Electrical Machines 9789339220051, 9339220056

Table of contents :
Title
Contents
1 Fundamental Concepts in Electrical Machines
2 Direct Current (DC) Machines
3 Single-phase Transformers
4 Three-Phase Transformers
5 Fundamentals of AC Rotating Machines
6 Synchronous Generator
7 Synchronous Motor
8 Three-phase Induction Motor
9 Single-phase Induction Motors
10 Special Machines
11 Review Problems (MCQs) with Solutions
Solution Manual
Index

Citation preview

Electrical Machines

ABOUT THE AUTHORS Abhijit Chakrabarti, PhD (Tech.) is Professor and former Head, Department of Electrical Engineering, Indian Institute of Engineering, Science and Technology (formerly BESU), Shibpur. He is also the former Vice-Chancellor of Jadavpur University and former Vice-Chairman and Chairman (acting), West Bengal State Council of Higher Education, West Bengal. Howrah. He has nearly eight years of industrial experience and around twenty-five years of teaching and research experience. He is a fellow of the Institution of Engineers (India) and has published/presented 129 research papers in international and national journals and conferences including IEEE Transaction, Elsevier, etc. He has published numerous books on Electrical Engineering. Professor Chakrabarti is the recipient of Pandit Madan Mohan Malviya award and the Power Medal, Merit and Best Paper awards (twice) from Central Electricity Authority, India, and an International Conference, China. His areas of interest include power systems, electric circuits, and power electronics and drives. Sudipta Debnath, PhD in Engineering, is Associate Professor, Department of Electrical Engineering, Jadavpur University, Kolkata. She has received the Career Award from AICTE, Govt. of India, and Young Scientist recognition from DST, Govt. of India. She is also the recipient of a certificate of merit from Institution of Engineers, India. She has published/presented 35 research papers in international and national journals and conferences. She has nearly one year of industrial experience and 18 years of teaching and research experience. Her areas of interest include power systems, electrical machines and soft computing techniques.

Electrical Machines

Abhijit Chakrabarti Professor Department of Electrical Engineering Indian Institute of Engineering, Science and Technology Shibpur, Howrah, West Bengal and Former Vice-Chancellor of Jadavpur University Former Vice-Chairman and Chairman (acting), West Bengal State Council of Higher Education, West Bengal. Sudipta Debnath Associate Professor, Department of Electrical Engineering Jadavpur University, Kolkata

McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

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CONTENTS

Preface 1. Fundamental Concepts in Electrical Machines 1.1 Introduction 1.1 1.2 Classification of Electrical Machines and their Essential Features 1.3 Important Terminologies 1.8 1.4 Magnetic Circuit 1.13 1.5 Electromagnetic Induction 1.18 1.6 Induced emf 1.21 1.7 Fleming’s Rules 1.24 1.8 Force on a Current-carrying Conductor 1.25 1.9 Torque in a Current-carrying Coil 1.26 1.10 Fundamental Concepts of Generator and Motor Action 1.44 1.11 Steady-state Voltage Equations 1.48 1.12 Magnetic Hysteresis 1.54 1.13 Eddy Current 1.57 1.14 Efficiency of an Electrical Machine 1.60 1.15 Electromechanical Energy-conversion Principles 1.61 1.16 Energy in a Singly Excited Magnetic Field System 1.62 1.17 Forces and Torques in a Magnetic Field 1.63 1.18 Energy in a Doubly Excited Magnetic Field System 1.69 Review Questions 1.92 Problems 1.92 Multiple-Choice Questions 1.93 Answers 1.98

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1.1–1.98 1.2

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2. Direct Current (dc) Machines 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29 2.30 2.31

Introduction 2.1 Principle of Operation of a dc Machine 2.1 Construction of a dc Machine 2.4 Armature Winding 2.7 Magnetic Circuit and Flux Path in a dc Machine 2.17 Equivalent Circuit of a dc Machine 2.17 Classification of dc Machines with respect to Types of Excitation 2.18 Emf Equation of a dc Machine 2.19 Relationship between Voltages and Currents in a dc Machine 2.27 Armature Reaction 2.37 Commutation 2.44 Characteristics of dc Generators 2.50 Voltage Regulation 2.58 Parallel Operation of dc Generator 2.58 dc Generators in Series 2.62 Principle of Working of a dc Motor 2.70 Electromagnetic Torque Equation of a dc Motor 2.70 Torque–current Characteristic of a dc motor 2.89 Speed Equation of a dc Motor 2.91 Speed Regulation of a dc Motor 2.91 Speed vs Armature Current Characteristic of a dc Motor 2.92 Speed vs Torque Characteristic of a dc Motor 2.94 Dynamic Behaviour of a dc Motor on Loading and Unloading 2.95 Speed Control of a dc Motor 2.96 Dynamic Behaviour During Speed Adjustment 2.107 Efficiency of dc Machines 2.139 Starting of a dc Motor 2.164 dc Motor Starting Circuits 2.171 Electric Braking of dc Motors 2.175 Testing of dc Machines 2.176 Applications of dc Machines 2.181 Review Questions 2.187 Problems 2.188 Multiple-Choice Questions 2.194 Answers 2.198

2.1–2.198

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3. Single-phase Transformers 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.21 3.22 3.23 3.24 3.25 3.26 3.27 3.28 3.29 3.30 3.31 3.32 3.33

Introduction 3.1 Transformer as a Coupled Circuit 3.1 Principle of Operation 3.4 EMF Equation 3.4 Construction of a Single-phase Transformer 3.5 Ideal Transformer 3.7 Distribution and Power Transformer 3.8 Transformation Ratio 3.8 Impedance Transformation 3.9 No-load Operation of a Transformer 3.22 Concept of Leakage Reactance 3.23 Operation of a Transformer on Load 3.24 Transient Behaviour when Loading and Unloading 3.27 Equivalent Circuit of a Transformer 3.28 Approximate Equivalent Circuit of a Transformer 3.31 Magnetization Current in a Real Transformer 3.32 Voltage Regulation of a Transformer 3.63 Expression of Voltage Regulation 3.64 Per-unit Expression for Resistance, Leakage Reactance and Impedance Voltage Drops 3.67 Approximate Per-unit Voltage Regulation 3.68 Condition for Zero Voltage Regulation 3.68 Condition for Maximum Voltage Regulation 3.68 Losses and Efficiency of Transformer 3.79 Condition for Maximum Efficiency 3.81 Testing of Transformers 3.104 All-day Efficiency 3.108 Parallel Operation of Two Single-phase Transformers 3.136 Auto-transformer 3.139 Step-down Auto-transformer 3.139 Power Relation in an Auto-transformer 3.141 Step-up Auto-transformer 3.143 Saving in Conductor Material in Auto-transformer 3.144 Advantages and Disadvantages of an Auto-transformer 3.145

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3.1–3.199

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3.34 3.35 3.36 3.37 3.38 3.39 3.40

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Application of an Auto-transformer 3.146 Pulse Transformer 3.146 Welding Transformer 3.147 Instrument Transformer 3.150 Transformer Cooling 3.152 Conservators and Breathers 3.153 Nameplate and Rating of a Single-phase Transformer Review Questions 3.173 Problems 3.174 Multiple-Choice Questions 3.195 Answers 3.199

3.153

4. Three-phase Transformers 4.1 Introduction 4.1 4.2 Three Single-phase Transformer Banks to Form One Three-phase Transformer 4.1 4.3 Three-phase Transformer as a Single Unit 4.2 4.4 Three-phase Transformer Connections 4.3 4.5 Salient Features of Commonly Used Transformer Connections 4.6 Open-delta or V–V Connection 4.10 4.7 Scott Connection or T-connection 4.12 4.8 Three-phase to Six-phase Conversion 4.19 4.9 Three-phase to Twelve-phase Conversion 4.22 4.10 Parallel Operation of Three-phase Transformers 4.40 4.11 Harmonic Phenomena in a Three-phase Transformer 4.40 4.12 Inrush Current 4.42 4.13 Transformer Tappings 4.43 4.14 Induction Regulator 4.45 4.15 Three-phase Auto-transformers 4.47 4.16 Three-winding Transformers 4.48 4.17 Audio Frequency Transformer 4.52 4.18 Rectifier Transformer 4.53 4.19 Grounding Transformer 4.53 4.20 Transformer Nameplate 4.53

4.1–4.91

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Review Questions 4.86 Problems 4.87 Multiple-Choice Questions 4.87 Answers 4.91

5. Fundamentals of ac Rotating Machines 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

Introduction 5.1 ac Windings 5.1 Concept of Electrical and Mechanical Degrees 5.6 Breadth Factor 5.7 Pitch Factor 5.9 Mmf of ac Windings 5.15 Emf Generated in Electrical Machines 5.21 Torque in ac Machines 5.25 Slot or Tooth Harmonics 5.33 Review Questions 5.40 Problems 5.41 Multiple-Choice Questions 5.41 Answers 5.47

6. Synchronous Generator 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14

5.1–5.47

Introduction 6.1 Principle of Operation 6.1 Construction 6.2 Different Types of Excitation Systems 6.5 Cooling 6.9 Emf Equation in a Synchronous Generator 6.10 Flux and mmf Diagram of Cylindrical-rotor Synchronous Generator 6.17 Concept of Synchronous Impedance 6.19 Voltage Regulation 6.20 Short-circuit Ratio 6.27 External Characteristic of a Synchronous Generator 6.45 Power-angle Characteristic of Cylindrical-rotor Synchronous Generator 6.47 Parallel Operation of Two Synchronous Generators 6.58 Synchronous Generator Connected to Infinite Bus 6.67

6.1–6.123

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6.15 6.16 6.17 6.18 6.19 6.20 6.21 6.22 6.23 6.24 6.25

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V Curves 6.72 Inverted V Curves 6.74 Operational Aspects 6.74 Salient-pole Synchronous Generator 6.77 Two-reaction Theory and Phasor Diagram of a Salient-pole Synchronous Generator 6.77 Methods of Determining Xd and Xq 6.79 Power-angle Characteristic of Salient-pole Synchronous Generator Synchronizing Power and Synchronizing Torque 6.84 Role of Damper Winding 6.91 Symmetrical and Unsymetrical Short Circuit in a Three-phase Synchronous Generator 6.91 Losses and Efficiency 6.94 Review Questions 6.109 Problems 6.110 Multiple-Choice Questions 6.118 Answers 6.123

6.82

7. Synchronous Motor 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13

Introduction 7.1 Principle of Operation 7.1 Equivalent Circuit and Phasor Diagram of a Synchronous Motor 7.2 Starting of Synchronous Motors 7.5 Power and Torque Developed in a Cylindrical Rotor Motor 7.11 Power and Torque Developed in a Salient-pole Motor 7.13 Effect of Change in Load 7.22 Effect of Change in Excitation 7.23 V Curves of a Synchronous Motor 7.24 Hunting of a Synchronous Machine 7.25 Methods of Reducing Hunting 7.26 Comparison between Synchronous and Induction Motors 7.26 Applications of Synchronous Motor 7.26 Review Questions 7.40 Problems 7.40 Multiple-Choice Questions 7.45 Answers 7.47

7.1–7.47

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8. Three-Phase Induction Motor 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16 8.17 8.18 8.19 8.20 8.21 8.22 8.23 8.24 8.25 8.26 8.27 8.28 8.29 8.30 8.31 8.32

Introduction 8.1 Construction of Induction Machines 8.1 Comparison of Squirrel-cage and Wound Rotors 8.4 Advantages and Disadvantages of a Three-phase Induction Motor 8.5 Principle of Operation 8.5 Induction Machine as a Generalized Transformer 8.6 Concept of Slip 8.7 Frequency of Rotor Voltage and Current 8.9 Emfs in Windings of Induction Motor 8.10 Equivalent Circuit of an Induction Motor 8.15 Power Relations in a Three-phase Induction Motor 8.20 Losses and Efficiency of Induction Motor 8.21 Torque Expression of Induction Motor 8.51 Torque–slip Characteristics of a Three-phase Induction Motor 8.55 Torque Condition During Loading, Breakdown and No-load 8.57 Determination of Motor Efficiency 8.83 Starting of a Three-phase Induction Motor 8.87 Reversal of Rotation 8.93 Speed Control of a Three-phase Induction Motor 8.102 Power Factor at Different Loading 8.107 Operation of an Induction Motor at Different Frequencies 8.108 Crawling and Cogging of Induction Motor 8.108 Skewing of Rotor Slots in a Three-phase Induction Motor 8.109 Operation on Unbalanced Voltage and Single Phasing 8.110 Circle Diagram 8.110 High-torque Induction Motors 8.114 Different Operating Zones of an Induction Machine 8.120 Induction Regulator 8.121 Frequency Changer 8.122 Induction Generator 8.123 Comparison of Induction Motor and Synchronous Motor 8.125 Application of Induction Motors 8.126 Review Questions 8.135 Problems 8.136

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Multiple-Choice Question Answers 8.149

8.146

9. Single-phase Induction Motors 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11

Introduction 9.1 Production of Torque 9.1 Equivalent Circuit 9.5 Determination of Parameters of Equivalent Circuit 9.10 Starting of Single-phase Induction Motors 9.14 Split-phase Induction Motors 9.15 Capacitor-start Motor 9.16 Capacitor-start Capacitor-run Motor 9.17 Shaded-pole Motors 9.19 Speed Control of a Single-phase Induction Motor 9.21 Comparison between Single-phase and Three-phase Induction Motors Review Questions 9.27 Problems 9.27 Multiple-Choice Questions 9.28 Answers 9.32

10. Special Machines 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11 10.12 10.13 10.14 10.15

Universal Motors 10.1 Repulsion Motor 10.5 Single-phase Series Motor 10.7 Single-phase Synchronous Motors 10.9 Synchronous Induction Motor 10.10 Linear Induction Motor 10.11 Reluctance Motor 10.13 Switched Reluctance Motor 10.14 Servomotor 10.15 Tachometer 10.16 Hysteresis Motor 10.17 Brushless DC Motors 10.19 Stepper Motors 10.22 Permanent Magnet DC (PMDC) Motors 10.26 Permanent Magnet Generators 10.27

9.1–9.32

9.22

10.1–10.33

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Review Questions 10.30 Multiple-Choice Questions Answers 10.33

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10.31

11. Review Problems (MCQs) with Solutions

11.1–11.108

Section A: DC Machines 11.1 Section B: Transformers 11.35 Section C: Induction Motors 11.67 Section D: Synchronous Generator 11.93

Index

I.1–I.7

PREFACE

In any mechanized industry, electrical machines are the backbone of the system. A course in electrical machines is usually a compulsory component in the syllabus of electrical engineering. This book is intended to serve as a textbook for the subject of electrical machines for graduate and postgraduate curricula in electrical engineering. It will also serve as a text for students of diploma level in electrical engineering. This book on electrical machines will be useful to the candidates appearing for GATE, UPSC Engineering Services, IAS Entrance Examinations, AMIE, IETE and other competitive examinations. The text of this book is arranged along with numerical problems in such a way that this book proves helpful to practicing engineers to understand the basics or to brush up theoretical application aspects of electrical machines. The study of electrical machines will continue to remain a core subject in electrical engineering, both at degree and diploma levels. The syllabi of electrical engineering in universities and institutes are mostly centred around rotating machines, transformers and some special machines. This book presents simple, explicit, rigorous and comprehensive treatment of transformers and electrical machines in a single volume. Considerable emphasis has been given to fundamental aspects, physical concepts, principles, circuit models, derivations and applications of transformers and electrical machines. The material presented in each chapter progresses from established principles to advanced topics. A large number of review problems have been added at the end of each chapter. In order to motivate the students, the subject material and worked-out problems have been presented in a systematic order. Once a student acquires a clear understanding of a machine, there is a need to develop the necessary equations associated with numerical problems. This part has been given utmost care in the present book. This book covers all the essential ingredients of electrical-machine learning expected in a graduate and undergraduate course in electrical engineering.

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Tbmjfou!Gfbuvsft The main highlights of this book are the following: � � � �

Focus on constructional details of machines Thorough coverage of special machines Application-based approach to prepare for a course on drives Theoretical treatise supplemented with numerous examples of descriptive, analytical, and mathematical types � One complete chapter dedicated to competitive examinations preparatory material � Rich pedagogy Illustrations: 300 Solved examples: 400 Practice questions: 250 Total problems: 881

Dibqufs!Pshboj{bujpo The content is organized in eleven chapters. Chapter 1 presents fundamental concepts of electromagnetism related to electrical machines. It explains the basic principles of generating and motoring actions. It also explains the working principle of all common types of rotating electrical machines from the view of magnetic field alignment. Basic principles of electromechanical energy-conversion devices are examined elaborately. This chapter also includes practical issues common to all electrical machines such as losses, efficiency, cooling, etc. Chapter 2 is devoted to the study of dc generators and dc motors. In-depth discussion on construction, operation, characteristics, testing, starting and braking phenomenon of dc machines have been carried out in this chapter. Explanation of the phenomena such as armature reaction and commutation along with winding details of dc machines are presented here. The dynamic behaviour of a dc motor during loading, unloading and speed adjustment have also been included in this chapter. Chapters 3 and 4 describe the construction, principle and different operational aspects of singlephase and three-phase transformers respectively. The phasor diagrams, equivalent circuit, tests for determination of equivalent circuit parameters, losses, voltage regulation and cooling phenomenon have been elaborately discussed. The harmonic phenomena in a three-phase transformer and vector grouping have also been presented. Operating principles of both single-phase and three-phase autotransformers have been included here. Chapter 5 provides in-depth discussion on fundamental aspects of ac rotating machines which are common to all types of rotating ac machines. General expressions for emf and torque are derived. The mmf produced by different types of windings in ac rotating machines has been explained with appropriate diagrams.

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Chapters 6 and 7 provide elaborate information on synchronous generators and synchronous motors. In addition to explaining the construction and different excitation systems, the principles of operation of these machines have been explained with the help of flux and mmf diagrams. The operation of a synchronous generator under symmetrical and unsymmetrical short-circuit condition and the effect of change in excitation and load in synchronous motors have also been included in these chapters. The operation of a three-phase induction motor is examined in Chapter 8. Emphasis has been given to develop the equivalent circuit and torque equation of the induction motor along with torque-slip characteristics of the motor. The determination of different performance data from the circle diagram is explained. The operation of an induction motor as induction regulator and frequency changer is also presented here. Chapter 9 deals with important concepts of a single-phase induction motor. The starting phenomenon, production of torque, equivalent circuit and circuit-parameter determination along with speed control has been discussed elaborately. The construction and operating principles along with application of different types of special machines have been presented in Chapter 10. The performance characteristics of different motors are also shown. Chapter 11 presents review problems on basic types of electrical machines and transformers in the form of multiple-choice questions. This chapter will help students in preparation for different competitive examinations.

Pomjof!Mfbsojoh!Dfoufs The Online Learning Center can be accessed at http://www.mhhe.com/chakrabarti/em and contains the following material: • Solutions Manual for Instructor’s The authors are confident that this book will certainly serve students as the most useful textbook for electrical machines. The authors have used their experience in presenting the basic concepts involved in the subject in such a way so that students can acquire them thoroughly.

Bdlopxmfehfnfout The authors acknowledge the support and comments received from their scholars and colleagues while writing the book. Sankar Narayan Mahato Indian Institute of Technology, Banaras Hindu University, Varanasi, Uttar Pradesh S N Mahendra Indian Institute of Technology, Banaras Hindu University, Varanasi, Uttar Pradesh

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R K Jarial Ajay Srivastava Mukhtar Ahmed Yaduvir Singh Rajesh Kumar Dubey Sanjay Parida Nitai Pal Sankari Bose Pramathes Das Ajit K Chattopadhyay Vaiju Kalkhambar A Subramanian U Sreenivas V Rajnikanth

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National Institute of Technology, Hamirpur, Himachal Pradesh Govind Ballabh Pant University of Agriculture and Technology, Pantnagar, Uttarakhand Aligarh Muslim University, Aligarh, Uttar Pradesh Harcourt Butler Technological Institute, Kanpur, Uttar Pradesh Jaypee Institute of Information Technology, Noida, Uttar Pradesh Indian Institute of Technology, Patna, Bihar Indian School of Mines, Dhanbad, Jharkhand KANKSA Academy of Technology & Management, Burdwan, West Bengal B P Poddar Institute of Management and Technology, Kolkata, West Bengal Bengal Engineering and Science University, Shibpur, West Bengal Walchand College of Engineering, Sangli, Maharashtra VRS College of Engineering and Technology, Arasur, Tamil Nadu Srinivasa Ramanujan Institute of Technology, Anatapur, Andhra Pradesh Sri Kalahastheeswara Institute of Technology, Srikalahasti, Andhra Pradesh

The authors also acknowledge the support from their family members and the publishers in bringing out the book.

Gffecbdl!Sfrvftu The authors have endeavoured to write and publish an accurate book. There may still be some minor errors in the book. It is also possible that some sections need further revision or clarification. The authors welcome any suggestion in this regard for future revisions. Abhijit Chakrabarti Sudipta Debnath Kolkata

QvcmjtifsÕt!Opuf McGraw Hill Education (India) invites suggestions and comments from you, all of which can be sent to [email protected] (kindly mention the title and author name in the subject line). Piracy-related issues may also be reported.

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An electrical machine is a device that can convert either mechanical energy to electrical energy or electrical energy to mechanical energy. When such a device is used to convert mechanical energy to electrical energy, it is called a generator. When it converts electrical energy to mechanical energy, it is called a motor. Almost all practical motors and generators convert energy from one form to another through the action of a magnetic field, and only machines using magnetic fields to perform such conversions are considered in this book. Since Thomas Alva Edison developed an electric generator, more than a hundred years ago, engineers have continually strived and successfully reduced the size and improved the efficiencies of electric machines by the use of improved materials and optimal design strategies. The transformer is an electrical device that is closely related to electrical machines. It converts ac electrical energy at one voltage level to ac electrical energy at another voltage level. Since transformers operate on the same principles as generators and motors, depending on the action of a magnetic field to accomplish the change in voltage level, they are usually studied together with generators and motors. These three types of electric devices are ubiquitous in modern daily life. Electric motors are used in home-run refrigerators, freezers, vacuum cleaners, blenders, air conditioners, fans and many similar appliances. In the workplace, motors provide the motive power for almost all tools. Of course, generators are necessary to supply the power used by all these motors. Apart from the application of the transformers in changing the voltage levels during power transmission and distribution, the transformers are also used extensively in electronic circuits to step down the voltage levels suitable for the low-voltage circuit.

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There are three basic rotating electric machine types, namely 1. the dc machine, 2. the polyphase synchronous machine (ac), and 3. the polyphase induction machine (ac). Three materials are mainly used in machine manufacture: steel to conduct magnetic flux, copper (or aluminium) to conduct electric current and insulation to insulate the voltage induced in conductors confining currents to them. All electric machines comprise two parts: the cylindrical rotating member called the rotor and the annular stationary member called the stator with the intervening air gap as illustrated in Fig. 1.1. The rotor has an axial shaft which is carried on bearings located in end covers bolted to the stator. The shaft extends out of the end cover usually at one end and is coupled to either the prime mover or the load. Air gap Rotor Shaft Stator

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2/3/2! Uif!ed!Nbdijof In a dc machine, the field poles are on the stator while the rotor is the armature as shown in the cross-sectional view of Fig. 1.2. The field poles are symmetrical and are even in number, alternately north and south. As the armature rotates, alternating emf and current induced in the armature winding are rectified to dc form by rotating mechanical switch called the commutator, which is tapped by means of stationary carbon brushes. The commutator is cylindrical in shape and comprises several wedge-shaped copper segments bound together while they are insulated from each other.

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The armature is made of laminated steel with slots cut out on the periphery to accommodate the insulated armature winding. The ends of each armature coil are connected to the commutator segments to form a closed winding. The armature, when carrying current, produces stationary poles (same as the number of field poles) which interact with the field poles to produce the electromagnetic torque. Main pole Field winding Pole shoe N

Yoke

Armature winding +

Ia

Va (dc)

S

S

– I

If

Vf (dc) N Commutator

Armature Brushes

Gjh/!2/3! Dsptt.tfdujpobm!wjfx!pg!b!ed!nbdijof

There are two methods of exciting the field windings of dc machines. They are separate excitation and self-excitation. The separately excited field winding consists of several turns of fine wire and is connected to a separate or external dc source. In self-excitation, the field winding of the machine is excited by its own armature. In these machines, the field poles must have residual magnetism. A self-excited machine can have shunt field, series field or both shunt and series fields. In shunt excitation, the field winding is in parallel with the armature and the field excitation is obtained from the armature voltage. The field winding resistance is high and hence the field current is low here. In series excitation, the field resistance is low and it is excited in series from the armature current. In compound excitation, both the series and shunt field are present.

2/3/3! Tzodispopvt!Nbdijof! A synchronous machine is known as doubly excited ac machine. The field winding of a synchronous machine is always excited from a dc source. In a synchronous generator, armature winding exports ac power and in a synchronous motor, the armature winding imports ac power. A synchronous generator is the most commonly used machine for the generation of electrical power and it is known

!

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Fmfdusjdbm!Nbdijoft

as alternator. It generates alternating voltage which is stepped up and transmitted. The generation of alternating voltage depends on the relative motion between field flux and armature winding. Like other electrical rotating machines, a synchronous machine has two main parts: stator and rotor. The high-power armature winding is placed on the stator slots in the stator core. The low-power field winding is constructed on the rotor. The construction of a synchronous machine depends upon the type of the prime mover used to rotate the rotor. The cross-sectional view of a synchronous machine is shown in Fig. 1.3. The field poles are excited by direct current. The stator forms the armature carrying a three-phase winding wound for the same number of poles as the rotor. All the three phases have identical windings with the same angular displacement between any pair of phases. When the rotor rotates, it produces alternating emf in each phase forming a balanced set with frequency given by f= where

nP 120

(1.1)

f = Frequency in Hz n = Rotor speed in rpm P = Number of field poles Armature winding

Field winding N

Ia Ib Ic

3-phase ac

S

S Stator

If Vf (dc)

N Rotor

Main pole

Pole shoe

Gjh/!2/4! Dsptt.tfdujpobm!wjfx!pg!b!tzodispopvt!nbdijof

For a given number of poles, there is a fixed correspondence between the rotor speed and the stator frequency; the rotor speed is, therefore, called the synchronous speed. When balanced threephase currents are allowed to flow in the armature winding, these produce a synchronously rotating

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field, stationary with respect to the rotor field as a result of which the machine produces torque of electromagnetic origin. The synchronous motor is, however, nonself-starting. Depending upon the construction of the rotor, a synchronous machine can be of salient-pole type and cylindrical-rotor type. In a salient-pole machine, the rotor has projecting poles and they are much larger in diameter. These machines are suitable for medium speed. A cylindrical rotor consists of a smooth cylinder having a number of slots for accommodating field coils. These machines are smaller in diameter and they are larger in axial length. The number of poles on the rotor is generally two and they can run at very high speed.

2/3/4! Joevdujpo!Nbdijof! Induction machines can be of two types: induction generator and induction motor. Induction generator is rarely used as its performance is unsatisfactory and hence in this book, a thorough discussion on induction motor has been done. The induction motor is the most widely used ac motor in the industry. Similar to other rotating electrical machines, a three-phase induction motor consists of two main parts: the stator and the rotor. The stator winding is three phase and they are connected either in star and/or in delta. When the stator winding is excited, it produces a synchronously rotating field. Depending on the construction of the rotor, an induction motor can be of two types: squirrel cage or simply cage rotor and phase wound or wound rotor or slip-ring rotors. 2/!Trvjssfm.dbhf!Npups! Here, the rotor has copper (or aluminium) bars embedded in slots which are short-circuited at each end as shown in Fig. 1.4(a). It is rugged and economic in construction but develops low starting torque. 3/!Tmjq.sjoh!)ps!Xpvoe.spups*!Npups The rotor has a proper three-phase winding with three leads brought out through slip rings and brushes as shown in Fig. 1.4(b). These leads are normally short-circuited when the motor is running. Resistances are introduced in the rotor circuit via the slip rings at the time of starting to improve the starting torque.

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Fmfdusjdbm!Nbdijoft

End rings

Conducting bar embedded in slots and shorted at both ends by end rings

Brushes

(a) A squirrel-cage rotor (schematic diagram)

Slip rings

Winding (details not shown) embedded in slots, leads brought out to slip rings (b) A wound rotor (schematic diagram)

Gjh/!2/5! Joevdujpo.npups!spupst

The rotating field created by the stator winding moves past the shorted rotor conductors inducing currents in the latter. These induced currents produce their own field which rotates at the same speed (synchronous) with respect to the stator-produced field. Torque is developed by the interaction of these two relatively stationary fields. The rotor runs at a speed close to synchronous speed but always slightly lower than it. At the synchronous speed, no torque can be developed as zero relative speed between the stator field and the rotor implies no induced rotor currents and, therefore, no torque. Single-phase ac motors are employed for low-voltage, low-power applications. They are the fractional kW motors. They operate on the same basic principles as the three-phase motor, but the pulsating single-phase field produces additional losses, reducing motor torque and the pulsating torque component increases the noise level of the motor. An induction machine connected to the mains when driven at super synchronous speed behaves as a generator feeding power into the electric system. It is used in small hydroelectric stations and aerospace applications.

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2/3/5! Usbotgpsnfs A transformer is a static electric device comprising two or more electric circuits coupled by a common magnetic circuit. One electric circuit is connected to alternating voltage source and the other circuits receive electrical energy through the magnetic circuit. The circuit to which the supply is connected is known as the primary winding of the transformer. The mutual flux generated by the primary winding links the other winding known as the secondary winding and induces voltage in it. The magnitude of the voltage depends upon the number of turns of winding in the primary and secondary winding. Hence, a transformer is essentially an electric device that transfers electrical energy from one circuit to another circuit through magnetic medium. Though the voltage and current levels in the two circuits are different, the frequency remains constant. The schematic diagram of a single-phase transformer is shown in Fig. 1.5. f

V1

V2

Gjh/!2/6! Tdifnbujd!ejbhsbn!pg!b!tjohmf.qibtf!usbotgpsnfs

Though a transformer is not an energy-conversion device, it is an indispensable component in many energy-conversion systems. It is widely used in low-power, low-current electronic and control circuits or isolating one circuit from another. It is an integral component of ac power system network and has made electric generation possible at low and economical voltage, power transfer through transmission line at most economical transmission voltage and power utilization at the most suitable voltage for the consumers. When the primary winding of the transformer is connected to a source, the emfs induced in both the primary and secondary windings are directly proportional to the number of turns in the respective windings. If the number of turns in the primary winding is greater than that in the secondary then secondary voltage will be less than the primary and the transformer may be treated as step down transformer. If the number of turns in the secondary winding is greater than that in the primary then the transformer is known as step up transformer. Transformers used in a power-system network can be classified as power transformers and distribution transformers. Power transformers are used in big substations and distribution transformers are used to supply loads to consumers at distribution voltage levels. Transformer action can occur even when two windings are coupled through air but coupling between the two windings can be made more effective when they are coupled through an iron core. Depending upon the construction of core, a transformer can be of core type or shell type. In the core-type transformer, the windings are wound around two legs of a rectangular magnetic core. In the shell type, the windings are wound around the centre leg of a three-legged core.

!

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Fmfdusjdbm!Nbdijoft

The insulation of a machine (or transformer) is its most vulnerable part because it cannot be stressed beyond a certain temperature. For a given frame size, the steady temperature rise is determined by the machine loading, the associated power loss (this appears in the form of heat) and the cooling provided. Thus, the maximum machine loading, called its rating for a given frame size, is limited by the permissible temperature rise which is dependent upon the class of insulation used. In the case of high-speed dc machines, poor commutation (reversal of current in armature coils) may become a limiting factor on account of the centrifugal forces developed. This limit is more stringent in dc machines with complicated armature construction than in the rugged rotor induction motor. Because of their high thermal capacity, machines are quite capable of withstanding a fair amount of overloads for short durations.

JNQPSUBOU!UFSNJOPMPHJFT!

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2/!Bohvmbs!Qptjujpo!q The angular position q of an object is the angle at which it is oriented, measured from some arbitrary reference point. Angular position is usually measured in radians or degrees. It corresponds to the linear concept of distance along a line. 3/!Bohvmbs!Wfmpdjuz!w Angular velocity (or speed) is the rate of change in angular position with respect to time. It is assumed positive if the rotation is in a counter-clockwise direction. Angular velocity is the rotational analog of the concept of velocity on a line. One-dimensional linear velocity along a line is defined as the rate of change of the displacement along the line (r) with respect to time. V=

dr dt

(1.2a)

Similarly, angular velocity w is defined as the rate of change of the angular displacement q with respect to time. w=

dq dt

(1.2b)

If the units of angular position are radians then angular velocity is measured in radians per second. In dealing with ordinary electric machines, engineers often use units other than radians per second to describe shaft speed. Frequently, the speed is given in revolutions per second or revolutions per minute. Because speed is such an important quantity in the study of machines, it is customary

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/:

to use different symbols for speed when it is expressed in different units. By using these different symbols, any possible confusion as to the units intended is minimized. The following symbols are used in this book to describe angular velocity: wm = Angular velocity expressed in radians per second fm = Angular velocity expressed in revolutions per second hm = Angular velocity expressed in revolutions per minute The subscript m on these symbols indicates a mechanical quantity, as opposed to an electrical quantity. If there is no possibility of confusion between mechanical and electrical quantities, the subscript is often left out. These measures of shaft speed are related to each other by the following equations: hm = 60 fm w fm = m 2p

(1.3a) (1.3b)

4/!Bohvmbs!Bddfmfsbujpo!a Angular acceleration is the rate of change in angular velocity with respect to time. It is assumed positive if the angular velocity is increasing in an algebraic sense. Angular acceleration is the rotational analog of the concept of acceleration on a line. Just as one-dimensional linear acceleration is defined by the equation a=

dv dt

(1.4)

a=

dw dt

(1.5)

angular acceleration is defined by

If the units of angular velocity are radians per second then angular acceleration is measured in radians per second square. 5/!Upsrvf!t!ps!U In linear motion, a force applied to an object causes its velocity to change. In the absence of a net force on the object, its velocity is constant. The greater the force applied to the object, the more rapidly its velocity changes. There exists a similar concept of rotation. When an object is rotating, its angular velocity is constant unless a torque is present on it. The greater the torque on the object, the more rapidly the angular velocity of the object changes.

!

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Torque can loosely be called the “twisting force” on an object. Intuitively, torque is fairly easy to understand. Imagine a cylinder that is free to rotate about its axis. If a force is applied to the cylinder in such a way that its line of action passes through the axis [Fig. 1.6(a)], then the cylinder will not rotate. However, if the same force is placed so that its line of action passes to the right of the axis [Fig. 1.6(b)], then the cylinder will tend to rotate in a counter-clockwise direction. The torque or twisting action on the cylinder depends on (a) the magnitude of the applied force, and (b) the distance between the axis of rotation and the line of action of the force.

F

t

F

t=0

Torque is counter-clockwise

Torque is zero (a)

(b)

Gjh/!2/7! Upsrvf!jo!dzmjoefs;!)b*!B!gpsdf!bqqmjfe!up!b!dzmjoefs!tp!uibu!ju!qbttft!uispvhi!uif!byjt!pg! spubujpo/!t!>!1/!)c*!B!gpsdf!bqqmjfe!up!b!dzmjoefs!tp!uibu!jut!mjof!pg!bdujpo!njttft!uif!byjt!pg! spubujpo/!Ifsf-!t!jt!dpvoufs.dmpdlxjtf/

The torque on an object is defined as the product of the force applied to the object and the smallest distance between the line of action of the force and the object’s axis of rotation. If r is a vector pointing from the axis of rotation to the point of application of the force, and if F is the applied force then the torque can be described as t = (Force applied) (Perpendicular distance) = (F) (r sin q) = r F sin q

(1.6)

where q is angle between the vector r and the F. The direction of the torque is clockwise if it would tend to cause a counter-clockwise rotation (Fig. 1.7). The units of torque are Newton-metres in SI units and pound-feet in the English system.

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r sin (180° – q) = r sin q

r 180° – q q

t = (perpendicular distance) (force) t = (r sin q) F, counter-clockwise

F

Gjh/!2/8! Efsjwbujpo!pg!uif!frvbujpo!gps!uif!upsrvf!po!bo!pckfdu

6/!OfxupoÕt!Mbx!pg!Spubujpo! Newton’s law of objects moving along a straight line describes the relationship between the force applied to an object and its resulting acceleration. This relationship is given by the equation F = ma

(1.7)

where F = Net force applied to an object m = Mass of object a = Resulting acceleration In SI units, force is measured in newton, mass in kilogram, and acceleration in metre per second square. In the English system, force is measured in pound, mass in slug and acceleration in foot per second square. A similar equation describes the relationship between the torque applied to an object and its resulting angular acceleration. This relationship, called Newton’s law of rotation, is given by the equation t = Ja

(1.8)

where t is the net applied torque in newton-metre or pound-feet and a is the resulting angular acceleration in radian per second square. The term J serves the same purpose as an object’s mass in linear motion. It is called the moment of inertia of the object and is measured in kilogram-metre square or slug-foot square.

!

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Fmfdusjdbm!Nbdijoft

7/!Xpsl!X For linear motion, work is defined as the application of a force through a distance. In equation form, W = Ú Fdr

(1.9)

where it is assumed that the force is collinear with the direction of motion. For the special case of a constant force applied collinearly with the direction of motion, this equation becomes just W = Fr

(1.10)

The units of work are joules in SI and foot-pound in the English system. For rotational motion, work is the application of a torque through an angle. Here, the equation for work is W =Útdq

(1.11)

W = tq

(1.12)

and if the torque is constant,

8/!Qpxfs!Q Power is the rate of doing work, or the increase in work per unit time. The equation for power is P=

dW dt

(1.13)

It is usually measured in joule per second (watts), but also can be measured in foot-pound per second or in horsepower. By this definition, and assuming that force is constant and collinear with the direction of motion, power is given by P= or,

dW d Ê dq ˆ = (tq ) = t Á ˜ = tw Ë dt ¯ dt dt

P = tw

(1.14) (1.15)

Equation (1.15) is very important in the study of electric machinery, because it can describe the mechanical power on the shaft of a motor or generator. Equation (1.15) is the correct relationship among power, torque and speed if power is measured in watt, torque in newton-metre and speed in radian per second. If other units are used to measure any of the above quantities then a constant must be introduced into the equation for unit conversion factors. It is still common in US engineering practice to measure torque in pound-feet, speed in

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

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revolution per minutes and power in either watt or horsepower. If the appropriate conversion factors are included in each term then Eq. (1.15) becomes

P(watts) =

t (lb - ft ) ◊ h ◊ ( r /min) 7.04

(1.16)

P(horsepower) =

t (lb - ft ) ◊ h ◊ ( r /min) 5252

(1.17)

where torque in measured in pound-foot and speed is measured in revolution per minute.

NBHOFUJD!DJSDVJU!

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There are many basic principles or laws related to electric machines that must be known and well understood before studying electric machines. This section briefly reviews some of the fundamentals of electricity and magnetism. Each magnetic circuit, shown in Fig. 1.8 is an arrangement of ferromagnetic materials called a core that forms a path to contain and guide the magnetic flux in a specific direction. The core shape shown in Fig. 1.8(a) is used in transformers. Fig. 1.8(b) shows the magnetic circuit of a simple two-pole motor; it includes a stator core, a rotor core, and air gap. Note that the flux always takes the shortest path across an air gap. Rotor iron f

l A

f

S

N

Air gaps

f Stator iron

I +

E

I

–

+

(a)

(b)

E

–

Gjh/!2/9! Nbhofujd!djsdvju;!)b*!Gps!b!usbotgpsnfs!)c*!Gps!b!tjnqmf!uxp.qpmf!npups

!

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Fmfdusjdbm!Nbdijoft

2/5/2! Dpodfqu!pg!Nbhofupnpujwf!Gpsdf!)NNG* The ampere-turns (A-t) of the respective coils in Fig. 1.8 represent the driving force, called magnetomotive force or mmf, that causes a magnetic field to appear in the corresponding magnetic circuits. Expressed in equation form, F = N◊I where

(1.18)

F = Magnetomotive force (mmf) in ampere-turns (A-t) N = Number of turns in coil I = Current in coil (A)

2/5/3! Dpodfqu!pg!Nbhofujd!Gjfme!Joufotjuz! Magnetic field intensity, also called mmf gradient, is defined as the magnetomotive force per unit length of magnetic circuit. The average magnitude of the field intensity in a homogeneous section of a magnetic circuit is numerically equal to the mmf across the section divided by the effective length of the magnetic section. That is, H= where

F N ◊I = l l

(1.19)

H = Magnetic field intensity (A-t/m) l = Mean length of the magnetic circuit, or section (m) F = mmf (A-t)

Note that in a homogeneous magnetic circuit of uniform cross section, the field intensity is the same at all points in the magnetic circuit. In composite magnetic circuits, consisting of sections of different materials and/or different cross-sectional areas, however, the magnetic field intensity differs from section to section. Magnetic field intensity has many useful applications in magnetic circuit calculations. One specific application is calculating the magnetic-potential difference, also called magnetic drop or mmf drop, across a section of a magnetic circuit. The magnetic drop in ampere-turns per metre of magnetic core length in a magnetic circuit is analogous to the voltage drop in volts per metre of conductor length in an electric circuit.

2/5/4! Dpodfqu!pg!Gmvy!Efotjuz The flux density is a measure of the concentration of lines of flux in a particular section of a magnetic circuit.

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Expressed mathematically, and referring to the homogeneous core in Fig. 1.8(a), B= where

f A

(1.20)

f = Flux, weber (Wb) A = Cross-sectional area (m2) B = Flux density (Wb/m2), or tesla (T)

2/5/5! Dpodfqu!pg!Sfmvdubodf A very useful equation that expresses the relationship between magnetic flux, mmf and the reluctance of the magnetic circuit is f= where

F N ◊I = R R

(1.21)

f = Magnetic flux (Wb) F = Magnetomotive force (A-t) R = Reluctance of magnetic circuit (A-t/Wb)

Reluctance R is a measure of the opposition the magnetic circuit offers to the flux and is analogous to resistance in an electric circuit. The reluctance of a magnetic circuit, or section of a magnetic circuit, is related to its length, cross-sectional area, and permeability. Solving Eq. (1.21) for R, dividing numerator and denominator by l, and rearranging terms, R=

N ◊ I N ◊ I/l H l = = = f f /l BA/ l ( B / H ) A

Defining B H l R= mA m=

where

B H l A m

(1.22) (1.23)

= Flux density (Wb/m2), or tesla (T) = Magnetic field intensity (A-t/m) = Mean length of magnetic circuit (m) = Cross-sectional area (m2) = Permeability of material (Wb/A-t.m)

Equation (1.23) applies to a homogeneous section of a magnetic circuit of uniform cross section. The ratio m = B/H is called magnetic permeability and has different values for different degrees of magnetization of a specific magnetic core material.

!

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Fmfdusjdbm!Nbdijoft

2/5/6! Sfmbujwf!Qfsnfbcjmjuz!boe!Nbhofuj{bujpo!Dibsbdufsjtujdt Relative permeability is the ratio of the permeability of a material to the permeability of free space; it is, in effect, a figure of merit that is very useful for comparing the magnetizability of different magnetic material whose relative permeability are known. Expressed in equation form, mr =

m mo

(1.24)

mo = Permeability of free space = 4p ¥ 10–7 (Wb/A-t.m) mr = Relative permeability, a dimensionless constant m = Permeability of material (Wb/A-t.m)

where

Representative graphs of Eq. (1.22) for some commonly used ferromagnetic materials are shown in Fig. 1.9. The graphs, called B-H curves, magnetization curves, or saturation curves, are very useful in design, and in the analysis of machine and transformer behaviour. The four principal sections of a typical magnetization curve are illustrated in Fig. 1.10. The curve is concave up for “low” values of magnetic field intensity, exhibits a somewhat (but not always) linear characteristic for “medium” field intensities, and then is concave down for “high” field intensities, eventually flattening to an almost horizontal line for “very high” intensities. The part of the curve that is concave down is known as the knee of the curve, and the “flattened” section is the saturation region. 1.50

l

t stee

Shee

el

t ste

Cas

Flux Density (B, T)

1.00

on

Cast ir 0.50

0

500

1000 1500 Magnetic Field Intensity (H, A-t/m)

2000

Gjh/!2/:! Sfqsftfoubujwf!C.I!dvswft!gps!tpnf!dpnnpomz!vtfe!gfsspnbhofujd!nbufsjbmt

2500

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

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Magnetic saturation is complete when all of the magnetic domains of the material are oriented in the direction of the applied magnetomotive force. Saturation begins at the start of the knee region and is essentially complete when the curve starts to flatten. Depending on the specific application, the magnetic core of an apparatus may be operated in the linear region, and/or the saturation region. For example, transformers and ac machines are operated in the linear region and lower end of the knee; self-excited dc generators and dc motors are operated in the upper end of the knee region, extending into the saturation region; separately excited dc generators are operated in the linear and lower end of the knee region.

Flux Density (B, T)

Saturation

Lin

ea r

ee Kn

Low

Medium

Very high Magnetic Field Intensity (H, A-t/m)

Gjh/!2/21! Fybhhfsbufe!nbhofuj{bujpo!dvswf!jmmvtusbujoh!uif!gpvs!qsjodjqbm!tfdujpot

The relationship between the relative permeability and the reluctance of a magnetic core is obtained by solving Eq. (1.24) for m, and then substituting into Eq. (1.23). The result is l l R = = (1.25) m A m r mo A Equation (1.25) indicates that the reluctance of a magnetic circuit is affected by the relative permeability of the material, which is dependent on the magnetization, and hence is not constant.

!

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Fmfdusjdbm!Nbdijoft

FMFDUSPNBHOFUJD!JOEVDUJPO!

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Oersted’s discovery in 1820 revealed that a current flowing in a wire deflects a compass needle placed near it and thereby proved that a magnetic field is associated with an electric current. The discovery provided the interaction between electricity and magnetism. The strength of the magnetic field can be increased by increasing the current in the conductor. In 1931, Michael Faraday discovered that a current is generated whenever the magnetic flux linked with the circuit changes. This current is called induced current and the emf in the circuit responsible for the production of the current is called induced emf. The phenomenon of the production of induced emf in a circuit due to change in the number of magnetic lines of force in it is called electromagnetic induction.

2/6/2! GbsbebzÕt!Mbxt!pg!Fmfduspnbhofujd!Joevdujpo! Faraday’s first law states that “whenever the number of magnetic lines of force (or magnetic flux) passing through a circuit changes an induced emf is set up in the circuit”. Faraday’s second law states that “the magnitude of the induced emf is proportional to the rate of change of magnetic lines of force”. Hence, according to Faraday’s laws of electromagnetic induction, eind =

d(Nf) df =N dt dt

(1.26)

where N is the number of turns and eind is the induced emf.

2/6/3! Mfo{Õt!Mbx Though Faraday’s laws give no idea about the direction of the induced emf, Lenz’s law gives the direction of induced emf which is based on the law of conservation of energy, It states that “the direction of the induced current (or emf) is such that it opposes the very cause producing this current (or emf), i.e. it opposes the change in magnetic flux.” Hence, from Lenz’s law, we get eind = –N

df dt

(1.27)

There is one major difficulty involved in using Eq. (1.27) in practical problems. The equation assumes that exactly the same flux is present in each turn of the coil. Unfortunately, the flux leaking out of the core into the surrounding air prevents this from being true. If the windings are tightly coupled, so that the vast majority of the flux passing through one turn of the coil does indeed pass through all of them then Eq. (1.27) will give valid answers. But if leakage is quite high or if extreme accuracy is required, a different expression that does not make that assumption will be needed.

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

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The magnitude of the voltage in the ith turn of the coil is always given by eind =

d (fi ) dt

(1.28)

If there are N turns in the coil of wire, the total voltage on the coil is N

eind =

 ei

(1.29a)

i =1 N

=

Â

d (fi ) i = 1 dt

(1.29b)

=

dÊN ˆ Âf dt ÁË i = 1 i ˜¯

(1.29c)

The term in parentheses in Eq. (1.29c) is called the flux linkage l of the coil, and Faraday’s law can be rewritten in terms of flux linkage as eind =

dl dt

(1.30)

N

where

l=

 fi

(1.31)

i =1

The units of flux linkage are weber-turns. Faraday’s law is the fundamental property of magnetic fields involved in transformer operation. The effect of Lenz’s law in transformers is to predict the polarity of the voltages induced in transformer windings.

2/6/4! Dpodfqu!pg!Tfmg.!boe!Nvuvbm!Joevdubodf! The change in current through a coil causes a change in flux. This change in flux induces an emf in the coil. This self-induced emf can be written as e =L where

di dt

(1.32)

e = emf induced, volts di = Rate of change of current, A/s dt L = Coefficient of self-inductance, H.

L, the coefficient of self-inductance or simply inductance has the units of henry (symbol H).

!

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Fmfdusjdbm!Nbdijoft

The inductance of a coil is 1 H if an emf of one volt is induced in it when the current through it changes at the rate of 1 A/s. By Lenz’s law, this emf is in a direction so as to oppose the external emf, which is driving current through the coil. From Eq. (1.26) and Eq. (1.32), e =L or

di df =N dt dt

L =N

df di

(1.33)

If rate of change of flux is constant, then df f = di i and Since

Therefore,

Nf i MMF Ni = f= reluctance 1/ma

L=

È Ni ˘ Í l /m a ˙ N 2 ma Î ˚ L =N = i l

(1.34)

When the flux of one coil links another coil, a mutually induced emf appears across the second coil. By Lenz’s law, this is also a counter emf. The mutually induced emf can be written as e=M

di dt

(1.35)

where e is the emf induced in the second coil, di/dt is the rate of change of current in the first coil and M is the coefficient of mutual inductance (units henry). The mutual inductance between two coils is 1 H if a current changing at the rate of 1 A/s in one coil induces an emf of 1 volt in the second coil. Let us consider two coils having turns NA and NB and assume that they are coupled magnetically. Let the current IA flowing in the first coil produce a flux of fA. Assuming that this whole flux is linkN f ing coil B, the flux linkage in coil B for unit current in the coil A is B A . IA The mutual inductance between the two coils is defined as the weber turns in one coil due to one ampere current in the other.

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

Hence,

M=

N Bf A IA

2/32

(1.36)

Now let us deduce an expression for coefficient of mutual inductance in terms of the dimensions of the two coils. Flux in coil A is fA = \ Flux per ampere

NA IA l /m A

fA N A ◊ m ◊ A = IA l

Assuming whole of flux fA is linking the coil B M=

N Bf A N B ◊ N A ◊ m ◊ A N A N B = = IA l l/m A

(1.37)

where m is the permeability of the medium between coil A and coil B. M

where M is the coefficient LA LB of mutual inductance between two coils having self-inductance LA and LB. When coefficient of coupling is nearly unity the coils are said to be closely coupled. When K is much less than unity the coils are said to be loosely coupled. Co-efficient of coupling may be defined as the ratio of mutual inductance present between the two coils to the maximum possible mutual inductance that can be present in the two coils. Hence, coefficient of coupling is unity when the flux produced by one coil completely links the other. It is zero when the flux produced by one coil does not link the other coil at all. The coefficient of coupling between two coils A and B is K =

JOEVDFE!FNG!

2/7

Emf is induced in a conductor in two ways. When the conductor remains stationary and the flux linked with it is changed by increasing or decreasing the current producing this flux then an emf is induced in the conductor which is known as statically induced emf. In the second case, the field is stationary but the conductor moves in the magnetic field and the flux linking the conductor changes. The emf induced in the conductor is called dynamically induced emf.

2/7/2! Tubujdbmmz!Joevdfe!fng!)Usbotgpsnfs!fng* The statically induced emf may be of two types: 1. Mutually induced emf 2. Self-induced emf

!

2/33

Fmfdusjdbm!Nbdijoft

2/!Nvuvbmmz!Joevdfe!fng Consider two coils A and B lying close to each other. If the flux in the coil A is varied by varying the current with the help of the rheostat connected to it then flux linked with the coil B also changes as shown Fig. 1.11. According to Faraday’s law, an emf will be induced in the coil B, the direction of which is given by Lenz’s law. E is the emf of the dc source connected in coil A and V is a voltmeter connected in the coil B. The voltmeter will indicate the emf induced in the coil B. In this case, both the coils are stationary and this emf induced in the coil B is called mutually induced emf.

Coil A

Coil B V

R

E

Gjh/!2/22! Nvuvbmmz!joevdfe!fng

3/!Tfmg.joevdfe!fng In this type, the emf is induced in a coil due to changes of its own flux linkage. If current through a coil changes, the flux linked with it also changes. This change in flux produces self-induced emf in the coil. The direction of the induced emf in the coil is given by Lenz’s law and the direction of the induced emf is such that it will oppose any change of flux.

2/7/3! Ezobnjdbmmz!Joevdfe!fng!)Npujpobm!fng* Figure 1.12 shows three conductors a, b, c moving in a magnetic field of flux density B in the directions indicated by arrows. Conductor a is moving in a direction perpendicular to its length and perpendicular to the flux lines. Therefore, it cuts the lines of force and a motional emf is induced in it. Let us consider that the conductor moves by a distance dx in a time dt. If the length of the conductor is l, the area swept by the conductor is l dx. Then change in flux linking the coil = df = B.l.dx Since there is only one conductor, e=

df Bl dx = dt dt

Since dx/dt is v, i.e. velocity of conductor, e = Blv volts

(1.38)

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

a

2/34

b

c

q

(a)

(b)

(c)

Gjh/!2/23! Npujpo!pg!b!dpoevdups!jo!b!nbhofujd!gjfme

where

e B v l

= emf induced, volts = Flux density, tesla = Velocity of conductor, metres/second = Length of conductor, metres

The motion of the conductor b (Fig. 1.12b) is at an angle q to the direction of the field. If the conductor moves by a distance dx, the component of distance traveled at right angles to the field is (dx) (sin q), and, proceeding as above, the induced emf is e = Blv sin q volts

(1.39)

Equation (1.39) includes Eq. (1.38) because when q = 90°, the two equations become identical. In Fig. 1.12(c), the motion of the conductor c is parallel to the field. Therefore, in this case, no flux is cut, q is zero and induced emf is also zero. Dynamically induced emf is also known as speed emf or motional emf or rotational emf. Equation (1.38) can also be written in a more general vectorial form: The force F on a particle of charge Q moving with a velocity v in a magnetic field B is F = Q(v ¥ B) N

(1.40)

Dividing F by Q, we get the force per unit charge, i.e. electric field E, as E=

F = v ¥ B volts/m Q

(1.41)

The electric field E is in a direction normal to the plane containing v and B. If the charged particle is one of the many electrons in a conductor moving across the magnetic field, the emf e between the end points of conductors is line integral of electric field E, or e =ã where

e = emf induced, volts

E.dl =

(v ¥ B).dl

(1.42)

!

2/35

Fmfdusjdbm!Nbdijoft

E dl v B

= Electric fields, volts/m = Elemental length of conductor, m = Velocity of conductor, metres/second = Flux density, tesla.

Equation (1.42) is the same as Eq. (1.39), but written in a more general form. If v, B and dl are mutually perpendicular, Eq. (1.42) reduces to Eq. (1.38).

GMFNJOHÕT!SVMFT!

2/8

The direction of the mechanical force experienced by a current-carrying conductor lying in a magnetic field is given by Fleming’s left-hand rule. According to this rule, the thumb, forefinger and middle finger are extended in three directions at right angles to one another. If the forefinger represents the direction of the field, middle finger represents the direction of the current then the thumb gives the direction of motion as shown in Fig. 1.13.

Motion Motion Field B Field

Cu

rre

nt

Current Gjh/!2/24! GmfnjohÕt!mfgu.iboe!svmf

The dc motor is based on the principle that when a current carrying conductor is placed in a magnetic field it experiences a mechanical force and starts rotating. Thus, the operation of a dc motor follows Fleming’s left-hand rule. The direction of the emf induced in a conductor may be found by applying Fleming’s right-hand rule. If the middle finger, forefinger and thumb of right hand are held perpendicular to each other and if the forefinger and thumb represents the direction of magnetic flux and direction of motion of the conductor respectively then the middle finger represents the direction of the emf induced in the conductor as shown in Fig. 1.14.

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/36

Motion Motion

Field

Field Emf Emf

Gjh/!2/25! GmfnjohÕt!sjhiu.iboe!svmf

Fleming’s right-hand rule is used when the induced emf is due to flux cutting, i.e. the emf is induced in a conductor when it moves in a magnetic field. This type of induced emf is called dynamically induced emf. When the emf is statically induced emf, i.e. emf is induced in a conductor due to change of flux linkage, Lenz’s law is applied to find the direction of the induced emf.

GPSDF!PO!B!DVSSFOU.DBSSZJOH!DPOEVDUPS!

2/9

Figure 1.15(a) shows a conductor lying in a magnetic field of flux density B. The conductor is carrying a current (entering the page). This current sets up a flux in clockwise direction. The external field is in the downward direction. As seen in Fig. 1.15(a), the field of the conductor assists the external field on the right-hand side of the conductor and opposes it on the left-hand side. This produces a force on the conductor towards left. If the direction of the current is reversed [Fig. 1.15(b)], the flux due to this current assumes counter-clockwise direction and the force on the conductor is towards right. In both cases, the force is in a direction perpendicular to both the conductor and the field and is maximum if the conductor is at right angles to the field. The magnitude of this force is F = BIl newtons

(1.43)

where B is flux density in tesla, I is current in amperes and l is the length of conductor in metres.

!

2/37

Fmfdusjdbm!Nbdijoft

Magnetic field

Magnetic field

Force

+

+

Flux due to current in conductor

Force

Flux due to current in conductor

(a)

(B)

Gjh/!2/26! Gpsdf!po!b!dpoevdups!jo!b!nbhofujd!gjfme;!)b*!Dvssfou!joup!uif!qbhf!)c*!Dvssfou!pvu!pg!uif! qbhf

If the conductor is inclined at an angle q to the magnetic field, the force is F = BIl sin q newtons

UPSRVF!JO!B!DVSSFOU.DBSSZJOH!DPJM!

(1.44)

2/:

Figure 1.16 shows a coil carrying current I and lying in a magnetic field of flux density B. From the discussion in Sec 1.7, it is seen that an upward force is exerted on the right-hand conductor and a downward force on the left-hand conductor. Equation (1.43) gives the force on each conductor and the total force is F = 2BIl newtons (1.45) I

Force l

N r

S

Force Flux

Gjh/!2/27! Upsrvf!po!b!dpjm!jo!b!nbhofujd!gjfme

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/38

If the coil has N turns, the total force is F = 2NBIl newtons

(1.46)

The torque is acting at a radius of r metres and is given by Torque = 2NBIl r

newton-metres

(1.47)

The configuration of Fig. 1.16 is the basic moving part in an electrical measuring instrument. An electric motor also works on this principle.

! Qspcmfn!2/2 B!dpoevdups!pg!36!n!mfohui!npwft!bu!sjhiu!bohmf!xjui!b!vojgpsn!wfmpdjuz!pg!6!n0t!jo!b!vojgpsn!nbh. ofujd!gjfme!pg!2!Xc0n3!gmvy!efotjuz/!Efufsnjof!uif!fng!joevdfe!jo!uif!dpoevdups/!Bmtp!gjoe!uif!fng! joevdfe!jo!uif!dpoevdups!xifo!uif!dpoevdups!npwft!bu!bo!bohmf!pg!71°!up!uif!ejsfdujpo!pg!uif!gjfme/

Solution Length of the conductor l = 25 cm = 0.25 m Velocity of the conductor v = 5 m/s Flux density B = 1 Wb/m2 \ emf induced in the conductor when it moves at right angles to the magnetic field = Blv sin 90° = 1 ¥ 0.25 ¥ 5 = 1.25 V When the conductor moves at an angle of 60° to the direction of the field, the emf induced is = Blv sin 60° = 1 ¥ 0.25 ¥ 5 ¥

3 = 1.08 V. 2

! Qspcmfn!2/3 B!dpoevdups!jt!spubujoh!jo!b!nbhofujd!gjfme!pg!1/6!Xc!qspevdfe!cz!3!qpmft!xjui!b!tqffe!pg!2611!sqn/! Dbmdvmbuf!uif!bwfsbhf!fng!joevdfe!jo!uif!dpoevdups/

Solution Average emf induced = N

df dt

Flux cut by the conductor in one revolution is 0.5 Wb Time taken to complete 1 revolution =

60 1 s = s = 0.04 s 1500 25

!

2/39

Fmfdusjdbm!Nbdijoft

\ d f = 0.5 Wb and dt = 0.04 s Hence, average emf induced = 1 ¥

0.5 = 12.5 V. 0.04

! Qspcmfn!2/4 B!dpjm!pg!611!W!sftjtubodf!jt!mzjoh!jo!b!nbhofujd!gjfme!pg!2/6!n!Xc/!Gjoe!uif!bwfsbhf!fng!joevdfe!jo! uif!dpjm!boe!uif!dvssfou!jo!uif!dpjm!jg!uif!dpjm!jt!npwfe!gspn!uif!hjwfo!gjfme!up!b!gjfme!pg!1/6!n!Xc!jo! 1/36/!Uif!ovncfs!pg!uvsot!jo!uif!dpjm!jt!611/

Solution Induced emf = N

df dt df = (1.5 – 0.5)m Wb = 1 mWb = 1 ¥ 10–3 Wb dt = 0.2 N = 500

1 ¥ 10 -3 = 2.5 V 0.2 2.5 Current in the coil = A = 5 mA. 500 \ induced emf = 500 ¥

! Qspcmfn!2/5 B!dpjm!dbsszjoh!3!B!dvssfou!jt!mzjoh!jo!b!nbhofujd!gjfme!pg!2!Xc0n3!gmvy!efotjuz/!Uif!dpjm!ibt!311! uvsot/!Efufsnjof!uif!gpsdf!fyqfsjfodfe!cz!21!dn!pg!uif!dpjm/

Solution Current I = 2 A Flux density B = 1 Wb/m2 Number of turns N = 200 Length of the coil l = 10 cm = 0.1 m \ force experienced by 10 cm of the coil F = 2 N BI l = 2 ¥ 200 ¥ 1 ¥ 2 ¥ 0.1 = 80 N.

! Qspcmfn!2/6 Efufsnjof!uif!fng!joevdfe!jo!b!dpjm!pg!6!nI!xifo!b!dvssfou!pg!21!B!jt!sfwfstfe!jo!21!njmmjtfdpoet/

Solution Self-inductance in the coil L = 5 mH = 5 ¥ 10–3 H

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

Change of current Time taken \ emf induced

2/3:

d I = {10 – (–10)} A = 20A dt = 10 ms = 0.01 s dI 20 = 5 ¥ 10–3 ¥ V dt 0.01 = 10 V.

eind = L

! Qspcmfn!2/7 B!dvssfou!pg!31!B!xifo!gmpxjoh!uispvhi!b!dpjm!pg!611!uvsot!hfofsbuft!b!gmvy!pg!1/6!nX/!Efufsnjof! uif!joevdubodf!pg!uif!dpjm/

Solution Current I = 20 A Number of turns N = 500 Flux f = 0.5 ¥ 10–3 Wb \ inductance of the coil

Nf I 500 ¥ 0.5 ¥ 10 -3 = H 20 = 0.0125 H = 12.5 mH.

L=

! Qspcmfn!2/8 B!tuffm!sjoh!ibwjoh!b!nfbo!djsdvngfsfodf!pg!61!dn!boe!21!dn3!dsptt.tfdujpobm!bsfb!ibt!b!dpjm!pg! 411!uvsot!xpvoe!vojgpsnmz!bspvoe!ju/!Efufsnjof!uif!sfmvdubodf!pg!uif!sjoh!boe!uif!dvssfou!sfrvjsfe! up!qspevdf!b!gmvy!pg!2!njmmj!Xc!jo!uif!sjoh/!Hjwfo!ms!>!411/

Solution Length of the ring l = 50 cm = 0.5 m Area A = 10 cm2 = 10 ¥ 10–4 m2 Turns N = 300 The reluctance of the ring R=

l 0.5 = -7 mo m r A 4p ¥ 10 ¥ 300 ¥ 10 ¥ 10 -4

= 1.327 ¥ 106 AT/Wb Now, flux f = 1 mWb = 1 ¥ 10–3 Wb

!

2/41

Fmfdusjdbm!Nbdijoft

Mmf = Flux ¥ Reluctance = 1 ¥ 10–3 ¥ 1.327 ¥ 106 = 1327 AT

\

\ the required current

Mmf 1327 = N 300 = 4.42 A.

I=

! Qspcmfn!2/9 Uxp!dpjmt!ibwjoh!211!boe!311!uvsot!bsf!xpvoe!po!b!dmptfe!jspo!djsdvju!pg!4!n!mfohui!xjui!611!dn3! dsptt.tfdujpobm!bsfb/!Efufsnjof!uif!nvuvbm!joevdubodf!cfuxffo!uif!dpjmt/!Hjwfo!ms!pg!jspo!bt!4111/

Solution N1 = 100 N2 = 200 l =3m A = 500 cm2 = 0.05 m2 mr = 3000 mo = 4p ¥ 10–7 H/m Mutual inductance

N1 N 2 Reluctance N1 N 2 = l /mo m r A 100 ¥ 200 ¥ 4p ¥ 10 -7 ¥ 3000 ¥ 0.05 = 3 = 1.256 H.

M=

! Qspcmfn!2/: B!dpjm!pg!261!uvsot!jt!spubufe!bu!2611!sqn!jotjef!b!nbhofujd!gjfme!pg!1/26!Xc0n3!gmvy!efotjuz/!Uif! ejbnfufs!pg!uif!dpjm!jt!41!dn!boe!byjbm!mfohui!jt!36!dn/!Efufsnjof!uif!wpmubhf!joevdfe!jo!uif!dpjm/!Jg! uif!dpjm!dbssjft!b!dvssfou!pg!31!B-!gjoe!uif!gpsdf!po!fbdi!dpoevdups!boe!uif!upsrvf!bdujoh!po!uif!dpjm/

Solution The velocity of the coil v=p¥

30 1500 ¥ = 23.55 m/s 100 60

Voltage induced in the coil = N Blv = 150 ¥ 0.15 ¥ 0.25 ¥ 23.55 = 132.468

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/42

\ force on each conductor = N BIl = 150 ¥ 0.15 ¥ 20 ¥ 0.25 = 112.5 N Torque acting on the coil T = Force ¥ Radius 0.3 = 112.5 ¥ = 16.875 Nm. 2

! Qspcmfn!2/21 B!sfdubohvmbs!jspo!dpsf!jt!tipxo!jo!Gjh/!2/28/!Ju!ibt!b!nfbo!mfohui!pg!nbhofujd!qbui!pg!211!dn-! dsptt!tfdujpo!pg!)3!dn!¥!3!dn*-!sfmbujwf!qfsnfbcjmjuz!pg!2511!boe!bo!bjs!hbq!pg!6!nn!jt!dvu!jo!uif! dpsf/!Uif!uisff!dpjmt!dbssjfe!cz!uif!dpsf!ibwf!ovncfs!pg!uvsot!O2!>!446-!O3!>!711!boe!O4!>!711-! boe!uif!sftqfdujwf!dvssfout!bsf!J2!>!2/7!B-!J3!>!5/1!B!boe!J4!>!4/1!B/!Uif!ejsfdujpot!pg!uif!dvssfout! bsf!bt!tipxo/!Dbmdvmbuf!uif!gmvy!jo!uif!bjs!hbq/ 5 mm Mean length = 100 cm

I2

l1 N1 = 335

N2 = 600

turns

turns

I3

N3 = 600 turns

Gjh/!2/28! B!sfdubohvmbs!jspo!dpsf!)Qspc/!2/21*

Solution The mmf acting in the magnetic circuit (considering current in the clockwise direction) = SNI = –335 ¥ 1.6 + 600 ¥ 4 – 600 ¥ 3 = 64 AT \ where

64 AT = li = Mean length, mr = Relative permeability lg = Air-gap cut length

˘ f È li Í + lg ˙ mo A Î m r ˚

!

2/43

\

Fmfdusjdbm!Nbdijoft

64 =

\ flux,

f -7

-4

È 1 7 -3 ˘ Í1400 + 5 ¥ 10 ˙ = 1.136 ¥ 10 f ˚ Î

4p ¥ 10 ¥ ( 2 ¥ 2) ¥ 10 64 f= = 5.63 ¥ 10–6 Wb 1.136 ¥ 107 = 5.63 mWb.

! Qspcmfn!2/22 Bo!bfspqmbof!xjui!b!31!n!xjohtqsfbe!jt!gmzjoh!bu!361!n0t!qbsbmmfm!up!uif!fbsuiÕt!tvsgbdf!bu!b!qmbof! xifsf!uif!ipsj{poubm!dpnqpofou!pg!uif!fbsuiÕt!nbhofujd!gjfme!jt!3!¥!21Ð6!uftmb!boe!bohmf!pg!ejq!jt! 71°/!Dbmdvmbuf!uif!nbhojuvef!pg!uif!joevdfe!fng!cfuxffo!uif!ujqt!pg!uif!xjoht/

Solution As the aeroplane is flying horizontally parallel to the earth’s surface, the flux linked with it will be due to the vertical component Bv on the earth’s field. Bv = BH tan q = 2 ¥ 10–5 ¥ tan 60°

\

= 2 3 ¥ 10–5 Wb/m2 \ induced emf is

|e| = Bv l.v sin q = 2 3 ¥ 10–5 ¥ 20 ¥ 250 ¥ sin 90°

or

l=

3 V = 0.173 V. 10

! Qspcmfn!2/23 B!sfdubohvmbs!mppq!pg!tjeft!36!dn!¥!21!dn!dbssjft!b!dvssfou!pg!26!B/!Ju!jt!qmbdfe!xjui!jut!mpohfs!tjef! qbsbmmfm!up!b!mpoh!tusbjhiu!dpoevdups!3/1!dn!bqbsu!boe!dbsszjoh!b!dvssfou!bu!36!B/!Gjoe!uif!ofu!gpsdf! po!uif!mppq/!Xibu!xjmm!cf!uif!ejggfsfodf!jo!gpsdf!jg!uif!dvssfou!jo!uif!mppq!cf!sfwfstfe@

Solution Let ABCD be the (length l, width b and current i1), with its longer side AB placed parallel and at a distance d from a long conductor XY, carrying current i2 as shown in Fig. 1.18.

Gjh/!2/29! Sfdubohvmbs!mppq!)Qspc/!2/3*

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/44

The attractive force on the side AB of the loop, due to current i2, is mo i1 ◊ i2 ◊ l ◊ towards XY 2p d Similarly, the (repulsive) force in the side CD of the loop is F1 =

F2 =

mo i1 ◊ i2 ◊ l ◊ away from XY 2p ( d + b)

\ the forces on the sides AD and BC of the loop; being equal, opposite and collinear, cancel each other. \ net force on the loop is F1 – F2 = =

mo È1 1 ˘ i1 ◊ i2 ◊ l Í ˙ 2p Î d ( d + b) ˚ mo i1 i2 ◊ l ◊ b ◊ towards XY 2p d ( d + b)

Substituting the values, i1 = 15 A, i2 = 25 A l = 0.25, b = 0.10 m d = 0.02 m, d + b = 0.12 m mo = 2 ¥ 10–7 NA–2, we get 2p

and

15 ¥ 25 ¥ 0.25 ¥ 0.10 0.02 ¥ 0.12 = 7.8 ¥ 10–4 N.

F1 – F2 = (2 ¥ 10–7) ¥

The net force is directed towards the long conductor. If the current in the loop, or in the long conductor, be reversed, the net force will remain same in magnitude but will then be directed away from the long conductor.

! Qspcmfn!2/24 B!dpjm!pg!711!uvsot!boe!pg!sftjtubodf!pg!31!W!jt!xpvoe!vojgpsnmz!pwfs!b!tuffm!sjoh!pg!41!dn!nfbo! djsdvngfsfodf!boe!:!dn3!dsptt.tfdujpobm!bsfb/!Ju!jt!dpoofdufe!up!b!tvqqmz!pg!31!W!)ed*/!Jg!uif!sfmb. ujwf!qfsnfbcjmjuz!pg!uif!sjoh!jt!2-!711!gjoe!)b*!uif!sfmvdubodf-!)c*!uif!nbhofujd!gjfme!joufotjuz-!)d*!uif! nng-!boe!)e*!uif!gmvy/

Solution Here, N = 600 turns, resistance of the coil is 20 W, l = 30 cm = 0.3 m, A = 9 cm2 = 9 ¥ 10–4 m2, relative permeability mr = 1600 and mo = 4p ¥ 10–7

!

2/45

Fmfdusjdbm!Nbdijoft

(a) Reluctance

R=

l 0.3 = -7 mo m r A 4p ¥ 10 ¥ 1600 ¥ 9 ¥ 10 -4

= 1.657 ¥ 105 At/wb (b) The magnetic field intensity Ê NI ˆ 600 ¥ 1 H=Á ˜ = = 2000 AT Ë l ¯ 0.3 V 20 Ê ˆ ÁË where I = Resistance of the coil = 20 = 1 Amp˜¯ (c) mmf = (NI) = 600 ¥ 1 = 600 AT mmf 600 ¥ 1 600 = (d) Flux (f) = = 5 reluctance 1.657 ¥ 10 1.65 ¥ 105 = 3.62 ¥ 10–3 Wb = 3.62 m Wb.

! Qspcmfn!2/25 B!njme.tuffm!sjoh!ibwjoh!b!dsptt.tfdujpobm!bsfb!pg!511!nn3!boe!b!nfbo!djsdvngfsfodf!pg!511!nn! ibt!b!dpjm!pg!311!uvsot!xpvoe!vojgpsnmz!bspvoe!ju/!\Hjwfo!ms!>!411^/ Efufsnjof ! )b*! Sfmvdubodf!pg!uif!sjoh ! )c*! Dvssfou!sfrvjsfe!up!qspevdf!b!gmvy!pg!911!m!Xc!jo!uif!sjoh

Solution (a) Flux density B in the ring is \

800 ¥ 10 -6 400 ¥ 10 -6

The reluctance R of the ring is

0.4

300 ¥ 4p ¥ 10 -7 ¥ 0.4 ¥ 10 -3 mmf Again we know f = Reluctance mmf 800 ¥ 10–6 = Reluctance mmf = 800 ¥ 10–6 ¥ 2.65 ¥ 106 = 2.122 ¥ 103 AT

\

and

= 2 Wb/m2.

magnetizing current is

mmf 2.122 ¥ 103 = Number of turns 200 = 10.6 A.

= 2.65 ¥ 106 A/Wb

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/46

! Qspcmfn!2/26 B!nbhofujd!djsdvju!ibt!261!uvsot!pg!dpjmt!boe!uif!dsptt.tfdujpobm!bsfb!boe!mfohui!pg!uif!nbhofujd! djsdvju!bsf!6!¥!21Ð5!n3!boe!36!¥!21Ð3!n!sftqfdujwfmz/!Efufsnjof!I!boe!uif!sfmbujwf!qfsnfbcjmjuz!ms! pg!uif!dpsf!xifo!uif!dvssfou!jt!3!B!boe!uif!upubm!gmvy!jt!1/4!¥!21Ð4!Xc/

Solution When I = 2 A, mmf = NI = 150 ¥ 2 = 300 A/T \

H=

NI 300 = 1200 A/m = l 25 ¥ 10 -2

B=

Flux 0.3 ¥ 10 -3 = = 0.6 T Area 5 ¥ 10 -4

m=

B 0.6 = = 500 ¥ 10–6 H/m H 1200

mr =

m 500 ¥ 10 -6 = = 3.9788 ¥ 102 = 397.88. -7 mo 4p ¥ 10

! Qspcmfn!2/27 Bo!bjs.dpsfe!dpjm!ibt!611!uvsot/!Uif!nfbo!mfohui!pg!nbhofujd!gmvy!qbui!jt!61!dn!boe!uif!bsfb!pg! dsptt!tfdujpo!jt!6!¥!21Ð5!n3/!Jg!uif!fydjujoh!dvssfou!jt!6!B-!efufsnjof!)b*!I-!)c*!uif!gmvy!efotjuz-!boe! )d*!uif!gmvy!)f*/

Solution mmf = NI = 500 ¥ 5 = 2500 A Given, l = 50 cm = 0.5 m a = 5 ¥ 10–4 m2 NI 2500 = = 5000 A/m l 0.5 (b) B = Flux density = m.H = mr. m0. H (a) H =

= m0. H = 4p ¥ 10–7 ¥ 5000 = 6.283 ¥ 10–3 T (c) Flux

(f) = B ¥ a = 6.283 ¥ 10–3 ¥ 5 ¥ 10–4 = 3.1415 ¥ 10–6 Wb.

[\mr = 1]

!

2/47

Fmfdusjdbm!Nbdijoft

! Qspcmfn!2/28 Bo!jspo!sjoh!pg!21!dn!nfbo!djsdvngfsfodf!jt!nbef!gspn!b!spvoe!jspo!pg!21Ð4!n3!dsptt!tfdujpo/!Jut! sfmbujwf!qfsnfbcjmjuz!jt!611/!Jg!ju!jt!xpvoe!xjui!361!uvsot-!xibu!dvssfou!xjmm!cf!sfrvjsfe!up!qspevdf! b!gmvy!pg!3!¥!21Ð4!Xc@

Solution The lines of magnetic flux follow the circular path of the iron so that l = 100 cm = 1 m a(area) = 10–3 m2 \

Reluctance R =

1 1 = mo m r a (500 ¥ 4p ¥ 10 -3 ¥ 10 -3 ) = 1.59 ¥ 106 A/Wb

Given flux (f) = 2 ¥ 10–3 Wb H = f . R = 2 ¥ 10–3 ¥ 1.59 ¥ 106 = 3.1847 ¥ 103 AT

\ As we know, H = NI \

l=

H 3.1847 ¥ 103 = N 250 = 12.738 A.

! Qspcmfn!2/29 Bo!jspo!sjoh!pg!221!dn!nfbo!mfohui!xjui!bo!bjs!hbq!pg!2/6!nn!ibt!b!xjoejoh!pg!711!uvsot/!Uif! sfmbujwf!qfsnfbcjmjuz!pg!jspo!jt!711/!Xifo!b!dvssfou!pg!5!B!gmpxt!jo!uif!xjoejoh-!dbmdvmbuf!uif!gmvy! efotjuz!C/!Ep!opu!dpotjefs!gsjohjoh/

Solution Given that li = 110 cm – 0.15 cm = 109.85 cm = 1.0985 m lg = 1.5 mm = 1.5 ¥ 10–3 m. N = 600 turns, mr = 600 (given) l =4A \ flux density

B=

m0 ◊ NI ◊ m r mo NI ◊ l + li lg

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/48

Êm lˆ = m0 NI Á r + ˜ lg ¯ Ë li È 600 ˘ 1 2 + = 4p ¥ 10–7 ¥ 600 ¥ 4 ¥ Í -3 ˙ Wb/m . 1 0985 1 5 10 . ¥ Î ˚ –3 2 2 = 3.0159 ¥ 10 ¥ (5.46 ¥ 10 + 6.666 ¥ 10 ) = 3.657 Wb/m2.

! Qspcmfn!2/2: Bo!jspo!sjoh!ibt!b!nfbo!djsdvngfsfodf!pg!91!dn-!boe!ibwjoh!b!dsptt.tfdujpobm!bsfb!pg!6!dn3!boe!b! dpjm!pg!261!uvsot/!Vtjoh!uif!gpmmpxjoh!ebub-!dbmdvmbuf!uif!fyjtujoh!dvssfou!gps!b!gmvy!pg!7/5!¥!21Ð5!Xc/! Bmtp!dbmdvmbuf!uif!sfmbujwf!qfsnfbcjmjuz!)ms*/ C)Xc0n3*;!!!!!!1/:!!!!! I)B0n*;!!!!!!!!!371!!!!!

2/2! 561!

2/3! 711!

2/4 931

Solution f = 6.4 ¥ 10–4 Wb

It is given that \ flux density

B=

f 6.4 ¥ 10 -4 = = 1.28 Wb/m2 area 5 ¥ 10 -4

Assuming B-H curve to be linear in the range from 1.2 to 1.3 Wb/m2, 820 – 600 ¥ (1.28 – 1.2) 1.3 – 1.2 = 776 A/m.

H = 600 +

\

mmf = 776 ¥

\

I= We know

\

80 = 620.8 A 100

620.8 = 4.138 A 150.0

B = m . H = m0 . mr . H B 1.28 = mr = = 1312.089. m0 ¥ H 4p ¥ 10 -7 ¥ 776

! Qspcmfn!2/31 Jo!uif!nbhofujd!djsdvju!tipxo!jo!Gjh/!2/2:-!uif!dsptt.tfdujpobm!bsfb!pg!mjnct!R!boe!S!bsf!1/12!n3!boe! 1/13!n3!sftqfdujwfmz!boe!uif!mfohuit!pg!uif!bjs!hbq!2/2!nn!boe!3/2!nn!sftqfdujwfmz-!bsf!dvu!jo!uif!

!

2/49

Fmfdusjdbm!Nbdijoft

mjnct!R!boe!S/!Jg!uif!nbhofujd!nfejvn!dbo!cf!bttvnfe!up!ibwf!jogjojuf!qfsnfbcjmjuz!boe!uif!gmvy!jo! uif!mjnc!jt!2/6!Xc-!dbmdvmbuf!uif!gmvy!jo!uif!mjnc!Q/ P

Q

R 1.1 mm

2.1 mm

Gjh/!2/2:! Nbhofujd!djsdvju!pg!Qspc/!2/31

Solution It is given that Area of cross section of limb Q = 0.01 m2 Area of cross section of limb R = 0.02 m2 Length of air gap = 1.1 mm for the limb Q and 2.1 mm for the limb R Flux in limb Q = 1.5 Wb As because the permeability of the magnetic medium is infinity, reluctance of the given iron path is zero. The electrical equivalent is shown in Fig. 1.20. f f1

R1

f2

R2

Gjh/!2/31! Fmfdusjdbm!frvjwbmfou!djsdvju!pg!Gjh/!2/2:

Now if we assume R1 is the reluctance of air gap of limb Q and R2 is the reluctance of the air gap of limb R respectively. Let f1 is the flux across the air gap of the limb Q and f2 is the flux across the air gap of the limb R. \

R1 ¥ f1 = R2 ¥ f2 l1 l2 ¥ f1 = ¥ f2 m0 ¥ a1 m0 ¥ a2

\

f2 = =

a2 l1 ¥ ¥ f2 a1 l2 0.02 1.1 ¥ ¥ 1.5 0.01 2.1

= 1.5714 Wb

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

\ flux in the limb

2/4:

P = f1 + f2 = 1.5 + 1.5714 = 3.0714 Wb.

! Qspcmfn!2/32 B! sjoh-! nbef! pg! tuffm-! ibt! b! sfdubohvmbs! dsptt.tfdujpobm! bsfb/! Uif! pvufs! ejbnfufs! pg! uif! sjoh! jt! 36!dn!xijmf!uif!joofs!ejbnfufs!jt!31!dn-!uif!uijdloftt!cfjoh!3!dn/!Uif!sjoh!ibt!b!xjoejoh!pg!611! uvsot!boe!xifo!dbsszjoh!b!dvssfou!pg!4!B-!qspevdft!b!gmvy!efotjuz!pg!2/3!U!jo!uif!bjs!hbq!qspevdfe! xifo!uif!sjoh!jt!dvu!up!ibwf!bo!bjs!hbq!pg!2!nn!mfohui!)Gjh/!2/32*/!Gjoe!)b*!uif!nbhofujd!gjfme!joufo. tjuz!pg!uif!tuffm!sjoh!boe!jo!uif!bjs!hbq-!)c*!sfmbujwf!qfsnfbcjmjuz!pg!uif!nbhofujd!nbufsjbm-!)d*!upubm! sfmvdubodf!pg!uif!nbhofujd!djsdvju-!)e*!joevdubodf!pg!uif!dpjm-!boe!)f*!fng!joevdfe!jo!uif!dpjm!xifo! uif!dpjm!dbssjft!b!dvssfou!pg!j)bd*!>!6!tjo!425!u/ Thickness (2 mm) cm .25 11

3A

2 cm

10 cm

1 mm X

N = 500

X 2.5 mm

X X 2.5 cm

Gjh/!2/32! Tuffm!sjoh!pg!Qspc/!2/32

Solution NI = 500 ¥ 3 = 1500 AT Bsteel = Bgap = 1.2 T (a) \

Hgap =

4p ¥ 10 -7

= 9.55 ¥ 105 AT/m

NI = Hgap ¥ Ig + Hcore ¥ Icore,

Since where

1.2

lgap(mean length of gap) = 1 mm = 1 ¥ 10–3 m, 2.5 ˆ Ê lcore = 2p ¥ Á10 + = (2p ¥ 11.25) cm Ë 2 ˜¯ = 2p ¥ 11.25 ¥ 10–2 m.

We can write, i.e.

1500 = 9.55 ¥ 105 ¥ 10–3 + Hcore ¥ 2p ¥ 11.25 ¥ 10–2 Hcore = 771.20 AT/m.

!

2/51

Fmfdusjdbm!Nbdijoft

(b) Also, Hcore = \

mr =

(c) R = R1 + R2 =

Bcore ; mo m r 1.2 -7

4p ¥ 10 ¥ 771.20

= 1238.2.

1 ¥ 10 -3 4p ¥ 10 -7 ¥ 1238.2 ¥ 2 ¥ 2.5 ¥ 10 -4

+

2p ¥ 11.25 ¥ 10 -2 4p ¥ 10 -7 ¥ 1238.2 2 ¥ 2 ¥ 2.5 ¥ 10 -4

= 2.5 ¥ 106 AT/Wb. N f 500 ¥ ( B ¥ a) 500 ¥ (1.2 ¥ 2.5 ¥ 2 ¥ 10 -4 ) = = = 0.1 H. I 3 3

(d)

L=

(e)

E=L

di d d = 0.1 ¥ (5 sin 314 t) = 0.5 sin (314 t) dt dt dt = 157 cos 314 t.

! Qspcmfn!2/33 Bo!jspo!sjoh!pg!djsdvmbs!dsptt!tfdujpo!pg!6!¥!21Ð5!n3!ibt!b!nfbo!djsdvngfsfodf!pg!3!n/!Ju!ibt!b!tbx. dvu!pg!3!¥!21Ð4!n!mfohui!boe!jt!xpvoe!xjui!911!uvsot!pg!xjsf/!Efufsnjof!uif!fydjujoh!dvssfou!xifo! uif!gmvy!jo!uif!bjs!hbq!jt!1/6!¥!21Ð4!Xc/!\Hjwfo;!ms!pg!jspo!>!711!boe!mfblbhf!gbdups!jt!2/3^!Bttvnf! bsfbt!pg!bjs!hbq!boe!jspo!bsf!tbnf/

Solution The flux linking with the iron ring is firon = fair gap ¥ Leakage factor = 0.5 ¥ 10–3 ¥ 1.2 = 0.6 ¥ 10–3 Wb Again we know, Ampere turns required = NI fair gap ¥ lair ˘ È firon ¥ liron + =Í ˙ Î m r ¥ m0 of iron ¥ area m0 of air ¥ area ˚

\

l=

[\ mr for air = 1]

1 È 0.6 ¥ 10 -3 ¥ 2 0.5 ¥ 10 -3 ¥ 2 ¥ 10 -3 ˘ + Í ˙ 800 ÍÎ 4 ¥ p ¥ 10 -7 ¥ 600 ¥ 5 ¥ 10 -4 4p ¥ 10 -7 ¥ 5 ¥ 10 -4 ˙˚

= 5.95 A.

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/52

! Qspcmfn!2/34 Bo!jspo!sjoh!pg!211!dn!nfbo!djsdvngfsfodf!jt!nbef!gspn!spvoe!jspo!pg!21!dn3! dsptt.tfdujpo/!JuÕt! sfmbujwf!qfsnfbcjmjuz!jt!611/!Opx!b!tbx.dvu!pg!3!nn!xjef!ibt!cffo!nbef!po!ju/!Ju!jt!xpvoe!xjui! 311!uvsot/!Efufsnjof!uif!ofx!dvssfou!sfrvjsfe!up!qspevdf!b!gmvy!pg!1/23!¥!21Ð3!Xc!jo!uif!bjs!hbq-! hjwfo!uibu!uif!mfblbhf!gbdups!y!jt!2/35-!boe!uibu!uif!sfmbujwf!qfsnfbcjmjuz!pg!uif!jspo!voefs!uif!ofx! dpoejujpo!jt!461/

Solution Given fair gap lair gap aair gap mair

= 1.2 ¥ 10–3 Wb = 0.2 ¥ 10–2 = 2 ¥ 10–3 m = 10–3 m2 =1

\ reluctance in the air gap is 2 ¥ 10 -3

= 1.325 ¥ 106 AT/Wb ( 4p ¥ 10 -7 ¥ 1 ¥ 1.2 ¥ 10 -3 ) firon path = 1.25 ¥ 1.2 ¥ 10–3 Wb = 1.5 ¥ 10–3 Wb liron path = 0.998 m airon path = 10–3 m2 miron path (new) = 350. \

reluctance in the air gap is =

998 ¥ 10 -3 (350 ¥ 4p ¥ 10 -7 ¥ 1.2 ¥ 10 -3 )

= 1890912.3 AT/Wb = 1.89 ¥ 106 AT/Wb As we cannot add up the value of air gap reluctance and iron path reluctance to get the total reluctance, we therefore calculate in this way. H = Hair gap + Hiron path = (Reluctance of air gap ¥ Flux in this path) + (Reluctance of iron path ¥ Flux in this path) = (1.325 ¥ 106 ¥ 1.2 ¥ 10–3) + (1.89 ¥ 106 ¥ 1.5 ¥ 10–3) = 4.425 ¥ 103 = 4425 \

current (= I) =

4, 425 = 22.125 A. 200

!

2/53

Fmfdusjdbm!Nbdijoft

! Qspcmfn!2/35 B!nbhofujd!djsdvju!tipxo!jo!Gjh/!2/33!jt!dpotusvdufe!pg!xspvhiu!jspo/

Gjh/!2/33! Nbhofujd!djsdvju!pg!Qspc/!2/35 Uif!dsptt!tfdujpo!pg!uif!dfousbm!mjnc!jt!7!dn3-!boe!fbdi!pvufs!mjnc!jt!5!dn3/!Jg!uif!dpjm!jt!xpvoe!xjui! 611!uvsot-!efufsnjof!uif!fydjujoh!dvssfou!sfrvjsfe!up!tfu!vq!b!gmvy!pg!2/1!n!Xc!jo!uif!dfousbm!mjnc/ C.I!dvswft!pg!xspvhiu!jspo!bsf C!)Xc0n3*! I!)BU0n*!

2/36! 711!

! 2/78 ! 3-211

Solution Given that flux (f1) in the central limb = 1.0 ¥ 10–3 Wb Area (a1) of the central limb = 6 ¥ 10–4 m2 l1 = 15 cm = 0.15 m B1 =

f1 1.0 ¥ 10 -3 = = 1.667 Wb/m2 a1 6.0 ¥ 10 -4

\ AT required = H1l1 = 2100 ¥ 0.15 = 315 AT For outer limb flux (f2) =

1 ¥ 1.0 ¥ 10–3 Wb 2

Area (a2) = 4 ¥ 10–4 m2, Length l2 = 25 cm = 0.25 cm \ From B-H curve, H2 = 600 AT/m \ AT required = H2 l2 = 600 ¥ 0.25 = 150 AT

B2 =

1/ 2 ¥ 1.0 ¥ 10 -3 4 ¥ 10

-4

= 1.25 Wb/m2

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/54

Air gap: Bg = 1.25 Wb/m2 lg = 1 ¥ 10–3 m \ AT required =

Bg lg mo

=

1.25 ¥ 1 ¥ 10 -3 4p ¥ 10 -3

= 994.45 AT

\ total AT required = 315 + 150 + 994.45 = 1,459.45 AT \ the exciting current I=

NI 1459.45 = = 2.92 A. N 500

! Qspcmfn!2/36 Bo!jspo!sjoh!nbef!vq!pg!uisff!qbsut-!l2!>!23!dn-!b2!>!7!dn3!21-!b3!>!6!dn3-!l4!>!9!dn!boe!b4!>! 5!dn3/!Ju!jt!tvsspvoefe!cz!b!dpjm!pg!311!uvsot/!Efufsnjof!uif!fydjujoh!dvssfou!sfrvjsfe!up!dsfbuf!b! gmvy!pg!1/6!n!Xc!jo!uif!jspo!sjoh/! \Hjwfo!m2!>!3781-!m3!>!2166-!m4!>!791/^

Solution Total reluctance

R = R1 + R2 + R3 3

=

Âm

a =1

=

=

l3 l1 l2 l = + + m0 m r a1 m0 m r a2 m0 m r a3 0 mr a

1 4p ¥ 10

-7

1 4p ¥ 10 -7

0.12 0.1 0.08 È ˘ + + Í -4 -4 -4 ˙ 1055 ¥ 5 ¥ 10 680 ¥ 4 ¥ 10 ˚ Î 2670 ¥ 6 ¥ 10 [0.074906 + 189573 + 0.294117]

= 4.445 ¥ 105 AT/Wb \

Flux (f) =

\

I=

mmf NI = Reluctance 4.445 ¥ 105 flux ¥ 4.45 ¥ 105 N

!

2/55

Fmfdusjdbm!Nbdijoft

=

0.5 ¥ 10 -3 ¥ 4.45 ¥ 105 200

= 1.11125 A = 1111.25 ¥ 10–3 Amps = 1111.25 mA.

GVOEBNFOUBM!DPODFQUT!PG!HFOFSBUPS!

2/21

BOE!NPUPS!BDUJPO It is seen from Blv and BlI equations that generator and motor actions are based on the physical reactions on conductors situated in magnetic fields. When a relative motion between conductor and field exists, an emf is generated in the conductor and when a conductor carries current and is situated in a magnetic field, a force is exerted on the conductor. Both generator and motor actions take place simultaneously in the winding of a rotating machine. Both generators and motors have current-carrying conductors in a magnetic field. Thus, both torque and speed voltage are produced. Within the winding it is not possible to distinguish between the generator and motor action without finding the direction of power flow. Constructionally, a generator and motor of one category are basically identical and differ only in details necessary for its best operation for intended service. Any generator or motor can be used for energy conversion in either direction. A generator converts mechanical energy into electrical energy. The torque produced, in a generator, is a counter torque opposing rotation. The prime mover must overcome this counter torque. An increase (or decrease) in electrical power output means an increase (or decrease) in counter torque, which finally results in an increase (or decrease) in torque supplied by the prime mover to the generator. A motor converts electrical energy into mechanical energy. The voltage generated in the conductors is a counter or back emf, which opposes the applied voltage. It is through the mechanism of back emf that a motor adjusts its electrical input to meet an increase (or decrease) in mechanical load on the shaft.

2/21/2! Bo!Fmfnfoubsz!Hfofsbups! Consider a coil being rotated in a magnetic field by applying a mechanical torque, Tm as shown in Fig. 1.23. An emf will be induced in the coil due to the change of flux linked by the coil. The direction of induced emf in the coil sides is shown in Fig. 1.23. The direction of induced emf is determined by applying either Lenz’s law or Fleming’s right-hand rule. If the coil ends are connected to an external resistance RL, a current i will flow in the direction shown in Fig. 1.23. The current-carrying coil will now experience a torque Te in the direction shown (as it is known that when a current-carrying conductor is placed in a magnetic field it experiences a force). The torque developed by the current-carrying coil is the electromagnetic torque, Te.

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/56

i

N

e

Te Tf

i

RL

e

Tm w

i

i S

Gjh/!2/34! Bo!fmfnfoubsz!hfofsbups

The electromagnetic torque developed will act in a direction opposite to the applied mechanical torque, Tm, responsible for causing rotation of coil in the magnetic field. If there is no external resistance RL connected across the coil, no current will flow through the circuit and hence no torque will be developed. The value of Tm will then be utilized in overcoming the frictional torque, Tf . When the circuit of the coil is closed through an external resistance, Te will be set up by the coil in a direction opposite to Tm. The value of Tm must, therefore, be increased to overcome Tf as well as Te. Thus, Tm = Te + Tf If Tf is considered negligible, Tm = Te The magnitude of current through the coil is i= where

e RL + r

r = Resistance of the coil RL = Load resistance

Thus, e = iRL + ir Multiplying both sides by i, ei = i2RL + i2r = vi + i2r where

and

ei = Electrical power developed i2RL = Electrical power used in the electrical load = vi v = Terminal voltage = iRL i2r = Power lost in the resistance of the coil.

(1.48)

!

2/57

Fmfdusjdbm!Nbdijoft

The electrical power developed ei = wTe, where Te is the portion of Tm used for conversion into electrical power. The relation ei = wTe can be proved as follows: We can write

ei = 2 B lvi = 2 Bil v v = 2 Fr r = Te w

Thus,

ei = wTe

(1.49)

Of the total mechanical power wTm supplied, wTe is converted into electrical power, the remaining power wTf is wasted as frictional losses. Of the electrical power developed ei, an amount of power i2r is wasted in the winding, the remaining power, vi, is available across the load. This is shown diagrammically in Fig. 1.24. Mechanical power input = wTm

Frictional losses = wTf

wTe = ei = Electrical power developed in the machine

Electrical power output = v. i

Copper loss in the winding = i2 r

Gjh/!2/35! Qpxfs!gmpx!ejbhsbn!pg!bo!fmfnfoubsz!hfofsbups

In a practical generator, instead of one coil, a number of coils connected together are housed inside the slots of a cylinder called the armature so that an emf of the desired magnitude is induced and can be made available for commercial use. The details regarding the various types of generators will be discussed at a later stage.

2/21/3! Bo!Fmfnfoubsz!Npups The same arrangement of Fig. 1.23 is shown in Fig. 1.25. Here, instead of connecting a load across the coil, a source of supply, a battery has been connected across it and the coil is not being rotated by a prime mover. The current flowing through the armature coil will produce a torque on the coil which will cause the coil to turn. For the position of the coil shown in Fig. 1.25, the magnetic field produced by the coil carrying current is at 90° with the main field axis.

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/58

Its north pole will be on the left-hand side and the south pole will be on the right-hand side. Thus, the coil will turn in the anti-clockwise direction by 90°. The two magnetic fields will then align with each other. Therefore, continuous rotation of the coil will not be obtained. Some special arrangement (brush and commutator arrangement) is made to achieve continuous rotation of the rotor which will be discussed in the sections to follow. For the time being, it is assumed that the coil is rotating continuously. The direction of emf induced in the coil for its anti-clockwise direction is shown in Fig. 1.25. It may be noted that in the case of the motor, the directions of induced emf and current through the armature coil are opposite to each other, whereas in the case of the generator they are in the same direction. See Figures 1.25 and 1.23 for comparison.

i

N e i

Te w

V Tm

e

Tf

i

i S

Gjh/!2/36! Bo!fmfnfoubsz!npups

Considering that the coil is rotating continuously at an angular velocity w, the following equations can be derived: V – e = ir where induced emf e is in opposite direction to that of applied voltage V. Multiplying both sides by i, Vi = ei + i2r

or,

Power supplied = Electrical power being available for conversion to mechanical power + Copper loss. Electrical power,

ei = wTe

A portion of wTe is lost to overcome friction. The rest is available for supplying mechanical output, i.e. to carry the mechanical load on the motor. or

Te = Tm + Tf wTe = wTm + wTf

Thus, the power-flow diagram for an electrical motor can be represented as shown in Fig. 1.26.

!

2/59

Fmfdusjdbm!Nbdijoft

Electrical power input = vi

Copper loss in the winding = i2 r

ei = wTe = electrical power converted to mechanical power

Mechanical power output = wTm

Frictional loss = wTf

Gjh/!2/37! Qpxfs!gmpx!ejbhsbn!pg!bo!fmfnfoubsz!npups

The following differences are found between a generator and a motor: 1. In the case of a generator, the directions of e and i are the same, whereas in the case of motor they are opposite. Thus, it may be concluded that a circuit in which e and i are in the same direction acts as a source of electrical energy, whereas a circuit in which e and i are in opposition to each other acts as a consumer of electrical energy. 2. The direction of frictional torque Tf is opposite to the direction of rotation, both in case of a generator and a motor. 3. For a generator, w and Te are in the opposite directions whereas they are in the same direction in the case of a motor.

TUFBEZ.TUBUF!WPMUBHF!FRVBUJPOT!

2/22

1. In dc machines, (1.50) V – E + Ia Ra = 0 (+ sign in case of generator and – sign in case of motor) where V is the terminal voltage, E is the dynamically induced emf and Ra is the armature resistance drop. 2. In case of polyphase ac machines where the magnitude of the flux generally remains constant, V + E + I ◊Z = 0

(1.51)

where V is the vector of the terminal voltage, E is vector of the dynamically induced voltage and I ◊ Z is the internal impedance drop vector of the machine. 3. In case of ac machines where the magnitude of the flux is also periodically changing, V + E + Es + I ◊ Z = 0

(1.52)

V , E and I ◊ Z have the some meaning as in Eq. (1.51) and Es is the vector of transformer emf induced.

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/5:

! Qspcmfn!2/37 B!ed!npups!ibwjoh!bsnbuvsf!djsdvju!sftjtubodf!pg!1/4!W!jt!svoojoh!bu!2111!sqn/!Uif!npups!esbxt!b! dvssfou!pg!51!B!gspn!511!W!tvqqmz/!Efufsnjof!uif!fmfduspnbhofujd!upsrvf!efwfmpqfe!cz!uif!npups/

Solution Emf induced in the rotor E = 400 – (40 ¥ 0.3) = 388 V The developed electromagnetic torque =

EI 388 ¥ 40 = N = 148.28 Nm. 2 p ¥ 1000 m w 60

! Qspcmfn!2/38 B!npups!ibt!611!dpoevdupst!jo!uif!bsnbuvsf-!fbdi!dpoevdups!dbsszjoh!41!B/!Uif!bsnbuvsf!dpsf! ibt!b!ejbnfufs!pg!411!nn!boe!b!mfohui!pg!311!nn/!Jg!uif!gmvy!efotjuz!jt!1/2!Xc0n3-!efufsnjof!uif! upsrvf!boe!uif!nfdibojdbm!qpxfs!efwfmpqfe!jg!uif!tqffe!jt!:11!sqn/

Solution Force on 500 conductors = N B Il = 500 ¥ 0.1 ¥ 30 ¥ 0.2 = 300 N Torque = Force ¥ Radius 0.3 = 300 ¥ 2 = 45 Nm The angular velocity

2p N 2p ¥ 500 rad/s = 60 60 = 52.33 rad/s.

w=

\ mechanical power developed = wT = 52.33 ¥ 45 = 2355 W = 2.355 Kw.

! Qspcmfn!2/39 B!311!W!tivou!npups!ublft!b!dvssfou!pg!26!B/!Uif!bsnbuvsf!sftjtubodf!boe!uif!tivou!gjfme!sftjtubodf! bsf!2!W!boe!611!W!sftqfdujwfmz/!Uif!dpsf!mptt!boe!gsjdujpobm!mptt!bnpvou!up!311!X/!Efufsnjof!uif! pvuqvu!qpxfs!boe!fggjdjfodz!pg!uif!npups!bu!uijt!dpoejujpo/

!

2/61

Fmfdusjdbm!Nbdijoft

Solution Supply voltage V = 200 V Line current IL = 15 A If =

Shunt field current

V 200 = A rf 500

= 0.4 A \ shunt field copper loss

If 2 rf = (0.4)2 ¥ 500 = 80 W

Armature current

Ia = IL – If = 15 – 0.4 = 14.6 A

\ armature copper loss = Ia2 ra = (14.6)2 ¥ 1 = 213.16 W \ total loss = Total copper loss + Core loss and frictional loss = 80 + 213.16 + 200 = 493.16 W Output power = Input – Loss = 200 ¥ 15 – 493.16 = 2506.84 W \

Efficiency =

Output 2506.84 = ¥ 100 % = 83.56 %. Input 200 ¥ 15

! Qspcmfn!2/3: B!tfsjft!ed!hfofsbups!efmjwfst!b!mpbe!dvssfou!pg!61!B!bu!511!W!boe!ibt!bsnbuvsf!boe!tfsjft!gjfme!sf. tjtubodf!pg!1/16!W!boe!1/15!W!sftqfdujwfmz/!Efufsnjof!uif!joevdfe!fng!jo!uif!bsnbuvsf!jg!uif!csvti! dpoubdu!espq!jt!2!W!qfs!csvti/

Solution In a series generator, the load current flows through the armature and the series field \ armature voltage drop = 0.05 ¥ 50 = 2.5 V series field voltage drop = 0.04 ¥ 50 = 2 V There are two brush contacts. Hence, total brush contact drop is 1 ¥ 2 = 2 V Load voltage = 400 V Hence, induced emf in the armature = 400 + 2.5 + 2 + 2 = 406.5 V.

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/62

! Qspcmfn!2/41 W

!

Solution The back emf developed without external resistance Eb1 = V – Ia ra = 400 – 40 ¥ 0.2 = 392 V When external resistance of R W is connected in series with the armature then back emf Eb2 = V – Ia (ra + R) = 400 – 40(0.2 + R) = 392 – 40 R. If N1 and N2 be the speed when the machine operates without and with the external resistance respectively then Eb1 N1 = where N1 = 1000 rpm and N2 = 700 rpm Eb2 N 2 392 1000 = 392 - 40 R 700 R = 2.94 W.

\ \

! Qspcmfn!2/42 !

" W #

$ %

!

$

Solution Speed of motor N1 = 1000 rpm V = 220 V Input power = P = 700 W 700 = 3.18 A 200 Shunt field current = 1 A

Line current of motor I1 = \ armature current

Ia = 3.18 – 1 = 2.18 A 1

%$

!

2/63

Fmfdusjdbm!Nbdijoft

Eb = V – Ia ra = 220 – 2.18 ¥ 0.2 1 = 219.56 V

Hence, back emf

When the machine operates as a generator then Armature current Generated emf

Ia = 3.18 + 1 = 4.18 A 2 Eg = V + Ia ra = 220 + 4.18 ¥ 0.2 2 = 220.836 V

If N2 be the speed of the generator then Eb N 1 219.56 1000 = or, or, N2 = 1006 rpm. = Eg N 2 220.836 N2

! Qspcmfn!2/43 & W

' '

(

)

Solution Ia = 50 A 1

V = 200 V N1 = 1000 rpm ra = 0.1 W T • Ia T2 = 0.7 T1 Ia = 0.7 Ia = 0.7 ¥ 50 = 35 A

Torque As \

2

1

Eb = V – Ia ra = 200 – 50 ¥ 0.1 1

1

1

= 195 V Eb = V – Ia ra = 200 – 35 ¥ 0.1 = 196.5 V 2

\

Eb1 Eb2

=

\

2

2

N1 where N2 is the new speed. N2 Eb 196.5 N2 = 2 N1 = ¥ 1000 = 1008 rpm. Eb1 195

! Qspcmfn!2/44 "

* ,

" -

% .

) $+

% &

,

$ $

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

" -

%

)

$ )

2/64

$

Solution Total number of terms N = 8 ¥ 300 = 2400 Initial flux = 0.04 ¥ 8 = 0.32 Wb Residual flux = 0.002 ¥ 8 = 0.016 Wb \ Change in flux

d f = (0.32 – 0.016) Wb = 0.304 Wb

Time taken to open the field circuit dt = 0.05 s \ average voltage induced across the field terminals =N

df 0.304 = 2400 ¥ V = 14,592 V. dt 0.05

! Qspcmfn!2/45 )

#

'

$

'

& /

+W &

W

Solution Torque developed in a dc motor T=

Ea I a where Eb is the back emf, Ia is the armature current and w is the angular velocity w 2p ¥ 1500 rad/s 60 Eb = V – Ia(ra + rse) = 240 – Ia (0.3 + 0.2) w=

Here,

= 240 – 0.5 Ia \

50 =

( 240 - 0.5 I a ) I a 240 I a - 0.5 I a2 = 2p ¥ 1500 157 60

0.5 I a2 – 240 Ia + 7850 = 0 Ia = 35.3 A.

or, \

! Qspcmfn!2/46 &W

$ 0 )

)

$

.

! + (%

!

2/65

Fmfdusjdbm!Nbdijoft

Solution Armature resistance ra = 0.15 W When speed N1 = 1200 rpm Eb = V – Ia ra = 220 – 60 ¥ 0.15 1 = 211 V Let the initial load be P1 \ final load P2 = 1.3 P1 If Eb and N2 be the final back emf and speed then 2

Eb = 220 – Ia ¥ 0.15 where Ia is the final armature current or, 2

2

Ia = 2

As \ \ or, \

2

220 - Eb2 0.15

P2 = 1.3 P1 Eb Ia = 1.3 Eb Ia = 1.3 ¥ 211 ¥ 60 = 16458 W 2

2

1

1

Ê 220 - Eb2 ˆ Eb2 Á ˜ = 16458 Ë 0.15 ¯ Eb22 – 220 Eb + 2468.7 = 0 Eb = 220 + ( 220) 2 - 4 ¥ 2468.7 2

= 208.14 V \ Speed

N2 =

Eb2 Eb1

¥ N1

208.14 ¥ 1200 rpm 211 = 1184 rpm. =

NBHOFUJD!IZTUFSFTJT!

2/23

Magnetic hysteresis may be defined as the lagging of magnetisation or flux density (B) behind the magnetisation force (H). When an unmagnetized bar of iron is magnetized by placing it in a magnetic field and varying the magnetic field, the flux density B is measured for different values of magnetizing field H, the graph of B vs. H is obtained which is called the B-H curve or magnetization curve. When the magnetizing field is increased from O up to a certain maximum value the corresponding values of flux density are noted. If we plot B vs. H curve, OP is obtained as shown in Fig. 1.27.

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/66

The maximum value of H i.e., Hm is called the saturation value of H and the corresponding flux density is Bm. B

Bm P

A

– Hm D

C

O

F

Hm H

G

E

– Bm B–H curve

Gjh/!2/38 1 2

When H is gradually decreased to O it is seen that B reduces but it is not zero when H = 0. When H is zero, B = OA. Thus, on removing the magnetizing force, H the iron bar is not completely demagnetized. This value of B(= OA) is known as the residual magnetic flux density. When a magnetic field H is reversed then B is reduced to zero at the point C where H = OC. This value of H required to make the flux density zero is known as coercive force. If now H is further increased in the negative direction the iron bar reaches a state of magnetic saturation at the point D. Now if H is increased in the positive direction till its saturation value Hm is reached it is observed that it follows the curve EGFP. The closed-loop PACEGFP which is obtained when the iron bar is taken through one complete cycle of magnetization is called hysteresis loop.

2/23/2! Iztufsftjt!Mptt! This loss occurs due to the B-H magnetization curve which swings to positive and negative maximum Bm before returning to zero. Ideally, the energy absorbed during the positive swing should be returned to the source during reversal of the magnetizing cycle. But in reality there is only a partial return to source, the rest being dissipated as heat. Figure 1.28 represents the hysteresis loop obtained from a steel ring of mean circumference (l) metres and cross-sectional area (a) square metres. Let (N) be the number of turns on the magnetizing coil.

!

2/67

Fmfdusjdbm!Nbdijoft

Gjh/!2/39!

Let (dB) = Increase of flux density when the magnetic field intensity is increased by a very small N amount dH (say) in dt seconds, and i = Current in amperes corresponding to om, i.e., om = i . l N Instantaneous emf induced in the winding is a ¥ dB ¥ V. dt and component of applied voltage to neutralize this emf equals dB ˘ È Ían ¥ dt ˙ V ˚ Î Therefore, instantaneous power supplied to the magnetic field is dB ˆ Ê ÁË i ¥ an ¥ dt ˜¯ W and energy supplied to the magnetic field in time dt second is (i ¥ an ¥ dB)J Since,

om =

Ni l

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

i =l¥

2/68

om N

Hence, energy supplied to magnetic field in time dt is l¥

om ¥ an ¥ dB J = (om ¥ dB ¥ lA)J N = Area of shaded strip, J/m3

Thus, energy supplied to the magnetic field when H is increased from zero to oa is equal to area f g b Bm f J/m3. Similarly, energy returned from the magnetic field when H is reduced from oa to zero is area b Bm cb J/m3. Then net energy absorbed by the magnetic field is Area f g b Bm f J/m3 Hence, hysteresis loss for a complete cycle is area of e f bc e joule per m3. If we define hysteresis loss as Ph, \

Ph = v . f . (Kh ¥ Bmn )W

(1.53)

where v is volume of core material and f is the frequency of variation of H in Hz. The value of n is between 1.5 and 2.5 (1.5 £ n £ 2.5) but mostly it is 1.6. Bm is the maximum flux density in Tesla and Kh is a constant and n is the exponent which depends on the material. The constant Kh (depending on the chemical properties of the material and the heat treatment and mechanical treatment the material has been subjected to) may have value as low as 5 ¥ 10–7 for permalloys and as high as 6 ¥ 10–5 for cast iron. However Kh for electrical sheet steel is generally 4 ¥ 10–5. The exponent n has been found by Steinmetz as 1.6 and does not have any theoretical basis. This value suits most material at flux densities generally not exceeding 1 Wb/m2. However, for higher value of flux densities, the value may be as great as 2.5.

FEEZ!DVSSFOU!2/24 Eddy currents are closed loops of induced current circulating in planes perpendicular to the magnetic flux. They flow parallel to the coil’s winding and the flow is limited to the area of the inducing magnetic field. Eddy currents concentrate near the surface adjacent to the excitation coil and their strength decreases with distance from the coil. A magnetic ckt is shown in Fig. 1.29 where a rectangular core of magnetic material is wrapped by an exciting coil.

!

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Fmfdusjdbm!Nbdijoft

Gjh/!2/3:

When the coil is excited from a sinusoidal source, an emf will be induced in an arbitrarily chosen closed path abcd. Hence, a circulating current will flow. The emf induced in the closed path is dependent on the flux enclosed by the path, i.e. fm = Bm ¥ area of the loop abcd. Neglecting the reactance, the magnitude of the eddy current is limited by the path resistance. It causes power loss and heating in the core. To evaluate the total eddy current loss, we have to add all the power losses of different eddy current paths covering the whole cross section. The power loss due to eddy current causes heating of the core and reduces the efficiency of the machine or apparatus. Hence, eddy current loss should be minimized. If the cross-sectional path of the eddy current is reduced then eddy voltage induced will be reduced as eddy voltage induced is proportional to the area and hence eddy current loss will be less. This is achieved by using several thin electrically insulated plates stacked together to form the core instead of a solid block of iron. These electrically insulated plates are called laminations.

2/24/2! Fyqsfttjpo!gps!Feez.Dvssfou!Mptt! Let us consider a thin magnetic plate of length L, height h and thickness t such that t is very small compared to both L and h. Also let us assume that sinusoidally varying field Bm sin wt exists perpendicular to the rectangular area formed by t and h as shown in Fig. 1.30. Let us now consider a small elemental rectangular closed path of thickness dx and at a distance x from the origin. The closed path may be considered a single coil through which the flux is crossing. Hence, there will be a voltage induced in it. The area of the closed path = 2hx Flux crossing the loop = Bm sin wt ¥ 2hx Rms voltage induced in the loop, E = 2p f Bm 2hx Resistance of the path through which eddy current flows is r( 2h + 4 x ) R= Ldx

Gjh/!2/41

Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft

2/6:

Now the power loss in the elementary strip dP =

E2 E 2 Ldx = R r( 2h + 4 x )

=

E 2 L dx 2r h

=

2p 2 f 2 Bm2 4 h2 x 2 Ldx 4p 2 f 2 Bm2 x 2 Lh = dx 2rh r

(\ x !5!W! S3!>!1/15!W! Y2!>!6!W! Y3!>!1/16!W ! )b*! Efufsnjof!upubm!mfblbhf!jnqfebodf!sfgfssfe!up!uif!IW!tjef!boe!MW!tjef/ ! )c*! Gjoe!IW!ufsnjobm!wpmubhf!boe!qfsdfoubhf!wpmubhf!sfhvmbujpo!xifo!uif!usbotgpsnfs!hjwft!jut! sbufe!lWB!bu!b!q/g/!pg!1/9!mbhhjoh!boe!1/9!mfbejoh!up!mpbet!bu!sbufe!wpmubhf/ ! )d*! Jg!uif!dpsf!mptt!jt!91!X-!gjoe!fggjdjfodz!gps!cpui!mbhhjoh!boe!mfbejoh!q/g/ ! )e*! Jg!uif!mpbe!jt!tipsu!djsdvjufe!gjoe!uif!dvssfou!jo!IW!tjeft/

Solution Transformer ratio =

2200 = 10 220

(a) Z1 (HV side) = (4 + 0.04 ¥ 102) + j (5 + 0.05 ¥ 102) Ê 4 ˆ Ê 5 ˆ Z2 (LV side) = Á 2 + 0.04˜ + j Á 2 + 0.05˜ Ë 10 ¯ Ë 10 ¯ = 0.08 + j 0.1

!

4/:3

Fmfdusjdbm!Nbdijoft

(b) Load current =

20, 000 = 9.09 A 2200 V1 = 2200 + 9.09 (8 ¥ 0.8 + 10 ¥ 0.6) = 2312.72 V

\

% regulation =

2312.72 - 2200 ¥ 100 = 5.12% 2200

For leading p.f., V1 = 2200 + 9.09 [8 ¥ 0.8 – 10 ¥ 0.6] = 2203.64 V Percentage regulation = 0.16% (c) Core loss Pi = 80 W Copper loss Pc = (9.09)2 ¥ 8 = 661.02 W Efficiency =

(d) Isc (HV)

20 ¥ 0.8 = 0.9557 p.u. = 95.57% 80 + 661.02 20 ¥ 0.8 + 1000

2200 = 171.79 A. 8 + j10

! Qspcmfn!4/98 B!211!lWB-!71110711!W-!61!I{-!tjohmf.qibtf!usbotgpsnfs!ibt!uif!gpmmpxjoh!xjoejoh!sftjtubodft!boe! mfblbhf!sfbdubodft; ! !

S2! >!3/4!W!!!!S3!>!1/13!W Y2! >!5!W!!!!!!!Y3!>!1/6!W

Dbmdvmbuf!uif!gpmmpxjoh;! ! )b*! Frvjwbmfou!sftjtubodf-!mfblbhf!sfbdubodf!boe!jnqfebodf!sfgfssfe!up!ijhi.wpmubhf!tjef ! )c*! Frvjwbmfou!sftjtubodf-!mfblbhf!sfbdubodf!boe!jnqfebodf!sfgfssfe!up!mpx.wpmubhf!tjef ! )d*! Upubm!Dv.mptt!pg!uif!usbotgpsnfs!bu!gvmm.mpbe!dpoejujpo

Solution R1 = 2.3 W X1 = 4 W Turns ratio =

R2 = 0.02 W X2 = 0.5 W

6000 = 10 600

(a) Equivalent resistance referred to HV side = (2.3 + 102 ¥ 0.02) W = 4.3 W

Tjohmf.qibtf!Usbotgpsnfst

4/:4

Equivalent reactance referred to HV side = (4 + 102 ¥ 0.5) = 54 W Equivalent impedance referred to HV side =

4.32 + 54 2 = 54.17 W

(b) Equivalent resistance referred to LV side 2.3 ˆ Ê = Á 0.02 + 2 ˜ W Ë 10 ¯ = 0.043 W Equivalent reactance referred to HV side 4 ˆ Ê = Á 0.5 + 2 ˜ Ë 10 ¯ = 0.54 W Equivalent impedance referred to HV side = 0.0432 + 0.54 2 = 0.5417 W 100 ¥ 103 6000 = 16.67 A

(c) Primary current (I1) =

Total Cu loss = (16.67)2 ¥ 4.3 = 1194.9 W.

! Qspcmfn!4/99 Uif!iztufsftjt!boe!feez.dvssfou!mptt!pg!b!usbotgpsnfs!bsf!361!X!boe!211!X!sftqfdujwfmz/!Jg!uif!tvq. qmz!wpmubhf!jt!611!W!bu!61!I{-!gjoe!uif!iztufsftjt!mptt!boe!feez.dvssfou!mptt!jg!uif!tvqqmz!wpmubhf!jt! nbef!epvcmf!bu!211!I{!gsfrvfodz/

Solution Hysteresis loss at 500 V, 50 Hz = 250 W Eddy-current loss at 500 V, 50 Hz = 100 W Induced emf (E) = 4.44 f fm N = 4.44 f Bm AN E  Bm f fi

Bm 

E f

!

4/:5

Fmfdusjdbm!Nbdijoft

Bm = a For the first case,

\

E f

[a in the constant]

Bm1 = 10 a

È E 500 ˘ ÍAs f = 50 = 10 ˙ Î ˚

Bm2 = 10 a

È E 1000 ˘ Í f = 100 = 10 ˙ Î ˚

Bm1 = Bm2 Ph = Cf and Pc = Df 2 Ph 250 = =5 50 f 100 D= = 0.04 50 ¥ 50 C=

Now, f = 100 Hz Ph = 5 ¥ 100 = 500 W Pe = 0.04 ¥ 1002 = 400 W.

! Qspcmfn!4/9: Ijhi.wpmubhf!xjoejoh!pg!b!tjohmf.qibtf!61!I{!usbotgpsnfs!jt!dpoofdufe!xjui!311!W!tvqqmz!boe!MW! xjoejoh!jt!tipsu!djsdvjufe/!Uif!gpmmpxjoh!ebub!bsf!hjwfo; ! !

S2! >!2/5!W!!!!!Y2!>!7!W S3! >!1/17!W!!!Y3!>!1/15!W

Uvsot!sbujp!>!5 Efufsnjof!dvssfou!jo!MW!xjoejoh-!q/g/!boe!dpqqfs!mptt!pg!uif!usbotgpsnfs/

Solution Ro1 = 1.4 + 42 ¥ 0.06 = 2.36 W Xo1 = 6 + 42 ¥ 0.04 = 6.64 W Zo1 = 2.36 2 + 6.64 2 = 7.047 W Isc =

200 A = 28.38 A 7.047

I1 = I¢2 = 28.38 A [Neglecting Io] I2 = 28.38 ¥ 4 = 113.52 A Total Cu loss = (28.38)2 ¥ 2.36 = 1900.8 W

Tjohmf.qibtf!Usbotgpsnfst

4/:6

1900.8 200 ¥ 28.38 = 0.335 (lagging).

Short-circuit power factor =

! Qspcmfn!4/:1 B!211!lWB!tjohmf.qibtf!tufq.epxo!usbotgpsnfs!ibt!b!gvmm.mpbe!tfdpoebsz!dvssfou!pg!511!B!boe!uif! upubm!sftjtubodf!sfgfssfe!up!uif!tfdpoebsz!jt!1/116!W/!Gjoe!uif!fggjdjfodz!pg!uif!usbotgpsnfs!bu!gvmm. 3 mpbe!boe! !uif!gvmm!mpbe!bu!)b*!vojuz!qpxfs!gbdups-!boe!)c*!1/9!q/g/!jg!jspo!mptt!pg!uif!usbotgpsnfs!jt! 311!X/! 4

Solution At full-load, I2(fL) = 400 A \ full-load Cu loss = 4002 ¥ 0.005 = 800 W Total losses = (200 + 800) W = 1000 W Output at unity p.f. = 100 ¥ 103 ¥ 1 = 105 W Efficiency =

105 105 + 1000

¥ 100% = 99%

Output at 0.8 p.f. = 100 ¥ 103 ¥ 0.8 W = 80000 W Efficiency =

80000 ¥ 100% 80000 + 1000

= 98.765% (Am) th

At

3 load 4

2

Ê 3ˆ Cu loss = Á ˜ ¥ 800 W = 450 W Ë 4¯ Output at unity p.f. = Efficiency =

3 ¥ 105 W = 75000 W 4

75000 ¥ 100% = 99.14% 75000 + 450 + 200

Output at 0.8 p.f. = 8000 ¥ Efficiency =

3 = 60000 W 4

60000 ¥ 100% = 98.9%. 60000 + 200 + 450

!

4/:7

Fmfdusjdbm!Nbdijoft

! Qspcmfn!4/:2 B!tjohmf.qibtf-!71110711!W!usbotgpsnfs!ibt!qsjnbsz!boe!tfdpoebsz!sftjtubodft!1/36!boe!1/129!W! sftqfdujwfmz/!Jg!jspo!mptt!pg!uif!usbotgpsnfs!jt!311!X-!dbmdvmbuf!uif!tfdpoebsz!dvssfou!bu!xijdi!nbyj. nvn!fggjdjfodz!pddvst/!Bmtp!dbmdvmbuf!uif!nbyjnvn!fggjdjfodz!bu!1/:!q/g/!mbhhjoh/

Solution 6000 = 10 600 R1 = 0.25 W R2 = 0.018 W

Turns ratio =

Ro2 = 0.018 +

0.25 10 2

= 0.021 W

At maximum efficiency, Copper-loss = Iron loss I22 Ro2 = 200 \

I2 =

200 = 97.59 A 0.021

Output at maximum efficiency = 600 ¥ 97.59 ¥ 0.9 W = 52698.6 W = 52.69 kW Total loss = 2 ¥ 200 = 400 W = 0.4 kW Maximum efficiency =

52.69 ¥ 100% = 99.25%. 52.69 + 0.4

! Qspcmfn!4/:3 B!tjohmf.qibtf-!261!lWB!usbotgpsnfs!ibt!bo!fggjdjfodz!pg!:9&!bu!gvmm.mpbe!po!1/9!q/g/!boe!po!ibmg! mpbe!bu!1/9!q/g/!mbhhjoh/!Gjoe!uif!gpmmpxjoh;! ! ! ! !

)b*! )c*! )d*! )e*!

Jspo!mptt Dpqqfs!mptt!bu!gvmm.mpbe Mpbe!lWB!bu!xijdi!nbyjnvn!fggjdjfodz!pddvst Nbyjnvn!fggjdjfodz!pg!uif!usbotgpsnfs!bu!1/9!q/g/!mbhhjoh

Solution At full-load, 150 ¥ 103 ¥ 0.8 150 ¥ 103 ¥ 0.8 + Pi + Pcu

= 0.98 [Pi = iron loss and Pcu = Cu loss at full-load]

Tjohmf.qibtf!Usbotgpsnfst

4/:8

Ê 1 ˆ Pi + Pcu = 150 ¥ 103 ¥ 0.8 Á - 1˜ Ë 0.98 ¯ = 2448.98 W

fi

(i)

At half-load, 1 ¥ 150 ¥ 103 ¥ 0.8 2

= 0.98

1 1 ¥ 150 ¥ 103 ¥ 0.8 + Pi + Pcu 2 4 fi

Pi +

1 Pcu = 75 ¥ 103 ¥ 0.8 4 = 1224.49 W

Ê 1 ˆ ÁË 0.98 - 1˜¯ (ii)

Solving Eqs (i) and (ii), Pi = 816.33 W

Pcu = 1632.7 W

Say, x be the fraction of full-load to get maximum efficiency of the transformer. x=

Pi = Pcu

816.33 = 0.7071 1632.7

\ maximum efficiency occurs at 70.71% of full-load. 0.7071 ¥ 150 ¥ 103 ¥ 0.8

¥ 100% 0.7071 ¥ 150 ¥ 103 ¥ 0.8 + 2 ¥ 816.33 [As at maximum efficiency, Pi = Pcu] = 98.11%.

Maximum efficiency =

! Qspcmfn!4/:4 Gjoe!pvu!uif!mpbe!)bu!nbyjnvn!fggjdjfodz*!boe!nbyjnvn!fggjdjfodz!pg!b!31!lWB-!tjohmf.qibtf!usbot. gpsnfs! bu! vojuz! q/g/! Jspo! mptt! boe! gvmm.mpbe! dpqqfs.mptt! bsf! 361! X! boe! 611! X! sftqfdujwfmz/! Jg! uif! nbyjnvn!fggjdjfodz!pddvst!bu!96&!pg!gvmm.mpbe-!dbmdvmbuf!uif!ofx!dpsf!mptt!boe!gvmm.mpbe!Dv!mptt!bt. tvnjoh!uibu!upubm!gvmm.mpbe!mptt!jt!b!dpotubou/

Solution Say, x is the fraction of full-load when maximum efficiency occurs. \

x=

Pi 250 = = 0.707 500 Pcu

\ maximum efficiency occurs at 70.7% of full-load.

[Pi = iron loss Pcu = Cu loss]

!

4/:9

Fmfdusjdbm!Nbdijoft

Output at unity p.f. = 0.707 ¥ 20 ¥ 1 = 14.14 kW Total losses = 2 ¥ 250 = 500 W = 0.5 kW [As at maximum efficiency Pi = Pcu] 14.14 ¥ 100% 14.14 + 0.5 = 96.58%

\ maximum efficiency =

When maximum efficiency occurs at 85% of full-load then, [As total loss contained] P¢i + P¢cu = Pi + Pcu = 750 W [Pi¢ and Pcu¢ new iron loss and Cu loss] Pi ¢ = 0.85 Pcu¢

Again,

Pi ¢ = 0.7225 Pcu¢

fi

Pi ¢+ Pcu¢ = 1.7225 Pcu¢ 750 = 1.7225 Pcu¢

fi fi fi

Pcu ¢ = 435.41 W Pi¢ = 314.58 W.

! Qspcmfn!4/:5 B!21!lWB-!5510311W-!61!I{!tjohmf.qibtf!usbotgpsnfs!sfrvjsft!211!W!po!IW!tjef!up!djsdvmbuf!gvmm. mpbe!dvssfou!xjui!MW!tipsu!djsdvjufe/!Uif!qpxfs!joqvu!jt!311!X/!Gjoe!uif!nbyjnvn!qpttjcmf!wpmubhf! sfhvmbujpo!boe!q/g/!bu!xijdi!ju!pddvst/!Bmtp!gjoe!uif!tfdpoebsz!ufsnjobm!wpmubhf!voefs!uijt!dpoejujpo/

Solution 10 ¥ 103 A = 22.73 A 440 Power input = 200 W = (22.73)2 Re1 Full-load HV current =

Hence, Also,

Re1 = 0.387 W 100 Ze1 = W = 4.4 W, hence Xe1 = Ze21 - Re21 = 4.38 W 22.73

Voltage regulation is (Rp.u. cos q2 + Rp.u. sin q2).

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4/::

Maximum voltage regulation occurs when d (Rp.u cos q2 + Xp.u sin q2) = 0 dq 2 or, or, Now and Hence,

–Rp.u. sin q2 + Xp.u. cos q2 = 0 X p.u. tan q2 = Rp.u. 0.387 ¥ 22.73 Rp.u. = = 0.02 440 4.38 ¥ 22.73 = 0.226 Xp.u. = 440 0.226 = 11.31 or, cos q2 is 0.088 tan q2 = 0.02

i.e., power factor is 0.088 lagging. Hence, maximum voltage regulation = 0.02 ¥ 0.088 + 0.226 ¥ 0.996 = 0.227 p.u. or, 22.7%. If (V2) be the terminal voltage then 1– or,

V2 = 0. 227 E2 V2 = (1 – 0.227) ¥ 200 = 154.6 V.

! Qspcmfn!4/:6 B!211!lWB-!211110311!W-!61!I{!tjohmf.qibtf!usbotgpsnfs!offet!411!W!po!IW!tjef!up!djsdvmbuf!gvmm. mpbe!dvssfou!xjui!MW!xjoejoh!tipsu.djsdvjufe/!Uif!qpxfs!joqvu!jt!2111!X/!Gjoe!nbyjnvn!qpttjcmf! sfhvmbujpo!boe!qpxfs!gbdups!bu!xijdi!ju!pddvst/!Xibu!jt!W3!voefs!uijt!dpoejujpo@

Solution 100 ¥ 103 = 10 A 10000 Taking 100 kVA and 10000 V as base values for high-voltage side, 10000 Base impedance = = 1000 W 10 10000 Resistance as referred to HV side = = 10 W 10 ¥ 10 300 = 30 W Impedance as referred to HV side = 10 Full-load current on HV side =

Reactance as referred to HV side = 30 2 - 10 2 = 28.28 W

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10 = 0.01 p.u. 1000 28.28 = 0.02828 p.u. = 1000

Rp.u. = Xp.u. Regulation will be maximum, when

d (Regulation ) =0 dq 2 Now, Regulation = Rp.u. cos q2 + Xp.u. sin q2 \ fi

d (Regulation ) = –Rp.u. sin q2 + Xp.u. cos q2 = 0 dq 2 0.02828 tan q2 = = 2.828 0.01 Power factor = cos q2 = 0.333 lagging sin q2 = 0.943 Regulation = (0.01) (0.333) + (0.02828) (0.943) = 0.03 p.u. or 3% 200 V2 = = 194.17 V. 1.03

! Qspcmfn!4/:7 B!tjohmf.qibtf!usbotgpsnfs!tvqqmjft!b!mpbe!pg!361!B!bu!1/9!q/g/!xifsfbt!joqvu!wpmubhf!pg!uif!usbot. gpsnfs!jt!6111!W-!boe!uvsot!sbujp!jt!21;2/!Ofhmfdujoh!op.mpbe!dvssfou-!dbmdvmbuf!tfdpoebsz!ufsnjobm! wpmubhf!boe!pvuqvu!qpxfs/!Hjwfo![2!>!2/7!,!25/4!W!boe![3!>!1/12:!,!k1/159!W

Solution Secondary terminal voltage (V2 –0°) is taken as reference

\

E2 = V2 –0° + I2 Z2 = V2 + 250 ––36.87° (0.019 + j 0.048) [as cos f = 0.8 lagging so f = 36.87°] = V2 + 11 + j 6.75 E1 = 10 fi E1 = 10 E2 E2 E1 = 10 V2 + 110 + j 67.5 V1 = E1 + I1 Z1 = 10 V2 + 110 + j 67.5 + I1 (1.6 + j 4.3)

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250 – - 36.87∞ = 25 (0.8 – j 0.6) 10 V1 = 10 V2 + 110 + j 67.5 + 25(0.8 – j 0.6) (1.6 + j 4.3) = 10 V2 + 206.5 + j 129.5 I1 =

\

| V1 |2 = (10 V2 + 206.5)2 + (129.5)2 | V1 | = 5000 | 5000 |2 = (10 V2 + 206.5)2 + (129.5)2 V2 = 4791.82 V

fi

Output power = 4791.82 ¥ 250 ¥ 0.8 kW = 95.836 kW.

! Qspcmfn!4/:8 Uif!sftjtubodf!boe!mfblbhf!sfbdubodf!pg!b!23!lWB-!61!I{!33110331!W!ejtusjcvujpo!usbotgpsnfs!bsf! bt!gpmmpxt;! s2!>!3!W! s3!>!1/13!W! y2!>!5!W!boe!y3!>!1/15!W Fbdi!rvboujuz!jt!sfgfssfe!up!jut!pxo!tjef!pg!usbotgpsnfs!)tvggjy!Ô2Õ!tuboet!gps!IW!boe!Ô3Õ!gps!MW*/ ! )b*! Gjoe!uif!upubm!mfblbhf!jnqfebodf!sfgfssfe!up!)j*!uif!IW!tjef-!boe!)jj*!uif!MW!tjef/ ! )c*! Dpotjefs!uif!usbotgpsnfs!up!hjwf!jut!sbufe!lWB!bu!b!q/g/!pg!1/9!mbhhjoh!up!b!mpbe!bu!sbufe!wpmu. bhf/!Gjoe!uif!IW!ufsnjobm!wpmubhf!boe!qfsdfoubhf!wpmubhf!sfhvmbujpo/ ! )d*! Sfqfbu!)c*!gps!b!q/g/!pg!1/9!mfbejoh/ ! )e*! Dpotjefs!uif!dpsf!mptt!up!cf!81!X/!Gjoe!uif!fggjdjfodz!voefs!uif!dpoejujpot!pg!qbsu!)c*/!Xjmm!ju! cf!ejggfsfou!gps!uif!dpoejujpot!voefs!qbsu!)d*@ ! )f*! Jg!uif!mpbe!jo!qbsu!)c*!hfut!tipsu.djsdvjufe-!gjoe!uif!tufbez.tubuf!dvssfou!jo!uif!IW!mjoft-!bttvn. joh!uibu!uif!wpmubhf!bqqmjfe!up!uif!usbotgpsnfst!sfnbjot!vodibohfe/

Solution a=

2200 = 10 220

(a) ZL1 = (2 + 0.2 ¥ 100) + j (4 + 0.04 ¥ 100) = 4 + j 8 Ê 2 ˆ Ê 4 ˆ + 0.02˜ + j Á + 0.04˜ = 0.04 + j 0.08 ZL2 = Á Ë 100 ¯ Ë 100 ¯ 12000 (b) I¢L = = 5.45 Amp 2200 VH = 2200 + 5.45 (4 ¥ 0.8 – 8 ¥ 0.6) = 2243.6 V % Regulation =

43.6 ¥ 100 = 1.98 2200

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Fmfdusjdbm!Nbdijoft

(c) VH = 2200 + 5.45 (4 ¥ 0.8 – 8 ¥ 0.6) = 2191.28 % Regulation = –

8.72 ¥ 100 = – 0.4 2200

(d) Pi = 70 W, Pc = (5.45)2 ¥ 4 = 118.81 W h=

(e) Isc (HV) =

10 ¥ 0.8 ¥ 100 = 97.13% 70 + 118.81 10 ¥ 0.8 + 1000

2200 = 245.97 A. 14 + j 81

! Qspcmfn!4/:9 Uif!ebub!pg!b!tjohmf.qibtf!usbotgpsnfs!bsf!hjwfo!cfmpx; Uvsot!sbujp!>!21-!S2!>!6!W-!Y2!>!9!W-!S3!>!1/3!W-!Y3!>!1/5!W-!op.mpbe!dvssfou!)2/5!B*!mfbet!uif!gmvy!cz! bo!bohmf!pg!41¡/!B!mpbe!pg!1/:!mbhhjoh!qpxfs!gbdups!jt!dpoofdufe!up!uif!tfdpoebsz!tp!uibu!uif!mpbe! dvssfou!jt!211!B!bu!b!ufsnjobm!wpmubhf!pg!461!W/!Dbmdvmbuf!uif!gpmmpxjoh!xjui!uif!ifmq!pg!b!qibtps! ejbhsbn; ! )b*! Qsjnbsz!bqqmjfe!wpmubhf ! )c*! Uif!qsjnbsz!qpxfs!gbdups ! )d*! Uif!fggjdjfodz!pg!uif!usbotgpsnfs

Solution Refer Fig. 3.11(a). V2 = 350 –0° V = 350 + j 0 I2 = 100 (0.9 – j 0.4) = 90 – j 40 = 98.488 ––23.96° Z2 = 0.2 + j 0.4 E2 = V2 + I2 Z2 = (350 + j 0) + (90 – j 40) (0.2 + j 0.4) = 350 + j 0 + 18 – j 8 + j 36 + 16 = 384 + j 28 = 385.02 –4.17° V \ angle between E2 and V2 a = 4.17° E1 = 10 (384 + j 28) = 3840 + j 280 –E1 = – 3840 – j 280 = 3850.19 ––175.83° -I -90 - j 40 I1¢ = 2 = = –9 – j 4 10 10 The angle between Io and V2 is (4.17° + 90° + 30°) = 124.17°

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\ \

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Io = 1.4 –124.17° = – 0.786 + j 1.158 I1 = I1¢ + Io = –9 – j 4 – 0.786 + j 1.158 = –9.786 – j 2.842 = 10.19 ––163.8° V1 = – E1 + I1 Z1 = – 3840 – j 280 + 10.19 ––163.8° (5 + j 8) = – 3840 – j 280 + 10.19 ––163.8° ¥ 9.434 –57.99° = – 3840 – j 280 + 96.132 ––105.81° = – 3840 – j 280 – 26.19 – j 92.49 = – 3866.19 – j 372.49 = 3884.09 ––174.49°

(a) The primary applied voltage = |V1 | = 3884.09 V (b) The primary power factor = cos f1 = cos (–174.49° + 163.8°) = 0.98 (lagging) (c) No-load primary input = Wo = V1 Io cos fo = 3884.09 ¥ 1.4 ¥ 1030° = 4709.21 W Wo is the iron loss of the transformer. Total Cu-loss of the transformer = 10.192 ¥ 5 + 98.4882 ¥ 0.2 = 519.18 + 1939.977 = 2459.157 W Total losses = (4709.21 + 2459.157) W = 7168.367 W Output = 350 ¥ 98.488 ¥ 0.9 W = 31023.72 W 31023.72 Efficiency = ¥ 100% = 81.23%. 31023.72 + 7168.367

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Solution If the linear dimensions of the second transformer are 2 times that of the first then its volume is 2 ¥ 2 ¥ 2 , i.e. 2 2 times the first and area is twice the first transformer. Since the voltage

!

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Fmfdusjdbm!Nbdijoft

applied to the second transformer is twice that of the first (2 ¥ 13.8 kV = 27.6 kV), the flux densities in both the windings are same. Hence, core loss is proportional to the volume. \ core loss of second transformer = 2 2 ¥ 3420 = 9673 W \ power drawn by the second transformer is 9673 W or 9.673 kW. Magnetizing reactance is inversely proportional to the length of core and directly proportional to Ê m AN 2 ˆ the cross-sectional area ÁE L = ˜ Ë l ¯ 2 \ no-load current of the second transformer is 4.93 ¥ = 6.97 A. 2

UFTUJOH!PG!USBOTGPSNFST!

4/36

The efficiency and regulation of a transformer are calculated by two types of tests, called opencircuit test and short-circuit test.

4/36/2! Pqfo.Djsdvju!Uftu! This test is performed to measure the iron losses. Also, the core-loss resistance and magnetizing reactance can be calculated from this test. In this test, one of the windings of the transformer is open circuited. The rated voltage at rated frequency is applied to the other winding. Generally, the high-voltage side is open circuited and the rated voltage is applied at the low-voltage side. The connections made are shown in Fig. 3.26; the rated voltage is applied through an autotransformer (also called variance). The readings of the wattmeter Wo, voltmeter V1 and ammeter Io are recorded. Since the secondary is open circuited, a very small current called the no-load current flows in the primary. This current Io is 2% to 6% of the rated current. Hence, the primary leakage impedance drop is negligible and applied voltage V1 can be considered equal to V1¢. The equivalent circuit gets modified to that shown in Fig. 3.27.

Gjh/!4/37! Djsdvju!ejbhsbn!gps!pqfo.djsdvju!uftu

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! Gjh/!4/38! Frvjwbmfou!djsdvju!voefs!pqfo.djsdvju!uftu

The wattmeter reading gives the total losses, i.e. core loss and copper loss of the transformer. The no-load current being negligible, the copper loss under no-load condition varies from 0.04% (0.02 ¥ 0.02 ¥ 100) to 0.36% (0.06 ¥ 0.06 ¥ 100) of the rated copper loss. Hence, copper loss in open-circuit test is negligible compared to core loss. Hence, the wattmeter reading is considered equal to core loss. \

Wo = Core loss The no-load power factor cos qo =

Wo V1I o

(3.47)

The core loss component of no-load current IW = Io cos qo V Ro = 1 IW

\ core loss resistance

(3.48)

The magnetizing component of no-load current

\ magnetizing reactance

Im = Io sin qo V Xo = 1 Im

(3.49)

Ro and Xo are referred to that side of the transformer where instruments are kept, generally they are referred to low-voltage side.

4/36/3! Tipsu.djsdvju!Uftu This test is performed to measure the copper loss of a transformer. In addition, it gives the equivalent resistance and leakage reactance of the transformer. As rated current is to be supplied in one winding of the transformer keeping the other side short circuited, this test is generally performed in the high-voltage side keeping the low-voltage side short circuited. The connections are made as shown in Fig. 3.28. All the instruments are placed on the high-voltage side and a low-voltage is applied using an autotransformer. This applied voltage is adjusted in such a way that the full-load

!

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Fmfdusjdbm!Nbdijoft

Wsc A

Isc

Vsc V

A

HV

LV

!

Gjh/!4/39! Djsdvju!ejbhsbn!gps!tipsu.djsdvju!uftu

current flows through both the windings. It has been observed that a primary voltage of 2% to 12% of its rated value is sufficient to circulate rated currents. Hence, about 1% to 6% of the rated voltage appears across the magnetizing branch and the core flux is also 1% to 6% of its rated value. Since core loss is proportional to the square of the core flux, the core loss is 0.01% (0.01 ¥ 0.01 ¥ 100) to 0.36% (0.06 ¥ 0.06 ¥ 100) of the rated core loss. Hence, the core loss is negligible and the wattmeter reading Wsc can be considered to be equal to copper loss only. Again at rated voltage, the exciting current Io is 2% to 6% of the full-load current. When the voltage across exciting branch is 1% to 6% of its rated value, Io is 0.02% to 0.36% of full-load current and hence can be neglected. The equivalent circuit can be drawn as shown in Fig. 3.29. R1

X1

X2

R2

Isc Vsc

! Gjh/!4/3:! Frvjwbmfou!djsdvju!voefs!tipsu.djsdvju!uftu

If Vsc, Isc and Wsc be the voltmeter, ammeter and wattmeter reading then ReH = ZeH = XeH =

Wsc I sc2 Vsc I sc 2 2 ZeH - ReH

ReH and XeH are equivalent resistance and reactance referred to high-voltage side.

(3.50) (3.51) (3.52)

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If the primary and secondary resistances and leakage reactances are required then R1 = R¢2 =

1 ReH and 2

X1 = X2¢ =

1 XeH 2

4/36/4! TvnqofsÕt!Uftu!)Cbdl!up!Cbdl!uftu!ps!Sfhfofsbujwf!Uftu* To determine the maximum temperature rise of a transformer, Sumpner’s test is performed. This test can also be performed to find out the efficiency of a transformer. Sumpner’s test is essentially a load test. It requires two identical transformers whose primaries are connected in parallel. The two secondaries are connected in series with their polarities in phase opposition. The primary windings are supplied at rated voltage and frequency. A voltmeter, ammeter and wattmeter are connected to the input as shown in Fig. 3.30. The range of the voltmeter V2 connected across the two secondaries should be double the rated voltage of either transformer secondary. As the two secondaries are connected in phase opposition, the two secondary emfs oppose each other and no current can flow in the secondary circuit. A regulating transformer excited by an ac mains supply is used to inject voltage in the secondary winding. The injected voltage is adjusted till the ammeter A2 reads fullload secondary current. The secondary current causes full-load current to flow through the primary windings whereas the primary current remains confined to the dotted path as shown in Fig. 3.30. The wattmeter W1 indicates total core losses, W2 indicates total copper losses and ammeter A1 indicates total no-load current of the two transformers. Thus, by this method, we can load the transformer to full-load but the supplying energy is only equal to that required for the losses only. This test can be continued for a long time to determine the maximum temperature rise of a transformer. W1 A1 ac supply

V1

V2

Regulating Transformers

A2 W2

! Gjh/!4/41! TvnqofsÕt!uftu

!

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Fmfdusjdbm!Nbdijoft

4/36/5! Qpmbsjuz!Uftu Polarities of a transformer can be checked by a simple test requiring only voltage measurements with a transformer on no-load. At any instant, one terminal of the primary winding is positive with respect to the other. At the same instant, one terminal of the secondary winding is positive with respect to the other. The polarities of the windings at any instant should be known in order to identify the like terminals and unlike terminals. Polarity test is required for parallel operation of transformers. In this test, the rated voltage is applied on one winding and one V terminal of the primary winding is connected to one terminal of the A C secondary winding through a voltmeter. The other two terminals of the windings are short circuited as shown in Fig. 3.31. ~ E1 E2 Suppose the voltages induced in the primary and secondary windings are E1 and E2 respectively. If the voltmeter reading is equal to E1 – E2 then the terminals A and C are of same polarity at any one inD B stant. Hence, terminals B and D will be of opposite polarities. A and ! C are like terminals and B and D are unlike terminals. If the voltmeter Gjh/!4/42! Qpmbsjuz!uftu reading is equal to E1 + E2 then A and C are of opposite polarities. In that case if terminal A is positive at any instant then terminal D will be positive at that instant. At the same instant, terminals B and C will have negative polarities.

BMM.EBZ!FGGJDJFODZ!

4/37

It is usual for the primary of a transformer to be connected permanently to the supply and for the switching of load to be carried out in the secondary circuit. Since the copper loss varies with load but iron loss is constant, the efficiency depending on loading and losses varies throughout the day. For transformers which are continuously excited but supply loads only intermittently, a low iron loss is particularly desirable, but a low copper loss is specially important where the load factor is high. Again, for a transformer working on full-load for the greater part of the day, maximum efficiency should be arranged to occur somewhere around the full-load value but for a transformer whose full-load value may be supplied for only ¼ of the day and the unit is only lightly loaded for the rest of the time, it would be desirable to arrange maximum efficiency to occur at about ½ fullload value. Considering the above factors, the efficiency of a transformer is better estimated on an energy rather than a power ratio and thus we have the term all-day efficiency. All-day efficiency =

Output in kWh for 24 hours Input in kWh for 24 hours

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4/21:

! Qspcmfn!4/211 Ipx!epft!b!xbuunfufs!dpoofdufe!up!uif!qsjnbsz!xjoejoh!pg!b!usbotgpsnfs!nfbtvsf!dpqqfs!mptt!jo! uif!tfdpoebsz!bmtp@

Solution When rated current flows in the primary winding, the secondary winding also carries rated current as N1 I1 = N2 I2. The copper losses in the secondary winding must be supplied from the primary as there is no source in the secondary. Hence, the entire power is recorded by the wattmeter connected to the primary which includes copper losses of the secondary windings also.

! Qspcmfn!4/212 Uif!gpmmpxjoh!sftvmut!xfsf!pcubjofe!jo!uftut!po!b!61!lWB-!tjohmf.qibtf-!44110511!W!usbotgpsnfs/ Pqfo.djsdvju!uftu;!4411!W-!541!X Tipsu.djsdvju!uftu;!235!W-!26/4!B-!646!X/ )tvqqmz!hjwfo!po!IW!tjef* Dbmdvmbuf!)b*!uif!fggjdjfodz!bu!gvmm.mpbe!boe!ibmg!gvmm.mpbe!bu!1/818!q/g/!mbhhjoh-!)c*!uif!sfhvmbujpo!bu! gvmm.mpbe!gps!q/g/!pg!1/818!)mbhhjoh!boe!mfbejoh*-!boe!)d*!gvmm.mpbe!ufsnjobm!wpmubhf!voefs!uif!dpoej. ujpo!pg!1/818!q/g/!)mbhhjoh*/

Solution For short-circuit test data, Ze1 =

124 535 W = 2.285 W W = 8.10 W, Re1 = 15.3 (15.3)2

Xe1 = (8.1)2 - ( 2.285) 2 = 7.77 W 50, 000 A = 15.15 A 3300 2 Ê 15.15 ˆ So, rated copper loss = 535 ¥ Á = 524.56 W Ë 15.3 ˜¯

(a) Rated current on the HV side =

Iron loss = 430 W Efficiency at full-load 0.707 p.f. lagging =

50 ¥ 103 ¥ 0.707 50 ¥ 103 ¥ 0.707 + 524.56 + 430

= 0.9737 or 97.37%

Efficiency at half-load and 0.707 p.f. lagging is 1 50 ¥103 ¥ 0.707 ¥ 2 = 0.9623 or 96.93% = 2 1 Ê 1ˆ 3 50 ¥ 10 ¥ (0.707) ¥ + Á ˜ ¥ 524.56 + 430 2 Ë 2¯

!

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(b) Voltage regulation at full-load and 0.707 lagging p.f. is =

I1Re1 cos q + I1 X e1 sin q 15.15 ( 2.285 ¥ 0.707 + 7.77 ¥ 0.707) = E1 3300 = 0.0326 = 3.26%

Voltage regulation at full-load and 0.707 p.f. leading is =

I1Re1 cos q - I1 X e1 sin q 15.15 ( 2.285 ¥ 0.707 - 7.77 ¥ 0.707) = E1 3300 = (–0.0178) or (–1.78%)

(c) If V2 is the terminal voltage at 0.707 p.f. leading È 400 - V2 ˘ ÍÎ 400 ˙˚ = 0.0326 or, V2 = 400 (1 – 0.0326) V = 386.96 V.

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Solution (a) From the short-circuit test, the equivalent impedance referred to the low-voltage side is ZeL =

674 W = 0.10784 W = 107.84 mW 6250

Equivalent resistance referred to the low-voltage side ReL = \

187000 (6250) 2

W = 4.78 mW

XeL = 107.73 mW

(b) The efficiency of the transformer at unity power factor h=

50 ¥ 106 ¥ 1 50 ¥ 106 ¥ 1 + 206 + 187

¥ 100% = 99.2%

Tjohmf.qibtf!Usbotgpsnfst

Voltage regulation =

=

I L ( ReL cos q 2 + X eL sin q 2 ) 8 ¥ 103 6.25 ¥ 103 {4.78 ¥ 10 -3 ¥ 1} 8 ¥ 103

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¥ 100% ¥ 100%

= 0.373% (c) Efficiency at 0.9 p.f. leading 50 ¥ 106 ¥ 0.9

h=

\

50 ¥ 106 ¥ 0.9 + 206 + 187 cos q = 0.9 sin q = 0.436 Voltage regulation =

¥ 100% = 99.13%

6.25 ¥ 103 {4.78 ¥ 10 -3 ¥ 0.9 - 107.73 ¥ 10 -3 ¥ 0.436} 8 ¥ 103

= –0.0333 or –3.33%.

! Qspcmfn!4/214 B!6!lWB-!5110311!W-!61!I{-!tjohmf.qibtf!usbotgpsnfs!ibt!uif!gpmmpxjoh!uftu!ebub; PD!Uftu;!311!W-!3!B-!91!X !!TD!Uftu;!31!W-!21!W-!261!X Uif!nfufst!bsf!qmbdfe!evsjoh!TD!uftu!po!uif!tjef!pqqptjuf!up!uibu!pg!PD!uftu/!Gjoe!frvjwbmfou!djsdvju! sfgfssfe!up!qsjnbsz!ijhi.wpmubhf!tjef/

Solution OC Test Since rated voltage is applied during OC test, meters are placed on the low-voltage side. 400 =2 200 No-load current (Io) = 2 A = (Io)LV Turns ratio =

(Io)HV = 2 ¥ Applied voltage at no-load = 400 Vo Io cos fo = 80 fi 400 ¥ 1 ¥ cos fo = 80 80 fi cos fo = = 0.2 400

1 =1A 2

!

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sin fo = 0.979 Iw = 1 ¥ 0.2 = 0.2 A Im = 1 ¥ 0.979 = 0.979 A

\

RoL =

Vo 400 = 2000 W = 0.2 IW

XoL =

Vo 400 = 408.58 W = I m 0.979

RoH = 2000 ¥ 22 = 8000 W

\

XoH = 408.58 ¥ 22 = 1634.32 W SC Test As per the given statement, the meters are placed on high-voltage side. Vsc = 20 V Isc = 10 A Psc = 150 W ZeH = ReH =

20 =2W 10 150 10 2

= 1.5 W

XeH = 22 - 1.52 = 1.323 W.

! Qspcmfn!4/215 B! tjohmf.qibtf! 4! lWB-! 5110311! W-! 61! I{! usbotgpsnfs! ibt! qsjnbsz! boe! tfdpoebsz! sftjtubodft! pg! 1/5!W!boe!1/2!W-!qsjnbsz!boe!tfdpoebsz!mfblbhf!sfbdubodft!pg!1/6!W!boe!1/26!W-!sftjtubodf!boe! sfbdubodf!pg!fydjubujpo!djsdvju!sfgfssfe!up!qsjnbsz!bt!861!W!boe!311!W!sftqfdujwfmz/!Gps!PD!boe! TD!uftut-!dbmdvmbuf!uif!nfufs!sfbejoh-!xifo!gps!cpui!uif!uftut-!tvqqmz!jt!hjwfo!po!ijhi.wpmubhf!tjef/

Solution R1 = 0.4 W

X1= 0.5 W

R2 = 0.1 W

X2 = 0.15 W

OC Test V1 = 400 V 400 Iw = A = 0.53 A 750 400 Im = =2A 200

Ro = 750 W

Xo = 200 W

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I m2 + I w2 = 22 + 0.532 = 2.069 A

Io =

Wo = 400 ¥ 0.53 W = 212 W The meter readings are 400 V, 2.069 A, 212 W SC Test ReH = 0.4 + 22 ¥ 0.1 = 0.8 W XeH = 0.5 + 22 ¥ 0.15 = 1.1 W

400 È ˘ ÍÎAs turn ratio = 200 = 2˙˚

ZeH = 0.82 + 1.12 = 1.36 W 3 ¥ 103 A 400 = 7.5 A

Full-load current in high-voltage winding (I1) =

Vsc = 7.5 ¥ 1.36 V = 10.2 V Psc = (7.5)2 ¥ 0.8 = 45 W Meter readings are 10.2 V, 7.5 A, 45 W.

! Qspcmfn!4/216 B!311!lWB!tjohmf.qibtf!usbotgpsnfs!tvqqmjft!mpbe!uispvhipvu!35!ipvst/!Gps!8!ipvst!jo!b!ebz-!uif! mpbe!jt!251!lX!bu!1/9!q/g/!mbhhjoh!boe!gps!8!ipvst-!uif!mpbe!jt!:1!lX!bu!1/:!q/g/!Gps!sfnbjojoh!ujnf! po!uif!sftu!qfsjpe-!ju!jt!bu!op.mpbe!dpoejujpo/!Gvmm.mpbe!Dv.mptt!jt!6!lX!boe!uif!jspo!mptt!jt!3!lX/!Dbm. dvmbuf!uif!bmm.ebz!fggjdjfodz!pg!uif!usbotgpsnfs/

Solution Full-load output = 200 kVA Full-load Cu-loss = 5 KW Iron loss = 2 KW All-day output = (140 ¥ 7) + (90 ¥ 7) = 1610 kWh 2

2

Ê 90 / 0.9 ˆ Ê 140 / 0.8 ˆ ¥5¥7+Á ¥5¥7 Cu-loss for 24 hours = Á Ë 200 ˜¯ Ë 200 ˜¯ = 26.79 + 8.75 = 35.54 Iron loss for 24 hours = 2 ¥ 24 = 48 kWh

!

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All-day input = 1610 + 35.54 + 48 = 1693.54 kWh 1610 All-day efficiency = ¥ 100% 1693.54 = 95.06%.

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Solution OC Test Meters placed to LV side and HV side are kept open. V2 = 250 V, Io = 1.0 A, Wo = 80 W cos fo =

80 = 0.32 250 ¥ 1.0

sin fo = 1 - 0.322 = 0.947 Iw = 1 ¥ 0.32 = 0.32 A Im = 1 ¥ 0.947 = 0.947 A \

250 W = 781.25 W 0.32 250 = W = 263.99 W 0.947

RoL = XoL

SC Test Meters placed on HV side and LV side are short circuited. Vsc = 25 V Isc = 12 A Psc = 100 W ZeH =

25 W = 2.08 W 12

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ReH =

100 122

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W = 0.69 W

XeH = 2.082 - 0.692 W = 1.96 W Turns ratio =

0.69

= 0.1725 W 22 1.96 = = 0.49 W. 4

ReL = XeL

500 =2 250

10 ¥ 103 A = 40 A 250 Secondary load voltage at 0.8 p.f. lagging with full-load current

(b) (I2)full-load =

= [250 – I2 (ReL cos f + XeL sin f)] = [250 – 40 (0.1725 ¥ 0.8 + 0.49 ¥ 0.6)] = [250 – 17.28] V = 232.72 V.

! Qspcmfn!4/218 B!31!lWB-!5110311!W-!61!I{-!tjohmf.qibtf!usbotgpsnfs!ibt!uif!gpmmpxjoh!uftu!sftvmut; ! ! ! ! ! ! ! ! ! ! ! ! PD!Uftu;!511!W-!2/3!B-!91!X!)MW!tjef!lfqu!pqfo* ! ! ! ! ! ! ! ! ! ! ! ! TD!Uftu;!61!W-!26!B-!:1!X!)MW!tjef!tipsufe* Efufsnjof!uif!sfhvmbujpo!boe!fggjdjfodz!pg!uif!usbotgpsnfs!bu!gvmm.mpbe!boe!bu!1/:!q/g/!mbhhjoh/

Solution From OC test of the transformer, iron loss of the transformer = 80 W SC Test Vsc = 50 V, Isc = 15 A, Psc = 90 W 50 W = 3.33 W 15 90 = 2 W = 0.4 W 15

Ze1 = Re1

Xe1 = 3.332 - 0.4 2 = 3.3 W Turns ratio =

400 =2 200

!

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Re2 = Xe2 =

0.4 22 3.3 22

= 0.1 W = 0.825 W

20 ¥ 103 A 200 = 100 A

Full-load secondary current =

Approximate voltage drop of the transformer = 100 (0.1 ¥ 0.9 + 0.825 ¥ 0.4) = 42 V 42 ¥ 100% 200 = 21%

Regulation of the transformer =

Full-load primary current =

20 ¥ 103 = 50 A 400

Cu-loss of the transformer at full-load Ê 50 ˆ = 90 ¥ Á ˜ Ë 15 ¯

2

= 1000 W = 1 kW Total full-load loss of the transformer (1000 + 80) W = 1080 W Secondary terminal voltage at full-load = (200 – 42) V = 158 V Output power = 158 ¥ 100 ¥ 0.9 = 14220 W 14220 Efficiency of the transformer = ¥ 100% 14220 + 1080 = 92.9%.

! Qspcmfn!4/219 Uif!PD!uftu!)MW!tjef*!boe!TD!uftu!)IW!tjef*!sftvmut!pg!b!tjohmf.qibtf!7!lWB-!3610611!W!usbotgpsnfs! bsf!361!W-!2/3!B-!91!X!boe!36!W-!21!B-!:6!X!sftqfdujwfmz/!Efufsnjof!uif!djsdvju!qbsbnfufst!sfgfssfe! up!MW!tjef!boe!bmtp!dbmdvmbuf!uif!sfhvmbujpo!boe!fggjdjfodz!pg!uif!usbotgpsnfs!bu!gvmm.mpbe!boe!ibmg. mpbe!bu!1/6!q/g/!mbhhjoh/

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Solution Turns ratio =

250 1 = 500 2

OC Test V1 = 250 V, Io = 1.2 A, Wo = 80 W 80 = 0.267 250 ¥ 1.2 sin fo = 0.964 Iw = 1.2 ¥ 0.267 = 0.32 Im = 1.2 ¥ 0.964 = 1.157

cos fo =

250 W = 781.25 W 0.32 250 = W = 216.07 W. 1.157

RoL = XoL SC Test Vsc = 25 V, Isc = 10 A, Psc = 95 W

25 W = 2.5 W 10 95 = 2 = 0.95 W 10

Ze2 = Re2

Xe2 = 2.52 - 0.952 = 2.312 W 2

Ê 1ˆ Re1 = Á ˜ ¥ 0.95 = 0.2395 W Ë 2¯ 2

Ê 1ˆ Xe1 = Á ˜ ¥ 2.312 = 0.578 W Ë 2¯ Regulation (I2)fl =

6 ¥ 103 = 12 A 500

DV = 12 (0.95 ¥ 0.5 + 2.312 ¥ 0.866) [p.f. = 0.5, i.e. cos f = 0.5, sin f = 0.866] = 29.73 V. Voltage regulation =

29.73 ¥ 100% = 5.95%. 500

!

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Output power = (500 – 29.73) ¥ 12 ¥ 0.5 W = 2821.62 W Iron loss = 80 W 2

Ê 12 ˆ Full-load Cu-loss = Á ˜ ¥ 95 = 136.8 W Ë 10 ¯ 2821.62 2821.62 + 80 + 136.8 = 92.86%.

Full-load efficiency =

Half-load Voltage drop at half-load =

29.73 V = 14.865 V 2

14.865 ¥ 100% = 2.97% 500 1 Output power = ¥ (500 – 14.865) ¥ 12 ¥ 0.5 W 2 = 1455.4 W Regulation =

2

Ê 1ˆ Half-load Cu-loss = Á ˜ ¥ 136.8 = 34.2 W Ë 2¯ 1455.4 ¥ 100% 1455.4 + 80 + 34.2 = 92.7%.

Half-load efficiency =

! Qspcmfn!4/21: Uif!PD!boe!TD!uftut!po!b!3310551!W-!61!I{-!tjohmf.qibtf!usbotgpsnfs!hbwf!uif!gpmmpxjoh!sftvmut; PD!Uftu;!)MW!tjef*;!331!W-!1/9!B-!91!X TD!Uftu;!)IW!tjef*;!31!W-!21!B-!236!X Gjoe!uif!frvjwbmfou!djsdvju!pg!uif!usbotgpsnfs!sfgfssfe!up!MW!tjef!boe!bmtp!dbmdvmbuf!uif!tfdpoebsz! wpmubhf!xifo!efmjwfsjoh!6!lX!bu!1/9!q/g/!mbhhjoh/

Solution OC Test Meters placed on LV side and HV side are kept open. V1 = 220 V, Io = 0.8 A, Wo = 80 W 80 cos fo = = 0.45 220 ¥ 0.8 sin fo = 1 - 0.452 = 0.89

Tjohmf.qibtf!Usbotgpsnfst

Iw = 0.8 ¥ 0.45 = 0.36 A Im = 0.8 ¥ 0.89 = 0.712 A 220 W = 611.11 W 0.36 220 = W = 308.98 W 0.712

ROL = XoL

SC Test Meters placed on HV side and LV side are kept open.

Vsc = 20 V, Isc = 10 A, Psc = 125 W ZeH = ReH =

20 W=2W 10 125 10 2

W = 1.25 W

XeH = 22 - 1.252 W = 1.56 W Turns ratio =

220 1 = 440 2 2

Ê 1ˆ ROL = Á ˜ ¥ 1.25 = 0.3125 W Ë 2¯

\

2

Ê 1ˆ XoL = Á ˜ ¥ 1.56 = 0.39 W Ë 2¯ Secondary current (I2) =

5 ¥ 103 A = 14.2 A 440 ¥ 0.8

Approximate voltage drop referred to secondary for 0.8 p.f. lagging = I2 (ReH cos f + XeH sin f) = 14.2 (1.25 ¥ 0.8 + 1.56 ¥ 0.6) V = 27.49 V \

secondary terminal voltage = (440 – 27.49) V = 412.51 V.

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4/22:

!

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Dbmdvmbuf!uif!gpmmpxjoh; ! )b*! Uif!fggjdjfodz!pg!uif!usbotgpsnfs!bu!ibmg.mpbe!boe!1/9!q/g/!mbhhjoh/ ! )c*! Uif!mpbe!lWB!bu!xijdi!nbyjnvn!fggjdjfodz!pddvst!boe!bmtp!uif!nbyjnvn!fggjdjfodz!bu!1/:!q/g/! mbhhjoh/ ! )d*! Uif!wpmubhf!sfhvmbujpo!bu!1/:!q/g/!mfbejoh!po!gvmm.mpbe!dpoejujpo/

Solution From OC test, iron loss of the transformer = 80 W 15 ¥ 103 Full-load primary current (I1) = = 7.5 A 2000 From SC test, 40 Ze1 = W=8W 5 100 Ro1 = 2 = 4 W 5 Xe1 = 82 - 4 2 W = 6.93 W 2000 Turns ratio = = 10 200 4 Re2 = 2 = 0.04 W 10 6.93 Xe2 = 2 W = 0.0069 W 10 Approximate voltage drop at secondary DV = I2 (Re2 cos f +Xe2 sin f) 15 ¥ 103 I2 at half-load = A 200 ¥ 2 = 37.5 A DV = 37.5 (0.04 ¥ 0.9 + 0.0069 ¥ 0.44) V = 1.464 V V2¢ = (200 – 1.464)V = 198.536 V Output power = 198.536 ¥ 37.5 ¥ 0.9 = 6700.59 W at half-load 2

Ê 1ˆ Cu-loss at half-load = Á ˜ ¥ 100 = 25 W Ë 2¯ (a) Efficiency of the transformer at half-load 6700.59 = ¥ 100% 6700.59 + 25 + 80 = 98.46%

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(b) At maximum efficiency, Cu-loss = Iron loss = 80 W I1 =

80 A = 20 A 4

(c) Voltage regulation at 0.9 p.f. leading DV = 37.5 ¥ 2 (0.04 ¥ 0.9 – 0.0069 ¥ 0.44) V = 2.475 V Regulation =

2.475 ¥ 100% = 1.24%. 200

! Qspcmfn!4/222 B!tjohmf.qibtf!51104111!W-!7!lWB-!61!I{!usbotgpsnfs!hjwft!uif!gpmmpxjoh!uftu!sftvmut; PD!Uftu!)MW!tjef*;!511-!1/9!B-!211!X TD!Uftu!)IW!tjef*;!31!W-!21!B-!261!X Efufsnjof!uif!gpmmpxjoh; ! )b*! Uif!djsdvju!dpotubout!sfgfssfe!up!MW!tjef/ ! )c*! Uif! bqqmjfe! wpmubhf! boe! fggjdjfodz! jg! uif! pvuqvu! dvssfou! jt! 26! B! bu! uif! ufsnjobm! wpmubhf! pg! 4111!W!bu!1/:!qpxfs!gbdups!mbhhjoh/

Solution Turns ratio =

400 4 2 = = . 3000 30 15

(a) OC Test Meters placed on LV side. V1 = 400 V, Io = 0.8 A, Wo = 100 W V1 Io cos fo = Wo fi \

100 = 0.3125 400 ¥ 0.8 sin fo = 0.9499 Iw = Io cos fo = 0.25 A Im = Io sin fo = 0.76 A

cos fo =

400 W = 1600 W 0.25 400 = W = 526.316 W. 0.76

Ro1 = Xo1

!

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SC Test Meters placed on HV side. Vsc = 20 V Isc = 10 A Psc = 150 W 20 =2W 10 150 = 2 = 1.5 W 10

Ze2 = Re2

Xe2 = 22 - 1.52 W = 1.323 W 2

Re1

Ê 2ˆ = Á ˜ ¥ 1.5 W = 0.027 W Ë 15 ¯ 2

Ê 2ˆ Xe1 = Á ˜ ¥ 1.323 W = 0.024 W. Ë 15 ¯ (b) To maintain the output voltage at 3000 V, the applied voltage can be calculated from Fig. 3.22. V¢1

I1 Xo1

V1 I1 Ro1 V1 sin f1 f1 I1

V1 cos f1

!

Gjh/!4/43! Qibtps!ejbhsbn!pg!Fybnqmf!4/222

V1¢2 = (V1 cos f1 + I1 Re1)2 + (V1 sin f1 + I1 Xe1)2 = (400 ¥ 0.9 + 112.5 ¥ 0.027)2 + (400 ¥ 0.436 + 112.5 ¥ 0.024)2 15 È ˘ Í I 2 = 15 A , I1 = 2 /15 = 112.5 A ˙ Î ˚ = 131796.23 + 31364.41 = 163160.64 V1¢ = 403.93 V

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To maintain the output voltage at 3000 V with given load, the applied voltage should be 403.93 V. From OC test, iron loss = 100 W From SC test, Cu-loss = 150 W at 10 A 2

Ê 15 ˆ \ full-load Cu-loss = Á ˜ ¥ 150 W Ë 10 ¯ = 337.5 W \ total loss of the transformer = (100 + 337.5) W = 437.5 W \

Efficiency =

6 ¥ 103 ¥ 0.9 6 ¥ 103 ¥ 0.9 + 437.5

¥ 100%

5400 ¥ 100% 5837.5 = 92.5%. =

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Solution 2000 = 10 200 (a) Iron loss or core-loss of the transformer = 85 W (b) From SC test, Vsc = 50 V, Isc = 15 A, Psc = 100 W Turns ratio =

50 W = 3.33 W 15 100 = 2 W = 0.44 W 15

Ze1 = Re1

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Xe1 = 3.332 - 0.44 2 = 3.3 W (c) Re2 =

Re1 10

2

=

0.44 = 0.0044 W 100 Xe2 =

3.3 10 2

= 0.033 W

15 ¥ 103 A = 75 A 200 \ approximate voltage drop = I2 (Re2 cos f + Xe2 sin f) = 75 (0.0044 ¥ 0.8 + 0.033 ¥ 0.6) = 1.749 V

(d) (I2)fl =

1.749 ¥ 100% 200 = 0.87%

\ regulation of the transformer =

(e) At full-load, transformer terminal voltage = (200 – 1.749) V = 198.25 V (f) Output = 198.25 ¥ 75 ¥ 0.8 = 11895.06 W 2

Ê 75 ˆ Cu-loss of the transformer at full-load = Á ˜ ¥ 100 Ë 15 ¯ = 2500 W Total loss = (2500 + 85) W = 2585 W 11895.06 ¥ 100% 11895.06 + 2585 = 82.15%

Efficiency at full-load =

75 A = 37.5 A 2 Now, approximate voltage drop = 37.5 (0.0044 ¥ 0.8 + 0.033 ¥ 0.6) V = 0.875 V At half-load, I2 (fl/2) =

Terminal voltage = (200 – 0.875) V = 199.125 V

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Output power = 199.125 ¥ 37.5 ¥ 0.8 = 5973.75 W 2

Ê 1ˆ Cu-loss at half-load = Á ˜ ¥ 2500 W = 625 W Ë 2¯ Total losses = (85 + 625) W = 710 W 5973.75 ¥ 100% 5973.75 + 710 = 89.377%.

Efficiency at half-load =

! Qspcmfn!4/224 B!6!lWB-!6110361!W-!61!I{-!tjohmf.qibtf!usbotgpsnfs!ibt!uif!gpmmpxjoh!uftu!sftvmut; PD!Uftu!)MW!tjef*;!361!W-!2/3!B-!91!X TD!Uftu!)IW!tjef*;!31!W-!21!B-!211!X Efufsnjof!uif!gpmmpxjoh; ! )b*! Bqqspyjnbuf!frvjwbmfou!djsdvju!sfgfssfe!up!IW!tjef! ! )c*! Wpmubhf!sfhvmbujpo!boe!fggjdjfodz!bu!gvmm.mpbe!boe!1/9!q/g/!mbhhjoh!mpbe ! )d*! Fggjdjfodz!pg!uif!usbotgpsnfs!bu!ibmg!pg!gvmm.mpbe!boe!1/9!q/g/!mbhhjoh!mpbe

Solution Turns ratio =

500 =2 250

OC Test (Io)LV = 1.2 A ( I o ) HV 1 = ( I o ) LV 2 1.2 = 0.6 A (Io)HV = 2

\

Meter reading when they are placed on HV side V1 = 500 V, Io = 0.6 A, Wo = 80 W 80 = 0.26 500 ¥ 0.6 Iw = 0.6 ¥ 0.26 A = 0.16 A Im = 0.6 ¥ 0.97 A = 0.579 A [cos fo = 0.26, i.e. sin fo = 0.97] 500 ROH = W = 3125 W 0.16 500 XoH = W = 863.56 W (referred to HV side). 0.579

cos fo =

!

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SC Test Meters placed on HV side. Vsc = 20 V, Isc = 10 A, Psc = 100 W 20 W=2W 10 100 = 2 W=1W 10

Ze1 = Re1

Xe1 = 22 - 12 W = 1.732 W.

\

(a) Full-load secondary current of the transformer 5 ¥ 103 I2 = A = 20 A 250 Approximate voltage drop of the transformer referred to secondary DV = I2 (Re2 cos f2 + Xe2 sin f2) Re2 =

Re1 2

=

1 W = 0.25 W 4

2 1.732 Xe2 = = 0.433 W 4 cos f2 = 0.8 (lagging) sin f2 = 1 - 0.82 = 0.6

DV = 20 (0.25 ¥ 0.8 + 0.433 ¥ 0.6) V = 9.196 V 9.196 ¥ 100% 250 = 3.678%

\ regulation of the transformer =

From OC Test

From SC Test

Iron loss = 80 W

Cu-loss at 10 A = 100 W \ full-load Cu-loss 2

Ê 20 ˆ = Á ˜ ¥ 100 W = 400 W Ë 10 ¯ \ total losses of the transformer at full-load = 80 + 400 = 480 W

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Output of the transformer = 5000 ¥ 0.8 W = 4000 W \ efficiency =

4000 ¥ 100% = 89.28%. 4000 + 480

(iii) At half-load Output =

4000 W = 2000 W 2 2

Ê 1ˆ Cu loss = Á ˜ ¥ 400 = 100 W Ë 2¯ \ efficiency =

2000 ¥ 100% = 91.74%. 2000 + 100 + 80

! Qspcmfn!4/225 Uif!bqqspyjnbuf!frvjwbmfou!djsdvju!pg!b!3!lWB-!3110511!W!tjohmf.qibtf!usbotgpsnfs-!sfgfs!up!MW!tjef! jt!tipxo!jo!Gjh/!4/44)b*/ ! )b*! Bo!pqfo.djsdvju!uftu!jt!dpoevdufe!cz!bqqmzjoh!311!W!up!uif!MW!tjef/!Lffqjoh!uif!IW!tjef!pqfo-! dbmdvmbuf!uif!qpxfs!joqvu-!qpxfs!gbdups!boe!dvssfou!esbxo!cz!uif!usbotgpsnfs/ ! )c*! B!tipsu.djsdvju!uftu!jt!dpoevdufe!cz!qbttjoh!gvmm.mpbe!dvssfou!gspn!uif!IW!tjef!lffqjoh!uif!MW! tjef!tipsufe/!Dbmdvmbuf!uif!wpmubhf!up!cf!bqqmjfe!up!uif!usbotgpsnfs!boe!uif!qpxfs!joqvu!boe! qpxfs!gbdups/ 0.15 W

V1 800 W

0.4 W

400 W

Isc

V¢2

0.6 W

1.6 W

Vsc

! ! !

Gjh/!4/44)b*! Djsdvju!ejbhsbn!pg!Fybnqmf!4/225! ! Gjh/!4/44)c*! Djsdvju!ejbhsbn!sfgfssjoh!up! ! IW!tjef

Solution (a) V1 = 200

( 200) 2 = 50 W 800 200 = 0.25 A IW = 800 200 = 0.5 A Im = 400 Po =

!

4/239

Fmfdusjdbm!Nbdijoft

Io = 0.25 + j 0.5 Io = 0.56 A, cos fo = 0.4976 (b) Referring to HV side and neglecting the magnetizing branch [see Fig. 3.33(b)], 2 ¥ 1000 =5A 400 = 0.6 + j 1.6 = 1.71 –69.4° W = 5 ¥ 1.71 = 8.55 V = cos 69.4° = 0.352 lag = 8.55 ¥ 5 ¥ 0.352 = 15.048 W.

Ifl(HV) = Z Vsc P fsc Psc

! Qspcmfn!4/226 B!61!lWB-!22110221!W!usbotgpsnfs!jt!uftufe-!boe!uif!gpmmpxjoh!sftvmu!jt!pcubjofe;.! PD!uftu-!nfbtvsfnfout!po!MW!tjef;!411!X-!9!B-!221!W TD!uftu-!nfbtvsfnfout!po!IW!tjef;!861!X-!33!B-!:1!W Dpnqvuf!bmm!uif!qbsbnfufst!pg!uif!frvjwbmfou!djsdvju!sfgfssfe!up!uif!IW!boe!MW!tjef!pg!uif!usbot. gpsnfs/

Solution OC Test (LV Side) YOL = GOL =

8 = 0.072 � 110 300 (110) 2

= 0.025 �

BOL = Yo2 - Go2 = 0.0675 �. SC Test (HV Side) 90 = 4.09 W 22 750 = = 1.55 W ( 22) 2

ZeH = ReH

XeH = ( 4.09) 2 - (1.55) 2 = 3.785 W. Transformer ratio =

1100 = 10 110

Tjohmf.qibtf!Usbotgpsnfst

4/23:

Equivalent circuit referred to HV side, 2

Ê 1ˆ GOH = 0.072 ¥ Á ˜ = 0.072 ¥ 10–2 � Ë 10 ¯ 2

Ê 1ˆ BOH = 0.025 ¥ Á ˜ = 0.025 ¥ 10–2 � Ë 10 ¯ For equivalent circuit referred to LV side, 2

ReL

Ê 1ˆ = 4.09 ¥ Á ˜ = 4.09 ¥ 10–2 W Ë 10 ¯

XeL

Ê 1ˆ = 3.785 ¥ Á ˜ = 3.785 ¥ 10–2 W Ë 10 ¯

2

IH

I¢L IOH 1.55 W

0.072 ¥

VH 10–2 �

I¢OH

4.09 W

0.025 ¥ 10–1 �

(4.09 ¥ 10–2) W

I¢H

Vi

V¢H 0.072 �

IL

(3.785 ¥ 10–2) W 0.025 �

V¢I

! Gjh/!4/45)b*! Frvjwbmfou!djsdvju!sfgfssfe! ! up!IW!tjef!

Gjh/!4/45)c*! Frvjwbmfou!djsdvju!sfgfssfe ! up!MW!tjef

The equivalent circuits referred to both sides of the transformer are shown is Fig. 3.34(a) and Fig. 3.34(b).

! Qspcmfn!4/227 B!usbotgpsnfs!ibt!jut!nbyjnvn!fggjdjfodz!pg!:9/6&!bu!21!lWB!bu!vojuz!q/g/!Evsjoh!uif!ebz-!ju!jt! mpbefe!bt!gpmmpxt; 23!ipvst;!3!lX!bu!q/g/!1/7! 7!ipvst;!21!lX!bu!q/g/!1/9 7!ipvst;!31!lX!bu!q/g/!1/: Gjoe!uif!Ôbmm.ebzÕ!fggjdjfodz!pg!uif!usbotgpsnfs/!

Solution 10 ¥ 1000 ¥ 1 = 0.985 10 ¥ 1000 ¥ 1 + 2 Pi Pi = 76.14 W = Pc (76.14 kVA) 2 kW, 0.6 p.f, 3.33 kVA, 12 A, 2 ¥ 12 = 24 kWh/output) È Ê 3.33 ˆ 2 ˘ 76.14 Í1 + Á ˜ ˙ ¥ 12 = 1.01 kWh (loss) Î Ë 10 ¯ ˚ hmax =

!

4/241

Fmfdusjdbm!Nbdijoft

10 kW, 0.8, 12.5 kVA, 6 h, 10 ¥ 6 = 60 kWh (output) È Ê 12.5 ˆ 2 ˘ 76.14 Í1 + Á ˜ ˙ ¥ 6 = 1.17 kWh (loss) Î Ë 10 ¯ ˚ 20 kW, 0.9 p.f, 22.22 kVA, 6 h, 20 ¥ 6 = 120 kwh (output) È Ê 22.22 ˆ 2 ˘ 76.14 Í1 + Á ˜ ˙ ¥ 6 = 2.71 kWh (loss) ÍÎ Ë 10 ¯ ˙˚ 204 kWh (output), 4.89 kWh (loss) henergy (all day) =

204 ¥ 100% = 97.65%. 204 + 4.89

! Qspcmfn!4/228 Uif!gpmmpxjoh!uftu!sftvmut!xfsf!pcubjofe!gps!b!21!lWB-!61!I{-!33110331!W!ejtusjcvujpo!usbotgpsnfs;! ! !

PD!uftu!)MW*;!331!W-!2/17!B-!236/6!X TD!uftu!)IW*;!66!W-!9/3!B-!386!X

! )b*! Xifo!uif!usbotgpsnfs!jt!pqfsbufe!bt!b!tufq.epxo!usbotgpsnfs!xjui!uif!pvuqvu!wpmubhf!frvbm! up!331!W-!tvqqmzjoh!b!mpbe!bu!v/q/g/-!efufsnjof!uif!nbyjnvn!fggjdjfodz!boe!uif!v/q/g/!mpbe!bu! xijdi!ju!pddvst/ ! )c*! Efufsnjof!uif!q/g/!pg!uif!sbufe!mpbe-!tvqqmjfe!bu!351!W-!tvdi!uibu!uif!ufsnjobm!wpmubhf!pc. tfswfe!po!sfevdjoh!uif!mpbe!up!{fsp!jt!tujmm!331!W/

Solution OC Test (LV) YOL = GOL =

1.06 = 0.0048 � 220 125.5 ( 220) 2

= 0.0026 �

BOL = [(0.0048)2 – (0.0026)2]1/2 = 0.004 �. SC Test (HV) 55 275 = 6.71 W, R = = 4.09 W 8.2 (8.2) 2 X = 5.32 W Z=

\ (a) Pi = 125.5 W. Ifl (HV) =

10 ¥ 1000 = 4.55 A 2200 Pc, fl = (4.55)2 ¥ 4.09 = 84.67 W

Tjohmf.qibtf!Usbotgpsnfst

Load at maximum efficiency = 10 ¥

hmax (up.f.) =

4/242

125.5 = 12.17 KVA 84.67 12.17 ¥ 1 = 97.98%. 12.17 ¥ 1 + 2 ¥ 0.1255

(b) Voltage regulation = 0% cos f = cos tan–1

4.09 = 0.79 leading. 5.32

! Qspcmfn!4/229 B!mjhiujoh!usbotgpsnfs!sbufe!bu!21!lWB!ibt!gvmm.mpbe!mpttft!pg!1/4!lX!xijdi!jt!nbef!vq!frvbmmz!gspn! uif!jspo!mpttft!boe!uif!dpqqfs!mpttft/!Uif!evuz!dzdmf!dpotjtut!pg!gvmm.mpbe!gps!4!ipvst-!ibmg!gvmm.mpbe! gps!5!ipvst!boe!op.mpbe!gps!uif!sfnbjoefs!pg!b!35.ipvs!qfsjpe/!Jg!uif!mpbe!pqfsbuft!bu!vojuz!qpxfs! gbdups-!dbmdvmbuf!uif!bmm.ebz!fggjdjfodz/!

Solution The load operates at unity power factor. For the first three hours, Energy output = 10 ¥ 1 ¥ 3 kWh = 30 kWh For the next four hours, 1 Energy output = ¥ 10 ¥ 1 ¥ 4 = 20 kWh 2 Total energy output = (30 + 20) kWh = 50 kWh Full-load losses = 0.3 kW So,

Ê 0.3 ˆ iron loss = Á ˜ kW = 0.15 kW Ë 2 ¯ 0.3 kW = 0.15 kW Full-load copper loss = 2 Iron loss energy = (0.15 ¥ 24) = 3.6 kWh 0.15 Ê ˆ Copper loss energy = Á 0.15 ¥ 3 + ¥ 4˜ kWh = (0.45 + 0.15) kWh = 0.6 kWh 2 ( 2) Ë ¯ Energy loss = (3.6 + 0.6) kWh = 4.2 kWh

\

all-day efficiency =

50 = 0.922 or 92.2%. 50 + 4.2

!

4/243

Fmfdusjdbm!Nbdijoft

! Qspcmfn!4/22: Uif!pqfo.djsdvju!boe!tipsu.djsdvju!uftut!po!b!31!lWB-!61!I{-!35110351!W!usbotgpsnfs!hbwf!uif!gpm. mpxjoh!sftvmut; Ijhi.wpmubhf!xjoejoh!pqfo.djsdvjufe;!351!W-!2/149!B-!233!X Mpx.wpmubhf!xjoejoh!tipsu.djsdvjufe;!72/4!W-!9/44!B-!368!X ! )b*! Dpnqvuf!uif!fggjdjfodz!bu!gvmm.mpbe!boe!sbufe!wpmubhf!bu!1/9!q/g/! ! )c*! Bttvnjoh!uibu!uif!mpbe!qpxfs!gbdups!jt!wbsjfe!xijmf!uif!mpbe!dvssfou!boe!tfdpoebsz!ufsnjobm! wpmubhf!bsf!ifme!dpotubou-!efufsnjof!uif!qpxfs!gbdups!bu!xijdi!sfhvmbujpo!jt!nbyjnvn/

Solution Rated current of high-voltage side, IH =

20 ¥ 103 = 8. 33 A 2400

\ rated copper loss = 257 W Rated core loss = 122 W Total loss = 379 W (a) Efficiency at full-load and 0.8 p.f. h=

20 ¥ 103 ¥ 0.8 20 ¥ 103 ¥ 0.8 + 379

= 7.368 = 97.68%

(b) From the short-circuit data, the equivalent impedance ZeH =

61.3 W = 7.36 W 8.33

Equivalent resistance ReH = 3.7 W XeH = (7.36) 2 - (3.7) 2 = 6.362 W When regulation is maximum, the primary and secondary voltages of the transformer are in phase. The voltage drop across the transformer is 8.33 ¥ 7.36 = 61.3088 V \ voltage regulation =

61.3088 ¥ 100% = 2.55%. 2400

Tjohmf.qibtf!Usbotgpsnfst

4/244

! Qspcmfn!4/231 B! usbotgpsnfs! ibt! jut! nbyjnvn! fggjdjfodz! pg! 1/:86! bu! 31! lWB! bu! vojuz! q/g/! Evsjoh! uif! ebz-! ju! jt! mpbefe!bt!gpmmpxt; 21!ipvst;!4!lX!bu!1/7!q/g/ 9!ipvst;!21!lX!bu!1/9!q/g/ 7!ipvst;!31!lX!bu!1/:!q/g/ Gjoe!uif!bmm.ebz!fggjdjfodz/!

Solution kWh output = (10 ¥ 3) + (8 ¥ 10) + (6 ¥ 20) = 30 + 80 + 120 = 230 kWh As maximum efficiency is 0.975, so total losses under this condition are [1 – 0.975] = 0.025 of output power. At unity p.f. output power = 20 ¥ 1 = 20 kW Hence, losses = 0.025 ¥ 20.000 = 500 W 500 \ core losses = copper losses = W = 250 W 2 As core loss is constant for all p.f. so total core losses in 24 hours 250 ¥ 24 = kWh = 6 kWh 103 For the first 10 hours, 3 kVA load = =5 0.6 2 250 Ê 5ˆ kWh = 0.156 kWh Total copper losses = 10 ¥ Á ˜ ¥ Ë 20 ¯ 1000 For the next 8 hours, 10 kVA load = = 12.5 0.8 2 250 Ê 12.5 ˆ kWh = 0.781 kWh ¥ Total copper losses = 8 ¥ Á Ë 20 ˜¯ 1000 For the last 6 hours, 20 kVA load = = 22.22 0.9 2 250 Ê 22.22 ˆ Total copper losses = 6 ¥ Á ¥ kWh = 1.85 kWh Ë 20 ˜¯ 1000 Total copper losses = 0.156 + 0.781 + 1.85 = 2.79 kWh Total loss = 6 kWh + 2.79 kWh = 8.79 kWh 230 All-day efficiency = = 0.963 or 96.3%. 230 + 8.79

!

4/245

Fmfdusjdbm!Nbdijoft

! Qspcmfn!4/232 Jo!b! TvnqofsÕt!uftu-!uif!xbuunfufs!sfbejoh!po! IW! tjef! boe! MW! tjef! pg! uxp! jefoujdbm! tjohmf.qibtf! usbotgpsnfst!sbufe!611!lWB-!2201/5!lW-!61!I{-!jt!7111!X!boe!26111!X!sftqfdujwfmz/!Po!ibmg!gvmm. mpbe!boe!1/9!q/g/-!gjoe!pvu!uif!fggjdjfodz!pg!fbdi!usbotgpsnfs/!Bmtp!gjoe!pvu!uif!nbyjnvn!fggjdjfodz! bu!vojuz!qpxfs!gbdups/

Solution 15000 = 7500 W = 7.5 kW 2 6000 = 3000 W Copper losses = 2 Copper losses at half full-load = (0.5)2 (3000) = 750 W = 0.75 kW Iron losses =

0.5 ¥ 500 ¥ 0.8 ¥ 100 0.5 ¥ 500 ¥ 0.8 + 7.5 + 0.75 = 96.04%

Efficiency =

Let us consider x to be the fraction of full-load at which maximum efficiency occurs. Hence,

x= =

Iron loss Cu-loss 7500 = 1.58 3000

kVA for maximum efficiency = 1.58 ¥ 500 = 790 kVA 790 ¥ 1 ¥ 100 = 98.14%. \ maximum efficiency = 790 ¥ 1 + 7.5 + 7.5

! Qspcmfn!4/233 B!61!lWB-!33110221!W!usbotgpsnfs!xifo!uftufe!hbwf!uif!gpmmpxjoh!sftvmut; ! !

PD!uftu;!MW!tjef!Æ!511!X-!21!B-!221!W TD!uftu;!IW!tjef!Æ!2111!X!31/6!B!211!W

Dbmdvmbuf!bmm!uif!qbsbnfufst!pg!uif!frvjwbmfou!djsdvju!sfgfssfe!up!IW!boe!MW!tjeft!pg!uif!usbotgpsnfs/

Solution OC Test (LV Side) 400 = 0.3636 110 ¥ 10 Iw = 10 cos qo = 3.636 A

cos qo =

Tjohmf.qibtf!Usbotgpsnfst

4/246

110 110 = = 30.25 W I w 3.636 Im = 10 sin qo = 9.315 A 110 XOL = = 11.81 W. 9.315

\

ROL =

SC Test (HV Side) 100 = 4.87 W 20.5 1000 = = 2.38 W ( 20.5) 2

ZeH = ReH

XeH = ( 4.87) 2 - ( 2.38) 2 = 4.25 W For equivalent circuit referred to LV side 2

Ê 1ˆ ReL = 2.38 ¥ Á ˜ = 5.95 ¥ 10–3 W Ë 20 ¯ 2

Ê 1ˆ XeL = 4.25 ¥ Á ˜ = 10.6 ¥ 10–3 W Ë 20 ¯ [where, transformer ratio = Equivalent circuit referred to LV side is shown in Fig. 3.35(a). 5.95 ¥ 10–3 W

I¢H I¢O V¢H

30.25 W

IL

10.6 ¥ 10–3 W 11.81 W

VL

! Gjh/!4/46)b*! Frvjwbmfou!djsdvju!sfgfssfe!up!mpx.wpmubhf!tjef

For equivalent circuit referred to HV side 2

ROH

Ê 2200 ˆ = 12100 W = RoL ¥ Á Ë 110 ˜¯

XOH

Ê 2200 ˆ = 4724 W = XOL ¥ Á Ë 110 ˜¯

2

Equivalent circuit referred to high-voltage side is shown in Fig. 3.35(b).

2200 = 20 ] 110

!

4/247

Fmfdusjdbm!Nbdijoft

2.38 W

IH

4.25 W

I¢L

IO VH

12100 W

4724 W

V¢L

! Gjh/!4/46)c*! Frvjwbmfou!djsdvju!sfgfssfe!up!uif!ijhi.wpmubhf!tjef

QBSBMMFM!!PQFSBUJPO!!PG!UXP! TJOHMF.QIBTF!USBOTGPSNFST

4/38

For operating transformers in parallel, their primaries are connected to the supply bus and the secondaries are connected together to form a secondary bus to feed the common load. The following are the advantages of operating the transformers in parallel. 1. With two or more transformers feeding a load, the systems become more reliable. 2. Switching of transformers can be done depending upon the demand, so as to achieve economical and efficient operation of the system. 3. The size and cost of stand by transformer is much less. 4. There is always a scope of future expansion of a substation. The conditions which must be fulfilled for the satisfactory parallel operation of transformers are as follows: 1. The transformers must be connected in the same polarity. 2. The transformation ratios of the transformers should be same. 3. Equivalent impedances of the transformers should be divided in inverse proportion to the current ratings, that is the internal impedance drop of the transformers should be equal. 4. The ratio of equivalent leakage reactance to equivalent resistance should be equal for all the transformers. This condition ensures that the transformers share active and reactive power according to their ratings. Out of the conditions listed above, the first two conditions should be strictly fulfilled, otherwise a large circulating current will flow. Thus, the first two conditions are essential for the parallel operation of the transformers, whereas the last two conditions are desirable.

4/38/2! Mpbe!Tibsjoh!xjui!Frvbm!Uvsot!Sbujp Consider two single-phase transformers m and n connected in parallel fulfilling all the conditions mentioned above. The arrangement is shown in Fig. 3.36. According to condition (3), the ratio of currents given by

Tjohmf.qibtf!Usbotgpsnfst

4/248

I m Zn = I n Zm

(3.53)

where Im and In are the currents supplied by the transformers m and n respectively. If I is the total current supplied to the load then I = Im + In

(3.54)

Solving Eqs (3.53) and (3.54), Zn I Zm + Zn Zm In = I Zm + Zn

Im =

(3.55) (3.56)

Xm X and n are equal then the current Im and In will be in phase and the total Rm Rn current I will be the arithmetic sum of Im and In. When the ratios

Xm

Rm

Transformer m L

Im

I

Zm 1-f ac supply

L o a d

Em

N Rn

Transformer n

Xn

In

Zn En

! Gjh/!4/47! Uxp!tjohmf.qibtf!usbotgpsnfst!jo!qbsbmmfm

Xn X and m are not equal then the current Im and In will not be in phase although the magRn Rm nitude of currents can be inversely proportional to the magnitudes of impedances. At rated output, the phasor sum of Im and In will be the total current I under such condition. As the phasor sum is less than the arithmetic sum, the load kVA is less than sum of the ratings of the transformers. But if

!

4/249

Fmfdusjdbm!Nbdijoft

4/38/3! Mpbe!Tibsjoh!xjui!Vofrvbm!Uvsot!Sbujp When the turn ratios are not equal then the secondary induced emfs are unequal. This inequality will produce a circulating current which will flow under no-load condition. The undesirable effects of circulating currents are as follows: 1. Increases the copper losses 2. Overload any one transformer 3. Reduces the permissible output of the bank On load condition, this circulating current will be superimposed on the load current. Let Em and En be the secondary induced emf in transformers m and n. The circulating current Ic under no-load condition is given by Ic =

Em - En Zm + Zn

If the load impedance is ZL then Em = Im Zm + I ZL = Im Zm + (Im + In) ZL En = In Zn + I ZL = In Zn + (Im + In) ZL

(3.57) (3.58)

Solving Eqs (3.57) and (3.58), we get Im =

Em Z n + ( Em - En ) Z L Zm Zn + ( Zm + Zn )Z L

(3.59)

In =

En Z m - ( Em - En ) Z L Zm Zn + ( Zm + Zn )Z L

(3.60)

and the total current I is the summation of Im and In. I=

Em Z n + En Z m Zm Zn + ( Zm + Zn )Z L

(3.61)

Secondary terminal voltage across load V is given by, V = I ZL =

Em Z n + En Z m Zm Zn + Zm + Zn ZL

(3.62)

In general, the load sharing of any number of transformers in parallel can more readily be found out by applying the Millman’s theorem which states that the secondary terminal voltage V of a number of transformers connected in parallel across a load impedance of ZL and having respective

Tjohmf.qibtf!Usbotgpsnfst

4/24:

internal impedances Zm, Zn, Zp, etc., is given by E 1 1 È 1 ˘ E + + + º˙ = m + n + … (3.63) VÍ Î Z L Zm Zn ˚ Zm Zn Hence, in case of a large number of transformers operating in parallel on a system with different turns ratio, the terminal voltage across the load can be ascertained using Eq. (3.63).

BVUP.USBOTGPSNFS!

4/39

A single-phase auto-transformer is a one-winding transformer in which a part of the winding is common to both high-voltage and low-voltage sides. Consider a single winding xyz of Fig. 3.37. The terminals x and z are the high-voltage terminals where y and z are the low-voltage terminals. y is the suitable tapping point. The winding yz is called common winding as it is common for both high-voltage and low-voltage sides.

TUFQ.EPXO!BVUP.USBOTGPSNFS!

4/3:

In a step-down auto-transformer, the primary voltage is greater than the secondary voltage. The source voltage V1 is applied to the full winding xyz and the load is connected across the secondary terminal yz. This type of arrangement is known as step-down auto-transformer and shown in Fig. 3.37. I1 x I1 = Ixy Nxy = N1 – N2 V1

Nxz = N1 Nyz = N2 y

p I2 = IL = Iyp I = Izy VL

z

Load

!

Gjh/!4/48! Tufq.epxo!bvup.usbotgpsnfs

Since the transformer windings are physically connected, a different terminology is used for the auto-transformer than for other types of transformers. Let N1 = Nxz = Number of turns of full winding xyz = Number of turns of the HV side.

!

4/251

Fmfdusjdbm!Nbdijoft

N2 = Nyz = Number of turns of the common winding yz = Number of turns of the LV side Nxy = N1 – N2 = Number of turns of the series winding xy V1 = Input voltage on the HV side V2 = Output voltage on the LV side I1 = Input current in the HV side I2 = Output current in the LV side Current in the series windings = Ixy = I1 Current in the common winding = Izy = 1 Two voltage ratios of the auto-transformer are circuit voltage ratio and winding voltage ratio. The circuit voltage ratio V1 N1 = = aA (3.64) V2 N 2 The quantity aA is called the transformation ratio of the auto-transformer. It is seen from Eq. (3.64) that aA is always greater than 1. When the load is connected across the secondary terminals, a current I flows in the common winding yz. It has a tendency to reduce the main flux but the primary current I1 increases to such a value that the mmf in windings xy neutralizes the mmf in windings yz. That is, Ixy Nxy = Iyz Nyz I1 (N1 – N2) = I N2 N - N 2 N1 I = 1 –1 = aA – 1 = N2 N2 I1

(3.65) (3.66)

The winding voltage ratio is a=

Vxy V yz

=

N xy N yz

a = aA – 1

=

N1 - N 2 N2 (3.67)

As the right-hand side of Eq. (3.67) is a pure number, it follows that the current in windings xy and zy are in phase. By KCL at the point y, I2 = I1 + I (3.68) (3.69) I = I2 – I1 Equation (3.66) becomes I = aA – 1 I1 I 2 - I1 = aA – 1 I1

Tjohmf.qibtf!Usbotgpsnfst

I2 = aA I1 I I1 = 2 aA

and From Eqs (3.65) and (3.71),

Let

I I2 E1 E2 Exy E xy E yz

\ and

E xy E yz

aA -1 aA = Exz = Eyz = Exz – Eyz E xz - E yz E xz = = –1 E yz E yz E = 1 –1 E2 =

= aA –1

aA =

4/252

(3.70) (3.71)

(3.72)

(3.73) (3.74)

E xy

+1 E yz aA = a + 1

(3.75)

where a=

E xy E yz

= Two-winding transformation ratio Equation (3.75) shows that the transformation ratio of an auto-transformer is greater than that if the same set of windings were connected as a two-winding transformer. Equations (3.66) and (3.72) show that the ratio of voltages and currents in the windings xy and yz are the same as if turn Nxy formed the primary and the turn Nyz formed the secondary of an ordinary transformer having a ratio of transformation of (aA – 1). Thus, the transformer action present in an auto-transformer takes place in the windings xy and yz.

QPXFS!SFMBUJPO!JO!BO!BVUP.USBOTGPSNFS!

4/41

An auto-transformer transfers electrical power between primary and secondary circuits partly through induction and party by conduction. Thus, an auto-transformer has two different types of volt-amperes: the transformed volt-amperes and conducted volt-amperes. The transformer action in the auto-transformer takes place in the windings xy and yz (Fig. 3.37) and at balance condition, the volt-ampere in the respective winding should be equal.

!

4/253

Fmfdusjdbm!Nbdijoft

Exy Ixy = Eyz Iyz = ST

(3.76)

ST is known as transformed volt-ampere. As the winding resistances and leakage reactance are negligible, transformed VA can be modified as ST = Vxy Ixy = Vyz Iyz

(3.77)

The input volt-amperes of the auto-transformer Sin = Vxz Ixy = (Vxy + Vyz) Ixy = Vxy Ixy + Vyz Ixy or

Sin = ST + Vyz Ixy

(3.78)

Vyz Ixy is known as conducted volt-amperes (Sc) where,

Sc = Vyz Ixy Sin = ST + Sc

(3.79) (3.80)

VA (Sc) = V2 I1 VA (ST) = Vyz Iyz = V2 I

(3.81) (3.82)

In Fig. 3.37, Conducted Transformed

An auto-transformer is rated on the basis of output VA rather than the transformer’s VA. In Fig. 3.37, if the auto-transformer is used as two winding transformer, its VA rating is (VA)TW + (V1 – V2) I1 = V2 I = (I2 – I1) V2 For an auto-transformer, (VA)auto = V1 I1 = V2 I2 The ratio of output VA of an auto-transformer and equivalent two-winding transformer is (VA)auto V2 I 2 I 2 = = I (VA)TW V2 I

(3.83)

(VA)auto aA = (VA)TW a A - 1

(3.84)

(VA)auto > (VA)TW

(3.85)

From Eq. (3.72), \ From Eq. (3.84) shows that

This shows that a two-winding auto-transformer will have greater VA rating than a two-winding conventional transformer.

Tjohmf.qibtf!Usbotgpsnfst

4/254

TUFQ.VQ!BVUP.USBOTGPSNFS!

4/42

In case of a step-up auto-transformer, the secondary voltage is greater than primary voltage. Hence, the source is connected to the yz windings and the load is connected across the full windings xyz. The arrangement is shown in Fig. 3.38. By using balancing mmf condition in windings xy and yz, I2 Nxy = I Nyz

(3.86)

Let the transformation for the step-up auto-transformer be defined as aA =

N xz V2 = N yz V1

(3.87)

From Eq. (3.83), N xy N xz - N yz N xz I = = = – 1 = aA –1 I2 N yz N yz N yz I = aA –1 I2

\

(3.88)

I + I2 = aA I2 x

I2

I1

SOURCE

y

V2

LOAD

I

V1 z

!

Gjh/!4/49! Tufq.vq!bvup.usbotgpsnfs

Again, and

I + I2 = I1 I1 = aA I2

(3.89) (3.90)

Dividing Eq. (3.88) by Eq. (3.90),

I aA -1 = aA I1

(3.91)

!

4/255

Fmfdusjdbm!Nbdijoft

By using KVL in the high-voltage side, Exy + Eyz = Exz Exy = Exz – Eyz = aA Eyz – Eyz = (aA –1) Eyz

\

TBWJOH!JO!DPOEVDUPS!NBUFSJBM! JO!BO!BVUP.USBOTGPSNFS

4/43

The length of the conductor in a winding is proportional to the number of turns and cross section of a conductor is proportional to the current through it. So the total weight of the conductor material is proportional to the product of current and number of turns. For a two-winding transformer, the weight of the conductor material in primary is k I1 N1, the weight of the conductor material in secondary is k I2 N2. Hence, total weight of the conductor material is k (I1 N1 + I2 N2). For the auto-transformer (Fig 3.37), the portion xy has (N1 – N2) turns and the current through it is I1. Therefore, the weight of the conductor material in section xy is k I1 (N1 – N2). The portion yz has N2 turns and the current through it is I (= I2 – I1). Therefore, the weight of the conductor material in section yz K (I2 – I1) N2 Total weight of conductor material K [I1 (N1 – N2) + (I2 – I1) N2] Now, consider Mauto = Weight of conductor material in an auto-transformer M2w = Weight of conductor material in a two-winding transformer Therefore, M auto I1 ( N1 - N 2 ) + ( I 2 - I1 ) N 2 = I1 N1 + I 2 N 2 M 2w =

( I1 N1 + I 2 N 2 ) - 2 I1 N 2 I1 N1 + I 2 N 2

=

2 I1 N1 + 2 I1 N 2 2 I1 N1

=1–

N2 1 =1N1 aA

(3.92)

Tjohmf.qibtf!Usbotgpsnfst

\

M auto 1 =1– M 2w aA M 1 1 – auto = p.u. M 2w aA

4/256

(3.93)

Saving of conductor material in using an auto-transformer, 1 = M2w – Mauto = M2w aA 1 If = 0.1, saving of conductor material is 10 %. aA 1 = 0.9, saving is 90%. Hence, the use of an auto-transformer is more economical when If aA 1 aA

Ê N2 ˆ ÁË = N ˜¯ is close to unity. The ratios of transformations used in auto-transformers are 3:1 or 4:1. 1

V1 = 3, V2 V If 1 = 2, V2 V If 1 = 4, V2 If

M auto 1 2 =1- = 3 3 M 2w M auto 1 1 =1- = 2 2 M 2w M auto 1 3 =1- = 4 4 M 2w

If the ratio of transformation is greater than 4, the advantage in the reduction of conductor material is not much.

BEWBOUBHFT!BOE!EJTBEWBOUBHFT! PG!BO!BVUP.USBOTGPSNFS

4/44

Bewboubhft Auto-transformers have the following advantages: 1. Less amount of copper is required compared to a two-winding transformer. 2. Due to smaller size, auto-transformers are cheaper than the two-winding transformer of same output. 3. It is possible to get smooth and continuous variation of voltage. 4. The resistance and reactance are less due to the absence of one winding. Hence, the voltage regulation of the auto-transformer is superior. 5. Volt-ampere rating is more. 6. Due to reduction in core material and conductor, the ohmic losses and core losses are small which provides higher efficiency than the equivalent two-winding transformer.

!

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Fmfdusjdbm!Nbdijoft

Ejtbewboubhft In spite of various advantages of auto-transformers, a number of disadvantages are present. They are as follows: 1. Since there is no electrical isolation between primary and secondary, the risk factor appears at high-voltage level. 2. The effective p.u. impedance of an auto-transformer is smaller compared to a two-winding transformer. That is why, there is a high possibility of high short circuit currents for shortcircuits on the secondary side. 3. The full primary current will appear across the secondary causing higher voltage on secondary terminal resulting in danger of accidents if the common winding is open circuited.

BQQMJDBUJPO!PG!BO!BVUP.USBOTGPSNFS!

4/45

1. Auto-transformers with a number of tappings are used for starting induction motors and synchronous motors. 2. Auto-transformers are used as variable a.c. voltage source (variac). 3. Auto-transformers are used as boosters to give a small boost to a distribution cable for compensating the voltage drop. 4. They are used as interconnections of power systems of different voltage levels. 5. They can be used as furnace transformers to supply power to the furnaces at the required supply voltage.

QVMTF!USBOTGPSNFS!

4/46

Pulse amplitude

0

Time Pulse duration

!

Gjh/!4/4:! Jefbm!sfdubohvmbs!qvmtf!joqvu

Figure 3.39 shows an ideal rectangular pulse input to the pulse transformer which couples a source of pulses of electrical energy to the load. Its shape as well as other properties are maintained properly. The size of these transformers is small and they have few turns. The inter-winding

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4/258

capacitance of these transformers is low. The leakage reactance of the windings is small. Ferrites, or wound strips, of high-permeability alloys like permalloy are used to construct the cores. Therefore, these transformers have high magnetizing inductance. A high-voltage insulation is used between windings and ground. Figure 3.39 shows practical rectangular input pulse to such transformers. Small pulse transformers are used in computers, pulse generators, and so on, whereas large pulse transformers are used in radar systems, and so on. Vtbhf!pg!Qvmtf!Usbotgpsnfs Pulse transformers have the following usages: • SCR and switching transistors • Radar systems • Cathode ray tubes • Microwave tube circuits • Data-handling circuits • Analog switching applications • Transmission-line pulse transformers are also useful in fast pulse signal transmission applications • Pulse transformers are also useful in digital signal processing.

XFMEJOH!USBOTGPSNFS!

4/47

A welding transformer has thin primary windings with a large number of turns, whereas the secondary has less number of turns with less voltage and high current. One end of the secondary winding is connected to the pieces to be welded and another end, to the welding electrode as shown in Fig. 3.40. When high current is flowing, heat is produced due to the contact resistance between the electrode and the pieces. Highly reactive winding is used for this type of transformer, otherwise a separate reactor may be added in series with the secondary winding. Figure 3.41 shows the volt-ampere characteristic of a welding transformer. Core Electrode Arc

ac supply

Pieces to be welded

Welding transformer

Gjh/!4/51! Xfmejoh!usbotgpsnfs

!

!

4/259

Fmfdusjdbm!Nbdijoft

v Open circuit voltage Operating point Arc voltage

Short circuit current l

!

Gjh/!4/52! Wpmu.bnqfsf!dibsbdufsjtujd!pg!b!xfmejoh!usbotgpsnfs

Sfbdupst!Vtfe!xjui!Xfmejoh!Usbotgpsnfst! To control the arc, various reactors are used with welding transformers. Some methods to control the arc are given below: )b*!Ubqqfe!Sfbdups! With the help of taps on the reactor, the output current is regulated. This

has limited number of current settings as shown in Fig. 3.42. Selector Secondary Tappings

Primary

Reactor

Electrode Work

Welding transformer

! Gjh/!4/53! Ubqqfe!sfbdups

)c*!Npwjoh.dpjm!Sfbdups! Figure 3.43 shows a moving-coil reactor in which the reactive distance between primary and secondary is adjusted. The current becomes less if the distance between the coils is large.

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4/25:

Arrangement for moving coil

Primary Secondary

Welding transformer Electrode Work

! Gjh/!4/54! Npwjoh.dpjm!sfbdups

)d*! Npwjoh.tivou! Sfbdups! Figure 3.44 shows a moving shunt reactor in which the position

of the central magnetic shunt can be adjusted. Change of the output current is obtained due to the adjustment of the shunted flux. )e*! Dpoujovpvtmz! Wbsjbcmf! Sfbdups! Figure 3.45 shows a continuously variable reactor in

which the height of the reactor is continuously varied. Greater reactance is obtained due to greater core insertion and hence the output current is less. )f*! Tbuvsbcmf! Sfbdups! Figure 3.46 shows a saturable reactor. To adjust the reactance of the

reactor, the required dc excitation is obtained from a dc controlled transducer. Reactor approaches saturation if the dc excitation current is more. Therefore, changes of current is obtained due to the change of reactance.

Moving shunt

Primary Electrode Work ! Gjh/!4/55! Npwjoh!tivou!sfbdups

!

4/261

Fmfdusjdbm!Nbdijoft

Reactor Reactor coil Electrode Work

Primary Welding transformer

Gjh/!4/56! Dpoujovpvtmz!wbsjbcmf!sfbdups

dc supply – +

dc transducer

Secondary

Primary Welding transformer

Electrode Work ! Gjh/!4/57! Tbuvsbcmf!sfbdups

JOTUSVNFOU!USBOTGPSNFS!

4/48

There are two types of instrument transformers, one is current transformer and another is potential transformer.

4/48/2! Dvssfou!Usbotgpsnfs Since the large alternating current cannot be passed through normal ammeters and current coils of wattmeters, the current is reduced with the help of a current transformer. A current transformer is a device having two windings called primary and secondary. It transfers energy from one side to another with suitable change in level of current or voltage. It has a primary coil with a few turns having heavy cross-sectional area. This side is connected in series with the line carrying high

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current as shown in Figure 3.47. On the other hand, the secondary is made up of many turns of fine wire having a rating of 5 A current. This is connected to the coil of a low-range meter. Current transformers are basically step-up transformers, that is, stepping up the voltage and hence stepping down the current. If I1 and I2 be the currents in primary and secondary sides and N1 and N2 be the number of turns of primary and secondary windings, we can write I 2 N1 = I1 N 2 If the current ratio of of CT and meter reading is known, the actual line current value can be determined. The secondary of a CT should not be kept open. If it is kept open, secondary current will be zero and ampere turns produced by the secondary will be zero. Since the secondary ampere turns oppose the primary ampere turns, the counter mmf will be zero. Hence, unopposed primary produces heavy flux in the core which causes excessive core losses. Hence, large emf will be induced in primary and secondary. This will damage the insulation which is dangerous from the operating point of view. Core

Secondary

Heavy line current

To load

Bar primary

I1

Secondary

I1

CT

Primary

I2

0–5 A

I2

0.5 A

A

A

Ammeter

Ammeter (b) Bar primary

(a) Winding primary

! Gjh/!4/58! Dvssfou!usbotgpsnfs

4/48/3! Qpufoujbm!Usbotgpsnfs These transformers reduce the voltage and have the same basic principle as current transformers. Figure 3.48 shows a potential transformer. The primary winding of PT consists of a large number of turns and the secondary consists of a small number of turns. The secondary is always grounded for safety purposes.

!

4/263

Fmfdusjdbm!Nbdijoft

PT

High voltage

V

Primary of PT

0 – 110 V voltmeter

Load

Secondary of PT

! Gjh/!4/59! Qpufoujbm!usbotgpsnfs

USBOTGPSNFS!DPPMJOH!

4/49

The core and copper losses cause heating of transformers. It is necessary to ensure that temperature of the transformer does not exceed the maximum value, otherwise it may cause damage to the insulation. The following are the methods for cooling these types of transformers. )b*!Bjs!Obuvsbm!Dppmjoh! Small transformers up to 25 kVA are cooled by natural circulation of

air surrounding them. )c*!Bjs!Cmbtu!Dppmjoh! In this type of cooling, a continuous blast of filtered air is forced through

the core and windings for better cooling. )d*!Pjm!Obuvsbm!Dppmjoh! A majority of transformers have their core and windings immeresed

in oil. Oil is a good insulating material and provides better heat dissipation than air. Oil-immersed transformers are enclosed in a sheet-steel tank. The heat produced in the transformer is passed to the oil. The oil is heated and it becomes lighter and rises to the top and its place is taken by cool air from the bottom of the tank. The heat of the air is transferred to the tank by natural circulation of air. The heat is then transferred to the surrounding atmosphere. )e*!Pjm!Cmbtu!Dppmjoh! Here, forced air is passed over cooling elements of a transformer immersed

in oil. )f*!Gpsdfe!Pjm!boe!Gpsdfe!Bjs!Dppmjoh! Heated oil is taken from the top of the transformer

tank to a cooling plant. Cooled oil is then circulated through the bottom of the tank. )g*!Gpsdfe!Pjm!boe!Xbufs!Dppmjoh! In this type of cooling, metallic tubes are situated inside the

tank, below the oil level. Water is circulated through these tubes to extract heat from the oil.

Tjohmf.qibtf!Usbotgpsnfst

DPOTFSWBUPST!BOE!CSFBUIFST!

4/264

4/4:

A conservator is an air tight metal drum placed above the top of the oil tank in oil cooled transformers. It is connected with the tank through a pipe. It takes up the expansion of oil with the change of temperature. As the conservator is air tight it does not allow the oil to come in contact with atmospheric air which may contain moisture. The moisture destroys the insulating properties of the oil. The oil expands and contracts due to the change in temperature. When the transformer cools the oil level goes down and the air is drawn in. The incoming air is passed through a device called breather, which contains silica gel or calcium chloride to extract moisture from air. Silica gel is checked regularly and replaced regularly.

OBNFQMBUF!BOE!SBUJOH!PG!B!TJOHMF.! QIBTF!USBOTGPSNFS

4/51

The specifications of transformers are given by BIS (Bureau of Indian Standards) 2026. As per this standard, every transformer must be provided with the following specifications: Type (power, distribution, auto, etc.), year of manufacture, number of phases, rated kVA, rated frequency, rated voltage of each winding, connection symbol, percent impedance voltage at rated current, type of cooling, total mass, mass and volume of insulating oil.

! Qspcmfn!4/234 Uxp!311!lWB!tjohmf.qibtf!usbotgpsnfst!bsf!up!cf!pqfsbufe!jo!qbsbmmfm/!Uif!joufsobm!jnqfebodf!pg! usbotgpsnfs!2!jt!)1/117!,!k1/19*!q/v/!xijmf!usbotgpsnfs!3!ibt!bo!joufsobm!jnqfebodf!pg!)1/119!,! k1/16*!q/v/!Ipx!xjmm!uifz!tibsf!b!mpbe!pg!411!lX!bu!1/9!mbhhjoh!qpxfs!gbdups@

Solution Z1 = (0.006 + j 0.08) = 0.08 –85.71° Z2 = (0.008 + j 0.05) = 0.0506 –80.91° 300 SL = –cos–1 0.8 = 375 ––36.87° kVA 0.8

Load

Load shared by transformer 1 Z2 SL S1 = Z1 + Z 2 \

S1 = =

0.0506 –80.91∞ ¥ 375 ––36.87° (0.006 + 0.008) + j (0.08 + 0.05) 0.0506 –80.91∞ ¥ 375 –-36.87∞ = 145.12 ––39.81° kVA. 0.13075 –83.85∞

!

4/265

Fmfdusjdbm!Nbdijoft

Load shared by transformer 2

\

S2 =

Z1 SL Z1 + Z 2

S2 =

0.08 –85.71∞ ¥ 375 ––36.87° (0.006 + 0.008) + j (0.08 + 0.05)

= 229.44 ––35° kVA.

! Qspcmfn!4/235 Uxp!tjohmf.qibtf!usbotgpsnfst!sbufe!2111!lWB!boe!911!lWB!bsf!dpoofdufe!jo!qbsbmmfm!up!tvqqmz!b! mpbe!pg!2611!lWB!bu!1/9!mbhhjoh!q/g/!Uif!sftjtubodf!boe!sfbdubodf!pg!uif!gjstu!usbotgpsnfs!bsf!4&! boe!7&!sftqfdujwfmz!boe!pg!uif!tfdpoe!usbotgpsnfs!bsf!3&!boe!8&!sftqfdujwfmz/!Dbmdvmbuf!uif!lWB! mpbejoh!boe!q/g/!bu!xijdi!fbdi!usbotgpsnfs!pqfsbuft/

Solution Choose kVA base of 1000 kVA. Z1 (p.u.) = (0. 03 + j 0.06) ¥

1000 1000

= 0.03 + j 0.06 Z2 (p.u.) = (0.02 + j 0.07) ¥

1000 800

= 0.025 + j 0.0875 Z2 SL S1 = Z1 + Z 2 =

[0.025 + j 0.0875] ¥ 1500 [0.8 – j 0.6] [0.055 + j 0.1475]

=

0.09100 –74.05 [1200 – j 900] 0.15742 –69.55

= 0.5781 –4.5 ¥ 1500 ––36.86 = 867.15 ––32.36 kVA S2 = =

[0.03 + j 0.06] 1500 ––36.86 0.15742 –69.55 0.067082 –63.43 1500 ––36.86 0.15742 –69.55

= 639.20 ––42.98 kVA.

Tjohmf.qibtf!Usbotgpsnfst

4/266

! Qspcmfn!4/236 Uxp!tjohmf.qibtf!usbotgpsnfst-!sbufe!2111!lWB!boe!711!lWB!sftqfdujwfmz-!bsf!dpoofdufe!jo!qbsbmmfm! po!cpui!IW!boe!MW!tjeft/!Uifz!ibwf!frvbm!wpmubhf!sbujoht!pg!22!lW0551!W!boe!uifjs!qfs!voju!jnqfe. bodft!bsf!)1/13!,!k1/18*!boe!)1/136!,!k1/1986*!sftqfdujwfmz/!Xibu!jt!uif!mbshftu!wbmvf!pg!uif!vojuz! qpxfs!gbdups!mpbe!uibu!dbo!cf!efmjwfsfe!cz!uif!qbsbmmfm!dpncjobujpo!bu!uif!sbufe!wpmubhf@

Solution S1 (rated) = 1000 kVA; S2 (rated) = 600 kVA Choose a kVA base of 1000. Z1 = 0.02 + j 0.07 = 0.0728 –74° Z 2 = (0.025 + j 0.0875) ¥ 2 = 0.182 –74° \

Z1 + Z 2 = 0.255 –74° S1 =

Z2 SL Z1 + Z 2

(i)

S2 =

Z1 SL Z1 + Z 2

(ii)

From Eq. (i), SL = 1000 ¥

0.255 = 1400 kVA 0.182

From Eq. (ii), SL = 600 ¥

0.255 = 2101.65 kVA 0.0728

As total load is increased, the 1000 kVA transformer will be the first to reach its full-load. SL (max) = 1400 kVA.

! Qspcmfn!4/237 Uxp!tjohmf.qibtf!usbotgpsnfst!sbufe!811!lWB!boe!711!lWB!sftqfdujwfmz!bsf!dpoofdufe!jo!qbsbmmfm! up!tvqqmz!b!mpbe!pg!2111!lWB!bu!1/9!mbhhjoh!qpxfs!gbdups/!Uif!sftjtubodf!boe!sfbdubodf!pg!uif!gjstu! usbotgpsnfs!bsf!4&!boe!7/6&!sftqfdujwfmz-!boe!pg!uif!tfdpoe!usbotgpsnfs!bsf!2/6&!boe!9&!sftqfd. ujwfmz/!Dbmdvmbuf!lWB!mpbejoh!boe!uif!qpxfs!gbdups!bu!xijdi!fbdi!usbotgpsnfs!pqfsbuft/

Solution S1 = 700 kVA S2 = 600 kVA SL = 1000 KVA, 0.8 p.f. lagging

!

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Fmfdusjdbm!Nbdijoft

Choose a kVA base of 1000 kVA. Z1 (p.u.) = (0.03 + j 0.065) ¥

1000 700

= 0.043 + j 0.093 1000 = 0.025 + j 0.134 600 Z2 0.025 + j 0.134 SL = S1 = ¥ 1000 (0.8 – j 0.6) Z1 + Z 2 0.068 + j 0.227 = 136.39 ––64.2 = 136.39 kVA at 0.44 p.f. lagging Z1 0.043 + j 0.093 SL = S2 = ¥ 1000 (0.8 – j 0.6) Z1 + Z 2 0.068 + j 0.227 = 90.09 ––76.7 = 90.09 kVA at 0.23 p.f. lagging.

Z 2 (p.u.) = (0.015 + j 0.08) ¥

! Qspcmfn!4/238 B!31!lWB-!51110311!W-!61!I{!usbotgpsnfs!xjui!bo!frvjwbmfou!jnqfebodf!pg!1/13!W!jt!up!pqfsbuf!jo! qbsbmmfm!xjui!b!26!lWB-!51110311!W-!61!I{!usbotgpsnfs!xjui!bo!frvjwbmfou!jnqfebodf!pg!1/136!W/! Uif!uxp!usbotgpsnfst!bsf!dpoofdufe!jo!qbsbmmfm!boe!nbef!up!dbssz!b!mpbe!pg!36!lWB/!Bttvnf!cpui! uif!jnqfebodft!up!ibwf!uif!tbnf!bohmf/ ! )b*! Gjoe!uif!joejwjevbm!mpbe!dvssfout/ ! )c*! Xibu!qfsdfou!pg!uif!sbufe!dbqbdjuz!jt!vtfe!jo!fbdi!usbotgpsnfs@

Solution Z1 = 0.02 W,

Z2 = 0.025 W

Since the impedances have the same angle, Z1 + Z 2 = 0.02 + 0.025 = 0.045 W S1 =

Z2 0.025 SL = ¥ 25 = 13.89 kVA 0.045 Z1 + Z 2

S2 =

Z1 0.02 = ¥ 25 = 11.11 kVA. Z1 + Z 2 0.045

(a) I1 =

13.89 ¥ 1000 = 69.45 A 200

I2 =

11.11 ¥ 1000 = 55.55 A. 200

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13.89 = 66.95% 20 11.11 = 74.06%. Percentage rated capacity used in transformer 2 = 15

(b) Percentage rated capacity used in transformer 1 =

! Qspcmfn!4/239 Uxp!usbotgpsnfst!Q!boe!R!bsf!dpoofdufe!jo!qbsbmmfm!boe!tvqqmz!b!dpnnpo!mpbe-!pqfo.djsdvju!fng!pg! Q!boe!R!bsf!7111!W!boe!6911!W!sftqfdujwfmz/!Frvjwbmfou!jnqfebodf!jo!ufsnt!pg!tfdpoebsz!pg!Q!boe! R!bsf!)1/!5!,!k5*!W!boe!)1/3!,!k3*!W/!Uif!mpbe!jnqfebodf!jt!)31!,!k5*!W/!Gjoe!uif!dvssfou!tvqqmjfe!cz! fbdi!usbotgpsnfs/

Solution EP = 6000 V, EQ = 5800 V, ZL = (20 + j 4) W, ZP = (0.4 + j 4) W, ZQ = (0.2 + j 2) W IP = = =

E P Z Q + ( E P - EQ ) Z L Z P ZQ + Z L ( Z P + ZQ ) 6000 (0.2 + j 2) + (6000 - 5800) ( 20 + j 4) (0.4 + j 4) (0.2 + j 2) + ( 20 + j 4) (0.4 + j 4 + 0.2 + j 2) 1200 + j 12000 + 4000 + j 800 2 (0.22 + 4) + ( 20 + j 4) (0.6 + j 6)

=

5200 + j 12800 8.08 + 12 + j 2.4 + j 120 - 24

=

13815.9 –67.89∞ -3.92 + j 122.4

=

13815.9 –67.89∞ = 112.82 ––23.9° A 122.46 –91.8∞

IQ =

EQ Z P - ( E P - E Q ) Z L Z P ZQ + Z L ( Z P + ZQ )

=

5800 (0.4 + j 4) - (6000 - 5800) ( 20 + j 4) (0.4 + j 4) (0.2 + j 2) + ( 20 + j 4) (0.4 + j 4 + 0.2 + j 2)

=

2320 + j 23200 - 4000 - j 800 122.46 –91.8∞

=

-1680 + j 22400 22462.9 –94.289∞ = 122.46 –91.8∞ 122.46 –91.8∞ = 183.43 –2.489° A.

!

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! Qspcmfn!4/23: Uxp!tjohmf.qibtf!usbotgpsnfst!N!boe!O!pg!frvbm!wpmubhf!sbujp!bsf!svoojoh!jo!qbsbmmfm!boe!tvqqmz! b!mpbe!pg!2111!B!bu!1/9!q/g!mbhhjoh!ibwjoh!frvjwbmfou!jnqfebodft!)3!,!k5*!W!boe!)4!,!k6*!W!sftqfd. ujwfmz/! Gjoe! uif! dvssfou! tvqqmjfe! cz! fbdi! usbotgpsnfs! boe! uif! sbujp! pg! uif! lX! pvuqvu! pg! uif! uxp! usbotgpsnfst/

Solution ZM = 2 + j 4 ZN = 3 + j 5 Z IM 3 + j 5 5.83 –59.04∞ = N = = IN Z M 2 + j 4 4.47 –63.43∞ = 1.3 ––4.39° = 1.296 – j 0.09 IM = (1.296 – j 0.09) IN

\

Taking secondary terminal voltage as reference, I = 1000 (0.8 – j 0.6) = (800 – j 600) A [cos f = 0.8, sin f = 0.6] I = IM + IN = (1.296 – j 0.09) IN + IN = IN (2.296 – j 0.09) \

IN =

800 - j 600 1000 –- 36.87∞ A A= 2.296 - j 0.09 2.298 –- 2.24∞ = 435.16 ––34.63° A = 358.07 – j 247.29 A

IM = 1.3 ––4.39° ¥ 435.16 ––34.63° A = 565.71 ––39.02° A = 439.51 – j 356.17 A. The ratio of kW output is equal to the ratio of in-phase components of the two currents. Output of M 439.51 = = 1.227. Output of N 358.07

! Qspcmfn!4/241 Uxp!usbotgpsnfst!Q!boe!R!bsf!dpoofdufe!jo!qbsbmmfm!bdsptt!b!tvqqmz!pg!2111!W/!Cpui!usbotgpsn. fst!ibwf!op.mpbe!sbujp!pg!21110611!W/!Usbotgpsnfs!Q!jt!sbufe!bu!261!lWB-!jut!upubm!sftjtubodf!boe!

Tjohmf.qibtf!Usbotgpsnfst

4/26:

sfbdubodf!cfjoh!3&!boe!4&!sftqfdujwfmz/!Uif!sbujoh!pg!usbotgpsnfs!R!jt!311!lWB-!jut!upubm!sftjtubodf! boe!sfbdubodf!cfjoh!29&!boe!5&!sftqfdujwfmz/!Efufsnjof!uif!mpbe!po!fbdi!usbotgpsnfs!boe!tfdpoe. bsz!wpmubhf!gps!b!mpbe!pg!411!lWB!bu!1/9!q/g/!mbhhjoh/

Solution Considering base kVA to be 150, % ZP = (2 + j 3)

Total load

150 (1 + j 4) 200 = 0.75 + j 3

% ZQ =

S = 350 (0.8 – j 0.6) = 280 – j 210 = 350 ––36.87°

Load shared by transformer P Ê ZQ ˆ Ê 0.75 + j 3 ˆ SP = S Á ˜ = 350 ––36.87° ÁË 2.75 + j 6 ˜¯ + Z Z Ë P Q¯ Ê 3.09 –75.96∞ ˆ = 350 ––36.87° Á Ë 6.6 –65.38∞ ˜¯ = 350 ––36.87° ¥ 0.468 –10.58° = 163.8 ––26.29°kVA Load shared by transformer Q SQ = S

ZP Ê 2 + j3 ˆ = 350 -36.87∞ Á Z P + ZQ Ë 2.75 + j 6 ˜¯ 3.6 56.3∞ = 350 -36.87∞ ¥ 6.6 63.38∞ = 350 -36.87∞ ¥ 0.545 -7.08∞ = 190.75 -43.95∞

Current shared by transformer P =

163.8 = 1.092 p.u. 150

Current shared by transformers Q =

190.75 = 0.95375 p.u. 200

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For the transformer P, the voltage regulation at 1.092 p.u. current is = 1.092 (Rp.u. cos q2 + Xp.u. sin q2) = 1.092 (0.02 ¥ 0.8 + 0.03 ¥ 0.6) = 0.03713 E2 - V2 = 0.03713 E2

or,

V2 = (1 – 0.03713) ¥ 500 = 481.435 V.

or,

! Qspcmfn!4/242 Uif!sftvmut!pg!b!tipsu.djsdvju!uftu!po!uxp!tjohmf.qibtf!usbotgpsnfst!bsf!bt!gpmmpxt; 411!lWB;!3/6&!sbufe!wpmubhf!boe!sbufe!dvssfou!pg!1/4!q/g/!mbh 811!lWB;!6&!sbufe!wpmubhf!boe!sbufe!dvssfou!pg!1/36!q/g/!mbh Efufsnjof!ipx!uif!usbotgpsnfst!xpvme!tibsf!b!mpbe!pg!861!lWB!jg!uifz!bsf!pqfsbufe!jo!qbsbmmfm/!Jg! uifjs!wpmubhf!sbujoht!bsf!22!lW0511!W-!efufsnjof!uif!tfdpoebsz!ufsnjobm!wpmubhf/

Solution Considering Irated = Ibase = 1 p.u. and Vrated = Vbase = 1 p.u. Per unit impedance of the first transformer is 0.025 {0.3 + j sin (cos–1 0.3)} = 0.0075 + j 0.024 Per unit impedance of the second transformer is 0.5 {0.25 + j sin (cos–1 0.25)} = 0.0125 + j 0.968 Consider a common base of 700 kVA. Converting the impendaces into the common base, 700 (0.0075 + j 0.024) = 0.0175 + j 0. 056 300 Ze2 = 0.0125 + j 0.968 S = 750 kVA Ze1 =

Total load

Load shared by the first transformer S1 = S ¥

Ze2 Ze1 + Ze2

= 750 ¥

0.0125 + j 0.968 0.0175 + j 0.056 + 0.0125 + j 0.968

= 750 ¥

0.0125 + j 0.968 0.03 + j 1.024

= 750 ¥

0.968 89.26∞ 1.0244 88.32∞

= 708 0.94∞ kVA

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Load shared by the second transformer S2 = S ¥

Ze1 Ze1 + Ze2

= 750 ¥

0.0175 + j 0.056 0.03 + j 1.024

= 750 ¥

0.0587 72.64∞ 1.0244 88.32∞

= 42.976 -15.67∞ kVA Current shared by the first transformer =

708 = 2.36 p.u. 300

Voltage regulation at 2.36 p.u. current is = 2.36 (0.0075 ¥ cos 0.94° – 0.056 ¥ sin 0.94°) = 2.36 (0.007499 – 0.00092) = 0.0155 V2 = (1 – 0.0155) ¥ 400 = 393.8 V.

\

! Qspcmfn!4/243 Bo!bvup.usbotgpsnfs!tvqqmjft!b!mpbe!pg!21!lX!bu!231!W!boe!bu!vojuz!q/g/!Jg!uif!qsjnbsz!wpmubhf!jt! 351!W-!efufsnjof!)b*!usbotgpsnbujpo!sbujp-!)c*!tfdpoebsz!dvssfou!boe!qsjnbsz!dvssfou!)d*!ovncfs!pg! uvsot!jo!tfdpoebsz!jg!upubm!ovncfs!pg!uvsot!jt!511-!)e*!qpxfs!usbotgpsnfe-!boe!)f*!qpxfs!dpoevdufe! ejsfdumz!gspn!uif!tvqqmz!nbjot!up!uif!mpbe/

Solution (a) Transformation ratio of the auto-transformer V 240 = 2. aA = H = VL 120 (b) Power output PL = VL IL cos f \

IL =

and primary current

IH =

10 ¥ 103 = 83.33 A 120 ¥ 1

83.33 A 2 = 41.67 A.

TH = 2 where TH and TL are the number of turns in the high-voltage and low-voltage TL winding respectively. 400 \ TL = = 200. 2

(c) aA =

!

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(d) Power transformed

1 ˆ Ê ¥ Power output Ptrans = Á1 Ë a A ˜¯ = 0.5 ¥ 10 = 5 kW. (e) Power conducted Pcond =

1 1 ¥ Power output = ¥ 10 kW = 5 kW. 2 aA

! Qspcmfn!4/244 Uif!qsjnbsz!boe!tfdpoebsz!wpmubhft!pg!bo!bvup.usbotgpsnfs!bsf!711!W!boe!411!W!sftqfdujwfmz/!Xjui! uif!ifmq!pg!uif!ejbhsbn-!tipx!uif!dvssfou!ejtusjcvujpo!jo!uif!xjoejoh/!Xifo!uif!tfdpoebsz!dvssfou! jt!211!B-!bmtp!gjoe!uif!qfsdfoubhf!tbwjoht!jo!dpoevdups!nbufsjbm/

Solution Transformation ratio aA = \ So,

VH =2 VL

IL = aA IH I 100 = 50 A IH = L = aA 2

The current distribution in the winding is shown in Fig. 3.49.

! Gjh/!4/5:! Djsdvju!ejbhsbn!pg!Fybnqmf!4/244

Percentage savings in conductor material in using auto-transformer =

1 p.u. = 0.5 p.u. or 50%. aA

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! Qspcmfn!4/245 B!33110331!W!usbotgpsnfs!jt!sbufe!bu!21!lWB!bt!b!uxp.xjoejoh!usbotgpsnfs/!Ju!jt!dpoofdufe!bt!bo! bvup.usbotgpsnfs!xjui!mpx.wpmubhf!xjoejoh!dpoofdufe!beejujwfmz!jo!tfsjft!xjui!ijhi.wpmubhf!xjoejoh/! Uif!bvup.usbotgpsnfs!jt!fydjufe!gspn!b!3611!W!tpvsdf/!Uif!bvup.usbotgpsnfs!jt!mpbefe!tp!uibu!uif! sbufe!dvssfout!pg!uif!xjoejoht!bsf!opu!fydffefe/!Gjoe!)b*!uif!dvssfou!ejtusjcvujpo!jo!uif!xjoejoht-! )c*!lWB!pvuqvu-!)d*!lWB!usbotgfssfe!dpoevdujwfmz!boe!joevdujwfmz!gspn!joqvu!up!pvuqvu-!boe!)e*!tbw. joht!jo!dpoevdups!nbufsjbm!bt!dpnqbsfe!up!b!uxp.xjoejoh!usbotgpsnfs!pg!tbnf!WB!sbujoh/

Solution 10 ¥ 103 = 4.55 A 2200 10 ¥ 1000 Rated current of 220 V winding = = 45.45 A 220

Rated current of 2200 V winding =

(a) Output current of auto-transformer = (4.55 + 45.45) A = 50 A 2200 ¥ 50 = 110 kVA 1000 2200 ¥ 45.45 (c) kVA transferred conductively = = 100 kVA 1000

(b) kVA output =

kVA transferred inductively =

2200 ¥ 4.55 = 10 kVA 1000

(d) Saving in conductor material =

1 2200 p.u. = aA 2420 = 0.909 p.u. = 90.9%.

! Qspcmfn!4/246 B!31!lWB-!3110611!W-!61!I{!tjohmf.qibtf!usbotgpsnfs!jt!dpoofdufe!bt!bvup.usbotgpsnfs!xifsf!W2! >!311!W!boe!W3!>!811!W/!Dbmdvmbuf!jut!wpmubhf!sbujp!boe!lWB!sbujoh/!Tipx!uif!dvssfou!ejtusjcvujpo! xifo!ju!efmjwfst!uif!sbufe!lWB!up!mpbe/

Solution Voltage ratio =

700 = 3.5 200 I2 (rated) =

The auto-transformer is shown in Fig. 3.50.

20 ¥ 103 = 40 A 500

!

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I2 = 40 A 500 V

I1

Load

700 V

I1 – I2 = 16 A

V1 = 200 V

! Gjh/!4/61! Djsdvju!ejbhsbn!pg!Fybnqmf!4/246

- I 2 + I1 200 2 = = I2 500 5 \

I1 = 56 A

\

(kVA)auto = 700 ¥

40 = 28. 1000

! Qspcmfn!4/247 B!31!lWB-!31110311!W-!uxp.xjoejoh!usbotgpsnfs!jt!up!cf!vtfe!bt!bo!bvup.usbotgpsnfs-!xjui!b!dpo. tubou!tpvsdf!wpmubhf!pg!3111!W/!Bu!gvmm.mpbe!pg!vojuz!qpxfs!gbdups!dbmdvmbuf!uif!qpxfs!pvuqvu-!qpxfs! usbotgpsnfe!boe!qpxfs! dpoevdufe/! Jg!uif!fggjdjfodz! pg! uif! uxp.xjoejoh! usbotgpsnfs!bu! 1/8! q/g/! jt! :8&-!gjoe!uif!fggjdjfodz!pg!uif!bvup.usbotgpsnfs/

Solution The auto-transformer is shown in Fig. 3.51. I2

100

I1 V2 = 2200 V V1 = 2000 V

I1 – I2

! Gjh/!4/62! Djsdvju!ejbhsbn!pg!Fybnqmf!4/247

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I2 =

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20 ¥ 103 = 100 A 200

Power output = 2200 ¥ 100 ¥ 1 = 220 kW Power transformed = 200 ¥ 100 ¥ 1 = 20 kW Power conducted = 200 kW h(two-winding) = 0.97 = \

20 ¥ 103 ¥ 0.7 20 ¥ 103 ¥ 0.7 + PL

PL = 433 W h(auto-trans) =

220 ¥ 100 = 99.8%. 220 + 0.433

! Qspcmfn!4/248 Uif!ovncfs!pg!qsjnbsz!uvsot!pg!bo!jefbm!usbotgpsnfs!jt!311/!Po!uif!tfdpoebsz!tjef-!uif!ovncfs!pg! uvsot!cfuxffo!B!boe!C!jt!711!boe!cfuxffo!C!boe!D!jt!411-!uif!usbotgpsnfs!tvqqmjft!b!sftjtups!dpo. ofdufe!cfuxffo!B!boe!D!xijdi!esbxt!9!lX/!Gvsuifs-!b!mpbe!pg!2111!–56¡!W!jt!dpoofdufe!cfuxffo!B! boe!C/!Uif!qsjnbsz!wpmubhf!jt!3!lW/!Gjoe!uif!qsjnbsz!dvssfou/!

Solution The auto-transformer is shown in Fig. 3.52. VAC =

900 ¥ 2 = 9 kV 200

IL1 =

8 ––0° = 0.89 A 9

VAB =

600 = 6 kV 200

IL2 =

6 ¥ 1000 = 6 ––45° A 1000 – 45∞

IBA = IL1 + IL2 = 0.89 + 6 ––45° = 5.13 – j 4.24 A I LB = IL1 = 0.89 + j 0

!

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IL1

A

1000 –450°W

600 250

2 kV

B

8 kW

IL2

300 C

! Gjh/!4/63! Djsdvju!ejbhsbn!pg!Fybnqmf!4/248

Secondary AT = 600 (5.13 – j 4.24) + 300 ¥ 0.89 = 3345 – j 2544. \ primary current =

3345 - j 2544 200

= 16.725 – j 12.72 \

I1 = 21.01 A.

! Qspcmfn!4/249 B! 311! lWB-! 31110511! W-! tjohmf.qibtf! uxp.xjoejoh! usbotgpsnfs! jt! up! cf! vtfe! bt! bo! bvup.usbot. gpsnfs!gps!tufqqjoh!vq!uif!wpmubhf!gspn!3111!W!up!3611!W/!Jg!bu!sbufe!mpbe-!uif!uxp.xjoejoh!usbot. gpsnfs! ibt! 4&! mptt-! 4/6&! wpmubhf! sfhvmbujpo! boe! 6&! jnqfebodf-! dbmdvmbuf! uif! gpmmpxjoh! gps! bo! bvup.usbotgpsnfs; ! ! ! ! ! !

)b*! )c*! )d*! )e*! )f*! )g*!

Wpmubhf!boe!dvssfou!sbujoh lWB!sbujoh Fggjdjfodz Qfsdfoubhf!jnqfebodf Sfhvmbujpo! Tipsu.djsdvju!dvssfou!po!fbdi!tjef

Solution (a) For an auto-transformer, LV side is 2000 V and HV side is 2500 V. Rated current of 400 V side of two-winding transformer =

200 ¥ 103 = 500 A = I2 (say). 400

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Rated current on LV side of an auto-transformer I1 = Rated current of HV side of two-winding transformer 200 ¥ 103 = 500 + 2000 = 600 A (b) kVA rating as an auto-transformer 600 ¥ 2000 = kVA = 1200 kVA. 1000 (c) Transformer ratio of auto-transformer 2000 4 = = = 0.8 2500 5 Percent full-load losses as an auto-transformer = (1 – 0.8) ¥ 3% = 0.6% \ efficiency as an auto-transformer = (100 – 0.6)% = 99.4% (d) Percentage impedance as an auto-transformer = (1 – 0.8) ¥ 5 =1% (e) Regulation = (1 – 0.8) ¥ 3.5 = 0.7% (f) Short circuit current as an auto-transformer 1 Ê 1 ˆ =Á ¥ p.u. ˜ Ë 1 - 0.8 ¯ 0.05 = 100 p.u.

! Qspcmfn!4/24: B!5510221!W-!21!lWB-!uxp.xjoejoh!usbotgpsnfs!jt!up!cf!fnqmpzfe!bt!bo!bvup.usbotgpsnfs!up!tvqqmz! b!551!W!djsdvju!gspn!b!611!W!tpvsdf-!xifo!uftufe!bt!b!uxp.xjoejoh!usbotgpsnfs!bu!sbufe!mpbe!1/9! q/g!mbhhjoh/!Jut!fggjdjfodz!jt!1/:9/ ! )b*! Efufsnjof!jut!lWB!sbujoh!bt!bo!bvup.usbotgpsnfs/ ! )c*! Gjoe!jut!fggjdjfodz!bt!bo!bvup.usbotgpsnfs/

Solution Refer Fig. 3.53. (a) I1 =

10 ¥ 1000 = 90.9 A 110

!

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! Gjh/!4/64! Djsdvju!ejbhsbn!pg!Fybnqmf!4/24:

(kVA)auto =

500 ¥ 90.9 1000

= 45.45 kVA - I1 + I 2 V1 110 1 = = = I1 V2 440 4 \

I2 =

5 I1 = 113.63 A. 4

(b) As a two-winding transformer, the efficiency is 10 ¥ 1000 ¥ 0.8 = 0.98 10 ¥ 1000 ¥ 0.8 + PL PL = 163.27 W h=

\

Full-load output as auto transformer (0. 8 p.f) = 45.45 ¥ 0.8 = 36.36 kW As auto-transformer efficiency is 36.36 ¥ 100 36.36 + 0.1633 = 99.55%.

hauto =

! Qspcmfn!4/251 B!21!lWB-!3110611!W-!61!I{/!B!tjohmf.qibtf!usbotgpsnfs!jt!dpoofdufe!bt!bo!pvuqvu!usbotgpsnfs! bt!tipxo!jo!Gjh/!4/65/!Efufsnjof!jut!wpmubhf!sbujp!boe!uif!lWB!sbujoh/!Nbsl!po!uif!ejbhsbn-!uif! nbhojuveft!boe!sfmbujwf!ejsfdujpot!pg!uif!dvssfout!jo!uif!xjoejoht!bt!xfmm!bt!jo!uif!joqvu!boe!pvuqvu! mjoft!xifo!efmjwfsjoh!uif!sbufe!lWB!up!mpbe/

Tjohmf.qibtf!Usbotgpsnfst

4/27:

Solution Refer Fig. 3.54(a). I2 = 20 A 500 V

500 V

Load

I1 = 28 A

Load

Input

200 V

V1 = 200 V

V2 = 700 V I1 – I2 =8A

! Gjh/!4/65! Djsdvju!ejbhsbn!pg!Fybnqmf!4/251! ! Gjh/!4/65)b*! Djsdvju!ejbhsbn!pg!Fybnqmf!4/251! ! !tipxjoh!ejsfdujpot!pg!dvssfout!!!!!!

V2 500 + 200 = = 3.5 V1 200 I2 (rated) =

10 ¥ 1000 = 20 A 500

\

- I 2 + I1 200 2 = = I2 500 5

or,

-20 + I1 2 = 20 5 I1 = 28 A

or, \

(kVA)auto = 700 ¥

20 = 14. 1000

! Qspcmfn!4/252 B!21!lWB-!21110211!W-!uxp.xjoejoh!usbotgpsnfs!jt!up!cf!vtfe!bt!bo!bvup.usbotgpsnfs-!xjui!b!dpo. tubou!tpvsdf!wpmubhf!pg!2111!W/!Bu!gvmm.mpbe!pg!v/q/g/-!dbmdvmbuf!uif!qpxfs!pvuqvu-!qpxfs!usbotgpsnfe! boe!qpxfs!dpoevdufe/!Jg!uif!fggjdjfodz!pg!uif!uxp.xjoejoh!usbotgpsnfs!bu!1/8!q/g/!jt!:8&-!gjoe!uif! fggjdjfodz!pg!uif!bvup.usbotgpsnfs/

!

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Solution Refer Fig. 3.55. I2

I1 V2 = 1100 V V1 = 1000 V I1 – I2

! Gjh/!4/66! Djsdvju!ejbhsbn!pg!Fybnqmf!4/252

I2 =

10 ¥ 1000 = 100 A 100

Power output = 1100 ¥ 100 ¥ 1 = 110 kW Power transferred = 100 ¥ 100 ¥ 1 = 10 kW Power conducted = 100 kW Let efficiency of two-winding transformer be hTW hTW = 0.97 = \

10 ¥ 1000 ¥ 0.7 10 ¥ 1000 ¥ 0.7 + PL

PL = 216.5 W hAuto =

110 ¥ 100 110 + 0.2165 = 99.8%.

! Qspcmfn!4/253 B!3310551!W-!31!lWB-!61!I{!usbotgpsnfs!jt!dpoofdufe!bt!bo!bvup.usbotgpsnfs!up!usbotgpsn!711!W! up!311!W/ ! )b*! Efufsnjof!uif!bvup.usbotgpsnfs!sbujp!b/ ! )c*! Efufsnjof!uif!lWB!sbujoh!pg!uif!bvup.usbotgpsnfs/! ! )d*! Xjui!b!mpbe!pg!31!lWB-!1/9!q/g!mbhhjoh!dpoofdufe!up!311!W!ufsnjobmt-!efufsnjof!uif!dvssfout! jo!uif!mpbe!boe!uxp!usbotgpsnfs!xjoejoht/

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Solution Refer Fig. 3.56. I1

V1 = 600 V

N1

I2

200 V

N2 I2 – I1

! Gjh/!4/67! Djsdvju!ejbhsbn!pg!Fybnqmf!4/253

(a) a = (b)

(a)

N1 600 = =3 N 2 200 I1 =

20 ¥ 1000 = 45.45 A 440

(kVA)auto =

600 ¥ 45.45 = 27.27 1000

20 ¥ 1000 = 90.9 A 220 \ I2 – I1 = 90.9 – 45.45 = 45.45 A. I2 =

! Qspcmfn!4/254 Uif!fggjdjfodz!pg!b!2310591!W-!21!lWB!usbotgpsnfs!bu!sbufe!mpbe!boe!vojuz!qpxfs!gbdups!jt!1/:8:/!Ju! jt!pqfsbufe!bt!bo!bvup.usbotgpsnfs!gspn!b!711!W!tvqqmz!up!qspwjef!b!591!W!djsdvju/!Efufsnjof!uif! lWB!sbujoh!bt!bo!bvup.usbotgpsnfs/!Bmtp!efufsnjof!uif!fggjdjfodz!pg!uif!bvup.usbotgpsnfs!bu!gvmm. mpbe!xjui!1/96!qpxfs!gbdups!mbhhjoh/

Solution The rated current of 120 V winding 10 ¥ 103 = 83.33 A 120 The circuit diagram of the transformer operated as an auto-transformer is shown in Fig. 3.57. I1 =

!

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! Gjh/!4/68! Djsdvju!ejbhsbn!pg!Fybnqmf!4/254

\ kVA rating of the auto-transformer is 600 ¥ 83.33 = 50 kVA 1000 At 0.85 p.f. the auto-transformer output is 50 ¥ 0.85 = 42.5 kW The efficiency of a two-winding transformer at unity p.f. is 0.979. \ loss at unity p.f. is (1 – 0.979) ¥ 10 ¥ 103 ¥ 1 = 210 W \ input of the auto-transformer = 42.5 ¥ 103 + 210 = 42710 W = 42.71 kW \ efficiency of the auto-transformer at full-load and 0.85 p.f. lagging is

42.5 ¥ 100% or 99.5%. 42.71

! Qspcmfn!4/255 B!61!NWB-!71!I{-!tjohmf.qibtf!usbotgpsnfs!ibwjoh!wpmubhf!sbujoh!pg!9089!lW!ibt!dpsf!mptt!pg!317!lX! boe!gvmm.mpbe!dpqqfs!mptt!pg!298!lX/!Jg!uijt!usbotgpsnfs!jt!up!cf!pqfsbufe!bt!bo!bvup.usbotgpsnfs-! efufsnjof!uif!wpmubhf!sbujoht!pg!uif!ijhi.wpmubhf!boe!mpx.wpmubhf!xjoejoht!gps!uijt!dpoofdujpo!boe! uif!lWB!sbujoh!pg!uif!bvup.usbotgpsnfs!dpoofdujpo/!Bmtp-!efufsnjof!uif!fggjdjfodz!pg!uif!usbotgpsnfs! jo!uijt!dpoofdujpo!xifo!ju!jt!tvqqmzjoh!sbufe!mpbe!bu!vojuz!qpxfs!gbdups/

Solution The auto-transformer connection is shown in Fig. 3.58.

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I1

8 kV 86 kV 78 kV

Load

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\ voltage rating of the auto-transformer is (78 + 8) kV/78 kV or 86/78 kV As a two-winding transformer, total loss is (206 + 187) kW = 393 kW The rated current in the 8 kV winding is I1 = \ kVA rating of the transformer is

50 ¥ 106 8 ¥ 103

= 6250 A

= 86 ¥ 6250 = 537500

\ efficiency of the auto-transformer at unity p.f. load is 537500 ¥ 1 = 0.9993 or 99.93%. 537500 ¥ 1 + 393

Sfwjfx!Rvftujpo! 1. Define a transformer. Explain the principle of operation of a transformer. 2. Distinguish between core-type and shell-type transformer. Why is the low-voltage winding placed near the core? Why is the core of a transformer laminated? 3. Derive an expression for the induced emf in the transformer winding. 4. What is an ideal transformer? Draw and explain the phasor diagram of an ideal transformer. 5. What is power transformer and distribution transformer? 6. Explain how the primary current increases as the current in the secondary side of the transformer is increased. 7. What is meant by ‘equivalent resistance referred to primary’ and ‘equivalent resistance referred to secondary’?

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8. Draw and explain the phasor diagram of a single-phase transformer under lagging p.f. 9. Draw the exact equivalent circuit of a transformer and describe the various parameters involved in it. 10. Define voltage regulation of a transformer. Derive an expression for voltage regulation under lagging p.f. load. 11. Derive the condition for zero voltage regulation and condition for maximum voltage regulation. What is the value of the maximum voltage regulation? 12. Describe briefly the various losses in a transformer. 13. Derive the condition for maximum efficiency of a transformer. 14. Explain why: (a) Open-circuit test is generally conducted on the low-voltage side and short-circuit test is conducted on the high-voltage side of the transformer. (b) Open-circuit test is conducted at rated voltage and short-circuit test is conducted at rated current. (c) Core loss is neglected in short-circuit test and copper loss is neglected in open circuit test. 15. Discuss how polarity of a transformer is determined from polarity test. 16. Explain how maximum temperature rise is obtained from Sumpner’s test. 17. What are the conditions for parallel operation of transformer? 18. What are the advantages of parallel operation of transformers? 19. Deduce expressions for load shared by two transformers connected in parallel (a) turns ratios of the transformers are same, and (b) turns ratios of the transformers are unequal. 20. Define an auto-transformer. State its merits and demerits over a two-winding transformer. What are the applications of an auto-transformer? 21. In an auto-transformer, the power transferred from primary to secondary circuit is partly by conduction and partly by induction. Explain. 22. Derive an expression for the rating of an auto-transformer as a fraction of the rating of a twowinding transformer when the auto-transformer has a transformation ratio of a. 23. What are the different methods of cooling of transformers? 24. Define all-day efficiency of transformer. 25. What do you mean by instrument transformers? Explain in brief each of them. 26. What is a winding transformer? What are the uses of a pulse transformer?

Qspcmfnt 1. The primary winding of a 50 Hz transformer is supplied from a 440 V, 50 Hz source and has 200 turns. Find the (a) peak value of flux, and (b) voltage induced in the secondary winding if it has 50 turns.

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Hints: f = 50 Hz E1 = 440 V N1 = 200 (a) If fm is the peak value of flux then E1 = 4.44 f fm N1 440 fm = Wb = 0.0099 Wb. 4.44 ¥ 50 ¥ 200

or (b) N2 = 50

Voltage induced in the secondary E2 = 4.44 f fm N2 = 4.44 ¥ 50 ¥ 0.0099 ¥ 50 V = 110 V. 2. A 200/50 V, 50 Hz transformer has a core area of 100 cm2. The maximum value of the flux density is 1 Wb/m2. Assuming 9% loss of area due to laminations, find the primary and secondary number of turns and transformation ratio. Hints A = 100 ¥ 10–4 m2 = 0.01 m2 E1 = 200 V; E2 = 50 V; Bm = 1 Wb /m2 Assuming 9% loss of area, net area of core = 0.01 ¥ 0.9 m2 = 0.009 m2 Primary turns N1 =

E1 200 = = 100 4.44 f Bm A 4.44 ¥ 50 ¥ 1 ¥ 0.009

Secondary turns N2 = Transformer ratio

E2 50 N1 = ¥ 100 = 25 E1 200

E1 N1 100 = = = 4. E2 N 2 25

3. A single-phase transformer has 200 and 100 turns respectively in its secondary and primary windings. The resistance of the primary winding is 0.05 W and that of the secondary is 0.3 W. Find the resistance of (a) the primary winding referred to the secondary, and (b) the secondary winding referred to the primary. Also find the equivalent resistance of the primary. Hints: Number of turns of primary winding N1 = 100 Number of turns of secondary winding N2 = 200 Resistance of primary winding R1 = 0.05 W Resistance of secondary winding R2 = 0.3 W

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(a) Resistance of primary winding referred to secondary 2

2

ÊN ˆ Ê 200 ˆ = 0.05 ¥ 4 = 0.2 W. = R1¢ = R1 ¥ Á 2 ˜ = 0.05 ¥ Á Ë 100 ˜¯ Ë N1 ¯ (b) Resistance of secondary winding referred to the primary 2

2 ÊN ˆ 0.3 Ê 100 ˆ W = 0.075 W R 2¢ = R2 ¥ Á 1 ˜ = 0.3 ¥ Á ˜¯ = Ë 200 4 N Ë 2¯

Equivalent resistance of the transformer referred to the primary R01 = R1 + R 2¢ = 0.05 + 0.075 = 0.125 W. [Also, R02 = R2 + R1¢ = 0.3 + 0.2 = 0.5 W]. 4. A 20 kVA, 1000/200 V single-phase transformer has a primary resistance of 1 W and a secondary resistance of 0.2 W. Find the equivalent resistance of the transformer referred to the secondary and the total resistance drop on full-load. Hints: If N1 and N2 be the number of turns of the primary and secondary winding then N1 1000 = N2 200 Resistance of the primary winding R1 = 1 W Resistance of the secondary winding R2 = 0.2 W Total equivalent resistance in terms of the secondary winding is (R1¢ + R2) 2

i.e.

2

ÊN ˆ Ê 200 ˆ + 0.2 = 0.04 + 0.2 = 0.24 W Ro2 = R1 Á 2 ˜ + R2 = 1 ¥ Á Ë 1000 ˜¯ Ë N1 ¯

Full-load secondary current I2 =

20 ¥ 103 = 100 A 200

Total resistance drop on full-load = I2 Ro2 = 100 ¥ 0.24 = 24 V. 5. The efficiency of a 10 kVA, 2000/400 V single-phase transformer at unity p.f. is 97% at rated load and also at half-rated load. Determine the transformer core losses and ohmic losses. Hints: Efficiency =

Output (Input - Loss) Losses = =1Input Input Input

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At full-load, 0.97 = 1 – or

0.97 = 1 –

Core losses + Copper losses osses Output + Core losses + Copper lo

Pc + Pcu 10 ¥ 103 ¥ 1 + Pc + Pcu

Pc + Pcu = 0.03 10000 ¥ Pc + Pcu

or

Pc + Pcu = 309.278 W

or

(i)

At half-load, Pc +

0.97 = 1 – 10 ¥ 103 ¥

1 P 4 cu

1 1 + Pc + Pcu 2 4

1 Ê ˆ ÁË Pc + Pcu ˜¯ 4

or

Pc +

1 Pcu = 150 + 0.03 4

or

Pc +

1 150 Pcu = = 154.693 W 4 0.97

(ii)

Solving Eqs (i) and (ii), Pcu = 206.185 W PC = 103.1 W.

and

6. A single-phase transformer supplies a load of 20 kVA at a p.f. of 0.81 (lagging). The iron loss of the transformer is 200 W and the copper losses at this load is 180 W. Calculate (a) the efficiency, (b) if the load is now changed to 30 kVA at a p.f. of 0.91 (lagging), calculate the new efficiency. Hints: Iron loss = 200 W Copper loss at a load of 20 kVA is = 180 W (a) Output at 0.81 p.f. (lag) = 20 ¥ 103 ¥ 0.81 = 16200 W Total losses = 200 + 180 = 380 W Input = Output + Losses = 16200 + 380 = 16580 W Efficiency =

Output 16200 = = 0.977 = 97.7% Input 16580

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(b) New load is 30 kVA at 0.91 p.f. (lag) Output = 30 ¥ 103 ¥ 0.91 = 27300 W 2

9 Ê 30 ˆ Copper losses at a load of 30 kVA is, Pcu = 180 ¥ Á ˜ = 180 ¥ = 405 W Ë 20 ¯ 4 Input = Output + Losses = 27300 + 200 + 405 = 27905 W Efficiency =

27300 = 0.978 = 97.8%. 27905

7. A 50 kVA, 3.3 kV/230 V single-phase transformer has an impedance of 4.2% and a copper loss of 1.8% at full-load. Calculate the ohmic value of resistance, reactance and impedance referred to the primary side. Estimate the primary short-circuit current, assuming the supply voltage to be maintained. Hints: 50, 000 A = 15.15 A Primary full-load current = 3300 4.2 I1Zo1 = 100 V1

Now, or

0.042 =

15.15 Zo1 300

Zo1 = 9.148 W

or

where Zo1 = equivalent impedance referred to the primary. Again, or

0.018 =

I12 Ro1 I1Ro1 15.15 Ro1 = = V1I1 V1 3300

Ro1 = 3.92 W,

where Ro1 = equivalent resistance referred to the primary. \ equivalent resistance referred to the primary = (9.148) 2 - (3.92) 2 W = 8.26 W. Under short circuit condition, the primary current =

Ê V1 ˆ 3300 A = 360.73 A. ÁË Z ˜¯ = 9.148 o1

8. Calculate the values of Ro, Xo, R1 and X1 in the diagram shown in Fig. 3.59 of a single-phase 8 kVA, 22/440 V, 50 Hz transformer of which the following are the test results: Open-circuit test 220 V, 0.9 A, 90 W on the low-voltage side

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Short-circuit test 20 V, 15 A, 100 W on the high-voltage side

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Hints: From the open-circuit test data, 90 = 0.4545 220 ¥ 0.9 sin qo = 0.89

cos qo =

No-load p.f. \ Core loss resistance Ro =

V1 220 = W = 537.83 W I o sin q o 0.9 ¥ 0.4545

Magnetizing reactance Xo =

V1 220 = = 274.65 W I o sin q o 0.89 ¥ 0.9

From short-circuit test data, ROH = Ro2 = ZOH = Zo2 =

100 (15) 2

W = 0.444 W

20 W = 1.33 W 15

where Ro2 and Zo2 are the equivalent resistance and impedance referred to the high-voltage side. Hence,

XOH = Xo2 = (1.33) 2 - (0.44) 2 = 1.257 W

Figure 3.59 shows the equivalent resistance R1 and reactance X1 referred to the low voltage side or primary side. 2

Hence,

Ê 220 ˆ = 0.111 W R1 = 0.444 ¥ Á Ë 440 ˜¯

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2

and

Ê 220 ˆ = 0.314 W X1 =1.257 ¥ Á Ë 440 ˜¯

Also

RO = 537.83 W and XO = 274.65 W.

9. A 8 kVA, 440/2000 V, 50 Hz single-phase transformer gave the following test results: No-load test: 440 V, 0.8 A, 80 W Short-circuit test: 50 V, 3 A, 20 W Calculate (a) the magnetizing current and the component corresponding to iron losses at normal voltage and frequency, (b) the efficiency on full-load at unity p.f., and (c) the secondary terminal voltage on full-load at unity p.f. Hints: (a) From no-load test data, No-load p.f. (cos qo) =

80 = 0.227 440 ¥ 0.8

Iron loss component current = 0.8 (cos qo) = 0.8 ¥ 0.227 = 0.182 A Magnetizing current = 0.8(sin qo) = 0.8 ¥ 0.974 = 0.779 A. (b) Iron loss = 80 W As the secondary side is the HV side, so a short-circuit test is performed on the secondary side. 8000 Rated current of the HV side I2 = =4A 2000 When current is 3 A, the wattmeter reading is 20 W. So, if rated current of 4 A flows through the high-voltage winding, the wattmeter read2

Ê 4ˆ ing = 20 ¥ Á ˜ = 35.55 W Ë 3¯ So rated copper losses = 35.55 W Efficiency on full-load at unity p.f. =

8 ¥ 103 ¥ 1 8 ¥ 103 ¥ 1 + 35.55 + 80

= 0.9857 or 98.57%.

10. A 17.5 kVA, 450/121 V, 50 Hz single-phase transformer gave the following data on test: Open-circuit test (OCT): 450 V, 1.5 A, 115 W Short-circuit test (SCT): 15.75 V, 38.9 A, 312 W Estimate the voltage on the secondary terminal and the efficiency of the transformer when supplying full-load current, at a p.f. of (0.8) lagging, from the secondary side. Assume the input voltage to be maintained at 450 V.

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Hints: 17500 A = 38.89 A 450 17500 Full-load secondary current = A = 144.63 A 121 Full-load primary current =

As 450 V is the applied voltage in OCT, so this test has been performed on the HV side and as the current in the SCT is 38.9 A which is the rated primary or low-voltage current, so this test has also been performed on the LV side. From short-circuit test data, Ro1 = Zo1 =

312 (38.9) 2

W = 0.206 W

15.75 W = 0.405 38.9

and

Xo1 = (0.405) 2 - (0.206) 2 = 0.3887 W

\

Ê 121 ˆ = 0.0149 W Ro2 = 0.206 ¥ Á Ë 450 ˜¯

2

2

and

Xo2

Ê 121 ˆ = 0.0252 W = 0.3487 ¥ Á Ë 450 ˜¯

If V2 be the secondary terminal voltage then or

[121 – V2] = 144.63 (0.0149 ¥ 0.8 + 0.0252 ¥ 0.6) = 3.91 V2 = 117.09 V

From OCT, iron loss = 115 W From SCT, full-load copper loss = 312 So efficiency at full-load and 0.8 p.f. lagging =

17.5 ¥ 103 ¥ 0.8 17.5 ¥ 103 ¥ 0.8 + 115 + 312

= 0.97 or 97%.

11. The maximum efficiency of a 100 kVA, single-phase transformer is 95% and occurs at 90% of full-load at 0.85 p.f. If the leakage impedance of the transformer is 5%, find the voltage regulation at rated load 0.8 p.f. lagging. Hints: Output at maximum efficiency = 100 ¥ 0.9 ¥ 0.85 = 76.5 kW Output 76.5 = Efficiency = (0.95) = Output + Losses 76.5 + Losses

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È 76.5 ˘ \ Losses = Í - 76.5˙ = 4.026 kW Î 0.95 ˚ At maximum efficiency, core losses = copper losses \ core losses = copper losses = 2.013 kW 2.013 kW is the copper losses at 90% of full-load. 2

Ê 1 ˆ = 2.485 kW So full-load ohmic losses = 2.013 ¥ Á Ë 0.9 ˜¯ If I2 be the full-load secondary current, (I22 RO2) = 2485, where RO2 is the equivalent resistance referred to as the secondary. \ or

ÊI R ˆ I2 V2 Á 2 O 2 ˜ = 2485 Ë V2 ¯ 100 ¥ 103 ¥ Rp.u = 2485, where Rp.u. is the p.u resistance

or Now

Rp.u. = 0.02485 Zp.u. = 0.05

\

Xp.u. = (0.05) 2 - (0.02485) 2 = 0.04338

Voltage regulation = (Rp.u. cos q2 + Xp.u. sin q2) = 0.02485 ¥ 0.8 + 0.04338 ¥ 0.6 = 0.0459 or, 4.59%. 12. A 10 kVA single-phase transformer has a core loss of 50 W and full-load ohmic loss of 120 W. The daily variation of load on the transformer is as follows: 6 a.m to 12 noon: 5 kW at 0.7 p.f. 12 noon to 6 p.m.: 4 kW at 0.8 p.f. 6 a.m to 1 a.m.: 8 kW at 0.9 p.f. 1 a.m to 7 a.m: No-load Find the all-day efficiency of the transformer. Hints: Core loss = 50 W; full-load ohmic loss = 120 W From 6 a.m. to 12 noon, Output = 5 ¥ 6 = 30 kWh 5 = 7.143 kVA load = 0.7 2

Ê 7.143 ˆ Ohmic losses for 6 hours = Á ¥ 120 = 61.22 W Ë 10 ˜¯

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Energy lost as ohmic loss = (61.22 ¥ 6) = 367.36 Wh From 12 noon to 6 p.m., Output = 4 ¥ 6 = 29 kWh 4 kVA load = =5 0.8 2

Ê 5ˆ Ohmic losses for 6 hours = Á ˜ ¥ 120 = 30 W Ë 10 ¯ Energy lost as ohmic loss = (30 ¥ 6) = 180 Wh From 6 p.m. to 1 a.m., Output 8 ¥ 7 = 56 kWh 8 kVA load = = 8.89 0.9 2

Ê 8.89 ˆ Ohmic losses for 7 hours = Á ¥ 120 = 94.84 W Ë 10 ˜¯ Energy lost as ohmic loss = 94.84 ¥ 7 = 663.87 Wh Daily energy lost as ohmic loss = (367.36 + 180 + 663.87) ¥ 10–3 kWh = 1.211 kWh Daily energy lost as core loss =

50 ¥ 24 103

kWh = 1.2 kWh

Total loss = (1.211 + 1.2) = 1.411 kWh Daily output = (30 + 24 + 56) = 110 kWh All-day efficiency =

110 = 0.9872 or 98.73%. 110 + 1.411

13. A 1000 kVA transformer with (0.02 + j0.1) p.u. impedance and a 500 kVA transformer with (0.015 + j0.05) p.u. impedance are operating in parallel. The no-load secondary voltages of the two transformers are equal. How will they share a load of 1500 kVA at unity p.f. load? Hints: Let base kVA = 100 kVA Z1 = (0.02 + j 0.1) = 0.102 –78.69° Converting impedance of second transformer to base kVA Z2 = Load

1000 (0.015 + j 0.05) = (0.03 + j 0.1) = 0.104 –13.3° 500

(SL) = 1500 –0° kVA

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Hence, load shared by transformer 1 S1 = 1500 –0° ¥ =

0.104 –73.3∞ (0.02 + 0.03) + j (0.1 + 0.1)

1500 ¥ 0.104 –73.3∞ = 757.28 ––2.66° kVA 0.206 –75.96∞

Load shared by transformer 2 S2 = 1500 –0° ¥

0.102 X –78.69∞ = 742.72 –2.73° kVA. (0.02 + 0.03) + j (0.1 + 0.1)

14. A 100 kVA transformer has iron losses of 1.2 kW and full-load copper losses of 1.5 kW. (a) Find the KVA for maximum efficiency, and (b) efficiency at the kVA found in part (a) for power factor of 1, 0.8 lagging and 0.8 leading. Hints: (a) Let x be the fraction of full-load kVA at which maximum efficiency occurs. Then, x2 (1.5) = 1.2 fi

x=

1.2 = 0.894 1.5

Load kVA for maximum efficiency = 0.894 (100) = 89.4 kVA. (b) (i) For power factor 1 Output = 89.4 ¥ 1 = 89.4 kW Iron losses = 1.2 kW Copper losses = 1.5 kW 89.4 ¥ 100 = 97.07% Efficiency = 89.4 + 1.2 + 1.5 0.8 lagging power factor output = 89.4 ¥ 0.8 = 71.52 kW 71.52 ¥ 100 = 96.36% Efficiency = 71.52 + 1.2 + 1.5 For 0.8 leading p.f., efficiency is 96.36% [as kW and losses are same for leading and lagging p.f.]. 15. A 200/400 V single-phase transformer has a core loss of 200 W. It has a resistance of 0.75 ohm and leakage reactance of 1.5 W, both referred to HV side. Find the p.f. at which regulation is zero. Full-load primary current is 30 A.

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Hints: Full-load secondary current = 30 ¥

200 = 15 A 400

Rp.u. =

Xp.u.

15 ¥ 0.75 400

= 0.028 p.u. 10 ¥ 1.5 = 0.0375 p.u. = 400

Regulation = Rp.u. cos q2 – Xp.u. sin q2 = 0 fi

tan q2 =

0.028 = 0.746 0.0375

Power factor = cos q2 = 0.8 leading. [regulation can be zero when p.f. is leading] 16. Two transformers have the following characteristics: A 150 kVA, 6600/225 V, IR = 1.5%, IX = 4% B 300 kVA, 6600/215 V, IR = 1%, IX = 5% They are connected in parallel. Find the no-load terminal voltage and current when 6600 V is applied at HV terminals. Hints: Full-load secondary current of transformer A =

150 ¥ 1000 225

= 666.67 A ZA = (0.015 + j 0.04)

225 666.67

= 0.005 + j 0.0135 W Full-load secondary current of transformer B 300 ¥ 1000 = 1395.35 A 215 Ê 215 ˆ ZB = (0.01 + j 0.04) Á Ë 1395.35 ˜¯ =

= 0.0015 + j 0.006 W

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Circulating current at no-load =

225 - 215 0.005 + j 0.0135 + 0.0015 + j 0.006

= 486. 5 ––71.56° A IA = 486.5 A IB = –486.5 A V = 225 – (486.5 ––71.56°) (0.005 + j 0.0135) = 225 – 7 ––1.88° = 218 V.

\

17. In a transformer if the load current is kept constant, calculate the power factor at which maximum efficiency occurs. Hints: V2 I 2 cos f2 h= V2 I 2 cos f2 + I 22 R2 + Pi As load current constant, I 22 R2 = constant \ I 22 R2 + Pi = constant = K 1

\ efficiency h = 1+

K V2 I 2 cosf2

So for maximum efficiency, denominator should be minimum. This is possible when cos f2 is maximum. Hence, maximum efficiency occurs when load p.f. is unity (resistive load). 18. A 100 kVA, 50 Hz, 440/11000 V, single-phase transformer has an efficiency of 98.5% when supplying full-load current at 0.8 p.f. lagging and an efficiency of 99% when supplying half full-load current at unity p.f. Find the core losses and copper losses corresponding to fullload current. At what value of load current will the maximum efficiency be attained? Hints: At full-load, 0.8 p.f., the efficiency = 98.5% \

Pi + Pe(f – L) = 1218

(1)

At half-load and unity p.f. efficiency = 99% \

Pi + 0.25 Pc (f – L) = 505

Solving Eqs (i) and (ii), Pi = 267.3 W

Pc (f – L) = 950.7 W

Full-load current at secondary side =

100 ¥ 103 = 9.09 A 11000

At maximum efficiency, I2M = 9.09

267.3 A = 4.82 A. 950.7

(ii)

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19. A single-phase transformer working at unity power factor has an efficiency of 90% at both half-load and at the full-load of 500 W. Determine the efficiency at 75% full-load and the maximum efficiency. Hints: h(f –L) = \

V2 I 2 cos f2 V2 I 2 cos f2 + Pi + Pc ( f - L ) Pi + Pc (f – L) = 55.56 h Ê1

ˆ ÁË f - L˜¯ 2

\

=

(i) ÊI ˆ V2 Á 2 ˜ cos f2 Ë 2¯ 2

ÊI ˆ Ê 1ˆ V2 Á 2 ˜ cos f2 + Pi + Á ˜ Pc ( f - L ) Ë 2¯ Ë 2¯

Pi + 0.25 Pc (f – L) = 27.78

Solving (i) and (ii), we get, Pi = 18.52 W Efficiency at 75% of full-load =

(ii) Pc (f – L) = 37.04 W.

500 ¥ 3/4 2

3 Ê 3ˆ 500 ¥ + Pi + Á ˜ Pc ( f - L ) Ë 4¯ 4

Output at maximum efficiency = 500

= 0.905 p.u.

18.52 = 353.55 W 37.04

\ at maximum efficiency Pc = Pi Maximum efficiency =

353.55 = 0.9051 p.u. 353.55 + 18.52 + 18.52

20. The maximum efficiency of a 100 kVA, single-phase transformer is 98% and occurs at 80% of full-load at 0.8 p.f. lagging. If the leakage impedance of the transformer is 5%, find the regulation at full-load. Hints: hmax = 0.98 \ iron loss (Pi) = 653 W At maximum efficiency, x = \ \

Pi Pc ( f - L )

= 0.8

Pc (f – L) = 1020 W Rep.u =

Pc ( f - L ) d

= 0.0102

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5 = 0.05 100 = 0.04895

Zep.u = Xep.u.

Voltage regulation for lagging p.f. = Rep.u. cos fr + Xep.u. sin fr = 0.03753 p.u. = 3.753%. 21. In a 25 kVA, 2000/200 V transformer, the iron and copper losses are 350 W and 400 W respectively. Determine the values of iron and copper losses which will give maximum efficiency and also find the value of maximum efficiency Hints: Power factor cos f2 = 1. At maximum efficiency, constant iron loss = variable copper loss. hmax =

V2 I 2 cos f2 25 ¥ 103 ¥ 1 = 0.973 p.u. = V2 I 2 cos f2 + Pi + Pi 25 ¥ 103 ¥ 1 + 2 ¥ 350 = 97.3%.

22. Open-circuit and short-circuit tests on a 5 kVA, 220/400 V, 50 Hz single-phase transformer gave the following results. OC test: 220 V, 2 A, 100 W (LV side) SC test: 40 V, 11.4 A, 200 W (HV side) Find the efficiency and approximate regulation of transformer at full-load 0.9 p.f. lagging. Hints: From OC test: core loss Pi = 100 W SC test: Copper loss Pc = 200 W and short-circuit current = 11.4 A 200 \ equivalent resistance (Re2) = = 1.54 W (11.4) 2 Again, \

V2 I2f L = kVA ¥ 103 I2fL =

5 ¥ 103 = 12.5 A 400

Copper loss at full-load (Pc fL) = I22 fL ¥ Re2 = 240. 6 W \ efficiency at full-load =

V2 I 2 cos f = 0.9296 p.u. V2 I 2 cos f + PcfL + Pi = 92.96%

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From short-circuit test, V2sc = 40 V \

Ze2 =

40 = 3.5 W 11.4

Re22 + Xe22 = Ze22 \

Xe2 = 3.15 W

Approximate voltage regulation =

I2 [R2 cos f + Xe2 sin f] V2

= 0.0862 p.u. = 8.62%. 23. A 400/100 V, 5 kVA, two-winding transformer is to be used as an auto-transformer to supply power at 400 V from 500 V source. Draw the connection diagram and determine the kVA output of the auto-transformer. Hints: For a two winding transformer, VH IH = VL IL = Sin = Sout 400 IH = 5 ¥ 103 or, IH = 12.5 A

\

IL = 50 A

Similarly,

Figure 3.60 shows the winding diagram of two-winding transformer working as an autotransformer.

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For an auto-transformer, aA =

VH 500 = = 1.25 VL 400

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\

IH =

IL I = L a A 1.25

Current through 400 V winding I = IL – IH = IL –

IL = 0.2 IL 1.25

Since the current rating of 400 V winding is 12.5 A or

0.2 IL = 12.5 IL = 62.5

The kVA output of the auto-transformer VL I L 400 ¥ 62.5 = = 25. 1000 1000 24. For a 2200/220 V, 50 Hz, single-phase transformer, the emf per turn is 12 V. Determine the number of primary and secondary turns and the net cross-sectional area of core for a maximum flux density of 1.5 T. Hints: V1 V2 = = 12 N1 N 2 2200 � 183 12

or,

N1 =

and

N2 = 18.33 � 18

We have

V1 � E1 = 4.44 f N1 fmax 12 4.44 ¥ 50 = 0.054 Wb

fmax =

\ cross-sectional area =

È V1 ˘ ÍE N = 12˙ 1 Î ˚

f max = 0.036 m2. Bmax

25. A 100 kVA, 1100/230 V, 50 Hz transformer has an HV winding resistance of 0.1 W and a leakage reactance of 0.4 W. The LV winding has a resistance of 0.006 W and a reactance of 0.01 W. Find the equivalent winding resistance, reactance and impedance referred to the HV and LV sides. Convert these to p.u. values. Hints: Z1 = r1 + j x1 = (0.1 + j 0.4) W

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Z2 = (0.006 + j 0.01) W Impedance referred to HV side 2 2 ˘ È ˘ È Ê 1100 ˆ Ê 1100 ˆ Z(HV) = Í0.1 + Á 0.006 ˙ + j Í0.4 + Á 0.01˙ ˜ ˜ Ë 230 ¯ Ë 230 ¯ ˚ Î ˚ Î = [0.237 + j 0.629] W

Similarly, impedance referred to LV sides ˘ ÈÊ 230 ˆ 2 ˘ ÈÊ 230 ˆ 2 . . j + + 0 1 0 006 0.4 + 0.01˙ Z(LV) = ÍÁ ÍÁ ˙ ˜ ˜ ˚ ÎË 1100 ¯ ˚ ÎË 1100 ¯ = [0.0104 + j 0.0275] W

Base impedance referred to HV winding ( kV ) 2 (1.1) 2 = = 12.1 W ( MVA ) 100 ¥ 10 -3 0.237 + j 0.629 Zp.u (HV) = = (0.019 + j 0.052) 12.1 Zp.u. (LV) = 0.0196 + j 0.052.

ZB (HV) =

[where ZB (LV) =

(0.23) 2 100 ¥ 10 -3

= 0.529]

26. A 20 kVA, 2000/200 V single-phase transformer has the following parameters. HV winding R1 = 3 W, X1 = 5.3 W LV winding R2 = 0.05, X2 = 0.05 W (a) Find the voltage regulation at 0.8 p.f. lagging u.p.f. and 0.707 p.f. leading. (b) Determine the secondary terminal voltage at 0.8 p.f. lagging u.p.f. and 0.707 p.f. leading when delivering full-load current with primary voltage fixed at 2 kV. Hints: Impedance referred to LV side. 2 2 ˘ È È Ê 1ˆ Ê 1ˆ ˘ Z(LV) = Í0.05 + Á ˜ 3˙ + j Í0.05 + Á ˜ 5.3˙ Ë 10 ¯ Ë 10 ¯ ˚ ˚ Î Î = 0.08 + j 0.103 W

\

I2 =

20 ¥ 103 = 100 A 200

!

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(a) Voltage drop at 0.8 p.f. lagging = 100 [0.08 ¥ 0.8 + 0.103 ¥ 0.6] = 12.58 V 12.58 ¥ 100 = 6.29% 200 Voltage drop at unity p.f. = 100 [0.08 ¥ 1] = 8 V \ voltage regulation = 4% Voltage drop at 0.707 leading p.f. = 100 [0.08 ¥ 0.707 – 0.103 ¥ 0.707] = –1.63 V \ voltage regulation = – 0.815%. (b) Secondary terminal voltage at 0.8 p.f. lagging \ voltage regulation =

V2 = (200 –12.58) = 187.4 V V2 (at) unity p.f. = (200 – 8) = 192 V V2 (at) 0.707 leading p.f. (200 + 1.63) = 201.63 V. 27. A 40 kVA, 2000/200 V transformer has nameplate leakage impedance of 10%. What voltage must be applied on the HV side to circulate full-load current with LV shorted? Hints: Z HV (W) I HV ( rated ) = 0.10 VHV ( rated ) ZHV (W) IHV (rated) = 0.10 VHV (rated) = 0.10 ¥ 4000 = 400 V = Vsc. 28. In a 25 kVA, 2000/200 V transformer, the iron and copper losses are 350 and 400 W respectively. (a) Calculate the efficiency on u.p.f. at full-load and half-load. (b) Determine the load for maximum efficiency and iron and copper loss in this case. Hints: h(f L, up.f.) =

25 ¥ 103 ¥ 1 25 ¥ 103 + 350 + 400

h Ê1

ˆ fL, upf ˜ ËÁ 2 ¯

(a) K =

=

= 97.08% 25 ¥ 103 ¥ 1/ 2 = 96.5% 1 1 3 25 ¥ 10 ¥ + 350 + ¥ 400 2 4

350 = 0.935 400

Load for max. h = 25 ¥ 0.935 = 23.385 kVA Pi = 350 W. Pc = (0.935)2 ¥ 400 = 350 W.

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29. The efficiency of a 1000 kVA, 110/220 V, 50 Hz, single-phase transformer is 98.5% at half full-load at 0.8 p.f. leading and 98.8% at full-load u.p.f. Calculate (a) iron loss, and (b) fullload copper loss, and (c) maximum efficiency at u.p.f. Hints: 0.985 =

500 ¥ 103 ¥ 0.8 1 P 4 c fL

500 ¥ 103 ¥ 0.8 + Pi + 0.988 =

1000 ¥ 1000 1000 ¥ 1000 + Pi + PcfL

Solving these, we get Pi = 4071 W and

PcfL = 8079 W

4071 = 0.71 8079 1000 ¥ 1000 ¥ 0.71 = = 98.9%. 1000 ¥ 1000 ¥ 0.71 + 2 ¥ 4071

K= hmax

30. A transformer has its maximum efficiency of 0.98 at 20 kVA at u.p.f. During the day, it is loaded as follows: 12 hours: 2 kW at 0.6 p.f. 6 hours: 10 kW at 0.8 p.f. 6 hours: 20 kW at 0.9 p.f. Find the all day efficiency of the transformer. Hints: hmax =

20 ¥ 103 ¥ 1 20 ¥ 103 ¥ 1 + 2 Pi

= 0.98

2 kW, 0.6 p.f (3.33 kVA), 12 h 2 ¥ 12 = 24 kWh (output) È Ê 3.33 ˆ 2 ˘ 200 Í1 + Á ˜ ˙ ¥ 12 = 2.47 kWh (loss) Î Ë 20 ¯ ˚ 10 kW, 0.8 p.f. (12.5 kVA), 6 h 10 ¥ 6 = 60 kWh È Ê 12.5 ˆ 2 ˘ 200 Í1 + Á ˜ ˙ ¥ 6 = 1.67 kWh (loss) Î Ë 20 ¯ ˚ 20 kW. 0.9 p.f. (22.22 kVA), 6 h 20 ¥ 6 = 120 kWh

!

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È Ê 22.22 ˆ 2 ˘ 200 Í1 + Á ˜ ˙ ¥ 6 = 2.68 kWh (loss) Î Ë 20 ¯ ˚ 204 h(all day energy) = = 96.77%. 204 + 6.82

\

31. A 400/100 V, 10 kVA, two-winding transformer is to be employed as an auto-transformer to supply a 400 V circuit from a 500 V source. When tested as a two-winding transformer at rated load, 0.85 p.f. lagging, its efficiency is 0.97. Find its kVA rating and efficiency as an auto-transformer. Hints: I1

100

I2 V1 = 500 V V2 = 400 V I2 – I1

! Gjh/!4/72! Djsdvju!ejbhsbn!pg!Qspc/!42

I1 = 10 ¥

1000 = 100 A 100 (kVA)auto = hTW =

500 ¥ 100 = 50 1000 10 ¥ 103 ¥ 0.85 10 ¥ 103 ¥ 0.85 + PL

= 0.97

PL = 262.9 W \ full-load output as auto (0.85 p.f.) = 50 ¥ 0.85 = 42.4 kW 42.5 hauto = = 99.38%. 42.5 + 0.2629 32. A 200/400 V, 20 kVA and 50 Hz transformer is connected as an auto-transformer to transform 600 V to 200 V. (a) Calculate the kVA rating of an auto-transformer. (b) With a load of 20 kVA, 0.8 p.f. lagging connected to 200 V terminals, determine the currents in the load and two-transformer winding.

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Hints:

! Gjh/!4/73! Djsdvju!ejbhsbn!pg!Qspc/!43

I1 = (kVA)auto =

20 ¥ 103 = 50 A 400 600 ¥ 50 = 30 1000

20 ¥ 103 = 100 A I2 = 200 \

I2 – I1 = 50 A.

33. A 20 kVA, 4400/220 V transformer with an equivalent impedance of 0.01 W is to operate in parallel with a 15 kVA, 4400/220 V transformer with an equivalent impedance of 0.015 W. The two transformers are connected in parallel and made to carry a load of 25 kVA. Assume both impedances to have the same angle. (a) Find the individual load currents. (b) What percent of the rated capacity is used in each transformer? Hints: S1 =

Z2 0.015 SL = ¥ 25 = 15 kVA Z1 + Z 2 0.025

S2 =

Z1 0.01 SL = 25 = 10 kVA Z1 + Z 2 0.025

!

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10 ¥ 1000 = 45.6 A 220 15 = 75% (b) Percentage rated capacity used in transformer 1 = 20 10 Percentage rated capacity used in transformer 2 = = 66.7%. 15 (a) I1 =

15 ¥ 1000 = 68.2 A. 220

I2 =

34. Two single-phase transformers, rated 1000 kVA and 500 kVA respectively, are connected in parallel on both HV and LV sides. They have equal voltage ratings of 11 kV/400V and their p.u. impedances are (0.02 + j0.07) and (0.025 + j0.0875) respectively. What is the largest value of the u.p.f. load that can be delivered by parallel combination at the rated voltage? Hints: S1 (rated) = 1000 kVA S2 (rated) = 500 kVA Z1 = 0.02 + j 0.07 = 0.0728 –74° Z2 = (0.025 + j 0.0875) ¥ 2 = 0.182 –74° S1 = \

Z2 SL Z1 + Z 2

SL = 1400 kVA

S2 =

Z1 SL Z1 + Z 2

SL = 3500 kVA

As total load is increased, the 1000 kVA transformer will be the first to reach its full-load. SL(max) = 1400 kVA.

Nvmujqmf.Dipjdf!Rvftujpot 1. A single-phase air core transformer, fed from a rated sinusoidal supply is operating at no load. The steady state magnetizing current drawn by the transformer from the supply will have the waveform I

I

(a)

t

(b)

t

I

I

(c)

(d) t

t

[GATE 2011]

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2. In a transformer, zero voltage regulation at full load is (a) not possible (b) possible at unity power factor load (c) possible at leading power factor load (d) possible at lagging power factor load [GATE 2007] 3. In transformers which of the following statements is valid? (a) In an open circuit test, copper losses are obtained while in short circuit test, core losses are obtained (b) In an open circuit test current is drawn at high power factor (c) In a short circuit test, current is drawn at zero power factor (d) In an open circuit test current is drawn at low power factor [GATE 2006] 4. The equivalent circuit of a transformer has leakage reactances X1, X2¢ and magnetizing reactance XM. Their magnitudes satisfy (a) X1 >> X2¢ >> XM

(b) X1 XM

(d) X1 ª X2¢ !31!lX/

Solution Primary phase voltage = 300 V 120 Primary current/phase = = 69.28 A 3 200 Z01 = = 2.88 W 69.28 Now,

I 12 R01 =

25 ¥ 103 = 8.33 ¥ 103 W = 8330 W 3

Uisff.qibtf!Usbotgpsnfst

R01 =

8330 (69.28) 2

5/3:

= 1.74 W

X01 = ( 2.88) 2 - (1.74) 2 = 8.2944 - 3.0276 = 2.92 W 69.28 ¥ 1.74 ¥ 100 = 1.82 6600 Percentage X = 2.40 Percentage R =

Percentage voltage regulation on full-load at 0.8 p.f. = 1.82 ¥ 0.8 + 2.40 ¥ 0.6 = 2.896.

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Solution (0.7 + j 0.2) W 2000 V(line)

D/Y

VL

Gjh/!5/42! Djsdvju!ejbhsbn!pg!Qspcmfn!5/:

According to the short-circuit test, 40 W = 1.33 W 30 800 = W = 0.889 W (30) 2

ZHV = RHV

XHV = 1.332 - 0.8892 = 0.989 W

!

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Fmfdusjdbm!Nbdijoft

On equivalent but star basis, transformer impedance on HV side \

ZT = (0.889 + j 0.989) W Z total = (0.7 + j 0.2) + (0.889 + j 0.989) = (1.589 + j 1.189) W = 1.985, –36.8° W f = 36.8°

\ (a) IL (HV) =

50 ¥ 1000 2000 / 3

= 43.3 0∞ A

Voltage drop = IL (R cos f + X sin f) = 43.3 (0.889 cos 36.8° + 0.989 sin 36.8°) = 56.47 N È Ê 2000 ˆ˘ - 56.47˜ ˙ VL (line to line) = Í 3 Á ¯˚ Î Ë 3 = 1902.19/11.55 V = 164.69 V (b) Three-phase short-circuit on secondary terminals I Fsc =

2000 100 3

2000 / 3 = 581.7 A 1.985

Isc (transformer primary) = 581.7 3 A = 1007.5 A (line current) Isc (transformer secondary) = 1007.5 ¥

2000 100 3

= 11633.6 A (line current).

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)b*! )c*! )d*! )e*! )f*!

Uif!qpxfs!dpotvnfe!cz!uif!voju!jo!lX Uif!upubm!lWB!vtfe! Uif!sbufe!mjof!dvssfout!bwbjmbcmf!gspn!uif!usbotgpsnfs!cbol! Uif!sbufe!usbotgpsnfs!qibtf!dvssfout!pg!uif!D.tfdpoebsjft Qfsdfou!pg!sbufe!mpbe!po!usbotgpsnfst!

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! )g*! Qsjnbsz!mjof!boe!qibtf!dvssfout ! )h*! Uif!lWB!sbujoh!pg!fbdi!joejwjevbm!usbotgpsnfs

Solution

2500 / 3 V

100 A

2500 V

10

:1

250 V

3

Gjh/!5/43! Djsdvju!ejbhsbn!pg!Qspcmfn!5/21

(a) The power consumed by the unit = 3 ¥ 100 ¥ 250 ¥ 0.8 W = 34641.02 W = 34.64 kW (b) Total kVA used = 3 ¥ 100 ¥ 250 = 43301.27 VA = 43.3 kVA (c) Rated line current =

60 ¥ 1000 3 ¥ 250

A = 138.56 A

(d) Rated phase current of D secondaries =

138.56 3

A = 80 A

43.3 ¥ 100% 60 = 72.167%

(e) Percent of rated load =

(f) Primary phase current =

100 ¥ 3 10

= 17.32 A Primary line current = 17.32 A

[turns ratio = 10]

!

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Fmfdusjdbm!Nbdijoft

(g) kVA rating of each individual transformer 60 kVA 3 = 20 kVA.

=

! Qspcmfn!5/22 Gjhvsf!5/44!tipxt!b!Tdpuu.dpoofdufe!usbotgpsnfs-!tvqqmjfe!gspn!22!lW-!uisff.qibtf-!61!I{!nbjot! xjui!tfdpoebsjft-!tfsjft!dpoofdufe!bt!tipxo/!Tvqqmz!611!B!bu!b!wpmubhf!pg!211! 3!up!b!sftjtujwf!mpbe/! Uif!qibtf!tfrvfodf!pg!uif!uisff.qibtf!tvqqmz!jt!BCD/ Teaser

500 A

A

Va

( 3 /2) N1 11 KV, 3-phase supply

N2

IA

Resistive load

100 2 V B

N1/2

M

N1/2

C

Vc

N2

Gjh/!5/44! Djsdvju!ejbhsbn!pg!Qspcmfn!5/22 ! )b*! Dbmdvmbuf!uif!uvsot.sbujp!pg!uif!ufbtfs!usbotgpsnfs/! ! )c*! Dbmdvmbuf!uif!mjof!dvssfou!JC!boe!jut!qibtf!bohmf!xjui!sftqfdu!up!uif!wpmubhf!pg!qibtf!B!up!ofv. usbm!po!uif!uisff.qibtf!tjef/

Solution (a)

Va = Vb =

100 2 2

= 100 V

N1 VBC 11000 = = = 110 Vb 100 N2

Uisff.qibtf!Usbotgpsnfst

Turns ratio = (b) IA = =

2 3 2 3

5/44

3 N1 ◊ = 3/2 ¥ 110 = 95.26 2 N2

¥

N2 ¥ 500 A N1

¥

100 ¥ 500 A 11000

= 5.249 A IBC =

N2 500 ¥ 500 A = A = 4.545 A N1 110

V and I are in-phase because of a resistive load. IB = IB C -

IA 2

= 4.545 –

5.249 = 1.92 A 2

So IB is lagging VAN by 45°.

! Qspcmfn!5/23 Bo!71110561!W-!61!I{-!uisff.qibtf!usbotgpsnfs!jt!efmub!dpoofdufe!po!uif!IW!tjef!boe!MW!xjoejoht! bsf!tubs!dpoofdufe/!Uifsf!bsf!up!cf!21!W!qfs!uvso!boe!uif!gmvy!efotjuz!jt!opu!up!fydffe!2/3!Xc0n3/! Dbmdvmbuf!uif!ovncfs!pg!uvsot!qfs!qibtf!po!fbdi!xjoejoh!boe!uif!ofu!jspo!dsptt.tfdujpobm!bsfb!pg! uif!dpsf/

Solution Ê Eph1 ˆ Induced voltage in the primary per turn Á ˜ = 10 V Ë Np1 ¯ HV side phase voltage (Eph1) = Line voltage (El1) = 6000 6000 = 10 Np1

\ \

NP1 = 600 Also, Now,

Eph 2 Np 2

= 10

Eph2 =

1 3

¥ 450 = 259.8 V

[As it is delta connected]

!

5/45

Fmfdusjdbm!Nbdijoft

259.8 = 25.98 ª 26 10 Eph2 = 4.44 Bm A f Np2 10 = 4.44 ¥ 1.2 ¥ A ¥ 50 10 A= = 0.037 m2. 4.44 ¥ 1.2 ¥ 50

\

Np2 =

\ \ \

! Qspcmfn!5/24 Uxp!211!lWB!tjohmf.qibtf!usbotgpsnfst!bsf!dpoofdufe!jo!pqfo.efmub!up!tvqqmz!b!341!W!cbmbodfe! uisff.qibtf!mpbe/! ! )b*! Xibu!jt!uif!upubm!mpbe!uibu!dbo!cf!tvqqmjfe!xjuipvu!pwfsmpbejoh!fjuifs!usbotgpsnfs@ ! )c*! Xifo!uif!efmub!jt!dmptfe!cz!uif!beejujpo!pg!b!uijse!211!lWB!usbotgpsnfs-!xibu!upubm!mpbe!dbo! opx!cf!tvqqmjfe@ ! )d*! Qfsdfou!jodsfbtf!jo!mpbe/

Solution (a) The rated secondary transformer current =

100 ¥ 103 = 434.78 A 230

Now load kVA = 3 ¥ 230 ¥ 434.78 ¥ 10–3 = 173.2 kVA [As the secondary transformer current = Load current] \

Load kVA 173.2 = Total capacity of two individual transformers 2 ¥ 100 = 0.866

(b) When the delta is closed by the addition of a third 100 kVA transformer then D–D bank will operate at full capacity of the individual transformer. \ now the lead kVA = 3 ¥ 100 = 300 kVA (c) Percent increase in load kVA 300 - 173.2 ¥ 100% 173.2 = 73.2%.

=

! Qspcmfn!5/25 Uxp!tjohmf.qibtf!gvsobdft!xpsljoh!bu!211!W!bsf!dpoofdufe!up!b!7711!W-!uisff.qibtf!tvqqmz!uispvhi! Tdpuu.dpoofdufe! usbotgpsnfst/! Efufsnjof! uif! dvssfout! jo! uif! uisff.qibtf! mjoft! xifo! uif! qpxfs! ublfo!cz!fbdi!gvsobdf!jt!611!lX!bu!b!q/g/!pg!1/6!mbhhjoh/!Ofhmfdu!usbotgpsnfs!mpttft/

Uisff.qibtf!Usbotgpsnfst

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Solution Np

6600 = 33 100 Ns For a teaser transformer, Turns ratio =

=

P2t = V2t I2t cos f 2t \

550 ¥ 103 A = 10000 A 100 ¥ 0.5 As the two-phase load is balanced, the three-phase side is also balanced. For the mmf balance of the teaser transformer, I2t =

IA ¥ \

\

3 N = I2t Ns 2 p N 2 IA = ¥ s ¥ I2 t 3 Np 2 1 = ¥ ¥ 10000 A = 349.9 A 3 33 | IA | = | IB | = | IC | = 349.9 A.

! Qspcmfn!5/26 B!efmub!tubs!2201/55!lW!uisff.qibtf!usbotgpsnfs!ibt!26!W!qfs!uvso!boe!nbyjnvn!gmvy!efotjuz!pg! 26!Xc0n3/!Efufsnjof!uif!ovncfs!pg!uvsot!qfs!qibtf!po!fbdi!xjoejoh!boe!uif!ofu!dsptt.tfdujpobm! bsfb/

Solution Voltage per turn is 15 V Line voltage on HV side = Phase voltage on HV side = 11 kV \ number of turns in the high-voltage side 11000 N1 = = 733 15 440 V Phase voltage on LV side = 3 \ number of turns on the LV side N2 =

440 3 ¥ 15

= 17

If A is the cross-sectional area of the core then

or,

E2 = 4.44 Bm A f N2 E 1 1 A= 2 ¥ = 15 ¥ = 0.045 m2. N2 4.44 Bm f 4.44 ¥ 1.5 ¥ 50

!

5/47

Fmfdusjdbm!Nbdijoft

! Qspcmfn!5/27 B!611!lWB!mpbe!bu!1/9!q/g/!mbhhjoh!jt!tvqqmjfe!cz!uisff!2f!usbotgpsnfst!dpoofdufe!jo!DÐD/!Fbdi!pg!uif! DÐD!usbotgpsnfst!jt!sbufe!bu!311!lWB-!36110361!W/!Jg!pof!efgfdujwf!usbotgpsnfs!jt!sfnpwfe!gspn! tfswjdf-!dbmdvmbuf!gps!uif!WÐW!dpoofdujpo/ ! ! ! ! !

)b*! )c*! )d*! )e*! )f*!

Uif!lWB!mpbe!dbssjfe!cz!fbdi!usbotgpsnfs! Qfsdfou!sbufe!mpbe!dbssjfe!cz!fbdi!usbotgpsnfs! Upubm!lWB!sbujoht!pg!uif!usbotgpsnfs!cbol!jo!WÐW Sbujp!pg!WÐW!cbol!up!DÐD!cbol!usbotgpsnfs!sbujoht Qfsdfou!jodsfbtf!jo!mpbe!po!fbdi!usbotgpsnfs!xifo!pof!usbotgpsnfs!jt!sfnpwfe

Solution (a) Load in kVA per transformer in V–V bank 1 = ¥ 500 = 288.67 kVA 3 288.67 (b) Percent rated load carried by each transformer = ¥ 100 = 144.335 percent 200 (c) Total kVA rating of the V–V bank = 3 ¥ 200 = 346.4 kVA (d)

346.4 V -V bank rating = = 57.7% D - D bank rating 3 ¥ 200

(e) Original load kVA per transformer in D–D 1 ¥ 500 = 166.67 kVA 3 Percent increase in load on each transformer when one transformer is removed =

288.67 - 166.67 ¥ 100% 166.67 = 73.198%.

=

! Qspcmfn!5/28 B!61!lWB-!249110319!W-!DÐZ!ejtusjcvujpo!usbotgpsnfs!ibt!b!sftjtubodf!pg!2/6&!boe!sfbdubodf!pg! 9&!qfs!voju/! Efufsnjof!uif!usbotgpsnfsÕt!jnqfebodf!sfgfssfe!up!uif!ijhiÐwpmubhf!tjef!boe!wpmubhf!sfhvmbujpo!bu! gvmm.mpbe!1/9!q/g/!mbhhjoh/

Solution The base impedance of the high-voltage side Zbase =

(13800) 2 = 11426 W 50, 000 / 3

(E high-voltage side is delta connected)

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\ now per-unit impedance is Zp.u. = 0.015 + j 0.08 \ transformer’s impedance referred to the high-voltage side is ZeH = (0.015 + j 0.08) ¥ 11426 = 171.39 + j 914 W The voltage regulation is IH reH cos q + IH xeH sin q VH Now rated current on the high-voltage side is 50, 000 / 3 = 1.21 A 13800 cos q = 0.8 lag sin q = 0.6 IH =

\

1.21 (171.39 ¥ 0.8 + 914 ¥ 0.6) 13800 = 0.06 or 6%.

voltage regulation =

! Qspcmfn!5/29 Uisff!tjohmf.qibtf!usbotgpsnfst-!fbdi!pg!sbujoh!26!lWB!-!33110221!W-!bsf!dpoofdufe!jo!efmub.efmub! gps!uisff.qibtf!cbmbodfe!pqfsbujpo/!Uif!qpxfs!gbdups!pg!fbdi!usbotgpsnfs!jt!1/9!mbh/!Jg!uif!bssbohf. nfou!jt!dpowfsufe!up!pqfo!efmub/! ! )b*! Efufsnjof!uif!mpbe!lWB!efmjwfsfe!jo!fbdi!dbtf!xjuipvu!pwfsmpbejoh!uif!usbotgpsnfst!boe!uif! pqfsbujoh!qpxfs!gbdups!pg!uif!usbotgpsnfst!pqfsbujoh!jo!pqfo!efmub/ ! )c*! Gjoe!uif!qpxfs!gbdups!jg!uif!usbotgpsnfst!xfsf!pqfsbujoh!bu!vojuz!qpxfs!gbdups!jo!efmub.efmub/ ! )d*! Xibu!xpvme!cf!uif!qpxfs!gbdups!gps!uif!usbotgpsnfst!jo!efmub.efmub-!jg!uif!qpxfs!gbdups!pg!pof! pg!uif!usbotgpsnfst!jt!up!cf!{fsp!jo!pqfo.efmub@!

Solution Rated secondary current I2 =

15000 = 136.36 A 110

(a) In delta-delta configuration, the transformer secondary currents are the phase currents of delta-connected secondary. \ line current in the secondary IL = 3 ¥ 136.36 = 236.2 A

!

5/49

Load kVA =

Fmfdusjdbm!Nbdijoft

3 VL I L

=

3 ¥ 110 ¥ 236.2

= 45 10 103 When the transformers are connected in open-delta, the line current is the rated current of the transformers. 3

\

IL = 136.36 A

\ load kVA = 3 ¥ 110 ¥ 136.36/103 = 25.98 The power factor of delta-delta transformer is 0.8 or q = cos–1 0.8 = 36.87° \ power factors of the two transformers operating in open delta are cos (36.87° + 30°) lag and cos (36.87° – 30°) lead or 0.39 lag and 0.99 lead (b) If the delta-delta transformers were operating at unity power factor then the power factors are cos (0° + 30°) lag and cos (0° – 30°) lead or 0.866 lag and 0.866 lead. (c) If the power factor of one of the transformers is zero in open delta then cos (f + 30°) = 0 or cos (f – 30°) = 0 where cos f is the power factor of the transformers operating in open-delta, \ Hence,

f = ± (90° – 30°) = ± 60° f = 60° lag or f = 60° lead

\ power factor is 0.5 lag or 0.5 lead.

! Qspcmfn!5/2: Uxp!221!W-!tjohmf.qibtf!gvsobdft!xjui!mpbet!pg!911!lX!boe!2111!lX!pqfsbuf!bu!qpxfs!gbdups!pg! 1/818!mbhhjoh!gspn!4/4!lW-!uisff.qibtf!tvqqmz!uispvhi!b!Tdpuu.dpoofdufe!usbotgpsnfs/!Efufsnjof! uif!dvssfout!jo!uif!uisff.qibtf!mjoft/

Solution N 3300 = 30 = 1 N2 110 For the main transformer, the secondary current Turns ratio a =

I2M =

1000 ¥ 103 = 12858 A 110 ¥ 0.707

I2T =

800 ¥ 103 = 10286 A 110 ¥ 0.707

For the teaser transformers,

Uisff.qibtf!Usbotgpsnfst

5/4:

The teaser primary current I1T = I2T ¥

N2 3 N 2 1

= 10286 ¥

2 3 ¥ 30

= 396 A

The main transformer primary current I1M =

I2 M 12858 = 428.6 A = a 30

The current in one phase IA = I1T = 396 A The current in the other two phase from Eq. (4.10), 2

2

2

2

ÊI ˆ ÊI ˆ Ê 396 ˆ Ê 12858 ˆ IB = IC = Á 1T ˜ + Á 2 M ˜ = Á + = 428.83 A. Ë 2 ˜¯ ÁË 30 ˜¯ Ë 2 ¯ Ë a ¯

! Qspcmfn!5/31 Uxp!tjohmf.qibtf!Tdpuu.dpoofdufe!usbotgpsnfst!tvqqmz!b!uisff.qibtf!mpbe!xjui!351!W!cfuxffo!mjoft! boe!ofvusbm/!Uif!qibtf!wpmubhf!jo!uif!uxp.qibtf!tztufn!jt!22!lW!boe!uijt!jt!uif!ijhi.wpmubhf!tjef/!Jg! uif!nbyjnvn!gmvy!efotjuz!jo!uif!dpsf!jt!2/6!Xc0n3!boe!uif!dsptt.tfdujpobm!bsfb!pg!uif!dpsf!jt!711! tr/!dn-!efufsnjof!uif!ovncfs!pg!uvsot!jo!cpui!uif!ijhi.!boe!mpx.wpmubhf!xjoejoht!boe!uif!qptjujpo! pg!uif!ofvusbm!qpjou/!

Solution If number of turns in the high-voltage side is NH, \

11000 = 4.44 ¥ 1.5 ¥ 600 ¥ 10–4 ¥ 50 NH NH = 550 Hence, the number of turns on the high-voltage side of both transformers is 550. The number of turns in the low-voltage side of the main transformer = 550 ¥ The number of turns in the high-voltage side of the teaser transformer =

From Fig. 4.24, the number of turns between S and N 1 TNS = ¥ 18 = 6 3 th \ the neutral point is located at the 6 turn from the point S.

3 ¥ 240 = 21 11000

3 ¥ 21 = 18 2

!

5/51

Fmfdusjdbm!Nbdijoft

QBSBMMFM!PQFSBUJPO!PG!UISFF.! QIBTF!USBOTGPSNFST

5/21

For parallel operation of three-phase transformers, the conditions for paralleling a single-phase transformer are required as well as the following additional conditions: 1. The ratio of voltage should refer to terminal voltage of primary and secondary. 2. The phase sequence must be identical. 3. The phase between primary and secondary of all transformers to be parallel must be identical. The main reason for operating transformers in parallel are as follows: 1. Using a single large transformer for a large load is uneconomical and impractical too. 2. In substations, the total load required may be supplied by an appropriate number of transformers of standard size. This reduces the spare capacity of the substation. 3. There is a scope of future expansion of a substation to supply a load beyond the capacity of the transformers already installed. 4. If there is a breakdown of a transformer in a system of transformers connected in parallel, there is no interruption of power supply for essential services. Similarly, when a transformer is taken out of service for its maintenance and inspection, the continuity of supply is maintained. If two star-star or delta-delta transformers are to be operated in parallel they must belong to the vector group, i.e. either they both should be connected for zero phase displacement or 180° phase displacement. A delta-delta transformer can also be operated with a star-star transformer if they belong to the same vector group. Similarly, DY transformer can be operated with another DY transformer and a YD transformer can be operated in parallel with another YD transformer if they belong to the same vector group. A DY transformer can also be operated in parallel with a YD transformer. But a DD or YY transformer cannot be put in parallel with a DY or YD transformer, as this would result in 30° phase displacement between primary and secondary line voltages and hence a large circulating current will flow.

IBSNPOJD!QIFOPNFOB!JO!B!UISFF.! QIBTF!USBOTGPSNFS

5/22

If the flux in a transformer core is sinusoidal, the induced emf is also sinusoidal but the magnetizing current is not sinusoidal due to non-linear B-H curve for core material. The magnetizing current contains third and higher harmonics necessary to produce sinusoidal flux. The amplitude of these harmonics increases with increase in maximum flux density in the core. If the phase voltage across each phase is to remain sinusoidal then the phase magnetizing currents are of the form as follows:

Uisff.qibtf!Usbotgpsnfst

Iam = Im sin wt + I3m sin (3wt + q3) + I5m sin (5m + q5) …

5/52

(4.11)

Ibm = Im sin (wt – 120°) + I3m sin [3 (wt – 120°) + q3] + I5m sin [5 (wt – 120°) + q5] = Im sin (wt – 120°) + I3m sin (3wt + q3) + I5m sin (5wt + 120° + q5) …

(4.12)

Icm = Im sin (wt – 240°) + I3m sin [3 (wt – 240°) + q3] + I5m sin [5 (wt – 240°) + q5] = Im sin (wt – 240°) + I3m sin (3wt + q3) + I5m sin (5wt + 240° + q5)…

(4.13)

Here, third and fifth harmonics components have been considered and higher order harmonics are neglected. From the equations (4.11) to (4.13), it is noted that third harmonic currents of all the three phases are in the same direction and hence they are cophasal. 2/!Efmub!Dpoofdujpo! If Iam, Ibm and Icm represent the magnetizing current in the phase of a delta-connected winding then the line current Iabm = Iam – Ibm = 3 Iam sin (wt + 30°) – 3 I5m sin (5wt – 30° + q5)…

(4.14)

Hence, from Eq. (4.14), it is seen that the third harmonic component does not exist in the line current of the delta-connected transformer though it exists in the phase current. The third harmonic current flows round the closed loop of delta-connected winding. For this reason, majority of threephase transformers have delta-connected winding and in case where it is not possible to have either primary and secondary connected in delta, a tertiary winding connected in delta is provided. The tertiary winding carries the circulating third harmonic current required by the sinusoidal flux in each limb of the core. A similar treatment for voltages is also applicable. Since the third harmonic current flows in the same direction, the third harmonic voltage round the closed delta is 3V3m sin (3wt + f3). This third harmonic voltage will circulate a third harmonic current round the closed loop of the delta. 3/!Tubs!Dpoofdujpo! In star connection, the neutral current IN = Iam + Ibm + Icm … Now,

Iam + Ibm + Icm = 3 I3m sin (3w t + q3) …

(4.15) (4.16)

Hence, under balanced condition, the current in the neutral wire is a third harmonic current having three times the magnitude of each third harmonic phase current. If neutral wire is not present (three-wire system), the neutral current is zero. Hence, I3m = 0. Hence, a three-phase three-wire star-connected system suppresses the flow of third harmonic magnetizing current.

!

5/53

Fmfdusjdbm!Nbdijoft

Now the line voltage in a star-connected system Vab = Vam – Vbm = 3 V5m sin (wt + 30° + f1) + 3 V5m sin (5wt – 30° + f5)…

(4.17)

Hence, the third harmonic component is not present in the line-to-line voltage of a star connection.

JOSVTI!DVSSFOU!

5/23

When a transformer is switched on, a heavy inrush of magnetizing current takes place. This transient current has distorted wave shape and the negative half-cycles may be wiped out. The transient current lasts for only a few cycles but during this time, it causes severe dip in the supply voltage and may cause tripping of overload and differential relays. I3

I1 B1

B3

Bm

B¢

B¢

t2

t1

I2 B2

Gjh/!5/45! Josvti!dvssfou!boe!gmvy!efotjuz

Consider Fig. 4.34; when the transformer operates under normal condition, the flux density follows curve B1 and magnetizing current follows curve I1. Suppose the transformer is switched off at time t1. The current is zero at this position but the flux density has some value, say B¢. If the transformer were not switched off, the flux density and current would have followed curve B2 and curve I2 respectively. The curve is switched on again at t2. Since flux cannot be created or destroyed instantly, the flux density wave starts from the initial value B¢ and traces curve B3 which is a displaced sinusoid. The magnetizing current corresponding to curve B3 is I3. Since the transformer

Uisff.qibtf!Usbotgpsnfst

5/54

is designed for flux density Bm, the peak value of the curve B3 will cause supersaturation of the magnetic circuit which results in very high value of magnetizing current. The maximum value of inrush current depends on the instant of switching on and the residual flux density. The residual flux density in transformers may be about 50 to 60% of the maximum flux density and is responsible for making the peak of inrush current about 2 to 5 times as high as it would be otherwise.

!USBOTGPSNFS!UBQQJOHT!

5/24

The taps in a transformer are provided to permit small changes in the turns ratio of the transformer A typical transformer installation might have four taps in addition to the nominal setting with spacings of 2.5 percent of full-load voltage. Such an arrangement provides for adjustments up to 5% above or below the nominal voltage rating of the transformer. The taps in a transformer permit the transformer to be adjusted in the field to accommodate variations in local voltages. Though taps may be provided on both the high- and low-voltage sides, the tappings are generally provided on the high-voltage side due to the following reasons: 1. The low-voltage winding is placed next to the core due to the insulation considerations. The high-voltage winding which is placed outside the low-voltage winding is easily accessible and tappings can be provided easily. 2. Tap changing gears if placed in the high-voltage side will have to handle lower current as against if they are placed in the low-voltage side. 3. As the low-voltage side has fewer number of turns, if the tappings are provided on the lowvoltage side, the variation in the output voltage will be limited. On the other hand, as the highvoltage side has more turns, smoother control of output voltage is possible if the tappings are provided in the high-voltage side. The tap changer is connected where the voltage to the neutral is minimum. In a star-connected transformer, the tapped end of the windings are connected to form the star point though physically the tapped coils are placed in the middle of the winding. In a delta-connected transformer, it is essential to provide the tapped coils in the middle so that the tap-changing gear is far removed from the line and lighting surges. Depending on whether the tap changer is designed to operate when the transformer is out of service or when the transformer is in operation, it can be of two types: off-load tap changer and on-load tap changer.

5/24/2! Pgg.Mpbe!Ubq!Dibohfs Off-load tap changers are used for seasonal voltage variations. A simple arrangement of off-load tap changer with six studs is shown Fig. 4.35. The high-voltage winding is tapped at 2.5% interval.

!

5/55

Fmfdusjdbm!Nbdijoft

LV

HV

2

3 S

4 5

1

6

Gjh/!5/46! Pgg.mpbe!ubq!dibohfs

S is a rotatable arm which can be rotatable with the help of a handle. When S short circuits studs 1 and 2, the output voltage is the full rated voltage of the high-voltage winding. When S short circuits studs 1 and 6, the output voltage is 97.5% of the full rated voltage. Similarly, when S short circuits studs 6 and 5, and 5 and 4, the output voltage is 95% and 92.5% respectively. When S connects studs 4 and 3 simultaneously, the output voltage is 90% of the rated voltage of the high-voltage winding. S cannot be rotated beyond 3 in anticlockwise direction due to mechanical stress limitations. The off-load tap changers are operated after switching off the transformer. If the transformer is operated with S at studs 4 and 5 and it is desired to operate the transformer with S at positions 6 and 5 then the transformer must first be de-energized, and S is rotated in clockwise direction to short circuit studs 6 and 5. After this, the transformer should be connected to the supply.

5/24/3! Po.Mpbe!Ubq!Dibohfs In this process of tap changing, the output voltage can be varied while the transformer is in operation. No part of the tapped winding should get short circuited and some resistors or reactors are introduced to prevent short circuit. Also, an alternate circuit is provided so that the load current can take the alternate path when the tap-changing selection process is carried out in the main circuit.

Uisff.qibtf!Usbotgpsnfst

5/56

HV

e LV

X d S c

X

b

a

Gjh/!5/47! Po.mpbe!ubq!dibohfs

When the transformer is delivering full rated voltage, the switch S is closed and switch a is connected as shown in Fig. 4.36. The mmfs generated in the two identical reactors X cancel each other as they carry equal currents in the opposite directions. When the tapping is required to be connected at position b, the switch S is opened. In this condition, the upper reactor carries the total current. Then switch b is closed, switch a is opened and again S is closed. This sequence of operation should be followed every time the tap setting is changed. In modern times, the on-load tap changers are designed for automatic operation.

JOEVDUJPO!SFHVMBUPS!

5/25

An induction regulator can be used for smooth variation of the output ac voltage. The output voltage is changed by varying the angle between the magnetic fluxes of primary and secondary windings unlike in a tap-changer transformer where the output voltage is regulated by altering the turns ratio. Also, in a tap-changer transformer, smooth variation of voltage is not possible, as voltage can be varied in discrete steps. An induction regulator is actually an induction machine constructed to be used as a variable voltage supplier. The rotor should be kept in a static position after turning it mechanically by the required angle, otherwise the rotor may try to rotate as a motor, specially in case of a three-phase regulator. The rotor rotation in an induction regulator is limited to a maximum of half revolution, hence the slip rings and brushes are not required.

!

5/57

Fmfdusjdbm!Nbdijoft

5/25/2! Tjohmf.qibtf!Joevdujpo!Sfhvmbups! Figure 4.37 shows the construction of a single-phase regulator. The primary winding is placed in rotor slots. It is designed to carry small current and has smaller conductor area. The secondary winding is housed in stator slots due to its larger cross-sectional area. In addition to the primary winding, the rotor also carries a short-circuited compensating winding whose magnetic axis is perpendicular to that of the primary winding. The secondary is connected in series with the output circuit. When the primary and secondary winding axes coincide, there is maximum magnetic coupling between them and zero magnetic coupling with the compensating winding. Hence, the secondary induced emf is maximum and the output voltage is the sum of primary voltage and the secondary induced emf. When the primary axis is perpendicular to the secondary axis, no emf is induced in the secondary and the output and input voltages are equal. If the primary is rotated through 180° from the secondary winding axis, the secondary induced emf is again maximum, but of reversed polarity. Hence, the output voltage is the difference of the primary voltage and the secondary induced emf. If V1 is the primary voltage and E2 be the secondary induced emf then the output voltage can be varied between V1 + E2 and V1 – E2 by moving the rotor through 180°. Input S P

C

P a S C

S

Output (a)

(b)

Gjh/!5/48! Tjohmf.qibtf!joevdujpo!sfhvmbups;!)b*!Dpotusvdujpo!)c*!Dpoofdujpo!ejbhsbn

The rating of a single-phase induction regulator is equal to the product of full-load output current and the maximum voltage variation from normal. The advantages of a voltage regulator over a tap-changer transformer is its smooth voltage control without interruption in the supply circuit. The disadvantages of induction regulators are its higher initial cost and higher magnetizing current due to the presence of air gap.

5/25/3! Uisff.qibtf!Joevdujpo!Sfhvmbups! The three-phase induction regulator is very similar to a three-phase slip-ring induction motor. As in a three-phase motor, the primary currents produce a constant amplitude rotating magnetic field

Uisff.qibtf!Usbotgpsnfst

5/58

when the primary windings are connected to a three-phase supply. This rotating magnetic field induces emfs in the secondary winding whose magnitudes depend on the primary to secondary turns and are independent of rotor position. The phase of the secondary induced emf changes relative to that of the primary, as the rotor is moved. Hence, a variation in the phase of the output voltage is obtained by altering the angular position of the rotor causing a phase shift of the secondary emf. If V1 and E2 be the input voltage and secondary induced emf respectively then the maximum output voltage V1 + E2 is obtained when E2 is in phase with V1 and minimum voltage V1 – E2 is obtained with E2 in phase opposition to V1. Figure 4.38 (a) shows the connection diagram of a three-phase induction motor. Figure 4.38 (b) shows the locus of output voltage V2 which is a circle with centre at extremity of V1 and of radius E2. When q = 0, the output voltage V2 = V1 + E2 and when q = 180°, V2 = V1 – E2. Figure 4.38(c) shows the phasor diagram for all the three phases. Primary

Secondary

R

E2

V1

R

V2

B Y

Y

B Input line voltage

E2 1 q

V1

V2

Output line voltage

(c)

Gjh/!5/49! Uisff.qibtf!joevdujpo!sfhvmbups;!)b*!Dpoofdujpo!ejbhsbn!)c*!Mpdvt!pg!pvuqvu!)d*!Qibtps! ejbhsbn

UISFF.QIBTF!BVUP.USBOTGPSNFST!

5/26

The working principle of a three-phase autotransformer is same as that of a single-phase autotransformer. Three auto-transformers are used for small ratios of transformations. Normally, star connections are used and delta connections are avoided in such transformers. These transformers

!

5/59

Fmfdusjdbm!Nbdijoft

are mainly used for interconnecting two power systems of different voltages, for example 220 kV to 400 kV, 132 kV to 220 kV, 66 kV to 132 kV system, etc. A three phase star-connected autotransformer is shown in Fig. 4.39.

Gjh/!5/4:! Uisff.qibtf!tubs.dpoofdufe!bvupusbotgpsnfs

UISFF.XJOEJOH!USBOTGPSNFST!

5/27

Three-winding transformers contain an additional winding, other than primary and secondary windings, called tertiary winding. Unlike two-winding transformers, the kVA rating of the three windings are different. The voltage rating of the primary is the highest and that of the tertiary is lowest. The voltage rating of the secondary winding lies in between primary and tertiary. The advanges of using tertiary winding are as follows: 1. The main advantage of using tertiary winding, which is always connected in delta, is that its delta connection suppresses any harmonic that may be generated in star-connected primary or secondary. 2. When the secondary load currents are unbalanced, the primary currents also become unbalanced and the circulating current is reduced in the tertiary winding. This tends to restore both primary and secondary phase voltages to their normal phase magnitudes and angles. 3. The tertiary winding can be used to feed a load having a voltage other than that of primary and secondary. For example, they are used to supply substation auxiliaries (lights, fans, pumps, etc.) at a voltage different from those of primary and secondary. 4. Synchronous capacitors or static high-voltage capacitors are connected across delta-connected output of the tertiary windings for reactive power injection into the system for power factor correction or voltage regulation. 5. The lower voltage rating of the tertiary winding makes it suitable for use in high-voltage transformer testing.

Uisff.qibtf!Usbotgpsnfst

5/5:

5/27/2! Frvjwbmfou!Djsdvju!pg!b!Uisff.xjoejoh!Usbotgpsnfs As in a two-winding transformer, a three-winding transformer can also be represented by equivalent resistance and reactance of each individual winding. Figure 4.40 shows the equivalent circuit of a three-winding transformer referred to the primary. R¢2 I¢2 R1

1

2

X1

I1

Io I¢3

V1

X¢2

Ro

3

R¢3

Xo

V¢2

X¢3 V¢3

Gjh/!5/51! Frvjwbmfou!djsdvju!pg!b!uisff.xjoejoh!usbotgpsnfs!sfgfssfe!up!qsjnbsz

R1 and X1 are the resistance and leakage reactance of the primary winding. R¢2 and X 2¢ are the resistance and leakage reactance of the secondary winding referred to the primary and R2¢ and X3¢ are the resistance leakage reactance of the tertiary winding referred to the primary. I1 is the primary winding current. I 2¢ and I 3¢ are the secondary and tertiary winding currents referred to the primary.

5/27/3! Efufsnjobujpo!pg!Uisff.Xjoejoh ! Usbotgpsnfs!Qbsbnfufst! Similar to the two-winding transformers, the parameters of the three-winding transformer can be determined from open-circuit and short-circuit tests. 2/!Tipsu.djsdvju!Uftu! The equivalent leakage impedances referred to a common base can be determined from shortcircuit test. This test is performed in three stages. At first, the instruments (ammeter, voltmeter and wattmeter) are connected in winding 1, and winding 3 is kept open circuited. A low voltage is applied in winding 1 so that rated current flows in winding 2 which is short circuited. The voltage, current and power rating of winding 1 are voted. The connection diagram is shown in Fig. 4.41.

!

5/61

Fmfdusjdbm!Nbdijoft

W 1

A

3

2

V

Gjh/!5/52! Dpoofdujpo!ejbhsbn!pg!tipsu.djsdvju!uftu

The equivalent circuit is shown in Fig. 4.42. 2

X2 R2 I1

R1

X1 R3 X3 3

Gjh/!5/53! Frvjwbmfou!djsdvju!ejbhsbn!pg!tipsu.djsdvju!uftu

If Z12 represents the short-circuit impedance of windings 1 and 2 with winding 3 open then Z12 = Z1 + Z2 …

(4.18)

Now if V1, I1, and W1 be the voltmeter, ammeter and wattmeter readings then Z12 = Equivalent resistance R12 =

W1 I12

V1 I1

2 2 and equivalent leakage reactance X12 = Z12 - R12

In the second stage, winding 3 is short circuited and winding 2 is open circuited. A low voltage is applied in winding 1 to circulate rated current in winding 3. If Z13 represents the short-circuit impedance of windings 1 and 3 with winding 2 open then Z13 = Z1 + Z3 Z13 can be found from the ratio of voltmeter and ammeter reading.

(4.19)

Uisff.qibtf!Usbotgpsnfst

5/62

In the third stage, all the instruments are connected in winding 2 keeping winding 1 open. Winding 3 is short circuited and a low voltage is applied to winding 2 to circulate rated current in winding 3. Z23 can be determined from this test where Z23 = Z2 + Z3

(4.20)

Solving equations (4.18), (4.19) and (4.20), 1 (Z12 + Z13 – Z23) 2 1 (Z23 + Z12 – Z13) Z2 = 2 1 Z3 = (Z13 + Z23 – Z12) 2 where Z1 = R1 + jX1, Z2 = R2 + jX2 and Z3 = R3 + jX3 Z1 =

(4.21) (4.22) (4.23)

It should be remembered that Z12 and Z13 are referred to winding 1 as the instruments are connected in winding 1. Z23 is referred to winding 2 as the instruments are connected in winding 2. Hence, Z23 must be referred to winding 1 to perform the calculations on a common base. 3/!Pqfo.djsdvju!Uftu! This test is performed on any one of the three windings to determine core losses, turns ratio and magnetizing impedance. Winding 1 may be excited with both windings 2 and 3 open-circuited. \

N12 =

V1 V2

N13 =

N V1 V V /V and N23 = 2 = 2 1 = 13 V3 V3 V3 /V3 N12

5/27/4! Wpmubhf!Sfhvmbujpo!pg!b!Uisff.xjoejoh!Usbotgpsnfs The different steps for calculating voltage regulation are as follows: 1. Determine kVA in each winding for the given load. The ratio of the magnitude of the actual kVA loading of the winding to the base kVA used are as follows: K1 =

Primary kVA loading base kVA

K2 =

Secondary kVA loading Tertiary kVA loading , K3 = base kVA base kVA

Similarly,

!

5/63

Fmfdusjdbm!Nbdijoft

2. Calculate voltage regulation for each winding at its operating p.f. If cos f1, cos f2, cos f3 are the operating power factors of three windings and Er1, Ex1, Er2, Ex2, Er3, Ex3 are the p.u. resistance and reactance drop of three windings respectively then p.u. voltage regulation for primary winding alone. E1 = K1 (Er1 cos f1 + Ex1 sin f1) The p.u. voltage regulation for secondary and tertiary windings are E2 = K2 (Er2 cos f2 + Ex2 sin f2) and E3 = K3 (Er3 cos f3 + Ex3 sin f3) respectively. 3. Now the voltage regulation for any pair of windings can be obtained by algebraic sum of individual voltage regulations. For example, if a three-winding transformer is energized from ac source, through primary and both secondary and tertiary are connected with load then voltage regulations are • From primary to secondary E12 = E1 + E2 • From primary to tertiary E13 = E1 + E3

5/27/5! Bqqmjdbujpot!pg!Uisff.xjoejoh!Usbotgpsnfst! 1. Supplying an auxiliary load at a voltage different from that of secondary 2. To feed a distribution system from two transmission circuits of different voltages 3. The secondary load can be subdivided into two secondary windings to reduce short-circuit current. 4. It can be used for load ratio control circuit. 5. A generating station feeding two outgoing transmission systems of different voltages.

BVEJP!GSFRVFODZ!USBOTGPSNFS!

5/28

These transformers are generally small transformers used in electronic or communication circuits for processing, control, measurement or testing with amplifiers, oscillators, etc. They are used in the audio frequency range of 20 to 20,000 Hz. These transformers step up or step down the voltage in high-frequency circuits and also act as an impedance transforming device for impedance matching or sometimes they may serve some other auxiliary function such as blocking dc to pass only ac to the next circuit. Certain larger transformers of medium power are needed for transmitters, induction and dielectric furnaces. In radio and television sets, these high-frequency transformers are used between different amplification stages. These transformers are called intermediate frequency transformers or IFT. In audio frequency transformers, the iron losses have to be kept under permissible limit by using special high-resistivity, low-hysteresis and high-permeability materials like soft ferrites or highpermeability silicon steel.

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Rectifier transformers are used with power rectifiers when a given ac supply is required to be rectified and used as rectified dc voltage in electric arc furnace, dc welding, battery charging, electrolysis, electroplating and other applications. There are centre- or neutral tapped and bridgetransformer rectifier circuits. Though a rectifier transformer has same general principles of power transformer, the secondary of rectifier transformers feed the rectifier in current pulses. This causes increased losses as well as leakage fluxes. Hence, compared to the power transformers, they have lower efficiency and higher percentage impedance.

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When neutral of the power transformer is not available for grounding, these transformers are used. For example, in delta-delta connection, there is no neutral point. In such cases, a star-delta transformer is used which is a step-down transformer. The system is connected to the star-connected primary of the grounding transformer whose neutral is grounded. The delta-connected secondary does not supply any load but provides a closed path for the triplen harmonics. Under normal conditions, the currents in a grounding transformer are only the exciting currents. But if a single line-to-ground fault occurs, large currents may flow through it. Hence, a grounding transformer should be of sufficient rating to withstand the effects of line-to-ground faults. Transformers with zigzag connections are also used as grounding transformers.

USBOTGPSNFS!OBNFQMBUF! The broad specifications of transformers are given below: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

kVA rating Rated voltage Number of phases (that is, 1f or 3f) Rated frequency Type of connections for 3f Tappings if they exist Type of core Power or distribution type Ambient temperature (it is generally 40°C) Nature of cooling Rise of temperature above ambient temperature

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12. Voltage regulation 13. No-load current 14. Efficiency The standard ratings of distribution transformers are 16, 25, 40, 50, 63, 80 and 100 kVA whereas standard power ratings are 25, 40, 63, 100, 125, 160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250, 1600, 2000, 2500, 3125, 4000, 6300, 8000, 10000, 12500 160000, 20000, 25000, 31500, 40000, 50000, 63000 and 80000 kVA.

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Solution For Transformer 1, È 1.5 + j 5 ˘ kVA rating = [1000 –– cos–1 0.8] Í ˙ Î 2.5 + j 8 ˚ È 5.22 –73.3∞ ˘ = (1000 ––36.87°) Í ˙ Î 8.38 –72.65∞ ˚ = 1000 ––36.87° ¥ 0.62 –0.65° = 620 ––36.22° kVA kW rating = 620 ¥ cos 36.22° kW = 500.187 kW Power factor = cos 36.22° = 0.8 lagging 620 = 32.54 A Current = 3 ¥ 11 For Transformer 2, È 1+ j3 ˘ kVA rating = [1000 –– cos–1 0.8] Í ˙ Î 2.5 + j 8 ˚ È 3.16 –71.56∞ ˘ = 1000 –– 36.87° Í ˙ Î 8.38 –72.63∞ ˚ = 1000 –– 36.87° ¥ 0.377 –– 1.09° = 377 –– 37.96° kVA kW rating = 377 (cos 37.96°) kW = 297.24 kW

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Power factor = cos 37.96° lagging = 0.788 lagging Current =

377 3 ¥ 11

= 19.79 A.

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Solution Here, the given per unit values refer to different ratings. They should be converted to the same base kVA, say 500 kVA. ZA = 0.02 + j 0.08 = 0.08 –75.96° p.u. 500 ZB = (0.015 + j 0.04) 750 = 0.01 + j 0.026 = 0.0278 –68.96° ZA + ZB = 0.02 + j 0.08 + 0.01 + j 0.026 = 0.03 + j 0.106 = 0.11 –74.197° p.u. Total load kVA = SL = 1000 –– cos–1 0.8° = 1000 ––36.9° SA = =

ZB SL ZA + ZB 0.0278 –68.96∞ ¥ 1000 –– 36.9° 0.11 –74.197∞

= 0.253 –– 5.237° ¥ 1000 –– 36.9° = 253 –– 42.137° [253 kVA at p.f. 0.7 lagging] SB =

ZA ¥ SL ZA + Z B

0.08 –75.96∞ ¥ 1000 ––36.9° 0.11 –74.197∞ = 0.727 –1.763° ¥ 1000 –– 36.9° = 727 –– 35.137° [727 kVA at p.f. 0.8 lagging].

=

!

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Solution Considering 1000 kVA as the base kVA, 1000 = 0.0375 800 1000 = 0.05 ¥ = 0.0625 800

R1p.u. = 0.03 ¥ X1p.u.

The load kVA is

1500 - cos -1 0.85 or 1765 - 31.8∞ 0.85

\ power output of the first transformer is S1 =

0.02 + j 0.07 ¥ 1765 - 31.8∞ 0.0375 + 0.02 + j (0.0625 + 0.07)

or,

S1 =

0.02 + j 0.07 ¥ 1765 - 31.8∞ 0.0575 + j 0.1325

or,

S1 = 892.3 - 24.3∞ kVA

Power factor is cos 24.3° or 0.911 lag \ power output of the second transformer is S2 = 1765 - 31.8∞ – 892.3 - 24.3∞ = 686.75 – j 563 = 888 - 39.34∞ kVA Power factor is 0.77 lag.

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Solution If E2 be the no-load voltage on the secondary side and V2 is the voltage under loaded condition, E2 – V2 = I2 Re2 cos q2 + I2 Xe2 sin q2 I R I X E2 = 1 + 2 e 2 cos q2 + 2 e 2 sin q2 V2 V2 V2 E2 = 1 + 0.02 ¥ 0.85 + 0.04 ¥ 0.5268 V2 E2 = 1.0381 V2

or, or, or,

Line terminal voltage on star side is 440 V. 440

\ phase voltage

V2 =

\

E2 = 1.0381 ¥

\ turns ratio is

3 440 3

= 263.7 V

E1 3300 or or 13. E2 263.7

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Solution To find the required tap setting, we need to know the change in the terminal voltage when it is loaded with respect to that under no-load condition. Voltage regulation = Secondary current

I 2 Re 2 cos q 2 + I 2 X e 2 sin q 2 V2 2 ¥ 103 I2 = = 43.74 A 3 ¥ 33 ¥ 103 ¥ 0.8 2

Re2

ÊN ˆ = R¢1 + R2 + RL = R1 Á 2 ˜ + R2 + RL ËN ¯ 1

Now

N2 33 = = 3 N1 3 ¥ 11

!

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( 3 ) + 2 + 15 = 17.9 W = 5 ¥ ( 3 ) + 15 + 5 = 35 W

\

Re2 = 0.3 ¥

and

Xe2

\

2

voltage regulation =

or, or,

2

43.74 33 ¥ 103

3

{17.9 ¥ 0.8 + 35 ¥ 0.6}

= 0.081 = 8.1% Hence, tap setting should be raised by 8.1%.

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Solution 100 ¥ 103 = 3 ¥ 340 ¥ IL ¥ 0.8 \

IL =

100 ¥ 103 3 ¥ 340 ¥ 0.8

= 212.26 A

Input volt-amperes per phase = Output volt amperes per phase 400

\

3

\

IH =

340 IL 3

IH =

340 ¥ 212.26 = 180.4 A 400

[IH = winding current]

Currents flowing from neutral to tapping points = (212.26 – 180.4) A = 31.86 A.

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Solution The secondary line current I2L =

800 ¥ 103 3 ¥ 3300

= 139.96 A

Since the secondary is star connected, the phase current I2p.u. = 139.96 A The secondary current transferred to the primary side I 2¢ = 139.96 ¥

3300 = 41.988 A 11000

Let the tertiary winding be operating at a power factor of cos q3 lag. \ the tertiary phase current I3 =

200 ¥ 103 151.51 A A= cos q 3 ¥ 3 ¥ 440 cos q 3

The tertiary phase current referred to the primary I 3¢ =

151.51 440 10.5 ¥ = A cos q 3 11000 3 cos q 3

Now primary current I1 = I 2 + I 3 + Im where Im is the magnetizing current The active component of primary current 10.5 ¥ cos q3 cos q 3 = 33.59 + 10.5 = 44.1 A

I1x = I 2¢ x + I3x ¢ = 41.988 ¥ 0.8 +

The reactive component of primary current I1y = I 2y ¢ + I 3y ¢ + 5 = 41.988 ¥ 0.6 + I 3y ¢ +5 = 30.2 + I 3y ¢ The primary power factor cos q = 0.8 lag \ \ \

tan q = 0.75 30.2 + I 3¢ y 0.75 = 44.1 I3y ¢ = 2.875

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!

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\ tertiary phase current

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I3 = (10.5) 2 + ( 2.875) 2 ¥

11000 3 ¥ 440

= 157.14 A I1y = 30.2 + 2.875 = 33.075 \ primary phase current

I1 = ( 44.1) 2 + (33.075) 2 = 55.125 A.

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Solution 600 = 1.5 W 400 800 =2W Z31 = X31 = 400 300 Z23 ¢ = X23 ¢ = = 0.2 W 1500 6.6 = 0.4 Z23 = X23 = 0.2 ¥ 3.3 1.5 + 2.5 + 0.4 Z123 = X123 = = 2.2 W 2 Z12 = X12 =

\ \

X1 = 2.2 – 0.4 = 1.8 W X2 = 2.2 – 2 = 0.2 W X3 = 2.2 – 1.5 = 0.7 W Base impedance for primary =

(Base kV) 2 (6.6)2 = = 7.26 W 6 Base MVA

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1.8 p.u. = 0.248 p.u. 7.26 0.2 p.u. = 0.0275 p.u. X2 = 7.26 0.7 p.u. = 0.0964 p.u. X3 = 7.26

\

X1 =

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Solution The active power supplied by the primary is equal to the active power supplied by the secondary and tertiary. \

3 ¥ 11 ¥ I1 cos q1 = 1000 ¥ 0.75 + 500 ¥ 0.85

where I1 and cos q1 are the primary current and primary power factor respectively. \

3 ¥ 11 ¥ I1 cos q1 = 1175

(1)

Similarly, equating for the reactive power (considering magnetizing current) 3 ¥ 11 ¥ I1 sin q1 = 3 ¥ 11 ¥ 8 + 1000 ¥ sin (cos–1 0.75) + 500 sin (cos–1 0.85) or,

3 ¥ 11 ¥ I1 sin q1 = 1077.25

(2)

Solving Eq. (1). and Eq. (2) tan q1 = \

1077.25 = 0.9168 1175

cos q1 = 0.737 1175 I1 = = 83.68 A 3 ¥ 11 ¥ 0.737

\

\ primary current is 83.68 A at 0.737 lag.

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!

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Solution Active component of primary current is 8 ¥ 0.3 or 2.4 A. Active component of secondary current referred to primary is 150 ¥ 0.75 ¥

5 or 56.25 A. 10

3 or 20.4 A. 10 \ total active component of current in the primary is 2.4 + 56.25 + 20.4 or 79.05 A. Similarly, considering the reactive component of all currents referred to the primary the total reactive component of current in the primary is Active component of tertiary current referred to primary is 80 ¥ 0.85 ¥

8 ¥ sin (cos–1 0.3) + 150 ¥ sin (cos–1 0.75) ¥

5 3 + 80 ¥ sin (cos–1 0.85) ¥ or, 69.88 A 10 10

\ primary current I1 = (79.05) 2 + (69.88) 2 = 105.51 A 69.88 ˆ Ê or, 0.749 lag. Power factor of primary current is cos Á tan-1 Ë 79.05 ˜¯

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Solution Power taken by the load is 3 ¥ 440 ¥ 200 ¥ 0.9 W or 137.18 kW The rating of the transformer 3 ¥ 440 ¥ 200 VA or 152.42 kVA Phase current = Line current in the HV side =

152.42 3 ¥ 11

=8A

Line current in the LV side is 200 A 200 \ Phase current in the LV side is A or 115.47 A. 3

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Solution Transformer rating =

50 kVA or 73.53 kVA 0.85 ¥ 0.8

Line current on the low-voltage side is

73.53 ¥ 103 3 ¥ 440

A or 96.48 A

As the secondary of the transformer is star connected, phase current on the low-voltage side is also 96.48 A. 73.53 Line current on the high-voltage side is A or 3.86 A. 3 ¥ 11 Since the HV side is delta connected, the phase current on the HV side is 3 ¥ 3.86 A or 6.68 A.

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Solution When efficiency is maximum, the output is 40 ¥ 0.9 kW or 36 kW \

Loss =

1 - efficiency 1 - 0.97 ¥ output = ¥ 36 efficiency 0.97

= 1.113 kW Under maximum efficiency condition, Core loss = Copper loss 1113 \ Core loss = Copper loss = W = 556.5 W 2 40 or 0.8 fraction of full-load. Maximum efficiency occurs when output is 40 kVA or 50 2

Ê 1 ˆ ¥ 556.5 W or 869.53 W. \ full-load copper loss is Á Ë 0.8 ˜¯ The efficiency for rated output is

50 ¥ 0.8 ¥ 103 50 ¥ 0.8 ¥ 103 + 869.53 + 556.5

¥ 100% or 96.56%.

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!

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Solution (a) The power rating of three-phase bank is 3 ¥ 100 or 300 MVA. The voltage rating of the primary of three-phase bank is 3 ¥ 13.8 or 23.9 kV. \ the required voltage rating is 23.9/115 kV (b) The equivalent impedance referred to the high-voltage side is 2

Ê 66.4 ˆ (0.0045 + j 0.19) ¥ Á W Ë 13.8 ˜¯ or, 0.104 + j 44 W/

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Solution The kVA rating of the three-phase bank is 4500. 4500 \ the rating of each single-phase transformer = kVA = 1500 kVA 3 (a) For YD connection 13.8 The voltage on the high-voltage side is kV or 7.97 kV 3 The voltage on the low-voltage side is 2300 V or 2.3 kV 1500 The current in each transformer in the high-voltage side is A or 188.2 A. 7.97 1500 A or 0.652 A The current in each transformer in the low-voltage side is 2300 Hence, rating of each transformer is 7.97/2.3 kV, 188.2/0.652 A, 1500 kVA (b) For DY connection The rating of each transformer is 13.8

or,

2.3 1500 1500 kV, A, 1500 kVA 13.8 2.3 3 3

13.8/1.32 kV, 108.7/1129.6 A, 1500 kVA

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(c) For YY connection The rating of each transformer is 13.8 3

2.3 1500 1500 kV, A, 1500 kVA 13.8 2.3 3 3 3 7.97/1.32 kV, 188.2/1129.6 A, 1500 kVA

or,

(d) For DD connection The rating of each transformer is 1500 1500 A, 1500 kVA 13.8 2.3 13.8/2.3 kV, 108.7/0.652 A, 1500 kVA. 13.8/2.3 kV,

or,

! Qspcmfn!5/47 B! uisff.qibtf! ZD! usbotgpsnfs! pg! sbujoh! 336035! lW-! 511! NWB! ibt! b! tfsjft! sfbdubodf! pg! 22/8! W! sfgfssfe! up! jut! ijhi.wpmubhf! tjef/! Uif! usbotgpsnfs! jt! tvqqmzjoh! b! mpbe! pg! 436! NWB! xjui! 1/:4! q/g/! mbhhjoh!bu!b!wpmubhf!pg!35!lW!)mjof!up!mjof*!po!jut!mpx.wpmubhf!tjef/!Ju!jt!tvqqmjfe!gspn!b!gffefs!xiptf! jnqfebodf!jt!1/22!,!k!3/3!W!dpoofdufe!up!jut!ijhi.wpmubhf!tjef/!Efufsnjof!uif!mjof.up.mjof!wpmubhf!bu! uif!ijhi.wpmubhf!tjef!pg!uif!usbotgpsnfs!boe!mjof.up.mjof!wpmubhf!bu!uif!tfoejoh!foe!pg!uif!gffefs/!

Solution The load current per phase =

325 ¥ 103 - cos -1 0.93 = 4514 21.56∞ A 3 ¥ 24

\ when referred to the high-voltage side, the load current IH = 4514 ¥

24 - 21.56∞ A = 833.97 - 21.56∞ A 225 3

\ phase voltage at the high-voltage side is 225 ¥ 103

+ 833.97 - 21.56∞ (j 11.7) 3 = 103 ¥ 130 + 9757.5 68.44

VHph =

= (133585 + j 9075) V = 133.89 kV \ the line voltage at the high-voltage side VHL = 133.89 ¥ 3 kV = 231.9 kV

!

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The phase voltage at the sending end of the feeder 225 × 103

+ 833.97 - 21.56∞ (0.11 + j 2.2 + j 11.7) 3 = 130 ¥ 103 + 833.97 - 21.56∞ ¥ 13.9 89.54∞

VFph =

= 130 ¥ 103 + 11592.18 67.98∞ = 134346.26 + j 10749.35 V = 134.77 4.57∞ kV \ the line voltage at the sending end of the feeder VFL = 3 ¥ 134.77 kV = 233.4 kV.

! Qspcmfn!5/48 B!DZ!dpoofdufe!cbol!pg!uisff!jefoujdbm!211!lWB-!35110231!W-!usbotgpsnfst!jt!tvqqmjfe!xjui!qpxfs! uispvhi!b!gffefs!xiptf!jnqfebodf!jt!1/176!,!k!1/98!W!qfs!qibtf/!Uif!tfoejoh.foe!wpmubhf!pg!uif! gffefs!jt!ifme!dpotubou!bu!3511!W!mjof!up!mjof/!Uif!sftvmut!pg!b!tjohmf.qibtf!tipsu.djsdvju!uftu!po!pof! pg!uif!usbotgpsnfst!xjui!jut!mpx.wpmubhf!ufsnjobmt!tipsu!djsdvjufe!bsf! WI!>!64/5!W-!JI!>!52/8!B-!Q!>!943!X Dbmdvmbuf!uif!mjof.up.mjof!wpmubhf!po!uif!mpx.wpmubhf!tjef!pg!uif!usbotgpsnfs!xifo!uif!cbol!efmjwfst! sbufe!dvssfou!up!b!cbmbodfe!uisff.qibtf!vojuz!qpxfs.gbdups!mpbe/!

Solution From the short-circuit test data, the equivalent impedance of the transformer referred to the highvoltage side 53.4 ZeH = W = 1.28 W 41.7 832 The equivalent resistance referred to the high-voltage side reH = W = 0.478 W ( 41.7) 2 \ equivalent reactance referred to the high-voltage side xeH = (1.28) 2 - (0.478) 2 = 1.187 W The rated current in the high-voltage side under unity power factor 100 ¥ 103 A = 41.67 A 2400 The voltage drop in the feeder and transformer winding in the HV side is IH =

41.67 {(0.478 + 0.065) + j (1.187 + 0.87)} = 41.67 {0.543 + j 2.057} = 88.65 75.21∞ V

Uisff.qibtf!Usbotgpsnfst

When the transformer is to the low-voltage side, the voltage drop is 88.65 ¥

5/78

120 75.21∞ 2400

4.4325 75.21∞ V

or,

\ the terminal phase voltage on the low voltage = 120 – 4.4325 75.21∞ = 118.95 V The line voltage on the low-voltage side = 118.95 ¥ 3 V = 206 V.

! Qspcmfn!5/49 Uif!sfbdubodf!pg!b!8/:8!lW0571!W-!86!lWB-!tjohmf.qibtf!usbotgpsnfs!ibt!tfsjft!sfbdubodf!pg!1/23! q/v/! ! )b*! Jg!uiftf!usbotgpsnfst!bsf!dpoofdufe!jo!ZZ!uisff.qibtf!dpoofdujpo-!efufsnjof!uif!uisff.qibtf! wpmubhf!boe!qpxfs!sbujoh-!qfs!voju!jnqfebodf!pg!uif!usbotgpsnfs!cbol-!tfsjft!sfbdubodf!sf. gfssfe!up!uif!ijhi.wpmubhf!boe!mpx.wpmubhf!ufsnjobmt/! ! )c*! Sfqfbu!qbsu!)b*!jg!uif!usbotgpsnfst!bsf!dpoofdufe!jo!Z!po!uif!ijhi.wpmubhf!tjef!boe!D!jo!uif! mpx.wpmubhf!tjef/!

Solution (a) For star-star connection, the voltage and power rating is 7.97 ¥ 3 kV/460 ¥ 3 V, 75 ¥ 3 kVA or,

13.8 kV/797 V, 225 kVA

Per unit impedance is same as the per unit reactance of the single-phase transformer, i.e. (13.8) 2 0.12 p.u. Base impedance of the high-voltage side is or 846.4 W 225 ¥ 10 -3 \ series reactance referred to HV side is 0.12 ¥ 846.4 or 102 W Base impedance of the low-voltage side is (797 ¥ 10 -3 ) 2

W or 2.823 W 225 ¥ 10 -3 \ series reactance referred to the low-voltage side 2.823 ¥ 0.12 W or 0.339 W (b) For YD connection, the voltage and power rating is

or,

7.97 ¥ 3 kV/460 V, 225 kVA 13.8 kV/460 V, 225 kVA

Series reactance referred to the high-voltage side is 102 W.

!

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Fmfdusjdbm!Nbdijoft

Series reactance referred to the low-voltage side is ( 460 ¥ 10 -3 ) 2 225 ¥ 10 -3

¥ 0.12 W or, 0.113 W/

! Qspcmfn!5/4: Jo! Qspcmfn! 5/49-! jg! uif! sbufe! wpmubhf! jt! bqqmjfe! up! uif! ijhi.wpmubhf! ufsnjobmt! boe! uif! uisff! mpx. wpmubhf!ufsnjobmt!bsf!tipsu!djsdvjufe-!dbmdvmbuf!uif!nbhojuvef!pg!uif!qibtf!dvssfou!jo!qfs!voju!boe! jo!bnqfsft!po!ijhi.wpmubhf!boe!mpx.wpmubhf!tjeft/!

Solution As the per unit reactance is 0.12 p.u., the per unit current in each case when the low-voltage side is short circuited is 1 or 8.33 0.12 For star-star connection Base current in the high-voltage side is 225 IH base = A 3 ¥ 13.8 \ phase current on high-voltage side is 225 3 ¥ 13.8

¥ 8.33 A or 78.4 A

Base current in the low-voltage side is IL base =

225 ¥ 103 3 ¥ 797

A

\ phase current on low-voltage side is 225 ¥ 103 3 ¥ 797

¥ 8.33 A or 1358 A

For star-delta connection Phase current on the high-voltage side is 78.4 A. Base current on the low-voltage side is IL base =

225 ¥ 103 3 ¥ 460

A

\ phase current on low-voltage side is 225 ¥ 103 3 ¥ 460

¥ 8.33 A or 2352 A.

Uisff.qibtf!Usbotgpsnfst

5/7:

! Qspcmfn!5/51 B!uisff.qibtf!usbotgpsnfs!sbufe!370456!lW-!961!NWB!ibt!b!tfsjft!jnqfebodf!pg!1/1146!,!k!1/198! q/v/!po!jut!cbtf/!Ju!jt!dpoofdufe!up!b!37!lW-!911!NWB!hfofsbups!xijdi!dbo!cf!sfqsftfoufe!bt!b!wpmu. bhf!tpvsdf!jo!tfsjft!xjui!b!sfbdubodf!pg!k!2/68!q/v/!po!uif!hfofsbups!cbtf/! ! )b*! Dpowfsu!uif!qfs!voju!hfofsbups!sfbdubodf!up!uif!tufq.vq!usbotgpsnfs!cbtf/ ! )c*! Uif! voju! jt! tvqqmzjoh! 811! NX! bu! 456! lW! boe! 1/:6! q/g/! mbhhjoh! up! uif! tztufn! bu! uif! usbot. gpsnfs!ijhi.wpmubhf!ufsnjobmt/!)j*!Efufsnjof!uif!usbotgpsnfs!wpmubhf!po!uif!mpx.wpmubhf!tjef! boe!uif!hfofsbups!joufsobm!wpmubhf!cfijoe!jut!sfbdubodf!jo!lW/!)jj*!Gjoe!uif!hfofsbups!pvuqvu! qpxfs!jo!NX!boe!uif!qpxfs!gbdups/!

Solution (a) The transformer base is 850 MVA and generator base is 800 MVA. \ the reactance of 1.57 p.u. on the generator base when transferred to the transformer base is 850 j 1.57 ¥ = j 1.67 p.u. 800 700 MVA or 736.84 MVA at transformer high-voltage terminals or the 0.95 736.84 unit is supplying a power of p.u. or 0.867 p.u. at 0.95 p.f. lagging 850

(b) The unit is supplying

\ per unit current

I=

0.867 - cos -1 0.95 1 0∞

= 0.867 -18.2∞ (i) The transformer voltage on the low-voltage side Vt = 1 0∞ + 0.867 -18.2∞ ¥ (0.0035 + j 0.087) = 1 0∞ + 0.0755 69.496∞ = 1.0744 + j 0.0707 = 1.0767 3.76∞ p.u. = 1.0767 ¥ 26 3.76∞ kV = 28 3.76∞ kV The generator internal voltage behind its reactance VG = 1 0∞ + 0.867 -18.2∞ (0.0035 + j 0.087 + j 1.57) = 1 0∞ + 0.867 -18.2∞ (0.0035 + j 1.657)

!

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= 1 0∞ + 1.4366 71.68∞ = 1.45 + j 1.364 = 2 43.25∞ p.u = 52 43.25∞ kV (ii) The generator terminal voltage = Transformer voltage on the low-voltage side = 1.0767 3.76∞ p.u. I = 0.867 -18.2∞ p.u. \ generator output power SG = Vt I* = 1.0767 3.76∞ ¥ 0.867 18.2∞ = 0.933 21.87∞ p.u. \ generator output power is or, or,

0.933 cos 21.87° p.u. 850 ¥ 0.933 cos 21.87° MW 736 MW

Power factor is cos 21.87° lagging or 0.928 lagging.

! Qspcmfn!5/52 B!uisff.qibtf!tubs.dpoofdufe!bvup.usbotgpsnfs!sfevdft!uisff.qibtf!wpmubhf!gspn!551!W!up!331!W! mjof!up!mjof/!Uif!mpbe!po!uif!mpx.wpmubhf!tjef!jt!23!lX!bu!1/96!q/g/!Ofhmfdujoh!nbhofuj{joh!dvssfou-! efufsnjof!uif!dvssfou!jo!uif!xjoejoh!tfdujpot!pg!uif!bvupusbotgpsnfs/!

Solution Since the load is balanced, the per phase power is 4 kW. 440 Per phase voltage on the high-voltage side = V = 254 V 3 220 Per phase voltage on the low-voltage side = V = 127 V 3 4000 Load current = A = 37 A 127 ¥ 0.85 220 ¥ 37 A = 18.5 A Input current = 440 The currents in the windings are 18.5, 37 A and (37 – 18.5) A or 18.5 A

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The currents in different sections are shown in Fig. 4.43. R

18.5 A 37 A

.5

18

A

18.5 A 18.5 A

18.5 A

37 A

3 Phase Load

37 A

B Y 18.5 A

Gjh/!5/54! Djsdvju!ejbhsbn!pg!Qspcmfn!5/52

! Qspcmfn!5/53 B!uisff.qibtf!tubs.dpoofdufe!bvup.usbotgpsnfs!tvqqmjft!411!W!gspn!b!551!W-!uisff.qibtf!tztufn/! Uif!mpbe!esbxt!91!lX!bu!1/9! q/g/!Efufsnjof!uif!dvssfout!jo!ejggfsfou!xjoejoht!pg!uif!bvupusbot. gpsnfs!boe!tipx!uifn!jo!uif!ejbhsbn/!

Solution The load current =

80 ¥ 103 3 ¥ 300 ¥ 0.8

A = 192.45 A

Live current drawn from source = 192.45 ¥

300 A 440

= 131.21 A The currents in the different sections of the windings are 192.45, 131.21 A and (192.45 – 131.21) A or 61.24 A. The currents are shown in Fig. 4.44. R

192.45 A

131.21 A 61.24 A

61.24 A 192.45 A

4

.2 61 A

B

192.45 A

131.21 A

Y 131.21 A

Gjh/!5/55! Djsdvju!ejbhsbn!pg!Qspc/!5/53

3 Phase Load

!

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Fmfdusjdbm!Nbdijoft

! Qspcmfn!5/54 Uif!sbujoht!pg!uif!qsjnbsz-!tfdpoebsz!boe!ufsujbsz!xjoejoht!pg!b!tjohmf.qibtf!uisff.xjoejoh!usbot. gpsnfs!bsf!bt!gpmmpxt; Qsjnbsz;!!!!!!!231!lWB-!4411!W! Tfdpoebsz;!!!!!91!lWB-!2211!W!!!!!!!!!!!! Ufsujbsz;!!!!!!!!!!91!lWB-!!551!W Uif!sftvmut!pg!uif!tipsu.djsdvju!uftut!bsf!bt!gpmmpxt; Qsjnbsz!fydjufe-!tfdpoebsz!tipsufe;!261!W-!41!B-!2111!X Qsjnbsz!fydjufe-!ufsujbsz!tipsufe;!251!W-!41!B-!2311!X! Tfdpoebsz!fydjufe-!ufsujbsz!tipsufe;!!61!W-!211!B-!2361!X Efufsnjof!uif!sftjtubodft!boe!mfblbhf!sfbdubodft!pg!uif!tubs!frvjwbmfou!djsdvju/!Bmtp!dbmdvmbuf!uifjs! wbmvft!gps!fbdi!xjoejoh/!

Solution The short-circuit impedances are 150 W=5W 30 140 = W = 4.67 W 30 50 = W = 0.5 W 100

Z12 = Z13 Z23

Z23 is referred to the secondary side. Hence, the value of Z23 when referred to the primary side 2

2 ÊN ˆ Ê 3300 ˆ Z¢23 = 0.5 ¥ Á 1 ˜ = 0.5 ¥ Á Ë 1100 ˜¯ ËN ¯ 2

= 0.5 ¥ 9 = 4.5 W \ the leakage impedances referred to the primary are 1 1 (Z12 + Z13 – Z 2¢ 3) = (5 + 4.67 – 4.5) W = 2.585 W 2 2 1 1 Z2 = (Z12 + Z¢23 – Z13) = (5 + 4.5 – 4.67) W = 2.415 W 2 2 1 1 Z3 = (Z13 + Z 23 ¢ – Z12) = (4.67 + 4.5 – 5) W = 2.085 W 2 2 The short-circuit resistances are Z1 =

R12 =

1000 (30) 2

= 1.11 W

Uisff.qibtf!Usbotgpsnfst

R13 = R23 =

1200 (30) 2

= 1.33 W

1250 (100) 2

= 0.1250 W

R23 which is referred to secondary side, when referred to the primary is 2

2 ÊN ˆ Ê 3300 ˆ R23 ¢ = R23 ¥ Á 1 ˜ = 0.1250 ¥ Á W = 1.125 W Ë 1100 ˜¯ ËN ¯ 2

\ the equivalent circuit resistances when referred to primary are R1 =

1 1 ¢ ) = (1.11 + 1.33 – 1.125) W = 0.6575 W (R12 + R13 – R 23 2 2

R2 =

1 1 (R12 + R¢23 – R13) = (1.11 + 1.125 – 1.33) W = 0.4525 W 2 2

R3 =

1 1 (R13 + R¢23 – R12) = (1.33 + 1.125 – 1.11) W = 0.6725 W 2 2

Hence, the leakage reactances referred to the primary are X1 = Z12 - R12 = ( 2.585) 2 - (0.6575) 2 = 2.5 W X2 = Z 22 - R22 = ( 2.415) 2 - (0.4525) 2 = 2.37 W X3 = Z32 - R32 = ( 2.085) 2 - (0.6725) 2 = 1.97 X The leakage impedance of primary winding Z1 = 0.6575 + j 2.5 W The leakage impedance of the secondary winding 2

Ê 1100 ˆ Z2 = (0.4525 + j 2.37) ¥ Á W = 0.05 + j 0.26 W Ë 3300 ˜¯ The leakage impedance of the tertiary winding 2

Ê 440 ˆ W = 0.0119 + j 0.035 W/ Z3 = (0.6725 + j 1.97) ¥ Á Ë 3300 ˜¯

5/84

!

5/85

Fmfdusjdbm!Nbdijoft

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Solution 1 (0.12 + 0.2 – 0.1) = 0.11 p.u. 2 1 X2 = (0.1 + 0.12 – 0.2) = 0.02 p.u. 2 1 X3 = (0.2 + 0.1 – 0.12) = 0.09 p.u. 2

X1 =

We assume that the phase angle of V3 with respect to the reference voltage V1 is 0°. It actually has a small angle which must otherwise be determined. The error caused is negligible. (a) Secondary current

I2 = 1 cos -1 0.8 = 1 - 36.86∞ A

Tertiary current

I3 = 1 0∞ A

\ primary current

I1 = 1 - 36.86∞ – 1 = 0.8 – j 0.6 + 1 = 1.8 – j 0.6 V1 = (1 + j 0.12) (1.8 – j 0.6) = 1.8 + j 0.216 – j 0.59 + 0.07 = 1.87 – j 0.374 = 1.9 –– 11.3° V1 = 1.9 ¥ 220 = 418 kV.

Uisff.qibtf!Usbotgpsnfst

I1

X1

I2

5/86

X2

X3 V1

V2

I3 V3

Gjh/!5/56! Djsdvju!ejbhsbn!pg!Qspc/!5/55

Referring Fig. 4.45, V1 = V2 + I2 X2 + I1X1 = 1 0∞ + 1 - 36.86∞ (j 0.02) + (1.8 – j 0.6) (j 0.11) = 1 0∞ + (0.8 – j 0.6) (j 0.02) + 0.066 + j 0.198 = 1+ 0.078 + j 0.214 = 1.078 + j 0.214 = 1.099 11.23∞ p.u \ primary line-to-line voltage is 1.099 ¥ 220 kV or 241.78 kV (b) From Fig. 4.45, the voltage at the tertiary terminals V3 = V2 + I2 X2 – I3 X3 = 1 0∞ + 1 - 36.86∞ ( j 0.02) – 1 ¥ j 0.09 = 1 + (0.8 – j 0.6) ¥ j 0.02 – j 0.09 = 1 + 0.012 – j 0.074 = 1.0147 - 4.18∞ p.u. \ line-to-line voltage at the tertiary terminals V3 = 1.0147 ¥ 11 kV = 11.16 kV (c) When the secondary load is removed, I2 = 0 and I1 = I3 \ or,

V1 = V3 + I3 Z3 + I1 Z1 = V3 + I3 (Z1 + Z3) V3 = V1 – I3 (Z1 + Z3) = 1.078 + j 0.214 – 1 (j 0.11 + j 0.09) = 1.078 + j 0.214 – j 0.2 = 1.078 + j 0.014 = 1.078 0.744∞ p.u.

\

V3 = 1.078 ¥ 11 kV = 11.858 kV

!

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Fmfdusjdbm!Nbdijoft

Hence, increase in the tertiary voltage when secondary load is removed is (11.858 – 11.16) kV or 0.698 kV.

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Solution Since the primary and secondary are connected in star, their line current is equal to their phase current 150 ¥ 103 \ R12 = = 1.25 W 3 ¥ ( 200) 2 R13 = R23 =

1.5 ¥ 103 3 ¥ (15) 2 2 ¥ 103 3 ¥ (50) 2

= 2.22 W = 0.267 W

As R23 is referred to secondary, when it is referred to primary, 2

2 Ê N1 ˆ Ê 33 ˆ R¢23 = R23 Á = 0.267 ¥ Á ˜ = 2.403 W Ë 11 ¯ Ë N ˜¯ 2

1 1 R1 = (R12 + R13 – R¢23) = (1.25 + 2.22 – 2.403) W = 0.5335 W 2 2 1 1 R2 = (R12 + R¢23 – R13) = (1.25 + 2.403 – 2.22) W = 0.7165 W 2 2 1 1 R3 = (R13 + R¢23 – R12) = (2.22 + 2.403 – 1.25) W = 1.6865 W 2 2

\

Now, Z12 =

300 3 W = 11.547 W 15 200 3 = W = 2.31 W 50

Z13 = Z23

Primary phase voltage 5000 3 = W = 14.43 W Primary phase current 200

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Z23 when referred to primary 2

Ê 33 ˆ Z¢23 = 2.31 ¥ Á ˜ = 20.79 W Ë 11 ¯ \

1 1 ¢ ) = (14.43 + 11.547 – 20.79) W = 2.59 W (Z12 + Z13 – Z 23 2 2 1 1 Z2 = (Z12 + Z 23 ¢ – Z13) = (14.43 + 20.79 – 11.547) W = 11.836 W 2 2 1 1 ¢ – Z12) = (11.547 + 20.79 – 14.43) W = 8.95 W Z3 = (Z13 + Z 23 2 2

\

X1 = ( 2.59) 2 - (0.5335) 2 = 2.53 W

Z1 =

X2 = (11.836) 2 - (0.7165) 2 = 11.81 W X3 = (8.95) 2 - (1.6865) 2 = 8.79 W Base impedance of the primary 2

Zb1 \

Z1 p.u. Z2 p.u. Z3 p.u.

Ê 33 ˆ ÁË ˜¯ 2 (Base kV) (33) 2 3 = = = = 72.6 W 15 Base MVA 15 3 2.59 = p.u. = 0.0357 p.u. 72.6 11.836 = p.u. = 0.163 p.u. 72.6 8.95 = p.u. = 0.123 p.u. 72.6

! Qspcmfn!5/57 B!uisff.qibtf-!uisff.xjoejoh!D0D0Z-!44!lW!02211!W!0551!W-!411!lWB!usbotgpsnfs!ibt!b!tfdpoebsz! mpbe!pg!311!lWB!bu!1/:!q/g/!mbhhjoh!boe!!b!ufsujbsz!mpbe!pg!211!lWB!bu!1/9!q/g/!mbhhjoh/!Uif!nbhofuj{. joh!dvssfou!jt!4&!pg!sbufe!mpbe!boe!uif!jspo!mptt!jt!2/6!lX/!Efufsnjof!uif!wbmvf!pg!qsjnbsz!dvssfou! xifo!uif!puifs!uxp!xjoejoht!bsf!efmjwfsjoh!uif!bcpwf!mpbet/!

Solution Secondary current per phase I2ph =

Secondary VA per phase Secondary voltage per phase

!

5/89

Fmfdusjdbm!Nbdijoft

300 ¥ 103 = 3 = 90.9 A 1100 Secondary current per phase referred to primary I 2¢ = 90.9 ¥

1100 - cos -1 0.9 = 3.03 - 25.84∞ 33000

= 2.727 – j 1.32 A Tertiary current per phase

I3ph

100 ¥ 103 Tertiary VA per phase = A = 3 Tertiary voltage per phase 440 3 = 131.216 A

Tertiary current per phase referred to primary 440 3 - cos -1 0.8 A I 3¢ = 131.216 ¥ 33000 = 1.01 (0.8 – j 0.6) A = 0.808 – j 0.606 A 330 ¥ 103 3 Primary rated current = = 3.03 A 33000 Magnetizing current = 4% of 3.03 A = 0.1212 A Core loss component of no-load current =

1500 / 3 = 0.015 A 33000

Hence, primary no-load current I0 = 0.015 – j 0.1212 A \ total primary current I1 = I0 + I 2¢ + I3¢ = 0.015 – j 0.1212 + 2.727 – j 1.32 + 0.808 – j 0.606 = 3.55 – j 2.0472 A = 4.098 - 30∞ A Hence, primary current is 4.098 A with a power factor of cos 30° lagging or 0.866 lagging.

Uisff.qibtf!Usbotgpsnfst

5/8:

! Qspcmfn!5/58 B!7/704/401/55!lW!tubs0tubs0efmub!usbotgpsnfs!ibt!qfs!voju!sftjtubodf!espqt!pg!1/119-!1/11:!boe! 1/12!boe!qfs!voju!sfbdubodf!espqt!1/16-!1/17!boe!1/176!gps!uif!qsjnbsz-!tfdpoebsz!boe!ufsujbsz! xjoejoht!sftqfdujwfmz!po!b!2111!lWB!cbtf/!Jg!uif!nbhofuj{joh!dvssfou!jt!8!B-!efufsnjof!uif!wpmubhf! sfhvmbujpo!gps!b!uisff.qibtf!cbmbodfe!mpbet!pg!2311!lWB!bu!1/:!mbhhjoh!q/g/!po!tfdpoebsz!boe!911! lWB!bu!1/9!q/g/!mbhhjoh!po!ufsujbsz/!

Solution Secondary kVA = 1200 (0.9 – j 0.436) Tertiary kVA = 800 (0.8 – j 0.6) Magnetizing kVA = – j 3 ¥ 6600 ¥ 7 ¥ 10–3 = – j 80.02 \ total load on the primary = 1200 (0.9 – j 0.436) + 800 (0.8 – j 0.6) – j 80.02 = 1720 – j 1083.22 = 2032.67 - 32.2∞ For primary

K1 =

Primary kVA loading 2032.67 = 2.033 = Base kVA 1000

Power factor of primary winding is cos 32.2° or 0.846 lagging. The per unit voltage regulation for the primary winding alone is E1 = K1 (Er1 cos f1 + Ex1 sin f1) = 2.033 (0.008 ¥ 0.846 + 0.05 ¥ 0.533) = 0.068 For secondary, 1200 = 1.2 1000 \ per unit voltage regulation for secondary winding alone is K2 =

E2 = K2 (Er2 cos f2 + Ex2 sin f2) = 1.2 (0.009 ¥ 0.9 + 0.06 ¥ 0.436) = 0.04 For tertiary,

K3 =

800 = 0.8 1000

\ per unit voltage regulation for tertiary winding alone is E3 = K3 (Er3 cos f3 + Ex3 sin f3) = 0.8 (0.01 ¥ 0.8 + 0.065 ¥ 0.6) = 0.0376.

!

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\ primary to secondary voltage regulation E12 = E1 + E2 = 0.068 + 0.04 = 0.108 p.u. or 10.8% Primary to tertiary voltage regulation E13 = E1 + E3 = 0.068 + 0.0376 = 0.1056 or 10.56% Secondary to tertiary voltage regulation E23 = 0.04 + 0.0376 = 0.0776 or 7.76%.

! Qspcmfn!5/59 B!22111044110551!W-!Z0Z0D-!uisff.xjoejoh!usbotgpsnfs!ibt!b!nbhofuj{joh!dvssfou!pg!6!B/!Bu!sbufe! wpmubhf-!uif!tfdpoebsz!tvqqmjft!b!cbmbodfe!mpbe!pg!411!lWB!bu!1/96!q/g/!mbh!boe!ufsujbsz!tvqqmjft!b! cbmbodfe!mpbe!pg!261!lX/!Uif!qsjnbsz!pqfsbuft!bu!b!qpxfs!gbdups!pg!1/:6!mbhhjoh/!Efufsnjof!uif! qsjnbsz!boe!ufsujbsz!qibtf!dvssfout!boe!bmtp!uif!qpxfs!gbdups!bu!xijdi!uif!ufsujbsz!pqfsbuft/!

Solution Magnetizing current IM = – j 5 A Secondary phase current I2 =

300 ¥ 103

= 52.486 A 3 ¥ 3300 Secondary phase current referred to primary I 2¢ = 52.486 ¥

3300 = 15.75 A 11000

Active component of tertiary current referred to primary = I 3¢ cos q3 =

150 ¥ 103 440 ¥ A = 7.87 A 11000 3 ¥ 440 3

Assuming lagging power factor for the tertiary load I 3¢ = 7.87 – j I3 sin q3 Total primary current I1 = IM + I 2¢ + I 3¢ = – j 0.5 + 15.75 - cos -1 0.85 + 7.87 – j I3 sin q3 or,

I1 = – j 0.5 + 13.387 – j 8.3 + 7.87 – j I3 sin q3 Now cos q1 = 0.95 lag where q1 is the primary power factor angle

\

tan q1 = 0.328

Uisff.qibtf!Usbotgpsnfst

\

0.328 = Hence,

5/92

0.5 + I 3 sin q 3 + 8.3 8.8 + I 3 sin q 3 = 13.387 + 7.87 21.257

I3 sin q3 = –1.83 A

As I3 sin q3 is negative, hence tertiary is operating at leading power factor and the magnitude of tertiary current referred to primary is I 3¢ = ( I 3 cos q 3 ) 2 + ( I 3 sin q 3 ) 2 = (7.87) 2 + (1.83) 2 = 8.08 A I3 = 8.08 ¥

\

1100 3 = 116.62 A 440

Power factor at which tertiary winding operates cos q3 = \

7.87 = 0.974 leading 8.08

I1 = – j 0.5 + 13.387 – j 8.3 + 7.87 + j 1.83 = 21.257 – j 6.97 = 22.37 -18.15∞ A

\ primary phase current is 22.37 A.

! Qspcmfn!5/5: Uxp!usbotgpsnfst-!fbdi!sbufe!361!lWB-!2206!lW!boe!61!I{-!bsf!dpoofdufe!jo!pqfo!efmub!po!cpui! uif!qsjnbsz!boe!tfdpoebsz/! ! )b*! Gjoe!uif!mpbe!lWB!uibu!dbo!cf!tvqqmjfe!gspn!uijt!usbotgpsnfs!dpoofdujpo/! ! )c*! B!efmub.dpoofdufe!uisff.qibtf!mpbe!pg!361!lWB-!1/9!q/g/-!6!lW!jt!dpoofdufe!up!uif!MW!ufsnjobmt! pg!uijt!pqfo.efmub!usbotgpsnfs/!Efufsnjof!uif!usbotgpsnfs!dvssfout!po!uif!22!lW!tjef!pg!uijt! dpoofdujpo/!

Solution

11 V

11 : 1 2

Gjh/!5/57! Djsdvju!ejbhsbn!pg!Qspcmfn!5/5:

2 KV

!

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250 = 50 A 5 Load kVA = Sopen delta = 3 ¥ 5 ¥ 50 = 433 kVA

(a) Iph (secondary) =

(b)

3 V Iph (secondary) = 250 Iph (secondary) =

250

= 28.87 A 3¥ 5 28.87 Iline (11 kV side) = ¥ 5 = 13.12 A 11 Iphase (11 kV side) = 13.12 A.

!Qspcmfn!5/61 Uxp!usbotgpsnfst!dpoofdufe!jo!pqfo.efmub!tvqqmz!xjui!b!411!lWB!cbmbodfe!mpbe!bsf!pqfsbujoh!bu! 1/977!q/g/!mbhhjoh/!Jg!uif!wpmubhf!jt!551!W-!dbmdvmbuf!uif!gpmmpxjoh;! ! )b*! lWB!tvqqmjfe!cz!fbdi!usbotgpsnfs ! )c*! lX!tvqqmjfe!cz!fbdi!usbotgpsnfs

Solution 300 / 2 0.866 = 173.21 kVA.

(a) kVA supplied by each transformer =

(b) cos f = 0.866 (lagging) f = 30° \ \

P1 = 173.21 ¥ cos (30° – f) = 173.21 kW P2 = 173.21 ¥ cos (30° + f) = 86.61 kW total = P1 + P2 = 259.82 kW.

! Qspcmfn!5/62 B!4!f-!DÐD!cbol!dpotjtut!pg!uisff!36!lWB-!44110411!W!usbotgpsnfst!boe!tvqqmjft!b!mpbe!pg!61!lWB/! Gps!WÐW!dpoofdujpo-!dbmdvmbuf! ! ! ! !

)b*! )c*! )d*! )e*!

lWB!mpbe!dbssjfe!cz!fbdi!usbotgpsnfs Qfsdfou!pg!mpbe!dbssjfe!cz!fbdi!usbotgpsnfs! Upubm!lWB!sbujoh! Sbujp!pg!DÐD!cbol!boe!WÐW!cbol!usbotgpsnfst!sbujoht/

Solution (a) kVA load supplied by each of two transformers =

50 3

= 28.86 kVA

Uisff.qibtf!Usbotgpsnfst

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28.86 = 1.7544 or 115.44% carried by each transformer 25 (c) kVA rating of V–V bank = 2 ¥ 25 ¥ 0.866 = 43.3 kVA D - D rating 75 (d) = = 1.7321 V - V rating 43.3 V - V rating = 0.577. or D - D rating (b) Percent of rated load =

! Qspcmfn!5/63 B!Tdpuu.dpoofdufe!usbotgpsnfs!jt!gfe!gspn!b!7711!W-!uxp.qibtf!ofuxpsl!boe!tvqqmjft!uisff.qibtf! qpxfs!bu!611!W!cfuxffo!mjoft!po!b!gpvs.xjsf!tztufn/!Jg!uifsf!bsf!611!uvsot!qfs!qibtf!po!uif!uisff. qibtf!tjef-!gjoe!uif!ovncfs!pg!uvsot!jo!uif!mpx.wpmubhf!xjoejoht!boe!uif!qptjujpo!pg!uif!ubqqjoh!pg! uif!ofvusbm!xjsf/!

Solution 6600 500 where NLV is the number of turns in the low voltage winding = 500 NLV 500 ¥ 500 = 38 6600 3 NPS = ¥ 38 = 33 2 2 NPN = ¥ 33 = 22 3 \ the neutral point is located at 22nd turn from the point A.

\

NLV =

(from Fig. 4.24a)

! Qspcmfn!5/64 B! :11! lWB! mpbe! jt! tvqqmjfe! cz! uisff! usbotgpsnfst! dpoofdufe! jo! efmub.efmub/! Uif! qsjnbsjft! bsf! dpoofdufe! up! b! 3411! W! tvqqmz! mjof-! xijmf! uif! tfdpoebsjft! bsf! dpoofdufe! up! b! 341! W! mpbe/! Jg! pof! usbotgpsnfs!jt!sfnpwfe!gps!sfqbjs-!xibu!mpbe!dbo!uif!sfnbjojoh!uxp!usbotgpsnfst!tvqqmz!xjuipvu! pwfsmpbejoh@!Xibu!bsf!uif!dvssfout!jo!uif!ijhi.!boe!mpx.wpmubhf!tjeft!pg!uif!usbotgpsnfs!xjoejoht! xifo!dpoofdufe!jo!pqfo!efmub@

Solution Delta-delta operation If VL1 and IL1 are the line voltage and line current respectively on HV side, SD – D = 3 VL1 IL1 \

900 ¥ 103 = 3 ¥ 2300 ¥ IL1 900 ¥ 103 IL1 = = 225.9 A 3 ¥ 2300

!

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Fmfdusjdbm!Nbdijoft

Transformer current per phase on HV side Ip1 =

1 3

¥ line current =

1 3

¥ 225.9 = 130.4 A

Transformer current per phase on LV side Ip2 = 130.4 ¥

2300 = 1304 A 230

Open-delta operation

When one transformer is removed from chosen delta, the currents through the phase windings of the transformers should not exceed the rated currents to avoid overloading. Therefore, the permitted phase winding currents on HV and LV sides are 130.4 A and 1304 A respectively. For open delta, the line is in series with the windings of the transformer and, therefore, the secondary line current is equal to the rated secondary current of the transformer. Therefore, the VA load that can be carried by the open-delta bank without overloading is SV–V = 3 V2B I2B = 3 VL1 Ip2 = 3 ¥ 230 ¥ 1304 VA = 519.5 kVA.

! Qspcmfn!5/65 B!7711044110511! W-!tubs0tubs0efmub!usbotgpsnfs!ibt!qfs!voju!sftjtubodf!espqt!pg!1/117-! 1/118-! 1/119!boe!qfs!voju!sfbdubodf!espqt!pg!1/14-!1/139-!1/145!gps!jut!qsjnbsz-!tfdpoebsz!boe!ufsujbsz! xjoejoht!sftqfdujwfmz!po!b!2611!lWB!sbufe!wpmubhf!cbtf/!Jut!nbhofuj{joh!dvssfou!jt!6!B/!Efufsnjof! uif!wpmubhf!sfhvmbujpot!gps!uisff.qibtf!cbmbodfe!mpbet!pg!2611!lWB!bu!qpxfs!gbdups!1/9!mbhhjoh!po! tfdpoebsz!boe!711!lWB!bu!qpxfs!gbdups!1/7!mfbejoh!ufsujbsz/!

Solution Magnetizing kVA = – j 57.16 Secondary load kVA = 1200 – j 900 Tertiary load kVA = 360 + j 480 Total load on primary = 1631.34 -17∞ kVA 1631.34 (0.006 ¥ 0.956 + 0.03 ¥ 0.2924) 1500 = 0.01578

Primary regulation =

Secondary regulation = 1 (0.007 ¥ 0.8 + 0.028 ¥ 0.6) = 0.0224 600 Tertiary regulation = (0.008 ¥ 0.6 – 0.034 ¥ 0.8) 1500 = – 0.00896

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Primary to secondary voltage regulation = 0.01578 + 0.0224 = 0.03818 p.u. Primary to tertiary voltage regulation = 0.01578 – 0.00896 = 0.00682 p.u. Secondary to tertiary voltage regulation = 0.0224 + 0.00896 = 0.03136.

! Qspcmfn!5/66 B!911!lWB-!2201/5!lW-!uisff.qibtf!efmub.tubs!usbotgpsnfs!ibt!pinjd!mptt!pg!4!lX!po!IW!tjef!boe!3! lX!po!MW!tjef!voefs!sbufe!mpbe/!Uif!upubm!mfblbhf!sfbdubodf!jt!1/19!q/v/!Efufsnjof!! ! )b*! Pinjd!wbmvft!pg!frvjwbmfou!sftjtubodf!boe!mfblbhf!sfbdubodf!po!cpui!efmub!boe!tubs!tjef! ! )c*! Qfs!voju!wbmvft!pg!sftjtubodft!po!cpui!IW!boe!MW!tjeft/!

Solution Total ohmic loss = 3 + 2 = 5 kW For HV side, Full-load current =

800 A 3 ¥ 11 re1 = re 1 p.u. =

5000 ¥ (33) 2 (800) 2

= 8.508 W

8.508 0.8 ¥ = 0.01875 p.u. 3 (11) 2

xe1 = 3 ¥

0.08 ¥ (11) 2 = 36.3 W 0.8

For LV side, Full-load current =

800 3 2000 = A 3 ¥ 0.4 3

re2p.u.

5000 ¥ 3

= 0.00375 W ( 2000) 2 0.00375 ¥ 0.8 = = 0.01875 (0.4) 2

re2 =

xe2 =

0.08 ¥ (11) 2 = 12.1 W/ 0.8

!

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Fmfdusjdbm!Nbdijoft

! Qspcmfn!5/67 Uxp!tjohmf.qibtf!Tdpuu.dpoofdufe!usbotgpsnfst!tvqqmz!b!uisff.qibtf-!gpvs.xjsf-!61!I{!ejtusjcvujpo! tztufn!xjui!361!W!cfuxffo!mjoft!boe!ofvusbm/!Uif!ijhi.wpmubhf!xjoejoht!bsf!dpoofdufe!up!b!uxp. qibtf!tztufn!xjui!b!qibtf!wpmubhf!pg!22111!W/!Bmmpx!b!nbyjnvn!gmvy!efotjuz!pg!2/3!Xc0n3!jo!b! hsptt!dpsf!tfdujpo!pg!661!dn3/!Efufsnjof!uif!ovncfs!pg!uvsot!jo!fbdi!tfdujpo!pg!uif!ijhi.wpmubhf! boe!mpx.wpmubhf!xjoejoht/

Solution Ai = 0.9 ¥ 550 ¥ 10–4 = 0.0495 m2 11000 TS = = 834 4.44 ¥ 1.2 ¥ 0.0495 ¥ 50 Number of turns on low-voltage side of main transformer 3 ¥ 250 = 33 11000 Number of turns on low-voltage side of teaser transformer 834 ¥

3 ¥ 33 = 30. 2

Sfwjfx!Rvftujpot 1. What are the advantages of a single three-phase transformer over three single-phase transformer banks of the same kVA rating? State the difference between a three-phase transformer bank and a three-phase transformer unit. 2. What is meant by the vector group of transformers? What are the distinguishing features of different groups? Mention their advantages and disadvantages. 3. How can the problems of unbalanced voltages and third harmonic currents be overcome in star-star connection? 4. Explain the open-delta connection with a suitable diagram. What are the uses of this connection? 5. What are the different schemes of three-phase to six-phase conversion? Explain with a diagram. 6. Explain with the help of connection and phasor diagrams how a Scott connection is used to obtain two-phase supply from three-phase supply. 7. What are the advantages of parallel operation of three-phase transformers? State the connections to be fulfilled for operating two three-phase transformers in parallel. 8. State briefly why all three-phase transformers cannot be operated in parallel. 9. Explain the phenomenon of inrush of magnetizing current. What factors contribute to the magnitude of inrush current? 10. Explain why it is essential to have one three-phase winding in delta for the transformers used in three-phase systems.

Uisff.qibtf!Usbotgpsnfst

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11. Why are tappings provided in transformers? Give reasons for tappings being generally provided on the high-voltage side of the transformer. 12. Explain with the help of connection diagrams the operation of off-load and on-load tap changers. 13. Explain how a three-phase transformer is operated as a three-phase induction regulator. 14. What are the applications of grounding transformers? 15. Explain the operation of audio frequency and rectifier transformers. 16. What are the advantages of using tertiary windings in transformers? What are its applications? 17. Draw and explain the equivalent circuit of a three-winding transformer.

Qspcmfnt 1. A three-phase step-down transformer is connected to 6600 V on the primary side. The ratio of turns per phase is 12 and the line current drawn from the mains is 20 A. Find the secondary line voltage, line current and output if the transformer is (a) YY, (b) YD, (c) DY, and (d) DD. [(a) 550 V, 240 A, 228.62 kVA (b) 317.55 V, 415.68 A, 228.62 kVA (c) 952.6 V, 138.56, 228.62 kVA (d) 550 V, 240 A, 228.62 kVA] 2. A 11000/440 V, three-phase transformer is delta connected on the HV side and the LV windings are star connected. There are 12 V per turn and the flux density is not to exceed 1.2 Wb/m2. Calculate the number of turns per phase on each winding and the net iron crosssectional area of the core. [917, 21, 450 cm2] 3. A 6600/400/110 V star/star/delta three-phase transformer has a magnetizing current of 5.5 A and balanced three-phase loads of 1000 kVA at 0.8 lagging power factor on the secondary and 200 kVA at 0.5 leading power factor on tertiary. Neglect losses. Find primary current, kVA and power factor. [90.1 A, 1030 kVA, 0.875 lag]

Nvmujqmf.Dipjdf!Rvftujpot 1. Three phase transformers are connected to form a 3-phase transformer bank. The transformers are connected in the following manner. The transformer connection will be represented by A1

A2

a2

a1

B1

B2

b2

b1

C1

C2

c2

c1

!

5/99

Fmfdusjdbm!Nbdijoft

(a) Yd0 (c) Yd6 [Hints:

(b) Yd1 (d) Yd11

[GATE 2009] c1

A2

a2

c2 b1

N

A1 B1 b2

C1 B2 C2

A2

a1

a2 30°

]

N

2. Which three phase connection can be used in a transformer to introduce a phase difference of 30° between its output and corresponding input time voltages? (a) Star-Star (b) Star-Delta (c) Delta-Delta (d) Delta-Zigzag [GATE 2005] 3. The star delta transformer shown below is excited on the star side with a balanced, 4 wire, 3 phase, sinusoidal voltage supply of rated magnitude. The transformer is under no load condition. a

A

B

b

C

c

N

S1

S1

with both S1 and S2 opened, the core flux waveform will be (a) sinusoidal at fundamental frequency (b) flat topped with third harmonic (c) peaky with third harmonic (d) none of the above

[GATE 2009]

Uisff.qibtf!Usbotgpsnfst

5/9:

4. In the previous question, with S2 closed and S1 opened, the current waveform in the delta winding will be (a) a sinusoidal at fundamental frequency (b) flat topped with third harmonic (c) only third harmonic (d) none of the above [GATE 2009] 5. Which of the following connections of three phase transformer will give the highest secondary voltage? (a) Star primary, star secondary (b) Star primary, delta secondary (c) Delta primary, delta secondary (d) Delta primary, star secondary 6. Two three phase delta star transformers are supplied from the same source. One transformer is DY1 connected and other is DY11 connected. The phase difference between the corresponding phase voltage of the secondaries would be (a) 30° (b) 0° (c) 60° (d) 120° 7. In which transformer connection tertiary winding is used? (a) YD (b) DD (c) DY (d) YY 8. A DY transformer is connected in parallel to a YD transformer. The turns ratio of the former is x times the latter. Hence, x is (a) 3 (b) 3 1 1 (c) (d) 3 3 9. Which of the following combinations of three phase transformers can operate successfully in parallel? (a) YY and DY (b) YD and DD (c) DY and DY (c) DD and DY 10. The magnitude of third harmonic in the wave shape of magnetizing current of a transformer is about (a) 1% (b) 20% (c) 50% (d) 5% 11. Which of the following three phase connections of a transformer causes interference to the nearly communication system? (a) Star star (b) Delta star (c) Star delta (d) Delta delta 12. A DY transformer has a phase to phase voltage transformation ratio of K Ê Phase voltage of delta connection ˆ . The line-to-line voltage ratio of YD connection is ÁË K = Phase voltage of star co onnection ˜¯

!

5/:1

(a)

Fmfdusjdbm!Nbdijoft

K

(b)

3

(c) K

(d)

3 K K 3

13. In a three-phase DY transformer shown in the figure the phase displacement of secondary line voltage with corresponding primary line voltage will be c2 a2

A1

C1 n

B1 b2

(a) 30° lag (b) 30° lead (c) 180° (d) zero 14. The figure below shows a DY connected three phase distribution transformer used to step down the voltage from 11,000 V to 415 V line to line. Under normal condition switch S1 is closed and switch S2 is open. Under special condition S1 is opened and S2 is closed. At this condition the magnitude of the voltage across the lv terminals a and c is HV A

B

LV

a

b

S2

c

C

S1

Uisff.qibtf!Usbotgpsnfst

(a) 240 V (c) 0 V

5/:2

(b) 480 V (d) 415 V

Botxfst 1. (b) 6. (c) 11. (a)

2. (b) 7. (c) 12. (b)

3. (b) 8. (a) 13. (b)

4. (a) 9. (c) 14. (c)

5. (d) 10. (b)

6 Gvoebnfoubmt!pg!BD! Spubujoh!Nbdijoft JOUSPEVDUJPO!

6/2

The objective of this chapter is to discuss how the basic principles and laws, as discussed in Chapter 1, are implemented in rotating machines. The general expressions for electromagnetic torque and generated emfs are developed. These basic torque and emf expressions are applicable to both ac and dc machines. Except some small motors used for domestic and specific purposes, ac machines are mostly three-phase machines. The most commonly used ac machines are synchronous machines and induction machines.

BD!XJOEJOHT!

6/3

As no commutator is needed in an ac winding it differs from a dc winding in its configuration. The ac winding need not be a closed one like a dc winding and hence ac windings are open windings. Since most ac machines are of three-phase type, the windings of all the three phases are identical and spaced 120 electrical degrees apart from each other. The windings are normally connected in star or delta connection. However, when the three windings are connected in delta, the combination forms a closed winding. The ac windings are characterized by the following: 1. Usually the number of phases are three. 2. Armature windings of synchronous generators and motors are usually connected in star. In the case of three-phase induction motors, both star and delta-connected windings may be used.

!

6/3

Fmfdusjdbm!Nbdijoft

3. The number of winding circuits in parallel per phase may be one or more than one. 4. The number of coil layers per slot may be one or two though two-layer windings are much common. 5. The angular spread of the consecutive conductors belongs to a phase belt and the arrangement of the end connections of ac windings differ from that of dc windings.

6/3/2! Tjohmf.!boe!Epvcmf.Mbzfs!Xjoejoht In a single-layer winding, one coil side occupies one slot completely. Therefore, the number of coils is equal to half the number of slots. In a double-layer winding, one coil side occupies the upper half of one slot, while the other coil side is placed in the lower half of another slot spaced about one pole pitch from the first one. In a single-layer winding, the coils are arranged in groups and the overhang of each group of coils is made to cross the overhang of other groups by adjusting the size and shape of different coil groups. However, in double-layer windings, all the coils are identical in shape and size. Hence, a double-layer winding results in a cheaper machine. All synchronous machines and most induction machines of medium and large capacity use double-layer windings. The advantages of double-layer winding over single-layer winding are 1. 2. 3. 4.

Simple manufacture and lower cost of the coils Fractional slot and chorded winding is possible Lower leakage reactance and hence better performance Better emf waveform

6/3/3! Qibtf!Tqsfbe! A group of adjacent slots belonging to any one phase under one pole pair is termed phase belt. The angle subtended by a phase belt is known as phase spread. Let us illustrate this arrangement by considering two slots per pole per phase (i.e. 12 slots per pole pair). The slot angle is obviously 30°. We can place the winding of phase R in slots 1–4, that of phase Y in slots 5–8 and that of phase B in slots 9–12. Each phase belt has a spread of 120°[Fig. 5.1(a)]. The induced emfs of conductors in adjacent slots have a phase difference equal to the slot angle. The resultant emf per phase is less than the sum of individual conductor emfs. [The phasor sum of induced emfs in conductors of 1–4 gives the phase emf of phase R.] In another possible arrangement, as shown in Fig. 5.1(b), the complete winding has been divided in six groups, each having a spread of 60°. Here, the conductors in slots 7 and 8 serve as return conductors for slots 1 and 2 (hence the induced conductor emfs for slots 7 and 8 have been marked negative). The phasor sum of conductor emfs gives the phase emf for the phase R. The magnitude of phase emf in Fig. 5.1(b) is more than that in Fig. 5.1(a). Hence, Three-phase windings are always designed for 60° phase

Gvoebnfoubmt!pg!BD!Spubujoh!Nbdijoft

6/4

spread. Moreover, the arrangement shown in Fig. 5.1(a) is not feasible to be used with a single-layer winding, because there are no conductors of one pole pitch apart to form return conductors. B 120°

120°

120°

2

3

1 1

3

2

4

5

6

R

7

8

9

10

11

12

120° 5 120° 6

B

Y (a)

Y

8

R

4

7 R –7

B 1 1

7

2 R

B¢

Y

8 R¢

B

–8 2

Y¢

(b) Y

Gjh/!6/2! Fggfdu!pg!qibtf!tqsfbe!po!hfofsbufe!fng;!)b*!231°!qibtf!tqsfbe!)c*!71°!qibtf!tqsfbe

6/3/4! Nvmujuvso!Dpjm!Xjoejoht! The ac machines with small number of poles and low value of flux per pole require a large number of conductors to have appreciable induced emf. Such machines have multiturn coils.

6/3/5! Gvmm.Qjudi!boe!Gsbdujpobm.Qjudi!Dpjmt! Figure 5.2(a) shows a two-pole machine which contains one coil housed in stator slots. The coil has a single turn only and one turn contains two conductors. AB and CD are the two coil sides. The peripheral distance between two adjacent poles is called pole pitch. A pole pitch is always equal to 180 electrical degrees. In Fig. 5.2(b), the coil span is equal to the pole pitch and in Fig. 5.2(c), the coil span is less than one pole pitch. Coil span is the distance between two coil sides of one coil. Hence, in Fig. 5.2(b), the coil span is equal to 180° and in Fig. 5.2(c), the coil span is less than 180° by an angle a. This angle a is called the chording angle and is defined as the angle by which the coil span is less than 180 electrical degrees. In a full-pitch coil, the coil sides are 180 electrical degrees, apart or the coil span is equal to one pole pitch as shown in Fig. 5.3(a). If the coil span is less than 180 electrical degrees, it is called a short pitch or chorded coil [Fig. 5.3(b)].

!

6/5

Fmfdusjdbm!Nbdijoft

Pole Pitch N

S

N

S

N B

Coil spare

C

B

C a

S A

(a)

D

A

(b)

D

(c)

Gjh/!6/3! )b*!B!uxp.qpmf!nbdijof;!)c*!Dpjm!tqbo!)d*!Dipsejoh!bohmf q = 0°

q = 180° Pole Pitch

(a) a

(b)

Gjh/!6/4! )b*!Gvmm.qjudi!dpjm!)c*!tipsu.qjudi!dpjm

Fractional-pitch windings are extensively used as the resulting voltage waveform is nearly sinusoidal in comparison to that of full-pitch winding. Also, because of saving in copper, we use fractional-pitch windings. It also offers greater stiffness of coils due to shorter end connections. The only disadvantage of fractional-pitch winding is that terminal voltage is comparatively less than that with full-pitch coils.

6/3/6! Joufhsbm!Tmpu!Xjoejoht Since the three phases in ac machines are identical, the total number of slots in an ac machine is always an integral multiple of three. However, the number of slots per pole phase may be an integer or a fraction. In integral slot windings, the number of slots per pole per phase is an integer. As for example, in a double-layer integral slot, full-pitch winding with 9 slots per pole, there are 3 slots per pole per phase. Here, the coil span is full pitch (i.e. 9 slots). For the same winding with a coil span of 8 slots, it gives a short pitch winding. Here, some slots hold coil sides of different phases [Fig. 5.4].

Gvoebnfoubmt!pg!BD!Spubujoh!Nbdijoft

6/6

6/3/7! Gsbdujpobm!Tmpu!Xjoejoht In these windings, the number of slots per pole phase is a fraction. A fractional slot winding may be single layer or double layer. Standard slotting arrangements over a wide range of pole numbers can be used in fractional slot windings. Pole Pitch

Pole Pitch

Coil Sides Upper Lower 1

2

3

4

5

6

7

8

9

10

11 12 13 14 15 16 17 18

19

R

R

R

B¢

B¢

B¢

Y

Y

Y

R¢

R¢

R¢

B

B

B

Y¢

Y¢

Y¢

R

R

R

R

B¢

B¢

B¢

Y

Y

Y

R¢

R¢

R¢

B

B

B

Y¢

Y¢

Y¢

R

R¢

R¢

R¢

B

B

B

Y¢

Y¢

Y¢

R

R¢

R¢

B

B

B

Y¢

Y¢

Y¢

R

R

Coil Sides (a)

R

R

R

B¢

B¢

B¢

Y

Y

Y

R

R

B¢

B¢

B¢

Y

Y

Y

R¢

Coil Sides (b)

Gjh/!6/5! Epvcmf.mbzfs!joufhsbm!tmpu!xjoejoh;!)b*!Gvmm.qjudi!dpjmt!)c*!Tipsu.qjudi!dpjmt

The total number of slots must be a multiple of 3, so that the windings of the three phases remain symmetrical. In a double-layer fractional slot winding, the arrangement of coil sides is repeated for the bottom layer. Here, the coil span is always less than one pole pitch.

6/3/8! Bewboubhft!pg!Ejtusjcvufe!Xjoejoh!boe ! Gsbdujpobm.Qjudi!Xjoejoh The advantages of distributed windings are 1. Reduction of the harmonic effect in the emf signal 2. Adding mechanical strength to the winding 3. Full utilization of armature iron and copper The advantages of fractional-pitch windings are 1. Reduction of harmonics in the generated emf signal 2. Reduction in the copper for overhang which results in reduction in the cost of the machine 3. Reduction in magnetic losses due to the elimination of high-frequency harmonics The only disadvantage of fractional-pitch winding is that the terminal voltage is a bit less than that with full-pitch coils.

!

6/7

Fmfdusjdbm!Nbdijoft

DPODFQU!PG!FMFDUSJDBM!BOE! NFDIBOJDBM!EFHSFFT

6/4

Let us consider a two-pole machine as shown in Fig. 5.5.

e

N 2

2 1

3

1

0

p 3

1

q

2p

4 S

4

(a)

(b)

Gjh/!6/6! )b*!Uxp.qpmf!nbdijof!xjui!tjohmf!spups!dpjm!)c*!Fng!wbsjbujpo!xjui!wu!gps!uxp!qpmf!nbdijof

We assume one rotor coil rotating at a uniform angular speed. When the rotor coil is at the position 1, the emf in the coil is zero (though flux cutting the coil is maximum but the rate of change of flux is minimum). Let us consider that 1 is the initial position of the rotor coil which is rotating in the clockwise direction. After 90° revolution, it reaches the position 2 where the emf in the coil becomes maximum (though flux cutting the coil is minimum at this position but the rate of change of flux is maximum). When the coil reaches the position 3, the emf is zero and at the position 4, the emf is maximum but the direction of the emf is opposite to that at the position 2. Thus, when the rotor coil completes one revolution and returns to the position 1, the emf in the coil is again zero. Hence, with one complete revolution of the rotor coil, one full cycle of the coil emf is generated. Now let us consider a four-pole machine as shown in Fig. 5.6. It is assumed that the rotor coil starts from the initial position 1 where the emf is 0. When it is at the position 2, the emf is maximum N 4

S

e

2

6

6

2

S 4p

1

7

0

1

3 p

2p 5

7 3p

1

8 N (a)

wt

8

4 (b)

Gjh/!6/7! )b*!Gpvs.qpmf!nbdijof!xjui!tjohmf!spups!dpjm!)c*!Fng!wbsjbujpo!xjui!wu!gps!gpvs.qpmf!nbdijof

Gvoebnfoubmt!pg!BD!Spubujoh!Nbdijoft

6/8

and again at the position 3, the emf is 0. At the position 4, the emf is maximum but in a direction opposite to that at the position 1 and hence the emf is considered negative maximum. Again, at the position 5, the emf is 0. Thus, one full cycle of emf is generated while the rotor coil completes only half revolution by moving from the position 1 to 5. In a similar way, when the rotor coil completes full revolution, another full cycle of emf will be generated. Hence, two cycles of emf, i.e. 720°, is generated in one revolution of 360 mechanical degrees for a four-pole machine. \ or, or,

720 electrical degrees = 2(360 mechanical degrees) P 720 electrical degrees = (360 mechanical degrees) 2 qelectrical =

P qmechanical 2

(5.1)

where P is the number of poles in the machine. In other words, in a P-pole machine, in one revolution of the rotor

P cycles of emf is generated. 2

CSFBEUI!GBDUPS!

6/5

When the coils of the same phase of the winding are placed in more than one slot per pole then the emfs of the windings in different slots will have different phase angles. Let the angle between the adjacent slots be g which is known as the slot angle. Hence, the emfs of the windings are displaced from each other by an angle g. Considering three slots per pole per phase, the component emfs E1, E2 and E3 of the coils are ab, bc and ca as shown in Fig. 5.7. The right bisectors of ab, bc and cd are go, fo and eo respectively and they meet at the point O. d

g c

e

g

f

p

g

g

b

g

g O

a

Gjh/!6/8! Efufsnjobujpo!pg!csfbeui!gbdups

!

6/9

In triangle aog,

Fmfdusjdbm!Nbdijoft

sin

g ag = 2 ao g 2

or,

ag = ao sin

Emf per coil is

ab = bc = cd = 2ag = 2 ao sin

g 2

Hence, the arithmetic sum of the coil emfs = ab + bc + cd = 3 ¥ 2 ao sin

g 2

g If there are q slots per pole phase then the arithmetic sum of the coil emf is q ¥ 2 ao sin 2 In the right-angled triangle apo, ap 3g = sin ao 2 3g or, ap = ao sin 2 The resultant emf ad which is the phasor sum of ab, bc and cd is given by ad = 2 ap = 2ao sin 3

g 2

If there are q slots per pole per phase then the resultant emf is ad = 2 ao sin

qg 2

The breadth or distribution factor is defined as the ratio of the phasor sum of the coil emfs to the arithmetic sum of the coil emfs. qg qg sin 2 = 2 Kd = g g 2ao q sin q sin 2 2 2ao sin

Hence, breadth factor

(5.2)

In other words, breadth factor is the ratio of the resultant emf with coils distributed in slots to the resultant emf with the coils concentrated in one slot. The flux-density wave may contain harmonics. Since the positive and negative halves of the fluxdensity wave are identical, only odd harmonics can be present. The harmonic poles of the nth space 1 harmonic have a pitch equal to of the fundamental pole pitch. n

Gvoebnfoubmt!pg!BD!Spubujoh!Nbdijoft

6/:

Therefore, the angle between adjacent slots is ng for the nth harmonic and breadth factor for the nth harmonic is Ê nqg ˆ sin Á Ë 2 ˜¯ Kdx = (5.3) Ê ng ˆ q sin Á ˜ Ë 2¯

QJUDI!GBDUPS! a

6/6

c a 2

a

b d

a¢

a¢

a 2

a (a)

(b)

a

Gjh/!6/9! Efufsnjobujpo!pg!qjudi!gbdups!gps!)b*!gvmm.qjudi!dpjm-!boe!)c*!tipsu.qjudi!dpjm

Figure 5.8(a) shows a full-pitch coil and Fig. 5.8(b) shows a short-pitch or fractional-pitch coil with chording angle a. In the full-pitch coil, the coil emf is twice the emf of each coil side. In the short-pitch coil, as the two coil sides are not in phase, the emf of each coil side should be added vectorially to get the resultant emf. Figure 5.8(b) shows emfs ab and bc on two coil sides of short-pitch coil. The resultant coil emf a ac = 2ad = 2ab cos 2 The arithmetic sum of coil emfs is 2ab. Now pitch factor is defined as the ratio of the resultant emf of a chorded coil to the resultant emf of a full-pitch coil. a 2ab cos 2 = cos a (5.4) Hence, pitch factor Kp = 2 ab 2 The product of breadth or distribution factor and pitch factor is known as the winding factor Kw where Kw = Kd ¥ Kp (5.5)

!

6/21

Fmfdusjdbm!Nbdijoft

The pitch factor given by Eq. (5.4) is for the fundamental frequency. If the flux-density distribution contains space harmonics, the pitch factor for nth harmonic is Kpn = cos (ng /2)

(5.6)

The nth harmonic emf is reduced to zero if the angle g is such that cos (ng /2) = 0 or

ng = 90° 2

(5.6a)

The flux-density distribution contains some odd harmonics (because of symmetry, even harmonics are absent). Therefore, phase voltage may contain third, fifth, seventh and higher-order harmonics. The components of triple harmonics are identical in the three phases and do not appear in the line-to-line voltage. Since the strength of harmonic components of voltage decreases with increasing frequency, only fifth and seventh harmonics are important. These are known as belt harmonics. It is evident from Eq. (5.6) that the pitch factor is different for different harmonics. By proper selection of coil span, the pitch factor for 5th and 7th harmonics can be made small and thus these harmonics can be nearly eliminated from the voltage wave. If the coil span is 5/6 of the pole pitch, the pitch factor is 0.259 for both 5th and 7th harmonics.

! Qspcmfn!6/2 Efufsnjof!uif!ejtusjcvujpo!gbdups!boe!uif!qjudi!gbdups!jo!b!4.qibtf!xjoejoh!ibwjoh!b!dpjm!tqbo!pg! 261¡!boe!qibtf!tqsfbe!pg!231¡/

Solution Phase spread = 120° Let number of slots per pole per phase be q and the angle between adjacent slots be g. Then qg = 120° qg sin 2 Distribution factor Kd = g q sin 2 qg sin g 2 [If g is small then sin = g, where g is in radians] = g 2 q 2 120∞ sin 2 = = 0.827 p 120∞ ¥ 2 180∞ Now coil span = 150°

Gvoebnfoubmt!pg!BD!Spubujoh!Nbdijoft

\ chording angle \ pitch factor

6/22

a = 180° – 150° = 30° a 30∞ Kp = cos = cos = 0.9659. 2 2

! Qspcmfn!6/3 Bo!bd!nbdijof!ibt!7!qpmft!xjui!:7!tmput/!Uif!dpjmt!bsf!xpvoe!xjui!24027!gsbdujpobm!qjudi/!Efufsnjof! uif!qjudi!gbdups/

Solution 13 fractional pitch. 16 \ the angle by which coils are short pitched The coils are wound with

3 Ê 13 ˆ = a = Á1 - ˜ ¥ 180° = ¥ 180° Ë 16 ¯ 16 \ pitch factor

Kp = cos

3 180∞ ¥ = 0.9569. 16 2

! Qspcmfn!6/4 B!tzodispopvt!nbdijof!ibt!65!tmput!boe!7!qpmft/!Ju!ibt!tjohmf.mbzfs!xjoejoh!xjui!gvmm.qjudi!dpjmt-! fbdi!dpjm!ibwjoh!9!uvsot/!Jg!uif!dpjmt!bsf!dpoofdufe!jo!tfsjft!jo!71¡!qibtf!hspvqt!boe!uif!gmvy!qfs!qpmf! jt!1/14!Xc-!efufsnjof!uif!qibtf!wpmubhf/

Solution Slots per pole per phase As slots per pole is \ slot angle \ distribution factor Pitch factor

q=

54 =3 6¥3

54 =9 6 180∞ = 20° g= 9 qg 3 ¥ 20∞ sin sin 2 = 2 Kd = = 0.9598 g 20∞ q sin 3 sin 2 2 Kp = 1

In a single-layer winding, the number of coils is half the number of slots. 54 \ number of coils = = 27 2 Each coil has 8 turns. Total number of turns = 27 ¥ 8 = 216

(∵ coil is full pitch)

!

6/23

Fmfdusjdbm!Nbdijoft

\ number of turns per phase =

216 = 72 3 E = 4.44 Kd f f N V = 4.44 ¥ 0.9598 ¥ 0.03 ¥ 50 ¥ 72 = 460.24 V.

The phase voltage

! Qspcmfn!6/5 Uif!upubm!ovncfs!pg!tmput!jo!b!epvcmf.mbzfs!bmufsobups!xjoejoh!jt!83/!Uif!nbdijof!ibt!9!qpmft!boe! fbdi!dpjm!ibt!7!uvsot/!Uif!xjoejoh!jt!tipsu!qjudife!cz!4!tmput/!Jg!uif!gmvy!qfs!qpmf!jt!1/13!Xc-!efufs. njof!uif!joevdfe!fng!qfs!qibtf/

Solution Number of slots per pole phase Slot angle

72 =3 8¥3 180∞ 180∞ g = = 20° = slots per pole 72 / 8 q=

3 ¥ 20∞ qg sin 20 = 0.9598 2 = Kd = g 20∞ q sin 3 sin 2 2 sin

Breadth factor

The winding is short pitched by 3 slots. \ chording angle = 3a = 3 ¥ 20° = 60° \ pitch factor

Kp = cos

a 60∞ = cos = 0.866 2 2

In double-layer winding, the number of coils is equal to the number of slots. \ total number of coils = 72 Each coil has 6 turns. \ total number of turns = 72 ¥ 6 = 432 432 = 144 \ number of turns per phase N = 3 Induced emf

E = 4.44 Kd Kp f f N = 4.44 ¥ 0.9598 ¥ 0.866 ¥ 0.02 ¥ 50 ¥ 144 = 531.428 V.

! Qspcmfn!6/6 B!2611!W-!5.qpmf!tzodispopvt!nbdijof!ibt!59!tmput/!Jg!uif!dpjm!qjudi!jt!57!tmput!boe!fbdi!tmpu!ibt!3! dpoevdupst-!efufsnjof!uif!gmvy!qfs!qpmf/

Gvoebnfoubmt!pg!BD!Spubujoh!Nbdijoft

6/24

Solution 48 4¥3 =4

Number of slots per pole per phase q =

180∞ = 15° 48/ 4 qg 4 ¥ 15∞ sin sin 2 = 2 Kd = g 15∞ q sin 4 sin 2 2 = 0.9576 g=

Slot angle \ distribution factor

Coil pitch = 46 slots Total number of slots = 48 slots \ the coils are short pitched by 2 slots. \ chording angle

a = 2 ¥ 15° = 30° 30∞ = 0.9659 Kp = cos 2

Pitch factor

Number of conductors = 48 ¥ 2 48 ¥ 2 \ total number of turns = = 48 2 48 = 16 Number of turns per phase N = 3 If f be the flux per pole then 1500

= 4.44 ¥ Kd Kp f f N = 4.44 ¥ 0.9576 ¥ 0.9659 ¥ 50 ¥ 16 f 3 f = 0.2636 Wb.

\

! Qspcmfn!6/7 Efufsnjof!uif!gvoebnfoubm-!uijse!boe!gjgui!ibsnpojd!ejtusjcvujpo!gbdupst!jo!b!31.qpmf!291.tmpu!tzo. dispopvt!nbdijof/

Solution Number of slots per pole per phase

Slot angular pitch

180 20 ¥ 3 =3 180∞ g= = 20° 180 / 20

q=

!

6/25

Fmfdusjdbm!Nbdijoft

If n is the harmonic number then nqg 2 Kdn = ng q sin 2 sin

For fundamental frequency, 3 ¥ 20∞ 2 Kd1 = = 0.9598 20∞ 3 sin 2 sin

Third harmonic distribution factor 3 ¥ 3 ¥ 20∞ sin 90∞ 2 = = 0.667 Kd3 = 3 ¥ 20∞ 3 sin 30∞ 3 sin 2 Fifth harmonic distribution factor 5 ¥ 3 ¥ 20∞ sin sin 150∞ 2 = = 0.2175. Kd5 = 5 ¥ 20∞ 3 sin 50∞ 3 sin 2 sin

! Qspcmfn!6/8 Uif!gmvy.efotjuz!ejtusjcvujpo!pg!b!tzodispopvt!nbdijof!jt!C!>!Cn!tjo!q!,!1/5Cn!tjo!4q!,!1/4Cn!tjo! 6q/!Jg!uif!dpjm!tqbo!jt!1/:!pg!qpmf!qjudi-!efufsnjof!uif!snt!fng!joevdfe!jo!pof!uvso/!Bttvnf!uif!gmvy! qfs!qpmf!up!cf!1/2!Xc/

Solution 1 ¥ 0.4 f1 = 0.133 f1 = 3 1 = ¥ 0.3 f 1 = 0.06 f 1 5 = f1 + f2 + f3 = f 1 + 0.133 f 1 + 0.06 f 1 = 1.193 f 1 = 0.1 Wb = 0.0838 Wb

The third harmonic flux

f3 =

The fifth harmonic flux

f5

Total flux per pole

As

f

f f1

As emf in one turn needs to be calculated, Kd1 = Kd3 = Kd5 = 1

1 Bn ˆ Ê ÁËEfm = n B f1 ˜¯ 1

Gvoebnfoubmt!pg!BD!Spubujoh!Nbdijoft

6/26

a = 180(1 – 0.9) = 18° 18∞ Kp1 = cos = 0.987 2

Chording angle \ The fundamental frequency emf

E1 = 4.44 f N f1 Kp1 Kd1 = 4.44 ¥ 50 ¥ 1 ¥ 0.0838 ¥ 0.987 ¥ 1 = 18.362 V Now and \

3a = cos 27° = 0.89 2 5a = cos 45° = 0.707 Kp5 = cos 2 K p3 B3 0.89 ¥ 0.4 E1 = ¥ 18.362 E3 = K p1 B1 0.987

Kp3 = cos

= 6.623 V E5 =

K p5 B5 K p1 B1

E1 =

0.707 ¥ 0.3 ¥ 18.362 0.987

= 3.946 V \ total emf induced

E=

E12 + E32 + E52

= (18.362) 2 + (6.623) 2 + (3.946) 2 = 19.915 V.

NNG!PG!BD!XJOEJOHT!

6/7

6/7/2!Nng!pg!Dpodfousbufe!Dpjm A full pitch coil on the stator of a 2 pole ac machine as shown in Fig. 5.9(a) is considered. The coil has N turns and each turn carries current i. The direction of the current in the two coil sides is shown by V and €. The magnetic flux set up by the current is shown by dotted lines. The flux completes its path through the air gap and the iron path. As the permeability of iron is much higher than that of air, we can assume infinite permeability for iron and hence the mmf can be considered to be consumed in the air gap only. Also it is assumed that the flux lines cross the air gap radially. The developed view of the coil with stator and rotor periphery laid out flat is shown in Fig. 5.9(b).

!

6/27

Fmfdusjdbm!Nbdijoft

Air gap

N

S

Coil magnetic axis Rotor

Stator (a)

e

S

rfac

r su

to Sta

N

Air gap

Rotor surface

(b)

Ni 2 –Ni 2

0°

180°

q

Pole pitch (c)

Pole pitch

Gjh/!6/:! )b*!Gvmm!qjudi!dpjm!po!tubups!)c*!Efwfmpqfe!wjfx!pg!uif!tubups!dpjm!)d*!Nng!ejtusjcvujpo!bmpoh! uif!bjs!hbq!qfsjqifsz

If H is the magnetic field intensity and lg is the uniform air gap then 1 2lg H = Ni as each flux line crosses the air gap twice \ lg H = Ni. The air gap mmf on the op2 posite sides of the rotor are equal in magnitude but opposite in direction. The variation of the mmf distribution along the air gap periphery is shown in Fig. 5.9(c). It is seen that at any instant the air 1 gap mmf wave is rectangular with magnitude Ni. If the coil current is dc, the magnitude of mmf 2

Gvoebnfoubmt!pg!BD!Spubujoh!Nbdijoft

6/28

wave does not vary with time and space. When the coil current is ac, the magnitude of mmf varies with time but not with space. 1 The rectangular mmf space wave of magnitude Ni can be resolved by Fourier series into a 2 fundamental component and a series of odd harmonic component. The fundamental component from Fourier series Ê 4 Ni ˆ cos q ˜ where q is the electrical angle measured from the magnetic axis of the coil Fa1 = Á Ëp 2 ¯ which coincides with the positive peak of the fundamental wave shown in Fig. 5.9(c). \

Ê 4 Ni ˆ Fa1 = (F1 peak cos q), where F1P = Á Ë p 2 ˜¯

which is the peak value of the sine mmf wave for a 2 pole machine. If the machine has P number of poles then Ê 4 Ni ˆ Fa1 = Á cos q ˜ …(5.7) Ëp P ¯

6/7/3!Nng!pg!Ejtusjcvufe!Dpjm The distributed coil of a three phase, 2 pole ac machine is shown in Fig. 5.10. The three coils

1¢

2¢

3¢ Stator

3

2

1

Gjh/!6/21! Ejtusjcvufe!dpjm!po!tubups

belong to a single phase. The mmf variation of coil 11¢ alone is a rectangular wave of magnitude 1 Ni as discussed in the mmf study for concentrated coil. Similarly the mmf distribution of coil 2 22¢ and 33¢ are identical except the fact that they are displaced from one another by a slot angle g. The mmf distribution of the three coils and the resultant of the three mmf distribution is shown in Fig. 5.11. It is observed that the resultant mmf wave is stepped unlike rectangular shape of individual coils.

!

6/29

Fmfdusjdbm!Nbdijoft

1

2

1¢

3

2¢

3¢

(a)

(b)

(c)

(d)

1 Ni 2 – 3 Ni 2

– 1 Ni 2

3 Ni 2

(e)

Gjh/!6/22! Nng!xbwfgpsn!pg!ejtusjcvufe!xjoejoh!

(a) Developed view of stator coil (c) Mmf waveform due to coil 22¢ (e) Resultant mmf waveform

(b) Mmf waveform due to coil 11¢ (d) Mmf waveform due to coil 33¢

When more than three slots per pole per phase are present the steps are neglected and the mmf waveform can be approximated to a smooth trapezoidal waveform as shown in Fig. 5.11(e). The fundamental component of air gap mmf in distributed winding can be obtained from Fourier series and fundamental component of mmf of phase a Fa1 =

N 4 K d K P Ph ia cos q p P

…(5.8)

where Kd and KP are distribution factor and pitch factor respectively, P is the number of poles, NPh is the number of turns per phase and ia is the current in phase a.

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6/2:

6/7/4!Nng!pg!Uisff.qibtf!Xjoejoht When a three-phase winding, displaced in space by 120°, is supplied by a three-phase current displaced in time by 120°, a magnetic flux is produced which rotates in space. This causes the rotor to rotate. Bobmzujdbm!Dpodfqu!pg!Uisff.qibtf!Spubujoh!Gjfme Let us consider a machine with three identical coils placed 120° apart in space with respect to each other, as shown in Fig. 5.12(a). The coils are supplied with balanced three-phase voltages varying sinusoidally in time. This will cause flow of sinusoidal current and each coil will produce an alternating flux along its own axis. Let the instantaneous flux of the three phase be given by f 1 = f m sin w t f 2 = f m sin (w t – 120°) f 3 = f m sin (w t – 240°)

(5.9a) (5.9b) (5.9c)

The resultant flux produced by this system is determined by resolving the components will respect to x and y axes, as shown in Fig. 5.12(b). Axis of phase 3

f3

1 .

30°

120°

2¢

120° . 120° 3

3¢ . 2

60° Axis of phase 1

f1

60° 30°

1¢

f2 Axis of phase 2 (a)

(b)

P

fp = 3 fm

fr

fv

2

fs = 3 f m 2 S

Q

fQ = 3 f m 2 fR = 3 fm

q fh

2

R (c)

(d)

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The resultant horizontal component of flux is then given by f h = f 1 – f 2 cos 60° – f 3 cos 60° = f 1 – (f 2 + f 3) cos 60° 1 = f 1 – (f 2 + f 3) ¥ 2 1 = f m sin wt – [fm sin (wt – 120°) + f m sin (wt – 240°)] 2 f Ê 1ˆ = f m sin wt – m ¥ (2 sin wt) Á - ˜ Ë 2¯ 2 3 = f m sin wt 2

(5.10a)

Similarly, the vertical component of flux is given by f v = 0 – f 2 cos 30° + f 3 cos 30° = =

3 [f 3 – f 2] 2

3 [f m sin (wt – 240°) – f m sin (wt – 120°)] 2

3 3 ¥ f m . 2 cos wt ¥ 2 2 3 = f m cos wt 2 =

(5.10b)

\ the resultant flux is obtained as [Fig. 5.9(d)] fr =

3 (fh ) 2 + (fv ) 2 = fm sin 2 w t + cos 2 w t 2

3 f m [∵ sin2 wt + cos2 wt = 1] 2 f tan q = v cot tw = tan (90° – wt) fh =

and It implies

q = (90° – wt)

(5.11)

(5.12)

The above equation shows that the resultant flux (f r) is free from time factor. It is a constant Ê 3ˆ magnitude flux of value equal to Á ˜ times the maximum flux per phase. However, q is dependent Ë 2¯ on time and we can calculate q at different values of (wt); when (wt = 0), q = p/2 corresponding to position P in Fig. 5.12(c).

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wt = p /2, q = 0°, (corresponding to position q), and wt = p, q = – p/2 (corresponding to position R), 3p with wt = , q = – p (corresponding to position S). 2 It is thus observed that the resultant flux f r rotates in space in the clockwise direction with anguPN s lar velocity of w radians per second. Since w = 2p f and f = , the resultant flux f r rotates with 120 synchronous speed (Ns). Thus, with application of a three-phase source at a three-phase apart space winding, we get a rotating field in the air gap of the machine. For when

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Let the space distribution of field flux density created by distributed stator winding be taken approximately as sinusoidal, i.e. bf = Bm sin q d-axis

1 q-axis

w q 1¢

Gjh/!6/24! Tubups!xjui!ejtusjcvufe!xjoejoh!boe!spups!xjui!tjohmf!dpjm!2Ð2¢

For simplicity, let a single coil 1–1¢ of N turns be placed in the rotor slots as shown in Fig. 5.10 and rotated at a speed of w. Let q be the angle measured from the quadrature axis. For an elemental angle dq of the rotor surface, the flux linkage dy by the coil is given by dy = N d f = N Bm d A = N Bm sin q l (rd q)

[∵ Flux = Flux density ¥ Area]

[∵ r is the radius of the coil, r d q is the arc and l is the axial length of the armature core] If P is the number of poles, angle d q is to be converted into equivalent mechanical degrees. However,

1° mechanical =

P electrical degrees 2

!

6/33

Or \

Fmfdusjdbm!Nbdijoft

2 degrees mechanical P 2 1 d y = N Bm sin q l (r d q) = 2N Bm lr sin q d q P P 1° electrical =

Let the angular position of the rotor coil 1–1¢ at time zero be at the Q-axis, i.e. at the origin, and after a time t, let the angular position be such that the coil makes an angle b with the Q-axis. Since a coil span is 180°, total flux y linked by the coil is given by p +b

Ú

y=

2 N Bm lr

b

1 sin q d q P

1 p +b - cos q ]b [ P 1 = 4 N Bm lr cos b P 4 = N Bm lr cos w t P = 2 N Bm lr

[∵ b = w t]

where w = 2p f, f being the frequency of alternation of induced voltage. We know that induced emf is the rate of change of flux linkage. Thus, instantaneous value of induced emf is dy d È4 ˘ = - Í NBm lr cos w t ˙ dt dt Î P ˚ 4 d [cos w t] = N Bm lr P dt 4 = N Bm lr 2p f sin wt P 8 = p f N Bm lr sin w t P

E=

(5.13a)

We can also calculate the flux per pole as, f = Flux density ¥ Area under a pole p

= Ú Bm sin q lrd q 0

\

f=

2 P

4 Bm lr P

Substituting the value of f in Eq. (5.13), the value of included emf becomes e = 2p f Nf sin w t

(5.13b)

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The equation is of the form e = Em sin w t where

Em = (2p f N f).

The rms value of induced emf,

E=

\

Em 2 2p

f Nf = 4.44 f f N 2 E = 4.44 f f N volts

E=

or

(5.14)

The magnitude of induced emf in a stator (or rotor) having distributed winding is affected by distribution factor or breadth factor (kd). The effect of use of short-pitch coils on the magnitude of induced emf is expressed by pitch factor (kp). The expression for induced emf is to be multiplied by these two factors since they directly affect the magnitude of the induced emf. The values of (kd) and (kp) are less than one. For a concentrated winding with full-pitch coils, values of (kd) and (kp) are unity. Thus, the emf equation for a distributed winding can be expressed as E = 4.44 f f N kd kp volts = 4.44 f f N Kw volts where Kw is the winding factor

(5.15a) (5.15b)

[It is interesting to note that Eq. (5.14) is same as that of the transformer shown in Fig. 5.11.] When harmonics are present in the flux waveform then for nth harmonic emf per phase, En = 4.44 n f N f fn Kwn where fn is the flux per pole and Kwn is the winding factor for nth harmonic. If B1 and Bn are the peak flux densities for fundamental and nth harmonics then fn =

1 Bn f1 n B1 Kw n Bn

\

En = Kw1 B1 E1

The pole pitch for nth harmonic is

1 of the pole pitch for fundamental frequency. n f = fm sin wt

l0 V~

N1

E1

E2

N2

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!

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The emf induced in the transformer winding is given by dy df =–N dt dt d = – N (fm sin wt) dt

E=

pˆ Ê = – Nw fm cos wt = 2 N p f fm sin Á w t - ˜ Ë 2¯ pˆ Ê e = Em sin Á w t - ˜ Ë 2¯

or, where, The rms value of induced emf, \

Em = 2p f fm N E=

Em 2 2p

f fm N = 4.44 fm f N 2 E1 = 4.44 fm f N1 E2 = 4.44 fm f N2 E=

or and

(5.16)

where E1 and E2 are the emfs induced in the primary and secondary windings of turns N1 and N2 respectively for the transformer shown in Fig. 5.14. Induced emf equation for a coil rotating in a magnetic field being equal to the emf induced in the windings of a transformer indicates that the motion of a coil under a stationary flux-density wave produces the same voltage as for a time-varying flux-density wave linking a stationary coil. It is noted that the voltage induced in the coil 1–1¢ in Fig. 5.13 is alternating in nature. In case of dc machines, the alternating voltage is induced in the armature coils which is rectified by means of brush and commutator arrangement. The commutator (a mechanical rectifier) reverses the negative half cycle of the alternating voltage such that the output voltage will be a pulsating unidirectional wave. In practical dc machines, due to large number of armature coils in distributed armature slots, the output voltage wave shape becomes more or less a dc wave. We can calculate the average value of the fluctuating dc wave as p

Edc =

1 E sin w t d (w t ) p Ú0 m

Substituting the value of Em from Eq. (5.16), we get p

1 Edc = Ú w N f sin w t d (w t ) p0 = 4f f N volts

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NP where P is the number of poles. If 120 the armature of a dc machine has Z conductors and A parallel paths then the number of conductors per parallel path is Z/A. Induced emf is the emf per parallel path. Here, f can be expressed in terms of speed N in rpm as f =

Number of turns being N =

or

Z , the emf equation becomes 2A NP Z Edc = 4f 120 2 A fZNP Edc = volts 60 A

UPSRVF!JO!BD!NBDIJOFT!

(5.17)

6/9

6/9/2! Qspevdujpo!pg!Upsrvf! The behaviour of any rotating electrical machine can be described in terms of electromechanical torque. In the beginning, physical concept of torque production in rotating electrical machines is discussed. There are mainly two types of torques developed in electrical machines. One type is due to the interaction of the magnetic fields produced by the currents in two windings which are in relative motion with respect to each other. This type of torque is known as electromagnetic or interaction torque. The second type is dependent on the current in one winding only and is produced due to variations in the reluctance of the air gap in the magnetic circuit carrying the flux which links that winding. 2/!Fmfduspnbhofujd!Upsrvf!

.

N

S

.

.

When stator coils and rotor coils both carry currents in an electrical machine then the flux produced by the rotor current interacts with the flux produced by the stator current and results in the net magnetic flux. Due to the resultant magnetic flux, the rotor conductors experience a force and this torque developed due to the interaction of the stator and rotor magnetic fields is called electromagnetic torque.

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!

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Figure 5.15 shows a current-carrying coil on the rotor. The direction of the current in the rotor coil under the influence of the stator north pole is opposite to the direction of the current in the other rotor coil which is under the influence of the south pole. The directions are indicated by dot and cross respectively. The stator north pole repels the rotor north pole and attracts the rotor south pole resulting in clockwise torque. Similarly, the stator south pole repels the rotor south pole and attracts rotor north pole again resulting in clockwise torque. The developed torque is known as electromagnetic torque. The magnitude of the electromagnetic torque developed in all rotating machines is given by Te μ Stator field strength ¥ Rotor field strength ¥ sin d where d is the angle between the stator field axis and the rotor field axis. 3/!Sfmvdubodf!Upsrvf! When a piece of magnetic material is free to move in a magnetic field, it tends to align itself with the magnetic field to minimize the reluctance of the magnetic circuit. In other words, the piece of magnetic material will orient itself toward the magnetic pole creating the field. The torque created in this way on the rotor is called reluctance torque or alignment torque.

N

S

Gjh/!6/27! Qspevdujpo!pg!sfmvdubodf!upsrvf

Figure 5.16 shows two salient stator poles carrying currents and producing magnetic flux. A ferromagnetic iron bar is placed in between the two stator poles. Since the magnetic flux has a tendency to follow minimum reluctance path or shorten the flux path, the rotor experiences a counterclockwise torque so that the rotor axis coincides with the stator magnetic axis and the reluctance offered to the magnetic flux is minimum. This torque developed to reduce the reluctance of the magnetic path is called reluctance torque or alignment torque. This torque is produced only when there is a change in the reluctance offered to the magnetic flux with the change in the position of the rotor. The reluctance torque becomes zero when the rotor axis coincides with stator polar axis.

6/9/3! Upsrvf!Frvbujpo! Let us assume a cylindrical rotor machine as shown in Fig. 5.17. The machine can be assumed to have two sets of windings; one on the stator and the other on the rotor, producing magnetic fields in the air gap. The torque is produced due to the interaction of these two magnetic fields.

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In deriving the torque equation, we assume that the tangential component of the magnetic field in the air gap is negligible as compared to the radial component (i.e. the flux goes straight across the gap). Also, the radial length of the air gap is small compared to the radius of the rotor or stator. The flux density at the rotor N surface (or at the stator surface) or at any intermediate radial Fs distance in the air gap is thus same. The resultant air-gap field S N is then the radial field (H) or (B) whose intensity varies with S the angle around the periphery. Ft Figure 5.17 shows the axes of the stator field (Fs) and rotor field (Fr). The angle between the two axes is (dsr). The phasor Gjh/!6/28! Dzmjoesjdbm!spups!tztufn diagram is shown in Fig. 5.18 where Fs and Fr represent the peak values of the stator and rotor mmf waves and the phasor Fsr represents the peak value of their resultant. Application of the trigonometric formula for the diagonal of a parallelogram yields Also, and

Fsr2 = Fs2 + Fr2 + 2Fs Fr cos dsr Fs sin dsr = Fsr sin dr Fr sin dsr = Fsr sin ds

Fs

90 – dsr

dr

sin

(5.18)

d sr

90 –

Fs

d sr

ds dsr

Fs

sin

d sr

Fr sin dsr Fsr

Fr r

Fs

sin

d sr

Gjh/!6/29! Qibtps!ejbhsbn!tipxjoh!sfmbujpotijq!pg!tubups!gjfme!Gt!spups!gjfme!Gs!boe!uifjs!sftvmubou!Gts

Now let us consider the magnetic field co-energy stored in the air gap. By definition, co-energy is the area under the BH curve and represents the energy converted into mechanical work. The coenergy density at the point of magnetic field intensity H, is given by 1 1 BH or mo H 2 (since B = moH) 2 2 The average value of energy density =

mo (average value of H 2) 2

!

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Fmfdusjdbm!Nbdijoft

The average value of the square of the sine wave is half its square of peak value. Since

Magnetic force Length of flux path F Hm = sr (neglecting reluctance of iron path) g H=

Average co-energy density =

mo H m2 mo Ê Fsr ˆ = 2 2 4 ÁË g ˜¯

2

Total co-energy = Average co-energy density ¥ Volume of air gap 2

Wf l d

m ÊF ˆ = o Á sr ˜ p Dl . g 4 Ë g ¯ m p Dl 2 = o Fsr 4g

(

mop Dl 2 Fsr + Fr2 + 2 Fs Fr cos d sr 4g

)

(5.19)

where D is the average diameter at the air-gap, l is axial length, g is the air-gap clearance and mo is the permeability of air. Equation (5.19) is the energy being converted into mechanical work. An expression for the electromagnetic torque T can now be obtained in terms of the interacting magnetic fields by taking the partial derivatives of the field co-energy with respect to the angle. For a two-pole machine, T=

∂(W fld ) ∂d sr

=

mop Dl Fs Fr sin dsr 2g

(5.20)

The torque for a P-pole machine is T=

P mop Dl Fs Fr sin dsr 2 2g

(5.21)

We know that the torque is produced by the interaction of rotor field and the resultant air-gap field. Substituting Fs sin dsr = Fsr sin dr in Eq. (5.21), we get P mop Dl Fs Fsr sin dr 2 2g F Bsr = m o Hsr = m o sr g B g Fsr = sr mo T=

We know that

(5.22)

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6/3:

Here, Bsr is the maximum value of flux density. Average value of flux density =

2 Bsr p f=

Flux or

Bsr =

2 p Dl Bsr ¥ P p f Pp f P 2p Dl 2 Dl

Substituting this value in the expression of Fsr, Fsr =

We get

fP g 2 Dl mo

Substituting Fsr in Eq. (5.22), 2

Ê Pˆ p f Fr sin d r T=Á ˜ Ë 2¯ 2

(5.23)

where f is the resultant air-gap flux produced by the combined effect of the stator and rotor mmfs. From the above equation, it is seen that torque is proportional to the interacting fields and the sine of the electric space angle between their magnetic axis. Equation (5.23) is the generalised torque equation. Let us now apply this equation to different types of machines and try to explain their behaviour. 2/!Upsrvf!Frvbujpo!gps!b!ed!Nbdijof! The essential features of a dc machine are shown in Fig. 5.19. The stator has salient poles with excited field coils. Fr (Q-axis) . . .

. . .

Armature

N

.

.

Pole

N

.

.

.

f (D-axis) S

Gjh/!6/2:! Fttfoujbm!gfbuvsft!pg!b!ed!nbdijof!tipxjoh!uif!gjfme!byjt!)E.byjt*!boe!uif!bsnbuvsf!nng! byjt!)R.byjt*

The air-gap flux distribution created by the field windings is symmetrical about the centre line of the field poles. This axis is called the field axis or direct axis. The brushes are located so that commutation occurs when the coil sides are in the neutral zone, midway between the field poles.

!

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Fmfdusjdbm!Nbdijoft

The axis of the armature mmf wave then is 90° electrical from the axis of the field poles, i.e. in the quadrature axis. The torque can be expressed in terms of the interaction of the direct-axis air-gap flux per pole f and the space fundamental component Fr1 of the armature mmf wave. With the brushes in the quadrature axis the angle between these fields is 90° electrical. Thus, referring to the generalized torque equation shown in Eq. (5.23), the torque equation for the dc machine becomes 2

p ÈP˘ T = Í ˙ f Fr1 sin 90∞ 2 Î2˚ 2

=

p ÈP˘ f Fr1 2 ÍÎ 2 ˙˚

(5.24)

The armature mmf of a distributed armature winding of a dc machine is triangular in shape and is shown in Fig. 5.20. A triangular mmf wave can be represented by fundamental sine wave and component harmonic waves. We will consider only a fundamental wave whose maximum value is 8/p 2 of the maximum value of the triangular wave. If Z is the total number of conductors, A is the number of parallel paths, P is the number of poles and i is the armature current then F = Ampere turns/Pole = The fundamental component

N

Fr1 =

8

Fr p2 8 Zi Fr1 = 2 p 2 PA

B 0

q N

S mmf

p Flux Density Wave Shape Due to Field Poles p

0 S

Zi 2 PA

2p

2p

(Armature mmf wave)

Gjh/!6/31! Gmvy.efotjuz! xbwf! boe! bsnbuvsf! nng! xbwf! pg! b! ejtusjcvufe! bsnbuvsf! xjoejoh! pg! b! ed! nbdijof!

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Substituting this value in Eq. (5.24), we get 2

8 Zi p Ê Pˆ T= Á ˜ f 2 Ë ¯ 2 2 p 2 PA T=

PZ fi = Kf i 2p A

(5.25)

Multiplying numerator and denominator of RHS of Eq. (5.25) by N/60, ÊNˆ ÁË ˜¯ PZfi T = 60 ÊNˆ ÁË ˜¯ 2p A 60 or

2p NT fZNP = .i 60 60 A 2p NT = ei 60

(5.26)

This is the power which is transferred through the air gap to the rotor. 3/!Upsrvf!Frvbujpo!gps!b!Qpmzqibtf!Tzodispopvt!Nbdijof A synchronous machine is one in which alternating current flows in the armature winding and dc excitation is supplied to the field winding. The armature winding is on the stator and has a three-phase winding. The field winding is on the rotor as shown in Fig. 5.21. The dc power required for excitation is supplied through slip rings. While carrying balanced polyphase currents, the ar120 f mature winding will produce a rotating magnetic field rotating at synchronous speed Ns = . P But the field produced by the dc rotor winding revolves with the rotor. For production of a steady unidirectional torque, the rotating field of the stator and rotor must be travelling at the same speed. A synchronous motor connected to a constant frequency voltage source, therefore, operates at a constant steady-state speed regardless of load. The behaviour of a synchronous motor under running conditions can be explained in terms of the torque equation. Y¢ O R O O B¢ DC O B

Y O O R¢

Gjh/!6/32! Spubujoh!gjfme!boe!tubujpobsz!bsnbuvsf!tztufn

!

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Fmfdusjdbm!Nbdijoft

2

T=

p Ê Pˆ Á ˜ f Fr sin dr 2Ë 2¯

The voltage generated by the air-gap flux wave is nearly equal to the terminal voltage Vt, Vt = 4.44 f Nf i.e. f =

Vt = constant, if Vt and f are constants. The rotor mmf Fr is determined by the dc field 4.44 fN

current and is also constant. T = K sin d r

Therefore,

(5.27)

The variation in the torque requirements of the load on the machine is taken care of entirely by variation of torque angle d r, as shown by the torque angle curve in Fig. 5.22 in which positive value of torque represents motor action and the positive value of d r represents angle of lag of the rotor mmf wave. With a light shaft load, relatively small electromagnetic torque is required and d r is small. When more shaft load is added, the rotor must drop back in space phase with respect to the rotating flux wave just enough so that d assumes the value required to supply the necessary torque. The readjustment process is actually a dynamic one accompanied by a temporary decrease in the instantaneous mechanical speed of the rotor and a damped mechanical oscillation, called hunting of the rotor, about its new position. When d r is 90°, the maximum possible torque or power, called the pull-out torque or pull-out power, for a fixed terminal voltage and field current is reached. If the load requirement exceeds this value, the motor slows down under the influence of the excess shaft torque and synchronous motor action is lost because the rotor and stator fields are no longer stationary with respect to each other. This phenomenon is known as pulling out of step or losing synchronism. Motor

T

Pullout Torque

–180°

–90°

0

90°

180°

Torque Angle, dr Generator

Gjh/!6/33! Upsrvf.bohmf!dibsbdufsjtujdt!pg!b!tzodispopvt!nbdijof

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4/!Upsrvf!Frvbujpo!gps!Qpmzqibtf!Joevdujpo!Nbdijoft Like a synchronous motor, a polyphase induction motor is supplied with three-phase supply to its three-phase stator windings. The rotor is a closed winding and gets its excitation through electromagnetic induction. When a three-phase supply is supplied across the three phase windings of the stator, a rotating magnetic field fsr is produced whose magnitude remains constant if the supply and frequency remain constant. The rotor mmf Fr is proportional to the rotor current Ir. Thus, the original equation 2

p Ê Pˆ T = Á ˜ fsr Fr sin dr gets reduced to T = kIr sin d r 2Ë 2¯

(5.28)

The motor rotates at a speed N, somewhat less than the speed of the rotating magnetic field, Ns, because at synchronous speed of the rotor, there will be no induced current in the rotor as there will be no relative speed between the rotating field and the rotor. The difference of Ns and Nr expressed as percentage of Ns is called slip S. S=

Ns - Nr ¥ 100 Ns

(5.29)

Slip is about 3 to 5 percent. Various types of machines mentioned so far will be dealt with in detail separately.

TMPU!PS!UPPUI!IBSNPOJDT!

6/:

The use of distributed windings creates an additional problem in ac machines. The reluctance of the flux path through teeth is lesser as compared to that through slots. This causes conversion and fringing of flux at the tooth tips as shown in Fig. 5.23 and consequent ripples in the flux wave. The effect is similar to an additional flux fs (which pulsates at the frequency with which the teeth pass the pole tips) superimposed on the main flux. A complete cycle of this tooth frequency takes place in the time required for the pole to move through an angle equivalent to one tooth pitch (or slot pitch). If a three-phase machine has q slots per pole per phase, there will be 2(3 ¥ q) cycles of tooth frequency for each cycle of fundamental frequency f.

Gjh/!6/34! Uvgujoh!pg!gmvy!bu!uppui!ujqt

!

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Fmfdusjdbm!Nbdijoft

If fs is the tooth or slot frequency and S is the total number of slots then fs = f (Number of slots per pole pair) = f (2S/P)

(5.30)

If the field flux per pole is f and additional flux due to slot harmonics is fs then ftotal = (f + fs sin 2p fs t) cos 2 p f t

(5.31)

The induced emf is then given by e = –N

df total f N = 2p f Nf sin 2p f t – s dt 2 [2p ( fs – f ) cos 2p ( fs – f )t + 2p ( fs + f ) cos 2p ( fs + f )t]

(5.32)

The emf thus contains a component corresponding to fundamental frequency and two other components due to tooth ripples. The frequencies of these components are ( fs – f ) and ( fs + f ) or f (2S/P + 1). For a two-pole machine having 3 slots per pole per phase, the order of these harmonics will be 11th and 13th, corresponding to 550 and 650 Hz. Slot harmonics may cause many problems, such as the following: 1. When the generated voltage of synchronous machines contain these harmonics, it causes interference in telecommunication lines. 2. They introduce vibration and noise in the machine. 3. The interaction of stator and rotor slot harmonics produces parasitic torques in induction motor which may have a serious effect on the torque speed curve of the machine. 4. An increase in core losses by introduction of high-frequency components of voltages and currents into the teeth of the stator. The use of fractional slot winding and skewed rotor conductors reduces slot harmonics. In a fractional slot winding, the space relation between teeth and slot opposite to the given pole face is not the same as in the next and succeeding poles. Therefore, the emfs of slot ripple frequencies (i.e. fs + f ) in different armature coils are out of phase and tend to cancel out. The skewed rotor conductors are used in induction motors. It is important to note that since the slot harmonics are set by the spacing between adjacent coil slots, variation in coil pitch and distribution cannot reduce these harmonics.

! Qspcmfn!6/9 Efufsnjof!uif!tqffe!pg!uif!nbhofujd!gjfme!qspevdfe!jo!b!4.qibtf!xjoejoh!ejtqmbdfe!jo!tqbdf!cz!231¡! xifo!tvqqmjfe!gspn!b!4.qibtf!61!I{!cbmbodfe!bd!tvqqmz/!Bttvnf!uif!ovncfs!pg!qpmft!up!cf!9/

Solution The speed of the rotating magnetic field = Synchronous speed =

120 f , where f = 50 Hz and P = 8 P

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6/46

\ speed of the magnetic field =

120 ¥ 50 = 750 rpm. 8

! Qspcmfn!6/: Jg!uif!spups!pg!b!7.qpmf!4.qibtf!joevdujpo!npups!svot!bu!b!tqffe!pg!:71!sqn-!efufsnjof!uif!tmjq!dpo. tjefsjoh!uif!tvqqmz!gsfrvfodz!up!cf!61!I{/

Solution P=6 f = 50 \ synchronous speed Ns =

120 f 120 ¥ 50 = = 1000 rpm P 6

Speed of the rotor Nr = 960 rpm \

1000 - 960 Ns - Nr ¥ 100% = ¥ 100% 1000 Ns = 4%.

slip =

! Qspcmfn!6/21 Jo!b!5.qpmf!tzodispopvt!nbdijof-!uif!spups!tqffe!jt!2611!sqn/!Uif!nbdijof!ibt!71!tmput!xjui!9! dpoevdupst!qfs!tmpu/!Jg!uif!upubm!gmvy!jt!1/15!Xc!boe!uif!xjoejoh!gbdups!jt!1/:6-!gjoe!uif!hfofsbufe! fng!qfs!qibtf/

Solution P=4 Synchronous speed Ns = 1500 rpm If the frequency be f then

or

120 f = 1500 4 f = 50 Hz

Total number of conductors = 60 ¥ 8 = 480 480 \ total number of turns = 2 480 = 80 Number of turns per phase N = 2¥3 Winding factor Kw = 0.95

!

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Generated emf per phase E = 4.44 Kw f f N = 4.44 ¥ 0.95 ¥ 0.04 ¥ 50 ¥ 80 = 674.88 V.

! Qspcmfn!6/22 Efufsnjof!uif!ovncfs!pg!qpmft!pg!b!61!I{!4.qibtf!joevdujpo!npups!xijdi!svot!bu!b!tqffe!pg!831! sqn/!

Solution f = 50 Nr = 720 rpm Considering Nr to be equal to synchronous speed, 120 ¥ 50 , where P is the number of poles P 120 ¥ 50 1 =8 P= 720 3

Nr = 720 = \

Since the number of poles is an even and whole number, it may be 8 or 10. If P = 8 then 120 ¥ 50 = 750 rpm which is greater than the speed of the induction motor. Ns = 8 If P = 10 then Ns = 600 rpm which is not possible as in an induction motor, the rotor speed is always less than the synchronous speed. \ number of poles is 8.

! Qspcmfn!6/23 B!4.qibtf!npups!jt!esjwjoh!b!qvnq/!Uif!tqffe!pg!uif!npups!voefs!mpbefe!dpoejujpo!jt!859!sfw0njo! boe!voefs!vompbefe!dpoejujpo!jt!831!sfw0njo/!Tubuf!xifuifs!uif!npups!jt!b!tzodispopvt!npups!ps! joevdujpo!npups!boe!gjoe!pvu!uif!ovncfs!pg!qpmft!jo!uif!npups/

Solution If 748 rpm is the synchronous speed then 120 f 748 = , where f = 50 Hz. and P is the number of poles. P 120 ¥ 50 = 8.02 \ P= 748 As the number of poles cannot be a fraction, hence it is not a synchronous motor, it is an induction motor.

Gvoebnfoubmt!pg!BD!Spubujoh!Nbdijoft

For

6/48

P=8

Synchronous speed =

120 ¥ 50 = 750 rpm 8

Under both loaded and unloaded condition, the motor is running below synchronous speed. The 748 ˆ Ê slip varies from 0.0027 Á = 1 ˜ to 0.04 Ë 740 ¯

720 ˆ Ê ÁË = 1 ˜. 750 ¯

! Qspcmfn!6/24 B!4.qibtf-!3.qpmf!nbdijof!jt!fydjufe!cz!b!cbmbodfe!4.qibtf!61!I{!wpmubhf/!Uif!nng!efwfmpqfe!jo! uif!b!qibtf!xjoejoh!jt!bt!gpmmpxt; Gb!>!jb!)L2!dpt!qb!,!L4!dpt!4qb!,!L6!dpt!6qb* Tjnjmbs!fyqsfttjpot!dbo!cf!xsjuufo!gps!qibtft!c!boe!d/!Dbmdvmbuf!uif!uisff.qibtf!nng/

Solution Fa = ia [K1 cos qa + K3 cos 3qa + K5 cos 5 qa] = Ia cos wt [K1 cos qa + K3 cos 3 qa + K5 cos 5 qa] Fb = ib [K1 cos (qa – 120°) + K3 cos 3 (qa – 120°) + K5 cos 5 (qa – 120°)] = Ia cos (wt – 120°) [K1 cos (qa – 120°) + K3 cos 3 (qa – 120°) + K5 cos 5 (qa – 120°)] Fc = ic [K1 cos (qa – 240°) + K3 cos 3 (qa – 240°) + K5 – cos 5 (qa – 240°)] = Ia cos (wt – 240°) [K1 cos (qa – 240°) + K3 cos 3 (qa – 240°) + K5 cos 5 (qa – 240)] Total mmf = Fa + Fb + Fc = Ia cos wt [K1 cos qa + K3 cos 3 qa + K5 cos 5 qa] + Ia cos (wt – 120°) [K1 cos (qa – 120°) + K3 cos 3 qa + K5 cos 5 (qa – 120°)] + Ia cos (wt + 120°) [K1 cos (qa + 120°) + K3 cos 3 qa + K5 cos 5 (qa + 120°] = Ia cos wt [K1 cos qa + K5 cos 5 qa] + Ia cos (wt – 120°) [K1 cos (qa – 120°) + K5 cos 5 (qa – 120°)] + Ia cos (wt + 120°) [K1 cos (qa + 120°) + K5 cos 5 (qa + 120°)] + K3 Ia cos 3qa [cos wt + cos (wt – 120°) + cos (wt + 120°)] Now cos wt + cos (wt – 120°) + cos (wt + 120°) = 0 \ total mmf 3 Ia [K1 cos (qa – wt) + K5 cos (5 qa + wt)] 2 wt ˆ ˘ 3 È Ê = Ia Í K1 cos (q a - w t ) + K5 cos 5 Á q a + ˜ ˙ . Ë 5 ¯˚ 2 Î =

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! Qspcmfn!6/25 B!ed!nbdijof!qspevdft!35!W!xifo!pqfsbufe!bu!b!tqffe!pg!2311!sqn/!Cz!xibu!gbdups!tipvme!uif!ovn. cfs!pg!bsnbuvsf!uvsot!cf!dibohfe!tvdi!uibu!gps!uif!tbnf!gjfme!gmvy!qfs!qpmf-!uif!hfofsbups!pvuqvu! wpmubhf!xjmm!cf!29!W!bu!2611!sqn!tqffe@

Solution From Eq. (5.17), the voltage output of a dc machine is V=

Pf ZN 60 A

The number of armature turns is proportional to the number of conductors Z. \ V = KTN where T is the number of turns and K is a constant. Now, V1 = 24 V V2 = 18 V

N1 = 1200 rpm N2 = 1500 rpm

\

T1 =

V1 V 24 18 = and T2 = 2 = KN1 1200 K KN 2 1500 K

\

T2 =

18 1200 3 ¥ T2 = T2 24 1500 5

3 Hence, the armature turns should be changed by a factor of . 5

! Qspcmfn!6/26 Uif!upubm!ovncfs!pg!bsnbuvsf!dpoevdupst!pg!b!3.qpmf!ed!hfofsbups!jt!431/!Uif!pqfo.djsdvju!wpmubhf! hfofsbufe!jt!351!W!xifo!uif!tqffe!pg!uif!bsnbuvsf!jt!2911!sqn/!Efufsnjof!uif!bjs.hbq!gmvy!qfs! qpmf/

Solution Generated voltage in a dc machine E=

Pf ZN where f is the flux per pole 60 A

As P = 2, A = 2 for either lap-connected or wave-connected machine. \

240 =

\

f=

2f ¥ 320 ¥ 1800 60 ¥ 2 240 ¥ 60 ¥ 2 = 0.025 Wb = 25 mWb. 2 ¥ 320 ¥ 1800

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6/4:

! Qspcmfn!6/27 B!7.qpmf!61!I{!tzodispopvt!nbdijof!ibt!b!spups!mfohui!pg!2/:8!n!boe!b!spups!sbejvt!pg!69!dn/! Efufsnjof!uif!sbufe!pqfsbujoh!tqffe!pg!uif!nbdijof/!Bmtp!dbmdvmbuf!uif!gmvy!qfs!qpmf!jg!uif!gvoeb. nfoubm!bjs.hbq!gmvy!efotjuz!jt!2/34!Xc0n3/

Solution Rated operating speed of the machine = Synchronous speed 120 ¥ 50 = = 1000 rpm 6 From Eq. 5.13(a), Flux per pole

f=

4 Bm lr P

where P = 6, Bm = 1.23 Wb/m2, l = 1.97 m, r = 0.58 m \

f=

4 ¥ 1.23 ¥ 1.97 ¥ 0.58 Wb = 0.937 Wb. 6

! Qspcmfn!6/28 Bttvnf!uibu!uif!dpjmt!jo!uif!tzodispopvt!nbdijof!pg!Qspcmfn!6/27!bsf!gvmm.qjudi-!22.uvso!dpjm!qfs! qpmf!qbjs!xjui!uif!dpjmt!dpoofdufe!jo!tfsjft/!Dbmdvmbuf!uif!hfofsbufe!wpmubhf!qfs!qibtf/

Solution There are 6 poles in the machine, i.e. 3 pole pairs. \ total number of turns N = 11 ¥ 3 = 33 f = 0.937 Wb f = 50 Hz From Eq. (5.14), Generated voltage per phase E = 4.44 f f N = 4.44 ¥ 0.937 ¥ 50 ¥ 33 = 6864 V = 6.864 kV.

! Qspcmfn!6/29 Uif!tzodispopvt!nbdijof!pg!Qspcmfn!6/27!ibt!uisff!qibtf!xjoejoh!xjui!56!tfsjft!uvsot!qfs!qibtf! boe!xjoejoh!gbdups!Lw!>!1/:39/!Gps!uif!tbnf!gmvy!dpoejujpo!boe!sbufe!tqffe!pg!Qspcmfn!6/27-!efufs. njof!uif!snt!wpmubhf!qfs!qibtf/

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Solution From Eq. (5.15a) generated emf, E = 4.44 f f N Kw Here, Kw = 0.928, f = 50, N = 33, f = 0.937 Wb \ E = 4.44 ¥ 0.937 ¥ 50 ¥ 33 ¥ 0.928 = 6370 V = 6.37 kV.

! Qspcmfn!6/2: Uif!4.qibtf!tzodispopvt!nbdijof!pg!Qspcmfn!6/27!jt!opx!pqfsbufe!gspn!71!I{!tvqqmz/!Ju!jt!hjwfo! uibu!gps!uif!pqfsbujoh!dpoejujpo!dpotjefsfe!jo!Qspcmfn!6/27!uif!snt!hfofsbufe!wpmubhf!jt!frvbm!up! 24!lW!mjof!up!mjof/!Ifodf-!uif!nbdijof!bsnbuvsf!nvtu!cf!sfxpvoe!xjui!b!ejggfsfou!ovncfs!pg!uvsot/! Dpotjefsjoh!xjoejoh!gbdups!pg!Lw!>!1/:39-!efufsnjof!uif!sfrvjsfe!ovncfs!pg!tfsjft!uvsot!qfs!qibtf/

Solution The rms phase voltage

E=

13, 000 3

V = 7505.77 V

F = 60 Hz Kw = 0.928 f = 0.937 Wb \

E = 4.44 f f N Kw

or,

N=

7505.77 = 144 0.937 ¥ 60 ¥ 0.928

\ number of series turns per phase is 144.

Sfwjfx!Rvftujpot 1. Define distribution factor and pitch factor. Derive the expression for these two factors. 2. What is the difference between (i) single-layer and double-layer winding, and (ii) integral slot and fractional slot winding? 3. Derive an expression for the emf generated in an N-turn full-pitch coil when the coil is rotated in a magnetic field of sinusoidally distributed flux density. 4. Explain how a rotating magnetic field is produced when a Three-phase winding is supplied from a three-phase balanced voltage. 5. Derive the relation between electrical and mechanical degree. 6. What are the advantages of distributed winding and fractional-pitch winding? 7. Explain electromagnetic torque and reluctance torque with suitable diagrams.

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8. Show that the variation of torque in a synchronous machine is entirely dependent on the variation of the torque angle. 9. What are slot harmonics? What are the problems caused by slot harmonics? How can they be reduced?

Qspcmfnt 1. A 3-phase 50 Hz, 4-pole star-connected synchronous machine has a single-layer winding in 36 slots with 30 conductors per slot. Determine the line voltage and the synchronous speed if the winding is full pitched and flux per pole is 0.05 Wb. [3322 V, 1500 rpm] 2. A 4-pole ac machine has a 3-phase winding wound in 60 slots. The coils are short pitched in such a way that if one coil side lies in slot 1, the other side of the same coil lies in slot 13. Determine the winding factor for the fundamental and third harmonics. [0.91 and 0.38] 3. A 16-pole, 3-phase synchronous machine has an air-gap flux of 0.06 Wb per pole. The stator has 2 slots per pole per phase and 4 conductors per slot are accommodated in two layers. The coil span is 150° electrical. Determine the phase voltage generated in the machine when the machine runs at 375 rpm. [795.3 V] 4. A 6-pole synchronous machine rotating at 1000 rpm has a single-phase winding housed in 3 slots per pole, the slots in groups of three being 20° apart. If each slot contains 10 conductors and the flux per pole is 20 mWb, determine the voltage generated in the machine. [383.5 V] 5. A 3-phase 10 pole synchronous machine with full-pitch coils has 16 slots per pole. Each coil has 20 turns and the machine has a single-layer winding. The current per conductor is 75 A and the flux per pole is 25 mWb. Determine the kVA output of the stator winding if the machine runs at 600 rpm. [636.5 kVA] 6. The line voltage of a star-connected three-phase machine is 400 V. Due to the presence of third harmonic voltage, the phase voltage is 244 V. Determine the value of the third harmonic voltage in the machine. [78.73 V]

Nvmujqmf.Dipjdf!Rvftujpot 1. Distributed winding and short chording winding employed in ac machines will result in (a) increase in emf and reduction in harmonics (b) reduction in emf and increase in harmonics (c) increase in both emf and harmonics (d) reduction in both emf and harmonics [GATE 2009] 2. It is desirable to eliminate 5th harmonic voltage from the phase voltage of an alternator. The coils should be short pitched by an electrical angle of (a) 30° (b) 36° (c) 72° (d) 18° [GATE 2001]

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[Hints: In order to eliminate any harmonic the pitch factor corresponding to that harmonic should be 0. 5a \ pitch factor corresponding to 5th harmonic KP = cos = 0 = cos 90° 2 2 \ a = 90° ¥ = 36°] 5 3. A rotating electrical machine having its self inductances of both the stator and the rotor windings, independent of the rotor position will definitely not develop (a) starting torque (b) synchronizing torque (c) hysteresis torque (d) reluctance torque [GATE 2004] 4. An electric motor with ‘constant output power’ will have a torque-speed characteristic in the form of a (a) straight line through the origin (b) straight line parallel to the speed axis (c) circle about the origin (d) rectangular hyperbola [GATE 2001] [Hints: P = Eb Ia = Tw = constant. Hence TN = constant] 5. Following are some of the properties of rotating electrical machines: P. Stator winding current is dc, rotor winding current is ac Q. Stator winding current is ac, rotor winding current is dc R. Stator winding current is ac, rotor winding current is ac S. Stator has salient poles and rotor has commutator T. Rotor has salient poles and slip rings, and stator is cylindrical m. Both stator and rotor have polyphase windings. Synchronous machines, dc machines and induction machines exhibit some of the above properties as given in the following table. Indicate the correct combination from this table. DC machines

Synchronous machines

Induction machines

(a)

P–S

Q–T

R–U

(b)

Q–U

P–T

R–S

(c)

P–S

R–U

Q–T

(d)

R–S

Q–U

P–T

[Hints: In dc machine armature or rotor winding current is ac and stator or field winding current is dc. The rotor has commutator and stator has salient poles. In synchronous machines field is the rotor part which carries dc current and stator or armature carries ac current. Stator is cylindrical and rotor has salient poles and slip rings. In induction machine both stator and rotor current is ac and both have three phase ac winding]

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6. A 4-pole, three-phase, double-layer winding is housed in a 36-slot stator for an ac machine with 60° phase spread. Coil span is 7 slot pitches. Number of slots in which top and bottom layers belong to different phases is (a) 24 (b) 18 (c) 12 (d) zero [GATE 2003] [Hints: S = 36 36 Slots/pole = =9 4 \ the coils are short pitched by 9 – 7 = 2 slots Number of slots/pole/phase =

36 =3 4¥3

In 3 slots, number of slots containing different phases = 2 \ in 36 slots, the number is

2 ¥ 36 = 24] 3

7. A 500 MW, 3-phase, Y-connected synchronous generator has a rated voltage of 21.5 KV at 0.85 p.f. The line current when operating at full load rated conditions will be (a) 13.43 kA (b) 15.79 kA (c) 23.25 kA (d) 27.36 kA [GATE 2004] IL =

[Hints:

500 ¥ 106 3 ¥ 21.5 ¥ 103 ¥ 0.85

= 15796 A = 15.79 kA]

8. Match the following List I with List II and select the correct answer using the codes given below the lists List I

List II

P.

Armature emf (E)

1.

Flux (f), speed (w) and armature current (Ia)

Q.

Developed torque (T)

2.

f and w

R.

Developed power (P)

3.

f and Ia

4.

Ia and w

5.

Ia only

(a) (b) (c) (d)

P 3 2 3 2

Q 3 5 5 3

R 1 4 4 1

[GATE 2005]

!

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[Hints:

E=

Pf ZN Pf Zw = Kfw = 2p A 60 A

Eb I a K fw I a = = K¢f Ia w w P = Eb Ia = Kfw Ia] T=

9. When stator and rotor windings of a 2-pole rotating electrical machine are excited, each would produce a sinusoidal mmf distribution in the air gap with peak values F1 and F2 respectively. The rotor mmf lags stator mmf by a space angle d at any instant as shown in figure. Thus, half of stator and rotor surfaces will form one pole with the other half forming the second pole. Further, the direction of torque acting on the rotor can be clockwise or counter-clockwise. The table given below gives four sets of statements as regards to poles and torque. Select the correct set corresponding to the mmf axes as shown in the figure. Stator

C

Air gap

c B

Rotor

D F2

b

d a

d

F1 Rotor mmf axis

A

Stator surface ABC forms

Stator mmf axis

Stator surface CDA forms

Rotor surface CDA forms

Rotor surface CDA forms

Torque is

(a)

North pole

South pole

North pole

South pole

Clockwise

(b)

South pole

North pole

North pole

South pole

Counter clockwise

(c)

North pole

South pole

South pole

North pole

Counter clockwise

(d)

South pole

North pole

South pole

North pole

Clockwise

[GATE 2003] Common data for Questions 10 and 11 A 4-pole, 50 Hz, synchronous generator has 48 slots in which a double-layer winding is housed. Each coil has 10 turns and is short pitched by an angle to 36° electrical. The fundamental flux per pole is 0.025 Wb. [GATE 2006] 10. The line-to-line induced emf (in volt), for a three-phase star connection is approximately (a) 808 (b) 888 (c) 1400 (d) 1538 [Hints:

q=

48 =4 4¥3

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y=

6/56

180∞ 180∞ = 15° = Slots/pole 48/ 4

4 ¥ 15∞ q/ y sin 2 = 2 = 0.9576 Kd = 15∞ y 4 sin q sin 2 2 sin

KP = cos

36∞ = 0.95 2

In double-layer winding, the number of coils is equal to number of slots \

10 ¥ 48 = 160 3 = 4.44 Kd KP f f N = 4.44 ¥ 0.9576 ¥ 0.95 ¥ 0.025 ¥ 50 ¥ 160

N= EPh

= 807.83 V \

EL = 3EPh = 1400 V]

11. The fifth harmonic component of phase emf (in volt) for a three-phase star connection is (a) Zero (b) 269 (c) 281 (d) 808 na 5 ¥ 36∞ ˘ È Í Hints: cos 2 = cos 2 = cos 90∞ = 0 ˙ ˚ Î 12. In synchronous machines, the phase spread is (a) 360° (b) 120° (c) 60° (d) 30° 13. The distribution factor for a three-phase ac winding is about (a) 0.85 (b) 0.75 (c) 0.95 (c) 0.5 14. The full-pitch coil in a synchronous machine, has a span of 18 slots. To eliminate third harmonic, the coil span should be (a) 18 slots (b) 15 slots (c) 12 slots (d) 9 slots [Hints:

a=

180∞ = 10° for full pitch coil 18

To eliminate third harmonic, chording angle is \ coil span should be reduced by \ coil span is 18 – 6 = 12 slots]

1 ¥ 180° = 60° 3

60∞ or 6 slots 10∞

!

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15. An ac winding has 2 slots per pole per phase. The slot harmonics will be (a) 17th and 19th (b) 5th and 7th (c) 11th and 13th (d) none of the above S =2¥3=6 P 2S ± 1 = (2 ¥ 6) ± 1 = 12 ± 1 i.e. 11 or 13] Slot harmonics = P

[Hints:

16. For a P pole machine, the relation between electrical and mechanical degrees is given by (a) qelec =

2 qmech P

(b) qelec =

4 qmech P

P qelec 2

(d) qelec =

P qmech 2

(c) qmech =

17. For eliminating nth harmonic from the emf generated in the phase of a 3 phase alternator, the chording angle should be 2 (a) n ¥ full pitch (b) ¥ full pitch n 3 1 (c) ¥ full pitch (c) ¥ full pitch n n th 18. For eliminating 5 harmonic from the phase emf generated in an alternator, the coil span in terms of full pitch (or pitch fraction) would be 4 5 5 (c) 6

5 4 6 (d) 5

(a)

[Hints: \

(b)

5a = 0 = cos 90° or, a = 36° 2 Coil span = 180° – 36° = 144° cos

144∞ 4 ˘ = 180∞ 5 ˙˚ 19. A uniformly distributed winding on the stator has three full pitched coils, each coil having N turns and each turn carrying a current i. The mmf produced by this winding is (a) sinusoidal in waveform with an amplitude 3Ni/2 (b) sinusoidal in waveform with an amplitude 3Ni (c) trapezoidal in waveform with an amplitude 3Ni/2 (d) trapezoidal in wave form with an amplitude 3Ni or

coil span =

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20. Mmf produced by one N turn coil carrying a current i is (a) rectangular of amplitude Ni Ni (b) rectangular of amplitude 2 (c) trapezoidal of amplitude Ni Ni (d) trapezoidal of amplitude 2 21. For eliminating nth harmonic from the emf generated in one phase of a 3 phase alternator, the coil span (pitch of the coils) must be n -1 ¥ full pitch n 2n (c) ¥ full pitch n +1

n ¥ full pitch n +1 2n - 1 ¥ full pitch (d) n

(a)

[Hints:

\

(b)

For eliminating nth harmonic, the chording angle 1 a = ¥ full pitch n 1 Coil span = Full pitch – full pitch n 1 Ê ˆ = Á1 - ˜ ¥ full pitch] Ë n¯

22. In all rotating electrical machines, electrical torque is developed when relative speed between stator field and rotor field is (a) equal to rotor speed (b) zero (c) equal and opposite to rotor speed (d) dependent upon the type of electrical machine

Botxfst 1. 5. 9. 13. 17. 21.

(d) (a) (c) (c) (d) (a)

2. 6. 10. 14. 18. 22.

(b) (a) (c) (c) (a) (b)

3. 7. 11. 15. 19.

(b) (b) (a) (c) (c)

4. 8. 12. 16. 20.

(d) (d) (c) (d) (b)

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A synchronous generator, commonly known as alternator, is the most commonly used machine for the generation of electrical power throughout the world. It generates alternating voltage which is stepped up and transmitted. Like other electrical rotating machines, a synchronous generator has two main components, viz. stator and rotor. The armature winding is placed on the stator and the field poles are placed on the rotor which are excited from a dc source. When the rotor is supplied with mechanical energy by coupling it to a prime mover engine or turbine, it generates electrical energy in the stator. A synchronous generator rotates at a speed fixed by the supply frequency and the number of poles, which is known as the synchronous speed. It is also called doubly excited ac machine because its field winding is always energized from a dc source in addition to the mechanical energy input to the rotor.

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A synchronous generator has dc winding on its rotor. Pairs of poles are formed on the rotor as shown in Fig. 6.1(a). Hence, there may be 2, 4, 6, etc. pole machines. They have three-phase windings A, B and C placed on the stator slots. These identical coils are with their axis at 120° to each other. During rotation of the rotor, its north and south pole pass by the stator windings A, B, C alternately. This gives the effect of a rotating magnetic field cutting the stator conductors. This induces a single-phase alternating voltage in each winding. These three alternating voltages induced have a time pattern as shown in Fig. 6.1(b). They are displaced from each other in time by one-third of the time in which one pole pair passes through one winding. When N pole is crossing a winding,

!

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Fmfdusjdbm!Nbdijoft

one polarity of emf is induced, and when S pole is crossing, the opposite polarity of emf is induced. Since one cycle time is represented by 360°, hence one-third of it is 120°. So the voltages in the three windings A, B and C are each shifted by 120°. A pole pair crosses the winding B 120° after it has crossed A and crosses C after 120° after B. This is known as the phase difference between voltages of the windings as depicted in Fig. 6.1(c). 120° VA

VB

B

VC

N

V

120° 120° t

120°

VC (a)

VC

12

0° 12

C

VB

0°

S

A

VA

(b) Time pattern of voltages of a three-phase system

VA 120°

VB (c) Phasor diagram of stator voltages

Gjh/!7/2! Qsjodjqmft!pg!uisff.qibtf!tzodispopvt!hfofsbups

Since crossing of one rotor pole pair across one stator phase produces one cycle of emf in that P phase, if there are P poles in a rotor then cycles of emf are induced in each phase per rotation 2 of the rotor. If rotor speed is Ns rpm or ns rps then the number of emf cycles induced per second in each phase P N P N P f = ns ¥ = s ¥ = s …(6.1) 2 60 2 120 or,

Ns =

120 f P

where f is supply frequency and Ns is the speed of synchronous machine in revolutions per minute. A low-speed machine has a large number of poles while a high-speed machine has less number of poles.

DPOTUSVDUJPO!

7/4

A synchronous machine has basically the following parts.

7/4/2! Gsbnf! Frame is the protective covering for the whole machine. It is usually built from cast iron or steel to which the stator core is damped. The frame is designed not to carry the flux but to provide mechanical support to the generator.

Tzodispopvt!Hfofsbups

7/4

7/4/3! Tubups!Dpsf! The stator core is made up of thin laminations of highly permeable steel in order to reduce iron losses. It carries the alternating flux and houses the stator windings. The axial length of the stator core is comparitively short for slow-speed, large-diameter generators. The generators have many poles and are left open on both ends for self-cooling. The axial length of high-speed generators having 2 or 4 poles can be many times its diameter. These generators require forced air circulation for cooling and are totally enclosed. Tubups!Xjoejoh! The stators of synchronous generators are wound with three distinct and independent windings to generate three-phase power. Usually, the stator winding material is made of high-conductivity copper. The three stator windings are exactly alike in shape but are displaced from each other by exactly 120° electrical so that the induced emfs in the windings are exactly 120° in time phase. The three-phase windings are either connected in star or delta. In star connection, the neutral point is brought out so that it can be properly grounded. The double-layer winding is often used to wind the armature of a synchronous generator. However, full pitch coils are rarely used and generators are wound with fractional pitch coils. The winding is distributed in nature. If Nph is the total number of series turns per phase of the polyphase stator winding then the emf generated in any one phase is 2p Kw f Nph f where Kw is the winding factor, f is the supply frequency and f is the flux.

7/4/4! Spups Stator

Rotor

N

Motion

Depending upon the type of rotor, synchronous machines can be of two types—cylindrical rotor machine and the salient-pole machine. A cylindrical rotor or non-salient type rotor consists of a smooth solid forged steel cylinder having a number of slots for accomodating field coils. These rotors are used in high-speed turbogenerators and have less number of poles. The rotor is built from solid steel forging. Chromium nickel-steel or chrome-nickelmolybdenum steel is used for these types of rotors. The forging has radial slots where the field copper in striped form is placed. Generally, two thirds of the rotor is slotted for the field winding and one third is left without slots to form pole faces Figure 6.2 shows a cylindrical rotor machine. The salient features of cylindrical-rotor-type synchronous machines are as follows:

S

Gjh/! 7/3! Dzmjoesjdbm.spups! tzodispopvt! hfofsbups

!

7/5

Fmfdusjdbm!Nbdijoft

1. 2. 3. 4. 5. 6.

They are smaller in diameter and larger in axial length. Usually the number of poles in the rotor is two or four. They have high operating speed. Windage loss is less. Rotor is subjected to high mechanical stresses due to high peripheral speed. They have nearly sinusoidal flux distribution around the periphery and hence, give a better emf waveform than that obtained from salient-pole machines. 7. They are mechanically more balanced than the salient-pole type. 8. They do not require damper windings. In a salient-pole-type machine, the rotor has a large number of poles projecting from the rotor surface having their cores bolted to a heavy magnetic core of cast iron or CRGO steel of good magnetic quality. The salient-type rotors are suitable for low and medium-speed machines. If run at high speeds, there are excessive windage loss and more noise. Also they do not have adequate mechanical strength to run at high speeds. They require damper windings on the rotor to damp the rotor oscillations. Figure 6.3 shows a salient-pole machine. The salient features of salient rotor-type synchronous machines are as follows:

a¢1 c1

b1 c¢1

a1

b¢1

N

S

c¢2

N

b¢2

a2

S

c2

b2 a¢2

1. They are larger in diameter and smaller in axial length. Tbmjfou.qpmf! tzodispopvt! 2. The pole shoes are wide and usually cover two- Gjh/! 7/4! hfofsbups thirds of the pole pitch. 3. Poles of these rotors are laminated in order to reduce the eddy-current loss. 4. These machines are suitable for low and medium speeds. 5. Damper windings are required to damp rotor oscillations Synchronous machines have high power armature winding on the stator and low power field winding on the rotor. The advantages of providing armature winding on the stator and low power winding on the rotor are the following: (a) Modern synchronous machines are rated for 6.6 kV or 11 kV or 33 kV. If armature winding is placed on the rotor, three slip rings, each capable of handling huge current, are required. Also, each slip ring must be properly insulated from the shaft of high voltage. Moreover, the shaft of every machine is electrically earthed, through metallic bearings, to the stationary frame of the machine.

Tzodispopvt!Hfofsbups

(b) (c) (d) (e)

(f) (g)

7/6

If the low power field winding is placed on the rotor, only two slips capable of handing lower current is required. Also each slip ring should be insulated from the shaft of low-voltage. It is easier to insulate low-voltage dc field winding on rotor than a high-voltage ac winding. It is easier to collect large armature current if the armature is stationary. The collection of large currents through slip rings is difficult. With field winding on the rotor, only two slip rings are required instead of three slip rings. Hence, slip-ring losses are reduced which results in better efficiency. Stationary armature can be cooled more efficiency resulting in use of high-power synchronous machines. As the rotor has low-voltage field winding, it requires less amount of copper and insulation. Hence, the overall weight and inertia of the rotor is reduced. Due to less weight, it requires cheaper bearings and efficient high-speed operation is possible. Due to reduced inertia, the machine gains full speed from rest within a very short time. Due to light and robust construction of the rotor, higher rotor speeds are permissible and this results in increased output of the synchronous machine. The high-voltage and high-current armature winding of synchronous machine is very complex and requires many connections and end connections. This winding remains more secure when housed on the stator.

EJGGFSFOU!UZQFT!PG!FYDJUBUJPO!TZTUFNT!

7/5

The excitation system in a synchronous machine provides dc excitation to the field winding which is placed on the rotor. The excitation systems may be classified as dc excitation system, ac excitation system and static excitation system.

7/5/2! ed!Fydjubujpo!Tztufn The dc excitation systems were used in earlier days. They have been replaced by ac or static excitation systems. This excitation system utilizes a dc generator as the source of excitation power. It provides current to the synchronous machine rotor through slip rings. The dc generator may be driven by a motor or the shaft of the main generator. The exciter generator may be either self-excited or separately excited. For a separately excited generator, the exciter field is supplied by a pilot exciter comprising a permanent magnet generator. Figure 6.4 shows a typical dc excitation system with amplidyne as voltage regulator.

!

7/7

Fmfdusjdbm!Nbdijoft

Main generator

dc exciter Amplidyne

Field

Field

Armature

Armature

Slip ring CT

PT

Exc. field rheostat

Voltage regulator

Gjh/!7/5! ed!fydjubujpo!tztufn

7/5/3! bd!Fydjubujpo!Tztufn In this type of excitation system, the source of excitation power to the main generator is another alternator which is on the same shaft as that of the main generator. The ac output of the exciter is rectified to produce direct current for the main generator field. Depending upon the type of rectifiers used, this type of excitation system is classified as stationary rectifier system and rotating rectifier system. 2/!Tubujpobsz!Sfdujgjfs!Tztufn In this excitation system, stationary rectifiers are used which supply dc power to the field winding of the main generator through slip rings. Figure 6.5 shows the schematic diagram of a fieldcontrolled alternator rectifier excitation system. The exciter which is an alternator, is driven by the main generator rotor. The exciter field is supplied through thyristor rectifiers and it is self-excited. In this system, an ac regulator is provided to automatically maintain the main generator stator terminal voltage at the desired value corresponding to the ac reference. The dc regulator is provided to maintain a constant generator field voltage as determined by the dc reference. The input signals to the ac regulator include auxiliary inputs which provide additional control and protective functions.

Tzodispopvt!Hfofsbups

ac exciter Field

7/8

Main generator

Armature

Stationary diode

Field

Armature

Slip ring CT dc ref.

dc regulator

Controlled rectifier

PT

ac ref. ac regulator Aux. inputs

Gjh/!7/6! Gjfme.dpouspmmfe!fydjubujpo!tztufn

3/!Spubujoh!Sfdujgjfs!Tztufn This system is a brushless excitation system as rotating rectifiers are used to feed the dc output directly to the main generator field. This system has been developed to avoid problems with the use of brushes for supplying high field currents to large generators. Figure 6.6 shows the schematic diagram of this type of excitation system. The armature of the ac exciter and diode rectifiers rotate with the main generator field. A small pilot exciter with a permanent magnet rotor (shown as NS in Fig. 6.6) rotates with the exciter armature and diode rectifiers. The stationary field of the ac exciter is energized from the rectified output of the pilot exciter stator. The voltage regulator controls the ac exciter field which in turn controls the field of the main generator. Rotating structure

Pilot exciter Armature

Threephase ac

Field Armature

ac exciter Field N S

Main generator

Armature CT

PT

Field

Regulator

Gjh/!7/7! Spubujoh.sfdujgjfs!fydjubujpo!tztufn!

Manual control Aux. inputs

!

7/9

Fmfdusjdbm!Nbdijoft

7/5/4! Tubujd!Fydjubujpo!Tztufn As the name suggests, all components in this system are stationary. Stationary rectifiers supply excitation current directly to the field of the main synchronous generator through slip rings. The rectifiers are supplied from the main generator through a transformer which steps down the voltage to a suitable level. This type of excitation system can be classified as potential-source-controlled rectifier system and compound-source rectifier system. 2/!Qpufoujbm.Tpvsdf.Dpouspmmfe!Sfdujgjfs!Tztufn In this system, the excitation power is supplied through a transformer from the generator terminals and is regulated by a controlled rectifier. The maximum exciter output voltage is dependent on the input ac voltage. But this system is inexpensive and is easy to maintain. This excitation system operates satisfactorily for generators connected to large power systems. Figure 6.7 shows the schematic diagram of this type of excitation system. Exciter transformer

Main generator Field Armature

Controlled rectifier Slip ring

Threephase ac

CT

PT

Field dc ref. dc regulator ac ref. ac regulator Aux. inputs

Gjh/!7/8! Qpufoujbm.tpvsdf.dpouspmmfe!sfdujgjfs!fydjubujpo!tztufn

3/!Dpnqpvoe.Tpvsdf!Sfdujgjfs!Tztufn! The power in this excitation system is fed both from current and voltage of the main generator. A power potential transformer and a saturable current transformer are used to fed power. The regulator controls the exciter output through controlled saturation of the excitation transformer. Under noload condition, the generator armature current is zero and the potential source suppliers the entire excitation power. Under loaded condition, part of the excitation power is supplied by the generator current. Figure 6.8 shows the schematic diagram of a compound-source rectifier system.

Tzodispopvt!Hfofsbups

7/:

Main generator

Saturable current transformer

Armature CT

To neutral grounding

Field Slip rings

Power rectifier

Current source

PT

Power potential transformer

Voltage source Linear reactor

Regulator Aux. inputs

Gjh/!7/9! Dpnqpvoe.tpvsdf!sfdujgjfs!fydjubujpo!tztufn

DPPMJOH!

7/6

The cooling medium in case of a synchronous machine is either air or hydrogen. The slotted part of the stator core which carries the armature conductors can only be reached through the air gap. Circulating cooling air through the air gap of nonsalient-pole-type machine is a major issue as it has long stator core length. In such cases, radial ducts are provided by placing stator core plates in packets so as to have open spaces in between. Axial ducts are also provided for air cooling which are parallel to the shaft axis. These ducts are formed by aligning the circular punching of core laminations to form tubular ducts. For cooling synchronous machines, either natural or forced air circulation is used. As salientpole machines have large diameter, they can be cooled by natural circulation of air through the ducts. But the nonsalient-type-machines require forced cooling which require total enclosed stator and fans to circulate the air with high velocity. Hydrogen as a cooling medium has certain advantages over air. It requires less volume as its specific heat is much greater than that of air. The thermal conductivity of hydrogen is also much larger than air. Hence, hydrogen cooling results in more efficient and less noisy system. The main disadvantage of using hydrogen as the cooling medium (hydrogen cobled generators) is that it is combustible when air-hydrogen ratio is less than 3. Hence, there is potential for fire hazard. This is minimized by giving a pressure gradient from the machine towards the atomsphere so that leakage is outwards, if any. Moreover, a number of auxiliary equipment is required for a hydrogen-cooled unit. While filling the generator casing with hydrogen, it is necessary to replace air by carbon dioxide, otherwise hydrogen-air mixture may be combustible. Hence, carbon dioxide cylinders, piping, valves, regulators, pressure monitors, etc. are required. Also, oil film gas seals become necessary and to avoid water vapour entering through seals, gas dryers are necessary.

!

7/21

Fmfdusjdbm!Nbdijoft

For deciding the actual cooling type to be used for synchronous machines, the parameters such as the rate of heat dissipation, cooling surface area, volume of air required, temperature difference between the cooling air and the part to be cooled, humidity, temperature, pressure of cooling air, etc. should be considered.

FNG!FRVBUJPO!JO!B!TZODISPOPVT!HFOFSBUPS!

7/7

Figure 6.9 shows the cross-sectional view of a Field winding axis synchronous generator. The stator has a two-pole, a b¢ three-phase winding. Full pitched coils are considq = wrt N ered with two coil sides a, a¢ of one coil located dic¢ ws ametrically opposite to each other. A salient-poletype rotor is shown rotating in clockwise direction. Stator coil axis The time origin is chosen at the instant when c the stator-coil axis coincides with the field-windS b ing axis. ws is the angular velocity of the rotor in radians per second which is the synchronous speed a¢ of the machine. Hence, from Fig. 6.9, it is evident that at q = 0°, i.e. when the stator-coil axis coin- Gjh/!7/:! Hfofsbujpo!pg!fng!jo!b!tzodispopvt! hfofsbups cides with the field-winding axis, the flux passing through coil aa¢ is maximum. When q = 90°, i.e. the stator-coil axis is in quadrature with the field-winding axis, the flux passing through the coil is minimum. Similarly, when q = 180°, the flux passing is again maximum, but reversed in direction. Hence, if the number of turns in the coil is N, the flux linkage y at any time t is y = N f cos ws t

…(6.2)

The emf induced according to Faraday’s law, e=–

dy df = N f ws sin wrt – N cos ws t dt dt

…(6.3)

df In Eq. (6.3), N f ws sin wr t is the speed voltage term and N cos wst is the transformer voltdt df age term. If field flux is time invariant then N cos wst is zero. Hence, the emf equation becomes dt e = N f ws sin wr t …(6.4) The maximum value of generated emf per phase Emax = N f ws = 2p f N f where f is the supply frequency.

…(6.5)

Tzodispopvt!Hfofsbups

7/22

The rms value of the generated emf per phase Emax

…(6.6) = 2 p f N f = 4.44 f N f 2 As the synchronous machine has distributed short-pitched winding, the emf equation for synchronous machine per phase is Eph =

Eph = 4.44 Kw f N f

…(6.7)

where Kw is the winding factor. Equation (6.4) can be written as pˆ Ê Êp ˆ e = N f ws cos Á - w s t ˜ = N f ws sin Á w s t - ˜ Ë Ë2 ¯ 2¯

…(6.8)

Hence, from Eq. (6.2) and Eq. (6.8), we can conclude that generated emf lags 90° behind the flux that generates it.

! Qspcmfn!7/2 B!uvscp!hfofsbups!ibt!5!qpmft/!Efufsnjof!uif!tqffe!pg!uif!tufbn!uvscjof!esjwjoh!ju!up!pcubjo!b!gsf. rvfodz!pg!61!I{/

Solution Number of poles P = 4 Frequency f = 50 Hz Hence, speed Ns =

120 f 120 ¥ 50 = 1500 rpm. = P 4

! Qspcmfn!7/3 B!xbufs!uvscjof!jt!dpvqmfe!up!b!261!NX!bmufsobups!jo!b!izefm!qpxfs!tubujpo!xijdi!hfofsbuft!qpxfs! bu!b!gsfrvfodz!pg!71!I{/!Xibu!tipvme!cf!uif!ovncfs!pg!qpmft!jg!uif!bmufsobups!tqffe!jt!:11!sqn@

Solution f = 60 Hz Ns = 900 rpm Hence, 900 = \P=

120 f where P is the number of poles. P

120 ¥ 60 = 8. 900

!

7/23

Fmfdusjdbm!Nbdijoft

! Qspcmfn!7/4 B!tzodispopvt!hfofsbups!dpoofdufe!up!b!61!I{!jogjojuf!cvt!jt!dpvqmfe!up!b!tzodispopvt!npups!xijdi! jt! tvqqmjfe! gspn! 36! I{! nbjot/! Xibu! tipvme! cf! uif! njojnvn! ovncfs! pg! qpmft! gps! uif! npups! boe! hfofsbups!boe!xibu!jt!uif!dpnnpo!tqffe@

Solution As the speeds of both the machines are equal, 120 ¥ 25 120 ¥ 25 = Pm Pg

where Pm and Pg are the number of poles of the motor and generator respectively. Pg

Hence,

Pm

=

50 =2 25

The number of poles of the generator is double the number of poles of the motor. As the minimum number of poles can be 2, if Pm = 2; then Pg = 4 Therefore, the minimum number of poles of the motor is 2 and that of the generator is 4. 120 ¥ 25 120 ¥ 50 = 2 4 = 1500 rpm.

The common speed Ns =

! Qspcmfn!7/5 B!tzodispopvt!dmpdl!xjui!opnjobm!gsfrvfodz!pg!61!I{!tipxt!dpssfdu!ujnf!bu!21!b/n/!Xibu!jt!uif! ujnf!joejdbufe!cz!uif!dmpdl!bu!4!q/n/!jg!evsjoh!uijt!qfsjpe!uif!bwfsbhf!gsfrvfodz!jt!5:/6!I{@

Solution Actual time between 10 a.m. and 3 p.m. is 5 hours, or 5 ¥ 3600 s At 50 Hz frequency in 1 s, the number of cycles completed is 50. Hence, in 5 ¥ 3600 s, the number of cycles = 50 ¥ 5 ¥ 3600 Now at 49.5 Hz frequency, 49.5 cycles are completed in 1 s. Hence, 50 ¥ 5 ¥ 3600 cycles are completed in

50 ¥ 5 ¥ 3600 s 49.5

= 18182 s = 5 h 3 min 2 s Hence, the clock will indicate (10 a.m. + 5 h 3 min 2 s) or 3 p.m. 3 min 2 s.

Tzodispopvt!Hfofsbups

7/24

! Qspcmfn!7/6 B!4.qibtf-!61!I{!tzodispopvt!hfofsbups!ibt!311!dpoevdupst!qfs!qibtf/!Uif!gmvy!qfs!qpmf!jt!41!nXc! boe!uif!xjoejoh!gbdups!jt!1/:7/!Xibu!jt!uif!wbmvf!pg!uif!mjof!wpmubhf!jg!uif!tubups!xjoejoh!jt!dpoofdufe! jo!)b*!efmub-!boe!)c*!tubs@

Solution From Eq. (6.7), Eph = 4.44 Kw f N f f = 50 Hz 200 = 100 (E 1 turn = 2 conductors) N= 2 f = 30 ¥ 10–3 Wb Kw = 0.96 Eph = 4.44 ¥ 0.96 ¥ 50 ¥ 100 ¥ 30 ¥ 10–3 = 639.36 V

Here,

\

(a) In delta connection, the line voltage is equal to phase voltage. \

Line voltage VL = 639.36 V

(b) In star connection, the line voltage is equal to \

times the phase voltage.

Line voltage VL = ¥ 639.36 V = 1107.4 V.

! Qspcmfn!7/7! B!5.qpmf-!61!I{!tubs.dpoofdufe!bmufsobups!ibt!7!tmput!qfs!qpmf!qfs!qibtf!boe!b!uxp.mbzfs!xjoejoh! xjui!5!dpoevdupst!qfs!tmpu/!Jg!uif!dpjm!tqbo!jt!261°-!gjoe!uif!op.mpbe!ufsnjobm!fng!jg!uif!gmvy!qfs!qpmf! jt!411!nXc/

Solution \

Total number of slots S = 6 ¥ 4 ¥ 3 = 72 Total number of conductors Z = 4 ¥ 6 ¥ 4 ¥ 3 = 288 Coil span = 150° Hence, pitch factor Kp = cos

180∞ - 150∞ = 0.966 2

= 48 2¥3 The number of slots per pole per phase q = 6 Number of turns per phase N =

!

7/25

Fmfdusjdbm!Nbdijoft

x 2 Distribution factor Kd = x q sin 2 sin q

180∞ 180∞ 180∞ = = = 10° 72 slots per pole 18 4 6 ¥ 10∞ sin 2 Kd = = 0.956 10∞ 6 sin 2

a=

where

Hence,

Winding factor Kw = Kp Kd = 0.966 ¥ 0.956

\

= 0.9235 \ no-load terminal emf per phase E = 4.44 Kw f Nf = 4.44 ¥ 0.9235 ¥ 50 ¥ 48 ¥ 300 ¥ 10–3 = 2952.24 V ¥ 2952.24 V The terminal voltage VL = = 5.113 kV.

! Qspcmfn!7/8 B!efmub.dpoofdufe!uisff.qibtf!26!lW-!711!sqn-!61!I{-!5.qpmf!hfofsbups!ibt!5!tmput!qfs!qpmf!qfs! qibtf/!Uif!dpjm!tqbo!jt!21!tmput!boe!uifsf!bsf!31!uvsot!qfs!dpjm/!Efufsnjof!uif!gmvy!qfs!qpmf!jg!uif! hfofsbups!ibt!epvcmf.mbzfs!xjoejoh/

Solution Number of slots per pole = 4 ¥ 3 = 12 Slot angle =

180∞ = 15° 12

Coil span is 10 slots Hence, the short pitch angle g = 15° ¥ (12 – 10) = 15° ¥ 2 = 30° g \ pitch factor Kp = cos = cos 15° = 0.9659 Number of slots per pole per phase q = 4 a=

180∞ 180∞ = 15° = Slots per pole 12

Tzodispopvt!Hfofsbups

7/26

a 4 ¥ 15∞ sin 2 = 2 a 15∞ sin 4 ¥ sin 2 2

sin

\ distribution factor

Kd =

\ winding factor

= 0.957 Kw = 0.9659 ¥ 0.957

In double-layer winding, number of coils = number of slots = 4 ¥ 4 ¥ 3 = 48 \

Number of turns per phase N =

20 ¥ 48 3

E = 15 ¥ 103 = 4.44 Kw fN f

Hence,

= 4.44 ¥ 0.9659 ¥ 0.957 ¥ 50 ¥ \

20 ¥ 48 ¥f 3

15 ¥ 103

f=

4.44 ¥ 0.9659 ¥ 0.957 ¥ 50 ¥

20 ¥ 48 3

= 0.2284 Wb.

! Qspcmfn!7/9 Uif!usbqf{pjebm!ejtusjcvujpo!pg!joevdfe!fng!jo!b!gvmm!qjudife!dpjm!jt!tipxo!jo!Gjh/!7/21/!Jg!uif!ovncfs! pg!uvsot!qfs!qibtf!jt!71-!efufsnjof!uif!gmvy!qfs!qpmf/ emf

E = 200 V

0

3

9

12

15

21

24

t(ms)

–E

Gjh/!7/21! Uif!fng!ejtusjcvujpo!pg!Qspc/!7/9

Solution From Fig. 6.10, the rms value of induced emf =

3 9 12 2 2 ˘ 1 È ÊE ˆ ÊE ˆ Í Ú Á t ˜ dt + Ú E 2 dt + Ú Á t ˜ dt ˙ Ë3 ¯ 12 ÍÎ 0 Ë 3 ¯ 3 9 ˚˙

!

7/27

Fmfdusjdbm!Nbdijoft

3 3 12 ˘ 1 ÈÍ 1 È t 3 ˘ 9 1 Èt ˘ ˙ [ ] + + =E t3 Í ˙ Í ˙ 12 ÍÎ 9 Î 3 ˚0 9 Î 3 ˚9 ˙˚

=

=

1 È 27 999 ˘ +6+ Í 12 Î 9 ¥ 3 9 ¥ 3 ˙˚ 1 [1 + 6 + 37] = 1.915 E 12 = 1.915 ¥ 200 V = 383 V

If f be the flux per pole then 383 = 4.44 ¥ 50 ¥ 60 ¥ f f = 0.029 Wb.

or

! Qspcmfn!7/: Uif!dpsf!mfohui!pg!b!uisff.qibtf!7.qpmf-!61!I{!bmufsobups!jt!frvbm!up!uif!qpmf!qjudi/!Uif!tmpu!qjudi!jt! 4!dn-!tmput!qfs!qpmf!qfs!qibtf!jt!7!boe!uif!ovncfs!pg!uvsot!qfs!dpjm!jo!uif!epvcmf.mbzfs!xjoejoh!jt!6/! )b*!Efufsnjof!uif!upubm!ovncfs!pg!tmput-!dpsf!ejbnfufs!boe!dpsf!mfohui/!)c*!Xibu!jt!uif!fng!hfofs. bufe!jg!uif!gmvy!qfs!qpmf!jt!1/14!Xc@

Solution Slot pitch tp =

pD where D is the core diameter and S is the number of slots. S p p = tp = 3 = 6 ¥ 6 ¥ 3 108

\

D=

108 ¥ 3 = 103.185 cm p

(a) If core length is L then L = pole pitch Now, pole pitch =

p D p ¥ 103.185 = = 54 P 6

\ L = 54 cm (b) E = 4.44 f N f In double-layer winding, number of coils = number of slots = 6 ¥ 6 ¥ 3 = 108 \ N = 5 ¥ 108 Hence, E = 4.44 ¥ 50 ¥ 5 ¥ 108 ¥ 0.03 = 3596 V.

Tzodispopvt!Hfofsbups

7/28

GMVY!BOE!NNG!EJBHSBN!PG!DZMJOESJDBM.SPUPS! 7/8 !TZODISPOPVT!HFOFSBUPS!!! For developing the phasor diagram of a cylindrical rotor alternator, first the no-load operation of the machine is considered. Then the armature reaction is considered and finally the effect of armature resistance and leakage reactance is considered. f

7/8/2! Op.Mpbe!Pqfsbujpo!

F

When the synchronous generator is operating at no-load condition with the field winding excited from the dc source, the field flux f and find mmf F is set up. The per phase generated emf E is given by Eq. (6.7). It has already been discussed in Article 6.6 that E lags behind f by 90°. The field mmf F, which is the product of field current and the number of turns in the field winding and the field f are in phase as shown in Fig. 6.11.

90° E

Gjh/!7/22! Qibtps!ejbhsbn! ! ! ! ! ! !voefs!op!mpbe

7/8/3! Fggfdu!pg!Bsnbuvsf!Sfbdujpo! When a balanced three-phase load is connected to the synchronous generator, an armature current Ia fr f flows in each phase of the armature winding. The three-phase currents produce a rotating magnetic field Fa which rotates at synchronous speed. As the F Fr rotor also rotates at synchronous speed, the field mmf F and armature mmf Fa are stationary with respect to each other. Let us assume that armature current Ia is in phase with E or the alternator is operating under unity power factor load. Hence, Fa is perpendicular to F and is in phase with Ia. The phasor sum of F and Fa gives the resultant air gap fa Fa Ia E mmf Fr. The phasor diagram under unity power factor load is shown in Fig. 6.12. The effect of ar- Gjh/! 7/23! Qibtps! ejbhsbn! voefs! vojuz! qpxfs! gbdups! mpbe! tipxjoh! mature reaction under unity power factor load is fggfdu!pg!bsnbuvsf!sfbdujpo cross magnetizing as Fa acts at 90° to the direction of field mmf F. Also the direction of armature flux fa, created by Fa, is along the direction of Fa. In a cylindrical rotor machine, as the air gap is uniform, the reluctance is uniform \

Fa + F = Fr

…(6.9)

!

or, or,

7/29

Fmfdusjdbm!Nbdijoft

Fa Fr F = + Reluctance Reluctance Reluctance

…(6.10)

f +f = f

…(6.11) f

\ the resultant fr is the vector sum of f and fa and is shown in Fig. 6.12. Now the operation of a synchronous generator under lagging power factor load is examined. Let us consider the armature current Ia is lagging the excitation voltage E by an angle q. Hence, Fa will be lagging with respect to F by an angle of 90° + q. The resultant mmf Fr under this condition is shown in Fig. 6.13. The armature reaction flux is constant in magnitude and rotates at synchronous speed. It is purely cross magnetizing in nature when the synchronous generator supplies load at unity power factor. It is partly demagnetizing and partly cross magnetizing in nature when the machine supplies load at lagging power factor. The armature reaction is partly magnetizing and partly demagnetizing in nature when it supplies load at leading power factor.

F fr Fr

E

q

Fa fa Ia

Gjh/! 7/24! Qibtps! ejbhsbn! voefs! mbhhjoh! qpxfs! gbdups! mpbe!tipxjoh!fggfdu!pg! bsnbuvsf!sfbdujpo

7/8/4! Fggfdu!pg!Bsnbuvsf!Sftjtubodf!boe!Mfblbhf!Sfbdubodf! F

Fr

90° E

y

d q

V

t

Fa

Ea

I ar a

When the armature of the synchronous generator is loaded, voltage drop occurs in the stator winding due to its resistance ra and leakage reactance xl. To draw the phasor diagram considering the effects of armature resistance and leakage reactance, the terminal voltage Vt is drawn with armature current Ia lagging behind it by an angle q as shown in Fig. 6.14. The armature resistance drop Iara and armature leakage reactance drop Iaxl is added with Vt to get the per phase emf Ea which is called the air-gap voltage. The air-gap voltage is induced by the resultant flux fR which leads it by 90°. The angle y between E and Ia is called the inner displacement angle or the internal power-factor angle. The angle between E and Vt is called the load angle or the power angle.

l I ax

Ia

Gjh/!7/25! Qibtps!ejbhsbn!pg!dzmjoesjdbm. spups!tzodispopvt!hfofsbups

Tzodispopvt!Hfofsbups

7/2:

DPODFQU!PG!TZODISPOPVT!JNQFEBODF!

Hence, and

E + Ear = Ea Ear = – j Ia xa

F

Fr

=

–I a

xa

E

Figure 6.16 shows the equivalent circuit of a synchronous generator. xa

Ia

ra

Vt Fa

Iaxl Iara Ia

Gjh/! 7/26! Qibtps! ejbhsbn! tipxjoh! bsnbuvsf!sfbdujpo!fng xe

Ea

E

ar

Ea

…(6.12) …(6.13)

E

The terminal voltage of a synchronous generator under loaded condition decreases due to the effect of armature reaction, armature resistance and armature leakage reactance. The effect of armature reaction can be represented by a fictitious reactance xa. As the field mmf F generates an emf E lagging it by 90°, the resultant mmf Fr generates an emf Ez, the armature reaction mmf Fa may also be considered to induce armature reaction emf Ear lagging Fa by 90° as shown in Fig. 6.15. As mmfs F and Fa give rise to the resultant mmf Fr, similarly, emfs E and Ear may be considered to produce the emf Ea.

7/9

Vt

Gjh/!7/27! Frvjwbmfou!djsdvju!pg!tzodispopvt!hfofsbups

From Fig. 6.16, or, or, \

E E E E

= Ea + j Ia xa = Vt + Ia ra + jIa xl + jIa xa = Vt + Ia ra + jIa (xl + xa) = Vt + Ia ra + jIa xs = Vt + Ia Zs

…(6.14) …(6.15)

The quantity xs being the sum of leakage reactance xl and fictitious armature reaction reactance xa is known as the synchronous reactance. Zs is known as the synchronous impedance. The synchronous reactance xs is the fictitious reactance considered to account for the voltage effects in the armature circuit produced by armature leakage reactance and by change in air-gap flux, produced by armature reaction.

!

7/31

Fmfdusjdbm!Nbdijoft

The synchronous impedance Zs is a fictitious impedance considered to account for the voltage effects in the armature circuit produced by the actual armature resistance, armature leakage reactance and the change in air-gap flux caused by armature reaction.

WPMUBHF!SFHVMBUJPO!

7/:

Voltage regulation of a generator is defined as the percentage rise of voltage from full load to no load when the excitation field and the speed remain constant. Voltage regulation =

E - V fl

…(6.16)

V fl

Vf l is the full load voltage and E is the no-load excitation voltage. For lagging power factor load, E is always greater than Vfl, hence voltage regulation is always positive. For leading power factor load, E may be greater or less than Vfl. Hence, the voltage regulation may be positive or negative. The terminal voltage of a synchronous generator changes with the application of load across its output terminals. The change is due to voltage drops in the windings and the effect of armature reaction. Figure 6.17 shows the variation of terminal voltage with armature current at different power factors.

Q

Leading pf

P A

Unit

y pf

Vt

La

gg ing

R

O

B

pf

Ia

Gjh/!7/28! Wbsjbujpo!pg!Wu!xjui!Jb!bu!ejggfsfou!qpxfs!gbdups!mpbet

Here, OB is the rated load current of the generator at which the terminal voltage is OA. If this load at unity power factor is removed, keeping the speed and excitation of the alternator constant, the terminal voltage will rise to OP. If the load at lagging power factor is removed, the terminal voltage rises to OQ, whereas when the load at leading power factor is removed, the terminal voltage drops to OR. The change of terminal voltage from full load to no load is more in case of lagging and leading power factor as compared to unity power-factor load.

Tzodispopvt!Hfofsbups

7/32

In synchronous generators, a constant output voltage at different power factor loads is maintained by using an automatic voltage regulator which automatically increases or decreases the field excitation, depending upon, the magnitude of the load power factor. To determine the expression for voltage regulation, the phasor diagram of a cylindrical-rotor synchronous generator operating at lagging, leading and unity power factor are shown in Fig. 6.18(a), Fig. 6.18(b) and Fig. 6.18(c) respectively. E

I az s d O

A

Iaxs

Ia ra

q

Vt

Ia

B

(a) Ia E Iaxs Iazs q

d Iara Vt (b) Iaxs E Iazs

d Ia

Vt

Iara

(c)

Gjh/! 7/29! Qibtps! ejbhsbn! pg! dzmjoesjdbm.spups! tzodispopvt! hfofsbups! bu! )b*! mbhhjoh! qpxfs.gbdups! mpbe-!)c*!mfbejoh!qpxfs.gbdups!mpbe-!boe!)d*!vojuz!qpxfs.gbdups!mpbe!sftqfdujwfmz

In Fig. 6.18(a), from DOAB, or,

OA2 = OB2 + BA2 E 2 = (Vt cos q + Ia ra)2 + (Vt sin q + Ia xs)2

\

E = (Vt cos q + I a ra )2 + (Vt sin q + I a xs )2

…(6.17)

!

7/33

Fmfdusjdbm!Nbdijoft

The no-load voltage E at any load Ia and any power factor cos q can be calculated if the terminal voltage Vt, armature resistance ra and synchronous reactance xs is known. Hence, voltage regulation can be calculated. From Fig. 6.18(b), at leading power factor, E = (Vt cos q + I a ra )2 + (Vt sin q - I a xs )2

…(6.18)

From Fig. 6.18(c), at unity power factor, E = (Vt + I a ra )2 + ( I a xs )2

…(6.19)

The voltage regulation in small machines can be obtained by directly loading the machine. But in large machines, indirect methods are used to determine the voltage regulation since direct loading of these machines require large input and due to the cost of dissipating the huge output. For large cylindrical-rotor synchronous generators, three indirect methods are used to determine the voltage regulation. They are 1. Synchronous impedance method or emf method 2. Ampere-turn method or mmf method 3. Zero power-factor method or Potier method

7/:/2! Tzodispopvt!Jnqfebodf!Nfuipe-!ps!fng!Nfuipe! It has been mentioned earlier that voltage regulation can be calculated if ra and xs is known. The following tests are performed to determine ra and xs or synchronous impedance Zs. 1. dc resistance test 2. Open-circuit test 3. Short-circuit test In dc resistance test, the armature resistance ra is measured by voltmeter-ammeter method using a battery, a dc voltmeter and a dc ammeter. The value of the resistance obtained is multiplied by a suitable factor to take into account the rise of resistance due to skin effect. For example, if the armature is star connected with dc field winding open, the dc resistance between each pair of terminals is measured. This resistance is divided by 2 to obtained the dc resistance per phase. Then it is multiplied by a factor 1.2 to 1.7 depending upon the size of the machine to consider the skin effect, and the armature resistance ra is obtained. If the armature winding is delta connected, the dc 3 armature resistance is ¥ dc resistance between each pair of terminals. Then it is multiplied by the 2 skin factor to obtain the actual ra. In open-circuit test, the loads in the synchronous generator are disconnected and the field current is made zero. The alternator is then run at rated synchronous speed. The field current is gradually increased in steps and the corresponding terminal voltage is measured. The field current may be increased to get the terminal voltage 25% more than the rated voltage. The open-circuit characteristic

Tzodispopvt!Hfofsbups

7/34

Open-circuit voltage

C OC

P

C

E

SC

Short-circuit armature current Ia sc

(Fig. 6.19) shows the relation between open-circuit voltage per phase and the field current. For low values of field current, the Open-Circuit Characteristic (OCC) is a straight line. For high values of field current, the characteristic departs from the straight line due to saturation.

Q

R O

Field current If

Gjh/!7/2:! Pqfo.djsdvju!boe!tipsu.djsdvju!dibsbdufsjtujd

In short-circuit test, the armature terminals are shorted and three ammeters are connected to measure the short-circuit current. The field current is reduced to zero and the synchronous machine is run at synchronous speed. The field current is gradually increased and the short-circuit armature current is measured. The field current may be increased to get armature current of 50% more than the rated value. The average of the three armature currents is considered and a graph is plotted between this current and field current. This characteristic of short-circuit armature current Iasc and field current If is known as short-circuit characteristic as shown in Fig. 6.19. From Fig. 6.19, the field current OR produces the rated open-circuit voltage of PR and the same field current OR produces the short-circuit armature current of QR. The synchronous impedance Zs may be calculated as follows: Zs = =

Rated open-circuit phase voltage per phase for the same value of field current Short-circuit armaature current

PR (in volts) QR (in amperes)

…(6.20)

Hence, synchronous impedance Zs is the ratio of the open-circuit phase voltage for a certain excitation to the short-circuit armature current for the same excitation. The synchronous reactance xs is obtained as xs =

Z s - ra

…(6.21)

!

7/35

Fmfdusjdbm!Nbdijoft

As ra is very small compared to xs , hence xs can be considered nearly equal to Zs. After getting the values of ra and xs, E can be calculated from Eq. (6.17) and voltage regulation E - Vt ¥ 100% can be measured. Vt

7/:/3! Bnqfsf.uvso!Nfuipe-!ps!nng!Nfuipe! The ampere-turn method, or mmf method, is based on the concept of replacing the effect of armature leakage reactance by an equivalent additional armature reaction mmf. The opencircuit characteristic and short-circuit characteristic along with stator resistance Ra should be available for finding the voltage regulation using If mmf method. The armature terminal voltage V t is considered as reference, and armature current If1 Ia is drawn lagging Vt by an angle q for which If2 regulation is to be calculated as shown in Fig. 6.20. The armature resistance drop Ia ra is drawn parallel to Ia and the resultant of Vt and Ia ra is represented Vt by E 1. From the open-circuit characteristic of Iara q Fig. 6.19, the field current If1 corresponding to E1 is E1 noted. This current If1 is drawn leading E1 by 90°. From the short-circuit characteristic, the field current Ia If2 corresponding to rated short-circuit armature Gjh/! 7/31! Qibtps! ejbhsbn! gps! wpmubhf. current is noted. If2 is required to overcome the sfhvmbujpo! dbmdvmbujpo! vtjoh! synchronous reactance drop Ia xs. If2 is drawn in phase bnqfsf.uvso!nfuipe opposition to Ia. The resultant of If1 and If2 is If which would generate a voltage E under no-load condition. The open-circuit voltage E corresponding to field current If is determined from open-circuit characteristic. Then the voltage regulation is E - Vt ¥ 100%. calculated as Vt

7/:/4! [fsp!Qpxfs.Gbdups!Nfuipe-!ps!Qpujfs!Nfuipe! Zero power-factor characteristic shows the variation of armature terminal voltage with field current when the alternator is delivering full rated current to zero power-factor load. For obtaining zero power-factor characteristic the synchronous machine is run at rated synchronous speed after connecting a purely inductive load across the armature terminals. The field current is increased till full load current flows across the armature winding. The armature terminal voltage is recorded at each step corresponding to the field current to obtain the zero power-factor characteristic. Figure 6.22 shows the phasor diagram corresponding to zero power-factor lagging load where the armature current Ia lags the terminal voltage Vt by 90°. The armature reaction mmf Fa, field mmf F and resultant mmf Fr is also shown. If the armature resistance ra is neglected then the corresponding phasor diagram is shown in Fig. 6.22.

Tzodispopvt!Hfofsbups

7/36

F

F Fa

Fa

Fr

Fr Vt

Vt

E Iaxl

E Iara

Fa

Iaxl

Fa Ia

! ! ! !

Ia

Gjh/!7/32! Qibtps!ejbhsbn!bu!{fsp! ! qpxfs.gbdups!mbhhjoh!mpbe!

Gjh/!7/33! Qibtps!ejbhsbn!bu!{fsp!qpxfs.gbdups ! mbhhjoh!ofhmfdujoh!bsnbuvsf!sftjtubodf

From Fig. 6.22, the magnitude of terminal voltage Vt = E – Ia xl Also, the magnitudes of the mmfs are related by the equation F = Fr + Fa Hence, the magnitudes of currents are related by If = Ir + Iar

…(6.22) …(6.23) …(6.24)

where If is the field current, Iar is the field current required to overcome armature reaction and Ir is the resultant current. The open-circuit characteristic and zero power-factor characteristic are shown in Fig. 6.23. Voltage y

E Iaxal Vt

p

x

z

y¢

E¢

V¢t

x¢

z¢ y≤

O

x≤

z≤

b

Ir If

Iar

Gjh/!7/34! Qpujfs!usjbohmf!

a

I

!

7/37

Fmfdusjdbm!Nbdijoft

Let us consider a point x on the zero power-factor characteristic curve corresponding to the rated terminal voltage Vt. The corresponding field current is If where If = oa. On the open-circuit characteristic, consider the point y corresponding to the generated emf E. As E = Vt + Ia xal, consider a point z such that yz = Ia xal = E – Vt. The triangle formed by the vertices x, y, z is called the Potier triangle. Now let us consider the zero power-factor operation of the synchronous generator at any other terminal voltage Vt¢ with rated armature current. Hence, the leakage reactance voltage drop will have the same value as that when the machine operates at terminal voltage Vt. Hence, if the Potier triangle xyz is shifted downward so that xz is kept horizontal and point x lies on the zero power-factor characteristic curve, the new Potier triangle at terminal voltage V¢t is triangle x¢, y¢, z¢. The point y¢ on the OCC will determine the generated voltage E ¢. Now if the triangle is still shifted downwards such that the terminal voltage Vt is zero, the Potier triangle is represented by triangle x≤, y≤, z≤. This triangle represents the limiting position which corresponds to the short-circuit condition. As the initial part of OCC is almost linear, another triangle y≤x≤ is formed when terminal voltage is zero. A similar triangle xyp can be formed from the Potier triangle at any terminal voltage by drawing the line py parallel to oy≤ from the vertex y of the Potier triangle such that px = ox≤. To determine the voltage regulation by zero power factor method at the terminal voltage Vt at lagging power factor of cos q, the phasor diagram shown in Fig. 6.24 is constructed considering Vt as reference. Ia is drawn making an angle q with Vt. Ia ra is drawn parallel to Ia and the line BC perpendicular to Iara represents Ia xl. OC is joined which represents the generated emf E. From the OCC, the field excitation current Ir corresponding to generated emf E is determined. OD is drawn equal to Ir perpendicular to OC. GD is drawn equal in magnitude to Ia and parallel to Ia. OG is joined to get the total field current If. G If

Ef

Ia

H

D Iaxa

Ir

E A

O

Vt Ia

C Iaxl

Iara B

Gjh/!7/35! Wpmubhf!sfhvmbujpo!gspn!{fsp!qpxfs.gbdups!nfuipe!

Now corresponding to the field current If (= OG), the emf Ef is determined from the open-circuit characteristic. Ef is represented by OH which is drawn perpendicular to OG and lagging by 90°. CH represents the voltage drop due to armature reaction. E f - Vt Hence, voltage regulation = ¥ 100% Vt

Tzodispopvt!Hfofsbups

TIPSU.DJSDVJU!SBUJP!

7/38

7/21

Short-circuit ratio of a synchronous generator is defined as the ratio of the field current required to produce rated voltage on open circuit to the field current required to produce rated current on three-phase short circuit. The short-circuit ratio is the reciprocal of the per unit value of saturated synchronous reactance. The short-circuit ratio determines the current through the armature during fault. Modern synchronous machines are built with short-circuit ratios between 0.5 and 1.5.

! Qspcmfn!7/21 B!211!lWB-!uisff.qibtf-!tubs.dpoofdufe!bmufsobups!ibt!b!sbufe!mjof!up!mjof!wpmubhf!pg!2211!W/!Uif! sftjtubodf!boe!tzodispopvt!sfbdubodf!qfs!qibtf!bsf!1/6!W!boe!7!W!sftqfdujwfmz/!Efufsnjof!uif!wpmu. bhf!sfhvmbujpo!bu!gvmm!mpbe!boe!1/96!qpxfs!gbdups!mbhhjoh/

Solution Line current IL =

100 ¥ 103 3 ¥ 1100

A = 52.486 A

Since the machine is star connected, armature current Ia = 52.486 A From Eq. (6.17), the no-load voltage E = (Vt cos q + I a ra )2 + (Vt sin q + I a xs )2 Vt =

1100 3

V = 635.1 V

cos q = 0.85, sin q = sin (cos–1 0.85) = 0.5268 ra = 0.5 W xs = 6 W \

E = (635.1 ¥ 0.85 + 52.486 ¥ 0.5)2 + (635.1 ¥ 0.5268 + 52.486 ¥ 6)2 = 320444.3 + 421832.95 = 861.55 V From Eq. (6.16), voltage regulation is E - Vt 861.55 - 635.1 ¥ 100% = ¥ 100% Vt 635.1

= 35.66%.

! Qspcmfn!7/22 B!gjfme!fydjubujpo!pg!9!B!jo!b!tzodispopvt!hfofsbups!qspevdft!b!dvssfou!pg!211!B!po!tipsu!djsdvju! boe!b!ufsnjobm!qibtf!wpmubhf!pg!811!W!po!pqfo!djsdvju/!Efufsnjof!uif!joufsobm!wpmubhf!espq!xjui!b! mpbe!dvssfou!pg!61!B/

!

7/39

Fmfdusjdbm!Nbdijoft

Solution From Eq. (6.20), synchronous impedance Zs =

Open-circuit voltage 700 W=7W = Short-circuit curent 100

Hence, the internal voltage drop at a load current of 50 A is 50 ¥ Zs or 50 ¥ 7 V or 350 V.

! Qspcmfn!7/23 Jo!b!2611!lWB-!4411!W-!61!I{-!uisff.qibtf-!tubs.dpoofdufe!tzodispopvt!hfofsbups-!b!gjfme!dvssfou! pg!61!B!qspevdft!b!tipsu.djsdvju!dvssfou!pg!361!B!boe!bo!pqfo.djsdvju!wpmubhf!pg!2211!W!mjof!up!mjof/ Efufsnjof!uif!wpmubhf!sfhvmbujpo!bu!gvmm!mpbe!boe!)b*!1/9!qpxfs!gbdups!mbhhjoh-!boe!)c*!1/9!qpxfs! gbdups!mfbejoh/!Dpotjefs!uif!bsnbuvsf!sftjtubodf!up!cf!1/4!W/

Solution Ia = Vt =

1500 ¥ 103 3 ¥ 3300 3300 3

A = 262.43 A

V = 1905.25 V

1100 Zs = 3 W = 2.54 W 250

ra = 0.3 W \

xs = (2.54)2 - (0.3)2 = 2.52 W

(a) From Eq. (6.17), no-load emf for lagging load E = (Vt cos q + I a ra )2 + (Vt sin q + I a xs )2 = (1905.25 ¥ 0.8 + 262.43 ¥ 0.3)2 + (1905.25 ¥ 0.6 + 262.43 ¥ 2.52)2 = 2569381 + 3256125 = 2413.6 V \ voltage regulation =

2413.6 - 1905.25 ¥ 100% = 26.68% 1905.25

(b) From Eq. (6.18), no-load emf for leading load E = (Vt cos q + I a ra )2 + (Vt sin q + I a xs )2 = (1905.25 ¥ 0.8 + 262.43 ¥ 0.3)2 + (1905.25 ¥ 0.6 - 262.43 ¥ 2.52)2 = 2569381 + 232157 = 1673.78 V

Tzodispopvt!Hfofsbups

\ Voltage regulation =

7/3:

1673.78 - 1905.25 ¥ 100% 1905.25

= –12.15%.

! Qspcmfn!7/24 B!22111!lWB-!7711!W-!4.qibtf!tubs.dpoofdufe!bmufsobups!xjui!bsnbuvsf!sftjtubodf!pg!1/4!W!boe 6!W!tzodispopvt!sfbdubodf!qfs!qibtf!efmjwfst!gvmm!mpbe!dvssfou!bu!1/9!qpxfs!gbdups!mbhhjoh!bu!sbufe! wpmubhf/!Efufsnjof!uif!ufsnjobm!wpmubhf!bu!1/9!qpxfs!gbdups!mfbejoh!mpbe!dvssfou!gps!uif!tbnf!fy. djubujpo/

Solution Ia = Vt =

11000 ¥ 103 3 ¥ 6600 6600 3

A = 962.25 A

V = 3810.5 V

The no-load voltage at 0.8 power factor lagging load E = (3810.5 ¥ 0.8 + 962.25 ¥ 0.3)2 + (3810.5 ¥ 0.6 + 962.25 ¥ 5)2 = 11136069.56 + 50375216 = 7842.9 V At constant excitation, E remains constant If V ¢t be the terminal voltage at 0.8 power factor leading current then 7842.9 = (Vt ¢ cos q + I a ra )2 + (Vt ¢ sin q - I a xs )2 = (Vt ¢ ¥ 0.8 + 962.25 ¥ 0.3)2 + (Vt ¢ ¥ 0.6 + 962.25 ¥ 5)2 or, or, or, or, \

(0.64 + 0.36) Vt¢ 2 + 2(0.8 ¥ 0.3 + 0.6 ¥ 5) ¥ 962.25 V ¢ + (0.3 ¥ 962.25)2 + (5 ¥ 962.25)2 = 61511080 Vt¢ 2 + 6235.38 Vt + 23231459.82 = 61511080 Vt¢2 + 6235.38 Vt – 38279620 = 0 V¢t = 3810.48 V Terminal line voltage = ¥ 3810.48 = 6600 V.

!

7/41

Fmfdusjdbm!Nbdijoft

! Qspcmfn!7/25 B!551!W-!31!lWB-!uisff.qibtf!tubs.dpoofdufe!bmufsobups!ibt!tubups!sftjtubodf!pg!1/4!W/!B!gjfme!dvs. sfou!pg!2/6!B!djsdvmbuft!bo!bsnbuvsf!dvssfou!pg!26!B!evsjoh!tipsu.djsdvju!uftu/!Po!pqfo!djsdvju-!uif! nbdijoft!hjwf!261!W!gps!uif!tbnf!gjfme!dvssfou/!Efufsnjof!uif!tzodispopvt!sfbdubodf/!Jg!uif!bmufs. obups!jt!tvqqmzjoh!b!mpbe!dvssfou!pg!36!B!bu!vojuz!qpxfs!gbdups-!gjoe!uif!ufsnjobm!wpmubhf!xifo!uif! mpbe!jt!tveefomz!uispxo!pgg/!Xibu!jt!uif!wpmubhf!sfhvmbujpo@

Solution ra = 0.3 W 150 Zs = W = 10 W 15 \

xs =

Z s2 - ra2 = (10)2 - (0.3) 2

= 9.995 W Terminal voltage per phase Vt =

440 3

V = 254 V

Load current Ia = 25 A At unity power factor, the no-load emf E = (Vt + I a ra )2 + ( I a xs )2 = (254 + 15 ¥ 0.3)2 + (15 ¥ 9.995) 2 = 298.83 V When the load is thrown off, the terminal voltage rises from 254 V to 298.83 V. \

Voltage regulation =

298.83 - 254 ¥ 100% = 17.65%. 254

! Qspcmfn!7/26 B!22-111!lWB-!44!lW-!tubs.dpoofdufe!tzodispopvt!hfofsbups!djsdvmbuft!gvmm.mpbe!tipsu!djsdvju!dvs. sfou!gps!b!gjfme!dvssfou!pg!31!B/!Xjui!uif!tbnf!gjfme!dvssfou-!uif!pqfo.djsdvju!wpmubhf!jt!2/6!lW/!Uif! bsnbuvsf!sftjtubodf!jt!nfbtvsfe!cz!b!wpmunfufs!boe!bo!bnnfufs!xijdi!hjwf!sfbejoht!pg!26!W!cf. uxffo!mjoft!boe!41!B/!Efufsnjof!uif!bsnbuvsf!fggfdujwf!sftjtubodf!boe!tzodispopvt!jnqfebodf!qfs! qibtf/

Solution As the voltage is measured between lines and the armature winding is star connected, the dc armature resistance radc =

1 Ê V ˆ 1 15 = 0.25 W Á ˜= ¥ 2 Ë I ¯ 2 30

Tzodispopvt!Hfofsbups

7/42

\ effective armature resistance ra = 1.5 ¥ 0.25 W = 0.375 W Full-load current = \

11000 3 ¥ 33

A = 192.45 A

Short-circuit current = 192.45 A

\

Zs =

1.5 ¥ 103 / 3 W = 4.5 W 192.45

! Qspcmfn!7/27 Jo!Qspcmfn!7/26-!dpotjefs!uibu!uif!bmufsobups!jt!dpoofdufe!jo!efmub/!Uif!sfbejoht!pcubjofe!bsf!tbnf/! Gjoe!uif!wpmubhf!sfhvmbujpo!bu!vojuz!qpxfs!gbdups/

Solution The resistance measured by voltmeter and ammeter ra (ra + ra ) 15 = ra + ra + ra 30 2 ra 15 = 3 30

or, or,

ra = Full-load line current IL =

15 3 ¥ = 0.75 W 30 2 11000 3 ¥ 33 192.45

\

Full-load phase current =

\

Zs =

\

xs = (13.5)2 - (0.75)2 = 13.48 W

3

= 192.45 A

A = 111.11 A

1.5 ¥ 103 W = 13.5 W 111.11

Vt = 33000 V

(E delta-connected alternator)

No-load voltage E = (Vt + I a ra )2 + ( I a xs )2 = (33000 + 111.11 ¥ 0.75)2 + (111.11 ¥ 13.48)2 = 13117 V \ voltage regulation at unity power factor =

33117 - 33000 ¥ 100% = 0.35%. 33000

!

7/43

Fmfdusjdbm!Nbdijoft

! Qspcmfn!7/28 B!4.qibtf-!2311!lWB-!3411!W-!tubs.dpoofdufe!tzodispopvt!hfofsbups!ibt!sftjtubodf!cfuxffo!fbdi! qbjs!pg!ufsnjobmt!bt!1/29!W!xifo!nfbtvsfe!cz!bqqmzjoh!ed!wpmubhf/!Dpotjefs!uif!fggfdujwf!sftjtubodf! up!cf!2/3!ujnft!uif!ed!sftjtubodf/!B!gjfme!dvssfou!pg!61!B!djsdvmbuft!tipsu.djsdvju!bsnbuvsf!dvssfou! frvbm!up!gvmm.mpbe!dvssfou!pg!411!B/!Uif!tbnf!gjfme!dvssfou!pg!61!B!qspevdft!bo!fng!pg!711!W!po!pqfo! djsdvju/!Dbmdvmbuf!uif!tzodispopvt!sfbdubodf!pg!uif!nbdijof!boe!uif!wpmubhf!sfhvmbujpo!bu!gvmm!mpbe! boe!1/96!qpxfs!gbdups!mbhhjoh/!

Solution Open-circuit voltage per phase =

600 3

V

\ synchronous impedance 600 Open-circuit voltage per phase = 3 W Zs = Short-circuit armature current 300

= 1.1547 W Armature resistance per phase ra = 1.2 ¥ \

0.18 W = 0.108 W 2

Synchronous reactance xs = (1.1547)2 - (0.108)2 W = 1.1496 W Now Zs = ra + jxs = 0.108 + j1.1496 = 1.1547 84.63∞ W Armature current Ia = 300 - cos - 1 0.85 A = 300 -31.79∞ A

\ generated voltage E = Vt + Ia Zs =

2500 3

0∞ + 300 -31.79∞ ¥ 1.1547 84.63∞

= 1443.37 + 346.41 52.84∞ or, Hence,

E = 1652.616 + j 276 = 1675.5 9.48∞ V voltage regulation = =

E - Vt ¥ 100% Vt 1675.5 - 1443.37 ¥ 100% 1443.37

= 16.08%.

Tzodispopvt!Hfofsbups

7/44

! Qspcmfn!7/29 B!23!lWB-!551!W-!61!I{-!tubs.dpoofdufe!tzodispopvt!hfofsbups!tvqqmjft!sbufe!mpbe!bu!1/9!qpxfs! gbdups!mbhhjoh/!Uif!bsnbuvsf!sftjtubodf!boe!tzodispopvt!sfbdubodf!bsf!1/4!W!boe!9!W!sftqfdujwfmz/! Efufsnjof!uif!upsrvf!bohmf!boe!uif!wpmubhf!sfhvmbujpo/

Solution Armature current Ia = Terminal voltage per phase Vt =

12 ¥ 103 3 ¥ 440 440 3

A = 15.746 A

V

ra = 0.3 W xs = 8 W E = Vt + Ia (ra + jxs)

\

=

440 3

+ 15.746 - cos -1 0.8 (0.3 + j8)

= 254 + 126.05 87.85∞ - 36.87∞ = 254 + 126.05 50.98∞ E = 333.36 + j 97.93 = 347.44 16.37∞ V

or, Hence, torque angle is 16.37°.

Voltage regulation =

347.44 - 254 ¥ 100% = 36.78%. 254

! Qspcmfn!7/2: B!4.qibtf-!tubs.dpoofdufe!bmufsobups!ibt!bo!pqfo.djsdvju!mjof!wpmubhf!pg!7/7!lW/!Uif!bsnbuvsf!sftjt. ubodf!boe!tzodispopvt!sfbdubodf!bsf!1/4!W!boe!6!W!sftqfdujwfmz/!Efufsnjof!uif!ufsnjobm!wpmubhf-! wpmubhf!sfhvmbujpo!boe!upsrvf!bohmf!jg!uif!mpbe!dvssfou!jt!311!B!bu!1/9!qpxfs!gbdups!mbh/

Solution Armature current Ia = 200 -36.87∞ A Let the torque angle be d. \

\ or,

6.6 ¥ 103

d = 3810.5 d V 3 Again, E = Vt + Ia (ra + jxs)

E =

3810.5 (cos d + j sin d) = Vt 0∞ + 200 -36.87∞ (0.3 + j5) 3810.5 cos d + j 3810.5 sin d = Vt + 1001.8 49.696∞ = (Vt + 648) + j 764

!

7/45

Fmfdusjdbm!Nbdijoft

Equating the imaginary terms, 3810.5 sin d = 764 sin d = 0.2005 or d = 11.57°

\ Equating the real terms,

3810.5 cos d = Vt + 648 Vt = 3810.5 cos 11.57° – 648 = 3085 V

or,

¥ 3085 V or 5.343 kV

\ terminal line voltage is

3810.5 - 3085 ¥ 100% 3085 = 23.52%.

Voltage regulation =

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Solution ra = 0.3 W xe = 0.8 W Vt =

440

V = 254 V 3 As armature reaction has twice the effect of armature reactance, xa = 2xl = 2 ¥ 0.8 = 1.6 W \ xs = xa + xl = 2.4 W (a) Ia =

15 ¥ 103 3 ¥ 440

A = 19.68 A

cos q = 0.6; sin q = 0.8 No-load phase voltage E = (Vt cos q + I a ra )2 + (Vt sin q + I a xs )2 = (254 ¥ 0.6 + 19.68 ¥ 0.3)2 + (254 ¥ 0.8 + 19.68 ¥ 2.4)2 = 25060 + 62716 = 296.27 V Line voltage when load is disconnected is ¥ 296.27 V or 513.15 V

Tzodispopvt!Hfofsbups

7/46

(b) Zs = ra2 + xs2 = (0.3)2 + (2.54)2 = 2.557 W When the alternator terminals are disconnected, the no-load voltage required to circulate rated current is Ia Zs or 19.68 ¥ 2.557 or 50.32 V per phase.

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Pqfo.djsdvju!Wpmubhf!)W* 3111 5111 6811 7511 8461 8911

[fsp!Qpxfs.gbdups!Wpmubhf!)W* 1 3211 5361 6511 7411 7:11

Solution The open-circuit characteristic and zero power-factor characteristic are plotted as shown in Fig. 6.25. c isti

r cte ara

8000

h

it c ircu nc c

e

Op

7000

b d

f 6000

Terminal voltage (V)

a

r tic we ris po cte ro ra Ze cha r cto

fa

5000

4000

3000

2000

1000

O

10

20

30

40

50

60

70

80

90 100 110 120 130 Field current (A)

Gjh/!7/36! Pqfo.djsdvju!boe!{fsp!qpxfs.gbdups!dibsbdufsjtujd!pg!Qspc/!7/32

!

7/47

Fmfdusjdbm!Nbdijoft

A horizontal line is drawn at rated voltage of 6.6 kV which meets the zero power-factor curve at a. On this line, a point b is taken so that ab = 20 A. From b, a line bc is drawn parallel to the line of which is the initial slope of the open-circuit characteristic. From c, a line cd is drawn perpendicular to ab. Here, adc is the Potier triangle and ad is the field current required to overcome armature reaction on no load which is 16 A or Iar = 16 A. Now dc measures 400 V. Hence, leakage impedance voltage drop per phase Iaxl =

400 3

V

Now,

Ia = 300 A

\

xl =

400 3 ¥ 300

W = 0.77 W

(a) At unity power factor load, Vt = \

6600 3

0∞ V (considering Ia as reference)

E = Vt + Ia (ra + jxl) =

6600 3

+ 300 (0.4 + j 0.77)

= 3810.5 + 120 + j231 = 3937.3 3.35∞ V per phase or,

E=

¥ 3937.3 V = 6820 V (line to line)

From open-circuit characteristic of Fig. 6.25, the field current corresponding to line voltage of 6820 V is 85 A. This current leads E by 90°. \

Ir = 85 3.35∞ + 90∞ = 85 93.35∞ A

Now Iar is in phase with Ia . \

Iar = 16

∞A

Hence, resultant current Ir = If + Iar or,

If = 85 93.35∞ – 16

∞ = 87.4 103.88∞ A

Tzodispopvt!Hfofsbups

7/48

From the open circuit characteristic, corresponding to 87.4 A current, the line value of voltage E = 6900 V \

Voltage regulation =

6900 - 6600 ¥ 100% 6600

= 4.5% (b) cos q = 0.8 lag = cos 36.87° If Ia is reference then Ia = 300 \

E=

∞ A and Vt =

6600 3

6600 3

36.87∞ V

36.87∞ + 300 (0.4 + j0.77)

= 3168.41 + j2517.3 = 4046.68 38.47∞ V per phase or,

E=

¥ 4046.68 V = 7009 V(line)

From open-circuit characteristic, the field current corresponding to 7009 V is 92 A. \

Ir = 92 90∞ + 38.47∞ A = 92 128.47∞ A

\

If = Ir – Iar = 92 128.47∞ - 16 0∞ = –73.23 + j72.03 = 102.72 135.47∞ A

From open-circuit characteristic, the open circuit voltage corresponding to 102.72 A. E = 7300 V Hence,

voltage regulation =

7300 - 6600 ¥ 100% 6600

= 10.6% (c) cos q = 0.8 lead = cos 36.87° \

Vt =

\

E=

6600 3 6600 3

-36.87∞ V -36.87∞ + 300(0.4 + j 0.77)

= 3168.41 – j 2055.3 = 3776.65 -32.97∞ V per phase \

E=

¥ 3776.65 V = 6541.35 V (line)

From open-circuit characteristic, the field current corresponding to 6541.35 V is 82 A.

!

7/49

Fmfdusjdbm!Nbdijoft

\

Ir = 82 90∞ - 32.97∞ A = 82 57.03∞ A

\

If = 82 57.03∞ – 16

∞ = 28.62 + j68.79°

= 74.5 67.41∞ A From open-circuit characteristic, the open-circuit voltage corresponding to 74.5 A E = 6250 V \

voltage regulation =

6250 - 6600 ¥ 100% 6600

= –5.3%.

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2

Pqfo.djsdvju!ufsnjobm!wpmubhf!)W*

4

5

6

7

211 421 491 571 621

Tipsu.djsdvju!dvssfou!)B*

7

31

37

[fsp!qpxfs.gbdups!ufsnjobm!wpmubhf!)W*

Ð

Ð

1

44

49

221 326

9

21

22

691 756 771 62

75

24

25

Ð

Ð

83

471 571 611

666 676

Solution The armature resistance per phase ra =

1 ¥ 0.4 W = 0.2 W 2

Terminal voltage per phase Vt =

460

V = 265.58 V 3 The open-circuit characteristic, short-circuit characteristic and zero power-factor characteristic is shown in Fig. 6.26. (a) The open-circuit voltage of 460 V (line) or 265.58 V (phase) is produced by the field current of 5 A. The same field current of 5 A produces a current of 33 A under short circuit condition.

Tzodispopvt!Hfofsbups

7/4:

Hence, synchronous impedance Zs =

460/ 3 W = 8.05 W 33

\ Synchronous reactance xs = Z s - ra = 8.047 W 700

600 f

500

100 b

a

d

90

400

80 70

300

60 50

200

40

Short-circuit current

Open-circuit terminal voltage

c

30 20

100

10

O

1

2

3

4

5

6

7

8

9

10

11

12

13

14

Gjh/!7/37! Pqfo.djsdvju-!tipsu.djsdvju!boe!{fsp!qpxfs.gbdups!dibsbdufsjtujd!pg!Qspc/!7/33

\

E = Vt + IaZs = 265.58 + 33 -31.79∞ ¥ (0.2 + j8.047) = 265.58 + 265.65 56.78∞ = 411.12 + j 222.23 = 467.34 28.39∞ V

\

467.34 - 460 ¥ 100% 460 = 1.595%

voltage regulation =

!

7/51

Fmfdusjdbm!Nbdijoft

(b) The field current which produces rated voltage of 460 V is 5 A. This current I f1 leads E by 90°. \ I f1 = 5 90∞ + 28.39∞ A = 5 118.39∞ A From Fig. 6.20, I f 2 is in phase opposition to Ia \

I f2 = 5 180∞ - 31.79∞ = 5 148.21∞ A

Now

If = I f1 + I f2 = 5(–1.325 + j1.4) = 9.64 133.42∞ A

From the open-circuit characteristic, the phase voltage corresponding to 9.64 A is 470 V. \

voltage regulation =

470 - 460 ¥ 100% 460

= 2.17% (c) From the rated voltage of 460 V, a horizontal line is drawn which meets the zero power factor curve at a. On this line, a point b is taken such that ab = 4 A. From b, a line bc is drawn parallel to the line of, which is the initial slope of the open circuit characteristic. From c, a line cd is drawn perpendicular on ab. Triangle adc is the Potier triangle and ad is the field current required to overcome armature reaction. \ ad = Iar = 3.3 A Now,

dc = 75 V

\

I a XL =

75 3

V

Ia = 33 A 75 XL = W = 1.31 W 3 ¥ 33

Now, \ Considering Ia as reference,

E = Vt + Ia (ra + jXL) =

460 3

31.79∞ + 33 (0.2 + j1.31)

= 265.58 31.79∞ + 43.73 81.32∞ = 232.34 + j183.14 = 295.84 38.25∞ V (phase) \

E=

¥ 295.84 V = 512.34 V (line)

From open-circuit characteristic, the field current corresponding to this voltage is 6.1 which leads E by 90°.

Tzodispopvt!Hfofsbups

\

7/52

Ir = 6.1 90∞ + 38.25∞ A = 6.1 128.25∞ A

Now Iar is in phase with Ia \

Iar = 3.3

\

∞A

If = Ir – Iar = 6.1 128.25∞ - 3.3 0∞ = –7.08 + j4.79 = 8.55 145.92∞ A

From open-circuit characteristic, the open-circuit voltage corresponding to 8.55 A E = 600 V \

voltage regulation =

600 - 460 ¥ 100% 460

= 30.4%.

! Qspcmfn!7/34 B!2611!lWB-!22!lW-!uisff.qibtf-!tubs.dpoofdufe!bmufsobups!ibt!tzodispopvt!jnqfebodf!pg![t!>!)1/6! ,!k9*!W!qfs!qibtf/!Bu!sbufe!wpmubhf-!ju!efmjwfst!gvmm!mpbe!dvssfou!bu!b!qpxfs!gbdups!pg!1/:!mbhhjoh/! Efufsnjof!uif!ufsnjobm!wpmubhf!gps!uif!tbnf!fydjubujpo!boe!dvssfou!bu!1/:!q/g/!mfbejoh/

Solution Vt = Ia =

\

11000 3

V

1500 3 ¥ 11

- cos -1 0.9 A

ra = 0.5 W xs = 8 W E = Vt + Ia (ra + jxs) =

11000 3

+

1500 3 ¥ 11

-25.84∞ (0.5 + j8)

= 6350.85 + 78.73 -25.84∞ ¥ 8.016 86.42∞ = 6350.85 + 631.1 60.58∞ = 6660.85 + j549.7 = 6683.49 4.72∞ V

!

7/53

Fmfdusjdbm!Nbdijoft

Now, with E = 6683.49 V and p.f. 0.9 leading, if the terminal voltage is V¢t then = Vt¢ +

1500 3 ¥ 11

25.84∞ (0.5 + j8)

= V¢t + 631.1 112.26∞ 2

E = (Vt¢ + 631.1 cos 112.26°)2 + (631.1 sin 112.26°)2 (Vt¢ + 631.1 cos 112.26°)2 = (6683.49)2 – (631.1 sin 112.26°)2 (Vt¢ – 239)2 = 44327904 Vt¢ = 6896.92 V.

\ or, or, \

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Solution Vt = 1 p.u., xs = 0.8 p.u. As

ra = 0.01 p.u.

= Vt + I a Z s, considering Ia as reference E2 (1.3)2 1.69 0.01 cos q + 0.8 sin q

\ \ or, or,

= (Vt cos q + Iara)2 + (V sin q + Iaxs)2 = Vt2 + 2Vt Ia (ra cos q + xs sin q) = 1 + 2 (0.01 cos q + 0.8 sin q) = 0.345

0.01ˆ Ê (0.01)2 + (0.8) 2 sin Á q + tan -1 ˜ = 0.345 Ë 0.8 ¯

or, or, or, \

Ia = 1 p.u., E = 1.3 p.u.

sin (q + 0.716°) = 0.4312 = sin 25.54° q = 24.83° Power factor = cos 24.83° lagging or, 0.91 lagging 2 2 Also, from Fig. 6.18(a), E + Vt – 2 EVt cos d = (Iazs)2 \

cos d =

(1.3) 2 + (1)2 - 1(0.012 + 0.82 ) 2 ¥ 1.3 ¥ 1

= 0.7884 = cos 37.96° \

d = 37.96°.

Tzodispopvt!Hfofsbups

7/54

! Qspcmfn!7/36 B! 6! NWB-! 22! lW-! 4.qibtf-! tubs.dpoofdufe! tzodispopvt! hfofsbups! ibwjoh! bsnbuvsf! sftjtubodf! pg! 1/6!W!qfs!qibtf-!tvqqmjft!:1!B!bu!{fsp!qpxfs.gbdups!mfbejoh!boe!ufsnjobm!wpmubhf!pg!23/6!lW/!Xifo! uif!mpbe!jt!sfnpwfe-!uif!ufsnjobm!wpmubhf!gbmmt!up!22!lW/!Efufsnjof!uif!wpmubhf!sfhvmbujpo!xifo!uif! bmufsobups!pqfsbuft!bu!gvmm!mpbe!boe!1/9!q/g/!mbh/

Solution E= Vt = Ia =

11000 3 12500 3

V = 6351 V = 7217 V

5 ¥ 103 3 ¥ 11

A = 262.43 A

Taking Ia as reference, = Vt -90∞ + Ia (0.5 + jxs) = 0.5 Ia – jVt + jIaxs E = (0.5 ¥ 262.43)2 + (Vt – 262.43 xs)2 (6351)2 = 17217.376 + (7217 – 262.43 xs)2 xs = 3.3 W 2

\ or, or,

For 0.8 lagging power factor operation, Vt = 6351 V \ the excitation emf E2 = (Vt cos q + Iara)2 + (Vt sin q + Iaxs)2 = (6351 ¥ 0.8 + 262.43 ¥ 0.5)2 + (6351 ¥ 0.6 + 262.43 ¥ 3.3)2 = (5212.015)2 + (4676.619)2 E = 7002.56 V

\ Hence,

voltage regulation =

7002.56 - 6351 ¥ 100% 6351

= 10.26%.

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!

7/55

Fmfdusjdbm!Nbdijoft

Solution Vt =

11, 000 3

V = 6351 V

P = 20 ¥ 106 W cos q = 0.8 lag \

20 ¥ 103

Ia = E E2 (7621.2)2 xs2 + 5.81 xs xs

\ or, or, or,

3 ¥ 11 ¥ 0.8

A = 1312.16 A

= 1.2 ¥ 6351 V = 7621.2 V = (Vt cos q)2 + (Vt sin q + Iaxs)2 = (6351)2 + (1312.16)2 xs2 + 2 ¥ 1312.16 ¥ 6351 ¥ 0.6 xs = 10.3077 = 1.42 W

! Qspcmfn!7/38 Bo!bmufsobups!xijdi!jt!pqfsbujoh!jo!qbsbmmfm!xjui!puifs!bmufsobupst!bu!sbufe!ufsnjobm!wpmubhf!ibt! bsnbuvsf!sftjtubodf!pg!2/6&!boe!tzodispopvt!sfbdubodf!46&/!Ju!efmjwfst!b!qpxfs!pvuqvu!jo!lX!frvbm! up!91&!pg!jut!sbufe!lWB!boe!uif!fng!jt!frvbm!up!2/26!ujnft!uif!ufsnjobm!wpmubhf/!Efufsnjof!uif!qpxfs! gbdups!bu!xijdi!uif!nbdijof!jt!pqfsbujoh/

Solution Considering Ia as reference for lagging power factor, = Vt q + Ia (ra + jxs) = Vt cos q + j Vt sin q + Iara + jIaxs 2

\ Now \ or, or, or,

E = (Vt cos q + Ia ra)2 + (Vt sin q + Ia xs)2 I r I x E = 1.3 Vt, a a = 0.015, a s = 0.35, Ia = 0.8/Irated Vt Vt 2 (1.15 Vt) = (Vt cos q + 0.015 Vt ¥ 0.8)2 + (Vt sin q + 0.35 Vt ¥ 0.8)2 1.3225 = (cos q + 0.012)2 + (sin q + 0.28)2 1.3225 = 1 + 0.024 cos q + 0.56 sin + 0.0785 0.024 cos q + 0.56 sin q = 0.244

We know that K1 cos q + K2 sin q = \ or, or, \

0.5605 sin (q + 2.45°) sin (q + 2.45°) q Power factor

2 1 +

2 2

Ê sin Á q + tan -1 Ë

= 0.244 = 0.435 = 23.335° = cos q = cos 23.335° = 0.918 lagging.

1ˆ

˜

2¯

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FYUFSOBM!DIBSBDUFSJTUJD!PG! B!TZODISPOPVT!HFOFSBUPS

7/22!

The external characteristic of a synchronous generator shows the variation of the terminal voltage with load current. The terminal voltage varies with the load current. If the excitation current is held constant, the generated emf per phase is constant. Neglecting the armature resistance, Vt = E – j Ia xs

…(6.25)

Since E and xs are constant, the terminal voltage Vt depends upon the magnitude of the load current and power factor. Though the magnitude of E is constant, its phase (the power angle) is free to change. Hence, the locus of E must be a circle. Let us now find the changes in Vt as a function of load with unity, lagging and leading power factor. Figure 6.27 shows the external characteristic for lagging, unity and leading power factor. Vt

cos q2 lead

q1 < q2

cos q1 lead E

cos q = 1 cos q1 lag cos q2 lag

q1 < q2

Ia

Gjh/!7/38! Fyufsobm!dibsbdufsjtujd!pg!b!tzodispopvt!hfofsbups!voefs!ejggfsfou!mpbe!dpoejujpot

For unity and lagging power factor, Vt decreases as Ia increases. However, the decrease in Vt is more in case of lagging power factor. For leading power factor, Vt increases as Ia increases. When the power factor is lagging Vt decreases further with the decrease in power factor or increase in power factor angle. For leading power factor the terminal voltage decreases with the increase in power factor.

7/22/2! Vojuz!Qpxfs.gbdups!Mpbe When the load is purely resistive, Vt and Ia are in phase. If the load current increases, the voltage drop across the synchronous reactance increases, power angle increases and the terminal voltage decreases. This is illustrated in Fig. 6.28.

!

7/57

Fmfdusjdbm!Nbdijoft

j Iaxs

E

j Iaxs

E

d

d Ia

Vt

(a)

Ia

Vt

(b)

Gjh/! 7/39! Fggfdu! pg! mpbe! dvssfou! dibohf! po! ufsnjobm! wpmubhf! bu! vojuz! qpxfs! gbdups;! )b*! Jb! tnbmm )c*!Jb!mbshf

7/22/3! Mbhhjoh!Qpxfs.gbdups!Mpbe For inductive load, Ia lags Vt by an angle q. When Ia increases, Vt decreases as shown in Fig. 6.29. If the power factor is decreased, keeping magnitude of Ia constant then also Vt decreases as shown in Fig. 6.30. E

j Ia x s

E j Ia x s q

d

d Vt

Vt

q

Ia Ia

(a)

(b)

Gjh/!7/3:! Fggfdu!pg!bsnbuvsf!dvssfou!po!Wu!voefs!mbhhjoh!qpxfs!gbdups;!)b*!Jb!tnbmm!)c*!Jb!mbshf!

E

j Iaxs E j Iaxs

d

d

q

q Ia Ia

(a)

(b)

Gjh/!7/41! Fggfdu!pg!mbhhjoh!qpxfs!gbdups!po!Wu;!)b*!dpt!q!ijhi!)c*!dpt!q!mpx

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7/58

7/22/4! Mfbejoh!Qpxfs.gbdups!Mpbe! For leading power-factor load, or for capacitive load, the load current Ia leads the terminal voltage Vt. The terminal voltage Vt increase with the increase in Ia as shown in Fig. 6.31. Ia E Ia q

E

j Iaxs

j Iaxs q

d

d

Vt

(a)

Vt

(b)

Gjh/!7/42! Fggfdu!pg!bsnbuvsf!dvssfou!po!Wu!voefs!mfbejoh!qpxfs!gbdups;!)b*!Jb!tnbmm!)c*!Jb!mbshf!

For leading power factor, the terminal voltage increases with decrease in power factor as shown in Fig. 6.32. Ia

Ia q d

j Iaxs E

E j I ax s

q d

Vt

Vt

(a)

(b)

Gjh/!7/43! Fggfdu!pg!mfbejoh!qpxfs!gbdups!po!Wu!;!)b*!dpt!q!ijhi!)c*!dpt!q!mpx

QPXFS.BOHMF!DIBSBDUFSJTUJD!PG!DZMJOESJDBM.! SPUPS!TZODISPOPVT!HFOFSBUPS

7/23!

The three-phase complex power delivered by a synchronous generator is S = P + jQ = 3 Vt Ia* where Vt is the terminal voltage per phase and Ia the armature current per phase.

…(6.26)

!

7/59

Fmfdusjdbm!Nbdijoft

From the phasor diagram shown in Fig. 6.18(a), Ia =

E d - Vt 0∞ E V = d - f s - t -f s Z s fs Zs Zs

Hence

V ÈE ˘ S = 3Vt Í f s - d - t f s ˙ Zs ˚ Î Zs

or,

S=

Vt [E cos (fs – d) + j E sin (fs – d) – Vt cos fs – j Vt cos fs] Zs

…(6.27)

…(6.28) …(6.29)

Equating the real and imaginary terms, real power P=

3Vt E 3V 2 cos (f s - d ) - t cos f s Zs Zs

…(6.30)

Q=

3Vt E 3V 2 sin (f s - d ) - t cos f s Zs Zs

…(6.31)

and reactive power

Here,

fs = tan -1

xs ra

As armature resistance is very small compared to the synchronous reactance, if we neglect armature resistance then VE P = t sin d (\ fs = 90°) …(6.32) Zs and

Q=

Vt E 3V 2 cos d – t Zs Zs

…(6.33)

or,

P=

Vt E sin d xs

…(6.34)

and

Q=

Vt E 3V 2 cos d – t xs xs

…(6.35)

As xs is constant in a synchronous generator, if E and Vt are held constant then the real power varies as sin d. Hence, d is known as the power angle or load angle. The graph of P vs. d is shown in Fig. 6.33 and this curve is known as the power-angle curve. The negative values of P indicate the operation of the machine as motor. The maximum power is transferred when d = 90°. This maximum power is known as steady-state stability limit. When the machine delivers this maximum power, the armature current becomes many times the rated current and under this condition, any disturbance may lead to loss of synchronism. Hence, in practice, d is kept well below 90°.

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P Pmax Generator –180°

–190°

0°

90°

180°

Motor

270° Motor

360°

d

–Pmax

Gjh/!7/44! Qpxfs.bohmf!dvswf!pg!dzmjoesjdbm.spups!tzodispopvt!nbdijof!

The real output power P is maximum when dP =0 dd 2 dP = d È 3Vt E cos(f - d ) - 3Vt cos f ˘ = 0 Í s s˙ dd Î Z s Zs dd ˚

Hence,

or, or,

Vt E sin (f s – d) = 0 Zs fs = d

…(6.36)

Therefore, the maximum power output Pmax =

3Vt E 3Vt 2 cos f s Zs Zs

…(6.37)

When real power is maximum, the reactive power Q =–

3Vt 2 sin f s Zs

…(6.38)

Neglecting armature resistance, Pmax =

Vt E xs

and the reactive power when real power is maximum is 3V 2 Q =– t xs The three-phase complex power input to the generator input

Ê E d - Vt ∞ ˆ * = 3 E Ia* = 3E d Á ˜ Ë Z s fs ¯

…(6.39)

…(6.40)

!

or,

7/61

Fmfdusjdbm!Nbdijoft

3E d 3E E f s - d - Vt f s = E f s - Vt f s + d Zs Zs Hence, the three-phase real input power input

=

(

)

(

Pinput =

E [E cos q – Vt cos (fs + d)] Zs

) …(6.41)

The three-phase reactive power input Qinput =

E [E sin fs – Vt sin (fs + d)] Zs

…(6.42)

The input real power is maximum when d Pinput dd

=

EVt sin (fs + d) = 0 Zs

fs + d = 0 or 180° d = – fs or d = 180° – fs d = 180° – fs

or, Hence, Putting

Pinput max =

3EVt 3E 2 cosf s + Zs Zs

…(6.43)

…(6.44)

Neglecting armature resistance, Pinput max =

EVt . xs

…(6.45)

! Qspcmfn!7/39 B!uisff.qibtf!221!NWB-!22!lW-!61!I{!tzodispopvt!hfofsbups!ibt!tzodispopvt!sfbdubodf!pg!2/6!q/v/! Efufsnjof!uif!nbyjnvn!qpxfs!efwfmpqfe!jg!uif!joufsobm!wpmubhf!pg!uif!hfofsbups!jt!lfqu!dpotubou!bu! 2/7!q/v/!Efufsnjof!uif!qpxfs!bohmf!xifo!uif!tzodispopvt!hfofsbups!tvqqmjft!81!NX/

Solution Maximum power developed from Eq. (6.45) is Pmax =

EVt 1.6 ¥ 1 = xs 1.5

= 1.06 p.u. = 1.06 ¥ 110 MW = 116.6 MW

(E Vt = 1 p.u.)

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7/62

When the generator supplies 70 MW power or 70 p.u. or 0.636 p.u. power then 110 1.6 ¥ 1 sin d 1.5 sin d = 0.596 or d = 36.58°.

0.636 = or,

! Qspcmfn!7/3: B!411!NWB-!22!lW-!61!I{-!tubs.dpoofdufe!tzodispopvt!hfofsbups!djsdvmbuft!b!tipsu.djsdvju!dvssfou! pg!811!B!jo!uif!bsnbuvsf!xjui!gjfme!dvssfou!pg!26!B/!Uif!pqfo.djsdvju!wpmubhf!jt!2311!W!xjui!uif! tbnf!gjfme!dvssfou/!Jg!uif!hfofsbups!tvqqmjft!b!mpbe!pg!4/6!lB!bu!1/9!qpxfs.gbdups!mbhhjoh-!efufsnjof! )b*!qpxfs!efwfmpqfe!cz!uif!hfofsbups-!)c*!nbyjnvn!qpxfs!uibu!dbo!cf!efwfmpqfe-!boe!)d*!uif!qpxfs! bohmf!jg!uif!hfofsbups!efwfmpqt!261!NX!gps!uif!tbnf!dpoejujpo/

Solution Neglecting armature resistance, xs = Zs =

1200 3 ¥ 700

W = 0.99 W

At 0.8 lagging power factor, 2

E=

2

Ê 11 ˆ Ê 11 ˆ ¥ 0.8˜ + Á ¥ 0.6 + 3.5 ¥ 0.99˜ kV ÁË ¯ Ë 3 ¯ 3

= 8.87 kV (a) Power developed =

¥ 11 ¥ 3.5 ¥ 0.8 MW = 53.35 MW

(b) Maximum power that can be developed =3¥

EVt = xs

3 ¥ 8.87 ¥ 0.99

11 3 MW

= 170.7 MW (c) If the power angle is d then 3 ¥ 8.87 ¥

150 =

or \

0.99 sin d = 0.8787 d = 61.49°.

11 3

sin d

! Qspcmfn!7/41 Bo!22!lW!uisff.qibtf-!tubs.dpoofdufe!tzodispopvt!hfofsbups!ibt!tzodispopvt!sfbdubodf!pg!2!W!qfs! qibtf!boe!uif!pvuqvu!qpxfs!jt!261!NX!bu!vojuz!qpxfs!gbdups/!Jg!uif!hfofsbups!fydjubujpo!jt!jodsfbtfe!

!

7/63

Fmfdusjdbm!Nbdijoft

cz!26&!xjui!uif!pvuqvu!sfnbjojoh!dpotubou-!efufsnjof!uif!bsnbuvsf!dvssfou!boe!qpxfs!gbdups/!Xjui! uijt!fydjubujpo-!jg!uif!uvscjof!pvuqvu!jt!jodsfbtfe!up!311!NX-!efufsnjof!uif!hfofsbups!dvssfou!boe! qpxfs!gbdups/!Ofhmfdu!mpttft/

Solution xs = 1 W The armature current at 150 MW output Ia =

150 ¥ 106 3 ¥ 11 ¥ 103 ¥ 1

A = 7872.96 A

The terminal voltage per phase Vt =

11 ¥ 103 3

V = 6350.85 V

The no-load emf E = Vt + ( I a xs ) (neglecting ra) = (6350.85)2 + (7872.96 ¥ 1)2 = 10115.17 V = 10.115 kV Neglecting losses, the input power per phase =

150 ¥ 103 kW 3

As the excitation is increased by 15%, E¢ = 1.15 ¥ 10.11 kV = 11.63 kV \ or, or

E ¢Vt 50 11.63 ¥ 6.35 = sin d sin d = xs 3 1

sin d =

d = 42.61° Armature current Ia¢ =

or,

150 = 0.677 3 ¥ 11.63 ¥ 6.35

11.63 42.61∞ - 6.35 1

Ia¢ = (7.87 – j 2.21) kA = 8.17 -15.68∞ kA Hence, the power factor is cos 15.68° lag or 0.96 lag With this excitation, the power is increased to 200 MW.

Tzodispopvt!Hfofsbups

7/64

Hence, 200 11.63 ¥ 6.35 = sin d 3 1

sin d = 0.903 d = 64.55°

or, or, The new armature current

=

11.63 64.55∞ - 6.35 kA 1

= (10.5 + j1.35)kA = 10.586 7.33∞ kA \

Power factor = cos 7.33° lead or 0.99 lead.

! Qspcmfn!7/42 B!tzodispopvt!hfofsbups!ibwjoh!tzodispopvt!sfbdubodf!pg!1/8!q/v/!jt!dpoofdufe!up!bo!jogjojuf!cvt! bu! sbufe! wpmubhf/! Uif! nbdijof! efmjwfst! bo! pvuqvu! pg! 1/9! q/v/! xifo! uif! fydjubujpo! fng! jt! 2/2! q/v/! Ofhmfdujoh!bmm!mpttft-!efufsnjof!uif!mpbe!bohmf-!bsnbuvsf!dvssfou!jo!qfs!voju!boe!uif!qpxfs!gbdups/

Solution xs = 0.7 p.u., Po = 0.8 p.u., E = 1.1 p.u. and Vt = 1 p.u. From Eq. (6.32), EVt sin d = Po where d is the load angle and Po is per phase power xs \ or, or,

1.1 ¥ 1 sin d = 0.8 0.7

sin d = 0.51 d = 30.6° From the phasor diagram shown in Fig. 6.18, neglecting ra, (Ia xs)2 = E 2 + Vt2 – 2 E Vt cos d

\

Ia =

(1.1)2 + 1 - 2 ¥ 1.1 ¥ 1 ¥ cos 30.6∞ 0.7

Again, Po = Vt Ia cos q \ \ power factor is unity.

cos q =

0.8 =1 1 ¥ 0.8

= 0.8 p.u.

!

7/65

Fmfdusjdbm!Nbdijoft

! Qspcmfn!7/43 Bo!22!lW-!tubs.dpoofdufe!tzodispopvt!hfofsbups!xjui!tzodispopvt!jnqfebodf!pg!)1/6!,!k5*!W!qfs! qibtf!jt!dpoofdufe!up!bo!jogjojuf!cvt!bu!sbufe!wpmubhf/!Uif!nbdijof!efmjwfst!91!B!bu!vojuz!qpxfs!gbd. ups/!Jg!uif!hfofsbups!fydjubujpo!jt!jodsfbtfe!cz!31&-!xjui!uif!pvuqvu!sfnbjojoh!dpotubou-!efufsnjof! uif!bsnbuvsf!dvssfou-!mpbe!bohmf!boe!qpxfs!gbdups/!Xjui!uif!jodsfbtfe!wbmvf!pg!fydjubujpo-!jg!uif! hfofsbups!jt!nbef!up!pqfsbuf!bu!vojuz!qpxfs!gbdups-!gjoe!uif!ofx!wbmvft!pg!bsnbuvsf!dvssfou-!mpbe! bohmf!boe!qpxfs!efmjwfsfe!up!uif!jogjojuf!cvt/

Solution Vt =

11, 000 3

V, Ia = 80 A, ra = 0.5 W, xs = 4 W, Zs = 4.03 82.87∞ W

At unity power factor from Eq. (6.19), E2 = (Vt + Ia ra)2 + (Ia xs)2 2

\

E=

Ê 11, 000 ˆ + 80 ¥ 0.5˜ + (80 ¥ 4) 2 = 6398.8 V ÁË ¯ 3

From Fig. 6.18(c), I a xs 80 ¥ 4 = = 0.05 Vt + I a ra 11000 + 80 ¥ 0.5 3 or, d = 2.86° From Eq. (6.30), power output per phase

tan d =

P=

11000 3

¥

(11000)2 cos 82.87∞ 6398.8 cos (82.87∞ - 2.86∞) 4.03 3 ¥ 4.03

= 1749305.58 – 1242237.08 = 507068.5 W Now generator excitation E1 = 1.2 ¥ 6398.8 V = 7678.56 V If d1 be the new value of the load angle then 507068.5 = or, \

11000 3

¥

(11000)2 cos 82.87∞ 7678.56 cos(82.87∞ - d1 ) 4.03 3 ¥ 4.03

cos (82.87° – d1) = 0.14456 = cos 81.688° d1 = 1.18° Now, E12 + Vt2 – 2E1 Vt cos d1 = (Ia Zs)2 1

Tzodispopvt!Hfofsbups

(7678.56) 2 +

\

Ia1 =

7/66

11000 (11000) 2 - 2 ¥ 7678.56 ¥ cos1.18∞ 3 3 4.03

= 331.38 A Again P = Vt Ia1 cos q1 \

Power factor cos q1 =

507068.5 11000 ¥ 331.38 3

= 0.241 lagging Now power factor cos q2 = 1 E2 = E1 = 7678.56 V E22 = (Vt cos q2 + Ia2 ra)2 + (Vt sin q2 + Ia2 xs)2

\

2

Ê 11000 ˆ (7678.56) = Á + I a 2 ¥ 0.5˜ + (0 + Ia2 ¥ 4)2 Ë 3 ¯ 2

or, or, or, \

16Ia22 + 0.25 Ia2 + 6351 Ia2 – 18625082 = 0 Ia22 + 391Ia2 – 1146159 = 0 Ia2 = 892.8 A Power delivered to the infinite bus per phase = Vt Ia2 cos q2 =

11000 3

¥ 892.8 ¥ 1 = 5670041 W = 5670 kW

\

total power delivered = 3 ¥ 5670 kW = 17010 kW Again, from Eq. (6.30), 5670041 =

\

11000 3

(11000) 7678.56 cos (82.87∞ - d 2 ) cos 82.87∞ 4.03 3 2

¥

d2 = 54.51°.

! Qspcmfn!7/44 B!tzodispopvt!hfofsbups!jt!dpoofdufe!up!bo!jogjojuf!cvt!cbs!bu!sbufe!wpmubhf/!Uif!joqvu!up!uif!hfo. fsbups!jt!jodsfbtfe!ujmm!uif!nbdijof!pqfsbuft!bu!sbufe!lWB/!Efufsnjof!uif!pqfsbujoh!qpxfs!gbdups! pg!uif!nbdijof!boe!jut!mpbe!bohmf!jg!uif!nbdijof!ibt!tzodispopvt!jnqfebodf!pg!)1/14!,!k1/:*!q/v/

Solution Let cos q be the lagging power factor at which the machine operates.

!

7/67

Fmfdusjdbm!Nbdijoft

From Eq. (6.18), E2 = (Vt cos q + Ia ra)2 + (Vt sin q + Ia xs)2 2

I x ˆ I r ˆ Ê Ê E = Vt Á cos q + a a ˜ + Vt 2 Á sin q + a s ˜ Vt ¯ Vt ¯ Ë Ë 2

or,

2

2

1 = 1(cos q + 0.03)2 + 1(sin q + 0.9)2 1 = cos2 q + sin2 q + 0.06 cos q + 1.8 sin q + 0.8109 0.06 cos q + 1.8 sin q + 0.8109 = 0

or, or, or,

0.06 ˆ Ê (0.06) 2 + (1.8) 2 sin Á q + tan -1 ˜ + 0.8109 = 0 Ë 1.8 ¯

or,

or, sin (q + 1.909°) = –0.45 or, sin (q + 1.909°) = sin (–26.74°) \ q = –28.649° \ power factor is cos 28.649° leading or, 0.877 leading Zs = 0.03 + j0.9 = 0.9005 88.09∞ p.u. From Eq. (6.30), power output per phase P=

Vt E V2 cos (88.09∞ - d ) - t cos 88.09∞ Zs Zs

Again, P = Vt Ia cos q \

1¥1 1 ¥ 0.033 = 1 ¥ 1 cos 88.09° cos (88.09∞ - d ) 0.9005 0.9005

\ or, \ load angle is 1.702°.

cos (88.09° – d) = 0.063 = cos 86.388° d = 1.702°

! Qspcmfn!7/45 B!26!NWB-!4.qibtf-!tubs.dpoofdufe!tzodispopvt!hfofsbups!ibt!tzodispopvt!sfbdubodf!pg!6!W!boe! sftjtubodf!pg!1/9!W/!Ju!jt!dpoofdufe!xjui!bo!22!lW!jogjojuf!cvt!cbs/!Jg!uif!fydjubujpo!wpmubhf!jt!22/9!lW-! efufsnjof!uif!nbyjnvn!qpxfs!pvuqvu!pg!uif!hfofsbups!boe!uif!bsnbuvsf!dvssfou!voefs!nbyjnvn! qpxfs!dpoejujpo/

Solution Vt =

11000 3

,E=

11800 3

, Zs = 52 + 0.82 80.9∞ W = 5.06 80.9∞ W

Tzodispopvt!Hfofsbups

7/68

From Eq. (6.45), maximum power per phase 2

Pmax

11000 11800 Ê 11000 ˆ ¥ Á ˜ 3 3 -Ë 3 ¯ = cos 80.9° 5.06 5.06

= 8550724.64 – 1260680.24 = 7290044.4 W = 7.29 MW total maximum power output = 3 ¥ 7.29 MW = 21.87 MW

\

Now (Ia Zs)2 = E 2 + Vt2 – 2E Vt cos 80.9° 2

or,

Ia =

2

11800 ¥ 11000 Ê 11000 ˆ Ê 11800 ˆ cos 80.9∞ ˜¯ - 2 ¥ ÁË ˜¯ ¥ ÁË 3 3 3 5.06

= 1189.24 A.

! Qspcmfn!7/46 B!uisff.qibtf!551!W-!tubs.dpoofdufe!tzodispopvt!hfofsbups!jt!svoojoh!pwfs.fydjufe!xjui!fydjubujpo! wpmubhf!pg!2/4!q/v/!Uif!nbdijof!jt!efmjwfsjoh!1/7!q/v/!qpxfs!up!bo!jogjojuf!cvt!boe!uif!tzodispopvt! sfbdubodf!jt!2/2!q/v/!Jg!uif!joqvu!upsrvf!jt!jodsfbtfe!cz!3&-!efufsnjof!uif!dibohf!jo!uif!sfbm!qpxfs! boe!sfbdujwf!qpxfs/

Solution P= \

0.6 =

\

EVt sin d xs

…(i)

1.3 ¥ 1 sin d 1

d = 30.5° As the torque is increased by 2%, the real power is increased by 2%. \ change in real power dP = 0.02 ¥ 0.6 = 0.012 Now Q =

Vt E V2 cosd - t xs xs

dQ VE = – t sin d dd xs

\ From Eq. (i),

EV dP = t cos d xs dd

…(ii)

!

\ or, \

7/69

Fmfdusjdbm!Nbdijoft

dQ V E sin d =– t = – tan d = – tan 30.5° dP EVt E cos d dQ = – 0.589 dP

dQ = – 0.589 ¥ 0.02 = – 0.01178 or, –1.178% Hence, the reactive power is decreased by 1.178% while the real power is increased by 2%.

QBSBMMFM!PQFSBUJPO!PG!UXP! TZODISPOPVT!HFOFSBUPST!

7/24!

When a single synchronous generator is supplying a load, problems may arise due to the following reasons: 1. If the generator is delivering full load and it becomes necessary to supply a larger load. 2. A fault develops in the generator supplying load which needs to be shut down causing discontinuity of power to the load. 3. The generator needs to be disconnected for maintenance purpose. 4. The load on the generator reduces from full load value, reducing the efficiency of the generator. To overcome the above difficulties, at least one more generator is connected in parallel with the existing generator. The parallel operation of generators provide the following advantages: 1. Can meet the additional load demand 2. Maintains continuity of service, i.e. load receives uninterrupted power supply 3. Can supply power to the load at higher efficiency. The basic idea of the parallel operation of generators is to make the generators meet the load collectively and there should be no flow of current and power between the generators themselves, i.e. when the two generators are connected in parallel and the external load is zero, there should be no current in the generators. Actually, the generator voltages are in phase opposition with respect to the local circuit but in phase agreement with respect to the external circuit. Hence, when load is connected, both the generators supply the load.

7/24/2! Dpoejujpot!gps!Qbsbmmfm!Pqfsbujpo! Let us consider that one generator is running under no-load condition and the other generator is to be connected in parallel with it. As the connection of equipotential points does not result in flow of current, hence, when two generators are connected in parallel, the actual terminals to be connected should have equal potentials so that there is no circulating current between the generators. As the voltage of the synchronous generator is alternating, the potentials of the terminals keep changing every instant with regular periodicity. Therefore, to connect another generator, which is

Tzodispopvt!Hfofsbups

7/6:

usually referred as the incoming generator, in parallel with a running generator, the essential requirement is that the instantaneous potential variation of the terminals of the incoming generator should be exactly same as that of the respective terminals of the running generator with which they are to be connected in parallel. During the process of putting an incoming alternator with a running generator, the terminals of the two machines are connected to an indicator which senses the difference in voltages across the pairs of terminals to be connected together. When voltages of the two generators have the same waveform, frequency, magnitude and phase sequence then the indicator reads zero, suggesting that the incoming alternator may be connected in parallel. Some allowance can be given in the case of frequencies being exactly identical. If the incoming generator is driven at a slightly higher or lower speed than required, there is an in-built mechanism which makes the incoming generator attain the same speed as that of the running generator. This phenomenon is due to the development of synchronizing power which slows down the slightly faster generator and accelerates the slightly slower generator. The process of connecting an incoming generator in parallel with a running generator is called synchronizing of generators. In earlier days, before the advent of modern electronic indicators, methods called dark-lamp method and bright-lamp method were used to identify the instant of synchronization. Hence, the conditions that must be fulfilled for parallel operation of two generators are the following: 1. The terminal voltage of the incoming generator must be same as that of the running generator. 2. The frequency of the incoming generator must be same as that of the running generator. 3. The phase of the voltage of the incoming generator must be identical to the phase of the running generator, with respect to the external circuit, i.e. opposite in phase with respect to the local circuit. 4. The phase sequence of the voltage of the incoming alternator must be the same as that of the running generator.

7/24/3! Tibsjoh!pg!Mpbe!Dvssfout! Let us consider that E1 and E2 be the induced emfs per phase, Zs1 and Zs2 be the synchronous impedances per phase of the two generators, Z be the load impedance and Vt be the voltage across the load or the terminal voltage. Hence,

E1 = I1Zs1 + (I1 + I2) Z

…(6.46a)

and

E2 = I2 Zs2 + (I1 + I2) Z

…(6.46b)

Solving equations (6.46a) and (6.46b), I1 =

and

I2 =

E1Z s2 + ( E1 - E2 ) Z

Z s1 Z s2 + ( Z s2 + Z s2 ) Z E2 Z s1 + ( E2 - E1 ) Z

Z s1 Z s2 + ( Z s1 + Z s2 ) Z

…(6.47a)

…(6.47b)

!

7/71

Fmfdusjdbm!Nbdijoft

The load current I = I1 + I2 =

Hence,

V = IZ =

E2 Z s2 + E2 Z s1

Z s1 Z s2 + ( Z s1 + Z s2 ) Z

…(6.48)

E2 Z s2 + E2 Z s1 Z s1 Z s2 + Z s1 + Z s2 Z

Dividing both numerator and denominator of Eq. (6.51) by Zs1, Zs2, E1 E + 2 Z s1 Z s2 V= 1 1 1 + + Z Z s1 Z s2

…(6.49)

If more than two generators are operated in parallel then E1 E + 2 +º Z s1 Z s2 V= 1 1 1 + + +º Z Z s1 Z s2

…(6.50)

7/24/4! Tibsjoh!pg!Qpxfs! Let us consider two machines with full load power rating of W1 and W2 operating in parallel. Let P1 and P2 be the power shared by the two machines when the total power is P. If the no-load and full-load frequency of the machines are f o1 and fl1 respectively then Drop of frequency from no-load to full load = fo1 – fl1 Drop of frequency per unit rating =

f o - fl W

f o - fl P1 W Operating frequency of Machine 1 = (no load frequency) – (drop in frequency)

Hence, drop of frequency for a load of P1 is

f = fo1 –

f o - fl P1 W

…(6.51)

f = fo2 –

f o - fl P2 W

…(6.52)

Similarly, for Machine 2,

where fo2 and f l2 are the no-load and full-load frequencies of Machine 2. \

f = fo1 –

f o - fl f - fl P1 = f o2 – o P2 W W

…(6.53)

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7/72

7/24/5! Tzodispoj{joh!Qpxfs!Evsjoh!Qbsbmmfm!Pqfsbujpo! The power that comes into play at the instant of synchronizing two generators is called synchronizing power. It provides a corrective action for slight imbalances in the machine frequencies ensuring the running of the generators in synchronism.

Zs1

Zs2 K

E1

ZL

E2

G1

G2

Gjh/!7/45! Jmmvtusbujpo!pg!tzodispoj{joh!qpxfs!evsjoh!qbsbmmfm!pqfsbujpo!

Figure 6.34 shows two synchronous generators G1 and G2 connected in parallel. Let E1 = E2 = E be the internal emfs of the two generators and Zs1 = Zs2 = Zs be the synchronous impedances. Initially, the load switch K is open. Let G1 is running at a speed slightly higher than the synchronous speed. Hence, the frequency of G1 is slightly higher than that of G2. Figure 6.35 shows the phasor diagram under this condition. E1 which is supposed to be in exact phase opposition to E2 is now ahead of its position of exact phase opposition. The resultant Er E1 of the voltages E1 and E2 is Er which circulates the Ir current Ir in the local circuit consisting of Zs1 and Zs2 in series. As Zs1 and Zs2 are predominantly inductive, E2 Ir lags Er by almost 90°. Ir has two components, one Gjh/!7/46! Qibtps!ejbhsbn!xifo!gsfrvfodz! in phase with E1 and another in phase opposition to pg!H2!jt!tmjhiumz!hsfbufs!uibo!H3 E2. Hence, G1 acts as a generator supplying power to G2 which acts as a motor as it receives power from G1. Hence, G1 develops electromagnetic torque opposing the driving primer torque, retarding G2. G2 develops an electromagnetic torque, motoring in nature, which accelerates G2. Thus, G1 and G2 attain same speed. The power due to current Ir is called synchronizing power. The synchronizing power Ps = EIr = where

Isc =

Er ˆ E Er Er I sc Ê = ÁËE I r = 2 Z ˜¯ 2 Zs 2 s

…(6.54)

E is the short-circuit armature current. Zs

Let the two generators with synchronous impedances Zs1 and Zs2 deliver a load current of I to a load impedance ZL and terminal voltage Vt. The currents shared by G1 and G2 are I1 and I2 respectively.

!

7/73

Fmfdusjdbm!Nbdijoft

E1 = V + I1 Zs1 = I ZL + I1 Zs1 Hence,

I1 =

or,

E - I ZL Zs

I = Isc1 –

V Zs

…(6.55)

where Isc1 is the short-circuit current of the generator G1. Similarly, for the generator G2, V I2 = Isc2 – Zs As I1 + I2 = IL =

…(6.56)

V ZL

V V V + I sc2 = Isc1 – Z Z ZL s1 s2

or, or,

V

1 1 ˆ Ê 1 = Isc1 + Isc2 + + ÁË Z Z s1 Z s2 ˜¯ L

V = (Isc1 + Isc2)

E1 -

Hence,

I1 =

I2 =

I sc1 + I sc2 1 1 1 + + Z L Z s1 Z s2 Z s1

E2 -

and

1 Ê Á 1 1 1 + + Á Ë Z L Z s1 Z s2

I sc1 + I sc2 1 1 1 + + Z L Z s1 Z s2 Z s2

ˆ ˜ ˜ ¯

…(6.57)

…(6.58)

…(6.59)

7/24/6! Fggfdu!pg!Dibohf!pg!Fydjubujpo! Let us consider two identical generators operating in parallel and sharing a given load current IL I equally. Let internal emfs E1 = E2 = E and currents supplied by generators I1 = I2 = L at the terminal voltage of Vt. The phasor diagram is shown in Fig. 6.36(a). E1 and E2 are in the same phase with

Tzodispopvt!Hfofsbups

7/74

respect to the external circuit. Let E1 be increased. The difference voltage E ¢ = E1 – E2 circulates a current I ¢ in the local circuit such that E¢ I¢ = Zs I ¢ lags E ¢ by an angle q which is nearly 90°. I ¢ gets added to I1 to make the current of G1 as I1¢ and I 1¢ gets subtracted from I2 to make the current of G2 as I¢2. Hence, the magnitude of the total current of G1 increases with the decrease in its power factor, whereas the magnitude of the current of G2 decreases with the increase in its power factor. The phasor diagram is shown in Fig. 6.36(b). As q is nearly 90°, the contribution of I¢ to the active components of generator currents is negligibly small. The reactive component of I¢ changes the reactive components of generator currents causing the change in the power factors. E¢ E1 E2

q

I¢2 –I¢

q1

q

E1

E2

q2

I¢

I1, I 2

V I1, I2

I¢ I¢1

(a)

(b)

Gjh/!7/47! Fggfdu!pg!dibohf!pg!fydjubujpo!po!qpxfs!gbdupst!pg!hfofsbupst

7/24/7! Fggfdu!pg!Dibohf!jo!Qsjnf.npwfs!Pvuqvu! Consider two generators with internal emfs E1 = E2 = E operating in parallel and sharing load equally. Consider the external circuit E1 and E2 are in phase and shown in Fig. 6.37(a). The currents shared by each generator I1 = I2 = I. Let the mechanical input to G1 be increased without changing load. For steady conditions, the generators continue to run in synchronism at a common frequency. But due to increase of mechanical input, G1 shifts its rotor position with respect to G2, ahead of its previous position by an angle a. E1 E¢

E¢

I¢

E1 E2

a

E2 I¢1 I¢ I

I

(a)

I¢2

–I¢

(b)

Gjh/!7/48! Fggfdu!pg!dibohf!jo!qsjnf.npwfs!pvuqvu

!

7/75

Fmfdusjdbm!Nbdijoft

The resultant voltage in the local circuit E¢ = 2E cos

a

circulates current I¢ given by I¢ =

E¢ which lags nearly 90° with respect to E ¢. Zs

The phase of I ¢ is such that it is practically in phase with the active component of the load current of G1 and in phase opposition to that of the generator G2. Hence, total active component of the current G1 is increased and that of G2 is decreased as shown in Fig. 6.37(b). It can be concluded that an increase of mechanical input to G1 increases the load shared by it and decreases the load shared by G2. Thus, by changing the mechanical input, the generators can be made to have a desired load sharing for a given load.

! Qspcmfn!7/47 Uxp!uisff.qibtf!tzodispopvt!hfofsbupst!bsf!pqfsbujoh!jo!qbsbmmfm/!Uif!sbujoht!pg!uif!nbdijoft!bsf! 211!lX!boe!261!lX/!Cpui!uif!nbdijoft!bsf!gjuufe!xjui!hpwfsopst!ibwjoh!b!esppq!pg!6&/!Efufsnjof! ipx!uif!nbdijoft!xjmm!tibsf!b!mpbe!pg!311!lX@

Solution Let the original frequency be f and P1 and P2 are the loads shared by Machine 1 and Machine 2 respectively when total load is 200 kW. For a load of 100 kW, the drop in frequency = 0.05 f For a load of P1 kW, the drop in frequency =

0.05 f P1 100

Operating frequency of the first machine = f –

0.05 f P1 100

Similarly, for Machine 2, the operating frequency = f – \

f–

0.05 f P2 150

0.05 0.05 f P1 = f – f P2 100 150

P1 =

2 P2 3

Again, P1 + P2 = 200 kW \

P1 = 200 ¥

3 = 120 kW and P2 = 200 – 120 = 80 W. 5

! Qspcmfn!7/48 Uxp!tjnjmbs!6111!lWB!tzodispopvt!hfofsbupst!bsf!pqfsbujoh!jo!qbsbmmfm/!Jo!uif!gjstu!hfofsbups-!uif! gsfrvfodz!espqt!gspn!61!I{!up!59/6!I{!boe!jo!uif!tfdpoe!pof-!uif!gsfrvfodz!espqt!gspn!61!I{!up! 58/6!I{/!Efufsnjof!uif!mpbe!tibsfe!cz!fbdi!xifo!uif!upubm!mpbe!jt!8111!lX/!

Tzodispopvt!Hfofsbups

7/76

Solution If the load shared by the first machine is P1, the load shared by the second one is (7000 – P1) kW From Eq. (6.53), 50 –

50 - 48.5 50 - 47.5 P1 = 50 – (7000 – P1) 5000 5000

1.5 P1 = 2.5 (7000 – P1) P1 = 5000 kW P2 = 7000 – 5000 = 2000 kW.

or, \ \

! Qspcmfn!7/49 Uxp!2611!lWB-!uisff.qibtf!tzodispopvt!hfofsbupst!bsf!pqfsbujoh!jo!qbsbmmfm!boe!tvqqmz!b!mpbe!pg! 3111!lWB!bu!1/9!mbhhjoh!qpxfs!gbdups/!Pof!nbdijof!jt!pqfsbujoh!bu!1/8!mbhhjoh!qpxfs!gbdups!boe! tvqqmzjoh!2311!lWB/!Efufsnjof!uif!qpxfs!pvuqvu!pg!uif!tfdpoe!nbdijof!boe!uif!qpxfs!gbdups!bu! xijdi!ju!jt!pqfsbujoh/

Solution Total load S = 2000 - cos -1 0.8 kVA = 1600 – j 1200 kVA Load supplied by the first machine S1 = 1200 - cos -1 0.7 = 840 – j 856.97 kVA Now S = S1 + S2 where S2 is the load supplied by second machine \

S2 = 1600 – j 1200 – 840 + j 856.9 = 760 – j 343.1 = 833.86 -24.3∞ kVA

Hence, power output of the second machine is 833.86 kVA and power factor is cos 24.3° lagging or, 0.91 lagging.

! Qspcmfn!7/4: Uxp!jefoujdbm!uisff.qibtf!tzodispopvt!hfofsbupst!pqfsbujoh!jo!qbsbmmfm!tibsf!frvbm!mpbet!pg!3111! lX!bu!22!lW!boe!1/9!mbhhjoh!qpxfs!gbdups/!Uif!bsnbuvsf!dvssfou!pg!uif!gjstu!hfofsbups!jt!211!B!bu! mbhhjoh!qpxfs!gbdups/!Efufsnjof!uif!bsnbuvsf!dvssfou!pg!uif!tfdpoe!hfofsbups!boe!uif!qpxfs!gbd. ups!bu!xijdi!fbdi!nbdijof!pqfsbuft/

Solution P1 = P2 = ¥ 11 ¥ Ia1 cos q1 = 1000

2000 = 1000 kW 2

!

7/77

Fmfdusjdbm!Nbdijoft

\

Ia1 cos q1 =

Now

1000 3 ¥ 11

A

Ia1 = 100 A

\

cos q1 =

1000 3 ¥ 11 ¥ 100

= 0.525 lagging

Ia1 = 100 - cos -1 0.525 = 100 -58.33∞ A

\ If total current is I then

¥ 11 ¥ I ¥ 0.8 = 2000 I = 131.22 A

\ or,

I = 131.22 - cos -1 0.8 = 131.22 -36.87∞ A

Now

I = Ia1 + Ia2

\

Ia2 = 131.22 -36.87∞ – 100 -58.33∞ = (104.97 – 52.5) – j (78.73 – 85.12) = 52.47 + j 6.39 = 52.86 -6.94∞ A

\ power factor of the second machine is cos 6.94° or 0.99 leading.

! Qspcmfn!7/51 Uxp!22!lW-!4.qibtf-!61!I{-!tubs.dpoofdufe!tzodispopvt!hfofsbupst!tvqqmz!b!mpbe!pg!6111!lX!bu! 1/9!mbhhjoh!qpxfs!gbdups/!Uif!tzodispopvt!jnqfebodf!pg!uif!uxp!nbdijoft!bsf!1/9!,!k!23!W!boe! 1/6!,!k!21!W!sftqfdujwfmz/!Uif!uxp!nbdijoft!tibsf!frvbm!mpbet!boe!uif!gjstu!nbdijof!efmjwfst!361!!B! bu!mbhhjoh!qpxfs!gbdups/!Efufsnjof!uif!dvssfou-!qpxfs!gbdups-!joevdfe!fng!boe!mpbe!bohmf!pg!fbdi! nbdijof/

Solution Ia1 = 250 A \ or, \

¥ 11 ¥ 250 cos q1 =

5000 2

cos q1 = 0.525 = cos 58.33 lagging Ia1 = 250 -58.33∞ A Total current I =

5000 3 ¥ 11 ¥ 0.8

A = 328.04 A

Tzodispopvt!Hfofsbups

or,

7/78

I = 328.04 -36.87∞ A Ia2 = I – Ia2 = 328.04 -36.87∞ – 250 -58.33∞ = 262.43 – j 196.82 – 131.256 + j 212.77 = 131.174 + j 15.95 = 132.14 6.93∞ A

\ power factor of second machine is cos 6.93° or 0.99 leading. E1 = Vt + Ia1 Z1 =

11000 3

+ 250 -58.33∞ ¥ (0.8 + j12)

= 6351 + 3006.66 27.86∞ = 9009 + j 1405 = 9117.9 8.86∞ V \ induced emf and load angle of Machine 1 are 9117.9 V and 8.86° respectively. Now E2 = Vt + Ia2 Z2 =

11000 3

+ 132.14 6.93∞ ¥ (0.5 + j10)

= 6351 + 1323 94∞ = 6258.7 + j1319.78 = 6396.34 11.9∞ V Hence, induced emf and load angle of Machine 2 are 6396.34 V and 11.9° respectively.

TZODISPOPVT!HFOFSBUPS!DPOOFDUFE! UP!JOGJOJUF!CVT

7/25

7/25/2! Fggfdu!pg!Dibohf!pg!Fydjubujpo!bu!Dpotubou!Pvuqvu! Let a synchronous generator with constant synchronous impedance Zs and delivering current Ig is connected to an infinite bus having voltage V. The internal impedance E = V + Zs Ig and Ig = Ia + jIr where Ia = Ig cos f and Ir = Ig sin f, f being the angle of Ig with respect to V. E = V + Zs Ia + j Zs Ir = V + Ia (ra + j xs) + j Ir (ra + j xs) or,

E = V + Ia ra + j Ia xs + j Ir ra – Ir xs

…(6.60)

!

7/79

Fmfdusjdbm!Nbdijoft

For constant real power output, V + Ia ra + j Ia xs = E1 = constant Hence, E = E1 + j Ir ra – Ir xs

…(6.61) E

At unity power factor, E = E1 = V + Ia ra + j Ia xs

…(6.62) j Iaxs

This excitation E is called normal excitation. Figure 6.38 shows the phasor diagram under this condition. If the generator excitation is increased above normal excitation, it is said to be over-excited. If the generator excitation is reduced below normal excitation, it is said to be under-excited.

q

d Ia

V

Iara

Gjh/! 7/49! Qibtps! ejbhsbn! voefs! opsnbm! fydjubujpo

2/!Pwfs.fydjubujpo! When the synchronous generator is over-excited, the additional field ampere turns provided has to be balanced by a lagging reactive component of armature current, since a lagging reactive armature current directly demagnetizes the main field. Hence, armature current Ig = Ia – j Ir Therefore, E = E1 – j Ir ra + Ir xs = V + Ia ra + j Ia xs – j Ir ra + Ir xs or,

|E| =

(V + I a ra + I r xs ) + ( I a xs - I r ra )

d = tan–1

and

…(6.63)

I a xs - I r ra V + I a ra + I r rs

…(6.64)

Ia z

s

No ex rma cit l ati on

If the magnitude of the over-excitation increases, the magnitude of E and d increases. The phasor diagram under over-excited condition is shown in Fig. 6.39.

Centre of constant excitation circle

f V

d f

Ir

Ia

Ig

Centre of constant current circle

Ir z s Irra jIrxs s I gz

Ov ex er cit ati o E C on

n

sta

nt

po w

er

jIaxs

q Iara Constant excitation circle

Constant current circle

Gjh/!7/4:! Qibtps!ejbhsbn!gps!pwfs.fydjufe!dpoejujpo

line

Tzodispopvt!Hfofsbups

7/7:

Z ÊZ ˆ Now Zs Ia = Á s ˜ V Ia = Po VIa where Po is the real power output of the generator and s is constant. ËV ¯ V

Hence, Zs Ia • Po and f = tan–1

Zs Ir I = tan–1 r Ia Zs Ia

…(6.65)

Hence, a reference line, marked as ‘Normal Excitation’, can be drawn from the tip of the reference phasor V at an angle q, which is the phasor angle of Zs as shown in Fig. 6.39. With respect to the ‘Normal Excitation’ line, the following observations can be made: 1. The distance of a point on the reference line from the tip of the reference phasor V represents ˆ ÊV the real power output of the generator to a suitable scale Á times the voltage scale˜ . ¯ Ë Zs

2. The angle of the synchronous impedance voltage drop Ig Zs with respect to the reference line gives the lagging power-factor angle. 3. The magnitude of Ig Zs represents the total armature current to a suitable scale Ê 1 ˆ ÁË Z times the voltage scale˜¯ . s

The locus traced by the tip of the phasor E for increasing over-excitation is a straight line which is indicated as ‘Constant Power Line’, because for all points on this locus, the power output is constant at the value determined by Ia. If the armature current has constant magnitude but varying phase angle, then the magnitude of Ig Zs is constant but has different phase angle. Hence, the locus traced by the tip of Ig Zs phasor at constant magnitude of armature current and variable over-excitation is a circle with the tip of V taken as its centre and radius equal to Ig Zs which is indicated as ‘Constant Current Circle’. When the excitation is held constant and the power output is varied, the tip of E follows a ‘Constant Excitation Circle’ as shown in Fig. 6.39. Constant Power Line, Constant Current Circle and Constant Excitation Circle are unique to the operation of synchronous generator connected to infinite bus. 3/!Voefs.Fydjubujpo! When the generator is under-excited, the reduction of field-ampere turns is balanced by a leading reactive component of the armature current. The leading reactive armature current causes magnetizing armature reaction in the direction of the main field.

!

7/81

Fmfdusjdbm!Nbdijoft

Hence,

E = E1 + j ra Ir – xs Ir = V + ra Ia + j xs Ia + j ra Ir – xs Ir | E | = (V + ra I a - xs I r )2 + ( xs I a - ra I r )

or,

2

…(6.66)

and the phase angle of E d = tan–1

I a xs + I r ra

…(6.67)

V + I a ra - I r xs

The phasor diagram for under-excited condition is shown in Fig. 6.40. If under-excitation becomes more pronounced, E decreases and d increases. Constant excitation circle

an

tp ow

er

line

E

N ex orma cita l tio n

nst

U exc nder itat ion

jIrxs

nd

er-

ex

cita

tio n

Co

fu

Ir z

it o

s

Igzs Ia z

s

Lim Centre of constant excitation circle

Ig

Ir q f

ex Over cita tio n

Irra

d Ia

Centre of constant current circle

f q V

jIaxs Constant current circle

I ara

Gjh/!7/51! Qibtps!ejbhsbn!gps!voefs.fydjufe!dpoejujpo

The tip of the phasor E traces a constant power line for more pronounced under-excitation at constant power output. Zs Ig represents armature current for leading power factor angle with respect to the reference line corresponding to normal excitation. In this case also, we have Constant Current Circle and Constant Excitation Circle. In both Fig. 6.39 and Fig. 6.40, the constant power line is the locus of synchronous impedance drop phasor due to reactive component of armature current. It is perpendicular to the locus of synchronous impedance drop phasor due to reactive component of armature current. The part of the locus to the right of the reference line of “Normal Excitation” corresponds to lagging power factor and the part to the left corresponds to leading power factor. The combined figure of Fig. 6.39 and Fig. 6.40 is shown in Fig. 6.41.

Tzodispopvt!Hfofsbups

U exc nderitat ion nst po an we tp r fa ow er cto line r

N Ex orma cit l (UP ation F)

Le ad ing

q

lg Z

f

O ex ver cita tio E Lagg n ing po we r fa

s

Lim

it o

fu

nd

er-

exc

itat

ion

Co

7/82

cto r

q

d f

V lg

Gjh/!7/52! Qibtps!ejbhsbn!tipxjoh!dpncjobujpo!pg!Gjh/!7/4:!boe!Gjh/!7/51

7/25/3! Fggfdu!pg!Dibohf!pg!Fydjubujpo!po!Op.Mpbe!

Ze

ro

pow

er

line

U ex nder cita tio n po we r fa cto r

Lim

it o

fu

nd er-

exc i

tat

ion

Le ad ing

N exc orma l it (UP ation F)

As the output of the generator is zero in this case, the active component of the current is zero. The constant-zero power line passes through the tip of the voltage phasor V. The normal excitation in this case corresponds to an excitation which produces an internal emf equal to the voltage of the infinite bus. The phasor diagram at normal excitation under no-load condition is shown in Fig. 6.42.

Ov cita ertio La n gg ing po we r ex

q

q V=E

fac

tor

Gjh/!7/53! Qibtps!ejbhsbn!gps!opsnbm!fydjubujpo!bu!op.mpbe!

During over-excited condition, the generator delivers lagging current I ro. The demagnetizing effect of Iro maintains the net excitation same as normal excitation. As the synchronous impedance takes into account the effect of armature reaction, the increased value of E due to increased excitation, provides terminal voltage equal to the infinite bus voltage and overcomes the synchronous impedance drop due to lagging wattless armature current. The phasor diagram during over-excited condition at no load is shown in Fig. 6.43.

7/83

Fmfdusjdbm!Nbdijoft

U exc nder itat ion ing po we Ze r ro fac po tor we r lin e

Le

cita -ex der it o f un Lim

N ex orma cita l (U tion PF )

ad

tion

!

q

V d

q

jlro

lrora

O exc veritat ion gg ing po we r fa

La zs

E

cto

jlroxs

lro

r

Gjh/!7/54! Qibtps!ejbhsbn!gps!pwfs.fydjubujpo!bu!op.mpbe

U ex nder cita ing tio n po we Ze r fa ro cto po we r r lin e Le

ad

lro

N exc orma l it (UP ation F)

Lim

it o

fu

nd

er-

exc

itat

ion

During under-excited condition, the generator delivers leading reactive current Iro. The magnetizing effect of Iro maintains net excitation same as normal. The decreased value of E due to decreased excitation provides terminal voltage equal to infinite bus voltage and overcomes synchronous impedance drop due to leading wattless armature current. The locus of the tip of E is the constant power line for zero output power as shown in Fig. 6.44.

jlroxs q

d

E lro lrora z s

V

q

O ex vercita tio La n gg ing po we r fa

cto

r

Gjh/!7/55! Qibtps!ejbhsbn!gps!voefs.fydjubujpo!bu!op.mpbe

W!DVSWFT!

7/26!

The plot of the magnitude of armature current versus main field current at constant output power is known as V curve. As the excitation is increased from zero, the armature current reduces from a large value, reaches a minimum value at a particular excitation and again increases with the increase of excitation. V curves can be obtained from Fig. 6.39 and Fig. 6.40. Figure 6.45 shows the combined constant power lines of Fig. 6.39 and Fig. 6.40 for lagging and leading power factors. In Fig. 6.45, four constant power output lines, viz. h1g1, h2 g2, h3 g3 and h4 g4 are shown for constant output powers Po1, Po2, Po3 and Po4 respectively such that Po4 > Po3 > Po2 > Po1. kf is the locus of synchronous impedance drop due to active component of armature current and is the reference line for armature current. The distance to a point on the constant power line from k is proportional to the total armature current at that point. The excitation is proportional to the distance from o to a point on the constant power line.

Tzodispopvt!Hfofsbups

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Let us consider the constant power line h1 g1. h1 corresponding to the operating point when the power angle is equal to the angle of synchronous impedance. Z1 corresponds to the operating point at which the power angle is zero, n1 is the operating point which corresponds to the normal excitation where there is no reactive component of armature current; n1 h1 is the region of underexcitation and n1 g1 is the region of over-excitation. At operating point h1, the excitation is oh1, the armature current is kh1 and the power-factor angle is f h1 leading with the increase of excitation from the operating point h1. The operating point moves towards z1 on the constant power line through a1, x1, b1 and g1. Figure 6.46 shows the plot of armature current versus field current at different output power condition. It is observed that as the operating point moves from h1 towards z1, the excitation increases, armature current reduces to a minimum value of kn1 at operating point n1 and again increases. The power-factor angle is initially leading, decreases to zero at n1 and again increases and becomes lagging. The V curves, as shown in Fig. 6.46 for increasing powers, line one above the other without intersecting each other. h4 h3 h2 h1

Inc

Po Po

4

Po 2 Po

rea

sin

g P Le

U.P .F. f

ad

o

3

1

La

n4 n3

a1

g

g4

n2

g3 g2

n1 b1

fh1 dmax = q

q

g1

z1

to z4 on x-axis to z3 on x-axis to z2 on x-axis

k

O

Gjh/!7/56! Dpotubou!qpxfs!mjoft! h4 |Ig|

h3 kh1

h1

h2

Locus of minimum current

V4 ka1

V3

a1

n4 z1

kz1

n3

V2

g1

kg1

V1

n2

kb1

b1 n1

kn1

oh1

oa1

on1

ob1

og1

oz1

Gjh/!7/57! W!dvswft!pg!b!tzodispopvt!hfofsbups!

If

!

7/85

Fmfdusjdbm!Nbdijoft

JOWFSUFE!W!DVSWFT!

7/27!

The plot of power factor of the armature current versus the main field current at constant output power is known as inverted V curve. It is known that at constant voltage, for constant output power, the product of current and its power factor is constant. Hence, the variation of power factor with main field excitation is inverse to the variation of armature current with main field excitation. Therefore the variation of power factor follows the inverse V curve. Figure 6.47 shows typical inverted V curves. Power factor n1 n2 n3 n4

1.0

U1 U2

h3

cos q cos fh1

O

U3

U4 h4 z1

h2

z2

z3

z4

h1 h1

n1

z1

If

Gjh/!7/58! Jowfsufe!W!dvswft!gps!b!tzodispopvt!hfofsbups!

The maximum power factor for all the power is unity and they occur at successively higher excitations for higher power factors. Let us consider the curve 1 on the inverted V curve in Fig. 6.47. The operating point h1 on this curve corresponds to the lowest possible excitation oh1 for which power factor cos f h1 is lowest. As the operating point moves from h1 to z1, the power factor increases with the increase in excitation, reaches 1 at n1 and decreases with lagging phase angle. At z1, the power angle is same as the synchronous impedance angle q. With the increase in output power, the minimum excitation increases and corresponding power factor increases. The locus of minimum power factor (leading) moves along h1, h2, h3 and h4 in Fig. 6.47. The locus of the lagging power factor cos f moves horizontally along z1, z2, z3 and z4. The unity power factor points on the inverted V curves are also shifted horizontally along n1, n2, n3 and n4. Hence, inverted V curves are a set of intersecting curves shifted horizontally for increasing output power.

PQFSBUJPOBM!BTQFDUT!

7/28

If the prime-mover input to the generator is increased, keeping the excitation constant, the operation can be treated equivalent to operating the generator in parallel with another. The other generator in this case is considered the constant voltage and constant frequency infinite bus bar. The change in prime-mover input does not affect the infinite bus bar but results in change in the output of the generator. But for constant excitation, the internal emf and terminal voltage of the generator is

Tzodispopvt!Hfofsbups

7/86

constant. Hence, the power-angle characteristic of the generator changes. Due to change in output power, the active component of armature current changes resulting in the change in power factor also. Figure 6.48 illustrates the effect of change in output power at constant excitation. A set of constant power lines Po1 to Po6 is shown in this figure. At constant output power Po1, g1 is the initial operating point, d 1 is the power angle, E1 is the internal emf and Ig1 is the armature current at a power factor of cos f1 lagging. If the excitation is held constant and the prime-mover input is increased gradually, the operating point moves from g1 over an arc of the circle with o as centre and og1 as radius. Hence, the operating points become g2, g3, g4, g5 and g6 at output powers of Po2, Po3, Po4, Po5 and Po6 respectively. The circular locus of three operating points is known as Constant Excitation Circle.

g6

Le

Po

ad

6

Po

U.P .F f

5

Ex1

Po

Ex2

4

Po

La

g

g5

3

Po

2

Po P

o0

1

g4 g3

=0

dmax = q O

E2 g2 f2 E g ZsIg2 1 1

d2 d1

k

f1 ZsIg1 g0

Gjh/!7/59! Fggfdu!pg!dibohf!jo!pvuqvu!qpxfs!bu!dpotubou!fydjubujpo!

Figure 6.48 shows the constant excitation circle. Ex1 intersects several constant power lines. At the operating point g2, the power angle increases to d 2, armature current is Ig2 at cos f2 power factor lagging and internal emf is E2 where E2 = E1. Similarly, at the operating point g3, the power angle is d 3 and armature current is Ig3 at cos f3 power factor lagging. Here, emf E3 = E1. At the operating point g4, the power factor is 1. For powers higher than Po4, the power factor is leading and starts reducing in magnitude. Power angle and magnitude of armature current keep increasing further but the internal emf is constant at each operating point. g6 is the limiting operating point at output power Po6, at which the power line is tangent to the excitation circle. At this point, the power angle is equal to the angle of the synchronous impedance. The other extreme limiting operating point is g 0 at which the output power is zero. The power factor here is zero lagging.

!

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Fmfdusjdbm!Nbdijoft

When the excitation is kept constant at the value equal to the voltage of infinite bus bar, the excitation circle is Ex 2 as shown in Fig. 6.48. In this case, with increase in prime-mover input, the armature current increases but with only leading power factors. At the instant of synchronizing the generator with infinite bus, the internal emf of the generator is made equal to the infinite bus-bar voltage. If the generator is made to deliver power to the infinite bus soon after synchronizing, keeping the excitation constant at the value just before synchronizing, the armature current operates only at leading power factors. A family of constant power lines, excitation circles and constant current circles drawn together is called an electrical load diagram. Knowing the limiting operating conditions of the generator, a boundary can be identified, the area within which is the permissible operating region of the generator. Such a boundary is known as the operating chart. In the electrical load diagram, the output power, excitation and armature currents are expressed in per unit. Figure 6.49 shows a typical electrical load diagram of a synchronous generator and the permissible operating area when excitation emf Ex should not exceed 1.5 p.u. and fall below 0.6 p.u., armature current should not exceed 1 p.u., active power should not exceed 1 p.u. and maximum lagging power-factor angle should not exceed synchronous impedance angle. The boundary line is the thick line a1 – a2 – a3 – a4 – a5 – a6 inside which none of the given specifications is exceeded. Ex = 1.5

0p .u.

Ex = 1.2 5

Ex = 1p Po Po

Po

=0

.25

=P

o0

f

.u. Po

=1

.25

.0 p .u.

=0 .75

Ex = 0 Po

=1

p.u .

p.u

.

p.u .

a4

.6 p .u. Po

p.u

=0

. a5 a6

=0 .5 p

.u.

Ig = 0.5 p Ig = 0.2

Ig = 0.7 5 .u.

a3

p.u

. a2 Ig = 1.0 p.u.

5p

dmax = q

q k

.u. a1

Gjh/!7/5:! Fmfdusjdbm!mpbe!ejbhsbn!pg!b!tzodispopvt!hfofsbups

Tzodispopvt!Hfofsbups

TBMJFOU.QPMF!TZODISPOPVT!HFOFSBUPS!

7/88

7/29!

As discussed earlier, the rotor of a cylindrical pole maq-axis chine has uniform air gap whereas in a salient pole machine, the air gap is non-uniform. Figure 6.50 shows a two-pole salient-pole synchronous generator. From the figure, it is clear that there are two axes of symmetry here. The axis of the rotor pole is called the direct or d d-axis axis and the axis perpendicular to the rotor pole is called the quadrature or q axis. The direct-axis flux path includes two small air gaps under pole faces only whereas quadrature axis flux path has two larger air gaps in the interpolar region. Hence, the direct-axis flux path has minimum reluctance and quadrature-axis flux path has maximum reluctance. The effect of armature reaction is less pronounced along the quadrature axis than along Gjh/!7/61! Uxp.qpmf!tbmjfou.qpmf!nbdijof the direct axis. Since the armature reaction reactance, which represents the effect of armature reaction, is a major component of synchronous reactance, a salient-pole machine may be considered to have two different armature reaction reactances, one representing the effect of armature reaction along the direct axis and the other along the quadrature axis. Consider xad and xaq as the armature reaction reactance along the direct axis and quadrature axis respectively. As inductance is flux linkage per unit ampere, xad > xaq. However, the leakage flux xl is unaffected by the difference in the air-gap length along the two axes. The net direct axis and net quadrature axis reactance are Xd = xad + xl and Xq = xaq + xl respectively. Hence, Xd > Xq. The typical ratio

Xq Xd

= 0.6.

UXP.SFBDUJPO!UIFPSZ!BOE!QIBTPS!EJBHSBN!PG! B!TBMJFOU.QPMF!TZODISPOPVT!HFOFSBUPS!

7/2:

The performance of a salient-pole synchronous generator can be explained by the two-reaction theory. According to the two reaction theory, the armature mmf Fa produced by armature current Ia is resolved into two components, one along the direct or d-axis and another along quadrature or q-axis. The d-axis component of Fa is Fad and the q-axis component is Faq. The effect of Fad is either magnetizing or demagnetizing (depending on whether p.f. is leading or lagging), whereas the effect of Faq is entirely cross magnetizing. Similarly, the flux fa produced by Ia is resolved into two components fad and faq. fad induces direct-axis armature reaction voltage Ead and faq induces quadrature-axis armature reaction voltage Eaq. If Id and Iq be the two components of Ia along direct and quadrature axis respectively then

!

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Fmfdusjdbm!Nbdijoft

and

Ead = – jXd Id

…(6.68)

Eaq = – jXqIq

…(6.69)

Figure 6.51 shows the phasor diagram of a salient-pole alternator based on two-reaction theory. The terminal voltage Vt is taken as reference. Ia is assumed to be lagging with respect to Vt by an angle q. The two perpendicular components of Ia along d-axis and q-axis are Id and Iq along d-axis and q-axis respectively. Similarly, the resultant flux f r has been resolved into d-axis and q-axis components fd and fq respectively. The daxis component of armature reaction voltage Ead lags behind fad by 90° and q-axis component of armature reaction voltage Eaq lags behind faq by 90° as shown in Fig. 6.51. The resultant air-gap voltage is given by Ea = E + Ead + Eaq Vt = Ea – Ia ra – j Ia xl

…(6.70) …(6.71)

Now Ia xl can be resolved into two components.

q-axis Eaq

E

Ead Ear Ea

Iaxl Iara

Vt

d Iq Faq faq

Ia q

Fa

fr

fad Fad

O

Id

d-axis

Ead

Gjh/! 7/62! Qibtps! ejbhsbn! pg! b! tbmjfou.qpmf bmufsobups! cbtfe! po! uxp.sfbdujpo! uifpsz

Ia xl = Id xl + Iq xl Again Id (xad + xl) = Id Xd Iq (xaq + xl) = Iq Xq

and

…(6.72) …(6.73) …(6.74)

where Xd and Xq are direct-axis and quadrature-axis synchronous reactance. The simplified phasor diagram is shown in Fig. 6.52. Let E1 be a part of E such that

E jIqxq

E1 = Vt + ra Id + ra Iq + jXq Id + jXq Iq = Vt + ra (Id + Iq) + jXq (Id + Iq)

jIdxd Iq

= Vt + (ra + jXq) Ia = Vt + Zsq Ia

Vt

…(6.75)

where Z sq is the quadrature-axis synchronous impedance. Figure 6.53 shows the phasor diagram of a salient-pole synchronous machine with the inclusion of E1 and Zsq Ia.

d

q

Iara Ia

Idra

Iqra

Id

Gjh/! 7/63! Tjnqmjgjfe! qibtps! ejbhsbn! pg! tbmjfou.qpmf! tzodispopvt! hfofsbups

Tzodispopvt!Hfofsbups

7/8:

E E1

q jI qX

sd

V t co

Iq

Vt

d

jI d(

Ia r

I dr a

q

jIaXq

sq I aZ

Ia

Xd

) – Xq

Xq jI d

a

Iqra

Id

Gjh/!7/64! Npejgjfe!qibtps!ejbhsbn!pg!b!tbmjfou.qpmf!tzodispopvt!hfofsbups!

From Fig. 6.53, E1 = Vt + ra Ia cos q + Xq Ia sin q + j (Xq Ia cos q – ra Ia sin q) E1 =

or, and

…(6.76)

(Vt + ra I a cos q + X q I a sin q )2 + ( X q I a cos q - ra I a sin q )2

d = tan–1

…(6.77)

X q I a cos q - ra I a sin q

…(6.78)

Vt + ra I a cos q - X q I a sin q

Also, | E | = | E1 | + (Xd – Xq) | Id |

…(6.79)

Similarly, for leading power factor d = tan–1

and

E1 =

X q I a cos q + ra I a sin q

…(6.80)

Vt + ra I a cos q - X q I a sin q

(Vt + ra I a cos q + X q I a sin q )2 + ( X q I a cos q - ra I a sin q )2

…(6.81)

NFUIPET!PG!EFUFSNJOJOH!Ye!BOE!Yr! Xd and Xq are the two reactances which exist due to different effects of armature mmf when acting along the direct axis and along the quadrature axis. The main problem in determining Xd and Xq is that during normal operation, the air-gap flux is due to the resultant of main field and armature field mmfs whereas it is required to evaluate the effect due to armature mmf alone. To understant the basic concept of slip test, a magnetic circuit, as shown in Fig. 6.54, is considered.

Core

N Turns

7/31

q

x Rotor

Gjh/!7/65! Djsdvju!gps!fyqmbobujpo!pg!tmjq!uftu!

!

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Here, the magnetic core is wound with N turns of a coil. The rotor rotates freely about a horiNf where f is the I flux produced by the winding when it carries a current I. The flux produced by the current I through the coil depends on the reluctance offered by the magnetic circuit to the mmf NI produced by the winding. In Fig. 6.54, when q = 0°, the length of the air gap is minimum and, therefore, the reluctance is minimum. The flux is maximum and let in this position the winding inductance be L1. If the rotor is rotated by 90°, i.e. when q = 90°, the air gap is maximum and the reluctance is maximum. Assuming that the voltage source drives the same current as before, the flux produced is minimum and let in this condition the inductance be L2. Hence, for frequency f, the maximum and minimum reactances are Xmax = 2pfL1 and Xmin = 2pfL2. On the other hand, for a constant voltage source, the current through the winding is minimum corresponding to the position of maximum flux and vice versa. Therefore, by noting the minimum and maximum currents due to shift of the position of rotor, the maximum and minimum values of reactances can be evaluated. The maximum reactance is the ratio of the applied voltage to the minimum current and minimum reactance is the ratio of the applied voltage to the maximum current. The method for evaluation of Xd and Xq is based on this principle. The circuit diagram shown in Fig. 6.54 is used for conducting test to measure Xd and Xq. The armature windings of the synchronous machine are supplied with balanced three-phase supply from a variac with the field windings kept open. A voltmeter measures the voltage across the field winding. Oscillograms of armature current, armature voltage and voltage across field winding may also be obtained by connecting suitable instruments. While conducting test on the synchronous machine, a balanced three-phase voltage is applied across armature winding in such a way that rated armature current is not exceeded. The threephase armature currents produce a magnetic field in the air gap rotating at synchronous speed. At stationary condition due to relative speed between the armature field and the field winding, a voltage is induced in the field winding at normal frequency. Now the rotor is rotated by an adjustable prime mover in the direction of armature field. Whether the rotor is rotating in the direction of the magnetic field or in the opposite direction can be determined from the reading of the voltmeter connected across the field winding. A decrease in voltage in the voltmeter indicates that the speed between the armature field and rotor is decreasing, which indicates that the direction of rotation of the rotor is same as that of the armature field. The voltage across the voltmeter increases if the speed of rotation of the rotor is opposite to that of the magnetic field. If the rotor is rotated at synchronous speed, there will be no voltage induced in the field winding since at this condition, the magnetic field is stationary with respect to the rotor. If the rotor is driven at a speed slightly less or slightly more than the synchronous speed in the direction of armature magnetic field, the armature field slips past the rotor at slip speed inducing slip-frequency voltage in the field winding. As the armature field is running at a speed slightly different than that of the rotor, the armature flux acts along the direct axis and quadrature axis of the rotor at equal intervals of time which is determined by the slip speed. If the slip is large, the time interval is small and vice versa. For example, a 6-pole,

zontal axis X. The inductance of the winding is the flux linkage per unit ampere,

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7/92

50 Hz generator has a synchronous speed of 1200 rpm. when driven at 1194 rpm, the slip is 0.5%. The slip speed is (1200 – 1194) rpm, i.e. 6 rpm. The time taken for one revolution of the armature field with respect to rotor is 10 s. As the rotor has 6 poles, the armature field acts along the direct 10 axis and quadrature axis at intervals of s or 1.67 s. When the armature field acts along the direct 6 axis, the situation is same for q = 0° in Fig. 6.54. The armature reactance is maximum and this is the direct-axis synchronous reactance Xd. After 1.67 s, the armature field acts along the quadrature axis and the armature reactance is minimum, i.e. Xq. Hence, for a constant voltage applied to the armature winding, the armature current swings from a maximum value corresponding to Xq to a maximum value corresponding to Xd. The time interval between the instant of maximum and minimum current is 1.67 s in this example. The record of maximum and minimum current will be better if the slip frequency is lower. Now at constant supply voltage, Applied armature voltage per phase Xd = …(6.82a) Minimum armature current per phase and

Xq =

Applied armature voltage per phase Maximum armature current per phase

…(6.82b)

Sometimes the voltage applied to the armature may not be constant but varies from a maximum value when the armature current is minimum to a minimum when the armature current is maximum. In this case, Maximum value of applied voltage per phase Xd = …(6.83) Minimum value of armature current per phase and

Xq =

Minimum value of applied voltage per phase Maximum value of armature current per phase

…(6.84)

As discussed earlier, the open-circuit field winding has a slip-frequency induced voltage reaching a maximum when the axes of the field winding and the armature field are in quadrature and zero when the two axes coincides. Hence, the instant at which the field voltage is zero indicates the coincidence of the axis of the armature field with the direct axis of the rotor and the instant when the field voltage is maximum indicates the coincidence of the axis of the armature field with the quadrature axis. The instants at while the armature flux is acting along the d-axis and q-axis can be determined precisely if we take the oscillogram of the open circuit voltage across the field winding. Typical oscillograms of the voltage across field winding, armature current and the voltage applied to armature are shown in Fig. 6.55. The envelope of armature current and armature voltage have a cyclic variation at twice the slip frequency. At the instant t1, the field voltage is zero, the armature current is minimum and armature voltage is maximum which correspond to the coincidence of the quadrature axis of the rotor with the axis of armature mmf. At the instant t2, the field voltage is maximum, armature current is maximum and the armature voltage is minimum which corresponds to the coincidence of the direct axis of the rotor with the axis of the armature mmf.

!

7/93

Fmfdusjdbm!Nbdijoft

Field voltage t1

t2

t

Armature current

t

Armature voltage

t

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QPXFS.BOHMF!DIBSBDUFSJTUJD!PG!TBMJFOU.! QPMF!TZODISPOPVT!HFOFSBUPS! The armature resistance ra has negligible effect on the relationship between the power output of a synchronous machine and its torque angle d. Hence, ra is neglected for determining power angle characteristic of synchronous generator. From the simplified phasor diagram of Fig. 6.52, neglecting ra, Fig. 6.56 is obtained. Taking E as reference,

E

Xq jI q

B

Xd jI d

Iq O

C

d q

Vt

A

Ia

Ia = Iq – j Id \

Id

Ia* = Iq + j Id Hence, complex power output per phase S = Vt I a* = Vt -d ¥ (Iq + j Id)

Gjh/!7/67! Tjnqmjgjfe!qibtps!ejbhsbn! pg!tbmjfou.qpmf!tzodispopvt! hfofsbups!ofhmfdujoh!sb

= (Vt cos d – j Vt sin d) (Iq + j Id) From Fig. 6.56, Iq Xq = AB = Vt sin d

…(6.85)

Tzodispopvt!Hfofsbups

\

Iq =

7/94

Vt sin d Xq

…(6.86)

Id Xd = OC – OB = E – Vt cos d \

Id =

E - Vt cosd Xd

…(6.87)

Substituting the value of Iq and Id in Eq. (6.85), E - Vt cos d ˆ Ê V sin d S = (Vt cos d – jVt sin d) Á t + j ˜¯ Xd Ë Xq ÊV2 ÊV E ˆ ˆ VE V2 V2 V2 = Á t sin d cos d + t sin d - t sin d cos d ˜ + j Á t cos d - t cos 2 d - t sin 2 d ˜ …(6.88) Xd Xq Xd Xq Ë Xq Ë Xd ¯ ¯ ÏÔV E ÏV E V2 V2 V2 Ê 1 1 ˆ Ô¸ Ô¸ sin 2d ˝ + j Ì t cos d - t (1 + cos 2d ) - t (1 - cos 2d ) ˝ = ÔÌ t sin d + Á ˜ X 2Xq 2 Ë Xq Xd ¯ 2Xd ÔÓ X d ˛Ô ÓÔ d ˛Ô ¸Ô ÏÔV E V2 Ê 1 1 ˆ = Ì t sin d + t sin 2d ˝ + X 2 ÁË X q X d ˜¯ ˛Ô ÓÔ d

ÏÔV E ¸Ô Vt 2 j Ì t cos d ÈÎ( X d + X q ) - ( X d - X q ) cos 2d ˘˚ ˝ 2X Xd Xq ÓÔ X d ˛Ô …(6.89)

\ real power output per phase P=

Vt E V2 Ê 1 1 ˆ sin d + t sin 2d Á Xd 2 Ë X q X d ˜¯

…(6.90)

Reactive power output per phase Q=

Vt E Vt 2 cos d È( X d + X q ) - ( X d - X q ) cos 2d ˘˚ Xd 2Xd Xq Î

…(6.91)

From an infinite bus system, Vt is constant and E can be made constant by keeping the excitation constant. Hence, real power P is a function of d only. The first term on the right-hand side of Eq. (6.90) is called excitation power and the second term is the reluctance power. Hence, Excitation power =

Reluctance power =

Vt E sin d Xd

…(6.92)

Vt 2 Ê 1 1 ˆ sin 2d 2 ÁË X q X d ˜¯

…(6.93)

The reluctance power depends on the saliency defined by the quantity

Ê ˆ . For cylinÁË X q X d ˜¯

drical rotor machine, Xd = Xq and hence, the reluctance power becomes zero for cylindrical rotor

!

7/95

Fmfdusjdbm!Nbdijoft

machine. The excitation power becomes zero if there is no field excitation. But the reluctance power exists in a salient-pole machine even when there is no field excitation. Hence, the machine has the same real power generator capability even in the absence of field excitation. But it is impractical to operate synchronous generator without field excitation because it would supply only about 25% or less of its real power rating. Also, it would absorb an excessive amount of reactive power. The reluctance power is 10 to 20 percent of the excitation power. Equation (6.92) shows that the excitation power is proportional to sin d and Eq. (6.93) shows that the reluctance power is proportional to sin 2d. The power-angle curve for salient-pole machine is shown in Fig. 6.57. Motor

Generator P = wst Pmax

PT

3 2È 1 1 ˘ V Í ˙ 2 ÍÎ X q X d ˙˚

3VEf Xd Pr

–dmax –180°

PE

90° O

–90°

dmax

d 180°

Reluctance power

–Pmax

PT = Total power PE = Excitation power Pr = Reluctance power

Gjh/!7/68! Qpxfs.bohmf!dvswf!gps!tbmjfou.qpmf!hfofsbups!

TZODISPOJ[JOH!QPXFS!BOE!

7/33

TZODISPOJ[JOH!UPSRVF! For a cylindrical rotor synchronous generator, P=

VE sin d xs

If the load angle is increased by a, the increased power, P¢ =

VE sin (d + a) xs

…(6.94)

Tzodispopvt!Hfofsbups

7/96

The difference between P¢ and P is known as the synchronizing power Ps \

Ps = =

3VE {sin(d + a ) - sin d } xs 3VE {sin d cos a + cos d sin a - sin d } xs

…(6.95)

As a is very small, cos a = 1 and sin a = a \

Ps =

VEa cos d xs

…(6.96)

Synchronizing power per electrical radian VE cos d …(6.97) xs Alternatively, the rate at which synchronous power P varies with d is called synchronizing power co-efficient Psy. dP EVt \ Psy = cos d watts/electrical radian …(6.98) dd xs

=

or,

Psy =

p EVt cos d watts/electrical degree 180 xs

…(6.99)

Psy is also known as the stiffness ratio of the machine. Similarly, for a salient-pole machine, Psy =

dP EVt 1 ˆ Ê 1 = cos d + Vt 2 cos 2d watts/electrical radian ÁË X q X d ˜¯ dd Xd

…(6.100)

Synchronizing power co-efficient gives rise to synchronizing torque co-efficient at synchronous speed. The synchronizing torque co-efficient dP Nm/electrical radian w s dd

…(6.101)

=

1 dP p Nm/electrical degree w s d d 180

…(6.102)

=

1 dP p P Nm/mechanical degree w s d d 360

…(6.103)

Tsy =

From Eq. (6.96), if the change in load angle is represented by Dd then synchronizing power Ps =

EVt dP Dd = cos d Dd for cylindrical rotor xs dd

…(6.104)

!

7/97

Fmfdusjdbm!Nbdijoft

For a salient-pole rotor, ¸ 1 ˆ Ï EV Ê 1 Ps = ÔÌ t cos d + Vt 2 cos 2d Ô˝ Dd Á ˜ X X X Ë q d¯ Ô˛ ÔÓ d At no load d @ 0 and Vt = E. Hence, synchronizing power at no load Vt Dd xs E Vt As = = steady-state short-circuit current Isc xs xs Ps = Vt Isc Dd watts

Ps =

…(6.105)

…(6.106)

…(6.107)

The synchronizing power gives rise to synchronizing torque Ts. Ts =

ws

Ps =

dp Dd = Tsy Dd. ws d d

…(6.108)

! Qspcmfn!7/52 Jo!b!tmjq!uftu!pg!b!551!W-!uisff.qibtf-!61!I{!tbmjfou.qpmf!tzodispopvt!nbdijof-!uif!gpmmpxjoh!sftvmut! bsf!pcubjofe; Wpmunfufs!sfbejoh!)Wpmubhf0qibtf*;!281!W-!261!W-!271!W!boe!251!W Bnnfufs!sfbejoh;! ! ! ! ! ! ! ! ! 26!B-!27!B-!26!B-!28/6!B! Efufsnjof!uif!wbmvft!pg!Ye!boe!Yr/

Solution Xd = = Xq = =

Maximum voltmeter reading Minimum ammeter reading 170 W = 11.33 W 15 Minimum voltmeter reading Maximum ammeter reading 140 W = 8 W. 17.5

! Qspcmfn!7/53 B! 26! NWB-! 22! lW-! 61! I{-! uisff.qibtf! tubs.dpoofdufe! tzodispopvt! hfofsbups! pqfsbujoh! bu! vojuz! qpxfs!gbdups!hjwft!uif!gpmmpxjoh!wpmunfufs!boe!bnnfufs!sfbejoht!jo!uif!tmjq!uftu;!46!W-!36!W-!26!B! boe!9!B/!Efufsnjof!uif!fydjubujpo!wpmubhf!boe!sfmvdubodf!qpxfs!efwfmpqfe!cz!uif!nbdijof/

Tzodispopvt!Hfofsbups

7/98

Solution From Eq. (6.83), Xd =

35 W = 4.375 W 8

Xq =

25 W = 1.67 W 15

From Eq. (6.84),

Full-load armature current 15 ¥ 103

Ia = Vt =

3 ¥ 11 ¥ 1 1100 3

A = 787.3 A

V = 6351 V

From Eq. (6.75), E1 = Vt + Ia (ra + j X q) Neglecting ra E1 = 6351 + j 787.3 ¥ 1.67 = 6351 + j 1314.79 = 6485.67 11.69∞ \ \

d = 11.69° Id = Ia sin d = 787.3 sin 11.69° = 159.52 A Per phase excitation emf per phase from Eq. (6.79), E = 6485.67 + (4.375 – 1.67) ¥ 159.52 = 6917.17 V

\ line value of excitation voltage E=

¥ 6917.17 = 11980.9 V = 11.98 kV

From Eq. (6.93), reluctance power is

=

Vt 2 Ê 1 1 ˆ sin 2d Á 2 Ë X q X d ˜¯

(6351) 2 Ê 1 1 ˆ ÁË ˜ sin (2 ¥ 11.69°) 2 1.67 4.375 ¯

= 8003058.682 (0.5988 – 0.22857) = 2962972 W = 2962 kW per phase Total reluctance power = 2962 ¥ 3 kW = 8.886 MW.

!

7/99

Fmfdusjdbm!Nbdijoft

! Qspcmfn!7/54 A three-phase star-connected synchronous generator is delivering a power of 0.8 p.u. to an infinite bus at rated voltage and at 0.9 power factor lagging. Determine the load angle and the excitation voltage if Xd = 0.85 p.u. and Xq = 0.5 p.u. Solution = Vt Ia cos q = 1 ¥ Ia ¥ 0.9 = 0.89 p.u. = 0.9, sin q = 0.436

P 0.8 Ia cos q

\ or, From Eq. (6.78),

d = tan–1 = tan–1 \

X q I a cos q Vt + X q I a sin q

(neglecting ra)

0.5 ¥ 0.89 ¥ 0.9 1 + 0.5 ¥ 0.89 ¥ 0.436

d = 18.54° From Fig. 6.56, Id = Ia sin (q + d) Now Id Xd = E – Vt cos d

\

E = 1 cos 18.54° + 0.89 sin (cos–1 0.9 + 18.54°) = 1.57 p.u.

! Qspcmfn!7/55 B!3111!lWB-!tubs.dpoofdufe-!3611!W-!uisff.qibtf!tbmjfou.qpmf!tzodispopvt!hfofsbups!ibt!Ye!>!3!W! boe!Yr!>!2/3!W!qfs!qibtf/!Ofhmfdujoh!mpttft-!efufsnjof!uif!fydjubujpo!wpmubhf!pg!sbufe!LWB!pqfsbujpo! boe!bu!1/9!qpxfs!gbdups!mbhhjoh/

Solution Vt = Ia = \

2500 3

V = 1443.37 V

2000 ¥ 103 3 ¥ 2500

A = 462 A at 0.8 p.f. lag

Ia = 462 - 36.87∞ A From Eq. (6.75), neglecting ra E1 = Vt + j Ia Xq = 1443.37 + j 462 - 36.87∞ ¥ 1.2 = 1443.37 + 554.4 53.13∞

Tzodispopvt!Hfofsbups

7/9:

= 1776 + j 443.52 = 1830.54 14∞ V From Fig. (6.56), Id = Ia sin (q + d) = 462 sin (36.87° + 14°) = 358.38 A From Eq. (6.79), E = 1830.54 + (2 – 1.2) 358.38 = 2117.24 V.

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Solution Vt = 440 V Xd = 0.5 W Xq = 0.1 W Ia =

800 3

-36.87∞ A

From Eq. (6.78), neglecting ra 800 ¥ 0.8 3 = tan–1 0.079 800 440 + 0.1 ¥ ¥ 0.6 3 0.1 ¥

d = tan–1 = 4.520 From Fig. 6.56,

Id = Ia sin (q + d) = and \

800 3

sin (36.87° + 4.52°) = 305.39 A

E = Id Xd + Vt cos d E = 305.39 ¥ 0.5 + 440 cos 4.52° = 591.33 V.

! Qspcmfn!7/57 B!3111!lWB-!4.qibtf-!tubs.dpoofdufe!22!lW-!5.qpmf-!61!I{!tzodispopvt!hfofsbups!ibt!tzodispopvt! sfbdubodf!pg!26&!boe!jt!dpoofdufe!up!bo!jogjojuf!cvt/!Efufsnjof!uif!tzodispoj{joh!qpxfs!boe!uif! tzodispoj{joh!upsrvf!qfs!nfdibojdbm!efhsff!pg!qibtf!ejtqmbdfnfou!)j*!bu!op!mpbe-!boe!)jj*!bu!gvmm!mpbe! 1/9!qpxfs!gbdups!mbhhjoh/

!

7/:1

Fmfdusjdbm!Nbdijoft

Solution Vt =

11, 000

Ia =

3

V = 3651 V

2000 3 ¥ 11

xs = 0.15 ¥

= 104.97 A

(11)2 (kV) 2 = 0.15 ¥ W 2 MVA = 9.075 W

(i) At no load, d = 0 and E = Vt = 11 kV \ synchronizing power per phase Ps = =

Vt E pP cosq W/mechanical degree xs 360 (3651) 2 p ¥ 4 ¥ W = 51.246 kW 9.075 360

\ synchronizing torque per mechanical degree Ts = =

w

¥ 51.246 ¥ 103 Nm

1 ¥ 51.246 ¥ 103 Nm 2 ¥ 50 2p ¥ 4

= 326.41 Nm/per phase (ii) At full-load, E = 3651 + 104.97 -36.87∞ ¥ j 9.05 = 3651 + 949.98 53.13∞ = 4221 + j 760 = 4289 10.21∞ V Hence, synchronizing power per mechanical degree 3651 ¥ 4289 p ¥ 4 cos 10.21° ¥ 9.05 360 = 59412 W/phase Similarly, the synchronizing torque per mechanical degree

Ps =

1 ¥ 59412 Nm 2 ¥ 50 2p ¥ 4 = 378.42 Nm/phase.

=

Tzodispopvt!Hfofsbups

SPMF!PG!EBNQFS!XJOEJOH!

7/:2

7/34

Damper windings are used in salient-pole machines which help in maintaining synchronism by providing additional damping. The damper winding consists of low-resistance copper or aluminium bars embedded in notches carved in rotor pole shoes and short-circuited at both ends by rings. This winding is also known as amortisseur winding. During normal operating condition, the relative speed between the damper winding and the airgap magnetic field is zero as the copper bars of the damper winding are embedded on the pole face of the rotor. Under this condition, there is no induced voltage in the damper windings. But when a disturbance occurs and the rotor deviates from synchronism, the damper bars have relative motion with respect to the air-gap field. Hence, a voltage is induced in the damper bars. Since the damper bars are short-circuited, a current flows in the bars. This induced current in the bars reacts with the air-gap field and produces an effect so as to restore synchronism between the rotor and the air-gap magnetic field. The torque due to the damper winding is the induction torque. The induced emf in the damper bar is proportional to the relative velocity between the rotor and the air-gap magnetic field. Let the initial load angle during normal operating condition be do. Let the incremental change in load angle be Dd when a disturbance occurs. Hence, the load angle becomes d = do + Dd and

d d d ( Dd ) = Dw = dt dt

where Dw is the incremental speed deviation of the rotor from synchronous speed. The restoring dd dd i.e., Td = Kd = Kd Dw where Kd is torque Td, known as damping torque, is proportional to dt dt the constant of proportionality in Nm/rad/s. The restoring torque is also called induction torque. The damper winding also serves as a starter for synchronous motor which is not self-starting. The cylindrical rotor machine does not require to use the damper winding. The solid-steel rotor cores of these machines provide the path for eddy currents, which provide damping torque.

TZNNFUSJDBM!BOE!VOTZNFUSJDBM!TIPSU!DJSDVJUT!

7/35

!JO!B!UISFF.QIBTF!TZODISPOPVT!HFOFSBUPS! The most severe transient condition that can occur in a synchronous generator is three-phase short circuit at the armature terminals. The machine is assumed to be initially unloaded and operates at synchronous speed after short circuit occurs. Figure 6.58 shows the nature of current variation in any one of the three phases when the armature terminals of the generator are suddenly short circuited. During the first few cycles, the waveform of armature current will depend upon the point in the emf cycle at which short circuit occurs. The current in each phase consists of an ac component and a dc component. The ac component of the symmetrical short-circuit current can be divided into three periods—subtransient, transient and steady-state period.

!

7/:3

Fmfdusjdbm!Nbdijoft

Subtransient period

Steady-state period

Short-circuit current

Transient period

O

Time

Actual envelope Extrapolation of steady value Extrapolation of transient envelope

Gjh/!7/69! Uif!bd!dpnqpofou!pg!tznnfusjdbm!tipsu!djsdvju!dvssfou!jo!b!tzodispopvt!hfofsbups

The subtransient period lasts for about 2 cycles after the fault occurs and the current is very high during this period, about 5 to 10 times the rated current. The current drops rapidly during this period. The rms value of initial current is called subtransient current and is denoted by I ≤. The reE where E is the actance corresponding to I≤ is known as subtransient reactance Xd≤. Hence Xd≤ = I ¢¢ rms value of the open circuit voltage. After the subtransient period, the current decreases at a slower rate. This period is known as transient period which lasts for about 30 cycles. The rms value of the current during this period is known as transient current I ¢. I ¢ is about five times the steady-state fault current. The reactance of the armature winding corresponding to I ¢ is known as transient reactance X¢d where

Xd¢ =

E I¢

After the transient period, the fault current reaches its steady-state value. The rms value of current during this period is Isc and the synchronous reactance Xd =

E I sc

If a three-phase star-connected synchronous generator is driven at rated speed with an excitation sufficient to develop its voltage on open circuit, a sudden short circuit across any two terminals or between one line and neutral will cause it to function as a single-phase machine. The single-phase

Tzodispopvt!Hfofsbups

7/:4

armature current produces an mmf which alternates at normal frequency and is stationary with respect to the armature winding. This alternating but stationary mmf may be resolved into two components each having half the amplitude of the alternating original mmf. One component is revolving at synchronous speed in the same direction as the rotor and the other at equal speed in the opposite direction. The unsymmetrical short-circuit fault condition is most effectively handled by resolving the unbalanced current by the method of symmetrical components, i.e. positive, negative and zero sequence currents. The positive sequence components constitute a balanced set of three-phase currents which produce an mmf moving in step with the rotor field. The negative sequence components produce an armature mmf revolving at synchronous speed in the direction opposite to that of the rotor and, hence, set up double-frequency emfs and currents in the field winding, in the damper winding and in the eddy-current paths. The zero sequence components of armature current all have the same time phase, as a result of which the mmfs due to zero sequence phase currents are mutually 120° apart in space phase.

7/35/2! Nfbtvsfnfou!pg!Ejsfdu.byjt!Tvcusbotjfou!Sfbdubodf For measuring direct-axis subtransient reactance, the field winding is short-circuited and a single-phase voltage is impressed across two armature phase windings in series (connected for 120° phase displacement) as shown in Fig. 6.59. The rotor is blocked in the angular position for maximum induced field current. The subtransient reactance X≤d is then equal to half the applied voltage divided by the resulting armature current.

Amp.

Watts

If E

Gjh/!7/6:! Djsdvju!gps!nfbtvsfnfou! pg!Y e≤

7/35/3! Nfbtvsfnfou!pg!Ofhbujwf!Tfrvfodf!Sfbdubodf! a

The negative sequence impedance can be determined experimentally by applying balanced negative sequence voltage to the arE mature terminals with the machine driven at rated speed with its field winding short-circuited. The negative sequence impedance Watt. b is obtained by dividing the impressed voltage per phase by the Am. c current per phase. Figure 6.60 shows a star-connected machine which is driven at rated speed with its two terminals, b and c Gjh/!7/71! Djsdvju!gps!nfbtvsfnfou! short circuited. Voltage E is applied between the terminal a and pg! ofhbujwf! tfrvfodf! sfbdubodf the junction of terminals b and c. An ammeter and current coil of a wattmeter are connected in the short-circuited phases and the potential coil of the wattmeter is connected across the voltage E. The negative sequence impedance is Z2 =

E I

!

7/:5

Fmfdusjdbm!Nbdijoft

If P be the reading of the wattmeter then the negative sequence reactance is x2 = Z2

P EI

7/35/4! Nfbtvsfnfou!pg![fsp!Tfrvfodf!Sfbdubodf I

Field

To measure zero sequence reactance, the machine is driven at rated speed with its field winding short-circuited. All the three phases are connected in series and a single-phase current is circulated by impressing a voltage upon the open terminals as shown in Fig. 6.61. The zero sequence reactance is then E xo = I

E

Gjh/!7/72! Djsdvju!gps!nfbtvsfnfou!pg!

{fsp!tfrvfodf!sfbdubodf Sometimes it is convenient to connect the phases in parallel instead of in series. In this case, the voltage required will be one third of that required for series connection and the current will be three times larger. It should be checked that the currents flow simultaneously in the same direction from the line terminals

MPTTFT!BOE!FGGJDJFODZ!

7/36

The losses in a synchronous generator include 1. I 2R loss (a) in the armature winding, (b) in the field winding, and (c) at the contact between brushes and slip rings 2. Core loss 3. Friction and windage loss, including the loss due to circulation of cooling air through a closed ventilating system and also the brush-friction loss at the slip rings 4. The stray-load loss caused by eddy currents in the armature conductors and by additional core loss due to the distortion of the magnetic field under load condition. For large machines, there are some additional losses such as (i) loss in the field regulating rheostat, and (ii) the loss in ventilating ducts external to the machine itself. The copper losses in the armature and field windings are calculated by using the dc values of resistance connected, at the working temperature of 75°C. The core loss, including the hysteresis and eddy-current loss, is that which corresponds to the main flux at rated frequency when the machine develops an open-circuit voltage per phase equal to the phasor sum of its rated voltage per phase and the ohmic drop per phase. Brush-contact losses of synchronous machines are very small and can be neglected. The strayload loss per phase is added to the armature copper loss per phase and then it is divided by the

Tzodispopvt!Hfofsbups

7/:6

square of the armature current to get the effective value of the armature resistance. The combined value of friction and windage loss is either measured or estimated on the basis of experience with similar machines and in both cases, this loss is treated as constant loss. The efficiency of a synchronous generator is given by h=

output output 3 VL I L cos f = = input output + losses 3 VL I L cos f + losses

¥ 100%

[losses in denominator includes armature loss, core loss, frictional windage loss, stray loss and other small losses, as applicable]

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Solution Ia = Synchronous impedance Zs =

50 ¥ 103 A = 113.64 A 440 400 W = 1.33 W 300

ra = 0.4 \

Synchronous reactance xs = (1.33)2 - (0.4)2 = 1.268 W Generated emf for 0.8 power factor lagging E = (Vt cos q + I a ra )2 + (Vt sin q + I a xs ) =

2

(440 ¥ 0.8 + 113.64 ¥ 0.4)2 + (440 ¥ 0.6 + 113.64 ¥ 1.268) 2

= 157971 + 166542 = 569.66 V \

569.66 - 440 ¥ 100% 440 = 0.294%.

voltage regulation =

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!

7/:7

Fmfdusjdbm!Nbdijoft

Solution Vt =

11, 000 3

V = 6351 V

E = 6351 + 250 -36.87∞ (0.3 + j7) = 6351 + 1751.6 50.68∞ = 7460.9 + j 1355.07 = 7582.96 10.29∞ V As the input is constant and losses are neglected, the output is also constant. The power output at unity power factor and current Ia2 is \

Vt Ia2 =

\

Vt Ia2 cos q1 =

Vt Ia2 X1

Vt ¥ 250 ¥ 0.8

Ia2 = 200 A

The new value of induced emf E¢ = 6351 + 200 ∞ (0.3 + 7) = 6351 + 1401.28 87.55∞ = 6411 + j 1400 = 6562.08 12.32∞ V \ percentage change in induced emf 7582.96 - 6562.08 ¥ 100% 7582.96 = 13.46%.

=

! Qspcmfn!7/5: B! 21! lWB-! 351! W-! 2611! sqn-! uisff.qibtf-! 61! I{-! tubs.dpoofdufe! tzodispopvt! hfofsbups! ibt! bs. nbuvsf!xjoejoh!jnqfebodf!pg!1/5!,!k!7!W!qfs!qibtf!boe!gjfme!xjoejoh!sftjtubodf!pg!6!W/!Xifo!uif! hfofsbups!pqfsbuft!bu!gvmm.mpbe!bu!1/9!q/g/!mbhhjoh-!uif!gjfme!xjoejoh!dvssfou!jt!9!B/!Jg!uif!spubujpobm! mptt!jt!811!X-!efufsnjof!gvmm.mpbe!sfhvmbujpo-!fggjdjfodz!boe!upsrvf!bqqmjfe!cz!uif!qsjnf!npwfs/

Solution Vt = Ia =

240 3

V = 138.56 V

10 ¥ 103 3 ¥ 240

-36.87∞ A = 24.056 -36.87∞ A

\ generated voltage per phase E = 138.56 + 24.056 -36.87∞ ¥ (0.4 + j 6) = 138.56 + 144.656 49.316∞ = 232.8 + j 109.7 = 257.35 25.23∞ V

Tzodispopvt!Hfofsbups

7/:8

257.35 - 240 ¥ 100% = 7.23% 240 Power output = 3 ¥ 138.56 ¥ 24.056 ¥ 0.8 = 8000 W Armature copper loss = 3 ¥ (24.056)2 ¥ 0.4 = 694.43 W Field copper loss = (8)2 ¥ 5 = 320 W \ total loss = 700 + 694.43 + 320 = 1714.43 W 8000 ¥ 100% \ Efficiency = 8000 + 1714.43 = 82.35% The angular velocity of the prime mover is

\

Voltage regulation =

w s = 2p ¥

1500 rad/s = 50p rad/s 60

The torque applied by the prime mover =

8000 + 700 + 694.43 Nm = 59.84 Nm. 50p

! Qspcmfn!7/61 B!uisff.qibtf-!44!lW-!:11!lWB-!61!I{-!tubs.dpoofdufe-!tbmjfou.qpmf!tzodispopvt!nbdijof!ibt!ejsfdu! byjt!sfbdubodf!pg!1/9!q/v/!boe!rvbesbuvsf.byjt!sfbdubodf!pg!1/5!q/v/!Efufsnjof!uif!joevdfe!fng!boe! uif!qpxfs!bohmf!xifo!uif!bmufsobups!jt!tvqqmzjoh!sbufe!mpbe!bu!vojuz!qpxfs!gbdups/

Solution Xd = 0.8 p.u. Xq = 0.4 p.u. Vt = Ia =

KV 900

A = 15.75 A 3 ¥ 33 cos q = 1, q = 0° In per unit, Vt = 1 and Ia = 1 From Eq. (6.78), d = tan–1

X q I a cos q Vt + X q I a sin q

= tan -1

0.4 ¥ 1 ¥ 1 1 + 0.4 ¥ 1 ¥ 0

= tan–1 0.4 = 21.8° From Fig. 6.56, \

Id = Ia sin (q + d) = 1 sin 21.8° = 0.371 p.u. E = Vt cos d + Id Xd

!

7/:9

Fmfdusjdbm!Nbdijoft

= cos 21.8° + 0.371 ¥ 0.8 = 1.225 V \ induced emf is 1.225 ¥ 11 kV or 13.475 kV (line to line).

! Qspcmfn!7/62 Bo! 22! lW-! uisff.qibtf-! tubs.dpoofdufe! dzmjoesjdbm! spups! tzodispopvt! hfofsbups! ibt! tzodispopvt! sfbdubodf! pg! 2! W! qfs! qibtf! boe! jt! tvqqmzjoh!b! qpxfs! pg! 961! NX! bu! 1/9! qpxfs! gbdups! mbhhjoh/! Jg! uif!fydjubujpo!jt!jodsfbtfe!cz!26&!xjui!uif!pvuqvu!qpxfs!sfnbjojoh!dpotubou-!gjoe!uif!bmufsobups! dvssfou!boe!qpxfs!gbdups/!Xjui!uijt!fydjubujpo-!jg!uif!joqvu!qpxfs!up!uif!hfofsbups!jt!jodsfbtfe!up! 86!NX-!gjoe!uif!bmufsobups!dvssfou!boe!qpxfs!gbdups!

Solution 50 ¥ 103

Ia = Vt =

3 ¥ 11 ¥ 0.8 11000 3

-36.87∞ A = 3280 -36.87∞ A

V = 6351 V

\

E = Vt + Ia jxs neglecting ra

or,

E = 6350 + 3280 -36.87∞ ¥ 1 90∞ = 6350 + 3280 53.13∞ = 8318 + j 2624 = 8722 17.51∞ V

Now the increased excitation E¢ = 1.15 ¥ 8722 = 10030 V From Eq. (6.34), 11 3¥ ¥ 103 ¥ 10030 3 sin d 50 ¥ 106 = 1 \ sin d = 0.2616 or d = 15.165° The armature current Ia¢ =

10030 15.165∞ - 6351 1

= 10030 -74.835∞ + j 6351 = 2623.84 – j 3329.7 = 4239.27 - 51.76∞ A Power factor is cos 51.76° lag or 0.619 lag. Now input power is 75 MW.

Tzodispopvt!Hfofsbups

75 ¥ 106 =

\ \

7/::

3 ¥ 10030 ¥ 6351 sin d 1

sin d = 0.39 or d = 22.95° 10030 22.95∞ - 6351 Ia = A = 10030 - 67∞ + j 6351 1 = 3919 – j 2882 = 4864.6 - 36.33∞ A

\

\ power factor is cos 36.33° lag or 0.8056 lag.

! Qspcmfn!7/63 B!36!NWB-!uisff.qibtf-!7/7!lW-!21.qpmf-!61!I{-!tbmjfou.qpmf!tzodispopvt!hfofsbups!ibt!sfbdubodft! pg!Ye!>!4!W!boe!Yr!>!2!W!xjui!ofhmjhjcmf!bsnbuvsf!sftjtubodf/!Efufsnjof!uif!gpmmpxjoh!bu!gvmm!mpbe-! vojuz!qpxfs!gbdups!boe!sbufe!wpmubhf; ! ! ! !

)b*! )c*! )d*! )e*!

Fydjubujpo!wpmubhf! Qpxfs! Tzodispoj{joh!qpxfs!qfs!fmfdusjdbm!efhsff!boe!dpssftqpoejoh!upsrvf Tzodispoj{joh!qpxfs!qfs!nfdibojdbm!efhsff!boe!dpssftqpoejoh!upsrvf

Solution Vt = Ia =

6600

V = 3810.5 V 3 25 ¥ 103 A = 2187 A 3 ¥ 6.6 ¥ 1

From Eq. (6.78), neglecting ra d = tan–1

X q I a cos q Vt X q I a sin q

= tan -1

1 ¥ 2187 ¥ 1 3810.5 + 1 ¥ 2187 ¥ 0

= 29.85° (a) From Fig. 6.56, Id = Ia sin (q + d) = 2187 sin (0° + 29.85°) = 1088.54 A Excitation voltage E = Id Xd + Vt cos d = 1088.54 ¥ 3 + 3810.5 cos 29.85° = 6570.58 V (b) P = =

EVt V2 Ê 1 1 ˆ sin d + t sin 2d Á Xd 2 Ë X q X d ˜¯ 6570.58 ¥ 3810.5 (3810.5)2 Ê 1ˆ sin 29.85∞ + ÁË1 - ˜¯ sin ( 2 ¥ 29.85∞) 3 2 3

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= 8332738 W per phase \ total power = 8332738 ¥ 3 W = 24998 KW (c)

dP EVt 1 ˆ Ê 1 = cos d + Vt 2 cos 2d ÁË X q X d ˜¯ dd Xd

=

6570.58 ¥ 3810.5 Ê 1 1ˆ cos 29.85∞ + (3810.5)2 Á - ˜ cos ( 2 ¥ 29.85∞) Ë 1 3¯ 3

= 12122312 W per phase = 36367 kW for 3 phases \ synchronizing power per electrical degree p ¥ 36367 kW = 634.72 kW 180 Synchronizing torque per electrical degree

Psy =

Tsy =

1 634.72 ¥ 103 Psy = Nm = 10107 Nm 2 ¥ 50 ws 2p ¥ 10

(d) Synchronizing power per mechanical degree p ¥ 36367 kW 360

Psy = =

p ¥ 10 ¥ 36367 kW = 3172 kW 360

Synchronizing torque per mechanical degree Tsy =

3172 ¥ 103 Nm = 50509 Nm. 2 ¥ 50 2p ¥ 10

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Solution Ia =

80 ¥ 103 3 ¥ 400

A = 115.47 - 36.87 A

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At full load, armature copper loss = 3 Ia2 ra = 3 ¥ (115.47)2 ¥ 0.03 W = 1200 W At full load, field-circuit loss = \ total losses at full load

(200)2 W = 266.67 W 150

= 450 + 300 + 1200 + 266.67 = 2216.67 W Output at full load and 0.8 power factor lagging = 80 ¥ 103 ¥ 0.8 W = 64000 W 2216.67 Ê ˆ Efficiency = Á1 ¥ 100% 64000 + 2216.67 ˜¯ Ë

\

= 96.65% 1 At one-third load, output at 0.8 power factor lagging = ¥ 64000 = 21333 W 3 Armature copper loss at one third load 2

2

ÊI ˆ Ê 115.47 ˆ = 3 Á a ˜ ra = 3 ¥ Á ¥ 0.03 = 133.33 W Ë 3¯ Ë 3 ˜¯

\ total loss at one-third load = 450 + 300 + 133.33 + 266.67 = 1150 W \

1150 Ê ˆ ¥ 100% Efficiency = Á1 21333 + 1150 ˜¯ Ë

= 94.88%.

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Solution Vt =

I=

220 3

V = 127 V

5 ¥ 103 3 ¥ 220

A = 13.12 A

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d = tan–1

X q I a cos q Vt + X q I a sin q

= tan -1

2 ¥ 13.12 ¥ 0.8 127 + 2 ¥ 13.12 ¥ 0.6

= 8.36° Id = Ia sin (q + d) = 13.12 sin (36.87° + 8.36°) = 9.31 A \ E = Id Xd + Vt cos d = 9.31 ¥ 5 + 127 cos 8.36° = 172.2 V per phase Output power per phase VE V2 Ê 1 1 ˆ sin 2d P = t sin d + t Xd 2 ÁË X q X d ˜¯ \

=

127 ¥ 172.2 (127)2 Ê 1 1 ˆ sin d + Á - ˜ sin 2d 5 2 Ë 2 5¯

= 4373.88 sin d + 2419.35 sin 2d For maximum power, dP =0 dd

\ or, or, or, \ or, \

4373.88 cos d + 2 ¥ 2419.35 cos 2d 4373.88 cos d + 4838.7(2 cos2 d – 1) 9677.4 cos2 d + 4373.88 cos d – 4838.7 cos2 d + 0.452 cos d – 0.5 cos d d

=0 =0 =0 =0 = 0.514 = 59°

Maximum power per phase = 4373.88 sin 59° + 2419.35 sin 2 ¥ 59° = 5885 W Total maximum power = 3 ¥ 5885 W = 17655 W = 17.655 kW.

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Solution (a) Let load on machine 1 be P1 kW \ load on machine 2 is (5 – P1) kW Operating frequency of Machine 1 is (52 – P1) Hz Operating frequency of Machine 2 is 51.5 – 1 (5 – P1) Hz Since the machines are operating in parallel, the operating frequencies should be same for both machines. \ or, or, or, \

52 – P1 52 – P1 2P1 P1 P2

= 51.5 – (5 – P1) = 51.5 – 5 + P1 = 52 – 51.5 + 5 = 5.5 = 2.75 kW = 5 – 2.75 = 2.25 kW

Operating frequency 52 – 2.75 = 49.25 Hz (b) New load is 7 kW. Let load shared by Machine 1 and Machine 2 be P¢1 kW and (7 – P¢1) kW respectively \ or, or,

52 – P¢1 = 51.5 – (7 – P¢1) 52 – P¢1 = 44.5 + P¢1 P¢1 = 3.75 kW

\ load shared by Machine 1 is 3.75 kW and that by Machine 2 is 3.25 kW. Operating frequency 52 – 3.75 or 48.25 Hz (c) Let the load shared by Machine 1 and Machine 2 be P≤1 kW and (7 – P≤1) kW respectively. No-load frequency of Machine 2 is 51.5 + 1 or 52.5 Hz \ or, or,

52 – P≤1 = 52.5 – (7 – P≤1) 52 – P≤1 = 45.5 + P≤1 P≤1 = 3.25 kW

\ load shared by Machine 1 is 3.25 kW and that of Machine 2 is 3.75 kW \ operating frequency is 52 – 3.25 or 48.75 Hz.

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Solution As the machines are identical, 1200 kW = 600 kW 2 If the armature current and power factor of the first machine are Ia1 and cos q1 respectively then

P1 = P2 =

Vt Ia1 cos q1 = P1 ¥ 11 ¥ Ia1 cos q1 = 600 600 \ Ia1 cos q1 = = 31.49 A 3 ¥ 11 Now Ia1 = 50 A \ cos q1 = 0.63 or, q1 = 50.95° If the total current is I then or,

¥ 11 ¥ I ¥ 0.8 = 1200 \ I = 78.73 A (a) As I = Ia1 + Ia2 where Ia2 is the current supplied by the second generator, Ia2 = 78.73 - 36.87∞ – 50 - 50.95∞ = 31.48 – j 8.4 = 32.58 - 14.94∞ A (b) The first machine operates at a power factor of 0.63 lagging and the second machine operates at a power factor of cos 14.94° lagging or 0.966 lagging. 11000 + 50 - 50.95∞ ¥ (3 + j 40) (c) The emf of the first machine = 3 =

11000 3

+ 2005.6 39∞

= 7909.5 + j 1262.16 = 8009.6 9.06∞ V \ line value of emf is

¥ 8009.6 V or 13.87 kV.

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Tzodispopvt!Hfofsbups

Solution E1 = 200 30∞ V E2 = 2000 15∞ V (a) From Eq. (6.49), the terminal voltage per phase 200 30∞ 200 15∞ + 8 10 Vt = 1 1 1 + + 8+ 6 8 10

or,

Vt =

200 (10 30∞ + 8 15∞) (8 + 6) 8 ¥ 10 + 10 (8 + 6) + 8 (8 + 6)

=

200 90∞ ¥ 10 36.87∞ ¥ 17.85 23.34 238.8 142.55∞

= 149.5 7.66∞ V \ line value of terminal voltage is (b) Armature current of Generator 1 =

¥ 149.5 or 259 V

200 30∞ - 149.5 7.66∞ 25.04 + 80.07 A= 8 8

= 10.487 -17.36 A Armature current of Generator 2 =

200 15∞ - 149.5 7.66∞ 45.02 + 31.84 = 10 10

= 5.51 - 55.59∞ A (c) The total power supplied by Generator 1 = 3 ¥ 149.5 ¥ 10.487 cos (7.66° + 17.36°) = 4262 W The total power supplied by Generator 2 = 3 ¥ 149.5 ¥ 5.51 cos (7.66° + 55.59°) = 1112.3 W \ total output power = 4262 W + 1112.3 W = 5374.3 W.

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Solution Full-load current If l = Vt = E=

3000 3 ¥ 6.6 ¥ 0.9 6600 3 7200 3

A = 291.6 A

V = 3810.5 V V = 4157 V

Voltage drop for current of 291.6 A is (4157 – 3810.5) V or 346.5 V 346.5 291.6 Let the machine be delivering a current I to the load at 0.9 p.f. lagging

Hence, voltage drop for 1 A current is

\ voltage drop for supplying current I is

346.5 I volts 291.6

346.5 ˆ Ê Hence, terminal voltage = Á 4157 ˜V Ë 291.6 ¯ Load terminal voltage = I ZL

= I 102 + 82 V 346.5 I = I 102 + 82 291.6 4157 – 1.188 I = 12.8 I I = 297.18 A

\

4157 –

or, or,

\ terminal voltage per phase is 297.18 102 + 82 V or, 3805.76 V Line value of terminal voltage is

¥ 3805.76 V or, 6592 V.

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Solution Ia = Vt =

15 ¥ 103 3 ¥ 240 240 3

= 36.08 A

= 138.56 V

E = Vt + Ia Zs = 138.56 + 36.08 36.87∞ (0.4 + j 0.75) = 138.56 + 30.668 98.8∞ = 133.86 + j 30.3 = 137.24 12.75∞ V Voltage requlation =

137.24 - 138.56 ¥ 100% 138.56

= –0.95% Let at a power factor of cos q lagging, the voltage regulation is zero at 36.08 A load current. E = 138.56 + 36.08 -q (0.4 + j 0.75) = 138.56 + 30.67 61.93∞ - q = 138.56 + 30.67 cos (61.93° – q) + j 30.67 sin (61.93° – q) Now, | E | = |Vt | as voltage regulation is zero. \ {138.56 + 30.67 cos (61.93° – q)}2 + {30.67 sin (61.93° – q)}2 = (138.56)2 or (138.56)2 + (30.67)2 cos2 (61.93° – q) + 2 ¥ 138.56 ¥ 30.67 cos (61.93° – q) + (30.67)2 sin2 (61.93° – q) = (138.56)2 or, (30.67)2 + 8499.27 cos (61.93° – q) = 0 or, cos (61.93° – q) = – 0.11 = cos 96.31° or, q = – 34.38° \ power factor is cos 34.38° leading or, 0.825 leading.

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Solution (a) From Eq. (6.43), for maximum real input power, x d = 180° – fs = 180° – tan–1 s ra 8 = 180° – tan–1 0.5 = 180° – 86.424 = 93.576° From Eq. (6.30), the output power 3¥

P=

11000 12500 ¥ 3 3 (0.5) 2 + (8) 2

cos (86.424° – 93.576°) –

Ê 11000 ˆ 3¥Á Ë 3 ˜¯

2

(0.5) 2 + (8) 2

¥ cos 86.424°

= 17020559 – 941546 = 16079013 W = 16079 kW (b) If power factor is cos q and armature current is I then 3 Vt I cos q = 16079 ¥ 103 \

I cos q =

16079 ¥ 103 3

11 ¥ 10 3¥ 3 (c) At E > Vt, power factor is lagging. Now, E = Vt + I (ra + j xs)

or,

12.5 ¥ 103 3

93.576∞ =

11 ¥ 103 3

A=

16079 3 ¥ 11

A

…(i)

I (cos q – j sin q) ¥ (0.5 + j 8)

or, –450.13 + j 7202.8 = 6351 I (0.5 cos q + 8 sin q – j 0.5 sin q + j 8 cos q) Equating real parts, – 450.13 = 6351 I (0.5 cos q + 8 sin q) – 450.13 = 6351 ¥ 0.5 ¥

16079 3 ¥ 11

or, I sin q = – 52.75 Dividing Eq. (ii) by Eq. (i), tan q = –

+ 8 ¥ 6351 I sin q

16079 ˆ Ê ÁËE cos q = 3 ¥ 11 form Eq. (i)˜¯ (ii)

52.75 ¥ 3 ¥ 11 = – 0.0625 16079

\ q = –3.576° \ power factor is cos 3.576° or 0.998 lag. From Eq. (i), 16079 A = 845.62 A. I= 3 ¥ 11 ¥ 0.998

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Sfwjfx!Rvftujpot! 1. Explain the differences between cylindrical rotor and salient pole rotor used in large synchronous generators. 2. Describe the various schemes used for exciting large synchronous machines. 3. With the help of a neat diagram, describe the main parts of an alternator with their functions. 4. Derive e.m.f. equation for an alternator. 5. Define voltage regulation of an alternator. Draw the phasor diagram of a loaded alternator for the following conditions: (a) unity power factor load, (b) lagging power factor load and (c) leading power factor. 6. What is armature reaction. Explain the effect of armature reaction on the terminal voltage of alternator at (a) unity power factor load (b) zero lagging power factor load and (c) zero leading power factor load. Draw relevant phasor diagram. 7. Discuss the synchronous impedance and mmf methods for calculating regulation of an alternator. 8. Discuss the Poteir triangle method of finding regulation of an alternator. 9. Explain the two reaction theory applicable to salient pole synchronous machine. 10. For a salient pole synchronous machine derive an expression for power developed as a function of load angle. Neglect the effect of armature resistance. 11. Draw and explain the phasor diagram of a salient pole synchronous generator supplying lagging power factor load. 12. Describe the slip test method for the measurement of Xd and Xq of synchronous machines. 13. Discuss the phenomenon of sudden three-phase short circuit of an alternator. Draw a typical wave shape of current and mark the different regions. 14. What is synchronizing power? Derive equations for synchronizing power of cylindrical rotor and salient pole alternators. 15. What is the necessity of parallel operation of alternators? State the conditions necessary for parallel operation of alternators. 16. Describe different methods of cooling of alternators. 17. Derive an expression for power developed in a cylindrical rotor alternator in terms of power angle and synchronous impedance. 18. Explain the effect of damper winding in alternators. 19. Enumerate the various losses in synchronous machines? 20. A cylindrical rotor alternator is operating at a lagging power factor. Show that E Ia cos q is EVt sin d where q is the internal power factor angle and d is the load angle. Neglect armature xs circuit resistance.

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Qspcmfnt 1. A 1500 kVA, 6.6 kV, three-phase, star connected synchronous generator has a resistance and synchronous reactance of 0.5 W and 5 W respectively. If the rated output of the generator at 0.8 p.f. is suddenly switch off what will be the voltage of the generator? What is the percentage rise in voltage? [7416 V, 12.4%] 1500

[Hints:

IL =

or

E2 = (3810 ¥ 0.8 + 131.2 ¥ 0.5)2 + (3810 ¥ 0.6 + 131.2 ¥ 5)2 E = 4283.6 V/phase or 7419 V line to line

% Rise in voltage =

3 ¥ 6.6

= 131.2 A

7419 - 6600 ¥ 100 = 12.4%] 6600

2. A three-phase 25 KVA, 400 V, 6-pole, star connected synchronous generator has a synchronous reactance of 12 W per phase. The machine is connect to infinite bus bar. Neglecting armature resistance determine (a) the emf and power angle when the machine is delivering rated kVA at 0.8 p.f. lagging (b) the machine excitation is increased by 25% while supplying the same power as in (a). Determine the power angle. (c) What is the maximum power the generator can develop under conditions rated in (b). [1037 V, 35.1°, 27.52°, 42 kW] Ia =

[Hints:

25 ¥ 103 3 ¥ 400

= 36 A

E2 = (231 ¥ 0.8)2 + (231 ¥ 0.6 + 432)2 = 598.8 V tan (d + q) =

138.6 + 432 = 3.09 184.8

q = 36.87°, d = 35.1°, E = 3 ¥ 598.8 = 1037 V E = 1.25 ¥ 598.8 = 748.5 V

Now,

Po = or

EVt sin d = 231 ¥ 36 ¥ 0.8 xs

748.5 ¥ 231 sin d = 231 ¥ 36 ¥ 0.8 12 d = 27.52°

748.5 ¥ 231 sin 90° = 14 kW/ph 12 \ total power = 14 ¥ 3 = 42 kW] Maximum power =

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3. A three-phase, 100 MVA, 50 Hz, 11 kV synchronous generator has synchronous reactance of 1.4 p.u. and is connected to a 11 kV bus. Determine the maximum power developed if the internal voltage of the generator is kept constant at 1.2 p.u. If the generator supplies 60 MW, find the power angle. [78 MW, 50.4°] [Hints: Vt = 1 p.u. 1 ¥ 1.2 \ Pm = = 0.857 p.u. = 0.857 ¥ 100 1.4 = 85.7 MW 60 1.2 ¥ 1 sin d = 1.4 100 d = sin–1 0.7 = 44.43°]

\

4. A three-phase 6.6 kV, 50 Hz, 500 kVA, star-connected alternator has an effective resistance of 0.5 W per phase and synchronous reactance of 5 W per phase. Determine the voltage regulation at full load at (a) unity power factor, (b) 0.8 p.f. lagging, and (c) 0.8 p.f. leading. [0.72%, 3.9%, –1.85%] [Hints:

Vt = Ia =

6600 3 500

= 3810.5 V

3 ¥ 66

= 43.73

(a) When cos q = 1, E = (3810 ¥ 1 + 43.73 ¥ 0.5) 2 + ( 43.73 ¥ 5) 2 = 3838 V 3838 - 3810.5 \ % regulation = ¥ 100 = 0.72. 3810.5 (b) E = (3810.5 ¥ 0.8 + 43.73 ¥ 0.5) 2 + (3810.5 ¥ 0.6 + 43.73 ¥ 5) 2 = 3962.8 V % regulation =

3962.8 - 3810.5 ¥ 100 = 3.9% 3810.5

(c) E = (3810.5 ¥ 0.8 + 43.73 ¥ 0.5) 2 + (3810.5 ¥ 6 - 43.73 ¥ 5) 2 = 3740.5 % Regulation =

3740.5 - 3810.5 ¥ 100 = –1.83] 3810.5

5. A 250 MVA, 11 kV, 50 Hz, star connected synchronous generator is tested under short circuit conditions and open circuit conditions. A field current of 12 A produces 800 A in the armature under short circuit condition and with the same field current the open circuit voltage is

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1500 V. If this machine supplies a load of 4000 A at 0.8 p.f., determine (a) power developed by the generator, (b) maximum power that can be obtained, and (c) power angle if the generator develops 100 MW for the same excitation. [60.97 MW, 169 MW, 36.28°] [Hints:

zs =

1500 3 ¥ 800

= 1.08 W 2

ˆ Ê 11000 ˆ Ê 11000 ¥ 0.6 + 4000 ¥ 1.08˜ ¥ 0.8˜ + Á E= Á ¯ Ë 3 ¯ Ë 3

2

= 9586 V (a) P = 3 ¥ 11000 ¥ 4000 ¥ 0.8 = 60.97 MW 11000 ¥ 9586 3 W = 169 MW (b) Pm = 3 1.08 11000 3 sin d 1.08

9586 ¥ (c) 100 ¥ 106 = 3 ¥ \ d = 36.28°]

6. A 2 MVA, three-phase, star connected, 8-pole, 750 rpm synchronous generator is operating on 6600 V bus bar. If xs is 6 W per phase, determine the synchronizing power and torque per mechanical degree of displacement for full-load, 0.8 p.f. lagging. [502.8 kW/mech. degree, 6400 Nm/mech/degree] [Hints:

I=

200 ¥ 103 3 ¥ 6000

= 192.45 A

6000

+ 192.45 -36.87∞ ¥ j 6 = 4258.32 12.53∞ V 3 6000 3¥ ¥ 4258.32 4p 3 Ps = cos 12.53∞ ¥ 6 180 E=

= 502.8 kW/mech. degree 502.8 ¥ 1000 = 6400 Nm/mech. degree] 750 2p ¥ 60 7. A three-phase, 400 V, star-connected synchronous machine is synchronised with an infinite bus at rated voltage. The synchronous machine is now made to deliver a shaft load of 9.5 kW. Ts =

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Xd = 5 W and Xq = 3.2 W and resistance is negligible. Friction, windage and core loss amounts to 500 W. Determine (a) power angle, armature current, power factor (b) maximum power output and the corresponding power angle. [11.623°, 14.572 A, 0.99 lag, 35.43 kW 67°] [Hints:

P = 9.5 kW + 500 W = 10,000 W 2 È ( 400) 2 ˘ 1 Ê 400 ˆ Ê 1 1 ˆ 10,000 = 3 Í - ˜ sin 2 d ˙ sin d + Á Á ˜ 2 Ë 3 ¯ Ë 3.2 5 ¯ ÍÎ 3 ¥ 5 ˙˚

or, 1 = 3.2 sin d + 0.9 sin2 d (a) d = 11.623° As E = Vt, cos q is leading 400 400 cos 11.623∞ E - Vt cos d 3 Id = = 3 = 0.947 A Xd 5 400 sin 11.623∞ Iq = 3 = 14.541 A 3.2 Ia = I d2 + I q2 = 14.572 A Iq = Ia cos (d + q) d + q = cos–1

Iq Ia

= 3.738°

\ q = –7.885 \ p.f. = cos 7.885° = 0.99 lag (b) For maximum power 230.95 ¥ 3.2 1 ± + (0.44) 2 4 ¥ 230.95 ¥ 1.8 2 d = 67°

cos d = – or,

Maximum power developed 3 ¥ ( 400) 2 ( 400) 2 sin 67∞ + 3 sin 134° = 35.93 kW 3¥5 3¥ 2 Maximum power output =

= 35.93 –

500 = 35.43 kW] 1000

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8. A 500 kVA, 11 kV, three-phase star connected synchronous generator has the following data: Friction and windage loss is 1500 W, open circuit core loss is 2500 W, effective armature resistance per phase is 4 W, field copper loss is 1000 W. Determine the alternator efficiency at half load and 0.8 p.f. lagging. Also find the maximum efficiency of the generator. [96.587%, 96.886%] 500

Ia =

[Hints:

3 ¥ 11

= 26.244 A

2

Ê 26.244 ˆ ¥4 (a) Armature copper loss = 3 ¥ Á Ë 2 ˜¯ = 2066.24 W Total loss at half load = 1500 + 2500 + 2066.24 + 1000 = 7066.24 W \

Ê ˆ 7066.24 Á ˜ h = Á1 ˜ ¥ 100% = 96.587% 1 Á 500, 000 ¥ ¥ 0.8 + 7066.24 ˜ Ë ¯ 2

(b) For maximum h, variable loss = constant loss \ \

3Im2 ¥ 4 = 1500 + 2500 + 1000 = 5000 W Im = 20.412 A

Output at maximum efficiency = 3 ¥

11000 3

¥ 20.412 ¥ 0.8

= 311, 111.54 W Total loss = 2 ¥ 5000 = 10,000 W Ê ˆ 10, 000 \ Max h = Á1 ˜¯ ¥ 100 = 96.886%] 311 111 54 + 10 000 , . , Ë 9. A 70 MVA, 13.8 kV, 60 Hz, star-connected, salient-pole, three-phase synchronous generator has Xd = 1.83 W and Xq = 1.21 W. It delivers rated load at 0.8 p.f. lagging. The armature resistance is negligible. Determine voltage regulation and power developed by the generator. [49.68%, 56 MW] [Hints:

Ia = tan d =

70 ¥ 106 3 ¥ 13800

= 2928.59 A

2928.59 ¥ 1.21 ¥ 0.8 = 0.28 13800 + 2928.59 ¥ 1.21 ¥ 0.6 3

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d = 15.69° Id = Ia sin (36.87° + 15.69°) = 2325.27 -74.31∞ A Iq = Ia cos (36.87° + 15.69°) = 1780.38 15.69∞ A E=

13800 3

+ 2325.27 -74.31∞ ¥ 1.83 +

13800 3

1569∞

= 11925.8 15.69∞ V \

regulation =

11925.8 - 7967.43 ¥ 100% = 49.68% 7967.43

P =3¥

13800

¥ 2928.59 ¥ 0.8 = 56 mW] 3 10. Two three-phase, star connected synchronous generator have per phase generator votlages of 120 10∞ V and 120 20∞ V under no load and reactances of j 5 W/phase and j 8 W/phase respectively. They are connected in parallel to a load impedance of 4 + j 3 W/phase. Determine per phase terminal voltage, armature current of each generator, power supplied by each generator and total power output. [82.17 V, 936 A, 7.31 A, 1624.68 W, 1617.82 W, 3242.5 W] [Hints:

Vt =

120 10∞ ¥ j 8 + 120 20∞ ¥ j 5 ¥ ( 4 + j 3) = 82.17 -5.93∞ V ( 4 + j 3) (5 + j8) + j 5 ¥ j8

I a1 =

120 20∞ - 82.17 -5.93∞ = 9.36 -51.17∞ A j5

120 20∞ - 82.17 -5.93∞ = 7.31 -32.06∞ A j8 P1 = 3 ¥ 82.17 ¥ 9.36 cos (51.17° – 5.93°) = 1624.68 W P2 = 3 ¥ 82.17 ¥ 7.31 cos (32.06° – 5.93°) = 1617.82 W Ia2 =

Total power P = P1 + P2 = 3242.5 W] 11. A three-phase, 50 Hz, 8-pole synchronous generator has a star-connected winding with 120 slots and 8 conductors per slot. The flux per pole is 0.05 Wb sinusoidally distributed. Determine the phase and line voltages. [1699 V and 2942.8 V] [Hints: Slots per pole phase =

120 =5 8¥3

Slot angle =

180∞ ¥ 8 = 12° 12

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5 ¥ 12∞ 2 Kd = = 0.9567 12∞ 5 sin 2 sin

\

Total no. of conductors = 8 ¥ 120 = 960 960 = 320 Conductors per phase = 3 320 = 160 Number of turns per phase = 2 Generated voltage per phase E = 2p ¥ 0.9567 ¥ 50 ¥ 0.05 ¥ 160 = 1699 V \ line value of E = 3 ¥ 1699 = 2942.8 V] 12. A three-phase, 1500 kVA, star-connected, 50 Hz, 2300 V synchronous generator has a resistance between each pair of terminals as measured by direct current is 0.16 W. Assume that the effective resistance is 1.5 times the ohmic resistance. A field current of 70 A produces a short circuit current equal to full load current of 376 A. The same field current produces an emf of 700 V on open circuit. Determine the synchronous reactance of the machine and its full load regulation at 0.8 power factor lagging. [1.068 W, 22.8%] 700 zs = 3 = 1.075 W 376 0.16 ra = 1.5 ¥ = 0.12 2

[Hints:

xs = (1.075) 2 - (0.12) 2 = 1.068 W Now, \

1500 ¥ 103 = 3 ¥ 2300 Ia Ia = 376 A E=

2300 3

+ 376 -36.87∞ ¥ (0.12 + j 1.068)

= 1631 10.39∞ V 2300 3 ¥ 100% = 22.8%] 2300 3

1631 \ regulation =

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13. Two synchronous generators are operating in parallel and supplying a load of 8 mW at 0.8 p.f. lagging. The output of generator A is adjusted to 5000 kW by changing its steam supply and its power factor is adjusted to 0.9 lagging by changing its excitation. Determine the power factor of generator B. [0.64 lagging] [Hints:

P = 8000 kW cos q = 0.8 8000 Q= ¥ 0.6 = 6000 KVAR 0.8 PA = 5000 kW cos qA = 0.9 5000 sin (cos–1 0.9) = 2421.6 KVAR 0.9 PB = P – PA = 3000 W QB = Q – QA = 3578.4 KVAR QA =

\ \ \

3578.4 = 1.1928 3000 cos qB = 0.64] tan qB =

14. Two identical three-phase synchronous generators are running in parallel sharing equally a total load of 10,000 kW at a lagging power factor of 0.8, the excitation of two machines remaining same. If the excitation of one machine is adjusted so that its armature current is 438 A, find the armature current and power factor of the second machine if the steam supply is not changed. [769.5 A, 0.568 lag] [Hints:

IL =

10, 000 ¥ 103 3 ¥ 6600 ¥ 0.8

= 1093.5 A

cos q = 0.8 sin q = 0.6 I1 = I2 =

1093.5 = 546.75 2

Active component of current = IL cos q = 874.8 A Reactive component of current = IL sin q = 656.1 A 874.8 = 437.4 A 2 Since steam supply is same, the active component remains same at 437.4 A

Active component of current supplied by each generator =

Now

I1 = 438 A

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\ reactive component of I1 = ( 438) 2 - ( 437.4) 2 = 23 A \ reactive component of I2 = 656.1 – 23 = 633.1 \

I2 = ( 437.4) 2 + (633.1) 2 = 769.5 A

\

tan q2 =

633.1 437.4

or, cos q2 = 0.568 lag]

15. A synchronous generator has a synchronous reactance of 1.2 p.u. It is running over excited with an excitation voltage of 1.5 p.u. and supplies a synchronous power of 0.6 p.u. to the bus. If the prime torque is increased by 1%, by how much will the synchronous power P and reactive power Q change? [1% and –0.125%] [Hints:

0.6 =

1.5 ¥ 1 sin d 1.2

sin d = 0.48, d = 28.68° dP = 1% of 0.6 = 0.006 p.u. EVt Vt 2 cos d Q= xs xs dQ -EVt = sin d dd xs

\

dP EVt = cos d dd xs dQ = –tan d = –tan 28.68° = –0.547 dP dQ = – 0.547 ¥ 0.006 = –0.328 ¥ 10–3

Also \ or,

\ % decrease change in Q =

0.328 ¥ 10 -3 1.5 ¥ 1 12 cos 28.68 1.2 1.2

¥ 100% = 0.125%]

Nvmujqmf.Dipjdf!Rvftujpot 1. The phase sequence of a three-phase alternator will reverse if (a) the field current is reversed keeping the direction of rotation same (b) the field current remains same but the direction of rotation is reversed

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(c) the field current is reversed and the number of poles doubled (d) the number of poles is doubled without reversing the field current [GATE 2000] 2. Xd, Xd¢ and X≤ d are steady state d-axis synchronous reactance, transient d-axis reactance and sub transient d-axis reactance of a synchronous machine respectively. Which of the following statements in true? (b) Xd≤ > X¢d > Xd (a) Xd > X¢d > X≤d (c) Xd¢ > Xd≤ > Xd (d) Xd > Xd≤ > Xd¢ [GATE 2001] 3. The Figure shows the magnetization curves of an alternator at rated armature current, unity power factor and also at no load. The magnetization curve for rated armature current, 0.8 power factor leading is given by (a) Curve A (b) Curve B (c) Curve C (d) Curve D [GATE 2001] Y

No load

Rated armature current unit p.f.

C A

B

D

Exciting current

4. Curves X and Y in the figure denote open circuit and full load zero power factor (ZPF) characteristics of a synchronous generator, Q is a point on the ZPF characteristics at 1 p.u. voltage. The vertical distance PQ in figure gives the voltage drop across Voltage (pm) P 1.0

Q

X Y

Field current

(a) synchronous reactance (b) magnetizing reactance (c) poteir reactance (d) leakage reactance [GATE 2003] 5. A standalone engine driven synchronous generator is feeding a partly inductive load. A capacitor is now connected across the load to completely multiply the inductive current. For this operating condition (a) the field current and fuel input have to be reduced (b) the field current and fuel input have to be increased

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(c) the field current has to be increased and fuel input left unaltered. (d) the field current has to be reduced and fuel input left unaltered [GATE 2003] 6. If field excitation of 20 A in a certain alternator results in an armature current of 400 A in short circuit and a terminal voltage of 2000 V on open circuit. The magnitude of the terminal voltage drop within the machine at a load current of 200 A is (a) 1 V (b) 10 V (c) 100 V (d) 1000 V [GATE 2010] [Hints: zs = \

2000 W=5W 400 voltage drop = IZs = 200 ¥ 5 = 1000 V]

7. Consider a stator winding of an alternator with an integral high resistance ground fault. The currents under the fault condition are as shown in the figure. The winding is protected using a differential current scheme with current transformers of ratio 400/5A as shown. The current through the operating coil is (a) 0.17875 A (b) 0.2 A (c) 0.375 A (d) 60 KA CT ratio 400/5

250 + j 0

CT ratio 400/5

(220 + j 0) A I1 I2 Operating coil

[GATE 2010] [Hints: Current through operating coil I1 - I 2 250 - 220 = 0.375 A] = CT ratio 400 / 5 8. When the real power of a cylindrical rotor alternator is maximum, the reactive power output is =

(a)

3Vtt2 xs

(b)

3Ett2 xs

(c)

3E 2 xs

(d)

-3Vt 2 xs

9. When an alternator is feeding an infinite bus bar, (a) the excitation controls the reactive power output and the governor setting controls the real power output

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10.

11.

12.

13.

14.

15.

16.

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(b) the excitation controls both the real and reactive power output (c) the excitation controls the terminal voltage and the governor setting controls the real power output (d) the governor setting controls the terminal voltage. In two-reaction theory, the armature current is decomposed into two components Id and Iq. If E and Vt are the generated emf and terminal voltage then (a) Id is in phase with E and Iq is perpendicular to E (b) Id is in phase with Vt and Iq is perpendicular to Vt (c) Id is perpendicular to E and Iq is in phase with E (d) Id is perpendicular to Vt and Iq is in phase with Vt. The function of damper bars in synchronous machine is to (a) prevent the rotor from running at super synchronous speed (b) prevent the rotor from running at subsynchronous speed (c) percent the rotor from running at synchronous speed (d) reduce the rotor oscillations about the operating point. A synchronous generator connected to an infinite bus is overexcited. Considering only the reactive power, from the point of view of the system, the machine acts as (a) an inductor (b) a capacitor (c) a resistor (d) as R-L circuit A three-phase synchronous generator is operating at constant load while the excitation is adjusted to give unity power factor current. If the excitation is now increased the power factor will (a) remain unity (b) become zero (c) become leading (d) become lagging Poteir reactance of an alternator is almost the same as (a) total armature reactance (b) field winding reactance (c) armature leakage reactance (d) leakage reactance of field winding Which of the following method gives more accurate results for determination of voltage regulation of an alternator? (a) mmf method (b) Potier triangle method (c) Synchronous impedance method (d) American Institution Standard method Synchronous reactance is (a) equal to armature leakage reactance (b) sum of armature leakage reactance and magnetization reactance (c) same as magnetization reactance (d) the difference of armature leakage reactance and magnetization reactance

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17. During ‘Slip Test’ for maximum current, the armature mmf aligns along (a) 45° to d-axis (b) 45° to q-axis (c) d-axis (d) q-axis. 18. Which of the following is not a necessary condition to be satisfied for synchronizing an incoming alternator to an already operating alternator? (a) Same frequency (b) Same voltage magnitude (c) Same phase sequence (d) Same prime mover speed 19. In a synchronous generator, the generated emf phasor (a) leads the flux phase by 90° (b) is in phase with flux phasor (c) lags behind the flux phasor by 90° (d) is in phase opposition to the flux phasor 20. A synchronous generator is feeding power to infinite bus bars at unity power factor. Its excitation is now increased. It will feed: (a) the same power but at a leading power factor (b) the same power but at a lagging power factor (c) more power at unity power factor (d) less power at unity power factor 21. A synchronous generator is feeding a zero power factor (lagging) load at rated current. The armature reaction is (a) magnetizing (b) demagnetizing (c) cross magnetizing (d) ineffective [GATE 2006] 22. In relation to the synchronous machines which of the following statements is false? [GATE 2005] (a) In salient pole machines, the direct axis synchronous reactance is greater than the quadrature axis synchronous reactance (b) The damper bars help the synchronous motor self-start (c) short circuit ratio is the ratio of the field current required to produce the rated voltage on open circuit to the rated current (d) The V curve of a synchronous motor represents the variation in the armature current with field excitation, at a given output power 23. Synchronous generator voltage obtained by the synchronous impedance method is (a) higher than actual as it does not account for magnetic saturation (b) lower than actual as it does not account for magnetic saturation (c) nearly accurate as it accounts for magnetic saturation (d) nearly accurate as the generator is normally operated in the unsaturated region of magnetization 24. Armature reaction AT of a synchronous generator at rated voltage zero power factor lagging is (a) magnetizing (b) demagnetizing (c) cross magnetizing (d) both magnetizing and cross magnetizing

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25. If the armature current in a generating synchronous machine is in phase with excitation emf, the armature mmf vector (a) lags behind the field mmf vector by 90° (b) is in the same direction as the field mmf vector (c) is in a direction opposite to the field mmf vector (d) leads the field mmf vector by 90°

Botxfst 1. 6. 11. 16. 21.

(b) (d) (d) (b) (b)

2. 7. 12. 17. 22.

(a) (c) (b) (d) (c)

3. 8. 13. 18. 23.

(d) (d) (d) (d) (a)

4. 9. 14. 19. 24.

(a) (a) (c) (c) (b)

5. 10. 15. 20. 25.

(d) (c) (b) (b) (a)

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Like other rotating machines, a synchronous generator can be operated as a synchronous motor and a synchronous motor can be operated as a synchronous generator. When a prime mover drives the synchronous machine, it operates as a synchronous generator converting mechanical energy to electrical energy. If the armature of the machine is supplied with electric power, it functions as a motor converting electric power to mechanical power. Synchronous motors run at a constant speed known as synchronous speed. It is a doubly fed motor. Three-phase alternating power is provided to the stator and the rotor is fed from a dc source. By varying the rotor excitation, the power factor of the machine is varied. Synchronous motors can operate both with lagging and leading power factors.

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When a three-phase stator winding of a synchronous machine is fed by a balanced three-phase ac supply then a magnetic flux of constant magnitude but rotating at synchronous speed is produced. Let us assume that the rotor is rotating round the stator at synchronous speed having the same number of poles as the stator. As the rotor is excited by an external dc source, the poles of the rotor retain the same polarity throughout but the polarity of the stator poles changes as it is connected to an ac supply, i.e. the polarity of the stator as poles are alternating. Two similar poles of the stator and rotor repel each other with the result that the rotor tends to rotate in a particular direction. After Ê 1ˆ (T/2) seconds, Á i.e. T = ˜ , the polarity of the stator poles is reversed but the polarity of the rotor f¯ Ë

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poles remains the same. Under this condition, stator N-poles attract rotor S-poles and stator S-poles attract the rotor N-poles and, hence, the torque produced will be in the reverse direction and thus the rotor starts to rotate in the reverse direction. For a frequency of 50 Hz., these changes will occur 100 times in one second; thus the torque acting on the rotor of the synchronous motor is pulsating and the rotor does not move in any direction and remains stationary. Therefore, the synchronous motor is not self-starting. Let us now consider that the rotor is rotated in a clockwise direction by some external means so that torque is clockwise. After half a period later, the stator N-pole and S-pole will become S-pole and N-pole. If the rotor speed is such that the N-pole of the rotor also turns by a pole pitch so that it is again under the N-pole then the torque acting on the rotor will again be clockwise. Hence, in order to obtain a continuous and unidirectional torque, the rotor must be rotated with such a speed that it advances 1-pole pitch by the time the stator poles interchange their polarity. This means the rotor must rotate at synchronous speed with the stator. At this instant, the stator and rotor poles get magnetically interlocked (i.e. N-pole of stator attracts S-pole of rotor and vice versa). It is because of this magnetic locking acting between the two that the motor rotates. The motor can rotate at synchronous speed only. When the mechanical load is applied to a synchronous motor, its speed cannot decrease since the rotor must operate at constant speed. Hence, speed is independent of load and can be varied only by varying the supply frequency.

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Xal Ra Ia The equivalent circuit and phasor diagram of a cylindrical rotor + Xar synchronous motor is similar to that of a synchronous generaXs E M V(–) tor. The effect of armature reaction is replaced by the fictitious reactance Xar, while the leakage reactance is Xal. The resultant of Xar and Xal is called the synchronous reactance Xs. The equiva- Gjh/!8/2! Frvjwbmfou!djsdvju!pg! lent circuit of a cylindrical rotor synchronous motor is shown in b!dzmjoesjdbm!spups! tzodispopvt!npups Fig. 7.1 Here, Vt is the terminal voltage and E is the counter emf

Vt = E + Ia Ra + j Ia Xs

Vt

(7.1) d

jIaXar

jIaXal

The corresponding phasor diagram is shown in Fig. 7.2. q E Iara The phasor diagram of a salient-pole synchronous motor is shown Ia in Fig. 7.3(a). From this phasor diagram, the terminal voltage is obGjh/!8/3! Qibtps!ejbhsbn!pg! tained as Vt = E + Ia Ra + j Id Xd + j Iq Xq

(7.2)

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Vt

jIdXd b

jIqXq

Id

jIqXq

E

q Iq

Idra Iara

Ia

IaRa Vt e

d

Iqra

d q I q

a

jIdXd

Id

d

E

Ia c

(a)

Ia ra (b)

Gjh/!8/4! Qibtps!ejbhsbn!pg!b!tbmjfou.qpmf!tzodispopvt!npups

The armature current Ia can be decomposed into two components: Id, which is lagging phasor E by 90° and Iq, which is in phase with E. Figure 7.3(b) shows the phasor diagram in terms of the known parameters. From this phasor diagram, it is also evident that Vt = E + j Iq Xq + j Id Xd + Ia Ra. Next, ac is drawn from point a perpendicular to Ia and ac = j Ia Xq. From point e, i.e. the terminal point of V, ed is drawn parallel to ac which meets the extended line of the vector E at d. Hence, Now, or or Hence,

cd ac j Ia Xq j Iq Xq Vt

= jId Xq = cd + da = j Id Xq + j Iq Xq = j Ia Xq – j Id Xq = E + j Iq Xq + j Id Xd + Ia Ra = E + Ia Ra + j Id Xd + j IaXq – j Id Xq = E + Ia Ra + j Ia Xq + j Id (Xd – Xq).

(7.3)

The equivalent circuit of the salient-pole synchronous motor is shown in Fig. 7.4. Ia ra For a cylindrical rotor synchronous motor, the excitation jXq voltage E and power angle d can be calculated for differjId(Xd – Xq) V ent power factors. If Vt is taken as the reference phasor and cos q is the lagging power factor (Fig. 7.2) then

E

E d = Vt 0∞ – Ia -q (ra + j xs) where ra is the armature Gjh/!8/5! Frvjwbmfou!djsdvju!pg!b! resistance and xs is the synchronous reactance.

tbmjfou.qpmf!tzodispopvt! npups

\

E = (Vt - I a ra cos q - I a xs sin q ) 2 + ( I a xs cos q - I a ra sin q ) 2

and

d = tan–1

I a ra sin q - I a xs cos q Vt - I a ra cos q - I a xs sin q

(7.4) (7.5)

When power factor is leading then E = (Vt - I a ra cos q + I a xs sin q ) 2 + ( I a xs cos q + I a ra sin q ) 2

(7.6)

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d = tan–1

and

I a xs cos q + I a ra sin q Vt - I a ra cos q + I a xs sin q

(7.7)

When power factor is unity, E = (Vt - I a ra ) 2 + ( I a xs ) 2

(7.8)

I a xs Vt - I a ra

(7.9)

d = tan–1

and

From Fig. 7.3, the excitation voltage and power angle for a salient-pole synchronous motor can be determined. From Fig. 7.3(a) for lagging power factor, Vt cos d Vt sin d Id Vt cos d Vt sin d

and Now, Hence, and

= E + Iq ra + Id Xd = Iq Xq – Id ra = Ia sin (q – d) and Iq = Ia cos (q – d) = E + Ia ra cos (q – d) + Ia Xd sin (q – d) = Ia Xq cos (q – d) – Ia ra sin (q – d)

(7.10) (7.11) (7.12) (7.13) (7.14)

From Eq. (7.14), Vt sin d = Ia Xq cos q cos d + Ia Xq sin q sin d – Ia ra sin q cos d + Ia ra cos q sin d Vt sin d = (Ia Xq cos q – Ia ra sin q) cos d + (Ia Xq sin q + Ia ra sin d)

or,

tan d =

or,

I a xq cos q - I a ra sin q

(7.17)

Vt - I a X q sin q - I a ra cos q

d = tan–1

or,

(7.15) (7.16)

I a X q cos q - I a ra sin q Vt - I a X q sin q - I a ra cos q

(7.18)

From Eq. (7.13), E = Vt cos d – Ia ra cos (q – d) – Ia Xd sin (q – d)

(7.19)

Similarly, for leading power factor, d = tan–1 and

I a X q cos q - I a ra sin q Vt + I a X q sin q - I a ra cos q

E = Vt cos d + Ia Xd sin (q + d) – Ia ra cos (q + d)

(7.20) (7.21)

For unity power factor, d = tan–1 and

Ia X q Vt - I a ra

E = Vt cos d + Ia Xd sin d

(7.22) (7.23)

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It is now evident that a synchronous motor needs an auxiliary starting arrangement. The methods of starting of a synchronous motor are as follows: 1. Starting with the help of a damper winding 2. Starting with the help of a separate small induction motor 3. Starting by using a dc motor coupled to the synchronous motor 4. Starting as induction motor and run as synchronous motor 2/!Tubsujoh!xjui!uif!Ifmq!pg!b!Ebnqfs!Xjoejoh! In this method, a synchronous motor is started independently using a damper winding. The damper winding is provided on the pole face slots in the field. Bars of aluminium, copper, bronze or similar alloys are inserted in slots of pole shoes. These bars are short-circuited by end-rings on each side of the poles. By short-circuiting of these bars, a squirrel cage winding is virtually formed. When a three-phase supply is given to the stator, a synchronous motor with damper winding will start as a three-phase induction motor with the speed of rotation near to synchronous speed. Now the dc excitation to the field winding of the rotor is applied and the rotor will be pulled into synchronism. A reduced supply voltage may be necessary, to limit the starting current drawn by the motor. In this method, since starting is done as an induction motor, the starting torque developed is rather low. Hence, a large capacity synchronous motor may not be able to start on full load if damper winding starting is employed. 3/!Tubsujoh!xjui!uif!Ifmq!pg!b!Tfqbsbuf!Tnbmm!Joevdujpo!Npups! In this method, a separate induction motor is used to bring the speed of the synchronous motor to synchronous speed. The number of poles of the synchronous motor needs to be more than that of poles of the induction motor to enable the induction motor to rotate at the synchronous speed of the synchronous motor. As the set attains synchronous speed, dc excitation is applied and as the rotor and stator of the synchronus motor are pulled in synchronism, the induction motor is switched off. 4/!Tubsujoh!cz!Vtjoh!b!ed!Npups!Dpvqmfe!up!b!Tzodispopvt!Npups! In this method, the dc motor drives the synchronous motor and brings it to synchronous speed. Then the synchronous motor is synchronised with the supply. 5/!Tubsujoh!bt!Joevdujpo!Npups In this method, the rotor winding is shorted at start and no dc excitation is given. The stator receives the applied voltage in steps and when near full speed is attained by the rotor, the rotor short circuit is

!

8/7

Fmfdusjdbm!Nbdijoft

removed and dc voltage is applied. The motor continues to operate as a synchronous motor. Instead of keeping the rotor winding shorted at start, sometimes there is one more additional winding which helps the machine to start as an induction motor. This winding remains open-circuited during the run of the machine as a synchronous motor. Out of these three methods, the method of using a damper winding for starting the synchronous motor is mostly used, because it requires no external motor.

! Qspcmfn!8/2 Uif!joqvu!up!bo!22111!W-!uisff.qibtf!tubs.dpoofdufe!tzodispopvt!npups!jt!71!B/!Uif!fggfdujwf!sftjt. ubodf!boe!tzodispopvt!sfbdubodf!qfs!qibtf!bsf!2!W!boe!41!W!sftqfdujwfmz/!Gjoe!uif!qpxfs!tvqqmjfe! up!uif!npups!boe!uif!joevdfe!fng!gps!b!q/g/!pg!1/9!mfbejoh/

Solution Per phase input voltage is

11000 3

V = 6351 V

Ia = 60 –cos–1 0.8 A = 60–36.87° A (Note that q is +ve as Ia is leading) Xs = 30 W and Ra = 1 W Power supplied to the motor is 3 VL Ia cos q = 3 ¥ 11000 ¥ 60 ¥ 0.8 W = 914522.82 W = 914.52 kW Induced emf per phase is obtained as E = V – Ia (Ra + j Xs) =

11000 3

– 60–36.87° (1 + j 30)

= 7383 – j 1476.03 = 7529.1 ––11.3° V Hence, line voltage of induced emf is 3 = 7529.1 V, i.e. 13040.38 V.

! Qspcmfn!8/3 B!uisff.qibtf!tzodispopvt!npups!pg!2111!lX!boe!7/7!lW!ibt!b!tzodispopvt!sfbdubodf!pg!21!W!qfs! qibtf/!Uif!fggjdjfodz!pg!uif!npups!jt!:1&/!Ofhmfdujoh!bsnbuvsf!sftjtubodf-!efufsnjof!uif!njojnvn! dvssfou!boe!uif!dpssftqpoejoh!joevdfe!fng!bu!gvmm!mpbe/

Solution At constant supply voltage and input power, the current is minimum when power factor is 1. Output power = 1000 kW

Tzodispopvt!Npups

8/8

Efficiency = 90% Hence, input power =

1000 = 1111.11 kW 0.9

At unity power factor, line current =

1111.11 ¥ 103 3 ¥ 6.6 ¥ 103 ¥ 1

= 97.2 A

Also, at unity power factor, induced emf E =

6600 3

–0° – j 97.2 –0° ¥ 10

= 3810.6–0° – j 972 = 3932.64 ––14.31° V. Hence, at full load, emf is 3932.64 V per phase.

! Qspcmfn!8/4 B!3611!W-!uisff.qibtf!tubs.dpoofdufe!tzodispopvt!npups!ibt!b!sftjtubodf!pg!1/46!W!qfs!qibtf!boe! tzodispopvt!sfbdubodf!pg!3/3!W!qfs!qibtf/!Uif!npups!jt!pqfsbujoh!bu!1/86!q/g/!mfbejoh!xjui!b!mjof! dvssfou!pg!361!B/!Efufsnjof!uif!fydjubujpo!wpmubhf!qfs!qibtf/

Solution Per phase voltage V =

2500 3

V = 1443.42 V

Resistance per phase Ra = 0.35 W Synchronous reactance per phase Xs = 2.2 W V Power factor is 0.75 leading jIXs I Hence, cos q = 0.75 and sin q = 0.66 q IRa d Line current I = 250 A E The phasor diagram of synchronous motor at leading p.f. is Gjh/!8/6! Qibtps!ejbhsbn!pg! shown in Fig. 7.5. uif!tzodispopvt! Hence, excitation voltage per phase can be obtained. npups!bu!mfbejoh!q/g/ E = (V cos q + IRa ) 2 + (V sin q - IX s ) 2 = (1443.42 ¥ 0.75 + 250 ¥ 0.35) 2 + (1443.42 ¥ 0.66 - 250 ¥ 2.2) 2 = 1369052 + 162132.82 = 1237.4 V.

!

8/9

Fmfdusjdbm!Nbdijoft

! Qspcmfn!8/5 B!551!W-!61!I{-!uisff.qibtf!tzodispopvt!npups!ublft!61!B!bu!b!qpxfs!gbdups!pg!1/96!mbhhjoh/!Uif! tzodispopvt!sfbdubodf!qfs!qibtf!jt!6!W!boe!uif!sftjtubodf!jt!ofhmjhjcmf/!Uif!ovncfs!pg!bsnbuvsf! dpoevdupst! qfs! qibtf! jt! 51/! Uif! ejtusjcvujpo! gbdups! jt! 1/:77! boe! uif! dpjmt! bsf! gvmm! qjudife! dpjmt/! Efufsnjof!uif!gmvy!qfs!qpmf/

Solution Terminal voltage per phase is Armature current Power factor Hence, Given,

440 3

I cos q sin q Xs Z Kd

Distribution factor

= 254 V = 50 = 0.85 = 0.5268 =5W = 40 = 0.966

V

E

q

V sin q

I

40 Tp = = 20 2 From the phasor diagram (Fig. 7.6).

jIXs

d

Hence, number of turns per phase is

V

co

sq

Gjh/!8/7! Qibtps!ejbhsbn!pg! Qspc/!8/5

E = (V cos q ) 2 + (V sin q - IX s ) 2 = ( 254 ¥ 0.85) 2 + ( 250 ¥ 0.5268 - 50 ¥ 5) 2 = 46612.8 + 13995 = 246.19 V We know,

E = 4.44 Kd Kp f f Tph

where f is the frequency, f is the flux per pole and Kp is the pitch factor which is 1 in this case. Hence, flux per pole is obtained f=

246.19 Wb = 57.4 mWb. 4.44 ¥ 0.996 ¥ 1 ¥ 50 ¥ 20

! Qspcmfn!8/6 B!3611!W-!uisff.qibtf!tubs.dpoofdufe!tzodispopvt!npups!ibt!b!tzodispopvt!sfbdubodf!pg!4!W!boe! bsnbuvsf!sftjtubodf!pg!1/5!W!sftqfdujwfmz/!Jg!uif!joevdfe!fng!jt!4111!W!boe!uif!joqvu!jt!2111!lX-! efufsnjof!uif!mjof!dvssfou!boe!qpxfs!gbdups/

Tzodispopvt!Npups

8/:

Solution Vt = E=

2500 3 3000 3

V V

3 ¥ VLine ¥ Ia cos q = 1000 ¥ 103 \

Ia cos q =

1000 ¥ 103 3 ¥ 2500

A = 230.94 A

Considering lagging power factor, E2 = (Vt – Ia ra cos q – Ia xs sin q)2 + (Ia xs cos q – Ia ra sin q)2 2

2

\

Ê 2500 ˆ Ê 3000 ˆ - 23.94 ¥ 0.4 - 3I a sin q ˜ + (3 ¥ 230.94 – 0.4 Ia sin q)2 ÁË ˜¯ = ÁË ¯ 3 3

or,

Ê 3000 ˆ 2 2 ÁË ˜ = (1433.8 – 3Ia sin q) + (692.82 – 0.4 Ia sin q) 3 ¯

2

Ia2 sin2 q – 112.83 Ia sin q – 5720 = 0 Ia sin q = 150.76 A 150.76 tan q = = 0.6528 230.94 cos q = 0.837 lag

or, \ \ \ power factor

Line current = Ia =

230.94 A = 276 A. 0.837

! Qspcmfn!8/7 Bo!22!lW-!uisff.qibtf!tubs.dpoofdufe!tzodispopvt!npups!esbxt!b!gvmm!mpbe!dvssfou!pg!261!B!bu!1/9! q/g/!mfbejoh/!Uif!bsnbuvsf!sftjtubodf!jt!2!W!boe!tzodispopvt!sfbdubodf!jt!9!W!qfs!qibtf/!Uif!tusbz! mpttft!pg!uif!nbdijof!bsf!5111!X/!Efufsnjof!uif!joevdfe!fng!boe!fggjdjfodz!pg!uif!nbdijof/

Solution Vt = Ia cos q ra xs

11000 3

V

= 150 A = 0.8 lead =1W =8W

!

8/21

Fmfdusjdbm!Nbdijoft

2

Ê 11000 ˆ E= Á - 150 ¥ 0.8 + 150 ¥ 8 ¥ 0.6˜ + (150 ¥ 8 ¥ 0.8 + 150 ¥ 1 ¥ 0.6) 2 Ë 3 ¯

\

= 7029.7 V \ line voltage of induced emf = 12176 V Input power = 3 VL Ia cos q = 3 ¥ 11000 ¥ 150 ¥ 0.8 W = 2286 kW Total copper loss = 3Ia2 ra = 3 ¥ (150)2 ¥ 1 W = 67.5 kW Stray loss = 4 kW Power output = 2286 – 67.5 – 4 = 2214.5 kW

\

Hence, efficiency =

2214.5 ¥ 100% = 96.87%. 2286

! Qspcmfn!8/8 B!uisff.qibtf!31!NWB-!7/7!lW-!61!I{-!tbmjfou.qpmf!tzodispopvt!npups!ibt!sfbdubodft!Ye!>!1/8!q/v/! boe!Yr!>!1/4!q/v/!Uif!npups!esbxt!sbufe!dvssfou!bu!b!qpxfs!gbdups!pg!1/96!mbhhjoh/!Efufsnjof!uif! fydjubujpo!wpmubhf/

Solution Vt = 1 0∞ p.u. Ia = 1 - cos -1 0.85 p.u. = 1 -31.78∞ p.u. cos q = 0.85, sin q = 0.527 From Eq. (7.17), neglecting armature resistance 1 ¥ 0.3 ¥ 0.85 Vt - I a X q sin q 1 0∞ - 1 ¥ 0.3 ¥ 0.527 = 0.303 d = 16.85°

tan d = \

I a X q cos q

=

From Eq. (7.19), E = Vt cos d – Ia Xd sin (q – d) = 1 ¥ 0.85 – 1 ¥ 0.7 sin (31.78° – 16.85°)

Tzodispopvt!Npups

8/22

= 0.85 – 0.18 = 0.67 p.u.

! Qspcmfn!8/9 B!41!NWB-!uisff.qibtf-!tubs.dpoofdufe-!22!lW-!61!I{-!tbmjfou.qpmf!tzodispopvt!npups!ibt!Ye!>!9!W-! Yr!>!5!W!boe!sb!>!2!W/!Efufsnjof!uif!fydjubujpo!wpmubhf!bu!gvmm!mpbe!boe!vojuz!qpxfs!gbdups/

Solution 30 ¥ 106

Ia =

3 ¥ 11 ¥ 103

= 1574.6 A

From Eq. (7.22) at unity power factor, d = tan–1

Ia X q Vt - I a ra

1574.6 ¥ 4

= tan -1

11 ¥ 103 - 1574.6 ¥ 1 3 = 52.82°

From Eq. (7.23), 11000

cos 52.82° + 1574.6 ¥ 8 sin 52.82° 3 = 13875.6 V = 13.87 kV.

E=

QPXFS!BOE!UPSRVF!EFWFMPQFE!JO! B!DZMJOESJDBM!SPUPS!NPUPS

8/6

Let S represent the per phase complex power drawn by the cylindrical rotor synchronous motor. *

S=

V ◊ I*a

V -E = V◊ , Zs

where Zs is the synchronous impedance. *

= V◊ Now, and

V E -V ◊ Zs Zs

*

E = | E | ––d Zs = | Zs | –f = Rs2 + X s2 – tan -1

Xs Rs

!

8/23

Hence,

Fmfdusjdbm!Nbdijoft

S = | V | –0° . =

| V | –0∞ | E | –d - | V | –0∞ ◊ | Z s | –- f | Z s | –- f

| V |2 | V || E | –f –d + f = P + jQ | Zs | | Zs |

\ active power P per phase is P=

| V |2 | V || E | cos (d + f) cos f | Zs | | Zs |

(7.24)

| V |2 | V || E | sin (d + f) sin f | Zs | | Zs |

(7.25)

and reactive power Q per phase is Q=

Neglecting stator resistance (i.e. f = 90°), P=

| V || E | sin d | Xs |

where Xs is the synchronous reactance. Total active power (or total mechanical power developed) by the motor is P=

3 | V || E | sin d | Xs |

(7.26)

= Pmax sin d, where Pmax =

3 | V || E | . | Xs |

Mechanical torque developed by the motor T=

Pmax sin d . ws

where (ws) is the synchronous speed \

T=

3 | V || E | sin d ws | X s |

= Tmax sin d where (Tmax) is the maximum torque which is also known as pull-out torque. The power angle characteristic is shown in Fig. 7.7.

(7.27)

Tzodispopvt!Npups

8/24

Pmax P –180°

–90°

0°

90°

180°

d

Generator

Motor

Gjh/!8/8! Qpxfs!bohmf!dibsbdufsjtujdt!pg!b!dzmjoesjdbm!spups!tzodispopvt!npups

The synchronous motor can be loaded up to a maximum value of Pmax which is called the static stability limit and after which it will lose synchronism. In order to increase the stability limit at fixed applied voltage V, the field current should be increased which in turn will increase the excitation voltage E.

QPXFS!BOE!UPSRVF!EFWFMPQFE!JO! B!TBMJFOU.QPMF!NPUPS

8/7

Neglecting the armature resistance, the simplified phasor diagram of salient-pole motor is shown in Fig. 7.8. From the phasor diagram, | V | cos d = | E | + Id Xd | V | sin d = Iq Xq | V | cos d - | E | Id = and Xd

and Hence,

Iq =

| V | sin d Xd

(7.28)

If Vd and Vq are the two components of V then

V

Vd = –| V | sin d [–ve sign appears as for motor d is –ve and sin (–d) = –sin d] Vd = | V | cos d Per phase active power P = Vd Id + Vq Iq = –| V | sin d . =

| V | cos d - | E | V sin d + | V | cos d Xd Xq

Ê 1 | V || E | sin d 1 ˆ + | V |2 sin d cos d Á ˜ Xd Ë Xq Xd ¯

a Id

d q

Iq

jIaXa b E jIdXd

jIqXq

Ia

Gjh/!8/9! Qibtps!ejbhsbn!pg! b!tbmjfou!qpmf!npups! )ofhmfdujoh!Sb*

!

8/25

Fmfdusjdbm!Nbdijoft

=

Xd - Xq | V || E | sin 2d sin d + | V |2 2Xd Xq Xd

(7.29)

Hence, total mechanical power developed Pm =

Xd - Xq 3 | V || E | sin 2d sin d + 3 | V |2 Xd 2Xd Xq

(7.30)

Total torque developed T=

Xd - Xq Pm 3 | V || E | = sin d + 3 | V |2 sin 2d ws ws X d 2w s X d X q

(7.31)

Ê 3 | V || E | ˆ sin d ˜ is the power of the synchronous motor due to excitation voltage E and ÁË X ¯ d

Ê 3 |V | ˆ ( X d - X q )sin 2d ˜ is the power due to saliency. Á Ë 2Xd Xq ¯ 2

Ê 3 | V |2 ˆ ( X d - X q )sin 2d ˜ is called Á Ë 2w s X d X q ¯

the reluctance torque and is independent of field excitation. The power angle curve of a salient-pole synchronous motor is shown in Fig. 7.9. Resultant power Power

3 | V || E | sin d Xd O

Motor

d

Generator

3 | V |2 (X – X ) sin 2d d q 2 Xd Xq

Gjh/!8/:! Qpxfs!bohmf!dibsbdufsjtujd!pg!b!tbmjfou.qpmf!tzodispopvt!npups

! Qspcmfn!8/: B!41!NWB-!22!lW-!uisff.qibtf!tubs.dpoofdufe!tzodispopvt!npups!ibt!b!ejsfdu.byjt!sfbdubodf!pg!7!W! boe!rvbesbuvsf!byjt!sfbdubodf!pg!5!W/!Efufsnjof!uif!fydjubujpo!wpmubhf!boe!qpxfs!bu!gvmm!mpbe!boe! vojuz!qpxfs!gbdups/!Ofhmfdu!uif!bsnbuvsf!sftjtubodf/

Solution Terminal voltage per phase V=

11000 3

= 6351 V

Tzodispopvt!Npups

8/26

Armature current per phase I=

30000 3 ¥ 11

A = 1574.638 A

Xd = 6 W and Xq = 4 W (given) From Eq. (7.3), the excitation voltage E = V – j IXq – j Id (Xd – Xq), neglecting armature resistance. From Fig. 7.8, ab = V – j IXq = 6351 – j 1574.638 –0° ¥ 4 = 6350 – j 6298.55 = 8944 ––44.767° \ load angle Now,

d = 44.767° Id = I sin d = 1574.638 sin 44.767° = 1108.9 A

Hence, excitation voltage E = V – j I Xq – jId (Xd – Xq) = 6351 – j 1574.638 ¥ 4 – j 1108.9(6 – 4) = 6351 – j 6298.55 – j 2217.8 = 6351 – j 8516.35 = 10623 ––53.3° V \ Excitation voltage per phase is 10623 V. From Eq. (7.30), total power at full load, P=

=

Xd - Xq 3VE sin d + V 2 3 sin 2d Xd 2Xd Xq 6-4 3 ¥ 6351 ¥ 10623 sin (44.767°) + (6351)2 ¥ 3 ¥ sin (2 ¥ 44.77°) 2¥6¥4 6

= 28797.61 kW.

! Qspcmfn!8/21 B!2111!lWB-!44!lW-!tubs.dpoofdufe!tbmjfou.qpmf!tzodispopvt!npups!ibt!Ye!>!31!W!boe!Yr!>!21!W! qfs!qibtf!sftqfdujwfmz/!Efufsnjof!uif!fydjubujpo!fng!xifo!uif!npups!jt!tvqqmzjoh!sbufe!mpbe!bu!1/9! mfbejoh! q/g/! Xibu! nbyjnvn!mpbe! dbo!uif!npups! tvqqmz! jg! uif! fydjubujpo! jt! dvu! pgg! xjuipvu! mptt! pg! tzodispojtn@!Ofhmfdu!bsnbuvsf!sftjtubodf!boe!bttvnf!mpbe!bohmf!up!cf!36/46°/

!

8/27

Fmfdusjdbm!Nbdijoft

Solution Armature current =

1000 3 ¥ 3.3

A = 175 A.

Terminal voltage per phase

V=

3.3 3

kV = 1.9053 kV = 1905.3 V.

Xd = 20 W and Xq = 10 W d = 25.35°. Id = I sin d = 74.92 A

Given, Here, load angle Now,

\ excitation voltage per phase is given as E = V – j IXq – jId (Xd – Xq) = 1905.3 – j 175 ¥ 10 – j 74.92 (20 – 10) = 1905.3 – j 1750 – j 749.2 = 1905.3 – j 2499.2 = 3142.6 ––52.68° V. When the excitation is cut off, output power is only the reluctance power. Power = 3V 2

Xd - Xq 2Xd Xq

= 3(1905.3)2

sin 2d

20 - 10 sin (2 ¥ 25.35°) 2 ¥ 20 ¥ 10

= 210687 W = 210.687 kW.

! Qspcmfn!8/22 B!7/7!lW-!tubs.dpoofdufe-!tzodispopvt!npups!jt!pqfsbujoh!bu!dpotubou!fydjubujpo/!Uif!tzodispopvt! jnqfebodf!jt!)2!,!k!9*!W/!Ju!pqfsbuft!bu!b!qpxfs!gbdups!pg!1/9!mbhhjoh!boe!esbxt!2111!lX!gspn!uif! tvqqmz/!Efufsnjof!jut!qpxfs!gbdups!xifo!uif!joqvu!jt!jodsfbtfe!up!2311!lX/

Solution Zs = 1 + j 8 = 8.06 –82.87° W V= P1 =

Hence,

6600 3

= 3810.62 V

1000 kW = 333.33 kW 3

cos q = 0.8 lagging Q1 = 333.33 ¥ tan q = 249.997 kVAR

Tzodispopvt!Npups

8/28

From Eqs (11.15) and (11.16), P1 = 333.33 ¥ 103 =

| V |2 | V || E | cos (d + q) cos q | Zs | | Zs |

(i)

| V |2 | V || E | sin (d + q) sin q | Zs | | Zs |

(ii)

Q1 = 249.997 ¥ 103 = From Eq. (i), 333.33 ¥ 103 =

(3810.62) 2 3810.62 | E | ¥ 0.8 cos (d + q) 8.06 8.06

E cos (d + q) = 2343.52

or

(iii)

From Eq. (ii), (3810.62) 2 3810.62 | E | 249.997 ¥ 10 = ¥ 0.6 sin (d + q) 8.06 8.06 3

E sin (d + q) = 1757.59

or

(iv)

Dividing Eq. (iv) by Eq. (iii), we get or Hence,

tan (d + q) = 0.75 d = 36.87° – 36.87° = 0 2343.456 E= = 2929.32 V cos q

Since

P2 =

1200 kW = 400 kW, we can write 3

(3810.62) 2 3810.62 ¥ 2929.32 cos q cos q 8.06 8.06 cos q = 0.96

400 ¥ 103 = or,

\ now power factor is 0.96 lagging.

! Qspcmfn!8/23 Uif!gvmm!mpbe!dvssfou!pg!b!7/7!lW-!tubs.dpoofdufe!npups!jt!311!B!bu!1/9!q/g/!mbhhjoh/!Uif!qfs!qibtf! sftjtubodf!boe!sfbdubodf!pg!uif!npups!jt!2!W!boe!8!W!sftqfdujwfmz/!Jg!uif!nfdibojdbm!mptt!jt!51!lX-! gjoe!uif!fydjubujpo!fng-!upsrvf!bohmf-!fggjdjfodz!boe!tibgu!pvuqvu!pg!uif!npups/

Solution Given, R =1W Xs = 7 W

!

8/29

Fmfdusjdbm!Nbdijoft

V=

6600 3

= 3810.6 V

I = 200 A, cos q = 0.85. Hence, excitation emf is E = 3810.6 – 200 ––cos–1 0.8 (1 + j 7) = 3810.6 – 200 ––36.87° ¥ 7.07 –81.87° = 3810.6 – 1414 –45° = 2810.75 – j 999.8 = 2983.29 ––19.58° V. \ excitation emf is 2983.29 V and torque angle is 19.58°. Mechanical power developed = 3 EI cos q, where q is the angle between E and I = 3 ¥ 2983.29 ¥ 200 cos (–36.87° + 19.58°) = 3 ¥ 2983.29 ¥ 200 cos (–17.29°) = 1709089.88 W = 1709 kW. The shaft output = 1709 – 40 = 1669 kW Power input = 3 VL I cos q = 3 ¥ 6600 ¥ 200 ¥ 0.8 = 1829045.65 W = 1829 kW Hence, efficiency =

1669 ¥ 100% = 91.25%. 1829

! Qspcmfn!8/24 B!211!NWB-!uisff.qibtf-!tubs.dpoofdufe-!22!lW-!7.qpmf-!61!I{-!tzodispopvt!npups!ibt!tzodispopvt! sfbdubodf!pg!1/3!q/v/!Efufsnjof!uif!tzodispoj{joh!qpxfs!qfs!nfdibojdbm!efhsff!jg!uif!npups!tvq. qmjft!gvmm!mpbe!bu!1/96!qpxfs!gbdups!mfbejoh/!Ofhmfdu!bsnbuvsf!sftjtubodf/

Solution Ia =

100 ¥ 106

3 ¥ 11 ¥ 103 cos q = 0.85 leading E = Vt – j Ia xs =

A = 5248.63 A

11000

– j 5248.63 cos -1 0.85 ¥ 0.2

3 = 6350.85 – 1049.7 121.79∞ = 6903.84 – j 892.2

Tzodispopvt!Npups

8/2:

= 6961.25 -7.41∞ V \

d = 7.41°

Total synchronizing power per mechanical degree Psyn = 3 =3 =

ˆ Pp dP p P d ÊV E ¥ ¥ = 3 Á t sin d ˜ dd 180 2 dd Ë x s ¯ 360 Vt E Pp cos d 360 xs

3 ¥ 11000 ¥ 6961.25 3 ¥ 0.2

cos 7.41∞ ¥

6¥p 360

= 34415 kW.

! Qspcmfn!8/25 B!61!NWB-!uisff.qibtf-!22!lW-!tubs.dpoofdufe!21.qpmf-!61!I{!tzodispopvt!npups!ibt!Ye!>!7!W!boe! Yr!>!4!W/!Bu!gvmm!mpbe!boe!vojuz!qpxfs!gbdups-!dbmdvmbuf! ! )b*! fydjubujpo!wpmubhf ! )c*! bdujwf!qpxfs ! )d*! tzodispoj{joh!qpxfs!qfs!fmfdusjdbm!efhsff!boe!uif!dpssftqpoejoh!upsrvf! ! )e*! tzodispoj{joh!qpxfs!qfs!nfdibojdbm!efhsff!boe!uif!dpssftqpoejoh!upsrvf

Solution 50 ¥ 106 = 3 ¥ 11 ¥ 103 ¥ Ia \

Ia = 2624.32 A From Eq. (7.22), d = tan–1

\

Ia X q Vt - I a ra

= tan -1

2624.32 ¥ 3 neglecting ra 11000 3

d = 51.1°

(a) From Eq. (7.23), E=

11000 3

cos 51.1° + 2624.32 ¥ 6 sin 51.1°

= 3988 + 12254 = 16242 V = 16.24 kV Line value of excitation voltage 3 ¥ 16.24 kV = 28.13 kV

!

8/31

Fmfdusjdbm!Nbdijoft

(b) Total active power from Eq. (7.30) P=

Xd - Xq 3Vt E sin d + 3Vt 2 sin 2d Xd 2Xd Xq 2

Ê 11000 ˆ 16240 6-3 =3¥ sin 2 ¥ 51.1° ¥ ¥ sin 51.1∞ + 3 Á ˜ Ë 3 ¯ 6 2¥6¥3 3 = 40133163 W + 9855610 W = 49988.77 kW dP p (c) Synchronizing power per electrical degree = W dd 180 11000

ÏÔ 3V E ¸Ô p Xd - Xq 2 ◊ cos 2d ˝ ¥ = Ì t cos d + 3 Vt 2 2Xd Xq ÔÓ X d Ô˛ 180 11000 ¸ Ï ¥ 16240 2 ÔÔ p ÔÔ 3 ¥ Ê ˆ 11000 6 3 3 2 2 51 . 1 ) cos 51.1∞ + 3 ¥ Á ¥ ¥ cos ( ¥ =Ì ˝¥ Ë 3 ˜¯ 6 2¥6¥3 Ô 180 Ô Ô˛ ÔÓ p = (32383380 – 4261716.7) 180 = 490.56 kW/electrical degree \ synchronizing torque =

=

Synchronizing power ws 490.56 ¥ 103 Nm/electrical degree 2 ¥ 50 2p ¥ 10

= 7811.46 Nm/electrical degree (d) Synchronizing power per mechanical degree =

dP p P 10 kW/mechanical degree ¥ = 490.50 ¥ 2 dd 180 2

= 2452.8 kW/mechanical degree 2452.8 ¥ 103 Torque = Nm/mechanical degree 2 ¥ 50 2p ¥ 10 = 39057 Nm/mechanical degree.

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8/32

! Qspcmfn!8/26 Bo!22!LW-!uisff.qibtf-!tubs.dpoofdufe!tzodispopvt!npups!ibt!Ye!>!23!W!boe!Yr!>!7!W/!Ju!jt!dpo. ofdufe!up!bo!jogjojuf!cvt/!Jg!uif!gjfme!dvssfou!jt!sfevdfe!up!{fsp-!efufsnjof!uif!nbyjnvn!qpttjcmf! mpbe!po!uif!tzodispopvt!npups/!Bmtp!dbmdvmbuf!uif!bsnbuvsf!dvssfou/

Solution From Eq. (7.29), per phase active power P=

Xd - Xq Vt E sin 2d sin d + Vt 2 2Xd Xq Xd

When field current is zero, E = 0 \

P = Vt2

Xd - Xq 2Xd Xq

sin 2d

2

Ê 11000 ˆ 12 - 6 =Á ¥ sin 2d W Ë 3 ˜¯ 2 ¥ 12 ¥ 6 For maximum power, sin 2d = 1 2

\

Ê 11000 ˆ 6 P=Á ¥ W = 1680555.56 W ˜ Ë 3 ¯ 2 ¥ 12 ¥ 6 = 1680.55 kW Total maximum load for all the three-phases = 1680.55 ¥ 3 kW = 5041.65 kW For maximum power sin 2d = 1 or d = 45° From Eq. (7.28), Id = = Iq = =

\

Vt cos d (E E = 0) Xd 11000 cos 45∞ 3 ¥ 12

A = 374.23 A

Vt sin d Xq 11000 sin 45∞ 3¥6

A = 748.45 A

Ia = (374.23) 2 + (748.45) 2 = 836.79 A.

!

8/33

Fmfdusjdbm!Nbdijoft

FGGFDU!PG!DIBOHF!JO!MPBE!

8/8

The synchronous motor always runs at synchronous speed. Hence, the change of load has no effect on the speed of the motor. In this section, we will examine the effect of change of load on the motor. Let us consider that the excitation is normal and is kept constant. The motor is operating at lagging power factor. Now if the load on the shaft of the motor is increased, the rotor slows down momentarily. It takes sometime to take more power from the line. In other words, the rotor poles slip back by a small angle with respect to the stator poles. Hence, the torque angle d becomes greater V E sin d increases. The increased torque increases the rotor speed and the and the induced torque t w xs motor picks up synchronous speed, with a larger value of torque angle d. With the excitation kept constant, the excitation emf E is constant. With increase in load as the power angle d increases, E moves on a circular locus keeping its centre at the origin and with a radius equal to E as shown in Fig. 7.10. The steady state stability limit is reached when d = 90°. A series of concentric circles can be drawn for different excitation values. With the increase of the load, the magnitude of Ia xs and d increases so that the relation V = E + jI a xs is satisfied. Also the power factor angle q changes. With the increase in load, the lagging power factor decreases or the power factor angle increases. Thus, when the load on a synchronous motor increases at constant excitation, the torque angle d and the armature current Ia increases, the phase angle q increases in the lagging direction, with the speed remaining constant at synchronous speed. Vt

O d1 d3

Ia1

d2

Ia1xs

Ia2

E1 Ia3 Ia2xs

E2

Ia3xs

E3 Excitation circles

Stability line

Gjh/!8/21! Fggfdu!pg!dibohf!jo!mpbe!xjui!fydjubujpo!dpotubou

If the load is increased, the torque angle increases till a point is reached when the rotor is pulled out of synchronism and the motor stalls. The maximum value of the torque that a synchronous

Tzodispopvt!Npups

8/34

motor can develop at rated voltage and frequency without losing synchronism is called the pull-out torque. It ranges from 1.5 to 3.5 times the full load torque.

FGGFDU!PG!DIBOHF!JO!FYDJUBUJPO!

8/9

The power developed by a synchronous motor P=

EVt sin d xs

If the load on the motor is kept constant, the power developed is constant and, hence, E sin d is constant. Hence, if the excitation of a synchronous motor is varied keeping the load constant, the excitation emf will change in such a way that E sin d is constant. On the other hand, as the power developed is constant, Vt Ia cos q is constant or Ia cos q is constant. The phasor diagram of a synchronous motor for varying excitation with constant load is shown in Fig. 7.11. Ia1, Ia2 and Ia3 are different armature currents for lagging, unity and leading power factors so that Ia1 cos q1 = Ia2 cos 0° = Ia3 cos q3 = constant Similarly, the varying excitation voltages corresponding to current Ia1, Ia2 and Ia3 are E1, E2, and E3 such that E1 sin d1 = E2 sin d2 = E3 sin d3 = constant Locus of Ia

I a3

Ia2

Ia1 E1

Ia2xs

Ia x 1 s

d2

E2

xs

Emin

d3

Vt

E sin d

3

q1 d1

Ia

q3

E3

Ia cos q Locus of E

Gjh/!8/22! Fggfdu!pg!dibohf!jo!fydjubujpo!xjui!mpbe!dpotubou!

The excitation which makes the armature current to be in phase with Vt, is the normal excitation. Here, the normal excitation is E2 at armature current Ia2 and unity power factor. In this case E2 cos d1 = Vt. When E cos d < Vt the motor has lagging current and considered to be under-excited.

!

8/35

Fmfdusjdbm!Nbdijoft

Here, at excitation emf is E1 at lagging armature current Ia1, E1 cos d1 < Vt and, hence, the motor is underexcited. When E cos d > Vt, the motor takes leading current and the field current is large in this case. The motor is said to be overexcited. In Fig. 7.11, during overexcited condition, excitation emf is E3 and E3 cos d3 > Vt when armature current is Ia3 at leading power factor angle d3. It is seen from Fig. 7.11 that if the excitation is reduced, the torque angle d increases. The minimum value of E occurs when d = 90°. Emin =

Hence,

Pxs Vt

W!DVSWFT!PG!B!TZODISPOPVT!NPUPS!

8/:

We have seen in the previous sections that keeping the motor load fixed, the armature current can be varied over a large range by varying the excitation. Let us assume that the motor is operating under Stability limit or minimum excitation

Full load 50% load

Armature current Ia

No load

Unity pf line

Lagging pf (a) V curves

Leading pf Field current If

Unity pf line

Power factor pf

Full load 50% load No load

Lagging pf (b) Inverted V curves

Leading pf Field current If

Gjh/!8/23! W!dvswft!boe!jowfsufe!W!dvswft!gps!ejggfsfou!mpbet

Tzodispopvt!Npups

8/36

no-load condition. If the field current is increased, the armature current under lagging power factor decreases until it becomes minimum. The lagging power factor goes on increasing until it becomes unity when the armature current is minimum. If the field current is further increased, the armature current again increases and the motor starts operating at leading power. The graph of armature current vs. field current takes the shape of letter V and is called V curve which is shown in Fig. 7.12(a). V curves are useful in adjusting the field current t. If the field current is increased beyond the level of minimum armature current, the motor will operate under leading power factor. Similarly, if the field current is decreased below the level of minimum armature current, the motor will start operating at lagging power factor. Hence, by controlling the field current, the reactive power supplied to or consumed from the power system network can be controlled. Inverted V curves are the characteristics showing the variation of power factor with the armature current for different excitations when the motor load is constant. The inverted V curves are shown in Fig. 7.12(b). The maximum point on each curve indicates unity power factor. The field current for unity power factor at full load is more than the field current for unity power factor at no load. If the synchronous motor at full load is operating at unity power factor then removal of shaft load causes the motor to operate at leading power factor.

IVOUJOH!PG!B!TZODISPOPVT!NBDIJOF!

8/21

When a synchronous motor is loaded mechanically at its shaft, the rotor falls back by a certain angle known as load angle d behind the poles of the forward rotating magnetic field. If the load is suddenly thrown off, the rotor poles are pulled into almost exact position to the poles of the forward field, but due to inertia of rotor, the rotor poles travel beyond. They are then pulled back again and this process may continue. Thus, an oscillation of the rotor is set up about the equilibrium position corresponding to new load. This oscillation of the rotor about its equilibrium position is known as hunting. Hunting may be caused due to the following reasons: (a) Change in load. (b) Change in excitation. (c) Change in supply frequency. Hunting is an undesirable characteristic of all synchronous motors. It produces severe mechanical stresses as well as great variations in current and power drawn by the motor. Hunting increases the possibility of resonance. Resonance takes place when the frequency of the torque component is equal to that of the transient oscillations of the synchronous motor. Hunting also increases the losses in motors and increases temperature of the machine.

!

8/37

Fmfdusjdbm!Nbdijoft

NFUIPET!PG!SFEVDJOH!IVOUJOH!

8/22

Hunting can be reduced by providing damper winding (or grids). These windings consist of shortcircuited copper bars embeded in the faces of the field poles of the synchronous motors. When the rotation at constant load is uniform, there is no relative motion between the rotor and stator forward rotating fields and, hence, no current is induced in these windings. But when hunting takes place, the relative motion of the rotor sets up eddy currents in these windings which flow such as to suppress the oscillations (as per Lenz’s Law). The dampers should have low resistance to be more effective. However, this method cannot eliminate hunting completely.

DPNQBSJTPO!CFUXFFO!TZODISPOPVT!BOE! JOEVDUJPO!NPUPST Synchronous motor

8/23

Induction motor

1. It is internally not a self-starting motor.

1.

It is a self-starting motor.

2. It runs at constant speed called synchronous speed and this speed is independent of the load.

2.

The speed of the motor is always less than the synchronous speed and its speed decreases as the shaft load increases.

3. It requires dc source for the field excitation.

3.

No dc exciter is needed.

4. It can be operated under a wide range of power factors including lagging and leading p.f.

4.

It runs with lagging p.f. only which may be very low at light loads.

5. It runs at synchronous speed only. The only way to change its speed is by varying the supply frequency.

5.

Many power electronic methods are available with which speed can be varied.

6. It is used to improve the p.f. and in that case it is called synchronous capacitor.

6.

It is used only to drive a mechanical load.

7. It is costlier and construction-wise more complicated. Maintenance is also more.

7.

It is cheaper and is commonly used in industry as a robust drive.

BQQMJDBUJPOT!PG!TZODISPOPVT!NPUPST!

8/24

An over-excited synchronous motor operates at leading p.f. and takes a leading current from the bus bar; so it can be used to raise the overall power factor of the bus-bar supplying load. When the motor is run without load with over-excitation for improving the voltage regulation of a transmission line, it is called a synchronous capacitor or synchronous condenser. Synchronous motors can be used in electrical clocks as they run at constant speed.

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8/38

! Qspcmfn!8/27 B!2311!lX!mpbe!ibt!b!qpxfs!gbdups!pg!1/7!mbhhjoh/!Gjoe!uif!sbujoh!pg!b!tzodispopvt!dpoefotfs!up! sbjtf!uif!qpxfs!gbdups!up!1/:!mbhhjoh/!Bmtp!gjoe!uif!upubm!lWB!tvqqmjfe!bu!uif!ofx!qpxfs!gbdups/

Solution The phasor diagram is shown in Fig. 7.13. P = 1200 kW At 0.3 p.f. lagging, S1 = OB =

Apprarent power is

1200 1200 = cos q1 0.6

B S1

= 2000 kVA. Reactive power

q1 O

Q1 = AB = S1 sin q1 = 2000 sin (cos–1 0.6) = 1600 kVAR

S2

q2 1200 kW

C A

Gjh/!8/24! Qibtps! ejbhsbn! pg!Qspc/!8/27

When power factor is raised to 0.9 by a synchronous condenser, real power remaining same, P = 1200 kW, 1200 1200 = = 1333.33 kVA cos q 2 0.9

Apparent power

S2 =

Reactive power

Q2 = AC = S2 sin q2 = 1333.33 sin (cos–1 0.9) = 581.18 kVAR

Hence, reactive power supplied by the condenser is BC = AB – AC = 1600 – 581.18 = 1018.8 kVAR. Therefore, rating of the synchronous condenser is 1018.8 kVAR and total kVA supplied is 1333.33 kVA.

! Qspcmfn!8/28 B!tvctubujpo!tvqqmjft!b!mpbe!pg!26!NWB!bu!1/9!q/g/!mbh/!Ju!jt!sfrvjsfe!up!sbjtf!uif!qpxfs!gbdups!up! )b*!1/:!mbh-!boe!)c*!vojuz!cz!vtjoh!bo!pwfs.fydjufe!tzodispopvt!npups/!Efufsnjof!uif!sbujoh!pg!uif! npups/

Solution Reactive power supplied by the substation = 15 sin (cos–1 0.8) = 15 ¥ 0.6 MVAR = 9 MVAR lag (a) The required reactive power = 15 sin (cos–1 0.9) = 6.538 MVAR lag

!

8/39

Fmfdusjdbm!Nbdijoft

Hence, the leading reactive power supplied by the motor = (9 – 6.538) MVAR = 2.462 MVAR \ rating of the motor is 2.462 MVAR (b) Required reactive power = 15 sin (cos–1 1) = 0 Hence, leading reactive power supplied by the motor = 9 – 0 = 9 MVAR \ rating of the motor is 9 MVAR.

! Qspcmfn!8/29 B!uisff.qibtf!21.qpmf-!551!W-!61!I{-!tubs.dpoofdufe!tzodispopvt!npups!pqfsbuft!po!op!mpbe!xjui! jut! joevdfe! fng! frvbm! up! uif! tvqqmz! wpmubhf/! Uif! tzodispopvt! sfbdubodf! jt! 21! W! boe! bsnbuvsf! sftjtubodf!jt!ofhmjhjcmf/!Jg!uif!npups!jt!mpbefe!boe!ju!tmjqt!cz!7°!gspn!jut!qptjujpo!pg!tzodispojtn-! efufsnjof!uif!qpxfs!joqvu/!

Solution Vt =

440 3

V=E

xs = 10 W P = 10 Power angle = 6 ¥ Now

P 10 =6¥ = 30° 2 2 Vt = E + j Ia xs Vt 0∞ = E -d + j Ia xs

\

j Ia xs =

or,

Ia =

or,

Ia =

440 3

-

440 3

440 -90∞ 3 ¥ 10

-

-30∞ 440 -30∞ - 90∞ 3 ¥ 10

44 Ê 1 3ˆ Á - j + 2 + j 2 ˜ = 12.7 – j 3.4 = 13.15 -15∞ A 3Ë ¯

Hence, power input = Vt Ia cos q =

440 3

¥ 13.15 ¥ cos 15° W = 3226.7 W per phase.

Tzodispopvt!Npups

8/3:

! Qspcmfn!8/2: B!7/7!lW-!uisff.qibtf-!tubs.dpoofdufe!tzodispopvt!npups!esbxt!2111!lX!bu!1/9!qpxfs!gbdups!mbh. hjoh/!Uif!tzodispopvt!sfbdubodf!pg!uif!npups!jt!6!W/!Ofhmfdujoh!uif!bsnbuvsf!sftjtubodf-!efufsnjof! uif!ofx!mjof!dvssfou!boe!qpxfs!gbdups!xifo!uif!cbdl!fng!pg!uif!nbdijof!jodsfbtft!cz!51&!boe!uif! npups!esbxt!tbnf!qpxfs!gspn!uif!tvqqmz/

Solution 3 VL IL cos q = P 3 ¥ 6.6 ¥ 103 ¥ IL ¥ 0.8 = 1000 ¥ 103 IL = 109.35 A

or, or,

E = Vt - I a Z s \

E =

6.6 ¥ 103 3

- 109.35 - cos -1 0.8 ¥ (j 5)

= 3810.5 – 546.75 -36.87∞ = 3373.1 + j 328 = 3389 5.55∞ V When E is increased by 40%, the new value of E is As

E = 1.4 ¥ 3389 V = 4744.6 V E > Vt

(E E = 4744.6 V and Vt = 3810.5 V)

\ power factor is leading 3 VL I¢ cos q ¢ = 1000 ¥ 103 \

I¢ cos q¢ = Again,

1000 ¥ 103 3 ¥ 6.6 ¥ 103

A = 87.5 A

E 2 = (Vt + Ia xs sin q)2 + (Ia xs cos q)2

When ra is neglected, (4744.6)2 = (3810.5 + 5 I¢ sin q ¢)2 + (5 ¥ 87.5)2 I¢ sin q ¢ = 182.78 A 182.78 \ tan q¢ = = 2.1 87.5 or, cos q ¢ = 0.43 87.5 \ I¢ = A = 203.49 A 0.43 The new line current and power factor are 203.49 A and 0.43 leading. \ or,

!

8/41

Fmfdusjdbm!Nbdijoft

! Qspcmfn!8/31 B! uisff.qibtf! 7/7! lW-! 61! I{-! tubs.dpoofdufe! tzodispopvt! npups! ibt! 1/2! W0qibtf! bsnbuvsf! sf. tjtubodf/!Uif!pvuqvu!pg!uif!npups!jt!211!lX!boe!ju!pqfsbuft!bu!b!qpxfs!gbdups!pg!1/9!mfbejoh/!Uif! jspo!boe!nfdibojdbm!mpttft!bsf!6111!X!boe!uif!fydjubujpo!dvssfou!jt!26!B!gspn!b!411!W!ed!tpvsdf/! Efufsnjof!uif!bsnbuvsf!dvssfou!boe!fggjdjfodz!pg!uif!npups/!

Solution 100 kW = 33333 W 3 5000 Per phase iron and mechanical loss = W 3 Per phase output power =

\ power developed by the motor = 33333 + 1667 = 35000 W Input power = Power developed + Copper losses in stator = 35000 + Ia2 ¥ 0.1 Again input power = Vt Ia cos q = 0.8 Vt Ia \

35000 + 0.1 Ia2 = 0.8 ¥

6600 3

Ia

0.1 Ia2 – 3048 Ia + 35000 = 0 \

Ia = =

3048 ± (3048) 2 - 4 ¥ 35000 ¥ 0.1 2 ¥ 0.1 3048 ± 3045.7 = 11.5 A 2 ¥ 0.1

Total input power per phase = Armature input power + Excitation input power =

6600 3

¥ 11.5 ¥ 0.8 +

h=

15 ¥ 300 = 36556 W 3

33333 ¥ 100% = 91.18%. 36556

! Qspcmfn!8/32 B!571!W-!61!lX-!71!I{-!uisff.qibtf!tzodispopvt!npups!ibt!b!tzodispopvt!sfbdubodf!pg!5/26!W!boe! bsnbuvsf!up!gjfme!nvuvbm!joevdubodf!pg!94!nI/!Uif!npups!jt!pqfsbujoh!bu!sbufe!ufsnjobm!wpmubhf!boe! bo!joqvu!qpxfs!pg!51!lX/!Efufsnjof!uif!nbhojuvef!boe!qibtf!bohmf!pg!mjof!up!ofvusbm!hfofsbufe! wpmubhf!boe!uif!gjfme!dvssfou!jg!uif!npups!jt!pqfsbujoh!bu!)b*!1/96!qpxfs!gbdups!mbhhjoh-!)c*!vojuz! qpxfs!gbdups-!boe!)d*!1/96!qpxfs!gbdups!mfbejoh/

Tzodispopvt!Npups

Solution (a) Armature current

Ia =

40 ¥ 103 3 ¥ 460 ¥ 0.85

A = 59.1 A

Ia = 59.1 - cos -1 0.85 A = 59.1 -31.78∞ A

or,

The line-to-neutral generated voltage E = Vt – j Ia xs =

460 3

- j 59.1 -31.78∞ ¥ 4.15

= 265.58 – 245.26 58.22∞ = 136.41 – j 208.49 = 249.15 -56.8∞ V If =

The field current

2E 2 ¥ 249.15 = A = 11.26 A w Laf 2p ¥ 60 ¥ 0.083

(b) At unity power factor, Ia = E=

40 ¥ 103 3 ¥ 460 460 3

0∞ A = 50.2 0∞ A

– j 50.2 ¥ 4.15

= 265.58 – j 208.33 = 337.54 -38.11∞ V If =

2 ¥ 337.54 A = 15.26 A 2p ¥ 60 ¥ 0.083

(c) At 0.85 leading power factor, Ia = 59.1 31.78∞ A \

E=

460 3

- j 59.1 31.78∞ ¥ 4.15

= 265.58 – 245.26 121.78∞ = 394.75 – j 208.49 = 446.42 -27.84∞ V \

If =

2 ¥ 446.42 A = 20.18 A. 2p ¥ 60 ¥ 0.083

8/42

!

8/43

Fmfdusjdbm!Nbdijoft

! Qspcmfn!8/33 B!61!I{-!uxp.qpmf-!861!lWB-!3411!W-!uisff.qibtf!tzodispopvt!nbdijof!ibt!b!tzodispopvt!sfbd. ubodf!pg!8/86!W!boe!bdijfwft!sbufe!pqfo.djsdvju!ufsnjobm!wpmubhf!bu!b!gjfme!dvssfou!pg!231!B/! ! )b*! Efufsnjof!uif!bsnbuvsf!up!gjfme!nvuvbm!joevdubodf! ! )c*! Uif!nbdijof!jt!up!cf!pqfsbufe!bt!b!npups!tvqqmz!pg!711!lX!mpbe!bu!jut!sbufe!ufsnjobm!wpmubhf/! Efufsnjof!uif!joufsobm!wpmubhf!boe!uif!dpssftqpoejoh!gjfme!dvssfou!jg!uif!npups!jt!pqfsbujoh! bu!vojuz!qpxfs!gbdups/

Solution Vt If f xs

= 2300 V = 120 A = 50 Hz = 7.75 W

(a) The armature to field mutual inductance Laf =

2 Vt 3w I f

=

2 ¥ 2300 3 ¥ 2p ¥ 50 ¥ 120

H = 49.8 mH

(b) Armature current at unity power factor Ia =

600 ¥ 103 3 ¥ 2300

0∞ A = 150.6 0∞ A

\ internal voltage 2300

– j 150.6 ¥ 7.75 3 = 1327.9 – j 1167.15 = 1767.9 -41.3∞ V

E = Vt – j Ia xs =

Field current

If =

2 ¥ 1767.9 2p ¥ 50 ¥ 49.8 ¥ 10 -3

A = 160 A.

! Qspcmfn!8/34 Uif!gpmmpxjoh!sfbejoht!bsf!ublfo!gspn!uif!sftvmut!pg!bo!pqfo!boe!tipsu.djsdvju!uftu!po!b!6111!lX-! 5271!W-!uisff.qibtf-!gpvs.qpmf-!2911!sqn!tzodispopvt!npups!esjwfo!bu!sbufe!tqffe/ Gjfme!dvssfou-!B Bsnbuvsf!dvssfou-!tipsu!djsdvju!uftu-!B Mjof!wpmubhf-!pqfo!djsdvju!dibsbdufsjtujd-!W Mjof!wpmubhf-!bjs!hbq!mjof-!W

27: 7:5 4:31 5751

2:3 8:1 5271 6381

Uif!bsnbuvsf!sftjtubodf!jt!22!nW0qibtf/!Uif!bsnbuvsf!mfblbhf!sfbdubodf!jt!ftujnbufe!up!cf! 1/23!qfs!voju!po!uif!npups!sbujoh!bt!cbtf/!Gjoe!)j*-!uif!tipsu.djsdvju!sbujp-!)jj*!uif!tbuvsbufe!tzodisp. opvt!sfbdubodf!jo!qfs!voju!boe!jo!pint!qfs!qibtf-!boe!)jjj*!uif!votbuvsbufe!wbmvf!pg!uif!tzodispopvt! sfbdubodf!jo!qfs!voju!boe!jo!pint!qfs!qibtf/

Tzodispopvt!Npups

8/44

Solution (a) Short-circuit ratio =

192 = 1.136 169

(b) Per unit saturated synchronous reactance =

1 = 0.88 1.136

Base impedance Zbase =

( 4160) 2 5000 ¥ 103

W = 3.46 W

\ saturated synchronous reactance = 0.88 ¥ 3.46 W = 3.0448 W (c) At field current of 169 A, the line-to-neutral voltage on the air-gap line is Vag =

4640

V 3 For the same field current, the armature current on short circuit is Isc = 694 A \ unsaturated value of synchronous reactance 4640 = 3 W = 3.86 W 694 Per unit value of unsaturated synchronous reactance =

3.86 = 1.11. 3.46

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Solution The armature current Ia = =

4200 ¥ 103 0.87 ¥ 3 ¥ 4160

cos -1 0.87 A

582.9 29.54∞ A = 670 29.54∞ A 0.87

xs = 0.12 p.u. = 0.12 ¥ 3.46 W = 0.4152 W The line-to-neutral value of excitation voltage | E | = Vt – Ia (ra + jxs)

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4160

- 670 29.54∞ (0.011 + j 0.4152) 3 = 2401.8 – 278.28 118∞ = 2532.44 – j 245.7 = 2544.33 V

=

Field current =

2544.33 ¥ 3 A = 310 A. 4160 3 ¥ 169

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Solution The armature current Ia =

5000 ¥ 103 3 ¥ 4160 ¥ 1

0∞ A = 694 0∞ A

\ armature copper loss = Ia2 ra = 3(694)2 ¥ 0.011 W = 3 ¥ 5298 W = 15894 W The excitation voltage |E| =

4160 3

– 694 (0.011 + j 0.4152)

= 2401.8 – 288.25 88.48∞ = 2394 – j 288 = 2411.26 V Field current

If =

2411.26 ¥ 3 A = 293.87 A 4160 3 ¥ 169

\ field copper loss = If2 Rf = (293.87)2 ¥ 0.279 = 24094 W Total loss = 37000 + 46000 + 15894 + 24094 = 122988 W = 122.988 kW Hence, output power = 5000 – 122.988 = 4877 kW 4877 \ efficiency = ¥ 100% = 97.54%. 5000

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Solution (a) Short-circuit ratio = 0.55 Hence, saturated synchronous reactance 1 = 1.82 p.u. Xs = 0.55 Now, Hence,

Xbase =

(11) 2 W = 4.03 W 30

Xs = 1.82 ¥ 4.03 W = 7.335 W

(b) If = 150 A \

E = 11 ¥

\

Ia =

150 kV = 8.25 kV (line to line) 200

Vt - E (11 - 8.25)103 = 217.5 A = Xs 3 ¥ 7.3

As the motor is operated under no load condition and losses are neglected, d = 0. As Vt > E, the motor is operating at a power factor of 0 lag. Hence, the motor acts as an inductor. If L be the equivalent inductance then wL = \

L=

11 ¥ 103 3 ¥ 217.5

W.

11 ¥ 103 3 ¥ 217.5 ¥ 2p ¥ 50

H = 0.093 H.

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!

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Solution Vt =

3300 3

V

cos q = 1 1200 ¥ 103

Ia =

A = 209.95 A 3 ¥ 3.3 ¥ 103 ¥ 1 Ia X q 209.95 ¥ 3 = tan -1 d = tan–1 = tan–1 0.33 = 18.26° Vt 3.3 ¥ 103 3

E=

3.3 ¥ 103 3

cos 18.26° + 5 ¥ 209.95 sin 18.26° = 2138 V

\ excitation emf = 3 ¥ 2138 V = 3.7 kV Mechanical power Pm =

Xd - Xq EVt sin 2d sin d + Vt 2 2Xd Xq Xd

For maximum mechanical power, dPm =0 dd Xd - Xq EVt cos d + Vt 2 cos 2d = 0 Xd Xd Xq

\

3300 2 3 cos d + Ê 3300 ˆ ¥ 5 - 3 cos 2d = 0 ÁË ˜ 5 5¥3 3 ¯

2138 ¥ \

427.6 cos d + 254 cos 2d 427.6 cos d + 254 (2 cos2 d – 1) cos2 d + 0.842 cos d – 0.5 cos d d

or, or, or, or, \

=0 =0 =0 = 0.402 = 66.3°

Hence, maximum mechanical power for all the three phases 3300 2 3 sin 66.3° + 3 ¥ Ê 3300 ˆ ¥ 5 - 3 sin (2 ¥ 66.3°) ÁË ˜ 5 5¥3 3 ¯

2138 ¥ 3¥

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= 2237.936 kW + 1068.8 kW = 3306.74 kW.

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Solution The equivalent circuit and phasor diagram is shown in Fig. 7.14(a) and Fig. 7.14(b) respectively. EG

jXSGI XSG

XSM

I

V

G

V jXSMI

M

EM (a)

(b)

Gjh/!8/25! Djsdvju!ejbhsbn!boe!qibtps!ejbhsbn!pg!Qspc/!8/39

If the current is I then

\

3 ¥ 3.3 ¥ I cos q = 1.5 ¥ 103 cos q = 1 1.5 ¥ 103 I= A = 262.4 A 3 ¥ 3.3 XSM = 2 W XSG = 3 W Let EM be the back emf of the motor and EG be the induced emf of the alternator From Fig. 7.14(b), 2

\

Ê 3300 ˆ EM = Vt 2 + ( X SM I ) 2 = Á + ( 2 ¥ 262.4) 2 = 1976 V per phase Ë 3 ˜¯

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Again from Fig. 7.14(b), 2

Ê 3300 ˆ EG = Vt 2 + ( X SG I ) 2 = Á + (3 ¥ 262.4) 2 = 2061.5 V per phase Ë 3 ˜¯ Maximum power transfer =3

E1E2 1976 ¥ 2061.5 =3 W = 2444 kW. X SG + X SM 2+3

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Solution Base impedance

Zbase =

(11) 2 = 24.2 W 5

ra = 0.05 ¥ 24.2 W = 1.21 W xs = 1 ¥ 24.2 W = 24.2 W Zs = 1.21 + j 24.2 W

\ Hence,

f = tan–1

\

24.2 = 87° 1.21

d = 180° – fs = 93°

For maximum power input

E = 1.5 ¥

11 3

kV = 9.526 kV per phase

From Eq. (7.24) per phase power, P=

=

=

Vt 2 VE cos f - t cos (d + f) ZS ZS Ê 11 ˆ ÁË ˜¯ 3

2

(1.21) 2 + ( 24.2) 2

cos 87∞ +

11 ¥ 9.526 3 (1.21) 2 + ( 24.2) 2

121 60.5 cos 87∞ + MW = 2.58 MW 3 ¥ 24.23 24.23

MW

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Maximum input power for all three phases = 2.58 ¥ 3 MW = 7.74 MW 3 Vt Ia cos q = 7.74 ¥ 106

\

Ia cos q =

or,

Vt = E + I a Z S

Now,

11 ¥ 103

\

7.74 ¥ 106 = 406.24 A 11 3¥ ¥ 103 3

3

= 9.526 ¥ 103 -93∞ + (406.24 + j Ia sin q) (1.21 + j 24.2)

or,

6351 = 9526 -93∞ + 491.55 – 24.2 Ia sin q + j (9831 + 1.21 Ia sin q)

or,

6351 = 9526 cos (–93°) + 491.55 – 24.2 Ia sin q + j {9526 sin (–93°) + 9831 + 1.21 Ia sin q}

\

Ia sin q = –262.7 Ia = 406.24 – j 262.7 = 483.8 -32.89∞ A

Hence,

Power factor cos 32.89° lag or 0.84 lag.

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Solution From Eq. (7.29), per phase output power P=

1.4 ¥ 1 (1) 2 (0.8 - 0.5) sin d + sin 2d 0.8 2 ¥ 0.8 ¥ 0.5

P = 1.75 sin d + 0.375 sin 2d

or, For P to be maximum,

dP dd 1.75 cos d + 0.75 cos 2d 1.75 cos d + 0.75 (2 cos2 d – 1) cos2 d + 1.167 cos d – 0.5 cos d

\ or, or, or, Hence,

=0 =0 =0 =0 = 0.33 or d = 70.73°

Pmax = 1.75 sin 70.73 + 0.375 sin (2 ¥ 70.73°) = 1.88 p.u.

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Sfwjfx!Rvftujpot! 1. Explain the principle of operation of a synchronous motor. 2. Draw and explain the phasor diagram of a cylindrical rotor synchronous motor operating at (a) lagging, (b) leading, and (c) unity power factor. 3. Draw and explain the phasor diagram of a salient pole synchronous motor operating at lagging power factor. 4. Explain why a synchronous motor does not have starting torque. 5. Describe the different methods of starting of synchronous motors. 6. Derive the power angle characteristic of a salient-pole synchronous motor. 7. What is a synchronous condenser? What is its application? 8. Explain the effect of change of excitation of a synchronous motor on its (a) armature current, and (b) power factor 9. Explain the effect of change of load of a synchronous motor on its (a) torque angle, and (b) armature current. 10. Explain V curves in a synchronous motor. 11. Compare between an induction motor and a synchronous motor. 12. What do you mean by hunting of a synchronous motor? What are the causes of hunting? How do you prevent hunting?

Qspcmfnt 1. An industrial plant has a load of 800 kW at power factor or 0.8 lagging. It is required to instal a synchronous motor to deliver a load of 200 kW and improve the overall power factor of a plant to 0.92. Determine the kVA rating of the synchronous motor and its power factor. The efficiency of the motor is 90%. [277.75 kVA, 0.8 lead] [Hints: \

\

cos q1 = 0.8 lag sin q1 = 0.6 P1 = 800 kW 800 S1 = = 1000 kVA 0.8 Q1 = 1000 ¥ 0.6 kVAR = 600 kVAR

200 kW = 222.2 kW 0.9 Total load = 800 + 222.2 = 1022.2 kW cos q2 = 0.92 P2 = 1022.2 kW sin q2 = 0.39 Input of motor =

Tzodispopvt!Npups

\

8/52

S2 =

1022.2 kVA 0.92

q2 =

1022.2 ¥ 0.39 kVAR = 433.3 kVAR 0.92

Reactive power supplied by motor = 600 – 433.3 = 166.7 kVAR

kVAR 166.7 = 0.75 = kW 222.2

\

tan q2 =

\

cos q2 = 0.8 lead

kVA of motor =

kW 222.2 = = 277.75] cos q 2 0.8

2. A 2500 HP, 2300 V, 20-pole, 50 Hz synchronous motor has a reactance of 1.85 W per phase. The field excitation is adjusted so that the power factor is unity at rated load. Determine the maximum torque which the motor can develop. [106.95 ¥ 103 Nm] [Hints: Total output =

2500 ¥ 735.5 kW = 1838.75 kW 1000 Ia =

1838.75 ¥ 103 3 ¥ 2300 ¥ 1

= 461.5 A

2

Ê 2300 ˆ E= Á + ( 461.5 ¥ 1.85) 2 = 1578.7 V Ë 3 ˜¯

Maximum power output =

Maximum torque =

3Vt E = xs

3¥

2300 ¥ 1578.7 3 = 3399.5 ¥ 103 W 1.85

3399.5 ¥ 103 Nm = 108.26 ¥ 103 Nm] 2 ¥ 50 2p ¥ 20

3. A 6.6 kV, three-phase, star-connected synchronous motor operates at constant voltage and constant excitation. The synchronous impedance is 2 + j 20 W per phase. When the input is 1000 kW, the power factor is 0.8 leading. Determine the power factor when the input is increased to 1500 kW. [0.935 lead] [Hints:

Ia =

1000 ¥ 103 3 ¥ 6.6 ¥ 103 ¥ 0.8

A = 109.35 A

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When p.f. is 0.8 leading, Ia = 109.35 (0.8 + j 0.6) = 87.47 + j 656 A 6600

\

= E -d ∞ + Ia (2 + j 20) 3 E sin d = 1880.6 E cos d = 4947.66 d = 20.8° and E = 5293.4 V

\

When input is 1500 kW, 6600 - 5293.4 -d ¢ 3 I¢a = 2 + j 20 I¢a cos q¢ = 18.86 – 263.35 cos (d ¢ + 84.29°) Again,

3 ¥ 6600 ¥ Ia¢ cos q¢ = 1500 ¥ 103

\

Ia¢ cos q¢ =

1500 ¥ 103 3 ¥ 6600

A = 131.21 A

\

d¢ = 30.96°

Now

Vt = E -d ¢ + (I¢ cos q ¢ + j I¢ sin q¢) ¥ (2 + j 20)

\

I¢ sin q¢ = 49.565 tan q¢ =

or,

49.565 = 0.377 131.2

Power factor cos q¢ = 0.935 load] 4. A salient-pole synchronous motor has Xd = 0.85 p.u. and Xq = 0.55 p.u. It is connected to an infinite bus of 1 p.u. voltage and the excitation is adjusted to 1.2 p.u. Determine the maximum power output that the motor can supply without loss of synchronism. [1.5332 p.u.] [Hints:

P=

=

EVt V2 Ê 1 1 ˆ sin d + t Á ˜ sin 2d 2 Ë Xq Xd ¯ Xd 1.2 ¥ 1 1 Ê 1 1 ˆ + Á sin 2d 0.85 2 Ë 0.55 0.85 ˜¯

= 1.412 sin d + 0.321 sin 2d For maximum power,

dP =0 dd

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dP = 1.412 cos d + 2 ¥ 0.321 cos 2d = 0 dd

\ \

d = 69.764° Pmax = 1.412 sin 69.764° + 2 ¥ 0.321 sin 2 ¥ 69.764° = 1.5332 p.u.]

5. For a three-phase, salient-pole synchronous motor running as a reluctance motor, show that under maximum power conditions Xd + Xq tan q = Xd - Xq Id = Ia sin (q – d) and Iq = Ia cos (q – d) Vt sin d = Iq Xq Vt cos d = Id Xd

[Hints:

\

tan d =

Iq X q Id X d

=

X q cos (q - d ) X d sin (q - d )

Under maximum power condition, d = 45° 1=

X q cos(q - 45∞) X d sin (q - 45∞)

=

X q (sin q + cos q ) X d (sin q - cos q )

X d sin q + cos q = X q sin q - cos q

\

Xd + Xq Xd - Xq

=

sin q = tan q] cos q

6. A 480 V, three-phase, star-connected, salient-pole, synchronous motor is operating at its full load and draws a current of 50 A at unity power factor. The d and q axis reactances are 3.5 W per phase and 2.5 W per phase respectively. The armature winding resistance is 0.5 W per phase. Determine excitation voltage of the motor. [564 V] [Hints: Ia = 50 0∞ A d = tan–1

E=

480 3

Ia X q Vt - I a ra

= tan -1

50 ¥ 2.5 = 26.37° 480 - 50 ¥ 0.5 3

cos 26.37° + 50 ¥ 3.5 sin 26.37° = 326 V

\ line value of emf = 3 ¥ 326 V = 564 V]

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7. A 6.6 kV, star-connected, three-phase synchronous motor operates at constant voltage and excitation. Its synchronous impedance is 2 + j 20 W per phase. The motor operates at 0.8 leading power factor and draws 800 kW from the main. Determine the power factor when the motor draws an increased power of 1200 kW from the mains. [0.9265 lead] [Hints:

E=

6600 3

800 ¥ 103

-

3 ¥ 6600 ¥ 0.8

36.87∞ ¥ (2 + j 20)

= (4.724 – j 1.504) kV = 4.96 -17.66∞ kV Zs = 2 + j 20 = 20.1 84.3∞ 2

Now \

Ê 6.6 ˆ 6.6 ¥ cos 84.3∞ ¥ 4.96 Á ˜ 1200 Ë 3 ¯ 3 cos (84.3° + d) = 20.1 20.1 3 d = 26.1° 6.6 - 4.96 -26.1∞ 3 = 113 22.1∞ Ia = 20.1 84.3∞ cos q = cos 22.1° = 0.9265 lead]

8. The full load current of a 3.3 kV, star-connected synchronous motor is 160 A at 0.8 power factor lagging. The synchronous impedance of the machine is 0.8 + j 5.5 W per phase. Determine the excitation emf, torque angle, efficiency and shaft output of the motor. Assume the mechanical stray load loss to be 30 kW. [2.46 kV, 26.2°, 87.5%, 640 kW] [Hints:

Zs = 0.8 + j 5 = 5.56 81.7∞ W E=

3.3 3

- 5.56 81.7∞ ¥

160 103

-36.87∞

= 1.42 -26.2∞ kV Line value of emf = 3 ¥ 1.42 kV = 2.46 kV d = 26.2° Power developed = 3 ¥ 1.42 ¥ 160 cos (–36.87° + 26.2°) = 670 kW Shaft power output = 670 – 30 = 640 kW Input power = 3 ¥ 3.3 ¥ 160 ¥ 0.8 = 731.5 kW h=

640 ¥ 100% = 87.5%] 731.5

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Nvmujqmf.Dipjdf!Rvftujpot 1. A three-phase salient-pole synchronous motor is connected to an infinite bus. It is operated at no load at normal excitation. The field excitation of the motor is first reduced to zero and then increased in the reversed direction gradually. Then the armature current (a) increases continuously (b) first increases and then decreases steeply (c) first decreases and then increases steeply (d) remains constant [GATE 2011] 2. A three-phase synchronous motor connected to ac mains is running at full load and unity power factor. If its shaft load is reduced by half, with field current held constant, its new power factor will be (a) unity (b) leading (c) lagging (d) dependent on machine parameters [GATE 2007] 3. An over-excited synchronous motor operates at (a) lagging power factor (b) leading power factor (c) unity power factor (d) lagging power factor at low loads and leading power factor at high loads 4. As the load on a synchronous motor increases, the torque angle (a) decreases (b) increases (c) remains same (d) depends on the power factor of the load 5. V curves of a synchronous motor show relation between (a) armature current and field current (b) armature current and terminal voltage (c) field current and terminal voltage (d) field current and excitation voltage 6. The function of amortisseur winding in a synchronous motor is (a) to improve power factor (b) to prevent hunting (c) to provide starting torque (d) to prevent hunting and provide starting torque 7. A salient-pole synchronous motor is running with normal excitation. If the excitation is reduced to zero, (a) it becomes an induction motor (b) it remains a synchronous motor

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(c) it becomes a reluctance motor (d) the motor would stop 8. Under the conditions of maximum load on a cylindrical rotor synchronous motor, the reactive power input to the motor is (a) – (c)

3Vt E Xs

3Vt 2 Xs

(b) –

3Vt 2 Xs

(d) –

3E2 Xs

9. A three-phase synchronous motor connected to an infinite bus is operating at half full load with normal excitation. When the load on the synchronous motor is suddenly increased, (a) its speed will fluctuate around synchronous speed and then become synchronous (b) its speed first increases and then become synchronous (c) its speed first decreases and then become synchronous (d) its speed will remain unchanged 10. A three-phase synchronous motor is connected to an infinite bus and operates at leading power factor. For constant load torque, if excitation is increased, (a) load angle and power factor angle both increases (b) load angle and power factor angle both decreases (c) load angle increases but power factor angle decreases (d) load angle decreases but power factor angle increases 11. A three-phase, 400 V, 50 Hz, synchronous motor is operating at a load angle of 20°. If load on the motor is doubled, keeping other parameters constant, the new load angle will be (a) 20° (b) 40° (c) 30° (d) 43.16° 12. With the damper winding, the synchronous motor starts as (a) synchronous motor (b) dc motor (c) induction motor (d) split phase motor 13. A synchronous motor draws minimum current at a given load when its power factor is (a) zero (b) lagging (c) leading (d) unity 14. A salient-pole synchronous motor running at synchronous speed suddenly losses its excitation. Then (a) it will run at a lower speed (b) it will not run (c) it will continue to run at synchronous speed with reduced output (d) it will continue to run at synchronous speed with same output

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15. A synchronous motor can operate over a range of power factors with both leading and lagging. This can be done by (a) varying load (b) varying voltage (c) varying frequency (c) varying the field excitation 16. A synchronous motor runs with a constant load. When its excitation is reduced, (a) the power angle decreases (b) the power factor becomes more lagging (c) the power factor improves (d) the speed increases 17. When a synchronous motor is run as a synchronous capacitor, it operates (a) with leading power factor without any load (b) with leading power factor supplying loads (c) with lagging power factor without any load (d) with lagging power factor supplying load 18. Armature reaction of a synchronous motor at rated voltage zero power factor leading is (a) cross magnetizing (b) magnetizing (c) demagnetizing (d) both demagnetizing and cross magnetizing 19. In a salient-pole synchronous motor, (a) Id and Iq are both in phase with E (b) Id and Iq are both at 90° to E (c) Iq is in phase with E and Id is at 90° to E (d) Id is in phase with E and Iq is at 90° to E 20. In a salient-pole synchronous motor, the reluctance power is given by (a) Vt2 (c) Vt2

Xd + Xq 2Xd Xq Xd - Xq 2Xd Xq

sin 2d

(b) Vt2

sin 2d

(d) Vt2

Xd + Xq 2Xd Xq Xd - Xq 2Xd Xq

sin d sin d

Botxfst 1. 6. 11. 16.

(b) (d) (d) (b)

2. 7. 12. 17.

(b) (c) (c) (a)

3. 8. 13. 18.

(b) (c) (d) (c)

4. 9. 14. 19.

(b) (a) (c) (c)

5. 10. 15. 20.

(a) (d) (d) (c)

9 Uisff.qibtf!Joevdujpo! Npups JOUSPEVDUJPO!

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In 1888, Ferrari introduced the concept of induction motor by publishing his revolutionary work on the rotating magnetic field produced by polyphase currents. Induction motors are versatile electric motors suited best for industrial applications as well as low-power application in domestic equipment because of their unique advantage compared to other types such as dc and synchronous motors. More than 80% of all the motors used in the world are induction motors. The dominance of induction motors is mainly due to the fact that they are the cheapest and most robust motors. In a three-phase induction motor, three-phase ac supply is given to the stator winding. The flux from the stator, flowing through the air gap links the rotor circuit. Voltages are induced in the shortcircuited rotor winding according to Faraday’s laws of electromagnetic induction causing currents to flow in them. The fact that the rotor current arises from induction rather than conduction, is the basis for the name of this class of machines. They are also called asynchronous motors because their operating speed is slightly less than the synchronous speed.

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Similar to other rotating electrical machines, a three-phase induction motor also consists of two main parts: the stator and the rotor (the stator is the stationary part and the rotor is the rotating part). Apart from these two main parts, a three-phase induction motor also requires bearings, bearing covers, end plates, etc., for its assembly.

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The stator of a three-phase induction motor has three main parts namely, stator frame, stator core and stator windings. The stator frame can either be casted or can be fabricated from rolled steel plates. The stator core is built up of high silicon sheet-steel laminations of 0.4 to 0.5 mm thickness. Each lamination is separated from the other by means of either varnish, paper or oxide coating. Each lamination is slotted on the inner periphery so as to house the winding. The laminations for small machines are in the form of complete rings, but for large machines these may be made in sections. The insulated stator conductors are connected to form a three-phase winding, the stator phase windings may be either star or delta connected. The rotor is also built up of laminations of the same material as the stator. The laminated cylindrical core is mounted directly on the shaft or a spider carried by the shaft. These laminations are slotted on their outer periphery to house the rotor conductors. There are two types of inductionmotor rotors. (a) Squirrel cage or simply cage rotor (b) Phase wound or wound rotor or slip-ring rotor In either case, the rotor windings are contained in slots in a laminated iron core which is mounted on the shaft. In small machines, the lamination stack is pressed directly on the shaft. In larger machines, the core is mechanically connected to the shaft through a set of spokes called a spider. The motor having the first type of rotor is known as a squirrel-cage induction motor. This type of rotor is cheap and has a simple and rugged construction, it is cylindrical in shape and is made of sheet-steel laminations. Here, the slots provided to accommodate the rotor conductors are not parallel to the shaft but are skewed. The purpose of skewing is (a) to reduce the magnetic hum, and (b) to reduce the magnetic locking. The rotor conductors are short-circuited at the ends by brazing the copper rings, resembling the cage of a squirrel and hence, the name squirrel-cage rotor. In present days, die-cast rotors have become very popular. The assembled rotor laminations are placed in a mound. The molten aluminium is forced under pressure to form the bars. Figure 8.1(a-c) shows a typical stator and rotor (both squirrel-cage type and slip-ring type) assembly. Figure 8.1(d) shows the schematic of a cage rotor separately. The motor having the second-type rotor, i.e. wound-type rotor, is named slip ring induction motor. In this motor, the rotor is wound for three-phase, similar to stator winding using open-type slots in the lamination. Rotor winding is always star connected and thus, only three remaining ends of the windings are brought out and connected to the slip rings as shown in Fig. 8.2. With the help of these slip rings and brushes, additional resistances can also be connected in series with each rotor phase (Fig. 8.2). This will increase the starting torque provided by the motor and will also help in reducing, the starting current. When running under normal condition, the external resistances are removed completely from the rotor by short circuiting these additional resistances from the rotor circuit and the rotor behaves just like a squirrel-cage rotor.

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End rings Copper bars iron rotor

Stator winding (a)

Squirrel cage rotor (b)

!! ! ! !

End ring

End ring

Rotor bars (slightly skewed) Schematic of a cage rotor

Slip ring rotor (c)

(d)

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Stator

Rotor

Slip rings

Starting resistance

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For an induction motor, the following things may be noted: 1. An induction motor is a singly-fed motor. Therefore, it does not require a commutator or brushes. In fact, there are no moving contacts between the stator and the rotor. This makes the motor rugged, reliable and almost maintenance free. 2. The absence of brushes eliminates the electrical loss at the brush and the brush voltage drop as well as the mechanical loss due to friction between the brushes and commutator or the slip rings. An induction motor thus, has a relatively high efficiency. 3. An induction motor carries alternating current in both the stator and the rotor windings.

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4. An induction motor is analogous to a rotating transformer in which the secondary winding receives energy by induction while it rotates. However, it may be remembered that the windings of a three-phase transformer and that of a three-phase induction motor are of different types. Owing to the widespread generation and transmission of three-phase power, most polyphase induction motors are of the three-phase type. In this chapter, we confine our discussion exclusively to three-phase induction motors. This theoretical development, however, can be easily extended to an n-phase induction motor where n ≥ 2.

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9/4

The advantages of a cage rotor induction motor are as follows: 1. 2. 3. 4.

The rotor is of robust construction and cheaper. The absence of brushes reduces the risk of sparking. Squirrel-cage rotors require lesser maintenance. Squirrel-cage induction motors have higher efficiency and better power factor.

On the other hand, wound rotors have the following merits: 1. High starting torque and low starting current. 2. Additional resistance can be connected in the rotor circuit to control speed. Ubcmf!9/2! B!dpnqbsjtpo!pg!trvjssfm.dbhf!boe!xpvoe!spups!npupst Squirrel Cage Motor

Wound Rotor Motor

1.

No rotor terminals available, hence, no external connections can be made to rotor circuit.

1. Rotor terminals are available, hence, an external impedance or device can be connected to the rotor circuit.

2.

No slip rings and brushes, so very robust construction and almost no maintenance required.

2. There are slip rings and brushes which require frequent maintenance and cause problems.

3.

Cheapest motor

3. Costlier than squirrel cage motor.

4.

Simplest starting accessories.

4. Comparatively complicated starting accessories.

5.

With the starters needed to limit the starting current, the starting torque becomes very low.

5. With the rotor resistance starter, starting current gets limited and starting torque becomes quite high.

6.

Used generally for all the low-capacity loads and for medium and large loads where starting is light which is the most common case.

6. Used normally with medium and large capacity loads where no load starting, i.e., high torque starting, is required.

7.

No speed control possible from the secondary side. 7. Speed can be controlled from the secondary side.

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9/6

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Bewboubhft 1. 2. 3. 4. 5. 6. 7.

It is very simple, robust, rugged and capable of withstanding rough use. It is quite cheap and reliable in operation. Its maintenance cost is low. The losses are reasonably small and hence, it has sufficiently high efficiency. It is mostly a trouble-free motor. Its power factor is reasonably good at full-load operation. It is simple to start (since it has a self-starting torque).

An induction motor is equivalent to a static transformer whose secondary is capable of rotating with respect to the primary. Usually, the stator is treated as the primary, while the rotor is treated as the secondary. The induction motor operation is electrically equal even if the rotor is primary and the stator operation is treated as secondary. Ejtbewboubhft! 1. 2. 3. 4. 5.

Its speed cannot be varied without sacrificing efficiency. Its speed decreases with an increase in load. Its starting torque is inferior to that of a dc shunt motor. For direct online starting, the starting current is usually 5 to 8 times the full-load rated current. It runs at a low lagging power factor when it is lightly loaded.

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A three-phase induction motor has a stator winding which is supplied by three-phase alternating balanced voltage and has balanced three-phase currents in the winding. The rotor is not excited from any source and has only magnetic coupling with the stator. Under normal running conditions, the rotor winding (cage or slip-ring) is always short circuited to allow induced currents to flow in the rotor winding. The flow of three-phase currents in the stator winding produces a rotating magnetic field of constant amplitude and rotates at a synchronous speed. Let us assume that the rotor is at standstill initially: the rotating stator field induces an emf in the rotor conductor by transformer action. Since the rotor circuit is a closed set of conductors, a current flows in the rotor circuit. This rotor current then produces a rotor field. The interaction of stator and rotor field produces a torque which causes the rotation of the rotor in the direction of the stator rotating field.

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As per Lenz’s law, the rotor field will try to oppose the very cause of its production. Thus, it speeds up in the direction of the stator field so that relative speed difference between these two fields is zero. In this way, the three-phase induction motor catches up the speed. When the rotor is at standstill, the relative motion between the stator field and rotor is maximum. Therefore, the emf induced in the rotor and rotor current are reduced. However, the rotor cannot attain the speed of the stator field which is equal to the synchronous speed. This is evidently due to the reason that if the rotor is moving at synchronous speed, there is no relative motion between the stator field and the rotor. Hence, the rotor-induced emf and current become zero and the torque becomes zero. This causes the rotor speed to decrease. As the rotor speed falls below the synchronous speed, the rotor emf and current continue to increase. Therefore, the electromagnetic torque continuous to increase. Finally, the rotor speed becomes constant at a value of speed slightly less than that of the stator field. When a mechanical load is put on the rotor, the automatic tendency of the load is to produce a braking action, i.e. the speed of the rotor is reduced which results in the increase of the relative speed between the rotor conductors and the rotating magnetic field. The direction of rotation of an induction motor is dependent on the direction of rotation of the stator flux, which in turn is dependent on the phase sequence of the applied voltage. Interchanging any two of the three line-leads to a three-phase induction motor will reverse the phase sequence, thus, reversing the rotation of the motor.

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An induction motor is very similar to a transformer as far as their principle of operation is concerned. The similarities between a transformer and an induction motor are as follows: 1. Both an induction motor and transformer work on the principle of electromagnetic induction. In a transformer, ac voltage is supplied to the primary winding. This voltage produces current in the primary winding which sets up flux f in the iron core. This flux links both the primary and the secondary winding. As the flux is alternating in nature, an emf is induced in the secondary winding. Hence, there is no electrical connection between the primary and secondary windings of a transformer. The secondary windings receives energy from the primary by electromagnetic induction. Similarly, in an induction motor, the three-phase ac supply is given to the stator winding which sets up a rotating magnetic field. This rotating magnetic field induces emf in the rotor conductors. Hence, the rotor receives energy from the stator by electromagnetic induction. 2. The primary and secondary winding of a transformer resembles the stator and rotor winding of an induction motor. The resistance and leakage reactance of stator and rotor winding of an induction motor are similar to the primary and secondary winding of a transformer.

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3. The operation of an induction motor under no-load condition is similar to a transformer under open circuit condition. Similarly a three-phase induction motor with its rotor blocked behaves similar to a transformer under short circuit condition. 4. The equation for the emfs induced in the primary and secondary windings of a transformer are similar to that of the emfs induced in the stator and rotor windings of an induction motor. The dissimilarities between an induction motor and a transformer are as follows: 1. The output of a transformer is electrical energy whereas the output of an induction motor is mechanical energy. 2. A transformer is a static device whereas an induction motor is a rotating device. 3. Frequency of the primary and secondary emf of a transformer are same whereas the frequency of the emf induced in the rotor winding of an induction motor depends upon the speed of the rotor. 4. As there is no rotating part in a transformer, the losses are lower compared to that of an induction motor and efficiency is higher. 5. The no-load current of an induction motor is very high compared to that of a transformer due to the fact that the mutual flux in a transformer completes its path through a low-reluctance iron core, whereas in an induction motor, this flux completes its path through an air gap. mmf As flux is equal to , a constant flux requires less mmf when it completes its path reluctance through a low-reluctance iron core in a transformer and hence, draws less magnetizing current. 6. In a transformer concentrated windings are used, whereas in induction motors distributed windings are used. Hence, voltage equations in induction motor involve winding factor.

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The speed of the rotating flux, called synchronous speed, is directly proportional to the frequency of the supply voltage and inversely proportional to the number of pairs of poles (poles only occur in pairs). Expressed mathematically, we have 2 ¥ fs fs = r/s P /2 P 120 ¥ f s ns = r/min P

ns =

(8.1)

where ns stands for the speed of the rotating stator field, P is the total number of poles and fs is the frequency of the supply voltage. The relationship between the speed of the rotating flux and the number of stator poles may be visualized by comparing the mechanical degrees of circular arc travelled by the flux in motors with

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different numbers of stator poles. Thus, the circular arc travelled by the rotating field of a four-pole motor is twice that of an eight-pole motor. Assuming the same frequency and same time period, the centre line of flux rotates 60°(phase A to phase B) for a four-pole winding, and 30°(A to B) for an eight-pole winding. Increasing the frequency of the supply voltage increases the frequency of the current in the stator circles, causing the flux to rotate at a proportionately higher speed. The synchronous speed of an induction motor operating from a fixed-frequency system can be changed by changing the number of poles in the stator. The magnitude and frequency of the rotor voltages depend on the speed of the relative motion between the rotor and the flux crossing the air gap. The difference between the synchronous speed and the rotor speed expressed as a fraction (or percent) of synchronous speed is known as slip, i.e. Slip speed = (ns – n) rev/s n -n Slip (s) = s p.u. ns n = ns (1 – s) rps

and or

(8.2) (8.3)

where ns = Synchronous speed (rev/s) n = Rotor speed (rev/s) s = Slip When the speed is expressed in rpm, we can write Ns - N N -N ¥ 100 (in %) p.u. = 1 Ns Ns N = Ns (1 – s) rpm s=

and

(8.4) (8.5)

The slip s is a very useful quantity in the study of induction motors. The value of slip at full-load is about 4 to 5% for small motors and about 2 to 2.5% for large motors. The slip at no-load is about 1%. Thus, the speed of an ac induction motor is almost constant from no-load to full-load. If the machine has P poles, the frequency of induced emf in the rotor, i.e. f2, is given by È Ê N - N ˆ˘ PN s P( N s - N ) Ns - N : f2 = and hence ( f 2 / f1 ) = Á s ¥ f1 ÍE f1 = ˙ 120 120 Ns Ë N s ˜¯ ˙˚ ÍÎ f2 = sf1 (8.6) f2 =

At standstill condition of the rotor, s = 1, i.e. the frequency of rotor currents is f1. Pole changing may be accomplished by using separate windings for each speed, or by reconnecting the windings of specially designed machines called consequent pole motors. When two separate windings are used, the machine is called a two-speed two-windings motor. Three separate windings, each arranged for a different number of poles, result in a three-speed three-winding motor.

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9/9

Let us consider a typical pair of rotor bars. As the rotor “slips” backward through the flux field, the flux linking these bars will vary cyclically. The voltage induced in the rotor circuit is composed of the voltages in these two bars and the end rings. It is at its peak at the instant when the rate of change of flux linkages is a maximum. Thus, one cycle of rotor voltage is generated as a given conductor slips past two poles of the air-gap flux field. In other words, one cycle of rotor voltage corresponds to 360 electrical degrees of “slips”. Then the frequency of the rotor voltages and currents is given by f2 = Pole-pairs slipped per second ( ns - n) P – ns. = s. f1 ns 2

PN s PN s Pns ˘ È ÍE f1 = 120 = 2 ¥ 60 = 2 ˙ Î ˚ Frequency of rotor voltage or current = Per unit slip ¥ Supply frequency At standstill, rotor speed is zero. =

(8.7)

( ns - n) ns - 0 = =1 ns ns f2 = f1

\

s=

and

\ Er (rotor induced voltage) = sE1 (stator voltage) 2/!Tuboetujmm!Dpoejujpot Let

E20 R2 X20 Z20 I20 Z20

= emf induced per phase of the rotor at standstill = Resistance per phase of the rotor = Reactance per phase of the rotor at standstill (= 2pf1L2) = Rotor impedance per phase at standstill (= R2 + jX20) = Rotor current per phase at standstill = R2 + jX20 E Hence, I20 = 20 Z 20 and power factor at standstill is given by R R20 cos f20 = 2 = 2 Z 20 R + X2 2

(8.8)

(8.9)

20

3/!Spups!Dvssfou!bu!Tmjq!t Induced emf per phase in the rotor winding at slip s is Er = sE2

(8.10)

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Rotor winding resistance per phase = R2 Rotor winding reactance per phase at slip (s) is X2s = 2pf2 L = 2p(sf1)L = sX20

(8.11)

Rotor winding impedance per phase at slip (s) is Z2s = R2 + jX2s = R2 + jsX20

(8.12)

\ rotor current at slip (s) is given by I2s =

Er Z2s

(8.13)

Power factor at slip (s) is cos f2s =

R2 Z2s

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(8.14)

9/:

Let suffixes 1 and 2 be used for stator and rotor quantities respectively V1 T1 T2 f E1 E2 Er R1 R2 L20 X20 f1 f2 X2s Kd1 Kd2 KP1 KP2

= Stator applied voltage per phase = Number of stator winding turns in series per phase = Number of rotor winding turns in series per phase = Flux per pole produced by the stator mmf = Resultant air-gap flux = Stator induced emf per phase = emf induced in the rotor per phase when the rotor is at standstill = emf induced in the rotor per phase when the rotor is rotating at a slip s = Resistance of stator winding per phase = Resistance of rotor winding per phase = Rotor inductance per phase at standstill due to leakage flux = Leakage reactance of the rotor winding per phase when the rotor is at standstill = Stator emf frequency (supply frequency) = Frequency of the induced emf in the rotor at a slip s = Leakage reactance of rotor winding per phase when the rotor is rotating at a slip s = Distribution factor of stator winding = Distribution factor of rotor winding = Pitch factor or coil-span factor of stator winding = Pitch factor or coil-span factor of rotor winding stator induced emf per phase E1 = 4.44 KP1 Kd1 f1 fT1

(8.15)

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Induced emf per phase in the rotor when the rotor is at standstill is given by E2 = 4.44 KP2 Kd2 f1 f T2

(8.16)

Induced emf per phase in the rotor when the rotor is rotating at a slip (s) is given by Er = sE20 = sE2 (say, assuming E2 same as E20) Er = 4.44 KP2 Kd2 sf1 f T2

\

(8.17) (8.18)

Let KP1 Kd1 = Kw1 = Winding factor of stator and KP2 Kd2 = Kw2 = Winding factor of rotor

(8.19) (8.20)

\

(8.21) (8.22)

E1 = 4.44 Kw1 f1 fT1 E2p = 4.44 Kw2 sf1 f T2 Here, we define Te1 ? Kw1 T1 Te2 ? Kw2 T2

(8.23) (8.24)

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Solution Ns = \

Slip =

120 f 120 ¥ 50 = 1500 rpm = P 4 N s - N 1500 - 1450 = 0.033 = 3.33%. = Ns 1500

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Solution (a) We know, Ns = (b) Slip (s) =

120 f 120 ¥ 50 = = 1000 rpm P 6

N s - N 1000 - 950 = 0.05 = Ns 1000

\ percentage of slip = 0.05 ¥ 100 = 5 (c) The frequency of rotor emf = sf1 = 0.05 ¥ 50 = 2.5 Hz.

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9/23

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Solution We know f2 = s. f1 f 2 s= 2 = = 0.04 = 4% f1 50

\

P ◊ Ns 120 120 ◊ f1 120 ¥ 50 = 1500 rpm = Ns = P 4 f1 =

Again \ Speed of the motor

N = (1 – s) . Ns = (1 – 0.04) ¥ 1500 = 1440 rpm.

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Solution For induction motor, Synchronous speed is given by 120 f 120 ¥ 70 = 840 rpm = P 10 N - N 840 - N s= s = Ns 840

Ns = Now slip, \

0.02 =

\

PN A 6 ¥ 1400 È ˘ ÍE f = 120 = 120 = 70 Hz ˙ Î ˚

840 - N 840

N = 823.2 rpm.

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Solution (a) Since the no-load slip of an induction motor is about one percent, the synchronous speed is slightly larger than the no-load speed of 890 rpm. For 60 Hz frequency, the number of poles and their corresponding synchronous speeds are P Ns (rpm)

2 3600

4 1800

6 1200

8 900

10 720

It is obvious that the synchronous speed can be only 900 rpm and therefore, the number of poles is 8. 900 - 890 ¥ 100 = 1.11% 900 900 - 855 (c) Full-load slip = ¥ 100 = 5% 900 1.11 (d) At no-load, f2 (= sf1) = ¥ 60 = 0.66 Hz 100 5 ¥ 60 = 3 Hz. (e) At full-load, f2 = 100

(b) No-load slip (s) =

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Solution Given the number of poles of alternator PA = 4 and the synchronous speed of the alternator is 1200 N ◊ PA 1200 ¥ 4 rpm, the frequency f is = 40 Hz = 120 120 \ frequency generated by the alternator is 40 Hz. For the given induction motor, P = 6. Speed at full-load N = 760 rpm, supply frequency from the alternator is f = 40 Hz. 120 f 120 ¥ 40 = 800 rpm. \ synchronous speed of the motor, Ns = = P 6 \ the percentage slip,

s=

800 - 760 Ns - N . 100 = ¥ 100 = 5%. 800 Ns

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!

9/25

Fmfdusjdbm!Nbdijoft

Solution (a) Given N = 900 rpm, f = 50 Hz and P = 60 \ \

120 ¥ f 120 ¥ 50 = = 1000 rpm P 6 N - N 1000 - 900 slip (s) = s = 0.1 or 10% = Ns 1000 Ns =

(b) Alternation of rotor voltage f ¢ = s ¥ f = 0.01 ¥ 50 = 0.5/sec or 30/min.

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Solution 120 f 120 ¥ 50 = = 1000 rpm P 6 Speed at no-load = (1 – slip at no-load) ¥ Ns = (1 – 0.008) ¥ 1000 = 992 rpm Speed at full-load = (1 – slip at full-load) ¥ Ns = (1 – 0.02) ¥ 1000 = 980 rpm Frequency of rotor current at standstill f2 = sf = 1 ¥ 50 = 50 Hz Frequency of rotor current at full-load, f2 = (slip at full-load) ¥ f = 0.02 ¥ 50 = 1.0 Hz.

(a) Ns = (b) (c) (d) (e)

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Solution 120 f 120 ¥ 50 = = 1500 rpm P 4 Rotor emf frequency, f2 = sf or 1.5 = s ¥ 50 1.5 \ slip (s) = = 0.03 or 3.0% 50

(a) Ns =

(b) Actual speed of motor is N = (1 – s). Ns = 1500 (1 – 0.03) = 1455 rpm.

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9/26

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)b*! )c*! )d*! )e*! !)f*!

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Solution (a) Ns =

120 ¥ f 120 ¥ 50 = = 1000 rpm P 6

3 ˆ Ê = 970 rpm (b) N = Ns (1 – s) = 100 Á1 Ë 100 ˜¯ (c) f2 = sf1 =

3 ¥ 50 = 1.5 Hz 100

120 ¥ f 2 120 ¥ 1.5 = 30 rpm = P 6 (e) Since the rotor is rotating at 970 rpm and the rotor mmf is revolving at 30 rpm with respect to rotor, therefore, speed of the rotor mmf relative to the stationary winding (stator) is (970 + 30) rpm = 1000 rpm.

(d) Speed of rotor mmf relative to rotor winding =

FRVJWBMFOU!DJSDVJU!PG! BO!JOEVDUJPO!NPUPS

9/21

When a balanced three-phase induction motor is excited by a balanced three-phase voltage source, the currents in the phase windings must be equal in magnitude and 120° electrical degrees apart in phase. The same must be true for the currents in the rotor windings as the energy transferred across the air gap from the stator to the rotor is by induction only. However, the frequency of the induced emf in the rotor is proportional to its slip. Since the stator and the rotor windings are coupled inductively, an induction motor resembles a three-phase transformer with a rotating secondary winding. The similarity becomes evident when the rotor is at rest (standstill condition s = 1). Thus, a threephase induction motor can be represented on a per-phase basis by an equivalent circuit at any slip (s) as depicted in Fig. 8.3. In this figure, V1 = Applied voltage (rms value) to the stator on per-phase basis R1 = Per-phase stator winding resistance L1 = Per-phase stator winding leakage inductance

!

9/27

Fmfdusjdbm!Nbdijoft

X1 (= 2pfL1) R2 L2 X2 (= 2pf L2) Xr (= 2psf Lb) Xm Rm T1 T2 Kw1 Kw2 f

= Per-phase stator winding leakage reactance = Per-phase rotor winding resistance = Per-phase rotor winding leakage inductance = Per-phase rotor winding leakage reactance under blocked rotor condition (s = 1) = sX2 = Per-phase rotor winding leakage reactance at slip (s) = Per-phase magnetization reactance = Per-phase equivalent core-loss resistance = Actual turns per-phase of the stator winding = Actual turns per-phase of the rotor winding = Winding factor for the stator winding = Winding factor for the rotor winding = Amplitude of the per-phase flux

E1 (= 4.44 f1T1 Kw1 f) = Per-phase induced emf in the stator winding R1

jX1 Im

I1 +

V1 –

I2

+

Ic

jXr

R2

+

E1 Xm

Rm Im

Er –

–

a:1

Gjh/!9/4! Qfs.qibtf!frvjwbmfou!djsdvju!pg!b!cbmbodfe!uisff.qibtf!joevdujpo!npups

E2 (= 4.44 f1T2 Kw2 f) = Per-phase induced emf in the rotor winding under blocked-rotor condition (s = 1) Er = sE2 = Per-phase induced emf in the rotor winding at slip (s) I2 = Per-phase rotor winding current I1 = Per-phase current supplied by the source Im = Ic + Im = Per-phase excitation current Ic = Per-phase core-loss current Im = Per-phase magnetization current a=

Kw1 T1 Kw 2 T2

Uisff.qibtf!Joevdujpo!Npups

9/28

From the per-phase equivalent circuit in Fig. 8.3, it is evident that the current in the rotor circuit is Er sE2 = R2 + jX r R2 + jsX 2 E2 = ( R2 / s) + jX 2

I2 =

(8.25)

Based upon the above equation, we can develop another circuit of an induction motor as given in Fig. 8.4. In this circuit, the hypothetical resistance (R2/s) in the rotor circuit is called the effective resistance. The effective resistance is the same as the actual rotor resistance when the rotor is at rest (standstill or blocked-rotor condition). On the other hand, when the slip approaches zero under no-load condition, the effective resistance is very high (R2/s Æ •). R1

R2

jX1 Im

I1

I2

+

–

+

E1

+

V1

Ic

jX2

s

jXm

Rm Im

E2 –

–

a:1

Gjh/!9/5! Npejgjfe!frvjwbmfou!djsdvju!pg!b!cbmbodfe!uisff.qibtf!npups!po!b!qfs.qibtf!cbtjt

We can represent the induction motor by its per-phase equivalent circuit as referred to the stator. Such an equivalent circuit is shown in Fig. 8.5, where R¢2 = a2 R2 X¢2 = a2 X2

(8.26) 2

i.e., when referred to primary, the secondary impedance is a times of its value at the secondary side. E1 K w1 T1 Te1 \ = =a (8.27) = E2 K w2 T2 Te2 where Te1 and Te2 are called effective stator and rotor turns per-phase respectively and a is effective turns ratio of an induction motor I1 Te2 1 = Also, (8.28) = I 2 Te1 a

!

9/29

Fmfdusjdbm!Nbdijoft

Equation (8.27) shows that the ratio between stator and rotor emfs is constant at standstill. The ratio depends on the turns ratio modified by distribution and pitch factors of the windings. The induction motor, therefore, behaves like a transformer. It is to be noted that the factors for stator and rotor windings are not the same, because the number of slots in them may be different. I2 a E¢2 = aE2 I2¢ =

Also, and For this equivalent circuit,

I 2 R2¢ + jI2 X2¢ s Im = Ic + Im

E1 = and where

Ic =

E1 E1 and Im = Rm jX m

The per-phase stator winding current and the applied voltage are I1 = Im + I¢2 and

V1 = E1 + I1 (R1 + jX1)

The equivalent circuit of the rotor in Fig. 8.4 is in terms of the hypothetical resistance R2/s. In this circuit, I22 R2/s represents the per-phase power delivered to the rotor. However, the per-phase copper loss in the rotor must be I22 R2. Thus, the per-phase power developed by the rotor is R¢2 jX1

R1

jX¢2

s

I¢2

Im

I1 +

V1

Ic –

jXm

Rm Im

! Gjh/!9/6! Qfs.qibtf!frvjwbmfou!djsdvju!pg!b!cbmbodfe!uisff.qibtf!joevdujpo!npups!bt!sfgfssfe!up!uif! tubups!tjef

I22

R2 È1 - s ˘ – I22 R2 = I22 R2 Í ˙ s Î s ˚

Uisff.qibtf!Joevdujpo!Npups

9/2:

R2 È1 - s ˘ (8.29) = R2 + R2 Í ˙ s Î s ˚ The above equation establishes the fact that the hypothetical resistance (R2/s) can be divided into two components: the actual resistance of the rotor (R2) and an additional resistance [R2[(1 – s)/s]]. This additional resistance is called load resistance or the dynamic resistance. The load resistance depends upon the speed of the motor and is said to represent the load on the motor because the mechanical power developed by the motor is proportional to it. In other words, the load resistance is the electrical equivalent of a mechanical load on the motor. An equivalent circuit of an induction motor in terms of the load resistance is given in Fig. 8.6. This circuit is said to be the exact equivalent circuit of a balanced three-phase induction motor on per-phase basis referred to the stator side. while

R1

R¢2

jX1

I¢2 I2 ˆ Ê ÁË I2¢ = a ˜¯

Im

I1 Ic

jXm

Rm

V1

jX¢2

E1 = E¢2

Im

R¢2 (1 – s) s

(E ¢2 = aE2)

Gjh/!9/7! Uif!frvjwbmfou!djsdvju!pg!Gjh/!9/6!npejgjfe!up!tipx!uif!spups!boe!mpbe!sftjtubodf

The approximate equivalent circuit of an induction motor is obtained by shifting the shunt branch to the input terminals. The approximate equivalent circuit is shown in Fig. 8.7. In an induction motor, as the no-load current is very high compared to that of the transformer (Article 8.6), the computation using approximate equivalent circuit of an induction motor involves considerable error. The phasor diagram of the induction motor is shown in Fig. 8.8. R1

I1 Im Ic V1

Im

R¢2

jX¢2

I¢2 jXm

Rm

jX1

V1 = E1 = E¢2

R¢2 (1 – s) s

Gjh/!9/8! Bqqspyjnbuf!frvjwbmfou!djsdvju!pg!b!uisff.qibtf!joevdujpo!npups

!

9/31

Fmfdusjdbm!Nbdijoft

fm E2 Im

I1

Im Ic

I1R1

Er

Im

)= (–E 1

I¢2 E2 =a E¢2

2 = sE

I2R2

I2sx2 I2

È-E1 = E2¢ = aE2; E1 / E2 = a; a > 1.˘ Í ˙ Í I2¢ = (I2 / a ); In = Ic2 + I m2 ; ˙ Í ˙ Í Er = sE2 = I2 (R2 + jX 2 ◊ s ) ˙ Í ˙ Î-E1 + I1(R1 + jx1) = V1 ˚

I1X1

V1

Gjh/!9/9! Qibtps!ejbhsbn!pg!b!uisff.qibtf!joevdujpo!npups!)pqfsbujoh!xjui!tmjqt*!po!qfs!qibtf!cbtjt

QPXFS!SFMBUJPOT!JO!B!UISFF.QIBTF! JOEVDUJPO!NPUPS

9/22

Since the load resistance varies with the slip and the slip adjusts itself to the mechanical load on the motor, the power delivered to the load resistance is equivalent to the power developed in the motor. Thus, the performance of the motor at any slip can be determined from its equivalent circuit, as given in Fig. 8.4. For a balanced three-phase induction motor, Pin = 3V1 I1 cos q

(8.30)

where q is the phase difference between the applied voltage V1 and the stator winding current I"1. The most important electrical loss that must be taken into consideration is the stator copper loss. The total stator copper loss is given by PL1 = 3 I12 R1

(8.31)

If the core loss is assumed to take phase in an equivalent core loss resistance (Rm), the total core loss (magnetic loss) is obtained as Pm = 3 Ic2 Rm

(8.32)

The electrical power that is crossing the air gap and is transported to the rotor by electromagnetic induction is called the air-gap power. In this case, the air-gap power is Pag =

3 I 22 R2 s

(8.33)

Uisff.qibtf!Joevdujpo!Npups

9/32

The electrical power loss in the rotor circuit or rotor copper loss is then PL2 = 3 I22 R2 = sPag

(8.34)

Hence, the power developed by the motor is

3 I 22 (1 - s) R2 (1 – s)Pag s The electromagnetic torque developed by the motor is Pe = Pag – PL2 =

Te =

Pe (1 - s) Pag Pag R = = = 3 I 22 2 w (1 - s) w s w s sw s

(8.35)

(8.36)

By subtracting the rotational loss (Prot) from the power developed Pe , we obtain the power output of the motor as Po = Pe – Rotor mechanical loss (Prot) (8.37a) Po = Pe – Pr Since the core loss has already been accounted for, the rotational loss includes the friction and windage loss Pf and Pw as well as the stray-load loss Pst. Thus,

Po = Pe – (Pf + Pw + Pst) = Te ¥ w

MPTTFT!BOE!FGGJDJFODZ!PG!JOEVDUJPO!NPUPS!

(8.37b)

9/23

At starting and during acceleration, the rotor core losses are high, and with the increase in speed these losses decrease to some extent. The friction and windage losses are zero at start and with increase in speed, these losses increase. However, the sum of friction windage and core losses is roughly constant for a motor even with variable speed. Hence, these categories of losses are often lumped together and called constant losses. \

P(constant loss) = Pcore loss + Pmechanical loss

Losses in a three-phase induction motor are of two types, mainly (a) fixed losses, and (b) variable losses (a) Fixed losses — Core loss — Bearing friction loss — Brush friction loss in wound rotors — Windage loss

!

9/33

Fmfdusjdbm!Nbdijoft

(b) Variable losses — Stator ohmic loss (I2 R loss in state) — Rotor ohmic loss (I2 R loss in rotor) — Brush contact loss for wound rotor motors only — Stray load loss The rotor output gives rise to the development of gross torque or electromagnetic torque Te which is partly wasted (in the form of winding and frictional losses in the rotor) and partly appears as useful shaft torque Tsh. \ Rotor output

Po = Te ¥ 2pn

(8.38)

Pag = Te ¥ 2pns

(8.39)

where n is the speed of the motor. From Eq. 8.36, Rotor input where ns is the synchronous speed. Hence, rotor efficiency h=

Output of rotor n (1 - s) ns =1–s = = Rotor input ns ns

(8.40)

From Eq. (8.34) and Eq. (8.35), Copper losses of rotor PL2 = sPag =

s Pe 1- s

\ Rotor input : Rotor copper loss : Mechanical power developed in the rotor = Pag : sPag : (1 – s) Pag =1:s:1–s

(8.41)

(8.42)

It is to be noted that air gap power Pag = Rotor input = Stator input – Stator copper loss – Stator core loss The shaft output torque Tsh is developed at the output of the rotor (i.e., at the shaft). The shaft torque is obtained from Eq. (8.37a) as Tsh =

Po Pe - Prot = w w

(8.43)

The torque of a polyphase induction motor may be expressed in synchronous watts. It is defined as the torque which develops a power of 1 W at the synchronous speed of the motor. Rotor input = Tg ¥ 2p ns

Uisff.qibtf!Joevdujpo!Npups

9/34

The motor efficiency h=

Po ¥ 100 % Pin

(8.44)

where Pin is the stator input power and Pin = Po + [Stator core loss + Stator copper loss] + [Rotor core loss + Rotor copper loss] + Mechanical loss Rotor core loss is negligible in induction motors. \ Fixed loss = Stator core loss + Mechanical loss Variable loss = Stator copper loss + Rotor copper loss As it is very difficult to measure stray load losses in induction motors, to account for this loss, the efficiency h is taken as 0.5% less than the calculated value on full-load and for other loads. The power input Pin can also be written as Pin = 3 V1 I1 cos f where V1 and I1 are stator input voltage and current and cos f is the stator power factor. From the equivalent circuit shown in Fig. 8.7, the electromagnetic power developed by the rotor Pe = 3 I 2¢ 2 R2¢ I2¢ =

\

1- s s V1 0.5

2 ÏÔÊ R¢2 ˆ 2¸ Ô R + + ( X1 + X 2¢ ) ˝ ÌÁ 1 ˜ Ë ¯ s ÔÓ Ô˛ 2 3 R2¢ V1 (1 - s) Pe = 2 ÏÔÊ R¢ ˆ 2¸ Ô s ÌÁ R1 + 2 ˜ + ( X1 + X 2¢ ) ˝ Ë ¯ s ÔÓ Ô˛

(8.45)

From Eq. (8.35), Pe 3 R2¢ V12 = Pag = 2 1- s ÏÔÊ R¢ ˆ 2¸ Ô s ÌÁ R1 + 2 ˜ + ( X 1 + X 2¢ ) ˝ Ë ¯ s ÔÓ Ô˛ If the circuit parameters of the induction motor are referred to the rotor side then Pe = 3 I22 R2

Er2 1 - s 3 R2 (1 - s) = 2 s s Ê R2 ˆ 2 ÁË s ˜¯ + X 2

(8.46)

!

9/35

Fmfdusjdbm!Nbdijoft

= \

Pag

3s (1 - s) Er2 R2

(8.47)

R22 + s 2 X 22 P 3s E 2 R = e = 2 r2 2 2 1 - s R2 + s X 2

(8.48)

The power stages of a three-phase induction motor are shown in Fig. 8.9. Air gap

Pe (mechanical power

developed in rotor)

Input power (Pin)

Shaft

Rotor

Stator

Pag

Stator copper loss + stator core loss (Pscu + Psc)

Rotor copper loss (Prcu)

Shaft output (Po) Shaft output torque (Tsh)

Rotor mechanical loss (Prot)

Te (gross mechanical torque developed in rotor)

[Pag = Pin – (Pscu + Psc); Pe = Pag – Prcu; Po = Pe – Prot.]

Gjh/!9/:! Qpxfs!tubhft!jo!b!uisff.qibtf!joevdujpo!npups

! Qspcmfn!9/22 B!uisff.qibtf-!6!IQ-!511!W-!61!I{!joevdujpo!npups!jt!xpsljoh!bu!gvmm.mpbe!xjui!bo!fggjdjfodz!pg!:1&! bu!b!qpxfs!gbdups!pg!1/9!mbhhjoh/ Dbmdvmbuf!)b*!uif!joqvu!qpxfs-!boe!)c*!uif!mjof!dvssfou/

Solution Rating of the motor = 5 HP = 5 ¥ 735.5 = 3677.50 watts; V = 400 V (line value); f = 50 Hz; fullload efficiency = 90% (= 0.9) and p.f = 0.8 (lagging). Output , Input Output 5 ¥ 735.5 Input power = = = 4.086 kW h 0.9

(a) E efficiency h =

(b) For a three-phase induction motor, Input power = 3 VL IL cos f or

4086 = 3 ¥ 400 ¥ IL ¥ 0.8

Hence, the line current (IL) =

4086 3 ¥ 400 ¥ 0.8

= 7.37 A.

Uisff.qibtf!Joevdujpo!Npups

9/36

! Qspcmfn!9/23 B!uisff.qibtf-!gpvs.qpmf!joevdujpo!npups!svot!bu!b!tqffe!pg!2551!sqn!po!611!W-!61!I{!nbjot/!Uif! nfdibojdbm!qpxfs!efwfmpqfe!cz!uif!spups!jt!31/4!IQ/!Uif!nfdibojdbm!mpttft!bsf!3/34!IQ/!Efufsnjof! )b*!uif!tmjq-!)c*!uif!spups!dpqqfs!mpttft-!boe!)d*!uif!fggjdjfodz/

Solution Ns =

(a) \

Slip =

120 ◊ f 120 ¥ 50 = = 1500 rpm P 4 N s - N 1500 - 1440 = 0.04 or 4% = Ns 1500

(b) Mechanical power developed in rotor = Power output + Rotor losses = 20.3 + 2.23 = 22.53 HP. = 22.53 ¥ 735.5 W = 16.571 kW 16571 16571 \ Power transferred from stator to rotor (Pag) = = = 17261.46 W (1 - s) (1 - 0.04) \ Rotor copper losses = 17261.46 – 16571.00 = 690.46 W. (c) Efficiency (h) =

Output 20.3 ¥ 735.5 = 0.865 = 86.5%. = Input 17261.46

! Qspcmfn!9/24 Uif!gvmm.mpbe!tmjq!pg!b!611!IQ-!61!I{!uisff.qibtf!joevdujpo!npups!jt!1/14/!Uif!spups!xjoejoh!ibt!b! sftjtubodf!pg!1/41!W0qibtf/!Efufsnjof!uif!tmjq!boe!uif!qpxfs!pvuqvu-!jg!bo!fyufsobm!sftjtubodf!pg 3!pint!jt!jotfsufe!jo!fbdi!spups!qibtf/!Bttvnf!uibu!uif!upsrvf!sfnbjot!tbnf/

Solution (a) R2 = 0.3 W, R¢2 = 2 + 0.3 = 2.3 W, s = 0.03 \

Slip s¢ =

R2¢ ◊ s 2.3 ¥ 0.03 = = 0.23 R2 0.3

(b) Let Ns be the synchronous speed; then and

N = Ns (1 – 0.03) = 0.97 Ns N¢ = Ns (1 – 0.23) = 0.77 Ns

[N¢ is the new speed when external resistance of 2 ohms is inserted in each rotor phase] Since the torque remains same, the output is directly proportional to speed. \ new motor output = 500 ¥

0.77 N s = 397 HP. 0.97 N s

!

9/37

Fmfdusjdbm!Nbdijoft

! Qspcmfn!9/25 B!uisff.qibtf-!61!I{-!gpvs.qpmf!joevdujpo!npups!ibt!b!tubs.dpoofdufe!xpvoe!spups/!Uif!spups!fng! jt!61!W!cfuxffo!uif!tmjq!sjoht!bu!tuboetujmm/!Uif!spups!sftjtubodf!boe!tuboetujmm!sfbdubodf!bsf!1/5!W! boe!3/1!W!sftqfdujwfmz/!Dbmdvmbuf ! )b*! uif!spups!dvssfou!qfs!qibtf!bu!tubsujoh!xjui!tmjq!sjoht!tipsu!djsdvjufe! )c*! uif!spups!dvssfou!qfs!qibtf!bu!tubsujoh!jg!61!W!qfs.qibtf!sftjtubodf!jt!dpoofdufe!cfuxffo!tmjq! sjoht! )d*! uif!spups!fng!xifo!uif!npups!jt!svoojoh!bu!gvmm.mpbe!bu!2551!sqn! )e*! uif!spups!dvssfou!bu!gvmm.mpbe-!boe! ! )f*! spups!qpxfs!gbdups!)q/g/*!bu!gvmm.mpbe/

Solution Ns = (a) Er =

50 3

120 ¥ 50 = 1500 rpm 4

= 28.867 V

At standstill with slip rings short-circuited, I2 =

Er ( R22

+

X 22 )0.5

=

28.867 {(0.4) 2 + 22 }0.5

= 14.15 A

(b) The total resistance in the rotor circuit is 5.4 ohms per phase. 28.867 \ I2 = = 5.01 A {(5.4) 2 + 22 }0.5 1500 - 1440 = 0.04 1500 \ Rotor emf = 28.87 ¥ 0.04 = 1.555 V/Ph sEr 1.155 (d) I2 = 2 = = 2.82 A. 2 0.5 2 [ R2 + ( sX 2 ) ] [0.4 + (0.04 ¥ 2) 2 ]0.5 (c) Full-load slip =

! Qspcmfn!9/26 B!fjhiu.qpmf-! 61! I{-!uisff.qibtf! joevdujpo!npups! ibt! spups! joqvu!pg! 211! lX! po! gvmm.mpbe/! Uif!sp. ups! fng! nblft! 231! dzdmft! qfs! njovuf/! Efufsnjof! )b*! spups! tqffe! jo! sqn-! )c*! spups! dpqqfs! mptt-! )d*! nfdibojdbm! qpxfs! efwfmpqfe-! boe! )e*! spups! sftjtubodf! qfs! qibtf! jg! spups! dvssfou! jt! 91! B! qfs! qibtf/

Solution No. of poles P = 8 Supply frequency f = 50 Hz

Uisff.qibtf!Joevdujpo!Npups

9/38

Input = 100 kW Rotor emf frequency = \

120 cycles/s = 2 cycles/s 60 = 2 Hz sf = 2 Hz where s is the slip.

\

s=

2 = 0.04 50

(a) Rotor speed Nr = (1 – s) Ns where Ns is the synchronous frequency Nr = (1 – 0.04) ¥

120 ¥ 50 8

= 720 rpm (b) Rotor input Pag = 100 kW \ Rotor copper loss = sPag = 0.04 ¥ 100 = 4 kW (c) Mechanical power developed Pe = (1 – s) Pag = (1 – 0.04) ¥ 100 = 96 kW (d) Rotor copper loss = I22 R2 where I2 is the rotor current and R2 is the rotor resistance per phase \

I22 R2 =

or,

R2 =

sPag 3

=

0.04 ¥ 100 0.04 ¥ 105 kW = W 3 3

0.04 ¥ 105 3 ¥ (80) 2

W = 0.21 W/

! Qspcmfn!9/27 B!xpvoe.spups!joevdujpo!npups!ibt!gvmm.mpbe!tqffe!pg!691!sqn/!Jg!uif!tvqqmz!gsfrvfodz!jt!61!I{-! efufsnjof!uif!ovncfs!pg!qpmft!boe!tmjq!bu!gvmm.mpbe/!Jg!spups!sftjtubodf!jt!epvcmfe-!efufsnjof!uif! dpqqfs!mptt!jg!uif!jojujbm!dpqqfs!mptt!jt!411!X/!

Solution Motor speed Nr = 580 rpm Supply frequency f = 50 Hz As slip lies between 0.03 to 0.05 at full-load, the synchronous speed Ns = 600 rpm If P be the number of poles then 120 f = Ns P

!

9/39

Fmfdusjdbm!Nbdijoft

120 f 120 ¥ 50 = = 10 Ns 600 580 \ Slip s = 1 – = 0.033 600 The copper loss is proportional to the rotor resistance. \ if rotor resistance is doubled, the copper loss is also doubled. \ copper loss = 300 ¥ 2 = 600 W. or,

P=

! Qspcmfn!9/28 B!uisff.qibtf!551!W-!tjy.qpmf-!61!I{!joevdujpo!npups!efwfmpqt!56!IQ!bu!::1!sqn!bu!b!qpxfs!gbdups! pg!1/:!mbhhjoh/!Efufsnjof!uif!spups!dpqqfs!mptt-!gsfrvfodz!pg!spups!fng!boe!upubm!qpxfs!joqvu/!Of. hmfdu!nfdibojdbm!boe!jspo!mpttft!pg!uif!spups!boe!bttvnf!tubups!mpttft!up!cf!2611!X/

Solution P=6 As mechanical losses are neglected, Output power = Mechanical power developed by the motor. Pe = 45 HP = 45 ¥ 735 W = 33075 W 990 990 Slip s = 1 – =1– = 0.01 120 ¥ 50 1000 6 If Pag is the rotor input power, (1 – s) Pag = 33075 \ Pag = 33409 W

\

\ copper loss = sPag = 334.1 W Frequency of rotor emf = sf = 0.01 ¥ 50 = 0.5 Hz Total power input = Pag + Stator losses = 33409 + 1500 = 34909 W.

! Qspcmfn!9/29 Uif! qbsbnfufst! pg! uif! frvjwbmfou! djsdvju! pg! b! uisff.qibtf! 61! IQ-! fjhiu.qpmf-! 551! W-! 61! I{! tubs. dpoofdufe!joevdujpo!npups!sfgfssfe!up!uif!tubups!bsf!bt!gpmmpxt; ! S2! >!1/9!W!Y2!>!2/6!W ! S¢3! >!1/6!W!Y¢3!>!1/6!W ! Yn! >!41!W Uif!tmjq!jt!4&/!Jg!uif!nfdibojdbm!boe!tubups!jspo!mpttft!bsf!2611!X-!boe!911!X-!efufsnjof!)b*!tqffe-! )c*!tubups!dvssfou-!)d*!qpxfs!gbdups-!)e*!bjs.hbq!qpxfs-!nfdibojdbm!qpxfs!efwfmpqfe!boe!qpxfs!pvu. qvu-!)f*!upsrvf!efwfmpqfe!boe!mpbe!upsrvf-!)g*!fggjdjfodz-!boe!)h*!spups!dpqqfs!mptt!qfs!qibtf/!Ofhmfdu! dpsf!mptt!sftjtubodf!Sn/

Uisff.qibtf!Joevdujpo!Npups

Solution Synchronous speed Ns=

120 f 120 ¥ 50 = = 750 rpm P 8

Slip s = 0.03 (a) Speed Nr = (1 – s) Ns = (1 – 0.03) ¥ 750 = 728 rpm 440 (b) Per phase supply voltage V1 = = 254 V 3 From Fig. 8.7, V1 (1 - s) R1 + R2¢ + R2¢ ◊ + j ( X 1 + X 2¢ ) s V1 = R2¢ ˆ Ê ÁË R1 + s ˜¯ + j ( X 1 + X 2¢ ) 254 = 0.5 ˆ Ê ÁË 0.8 + 0.03 ˜¯ + j (1.5 + 0.5) 254 = = 14.44 -6.53∞ A 17.47 + j 2 254 Im = A (Neglecting Rm) j 30 = 8.47 -90∞ A I¢2 =

\

\ Stator current

I1 = I¢2 + Im = 14.44 -6.53∞ + 8.47 -90∞ = 14.346 – j 10.11 = 17.55 -35.17∞ A

(c) Power factor cos 35.17° lag or 0.817 lag (d) Power input

Pi = 3 VL IL cos q = 3 ¥ 440 ¥ 17.55 ¥ 0.817 = 10927.3 W

Stator iron loss = 800 W Stator copper loss = 3 I12 R1 = 3 ¥ (17.55)2 ¥ 0.817 \ air-gap power Pag = 10927.3 – 800 – 754.9 = 9091.228 W

9/3:

!

9/41

Fmfdusjdbm!Nbdijoft

Mechanical power developed Pe = (1 – s) Pag = (1 – 0.03) ¥ 9372.4 = 9091.228 W Power output

Po = 9091.228 – 1500 = 7591.228 W

Pag

9372.4 = 119.39 Nm 750 ws 2p ¥ 60 Po 7591.228 Tsh = = = 99.63 Nm 728 w 2p ¥ 60 Te =

(e) Torque developed

Load torque

(f) Efficiency

h=

=

Po 7591.228 = ¥ 100% = 69.47% 10927.3 Pi

Rotor copper loss per phase =

sPag 3

=

0.03 ¥ 9372.4 W = 93.724 W. 3

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Solution The output of the motor

Input

Po = 30 HP = 30 ¥ 735 = 22050 W 22050 Pi = = 26250 W 0.84

\ total loss = Pi – Po = 4200 W Let stator iron loss be x. \ stator copper loss = Rotor copper loss = x No-load loss = Stator iron loss + Mechanical loss = x + Mechanical loss 1 \ total mechanical loss = (x + Mechanical loss) 5 x \ total mechanical loss = 4 \ Stator iron loss + Stator copper loss + Rotor copper loss + Mechanical loss = 4200

Uisff.qibtf!Joevdujpo!Npups

9/42

x = 4200 4 13 x = 4200 4 x = 1292.31 W

\

x+x+x+

or, or,

\ mechanical power produced by the motor Pe = Po + Mechanical loss 1292.31 = 22050 + = 22373 W 4 Power input to the rotor Pag = Pi – Stator losses = 26250 – 1292.31 ¥ 2 = 23665 W If s be the slip at full-load, (1 – s) Pag = Pe 22373 1–s = = 0.9454 23665 s = 0.055.

\ or,

! Qspcmfn!9/31 B!uisff.qibtf!joevdujpo!npups!svoojoh!bu!b!tmjq!pg!51&!jt!ubljoh!bo!joqvu!pg!71!lX/!Jg!uif!tubups! mpttft!bsf!2/6!lX!boe!gsjdujpo!boe!xjoebhf!mpttft!bsf!3/6!lX-!efufsnjof!)b*!qpxfs!joqvu!up!uif!sp. ups-!)c*!spups!dpqqfs!mptt!qfs.qibtf-!)d*!nfdibojdbm!qpxfs!efwfmpqfe-!boe!)e*!npups!fggjdjfodz/

Solution Slip s = 0.04 Input = 60 kW Stator loss = 1.5 kW (a) \ rotor input Pag = 60 – 1.5 = 58.5 kW (b) Rotor copper loss PL2 = sPag = 0.04 ¥ 58.5 = 2.34 kW (c) Mechanical power developed Pe = (1 – s) Pag = (1 – 0.04) ¥ 58.5 = 56.16 kW (d) Motor output \ efficiency =

Po = Pe – 2.5 = 56.16 – 2.5 = 53.66 kW 53.66 ¥ 100% = 89.43%. 60

!

9/43

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Solution (a) Motor speed, N = Ns (1 – s) =

120 f (1 - s) 120 ¥ 50 (1 - 0.04) = = 1440 rpm P 4

Gross power developed in rotor of motor Tshaft ¥ 2p N + Friction + Windage losses 60 160 ¥ 2p ¥ 1440 (Pc) = + 500 = 24615 W 60 (Pc) =

or, \ Rotor input

(Pg) =

Pm 24615 = 25640 W = (1 - s) (1 - 0.04)

(b) Motor input (Pin) = Rotor input (Pag) + Stator losses = 25640 + 1000 = 26640 W (c) Efficiency (h) =

Net motor output (Po ) = 0.9052 = 90.52%. Motor input (Pin )

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Solution (a) Ns =

120 f 120 ¥ 50 = 1500 rpm = P 4

Now,

slip (s) = 0.04 =

( N s - N ) (1500 - N ) = Ns 1500

\ speed of motor, N = 1440 rpm (b) Frequency of rotor emf, f2 (= sf1) = 0.04 ¥ 50 = 2 Hz = 120 rpm (c) (i) At standstill, N = 0, s = 1 \ Rotor reactance = 4 ¥ s = 4 ¥ 1 = 4 W

Uisff.qibtf!Joevdujpo!Npups

9/44

\ Rotor impedance = (1 + j4) ohms = 4.123 – 75.96° W and p.f. (cos f) = cos 75° 96¢ = 0.243 (lag). [Rotor resistance is independent of slip and hence, R2 = 1 W] (ii) Slip at 1400 rpm speed is given by s¢ = (1500 – 1400)/1500 = 0.067. (Z¢) = 1 + j (4 ¥ 0.067) = (1 + j 0.268) W

\ rotor impedance and p.f

(cos f) =

1 {1 + (0.268) 2 }0.5

= 0.966 lag.

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Solution (a) Ns = \

120 f 120 ¥ 50 = 1000 rpm = P 6 N - N 1000 - 960 Slip (s) = s = 4% = Ns 1000

40 = 54.38 BHP [E 1 HP = 735.5 W] 0.7355 (c) Motor input = 40 kW, stator loss = 1 kW \ Rotor input = 36 kW Rotor copper loss = Slip ¥ Rotor input = 0.04 ¥ 39 = 1.56 kW (d) Rotor gross output is (1 – s) ¥ rotor input = 39(1 – 0.04) = 37.44 kW \ rotor output power = (37.44 – 0.2) kW = 37.24 kW (b) BHP (Brake Horse Power) =

\ motor efficiency i.e.,

37.24 ¥ 100 40 h = 93% (app.).

(h) =

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Solution Motor output = 18650 W, friction and windage losses =

2.6 ¥ 18650 = 484.9 W. 100

!

9/45

Fmfdusjdbm!Nbdijoft

\ rotor gross power developed = 18650 + 484.9 = 19134.9 W (= Pm) Ê s ˆ (a) Rotor copper loss = Rotor gross power developed ¥ Á Ë 1 - s ˜¯ = 19134.9 ¥ 0.42/(1 – 0.042) = 838.89 (b) Rotor input

Pag =

Rotor copper loss 838.89 = 19973.5 W = Slip 0.042 [Alternatively, Rotor input = 19134.9 + 838.89 = 19973.79 W]

(c) Ns =

120 f 120 ¥ 50 = 1500 rpm = P 4 N = Ns (1 – s) = 1500 (1 – 0.042) = 1437 rpm Output in watts (18650 ¥ 60) Tsh = = 123.6 Nm = ( 2p N / 60) 2p ¥ 1437

Shaft torque,

(d) Gross mechanical torque Te Gross mechanical power developed in rotor in watts 19134.9 ¥ 60 = 2p N / 60 2p ¥ 1437 = 127.2 Nm. =

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Solution (a) Slip =

N s - N 1000 - 950 = 0.05 = Ns 1000

50 ¥ 120 Ê ˆ = 1000 rpm˜ ÁË as N s = ¯ 6

(b) Rotor copper loss = Slip ¥ Rotor input = 0.05 ¥ 48.8 kW = 2.44 kW [Rotor input = Input – Stator loss = 50 – 1.2 = 48.8 kW] (c) Rotor output = Rotor input – Rotor copper loss – Friction and windage loss = 48.8 – 2.44 – 1.5 = 44.86 kW (d) Efficiency (h) =

Motor output 44.86 ¥ 100 = ¥ 100 = 0.897 = 89.7%. Motor input 50

Uisff.qibtf!Joevdujpo!Npups

9/46

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Solution 0.2 W

I1 Io 415 3

V 50 W

0.1 W

0.5 W

0.5 W

I¢2 10 W

1 - sˆ Ë s ¯˜

RL = R¢2 ÊÁ

Gjh/!9/21! Djsdvju!ejbhsbn!pg!Qspc/!9/37

Let Z be the total impedance of the circuit (Fig. 8.10). Ê1 - sˆ Ê 1 - 0.04 ˆ RL = R¢2 Á = 0.1 Á = 2.4 W ˜ Ë s ¯ Ë 0.04 ˜¯

Load resistance

R1e = 2.4 + 0.2 + 0.1 = 2.7 W X1 = 0.5 + 0.5 = 1 W

\ total resistance and total reactance

Z1 = ( 2.7) 2 + 1 = 8.29 = 2.88 W

Impedance

Angle of Z1 is tan–1 is (1/2.7) i.e., 20.323° (lag) Given,

VL = 415 V

\

Vphase =

\

I¢2 =

415 3 Vph Z1

= 240 V =

240 –0∞ = 83.36 – – 20.323° A 2.88 – 20.325∞

i.e. rotor current (referred to stator) = 83.36 A [Ans. of (c)] I0 = IC + Im

E and and \

240 = 4.8 A 50 240 Im = = 24 A 10 I0 = Ic + I m = (4.8 – j 24) A

IC =

!

9/47

Thus, \ Again, or, i.e.

Fmfdusjdbm!Nbdijoft

I1 = I0 + I 2¢ = (4.8 – j 24) + (78.17 – j 28.95) = (82.97 – j 52.95) A | I1 | = (82.97) 2 + (52.95) 2 98.44 A

[Ans. of (a)]

52.95 = 0.63818 82.97 f1 = 32.545° cos f1 = cos (32.545°) = 0.843 (lagging)

[Ans. of (b)].

tan f1 =

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Solution Here, VL = 500 V, f = 50 Hz, s = 4% = 0.04 Given, BHP = 20, stator losses = 1000 W Mechanical loss = 1 HP = 735.5 W Power output = 20 BHP = 20 ¥ 735.5 W = 14710 W = Rotor net output. \ Rotor gross mechanical power developed = Rotor net output + Mechanical loss = 14710 + 735.5 = 15445.5 W Hence, Rotor input =

Rotor gross mechnial power developed 15445.5 = = 16089.06 W (1 - s) 1 - 0.04

\ Stator input = Rotor input + Stator losses = 16089.06 + 1000 = 17089.06 W.

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Solution Here, P = 6, N = 960 rpm, f = 50 Hz, VL = 550 V, p.f = cos f = 0.90 (a) Ns =

120 f 120 ¥ 50 = = 1000 rpm P 6 Slip s =

N s - N 1000 - 960 = 0.04 = 4% = Ns 1000

Uisff.qibtf!Joevdujpo!Npups

9/48

30 ¥ 735.5 30 ¥ 735.5 = = 22984.4 W 1- s (1 - 0.04) \ rotor copper loss = (22984.4 – 28 ¥ 735.5) = 2390.4 W (c) Total input = (30 ¥ 735.5 + 2000) = 24,065 W 30 ¥ 735.5 (d) Efficiency = = 0.96 % 22984.4 24065 24065 Inut = = = 28.06 A. (e) Line current (IL) = 3 ¥ 550 ¥ 0.9 857.365 3 VL cos f (b) Power transferred from stator to rotor (Pag) =

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)b*! )c*! )d*! )e*!

Xibu!jt!uif!npupsÕt!tmjq@ Xibu!jt!uif!joevdfe!upsrvf!jo!On@ Xibu!xjmm!cf!uif!pqfsbujoh!tqffe!pg!uif!npups!jg!jut!upsrvf!jt!epvcmfe@ Ipx!nvdi!qpxfs!xjmm!cf!tvqqmjfe!cz!uif!npups!jg!uif!upsrvf!jt!epvcmfe@

Solution 1440 = 0.04 120 ¥ 50 4 (b) Induced torque is assumed to be equal to the load torque and the load power is 10 kW (a) Slip s = 1 –

10 ¥ 103 Nm = 66.35 Nm 1440 2p ¥ 60 (c) For low values of slip, the torque speed curve is linear; hence, induced torque is directly proportional to slip. \ when torque is doubled, slip is doubled. \ Induced torque =

\

s = 0.04 ¥ 2 = 0.08 120 ¥ 50 = 1380 rpm Speed = (1 – 0.08) ¥ 4

\

(d) \ power supplied to the motor P = 2 ¥ (66.35) ¥ 2p ¥

1380 = 19167 W. 60

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!

9/49

Fmfdusjdbm!Nbdijoft

Efufsnjof!tubups!dvssfou!boe!q/g/!bu!sbufe!wpmubhf!boe!bu!t!>!1/14/!Ofhmfdu!Sn/

Solution Referring to the equivalent circuit of Fig. 8.9, rotor current I¢2 =

=

=

V1 R2¢ ˆ Ê ÁË R1 + s ˜¯ + j ( X 1 + X 2¢ ) 440 / 3 0 . 15 Ê ˆ ÁË 0.2 + 0.03 ˜¯ + j (0.5 + 0.6) 254 A 5.2 + j 1.1

= 47.79 -11.94∞ A Im = \ stator current

254 = – j 8.467 A j 30

I1 = I¢2 + Im = 47.79 -11.94∞ – j 8.467 = 46.756 – j 18.354 = 50.23 -12.43∞ A

Power factor cos 21.43° lag or 0.931 lag.

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Solution Full-load slip sf l = 0.04 Output = 50 ¥ 735.5 = 36775 W Rotor current

Er (neglecting rotor reactance) R2 s 80 ¥ 0.04 3.2 = = R2 R2

I2 =

Uisff.qibtf!Joevdujpo!Npups

Power output (Fig. 8.7) Po = or, or, Motor speed

3 I 22 R2 (1 - s) = 36775 s

3(3.2) 2 (1 - 0.04) = 36775 R2 0.04 R2 = 0.02 W Nr = (1 – 0.04) Ns = 0.96 Ns

where Ns is the synchronous speed. Now, the new motor speed Nr¢ = 0.9 Nr = 0.9 ¥ 0.96 Ns = 0.864 Ns As torque varies as cube of speed T¢ T¢ Po Po

Torque or, Power output \

μ Nr¢3 μ (0.864 Ns)3 μ T ¢ ¥ Ns μ (0.864 Ns)4

\ new power output = 36775 ¥ Now

(0.864 N s ) 4 (0.96 N s ) 4

= 24128 W

Nr¢ = (1 – s¢)Ns = 0.864 Ns

where s¢ is the new value of slip. \

s¢ = 1 – 0.864 = 0.136

If R¢2 be the new resistance added per phase then new rotor current I¢2 =

80 ¥ 0.136 10.88 = 0.02 + R¢2 0.02 + R2¢

New power output 2

Ê 10.88 ˆ (0.02 + R2¢ ) (1 - 0.136) ¥ = 24128 P¢o = 3 Á ˜ 0.136 Ë 0.02 + R ¢ ¯ 2

or, or, or,

3 ¥ (10.88) 2 (1 - 0.136) = 24128 (0.02 + R2¢ ) ¥ 0.136 (0.02 + R¢2) ¥ 3281.41 = 306.83 R¢2 = 0.0735 W/phase.

9/4:

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9/51

Fmfdusjdbm!Nbdijoft

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Solution Frictional and windage loss = 2 ¥ 735 = 1470 W Output power Po = 30 HP = 30 ¥ 735 = 22050 W \ mechanical power developed by the motor Pe = 22050 + 1470 = 23520 W (a) If Pag be the air gap power then Rotor copper loss = sPag Pe = (1 – s) Pag P Pag = e where s is the slip 1- s 980 s =1– = 0.02 120 ¥ 50 6

Again, or, \

\ rotor copper losses = 0.02 ¥

23520 = 480 W 1 - 0.02

(b) Stator loss = 1500 W 480 + 1500 = 25500 0.02 P 22050 \ efficiency = o ¥ 100% = ¥ 100% = 86.47% Pi 25500

\ power input Pi = Pag + 1500 =

(c) If IL be the line current then or, or,

3 VL IL cos q = Pi 3 ¥ 440 ¥ IL ¥ 0.85 = 25500 IL = 39.365 A.

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Uisff.qibtf!Joevdujpo!Npups

Solution Shaft power

9/52

Psh = 100 H.P. = 100 ¥ 735 = 73500 Watts

Efficiency

h = 0.91 73500 Pin = 0.91 = 80769 W

\ power input to the stator

Total losses = Pin – Psh = 7269 W Total stator loss = Stator core loss + Stator copper loss = 1697 + 2803 = 4500 W \ Air-gap power = Power input to the rotor (Pag) = 80769 – 4500 = 76269 W If s be the slip then Rotor copper loss = sPag \

sPag = 1549 1549 s= = 0.0203 76269

or,

\ electromagnetic power developed by the rotor Pe = (1 – s) Pag = 74720 W \ combined windage and friction

Now Rotor speed

loss = 74720 – 73500 = 1220 W 80769 Pin = 3 VL IL cos q = = 0.817 3 ¥ 230 ¥ 248 120 ¥ 50 Nr = (1 – s) Ns = (1 – 0.0203) ¥ 6 = 980 rpm

\ Shaft torque

73500 980 2p ¥ 60 = 716.56 Nm.

Tsh =

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!

9/53

Fmfdusjdbm!Nbdijoft

Solution Input power

Pin = 3 VL IL cos q = 3 ¥ 440 ¥ 50 ¥ 0.8 = 30484.1 W

Air-gap power

Pag = Pin – Stator core loss – Stator copper loss Pag = 30484.1 – 2000 – 3000 = 25484.1 W

\ Output power

Po = Pag – Rotor copper loss – Friction and windage loss = 25484.1 – 1000 – 800 = 23684.1 W

\ efficiency of the motor h=

23684.1 ¥ 100% = 77.69%. 30484.1

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! !

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Solution (a) s = 0.025 120 ¥ 50 = 1500 rpm 4 Motor’s speed = (1 – 0.025)1500 = 1463 rpm (b) From the equivalent circuit of Fig. 8.5, the total equivalent impedance Ns =

Ê R¢ ˆ jX m Á 2 + jX 2¢ ˜ Ë s ¯ Z = R1 + jX1 + (neglecting Rm) R2¢ jX m + + jX 2¢ s Ê 0.35 ˆ j 25 Á + j 0.5˜ Ë 0.025 ¯ = 0.65 + j 1.15 + 0.35 j 25 + + j 0.5 0.025

Uisff.qibtf!Joevdujpo!Npups

= 0.65 + j 1.15 +

25 90∞ (14 + 0.5 90∞) 14 + j 25.5

= 0.65 + j 1.15 +

350 90∞ + 12.5 180∞) 29.1 61.23∞

= 0.65 + j 1.15 + 1203 28.77∞ + 0.429 118.77∞ = 0.65 + j1.15 + 10.33 + j 6.16 = 10.98 + j 7.31 = 13.191 33.65∞ Stator current

I1 =

440 3 ¥ 13.191 33.65∞

= 19.26 -33.65∞ A

(c) Power factor = cos (–33.65°) = 0.833 lag (d) Input power to the motor Pin = 3 ¥ 440 ¥ 19.26 ¥ 0.833 = 12226.85 W Stator copper loss = 3 I 12 R1 = 3 ¥ (19.26)2 ¥ 0.65 = 723.35 W \ power input to the rotor Pag = 12226.85 – 723.35 = 11503.5 W Electromagnetic power developed by the rotor Pe = (1 – s) Pag = (1 – 0.025) ¥ 11503.5 = 11215.9 W \ power output Po = Pe – mechanical loss = 11215.9 – 1200 = 10016.9 W Po 10016.9 = = 65.41 Nm 1463 w 2p ¥ 60 Po 10016.9 (f) Efficiency h = = = 81.925%. Pin 12226.85

(e) Shaft torque Tsh =

9/54

!

9/55

Fmfdusjdbm!Nbdijoft

! Qspcmfn!9/47 B!tjy.qpmf-!uisff.qibtf!joevdujpo!npups!jt!svoojoh!bu!b!tqffe!pg!:61!sqn!xifo!uif!joqvu!jt!61!lX/! Bu!uijt!dpoejujpo-!uif!tubups!dpqqfs!mptt!jt!2/6!lX!boe!uif!spubujpobm!mptt!jt!2!lX/!Efufsnjof!uif! spups!dpqqfs!mptt-!fmfduspnbhofujd!qpxfs!efwfmpqfe!cz!uif!spups!boe!uif!nfdibojdbm!qpxfs!pvuqvu/

Solution 120 ¥ 50 = 1000 rpm 6 Rotor speed = 950 rpm Synchronous speed =

Hence, slip

950 = 0.05 1000 Input = 50 kW s =1–

Hence, air-gap power (Pag) = (50 – 1.5) = 48.5 kW \ rotor copper loss is (sPag) (= 0.05 ¥ 48.5) or 2.425 kW \ gross mechanical power developed by the rotor is [(1 – s)Pag] or, (1 – 0.05) ¥ 48.5 = 46.075 kW \ Shaft power output = Gross mechanical power – Rotational loss = 46.075 – 1 = 45.075 kW.

! Qspcmfn!9/48 B!uisff.qibtf-!526!W-!5!lX!efmub.dpoofdufe!joevdujpo!npups!ibt!b!tipsu.djsdvju!mjof!dvssfou!pg!31! B!bu!311!W/!Uif!npups!jt!tubsufe!cz!b!tubs.efmub!tubsufs/!Jg!uif!gvmm.mpbe!fggjdjfodz!boe!q/g/!bsf!1/96! boe!1/9!sftqfdujwfmz-!efufsnjof!uif!tubsujoh!dvssfou!esbxo!cz!uif!npups!boe!sbujp!pg!tubsujoh!up! gvmm.mpbe!dvssfou/

Solution Short-circuit line current of the motor at 200 V is 20 A. Hence, phase current of the motor is 20 A, i.e. 1155 A. 3 415 The phase voltage of the motor is = 239.6 V (when started by star–delta starter). 3 239.6 ˘ È A , i.e., 13.83 A. Hence, the starting current drawn by the motor is Í11.55 ¥ 200 ˙˚ Î At full-load condition, the motor is delta connected. 40000 Hence, full-load line current is = 8.184 A 3 ¥ 415 ¥ 0.85 ¥ 0.8 13.83 \ the ratio of starting to full-load current = 1.689. 8.184

Uisff.qibtf!Joevdujpo!Npups

9/56

! Qspcmfn!9/49 Uif!gpmmpxjoh!bsf!uif!qbsbnfufst!pg!uif!frvjwbmfou!djsdvju!pg!b!526!W-!uisff.qibtf-!tjy.qpmf!tubs. dpoofdufe!joevdujpo!npups; Tubups!jnqfebodf!>!)1/3!,!k!1/6* W Nbhofuj{joh!sfbdubodf!>!36!W Dpsf!mptt!sftjtubodf!>!261!W Frvjwbmfou!spups!jnqfebodf!sfgfssfe!up!uif!tubups!>!)1/4!,!k!1/8*!W Efufsnjof!uif!tubups!dvssfou-!spups!dvssfou-!nfdibojdbm!qpxfs!pvuqvu!boe!joqvu!qpxfs!bu!tmjq!pg!5&! vtjoh!uif!fybdu!frvjwbmfou!djsdvju/

Solution The per-phase exact equivalent circuit of the motor is shown in Fig. 8.5. Here,

\

R1 + jX1 Xm Rm R2¢ + jX2¢ s

= (0.2 + j 0.5) W = 25 W = 150 W; s = 0.04 0.3 = + j 0.7 = (7.5 + j 0.7) W 0.04

The parallel combination of Rm and Xm gives Zm =

Rm ( jX m ) 150 ( j 25) j 150 = = W Rm + jX m 150 + j 25 6 + j

È R¢ ˘ However, Zo is in parallel with Í 2 + jX 2¢ ˙ Î s ˚ È R¢ ˘ Z m Í 2 + jX 2¢ ˙ s Î ˚ Hence, total input impedance is Z = Z1 + È R2¢ ˘ Z m + Í + jX 2¢ ˙ Î s ˚ \

j 150 (7.5 + j 0.7) 6+ j Z = (0.2 ¥ j 0.5) + j 150 + (7.5 + j 0.7) 6+ j = (0.2 + j 0.5) +

24.67 –80.54∞ ¥ 7.53 –5.33∞ 24.67 –80.54∞ + 7.5 + j 0.7

= (0.2 + j 0.5) +

185.765 –85.87∞ 11.55 + j 25.03

!

9/57

Fmfdusjdbm!Nbdijoft

= (0.2 + j 0.5) +

185.765 –85.87∞ 27.57 –65.23∞

= 0.2 + j 0.5 + 6.74 – 20.64° = (6.51 + j 2.87) W. \ stator current I1 is obtained as 415 –0∞ E1 239.6 –0∞ 3 = = 33.7 ––23.79° A I1 = = Z 6.51 + j 2.87 7.11 – 23.79∞ Voltage across the magnetizing branch is obtained as E¢2 = E1 – I1(R1 + jX2) = 239.6 –0° – 33.7 ––23.79° (0.2 + j 0.5) = 239.6 –0° – 18.13 –44.4° = 226.65 – j12.68 = 227 ––3.2° V \ current through Rm is, Current through Xm is,

227 – - 3.2∞ = 1.51 ––3.2° A 150 227 – - 3.2 Im = = 9.08 ––93.2° A j 2.5

IC =

Hence, no-load current Im = 1.51 ––3.2° + 9.08 ––93.2° = (1.5 – 0.5) + j (–0.08 – 9.06) = 1 – j 9.146 = 9.2 ––83.76° A Also, the rotor current referred to stator is I 2¢ = I1 – Im = 33.7 ––23.79° – 9.2 ––83.76° = (30.8 – 1) + j (–13.59 + 9.145) = 29.8 – j 4.445 = 30.13 ––8.48° A Per-phase mechanical power output is Ê1 - sˆ Ê 1 - 0.04 ˆ I¢2 R¢2 Á = (30.13)2 ¥ 0.3 Á ˜ Ë s ¯ Ë 0.04 ˜¯ = 907.82 ¥ 0.3 ¥ 24 = 6536.3 W Per-phase input power is (3E1 I1 cos q), i.e.,

Pin = 3 ¥ 239.6 ¥ 33.7 cos 23.79° = 22165.286 W = 22.165 kW.

Uisff.qibtf!Joevdujpo!Npups

9/58

! Qspcmfn!9/4: B!uisff.qibtf-!61!I{-!211!lX!joevdujpo!npups!ibt!b!gvmm.mpbe!fggjdjfodz!pg!96&/!Uif!tubups!dpqqfs! mptt!boe!spups!dpqqfs!mptt!bsf!fbdi!frvbm!up!uif!tubups!dpsf!mptt!bu!gvmm.mpbe/!Uif!nfdibojdbm!mptt! jt!frvbm!up!pof!gpvsui!pg!uif!spups!dpqqfs!mptt/!Dbmdvmbuf!)b*!uif!spups!dpqqfs!mptt-!)c*!uif!bjs!hbq! qpxfs-!boe!)d*!uif!tmjq/

Solution 100 kW = 117.65 kW 0.85 Total loss = Input – Output = (117.65 – 100) kW = 17.65 kW Let stator copper loss = Rotor copper loss = Stator core loss = P 1 P Mechanical loss = Rotor copper loss = 4 4 Now, Total loss = Stator core loss + Stator copper loss + Rotor copper los + Mechanical loss P 13 P =P+P+P+ = 4 4 13 P = 17.65 Hence, 4 or P = 5.43 Input power of the motor is

\ rotor copper loss is 5.43 kW. Air gap power = Input power – Stator core loss – Stator copper loss = 117.65 – 5.43 – 5.43 = 106.788 kW But, Rotor copper loss = Slip ¥ Air gap power 5.43 Hence, slip = = 0.05. 106.788

! Qspcmfn!9/51 Bo!fjhiu.qpmf-!61!I{-!uisff.qibtf!joevdujpo!npups!ibt!b!gvmm.mpbe!upsrvf!pg!311!On!xifo!uif!gsf. rvfodz!pg!uif!spups!fng!jt!3/6!I{/!Jg!uif!nfdibojdbm!mptt!jt!26!On-!efufsnjof!uif!spups!dpqqfs!mptt! boe!uif!fggjdjfodz!pg!uif!npups/!Uif!upubm!tubups!mptt!jt!2111!X/

Solution If s be the slip at full-load then s.f = 2.5 (from the given data) 2.5 \ s= = 0.05 50 The speed of the motor at full-load N = (1 – s) Ns

!

9/59

Fmfdusjdbm!Nbdijoft

But synchronous speed

Ns =

120 f 120 ¥ 50 = = 750 rpm P 8

Hence,

N = (1 – 0.05) 750 = 713 rpm 713 ¥ 2p rpm = 74.665 rad/s or w= 60 Mechanical power developed by the rotor Pm = (200 + 15) Nm = 215 Nm = 215 ¥ 74.665 = 16052.975 W

[E P = ws ¥ T]

If Pag be the air-gap power then, (1 – s) Pag = Pm or

Pag =

16.053 kW = 16.898 kW 1 - 0.05

Rotor copper loss is (sPag) = 0.05 ¥ 16.898 = 0.845 kW Motor input = 16.052 + 0.845 + 1 = 17.897 kW Motor output = 200 ¥ 74.665 W = 14.933 kW 14.933 Hence, efficiency is = 0.8344 or 83.44%. 17.897

! Qspcmfn!9/52 B!tjy.qpmf-!uisff.qibtf!joevdujpo!npups!efwfmpqt!36!IQ!jodmvejoh!4!IQ!nfdibojdbm!mpttft!bu!b!tqffe! pg!:71!sqn!xifo!dpoofdufe!up!551!W-!uisff.qibtf!nbjot/!Uif!qpxfs!gbdups!jt!1/9/!Gjoe!)b*!uif!tmjq-! )c*!uif!spups!dpqqfs!mptt-!)d*!uif!upubm!joqvu!jg!tubups!mptt!jt!4!lX-!boe!)e*!uif!fggjdjfodz/

Solution Synchronous speed Speed of the motor

120 ¥ 50 = 1000 rpm 6 N = 960 rpm

Ns =

960 ˆ Ê (a) Slip = Á1 = 0.04 Ë 1000 ˜¯ (b) Gross mechanical power developed is 35 ¥ 735.5 W = 25742.5 W or \ air-gap power

Pm = 25742.5 W = 25.742 kW P 25.742 (Pag) = m = = 26.81 kW 1 - s 1 - 0.04

Hence, rotor copper loss is sPag, i.e. 0.04 ¥ 26.81 or 1.072 kW.

Uisff.qibtf!Joevdujpo!Npups

9/5:

(c) Stator loss is 3 kW (given). Hence, total input = 26.81 + 3 = 29.81 kW (d) Input = 29.81 kW \ Output = 35 – 3 = 32 HP = 32 ¥ 0.7355 = 23.536 kW 23.536 ¥ 100% = 78.95%. \ Efficiency = 29.81

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Solution The equivalent circuit of the induction motor is shown in Fig. 8.5. The per-phase applied stator voltage E1 =

440

V = 254 V 3 s = 0.05 (given)

Slip

Ê 0.18 ˆ j 30 Á + j 1.5˜ Ë 0.05 ¯ Neglecting Rm total impedance from input = (0.2 + j1) + W 0.18 j 30 + + j 1.5 0.05 -45 + j 108 W = 0.2 + j + 3.6 + j 31.5 117 –112.62∞ = 0.2 + j + 31.7 –83.48∞ = 0.2 + j + 3.69 – 29.14° W = 1.073 + j 1.487 + 1.834 – 54.186° W \ stator current

I1 =

254 = 138.49 – –54.186° 1.834 –54.186∞

Stator input power = 3 V1 I1 cos 54.186° = 3 ¥ 440 ¥ 138.49 cos 54.186° = 61.757 kW

!

9/61

Fmfdusjdbm!Nbdijoft

Air-gap power Pag = 61.757 –

3(138.49) 2 ¥ 0.2 = 50.25 kW 1000

[\ Pag = Stator input power – Stator copper loss, core loss being neglected in stator] Mechanical power developed Pm = (1 – s) Pg = (1 – 0.05) 50.25 = 47.737 kW 500 ˆ Ê Pout = Á 47.737 - 3 ˜ = 47.237 kW Ë 10 ¯

Power output

Output torque Tout =

Tout =

47.237 ¥ 103 Nm. [where N (rpm) is the speed of the motor, 2p ¥ N 60

Pout N 120 ¥ 50 , ws = 2pns, ns = = 712.5 rpm] rpm; N = (1 – 0.05) ¥ ws 60 8

Hence, output torque = Efficiency =

47.237 ¥ 103 ¥ 60 = 633.40 Nm 2p ¥ 712.5

Power output 47.237 = ¥ 100% = 76.48%. Power input 61.757

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Solution Synchronous speed Slip

120 f 120 ¥ 50 = = 750 rpm P 8 È 720 ˘ (s) = Í1 ˙ = 0.04 Î 750 ˚ Ns =

If R2 be the rotor resistance then the rotor copper loss is I22 R2, where I2 is the rotor current. If Pag be the air-gap power or power input to the rotor then sPag = I22 R2 or

0.04 =

I 22 R2 0.5 I 22 = Pag Pag

(i)

Uisff.qibtf!Joevdujpo!Npups

The new speed

9/62

N = 680 rpm 680 ˆ Ê = 0.093 s = Á1 Ë 750 ˜¯

\

(ii)

Let R be the total resistance of the rotor circuit. In order that the full-load torque remains same, Pag should have the same value as the previous one. Hence,

0.093 =

I 22 R Pag

From equations (i) and (ii), 0.04 . R 0.5 R = 1.1625 W

0.093 = or

Hence, the external resistance to be connected is (R – R2) or (1.1625 – 0.5), i.e., 0.6625 W per phase.

UPSRVF!FYQSFTTJPO!PG!JOEVDUJPO!NPUPS!

9/24

From Eq. (8.36) and Eq. (8.46), the electromagnetic torque developed by the induction motor Te =

Pag ws

=

3 R2¢ V12 2 ÏÔÊ ¸Ô R¢ ˆ w s s ÌÁ R1 + 2 ˜ + ( X 1 + X 2¢ ) 2 ˝ Ë a¯ ÓÔ ˛Ô

Nm

(8.49)

when the circuit parameters are referred to the stator side. From Eq. (8.48), Te =

3 R2 sEr2 w s ( R22 + s 2 X 22 )

(8.50)

when the circuit parameters are referred to the rotor side.

9/24/2! Tubsujoh!Upsrvf!)Ut* The expression for starting torque is obtained from Eq. (8.49) putting s = 1 \

Ts =

V12 3 R¢2 Nm ws ( R1 + R2¢ ) 2 + ( X 1 + X 2¢ ) 2

(8.51)

!

9/63

Fmfdusjdbm!Nbdijoft

From Eq. (8.50), Ts =

E2 3 R2 2 r 2 Nm ws R2 + X 2

(8.52)

9/24/3! Fggfdu!pg!Dibohf!pg!Tvqqmz!Wpmubhf!po!Upsrvf!boe!Tmjq! From Eq. (8.49), Te =

V12 s 3 R¢2 ws ( R1s + R2¢ ) 2 + ( sX 1 + sX 2¢ ) 2

As R1s 1 R¢2 and sX1 and sX 2¢ are negligible for low values of s.

or,

Te =

V2 s 3 R¢2 1 2 ws R2¢

Te μ

V12 s R2¢

(8.53)

Therefore, torque under normal conditions is proportional to the square of the supply voltage. With drop in supply voltage, running torque T decreases and to maintain the same torque, slip must be increased.

9/24/4! Dpoejujpo!gps!Nbyjnvn!Upsrvf!)Unby* To get the slip for maximum torque condition, we differentiate Eq. (8.49) with respect to slip. Since the numerator of this equation is independent of s, we differentiate the denominator w.r.t. s. d ds

\

2 ¸Ô ÏÔÊ R2¢ ˆ s ÌÁ R1 + ˜ + ( X 1 + X 2¢ ) 2 ˝ = 0 Ë s ¯ ˛Ô ÓÔ

or,

Ï Ê R¢ ˆ Ê R¢ ˆ ¸ Ê R¢ ˆ s Ì2 Á R1 + 22 ˜ Á - 22 ˜ ˝ + Á R1 + 2 ˜ + (X1 + X2¢)2 = 0 s ¯ s ¯ Ë s ¯˛ Ë Ó Ë

or,

–2

or, or,

R1R2¢ R¢ R¢ 2 2 R R¢ - 2 22 + R12 + 22 + 1 2 + (X1 + X2¢)2 = 0 s s s s R12 –

R2¢ 2 s

2

+ (X1 + X2¢)2 = 0 R2¢ 2 s

2

= R12 + (X1 + X 2¢ )2 = 0

Uisff.qibtf!Joevdujpo!Npups

9/64

\ slip at maximum torque R2¢

smax =

R12

(8.54)

+ ( X1 + X 2¢ ) 2

Neglecting the stator impedance, R2¢ R = 2 X 2¢ X 2

smax =

(8.55)

Thus, the slip at maximum torque is the ratio of rotor resistance and standstill rotor reactance. Substituting the value of smax in Eq. (8.49), the expression for maximum torque 3 R2¢ V12

Tmax = ws

=

=

=

=

ÏÊ R2¢ R12 + ( X1 + X 2¢ ) 2 Ô + R Á Ì 1 R2¢ R12 + ( X1 + X 2¢ ) 2 ÔÁË Ó R2¢

2 ¸ ˆ 2Ô ˜ + ( X 1 + X 2¢ ) ˝ ˜¯ Ô ˛

R12 + ( X 1 + X 2¢ ) 2

3 V12

w s {R12 + R12 + ( X1 + X 2¢ ) 2 + 2 R1 R12 + ( X 1 + X 2¢ ) 2 + ( X 1 + X 2¢ ) 2 } 3 V12

R12 + ( X 1 + X 2¢ ) 2

w s {2 R12 + 2 ( X 1 + X 2¢ ) 2 + 2 R1 R12 + ( X 1 + X 2¢ ) 2 } 3 V12

R12 + ( X 1 + X 2¢ ) 2

2 w s R12 + ( X 1 + X 21 ) 2 { R12 + ( X 1 + X 2¢ ) 2 + R1}

(

3 V12

2 w s R1 + R12 + ( X 1 + X 2¢ ) 2

)

(8.56)

Neglecting R1 and X1, 3 V12 Tmax = 2 w s X 2¢

(8.57)

Again maximum torque from Eq. (8.50), 3 R2 Er2 Tmax =

R2 X2

( )

w s 2 R22

=

3 Er2 2w s X 2

(8.58)

!

9/65

or,

Tmax =

where \

Fmfdusjdbm!Nbdijoft

K= Tmax μ

K Er2 X2 3 2 ws 1 X2

Thus, the maximum torque of an induction motor operating on constant applied voltage and constant supply frequency is inversely proportional to the standstill rotor reactance X2 but independent of rotor circuit resistance R2. From Eq. (8.50) and Eq. (8.58), we have Te 2s X R 2s 2s = 2 22 2 2 = = 2 R2 X Tmax R2 + s X 2 + s 2 2 smax + s X2 R2 smax 2 = smax s + s smax

(8.59)

Again from Eq. (8.59), putting s = 1, Ts = Tmax

2 smax +

1

(8.60)

smax

For squirrel-cage induction motor since the rotor is permanently short circuited, no resistance can be inserted in its rotor circuit. Thus, slip for maximum torque is constant and its value is considerably varied. Since external resistance can be inserted into the rotor circuit in a slip-ring induction motor, its slip can be varied for obtaining maximum torque. For rotor circuit without external resistances, R smax (1) = 2 X2 and if a resistance r per phase is inserted in the rotor circuit, smax (2) =

R2 + r X2

9/24/5! Dpoejujpo!gps!Nbyjnvn!Tubsujoh!Upsrvf! We obtain the condition for maximum starting torque by differentiating Eq. (8.52) with respect to R2.

Uisff.qibtf!Joevdujpo!Npups

\

d Ts d Ê 3 R2 Er2 ˆ =0 = dR2 dR2 ÁË w s ( R22 + X 22 ) ˜¯

or,

d Ê R2 ˆ =0 dR2 ÁË R22 + X 22 ˜¯

9/66

R2 (2 R2) – R22 – X22 = 0 R22 = X22 R2 = X2

or,

(8.61)

\ for a given applied voltage, the starting torque is maximum when the resistance of rotor R2 equals its reactance X2.

UPSRVFÐTMJQ!DIBSBDUFSJTUJDT!PG!B! UISFF.QIBTF!JOEVDUJPO!NPUPS

9/25!

From Eq. (8.50), we can write Tμ

s R22

+ X 22 s 2

At synchronous speed, slip s is zero, hence, torque T is zero. Consequently, the torque slip curve starts from origin (i.e. s = 0), and ends at s = 1. Dbtf!Tuvez!2 s When s (slip) is very low (at rotor speeds close to synchronous speed), sX2 1 R2 and T μ 2 (at R2 low slips). i.e. torque-slip curve at low values of slip is a straight line passing through the origin, and torque is R maximum when s = 2 (from Eq. 8.55). X2 Dbtf!Tuvez!3 When the load on the motor increases, the speed of the motor decreases. When slip s is large, compared to R2, sX2 is much large and hence, sX2 1 R2. Tμ

s ( sX 2 ) 2

μ

1 sX 22

μ

1 (at high slips) s

!

9/67

Fmfdusjdbm!Nbdijoft

i.e. the torque T slip s curve for larger values of slip is approximately a rectangular hyperbola. Consequently, any further increase in motor load, beyond the print of maximum torque, results in decrease of the torque developed by the motor. Eventually, the motor slows down. The maximum torque developed in an induction motor is called the pull-out torque or breakdown torque. This torque is a measure of the short-time overloading capability of the motor. Figure 8.11 shows the torque-slip characteristics of an induction motor operating with constant applied voltage and constant frequency. R +r Smax(3) = 2 2 X2

R +r Smax(2) = 2 1 X2 S

max(1) =

Tmax

R2 X2

Torque (T)

1 2

Max. torque

3

(s = 0) (N = Ns)

O s = 1(N = 0) Slip Speed

[r2 > r1]

Gjh/!9/22! Upsrvf.tmjq!dibsbdufsjtujdt!pg!bo!joevdujpo!npups

Dvswf!2! represents (T-s) characteristic of an induction motor having low rotor resistance or when no resistance is inserted in the rotor circuit. R Maximum torque is developed at Smax (1) = 2 . X2 Dvswf!3! represents the (T-s) characteristic of an induction motor. When an external resistance of r1 W/phase is inserted in the rotor circuit, the magnitude of the maximum torque remains unchanged, but the slip for maximum torque in Smax (2) = (R2 + r1)/X2. Dvswf!4! represents the (T-s) characteristic of an induction motor, when an external resistance of

r2 W/phase is inserted in the rotor circuit such that R2 + r2 = X2, a condition for maximum torque at starting. It may be noted here that (R2 + r2) > (R2 + r1) > R2 It is also seen that as the rotor resistance is increased, the pull-out speed of the motor decreases, but the maximum torque remains constant. However, for squirrel-cage rotors, it is not possible to insert any rotor resistance under normal operating conditions and hence, it is not easily possible to enhance the value of the starting or maximum torque for a squirrel-cage induction motor.

Uisff.qibtf!Joevdujpo!Npups

UPSRVF!DPOEJUJPO!EVSJOH!MPBEJOH-! CSFBLEPXO!BOE!OP.MPBE

9/68

9/26

Assume the induction motor is operating initially at rated load and then subsequently mechanical load increases. For this condition, the load torque becomes greater than the developed torque, and the motor slows down. The resultant increase in slip causes an increase in the developed torque. If the new developed torque (caused by the increase in slip) equals the load torque on the shaft plus windage, friction, and stray load, the motor will then operate at the steady-state speed and at a higher slip. Further increase in shaft load (overload) causes additional deceleration, accompanied by increase in input power and thus, increase in developed torque. If, however, the load torque on the shaft is increased to a value greater than the maximum torque that the machine can develop, the machine will “break down”. Increase in slip, due to increase in shaft load above the breakdown value, causes a rapid decrease in speed and may even stall the motor. There will be a very high input current (similar to short circuit of secondary of transformer) and this will burn out the motor windings unless protective devices remove the machine from the line. The breakdown torque is then defined as the maximum torque that a motor can develop while being loaded (at rated voltage and rated frequency) without suffering an abrupt drop in speed. Although an induction motor can be operated momentarily at overloads up to the breakdown point, it cannot do so continuously without overheating and causing severe damage, to both stator and rotor. If a sustained overload occurs, to prevent damage, a motor control circuits using overload relays and/or solid state devices should be used to trip the machine from the line. If there is no-load on the shaft, the rotor will run near synchronous speed and the rotor current will be very low and near zero value. Under such conditions, the line current drawn by the stator will only be enough to produce the rotating magnetic field and supply the friction, windage and iron losses. Thus, in a way, the no-load current drawn by the stator of an induction motor is similar to the exciting current of a transformer that supplies only the transformer flux and iron losses. The complete equivalent circuit of an induction motor, including stator and rotor windings, is similar to the equivalent circuit of a transformer. Neglecting the induction motor exciting current, the stator (primary) current will be directly proportional to the rotor (secondary) current. Increasing the shaft load increases the rotor current, causing a proportional increase in stator current (Istator μ Irotor).

9/26/2! Qbsbtjujd!Upsrvft The periodic variation of magnetic-circuit reluctance, caused by rotor and stator slots, results in a nonsinusoidal space distribution of the rotating flux. Analysis of this rotating flux pattern shows it to consist of a number of rotating fields of different speeds called space harmonics. The first harmonic, called the fundamental, runs at a speed corresponding to the number of poles in the actual winding. The fifth space harmonic rotates backward at one-fifth the speed of the fundamental, the

!

9/69

Fmfdusjdbm!Nbdijoft

seventh space harmonics rotates forward one-seventh the speed of the fundamental, and so forth. There are no even space harmonics and no third harmonics or its multiples. Although the fundamental dominates, the component torques produced by the fifth and seventh harmonics, called parasitic torques or harmonic torques, can cause undesirable bumps and dips in the motor torquespeed characteristic during acceleration, and may even cause the rotor to lock in at some subsynchronous speed and “crawl”. Figure 8.12 shows the effect of parasitic torques on the torque-speed characteristic. The presence of significant dips in the torque-speed characteristic of an induction motor may indicate a defective design, a damaged rotor, or improper repair of a damaged stator.

Developed Torque

Breakdown

Pull up Locked rotor

O

Speed

100%

Gjh/!9/23! Fggfdu!pg!qbsbtjujd!upsrvft!po!uif!upsrvf.tqffe!dibsbdufsjtujd!pg!bo!joevdujpo!npups

9/26/3! Qvmm.vq!Upsrvf! The pull-up torque of an induction motor is the minimum torque developed by the motor during the period of acceleration from rest to the speed at which breakdown torque occurs. For the torquespeed characteristic, the pull-up torque is the value of torque at the bottom of the dip caused by a parasitic torque. If the pull-up torque is less than the load torque on the shaft, the motor will not accelerate past the pull-up point.

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Uisff.qibtf!Joevdujpo!Npups

9/6:

Solution Ns =

120 f 120 ¥ 50 = = 1500 rpm P 4

smax =

For maximum torque slip,

R2 0.020 = = 0.04 X2 0.5

\ speed at maximum torque = Ns (1 – smax) = 1500 (1 – 0.04) = 1440 rpm.

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Solution Given:

f = 50 Hz, P = 8, s = 0.015, R2 = 0.001 W; X2 = 0.005 W, smax =

Now

T Tmax T

\

Tmax Now,

=

=

Ns =

R2 0.001 = 0.2 = X 2 0.005 2 ◊ s ◊ smax 2 s 2 + smax

=

4.02 ¥ 10 -2 6 ¥ 10 -3

2 ¥ 0.015 ¥ 0.2 (0.015) 2 + (0.2) 2

=

6 ¥ 10 -3 4.02 ¥ 10 -2

= 6.7

120 f 120 ¥ 50 = = 750 rpm P 8

\ speed at maximum torque = Ns (1 – smax) = 750(1 – 0.2) = 600 rpm.

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Solution Slip

(s) = 0.04 smax =

R2 0.01 = = 0.2 X 2 0.05

!

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Fmfdusjdbm!Nbdijoft

\ speed at maximum torque = Ns (1 – smax) = (1 – 0.2) ¥ 1000 = 800 rpm \

2 Tmax s 2 + smax (0.04) 2 + (0.2) 2 = = = 2.6 2 ◊ s ◊ smax 2 ¥ 0.04 ¥ 0.2 T

or

Tmax = 2.6 ¥ T

[T is full-load torque].

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Solution The synchronous speed Ns =

120 f 120 ¥ 50 = = 1500 rpm P 4

Speed at maximum torque = 1400 rpm (given) N s - Speed at maximum torque Ns 1500 - 1400 = = 0.067 1500

\ slip (smax) at maximum torque =

smax =

Also, \

X2 =

R2 X2 R2 0.25 = = 3.73 smax 0.067

From Eq. (8.60), Ts = Tmax \ or,

smax +

2 smax +

1 smax

1

=4 smax smax = 0.2679

If r be the external resistance inserted then

\

0.25 + r = 0.2679 3.73 r = 0.75 W.

=

1 2

Uisff.qibtf!Joevdujpo!Npups

9/72

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Solution Synchronous speed

Ns =

120 f 120 ¥ 50 = = 250 rpm P 24

\ slip

(s) =

N s - N 250 - 247 = 0.012 = Ns 250

Also,

smax =

R2 0.016 = 0.059 = X 2 0.270

T

2 ◊ smax ◊ s

We know,

Tmax T

Here,

Tmax T

or

Tmax

= =

=

2 s 2 + smax

2 ¥ 0.059 ¥ 0.012 (0.012) 2 + (0.059) 2 (0.012) 2 + (0.059) 2 = 2.56 2 ¥ 0.059 ¥ 0.012

Let N¢ be the intended speed at maximum torque. Then,

smax =

N s - N ¢ 250 - N ¢ = Ns 250

smax = 0.059 from calculations we have got earlier. 250 - N ¢ 250 N¢ = 235.25 rpm.

i.e.,

0.059 =

or

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Solution Given, f = 50 Hz, P = 6

!

9/73

Fmfdusjdbm!Nbdijoft

\

Ns =

Slip at maximum torque =

120 f 120 ¥ 50 = = 1000 rpm P 6

N s - Speed maximum torque 1000 - 960 = 0.04 (= smax) = Ns 1000 smax =

Also, \

X2 =

If T is the torque at slip s,

T Tmax

=

2 s ◊ smax 2 s 2 + smax

R2 X2 R2 0.6 = = 15 W smax 0.04

.

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Solution Ns =

120 f 120 ¥ 50 = 375 rpm = P 16

Speed at full-load = 350 rpm 375 - 350 = 0.06 375 Slip at maximum torque

\ slip at full-load =

smax =

R2 0.02 2 = = = 0.133 X 2 0.15 15

2ˆ Ê (a) Speed at which maximum torque occurs = (1 – smax) Ns = Á1 - ˜ ¥ 375 = 325 rpm Ë 15 ¯ 2 Ê 2ˆ (0.06) 2 + Á ˜ 2 Ë 15 ¯ + s2 smax T (b) = = 1.33 = 2 2 s ◊ smax Tmax 2 ¥ 0.06 ¥ 15

Uisff.qibtf!Joevdujpo!Npups

9/74

(c) Let the external resistance per phase added to the rotor circuit be ‘r’ W, so that rotor resistance per phase, R2 = (0.02 + r). The starting torque will be maximum when R2 = X2 \ or

0.02 + r = 0.15 r = 0.13 W per phase.

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Solution smax =

R2 0.05 = = 0.167 X2 0.3

i.e., at 16.7% slip, torque will be maximum. The full-load slip is obtained as sf l =

N s - N 600 - 585 = 0.025 i.e., 2.5% = Ns 600 [\ Ns =

120 f 120 ¥ 50 = = 600 rpm; N = 585 rpm] P 10

In Eq. 8.59, the ratio of full-load torque and maximum torque has been obtained. T Tmax

=

2 ◊ s ◊ smax 2 s 2 + smax

For the given problem, we find T Tmax

=

(0.025) 2 + (0.167) 2 = 3.414. 2 ¥ 0.025 ¥ 0.167

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Solution Synchronous speed

Ns =

120 ¥ 50 = 1000 rpm 6

!

9/75

Fmfdusjdbm!Nbdijoft

Slip

s =1–

900 = 0.1 1000

Torque is maximum when slip smax =

R2 = 0.1 X2

From Eq. (8.59), Te 2 = smax s Tmax + s smax where \

s = 0.04 Te = 500 ¥

= 500 ¥

2 0.1 0.04 + 0.04 0.1 2 = 96.154 Nm 10 + 0.4

Rotor resistance R2 per phase is 0.5 W. If X2 be the rotor reactance per phase R2 = smax = 0.1 X2 R 0.5 X2 = 2 = = 5 W. 0.1 0.1

\

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Solution Stator voltage per phase E1 = 440 V 200 Rotor voltage per phase Er = = 115.47 V 3

Uisff.qibtf!Joevdujpo!Npups

9/76

\ effective turns ratio a=

E1 440 = 3.81 = Er 115.47

Rotor resistance and standstill reactance referred to the stator R¢2 = 0.08 ¥ (3.81)2 = 1.1613 W X¢2 = 0.5 ¥ (3.81)2 = 7.258 W

and (a) Slip at maximum torque

smax =

R2¢ R12 + ( X 1 + X 2¢ )2

=

1.1613 1 + (3 + 7.258)2

= 0.1127 (b) Maximum torque Tmax =

=

=

3 V12

(

2 w s R1 + R12 + ( X 1 + X 2¢ ) 2

)

3 ¥ ( 440) 2 ¥ 50 2 ¥ 2p ¥ 1 + 12 + (3 + 7.258) 2 4 2

(

)

3 ¥ ( 440) 2 = 163.6 Nm 4p ¥ 25 ¥ 11.306

(c) Full-load torque Tf l =

1 Tmax = 54.53 Nm 3

From Eq. (8.49), 3 R2¢ V12

or,

or,

= 54.53 2 ÏÔÊ ¸ R2¢ ˆ 2Ô w s s ÌÁ R1 + ˜ + ( X 1 + X 2¢ ) ˝ Ë s ¯ ÓÔ ˛Ô 2 3 ¥ 1.1613 ( 440) = 54.53 2 ¸ 2 ¥ 50 ÏÔÊ 1.1613 ˆ 2Ô s ÌÁ1 + + (3 + 7.258) ˝ s ˜¯ 4 ÔÓË Ô˛ 2 ÔÏÊ 1.1613 ˆ Ô¸ s ÌÁ1 + + 105.226 ˝ = 496.76 ˜ s ¯ ÔÓË Ô˛

!

9/77

or, or, or,

Fmfdusjdbm!Nbdijoft

1.3486 + 2.3226 + 105.226s = 494.76 s 106.226s2 – 492.4374s + 1.3486 = 0 s2 – 4.6357s + 0.01269 = 0 s+

\ s = 0.00274 (neglecting the other value of s which is not feasible) \ full-load speed Nf l = (1 – 0.00274)

120 ¥ 50 = 1496 rpm 4

Power output 1496 Tf l 60 2p ¥ 1496 ¥ 54.53 = 60 = 8538.38 W.

Po = 2p

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Solution Slip at maximum torque smax = Synchronous speed

Ns =

R2 = 0.08 X 2¢

120 ¥ 50 = 1500 rpm 4

The slip of the motor \ speed The new speed \

s = 0.04 Nr = (1 – 0.04) 1500 = 1440 rpm Nr¢ = 1400 rpm 1400 New slip = 1 – = 0.0667 1500

As torque is proportional to the square of the speed, 2

T ¢ Ê N r¢ ˆ Ê 1400 ˆ = = T ÁË N r ˜¯ ÁË 1440 ˜¯

2

Uisff.qibtf!Joevdujpo!Npups

9/78

Now V12 3 neglecting stator impedance s w s Ê R2¢ ˆ 2 ÁË s ˜¯ + X 2¢

T=

If the new supply voltage is V¢1 then 2

2

Ê R2¢ ˆ 2 ÁË 0.04 ˜¯ + X 2¢

Ê V1¢ ˆ Ê 1400 ˆ =Á ÁË 400 ˜¯ 2 Ë 1440 ˜¯ Ê R2¢ ˆ 2 ÁË 0.0667 ˜¯ + X 2¢

2

2

Ê R2¢ ˆ Ê R2¢ ˆ ÁË 0.04 ˜¯ + ÁË 0.08 ˜¯

or,

V1¢ 2 Ê 140 ˆ ¥ =Á 2 2 Ë 144 ˜¯ 160000 Ê R2¢ ˆ Ê R2¢ ˆ ÁË 0.06667 ˜¯ + ÁË 0.08 ˜¯

2

V1¢ 2 781.25 Ê 140 ˆ ¥ = 160000 381.025 ÁË 144 ˜¯

2

or, \

V1¢ = 271.58 V.

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Solution Synchronous speed \

120 ¥ 50 = 1500 rpm 4 R¢ R 1440 =1– = 0.04 = 2 = 2 X 2¢ X 2 1500

Ns = smax

Now slip s = 0.03 From (Eq. 8.59), T Tmax \

=

2 smax s + s smax

=

2 0.04 0.03 + 0.03 0.04

T = 0.96 ¥ 300 = 288 Nm.

= 0.96

!

9/79

Fmfdusjdbm!Nbdijoft

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Solution R2 = 0.3 W Tm = 100 Nm 700 700 smax = 1 – =1– = 0.067 120 ¥ 50 750 8 R2 = 0.067 X2 0.3 = 4.4776 X2 = 0.067

\ or, From Eq. (8.58),

3 Er2 = 100 2w s X 2 Er2 2 2 ¥ 100 ¥ 4.4776 = ¥ 100 ¥ X2 = ws 3 3

\

Now from Eq. (8.52), the starting torque Ts =

3( R2 + r ) Er2 ws ( R2 + r ) 2 + X 22

where r is the external resistance. \

w s {(0.3 + r ) + ( 4.4776) } 0.3 + r = R

Let \ or, or, \

3 (0.3 + r )Er2

Ts =

R R + ( 4.4776) 2

2

R R + 20.05 2

= 60 ¥

2

ws 3 Er2

= 0.067

0.067 R2 – R + 1.343 = 0 R = 1.49 W

\ external resistance r = 1.49 – 0.3 = 1.19 W.

=

2

=

20 ¥ 3 2 ¥ 100 ¥ 4.477

3 ¥ 100 = 60 5

Uisff.qibtf!Joevdujpo!Npups

9/7:

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Solution From Eq. (8.56), Tmax =

(a) Maximum torque

3 V12 2 w s ( R1 + R12 + ( X 1 + X 2¢ ) 2 2

Ê 440 ˆ 3¥Á Ë 3 ˜¯ = 2 ¥ 50 2 ¥ 2p ¥ (0.5) 2 + ( 2 + 1.5) 2 4 = 152.78 Nm

{

}

From Eq. (8.54), the slip at maximum torque smax =

R2¢ R12

+ ( X1 + X 2¢ )

2

=

0.4 (0.5) + ( 2 + 1.5) 2 2

= 0.113 (b) From Eq. (8.51), the starting torque Ts =

3 V12 R2¢ w s {( R1 + R2¢ ) 2 + ( X 1 + X 2¢ ) 2 } 2

Ê 440 ˆ 3¥Á ¥ 0.4 Ë 3 ˜¯ = 2 ¥ 50 2p ¥ {(0.5 + 0.4) 2 + ( 2 + 1.5) 2 } 4 = 37.77 Nm When the rotor resistance is doubled

\

smax = 2 ¥ 0.113 = 0.226 120 ¥ 50 Speed = (1 – 0.226) ¥ = 1161 Nm 4

!

9/81

Fmfdusjdbm!Nbdijoft

2

Ê 440 ˆ 3¥Á ¥ 0.4 ¥ 2 Ë 3 ˜¯ Ts = 2 ¥ 50 2p ¥ {(0.5 + 0.4 ¥ 2) 2 + ( 2 + 1.5)2 } 4 = 70.767 Nm.

New starting torque

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Solution Synchronous speed

Ns =

120 ¥ 50 = 1500 rpm 4

From Eq. (8.55), smax =

R2 0.5 = = 0.1 X2 5

\ speed at maximum torque Nr = (1 – 0.1) ¥ 1500 = 1350 rpm 1 smax + smax Tmax (from Eq. 8.60) = 2 Ts 1 0.1 + 0.1 = 5.05 = 2 1 Tmax then 2 1 smax ¢ + smax ¢ where s¢max is the new slip 2= 2

Ts =

If

\ or, \

s¢max +

1 =4 smax ¢

2 – 4 s¢max + 1 = 0 s¢max

s¢max =

4 ± 16 - 4 2

Uisff.qibtf!Joevdujpo!Npups

9/82

As s¢max cannot be greater than 1, s¢max = 0.268 If R¢2 be the new value of rotor resistance R¢2 = 0.268 X2 = 0.268 ¥ 5 = 1.34 W.

! Qspcmfn!9/6: B! gpvs.qpmf-! 61! I{-! uisff.qibtf! joevdujpo! npups! ibt! b! tubsujoh! upsrvf! pg! 28/9&! pg! uif! gvmm.mpbe! upsrvf! pg! boe! 246&! nbyjnvn! upsrvf! pg! uif! gvmm.mpbe! upsrvf/! Efufsnjof! uif! gvmm.mpbe! tqffe! boe! tqffe!bu!nbyjnvn!upsrvf/

Solution T 17.8 Starting torque = 0.178 = s = Full-load torque T 100 Maximum torque Tmax 135 = = 1.35 = Full-load torque 100 T Ts 0.178 = = 0.1318 Tmax 1.35

Hence,

Now, at slip s if the torque be T then we have

T = Tm

2 s smax

torque.

+

smax s

where sm is the slip at maximum

È ˘ Í ˙ 2◊s◊s T 2 ÍEFrom Eq. (8.59), we have ˙ = 2 2max = smax ˙ s Tm s + smax Í + Í smax s ˙˚ Î At starting s = 1, hence, from the above relations we can write, 2 Ts = = 0.1318 s 1 Tm + m sm

or or

0.1318 s2max – 2 smax + 0.1318 = 0 smax = 0.066 Synchronous speed (Ns) =

120 ¥ 50 = 1500 rpm 4

1

!

9/83

Fmfdusjdbm!Nbdijoft

If N1 be the speed at maximum torque then

N1 = (1 – smax)Ns = (1 – 0.066) 1500 = 1401 rpm T 2 = sfl sm Tmax + sm sfl 1 2 = sfl 0.066 1.35 + sfl 0.066 2 ¥ dfl ¥ 0.066 1 = 2 1.35 sfl + 0.004356

E

Here,

or

s 2fl – 0.1782 sf l + 0.004356 = 0 sf l = 0.029

or \

Hence, full-load speed is [(1 – 0.029) 1500] or, 1456.5 rpm.

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Solution 120 f 120 ¥ 50 = = 1500 rpm P 4 N = 1450 rpm (given) 1450 sf l (full-load slip) = 1 – = 0.033 (= s) (i.e. 3.3%) 1500 Ns =

\

smax (slip at maximum torque) = T

E

Tmax Here,

\ Also,

T

R2 0.05 = 0.1 (i.e. 10%) = X2 0.5

Full-load torque 2 ◊ s ◊ smax = 2 Maximum torque s 2 + smax 2 ¥ 0.033 ¥ 0.1 0.0066 = = = 0.595 0.011 0.0332 + 0.12 =

Tmax T = 1.68 Tmax Tstarting T

=

2 ◊ smax 1+

2 smax

=

2 ¥ 0.1 1 + 0.12

= 0.198

Uisff.qibtf!Joevdujpo!Npups

9/84

We have seen in the text that Ts = And

T=

KE12 R2 R22 + X 22 KE12 ◊ sR2 R22 + ( sX 2 ) 2

Ts KE 2 R R 2 + ( sX ) 2 = 2 1 22 ¥ 2 2 2 T R2 + X 2 KE1 sR2

\

=

R22 + ( sX 2 ) 2 ( R22

+

X 22 ) s

=

(0.05) 2 + (0.033 ¥ 0.5) 2 0.033(0.05 + 0.5 ) 2

2

=

0.0028 = 0.336. 0.00833

Ts /T = 0.336.

i.e.,

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Solution 120 ¥ 50 = 1500 rpm 4 Ê 1350 ˆ = 0.1 Slip at maximum torque smax = Á1 Ë 1500 ˜¯ Synchronous speed Ns =

If P2 and X2 be the resistance and reactance of the rotor circuit, R2 R 0.2 = smax or, X2 = 2 = =2W X2 smax 0.1 Writing the expression for maximum torque using Eq. 8.58, we get Tmax =

3 Er2 3 Er2 3 E2 ◊ = = ◊ r w s 2 X 2 w1 ¥ 2 ¥ 2 4 w r

(i)

(a) When slip is 4%, we can write, s = 0.04. From torque equation (10.13) we have, for threephase torque, E 2 ◊ s ◊ R2 3 T= ◊ 2r Nm (ii) w s [ R2 + ( sX 2 ) 2 ] 3 . 2 Er from Eq. (ii) by 4Tmax (as obtained in Eq. (i)) we get Replacing ws

!

9/85

Fmfdusjdbm!Nbdijoft

T=

4 Tmax ◊ s ◊ R2 [ R22

+ ( sX 2 ) ] 2

=

4 ¥ 15 ¥ 0.04 ¥ 0.2 0.22 + (0.04 ¥ 2) 2

= 10.34 Nm

(b) Starting torque Tst = 0.7 ¥ 15 = 10.5 Nm (as per question) and at starting, slip is 1. If R be the net rotor resistance after addition of external resistance then Ts =

3 E12 ◊ R w s ( R 2 + X 22 )

Nm [using Eq. (ii) with R2 replaced by R and s = 1]

i.e., i.e., i.e.,

3 E12 R R ◊ 2 = 4 Tmax . 2 2 ws R + X 2 R + 22 2 10.5 R + 42 = 60 R R = either 4.9 W or 0.816 W 10.5 =

With R = 4.9 W, smax =

R 4.9 = = 2.45, which is an impossible value. X2 2

\ feasible value of R is 0.816 W. Thus, external resistance to be added is (R – R2), i.e., (0.816 – 0.2) or, 0.616 W/phase.

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Solution Ns =

Synchronous speed

120 ¥ 50 = 1000 rpm 6

È Ê 900 ˆ ˘ smax = Í1 - Á ˜ ˙ = 0.1 Î Ë 1000 ¯ ˚ R2 = 0.25 (given)

At maximum torque slip

Hence,

X2 =

R2 0.25 = = 2.5 W smax 0.1

Torque at any slip s is given by T=

E 2 sR 3 ◊ 2 r 2 2 Nm w s [ R2 + ( sX 2 ) ]

3 Er2 3 0.05 ¥ 0.25 2 ◊E ¥ = 0.16 = w s r (0.25) 2 + (0.05 ¥ 2.5) 2 ws

Uisff.qibtf!Joevdujpo!Npups

9/86

Maximum torque is given by Tmax =

3 Er2 3 Er2 1 1 3E2 ◊ = ◊ = ◊ r ws 2 X 2 w s 2 ¥ 2.5 5 w s

Using this expression of (Tmax) in the expression of (T), we get T = 5 Tmax 0.6 As Tmax = 200 Nm, we get T = 5 ¥ 200 ¥ 0.16 = 160 Nm.

! Qspcmfn!9/74 B!uisff.qibtf!joevdujpo!npups!ibt!b!261&!tubsujoh!upsrvf!pg!gvmm.mpbe!boe!311&!nbyjnvn!upsrvf! pg!gvmm.mpbe/!Efufsnjof!uif!tmjq!bu!nbyjnvn!upsrvf-!gvmm.mpbe!tmjq!boe!spups!tubsujoh!dvssfou!bt!b! qfsdfoubhf!pg!gvmm.mpbe!dvssfou/

Solution We know that in any slip s, the developed torque can be expressed in terms of maximum torque, i.e. T 2 = s s Tmax + max smax s where Tmax is the maximum torque at slip smax. Starting torque (Ts ) = 1.5 Full-load torque (T ) Tmax =2 T Ts 1.5 3 = = = Tmax 2.0 4

Now, as well as Hence,

2 1 smax

.

+ smax

During starting, the slip is 1. \ or or

Ê 1 ˆ 8 + smax ˜ = = 2.67 ÁË s ¯ 3 max s 2max – 2.67 smax + 1 = 0 smax =

2.67 ± ( 2.67)2 - 4 2

= 0.45 (the other value is rejected as it is greater than 1)

!

9/87

Fmfdusjdbm!Nbdijoft

T

Also,

Tmax

2

=

s + max sfl

sfl smax

=

1 2

s 2fl – 4 ¥ 0.45 . sf l + (0.45)2 = 0 s 2fl – 1.8 sf l + 0.2025 = 0

or or

1.8 ± (1.8) 2 - 4(0.2025) = 0.12 sf l = 2

or

At full-load, rotor current may be obtained as E2 R2 + jX 2 sfl

I2f l =

[refer approximate equivalent circuit neglecting the magnetizing branch and stator impedance] 2 I 2f l=

or

E2 2

Ê R2 ˆ 2 ÁË s ˜¯ + X 2 fl

Similarly, starting current 2 = I 2st 2

Hence,

Ê I 2fl ˆ ÁË I ˜¯ = 2st

Now,

smax =

E22 R22 + X 22 R22 + X 22 2

Ê R2 ˆ 2 ÁË 0.12 ˜¯ + X 2 R2 = 0.45 X 2¢ 2

2

Hence,

Ê I 2fl ˆ ÁË I ˜¯ = 2 fl

Ê R2 ˆ ÁË X ˜¯ + 1 2 2

Ê R2 ˆ ÁË 0.12 X ˜¯ + 1 2

Hence,

I 2fl = 0.0798 = 0.28 I 2st

or

I 2st = 3.54. I 2fl

=

(0.45) 2 + 1 2

Ê 0.45 ˆ ÁË 0.12 ˜¯ + 1

= 0.0798

Uisff.qibtf!Joevdujpo!Npups

9/88

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Solution (a) When slip rings are directly shorted Starting current

Is =

V1 ( R1 + R2¢ ) 2 + ( X 1 + X 2¢ ) 2 440 / 3

=

(0.3 + 0.1) 2 + (0.6 + 0.5) 2 = 217.04 A

Starting torque

Ts =

3 I s2 R2¢ ws

3 ¥ ( 217.04) 2 ¥ 0.1 Nm 2 ¥ 50 2p ¥ 8 = 180.02 Nm

=

(b) When external rotor resistance of 0.5 W is inserted Is =

440 / 3 (0.3 + 0.1 + 0.5) 2 + (0.6 + 0.5) 2

= 178.74 A Starting torque 3 ¥ (178.74) 2 ¥ (0.1 + 0.5) Nm 2 ¥ 50 2p ¥ 8 = 732.565 Nm.

Ts =

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!

9/89

Fmfdusjdbm!Nbdijoft

Solution Slip

s=1–

1440 = 0.04 1500

If the rated supply voltage is V1, the new voltage is 0.9 V1, from Eq. (8.53), Te μ V12 s if rotor resistance is constant. As load torque is constant, V12 s = constant 1 sμ 2 V1

or,

If s¢ be the slip when voltage drops by 90%, V12 s¢ = = 0.049 0.04 (0.9 V1 ) 2 For small values of slip, \ or,

R2¢ has much larger value compared to R1, X1 or X2¢. s V Vs I¢2 = 1 = 1 R2¢ R2¢ s I¢2 μ V1 s

Rotor copper loss is proportion to I¢22 \

Copper loss with reduced voltage (0.9 V1 ¥ 0.049) 2 = = 1.2155 Copper loss with normal volltage (V1 ¥ 0.04) 2

Hence, when voltage drops, slip increases or speed decreases and rotor copper loss increases. Therefore, due to reduction of supply voltage, temperature rises noticeably under constant loss condition.

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Solution Let the rotor leakage reactance at 50 Hz be X2.

Uisff.qibtf!Joevdujpo!Npups

\ at 40 Hz. the rotor reactance is

9/8:

40 X or, 0.8 X2 50 2

At starting, slip is 1. \

R2¢ is less as compared to X2¢ s Hence, starting current Is μ

V (neglecting stator impedance) X2 Starting current at 50 Hz 440 0.8 X 2¢ = 0.88 = Starting current at 40 Hz X 2¢ 400

\

Starting torque μ

V2 w s X 2¢ 2

Starting torque at 50 Hz = Starting torque at 40 Hz

(440)2 50 X 2¢ 2 ( 400) 2 40 ¥ (0.8) 2 X 2¢ 2 2

40 Ê 440 ˆ =Á ¥ ¥ (0.8)2 = 0.619 ˜ Ë 400 ¯ 50 From Eq. (8.57), Maximum torque at 50 Hz = Maximum torque at 40 Hz

(440)2 50 X 2¢ ( 400) 2 40 ¥ 0.8 X 2¢ 2

40 Ê 440 ˆ =Á ¥ ¥ 0.8 = 0.7744. Ë 400 ˜¯ 50

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!

9/91

Fmfdusjdbm!Nbdijoft

Solution At 50 Hz smax1 =

R2 = 0.2 X2

R2 μ 0.2 50

\ If s1 be the slip then s1 = At 40 Hz,

(E X2 μ f )

3 = 0.06 50 R2 μ smax2 40

\

smax2 = 0.2 ¥

50 = 0.25 40

From Eq. (8.57), Tmax μ Tmax μ

or,

V2 w s X 2¢ V2

(E ws μ f and X2 μ f )

f 2 X2

where X2 is the reactance at 50 Hz. Tmax1

\

Tmax 2

=

( 440) 2 ¥ ( 40) 2 (50) 2 ¥ ( 400) 2

= 0.7744

Let s2 be the slip when the motor is operating at 40 Hz, 400 V supply. Now from Eq. (8.59), Te1 Tmax1 Te2 Tmax 2

=

2 smax1 s1

=

or,

Te1 Te2

Te1 Te2

=

smax1

2 smax 2 s2

\

+

s1

+

=

s2 smax 2

=

2 0.2 0.06 + 0.06 0.2 2 s 0.25 + 2 0.25 s2

Tmax1 0.55 ¥ 2 Tmax 2 s2 0.25 + s2 0.25

Ê 0.25 s ˆ = 0.21296 ¥ Á + 2 ˜ 0.25 ¯ Ë s2

= 0.55

Uisff.qibtf!Joevdujpo!Npups

9/92

Now as the developed torque remains same, Te1 = Te2 0.05324 + 0.85184 s2 = 1 s2

\

s 22 + 0.0625 – 1.1739 s = 0 s2 = 0.0559

or, or

\ slip frequency is 0.0559 ¥ 40 Hz or 2.236 Hz.

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Solution Synchronous speed

Ns =

120 ¥ 50 = 1500 rpm 4

\ slip at three-fifth of rated torque 1470 = 0.02 1500 Neglecting leakage reactance and stator resistance from Eq. (8.49), s=1–

Te = Te μ

or,

3 V12 w s ◊ s ◊ ( R2¢ /s 2 ) V12 s f R2

(\ ws μ f )

where f is the frequency of supply voltage. \

Te1 Te2

=

s f ¢ R2 0.02 f ¢ = 50 R2 ¥ s¢ 50 s¢

where f ¢ and s¢ are the frequency and slip under new condition. As torque is same in both conditions,

\

0.02 f ¢ =1 50 s¢ f ¢ = 50 = 2500 0.02 s¢

(i)

!

9/93

Fmfdusjdbm!Nbdijoft

Synchronous speed at frequency f ¢ 120 f ¢ = 30 f ¢ 4 1570 s¢ = 1 – 30 f ¢

Ns¢ = \ Substituting s¢ in Eq. (i),

f¢ = 2500 1570 130 f ¢

\

f ¢ = 2500 – \

130833 f¢

f ¢2 – 2500 f ¢ + 130833 = 0 f ¢ = 53.477 Hz.

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Solution Slip at maximum torque smax = \

R2 = 0.25 X2

R2 = 0.25 X2

40 If the rotor leakage reactance at 50 Hz be X2, the rotor leakage reactance at 40 Hz is X2 or 50 0.8 X2. Starting currents: Ist1 = Ist2 =

\ \

I st1 I st2

=

440 R22 + X 22

=

440 (0.25 X 2 ) 2 + X 22

440 (0.25 ¥ 3 X 2 ) + (0.8 X 2 ) 2

(0.25 X 2 ) 2 + (0.8 X 2 ) 2

Ist2 = 1.23

(0.25 X 2 ) 2 + X 22

2

=

=

440 (0.25 X 2 ) 2 + (0.8 X 2 ) 2

0.838 = 0.813 1.031

Uisff.qibtf!Joevdujpo!Npups

9/94

Starting torques: Tst1 = Tst2 = Tst 2

\

Tst1

=

( 440) 2 R2 w s ( R22 + X 22 )

=

( 440) 2 ¥ 0.25 X 2 50 {(0.25 X 2 ) 2 + X 22 }

( 440) 2 ¥ 0.25 X 2 40 {(0.25 X 2 ) 2 + (0.8 X 2 ) 2 } 0.25 X 2 ¥ 50 {(0.25) 2 + 1} 2

2

40{(0.25 X 2 ) + (0.8) } ¥ 0.25

=

13.28 = 1.89 7.025

Maximum torques: Tmax1 =

3 ( 440) 2 2 ¥ 50 ¥ X 2

3 ( 440) 2 2 ¥ 40 ¥ 0.8 X 2 50 X 2 50 = = 1.56. = 40 ¥ 0.8 X 2 40 ¥ 0.8

Tmax2 = \

Tmax 2 Tmax1

EFUFSNJOBUJPO!PG!NPUPS!FGGJDJFODZ!

9/27

The efficiency of small induction motors can be determined by directly loading them and by measuring their input and output powers. For larger motors, it may be difficult to arrange loads for them. Moreover, the power loss will be large with direct loading tests. Therefore, indirect methods are used to determine the efficiency of three-phase induction motors. The following tests are performed on the motor: (a) No-load test (or open circuit test) (b) Blocked-rotor test The parameters of the equivalent circuit can be found from the no-load and blocked-rotor test also.

9/27/2! Pqfo.Djsdvju!ps!Op.mpbe!Uftu This test is similar to the open-circuit test on a transformer. A three-phase auto-transformer is used to supply rated voltage at the frequency. The motor is run at no-load. The power input is measured by two wattmeter method. The power factor under no-load condition is generally less than 0.5. Therefore, use of the wattmeters will show negative reading. It is, therefore, necessary to reverse the direction of current coil terminals to take the reading. Let Wo, i.e. the wattmeter reading, be equal to the sum of stator core losses and mechanical losses. Let Vo and Io be the per-phase values of voltage and current.

!

9/95

Fmfdusjdbm!Nbdijoft

Then no-load power factor is cos qo =

Wo 3 Vo I o

However, and

IC = Io cos qo Im = Io sin qo

\

Vo Vo = I c I o cos q o V Vo Xo = o = I m I o sin q o

(8.62a)

Ro =

(8.62b)

The circuit diagram for no-load test and the corresponding equivalent circuit is shown in Fig. 8.13(a) and Fig. 8.13(b). W1 (P ) 1

A

Threephase supply at rated voltage and frequency

V

Three-phase induction motor

W2

(no mechanical load) (P2 )

Gjh/!9/24)b*! Djsdvju!ejbhsbn!gps!op.mpbe!uftu!po!b!uisff.qibtf!joevdujpo!npups

Io Io E1

Ic

Im

Ro

Xo

3

Rotor circuit open

R2¢ Ê ˆ Áˁ s = 0, s = •˜¯

Gjh/!9/24)c*! Frvjwbmfou!djsdvju!bu!op.mpbe

9/27/3! Cmpdlfe.Spups!Uftu! The circuit for blocked rotor test is shown in Fig. 8.14. The motion of the rotor is blocked by a brake (or a belt). This test is analogous to the short-circuit test of a transformer because the rotor winding is short-circuited through slip rings, and in cage motors, the rotor bars are permanently short circuited. Only a reduced voltage needs to be applied to the stator at rated frequency. The voltage should be such that the ammeter reads rated current of the motor.

Uisff.qibtf!Joevdujpo!Npups

Isc

W1

A

9/96

(P1)

V

Reduced supply voltage

Three-phase induction motor

W2

Blocked rotor (P2 )

Gjh/!9/25)b*! Djsdvju!ejbhsbn!gps!cmpdlfe.spups!uftu

Is/c Rs

Xs

X¢2 R2¢

E(reduced) 3

= Vs /c

s

= R2¢

( s = 1)

Gjh/!9/25)c*! Frvjwbmfou!djsdvju!evsjoh!cmpdlfe.spups!uftu

The total power input on short circuit Ws/c is equal to the algebraic sum of the two wattmeter readings, i.e. equals the copper losses of the stator and rotor. Let Vs/c and Is/c be the voltage and current per phase. Then the power factor under blocked rotor condition is cos qs/c =

Ws / c 3 (Vs / c ) ( I s/ c )

[neglecting the core and mechanical losses] Since in a R-L circuit, R = Z cos q and X = Z sin q, here we can write Ê Vs / c ˆ (Rs + R¢2) = Á ˜ cos qs/c Ë Is / c ¯

(8.63a)

Ê Vs / c ˆ (8.63b) (Xs + X 2¢ ) = Á ˜ sin qs/c Ë Is / c ¯ The stator resistance Rs is measured separately by using a battery, ammeter and a voltmeter. Then R¢2 can be found from Eq. 8.63(a). The reactances (Xs) and (X2¢) are generally assumed equal.

9/27/4! Uif!ed!Uftu! The purpose of the dc test is to determine R1. This is accomplished by connecting any two stator leads to a variable-voltage dc source as shown in Fig. 8.15(a). The dc source is adjusted to provide approximately rated stator current, and the resistance between the two stator leads is determined from voltmeter and ammeter readings. Thus, from Fig. 8.15(a),

!

9/97

Fmfdusjdbm!Nbdijoft

Rdc =

Vdc I dc T1

A Adjustable dc source

V

Induction motor

T2 T3

(a)

T1

T1 R 1, D

R 1, y R 1, y

T2

T2 R 1, y

Vdc

R1 , D Vdc

T3

R1, D

T3 Idc

Idc

(b)

(c)

Gjh/!9/26! Djsdvjut!gps!ed!uftu!up!efufsnjof!qbsbnfufs!S2

If the stator is star connected, as shown in Fig. 8.15(b), Rdc = 2 R1, g R1,g =

Rdc 2

(8.64)

If the stator is delta connected, as shown in Fig. 8.15(c), Rdc =

R1D ◊ 2 R1D 2 = R1D R1D + 2 R1D 3

R1D = 1.5 Rdc

(8.65)

Uisff.qibtf!Joevdujpo!Npups

TUBSUJOH!PG!B!UISFF.QIBTF!JOEVDUJPO!NPUPS!

9/98

9/28

A three-phase induction motor has a definite positive starting torque. When switched on to supply, it starts itself but draws a high starting current. This is evident from the equivalent circuit. At the È R (1 - s) ˘ time of starting, slip s = 1 and hence, the resistance Í 2 ˙ becomes zero (the motor behaves s Î ˚ as a short-circuited transformer). The current in the rotor and the stator windings may be about five times more than full-load values. These high rotor and stator currents cause many problems. 1. High electromagnetic forces between the conductors on the same part. 2. High heat generation causing high temperature may damage the insulation. 3. High current (at low power factor) may cause an appreciable drop in supply voltage causing undesirable effect on other equipment. Therefore, suitable means must be provided with the motor at start, to limit the starting current up to safe value. The device which is used to start the three-phase induction motor is termed starter. The function of the starter is to limit the initial rush of current to a predetermined safe value. The various methods of starting the three-phase induction motor are 1. By Direct On-Line (DOL) starter 2. By star-delta starter 3. By auto-transformer starter

9/28/2! Ejsfdu!Po.Mjof!Tubsujoh!)EPM* For small size squirrel-cage (less than 2 HP) motors or for motors in a power system where inrush of high-starting current is permissible, direct start may be used. For these small motors, the starting torque is about twice the full-load torque and the starting period lasts only a few seconds. Figure 8.16 shows a starter for direct starting with in-built short circuit, overload and undervoltage protection. When the motor is to be started, the main switch is put on and the start button is pressed. This energizes the relay coil S causing the normally open contacts S1, S2, S3 to close. Power is supplied to the motor and it starts. The contact S4 also shuts, thus, shorting out the starting switch allowing the operator to release it without removing power from the S relay. When the stop button is pressed, the S relay is de-energised and the S contacts open, thus, stopping the motor. Short-circuit protection is provided by fuses F1, F2 and F3. A Thermal Overload Relay (OLC) protects the motor from sustained overloads opening the contact D.

!

9/99

Fmfdusjdbm!Nbdijoft

Three-phase supply

a

S

L1

L2

L3

F1

F2

F3 Fuses

S4

S1

S2

S3

b Local stop Start

OLC

Motor

Remote stop

Gjh/!9/27! Ejsfdu!po.mjof!tubsufs

9/28/3! Tubs.Efmub!Tubsufs! Figure 8.17 shows the diagram of the star-delta starter. A star-delta starter can be used only for those three-phase induction motors whose stator winding has been designed for delta connection. All the six terminals (of the three-phases) are brought out. For starting, the phases are connected in 1 of its normal value. start thereby reducing the voltage of each phase to 3 From Eq. (8.33), we have I2 R Pag = 2 2 s Pag 1 2 1 \ Tf l = = ◊ I 2 R2 ¥ ws ws sfl where Tf l is the full-load torque at slip sf l. When the circuit parameters are referred to the stator side, Tf l =

I 2¢ 2 R2¢ w s sfl

Uisff.qibtf!Joevdujpo!Npups

9/9:

Three-phase supply Main switch (with fuse)

Rotor Delta-run

Stator Three-phase induction motor

Star-start

Gjh/!9/28! Tubs.efmub!tubsufs

At s = 1 (i.e., at starting), I 2¢¢ 2 R2¢ 1 2 I2¢ ¥ R¢2 = Ts = ws ws

Ts Ê I 2¢¢ 2 ˆ =Á ˜ Tfl Ë I 2¢ ¯

\

If T represents full-load torque, If l the full-load rotor current reflected to primary, we have If l = I2¢ neglecting the magnetizing branch current. Similarly, I≤ 2 represents the starting current (Is) at stator, the magnetizing branch being neglected. 2

\ we can write,

Ts = Ê I s ˆ ¥ s fl Á ˜ Tfl Ë I fl ¯

(8.66)

The starting line current of the motor with star-delta starter is thus, also reduced to 2

1 3

full volt-

age starting-line current. The starting torque which is proportional to Ê E1 ˆ is reduced to 1/3 of ÁË ˜¯ 3 the full-load torque. Thus, for star-delta start, though we are able to reduce the starting current, we sacrifice the torque and the starting torque reduces to 1/3 of the full-load torque. Let us analyse the star-delta starting method to find the torque. We assume that the motor first operates with star connection [Fig. 8.18(a)] and when speeds up, it operates with delta connection of the stator [Fig. 8.18(b)].

!

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In Fig. 8.18(a), starting line (phase) current (Is(start)) is given by Is(start) =

E/ 3 1 E 1 = ◊ = ◊ I P ( start ) Zs 3 Zs 3

[IP(start) is the starting phase current] Is(delta)

Is(star) E E

3

Zs E

Zs

(a)

(b)

Gjh/!9/29! Tubs.efmub!tubsujoh

In Fig. 8.18(b), Starting phase current

IP(start) =

E Zs

\ starting line current

Is(delta) = 3 IP(start) I s( start ) Ê 1 ˆ 1 \ = ◊I ∏ 3 I P (start ) = I s(delta) ÁË 3 P ( start ) ˜¯ 3 1 Using the relation (8.66), we can write Ts(start)/Tf l = (IP(start)/If l)2 ¥ sf l. Thus, starting torque is 3 1 of that obtained in DOL starting. 3 This method is a bit economical one but for motors rated beyond 3 kV, this method is not applicable. Like other three-phase motor starters, in this starter also overload coil and no-voltage coils are provided for the protection of the motor (not shown in the star-delta figure). An automatic star-delta starter can also be made by using pushbuttons, contactors, Time Delay Relay (TDR), etc.

(

)

9/28/4! Bvup.Usbotgpsnfs!Tubsufs In this method, reduced voltage is obtained by some fixed tappings on the three-phase auto-transformer. Generally, 60 to 65% tappings can be used to obtain a safe value of starting current. The full rated voltage is applied to the motor by taking the auto-transformer out of the motor circuit when the motor has picked up speed up to 85% of its normal speed. Figure 8.19 shows the circuit. Let us assume that the input voltage E is reduced to xE using auto-transformer tappings. Therefore, the motor starting current is, Is = xI, where I is the motor starting current when full voltage E is applied. However, the current drawn from the supply Is (line) is obtained from the relation

Uisff.qibtf!Joevdujpo!Npups

I s( line ) I s( motor )

9/:2

=x

Is (line) = xIs(start) = x2 I

\ here we have

Is(line)

Three-phase AC

Auto-transformer

E Main switch

xE Start Run

Is(motor)

Stator

Rotor

Gjh/!9/2:! Bvup.usbotgpsnfs!tubsufs

Hence, from the relation (8.66), we get 2

Ts = x2 Ê I ˆ . s fl ÁË I ˜¯ Tfl fl It is found that while the starting torque is reduced by x2 of that of DOL starting, the line current is also reduced by the same fraction.

9/28/5! Dbmdvmbujpo!pg!Spups!Sftjtubodf!Tubsufs!Tufqt In any motor starter, the steps are so designed that the current during starting fluctuates between a maximum and a minimum value. It is customary to design the steps of a slip-ring induction motor rotor resistance starter following the same criterion. Let the rotor current at the start of each step be I2max and that at the end of each step be I2min. Also the rotor induced emf per phase at standstill be E2. If s1, s2, s3, … sn are the slips at steps 1, 2, 3, … n, then s1 = 1 and sn = final slip.

!

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We can now write I2max =

E2 2

E2

=

Ê r1 ˆ 2 ÁË s ˜¯ + X 2

Ê r2 ˆ 2 ÁË s ˜¯ + X 2

1

E2

=

2

2

Ê r3 ˆ 2 ÁË s ˜¯ + X 2

2

E2

=º=

2

Ê rn ˆ 2 ÁË s ˜¯ + X 2 n

3

where, r1, r2, r3, … , rn are the rotor resistance including corresponding starter step resistances at slips s1, s2, … , sn respectively, and X2 is the rotor reactance at standstill and E2 the rotor voltage per phase. r r1 r2 r3 = = =º= n s1 s2 s3 sn

Hence,

(8.67a)

where rn = R2, the rotor resistance. Also,

I2min =

E2 2

E2

=

Ê r1 ˆ 2 ÁË s ˜¯ + X 2 2

E2

=

2

Ê r2 ˆ 2 ÁË s ˜¯ + X 2 3

2

E2

=º=

Ê r3 ˆ 2 ÁË s ˜¯ + X 2 4

2

Ê rn - 1 ˆ 2 ÁË s ˜¯ + X 2 n

rn - 1 r r1 r = 2 = 3 =º= s2 s3 s4 sn s r s2 r2 = and 3 = 3 s1 r1 s2 r2

i.e., From (8.67a),

(8.67b) (8.67c)

and so on. From (8.67b), and so on. From (8.67c) and (8.67d), we have,

r r2 s3 s = and 3 = 4 r1 s2 r2 s3

s r r r s2 s3 s4 r = = =º= n = 2 = 3 = 4 = … n = a s1 s2 s3 sn - 1 r1 r2 r3 rn -1 where a is a constant. From (8.67a), we have, r1/s1 = rn/sn \ r1 = R2 /sn (E s1 = 1)

(8.67d)

(8.67e)

(8.68)

2

From (8.67e), we have, r2 = a r1 ; r3 = a r2 = a r1; r4 = a r3 = a2 r2 = a3 r1 and so on \ rn (= R2) = an–1 . r1 \ a = (R2/r1)1/(n – 1) = (sn)1/(n – 1)

(8.69) (8.70)

Thus, the value of a can be found from Eq. (8.70) and then the total resistance in the rotor at various steps can be obtained from Eq. (8.69).

Uisff.qibtf!Joevdujpo!Npups

9/:4

Ubcmf!9/3! Dpnqbsjtpo!pg!Ejggfsfou!Tubsujoh!Bssbohfnfou! DOL starter

Star-delta starter

Auto-transformer starter

1. Full voltage is applied to the motor at the time of starting.

1. Each winding gets 58% of the rated line voltage at the time of starting.

1. The starting voltage can be adjusted according to the requirement.

2. The starting current is 5–6 times the full-load current.

2. The starting current is reduced to 1 that of direct on-line starting. 3

2. The starting current can be reduced as desired.

3. The three windings are connected generally in star.

3. The three windings are connected 3. The three windings are generally in star at the time of starting, and connected in delta. then in delta at the time of running.

4. Only three wires are to be brought out from the motor.

4. Six wires to be brought out from the motor.

4. Only three wires are to be brought out from the motor.

5. Easy-to-connect motor with direct on line.

5. Identification of three starting leads and three end leads is not so easy.

5. Input and output connections of the auto-transformers are to be made properly.

6. Very easy operation.

6. It is required that connections are first to be made in star, and then in delta either manually or automatically.

6. Skilled operator is needed for connection and starting.

7. Low cost.

7. More cost.

7. High cost.

8. Less space required for installation.

8. More space required.

8. More space required.

9. Used for motor up to 5 HP.

9. Up to 10 HP.

9. Large motors.

SFWFSTBM!PG!SPUBUJPO!

9/29

The direction of rotation of a three-phase induction motor can be reversed by reversing the direction of the rotation of the magnetic field. This can be done by interchanging the connections of any two of the three wires of the three-phase power supply. This causes the currents in the phases to interchange their relative timings in going positive and negative with the result that the magnetic field produces reversal in direction of rotation.

! Qspcmfn!9/81 B!31!IQ!uisff.qibtf-!511!W!tubs.dpoofdufe!joevdujpo!npups!hbwf!uif!gpmmpxjoh!uftu!sftvmut; ed!uftu!xjui!uif!tubups!xjoejoht!pg!uxp!qibtft!jo!tfsjft;!32!W-!41!B/!

!

9/:5

Fmfdusjdbm!Nbdijoft

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Solution Since two phases of the stator windings are in series in the dc test, we have 21 = 0.70 W 30 R1 = 0.35 W

2R1 = or No-load test: Vo =

400

= 230.95 V; Io = 8 A 3 Wo = (W10 + W20) = 2360 – 1160 = 1200 W 1200 = 0.216 3 ¥ 230.95 ¥ 8 Vo 230.95 = 133.65 W Ro = = I o cos q o 8 ¥ 0.216 Vo 230.95 = Xo = = 29.57 W I o sin q o 8 ¥ 0.976

\

cos qo =

Short-circuit test: Vsc =

140

= 80.83 V 3 Isc = 33 Amps; Wsc = 2450 W 2450 = 0.306 3 ¥ 80.83 ¥ 33

\

cos q sc =

\

Vsc . 80.83 cos q sc = ¥ 0.306 = 0.745 W I sc 33 V 80.83 80.83 X1 + X 2¢ = sc . sin q sc = ¥ sin q sc = ¥ 0.9518 = 2.33 W I sc 33 33 X1 = X2¢ = 0.5 ¥ 2.33 = 1.666 W R¢2 = 0.745 – 0.35 = 0.395 W R1 + R¢2 =

and \ and

Hence, we can write R0 = 133.65 W X0 = 29.57 W

Uisff.qibtf!Joevdujpo!Npups

9/:6

R1 = 0.35 W R¢2 = 0.395 W X1 = X¢2 = 1.166 W.

! Qspcmfn!9/82 B!551!W-!61!I{!fjhiu.qpmf!tubs.dpoofdufe!uisff.qibtf!joevdujpo!npups!ibt!uif!gpmmpxjoh!uftu!sftvmut; Op.mpbe!uftu;!551!W-!36!B-!3611!X Cmpdlfe!spups!uftu;!261!W-!226!B-!:111!X Efufsnjof!uif!frvjwbmfou!djsdvju!qbsbnfufst!pg!uif!npups!xifo!uif!qfs.qibtf!tubups!sftjtubodf!jt! 1/3!W/!

Solution From no-load test we have, E0 =

Per phase voltage

440

V = 254.03 V 3 Per phase current Io = 25 A 2500 Per phase power W0 = W = 833.33 W 3 W Power factor under no-load condition (cos q0) = o Vo I o

833.33 = 0.1312 (lag) 254.03 ¥ 25 [usually the no-load p.f. is very low for induction motors] Core loss component of the no-load current Hence,

cos q0 =

IC = I0 cos q0 = 25 ¥ 0.1312 = 3.2804 A Magnetizing component of the no-load current If = I0 sin q0 = 25 sin (cos–1 0.1312) = 24.783 A Voltage across the magnetizing branch is obtained as [V0 – I0(Rs + jXs)], where (Rs + jXs) is the per phase stator impedance in ohms. Again from blocked-rotor test data, we have Per-phase voltage

Es/c =

Per-phase current

Is/c

Per-phase power

Ws/c =

150

V = 86.6 V 3 = 115 A 9000 W = 3000 W 3

!

9/:7

Fmfdusjdbm!Nbdijoft

\ per-phase impedance

Zs/c =

Es / c Is / c

Ws / c

=

=

86.6 W = 0.753 W 115

3000

W = 0.2268 W

Per-phase resistance

Rs/c =

Per-phase reactance

Xs/c = (0.753) 2 - (0.2268) 2 = 0.718 W

I s2/ c

(115) 2

Per-phase rotor reactance referred to the stator, R¢2 = Rs/c – Rs = 0.2268 – 0.2 = 0.0268 W We assume here that per-phase stator reactance Xs = Per-phase rotor reactance X2¢ \

X1 = X2¢ =

Xs/c 2

=

0.718 = 0.359 W 2

Voltage across the magnetizing branch is obtained from the formula [E0 – I0(Rs + jXs)]. This gives the required voltage as [254.03 – 25(0.2 + j 0.359)] V, (249.03 – j 8.975) V or (249.192 ––2.064° V)

i.e., \ core loss resistance Magnetizing reactance

RC =

249.192 249.192 = W = 75.963 W IC 3.2804

Xo =

249.192 249.192 = W = 10.054 W If 24.783

Hence, the equivalent circuit parameters are R0 = 75.963 W R¢2 = 0.0268 W

X0 = 10.054 W Xs = X2¢ = 0.359 W.

R1 = 0.2 W

! Qspcmfn!9/83 B!uisff.qibtf-!26!lX-!551!W-!61!I{-!tjy.qpmf!trvjssfm.dbhf!joevdujpo!npups!ibt!b!efmub.dpoofdufe! tubups!xjoejoh/!Uif!npups!evsjoh!cmpdlfe.spups!uftu!zjfmet!uif!gpmmpxjoh!sftvmut;!351!W-!36!B-!8!lX/ Uif!ed!sftjtubodf!nfbtvsfe!cfuxffo!boz!uxp!tubups!ufsnjobmt!jt!2!W/!Jg!uif!tubups!dpsf!mptt!bu!sbufe! wpmubhf!jt!511!X-!efufsnjof!uif!tubsujoh!upsrvf!xifo!sbufe!wpmubhf!jt!bqqmjfe/

Solution If the stator winding resistance per phase is r then the resistance between any two terminals (see Fig. 8.20) is r(r + r ) =1 r + (r + r)

Uisff.qibtf!Joevdujpo!Npups

9/:8

2r 2 =1 3r 3 r = W = 1.5 W 2

or or

r

r

r

Gjh/!9/31! Tubups!xjoejoh!)D* 2

Ê 440 ˆ kW, i.e., At the rated voltage, power input during the blocked rotor test will be 7 ¥ Á Ë 240 ˜¯ 440 ˆ Ê 23.52 kW. At rated voltage, stator current during blocked rotor test will be Á 25 ¥ or 45.83 A. Ë 240 ˜¯ È Ê 45.83 ˆ 2 ˘ Thus, at rated voltage the stator copper loss will be Í3 ¥ Á ¥ 1.5˙ or, 3150.768 W ˜ ÍÎ Ë 3 ¯ ˙˚ Air-gap power Pag = Power input – (Stator copper loss + Stator core loss) = (23520 – 3150.768 – 400) = 19.969 kW Synchronous speed

Ns =

120 f 120 ¥ 50 = 1000 rpm = P 6

Hence, starting torque when rated voltage is applied

Pag ws

=

19969 Nm = 190.79 Nm 2p ¥ 1000 60 Ns È ˘ ÍEw s = 2p ns = 2p ◊ 60 , N s being expressed in rpm ˙ . Î ˚

! Qspcmfn!9/84 B!uisff.qibtf!trvjssfm.dbhf!joevdujpo!npups!hjwft!b!cmpdlfe!spups!uftu!dvssfou!pg!311&!pg!uif!sbufe! dvssfou!xifo!41&!pg!uif!sbufe!wpmubhf!jt!bqqmjfe/!Uif!tubsujoh!upsrvf!jt!41&!pg!uif!sbufe!upsrvf/! Uif!npups!xifo!tubsufe!cz!bo!bvup.usbotgpsnfs!mjnjut!uif!tubsujoh!mjof!dvssfou!up!271&!pg!uif!sbufe! dvssfou/!Efufsnjof!uif!qfsdfoubhf!tubsujoh!upsrvf!xjui!bvup.usbotgpsnfs!tubsujoh/

!

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Fmfdusjdbm!Nbdijoft

Solution At 30% or rated voltage, blocked rotor current is 200% (given) At rated voltage, blocked rotor current is 2 . Isc = If l = 6.67 If l, where If l is the full-load current. 0.3 Now, if x be the fraction of the voltage applied to the stator during auto-transformer starting then per phase starting current would be Ist = x2 Isc Ist = x2 (6.67 If l) (Ist) = 1.6 If l

Here, Again, it is given that

(i) (ii) (ii)

From the equations (i) and (ii), x2 =

1.6 = 0.23988 6.67

Also, T μ voltage2, here, 0.3 Tf l μ (0.3 V)2, where V is the rated voltage, Tf l is the full-load torque T, 0.3 . i.e., V2 μ Tf l 0.09 From the text we know that Starting torque with auto-transformer starting = x2 que with direct on line starting Starting torq Hence, starting torque with auto-transformer starting is, 0.3 0.3 . Tf l = 0.1918 Tf l Tf l = (0.23988)2 ¥ 0.09 0.09 i.e., starting torque with auto-transformer starting is 19.18% of the full-load torque. x2 ¥

! Qspcmfn!9/85 B!dbhf!npups!ibt!b!tubsujoh!dvssfou!pg!51!B!xifo!txjudife!po!ejsfdumz/!Bvup.usbotgpsnfs!xjui!56&! ubqqjoh!jt!vtfe/ Efufsnjof!)b*!tubsujoh!dvssfou-!boe!)c*!sbujp!pg!tubsujoh!upsrvf!xjui!bvup.usbotgpsnfs!up!uif!tubsujoh! upsrvf!xjui!ejsfdu!txjudijoh/

Solution The ratio of transformation (x) is 0.45. (a) \ Starting current with auto-transformer = (0.45)2 ¥ 40 = 8.1 A Starting torque with auto-transformer (b) = (0.45)2 = 0.2025. Starting torque with direct starting

Uisff.qibtf!Joevdujpo!Npups

9/::

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Solution Given: Output = 10 kW No. of poles = 6 Frequency

f = 50 Hz N = 900 rpm h = 90%

Full-load p.f. = 0.88. Full-load line current drawn by a three-plane D-connected induction motor is given as (If l) = =

Output in watt 3 ◊ VL ¥ P.f. ¥ efficiency 10 ¥ 1000 3 ¥ 400 ¥ 0.88 ¥ 0.9

=

10000 = 18.22 A 548.71

Now, full-load current per phase (D-connection) If l =

18.22 3

= 10.52 A

On direct on-line start the current Isc drawn by the motor per phase is given as Isc = Synchronous speed Full-load slip

85 3

= 49.07 A

(Ns) =

120 f 120 ¥ 50 = 1000 rpm = P 6

(s) =

N s - N 1000 - 960 = 0.04 = Ns 1000 2

E

Ts = 1 Ê I sc ˆ ¥ s fl Á ˜ Tfl 3 Ë I fl ¯

\ Here

Ts = 1 Ê 49.07 ˆ ¥ 0.04 = 0.290. Á ˜ Tfl 3 Ë 10.52 ¯

2

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! Qspcmfn!9/87 B!uisff.qibtf!efmub.dpoofdufe!dbhf.uzqf!joevdujpo!npups!xifo!dpoofdufe!ejsfdumz!up!b!511!W-!61!I{! tvqqmz!ublft!b!tubsujoh!dvssfou!pg!216!B!jo!fbdi!tubups!qibtf/!Gjoe!pvu! ! )b*! uif!mjof!dvssfou!gps!EPM!tubsujoh! )c*! mjof!boe!qibtf!tubsujoh!dvssfout!gps!Z.D!tubsujoh-!boe! ! )d*! mjof!boe!qibtf!tubsujoh!dvssfout!gps!b!81&!ubqqjoh!po!bvup.usbotgpsnfs!tubsujoh/

Solution (a) Direct on line (DOL) starting current = 3 ¥ 105 = 181.86 A V 400 (b) In Y-D starting, phase voltage on starting = L = = 230.9 V 3 3 400 105 will produce = 60.62 A Since 400 V produce 150 A in phase winding, 3 3 \ starting phase current = 60.62 A In Y-connection, line current = phase current \ starting line current = 60.62 A (c) In auto-transformer start, with 70% tapping on auto-transformer, the line voltage across the D-connected motor = 0.7 ¥ 400 V. In D-connection, Phase voltage = Line voltage = 0.7 ¥ 400 = 280 V. Since 400 V produces 105 A in phase winding. (0.7 ¥ 400) V = 280 V produces 0.7 ¥ 105 = 73.5 A Hence, motor phase current is 73.5 A \ motor phase current = 3 ¥ 73.5 = 127.30 A supply line current motor applied voltage = 0.7 = supply voltage motor line current

But

\ supply line current = 0.7 ¥ 127.30 = 89.11 A.

! Qspcmfn!9/88 Dbmdvmbuf! uif! tvjubcmf! ubqqjoh! po! bo! bvup.usbotgpsnfs! tubsufs! gps! b! uisff.qibtf! joevdujpo! npups! sfrvjsfe!up!tubsu!uif!npups!xjui!61&!pg!gvmm.mpbe!upsrvf/!Uif!tipsu.djsdvju!dvssfou!pg!uif!npups!jt! 7!ujnft!uif!gvmm.mpbe!dvssfou!boe!gvmm.mpbe!tmjq!jt!1/146/!Bmtp!dbmdvmbuf!uif!dvssfou!esbxo!gspn!uif! nbjo!tvqqmz!bt!b!gsbdujpo!pg!gvmm.mpbe!dvssfou/

Solution 2

Ê I sc ˆ Starting torque = x . Á ˜ ¥ Slip at full-load ¥ Torque at full-load Ë I fl ¯ 2

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or 0.5 ¥ Torque at full-load = x2 ¥ 62 ¥ 0.035 ¥ Torque at full-load x2 =

\

0.5

6 ¥ 0.035 x = 0.629

i.e.,

2

= 0.396

\ current drawn from the supply mains = x2 Isc = 0.396 ¥ 6 ¥ Ifull-load = 2.736 If l.

! Qspcmfn!9/89 B!526!W-!61!I{-!fjhiu.qpmf!uisff.qibtf!efmub.dpoofdufe!trvjssfm.dbhf!joevdujpo!npups!ibt!b!tubsujoh! dvssfou!pg!41!B!xifo!dpoofdufe!ejsfdumz!up!uif!tvqqmz/!Gjoe!)b*!uif!mjof!boe!qibtf!dvssfou!esbxo!cz! uif!npups!xifo!dpoofdufe!ejsfdumz!po!mjof-!)c*!uif!mjof!dvssfou!xifo!tubsufe!cz!bo!bvup.usbotgpsnfs! xjui!81&!ubqqjoh-!boe!)d*!uif!mjof!dvssfou!xifo!tubsufe!cz!b!tubs.efmub!tubsufs/

Solution (a) Line current when connected directly to the supply is 30 A. As the motor is delta connected, the phase current under direct online supply is

30 3

A=

1732 A. (b) At 70% auto-transformer tapping the applied line voltage is 415 ¥ 0.7 V = 290.5 V. As the motor is delta-connected phase voltage is 290.5 V. When phase voltage is 415 V, the phase current is 17.32 A. 17.32 ¥ 290.5 When phase voltage is 290.5 V, the current supplied by the auto-transformer is 415 = 12.124 A. Hence, phase current of the motor = 0.7 ¥ 12.124 = 8.48 A Hence, the line current when started by auto-transformer starter is 8.48 ¥ 3 = 14.7 A (c) When the motor is started by a star-delta starter, the motor is connected in star at the instant of 415 V = 239.6 V. Phase current starting. Hence, phase voltage of the motor during stating is 3 239.6 at phase voltage of 239.6 V is 17.32 ¥ = 10 A. 415 \ Line current = Phase current = 10 A at start.

! Qspcmfn!9/8: B!uisff.qibtf!trvjssfm.dbhf!joevdujpo!npups!ibt!cmpdlfe!spups!dvssfou!pg!611&!boe!upsrvf!pg!311&! pg!jut!gvmm.mpbe!dvssfou!boe!upsrvf!sftqfdujwfmz/!Efufsnjof!uif!dvssfou!boe!upsrvf!bt!b!qfsdfoubhf! pg! gvmm.mpbe! wbmvft! xifo! tubsufe! cz! )b*! tubs.efmub! tubsufs-! )c*! bvup.usbotgpsnfs! tubsufs! xjui! 91&! ubqqjoh-!boe!)d*!qsjnbsz!sftjtubodf!tubsufs-!mjnjujoh!uif!tubsujoh!dvssfou!up!81&!pg!ejsfdu!po!mjof! dvssfou/

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Fmfdusjdbm!Nbdijoft

Solution (a) When started by star-delta starter 1 ¥ 500% = 166.67% 3 1 Starting torque = ¥ 200% = 66.67% 3

Starting current =

(b) With auto-transformer starter, Starting current = 500 ¥ (0.8)2 = 320% Starting torque = 200 ¥ (0.8)2 = 128% (c) With primary resistance starter, Starting current = 500 ¥ 0.7 = 350% Starting torque = 200 ¥ (0.7)2 = 9.8%.

TQFFE!DPOUSPM!PG!B!UISFF.QIBTF! JOEVDUJPO!NPUPS

9/2:

The synchronous speed (Ns) of a three-phase induction motor is given by 120 f N (1 – s) or N = Ns (1 – s) = P 1- s The speed N of an induction motor can be changed by the three basic methods. Ns =

)b*!Gsfrvfodz!Dpouspm! By changing the supply frequency f, the speed can be varied directly proportional to the supply frequency of ac supply. )c*!Qpmf!Dibohjoh! Speed control can also be obtained by changing the number of poles P on the stator (as speed is inversely proportional to the number of poles). This change can be incorporated by changing the stator winding connections with a suitable switch. The change in the number of stator poles P changes the synchronous speed Ns of the rotating flux, thereby the speed of the motor also changes. )d*! Cz! Dibohjoh! uif! Tmjq! This can be accomplished by introducing resistance in the rotor

circuit, which causes an increase in slip, thereby bringing down the speed of the motor.

9/2:/2! Tqffe!Dpouspm!cz!Dibohf!pg!Tvqqmz!Gsfrvfodz If the frequency of the supply to the stator of an induction motor is changed its synchronous speed is changed, depending on the frequency and hence, provides a direct method of speed control. To

Uisff.qibtf!Joevdujpo!Npups

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keep the magnetization current within limits, the applied voltage must be reduced in direct proportion to the frequency. Otherwise, the magnetic circuit will become saturated resulting in excessive magnetization current. The starting torque at reduced frequency is not reduced in the same proportion, because rotor power factor improves with reduction in frequency. The torque that can be produced by the maximum permissible rotor current is equal to that at rated conditions. Since power is the product of torque and speed, operation at reduced speed results in lesser permissible output. This method of speed control is not a common method and hence, this method would be used only as a special case. In earlier days, the variable frequency was obtained from a motor generator set or mercury arc inverter. In recent days, frequency control is used by SCR based inverters or by using IGBT inverters.

9/2:/3! Tqffe!Dpouspm!cz!Qpmf!Dibohjoh If an induction motor is to run at different speeds, one way is to have different windings for the motor so that it will have different synchronous speeds and the running speeds. Another method is used with suitable connections for a changeover to double the number of poles. The principle of formation of consequent poles is used. The method of changing the number of poles is accomplished by producing two sections of coils for each phase which can be reversed with respect to the other section. It is important to note in this connection that slot angle (i.e., electrical degrees), phase spread, breadth factor and pitch factor will be different for the low- and high-speed connections. The three phases can be connected in star or delta, thus, giving a number of connections. If 50% per pole pitch is used for a high-speed connection, a full-pitch winding is obtained for low-speed connection. The method of connecting coils of a four-pole motor is shown in Fig. 8.21 for one phase and also change over connections to obtain eight poles for the same machine with the same winding. 1

2

3

4

N

S

N

S

(a) 2

1 N

S

N

4

3 S

N

S

N

S

(b)

Gjh/!9/32! Gpvs.qpmf0fjhiu.qpmf!dpoofdujpot!gps!pof!qibtf!pg!joevdujpo!npups

The methods of speed control by pole changing are suitable for squirrel-cage motors only because, a cage rotor has as many poles induced in it as there are in the stator and can thus, adapt when the number of stator poles changes.

!

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Fmfdusjdbm!Nbdijoft

9/2:/4! Tqffe!Dpouspm!cz!Dibohf!pg!Tmjq Slip can be varied in three ways: 1. By varying resistance of the rotor circuit 2. By varying the stator applied voltage 3. By injecting an external slip emf in the rotor circuit 2/!Spups!Sftjtubodf!Dpouspm! This method is applicable to slip-ring induction motor as the slip rings make it possible to connect external resistance to the rotor circuit. When speed control in a limited range for a short period is required, this method is used. As the total resistance in the rotor circuit increases, the speed drops for constant torque as evident from the torque-slip characteristic of Fig. 8.11. For a constant stator voltage, the air-gap flux is constant. Therefore, for a constant torque, the rotor current and the stator current are constant. When the rotor resistance is increased, the decrease of speed increases the rotor emf because of relative increase of speed with respect to the rotating air-gap flux, tending to increase the rotor current. But the increase of rotor resistance due to increase of rotor frequency offsets this tendency. Hence, there is a built-in mechanism for limiting the current at constant torque. The increase of rotor resistance with the same rotor current increases the rotor copper loss resulting in reduced efficiency and inversed temperature rise. Hence, this method is well suited when speed control for short periods is required. For a constant rotor current, the torque at standstill can be made much higher than normal full-load torque by inserting a suitable external resistance, so that when started on full-load, the motor accelerates faster. 3/!Tubups!Wpmubhf!Dpouspm! Speed control by controlling the applied voltage to the stator is used only in small three-phase squirrel induction motors. The torque developed by an induction motor is proportional to the square of the applied voltage (as evident from Eq. 8.49). A reduction of stator voltage reduces the torque. But if the load torque is constant, the electromagnetic torque cannot decrease for a stable operation. Hence, the slip changes (increases) in such a way so as to make up the reduction of stator voltage, i.e., the speed reduces. In an induction motor, the electromagnetic torque is developed due to the interaction of rotor current and air-gap flux. The strength of the air-gap flux is proportional to the stator voltage due to the fact that the magnetizing current depends on the stator voltage. Therefore, a reduction of stator voltage reduces the air-gap flux so that for a fixed electromagnetic torque, the rotor current and, therefore, the stator current increase. This increase of current should be such that the full-load value is not exceeded. Now if the motor is operating at rated stator voltage and rated stator current, a reduction of stator voltage requires an increase of stator current above the rated value for the same

Uisff.qibtf!Joevdujpo!Npups

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load torque. As this is not possible, the speed control by stator voltage control is possible with a load torque only if it is less than the motor torque corresponding to rated voltage and rated current. 4/!Tqffe!Dpouspm!cz!fng!Jokfdujpo!jo!Tfdpoebsz!Djsdvju! This method of speed control is possible only in case of slip-ring induction motors. Let the standstill rotor voltage be E2. When the motor runs on load with slip rings shorted, the rotor voltage reduces to sE2 = I2 Z2s where Z2s is the rotor impedance. The current I2 provides the necessary torque. Now if the motor speed is to be decreased then an external emf Ef of slip frequency (that means the frequency of the externally injected emf should be exactly the same as that of sE2) is injected in the rotor circuit in each phase and in phase opposition to E2. If the load torque on the motor remains approximately the same, the current in the motor remains in the same direction and of the same value. This means I2Z2s remains approximately the same value. This is possible only when the induced emf in the rotor increases to a value s1E2 (s π s1). This means the slip must increase or speed must decrease. There are two voltages applied in the rotor circuit s1E2(emf induced) and Ef (emf injected). Since there are two voltages applied to the rotor circuit s1 E2 (emf induced) and Ef (emf injected), hence, the difference of their magnitudes should be equal I2 Z2s. \

s1 E2 – E2 = I2Z2s

(8.71)

Since Z2s (the rotor impedance) is normally small, hence, we may neglect I2Z2s. This gives \

s1E2 = Ef or s1 = Ef /E2

(8.72)

On the other hand, If the external slip emf is injected in the same phase with E2 then by the same reasoning we have, s1 E2 + Ef = 0 or s1 = –

Ef E2

(8.73)

Hence, the slip becomes negative, i.e. the motor runs at super-synchronous speed. Thus, we observe that if the external slip emf is injected in the rotor in phase with E2 the rotor runs above synchronous speed and if injected at 180° to E2, the motor runs below synchronous speed. More the magnitude of Ef , more the speed variation. For a wide variation of speed, by this method slip is increased to high values, so that the power absorbed in the secondary circuit (i.e. rotor copper loss) increases. Power absorbed in the external rotor circuit (neglecting the small power loss in rotor winding itself) = s ¥ Power transferred between stator and rotor. An external resistance in rotor can consume electrical power but cannot supply electrical power Hence, it can change motor speed in subsynchronous region only. Actually, it injects an external voltage (in the form of I2R2 drop across the external resistance in each phase) at 180° to E2 in the rotor circuit. Proper electronic thyristor circuitary, which can make controlled electrical power flow in both the directions and can generate slip frequency emf on one side and power frequency on the

!

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Fmfdusjdbm!Nbdijoft

other, when connected to secondary circuit of a slip trying induction motor, can control its speed below or above synchronous speed. If we connect capacitors or pure inductors in the rotor circuit, they do not absorb any real power or generate any electrical power and they cannot control the speed of the motor. The I2X2 drop which gets injected in the secondary circuit is at 90° to E2 and it simply changes the power factor of the motor current. If the regulating mechanism at rotor circuit is capable of handling the electrical power absorbed in such a way that it takes electrical power from supply system and feeds to rotor (–ve absorption) to cater for extra mechanical power developed then the speed control is done at high efficiency. The regulating power is not lost as heat. 5/!Dbtdbef!Dpoofdujpo Here, two induction motors are required, and one of them must be a slip ring induction motor. The connections are as shown in Fig. 8.22. The two motors are coupled electrically as well as mechanically. For motor A, it is external slip emf control from the secondary side. 3-f supply

A

B

Gjh/!9/33! Dbtdbef!dpouspm

Four speeds are possible: (a) Speed corresponding to poles P1 of the machine A when only this machine is connected to the line B is open. (b) Speed corresponding to poles P2 of the machine B when only this is connected to the line and the machine A is left open. (c) Speed corresponding to poles (P1 + P2) when the machines are connected as in Fig. 8.21. Machine A is working at a very high slip. The slip power is being fed to the machine B to get converted into mechanical power. The motor B runs at its normal slip w.r.t. its synchronous speed governed by frequency coming through slip rings of A. (d) Speed corresponding to number of poles is equal to (P1 – P2). This is possible if A is first run above its synchronous speed with supply B opened and then suddenly connecting the slip frequency supply to B with reserved phase sequence as compared to what was the phase sequence to B in (c).

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This method has not found wide application because of large weight of the two motors and comparatively lower power factor; however, sometimes it is used in drives like rolling mills and large ventilating fans.

QPXFS!GBDUPS!BU!EJGGFSFOU!MPBEJOHT!

9/31

Referring to the equivalent circuit of the induction motor as shown in Fig. 8.13(b), the power factor of the motor means the power factor of I1 with respect to V1. At no-load, I 2¢ is negligible. So the power factor is due to the current Im only. Im is the magnetizing component predominantly and hence, has very low power factor. At full-load or near full-load, I2 is predominant and I¢2 is governed by R¢2. Thus, the power factor of the motor is much better at full-load or close to full-load values, Im has little effect this time. However, at high value of slip, (R¢2/s) decreases and obviously the power factor starts deteriorating. At s = 1, blocked-rotor condition, the power factor is worst as the motor current is then highly inductive. Power factor of an induction motor on full-load is a design factor. Normally, less number of poles or higher speed motors have better power factors. Power of larger capacity motors is normally better than that of smaller capacity motors. Actually, power factor is a function of air gap, leakage reactance, core material, etc., in the machine.

Fggjdjfodz!wt!Mpbe!Dibsbdufsjtujd Efficiency of induction note is naturally very low at no-load and increases fast as load increases, and reaches maximum value when the variable losses become equal to constant losses. It then again starts decreasing slowly with further increase in load as shown in Fig. 8.23. Full-load efficiency of a machine is again a design factor and depends mainly upon the losses in the machine. Higher speed and higher capacity motors have generally better efficiencies.

h

Load

Gjh/!9/34! h!wt!mpbe!dibsbdufsjtujdt!pg!bo!joevdujpo!npups

!

9/219

Fmfdusjdbm!Nbdijoft

PQFSBUJPO!PG!BO!JOEVDUJPO!NPUPS!BU! EJGGFSFOU!GSFRVFODJFT

9/32

When a voltage V1 is applied to the primary of an induction motor, a voltage E1 (= V1) is induced in that primary such that E1 = 4.44 fm f N1. If the motor is to be operated at a higher frequency, the value of fm decreases as V1 is constant. It is not a problem as regards to no-load current, but the speed of the motor will increase. If the torque of the load remains constant, the power delivered will be more and load component of the current will become excessive. In order to bring down the current, either V1 has to be increased (to make V/f constant) or the stator must be rewound with less number of turns so that N1 f is constant and the flux is constant. The iron losses will be more because of the increased frequency. If the motor is to be run on a lower frequency than the normal frequency, the flux fm is required to be increased. Normally, the core and especially the teeth are operating near saturation region of the magnetization curve. Thus, Io may become very high. One solution to their problem is that the applied voltage may be decreased such that V/f is a constant. So whenever a machine is to be operated at a lower frequency, it should be ensured that the applied voltage is decreased in the same proportion. But with this measure, the starting torque, stalling torque, etc., also become less as torque is proportional to square of the applied voltage. If the applied voltage is to be kept same, the solution is to rewind the stator with increased number of turns such that N1 f is constant.

DSBXMJOH!BOE!DPHHJOH!PG!BO!JOEVDUJPO!NPUPS!

9/33

In a squirrel-cage induction motor, the starting torque is usually low. Sometimes, a squirrel-cage motor, after starting, run at a very low speed of about 1/7th of its full speed. This is known as crawling of the induction motor. The reason is as follows; Because of certain practical reasons, it is not possible to have more than 3 or 4 slots per pole per phase in the stator. The windings in these slots produce an approximately stepped mmf wave in place of a sine wave and this stepped wave is composed of a large number of harmonics. This can be proved by Fourier analysis of the mmf wave. Since the wave shape is almost perfectly symmetrical about the x-axis, the even harmonics are totally absent. Third harmonics and their multiples cancel out because of three-phase connection in the windings. Hence, only 5th, 7th, 11th, 13th, 17th, etc., harmonics are present. But the strength of these harmonics goes on diminishing with their increasing order. Thus, the 5th and 7th harmonics in the flux wave are of predominance. The fifth harmonic has negative phase sequence, so it does not produce a forward torque. The seventh harmonic has the same direction of rotation as the fundamental, and thus, its forward torque slip characteristics get superimposed over that due to fundamental. The seventh harmonic in the air-gap flow produces the effect of number of poles seven times more than the poles in fundamental. Thus, the synchronous speed for the seventh harmonic is 1/7th of the fundamental. That is why there is a hump and a cusp in the torque vs slip characteristics of the motor

Uisff.qibtf!Joevdujpo!Npups

9/21:

at about 1/7th of the synchronous speed as shown in Fig. 8.24. If the load characteristics intersects the motor characteristics somewhere on the torque descending line AB then the motor will keep on running at this low speed. These gap flux harmonics are produced not only by windings but also due to slotting, saturation, gap-length irregularities, harmonics or unbalance in supply voltage, etc. This phenomenon of operation of the induction motor at more than 1/7th of synchronous speed is called creating crawling of the motor.

T A

B

T/S curve for load

N s

Gjh/!9/35! Fggfdu!pg!uif!ibsnpojd!jo!gmvy!po!U0t!dvswf

When the number of the stator and the number of rotor slots are such that they are in the same position then the reluctance of the magnetic path is minimum. If the rotor leaves this position, the reluctance increases and a reluctance torque is produced which opposes the movement of the rotor. If the starting torque of the motor is less than the reluctance torque, the motor will fail to start. This phenomenon is known as cogging or magnetic locking and is due to equalization of number of stator and rotor slots.

TLFXJOH!PG!SPUPS!TMPUT!JO!B!UISFF.! QIBTF!JOEVDUJPO!NPUPS!

9/34

A larger number of induction motors are produced with skewed rotor slots. Placing the rotor bars in skewed slots results in reduction of cogging, crawling and other torque problems, minimizes the motor noise and increases the rotor resistance (because of the increase in the rotor bar length). The torque defects are minimized because the rotor and stator slots are no more parallel to each other to provide possibility of magnetic locking and one end of a rotor bar may be under the north pole of a harmonic while the other end of the same bar may be under south of the same harmonic. This results in no current flow in the bar due to the harmonics. Thus, skewing improves performance of the motor considerably.

!

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Fmfdusjdbm!Nbdijoft

PQFSBUJPO!PO!VOCBMBODFE!WPMUBHF! BOE!TJOHMF!QIBTJOH!

9/35

If the three line voltages supplied to a three-phase induction motor are not equal, they not only cause unequal phase currents in the rotor and stator windings but the percentage current unbalance may be 6 to 10 times larger than the percentage voltage unbalance. The resultant increase in I 2R losses will overheat the insulation, shortening its life. Unbalanced voltages also cause a decrease in locked-rotor and breakdown torque. Thus, in those applications where there is only a small margin between the locked-rotor torque and the load torque, severe voltage unbalance may prevent the motor from starting. The full-load speed of the running motor is reduced slightly by voltage unbalance. Percent voltage unbalance is defined by the maximum line voltage deviation from the average value of the three line voltages, times 100. Expressed as an equation, we have Percent unbalanced Vmax dev. . 100 voltage = (8.74) Vavg. where

Vavg = (V1 + V2 + V3)/3

and Vmax dev = maximum deviation of voltage between a line voltage and Vavg. Single phasing in a three-phase induction motor means that one of the three phases of the supply has been cut off due to any reason, or one phase fuse has blown off or removed or disconnected somewhere in one phase. The motor, when already running, keeps on running as a single-phase motor making a characteristic noise. This is a single-phasing condition because the current in both the remaining two lines now is the same single current. In a star-connected stator, the load is taken by the remaining two phases so the motor now can take a load of 1/ 3 times (57.7%) its rating. In single phasing condition, for the same current and voltage, P = VI cos f; while in three-phase operation, P = 3 VI cos f. Thus, if the motor is operating at a load near full-load, the healthy phases will be overloaded. If the stator is delta connected the current distribution in different phases is not the same. One of the windings gets overloaded even when the current in the healthy lines is normal. Under single-phasing condition, apart from the possibility of overloading of healthy windings, there is the danger of increased heating due to increased rotor losses. Hence, protection devices must be provided to disconnect the supply to the three-phase motor whenever single phasing occurs. If the motor is attempted to start in single-phasing condition from the stalling condition, it will not start as there is no three-phase rotating field to rotate the rotor.

DJSDMF!EJBHSBN!

9/36

To determine the performance characteristics of induction motors on load, it is not really easy to provide the loading facility. Especially, for large capacity motors it is almost impossible to load

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the motors on the manufacturing sites. So by simply conducting the no-load and the blocked-rotor tests and by drawing the circle diagram, one can compute all the performance characteristics of the motor on all possible loads. The performance characteristics of a motor can be theoretically predetermined, even before the production of the motor, from its design data using the equivalent circuit analysis. The circle diagram is an extension of the equivalent circuit analysis. It is faster in manually computing the characteristics of a motor at various loads. Besides the approximations of the equivalent circuit, it involves the approximation of construction and length measurement. The approximate equivalent circuit of an induction motor as referred to primary is as in Fig. 8.7. V I¢2 = Load current as referred to primary = Rt2 + X t2 where

Rt = R1 +

R2¢ and Xt = X1 + X2¢ s

Let f be the phase angle between I 2¢ and V. \

cos f =

\

I2¢ =

Rt Rt2 + X t2

and sin f =

Xt Rt2 + X t2

V sin f Xt

If the applied voltage and the leakage inductances of the stator and rotor windings are constant V then the above expression represents a circle of diameter passing through the origin. We assume Xt Xt as constant for all the normal motors. This assumption is very nearly true. But for deep-bar and double-cage motors, the diameter of the circle goes on changing with the changing Xt. The value of I2¢ changes as the load or Rt changes and it lies on the locus of a circle as shown in Fig. 8.25(a). If the voltage is taken on the y-axis and if Rt = • then sin f = 0, and lies on y-axis. If Rt = 0, sin f = 1, I2¢ = V/Xt at 90° to V, i.e. on x-axis.

Mpdvt!pg!J¢3!po!uif!Djsdmf! Since V/Xt is the maximum value of I¢2, therefore, it forms a semicircle above x-axis. I¢2 will be equal to V/Xt only if Rt = 0 which is practically impossible. So, it is only a theoretical value. Since I1 = Io + I¢2, therefore, Fig. 8.25(b) gives the primary current I1 for any load current I¢2. Io in the equivalent circuit is actually the open-circuit current, i.e. the core loss and magnetization current. But on performing the no-load test on an induction motor, it gives Io plus a small amount of I¢2 corresponding to the friction and windage losses in the motor. Thus, this also introduces a small approximation in these computations.

!

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Fmfdusjdbm!Nbdijoft

V

I¢2 Motoring

I1 I¢2

Io I¢2 =

V

Xt

Generating

(b)

(a)

Gjh/!9/36! Djsdmf!ejbhsbn

One knows Io and fo from no-load test, Is and fs from blocked-rotor test and the resistance of stator winding per phase (Rs) can be measured. Is is the blocked-rotor current with rated supply voltage. The construction of a circle diagram is described in Fig. 8.26. At first, a vertical line OY representing the input voltage V is drawn. Now a line OA is drawn making an angle fo with the vertical axis to represent the no-load current Io. From A, a line AB is drawn at an angle fs with the vertical axis with magnitude equal to Is. Join OB. Now a perpendicular bisector of AB is drawn which meets the horizontal line drawn from A, i.e., AC at E. With E as centre and EA as radius, a semicircle is drawn. From B on the semicircle, a vertical line is drawn which meets AC at H and the horizontal line drawn from O at G. The vertical line BG represents the input power under blocked-rotor condition with rated voltage applied across the stator. Now from A a perpendicular is drawn on OG at U. Hence, HG = AU Again, from D OAU AU = Io cos fo Hence, AU can be equated to the no-load input power which supplies mainly core loss, friction and windage loss. The power scale can be determined from AU as it is equal to the no-load power Wo.

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Wo Length of AU in cm Wo = Length of HG in cm

\ Power scale

1 cm =

As BG represents input power under blocked-rotor condition and HG represents no-load power, BH represents the copper loss in the stator and rotor as the output in the blocked-rotor condition is zero. BH is divided at the point J such that HJ represents the stator copper loss and JB represents the rotor copper loss. If stator and rotor copper losses are assumed to be equal then BJ = HJ

E

The line joining points A and J, i.e. the line AJ represents the torque line and line AB represents the output line. Now let us assume that the motor is taking a current represented by the line OK. The perpendicular line KL represents the input power. LM represents the fixed losses. MN and NP represent the stator and rotor copper loss. PK represents output power and NK represents the output torque. For the determination of maximum torque developed by the motor, a line parallel to the torque line AJ is drawn which makes a tangent to the circle at the point T. Then draw a vertical line TS to the torque line. The length of TS represents the maximum torque. Y V

K

T

Q B

e

t lin tpu

Ou

I1

I¢2

D

f

P

fs

J

line

S

N

fo

E

A O

Torque

R

Ia U

H

F

C

M L

G

Gjh/!9/37! Djsdmf!ejbhsbn

Similarly, to determine the maximum power developed by the motor, a line parallel to the output line AB is drawn which makes a tangent to the circle at the point Q. Then draw a vertical line QR on the output line. The length of QR represents the maximum output.

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It is advisable to draw as big a circle diagram as is reasonably possible. This will minimize the measurement errors.

IJHI.UPSRVF!JOEVDUJPO!NPUPST!

9/37

Conventional squirrel-cage motors suffer from the disadvantage of low starting torque because of low rotor resistance. The starting torque can be increased by using bar material of higher resistivity. A higher rotor resistance gives a higher starting torque and lower starting line current at a higher power factor. However, higher rotor resistance reduces the full-load speed and increases rotor ohmic losses resulting in lower efficiency. A low rotor resistance is required for normal operation, when running, so that the slip is low and the efficiency is high. Therefore, for good starting performances, the rotor resistance should be high at start, and under normal operating speeds, the rotor resistance should be low. In wound-rotor induction motors, these conditions are fulfilled by connecting external resistances in the rotor circuit at the time of starting. As the motor speeds up, the external resistance is cut out and the rotor windings are short-circuited through the slip rings. In order to obtain high rotor resistance at starting and low rotor resistance at running, two types of rotor connections are used in squirrel-cage motors: 1. Deep bar rotor 2. Double-cage rotor Figure 8.27 shows a cage rotor with deep and narrow bars. As such, a bar may be assumed to be made up of number of narrow layers connected in parallel. Figure 8.27 shows three such layers, X, Y and Z. It may be observed that the topmost layers, element X is linked with minimum leakage flux

Element X

Rotor bar

Element Y

Element Z

Leakage flux

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Uisff.qibtf!Joevdujpo!Npups

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and, therefore, its leakage inductance is minimum. On the other hand, the bottom layer Z links with maximum leakage flux and, therefore, its leakage inductance is maximum. At starting, the rotor frequency is equal to the supply frequency. The bottom-most layer element Z offers greater impedance to the flow of current than the topmost layer element X. Therefore, maximum current flows through the topmost layer and minimum current flows through the bottom layer. Because of the unequal distribution of current, the effective rotor resistance increases and the leakage reactance decreases. This causes a high rotor resistance at starting conditions; the starting torque is relatively higher while the starting current is relatively lower. Under normal operating conditions, the slip and the rotor frequency are very small. The reactance of all layers of the bar are very small in comparison to their resistances. The impedances of all layers of the bar are nearly equal and hence, current flowing through all parts of the bar is mostly identical. Also the resulting larger cross-sectional area makes the rotor resistance itself quite small, resulting in a good efficiency at low slips. An induction motor with two rotor windings (or cages) is also used for obtaining high starting torque at low starting current. The stator of a double-cage rotor induction motor is similar to that of an ordinary induction motor. In the double-cage rotor, there are two layers of bars as shown in Fig. 8.28.

Outer cage

Negligible outer cage leakage flux

Slit

Inner cage

Inner cage leakage flux

Mutual flux

Gjh/!9/39! Epvcmf.dbhf!spups!tmpu

Each layer is short-circuited by end rings. The outer cage bars have a smaller cross-sectional area than the inner bars and are made of high resistivity materials like brass, aluminum, bronze, etc. The inner-cage bars are made of low-resistance copper. Thus, the resistance of the outer cage is greater than the resistance of the inner cage. There is a slit between the top and bottom slots. The slit increases leakage flux around the inner-cage bars. Thus, the leakage flux linking the inner-cage winding is much greater than that of the outer-cage winding, and hence, the inner winding has a greater self-inductance.

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Fmfdusjdbm!Nbdijoft

At starting, the voltage induced in the rotor is same as the supply frequency voltage. The leakage reactance of the inner-cage winding (= 2pfL) is much larger than that of the outer-cage winding as frequency of rotor is same as the frequency of stator. Therefore, most of the starting current is flowing in the outer-cage winding which offers low-impedance to the flow of current. The highresistance outer-cage winding, therefore, develops a high starting torque. As the rotor speed increases, the frequency of the rotor emf (fr = sf ) decreases. At normal operating speed, the leakage reactances (= 2p sfL) of both the windings become negligibly small. The rotor current division between the two cages is governed mainly by their resistances. Since the resistance of the outer cage is about 5 to 6 times that of the inner cage, most of the rotor current flows through the inner cage. Hence, under normal operating speed, torque is developed mainly by the low-resistance inner cage. It may, thus, be noted that for low-starting torque requirements, an ordinary cage motor can be used. For higher torque requirements, a deep-bar motor may be used. A double-cage motor is used for still higher torques. For large-size motors with very large starting torques and long starting periods, slip ring motors are used. A single-cage motor and a double-cage motor of the same rating can be compared as follows: 1. A double-cage rotor has low starting current and high starting torque. Therefore, it is more suitable for direct on-line starting. 2. The high resistance of the outer cage increases the effective resistance of a double-cage motor. Therefore, full-load copper losses are increased and the efficiency of the double-cage motor is decreased. 3. Since effective rotor resistance of a double-cage motor is higher, there is a larger rotor heating at the time of starting as compared to that of a single-cage rotor. 4. A double-cage induction motor has higher effective leakage reactance due to additional reactance of the inner cage. Therefore, the full-load power factor is reduced. 5. The pull-out torque of a double-cage motor is smaller than that of a single-cage motor because the two cages produce the maximum torque at different speeds. 6. By a proper choice of resistances and reactance of the outer and inner cages, a wide range of torque-slip characteristics can be obtained with double-cage motors. This is not possible with a single-cage motor. 7. The cost of a double-cage motor is higher than that of a single-cage motor of the same rating. Let

R1 X1 R¢20 X2o ¢ R¢2i X¢2i s

= Resistance per phase of stator = Reactance per phase of stator = Resistance per phase of outer cage referred to stator = Standstill leakage reactance per phase of outer cage referred to stator = Resistance per phase of the inner cage referred to stator. = Standstill leakage reactance per phase of the inner cage referred to stator = Slip

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It is assumed that the main flux completely links both the cages and hence, the impedances of the two cages can be considered in parallel. I1

X1

R1

I¢2 Io Ic

I¢2i

I¢2o Im

X¢2i

X2o Ro

E¢2

Xo

V1

R¢2s s

R¢2i s

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The equivalent circuit of the double-cage induction motor at slip s is shown in Fig. 8.29(a). If the shunt branches containing Ro and Xo are neglected, the equivalent circuit becomes simplified and is shown in Fig. 8.29(b). I1

R1

X1 I¢2o

I¢2i

X¢2o

X¢2i

E¢2 V1

R¢2o s

R¢2i s

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At slip s, the outer-cage impedance, Z¢2o =

R2¢ o + jX2o ¢ s

Z 2i ¢ =

R2¢ i + jX 2i ¢ s

At slip s, the inner-cage impedance,

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The impedance of the stator, Z1 = R1 + jX1 Equivalent impedance per phase of the motor referred to stator Ze1 = Z1 + (Z¢2o || Z 2¢ i) \

Ze1 = R1 + jX1 +

1 1 1 + Z 2¢ o Z 2¢ i

= R1 + jX1 +

Z 2¢ o Z 2¢ i Z 2¢ o + Z 2¢ i

(8.75)

Current through the outer cage is given by I¢2o =

E2 Z 2¢ o

(8.76)

The current through the inner cage is obtained as I¢2i =

E2 Z 2¢ i

(8.77)

The rotor current (referred to the stator) is equal to the phasor sum of the currents through the outer and inner cages, i.e. I¢2 = I¢2o + I2i¢ (8.78) Since the two cages develop two separate torques, the total torque of the motor is equal to the sum of the two such torques. The torque-slip characteristics of the two cages are shown in Fig. 8.30. The total torque of the motor is also shown in Fig. 8.30. 4

Torque, current(Pm)

Input current

Starting torque Total torque

2

Torque cage 2

Torque cage 1 slip (pu) O

1

o Speed w

ws Synchronous speed

Gjh/!9/41! Upsrvf!boe!dvssfou!dibsbdufsjtujdt!pg!b!epvcmf.dbhf!joevdujpo!npups

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9/22:

The resultant torque-speed characteristics can be modified according to the requirement. This is done by modifying the individual cage resistances and leakage reactances. The resistances can be changed by changing the area of cross section of bars. The leakage reactance can be changed by changing the width of the slot openings and the depth of the inner cage. Power developed per phase by the outer cage is given by Pdo = I 22o¢

R2¢ o s

(8.79)

Power developed per phase by the inner cage is R2¢ i s \ power developed per phase by both the cages is given as Pdi = I 22i¢

Pd = Pdo + Pdi = (I 2¢ o)2

R2¢ o R¢ + ( I 2¢ i ) 2 2i s s

(8.80)

(8.81)

From the equivalent circuit of the double-cage motor, I2¢o =

E2¢ Z 2¢ o

(8.82)

I2i¢ =

E2¢ Z 2¢ i

(8.83) 2

Ê R¢ ˆ Z 2o ¢ = Á 2o ˜ + ( X 2¢ i ) 2 Ë s ¯

(8.84)

2

Ê R¢ ˆ Z2¢ i = Á 2i ˜ + ( X 2¢ i ) 2 Ë s ¯

(8.85)

Let Tdo = Torque developed by the outer cage Tdi = Torque developed by the inner cage Td = Total torque developed by the two cages \ where,

Pd = (2p ns) Td P 1 Td = d = 2p ns 2p ns

È 2 R2¢ o 2 R2i ˘ Í( I 2¢ o ) s + ( I 2¢ i ) s ˙ Î ˚

(8.86)

2

Also,

Ê R2¢ o ˆ 2 ÁË s ˜¯ + ( X 2¢ o )

Tdo = 2 Tdi Ê R2¢ i ˆ 2 ÁË s ˜¯ + ( X 2¢ i )

(8.87)

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9/38

The torque slip characteristic for a normal induction machine has been drawn for slips from +2 to –1 as shown in Fig. 8.31. Braking mode

+2

Motoring mode

Generating mode

1 S

0

ns

–1 Speed

Gjh/!!9/42! U0tmjq!dibsbdufsjtujdt!efqjdujoh!ejggfsfou!npeft!pg!pqfsbujpo

For slip 1 to 0, the machine is rotating in the direction of the field and the torque developed is in the same direction, so it is motoring action. For all the values of slips above +1, the machine is running in the opposite direction to the field causing braking action. When slip is negative, the mechanical torque developed is also negative. Negative mechanical torque developed by a machine means it is working as a generator. That means an induction machine works in the generating mode when its slip is negative. 2/!Npupsjoh!Npef When the slip is between 0 and 1, the mechanical power developed is positive and the power input to the rotor is also positive. Thus, it means electrical power is received by the rotor through stator and there it is converted into mechanical power. The machine is working as a motor. 3/!Hfofsbujoh!Npef When slip is negative, i.e. when rotor speed is higher than synchronous speed in the direction of the field, then there is a reverse flow of power. Mechanical power is not developed but is consumed and the rotor does not receive electrical power from the stator, instead, it delivers it. Hence, the machine is working as a generator.

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4/!Csbljoh!Npef When slip is more than unity, i.e., when the rotor is running in opposite direction to the direction of the field, power developed is negative while power input is positive. For this mode, slip is greater than 1. This is achieved by bringing the rotor to a quick stop by a braking action called plugging. In plugging, any two stator terminals are interchanged causing reversal of phase sequence. The direction of the rotating magnetic field becomes opposite to the direction of rotation of the rotor. The electromagnetic torque is developed in the opposite direction causing braking action. The motor is brought to rest quickly but the supply must be disconnected from the stator before the rotor starts rotating in the opposite direction.

JOEVDUJPO!SFHVMBUPS!

9/39

An induction machine to be used as a variable voltage supplier is known as induction regulator. In this mode of operation, it is in standstill condition. Hence, it works like an auto-transformer. Induction regulator is different from auto-transformer in the way that the latter supplies different voltages by changing tappings on its windings while the motor gives different voltages by changing its rotor position. The rotor has to be held after turning it by the required angle in the position by some mechanical means, otherwise the rotor may try to rotate as a motor (specially in case of a three-phase regulator). The primary is usually on the rotor. The principle of operation is described here by considering the primary supply to the stator. Mag. field axis Prime V1

V1

V0

V1

V0

V1

V0

V0

sec. comp.widg

(a)

V2 V1

V2

comp.widg

(b)

V0

V2 V1

V0

(c)

V1

V0

(d)

V1V2 V0

(a) Stator and rotor windings, axes same

(c) Stator and rotor (b) Stator and rotor windings axes at 90° windings, axes at 45°

(d) Stator and rotor axes at 180° to each other

Gjh/!!9/43! Qsjodjqmf!pg!b!tjohmf.qibtf!joevdujpo!sfhvmbups

The principle of operation of a single-phase induction regulator is quite clear from Fig. 8.32. The output voltage Vo = V1 ± V2 where the magnitude of V2 changes with rotor position. Thus, one can get any output voltage between the maximum value (V1 + V2) and minimum value (V1 – V2).

!

9/233

Fmfdusjdbm!Nbdijoft

A single-phase induction regulator is always provided with a short-circuited compensatory winding in quadratic with the primary. If this winding is not provided, the leakage reactance of the secondary is very large in the position of the rotor when the primary and secondary winding axes are not in the same line. It will cause a very heavy internal voltage drop in the machine. There is normally a handle for turning the rotor to different positions. So any output voltage may be set. The principle of operation is illustrated in Fig. 8.33. The only difference from the single-phase regulator is that whatever be the rotor position, the magnitude of V2 remains constant, only its phase changes. It is because the three-phase primary produces a constant-magnitude rotating field. Therefore, the magnitude of the emf induced in any coil coming under the influence of this field is same for any position of the coil. The difference is only that of the timings (timings of instantaneous value of emf attaining zero or maximum value). This decides the phase difference of V2 w.r.t. V1. Thus, the vectorial sum of V1 and V2 keeps on changing with different positions of the rotor. That is, the magnitude and also the phase of Vo depends upon the rotor position as illustrated by 8.33(c). For getting the designed output voltage, the locking latch is released, the rotor is turned to the required position and locked again. B C

V0 min q

Vo

A

V1 V1

(a) 3-phase connections

(b) Connections of each phase

(c) Phasor diagram

Gjh/!9/44! Uisff.qibtf!joevdujpo!sfhvmbups

GSFRVFODZ!DIBOHFS!

9/3:

This is a specially designed three-phase slip-ring induction machine to generate variable frequency. In the frequency changer as shown in Fig. 8.34, output frequency (f2) = sf1, where f1 is the frequency of the supply to its stator.

Uisff.qibtf!Joevdujpo!Npups

3-phase supply at of frequency f1

9/234

3-phase output at frequency

f2

Driving motor

Frequency changer

Gjh/!9/45! Joevdujpo!gsfrvfodz!dibohfs

This machine is generally used for getting higher frequency supplies. The driving motor is run in a direction opposite to that of the field of the frequency changing machine. The slip for the machine is then more than unity and output frequency is more than the input frequency. Ignoring losses, the power output share of the driving motor is the fraction (f2 – f1)/f2 of the higher frequency output.

JOEVDUJPO!HFOFSBUPS!

9/41

Figure 8.31 shows the complete torque-speed curve of an induction machine for all ranges of speeds. If the machine is driven at a speed greater than synchronous speed by a prime mover, the direction of torque reverses and it acts as an induction generator. There is a maximum torque in the generating region too (just as in motoring mode, the maximum torque or pull-out torque exists). If the prime mover applies a torque greater than this maximum value, the machine will have overspeed. The rotating magnetic field in an induction motor is set up by the magnetizing current drawn from the mains. Even when the speed of the machine is above synchronous speed and it is acting as a generator, this magnetizing current must be available. Thus, an induction generator is not self-exciting, but must be operated in parallel with another source which can supply its magnetizing power. An induction generator has many limitations. Since it does not have a field circuit, it cannot produce reactive power. It consumes reactive power from the magnetizing current must be supplied to it. It cannot control its output voltage and the generator’s output voltage is maintained by the external power system connected to it. It is possible to operate an isolated induction generator successfully if a capacitor bank is connected across its terminals as shown in Fig. 8.35. The capacitor bank supplies the reactive power of the generator and the other loads.

!

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Fmfdusjdbm!Nbdijoft

P

P 3-phase induction generator

Q

Q

Q

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The magnetizing current of an induction machine is a function of terminal voltage as shown in Fig. 8.36(a). The reactive current of a capacitor is proportional to the voltage as shown in C3 C2

Voltage

Terminal voltage

C1

C1> C2 > C3

Magnetizing current Capacitor current (a) Induction generator alone (b) Capacitor bank alone

C3

C2

C1

V1 V2 V3

O

I3

I2

I1

I

(c) Induction generator in parallel with capacitor bank.

Gjh/!9/47! Dibsbdufsjtujdt!pg!joevdujpo!hfofsbups!boe!dbqbdjups!cbol;

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Fig. 8.36(b). If a three-phase bank of capacitors is connected to the terminals of an induction generator, the no-load terminal voltage of generator will be given by the point of intersection of generator’s magnetization curve and capacitor’s load line [Fig. 8.36(c)]. When the induction generator is made to run, its residual magnetism produces a small terminal voltage which produces a capacitive current. This current increases the voltage which in turn increases the current till the voltage builds up to the point of intersection of capacitor load line and generator’s magnetization curve. The process is similar to that in a self-excited dc generator. If there is no residual flux in the rotor of induction generator, voltage build-up cannot take place. In that case, it must first be run as a motor to generate residual magnetism. The output voltage varies widely with changes in load, as shown in Fig. 8.37. For a lagging power factor load, the voltage collapses very rapidly which causes a serious disadvantage of an induction generator. The output frequency varies by about 5% from no-load to full-load. The induction generators were used in industry till the early twentieth century. Due to increase in fuel prices, there is a trend for new conventional sources of energy and the induction generator is ideal for such applications because of its simple control and maintenance. Because of its simplicity and small size per kW of power output, an induction generator is suitable for windmills. V

IL Gjh/!9/48! Wpmubhf!hsbqi

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9/42

A three-phase induction motor and synchronous motor have different characteristics in their own field of application. The main differences in characteristics and applications are as under: 1. A synchronous motor requires dc excitation (it is a doubly excited machine), whereas an induction motor is singly excited machine and does not require any dc excitation. 2. Induction motor is a self-starting motor, whereas a synchronous motor requires a special starting method (since it has no starting torque).

!

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3. An induction motor always operates at a lagging power factor. However, the power factor of a synchronous motor can be adjusted to be leading as well as lagging by variation of excitation current. 4. A synchronous motor runs at synchronous speed only. The speed of an induction motor at full-load is about 5% less than its synchronous speed. An induction motor does not develop any torque at synchronous speed and cannot run at this speed. 5. An induction motor is used only for supplying mechanical loads, whereas a synchronous motor is used for supplying mechanical loads as well as for power-factor control in power systems.

BQQMJDBUJPO!PG!JOEVDUJPO!NPUPST!

9/43

The three-phase induction motor is the most widely used motor in industry and workshops. It is manufactured in a variety of ratings from a few kW to 5000 kW or so. It is simple and of rugged construction; it has almost maintenance-free operation and does not require any dc excitation current. Typical application of three-phase induction motors include fans, blowers, centrifugal pumps, compressors, crushers, agitators, reciprocating pumps, shears, punch presses, bulldozers, boilers, hoists, cranes, etc.

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Solution From the no-load test data, the no-load power factor cos fo = \

1200 3 ¥ 400 ¥ 20

fo = 85°

= 0.0866

Uisff.qibtf!Joevdujpo!Npups

9/238

From the blocked-rotor test data cos fs = \

2800 3 ¥ 100 ¥ 45

= 0.359

fs = 69°

For constructing the circle diagram, the voltage V is represented on the vertical axis as shown in Fig. 8.38. The no-load current of 20 A is drawn making an angle of 85° to that of V. Considering current scale of 1 cm = 10 A, the no-load current is represented by OA whose length is 2 cm and lagging V by 85°. 45 Now under blocked-rotor condition, the input current at 400 V is ¥ 400 = 180 A. To represent 100 this current, a line AB is drawn with 18 A length and making an angle of 69° with the vertical line. Join OB and draw a perpendicular bisector of AB to cut the horizontal line drawn from A at E. With E as centre and EA as radius, draw a semicircle. From B on the semicircle, a vertical line is drawn which cuts the horizontal line from O at G. BG cuts the extended line AE at H. Since stator and rotor copper losses are assumed to be equal, BH is divided at J in two equal parts. Draw the line AJ, which represents the torque line and AB represents the output line. In the circle, diagram length of BG is 6.8 cm. BG represents the input at blocked rotor condition. 2

Ê 400 ˆ = 44800 W Now input power at blocked-rotor condition at 400 V is 2300 ¥ Á Ë 100 ˜¯ \

BG = 44800 W From this, the power scale can be derived which is 1 cm =

44800 = 6588 W 6.8

\ full-load output of the motor = 40 HP = 40 ¥ 735.5 = 29420 W 29420 = 4.46 mm 6588 which intersects the semicircle at K.OK represents the full-load current I1 which is 6.1 cm. \ full-load current I1 = 6.1 ¥ 10 = 61 A The angle made by the line OK with the vertical axis is the power factor at full-load. \ power factor at full-load cos f = cos 34° = 0.83 power input = KL = 5.1 ¥ 6488 W From the output line, PK is drawn parallel to the vertical line with length

!

9/239

Fmfdusjdbm!Nbdijoft

Power output = KP = 4.46 ¥ 648 W KP 4.46 \ efficiency = = = 87.45% KL 5.1 To determine the maximum power output, a line parallel to the output line AB and tangent to the semicircle is drawn at Q. The vertical line QR represents the maximum output. Length of QR is 6.7 cm. Using the power scale, Maximum output = 6.7 ¥ 6588 = 44139 W To determine the maximum torque a line is drawn parallel to the torque making tangent to the semicircle at point T. The vertical line TS represents the maximum torque and the length of line is 8.1 cm. \ maximum torque = 6588 ¥ 8.1 = 53363 W Now to determine the speed, Rotor copper loss = s ¥ Rotor input Rotor input PN 0.3 = 0.0645 = = \ Slip s = Rotor copper loss KN 4.65 Now synchronous speed, 120 f 120 ¥ 50 = = 1500 rpm P 4 Nr = (1 – s) Ns = 1403 rpm

Ns = \ motor speed

\ maximum torque =

53363 = 363.39 Nm. 1403 2p ¥ 60

V

T Q B

I1

f

K

e

ut lin

Outp P 85°

69°

R

I0

N M

A 0

J

e rque lin

S

To

H E

L

Gjh/!9/49! Djsdmf!ejbhsbn!pg!Qspc/!9/91

G

F

Uisff.qibtf!Joevdujpo!Npups

9/23:

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Solution (a) Neglecting the stator impedance, the equivalent impedance of the two cages at standstill are Ze2 =

(0.06 + j 0.6) (0.5 + j 0.15) 0.06 + j 0.6 + 0.5 + j 0.15

=

0.603 84.29∞ ¥ 0.522 16.7∞ 0.56 + j 0.75

=

0.315 101∞ = 0.3365 47.75∞ 0.936 53.25∞

\ power developed by the motor at standstill 2

Ê ˆ 440 Pe = 3 I 2 R2 = 3 ¥ Á ˜ ¥ 0.3365 cos 47.75 Ë 3 ¥ 0.3365 ¯ = 386.84 kW 2

The equivalent impedance of the two cages at 4% slip Ê 0.06 ˆ Ê 0.5 ˆ ÁË 0.04 + j 0.6˜¯ ÁË 0.04 + j 0.15˜¯ Ze2 = 0.06 0.5 + j 0.15 + j 0.6 + 0.04 0.04 =

(1.5 + j 0.6) (12.5 + j 0.15) 1.5 + j 0.6 + 12.5 + j 0.15

=

1.6155 21.8∞ ¥ 12.5 0.6875∞ 14 + j 0.75

=

20.19 22.49∞ = 1.44 19.49∞ W 14.02 3∞ 2

Ê 440 ˆ 1.44 cos19.49 \ power developed = 3 ¥ Á ˜ ¥ 0.04 Ë 3 ¥ 1.44 ¯ = 3168.5 kW.

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(b) Let at slip s the two cages develop equal power. Power developed by inner cage 2

Ê ˆ Á ˜ 440 / 3 Á ˜ ¥ 0.06 P1 = 3 ¥ Á ˜ 2 s Á Ê 0.06 ˆ + (0.6) 2 ˜ ÁË ÁË s ˜¯ ˜¯ Power developed by outer cage 2

Ê ˆ Á ˜ 440 / 3 Á ˜ ¥ 0.5 P2 = 3 ¥ Á ˜ 2 s Á Ê 0.5 ˆ + (0.15) 2 ˜ Á ˜ ÁË Ë s ¯ ˜¯ 2

\

or,

Ê 0.06 ˆ 2 ÁË s ˜¯ + (0.6) 0.06

2

=

Ê 0.5 ˆ 2 ÁË s ˜¯ + (0.15) 0.5

2 ÏÔÊ 0.5 ˆ 2 ¸Ô Ê 0.06 ˆ 2 + (0.15) 2 ˝ ÁË s ˜¯ + (0.6) = 0.12 ÌÁË ˜ ¯ ÓÔ s ˛Ô

or, or,

0.0264 = 0.3573 s s = 0.272 or 27.2.

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Solution The current in the outer cage

I21 =

E2 2

Ê 0.6 ˆ 2 ÁË s ˜¯ + (0.8)

Uisff.qibtf!Joevdujpo!Npups

The current in the inner cage

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E2

I22 =

2

Ê 0.3 ˆ 2 ÁË s ˜¯ + (1.4)

(a) At standstill s = 1 \

I21 = I22 =

(0.6) + (0.8)

2

E2 (0.3) + (1.4) 2

2

=

E2 = E2 1

=

E2 1.432

I 21 = 1.432 I 22 R 3 I 22 2 s Torque = wr

\

\

E2 2

Torque developed by the outer cage 0.6 = (1.432)2 ¥ = 4.1 Torque developed by the inner cage 0.3

(b) At s = 0.06 2

Ê 0.3 ˆ 2 ÁË 0.06 ˜¯ + (1.4)

I 21 = = 2 I 22 Ê 0.6 ˆ 2 ÁË 0.03 ˜¯ + (0.8)

26.96 = 0.259 400.64

Torque developed by the outer cage 0.6 = (0.259)2 ¥ = 0.134. Torque developed by the inner cage 0.3

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Solution Since stator resistance is neglected the rotor output is 12 kW. (a) Rotor copper loss = s ¥ 12 The slip in induction generator is negative and its magnitude is

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120 ¥ 50 1040 - 1000 6 = 0.04 = 120 ¥ 50 1000 6

1040 -

\ rotor copper loss = 0.04 ¥ 12 = 0.48 kW (b) If R2 be the rotor resistance per phase then 3 I 22

R2 E2 R 3 E 2 sR = 2 = 2 2 2 2 2 = 12000 =3 2 2 s s R2 R2 + s X 2 + X 22 2 s

2

or,

Ê 300 ˆ 3¥Á ¥ 0.04 R2 Ë 3 ˜¯ R22

or,

= 12000 (Neglecting rotor reactance)

R2 = 0.3 W

(c) Let R¢2 be the new rotor resistance 2

Ê 300 ˆ 3¥Á ¥ 0.04 R2¢ Ë 3 ˜¯ \

R2¢ 2

= 8000

R¢2 = 0.45 W

\ external resistance to be added is (0.45 – 0.3) W = 0.15 W.

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Solution (a) f = 50 Hz Ns =

120 ¥ 50 = 1000 rpm 6

If s be the slip then sf = 10 or

s=

10 = 0.2 50

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If N be the motor speed then N = (1 – s) Ns = (1 – 0.2) ¥ 1000 = 800 rpm (b) sf = 90 s=

90 = 1.8 50

\ motor speed N = (1 – 1.8) ¥ 1000 = –800 rpm Hence, the speed is 800 rpm and the direction of rotation of rotor is opposite to the direction of the rotating magnetic field.

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Solution Synchronous speed

Ns =

120 ¥ 50 = 750 rpm 8

(a) Speed is 1500 rpm in opposite direction \

Nr = –1500 rpm

\ Frequency of the machine

Ï Ê -1500 ˆ ¸ fr = sf = Ì1 - Á ˜ ˝¥ 5 Ó Ë 750 ¯ ˛ = (1 + 2) ¥ 50 = 150 Hz

Now speed is 350 rpm in the same direction as that of the rotating magnetic field. \

Nr = 350 rpm 375 ˆ Ê fr = sf = Á1 ¥ 50 = 25 Hz Ë 750 ˜¯

\ (b) (i) when fr = 30 Hz

N ˆ Ê 30 = Á1 - r ˜ ¥ 50 Ë 750 ¯ or,

Nr = 300 rpm in the same direction

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(ii) When fr = 500 Hz N ˆ Ê 500 = Á1 - r ˜ ¥ 50 Ë 750 ¯ Nr = –6750 rpm \ speed of the machine is 6750 rpm in a direction opposite to the rotating field. (iii) When fr = 150 Hz

\

N ˆ Ê 150 = Á1 - r ˜ 50 Ë 750 ¯ Nr = –1500 rpm

\ speed is 1500 rpm in a direction opposite to the rotating field.

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Solution From Eq. (8.68), r1 =

R2 0.1 = 3.33 = sn 0.03

From Eq. (8.70), a=

1 n -1 sn

1

= (0.03) 4 = 0.4162

From Eq. (8.69), r2 r3 r4 r5

= a r1 = 0.4162 ¥ 3.33 = 1.387 = a r2 = 0.5773 = a r3 = 0.24 = 0.1

Total external resistance is 3.33 – 0.1 = 3.23 W with tappings at (1.387 – 0.1) = 1.287 W, (0.5773 – 0.1) = 0.4773 and (0.24 – 0.1) = 0.14 W should be used.

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Solution Without feeder starting torque, Ts1 =

{

3V 2

w s (1.5) 2 + ( 4) 2

}

With feeder starting torque, Ts2 =

3V 2

{

w s ( R + 1.5) 2 + 4 2

}

.

where R is the resistance of the feeder. Starting torque with feeder should be at least 0.75 times the starting torque without feeder. \

Ts2 = 0.75 Ts1 1

\

( R + 1.5) + 4 2

\

2

=

0.75 (1.5) 2 + 4 2

R = 1.39 W l = 400 m f = 0.03 W/m/mm2

Length of feeder Resistivity If A be the cross-sectional area in mm2,

rl = 1.39 A 0.03 ¥ 400 = 8.63 mm2. A= 1.39

\

Sfwjfx!Rvftujpot! 1. Explain why the speed of an induction motor cannot be equal to the synchronous speed. 2. Explain the principle of operation of a three-phase induction motor. 3. Describe with a suitable diagram the constructional features of squirrel-cage and slip-ring induction motor. 4. What are the advantages and disadvantages of a squirrel-cage motor over a wound-rotor motor? 5. Why is an induction motor called a generalized transformer? In what respect is the operation of an induction motor different than that of a transformer? 6. Define slip. Deduce an expression for the rotor current frequency in terms of the supply frequency.

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7. Draw and describe the approximate equivalent circuit of an induction motor. Why does the computation using approximate equivalent circuit involve considerable error in induction motor? 8. Show that the rotor copper loss is slip times the power input to the rotor. 9. Derive the equation for the torque developed in an induction motor. 10. Draw and explain the torque-slip cause of a three-phase induction motor. 11. Deduce the expression for slip under maximum torque condition and also find the expression for the maximum torque. 2 where sm is the 12. Prove that the ratio of the starting torque to the maximum torque is 1 sm + sm 1 slip under maximum torque. 13. Prove that for a constant applied voltage, the magnitude of rotor resistance equals that of rotor reactance for the starting torque to be maximum. 14. What are the different losses of an induction motor? How do you find efficiency of such motors? 15. Why do we perform no-load test and blocked-rotor test on induction motors? Describe how we can find circuit parameters from these two tests. 16. Why are starters needed for starting three-phase induction motors? 17. Discuss the principle of operation of a star-delta starter. 18. What are the different methods of speed control of a three-phase induction motor? 19. Discuss the methods of speed control by change of slip. 20. Explain the phenomenon a of crawling and cogging in a three-phase induction motor. 21. What is induction generator? Discuss the principle of operation of an induction generator. 22. How do you compare the operation of a three-phase induction motor with that of a synchronous motor? 23. Explain the procedure of drawing the circle diagram of an induction motor. What information can be obtained from it and how? 24. Discuss the working principle of deep-bar rotor and double-cage induction motor. 25. Describe the operation of three-phase induction motor under single phasing condition.

Qspcmfnt 1. A six-pole, 60 Hz induction motor rotates at 3% slip. Find the speed of stator field, the rotor and the rotor field. What is the frequency of the rotor currents? [1200 rpm, 1164 rpm, 1200 rpm 1.8 Hz] 120 ¥ 60 [Hint: Ns = = 1200 rpm 6

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\ Stator field rotates at 1200 rpm. Rotor field rotates in the air gap in the same speed. N(rotor speed) = Ns (1 – s) = 1200 (1 – 0.03) = 1164 rpm The rotor speed is then 1164 rpm. If frequency of rotor current is fr, fr = sfs = 0.03 ¥ 60 = 1.8 Hz Since rotor rotates at 1164 rpm while the speed of the rotor field is 1200 rpm, hence, the field speed with respect to the rotor is (Ns – N) i.e., 36 rpm]. 2. A three-phase eight-pole squirrel-cage induction motor, connected to a 400 V (L – L) 50 Hz supply, rotates at 3% slip at full-load. The copper and iron losses at the stator are 2 kW and 0.5 kW respectively. If the motor takes 50 kW at full-load, find the full-load developed torque at the rotor. [605 Nm] [Hint:

Pag = Pin – Pscu – Psc = 50 – 2 – 0.5 = 47.5 kW

\

T=

Pag ws

=

47.5 ¥ 103 = 605 Nm] 120 ¥ 50 2p ¥ 8 ¥ 60

3. The power input to a three-phase induction motor is 50 kW. Stator loss is 1 kW. Find the gross mechanical power developed in the rotor and the rotor copper loss per phase when the motor has a full-load slip of 4%. [15.68 KW, 1.96 KW] [Hint: Pin = 50 kW; s = 0.04; Pscu = 1 kW \

Pag = Pin – Pscu = 49 kW Ê 1.96 ˆ Prcu = s ¥ Pag = 0.04 ¥ 49 = 1.96 kW Á = kW Per phase˜ Ë ¯ 3

\

Ê 49 ˆ 1.93 = 15.68 kW] Pm = Pag – Prcu = Á ˜ Ë 3¯ 3

4. The loss at the stator of a three-phase squirrel-cage 25 HP, 1500 rpm induction motor is 2 kW. What is rotor mechanical power if the rotor copper loss is 1 kW? What is the running slip? [15.65 kW; 6%] [Hint:

Pin = 25 ¥ 746 ¥ 10–3 = 18.65 kW

\

Pag = 18.65 – 2 = 16.65 kW Pm = Pag – Prcu = 16.65 – 1 = 15.65 kW

E

Prcu = s ¥ Pag; s =

Prcu 1000 = = 0.06, i.e. slip is 6%] Pag 16650

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5. The rotor of a six-pole, 50 Hz, slip-ring induction motor has a resistance of 0.3 W/phase and it runs at 960 rpm at full-load. How much external resistance/phase must be added to the rotor circuit to reduce the speed to 800 rpm, the torque remaining constant? [1.2 W] [Hint:

120 f = 1000 rpm P 1000 - 960 = 0.04 sf l (full-load slip) = 1000 Ns =

If r be the additional resistance per phase in rotor circuit, we can write

snew R2 + r . = R2 sfl

Since the power input to the rotor and rotor current remain constant for constant torque and Rotor Cu loss hence, from the relation, slip = , we have Rotor input snew 3I 22 ( R2 + r ) R2 + r = = . R2 sfl 3I 22 R2 Substitution of the values of sf l = 0.04 snew =

1000 - 800 = 0.2 and R2 = 0.3, yields r = 1.2 W] 1000

6. A three-phase, 50 Hz induction motor has an output rating of 500 HP, 3.3 kV (L – L). Calculate the approximate full-load current at 0.85 p.f., locked-rotor current and no-load current. What is the apparent power drawn under locked rotor condition? Assume the starting current to be 6 times full-load current and no-load current to be 30% of full-load current. [If l = 76.78 A; Ino-load = 23.093 A; Ilock rotor = 460.68 A; Plocked rotor = 2633 kVA] [Hint: \ E Hence,

If l = Ino-load Ilock rotor Istart

500 ¥ 746

= 76.78 A 3 ¥ 3300 ¥ 0.85 = 0.3 ¥ 76.78 = 23.03 A ∫ Istart = 6 ¥ If l = 460.68 A

Apparent power drawn during locked rotor condition is PA = 3 ¥ VL ¥ Ist = 3 ¥ 3300 ¥ 460.68 = 2633 kVA] 7. A four-pole, 60 Hz, 460 V, 5 HP induction motor has the following equivalent circuit parameters: Rs = 1.21 W; R¢2 = 0.742 W; X0 = 65.6 W

Xs = 3.10 W X¢2 = 2.41 W

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Find the starting and no-load current of the machine. [46.15 ––70.67° A; 3.87 ––89° A] [Hint: With reference to the equivalent circuit of the induction motor, the input impedance looking from the input side is Zin = (Rs + jXs) +

jXo ( R2¢ + jX 2¢ ) R2¢ + jX 2¢ + jXo

È j 65.6 ( 0.742 + j 2.41) ˘ = Í(1.21 + j 3.1) + W = 5.75 –70.72° W 0.742 + j 2.41 + j 65.6 ˙˚ Î At start, s = 1.0. This means the load resistor in equivalent circuit is shorted, since 1 – s = 0 \

Ist =

VL - L 3Zin

= 46.15 ––70.67° A

At no-load, s = 0, i.e., the load element in the equivalent circuit is open. \

Zin (no-load) = (Rs + jXs) + jX0 = (1.21 + j68.7) W = 68.71 –89° W

\

INL =

VL - L 3Zin (NL)

=

460 –0∞ 3 (68.71–89∞)

= 3.87 ––89° A]

8. A 30 HP, three-phase, six-pole, 50 Hz slip-ring induction motor runs at full-load at a speed of 960 rpm. The rotor current is 30 A. If the mechanical loss in the rotor is 1 kW while 200 W loss is being incurred by the rotor short-circuiting system, find the rotor resistance per phase. [R2 = 0.287 W] [Hint: E rotor copper loss = for this problem,

Ns = 1000 rpm; s =

Ns - N = 0.04 Ns

s ¥ Gross mechanical power developed in rotor Hence, we can write 1- s 0.04 (30 ¥ 746 + 1000) 1 - 0.04 3 ¥ 302 ¥ R2 = 774.17 R2 = 0.287 W]

3 I 22 R2 + 200 = or, or,

9. A 10 HP, 400 V(L – L), 50 Hz, three-phase induction motor has a full-load p.f. of 0.8 and efficiency of 0.9. The motor draws 7 A when a voltage of 160 V is applied directly across the live terminals, the motor being standstill. Determine the ratio of starting to full-load current when a star-delta starter is used to start the motor. [(Ist/If l) = 0.39] [Hint:

If l =

10 ¥ 746 3 ¥ 400 ¥ 0.8 ¥ 0.9

= 15 A

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Is/c at 160 V input = 7.0 A \

Is/c at 400 V(L – L) is

400 ¥ 7.0 = 17.5 A 160

With star-delta starter, Istarting = I starting

\

If l

=

1 1 ¥ Is/cf = ¥ 17.5 = 5.833 A 3 3 5.833 = 0.39] 15

10. A 75 kW, three-phase induction motor has 1500 rpm synchronous speed. It is connected across a 440 V(L – L) supply and rotates at 1440 rpm at full-load. The two-wattmeter method is applied to measure the power input which shows that the motor absorbs 70 kW while the line current is 80 A. Find (a) the power supplied to the rotor (b) the rotor copper loss (c) the mechanical power developed at shaft (d) the torque developed at rotor (e) the efficiency of the motor Assume stator resistance/phase = 0.2 W. [Ans: Pag = 64.16 kW; Prcu = 2.57 kW; Pm = 61.59 kW; T = 408.66 Nm; h = 85.84%] [Hint: (a) Pin = 70 kW; Pscu (stator Cu loss) = 3I 2fl ¥ Rs = 3 ¥ 802 ¥ 0.2 = 3.84 kW Psc (core loss in stator) = 2 kW Pag = Pin – Pscu – Psc = 64.16 kW

\

(b) Prcu = s ¥ Pag = 0.04 ¥ 64.16 = 2.57 kW (c) Pm = Pag – Prcu = 64.16 – 2.57 = 61.59 kW 64.16 ¥ 103 = 408.66 Nm 1500 ws 2p ¥ 60 P0 Pm - Mech ◊ loss in rotor = (e) h = Pin Pin

(d) T =

Pag

=

=

61.59 - 1.5 = 0.8584 i.e., 85.84%] 70

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11. A four-pole, three-phase 50 Hz induction motor develops a maximum torque of 20 Nm at 1440 rpm. Obtain the torque exerted by the motor at 5% slip. Assume rotor resistance to be 0.5 W. [19.51 Nm] 120 f = 1500 rpm P N = 1440 rpm

Ns =

[Hint:

\ But

Ns - N = 0.04 = smax Ns R = 2 X2

s= smax

\

X2 =

R2 0.5 = = 12.5 W smax 0.04

If s be the required slip (5%) then from the relation, T Tmax \

=

2 ◊ s ◊ smax s + 2

2 smax

=

2 ¥ 0.05 ¥ 0.04 (0.05) 2 + (0.04) 2

= 0.9756

T = 0.9756 ¥ 20 = 19.51 Nm]

12. You have a 50 HP, three-phase, 60 Hz, four-pole, 1765 rpm induction motor operating at 400 V (L – L). Find the shaft torque. If the mechanical loss in the rotor is 500 W, find the shaft and the developed torque. If the rotor copper loss is 800 W, what is the value of air-gap power? Also find the electromagnetic torque developed. If the total stator losses are 1 kW, find the motor input power. [Tsh = 201.91 Nm, Tm = 204.62 Nm, Pag = 38.6 kW; T = 204.89 Nm; Pin = 39.6 kW] P 50 ¥ 746 [Hint: Tsh = 0 = = 201.91 Nm 1765 w 2p ¥ 60 Pm = P0 + Mechanical loss in rotor = 50 ¥ 746 ¥ 10–3 + 500 ¥ 10–3 = 37.8 kW Pm 37.8 ¥ 103 = = 204.62 Nm 1765 w 2p ¥ 60 Pag = Pm + Rotor copper loss Tm =

= 37.8 ¥ 103 + 800 = 38600 W = 38.6 kW

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T=

Pag ws

=

38.6 ¥ 103 = 204.89 Nm 1800 2p ¥ 60

120 ¥ 60 Ê ˆ = 1800 rpm˜ ÁËE Ns ¯ 4

E stator copper loss is 1 kW. \ Pin = 1 + Pag = 1 + 38.6 = 39.6 kW] 13. A 400 V, four-pole, 50 Hz, three-phase squirrel-cage induction motor develops 25 HP at 4% slip on full-load. If the ratio of motor resistance to standstill reactance is 1:4, find the ratio of starting torque to full-load torque. [(Ts/T) = 1.51] [Hint: Ns = 1500 rpm; N = Ns(1 – s) = 1440 rpm (E s = 0.04) T(full-load) =

25 ¥ 735.5 = w

25 ¥ 735.5 = 0.122 Nm 1440 3 2p ¥ ¥ 10 60

At standstill the torque is Ts while at 4% slip it is T. s0 R2 E12 \

Ts R22 + ( s0 X 2 ) 2 = , (s0) being the slip at starting. T sR2 E12 R22 + ( sX 2 ) 2

With s0 = 1, 2

2 Ê R2 ˆ 2 Ê 1ˆ 2 + s ÁË 4 ˜¯ + (0.04) ÁË X ˜¯ Ts R22 + ( sX 2 ) 2 2 = = = 1.51] = 2 2 2 T ÈÊ 1 ˆ 2 ˘ È ˘ s [ R2 + X 2 ] Ê R2 ˆ + 1˙ 0.04 ÍÁ ˜ + 1˙ s ÍÁ ÍÎË 4 ¯ ÍË X 2 ˜¯ ˙˚ ˙ Î ˚

14. A 415 V, three-phase, six-pole, 50 Hz induction motor runs at a slip of 4% on full-load. The rotor resistance and reactance are 0.01 W and 0.05 W when the motor is at standstill. Find the ratio of full-load torque to maximum torque. Also obtain the speed at which the maximum torque occurs. [Nmax = 800 rpm; (T/Tmax) = 0.385] [Hint:

smax =

R2 0.01 = = 0.2. X 2 0.05

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The speed at a slip of 0.2 would then be 800 rpm as Ns = 1000 rpm. T

E

Tmax T

For this problem,

Tmax

= =

2 ◊ s ◊ smax 2 s 2 + smax

2 ¥ 0.04 ¥ 0.2 (0.04) 2 + (0.2) 2

= 0.385]

15. For a 30 HP, 440 V, 50 Hz, three-phase, six-pole star connected motor, no-load test and blocked rotor tests are conducted. The test results are as follows: No-load test No-load voltage V0 line to line: 440 V No-load current I0: 15 A No-load power Po: 1550 W Resistance measured for each stator phase winding = 0.5 W Locked rotor test Input voltage (L – L) : 160 V Input current : 60 A Input power : 7000 W Determine the equivalent circuit of the motor. Assume stator leakage reactance to be is equal to rotor stand still reactance referred to stator side. [R0 = 130.27 W, X0 = 17.08 W, Rs = 0.5 W; R¢2 = 0.148 W, Xs = 0.7 W, X¢2 = 0.7 W] [Hint:

Rs = 0.5 W. W0 = cos q0 = 3V0 I 0

1500 = 0.13 440 3¥ ¥ 15 3 IC = I0 cos q0 = 15 ¥ 0.13 = 1.95 A If = I0 sin q0 = 15 ¥ 0.99 = 14.873 A

\

R0 =

V0 440 / 3 = = 130.27 W Ic 1.95

X0 =

V0 440 / 3 = = 17.08 W If 14.873

Next, from locked rotor test, cos qs/c =

Ws / c = 3 (Vs / c )( Is / c )

7000 = 0.421 160 3¥ ¥ 60 3

!

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Fmfdusjdbm!Nbdijoft

Vs / c 160 / 3 cos qs/c = 0.421 I s/c 60 = 0.648 R¢2 = 0.648 – Rs = 0.148 W

\

Rs + R¢2 =

or

Xs + X¢2 =

Vs / c 160 / 3 sin qs/c = 0.91 I s/c 60 = 1.4 W

1.4 = 0.7 W. 2 The approximate equivalent circuit is shown below. Since Xs = X2¢, hence, Xs = X2¢ =

R0 = 130.27 W; X0 = 17.08 W; Rs = 0.5 W; R¢2 = 0.148 W, Xs = 0.7 W; X 2¢ = 0.7 W] 16. An induction motor has a short circuit current of 6 times the full-load current at normal supply voltage. It has a full-load slip of 5%. Calculate the starting torque in terms of the full-load torque if started by (a) star-delta starter (b) auto-transformer starter with 60% tapping Neglect magnetizing branch currents. [0.2; 0.648] [Hint: 2

Ts = Ê I s ˆ Á ˜ ◊ sfl Tf l Ë If l ¯

we have

Ê 1 ˆ (a) For Y-D starting: At start, winding is placed in star and only Á ˜ of normal voltage Ë 3¯ is applied. Again, short-circuit current being 6 times the full-load current (winding in delta connection), we can write short-circuit phase current as Is/c(ph) = \ as i.e., \

6 ¥ f .l. current

Istarting/phase = 1 3

3 1 3

¥ Is/c(ph)

of normal voltage is applied per phase, Ist/ph =

1 3 1

¥

6 3 6

¥ If l.

I st = ¥ =2 3 3 Ifl

Uisff.qibtf!Joevdujpo!Npups

Hence,

9/256

Ts = (2)2 ¥ 0.05 = 0.2 Tfl

(b) Auto-transformer starting with 60% tapping: At start, stator winding remain in delta connection. However, only 60% voltage is made available at stator. \ Hence,

Is =

60 ¥ 6 If l = 3.6 If l 100

Ts = (3.6)2 ¥ 0.05 = 0.648] Tfl

17. The rotor of a four-pole, 50 Hz slip-ring induction motor has a resistance of 0.2 W per phase and runs at 1440 rpm at full-load. Determine the external resistance per phase which must be added to reduce the speed to 1200 rpm, the torque remaining same in both the cases. [0.8 W] 18. An eight-pole 400 V, 50 Hz delta-connected, three-phase induction motor has stator to rotor turns ratio 2:1. The stator impedance per phase is (0.13 + j0.6) W and rotor circuit standstill impedance per phase is (0.035 + j0.15). Determine (a) the maximum torque developed and its corresponding slip, and (b) the starting torque. [2280.42 Nm, 11.57%, 520.62 Nm] 19. An eight-pole, 50 Hz, three-phase wound rotor induction motor has rotor winding resistance and standstill reactance per phase of 0.03 W and 0.15 W respectively. Determine (a) the speed at which the developed torque is maximum, and (b) the external resistance per phase to be added to the rotor circuit in order to develop three-fourth of maximum torque at starting. [600 rpm, 0.03 W] 20. Calculate the full-load efficiency of a 415 V, three-phase, 50 Hz delta connected induction motor from the following tests: No-load power intake is 1500 W at rated input voltage. Full-load line current is 50 A, power factor is 0.85 and slip is 0.04. Resistance of stator winding per phase is 0.5 W. Assume ratio of stator core loss to friction and windage loss as 3:2. [87.28%] [Hints: Motor input 3 ¥ 415 ¥ 50 ¥ 0.5 = 30548 W at full-load 2

Ê 50 ˆ Stator copper loss on full-load = 3 ¥ Á ˜ ¥ 0.5 Ë 3¯ = 1250 W No-load input power = Stator core loss + Friction and windage loss 3 \ stator core loss = ¥ 1500 = 900 W 5 2 Friction + Windage loss = ¥ 1500 = 600 W 5

!

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Fmfdusjdbm!Nbdijoft

Power input to rotor

Pag = 30548 – (1250 + 900) = 28398

\ rotor copper loss = sPag = 1136 W Shaft power = 28398 – (1136 + 600) = 26,662 W \

h=

26662 ¥ 100% = 87.28%] 30548

Nvmujqmf.Dipjdf!Rvftujpo! 1. The slip of an induction motor normally does not depend on (a) rotor speed (b) synchronous speed (c) shaft torque (d) core-loss component [GATE 2012] 2. A balanced three phase voltage is applied to a star connected induction motor, the phase to neutral voltage being V. The stator resistance, rotor resistance referred to the stator, stator leakage reactance, rotor leakage reactance referred to the stator, and the magnetizing reactance are denoted by rs, rr, xs, xr and xm respectively. The magnitude of the starting current of the motor is given by (a) (c)

V ( rs + rr ) + ( xs + xr ) 2

V ( rs + rr ) + ( xm + xr ) 2

V

(b)

2

rs2 +

V

(d)

2

( x s + xm ) 2

rs2 +

( xm + xr ) 2

[GATE 2010]

3. A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor (solid curve) and of the load (dotted curve) are shown. Of the two equilibrium points A and B, which of the following options correctly describes the stabilities of A and B?

B

Torque A

O

1

N Ns

[GATE 2009] (a) A is stable, B is unstable (b) A is unstable, B is stable (c) Both are stable (d) Both are unstable [Hints: A is unstable as at A slip is nearly equal to 0 and B is stable as at B slip is nearly equal to 1]

Uisff.qibtf!Joevdujpo!Npups

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4. On the torque/speed curve of induction motor shown in the figure four points of operation are marked as W, X, Y and Z. Which one of them represents the operation at a slip greater than 1? Torque X Y W

O

Speed Z

(a) W (c) Y

(b) X (d) Z

[GATE 2005]

[Hints: When speed = 0, slip = 1. When speed is negative slip > 1] 5. For an induction motor, operating at a slip s, the ratio of gross power output to air-gap power is equal to (b) (1 – s) (a) (1 – s)2 (c) (1 - s)

(d) (1 – s )

[GATE 2005]

6. The direction of rotation of a three phase induction motor is clockwise when it is supplied with three phase sinusoidal voltage having phase sequence A–B–C. For counterclockwise rotation of the motor, the phase sequence of the power supply should be (a) B – C – A (b) C – A – B (c) A – C – B (d) B – C – A or C – A – B [GATE 2004] [Hints: For reversing the direction of rotation any two phases should be interchanged] 7. A 3-phase, 4-pole squirrel cage induction motor has 36 stator and 28 rotor slots. The number of phases in the rotor is (a) 3 (b) 9 (c) 7 (d) 8 [GATE 2000] [Hints: The number of phases in a three phase induction motor is 3 in both stator and rotor.] 8. The following starting method for an induction motor is inferior view of the poor starting torque per ampere of the line current drawn: (a) Direct line starting (b) Auto transformer method of starting (c) Series induction method of starting (d) Star-delta method of starting 9. If an induction motor is run at above synchronous speed, it acts as (a) a synchronous motor (b) an induction generator (c) an induction motor (d) none of these

!

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Fmfdusjdbm!Nbdijoft

10. In an induction motor if the air gap is increased (a) efficiency will improve (b) speed will reduce (c) breakdown torque will reduce (d) power factor will be lowered 11. Unbalanced supply voltage given to a three phase, delta connected induction motor will cause (a) less heating of the rotor (b) negative sequence component current (c) zero sequence current (d) all of these 12. When the supply voltage to an induction motor is reduced by 10%, the maximum torque will decrease by approximately (a) 10% (b) 20% (c) 40% (d) 5% 13. The stator of a 4 pole, 3 phase induction motor is fed from 50 Hz three phase supply which contains pronounced 5th time harmonic. The speed of the 5th space harmonic field produced by the fifth time harmonic in the stator supply will be (a) 1500 rpm (b) 2000 rpm (c) 7500 rpm (d) 3000 rpm 14. The mmf produced by the rotor currents of a three phase induction motor (a) rotates at slip speed with respect to the stator mmf (b) rotates at synchronous speed with respect to the rotor (c) rotates at the speed of rotor in the air gap (d) is at standstill with respect to stator mmf 15. The variable resistance representing the mechanical load in the equivalent circuit of a three phase induction motor is given by (a) r2 ÊÁ1 - ˆ˜ Ë s¯ 1

(b) r2(1 – s)

(d) r2 (s – 1) (c) r2 ÊÁ - 1ˆ˜ Ës ¯ 16. The no load current of a three phase induction motor in terms of its rated current is (a) 30 to 50% (b) 20 to 30% (c) 10 to 20% (d) 3 to 5% 17. In a three phase induction motor the electromagnetic torque Te at any slip s and maximum torque Tm at slip sm are related as 1

(a)

2s Te = 2 m2 Tm s + sm

(b)

(c)

Te ss = 2 m2 Tm s + sm

(d)

2s s Te = 2 m2 Tm s + sm 2s s 2 + sm2

18. A star-delta starter is equivalent to an autotransformer starter with a tapping of (a) 58% (b) 57% (c) 86.6% (d) 57.73%

Uisff.qibtf!Joevdujpo!Npups

9/25:

19. Induction generator delivers power at (a) lagging power factor (b) leading power factor (c) zero power factor (d) unity power factor 20. The torque slip characteristics of a polyphase induction motor becomes almost linear at small vales of slips because in this range of slips (a) rotor reactance is equal to the stator reactance (b) effective rotor circuit resistance is very large compared to the rotor reactance (c) rotor resistance is equal to the stator resistance (d) rotor resistance is equal to the rotor reactance 21. If the stator impedance of a three phase induction motor is neglected, the maximum torque is (a)

3V12 w s r2

(b)

3V12 2 w s x2

(c)

3V12 w s x2

(d)

3V12 2 w s r2

22. In a polyphase induction motor starting torque can be increased by (a) increasing the number of poles (b) increasing the supply frequency (c) using double cage rotor (d) none of these. 23. Under blocked rotor condition the frequency of rotor current in a 50 Hz, three phase induction motor is (a) 50 (b) about 50 (c) very low (d) very high 24. The usual value of slip of a three phase induction motor at full load is about (a) 0.8 (b) 0.1 (c) 0.05 (d) 0.3 25. The speed regulation of a three phase induction motor from no load to full load is about (a) 20% (b) 40% (c) 3% (d) 10%

Botxfst 1. 5. 9. 13. 17. 21. 25.

(d) (b) (b) (c) (b) (b) (c)

2. 6. 10. 14. 18. 22.

(a) (c) (d) (d) (d) (c)

3. 7. 11. 15. 19. 23.

(b) (a) (b) (c) (b) (a)

4. 8. 12. 16. 20. 24.

(a) (c) (b) (a) (b) (c)

: Tjohmf.qibtf!Joevdujpo! Npupst JOUSPEVDUJPO!

:/2

Single-phase induction motors have numerous and diversified applications, both in homes and the industry. It is probably safe to say that single-phase induction motor applications far outweigh three-phase motor applications in the domestic sector. At homes, normally only single-phase power is provided, since power was originally generated and distributed to provide lighting. For this reason, early motor-driven appliances in homes depended on the development of the single-phase motor. Single-phase induction motors are usually small-sized motors of fractional kilowatt rating. They find wide applications in fans, washing machines, refrigerators, pumps, toys, hair dryers, etc. Single-phase induction motors operate at low power factors and are less efficient than three-phase induction motors.

QSPEVDUJPO!PG!UPSRVF!

:/3

From the study of three-phase induction motors, it is seen that the three-phase distributed stator winding sets up a rotating magnetic field which is fairly constant in magnitude and rotates at synchronous speed. In a single-phase induction motor, there is only single-field winding excited with alternating current and, therefore, it is not inherently self-starting since it does not have a true revolving field. Various methods have been devised to initiate rotation of the squirrel-cage rotor and the particular method employed to start the rotor of single-phase motor will designate the specific type of motor. Consider the behaviour of the magnetic field set up by an ac current in the single-phase winding. With reference to Fig 9.1, when the sinusoidal current is flowing in the field winding, neglecting the

!

:/3

Fmfdusjdbm!Nbdijoft

saturation effects of the magnetic iron circuit, the flux through the armature will vary sinusoidally with time. The magnetic field created at a particular instant of time will reverse during the next half-cycle of the ac supply voltage. Since the flux is pulsating, it will induce currents in the rotor bars which, in turn, will create a rotor flux which by Lenz’s law opposes that of the main field. The direction of the rotor current as well as the torque created can also be determined. It is apparent that the clockwise torque produced is counteracted by the counter-clockwise torque and so no motion results, i.e. the motor is at standstill. Squirrel cage rotor

Simple field wind

Main flux

Single phase ac

l

Direction of torque, shown by small arrows

Gjh/!:/2

Hence, a single-phase induction motor has no inherent self-starting torque. When the rotor is made to rotate by auxiliary means in any one direction, it continues running in that direction. For explaining the running performance of the single-phase induction motor, two different theories, viz. revolving-field theory and cross-field theory have been adopted.

:/3/2! Sfwpmwjoh.Gjfme!Uifpsz! According to this theory, the pulsating stationary mmf wave can be revolved into two counterrotating mmf waves of equal magnitudes and rotating at synchronous speed. If two sinusoidally distributed mmfs, each of magnitude Fmax/2, rotate in opposite directions, their combined effect is equivalent to one pulsating field Fmax cos wt varying between +Fmax and –Fmax which is shown in Fig. 9.2(a). A physical interpretation of the two oppositely rotating field components is depicted in Fig. 9.2(b). Assuming the stator mmf wave to be sinusoidally distributed in space and varying sinusoidally with time, it can be represented as Fs = Fs max sinwt cos a where Fs max is the peak value corresponding to maximum instantaneous alternating current in stator winding and a is the space-displacement angle measured from the stator main-winding axis. Here, sin wt indicates that mmf variation is sinusoidal with time and the term cos a indicates its co-sinusoidal distribution in space along the air-gap periphery.

Tjohmf.qibtf!Joevdujpo!Npupst

Now

:/4

Fs = Fs max sin wt cos a =

1 1 Fs max sin (wt – a) + Fs max sin (wt + a) 2 2

1 Fs max and they travel in opposite 2 directions. The first wave whose argument is (wt – a) travels in the forward direction and the other, whose argument is (wt + a), travels in the backward direction. Each field component acts independently on the rotor in a similar fashion as the rotating field in a three-phase induction motor. The only difference is that here there are two fields, one tending to rotate the rotor clockwise and the other tending to rotate the rotor anticlockwise. If the rotor runs at a speed nr the rotor speed relative to the forward field is (ns – nr) where ns is the synchronous speed. \ slip of the rotor current due to forward field Hence, the two mmf components have maximum value of

sf =

ns + nr =s ns

The rotor speed relative to the backward field is ns + nr. \ slip of the rotor current due to backward field is sb =

ns + nr =2–s ns

Hence, the forward field induces rotor currents of frequency sf and backward field of frequency (2 – s) f in the same rotor conductors. At standstill, both forward and backward fields rotate at synchronous speed ns with respect to rotor conductors. Hence, both these fields induce equal rotor emfs, equal rotor currents and produce equal rotor mmfs. These rotor mmfs react with their corresponding equal counter-rotating stator mmfs and hence both forward and backward rotating fields are equal with the rotor at standstill. When the rotor rotates at a speed of nr, the speed of forward flux wave is (ns – nr) with respect to rotor conductors. This small relative speed induces small rotor emf, small rotor current which, in turn, produces small rotor mmf as compared to their values at standstill. The rotor frequency sf results in less rotor leakage reactance and hence a better rotor power factor. The reduced rotor mmf component at a better power factor is much smaller than its magnitude at standstill and it opposes the constant forward mmf wave. Hence, the forward flux wave becomes higher than its value at standstill. On the other hand, at rotor speed nr, the speed of backward field becomes ns + nr with respect to the rotor conductors. Due to this high relative speed, the backward field induces large rotor emf, large rotor current and hence large rotor mmf, as compared to their standstill values. As the rotor frequency is (2 – s) f, the rotor power factor is poor. This large rotor mmf component at poor power factor is much greater than its value at standstill which opposes the constant backward rotating mmf wave. Hence, the backward flux wave is much reduced from its magnitude at standstill.

!

:/5

Fmfdusjdbm!Nbdijoft

From the above discussion, it is clear that as the speed increases, the backward flux wave decreases and the forward flux wave increases. Hence, in the under running condition, the forward field torque Tf is greater than the backward field torque Tb and the net torque (Tf – Tb) is in the direction of rotor rotation. In the normal operating region, the torque speed curve of a single-phase induction motor is not far inferior from the three-phase induction motor as the forward flux wave at small slips is several times greater than the backward flux wave and the resultant air-gap field is similar to three-phase induction motor. Fsmax

F = F1 + F2 F2 1 F smax = F1 2

ax

1 F sm = F2 2

F1

(a)

(b)

Gjh/!:/3

(forward)

Torque

The torque–slip curve of the motor can be obtained by applying the principle of superposition. The superposition of two torques Tf and Tb, as shown in Fig. 9.3, gives the required torque–slip curve of a single-phase induction motor. From Fig. 9.3, it can be inferred that when the rotor speed is zero and the rotor speed is slightly less than ns, the motor torque T is zero. Also, the motor can run in either direction.

Standstill

–ns

T = Tf – Tb nr < n s

1 0

Tb

(backward)

2.0

Tf

0 SLIP

ns

Gjh/!:/4

The single-phase induction motor having a simple winding, once started, will continue to run in the direction in which it is started. The manual self-starting is not a desirable practice and modifications are introduced to obtain the torque required to start. To accomplish this, a quadrature flux

Tjohmf.qibtf!Joevdujpo!Npupst

:/6

component in time and space with the stator flux must be provided at standstill. Auxiliary windings are normally placed on the stator to provide starting torque.

:/3/3! Dsptt.gjfme!Uifpsz! The cross-field theory is an application of two axes or generalized theory of electrical machines. The performance of induction motor under normal running condition can be explained by this theory. Hence the flux is resolved into two components—one component acting along the stator winding axis and the other at right angles to it as shown in Fig. 9.4. x1 x2 and y1 y2 represent the two groups of short-circuited rotor conductors. x1 x2 links with one component flux and y1 y2 links with the other component flux. fd

V

x1

y1

x2

fq

y2

Gjh/!:/5

When the motor is at standstill, the stator applied voltage V sets up an mmf in q-axis and this mmf causes flux fq in this axis. This flux links with turn y1 y2 and induces the transformer emf in this turn which causes a current Iq. This current sets up an opposing mmf along the q-axis. Actually the resultant of these two mmfs sets up the fq. As both these mmfs are along the same axis, no torque is produced. The axis of the coil x1x2 is 90° to q-axis. Hence, no emf is induced in this coil and no current flows. Therefore, there is no torque in either coils. When the motor is running, a back emf is induced in the stator winding which is equal to the voltage applied to the stator provided the voltage drops in the stator resistance and leakage reactances are neglected. The flux fq is constant. The flux fq sets up a motional emf in the coil x1x2. A current Id starts flowing in the coil as it is short-circuited. This current sets up an mmf and flux fd in the d-axis. The interaction of flux fq with current Id and of flux fd with current Iq produces torque and the motor continues to run.

FRVJWBMFOU!DJSDVJU!

:/4

At standstill, the equivalent circuit of a single-phase induction motor is exactly similar to that of a transformer on short circuit. The equivalent circuit at standstill condition is shown in Fig. 9.5(a). Rc and Xf represent the core loss and magnetizing reactance. R1 and x1 are the resistance and leakage

!

:/7

Fmfdusjdbm!Nbdijoft

reactance of the stator, r¢2 and x¢2 are the resistance and leakage reactance of the rotor referred to the stator. r1

x1

r¢2

Rc

Xf

x¢2

Gjh/!:/6)b*

The air-gap flux can be resolved into two oppositely rotating components. These components at standstill are equal in magnitude, each one contributing an equal share to the resistive and reactive voltage drops in the rotor circuit. Hence, r2 and x2 can be split into two parts, each one corresponding to the effects of one of the magnetic fields. Ef and Eb are the voltages set up by the two oppositely rotating fields, viz. forward and backward rotating fields respectively. The equivalent circuit considering the effects of forward and backward flux components is shown in Fig. 9.5(b). r1

x1 r¢2 Rc

Xf

2

2

2

x¢2

Ef (Forward)

2

V1 r¢2 Rc

Xf

2

2

2

x¢2

Eb (Backward)

2

Gjh/!:/6)c*

When the motor is running at a slip s, the slip for the forward field is s and for the backward Ê r¢ ˆ field is (2 – s). Hence, the resistance in the forward field becomes Á 2 ˜ and in the backward field Ë 2s ¯ Ê r2¢ ˆ Ê r2¢ ˆ Ê r¢ ˆ becomes Á . As s is normally very small, Á 2 ˜ is much higher than Á . Hence, Ef ˜ Ë ¯ 2s Ë 2( 2 - s) ¯ Ë 2( 2 - s) ˜¯ is much greater than Eb.

Tjohmf.qibtf!Joevdujpo!Npupst

:/8

Then equivalent circuit at any slip (s) is shown in Fig. 9.6. r1

x1 r¢2 Rc 2

Xf 2

2

Ef (Forward)

x¢2 2

r¢2 Rc

Xf

2

2

Eb (Backward)

2

x¢2 2

Gjh/!:/7

Pgf = Air-gap power of forward field = ( I 2¢ f )2

r2¢ W 2s

Pgb = Air-gap power of backward field = ( I 2¢ b )2 Tf = Torque due to forward field =

Pgf 2p ns

Tb = Torque due to backward field =

r2¢ W 2( 2 - s)

Nm

Pgb 2p ns

Nm

Net torque, T = Tf – Tb Rotor copper loss due to forward field (Pcu(rot.) f ) = sPgf Rotor copper loss due to backward field (Pcu(rot.)b) = (2 – s)Pgb Total rotor copper loss (Pcu(rot.) = sPgf + (2 – s)Pgb Mechanical power developed (=Pm) = (1 – s) (Pgf – Pgb)

! Qspcmfn!:/2 B!311!X-!351!W-!61!I{!tjohmf.qibtf!joevdujpo!npups!svot!po!sbufe!mpbe!xjui!b!tmjq!pg!1/16!q/v/!Uif! qbsbnfufst!bsf! ! !

s2! >!22/5!W-!y2!>!25/6!Ws¢3! >!24/9!W-!y3¢ !>!25/5!W-!Yf!>!381!W

Dbmdvmbuf!)b*!qpxfs!gbdups-!)c*!joqvu!qpxfs-!boe!)d*!fggjdjfodz/

!

:/9

Fmfdusjdbm!Nbdijoft

Solution From Fig. 9.6, neglecting the core loss resistance RC, the total series impedance X f Ê r2¢ x¢ ˆ X f Ê r2¢ x2¢ ˆ + j 2˜ j Á ÁË + j ˜¯ 2¯ 2 Ë 2( 2 - s) 2 2s 2 + Z = r1 + j x1 + Ê X f x2¢ ˆ Ê X f x2¢ ˆ r2¢ r2¢ + jÁ + ˜ + jÁ + ˜ Ë 2 Ë 2 2¯ 2s 2 ¯ 2( 2 - s) j

270 ¥ 14.4 270 ¥ 13.8 270 ¥ 14.4 + + j 4 4 4 ¥ 0.05 = 11.4 + j14.5 + + 13.8 13.8 Ê 270 14.4 ˆ + + jÁ + ˜¯ Ë 2( 2 - 0.05) 2 ¥ 0.05 2 2 -

= 11.4 + j14.5 +

-972 + j18630 -972 + j 477.69 + 138 + j142.2 3.538 + j142.2

= 11.4 + j14.5 +

18655 – 92.98∞ 1083 – 153.83∞ + 198 – 45.86∞ 142.24 – 88.57∞

270 ¥ 13.8 4( 2 - 0.05) 270 + 14.4 j 2 j

= 11.4 + j14.5 + 94.22 – 47.12° + 7.6 – 65.25° = (11.4 + 64.11 + 3.18) + j(14.5 + 69 + 6.9) = (78.69 + j90.4) W \ input current =

240 – 0∞ 240 – 0∞ = = 2 – – 48.96° A 78.69 + j 90.4 119.85 – 48.96∞

Hence, power factor is (cos 48.96°) lagging, i.e. 0.656 lagging. Input power = 240 ¥ 2 ¥ 0.656 W, i.e. 314.88 W. Output power is 200 W. Hence, efficiency =

Output 200 = = 0.638, i.e. 63.5%. Input 314.88

! Qspcmfn!:/3 B!341!W-!61!I{/!5.qpmf!tjohmf.qibtf!joevdujpo!npups!ibt!uif!gpmmpxjoh!qbsbnfufst;! ! ! ! ! s2!>!3/62!W-!y2!>!5/73!W-!s3¢ !>!8/92!W-!y3¢ !>!5/73!W boe! ! Yf!>!261/99!W Efufsnjof!uif!tubups!nbjo!xjoejoh!dvssfou!boe!qpxfs!gbdups!xifo!uif!npups!jt!svoojoh!bu!b!tmjq! pg!1/16/

Tjohmf.qibtf!Joevdujpo!Npupst

:/:

Solution The total series impedance is obtained as 150.88 Ê 7.81 4.62 ˆ 150.88 Ï 7.81 4.62 ¸ + j + j j Ì ˝ Á ˜ 2 Ë 2 ¥ 0.05 2 ¯ 2 Ó 2( 2 - 0.05) 2 ˛ Z = 2.51 + j4.62 + + 7.81 7.81 Ê 150.88 4.62 ˆ Ê 4.62 150.88 ˆ + + jÁ + + jÁ ˜¯ ˜ Ë 2 Ë 2 ¯ 2 ¥ 0.05 2 2 2( 2 - 0.05) j

= 2.51 + j4.62 +

-174.26 + j 5891.86 174.26 + j151.07 + 78.1 + j 77.75 2 + j 77.75

= 2.51 + j4.62 +

5894.40 – 91.69∞ 230.62 – 139.077∞ + 110.20 – 44.87∞ 77.77 – 88.50

= 2.51 + j4.62 + 53.48 – 46.82° + 2.965 – 50.58° = (2.51 + 36.596 + 1.88) + j (4.62 + 39 + 2.3) = 40.986 + j45.92 = 61.54 – 48.25° W 230 – 0∞ , i.e. 3.73 – – 48.25° A. 61.54 – 48.25∞ Hence, power factor is (cos 48.25°) i.e. 0.666 lagging. Stator main winding current is

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Solution Air-gap power of forward field Pgf = 160 W Air-gap power of backward field Pgb = 20 W Net power = Pgf – Pgb = 160 W – 20 W = 140 W 120 ¥ 50 Synchronous speed Ns = = 1000 rpm 6 Speed of motor Nr = 950 rpm Hence, slip s =

1000 - 950 = 0.05 1000

Power output is (1 – s) ¥ 140 – 75 = 58 W (= shaft power) Shaft torque =

58 shaft power = = 0.58 Nm. 950 95 2p ¥ 2p ¥ 60 6

!

:/21

Fmfdusjdbm!Nbdijoft

EFUFSNJOBUJPO!PG!QBSBNFUFST!PG! FRVJWBMFOU!DJSDVJU

:/5

The parameters of the equivalent circuit of a single-phase induction motor can be determined from the no-load and blocked rotor test.

:/5/2! Cmpdlfe.Spups!Uftu! In this test, a very small voltage is applied to the stator and the rotor is blocked (care is to be taken such that the stator current does not exceed the full-load current.) The voltage, current and power input to the stator are measured. When the rotor is blocked, Ê Xf ˆ x¢ ˘ È r¢ ÊR ˆ is much greater than Í 2 + j 2 ˙ s = 1 and hence parallel combination Á c ˜ and Á Ë 2 ˜¯ Ë 2¯ 2˚ Î2 (in Fig. 9.6). Therefore, under blocked rotor test, the equivalent circuit reduces to that shown in Fig. 9.7. Since (Rc/2) and (Xf/2) are of very high values, hence they can be neglected in the equivalent circuit. Let Vsc, Isc and Wsc be the input voltage, current and power during blocked rotor test. (r1 + r¢2) =

The total resistance Total impedance,

Zsc = Isc

r1

Wsc I sc2

= Rsc

Vsc I sc x1 r2¢ ¸ 2 ÔÔ ˝ (Forward) x2¢ Ô 2 Ô˛

Vsc

r2¢ ¸ 2 ÔÔ ˝ (Backward) x2¢ Ô 2 Ô˛

Gjh/!:/8

Hence, total reactance (x1 + x¢2) = Zsc2 - Rsc2 Generally, r1 = r¢2 and x1 = x¢2. Hence, r1, r¢2, x1 and x¢2 can be determined from this test.

Tjohmf.qibtf!Joevdujpo!Npupst

:/22

:/5/3! Op.Mpbe!Uftu In this test, the motor is run on no-load condition and voltage Vo, current Io and power Wo of the stator are measured. At no load, s is very small and core loss resistance Rc is neglected. Hence, Ê Xf ˆ r2¢ Ê r2¢ ˆ Ê r¢ ˆ . Also, from Fig. 9.6, Á 2 ˜ is much greater than Á Á = ˜ is much smaller than Xf /2. Ë 2 ˜¯ Ë s ¯ 2( 2 - s) Ë 4 ¯ Therefore, under no-load condition, the equivalent circuit can be reduced to that shown in Fig. 9.8. Ê Xf ˆ Ê r¢ ˆ are neglected in equivalent circuit. Here, Á 2 ˜ and Á Ë 2 ˜¯ Ë s¯ No load p.f. (cos f0) =

Wo Vo I o Io

r1

x1 xf 2

Vo

r¢2 4

x¢2 2

Gjh/!:/9

Ê Xf ˆ È ÏÊ r ˆ x ¢ ˆ ¸˘ Ê is ÍVo - I o – fo ÌÁ r1 + 2 ˜ + j Á x1 + 2 ˜ ˝˙ Now voltage across Á ˜ Ë 2 ¯ Ë 4¯ 2 ¯ ˛˚ ÓË Î

Hence,

ÈÊ r¢ ˆ x¢ ˆ ˘ Ê Vo - I o – - fo ÍÁ r1 + 2 ˜ + j Á x1 + 2 ˜ ˙ Ë ¯ Ë 4 2 ¯˚ Î = 2 Io

Xf

and Xf can thus be determined.

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!

:/23

Fmfdusjdbm!Nbdijoft

Solution From Fig. 9.5(a), neglecting the stator impedance and magnetizing reactance Z2 = r¢2 + jx¢2 =

240 = 4.8 50∞ W = 3.08 + j3.68 50 -50∞

r¢2 = 3.08 W and x¢2 = 3.68

\

From Fig. 9.6, at slip s = 0.04, r2¢ x¢ +j 2 2s 2 3.08 3.68 +j = 2 ¥ 0.04 2 = 38.5 + j1.84 W

Forward impedance

Zf = Rf + jXf =

Backward impedance

r2¢ x¢ + j 2 2( 2 - s) 2 3.08 3.68 = + j 2( 2 - 0.04) 2 = 0.786 + j 1.84 W

Zb = Rb + j ¥ b =

\ total input impedance when the motor is running at a slip of 4% is Z1 = Rf + jXf + Rb + jXb = 39.286 + j 3.68 W (neglecting stator impedance) The stator current

V 240 0∞ = A Z1 39.286 + j 3.68 240 0∞ A = 39.457 5.35∞

I1 =

= 6.08 -5.35∞ A Air-gap power of forward field Pgf = I12

r ¢2 = I 12 Rf 2s

Air-gap power of backward field Pgb = I 12

r ¢2 = I 12 Rb 2( 2 - s)

\ mechanical power developed Pm = (1 – s) (Pgf – Pgb) = (1 – 0.04) ¥ (6.08)2 ¥ (Rf – Rb)

Tjohmf.qibtf!Joevdujpo!Npupst

:/24

= 35.488 (38.5 – 0.786) = 1338.39 W Torque developed

T=

Pm Pm = w r (1 - s)w s 1338.39

=

2 ¥ 50 (1 - 0.04) ¥ 2p ¥ 8 = 17.76 Nm.

Nm

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Solution From blocked-rotor test, Zsc =

\

Now, \ and

Vsc 115 = = 19.17 W I se 6

Ê r ¢ ˆ 400 Rsc = r1 + 2 Á 2 ˜ = W = 11.11 W Ë 2 ¯ (6 ) 2 Ê x¢ ˆ Xsc = x1 + 2 Á 2 ˜ = (19.17) 2 - (11.11) 2 Ë 2¯ = 15.62 W r1 = 3 W r¢2 = 11.11 – 3 = 8.11 W 1 x1 = x¢2 = Xsc = 7.81 W 2

From no-load test, Znl = No-load power factor \

230 W = 76.67 W 3

140 = 0.2 230 ¥ 3 sin qnl = 0.98

cos qnl =

!

:/25

Fmfdusjdbm!Nbdijoft

\

Xnl = x1 +

xf 2

+

x2¢ = Znl sin q = 76.67 ¥ 0.98 2

= 75.1366 W Now x1 = x¢2 = 7.81 W \

7.81ˆ Ê Xf = 2 Á 75.1366 - 7.81 ˜ Ë 2 ¯ = 126.84 W

The power input to a single-phase induction motor at no load is equal to core, friction, windage and ohmic loss. From Fig. 9.8, the ohmic loss at no load is r¢ ˆ 8.11ˆ Ê Ê I nl2 Á r1 + 2 ˜ = 32 Á 3 + ˜ = 45.2475 W Ë Ë 4¯ 4 ¯ \ the core, friction and windage losses are 140 – 45.2475 = 94.7525 W.

TUBSUJOH!PG!TJOHMF.QIBTF!JOEVDUJPO!NPUPST!

:/6

Since a single-phase induction motor does not have a starting torque, it needs special methods of starting. The stator is provided with two windings, called main and auxiliary windings, whose axes are space displaced by 90 electrical degrees. The auxiliary winding is excited by a current which is out of phase with the current in the main winding, both currents derived from the same supply. If the phase difference between the two currents is 90° and the mmfs created by them are equal, maximum starting torque is produced. If the phase difference is not 90° and the mmfs are equal, the starting torque will be small, but in many applications, it is still sufficient to start the motor. The auxiliary winding may be disconnected by a centrifugal switch after the motor has achieved about 75% speed. Single-phase induction motors are usually classified according to the auxiliary means used to start the motors. They are classified as follows: 1. 2. 3. 4.

Split-phase motor Capacitor-start motor Capacitor-start capacitor-run motor Shaded-pole motor

Tjohmf.qibtf!Joevdujpo!Npupst

:/26

TQMJU.QIBTF!JOEVDUJPO!NPUPST!

:/7

One of the most widely used types of single-phase motors is the split-phase induction motor. Its service includes a wide variety of applications such as refrigerators, washing machines, portable hoists, small machine tools, blowers, fans, centrifugal pumps, etc. The essential parts of the split-phase motor is shown in Fig. 9.9(a). It shows auxiliary winding, also called the starting winding, in space quadrature, i.e., 90 electrical degrees displacement with the main stator winding. The rotor is normally of squirrel-cage type. The two stator windings are connected in parallel to the ac supply. A phase displacement between the winding currents is obtained by adjusting the winding impedances, either by inserting a resistor in series with the starting winding or as is generally the practice, by using a smaller gauge wire for the starting winding. A phase displacement of 30° between the currents of main winding Im and auxiliary winding Ia can be achieved at the instant of starting. A typical phasor diagram is shown in Fig. 9.9(b). la

l

lm Singlephase ac voltage

Main winding

V

Centrifugal switch

Rotor

la l

lm

Auxiliary (or starting winding) (a)

Gjh/!:/:

!

(b)

"# $!

# $

When the motor has come to about 70 to 75% of synchronous speed, the starting winding may be opened by a centrifugal switch and the motor will continue to operate as a single-phase motor. At the point where the starting winding is disconnected, the motor develops nearly as much torque with the main winding alone as it was with both windings connected. It can be observed from the typical torque–speed characteristic for this type of motor in Fig. 9.10. The starting winding is designed to take the minimum starting current from the required torque. The locked rotor starting current may be typically in the range 5 to 7 times the rated current while the starting torque is also about 1.5 to 2 times the rated torque. The high starting current is not objectionable since once started, it drops off almost instantly. The major disadvantages of this type of induction motor are relatively low starting torque and high slip. Moreover, the reversal of rotation can be made only when the motor is standstill (by reversing the line connections of either the main winding or the starting winding) but not while running. Also, the efficiency is lower.

!

:/27

Fmfdusjdbm!Nbdijoft

Torque

Torque for combined windings for starting

Switch operates

ly

g din

on

n

in Ma O

wi

Speed

Gjh/!:/21

DBQBDJUPS.TUBSU!NPUPS!

:/8

In the split-phase motor, the phase shift between the stator currents was accomplished by adjusting the impedances of the windings, i.e. by making the starting winding of a relatively higher resistance. This resulted in a phase shift of nearly 30°. Since the developed torque of any split-phase motor is proportional to the pole flux produced and the rotor current, it is also dependent on the angle between the winding currents. This implies that if a capacitor is connected in series with the starting winding, the starting torque will increase. By proper selection of the capacitor, the current in the starting winding will lead the voltage across it and a greater displacement between winding currents is obtained. Figure 9.11 shows the capacitor-start motor and its corresponding phasor diagram indicating a typical displacement between winding currents of about 80°–90°. The value of the capacitor needed to accomplish this is typically 135 pF or a 1/4 h.p. motor and 175 pF for a 1/3 h.p. motor. Contrary to the split-phase motor discussed earlier, the speed of the capacitor-start motor under running conditions is reversible. If temporarily disconnected from the supply line, its speed will drop allowing the centrifugal switch to close. The connections to the starting winding are reversed during this interval and the motor is reconnected to the supply with closed centrifugal switch. The resulting rotating field will now rotate opposite to the direction in which the motor rotates. Since the current displacement between the windings is much larger in this motor compared to the split-phase motor, the torque being proportional to this will also be much larger and exceed the torque produced by the rotor. Therefore, the motor will slow down, stop and reverse its direction of rotation. When the speed reaches to about 75 to 80% of synchronous speed, the centrifugal switch opens and the motor will reach speed as dictated by the load. Because of higher starting torques, capacitor-start motors are used in applications where not only higher starting torques are required but also where reversible motors are needed. Applications of capacitor motors are in washing machines, betted fans and blowers, dryers, pumps and compressors.

Tjohmf.qibtf!Joevdujpo!Npupst

lline

ls

ta

rt

lst

:/28

lrun

C Rotor Starting winding

V

Singlephase ac voltage

Main winding

lline Centrifugal switch

lrun Phase diagram (b)

Circuit diagram (a)

Combined winding

Torque

Switch operates

Main winding only

Speed Torque speed charge (c)

Gjh/!:/22

DBQBDJUPS.TUBSU!DBQBDJUPS.SVO!NPUPS!

:/9

The capacitor-start motor discussed above has still relatively low starting torque, although it is considerably better than the split-phase motor. In case where higher starting torques are required, best results will be obtained if a large value of capacitance is used at start which is then gradually decreased as the speed increases. In practice, two capacitors are used for starting and one is cut out of the circuit by a centrifugal switch once a certain speed is reached, usually about 75% of full-load speed. This starting or intermittent capacitor is of fairly high capacity (usually of the order of 10 to 15 times the value of the running capacitor, which remains in the circuit.) Figure 9.12 illustrates the connection diagrams for the capacitor motor showing two methods generally encountered. The first method shown in Fig. 9.12(a) uses an electrolytic capacitor in the starting circuit whose leakage is too high. The second capacitor is oil-filled which remains in the circuit always and has little leakage; it is therefore suitable for continuous operation.

!

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Fmfdusjdbm!Nbdijoft

The second circuit [Fig. 9.12(b)] uses an auto-transformer and one-oil filled high-voltage capacitor. This method utilizes the transformer principle of reflected impedance from the secondary to the primary. For instance, an auto-transformer with 180 turns tapped at the 30-turn point would reflect 2

Ê 180 ˆ ¥ 8 mF = 288 mF, representing an increase of an 8 mF running capacitor to the primary as Á Ë 30 ˜¯ about 36 times. Thus, running an oil-filled capacitor may be used for starting purposes as well, by eliminating one capacitor in lieu of the auto-transformer, which is of comparable cost. Care must be taken to ensure that the capacitor can withstand the stepped-up voltage which is 180/30 = 6 times the rated voltage at start. Like the capacitor-start motor, the capacitor-run motor may be damaged if the centrifugal switch fails to operate properly. The primary advantage of a capacitor-run motor or a two-value capacitor motor is its high starting torque, good running torque and quiet operation. Reversing the line leads to one of the windings in the usual manner causing motor operation in the opposite direction. It is, therefore, classified as a reversible-type motor. These motors are manufactured in a number of sizes from 1/8 to 3/4 hp and are used in compressors, conveyors, pumps and other high-torque loads. Single-phase ac Main

Cs C

Start

Centrifugal switch (a) Single-phase ac

Auto-transformer

Main Run

C Start

Start (b)

Gjh/!:/23

Tjohmf.qibtf!Joevdujpo!Npupst

:/2:

TIBEFE.QPMF!NPUPST!

:/:

Like any other induction motor, the shaded-pole motor is caused to run by the action of the magnetic field set up by the stator windings. There is, however, one extremely important difference between the polyphase induction motor and the single-phase induction motor discussed so far. As discussed, these motors have a truly rotating magnetic field, either circular, as in the three-phase machine, or of elliptical shape as encountered in most of the single-phase motors. In the shadedpole motor, the field merely shifts from one side of the pole to the other. In other words, it does not have a rotating field but one that sweeps across the pole faces. An elementary understanding of how the magnetic field is created may be gained from the simple circuit in Fig. 9.13, illustrating the shaded-pole motor. As can be seen, the poles are divided into two parts, one of which is “shaded”, i.e., around the smaller of the two areas formed by a slot cut across the laminations, a heavy copper short circuited ring, called the shading coil, is placed. That part of the iron around which the shading coil is placed is called the shaded part of the pole. When the excitation winding is connected to an ac source, the magnetic field will sweep across the pole face from the unshaded to the shaded portion. This, in effect, is equivalent to an actual physical motion of the pole, the result is that the squirrel-cage rotor will rotate in the same direction. Shading pole

Shading pole

Gjh/!:/24

!

To understand how this sweeping action of the field across the pole face occurs, let us consider the instant of time when the current flowing in the excitation winding is starting to increase positively from zero, as illustrated in Fig. 9.14(a). In the unshaded part of the pole, the flux will start to build up in phase with the current. Similarly, the flux f, in the shaded portion of the pole, will build up, but this flux change will induce a voltage in the shading coil which will cause current to flow. By Lenz’s law, this current flows in such a direction as to oppose the flux change that induces it. Thus, the building up of flux f, in the shaded portion is delayed. It has the overall effect of shifting the axis of the resultant magnetic flux into the unshaded portion of the pole. When the current in the excitation coil is at or near the maximum value as indicated in Fig. 9.14(b), the flux does not change appreciably. With an almost constant flux, no voltage is induced in the shading coil and, therefore, it, in turn, does not influence the total flux. The result is that the resulting magnetic flux shifts to the centre of the pole.

!

:/31

Fmfdusjdbm!Nbdijoft

f

t

Dt1 (a)

Main flux f

Axis of resultant flux f

t

Dt2 (b)

Main flux f

t

Dt3

Main flux f (c)

Gjh/!:/25

!

Figure 9.14(c) shows the current in the excitation coil decreasing. The flux in the unshaded portion of the pole decreases immediately. However, currents induced in the shading coil tend to oppose this decrease in flux; consequently, they try to maintain the flux. The result of this action translates into a movement of the magnetic flux axis towards the centre of the shaded portion of the pole. Hence, flux f continues to lag behind the flux axis during this part of the cycle. It can similarly be reasoned that at any instant of the current cycle, the flux f lags behind in time. The net effect of this time and space displacement is to produce a gliding flux across the pole face and consequently in the air gap, which is always directed towards the shaded part of the pole. Therefore, the direction of rotation of the shaded-pole motor is always from the unshaded towards the shaded part of the pole.

Tjohmf.qibtf!Joevdujpo!Npupst

:/32

Simple motors of this type cannot be reversed but must be assembled so that the rotor shaft extends from the correct end in order to drive the load in the proper direction. There are specially designed shaded-pole motors which are reversible. One form of design is to use two main windings and a shading coil. For one direction of rotation, one main winding is used and for the opposite rotation the other; such an arrangement is adaptable only to distributed windings, hence this necessitates a slotted stator. Another method employed is to use two sets of open-circuited shading coils, one set placed on each side of the pole. A switch is provided to short circuit the shading coil. Depending on the rotational direction desired, offsetting the simple construction and a low cost of this motor. This motor has a low starting torque, little overload capacity and low efficiencies (5 to 35%). These motors are built in sizes ranging from 1/250 hp up to about 1/20 hp. Typical applications of shaded-pole motors are where efficiencies are of minor concern such as in toys and fans.

TQFFE!DPOUSPM!PG!B!TJOHMF.QIBTF! JOEVDUJPO!NPUPS

:/21

The speed control of a single-phase induction motor can be obtained by change of line frequency, change of number of poles and change of voltage applied to the stator. But the only economical method is to change the voltage applied to the stator as these motors are small in size. The change of voltage applied to the stator can be achieved by any of the following methods: 1. The motor can be fed from the variable secondary voltage of the auto-transformer. This method is expensive and rarely used. 2. SCR or TRIAC circuit may be used to reduce the applied voltage by ac phase control. 3. A resistor in series with the motor circuit gives a very cheap method of voltage control. It has the disadvantage of reduced efficiency due to energy loss in the resistor. But it is the most commonly used method. Figure 9.15 shows a single-phase ac regulator. Two thyristors connected in antiparallel are used. Thyristor T1 is fired at an angle a and conducts from a to p. At p, the current through T1 falls to zero, and this thyristor is subjected to a reverse bias and turns off. Thyristor T2 is fired at (p + a) and conducts from (p + a) to 2p. At 2p, the thyristor T2 is turned off and this cycle is repeated. As a is varied from 0 to p, the rms value of the applied voltage changes from V to 0.

!

:/33

Fmfdusjdbm!Nbdijoft

T1

T2 ac

Gjh/!:/26

LOAD

!

DPNQBSJTPO!CFUXFFO!TJOHMF.QIBTF!BOE! UISFF.QIBTF!JOEVDUJPO!NPUPST

:/22

Single-phase induction motors, when compared with three-phase induction motors, have the following disadvantages: Rotor core loss and rotor ohmic loss are more in single-phase induction motors due to doublefrequency currents induced by backward rotating field. 1. For the same load torque, a single-phase induction motor requires more stator current and operates at higher slip. Hence, they have more stator copper losses. 2. For the same size, the output of a single-phase induction motor is less than that of three-phase induction motor because the backward field torque opposes the forward field torque. Hence, for the same size, a single-phase motor has higher temperature rise and lower efficiency as compared to three-phase induction motor. 3. As the stator winding of a single-phase induction motor carries magnetizing current for both the forward and backward fields, the ratio of magnetizing current to the active component of stator current is much greater than that in a three-phase motor. Hence, single-phase motors operate at a poor power factor compared to three-phase induction motors. 4. Single-phase induction motors require more iron than three-phase induction motors for the same power and speed rating. Moreover, the necessity of providing auxiliary windings makes them costlier as compared to three-phase induction motors. However, large-scale production of fractional kilowatt single-phase induction motors decrease their cost and the cost is comparable to the cost of three-phase induction motors of the same power rating and speed. However, single-phase induction motors are simple, robust, reliable and less expensive for small ratings.

Tjohmf.qibtf!Joevdujpo!Npupst

:/34

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Solution From Fig. 9.6, neglecting the core loss resistance RC (a)

r2¢ x¢ 15 16 + j 2 = + j = 187.5 + j 8 ohms 2s 2 2 ¥ 0.04 2 = 187.67 – 2.44° ohms j

Xf

275 = 137.5 – 90° ohms 2 2 (187.67 – 2.44∞) (137.5 –90∞) Zf = = 108.73 – 54.64° 187.67 – 2.44∞ + 137.5 –90∞ =j

= 62.92 + j 88.67 ohms (2 – s) = (2 – 0.04) = 1.96 r2¢ x¢ 15 16 + j 2 = + j = 3.83 + j8 ohms = 8.86 – 64.43° W 2( 2 - 5) 2 2 ¥ 1.96 2 (8.86 – 64.43∞) (137.5 – 90∞) Zb = = 8.37 – 65.93° ohm 8.86 – 64.43∞ + 137.5 – 90∞ = (3.41 + j 7.64) ohms Total impedance = r1 jx1 + Zf + Zb = 10 + j 16 + 62.92 + j 88.67 + 3.41 + j 7.64 = 76.33 + j 112.31 = 135.79 – 55.8° ohm 220 = 1.62 A Input current = 135.79 Input power factor = cos (55.8°) = 0.56 lagging (b) Ef = 1.62 ¥ 108.73 = 176.14 V Eb = 1.62 ¥ 8.37 = 13.56 V I2f ¢ =

Ef r2¢ x¢ + j 2 2s 2

=

176.14 = 0.94 A 187.5 + j8

!

:/35

Fmfdusjdbm!Nbdijoft

I2b ¢ =

Ns =

Eb

r2¢ x¢ + j 2 2( 2 - 5) 2

=

13.56 = 1.53 A 8.86 – 64.43∞

2 ¥ 50 50 = rps 6 3

15 ˆ 2Ê Ê r¢ ˆ I 2¢ f Á 2 ˜ (0.94) Á Ë 2 ¥ 0.04 ˜¯ Ë 2s ¯ = Torque due to forward field = = 1.582 Nm 2p ns Ê 50 ˆ 2¥pÁ ˜ Ë 3¯ È r2¢ ˘ ( I2¢ b ) 2 Í 2 ˙ Î 2( s - 2) ˚ = (1.53) ¥ 3.83 = 0.085 Nm Torque due to backward field = 2p ns Ê 50 ˆ 2¥p ¥Á ˜ Ë 3¯ Gross torque = 1.582 – 0.085 = 1.496 Nm (c) Gross output = (1.496) (2pn) = 2(1.496) (2pns) (1 – s) 2p.50 = 1.496 ¥ ¥ (1 – 0.04) 3 = 150.32 W Net output = (150.32 – 30) = 120.32 W (d) Input power = 220 ¥ 1.62 ¥ 0.56 = 199.58 W output 120.32 (e) Efficiency = = = 60.3%. input 199.58

! Qspcmfn!:/8 B!5.qpmf-!341!W!tjohmf.qibtf!joevdujpo!npups!ibt!gpsxbse.gjfme!bjs.hbq!qpxfs!pg!311!X!boe!cbdl. xbse.gjfme!bjs.hbq!qpxfs!pg!51!X!bu!b!tqffe!pg!2536!sqn/!Efufsnjof!uif!tibgu!upsrvf!jg!uif!op.mpbe! mpttft!bsf!61!X/

Solution Net air-gap power = 200 – 40 = 160 W Mechanical power developed Pm = (1 – s) ¥ 160 W Now synchronous speed Ns =

120 ¥ 50 = 1500 rpm 4

Tjohmf.qibtf!Joevdujpo!Npupst

:/36

1425 = 0.05 1500 Pm = 0.95 ¥ 160 = 152 W

\

slip s = 1 –

\

No-load losses = 90 W \ shaft power = 152 – 90 = 62 W \ shaft torque =

62 = (1 - s)w s

62 2f (1 - 0.95) ¥ 2p P 62 = 2 ¥ 50 0.95 ¥ 2p ¥ 4 62 = 0.95 ¥ p ¥ 50 = 0.4157 Nm.

! Qspcmfn!:/9 B!331!W!tjohmf.qibtf!joevdujpo!npups!hbwf!uif!gpmmpxjoh!uftu!sftvmut; Cmpdlfe.spups!uftu;! ! 211!W-! ! :!B-! ! 491!X Op.mpbe!uftu;! ! ! ! !!!331!W-! ! 6!B-! ! 231!X Gjoe!uif!qbsbnfufst!pg!uif!frvjwbmfou!djsdvju!ofhmfdujoh!uif!dpsf!mptt!sftjtubodf/!Bmtp!gjoe!uif!jspo-! gsjdujpo!boe!xjoebhf!mptt/

Solution From Art. 9.4.1, r1 + r¢2 =

380 92

W = 4.69 W

Ê V ˆ 100 Zsc Á = sc ˜ = = 11.11 W 9 Ë I sc ¯ Hence,

x1 + x¢2 = (11.11) 2 - ( 4.69) 2 = 10 W 4.69 = 2.345 W and 2 10 x1 = x¢2 = =5W 2 r1 = r¢2 =

120 Ê Wolc ˆ cos qo = Á = = 0.11 or, qo = 83.74° ˜ Ë Volc ¥ I olc ¯ 220 ¥ 5

!

:/37

Fmfdusjdbm!Nbdijoft

ÈÊ 2.345 ˆ 5ˆ ˘ Ê 220 - 5 – - 83.74∞ ÍÁ 2.345 + ˜¯ + j ÁË 5 + ˜¯ ˙ Ë 4 2 ˚ Î = 2 5 = 44 – 1.0 ––83.74° (2.93 + j7.5) = 44 – 1.0 ––83.74° ¥ 8 – 68.66° = 44 – 8 ––15° = 36.275 + j 2 = 36.33 – 3.156° W. Xf = 72.66 – 3.156° W

Xf

\

Hence

Iron, friction and windage loss (from Fig. 9.8) are given as r¢ ˆ Ê = 120 – 52 Á r1 + 2 ˜ Ë 4¯

È r2¢ ˆ ˘ 2Ê Íi.e.,Wolc - I o ÁË r1 + ˜¯ ˙ 4 ˚ Î

2.345 ˆ Ê = 120 – 25 Á 2.345 + ˜ = 46.72 W. Ë 4 ¯

! Qspcmfn!:/: B!311!W-!61!I{!dbqbdjups.tubsu!npups!ibt!uif!gpmmpxjoh!jnqfebodft!bu!tuboetujmm;! Nbjo!xjoejoh![n!>!)9!,!k!4*!W Bvyjmjbsz!xjoejoh![b!>!)21!,!k!9*!W Gjoe!uif!wbmvf!pg!dbqbdjubodf!up!cf!dpoofdufe!jo!tfsjft!xjui!bvyjmjbsz!xjoejoh!up!hjwf!qibtf!ejtqmbdf. nfou!pg!:1¡!cfuxffo!dvssfout!jo!uif!uxp!xjoebht/

Solution The phasor diagram is shown in Fig. 9.11(b). 3ˆ Ê Phase angle of current in main winding is Á tan -1 ˜ = 20.55° Ë 8¯ With capacitor C in the auxiliary winding, the phase angle of current in auxiliary winding is 1 ˆ Ê 8Á -1 wC ˜ . ÁË tan ˜ 10 ¯ To give a phase displacement of 90° between the two winding currents, we can write, –tan–1

i.e.

1 wC – (–20.55°) = 90° 10 1 8–1 wC –tan = 69.45° 10

8-

Tjohmf.qibtf!Joevdujpo!Npupst

or

8–

:/38

1 = 10 tan (–69.45°) = –26.67 wC 1 = 34.67 W wC

Hence or

C=

1 F = 91.84 mF. 2p ¥ 50 ¥ 34.67

Sfwjfx!Rvftujpot 1. Explain why single-phase induction motors do not have self-starting torque. 2. Explain the principle of operation of a capacitor-start capacitor-run induction motor with suitable diagrams. 3. Draw and explain torque–slip characteristic of a single-phase induction motor on the basis of double-revolving field theory. 4. Discuss the procedure for determination of the parameters of equivalent circuit of a singlephase induction motor. 5. What are the different methods of starting of a single-phase induction motor? 6. Explain why the backward flux wave decreases and forward flux wave increases when the speed of a single-phase induction motor increases. 7. Discuss in brief how the operation of a single-phase induction motor can be explained by cross-field theory under normal running condition. 8. Write short notes on (a) split-phase induction motor, and (b) shaded-pole induction motor. 9. Discuss the differences between capacitor-start and capacitor-start capacitor-run induction motors. Why is the auxiliary winding of a capacitor-start motor disconnected after the motor has picked up speed? 10. What are the disadvantages of a single-phase induction motor in comparison to a three-phase induction motor? 11. What are the different methods of speed control of a single-phase induction motor?

Qspcmfnt 1 HP, 4-pole, 120 V, 60 Hz, single-phase induction motor has the following equivalent 3 circuit parameters:

1. A

r1 = 2.5 W, x1 = 1.25 W, r¢2 = 3.75 W, x¢2 = 1.25 W

!

:/39

Fmfdusjdbm!Nbdijoft

and Xf = 65 W. The motor has a core loss of 25 W, and friction and windage loss is 2W. Determine the shaft torque and the efficiency of the motor. The motor runs at 1710 rpm. [1.295 Nm, 65.86%] 2. A 4-pole, 60 Hz, 115 V, single-phase induction motor is rotating in the clockwise direction at a speed of 1710 rpm. Determine its per unit slip (a) in the direction of rotation, and (b) in the opposite direction. Also determine the effective rotor resistance in the forward and backward branches if rotor resistance at standstill is 12.5 W. [0.05, 1.95, 125 W, 3.205 W] 3. The equivalent circuit parameters of a 4-pole single-phase induction motor are as follows: r¢2 = 4.5 W r1 = 2.2 W x¢2 = 2.6 W x1 = 3.1 W Xf = 80 W The friction, windage and core loss = 40 W For a slip of 0.03 pu, determine (a) input current, (b) power factor, (c) developed power, (d) output power, and (e) efficiency. [5.685 -60.96 , 0.4856 lag, 479.65 W, 439.65 W, 69.2%] 4. The test results of a 220 V, single-phase induction motor are as follows: Blocked-rotor test: 120 V, No-load test: 220 V,

9.6 A, 4.6 A,

460 W 125 W

The stator winding resistance is 1.5 W and during the blocked-rotor test, the starting winding is open. Determine the equivalent circuit parameters. Also find the core, friction and windage losses. [x1 = x¢2 = 5.73 W, r¢2 = 3.49 W, Xf = 47.46 W, 74.8 W] 5. The following are the equivalent circuit parameters of a single-phase, 4-pole, 230 V, 50 Hz induction motor: The stator resistance and leakage reactances are 2.3 W and 3.2 W respectively. The rotor resistance and leakage reactance referred to the stator are 4.2 W and 3.2 W respectively. Magnetizing reactance is 74 W. Determine the stator current, power factor, power output, torque and efficiency at a slip of 5% if the core loss is 98 W, and friction and windage loss is 30 W. [6.95 -50.4∞, 0.637 lag, 637 W, 4.27 Nm, 62.5%]

Nvmujqmf.Dipjdf!Rvftujpot 1. For a single-phase capacitor-start motor, which, of the following statements is valid? (a) The capacitor is used for power factor improvement. (b) The direction of rotation can be changed by reversing the main winding terminals. (c) The direction of rotation cannot be changed. (d) The direction of rotation can be changed by interchanging the supply terminals. [GATE 2006]

Tjohmf.qibtf!Joevdujpo!Npupst

:/3:

2. The type of single phase induction motor having the highest power factor at full load is (a) shaded-pole type (b) split-phase type (c) capacitor-start type (d) capacitor-run type [GATE 2004] 3. In a single phase induction motor driving a fan load, the reason for having a high resistance rotor is to achieve (a) low starting torque (b) quick acceleration (c) high efficiency (d) reduced size [GATE 2005] 4. A single phase 230 V, 50 Hz, 4 pole, capacitor start induction motor has the following standstill impedances. Main winding Zm = 6 + j4 W Auxiliary winding Za = 8 + j6 W The value of the starting capacitor required to produce 90° phase difference between the currents in the main and auxiliary windings will be (a) 176.84 mF (b) 167.24 mF (c) 265.26 mF (d) 280.86 mF [GATE 2004] Im =

[Hints:

Ia =

230 0∞ V = = | IM | -33.7∞ Zm 6 + j 4 V j Za wc

1 wc – (–33.7°) = 90° 8 1 6wc tan–1 = –56.3° 8 1 6– = –12 wc

=

230 0∞ j 8 + j6 wc

1 wc 8

6= | I a | - tan-1

6\

or, or, or, \

–tan–1

1 = 18 wc C=

1 F = 176.8 mF]. 2p ¥ 50 ¥ 18

5. A single-phase induction motor with only the main winding excited would exhibit the following response at synchronous speed

!

:/41

6.

7.

8.

9.

10.

11.

12.

13.

14.

Fmfdusjdbm!Nbdijoft

(a) rotor current is zero (b) rotor current is non zero and is at slip frequency (c) forward and backward rotating fields are equal (d) forward rotating field is more than the backward rotating field [GATE 2003] If the slip of a single phase motor with respect to its forward field is 1, the slip with respect to its backward field is (a) 2 (b) 0 (c) 1 (d) 0.01 In a single phase induction motor no starting torque will be developed if the phase difference between the currents of starting and running winding is (a) 90° (b) zero (c) less than 90° (d) more than 90° A single-phase induction motor has (a) two windings on the rotor (b) two windings on the stator (c) one winding on the rotor and the other on the stator (d) may have one or two windings on the rotor The rotor of a single-phase induction motor (a) is always a wound rotor (b) is always a cage rotor (c) has a double cage rotor (d) may be a wound or cage rotor In a capacitor-start motor, the capacitor is connected (a) in series with auxiliary winding (b) in series with main winding (c) in parallel with auxiliary winding (d) in series with both the windings A cooling fan uses (a) capacitor-start capacitor-run motor (b) split-phase motor (c) universal motor (d) capacitor start motor According to double revolving field theory, a pulsating field can be resolved into two revolving fields. The magnitude of each of these component fields is (a) half the amplitude of pulsating field (b) two-third the amplitude of the pulsating field (c) one-third the amplitude of the pulsating field (d) equal to the amplitude of the pulsating field In a split-phase motor, (a) X/R ratio of the main winding is lower than that of the auxiliary winding (b) X/R ratio of the two windings are exactly equal (c) X/R ratio of the two windings are nearly equal (d) X/R ratio of the main winding is higher than that of the auxiliary winding In a capacitor split motor, if C is the capacitance required for best starting torque and C¢ for best running performance then

Tjohmf.qibtf!Joevdujpo!Npupst

:/42

(a) C is much larger than C ¢ (b) C is much smaller than C¢ (c) C is almost equal to C ¢ (d) C and C¢ may have any values 15. In a single-phase induction motor, the torque developed is proportional to (a)

V1

(b) V1 V1 (d) V12

(c) V1

16. In a single-phase split-phase induction motor, Im and Ia are the main and auxiliary winding currents and a is the angle between them. The torque developed by the motor is proportional to (b) Ia Im sin a (a) Ia Im cos a a (c) Ia Im (d) Ia Im cos 2 17. A 6-pole, 50 Hz, single-phase induction motor runs at a speed of 900 rpm. The frequency/ frequencies of current in the cage rotor will be (a) 5 Hz (b) 55 Hz, 5 Hz (c) 5 Hz, 95 Hz (d) 55 Hz, 95 Hz [Hints: \

900 = 0.1 120 ¥ 50 6 f1 = sf = 0.1 ¥ 50 = 5 Hz f2 = (2 – s) f = 1.9 ¥ 50 = 95 Hz]. s =1–

18. A capacitor start motor is modified by replacing the capacitor by an inductor having the same reactance. The motor will (a) be damaged (b) start but run at lower speed (c) start and run at rated speed (d) not start 19. Single phase induction motors are more noisy than three phase induction motors due to (a) squirrel cage rotor design (b) inherently low power rating (c) pulsating electrical power input (d) pulsating electrical power input and constant mechanical power output. 20. A 220 V, 50 Hz, single phase induction motor has the following connection diagram and winding orientations shown. MM¢ is the axis of the main stator winding (M1 M2) and AA¢ is that of the auxiliary winding (A1 A2). Directions of the winding axes indicate direction of flux when currents in the windings are in the directions shown. Parameters of each winding are indicated.

!

:/43

Fmfdusjdbm!Nbdijoft

M1 rm = 0.1 W Lm = (0.1/p)H ra = 1 W La = (10/p)H S

M2

M

A

A¢

A2

A1

220 V M¢ 50 Hz

When the switch s is closed, the motor (a) rotates clockwise (b) rotates anticlockwise (c) does not rotate (d) rotates momentarily and comes to a halt

[GATE 2009]

0.1 = 0.1 + j10 = 10 89.43∞ W p 10 = 1 + j1000 = 1000 89.94∞ W Za = 1 + j 2p ¥ 50 ¥ p The torque is proportional to the difference of angle d between the main and auxiliary winding currents. Here, d is very small and hence the motor will not rotate]. 21. An 8-pole, 50 Hz single phase induction motor is running at 690 rpm. Its slip with respect to backward field is (a) 0.08 (b) 1.92 (c) –0.08 (d) 1 Zm = 0.1 + j 2p ¥ 50 ¥

[Hints:

Botxfst 1. 5. 9. 13. 17. 21.

(b) (d) (b) (d) (c) (b)

2. 6. 10. 14. 18.

(d) (c) (a) (a) (d)

3. 7. 11. 15. 19.

(b) (b) (d) (d) (c)

4. 8. 12. 16. 20.

(a) (b) (a) (b) (c)

21 Tqfdjbm!Nbdijoft

VOJWFSTBM!NPUPST!

21/2

A dc series motor can operate under ac supply also, though it does not show good performance when operated on ac. In a dc series motor, the field winding and the armature winding are in series and the same current flows through them. The direction of the torque developed in a dc series motor is determined by both field polarity and the direction of the current through the armature (T μ f ia). As the direction of the current changes simultaneously in both field and armature windings, the direction of the torque produced remains same as shown in Fig. 10.1. However, under ac supply, the motor runs at lower speed and operates at low power factor due to large armature voltage drop caused by large armature inductive reactance. Also, the efficiency is low due to hysteresis and eddy-current losses and the sparking at the brushes is excessive. Hence, to make a dc series motor work efficiently on ac supply, an armature inductance compensating winding is provided in the stator. Such a motor is called a universal motor. Thus, a universal motor is a dc series motor which can run on both ac and dc supply. Such motors are also known as ac commutator motors. Figure 10.2 depicts the speed–torque characteristics of a universal motor under dc and ac supplies. Main field

Armature field T clockwise

i

T clo

ckw

ise

t (a) ac current

Armature field (a) When ac current is positive

Main field (c) When ac current is negative

Gjh/!21/2! Efwfmpqnfou!pg!upsrvf!jo!b!vojwfstbm!npups

!

21/3

Fmfdusjdbm!Nbdijoft

10000

N in rpm

8000 6000

dc voltage supply

4000 2000

ac voltage supply 0 1

2

T in Nm

Gjh/!21/3! Pqfsbujoh!dibsbdufsjtujdt!pg!b!vojwfstbm!npups

21/2/2! Dpotusvdujpo! Since the motor is to operate on ac also, the field as well as frame structure are all laminated to minimize losses and heating, as shown in Fig. 10.3. The two types of distributed-field universal motors are the single-field compensated motor and the two-field compensated motor.

Gjh/!21/4! Gjfme!dpsf!pg!b!uxp.qpmf!vojwfstbm!npups

The field winding of a two-pole single-field compensated motor resembles the stator winding of a two-pole, split-phase induction motor. A two-pole single-field compensated motor has a stator which contains a main winding and a compensated winding spaced 90 electrical degrees apart. The compensating winding decreases the reactance voltage developed in the armature by the alternating flux when the motor is operated from an ac source. Figure 10.4 shows the schematic diagram of a compensated universal motor. The armature of a typical universal motor resembles the armature of

Tqfdjbm!Nbdijoft

21/4

a typical dc motor barring the fact that a universal motor armature is slightly larger for the same horsepower output. The frames of universal motors are made of aluminium, cast iron and rolled steel. The field poles are bolted to the frame. The field core consists of laminations pressed together and held by bolts. The armature core is also laminated and has a commutator and brushes. End plates resemble those of other motors except that in many universal motors, one end plate is cast as part of the frame. Both ball and steel bearings are used in universal motors. Main field

Compensating field

Armature

V

Gjh/!21/5! Tdifnbujd!ejbhsbn!pg!b!dpnqfotbufe!vojwfstbm!npups

21/2/3! Tqffe!Dpouspm Universal motors operate at approximately the same speed on dc or single-phase ac. As these motors are series wound, they operate at excessive speed under no-load condition. The speed of the motor is regulated by inserting a resistance in series with the motor. The resistance may be in the form of tapped resistors, rheostats, or tapped nichrome-wire coils wound over a single field pole. The speed may also be controlled by varying the inductance through taps on one of the field poles. Gear boxes are also used. Speed can also be varied over a wide range by varying the voltage with TRIAC control. The direction of rotation of a series-wound motor can be reversed by changing the direction of the current in either the field or the armature circuit. Universal motors are sensitive to brush position. When the direction of rotation is changed without shifting the brushes to the neutral plane, severe arcing at the brushes occurs.

21/2/4! Dpnqfotbujpo! The ac motors rated at more than ½ horsepower are used to drive loads requiring a high starting torque. Two methods are used to compensate for excessive armature reaction under loaded condition.

!

21/5

Fmfdusjdbm!Nbdijoft

In the conductively compensated type of motor, an additional compensating winding is placed in slots cut directly into the pole faces. The strength of this field increases with the increase in load current and thus minimizes the distortion of the main field flux caused by the armature flux. The compensating winding is connected in series with the series field winding and the armature as shown in Fig. 10.4. Although conductively compensated motors have high starting torque, the speed regulation is poor. But a wide range of speed control is possible with the use of resistor-type starter controllers. Armature reaction in ac series motor may also be compensated with an inductively coupled winding which acts as a short-circuited secondary winding of a transformer. This winding is placed in such a way that it links the cross-magnetizing flux of the armature which acts as the primary winding of a transformer. Figure 10.5 is the schematic diagram of an inductively compensated universal motor. The magnetomotive force of the secondary is nearly opposite in phase and equal in magnitude to the primary magnetomotive force. Therefore, compensating winding flux nearly neutralises the armature cross flux. This type of motor cannot be used on dc current. The operating characteristics of an inductively compensated motor are very nearly similar to those of the conductively compensated motor due to its dependency on induction. Inductive compensating winding Main field

Armature

V

Gjh/!21/6! Dpoofdujpot!gps!bo!joevdujwfmz!dpnqfotbufe!vojwfstbm!npups

21/2/5! Bqqmjdbujpot! Typical applications for universal motors are in electrically powered tools, like drills, etc., and in domestic appliances like mixies, sewing machines, etc.

Tqfdjbm!Nbdijoft

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SFQVMTJPO!NPUPS!

21/3

21/3/2! Dpotusvdujpo Repulsion motors have a single-phase stator winding similar to an induction motor (Fig. 10.6). The armature is like that of a direct-current motor. Carbon brushes, shorted across in pairs to match the number of pairs of poles, run on the commutator. L N

220 V, 50 Hz

Single phase stator winding M

(a)

dc armature winding

(b)

Gjh/!21/7! Sfqvmtjpo!npups

21/3/3! Qsjodjqmft!pg!Pqfsbujpo!boe!Tubsujoh! Figure 10.7 shows the operation of a repulsion motor. For the sake of simplicity, a ring winding is chosen for the armature for description. The brushes run directly on the wires of the winding. If an

Gjh/!21/8! Qsjodjqmf!pg!pqfsbujpo!pg!b!sfqvmtjpo!npups

!

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Fmfdusjdbm!Nbdijoft

alternating voltage is applied to the field winding terminals, an alternating voltage is induced in the armature winding, according to the transformer principle. Let us consider a point in time t. According to Lenz’s law, the current and voltage polarities will be as shown in Fig. 10.7. When the brushes are in the neutral zone [Fig. 10.7(a)], the voltages induced in the top and bottom halves of the armature winding cancel each other. No current flows, and therefore no torque is developed. If the brushes are now displaced by an angle a [Fig. 10.7(b)], the voltages cannot cancel each other out. Current flows in both halves of the winding and an alternating magnetic field is created. This is displaced with respect to the alternating magnetic field of the stator. Magnetic forces are set up between the two fields. A torque is exerted on the armature, causing it to turn in a direction opposite to the brush displacement. The polarity of the armature field remains unchanged. When the brushes are displaced in the opposite direction, the direction of rotation is reversed. The maximum torque is obtained with a brush-displacement angle of about 70°. If the brushes are displaced by 90°, the motor acts as a short-circuited transformer. Despite the large short-circuit current, no torque will be developed. The winding will be damaged and cannot be repaired. Therefore, a blocking mechanism is provided to prevent the brush being displaced by an angle of more than 70°. It is also essential that the brushes do not remain too long in the zero position, since they short circuit coils in the armature winding and the resulting currents overheat the motor. Repulsion motors are used where smooth starting and sensitive speed control is required, for example in spinning and printing machinery.

21/3/4! Dibsbdufsjtujdt The repulsion motor has series characteristics as represented in Fig. 10.8. Its speed–torque characteristics are highly drooping. It has quite a high starting torque. N in rpm

5000 4000 3000 2000 1000 T in Nm 0

1

2

Gjh/!21/9! Pqfsbujoh!dibsbdufsjtujdt!pg!b!sfqvmtjpo!npups

Tqfdjbm!Nbdijoft

21/8

TJOHMF.QIBTF!TFSJFT!NPUPS!

21/4

21/4/2! Dpotusvdujpo! If an ordinary dc series motor is fed from a single-phase ac supply, the torque which is produced is always positive though it is pulsating in nature. But the dc series motor does not show good performance when used on alternating current as discussed in Article 10.1. The efficiency is low, the power factor is poor and there is considerable sparking at the brushes. The sparking at the brushes is due to poor commutation. The voltage induced by transformer action in the coil undergoing commutation further intensifies commutation difficulties.

Compensating winding

ac Armature Field

Gjh/!21/:! Bo!bd!tfsjft!npups!xjui!dpnqfotbujoh!xjoejoh

To improve the efficiency, both the stator and rotor cores are laminated (the rotor or armature core is laminated in dc machines also). To reduce the reactance of armature winding, a compensating winding, is added in the pole faces and connected in series with armature winding, as shown in Fig. 10.9. The number of turns per pole in the compensating winding is so chosen that the mmf it produces is equal and opposite to that of the armature, i.e. Nc = where

Nc Z A P

Z 2 PA

(10.1)

= Number of turns per pole in compensating winding = Number of armature conductors = Number of parallel paths in armature = Number of poles

The compensating winding cannot neutralize the armature field in the commutation zone, because its space distribution is not identical with that of armature winding. Thus, there will be residual armature field owing to the armature conductors lying between the pole tips. To neutralize this flux in the commutating zone, interpoles are necessary.

!

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Fmfdusjdbm!Nbdijoft

The flux which is responsible for the reactance of the field winding is the main flux which is essential for the operation of the motor. Evidently, this flux should not be neutralized. Nevertheless, the field-winding reactance must be kept as low as possible, if the power Er IR¢ factor is to be maintained reasonably lra lrf lrc high. This requirement calls for the V minimum possible number of field lxc turns per pole, which means minimum flux per pole. Since the field flux is small, the number of armature lxf IX¢ conductors has to be high to develop the required torque. Thus, the ratio of lx a armature ampere turns per pole to the field ampere turns per pole is high in ac series motors. However, there is a limit to the number of armature am- Ef IX pere turns, which can be accommoq dated on a given core, this limit being fixed by the consideration of heating. fd I The only way to develop the desired torque in motors of considerable rat- Gjh/!21/21! Qibtps!ejbhsbn!pg!tjohmf.qibtf!tfsjft!npups ing is to use numerous poles, thus introducing an additional difference between ac series motor and dc series motor. The use of numerous poles, each with relatively few turns, improves the commutation and reduces the leakage inductance of the field winding. The power factor, efficiency and commutation are all improved by lowering the frequency of ac supply.

21/4/3! Qibtps!Ejbhsbn The current in a series field winding sets up a flux fd in the direct axis. This current also sets up a quadrature axis fq, but this flux is neutralized by the compensating winding. The current I and flux fd are in phase. The applied voltage V has to balance the following emfs and voltage drops: 1. The speed emf Er is produced by the flux fd. This emf appears across the brushes which are in neutral or q-axis. This emf Er is in phase with flux fd. 2. The self-induced emf (or transformer emf) Ef is induced in the field winding by the flux fd. This emf is induced because flux fd is an alternating flux. If the machine is operated from dc, this emf is absent. The emf Ef is in quadrature with the flux fd. 3. Voltage drops across ra, rf and rc, i.e. resistance of armature, field and compensating winding are in phase with the current I. 4. Voltage drops across xa, xf and xc, i.e. leakage reactance of armature, field and compensating windings are in quadrature with the current.

Tqfdjbm!Nbdijoft

21/:

Therefore, the phasor sum of Er, Ef and voltage drops in resistances and leakage reactances gives the applied voltage V. Figure 10.10 shows the approximate phasor diagram. The power factor is cos q. The effects of iron losses and current in the coil undergoing commutation have been neglected.

21/4/4! Bqqmjdbujpo The series motor has high starting torque and this feature makes it ideally suited for use in traction. The speed and torque characteristics are well suited to such a service. The speed is easily and efficiently regulated by varying the applied voltage. The ac series motors are, therefore, widely used 2 in electric locomotives. In the past, these motors were usually operated at 16 Hz and 25 Hz and 3 the railway electric-distribution network had to be specially supplied at these frequencies through frequency converters. Modern developments have enabled satisfactory operation at 50 Hz, so that railway electric networks can be fed directly from the power-system grid. However, the latest trend is to use dc series motors fed from the power system through solid-state rectifier circuits.

P.F.

T, h, p, f, Speed

h

T

Spe

ed

Current

Gjh/!21/22! Dibsbdufsjtujdt!pg!bd!npups!tfsjft

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The unique characteristic of a synchronous motor is that its speed remains exactly constant. The speed depends solely on the frequency of the supply voltage. For small loads, such motors are used when high running accuracy and extremely precise speed control are required.

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Figure 10.12 shows the construction of a synchronous fractional horsepower motor. The pole horns and the yoke are made from core sheet sections. The alternating field is generated in a coil. The rotor usually consists of soft magnetic iron, Main coils but sometimes of permanent magnets. The pole N horns and the rotor have teeth arranged so that a rotor tooth always exactly faces a pole-horn V tooth. When the motor is switched on, the polePole horns horn teeth attract the rotor teeth. When the rotor is started, it continues to rotate (Fig. 10.13). If there is a positive current flow in the coil, the pole-horn teeth (north poles) attract the rotor S teeth. While the current is passing through its Rotor Yoke zero point (no magnetic field) the rotor continues to turn due to the fly-wheel effect. When the Gjh/!21/23! Tzodispopvt! gsbdujpobm! ipstfqpxfs! npupst current flow reverses, the pole horns become south poles. The next rotor teeth are now attracted. This process is repeated at a frequency f of the applied alternating voltage. The rotor turns at a speed n = 2 f /z, where z is the number of rotor teeth. A similar process takes place in a permanent magnet rotor motor. A shaded-pole stator is often used to generate the rotating field. In permanent magnet synchronous motors the permanent magnets are embedded in the rotor. The rotor is of squirrel cage type. When the motor is connected to a single phase ac supply it starts as an induction motor, attains synchronous speed and locks into synchronism and operates as synchronous motor. One example of the use of synchronous fractional horsepower motors is the motor in an electric clock. It is also used in control apparatus and timing devices. i

t

t2 3 N

3

t

t

t1

t3

i

2 N 2

1 N 1

3

!!!!!!!!!!!!!

2 3

3 S

1 2

1

!!!!!!!!

2 S 3

1 S 2

Gjh/!21/24! Pqfsbujpo!pg!tzodispopvt!gsbdujpobm!ipstfqpxfs!npups

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A synchronous induction motor, as the name suggests, can run both as induction motor and a synchronous motor. This motor works as induction motor during starting and acts as synchronous motor during running condition.

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If the rotor of a slip ring induction motor is fed from dc supply it is possible to run this motor at a synchronous speed. The rotor of a slip ring induction motor becomes an electromagnet when it is fed from dc source. The fixed rotor poles get magnetically locked with the rotating magnetic field developed by the three phase stator windings carrying ac current and the motor runs at a constant speed equal to the synchronous speed. Though larger air gap results in poor power factor in ordinary induction motors, in synchronous induction motor large air gap is provided as the power factor can be controlled by variation of dc excitation. The larger air gap produces higher peak synchronous torque. These machines are provided with heavy rotor winding for having low slip which facilitates in pulling it into synchronism. Also in order that the induced emf in the field is not very high at starting, the field turns are kept few in number and the excitation voltage is kept low. As the exciter winding serves the purpose of damper winding, hence there is no need for providing separate damper winding with these machines. The synchronous induction motor is started as a slip ring induction motor by inserting resistances in the rotor circuit. The additional resistances are slowly cut out and the motor runs at low slip. The rotor is then disconnected from the starting resistances and connected to the dc excitation source which is mounted on the same shaft of induction motor. The motor now operates as a synchronous motor. Synchronous induction motors are used where high starting torque and constant speed operation is required. They are generally built for ratings above 25 KW because of the high cost of the exciter. They are used in fans, generators, blowers, pumps, air compressors, machinery and line shafting in industries such as in cement mills, rolling mills, flour mills, paper mills and textile mills.

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A linear induction motor is a special type of induction motor which gives linear or translational motion unlike conventional induction motor which gives rotational motion. If a conventional induction motor is cut and laid flat, a linear induction motor can be obtained. In this motor the stator is termed as primary and the rotor as secondary. The primary consists of a magnetic core with three phase winding and the secondary consists of a flat aluminium conductor as shown in Fig. 10.14. When the primary winding is energized from a three phase ac supply, a travelling flux wave is produced, like conventional induction motors which travels along the length of the primary. Due to the relative motion between the travelling flux wave and aluminium conductor, current is induced in the aluminium conductors. The interaction of the induced currents and the travelling flux produces a linear force or thurst. If the primary is free to move and the secondary is fixed, the force will move the primary in the direction of the travelling wave. A linear induction motor may have one or two primary. The motor shown in Fig. 10.14 contains one primary and is called single sided linear induction motor. The motor shown in Fig. 10.15 has primary on both sides of the secondary and is known as doubled sided linear induction motor.

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Primary The phase winding Aluminium conductor

Secondary

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Secondary Primary

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In order to maintain a constant force over a considerable distance, one side (either primary or secondary) is kept shorter than the other. For example in high speed ground transportation, a short primary and a long secondary is used. The primary forms an integral part of the vehicle, whereas the secondary forms the track. The linear synchronous speed vs = 2f ¥ pole pitch m/s where f is the supply frequency The linear velocity of the movable secondary vr = vs (1 – s) where s is the slip of the motor The linear force

F=

Air-gap power vs

The equivalent circuit of linear induction motor is similar to that of the rotary induction motor. The force velocity curve as shown in Fig. 10.16 is also similar to the torque speed curve of rotary induction motor. The linear induction motor has larger air gap than rotary induction motor. Hence it has larger magnetizing current, poor power fac- Force tor and lower efficiency than rotary induction motor. Again in rotary induction motor, the vs stator and rotor developments are of same Speed length due to smaller air gap and after one revolution the stator and rotor comes back to Gjh/!21/27! Upsrvf.tqffe!dvswf!pg!b!mjofbs!joevdujpo! npups the same position with respect to each other. In linear induction motor the shorter member continuously runs over a new part of the longer member.

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The advantages of linear induction motor are lower initial cost, low maintenance cost, simplicity, no limitation of maximum speed due to centrifugal forces and no over heating as the motor moves continuously over cool rotor plate leaving behind heated rotor portion. The disadvantages are its poor efficiency and low power factor, high capital cost of reaction rail fixed along the centre line of the track, difficulties encountered in maintaining adequate clearances at points and crossings. The main application of linear induction motor is in transportation. The primary is mounted on the vehicle and the secondary is laid along the track. It is also used in cranes for material handling, pumping of liquid metal, actuators for door movement and high voltage circuit breakers.

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Reluctance motors are basically synchronous motors, but they do not need dc excitation. They usually have a single-phase stator but may be polyphase also. They are self-starting and use reluctance torque for running. A reluctance torque is the torque produced in a motor in which the reluctance of the air gap is a function of angular position of the rotor, with respect to stator coils. In a salient-pole synchronous motor, the torque has two components, and one component is a function of (xd – xq), i.e. difference in the motor reactance in the direct and quadrature axis. This component is known as the reluctance torque. A motor which develops torque only due to the difference in reluctance in the two axes is known as a reluctance motor. The reluctance is proportional to sin 2d, where d is the angle between the rotor and stator magnetic fields. Therefore, reluctance torque is maximum when d = 45°. The reluctance torque develops due to the tendency of the rotor to align itself in the minimum reluctance position, with respect to the stator field revolving at synchronous speed. 600 Main and Auxiliary Winding

300 200 100

(a) Rotor punching 0

Switching Speed

400

Varies with Starting Position of Rotor

Percent Torque

500

Main Winding only 20

40

60

80

Percent Synchronous Speed (b) Torque-speed curve

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100

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Figure 10.17(a) shows the rotor punching of a self-starting reluctance motor. It is similar to the punching of a cage rotor, except that some of the teeth have been removed, leaving the bars and end rings intact. The stator is either single-phase (having main and auxiliary winding as in single-phase induction motor) or three-phase. The motor starts due to induction motor action. At speeds less than synchronous speed, the torque alternates in positive and negative half-cycles. If the rotor has small inertia and the motor is lightly loaded, the rotor will be pulled into synchronism during the positive half-cycle and continue to run at synchronous speed. Figure 10.17(b) shows the torque-speed characteristic of a reluctance motor. As mentioned above, it is basically a salient-pole synchronous motor, but has a split-phase stator winding for providing the starting torque. The starting torque is a function of the rotor position. This phenomenon is known as cogging. The starting torque is quite high, because to obtain satisfactory synchronous motor characteristics, it is necessary to use a frame bigger than that for an induction motor of the same rating. These motors are used for constant speed application such as electric clock timers, recording instruments, signalling devices, etc.

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The switched reluctance motor is a type of stepper motor that has less number of poles. The stator poles are wound with excitation windings but has no windings on its rotor which is made of soft magnetic material. The change in reluctance along the periphery of the stator forces the rotor poles to align with those of the stator. Hence the torque develops in the motor and the rotation takes place. These motors have one pole pair less in the rotor than the stator. Both the stator and rotor poles are projected type. The diametrically opposite stator poles are excited simultaneously in a sequence. Fig. 10.18 shows the schematic diagram of a switched reluctance motor. Magnetic axis or Phase-A

Soft iron rotor

Phase-B

Axis of rotor

Phase-A

d Magnetic axis of Phase-B

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Though the construction of switched reluctance motor is very simple, it requires complicated control mechanism. In order to develop torque in the motor the rotor position should be determined by sensors so that excitation timing of the phase windings is precise. The developed torque in the motor is independent of the direction of supply current as the torque is proportional to the square of the phase currents. The initial position of the rotor has a significant impact on the developed torque. Common uses of these motors are in the applications where the rotor must be held stationary for long periods and in potentially explosive environments such as mining as they do not have mechanical commutator. They are also used in some washing machines and control rod drive mechanisms of nuclear reactors.

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FROM CONTROLLER

Servomotors are generally known as control motors. Their power ratings vary from a fraction of a watt to a few hundred watts. They have a high speed of response and generally operate at very low speeds. Servomotors can operate from both dc and ac supplies. These motors are used in feedback control systems as output actuators. The dc servomotors are separately excited dc motors or permanent magnet dc motors. The speeds of dc servomotors are controlled by varying the armature voltage. The armature resistance is very high and a step change in armature voltage or current produces a quick change in the position or speed of the rotor. The ac servomotors are basically two-phase squirrel-cage induction motors used for low power applications. The rotor has high resistance and small diameter-to-length ratio to give good accelerating characteristics. They produce high torques at all speeds including zero speed. Control winding AMPLIFIER

vc ROTOR

Reference winding

vr FIXED ac VOLTAGE

Gjh/!21/2:! Uxp.qibtf!tfswpnpups

Figure 10.19 shows the connection diagram of a two-phase servomotor. One of the stator windings, called reference winding, is excited by a fixed ac voltage Vr, while the second winding, known

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as control winding, is excited by the voltage VC. The two voltages must be in synchronism, which is a necessary condition for production of torque. Therefore, the two voltages must be derived from the same source. The control voltage is fed to the motor through an amplifier. Moreover, the two voltages must have a phase difference of 90°. This phase shift is usually obtained by using a phase-shifting network in an amplifier circuit. The 90° phase shift can also be obtained by adding a capacitor in series with the reference winding. When VC leads Vr by 90°, rotation in one direction is obtained and when VC lags Vr by 90°, rotation in a reverse direction is obtained. Since the torque is a function of both Vr and VC, changing the magnitude of VC changes the torque developed. Esbh.dvq!Spups When the rating of a servomotor is only a few watts or less, inertia can be decreased by using a drag-cup rotor. This rotor (Fig. 10.20) is in the form of a thin metallic cup. The iron core is stationary. Since only the metallic cup is the rotating member, the inertia is very small. Stator Drag cup rotor

Stationary Rotor core

Shaft

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2/!ed!Ubdipnfufs! In many control systems, it is necessary to feed back a voltage proportional to the speed of the shaft. In a dc servo mechanism, this can be achieved by using a dc tachometer, which is a permanent magnet dc generator. The permanent magnet ensures a constant air-gap flux. Therefore, its output voltage is directly proportional to the speed of the shaft only. A dc tachometer can be used in ac servo mechanism by converting the dc output voltage to an ac voltage by using an inverter circuit.

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3/!bd!Ubdipnfufs An ac tachometer is used in feed back control systems, to feed back and ac voltage proportional to the speed of the shaft. It is basically a two-phase induction motor as shown in Fig. 10.21. One stator phase winding, known as reference winding, is excited by a suitable ac voltage of constant magnitude and frequency. A voltage of the same frequency is genr erated across the other winding known as auxiliary winding or control winding. This output voltage is applied to the Vp high-impedance input circuit of an amplifier, so that the control winding can be considered as an open circuit. It is necessary that the voltage developed across the control winding is linearly proportional to shaft speed and the phase of this Control winding voltage be fixed with respect to the voltage applied to the reference winding. To The operation of an ac tachometer can be explained by Amplifier the double revolving field theory. The tachometer is equivaGjh/!21/32! bd!ubdipnfufs lent to a small single-phase induction motor. At standstill, the forward and backward fields are equal and voltage developed across the control winding is zero. When the rotor is revolving, the impedance of the forward field increases and that of the backward field decreases, the difference between them being a function of speed. Therefore, the voltage developed across the control winding is a function of speed. Reversal of direction of rotation reverses the phase of output voltage. For a constant-phase angle of output voltage and linear relationship between output voltage and speed, the ratio x2/r2 of the rotor (where x2 is rotor reactance and r2 is rotor resistance) should be either small (i.e. less than 0.1) or high (i.e. more than 10). If x2/r2 is low, the sensitivity (i.e. volts per revolution per minute) is sacrificed but linear speed range is wide. If x2/r2 is high, the speed range is limited to a small fraction of synchronous speed to meet the condition of linearity of voltage and constancy of phase angle. Generally, an intermediate value of x2/r2 ratio gives satisfactory performance. An ac tachometer should have low inertia when rapid speed variations are encountered as in automatic control systems. The drag-cup construction (Fig. 10.16) gives low inertia and is used frequently.

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In a hysteresis motor, the torque is produced due to the hysteresis phenomenon. The stator has a three-phase or permanent split capacitor single-phase winding, thus producing a rotating field. It is necessary that the stator field has sinusoidal space distribution, so that the losses are minimum. That rotor is a smooth cylinder of hard steel and does not have any teeth or winding as shown in Fig. 10.22(a).

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Axis of Stator Field

ws

Axis of Rotor Field

d

Torque

Stator

Rotor

0

100 Percent Synchronous Speed

(a)

(b)

Gjh/!21/33! Iztufsftjt!npups;!)b*!Nbhofujd!gjfmet!)c*!UpsrvfÐtqffe!dvswf

At speeds less than synchronous speed, the torque is partly due to eddy currents, the component due to hysteresis being much greater than that due to eddy currents. When the magnetic fields of the stator sweep around the rotor surface, the rotor flux wave lags behind the stator flux wave due to hysteresis. The angle of lag is d. The greater the intrinsic hysteresis loss of rotor material, the greater is the angle by which the rotor magnetic field lags the stator field. The torque is proportional to sin d. In addition, the stator magnetic field induces eddy currents in the rotor. These eddy currents produce a magnetic field of their own and torque is produced. The greater the relative motion between the rotor and stator fields, the greater are the eddy-currents and eddy-current torque. When the motor reaches synchronous speed, the eddy-current torque reduces to zero and only hysteresis torque is present. The hysteresis torque is constant from zero to synchronous speed. Since the total torque at any speed less than synchronous speed is slightly more than the torque at synchronous speed, this motor can accelerate any load up to its full load. After reaching synchronous speed, the motor adjusts its torque angle, so as to develop torque required by the load and continues to run at synchronous speed. Since the eddy-current torque is much less than hysteresis torque, the total torque is nearly constant from starting to synchronous speed [Fig. 10.22(b)]. Since the rotor surface is smooth, without any slots the motor does not experience any magnetic or mechanical vibrations and is, therefore, extremely quite. Hence, it is used for electric clocks and other timing devices. These motors are also used for driving gyros. As these motors are noiseless, they are used for sound-playing instruments like record players, tape recorders, etc. There is a basic difference between hysteresis motor and reluctance motor. The reluctance motor starts as a single-phase induction motor and the rotor pulls into synchronism under favourable conditions. In a hysteresis motor, any load gets synchronized with stator poles provided hysteresis

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torque is able to accelerate it. Hence, in a reluctance motor, there is a tendency for the rotor to oscillate before synchronism, but in a hysteresis motor, the rotor and stator poles lock with each other without any oscillation.

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In conventional dc motors, the armature is on the rotor, and the field magnets are placed in the stator. The construction of modern brushless dc motors is very similar to the ac motor, known as the permanent magnet synchronous motor (Fig. 10.23). The armature windings are part of the stator, and the rotor is composed of one or more magnets. The windings in a brushless dc motor are similar to those in a polyphase ac motor and the most efficient motor has a set of three phase windings and is operated in bipolar excitation. Brushless dc motors are different from ac synchronous motors in that the former incorporates some means to detect the rotor position (or magnetic poles) to produce signals to control the electronic switches. The most common position/pole sensor is the Hall element, but some motors use optical sensors.

Permanent magnet rotor

Winding Hall elements

Gjh/!21/34! Ejtbttfncmfe!wjfx!pg!b!csvtimftt!ed!npups;!qfsnbofou!nbhofu!spups-!xjoejoh!boe!Ibmm! fmfnfou

By examining a simple three-phase unipolar operated motor, one can easily understand the basic principles of brushless dc motors. Figure 10.24 illustrates a motor of this type that uses optical sensors (phototransistors) as position detectors. Three phototransistors PT1, PT2 and PT3 are placed on the end plate at 120° intervals, and are exposed to light in sequence through a revolving shutter coupled to the motor shaft.

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Revolving shutter

I1 Phototransistors Pt1

W3

W1

Pt3

P1 N P3 S P2

Pt2

W2

Tr1

Tr2

I2

I3

I11 E11

Tr3

Gjh/!21/35! Uisff.qibtf!vojqpmbs!esjwfo!csvtimftt!ed!npups

As shown in Fig. 10.20, when the south pole of the rotor faces the salient pole P2 of the stator, the phototransistor Pt1 detects the light and turns transistor Tr1 on. In this state, the south pole which is created at the salient pole P1 by the electrical current flowing through the winding W1 is attracting the north pole of the rotor to move it in the direction of the arrow. When the south pole comes in the position to face the salient pole P1, the shutter, which is coupled to the rotor shaft, will shade Pt1, and Pt2 will be exposed to the light and a current will flow through the transistors Tr2. When a current flows through the winding W2, and creats a south pole on the salient pole P2, then the north pole in the rotor will revolve in the direction of the arrow and face the salient pole P2. At this moment, the shutter shades Pt2, and the phototransistor Pt3 is exposed to the light. These actions steer the current from the winding W2 to W3. Thus, the salient pole P2 is de-energized, while the salient pole P3 is energized and creats the south pole. Hence, the north pole on the rotor further travels from P2 to P3 without stopping. By repeating such a switching action in the sequence given in Fig. 10.21, the permanent magnetic rotor revolves continuously. Nfuipe!pg!Sfwfstjoh!uif!Ejsfdujpo!pg!Spubujpo In order to reverse the direction of rotation of a conventional dc motor, the terminal voltage simply has to be reversed. However, such an action will not reverse a brushless dc motor which uses semiconductor devices like transistors, because most semiconductor devices are unidirectional switches. Therefore, some additional circuit is necessary when the motor is to be driven in either direction.

Tqfdjbm!Nbdijoft

0

120

Pt1

21/32

240 Revolving angle (degree)

360

480

Pt2 Pt3 Output signals from phototransistors

I1 I2 I3

P3 P1

P1

P2 Time

Gjh/!21/36! Txjudijoh!tfrvfodf!boe!spubujpo!pg!tubupsÕt!nbhofujd!gjfme

In Fig. 10.25, the connections between the phototransistors (PT1, PT2, and PT3) and the transistors (Tr1, Tr2 and Tr3) are arranged as Pt1—Tr1 for controlling current through W1 Pt2—Tr2 for controlling current through W2 Pt3—Tr3 for controlling current through W3 These connections make the motor rotate counter-clockwise. If the connections are changed over to Pt1—Tr3 Pt2—Tr1 Pt3—Tr2 the rotational direction will be reversed.

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This type of motor is today one of the most popular small control motors used for accurate position controls. It is designed to rotate through a specific angle (called a step) for each electrical pulse received by its control unit. Stepper motors are controlled by current pulses generated in electronic circuits. The rotor turns in single steps or at a synchronous speed governed by the pulse frequency. The stator consists of a large number of independent separately controlled coils. Permanent magnet and soft magnetic iron rotors (reluctance step motor) are most frequently used. Stepper motors have wide range of applications. As they are digitally controlled i.e. controlled by using input pulses they can be used with computer controlled system. They are used in numerical control of machine tools, tape drivers, line printers, X – Y plotters, other computer peripheral devices, robotics, electric watches, etc. They are also used in space crafts, science fiction movies, various commercial, medical and military applications. Compared with other devices that can perform the same or similar functions, a control system using a stepper motor (Fig. 10.26) has several sigdc power nificant advantages as follows: supply 1. No feedback is normally required for either position control or speed control. 2. Positional error is noncumulative. 3. Stepping motors are compatible with modern digital equipment.

Integrated circuits or microprocessor

Transistorized driver

Generation of switching signals

Switching of power circuits

Motor

There are three types of stepper motors, per- Gjh/!21/37! Npefso! esjwjoh! tztufn! gps! b! tufqqfs!npups manent magnet stepper motor, variable reluctance stepper motor and hybrid stepper motor

21/24/2! Qfsnbofou!Nbhofu!Tufqqfs!Npups In permanent magnet stepper motor the stator poles have concentrated winding and they are of salient type. The rotor is cylindrical and consists of permanent magnet poles made of high retentivity steel. The concentrated windings on diametrically opposite poles are connected in series to form two phase winding on the stator. The rotor poles align with the stator teeth (or poles) depending on the excitation of the winding. Figure 10.27 explains how a stepper motor with a permanent magnet rotor works. Consider that 1 and 5 are fed from a positive pulse and at the same time coils 3 and 7 are fed from a negative pulse. The rotor position is shown in Fig. 10.27. Now the positive pulse is fed to coils 2 and 6 and the negative pulse to coils 4 and 8. The rotor turns through 45°. To make the rotor turn one step

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further, we have to feed a negative pulse to coils 1 and 5 and a positive pulse to coils 3 and 7. By feeding suitable pulses to coils 2 and 4 and coils 6 and 8, we can make the rotor turn yet another step. If we feed the coils with pulses which are periodi1 1 1¢ 2¢ 2 cally repeated, the rotor will turn step by step. If the 2 S 8 2¢ pulse frequency is high enough, it will turn at a con2 1 1¢ stant synchronous speed of rotation. The speed de7 N N 3 pends on the number of poles and the pulse frequency. 1 1¢ If there are enough coils, high speeds can be achieved. 2¢ 6 4 2 S The residual magnetism in the permanent magnet 2¢ 2 5 material produces detent torque or restraining torque 1 1¢ on the rotor when the stator coils are not energized. The detent torque can be defined as the maximum load torque that can be applied to the shaft of an unexcited Gjh/!21/38! Tufqqfs!npups!xjui!qfsnbofou! permanent magnet and hybrid stepper motor without nbhofu!spups causing continuous rotation.

21/24/3! Wbsjbcmf!Sfmvdubodf!Tufqqfs!Npups In variable reluctance stepper motor the stator and rotor gets aligned in such a way so that the magnetic reluctance becomes minimum. There are two types of variable reluctance stepper motor, single stack type and multistack type. A single stack variable reluctance motor has salient pole stator and concentrated winding. The rotor is a slotted structure which carries no windings. The exciting current required is very small as both the stator and rotor are made up of high quality magnetic materials having very high permeability. With the help of semiconductor switches the stator poles are excited from dc source in a proper sequence to create a magnetic field. The rotor aligns itself to the stator field axis to take the minimum reluctance position. The angle through which the rotor moves for each change in excitation sequence of the motor is called the step angle which is given by 360∞ h Number of stator phases ¥ number of rotor poles or teeth The step angle can also be expressed as Step angle =

Step angle = 360° ¥

…(10.2)

Difference of the stator and rotor poles or teeth …(10.3) Number of stator poles or teeth ¥ number of rotor poles or teeth

The cross section of a three phase motor having two stator poles per phase and four rotor poles is shown in Fig. 10.28. If the phase winding is excited in sequence ABCA the steps are clockwise and if the excitation sequence is ACBA the steps are anticlockwise A multistack variable reluctance motor consists of identical single stack reluctance motors with their rotors mounted on a single shaft. The stator and rotor have the same number of poles or teeth. For a m stack motor, the stator pole or teeth in all m stacks are aligned. The rotor poles or teeth are

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Fmfdusjdbm!Nbdijoft

A

A

C

C

C

C

B

B

B

B A

A

Gjh/!21/39! Dsptt!tfdujpo!pg!b!tjohmf.tubdl!wbsjbcmf.sfmvdubodf!tufqqfs!npups

1 of the pole pitch angle. As the stator pole windings in a given stack m are excited simultaneously, the stator winding of each stack forms one phase. Hence the motor has the same number of phases as the number of stacks. The step angle in multistack type stepper motor is given by 360∞ …(10.4) Step angle = number of stacks or phase ¥ number of rotor poles or teetth

displaced from each other by

The cross section of a three stack variable reluctance stepper motor is shown in Fig. 10.29. The stator teeth in each stack is aligned. When phase winding A is excited rotor teeth of stack A are aligned with the stator teeth. When phase B is energized and phase A is deenergized the rotor moves by one step angle and rotor teeth of stack B align with stator teeth. iB

iA

Stack A Rotor

Stator

iC

Stack B Rotor

Stack C Rotor

Stator

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21/24/4! Izcsje!Tufqqfs!Npups This is, in fact, a permanent magnet stepper motor with constructional features of toothed and stacked rotors adopted from the variable-reluctance motor. The stator has only one set of windingexcited poles which interact with the two rotor stacks. The permanent magnet is placed axially along the rotor in the form of an annular cylinder over the motor shaft. The stacks at each end of the rotor are toothed. So all the teeth on the stack at one end of the rotor acquire the same polarity while all the teeth of the stack at the other end of the rotor acquire the opposite polarity. The two sets of teeth are displaced from each other by one half of the tooth pitch (also called pole pitch). These constructional details are brought out by Fig .10.30(a) and Fig. 10.30(b) for the case PM of three teeth on each stack so that tooth pitch Yt = 360°/3 =120°. This motor has a S two-phase, four-pole stator. N Consider now that the stator phase a is excited such that the top stator pole acquires north polarity while the bottom stator pole acquires south polarity. As a result, the nearest tooth of the front stack (assumed to be of north polarity) is pulled into lock(a) ing position with the stator south pole (top), and the diametrically opposite tooth of the Phase a near stack (south polarity) is simultaneously locked into the stator north pole (bottom). The repulsive forces on the remaining two (S) front stack teeth balance out as these are symmetrically located with respect to the N bottom stator pole. This rotor position is S thus a stable position with net torque on roS tor being zero. Rotor If now the stator phase b is excited N instead of phase a such that the right- Phase b N hand pole becomes south and the leftS hand pole becomes north, this would cause the rotor to turn anti-clockwise by (N) Yt /4 = 30° into the new locking position. The rotor will turn clockwise by 30° if Stator phase b is oppositely excited. (b) If the stator excitation is removed, the rotor will continue to remain locked into the Gjh/!21/41! Tdifnbujd! ejbhsbn! pg! b! izcsje! tufqqfs! npups same position as it is prevented to move in either direction by torque because of the

!

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Fmfdusjdbm!Nbdijoft

permanent magnet excitation. This fact is in favour of hybrid (also PM) motors. Compared to a PM motor, finer steps for better resolution are easily obtained in a hybrid motor by increasing the number of stack teeth and also by adding additional stack pairs on the rotor. For example, for a seven-teeth stack, the step size is (360°/7)/4 = (90/7)°. Also compared to variable-reluctance motor, a hybrid motor requires less stator excitation current because of the PM excited rotor. Half-stepping can be achieved in a hybrid motor by exciting phase ‘a’ and then exciting phase ‘b’ before switching off the excitation of phase ‘a’, and so on. In fact, any fractional step can be obtained by suitably proportioning the excitation of the two phases. Such stepping is known as microstepping. Typical step angles for stepper motors are 15°, 7.5°, 2° and 0.72°. The choice of the angle depends upon the angular resolution required for application.

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The permanent magnet dc (PMDC) motor is a dc motor whose poles are made of permanent magnets. These motors have many advantages compared to dc shunt motors. They do not have field-circuit copper losses since they do not require external field circuit. This also makes them smaller than corresponding dc shunt motors and this fact partially offsets the high cost of permanent magnets. PMDC motors are especially common in smaller fractional and subfractional horsepower sizes. These motors offer shunt-type characteristics. The risk of permanent magnetism getting destroyed by armature reaction (at starting/reversing or heavy overloads) has been greatly reduced by the new PM materials like ceramics and rare-earth magnetic materials. There are many disadvantages of permanent magnet motors. As they cannot produce high flux density as an externally supplied shunt field, hence a PMDC motor has lower induced torque per ampere of armature current than a shunt motor of the same size and construction. Moreover, PMDC motors suffer from risk of demagnetization. In a dc machine, the net flux produced by the field poles is reduced due to the magnetic field developed by the armature current which is known as armature reaction. In PMDC motors, the pole flux is the residual flux in the permanent magnets only. When the armature current is large, there is some risk that the armature mmf may demagnetize the poles, permanently reducing and reorienting the residual flux in them. Demagnetization may also be caused by excessive heating which can occur during prolonged periods of overload. For normal dc machines, the ferromagnetic materials for stator and rotor should have low values of residual flux density and coercive, magnetizing intensity since such materials have low hysteresis losses. On the other hand, in PMDC motors the good materials for the poles are those which have large residual flux density and large coercive magnetizing intensity. The large residual flux density produces large flux in the machine and large coercive magnetizing intensity means that a very large current would be required to demagnetize the poles. Since the flux in a PMDC motor is fixed, hence it is not possible to control the speed of a PMDC motor by varying the field current or flux. The only method of speed control for PMDC motors are armature voltage control and armature resistance control.

Tqfdjbm!Nbdijoft

QFSNBOFOU!NBHOFU!HFOFSBUPST!

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21/26

A permanent magnet generator is a synchronous generator in which the excitation coil, which is in the rotor is replaced by a system made up of permanent magnets which provides a constant field excitation. The operation of a permanent magnet generator differs synchronous generator in the way that in normal synchronous generator the voltage is controlled by excitation whereas in permanent magnet generator the excitation is constant and hence when it is charged the voltage drops and there is no provision to regulate the voltage. Hence these generators find their applications where voltage drop upto a certain degree is allowable. Otherwise a constant voltage output electronic circuit is connected at the output of the generator which can convert a voltage range into continuous voltage of constant value and in this way the generator can be used for many purposes. These generators are used to feed the regulators which provides excitation to large synchronous generators. The advantage is that when there is a short circuit in the large synchronous generator, they can supply the energy necessary for the regulator so that it can overexcite the generator and maintain the voltage for sufficient time before the protection circuits activate. These generators also find their applications in wind power generation. The main advantage of permanent magnet generator is its simplicity. The manufacturing and assembly of the rotor is cheaper. It does not require any maintance as it does not have brushes. It does not need any excitation system and energy saving upto 20% can be obtained. Being an energy source independent from the generator, it can serve to supply energy to auxiliary systems of the main generator.

! Qspcmfn!21/2 Efufsnjof!uif!tufqqjoh!bohmf!gps!b!)b*!uisff!qibtf!uxfouz!qpmf!qfsnbofou!nbhofu!tufqqfs!npups-! boe!)c*!uisff.tubdl!uxfmwf.uppui!wbsjbcmf!sfmvdubodf!tufqqfs!npups/

Solution 360∞ = 6° (from eq. 10.2) 3 ¥ 20 360∞ (b) Stepping angle = = 10° (from eq. 10.4). 3 ¥ 12 (a) Stepping angle =

! Qspcmfn!21/3 B!wbsjbcmf!sfmvdubodf!tufqqfs!npups!ibt!5!qpmft!xjui!21!uffui!jo!fbdi/!Efufsnjof!uif!tufqqjoh!bohmf! jg!uif!spups!ibt!71!uffui/

Solution Number of stator teeth = 4 ¥ 10 = 40 Number of rotor teeth = 60

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Fmfdusjdbm!Nbdijoft

\ from eq. (10.3) Stepping angle = 360° ¥

60 - 40 = 3°. 60 ¥ 40

! Qspcmfn!21/4 Uif!sftjtubodf!boe!joevdubodf!pg!b!tjohmf!qibtf!bd!tfsjft!npups!bsf!51!W!boe!1/6!I!sftqfdujwfmz/! Xifo!dpoofdufe!up!341!W!ed!tvqqmz!uif!mjof!dvssfou!jt!2!B!boe!uif!npups!svot!bu!4111!sqn/!Efufs. njof!uif!tqffe!boe!qpxfs!gbdups!pg!uif!npups!xifo!dpoofdufe!up!341!W-!61!I{!bd!tvqqmz!boe!ublft! uif!tbnf!mpbe!dvssfou/

Solution When connected to dc supply the developed back emf is Eb = V – IR = 230 – 1 ¥ 20 = 210 V When connected to ac supply from the phasor diagram shown in Fig. 10.10 V 2 = (IX)2 + (Er + IR)2 where Er is the speed emf \

Er = V 2 - ( IX ) 2 – IR

or,

Er = ( 230) 2 - (1 ¥ 2p ¥ 50 ¥ 0.5) 2 – 1 ¥ 20 = 148 V If N1 and N2 be the speeds under dc and ac operating conditions respectively, then N Eb = 1 Er N 2 N2 = N1 ¥

or, Power factor

cos q =

Er 148 = 3000 ¥ = 2114 rpm Eb 210

Er + IR 148 + 1 ¥ 20 = V 230

= 0.73 lag.

! Qspcmfn!21/5 B!vojwfstbm!bd!pqfsbufe!npups!ibt!3!qpmf!bsnbuvsf!xjui!2111!dpoevdupst/!Uif!mpbe!dvssfou!jt!6!B! xifo!uif!npups!tqffe!jt!4111!sqn/!Uif!bsnbuvsf!sftjtubodf!jt!5!W/!Uif!tvqqmz!wpmubhf!jt!211!W! boe!uif!joqvu!qpxfs!jt!511!X/!Efufsnjof!uif!fggfdujwf!bsnbuvsf!sfbdubodf!boe!vtfgvm!gmvy!qfs!qpmf/

Solution P = VI cos q

Tqfdjbm!Nbdijoft

cos q =

\

21/3:

400 = 0.8 or q = 36.87° lag 100 ¥ 5

If Er be the rotational emf then V - Er =I r + jX Now I and Er are in the same phase and we take I as reference. Hence supply voltage V = 100 36.87∞ 100 36.87∞ - Er = 5 0∞ 4 + jX

\

50 + j60 – Er = 20 + j5X

or,

Equating the real and imaginary terms and Now

80 – Er = 20 60 = 5X Er = 60 V and X = 12 W Pf ZN 60 A E ¥ 60 A 60 ¥ 60 ¥ 2 f= r = = 1.2 mWb PZN 2 ¥ 1000 ¥ 3000

\

Er =

or,

The maximum value of useful flux is 1.2 ¥ 10–3 ¥

2 = 1.7 mWb.

! Qspcmfn!21/6 B!qfsnbofou!nbhofu!ed!npups!ibt!op!mpbe!tqffe!pg!4111!sqn!xifo!dpoofdufe!up!211!W!ed!tvqqmz/! Uif!bsnbuvsf!sftjtubodf!jt!4!W/!Ofhmfdujoh!puifs!mpttft!efufsnjof!uif!tqffe!pg!uif!npups!xifo!ju!jt! dpoofdufe!up!91!W!tvqqmz!boe!efwfmpqt!b!upsrvf!pg!1/9!On/

Solution In a permanent magnet motor, Again or,

Eb • N since f is constant T • f Ia T • Ia

At no-load, Ebo = 100 V Speed No = 3000 rpm Let Iao and Ia1 be the currents at no load and loaded condition.

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Ia Ia To = o or, To = 0.8 ¥ o I a1 I a1 T1

\

Again

To =

=

100 ¥ I ao

3000 2p ¥ 60 Ia To = 0.3185 Iao = 0.8 o I a1 Ia1 = 2.5 A

\ or,

Ebo

\

Eb1 N1 = No

or,

Ebo I ao

Eb1 Ebo

=

wo

= 0.3185 Iao

No N1

= 3000 ¥

80 - 2.5 ¥ 3 100

= 2175 rpm.

Sfwjfx!Rvftujpot 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

What are the changes made in dc series motor to operate it under ac supply? How does a hysteresis motor differ from a reluctance motor? Discuss the methods of compensation of armature reaction in a universal motor. Describe the construction of a universal motor. Explain the principle of operation of a repulsion motor. What are the uses of a repulsion motor? Draw and explain the phasor diagram of a single-phase series motor. Discuss the principle of operation of a single-phase synchronous motor. What is reluctance torque? Draw the torque–speed characteristic of a reluctance motor. What is a two-phase servomotor? Where is it used? Draw the connection diagram and explain the principle of operation. What is the advantage of drag-cup rotor? Discuss the principle of operation of an ac tachometer. Describe the torque-producing phenomenon in a hysteresis motor. Why are these motors noiseless? What are the uses of these motors? What is the difference between a permanent magnet type of stepper motor and a reluctancetype stepper motor? Describe the operation of each type of stepper motor. Why are stepper motors most popular control motors? Differentiate between brushless dc motor and dc motor. How is the direction of rotation of a brushless dc motor reversed?

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15. Explain the construction and principle of operation of a dc brushless motor. 16. What are the advantages and disadvantages of a permanent magnet dc motor over a dc shunt motor? What are the characteristics of the materials used in poles of PMDC motors?

Nvmujqmf.Dipjdf!Rvftujpot 1. In a stepper motor, the detent torque means (a) minimum of the static torque with phase winding excited (b) maximum of the static torque with phase winding excited (c) minimum of the static torque with phase winding unexcited (d) maximum of the static torque with phase winding unexcited [GATE 2009] 2. For a given stepper motor which of the following torque has the highest numerical value? (a) Detent torque (b) Pull-in torque (c) Pull-out torque (d) Holding torque [GATE 2004] 3. Which of the following motors definitely has a permanent magnet rotor? (a) DC commutator motor (b) Brushless dc motor (c) stepper motor (d) reluctance motor [GATE 2004] 4. Match the following List I with List II and select the correct answer using the codes given below: List I

List II

P.

Food mixer

1.

Permanent magnet dc motor

Q.

Cassette tape recorder

2.

Single-phase induction motor

R.

Domestic water pump

3.

Universal motor

S.

Escalator

4.

Three phase induction motor

5.

DC series motor

6.

Stepper motor

P Q R S (a) 3 6 4 5 (b) 1 3 2 4 (c) 3 1 2 4 (d) 3 2 1 4 [GATE 2003] 5. A 1.8° step, 4 phase stepper motor has a total of 40 teeth on 8 poles of stator. The number of rotor teeth for this motor will be (a) 40 (b) 50 (c) 100 (d) 80 [GATE 2000] [Hints: Number of rotor teeth =

360∞ = 50] 1.8∞ ¥ 4

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6. A three phase three stack variable reluctance step motor has 20 poles on each rotor and stator stack. The step angle of this step motor is (a) 3° (b) 6° (c) 9° (d) 18° [GATE 2007] 360∞ = 6°] 3 ¥ 20 7. For a 1.8°, 2 phase bipolar stepper motor, the stepping rate is 100 step/s. The rotational speed of the motor in rpm is (a) 15 (b) 30 (c) 60 (d) 90 [GATE 2004] 360∞ = 200 [Hints: Steps for 1 revolution = 1.8∞ 1 \ 100 steps is covered in revolution 2 1 Hence in 1 second number of revolution = 2 1 In 60 second or 1 min number of revolution = ¥ 60 = 30] 2 8. A permanent magnet dc commutator motor has a no load speed of 6000 rpm when connected to a 120 V dc supply. The armature resistance is 2.5 W and other losses may be neglected. The speed of the motor with supply voltage of 60 V developing a torque of 0.5 Nm is (a) 3000 rpm (b) 2673 rpm (c) 2836 rpm (d) 5346 rpm [GATE 2000] [Hints: Back emf Eb • N, since f is constant Again Torque T • Ia At no-load, Ebo = 120 V [Hints: Step angle =

\

\

Again \

Ia Ia Ia Eb I a 120 I ao To T = o or, o = o or, To = 0.5 o = o o = 6000 I a1 I a1 0.5 I a1 T1 wo 2p ¥ 60 0.5 120 = or, Ia1 = 2.62 A I a1 200 p Ebo Eb1

=

N1 =

No N1 Eb1 Ebo

No =

60 - 2.62 ¥ 2.5 ¥ 6000 120

= 2673 rpm]

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9. An ac series motor is switched on to dc supply of rated voltage. The speed of motor will be (a) lower than that under ac operation (b) higher than that under ac operation (c) equal or less than that under ac operation (d) equal to that under ac operation 10. The function of compensating winding in ac series motor is (a) to improve the efficiency of the machine (b) to convert it into a two phase motor (c) to provide starting torque (d) to reduce reactance of armature winding 11. A dc series motor is switched on to ac supply (a) the motor will not start (b) the motor will start and run (c) the motor will start and run but will show poor performance (d) the motor will be damaged 12. The most common application of a three phase induction generator can be in a (a) hydro power station (b) wind power station (c) nuclear power station (d) steam power station 13. Induction generators deliver power at (a) zero pf (b) lagging pf (c) leading pf (d) unity pf 14. In a single phase series motor as the load torque increases (a) both power factor and speed decreases (b) power factor decreases but speed increases (c) power factor increases but speed decreases (d) both power factor and speed increases 15. A single phase repulsion motor with one stator winding is started from (a) low impedance position and it has shunt type speed torque characteristics (b) low impedance position and it has series type speed torque characteristics (c) high impedance position and it has shunt type speed torque characteristics (d) high impedance position and it has series type speed torque characteristics

Botxfst 1. (c) 6. (b) 11. (c)

2. (c) 7. (b) 12. (b)

3. (c) 8. (b) 13. (c)

4. (c) 9. (b) 14. (a)

5. (b) 10. (d) 15. (b)

22 Sfwjfx!Qspcmfnt!)NDRt*!xjui! Tpmvujpot TFDUJPO!B;!ED!NBDIJOFT A1 A 220 V, 15 kW, 1000 rpm shunt motor with armature resistance of 0.25 W, has a rated line current of 68 A and a rated field current of 2.2 A. The change in field flux required to obtain a speed of 1600 rpm while drawing a line current of 52.8 A and a field current of 1.8 A is (a) 18.18% increase (c) 36.36% increase

(b) 18.18% decrease (d) 36.36% decrease

[GATE 2012]

Solution Eb1 = 220 – (68 – 2.2) ¥ 0.25 = 203.55 V Eb2 = 220 – (52.8 – 1.8) ¥ 0.25 = 207.25 V Eb1

\

Eb2

=

f1 f2

1 2

or,

203.55 f1 ¥ 1000 = 207.25 f2 ¥ 1600

\

f2 = 0.6363 f1

\ field flux should be reduced by (100 – 63.63)% or 36.36% Ans. (d) 36.36% decrease A2 A 220 V, dc shunt motor is operating at a speed of 1440 rpm. The armature resistance is 1 W and armature current is 10 A. If the excitation of the machine is reduced to 10%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will be (a) 1.79 W (b) 2.1 W (c) 3.1 W (d) 18.9 W [GATE 2011]

!

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Fmfdusjdbm!Nbdijoft

Solution As speed is same,

I2

Eb1 Eb2

If

=

1 f1 2 f2

=

f1 f2

Ra = 1 W

f 220 - 10 ¥ 1 = 1 Eb2 0.9 f1

\

220 V

Gjh/!B!3/2! Npups! djsdvju! pg! Qspc/!B3

Eb2 = 210 ¥ 0.9 = 189 V

or,

Ia

If external resistance is R and armature current Ia2 then Eb2 = 220 – Ia2 (1 + R) = 189 Now as torque remains same, 1

=

2

\

Ia2 =

\

220 – or

f1 I a1 f2 I a2

or

f1 ¥ 10 =1 0.9 f1 I a2

10 100 A = 0.9 9

100 (1 + R) = 189 9 R = 1.79 W

Ans. (a) 1.79 W A3 A separately excited dc motor runs at 1500 rpm under no-load with 200 V applied to the armature. The field voltage is maintained at its rated value. The speed of the motor when it delivers a torque of 5 Nm is 1400 rpm as shown in Fig. A3.1. The rotational losses and armature reaction are neglected. [GATE 2010] (i) The armature resistance of the motor is (a) 2 W (b) 3.4 W (c) 4.4 W (d) 7.7 W (ii) For the motor to deliver a torque of 2.5 Nm at 1400 rpm, the armature voltage to be applied is (a) 125.5 V (b) 193.3 V (c) 200 V (d) 241.7 V

Speed (rpm) 1500 1400

0

5

Tarque (Nm)

Gjh/!B4/2! Tqffe.upsrvf! dvswf! pg! npups!pg!Qspc/!B4

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

22/4

Solution (i) Eb1 = 200 = K N1 = K ¥ 1500 (E f constant) \

\

K=

Pf Z 200 = 60 A 1500

T=

Pf ZI a Pf Z 60 200 60 = Ia = ¥ Ia = 5 2p A 60 A 2p 1500 2p

Ia = 3.925 A Eb2 = 200 – 3.925 R = KN2 = 200 -

\

R=

200 ¥ 1400 1500

200 ¥ 1400 1500 = 3.397 W 3.925 / 3.4 W

Ans. (b) 3.4 W 60 200 60 (ii) T = 2.5 = K Ia¢ = Ia¢ ¥ 2p 1500 2p \ I¢a = 1.9625 A 200 Eb¢ = KN ¢ = ¥ 1400 = V – Ia¢ Ra 1500 2800 \ = V – 1.9625 ¥ 3.4 15 \ V = 193.3 V Ans. (b) 193.3 V A4 A 240 V, dc shunt motor draws 15 A while supplying the rated load at a speed of 80 rad/s. The armature resistance is 0.5 W and the field winding resistance is 80 W. (i) The net voltage across the armature resistance at the time of plugging will be (a) 6 V (b) 234 V (c) 240 V (d) 474 V (ii) The external resistance to be added in the armature circuit to limit the armature current to 125% of its rated value is (a) 31.1 W (b) 31.9 W (c) 15.1 W (d) 15.9 W [GATE 2008] Solution (i) Ish =

240 A=3A 80

\ Hence,

I2 = 15 A Ia = 15 – 3 A = 12 A Eb = V – Ia ra = 240 – 12 ¥ 0.5 = 234 V

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In plugging, the connections to the armature are reversed suddenly. The effective voltage across the armature initially is Eb + V. \ the net voltage across the armature resistance at the time of plugging is (240 + 234) V or 474 V Ans. (d) 474 V (ii) I a¢ = 1.25 ¥ 12 A = 15 A During plugging, –Eb = V – I a¢ (R + 0.5) or, – 234 = 240 – 15 (R + 0.5) \ R = 31.1 W Ans. (a) 31.1 W A5 A 220 V dc machine supplies 20 A at 200 V as a generator. The armature resistance is 0.2 W. If the machine is now operated as a motor at same terminal voltage and current but with the flux increased by 10% then ratio of motor speed to generator speed is (a) 0.87 (b) 0.95 (c) 0.96 (d) 1.06 [GATE 2006] Solution Eg = 200 + 20 ¥ 0.2 = 204 V Eb = 200 – 20 ¥ 0.2 = 196 V Eg

\

Eb

=

f Ng 204 = 196 1.1 f N m

Nm 196 = = 0.8734 Ng 204 ¥ 1.1

\

\ motor speed to generator speed is 0.87 Ans. (a) 0.87 A6 A 50 kW dc shunt motor is loaded to draw rated armature current at any given speed. When driven (i) at half the rated speed by armature voltage control, and (ii) at 1.5 times the rated speed by field control, the respective output powers delivered by the motor are approximately. (a) (b) (c) (d)

25 kW in (i) and 75 kW in (ii) 25 kW in (i) and 50 kW in (ii) 50 kW in (i) and 75 kW in (ii) 50 kW in (i) and 50 kW in (ii)

[GATE 2005]

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

22/6

Solution (i) Armature current Ia is constant Now output power \

P = Eb Ia (neglecting mechanical losses) P • Eb

For shunt motor, \

Eb • N (in armature voltage control method f is constant) P •N

Hence, output power at half speed will be half of rated power, i.e. 25 kW. (ii) In field control method, the voltage across the armature is constant. \ Eb is constant and Ia is also constant (E given) Hence, though the speed becomes 1.5 times the rated speed, the flux will be decreased by 1.5 times making Eb constant. \ output power will remain same, i.e. 50 kW Ans. (b) 25 kW in (i) and 50 kW in (ii) A7 Match the following List I with List II and select the correct answer using the codes given below the lists. List I List II P. Armature emf (E) 1. Flux (f), speed (w) and armature current (Ia) Q. Developed torque (T) 2. f and w R. Developed power (P) 3. f and Ia 4. Ia and w 5. Ia only P Q R (a) 3 3 1 (b) 2 5 4 (c) 3 5 4 (d) 2 3 1 [GATE 2005] Solution Armature emf

E=

Pf Z w = Kfw pA E • fw

=

\

Pf ZN 60 A

where K is constant

!

22/7

Fmfdusjdbm!Nbdijoft

Hence, E depends on f and w Developed torque in dc motor T=

Pf ZI a = Kf Ia pA

\ developed torque depends on f and Ia Developed power P = Eb Ia = Kfw Ia \ developed power depends on f, w and Ia Ans. (d)

P Q 2 3

R 1

A8 The armature resistance of a permanent magnet dc motor is 0.8 W. At no load, the motor draws 1.5 A from a supply voltage of 25 V and runs at 1500 rpm. The efficiency of the motor while it is operating on load at 1500 rpm drawing a current of 3.5 A from the same source will be (a) 48% (b) 57.1% (c) 59.2% (d) 88.8% [GATE 2004] Solution ra = 0.8 W At no load, the output is zero \ \

Loss = Input = 1.5 ¥ 25 W constant loss = 1.5 ¥ 25 – (1.5)2 ¥ 0.8 = 35.7 W When

Ia = 3.5 A Loss = (3.5)2 ¥ 0.8 + 35.7 = 45.5 W

\

h=

Output 3.5 ¥ 25 - 45.5 ¥ 100% = 48% = Input 3.5 ¥ 25

Ans. (a) 48% A9 An 8-pole, dc generator has a simplex wave wound armature containing 32 coils of 6 turns each. Its flux per pole is 0.06 Wb. The machine is running at 250 rpm. The induced armature voltage is (a) 96 V

(b) 192 V

(c) 384 V

Solution P =8 Z = 2 ¥ total number of turns = 2 ¥ 32 ¥ 6 f = 0.06 Wb

(d) 768 V

[GATE 2004]

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

22/8

N = 250 rpm Pf ZN E= 60 A For simplex wave wound, A =2 8 ¥ 0.06 ¥ 2 ¥ 32 ¥ 6 ¥ 250 E= = 384 V 60 ¥ 2

\ Ans. (c) 384 V

A10 To conduct load test on a dc shunt motor, it is coupled to a generator which is identical to the motor. The field of the generator is also connected to the same supply source as the motor. The armature of the generator is connected to a load resistance. The armature resistance is 0.02 p.u. Armature reaction and mechanical losses can be neglected. With rated voltage across the motor, the load resistance across the generator is adjusted to obtain rated armature current in both motor and generator. The p.u. value of this load resistance is (a) 1

(b) 0.98

(c) 0.96

(d) 0.94

[GATE 2003]

Solution

\ \ load resistance

Eb = V – Ia ra = 1 – 1 ¥ 0.02 = 0.98 p.u. Eg = Eb = 0.98 = V + Ia ra = V + 1 ¥ 0.02 V = 0.98 – 0.02 = 0.96 p.u. V 0.96 = 0.96 p.u. R= = 1 Ia

Ans. (c) 0.96 A11 A dc series motor driving an electric train faces a constant power load. It is running at rated speed and rated voltage. If the speed has to be brought down to 0.25 p.u., the supply voltage has to be approximately brought down to (a) 0.75 p.u. (b) 0.5 p.u. (c) 0.25 p.u. (d) 0.125 p.u. [GATE 2003] Solution In series motor T • f Ia and f • Ia T • Ia2

\ Power

(Eb Ia) = Tw 1 If speed decreases to 0.25 p.u, w decreases by th and to maintain constant power load, T should 4 be increased by 4 times.

!

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As T • Ia2, four times increase of T means twice the increase of Ia. Now if current increases by two times, the supply voltage should be halved in order to maintain constant power load. Ans. (b) 0.5 p.u. A12 A 200 V, 2000 rpm, 10 A, separately excited dc motor has an armature resistance of 2 W. Rated dc voltage is applied to both the armature and field winding of the motor. If the armature draws 5 A from the source, the torque developed by the motor is (a) 4.3 Nm Solution V N Ia ra Eb1

(b) 4.77 Nm

(c) 0.45 Nm

(d) 0.5 Nm

[GATE 2002]

= 200 V = 2000 rpm = 10 A =2W = 200 – 10 ¥ 2 = 180 V T1 =

Torque

Eb I a

=

w

180 ¥ 10 180 ¥ 10 ¥ 60 Nm = Nm 2000 2p ¥ 2000 2p ¥ 60

Eb2 = 200 – 5 ¥ 2 = 190 V Eb1

Now

Eb2

1

=

2

\

N2 = N1 ¥

\

T2 =

Eb2 Eb1

Eb2 I a2 w2

=

= 2000 ¥

190 rpm 180

190 ¥ 5 ¥ 60 = 4.3 Nm 190 2p ¥ 2000 ¥ 180

Ans. (a) 4.3 Nm A13 A 240 V dc series motor takes 40 A when giving its rated output at 1500 rpm. Its resistance is 0.3 W. The value of resistance which must be added to obtain rated torque at 1000 rpm is (a) 6 W

(b) 5.7 W

Solution V Ia N1 ra + rse

= 240 V = 40 A = 1500 rpm = 0.3 W

(c) 2.2 W

(d) 1.9 W

[GATE 2000]

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

Eb1 Eb2

=

22/:

I a1 N1 I a2 N 2

240 - 40 ¥ 0.3 40 ¥ 1500 = Eb2 I a2 ¥ 1000

or,

Now torque in the series motor T • I a2 As torque in both the conditions are same, Ia2 = Ia1 = 40 A \

228 60 60 = = Eb2 I a2 40

or,

Eb2 = 228 ¥

60 = 152 V 40

Now if R be the resistance added, Eb2 = 240 – 40 ¥ (0.3 + R) = 152 \

R=

240 - 152 – 0.3 = 1.9 W 40

Ans. (d) 1.9 W A14 A separately excited dc motor with armature resistance of 0.5 W draws an armature current of 20 A at 1500 rpm when connected to a 250 V supply. The torque developed when the machine draws an armature current of 10 A for the same field current is (a) 16.5 Nm Solution V I a1 ra N1

(b) 15.28 Nm

(c) 17 Nm

= 250 V = 20 A = 0.5 W = 1500 rpm

Torque

(250 - 20 ¥ 0.5) ¥ 20 Nm 1500 w 2p ¥ 60 = 30.57 Nm

T1 =

Eb I a

=

(d) 30.57 Nm

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Fmfdusjdbm!Nbdijoft

As field current is constant, when armature current is 10 A, the developed torque is Ia 10 = 15.28 Nm T2 = T1 ¥ 2 = 30.57 ¥ I a1 20 Ans. (b) 15.28 Nm A15 An 8-pole dc generator has a lap-wound armature winding containing 60 slots with 15 conductors in each slot. If the flux per pole is 0.01 Wb and it runs with a speed of 1800 rpm, the value of the generated voltage is (a) 1080 V (b) 18 V (c) 270 V (d) 4.5 V Solution E= Here, \

Pf ZN 60 A

P = 8, f = 0.01 Wb, Z = 60 ¥ 15 = 900, N = 1800 rpm, A = 8 (\ lap wound) 8 ¥ 0.01 ¥ 900 ¥ 1800 E= V = 270 V 60 ¥ 8

Ans. (c) 270 V A16 An 80 kW, 1500 rpm generator operating at rated load has a terminal voltage of 240 V. If the voltage regulation is 3%, the no-load voltage will be (a) 247.2 V (b) 232.8 V (c) 247.42 V (d) 240 V Solution Voltage regulation =

no load

-

rated

rated

\ or,

0.03 =

- 240 240

no load

Vno load = 240 ¥ 0.03 + 240 = 247.2 V

Ans. (a) 247.2 V A17 A 4-pole, wave-wound dc shunt generator supplies 80 lamps, each of 100 W at 240 V. If the armature and field resistances are 0.3 W and 50 W respectively, the current in each armature conductor is (a) 16.68 A Solution P =4

(b) 38.16 A

(c) 33.36 A

(d) 19.08 A

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

22/22

A =2 V = 240 V Current drawn by each lamp is

100 A or 0.417 A 240

\ total load current IL = 0.417 ¥ 80 = 33.36 A Ish =

Field current \ armature current

240 A = 4.8 A 50

Ia = IL + Ish = 38.16 A

As there are two parallel paths in a wave-wound machine, the current in each armature conductor 38.16 is A or 19.08 A 2 Ans. (d) 19.08 A A18 A separately excited dc generator running at a speed of 1200 rpm delivers 150 A at 400 V to a constant resistive load. The armature resistance is 0.12 W. If the current is reduced to 100 A, neglecting armature reaction, the generator speed is (a) 667 rpm (b) 1183 rpm (c) 1200 rpm (d) 800 rpm Solution N1 = 1200 rpm, Ia1 = IL1 = 150 A, V = 400 V, ra = 0.12 400 W = 2.67 W 150 E1 = 400 + Ia ra = 400 + 150 (0.12) = 418 V

RL =

Now IL2 = Ia2 = 100 A; the terminal voltage Vt = 100 ¥ 2.67 = 267 V \ \

E2 = 100 ¥ 0.12 = 279 V 1

=

2

or,

N2 =

1 2 2 1

N1 =

279 ¥ 1200 = 800.95 rpm 418

Ans. (d) 800 rpm A19 A shunt generator delivers 100 kW at 240 V at a speed of 750 rpm. The armature and field resistances are 0.05 W and 50 W respectively. If the machine is operated as a motor and drawing

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Fmfdusjdbm!Nbdijoft

80 kW from the 240 V source, the speed of the motor will be (a) 642 rpm (b) 750 rpm (c) 650 rpm

(d) 942 rpm.

Solution Generator V = 240 V Ng = 750 rpm Ia = IL + Ish =

100 ¥ 103 240 + 240 50

= 416.67 + 4.8 = 421.47 A Eg = V + Ia ra = 240 + 421.47 ¥ 0.05 = 261.07 V Motor Ia = IL – Ish =

80, 000 240 = 333.33 – 4.8 240 50 = 328.53 A

Em = V – Ia ra = 240 – 328.53 ¥ 0.05 = 223.57 V Eg

\

Em

=

Ng Nm

or Nm =

Em 223.57 Ng = ¥ 750 Eg 261.07

= 642.27 rpm Ans. (a) 642 rpm A20 The armature, shunt and series winding resistances of a 150 A, 240 V compound generator are 0.05 W, 50 W and 0.03 W respectively. The induced emfs of the generator when connected as long shunt and short shunt are respectively (a) 252.24 V, 252.38 V (b) 252.38 V, 252.24 V (c) 245.3 V, 245 V (d) 263.2 V, 265 V Solution When connected as long shunt (Fig. A20.1),

\

IL = 150 A V = 240 V 240 Ish = A = 4.8 A 50 Ia = IL + Ish = 150 + 4.8 = 154.8 A E = V + Ia (ra + rse) = 240 + 154.8 (0.05 + 0.03) = 252.384 V

Ia

150A Ish V

Gjh/!B31/2! Mpoh!tivou!djsdvju

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22/24

When connected as short shunt (Fig. A20.2), Voltage across shunt field,

150A

Vsh = 240 + 150 ¥ 0.03 = 244.5 V Ish =

244.5 A = 4.89 A 50

V

\

Ia = IL + Ish = 150 + 4.89 = 154.89 A

\

E = Vsh + Ia ra = 244.5 + 154.89 ¥ 0.05 = 252.24 V

Gjh/!B31/3! Tipsu!tivou!djsdvju

Ans. (b) 252.38 V, 252.24 V A21 A dc shunt generator delivers 200 A at a terminal voltage of 240 V. If the armature and field resistances are 0.05 W and 50 W respectively and stray losses are 1000 W, the efficiency of the machine is (a) 97.96%

(b) 93.66%

(c) 91.87%

(d) 93.94%

Solution IL = 200 A 240 Ish = W = 4.8 A 50 Ia = 204.8 A

\

Armature copper loss = Ia2 ra = (204.8)2 ¥ 0.05 = 2097.15 W 2 rsh = (4.8)2 ¥ 50 = 1152 W Field copper loss = Ish

\ total copper loss = 3249.15 W Stray loss = 1000 W \ total loss = 4249.15 W h=

Output 200 ¥ 240 ¥ 100% = 91.87% = Input 200 ¥ 240 + 4249.15

Ans. (c) 91.87% A22 A 25 kW, 125 V separately excited dc machine runs at a constant speed of 3000 rpm. The field current is maintained constant such that the open-circuit armature voltage is 125 V. The armature resistance is 0.02 W. If the terminal voltage is equal to 128 V, which of the following is true? (a) The machine operates as a motor and the electromagnetic torque is equal to 59.7 Nm. (b) The machine operates as a motor and the electromagnetic torque is equal to 61.1 Nm. (c) The machine operates as a generator and the electromagnetic torque is equal to 50.3 Nm. (d) The machine operates as a generator and the electromagnetic torque is equal to 24.8 Nm.

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22/25

Solution Open-circuit voltage Terminal voltage As

Fmfdusjdbm!Nbdijoft

Eo = 125 V V = 128 V V > Eo

the machine will operate as a motor. Ia = Now \

128 - 125 = 150 A 0.02

Eb = 125 V torque =

Eb I a 125 ¥ 150 = Nm = 59.7 Nm 3000 w 2p ¥ 60

Ans. (a) Motor and torque, 59.7 Nm A23 (a) (b) (a) (a)

In the previous problem, if the terminal voltage is 122 V, which of the following is true? The machine operates as a motor and the electromagnetic torque is 597.13 Nm. The machine operates as a generator and the electromagnetic torque is 597.13 Nm. The machine operates as a motor and the electromagnetic torque is 60.7 Nm. The machine operates as a generator and the electromagnetic torque is 60.7 Nm.

Solution As Eo > V, the machine operates as a generator. Ia = T=

125 - 122 A = 150 A 0.02 Ea I a 125 ¥ 150 = Nm = 597.13 Nm 3000 w 2p ¥ 60

Ans. (b) A24 Two shunt generators rated 40 kW, 400 V and 100 kW, 400 V operate in parallel and deliver a total load current of 300 A. The regulation of generators are 5% and 3%. The currents delivered by the generators are (a) 150 A, 150 A (b) 127.8 A, 172.2 A (c) 62.4 A, 237.6 A (d) 58 A, 242 A

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22/26

Solution Load current of the first generator ILfl =

40 ¥ 103 = 100 A 400

Voltage regulation = 0.05 \ full-load voltage drop = 400 ¥ 0.05 = 20 V If the load current contributed by the first generator is I1 when both the generators are operating in parallel then the voltage of the first generator at load current I1 V1 = 400 –

Full load voltage drop 20 I1 = 400 – I1 100 Full load current

Similarly, for the second generator, I Lfl =

100 ¥ 103 = 250 A 400

Voltage regulation = 0.03 \ full-load voltage drop = 400 ¥ 0.03 = 12 V If current contributed is I2 then voltage V2 = 400 –

12 I2 250

For parallel operation, V1 = V2 \

400 – Again,

20 12 I1 = 400 – I2 100 250

I1 + I2 = 300 A

Solving the above two equations, I1 = 58 A and I2 = 242 A Ans. (d) 58 A, 242 A A25 A 80 kW, 240 V, long shunt compound generator (Fig. A25.1) has armature, shunt field and series field resistances of 0.06 W, 50 W and 0.04 W respectively. The rotational loss is 3 kW. It supplies a load at its maximum efficiency and rated voltage. The maximum efficiency of the generator is (a) 93.2% (b) 95.6% (c) 85.2% (d) 88.3% Solution V = 240 V, ra = 0.06 W, rsh = 50 W, rse = 0.04 W 240 Ish = A = 4.8 A 50

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22/27

Fmfdusjdbm!Nbdijoft

For maximum efficiency, Variable loss = Constant loss I a2 max(ra + rse) I a2 max(0.1) Ia max IL max

\ \ or, \

2 = Pr + Ish rsh = 3000 + (4.8)2 ¥ 50 = 4152 = 203.76 A = 203.76 – 4.8 = 198.96 A

Long shunt

Power output at maximum efficiency = 240 ¥ 198.96 = 47.75 kW

Gjh/!B36/2! Mpoh! tivou! dpoofdujpo

Total loss under maximum efficiency condition = 2 ¥ {3000 + (4.8)2 ¥ 50} = 8304 W \

hmax =

47.75 ¥ 103 47.75 ¥ 103 + 8304

or Ans. (c) 85.2%

= 0.852 = 85.2%

A26 Two generators with ratings of 300 kW and 600 kW are operated in parallel at a rated voltage of 220 V for each. The no-load voltage for both the generators is 250 V. If the terminal voltage is 230 V then output of the two generators will be (a) 200 kW, 400 kW (c) 290 kW, 610 kW

(b) 450 kW, 450 kW (d) 209 kW, 418 kW

Solution For Generator 1, Pout = 300 kW E = 250 V Vrated = 220 V \

IL1 =

300 ¥ 103 A 220

\

E = V + Ia1 ra

or,

ra = =

E -V IL

(Neglecting shunt field current)

250 - 220 3

300 ¥ 10

¥ 220 =

30 ¥ 220 300 ¥ 103

W

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22/28

When V = 230 V then if, current is I¢L1 230 = 250 – IL1 ¥ \

I¢L1 =

30 ¥ 220 300 ¥ 103

20 ¥ 300 ¥ 103 = 909.1 A 30 ¥ 220

\ power output = 909.1 ¥ 230 = 209 kW Similarly, for Generator 2, 20 ¥ 600 ¥ 103 = 1818.18 A I¢L2 = 30 ¥ 220 \ power output = 1818.18 ¥ 230 = 418 kW Ans. (d) 209 kW, 418 kW A27 A 6-pole, 50 kW, dc machine operating at 1180 rpm has a generated emf of 136.8 V. If the speed is reduced to 75% of its initial value and the flux per pole is doubled then the induced emf is (a) 205.2 V

(b) 102.6 V

(c) 364.8 V

(d) 136.8 V

Solution P =6 N1 = 1180 rpm E1 = 136.8 V E= \

1 2

=

Pf ZN 60 A f1 f2

1 2

N2 = 0.75 N1 and f2 = 2f1 then E2 = E1 ¥ or, Ans. (a) 205.2 V

f2 f1

2 1

= 136.8 ¥

2 f1 ¥ 0.75 f1 1

1

E2 = 205.2 V

A28 An 8-pole, 240 V dc shunt generator is supplying full-load current when it is running at 1200 rpm. The armature and field resistances are 0.03 W and 200 W respectively. The machine is lap wound and it has 800 conductors. If the voltage across the armature resistance is 6.7 V, the value of load current and flux per pole are (a) 230.5 A, 20 mWb (b) 235 A, 10 mWb (c) 224.5 A, 15.5 mWb (d) 217.5 A, 12.5 mWb

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22/29

Fmfdusjdbm!Nbdijoft

Solution P =8 V = 240 V N = 1200 rpm ra = 0.03 W rsh = 200 W A =P Z = 800 Ia ra = 6.7 V \

6.7 A = 223.33 A 0.03 240 = = 1.2 A 200

Ia = Ish

\

I2 = Ia + Ish = 224.53 A E = V + Ia ra = 240 + 223.33 ¥ 0.03 = 246.7 V As E =

Pf ZN 60 AE 60 ¥ 8 ¥ 246.7 , hence f = Wb = 0.0154 Wb = 60 A PZN 8 ¥ 800 ¥ 1200

Ans. (c) 224.5 A, 15.4 mWb A29 A 220 V dc generator supplies 8 kW at a terminal voltage of 220 V. The armature resistance is 0.4 W and when the flux per pole increases by 20% the machine operates as a motor with the same terminal voltage and armature current. The ratio of generator speed to the motor speed is (a) 1.175

(b) 1.37

(c) 1.05

(d) 0.913

Solution For generator, Ia =

8 ¥ 103 A = 36.36 A 220

Eg = V + Ia ra = 220 + 36.36 ¥ 0.4 = 234.54 V For motor, Em = V – Ia ra = 220 – 36.36 ¥ 0.4 = 205.456 V Now, \

fm = 1.2 fg Eg Em

=

fg N g fm N m

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

Ng

or,

Nm

=

Eg Em

¥

22/2:

fm 234.54 = ¥ 1.2 f g 205.456

= 1.37 Ans. (b) 1.37 A30 Two generators are operating in parallel and deliver a total load current of 60 A. The terminal voltage of the first generator is 280 V on no load and 240 V while supplying 40 A current. The second generator has a voltage of 300 V when on no load and the terminal voltage is 240 V when it supplies 50 A current. The output voltage of each generator is (a) 240 V (b) 271.7 V (c) 256.4 V (d) 238.2 V Solution IL = I1 + I2 = 60 A fi I2 = 60 – I1 Vnl1 = 280 V VL1 = 240 V IL1 = 40 A Vnl - VL1 280 - 240 =1W ra1 = 1 = I L1 240

\ Similarly,

ra2 =

300 - 240 = 1.2 W 50

If V be the output voltage then V = Vnl1 – I1 ra1 = Vnl2 – (60 – I1) ra2 V = 280 – I1 and V = 300 – (60 – I2) ¥ 1.2 I1 = 23.63 A and I2 = 36.37 A V = 280 – 23.63 = 256.37 V

or, \ \ Ans. (c) 256.4 V

A31 In the previous problem, the output power generated by the first and second generators are respectively (a) 9.3 kW, 9.3 kW (b) 9.6 kW, 12 kW (c) 5.6 kW, 8.7 kW (d) 6 kW, 9.3 kW Solution V = 256.37 V I1 = 23.63 A \ output power of the first generator is 256.37 ¥ 23.63 W or 6 kW I2 = 36.37 A

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\ output power of second generator is 256.37 ¥ 36.37 W or 9.3 kW Ans. (d) 6 kW, 9.3 kW A32 A 240 V, 15 kW shunt generator is driven by a 25 HP motor to generate the rated output. The armature circuit resistance is 0.3 W and the field resistance is 100 W. The armature current for maximum efficiency is (a) 64.9 A (b) 62.5 A (c) 100 A (d) 84.14 A Solution Shunt field current Ish =

240 V A = 2.4 A = rsh 100

\ shunt field copper loss = I 2sh rsh = (2.4)2 ¥ 100 = 576 W Line current IL =

15 ¥ 103 A = 62.5 A 240

\ armature current Ia = IL + Ish = 64.9 A Generated emf E = 240 + 64.9 ¥ 0.3 = 259.47 V Power developed in the armature Pa = E Ia = 259.47 ¥ 64.9 = 16839.6 W Rotational loss = Input power – Pa = 25 ¥ 735.5 – 16839.6 = 1547.9 W \ Constant loss = Rotational loss + Shunt field copper loss = 1547.9 + 576 = 2123.9 If Ia be the armature current under maximum efficiency condition then I 2a ra = 2123.9 or

Ia =

2123.9 = 84.14 A 0.3

Ans. (d) 84.14 A A33 The relation between the induced emf and series field current in the linear region of a series generator is given as Ea = 0.5 Is. The armature and series field winding resistances are 0.04 W and 0.06 W respectively. The generator is used as a booster between a 300 V station bus bar and a feeder of 0.5 W resistance. If the current in the feeder is 200 A, the voltage between the far end of the

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

feeder and the bus bar is (a) 300 V (b) 200 V

(c) 500 V

22/32

(d) 100 V

Solution Feeder current = 200 A \ Is = 200 A \ Ea = 0.5 ¥ 200 = 100 V Bus-bar voltage is 300 V. As the generator is used a booster, the total voltage supplied to the feeder is = 300 + 200 – Is (ra + rs) = 500 – 200 (0.04 + 0.06) = 300 V \ voltage between the far end of the feeder and bus bar = 300 – feeder resistance drop = 300 – 200 ¥ 0.5 = 300 – 100 = 200 V Ans. (b) 200 V A34 A 2-pole wave-wound double-layer dc machine has 24 slots with 18 turns per coil. The effective length of the machine is 20 cm and the radius of the armature is 10 cm. The magnetic poles cover 80% of the armature periphery. The average flux density per pole is 1 T and the armature angular velocity is 183.2 rad/s. The induced emf per coil is (a) 105 V (b) 52.5 V (c) 230.4 V (d) 630 V Solution For double-layer winding, the number of coils is equal to the number of slots. \ total number of conductors Z = 2 ¥ 18 ¥ 24 = 864 Actual pole area =

2p rl 2p ¥ 0.1 ¥ 0.2 = 0.0628 sqm = P 2

Effective pole area = 0.8 ¥ 0.0628 = 0.05 sqm Flux per pole = 1 ¥ 0.05 = 0.05 Wb Induced emf E =

Pf ZN 2 ¥ 0.05 ¥ 864 ¥ 183.2 ¥ 60 = 1260 V = 60 A 2p ¥ 60 ¥ 2 Total number of coils 2 24 = = 12 2

Since there are two parallel paths, the number of coils in each path =

\ emf induced in each coil is Ans. (a) 105 V

1260 V = 105 V 12

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A35 An 80 kW, 240 V shunt generator with armature and field resistance of 0.03 W and 60 W respectively is operated at rated voltage. If the generator has a total mechanical and core loss of 2 kW, the output of the prime mover to drive the generator at full load is (a) 80.75 kW (b) 86.37 kW (c) 85 kW (d) 93.7 kW Solution V = 240 V Ish =

240 A=4A 60

IL =

80 ¥ 103 A = 333.33 A 240

Ia = Ish + IL = 337.33 A Armature copper loss = I a2 ra = (337.33)2 ¥ 0.03 W = 3413.75 W 2 Field copper loss = I sh rsh = (4)2 ¥ 60 W = 960 W Total loss = 3413.75 + 960 + 2000 = 6373.75 W = 6.37 kW Output = 80 kW \ Input = 80 + 6.37 = 86.37 kW

Ans. (b) 86.37 kW A36 What is the increase in torque expressed as percentage of initial torque, if the current drawn by a series motor is increased form 15 A to 20 A? Neglect saturation. (a) 78%

(b) 33.33%

(c) 15.47%

Solution For series motor, T • I 2a \

1 2

=

I a21 I a22

Ê 15 ˆ =Á ˜ Ë 20 ¯ 2

Ê 20 ˆ T2 = Á ˜ T1 = 1.78 T1 Ë 15 ¯

or \ increase in torque =

2

2

1

1

¥ 100%

(d) 5%

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22/34

1.78 - 1 ¥ 100% 1 = 78%

= Ans. (a) 78%

A37 The armature resistance of a permanent magnet dc motor is 3 W and the applied voltage is 10 V. If the motor draws an armature current of 15 mA at no-load speed of 12,000 rpm the motor torque constant in V/(rad/s) and the no-load rotational losses in mW are respectively. (a) 0.1321 ¥ 10–3, 149.3 (c) 0.1321 ¥ 10–3, 1.5

(b) 7.926 ¥ 10–3, 1.5 (d) 7.926 ¥ 10–3, 149.3

Solution Back emf at no load Eb = V – Ia ra = 10 – 15 ¥ 10–3 ¥ 3 = 9.955 V \

Torque constant =

Eb 9.955 = = 7.926 ¥ 10–3 V/(rad/s) 2 p ¥ 12 , 000 w 60

No-load rotational losses = Eb Ia = 9.955 ¥ 15 ¥ 10–3 = 149.325 mW –3 Ans. (d) 7.926 ¥ 10 , 149.3 A38 A 230 V, dc shunt motor draws an armature resistance of 0.3 W and armature current of 80 A while driving constant load torque. An external resistance of 1.5 W is inserted in series with the armature while the shunt field current is unchanged. The fractional change in motor speed neglecting the armature reaction and rotational losses is (a) 0.35 (b) 0.51 (c) 0.27 (d) 0.58 Solution V = 230 V Ia1 = 80 A ra = 0.3 W Eb1 = 230 – 80 ¥ 0.3 = 206 V

\

For constant load, torque Ia will remain same \

Eb2 = 230 – 80 (0.3 + 1.5) = 86 V If N2 be the speed then Eb2 Eb1

=

2 1

=

86 206

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\ change in motor speed =

1-

2

=

1-

1

86 206

1

= 0.58

1

Ans. (d) 0.58 A39 A 10 HP, 220 V series motor has armature resistance of 0.5 W, field resistance of 0.25 W and rotational losses of 1040 W. The armature current when the motor delivers rated load is (a) 41.88 A (b) 33.81 A (c) 50 A (d) 38.63 A Solution Output power Po = 10 ¥ 735.5 = 7355 W Power developed = 7355 + 1040 = 8395 W \

Eb Ia = 8395 or, Eb =

8395 Ia

Eb = V – Ia (ra + rse) = 220 – Ia (0.5 + 0.25) = 220 – 0.75 Ia

Back emf

…(i)

…(ii)

8395 = 220 – 0.75 Ia Ia

\

0.75 Ia2 – 220 Ia + 8395 = 0

or,

220 ± (220)2 - 4 ¥ 0.75 ¥ 8395 2 220 ± 152.36 = = 33.82 A or 186.2 A 2

Ia =

or,

Ans. (b) 33.8 A A40 A 250 V, dc shunt motor runs at a speed of 100 rpm under no-load condition. Under no load condition if an external resistance of 3 W is connected in series with the shunt field, the speed is observed to be 1100 rpm, the terminal voltage remaining same. The value of series resistance is (a) 50.3 W (b) 43.2 W (c) 30 W (d) 45.7 W Solution V = 250 V N1 = 1000 rpm Under no load, Now

V = Eb Eb • fN

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22/36

Eb • Ish N 250 = Eb1 = KIsh1 ¥ 1000

\ When 3 W is inserted,

250 = Eb2 = KIsh2 ¥ 1100 Ish1 ¥ 1000 = Ish2 ¥ 1100

\ or,

Ish2 =

10 Ish, 11

As the terminal voltages are same, Ish1 Rsh = Ish2 (3 + Rsh) 10 Ish (3 + Rsh) 11 1 3 11 +1 = 10 sh 3 = 0.1

Ish1 Rsh =

or, or, or,

sh

Rsh = 30 W

or, Ans. (c) 30 W

A41 The applied armature voltage to a dc motor is 220 V. At 1500 rpm, it draws full load armature current of 200 A. If the armature resistance is 0.3 W and rotational loss is 1500 W, the output torque is (a) 3.67 Nm (b) 219.75 Nm (c) 1380 Nm (d) 270.7 Nm Solution V = 220 V N = 1500 rpm Ia = 200 A ra = 0.2 W Pr = 1500 W Eb = V – Ia ra = 220 – 200 ¥ 0.2 = 220 – 40 = 180 V Power developed Power output \

Pd = Eb Ia = 180 ¥ 200 = 36,000 W Po = 36,000 – 1500 = 34500 W

Output torque =

Ans. (b) 219.75 Nm

Po 34500 Nm = 219.75 Nm = 2 p ¥ 1500 w 60

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Fmfdusjdbm!Nbdijoft

A42 The armature resistance of a permanent magnet dc motor is 1.5 W. At no load, the speed is 1800 rpm and it draws a current of 1.5 A from 50 V supply. Assuming the rotational loss to be constant, the power output of the motor when it operates at 1500 rpm from a 60 V source is (a) 535.7 W

(b) 232.5 W

(c) 464.34 W

(d) 375.45 W

Solution ra = 1.5 W NNL = 1800 rpm IaNL = 1.5 A VNL = 50 V EbNL = 50 – 1.5 ¥ 1.5 = 47.75 V Under no-load condition, power developed by the armature is equal to the rotational loss \ Now

Pr = 47.75 ¥ 1.5 = 71.625 W N = 1500 rpm V = 60 V

For a permanent magnet dc motor, Eb • N EbNL

\

Eb

=

N NL 1800 = N 1500

47.75 18 = Eb 15

or, or,

Eb = 39.79 V Armature current

Ia =

V - Eb 60 - 39.79 = ra 1.5

= 13.47 A \ power developed Pd = Eb Ia = 39.79 ¥ 13.47 = 535.97 W \ power output

Po = Pd – Pr = 535.97 – 71.625 = 464.34 W

Ans. (c) 464.34 W A43 A 230 V, dc series motor draws 100 A at a speed of 1000 rpm. It has armature and series field resistances of 0.15 W and 0.1 W respectively. Due to saturation, the flux at an armature current of 50 A is 50% of that at armature current of 100 A. The motor speed when the armature voltage is

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

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230 V and the armature current is 40 A is (a) 1768 rpm Solution V = 230 V Ia1 = 100 A N1 = 1000 rpm ra = 0.15 W rse = 0.1 W \

(b) 1500 rpm

Eb1 Ia2 f2 Eb2 Eb1

Now

Eb2

or,

(c) 874 rpm

(d) 1236 rpm

= 230 – 100 (0.15 + 0.1) = 205 V = 50 A = 0.6 f1 = 230 – 50 (0.15 + 0.1) = 217.5 V =

N2 =

f1 f2 Eb2 Eb1

1 2

¥

f1 217.5 1000 N1 = = 1768.3 rpm ¥ f2 205 0.6

Ans. (a) 1768 rpm Common Statement for problems 44 to 46 A 400 V shunt motor draws a rated current of 30 A at a speed of 100 rad/s. The armature and field resistances are 1 W and 200 W respectively. A44 The total voltage in the armature circuit at the time of plugging is (a) 28 V (b) 372 V (c) 400 V (d) 772 V Solution V = 400 V I2 = 30 A Ish =

V 400 = A=2A rsh 200

ra = 1 W Eb = V – Ia ra = 400 – (IL – Ish) ¥ 1 = 400 – (30 – 2) = 372 V Total voltage in the armature circuit at the time of plugging is = V + Eb = 400 + 372 = 772 V Ans. (d) 772 V

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A45 The external resistance to be inserted in series with the armature circuit to limit the armature current of 150% of its rated value when the motor is plugged is (a) 7.85 (b) 16.15 (c) 12.3 W (d) 17.38 W Solution Ia = IL – Ish = 30 – 2 = 28 A Iamax = 1.5 ¥ 28 = 42 A If the external resistance be R then Iamax (ra + R) = 772 \

R=

772 – 1 = 17.38 W 42

Ans. (d) 17.38 W A46 The braking torque at the instant of plugging is (a) 180.75 Nm

(b) 156.24 Nm

(c) 240.67 Nm

(d) 80.97 Nm

Solution With the external resistance, armature current at the time of plugging is 42 A. Eb = 372 \ braking power = 372 ¥ 42 W Braking torque =

372 ¥ 42 372 ¥ 42 = 100 w

= 156.24 Nm Ans. (b) 156.24 Nm A47 A 240 V, dc series motor runs at 800 rpm and takes 50 A while driving a load. The load varies as the square of the speed. If the armature circuit resistance is 1.5 W and the speed is raised to 1000 rpm, the supply voltage required is (a) 439.42 V Solution V1 = 240 V Ia1 = 50 A Ia • N2 N1 = 800 rpm Ia1 = KN 12

(b) 165 V

(c) 315 V

(d) 417.7 V

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

or

K=

50 (800)

2

=

22/3:

5 64000

ra = 1.5 W Eb1 = V1 – Ia1 ra = 240 – 50 ¥ 1.5 = 165 V

\ Now

N2 = 1000 rpm

If V be the new supply voltage, Eb2 = V – Ia2 ra = V – 1.5 Ia2 Ia2 = KN 22 =

5 (1000)2 = 78.125 A 64000

Eb2 = V – 78.125 ¥ 1.5 = V – 117.18

\ Now in a series motor, or \ \ or, or Ans. (a) 439.42 V

Eb • f N Eb • Ia N Eb2 Eb1

=

I a2 N 2 I a1 N1

=

78.125 ¥ 1000 = 1.953 50 ¥ 800

Eb2 = 1.953 ¥ 165 = 322.24 V V – 117.18 = 322.24 V = 439.42 V

A48 A 240 V, dc series motor operates at full rated load at 800 rpm and draws a current of 220 A. The armature and field resistances are 0.15 W and 0.05 respectively. If the line current drops to 120 A, the new developed torque is (a) 280.9 Nm

(b) 153.2 Nm

(c) 118.3 Nm

(d) 227.6 Nm

Solution V = 240 V Ia1 = 220 A N1 = 800 rpm ra = 0.15 W rse = 0.05 W Eb1 = V – Ia1 (ra + rse) = 240 – 220 (0.15 + 0.05) = 196 V

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Power developed in the armature = Eb1 ¥ Ia1 = 43120 W 43120 43120 = = 514.97 Nm 2p ¥ 800 w1 60 = 120 A

Developed torque

T1 =

Now In a series motor,

Ia2

T • I a2 \

1 2

Ê Ia ˆ =Á 1˜ Ë I a2 ¯

2

2

Ê 120 ˆ T2 = Á ¥ 514.97 = 153.21 Nm Ë 220 ˜¯

or, Ans. (b) 153.2 Nm

A49 A 400 V, dc shunt motor drives a load whose torque varies as the cube of the speed. The armature resistance is 0.4 W and the motor draws 50 A at a certain speed. The additional resistance required to be connected in series with the armature to reduce the speed to 60% of the original speed is (a) 15.3 W (b) 28 W (c) 40.7 W (d) 17 W Solution T • N3 ra = 0.4 W IL1 = 50 A As Ish is very small in shunt motors, Ia / IL and hence Ia1 = 50 A N2 = 0.6 N1 In a shunt motor, T • Ia \

1 2

\ Now,

=

I a1 I a2

3

50 N13 Ê N1 ˆ 1 = = 3 =Á = ˜ I a2 N 2 Ë 0.6 N1 ¯ 0.216

Ia2 = 0.216 Ia1 = 0.216 ¥ 50 = 10.8 A Eb1 = V – Ia1 ra = 440 – 50 ¥ 0.4 = 420 V Eb2 = 440 – (0.4 + R) 10.8

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

Eb1 Eb2

\

1

=

2 1

E b2 =

¥ Eb1 =

2

or, or, or, Ans. (d) 17 W

22/42

0.6

1

¥ 420

1

440 – (0.4 + R) ¥ 10.8 = 0.6 ¥ 420 = 252 435.68 – 10.8 R = 252 R = 17 W

A50 A 15 kW, 240 V shunt generator is delivering full load rated voltage at 1000 rpm. The armature and field resistance is 0.3 W and 300 W respectively. If the machine now runs as a motor taking 12 kW at 240 V, the speed of the motor is (a) 1200 rpm

(b) 870 rpm

(c) 850 rpm

(d) 900 rpm

Solution Generator IL =

15 ¥ 103 A = 62.5 A 240

N1 = 1000 rpm ra = 0.3 W rsh = 300 W 240 A = 0.8 A 300 Ia = IL + Ish = 63.3 A

Ish = \

Eg = V + Ia ra = 240 + 63.3 ¥ 0.3 = 259 V Motor 12 ¥ 103 A = 50 A 240 Ia = IL – Ish = 50 – 0.8 = 49.2 A

IL =

Eb = V – Ia ra = 240 – 49.2 ¥ 0.3 = 225.24 \

Eb = Eg

2 1

or N2 = N1

Eb 225.24 = 1000 ¥ = 870 rpm Eg 259

Ans. (b) 870 rpm A51 A 230 V, dc shunt motor running at 800 rpm draws 20 A current at rated voltage. The armature and field resistances are 0.2 W and 150 W respectively. An additional resistance is connected in

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series with the armature circuit to reduce the speed to 700 rpm. If the load torque is independent of speed, the value of the external resistance is (a) 3 W

(b) 2.7 W

(c) 1.3 W

(d) 1.53 W

Solution V = 230 V N1 = 800 A IL1 = 20 A Ish =

230 A = 1.53 A 150

Ia1 = IL1 – Ish = 18.47 A As the load torque is independent of speed, the torque remains constant at both speeds. T • f1 Ia1 • f2 Ia2 As f is constant, Ia1 = Ia2 = 18.47 Eb1 = 230 – 18.47 ¥ 0.2 = 226.306 Eb2 = 230 – 18.47 (0.2 + R) = 226.306 – 18.47 R \

Eb2 = Eb1 ¥

2

= 226.306 ¥

1

\ or,

700 = 198 V 800

226.306 – 18.47 R = 198 R = 1.53 W

Ans. (d) 1.53 W A52 A dc shunt motor has an armature resistance of 0.8 W. Under no-load condition, it draws an armature current of 1.5 A from 240 V supply and runs at 1500 rpm. A load is applied to the motor shaft which causes armature current to rise to 50 A and the speed falls to 1400 rpm. The percentage reduction in the flux per pole due to armature reduction is (a) 10.2%

(b) 3.5%

Solution V = 240 V ra = 0.8 W Ia1 = 1.5 A N1 = 1500 rpm Eb1 = 240 – 1.5 ¥ 0.8 = 238.8 V

(c) 11.3%

(d) 4.5%

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Ia2 = 50 A N2 = 1400 rpm Eb2 = 240 – 50 ¥ 0.8 = 200 V Eb1

=

Eb2

f1 f2

1 2

or,

238.8 f1 1500 = ¥ f2 1400 200

or,

f1 = 1.1144 f2

\

Reduction in flux =

f1 - f2 1.1144 f2 - f2 = 0.102 = 10.2% = f1 1.1144 f2

Ans. (a) 10.2% A53 A 220 V shunt motor draws an armature current of 30 A and the armature resistance is 0.3 W. If the developed torque is constant, by how much should the main flux be reduced to raise the speed by 60%? (a) 40% (b) 60% (c) 39% (d) 52.5% Solution V = 220 V Ia1 = 30 A ra = 0.3 W Torque

As

T T1 T2 N2 T1 f1 Ia1

• f Ia = K f1 Ia1 = K f2 Ia2 = 1.6 N1 = T2 = f2 Ia2

Ia Ia f1 = 2 = 2 f2 I a1 30

or, Now

Eb1 = 220 – 30 ¥ 0.3 = 211 V Eb2 = 220 – Ia2 ¥ 0.3 = 220 – 0.3 Ia2 Eb2 Eb1

=

f2 f1

2 1

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22/45

Fmfdusjdbm!Nbdijoft

220 - 0.3 I a2 211

=

30 ¥ 1.6 I a2

220 Ia2 – 0.3 I a22 = 10128

or,

0.3 Ia2 – 220 Ia2 + 10128 = 0

or, or,

Ia2 = =

220 ± (220) 2 - 4 ¥ 0.3 ¥ 10128 2 ¥ 0.3 220 ± 190.38 2 ¥ 0.3

\

Ia2 = 49.36 A

\

30 f2 = = 0.61 49.36 f1

\

Reduction in flux =

f1 - f2 ¥ 100% f1

f1 - 0.61f2 ¥ 100% f1 = 39%

=

Ans. (c) 39% A54 A 240 V, dc shunt motor takes 8 A current on no load while running at a speed of 1200 rpm. The armature and field resistances are 0.5 W and 100 W respectively. The motor is now driving a load drawing 50 A from the supply. If the armature reaction weakens the field by 3%. What would be the speed of the motor? (a) 1437 rpm

(b) 1128 rpm

(c) 1500 rpm

Solution V = 240 V I L1 = 8 A N1 = 1200 rpm ra = 0.5 W rsh = 100 W Ish = \

240 A = 2.4 A 100

Ia1 = 8 – 2.4 = 5.6 A Eb1 = 240 – 5.6 ¥ 0.5 = 237.2 V

(d) 900 rpm

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

I L2 Ia2 Eb2 f2 Eb2

\ Now \

Eb1

\ Ans. (b) 1128 rpm

22/46

= 50 A = 50 – 2.4 = 47.6 A = 240 – 47.6 ¥ 0.5 = 216.2 V = 0.97 f1 =

216.2 f2 = 237.2 f1

2 1

=

0.9 ¥ 1200

2

N2 = 1128 rpm

A55 A 220 V, dc shunt motor is running at 800 rpm with an armature current of 60 A. The armature resistance of the motor is 0.3 W. If the torque is doubled while the flux remains constant, the speed of the motor will be (a) 815 rpm (b) 750 rpm (c) 729 rpm (d) 825 rpm Solution V = 220 V N1 = 800 rpm Ia1 = 60 A ra = 0.3 W Torque T • Ia in shunt motor Hence, if torque is doubled, the armature current will also be doubled. \

T2 = 2T1 and Ia2 = 2Ia1 = 2 ¥ 60 = 120 A Eb1 = V – Ia1 ra = 220 – 60 ¥ 0.3 = 202 V Eb2 = V – Ia2 ra = 220 – 120 ¥ 0.3 = 184 V

\ or,

Eb2 Eb1

=

2 1

N2 = N1 ¥

Eb2 Eb1

= 800 ¥

184 = 729 rpm 202

Ans. (c) 729 rpm

TFDUJPO!C;!USBOTGPSNFST B1 A single-phase 10 kVA, 50 Hz transformer with 1 kV primary winding draws 0.5 A and 55 W at rated voltage and frequency on no load. A second transformer has a core with all its linear times the corresponding dimensions of the first transformer. The core material and dimension lamination thickness are the same in both transformers. The primary windings of both the trans-

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Fmfdusjdbm!Nbdijoft

formers have the same number of turns. If a rated voltage of 2 kV at 50 Hz is applied to the primary of the second transformer then the no-load current and power respectively are (a) 0.7 A, 77.8 W (c) 1 A, 110 W

(b) 0.7 A, 155.6 W (d) 1 A, 220 W

[GATE 2012]

Solution Inductance L =

m AN l

For first transformer, no-load current is 0.5. \

V = X m1

0.5 =

m AN l

\

1 ¥ 103 m AN 2 l

2p f

103 103 = 2p ¥ 0.5 p

=

For second transformer, l¢ = A¢ =

Length Area

l ¥

A=2A

\ no-load current Io =

\

Io¢ =

For first transformer, No-load power

V = Xm

V

(E number of turns are same)

m A¢ N pf l¢

V m 2 AN 2p f 2l

2

=

2 ¥ 103 3

10 2p f 2 ¥ pf

=

2 2 2

V1 Io cos q = 55 1 ¥ 10 ¥ 0.5 cos q = 55 3

or

cos q =

or

55 0.5 ¥ 103

For second transformer, No-load power = 2 ¥ 103 ¥ = 2¥ Ans. (b)

55 3

0.5 ¥ 10

¥

55 = 155.56 W 0.5

1 2

=

1 2

= 0.7 A

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B2 A single-phase transformer has a turns ratio of (1 : 2) and is connected to a purely resistive load (RL) as shown in Fig. B2.1. The magnetizing current drawn is 1 A and the secondary current is 1 A. If core losses and leakage reactances are neglected, the primary current is [GATE 2010] (a) 1.41 A (b) 2 A 1:2 1A (c) 2.24 A (d) 3 A I1 Solution As the secondary load is resistive, V2 and I2 are in phase. As the core loss is neglected, so there is no core-loss component of no load current, Ic = 0; I1 = primary current (Fig. B2.2) \

I1 =

m

V2

RL

Supply N1 T

Gjh/!C3/2! Ejbhsbn!pg!Qspc/!C3 I1

I¢2

+ ¢

Ê N ˆ = 1 + Á I2 2 ˜ Ë N1 ¯

2

Im

2

=

N2

12 + (1 ¥ 2)2 = 5 = 2.236 A

f

V2, I2

Gjh/!C3/3! Qibtps!ejbhsbn!pg Qspc/!C3

Ans. (c) 2.24 A

B3 The single-phase, 50 Hz iron-core transformer in the circuit (Fig. B3.1) has both the vertical arms of 20 cm2 cross-sectional area and both the horizontal arms of 10 cm2 cross-sectional area. If the two windings shown were wound instead on opposite horizontal arms, the mutual inductance will (a) double (b) remain same (c) be halved (d) become one quarter [GATE 2009] Solution N f NBA Self-inductance L = = Gjh/!C4/2! Jspo! dpsf! usbotgpsnfs! pg! I I Qspc/!C4 When wound on vertical arms L= Mutual inductance \

M= 1

=

2

or

M2 =

NB ¥ 20 ¥ 10-4 , or L • A I L1 L2 = 20 ¥ 20 10 ¥ 10 1 M1 2

A1 A2 =

20 10

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Fmfdusjdbm!Nbdijoft

\ mutual inductance will be halved. Ans. (c) be halved B4 It is desired to measure parameters of a 230 V/115 V, 2 kVA, single-phase transformer. The following wattmeters are available in a laboratory: W1: 250 V, 10 A, low power factor W2: 250 V, 5 A, low power factor W3: 150 V, 10 A, high power factor W4: 150 V, 5 A, high power factor The wattmeters used in open-circuit test and short-circuit test of the transformer will respectively be (b) W2 and W4 (c) W1 and W4 (d) W2 and W3 (a) W1 and W2 [GATE 2009] Solution Open-circuit test is preformed on low voltage side with rated voltage, i.e. 230 V. 2000 A or 17.4 A. The rated current on low voltage side is 115 The current in the low voltage side during open-circuit test is generally 2 to 6% of 17.4 A and power factor is very low. \ for open-circuit test, W2 is required. Short-circuit test is performed on high voltage side with rated current. 2000 \ rated current on the high voltage side A or 8.7 A. 230 The voltage required to circulate the rated current is very small. \ for short-circuit test, W3 is required Ans. (d) W2 and W3 B5 Three single-phase transformers are connected to form a three-phase transformer bank. The transformers are connected in the following manner (Fig. B5.1). The transformer connection will be represented by [GATE 2009] (b) Yd1 (c) Yd6 (d) Yd11 (a) Yd0 A1

B1

C1

A2

a2

a1

B2

b2

b1

c2

c1

C2

Gjh/!C6/2! Uisff.qibtf!usbotgpsnfs!cbol!pg!Qspc/!C6

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

22/4:

Solution A2

a 2, A1 C1 N

b1

B1

,c

2

a2

A2

c1

N

a1 ,

A1

b2

a1 N

B2

C2

Gjh/!C6/3! Wfdups!ejbhsbn!gps!Qspc/!C6

The vector diagrams are shown above which indicates that the connection is Yd1. Ans. (b) Yd1 B6 A 500 kVA, three-phase transformer has iron losses of 300 W and full load copper losses of 600 W. The percentage load at which the transformer is expected to have maximum efficiency is (a) 50%

(b) 70.7%

(c) 141.4%

(d) 200%

[GATE 2004]

Solution Maximum efficiency occurs when in a transformer, Copper loss = Iron loss Let at x fraction, the full load copper loss becomes equal to iron loss x2 ¥ 600 = 300

\ or,

x=

3 1 = 0.707 = 6 2

\ percentage load is 70.7% Ans. (b) 70.7% B7 A single-phase transformer has a maximum efficiency of 90% at full load and unity power factor efficiency at half load at the same power factor is (a) 86.7%

(b) 88.26%

(c) 88.9%

(d) 87.8%

[GATE 2003]

Solution Let at maximum efficiency condition, Iron loss = copper loss = x VI cos q VI = VI cos q + 2 x VI + 2 x

\

hmax =

or,

VI = 0.9 VI + x

(E cos q = 1)

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Fmfdusjdbm!Nbdijoft

or, or,

0.1 VI = 1.8x VI = 18x 1 load at unity p.f. 2 1 VI 2 hmax = 2 1 Ê 1ˆ VI + x + Á ˜ x Ë 2¯ 2

Efficiency at

=

1 VI 2 1 5 VI + x 2 4

=

2

(E copper loss at

VI VI +

5 x 2

=

1 Ê 1ˆ load is Á ˜ ¥ full load copper loss) Ë 2¯ 2

18 x 36 = = 87.8% 5 41 18 x + x 2

Ans. (d) 87.8% B8 A 400 V/200 V/200 V, 50 Hz three-winding transformer is connected as shown in Fig. B8.1. The reading of the voltmeter V will be [GATE 2002] (a) zero (b) 400 V (c) 600 V (d) 800 V 400 V 50 Hz V

400 : 200 : 200

Gjh/!C9/2! Uisff.xjoejoh!usbotgpsnfs!pg!Qspc/!C9

Solution The voltages in the three windings are 400 V, 200 V and 200 V. The two 200 V winding are connected in additive polarity and 400 V winding in subtractive polarity. \ reading of voltmeter is 400 – (200 + 200) or 0 V Ans. (a) zero B9 A balanced star-connected and purely resistive load is connected at the secondary of a stardelta transformer as shown in Fig. B9.1. The line-to-line voltage rating of the transformer is 110 V/200 V. Neglecting the non-idealities of the transformer, the impedance Z of the equivalent

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star-connected load, referred to the primary side of the transformer is (a) (3 + j 0) W (b) (0.866 – j 0.5) W (c) (0.866 + j 0.5) W

(d) (1 + j 0) W

[GATE 2010]

4W

4W

4W

(110/220) V

Gjh/!C:/2!Usbotgpsnfs!dpoofdujpo!pg!Qspc/!C:

Solution The secondary load of 4 W when transferred to the primary side is 2

1 Ê 110 ˆ = 4 ¥ = 1 W or (1 + j 0) W 4¥Á Ë 220 ˜¯ 4

Ans. (d) (1 + j 0) W B10 The circuit diagram shows a two-winding loss-less transformer with no leakage flux, excited from a current source I (t) whose waveform is also shown. The transformer has a magnetising inductance of

400 mH. p

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Fmfdusjdbm!Nbdijoft

I(t)

A

1:1

10 A S I(t)

N2

O 30 W

5ms

10ms 15ms

20ms 25ms

30ms

t

10 A B

Gjh/!C21/2! Uxp.xjoejoh!usbotgpsnfs!boe!uif!dvssfou!tpvsdf!pg!Qspc/!C21

(i) The peak voltage across A and B with S opened is 400 4000 800 V (b) 800 V (c) V (d) V p p p (ii) If the waveform of I (t) is changed to I (t) = 10 sin 100 p tA, the peak voltage across A and B with S closed is (a) 400 V (b) 240 V (c) 320 V (d) 160 V [GATE 2009]

(a)

Solution (i) L =

400 mH p

10 di 400 800 ¥ 10-3 ¥ = V Induced voltage e = L = -3 p dt p 5 ¥ 10 800 Ans. (d) V p

(ii) I (t) = 10 sin 100 pt A Induced voltage e = L

di 400 = ¥ 10–3 ¥ 10 ¥ 100 p cos 100 pt = 400 cos 100 pt V dt p

Peak voltage 400 V Ans. (a) 400 V B11 Figure B11.1 shows coils 1 and 2 with dot markings as shown, having 4000 and 6000 turns respectively. Both coils have a rated current of 25 A. Coil 1 is excited with a single phase, 400 V, 50 Hz supply. [GATE 2009] A C (i) The coils are to be connected to obtain a single-phase 400/1000 V autotransformer to drive a load of 10 kVA. Which Coil 1 Coil 2 of the options given should be exercised to realize the required autotransformer? (a) Connect A and D; Common B B D (b) Connect B and D; Common C Gjh/!C22/2! Usbotgpsnfs! pg! (c) Connect A and C; Common B Qspc/!C22 (d) Connect A and C; Common D

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22/54

(ii) In the autotransformer obtained in Question (i), the currents in Coil 1 and Coil 2 respectively are (a) 25 A and 10 A (b) 10 A and 25 A (c) 10 A and 15 A (d) 15 A and 10 A Solution (i) The connection will be as follows and the machine will act as a step-up autotransformer (Fig. B11.2). C Here, A and D are connected and the terminal B is common. AB acts as the primary with applied voltage 400 V. The voltage across windings C and D is 600 V (E AB and CD have 4000 and 6000 turns respectively). Hence, voltage across CB is (400 + 600) V = 1000 V. The autotransformer will have a rating of A D 400/1000 V. Ans. (a) connect A and D, common B (ii) Current in winding AB is 25 A (given). Output of the autotransformer is 10 kVA. B

10 ¥ 103 A or 10 A \ current in coil 2 is 1000

Gjh/!C22/3! Tufq.vq! bvupusbotgpsnfs! pg!Qspc/!C22

Ans. (a) 25 A and 10 A

B12 The core of a two-winding transformer is subjected to a magnetic flux variation as indicated in Fig. B12.1. f

p

f(Wb)

r

+

+ epq

100

ers

200

–

0.12

– q

s O

1

2

2.5

t(s)

Gjh/!C23/2! Uxp.xjoejoh!usbotgpsnfs!boe!hsbqi!pg!gmvy!wbsjbujpo!pg!Qspc/!C23

The induced emf (ers) in the secondary winding is a function of time and will be of the form (a)

ers 24 V O

–48 V

2 1

2.5 t(s)

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Fmfdusjdbm!Nbdijoft

ers

(b)

48 V 1

2

2.5

t(s)

–24 V

(c)

ers 48 V 24 V 1

(d)

2

2.5

t(s)

ers 1

2

2.5

[GATE 2008]

t(s) –24 V –48 V

Solution e = –N Here, From 0 to 1 s,

df dt

N = N2 = 200 d f 0.12 = Wb/s dt 1 ers = – 200 ¥ 0.12 = –24 V

\ From 1 to 2 s,

df =0 dt ers = 0

\ From 2 to 2.5 s,

\

df 0.12 =– Wb/s = – 0.24 Wb/s dt 0.5 ers = – 200 ¥ (– 0.24) = 48 V

\ the waveform of the secondary induced voltage will follow Figure (b). Ans. (b) B13 A single-phase 50 kVA, 250 V/500 V two-winding transformer has an efficiency of 95% at full load, unity power factor. If it is reconfigured as a 500 V/750 V autotransformer, its efficiency at

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

its new rated load at unity power factor will be (a) 95.752% (b) 97.851% (c) 98.276%

22/56

(d) 99.241%

[GATE 2007]

Solution In a two-winding transformer, 0.95 =

50 ¥ 103 50 ¥ 103 + Losses

Ê 50 ¥ 103 ˆ Losses = Á - 50 ¥ 103 ˜ W = 2632 W Ë 0.95 ¯

\

For a step-up autotransformer kVA of autotransformer = -a kVA of two winding transformer

a=

where

1 2

\ kVA of autotransformer = 50 ¥

1 1-

\ efficiency of autotransformer =

2 3

=

500 2 = 750 3

= 150

150 = 0.9827 150 + 2.632

Ans. (c) 98.276% B14 Two transformers are to be operated in parallel such that they share load in proportion to their kVA ratings. The rating of the first transformer is 500 kVA and its p.u. leakage impedance is 0.05 p.u. If the rating of the second transformer is 250 kVA then its p.u. leakage impedance is (a) 0.2

(b) 0.1

(c) 0.05

(d) 0.025

Solution

1 ˆ (per unit impedance) • ÊÁ Ë kVA ˜¯ \

1 p.u. 2 p.u.

\ Ans. (b) 0.1

=

kVA 2 kVA1

Z2 p.u. = Z1 p.u.

kVA1 = 0.05 ¥ 500 = 0.1 250 kVA 2

[GATE 2006]

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B15 A 300 kVA transformer has 95% efficiency at full load 0.8 power factor lagging and 96% efficiency at half load, unity power factor. (i) The iron loss (Pi) and copper loss (Pc) in kW, under full load operation are (b) Pc = 6.59, Pi = 9.21 (a) Pc = 4.12, Pi = 8.51 (a) Pc = 12.72, Pi = 3.07 (c) Pc = 8.51, Pi = 4.12 (ii) What is the maximum efficiency (in %) at unity power factory load? (a) 95.1

(b) 96.2

(c) 96.4

(d) 98.1

[GATE 2006]

Solution (i) 0.95 =

300 ¥ 0.8 300 ¥ 0.8 + Pi + Pc

or,

Pi + Pc = 12.63

…(i)

1 ¥1 2 0.96 = 2 1 Ê 1ˆ 300 ¥ ¥ 1 + Pi + Á ˜ Pc Ë 2¯ 2 1 Pi + Pc = 6.25 4 300 ¥

or,

…(ii)

Subtracting Eq. (ii) from Eq. (i), 3 Pc = 6.36 4 Pc = 8.51 kW and Pi = 4.12 kW

\

Ans. (c) Pc = 8.51, Pi = 4.12 (ii) Maximum efficiency occurs when copper loss is equal to iron loss. Let at x fraction of full load, maximum efficiency occurs. x2 Pc = Pi x2 ¥ 8.51 = 4.12

\ or or,

x=

4.12 = 0.696 8.51

\ maximum efficiency hmax = Ans. (b) 96.2

300 ¥ 0.696 ¥ 1 = 0.962 or 96.2% 300 ¥ 0.696 ¥ 1 + 2 ¥ 4.12

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N2

Vni = 3300 V

N1

Gjh/!C27/2! Usbotgpsnfs!dpoofdujpo!pg!Qspc/!C27

B16 A 50 kVA, 3300/230 V single phase transformer is connected as an autotransformer as shown in Fig. B16.1. The nominal rating of the autotransformer will be (a) 50 kVA (b) 53.5 kVA (c) 717.4 kVA (d) 767.4 kVA [GATE 2004] Solution Output voltage Vout = 3300 + 230 = 3530 V Current flowing in N2 turns I2 =

50 ¥ 103 A 230

\ rating of autotransformer is 3530 ¥

50 ¥ 103 ¥ 10–3 kVA or 767.4 kVA 230

Ans. (d) 767.4 kVA B17 The resistance and reactance of a 100 kVA, 11000/400 V, DY distribution transformer are 0.02 and 0.07 p.u. respectively. The phase impedance of the transformer referred to the primary is (a) (0.02 + j 0.07) W (c) (15.125 + j 52.94) W

(b) (0.55 + j 1.925) W (d) (72.6 + j 254.1) W

Solution If R be the resistance referred to the primary 0.02 =

R Z base primary

=

R 2

(Base KV) MVA per phase

=

R (11) 2 100 ¥ 10-3 3

[GATE 2004]

!

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Fmfdusjdbm!Nbdijoft

R = 0.02 ¥

or,

(11)2 100 ¥ 10-3 3

= 72.6 W

Similarly, reactance referred to the primary is X = 0.07 ¥

(11)2 100 ¥ 10-3 3

= 254.1 W

Ans. (d) (72.6 + j 254.1) W B18 Figure B18.1 shows an ideal three-winding transformer. The three windings 1, 2, 3 of the transformer are wound on the same core as shown. The turns ratio N1 : N2 : N3 is 4 : 2 : 1. A resistor of 10 W is connected across the winding 2. A capacitor of 2.5 W reactance is connected across the winding 3. Winding 1 is connected across a 400 V ac supply. If the supply voltage phasor V1 = 400 ∞ , the supply current phasor I1 is given by (a) (–10 + j 10) A (c) (10 + j 10) A

(b) (–10 – j 10) A (d) (10 – j 10) A

[GATE 2003]

I2

I1 V1 N2

N1

1

2

R = 10 W

N3 3 I3 Xc = 2.5 W

Gjh/!C29/2! Uisff.xjoejoh!usbotgpsnfs!pg!Qspc/!C29

Solution 1 2

\

Similarly,

=

1 2

Ê E2 = E1 ¥ Á Ë

2ˆ

Ê 1ˆ ˜¯ = 400 ¥ ÁË 2 ˜¯ = 200 V 1

Ê E3 = E1 ¥ Á Ë

3ˆ

Ê 1ˆ ˜¯ = 400 ¥ ÁË 4 ˜¯ = 100 V 1

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

22/5:

E2 200 = 20 A = R 10 E 100 I3 = 3 = = j 40 A - jXc - j 2.5

I2 =

When referred to the primary, Ê I¢2 = I2 ¥ Á Ë

2ˆ

2 ˜¯ = 20 ¥ 4 = 10 A 1

ÊN ˆ 1 I¢3 = –I3 ¥ Á 3 ˜ = - j 40 ¥ = –j 10 A 4 Ë N1 ¯

\ primary current I1 = I 2¢ + I 3¢ = (10 – j 10) A Ans. (d) B19 The hysteresis loop of a magnetic material has an area of 5 cm2 with the scales given as 1 cm = 2 AT and 1 cm = 50 mW. At 50 Hz, the total hysteresis loss is (a) 15 W

(b) 20 W

(c) 25 W

(d) 50 W

[GATE 2001]

Solution Hysteresis loss = 5 ¥ 2 ¥ 50 ¥ 10–3 f = 5 ¥ 2 ¥ 50 ¥ 10–3 ¥ 50 = 25 W Ans. (c) 25 W B20 A three-phase delta/star transformer is supplied at 6000 V on the delta-connected side. The terminal voltage on the secondary side when supplying full load at 0.8 lagging power factor is 415 V. The equivalent resistance and reactance drops for the transformer are 1% and 5% respectively. The turns ratio of the transformer is (a) 14 (b) 24 (c) 42 (d) 20 [GATE 2000] Solution E -V = Er cos q2 + Ex sin q2 V

or, or,

E – 1 = 0.01 ¥ 0.8 + 0.05 ¥ 0.6 V 2

415

= 1 + (0.008 + 0.03) = 1.038

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\

E2 = 415 ¥ 1.038 = 430.77 (line-line) Per phase value of

\ turns ratio

E2 =

430.77 3

V

6000 6000 ¥ 3 = = 24 430.77 430.77 3

Ans. (b) 24 B21 A three-phase transformer has rating of 20 MVA, 220 kV (star) –33 kV (delta) with leakage reactance of 12%. The transformer reactance (in ohm) referred to each phase of the LV delta connected side is (a) 23.5 (b) 19.6 (c) 18.5 (d) 8.7 [GATE 2001] Solution Base impedance of low voltage side Zbase = =

(Base kV)2 Base MVA (33)2 Ê 2 3 ˆ = Á 33 ¥ ˜ (20 / 3) Ë 20 ¯

Xp.u. = 0.12 \ transformer reactance referred to low voltage side is 0.12 ¥

(33)2 ¥ 3 = 19.6 W 20

Ans. (b) 19.6 W B22 A 2500/250 V, 100 kVA, Y–D distribution transformer bank delivers 200 A at 0.8 p.f. lagging to an industrial load from its secondary side. The power consumed by the load and the primary line current drawn by each transformer are respectively (a) 40 kW, 200 A (c) 23.1 kW, 11.55 A

(b) 69.28 kW, 20 A (d) 86.6 kW, 34.64 A

Solution Power consumed by load = I1 ph I 2 ph

¥ 250 ¥ 200 ¥ 0.8 = 69.281 kW =

2 1

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

\

I1ph = I2ph

2

=

200 3

1

¥

22/62

250 = 20 A 2500 3

Ans. (b) 69.28 kW, 20 A B23 An impedance of 3 + j 5 is connected in series with the secondary winding of an ideal transformer as shown in Fig. B23.1 2W j5W When referred to the primary side, the value of the se3:1 + + ries impedance is (a) (18 + j 45) W V1 V2 (b) (0.22 + j 0.55) W (c) (2 + j 5) W – – (d) (6 + j 15) W Gjh/!C34/2! Usbotgpsnfs!pg!Qspc/!C34

Solution

Ê The series impedance when referred to the primary side = (2 + j 5) ¥ Á Ë

1ˆ

2

˜

2¯

= (2 + j 5) ¥ 32 = (18 + j 45) W Ans. (a) (18 + j 45) W B24 A magnetic circuit having cross-sectional area of 50 cm2 is to be operated at 50 Hz from a 240 V supply. To achieve a peak flux density of 1 T in the core, the number of turns required is (a) 1356 (b) 959 (c) 153 (d) 216 Solution V = 4.44 f Bm N A \

N= =

V 4.44 f Bm A 240

4.44 ¥ 50 ¥ 1 ¥ 50 ¥ 10-4 = 216

Ans. (d) 216 B25 A 4.4 kV/440 V, 100 kVA single-phase transformer has a series reactance of 0.15 p.u. The reactance in ohms referred to LV and HV sides are to be (a) 290.4 W, 29 kW (c) 0.077 W, 0.75 mW

(b) 0.2904 W, 29.04 W (d) 0.66 W, 6.6 W

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Solution Per unit reactance =

Reactance in W ¥ MVA (kV)2

When referred to LV side, Reactance = 0.15 ¥

(440 ¥ 10-3 ) 2 100 ¥ 10-3

W = 0.2904 W

When referred to HV side, Reactance = 0.15 ¥

(4.4) 2 100 ¥ 10-3

W = 29.04 W

Ans. (b) 0.2904 W, 29.04 W B26 A 40 kVA, single-phase step-down transformer has a full load secondary current of 200 A. The effective resistance referred to secondary is 0.008 W and the iron loss of the transformer is 190 W. (i) The efficiency of the transformer at full load, 0.8 p.f. is (a) 98.74% (c) 99.4%

(b) 96.50% (d) 98.43%

(ii) The efficiency of the transformer at (a) 99.08% (c) 96.5%

3 th of the full load, unity power factor will be 4 (b) 98.78% (d) 98.48%

Solution (i) Full-load copper loss

I 22 re2 = (200)2 ¥ 0.008 W = 320 W

Iron loss = 190 W Total loss at full load = 320 W + 190 W = 510 W Output power at full load = 40 ¥ 103 ¥ 0.8 W = 32000 W 32000 Efficiency at full load and 0.8 p.f. = % = 98.43% 32000 + 510 Ans. (d) 98.43% 3 (ii) At th of full load, the copper loss 4 2 Ê 3ˆ = Á ˜ ¥ 320 W = 180 W Ë 4¯ Total loss at

3 th of full load = 190 + 180 = 370 W 4

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22/64

3 ¥ 40 ¥ 103 ¥ 1 3 4 \ efficiency at th of full load and unity power factor = 3 4 ¥ 40 ¥ 103 ¥ 1 + 370 4

= 98.78% Ans. (b) 98.78% B27 A transformer under no-load condition draws a current of 4.5 A at a power factor of 0.25 lagging when connected to a 230 V, 50 Hz supply. The number of turns of the primary winding is 250. The magnetizing current (Im) and iron loss (Pi) are (a) Im = 1.125 A, Pi = 1 kW

(b) Im = 4.35 A, Pi = 1 kW

(c) Im = 4.5 A, Pi = 259 W

(d) Im = 4.22 A, Pi = 259 W

Solution cos qo = 0.25 Io = 4.5 A Im = Io sin qo = 4.5 ¥ 1 - (0.25)2 = 4.22 A Pi = V1 Io cos qo = 230 ¥ 4.5 ¥ 0.25 W = 258.75 W Ans. (d) In = 4.22 A, Pi = 259 W B28 An ac source of 10 ∞ V having an internal impedance of 1 kW is applied to the primary of a single-phase transformer. If the secondary is connected to a purely resistive load of 10 W, the maximum power dissipated in the load is (a) 2.5 W (b) 25 W (c) 1 W (d) 250 W Solution Ê 1 kW when transferred to the secondary side is 1 ¥ 103 ¥ Á Ë

2ˆ

2

˜ W

1¯

Ê For maximum power dissipation, the load resistance is also 1 ¥ 10 Á Ë 3

2ˆ

\ current through the load under maximum power transfer condition is 10 ¥

IL =

2 1

Ê 2 ¥ 1 ¥ 10 ¥ Á Ë 3

2ˆ

˜

1¯

2

A

˜

1¯

2

!

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Fmfdusjdbm!Nbdijoft

\ power dissipated in the load = I L2 RL

=

=

Ê ÁË10 ¥

2ˆ

˜

1¯

Ê 4 ¥ 106 ¥ Á Ë 100 ¥ 103 4 ¥ 10

6

2

=

Ê ¥ 1 ¥ 10 ¥ Á Ë 3

2ˆ

4

2ˆ

2

˜

1¯

˜

1¯

1 = 0.025 W 40

Ans. (b) 25 mN B29 A 24 kVA, 2400/240 V single-phase two winding is to be connected as an autotransformer in two different ways shown in Fig. B29.1. What are the nominal ratings of the autotransformer in Fig. B29.1 (a) and Fig. B29.1 (b) respectively? + N1

N1 +

Vin N2

Load

–

Load

Vin N2 –

)b*! ! ! ! ! ! ! ! ! ! ! ! ! ! )c* Gjh/!C3:/2! Ejbhsbn!pg!Qspc/!C3:

(a) 264 kVA, 264 kVA

(b) 26.4 kVA, 264 kVA

(c) 26.4 kVA, 26.4 kVA

(d) 26.4 kVA, 290.4 kVA

Solution In Fig. B29.1(a), Vin = 2400 V + 240 V = 2640 V I1 = \ Input kVA =

Vin I1 3

10

=

2640 ¥ 10 103

24 ¥ 103 A = 10 A 2400

= 26.4.

In Fig. B29.1(b), Vin = 2400 V

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22/66

Current through N1 turns I1 =

24 ¥ 103 A = 10 A 2400

I2 =

24 ¥ 103 A = 100 A 240

Current through N2 turns

\ total input current is I1 + I2, i.e., 110 A \ input kVA =

2400 ¥ 110 103

= 264

Ans. (b) 26.4 kVA, 264 kVA B30 A single-phase 480 kVA, 480 V/8 kV two-winding transformer has an efficiency of 97.8% at full-load unity power factor. If it is connected as an 8000 V/9000 V autotransformer then what is its rated kVA and efficiency at full load and unity p.f.? (a) 8 MVA, 88.9% (c) 9 MVA, 99.8%

(b) 8 MVA, 99.8% (d) 10.125 MVA, 86.9%

Solution For a two-winding transformer, I1 = I2 =

480 ¥ 103 = 1000 A 480 480 ¥ 103 8 ¥ 103

I2a N1

= 60 A I1a +

For a 8000/9000 V autotransformer V1a = 8000 V

+

V2a = 9000 V

V2a V1a

N2

The connection is shown in Fig. B30.1. – – Rating of autotransformer is 9000 ¥ 1000 VA or 9 MVA Gjh/!C41/2! Dpoofdujpo! pg! Efficiency of the two-winding transformer at full load and unity usbotgpsnfs! jo! Qspc/!C41 p.f. is 97.8% 3 480 ¥ 10 ¥ 1 \ 0.978 = 480 ¥ 103 + Loss \

Loss = 10800 W

!

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As full-load loss remains same in the autotransformer, the efficiency of the autotransformer is h=

9 ¥ 106 ¥ 1 9 ¥ 106 + 10800

= 99.88%

Ans. (c) 9 MVA, 99.8% B31 A single-phase transformer has 400 and 600 turns in the primary and secondary windings. If the primary is connected to 440 V, 50 Hz ac supply and the net cross-sectional area of the core is 80 cm2, the maximum flux density in the core is (a) 2.75 T

(b) 137.5 T

(d) 2.75 ¥ 10–4 T

(c) 220 T

Solution E1 = 4.44 f N1 fm = 4.44 f N1 Bm A \

Bm =

440 50 ¥ 400 ¥ 80 ¥ 10-4

Wb/m2

= 2.75 Wb/m2 = 2.75 T Ans. (a) 2.75 T B32 A single-phase transformer takes a no-load current of 1.5 A when high-voltage winding is kept open. If the core loss component of no-load current is 0.8 A, the magnetizing component of no load current is (a) 1 A (b) 0.7 A (c) 1.27 A (d) 2.3 A Solution Io = 1.5 A Ic = 0.8 A 0.8 cos qo = = 0.53 \ sin qo = 0.85 1.5 Im = Io sin qo = 1.27 A Ans. (c) 1.27 A B33 A 400/2400 V transformer has a series leakage reactance of 40 W referred to the high voltage side. A load connected to the low voltage side absorbs 30 kW at unity power factor. If the voltage on the low voltage side is 420 V then the voltage and power factor at the high voltage side is (a) 2520 V, 1

(b) 2569 V, 1

(c) 2520 V, 0.8

Solution IL =

30 ¥ 103 A = 75 A 400

IH =

NL 400 IL = ¥ 75 A = 12.5 A NH 2400

(d) 2569 V, 0.98

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

when

VL = 420 V, VH = 420 ¥

22/68

2400 V = 2520 V 400

Considering the impedance drop voltage on the high voltage side is 2520 + j 40 ¥ 12.5 = 2569 11.22∞ V Power factor is cos 11.22° or 0.98 Ans. (d) 2569 V, 0.98 B34 A 25 kVA, single-phase transformer has core loss of 300 W and full-load copper loss of 600 W respectively. If the maximum efficiency occurs at 90% of full load, the new iron loss and copper loss under full load assuming that full load is constant is (a) 450 W, 450 W (b) 300 W, 600 W (c) 402.76 W, 497.24 W (d) 300 W, 486 W Solution Pc = 300 W Pcu = 600 W Let the new losses be P¢c and P¢cu under maximum efficiency condition (0.9)2 ¥ P¢cu = Pc¢

\ Again, as full load is constant Pc¢ + P¢cu or, (0.9)2 P¢cu + P¢cu or, P¢cu \ P¢c Ans. (c) 402.76 W, 497.24 W

= 600 + 300 = 900 W = 900 W = 497.24 W = 900 – 497.24 = 402.76 W

B35 A 4400/440 V, single-phase transformer has primary and secondary resistances of 0.3 W and 0.02 W respectively. The iron loss in the transformer is equal to 300 W. The value of secondary current at which efficiency is maximum is (a) 114.21 A (b) 100 A (c) 300 A (d) 1000 A Solution Equivalent resistance referred to the secondary side 2

Ê 440 ˆ + 0.02 = 0.023 W 0.3 ¥ Á Ë 4400 ˜¯

At maximum efficiency, iron loss is equal to the copper loss. If I2 be the secondary current, I 22 ¥ 0.023 = 300

!

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\ Ans. (a) 114.21 A

I2 = 114.21 A

B36 Refer to the previous problem. What is the maximum efficiency at a lagging power factor of 0.85? (a) 99.3% (b) 98.6% (c) 90% (d) 93.7% Solution At 0.85 power factor, the output under maximum efficiency Po = V2 I2 cos q2 = 440 ¥ 114.21 ¥ 0.85 = 42714.5 W Total loss = 2 ¥ Iron loss = 2 ¥ 300 = 600 W 42714.5 \ maximum efficiency hmax = = 0.9861 or 98.61% 42714.5 + 600 Ans. (b) 98.6% B37 A single-phase, 300 kVA transformer has an efficiency of 97% at full load on 0.9 power factor. Same efficiency is achieved on half load 0.9 power factor lagging. The copper loss and iron loss in kW under full-load operation respectively are (a) 4.56 kW, 3.78 kW (b) 4.17 kW, 4.17 kW (c) 5.56 kW, 2.78 kW (d) 6.56 kW, 1.78 kW Solution 0.97 = \

300 ¥ 103 ¥ 0.9 300 ¥ 103 ¥ 0.9 + Pc + Pcu

Ê 1 ˆ Pc + Pcu = 300 ¥ 103 ¥ 0.9 Á - 1 = 8350.515 W Ë 0.97 ˜¯ 300 ¥ 103 ¥

0.97 = 300 ¥ 103 ¥

\

Pc +

1 ¥ 0.9 2

1 1 ¥ 0.9 + Pc + Pcu 2 4

1 1 Ê 1 ˆ Pcu = 300 ¥ 103 ¥ ¥ 0.9 Á -1 Ë 0.97 ˜¯ 4 2

= 4175.26 W 3 Pcu = 4175.255 4

or,

Pcu = 5567 W = 5.567 kW

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

\

22/6:

Pc = 2783.51 W = 2.78 kW

Ans. (c) 5.56 kW, 2.78 kW B38 A 10 kVA transformer has an efficiency of 97% at full load and also at 70% of full load and 3 unity power factor. The efficiency of the transformer at th of full load and 0.8 power factor is 4 (a) 91%

(b) 95%

(c) 98%

(d) 97%

Solution 0.97 = \

10 ¥ 103 ¥ 1 10 ¥ 103 ¥ 1 + Pc + Pcu

Pc + Pcu = 309.28 W

where Pc and Pcu are the core and copper losses respectively at full load. Again, \ \ or, and

0.97 = Pc + 0.49 Pcu 0.51 Pcu Pcu Pc

3 \ efficiency at th of full load is 4

10 ¥ 103 ¥ 0.7 10 ¥ 103 ¥ 0.7 + Pc + (0.7) 2 Pcu

= 216.5 W = 92.785 = 181.93 W = 309.28 – 181.93 = 127.35 W

3 ¥1 4 or, 97% 2 3 Ê 3ˆ 3 10 ¥ 10 ¥ ¥ 1 + 127.35 + Á ˜ ¥ 181.93 Ë 4¯ 4 10 ¥ 103 ¥

Ans. (d) 97% B39 A 500 kVA, 50 Hz transformer having 3.3 kV in the primary winding draws a current of 6 A at no load, at rated voltage and frequency. Another transformer having core with all its linear times of the first transformer has 6.6 kV at 50 Hz applied to its primary winding. dimensions The primary winding of both the transformers have same number of turns. The no load current drawn by the second transformer is (a) 8.48 A (b) 4.24 A (c) 12 A (d) 3 A Solution As the linear dimensions of the second transformer is the second transformer is twice that of the first. \ Ac2 = 2 Ac1

times that of the first, the core area of

!

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Now V1 = 3.3 kV and V2 = 6.6 kV Again, Bm =

E E =K where K is constant 4.44 f NAc Ac

\

V V1 = K 2 = Bm 2 Ac2 Ac1

Bm1 = K

\ flux densities in both the transformers are same. m AN l Ac L1 = K l

Now, magnetizing inductance L = \

Ac

and

L2 = K

\

L2 = K

\

L2 = 2L1 No-load current

Io •

l 2 Ac1

=K 2

2 l1

Ac1 Ac2

V L

\ no-load current in the second transformer Io2 •

V L 2V1

\

Io2 •

or,

Io2 =

Io1

\ Ans. (a) 8.48 A

Io2 =

¥ 6 A = 8.485 A

2 L1

B40 A single-phase, 250 kVA transformer has primary and secondary voltage of 11 kV and 440 V respectively. The resistance of primary and secondary windings 0.5 W and 0.005 W respectively. If the iron loss is 5 kW, the efficiency on full load at 0.8 power factor lagging is (a) 92.3%

(b) 95.7%

(c) 98.7%

Solution Primary current

I1 =

250 A = 22.73 A 11

(d) 96.7%

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

Secondary current

I2 =

22/72

250 ¥ 103 A = 568.18 A 440

Total copper losses = (22.73)2 ¥ 0.5 + (568.18)2 ¥ 0.005 = 1872.47 W. Efficiency on full load at 0.8 power factor lagging is 250 ¥ 103 ¥ 0.8 250 ¥ 103 ¥ 0.8 + 5000 + 1872.47

or

0.967 or 96.7%

Ans. (d) 96.7% B41 An 11 kV/440 V, 100 kVA single-phase transformer has equivalent impedance of 1.5 + j 2 W referred to the high voltage side. The per unit equivalent impedance referred to the low voltage side is (a) 0.124 + j 0.165 (b) 0.00124 + j 0.00165 (c) 0.0024 + j 0.0032 (d) 0.024 + j 0.032 Solution Base impedance Zbase =

(kVbase ) 2 MVA base

Base impedance referred to low voltage side ZL base =

(0.44) 2 100 ¥ 10-3

W = 1.936 W

The equivalent impedance referred to low voltage side is ÊN ˆ (1.5 + j 2) ¥ Á L ˜ or (1.5 + j 2) ËN ¯ H

Ê 440 ˆ ÁË ˜ 11000 ¯

(0.0024 + j 0.0032) W

or,

\ per unit equivalent impedance referred to low voltage side is 0.0024 + 0.0032 1.936

or, Ans. (b) 0.00124 + j 0.00165

(0.00124 + j 0.00165)

2

!

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B42 A 230/115 kV, delta delta, 100 MVA three-phase transformer has a per unit resistance of 0.02 and a per unit reactance of 0.055. What is the percentage regulation of a transformer if it supplies a load of 80 MVA at 0.85 p.f. lagging? (a) 3.7% (b) 5.7% (c) 4.6% (d) 2.1% Solution The line current on the high-voltage side when it supplies a load of 80 MVA 100 ¥ 106

IH =

3 ¥ 115 ¥ 103

W = 401.65 A

The base value of current on the high voltage side 100 ¥ 106

IH base =

3 ¥ 115 ¥ 103

A = 502.06 A

\ per unit current IH p.u. =

401.65 = 0.8 502.06

Per unit voltage on the primary side Vp = 1 + 0.8 - cos -1 0.85 (0.02 + j 0.055) = 1 + 0.8 -31.78∞ ¥ 0.058 70.017∞ = 1 + 0.0464 38.23∞ = 1.0364 + j 0.0287 = 1.037 1.586∞ Voltage regulation =

1.037 - 1 = 0.037 or 3.7% 1

Ans. (a) 3.7% B43 The core of a two-winding transformer is subjected to a magnetic flux variation as shown in Fig. B43.1. f

c

a

–

+ eab

400

1000

ecd

–

0.009

+ b

d O

0.06

0.1

Gjh/!C54/2! Usbotgpsnfs!boe!gmvy!wbsjbujpo!hsbqi!pg!Qspc/!C54

0.12

t

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

22/74

The variation of e1 will follow the curve (a)

eab 450

(b)

eab 60 0.1 0

0

0.06 0.1

–180

eab

(d)

150

eab 180

0.1 0

t

t

0.12

–150

(c)

0.12

0.06

0.12

t

0.06

0

0.06 0.1

0.12

t

–60

–450

Solution For time between 0 s and 0.06 s, 0.009 t = 0.15 t Wb 0.06 df = –400 ¥ 0.15 = –60 V \ eab = – Nab dt Between 0.06 s and 0.1 s, the flux f is constant.

f=

Induced emf being proportional to

df , eab is 0 between 0.06 s and 0.1 s. dt

Between 0.1s and 0.12s 0.009 t = –0.45 t 0.02 = –400 ¥ (0.45 t) = 180 V

f =– \ Ans. (d) B44 (a)

eab

In the previous problem, the induced emf ecd in the secondary winding will follow the curve ecd

(b)

ecd 90

150 0.1 0

0.06

0.12

t 0 –60

–450

0.06 0.1

0.12

t

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(c)

Fmfdusjdbm!Nbdijoft

ecd

(d)

ecd 60

450

0.1 0

0

0.12

0.06

t

0.06 0.1

t

0.12

–90

–150

Solution For time between 0 s and 0.06 s, df ˆ Ê ecd = –edc = – Á - N cd ˜ = 1000 ¥ 0.15 V = 150 V Ë dt ¯

Between 0.06 s and 0.1 s, the emf is zero. For time between 0.1 s and 0.012 s, ecd = –edc = –{–1000 ¥ (–0.45)} = –450 V Ans. (a) B45 A three-winding transformer with turns ratio of 9 : 3 : 5 has resistive loads of 50 W and 20 W connected to the second and third windings respectively. What is the power drawn by the primary at unity power factor?

240 V

N2

50 W

N3

20 W

N1

Gjh/!C56/2! Usbotgpsnfs!dpoofdujpo!pg!Qspc/!C56

(a) 112 W

(b) 2036 W

(c) 1018 W

(d) 823 W

Solution The resistance connected to the second winding when referred to the primary Ê Z¢2 = Z2 Á Ë

1ˆ

2

2

Similarly,

2

Ê 9ˆ ˜¯ = 50 ¥ ÁË 3 ˜¯ = 450 W 2

Ê 9ˆ Z¢3 = 20 ¥ Á ˜ = 64.8 W Ë 5¯

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\ total impedance in the primary Z1 = Z¢2 || Z¢3 = \

450 ¥ 64.8 W = 56.64 W 450 + 64.8

240 A = 4.24 A 56.64 Power drawn by the primary at unity power factor is 240 ¥ 4.24 ¥ 1 = 1017.6 W

I1 =

Ans. (c) 1018 W B46 A 50 kVA, 13.8 kV/208 V, delta star distribution transformer has per unit resistance of 1% and per unit reactance of 7%. The transformer’s impedance referred to the high voltage side is (a) (0.036 + j 0.25) W (c) (0.026 + j 0.18) W

(b) (38 + j 266.6) W (d) (114.2 + j 800) W

Solution The base impedance referred to the high-voltage side is ZH base =

(13.8) 2 W = 11426 W 0.05 3

Impedance in ohms referred to high-voltage side ZH = 11426 (0.01 + j 0.07) = (114.2 + j 800) W Ans. (d) (114.2 + j 800) W B47

Referring to the previous question, the voltage regulation at full load and 0.8 p.f. lagging is

(a) 5.1% (c) 4.85%

(b) 1.24% (d) 3.3%

Solution Considering the per unit quantities, the output voltage is 1

∞

the output current is 1 - cos -1 0.8 or 1 -36.87° the input voltage V1 = 1

∞ + (0.01 + j 0.07) ¥ 1 -36.87∞

= 1 + 0.071 21.87∞ ¥ 1 -36.87∞ = 1 + 0.071 45∞ = 1.05 + j 0.05 = 1.051 2.726∞ \ voltage regulation = Ans. (a) 5.1%

1.051 - 1 or 0.051 or 5.1% 1

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B48 A single-phase transformer having 150 turns in the primary and 750 turns in the secondary is connected to a 240 V, 50 Hz supply on its primary side. The secondary winding supplies a load of 4 A at a lagging power factor of 0.08. The maximum flux in the core and power supplied to the load is (a) 7.21 mWb, 3840 W (c) 3.840 mW, 7.21 kW

(b) 7.21 Wb, 3.8 kW (d) 3.840 W, 7 kW

Solution fm =

Maximum flux

E1 240 = Wb 4.44 fN1 4.44 ¥ 50 ¥ 150

= 7.21 ¥ 10–3 Wb = 7.21 mWb Power supplied to the load is V2 I2 cos q2 = 240 ¥ (a) 7.21 mWb, 3840 W (c) 1.44 mWb, 153.6 W

750 ¥ 4 ¥ 0.8 W = 3840 W 150

(b) 7.21 mWb, 153.6 W (d) 36 mWb, 3840 W

Ans. (a) 7.21 mWb, 3840 W B49 A three-phase YY autotransformer interconnects a 13.2 kV distribution line and a 13.8 kV distribution line. The turns ratio N1/N2 is (a) 1 (b) 22 (c) 44 (d) 88 Solution VH 13.8 = = VL 13.2

(As the transformer has YY connection \ or, or

1+

2

1

VH 13.8 / 3 = ) VL 13.2 / 3

13.8 N1 = 13.2 N1 + 13.2 N2 0.6 N1 = 13.2 N2 1 2

=

132 = 22 6

Ans. (b) 22 B50 A single-phase transformer operating at no load draws an exciting current Io = 6 A when the primary is connected to 240 V, 50 Hz supply. If the core loss is 200 W, the magnetizing reactance is (a) 288 W (b) 50 W (c) 100 W (d) 40.4 W

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Solution If the no-load p.f. is cos qo then 240 ¥ 6 ¥ cos qo = 200 cos qo = 0.14 sin qo = 0.99

\ \

Magnetizing current is Io sin qo or 6 ¥ 0.99 A or 5.94 A \ magnetizing reactance Xm =

240 W = 40.4 W 5.94

Ans. (d) 40.4 W

TFDUJPO!D;!JOEVDUJPO!NPUPST C1 If a 400 V, 50 Hz, star-connected, three-phase squirrel-cage induction motor is operated from a 400 V, 75 Hz supply, the torque that the motor can now provide while drawing rated current from the supply (a) (b) (c) (d)

reduces increases remains the same increases or reduces depending upon the rotor resistance

Solution 3I 22

Torque Te =

r2 s

ws

Êw ˆ ÊN ˆ Slip (s) = 1 – Á r ˜ = - Á r ˜ N Ë ws ¯ Ë s¯

\

Te =

3 I 22 r2 3 I 22 r2 = ws - wr Ê wr ˆ ÁË1 - w ˜¯ w s s

Ê fˆ Now if frequency increases, ws increases, as ws = 2p Á ˜ Ë P¯ Hence, the denominator ws – wr increases and Te reduces.

Ans. (a) reduces

[GATE 2002]

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C2 A three-phase 440 V, 6-pole, 50 Hz squirrel-cage induction motor is running at a slip of 5%. The speed of the stator magnetic field with respect to rotor magnetic field and speed of rotor with respect to stator magnetic field are (a) zero, –50 rpm (b) zero, 950 rpm (c) 1000 rpm, –50 rpm (d) 1000 rpm, 950 rpm [GATE 2011] Solution The speed of both the stator magnetic field and rotor magnetic field is at synchronous speed (Ns). Hence, their relative speed is 0. 120 f 120 ¥ 50 = Ns = = 1000 rpm P 6 s = 0.05 \ Nr = (1 – 0.05) ¥ 1000 = 950 rpm \ Relative speed between rotor and stator magnetic field is (Nr – Ns) i.e., –50 rpm Ans. (a) zero, –50 rpm C3 A 400 V, 50 Hz, 4-pole, 1400 rpm, star-connected squirrel-cage induction motor has the following parameters referred to the stator: R¢r = 1 W,

Xr = X r¢ = 1.5 W

Neglect stator resistance and core and rotational losses of the motor. The motor is controlled from a 3-phase voltage source inverter with constant (V/f ) control. The stator line-to-line voltage (rms) and frequency to obtain the maximum torque at starting will be (a) 20.6 V, 2.7 Hz

(b) 133.3 V, 16.7 Hz

(c) 266.6 V, 33.3 Hz

(d) 323.3 V, 40.3 Hz

Solution V V = f f

or,

V2 =

f2 Ê 400 ˆ = 8f2 ¥ V1 = f 2 Á Ë 50 ˜¯ f1

Ê 50 ˆ Ns = 120 ¥ Á ˜ = 1500 rpm Ë 4¯ Nr = 1400 rpm

\ given,

Ê 1400 ˆ s = Á1 = 0.067 Ë 1500 ˜¯

r¢r = 1

[GATE 2008]

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x¢r = 1.5 W 2p ¥ 50 ¥ L¢r = 1.5 W

or,

1.5 = (4.777 ¥ 10–3) H 100 p

\

L¢r =

\

Ls = L¢r = (4.777 ¥ 10–3) H Now if Tst = Tm, Tst =1= Tm

At max torque, r2 = sx2: sm =

2 1 sm + sm

rs + rr¢ xs + xr¢

sm =

rr xr¢ + xs

(Neglecting stator resistance)

1 =1 2p f 2 ( Lr¢ + Ls )

i.e., \

; or, sm = 1

1 2p

2 ( 4.777

¥ 10-3 + 4.777 ¥ 10-3 )

=1

f2 = 16.67 Hz. V2 = 8 f2 = 8 ¥ 16.67 = 133.36 V

or, \ Ans. (b) 133.3 V, 16.7 Hz.

C4 A 400 V, 50 Hz, 30 HP, three-phase induction motor is drawing 50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be (a) 23.06 kW Solution Input power \

Ans. (c) 25.01 kW

(b) 24.11 kW

(c) 25.01 kW

(d) 26.21 kW [GATE 2008]

¥ 400 ¥ 50 ¥ 0.8 W (Pin) = = 27712.81 W Pg = (Pin – stator copper loss – core loss) = 27712.81 – 1500 – 1200 = 25012.81 W = 25.01 KW

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Common Data for Problems C5 and C6. A three-phase, 440 V, 50 Hz, 4-pole slip-ring induction motor is fed from the rotor side through an autotransformer and the stator is connected to a variable resistance as shown in the Fig. C5.1.

Induction motor

Rex

3-phase, 50 Hz, Supply +

Autotransformer

220 V

–

Gjh/!D6/2! Ejbhsbn!gps!Qspcmfnt!D6-!D7

The motor is coupled to a 220 V, separately excited dc generator feeding power to fixed resistance of 10 W. The two-wattmeter method is used to measure the input power to induction motor. The variable resistance is adjusted such that the motor runs at 1410 rpm and the following readings were recorded W1 = 1800 W, W2 = – 200 W. C5 The speed of rotation of stator magnetic field with respect to rotor structure will be (a) (b) (c) (d)

90 rpm in the direction of rotation 90 rpm in the opposite direction of rotation 1500 rpm in the direction of rotation 1500 rpm in the opposite direction of rotation

[GATE 2008]

Solution Speed of stator magnetic field Ns =

120 ¥ 50 = 1500 rpm 4

Speed of rotor = 1410 rpm \ speed of rotation of stator magnetic field with respect to rotor structure is Ns – Nr or (1500 – 1410), or 90 rpm in the direction of rotation as the rotor rotates in the direction of stator magnetic field. Ans. (a) 90 rpm in the direction of rotation

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C6 Neglecting all losses of both the machines, the dc generator power output and the current through resistance (Rex) will respectively be (a) 96 W, 3.10 A (c) 1504 W, 12.26 A

(b) 120 W, 3.46 A (d) 1880 W, 13.71 A

[GATE 2008]

Solution From wattmeter readings, input to the induction motor Pin = 1800 – 200 = 1600 W Neglecting stator losses, the air-gap power Pg = 1600 W If Pm be the power developed by the motor then Pm = (1 – s) Pg 1410 = 0.06 1500 Pm = (1 – 0.06) ¥ 1600 = 1504 W.

Now

s =1–

\

Neglecting the friction and windage losses, output power Po = Pm = 1504 W. \ Input to the dc generator is 1504 W Neglecting generator losses, output of generator is 1504 W which is dissipated in a 10 W resistor. If I be the current in the resistor then I 2 ¥ 10 = 1504 or, I = 12.26 A Ans. (c) 1504 W, 12.26 A C7 A three-phase squirrel-cage induction motor has a starting torque of 150% and a maximum torque of 300% with respect to rated torque at rated voltage and rated frequency. Neglect the stator resistance and rotational losses. The value of slip for maximum torque is (a) 13.48%

(b) 16.24%

Solution Tst = 1.5 Tfl Tm =3 Tfl Tst 1.5 = = 0.5 Tm 3

\ Now

Tfl 2 = sm sfl Tm + sfl sm

(c) 18.92%

(d) 26.79%

[GATE 2007]

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Tst = Tm

and

2 sm +

1 sm

= 0.5

sm2 – 4sm + 1 = 0 sm = 0.2679, neglecting the negative value sm = 26.79%

or, or Ans. (d) 26.79%

Common Data for Problems C8, C9 and C10 A three-phase squirrel cage induction motor has a starting current of seven times the full-load current and full-load slip of 5%. [GATE 2007] C8 If an autotransformer is used for reduced voltage starting to provide 1.5 p.u. starting torque, the autotransformer ratio should be (a) 57.77% (b) 72.56% (c) 78.25% (d) 81.33% Solution Ist = 7 Ifl = Isc, when directly switched on to the supply sfl = 0.05 If x be the autotransformer tapping, 2

Ê xI ˆ Tst = Á sc ˜ sfl Tfl Ë I fl ¯

1.5 = x2 (7)2 ¥ 0.05 x = 0.7825 or 78.25%

or, or, Ans. (c) 78.25%

C9 If a star-delta starter is used to start this induction motor, the per unit starting torque will be (a) 0.607 (b) 0.816 (c) 1.225 (d) 1.616 Solution For the star-delta starter, Ist =

sc

3 2

2

\

Ê I ˆ ÊI ˆ Tst = Á st ˜ sfl = Á sc ˜ ¥ sfl Tfl Ë I fl ¯ Ë 3 I fl ¯

\

Tst Ê 7 ˆ = Á ˜ ¥ 0.05 = 0.816 Ë 3¯ Tfl

2

Ans. (b) 0.816

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C10 If a starting torque of 0.5 p.u. is required then the per unit starting current should be (a) 4.65 (b) 3.75 (c) 3.16 (d) 2.13 Solution Tst = 0.5 Tfl 2

As

ÊT ˆ Tst = Á st ˜ sfl Tfl Ë Tfl ¯ Tst = Tfl

\

0.5 = 10 = 3.16 0.05

Ans. (c) 3.16 C11 The speed of a 4-pole induction motor is controlled by varying the supply frequency while maintaining the ratio of supply voltage to supply frequency (V/f ) constant. At rated frequency of 50 Hz and rated voltage of 400 V, its speed is 1440 rpm. Find the speed at 30 Hz, if the load torque is constant. (a) 882 rpm (b) 864 rpm (c) 840 rpm (d) 828 rpm [GATE 2006] Solution f1 = 50 Hz. Ns1 =

120 ¥ 50 = 1500 rpm 4

Nr1 = 1440 rpm 1440 s1 = 1 – = 0.04 1500 f2 = 30 Hz 120 ¥ 30 Ns2 = = 900 rpm 4

\

If the load torque is constant, s is constant \

Nr2 = (1 – s) Ns2 = (1 – 0.04) ¥ 900 = 864 rpm

Ans. (b) 864 rpm C12 A three-phase, 4-pole, 400 V, 50 Hz star-connected induction motor has the following circuit parameters r1 = 1 W, r¢2 = 0.5 W, x1 = x¢2 = 1.2 W, xm = 35 W. The starting torque when the motor is started direct online is (use approximate equivalent circuit model) (a) 63.6 Nm (c) 190.8 Nm

(b) 74.3 Nm (d) 222.9 Nm

[GATE 2006]

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Solution T=

Pg ws

=

3I 2¢ 2

r2¢ s

ws 2

¸ Ô r2¢ Ô ˝ ¥ Ô sw s Ô ˛

Ï Ô V1 Ô = Ì 2 Ô Ê r2¢ ˆ 2 + r Á ˜ Ô Ë 1 s ¯ + ( x1 + x2¢ ) Ó

At starting s = 1,

\

Tst =

Ê 400 ˆ 3¥Á Ë 3 ˜¯

2

(1 + 0.5) + (1.2 + 1.2) 2

= 63.61 Nm

2

¥

0.5 2 ¥ 50 2p ¥ 4

Ans. (a) 63.6 Nm C13 A three-phase, 10 kW, 400 V, 4-pole, 50 Hz star-connected induction motor draws 20 A on full load. Its no load and blocked rotor test data are given below: No-load test

400 V

6A

1002 W

Blocked rotor test

90 V

15 A

762 W

Neglecting copper loss in no-load test and core loss in blocked rotor test, estimate the motor’s full load efficiency. (a) 76% (b) 81% (c) 82.4% (d) 85% [GATE 2006] Solution Full-load current of 20 A 2

Ê 20 ˆ \ from blocked rotor test data, the full-load copper loss is Á ˜ ¥ 762 = 1354.67 W Ë 15 ¯ Total loss = 1002 + 1354.67 = 2356.67 W Total input = 10,000 + 2356.67 = 12356.67 W

\

h=1–

Loss 2356.67 =1= 0.809 or 81% Input 12356.67

Ans. (b) 81% C14 A three-phase squirrel-cage induction motor is started by direct online switching at the rated voltage. If the starting current drawn is 6 times the full load current and the full load slip is 4% then

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ratio of the starting torque to the full-load torque is approximately equal to (a) 0.24 (b) 1.44 (c) 2.4 (d) 6

[GATE 2005]

Solution Ist = 6 Ifl sfl = 0.04 2

ÊI ˆ Tst = Á st ˜ sfl = (6)2 ¥ 0.04 = 1.44 Tfl Ë I fl ¯

Ans. (b) 1.44 C15 A 400 V, 15 kW, 4-pole, 50 Hz, Y-connected induction motor has full slip of 4%. The output torque of the machine at full load is (a) 1.66 Nm

(b) 95.5 Nm

(c) 99.47 Nm

(d) 624.73 Nm [GATE 2004]

Solution sfl = 0.04 Neglecting the frictional and windage losses Pm = 15 kW \

T=

=

Pm 15 ¥ 103 15 ¥ 103 = = N 2p wr (1 - sfl ) N s 2p r 60 60 15 ¥ 103 120 ¥ 50 2p (1 - 0.04) ¥ 60 4

= 99.5 Nm Ans. (c) 99.5 Nm C16 The rotor of a three-phase, 5 kW, 400 V, 50 Hz, slip-ring induction motor is wound for 6 poles while its stator is wound for 4 poles. The approximate average no-load steady-state speed when this motor is connected to a 400 V, 50 Hz supply is (a) 1500 rpm (b) 500 rpm (c) zero (d) 1000 rpm [GATE 2002] Solution In a three-phase induction motor, the rotor will rotate when the number of stator and rotor poles are equal. Here, as the stator and rotor poles are not equal, the rotor will not rotate and the speed will be 0. Ans. (c) zero

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C17 The power input to a 415 V, 50 Hz, 6-pole, 3-phase induction motor running at 975 rpm is 40 kW. The stator losses are 1 kW and friction and windage losses total 2 kW. The efficiency of the motor is (a) 92.5%

(b) 98%

(c) 90%

(d) 88%

[GATE 2000]

Solution Ns =

120 ¥ 50 = 1000 rpm 6

Nr = 975 rpm s =1–

975 = 0.025 1000

Pin = 40 kW Pg = 40 ¥ 103 – 1 ¥ 103 = 39 ¥ 103 W Pm = (1 – s) Pg = (1 – 0.025) ¥ 39 ¥ 103 = 38025 W

\

\ Power output = Pm – Friction and windage losses = 38025 – 2000 = 36025 W h=

36025 ¥ 100% = 90.06% 40, 000

Ans. (c) 90% C18 A three-phase, 4-pole, 50 Hz induction motor runs at a speed of 1440 rpm. The stator winding resistance is 0.3 W per phase. The terminal current is 10 A and the input power is 12 kW. The power dissipated in the rotor is (a) 90 W

(b) 1 kW

(c) 842 W

Solution Stator copper loss = 3 ¥ (10)2 ¥ 0.3 = 90 W Neglecting stator core loss, the air-gap power Pag = 12 ¥ 103 – 90 = 11910 W Synchronous speed Ns = Rotor speed = 1440 rpm \ slip s = 1 –

1440 = 0.04 1500

120 f 120 ¥ 50 = = 1500 rpm P 4

(d) 476 W

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Power dissipated in rotor sPag = 0.04 ¥ 11910 = 476.4 W Ans. (d) 476 W C19 A 440 V, 50 Hz, three-phase induction motor has line current of 40 A at 0.8 p.f. lagging. The stator and rotor copper losses are 3000 W and 1000 W respectively. The friction and windage losses are 800 W and core losses are 200 W. The air-gap power is (a) 21.2 kW

(b) 21.39 kW

(c) 24.2 kW

(d) 23.38 kW

Solution ¥ 440 ¥ 40 ¥ 0.8 = 24387 W

Pin =

Power input Air-gap power

Pag = Pin – Stator core loss – stator copper loss = 24387 – 200 – 3000 = 21187 W = 21.2 kW Ans. (a) 21.2 kW C20 A 220 V, 50 Hz, 4-pole delta-connected three-phase induction motor operates at a full load speed of 1400 rpm. The output power is 500 W at this speed. If the supply voltage decreases by 10%, the output torque of the motor is (a) 3 Nm

(b) 2.76 Nm

(c) 4.12 Nm

(d) 3.41 Nm

Solution Full-load output torque Tfl =

500 Nm = 3.41 Nm 1400 2p ¥ 60

Now T • V 2 If voltage decreases by 10%, V2 = 0.9 V1 \

Ê Torque T2 = T1 ¥ Á Ë

2ˆ

2

2 ˜¯ = 3.41 ¥ (0.9) = 2.762 Nm 1

Ans. (b) 2.76 Nm C21 A 3-phase six-pole wound rotor induction motor is rotating in the direction opposite to the direction of the rotating magnetic field at a speed of 800 rpm. The frequency of the rotor current at this condition is (a) 10 Hz (b) 5 Hz (c) 50 Hz (d) 90 Hz

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Solution 120 ¥ 50 = 1000 rpm 6 The rotor speed with respect to the rotating magnetic field is 1000 + 800 = 1800 rpm

Speed of the rotating magnetic field =

\

Relative speed between rotor and rotating magnetic field Synnchronous speed 1800 = = 1.8 1000

Slip =

\ frequency of rotor current is s f = 1.8 ¥ 50 Hz = 90 Hz Ans. (d) 90 Hz. C22 A three-phase squirrel-cage induction has starting current five times that of full-load current and the full-load slip is 3%. The ratio of starting torque to full-load torque is (a) 0.75 (b) 0.8 (c) 1 (d) 0.3 Solution Ist = 5 Ifl sfl = 0.03 sst = 1 3I 22

T=

R2 s

ws 2

\

ÊI ˆ s Tst 0.03 = Á st ˜ fl = (5)2 ¥ = 0.75 Tfl 1 Ë I fl ¯ sst

Ans. (a) 0.75 C23 Assuming negligible stator resistance in the previous problem, the slip at which maximum torque occurs is (a) 0.03 (b) 0.05 (c) 0.04 (d) 0.02 Solution Under the condition for maximum power transfer, smax = I=

R X

neglecting stator resistance V12

R2 + X2 s

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

\

Ifl =

Ist =

22/8:

V1 2

Ê R2 ˆ 2 ÁË 0.03 ˜¯ + X 2 V1 R22

+ X 22 2

2

Ê I st ˆ ÁË I ˜¯ =

Ê R2 ˆ 2 ÁË 0.03 ˜¯ + X 2

fl

2 2

\

0.0009

R22

R22 + X 22

= 25

+ X22 = 25 R22 + 25 X22

(1111.11 – 25) = X22 (25 – 1) R2 = 0.0221 X2 smax = 0.0221

or, \ Ans. (d) 0.02

C24 A three-phase, 6-pole, 50 Hz star-connected induction motor has rotor resistance and reactance of 0.04 W and 0.6 W per phase respectively. Neglecting the stator resistance, the speed at maximum torque is equal to (a) 900 rpm

(b) 1000 rpm

(c) 933 rpm

(d) 960 rpm

Solution r2 = 0.04 W x2 = 0.6 W Slip at maximum torque sm =

r2 0.04 = 0.067 = x2 0.6

\ rotor speed at maximum torque N = (1 – sm) Ns = (1 – 0.067) ¥

120 ¥ 50 60

= 933 rpm Ans. (c) 933 rpm C25 A 440 V, 50 Hz, three-phase induction motor is drawing 80 A at 0.85 p.f. lagging from the supply. The stator and rotor copper losses are 1800 W and 1100 W respectively. The core losses are 1000 W and friction and windage losses are 900 W. How much power is converted from electrical to mechanical form?

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(a) 47.9 kW

(b) 50 kW

(c) 47 kW

(d) 48.9 kW

Solution ¥ 440 ¥ 80 ¥ 0.85 = 51822.96 W

Power input =

Mechanical power developed = Power input – Copper losses – Core losses = 51822.96 – (1800 + 1100) – 1000 = 47922.96 W = 47.92 KW Ans. (a) 47.9 kW C26 A 100 kW, 4-pole induction motor has full load slip of 5%. At full-load, the friction and windage losses are 500 W and the core losses are 800 W. The load torque is (a) 670.5 Nm

(b) 637 Nm

(c) 650 Nm

(d) 720 Nm

Solution Pout = 100 kW Nr = (1 – s) Ns = (1 – 0.05) ¥ \

Load torque =

out

w

=

120 ¥ 50 = 1425 rpm 4

100 ¥ 103 Nm = 670.5 Nm 1425 2p ¥ 60

Ans. (a) 670.5 Nm C27 In the previous problem, the developed torque is equal to (a) 673.8 Nm

(b) 670.5 Nm

(c) 600 Nm

(d) 720 Nm

Solution Developed power in the rotor = 100 ¥ 103 + 500 = (100500) W \

Developed torque =

100500 = 673.8 Nm 1425 2p ¥ 60

Ans. (a) 673.8 Nm C28 A three-phase star-connected induction motor has rotor resistance and standstill reactance of 0.06 W and 0.2 W respectively per phase. When the rotor is at standstill, the induced emf between the slip ring terminals is 150 V. The value of the rotor current at standstill is (a) 624.5 A

(b) 718.37 A

(c) 414.75 A

(d) 400.8 A

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Solution r2 = 0.06 W x2 = 0.2 W Rotor current

I2 =

150 3 2

150 3

=

= 414.75 A

2

Ê 0.06 ˆ 1 ˜ + (0.2) ÁË 1 ¯

Ê r2 ˆ 2 ÁË s ˜¯ + x2

Ans. (b) 414.75 A C29 In the previous problem, the phase angle between rotor voltage and rotor current at 5% slip is (a) 36.86° (b) 9.6° (c) 78° (d) 60° Solution Phase angle cos q2 =

=

r s

=

Êr ˆ ÁË s ˜¯ + x

r2 r22

+ ( sx2 )2

0.06 (0.06)2 + (0.05 ¥ 0.2) 2

= 0.986

q2 = cos–1 0.986 = 9.6°

\ Ans. (b) 9.6°

C30 A 230 V, 60 Hz, 6-pole, three-phase induction motor driving a constant load torque at rated voltage and frequency has speed of 1175 rpm. If the system disturbance causes 10% drop in voltage and 6% drop in frequency assuming constant friction and windage losses, the new operating speed is (a) 1101 rpm (b) 1128 rpm (c) 1200 rpm (d) 1170 rpm Solution T•

Vs Ns

V2 = 0.9 V1 120 ¥ 60 = 1200 rpm 6 1175 = 0.021 s1 = 1 – 1200

N s1 =

!

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f2 = 60 – 60 ¥ 0.06 = 56.4 Hz \

N s2 =

120 ¥ 56.4 = 1128 rpm 6

As torque is constant, V12 s1 V s = N s1 Ns 2

Ê V ˆ Ns

2 s2 = Á 1 ˜ ¥ s1 Ë V2 ¯ Ns1

2

1128 Ê 1 ˆ =Á ˜ ¥ ¥ 0.021 = 0.024 Ë 0.9 ¯ 1200

New operating speed is (1 – 0.024) ¥ 1128 = 1101 rpm Ans. (a) 1101 rpm C31 A 12-pole, three-phase, 50 Hz, 440 V star-connected induction motor has rotor resistance and standstill reactance of 0.02 W and 0.4 W respectively. If the motor speed is 495 rpm, the ratio of full load torque to maximum torque is (a) 0.2

(b) 0.38

(c) 0.5

Solution 120 ¥ 50 = 500 rpm 12 r2 = 0.02 W x2 = 0.4 W

Ns =

\ slip at maximum torque sm =

r2 0.02 = 0.05 = x2 0.4

Nr = 495 rpm at full load \ full-load slip sfl = 1 – Now

Ans. (b) 0.38

Te fl Tem

=

495 = 0.01 500

2 2 = = 0.38 sm sfl 0.05 0.01 + + sfl sm 0.01 0.05

(d) 1

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C32 A 4-pole, 50 Hz, 3-phase slip-ring induction motor has rotor resistance of 0.1 W per phase. The speed under maximum torque condition is 1400 rpm. How much resistance must be inserted in the rotor phase to obtain the maximum torque at starting? Ignore magnetizing current and stator leakage impedance. (a) 1.49 W (b) 0.75 W (c) 1.39 W (d) 0.23 W Solution 120 ¥ 50 = 1500 rpm 4 = 1400 rpm

Ns = Speed at maximum torque \

Nm

sm = 1 – Now As

Now,

sm =

1400 = 0.067 1500

r x

r2 = 0.1 W r 0.1 x2 = 2 = = 1.49 W sm 0.067 Test = Tem

Let at slip s¢m, maximum torque occurs at starting and let r¢2 be the resistance inserted to obtain maximum torque at starting. Test 2 2 =1= = s s 1 ¢ Tem m + st sm¢ + sm¢ sst sm¢

\

or, or, Now or, or,

s¢m2 – 2s¢m + 1 s¢m r + r¢ x 0.1 + r2¢ x2 r¢2

=0 =1 =1 =1 = 1.49 – 0.1 = 1.39 W

Ans. (c) 1.39 W C33 An 8-pole, 50 Hz, three-phase induction motor running on full load develops useful torque of 100 Nm when the rotor emf makes 120 complete cycles per minute. The friction and windage losses are 1 kW. Determine the air-gap power and the full-load copper loss. (a) 8536 W, 356 W (b) 7536 W, 1000 W (c) 8892 W, 1000 W (d) 8892 W, 356 W

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Solution Ns =

120 ¥ 50 = 750 rpm 8

sf =

120 =2 60

s=

2 = 0.04 at full load 50

\ \ rotor speed

Nr = (1 – 0.04) ¥ 750 = 720 rpm Tsh = 100 Nm

\

Psh = 100 ¥

2p ¥ 720 = 7536 W 60

\ mechanical power developed Pm = 7536 + 1000 = 8536 W Now if Pag is the air-gap power Pm = (1 – s) Pag or

Pag =

8536 = 8892 W 1 - 0.04

The rotor copper loss = sPag = 0.04 ¥ 8892 = 356 W Ans. (d) 8892 W, 356 W Common Data for Problems C34, C35 and C36 A squirrel-cage induction motor has a starting current five times the full load current at slip 0.06. Determine starting torque in p.u. of full load values in the following methods of starting. C34 Direct switching (a) 1.8 (b) 0.84

(c) 1.5

(d) 0.5

Solution I2 sc = I2 st = 5I2 fl sfl = 0.06 As

Te =

I r sw s

per phase 2

2 Ê I 2 st ˆ ÊI ˆ s Tst 2 = Á 2 st ˜ fl = Á ˜ ¥ 0.06 = (5) ¥ 0.06 = 1.5 Tfl I Ë I 2fl ¯ sst Ë 2 fl ¯

Ans. (c) 1.5

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C35 Autotransformer starting with 75% tapping (a) 0.75 (b) 0.84 (c) 1.5

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(d) 0.5

Solution x = 0.75 sc

=5

fl 2

Ê xI ˆ Tst = Á sc ˜ rfl = (0.75 ¥ 5)2 ¥ 0.06 = 0.84 Tfl Ë I fl ¯

Ans. (b) 0.84 C36 Star-delta starting (a) 0.75 (b) 0.84

(c) 1.5

(d) 0.5

Solution In star-delta starting, Ist =

Starting current

sc

sc

3

=5

fl 2

2

Ê ˆ ÊI ˆ Tst = Á st ˜ ¥ sfl = Á sc ˜ ¥ 0.06 Tfl Ë I fl ¯ Ë 3 fl ¯ 2

25 ¥ 0.06 Ê 5 ˆ = Á ˜ ¥ 0.06 = = 0.5 Ë 3¯ 3

Ans. (d) 0.5 C37 A 4-pole, 50 Hz, slip-ring induction motor has rotor resistance of 0.3 W and rotor reactance of 1 W. It runs at 1440 rpm at full load. The external resistance per phase to be inserted in the rotor circuit to lower the speed to 1300 rpm, the torque remaining same is (a) 2.5 W (b) 1.3 W (c) 0.3 W (d) 3 W Solution T•

r Êr ˆ ÁË s ˜¯ + x

r2 = 0.3 W

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x2 = 1 W N1 = 1440 rpm Ns =

120 ¥ 50 = 1500 rpm 4

s1 = 1 –

1440 = 0.04 1500

N2 = 1300 rpm s2 = 1 –

1300 = 0.133 1500

T1 = T2

As r2 2

Ê r2 ˆ 2 ÁË s ˜¯ + x2 1

=

r + r¢ Ê r + r¢ ˆ ÁË s ˜¯ + x

where r¢ is the external resistance per phase inserted in the rotor circuit 0.3

\

2

Ê 0.3 ˆ 2 ˜ +1 ÁË 0.04 ¯

=

5.24 ¥ 10–3 =

or, or, \ Ans. (d) 3 W

0.3 + ¢ (0.3 + ¢ ) 2 + 12 (0.133) 2 0.3 + ¢ 918.27 (0.3 + ¢ ) 2 + 1

918.27 (0.3 + r¢)2 + 1 = 190.84 (0.3 + r¢) r¢ = 0.028 W or 3 W

C38 An induction motor delivering an output of 40 kW has an efficiency of 90%. At this load, the stator copper loss and rotor copper loss each equals the iron loss. The mechanical losses are one third of the no-load loss. The mechanical loss is (a) 1269 W (b) 634 W (c) 4440 W (d) 440 W Solution Po = 40 kW h = 90% 40 Input = kW = 44.44 kW 0.9

\ Total losses = 4.44 kW

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Stator copper loss = Rotor copper loss = Iron loss = P If x be the mechanical loss then x= or,

1 (P + x) 3

3x = P + x

or,

x=

\

Total loss = P + P + P +

or,

2

=

7 = 4440 2

P = 1268.57 W

\

Mechanical loss =

1268.57 W = 634.3 W 2

Ans. (b) 634 W C39 In the previous problem, determine the value of slip. (a) 0.05 (b) 0.04 (c) 0.06 Solution Air-gap power

(d) 0.03

Pg = Motor input – Stator copper loss – Stator iron loss = 44.44 – 1.269 – 1.269 = 41.9 kW

Rotor copper loss = sPg = 1.269 or

s=

1.269 = 0.03 41.9

Ans. (d) 0.03 C40 A three-phase, 4-pole, 50 Hz induction motor has rotor copper loss equal to 400 W. The torque developed by the motor at 3% slip is (a) 84.9 Nm (b) 90 Nm (c) 42.5 Nm (d) 50 Nm Solution Ns =

120 ¥ 50 = 1500 rpm 4

Copper loss = 400 = sPg where Pg is the air-gap power When s = 0.03 400 Pg = = 13333.33 W 0.03

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\ developed torque T=

Pg

=

ws

13333.33 Nm = 84.92 Nm 1500 2p ¥ 60

Ans. (a) 84.9 Nm C41 The full-load shaft torque of a three-phase, 6-pole, 50 Hz induction motor is 200 Nm. If the rotor emf makes 90 cycles per minute, the motor output is (a) 20.3 kW

(b) 200 kW

(c) 50.4 kW

(d) 20.9 kW

Solution Ns =

120 ¥ 50 = 1000 rpm 6

Frequency of rotor emf fr = sf 90 f r 60 1.5 s= = = = 0.03 50 50 f

\

Tsh = 200 Nm Motor output

Po = Tsh ¥ wr

Now,

Nr = (1 – s) Ns = (1 – 0.03) ¥ 1000 = 970 rpm Po = 200 ¥ 2p ¥

\

970 W = 20305 W = 20.305 kW 60

Ans. (a) 20.3 kW C42 A three-phase, 4-pole, 50 Hz, 240 V star-connected induction motor has rotor resistance and standstill reactance of 0.15 W and 1 W respectively. If the ratio of stator to rotor turns is 5, the airgap power at 5% slip is (a) 300 W (b) 109 W (c) 440 W (d) 623.8 W Solution Stator emf

E1 =

Standstill rotor emf

E2 =

Rotor current

I2 =

240 3 1

5

V

=

240 5 3

= 27.71 V

sE2 R22 + ( sX 2 )2

=

0.05 ¥ 27.71 (0.15) 2 + (0.05 ¥ 1) 2

= 3.48 A

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Rotor copper loss = 3I22 R2 = 3 ¥ (3.48)2 ¥ 0.15 W Air-gap power =

Rotor copper loss

=

3 ¥ (3.48)2 ¥ 0.15 = 109 W 0.05

Ans. (b) 109 W C43 A six-pole, three-phase, 50 Hz induction motor has a rotor resistance and rotor reactance of 0.025 W and 0.12 W respectively at standstill. The external resistance to be added in the rotor circuit to get 70% of maximum torque at starting is (a) 0.05 W (b) 0.12 W (c) 0.035 W (d) 0.085 W Solution Tst = 0.7 Tm Tst = Tm 2 sm

\

sm2 + 1

2 sm +

1 sm

= 0.7

= 0.7

sm2 – 2.857 sm + 1 = 0 sm = 0.4

or, or

If R is the external resistance, 0.025 + 0.12 R = 0.023 W

0.4 = or, Ans. (c) 0.023 W

C44 A 240 V, 6-pole, 50 Hz three-phase star-connected wound rotor induction motor has a rotor impedance of 0.03 + j 0.1 W per phase and negligible stator impedance at standstill. Consider the following statements: (i) The slip at which maximum torque occurs is 0.04. (ii) The starting torque is 55% of the maximum torque. (iii) The maximum torque developed by the motor is 2742.85 Nm. Which of the above statements is/are correct? (a) (i) and (ii)

(b) (i) only

(c) (ii) and (iii)

Solution sm =

r2 0.03 = 0.3 = x2 0.1

(d) (i), (ii) and (iii)

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Hence, the statement (i) is false. Tst = Tm

Now

or,

2 1 sm + sm

2

=

0.3 +

1 0.3

= 0.55

Tst = 55% of Tm Hence, the statement (ii) is correct. 2 ¥ 50 = 104.67 rad/s 6

ws = 2p ¥

Now

Neglecting the stator impedance, I2 =

V1 2

Ê r2 ˆ 2 ÁË s ˜¯ + x2

\ maximum torque developed 3I 22

Tm =

ws

r2 s =

r2 sm 2 ¸ ÏÔÊ r ˆ 2Ô 2 ÌÁ ˜ + x2 ˝ ¥ w s Ô˛ ÔÓË sm ¯ 3V12

2

0.03 Ê 240 ˆ ¥ 3¥Á Ë 3 ˜¯ 0.3 = 2 ÏÔÊ 0.03 ˆ ¸ 2Ô ÌÁË ˜¯ + (0.1) ˝ ¥ 104.67 ÔÓ 0.3 Ô˛

=

5760 Nm = 2742.85 Nm 2.1

Hence, the statement (iii) is correct. Ans. (c) (ii) and (iii) C45 A 15 HP, 6-pole, 50 Hz, three-phase induction motor is operating at 950 rpm on full load. The stator loss is 300 W and the rotational loss is 400 W. The efficiency of the motor is (a) 87.3%

(b) 91.67%

(c) 93.7%

Solution Ns =

120 ¥ 50 = 1000 rpm 6

(d) 89.5%

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\

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950 = 0.05 1000 = 15 ¥ 735.5 = 11032.5 W

s =1– Pout Power developed by rotor

Pm = Pout + Rotational losses = 11032.5 + 400 = 11432.5 W \ air-gap power Input power

\

Pg =

Pm 11432.5 = = 12034.2 W - s 1 - 0.05

Pi = Pg + Stator loss = 12034.2 + 300 = 12334.2 W h=

Po 11032.5 = = 0.8945 or 89.45% Pin 12334.2

Ans. (d) 89.5% C46 A 240 V, 4-pole, 50 Hz delta-connected three-phase induction motor has stator impedance of 0.3 + j 0.5 W per phase and an equivalent rotor impedance of 0.5 + j 0.8 W per phase at standstill. The motor speed at which the developed power is maximum is (a) 938 rpm

(b) 1440 rpm

(c) 875 rpm

(d) 1275 rpm

Solution Considering both the stator and rotor impedance, r2 0.5 = = 0.375 sm = 2 2 2 r1 + ( x1 + x2 ) (0.3) + (0.5 + 0.8) 2 120 ¥ 50 = 1500 rpm 4 \ motor speed at maximum torque

Ns =

Nr = (1 – sm) Ns = (1 – 0.375) ¥ 1500 = 938 rpm Ans. (a) 938 rpm C47 A three-phase, 6-pole, 50 Hz, 440 V, 12 kW, slip-ring induction motor has full load slip of 4%. The starting torque is equal to the full-load torque. If the applied voltage is reduced to 400 V, the new value of starting torque is (a) 120 Nm (b) 98.7 Nm (c) 100 Nm (d) 90.3 Nm Solution Ns =

120 ¥ 50 = 1000 rpm 6

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s = 0.04 Nr = (1 – s) Ns = 960 rpm Full-load output power Po = 12 kW \

Tfl =

Po 12 ¥ 103 = Nm = 119.43 Nm 960 wr 2p ¥ 60

Tst = Tfl = 119.43 Nm As

T • V12

\ when voltage is reduced to 400 V from 440 V, 2

2

Ê 400 ˆ Ê 400 ˆ = 119.43 ¥ Á = 98.7 Nm T¢st = Tst ¥ Á ˜ Ë 440 ¯ Ë 440 ˜¯

Ans. (b) 98.7 Nm C48 A 6-pole, three-phase, 50 Hz induction motor has maximum torque of 250 Nm which occurs at a speed of 865 rpm. Neglecting stator resistance, the torque at 3% slip is (a) 237.6 Nm

(b) 105.9 Nm

(c) 157.4 Nm

Solution Ns =

120 ¥ 50 = 1000 rpm 6

r2 = 0.5 Nr = 865 rpm \

sm = 1 –

865 = 0.135 1000

If torque at slip s = 0.03 is T then T 2 2 = = . s 0 03 0.135 s Tm + + m 0.135 0.03 sm s

or, Ans. (b) 105.9 Nm

T = 250 ¥

2 = 105.9 Nm 0.222 + 4.5

(d) 250 Nm

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TFDUJPO!E;!TZODISPOPVT!HFOFSBUPS D1 The direct-axis and quadrature-axis reactances of a salient-pole alternator are 1.2 p.u. and 1 p.u. respectively. The armature resistance is negligible. If this alternator is delivering rated kVA at unity power factor and at rated voltage then its power angle is (a) 30°

(b) 45°

(c) 60°

(d) 90°

[GATE 2011]

Solution Xd Xq Ia Vt cos q

= 1.2 p.u. = 1 p.u. = 1 p.u. = 1 p.u. =1 tan d =

\ Ans. (b) 45°

I a X q cos q + I a ra sin q Vt + I a X q sin q - I a ra cos q

=

¥ ¥

=1

d = 45°

D2 A 100 kVA, 415 V (line) star-connected synchronous machine generates rated open-circuit voltage of 415 V at a field current of 15 A. The short-circuit armature current at a field current of 10 A is equal to the rated armature current. The per unit saturated synchronous reactance is (a) 1.731

(b) 1.5

(c) 0.666

(d) 0.577

Solution Synchronous impedance =

Open-circuit phase voltage for a particular field current Sh hort-circuit phase current for same field current

415 3 = W = 1.148 W 100 ¥ 103 15 ¥ 3 ¥ 415 10 15 (Short-circuit current for 15 A field current = ¥ Rated current) 10 Neglecting armature resistance, synchronous reactance is 1.148 W \ synchronous reactance 1.148 ¥ 0.1 = p.u. = 0.666 p.u. (0.415) 2 Ans. (c) 0.666

[GATE 2007]

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Common Data for Problems D3 and D4 A 1000 kVA, 6.6 kV, three-phase star-connected cylindrical-pole synchronous generator has a synchronous reactance of 20 W. Neglect the armature resistance and consider operation at full load and unity power factor. [GATE 2005] D3 The induced emf (line to line) is close to (a) 5.5 kV (b) 7.2 kV (c) 9.6 kV (d) 12.5 kV Solution Ia =

1000 3 ¥ 6.6

A = 87.48 A

Induced emf per phase 2

Ê 6600 ˆ + (87.48 ¥ 20) 2 = 4192.98 V E = Vt 2 + ( I a xs )2 = Á Ë 3 ˜¯

Hence, induced emf (line to line) is 4192.98 ¥

g or, 7.2 kV.

Ans. (b) 7.2 kV D4 The power (or torque) angle is close to (a) 13.9° (b) 18.3° (c) 24.6°

(d) 33°

Solution Torque angle d = tan–1

I a xs 87.48 ¥ 20 = tan–1 = 24.6° 6600 Vt 3

Ans. (c) 24.6° D5 A 400 V, 50 kVA, 0.8 p.f. leading D connected, 50 Hz synchronous machine has a synchronous reactance of 2 W and negligible armature resistance. The friction and windage losses are 2 kW and the core loss is 0.8 kW. The shaft is supplying 9 kW load at a power factor of 0.8 leading. The line current drawn is [GATE 2004] (a) 12.29 A (b) 16.24 A (c) 21.29 A (d) 36.88 A Solution Power input

Pin = (9 + 2 + 0.8) kW = 11.8 kW

If I1 be the input current then VL IL cos q = 11.8 ¥ 103

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¥ 400 ¥ IL ¥ 0.8 = 11.8 ¥ 103 IL = 21.29 A

or, \ Ans. (c) 21.29 A

D6 Two three-phase, Y connected alternators are to be paralleled to a set of common busbars. The armature has a per phase synchronous reactance of 1.7 W and negligible armature reactance. The line voltage of the first machine is adjusted to 3300 V and that of the second machine is adjusted to 3200 V. The machine voltages are in phase at the instant they are paralleled. Under this condition, the synchronizing current per phase will be (a) 16.98 A (b) 29.41 A (c) 33.96 A (d) 58.82 A [GATE 2004] Solution 3300 3200 3 3 A Synchronizing current = 1.7 + 1.7 = 16.98 A Ans. (a) 16.98 D7 A 500 MW, 3-phase Y connected synchronous generator has a rated voltage of 21.5 kV at 0.85 p.f. The line current when operating at full load rated conditions will be (a) 13.43 KA (b) 15.79 KA (c) 23.25 KA (d) 27.36 KA [GATE 2004] Solution P = 500 ¥ 106 W VL = 21.5 ¥ 103 V cos q = 0.85 If IL be the line current then 3 VL IL cos q = P \

IL =

P 3 VL cos q

=

500 ¥ 106 3 ¥ 21.5 ¥ 0.85

A = 15796 A

= 15.79 KA Ans. (b) 15.79 KA D8 A field excitation of 20 A in a certain alternator results in an armature current of 400 A in short circuit and a terminal voltage of 2000 V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200 A is (a) 1 V (b) 10 V (c) 100 V (d) 1000 V. [GATE 2009]

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Solution 2000 W=5W 400 \ internal voltage drop at a load current of 200 A is Synchronous impedance =

5 ¥ 200 or, 1000 V Ans. (d) 1000 V D9 A 5 kW three-phase, permanent magnet synchronous generator produces an open-circuit voltage of 208 V line to line, 60 Hz, when driven at a speed of 1800 r/min. When operating at rated speed and supplying a resistive load, its terminal voltage is observed to be 192 V line to line for a power output of 4.5 kW. Assuming the generator armature resistance to be negligible, the generator synchronous reactance is (a) 5.91 W (b) 1.97 W (c) 3 W (d) 3.41 W Solution Armature current

Ia =

4.5 ¥ 103 3 ¥ 192

A = 13.53 A

E = Vt 2 + ( I a xs )2 \

xs = E=

\

xs =

E - Vt Ia 208 3

V and Vt =

192

(208) 2 - (192) 2 3 ¥ 13.53

3

V

W = 3.41 W

Ans. (d) 3.41 W D10 What is the maximum per unit reactive power that can be supplied by a synchronous machine operating at its rated terminal voltage whose synchronous reactance is 1.6 per unit and whose maximum field current is limited to 2.4 times that required to achieve rated terminal voltage under open-circuit conditions? (a) 1.4 p.u. (b) 1.5 p.u. (c) 0.875 p.u. (d) 1 p.u. Solution Vt = 1 p.u. Emax = 2.4 p.u. xs = 1.6 p.u.

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

\ maximum reactive power =

22/:8

Emax - Vt 2.4 - 1 = 0.875 p.u. = xs 1.6

Ans. (c) 0.875 p.u. D11 The manufacturer’s data sheet for a 26 kV, 750 MVA, 60 Hz, three-phase synchronous generator indicates that it has a synchronous reactance xs = 2.04 and leakage reactance xal = 0.18, both in per unit on the generator base. The synchronous inductance and armature leakage inductance are (a) 1.838 H, 0.43 H (b) 4.88 mH, 0.43 mH (c) 4.88 mH, 0.162 H (d) 0.901 H, 0.162 H Solution (Base kV)2 (26) 2 = 0.901 W = Base MVA 750 xs = 2.04 ¥ 0.901 W = 1.838 W

Zbase = \ or, synchronous inductance =

1.838 H = 4.88 mH 2p ¥ 60

xal = 0.18 ¥ 0.901 W = 0.162

Similarly,

0.162 H 2p ¥ 60 = 0.43 mH

\ armature leakage inductance = Ans. (b) 4.88 mH, 0.43 mH

D12 The open-circuit characteristic of a 45 kVA, three-phase, star-connected, 220 V, 6-pole, 60 Hz. synchronous machine gives the following data: Line to line voltage = 220; Field current = 2.84 A Short-circuit characteristic gives the following data: Armature current: 118 A 152 A Field current: 2.20 A 2.84 A From the air-gap line: Field current = 2.20 A, line to line voltage = 202 V The unsaturated synchronous reactance, saturated synchronous reactance and the short-circuit ratio are (a) 0.987 W, 0.987 W, 1 (b) 0.987 W, 0.836 W, 1.29 (c) 1.44 W, 1.87 W, 0.77 (d) 1.44 W, 1.87 W, 1.29 Solution At a field current of 2.20 A, the phase voltage on the air-gap line is

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Vag =

202 3

= 116.7 V

For the same field current, the armature current on short circuit is Isc = 118 A Hence, unsaturated synchronous reactance xs unsaturated =

116.7 = 0.987 W 118

Saturated synchronous reactance 220 xs saturated = 3 W = 0.836 W 152

Short-circuit ratio =

2.84 = 1.29 2.20

Ans. (b) 0.987 W, 0.836 W, 1.29 Common Data for Problems D13, D14, D15 and D16 The full-load torque angle of a synchronous machine at rated voltage and frequency is 35 electrical degrees. Neglect the effects of armature resistance and leakage reactance. The field current is held constant. Determine the full-load torque angle due to the following changes in operating condition. D13 (a) D14 (a) D15 (a) D16 (a)

Frequency reduced by 10%, load power and applied voltage constant: 39.6° (d) 38.5° (c) 31.07° (d) 35° Both frequency and applied voltage reduced by 10%, load power constant: 39.6° (b) 38.5° (c) 31.07° (d) 35° Frequency reduced by 10%, load torque and applied voltage constant: 39.6° (b) 38.5° (c) 31.07° (d) 35° Both frequency and applied voltage reduced by 10%, load torque constant: 39.6° (b) 38.5° (c) 31.07° (d) 35°

Solution D13 P • Vt sin d Po = K Vt sin 35°

…(i)

As power is independent of frequency, hence, torque angle will remain unchanged if frequency is reduced by 10%. Hence, torque angle is 35 electrical degrees. Ans. (d) 35°

Sfwjfx!Qspcmfnt!)NDRt*!xjui!Tpmvujpot

Solution D14 If applied voltage is reduced by 10% then P = K(0.9 Vt) sin d From Eqs. (i) and (ii),

22/::

…(ii)

K Vt sin 35° = K(0.9 Vt) sin d sin 35∞ 0.9 d = 39.59°

sin d =

or, or, Ans. (a) 39.6° Solution D15

\

To =

Po P • o f ws

To •

V sin 35∞ Vt sin d or To • t f f

…(iii)

If frequency is reduced by 10% then T•

Vt sin d 0.9 f

From Eq. (iii) and Eq. (iv), Vt sin d Vt sin 35∞ = 0.9 f f

sin d = 0.9 sin 35° or, d = 31.07° Ans. (c) 31.07° Solution D16 If both frequency and applied voltage is reduced by 10% then

or

T•

0.9Vt sin d 0.9 f

T•

Vt sin d f

Hence, d remains unchanged at 35 electrical degrees. Ans. (d) 35°

…(iv)

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D17 The open-circuit terminal voltage of a three-phase, 60 Hz synchronous generator is found to be 15.4 kV rms line to line. Determine the open-circuit terminal voltage if the field current is held constant while the generator speed is reduced so that the frequency of the generated voltage is 50 Hz. (a) 15.4 kV (b) 12.83 kV (c) 18.48 kV (d) 8.89 kV Solution As

p fNf

E=

\ E • f, other quantities remaining constant \ at 50 Hz,

E = 15.4 ¥

50 kV = 12.83 kV 60

Ans. (b) 12.83 kV Common Data for Problems D18, D19 and D20 A 460 V, 50 kW, 60 Hz, three-phase synchronous machine has a synchronous reactance of xs = 4.15 W and an armature to field mutual inductance, Laf = 83 mH. The machine is operating at rated terminal voltage. Calculate the magnitude and phase angle of the line-to-neutral generated voltage and field current if the machine acts as a generator and operating at D18

0.85 power factor lagging:

(a) 500 V, 22.6 A D19

(b) 280.5 V, 12.68 A

(c) 371.96 V, 16.81 A (d) 460 V, 73.8 A

Unity power factor:

(a) 500 V, 22.6 A

(b) 280.5 V, 12.68 A

(c) 371.96 V, 16.81 A (d) 460 V, 73.8 A

D20 0.85 power factor leading: (a) 500 V, 22.6 A

(b) 280.5 V, 12.68 A

(c) 371.96 V, 16.81 A (d) 460 V, 73.8 A

Solution D18 Ia = Generated voltage

50 ¥ 103 3 ¥ 460 ¥ 0.85

A = 73.83 A

E = Vt + j Ia xs =

460 3

+ 73.83 - cos -1 0.85 ¥ 4.15

= 265.58 + 306.39 58.21∞

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= 426.99 + j 260.43 = 500 31.38∞ V The field current can be calculated from the magnitude of generator voltage. If =

E w Laf

where Laf is the armature to field mutual inductance

\

If =

2 ¥ 500 A = 22.6 A 2p ¥ 60 ¥ 0.083

Ans. (a) 500 V, 22.6 A Solution D19 50 ¥ 103

Ia = E=

3 ¥ 460 ¥ 1 460 3

= 62.755 A

+ j 62.755

∞ ¥ j 4.15

= 265.58 + 260.433 90∞ = 371.96 44.44∞ V If =

2 ¥ 371.96 A = 16.81 A 2p ¥ 60 ¥ 0.083

Ans. (c) 371.96 A, 16.81 A Solution D20 Ia = E=

50 ¥ 103 3 ¥ 460 ¥ 0.85 460 3

A = 73.83 A

+ j 73.83 cos -1 0.85 ¥ 4.15

= 265.58 + 306.39 121.788∞ = 104.18 + j 260.43 = 280.49 68.2∞ V If =

2 ¥ 280.49 A = 12.68 A 2p ¥ 60 ¥ 0.083

Ans. (b) 280.5 V, 12.68 A Common Data for Problems D21, D22 and D23 The following readings are taken from the results of an open and a short-circuit test on an 800 MVA, three-phase, star connected, 26 kV, two-pole, 60 Hz turbine generator driven at synchronous speed.

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Field current, A Armature current, short-circuit test, kA Line voltage, open circuit characteristic, kV Line voltage, air-gap line kV

1540 9.26 26 29.6

2960 17.8 (31.8) (56.9)

The number in parentheses are extrapolations based upon the measured data. D21 The short-circuit ratio is (a) 1.92

(b) 0.52

(c) 1

(d) 1.5

Solution Short circuit ratio = =

Field current required to achieve rated open-circuit voltagge urrent Field current required to achieve rated short-circuit cu

1540 = 0.52 2960

Ans. (b) 0.52 D22 The unsaturated value of synchronous reactance in per unit is (a) 0.85

(b) 1.85

(c) 2.19

(d) 1

Solution Zbase =

(Base kV) 2 (26) 2 W = 0.845 W = Base MVA 800

The unsaturated synchronous reactance

xsu \

29.6 = 3 W = 1.85 W 9.26

xsu per unit =

1.85 = 2.19 p.u.t 0.845

Ans. (c) 2.19 D23 The saturated synchronous reactance in ohms per phase (a) 0.44 W

(b) 2 W

(c) 1.5 W

(d) 1.62 W

Solution Saturated value of synchronous reactance xs =

1 1 = = 1.92 p.u. Short-circuit ratio 0.52

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= 1.92 ¥ 0.845 W = 1.62 W Ans. (d) 1.62 W Common Data for Problems D24, D25 and D26 Consider a two-pole, 60 Hz, 13.8 kV, 10 MVA synchronous generator which achieves rated opencircuit armature voltage at a field current of 842 A. It achieves rated armature current under threephase terminal short-circuit for a field current of 226 A. D24 The per unit synchronous reactance is (a) 3.7

(b) 0.268

(c) 0.875

(d) 0.5

D25 Consider the situation in which this generator is connected to a 13.8 kV distribution feeder of negligible impedance and operating at an output power of 8.75 MW at 0.9 p.f. lagging. The field current in amperes, the reactive power output in MVAR, and the rotor angle for this operating condition is (a) 960 A, 4.233 MVAR, 11.89° (c) 842 A, 8.75 MVAR, 11.89°

(b) 842 A, 8.75 MVAR, 25.84° (d) 960 A, 4.23 MVAR, 25.84

D26 The resultant rotor angle and reactive power output in MVAR if the field current is reduced to 842 A while the shaft power supplied by the prime mover to the generator remains constant is (a) 25.84°, 4.3 MVAR (c) 11.89°, 1 MVAR

(b) 13.56°, 1 MVAR (d) 13.56°, 4.3 MVAR

Solution D24 Synchronous reactance xs =

226 = 0.268 p.u. 842

Ans. (b) 0.268 Solution D25 8.75 p.u. = 0.875 p.u. 10 As power factor is 0.9 lagging, the apparent power

Output power

Po = 8.75 MW =

(s) =

0.875 p.u. or, 0.972 - cos -1 0.9 p.u. 0.9

\ Armature current Ia = 0.972 - cos -1 0.9 p.u.

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Generated voltage E = Vt + j Ia xs = 1 + j 0.972 -25.84∞ ¥ 0.268 = 1.14 11.89∞ p.u. The field current required to achieve rated open-circuit armature voltage of 1 p.u. is 842 A. Hence, field current required to achieve 1.14 p.u. voltage is 842 ¥ 1.14 A or 960 A. Reactive power output = S - P = (0.972)2 - (0.875)2 = 0.4233 p.u. = 4.233 MVAR

or, Rotor angle is 11.89°

Ans. (a) 960 A, 4.233 MVAR, 11.89° Solution D26 If field current is 842 A then E = 1 d p.u. P = 0.875 p.u. \

EV sin d = 0.875 xs

or,

sin d =

0.875 ¥ 0.268 1¥1

or,

d = 13.56°

\

E = 1 13.56∞

\

Ia =

E - Vt 1 13.56∞ - 1 = jxs ¥ 0.268

= 3.73 -76.44∞ + 3.73 90∞ = 0.875 – j 3.63 + j 3.73 = 0.875 + j 0.1 = 0.88 6.52∞ A \ reactive power

or, Ans. (b) 13.56°, 1 MVAR

Q = VI sin q = 1 ¥ 0.88 sin 6.52 = 0.1 p.u. leading 1 MVAR leading

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Common Data for Problems 27 and 28 A synchronous generator having synchronous reactance of 1 p.u. is connected to infinite bus bars of 1 p.u. voltage through two parallel lines each of 0.5 p.u. reactance. The generator delivers rated current (1 p.u.) at unity power factor at its terminals. D27 The generator excitation and terminal voltage are (a) 1.25 p.u., 0.968 p.u. (c) 1.25 p.u., 1 p.u.

(b) 1.39 p.u., 0.968 p.u. (d) 1 p.u., 0.75 p.u.

Solution 2

Ê 0.5 ˆ ¥ 1˜ = 0.968 p.u. Vt = 1 - Á ¯ Ë 2

E = Vt 2 + 12 = (0.968)2 + 12 = 1.39 p.u. Ans. (b) 1.39 p.u., 0.968 p.u. D28 The active power delivered to the infinite bus bar is (a) 0.968 p.u.

(b) 0.937 p.u.

(c) 1 p.u.

(d) 0.25 p.u.

Solution Load angle d = sin–1 0.25 = 14.48° Active power delivered to bus bar is Vt Ia cos d = 1 ¥ 1 ¥ cos 14.48° = 0.968 p.u. Ans. (a) 0.968 p.u. D29 A three-phase synchronous generator has a direct-axis synchronous reactance of 0.8 p.u. and a quadrature-axis synchronous reactance of 0.5 p.u. The generator is supplying full load at 0.8 lagging power factor at 1 p.u. terminal voltage. The power angle and the no-load voltage if the excitation remains unchanged are (a) 36.8°, 1 p.u. (c) 17.17°, 1.6 p.u.

(b) 17.17°, 1 p.u. (d) 36.8°, 1.6 p.u.

Solution Vt = 1 p.u. Ia = 1 -36.8∞ p.u. Xd = 0.8 p.u. and Xq = 0.5 p.u.

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Power angle

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d = tan–1 = tan–1

Vt sin q + I a X q Vt cos q + I a ra

–q

1 ¥ 0.6 + 1 ¥ 0.5 – 36.8° = 17.17° 1 ¥ 0.8

Ê 1 ¥ 0.6 + 1 ¥ 0.5 ˆ E = Vt cos d + Id Xd = Vt cos d + Ia sin Á tan -1 ˜¯ Xd 1 ¥ 0.8 Ë

= 1 cos 17.17° + 1 ¥ sin 54° ¥ 0.8 = 1.6 p.u. Ans. (c) 17.17°, 1.6 p.u. Common Data for Problems 30 and 31 A three-phase 1200 kVA, 50 Hz, star-connected, 2.5 kV synchronous generator has resistance of 0.2 W between each pair of terminals measured by direct current. The effective resistance is 1.2 times the ohmic resistance. A field current of 50 A, produces a short-circuit current of 300 A in each line. The same field current produces an emf of 750 V on open circuit. D30 The synchronous reactance of the machine is (a) 1.42 W

(b) 2.5 W

(c) 1.5 W

(d) 1.435 W

Solution 450 Synchronous impedance Zs = 3 W = 1.44 W 300

ra = 1.2 ¥

0.2 W = 0.12 W 2

\ synchronous reactance Xs = (1.44)2 - (0.12)2 W = 1.435 W Ans. (d) 1.435 W D31 The full-load voltage regulation at 0.8 p.f. lagging (a) 20.15%

(b) 15%

(c) 25%

(d) 6%

Solution Ia = E=

1200 3 ¥ 2.5 2500 3

A = 277 A

+ 277 -36.87∞ ¥ (0.12 + j 1.435)

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= 1443.37 + 398.88 48.35∞ = 1708.46 + j 298.05 = 1734.26 V 1734.26 -

\ voltage regulation =

2500 3

2500 3

¥ 100% = 20.15%

Ans. (a) 20.15% D32 A three-phase 12-pole alternator has star-connected winding with 144 slots and 10 conductors per slot. The flux per pole is 0.05 Wb and speed of the machine is 500 rpm. The line induced emf is (a) 4420 V (b) 2552 V (c) 4615 V (d) 2665 V Solution Ns = \ Slots per pole per phase Slot angle

120 f P

f=

12 ¥ 500 = 50 Hz 120

q=

144 =4 12 ¥ 3

(g) =

180∞ 180∞ = 15° = Slots/pole 144 /12 g 4 ¥ 15∞ sin 2 = 2 = 0.9577 g 15∞ sin 4 sin 2 2

sin

Kd =

Distribution factor Number of turns per phase

N= Emf = Line value of induced emf = Ans. (a) 4420 V

144 ¥ 10 = 240 2¥3

pf f N Kd = ¥ p ¥ 0.05 ¥ 50 ¥ 240 ¥ 0.9577 = 2551.67 V

¥ 2551.67 V = 4419.6 V / 4420 V

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D33 1200 kVA, 11 kV, three-phase star-connected synchronous generator delivers full-load current at 0.8 power factor lagging. The armature resistance is 0.6 W and synchronous reactance is 8 W per phase. The terminal voltage for same excitation and load current at 0.8 power factor leading is (a) 6694 V

(b) 11594.6 V

(c) 11000 V

(d) 12042 V

Solution Ia =

1200 3 ¥ 11

= 62.98 A 2

Ê 11000 ˆ Ê 11000 ˆ ¥ 0.8 + 62.98 ¥ 0.6˜ + Á ¥ 0.6 + 62.98 ¥ 8˜ E= Á Ë 3 ¯ Ë 3 ¯

2

= 6694.19 V As excitation is constant, E remains constant. If the terminal voltage at 0.8 power factor leading is Vt then 6694.19 = (Vt ¥ 0.8 + 62.98 ¥ 0.6)2 + (Vt ¥ 0.6 - 62.98 ¥ 8)2 or,

Vt = 6952.6 V Line voltage =

¥ 6952.6 = 12042 V

Ans. (d) 12042 V

TFDUJPO!F;!TZODISPOPVT!NPUPS Statement common to Question no. 1 and 2: A synchronous motor is connected to an infinite bus at 1 p.u. voltage and draws 0.6 p.u. current at unity power factor. Its synchronous reactance is 1 p.u. and resistance is negligible. [GATE 2008] E1 The excitation voltage E and load angle d will respectively be (a) 0.8 p.u. and 36.86° lag (c) 1.17 p.u. and 30.96° lead

(b) 0.8 p.u. and 36.86° lead (d) 1.17 p.u. and 30.96° lag

Solution E = Vt – j Ia xs Vt = 1 0∞, Ia = 0.6 0∞, xs = 1 \

E = 1 – j 0.6 ¥ 1 = 1 – j 0.6 = 1.17 -30.96∞

Ans. (d) 1.17 p.u. and 30.96° lag

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E2 Keeping the excitation voltage same, the load on the motor is increased such that the motor current increases by 20%. The operating power factor will become (a) 0.995 lagging (b) 0.995 leading (c) 0.791 lagging (d) 0.848 leading Solution

\ or, \

Ia E2 + Vt2 – 2EVt cos d (1.17)2 + 12 – 2 ¥ 1.17 ¥ 1 cos d cos d sin d

= 12 ¥ 0.6 = 0.72 p.u. = (Ia xs)2 = (0.72 ¥ 1)2 = 0.79 = 0.6123

Vt E sin d = Vt Ia cos q xs 1.17 ¥ 0.6123 cos q = = 0.995 1 ¥ 0.72 E cos d = 1.17 ¥ 0.79 = 0.924 Vt > E cos d

\ \

Hence power factor is lagging i.e., power factor is 0.995 lagging Ans. (a) 0.995 lagging E3 A three-phase, 400 V, 5 kW, star connected synchronous motor having an internal reactance of 10 W is operating at 50% load, unity power factor. Now the excitation is increased by 1%. What will be the new load in per cent if the power factor is to be kept same? Neglect all losses and consider linear magnetic circuit. [GATE 2006] (a) 67.9%

(b) 56.9%

(c) 51%

(d) 50%

Solution Ia =

5000 3 ¥ 400 ¥ 1

= 7.217 0∞ A

E = (Vt cos q - I a ra ) 2 + (Vt sin q - I a xs ) 2 2

Ê 400 ˆ Ê 400 ˆ 7.217 = Á ¥ 1 - 0˜ + Á ¥0¥ 10˜ Ë 3 ¯ Ë 3 ¯ 2 = 53333.33 + 1302.13 = 233.74 V \

E = 1.01 ¥ 233.74 = 236 V E2 = (Vt cos q – Ia ra)2 + (Vt sin q – Ia xs)2

2

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E2 = Vt2 + Ia2 xs2

or,

or,

Ia = = New load is

E 2 - Vt 2 xs2

=

Ê 400 ˆ ( 236) 2 - Á Ë 3 ˜¯

2

(10) 2

48.607 A = 4.86 A 10

4.86 % or 67.4% 7.21

Ans. (a) 67.9% E4 A three-phase star connected synchronous motor has power input of 5000 W at rated voltage. The synchronous reactance is 12 W and armature resistance is negligible. If the excitation voltage is adjusted equal to the rated voltage of 400 V, the power angle is (a) 7.18°

(b) 30°

(c) 22°

(d) 10°

Solution E =V=

400 3

3VE sin d = 5000 xs 5000 ¥ 12 = 0.375 sin d = 400 400 ¥ ¥3 3 3

or,

d = 22° E5 A 2 MVA, 6.6 KV, 50 Hz, 6 pole, 3-phase synchronous motor has a per unit resistance of 0.02 and per unit synchronous reactance of 0.8. If the excitation voltage is 1.5 p.u. the value of maximum power input is (a) 3.8 MW

(b) 0.6 MW

(c) 6.2 MW

Solution ( Base kV ) 2 (6.6) 2 = Base MVA 2 = 21.78

Base impedance =

Armature resistance Synchronous reactance

ra = 0.02 ¥ 21.78 = 0.4356 W xs = 0.8 ¥ 21.78 = 17.424 W

(d) 2.5 MW

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zs = 0.4356 + j17.424 = 17.43 88.56∞

Maximum power input Pinput =

3V 2 3 EV cos fs + zs zs 2

Ê 6600 ˆ 3¥Á Ë 3 ˜¯ 3 ¥ 1.5 ¥ 6600 6600 cos 88.56∞ + ¥ = 17.43 17.43 ¥ 3 3 = 62803.6 + 3748709 = 3811512.6 W = 3.8 MW Ans. (a) E6 A 400 V, 3-phase system feeds a 500 KVA load at 0.8 lagging p.f. A synchronous motor supplying 100 H.P. mechanical load is connected to improve the power factor. If the synchronous motor is operating at 0.8 p.f. leading, the total KVA and overall power factor is (a) 495, 0.91 lead (b) 495, 0.91 lag (c) 544.25, 0.91 lead (d) 544.25, 0.91 lag Solution System cos q1 = 0.8 lag sin q1 = 0.6

KW1 = 500 ¥ 0.8 = 400 KVAR1 = 500 ¥ 0.6 = 300

Motor cos q2 = 0.8 lead sin q2 = 0.6 \ \

KVAR2 = –116.56 ¥ 0.6 = –69.93 Total KW = 400 + 93.25 = 493.25 Total KVAR = 300 – 69.93 = 230 Total KVA = ( 493.25) 2 + ( 230) 2 = 544.24

Overall power factor = Ans. (a)

100 ¥ 0.746 = 93.25 0.8 93.25 KVA2 = = 116.56 0.8 KW2 =

493.25 = 0.906 lag 544.24

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E7 A 6.6 kV, 3-phase star connected synchronous motor draws a full load current of 100 A at 0.8 power factor leading. The armature resistance is 3 W. If the stray loss is 3000 W, the output power is (a) 93 kW

(b) 435 kW

(c) 400 kW

(d) 530 kW

Solution Power input = 3 Vt Ia cos q = 3 ¥

6600 3

¥ 100 ¥ 0.8

= 528000 W = 528 kW Total copper loss = 3 Ia2 ra = 3 ¥ (100)2 ¥ 3 = 90000 W = 90 kW Total loss = 90 + 3 = 93 kW \ Output power = 528 – 93 = 435 kW Ans. (b) Common data for Questions 8, 9 and 10. A synchronous generator is feeding a synchronous motor through a transmission line as shown in the Fig. Both the motor and generator are rated as 20 KVA, 440 V, 50 Hz. The field currents of the two machines are maintained such that the motor terminal voltage Vt = 440 V and the power factor is 0.8 leading. 0.2 p.u.

Ia

0.8 p.u. 0.8 p.u. Vt Eg

+ –

+ –

Em

E8 The magnitude of the induced voltage of the motor (a) 0.9 p.u. Solution Motor induced emf

(b) 0.61 p.u.

(c) 1.8 p.u.

(d) 0.894 p.u.

Em = Vt – Ia zs Vt = 1 0∞ p.u. =

Pe = 20 ¥ 0.8 = 16 kW

16 p.u. = 0.8 p.u. 20

(E KVABase = 20 KVA)

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\

Pe = Vt Ia cos q 0.8 = 1 ¥ Ia ¥ 0.8 or, Ia = 1 36.87∞ p.u.

\

Em = 1 – 1 36.87∞ Xj 0.8 = 1 – j 0.8 36.87∞

Now

= 1 + 0.8 -53.13∞ = 1.48 – j 0.64 = 1.612 -23.38∞ p.u. Ans. (b) 0.61 p.u. E9 The magnitude of the internal generated voltage of the generator is (a) 0.894 p.u.

(b) 0.61 p.u.

(c) 0.9 p.u.

(d) 1.8 p.u.

Solution Eg = Vt + Ia zs = 1 0∞ + 1 36.87∞ j (0.8 + 0.2) = 1 + 1 126.87∞ = 0.4 + j 0.8 = 0.894 63.4∞ p.u. Ans. (a) E10 The minimum value of the induced emf Em for the machine to remain in synchronism is (a) 300 V

(b) 358 V

(c) 400 V

(d) 256 V

Solution For stability maximum value of d is 90° Minimum value of

Em =

0.8 ¥ 0.8 = 0.64 p.u. 1

= 0.64 ¥ 400 = 256 V Ans. (d) Common data for Questions 11 and 12 A 240 V, three-phase star connected synchronous motor draws 8 kW at 0.8 p.f. at a certain load. The synchronous reactance of the motor is 2 W. E11 If the excitation voltage is increased by 50% by raising the field excitation, developed power remaining same, the new value of torque angle will be (a) 20.7°

(b) 15.9°

(c) 11.9°

(d) 30°

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Solution Vt = Ia =

240 3

V

8000 A = 13.89 A 3 ¥ 240 ¥ 0.8

E = (Vt cos q - I a ra ) 2 + (Vt sin q - I a xs ) 2 2

Ê 240 ˆ Ê 240 ˆ = Á ¥ 0.8˜ + Á ¥ 0.6 - 13.89 ¥ 2˜ Ë 3 ¯ Ë 3 ¯

2

= 12288 + 3064.56 = 123.9 V E¢ = 1.5 ¥ 123.9 V = 185.85 V 240 3 ¥ 185.85 ¥ 3 sin d 8000 = 2

New excitation voltage

Now \

sin d =

8000 ¥ 2

240 3 ¥ 185.85 ¥ 3 d = 11.94°

\

= 0.207

Ans. (c) E12 The value of new armature current (a) 28.9 A

(b) 13.9 A

(c) 30 A

(d) 40.3 A

Solution V -E = Ia = t jxs

240 - 185.85 -11.94∞ 3 j2

= 69.28 -90∞ – 92.925 -101.94∞ = – j 69.28 + 19.22 + j 90.91 = 19.22 + j 21.63 = 28.93 48.37∞ Ans. (a) E13 A 440 V, 50 Hz. six-pole synchronous motor is drawing 80 A of current at full load and unity power factor. Assuming the motor to be lossless, the output torque of the motor is (a) 735 Nm

(b) 185.7 Nm

(c) 384.2 Nm

(d) 582.5 Nm

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Solution Pi = 3 Vt Ia cos q = 3 ¥ 440 ¥ 80 ¥ 1

Input power

= 60968.2 W As the motor is lossless, output power = Pi = 60968.2 W Speed of motor = Synchronous speed =

120 ¥ 50 6

= 1000 rpm Hence output torque =

60968.2 Nm = 582.5 Nm 1000 2p ¥ 60

Ans. (d) 582.5 Nm E14 A plant operates at 0.8 lagging power factor and takes 3000 kW. A synchronous capacitor is used to raise the overall power factor to unity. The synchronous capacitor losses are 300 kW. The KVA rating and power factor of the synchronous capacitor are respectively (a) 3750, 0.8

(b) 2270, 0.132

(c) 2250, 0.25

(d) 3000, 0.6

Solution kW rating of the plant = 3000 Power factor = 0.8 lag 3000 ¥ 0.6 KVAR rating of the plant = 0.8 = 2250 kW rating of the capacitor = 300 To make the overall power factor unity, the KVAR of the capacitor should be –2250 Hence KVA rating of the capacitor = 300 – j 2250 KVA = 2270 -824∞ KVA \ Power factor = cos 82.4° = 0.132 leading Ans. (b) E15 A 3-phase, 6.6 kV, 300 rpm, 50 Hz. star connected motor has a synchronous reactance of 1.5 W. The generated voltage is equal to 3600 V. At full load the rotor poles are displaced by a mechanical angle of 2° from the no load position. The total developed mechanical power is (a) 27435 kW

(b) 7825 kW

(c) 9384 kW

(d) 30 kW

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Solution 300 =

120 ¥ 50 P

Number of poles 120 ¥ 50 = 20 300

P= The electrical torque angle is d= Vt =

Pa 20 ¥ 2∞ = = 20° 2 2 6600

V

3

Mechanical power developed P=

3Vt E sin d xs 3¥

=

6600 ¥ 3600 3 sin 20° 1.5

= 9383556.8 W = 9383.5 kW Ans. (c) E16 A 440 V, three-phase, star connected salient pole synchronous motor is drawing 80 A current at full load unity power factor. The d axis and q axis reactance per phase are 3 W and 2 W respectively. The power angle is equal to (a) 32.2°

(b) 25.1°

(c) 16.9°

Solution Vt =

440 3

Ia = 80 0∞ A E = Vt – Ia ra – j Ia Xq =

440 3

– 80 ¥ 0 – j 80 ¥ 2

(d) 0°

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– j 160 = 300 -32.2∞

Ans. (a) Common data for Questions 17 and 18 A cylindrical rotor synchronous motor having a synchronous reactance of 1.5 p.u. is connected to an infinite bus whose voltage is 1 p.u. through an equivalent reactance of 0.5 p.u. For a power output of 0.8 p.u., the power angle is found to be 30°. E17 The excitation voltage of the motor is (a) 1.5 p.u.

(b) 2.7 p.u.

(c) 1.6 p.u.

(d) 3.2 p.u.

Solution Pe = Pmax sin d 0.8 = Pmax sin 30° Pmax = 1.6 p.u.

\ Now \

1.6 =

EVt E ¥1 E = = 1.5 + 0.5 2 x

E = 2 ¥ 1.6 = 3.2.

Ans. (d) E18 Keeping the output voltage same the power angle is now reduced to 25°. The value of the equivalent reactance which connects the machine to the bus will be (a) 1.69 p.u.

(b) 0.19 p.u.

(c) 1 p.u.

(d) 3.2 p.u.

Solution d = 25° 0.8 = Pmax sin 25° \

Pmax = 1.89 EVt 3.2 ¥ 1 = x¢ x¢ 3.2 x¢ = = 1.69 1.89

1.89 = \

Hence equivalent reactance which connects the machine to the bus is 1.69 – 1.5 = 0.19 p.u. Ans. (b)

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Fmfdusjdbm!Nbdijoft

E19 A factory has a load of 1500 kW at power factor of 0.75 lagging. It is derived to improve the factory power factor to 0.9 lagging with the installation of a synchronous condenstor. The kVAR rating of the synchronous condenser is (a) 1322.87

(b) 726.5

(c) 924

(d) 596.4

Solution 1500 = 2000 0.75 Factory kVAR = 2000 sin (cos–1 0.75) = 1322.87 Final power factor = 0.9 \ combined kVAR = 1500 tan (cos–1 0.9) = 726.48 Factory kVA =

VA 0K

1322.87 KVAR

0 20

–1 0.75

cos

cos–1 0.9

1500 kW

Hence kVAR supplied by condenser = 1322.87 – 726.48 = 596.4 Ans. (d) E20 A 440 V, 100 KW, 50 Hz. star connected synchronous motor operating at 0.8 power factor leading has full load efficiency of 90%. The armature resistance is 0.1 W. How much electrical power is connected to mechanical power? (a) 10 kW

(b) 9.96 kW

(c) 101.15 kW

(d) 111.11 kW

Solution 100 = 111.11 kW 0.9 111.11 ¥ 103 = 182.24 A Ia = 3 ¥ 440 ¥ 0.8

Pin =

Copper losses

Pcu = 3 Ia2 ra = 3 ¥ (182.24)2 ¥ 0.1 = 9963.42 W

Neglecting other losses the power converted from electrical to mechanical form is Pin – Pcu = 111.11 – 9.963 = 101.147 kW. Ans. (c)

Tpmvujpo!Nbovbm

Dibqufs!2 Gvoebnfoubm!Dpodfqut!jo!Fmfdusjdbm!Nbdijoft 1.

I B l sin q

=5A = 1 Wb/m2 = 10 cm = 0.1 m = sin 90° = 1

Force on the conductor = B I l sin q = 1 ¥ 5 ¥ 0.1 ¥ 1 = 0.5 N 2.

l = 10 cm = 0.1 m A = 10 cm2 = 10 ¥ 10–4 m2 NA = 100 NB = 500 mr = 300 Coefficient of mutual inductance M= =

N1 N 2 l /mo m r A 100 ¥ 500 ¥ 4p ¥ 10 -7 ¥ 300 ¥ 10 -3 = 0.1884 H 0.1

Change in current di = 10 A

!

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Fmfdusjdbm!Nbdijoft

Time dt = 0.1 s Hence emf induced in coil B is M

di 10 V = 18.84 V = 0.1884 ¥ dt 0.1

3. f = 5 ¥ 10–4 Wb I = 10 A N = 150 Self inductance L=

N f 150 ¥ 5 ¥ 10 -4 H = 7.5 mH = I 10

Change of current di = 10 – (– 10) = 20 A Time dt = 0.2 s di 20 Hence induced emf e = L = 7.5 ¥ 10–3 ¥ = 0.75 V dt 0.2 4. N = 300 L = 8 cm = 0.08 m r = 2 cm = 0.02 m B = 1.1 Wb/m2 I =2A Force experienced by the coil F =2NBIl = 2 ¥ 300 ¥ 1.1 ¥ 2 ¥ 0.08 N = 105.6 N \ Torque T = F ¥ r = 105.6 ¥ 0.02 Nm = 2.112 Nm 5.

N = 500 d = 300 mm = 300 ¥ 10–3 m = 0.3 m (50) 2 p A =p¥ sqmm = ¥ 2500 ¥ 10–6 sqm 4 4 \ l = p ¥ 0.3 m Inductance -7 p -6 2 m AN 2 4p ¥ 10 ¥ 4 ¥ 2500 ¥ 10 ¥ (500) = L= p ¥ 0.3 l

p ¥ 10 -7 ¥ 2500 ¥ 10 -6 ¥ 250000 0.3 = 0.654 mH =

Tpmvujpo!Nbovbm

6.

f1 = 600 Hz Bm1 = 0.6 Wb/m2 Pe1 = 16 W f2 = 800 Hz Bm2 = 0.5 Wb/m2 Pe • f 2 Bm2 \

Pe2 = Pe1 ¥

f 22 Bm2 2 f12 Bm21

= 16 ¥

(800) 2 ¥ (0.5) 2 (600) 2 ¥ (0.6) 2

= 19.75 W 7. Output power = 100 ¥ 0.7 kW = 70 kW Core loss = 800 W Copper loss = 1000 W Efficiency =

Output Output = Input Output + Loss

70 ¥ 103 ¥ 100% 70 ¥ 103 + 1800 = 97.49% =

8. P = 8 Ns = 1500 rpm 120 f Ns = P \

f=

PN s 8 ¥ 1500 = 100 Hz = 120 120

9. P = 6 Output = (35 – 3) HP = 32 HP = 32 ¥ 735.5 W = 23.536 kW Total loss = (3 ¥ 735.5 ¥ 10–3 + 3 + 1.07) kW = 6.2765 m= 10. V N Ia ra Eb

Output 23.536 ¥ 100% = 78.95% = Output + Loss 23.536 + 6.2765

= 400 V = 795 rpm =8A = 0.3 W = V – Ia ra = 400 – 8 ¥ 0.3 = 397.6 V

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2/5

Fmfdusjdbm!Nbdijoft

Torque T =

=

Eb I a Eb I a = 2p N w 60 397.6 ¥ 8 Nm 795 2p ¥ 60

= 38.23 Nm

Tpmvujpo!Nbovbm

Dibqufs!3 Ejsfdu!Dvssfou!)ed*!Nbdijoft 1. V Ia ra Z

= 250 V = 100 A = 0.2 W = 272

Brush drop = 2 V f = 0.05 Wb For lap winding A=P (a) For generator, generated emf E = V + Ia ra + Brush drop = 250 + 100 ¥ 0.2 + 2 = 272 V If N be the speed in rpm then Pf ZN 60 A 60 AE 60 ¥ P ¥ 272 N= = 1200 rpm = Pf Z P ¥ 0.05 ¥ 272 E=

or, (b) For motor, back emf

E = V – Ia ra – Brush drop = 250 – 100 ¥ 0.2 – 2 = 228 V \ or,

P ¥ 0.05 ¥ 272 N 60 P 228 ¥ 60 = 1006 rpm N= 272 ¥ 0.05

228 =

2. E1 = 400 V Let initial flux be f1 and initial speed be N1 (a)

E2 f 2 = E1 f1 f2 = 0.8 f1 \

E2 = E1

f2 = 400 ¥ 0.8 = 320 V f1

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2/7

Fmfdusjdbm!Nbdijoft

(b) N2 = 1.25 N1 E2 N 2 = E1 N1 \

E2 = 400 ¥ 1.25 = 500 V

(c) N2 = 0.25 N1 \

E2 = 400 ¥ 0.25 = 100 V

(d) N2 = 1.25 N1 f2 = 0.8 f1 E2 f 2 N 2 = E1 f1 N1 \ (e) N2 = 0.8 N1

E2 = 400 ¥ 0.8 ¥ 1.25 = 400 V f2 = 0.8 f1

\

E2 = E1 ¥

f2 N 2 = 400 ¥ 0.8 ¥ 0.8 f1 N1 = 256 V

3. V N P A

= 1000 V = 1500 rpm = 10 = 2 (for wave wound)

Number of turns T = 200 ¥ 5 = 1000 Number of conductors Z = 2T = 2000 Each path has resistance = 0.006 ¥ 1000 = 3 W As there are two parallel paths, total armature resistance is Ia = 10 ¥ 2 A = 20 A (a) Number of commutator bars = 200 = number of coils (b) Generated emf E = V + Ia ra = 1000 + 20 ¥ \

E=

Pf ZN 60 A

3 = 1030 V 2

3 W 2

Tpmvujpo!Nbovbm

60 AE 60 ¥ 2 ¥ 1030 = = 4.12 ¥ 10–3 Wb PZN 10 ¥ 2000 ¥ 1500 = 0.004 Wb EIa E Ia 1030 ¥ 20 (c) Torque = = 131 Nm = = N 1500 w 2p 2p ¥ 60 60 f=

or,

4. rsh Rse ra V IL

= 40 W = 0.03 W = 0.06 W = 250 V = 100 A

(a) Ish =

V 250 = 6.25 A = rsh 40

\ \

Ia = 100 + 6.25 = 106.25 A E = V + Ia ra = 250 + 106.25 ¥ 0.06 = 256.375

(b) Voltage across shunt winding is V + IL rse = 250 + 100 ¥ 0.03 = 253 V Ish =

253 A = 6.325 A 40

\

Ia = IL + Ish = 100 + 6.325 = 106.325 A

\

E = 253 + 106.325 ¥ 0.06 + 2 = 261.38 V

(c) Ish =

250 A = 6.25 A 40

\

Ia = 100 + 6.25 = 106.25 A

\

E = V + Ia (ra + rse) = 250 + 106.25 (0.03 + 0.06) + 2 = 261.56 V

(d) E = 250 + 100 (0.03 + 0.06) + 2 = 261 V 5. V = 240 V Voltage regulation = \

2.5 100 Vnl - V 2.5 = V 100

2/8

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Fmfdusjdbm!Nbdijoft

Vnl 2.5 –1 = V 100 2.5 ˆ Ê Vnl = 240 Á1 + = 246 V Ë 100 ˜¯ 6.

V N1 I L1 ra rsh

= 240 V = 850 rpm = 72 A = 0.242 W = 95.2 W 240 A = 2.52 95.2 I a1 = 72 – 2.52 = 69.48 A Eb1 = V – I a1 ra = 240 – 69.48 ¥ 0.242 = 223.18 V Ish =

N2 = 1650 rpm I a2 = 50.4 A Eb2 = V – I a2 ra = 240 – 50.4 ¥ 0.242 = 227.8 V Eb2 Eb1

=

f2 N 2 f1 N1

f2 Eb2 N1 227.8 850 = ¥ = ¥ = 0.5258 f1 Eb1 N 2 223.18 1650

\ Reduction in flux =

f1 - f2 ¥ 100% f1

Ê f ˆ = Á1 - 2 ˜ ¥ 100% f1 ¯ Ë = (1 – 0.5258) ¥ 100% = 47.4% 7.

V N1 IL N2 ra rsh

= 240 V = 850 rpm = 91 A = 634 rpm = 0.221 W = 120 W

Resistance in series with armature R = 2.14 W

Tpmvujpo!Nbovbm

Ish =

240 = 2 A; I a1 = 91 – 2 = 89 A 120

Eb1 = V – I a1 ra = 240 – 89 ¥ 0.221 = 220.331 Eb2 = V – I a2 (ra + R) = 240 – I a2 (0.221 + 2.14) = 240 – 2.36 Ia2 Eb1 Eb2 \

V N IL ra rsh

N1 fi Eb2 = Eb1 ¥ N 2 N2 N1

240 – 2.36 I a2 = 220.331 ¥

634 = 164.34 850

I a2 = 32.05 A

or, 8.

=

= 240 V = 2500 rpm = 140 A = 0.0873 W = 95.3 W Ish =

240 A 95.3

Ia = IL – Ish = 140 –

240 = 137.48 A 95.3

Eb = V – Ia ra = 240 – 137.48 ¥ 0.0873 = 228 V Torque =

Eb I a 228 ¥ 137.48 Nm = 119.8 Nm = 2500 w 2p ¥ 60

9. Area of pole face = Pole arc ¥ axial length Pole arc = 0.7 Pole pitch Length of pole shoe = 0.2 m Diameter of pole shoe circle = 0.35 m p ¥ 0.35 ¥ 0.2 = 0.03846 m2 Area of pole face = 0.7 ¥ 4 Flux density =

Flux area of pole shoe

2/:

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Fmfdusjdbm!Nbdijoft

E=

Pf ZN 60 AE 60 ¥ P ¥ 250 fif= = 60 A PZN P ¥ 1200 ¥ 500 = 0.025 Wb

\ Flux density =

0.025 Wb/m2 = 0.65 Wb/m2 0.03846

10. P = 50 kW V = 250 V N = 400 rpm IL =

50, 000 = 200 A 250

ra = 0.02 W rsh = 50 W 250 Ish = =5A 50 Ia = IL + Ish = 205 A E1 = V + Ia ra + 2 = 250 + 205 ¥ 0.02 + 2 = 256.1 V When the machine runs as motor IL =

50, 000 A = 200 A 250

Ia = IL – Ish = 195 A E2 = V – Ia ra – 2 = 250 – 195 ¥ 0.02 – 2 = 244.1 V E2 N 2 = N1 E1 or, 11.

N2 = N1 ¥

E2 244.1 = 400 ¥ = 381.3 rpm E1 256.1

r = 0.5 W Ia = 60 A V = 500 V I • N3 If N1 = N then N2 = 0.75 N T2 N 23 (0.75 N )3 = = = 0.4218 T1 N13 N3

Tpmvujpo!Nbovbm

2/22

For series motor T • Ia2 I a22 I a22 T2 = 0.4218 = 2 = T1 I a1 (60) 2 I a2 = 60 0.4218 = 38.97 A E2 N 2 f2 N 2 I a2 = = E1 N1 f1 N1 I a1

E1 = V – Ia r = 500 – 60 ¥ 0.5 = 470 V E2 = 500 – 38.97 ¥ (0.5 + R)

Now,

500 – 38.97( R + 0.5) 0.75 ¥ 38.97 = 470 60

\

12.

or, V P A ra rsh Z f IL

(E f • Ia in series motor)

R + 0.5 = 6.955 R = 6.455 W = 250 =4 =2 = 0.25 W = 125 W = 500 = 0.02 Wb = 14 A

Rotational loss = 300 W 250 ˆ Ê Eb = V – Ia ra = 250 – Á14 – ¥ 0.25 Ë 125 ˜¯ = 250 – 12 ¥ 0.25 = 247 V Eb = \

T=

Pf ZN 247 ¥ 60 ¥ 2 fiN= = 741 rpm 60 A 4 ¥ 0.02 ¥ 500 Eb I a 247 ¥ 12 = Nm = 38.2 Nm 741 w 2p ¥ 60

Output = Eb Ia – 300 = 247 ¥ 12 – 300 = 2664 W 2664 \ Efficiency = ¥ 100% = 76.11% 250 ¥ 14

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Fmfdusjdbm!Nbdijoft

13. P = 10 kW V = 250 V Rotational loss = 400 W ra = 0.5 W rsh = 250 W 10, 000 IL = A = 40 A 250 250 Ish = A=1A 250 Ia = IL + Ish = 41 A Shunt field copper loss = I 2sh rsh = 12 ¥ 250 = 250 W Armature copper loss = Ia2 ra = (41)2 ¥ 0.5 = 840.5 W Total loss = 250 + 840.5 + 400 = 1490.5 Loss 1490.5 Efficiency = 1 – =1– = 87.02% Input 10,000 + 1490.5 For maximum efficiency, armature copper loss = constant loss Constant loss = I 2sh rsh + 400 = 250 + 400 = 650 W Ia2 ra = 650 fi Ia =

\

650 = 36 A 0.5

Output = 250 ¥ IL = 250 (36 – 1) = 250 ¥ 35 \ Efficiency = 14. V N I ra rse

Output 250 ¥ 35 = = 87.06% Output + Loss 250 ¥ 35 + 2 ¥ 650

= 230 V = 1500 rpm = 20 A = 0.3 W = 0.2 W Eb = V – Ia (ra + rse) = 230 – 20 ¥ 0.5 = 220 V

(i) At starting Eb = 0 \

V – Ia (R + 0.5) = 0 where R is the external resistance 230 \ R= – 0.5 = 11 W 20 (ii) N2 = 1000 rpm As torque is constant I a2 = Ia

Tpmvujpo!Nbovbm

2/24

Eb2 = V – Ia (0.5 + R¢) = 230 – 20 (R¢ + 0.5) = 220 – 20 R¢ Eb I N fN N 1500 = = a = = f2 N 2 I a2 N 2 N 2 1000 Eb2 220 3 = or, 440 = 660 – 60 R¢ 220 – 20 R¢ 2 or,

R¢ =

660 - 440 = 3.67 W 60

15. Armature mmf = 12000 AT/pole Pole arc = 0.7 Pole pitch Flux density in interpolar gap is 0.3 Wb/m2 Length of interpolar gap 1.25 cm. Mmf in air gap =

0.3 1.25 ¥ = 2985 AT/pole -7 100 4p ¥ 10

Total mmf = 12000 + 2985 = 14985 AT/pole Mmf of compensating winding = 12000 ¥ 0.7 = 8400 AT/pole Mmf of interpoles = 14985 – 8400 = 6585 AT/pole 16.

P V ra rsh

= 800 kW = 500 V = 0.005 W = 50 W

Mechanical loss = 10,000 W Iron loss = 11,000 W Stray loss = 1% of 800 kW = 8000 W Shunt field copper loss =

V 2 500 ¥ 500 = = 5000 W rsh 50 2

Ê 800, 000 500 ˆ ¥ 0.005 = 12960.5 W Armature copper loss = Ia2 ra = Á + Ë 500 50 ˜¯ Brush loss = Ia ¥ 2 = 1610 ¥ 2 = 3210 W Total loss = 10 + 11 + 8 + 5 + 12.9605 + 3.210 = 50.1705 kW

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Fmfdusjdbm!Nbdijoft

(a) Efficiency at full load =

Output 800 = ¥ 100% = 94% Input 800 + 50.1705

(b) At half load Stray loss = 4 kW Ia =

400, 000 500 = 800 + 10 = 810 A + 500 50

Armature copper loss = (810)2 ¥ 0.005 = 3280.5 W Brush loss = 810 ¥ 2 = 1620 W Total loss = 10 + 11 + 4 + 3.2805 + 1.620 + 5 = 34.9005 kW Efficiency =

400 = 91.97% 400 + 34.9005

(c) N = 1500 rpm Input torque = (b) Induced torque

Input power 850.1705 ¥ 103 = Nm = 5415 Nm 1500 w 2p ¥ 60

Input power - Iron loss - Mechanical loss - Stray loss w 850.1705 - 10 - 11 - 8 = Nm = 5230 Nm 2p ¥ 1500 60

=

17.

V = 250 V ra = 0.6 W rsh = 150 W Ianl = 5 A Nnl = 1000 rpm Ishnl =

250 W 150

Ebnl = V – Ia ra = 250 – 5 ¥ 0.6 = 247 V Now, T = 100 Nm =

Eb I a 100 ¥ 2p ¥ N fi Eb = w 60 ¥ I a Eb nl N nl = Eb N

Tpmvujpo!Nbovbm

fi \

Eb =

N N ¥ 247 Eb nl = 1000 N nl

100 ¥ 2p N N = ¥ 247 60 I a 1000

or,

Ia =

100 ¥ 2p ¥ 1000 = 42.37 A 60 ¥ 247

Now

Eb =

100 ¥ 2p ¥ N = V – Ia ra = 250 – 42.37 ¥ 0.6 60 ¥ 42.37 = 224.578

\ Now

N = 909 rpm P = 10 kW N = 1200 rpm 10,000 = Eb Ia = (V – Ia ra) Ia = (250 – Ia ¥ 0.6) Ia

or, 0.6 Ia2 – 250 Ia + 10,000 = 0 250 ± ( 250)2 - 4 ¥ 0.6 ¥ 10, 000 2 ¥ 0.6 250 ± 196.21 = = 44.825 A 1.2

\

Ia =

\

Eb = 250 – 44.825 ¥ 0.6 = 223.105

Now

Eb I N fN = = sh Eb nl f nl N nl I sh nl N nl I sh ¥ 1200 = 223.105 250 ¥ 1000 150 250 1 = 223.105 ¥ ¥ 1000 ¥ 150 247 ¥ 1200

\

Eb = 247 ¥

\

Ish

= 1.2545 A If R is the external resistance then 1.2545 = or,

R=

250 150 + R 250 – 150 = 49 W 1.2545

2/26

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18.

V T N1 I L1

Fmfdusjdbm!Nbdijoft

= 200 V • N2 = 600 rpm = 30 A

Neglecting field current I a1 = 30 A Now, for shunt motor T • Ia I a1 I a2

=

T1 N12 = T2 N 22

\

30 (600) 2 = I a2 N 22

\

I a2 = Eb1 Eb2

\

30 N2 3600 2 200 (Neglecting ra) = 200 – I a2 ¥ 20

200 600 = N2 200 - 20 I a2

Solving Eq. (i) and Eq. (ii) N = 33.65 rpm Ia = 9.436 A 19.

V = 250 V P = 15 kW hmax = 0.88 N = 700 rpm At maximum efficiency output power Pmax = 0.8 ¥ 15 kW = 12 kW rsh = 100 W Under maximum efficiency

\

0.88 =

12 12 + Loss

Loss =

12 – 12 = 1.636 kW = 1636 W 0.88

…(i)

…(ii)

Tpmvujpo!Nbovbm

Constant loss = Variable loss =

1636 W = 818 W 2 Ia2 ra = 818

\

Now, Input = 12 + 1.636 = 13.636 kW \

13636 = V IL = 250 ¥ IL

or,

IL = 54.544 A Ish =

\

250 A 100

Ia = IL – Ish = 52 A Ia2 ra = 818

Now,

818 = 0.3 W (52) 2

\

ra =

Now, \

IL = 78 A Ia = 78 – 2.5 = 75.5 A

Variable loss = Ia2 ra = (75.5)2 ¥ 0.3 = 1710.075 W Total loss = 818 + 1710.075 = 2528.075 W \

h =1–

Again

Eb2

=

250 - 52 ¥ 0.3 700 = 250 - 75.5 ¥ 0.3 N 2

N2 = 700 ¥

or, 20. P V N Ia ra

Eb1

Loss 2528.075 =1– = 87% Input 250 ¥ 78

227.35 = 678 rpm 234.4

= 50 kW = 400 V = 800 rpm = 140 A = 0.1 W Eb = 400 – 140 ¥ 0.1 = 386 V Tfl =

50 ¥ 103 = 597 Nm 800 2p ¥ 60

2/28

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Fmfdusjdbm!Nbdijoft

Initial plugging current = 200 A If initial plugging torque is T then T 200 = Tfl 140 or Now,

(E T • Ia in shunt motor)

200 = 853 Nm 140 N¢ = 400 rpm T = 597 ¥

E¢b = Eb ¥ Ia =

N¢ 400 = 386 ¥ = 193 V N 800

V - Eb¢ 400 - 193 = 2070 = ra 0.1

Plugging torque at half speed = 597 ¥

2070 Nm = 8827 Nm. 140

Tpmvujpo!Nbovbm

Dibqufs!4 Tjohmf.Qibtf!Usbotgpsnfst 1.

f = 50 Hz E1 = 440 V N1 = 200 (a) If fm is the peak value of flux then E1 = 4.44 f fm N1 fm =

or

440 Wb = 0.0099 Wb. 4.44 ¥ 50 ¥ 200

(b) N2 = 50 Voltage induced in the secondary E2 = 4.44 f fm N2 = 4.44 ¥ 50 ¥ 0.0099 ¥ 50 V = 110 V. 2. A = 100 ¥ 10–4 m2 = 0.01 m2 E1 = 200 V; E2 = 50 V; Bm = 1 Wb/m2 Assuming 9% loss of area, net area of core = 0.01 ¥ 0.9 m2 = 0.009 m2 Primary turns N1 =

E1 200 = 100 = 4.44 f Bm A 4.44 ¥ 50 ¥ 1 ¥ 0.009

Secondary turns N2 = Transformer ratio

E2 50 ¥ 100 = 25 N1 = E1 200

E1 N1 100 = = = 4. E2 N 2 25

3. Number of turns of primary winding N1 = 100 Number of turns of secondary winding N2 = 200 Resistance of primary winding R1 = 0.05 W Resistance of secondary winding R2 = 0.3 W (a) Resistance of primary winding referred to secondary 2

2 ÊN ˆ Ê 200 ˆ R1¢ = R1 ¥ Á 2 ˜ = 0.05 ¥ Á = 0.05 ¥ 4 = 0.2 W. Ë 100 ˜¯ Ë N1 ¯

(b) Resistance of secondary winding referred to the primary 2

2 Ê N1 ˆ 0.3 Ê 100 ˆ R2¢ = R2 ¥ Á = 0.3 ¥ Á = W = 0.075 W ˜ ˜ Ë 200 ¯ 4 Ë N2 ¯

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Fmfdusjdbm!Nbdijoft

Equivalent resistance of the transformer referred to the primary R01 = R1 + R¢2 = 0.05 + 0.075 = 0.125 W. [Also, R02 = R2 + R¢1 = 0.3 + 0.2 = 0.5 W]. 4. If N1 and N2 be the number of turns of the primary and secondary winding then N1 1000 = N2 200 Resistance of the primary winding R1 = 1 W Resistance of the secondary winding R2 = 0.2 W Total equivalent resistance in terms of the secondary winding is (R¢1 + R2) 2

i.e.

2

ÊN ˆ Ê 200 ˆ Ro2 = R1 Á 2 ˜ + R2 = 1 ¥ Á + 0.2 = 0.04 + 0.2 = 0.24 W Ë 1000 ˜¯ ËN ¯ 1

Full-load secondary current I2 =

20 ¥ 103 = 100 A 200

Total resistance drop on full-load = I2 Ro2 = 100 ¥ 0.24 = 24 V. 5. Efficiency =

Output (Input - Loss) Losses = =1Input Input Input

At full-load,

or or or

0.97 = 1 –

Core losses + Copper losses osses Output + Core losses + Copper lo

0.97 = 1 –

Pc + Pcu 10 ¥ 103 ¥ 1 + Pc + Pcu

Pc + Pcu = 0.03 10000 ¥ Pc + Pcu Pc + Pcu = 309.278 W

(i)

At half-load, Pc + 0.97 = 1 – 10 ¥ 103 ¥

1 P 4 cu

1 1 + Pc + Pcu 2 4

Tpmvujpo!Nbovbm

2/32

or

Pc +

1 1 ˆ Ê Pcu = 150 + 0.03 Á Pc + Pcu ˜ Ë 4 4 ¯

or

Pc +

1 150 Pcu = = 154.693 W 4 0.97

(ii)

Solving Eqs. (i) and (ii), Pcu = 206.185 W Pc = 103.1 W.

and 6. Iron loss = 200 W Copper loss at a load of 20 kVA is = 180 W

(a) Output at 0.81 p.f. (lag) = 20 ¥ 103 ¥ 0.81 = 16200 W Total losses = 200 + 180 = 380 W Input = Output + Losses = 16200 + 380 = 16580 W Efficiency =

Output 16200 = = 0.977 = 97.7% Input 16580

(b) New load is 30 kVA at 0.91 p.f. (lag) Output = 30 ¥ 103 ¥ 0.91 = 27300 W 2

9 Ê 30 ˆ Copper losses at a load of 30 kVA is, Pcu = 180 ¥ Á ˜ = 180 ¥ = 405 W Ë 20 ¯ 4 Input = Output + Losses = 27300 + 200 + 405 = 27905 W 27300 = 0.978 = 97.8%. 27905 50, 000 A = 15.15 A 7. Primary full-load current = 3300 4.2 I1Zo1 = Now, V1 100 Efficiency =

or

0.042 =

15.15 Zo1 300

Zol = 9.148 W

or

where Zo1 = equivalent impedance referred to the primary. Again, or

0.018 =

I12 Ro1 I1 Ro1 15.15 Ro1 = = V1 I1 V1 3300

Ro1 = 3.92 W,

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Fmfdusjdbm!Nbdijoft

where Ro1 = equivalent resistance referred to the primary. \ equivalent resistance referred to the primary = (9.148) 2 - (3.92) 2 W = 8.26 W. Ê V ˆ 3300 Under short circuit condition, the primary current = Á 1 ˜ = A = 360.73 A. Ë Zo1 ¯ 9.148 8. From the open-circuit test data, 90 No-load p.f. cos qo = = 0.4545 220 ¥ 0.9 \ Core loss resistance

sin qo = 0.89 V1 220 Ro = W = 537.83 W = I o sin q o 0.9 ¥ 0.4545 Xo =

Magnetizing reactance

V1 220 = = 274.65 W I o sin fo 0.89 ¥ 0.9

From short-circuit test data, ROH = Ro2 =

100 W = 0.444 W (15) 2

ZOH = Zo2 =

20 W = 1.33 W 15

where Ro2 and Zo2 are the equivalent resistance and impedance referred to the high-voltage side. Hence,

XOH = Xo2 = (1.33) 2 - (0.44) 2 = 1.257 W

Figure 3.59 shows the equivalent resistance R1 and reactance X1 referred to the low voltage side or primary side. 2

Hence,

Ê 220 ˆ R1 = 0.444 ¥ Á = 0.111 W Ë 440 ˜¯

and

Ê 220 ˆ = 0.314 W X1 = 1.257 ¥ Á Ë 440 ˜¯

Also

RO = 537.83 W and XO = 274.65 W.

2

9. (a) From no-load test data, No-load p.f. (cos qo) =

80 = 0.227 440 ¥ 0.8

Iron loss component current = 0.8 (cos qo) = 0.8 ¥ 0.227 = 0.182 A Magnetizing current = 0.8 (sin qo) = 0.8 ¥ 0.974 = 0.779 A.

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(b) Iron loss = 80 W As the secondary side is the HV side, so a short-circuit test is performed on the secondary side. 8000 Rated current of the HV side I2 = =4A 2000 When current is 3 A, the wattmeter reading is 20 W. So, if rated current of 4 A flows through the high-voltage winding, the wattmeter reading 2

4 = 20 ¥ ÊÁ ˆ˜ = 35.55 W Ë 3¯ So rated copper losses = 35.55 W Efficiency on full-load at unity p.f. = 10. Full-load primary current =

8 ¥ 103 ¥ 1 = 0.9857 or 98.57%. 8 ¥ 103 ¥ 1 + 35.55 + 80

17500 A = 38.89 A 450

17500 A = 144.63 A 121 As 450 V is the applied voltage in OCT, so this test has been performed on the HV side and as the current in the SCT is 38.9 A which is the rated primary or low-voltage current, so this test has also been performed on the LV side. From short-circuit test data, Full-load secondary current =

Ro1 =

312 W = 0.206 W (38.9) 2

Zo1 =

15.75 W = 0.405 38.9

and

Xo1 = (0.405) 2 - (0.206) 2 = 0.3887 W

\

Ê 121 ˆ Ro2 = 0.206 ¥ Á = 0.0149 W Ë 450 ˜¯

and

Ê 121 ˆ = 0.0252 W Xo2 = 0.3487 ¥ Á Ë 450 ˜¯

2

2

If V2 be the secondary terminal voltage then or

[121 – V2] = 144.63 (0.0149 ¥ 0.8 + 0.0252 ¥ 0.6) = 3.91 V2 = 117.09 V

From OCT, iron loss = 115 W

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Fmfdusjdbm!Nbdijoft

From SCT, full-load copper loss = 312 So efficiency at full-load and 0.8 p.f. lagging =

17.5 ¥ 103 ¥ 0.8 = 0.97 or 97%. 17.5 ¥ 103 ¥ 0.8 + 115 + 312

11. Output at maximum efficiency = 100 ¥ 0.9 ¥ 0.85 = 76.5 kW Efficiency = (0.95) = \

Output 76.5 = Output + Losses 76.5 + Losses

˘ È 76.5 Losses = Í - 76.5˙ = 4.026 kW ˚ Î 0.95

At maximum efficiency, core losses = copper losses \ core losses = copper losses = 2.013 kW 2.013 kW is the copper losses at 90% of full-load. 2

Ê 1 ˆ = 2.485 kW So full-load ohmic losses = 2.013 ¥ Á Ë 0.9 ˜¯ If I2 be the full-load secondary current, ( I 22 RO 2 ) = 2485, where RO2 is the equivalent resistance referred to as the secondary. \ or

ÊI R ˆ I2 V2 Á 1 O 2 ˜ = 2485 Ë V2 ¯ 100 ¥ 103 ¥ Rp.u. = 2485, where Rp.u. is the p.u. resistance

or

Rp.u. = 0.02485

Now

Zp.u. = 0.05

\

Xp.u. = (0.05) 2 - (0.02485) 2 = 0.04338

Voltage regulation = (Rp.u. cos q2 + Xp.u. sin q2) = 0.02485 ¥ 0.8 + 0.04338 ¥ 0.6 = 0.0459 or, 4.59%. 12. Core loss = 50 W; full-load ohmic loss = 120 W From 6 a.m. to 12 noon, Output = 5 ¥ 6 = 30 kWh 5 kVA load = = 7.143 0.7 2 Ê 7.143 ˆ ¥ 120 = 61.22 W Ohmic losses for 6 hours = Á Ë 10 ˜¯

Tpmvujpo!Nbovbm

Energy lost as ohmic loss = (61.22 ¥ 6) = 367.36 Wh From 12 noon to 6 p.m., Output = 4 ¥ 6 = 29 kWh 4 kVA load = =5 0.8 2 Ê 5ˆ Ohmic losses for 6 hours = Á ˜ ¥ 120 = 30 W Ë 10 ¯ Energy lost as ohmic loss = (30 ¥ 6) = 180 Wh From 6 p.m. to 1 a.m., Output 8 ¥ 7 = 56 kWh 8 kVA load = = 8.89 0.9 2 Ê 8.89 ˆ ¥ 120 = 94.84 W Ohmic losses for 7 hours = Á Ë 10 ˜¯ Energy lost as ohmic loss = 94.84 ¥ 7 = 663.87 Wh Daily energy lost as ohmic loss = (367.36 + 180 + 663.87) ¥ 10–3 kWh = 1.211 kWh Daily energy lost as core loss =

50 ¥ 24 kWh = 1.2 kWh 103

Total loss = (1.211 + 1.2) = 1.411 kWh Daily output = (30 + 24 + 56) = 110 kWh 110 All-day efficiency = = 0.9872 or 98.73%. 110 + 1.411 13. Let base kVA = 100 kVA Z1 = (0.02 + j 0.1) = 0.102 –78.69° Converting impedance of second transformer to base kVA Z2 =

1000 (0.015 + j 0.05) = (0.03 + j 0.1) = 0.104 –13.3° 500

Load (SL) = 1500 –0° kVA Hence, load shared by transformer 1 S1 = 1500 –0° ¥ =

0.104 –73.3∞ (0.02 + 0.03) + j (0.1 + 0.1)

1500 ¥ 0.104 –73.3∞ = 757.28 ––2.66° kVA 0.206 –75.96∞

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Load shared by transformer 2 S2 = 1500 –0° ¥

0.102 X –78.69∞ = 742.72 –2.73° kVA. (0.02 + 0.03) + j (0.1 + 0.1)

14. (a) Let x be the fraction of full-load kVA at which maximum efficiency occurs.. Then, x2 (1.5) = 1.2 fi

x=

1.2 = 0.894 1.5

Load kVA for maximum efficiency = 0.894 (100) = 89.4 kVA. (b) (i) For power factor 1 Output = 89.4 ¥ 1 = 89.4 kW Iron losses = 1.2 kW Copper losses = 1.5 kW 89.4 ¥ 100 = 97.07% Efficiency = 89.4 + 1.2 + 1.5 0.8 lagging power factor Output = 89.4 ¥ 0.8 = 71.52 kW 71.52 ¥ 100 = 96.36% Efficiency = 71.52 + 1.2 + 1.5 For 0.8 leading p.f., efficiency is 96.36% [as kW and losses are same for leading and lagging p.f.]. 15. Full-load secondary current = 30 ¥

200 = 15 A 400

Rp.u. =

15 ¥ 0.75 400

= 0.028 p.u. Xp.u. =

10 ¥ 1.5 = 0.0375 p.u. 400

Regulation = Rp.u. cos q2 – Xp.u. sin q2 = 0 fi

tan q2 =

0.028 = 0.746 0.0375

Power factor = cos q2 = 0.8 leading. [regulation can be zero when p.f. is leading]

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16. Full-load secondary current of transformer A =

150 ¥ 1000 225

= 666.67 A 225 666.67

ZA = (0.015 + j0.04)

= 0.005 + j0.0135 W Full-load secondary current of transformer B =

300 ¥ 1000 = 1395.35 A 215

Ê 215 ˆ ZB = (0.01 + j 0.04) Á Ë 1395.35 ˜¯ = 0.0015 + j 0.006 W Circulating current at no-load =

225 - 215 0.005 + j 0.0135 + 0.0015 + j 0.006

= 486.5 ––71.56° A \

IA = 486.5 A IB = –486.5 A V = 225 – (486.5 ––71.56°) (0.005 + j0.0135) = 225 – 7 ––1.88° = 218 V.

17. h =

V2 I 2 cos f2 V2 I 2 cos f2 + I 22 R2 + Pi

As load current constant, I22 R2 = constant \ \

I 22 R2 + Pi = constant = K 1

efficiency h = 1+

K V2 I 2 cos f2

So for maximum efficiency, denominator should be minimum. This is possible when cos f2 is maximum. Hence, maximum efficiency occurs when load p.f. is unity (resistive load). 18. At full-load, 0.8 p.f., the efficiency = 98.5% \

Pi + Pe (f – L) = 1218

(i)

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At half-load and unity p.f. efficiency = 99% \

Pi + 0.25 Pc (f – L) = 505

(ii)

Solving Eqs (i) and (ii), Pi = 267.3 W

Pc (f – L) = 950.7 W

Full-load current at secondary side =

100 ¥ 103 = 9.09 A 11000

At maximum efficiency, I2M = 9.09

267.3 A = 4.82 A. 950.7

19. h(f – L) = \

V2 I 2 cos f2 V2 I 2 cos f2 + Pi + Pc ( f - L ) Pi + Pc (f – L) = 55.56 hÊ 1

ˆ ÁË 2 f - L˜¯

\

=

(i) ÊI ˆ V2 Á 2 ˜ cos f2 Ë 2¯ 2

ÊI ˆ Ê 1ˆ V2 Á 2 ˜ cos f2 + Pi + Á ˜ Pc ( f - L ) Ë 2¯ Ë 2¯

Pi + 0.25 Pc (f – L) = 27.78

Solving (i) and (ii), we get, Pi = 18.52 W Efficiency at 75% of full-load =

(ii) Pc (f – L) = 37.04 W.

500 ¥ 3/ 4 2

3 Ê 3ˆ 500 ¥ + Pi + Á ˜ Pc ( f - L ) Ë 4¯ 4

Output at maximum efficiency = 500

= 0.905 p.u.

18.52 = 353.55 W 37.04

\ at maximum efficiency Pc = Pi Maximum efficiency =

353.55 = 0.9051 p.u. 353.55 + 18.52 + 18.52

20. hmax = 0.98 \ iron loss (Pi) = 653 W At maximum efficiency, x =

Pi Pc ( f - L )

= 0.8

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\

2/3:

Pc (f – L) = 1020 W

\

Rep.u. = Zep.u. =

Pc ( f - L ) d

= 0.0102

5 = 0.05 100

Xep.u. = 0.04895 Voltage regulation for lagging p.f. = Rep.u. cos fr + Xep.u. sin fr = 0.03753 p.u. = 3.753%. 21. Power factor cos f2 = 1. At maximum efficiency, Constant iron loss = variable copper loss. hmax =

V2 I 2 cos f2 25 ¥ 103 ¥ 1 = = 0.973 p.u. V2 I 2 cos f2 + Pi + Pi 25 ¥ 103 ¥ 1 + 2 ¥ 350 = 97.3%.

22. From OC test: core loss Pi = 100 W SC test: copper loss Pc = 200 W and short-circuit current = 11.4 A \ equivalent resistance (Re2) = Again, \

200 = 1.54 W (11.4) 2

V2 I2fL = kVA ¥ 103 I2fL =

5 ¥ 103 = 12.5 A 400

Copper loss at full-load (PcfL) I 22 fL ¥ Re2 = 240.6 W \ efficiency at full-load =

From short-circuit test, \

V2 I 2 cos f = 0.9296 p.u. V2 I 2 cos f + PcfL + Pi = 92.96% V2sc = 40 V 40 Ze2 = = 3.5 W 11.4 Re22 + Xe22 = Ze22

\

Xe2 = 3.15 W

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Approximate voltage regulation =

I2 [R2 cos f + Xe2 sin f] V2

= 0.0862 p.u. = 8.62%. 23. For a two winding transformer, VH IH = VL IL = Sin = Sout 400 IH = 5 ¥ 103 or, IH = 12.5 A

\ Similarly,

IL = 50 A

Figure 3.60 shows the winding diagram of two-winding transformer working as an autotransformer. IH 500 A

IL = 62.5 A VH = 500 V VL = 400 V

12.5 A

Gjh/! Djsdvju!ejbhsbn!pg!Qspc/!34

For an auto-transformer,

\

aA =

VH 500 = 1.25 = VL 400

IH =

IL I = L a A 1.25

Current through 400 V winding I = IL – IH = IL –

IL = 0.2 IL 1.25

Since the current rating of 400 V winding is 12.5 A

or

0.2 IL = 12.5 IL = 62.5

Load

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The kVA output of the auto-transformer VL I L 400 ¥ 62.5 = 25. = 1000 1000 24.

V1 V2 = = 12 N1 N 2 2200 / 183 12

or,

N1 =

and

N2 = 18.33 / 18

We have

V1 / E1 = 4.44 f N1 fmax fmax =

È V1 ˘ = 12˙ ÍE Î N1 ˚

12 4.44 ¥ 50

= 0.054 Wb \ cross-sectional area =

f max = 0.036 m2. Bmax

25. Z1 = r1 + j x1 = (0.1 + j 0.4) W Z2 = (0.006 + j 0.01) W Impedance referred to HV side 2 È ˘ Ê 1100 ˆ Z(HV) = Í0.1 + Á 0.006 ˙ + ˜ Ë 230 ¯ ÍÎ ˙˚

2 È ˘ Ê 1100 ˆ 0.01˙ j Í0.4 + Á ˜ Ë 230 ¯ ÍÎ ˙˚

= [0.237 + j 0.629] W Similarly, impedance referred to LV sides ÈÊ 230 ˆ 2 ˘ 0.1 + 0.006 ˙ + Z(LV) = ÍÁ ˜ ÍÎË 1100 ¯ ˙˚

ÈÊ 230 ˆ 2 ˘ 0.4 + 0.01˙ j ÍÁ ˜ ÍÎË 1100 ¯ ˙˚

= [0.0104 + j 0.0275] W Base impedance referred to HV winding ZB (HV) =

( kV ) 2 (1.1) 2 = 12.1 W = ( MVA ) 100 ¥ 10 -3

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Zp.u. (HV) =

0.237 + j 0.629 = (0.019 + j 0.052) 12.1 È ˘ (0.23)2 = 0.529˙ Íwhere Z B ( LV) = -3 100 ¥ 10 Î ˚

Zp.u. (LV) = 0.0196 + j 0.052. 26. Impedance referred to LV side. 2 È Ê 1ˆ ˘ Z(LV) = Í0.05 + Á ˜ 3˙ + Ë 10 ¯ ˙ ÍÎ ˚

2 È ˘ Ê 1ˆ j Í0.05 + Á ˜ 5.3˙ Ë 10 ¯ ÍÎ ˙˚

= 0.08 + j 0.103 W \

I2 =

20 ¥ 103 = 100 A 200

(a) Voltage drop at 0.8 p.f. lagging = 100 [0.08 ¥ 0.8 + 0.103 ¥ 0.6] = 12.58 V \ voltage regulation =

12.58 ¥ 100 = 6.29% 200

Voltage drop at unity p.f. = 100 [0.08 ¥ 1] = 8 V \ voltage regulation = 4% Voltage drop at 0.707 leading p.f. = 100 [0.08 ¥ 0.707 – 0.103 ¥ 0.707] = – 1.63 V \ voltage regulation = –0.815% (b) Secondary terminal voltage at 0.8 p.f. lagging V2 = (200 – 12.58) = 187.4 V V2 (at) unity p.f. = (200 – 8) = 192 V V2 (at) 0.707 leading p.f. (200 + 1.63) = 201.63 V. 27.

Z HV (W) I HV ( rated ) = 0.10 VHV ( rated ) ZHV (W) IHV (rated) = 0.10 VHV (rated) = 0.10 ¥ 4000 = 400 V = Vsc.

28. h(fL, up.f.) =

25 ¥ 103 ¥ 1 = 97.08% 25 ¥ 103 + 350 + 400 hÊ 1

ˆ ÁË 2 fL, upf ˜¯

=

25 ¥ 103 ¥ 1 / 2 = 96.5% 1 1 25 ¥ 103 ¥ + 350 + ¥ 400 2 4

Tpmvujpo!Nbovbm

(a) K =

350 = 0.935 400

Load for max.

h = 25 ¥ 0.935 = 23.385 kVA Pi = 350 W. Pc = (0.935)2 ¥ 400 = 350 W.

29. 0.985 =

500 ¥ 103 ¥ 0.8 500 ¥ 103 ¥ 0.8 + Pi +

0.988 =

1 P 4 cfL

1000 ¥ 1000 1000 ¥ 1000 + Pi + PcfL

Solving these, we get Pi = 4071 W PcfL = 8079 W and

K= hmax =

30. hmax =

4071 = 0.71 8079 1000 ¥ 1000 ¥ 0.71 = 98.9%. 1000 ¥ 1000 ¥ 0.71 + 2 ¥ 4071

20 ¥ 103 ¥ 1 = 0.98 20 ¥ 103 ¥ 1 + 2 Pi 2 kW, 0.6 p.f. (3.33 kVA), 12 h 2 ¥ 12 = 24 kWh (output) È Ê 3.33 ˆ 2 ˘ 200 Í1 + Á ˜ ˙ ¥ 12 = 2.47 kWh (loss) ÍÎ Ë 20 ¯ ˙˚ 10 kW, 0.8 p.f. (12.5 kVA), 6 h 10 ¥ 6 = 60 kWh È Ê 12.5 ˆ 2 ˘ 200 Í1 + Á ˜ ˙ ¥ 6 = 1.67 kWh (loss) ÍÎ Ë 20 ¯ ˙˚ 20 kW. 0.9 p.f. (22.22 kVA), 6 h 20 ¥ 6 = 120 kWh È Ê 22.22 ˆ 2 ˘ 200 Í1 + Á ˜ ˙ ¥ 6 = 2.68 kWh (loss) ÍÎ Ë 20 ¯ ˙˚

\

h(all day energy) =

204 = 96.77%. 204 + 6.82

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31. I1

100 I2 V1 = 500 V V2 = 400 V I2 – I1

Gjh/! Djsdvju!ejbhsbn!pg!Qspc/!42

1000 = 100 A 100 500 ¥ 100 = 50 = 1000

I1 = 10 ¥ (kVA)auto

hTW =

10 ¥ 103 ¥ 0.85 = 0.97 10 ¥ 103 ¥ 0.85 + PL

PL = 262.9 W \ full-load output as auto (0.85 p.f.) = 50 ¥ 0.85 = 42.4 kW 42.5 hauto = = 99.38%. 42.5 + 0.2629 32. I1

I2

V1 = 600 V

V2 = 200 V

N2 I2 – I1

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20 ¥ 103 = 50 A 400 600 ¥ 50 = 30 = 1000

I1 = (KVA)auto

I2 = \ 33. S1 = S2 =

20 ¥ 103 = 100 A 200

I2 – I1 = 50 A. Z2 0.015 SL = ¥ 25 = 15 kVA Z1 + Z 2 0.025 Z1 0.01 SL = 25 = 10 kVA Z1 + Z 2 0.025

(a) I1 =

15 ¥ 1000 10 ¥ 1000 = 68.2 A. I2 = = 45.6 A 220 220

15 = 75% 20 10 Percentage rated capacity used in transformer 2 = = 66.7%. 15

(b) Percentage rated capacity used in transformer 1 =

34. S1 (rated) = 1000 kVA S2 (rated) = 500 kVA Z1 = 0.02 + j 0.07 = 0.0728 –74° Z2 = (0.025 + j 0.0875) ¥ 2 = 0.182 –74° S1 =

Z2 SL Z1 + Z 2

SL = 1400KVA

S2 =

Z1 SL Z1 + Z 2

S2 = 3500KVA

As total load is increased the 1000KVA transformer will be the first to reach its full value. SL (max) = 1400KVA.

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Dibqufs!5 Uisff.qibtf!Usbotgpsnfst 1. V1L = 6600 V N1 = 12 N2 I1L = 20 A (a) For YY connection 6600 V 3 N = 1 = 12 N2

V1P = V1P V2 P \ Hence

V2P =

6600 V 12 3

V2L = 3 V2P =

6600 = 550 V 12

I1P = I1L = 20 A I1P N = 2 I 2P N1 or,

I2P = I1P

N1 = 20 ¥ 12 = 240 A N2

I2L = I2P = 240 A Output = 3V2L I2L = 3 ¥ 550 ¥ 240 VA = 228.63 kVA (b) For YD connection V1P =

6600 V 3

V1P N = 1 V2 P N 2 or,

V2P =

N2 6600 V V1P = N1 3 ¥ 12

V2L = V2P = 317.54 V

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I1L = IIP = 20 A N I2P = I1P ¥ 1 = 20 ¥ 12 A N2

\

I2L = 3I2P = 3 ¥ 240 = 415.69 A Output = 3 V2L I2L = 3¥ 317.54 ¥ 415.69 VA = 228.63 KVA (c) For DY connection V1P = V1L = 6600 V N 6600 V2P = V1P ¥ 2 = V N1 12 \

V2L = 3 V2P = 3 ¥ I2P = I1P

6600 = 952.6 V 12

N1 I1L N1 20 ¥ 12 A = = N2 3 N2 3

I2L = I2P =

240 A = 138.56 A 3

Output = 3V2L ¥ I2L = 3 ¥ 3 ¥ = 3¥

6600 240 VA ¥ 12 3

6600 ¥ 240 VA 12

= 228.63 KVA. (d) For DD connection V1L = V1P = 6600 V V2L = V2P = V1P ¥ I1P =

I1L 20 A = 3 3

I2P = I1P ¥ I2L =

N 2 6600 = V = 550 V N1 12

N1 20 ¥ 12 A = N2 3

3 I2P =

3¥

20 ¥ 12 = 240 A 3

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Output =

3 V2L I2L =

3 ¥ 550 ¥ 240 VA

= 228.63 KVA 2. V1L = 11000 V V2L = 440 V The transformer is DY. V1P V2 P = = 12 N1 N 2 N1 =

V1P V1L 11000 = 917 = = 12 12 12

V2 L V 440 = 21 N2 = 2 P = 3 = 12 12 3 ¥ 12 Flux density B = 1.2 Wb/m2 V1P = 4.44 f f N1 or, 11000 = 4.44 ¥ 1.2 A ¥ 50 ¥ 915 \

A = 0.0450 m2 = 450 cm2

3. The rating of a YYD connected transformer is 6600/400/110 V. Magnetizing current 5.5 A Load on secondary is 1000 kVA at 0.8 lagging power factor Load on tertiary is 200 kVA at 0.5 leading power factor Secondary line current I2L =

1000 ¥ 103 A 3 ¥ 400

Since the secondary is star connected, phase current I2P = I2L =

1000 ¥ 103 A 3 ¥ 400

400 1000 ¥ 103 ¥ 3 The secondary current referred to the primary I ¢2 = 3 ¥ 400 6600 3 =

1000 ¥ 103 A = 87.47 A 3 ¥ 6600

Tpmvujpo!Nbovbm

Tertiary line current I3L =

200 ¥ 103 A 3 ¥ 110

Since the tertiary is delta connected, phase current I3P =

I 3 L 200 ¥ 103 A = 3 ¥ 110 3

Tertiary current referred to the primary I¢3 =

200 ¥ 103 110 200 ¥ 103 A = 17.49 A ¥ A= 6600 3 ¥ 110 3 ¥ 6600 3

Primary current = 87.47 – cos -1 0.8 + 17.49 cos -1 0.5 + 5.5 90∞ = 78.721 – j 42.832 = 90 -28.55∞ A \ Power factor is cos 28.55° lag or 0.87 lag Primary KVA = 3 ¥ 6600 ¥ 90 ¥ 10–3 = 1030

2/4:

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Dibqufs!6 Gvoebnfoubmt!pg!bd!Spubujoh!Nbdijoft 1. P = 4 f = 50 Hz Number of slots S = 36 Number of conductors per slot = 30 \ Number of conductors = 36 ¥ 30 = 1080 1080 = 540 2 540 = 180 \ Number of turns per phase N = 3 36 Number of slots per pole per phase q = =3 4¥3 Total Number of turns =

Slot angle y =

180∞ 180∞ = = 20° Slots per pole 36/4

Hence, distribution factor 3 ¥ 20∞ qy sin 2 = 2 = 0.9598 Kd = y 20∞ q sin 3 sin 2 2 sin

Phase voltage E = 4.44 Kd f f N = 4.44 ¥ 0.9598 ¥ 0.05 ¥ 50 ¥ 180 = 1917.68 V \ Line voltage = 3E = 3322 V 120 f 120 ¥ 50 Synchronous speed Ns = = = 1500 rpm P 4 2. Number of slots S = 60 Number of poles P = 4 60 =5 4¥3 180∞ 180∞ = Slots angle y = = 12° Slots per pole 60 / 4 Slots per pole per phase q =

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Fundamental distribution factor 5 ¥ 12∞ qy sin 2 = 2 K d1 = = 0.9567 y 12∞ q sin 5 sin 2 2 sin

Slots per pole =

60 = 15 4

\ For full pitched coils if one coil lies in slot 1 the other side of the coil should be in slot 16. But here the other side lies in slot 13. Hence the coil is short pitched by 3 slots. \ Chording angle a = 3 ¥ 12° = 36° \ Pitch factor K P1 = cos

36∞ = 0.951 2

\ Fundamental winding factor Kw1 = K d1 K P1 = 0.9567 ¥ 0.951 = 0.91 For 3rd harmonic 3 ¥ 5 ¥ 12∞ 1 2 = = 0.647 = 3 ¥ 12∞ 1.545 5 sin 2 3a 3 ¥ 36∞ = cos = 0.5878 = cos 2 2 sin

K d3

K P3 \

Kw3 = K d3 K P3 = 0.647 ¥ 0.5878 = 0.38

3. P = 16 f = 0.06 Wb q =2 Conductors per slot = 4 Coil span = 150° Total number of slots s = 2 ¥ 16 ¥ 3 = 96 \ Total number of conductors Z = 4 ¥ 96 \ Number of turns per phase N=

4 ¥ 96 = 64 2¥3

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Slot angle r =

180∞ 180∞ = = 30° Slots per pole 2 ¥ 3

qy 2 ¥ 30∞ sin 0.5 2 = 2 = 0.966 Kd = = y 30∞ 0.5176 q sin 2 sin 2 2 sin

\

As coil span is 150°, chording angle a = 180° – 150° = 30° \ \

KP cos

30∞ = 0.9659 2

Phase voltage E = 4.44 Kd KP f f N = 4.44 ¥ 0.966 ¥ 0.9659 ¥ 0.06 ¥ 50 ¥ 64 = 795.4 V

4. P = 6 Ns = 1000 Ns = \

f=

120 f P P N s 6 ¥ 1000 = = 50 Hz. 120 120

Number of slots per pole per phase q = 3 Slots angle y = 20° 3 ¥ 20∞ qy sin sin 2 = 2 = 0.9598 \ Kd = y 20∞ q sin 3 sin 2 2 Number of slots in each phase S = 3 ¥ 6 = 18 Conductors per slot = 10 \ Total number of conductor in each phase = 18 ¥ 10 = 180 180 Number of turns per phase N = 2 Phase voltage E = 4.44 f f N Kd = 4.44 ¥ 20 ¥ 10–3 ¥ 50 ¥ = 383.5 V

180 ¥ 0.9598 2

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5. No. of poles P = 10 Slots per pole = 16 \ Total number of slots = 16 ¥ 10 = 160 In single layer winding machine the number of coils = \ Total number of coils =

1 (number of slots) 2

1 ¥ 160 = 80 2

Each coil has 20 turns \ Total number of turns = 80 ¥ 20 = 1600 1600 Number of turns per phase N = = 533 3 16 Number of slots per pole per phase q = 3 Slot angle y =

180∞ 180∞ = = 11.25° Slots/pole 16

16 11.25∞ ¥ 3 2 = 0.5 = 0.9566 Kd = 16 11.25∞ 0.5227 sin 3 2 sin

\

E = 4.44 Kd f f N

Phase voltage

= 4.44 ¥ 0.9566 ¥ 50 ¥ 0.025 ¥ 533 = 2829.77 V Total KVA = 3 ¥ 2829.77 ¥ 75 ¥ 10–3 = 636.5 6. Phase voltage E1 =

400 V 3

If E3 be the third harmonic voltage component then E12 + E32 = 244 2

\

Ê 400 ˆ E3 = ( 244) 2 - Á = 59536 - 53333 = 78.75 V Ë 3 ˜¯

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Dibqufs!7 Tzodispopvt!Hfofsbups 1500 = 131.2 A 3 ¥ 6.6

1. IL =

E2 = (3810 ¥ 0.8 + 131.2 ¥ 0.5)2 + (3810 ¥ 0.6 + 131.2 ¥ 5)2 or

E = 4283.6 V/phase or 7419 V line to line

% Rise in voltage = 2. Ia =

7419 - 6600 ¥ 100 = 12.4%. 6600

25 ¥ 103 = 36 A 3 ¥ 400 E2 = (231 ¥ 0.8)2 + (231 ¥ 0.6 + 432)2 = 598.8 V tan (d + q) =

138.6 + 432 = 3.09 184.8

q = 36.87°, d = 35.1°, E = 3 ¥ 598.8 = 1037 V E = 1.25 ¥ 598.8 = 748.5 V

Now,

Po = or

EVt sin d = 231 ¥ 36 ¥ 0.8 xs

748.5 ¥ 231 sin d = 231 ¥ 36 ¥ 0.8 12 d = 27.52° Maximum power =

\

748.5 ¥ 231 sin 90° = 14 kW/ph 12

total power = 14 ¥ 3 = 42 kW.

3. Vt = 1 p.u. \

Pm =

1 ¥ 1.2 = 0.857 p.u. = 0.857 ¥ 100 = 85.7 MW 1.4

60 1.2 ¥ 1 sin d = 100 1.4 \

d = sin–1 0.7 = 44.43°.

Tpmvujpo!Nbovbm

4. Vt = Ia =

2/56

6600 = 3810.5 V 3 500 = 43.73 3 ¥ 66

(a) When cos q = 1, E = (3810 ¥ 1 + 43.73 ¥ 0.5) 2 + ( 43.73 ¥ 5) 2 = 3838 V \ % regulation =

3838 - 3810.5 ¥ 100 = 0.72. 3810.5

(b) E = (3810.5 ¥ 0.8 + 43.73 ¥ 0.5) 2 + (3810.5 ¥ 0.6 + 43.73 ¥ 5) 2 = 3962.8 V 3962.8 - 3810.5 ¥ 100 = 3.9% 3810.5

% regulation =

(c) E = (3810.5 ¥ 0.8 + 43.73 ¥ 0.5) 2 + (3810.5 ¥ 6 - 43.73 ¥ 5) 2 = 3740.5 % regulation = 5. zs =

3740.5 - 3810.5 ¥ 100 = –1.83. 3810.5

1500 = 1.08 W 3 ¥ 800 2

Ê 11000 ˆ Ê 11000 ˆ E= Á ¥ 0.8˜ + Á ¥ 0.6 + 4000 ¥ 1.08˜ Ë 3 ¯ Ë 3 ¯ = 9586 V (a) P = 3 ¥ 11000 ¥ 4000 ¥ 0.8 = 60.97 MW 1100 ¥ 9586 3 W = 169 MW (b) Pm = 3 1.08 11000 3 sin d 1.08

9586 ¥ (c) 100 ¥ 106 = 3 ¥ \ d = 36.28°.

2

!

2/57

6. I =

Fmfdusjdbm!Nbdijoft

200 ¥ 103 = 192.45 A 3 ¥ 6000 6000 E= + 192.45 -36.87∞ ¥ j 6 = 4258.32 12.53∞ V 3 6000 3¥ ¥ 4258.32 4p 3 Ps = cos12.53∞ ¥ 6 180 = 502.8 kW/mech. degree Ts =

502.8 ¥ 1000 = 6400 Nm/mech. degree 750 2p ¥ 60

7. P = 9.5 kW + 500 W = 10,000 W 2 È ( 400) 2 ˘ 1 Ê 400 ˆ Ê 1 1 ˆ 2 ˙ 10,000 = 3 Í sin d + Á sin d 2 Ë 3 ˜¯ ÁË 3.2 5 ˜¯ ÍÎ 3 ¥ 5 ˙˚

1 = 3.2 sin d + 0.9 sin2 d

or,

(a) d = 11.623° As E = Vt, cos q is leading 400 400 cos 11.623∞ E - Vt cos d 3 = 0.947 A Id = = 3 5 Xd 400 sin 11.623∞ 3 Iq = = 14.541 A 3.2 Ia = I d2 + I q2 = 14.572 A Iq = Ia cos (d + q) d + q = cos–1 \

Iq Ia

= 3.738°

q = –7.885 \ p.f. = cos 7.885° = 0.99 lag

(b) For maximum power cos d = –

230.95 ¥ 3.2 ± 4 ¥ 230.95 ¥ 1.8

1 + (0.44) 2 2

Tpmvujpo!Nbovbm

d = 67°

or,

Maximum power developed =

3 ¥ ( 400) 2 ( 400) 2 sin 134° = 35.93 kW sin 67∞ + 3 3¥5 3¥ 2

Maximum power output = 35.93 – 8. Ia =

500 = 26.244 A 3 ¥ 11

500 = 35.43 kW 1000

2

Ê 26.244 ˆ (a) Armature coper loss = 3 ¥ Á ¥4 Ë 2 ˜¯ = 2066.24 W Total loss at half load = 1500 + 2500 + 2066.24 + 1000 = 7066.24 W \

Ê ˆ 7066.24 Á ˜ h = Á1 ˜ ¥ 100% = 96.587% 1 ÁË 500, 000 ¥ ¥ 0.8 + 7066.24 ˜¯ 2

(b) For maximum h, variable loss = constant loss 3I m2 ¥ 4 = 1500 + 2500 + 1000 = 5000 W

\ \

Im = 20.412 A

Output at maximum efficiency = 3 ¥

11000 ¥ 20.412 ¥ 0.8 3

= 311,111.54 W Total loss = 2 ¥ 5000 = 10,000 W \ 9. Ia =

Ê ˆ 10, 000 Max h = Á1 ¥ 100 = 96.886%. Ë 311,111.54 + 10, 000 ˜¯

70 ¥ 106 = 2928.59 A 3 ¥ 13800 tan d =

2928.59 ¥ 1.21 ¥ 0.8 = 0.28 13800 + 2928.59 ¥ 1.21 ¥ 0.6 3

2/58

!

2/59

Fmfdusjdbm!Nbdijoft

\

d = 15.69° Id = Ia sin (36.87° + 15.69°) = 2325.27 –74.31∞ A Iq = Ia cos (36.87° + 15.69°) = 1780.38 15.69∞ A E=

13800 13800 + 2325.27 –74.31∞ ¥ 1.83 + 1569∞ 3 3

= 11925.8 15.69∞ V \

regulation =

11925.8 - 7967.43 ¥ 100% = 49.68% 7967.43

P =3¥

10. Vt =

13800 ¥ 2928.59 ¥ 0.8 = 56 mW 3

120 10∞ ¥ j8 + 120 20∞ ¥ j 5 ¥ (4 + j3) = 82.17 -5.93∞ V ( 4 + j 3) (5 + j8) + j 5 ¥ j8 I a1 =

120 20∞ - 82.17 -5.93∞ = 9.36 -51.17∞ A j5

I a2 =

120 20∞ - 82.17 -5.93∞ = 7.31 -32.06∞ A j8

P1 = 3 ¥ 82.17 ¥ 9.36 cos (51.17° – 5.93°) = 1624.68 W P2 = 3 ¥ 82.17 ¥ 7.31 cos (32.06° – 5.93°) = 1617.82 W Total power P = P1 + P2 = 3242.5 W. 11. Slot per pole phase = Slot angle =

120 =5 8¥3

180∞ ¥ 8 = 12° 12

\

5 ¥ 12∞ 2 Kd = = 0.9567 12∞ 5 sin 2 sin

Total no. of conductors = 8 ¥ 120 = 960 Conductors per phase =

960 = 320 3

Tpmvujpo!Nbovbm

Number of turns per phase =

320 = 160 2

Generated voltage per phase E = 2 p ¥ 0.9567 ¥ 50 ¥ 0.05 ¥ 160 = 1699 V \ line value of E = 3 ¥ 1699 = 2942.8 V 700 12. zs = 3 = 1.075 W 376 ra = 1.5 ¥

0.16 = 0.12 2

xs = (1.075) 2 - (0.12) 2 = 1.068 W Now, \

1500 ¥ 103 = 3 ¥ 2300 Ia Ia = 376 A E=

2300 + 376 -36.87∞ ¥ (0.12 + j 1.068) 3

= 1631 10.39∞ V 2300 3 ¥ 100% = 22.8% 2300 3

1631 \

13.

regulation =

P = 8000 kW cos q = 0.8 Q=

8000 ¥ 0.6 = 6000 KVAR 0.8

PA = 5000 kW cos qA = 0.9 QA =

5000 sin (cos–1 0.9) = 2421.6 KVAR 0.9

2/5:

!

2/61

Fmfdusjdbm!Nbdijoft

\

PB = P – PA = 3000 W QB = Q – QA = 3578.4 KVAR 3578.4 = 1.1928 3000

\

tan qB =

\

cos qB = 0.64.

14. IL =

10, 000 ¥ 103 = 1093.5 A 3 ¥ 6600 ¥ 0.8

cos q = 0.8 sin q = 0.6 I1 = I2 =

1093.5 = 546.75 2

Active component of current = IL cos q = 874.8 A Reactive component of current = IL sin q = 656.1 A 874.8 = 437.4 A 2 Since steam supply is same, the active component remains same at 437.4 A Now I1 = 438 A Active component of current supplied by each generator =

\ reactive component of

I1 = ( 438) 2 - ( 437.4) 2 = 23 A

\ reactive component of

I2 = 656.1 – 23 = 633.1

\

I2 = ( 437.4) 2 + (633.1) 2 = 769.5 A

\ 15. 0.6 =

tan q2 =

633.1 or, cos q2 = 0.568 lag. 437.4

1.5 ¥ 1 sin d 1.2 sin d = 0.48, d = 28.68° dP = 1% of 0.6 = 0.006 p.u. Q=

\

EVt V2 cos d - t xs xs

dQ -EVt sin d = xs dd

Tpmvujpo!Nbovbm

Also

dP EVt cos d = xs dd

\

dQ = –tan d = –tan 28.68° = –0.547 dd

or,

dQ = –0.547 ¥ 0.006 = –0.328 ¥ 10–3

\ % decrease change in Q =

0.328 ¥ 10 -3 ¥ 100% = 0.125%. 1.5 ¥ 1 12 cos 28.68 1.2 1.2

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!

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Fmfdusjdbm!Nbdijoft

Dibqufs!8 Tzodispopvt!Npups cos q1 = 0.8 lag sin q1 = 0.6 P1 = 800 kW 800 S1 = = 1000 kVA 0.8 Q1 = 1000 ¥ 0.6 kVAR = 600 kVAR

1. \

\

200 kW = 222.2 kW 0.9 Total load = 800 + 222.2 = 1022.2 kW Input of motor =

cos q2 = 0.92 P2 = 1022.2 kW sin q2 = 0.39

\

S2 =

1022.2 kVA 0.92

q2 =

1022.2 ¥ 0.39 kVAR = 433.3 kVAR 0.92

Reactive power supplied by motor = 600 – 433.3 = 166.7 kVAR

kVAR 166.7 = 0.75 = kW 222.2

\

tan q2 =

\

cos q2 = 0.8 lead

kVA of motor =

2. Total output =

kW 222.2 = 277.75] = cos q 2 0.8

2500 ¥ 735.5 kW = 1838.75 kW 1000 Ia =

1838.75 ¥ 103 3 ¥ 2300 ¥ 1 2

= 461.5 A

Ê 2300 ˆ E= Á + ( 461.5 ¥ 1.85) 2 = 1578.7 V Ë 3 ˜¯

Tpmvujpo!Nbovbm

Maximum power output =

Maximum torque =

3Vt E = xs

3¥

2/64

2300 ¥ 1578.7 3 = 3399.5 ¥ 103 W 1.85

3399.5 ¥ 103 Nm = 108.26 ¥ 103 Nm] 2 ¥ 50 2p ¥ 20 1000 ¥ 103

Ia =

3.

3 ¥ 6.6 ¥ 103 ¥ 0.8

A = 109.35 A

When p.f. is 0.8 leading, Ia = 109.35 (0.8 + j 0.6) = 87.47 + j 656 A \

\

6600

= E -d ∞ + Ia (2 + j 20) 3 E sin d = 1880.6 E cos d = 4947.66 d = 20.8° and E = 5293.4 V

When input is 1500 kW, 6600 - 5293.4 -d ¢ 3 I¢a = 2 + j 20 I¢a cos q¢ = 18.86 – 263.35 cos (d ¢ + 84.29°) Again, \

3 ¥ 6600 ¥ Ia¢ cos q¢ = 1500 ¥ 103 Ia¢ cos q¢ =

1500 ¥ 103 3 ¥ 6600

A = 131.21 A

\

d¢ = 30.96°

Now

Vt = E -d ¢ + (I¢ cos q ¢ + j I¢ sin q¢) ¥ (2 + j 20)

\

I¢ sin q¢ = 49.565

or, Power factor cos q¢ = 0.935 lead]

tan q¢ =

49.565 = 0.377 131.2

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Fmfdusjdbm!Nbdijoft

EVt Vt 2 Ê 1 1 ˆ P= sin d + Á ˜ sin 2d 2 Ë Xq Xd ¯ Xd

4.

=

1.2 ¥ 1 1 Ê 1 1 ˆ sin 2d + Á 0.85 2 Ë 0.55 0.85 ˜¯

= 1.412 sin d + 0.321 sin 2d For maximum power,

dP =0 dd

\

dP = 1.412 cos d + 2 ¥ 0.321 cos 2d = 0 dd

\

d = 69.764° Pmax = 1.412 sin 69.764° + 2 ¥ 0.321 sin 2 ¥ 69.764° = 1.5332 p.u.] Id = Ia sin (q – d) and Iq = Ia cos (q – d)

5.

Vt sin d = Iq Xq Vt cos d = Id Xd \

tan d =

Iq X q Id X d

=

X q cos (q - d ) X d sin (q - d )

Under maximum power condition, d = 45° 1= \

X q cos(q - 45∞) X d sin (q - 45∞)

=

X q (sin q + cos q ) X d (sin q - cos q )

X d sin q + cos q = X q sin q - cos q Xd + Xq Xd - Xq

=

sin q = tan q] cos q

6. Ia = 50 0∞ A d = tan–1

Ia X q Vt - I a ra

= tan -1

50 ¥ 2.5 = 26.37° 480 - 50 ¥ 0.5 3

Tpmvujpo!Nbovbm

E=

480 3

2/66

cos 26.37° + 50 ¥ 3.5 sin 26.37° = 326 V

\ line value of emf = 3 ¥ 326 V = 564 V] E=

7.

6600 3

800 ¥ 103

-

3 ¥ 6600 ¥ 0.8

36.87∞ ¥ (2 + j 20)

= (4.724 – j 1.504) kV = 4.96 -17.66∞ kV Zs = 2 + j 20 = 20.1 84.3∞ 2

Now \

Ê 6.6 ˆ 6.6 ¥ cos 84.3∞ ¥ 4.96 Á ˜ 1200 Ë 3 ¯ 3 cos (84.3° + d) = 20.1 20.1 3 d = 26.1° 6.6 - 4.96 -26.1∞ 3 = 113 22.1∞ Ia = 20.1 84.3∞ cos q = cos 22.1° = 0.9265 lead]

8.

Zs = 0.8 + j 5 = 5.56 81.7∞ W E=

3.3 3

- 5.56 81.7∞ ¥

160 103

-36.87∞

= 1.42 -26.2∞ kV Line value of emf = 3 ¥ 1.42 kV = 2.46 kV d = 26.2° Power developed = 3 ¥ 1.42 ¥ 160 cos (–36.87° + 26.2°) = 670 kW Shaft power output = 670 – 30 = 640 kW Input power = 3 ¥ 3.3 ¥ 160 ¥ 0.8 = 731.5 kW h=

640 ¥ 100% = 87.5%] 731.5

!

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Fmfdusjdbm!Nbdijoft

Dibqufs!9 Uisff!Qibtf!Joevdujpo!Npups 120 ¥ 60 = 1200 rpm 6 \ Stator field rotates at 1200 rpm. Rotor field rotates in the air gap in the same speed.

1. Ns =

N(rotor speed) = Ns (1 – s) = 1200 (1 – 0.03) = 1164 rpm The rotor speed is then 1164 rpm. If frequency of rotor current is fr, fr = sfs = 0.03 ¥ 60 = 1.8 Hz Since rotor rotates at 1164 rpm while the speed of the rotor field is 1200 rpm, hence, the field speed with respect to the rotor is (Ns – N) i.e., 36 rpm]. 2. Pag = Pin – Pscu – Psc = 50 – 2 – 0.5 = 47.5 kW \

T=

Pag ws

=

47.5 ¥ 103 = 605 Nm. 120 ¥ 50 2p ¥ 8 ¥ 60

3. Pin = 50 kW; s = 0.04; Pscu = 1 kW \

Pag = Pin – Pscu = 49 kW Ê 1.96 ˆ kW per phase˜ Prcu = s ¥ Pag = 0.04 ¥ 49 = 1.96 kW Á = Ë ¯ 3

\

Ê 49 ˆ 1.93 Pm = Pag – Prcu = Á ˜ = 15.68 kW. Ë 3¯ 3

4. Pin = 25 ¥ 746 ¥ 10–3 = 18.65 kW \

Pag = 18.65 – 2 = 16.65 kW Pin = Pag – Prcu = 16.65 – 1 = 15.65 kW

E 5. Ns =

Prcu = s ¥ Pag; s =

Prcu 1000 = = 0.06, i.e. slip is 6% Pag 16650

120 f = 1000 rpm P Sfl (full-load slip) =

1000 - 960 = 0.04 1000

Tpmvujpo!Nbovbm

If r be the additional resistance per phase in rotor circuit, we can write

2/68

snew R2 + r = . sfl R2

Since the power input to the rotor and rotor current remain constant for constant torque and hence, from the relation, Slip =

Rotor Cu loss , we have Rotor input

snew 3I 22 ( R2 + r ) R2 + r = = . R2 sfl 3I 22 R2 Substitution of the values of sfl = 0.04 Snew = 6. Ifl =

1000 - 800 = 0.2 and R2 = 0.3, yields r = 1.2 W 1000

500 ¥ 746 = 76.78 A 3 ¥ 3300 ¥ 0.85

\ E Hence,

Ino-load = 0.3 ¥ 76.78 = 23.03 A Ilock rotor ∫ Istart Istart = 6 ¥ Ifl = 460.68 A

Apparent power drawn during locked rotor condition is PA = 3 ¥ VL ¥ Ist = 3 ¥ 3300 ¥ 460.68 = 2633 kVA 7. With reference to the equivalent circuit of the induction motor, the input impedance looking from the input side is jX o ( R2¢ + jX 2¢ ) Zin = (Rs + jXs) + R2¢ + jX 2¢ + jX o È j 65.6(0.742 + j 2.41) ˘ = Í(1.21 + j 3.1) + W = 5.75 –70.72° W 0.742 + j 2.41 + j 65.6 ˙˚ Î At start, s = 1.0. This means the load resistor in equivalent circuit is shorted, since 1 – s = 0 VL – L = 46.15 ––70.67° A \ Ist = 3 Zin At no-load, s = 0, i.e., the load element in the equivalent circuit is open. \ \

Zin (no-load) = (Rs + jXs) + jX0 = (1.21 + j68.7) W = 68.71 –89° W VL – L 460 –0∞ INL = = 3.87 ––89° A = 3 Zin (NL) 3 (68.71 –89∞)

!

2/69

Fmfdusjdbm!Nbdijoft

8. Ns = 1000 rpm; s =

Ns - N = 0.04 Ns

E rotor copper loss =

s ¥ Gross mechanical power developed in rotor hence, we can write 1- s

for this problem, 3I22 R2 + 200 =

3 ¥ 302 ¥ R2 = 774.17 R2 = 0.287 W

or, or, 9. Ifl =

0.04 (30 ¥ 746 + 1000) 1 - 0.04

10 ¥ 746 = 15 A 3 ¥ 400 ¥ 0.8 ¥ 0.9 Is/c at 160 V input = 7.0 A

\

Is/c at 400 V(L – L) is

400 ¥ 7.0 = 17.5 A 160

With star-delta starter, Istarting = I starting

\

I fl

=

1 1 ¥ I s/cf = ¥ 17.5 = 5.833 A 3 3

5.833 = 0.39 15

10. (a) Pin = 70 kW; Pscu (stator Cu loss) = 3Ifl2 ¥ Rs = 3 ¥ 802 ¥ 0.2 = 3.84 kW Psc (core loss in stator) = 2 kW \

Pag = Pin – Pscu – Psc = 64.16 kW

(b) Prcu = s ¥ Pag = 0.04 ¥ 64.16 = 2.57 kW (c) Pm = Pag – Prcu = 64.16 – 2.57 = 61.59 kW (d) T =

(e) h = =

Pag ws

=

64.16 ¥ 103 = 408.66 Nm 1500 2p ¥ 60

P0 Pin - Mech ◊ loss in rotor = Pin Pin 61.59 1.5 = 0.8584, i.e., 85.84% 70

Tpmvujpo!Nbovbm

11. Ns =

2/6:

120 f = 1500 rpm P N = 1440 rpm N -N = 0.04 = smax s= s Ns R smax = 2 X2

\ But \

X2 =

R2 0.5 = 12.5 W = smax 0.04

If s be the required slip (5%) then from the relation, T Tmax \ 12. Tsh =

=

2 ◊ s ◊ smax 2 ¥ 0.05 ¥ 0.04 = 0.9756 = 2 2 s + smax (0.05) 2 + (0.04) 2

T = 0.9756 ¥ 20 = 19.51 Nm P0 50 ¥ 746 = 201.91 Nm = w 2p ¥ 1765 60 Pm = P0 + Mechanical loss in rotor = 50 ¥ 746 ¥ 10–3 + 500 ¥ 10–3 = 37.8 kW Pm 37.8 ¥ 103 = 204.62 Nm = w 2p ¥ 1765 60 = Pm + Rotor copper loss

Tm = Pag

= 37.8 ¥ 103 + 800 = 38600 W = 38.6 kW T=

Pag ws

=

38.6 ¥ 103 = 204.89 Nm 1800 2p ¥ 60

120 ¥ 60 Ê ˆ = 1800 rpm˜ ÁËE N s ¯ 4

E stator copper loss is 1 kW \

Pin = 1 + Pag = 1 + 38.6 = 39.6 kW

13. Ns = 1500 rpm; N = Ns (1 – s) = 1440 rpm (E s = 0.04) T(full-load) =

25 ¥ 735.5 25 ¥ 735.5 = = 0.122 Nm 1440 w 2p ¥ ¥ 103 60

!

2/71

Fmfdusjdbm!Nbdijoft

At standstill the torque is Ts while at 4% slip it is T.

\

s0 R2 E12 R 2 + ( s0 X 2 ) 2 Ts , (s0) being the slip at starting. = 2 T sR2 E12 R22 + ( sX 2 )

With

s0 = 1, 2

2 Ê R2 ˆ 2 Ê 1ˆ 2 + s ÁË X ˜¯ ÁË 4 ˜¯ + (0.04) Ts R22 + ( sX 2 ) 2 2 = = 1.51 = = 2 2 2 T È ÈÊ 1 ˆ 2 ˘ ˘ s[ R2 + X 2 ] Ê R2 ˆ + 1˙ 0.04 ÍÁ ˜ + 1˙ s ÍÁ ÍË X 2 ˜¯ ÍÎË 4 ¯ ˙ ˙˚ Î ˚ R2 0.01 = 14. smax = = 0.2. X 2 0.05

The speed at a slip of 0.2 would then be 800 rpm as Ns = 1000 rpm. T

E

Tmax T

For this problem,

Tmax

=

2 ◊ s ◊ smax 2 s 2 + smax

=

2 ¥ 0.04 ¥ 0.2 = 0.385 (0.04) 2 + (0.2) 2

15. Rs = 0.5 W. W0 1500 = 0.13 = 3V0 I 0 3 ¥ 440 ¥ 15 3 Ic = I0 cos q0 = 15 ¥ 0.13 = 1.95 A

cos q0 =

If = I0 sin q0 = 15 ¥ 0.99 = 14.873 A \

R0 =

V0 440 / 3 = = 130.27 W Ic 1.95

X0 =

V0 440 / 3 = = 17.08 W Ic 14.873

Next, from locked rotor test, cos qs/c =

Ws/c 7000 = 0.421 = 160 3(Vs/c )(I s/c ) 3 ¥ ¥ 60 3

Tpmvujpo!Nbovbm

\

Rs + R¢2 =

2/72

Vs/c 160 / 3 0.421 = 0.648 cos qs/c = I s/c 60

R¢ = 0.648 – Rs = 0.148 W

or

Xs + X2¢ =

Vs/c 160 / 3 0.91 = 1.4 W sin qs/c = I s/c 60

Xs = X2¢, hence, Xs = X 2¢ =

Since

1.4 = 0.7 W. 2

The approximate equivalent circuit is shown below. R0 = 130.27 W; X0 = 17.08 W; Rs = 0.5 W, R¢2 = 0.148 W, Xs = 0.7 W; X 2¢ = 0.7 W 16. We have 2

Ts Ê I s ˆ = ◊s Tfl ÁË I fl ˜¯ fl Ê 1 ˆ (a) For Y-D starting: At start, winding is placed in star and only Á ˜ of normal voltage is Ë 3¯ applied. Again, short-circuit current being 6 times the full-load current (winding in delta connection), we can write short-circuit phase current as Is/c = \

Istarting/phse =

6 ¥ f.l. current 3 1 ¥ Is/c(ph) 3

1 of normal voltage is applied per phase, 3 1 6 i.e., Ist/ph = ¥ ¥ Ifl 3 3 as

\

I st 1 6 ¥ =2 = I fl 3 3

Hence,

Ts = (2)2 ¥ 0.05 = 0.2 Tfl

(b) Auto-transformer starting with 60% tapping: At start, stator winding remain in delta connection. However, only 60% voltage is made available at stator.

!

2/73

Fmfdusjdbm!Nbdijoft

\

Is =

Ts = (3.6)2 ¥ 0.05 = 0.648 Tfl

Hence, 17. P = 4 f = 50 Hz

60 ¥ 6 Ifl = 3.6 Ifl 100

r2 = 0.2 W Ns =

\

N1 = 1440 rpm s1 = 1 –

N1 Ns

120 ¥ 50 = 1500 rpm 4

s1 = 1 –

1440 = 0.04 1500

If Pag is the air gap power and I2 the rotor current then s1 Pag = I22 r2 I 22 ¥ 0.2 = 5I22 0.04

\

Pag =

New speed

N2 = 1200 rpm 1200 s2 = 1 – = 0.2 1500

New slip

Let R be the total resistance of the rotor circuit As torque is same Pag will have the same value s2 Pag = I22 R

\

0.2 ¥ 5 I22 = I22 R or,

R =1

Hence external resistance is (1 – 0.2) W or 0.8 W 18. r1 = 0.13 W x1 = 0.6 W

r2 = 0.035 ¥ 22 = 0.14 W

x2 = 0.15 ¥ 4 = 0.6 (a) Slip at maximum torque sm =

r2 r12 + ( x2 + x1 ) 2

= =

0.14 (0.13) 2 + (0.6 + 0.6) 2 0.14 1.21

= 0.1157

Tpmvujpo!Nbovbm

2/74

Hence slip at maximum torque is 11.57% Maximum torque

Tem =

3V 2 ws

r2 s

2

Ê r2 ˆ 2 ÁË r1 + s ˜¯ + ( x1 + x2 ) m

3 ¥ ( 400) 2 2p 120 ¥ 50 ¥ 0.14 60 8 = ¥ 2 0.1157 0.14 ˆ Ê 2 ÁË 0.13 + 0.1157 ˜¯ + (0.6 + 0.6) =

3 ¥ ( 400) 2 ¥ 0.373 = 2280 Nm 25p

(b) If Test be the starting torque then Test 2 = Tem s + 1 m sm \

Test =

2 ¥ 2280 = 520.62 Nm 1 0.1157 + 0.1157

19. P = 8 r2 = 0.03 W x2 = 0.15 W (a) Slip at maximum torque sm =

r2 0.03 = = 0.2 x2 0.15

Hence speed at maximum torque N = (1 – sm) Ns = (1 – 0.2) ¥

120 ¥ 50 = 600 rpm 8

(b) If Ts and Tm are the starting and maximum torque then Ts = Tm

2 sm¢ +

1 sm¢

=

3 4

!

2/75

Fmfdusjdbm!Nbdijoft

3s¢m2 – 8s¢m + 3 = 0 s¢m = 0.4517 If r be the external resistance then 0.03 + r = 0.4517 0.15 or, or,

0.03 + r = 0.06 r = 0.03 W

20. Motor input 3 ¥ 415 ¥ 50 ¥ 0.5 = 30548 W at full-load 2

Ê 50 ˆ Stator copper loss on full-load = 3 ¥ Á ˜ ¥ 0.5 Ë 3¯ = 1250 W No-load input power = Stator core loss + Friction and windage loss 3 ¥ 1500 = 900 W 5 2 Friction + Windage loss = ¥ 1500 = 600 W 5

\ stator core loss =

Power input to rotor Pag = 30548 – (1250 + 900) = 28398 \ rotor copper loss = sPag = 1136 W Shaft power = 28398 – (1136 + 600) = 26,662 W \

h=

26662 ¥ 100% = 87.28% 30548

Tpmvujpo!Nbovbm

2/76

Dibqufs!: Tjohmf!Qibtf!Joevdujpo!Npupst 1. r1 = 2.5 W

x1 = 1.25 W

r¢2 = 3.75 W

x¢2 = 1.25 W

Xf = 65 W

Core loss = 25 W Friction and windage loss = 2 W Ns = s=

120 ¥ 60 = 1800 rpm 4 1800 – 1710 = 0.05 1800

Ê 3.75 1.25 ¸ 1.25 ˆ 65 Ï 3.75 + j j 65 Á + j 2 j Ì ˝ ˜ 2 Ó 2( 2 - 0.5) 2 ˛ 2 ¯ Ë 0.05 ¥ 2 Z = 2.5 + j 1.25 + + 3.75 3.75 Ê 65 1.25 ˆ Ê 65 1.25 ˆ + jÁ + + jÁ + ˜ Ë 2 Ë 2 2( 2 - 0.05) 2 ˜¯ 2 ¥ 0.05 2 ¯ = 2.5 + j 1.25 + 15.822 + j18.524 + 0.925 + j 0.64 = 19.247 + j 20.414 W Input current

I1 =

120 = 4.277 -46.69∞ 19.247 + j 20.414

I2b

j 65

¥ 4.277 = 2.778 3.75 + j (1.25 + 65) 0.05 j 65 = ¥ 4.277 = 4.195 3.75 + j (1.25 + 65) 1.95

I2f =

Air gap power of forward field Pag f = (2.778)2 ¥

3.75 = 289.4 W 2 ¥ 0.05

Air gap power of backward field Pag b = (4.195)2 ¥

3.75 = 16.9 W 2 ¥ 1.95

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Fmfdusjdbm!Nbdijoft

The net air gap power is Pag = 289.4 – 16.9 = 272.5 W Gross power developed Pd = (1 – 0.05) ¥ 272.5 = 258.9 W Net output power Po = 258.9 – 25.2 = 231.9 W Efficiency =

231.9 = 0.6586 or 65.86% 352.1

Shaft torque =

2. Ns =

231.9 = 1.295 Nm 2p ¥ 1710 60

1200 ¥ 60 = 1800 rpm 4

(i) Slip in the forward direction s=

1800 - 1710 = 0.05 1800

(ii) Slip in the backward direction sb = 2 – 0.05 = 1.95 Effective rotor resistance in the forward branch is r2 12.5 = = 125 W 2 s 2 ¥ 0.05 Effective rotor resistance in the backward branch is 3. r1 = 2.2 W

r¢2 = 4.5 W

x1 = 3.1 W

x¢2 = 2.6 W

r2 12.5 = = 3.205 W 2( 2 - s) 2 ¥ 1.95

Xf = 80 W The friction, windage and core loss = 40 W s = 0.03 Ê 4.5 2.6 ˆ ÁË 2 ¥ 0.03 + j 2 ˜¯ Z = 2.2 + j 3.1 + 4.5 2.6 + j + 2 ¥ 0.03 2

80 Ê 2.6 ˆ 80 4.5 + j j 2 ÁË 2( 2 - 0.03) 2 ˜¯ 2 + 80 4.5 2.6 80 + j + j j 2 2( 2 - 0.03) 2 2 j

Tpmvujpo!Nbovbm

= 2.2 + j 3.1 + 16.37 + j 30.98 + 1.07 + j 1.29 = 19.64 + j 35.37 = 40.457 60.96∞ W (a) Input current I1 =

230 0∞ = 5.685 -60.96∞ 40.457 60.96∞

(b) Power factor = cos (–60.956°) = 0.4856 lagging (c) Developed power = I12 (Rf – Rb) (1 – s) = (5.685)2 (16.37 – 1.07) (1 – 0.03) = 479.65 W (d) Output power = 479.65 – 40 = 439.65 W Input power = VI1 cos q = 230 ¥ 5.685 ¥ 0.4856 = 634.9 W output 439.65 = 0.692 p.u. = input 634.9

(e) Efficiency =

4. From blocked rotor test Vsc = 120 V, Isc = 9.6 A, Psc = 460 W Zsc =

Vsc 120 = = 12.5 W I sc 9.6

Rsc =

Psc 460 = 4.99 W = 2 I sc (9.6) 2

Xsc = Zsc2 - Rsc2 = 11.46 W x1 = x¢2 =

1 1 Xsc = ¥ 11.46 W = 5.73 W 2 2

r1 = 1.5 r¢2 = Rsc – r1 = 3.49 W

\ From no load test

Vnl = 220 V Inl = 4.6 A Pnl = 125 W No load power factor cos qnl =

125 = 0.1235 220 ¥ 4.6

2/78

!

2/79

Fmfdusjdbm!Nbdijoft

sin qnl = 0.9923 Znl =

Vnl 220 = = 47.83 W I nl 4.6

Xnl = Znl sin qnl = 47.83 ¥ 0.9923 = 47.46 W = Xm Core, friction and windage losses = Power input at no load – no load copper loss r¢ ˆ 3.49 ˆ Ê Ê = Pnl – Inl2 Á r1 + 2 ˜ = 125 – (4.6)2 Á1.5 + = 74.8 W Ë Ë ¯ 4 4 ˜¯ 5. r1 = 2.3 W

x1 = 3.2 W,

r¢2 = 4.2 W,

x¢2 = 3.2 W,

Xm 2 Zf = Rf + jXf = r2¢ + 2s j

Ê r2¢ ÁË 2 s + x¢ j 2+ 2

Xm = 74 W,

x2¢ ˆ 2 ˜¯ X j m 2 j

37 Ê 4.2 3.2 ˆ + j Á 2 Ë 2 ¥ 0.05 2 ˜¯ = 27.23 49.64 = 4.2 3.2 74 + j + j 2 ¥ 0.05 2 2 j

= 17.66 + j20.75 W. Backward field impedance is X m Ï r2¢ x¢ ¸ + j 2˝ Ì 2 Ó 2( 2 - s) 2˛ Zb = Rb + jXb = Xm r2¢ x2¢ + j + j 2( 2 - s) 2 2 j

37 Ï 4.2 3.2 ¸ + j Ì ˝ 2 Ó 2 ( 2 0.05) 2 ˛ = 4.2 3.2 74 + j + j 2 ( 2 - 0.05) 2 2 j

= 1.85 57.6∞ = 0.99 + j 1.562 W Total motor impedance = 2.3 + j 3.2 + 17.66 + j20.75 + 0.99 + j1.562 = 20.95 + j 25.5 = 33.1 50.4∞ W

s = 0.05

Tpmvujpo!Nbovbm

Stator current I1 =

230 A = 6.95 –50.4∞ A 33.1 50.4∞

Power factor cos 50.4° = 0.637 lagging Power input = V1 I1 cos q = 230 ¥ 6.95 ¥ 0.637 = 1020 W Power delivered to forward field Pgf = I12 Rf = (6.95)2 ¥ 17.66 = 854 watts Power delivered to the backward field Pgb = I12 Rb = (6.95)2 ¥ 0.99 = 47.8 watts Total electrical power developed = 854 – 47.8 = 806.2 watts Total mechanical power developed = (1 – s) (Pgf – Pgb) = 0.95 ¥ 806.2 = 765 watts Power output = 765 – 98 – 30 = 637 watts Torque =

Power output = Rotor speed

= 4.27 Nm Efficiency =

637 120 ¥ 50 2p (1 - 0.05) ¥ ¥ 4 60

637 ¥ 100% = 62.5% 1020

2/7:

Joefy A

B

Accelerated commutation 2.47 ac commutator motors 10.1 ac excitation system 6.6 Air-gap power 8.20, 8.28, 8.40, 8.41, 8.44 All-day efficiency 3.108 Alternator 1.4, 6.1, 6.6, 6.11, 6.13, 6.16, 6.17, 6.20, 6.22, 6.24, 6.27, 6.29–6.31, 6.33–6.35, 6.41, 6.43, 6.44, 6.59, 6.78, 6.97, 6.98, 6.106, 6.109, 6.111, 6.114, 6.118–6.122 Amortisseur winding 6.91 Ampere-turn method 6.24 Armature 1.2–1.4, 1.8, 1.46–1.50, 2.5–2.27, 2.29–2.44, 2.48–2.55, 2.57, 2.58, 2.60–2.63, 2.65–2.67, 2.69–2.94 Armature control method 2.96 Armature reaction 6.17–6.20, 6.24–6.26, 6.34, 6.36, 6.40, 6.69, 6.71, 6.77, 6.78, 6.109, 6.122 Armature reaction 2.37, 2.39–2.41, 2.55 Armature winding 6.1, 6.4, 6.5, 6.17, 6.22, 6.24, 6.30, 6.80, 6.81, 6.92–6.94, 6.96 Asynchronous motors 8.1 Audio frequency transformer 4.52 Auto-transformer 3.139, 3.143–3.146 Autotransformers 4.47 Auto-transformer starter 8.90 Auxiliary windings 9.14, 9.22, 9.29

Back pitch 2.9, 2.14 Back-to-back 2.178 Back to back test 3.107 Backward field 9.3, 9.4, 9.6, 9.7, 9.9, 9.12, 9.22– 9.24, 9.30, 9.32 Belt harmonics 5.10 Blocked-rotor test 8.83 Blocked-rotor test 8.84, 9.10 Brake test 2.180 Breadth factor 5.7, 5.8, 5.23 Breadth factor 5.7, 5.12 Breakdown torque 8.55, 8.57, 8.58 Breathers 3.153 Bright-lamp method 6.59 Brushless dc motors 10.19 Brushless excitation system 6.7 Brush shifting 2.42

C Capacitor-start capacitor-run motor 9.14, 9.17 Capacitor-start motor 9.16 Cascade connection 8.106 Change of slip 8.104 Change of supply frequency 8.102 Chorded coil 5.3, 5.9 Chording angle 5.3, 5.9, 5.10, 5.12, 5.13, 5.45–5.47

!

J/3

Joefy

Circle diagram 8.110 Coenergy 1.66–1.68, 1.70, 1.71, 1.90, 1.96 Coercive force 1.55 Cogging 10.14, 8.108 Coil pitch 2.7 Coil span 5.3, 5.4, 5.43, 5.45–5.47 Commutating poles 2.5, 2.49 Commutation 2.44, 2.45, 2.46, 2.47, 2.48 Commutator 1.2, 1.3, 1.47, 2.2, 2.3, 2.5–2.10, 2.14–2.16, 2.39, 2.42, 2.44–2.46, 2.48, 2.50, 2.54 Commutator pitch 2.10, 2.15 Compoles 2.49 Compound excitation 1.3 Compound excitation 2.18 Compound-source rectifier system 6.8 Conductively compensated type of motor 10.4 Consequent pole motors 8.8 Conservators 3.153 Control winding 10.16, 10.17 Copper loss 1.60 Copper losses 3.79, 3.81, 3.84, 3.88 Core-type transformer 4.2, 4.3 Core-type transformer 3.5 Crawling 8.108 Critical field resistance 2.52, 2.69 Critical speed 2.53 Cross-field theory 9.5 Cross-magnetizing armature reaction 2.39, 2.40 Cross magnetizing component 2.39 Current transformer 3.150 Cylindrical-rotor 1.5 Cylindrical rotor machine 6.3, 6.17, 6.83, 6.91

D Damper winding 7.5, 7.6, 7.26, 7.46 Damper winding 6.91 Damping torque 6.91 Dark-lamp method 6.59 d axis 6.77 dc excitation systems 6.5 dc generator 2.1, 2.2, 2.6, 2.17, 2.23, 2.26, 2.27, 2.30, 2.33, 2.37, 2.44, 2.50–2.53, 2.70

dc machine 1.2 dc motor 2.1, 2.4, 2.17, 2.29, 2.70–2.72, 2.89, 2.90 Deep bar rotor 8.114 Delayed commutation 2.46 Delayed commutation 2.47 Delta-delta connection 4.10 Demagnetizing armature reaction 2.39, 2.87 Diametrical connection 4.19 Direct-axis subtransient reactance 6.93 Direct on-line starting 8.87 Distributed winding 2.7 Distributed winding 5.5 Distribution factor 5.8–5.11, 5.13, 5.14, 5.18, 5.23, 5.40, 5.45 Distribution transformer 3.8 Distribution transformers 1.7 Double-cage rotor 8.114, 8.115 Double-delta connection 4.19, 4.21 Double-layer winding 2.8, 2.9, 5.2, 5.12, 5.40 Double-star connection 4.19, 4.20 Doubly excited 1.3, 6.1 Drag-cup rotor 10.16 Dummy coil 2.15 Duplex winding 2.9 Dynamically induced emf 1.22 Dynamic braking 2.176

E Eddy current 1.57, 1.58 Eddy current loss 1.58 Eddy current losses 3.79 Efficiency 3.8, 3.79, 3.80–3.91, 3.95, 3.96, 6.5, 6.58, 6.95, 6.96, 6.100, 6.114 Efficiency of electrical machine 1.60 Electrical degrees 5.1, 5.3, 5.7, 5.21 Electromagnetic torque 5.25 Emf equation 6.10, 6.11 Emf injection in secondary circuit 8.105 Emf method 6.22 Equalizer connection 2.12, 2.13, 2.16 Equalizer rings 2.13, 2.16 Equivalent circuit of an induction motor 8.15

Joefy

Equivalent circuit of a transformer 3.28 Excitation circle 6.75, 6.76 Excitation power 6.5, 6.6, 6.8, 6.83, 6.84 External characteristic 6.45

J/4

Hysteresis loss 1.55 Hysteresis losses 3.79, 3.80 Hysteresis motor 10.17

I F Faraday’s laws 1.18 Field flux control method 2.100 Field poles 1.2–1.4 Fleming’s left-hand rule 1.24 Fleming’s right-hand rule 1.24, 1.25, 1.44 Fleming’s rules 1.24 Flux density 1.14 Forward field 9.3, 9.4, 9.6, 9.7, 9.9, 9.12, 9.22, 9.24, 9.30 Four-point starter 2.167 Fractional-pitch windings 5.4 Fractional slot windings 5.5 Frame 6.2 Frequency changer 8.122 Frog-leg winding 2.16 Frog-leg winding 2.16 Front pitch 2.9, 2.14 Full-pitch coil 5.3, 5.9, 5.40

Ideal transformer 3.7 Impedance transformation 3.9 Induced current 1.18 Induced emf 1.18–1.23, 1.25, 1.32, 1.44, 1.47, 1.48, 1.50 Induced emf 1.21, 1.22, 1.28, 1.32 Induction generator 8.123 Induction machine 1.5 Induction motors 8.1 Induction regulator 4.45, 4.46, 8.121 Induction torque 6.91 Inductively compensated universal motor 10.4 Inner displacement angle 6.18 Inrush current 4.42 Instrument transformer 3.150 Integral slot windings 5.4 Internal power-factor angle 6.18 Interpoles 2.5, 2.42, 2.49, 2.50, 2.80 Inverted V curve 6.74 Iron losses 1.60, 3.79, 3.84, 3.85

G Generalized transformer 8.6 Generator 1.1, 1.3, 1.5, 1.6, 1.12, 1.44–1.48, 1.50 Geometrical neutral axis 2.37 Grounding transformer 4.53

H Harmonic phenomena 4.40 Harmonic torques 8.58 High-torque induction motors 8.114 Hopkinson’s test 2.178 Hunting 5.32, 7.25, 7.26, 7.40, 7.45 Hybrid stepper motor 10.25 Hydrogen cobled generators 6.9 Hysteresis coefficient 3.79 Hysteresis loop 1.55, 1.72, 1.73, 1.97

L L 4.10 Lap winding 2.5, 2.10–2.14, 2.16, 2.21, 2.24, 2.25, 2.72 Lap winding 2.10, 2.11, 2.12 Leakage flux 2.17 Leakage reactance 3.23 Lenz’s law 1.18 Level or flat-compounded generator 2.19 Linear commutation 2.46 Linear induction motor 10.11 Load angle 7.15, 7.16, 7.25, 7.46 Long-shunt 2.28, 2.29 Losses in a synchronous generator 6.94 Low-voltage release coil 2.166

!

J/5

M Magnetic field intensity 1.14 Magnetic field system 2.4 Magnetic hysteresis 1.54 Magnetic locking 8.109 Magnetic neutral axis 2.37 Magnetic permeability 1.15 Magnetization current 3.32 Magnetizing branch 3.29, 3.31, 3.58 Magnetomotive force 1.14 Main 9.2, 9.8, 9.9, 9.11, 9.14, 9.15, 9.21, 9.26, 9.28–9.32 Maximum torque 7.12, 7.41 Maximum torque 8.52 Mechanical degrees 5.6, 5.7, 5.21, 5.46 Mmf method 6.22, 6.24, 6.38, 6.121 Mmf of ac windings 5.15 Mmf of concentrated coil 5.15 Mmf of distributed coil 5.17 Mmf of three-phase windings 5.19 Moment of inertia 1.11 Motor 1.1, 1.3, 1.5, 1.6, 1.8, 1.12, 1.13, 1.24, 1.27, 1.44, 1.47–1.49 Multiplex winding 2.9, 2.10 Mutually induced emf 1.21, 1.22

N No-load operation 6.17 No-load operation 3.22 No-load test 8.83 No-load test 9.11 Normal excitation 6.68, 6.70, 6.71, 6.73 Normally closed contacts 2.171 Normally open contacts 2.171 No-volt release 2.166, 2.167, 2.168

O Off-load tap changer 4.43 On-load tap changer 4.44 Open-circuit characteristic (OCC) 6.23 Open-circuit test 4.51

Joefy

Open-circuit test 3.104 Operating chart 6.76 Over-compounded generator 2.19 Over-excited 6.57, 6.68, 6.71 Overload release 2.166, 2.167, 2.188

P Parallel operation 3.136 Parallel operation 6.58, 6.61 Parallel operation of dc generator 2.58 Parallel operation of three-phase transformers 4.40 Parasitic torques 8.57 Permanent magnet dc (PMDC) motors 10.26 Permanent magnet generators 10.27 Permanent magnet stepper motor 10.22 Per-unit expression 3.67 Phase belt 5.2 Phase spread 5.2 Phasor groups 4.4 Pilot exciter 6.5, 6.7 Pitch factor 5.9, 5.11, 5.13, 5.42 Plugging 2.175, 2.194 Polarity test 3.108 Pole changing 8.102, 8.103 Pole pitch 5.2, 5.3, 5.5, 5.8, 5.10, 5.14, 5.23 Pole shoe 2.4, 2.8, 2.25, 2.26 Positive, negative and zero sequence currents 6.93 Potential-source-controlled rectifier system 6.8 Potential transformer 3.151 Potier method 6.24 Potier triangle 6.25, 6.26, 6.36, 6.38, 6.40, 6.121 Power angle characteristic 7.12, 7.40 Power-angle curve 6.48, 6.84 Power angle or load angle 6.48 Power transformer 3.8 Power transformers 1.7 Primary winding 1.7 Progressive lap winding 2.10 Pulling out of step 5.32 Pull-out torque 5.32, 7.12, 7.23, 8.55 Pull-up torque 8.58 Pulse transformer 3.146, 3.147

Joefy

Q q axis 6.77

R Reactance voltage 2.44, 2.49 Rectifier transformer 4.53 Reference winding 10.15–10.17 Regenerative braking 2.176 Regenerative test 2.178, 3.107 Reluctance 1.15, 1.30, 1.34, 1.36, 1.41 Reluctance motor 10.13 Reluctance power 6.83, 6.84, 6.86, 6.87 Reluctance torque 10.13, 10.30 Reluctance torque 5.26 Repulsion motor 10.5 Resistive commutation 2.46 Retrogressive lap winding 2.10 Reversal of rotation 8.93 Reversible-type motor 9.18 Revolving-field theory 9.2 Rheostatic braking 2.61 Rotating rectifier system 6.6 Rotational emf 1.23 Rotor 1.2, 1.4–1.6, 1.8, 1.13, 1.47, 1.49, 6.1–6.7, 6.10, 6.17, 6.18, 6.21, 6.22, 6.49, 6.63, 6.77, 6.80, 6.81, 6.83–6.86, 6.91, 6.93, 6.98, 6.106, 6.107, 6.109, 6.120, 6.121, 8.1–8.23, 8.25–8.28, 8.30–8.48, 8.50, 8.51, 8.53–8.64, 8.67–8.79, 8.81–8.83 Rotor resistance control 8.104 Rotor resistance starter steps 8.91

S Salient-pole 1.5 Salient-pole machine 6.3, 6.4, 6.77, 6.84, 6.85 Salient poles 2.4 Scott connection 4.12, 4.16 Secondary winding 1.7 Self-excitation 1.3 Self-excitation 2.18 Self-induced emf 1.21, 1.22

J/6

Separate excitation 1.3 Separate excitation 2.18 Separately excited dc motor 2.29 Series excitation 2.18 Series excited dc motor 2.29 Series field 1.3, 1.50 Series motor 10.1, 10.4, 10.7–10.9, 10.28, 10.30, 10.31, 10.33 Servomotor 10.15 Shaded-pole motors 9.19 Shading coil 9.19, 9.20, 9.21 Shell-type transformer 4.2, 4.3 Shell-type transformer 3.6, 3.7 Short-circuit ratio 6.27 Short-circuit test 3.105, 4.49 Short pitch 5.3, 5.4 Short-shunt 2.27, 2.29 Shunt excitation 2.18 Shunt-excited dc motor 2.29 Shunt field 1.3, 1.49, 1.50 Simplex winding 2.9, 2.10, 2.14 Single-layer winding 2.8, 5.2, 5.3, 5.11, 5.41 Single-phase induction motors 9.1, 9.14, 9.22 Single-phase synchronous motors 10.9 Six-phase star connection 4.19, 4.22 Skewing 8.109 Slip 8.7–8.9, 8.11, 8.25, 8.28, 8.29, 8.31, 8.33, 8.34, 8.36, 8.37, 8.47–8.49, 8.50, 8.52, 8.55, 8.59, 8.62, 8.64–8.66, 8.73, 8.78, 8.82 Slip-ring 1.5 Slip ring induction motor 8.2, 8.69 Slip speed 8.8 Slip test 6.79, 6.82, 6.86, 6.109 Slot angle 5.2, 5.7, 5.11, 5.17 Speed control of dc motor 2.96 Speed emf 1.23 Split-phase induction motors 9.15 Squirrel cage 1.5 Squirrel-cage induction motor 8.2, 8.49, 8.54, 8.56 Star-delta or delta-star connection 4.9 Star-delta starter 8.88

!

J/7

Star-star connection 4.10 Starter 2.164, 2.165, 2.167, 2.168, 2.170 Starting torque 8.51, 8.54 Statically induced emf 1.21 Stationary rectifier system 6.6 Stator 1.2, 1.4–1.6, 1.13, 8.1–8.3, 8.5–8.12, 8.14– 8.20, 8.23, 8.25, 8.28, 8.30–8.38, 8.40–8.42, 8.44–8.51, 8.53, 8.57, 8.58, 8.61, 8.64, 8.65, 8.67–8.69, 8.74, 8.76, 8.77, 8.79, 8.81–8.83 Stator core 6.2, 6.3, 6.9 Stator voltage control 8.104 Steady-state stability limit 6.48 Step angle 10.23, 10.24, 10.32 Step-down auto-transformer 3.139 Step down transformer 1.7 Stepper motors 10.22 Step-up auto-transformer 3.143 Step up transformer 1.7 Stiffness ratio of the machine 6.85 Stray load losses 1.60 Subtransient current 6.92 Subtransient reactance 6.92, 6.93 Sumpner’s test 3.107 Swinburne’s test 2.176, 2.178 Switched reluctance motor 10.14 Synchronizing of generators 6.59 Synchronizing power 6.59, 6.61, 6.85, 6.86, 6.89, 6.90, 6.100, 6.109, 6.112 Synchronizing power co-efficient Psy 6.85 Synchronous capacitor 7.26, 7.47 Synchronous condenser 7.26, 7.27, 7.40 Synchronous generator 6.1–6.4, 6.8, 6.10, 6.12, 6.13, 6.17–6.22, 6.26–6.28, 6.30, 6.32, 6.33– 6.35, 6.38, 6.42, 6.43, 6.45, 6.47, 6.48, 6.50, 6.51, 6.53–6.58, 6.67–6.69, 6.73, 6.74, 6.76– 6.79, 6.82, 6.84, 6.86, 6.88, 6.89, 6.91, 6.92, 6.94–6.96, 6.98–6.101, 6.106, 6.107, 6.109– 6.112, 6.114–6.116, 6.118, 6.119, 6.121, 6.122 Synchronous impedance 6.19, 6.20, 6.22, 6.23, 6.28, 6.30, 6.32, 6.38, 6.39, 6.41, 6.54, 6.55, 6.66, 6.67, 6.69–6.76, 6.78, 6.95, 6.103, 6.106, 6.107, 6.109, 6.122

Joefy

Synchronous impedance method 6.22 Synchronous induction motor 10.10 Synchronous machine 1.3 Synchronous motor 7.1–7.11, 7.13–7.16, 7.18, 7.19, 7.21–7.23, 7.25–7.30, 7.32, 7.35, 7.37–7.47 Synchronous reactance 6.19, 6.22–6.24, 6.27, 6.29, 6.30, 6.32, 6.33, 6.43–6.45, 6.48, 6.50, 6.51, 6.53, 6.56, 6.57, 6.77, 6.78, 6.81, 6.89, 6.92, 6.95, 6.98, 6.110, 6.111, 6.116, 6.118, 6.119, 6.122 Synchronous speed 1.4, 1.6, 6.1, 6.10, 6.17, 6.18, 6.22–6.24, 6.61, 6.80, 6.81, 6.85, 6.91, 6.93, 6.121

T Tachometer 10.16, 10.17 T-connection 4.12 Teaser transformer 4.12–4.15, 4.18, 4.20, 4.32, 4.35, 4.39, 4.86 Tertiary winding 4.10, 4.23, 4.41, 4.48, 4.49, 4.58, 4.59, 4.73, 4.79, 4.81, 4.89 The dc test 8.85 Three-phase to six-phase conversion 4.19 Three-phase to twelve-phase conversion 4.22, 4.23 Three-phase to two-phase conversion 4.12 Three-phase transformer 4.1, 4.3 Three point starter 2.165 Three-winding transformers 4.48 Tooth harmonics 5.33 Torque 1.6, 1.9, 1.10, 1.26, 1.27, 1.31, 1.49 Torque expression of induction motor 8.51 Transformer 3.1, 3.4–3.53, 3.55–3.67, 3.69–3.77, 3.79–3.96 Transformer 1.7, 1.21 Transformer cooling 3.152 Transformers 1.1, 1.7, 1.13, 1.17, 1.19 Transformer tappings 4.43 Transient current 6.92 Transient reactance 6.92 Two-reaction theory 6.77

Joefy

J/8

U

W

Under-compounded generator 2.19 Under-excited 6.68–6.70, 6.72 Universal motors 10.1, 10.3

Ward leonard method 2.102 Wave winding 2.5, 2.10, 2.12–2.16, 2.20, 2.24, 2.25, 2.72 Wave winding 2.14, 2.15 Welding transformer 3.147 Windage losses 1.60 Wound-type rotor 8.2

V Variable reluctance stepper motor 10.23 V curve 7.25 V curves 6.72, 6.73, 6.74 Voltage commutation 2.50 Voltage regulation 6.20, 6.26, 6.29, 6.30, 6.33, 6.34, 6.37, 6.97 Voltage regulation 2.58, 3.63, 3.64, 4.51 Voltage regulation down 3.64 Voltage regulation up 3.64 V–V connection 4.10

Y Yoke 2.6

Z Zero power-factor method 6.24