Gas Turbines [3 ed.] 9780070681927

Table of contents :
Cover
Contents
Chapter 1 Introduction
1.1 Prime Movers
1.2 Simple Gas Turbine
Review Questions
Multiple Choice Questions
Chapter 2 Review of Basic Principles
2.1 Definitions and Laws
2.2 Energy Equation
2.3 Fluid Dynamics
2.4 Basic Definitions
2.5 Streamtube Area-Velocity Relation
2.6 Normal Shock Waves
2.7 Equations of Motion for a Normal Shock Wave
2.8 Oblique Shock and Expansion Waves
2.9 Flow with Friction and Heat Transfer
2.10 Flow in Constant-Area Duct with Friction
Review Questions
Multiple Choice Questions
Chapter 3 Fundamentals of Rotating Machines
3.1 General Fluid Dynamics Analysis
3.2 The Physical Meaning of the Energy Equation
3.3 Classification of Machines
3.4 General Thermodynamic Analysis
3.5 Efficiency of Rotating Machines
3.6 Dimensional Analysis of Rotating Machines
3.7 Elementary Airfoil Theory
Review Questions
Multiple Choice Questions
Chapter 4 Cycle Arrangements
4.1 Open-Cycle Arrangements
4.2 The Closed-Cycle
4.3 Basic Requirements of the Working Medium
4.4 Properties of Various Working Media
4.5 Applications
4.6 Comparison of Gas Turbines with Reciprocating Engines
Review Questions
Multiple Choice Questions
Chapter 5 Ideal Cycles and their Analysis
5.1 Assumptions in Ideal Cycle Analysis
5.2 The Simple Gas Turbine Cycle
5.3 The Heat Exchange Cycle
5.4 The Reheat Cycle
5.5 The Reheat and Heat Exchange Cycle
5.6 The Intercooled Cycle
5.7 The Intercooled Cycle with Heat Exchanger
5.8 The Intercooled and Reheat Cycle
5.9 Intercooled Cycle with Heat Exchange and Reheat
5.10 Comparison of Various Cycles
5.11 Ericsson Cycle
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
Chapter 6 Practical Cycles and their Analysis
6.1 Assumptions
6.2 Stagnation Properties
6.3 Compressor and Turbine Efficiency
6.4 Pressure or Flow Losses
6.5 Heat Exchanger Effectiveness
6.6 Effect of Varying Mass Flow
6.7 Effect of Variable Specific Heat
6.8 Mechanical Losses
6.9 Loss due to Incomplete Combustion
6.10 Cycle Efficiency
6.11 Polytropic Efficiency
6.12 Performance of Actual Cycle
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
Chapter 7 Jet Propulsion Cycles and Their Analysis
7.1 Reciprocating or Propeller Engines
7.2 Gas Turbine Engines
7.3 The Ramjet Engine
7.4 The Pulse Jet Engine
7.5 The Turboprop Engine
7.6 The Turbojet Engine
7.7 Thrust and Thrust Equation
7.8 Specific Thrust of the Turbojet Engine
7.9 Efficiencies
7.10 Parameters Affecting Flight Performance
7.11 Thrust Augmentation
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
Chapter 8 Centrifugal Compressors
8.1 Essential Parts of a Centrifugal Compressor
8.2 Principle of Operation
8.3 Ideal Energy Transfer
8.4 Blade Shapes and Velocity Triangles
8.5 Analysis of Flow through the Compressor
8.6 Diffuser
8.7 Volute Casing
8.8 Performance Parameter
8.9 Losses in Centrifugal Compressors
8.10 Compressor Characteristics
8.11 Surging and Choking
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
Chapter 9 Axial Flow Compressors
9.1 Historical Background
9.2 Geometry and Working Principle
9.3 Stage Velocity Triangles
9.4 Work Done Factor
9.5 Enthalpy–Entropy Diagram
9.6 Compressor Stage Efficiency
9.7 Performance Coefficients
9.8 Degree of Reaction
9.9 Flow through Blade Rows
9.10 Flow Losses
9.11 Stage Losses
9.12 Pressure Rise Calculation in a Blade Ring
9.13 Performance Characteristics
9.14 Comparison of Axial and Centrifugal Compressors
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
Chapter 10 Combustion Systems
10.1 Combustion Theory Applied to Gas Turbine Combustor
10.2 Factors Affecting Combustion Chamber Design
10.3 Factors Affecting Combustion Chamber Performance
10.4 Form of Combustion System
10.5 Requirements of the Combustion Chamber
10.6 The Process of Combustion in a Gas Turbine
10.7 Combustion Chamber Geometry
10.8 Mixing and Dilution
10.9 Combustion Chamber Arrangements
10.10Some Practical Problems
Review Questions
Multiple Choice Questions
Chapter 11 Impulse and Reaction Turbines
11.1 A Single Impulse Stage
11.2 A Single Reaction Stage
11.3 Multistage Machines
11.4 Velocity Triangles of a Single Stage Machine
11.5 Expression for Work Output
11.6 Blade loading and flow coefficients
11.7 Blade and Stage Efficiencies
11.8 Maximum Utilization Factor for a Single Impulse Stage
11.9 Velocity-compounding of Multistage Impulse Turbine
11.10Pressure Compounding of Multistage Impulse Turbine
11.11The Reaction Turbine
11.12Multistage Reaction Turbines
11.13Blade-to-Gas Speed Ratio
11.14Losses and Efficiencies
11.15Performance Graphs
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
Chapter 12 Transonic and Supersonic Compressors and Turbines
12.1 The Supersonic Compressor
12.2 Supersonic Axial Flow Compressors
12.3 Supersonic Radial Compressors
12.4 Supersonic Axial Flow Turbine Stages
Review Questions
Multiple Choice Questions
Chapter 13 Inlets and Nozzles
13.1 Inlets
13.2 Subsonic Inlets
13.3 Diffuser
13.4 Supersonic Inlets
13.5 Exhaust Nozzles
Review Questions
Multiple Choice Questions
Chapter 14 Blades
14.1 Blade Materials
14.2 Manufacturing Techniques
14.3 Blade Fixing
14.4 Problems of High Temperature Operation
14.5 Blade Cooling
14.6 Liquid Cooling
14.7 Air Cooling
14.8 Practical Air Cooled Blades
Review Questions
Multiple Choice Questions
Chapter 15 Component Matching and Performance Evaluation
15.1 Performance Characteristics
15.2 Equilibrium Running Diagram
15.3 To Find the Equilibrium Points
15.4 Procedure to find Equilibrium Point
15.5 Performance Evaluation of Single-spool Turbojet Engine
15.6 Operating Line
15.7 General Matching Procedure
15.8 Transient Operation
Review Questions
Multiple Choice Questions
Chapter 16 Environmental Considerations and Applications
16.1 Air Pollution
16.2 Aircraft Emission Standards
16.3 Stationary Engine Emission Standards
16.4 NOx Formation
16.5 NOx Reduction in Stationary Engines
16.6 Noise
16.7 Noise Standards
16.8 Noise Reduction
16.9 Assessment of the Gas Turbine
16.10Typical Applications of Gas Turbines
16.11The Small Gas Turbine Applications
16.12Electric Power Generation Applications
16.13Marine Application
16.14Gas Pumping Applications
16.15Locomotive Applications
16.16Automotive Applications
16.17Aircraft Applications
16.18Process Applications
16.19Additional Features of Gas Turbine Engines
16.20Trends in the Future Development
Review Questions
Multiple Choice Questions
Chapter 17 Rocket Propulsion
17.1 Classification of Rockets
17.2 Principle of Rocket Propulsion
17.3 Analysis of an Ideal Chemical Rocket
17.4 Optimum Expansion Ratio for Rocket
17.5 The Chemical Rocket
17.6 Advantages of Liquid Propellant Rockets over Solid Propel-lant Rockets
17.7 Free Radical Propulsion
17.8 Nuclear Propulsion
17.9 Electro Dynamic Propulsion
17.10Photon Propulsion
17.11Comparison of various types of rockets
17.12Staging
17.13Multistage Rocket
17.14Comparison of Various Propulsion Systems
17.15Propulsive Efficiency
Review Questions
Multiple Choice Questions
Appendix
Index

Citation preview

Gas Turbines Third Edition

About the Author V Ganesan, currently working as Professor of Mechanical Engineering, Indian Institute of Technology Madras, is the recipient of Anna University National Award for Outstanding Academic for the Year 1997. He was the Head of the Department of Mechanical Engineering, between October 2000 and June 2002 and was the Dean (Academic Research) between January 1998 and October 2000. He has so far published more than 300 research papers in national and international conferences and journals, and has guided 20 MS and 40 PhDs. Among other awards received by him are the Babcock Power Award for the best fundamental scientific paper of Journal of Energy (1987), the Institution of Engineers Merit Prize and Citation (1993), SVRCET Surat Prize (1995), Sri Rajendra Nath Mookerjee Memorial Medal (1996), Automobile Engineer of the Year by the Institution of Automobile Engineers (India) (2001), Institution of Engineers (India), Tamil Nadu Scientist Award (TANSA) – 2003 by Tamil Nadu State Council for Science and Technology, ISTE Periyar Award for Best Engineering College Teacher (2004), N K Iyengar Memorial Prize (2004) by Institution of Engineers (India) SVRCET Surat Prize (2004), Khosla National Award (2004), Bharat Jyoti Award (2006) UWA Outstanding Intellectuals of the 21st Century Award by United Writers Association, Chennai (2006), 2006 SAE Cliff Garrett Turbomachinery Engineering Award by SAE International, USA, Sir Rajendra Nath Mookerjee Memorial Prize (2006) by Institution of Engineers, Environmental Engineering Design Award 2006 by The Institution of Engineers (India), 2006 SAE Cliff Garrett Turbomachinery Engineering Award (2007) and Excellence in Engineering Education (Triple “E”) Award by SAE International, USA (2007). He has received the Best Faculty Award from Nehru Group of Institution in Coimbatore in 2009. He is the Fellow of Indian National Academy of Engineering, National Environmental Science Academy, Fellow of SAE International, USA, and Fellow of Institution of Engineers (India). He has authored several books including the first edition of Internal Combustion Engines, McGraw Hill International Publications, Gas Turbines, Computer Simulation of Four-Stroke Spark-Ignition Engines, Computer Simulation of Four-Stroke Compression-Ignition Engines, and Heat Transfer and has edited several proceedings. He was formerly the Chairman of Combustion Institute (Indian Section) and the Chairman of Engineering Education Board of SAE (India), besides being a member of many other professional societies. Dr Ganesan is actively engaged in a number of sponsored research projects and is a consultant for various industries and R&D organizations.

Gas Turbines Third Edition

V Ganesan Professor Department of Mechanical Engineering Indian Institute of Technology Madras

Tata McGraw Hill Education Private Limited NEW DELHI McGraw-Hill 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

Published by Tata McGraw Hill Education Private Limited, 7 West Patel Nagar, New Delhi 110 008 Gas Turbine, 3/e Copyright © 2010, 1990, 1967 by Tata McGraw Hill Education Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, Tata McGraw Hill Education Private Limited ISBN (13 digits): 978-0-07-068192-1 ISBN (10 digits): 0-07-068192-9 Managing Director: Ajay Shukla Head—Higher Education Publishing: Vibha Mahajan Manager—Sponsoring—SEM & Tech Ed: Shalini Jha Assoc. Sponsoring Editor: Suman Sen Development Editor: Devshree Lohchab Executive—Editional Services: Sohini Mukherjee Sr Production Manager: PL Pandita General Manager: Marketing—Higher Education: Michael J Cruz Dy Marketing Manager—SEM & Tech Ed: Biju Ganesan General Manager—Production: Rajender P Ghansela Asst. General Manager—Production: B L Dogra Information contained in this work has been obtained by Tata McGraw Hill, from sources believed to be reliable. However, neither Tata McGraw Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither Tata McGraw Hill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Tata McGraw Hill and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Published by Tata McGraw Hill Education Private Limited and printed at Adarsh Printers, C-50/51, Mohan Park, Shahdara, Delhi 110 032ahdara, Delhi 110 094 Cover: Rashtriya Printers RAZDRRCZRYYCC

Dedicated to Him

Preface This text on gas turbines has been prepared keeping in mind the need of a first-level textbook for the undergraduate, graduate and a professional reference for practicing engineers. When I started writing the manuscript a few years ago, to my knowledge, suitable text material was practically non-existent, especially in SI units, and it seems to be true even today. It is my earnest hope that this textbook will simplify the teaching of this important subject.

Target Audience This book is primarily intended for undergraduate/postgraduate students. Hence, I have endeavoured to explain various topics right from the fundamentals so that even a beginner can understand the exposition. Besides, topics such as inlets, blades, environmental consideration, etc., are included to help the practising engineers as well. Professionals who are engaged in the design and development of various power plants, who are often called upon to undertake the work related to gas turbines, will find this book extremely useful.

New to this Edition It is gratifying to note that this book on Gas Turbines is being well received by the academic community. Based on the suggestions and comments received, all the chapters have been thoroughly revised. A set of multiplechoice questions has been introduced in each chapter. New sections and subsections have been added in various chapters. A new chapter on Transonic and Supersonic Compressors and Turbines has been included. Further, minor errors brought to my notice have been rectified.

Organisation of the Book SI units have been used consistently throughout the book. The book includes a large number of typical worked-out examples and several illustrative figures for an easier understanding of the subject. Exercises have been given in various chapters so that the inquisitive students may solve these problems and compare their answers with the answers provided. Quite a lot of technical literature is available on this subject and owing to the rate at which these technical papers now appear, a comprehensive

viii Preface bibliography has not been included. However, a few references have been selected and are appended to appropriate chapters. The text material is divided into eight parts. ∑ Introduction and Fundamentals (Chapters 1 to 4) ∑ Cycle Analysis (Chapters 5 to 7) ∑ Compressors (Chapters 8 and 9) ∑ Combustion Systems (Chapter 10) ∑ Turbines (Chapter 11) ∑ Advanced Topics (Chapters 12 to 15) ∑ Environmental Considerations (Chapter 16) ∑ Rocket Propulsion (Chapter 17) By dividing the chapters in the above manner, it may be seen that the work has been organised to form a continuous logical narrative.

Web Supplements The web supplements can be accessed at http://www.mhhe.com/ganesan/ gt3e and contain Solution Manual and PowerPoint Lecture Slides for Instructors, and a Sample Chapter and Links to Reference Material for students.

Acknowledgements It would be impossible to refer in detail to all those who have been consulted in the compilation of this work. A special note of appreciation is due to my gas-turbines course students of the 1999 batch, for their help in checking the worked out as well as exercise problems. I am thankful for the help and support received from colleagues of the department and sister departments of IIT Madras. I am particularly grateful to Prof. P Srinivasa Rao and Prof. S Sampath who spared a great amount of their valuable time in correcting the entire manuscript. Their critical comments and suggestions were of immense help in improving the presentation. In this context, I would also like to mention the names of all the reviewers of the book. Avinash Kumar Agarwal

Indian Institute of Technology (IIT) Kanpur Kanpur, Uttar Pradesh

V K Bajpai

National Institute of Technology (NIT) Kurukshetra Kurukshetra, Haryana

L D Garg

Punjab Engineering College (PEC) University of Technology Chandigarh, Punjab

Preface

ix

Souvik Bhattacharyya

Indian Institute of Technology (IIT) Kharagpur Kharagpur, West Bengal

Dipankar Chaterjee

Haldia Institute of Technology (HIT) Haldia, West Bengal

Ranjan Basak

Sikkim Manipal Institute of Technology Majitar, Sikkim

Mahesh Rathore

SNJB’s KBJ College of Engineering Nashik, Maharashtra

Venkatraj Varadharajulu

Bharati Vidyapeeth College of Engineering and Technology Mumbai, Maharashtra

M Udayakumar

National Institute of Technolgy (NIT) Tiruchirapalli Tiruchirapalli, Tamil Nadu

P Chitambarnathan

Dr Sivanthi Adithanar College of Engineering Thiruchendur, Tamil Nadu

B Srinivas Reddy

G Pulla Reddy Engineering College Kurnool, Andhra Pradesh

I wish to express my thanks to the Centre for Continuing Education for encouraging me to write this book and helping me in the preparation of this manuscript. I will be failing in my duty if I do not acknowledge the help of my family members, Ms P Rajalakshmi, Ms Vijayashree, Mr Venkatasubramanian and Ms Aparna. Words are not sufficient to express my gratitude to Ms Vijayashree who has put her heart and soul in the preparation of the book. My sincere thanks are due to all of them.

Feedback Although maximum care has been taken to minimize the errors, I will be grateful for any constructive criticism for further improvement of the book. I hope and wish the present edition also would be well received. V GANESAN

x Preface

Publisher’s Note Do you have a feature request? A suggestion? We are always open to new ideas (the best ideas come from you!). You may send your comments to [email protected] (kindly mention the title and author name in the subject line). Piracy related issues may also be reported.

Contents Preface Preface to the Second Edition Preface to the First Edition Nomenclature 1 Introduction 1.1 Prime Movers 1.2 Simple Gas Turbine Review Questions Multiple Choice Questions

ii viii ix xix 1 1 2 4 5

2 Review of Basic Principles 2.1 Definitions and Laws 2.2 Energy Equation 2.3 Fluid Dynamics 2.4 Basic Definitions 2.5 Streamtube Area-Velocity Relation 2.6 Normal Shock Waves 2.7 Equations of Motion for a Normal Shock Wave 2.8 Oblique Shock and Expansion Waves 2.9 Flow with Friction and Heat Transfer 2.10 Flow in Constant-Area Duct with Friction Review Questions Multiple Choice Questions

7 7 15 19 19 25 28 28 30 32 32 36 37

3 Fundamentals of Rotating Machines 3.1 General Fluid Dynamics Analysis 3.2 The Physical Meaning of the Energy Equation 3.3 Classification of Machines 3.4 General Thermodynamic Analysis

39 39 42 44 45

xii

Contents

3.5 Efficiency of Rotating Machines 3.6 Dimensional Analysis of Rotating Machines 3.7 Elementary Airfoil Theory Review Questions Multiple Choice Questions 4 Cycle Arrangements 4.1 Open-Cycle Arrangements 4.2 The Closed-Cycle 4.3 Basic Requirements of the Working Medium 4.4 Properties of Various Working Media 4.5 Applications 4.6 Comparison of Gas Turbines with Reciprocating Engines Review Questions Multiple Choice Questions

47 48 54 59 60 63 63 68 71 72 72 73 75 76

5 Ideal Cycles and their Analysis 5.1 Assumptions in Ideal Cycle Analysis 5.2 The Simple Gas Turbine Cycle 5.3 The Heat Exchange Cycle 5.4 The Reheat Cycle 5.5 The Reheat and Heat Exchange Cycle 5.6 The Intercooled Cycle 5.7 The Intercooled Cycle with Heat Exchanger 5.8 The Intercooled and Reheat Cycle 5.9 Intercooled Cycle with Heat Exchange and Reheat 5.10 Comparison of Various Cycles 5.11 Ericsson Cycle Worked out Examples Review Questions Exercise Multiple Choice Questions

79 80 80 83 86 89 92 94 96 99 101 103 104 128 130 134

6 Practical Cycles and their Analysis 6.1 Assumptions 6.2 Stagnation Properties 6.3 Compressor and Turbine Efficiency 6.4 Pressure or Flow Losses 6.5 Heat Exchanger Effectiveness 6.6 Effect of Varying Mass Flow 6.7 Effect of Variable Specific Heat 6.8 Mechanical Losses

137 137 138 140 143 147 148 149 150

Contents

6.9 Loss due to Incomplete Combustion 6.10 Cycle Efficiency 6.11 Polytropic Efficiency 6.12 Performance of Actual Cycle Worked out Examples Review Questions Exercise Multiple Choice Questions

xiii

150 151 152 156 165 203 203 210

7 Jet Propulsion Cycles and Their Analysis 7.1 Reciprocating or Propeller Engines 7.2 Gas Turbine Engines 7.3 The Ramjet Engine 7.4 The Pulse Jet Engine 7.5 The Turboprop Engine 7.6 The Turbojet Engine 7.7 Thrust and Thrust Equation 7.8 Specific Thrust of the Turbojet Engine 7.9 Efficiencies 7.10 Parameters Affecting Flight Performance 7.11 Thrust Augmentation Worked out Examples Review Questions Exercise Multiple Choice Questions

213 214 215 216 220 224 227 235 237 238 246 246 252 271 272 276

8 Centrifugal Compressors 8.1 Essential Parts of a Centrifugal Compressor 8.2 Principle of Operation 8.3 Ideal Energy Transfer 8.4 Blade Shapes and Velocity Triangles 8.5 Analysis of Flow through the Compressor 8.6 Diffuser 8.7 Volute Casing 8.8 Performance Parameter 8.9 Losses in Centrifugal Compressors 8.10 Compressor Characteristics 8.11 Surging and Choking Worked out Examples Review Questions Exercise Multiple Choice Questions

279 280 282 286 287 290 302 306 307 310 310 311 313 326 327 331

xiv

Contents

9 Axial Flow Compressors 9.1 Historical Background 9.2 Geometry and Working Principle 9.3 Stage Velocity Triangles 9.4 Work Done Factor 9.5 Enthalpy–Entropy Diagram 9.6 Compressor Stage Efficiency 9.7 Performance Coefficients 9.8 Degree of Reaction 9.9 Flow through Blade Rows 9.10 Flow Losses 9.11 Stage Losses 9.12 Pressure Rise Calculation in a Blade Ring 9.13 Performance Characteristics 9.14 Comparison of Axial and Centrifugal Compressors Worked out Examples Review Questions Exercise Multiple Choice Questions

333 333 335 336 340 342 343 344 345 351 353 356 357 359 365 366 386 387 390

10 Combustion Systems 10.1 Combustion Theory Applied to Gas Turbine Combustor 10.2 Factors Affecting Combustion Chamber Design 10.3 Factors Affecting Combustion Chamber Performance 10.4 Form of Combustion System 10.5 Requirements of the Combustion Chamber 10.6 The Process of Combustion in a Gas Turbine 10.7 Combustion Chamber Geometry 10.8 Mixing and Dilution 10.9 Combustion Chamber Arrangements 10.10Some Practical Problems Review Questions Multiple Choice Questions

393 394 397 398 402 403 405 406 409 409 411 418 419

11 Impulse and Reaction Turbines 11.1 A Single Impulse Stage 11.2 A Single Reaction Stage 11.3 Multistage Machines 11.4 Velocity Triangles of a Single Stage Machine 11.5 Expression for Work Output 11.6 Blade loading and flow coefficients 11.7 Blade and Stage Efficiencies

421 422 422 423 424 426 427 427

Contents

xv

11.8 Maximum Utilization Factor for a Single Impulse Stage 11.9 Velocity-compounding of Multistage Impulse Turbine 11.10Pressure Compounding of Multistage Impulse Turbine 11.11The Reaction Turbine 11.12Multistage Reaction Turbines 11.13Blade-to-Gas Speed Ratio 11.14Losses and Efficiencies 11.15Performance Graphs Worked out Examples Review Questions Exercise Multiple Choice Questions

428 430 435 435 436 450 451 451 454 463 464 465

12 Transonic and Supersonic Compressors and Turbines 12.1 The Supersonic Compressor 12.2 Supersonic Axial Flow Compressors 12.3 Supersonic Radial Compressors 12.4 Supersonic Axial Flow Turbine Stages Review Questions Multiple Choice Questions

467 468 469 474 475 478 478

13 Inlets and Nozzles 13.1 Inlets 13.2 Subsonic Inlets 13.3 Diffuser 13.4 Supersonic Inlets 13.5 Exhaust Nozzles Review Questions Multiple Choice Questions

481 481 482 484 486 488 498 498

14 Blades 14.1 Blade Materials 14.2 Manufacturing Techniques 14.3 Blade Fixing 14.4 Problems of High Temperature Operation 14.5 Blade Cooling 14.6 Liquid Cooling 14.7 Air Cooling 14.8 Practical Air Cooled Blades Review Questions Multiple Choice Questions

501 501 503 508 510 513 515 515 518 521 521

xvi

Contents

15 Component Matching and Performance Evaluation 15.1 Performance Characteristics 15.2 Equilibrium Running Diagram 15.3 To Find the Equilibrium Points 15.4 Procedure to find Equilibrium Point 15.5 Performance Evaluation of Single-spool Turbojet Engine 15.6 Operating Line 15.7 General Matching Procedure 15.8 Transient Operation Review Questions Multiple Choice Questions

523 526 526 527 530 531 537 543 545 546 547

16 Environmental Considerations and Applications 16.1 Air Pollution 16.2 Aircraft Emission Standards 16.3 Stationary Engine Emission Standards 16.4 NOx Formation 16.5 NOx Reduction in Stationary Engines 16.6 Noise 16.7 Noise Standards 16.8 Noise Reduction 16.9 Assessment of the Gas Turbine 16.10Typical Applications of Gas Turbines 16.11The Small Gas Turbine Applications 16.12Electric Power Generation Applications 16.13Marine Application 16.14Gas Pumping Applications 16.15Locomotive Applications 16.16Automotive Applications 16.17Aircraft Applications 16.18Process Applications 16.19Additional Features of Gas Turbine Engines 16.20Trends in the Future Development Review Questions Multiple Choice Questions

549 549 551 555 557 559 560 562 565 567 569 569 570 570 571 572 572 573 573 574 578 578 579

17 Rocket Propulsion 17.1 Classification of Rockets 17.2 Principle of Rocket Propulsion 17.3 Analysis of an Ideal Chemical Rocket 17.4 Optimum Expansion Ratio for Rocket 17.5 The Chemical Rocket

581 582 583 584 587 589

Contents

xvii

17.6 Advantages of Liquid Propellant Rockets over Solid Propellant Rockets 17.7 Free Radical Propulsion 17.8 Nuclear Propulsion 17.9 Electro Dynamic Propulsion 17.10Photon Propulsion 17.11Comparison of various types of rockets 17.12Staging 17.13Multistage Rocket 17.14Comparison of Various Propulsion Systems 17.15Propulsive Efficiency Review Questions Multiple Choice Questions

600 601 601 602 606 606 607 608 609 609 611 613

Appendix

615

Index

619

xviii

Contents

1 INTRODUCTION 1.1

PRIME MOVERS

The distinctive feature of our civilization, one that makes it different from all others, is the wide use of power from mechanical means. At one time the primary source of power, or prime mover, for the work of peace and war was chiefly man’s muscles. Even after animals had been trained to help and after the wind and running streams had been harnessed, man was mainly depending on his muscle power. But when he learned to convert the heat of chemical reactions into mechanical energy, it revolutionized the world. Machines which serve this purpose are known as heat engines. The first heat engine to have a revolutionary effect was the gun. It is perhaps hard to realize that the vital difference between the bow and the gun was the substitution of gunpowder for the bowman’s muscles, because the musket was much inferior to the bow in range and accuracy. In principle, the gun is the ancestor of our internal combustion piston engines. Another revolution began when Watt perfected the steam piston engine. Here, an intermediate working fluid was used, so that the products of combustion did not act directly on the moving parts of the mechanism. Later, the ancient principle of the turbine was adopted to steam, and the piston engine took the back seat. Among engines for the production of mechanical power there were, a few years ago, three principal competitors in the field: (i) the steam turbine plant, (ii) the diesel piston engine, and (iii) the gasoline piston engine. The steam turbine, since the beginning of its career around the turn of the nineteenth century, has become the most important prime mover for power generation and a widely used power plant for marine application. However, it has an inherent disadvantage of the need to produce highpressure, high-temperature steam. This involves the installation of a bulky

2

Gas Turbines

and expensive steam-generating equipment, a boiler or a nuclear reactor. The hot gases produced in the boiler furnace or reactor core never reach the turbine. They, merely are used to produce an intermediate working fluid, namely steam. Clearly, a much more compact power plant results when the water to steam step is eliminated and the hot gases themselves are used to drive the turbine. Diesel piston engines started replacing bulky steam power plants for power generation. Gasoline engines were used in early days for aircraft propulsion. A device known as the gas turbine came into existence to a limited extent in certain types of superchargers used with piston engines. The earliest patent on gas turbine was that of the Englishman, John Barber, in 1879. Early designs were unsuccessful, largely due to two factors: (i) the low efficiency of the compressors, and (ii) the combustion temperature limitations imposed by the materials then available. Serious development of gas turbine began only after the second world war with shaft power in mind, but attention was soon shifted to the turbojet engine for aircraft propulsion. Since then, the gas turbine made a progressively greater impact in an increasing variety of applications. However, only in the recent past much research effort has been focused on the design and development of efficient gas turbine units. Of the various means of producing mechanical power available today the gas turbine, in many respects, seems to be the most satisfactory power plant. It is mainly due to (i) the absence of reciprocating and rubbing members which reduces the vibration and balancing problems, (ii) high reliability, (iii) low lubricating-oil consumption, and (iv) high power-to-weight ratio. 1.2

SIMPLE GAS TURBINE

In order to produce an expansion through a turbine a pressure ratio must be provided. Hence, the first necessary step in the cycle of a gas turbine plant must be compression of the working fluid. If, after the compression, the working fluid is to be expanded directly in the turbine and there were no losses in either component, the power developed by the turbine would just equal that absorbed by the compressor. Thus, if the two were coupled together, the combination would do no more than turn itself round. The power developed by the turbine can be increased by the addition of energy to raise the temperature of the fluid prior to expansion. When the working fluid is air, a very suitable means of doing this is by the combustion of fuel in the air which has been compressed. Expansion of hot working fluid

Introduction

3

then produces a greater power output from the turbine than the power necessary to drive the compressor. The three main components are, therefore, a compressor, a combustion chamber and a turbine, connected together as shown in Fig. 1.1. Fuel

Combustion chamber Products of combustion

Air

Power output Compressor

Turbine

Fig. 1.1 A simple gas turbine In a practical cycle, losses do occur in both the compressor and the turbine which increase the power absorbed by the compressor and decrease the power output of the turbine. A certain minimum addition of energy in the form of fuel to the fluid will therefore be required before one component can drive the other. This fuel produces no useful power, due to component losses and lowering of efficiency of the machine. Further addition of fuel will result in a useful power output. However, there is a limit to the amount of fuel that can be added per unit mass of air and therefore to the net power output. The fuel-air ratio that may be used is governed by the working temperature of the highly stressed turbine blades. This temperature is limited by the creep strength of the materials used in the turbine blades and the working life required. The two main factors which affect the performance of gas turbines are the efficiencies of various components and turbine working temperature. The higher they are made, the better is the all-round performance of the plant. It was, in fact, low efficiencies and poor turbine materials which caused the failure of a number of early attempts. For example, in 1904 two French engineers built a unit which did little more than turn itself over, with compressor efficiency of about 60% and the maximum gas temperature of about 740 K. The overall efficiency of the gas turbine cycle mainly depends upon the pressure ratio of the compressor. The development of science of aerodynamics and that of metallurgy made it possible to employ very high pressure ratios (20:1) with an adequate compressor efficiency (85-90%) and high turbine inlet temperatures, up to 1500 K. Two possible combustion systems were proposed: one at constant-pressure and the other at constant-volume. Theoretically, the thermal efficiency of the constant-volume cycle is higher than that of constant-pressure cycle. However, constant-volume combustion involves mechanical difficulties requiring valves to isolate the combustion chamber from the compressor and turbine. Combustion is intermittent impairing the smooth running of the

4

Gas Turbines

machine. After certain initial attempts, constant-volume type combustion was discontinued. In the constant-pressure gas turbine, combustion is continuous and valves are not necessary. It was soon accepted that the constant-pressure cycle had the greater possibilities for future development. It is important to realize that in the gas turbine the process of compression, combustion and expansion do not occur in a single component as they do in a reciprocating engine. They occur in components which are separate, in the sense that they can be designed, tested and developed individually and these components can be linked together to form a gas turbine unit in a variety of ways. The possible number of components is not limited to those already mentioned. More compressors and turbines can be added with intercoolers between the compressors, and reheat combustion chambers between turbines. Water coolers and heat exchangers can be additionally fitted. These refinements are used to increase the power output and efficiency of the plant at the expense of added complexity, weight and cost. The way these components are added not only affects the maximum overall thermal efficiency but also the variation of thermal efficiency. Each arrangement is to be chosen depending on its suitability for a given application. Thus, it is seen that a simple gas turbine consists mainly of three components. Of the three, two are rotating machines and the third one is a heat addition device. However, more complex system is possible with the addition of auxiliary devices such as heat exchanger, intercooler and reheater. In order to understand the working principle and cycle arrangements one should first have some fundamental knowledge of rotating machines. This will be taken up in Chapter 3 after the review of some basic principles in Chapter 2. Review Questions 1.1 What is the difference between our civilization and the ancient ones and what action of the man revolutionized the world? 1.2 What is a heat engine and what is its origin? 1.3 What is a steam engine and what are its characteristics? 1.4 What is the inherent disadvantage of a steam power plant? 1.5 Which were the three contemporary power plants for the production of power? 1.6 When was the earliest gas turbine design attempted and what was the result? 1.7 What are the latest research efforts in gas turbines and why is this power plant considered to be satisfactory? 1.8 Explain with a neat sketch the details of a simple gas turbine power plant.

Introduction

5

1.9 Compare the steam and gas turbine power plants. 1.10 Which parameter is most important to determine the overall efficiency of a gas turbine cycle? 1.11 What are the two factors which affect the performance of a gas turbine? 1.12 What are the two types of combustion systems? 1.13 Why is constant-pressure heat addition is more advantageous in a gas turbine? 1.14 What advantage does a gas turbine power plant have over a reciprocating engine from the point of view of component developments? 1.15 How many components are there in a simple gas turbine power plant? Mention the additional components that can be added to improve the power output and efficiency. Multiple Choice Questions (choose the most appropriate answer) 1. Around the turn of nineteenth century the most important prime mover was (a) gas turbines (b) diesel engines (c) steam turbines (d) gasoline engines 2. Early aircraft engines used (a) gas turbine engines (b) gasoline engines (c) Diesel engines (d) steam turbines 3. The gas turbine was invented by (a) John Barber (b) Brayton (c) Otto (d) Atkinson 4. Early designs of the gas turbine were unsuccessful due to (a) low efficiency of the compressor (b) low efficiency of the turbine

6

Gas Turbines

(c) combustion temperature limitation (d) all of the above 5. Reciprocating engines are preferred over gas turbines because of (a) (b) (c) (d)

high reliability high power to weight ratio all of the above none of the above

6. A simple open-cycle gas turbine power plant consists of (a) (b) (c) (d)

turbine, turbine, turbine, turbine,

combustion chamber combustion chamber combustion chamber compressor and heat

and heat exchanger and charge cooler and compressor exchanger

7. The combustion in a gas turbine is at (a) (b) (c) (d)

constant-pressure constant-volume partly constant-pressure and partly constant-volume constant temperature

8. The performance of a simple gas turbine depends on (a) (b) (c) (d)

efficiency of the compressor efficiency of the turbine efficiency of the compressor and turbine none of the above

9. The pressure ratio of the modern gas turbine power plant is (a) (b) (c) (d)

5:1 10:1 15:1 20:1

10. The highest turbine inlet temperature is (a) (b) (c) (d)

1000 1200 1500 1800

K K K K

Ans:

1. – (c) 6. – (c)

2. – (b) 7. – (a)

3. – (a) 8. – (c)

4. – (d) 9. – (d)

5. – (d) 10. – (c)

2 REVIEW OF BASIC PRINCIPLES INTRODUCTION Turbines and compressors are usually analyzed using thermodynamic and fluid dynamic equations. The thermodynamic equations relate temperature, pressure and volume whereas the fluid dynamic equations relate force, mass and velocity. The following are the laws that are frequently used in dealing with problems of and operation of these machines: (i) energy equation in its various forms from the first law of thermodynamics, (ii) temperature, entropy and gas relations from the second law of thermodynamics, (iii) continuity relationships from the law of conservation of mass, and (iv) momentum equation from Newton’s second law of motion. 2.1

DEFINITIONS AND LAWS

Before discussing the various aero-thermodynamic aspects of compressors and turbines, let us review some important definitions used in the analysis of compressible flow useful for rotating machines. However, the reader is advised to refer to standard textbooks on thermodynamics and fluid dynamics for more details. 2.1.1

System

A fixed identity with an arbitrary collection of matter is known as a system. The boundary is an imaginary surface which separates the system from its surroundings. Surroundings are those which are outside the system. System can be classified as either an open system or a closed system.

8

Gas Turbines

Open System When there is a continuous flow of matter, it is called an open system. Such a system is usually depicted by a control volume. It has a fixed space but does not contain a fixed mass of matter; instead there is a continuous flow of mass through it. The properties of the matter occupying the control volume can vary with time. The surface which encloses a control volume is called control surface. Closed System When there is a fixed quantity of matter (fluid or gas), it is called a closed system. There is no inflow or outflow of matter to and from a closed system. However, a closed system can interact with its surroundings through work and heat transfers. The boundaries of a closed system containing the fixed mass of matter can change. Expanding gas in a reciprocating internal combustion engine is one such example. 2.1.2

State

Condition of a system, defined by its properties, is known as the state of a system. 2.1.3

Process

A change or a series of changes in the state of a system is known as a process. 2.1.4

Cycle

If the initial and final states of a system experiencing a series of processes are identical, it is said to have executed a cycle. 2.1.5

Pressure

Pressure at a point surrounded by an infinitesimal area, ΔA, is the force per unit area. Pressure is usually designated by Pascal in SI units. It may also be expressed as N/m2 or bar. In this book, we will follow ’bar’ as the unit of pressure. 2.1.6

Density

The density of a medium is the mass of the matter (gas) per unit volume. Density is expressed in kg/m3 . 2.1.7

Temperature

When two systems are in contact with each other and are in thermal equilibrium, the property common to both the systems having the same value is called temperature. Thus temperature is a measure of the thermal potential of a system.

Review of Basic Principles

2.1.8

9

Energy

Energy is the capacity to do work. The state of a system can be changed by adding or removing energy. Heat and work are different forms of energy in transit. They are not contained in any system. Heat is the form of energy which transfers between two systems by virtue of the temperature difference between them. Heat transfer to or from a system changes its state. Work is said to be done by a system on its surroundings when they are moved through a distance by the action of a force; this is exerted by the system on the surroundings in the direction of the displacement of the surroundings. The magnitude of mechanical work is given by Work done

=

Force × Distance in the direction of force

Both heat and work are path functions and they depend on the type of process and therefore, are not properties of a system. In SI units, energy, heat and work are all expressed in joules (J), kilojoules (kJ) or Newton metres (Nm). 2.1.9

First Law of Thermodynamics

It states that when a system executes a cyclic process, the algebraic sum of the work transfers is proportional to the algebraic sum of the heat transfers. dW

∝

dQ

=

J

dQ

(2.1)

When heat and work are expressed in the same units, then dQ −

dW

= 0

(2.2)

It can easily be shown that the quantity (dQ − dW ) is independent of the path of the process and hence it represents a change in the property of the system. This property is referred to as energy, denoted here by the symbol E. Thus, dE

=

dQ − dW

(2.3)

Equation 2.3 for the two states of a system can be written as E2 − E1

=

Q−W

Q

=

W + (E2 − E1 )

Heat transfer = 2.1.10

(2.4)

Work done + Change in energy

Specific Heats of Gases

The specific heat of a gas is the heat carrying capacity in a process. It is the amount of heat that is required to raise the temperature of a unit mass of the gas by one degree.

10

Gas Turbines

In thermodynamic analysis two different types of specific heats are used: (i) Specific heat at constant-volume, and (ii) Specific heat at constant-pressure. The specific heat at constant-volume (Cv ) is the amount of heat required to raise the temperature of a unit mass of the gas by one degree at constantvolume. It is given by Cv

∂q ∂T

=

= v

∂U ∂T

(2.5) v

The specific heat at constant-pressure (Cp ) is the amount of heat required to raise the temperature of a unit mass of the gas by one degree at constantpressure . It is given by Cp

∂q ∂T

=

= p

∂h ∂T

(2.6) p

The specific heats of actual gases are a function of temperature and vary with temperature Cp , Cv = f (T ) (2.7) The ratio of the two specific heats, (γ), is an important parameter in compressible flow problems of turbomachines and is defined as γ 2.1.11

=

Cp Cv

(2.8)

Internal Energy

The internal energy of a gas is the energy stored in it by virtue of its molecular motion. If it is assumed that the internal energy of a perfect gas is zero at the absolute zero temperature, its value at a temperature T is given by U = Cv T (2.9) 2.1.12

Enthalpy

The heat supplied to or rejected by a system at constant-pressure is the change of enthalpy during the process. The value of enthalpy at a given state is given by h =

U + pv

h =

Cp T

=

U+

p ρ

(2.10)

and for an ideal gas (2.11)

Review of Basic Principles

2.1.13

11

Ideal Gas

An ideal gas obeys both Boyle’s and Charle’s laws, i.e., (pv)T v T

p

=

constant (Boyle’s law)

(2.12)

=

constant (Charle’s law)

(2.13)

Thus, an ideal gas obeys the simple equation of state pv = RT p =

ρRT

(2.14) (2.15)

The two specific heats and the gas constant for an ideal gas are related by the following equation: Cp − Cv

=

R

(2.16)

Substituting Eq. 2.8 in 2.16, we get 1 R Cv = γ−1

(2.17)

γ R γ−1

(2.18)

Cp 2.1.14

=

Perfect Gas

A perfect gas is an ideal gas whose specific heats remain constant at all temperatures. d (Cv ) dT

= 0

(2.19)

Cv = constant with temperature and d (Cp ) dT

= 0

(2.20)

Cp = constant with temperature. Unless otherwise stated, analyses of compressible flow given in this book assume perfect gas relations.

2.1.15

Semi-perfect Gas

A semi-perfect gas is an ideal gas whose specific heats vary with temperature. Cv = f (T ) and Cp = f (T ) (2.21) 2.1.16

Real Gas

The real gas behaviour deviates from that of an ideal gas. It does not obey the equation of state (Eqs. 2.14, 2.15, etc.,). Different equations of state are used for real gases.

12

Gas Turbines

2.1.17

Second Law of Thermodynamics

The second law of thermodynamics can be stated in a number of ways. Some of them are as follows: (i) Clausius statement Heat cannot, on its own, flow from a body at lower temperature to a body at higher temperature. (ii) Kelvin-Planck’s statement It is impossible to construct a heat engine which performs a complete cycle and delivers work exchanging heat from a single source. The following relations can be derived from the second law of thermodynamics: Definition of entropy 2

s2 − s1 Clausius inequality

dQ T

= 1

dQR T

≤ 0

(2.22)

(2.23)

In any irreversible process, 2

s2 − s1

> 1

dQ T

In an irreversible adiabatic process, s2 − s1 > 0

(2.24)

(2.25)

In any reversible cycle, dQR T

= 0

(2.26)

In an isentropic, i.e., a reversible adiabatic process, s 2 − s1 2.1.18

= 0

(2.27)

Reversible Process

A process is reversible if the system and its surroundings can be restored to their initial states by reversing the process. A reversible process in a flow machine is possible only in the absence of fluid friction and heat transfer with finite temperature difference. Since these conditions are impossible to achieve in actual processes, all real flows in compressors and turbines are irreversible. The reversible process is used only as an ideal reference process for comparison with its equivalent actual process.

Review of Basic Principles

2.1.19

13

Irreversible Process

An irreversible process is one which does not satisfy the above conditions of reversibility. 2.1.20

Adiabatic Process

During a process if there is no heat transfer between the system and the surroundings, it is known as an adiabatic process. All the rotating machines discussed in this book are assumed to follow only adiabatic processes. 2.1.21

Isentropic Process

An adiabatic process in which entropy remains constant is known as a reversibly adiabatic or isentropic process. For unit mass, this is governed by the following relations: pv γ T1 T2

= =

T ds = 2.1.22

constant p1 p2

(2.28)

γ−1 γ

=

dh − vdp

=

v2 v1

γ−1

ρ1 ρ2

=

1 dh − dp ρ

=

0

γ−1

(2.29) (2.30)

Non-flow Process

A process that occurs in a closed system is a non-flow process. One such example is shown in Fig. 2.1. It represents the expansion of a fixed mass of gas inside the cylinders of a reciprocating engine. 1

pV = constant

dV

p dV

Volume

2

14

Gas Turbines 2

Wnf p

=

pdv

(2.32)

1

For an adiabatic process, assuming perfect gas relations Wnf p

2.1.23

=

1 (p1 v1 − p2 v2 ) γ−1

(2.33)

=

Cv (T1 − T2 ) = u1 − u2

(2.34)

Flow Process

A process that occurs in an open system or in a control volume is a flow process. Processes occurring in all turbomachines are of this type. Figure 2.2 represents such a process.

Pressure

1

dp γ

pV = constant V dp 2 Volume

Fig. 2.2 Illustration of a flow process The infinitesimal work done in a reversible process is given by dWf p

=

−vdp

Wf p

=

−

For a finite process

(2.35)

2

vdp

(2.36)

1

For an adiabatic process in a perfect gas, γ Wf p = (p1 v1 − p2 v2 ) γ−1

(2.37)

Wf p

=

Cp (T1 − T2 )

(2.38)

Wf p

=

h1 − h2

(2.39)

Review of Basic Principles

2.2

15

ENERGY EQUATION

The energy equation (Eq. 2.4) is basically derived from the first law of thermodynamics as given in Section 2.1.9. It is written as Q

=

W + (E2 − E1 )

(2.40)

For application in turbomachines, the energy terms will include internal energy, gravitational potential energy and kinetic energy. Other forms of energy which can be included but are not relevant here are strain energy, magnetic energy, etc., 1 E = U + m(gZ) + mc2 (2.41) 2 dE

=

dU + mg dZ + m d

1 2 c 2

(2.42)

where U is the internal energy and Z is the potential energy and c is the velocity of the fluid. The change in the energy in a finite process between two states is given by 1 E2 − E1 = (U2 − U1 ) + mg(Z2 − Z1 ) + m c22 − c21 (2.43) 2 Substituting Eq. 2.43 in Eq. 2.4, a general form of the energy equation can be obtained. 1 Q = W + (U2 − U1 ) + mg(Z2 − Z1 ) + m c22 − c21 (2.44) 2 Dividing throughout by m 1 q = w + (u2 − u1 ) + g(Z2 − Z1 ) + c22 − c21 (2.45) 2 2.2.1

Steady-flow Energy Equation

For steady flow processes through turbomachines, the work term in Eqs. 2.44 and 2.45 contains shaft work and flow work. Thus, W

=

Ws + (p2 V2 − p1 V1 )

(2.46)

Substituting Eq. 2.46 in Eq. 2.44 and rearranging, we get 1 Q = Ws + (U2 + p2 V2 ) − (U1 + p1 V1 ) + mg(Z2 − Z1 ) + m c22 − c21 (2.47) 2 Writing enthalpy H for the quantity U + pV , 1 1 H1 + mgZ1 + mc21 + Q = H2 + mgZ2 + mc22 + Ws (2.48) 2 2 In terms of specific quantities, 1 1 h1 + gZ1 + c21 + q = h2 + gZ2 + c22 + Ws (2.49) 2 2 Equation 2.48 or 2.49 is the steady-flow energy equation for a control volume or an open system. This will now be rewritten for processes in various turbomachines and their components.

16 2.2.2

Gas Turbines

Hydro-turbomachines

In hydro-turbomachines ρ

=

1 v

u1

≈

u2

q

≈

0

=

constant

Therefore, from Eq. 2.49, shaft work is given by 1 Ws = g(Z1 − Z2 ) + c21 − c22 + (p1 − p2 )V 2

(2.50)

In a stationary component, such as guide vanes or draught tubes, shaft work is absent. Therefore, Eq. 2.50 gives c22 − c21 2.2.3

=

2[g(Z1 − Z2 ) + (p1 − p2 )V ]

(2.51)

Compressible Flow Machines

Most of the compressible flow turbomachines such as turbines and compressors are considered as adiabatic machines, i.e., q ≈ 0. In these machines, change in potential energy (Z1 − Z2 ) is also negligible as compared c2 c2 to changes in enthalpy (h1 − h2 ) and kinetic energy 21 − 22 . Therefore, Eq. 2.49 yields 1 h1 + c21 2

=

1 h2 + c22 + Ws 2

(2.52)

The shaft work is given by Ws

=

1 h1 + c21 2

1 − h2 + c22 2

(2.53)

If the entry and exit velocities are small or the difference between them is negligible, then shaft work is given by the difference between the static enthalpies at the two states Ws 2.2.4

=

h1 − h2

(2.54)

Energy Transformation

It may be noted that energy transfer (shaft work input or output) in a turbomachine stage is possible only in the rotor, whereas energy transformation can occur both in moving and fixed blades. An application of the energy equation for stationary components of compressors and turbines such as nozzle blade rings, diffusers and volute casings can be made. The shaft work is absent in these components and the flow is almost adiabatic (q ≈ 0). Therefore Eq. 2.52 gives 1 h1 + c21 2

=

1 h2 + c22 2

=

constant

(2.55)

Review of Basic Principles

2.2.5

17

Stagnation Enthalpy

In an adiabatic energy transformation process, if the initial state is represented by h, T, c, etc., and the final gas velocity is zero, the resulting value of the enthalpy (h2 = h0 ) has a special significance. Under these conditions, Eq. 2.55 yields 1 h 0 = h + c2 (2.56) 2 Since the gas is stagnant or stationary in the final state, the quantity h0 in Eq. 2.56 is known as the stagnation enthalpy. Thus, the stagnation enthalpy can be defined as the enthalpy of a gas or vapour when it is adiabatically brought to rest. It may be observed that the definition of stagnation enthalpy in Eq. 2.56 is only another form of the energy equation. 2.2.6

Stagnation Temperature

For a perfect gas, a stagnation temperature is defined through stagnation enthalpy. From Eq. 2.56, Cp T0 T0

1 = Cp T + c2 2 = T+

c2 2Cp

(2.57)

T0 is known as the stagnation temperature whereas T is the static temperature and c2 /2Cp is the equivalent of kinetic energy temperature (Tc ). c2 2Cp

Tc

=

(2.58)

T0

= T + Tc

(2.59)

Equation 2.57 can be used to obtain an important relation for compressible flow machines. T0 T

= 1+

c2 2Cp T

(2.60)

Using Eq. 2.18, T0 T

= 1 + c2

γ−1 2γRT

The velocity of sound in a gas at a local temperature T is given by a

=

γRT

(2.61)

The Mach number of the flow is defined as the ratio of the local velocity of the gas and of sound

18

Gas Turbines

c a

=

M

=

T0 T

= 1+

c √ γRT

Therefore,

2.2.7

γ−1 2

c2 a2

(2.62)

=

1+

γ−1 M2 2

(2.63)

Stagnation Velocity of Sound

Stagnation values of various flow parameters are used as reference values in the analysis of compressible flow machines. Therefore, an expression for the stagnation velocity of sound is derived here. By definition, a0

=

γRT0

(2.64)

a0

=

(γ − 1)Cp T0

(2.65)

a0

=

(γ − 1)h0

(2.66)

Substituting for R,

Since Cp T0 = h0

2.2.8

Stagnation Pressure

The stagnation pressure is the pressure of the gas or fluid when it is brought to rest adiabatically and reversibly (i.e., isentropically). The ratio of the stagnation and static pressures can be obtained from Eq. 2.63 p0 p

=

T0 T

γ γ−1

=

γ−1 2 M 1+ 2

γ γ−1

(2.67)

When the pressure changes are small, the process can be assumed to be incompressible (ρ ≈ constant). Then the stagnation pressure can be determined from 1 p0 = p + ρc2 (2.68) 2 2.2.9

Stagnation Density

The density of a stationary gas or vapour is the stagnation density. For an ideal gas, its value at known values of stagnation temperature and pressure is given by ρ0

=

p0 RT0

(2.69)

For an isentropic process from Eq. 2.29, ρ0 ρ

=

T0 T

1 γ−1

=

γ−1 2 M 1+ 2

1 γ−1

(2.70)

Review of Basic Principles

2.2.10

19

Stagnation State

The concept of a reference state of the gas in a compressible flow machine is very useful. The stagnation state of a gas is often used as a reference state. A state defined by the stagnation temperature and pressure is the stagnation state of the gas. This state is obtained by decelerating a gas isentropically to zero velocity. It should be observed that it is necessary here to qualify the deceleration process as an isentropic process. This is not necessary in defining stagnation enthalpy and temperature.

2.3

FLUID DYNAMICS

In this section we shall discuss some basic definitions used in fluid dynamics. The analysis of flow in turbomachines requires the application of Newton’s second law of motion along with the equations of continuity and energy which will be discussed in this section. The application of Newton’s second law of motion provides the equations of motion which are also known as Euler’s momentum equations. These, and the continuity equation will be used in the subsequent chapters. We will summarize the various forms of these equations in the next section.

2.4

BASIC DEFINITIONS

First of all we will define various terms associated with fluid dynamics. These definitions will help us in understanding the various equations used in the fluid dynamics which will be applied to analyze the flow in turbomachines.

2.4.1

Fluid

A fluid is a substance which gets deformed continuously when shearing forces are applied. Liquids, gases and vapours are all fluids. A non-viscous or inviscid fluid is referred to as an ideal fluid.

2.4.2

Fluid Velocity

The instantaneous velocity of the fluid particle passing through a point is known as the fluid velocity at that point.

2.4.3

Streamline

A curve in a flow field which is always tangential to the direction of flow is referred to as a streamline. These are shown in Fig. 2.3.

20

Gas Turbines Streamlines

Stream tube

Fig. 2.3 One-dimensional flow through a stream tube 2.4.4

Stream Tube

A stream tube (Fig. 2.3) is an infinitesimal portion of the flow field. It is a collection of a number of streamlines forming an imaginary tube. There is no flow through the walls of a stream tube. The properties of the flow are constant across the section of a stream tube. Therefore, the flow in a stream tube is one-dimensional. 2.4.5

Incompressible Flow

If the relative change in the density of a fluid in a process is negligible, it is referred to as an incompressible process. In such a flow (or process) the fluid velocity is much smaller than the local velocity of sound in it. The flow of gases and vapours at Mach numbers less than 0.30 can be assumed to be incompressible without much sacrifice in accuracy. 2.4.6

Compressible Flow

In compressible flows the relative changes in the fluid density are considerable and cannot be neglected. The fluid velocities in such flows are appreciable compared to the local velocity of sound. If the Mach number in a flow is higher than 0.3, it is considered to be compressible. 2.4.7

Steady Flow

A flow is known to be steady if its properties do not change with time. The shape of the stream tube does not change in a steady flow. For such a flow ∂c ∂t 2.4.8

=

∂p ∂t

=

∂T ∂t

=

∂ρ ∂t

=

∂m ∂t

=

0

(2.71)

Unsteady Flow

If one or more parameters (c, p, T, ρ, m etc) in a flow change with time, it is known as unsteady flow.

Review of Basic Principles

2.4.9

21

Viscosity

Viscosity is the property which resists the shearing motion of two adjacent layers of the fluid. A fluid is known as a Newtonian fluid if the relation between the shear stress and the angular deformation is linear. The shear stress is given by τ

∝

dc dy

=

μ

dc dy

(2.72)

The constant of proportionality, μ, is known as the coefficient of viscosity or dynamic viscosity. The kinematic viscosity, ν, is the ratio of the dynamic viscosity and the density of the fluid. ν

=

μ ρ

(2.73)

All real flows experience fluid viscosity. Therefore, their behaviour is influenced by the viscous forces.

2.4.10

Inviscid Flow

If the viscosity of the fluid is assumed to be absent, the flow is referred to as inviscid flow. Such a flow glides freely over its boundaries without experiencing viscous forces.

2.4.11

Reynolds Number

The Reynolds number is the ratio of inertia forces to viscous forces. Inertia force = Viscous force =

ρAc2 μcl

Therefore, the Reynolds number is given by Re =

ρAc2 μcl

where l= characteristic length, A = l2 and μρ = ν. Therefore, cl ρcl = Re = μ ν

(2.74)

The value of the Reynolds number in a flow gives an idea about its nature. For example, at higher Reynolds numbers the magnitude of viscous forces is small compared to the inertia forces.

22

Gas Turbines

2.4.12

Mach Number

The Mach number is an index of the ratio of inertia and elastic forces. This is defined by M2

=

Inertia force Elastic force

=

ρAc2 KA

where K is the bulk modulus of elasticity and A is the flow area. However, K = ρa2 and therefore, M = ac . This relation gives another important definition of the Mach number as the ratio of the fluid velocity to the local velocity of sound. Thermodynamic relations derived in sections 2.2.6, 2.2.7, 2.2.8 and 2.2.9 demonstrate its application. 2.4.13

Mach Angle and Mach Cone

In a flow field if a small disturbance is introduced, it will be felt throughout the field in the form of a wave traveling at the local velocity of sound relative to the medium. Consider the propagation of pressure disturbance shown in Fig.2.4. Assume that an object (say, an aircraft) is moving with velocity c = 0, c = a/2, c = a and c > a as shown in Figs. 2.4(a – d). The moving object creates disturbance waves. As in Figs. 2.4(a and b) the disturbance waves will reach a stationary observer before the source of disturbance (moving object) could reach him in subsonic flow. In supersonic flows, Fig. 2.4(d) it takes considerable amount of time for an observer to perceive the pressure disturbance, after the source has passed him. This is one of the fundamental differences between subsonic and supersonic flows. Therefore, in a subsonic flow, the streamlines sense the presence of any obstacle in the flow field and adjust themselves ahead of it and flow around it smoothly. But in the supersonic flow, the streamlines feel the obstacle only when they hit it. The obstacle acts as a source of disturbance and so the streamlines deviate as the Mach cone as shown in Fig. 2.4(d). It should be noted that the disturbance due to obstacle is sudden and therefore, the flow behind the obstacle has to change abruptly. In Fig. 2.4(d), it is shown that for supersonic motion of an object, there is a well-defined conical zone in the flow field with the object located at the nose of the cone, and the disturbance created by the moving object is confined only to the field included inside the cone. The flow field zone outside the cone does not even feel the disturbance. For this reason, von Karman termed the region inside the cone as the zone of action, and the region outside the cone as the zone of silence. The lines at which the pressure disturbance is concentrated and which generate the cone are called Mach waves or Mach lines. The angle between the Mach line and the direction of motion of the body is called the Mach angle, μ.

Review of Basic Principles

(a) c = 0

23

(c) c = a

(b) c = a /2

Mac

h con

e Zone of action

at μ

ct Zone of silence

(d) c > a

Fig. 2.4 Propagation of disturbance wave From Fig. 2.4(d), it can be written as sin μ =

at ct

sin μ =

1 M

=

a c

(2.75) (2.76)

From the disturbance waves propagation shown in Fig. 2.4, we can infer the following features of the flow regimes. (i) When the medium is incompressible [M = 0, Fig. 2.4(a)] or when the speed of the moving disturbance is negligibly small compared to the local sound speed, the pressure pulse created by the disturbance will spread uniformly in all directions. (ii) When the disturbance source moves with a subsonic speed [M < 1, Fig. 2.4(b)], the pressure disturbance will be felt in all directions. At all points in space (neglecting viscous dissipation), the pressure pattern will not be symmetrical. (iii) For sonic velocity [M = 1, Fig. 2.4(c)] the pressure pulse will be at the boundary between subsonic and supersonic flow and the wave front will be in a plane. (iv) For supersonic speeds [M > 1, Fig. 2.4(d)] the disturbance wave phenomena are totally different from those at subsonic speeds. All the

24

Gas Turbines

pressure disturbances are included in a cone which has the disturbance source at its apex. The effect of the disturbance is not felt upstream of the disturbance source.

Sho

ck

Flow around a wedge shown in Figs. 2.5(a) and (b)can be seen with smooth change and abrupt change in flow direction for subsonic and supersonic flow, respectively.

Moo< 1

M oo >1

(a) Subsonic flow

(b) Supersonic flow

Fig. 2.5 Flow around a wedge For M∞ < 1, the flow changes its direction smoothly and pressure decreases with acceleration; for M∞ > 1; there is sudden change in flow direction at the body and pressure increases downstream of the shock.

2.4.14

Small Disturbance

When the apex angle of wedge, δ, is extremely small, the disturbances will be small. Then, we can consider these to be identical to sound pulses. In such a case, the deviation of streamlines will be small. There will be infinitesimally small increase in pressure across the Mach cone as shown in Fig. 2.6. e av

w ch

a

M

μ δ

Fig. 2.6 Mach cone

Review of Basic Principles

2.4.15

25

Finite Disturbance

Sh

oc k

When the wedge angle, δ, is finite, the disturbances introduced are finite, and then the wave is not called Mach wave but a shock or shock wave (Fig. 2.7). The angle of shock, β, is always smaller than the Mach angle.

β δ

Fig. 2.7 Shock wave The deviation of streamline is finite and there is finite pressure increase across the shock wave. 2.4.16

Transonic Flow

When a body is kept in transonic flow (Mach number between 0.8 and 1.2), it experiences subsonic flow over some portions of its surface and supersonic flow over other portions. There is also a possibility of shock formation on the body. It is this mixed nature of the flow field which makes the study of transonic flows complicated. 2.4.17

Hypersonic Flow

The temperature at stagnation point and over the surface of an object in the hypersonic flow becomes very high and, therefore, it requires special treatment. That is, we must consider the thermodynamic aspects of the flow along with gas dynamic aspects. That is why hypersonic flow theory is also called aero-thermodynamic theory. Besides, because of high temperature, the specific heats become functions of temperature and hence the gas cannot be treated as perfect gas. If the temperature is quite high (of the order of more than 2000 K), even dissociation of gas can take place. The complexities due to high temperatures associated with hypersonic flow makes its study complicated. 2.5

STREAMTUBE AREA-VELOCITY RELATION

In this section, let us consider quasi-one-dimensional flow, allowing the streamtube area A to vary with distance x, as shown in Fig. 2.8.

26

Gas Turbines A = A(x) p = p(x)

y

ρ = ρ(x) T = T(x) V = V(x) x

z

Fig. 2.8 Quasi-one-dimensional flow Let us continue to assume that all flow properties are uniform across any given cross-section of the streamtube, and hence are functions of x only for steady flows. Such a flow, where A = A(x), p = p(x), ρ = ρ(x), and c = c(x) for steady flow, is defined as quasi-one-dimensional flow. Algebraic equations for steady quasi-one-dimensional flow can be obtained by applying the integral form of the conservation equations. For any streamtube of area A, the continuity equation is given by ρAc

= constant

(2.77)

Differentiating with respect to c, we obtain d(ρAc) dA d(ρc) = ρc +A =0 dc dc dc A Since,

d(ρc) dc dp dc

= A ρ+c

dρ dp dp dc

= Aρ 1 − M 2

= −ρc

and from the Laplace equation, we have dp = a2 dρ Therefore, ρc

dA + Aρ 1 − M 2 dc dA dc

= 0 = −

A 1 − M2 c

(2.78)

Equation 2.78 is an important result. It is called the area-velocity relation. The following information can be derived from the area-velocity relation: (i) For incompressible flow limit, i.e., for M → 0, Eq. 2.78 shows that Ac = constant. This is the famous volume-conservation equation or continuity equation for incompressible flow.

Review of Basic Principles

27

(ii) For 0 ≤ M ≤ 1, a decrease in area results in increase of velocity, and vice versa. Therefore, the velocity increases in a convergent duct and decreases in a divergent duct. This result for compressible subsonic flows is the same as that for incompressible flow. (iii) For M > 1, an increase in area results in increase of velocity and vice versa, i.e., the velocity increases in a divergent duct and decreases in a convergent duct. This is directly opposite to the behaviour of subsonic flow in convergent and divergent ducts. (iv) For M = 1, by Eq. 2.78, dA/A = 0, which implies that the location where the Mach number is unity, the area of the passage is either minimum or maximum. We can easily show that the minimum in area is the only physically realistic solution. The above results are shown schematically in Fig. 2.9. Velocity, c

c2 > c1 Increases

c2 < c1

Subsonic, M < 1

Supersonic, M > 1

c1

c2

c1

c1

c2

c1

c2

c2

Decreases

Fig. 2.9 Flow in convergent and divergent ducts From the above discussions, it is c1ear that for a gas to expand isentropically from subsonic to supersonic speeds, it must flow through a convergentdivergent duct, as shown in Fig.2.10. The minimum area that divides the convergent and divergent sections of the duct is called the throat. From the point (iv) above, we know that the flow at the throat must be sonic with M = 1. Conversely, for a gas to get compressed isentropically from supersonic to subsonic speeds, it must again flow through a convergent-divergent duct, with a throat where sonic flow occurs.

28

Gas Turbines

V increasing M1

Throat� M =1

Fig. 2.10 Flow in a convergent-divergent duct 2.6

NORMAL SHOCK WAVES

The shock may be described as a compression front in a supersonic flow field and the flow process across the front results in an abrupt change in fluid properties. The thickness of the shocks is comparable to the mean free path of the gas molecules in the flow field. To have some physical feel about the formation of such shock waves, consider a cylinder placed in a flow as shown in Fig. 2.11. From the kinetic theory of gases it is known that the flow consists of a large number of fluid molecules in unit volume. The transport of mass, momentum and energy takes place through the motion of these molecules. Also, the molecules carry the signals about the presence of the cylinder around the flow field at a speed equal to the speed of sound. In Fig. 2.11(a), the incoming stream is subsonic, c∞ < a∞ , and the molecules far upstream of the cylinder get the information about the presence of the body through the signals which travel with speed, a∞ , well in advance before reaching the cylinder. Therefore, the molecules orient themselves in order to flow around the cylinder as shown in Fig. 2.11(a). But when the incoming stream is supersonic, the molecules travel faster than the signals. Therefore, there is no possibility that they will be informed of the presence of the body, before they reach the cylinder. Also, the reflected signals from the face of the cylinder tend to coalesce a short distance ahead of the body. Their coalescence forms a thin compression front called shock wave, as shown in Fig. 2.11(b). Upstream of the shock, the flow has no information about the presence of the body. However, the streamlines behind the normal shock quickly compensate for the obstruction, since the flow is subsonic after a normal shock. Although the shock formation discussed above is for a specific situation, the mechanism described is, in general, valid. However, we should realize that, when the flow just starts, there is no shock. The formation of shock takes place after the fluid molecules impinge on to the face of the cylinder and rebound. 2.7

EQUATIONS OF MOTION FOR A NORMAL SHOCK WAVE

For a quantitative analysis of changes across a normal shock wave, let us consider an adiabatic, constant-area flow through a non-equilibrium region,

Review of Basic Principles

29

Voo < a oo Moo< 1

(a) Subsonic flow Voo > a oo Moo> 1

Shoc k

(b) Supersonic flow Fig. 2.11 Streamlines around a blunt-faced cylinder in subsonic and supersonic flows as shown in Fig. 2.12(a). Let we assume that sections 1 and 2 be sufficiently away from the nonequilibrium regions so that we can define flow properties at these stations, as shown in Fig. 2.12(a). Now we can write the equations of motion for the flow considered, as follows: By continuity, ρ1 c1

=

ρ2 c2

(2.79)

The momentum equation p1 + ρ1 c21

=

p2 + ρ2 c22

(2.80)

=

1 h2 + c22 2

(2.81)

The energy equation

1 h1 + c21 2

Equations 2.79 to 2.81 are general – they apply to all gases. Also, there is no restriction on the size or details of the nonequilibrium region as long as the reference sections 1 and 2 are outside of it. The solution of these equations will provide the relations that must exist between the flow parameters at these two sections. Since, there is no restriction on the size or details of the nonequilibrium region, let us idealize it by a vanishingly thin region, as shown in Fig. 2.12(b), across which the flow parameters are said to jump. The control sections 1 and 2 may also be brought arbitrarily close to it.

30

Gas Turbines

p

p

V1

V2

T1

T2

ρ

ρ

1

2

1

2

1

2 Non-equilibrium region

(a)

Shock p

p

V1

V2

T1

T2

ρ

ρ

2

1

2

1

1

2 (b)

Fig. 2.12 Flow through a normal shock Such a front or discontinuity across which there is sudden change in flow properties is called a shock wave. There is no heat added or taken away from the flow as it traverses the shock wave; hence the flow across the shock wave is adiabatic. At this stage, one question obviously arises. In a real fluid, is it possible to have a discontinuity in a continuum flow field? It is to be clearly understood that the above consideration is only an idealization. Actually very high gradients occur in a shock wave, in the transition from state 1 to state 2. These severe gradients produce viscous stress and heat transfer, i.e., nonequilibrium conditions inside the shock. The processes taking place inside the shock wave are extremely complex. It cannot be explained on the basis of equilibrium thermodynamics. Temperature and velocity gradients internal to the shock provide heat conduction and viscous dissipation. This makes the process in the shock wave internally irreversible. In most of the practical applications, the primary interest is not generally focused on the internal mechanism of the shock wave. We are more interested on the net changes in fluid properties taking place across the wave. However, there are situations where the detailed information about the flow mechanism inside the shock describing its structure is essential for studying practical problems! However, these are beyond the scope of the book. 2.8

OBLIQUE SHOCK AND EXPANSION WAVES

It should be understood that in the normal shock, a compression wave normal to the flow direction is generated. However, in a majority of physical

Review of Basic Principles

31

situations, a compression wave inclined at an angle to the flow occurs. Such a wave is called an oblique shock. Hence, the normal shock wave can be considered as a special case of oblique shock wave. If we superimpose a uniform velocity normal to the upstream flow of the normal shock, then, it will result in a flow filed of oblique shock wave. This phenomenon can be used to get the oblique shock relations. Oblique shocks usually occur when a supersonic flow is turned into itself. The opposite of this, i.e., when a supersonic flow is turned away from itself, results in the formation of an expansion fan. These two families of waves play a dominant role in all flow fields involving supersonic velocities. Typical flows with oblique shock and expansion fan are illustrated in Fig. 2.13.

Oblique shock 1

Expansion fan

2 M 2 < M1

M 2 > M1

1 M1

M1

2 M2 θ

θ (a) Compressor corner

(b) Expansion corner

Fig. 2.13 Supersonic flow over corners

In Fig. 2.13(a) the flow is deflected into itself by the oblique shock. All the streamlines are deflected to the same angle θ at the shock, resulting in uniform parallel flow downstream of shock. The angle θ is referred to as flow deflection angle. Across the shock wave, the Mach number decreases, and the pressure, density, and temperature increase. The corner which turns the flow into itself is called compression or concave corner. In contrast, in an expansion or convex corner, the flow is turned away from itself through an expansion fan. All the streamlines are deflected to the same angle θ after the expansion fan, resulting in uniform parallel flow downstream of the fan. Across the expansion wave, the Mach number increases, and the pressure, density and temperature decrease. From Fig. 2.13, it is seen that the flow turns suddenly across the shock and the turning is gradual across the expansion fan, and hence all flow properties through the expansion fan change smoothly, with the exception of the wall streamline which changes suddenly. Oblique shock and expansion waves prevail in two- and three-dimensional supersonic flows, in contrast to normal shock waves, which are one-dimensional. In this chapter, we shall focus our attention on steady, two-dimensional (plane) supersonic flow.

32

Gas Turbines

2.9

FLOW WITH FRICTION AND HEAT TRANSFER

So far, we have discussed compressible flow of gases in ducts, where changes in flow properties were brought about solely by area change, i.e., where effects of viscosity are neglected. But, in a real flow situation like stationary power plants, aircraft propulsion engines, high-vacuum technology, transport of natural gas in long pipelines, transport of fluids in chemical process plants, and various types of flow systems, the high-speed flow travels through passages of sufficient length, the effects of viscosity (friction) cannot be neglected. In many practical flow situations, friction may even have a decisive effect on the resultant flow characteristics. The inclusion of friction terms in the equations of motion makes the analysis of the problem far more complex. 2.10

FLOW IN CONSTANT-AREA DUCT WITH FRICTION

Consider one-dimensional steady flow of a perfect gas, with constant specific heats, through a constant-area duct. Also, let there be neither external heat exchange nor external shaft work and let differences in elevation produce negligible changes compared to frictional effects. The flow with the above mentioned conditions, namely, adiabatic flow with no external work, is called Fanno line flow. Let the wall friction (due to viscosity) be the chief factor bringing about changes in fluid properties, for the adiabatic compressible flow through ducts of constant-area,under consideration. The energy equation of steady flow under the above assumptions may be written as h+

c2 2

=

h0

(2.82)

where h and c are respectively the corresponding values of the enthalpy and velocity at an arbitrary section of the duct, and h0 (the stagnation enthalpy) has a constant value for all sections of the duct. By equation of continuity, m ˙ = ρc = G (2.83) A where ρ is the density at the section where c and h are measured, and G is called the mass velocity, which has a constant value for all sections of the duct. Combining Eqs. 2.82 and 2.83, we get the equation of the Fanno line in terms of the enthalpy and density as G2 h = h0 − 2 (2.84) 2ρ where h and h0 are the static and stagnation enthalpy, ρ is density and G is mass velocity. This equation shows that for a given initial condition,

Review of Basic Principles

33

the relation between the local density ρ and local enthalpy h is fixed. This implies that the relation between any two properties of the flowing gas is also fixed. The friction coefficient f is defined as f

=

wall shearing stress dynamic head of the stream

The hydraulic diameter, D is defined as 4 (cross-sectional area) D = wetted perimeter

(2.85)

(2.86)

The advantage of using hydraulic diameter is that the equations, in terms of hydraulic diameter, are valid even for ducts with non-circular cross-section. The maximum length of the duct required for the flow to choke for a given initial Mach number is given by 4f

Lmax D

=

1 − M2 (γ + 1)M 2 γ+1 ln + 2 γM 2 2γ 2 1 + γ−1 2 M

(2.87)

where f is the mean friction coefficient with respect to duct length, defined by f

=

Lmax

1 Lmax

f dx

(2.88)

0

The duct length required for the flow to pass from a given initial Mach number M1 to a given final Mach number M2 can be obtained from the expression 4f

L D

=

4f

Lmax D

M1

− 4f

Lmax D

(2.89) M2

It should be understood that Fanno flow friction always drives the Mach number towards unity, decelerating a supersonic flow and accelerating a subsonic flow. For any given initial Mach number, for a certain value of L the flow becomes sonic. For this condition the flow is said to be choked, since any further increase in L is not possible without causing a drastic change of the inlet conditions. For instance, if the inlet conditions were achieved by expansion through a supersonic nozzle, and if L were larger than that allowed for attaining Mach 1 at the exit, then a normal shock would form inside the supersonic nozzle and the duct inlet conditions would suddenly become subsonic. It is important to note that friction always causes the total pressure to decrease whether the inlet flow is subsonic or supersonic. Further, unlike the Rayleigh curve for flow with heating or cooling, the upper and lower portions of the Fanno curve cannot be traversed by the same one-dimensional flow. In other words, it is not possible to first decelerate a supersonic flow to sonic condition by friction, and then further retard it to subsonic speeds

34

Gas Turbines

also by friction, since such a subsonic deceleration violates the second law of thermodynamics. In summary, it can be stated that change of state in flow properties is achieved by the following three means: (i) with area change, treating the fluid to be inviscid and passage to be frictionless, (ii) with friction, considering the heat transfer between the surrounding and system to be negligible, and (iii) with heat transfer, assuming the fluid to be inviscid and passage to be frictionless. These three types of flows are called isentropic flow, frictional or Fannotype flow , and Rayleigh-type flow , respectively. All gas dynamic problems encountered in practice can be classified under these three flow processes, of course with appropriate assumptions. Although it is impossible to have a flow process which is purely isentropic or Fanno-type or Rayleigh-type. In practice, it is justified in assuming so, since the results obtained from these processes prove to be accurate enough for most of the practical situations in gas dynamics. Flows in which wall friction is the chief factor bringing about changes in fluid properties, assuming that no heat is transferred to or from the fluid stream are termed Fanno-type flow. When the ducts are short, the flow is approximately adiabatic. However, when the ducts are extremely long, as in the case of natural gas pipe lines, there is sufficient area for heat transfer to make the flow non-adiabatic and approximately isothermal. 2.10.1

Laminar Flow

In laminar flow the fluid flows over a body in orderly parallel layers with no components of fluctuations in any of the three directions (x, y and z directions). In such a flow, the surface friction force predominates and keeps the flow parallel to the surface. Other layers of flow slide on top of the other. The values of the Reynolds number in such flows are comparatively lower. 2.10.2

Turbulent Flow

At higher values of the Reynolds number, the inertia force becomes predominant and the fluid particles are no longer constrained to move in parallel layers. Such a flow experiences small fluctuation components cx , cy and cz in the three reference directions. These fluctuations cause continuous mixing of various layers of the flow leading to flow equalization in the major part of the flow field. On account of different flow patterns in laminar and turbulent flow, the velocity profiles (shown in Fig. 2.14) in these are different. The nature of flow in blade passages in a turbine can be identified to a great extent by the velocity profiles and the values of Reynolds number.

Review of Basic Principles

Flow passage Laminar

Turbulent

35

36

Gas Turbines

2.10.5

Friction Factor

Friction factor or the coefficient of skin friction is a measure of the frictional resistance offered to the flow. This is defined by f

=

τw 1 2 2 ρc

(2.91)

where f is the Fanning’s coefficient of skin friction. It may be noted that Darcy’s friction factor is four times the Fanning’s coefficient. 2.10.6

Boundary Layer Separation

The boundary layer is the slow-moving or tired layer of the flow near a solid surface. When the flow occurs in the direction of static pressure rise (adverse pressure gradient), the boundary layer becomes thicker and reverses if this static pressure gradient (or the pressure hill) is too high. The leaving of the boundary layer from the surface and its reversal is known as separation. This leads to chaotic flow, large drag and high energy losses. In an accelerating flow, on account of the continuously decreasing static pressure the thickening of the boundary layer is prevented; in fact, the higher inertia forces make it thinner. The available pressure drop helps in washing down any localized thickening of the boundary layer or its separation. The separation of boundary layer and the point of separation depend on the geometry and roughness of the surface, nature of the flow (turbulent or laminar) and Reynolds number. The laminar boundary layer gets separated earlier than the turbulent boundary layer. In order to achieve high lift and good performance it is necessary to prevent or delay the separation of the boundary layer. Two of the methods to achieve this are (i) sucking away the decelerated layer, and (ii) energizing it by injecting high energy fluid parallel to the surface. Separation can also be delayed by achieving transition of the laminar flow into the turbulent earlier. In all turbomachines boundary layer separation should be avoided. Review Questions 2.1 State the various laws used in the design of turbomachines. 2.2 Explain (i) open system, and (ii) closed system. 2.3 State the relationship between enthalpy and internal energy? 2.4 State Clausius and Kelvin-Planck’s statement of second law of thermodynamic. 2.5 Explain the flow and non-flow process. 2.6 Derive steady-state energy equation. 2.7 Explain the terms ’stagnation pressure’ and ’stagnation temperature’.

Review of Basic Principles

37

2.8 What are streamlines and streamtubes? Explain with a sketch. 2.9 Explain the difference between laminar and turbulent flow. 2.10 What is a boundary layer? Give a brief account of boundary layer separation. Multiple Choice Questions (choose the most appropriate answer) 1. Pressure can be expressed in units (a) Pascal (b) N/m2 (c) bar (d) none of the above 2. Cv can be expressed as (a) Cv =

∂q ∂T

(b) Cv =

∂T ∂q

(c) Cv = (d) Cv =

v

v ∂T ∂v v ∂h ∂P v

3. Cp can be expressed as (a) Cp =

∂T ∂h v

(b) Cp =

∂T ∂q

(c) Cp =

∂q ∂T

(d) Cp =

v

v ∂h ∂T v

4. A process that occurs in a closed system is (a) closed flow process (b) non-flow process (c) open flow process (d) none of the above 5. A process that occurs in a open system is (a) open flow process (b) non-flow process (c) flow process (d) none of the above

38

Gas Turbines

6. Energy transformation can occur (a) in moving blades (b) in fixed blades (c) both moving and fixed blades (d) none of the above 7. If the relative change in the density of a fluid in a process is negligible then it is called (a) compressible flow (b) incompressible flow (c) steady flow (d) unsteady flow 8. Reynolds number is the ratio of (a) elastic force / viscous force (b) inertia force / elastic force (c) viscous force / inertia force (d) inertia force / viscous force 9. Mach number is the ratio of (a) inertia force / viscous force (b) inertia force / elastic force (c) viscous force / inertia force (d) elastic force / viscous force 10. In a pipe flow if the Reynolds number is greater than 2000 then it is considered as (a) laminar (b) steady (c) turbulent (d) unsteady Ans:

1. – (d) 6. – (c)

2. – (a) 7. – (b)

3. – (d) 8. – (d)

4. – (b) 9. – (b)

5. – (c) 10. – (c)

3 FUNDAMENTALS OF ROTATING MACHINES INTRODUCTION In the last chapter we reviewed some basic principles useful for the gas turbine cycle analysis. In this chapter we will deal with the fundamental principle of operation of rotating machines. Rotating machines are usually called turbomachines. These machines work on the principle of work addition or extraction. When a fluid passes through a rotating machine two things happen, viz., energy transfer and energy transformation. The energy transfer means transfer of available energy from one part (rotor) to the medium (fluid) or vice versa. Energy transformation means change of one form of energy into another form, for example, change of kinetic energy to pressure energy in a compressor. The energy transfer can occur only in its moving or rotating elements whereas the energy transformation can occur in both stationary and rotating elements. As both the compressors and turbines are concerned with energy transfer, we will consider their basic performance together. In compressors the energy is transferred from the rotor to the fluid while in turbines it is from fluid to the rotor. The effectiveness of this transfer of energy in a fluid machine is governed mainly by the fluid dynamics of the system. 3.1

GENERAL FLUID DYNAMICS ANALYSIS

Figure 3.1 shows the details of the passage of a fluid through a rotor of any shape. The rotor has an axis A − A and rotates at a steady angular velocity of ω radians per second. Let us assume that at point 1 the fluid enters with a velocity c1 and leaves at point 2 with velocity c2 . The radial distance of these points from the axis A − A is r1 and r2 respectively. The velocity c1 can be represented by three velocity components, viz., (i)

ca1

: axial velocity in a direction parallel to the axis A − A

40

Gas Turbines

c r2 ca2 c2

A

2 r2

ct2

w rad/s r1

A

1 c t1

c a1

c r1 c 1

Fig. 3.1 Fluid flow through a rotor (ii) cr1 (iii) ct1

: radial velocity in the direction normal to A − A : tangential velocity in the direction normal to any radius

Similarly, at the exit point 2, the velocity c2 will have the three components, ca2 , cr2 and ct2 respectively. Figure 3.2 shows the velocity triangles at the entry as well as at the exit of a general rotating machine. All velocity vectors shown are in the same plane and assumed to remain constant over the entire entry and exit section.

Exit

Blade row

Entry

c

1

c2

c

r2

w2

c t2

u2

c r1 w1

ct1

w

u1

r1 r2

Fig. 3.2 Energy transfer in a turbomachine The angular speed of the rotor is ω radians per second. ω

=

2πN 60

Fundamentals of Rotating Machines

41

The peripheral velocities of the blades at the entry and exit corresponding to diameters d1 and d2 are u1

=

πN d1 60

u2

=

πN d2 60

The directions of the relative velocity vectors correspond to the rotor blade angles. The absolute velocity vectors, c, are those which will be observed by a person standing outside the rotor, whereas, the relative velocity vectors, w are the ones which will be observed by an observer positioned on the rotor. The three velocity vectors c, w and u at a section (or station) are related by the simple vector equation c

=

u+w

The absolute velocity c at both the entry and exit has a tangential component ct and a radial component cr . The torque exerted by the rotor or by the fluid is obtained by employing Newton’s second law of motion for the change of moment of momentum. Torque

=

Rate of change in moment of momentum

Considering the interaction of fluid and rotor it will be seen that axial and radial components do not contribute to the rotation. The axial component, ca , produces thrust and the radial component, cr , the radial force. The tangential component, ct , produces rotational effects. Considering unit mass of fluid, entering and leaving in unit time, we can write The angular moment of momentum at inlet

= ct1 r1

(3.1)

The angular moment of momentum at exit = ct2 r2

(3.2)

As per Newton’s law, the rate of change of angular moment of momentum is equal to the torque produced. Therefore, T

= ct1 r1 − ct2 r2

(3.3)

and hence the rate of energy transfer for unit mass flow per unit time will be given by E

=

Tω

= ω(ct1 r1 − ct2 r2 )

=

ct1 r1 ω − ct2 r2 ω

(3.4)

But, ωr1 = u1 and ωr2 = u2 . Hence, E

= ct1 u1 − ct2 u2

(3.5)

The above equation is called Euler’s energy equation. The equation can be analyzed under two conditions, viz.,

42

Gas Turbines

(i) For ct1 u1 > ct2 u2 , in which the energy transfer is positive (ii) For ct1 u1 < ct2 u2 , in which the energy transfer is negative Positive energy transfer means that the energy is transferred from fluid to rotor and therefore it is called turbine rotor. Negative energy transfer means that the energy is transferred from rotor to the fluid and therefore it is called compressor rotor. Therefore, the energy transfer equation can be written now separately as ET

= ct1 u1 − ct2 u2

(3.6)

EC

= ct2 u2 − ct1 u1

(3.7)

Equations 3.6 and 3.7 can be used freely for any type of turbine or compressor respectively which satisfy the following few conditions. (i) The flow must be steady, i.e., there should not be any change in angular velocity, flow rate, fluid properties and heat transfer rate with respect to time. (ii) The relationship applies strictly to every infinitesimal stream line, i.e., if the velocity is not uniform over the inlet and exit areas then an integration over each area must be done. (iii) There should not be any discontinuity of pressure, i.e., a choked nozzle at the rotor discharge. In almost all applications, conditions (i), (ii) and (iii) are mostly satisfied. 3.2

THE PHYSICAL MEANING OF THE ENERGY EQUATION

Consider Fig. 3.3 which is the velocity diagram at the exit from a rotor. From geometry of the velocity triangle we have c2r2

=

c22 − c2t2

c2r2

=

w22 − (u2 − ct2 )

and also

(3.8) 2

(3.9)

where cr2 is called meridional component. Equating the values of c2r2 and expanding c22 − c2t2

=

w22 − u22 + 2ct2 u2 − c2t2

(3.10)

ct2 u2

=

1 2 c + u22 − w22 2 2

(3.11)

hence,

Fundamentals of Rotating Machines

43

A similar expression can be obtained for inlet, thus ct1 u1

=

1 2 c + u21 − w12 2 1

(3.12)

Now the energy transfer as per Eq. 3.5 can be written as E = ct1 u1 − ct2 u2 hence, E

=

⎤

⎡

1⎢ 2 ⎥ ⎣ c1 − c22 + u21 − u22 + w22 − w12 ⎦ 2 I

II

c2

(3.13)

III

w2 cr2

c t2

u2

Fig. 3.3 Velocity diagram at the exit of a rotor Let us consider the three terms (I), (II) and (III) separately, in order to understand their significance. c2 −c2

The first term 1 2 2 represents the energy transfer due to change of absolute kinetic energy of the fluid between the entrance and the exit sections. This effect is also known as impulse effect. Since it is concerned with the absolute velocity, it represents a virtual pressure rise which can be accomplished if c2 is reduced in a diffuser so as to bring about a rise in pressure. The rotor of a compressor will usually increase the absolute velocity of the fluid but it does not necessarily increase the fluid pressure. Therefore, an additional device is needed in order to slow down the absolute velocity to some lower magnitude so that a static pressure rise may be accomplished. All compressors have either a scroll or a diffuser or some type of vanes which follow the rotating passage in order to reduce the exit velocity of the fluid. As this absolute kinetic energy change may be used to accomplish a rise in pressure, it can be called a virtual pressure rise or a pressure rise which is theoretically possible to attain. If we can slow down the air velocity by some means from c2 to c1 we can attain the ideal head provided that the process is isentropic. Since this pressure rise comes from some device c2 −c2 which is external to the rotor, the term, 2 2 1 is sometimes called external effect.

44

Gas Turbines

The other two terms, viz., the second and the third in Eq. 3.13 are factors which produce pressure rise within the rotor itself and therefore they are called internal effect. u2 −u2

The term 1 2 2 represents the energy transfer due to centrifugal effect. It is well known from fluid mechanics that a rotating fluid in equilibrium has a pressure at any radius proportional to the product of the square of radius and the square of the angular velocity, ω 2 r2 , i.e., u2 . w 2 −w 2

The third term 2 2 1 represents the energy transfer due to the change of the relative kinetic energy of the fluid. This is nothing but a reaction effect and gives rise to the change in static pressure. From the above discussion, it is apparent that a turbomachine derives its pressure change due to three distinct effects. The first term represents energy transfer due to the change of kinetic energy. The last two terms represent energy change involving change of static pressure. Thus, energy transfer is dependent on a change of stagnation or total pressure, of which dynamic component may be important. In some rotating machines it may happen that only one of the three individual types of energy transfer is present. In particular, the second term of the Eq. 3.13 shows that a radial path of the fluid can be effective. That is to say that substantial effect may be produced by allowing the fluid to enter and leave the rotor at different radii. It is also possible for one of the components to be acting in the opposite direction to the other two so that the net effect is as desired. 3.3

CLASSIFICATION OF MACHINES

The compressor and turbine can be classified based upon (i) the flow direction, and (ii) the degree of reaction. The details of these two parameters are discussed in the following sections. 3.3.1

Classification Based upon Flow Direction

Based on the flow direction, machines can be divided into axial, radial or mixed flow type. (i) Axial Flow The machines which have no significant change of radius between flow entry and exit, i.e., u1 ≈ u2 are called axial machines. (ii) Radial Flow In this type there is a substantial change of energy due to change of radius between the entry and the exit of flow. These are often called as centrifugal machines in case of compressors and centripetal machines in case of turbines. (iii) Mixed Flow Although the majority of compressors and turbines in present day use may be classified as axial or radial, it is possible to have mixed-flow types in which neither effect clearly dominates.

Fundamentals of Rotating Machines

3.3.2

45

Classification Based upon the Degree of Reaction

Degree of reaction [R] is usually defined as the rate of energy transfer by virtue of change of static pressure to the total energy transfer. R =

1 2

u21 − u22 + w22 − w12 E

(3.14)

where E may be expressed either by Eq. 3.6 or by Eq. 3.7 depending on whether it is for turbine or compressor respectively. Another way of defining R is as follows: Change of enthalpy in rotor R = (3.15) Change of enthalpy in stage where stage is the combination of stator and rotor. Hence, for compressor R =

Enthalpy increase in rotor Enthalpy increase in stage

R =

Enthalpy drop in rotor Enthalpy drop in stage

and for turbine (3.16)

Based on the above definitions the machines can be classified as (i) Impulse Machine In impulse machines there is no change of static pressure in the rotor. Therefore these machines have a degree of reaction (R) equal to zero. In the case of a turbine of this type the energy transfer is wholly effected by a jet of fluid striking the blade. A simple example of pure impulse machine is a paddle wheel or a Pelton wheel. (ii) Reaction Machine The reaction principle is best illustrated by rocket propulsion or ordinary lawn sprinkler. In a pure reaction machine, R = 1, all energy transfer occurs by virtue of change of static pressure in the rotor. A reaction of unity for a compressor means that the fluid enters and leaves with the same absolute velocity. 3.4

GENERAL THERMODYNAMIC ANALYSIS

So far we have talked about the energy transfer by the application of the principle of fluid dynamics, i.e., in terms of rotor and fluid velocities. The work output or the energy transfer can also be determined based on thermodynamic considerations. The two equations, viz., the fluid dynamics and thermodynamics equations can be linked to determine the unknown quantities. The process of expansion and compression taking place in a gas turbine engine will fall into the category of steady flow process. Considering unit

46

Gas Turbines

Q

W

1

2

c1

c2

.

.

.

.

.

.

.

.

Z1

.

Z2 = Z 1

Fig. 3.4 Thermodynamic analysis of fluid flow mass of the working fluid and writing the general energy equation (Fig. 3.4), we get U1 + p1 V1 +

c21 c2 + gZ1 + Q + W = U2 + p2 V2 + 2 + gZ2 2 2

(3.17)

where the symbols denote U pV c Z Q&W

: : : : :

Internal energy Flow work done on or by the fluid Velocity of the fluid Potential energy Heat and work supplied from external source

In the case of gas turbine, the rate of flow of working fluid is quite large, and the surface area available for transfer of heat is quite small. Therefore, the process may be assumed to be adiabatic, i.e., Q

=

0

Equation 3.17 can now be written as U1 + p1 V1 +

c21 + gZ1 + W 2

=

U2 + p2 V2 +

c22 + gZ2 2

If Z2 ≈ Z1 as in Fig. 3.4, Eq. 3.18 reduces to W

= (U2 + p2 V2 ) − (U1 + p1 V1 ) −

c2 c21 − 2 2 2

= (Cv T2 + RT2 ) − (Cv T1 + RT1 ) −

c2 c21 + 2 2 2

(3.18)

Fundamentals of Rotating Machines

= (T2 − T1 )(Cv + R) − W

= Cp (T2 − T1 ) − =

W

h2 +

c22 2

c2 c21 + 2 2 2

47

(3.19)

c2 c2 c21 c2 + 2 = (h2 − h1 ) − 1 + 2 2 2 2 2

− h1 +

c21 2

(3.20)

= h02 − h01

(3.21)

= Δh0

(3.22)

and if we neglect the change in kinetic energy then W

= Δh

(3.23)

[Suffix 0 denotes total head or stagnation condition]. If in Eq. 3.22 or 3.23, change of enthalpy is positive then the work input is considered positive and the case is that of a compressor. If it is negative, that means the work input is negative, i.e., there is a work output and the case is that of a turbine. (3.24) 3.5

EFFICIENCY OF ROTATING MACHINES

An actual machine receiving gas at one particular condition and exhausting at another condition will, because of losses, do less work than represented by Eq. 3.23 for an expansion process and will absorb more amount of work for a compression process. This action gives rise to the term efficiency commonly called isentropic efficiency of the machine. For a compression machine Isentropic efficiency (ηC )

=

Isentropic work input Actual work input

ηC

=

h2 − h1 h2 − h1

(3.25)

ηC

=

T2 − T1 T2 − T1

(3.26)

Isentropic efficiency (ηT )

=

Actual work output Isentropic work output

ηT

=

h3 − h4 h3 − h4

If Cp is taken to be constant, then

For an expansion machine

If Cp is taken to be constant, then

(3.27)

48

Gas Turbines

ηT

=

T3 − T4 T3 − T4

(3.28)

ta nt

The isentropic efficiencies of compressor and turbine can be evaluated from the enthalpy-entropy (h-s) diagrams, the details of which are given in Figs.3.5(a) and (b) respectively.

co

ns

3

p

=

nt

h 2’

ta ns

2

2

=

co

p1

nt

h

a nst

o

=c p2

p 1=

ant

st con

4 4’

1 s

s

(a) Compressor

(b) Turbine

Fig. 3.5 h-s diagram for compressor and turbine

3.6

DIMENSIONAL ANALYSIS OF ROTATING MACHINES

Basically, dimensional analysis is a method for reducing the number and complexity of experimental variables which affect a given physical phenomenon, using a sort of compacting technique. If a phenomenon depends upon n dimensional variables, dimensional analysis will reduce the problem to only k dimensionless variables, where the reduction n − k = 1, 2, 3 or 4 depending upon the problem complexity. Generally n − k equals the number of different dimensions (sometimes called basic or primary or fundamental dimensions) which govern the problem. In fluid flow problems, the four basic dimensions are usually taken to be mass M , length L, time T and temperature Θ or an M LT Θ system for short. Sometimes one uses an F LT Θ system, with force F replacing mass, M . In SI system of units (System Internationale d’units) the units and dimensions used for fluid mechanics properties are given in Table 3.1. Dimensional analysis of problems in turbomachines identifies the variables involved and groups them into non-dimensional quantities much lesser in number than the variables themselves. In a design problem or performance test these non-dimensional quantities (or dimensional parameters or numbers as they are sometimes called) are varied instead of the large number of parameters forming these groups. While a great convenience and economy in test runs provided by employing this technique is obvious, the design procedure uses these non-dimensional numbers to obtain optimum performance. Some non-dimensional numbers give an idea of the type of

Fundamentals of Rotating Machines

Table 3.1 Units and Dimensions of Fluid-Mechanics Properties Quantity

Units

Symbol

Dimensions M LT Θ F LT Θ

Length

m

L

L

L

Area

m2

A

L2

L2

Volume

m3

V

L3

L3

Velocity

m/s

c or u

LT −1

LT −1

Speed of sound

m/s

a

LT −1

LT −1

Volume flow

m3 /s

Q

L3 T −1

L3 T −1

Mass flow

kg/s

m ˙

M T −1

F T L−1

Pressure, stress

kg/ms2

p, σ

M L−1 T −2

F L−2

Strain rate

1/s

T −1

T −1

Angle

None

θ

None

None

Angular velocity

1/s

ω

T −1

T −1

Viscosity

kg/m s

μ

M L−1 T −1

F T L−2

Kinematic viscosity

m2 /s

ν

L2 T −1

L2 T −1

Surface tension

kg/s2

γ

M T −2

F L−1

Force

kg m/s2

F

M LT −2

F

Moment, torque

kg m2 /s2

M

M L2 T −2

FL

Power

kg m2 /s

P

M L2 T −1

F LT −1

Work, energy

kg m2 /s2

W, E

M L2 T −2

FL

Density

kg/m3

ρ

M L−3

F T 2 L−4

Temperature

K

T

Θ

Θ

Specific heat

m2 /s2 K

Cp , Cv

L2 T −2 Θ−1

L2 T −2 Θ−1

Thermal conductivity

kg m/s3 K

k

M LT −3 Θ−1

F T −1 Θ−1

Expansion coefficient

1/K

β

Θ−1

Θ−1

49

50

Gas Turbines

machine and its range of operation. The presentation of the performance of a machine is also considerably simplified by adopting non-dimensional numbers. 3.6.1

The pi Theorem

There are several methods of reducing a number of dimensional variables into a smaller number of dimensionless groups. The scheme given here was proposed in 1914 by Buckingham and is now called the Buckingham pi theorem. The name pi comes from the mathematical notation Π, meaning a product of variables. The dimensionless groups found from the theorem are power products denoted by Π1 , Π2 , Π3 , etc., The method allows the pis to be found in sequential order without resorting to free exponents. The first part of the pi theorem explains what reduction in variables to expect: If a physical process satisfies the Principle of Dimensional Homogeneity (PDH) and involves n dimensionless variables, it can be reduced to a relation between only k dimensionless variables or Π’s. The reduction j = n−k equals the maximum number of variables which do not form a pi among themselves and is always less than or equal to the number of dimensions describing the variables. The second part of the theorem shows how to find the pis one at a time: Find the reduction j, then select j variables which do not form a pi among themselves† . Each desired pi group will be a power product of these j variables plus one additional variable which is assigned any convenient nonzero exponent. Each pi group, thus, found is independent. To be specific, suppose that the process involves n variables and assuming n = 5, we have, v1 = f (v2 , v3 , v4 , v5 ) If the five variables can be expressed in terms of three (j) dimensions (M LT ), then the number of dimensionless variables (k) or Πs will be k = 5 − 3 = 2. That is, we can expect from the theorem, two and only two pi groups. Now let us pick out three convenient variables which do not form a pi and suppose, these turn out to be v2 , v3 and v4 . Then the two pi groups are formed by power products of these three plus one additional variable Π1

=

M 0 L0 T 0

=

(v2 )a (v3 )b (v4 )c v1

Π2

=

M 0 L0 T 0

=

(v2 )a (v3 )b (v4 )c v5

Here, we have arbitrarily chosen the added variables, v1 and v5 , whose exponents is taken as unity. Equating exponents of the various dimensions is guaranteed by the theorem to give unique values of a, b and c for each pi. And they are independent because only Π1 contains v1 and only Π2 contains v5 . It is a very neat system once you get used to the procedure. † Make a clever choice here because all pis will contain these j variables in various groupings

Fundamentals of Rotating Machines

51

Typically, there are six steps involved: 1. List and count the n variables involved in the problem. If any important variables are missing, dimensional analysis will fail. 2. List the dimensions of each variable according to M LT Θ or F LT Θ. A list is given in Table 3.1. 3. Find j, initially guess j equal to the number of different dimensions present and look for j variables which do not form a pi product. If no luck, reduce j by 1 and look again. With practice, you will find j rapidly. 4. Select j variables which do not form a pi product. Make sure they please you and have some generality if possible, because they will then appear in every one of your pi groups. 5. Add one additional variable to your j variables and form a power product. Algebraically find the exponents which make the product dimensionless. Try to arrange for your output or dependent variables (force, pressure drop, torque, power) to appear in the numerator and your plots will look better. Do this sequentially, adding one new variable each time, and you will find all n − j = k desired pi products. 6. Write the final dimensionless function and check your work to make sure all pi groups are dimensionless. 3.6.2

Non-dimensional Parameters for Rotating Machines

The technique of dimensional analysis is used to reduce the number of variables into a few number of dimensionless groups. The dimensionless groups involved in rotating machines for plotting the necessary performance curves can be obtained in the following way. Step 1 Write the function and count the variables. Let us express the outlet pressure from the compressor, p2 , as function of other variables, viz., p2 where

p2 p1 D ρ1 m ˙ N

: : : : : :

=

f (p1 , ρ1 , D, m, ˙ N)

(3.29)

outlet pressure from the compressor inlet pressure to the compressor characteristic linear dimension density of the fluid mass flow rate of the fluid rotational speed

Note that in this problem there are six variables. Step 2 Let us now write the dimensions of each of the variables (refer Table 3.1). p : M L−1 T −2 ρ : M L−3 D : L m ˙ : M T −1 N : T −1 Step 3 Find j. No variable contains the dimension Θ and so j is less or equal to 3 (M LT ). On inspection of the list it can be seen that p1 , ρ and

52

Gas Turbines

D cannot form a pi group. Therefore, j equals 3, and n − j = 6 − 3 = 3. The pi theorem guarantees for this problem that there will be exactly three independent dimensionless groups. Step 4 Select j variables. The group p1 , ρ, D will do the job. Step 5 Combine p1 , ρ and D with one additional variable, in sequence to find the three pi products. First add p2 to find Π1 . You may select any exponent on this additional term as you please, to place it in the numerator or denominator to any power. Since p2 is the output, or dependent variable, we select it to appear to the first power in the numerator. Π1

=

pa1 ρb Dc p2

Π1

=

M 0 L0 T 0 M a L−a T −2a M b L−3b [Lc ] M L−1 T −2

= =

(3.30)

M a+b+1 L−a−3b+c−1 T −2a−2

(3.31)

Equating exponents, Mass Length Time

: a+b+1 : −a − 3b + c − 1 : −2a − 2

= = =

0 0 0

On solving the above three simultaneous equations, we get a

=

−1;

b

=

0;

c

=

0

Π1

=

p2 p1

(3.32)

Π2

=

pa1 ρb Dc N

(3.33)

Π2

=

M 0 L0 T 0

Hence, from Eq. 3.30,

M a L−a T −2a M b L−3b [Lc ] T −1

=

M a+b L−a−3b+c T −2a−1

= Equating exponents, Mass : a + b Length : −a − 3b + c Time : −2a − 1

= = =

0 0 0

(3.34)

Fundamentals of Rotating Machines

53

On solving the above simultaneous equations, we have a

1 = − ; 2

b

=

1 ; 2

c

=

1

√ ND ρ √ p1

Π2

=

Π3

= pa1 ρb Dc m ˙

Π3

= M 0 L0 T 0

(3.35) (3.36)

M a L−a T −2a M b L−3b [Lc ] M T −1

=

= M a+b+1 L−a−3b+c T −2a−1

(3.37) (3.38)

Equating exponents Mass : a + b + 1 Length : −a − 3b + c Time : −2a − 1

= = =

0 0 0

On solving the above simultaneous equations, we have a

1 − ; 2

=

b Π3

1 = − ; 2 =

c

=

m ˙ √ D 2 p1 ρ

−2 (3.39)

Step 6 Π1

= f (Π2 , Π3 )

p2 p1

= f

Therefore,

√ N D ρ1 m ˙ , 2√ √ p1 D p1 ρ

For a given compressor and working fluid, D and the gas constant, R, are fixed. Writing p in terms of ρRT , the pressure ratio is given by √ p2 N m ˙ T1 (3.40) = f √ , p1 p1 T1 The process may be repeated, letting the outlet temperature T2 be a function of the same variables, in which case TT21 will again be found to be the √

function of √NT and m˙ p1T1 . If the stagnation pressures and temperatures 1 are considered (as in the case of a gas turbine), then again Eq. 3.40 will result except that stagnation temperatures and pressures are substituted for their static parameters in the equations. The experimentally determined

54

Gas Turbines

N

Surge line

= constant

T 01

p

02

p

01

m T 01 p

01

Fig. 3.6 Performance map of a compressor performance map for a compressor based on the Eq. 3.40 will be as shown in Fig. 3.6. In the above analysis, it may be noted that the viscosity was not considered. Over the operating range of compressors and turbines the variation of μ has a negligible effect on their performance and hence the exclusion of viscosity from the analysis may be justified. 3.7

ELEMENTARY AIRFOIL THEORY

An airfoil may be defined as a streamlined body bounded principally by two flattened curves and whose length and width are very large in comparison with the thickness. It has a thick, rounded leading edge and a thin (sometimes sharp) trailing edge. Its maximum thickness occurs somewhere near the midpoint of the chord. The backbone line lying midway between the upper and lower surfaces is known as the camber line. When such a blade is suitably shaped and properly oriented in the flow, the force acting on it - normal to the direction of flow - is considerably larger than the force resisting its motion. Aerofoil shapes are used for aircraft wing sections and the blades of various turbomachines. 3.7.1

Nomenclature

A blade section of infinitesimal thickness is a curved line known as “camber line” (Fig. 3.7). This forms the backbone line of a blade of finite thickness. Its shape is specified by x-y coordinates which are conventionally presented in the non-dimensional form x/ and y/ as shown in the figure. They are applicable to a blade section of any chord length ( ). Maximum camber (y/ )max and its position (a/ ) are important properties.

Fundamentals of Rotating Machines

55

y/l 0.1 Maximum camber

Camber line

y 0

0.5

x

x/l

1.0

a’ Fig. 3.7 Camber line profile The shapes of profiles of the upper and lower surfaces of the blades are specified by the thickness distribution along the chord or the camber line. This may be symmetrical or unsymmetrical giving a symmetrical or an unsymmetrical profile. Figure 3.8 shows the base profile of a symmetrical blade. This may by used for an actual curved blade in a cascade of an axial flow compressor. The maximum thickness (tmax ) of the profile and its distance (a ) from the leading edge, and the shapes of the leading and trailing edges (LE and TE) are important parameters which govern the flow pattern and losses.

0.1 x t/2 t max

LE 0

TE

a’

-0.1

Chord (l) or length of the camber line 0

0.5

1.0

Fig. 3.8 Thickness distribution of an aerofoil (base profile)

3.7.2

Nature of Flow

When a flat plate is moved through a fluid, it experiences certain resistance to its motion on account of the fluid friction on its surface. If the flow direction is parallel to its length, the force normal to the plate will be zero. However, when the plate is inclined (at an angle i) to its direction of motion, it will experience a resultant force R normal to its length. This force has a drag force component (D) parallel to the flow and another lift

56

Gas Turbines

D L

R

Flat plate

Free stream +

i

+

-

-

-

+

+

+

Fig. 3.9 Lift force on an inclined flat plate force component (L) perpendicular to it. The drag and lift forces are as shown in Fig. 3.9. As can be seen in Fig. 3.9 the lift force arises on account of the negative and positive pressures prevailing on the upper and lower surfaces respectively. The figure shows only constant average values of pressures. In actual practice the pressures on the two sides will vary from the leading to the trailing edge. When the angle of incidence (i) as shown in Fig. 3.10 is zero, the lift force will be zero though CL = 0. The lift will increase up to an optimum angle of attack and decrease as can be seen from Fig. 3.10. Along with lift the drag also increases. Beyond the optimum angle of attack the drag force starts dominating and increasing very rapidly followed by a decrease in the lift force. This is obviously undesirable. The maximum drag occurs when the plate is normal (i = 90◦ ) to the flow direction; the lift is zero for this position. In practical applications for aircraft wings and turbomachine blades, the flat plate (if used) will have to be of finite thickness. In order to achieve a high lift-to-drag ratio, the leading edge is rounded and the blade section is tapered towards the thin trailing edge. To obtain further increase in the value of L/D, the blade is slightly curved, thus giving a curved camber line. It may be seen here that such a blade approaches a cambered aerofoil shape. Figure 3.11 shows (a) an uncambered aerofoil blade with zero angle of attack, (b) an uncambered blade with angle of attack i, and (c) a cambered blade with an angle of attack i. Static pressure distribution around a cambered aerofoil blade is shown in Fig. 3.12. The centrifugal force on the fluid particles on the upper (convex) side tries to move them away from the surface. This reduces the static pressure on this side below the free-stream pressure. On account of this suction effect, the convex surface on the blade is known as the suction side. In contrast to this, the centrifugal force on the lower (concave) side presses the fluid harder on the blade surface, thus increasing the static pressure

Fundamentals of Rotating Machines

57

0.7 0.6

CL

C L or C D

0.5

CD

0.4

0 -0.1

i

-0.2 -0.3 -4

-2

0

2

4

6

8

10

Angle of incidence Fig. 3.10 Variation of lift and drag coefficients with incidence above that of the free stream. Therefore, this side of the blade is known as the pressure side. Due to the above phenomenon, the flow on the suction side begins accelerating along the blade chord accompanied by a deceleration on the pressure side. However, the common boundary conditions at the trailing edge require the flows on the two sides to equalize. Therefore, the accelerated flow on the suction side experiences deceleration of the flow as it approaches the trailing edge. The already decelerated flow in the pressure side starts accelerating at some point along the chord. In spite of these processes the flows coming from the two sides of the blade surface may fail to equalize. The resultant upward force on the blade (Fig. 3.12) is the result of the cumulative effect of the positive static pressure on the pressure side and the negative pressure on the suction side. Now the total upward force acting on the aerofoil is equal to the projected area times the pressure difference between the two sides. On an aircraft wing there is a large area available for the production of lift force. Therefore, only a small pressure difference over its aerofoil wing section will provide the required lift. This requires only a slight deflection of the approaching flow over the aircraft wings which is achieved by only slightly cambered sections. In contrast to this, the projected areas of turbomachine blades are much smaller. Therefore, a considerable difference of static pressure between the pressure and suction sides is required to provide the necessary lift or the tangential force. This can only be achieved by providing highly cambered blade sections.

58

Gas Turbines

Trailing edge

Leading edge

i=0 Chord = l (a) Uncambered aerofoil with zero incidence

i

Chord line

(b) Uncambered aerofoil with incidence Camber line

i Camber

Chord line

(c) Cambered aerofoil with incidence Fig. 3.11 Flow around aerofoil blades

Suction side

-

-

Aerofoil

-

+

+

+

Pressure side

Fig. 3.12 Pressure distribution around a cambered aerofoil blade

Fundamentals of Rotating Machines

59

The lift force is exerted by the fluid on the aircraft wings and on the turbine rotor blades, whereas in power-absorbing machines like compressors, this lift force (exerted by their rotor blades on the fluid) is supplied by the torque input. 3.7.3

Coefficients of Lift and Drag

The resultant force due to the flow around an aerofoil blade acts at its centre of pressure. It has two components – lift force, normal to the flow direction (or blade chord) and the drag force parallel to the flow. These forces depend only on the density and velocity of the fluid and the blade chord. Thus, L and D are functions of density, velocity and chord length. L, D

=

f (ρ, c, )

The projected area per unit length of the blade is A =

×1

The lift and drag coefficients based on this area relate the dynamic pressure 12 ρc2 to the lift and drag forces. L =

CL A

1 ρ c2 2

D

CD A

1 ρ c2 2

=

Conventionally, these coefficients are expressed as CL

=

CD

=

L 1 2ρ

c2

D c2

1 2ρ

(3.41)

(3.42)

As already stated, the lift and drag coefficients are functions of angle of attack and the variation of CL and CD with i for an aerofoil blade will be similar to Fig. 3.10. Review Questions 3.1 What is the principle of operation of a rotating machine and what factor governs the energy transfer? 3.2 What is the general energy transfer equation for a rotating machine? Derive the same. 3.3 How do you differentiate between a turbine rotor and a compressor rotor from the point of view of energy equation?

60

Gas Turbines

3.4 What are the conditions to be satisfied for the Euler’s energy equation to be valid? 3.5 Discuss the physical meaning of the Euler’s energy equation. 3.6 What is meant by internal effect and external effect? 3.7 What are the two ways by which the compressor and turbine can be classified? Discuss them briefly. 3.8 By means of a thermodynamic analysis develop an expression for the energy transfer in a rotating machine. 3.9 Define the isentropic efficiency of a rotating machine (compressor and turbine) and show the details on an h-s or a T -s diagram. 3.10 What is meant by dimensional analysis? Derive the dimensionless parameters for a rotating machine. 3.11 Discuss briefly the theory of airfoil. 3.12 What is meant by lift and drag? Show the effect of angle of attack on CL and CD on a graph. Multiple Choice Questions (choose the most appropriate answer) 1. When the fluid passes through rotating machine, there is (a) only energy transfer (b) only energy transformation (c) both energy transfer and transformation (d) none of the above 2. The energy transfer in a rotating machine is given by (a) steady flow energy equation (b) unsteady flow energy equation (c) Euler’s energy equation (d) all of the above 3. The Euler’s equation is given by (a) E = Ct1 u1 × Ct2 u2

(b) E = Ct1 u1 − Ct2 u2 (c) E = Ct1 u1 + Ct2 u2

(d) E = Ct1 u1 /Ct2 u2

Fundamentals of Rotating Machines

4. The equation for energy transfer in a rotating flow is given by (a) 0.5 ×

c21 − c22 + u21 − u22 w12 − w22

(b) 0.5 ×

c21 − c22 + u21 − u22 w22 − w12

(d) 0.5 ×

c22 − c21 + u22 − u21 w22 − w12

(c) 0.5 ×

c21 − c22 + u22 − u21 w22 − w12

5. In the energy transfer equation, Δc = c21 − c22 represents (a) impulse effect (b) reaction effect (c) centrifugal effect (d) centripetal effect 6. In the energy transfer equation Δu = u21 − u22 represents (a) impulse effect (b) reaction effect (c) centrifugal effect (d) centripetal effect 7. In the energy transfer equation Δw = w22 − w12 represents (a) impulse effect (b) reaction effect (c) centrifugal effect (d) centripetal effect 8. The degree of reaction for a rotating machine is defined as (a)

Enthalpy change in the rotor Enthalpy change in the stator

(b)

Enthalpy change in the stator Enthalpy change in the rotor

(c)

Enthalpy change in the stator Enthalpy change in the stage

(d)

Enthalpy change in the rotor Enthalpy change in the stage

9. Isentropic efficiency of a turbine is (a)

Actual temperature drop Isentropic temperature drop

(b)

Isentropic temperature drop Actual temperature drop

61

62

Gas Turbines

Actual pressure drop Isentropic pressure drop Isentropic pressure drop (d) Actual pressure drop (c)

10. The efficiency of the modern compressors is (a) 70% (b) 75% (c) 80% (d) 85% Ans:

1. – (c) 6. – (c)

2. – (c) 7. – (b)

3. – (b) 8. – (d)

4. – (b) 9. – (a)

5. – (a) 10. – (d)

4 CYCLE ARRANGEMENTS INTRODUCTION As already seen in Chapter 1, a simple gas turbine unit consists of three components, viz., a compressor, a heat addition device and a turbine. These three components can be arranged either in an open or a closed form. Accordingly, a gas turbine cycle can be classified into two categories: (i) open-cycle arrangement (ii) closed-cycle arrangement Of the two, open-cycle arrangements are much more common. In this arrangement fresh atmospheric air is drawn into the system continuously and energy is added by combustion of fuel in the working fluid itself. The products of combustion are expanded through the turbine and exhausted into the atmosphere. In the closed-cycle, the same working fluid, be it air or some other gas, is repeatedly circulated through the system. It may be noted that in this type of plant whether the working fluid is air or some other gas, fuel cannot be burnt directly in the working fluid and the necessary energy must be added in a heater or gas boiler. 4.1

OPEN-CYCLE ARRANGEMENTS

If a gas turbine is to be operated at a fixed speed and fixed load condition such as peak-load power generation, a single shaft arrangement as shown in Fig. 4.1 may be suitable. Flexibility of operation, i.e., the rapidity with which the machine can accommodate itself to changes of load and speed, and efficiency at part load, are in this case considered unimportant. A heat exchanger can be added as shown in Fig. 4.2 to improve the thermal efficiency, although for a given size of the plant, power output may be reduced by 10% due to pressure losses in the heat exchanger.

64

Gas Turbines

Combustion chamber

Product of combustion

1 3

2

Air

4

Fuel

Power output

Compressor

Turbine

Fig. 4.1 A simple gas turbine with single-shaft arrangement

Heat exchanger 6

Combustion chamber

5

4

1 3

2 Fuel

Power output

Compressor

Turbine

Fig. 4.2 A simple gas turbine with a heat exchanger

The modified form, as shown in Fig. 4.3, is more suitable for fuels whose products of combustion contain constituents which may corrode or erode the turbine blades. It is much less efficient than the simple cycle power plant because the heat exchanger, inevitably less than perfect, in transferring the energy input because of the effectiveness of the heat exchanger. Such a cycle may be considered only if inferior fuels are to be used. When flexibility of operation is of paramount importance, such as in road, rail and marine applications, a mechanically independent power turbine is used. In this twin-shaft arrangement (Fig. 4.4), the compressor and highpressure turbine combination acts as a gas generator for the low pressure turbine. Fuel flow to the combustion chamber is controlled to achieve variation of power. It should be noted that this will cause a decrease in cycle pressure ratio and maximum temperature. At off-design conditions the power output reduces with the result that the thermal efficiency deteriorates considerably at part loads. Alternative arrangements to overcome the above disadvantages are the series flow and parallel flow gas turbines (Figs. 4.5 and 4.6). In these arrangements power output is controlled by the adjustment of fuel supply to the combustion chamber in the power turbine line.

Cycle Arrangements

65

Combustion chamber Heat exchanger 4

6 1

Fuel 3

5

2

Power output Compressor

Turbine

Fig. 4.3 A simple gas turbine with heat exchanger – an alternative arrangement

Heat exchanger Combustion chamber

Air Fuel HPC

HPT

Compressor

Turbine Power LPT output Separate power turbine

Fig. 4.4 A simple twin-shaft arrangement

66

Gas Turbines

Heat exchanger Combustion chamber

Air Fuel HPC

HPT Turbine

Compressor

Fuel Combustion chamber Power LPT

output

Separate power turbine

Fig. 4.5 Series flow twin-shaft arrangement Heat exchanger Fuel Power output Compressor

Turbine

Separate power turbine

Fig. 4.6 Parallel flow twin-shaft arrangement The performance of a gas turbine may be improved substantially by reducing the work of compression and/or increasing the work of expansion. For any given pressure ratio, the power required for compression per kg of working fluid is directly proportional to the inlet temperature. If, therefore, the compression process is carried out in two or more stages with intercooling, the work of compression will be reduced. This arrangement is shown in (Fig. 4.7). Similarly, the turbine output can be increased by dividing the expansion into two or more stages and reheating the gas to the maximum permissible temperature between the stages (Fig. 4.8). By employing a heat exchanger, the cost of additional fuel can be minimized. Complex cycles offer good part load performance and high flexibility but it is to be noted that the inherent simplicity and compactness of the power plant are lost. To obtain higher thermal efficiencies without a heat exchanger, a high pressure ratio is essential. Axial compressors are normally preferred, particularly for large units, as its efficiency is appreciably higher than that

Cycle Arrangements

Coolant Out

Coolant In

67

Heat exchanger Combustion chamber

Air Intercooler

Fuel HPT

HPC

LPC

Compressor

Compressor

Turbine Fuel

Combustion chamber

Power LPT

Separate power turbine

output

Fig. 4.7 Series flow with intercooling

Heat exchanger Combustion chamber

Fuel Compressor

Turbine

Reheat combustion chamber

Fuel

Power

Fuel

output Turbine

Turbine

Fig. 4.8 Series flow with reheating

of the centrifugal compressor. Unfortunately, axial compressors are more prone to instability when operating at off-design conditions. The unstable operation, manifested by violent aerodynamic vibration, is likely when a gas turbine is started up or operated at low power. The problem is particularly severe if an attempt is made to obtain a pressure ratio of more than 8:1 in one compressor. One way of overcoming this difficulty is to divide the compressor into two or more sections. This mechanical separation permits each section to run at different rotational speed. When the compressors are mechanically independent, each will have its own turbine. The LP compressor is driven by the LP turbine and the HP compressor by the HP turbine. This arrangement is called straight compounding. Power is normally taken either from the low pressure turbine shaft or from an additional free power turbine. This configuration is generally referred to as

68

Gas Turbines

a twin-spool engine (Fig. 4.9). This arrangement is widely used both for shaft power units and for the turbojet aircraft engines, employing pressure ratios in the range of 10:1 to 20:1.

Heat exchanger Combustion chamber Fuel HPC Compressor

HPT Turbine

Power LPC Compressor

LPT

output

Turbine

Fig. 4.9 Straight compounded twin-spool arrangement An arrangement, known as cross compounding wherein the LP compressor is driven by the HP turbine and the HP compressor by the LP turbine, is claimed to give better efficiency at part load. Unfortunately, the effect on stability of operation is the opposite of straight compounding, i.e., it makes the problem worse instead of better. 4.2

THE CLOSED-CYCLE

In all the arrangements discussed so far, atmospheric pressure and temperature have been considered as the datum, and the exhaust gases were discharged at atmospheric pressure. The average exhaust temperature will be around 700 K for an average maximum temperature of about 1000 K. It follows that 1 kg of gas will occupy a volume according to the law P V = mRT which is about of 2.34 m3 . At low pressure-ratios the constant-pressure cycle requires large mass flow rate, thus the total volume flow rate for any given power output will be quite large. In order to accommodate such large flow, a large rotor diameter and long blades are required. This results in excessive stress at the root of the blades due to large centrifugal force. Hence, there is an upper limit on the size and power output of a turbine or compressor. It will be shown in the next chapter that the efficiency expression of various constant-pressure cycles is a function of pressure-ratio and the gas temperatures;, i.e., the efficiency depends not upon the magnitude of the

Cycle Arrangements

69

pressures, but only upon their ratio. However, the volume occupied by a gas depends upon the magnitude of the pressure. When the volume to be circulated becomes large, the pressure level can be raised so that the volume per kg of fluid, i.e., the specific volume, is reduced to the level desired. However, the same pressure ratio (i.e., ratio of maximum to minimum pressures) should be maintained as before for attaining the same efficiency. It is true that the maximum pressure will increase, but only in proportion to the change of absolute pressures on the system. Thus, when the system is closed from the atmosphere, the same fluid will circulate again and again. It follows that if the fuel is burnt directly in the circulating air, the oxygen will soon deplete and combustion will fail. It is, thus necessary to supply a certain amount of fresh air to the closed circuit. The normal overall air-fuel ratio of a gas turbine is between 60:1 to 100:1. It follows that only a small fraction of burnt oxygen is to be supplied each time through the combustion chamber in order to have continuous operation. Thus, majority of the exhaust gas leaving the turbine would return to the compressor entrance;, i.e., the inlet temperature to the compressor would be greatly increased. It may be noted that the work of compression per kg can be shown to be WC

Cp T1 γ−1 r γ −1 ηC

=

which is seen to depend upon the inlet temperature, T1 and the pressure ratio, r. Thus, in a closed-cycle, the net output would be much reduced unless a gas cooler is added between the turbine exhaust and the compressor inlet to cool the gas which is being recirculated, down to approximately normal inlet temperature. The details of such a system with internal firing is shown in Fig. 4.10. Compressor

Main turbine Generator p

p p

1

Gas cooler

p

2 1

Regenerator

In Out

2

2

p

p

1

Auxiliary turbine

2

p

Combustion chamber

2

Fuel p

Compressor

atm

Make up air inlet

p

atm

Exhaust

Fig. 4.10 A closed-cycle arrangement with air as working medium The compressor draws gas from turbine exhaust at point (1) through a regenerator and gas cooler under pressure p1 , which is greater than atmospheric pressure. After compression, the gas at pressure p2 is delivered

70

Gas Turbines

to combustion chamber through regenerator. In the regenerator the compressed gas is heated to higher temperature by the main turbine exhaust. In the combustion chamber, fuel is added to obtain the desired maximum gas temperature. From combustion chamber the gases pass on to main turbine, where they expand from pressure p2 to pressure p1 . The turbine provides net work to the load after meeting the compression work of the cycle. A small portion of the gases leaving the combustion chamber is by-passed to auxiliary turbine, where expansion takes place from p2 to patm . The auxiliary turbine develops enough work to run a compressor, which draws air from atmosphere and delivers it at pressure p2 of the combustion chamber. The amount of fresh air supplied by this compressor must be equal to the amount of gases by-passed after combustion to atmosphere via auxiliary turbine and also to the amount of air which is sufficient to burn completely the fuel that is added to maintain the maximum temperature of the cycle and thereby the maximum output of the main turbine. Figure 4.11 illustrates another method on the closed-cycle concept, where instead of supplying the fuel directly to, and burning it with, the air inside the system, the fuel could be burned in a separate combustion chamber built like a regenerator. The heat of combustion is transferred through the containing walls of the furnace to the air flow in the turbines as shown in Fig. 4.11. Thus, the air or gas contained in the closed system can be used over and over again without the necessity for make-up air. This results in that the products of combustion do not come in contact with the moving parts and no deposit will accumulate on the turbine blades. In such type of closed circuit every kind of solid fuels like coal, can also be burned in the furnace. Further the working medium other than air having the desired properties can be used. The details of working of such a system is shown in Fig. 4.11. Advantages and disadvantages of a closed-cycle system are enumerated in the following sections.

Compressor

Main turbine Generator

Regenerator Gas precooler Gas heater In Out Water

Air preheater Fuel

Air Exhaust

Fig. 4.11 Another closed-cycle arrangement with working medium other than air

Cycle Arrangements

4.2.1

71

Advantages

(i) Use of high pressure (and hence gas density) level throughout the cycle would result in a reduced size of the plant for a given output. (ii) Wide range of load variation is possible by varying the pressure levels without altering the maximum cycle temperature. Hence, there will be almost no variation of overall efficiency. (iii) Erosion of the turbine blades due to the products of combustion is eliminated. (iv) Filtration of working medium is not required except charging for the first time. (v) High density of the working medium improves the effectiveness of the heat exchanger. (vi) Gases other than air having more desirable thermal properties, such as helium etc., with γ = 1.66 can be used to increase the power output and thermal efficiency. (vii) Cheaper fuels can be used. 4.2.2

Disadvantages

(i) The heating system is quite bulky. (ii) It is quite difficult to make the system absolutely leak proof. (iii) Large capacity cooler is necessary. (iv) Useful only for stationary power plants. 4.3

BASIC REQUIREMENTS OF THE WORKING MEDIUM

In a closed-cycle arrangement, the operating medium can be other than air and it must satisfy the following requirements. (i) Availability as well as cheapness of the working medium. (ii) The circulating working medium must be stable, non-explosive and non-corrosive. (iii) It should be non-toxic and non-inflammable. (iv) It should have high specific heat value Cp and high specific heat ratio, γ. (v) It must have a higher thermal conductivity, k.

72

Gas Turbines

Table 4.1 Properties of Various Gases Gas Air Argon Helium Hydrogen

4.4

Cp kJ/kg K 1.005 0.515 5.191 14.186

γ 1.400 1.660 1.660 1.408

k at 0◦ C Watt/mK 0.0241 0.0163 0.1425 0.1743

R kJ/kg K 0.287 0.151 10.050 59.934

PROPERTIES OF VARIOUS WORKING MEDIA

Table 4.1 shows the properties of various gases which can be used as a working medium in the gas turbine. An examination of the properties of some representative gases shown in the table would indicate that when desirable values of Cp and γ are available, they are accompanied by high values of R. The work done during compression per kg of flow of any gas is given by WC

= Cp (T2 − T1 )

and the temperatures after compression, T2 , is given by T2

= T1 r

γ−1 γ

It is clear, therefore, that low values of γ and Cp are conducive to lower compression work. On the other hand, a high value of γ and Cp are essential for higher turbine output, assuming the same R. It may be noted that the net output of the cycle will always increase with high values of Cp and γ. Then the increased output per kg of working medium would permit a reduction in mass flow rate. However, high values of thermal conductivity would result in appreciable saving in heat exchangers, i.e., regenerators and precoolers. The values given in the Table 4.1 for k are based on 0◦ C for comparative purpose. Actually, these values would be considerably higher at the temperatures encountered in regenerators. Therefore, it may be concluded that any serious reduction in size of the gas turbine plant by using a gas other than air would have to be realized in heat exchangers. Of course, a gas other than air could be used in an externally-fired closed-cycle. But it should be the aim to have a reduction in the size of heat exchanger and is particularly important. 4.5

APPLICATIONS

Gas turbines can be classified into aircraft and industrial gas turbines, the second term meaning all those gas turbine power plants which are not included in the first category. The aircraft gas turbines differ from the industrial gas turbines in three main aspects.

Cycle Arrangements

73

(i) The life of the industrial gas turbine is expected to be of the order of 120,000 hours without major overhaul as against 600-1200 hours for aircraft gas turbines. (ii) Size and the weight of an aircraft power plant is very crucial compared to industrial units. (iii) The aircraft power plant can make use of kinetic energy of the gases leaving the exhaust whereas it is wasted in other types and consequently, this energy loss must be kept as minimum as possible. These differences in the requirements have considerable effect on design, although fundamental theory is same for both the categories. Industrial gas turbines are rugged in construction, with many auxiliary equipments. They often employ a single, large cylindrical combustion chamber. They are also designed for multifuel capability. Apart from the aircraft market, the widest application of gas turbines have been in pump sets for oil and gas transmission pipe lines and generation of electricity. So far gas turbines have made no inroads into the world of merchant shipping but it is extensively used in naval operations. A major disadvantage of the gas turbine in naval use is its poor part load performance and higher specific fuel consumption. To overcome this problem, combined power plants consisting of gas turbines in conjunction with steam turbines, diesel engines and other gas turbines have been considered. To date, little impact has been made in the field of rail transport. Experimental trains have been operating in some countries. The high speed passenger train with gas turbine power is an attractive concept for the future. Maybe in the near future, a long haul truck market will provide a major application for the gas turbine. Major automobile industries are active in developing engines in the range of 200 – 300 kW. These vehicular engines employ low pressure ratio, centrifugal compressor, free power turbine and a rotary heat exchanger. Concern with exhaust pollution will be a critical factor in favour of gas turbine. The major problem is still with high part-load fuel consumption. Another concept of potentially great importance is the so-called Total Energy Plant, where exhaust heat is used to provide building heating in winter and refrigeration/air conditioning in the summer. Other uses for energy in a gas turbine’s exhaust are found in process industries. The gas turbine can also be used as a compact air compressor suitable for supplying large quantities of air at moderate pressures. 4.6

COMPARISON OF GAS TURBINES WITH RECIPROCATING ENGINES

Gas turbine is also an internal combustion engine. Its competitor in early stages was the reciprocating internal combustion engines. Let us compare them.

74 4.6.1

Gas Turbines

Advantages of Gas Turbines over Reciprocating Engines

(i) Mechanical efficiency Mechanical efficiency of the gas turbine is considerably higher than that of the best reciprocating engine. For simple gas turbine design mechanical efficiency of 90% to 95% has been claimed while for reciprocating engine it is from 85 to 90% under full load conditions. It is due to more frictional losses in reciprocating engines. (ii) Balancing Due to absence of any reciprocating mass in gas turbine engine, balancing can be near perfect. Torsional vibrations are absent because gas turbine is a steady flow machine. (iii) Cost In case of larger output gas turbine units of 2500 kW, it can be built at an appreciably lower cost and in a shorter time than the corresponding multicylinder petrol or diesel engines. (iv) Weight The fuel consumption per kW hour of best available aircraft gas turbine is almost twice that of the normal petrol engine. However, it has much lighter weight per kW so that the total weight of turbine plus fuel does not compare unfavourably with reciprocating type of engine and its fuel. To give quantitative example, the specific weight of (a) steam turbine is about 53 kg/kW, (b) diesel engine is about 115 kg/kW and (c) gas turbine is about 20 kg/kW. (v) External shape and size The basic cylindrical shape of turbine and compressor unit renders the gas turbine more convenient to start, especially in aircraft and locomotives. (vi) Fuel The turbine can be designed to operate with cheaper and more readily available fuels such as benzene, powdered coal, and heavy graded hydrocarbons. Promising results have been obtained using furnace oil and also pulverized coal as fuel. (vii) Lubrication Compared with reciprocating engines the lubrication of gas turbines is comparatively simpler. The requirement is chiefly to lubricate the main bearing, compressor shaft and bearings of auxiliaries. (viii) Maintenance The fact that the gas turbine consists of essentially a single turbine and compressor unit with a common or coupled shaft running in a relatively smaller number of main bearings, only minimum maintenance is necessary as compared to the reciprocating internal combustion engines. (ix) Low operating pressures The gas turbine generally operates at relatively low pressures so that the parts exposed to these pressures can be made light although the effects of thermal expansion and contraction must be taken into account. The maximum combustion pressure is much lower than that in reciprocating engines so that the pressure joints and piping do not pose any difficulty.

Cycle Arrangements

75

(x) Silent operation Since the exhaust from a gas turbine occurs under practically constant-pressure conditions unlike the pulsating nature of reciprocating engine exhaust, the turbine and compressor, if dynamically balanced, can run very smoothly. The usual vibrational noises as in the case of reciprocating engine are almost absent. (xi) Smokeless exhaust With the present tendency to use relatively large surplus air for combustion in order to reduce temperature of gases, the exhaust from the turbine is almost smokeless and generally free from pungent odour associated with optimum and rich fuel mixture which is characteristic of reciprocating engines. (xii) High operational speed Turbine can be made lighter than the reciprocating engine of similar output. It can be run at much higher speed than reciprocating engines. The output of any engine varies directly as the product of the driving shaft torque and its rpm. Therefore, for a given output and higher speed the torque will be lower. It may be noted that the torque characteristics of the gas turbine is much better than that of reciprocating engine, since the former gives a high initial torque and its variation with speed is comparatively less. 4.6.2

Advantages of Reciprocating Engines over Gas Turbines

(i) Efficiency The overall efficiency of the turbine is much less than the reciprocating engine since 70% of the output of the turbine is to be fed to the compressor and other accessories and auxiliary parts. (ii) Temperature limitation The maximum temperature in gas turbine cannot exceed 1500 K because of the material consideration of the blade while in reciprocating engines with complete combustion of the fuel the maximum temperature can be raised to 2000 K. This high temperature is permitted since the piston and cylinder head are subjected to this high temperature only for a fraction of a second. (iii) Cooling We can achieve very good results by efficient cooling in reciprocating engine by which the heat of the cylinder walls is taken away, which enables to keep the wall temperature only around 500 K but in gas turbine, cooling is complicated, and, therefore, much higher temperature cannot be allowed to reach. (iv) Starting difficulties It is more difficult to start a gas turbine than a reciprocating engine as it requires compressed air or some suitable starter mechanism which are complicated. Review Questions 4.1 What are the two types of cycle arrangements possible for a gas turbine system and what is the basic difference between the two arrangements?

76

Gas Turbines

4.2 Explain the details of single-shaft arrangement for fixed load conditions. 4.3 What is the purpose of adding a heat exchanger? Explain the details with suitable sketches. 4.4 With neat sketches explain the details of a gas turbine system when flexibility of operation is of paramount importance. 4.5 Explain with neat sketches the cycle arrangements of a series flow with intercooling and series flow with reheating. 4.6 With a neat sketch explain the straight compounded twin-spool arrangement. 4.7 With neat sketches explain the working of closed-cycle arrangements. 4.8 Discuss the advantages and disadvantages of closed-cycle system over open-cycle system. 4.9 What are important factors to be considered if working media other than air is to be used? 4.10 Briefly discuss the applications of gas turbines. 4.11 Discuss the advantages of gas turbines over reciprocating engines and vice-versa. 4.12 Discuss the advantages of reciprocating engines over gas turbines. Multiple Choice Questions (choose the most appropriate answer) 1. A gas turbine cycle can be operated (a) only as an open-cycle arrangement (b) only as a closed-cycle arrangement (c) both as an open-cycle and closed-cycle arrangement (d) none of the above 2. For the same pressure ratio and cycle peak temperature (a) open-cycle is more efficient than closed-cycle (b) open-cycle is less efficient than closed-cycle (c) both open-cycle and closed-cycle will have the same efficiency (d) none of the above 3. Adding heat exchanger to a simple ideal cycle (a) improves work output (b) reduces work output

Cycle Arrangements

77

(c) improves efficiency (d) improves both work output and efficiency 4. When flexibility of operation is important (a) single-shaft arrangement is better (b) twin-shaft arrangement is better (c) multiple compression is better (d) none of the above 5. Major applications of gas turbines is for (a) aircraft (b) locomotive (c) automotive (d) all of the above 6. The specific weight of a gas turbine power plant is (a) 50 kg/kW (b) 20 kg/kW (c) 100 kg/kW (d) 80 kg/kW 7. The selection of the working medium of a closed-cycle gas turbine power plant is based on (a) the availability and cost (b) non-explosive and non-corrosive properties (c) high specific heat value and specific heat ratio (d) all of the above 8. The life of an industrial gas turbine without major overhaul is (a) 120,000 h (b) 12,000 h (c) 1200 h (d) 120 h 9. The life of a aircraft gas turbine without major overhaul is (a) 120,000 h (b) 12,000 h (c) 1200 h (d) 120 h

78

Gas Turbines

10. The overall efficiency of a gas turbine engine compared to reciprocating engine for the same power output (a) more (b) less (c) same as reciprocating engine (d) none of the above Ans:

1. – (c) 6. – (b)

2. – (c) 7. – (d)

3. – (c) 8. – (a)

4. – (b) 9. – (c)

5. – (a) 10. – (b)

5 IDEAL CYCLES AND THEIR ANALYSIS INTRODUCTION In the previous chapter, we have seen, how large is the number of possible varieties of gas turbine cycle arrangements, when multistage compression and expansion, heat exchange, reheat and intercooling are incorporated. A comprehensive study of the performance of all such cycles allowing for the various inefficiencies of the components, would result in a very large number of performance curves. This chapter is mainly concerned with describing methods for calculating cycle performance under ideal conditions. For convenience, we will treat cycles in two groups: (i) shaft power cycle, and (ii) aircraft propulsion cycle. An important distinction between the two groups arises from the fact that the performance of aircraft propulsion cycles depend very significantly upon forward speed of the aircraft and altitude at which it is flying. However, these two variables do not enter into performance calculations of shaft power cycle used in marine and land based power plants to which this chapter is confined. Before proceeding with the main task, it will be useful to review the performance of ideal gas turbine cycles in which perfection of the individual components is assumed. The specific work output and cycle efficiency will only be a function of pressure ratio and maximum cycle temperature. The limited number of performance curves so obtained enables one to analyze the major effects of various additions to the simple cycle. To be seen clearly, such curves also show the upper limit of performance. In the limit this is the maximum that can be expected of real cycles as the efficiency of the gas turbine component is improved. Before going into the details of performance analysis let us list the various assumptions we make in the analysis of ideal cycle.

80

Gas Turbines

5.1

ASSUMPTIONS IN IDEAL CYCLE ANALYSIS

The assumption of ideal gas turbine cycle will be taken to imply the following: (i) The change of kinetic energy of the working fluid between inlet and outlet of each component is negligible. (ii) Compression and expansion are reversible and adiabatic, i.e., isentropic. (iii) There are no pressure losses in the inlet ducting, combustion chambers, heat exchangers, intercoolers, exhaust ducting and ducts connecting the components. (iv) Heat transfer in a heat exchanger is assumed to be perfect (100% effectiveness). (v) The mass flow of gas is constant throughout the cycle. (vi) Working fluid has the same composition throughout the cycle and is a perfect gas with constant specific heats. (vii) Bearing and windage friction, etc., are neglected. 5.2

THE SIMPLE GAS TURBINE CYCLE

The schematic details of a simple gas turbine are shown in Fig. 5.1. Figure 5.2 shows the various processes on a p-V diagram whereas Fig. 5.3 gives the details on a T -s diagram. Figures 5.4 and 5.5 show the performance curves of the cycle. From the thermodynamic analysis the relevant steady flow energy equation has been shown to be (refer Eq. 2.54) ws

=

Δh

=

h2 − h1

(5.1)

where ws is the work transfer per unit mass flow, and h stands for enthalpies. Applying Eq. 5.1 to each component and assuming unit mass flow of the working fluid, we can write the work input to the compressor (process 1→2) as Compressor work [WC ] W12

= (h2 − h1 )

=

Cp (T2 − T1 )

(5.2)

= (h3 − h2 )

=

Cp (T3 − T2 )

(5.3)

Heat addition [Q] Q23 Turbine work [WT ]

Ideal Cycles and their Analysis

Combustion chamber

Product of combustion

1 3

2

Air

81

4

Fuel

Power output

Compressor

Turbine

Fig. 5.1 Schematic arrangement of a simple gas turbine

2

10

1500

3

8

3

1000 T

p6 4

500

2 0

1 1

0

4

1

V

0

2

Fig. 5.2 p-V diagram

20

30

40 s

50

60

Fig. 5.3 T -s diagram

0.8

2

t=5

1.5

WN /CpT1

4

2

0.6

t=4

1

η 0.4

t=3

0.5

6 γ =1.6 0 γ =1.4

0.2

t=2 0

0

5

Fig. 5.4 r vs

10 r WN Cp T1

15

20

0

0

5

10

r Fig. 5.5 r vs efficiency

15

20

82

Gas Turbines

W34

= (h3 − h4 )

=

Cp (T3 − T4 )

(5.4)

Net work output [WN ] = WT − WC = Cp (T3 − T4 ) − Cp (T2 − T1 ) T3 T4 T2 − − +1 T1 T1 T1

= Cp T1 Let

then

Let r

γ−1 γ

(5.5)

T3 T1

= t and

p2 p1

T2 T1

=

T3 T4

r

=

=

(5.6)

r

(5.7)

γ−1 γ

= c. From Eq. 5.6 WN Cp T1

T3 T4 T3 T2 − − +1 T1 T3 T1 T1

=

= t− WN Cp T1

t −c+1 c

= t 1−

1 c

(5.8)

− (c − 1)

(5.9)

N Equation 5.9 shows that the specific work output CW , upon which p T1 the size of the plant depends is a function of not only the pressure ratio, (r), but also of maximum cycle temperature T3 . Now,

η

=

Net work output Heat input

=

Cp T1 t 1 − 1c − (c − 1) Cp (T3 − T2 )

=

Cp T1 (t − c) − Cp T1

η

=

1−

1 c

T3 T1

=

−

=

t−c c T2 T1

1−

1 r

γ−1 γ

WN Q

(5.10)

(5.11)

=

(t − c) − t−c

t−c c

(5.12)

The efficiency of a simple gas turbine thus depends only on the pressure ratio and the nature of the gas, (γ). Figure 5.4 shows the plot of work output in non-dimensional form CWpN T1 as a function of r and t. It may be noted that the value of t depends on the

Ideal Cycles and their Analysis

83

maximum cycle temperature, known as the metallurgical limit. The highly stressed parts of the turbine should withstand this temperature during the required working life. For an industrial plant it might be between 3.5 and 4.0, whereas a value of 5.0 to 5.5 is permissible for an aircraft engine with cooled turbine blades. A glance at the specific output curves (Fig. 5.4) show that a constant t curve exhibits a maximum at a certain pressure ratio; W = 0 at r = 1 and also at the value of the pressure ratio, r = tγ/(γ−1) (i.e., c = t) for which the compression and expansion processes coincide (refer Eq. 5.9). For any given value of t, there must be an optimum pressure ratio to give a maximum specific output. This can be found by differentiating the Eq. 5.9 with respect to c and equating it to zero; the result is, √ (γ−1)/γ ropt = t (5.13) Squaring both sides t =

r

2

γ−1 γ

since, r(γ−1)/γ

=

T2 T1

=

T3 T4

(5.14)

it can be written as t =

T2 T3 T1 T4

=

T2 t T4

(5.15)

If Eq. 5.15 is to be true then it follows that T2 must be equal to T4 , i.e., T2 = T4 . This means that the specific work output is maximum when the pressure ratio is such that the compressor and turbine outlet temperatures are equal. For all values of r between 1 and tγ/2(γ−1) , T4 will be greater than T2 and a heat exchanger can be incorporated to utilize the energy in the exhaust gas to heat the air coming out of the compressor and thereby, the efficiency improvement can be achieved. Figure 5.5 shows the relation between η and r when the working fluid is air (γ = 1.4), or a mono-atomic gas such as argon (γ = 1.66) is used. It can be seen that the efficiency increases with pressure ratio but the rate of increase reduces with the increase in pressure ratio. 5.3

THE HEAT EXCHANGE CYCLE

A schematic arrangement of a simple gas turbine cycle incorporating exhaust heat exchanger is shown in Fig. 5.6. The corresponding p-V and T -s diagrams are shown in Fig. 5.7 and Fig. 5.8 respectively. Figures 5.9 and 5.10 show the performance curves of the cycle.

84

Gas Turbines

Heat exchanger Combustion chamber 4

6 1

5

2

3 Fuel

Power output

Compressor

Turbine

Fig. 5.6 Schematic arrangement of a heat exchange cycle

12 2

10

5

3

1500 3

8 p6

1000 T

4

500

5 6

2 0

1

0

1

6

1

2

V

4

0 3

Fig. 5.7 p-V diagram

20

30

50

60

0.8

t=5

1.5

0.6

0.2

t=2

0

5

Fig. 5.9 r vs

10 r WN Cp T1

3

2

η 0.4

t=3

0.5

t=

t=5 t=4

t=

t=4

1

0

40 s

Fig. 5.8 T -s diagram

2

WN /CpT1

4

2

Simple cycle η

15

20

0

0

5

10

r Fig. 5.10 r vs efficiency

15

20

Ideal Cycles and their Analysis

85

It may be noticed that the T -s diagram is unchanged in outline from that of the simple gas turbine cycle, as can be seen from Figs. 5.3 and 5.8, except for the presence of the heat exchanger as indicated by the two dotted lines 4-5 and 2-6. It may be seen that the temperature of the compressed air has been raised from T2 to T5 in the heat exchanger resulting in the fall in temperature of the exhaust gases from T4 to T6 . Now, for a unit mass flow rate Compressor work [WC ] W12

=

(h2 − h1 )

=

Cp (T2 − T1 )

(5.16)

=

(h3 − h5 )

=

Cp (T3 − T5 )

(5.17)

(h3 − h4 )

=

Cp (T3 − T4 )

(5.18)

Heat addition [Q] Q53

Turbine work [WT ] W34

=

Since, the expressions WN = WT −WC is identical compared to a simple cycle (Eq.5.5), we have, WN Cp T1

=

t 1−

1 c

− (c − 1)

However, the expression for η will be different WN Q Since, T5

=

=

η

Cp T1 t 1 − 1c − (c − 1) Cp (T3 − T5 )

=

T4 η

=

Cp T1 t 1 − Cp T1

η

=

t 1−

=

1−

c t

1 c

T3 T1

1 c

−

− (c − 1)

T4 T3

− (c − 1) t − ct

T3 T1

=

t−c t (5.19)

As can be seen from Eq. 5.19 the efficiency of the heat exchange cycle is not independent of maximum cycle temperature and clearly increases as t is increased. Further, it is evident that for a given value of t the efficiency increases with decrease in pressure ratio and not with increase in pressure ratio as for the simple cycle. In Fig. 5.10 the solid line curves represent the above equation (Eq. 5.19). Each curve for t starts at r = 1 with a value of η = 1− 1t , which is the Carnot efficiency. This is to be expected because in this limiting case the Carnot requirement of complex external heat reception and rejection at the upper

86

Gas Turbines

and lower cycle temperatures is satisfied. The curves fall with √ increasing pressure ratio until a value corresponding to the value of t equals c is reached, and at this point the efficiency becomes that of the simple cycle. At this pressure ratio T4 = T2 and the output is maximum. For higher values of r a heat exchanger would cool the air leaving the compressor reducing the efficiency. The specific output is unchanged by the additions of a heat exchanger. Compare the curves of work output of Fig. 5.4 and Fig. 5.9. From the curves of efficiency for heat exchange cycle (Fig. 5.10) it can be concluded that to obtain an appreciable improvement in efficiency by heat exchange cycle we must have a value of r appreciably less than that for the optimum specific work output. It may be noted that it is not necessary to use a higher cycle pressure ratio as the maximum cycle temperature is increased. The following observations can be made from the performance curves: (i) With heat exchanger cycle, the cycle efficiency reduces as the pressure ratio increases, which is opposite to that of a simple cycle. This is due to the fact that as the pressure ratio increases the delivery temperature from the compressor increases and ultimately will exceed that of the exhaust gas from the turbine. Then heat in the heat exchanger will be lost from the air to the exhaust gases instead of desired gain. The efficiency with lower temperatures, say at t = 2, is seen to become negative soon after the pressure ratio 11.3 is exceeded (refer Fig. 5.10 WN < 0). The reason is that the temperature at compressor outlet actually exceeds the assumed combustion temperature in this case. (ii) In many cases, regeneration is not desirable. With high pressure ratios, efficiencies are higher without regeneration, again because loss of heat from the compressed air to the exhaust gases. (iii) Efficiency with heat exchanger cycle rises very rapidly with increase in maximum temperature of the cycle. (iv) Lower pressure ratios and high cycle temperatures are favourable for the regenerative cycle, since a large heat recovery is then possible. (v) For a given temperature-ratio, the curve falls with increasing value of pressure ratio until a value of c given by c2 = t is reached (Fig. 5.10). After this, the efficiency is equal to the ideal cycle without regeneration. Any further increase of pressure ratio will yield an efficiency which is lower than this value and is of no interest. 5.4

THE REHEAT CYCLE

A good improvement in specific work output can be obtained by splitting the expansion and reheating the gas between the high pressure and low pressure turbines. Figure 5.11 shows the schematic arrangement of the reheat cycle along with the p-V (Fig. 5.12) and T -s (Fig. 5.13) diagrams. Figures 5.14 and 5.15 show the performance curves of the cycle. The turbine

Ideal Cycles and their Analysis

87

work increase is obvious from the fact that the vertical distance between any pair of constant-pressure lines increases as the entropy increases. Thus (T3 − T4 ) + (T5 − T6 ) > (T3 − T4 ) Consider the T -s diagram in Fig. 5.13 in which the expansion is carried out in two stages, reheating of the working fluid to the upper limit of temperature T3 taking place between the stages. Let the pressure ratio in compression be r and the pressure ratios of the expansion stages be r3 and r4 so that r

=

c =

r3 × r4

(5.20)

r(γ−1)/γ

=

T2 T1

(5.21)

(γ−1)/γ

=

T3 T4

(5.22)

(γ−1)/γ

=

T5 T6

(5.23)

c3

=

r3

c4

=

r4

Therefore, c =

c3 × c4

(5.24)

For a unit quantity of fluid flow Compressor work [WC ] W12

=

(h2 − h1 )

=

Cp (T2 − T1 )

(5.25)

Heat addition [Q] Q23

=

(h3 − h2 )

=

Cp (T3 − T2 )

(5.26)

Q45

=

(h5 − h4 )

=

Cp (T5 − T4 )

(5.27)

Turbine work [WT ] W34

=

(h3 − h4 )

=

Cp (T3 − T4 )

(5.28)

W56

=

(h5 − h6 )

=

Cp (T5 − T6 )

(5.29)

Net work output [WN ]

WN Cp T1

=

WT − WC

=

Cp (T3 − T4 ) + Cp (T5 − T6 ) − Cp (T2 − T1 )

=

Cp T1

=

T3 T4 T3 T5 T6 T5 T2 − + − − +1 T1 T3 T1 T1 T5 T1 T1

T3 T4 − T1 T1

+

T5 T6 − T1 T1

−

T2 −1 T1 (5.30)

88

Gas Turbines

Combustion chamber 1

2

Air

Fuel

Reheat combustion chamber 5 4 Fuel

3

Product of combustion 6 Power output

Compressor

Turbine

Turbine

Fig. 5.11 Schematic arrangement of a reheat cycle

10

2

1600

3

8

1100 T

p6 4

4

5

4

6

4’

2

600

5

3

2 0

1

0

6

1

2 V

3

100 4

Fig. 5.12 p-V diagram

30

40

s

50

60

70

0.6

t=5

2

Wmax /CpT1

20

Fig. 5.13 T -s diagram

2.5

1.5

t=4

1

t=3

t=5 0.4

t=3

η

0.5 0

1

t= Simple cycle

t=2

0

5

Fig. 5.14 r vs

10 r WN Cp T1

15

2

0.2

20

0

0

5

10

r Fig. 5.15 r vs efficiency

15

20

Ideal Cycles and their Analysis

Since T5 = T3 and c3 · c4 = c, the above equation simplifies to WN t t = 2t − − −c+1 Cp T1 c3 c4

89

(5.31)

For maximum output d dc3

W Cp T1

=

0

(5.32)

writing c4 in terms of c and c3 and treating c, T and t as constants with respect to c3 t t = 0 − c23 c √ c3 = c = c4 (5.33) Equation 5.33 shows that to obtain maximum output the stage pressure ratios must be same and this should be the square root of the overall pressure ratio for two stage expansion. Now, Wmax 2t (5.34) = 2t − √ − c + 1 Cp T1 c

ηmax

ηmax

=

1 2t 1 − √ c

=

Wmax Q

=

2t 1 −

=

2t 1 −

T3 T1

−

− (c − 1) Cp T1 2t − c + 1 −

=

(5.35) 2t √ c

Cp [T3 − T2 ] + Cp [T5 − T4 ]

√1 − (c − 1) c T2 T5 T4 T1 + T1 − T1 √1 c

− (c − 1)

2t − c −

√t c

(5.36)

Comparisons of the CWp N T1 curves (Fig. 5.14) with the simple cycle (Fig. 5.4) indicates that reheat markedly increase the specific output, but the curves for efficiency indicate that this is achieved at the expense of efficiency (Fig. 5.15 ). This is to be expected because a less efficient cycle (4 456) (refer Fig. 5.13) is added to the simple cycle – less efficient because it operates over a smaller temperature range. Note that the rate of reduction in η becomes less as maximum cycle temperature is increased. 5.5

THE REHEAT AND HEAT EXCHANGE CYCLE

It is seen that improvement in specific power output is achieved in reheat cycle at the expense of the efficiency. This can be overcome by adding a

90

Gas Turbines

heat exchanger to the reheat cycle. The schematic arrangements of the reheat cycle with heat exchanger is given in Fig. 5.16. The corresponding p-V and T -s diagrams are shown in Figs. 5.17 and 5.18 respectively. Figures 5.19 and 5.20 show the performance curves of the cycle. For a unit quantity of mass flow, Compressor work [WC ] W12

=

Cp (T2 − T1 )

(5.37)

Cp (T3 − T7 ) + Cp (T5 − T4 )

(5.38)

Cp (T3 − T4 ) + Cp (T5 − T6 )

(5.39)

Heat addition [Q] Q73 + Q45

=

Turbine work [WT ] W34 + W56

=

Net work output [WN ] =

WT − WC

=

Cp (T3 − T4 ) + Cp (T5 − T6 ) − Cp (T2 − T1 )

=

Cp T1

T3 T4 T5 T6 T2 − + − − +1 T1 T1 T1 T1 T1

(5.40)

The maximum work output expression will be same as Eq. 5.34 since heat exchanger will improve only the efficiency and not the work output. Hence, Wmax Cp T1

ηmax

=

2t 2t − c + 1 − √ c

=

Wmax Q

=

Cp T1 2t − c + 1 −

=

Cp [T3 − T7 ] + Cp [T5 − T4 ]

2t − c + 1 −

T3 T1

−

T7 T1

+

2t √ c

T5 T1

(5.41)

2t √ c − TT41

Since T7 = T4 and T5 = T3 , we have ηmax

=

=

ηmax

=

2t − c + 1 −

T3 T1

−

T4 T1

+

T3 T1

2t − c + 1 − 2t −

1−

2t √ c − TT41

=

2t − c + 1 − 2

T3 T1

−

2t √ c

T4 T1

2t √ c

2t √ c

c−1 2t 2t − √ c

(5.42)

Ideal Cycles and their Analysis

91

Heat exchanger Combustion chamber

8

7

1 Air

2

Compressor

6

5

4

3

Fuel

Reheat combustion chamber Fuel

Turbine

Power output

Turbine

Fig. 5.16 Schematic arrangement of a reheat cycle with heat exchanger

10

2

1600

3

5

3

7

8

7

1100 T

p6 4

4

2

600

5

6

4

8

2 0

1

0

6

8

1

2 V

3

100 4

Fig. 5.17 p-V diagram

0.8

30

40

s

50

60

0.6

1.5

t=4

1

t=3

t=

η 0.4

t=5

2

t=3 t= 2

0.2

0.5

t=2

0

5

Fig. 5.19 r vs

10 r Wmax Cp T1

70

t=5 t=4 t=3

t=5

2

Wmax /Cp T1

20

Fig. 5.18 T -s diagram

2.5

0

1

15

Reheat cycle

20

0

0

5

10

r Fig. 5.20 r vs efficiency

15

20

92

Gas Turbines

The higher exhaust gas temperature is now fully utilized in the heat exchanger. In fact, when a heat exchanger is employed, the efficiency is higher with reheat than without. The family of constant t lines exhibit the same features as those for simple cycle with heat exchanger. Each curve starts with the Carnot values at r = 1 and falls with increasing r to meet the corresponding curve of the reheat cycle without heat exchanger at a value of r corresponding to maximum specific output. 5.6

THE INTERCOOLED CYCLE

We have seen in Section 5.4 that specific output of the cycle can be improved by increasing the turbine work output incorporating the reheat cycle. Another way of achieving the same is reducing the work of compression, i.e., compression in more than one stage and using an intercooler in between. That is, the compression of the working fluid is cut off at some intermediate pressure and the fluid is cooled by passing it through a heat exchanger supplied with coolant from some external source before being compressed in a second stage to the required pressure ratio, a certain improvement in overall output can be achieved. The details of such an arrangement are shown in Fig. 5.21. The corresponding p-V and T -s diagrams are as shown in Figs. 5.22 and 5.23 respectively. Figures 5.24 and 5.25 show the performance curves of the cycle. Now, for a unit quantity of mass flow Compressor work [WC ] W12 + W34

=

Cp (T2 − T1 ) + Cp (T4 − T3 )

(5.43)

Cp (T5 − T4 )

(5.44)

Cp (T5 − T6 )

(5.45)

Heat addition [Q] Q45

=

Turbine work [WT ] W56

=

Net work output [WN ]

Let,

=

WT − WC

=

Cp (T5 − T6 ) − Cp (T2 − T1 ) − Cp (T4 − T3 )

=

Cp T1

T5 T6 T2 T4 T3 − − +1− + T1 T1 T1 T1 T1

T5 T1

=

t;

T5 T6

T2 T1

=

c1 ;

T4 T3

= =

(5.46)

c c2

(5.47)

Ideal Cycles and their Analysis

93

Combustion chamber 1

2

Air

Intercooler

Compressor

4

3

Fuel

Compressor

5

Product of combustion 6 Power output

Turbine

Fig. 5.21 Schematic arrangement of a intercooled cycle

4

10

1600

5

8

5

1100 T

p6

6

4

3

600

2

0

0

1

3

6

1

V

2

100 3

Fig. 5.22 p-V diagram

8

1

18

28

s

38

48

58

Fig. 5.23 T -s diagram

0.6

2

t=5

t=5

t=4

1.5

Wmax /CpT1

2

4

2

t=4

0.4

t=3

η

1

t=3

t=

0.2

0.5

2

t=2 0

0

5

Fig. 5.24 r vs

10 r Wmax Cp T1

15

20

0

0

5

10

r Fig. 5.25 r vs efficiency

15

20

94

Gas Turbines

It can be shown that for maximum power output and perfect intercooling, (following the similar procedure as we have done for reheat cycle) √ c1 = c = c2 (5.48) =

T5 T6 T5 T2 T4 − − +1− +1 T1 T5 T1 T1 T3

=

t−

√ t √ − c+1− c+1 c

Wmax Cp T1

=

t−

√ t −2 c+2 c

ηmax

=

Wmax Q

=

t−

Wmax Cp T1

Since T3 = T1 ηmax

=

=

t c

(5.49)

√ Cp T1 t − ct − 2( c − 1) Cp (T5 − T4 )

=

√ − 2( c − 1)

T5 T1

−

T4 T3

√ √ c − ct − c + 2 √ t− c √ t c−2 c + √ 1− t− c t−

(5.50)

(5.51)

The specific work output and the efficiency curves are as shown in Figs. 5.24 and 5.25 respectively. Intercooling will help to increase the net work output of the cycle. Because of the lower compressor outlet temperature, the fuel flow rate to obtain a given turbine inlet temperature will increase. Therefore, the thermal efficiency of the intercooled cycle will be less than that for a simple cycle. Intercooling is useful when the pressure ratios are high and the efficiency of the compressor is low. Due to lower compressor outlet temperature there will be more scope for adding a heat exchanger. Due to regeneration, a substantial amount of heat from exhaust gases can be recovered. This will be seen in the next section. 5.7

THE INTERCOOLED CYCLE WITH HEAT EXCHANGER

As seen in the previous section, by intercooling we can improve the work output of the cycle. Suppose we have to improve the efficiency also, then we can add a heat exchanger to the intercooled cycle. The basic schematic arrangements along with the p-V and T -s diagrams are shown in Figs. 5.26, 5.27 and 5.28 respectively. Figures 5.29 and 5.30 show the performance curves of the cycle. Now, for an unit quantity of mass flow Compressor work [WC ]

Ideal Cycles and their Analysis

Products of combustion 1 Air

8 2

95

Heat exchanger 3

Intercooler

Compressor

7

Combustion chamber

5

4 Fuel

6 Power output

Compressor Turbine

Fig. 5.26 Schematic arrangement of an intercooled cycle with heat exchanger

4

10

1600

5

5

7

8

1100 T

p6 4

3

7

600

2

0

1

0

1

6

8

V

2

8

3

100 3

Fig. 5.27 p–V diagram

8

1

18

28

t=5

1.5

s

38

48

t=

0.6

t=4

t=

η 0.4

1

t=5 t=4

3

t=5

2

t=3 t= 2

t=3 0.5

0.2

t=2 0

0

5

Fig. 5.29 r vs

10 r Wmax Cp T1

58

Fig. 5.28 T -s diagram

0.8

2

Wmax /CpT1

2

4

2

6

15

Intercooled cycle

20

0

0

5

10

r Fig. 5.30 r vs efficiency

15

20

96

Gas Turbines

W12 + W34

=

Cp (T4 − T3 ) + Cp (T2 − T1 )

(5.52)

Cp (T5 − T7 )

(5.53)

Cp (T5 − T6 )

(5.54)

Heat addition [Q] Q75

=

Turbine work [WT ] W56

=

Net work output [WN ] =

WT − WC

=

Cp (T5 − T6 ) − Cp (T2 − T1 ) − Cp (T4 − T3 )

The above expression is identical to equation in a simple cycle with intercooling, Hence, the maximum specific work output is given by Wmax Cp T1 ηmax

√ t −2 c+2 c

=

t−

=

Wmax Q

=

t−

=

t−

t c

ηmax

=

1−

√ 2( c − 1) t 1 − 1c

=

√ t c −2 c T5 T7 T1 − T1

Since T7 = T6 T5 T1

√ Cp T1 t − ct − 2 c + 2 Cp (T5 − T7 ) +2

√ −2 c+2

−

T6 T5

(5.55)

T5 T1

=

t−

t c

√ −2 c+2 t − ct (5.56)

Figures 5.29 and 5.30 give the variation of specific power output and efficiency respectively with r for various t. As already discussed the specific power output increases over a simple cycle because of intercooling. Because of the addition of heat exchanger, substantial recovery of heat from the exhaust gas is possible which increases the efficiency also. It may be noted that the improvement in efficiency is substantial at lower pressure ratios and higher turbine inlet temperatures. 5.8

THE INTERCOOLED AND REHEAT CYCLE

We have seen – in the previous section – how to improve the efficiency of an intercooled cycle by adding an heat exchanger. We can further improve the specific work output of the intercooled cycle by adding reheat. The schematic diagram of the cycle is as shown in Fig. 5.31. The corresponding

Ideal Cycles and their Analysis

97

p-V and T -s diagrams are as shown in Figs. 5.32 and 5.33 respectively. Figures 5.34 and 5.35 show the performance curves of the cycle. For a unit quantity of mass flow, Compressor work [WC ] W34 + W12

= Cp (T4 − T3 ) + Cp (T2 − T1 )

(5.57)

Heat addition [Q] Q45 + Q67

= Cp (T5 − T4 ) + Cp (T7 − T6 )

(5.58)

Turbine work [WT ] W56 + W78

= Cp (T5 − T6 ) + Cp (T7 − T8 )

(5.59)

Net work output [WN ] = WT − WC = Cp (T5 − T6 + T7 − T8 ) − Cp (T4 − T3 + T2 − T1 ) = Cp T1 WN Cp T1

=

(5.60)

T5 T6 T7 T8 T4 T3 T2 − + − − + − +1 T1 T1 T1 T1 T1 T1 T1

T5 T6 T5 T5 T8 T5 T4 T1 T2 − + − − + − +1 T1 T5 T1 T1 T7 T1 T3 T1 T1

By following the usual procedure to get the maximum power output, we can show that √ √ Wmax t t = t− √ +t− √ − c+1− c+1 Cp T1 c c √ 2t = 2t − √ − 2 c + 2 c Wmax Cp T1 ηmax

√ t = 2 t− √ − c+1 c =

=

=

(5.61)

Wmax Q 2Cp T1 t −

√t c

−

√ c+1

Cp (T5 − T4 ) + Cp (T7 − T6 ) 2Cp T1 t − Cp T1

T5 T1

−

√t c

−

T4 T1

+

√ c+1 T7 T1

−

T6 T1

(5.62)

98

Gas Turbines

Products of combustion Reheat combustion

Combustion chamber

1 2 3 4 Intercooler Air LP HP Compressor

Fuel

chamber

5

6

Fuel

8 Power output

LP

HP

Compressor

7

Turbine

Turbine

Fig. 5.31 Schematic arrangement of an intercooled and reheat cycle

4

10

5

7

5

1500

8 1000 T

p6 4

3

2 0

2

7

500

6

8

1

0

1

2 V

0

3

Fig. 5.32 p–V diagram

2.5

Wmax /Cp T1

2

3

1

10 20 30 40 50 60 70 s

Fig. 5.33 T -s diagram

t=5

0.4

t=4

t=3

1.5

η

t=3

1

t=2 0.2

t=2

0.5 0

4

t=5

2

8

6

0

5

Fig. 5.34 r vs

10 r Wmax Cp T1

15

20

0

0

5

10

r Fig. 5.35 r vs efficiency

15

20

Ideal Cycles and their Analysis

99

√ − c+1 √ 2t − c − √tc

=

2 t−

=

2(t −

√t c

√

c) − 2 √tc − 1 √ 2t − c − √tc

√ + c−2 = 1− √ 2t − c − √tc √t c

(5.63)

(5.64)

The specific work output and efficiency curves are shown in Figs. 5.34 and 5.35 respectively. 5.9

INTERCOOLED CYCLE WITH HEAT EXCHANGE AND REHEAT

This is the most complex cycle arrangement so far discussed. Even though the specific work output and the efficiency of this cycle will be the maximum, it is at the expense of the simplicity. The schematic arrangement and the corresponding p-V and T -s diagrams are given in Figs. 5.36, 5.37 and 5.38 respectively. Figures 5.39 and 5.40 show the performance curves of the cycle. For unit quantity of mass flow Compressor work [WC ] W34 + W12

=

Cp (T4 − T3 ) + Cp (T2 − T1 )

(5.65)

Cp (T5 − T9 ) + Cp (T7 − T8 )

(5.66)

Heat addition [Q] Q95 + Q67

=

Turbine work, [WT ] W56 + W78

=

Cp (T5 − T6 ) + Cp (T7 − T8 )

(5.67)

Net work output [WN ] =

WT − WC

=

[Cp (T5 − T6 ) + Cp (T7 − T8 )] − [Cp (T4 − T3 ) + Cp (T2 − T1 )]

(5.68)

This expression is exactly identical to the expression for the simple cycle with intercooling and reheat (Eq. 5.60). Hence, Wmax Cp T1

=

√ t 2 t− √ − c+1 c

100

Gas Turbines

Heat exchanger

Products of combustion

Combustion Reheat Combustion chamber chamber

10 1

Air

9

2

Intercooler

3 4

LP

HP

Compressor

6

5

Fuel

Fuel

Power output

LP

HP

Compressor

8

7

Turbine

Turbine

Fig. 5.36 Schematic arrangement of intercooled cycle with heat exchanger and reheat

4

10

5

8

4

3

2 0

2

7 10

8

1

0

1

2 V

0

3

2.5

0

η

t=3 t=2

0

5

Fig. 5.39 r vs

10 r Wmax Cp T1

3

1

10

10 20 30 40 50 60 70 s

15

t=5

t= 0.4

t=5

2

t=3 t=2

0.2

20

0

t=4

t=3

0.6

1.5

0.5

2

0.8

t=4

1

4

Fig. 5.38 T -s diagram

t=5

2

8

6

500

6

Fig. 5.37 p–V diagram

Wmax /Cp T1

9

1000 T

p6

7

5

1500

9

Intercooled with reheat

0

5

10

r Fig. 5.40 r vs efficiency

15

20

Ideal Cycles and their Analysis

ηmax

=

=

101

Wmax Q 2Cp T1 t −

√t c

−

√ c+1 (5.69)

Cp (T5 − T9 ) + Cp (T7 − T6 )

Since T6 = T8 ηmax

ηmax

=

2 t−

=

2t −

=

T5 T1

−

√ √t − c+1 c T8 T7 T8 T1 + T1 − T1

2t √ c

√ −2 c+2

2t −

2t √ c

=

2t −

t−

2t √ c √t c

√ −2 c+2 +t−

√t c

√ c−1 1− t √ √ ( c − 1) c

=

√ c 1− t

(5.70)

The specific power output and the efficiency curves are as shown in Figs. 5.39 and 5.40 respectively. 5.10

COMPARISON OF VARIOUS CYCLES

However, the purpose of the foregoing sections has been to illustrate the individual effects of the most important modifications which can be made to a simple gas turbine cycle. Using the various expressions developed in previous sections, Table 5.1 has been compiled for r = 4 and t = 3 from which the percentage gains or losses over the performance of a typical simple cycle, can be ascertained. Table 5.1 Comparison of Various Cycles S.No. 1. 2. 3. 4. 5. 6. 7.

Addition to simple cycle Heat exchange Intercooling Reheat Reheat and heat exchange Intercooling and heat exchange Reheat and intercooling Reheat, intercooling and heat exchange

Effect on W η Cp T1 +50.0 No change − 6.50 +10.2 −10.4 +24.5 +66.7 +24.5 +68.0 +10.2 −18.2 +80.0

+34.7 +34.7

These values of pressure ratio and the maximum temperature ratio are used in all the cycles, and equal division of the overall pressure ratio in compression or expansion – corresponding to maximum work output, is assumed when intercooling and/or reheat are used.

102

Gas Turbines

2

(a) Simple cycle (b) Reheat (c) Intercooling (d) Intercooling & reheat

1.5

d c

W/CpT1

b d

a

1

c b

0.5 t=4 t=3

0

0

5

a

10

15

20

r Fig. 5.41 Comparison of various cycles It should be noted that this is not an exhaustive comparison, because the value of the overall pressure ratio for maximum work output is also affected by the various additions to the simple cycle. The effect may be seen from the curves of Fig. 5.41, which show the variation of work output with r for some of the cycles under two conditions of turbine inlet temperature. The chain-dotted curves correspond to t = 3 so that, if the compressor inlet temperature is taken as 288 K, the turbine inlet temperature will be around 865 K, a value suitable for a long life unit. The full line curves are for t = 4, which gives T3 ≈ 1150 K a figure somewhat reasonable for the power plants used at present in short life units such as jet engines for fighter aircraft. No mention is made of heat exchange in Fig. 5.41, as we have seen that the addition of this process to an ideal cycle makes no difference to the work output. It would of course be possible to make a comparison of the performance of cycles without heat exchange using, for each cycle, the overall pressure ratio giving maximum work output. This might, however, give a misleading impression because, as will be shown in the next Chapter, the efficiency of compression is not only less than 100% in practice, but also decrease with increase of pressure ratio. The gain in cycle efficiency with increase of pressure ratio is not, therefore, so great as would appear from calculations based on ideal cycles. It is, of course, possible to incorporate in a cycle multiple arrangements of the above modification, such as several compression stages with intercooling, several expansion stages with reheating, and heat exchanger. A study of any such cycle will reveal that improvements in performance will follow the form of those realized with each individual

Ideal Cycles and their Analysis

103

addition as shown above. Appendix 1 provides in a nutshell the complete details of the 10 cycles we have discussed in the previous sections. It is intended as a ready reckoner for the reader. 5.11

ERICSSON CYCLE

A discussion of ideal cycles for the gas turbine would not be complete without some reference to the theoretical cycle of Ericsson. In introducing multi-stage compression with intercooling, and multistage expansion with reheating, it is immediately seen from the T -s diagram that an approach is being made towards isothermal processes. This would, of course, be achieved only if the number of stages were infinite. In this case, the cycle would follow the outline 1234 given in Fig. 5.42. With a perfect heat ex1000 4

3

800

T 600

400 1

2

200

0

10

20

30 s

40

50

60

Fig. 5.42 Ericsson cycle changer, the heat supplied from 2 to 3 would be completely furnished by the heat rejected from 4 to 1, so that the only heat supplied is that during the isothermal expansion from 3 to 4. Similarly, the only heat rejected is that during the isothermal compression from 1 to 2. These heat quantities P2 2 are easily shown to be equal to RT3 loge P P1 and RT1 loge P1 respectively, so that the work done is equal to their difference, which can be written as 2 R(T3 − T1 ) loge P P1 . The efficiency is therefore given by η

=

T3 − T1 T3

since the pressure ratio is the same in compression and expansion. Here R refers to gas constant. The cycle is therefore of optimum efficiency like the Carnot cycle, all heat received at the upper temperature and rejected at the lower temperature. However, this cycle is really an ideal one and is of theoretical interest only.

104

Gas Turbines

Worked out Examples 5.1 Compute the indicated mean effective pressure and efficiency of a Joule cycle if the temperature at the end of combustion is 2000 K and the temperature and pressure before compression is 350 K and 1 bar. The pressure ratio is 1.3. Assume Cp = 1.005 kJ/kg K. Solution 2

10

3

8 p6 4 2 0

1

0

1

4

2

V

Fig. 5.43

Consider the isentropic compression process 1→ 2, γ−1 p2 γ = T1 = 350 × (1.3)0.286 = 377.27 K T2 p1 Consider the isentropic expansion process 3→ 4 T4

=

T3

p4 p3

γ−1 γ

= 2000 ×

1 1.3

0.286

= 1855.42 K

Work done during compression, (WC ) WC

=

Cp (T2 − T1 ) = 1.005 × (377.27 − 350) = 27.40 kJ/kg

Work done during expansion, (WT ) WT

WN

=

Cp (T3 − T4 )

=

1.005 × (2000 − 1855.42)

=

WT − WC

Volume at various state points:

=

=

145.30 kJ/kg

145.30 − 27.40 = 117.90 kJ/kg

Ideal Cycles and their Analysis

0.287 × 103 × 350 = 1.0045 m3 /kg 1 × 105

V1

=

RT1 p1

V2

=

0.287 × 103 × 377.27 1.3 × 105

V3

=

0.287 × 103 × 2000 1.3 × 105

V4

=

0.287 × 103 × 1855.42 1.0 × 105

imep

=

net work done change in volume

=

117.90 × 103 × 10−5 5.325 − 0.833

=

Cp (T3 − T2 ) = 1.005 × (2000 − 377.27)

=

1630.84 kJ/kg

=

WN q

q

η

=

=

105

0.833 m3 /kg

=

4.415 m3 /kg

=

=

=

5.325 m3 /kg

WN V4 − V2 =

117.9 × 100 1630.84

0.2625 bar

=

7.23%

Ans

⇐=

Ans

⇐=

Since the pressure ratio is quite small in spite of higher peak temperature, the efficiency is very low. This example illustrates that the efficiency of a simple cycle depends mainly on the pressure ratio and not on the peak temperature. 5.2 Calculate the improvement in the efficiency when a heat exchanger is added to the simple cycle given in the previous example. Solution The T -s diagram of the cycle is shown in Fig. 5.44. As can be seen the heat input is q = Cp (T3 − T5 ) For a perfect heat exchange, T5 = T4 q

=

1.005 × (2000 − 1855.42) = 145.30 kJ/kg

η

=

117.90 WN = q 145.30

=

0.81

=

81%

Ans

⇐=

Note the tremendous improvement in the efficiency when a heat exchanger is added to a simple cycle with a lower pressure ratio. This example amply

106

Gas Turbines

3

1500 1000 T

5 4

2

500

6 1

0

20

30

40 s

50

60

Fig. 5.44

illustrates that for a heat exchange cycle lower pressure ratio and higher peak temperature is advantageous. η2 η1

81 7.23

=

=

Ans

11.20

⇐=

More than 11 times efficiency improvement is achieved because of the addition of heat exchanger.

5.3 A gas turbine operates on a pressure ratio of 6. The inlet air temperature to the compressor is 300 K and the air entering the turbine is at a temperature of 577◦ C. If the volume rate of air entering the compressor is 240 m3 /s. Calculate the net power output of the cycle in MW. Also compute its efficiency. Assume that the cycle operates under ideal conditions. Solution Consider the isentropic compression process 1→ 2 T2

=

T1

p2 p1

γ−1 γ

0.286

= 500.81 K

=

509.18 K

= 300 × (6)

Consider the isentropic expansion process 3→ 4 T3 850 T4 = = γ−1 0.286 6 γ p3 p4

Work input to the compressor, (WC ), WC = Cp (T2 − T1 )

=

1.005 × (500.81 − 300) = 201.81 kJ/kg

Ideal Cycles and their Analysis

1500

107

3

1000 T

4

2

500 1

0

20

30

40 s

50

60

Fig. 5.45

Work done by the turbine, (WT ), =

Cp (T3 − T4 )

=

1.005 × (850 − 509.18) = 342.52 kJ/kg

=

WT − WC

=

140.72 kJ/kg

=

Cp (T3 − T2 )

=

1.005 × (850 − 500.81) = 350.94 kJ/kg

ρ1

=

p1 1 × 105 3 = 1.161 kg/m = RT1 287 × 300

Power output

=

140.72 × 240 × 1.161 = 39210.22 kW

=

39.2102 MW

=

WN q

WT

WN

q

η

=

=

342.52 − 201.80

140.72 × 100 = 40.16% 350.94

Ans

⇐= Ans

⇐=

5.4 A gas turbine cycle has a perfect heat exchanger. Air enters the compressor at a temperature and pressure of 300 K and 1 bar and discharges at 475 K and 5 bar. After passing through the heat exchanger the air temperature increases to 655 K. The temperature of air entering and leaving the turbine are 870◦C and 450◦ C. Assuming no pressure drop through the heat exchanger, compute (i) the output per kg of air, (ii) the efficiency of the cycle, and (iii) the work required to drive the compressor.

108

Gas Turbines

Solution

12 10

2

5

3

1500

3

8 p6

1000 T

5 4

2

4

500

6

2 0

1

0

6

1

V

2

1

4

0

3

20

30

40 s

50

60

Fig. 5.46 Work input to the compressor, (WC ), WC

=

Cp (T2 − T1 )

=

175.9 kJ/kg

=

1.005 × (475 − 300)

=

1.005 × (1143 − 723)

=

1.005 × (1143 − 655)

Work done by the turbine, (WT ), WT

Heat added, q

=

Cp (T3 − T4 )

=

422.1 kJ/kg

=

Cp (T3 − T5 )

=

490.44 kJ/kg

The output per kg of air WN

=

WT − WC

=

246.2 kJ/kg

=

422.1 − 175.9 Ans

⇐=

The efficiency of the cycle, η

=

Work output Heat supplied

=

246.2 × 100 490.44

=

50.2%

Ans

⇐=

The work required to drive the compressor =

175.9 kJ/kg

Ans

⇐=

Ideal Cycles and their Analysis

109

5.5 A closed-cycle regenerative gas turbine operating with air as the working medium. Assume the following data: p1 = 1.4 bar, T1 = 310 K, p2 /p1 = 5, Tmax = 1050 K, effectiveness of the regenerator is 100%, net output = 3000 kW. Assuming the compression and expansion to be isentropic, calculate (i) thermal efficiency, and (ii) mass flow rate of air per minute.

Solution 3

1500 1000 T

5 4

2

500

6 1

0

20

30

40 s

50

60

Fig. 5.47

T2

=

T4

=

p2 p1

T1

γ−1 γ

T3 (rp )

310 × 50.286

=

1050 50.286

=

γ−1 γ

=

=

491.2 K

662.65 K

As regenerator effectiveness is 100%, T4

=

T5

WN

=

mC ˙ p (T3 − T4 ) − mC ˙ p (T2 − T1 )

3000

=

m ˙ × 1.005 × [(1050 − 662.65) − (491.2 − 310)]

3000

=

m ˙ × 207.18

m ˙

=

14.48 kg/s

ηth

=

WN Heat added

=

662.65 K

=

868.8 kg/min

Ans

⇐=

110

Gas Turbines

=

mC ˙ p [(T3 − T4 ) − (T2 − T1 )] mC ˙ p (T3 − T5 )

=

(1050 − 662.65) − (491.2 − 310) × 100 (1050 − 662.65)

=

53.22%

Ans

⇐=

5.6 A gas turbine with a regenerator has got the following data: Compressor inlet temperature : 290 K Compressor outlet temperature : 460 K Inlet temperature of the turbine : 900◦C Outlet temperature to the turbine : 467◦C Assuming no pressure drop in the heat exchanger, calculate (i) the pressure ratio of the compressor and turbine, (ii) the specific power output, (iii) the overall efficiency of the cycle, and (iv) the work required to drive the compressor. Assume 100% mechanical efficiency. Solution 3

1500 1000 T

5 4

2

500

6 1

0

20

30

40 s

50

60

Fig. 5.48

T2 T1

=

c

=

460 290

=

1.586

Compressor pressure ratio, r

=

γ

c γ−1

=

1.5863.5

Specific power output (per kg of air)

=

5

Ans

⇐=

Ideal Cycles and their Analysis

WN

=

Cp (T3 − T4 ) − Cp (T2 − T1 )

=

1.005 × (1173 − 740) − 1.005 × (460 − 290)

=

264.3 kJ/kg

Overall efficiency of the cycle = η

111

Ans

⇐=

Specific power output Heat supplied per kg of air

264.3 Cp (T3 − T5 )

=

Assuming regenerator effectiveness to be 100%, T5 = T4 . 264.3 × 100 1.005 × (1173 − 740)

=

=

60.7%

Ans

⇐=

Work required to drive the compressor

T3 T4

=

Cp (T2 − T1 )

=

170.85 kJ/kg

=

c

=

=

1173 740

1.005 × (460 − 290) Ans

⇐= =

1.585

Turbine pressure ratio, r

γ

=

c γ−1

=

1.5853.5

=

Ans

5

⇐=

5.7 The ratio of net work to turbine work of an ideal gas turbine plant is 0.563. Take the inlet temperature to the compressor as 300 K. Calculate the temperature drop across the turbine if the thermal efficiency of the unit is 35%. Assume a mass flow rate of 10 kg/s, Cp = 1 kJ/kg K and γ = 1.4. Solution For ideal simple cycle, 1 c

η

=

1−

c

=

1.5384

T2

=

T1 × c

=

300 × 1.5384

=

Cp (T2 − T1 )

=

1 × (461.52 − 300)

WC

=

0.35

=

461.52 K

=

161.52 kJ/kg

112

Gas Turbines

1500

3

1000 T

4

2

500 1

0

20

30

40 s

50

60

Fig. 5.49

Net work Turbine work

=

0.563

WC WT

=

0.563

WT

=

WC 1 − 0.563

=

369.61 kJ/kg of air

=

WT − WC

=

208.09 kJ/kg of air

1−

WN

161.52 0.437

=

=

369.61 − 161.52

Net heat supplied per kg of air =

1 × WN ηth

=

594.54 kJ/kg of air

q

=

Cp (T3 − T2 )

594.54

=

1 × (T3 − 461.52)

T3

=

1056.06 K

c

=

T3 T4

T4

=

T3 c

=

1056.06 1.5384

q

=

208.09 0.35

=

1.5384

=

686.47 K

Ideal Cycles and their Analysis

113

Hence, temperature drop across the turbine T3 − T4

=

1056.06 − 686.47 = 369.59 K

Ans

⇐=

5.8 The specific power output of a turbine is 336.5 kW/kg and the exhaust gases leave the turbine at 700 K. Calculate the turbine pressure ratio if the value of Cp and Cv are 1 kJ/kg K and 0.717 kJ/kg K respectively. Round off the pressure ratio to the nearest integer. Solution γ

=

1 Cp = Cv 0.717

Cp (T3 − T4 )

=

336.5

T3 − 700

=

336.5

T3

=

1036.5 K

p3 p4

γ γ−1

T3 T4

= =

=

=

1.3947

1036.5 700

1.3947 0.3947

Ans

4

⇐=

5.9 A gas turbine plant works between temperature limits of 300 K and 900 K. The pressure limits are 1 bar and 4 bar. Estimate the thermal efficiency of the plant and the shaft power available for the external load in kW. Assume mass rate of flow of air to the compressor as 1600 kg/min. Solution

1500

3

1000 T

4

2

500 1

0

20

30

40 s

Fig. 5.50

50

60

114

Gas Turbines

For simple cycle gas turbine plant η

1 c

=

1−

c

=

p2 p1

η

=

1−

W Cp T1

=

t 1−

1 c

t

=

900 300

=

W 1.005 × 300

=

3× 1−

=

0.4952 kJ/kg of air

=

0.4952 × 1.005 × 300

=

1600 60

W Mass flow of air,

Shaft power available =

γ−1 γ

40.286

= 1 1.486

× 100

=

=

1.486 Ans

32.7%

⇐=

− (c − 1) 3 1 1.486

=

− (1.486 − 1)

=

149.3 kJ/kg of air

26.66 kg/s

149.3 × 26.66

=

3980.34 kW

Ans

⇐=

5.10 In a gas turbine, the pressure ratio to which air at 15◦ C is compressed is 6. The same air is then heated to a maximum permissible temperature of 750◦ C. First in a heat exchanger and then combustion chamber. It is then expanded in two stages such that the expansion work is maximum. The air is reheated to 750◦ C after the first stage. N Determine the cycle thermal efficiency, the work ratio W and net WT shaft work per kg of air. Solution c

=

60.286

T2

=

T1 × 1.669

For maximum expansion work p3 p5 = = p4 p6

=

1.669 =

√ rc

288 × 1.669 =

√

6

=

=

2.45

480.67 K

Ideal Cycles and their Analysis

1600 7

6

4 2

600 100

5

3

1100 T

115

8

1

20

30

40

s

50

60

70

Fig. 5.51

T4

=

T3 p3 p4

γ−1 γ

1023

=

=

(2.45)0.286

791.7 K

As pressure ratio is same, T6 = T4 = 791.7 K. WC

WT

q

=

Cp (T2 − T1 )

=

1.005 × (480.67 − 288)

=

193.63 kJ/kg of air

=

Cp (T3 − T4 ) + Cp (T5 − T6 )

=

2 × 1.005 × (1023 − 791.7) = 464.9 kJ/kg of air

=

Cp (T3 − T7 ) + Cp (T5 − T4 )

Because of 100% regeneration, q

=

2 × 1.005 × (1023 − 791.7)

=

464.9 kJ/kg of air

=

WT − WC

=

271.34 kW

η

=

WN q

=

271.34 × 100 464.9

Wratio

=

WN WT

=

271.34 464.9

WN

=

Ans

⇐=

464.9 − 193.56 Ans

⇐=

=

=

58.36%

0.5836

Ans

⇐= Ans

⇐=

116

Gas Turbines

5.11 A gas turbine plant operates between 5◦ C and 839◦ C. Find (i) pressure ratio at which cycle efficiency equals Carnot cycle efficiency, (ii) pressure ratio at which maximum work is obtained, and (iii) efficiency under conditions giving maximum work. Solution

1500

3

1000 T

4

2

500 1

0

20

30

40 s

50

60

Fig. 5.52

ηcarnot

=

1−

Tmin Tmax

=

1−

1 c

=

1−

278 1112

=

0.75

For a simple cycle η

c

=

4

p2 p1

=

43.5

=

0.75 p2 p1

= =

γ−1 γ

128

Therefore, pressure ratio at which cycle efficiency equals Carnot efficiency Ans = 128 ⇐= For a simple cycle, WN Cp T1

=

t 1−

1 c

− (c − 1)

Pressure ratio for maximum work is obtained when

Ideal Cycles and their Analysis

dWN dc

=

0

0

=

t 0+

t −1 c2

=

0

c

=

√ t

=

2

=

23.5

p2 p1

1 c2

117

− (1 − 0)

√ 4

=

=

2

γ−1 γ

p2 p1

=

11.31

Pressure ratio for maximum work =

Ans

11.31

⇐=

Efficiency at maximum work output 1 = = 1− c

1−

1 2

=

50%

Ans

⇐=

5.12 An ideal open-cycle gas turbine plant using air operates in an overall pressure ratio of 4 and between temperature limits of 300 K and 1000 K. Assuming the constant value of specific heat Cp = 1 kJ/kg K and Cv = 0.717 kJ/kg K, evaluate the specific work output and thermal efficiency for each of the modification below and state the percentage change from the basic cycle. Assuming optimum stage pressure ratios, perfect intercooling and perfect regeneration, find (i) basic cycle, (ii) basic cycle with heat exchanger, (iii) basic cycle with two stage inter cooled compressor, and (iv) basic cycle with heat exchanger and two-stage intercooled compressor. Solution Basic cycle: (refer Fig. 5.3) k

=

Cp Cv

c

=

(rp )

t

=

T3 T1

= γ−1 γ

=

1 0.717 =

=

40.283

1000 300

=

1.3947 =

1.4804

3.333

118

Gas Turbines

WN Cp T1

WN

η

1 c

− (c − 1)

=

t 1−

=

3.333 × 1 −

=

0.6012

=

Cp × T1 × 0.6012

=

1 × 300 × 0.6012

=

1−

1 c

1 1.4804

=

1−

− (1.4804 − 1)

=

180.36 kJ/kg

1 1.4804

× 100

=

32.45%

=

55.58%

Basic cycle with heat exchanger: (refer Fig. 5.8) WN

=

180.36 kW

η

=

1−

c t

=

1−

1.4804 3.333

× 100

Basic cycle with intercooled compressor: (refer Fig. 5.23) WN Cp T1

1 c

−2

√ c−1

=

t 1−

=

3.333 × 1 −

=

0.64815

WN

=

η

=

1 × 300 × 0.64815 √ t c−2 c + √ 1− t− c

=

1−

1 1.4804

−2×

=

194.44 kJ/kg

√ 1.4804 − 2 √ 3.333 − 1.4804

3.333 1.4804

+

√ 1.4804 − 1

× 100

=

30.63%

Basic cycle with heat exchanger and intercooled compressor: (refer Fig. 5.28) WN η

=

194.47 kJ/kg

=

√ √ 2× 1.4804 − 1 2( c − 1) = 1− 1− 1 1 t 1− c 3.333 × 1 − 1.4804

=

59.92%

× 100 Ans

⇐=

Ideal Cycles and their Analysis

Cycle

WN (kJ/kg) 180.36 180.36 194.47 194.47

Basic cycle With heat exchanger With intercooling With heat exchanger & intercooling

η (%) 32.45 55.58 30.63 59.92

% change in WN – 0 7.8 7.8

119

% change in η – 71.28 −5.61 84.65

5.13 Prove that the efficiency corresponding to the maximum work done in a Brayton cycle is given by the relation 1 ηwmax = 1 − √ t where t is the ratio of the maximum and minimum temperatures of the cycle. A gas turbine operating on Brayton cycle between 27◦ C and 827◦C. Determine the maximum net work per kg and cycle efficiency. Also compare Carnot efficiency with Brayton efficiency for these temperature limits. Solution

1500

3

1000 T

4

2

500 1

0

20

30

40 s

50

60

Fig. 5.53

WT

=

Cp (T3 − T4 )

WC

=

Cp (T2 − T1 )

WN

=

WT − WC

=

Cp T3 1 −

= T4 T3

Cp [(T3 − T4 ) − (T2 − T1 )] − T1

T2 −1 T1

120

Gas Turbines

Let rp be the pressure ratio, ! WN

Let c = (rp ) WN

=

γ−1 γ

Cp T3 1 −

(rp )

− T1 (rp )

γ−1 γ

γ−1 γ

−1

"

T3 T1

and t = =

1

1 c

Cp T3 1 −

− T1 (c − 1)

For maximum work dWN /dc = 0. 0

=

Cp T3

T1

=

T3

c2

=

c

=

η

=

=

1 − T1 c2

1 c2

T3 T1 √ t

=

t

(T3 − T4 ) − (T2 − T1 ) (T3 − T2 ) T3 1 −

T4 T3

T3 T1

T1 T3 T1

=

=

=

=

1−

t

(rp

−1

T2 T1

− (rp )

γ−1 ) γ

− (rp )

γ−1 γ

−1

γ−1 γ

1 c

− (c − 1) (t − c)

c−1 c

− (c − 1) (t − c)

(c − 1) t−c c (t − c)

For maximum work c =

−

1

T3 T1

t 1−

T2 T1

− T1

=

=

(c − 1) ct − 1 (t − c)

c−1 c

=

1−

1 c

√ t

ηwmax

=

1 1− √ t

T1

=

27 + 273

Ans

⇐= =

300 K

Ideal Cycles and their Analysis

T2

=

1100 K

t

=

1100 300

Wmax

=

1 Cp T3 1 − √ t

=

1.005 ×

=

3.666 − T1

1 1100 × 1 − √ 3.666 =

121

√ t−1

− 300 ×

√

3.666 − 1 Ans

⇐=

252.34 kJ/kg of air

ηwmax

=

1 1− √ 3.666

ηcarnot

=

Tmax − Tmin Tmax

=

0.7272

=

× 100

=

47.77%

Ans

⇐=

1100 − 300 1100

=

Ans

72.72%

⇐=

5.14 In an ideal gas turbine cycle with reheat, air at state (p1 , T1 ) is compressed to pressure rp1 and heated to T3 . The air is then expanded in two stages, each turbine having the same pressure ratio, with reheat to T3 between the stages. Assuming the working fluid to be a perfect gas with constant specific heats, and that the compression and expansion are isentropic, show that the specific work output will be a maximum when r is given by r where t =

T3 T1

and a =

=

t(2/3a)

=

c

γ−1 γ .

Solution

T3 T4

=

T2 T1

=

T5 T6

=

ra √ r

a

=

√

c

Assume unit mass flow of air and given T5 = T3 . Total turbine work output, WT

=

Cp (T3 − T4 ) + Cp (T5 − T6 )

=

Cp (T3 − T4 ) + Cp (T3 − T6 )

122

Gas Turbines

1600 1100 T

5

4

6

4’

2

600 100

3

1

20

30

40

s

50

60

70

Fig. 5.54

=

1 2Cp (T3 − T4 ) = 2Cp T3 1 − √ c

=

Cp (T2 − T1 )

=

Cp T1 (c − 1)

=

1 2Cp T3 1 − √ c

=

2

=

2t 1 − c− 2 − (c − 1)

Compressor work input, WC

=

Cp T1

T2 −1 T1

Net work output, WN

− Cp T1 (c − 1)

Specific work output, WN Cp T1

1 T3 1− √ T1 c 1

Specific work output is maximum when entiating Eq.(1) and equating it to zero, 2t −1 × −

1 2

3

× c− 2 − 1

r(

γ−1 γ

WN Cp T1

= 0. Therefore, differ-

0

=

1

)

=

ra

r

=

t(2/3a)

3

=

d dc

(1)

=

t × c− 2 c

− (c − 1)

=

2

t3 Ans

⇐=

Ideal Cycles and their Analysis

123

5.15 A Brayton cycle works between 1 bar, 300 K and 5 bar, 1250 K. There are two stages of compression with perfect intercooling and two stages of expansion. The work out of first expansion stage being used to drive the two compressors, where the interstage pressure is optimized for the compressor. The air from the first stage turbine is again heated to 1250 K and expanded. Calculate the power output of free power turbine and cycle efficiency without and with a perfect heat exchanger and compare them. Also calculate the percentage improvement in the efficiency because of the addition of heat exchangers. Solution 5

1500 1000 T 500 0

7

4

2

3

1

9

8

6

5

7

6

8

T 2

4

10 1

3

10 20 30 40 50 60 70 s

s

Fig. 5.55 Because of perfect intercooling, √ p2 = r =

√ 5

=

2.236 bar

Further, T4 = T2 and T3 = T1 . c1

=

2.2360.286

T2

=

T 1 × c1

=

T4

=

T2

377.7 K

WC

=

Wc1 + Wc2

=

2 × 1.005 × (377.7 − 300)

WT 1

=

WC

Cp (T5 − T6 )

=

156.2

T6

=

1250 −

=

T5 T6

p5 p6

=

=

156.2 1.005

γ γ−1

=

1.259 300 × 1.259

=

2Wc1

=

=

= =

377.7 K

2Cp (T2 − T1 ) 156.2 kJ/kg

1094.6 K

1250 1094.6

3.5

= 1.591

124

Gas Turbines

=

5 1.591

T7 T8

=

p7 p8

T8

=

1250 ×

WT 2

=

Cp (T7 − T8 )

=

1.005 × (1250 − 900.9)

η

=

WN q

q

=

Cp (T5 − T4 ) + Cp (T7 − T6 )

=

1.005 × [(1250 − 377.7) + (1250 − 1094.6)]

=

1032.84 kJ/kg

=

350.8 × 100 1032.84

=

Cp × [(T5 − T9 ) + (T7 − T6 )]

=

1.005 × [(1250 − 900.9) + (1250 − 1094.6)]

=

507.02 kJ/kg

=

350.8 × 100 507.02

p6

η

=

3.143 bar

γ−1 γ

1 3.143

0.286

=

900.9 K

=

350.8 kJ/kg

WT 2 q

=

=

33.96%

With regenerator q

η

=

69.2%

=

50.9%

Percentage improvement in efficiency =

69.2 − 33.96 69.2

Ans

⇐=

5.16 The p-V diagrams of two gas turbine power plants operating on identical inlet conditions of 1 bar and 27◦ C are shown in the Fig. 5.56. The maximum turbine inlet temperature is 1200 K for both the plants. Assuming that the pressure ratio and temperature ratio are the same for both the plants, calculate the efficiency ratio of the power plants. Also find the pressure ratio of LPC, HPC, HPT and LPT. Solution η1

=

1−

c (Eq. 5.19) t

Ideal Cycles and their Analysis

12 10

2

5

5 9

3

8

8 p6

p6 4

4

3

2

7 6

2

2 0

4

10

125

1

0

1

4

6

V

2

0 3

10

8

1

0

1

2 V

3

Fig. 5.56

r1

=

√ c (Eq. 5.70) t √ 1 − c/t = 1 − c/t

η2

=

1−

η2 η1

=

x

t

=

r

c

=

r

=

40.286

√ c

=

1.219

x

=

1 − (1.219/4) 1 − (1.487/4)

η2 η1

=

1.107

r2

=

r3

=

2

=

4

γ−1 γ

=

=

1.487

=

1.107 Ans

⇐= r4

=

√ r

=

√ 4 Ans

⇐=

5.17 An ideal gas turbine operates with m number of compressor stages and n number of turbine stages with a overall pressure ratio, r. The maximum temperature is Tmax and the minimum temperature is Tmin . Assume pressure ratios in compressor is same in all m stages and perfect intercooling and reheating. Also assume that pressure ratios in all n stages in turbine are same. Show that t

=

texit × (ra )

m+n mn

T4 t = TTmax , a = γ−1 γ , and texit = T2 . where T4 and T2 are the exit min temperatures of the turbine and compressor respectively.

126

Gas Turbines

Solution

Stage n

Stage 3

Tmin

4

Stage 1

Stage 2

Stage 1 Stage 3

2

Stage m

T

Stage 2

3

Tmax

1 s Fig. 5.57

Let ric and rit be the stage pressure of the compressor and turbine respectively. m n r = ric and r = rit Now referring to the Fig. 5.57, T2 a = (r) Tmin

(i)

and

Tmax a = (r) T4

(i) × (ii) gives

Tmax T2 × Tmin T4

=

(ric ) × (rit ) = r m × r n

t

=

texit × (ra )

a

a

1

m+n mn

1

a

(ii)

= r

m+n mn

a

Ans

⇐=

5.18 In the above problem, if the gas turbine is working on optimum pressure ratio to give the maximum specific work output, establish the relation between t and c. Solution Let ri be the pressure ratio of each stage, where i = 1, 2, . . . m for compressor and i = 1, 2, . . . n for turbine. For compressor r

=

r1 × r2 × r3 × . . . × rm

=

rim

The exit temperature of the compressor is T2 and that of turbine is T4 . Since, the stage pressure ratios are equal, the temperature ratio will also be equal.

Ideal Cycles and their Analysis

127

Tmin

Stage n

Stage 3

4

Stage 1

Stage 1 Stage 2

Stage 3

Stage m

2

T

Stage 2

3

Tmax

1 s Fig. 5.58

r

r

γ−1 γ

=

rim

T2 Tmin

=

T2 Tmin

=

T2m

m Tmin

=

mn Tmin

=

T2mn cn

n Tmax

=

T4n r

mn Tmax

=

T4mn cm

=

T4 T2

r

γ−1 γ

=

m

=

γ γ−1

Tmax T4

m

n

T2m c (i)

Similarly, γ−1 γ

From (i) and (ii), tmn

(ii)

mn

cm+n

Since, for the given condition T4 = T2 t

=

c

m+n mn

Ans

⇐=

5.19 A gas turbine unit operates at a mass flow of 30 kg/s. Air enters the compressor at a pressure of 1 bar and temperature 15◦ C and is discharged from the compressor at a pressure of 10.5 bar. Combustion occurs at constant-pressure and results in a temperature rise of 420 K. If the flow leaves the turbine at a pressure of 1.2 bar, determine the net power output from the unit and also the thermal efficiency. Take Cp = 1.005 kJ/kg K and γ = 1.4.

128

Gas Turbines

Solution p2 p1

γ−1 γ

T2

=

T1

T3

=

564.22 + 420

T4

=

T3 p3 p4

γ−1 γ

= 288 × 10.50.286 = 564.22 K =

984.22 K

984.22

=

10.5 0.286 1.2

= 529.27 K

=

mC ˙ p (T2 − T1 )

=

30 × 1.005 × (564.22 − 288) = 8328 kW

=

mC ˙ p (T3 − T4 )

=

30 × 1.005 × (984.22 − 529.27)

=

13716.74 kW

WN

=

W T − WC

Q

=

mC ˙ p (T3 − T2 )

=

12663 kW

=

Work done Heat supplied

=

42.55%

⇐=

=

5388.74 kW

⇐=

WC

WT

ηth

Power output

=

5388.74 kW =

30 × 1.005 × 420 Ans

⇐= =

5388.74 × 100 12663 Ans

Ans

Review Questions 5.1 In what way the gas turbine cycles can be grouped and what are the important distinctions between them? 5.2 State the assumptions made in an ideal cycle analysis of gas turbines. 5.3 Draw the schematic diagram of a simple cycle and explain briefly the working of the cycle. Draw the p-V and T -s diagrams of the cycle. 5.4 Derive the expression for specific work output and the efficiency of a simple cycle. Draw their trends as a function of pressure ratio. 5.5 Show that the specific work output is maximum when the pressure ratio is such that the compressor outlet and turbine outlet temperatures are equal.

Ideal Cycles and their Analysis

129

5.6 Find a suitable equation for the two pressure ratios where the specific output becomes zero for a constant temperature ratio. 5.7 Explain why the specific power output becomes zero at these two pressure ratios for the above cycle. 5.8 Draw the schematic diagram of a simple cycle with a heat exchanger and explain briefly the working principle. Draw also the p-V and T -s diagrams of the cycle. 5.9 Derive the expression for specific work output and the efficiency of a simple cycle with a heat exchanger. Draw their trends as a function of pressure ratio. 5.10 Explain the important observations from the specific work output and efficiency variation as a function of pressure ratio for the above cycle. 5.11 Draw the schematic diagram of a simple cycle with reheat and explain briefly the working principle. Draw also the p-V and T -s diagrams of the cycle. 5.12 Derive the expression for specific work output and the efficiency of a simple cycle with reheat. Draw their trends as a function of pressure ratio. 5.13 Explain the important observations from the specific work output and efficiency variation as a function of pressure ratio for the above cycle. 5.14 Draw the schematic diagram of a simple cycle with reheat and heat exchange and explain briefly the working principle. Draw also the p-V and T -s diagrams of the cycle. 5.15 Derive the expression for specific work output and the efficiency of a simple cycle with reheat and heat exchange. Draw their trends as a function of pressure ratio. 5.16 Explain the important observations from the specific work output and efficiency variation as a function of pressure ratio for the above cycle. 5.17 Draw the schematic diagram of a simple cycle with intercooler and explain briefly the working principle. Draw also the p-V and T -s diagrams of the cycle. 5.18 Derive the expression for specific work output and the efficiency of a simple cycle with intercooler. Draw their trends as a function of pressure ratio. 5.19 Explain the important observations from the specific work output and efficiency variation as a function of pressure ratio for the above cycle. 5.20 Draw the schematic diagram of a simple cycle with intercooled and heat exchanger and explain briefly the working principle. Draw also the p-V and T -s diagrams of the cycle.

130

Gas Turbines

5.21 Derive the expression for specific work output and the efficiency of a simple cycle with intercooled and heat exchanger. Draw their trends as a function of pressure ratio. 5.22 Explain the important observations from the specific work output and efficiency variation as a function of pressure ratio for the above cycle. 5.23 Draw the schematic diagram of a simple cycle with intercooled and reheat and explain briefly the working principle. Draw also the p-V and T -s diagrams of the cycle. 5.24 Derive the expression for specific work output and the efficiency of a simple cycle with intercooled and reheat. Draw their trends as a function of pressure ratio. 5.25 Explain the important observations from the specific work output and efficiency variation as a function of pressure ratio for the above cycle. 5.26 Draw the schematic diagram of a simple cycle with intercooled, heat exchange and reheat and explain briefly the working principle. Draw also the p-V and T -s diagrams of the cycle. 5.27 Derive the expression for specific work output and the efficiency of a simple cycle with intercooled, heat exchange and reheat. Draw their trends as a function of pressure ratio. 5.28 Explain the important observations from the specific work output and efficiency variation as a function of pressure ratio for the above cycle. 5.29 By means of suitable graphs compare the specific work output of various cycles without heat exchanger. 5.30 With a T -s diagram briefly explain the Ericsson cycle. Exercise [Note: Take γ−1 γ = 0.286 and Cp = 1.005 kJ/kg K for all problems, unless stated otherwise.] 5.1 In a gas turbine plant, air enters the compressor at 1 bar and 27◦ C. The pressure ratio is 6. The temperature at turbine inlet is 1000 K. The mass flow rate of air is 10 kg/s. Determine (i) power required to drive the compressor and the turbine power output, (ii) the ratio of the turbine to compressor work, (iii) net power developed by the plant, and (iv) the thermal efficiency. Ans: (i) 2018.14 kW; 4029.75 kW (ii) 1.997 (iii) 2011.61 kW (iv) 40.10%

Ideal Cycles and their Analysis

131

5.2 A gas turbine is supplied with 60 kg/s of gas at 5 bar and 800◦ C and expands it isentropically to 1 bar. Take the mean specific heats of the gas at constant-pressure and constant-volume to be 1 kJ/kg K and 0.717 kJ/kg K respectively. Calculate the exhaust gas temperature and the power developed in MW. Ans: (i) 680.44 K (ii) 23.55 MW 5.3 An open-cycle gas turbine receives air at 0.98 bar and 23◦ C. The air is compressed to 5.25 bar and reaches a maximum temperature of 650◦C before entering into the turbine. The hot air expands back to 0.98 bar. Assuming air-standard cycle and for unit mass flow rate, compute the thermal efficiency of the plant if the compression and expansion processes are isentropic. What is the ratio of the work required to drive the compressor to the work developed by the turbine. Ans: (i) 38.13% (ii) 0.5183 5.4 In a gas turbine plant the air enters the compressor at 1 bar and 300 K. The pressure ratio is 5. The temperature at the turbine inlet is 1200 K. The mass rate of flow is 12 kg/s. Sketch the cycle on p-V and T-s planes and indicate the area representing the heat supply, heat rejection and net work of the cycle. Determine, (i) (ii) (iii) (iv)

compressor and turbine work, net work developed, the ratio of turbine work to compressor work, and the thermal efficiency. Ans: (i) 2116.53 kW; 5341.37 kW (ii) 3224.84 kW (iii) 2.5236 (iv) 36.9%

5.5 In an air-standard cycle heat supply is at constant-volume and the heat rejection is at constant-pressure. The compression and expansion are isentropic and the air at the start of the compression is at 30◦ C and 1 bar. The pressure ratio is 6. The heat supply is 860 kJ/kg of air and air flow is 2.0 kg/s. Assume Cp = 1.005 kJ/kg K and Cv = 0.717 kJ/kg K. Calculate (i) temperature at the end of each processes, (ii) the power developed, and (iii) the thermal efficiency. Ans: (i) 505.82 K; 1705.26 K; 721.56 K (ii) 1569.6 kW (iii) 91.25% 5.6 The inlet pressure and temperature before compression in a Joule cycle are 1 atm and 300 K respectively. What will be the imep of the cycle when the pressure ratio is 2.0 and peak temperature at the end of combustion is 2200 K. Ans: 0.7218 bar

132

Gas Turbines

5.7 If an ideal regenerator is added to the above cycle, for the same temperature and pressure ranges as for the previous problem, calculate the work done and efficiency. Ans: (i) 331.8 kJ/kg (ii) 83.4% 5.8 A simple ideal gas turbine works with a pressure ratio of 8. The compressor and turbine inlet temperatures are 300 K and 800 K respectively. If the volume flow rate is 250 m3 /s, compute the net power output and cycle efficiency. Ans: (i) 33.557 MW (ii) 44.75% 5.9 A gas turbine power plant, working on an air-standard cycle, the heat supply is at constant-volume and heat rejection is at constantpressure. The compression and expansion are isentropic. The atmospheric temperature and pressure are 27◦ C and 1 atm respectively. The pressure ratio is 9. The heat supply is 600 kJ/kg of air. For an air flow of 3 kg/s calculate, (i) temperature at the end of each process, (ii) the net power developed and (iii) the thermal efficiency. Draw the p-V and T -s diagrams. Ans: (i) 562 K; 1398.8 K; 574.9 K (ii) 1694.12 kW (iii) 94.1% 5.10 A turbine supplied with gas at 5.15 bar and 800◦ C and expands it isentropically to 1.03 bar. If the mean specific heat of the gas at constant-pressure and constant-volume are 1 kJ/kg K and 0.717 kJ/kg K respectively. The air inlet temperature to compressor is at 30◦ C. Calculate (i) the exhaust temperature and (ii)the power developed in kJ per kg of gas per minute. Ans: (i) 680.4 K (ii) 13068 kJ/kg 5.11 An open-cycle gas turbine plant receives air at 1 bar and 23◦ C. The air is compressed to 5.5 bar and reaches a maximum temperature of 700◦C in the cycle. The hot air expands back to 1 bar. Assuming air-standard cycle, compute thermal efficiency of the plant if the compression and expansion are isentropic. What is the ratio of work required to drive the compressor to the work developed by the turbine. Ans: (i) 38.6% (ii) 0.495 5.12 In a gas turbine plant the air at 10◦ C and 1 bar is compressed to 4 bar with compressor efficiency of 100%. The air is heated in the regenerator having 100 per cent effectiveness and the combustion chamber till its temperature is raised to 700◦C and has a pressure drop of 0.14 bar. Determine the thermal efficiency of the plant. Ans: (i) 55.9% 5.13 A gas turbine operating between pressure limits of 1.5 bar and 5.5 bar. The inlet air temperature of the compressor is 20◦ C and the air entering the turbine is at a temperature of 560◦ C. If the volume rate of air entering the compressor is 1600 m3 /min, determine the available power output for the cycle. Assume that the cycle operates under ideal conditions. Ans: (i) 6057.57 kW

Ideal Cycles and their Analysis

133

5.14 In a regenerator gas turbine cycle, air enters the compressor at a temperature and pressure of 30◦ C and 1.5 bar and discharges at 220◦C and 5.2 bar. After passing through the regenerator the air temperature is 395◦ C. The temperature of air entering and leaving the gas turbine are 900◦ C and 510◦C. Assuming no pressure drop through the regenerator, determine (i) the output per kg of air, (ii) the efficiency of the cycle and (iii) the work required to drive the compressor. Ans: (i) 201 kJ/kg (ii) 39.6% (iii) 190.95 kJ/kg 5.15 Compare the maximum work delivered by an aircraft gas turbine which works in the following two atmospheric conditions (two-stage compression with perfect intercooling, but without reheat and regeneration). Compressor pressure ratio is 4 and metallurgical temperature limit is 1000 K. At ambient conditions: pressure = 1 atm and temperature = 28◦ C and at 6000 m altitude : pressure = 0.5 atm and temperature = −25◦ C. Find the percentage change in Net work output, efficiency and exhaust temperature if the volume flow rate of air is 2.5 m3 /s. Ans: (i) 32.11% decrease (ii) 1.34% increase (iii) Nil 5.16 A closed-cycle gas turbine (with reheat) power plant operates using helium as the working medium. The pressure ratio is 10. The maximum permitted temperature is 1000 K. Assuming the work output to be maximum, calculate the efficiency. If air is used instead of helium, calculate the efficiency and difference in heat added. Assume ideal Brayton cycle. Temperature at the inlet of compressor = 27◦ C, Cp of helium = 5.204 kJ/kg K and γ of helium = 1.67. Ans: (i) 40.16%; 46.23% (ii) 2490.22 kJ/kg 5.17 A Brayton cycle operates with ideal air between 1 bar, 300 K and 5 bar, 1000 K. The air is compressed in two stages with perfect intercooling. Similarly in the turbine expansion occurs in two stages with perfect reheating. Calculate the optimum pressure in bar, net work output and the fraction of turbine output that has to be put back to compressor (WC /WT ) Ans: (i) 2.236 bar (ii) 257.3 kJ/kg (iii) 0.378 5.18 A gas turbine unit operates at a mass flow of 30 kg/s. Air enters the compressor at a pressure of 1 bar and temperature 15◦ C and is discharged from the compressor at a pressure of 10.5 bar. Combustion occurs at constant-pressure and results in a temperature rise of 420 K. If the flow leaves the turbine at a pressure of 1.2 bar, determine the net power output from the unit and also the thermal efficiency. Ans: (i) 5388.74 kW (ii) 42.55% 5.19 An ideal open-cycle gas turbine plant using air operates on an overall pressure ratio of 4 and between the temperature limits of 300 K and 1000 K. Assuming constant specific heats, Cp = 1.005 kJ/kg K and

134

Gas Turbines

Cv = 0.717 kJ/kg K, evaluate the specific work output and thermal efficiency for each of the modifications below and state the percentage change from the basic cycle. Assuming optimum stage pressure ratios, perfect intercooling and perfect regeneration, find (i) basic cycle, (ii) basic cycle with heat exchanger, (iii) basic cycle with two-stage intercooled compressor, and (iv) basic cycle with heat exchanger and intercooled compressor. ηth

% change

% 32.7 55.4 30.9 59.8

in

WN Cp T01

Cycle Basic Heat exchange Intercooled Heat exchange & intercooled

0.6 0.6 0.653 0.653

W Cp T01

– 0 8.8 8.8

% change in ηth – 69.4 −5.5 82.9

Multiple Choice Questions (choose the most appropriate answer) 1. Performance of an ideal cycle power plant pertains to (a) work output (b) efficiency (c) specific fuel consumption (d) all of the above 2. A simple ideal cycle consists of (a) two adiabatic and two isentropic (b) two adiabatic and two isothermal (c) two isothermal and two constant-pressure (d) two isentropic, one constant-pressure, one constant-volume 3. For the maximum specific output, for any given value of t, the optimum pressure ratio is given by (a) c = t (b) c = t2 √ (c) c = t (d) c =

1 t

Ideal Cycles and their Analysis

135

4. Power output of simple cycle is a function of (a) only pressure ratio (b) only temperature ratio (c) both pressure ratio and temperature ratio (d) all of the above 5. The efficiency of the simple ideal cycle is a function of (a) only the γ of the working fluid (b) only the pressure ratio (c) both pressure ratio and γ (d) inlet temperature of the turbine 6. Maximum power output is achieved for a simple ideal cycle with respect to pressure ratio when (a) outlet temperature of the compressor is equal to outlet temperature of the turbine (b) outlet temperature of the compressor is lower than the outlet temperature of the turbine (c) outlet temperature of the compressor is higher than the outlet temperature of the turbine (d) It has nothing to do with outlet temperatures 7. When a heat exchanger is added to an ideal simple cycle, (a) power output decreases but the efficiency increases (b) power output increases but the efficiency decreases (c) both remain the same (d) power output remains the same but the efficiency increases 8. For better performance, an ideal heat exchange cycle should be operated with (a) lower t and higher r (b) higher t and higher r (c) higher t and lower r (d) r and t should be the same 9. Multistage compression with intercooling and multistage expansion with reheating improves (a) efficiency (b) power output

136

Gas Turbines

(c) both power output and efficiency (d) none of the above 10. An ideal cycle with reheat, intercooling and heat exchange will increase (a) efficiency (b) work output (c) both efficiency and work output (d) none of the above Ans:

1. – (d) 6. – (a)

2. – (c) 7. – (d)

3. – (c) 8. – (c)

4. – (c) 9. – (b)

5. – (c) 10. – (c)

6 PRACTICAL CYCLES AND THEIR ANALYSIS INTRODUCTION In the last chapter, ideal gas turbine cycles have been analyzed. In practice, it is difficult to achieve those ideal conditions. Therefore, in this chapter, we will discuss the performance calculations of practical gas turbine cycles by taking into account the various losses in different components. Before going into the details, various assumptions made are enumerated below. 6.1

ASSUMPTIONS

The practical gas turbine cycles differ from ideal cycles in the following main respects: (i) As the fluid velocities are high in turbo machinery, the change in kinetic energy between inlet and outlet of each component will be taken into account. (ii) The compression and expansion processes are irreversible adiabatic involving increase in entropy. (iii) Fluid friction results in pressure losses in combustion chamber and heat exchangers and also in the inlet and exhaust ductings. (iv) Complete heat exchange is not possible in a heat exchanger. (v) The mass flow is assumed to be the same in spite of the addition of the fuel. This is justified because the bleeding of air from the compressor (which is around 1 to 2%) for cooling the turbine discs and blade roots is compensated by the addition of fuel.

138

Gas Turbines

(vi) The values of Cp and γ of the working fluid vary throughout the cycle due to change of temperature and due to changes in chemical composition of the working medium. (vii) Slightly more work than that required for the compression process will be necessary to overcome bearing and windage friction in the transmission and to drive ancillary components. The implication of the above assumptions are discussed in the following sections. 6.2

STAGNATION PROPERTIES

The first assumption brings out the effect of fluid velocity. The kinetic energy terms in the steady flow energy equation can be accounted for, implicitly by making use of stagnation (or total) enthalpy. The energy equation is 1 2 h0 − h = c −0 (6.1) 2 where, the stagnation enthalpy h0 is the enthalpy which a gas stream of enthalpy h and velocity c would possess when brought to rest adiabatically and without work transfer. From Eq. 6.1, c2 h0 = h + (6.2) 2 when the fluid is a perfect gas, Cp T can be substituted for h, and the corresponding concept of stagnation temperature T0 is defined by T0

=

T+

c2 2Cp

(6.3)

In the above expression, c2 /2Cp is called the dynamic temperature and, when it is necessary to emphasize the distinctions, T is referred to as the static temperature. An idea of the order of magnitude of the difference between T0 and T can be obtained by considering air at atmospheric temperature for which Cp = 1.005 kJ/kg K, flowing at 100 m/s, then T0 − T

=

1002 2 × 1.005 × 103

≈

5K

It follows from the energy equation that, if there is no heat or work transfer, T0 will remain constant. If the duct is varying in cross-sectional area or friction is degrading the directed kinetic energy into random molecular energy, the static temperature will change, but T0 will not. Applying

Practical Cycles and their Analysis

139

the concept of an adiabatic compression, the energy equation becomes ⎤ ⎡ W

=

=

⎢ ⎢ (T2 − T1 ) + − Cp ⎢ ⎢ ⎣ static component

c22 2Cp

−

c21 2Cp

⎥ ⎥ ⎥ ⎥ ⎦

dynamic component

− Cp (T02 − T01 )

(6.4)

stagnation Similarly, for a heating process without work transfer Q

= Cp (T02 − T01 )

(6.5)

Thus, if stagnation temperatures are employed there is no need to refer explicitly to the kinetic energy term. A practical advantage is that it is easier to measure the stagnation temperature of a high velocity stream than the static temperature. When a gas is brought to rest, not only adiabatically but also reversibly, i.e., isentropically, the stagnation pressure p0 is defined by p0 p

γ/(γ−1)

T0 T

=

(6.6)

p0 and T0 can be used in the same way as static values. The stagnation and static states at the inlet and outlet of a compression process are shown in Fig. 6.1. p

02

02 p

2

02’

c 22 / 2 C p

2 2’

p

01

T

p

1

01

c21 / 2C p

1

s

Fig. 6.1 Stagnation and static properties

140

Gas Turbines

Thus the implication of the first assumption, viz., the effect of the inclusion of velocity in the calculation is that we shall refer to stagnation conditions and not static conditions as we did in the ideal cycle analysis. 6.3

COMPRESSOR AND TURBINE EFFICIENCY

The effect of compressor and turbine efficiency is discussed in this section. The actual simple gas turbine cycle is shown in Fig. 6.2, on a T -s diagram. During compression process a considerable amount of energy supplied to the compressor is wasted in churning up the working fluid. This energy does not contribute to the pressure rise but is converted into heat by friction. The outcome is that the temperature of the working fluid is higher at the end of the compression than it would have been had the process been fully and truly isentropic. The compression process on the T -s diagram will, therefore be represented by the line 1 – 2, the temperature at 2 being higher than at 2 , which is the temperature that would have been reached by isentropic compression over the same pressure ratio. Because of this, more work input to the compressor is required and thereby the efficiency of the compressor comes into picture. 03

04 T

04’

02 02’

01

01-02’-03-04’ Ideal cycle 01-02-03-04 Actual cycle s

Fig. 6.2 Actual gas turbine cycle on a T -s diagram The compressor efficiency may be defined as the ratio of the work required for isentropic compression to the actual work input. In ideal cycle analysis, it was assumed that the kinetic energy is the same before and after the process. But in actual gas turbine cycles the changes in kinetic energy are appreciable and therefore this assumption is not valid. Thus, under actual conditions we must consider the stagnation properties for calculating work done in various components. The compressor efficiency is given by ηC

=

Isentropic compression work Actual compression work

Practical Cycles and their Analysis

h02 − h01 h02 − h01

=

T02 − T01 T02 − T01

=

141

(6.7)

for the constant value of Cp . This ratio is also known as total-head isentropic efficiency of the compression process. Let rc be the total-head pressure ratio (or stagnation pressure ratio) during compression, then in the isentropic process (1 − 2 ) T02 T01

= rc(γ−1)/γ

and ηC

=

T02 T01

T01

(6.8)

(γ−1)/γ

−1

=

T02 − T01

T01 rc

−1

T02 − T01

(6.9)

or in the most suitable form for cycle calculations T02 − T01

=

T01 ηC

rc(γ−1)/γ − 1

(6.10)

The above equation gives the actual stagnation temperature rise during compression process. Now referring to the expansion process of the gas turbine cycle, (Fig. 6.2) a similar effect of wasteful heating of the working fluid will occur, as in the case of compressor. This will result in the expansion following the line 3–4 in the Fig. 6.2, instead of ideal isentropic line 3–4 . The effect is therefore to reduce the work output of the turbine. The efficiency of the turbine can be written as ηT

=

Actual turbine work output Isentropic turbine work output

=

h03 − h04 h03 − h04

=

(T03 − T04 ) (T03 − T04 )

(6.11)

for the constant value of Cp . If rt is the total head pressure ratio during the expansion process (3−4 ) T03 T04 ηT

(γ−1)/γ

= rt =

(6.12)

T03 − T04 T03 1 −

(6.13)

1 (γ−1)/γ

rt

or T03 − T04

= ηT T03 1 −

1 (γ−1)/γ rt

(6.14)

When the turbine is exhausting direct to the atmosphere, the kinetic energy of the exhaust is wasted. Then, the efficiency will be given by the

142

Gas Turbines

ratio of work output to the isentropic enthalpy drop from stagnation inlet to static outlet conditions. This efficiency may be denoted by ηT and is given by T03 − T04 ηT = (6.15) T03 1 − (P /P 1)(γ−1)/γ 03

4

It should be noted that the extra work supplied to the compressor, because of its inefficiency, is not entirely wasteful. The additional heating effect, resulting from non-isentropic compression, means that the heat to be supplied in the combustion chamber is reduced. The gain in overall efficiency so derived is not sufficient, however, to overcome the loss in efficiency due to the increased energy required to drive the compressor. The higher exhaust temperature resulting from turbine inefficiency can be, to some extent, utilized if an exhaust heat exchanger is included in the cycle. The actual work of compression for unit mass flow rate is given by Wcact

Cp T01 ηC

=

rc(γ−1)/γ − 1

(6.16)

and similarly, the actual work of expansion for unit mass flow rate is given by Wtact

ηT Cp T03 1 −

=

1 (γ−1)/γ rt

(6.17)

Net work done is given by WN = ηT Cp T03 1 −

1 (γ−1)/γ rt

Cp T01 ηC

−

rc(γ−1)/γ − 1

(6.18)

If the works of compression and expansion are equated, the minimum temperature ratio TT03 = tmin can be calculated for which no net work output 01 is possible. Thus ηT Cp T03 1 −

1

=

(γ−1)/γ rt

Cp T01 ηC

rc(γ−1)/γ − 1

(6.19)

Assuming the specific heats in both processes are same and also rc = rt = r T03 T01

=

r(γ−1)/γ ηT ηC

tmin

=

T03 T01

=

(6.20) r(γ−1)/γ ηT ηC

(6.21)

For r = 5 and ηC = ηT = 0.85, with γ = 1.4, with minimum temperature ratio about 2.19. Thus with T01 = 300 K, T03 must be about 657 K. The Eq. 6.21 explains why the gas turbine has not been developed into a satisfactory prime mover, even many years after it had been first demonstrated.

Practical Cycles and their Analysis

143

Neither materials suitable for high temperature nor compressors of high efficiency were available until about the middle 1930’s. Thus for r = 5 and ηT = ηC = 0.7, T03 must be 970 K before the unit is self-driving and must be considerably higher than this, for developing power to take any extra load on the shaft. The difference between the efficiencies of steam turbine plants and gas turbine plants operating at the same temperature can be comprehended by consideration of the net work expression. For steam turbines, the compression work is very small, since the working fluid is in liquid state, and thus the compression efficiency is a negligible factor. For the gas turbine, the negative work of compression is a considerable fraction of the total turbine work and so the compression efficiency is very significant. Small variations in the compressor and turbine efficiencies can thus have a considerable effect on the overall performance. The magnitude of this effect can be explained by a parameter known as the ‘work ratio’, defined as the ratio of net work to the total turbine work. Higher the value of work ratio, less is the effect on the variation of compressor efficiency and the performance. Work ratio = WN WT

=

WN WT − WC WC = =1− WT WT WT 1−

Cp T01 ηC

(γ−1)/γ

rc

ηT Cp T03 1 −

−1

(6.22)

1 (γ−1)/γ

rt

If rc = rt = r Work ratio =

1−

r(γ−1)/γ T01 ηC ηT T03

=

1−

c 1 t ηC ηT

(6.23)

. where c = r(γ−1)/γ and t = TT03 01 From this expression, it will be noted that the work ratio is increased by high temperature ratio, t, and by low pressure ratio, r. In general, the value of work ratio is not used as any major criterion but it may be useful as a deciding factor. Thus, the second assumption brings into light how important the component efficiencies are to achieve the best performance from the power plant. It may be noted that the higher values of efficiencies culminate in better performance. 6.4

PRESSURE OR FLOW LOSSES

In this section, effect of pressure losses will be considered. Although the irreversibility occurring in the compression and turbine has the major effect on reducing the cycle performance from the ideal values, loss due to friction

144

Gas Turbines

and turbulence occurs throughout the whole plant. It is due to the fact that no fluid flow process can be completely reversible. This overall loss may conveniently be divided into the following losses. (i) air-side intercooler loss, (ii) air-side heat exchanger loss, (iii) combustion chamber loss (both main and reheat), (iv) gas-side heat exchanger loss, and (v) duct losses between components and at intake and exhaust. These losses are measured as differences of pressure from the ideal value. All the losses up to the turbine inlet may be considered as being equivalent to a reduction of pressure ratio during the compression and all those following the turbine being equivalent to a reduction of pressure ratio during the expansion. The various losses are indicated in Fig. 6.3. p 03

04 04’

cc

05

Δp

heg

Δp

06

hea

p

p

04

Δp

T

03

02’ 02 p

02

01

01 s

Fig. 6.3 T –s diagram showing various losses It can be seen that because of the losses, the pressure ratios for compression and expansion are not equal. These fluid flow losses sometimes called parasitic losses, are usually small individually compared with those of compressor and turbine but are controllable to a large extent by suitably designing the components and ducts. The loss in pressure due to flow is proportional to ρc2 . From the continuity equation m ˙ = therefore,

Acρ

(6.24)

Practical Cycles and their Analysis

c =

m ˙ Aρ

∝

ρc2

(6.25)

and Δp

145

m ˙2 A2 ρ

∝

(6.26)

Thus, for a given mass flow rate m ˙ and density ρ, the pressure loss is inversely proportional to the square of flow area, or assuming circular duct of diameter d, 1 Δp ∝ (6.27) d4 Losses, thus can be minimized by increasing cross-sectional area. But the problem is not that simple, since for components like diffusers and heat exchangers, the length must be comparatively increased if low velocities are to be achieved without higher flow losses. Thus, the overall bulk may increase. The greatest difficulty is often on the exhaust side, i.e., following the turbine, because the gases have a very low pressure and high temperature thereby the specific volume will be large. Hence, the gas-side of the heat exchanger and the final exhaust duct is quite critical as far as design is concerned. It may be noted that the individual flow losses are small compared with those due to irreversibility in compressor and turbine. Nevertheless, they are extremely important because of their cumulative effect. Further, there is little point in making attempts to get the best component efficiencies, if their effect is nullified by parasitic losses. Because, flow losses are so dependent on size and shape, the actual layout of a gas turbine plant has a great influence on its overall performance. The effect of varying flow losses on cycle performance can be generalized in a manner convenient for calculation. The analysis consists of finding the difference in performance between the cycle without losses and the cycle with losses. Without pressure losses p03 = p02 and p04 = p01 , hence, the turbine work is ⎡ ⎤ (γ−1)/γ p02 − 1 ⎢ p01 ⎥ WT = ηT Cp T03 ⎣ (6.28) ⎦ (γ−1)/γ p02 p01

Let

γ−1 γ

= a, then ⎡

p02 p01

a

−1

⎤

WT

=

ηT Cp T03 ⎣

with pressure losses p03

=

p02 − Δpc

(6.30)

p04

=

p01 + Δpt

(6.31)

p02 p01

a

⎦

(6.29)

146

Gas Turbines

Wt

⎡

=

a

p02 −Δpc p01 +Δpt

ηT Cp T03 ⎣

−1 a

p02 −Δpc p01 +Δpt

⎤

(6.32)

⎦

where Δpc is the pressure loss between the compressor delivery, (state point 2) and turbine inlet, (state point 3) and Δpt between turbine discharge, (state point 4) and the atmosphere ⎛ a ⎞ a Δp p02 1−

Wt

c

p02 ⎜ pa 1+ Δpt a − 1 ⎟ p01 ⎜ 01 ⎟ ηT Cp T03 ⎜ ⎟ a c ⎝ pa02 1− Δp ⎠ p02

=

Δpt

pa 01 1+ p

The term 1 ±

Δp p0

a

01

can be expanded as a series 1±a

Δp ± ... p0

By omitting higher order terms of work with losses becomes

Wt

Δp p0

⎛

which by itself is small, the turbine

− 1⎟ ⎟ ⎟ c 1−a Δp ⎠ p02

02 Δp 1+a p t 01

⎜ pa ⎜ 01 ηT Cp T03 ⎜ ⎝ pa02

=

=

⎡

ηT Cp T03 ⎣

p02 p01

a

Wloss

⎡

=

(6.34)

Δpt 01

c t 1 − a Δp − 1 + a Δp p02 p01

p02 p01

Wt − Wt gives Wloss

⎞

Δpc pa 02 1−a p

pa 01 1+a p

Wt

(6.33)

a

ηT Cp T03 ⎣

a

c 1 − a Δp p02

Δpc t a Δp p01 + a p02 p02 p01

a

1−

c a Δp p02

⎤ ⎦

⎤ ⎦

(6.35)

(6.36)

Δpc c In the term 1 − a Δp p02 , the value of a p02 will be close to zero since

a

=

γ−1 γ

=

0.286

c for exhaust gas and Δp p02 should not exceed 0.10. Thus, the loss of output can be simplified to

Wloss

=

ηT Cp T03 p02 p01

a

a

Δpt Δpc + p01 p02

(6.37)

Practical Cycles and their Analysis

147

In the demonstration above, only two pressure losses were used, but we can generalize on the basis of effect of a loss being inversely proportional to the pressure level at which it occurs so that ) Δp p

ΔpA pA

=

+

ΔpB pB

+

ΔpC pC

+ ...

(6.38)

The above equation is not exact but is good enough for initial calculation and particularly for estimating the effect of varying amount of losses in the whole cycle. 6.5

HEAT EXCHANGER EFFECTIVENESS

The implication of the fourth assumptions, viz., the effectiveness of heat exchanger is discussed in this section. For ideal cycle calculations the regeneration was taken as 100%, i.e., the air from the compressor was assumed to be heated up to the turbine discharge temperature. Thus in Fig. 6.4, T05 = T04 and as the fluid has constant specific heat, mass flow rate is the same everywhere. Therefore, T04 − T06

=

T05 − T02

T02

=

T06

(6.39)

i.e.,

1620

03

1420 1220 1020 05

T 820

04

02

620

06

420

20

1.0 0.8 0.6

01

220

20

30

40 s

50

Fig. 6.4 Effect of heat exchanger effectiveness

60

148

Gas Turbines

It is a well-known fact that heat transfer requires a temperature difference and when this difference is very small, as it is in the ideal heat exchanger, then, infinite area is required. For actual heat exchanger the required surface area increases very rapidly as the temperature difference available decreases and thus for economic reasons the amount of regeneration is limited (Fig. 6.4) Thus, T05 is less than T04 and correspondingly T06 > T02 . The mass flow rate of expanded gas in an actual gas turbine is not quite the same as the mass flow rate of compressed air. It is usually slightly greater by the amount of the fuel injected in the combustion chamber provided there is no bleeding of air from compressor outlet to cool the turbine blades. Normally the mass of fuel injected and the amount of bleed air is almost same. The specific heat of the hot gases, Cpg , is greater than that of the comparatively cool air by virtue of their higher average temperature. Thus, for a heat balance m ˙ a Cpa (T05 − T02 ) or T05 − T02

=

=

m ˙ g Cpg (T04 − T06 )

m ˙ g Cpg (T04 − T06 ) > T04 − T06 m ˙ a Cpa

(6.40)

(6.41)

The effectiveness of the heat exchanger is defined as the ratio of the temperature rise of the air (T05 − T02 ) to the maximum temperature difference available (T04 − T02 ). This ratio is called the effectiveness or thermal ratio and is denoted by . Thus =

T05 − T02 T04 − T02

(6.42)

For ideal cycle with 100% effectiveness of the heat exchanger, the cycle efficiency increases with decrease of pressure ratio and this holds true for the actual cycle with < 1.0. However, it is modified by the effect of component efficiency to produce an optimum pressure ratio for given values of , ηC and ηT . Thus the implication of the fourth assumption is that effectiveness of the heat exchanger has a say in the performance of the cycle. 6.6

EFFECT OF VARYING MASS FLOW

The compressor handles a mass flow rate, m ˙ a , taken from the atmosphere and sends it to the combustion chamber where a fuel flow rate of m ˙ f is added. Because of this the turbine gas flow rate, m ˙ g , is greater than the compressor air flow rate, m ˙ a , by (1 + f ), where f is the fuel-air ratio by mass, i.e.,

and

m ˙g

=

m ˙ a+m ˙f

m ˙g m ˙a

=

1+

m ˙f m ˙a

(6.43)

=

1+f

(6.44)

Practical Cycles and their Analysis

149

Although, the fuel-air ratio is convenient to work with the calculations, it is usually more meaningful to think in terms of its reciprocal, the airfuel ratio, AF in gas turbine calculations. For most hydrocarbon fuels, the stoichiometric air-fuel ratio is about 15:1 in whole numbers so that the 200% and 400% of theoretical air represent air-fuel ratios of about 30:1 and 60:1 respectively. However, there is always nearly a compensating effect due to the bleeding of air from the compressor, either at discharge or intermediately, for a variety of ancillary purposes, such as cooling air for bearing, turbine wheel, cabin cooling etc., The quantity of bleed air is not known exactly when design cycle calculations are being made. It is normally of the order of 1 to 2% of the total air. It is reasonable to assume that the air and gas mass flows are equal. For final calculations, it is necessary to know or to estimate as exactly as possible, the amount of bleed air because it can be of such proportions as to reduce the nominal performance significantly. Because of the higher speeds of military aircraft and use of higher combustion temperature, with correspondingly greater demand of cooling air, the demand for compressor bleed air increases. 6.7

EFFECT OF VARIABLE SPECIFIC HEAT

The effect of variable specific heat is discussed in this section. The specific heat of air is independent of pressure within the operating limits of gas turbine. However, it varies considerably with temperature. At 300 K the constant-pressure specific heat of air is 1.005 kJ/kg K but at 1000 K, it is 1.140 kJ/kg K – an increase of about 13.4%. Further, the internal combustion of fuel causes the expanding exhaust gas which contains products of combustion, principally CO2 and H2 O vapour, both having higher value of specific heat than that of pure air. At 1000 K, a typical value of Cp for gases after combustion is 1.147. In ideal cycle calculations the specific heat has been taken to be constant throughout the cycle, with a value of 1.005 kJ/kg K. Thus, this assumption would seem to introduce considerable error, because the difference between the cold air and hot gas values given above is about 15%. Although there is an error, it is much less than this value, because of the compensating effect of the decrease in value of γ with temperature. As γ decreases, the exponent (γ−1) decreases and thus the change γ of temperature ΔT for a given pressure ratio and initial temperature decreases. Because in the calculation of work input or output the change of enthalpy, Cp ΔT , is involved. The effect of increased Cp is neutralized to some extent. Actually the specific heat for air and gases, changes continuously during compression and expansion. For precise calculation, an integration process is required. Keenan and Kaye∗ have provided tables of air and gas properties at various air-fuel ratios in their ‘Gas Tables’ which can be used for precise calculations. ∗ Keenan,

J.H. and Kaye, J. Gas Tables, Wiley, 1948

150

Gas Turbines

It has been shown that use of average value of specific heat over typical ranges of conditions existing in gas turbine is very suitable and convenient method. Adoption of a fixed value of Cp for the compression process and another value for heating and expansion process is most appropriate. Corresponding values of γ can also be chosen. Numerical values commonly used for gas turbines with air as working fluid are Cpa = 1.005 kJ/kg K and γ = 1.40 during compression Cpg = 1.147 kJ/kg K and γ = 1.33 during heating and expansion The point to note is that even though Cp and γ values vary during a cycle, it is recommended that a constant value for compressor (Cpa = 1.005 kJ/kg K) and another constant value for heating and expansion (Cpg = 1.147 kJ/kg K) will be more appropriate as this does not introduce any significant error in the calculations. 6.8

MECHANICAL LOSSES

Let us see the last assumption in the practical cycle analysis. In most of the gas turbine plants the power required to drive the compressor is transmitted directly from the turbine without any intermediate gearing. Any loss that occurs is therefore, only due to bearing friction, and windage. The loss is very small and is usual to assume that it amounts to about 10% of the power necessary to drive the compressor. If any power is used to drive ancillary components such as oil pump, fuel pump, etc., then this is accounted for by simply subtracting it from the net output of the turbine unit. If all these are taken into account then it is the practice to club it under ηmech so that the work required to drive the compressor is calculated using the relation WC 6.9

=

1 Cpa (T02 − T01 ) kJ/kg of air ηmech

(6.45)

LOSS DUE TO INCOMPLETE COMBUSTION

At the design operating conditions combustion efficiency in a gas turbine limit is quite high, close to 100%. It is defined as the ratio of enthalpy released to available enthalpy of fuel. It may be noted that the combustion efficiency is extremely difficult to measure accurately and it is the conservative practice to assume it as 98% for purpose of cycle calculations. It is taken into account in the performance calculation by dividing the theoretical amount of fuel required by 0.98 and it affects only cycle efficiency and not work output. It is extremely difficult to maintain combustion efficiency at this value in very lean mixtures especially at low loads or very high altitude in a turbo jet engine. It may be necessary to use lower values if operational

Practical Cycles and their Analysis

151

requirements dictate considerable periods at this condition. However, since both part load and high altitude imply that the total fuel consumption is relatively small, even a lower value of combustion efficiency may not be of very great importance. 6.10

CYCLE EFFICIENCY

Air-fuel ratio has considerable effect on cycle efficiency since gas turbine operate on an overall air-fuel ratio of the order of 60:1 to 100:1. In order to calculate the cycle efficiency an accurate determination of air-fuel ratio is essential. Assume m ˙ f and m ˙ a are the mass flow rates of fuel and air respectively. Equating heat supplied to the heat required to raise the temperature of the gas to T03 we have m ˙ f ηb ΔHrp

=

(m ˙ f +m ˙ a )Cpg (T03 − T02 )

(6.46)

where ηb is the combustion efficiency. ΔHrp is the enthalpy of reaction for unit mass at 25◦ C (lower calorific value). By dividing the above equation with m ˙a f ηb ΔHrp

= (1 + f )Cpg (T03 − T02 )

(6.47)

= Cpg (T03 − T02 )

(6.48)

f is the fuel-air ratio f ηb ΔHrp

since f is small compared to 1.0. Consider Eq. 6.48. There are two unknowns, viz., f and Cpg . Note that Cpg is a function of temperature and the species concentration in the exhaust. The species concentration depends upon the air-fuel ratio. In order to solve the Eq. 6.48, start with a mean value of Cpg = 1.147 kJ/kg K; ηb = 98% and the lower calorific value of the fuel as 43 MJ/kg. Find the value of f . For this value of f , we can find the species concentration and more accurate value of Cpg . With this new value of Cpg find f and the process is repeated until the value of Cpg and f do not vary by a percentage, say, 0.1%. A set of standard curves, connecting the ideal temperature rise with air-fuel ratio for a range of initial temperatures are also available in the literature to find the value of fuel-air ratio. Specific fuel consumption can be found directly from sf c

=

3600 m ˙f WN m ˙a

=

3600 f kg/kW h WN

(6.49)

where WN is the net work output in kW/kg/s of air flow. The cycle η may be defined as η

=

WN f CV

=

3600 sf c CV

(6.50)

where the heating value, CV = 43 MJ/kg. Thus knowing the accurate air-fuel ratio actual cycle efficiency can be determined.

152

Gas Turbines

6.11

POLYTROPIC EFFICIENCY

It should be noted that isentropic efficiencies of compression and expansion process vary with pressure. Therefore, it is necessary to take this variation into account while calculating the performance of actual cycles. This variation can be taken into account by defining a new parameter called polytropic or small-stage efficiency. Let us consider an infinitesimal compressor stage (small stage with pressure and temperature rise of dp and dT respectively) in which the isentropic efficiency remains constant. Then, the small stage efficiency is defined as the isentropic efficiency of an elemental stage of the compression which is constant throughout the whole process. To understand the effect of overall pressure ratio on the relationship between small-stage efficiency and overall efficiency, let us imagine the process on a T -s diagram (Fig. 6.5).

1020

05 04’

820

04 05’’

T 620

03’

02’

420

02 01

220 20

03

0

10

20

30 s 40

50

60

70

Fig. 6.5 Multi-stage compressor process in a T –s diagram Since the process is not isentropic, the starting point for any elemental stage will be at higher level than the starting point of the whole process. Thus, since the vertical distance between two constant-pressure lines increases with entropy, the isentropic temperature rise for the elemental or small stage is greater than that for the corresponding stage in an isentropic compression. Hence, the sum of isentropic temperature * rises for all the elemental stages making up the complete compression, ΔT s, is greater than the single overall temperature rise which would result from an isentropic compression ΔT and this effect is magnified as the overall pressure ratio increases. Now, the actual temperature-rise is found by dividing each ΔT by the polytropic efficiency. It follows that the overall efficiency is less than the polytropic efficiency, and that it decreases with the increase in overall pressure ratio. A similar argument holds good for an expansion process also. Here the overall efficiency is always greater than the polytropic efficiency and increases with increase in pressure ratio.

Practical Cycles and their Analysis

153

Now, a relationship between overall efficiency and polytropic efficiency for varying pressure ratio can be obtained by assuming a polytropic law for the process, viz., pV n

=

constant

(6.51)

pV

=

RT

(6.52)

From Eqs. 6.51 and 6.52, we can obtain p1−n T n

=

constant

(6.53)

=

log c

(6.54)

dp dT +n p T

=

0

(6.55)

1 − n dp n p

=

n − 1 dp n p

(6.56)

Also, for a perfect gas,

Taking the logarithm on both sides, we get (1 − n) log p + n log T Differentiating this equation, we get (1 − n) Therefore,

dT T

=

−

Thus the temperature rise in an elemental stage is given by dT

=

n − 1 dp T n p

(6.57)

Had the process been isentropic, temperature rise in the elemental stage would have been γ − 1 dp dT = T (6.58) γ p Hence, the efficiency of small stage, which is defined as the ratio of the isentropic temperature rise to the actual temperature rise, i.e., ηpc

=

dT dT

=

γ−1 n γ n−1

(6.59)

This expression has been developed on the basis of static temperatures and pressures. The same result will apply to stagnation conditions also. The index n refers to an assumed polytropic law connecting stagnation pressure and temperature. ηpc is the small-stage total head isentropic efficiency or it is called polytropic efficiency. Similarly, the polytropic efficiency in expansion process can be shown to be γ n−1 ηpe = (6.60) γ−1 n ηpc and ηpe enable us to calculate overall efficiency in terms of the polytropic efficiency.

154

Gas Turbines

Let rc be total head pressure ratio in compression and (T02 − T01 ) be the total head temperature rise, then

But,

ηC

=

T02 − T01 T02 − T01

T02 T01

=

rc γ

T02 T01 T02 T01

=

−1

−1

(6.61)

γ−1

Since T02 − T01 , is the temperature rise during isentropic process, and T02 T01

γ−1 1 γ ηpc

n−1

=

rc n

=

rc

(6.62)

Therefore, γ−1

ηC

=

rc γ − 1

γ−1 γ

1 ηpc

rc

(6.63)

−1

and similarly ηT

=

γ−1 γ

1 rt

1− 1−

1 rt

ηpe

(6.64)

γ−1 γ

For any given value of the polytropic efficiency, curves can be drawn showing the variation of overall efficiency with pressure ratio. Figure 6.6 gives curves for both compression and expansion for the polytropic efficiencies of 80, 85 and 90%. For polytropic process, the temperature rise across the compressor is given by

T02 − T01 where

n−1 n

⎡

= T01 ⎣ =

p02 p01

(n−1) n

⎤

− 1⎦

1 γ−1 ηpc γ

(6.65)

(6.66)

and the temperature drop across the turbine will be given by ⎡ (n−1) ⎤ n 1 ⎦ T03 − T04 = T03 ⎣1 − p03

(6.67)

p04

where

n−1 n

=

γ−1 ηpe γ

(6.68)

Practical Cycles and their Analysis

155

0.95 ηpe = 0.9 ηpe = 0.85

Isentropic efficiency

0.9

ηpc = 0 .9

0.85

η pe = 0.8

ηpc = 0.8

5

0.8 0.75

ηpc = 0 .8

0.7 0.65

0

5

10

15

20

25

r

Fig. 6.6 Variation of isentropic η with pressure ratio for different polytropic efficiency

Another formula for polytropic efficiency can be found in terms of pressure and temperature. For a small stage compression

ηpc

ηpc

=

dT dT

dT T

=

γ − 1 dp γ p

=

γ − 1 dp T γ p dT

(6.69) (6.70)

Since, ηpc is constant, the equation can be integrated. On integration, we will get ηpc ln

T2 T1

ηpc

= ln

p2 p1

ln

p2 p1

= ln

γ−1 γ

(6.71) γ−1 γ

(6.72)

T2 T1

when stagnation conditions are considered ln ηpc

= ln

p02 p01 T02 T01

γ−1 γ

(6.73)

This expression can be used to calculate polytropic efficiency from measured values of stagnation pressures and temperatures.

156

Gas Turbines

The corresponding expression for expansion process is T03 T04

ln ηpe

=

p03 p04

ln 6.12

(6.74)

γ−1 γ

PERFORMANCE OF ACTUAL CYCLE

As per the simple cycle on T -s diagram, the temperature rise during the compression is given by T02 − T01

=

T01 ( γ−1 ) rc γ a − 1 ηC

(6.75)

and during expansion, T03 − T04

⎡

= ηT T03 ⎣1 −

1 ( γ−1 γ )g

rt

Now the net work output is given by WN

=

⎤ ⎦

n−1 1 ( n )e 1 ( n−1 ) − Cpa T01 re n c − 1 rt ηmech

Cpg T03 1 −

(6.76)

(6.77)

where rc and rt are the pressure ratios of compression and expansion respectively. Then overall cycle efficiency η=

1 Cpg (T03 − T04 ) − ηmech Cpa (T02 − T01 ) Net work = Heat supplied Cpg (T03 − T02 ) Cpg ηT T03

1−

= Cpg

( γ−1 ) ( r1t ) γ g + T03 −T01

1 ηC

−

T01 1 C ηmech pa ηC

( γ−1 γ )a −1

rc

( γ−1 γ )a −1

rc

,

(6.78)

+1

where ηmech is the mechanical transmission efficiency and Cpa and Cpg are the mean specific heats for air and gas respectively. Also, the formula for overall efficiency can be expressed in terms of index of polytropic law for both compression and expansion which may be determined from equations already derived for known values of polytropic efficiencies. Then,

η=

Cpg T03 1 −

1 rt

( n−1 n )e

−

1 ηmech Cpa T01

( n−1 ) Cpg T03 − T01 rc n c

( n−1 ) rc n c − 1

(6.79)

The suffixes e and c on the n−1 n powers distinguish between expansion and compression processes respectively.

Practical Cycles and their Analysis

6.12.1

157

The Simple Cycle

The above general expressions (Eqs. 6.78 and 6.79) enable us to draw curves showing the variation of work output and cycle efficiency with pressure ratio and turbine inlet temperature and for a given compressor inlet temperature T01 . The various input values are given in Table 6.1. A set of such curves using the values in Table 6.1 are shown in Fig. 6.7 for work output and Fig. 6.8 for efficiency for a simple cycle. Table 6.1 Input Values ηpc

ηpe

ηmech

γair

γgas

Cpa

Cpg

γa −1 γa

γg −1 γg

T01

85%

87%

98%

1.4

1.33

1.005

1.147

0.286

0.248

300 K

The specific output curves for the conditions stated above are shown in Fig. 6.7 for four values of t, where t is the ratio of turbine inlet temperature to the compressor inlet temperature. For comparison, with the ideal cycles, the work output is divided by Cp T01 . Both WN and η show a peak at certain values of pressure ratio, r, the value of r being lower for the WN than for η for a given temperature ratio. The effect of increasing t is to increase the pressure ratio at which the peak value occur. Increase of turbine inlet temperature has a comparatively greater effect on work output than on efficiency. The importance of a higher value of t in order to obtain reasonable performance is clearly demonstrated in these figures. The simple cycle gas turbine, therefore, must use cheap fuels and have a high working temperature in order to compete in fuel cost with reciprocating combustion engines. It should be noted that, gas turbine power plants offer considerable advantages in size and weight. The effect of different values of compressor and turbine isentropic efficiencies on the performance of the gas turbine can now be easily studied. This will show that, effect of reducing either or both, will reduce W and η. On the other hand, the increased component efficiency increases markedly both W and η and moves the maximum values to a higher pressure ratios. It will also be noted that if low component efficiencies are expected then it will not be worthwhile to use higher pressure ratio, since the maximum to both W and η occur relatively at lower values of pressure ratio. While plotting curves of Figs. 6.7 and 6.8, inlet air temperature was assumed to be constant at 300 K. Since, the compressor work is directly proportional to T01 , the specific work output and cycle efficiency will decrease as inlet air temperature increases. This effect is usually of little importance in industrial applications. However, it has considerable importance in aircraft applications and results in improved performance at higher altitudes due to reduced ambient temperatures. From the study of these curves, it can be concluded that a comparatively lower turbine temperature may be advantageous in conjunction with high component efficiencies in order to achieve better performance. The use of lower temperatures allow

Gas Turbines

1.5

t=5

1

W/CpT01

158

t=4

0.5 t=3 t=2

0

0

5

10

15

20

r W Cp T01

Fig. 6.7

vs r for a simple cycle

0.6 Practical Ideal

0.4

t=5 t=4

η 0.2

t=

3

t= 2

0

0

4

8

12

16

r Fig. 6.8 η vs r for a simple cycle

20

Practical Cycles and their Analysis

159

less expensive materials to be used and avoids possible fouling of turbine blades. 6.12.2

The Heat Exchange Cycle

For better specific fuel consumption, it is necessary to add a heat exchanger to the simple gas turbine cycle in order to approach the regenerative cycle. Figures 6.9 and 6.10 show the effect on performance of adding heat exchanger of 0.80 effectiveness to the simple cycle. It can be seen that addition of heat exchanger has almost negligible effect on net work output since we have not assumed any pressure loss in the heat exchanger. If one assumes pressure losses, the reduction in work output can be attributed to the increased pressure losses and hence the reduced pressure ratio across the turbine. However, the efficiency increase is quite notable. It may be noted that with a heat exchanger effectiveness of 0.80 and t = 3, a peak efficiency of about 29.0% is achieved. An almost equally important effect is that regeneration lowers the pressure ratio at which the peak efficiency occurs. This holds the promise for the possibility of simple and less expensive compressor so as to obtain high efficiency with regeneration. Further, it is seen that the optimum pressure ratio for efficiency is lower than that for specific work output. The range of possible heat exchanger effectiveness is very wide and higher values can be achieved practically only by the regenerative type heat exchanger. This, however, represents a more difficult problem of design and construction. It should however be noted that overall efficiency of over 30% is possible with standard values for the cycle, which makes the gas turbine competitive in fuel consumption with other combustion engines. 6.12.3

The Reheat Cycle

The next variation to the simple cycle is the addition of reheat. Reheating is undertaken in order to achieve higher outputs. It may be noted that reheat will not drastically alter the efficiency. The variation of specific power output and efficiency with respect to the pressure ratio is shown in Figs. 6.11 and 6.12. It is seen that there is a marked improvement in the work output with increase in t whereas the efficiency slightly decreases for t = 4 and t = 5. This is because of the less efficient cycle is included due to reheat with lower expansion ratio. 6.12.4

The Reheat and Heat Exchange Cycle

Next variation to the simple cycle is to add both reheat and heat exchange in order to improve the work output as well as the efficiency. The effect of such addition is illustrated in Figs. 6.13 and 6.14. With heat exchange, addition of reheat improves the specific output considerably without loss of efficiency (Figs. 6.13 and 6.14). The curves of Fig. 6.13 are based on the assumption that the gas is reheated to the maximum cycle temperature at the point in the expansion giving equal

Gas Turbines

1.5

t=5

1

W/CpT01

160

t=4

0.5 t=3 t=2

0

0

4

8

12

16

20

r Fig. 6.9

0.6

W Cp T01

vs r for a heat exchange cycle

Without heat exchange With heat exchange

t=5 t=4

0.4

t=5 t=

η

t=4

3

0.2

t=

3

t= 2

0

0

4

8

12

16

r Fig. 6.10 η vs r for a heat exchange cycle

20

Practical Cycles and their Analysis

1.5

With reheat Without reheat

t=5

t=5

1

W/CpT01

t=4 t=4

0.5 t=3 t=3

t=2

0

0

4

8

12

16

20

r W Cp T01

Fig. 6.11

vs r for a reheat cycle

0.4

t=5

With reheat Without reheat

t=5

0.3

t=4

t=4

η 0.2

t=

3

t= 3

0.1

0

t=

0

2

4

8

12

16

r Fig. 6.12 η vs r for a reheat cycle

20

161

Gas Turbines

2 With reheat and heat exchange Without reheat and heat exchange

1.5

W/CpT01

162

t=

5 t=5

1 t=

4 t=4

0.5

0

t=3 t=3

t=2

0

4

8

12

16

20

r Fig. 6.13

W Cp T01

0.6

vs r for a reheat and heat exchange cycle

Without reheat and heat exchange With reheat and heat exchange

t=5 t=4

0.4

t=5

t=3

η

t=4

0.2 t=3 t= 2

0

0

4

8

12

16

20

r Fig. 6.14 η vs r for a reheat and heat exchange cycle

Practical Cycles and their Analysis

163

pressure ratios for the two turbines. The gain in efficiency due to reheat obtained with the ideal cycle is not realized in practice, partly because of the additional pressure loss in the reheat chamber and the inefficiency of the expansion process. However, it may be pointed out that the improvement is not realized mainly because, the effectiveness of the heat exchanger is well short of unity and the additional energy in the exhaust gas is not fully recovered. It is important to use a pressure ratio not less than the optimum value for maximum efficiency, because at lower pressure ratios the addition of reheat can actually reduce the efficiency as indicated by the curves. It is worthwhile to note that reheat has not been widely used in practice because of the additional combustion chamber, and the associated control problems. It can offset the advantage gained from the decrease in size of the main components consequent upon the increase in specific output. With some possible exception of the application, reheat would be considered only (i) if the expansion had to be split between two turbines for other reasons and (ii) if the additional flexibility of control provided by the reheat fuel supply was thought to be desirable. With regard to (i), it must be noted that the natural division of expansion between a compressor turbine and power turbine may not be at the optimum point. If the expansion is not done optimally then the full advantage of reheat will not be realized. Finally, readers familiar with steam turbine design may appreciate that reheat introduces additional mechanical problems arising from the decrease in gas density. Hence, the need for longer blading, in the low pressure stages will become necessary.

6.12.5

The Intercooled Cycle

The next variation to the simple cycle is the introduction of intercooling. The performance of such a cycle is as shown in Fig. 6.15 and 6.16. From cycle analysis it will be known that intercooling improves the work output but decreases the efficiency at higher temperature ratios provided there are no pressure losses. However, with pressure losses the efficiency will decrease irrespective of temperature ratios. The analysis of actual cycle will show that the work output increases considerably and peaks at much higher pressure ratio. The efficiency, however, is but little changed at low pressure ratios and increases as the pressure ratio increases. This is because with irreversible compression the saving in negative work with intercooling outweighs the additional fuel necessary for heating the air delivered by the compressor at a lower temperature. The change in efficiency is not large and the main use of intercooling is to achieve a greater work output per kg of air. For high pressure ratios, intercooling is almost a necessity, because as the air temperature increases, the increment of pressure for given amount of work decreases. However, it is worthwhile to point out here that intercoolers

Gas Turbines

2 With intercooling Without intercooling

1.5

W/CpT01

164

t=5 t=5

1

t=4 t=4

0.5 t=3 t=3

t=2

0

0

4

8

12

16

20

r Fig. 6.15

W Cp T01

vs r for an intercooled cycle

0.4

t=5

With intercooling Without intercooling

t=4

0.3

t=4

t=3

η 0.2 t= 3

0.1 t= t=2

0

0

4

2

8

12

16

r Fig. 6.16 η vs r for an intercooled cycle

20

Practical Cycles and their Analysis

165

tend to be bulky and if they require cooling water, the self-contained nature of the gas turbine is lost. Figures 6.17 and 6.18 show the performance characteristics of intercooling combined with heat exchange and this provides one of the most effective gas turbine cycles. Intercooling provides increased specific output, while both intercooling and regeneration improve the efficiency. The heat exchange modifies the effect of intercooling moving the pressure ratio up for peak efficiency. 6.12.6

Intercooled Cycle with Reheat

Figures 6.19 and 6.20 show the results of intercooling with reheat which are somewhat similar to these for heat exchange in increasing power and efficiency. Reheat has been used in practice only on experimental units, as the introduction of second combustion chamber leads to more complex installations and control problems. 6.12.7

Intercooled Cycle with Heat Exchange and Reheat

Figures 6.21 and 6.22 show the complex cycle, i.e., the cycle with intercooling, reheat and heat exchange, representing the closest practical approach to the ideal Ericsson cycle. It results in high efficiencies over a wide range of pressure ratios and quite a high power outputs. It has been used only in experimental plants, because in spite of its good performance, it requires an elaborate plant which except for its more moderate use of cooling water, offers little advantage over a conventional steam plant. From the above analysis it may be concluded that in practice gas turbines utilize either higher pressure ratio simple cycle or a lower pressure ratio heat exchange cycle. The other additions mentioned do not normally show sufficient advantage to the offset increased complexity and capital cost. Worked out Examples 6.1 An oil gas turbine installation consists of a compressor, a combustion chamber and turbine. The air taken in at a pressure of 1 bar and temperature of 30◦ C is compressed to 6 bar, with an isentropic efficiency of 87%. Heat is added by the combustion of fuel in combustion chamber to raise the temperature to 700◦ C. The efficiency of the turbine is 85%. The calorific value of the oil used is 43.1 MJ/kg. Calculate for an air flow of 80 kg/min. Neglect the effect of fuel in the mass flow rate. Calculate (i) the air/fuel ratio of the turbine gases, (ii) the final temperature of exhaust gases, (iii) the net power of installation, and (iv) the overall thermal efficiency of the installation.

Gas Turbines

2 With intercooling and heat exchange Without intercooling and heat exchange

1.5

W/CpT01

166

t=

5

t=5

1 t=

4 t=4

0.5

t=3 t=3

t=2

0

0

4

8

12

16

20

r Fig. 6.17

W Cp T01

0.6

vs r for an intercooled with heat exchangers With intercooling and heat exchange Without intercooling and heat exchange

t=5

0.4

t=4

η

t=5 t=4

t=3

0.2

t=3

t= 2

0

0

4

8

12

16

20

r Fig. 6.18 η vs r for an intercooled with heat exchangers

Practical Cycles and their Analysis

2 With intercooling and reheat Without intercooling and reheat

t=

1.5

5

t=5

W/CpT01

t=4

1

t=4 t=3

0.5

t=3 t=2

0

0

4

8

12

16

20

r Fig. 6.19

0.5

W Cp T01

vs r for an intercooled with reheat cycle

With intercooling and reheat Without intercooling and reheat

0.4

t=5

t=4 t=4 t=3

0.3

η

t=3

0.2 t=2

0.1

t=

0

0

4

2

8

12

16

20

r Fig. 6.20 η vs r for an intercooled with reheat cycle

167

Gas Turbines

2.5

t=5

With pressure losses Without pressure losses Ideal cycle

2 t=5 t=4

W/CpT01

168

1.5

t=5

1

t=4 t=3 t=4 t=3

0.5 t=3

0

0

4

8

12

16

20

r Fig. 6.21

W Cp T01

vs r for a complex cycle

0.8 t=5 t= 4

0.6

t=

3

t=5 t=4 t=5

η 0.4

t=4

t=3 t=3

0.2 With pressure losses Without pressure losses Ideal cycle

0

0

4

8

12

16

r Fig. 6.22 η vs r for a complex cycle

20

Practical Cycles and their Analysis

169

Assume Cpa = 1.005 kJ/kg K, γa = 1.4, Cpg = 1.147 kJ/kg K, γg = 1.33. Solution 03

04 04’

02

T

02’ 01

s Fig. 6.23

The T -s diagram is as shown in the figure above. p02 p01

T02

=

T01

T02

=

T01 +

γ−1 γ

303 × 60.2857

=

=

505.5 K

T02 − T01 505.5 − 303 = 535.8 K = 303 + ηC 0.87

Neglecting the addition of fuel in the combustion chamber, we have m ˙f+ m ˙a≈m ˙ a. =

m ˙ a Cpg (T03 − T02 ) CV

=

80 1.147 × (973 − 535.8) × = 0.0155 kg/s 60 43100

m ˙a m ˙f

=

1 80 × 60 0.0155

A/F

=

86

T04

=

T03

T04

=

T03 − ηT (T03 − T04 )

=

973 − 0.85 × (973 − 623.8)

m ˙f

=

86 Ans

⇐= p04 p03

γ−1 γ

= 973 ×

1 6

The net power of installation, WN (neglect fuel)

0.2481

= 623.8 K

=

676.2 K

Ans

⇐=

170

Gas Turbines

WN

=

m ˙ a Cpg (T03 − T04 ) − m ˙ a Cpa (T02 − T01 )

=

1.3333 × 1.147 × (973 − 676.2) − 1.3333 × 1.005 × (535.8 − 303)

=

Ans

142 kW

⇐=

The overall thermal efficiency, ηth ηth

=

Net power Heat input

=

21.25%

142 × 100 0.0155 × 43100

=

Ans

⇐=

6.2 In a gas turbine plant air enters the compressor at 1 bar and 7◦ C. It is compressed to 4 bar with an isentropic efficiency of 82%. The maximum temperature at the inlet to the turbine is 800◦ C. The isentropic efficiency of the turbine is 85%. The calorific value of the fuel used is 43.1 MJ/kg. The heat losses are 15% of the calorific value. Calculate the following: (i) Compressor work in KJ/kg (ii) Heat supplied in KJ/kg (iii) Turbine work in KJ/kg (iv) Net work in KJ/kg (v) Thermal efficiency (vi) Air/fuel ratio (vii) Specific fuel consumption in kg/kW h (viii) Ratio of compressor work to turbine work Assume Cpa = 1.005 kJ/kg K, γa = 1.4, Cpg = 1.147 kJ/kg K, γg = 1.33. Solution T02

T02

WC

p02 p01

γ−1 γ

=

T01

=

280 × 40.2857

=

T01 +

T02 − T01 ηC

=

280 +

416.1 − 280 0.82

=

Cpa (T02 − T01 )

=

416.1 K

=

446 K

Practical Cycles and their Analysis

171

03

T

04 04’

02

02’ 01

s Fig. 6.24

Q

T04

=

1.005 × (446 − 280)

=

166.83 kJ/kg of air

=

Cpg (T03 − T02 ) = 1.147 × (1073 − 446)

=

719.17 kJ/kg of air p04 p03

γ−1 γ

Ans

⇐= 1 4

0.2481

=

T03

=

760.7 K

=

T03 − ηT (T03 − T04 )

=

1073 − 0.85 × (1073 − 760.7) = 807.5 K

=

Cpg (T03 − T04 ) = 1.147 × (1073 − 807.5)

=

304.53 kJ/kg of air

=

WT − WC

=

137.7 kJ/kg of air

=

WN Q/0.85

=

16.3%

m ˙ f CV (1 − 0.15)

=

(m ˙ a+m ˙ f )Q

m ˙a m ˙f

=

0.85 CV −1 Q

T04

WT

Net work, WN

ηth

= 1073 ×

Ans

⇐=

=

=

Ans

⇐=

304.53 − 166.83 Ans

⇐=

137.7 × 100 719.19/0.85 Ans

⇐=

172

Gas Turbines

0.85 × 43100 −1 719.17

= sf c

WC WT

=

49.94

Ans

⇐=

Fuel consumed/hour Net power

= =

m ˙ f × 3600 m ˙ a WN

=

0.524 kg/kW h

=

166.83 × m ˙a 304.53 × (m ˙ a+m ˙ f)

=

166.83 × 49.94 304.53 × 50.94

=

3600 49.94 × 137.7 Ans

⇐=

=

0.537

Ans

⇐=

6.3 The following data refer to a gas turbine set employing a regenerator: Isentropic efficiency of the compressor Isentropic efficiency of the turbine Mechanical transmission efficiency Pressure ratio Maximum cycle temperature Combustion efficiency Calorific value of the fuel Air mass flow Regenerator effectiveness Regenerator gas-side pressure loss Ambient temperature and pressure

: : : : : : : : : : :

82% 85% 99% 7:1 1000 K 97% 43.1 MJ/kg 20 kg/s 75% 0.1 bar 327 K; 1 bar

Calculate the output, specific fuel consumption and overall thermal efficiency. Assume that the pressure losses in the air-side of the regenerator and combustion chamber are accounted for in the compressor efficiency. Compare these results with those obtained for the same plant without the regenerator, and with regenerator but without pressure loss and also comment on the results. Neglect the effect of fuel mass addition in the heat balance in the combustion chamber but include in the turbine calculation. Solution (i) With regeneration p02 p01

T02

=

T01

T02

=

T01 +

γ−1 γ

= 327 × 70.2857 = 570.2 K

T02 − T01 ηC

Practical Cycles and their Analysis

173

03

04 04’

05 02

T

02’ 01

s Fig. 6.25

p04

570.2 − 327 0.82

=

327 +

=

p01 + Δp p04 p03

=

=

1 + 0.1

γ−1 γ

=

1.1 bar

1.1 7

0.2481

T04

=

T03

T04

=

T03 − ηT (T03 − T04 )

=

1000 − 0.85 × (1000 − 631.8)

=

T02 + (T04 − T02 ) = 623.6 + 0.75 × (687 − 623.6)

=

671.2 K

T05

= 1000 ×

623.6 K

= 631.8 K

=

687 K

Neglecting the effect of change in mass flow rate due to m ˙ f in combustion chamber, the heat balance is given by m ˙ f CV ηcomb

=

m ˙ a Cpg (T03 − T05 )

m ˙f

=

20 × 1.147 × (1000 − 671.2) = 0.18 kg/s 43100 × 0.97

WT − WC

=

(m ˙ a+m ˙ f )Cpg (T03 − T04 ) − m ˙ a Cpa (T02 − T01 ) ηm

WN

=

20.18 × 1.147 × (1000 − 687) − 20 × 1.005 × (623.6 − 327) 0.99

=

7244.84 − 6021.88

=

1222.96 kW

Ans

⇐=

174

Gas Turbines

sf c

ηth

=

m ˙ f × 3600 WN

=

0.53 kg/kW h

=

1222.96 × 100 0.18 × 43100

0.18 × 3600 1222.96

=

=

Ans

⇐=

Ans

15.8%

⇐=

(ii) Without regenerator p04 T04

=

1 bar

=

p04 p03

p01 γ−1 γ

=

617.1 K

=

T03 − ηT (T03 − T04 )

=

1000 − 0.85 × (1000 − 617.1)

=

m ˙ a Cpg (T03 − T02 ) CV ηcomb

=

20 × 1.147 × (1000 − 623.6) = 0.207 kg/s 43100 × 0.97

=

WT − WC

=

20.207 × 1.147 × (1000 − 674.5) − 6021.88

=

1522.37 kW

⇐=

sf c

=

0.207 × 3600 = 0.49 kg/kW h 1522.37

⇐=

ηth

=

1522.37 × 100 0.207 × 43100

⇐=

m ˙f

WN

1000 ×

0.2481

T03

T04

=

1 7

=

=

674.5 K

Ans

=

17.06%

Ans

Ans

Similarly, we can work out the results with regenerator but without pressure loss. The following table gives the details. Quantities p03 p04

T04 (K) m ˙ f (kg/s) WT (kW) sf c (kg/kW h) ηth (%)

Regenerator With Δp Without Δp 6.364 7 687 674.5 0.18 0.186 7244.84 7536.41 0.53 0.44 15.8 18.9

Without regenerator & Δp 7 674.5 0.207 7544.25 0.49 17.06

Practical Cycles and their Analysis

175

Remarks : As can be seen from the table that pressure loss plays a major role in the efficiency than the regenerator. Hence, more care should be taken in the design to have minimum pressure loss. 6.4 The following data apply to gas turbine set employing a separate power turbine, regenerator, and intercooler between two-stage compression: Isentropic efficiency of compression each stage : 80% Isentropic efficiency of compressor turbine : 88% Isentropic efficiency of power turbine : 88% Turbine to compressor transmission efficiency : 98% Pressure ratio in each stage of compression : 3:1 Temperature after intercooler : 297 K Air mass flow : 15 kg/s Regenerator effectiveness : 80% Regenerator gas-side pressure loss : 0.1 bar Maximum turbine temperature : 1000 K Ambient temperature : 327 K Ambient pressure : 1 bar Calorific value of the fuel : 43.1 MJ/kg Calculate the net power output, specific fuel consumption, and overall thermal efficiency. Assume that the pressure losses in the air-side of the regenerator and combustion chamber are accounted for in the compressor efficiency. Assume Cpa = 1.005 kJ/kg K, γa = 1.4, Cpg = 1.147 kJ/kg K, γg = 1.33. Solution 03 09’

05

T

02

06

07 02’ 07’ 08

09 04 04’

01

s Fig. 6.26

p07 p01

T07

=

T01

T07

=

T01 +

=

477.8 K

γ−1 γ

T07 − T01 ηC

327 × 30.2857

= =

327 +

=

447.6 K

447.6 − 327 0.8

176

Gas Turbines

WLP C

=

15 × 1.005 × (477.8 − 327) p02 p08

γ−1 γ

=

2273.31 kW

297 × 30.2857

T02

=

T08

=

T02

=

T08 +

=

433.9 K

WHP C

=

15 × 1.005 × (433.9 − 297) = 2063.77 kW

WC

=

2063.77 + 2273.31

T02 − T08 ηC

=

297 +

=

Actual work input to compressor, Wca WC 4337.08 WCa = = ηtrans 0.98

=

406.5 K

406.5 − 297 0.8

4337.08 kW

=

4425.59 kW

Neglecting the effect of m ˙ f, WCT

=

m ˙ a Cpg (T03 − T09 )

T09

=

1000 −

T09

=

T03 −

=

707.7 K

p09

T04

T03 − T09 ηT

p03

=

T03 T09

T04

4425.59 15 × 1.147

γ γ−1

p04 p09

=

= =

742.8 K 1000 −

9 1000 4.0303 707.7

= 2.234 bar

γ−1 γ

=

1000 − 742.8 0.88

=

T09

742.8 ×

=

623.1 K

=

T09 − ηT (T09 − T04 )

=

742.8 − 0.88 × (742.8 − 623.1)

1.1 2.234

0.2481

=

637.5 K

Work output of the power turbine, WT P WT P

=

15 × 1.147 × (742.8 − 637.5) = 1811.69 kW

(ii) Specific fuel consumption, sf c T05

=

T02 + (T04 − T02 )

=

433.9 + 0.8 × (637.5 − 433.9) = 596.8 K

Practical Cycles and their Analysis

177

=

m ˙ a Cpg (T03 − T05 ) CV

=

15 × 1.147 × (1000 − 596.8) = 0.161 kg/s 43100

sf c

=

0.161 × 3600 = 0.32 kg of fuel/kW h 1811.7

⇐=

ηth

=

1811.7 × 100 0.161 × 43100

⇐=

m ˙f

=

26.1%

Ans

Ans

6.5 A 1850 kW open-cycle stationary plant is to have one intercooler, one reheater and a regenerator. On one shaft the high pressure turbine drives the low pressure compressor. On another shaft, the low pressure turbine drives the high pressure compressor and the load. The following data may be assumed (Neglect the weight of the fuel and the mechanical efficiencies). Ambient pressure : 1 bar Ambient temperature : 27◦ C Maximum cycle temperature : 720◦C Pressure ratio in each stage of compression : 2.5 Turbine and compressor efficiencies : 80% Pressure drop in each heater and : 3% each side of regenerator Regenerator effectiveness : 75% Assume intercooling to ambient temperature and reheating to maximum cycle temperature. Sketch the T -s diagram and index all state points. Calculate these state points and the necessary ideal states. Calculate the thermal efficiency and air rate. Find the power output of each turbine and compressor. If the fuel used has 43.1 MJ/kg heating value and that the combustion efficiency is 98%, calculate the fuel used per hour at rated load and specific fuel consumption. Solution 03 10 09’

09

05

T

06

02 02’ 08

07’

07

01

s Fig. 6.27

04 04’

178

Gas Turbines

T07

γ−1 γ

p07 p01

300 × 2.50.2857

=

T01

=

389.8 K

=

T01 +

=

412.3 K

WLP C

=

WHP T = 1.005 × (412.3 − 300) = 112.86 kJ/kg of air

WHP T

=

WLP C

T09

=

T03 −

WHP T 112.86 = 894.6 K = 993 − Cpg 1.147

⇐=

T09

=

T03 −

T03 − T09 993 − 894.6 = 870 K = 993 − ηT 0.8

⇐=

p03

=

0.97 × p02 × (1 − 0.03)

=

0.97 × 6.25 × 0.97

T07

p09

=

=

=

=

300 +

389.8 − 300 0.8 Ans

T02

⇐=

Cpg (T03 − T09 )

=

5.88

=

= 3.45 bar

1.03 3.347

Ans

0.2481

=

T10

T04

=

T10 − ηT (T10 − T04 )

=

993 − 0.8 × (993 − 741.2)

=

Cpg (T10 − T04 )

=

231 kJ/kg of air

WN

=

231 − 112.86

=

m ˙a

=

Wplant WN

1850 118.14

T05

=

T02 + (T04 − T02 ) = 412.3 + 0.75 × (791.6 − 412.3)

=

696.8 K

WLP T

=

=

=

Ans

⇐=

⇐=

T04

= 993 ×

Ans

⇐=

3.347 bar

γ−1 γ

Ans

Ans

5.88 bar

993 4.0303 870

3.45 × (1 − 0.03) p04 p10

⇐=

=

=

γ γ−1

Ans

T02

T07 − T01 ηC

p03

=

T03 T9

p10

=

Ans

= 741.2 K ⇐=

791.6 K

Ans

⇐=

1.147 × (993 − 791.6)

118.14 kJ/kg of air =

15.7 kg/s

Ans

⇐=

Ans

⇐=

Practical Cycles and their Analysis

Q

=

Cpg T03 − T05 + T10 − T09

=

1.147 × (993 − 696.8 + 993 − 894.6)

=

452.61 kJ/kg of air

ηth

=

118.14 × 100 452.61

WLP C

=

WHP C

=

15.7 × 112.86

WLP T

=

15.7 × 231

m ˙f

=

15.7 × 452.61 × 3600 m ˙ a × Heat input = ηcomb × CV 0.98 × 43100

=

605.65 kg of fuel/h

=

605.65 1850

sf c

=

=

26.1%

179

Ans

⇐=

WHP T =

=

=

1771.9 kW 3626.7 kW

0.327 kg of fuel/kW h

Ans

⇐=

6.6 A gas turbine operating at a pressure ratio of 11.3137 produces zero net work output when 476.354 kJ of heat is added per kg of air mass. If the inlet air total temperature is 300 K and the turbine efficiency is 71%. Find the compressor efficiency and the temperature ratio. Assume γ = 1.4 and Cp = 1.005 kJ/kg K for the whole cycle. Solution 03

T

04 04’

02

02’ 01

s Fig. 6.28

Since WN = 0, ΔTC = ΔTT and assuming Cp to be constant, T02 − T01 From the above,

=

T03 − T04

180

Gas Turbines

T03 − T02

=

T04 − T01

(1)

γ−1 γ

T02 T01

=

T02

=

300 × (11.3137)0.4/1.4

ηT

=

T03 − T04 T03 − T04

T03 T04

=

T03 T04

=

p02 p01

0.4/1.4

=

(11.3137) =

(2)

γ−1 γ

p03 p04

600 K

0.4/1.4

=

(11.3137)

2

(3)

Given that heat added = 476.354 kJ/kg of air, Cp (T03 − T02 )

=

476.354

T03 − T02

=

476.354 1.005

=

474 K

(4)

From (2), 0.71

=

T03 1 −

T04 T03

T03 T04

T04

−1

T04 T03

1−

0.71

=

2×

T04 T03

=

1−

T04

=

T01 + (T03 − T02 ) = 300 + 474 = 774 K

T03

=

774 0.645

=

1200 K

t

=

1200 300

=

4

T02

=

1200 − 474

=

ηC

=

T02 − T01 T02 − T01

=

=

70.4%

2−1

0.71 2

=

Ans

0.645

⇐=

From (1) and (4)

Ans

⇐= 726 K 600 − 300 726 − 300

× 100 Ans

⇐=

Practical Cycles and their Analysis

181

6.7 The efficiencies of the compressor and turbine of a gas turbine are 70.42% and 71% respectively. The heat added in the combustion chamber per kg of air is 476.354 kJ/kg. Find a suitable pressure ratio such that the work ratio is 0.0544. Also find the corresponding temperature ratio. The inlet total temperature of air 300 K. Solution 03

T

04 04’

02

02’ 01

s Fig. 6.29

Work ratio

=

1−

c 1 t ηT ηC

0.0544

=

1−

c t

c 1 × t 0.7042 × 0.71

=

0.4728

Q

=

476.334 kJ/kg

Cpg (T03 − T02 )

=

476.334

T02

=

T01 +

476.334

=

1.147 × T03 − T01 +

476.334 1.147

=

T01

415.286

=

300 × [t − (1 + 0.6714t − 1.42)]

1.384

=

0.3286t + 0.42

0.3286t

=

0.964

t

=

2.934

T01 (c − 1) ηC T01 T01 c− ηC ηC

T03 1 1 − 1+ × 0.4728 × t − T01 ηC ηC

Ans

⇐=

182

Gas Turbines

(rp )

T03 T01

=

2.934

c t

=

0.4728

c

=

2.934 × 0.4728

=

1.3872

=

(1.3872)3.5

γ−1 γ

rp

=

=

1.3872

Ans

3.144

⇐=

6.8 A practical gas turbine cycle having the isentropic compressor efficiency of ηC and isentropic turbine efficiency of ηT with a work ratio of 0.30. If the pressure ratio r and temperature ratio t are 12 and 4 respectively, calculate the minimum temperature ratio required to drive the compressor. If the above unit works at ideal condition what will be its efficiency? Solution 03

04 04’

02

T

02’ 01

s Fig. 6.30

Wratio

=

1−

c 1 · t ηC ηT

0.3

=

1−

1 120.286 · 4 ηC ηT

ηC ηT

=

0.509 (1 − 0.3)

tmin

=

c ηC ηT

=

120.286 = 2.79 = 2.8 0.73

η

=

1−

1 c

=

1−

=

=

1−

0.509 ηC ηT

0.73

1 120.286

=

0.508

Ans

⇐= Ans

⇐=

Practical Cycles and their Analysis

183

6.9 The polytropic efficiency of the compressor is 85%. If the ideal outlet temperature of the compressor is twice that of the inlet, calculate the isentropic efficiency of the compressor. If the polytropic efficiency of the turbine is same as the compressor, calculate the isentropic efficiency of the turbine assuming that there is no pressure loss. Take γa = 1.4 and γg = 1.33. Solution 03

04 04’

02

T

02’ 01

s Fig. 6.31

γ−1

ηC

=

rc γ − 1

γ−1

rc γ · rc

=

ηC

=

ηT

=

−1

γ γ−1

T02 T01

=

23.5

=

11.31

1 2−1 = = 0.794 = 79.4% 1.26 11.310.286/0.85 − 1 1− 1−

=

1 ηpc

0.400 0.452

ηpe · γ−1 γ

1 rt 1 rt

=

γ−1 γ

=

1−

0.885

1

=

Ans

⇐=

0.85×0.248 1 11.31 0.248 1 − 11.31 Ans

88.5%

⇐=

6.10 A compressor has an isentropic efficiency of 0.85 at a pressure ratio of 4.0. Calculate the corresponding polytropic efficiency, and thence plot the variation of isentropic efficiency over a range of pressure ratio from 2.0 to 10.0. Solution ηC

=

( γ−1 ) rc γ − 1 rc

γ−1 1 γ ηpc

−1

184

Gas Turbines

0.85

40.286 − 1

= 4

rc ηc

−1

=

1.572

1 × ln 4 ηpc

=

ln 1.572

ηpc

=

0.877

⇐=

ηC

=

−1 r0.286 c r0.326 −1 c

⇐=

4 0.286 ×

1 0.286× ηpc

1 0.286× ηpc

2 0.863

4 0.850

6 0.842

4

6

8 0.836

Ans

Ans

10 0.832

0.9

ηc

0.85

0.8

2

8

10

r

Fig. 6.32

6.11 A closed-cycle gas turbine is to be used in conjunction with a gas cooled nuclear reactor. The working fluid is helium (Cp = 5.19 kJ/kg K and γ = 1.66). The layout of the plant consists of two-stage compression with intercooling followed by a heat-exchanger; after leaving the cold side of the heat-exchanger the helium passes through the reactor channels and on to the turbine; from the turbine it passes through the hot-side of the heat-exchanger and then a pre-cooler before returning to the compressor inlet. The following data are applicable: Compressor and turbine polytropic efficiencies : 0.88 Temperature at LP compressor inlet : 310 K Pressure at LP compressor inlet : 14.0 bar Compressor pressure ratios (LP and HP) : 2.0 Temperature at HP compressor inlet : 300 K Mass flow of helium : 180 kg/s Reactor thermal output (heat input to gas turbine) : 500 MW Pressure loss in pre-cooler and intercooler (each) : 0.34 bar Pressure loss in heat-exchanger (each side) : 0.27 bar

Practical Cycles and their Analysis

Pressure loss in reactor channels Helium temperature at entry to reactor channels

: :

185

1.03 bar 700 K

Calculate the power output and thermal efficiency, and the heatexchanger effectiveness implied by the data. Solution 05 06

07

T

06’

04 02 04’ 02’ 01

03

s Fig. 6.33

ηpc

=

γ−1 n γ n−1

0.88

=

0.66 n × 1.66 n − 1

n−1 n

=

0.4518 for compressor

T02 T01

=

(rc )

T02

=

310 × 20.4518

T04

=

T03 (rc )

Q

=

mC ˙ p (T05 − T07 )

500 × 103

=

180 × 5.19 × (T05 − 700)

T05

=

1235.217 K

Total pressure loss, =

n−1 n

n−1 n

= =

424 K 300 × 20.4518

0.34 + 0.34 + 0.27 + 1.03

p05

=

56.00 − 1.98

p06

=

14.00 + 0.27 + 0.34

ηpt

=

γ n−1 γ−1 n

=

=

=

1.98 bar

54.02 bar =

410.32 K

14.61 bar

186

Gas Turbines

n−1 n

=

T06

=

T05 p05 p06

WC

WT

WN

η

0.66 1.66

0.88 ×

n−1 n

=

0.3498 1235.217

=

54.02 0.3498 14.61

= 781.790 K

=

mC ˙ p [(T02 − T01 ) + (T04 − T03 )]

=

180 × 5.19 × [(424 − 310) + (410.32 − 300)]

=

209559 kW

=

mC ˙ p (T05 − T06 ) = 180 × 5.19 × (1235.217 − 781.79)

=

423591 kW

=

WT − WC

=

214.032 MW

⇐=

=

214.032 WN × 100 = × 100 = 42.80% q 500

⇐=

=

T07 − T04 T06 − T04

=

289.68 × 100 371.47

=

209.559 MW

= =

=

423.591 MW 423.591 − 209.559 Ans

Ans

700 − 410.32 781.79 − 410.32 =

78%

Ans

⇐=

6.12 An open-cycle gas turbine plant consists of a compressor, combustion chamber and turbine. The isentropic efficiencies of compressor and turbine are ηC and ηT , the maximum and minimum cycle temperatures are T03 and T01 respectively, and rp is the pressure ratio for both the compression and expansion. Neglecting pressure losses and assuming that the working substance is a perfect gas, show that the necessary condition for positive power output is n−1

ηC ηT T03 > T01 rp n Such a plant delivers 1500 kW and operates such that inlet pressure and temperature at the compressor is 1 bar and 25◦ C and that at turbine is 4 bar and 700◦ C. Calculate the isentropic efficiency of the turbine and the requisite mass flow of air in kg/s if the compressor efficiency is 85% and overall thermal efficiency is 21%. Assume Cp to be 1.005 kJ/kg K throughout. Solution n−1

WC

=

Cp T01 rp n − 1 ηC

Practical Cycles and their Analysis

187

03

04 04’

02

T

02’ 01

s Fig. 6.34

WT

=

Cp ηT 1 −

1

T03

n−1 n

rp

If WT = WC , then n−1

Cp ηT

n−1

rp n − 1

rp n − 1

T03

=

Cp T01

ηC ηT T03

=

T01 rp n

n−1

ηC

rp n n−1

This means there will not be any power output. Hence for a positive power output, n−1

ηC ηT T03 T02

Given η0 = 0.21 =

>

T01 rp n T01 r

γ−1 γ

−1

=

T01 +

=

298 +

Q

=

WN 1500 = 7142.86 kW = η0 0.21

m ˙

=

Q Cp (T03 − T02 )

=

7142.86 1.005 × (973 − 468.59)

=

14.09 kg/s

ηC 298 × 40.286 − 1 = 468.59 K 0.85

1500 Q

Ans

⇐=

188

Gas Turbines

WN

=

1500 14.09

WN

=

Cp (T03 − T04 ) − Cp (T02 − T01 )

=

1.005 × (973 − T04 − 468.59 + 298)

T04

=

696.48 K

T04

=

T03 973 = 0.286 = 654.52 r0.286 4

ηT

=

T03 − T04 T03 − T04

=

=

0.868

86.8%

=

=

106.46 kJ/kg

973 − 696.48 973 − 654.52 Ans

⇐=

6.13 In a compound gas turbine the air from the compressor passes through a heat exchanger heated by the exhaust gases from the low-pressure turbine, and then into the high pressure combustion chamber. The high-pressure turbine drives the compressor only. The exhaust gases from the high pressure turbine pass through the low-pressure combustion chamber to the low pressure turbine which is coupled to an external load. The following data refer to the plant: Pressure ratio of the compressor : 4:1 Isentropic efficiency of compressor : 0.86 Isentropic efficiency of HP turbine : 0.84 Isentropic efficiency of LP turbine : 80.0 Mechanical efficiency of drive to compressor : 0.92 Temperature of gases entering HP turbine Temperature of gases entering LP turbine Atmospheric temperature Atmospheric pressure

: : : :

660◦ C 625◦ C 15◦ C 1 bar

In the heat exchanger 75% of the available heat is transferred to the air. Assuming that the specific heat Cp of air and gas is 1.005, determine (i) the pressure of the gases entering the low pressure turbine, and (ii) the overall efficiency. Solution p02 p01

γ−1 γ

T02

=

T01

ηC

=

T02 − T01 T02 − T01

0.86

=

428.13 − 288 T02 − 288

T02

=

450.94 K

= 288 × 40.286 = 428.13 K

Practical Cycles and their Analysis

T05 < T03

03 05 07 04’

T

189

02

06 06’ 08 04

02’

01 s Fig. 6.35

Work done in compressor = ηm × Work done in turbine Cp (T02 − T01 )

=

0.92 × Cp (T03 − T04 )

450.94 − 288

=

0.92 × (933 − T04 )

T04

=

755.9 K

ηT

=

T03 − T04 T03 − T04

0.84

=

933 − 755.9 933 − T04

T04

=

722.16 K

T03 T04

=

p04

=

In HP turbine,

p03 p04

γ−1 γ

4 3.5 933 722.16

In LP turbine, γ−1 γ

T05 T06

=

p05 p06

898 T06

=

1.63 1

T06

=

780.90 K

ηT

=

T05 − T06 T05 − T06

0.286

=

1.63 bar

Ans

⇐=

190

Gas Turbines

0.8

=

898 − T06 898 − 780.9

T06

=

804.3 K

=

T07 − T02 T06 − T02

0.75

=

T07 − 450.94 804.3 − 450.94

T07

=

716 K

Q

=

Cp (T03 − T07 + T05 − T04 )

=

1.005 × (933 − 716 + 898 − 755.9)

=

360.9 kJ/kg

=

Cp (T02 − T01 )

=

163.75 kJ/kg

=

Cp (T03 − T04 ) + Cp (T05 − T06 )

=

1.005 × [(933 − 755.9) + (898 − 804.3)]

=

272.15 kJ/kg

=

272.15 − 163.75 WT − WC = = 30% Q 360.89

WC

WT

ηth

=

1.005 × (450.94 − 288)

Ans

⇐=

6.14 In a closed-cycle turbine plant the working fluid at 38◦ C is compressed with an adiabatic efficiency of 82%. It is then heated at constantpressure to 650◦ C. The fluid then expands down to initial pressure in a turbine with an adiabatic efficiency of 80%. The fluid after expansion is cooled to 38◦ C. The pressure ratio is such that work done per kg is maximum. For the working fluid take Cpa = 1.005 kJ/kg K and Cpg = 1.147. Calculate the pressure ratio and cycle efficiency. Solution t

=

T03 T01

=

923 311

=

2.96

For maximum specific work we know that γ−1 √ √ γ ropt = tηC ηT = 2.96 × 0.82 × 0.8 ropt

=

3.194

=

1.3935 Ans

⇐=

Practical Cycles and their Analysis

191

03

04 04’

02

T

02’ 01

s Fig. 6.36

p02 p01

γ−1 γ

= 311 × 3.1940.286 = 433.51 K

T02

=

T01

T02

=

T01 +

T04

=

T03 3.1940.248

ηT

=

T03 − T04 T03 − T04

0.80

=

923 − T04 923 − 692.03

T04

=

738.22 K

ηth

=

1.147 × (923 − 738.22) − 1.005 × (460.40 − 311) 1.147 × (923 − 460.40)

=

211.43 − 150.15 × 100 530.60

T02 − T01 433.51 − 311 = 460.40 K = 311 + ηC 0.82 =

923 3.1940.248

=

=

692.03 K

11.55%

Ans

⇐=

6.15 The following data relate to a test on simple gas turbine plant. Stagnation pressure at entry is 1 bar, static pressure and temperature at compressor entry are 0.93 bar and 10◦ C respectively; compressor delivery total head pressure and temperature, 6 bar and 230◦ C respectively; turbine exhaust pipe temperature, 460◦ C, turbine horsepower 5100 kW. Calculate the total head isentropic efficiency of the compressor, and taking the compressor entry area as 0.10 m2 , calculate the air mass flow and estimate the temperature of the gases at entry to the turbine. It may be assumed that there are no losses at compressor entry, that the entry velocity distribution is uniform and that increase of mass flow due to fuel addition is negligible.

192

Gas Turbines

Take Cp = 1.005 kJ/kg K and γ = 1.4 for compression and Cp = 1.147 kJ/kg K, γ = 1.333 for expansion. Solution 03 04 02’ 02

T

04’

01’ 01

s Fig. 6.37

γ γ−1

p01 pa

=

1+

γ−1 2 M 2

1 0.93

=

1+

1.4 − 1 2 M 2

-

1 3.5

1 0.93

−1

3.5

=

0.324

M

=

T01 T1

=

1 + 0.2 × 0.3242

T01

=

283 × 1 + 0.2 × 0.3242

0.2

p02 p01

γ−1 γ

=

288.94 K

= 288.94 × 60.286 =

T02

=

T01

ηC

=

T02 − T01 482.35 − 288.94 = 90.35% = T02 − T01 503 − 288.94

= 482.35 K Ans

⇐=

We know m ˙ = ρAV . By equation of state p

=

ρRT

ρs

=

ps RTs

a

=

V

=

=

γRT =

0.93 × 105 287 × 283

=

3

1.145 kg/m

√ 1.4 × 0.287 × 283 × 1000 = 337.21 m/s

M × a = 0.324 × 337.21 = 109.26 m/s

Practical Cycles and their Analysis

m ˙

=

ρAV = 1.145 × 0.1 × 109.26 = 12.51 kg/s

193 Ans

⇐=

We know turbine output power = 5100 kW 5100

=

mC ˙ p (T03 − T04 )

5100

=

12.51 × 1.147 × (T03 − 733)

Turbine inlet stagnation temperature T03

=

Ans

1088.43 K

⇐=

6.16 In the last seventy years, the gas turbine, inlet temperature has increased from 700◦C to 1000◦C, the turbine efficiencies have increased from 70 to 90 per cent, and the compressor efficiencies from 65 to 85 per cent. For a pressure ratio of 3 calculate: (i) the efficiency and work ratio of turbines of 70 years ago and (ii) the efficiency and work ratio of modern turbines. Assume an inlet pressure of 1 bar and temperature of 27◦ . Solution 03

04 04’

02

T

02’ 01

s Fig. 6.38

T02

=

T01 r

γ−1 γ

=

300 × 30.286

=

410.75 K

Turbines 70 years ago T02

!

1 1+ ηC

=

T01

=

300 × 1 +

p02 p01

γ−1 γ

−1

1 × 30.286 − 1 0.65

" =

470.40 K

194

Gas Turbines

T04

ηth

1

=

T03 1 − ηT 1 −

=

973 × 1 − 0.7 × 1 −

=

1.147 × (973 − 810.56) − 1.005 × (470.40 − 300) 1.147 × (973 − 470.4)

=

186.315 − 171.24 × 100 = 2.62% 576.48

=

186.32 − 171.24 186.32

Work ratio

Modern turbines T02 =

T01 +

r

γ−1 γ

1 30.248

=

=

810.56 K

Ans

⇐=

0.0809

T02 − T01 410.75 − 300 = 430.29 K = 300 + ηC 0.85

T04

=

1273 30.248

T04

=

T03 − ηT (T03 − T04 )

=

1273 − 0.9 × (1273 − 969.398)

WC

=

1.005 × (430.29 − 300)

WT

=

1.147 × (1273 − 999.758)

ηth

=

313.4 − 130.941 = 18.88% 1.147 × (1273 − 430.29)

⇐=

=

WN WT

⇐=

Work ratio

=

=

969.398 K

182.46 313.4

=

=

=

999.758 K

130.941 kJ/kg =

0.58

313.4 kJ/kg Ans

Ans

6.17 A gas turbine operates at a pressure ratio of 7 and maximum temperature is limited to 1000 K. The isentropic efficiency of the compressor is 85 per cent and that of the turbine is 90 per cent. If the air enters the compressor at a temperature of 288 K, calculate the specific output and efficiency of the plant and compare these values with those achieved by the ideal Joule cycle. If the unit under actual condition is required to produce a power output of 750 kW, determine the necessary mass flow rate. Take Cp equal to 1.005 for ideal and actual cycles. Solution T02

=

T01 1 +

γ−1 1 r γ −1 ηC

Practical Cycles and their Analysis

195

03

04 04’

02

T

02’ 01

s Fig. 6.39

T04

WN

ηth

1 70.286 − 1 0.85

=

540.29 K

=

288 × 1 +

=

T03 1 − ηT 1 −

=

1000 × 1 − 0.9 1 −

=

Cpg (T03 − T04 ) − Cpa (T02 − T01 )

=

1.005 × (1000 − 615.87) − 1.005 × (540.29 − 288)

=

132.499 kJ/kg

=

WN Cpg (T03 − T02 )

=

0.2867

=

1 r

γ−1 γ

1 70.286

=

615.87 K

Ans

⇐= 132.499 1.005 × (1000 − 540.29)

=

Ans

⇐=

28.68%

Ideal Joule cycle WN

1

=

Cp T03 1 −

=

1.005 × 1000 × 1 −

=

213.42 kJ/kg

ηth

=

1−

m ˙a

=

pN WN

1 r

γ−1 γ

=

r

γ−1 γ

=1−

− T01 r 1 70.286

γ−1 γ

−1

− 288 × 70.286 − 1 Ans

⇐= 1 = 0.427 = 42.7% 70.286

750 132.499

=

5.60 kg/s

Ans

⇐= Ans

⇐=

196

Gas Turbines

6.18 The layout of a gas turbine is shown in the diagram. The compressor is driven by the HP stage of a two-stage turbine and compresses 5 kg of air per second from 1 bar to 5 bar with an isentropic efficiency of 85%. The H.P. stage has an isentropic efficiency of 87% and its inlet temperature is 675◦ C. The L.P. stage, which is mechanically independent has an isentropic efficiency of 82%. The expansion pressure ratios of the two turbines are not equal. The exhaust gases from the L.P. stage pass to a heat exchanger which transfers 70% of the heat available in cooling the exhaust to raise the compressor temperature at delivery. Assuming the working fluid to be air throughout, of constant specific heat, and neglecting pressure losses, estimate the intermediate pressure p04 and the temperature T04 between the two turbine stages, the power output of the L.P. stage and the overall plant efficiency. Assume an inlet pressure of 1 bar and temperature of 15◦ .

Solution

Heat exchanger Combustion chamber

Air Fuel HPC

HPT

Compressor

Turbine Power LPT output Separate power turbine Fig. 6.40

p02 p01

γ−1 γ

T02

=

T01

ηC

=

T02 − T01 T02 − T01

0.85

=

456.35 − 288 T02 − 288

= 288 × 50.286 = 456.35 K

Practical Cycles and their Analysis

197

03 04

04’

0x

T

02

05 05’

02’

01

s Fig. 6.41

T02

=

486.06 K

As LPT mechanically independent implies HPT power = Compressor power. Now, Cp (T02 − T01 )

=

Cp (T03 − T04 )

T04

=

T01 − T02 + T03 = 288 − 486.06 + 948

T04

=

749.94 K

ηHP T

=

T03 − T04 T03 − T04

0.87

=

948 − 749.94 948 − T04

T04

=

720.34 K γ γ−1

p03 p04

=

T03 T04

5 p04

=

948 720.34

p04

=

1.91 bar

T04 T05

=

T05

=

ηLP T

=

p04 p05

Ans

⇐=

3.5

Ans

⇐=

γ−1 γ

749.94 1.91 0.286 1

T04 − T05 T04 − T05

=

623.23 K

198

Gas Turbines

0.82

=

749.94 − T05 749.94 − 623.23

T05

=

646.04 K

Heat exchanger effectiveness 0.7

=

T0x − 486.06 T0x − T02 = T05 − T02 646.04 − 486.06

T0x

=

598.046 K

WLP T

=

Cp (T04 − T05 )

=

1.005 × (749.94 − 646.04) = 104.42 kJ/kg

PLP T

=

104.42 × 5 = 522.10 kW

Q

=

Cp (T03 − T0x ) = 1.005 × (948 − 598.04)

=

351.71 kJ/kg

=

WLP T Q

ηth

=

Ans

⇐=

Ans

⇐= 104.42 = 29.69% 351.71

Ans

⇐=

6.19 In the gas turbine plant shown, each compressor operates on a pressure ratio of 3 and an isentropic efficiency of 82%. After the low pressure compressor, some of the air is extracted and passed to a combustion chamber from which the products leave at a temperature of 650◦ C and expand in power turbine. The remainder of the air passes through the high pressure compressor and into a combustion chamber from which it leaves at a temperature of 540◦C and expands in a turbine which drives both the compressors. The isentropic efficiency of each turbine is 87%. If the temperature of the air at inlet to the low pressure compressor is 15◦ C, determine the percentage of the total air intake that passes to the power turbine and the thermal efficiency of the plant. For compression assume Cp = 1.005 kJ/kg K and γ = 1.4 and for heating and expansion Cp = 1.147 kJ/kg K and γ = 1.33. Solution T02

T03

γ−1 1 rLPγ − 1 ηC

=

T01 1 +

=

288 × 1 +

=

T02 1 +

1 30.286 − 1 0.82

γ−1 1 γ rHP −1 ηC

= 417.66 K

Practical Cycles and their Analysis

199

04

05

07 07' 06 03

06'

03' 02'

02

01

Fig. 6.42 CC

07

04 PT x 1-x CC

01

03

02

HPC

LPC

05

06

CT

Fig. 6.43

1 30.286 − 1 0.82

=

417.66 × 1 +

T06

=

T05 90.248

T06

=

T05 − ηT (T05 − T06 )

=

813 − 0.87 × (813 − 471.45) = 515.85 K

=

= 605.7 K

813 = 471.45 K 90.248

Let x be the mass of air passing through the power turbine, then m ˙ a Cpa (T02 − T01 ) + m ˙ a (1 − x)Cpa (T03 − T02 ) = m ˙ a (1 − x)Cpg (T05 − T06 )

(1 − x)

Cpg (T05 − T06 ) − (T03 − T02 ) = T02 − T01 Cpa

200

Gas Turbines

x

=

1−

=

1−

=

0.142

Cpg Cp (T05

1.147 1.005

T02 − T01

− T06 ) − (T03 − T02 )

417.66 − 288 × (813 − 515.85) − (605.7 − 417.66)

Percentage of the total air intake that passes to the power turbine

T07

ηth

Ans

=

14.2%

=

T04 1 − ηT 1 −

⇐= 1 3

γg −1 γg

1

= 731.50 K

=

923 × 1 − 0.87 × 1 −

=

xCpg (T04 − T07 ) (1 − x)Cpg (T05 − T03 ) + xCpg (T04 − T02 )

3

1.33−1 1.33

=

0.142×1.147×(923−731.50) (1−0.142)×1.147×(813−605.7)+0.142×1.147×(923−417.66)

=

0.109

=

10.9%

Ans

⇐=

6.20 In an open-cycle gas turbine plant the air is compressed in a two-stage compressor with complete intercooling to the initial temperature. After passing through an exhaust heat exchanger and combustion chamber, the gases are expanded in a two-stage turbine with reheating in a second combustion chamber between the stages. Both the compressor stages are driven by high pressure turbine stage and the power output of the plant is taken from the mechanically independent low pressure turbine stage. After expansion in the low pressure turbine, the gases pass through the heat exchanger to atmosphere. The pressure ratio of each compressor is 2:1, the air inlet pressure and temperature are 1 bar and 15◦ C and the gas inlet temperature to both turbines is 700◦ C. If the isentropic efficiency in each compressor and turbine stage is 85% and the thermal ratio of heat exchanger is 50% determine the output per kg of air per second and the thermal efficiency of the plant, neglecting pressure losses in the heat exchanger, intercooler and combustion chambers and any variation in the mass flow of the working fluid due to addition of fuel. For gases during heating and expansion take Cp = 1.147 and γ = 1.333. Solution T01

=

T03

Practical Cycles and their Analysis

05

08 08� 06� 011 011�

09

T

07 06

04 02 04� 02�

010

01

03

s Fig. 6.44

p02 p01

=

2

p04 p01

=

p04 p02 × p03 p01

p04 p01

=

4

p04 p03

=

=

2×2

=

4

p05 p08

=

γ−1 γ

p02 p01

= 288 × 20.286 = 351.15 K

T02

=

T01

T02

=

T01 +

=

362.3 K

T05 T06

=

20.248

T06

=

973 1.1876

T06

=

T05 − ηT (T05 − T06 ) = 973 − 0.85 × (973 − 819.3)

=

842.36 K

T02 − T01 351.15 − 288 = 288 + ηC 0.85 = =

T04 1.1876

=

819.3 K

=

T08

Heat exchanger effectiveness

T09

WN

=

T09 − T04 T08 − T04

=

T04 + (T08 − T04 )

=

362.3 + 0.5 × (842.36 − 362.3)

=

WLP T

=

Cp (T07 − T08 )

=

602.33 K

201

202

Gas Turbines

=

1.147 × (973 − 842.36) = 149.84 kJ/kg

Ans

⇐=

Heat addition Q

ηth

=

Cpg (T05 − T09 ) + Cpg (T07 − T06 )

=

Cpg (2T05 − T09 − T06 )

=

1.147 × (2 × 973 − 602.33 − 842.36) = 575 kJ/kg

=

WN Qs

149.84 575

=

=

26.06%

Ans

⇐=

6.21 The following data refer to a closed-cycle gas turbine plant using helium as working fluid and incorporating two-stage compression with intercooling and two-stage expansion with reheating; temperature at entry to each compression stage is 270◦C; pressure at entry to first compression stage and exit from the second turbine stage is 1 bar; first compression stage pressure ratio is 6; each compressor stage isentropic efficiency is 0.85; temperature at inlet to each expansion stage is 1150◦C; isentropic efficiency of each expansion stage is 0.9; reheat pressure is 6 bar; for helium polytropic index n is 1.24 and R is 10.05 kJ/kg K. Calculate the cycle thermal efficiency. 05 07 06’

T

06

08 08’

04 02 04’ 02’ 01

03

s Fig. 6.45

Solution p02 p01

T02

=

T01

T02

=

T01 +

Cp Cv

=

1.24

Cp − Cv

=

10.05

γ−1 γ

= 543 × 60.1935 = 768 K

T02 − T01 768 − 543 = 807.81 K = 543 + ηC 0.85

Practical Cycles and their Analysis

Cp

=

51.925 kJ/kg K

Cv

=

41.875 kJ/kg K

WC

=

2 × (807.81 − 543) × 51.925 = 27500.5 kJ/kg

T05 T06

=

60.24/1.24 = 1.4144

T06

=

1423 1.4144

T06

=

1423 − 0.9 × (1423 − 1006)

WT

=

2 × 51.925 × (1423 − 1047.7) = 38974.905 kJ/kg

Q

=

q1 + q2

=

51.925 × (1423 − 807.81) +

=

203

1006 K =

1047.7 K

51.925 × (1423 − 1047.7) = 51431.193 kJ/kg WN

=

WT − WC = 38974.905 − 27500.5 = 11474.405 kJ/kg

η

=

11474.405 51431.193

=

22.31%

Ans

⇐=

Review Questions 6.1 What are the various assumptions made in practical cycle analysis? 6.2 Bring out clearly the implications of the various assumptions. 6.3 Define polytropic efficiency. Derive suitable expressions for polytropic efficiency and bring out the relation between the polytropic efficiency and isentropic efficiency. 6.4 Show the variation of isentropic efficiency for various polytropic efficiency as a function of pressure ratio. 6.5 With suitable graphs, compare the performance of various practical cycles with respect to a simple cycle. Exercise [Note: Take γaγ−1 = 0.286 and a stated otherwise]

γg −1 γg

= 0.248 for all problems unless

6.1 In a gas turbine the compressor takes in air at a temperature of 27◦ , and compresses it to five times the initial pressure with an isentropic efficiency of 85%. The air is then passed through a regenerator heated

204

Gas Turbines

by the turbine exhaust before reaching the combustion chamber. The effectiveness of the regenerator is 80%. The maximum temperature after constant-pressure combustion is 677◦ C and the efficiency of the turbine is 80%. Neglecting all losses except mentioned, and assuming the working fluid throughout the cycle to have the characteristics of air (i) Sketch the cycle on the T -s diagram. (ii) Calculate the efficiency of the cycle. Ans: (i) 23.7% 6.2 A simple gas turbine takes in air at 1.0 bar and 27◦ C and compresses to a pressure of 6 bar with the isentropic efficiency of compression being 85%. The air passes to the combustion chamber, and after combustion the gases enter the turbine at a temperature of 560◦C and expand to 1.00 bar, the turbine efficiency being 80%. Neglecting the change of mass flow rate due to fuel, calculate the flow of air in kg per second for a net output of 1500 kW making the following assumptions: Loss of pressure in combustion chamber = 0.08 bar. Ans: (i) 42.67 kg/s 6.3 The axial compressor of a gas turbine delivers 20 kg/s of air at a pressure ratio of 5.0 with an isentropic efficiency of 80%. Inlet temperature and pressure are 22◦ C and 1 bar. Calculate the power required by the compressor. After heating at constant-pressure to 870◦C the gas is expanded to 1 bar through a turbine with efficiency of 85%. The turbine is direct coupled to the compressor and to an airscrew reduction gear which has an efficiency of 95%. What power is available for the airscrew? Ans: (i) 4332.56 kW (ii) 2852.17 kW 6.4 For a gas turbine operating at a pressure ratio of 8.7 the maximum temperature ratio to be maintained such that the turbine just supports the compressor, which is given by tmin = 3.0. If the compressor inlet total temperature and the turbine efficiency are respectively 300 K and 0.75, find the following: (i) the compressor efficiency (ii) the temperature ratio at which the compressor work is 80% of the power produced by turbine. Also find the corresponding heat addition and Net work output per unit mass flow in the gas turbine. Take Cpa = 1.005 kJ/kg K and Cpg = 1.147 kJ/kg K. (iii) for the temperature ratio found at (ii) find the required compressor pressure ratio at which the compressor work and turbine work are equal. Take γg = γa and Cpg = Cpa . Ans: (i) 0.8 (ii) (a) 3 (b) 396.06 kW (c) 64.29 kW (iii) 12.33

Practical Cycles and their Analysis

205

6.5 In a gas turbine the pressure ratio of r is achieved by two stage compression with intercooling. If η1 is the efficiency of LP compressor, η2 is the efficiency of the HP compressor and is the effectiveness of the intercooler. Derive an expression for the above for stage pressure ratios in terms of overall pressure ratio between first stage and second stage such that work of compressor is minimum. If the efficiencies of the HP and LP compressors are respectively 0.75 and 0.85, and if the effectiveness of the intercooler is 80%, find the work of compression assuming the inlet temperature at LP compressor as 300 K. The total pressure ratio for overall compression ratio is 11.3137.

−1 Ans: (i) c01 = c ηη12 + + −1 (iii) 327.368 kJ/kg

1 2

(ii) c02 = c

η2 + −1 η1 + −1

1 2

6.6 A simple gas turbine with heat-exchanger has a compressor and turbine having respective isentropic efficiencies ηC and ηT . Show that the combined effect of small pressure drops Δphg (in gas-side of heatexchanger) and Δp (total in combustion chamber and air-side of heatexchanger) is to reduce the specific work output by an amount given by γ−1 Cp T03 ηT Δp × Δphg + γ−1 γ r γ r p01 where T03 is turbine inlet temperature, p01 is compressor inlet pressure and r, the compressor pressure ratio. Assume that Cp and γ are constant throughout the cycle. 6.7 In a gas turbine plant, air is compressed from state (p01 , T01 ) to a pressure rp01 and heated to temperature T03 . The air is then expanded in two stages with reheat to T03 between the turbines. The isentropic efficiencies of the compressor and each turbine are ηC and ηT . If xp01 is the intermediate pressure between the turbines, show that, for given values p01 , T01 , √ T03 , ηC , ηT and r, the specific work output is a maximum when x = r. If this division of the expansion between the turbines is maintained, show that: (i) when r is varied, the specific work output is a maximum with r given by γ/(γ−1) ηC ηT T03 r3/2 = T01 (ii) when a perfect heat-exchanger is added, the cycle efficiency is given by T01 r(γ−1)/2γ r(γ−1)/2γ + 1 η =1− 2ηC ηT T03 Assume that the working fluid is a perfect gas with constant specific heats, and that pressure losses in the heater, reheater, and heatexchanger are negligible.

206

Gas Turbines

6.8 A peak load generator is to be powered by a simple gas turbine with free power turbine delivering 20 MW of shaft power. The following data are applicable: Compressor pressure ratio Compressor isentropic efficiency Combustion chamber pressure loss Combustion efficiency Turbine inlet temperature Power turbine isentropic efficiency Mechanical efficiency (each shaft) Ambient conditions pa , Ta

: : : : : : : :

11.0 0.82 0.4 bar 0.99 1150 K 0.89 0.98 1 bar, 300 K

Assume gas generator turbine isentropic efficiency = 0.87, Draw the schematic diagram of the cycle. If the calorific value of the fuel = 43 MJ/kg, calculate the air mass flow required and the specific fuel consumption. Ans: (i) 130.51 kg/s (ii) 0.3726 kg/kW h 6.9 A gas turbine is to consist of a compressor, combustion chamber, turbine and heat-exchanger. It is proposed to examine the advantage of bleeding off a fraction Δm of the air delivered by the compressor m and using it to cool the turbine blades. By so doing the maximum permissible cycle temperature may be increased from T to (T + ΔT ). The gain in efficiency due to the increase of temperature will be offset by a loss due to the decrease in effective air flow through the turbine. Show that, on the following assumptions, there is no net gain in efficiency when ΔT Δm T = m 1 + ΔT T and that this result is independent of the compressor and turbine efficiencies. Assumptions: (i) No pressure loss in combustion chamber or heat exchanger. (ii) Working fluid is air throughout and the specific heats are constant. (iii) Air bled for cooling purposes does no work in the turbine. (iv) Temperature of the air entering the combustion chamber is equal to that of the turbine exhaust. A plant of this kind operates with an inlet temperature of 288 K, a pressure ratio of 6.0, a turbine isentropic efficiency of 90 per cent and a compressor isentropic efficiency of 87 per cent. Heat transfer calculations indicate that if 5 per cent of the compressor delivery is bled off for cooling purposes, the maximum temperature of the cycle can be raised from 1000 to 1250 K. Find the percentage increase in (a) efficiency, and (b) specific work output, which is achieved by the combined bleeding and cooling process. Make the same assumptions as before and take γ = 1.4 throughout. Ans: (i)25.1% (ii) 48.55%

Practical Cycles and their Analysis

207

6.10 The efficiencies of the compressor and turbine of a gas turbine are ηC = 0.80 and ηT = 0.75 respectively. If the temperature ratio is 4 and the inlet total temperature of air 300 K, find the optimum pressure ratio, and corresponding Net work output, work ratio, heat addition and efficiency. Ans: (i) 4.626 (ii) 158.96 kJ/kg (iii) 0.43 (iv) 796.16 kJ/kg (v) 19.96% 6.11 An auxiliary gas turbine for use on a large airliner uses a single-shaft configuration with air bled from the compressor discharge for aircraft services. The unit must provide 1.5 kg/s bleed air and a shaft power of 200 kW. Calculate (a) the total compressor air mass flow and (b) the power available with no bleed flow, assuming the following: Compressor pressure ratio Compressor isentropic efficiency Combustion pressure loss Turbine inlet temperature Turbine isentropic efficiency Mechanical efficiency (compressor rotor) Mechanical efficiency (driven load) Ambient conditions

: : : : : : : :

3.80 0.85 0.12 bar 1050 K 0.88 0.99 0.98 1 bar, 288 K

Ans: (i) 4.97 kg/s (ii) 630.9 kW 6.12 In a simple gas turbine plant air enters the compressor at 1 bar and 15◦ C and leaves at 6 bar. It is then heated in combustion chamber to 700◦ C and then enters the turbine and expands to atmospheric pressure. The isentropic efficiency of compressor and turbine are 0.80 and 0.85 respectively and the combustion efficiency is 0.90. The fall in pressure in the combustion chamber is 0.1 bar. Determine (i) air-fuel ratio, (ii) work ratio, (iii) thermal efficiency, (iv) air rate in kg per shaft kilowatt power, and (v) specific fuel consumption. Take calorific value of the fuel equal 42000 kJ/kg. Ans: (i) 73.219:1 (ii) 0.2831 (iii) 16.89% (iv) 0.0105 kg/kW (v) 0.5158 kg/kW h 6.13 An industrial gas turbine takes in air at 1 bar and 27◦ C and compresses it to 5.5 times the original pressure. The temperatures at the salient points are compressor outlet, 251◦ C, turbine inlet 760◦ C and turbine outlet 447◦C. Calculate (i) the compressor and (ii) turbine efficiencies. Compare the ideal cycle and the practical cycle considering component efficiencies for the following (iii) work ratio,

208

Gas Turbines

(iv) thermal efficiency, (v) optimum pressure ratio for maximum output. Ans: (i) 0.842 (ii) 0.88 (iii) 0.47; 0.37 (iv) 0.308; 0.229 (v) 8.7; 5.7 6.14 Deduce an expression for the specific output (kJ/kg of working fluid) of a simple constant-pressure gas turbine in terms of temperatures at the beginning of compression and at the beginning of expansion, the isentropic efficiencies of the compressor and turbine, the pressure ratio and the isentropic index. The specific heat at constant-pressure may be assumed constant, and the weight of the fuel added may be neglected. Hence determine the pressure ratio at which the specific output is maximum for the following operating conditions. Temperature at compressor inlet is 15◦ C, temperature at turbine inlet is 630◦C and the isentropic efficiency for compressor and turbine is 0.85 and 0.90 respectively. Neglect the effect of fuel flow and assume constant Cp of air. Determine in kg/kW-h the air flow and the specific fuel consumption, if 1 kg/s of fuel is flowing with the calorific value of 42000 kJ/kg. ⎡ ⎤ ⎢ Ans: (i) WN = Cp ηT T03 ⎢ ⎣1 −

1

γ−1 rp γ

⎥ ⎥− ⎦

Cp ηC T01

γ−1

rp γ − 1

(ii) 4.62 (iii) 35.1 kg/kW h (iv) 0.362 kg/kW h

6.15 In a gas turbine plant, comprising a single stage compressor, combustion chamber and turbine, the compressor takes in air at 15◦ C and compresses it to 4 times the initial pressure with an isentropic efficiency of 85 per cent. The fuel-air ratio is 0.0125 and the calorific value of the fuel is 42000 kJ/kg. If the isentropic efficiency of the turbine is 82 per cent, find the amount of air intake for a power output of 260 kW and also the overall thermal efficiency. Take the mass of the fuel into account. The turbine inlet temperature is 1000 K. Ans: (i) 2.33 kg/s (ii) 21.3% 6.16 At design speed the following data apply to a gas turbine set employing a separate power turbine, heat exchanger, reheater and intercooler between two-stage compression. Efficiency of compression in each stage : 80% Isentropic efficiency of compressor turbine : 87% Isentropic efficiency of power turbine : 80% Transmission efficiency : 99% Pressure ratio in each stage of compression : 2:1 Pressure loss in intercooler : 0.07 bar Temperature after intercooling : 300 K Thermal ratio of heat exchanger : 0.75 Pressure loss in combustion chamber : 0.15 bar Combustion efficiency of reheater : 98%

Practical Cycles and their Analysis

Maximum cycle temperature Temperature after reheating Air mass flow Ambient air temperature Ambient air pressure

: : : : :

209

1000 K 1000 K 25 kg/s 15◦ C 1 bar

Take the calorific value of fuel as 42 MJ/kg and pressure loss in each side of heat exchanger as 0.1 bar. Find the net power output, overall thermal efficiency, specific fuel consumption. Neglect the kinetic energy of the gases leaving the system. Ans: (i) 2.457 MW (ii) 23.72% (iii) 0.3735 kg/kW h 6.17 The following diagram refers to an automotive gas turbine and the data refer to this turbine. Rated power (kW) : 220 Pressure ratio : 16:1 Turbine inlet temperature,◦ C : 925 Reheat turbine inlet temperature,◦ C : 925 bsf c, kg/kW h : 0.3465 Exhaust temperature,◦ C : 316 Regenerator efficiency, % : 75 Burner efficiency, % : 96 Compressor efficiency, % : 80 Turbine efficiency, % : 85 Ambient temperature,◦ C : 27 The schematic diagram is as shown below. 07 C2 03

T2

04

06 CC

Air cooler

Air

08

01 02

011

010

C

CC

05

012

T1

T4

T3

09

Fig. 6.46

Calculate the (i) outlet temperature of all turbines, (ii) net work output, (iii) thermal efficiency,

Transmissio

210

Gas Turbines

(iv) work ratio, and (v) mass flow rate. Ans: (i) T07 = 1038 K; T09 = 1057.5 K; T010 = 953.1 K; T011 = 793.1 K; (ii) 280.9 kJ/kg (iii) 36.6% (iv) 0.434 (v) 0.783 kg/s 6.18 The maximum and minimum temperatures in a simple gas turbine plant working on the Joule cycle are 1000 K and 288 K respectively. The pressure ratio is 6, and the isentropic efficiencies of the compressor and turbine are 85 and 90 per cent respectively. Calculate work output per kg and the efficiency of the plant. Ans: (i) 143.041 kJ/kg (ii) 25.7% 6.19 In a gas turbine plant the air enters the compressor at 1 bar and 27◦ C. The pressure leaving the compressor is 5 bar and the maximum temperature in the cycle is 850◦C. Compare the compressor work, turbine work, work ratio and cycle efficiency for the following two cases: (i) the cycle is ideal (ii) the cycle is actual with compressor efficiency of 80%, turbine efficiency of 85% and the pressure loss in the combustion chamber is 0.15 bar. If in the above cycle the air expands in the turbine to such a pressure that the turbine work is just equal to compressor work, and further expansion of air upto atmospheric pressure takes place in a nozzle; calculate the velocity of air leaving the nozzle. Assume a nozzle efficiency of 90%. If an ideal regenerator is incorporated in the ideal cycle, what is the thermal efficiency of the cycle? Ans: For an ideal cycle: (i) 176.24 kJ/kg (ii) 416.35 kJ/kg (iii) 0.5767 (iv) 36.89% For an actual cycle: (i) 220.3 kJ/kg (ii) 354.76 kJ/kg (iii) 0.379 (iv) 19.41% Ve = 543.52 m/s; ηth = 57.67% Multiple Choice Questions (choose the most appropriate answer) 1. The important assumptions in practical cycle analysis are (a) the compressor and expansion process are irreversible adiabatic (b) complete heat exchange is possible in a heat exchanger (c) the pressure losses are negligibly small (d) none of the above

Practical Cycles and their Analysis

211

2. The isentropic efficiency of the modern compressor is (a) 65% (b) 75% (c) 85% (d) 90% 3. Isentropic efficiency of a compressor is defined as the ratio of Isentropic work input Stoichiometric work input Isentropic work input (b) Actual work input Actual work input (c) Isentropic work input Stoichiometric work input (d) Isentropic work input (a)

4. The isentropic efficiency of a turbine is defined as the ratio of Actual work output Isentropic work output Isentropic work output (b) Actual work output Actual work input (c) Isentropic work input Isentropic work input (d) Actual work input (a)

5. The work ratio is defined as (a)

WT −WC WT

(b) 1 − (c)

WC WT

WN WT

(d) all of the above 6. The effectiveness of the modern heat exchanger is (a)

= 55%

(b)

= 65%

(c)

= 75%

(d)

= 90%

7. The assumption of mass flow rate to be the same in spite of the fuel addition is due to (a) the gas turbine operates at a very lean air-fuel ratio of the order of 60 to 100

212

Gas Turbines

(b) fuel flow rate compared to air flow rate is very small (c) the bleed air from the compressor to cool the turbine blades which compensates for the fuel addition (d) none of the above 8. The mechanical losses are accounted by ηmech on (a) compressor side (b) combustion chamber side (c) heat exchanger side (d) turbine side 9. Cycle efficiency can be defined as (a) (b) (c)

WT −WC f ×CV

WN f ×CV

3600 sf c×CV

(d) any of the above equation 10. For a constant value of ηpc (say 85%) ηc (a) increases with increase in pressure ratio (b) decreases with increase in pressure ratio (c) remains constant with increase in pressure ratio (d) none of the above Ans:

1. – (d) 6. – (c)

2. – (c) 7. – (c)

3. – (b) 8. – (a)

4. – (a) 9. – (d)

5. – (d) 10. – (b)

7 JET PROPULSION CYCLES AND THEIR ANALYSIS INTRODUCTION The last two chapters were dealt with the ideal and practical shaft power cycles respectively. In this chapter, the analysis of jet propulsion cycles will be taken up. Gas turbine cycles for jet propulsion differ from shaft power cycles because of the fact that the useful power output for jet propulsion is produced, wholly or partially, as a result of expansion of gas in a propelling nozzle; wholly in turbojet engines and partially in turboprop engines. A second distinguishing feature is the need to consider the effect of forward speed and altitude on the performance of propulsion engines. The beneficial effect of those parameters, together with an inherently high power/weight ratio have enabled the gas turbine to replace the reciprocating engine for aircraft propulsion except for the very low power levels. The principle of jet propulsion is obtained from the application of Newton’s laws of motion.∗ We know that when a fluid is to be accelerated, a force is required to produce this acceleration in the fluid. At the same time, there is an equal and opposite reaction force of the fluid on the engine which is known as the thrust. Hence, it may be stated that the working of jet propulsion is based on the reaction principle. Thus all devices that move through fluids must follow this basic principle. In principle, any fluid can be used to achieve the jet propulsion. Thus water, steam or combustion gases can be used to propel a body in a fluid. But there are limitations in the choice of the fluid when the bodies are to be propelled in the atmosphere. Experience shows that only two types of fluids are particularly suitable for jet propulsion. ∗ Newton’s Second Law: Rate of change of momentum in any direction is proportional to the force acting in that direction. Newton’s Third Law : For every action there is equal and opposite reaction.

214

Gas Turbines

(i) A heated and compressed atmospheric air – admixed with the products of combustion produced by burning fuel in that air can be used for jet propulsion. The thermochemical energy of the fuel is utilized for increasing the temperature of the air to the desired value. The jet of this character is called a thermal jet and the jet propulsion engine using atmospheric air is called air breathing engines. (ii) Another class of jet-propulsion engines use a jet of gas produced by the chemical reactions of fuel and oxidizer. Each of them is carried with the system itself. The fuel-oxidant mixture is called the propellant. No atmospheric air is used for the formation of the jet. But the oxidant in the propellant is used for generating the thermal jet. A jet produced in this way is known as rocket jet and the equipment wherein the chemical reaction takes place is called a rocket motor. The complete unit including the propellant is called a rocket engine. From the above discussion it is clear that jet-propulsion engines may be classified broadly into two groups. (i) air breathing engines and (ii) rocket engines In this chapter, we will discuss the air breathing engines. Air breathing engines can be further classified as follows: (i) reciprocating or propeller engines (ii) gas turbine engines. 7.1

RECIPROCATING OR PROPELLER ENGINES

In early days, the source of power for an air breathing engine was a reciprocating internal combustion engine which used to drive a propeller connected to it. The propeller displaces rearwards, a large mass of air, accelerating it in the process. Due to this acceleration of the fluid a propulsive force is produced which drives the aircraft. The extensive use of aircraft for military purposes led to a very rapid development of reciprocating internal combustion engines during the two world wars and it is now a highly developed piece of equipment as compared to its industrial counterpart. Further development resulted in the use of highly supercharged and turbocharged engines. All these engines were gasoline engines. Diesel engines in spite of their good fuel economy and reliability was not used due to higher weight. For small aircraft flying at velocities less than about 500 km/h reciprocating engine is in an enviable position due to its excellent fuel economy and good take-off characteristics. However, due to comparatively large drop in power with altitude operation and the need of using high octane fuels, along

Jet Propulsion Cycles and Their Analysis

215

with the difficult cooling and lubrication problems, high weight/power ratio, and larger frontal area these are being replaced by turbojets in higher speed ranges. Rapid developments in design of turbojet and turboprop engines have started exploding the best fuel economy myth of the reciprocating engines. They are nearing the specific fuel consumption value of reciprocating engines. Still the reciprocating engine is likely to be used for small aircraft needing only a few hundred kilowatts both because of good take off characteristics and due to difficulties in the development of smaller gas turbine engines giving reasonable fuel economy and cost. However, the use of reciprocating engines is continuously on the decline because its development has reached almost a saturation stage as far as maximum power is concerned. The demand of present day aircraft, in terms of high flight speeds, long distance travels and high load carrying capacities, is soaring to new heights. A power output more than 4000 kW is difficult to obtain without modifications in the present reciprocating engine power plant. The output can be increased by increasing the cylinder sizes, installing large number of cylinders or by running the engine at higher speeds. Unfortunately all these methods of raising the output of the engine increase the engine size, frontal area of the aircraft, complexity and cost of the plant. The drag of the plane will also increase to critical values with increase in engine size. Hence, for aircraft propulsion, gas turbine engines are the ideal power plant. 7.2

GAS TURBINE ENGINES

World War II was the turning point for the development of gas turbine technology. All modern aircrafts are fitted with gas turbines. Gas turbine engines can be classified into (i) ramjet engines, (ii) pulse jet engines, (iii) turbojet engines, (iv) turboprop engines, and (v) turbofan engines. Taken in the above order they provide propulsive jets of increasing mass flow and decreasing jet velocity. Therefore, in that order, it will be seen that ramjet can be used for highest cruising speed whereas the turboprop engine will be useful for the lower cruising speed at low altitudes. In practice, the choice of the power plant will depend on the required cruising speed, desired range of the aircraft and maximum rate of climb. The details of various gas turbine engines mentioned above are discussed under two categories: (i) pilotless operation, and (ii) piloted operation. The ramjet and pulse jet engines come under the category of pilotless operation whereas the turboprop and turbojet engines are used for piloted operation.

216 7.3

Gas Turbines

THE RAMJET ENGINE

The fact of obtaining very high pressure ratios of about 8 to 10 by ram compression has made it possible to design a jet engine without a mechanical compressor. A deceleration of the air from Mach number 3 at diffuser inlet to Mach number 0.3 in combustion chamber would cause pressure ratio of more than 30. Due to shock and other losses inevitable at such velocities all of this pressure rise is not available; still whatever we get is more than sufficient for raising the air pressure to the required combustion pressure. This principle of ram pressure rise is used in the ramjet engines. The ram pressure rise can be achieved in diffusers. It may be noted that the simplest types of air breathing engine is the ramjet engine and a simplified sketch of the engine is illustrated in Fig. 7.1. The engine consists of (i) supersonic diffuser (1–2), (ii) subsonic diffuser section (2–3), (iii) combustion chamber (3–4), and (iv) discharge nozzle section (4–5).

Air inlet and supersonic diffuser Subsonic diffuser Combustor Nozzle F

A ma

B m +m

A P

c A 0 1 Oblique shock

2

3

mf

Normal shock Central body housing accessories

4

c B 5

Jet Propulsion Cycles and Their Analysis

217

the fuel is injected by suitable injectors and mixed with the unburnt air. The air is heated to a temperature of the order of 1500 – 2000 K by the continuous combustion of fuel. The fresh supply of air to the diffuser builds up pressure at the diffuser end so that these gases cannot expand towards the diffuser. Instead, the gases are made to expand in the combustion chamber towards the tail pipe. Further, they are allowed to expand in the exhaust nozzle section. The products will leave the engine with a speed exceeding that of the entering air. Because of the rate of increase in the momentum of the working fluid, a thrust, F , is developed in the direction of flight. Normally, the air enters the engine with a supersonic speed which must be reduced to a subsonic value. This is necessary to prevent the blow out of the flame in the combustion chamber. The velocity must be small enough to make it possible to add the required quantity of fuel for stable combustion. Both theory and experiment indicate that the speed of the air entering the combustion chamber should not be higher than that corresponding to a local Mach number of 0.2 approximately. The cycle pressure ratio of a ramjet engine depends upon its flight velocity. The higher the flight velocity the larger is the ram pressure, and consequently larger will be the thrust. This is true until a condition is reached where the discharge nozzle becomes chocked. Thereafter, the nozzle operates with a constant Mach number of 1 at its throat. Therefore, a ramjet having fixed geometry is designed for a specific Mach number and altitude, and at the design point, will give the best performance. Since the ramjet engine cannot operate under static conditions, as there will be no pressure rise in the diffuser, it is not self-propelling at zero flight velocity. To initiate its operation, the ramjet must be either launched from an airplane in flight or be given an initial velocity by some auxiliary means, such as launching rockets. Since the ramjet is an air breathing engine, its maximum altitude is limited. Its field of operations is inherently in speed ranges above those of the other air breathing engines. However, it has a limited use in the high subsonic speed range. Its best performance capabilities, however, are in supersonic speed range of Mach numbers between 2 and 5. The upper speed is limited by the problem of cooling of the outer skin of the engine body at the high flight Mach numbers. 7.3.1

Thermodynamic Cycle

The ramjet works on the thermodynamic cycle, viz., the Brayton cycle. Figure 7.2 shows this cycle on a T -s diagram. In an ideal cycle process, 1 → 2 is isentropic ram compression in both the diffusers and process 3 → 4 is the isentropic expansion in the nozzle. In actual practice there will be losses due to shock, friction and mixing at the diffuser and losses in the nozzle. The actual compression and expansion is shown by the processes 1 → 2 and 3 → 4 respectively. Combustion is represented by the process 2 → 3. Since the expansion and compression is assumed isentropic, in an ideal ramjet cycle the stagnation pressure must remain constant and the pressure ratio of ram compression and nozzle expansion ratio must be same.

218

Gas Turbines

3 Nozzle

4 T

4’

2 2’

Ram

1 s Fig. 7.2 Ramjet cycle on T -s diagram p02 p1

=

1+

γ−1 2 Mi 2

p03 p4

=

1+

γ−1 2 Mj 2

and

γ γ−1

γ γ−1

where Mi and Mj are inlet and exhaust Mach number. We have, p02 p03 = p1 p4 which gives

or

Mi

=

Mj

c √ i γRT1

=

c √ j γRT4

cj

=

ci

or T4 T1

(7.1)

Knowing cj and ci , thrust can be calculated. 7.3.2

Performance

Figure 7.3 shows the net thrust and thrust specific fuel consumption per unit weight of a ramjet engine. It can be seen that at low flight speeds

Jet Propulsion Cycles and Their Analysis

219

l leve Sea 0m 700 00 m 130 00 m 200 00 m 300

Mach number

Net specific fuel consumption (kg/N thrust)

Net thrust (kN)

the thrust fuel consumption is quite high, while at high speeds it reduces considerably. This is because at low speeds the compression is poor. As can be seen from the Fig.7.3 the thrust is maximum at sea level and decreases as the altitude increases, due to reduction in air density with altitude. However, the velocity of the ramjet increases due to lower resistance at high altitude, thereby, improving specific fuel consumption. If the increase in thrust due to increased jet velocity is more than the decrease in thrust due to reduced density; the specific fuel consumption will improve. Otherwise, the specific fuel consumption will be independent of altitude. The thrust, at a given altitude, rises rapidly with an increase in the flight speed due to increased pressure ratio but at too high Mach numbers the losses in inlet diffuser reduce the thrust. Ramjets have highest thrust per unit weight amongst air breathing engines and is only next to rockets in this respect. Though a ramjet can operate at subsonic velocities just below the sonic velocity, it is most efficient at high velocities of about 2400–6000 km/h and at very high altitudes. Max. thrust

Cruise thrust

Mach number

Fig. 7.3 Performance of a ramjet engine

7.3.3

Advantages, Disadvantages and Characteristics

In this section, the various advantages and disadvantages of ramjet are enumerated. Advantages of ramjet (i) Ramjet is very simple and does not have any moving part. It is very cheap to produce and requires almost no maintenance. (ii) Due to the fact that a turbine is not used to drive the mechanical compressor, the maximum temperature which can be allowed in ramjet is very high, about 2000◦ C as compared to about 900◦ C in turbojets. This allows a greater thrust to be obtained by burning fuel at air-fuel ratio of about 13:1, which gives higher temperatures.

220

Gas Turbines

(iii) The specific fuel consumption is better than other gas turbine power plants at high speed and high altitudes. (iv) Theoretically there seems to be no upper limit to the flight speed of the ramjet. Disadvantages of ramjet (i) Since the compression of air is obtained by virtue of its speed relative to the engine, the take-off thrust is zero and it is not possible to start a ramjet without an external launching device. (ii) The engine heavily relies on the diffuser and it is very difficult to design a diffuser which will give good pressure recovery over a wide range of speeds. (iii) Due to high air speed, the combustion chamber requires flame holder to stabilize the combustion. (iv) At very high temperatures of about 2000◦C dissociation of products of combustion occurs which will reduce the efficiency of the plant if not recovered in nozzle during expansion. 7.3.4

Basic Characteristics and Applications

The basic characteristics of the ramjet engine can be summarized as follows: (i) It is a simple engine and should be adaptable for mass production at relatively low cost. (ii) It is independent of fuel technology and a wide range of liquid, and even solid fuels can be used. (iii) Its fuel consumption is comparatively very large for its application in aircraft propulsion or in missiles at low and moderate speeds. (iv) Its fuel consumption decreases with flight speed and approaches reasonable values when the flight Mach number is between 2 and 5 and therefore, it is suitable for propelling supersonic missiles. Due to its high thrust at high operational speed, it is widely used in high-speed military aircrafts and missiles. Subsonic ramjets are used in target weapons, in conjunction with turbojets or rockets for getting the starting torque. 7.4

THE PULSE JET ENGINE

Pulse jet engine (Fig. 7.4) is very similar to ramjet engine in construction except that in addition to the diffuser at intake, combustion chamber and exhaust nozzle, it has mechanically operated flapper valve grids which can allow or stop air flow in the combustion chamber. Thus pulse jet is an

Jet Propulsion Cycles and Their Analysis

221

intermittent flow, compressorless type of device with minimum number of moving parts. Pulse jet was the power plant of German V-1 bomb popularly known as ‘Buzz Bomb’ first used in World War II in 1944. Inlet diffuser Valve grid

Combustion chamber Tail piece

c

Fuel

Spark plug

Fig. 7.4 The pulse jet engine The basic features of the pulse jet engine are illustrated in Fig. 7.4. It consists essentially of the following parts: (i) a diffuser, (ii) a valve grid which contains springs that close on their own spring pressure, (iii) a combustion chamber, (iv) a spark plug, and (v) a tail pipe or discharge nozzle. The theoretical and actual p-V diagrams of the pulse jet engine are shown in Fig. 7.5 and Fig. 7.6 respectively. The operation of the pulse jet is as follows: During starting compressed air is forced into the inlet which opens the spring loaded flapper valve grid; the air enters combustion chamber into which fuel is injected and burnt with the help of a spark plug. Combustion occurs with a sudden explosion process 2–3 in Fig. 7.5, i.e., the combustion is at constant-volume instead of at constant-pressure as in other propulsive devices. The pulse jet cycle is more near to Otto cycle. Ram action can also be used to increase the pressure of the cycle (Fig. 7.5). The function of the diffuser is to convert the kinetic energy of the entering air into static pressure rise by slowing down the air velocity. When a certain pressure difference builds up across the valve grid, the valves will open. This makes the fresh air to enter the combustion chamber, where fuel is mixed with the air and combustion starts. To start the combustion initially the spark plug is used. Once the combustion starts it proceeds at constant-volume. Thereby, there is a rapid increase in pressure, which causes the valve to close rapidly. The products of combustion surges towards the nozzle. They expand in the nozzle and escape into the atmosphere

222

Gas Turbines

3 3’ p

Isentropic

2 4

1 V

Fig. 7.5 Theoretical pulse jet cycle on p-V diagram

3 p 1

Expansion and acceleration of gases

2

4 Pulsating gas column leaving the system V

Fig. 7.6 Actual pulse jet cycle on p-V diagram

with a higher velocity so that the exit velocity is much higher than the inlet velocity. Thus, the rate of momentum of the working fluid is changed so as to cause a propulsive thrust. Since, the combustion process causes the pressure to increase, the engine can operate even at static conditions once it gets started. When the combustion products accelerate from the chamber, they leave a slight vacuum in the combustion chamber. This, in turn, produces sufficient pressure drop across the valve grid, allowing the valves to open again. A new charge of air enters the combustion chamber which is mixed with fuel that flows continuously. The fresh fuel-air mixture is ignited by the charge leaving and/or by residual charge. New charge need not be ignited with a spark plug again. Proper design allows the duct to fire at a given pulse rate when the fuel flows continuously. The frequency of pulsation is determined by the duct shape and working temperature and may be as high as 500 cycles per second in very small units. The thrust of the pulse jet engine is proportional to the average mass flow rate of gases through the engine multiplied by its increase in velocity. Like ramjet engines, the maximum operating altitude of the pulsejet is also limited by air density consideration. Unlike the ramjet, the pulse jet engine develops thrust at zero speed. A high initial launching velocity,

Jet Propulsion Cycles and Their Analysis

223

however, improves its performance. The thrust of the engine, of course, decreases with altitude and does not continue to increase with increasing flight speeds up to supersonic range as is true of ramjet. The maximum flight speed of the pulse jet engine is limited by aerodynamic consideration to below 800 km/h. The pulse jet engine is simple and cheap for subsonic flight and well adapted to pilotless aircraft. The use of the pulse jet engine is restricted to pilotless aircraft due to its severe vibration and high intensity noise. The pulse jet has low thermal efficiency and limited speed range. In early designs the efficiency obtained was about 2 to 3 percent with a total flight life of 30 to 60 minutes. The maximum operating speed of the pulse jet is seriously limited by two factors: (i) It is not possible to design a good diffuser at high speeds. (ii) The flapper valves, the only mechanical part in the pulse jet, also have certain natural frequency and if it coincides with the cycle frequency resonance occurs and the valve may remain open and no compression will take place. Also, as the speed increases it is difficult for the air to flow back. This reduces the total compression pressure as well as the mass flow of air which results in inefficient combustion and lower thrust. The reduction in thrust and efficiency is quite sharp as the speed increases. At subsonic speeds it might not operate as the speed is not sufficient to raise the air pressure to the required combustion pressure. 7.4.1

Advantages and Disadvantages

Advantages of pulse jet (i) This is a very simple device next to ramjet and is light in weight. It requires very small and occasional maintenance. (ii) Unlike ramjet, it has static thrust because of the compressed air starting ; thus it does not need a device for initial propulsion. The static thrust is even more than the cruise thrust. (iii) It can run on almost any types of liquid fuels without much effect on the performance. It can also operate on gaseous fuel with a little modifications. (iv) Pulsejet engine is relatively cheap. Disadvantages of pulse jet (i) The biggest disadvantage is very short life of flapper valves and high rates of fuel consumption. The specific fuel consumption is as high as that of ramjet.

224

Gas Turbines

(ii) The speed of the pulse jet is limited to a very narrow range of about 650–800 km/h because of the limitations in the aerodynamic design of an efficient diffuser suitable for a wide speed range. (iii) The operational range of the pulse jet is also limited in altitude range. (iv) The high degree of vibrations due to intermittent nature of the cycle and the buzzing noise has made it suitable for pilotless crafts only. (v) It has lower propulsive efficiency than turbojet engines. 7.4.2

Applications

Pulse jet is highly suited for bombers like the German V-1 and it has also been used in some helicopters, target aircrafts missiles, etc. 7.5

THE TURBOPROP ENGINE

It is a known fact that an higher thrust per unit mass flow of fuel can be obtained by increasing the mass flow of air which results in better fuel economy. This fact is utilized in a turboprop engine which is an intermediate between a pure jet engine and a propeller engine. Turboprop engine attempts to increase the air flow by using a propeller driven by the turbine in addition to a small thrust produced by the exhaust nozzle. The details of this engine is illustrated in Fig. 7.7. In this engine the turbine is designed so as to develop enough shaft power for driving a propeller which provides most of the propulsive thrust. It may be noted that the thrust produced due to jet action is quite small.

Diffuser Compressor

Fuel Turbine Nozzle

G

Reduction gear Propeller Shaft

Fuel

Combustion chamber

Fig. 7.7 The turboprop engine The engine consists of the following components: (i) a diffuser, (ii) a compressor, (iii) a combustion chamber, (iv) a turbine,

Jet Propulsion Cycles and Their Analysis

225

(v) an exhaust nozzle, (vi) a reduction gear, and (vii) a propeller. In the turboprop engine, the turbine extracts much more power because the turbine is to provide power for both the compressor and the propeller. When all of this energy is extracted from the high temperature gases, there is still a little energy left for producing jet thrust. Thus the turboprop engine derives most of its propulsive thrust from the propeller and derives only a small portion (10 to 20% depending upon the flight velocity) from the exhaust nozzle. Since the shaft speed of gas turbine engine is very much higher than that of a propeller, a reduction gear must be placed between the turbine shaft and the propeller to enable the propeller to operate efficiently. Still, as flight speed increases, the ratio of nozzle power to propeller power for maximum thrust tends to increase. It may be noted that the propulsive thrust developed is due to the following: (i) the propeller increases the air momentum, and (ii) the overall engine – from diffuser to nozzle – provides an internal momentum increase. The sum of these two thrusts is the total thrust developed by the engine. 7.5.1

Thermodynamic Cycle

The thermodynamic cycle of turboprop is similar to that of a pure jet engine except that now more energy is used in the turbine as seen from the T -s diagram of Fig. 7.8. About 80 to 90 per cent of the total energy is used in the turbine and only about 10 to 20 per cent is used in exhaust nozzle. The propeller produces its own thrust and thus the turboprop engine is essentially a two fluid stream engine. The optimum power ratio for turboprop engine can be determined as follows: Let the total expansion be divided into parts such that Total thrust = Nozzle thrust + Propeller thrust = m ˙ a (cj − ci ) + m ˙ a ci

where Δhnoz ΔhT ηtr ηT ηnoz m ˙a

= = = = = =

= m ˙a

˙a 2Δhnoz ηnoz − ci + m

= m ˙a

2Δhnoz ηnoz − ci +

ηT ΔhT ηtr

ηT ΔhT ηtr

enthalpy drop in the nozzle enthalpy drop in the turbine transmission efficiency of the propeller and gears turbine efficiency nozzle efficiency mass flow rate in turbine

(7.2)

226

Gas Turbines

3

4 T

4’

2 2’

1 s

Fig. 7.8 T -s diagram for a turboprop engine By assigning suitable values to ηT , ηnoz and ηtr this equation can be optimized for maximum thrust. 7.5.2

Performance

Turboprop engines combine in them the high take-off thrust and good propeller efficiency of the propeller engines at speeds lower than 800 km/h and the small weight, lower frontal area, and reduced vibration and noise of the pure jet engine. Its operational range is between that of propeller engines and turbojets though it can operate in any speed upto 800 km/h. The power developed by the turboprop remains almost same at high altitudes and high speeds as that under sea-level and take-off conditions because as speed increases ram effect also increases. The specific fuel consumption increases with increase in speed and altitude. The thrust developed is high at take-off and reduces at increased speed. 7.5.3

Advantages and Disadvantages

Advantages of Turboprop (i) Turboprop engines have a higher thrust at take-off and better fuel economy. (ii) The frontal area is less than propeller engines so that the drag is reduced. (iii) The turboprop can operate economically over a wide range of speeds ranging from low speeds where pure jet engine is uneconomical to high speeds of about 800 km/h where the propeller engine efficiency is low.

Jet Propulsion Cycles and Their Analysis

227

(iv) It is easy to maintain and has lower vibrations and noise. (v) The power output is not limited as in the case of propeller engines. (vi) The multishaft arrangement allows a great flexibility of operation over a wide range of speeds. Disadvantages of Turboprop (i) The main disadvantage is that at high speeds, due to shocks and flow separation, the propeller efficiency decreases rapidly, thereby, putting up a maximum speed limit on the engine. (ii) It requires a reduction gear which increases the cost and also consumes certain amount of energy developed by the turbine in addition to requiring more space. 7.5.4

Applications

(i) The turboprop engine is widely used in commercial and military aircraft due to its high flexibility of operation and good fuel economy. It is likely to eliminate propeller engines from moderate power and speed aircraft. 7.6

THE TURBOJET ENGINE

The two pilotless air breathing engines described, viz., ramjet and pulsejet are simple in construction. However, their application is limited and, to date, they have not been used very extensively. The most common type of air breathing engine apart from turboprop is the turbojet engine. The important features are shown in Fig. 7.9. Diffuser Compressor

0

1

Fuel

Turbine

2

Nozzle

4 3 Fuel Shaft Combustion chamber

Fig. 7.9 The turbojet engine This engine consists of the following components: (i) a diffuser, (ii) a mechanical compressor, (iii) a combustion chamber,

5

228

Gas Turbines

(iv) a mechanical turbine, and (v) an exhaust nozzle. The function of the diffuser is to convert the kinetic energy of the entering air into a static pressure rise which is achieved by the ram effect. After this air enters the mechanical compressor. The compressor used in a turbojet can be either centrifugal type or axial flow type. The use of a particular type of compressor gives the turbojet typical characteristics. The centrifugal compressor produces a high pressure ratio of about 4:1 to 5:1 in a single stage and usually a doublesided rotor is used to reduce the engine diameter. The turbojet using a centrifugal compressor has a short and sturdy appearance. The advantages of centrifugal compressor are high durability, ease of manufacture and low cost, and good operation under adverse circumstances such as icing and when sand and small foreign particles are inhaled in inlet duct. The axial flow compressor is more efficient than the centrifugal type and gives the turbojet a long, slim, streamlined appearance. The engine diameter is reduced which results in low aircraft drag. A multistage axial flow compressor can develop a pressure ratio as high as 6:1 or more. The air handled by it is more than that handled by a centrifugal compressor of the same diameter. The demand for increased power has led to the use of split or twospool axial flow compressor. A very high pressure ratio of about 9:1 to 13:1 is obtained by using a high pressure and a low pressure rotor driven by separate shafts. The use of high pressure ratio gives very good specific fuel consumption (0.75 kg/kg thrust per hour) and use of two rotors allows greater efficiency because firstly, the high pressure rotor can be governed for speed and secondly, the low pressure rotor can be allowed to run at a speed giving maximum efficiency. The turbojets having centrifugal compressor have about 20 per cent weight advantage over the axial flow turbojets. Thrust per unit weight is more for the first type while thrust per unit diameter is more the second type. The axial flow turbojets have about 6 to 8 per cent less specific fuel consumption. After the compressor air enters to the combustion chamber, the fuel nozzles feed fuel continuously and continuous combustion takes place at constant-pressure. The high pressure, high temperature gases then enter the turbine, where they expand to provide enough power output from the turbine. The turbine is directly connected to the compressor and all the power developed by the turbine is absorbed by the compressor and the auxiliaries. The main function of the turbine is to provide power, to drive the compressor. After the gases leave the turbine they expand further in the exhaust nozzle, and are ejected into the atmosphere with a velocity greater than the flight velocity thereby producing thrust for propulsion. Current turbojet engines operate with compressor pressure ratios between 6 and 16, and with turbine inlet temperatures of the order of 1200 K.

Jet Propulsion Cycles and Their Analysis

229

The corresponding speed of the exhaust jet when propelling an aircraft at 900 km per hour (250 m/s) is of the order of 500 m/s. Like the ramjet engine, the turbojet engine is a continuous flow engine. Here a compressor run by a turbine is used to provide additional pressure rise which is not available in a ramjet engine. Since this engine has a separate mechanical compressor, it is capable of operating even under static conditions. However, increase in flight velocity improves its performance because of the benefit of ram pressure achieved by the diffuser. Because of turbine material limitations, only a limited amount of fuel can be burnt in the combustion chamber. The exhaust products downstream of the turbine still contain a considerable amount of excess oxygen. Additional thrust augmentation can be achieved from the turbojet engine by providing an afterburner in which additional fuel can be burnt. Properly designed afterburner can greatly increase the temperature and hence the velocity of exhaust gases – providing thrust augmentation. Of the four air-breathing engines, discussed so far only the turbojet and turboprop engines have been applied as the propulsion device for piloted aircraft. Turbojet engine is eminently suited for propelling aircraft at speeds above 800 km/h. As the flight speed is increased, the ram pressure increases rapidly and at supersonic speeds, (Mach number = 3) the characteristics of turbojet engine tend to merge with those of the ramjet engine. 7.6.1

Thermodynamic Cycle

Figures 7.10 and 7.11 show the basic thermodynamic cycle of a turbojet engine of p-V and T -s diagrams. This is Joule or Brayton cycle. In the analysis of turbojet cycle following assumptions are made: (i) There is no loss of pressure in the combustion chamber. (ii) The specific heat is constant. (iii) Power developed by the turbine is just sufficient to drive the compressor. At the inlet to the diffuser air enters with a velocity equal to the forward speed of the aircraft. In diffuser air velocity is decreased and pressure is increased. In the ideal case the pressure will rise such that the velocity at the exit of the diffuser is zero. However, in actual practice the air will have a velocity of about 60–90 m/s at diffuser exit. If ηd is the efficiency of the diffuser then the total pressure at the end of diffusion process is given by p01 p0

=

γ−1 2 M 1 + ηd 2

γ γ−1

From the diffuser air goes into a compressor. If ηC is the compressor efficiency and pp02 the pressure ratio, we get 01

Gas Turbines

Combustion

2

3 Pressure loss in combustion chamber

ine

r esso

rb Tu

pr Com

230

1

4 No

Diffuser

zzl e

5

0

V Fig. 7.10 p-V diagram of a turbojet engine

03 n

bustio

02’ 02 T

Com

04

Compressor

p

Turbine Nozzle

04’

01’

01

5 5’

Diffuser 0

s Fig. 7.11 T -s diagram of a turbojet engine

Jet Propulsion Cycles and Their Analysis

ηC

=

h02 − h01 h02 − h01

h02 − h01

=

1 (h02 − h01 ) ηC

=

Cp T01 ηC

or

=

T02 −1 T01

231

Cp (T02 − T01 ) ηC =

Cp T01 ηC

p02 p01

γ−1 γ

−1

The heat input per kg of air in the combustion chamber is given by q

=

h03 − h02

We assume that there is no pressure loss in the combustion chamber so that the pressure remains constant and full pressure is available for expansion in the turbine and nozzle. Since we have assumed that turbine and compressor work are same. h02 − h01

=

h03 − h04

If ηT is the turbine efficiency and h03 − h04

=

p04 p03

p04 p03

ηT Cp T03 1 −

is the turbine pressure ratio, γ−1 γ

By equating the turbine and compressor work, we get Cp T01 ηC

p02 p01

γ−1 γ

− 1 = ηT Cp T03 1 −

p04 p03

γ−1 γ

(7.3)

From Eq. 7.3, knowing the ram compression and the compressor pressure ratio, the required turbine pressure ratio to produce a power equal to that absorbed by the compressor can be obtained and from the turbine pressure ratio the nozzle pressure can be obtained. If ηnoz is the nozzle efficiency then, ηnoz

=

h04 − h5 h04 − h5

(7.4)

It is assumed that there is no loss in passing the gas from turbine exhaust to the nozzle. It should be noted that in Eq. 7.4, h5 is used instead of total enthalpy h05 because the exhaust nozzle efficiency is an indication of the percentage of total energy converted into velocity energy. Therefore, h04 − h5

ηnoz Cp T04 1 −

=

p5 p04

γ−1 γ

The exit velocity at the nozzle end can be obtained by writing energy balance equation. h04 − h5

=

c2j 2

232

Gas Turbines

c2j 2

cj

γ−1 γ

p5 p04

=

ηnoz Cp T04 1 −

=

. / / 02η C T 1 − noz p 04

p5 p04

γ−1 γ

Thrust developed =

=

m ˙ a [(1 + f )cj − ci ] . ⎧ / ⎨ / m ˙ a (1 + f )02ηnoz Cp T04 1 − ⎩

p5 p04

γ−1 γ

− ci

⎫ ⎬

(7.5)

⎭

From the above, thrust specific fuel consumption and also the thermal efficiency can be calculated. 7.6.2

Performance of a Turbojet Engine

With the help of the above analysis it is possible to estimate the performance of a turbojet engine taking into account the component efficiencies and other parameters. Figure 7.12 shows the thrust specific fuel consumption for various compressor pressure ratios at two different Mach numbers. It is evident that for a given Mach number there is only one compressor pressure ratio which gives best fuel economy for given values of component efficiencies and maximum allowable temperature. As the pressure ratio increases with a given maximum temperature fuel consumption decreases to a minimum. After that further increase in pressure ratio will not improve the fuel economy until maximum temperature is not raised. For a given pressure ratio higher maximum temperature will result in more thrust. Maximum thrust per unit mass of fuel is achieved at a lower pressure ratio than the one which produces minimum specific fuel consumption at the same turbine inlet temperature. In addition to the above parameters, three more variables – flight speed, altitude (inlet temperature and pressure), and fuel flow rate – greatly affect the performance of a jet engine. The turbojet is almost a constant thrust engine. The specific fuel consumption based on thrust power reduces because with almost constant thrust, the thrust power increases as shown in Fig. 7.13. Therefore, the maximum speed is the most efficient operational point for the turbojet. The static thrust of such engines is very low as compared to propeller engine aircraft for which cruise thrust is about 60 per cent of the takeoff thrust. The thrust at first decreases with increase in speed because the velocity cj in the thrust equation increases. After a minimum value, the thrust starts increasing (see Figs. 7.14 and 7.15) due to increased ram compression at higher speeds. As the altitude increases, the thrust decreases due to decrease in density, pressure and temperature of the air (see Fig. 7.15). However, the rate of

Jet Propulsion Cycles and Their Analysis

233

TSFC

M = 0.66

M = 0.33

r

Fig. 7.12 Thrust specific fuel consumption vs compressor pressure ratio for a turbojet engine

TP

TP

TPSFC

TPSFC

Flight speed Fig. 7.13 Thrust specific fuel consumption and thrust power vs flight speed

decrease of thrust is less than the rate of decrease of density with altitude because some loss due to reduced density is compensated by lesser drag. The thrust is maximum at sea level. Due to considerable reduction in drag (at an altitude of 8000 m the drag is reduced to less than 25 per cent of sea level drag), the turbojet is most efficient when flown at high altitudes and at relatively high speeds. The fuel consumption on fuel mass per km of travel increases with speed as power output also increases with speed. The operational range of turbojet engine is about 800 to 1100 km/h and the specific fuel consumption is about 1.0 to 1.5 kg/thrust h at cruising speeds and are still greater at lower speeds. The altitude limit is about 10000 m.

234

Gas Turbines

r lle

pe ro hp

itc

ep

bl

Relative thrust

sta ju

Ad

Jet engine

Flight speed (km/h)

Fig. 7.14 Thrust vs speed for propeller driven and turbojet engine

Thrust

Sea level 3000 m 6000 m 9000 m 12000 m

Flight speed

Fig. 7.15 Effect of altitude on thrust at maximum rpm 7.6.3

Advantages and Disadvantages of Turbojet

Advantages of Turbojet (i) The power to weight ratio of a turbojet is about 4 times that of a propeller system having reciprocating engine. (ii) It is simple, easy to maintain and requires lower lubricating oil consumption. Furthermore, complete absence of liquid cooling results in reduced frontal area. (iii) There is no limit to the power output which can be obtained from a turbojet while the piston engines have reached almost their peak power and further increase will be at the cost of complexity and greater engine weight and frontal area of the aircraft.

Jet Propulsion Cycles and Their Analysis

235

(iv) The speed of a turbojet is not limited by the propeller and it can attain higher flight speeds than engine propeller aircrafts. Disadvantages of Turbojet (i) The fuel economy at low operational speeds is extremely poor. (ii) It has low take-off thrust and hence poor starting characteristics. 7.7

THRUST AND THRUST EQUATION

Having gone through the details of various propelling devices, their thermodynamic cycles and performances, it is worthwhile to discuss some of the basic laws of thrust production and the factors which affect the performance of the engine. Let us consider the control volume of a schematic propulsive device shown in Fig. 7.16. A mass m ˙ i of air enters the control volume with a velocity ci and pressure pi and the products of combustion of mass m ˙j leaves the control volume with a velocity cj and pressure pj . The flow is assumed to be steady and reversible outside the control volume, the pressure and velocity being constant over the entire control volume except that at the exhaust area Aj . Force F is the force necessary to balance the thrust produced due to change in momentum of the fluid as it passes through the control volume. Fuel

Aj p j cj mj

Ai p i ci mi

Fig. 7.16 General schematic diagram of a propulsive device If pa is the atmospheric pressure, then writing the momentum equation, we get m ˙ j cj − m ˙ i ci

= F + (pi − pa )Ai − (pj − pa )Aj

or thrust, F

= (m ˙ j cj − m ˙ i ci ) + (pj − pa )Aj − (pi − pa )Ai

We have by mass balance,

(7.6)

236

Gas Turbines

m ˙j

= m ˙ i+m ˙f

(7.7)

where m ˙ j, m ˙ i and m ˙ f are the mass flow rates of exhaust gases, air, and fuel respectively. If f m ˙j

= Fuel air ratio

=

m ˙f m ˙i

= m ˙ i (1 + f )

Thrust = m ˙ i [(1 + f )cj − ci ] + (pj − pa )Aj − (pi − pa )Ai Momentum thrust

(7.8)

Pressure thrust

From Eq. 7.8 it is clear that the net thrust produced is made up two parts, viz., momentum thrust and the pressure thrust . If the exhaust velocity cj from the control volume is subsonic, then pj ≈ pa and also pi ≈ pa so that the pressure thrust is quite small. Similar is the case for propeller engines. For supersonic exhaust velocity the pressure pj may differ from pa . However, the pressure thrust developed is so small as compared to the momentum thrust that it can safely be neglected for simple calculations and the net thrust is given by Thrust

=

m ˙ i [(1 + f )cj − ci ]

(7.9)

The thrust, given by Eq. 7.9 can be increased by increasing the mass flow or increasing the velocity of the exhaust jet for a given ci . Equation 7.9 has been derived for a simple control volume shown in Fig. 7.16, but is equally applicable to an aircraft flying at a forward speed ci . In the latter case the velocities are considered relative to the aircraft. The air has a velocity ci equal to aircraft forward speed, relative to the engine. Thus, we see that a large amount of thrust can be obtained either by propelling a large mass of air and increasing its velocity by a small amount, or by increasing the velocity of a small mass of air to a high value. In the case of aircraft power plants the fuel-air ratio f is very small (about 0.01 to 0.02) and hence the mass of fuel can be neglected safely without causing much error in the performance calculations. The thrust is, then, given by F

=

m ˙ i (cj − ci )

the effective speed ratio, α, is given by ci α ≡ cj and F

=

m ˙ i ci

(7.10)

(7.11)

1 −1 α

(7.12)

Jet Propulsion Cycles and Their Analysis

237

The product m ˙ i cj is called the gross thrust and m ˙ i ci is called the inlet drag or inlet momentum. Equation 7.12 is also applicable to turboprop engines. The thrust developed by the engine overcomes the drag on the aircraft and in doing so it develops power, called the thrust power which is given by PT

=

F ci

=

m ˙ i (cj − ci )ci

(7.13)

Note that turboprop engine is rated in kilowatts while turbojet and ramjet are rated on the basis of thrust developed.

7.8

SPECIFIC THRUST OF THE TURBOJET ENGINE

The thrust per kg of air flow is known as specific thrust or specific impulse. Assuming m ˙ a is the mass flow rate at inlet, the specific impulse, Isp , is given by Isp

=

F m ˙a

=

(cj − ci )

In terms of the effective speed ratio, α = Isp

=

ci cj ,

cj (1 − α)

(7.14)

Eq. 7.14 becomes (7.15)

The specific thrust is a criterion of the size of engine required for producing a given total thrust. It is apparent from Eq. 7.15 that, to achieve a large specific thrust, the speed ratio α must be kept small. Also this equation shows that if the effective jet velocity cj is maintained constant, the specific thrust, Isp , decreases with increase in the flight speed ci . The maximum value of Isp under the above mentioned conditions is realized when the flight speed is zero. The magnitude of the specific thrust at ci = 0 is of particular significance because it is the measure of the thrust available for take off. The specific static thrust, (Isp )st is given by (Isp )st

=

cj

(7.16)

Hence to develop a large specific thrust at take off the effective jet speed should be as large as possible. Let Δhnoz be the enthalpy drop in the nozzle corresponding to the velocity cj , then c2j 2

= Δhnoz

cj

=

2Cp ΔT

(7.17)

238 7.9

Gas Turbines

EFFICIENCIES

To evaluate the performance of a turbojet engine the efficiencies of its components must be known. Here, we shall define the various efficiencies associated with jet propulsion plants which in turn will indicate the state of development of the components of the plant. 7.9.1

Inlet Diffuser or Ram Efficiency

h

p

h01= h 01’

01

p

01

’

=

=

co n

co ns tan t

sta n

t

In all jet propulsion plants, except turboprop and rocket engines, a diffuser is used at the inlet to convert kinetic energy into static pressure rise. This compression process is essentially adiabatic but it cannot be considered as reversible since fluid friction is present. The most widely used efficiency of the inlet jet process is based upon the pressure rise that actually takes place; compared to the pressure rise which would have taken place had the process been isentropic. By definition, this efficiency is (refer Fig. 7.17)

c 21

s

on

2 pa

Ideal

=c

Actual t tan

s

Fig. 7.17 h–s diagram for inlet diffuser

ηram

=

p01 − pa p01 − pa

=

p01 pa p01 pa

−1

−1

(7.18)

In an ideal diffuser, the process will be isentropic and the final pressure will be given by p01 pa But

=

T01 Ta

γ γ−1

(7.19)

Jet Propulsion Cycles and Their Analysis

T01

= Ta +

c2i 2Cp

239

(7.20)

where ci is the forward speed of the engine. γ ⎡ ⎤ γ−1 c2i T + a p01 2Cp ⎦ = ⎣ = pa Ta

c2i 1+ 2Cp Ta

γ γ−1

(7.21)

or

p01

= pa

c2i 1+ 2Cp Ta

γ γ−1

(7.22)

But due to losses in friction and shock, the actual pressure will not rise to the value given by Eq. 7.22. From Eq. 7.18 actual pressure rise is p01 − pa

=

ηram (p01 − pa )

(7.23)

or p01

=

pa + ηram pa

γ γ−1

c2i 1+ 2Cp Ta

− pa

or p01

=

pa 1 + ηram

!

γ γ−1

c2i 1+ 2Cp Ta

−1

"

(7.24)

Efficiencies of the order of 85 to 90 percent are obtained with well designed diffusers at subsonic speeds. The diffusion process is dependent more on the type of aircraft and ducting than it is on the engine itself. Hence, it should not be considered as an engine characteristic, although it affects the overall engine cycle. Equation 7.18 has one serious disadvantage. At low flight speeds p01 −pa might actually become negative as a result of inlet duct pressure drop, whereas if the airplane is standing still or if the engine is on test bed, p01 − pa is equal to zero causing Eq. 7.18 to become negatively infinite. A more consistent method of expressing diffuser friction losses is in the form of a percentage pressure drop or pressure coefficient. Referring to Fig. 7.17, the diffuser pressure drop coefficient is expressed by pdc 7.9.2

=

p01 − p01 p01

=

1−

p01 p01

(7.25)

Thermal Efficiency of the Turbojet Engine

Thermal efficiency of a propulsive device is an indication of the degree of utilization of energy in fuel in accelerating the fluid flow and is defined as the ratio of propulsive power furnished to exhaust nozzle to the heat

240

Gas Turbines

supplied and is given by Thermal efficiency =

Propulsive power Fuel flow rate × Calorific value of fuel ηth

=

P m ˙ a Qi

(7.26)

where Qi = f CV = heat supplied to the engine per kg of air and f = m ˙ f /m ˙ a is the fuel-air ratio and CV is the calorific value of the fuel. The propulsive power, P , is given by the net increase in the energy of the working fluid between inlet and exit, viz., stations 0 and 5 (refer Fig. 7.9). Hence, P

=

1 (m ˙ a+m ˙ f )c2j − m ˙ a c2i 2

(7.27)

Assuming m ˙ =m ˙ a+m ˙f ≈m ˙ a , the Eq. 7.27 reduces to P

=

m ˙a 2 c − c2i 2 j

(7.28)

Introducing α = ci /cj and substituting P in Eq. 7.26, we get ηth

=

m ˙ a c2j (1 − α2 ) 2m ˙ f CV

=

c2j 1 − α2 2f CV

(7.29)

If h02 and h03 are specific stagnation enthalpies of the working fluid entering and leaving the combustion system, and ηcomb denotes the combustion efficiency, then by referring Fig. 7.18,

Qi

= f CV

=

h03 − h02 ηcomb

(7.30)

Substituting the value of Qi in Eq. 7.26 we get ηth

=

ηcomb c2j 1 − α2 2(h03 − h02 )

(7.31)

If we assume that the mean specific heat remains constant during process 2-3, then ηth

=

ηcomb c2j 1 − α2 2Cp (T03 − T02 )

(7.32)

For constant values of altitude and turbine inlet temperature T03 , the enthalpy added by the combustion process (h03 − h02 ) is also constant. It is seen from the Eq. 7.32, that the thermal efficiency, ηth , reduces to zero

Jet Propulsion Cycles and Their Analysis

Turbine 03

h03

WT

04 h

Burner

q

Compressor

241

p 04

c 2j / 2

04’ 5’

5

Nozzle (jet)

02

p

a

p=

’

p

01

p 01

01

c2i /2

01’

5

02’

WC

0

Diffuser (ram) s

Fig. 7.18 h–s diagram for turbojet engine when α = 1, i.e., when cj = ci . That conclusion is logical because when α = 1, the engine develops no thrust and no energy is added to the fluid flowing through the engine. On the other hand when α = 0, i.e., when ci = 0 the thermal efficiency of the engine depends entirely on the jet velocity cj . It should be noted that Eq. 7.32 applies only to turbojet, ramjet and pulse jet engines and does not apply to propeller engines and rocket engines. For propeller engines thermal efficiency is defined as ηth =

Brake power Fuel rate × CV of fuel

(7.33)

In the case of turboprop engine shaft power as well as thrust power both are used for propulsion. However, the shaft power is considerably larger than the thrust power and it is usual to define thermal efficiency on the equivalent shaft power basis such that the thrust power due to exhaust jet is also included. For calculating thrust power some suitable velocity for the aircraft is selected. 7.9.3

Propeller Efficiency

The propeller produces thrust power by accelerating the air. The propeller itself is driven by the engine. The efficiency of the propeller is defined as the ratio of the thrust power to the shaft power. Propeller efficiency

=

Thrust power Shaft power

=

F × ci sp

(7.34)

In the case of turboprop engine the thrust power developed by the exhaust is also considered.

242 7.9.4

Gas Turbines

Transmission Efficiency

In many cases the engine or the turbine output cannot be directly applied to the propeller some form of transmission is involved between the engine and the propeller in the form of a reduction gear. The main reason for providing reduction gear in the case of a turboprop engine is high rotational speed of the turbine at which the propeller cannot be rotated efficiently. In addition to this some layout problems always occur. Due to friction and other losses the output from the transmission system is always less than input to it and the transmission efficiency is defined as Efficiency of transmission =

Output of the transmission Input to the transmission

(7.35)

Measures of the thermal efficiency of an engine for a given air speed and jet speed are; the thrust specific fuel consumption (T SF C) for cycle efficiency and the thrust power specific fuel consumption for overall thermal efficiency (T P SF C) by definition, the thrust specific fuel consumption is T SF C

=

3600 × m ˙f kg of fuel/N thrust-hour F

(7.36)

where m ˙ f is in kg/s. Dividing Eq. 7.36 by m ˙ a , we get T SF C

=

3600 × f Isp

(7.37)

Similarly thrust power specific fuel consumption is T P SF C

=

3600 × m ˙f kg/thrust kW h TP

(7.38)

=

3600 × f (T P )s

(7.39)

The cycle efficiency of a turbojet engine remains relatively constant with an increase in air speed, whereas the propulsive efficiency increases. Thus, it may be expected that the thrust power specific fuel consumption would improve with air speed. Typical curves showing the trend of thrust, thrust power and fuel consumption are shown in Fig. 7.19. 7.9.5

Propulsive Efficiency

During the forward motion, the specific thrust, cj , is reduced by the inlet drag ci . The net thrust thus, is dependent not only on the power plant but also on the flight speed. The utilization of the gross thrust may be considered in terms of the propulsive efficiency. Thus, propulsive efficiency is the measure of the effectiveness with which the kinetic energy imparted to the fluid is transferred into useful work. The useful work done by the system is the product of the thrust and flight velocity, i.e., F × ci which is called the thrust power. The kinetic energy

243

Th

rus

t po

we

r

Jet Propulsion Cycles and Their Analysis

Thrust

TSFC TP

SF C

Air speed or flight speed (km/h)

Fig. 7.19 Typical performance curves for a jet engine imparted to the fluid is the difference between the kinetic energy at exit and the kinetic energy at inlet and is called propulsive power. The difference between propulsive power and thrust power is called the leaving losses. Now, Kinetic energy of air at inlet

c2i 2

=

m ˙a

=

m ˙ a (1 + f )

c2j 2

(7.41)

Propulsive power =

m ˙ a (1 + f )

c2j c2 −m ˙a i 2 2

(7.42)

(7.40)

Kinetic energy of gases at exit

Thrust power = Propulsive efficiency = =

(7.43)

F × ci Thrust power Propulsive power m ˙ a [(1 + f )cj − ci ]ci c2

m ˙ a (1 + f ) 2j − =

[(1 + f )cj − ci ]ci c2

(1 + f ) 2j −

c2i 2

c2i 2

(7.44)

As the turbojet engines operate at very low fuel-air ratios, f , (i.e., very high air-fuel ratios), f may be neglected, then ηp

=

2(cj − ci )ci c2j − c2i

244

Gas Turbines

=

ηp =

2ci c j + ci

2(cj − ci )ci (cj − ci )(cj + ci ) =

2α 1+α

(7.45)

(7.46)

ci . cj Differentiating 7.46 with respect to the speed ratio α, and equating it to zero, the conditions for the best propulsive efficiency can be obtained, which is α = 1. This means that the velocity of the gas jet, cj , is equal to the velocity of the aircraft. At this point, the propulsive efficiency becomes 100 per cent. However, from Eq.7.12, it is clear that at the condition of maximum propulsive efficiency, the thrust developed is zero and no useful work output can ever be obtained from the power plant. Hence, it is not practical to maximize propulsive efficiency of a jet engine. Other parameter such as thrust etc., must be considered for performance evaluation. Now from Eq.7.13, thrust power where α =

PT

=

m ˙ a c2j α(1 − α)

(7.47)

By differentiating the Eq. 7.47 with respect to α and equating to zero, it can be easily seen that the maximum thrust occurs at α = 0.5, i.e., when the speed of the gas jet at the exhaust is twice that of the air at inlet. The corresponding value of the propulsive efficiency is given by ηp

=

2α 1+α

=

2 × 0.5 1 + 0.5

=

0.667

(7.48)

Thus it is evident from the above analysis that for a turbojet, ramjet and pulsejet, the points of maximum propulsive efficiency and maximum thrust are different. Some compromise must be made to get reasonable thrust with a good propulsive efficiency. Turboprop engines are essentially two-fluid stream engines and this analysis cannot be applied to them. For each stream a separate equation must be written. In the case of propeller driven aircraft engine, the propulsive efficiency is based on the brake power (bp) of the engine ηp

=

Thrust power Brake power

(7.49)

It can be noted that ηp , of an engine increases with the aircraft speed. To put it in a nutshell large thrust per unit flow rate of working fluid cannot be obtained from a small light weight propulsion system unless the jet velocities are very large. To realize reasonable propulsive efficiencies the flight speed must be high. These are met usually in turbojet engines.

Jet Propulsion Cycles and Their Analysis

7.9.6

245

Overall Efficiency of a Propulsive System

The performance of the propulsion system is normally evaluated in terms of overall efficiency, η0 is defined as the ratio of rate at which useful propulsion work is done to the rate at which energy is supplied to the system. In other words the overall efficiency ηo of a propulsive device is the ratio of the useful work done to the chemical energy supplied in the form of fuel. ηo

=

Useful propulsive work Chemical energy supplied

=

Transmission output Engine output × × Energy input Transmission input (engine output) Useful propulsive work Input to the propulsive device (transmission output)

ηo

=

Thermal efficiency × Transmission efficiency × Propulsive efficiency

ηo

=

ηth × ηtr × ηp

(7.50)

It follows from the above that the overall effect of the propulsive device is made of a power plant converting some percentage of the chemical energy supplied to it into a form useful for propulsion (shaft power in case of propeller devices and acceleration of mass flow for jet devices). This includes the losses in the transmission, if used, (the transmission converting the energy in a more useful form, i.e., in acceleration of the fluid); the final propulsive device converting this energy into useful work output. This is shown in Fig. 7.20. Fuel Engine

Engine thermal efficiency

Transmission system

Propulsive device

Output of engine = Input to transmission

Output of transmission = Input to propulsion

Transmission efficiency

Propulsion efficiency

Overall efficiency =

Useful propulsion work Energy in fuel

Fig. 7.20 Block diagram for aircraft engine efficiencies

Useful propulsive work

246

Gas Turbines

7.10

PARAMETERS AFFECTING FLIGHT PERFORMANCE

As has been already mentioned that two parameters which affect the performance of the jet propulsion cycle are – the forward speed of the aircraft and the altitude at which the aircraft flies. These two do not enter into the analysis of shaft power cycle. The effect of the two parameters are discussed in the following sections. 7.10.1

Effect of Forward Speed

The forward speed affects the inlet pressure and temperature of the compressor. The inlet duct to the compressor acts as a diffuser. The air which enters the diffuser at flight speed is slowed down to the speed acceptable to compressor, and at the same time raising its pressure and temperature. This increase of pressure and temperature due to aircraft speed is known as ram effect, or simply ram. It becomes more and more prominent as the flight speed increases. For a given aircraft speed and ram efficiency, the ram pressure ratio increases as the ambient temperature decreases at high altitudes. The second effect of aircraft forward speed is in relation to propulsive efficiency. As flight velocity increases, the inlet drag also increases. If there were no ram effect, the net specific thrust, Isp , would decrease. This is because the jet velocity remains the same. With ram, the increase of inlet temperature reduces the gross-thrust to some extent. However, the increase of inlet pressure more than compensates for this. Because of this, the cycle pressure ratio increases without shaft work being necessary. The overall effect of forward speed on inlet drag and ram is to reduce somewhat the net specific thrust. 7.10.2

Effect of Altitude

The effect of altitude on a turbojet is by virtue of reduction of ambient pressure and temperature. The temperature of atmosphere varies considerably and continuously with location and time, so that the standard atmosphere is used for calculating the performance at various altitudes. The most general data used are those for International Standard Atmosphere or ICON atmosphere (International Commission On Navigation). It corresponds approximately to average values found in the middle latitudes. It is normalized by using a linear decrease of temperature or lapse rate of 1.98◦C per 300 m of altitude, starting with a ground level temperature of 15◦ C. With the temperature fixed, the pressure can be calculated according to the principles of hydrostatics using 1.03 bar as the ground level pressure. Table 7.1 provides values of International Standard Atmosphere. 7.11

THRUST AUGMENTATION

The poor take-off characteristic of the turbojet engine can be improved by augmenting the thrust. The thrust from a turbojet is given by

Jet Propulsion Cycles and Their Analysis

247

Table 7.1 International Standard Atmosphere. Altitude [m] 0 300 1500 3000 4500 6000 7500 9000 12000 15000 18000 21000

F

Temperature [◦ C] 15.0 13.0 5.1 − 6.8 − 16.7 − 26.6 − 36.5 − 46.6 − 56.3 − 56.3 − 56.3 − 56.3

=

Pressure [bar] 1.030 0.992 0.856 0.708 0.579 0.473 0.382 0.306 0.190 0.118 0.073 0.066

m ˙ a [(1 + f )cj − ci ]

in which the exhaust cj , is the function of the maximum temperature in the cycle. Higher the maximum temperature higher is the value of cj . Another method of increasing thrust is to increase the mass flow rate. Improved thrust results in shorter take-off distances, high climb rate and good manoeuvrability at high altitudes. The thrust augmentation can be effected by the following methods: (i) Burning of additional fuel in the tail pipe between the turbine exhaust section and entrance section of the exhaust nozzle. This method of thrust augmentation increases the jet velocity and is known as afterburning. The device used is called the afterburner. (ii) Injecting refrigerants, water or water-alcohol mixture at some point between inlet and exit sections of the air compressor. This method of thrust augmentation increases the mass flow rate and decreases the work of compression. (iii) Bleeding off air in excess of that required for stoichiometric combustion in the main combustion chamber – at the entrance section of the combustion chamber, and burning it with the stoichiometric fuel-air ratio in a separate one. The combustion products from the latter combustor are expanded in a separate auxiliary exhaust nozzle. The bled air is replaced by water which is injected in the main combustors. This method of thrust augmentation is known as the bleed burn cycle. 7.11.1

The Afterburner

This method of thrust augmentation is being widely applied for obtaining high thrust for short duration. It is known that turbine blade material

248

Gas Turbines

considerations limit the combustion chamber temperature rise. This in turn, limits the basic engine fuel- air ratio to values of about 0.017. As a result, the products of combustion leaving the turbine contain enough unutilized oxygen to support further combustion. Thus if a suitable burner is installed between the turbine and exhaust nozzle, a considerable amount of fuel can be burned in this section to produce temperatures entering the nozzle as high as 2000◦C. This increases the gas velocity, and hence provides a thrust increase. A boost of about 30 per cent can be obtained in this manner. However, the fuel consumption increases rapidly. For about 20 per cent thrust increase by use of reheat the overall fuel consumption may be increased by more than 100 per cent and this additional mass of fuel has to be carried by the turbojet. Therefore, it is used only for take-off or for high climbing rates and for a very short duration. Because of the temperature rise in the afterburner, there is a large increase in specific volume of gases, and to keep the pressure drop as small as possible, the tail pipe of the afterburning area is more than that of the normal engine. Furthermore, the afterburning engine must be equipped with a variable exit area exhaust nozzle so that by varying its area with the afterburner operating, the normal conditions at inlet to the afterburner will be unaffected. Another way of seeing the need for an increase of nozzle exit area is to examine the compressor performance chart. Figure 7.21 shows the sketch of the typical thrust variation with speed for an afterburner engine.

On Off

Thrust

After burner

Engine speed

100%

Fig. 7.21 Variation of thrust with speed As shown by the thrust versus rpm plot, the augmented thrust occurs at a constant 100% rpm; therefore, on the compressor performance chart (Fig. 7.22) the augmented thrust should lie on 100% rpm line. Point X represents operating point for 100% rpm operation without afterburner. Now, when the afterburner is tuned on without increasing the nozzle area, the flow resistance felt in the compressor will increase and the mass flow rate will decrease along the constant rpm line to produce operating point Y , which is in the stalled region. This type of operation of course, cannot be tolerated. For optimum conditions, the added flow resistance, caused by the specific volume increase in the afterburner, must be exactly balanced by

Jet Propulsion Cycles and Their Analysis

249

Operating line Surge line

p

Y

03

X

100%

p

02

95% N = 90% ma Fig. 7.22 Performance chart for an afterburner engine

the decrease in flow resistance brought about by opening the nozzle by the proper amount. Under these conditions, the operating point for afterburner operation will coincide with point X. Figure 7.23 shows the thermodynamics cycle of a turbojet engine fitted with afterburner on h-s plane. The process lines up to the turbine section are the same as for a basic engine. The afterburner process 5–6 is ideally a constant-pressure process, but the internal drag and momentum pressure loss produce a total pressure drop of about 5%. It is apparent that more thrust can be realized per kg of air by examining the relative magnitude of enthalpy drops across the normal nozzle and across the afterburner nozzle.

About 1900 o C

6

2 0

a

p

p

e

p

7’

4 5

3

05

p

03

or T About 850 o C

7

Nozzle process without afterburne

p 02 Atmospheric pressure s

Fig. 7.23 h–s diagram of a turbojet engine with afterburner

250 7.11.2

Gas Turbines

Injection of Water-Alcohol Mixture

This method of thrust augmentation is probably the simplest one to achieve. Mixture of water and alcohol or just water is injected at the combustion chamber or compressor inlet section through a series of suitable spray nozzles to produce an increase in thrust. The effect which produces the greater gain in thrust is the cooling effect within the compressor through water evaporation, which brings about rise in compressor discharge pressure. The rise in pressure can be seen from the compressor h-s diagram shown in Fig. 7.24. Process 1–2 represents normal dry compression process which is achieved by the turbine work ΔhT , being delivered to the compressor. Now, when the water alcohol injection system is turned on (usually at 100% rpm) the mixture as it evaporates within the compressor, cools the air to produce the compression process line 1–2 . The amount of work delivered by the turbine, with or without water injection, remains essentially the same, therefore, points 2 and 2 must line on the same enthalpy line. It is apparent that water alcohol injection process produces a higher compressor discharge pressure which in turn produces an increase in thrust. 2" p 02

p 02"

h or T

2

2’ Δ hT = Δ h C

1 s

Fig. 7.24 Compressor h–s diagram The factors which contribute to thrust augmentation by water-alcohol mixture injection can be summarized such as: (i) evaporative cooling which produces higher pressure and higher mass flow; (ii) additional mass of injected fluid; and (iii) possibility of burning of alcohol.

Jet Propulsion Cycles and Their Analysis

251

Mass flow rate (a) Axial compressor

e in

e

gl

lin

tin era

N/ T

0 all

0

Δp

N/ T

Op

Δp

X

St

Pressure ratio St all lin Op e era tin gl in e

X

Pressure ratio

The first factor determines whether a water injection system is practical on a given engine installation. The first factor also provides the key to determine where water injection produces the higher augmentation thrust ratios. Cooling will be accomplished by the injected fluid until the air at compressor discharge is saturated. Thus to get this saturation point, the amount of cooling increases as the thrust ratio is greater on a hot day than on a cold day and the augmented thrust ratio decreases with altitude and increases the flight Mach number. The water injected into an axial flow compressor tends to be centrifugally separated from the air. To eliminate this problem, water injection into the combustion chamber has been developed. The principle of operation of this method is illustrated on the compressor-performance diagram (Fig. 7.25).

Mass flow rate (b) Centrifugal compressor

Fig. 7.25 Compressor performance charts for typical axial and centrifugal flow machines The

√N T0

lines for an axial flow compressor are steeper near the surge

line than the √NT lines of the centrifugal compressor. Point 0 in the figures 0 represents the normal operating point at 100% speed. At this operating conditions the gas-flow is normally chocked, thus the only way to increase the thrust output of the engine is to raise the pressure upstream to turbine or nozzle. Now, when the water is injected into the combustion chamber, the flow resistance on the compressor increases, and since the engine speed is fixed, the compressor operating point shifts upward of constant √NT line until a 0 stabilized operating point X is reached. This point occurs when the pressure required by the new flow system is achieved. It is clear that if too much flow resistance is added, compressor stall will occur, therefore, it is necessary to maintain close control over the rate of water injection. The increase in flow resistance causes the pressure level of the engine, and consequently the thrust, to increase, but at the same time, the added compressor flow resistance produces a tendency to decrease the thrust by reducing the mass flow through the compressor. However, since the pressure is higher

252

Gas Turbines

at the turbine and nozzle, these components will handle more mass, the additional mass being the water injected into the burner. Effectively water injection into the combustion chamber produces a thrust increase by (i) increasing the compressor pressure ratio due to reduced compressor air flow, and (ii) increasing the total mass flow through turbine and exhaust nozzle. The magnitude of thrust increase is entirely dependent upon compressor operating characteristics. 7.11.3

Bleed Burn Cycle

Since in a turbine excess air is also present, a small percentage of highpressure air from the compressor is bled to an auxiliary combustion chamber by by-passing the turbine. In auxiliary combustion chamber the bled air is heated by an additional fuel supply to a higher temperature than would be permissible in the main engine on account of the limiting temperature at the turbine blades. The hot gases are then discharged forming an additional jet. A shut off valve is used to bring the engine to normal position. Water is injected into main combustion chamber to replace the mass of the extracted air, thus maintaining the discharge of main jet at the same level. This method is usually used for take-offs only due to high rate of liquid consumption which cannot be carried with engine during its flight. The augmented thrust ratio is highest for this method among the three methods. Of the three methods discussed, afterburning seems to be the only practical method for thrust augmentation during flight. The air bleed-off system gives maximum thrust augmentation but at the expense of large fuel consumption. This is used only when a large take-off thrust is needed. For smaller thrust augmentation ratio water injection is used because of simplicity and light weight. Afterburner is used for medium thrust augmentation ratio. Afterburner combined with water injection can also be used. Worked out Examples [Note : Take

γa −1 γa

= 0.286;

γg −1 γg

= 0.248]

7.1 A turbojet power plant uses aviation kerosene having a calorific value of 43 MJ/kg. The fuel consumption is 0.18 kg per hour per N of thrust, when the thrust is 9 kN. The aircraft velocity is 500 m/s the mass of air passing through the compressor is 27 kg/s. Calculate the air-fuel ratio and overall efficiency. Solution m ˙f

=

0.18 × 9000 3600

=

0.45 kg/s

Jet Propulsion Cycles and Their Analysis

Air-fuel ratio

=

27 0.45

Thrust power, PT

=

F × ci

=

9 × 500

Heat input, Q

=

0.45 × 43000

η

=

PT Q

=

4500 × 100 19350

=

23.26%

=

Ans

60 : 1

=

253

⇐=

4500 kW =

19350 kW

Ans

⇐=

7.2 A simple turbojet unit operates with a maximum turbine inlet temperature of 1200 K, a pressure ratio of 4.25:1 and a mass flow of 25 kg/s under design conditions, the following component efficiencies may be assumed. Isentropic efficiency of compressor Isentropic efficiency of turbine Propelling nozzle efficiency Transmission efficiency Combustion chamber pressure loss

: : : : :

87% 91.5% 96.5% 98.5% 0.21 bar

Assume Cpa = 1.005 kJ/kg K and γ = 1.4, Cpg = 1.147 kJ/kg K and γ = 1.33. Calculate the total design thrust. Also calculate the total thrust and specific fuel consumption taking into consideration the nozzle choking condition. Assume that the unit is stationary and at sea level, where the ambient conditions may be taken as 1 bar and 293 K. Assume air-fuel ratio of 50. Solution 03 04

04’

T

02 02’

01 s Fig. 7.26

5 5’

254

Gas Turbines

Actual temperature rise in the compressor T02 − T01

T02

=

T01 (γa −1)/γa r −1 ηC C

=

293 × 4.250.286 − 1 = 172.63 ◦ C 0.87

=

172.63 + 293

=

465.63 K

Since the work output of the turbine must be just sufficient to drive the compressor we have, ηtr Cpg (T03 − T04 )

=

Cpa (T02 − T01 )

T03 − T04

=

1.005 × 172.63 0.985 × 1.147

T04

=

1200 − 153.56

=

153.56 K

=

1046.44 K

Thus, the total head pressure ratio, rt across the turbine is, 1

T03 − T04

=

ηT T03 1 −

153.56

=

0.915 × 1200 × 1 −

=

1−

1 rt0.248 rt

=

(γ −1)/γg rt g

153.56 0.915 × 1200

1 0.8602

1 rt0.248

=

0.8602

=

1.835

1/0.248

The total head pressure at exit from the turbine is, p04

p02 − Δpcomb rt

=

p03 rt

=

4.25 − 0.21 1.835

=

=

2.20 bar

For isentropic flow through the nozzle, T04 T5

T5

γg −1 γg

=

p04 p5

=

2.20 1.0

=

1046.47 1.216

0.248

= 1.216 =

860.56 K

Jet Propulsion Cycles and Their Analysis

255

Now, the propelling nozzle efficiency is given by ηn

=

T04 − T5 T04 − T5

T04 − T5

=

ηn (T04 − T5 )

=

0.965 × (1046.47 − 860.56) = 179.38 K

T5

=

1046.47 − 179.38 = 867.07 K

c2j 2Cp

=

T04 − T5 = 179.38

c5

=

√ 2 × 1.147 × 1000 × 179.38

=

641.46 m/s

=

25 × 641.46 = 16036.6 N

Total design thrust/s

Ans

⇐=

Now, we should compare the operating pressure ratio, pp045 , of nozzle with the corresponding critical pressure ratio, pp04 . If pp045 > pp04c , then the nozzle c will choke and a convergent-divergent nozzle will be required. If pp04 < pp04c , 5 then the nozzle will not choke and only a simple convergent nozzle will be sufficient. In practice, a convergent nozzle is used, in which expansion takes place to the critical pressure, pc , which is greater than the ambient pressure, pa = p5 . Then in addition to momentum thrust there will be a pressure thrust, which is given by A(pc − pa ) where A is the cross-sectional area of the nozzle outlet. Thus, in this problem, p04 p5

=

p04 pa

=

2.2

Since the nozzle efficiency is given, critical pressure ratio is given by p04 pc

1

= 1−

1 ηn

γg −1 γg +1

γg γg −1

1

= 1−

1 0.965

1.33−1 1.33+1

4.03

= 1.895

p Therefore, in this problem, p04 is greater than the critical pressure ratio of a the propelling nozzle and as such the complete expansion within nozzle is not possible. Then, the static pressure at the nozzle exit will be

256

Gas Turbines

p5 = pc

=

p04

1 p04 /pc pc p04

γg −1 γg

T5 = Tc

=

T04

Tc

=

892.86 K

R

=

Cp (γ − 1) γ

=

284.6 J/kg K

cj

ρc

= 2.2 ×

1 = 1.160 bar 1.895 1 1.895

= 1046.44 ×

0.33 1.33

1147 × 0.33 1.33

=

√ 1.33 × 284.6 × 892.86

=

γRTc

=

=

581.34 m/s

=

1.160 × 105 284.6 × 892.86

3

=

0.456 kg/m

Let A be the required area of the propelling nozzle, then, 25 m ˙ = 0.0943 m2 = A = ρ c cj 0.456 × 581.34 If the nozzle is circular, then π 2 d = 0.0943 4 d

=

4 × 0.0943 π

=

0.3466

=

34.66 cm

Actual diameter would have to be slightly greater than this owing to the effect of the boundary layer at the nozzle outlet. Now, Pressure thrust

=

(pc − pa )A

=

(1.16 − 1) × 105 × 0.0943

=

25 × 583.11

=

1508.8 N

Momentum thrust =

14577.75 N

Therefore, total thrust per second, = sf c

=

=

14577.75 + 1508.8

=

16086.55 N

Ans

⇐=

m ˙a A/F Total thrust 25 × 3600 50 = 0.111 kg/N thrust h 16086.55

Ans

⇐=

Jet Propulsion Cycles and Their Analysis

257

7.3 In an aircraft power plant, the gas expands through a turbine to an intermediate pressure and on leaving the turbine it expands from intermediate pressure to the back pressure, generating kinetic energy for jet. All the power of turbine is absorbed in driving the associated compressor. In such a unit gas enters the turbine at 4.5 bar and 800◦ C and expands therein to 1.75 bar. Turbine absorbs 75% of the available enthalpy drop. Expansion occurs through the jet from exhaust condition to 1.03 bar. Assume that the velocity of gas entering the turbine and jet (nozzle) is negligible. There is no heat loss and conversion of kinetic energy is 100% of the available adiabatic enthalpy drop. Calculate (i) temperature of the gas entering the jet (nozzle), and (ii) velocity of the gas leaving the jet. Assume Cp = 1.05 kJ/kg K and γ = 1.38. Solution 03 04

04’

T

02

5 5’

02’

01 s Fig. 7.27

Temperature of gas entering the nozzle, T04

=

T03 1 − ηT 1 −

=

1073 ×

!

=

888.92 K

1 r

γ−1 γ

1 − 0.75 1 −

"

1 4.5 0.275 1.75

Ans

⇐=

Velocity leaving the nozzle, cj

=

2Cp (T04 − T5 )

=

. / / 02 × 1.05 × 888.92 × 1 −

=

503.2 m/s

1 1.75 0.2754 1.03 Ans

⇐=

258

Gas Turbines

7.4 The effective jet exit velocity from a jet engine is 2700 m/s. The forward flight velocity is 1350 m/s and the air flow rate is 78.6 kg/s. Calculate (i) thrust, (ii) thrust power, and (iii) propulsive efficiency. Solution α

=

ci cj

Thrust, F

=

m ˙ a (cj − ci )

=

78.6 × (2700 − 1350)

=

106110 N

=

F × ci

=

106110 × 1350

=

143.25 × 106 W

=

2α α+1

=

2 × 0.5 1.5

=

66.67%

Thrust power

ηp

=

1350 2700

=

=

0.5

Ans

⇐=

0.6667 Ans

⇐=

7.5 The following data apply to a turbojet aircraft flying at an altitude of 6.1 km where the ambient conditions are 0.458 bar and 248 K. Speed of aircraft : 805 km/h Pressure ratio of compressor : 4:1 Combustion chamber pressure loss : 0.21 bar Turbine inlet temperature : 1100 K Intake duct efficiency : 95% Isentropic efficiency of compressor : 0.85 Isentropic efficiency of turbine : 0.90 Mechanical efficiency of transmission : 99% Nozzle efficiency : 95% Nozzle outlet area : 0.0935 m2 L.C.V. of fuel : 43 MJ/kg Find the thrust and specific fuel consumption in kg/Nh of thrust. Assume convergent nozzle. Take Cpa = 1.005 kJ/kg K and γ = 1.4, Cpg = 1.147 kJ/kg K and γ = 1.33.

Jet Propulsion Cycles and Their Analysis

03 04’

T

02 02’ 01’

04 5 5’

01

a

s Fig. 7.28

Solution Speed of aircraft =

p01

805 × 1000 3600 !

=

223.6 m/s γ γ−1

Ci2 1+ 2Cp Ta

=

pa 1 + ηram

=

0.458× 1+0.95×

=

0.631 bar

p02

=

r × p01

T01

=

Ta +

=

272.87 K

T02

=

T01 +

429.08

=

272.87 +

T03 − T04

=

Cpa (T02 − T01 ) Cpg ηm

=

1.005 × (429.08 − 272.87) 1.147 × 0.99

=

138.25 K

T04

=

1100 − 138.25

p03

=

p02 − 0.21 = 2.52 − 0.21 = 2.31 bar

7

Ci2 2Cp

=

−1 2

223.6 1+ 2×1.005×248×1000

4 × 0.631 =

248 +

T01 γ−1 r γ −1 ηC

=

3.5

−1

223.62 2 × 1005

=

8

2.524 bar

429.08 K

272.87 × 40.286 − 1 0.85

=

"

961.75 K

259

260

Gas Turbines

T04

p04

T03 − T04 ηT

=

T03 −

=

946.389 K

=

p03

=

1.26 bar

=

1100 −

γ γ−1

T04 T03

= 2.31 ×

138.25 0.90

946.389 1100

4.03

The nozzle pressure ratio is p04 p04 = p5 pa

=

1.26 0.458

=

2.751

The critical pressure ratio 1

= 1 ηn

1−

1

= 1 0.95

1− p04 pc

=

γg γg −1

γg −1 γg +1

4.03

1.33−1 1.33+1

1.917

Thus nozzle is choking. γg −1 γg

pc p04

T5 = Tc

=

T04

Tc

=

818.35 K

p5 = pc

=

1.26 p04 = = 0.657 bar 1.917 1.917

ρ5

=

ρc

=

0.282 kg/m

=

γg RTc

cj

m ˙

=

pc RTc

= 961.75 ×

=

1 1.917

0.33 1.33

0.657 × 105 284.6 × 818.35

3

=

√ 1.33 × 0.2846 × 1000 × 818.35

=

556.56 m/s

=

ρc Ac cj

=

14.67 kg/s

=

0.282 × 0.0935 × 556.56

Jet Propulsion Cycles and Their Analysis

261

Total thrust F

=

m ˙ a (cj − ci ) + A5 (pc − pa )

=

14.67 × (556.56 − 223.6) + 0.0935 × (0.657 − 0.458) × 105

m ˙f

sf c

Ans

=

6745.17 N

=

mC ˙ p (T03 − T02 ) CV

=

14.67 × 1.147 × (1100 − 429.08) 43000

=

0.263 kg/s

=

m ˙f F

=

0.140 kg/N h

=

⇐=

0.263 × 3600 6745.17 Ans

⇐=

7.6 A turboprop aircraft is flying at 600 km/h at an altitude where the ambient conditions are 0.458 bar and −15◦C. Compressor pressure ratio 9:1. Maximum gas temperature 1200 K. The intake duct efficiency is 0.9 and total head isentropic efficiency of compressor and turbine is 0.89 and 0.93 respectively. Calculate the specific power output in kJ/kg, thermal efficiency of the unit taking mechanical efficiency of transmission as 98% and neglecting the losses other than specified. Assume that exhaust gases leave the aircraft at 600 km/h relative to the aircraft. (Hint : No thrust as the unit is turboprop) Solution 03 04’

T

02 02’ 01’

04 5 5’

01

a

s Fig. 7.29

ci

=

600 ×

5 18

=

166.67 m/s

262

Gas Turbines

5 18

=

166.67 m/s

c2i 2Cp

=

258 +

cj

=

600 ×

T01

=

Ta +

=

271.82 K

p01

=

166.672 2 × 1005

T01

γ γ−1

T01 Ta

pa

=

0.5497 bar

=

0.9 × (0.5497 − 0.458) + 0.458

=

0.5405 bar

p02

=

9 × 0.5405

T02

=

T01 (r)

=

271.82 × 90.286

=

271.82 +

T04

=

T03

p04 p03

T04

=

1200 − 0.93 × (1200 − 695.87)

WN

=

WT −

=

Cpg (T03 − T04 ) −

=

1.147 × (1200 − 731.16) −

=

263.76 kJ/kg

=

263.76 × 100 1.147 × (1200 − 539)

p01

T02

ηth

=

271.82 258

=

=

0.458 ×

3.5

4.865 bar

γ−1 γ

=

509.60 K

509.60 − 271.82 0.89 γ−1 γ

= 1200 ×

= 1 9

539 K

0.248

= 695.87 K =

731.16 K

WC ηm Cpa (T02 − T01 ) ηm 1.005 × (539 − 271.82) 0.98 Ans

⇐= =

34.79%

Ans

⇐=

7.7 In a turbojet unit with forward facing ram intake, the jet velocity relative to the propelling nozzle at exit is twice the flight velocity. Determine the rate of fuel consumption in kg/s, when developing a thrust of 25000 N under the following conditions.

Jet Propulsion Cycles and Their Analysis

Ambient pressure and temperature Compression total head pressure ratio Flight speed CV of fuel Ram efficiency Isentropic efficiency of compressor Isentropic efficiency of turbine Isentropic efficiency of nozzle Combustion efficiency Turbine pressure ratio

: : : : : : : : : :

263

0.7 bar; 1◦ C 5:1 800 km/h 42000 kJ/kg 100% 85% 90% 95% 98% 2.23

Assume the mass flow of fuel is small compared with the mass flow of air and that the working fluid throughout has the properties of air at low temperature. Neglect the extraneous pressure drop. Assume Cpg = Cpa = 1.005 kJ/kg K. Solution 03 04’

T

02 02’ 01’

04 5 5’

01

a

s Fig. 7.30

ci

=

800 km/h

cj

=

2 × ci = 2 × 222.22 = 444.44 m/s

T01

=

Ta +

=

298.55 K

T02 − T01

=

T01 (γ−1)/γ r −1 ηC c

T02 − 298.55

=

298.55 × 50.286 − 1 0.85

T02

=

503.9 K

p01

=

pa

c2i 2Cp

=

222.22 m/s

= =

c2i 1+ 2Cp Ta

274 +

222.222 2 × 1005

T01

γ γ−1

=

205.31

264

Gas Turbines

222.222 2 × 274 × 1005

3.5

=

0.7 × 1 +

ηr

=

p01 − pa p01 − pa

p01

=

1 × (0.945 − 0.7) + 0.7

=

0.945 bar

p02

=

5 × 0.945

T03 − T04

=

T02 − T01 = 503.9 − 298.6 = 205.3 K

T03 − T04

=

ηT T03 1 −

=

0.9 × T03 × 1 −

205.3

=

0.1845 × T03

T03

=

1112.73 K

F

=

m(c ˙ j − ci )

25000

=

m ˙ × (444.44 − 222.22)

m ˙a

=

112.5 kg/s

=

= 0.945 bar

4.725 bar

1 rt0.286 1 2.230.286

Ans

⇐=

Neglecting the effect of mass addition of fuel on the right hand side. m ˙ f CV

=

m ˙ a Cp (T03 − T02 )

m ˙f

=

112.5 × 1.005 × (1112.73 − 503.9) 42000

=

1.64 kg/s

Ans

⇐=

7.8 Under take-off conditions when the ambient pressure and temperature are 1.01 bar and 288 K, the stagnation pressure and temperature in the jet pipe of a turbojet engine are 2.4 bar and 1000 K, and the mass flow is 23 kg/s. Assuming that the expansion in the converging propelling nozzle is isentropic, calculate the exit area required and the thrust produced. For a new version of the engine the thrust is to be increased by the addition of an aft fan which provides a separate cold exhaust stream. The fan has a by-pass ratio of 2.0 and a pressure ratio of 1.75, the isentropic efficiencies of the fan and fan-turbine sections being 0.88 and 0.90 respectively. Calculate the take-off thrust assuming that the

Jet Propulsion Cycles and Their Analysis

265

expansion in the cold nozzle is also isentropic, and that the hot nozzle area is adjusted so that the hot mass flow remains at 23 kg/s. Solution As expansion in the nozzle is isentropic, the nozzle efficiency is 100%. Then, pc

γ γ−1

2 γ+1

=

p04

=

1.2969 bar

=

2.4 ×

2 2.33

4.03

Since pc > pa the nozzle will choke. Therefore, all the variables are calculated with respect to throat conditions. Tc

=

T04

2 γ+1

ρc

=

pc RTc

=

=

0.531 kg/m3

cj

=

1000 ×

2 = 858.37 K 2.33

1.2969 284.6 × 858.37 √ 1.33 × 284.6 × 858.37

=

γRT

=

=

570 m/s

=

m ˙ ρ c cj

=

0.076 m2

=

mc ˙ j + Ac (pc − p1 )

=

23 × 570 + 0.076 × 1.2969 × 105 − 1.01 × 105

=

15290.47 N

T02

=

T01

p02 p1

T02

=

T01 +

T02 − T01 ηf

=

288 +

338 − 288 0.88

At nozzle exit, A

Total thrust

=

23 0.9824 × 570 Ans

⇐=

Ans

⇐=

Aft fan engine

Cold nozzle

γ−1 γ

= 288 × 1.750.286

=

344.8 K

=

338 K

266

Gas Turbines

Addition of an aft fan does not affect the main flow. Cold nozzle thrust adds to the hot nozzle thrust. Flow through cold nozzle =

2 × 23

=

46 kg/s

Assuming nozzle efficiency to be 100%, pc

Thrust (cold)

2 γ+1

γ γ−1

=

p02

=

(1.01 × 1.75) ×

=

0.93374 bar

=

46 ×

=

15542.8 N

2 1.4 + 1

3.5

(below ambient)

2 × 1005 × (344.8 − 288)

Fan turbine 23 × 1.147 × (1000 − T05 )

=

46 × 1.005 × (344.8 − 288)

T05

=

900.5 K

T05

=

1000 −

=

889.44 K

p05

=

2.4 ×

889.44 1000

pc

=

1.5 ×

2 1.33 + 1

=

0.811 bar (subcritical)

=

m ˙

=

23 ×

=

27263.25 N

Fcold

=

15542.8 N

Take-off thrust

=

15542.8 + 27263.25

=

42806.05 N

Fhot

1000 − 900.5 0.9

4.03

=

1.5 bar

4.03

2Cp (T05 − T06 ) 2 × 1147 × (900.5 − 288)

Ans

⇐=

Jet Propulsion Cycles and Their Analysis

267

7.9 A jet propelled plane consuming air at the rate of 18.2 kg/s is to fly at Mach number 0.6 at an altitude of 4500 m (pa = 0.55 bar, Ta = 255 K). The diffuser which has a pressure coefficient of 0.9, decreases the flow to a negligible velocity. The compressor pressure ratio is 5 and maximum temperature in the combustion chamber is 1273 K. After expanding in the turbine, the gases continue to expand in the nozzle to a pressure of 0.69 bar. The isentropic efficiencies of compressor, turbine and nozzle are 0.81, 0.85 and 0.915 respectively. The heating value of the fuel is 45870 kJ/kg. Assuming Cp = 1.005 kJ/kg K, Cpg = 1.147 kJ/kg K. γair = 1.4, γgas = 1.33, calculate (i) power input to the compressor (ii) power output of the turbine (iii) the fuel-air ratio (iv) the thrust provided by the engine and (v) the thrust power developed. Solution 03 04’

T

02 02’ 01’

04 5 5’

01

a

s Fig. 7.31

=

Mi

=

192.05 m/s

=

T01

=

Ta +

c2i 2Cp

=

255 +

192.052 2 × 1005

=

p01 pa

=

T01 Ta

p01

=

0.55 ×

ci

T01

γRTa

=

0.6 ×

√ 1.4 × 287 × 255

273.35 K

γ γ−1

273.35 255

3.5

=

0.701 bar

268

Gas Turbines

ηram

=

p01 − pa p01 − pa

p01

=

0.9 × (p01 − pa ) + pa

=

0.9 × (0.701 − 0.55) + 0.55 = 0.686 bar

p02 p01

=

rc

p02

=

5 × 0.686

T02

=

T01 1 +

=

273.35 × 1 +

=

m ˙ a Cp (T02 − T01 )

=

18.2 × 1.005 × (470.62 − 273.35)

=

3608.26 kW

WC

=

=

0.9

5 =

3.43 bar

γ−1 1 rc γ − 1 ηC

1 50.286 − 1 0.81

= 470.62 K

Ans

⇐=

Power output of the turbine = Power input to the compressor =

WC

Fuel-air ratio

=

(m ˙ a+m ˙ f )Cpg (T03 − T02 )

=

(18.2 + m ˙ f ) × 1.147 × (1273 − 470.62)

=

m ˙ f × 45870

m ˙f

=

0.373 kg/s

f

=

0.373 18.2

WT

=

(m ˙ a+m ˙ f )Cpg (T03 − T04 )

3608.26

=

(18.2 + 0.373) × 1.147 × (1273 − T04 )

T04

=

1103.62 K ⎡ ⎛

T04

=

=

Ans

WT

3608.26 kW

=

⇐= =

m ˙ f CV

Ans

0.0205

T03 ⎣1 − ηT ⎝1 −

⎞⎤

1 ⎠⎦

γ−1 γ

rt

⇐=

1103.62

=

1273 × 1 − 0.85 × 1 −

rt

=

1.99

1 rt0.248

Jet Propulsion Cycles and Their Analysis

p03 p04

=

1.99

p04

=

1.724 bar

p04 pa

=

1.724 0.55

=

269

3.13

As nozzle efficiency is given, the critical pressure ratio is given by p04 pc

1

=

=

γ γ−1

γ−1 γ+1

1−

1 ηn

1−

1 0.915

1 ×

4.03

0.33 2.33

=

1.97

Critical pressure, pc =

1.724 1.97

=

0.875 bar

The nozzle will be choking Tc

cj

pc p04

=

T04

=

932.73 K

γg −1 γg

0.33 1.33

√ 1.33 × 284.6 × 932.73

=

γRTC

=

594.18 m/s

=

pc RTC

=

0.875 × 105 284.6 × 932.73

m ˙ a+m ˙f

=

ρc An cj

18.2 + 0.373

=

0.33 × An × 594.18

An

=

0.0947 m2

Thrust, F

=

[(m ˙ a+m ˙ f )cj − m ˙ a ci ] + [An (pc − pa )]

=

[(18.2 + 0.373) × 594.18 − 18.2 × 192.05] +

ρc

=

= 1103.62 ×

1 1.97

=

3

0.33 kg/m

0.0947 × (0.875 − 0.55) × 105 =

10618 N

Ans

⇐=

270

Gas Turbines

Thrust power, Fp

=

F × ci

=

2060 kW

10726.7 × 192.1 × 10−3

=

Ans

⇐=

7.10 Air enters a turbojet engine at a rate of 12×104 kg/h at 15◦ C and 1.03 bar and is compressed adiabatically to 182◦ C and four times the pressure. Products of combustion enter the turbine at 815◦ C and leave it at 650◦C to enter the nozzle. Calculate the isentropic efficiency of the compressor, the power required to drive the compressor, the exit speed of gases and thrust developed when flying at 800 km/h. Assume the isentropic efficiency of turbine is same as that of the compressor and the nozzle efficiency 90%. Solution 03 04

04’

T

5 5’

02 02’

01 s Fig. 7.32

T02 − T01

=

T01 ( γ−1 ) rc γ − 1 ηC

455 − 288

=

288 × 40.286 − 1 ηC

ηC

=

84%

m ˙a

=

12 × 104 3600

WC

=

m ˙ a Cp (T02 − T01 ) = 33.33 × 1.005 × (455 − 288)

=

5594 kW

T03 − T04

=

ηT T03 1 −

1088 − 923

=

1088 × 0.84 × 1 −

Compressor work

=

Ans

ηT

=

⇐=

33.33 kg/s

Ans

⇐= 1 rt0.248 1 rt0.248

Jet Propulsion Cycles and Their Analysis

rt

=

2.23

p04

=

p03 rt

p04 pa

=

1.8475 = 1.79 1.03

p04 pc

=

=

4.12 2.23

=

=

271

1.8475 bar

1 1−

1 ηn

1−

1 0.9

γ γ−1

γ−1 γ+1

1 ×

0.33 2.33

4.03

=

1.99

Nozzle will not choke. T5 T04

=

p5 p04

γ−1 γ

1.03 1.8478

0.33 1.33

T5

=

923 ×

ηn

=

T04 − T5 T04 − T5

T5

=

T04 − ηn (T04 − T5 )

=

923 − 0.9 × (923 − 798.46)

cj

F

=

=

798.46 K

=

810.91 K

2 × 1147 × (923 − 810.91)

=

507.1 m/s

=

m(c ˙ j − ci )

=

9495.72 N

Ans

⇐= =

33.33 × (507.1 − 222.2) Ans

⇐=

Review Questions 7.1 What is meant by jet propulsion? What are the basic differences between jet propulsion cycle and shaft power cycle. 7.2 Explain the principle of jet propulsion and mention how the jet propulsion engines are classified. 7.3 What do you understand by the term ’air-breathing engines’? How are they classified? 7.4 With the aid of a neat diagram, explain the working principle of a ramjet engine.

272

Gas Turbines

7.5 Draw the thermodynamic cycle of the ramjet engine and derive the equation for thrust. 7.6 What are the advantages and disadvantages of a ramjet engine and what are its applications? 7.7 With the aid of a schematic diagram, explain the working principle of pulse jet engine and also draw the ideal and actual p-V diagram. 7.8 Mention the various advantages and disadvantages of the pulse jet engine. 7.9 With the aid of the schematic diagram and thermodynamic cycle, explain the working of a turboprop engine. 7.10 Mention the various advantages and disadvantages of a turboprop engine and also bring out the applications. 7.11 With a neat sketch and T -s diagram, explain the working of turbojet engine and also derive the expression for the thrust developed. 7.12 Explain with suitable graphs the performance of a turbojet engine. What are the advantages and disadvantages of a turbojet engine? 7.13 What is meant by thrust? Derive the thrust equation for a general propulsion system. 7.14 Explain the various efficiencies associated with a propulsion device. 7.15 Explain clearly the various factors affecting the performance of a propulsion device. 7.16 What is meant by thrust augmentation and explain how it is effected. 7.17 Explain the principle of afterburner in thrust augmentation. 7.18 Draw the performance chart of an afterburner engine and explain. 7.19 How does water-alcohol mixture injection help thrust augmentation? Explain. 7.20 Explain the principle of bleed burn cycle for thrust augmentation. Exercise [Note: Take γaγ−1 = 0.286 and a stated otherwise]

γg −1 γg

= 0.248 for all problems unless

7.1 The exit velocity from a jet unit is 650 m/s for an air flow of 40 kg/s through the unit. The aircraft is flying at 250 km/h. Calculate the thrust developed, the thrust power and the propulsion efficiency. Neglect the effect of fuel. Ans: (i) 23222.4 N (ii) 1612.56 kW (iii) 19.3%

Jet Propulsion Cycles and Their Analysis

273

7.2 A simple jet engine has compressor directly coupled to the turbine mounted in an aircraft with forward facing intake and rearward convergent propelling nozzle. Calculate the total thrust when the aircraft flies at true air speed of 300 m/s in the ambient total conditions of −10◦ C and 0.58 bar. Air mass flow rate Compressor stagnation pressure ratio Turbine inlet stagnation temperature Combustion chamber loss in stagnation Compressor stage efficiency Turbine stage efficiency Combustion efficiency Ram efficiency Nozzle efficiency Mechanical efficiency

: : : : : : : : : :

39 kg/s 7.5 : 1 650◦ C 4% 82% 85% 100% 90% 100% 100% Ans: 9683.96 N

7.3 In a turbojet unit with forward facing ram intake, the jet velocity relative to the propelling nozzle at exit is twice the flight velocity. Determine the rate of fuel consumption in kg/s, when developing a thrust of 25000 N under the following conditions. Ambient pressure Ambient temperature Compression total head pressure ratio Pressure at exit Flight speed L.C.V. of fuel Ram efficiency Isentropic efficiency of compressor Isentropic efficiency of turbine Isentropic efficiency of nozzle Combustion efficiency

: : : : : : : : : : :

0.7 bar 1◦ C 5:1 0.7 bar 233 m/s 43 MJ/kg 100% 85% 90% 95% 98%

Neglect the effect of fuel in total mass flow rate. Ans: 0.794 kg/s

7.4 A naval aircraft is powered by a turbojet engine, with provision for flap blowing. When landing at 55 m/s, 15 % of the compressor delivery is bled off for flap blowing and it can be assumed to be discharged perpendicularly to the direction of flight. If a propelling nozzle area of 0.13 m2 is used, calculate the net thrust during landing given that the engine operating conditions are as follows

274

Gas Turbines

Compressor pressure ratio Compressor isentropic efficiency Turbine inlet temperature Turbine isentropic efficiency Combustion pressure loss Nozzle isentropic efficiency Mechanical efficiency Ambient conditions

: : : : : : : :

9.0 0.82 1275 K 0.87 0.45 bar 0.95 0.98 1 bar, 288 K

The ram pressure and temperature rise can be regarded as negligible. Ans: 18489.6 N 7.5 The following data apply to a twin-spool turbofan engine, with the fan driven by the LP turbine and the compressor by the HP turbine. Separate cold and hot nozzles are used. Overall pressure ratio Fan pressure ratio By-pass ratio mc /mh Turbine inlet temperature Mechanical efficiency of each spool Combustion pressure loss Total air mass flow

: : : : : : :

19.0 1.65 3.0 1300 K 0.99 1.25 bar 115 kg/s

It is required to find the thrust under sea level static conditions where the ambient pressure and temperature are 1.0 bar and 288 K. Assume the fan, compressor and turbine polytropic efficiency as 0.90 and the isentropic efficiency of each propelling nozzle as 0.95. The values of n−1 n for the polytropic compression and expansion are For compression, n−1 n

=

1 ηpc

γ−1 γ

a

ηpt

γ −1 γ

g

=

0.3175

=

0.225

For expansion, n−1 n

=

Draw the schematic as well as T -s diagrams. Ans: 38892.3 N 7.6 Extend the above example on the turbofan, with the additional information that the combustion efficiency is 0.99, determine the sf c. Also, calculate the thrust and sf c when a combustion chamber is incorporated in the by-pass duct and the ‘cold’ stream is heated to 1000 K. The combustion efficiency and pressure loss for this process may be assumed to be 0.97 and 0.05 bar respectively. Assume CV = 42 MJ/kg. Ans: (i) 1.135 ×10−5 kg/N s (ii) 55823.17 N (iii) 3.672 ×10−5 kg/N s

Jet Propulsion Cycles and Their Analysis

275

7.7 A turbojet engine is traveling at 920 km/h at standard sea-level conditions. The ram efficiency is 87%, the compressor pressure ratio is 4.3:1, the compressor efficiency is 82%, the burner pressure loss is 2%, the fuel-air ratio is 0.0119, the turbine inlet total temperature is 688◦ C. The turbine efficiency is 83.5%. Calculate (i) the specific gross thrust and (ii) the thrust specific fuel consumption. Ans: (i) 255.05 N (ii) 0.168 kg/N h 7.8 A turbojet engine inducts 51 kg of air per second and propels an aircraft with an uniform flight speed of 912 km/h. The isentropic enthalpy change for the nozzle is 200 kJ/kg and its velocity coefficient is 0.96. The fuel-air ratio is 0.0119, the combustion efficiency is 0.96 and the lower heating value of the fuel is 42 MJ/kg. Calculate (i) the thermal efficiency of the engine, (ii) the fuel flow rate in kg/h and T SF C, (iii) the propulsive power in kW (iv) the thrust power, and (v) the propulsive efficiency. Ans: (i) 30.9% (ii) 2184.84 kg/h; 0.1186 kg/N h (iii) 7876.16 kW (iv) 4664.6 kW (v) 59.22% 7.9 A turbojet engine is being used to propel an aeroplane. The drag is 3900 N. The coefficient of drag is 0.01835. The wing area is 21.25 m2 . The air consumption per second of the engine is 14.5 kg/s and the thrust developed is 8900 N. Calculate the flight velocity and effective jet velocity. Also calculate the specific thrust where the Cp = 1.005 kJ/kg K. What is the density ratio at this altitude of 10000 m. Take ρ = 0.5 kg/m3 at this altitude. Ans: (i) 200 m/s (ii) 813.8 m/s (iii) 613.8 N (iv) 0.41 7.10 In a theoretical cycle for jet propulsion unit both the compression and expansion are considered isentropic, and the heat is supplied at constant-pressure. Show that the thrust developed per kg of air per second when the velocity of approach is neglected is γ−1 γ

2Cp T (q − 1) rp

1 2

−1

where q is the ratio of absolute temperatures after and before combustion, rp is the compressor pressure ratio and T is the absolute atmospheric temperature.

276

Gas Turbines

7.11 A simple turbojet is operating with a compressor pressure ratio of 8.0, a turbine inlet temperature of 1200 K, polytropic efficiencies of compressor and turbine of 0.87 and a mass flow rate of 15 kg/s, when the aircraft is flying at 260 m/s at an altitude of 7000 m. Assuming the following component efficiencies, and I.S.A. conditions, calculate the propelling nozzle area required, the net thrust developed and the thrust specific fuel consumption. Isentropic efficiency of intake Isentropic efficiency of propelling nozzle Mechanical efficiency Combustion chamber pressure loss Calorific value of the fuel Combustion efficiency

: : : : : :

0.95 0.95 0.99 6 % of p02 43 MJ/kg 0.97

Ans: (i) 0.069 m2 (ii) 7893.4 N (iii) 0.124 kg/h N 7.12 A turboprop aircraft is flying at 800 km/h at an altitude where the ambient conditions are 0.567 bar and −20◦C. Compressor pressure ratio 8:1. Maximum gas temperature 1100 K. The intake duct efficiency is 0.95 and total head isentropic efficiency of compressor and turbine is 0.92 and 0.95 respectively. Calculate the specific power output in kJ/kg of air and the thermal efficiency of the unit taking mechanical efficiency of transmission as 0.96 and neglecting the losses other than specified. Assume that exhaust gases leave the aircraft at 800 km/h relative to the aircraft. (Hint : No thrust as the unit is turboprop) Ans: (i) 226.23 kJ/kg (ii) 34.17% Multiple Choice Questions (choose the most appropriate answer) 1. The useful power output for the propulsion cycle is from (a) inlet diffuser (b) outlet nozzle (c) turbine (d) compressor 2. The working of jet propulsion is based on (a) impulse principle (b) centrifugal principle (c) reaction principle (d) centripetal principle 3. The difference between air breathing engine and rocket engine are (a) there is altitude limit (b) there is temperature limit

Jet Propulsion Cycles and Their Analysis

277

(c) atmospheric air is used for formation of the jet (d) all of the above 4. Reciprocating engines are not used in modern aircrafts due to (a) high specific weight (b) high speed requirement (c) large drop in power with altitude (d) all of the above 5. Ramjet engines are highly suited for (a) subsonic application (b) supersonic application (c) hypersonic application (d) none of the above 6. Pressure rise in a ramjet is achieved by a (a) diffuser (b) centrifugal compressor (c) axial flow compressor (d) none of the above 7. Compared to a ramjet engine, a pulse jet engine has (a) higher propulsive efficiency (b) no altitude limitation (c) larger life (d) none of the above 8. Turbo prop engines have (a) lower thrust at take-off (b) less frontal area (c) difficult to maintain (d) none of the above 9. In a turbojet engine compared to momentum thrust, pressure thrust is (a) quite low (b) quite high (c) almost equal (d) none of the above

278

Gas Turbines

10. The propulsive efficiency of a turbo jet engine as a function of speed ration, α is given by (a) (b) (c) (d)

1+α 2α 1−α 2α 2α 1+α 2α 1−α

Ans:

1. – (b) 6. – (a)

2. – (c) 7. – (d)

3. – (d) 8. – (b)

4. – (d) 9. – (a)

5. – (b) 10. – (c)

8 CENTRIFUGAL COMPRESSORS INTRODUCTION The second World War was the turning point for the development of gas turbine technology. The most rapid progress in the development of gas turbines was made during this period with the use of the centrifugal compressors. Attention was focused on the simple turbojet unit and the power to weight ratio was one of the most important considerations. Centrifugal compressor was the best possible type available at that time. Development time was another factor in the perfection of gas turbine design. Much expertise was derived from the experience gained on the design of small high-speed centrifugal compressors for supercharging reciprocating engines. Since the war, however, the axial compressor has been developed to the point where it has an appreciably higher isentropic efficiency. We have already seen in the analysis of the practical cycles (Chapter 6) the importance of efficiency of each component from the point of view of overall performance of the power plant. A centrifugal compressor is one of its class of machines in producing pressure rise and is known as turbo-compressors. In this type, energy is transferred by dynamic means from a rotating member (or impeller) to the continuously flowing working fluid. The main feature of the centrifugal compressor is that, the angular momentum of the fluid flowing through the impeller is increased partly by virtue of the impeller’s outlet diameter being significantly larger than its inlet diameter. The centrifugal compressor may be known as a fan, blower, supercharger, booster, exhauster or compressor; the distinction between these types being very vague. Broadly speaking, fans are classified as low-pressure compressors and blowers as mediumpressure compressors. Boosters, exhausters and superchargers are named from their point of view of applications. Although the centrifugal compressor is unlikely to be used in gas turbine power plants where specific fuel consumption is the main criterion, it still has certain advantages for some applications.

280

Gas Turbines

(i) It occupies a smaller length than the equivalent axial flow compressor. (ii) It is not so liable to loss of performance by build up of deposits on the surfaces of the air channels. (iii) It can work reasonably well in a contaminated atmosphere compared to axial flow machine. (iv) It is able to operate efficiently over a wider range of mass flow rate at any particular rotational speed. This latter feature, viz., a wider range of mass flow rate matching a wide range of operating conditions with turbine, makes centrifugal compressors attractive compared to axial flow machines. A pressure ratio of the order of 4:1 can be obtained from a single-stage, manufactured using conventional materials. It has already been shown in chapters 5 and 6 that this is adequate for a heat-exchange cycle when the turbine inlet temperature is in the region of 1000-1200 K. Certainly, it can find an application in small power units. It is mainly because the higher isentropic efficiency of axial compressors cannot be maintained for very small sizes of machines. Most current proposals for vehicular gas turbines utilize a centrifugal compressor in a heat-exchange cycle. Materials such as titanium now enable pressure ratios of over 6:1 to be used. There is a renewed interest in the centrifugal stage, used in conjunction with one or more axial stages, for small turbofan and turboprop aircraft engines. The centrifugal compressor is less suitable when the cycle pressure ratio requires the use of more than one stage in series because of the tortuous path the air must follow between the stages. Nevertheless, a single stage centrifugal compressor has been used successfully in turboprop engines. 8.1

ESSENTIAL PARTS OF A CENTRIFUGAL COMPRESSOR

The compression process is carried out in a centrifugal compressor, which comprises mainly of four elements (Fig. 8.1): (i) The inlet casing with converging nozzle, whose function is to accelerate the fluid to the impeller inlet. The outlet of the inlet casing is known as the eye. (ii) The impeller, in which the energy transfer takes place, resulting in a rise of fluid kinetic energy and static pressure. (iii) The diffuser, whose function is to transform the high kinetic energy of the fluid at the impeller outlet into static pressure. (iv) The outlet casing, which comprises a fluid collector known as a volute or scroll. Further definitions are required to describe the impeller (Fig. 8.2) and diffuser (Fig. 8.3). The various impeller components shown in Fig. 8.2 are the following:

Centrifugal Compressors

281

Volute 3 2

Diffuser

Inlet casing with converging nozzle Impeller 0

1

Fig. 8.1 A centrifugal compressor stage (i) The impeller vanes, help to transfer the energy from the impeller to the fluid. (ii) The hub, which is surface AB. [Fig. 8.2(a)]. (iii) The shroud, which is surface CD [Fig. 8.2(a)]. Impellers enclosed on the surface CD are known as shrouded impellers, but the surface CD is referred to as the shroud in descriptions of impeller geometry whether the impeller is enclosed or not. (iv) The inducer, the section EF in impellers of the form shown in Fig. 8.2(b) whose function is to increase the angular momentum of the fluid without increasing its radius of rotation. The diffuser may consist of any annular space [Fig. 8.3(a)] known as a vaneless diffuser or may be in the form of a set of guide vanes, when it is known as a vaned diffuser [Fig. 8.3(b)]. The main aim of providing diffusers is to increase the static pressure by reducing the kinetic energy. Diffusers will be discussed in greater detail in a later section. To put it in a nutshell, a centrifugal compressor has essential two parts of energy transformation: (i) Rotating impeller which imparts a high velocity to the fluid and at the same time increases the static pressure. Impellers are housed inside a stationary casing. (ii) A number of fixed diverging passages in which the air is decelerated increasing the static pressure.

282

Gas Turbines

Impeller vanes

Shroud D B C

Inducer hub

E

F

A A C

hub

Shroud D B (a)

(b) Fig. 8.2 The impeller

Outlet diffuser

Vaneless diffuser Volute

Impeller

Diffuser vanes (a)

(b)

Fig. 8.3 Diffusers Figure 8.4 is a schematic diagram of a centrifugal compressor. The impeller may be designed as single-sided or double-sided as in Fig. 8.5 (a) or 8.5 (b), but both operates on the same principle. The double-sided impeller was required in early aero-engines because of the relatively small flow capacity of the centrifugal compressor for a given overall diameter. 8.2

PRINCIPLE OF OPERATION

Air is sucked into the impeller eye through an accelerating nozzle and whirled round at high speed by the vanes on the impeller disc (refer Fig. 8.4). At any point, in the impeller, the flow experiences a centripetal acceleration due to a pressure head. Hence, the static pressure of the air increases from the eye to the tip of the impeller. The remainder of the static pressure rise is obtained in the diffuser. It may be noted that air enters the impeller eye with a very high velocity. The friction in the diffuser will cause some loss in stagnation pressure. It

Centrifugal Compressors o

90 bend taking air to combustion chambers

283

284

Gas Turbines

Depth of diffuser

Centrifugal Compressors

3

2

p 0

Inlet casing

1 Impeller

Diffuser

Channel Fig. 8.6 Pressure rise across centrifugal compressor

p

02

02

p

03

03 p

3

03’ 2 C2

h03 , h02 2

C3 2

3 3’

2

p

h

2

2’

2 p

01

h00 , h 01

00 p

01

00

2 C1

p

1

2

1’ 1 s

Fig. 8.7 Enthalpy–entropy diagram of a centrifugal compressor

285

286

Gas Turbines

converted to pressure (isentropically), the delivery pressure could be p02 . Since the diffusion process is not accomplished isentropically, the process (2–3) results, and some kinetic energy remains at diffuser exit (velocity c3 ), the static delivery pressure at point 3 is p3 . The final state, in the collector, has static pressure p3 , low kinetic energy c23 /2 , and a stagnation pressure p03 which is less than p02 since the diffusion process is incomplete as well as irreversible. 8.3

IDEAL ENERGY TRANSFER

Let us first consider the case of an ideal compressor with the following assumptions for a radial vaned impeller: (i) Losses due to friction are negligible. (ii) Energy loss or gain due to heat transfer to or from the gas is considered very small. (iii) The gas leaves the impeller with a tangential velocity equal to the impeller velocity (i.e., ct2 = u2 ), no slip condition is assumed. (iv) The air enters the rotor directly from the atmosphere without any tangential component, i.e., ct1 = 0. Applying these assumptions the Euler’s energy equation (Eq. 8.1) under ideal conditions becomes E

=

u22

(8.2)

This is the maximum energy transfer that is possible. Therefore, the work done by the impeller on unit quantity of air is given by W

=

E

=

u22

(8.3)

Equation 8.3 represents the maximum work capacity of a radial vaned impeller for ideal operation. It may be noted that air cannot leave the impeller with a tangential velocity greater than u2 , and in practical machines the work or energy absorbed by the fluid will be less than that given by Eq. 8.3. Also from thermodynamic analysis we have the equation (refer Eq. 3.21) for the energy transfer, E, as W

=

E

=

h02 − h01

If rc is the pressure ratio based on total pressure

(8.4) p02 p01

; we shall have

γ−1

W

=

Cp T01 rc γ − 1

(8.5)

If c1 ∼ = c2 , i.e., when there is no kinetic energy change between inlet and exit we can refer to static conditions so that

Centrifugal Compressors

287

γ−1

W

Cp T1 rc γ − 1

=

p2max p1

Here, rc is the static pressure ratio u22

(8.6)

. From Eq. 8.3 and 8.5 we have γ−1

Cp T01 rc γ − 1

=

(8.7)

where Cp is in kJ/kg K.

8.4

BLADE SHAPES AND VELOCITY TRIANGLES

In the previous section, we dealt with ideal energy transfer. In order to understand the actual energy transfer and flow through the compressor, we will use two velocity triangles, viz., entry velocity triangle and exit velocity triangle. The notations used here correspond to the x, r and θ coordinate system. As per the convention for radial machines, at a given point the angles are measured from the tangential direction. The absolute and relative air angles at the entry and exit of the impeller are denoted by α1 , α2 and β1 , β2 respectively. Based on the value of β2 the blade shapes are given the name as forward curved blades (β2 > 90◦ ), radial blades (β2 = 90◦ ) and backward curved blades (β2 < 90◦ ) (refer Fig. 8.8). Note that angle β2 is measured with respect to negative of u (i.e., −u). The impeller linear velocities at inlet and exit are given by

β > 90

u1

=

2πr1 N 60

u2

=

2πr2 N 60

o

β = 90

2

o

2

Forward

Radial

β < 90

o

2

Backward

Fig. 8.8 Blade shapes 8.4.1

Entry Velocity Triangle

Figure 8.9(a) shows the flow at the entry of the inducer section of the impeller without Inlet Guide Vanes (IGV). The absolute velocity (c1 ) of

288

Gas Turbines

the flow is axial (α1 = 90◦ ) and the relative velocity (w1 ) is at an angle β1 from the tangential direction. Thus the swirl or whirl component ct1 = 0. c1 ca1 = (8.8) tan β1 = u1 u1

Flow Inducer section of the impeller

u1 ca1= c1

w1 α1

ct1 = 0

u1

β1

(a) Without inlet guide vanes Inducer c1 α1 u1 = ct1

β1

w 1 = c a1

Inlet guide vanes Entry (b) With inlet guide vanes Fig. 8.9 Flow through the inducer section Figure 8.9(b) shows the flow through axially straight inducer blades in the presence of IGVs. The air angle (α1 ) at the exit of the IGVs is such that it gives the direction of the relative velocity vector (w1 ) as axial, i.e., β1 = 90◦ . This configuration seems to offer some manufacturing and aerodynamic advantages, viz., (i) centrifugal impellers with straight blades are much easier and cheaper to manufacture and (ii) the relative velocity (w1 ) approaching the impeller is considerably reduced. In this case β1 = 90◦ and the positive swirl component is ct1 tan α1

= u1 =

w1 u1

(8.9) =

ca1 u1

(8.10)

Centrifugal Compressors

289

Figure 8.10 shows the entry and exit velocity triangles for impeller blades located only in the radial section. For the sake of generality, the absolute velocity vector c1 is shown to have a swirl component ct1 . However, if there are no guide vanes, c1 will be radial (c1 = cr1 ) and α1 = 90◦ , ct1 = 0. This particular condition is expressed by zero whirl or swirl at the entry and would be assumed in this chapter unless otherwise mentioned. c2 α2 c t2

c r2

w2 β2

u2

c1 α1

Impeller blade ring

c t1

cr

1

u1

β1

w1

Fig. 8.10 Velocity triangles for backward swept impeller blades (β2 < 90◦ )

8.4.2

Exit Velocity Triangle

The impeller shown in Fig. 8.10 are having backward swept blades, i.e., β2 < 90◦ . The exit velocity triangle is as shown in the figure. The flow leaves the blades with a relative velocity w2 and at an air angle, β2 . The absolute velocity of flow leaving the impeller is c2 at an air angle α2 . Its tangential (swirl or whirl) component is ct2 and the radial component cr2 . The following relations can be obtained from the velocity triangles as shown in Fig. 8.10 at the entry and exit. cr1

=

c1 sin α1

= w1 sin β1

ct1

=

c1 cos α1

= cr1 cot α1

cr2

=

c2 sin α2

= w2 sin β2

ct2

=

c2 cos α2

= cr2 cot α2

(8.11) =

u1 − cr1 cot β1

(8.12) (8.13)

=

u2 − cr2 cot β2

(8.14)

Figure 8.11 shows the velocity triangles at the entry and exit of a radialtipped blade extending into the inducer section. The velocity triangle at the entry is similar to that in Fig. 8.10; here ca1 replaces the velocity component cr1 . The exit velocity triangle here is only a special case of the triangle in Fig. 8.10 with β2 = 90◦ . This condition when applied in Eqs. 8.13 and 8.14, gives

290

Gas Turbines

c2

(at exit)

w 2 = c r2

β2

α2 c t2 = u 2

c1 α1

ca1 w1 β1

c t1

u1 (at inlet)

Impeller blade ring

Fig. 8.11 Velocity triangles for radial-tipped impeller with inducer blades, (β2 = 90◦ ) cr2 ct2

=

=

w2

c2 cos α2

= c2 sin α2 = cr2 cot α2

(8.15) =

u2

(8.16)

Figure 8.12 shows the velocity triangles for forward swept blades, (β2 > 90◦ ) with zero swirl at the entry. It may be observed that such blades have large fluid deflection and give ct2 > u2 . This increases the work capacity of the impeller and the pressure rise across it. But in practice, this configuration is unsuitable for higher speeds and leads to higher losses. 8.5

ANALYSIS OF FLOW THROUGH THE COMPRESSOR

In the previous section we have seen various blade shapes and the corresponding inlet and exit velocity triangles to understand flow conditions. In this section we will analyze the flow through the compressor from inlet to exit. Ideal conditions cannot be achieved in practice and various losses do occur in an actual compressor. The losses are the major factors responsible for the decrease of efficiency in the centrifugal compressors. In order to understand the various losses it is necessary to understand the flow through the compressor first. In the early days of radial flow compressor development, the design and theoretical analysis was mainly based on the assumption that the air flow through the compressor may be considered one-dimensional. However, there are complications due to flow separation and the appearance of shocks. Still, one-dimensional analysis provides the base for the initial design of the components. The technique has been further refined over the years by the introduction of design parameters or design criteria gained from experimental results. Finally, the design can be modified with the

Centrifugal Compressors

291

c2

α2

u

2

β2

c

w2

cr2

t2

c1 Impeller blade ring

w1 u1

β1

Fig. 8.12 Velocity triangles for forward swept blades, with zero swirl at entry (β2 > 90◦ )

help of three-dimensional flow analysis to obtain optimum compressor performance. With the aid of the momentum equation, steady-flow energy equation, continuity equation and thermodynamic equation of state, it is possible to estimate the representative state of the air (that is, pressure, temperature, velocity, Mach number etc.,) at any point in the compressor channel. The assumptions in the estimation are that the fluid is a perfect gas and the flow is isentropic and one-dimensional. This will help to determine the shape of the compressor components (channel geometry) to produce desirable velocity or Mach number profiles from inlet to exit. When air flows from inlet to exit, it passes through the following, viz., (1) the inlet casing, with the accelerating nozzle, (2) the impeller, (3) the diffuser and (4) the volute. The details of various elements through which air flows with the representative dimensions are shown in Fig. 8.13.

8.5.1

The Inlet Casing

The inlet casing consists of an accelerating nozzle with or without inlet guide vanes. The function of the inlet casing is to deliver air to the impeller eye with minimum loss and to provide a uniform velocity profile at the eye. The inlet flange is axisymmetric and the inlet duct takes the form of a simple accelerating convergent nozzle. Since there is no energy transfer, stagnation enthalpy remains constant. Hence, from the h-s diagram (Fig. 8.7), T1

=

T01 −

c21 2Cp

(8.17)

292

Gas Turbines

Volute (4) Diffuser (3) Casing

Shroud

Impeller eye

d2

Impeller (2)

2

Hub dt 2

d1 2

Driving shaft dh Inducer section

2 IGV

Inlet casing with accelerating nozzle (1)

Fig. 8.13 Details of airflow passage of a centrifugal compressor stage With an air filter fitted to the compressor inlet casing, a pressure drop will occur which must be kept to a minimum. However, it should be accounted for, in the flow analysis and prediction of compressor performance. Significant losses can occur in inlet casings fitted with silencing baffles. Under this condition an inlet casing efficiency (ηic ) may be used. Similar to the nozzle efficiency, ηic , can be defined as (Fig. 8.7) ηic

=

h00 − h1 h00 − h1

(8.18)

where h00 − h1 is the enthalpy change in isentropic expansion from p00 to p1 . For a perfect gas assumption (Cp = constant) Eq. 8.18 becomes ηic

=

T00 − T1 T00 − T1

=

T00 − T1 T00

=

1 1− ⎡

T00 − T1 ⎢ ⎣ T00

T1 T00

1 1−

p1 p00

T00 = T01 since h00 = h01 (refer Fig. 8.7). Now, T01 − T1

=

c21 2Cp

Rearranging and substituting in Eq. 8.19 gives

(γ−1)/γ

⎤ ⎥ ⎦

(8.19)

Centrifugal Compressors

p1 p00

=

1−

c21 2Cp ηic T00

293

γ/(γ−1)

From values of p1 and T1 = T00 − c21 /2Cp , the density at the impeller eye can be calculated. p1 ρ1 = RT1 and hence the mass flow rate entering the impeller eye can be estimated π 2 m ˙ = d − d2h1 ca1 ρ1 (8.20) 4 t1 where dt1 and dh1 are the impeller eye tip and eye hub diameters respectively, and ca1 is the axial component of the absolute velocity of the air entering the impeller eye. In the case of a swirl-free intake ca1

=

c1

(8.21)

For compressor intakes, where the air must be turned from a radial to an axial direction, the air incidence angle at the impeller eye (inducer) has to be chosen very carefully. This is mainly because of the variation in the axial velocity distribution caused by the free vortex flow effect in the bend of the inlet ducting. 8.5.2

The Inducer

The inducer, or entry section of the impeller has a very pronounced effect on the impeller performance and hence on over-all compressor efficiency. The eye hub diameter (dh1 ) of the inducer is determined by a number of design considerations such as inducer stress/vibration, impeller manufacturing techniques and the number of blades. The tip of the inducer eye is the point where the highest inlet relative Mach number occurs, and therefore careful attention should be paid to the choice of the inducer eye tip diameter (dt1 ). For a given design mass flow rate and impeller speed, dt1 should be chosen so as to obtain a minimum relative Mach number (Fig. 8.14). To further reduce the relative Mach number (Mt1 ), pre-whirl (inlet guide vanes) may be used. However, the penalty will be a reduction in energy transfer in the impeller. Thus the technique is usually used only on highpressure ratio compressors, where the inlet Mach number exceeds unity and shock waves reduce impeller efficiency. The use of inlet guide vanes has to be carefully evaluated for each application. Since there is a trade-off between improved stage performance from the impeller and reduced stage performance from the diffuser. 8.5.3

The Impeller

Energy transfer occurs in the impeller of the compressor, hence there is an increase in stagnation enthalpy and pressure. To attain peak compressor

Gas Turbines

Mach number at eye tip

294

Diameter of eye tip

Fig. 8.14 Effect of eye tip diameter efficiency great care must be taken so that very efficient diffusion processes is achieved in the impeller and the diffuser. Since the diffusion processes are related to the flow, a reasonable relative Mach number at the impeller inlet, and a minimum absolute Mach number at the impeller outlet are to be ensured. 8.5.4

The Effect of Impeller Blade Shape on Performance

As already mentioned in section 8.4 the various blade shapes utilized in impellers of centrifugal compressors can be classified as (i) forward-curved blades (β2 > 90◦ ), (ii) radial-curved blades (β2 = 90◦ ), (iii) backward-curved blades (β2 < 90◦ ). Figure 8.15 represents the variation of pressure-ratio that can be obtained with respect to mass flow rate for the above mentioned various shapes for operation at a given rpm. The basic velocity diagram at exit is shown for each type of blade is Fig. 8.16. For example, the Euler energy equation without inlet swirl (ct1 = 0) is written as E

=

u2 ct2

=

u2 (u2 − cr2 cot β2 )

(8.22)

For any particular impeller running at a constant speed, energy equation can be written as E

=

K1 − K2 Q

(8.23)

where Q is the volume flow rate which is proportional to flow velocity for given impeller, and K1 and K2 are constants as given by K1

=

u22

(8.24)

K2

=

u2 cot β2 πd2 b2

(8.25)

and

Centrifugal Compressors

295

Pressure ratio or head

Forward

Radial

Backward

Mass flow rate

Fig. 8.15 The effect of impeller blade shape on performance

c2

c2

w2 β2

u2

c t2

Forward

w2 = cr2

c2 cr2

cr2 β2

c t2 = u2

Radial

w2 β2

c t2

u2

Backward

Fig. 8.16 Exit velocity diagrams for different shapes of blades

The comparison of performance for the three types of vanes are made for the same volume flow rate, each blade having unit depth, and for the same vector value of cr2 . Figure 8.16 shows the exit velocity triangles for three types of impellers from which the relative performance of the blades can be evaluated. Centrifugal effects of the curved blades create a bending moment and produce increased stresses which reduce the maximum speed at which the impeller can be run. Good performance can be obtained with radial impeller blades. Backward-curved blades are slightly better in efficiency and are stable over a wider range of flows than either radial or forward-curved blades. The forward-curved impeller can produce the highest pressure ratio for a given blade tip speed; but is inherently less stable and has a narrow operating range. Its efficiencies are lower than that are possible with the backward-curved or radial-curved blades. Although all the three types can be used in compressors, the radial blade is used almost exclusively in turbojet engine applications.

296 8.5.5

Gas Turbines

The Impeller Channel

The relative velocity of the air in the impeller channels undergoes a rapid deceleration from the impeller eye to the impeller tip. In addition to the function of transferring energy to the air, the impeller should act as an efficient diffuser. A badly shaped channel will interfere with the diffusion process, causing flow separation at the impeller walls; leading to higher losses. To obtain gradual deceleration of the flow in the channel, the geometry of the impeller channels should be such that it can provide a smooth change in the relative Mach number along the mean flow path. The flow in an impeller is not completely guided by the impeller vanes and hence the effective fluid outlet angle does not equal to the impeller vane outlet angle. To account for this deviation, a factor known as the slip factor is used to correct the energy transfer calculated from simple one-dimensional theory. It may be kept in mind that impeller flow is not one-dimensional. Many analyses have been made of the velocity distribution within an impeller assuming the flow to be isentropic. Although such solutions may depart from experimental velocity distributions, they are useful in predicting regions in an impeller channel where losses may occur. No quantitative criteria for boundary layer separation in impellers are at present available and much more work is necessary to take these factors into account. It is a known fact that the vorticity of frictionless fluid does not change with time. Hence, if the flow at the inlet to an impeller is irrotational, the absolute flow must remain irrotational throughout the impeller. As the impeller has an angular velocity, ω, the fluid must also have an angular velocity −ω, relative to the impeller which is known as the relative eddy. Thus if there were no flow through the impeller the fluid in the impeller channels would rotate with an angular velocity equal and opposite to the impeller’s angular velocity [Fig. 8.17(a)].

(a) Relative eddy

(b) Relative flow through the impeller

Fig. 8.17 Relative eddy and flow through the impeller The flow through a rotating impeller can be considered as the vector sum of the flow through the impeller passages – when the impeller is stationary – and the flow produced in the fluid by the rotation of the impeller. It may be

Centrifugal Compressors

297

noted that the large amount of the mass of air flowing through the impeller has certain inertia and due to the formation relative eddies, the velocity of air at the tip is always less than the blade speed. Further, the relative flow through the impeller is not perfectly guided by the impeller vanes but is deflected away from the direction of rotation of the impeller; because of which air leaves at an angle smaller than the vane angle [Fig. 8.17(b)]. The resultant velocity triangle at impeller outlet will be as shown in Fig. 8.18. It may be seen that the actual tangential component of absolute velocity (ct2 ) of the fluid is less than that of the perfectly guided value (ct2 ). The perfectly guided value will be obtained if the fluid leaves the impeller at the vane angle (β2 ). The difference between the value of perfectly guided and actual tangential component (ct2 − ct2 ) is called slip velocity (cs ). Hence, the energy transfer can be written as E

=

u2 ct2

(8.26)

It follows that the actual energy transfer in the impeller is less than the perfectly guided value as the fluid does not come out at an angle β2 , the exit blade angle. Actual velocity triangle cs

Perfectly guided velocity triangle

c2

w2

w2'

c 2’

cr

β2 β2’

ct2

u2 ct2’

Fig. 8.18 Effect of relative eddy on impeller outlet velocity

8.5.6

Slip Factor

The actual velocity profiles at the impeller tip due to real flow behaviour are shown in Figs. 8.19 and 8.20. Figure 8.19 corresponds to the velocity profile at exit in the meridional plane whereas Fig. 8.20 depicts the details between two vanes. The energy transfer occurring in the impeller corresponding to these velocity profiles is less than the one that would have been obtained with one-dimensional flow. The relative eddy [Fig. 8.17(a)] mentioned earlier causes the flow in the impeller passages to deviate (Fig. 8.21) from the blade angle (β2 ) at the exit to an angle (β2 ), the difference being larger for a larger blade pitch or smaller number of impeller blades.

298

Gas Turbines

cr= w

Exit velocity profile

Hub Shroud Meridional plane*

* Meridional plane is a plane which contains the axis of rotation and is swept around that axis

Fig. 8.19 Velocity profile at the impeller tip in the meridional plane

Exit velocity profile

w

Vane to vane plane

ω

Fig. 8.20 Vane to vane velocity profile at the exit of the impeller

Centrifugal Compressors

Actual Ideal

299

cs

cr

2

2

w

2

cr

2’

c

w 2’

c2 c 2’ ct

β2’ β

t2’

2

u

2

ω

Fig. 8.21 Exit velocity triangles with and without slip

On account of the aforementioned effects, the apex of the actual velocity triangle at the impeller exit is shifted away (opposite to the direction of rotation) from the apex of the perfectly guided velocity triangle as shown in Fig. 8.21. This phenomenon is known as slip and the shift of the apex is the slip velocity (cs ). It may be noted that on account of the slip, the whirl component is reduced which in turn decreases the energy transfer and the pressure developed. The ratio of the actual and perfectly guided values of the whirl components at the exit is known as slip factor (μ) μ =

ct2 ct2

(8.27)

(1 − μ)ct2

(8.28)

The slip velocity is given by cs

8.5.7

=

ct2 − ct2

=

Determination of Slip Factor

Several theoretical and empirical equations have been derived for the slip factor to allow calculation of energy transfer during the design of a new impeller. The most widely used of these are theoretically based equations

300

Gas Turbines

derived by Stodola∗ , Stanitz† and Balje‡ , all of which assume the flow of an inviscid fluid through the impeller. 8.5.8

Stodola’s Formula

Stodola∗ assumed that the relative eddy was equivalent to the rotation of cylinder with an angular velocity equal in magnitude and opposite in direction to that of the impeller. Figure 8.22 depicts the model of flow with slip as suggested by Stodola. The relative eddy is assumed to fill almost the entire exit section of the impeller passage. It is considered equivalent to the rotation of a cylinder of diameter d = 2r at an angular velocity ω which is equal and opposite to that of the impeller as shown in the figure. The diameter, and hence, the tangential velocity of the cylinder is approximately determined as follows: s = 2 π r2 / z

β

2’

ω

2r

ω

Fig. 8.22 Stodola’s model of flow with slip The blade pitch at the outer radius (r2 ) of the impeller with z number of blades is s =

2πr2 z

The diameter of the cylinder is 2r

≈ s sin β2

=

2πr2 sin β2 z

(8.29)

∗ Stodola, A., Steam and Gas Turbines, McGraw Hill Book Co., New York, Vol.I & II, 1927, Reprint Peter Smith, New York, 1945. † Stanitz, J.D., ’Some theoretical aerodynamic investigations of impellers in radial and mixed flow centrifugal compressors’, Trans ASME, 74, 4, 1952 ‡ Balje, O.E., ’A study of design criteria and matching of turbomachines. Pt. B – Compressor and pump performance and matching of turbo components’, ASME paper No.60–W A–231, 1960

Centrifugal Compressors

301

The slip velocity, is assumed to be due to the rotation of the cylinder and can therefore be taken as, cs

= ωr

Substituting in the above equation for r from Eq. 8.29, we have ωπr2 sin β2 cs = z

(8.30)

However, ωr2 = u2 . Therefore, cs

π u2 sin β2 z

=

(8.31)

Substituting Eq. 8.31 in Eq. 8.28, we have π u2 sin β2 (1 − μ)ct2 = z μ = 1−

π u2 sin β2 z ct2

From Fig. 8.21, we have ct2

= u2 − cr2 cot β2 ,

and μ = 1−

π z

sin β2 1 − φ2 cot β2

(8.32)

where the flow coefficient, φ2 = cr2 /u2 For a radial-tipped impeller (β2 = 90◦ ) π μ = 1− z

(8.33)

The above expressions for μ show that for a given geometry of flow, (blade exit angle) the slip factor increases with the number of impeller vanes. When the slip factor approaches 1, the slip velocity tends to be zero (i.e., ct2 = ct2 ). Hence, larger the number of blades lower will be the slip velocity. However, there is a limitation from the manufacturing point of view to decide about the number of impeller vanes. Thus, the number of impeller vanes is one of the governing parameters in the calculation of losses and this should be given due consideration. It should be noted that because of the phenomenon of the slip the actual energy transfer is given by E 8.5.9

=

μu2

(8.34)

Stanitz’s Formula

A method based on the solution of potential flow in the impeller passages is suggested by Stanitz for β2 between 45◦ and 90◦ . The slip velocity is

302

Gas Turbines

found to be independent of the blade exit angle and the compressibility. This is given by cs

=

1.98 u2 z

(1 − μ)ct2

=

1.98 u2 z

(8.35)

μ =

1−

1.98 u2 z ct2

μ =

1−

1.98 z

For β2 = 90◦

=

1−

1.98 z(1 − φ2 cot β2 )

(8.36)

(8.37)

Equations 8.33 and 8.37 are of identical form.

8.5.10

Balje’s Formula

Balje suggests an equation for radial-tipped (β2 = 90◦ ) blade impellers: μ = where n = 8.6

1+

6.2 zn2/3

−1

(8.38)

Impeller tip diameter Eye tip diameter

DIFFUSER

Centrifugal compressors are usually fitted with either a vaneless or a vaned diffuser, although in some low-speed applications a volute is fitted directly around the impeller. The influence of the diffuser upon compressor performance cannot be over emphasized: a considerable proportion of the fluid energy at the impeller tip is kinetic energy (especially in radial-vaned impellers) and its efficient transformation into static pressure is important. Losses may be high in the diffuser as the fluid is flowing against an adverse pressure gradient. Hence, careful design of diffuser is a must. 8.6.1

Vaneless Diffuser

As the name indicates, the gas in a vaneless diffuser is diffused in the vaneless space around the impeller before it leaves the stage through a volute casing. In some applications the volute casing is omitted. The gas in the vaneless diffuser gains static pressure rise simply due to the diffusion process from a smaller diameter (d2 ) to a larger diameter (d3 ). The corresponding area of cross-section in the radial direction are A2

=

πd2 b2

=

2πr2 b2

(8.39)

Centrifugal Compressors

A3

=

πd3 b3

=

2πr3 b3

303

(8.40)

Such a flow in the vaneless space is a free-vortex flow in which the angular momentum remains constant. It can be shown that for a parallel walled (constant width) diffuser, the ratio of tangential velocity at the exit to that of inlet is given by ct3 cr3 c3 r2 = = = (8.41) ct2 cr2 c2 r3 Further, it can be shown that for a constant width diffuser with compressible flow cr2 cr3 α2 = α3 = tan−1 = tan−1 (8.42) ct2 ct3 Equation 8.41 clearly shows that the diffusion is directly proportional to the diameter ratio (d3 /d2 ). This leads to a relatively large-sized diffuser which is a serious disadvantage of the vaneless type. In some cases the overall diameter of the compressor may be impractically large. This is a serious limitation which prohibits the use of vaneless diffusers in aeronautical applications. Besides this, the vaneless diffuser has a lower efficiency and can be used only for a small pressure rise. However, for industrial applications where large-sized compressors are needed, the vaneless diffuser is economical and provides a wider range of operation. Besides this, it does not suffer from blade stalling and shock waves. 8.6.2

Vaned Diffuser

For a higher pressure ratio across the radial diffuser, the diffusion process has to be achieved across a relatively shorter radial distance. This requires the application of vanes which provide greater guidance to the flow in the diffusing passage. Diffuser blade rings can be fabricated from sheet metal or casting can be made as cambered and uncambered shapes of uniform thickness (see Figs. 8.23 and 8.24). Figure 8.25 shows a diffuser ring made up of cambered aerofoil blades. To avoid flow separation, the divergence of the diffuser blade passages in the vaned diffuser ring can be kept small by employing a large number of vanes which will lead to higher friction losses. Thus an optimum number of diffuser vanes must be employed. It is the normal practice to have the divergence of the flow passages not more than 12◦ . The flow leaving the impeller has jets and wakes. When such a flow enter a large number of diffuser passages, the quality of flow entering different diffuser blade passages differs widely and some of the blades may experience flow separation leading to rotating stall and poor performance. To avoid such a possibility it is safer to provide a smaller number of diffuser blades than that of the impeller. In some designs the number of diffuser blades is kept one-third of the number of impeller blades. This arrangement provides a diffuser passage with flows from a number of impeller blade channels.

304

Gas Turbines

Flow

Fig. 8.23 Diffuser ring with cambered blades

α3 (90- α 3)

α2

Straight flat blade

r3 r2

Fig. 8.24 Diffuser ring with straight (uncambered) flat blades

il ofo Aer

es

blad

Flow

Fig. 8.25 Diffuser ring with cambered aerofoil blades

Centrifugal Compressors

305

Thus the nature of flow entering various diffuser passages does not differ significantly. Another method to prevent steep velocity gradients at the diffuser entry is to provide a small (0.05d2 − 0.1d2 ) vaneless space between the impeller exit and the diffuser entry as shown in Fig. 8.26 and Fig.8.27. This allows the non-uniform impeller flow to mix out and enter the diffuser with less steep velocity profiles. Besides this the absolute velocity (Mach number) of the flow is reduced at the diffuser entry. This is a great advantage, especially, if the absolute Mach number at the impeller exit is greater than unity. The supersonic flow at the impeller exit is decelerated in this vaneless space at constant angular momentum without shock. Diffuser blades Flow

ω d2 2

Impeller blades

dh 2

Vaneless space

d1 2

Fig. 8.26 Vaneless space between impeller exit and diffuser entry

b3

Diffuser

θ

b2

r3

Vaneless space Impeller

r2

Fig. 8.27 Radial diffuser passage with diverging walls Every diffuser blade ring is designed for given flow conditions at the entry at which optimum performance is obtained. Therefore, at off-design

306

Gas Turbines

operations the diffuser will give poor performance on account of mismatching of the flow. In this respect a vaneless diffuser or a vaned diffuser with aerofoil blades (Fig. 8.25) is better. For some applications it is possible to provide movable diffuser blades whose directions can be adjusted to suit the changed conditions at the entry. In some designs for industrial applications, a vaneless diffuser supplies the air or gas direct to the scroll casing, whereas for aeronautical applications, various sectors of the vaned diffuser are connected to separate combustion chambers placed around the main shaft.

8.7

VOLUTE CASING

The volute or scroll casing collects and guides the flow from the diffuser or the impeller (in the absence of a diffuser). The flow is finally discharged from the volute through the delivery pipe. For high pressure centrifugal compressors or blowers, the gas from the impeller is discharged through a vaned diffuser whereas for low pressure fans and blowers, the impeller flow is invariably collected directly by the volute; since a diffuser is not required owing to the relatively low pressures. Figure 8.28 shows a volute casing along with impeller, diffuser and vaneless spaces. The volute base circle radius (r) is a little larger (0.05 to 0.10 times the diffuser or impeller radius) than the impeller or diffuser exit radius. The vaneless space before volute decreases the non-uniformities and turbulence of flow entering the volute as well as noise level. Some degree of diffusion in the volute passage is also achieved in some designs, while others operate at constant static pressure.

Delivery pipe Exit

Throat

Vaneless spaces

Tongue Flow

Impeller r Di

Volute passage

r

ffu s

er

Volute section dr

Fig. 8.28 Flow through volute casing of a centrifugal compressor

Centrifugal Compressors

307

Different cross-sections are employed for the volute passage as shown in Fig. 8.29. The rectangular section is simple and convenient when the volute casing is fabricated from sheet metal by welding the curved wall to the two parallel side walls. While the rectangular section is very common in centrifugal blowers, the circular section is widely used in compressor practice. (a)

(b)

(c)

r4

r4

r4

Circular

Trapozoidal

Rectangular

axis

Fig. 8.29 Different cross-section of the volute passage While, on the one hand, the volute performance is dependent on the quality of flow passed on to it from the impeller or diffuser, the performance of the impeller or the diffuser also depends on the environment created by the volute around them. The non-uniform pressure distribution around the impeller provided by its volute gives rise to the undesirable radial thrust and bearing pressures. 8.8

PERFORMANCE PARAMETER

As ideal energy transfer given by Eq. 8.2 will not hold good for an actual compressor, in order to assess the performance, certain parameters are used. In this section we will describe them briefly. 8.8.1

Power Input Factor

The ideal energy transfer is given by Eq. 8.2, which is E = u22 . In practice, the actual energy transfer to the air from the impeller is lower than that given by Eq. 8.34, which is μu22 . Further, some energy is lost in friction between the casing and the air carried round by the vanes, and in disc friction or windage. In order to take this into account, power input factor, Pif is introduced, so that the work input to the compressor becomes Wc

=

Pif μ u22

(8.43)

The normal value used for Pif is between 1.035 to 1.04. Let ΔTc = T02 − T01 be the total head temperature rise across the compressor. As there is no energy addition in the diffuser this temperature rise is also the temperature rise across the impeller. Thus, ΔTc

=

T02 − T01

=

Pif μu22 Cp

(8.44)

308 8.8.2

Gas Turbines

Pressure Coefficients

This is a performance parameter which is useful in comparing various centrifugal compressors. Each impeller has a definite maximum work capacity limited by the maximum tangential velocity at the exit. If this maximum work is utilized to the maximum advantage, an isentropic compressor will result, and a delivery pressure of p02 max will be obtained. Now referring to Fig. 8.30 we have,

02’ max

T02’ max T02

02 02’

T 02’

T

p

p

02

02max

p

01

04 03

T01

01

s

Fig. 8.30 T –s diagram of the compression process

p02max p01

γ−1 γ

T02 max

=

T01

(8.45)

Wisen

=

Wmax = Cp T01

p02max p01

γ−1 γ

−1

(8.46)

The actual compressor produces only p02 pressure after an expenditure of adiabatic work given by

Wadia

=

Cp T01

p02 p01

γ−1 γ

Now pressure coefficient ψp is defined as

−1

(8.47)

Centrifugal Compressors

ψp

=

p02 p01

Cp T01

Wadia Wisen

γ−1 γ

= p02max p01

Cp T01

−1

γ−1 γ

309

(8.48)

−1

and from Eq. 8.7, the denominator is u22 and therefore ψp

8.8.3

=

Cp T01 (rc )

γ−1 γ

u22

−1

(8.49)

Compressor Efficiency

We have already dealt with the compressor efficiency. However, it is better to recall it once again here. The overall stagnation isentropic efficiency (Fig. 8.31) in terms of pressure ratio can be written as

p

02

2

p

01

h

2’

Δ h0 Δ h 0’

1 s

Fig. 8.31 h–s diagram of the compressor

ηc

=

Δh0 Δh0

In terms of temperature and pressure ratios γ−1

ηc From Eq. 8.44,

=

T01 rc γ − 1 (T02 − T01 )

(8.50)

310

Gas Turbines

T02 − T01

=

Pif μu22 Cp

(8.51)

=

ψp u22 Cp

(8.52)

=

ψp Pif μ

(8.53)

From Eq. 8.49, γ−1

T01 rc γ − 1 Substituting in the Eq.8.50 ηc

8.9

LOSSES IN CENTRIFUGAL COMPRESSORS

Total losses in a centrifugal compressor may be divided into two groups: (i) Frictional losses These are proportional to c2 and hence proportional to m ˙ 2. (ii) Incidence losses These loses in terms of drag coefficient CD are proportional to CD c2 . Figure 8.32 shows the variation of the above two losses with respect to the mass flow rate.

Frictional losses

Losses

Total losses

Incidence losses Mass flow rate

Fig. 8.32 Variation of losses with respect to the mass flow rate

8.10

COMPRESSOR CHARACTERISTICS

The ideal performance characteristics as shown in Fig. 8.15 for a radialvane compressor are to be modified because of the losses mentioned above. If these losses are subtracted from the ideal energy transfer for a radialvaned impeller, then the constant-pressure ratio straight line characteristics becomes curved, with a maximum value of energy at some particular value of the mass flow rate as shown in Fig. 8.33. Radial-vaned impellers are

Centrifugal Compressors

311

mostly used with gas turbine compressors. Therefore, a characteristic which will have a point of maximum pressure ratio with a positive and a negative slope is shown in Fig. 8.33. This particular variation decides upon the range of compressor operation.

Ideal Actual

For

= 90

Mass flow rate

Loss

Gas Turbines

6

N

Pressure ratio

Relative to design value

T 01

5 4

Locus of points of maximum efficiency

3 Surge line 2 1 0

0.2

0.4

0.6

0.8

1.0

1.2

m T 01 Relative to design value p

01

Fig. 8.35 Actual characteristics of a centrifugal compressor

100

Total head efficiency (%)

312

80

60 0.6 0.7 0.8 0.9 1.0

40

N T

Relative to design value

01

20

0

0.2 m

0.4 T 01

0.6

0.8

1.0

1.2

Relative to design value

p01

Fig. 8.36 Total head efficiency of a centrifugal compressor

Centrifugal Compressors

313

pressure ratio starts reducing. At this moment, there is a higher pressure in the downstream of the system (near exit) than at compressor delivery and the flow stops momentarily and may even reverse its direction. This reduces the pressure downstream. After a short interval of time, compressor again starts to deliver fluid and the operating point shifts to C again. Again the pressure starts increasing and the operating point moves from right to left. If the downstream conditions are unchanged then once again the flow will break down after point A and the cycle will be repeated with a high frequency. This phenomenon is known as surging or pumping. This will be again explained in greater detail with respect to axial flow compressor in the next chapter. This instability will be severe in compressors producing high pressure ratios, which may ultimately lead to physical damage due to impact loads and high-frequency vibration. Because of this particular phenomenon of surging or pumping at low-mass flow rates, the compressor cannot be operated at any point to the left of the maximum pressure ratio point A, i.e., it cannot be operated on the positive slope of the characteristic. At higher mass flow rate points on the characteristic a different situation occurs. At a constant rotor speed the tangential velocity component at the impeller tip remains constant. As can be seen from Fig. 8.34, with the increase in mass flow rate the pressure ratio decreases (ABCD) and hence the density also decreases. These effects result in a considerably increased velocity which increases the absolute velocity and the incidence angle at the diffuser vane top. Thus there is a rapid progress towards a choking state. The slope of the characteristic therefore steepens and finally after point D mass flows cannot be increased any further. The characteristic finally becomes vertical. The point D on the characteristic curve is called a choking point. The actual characteristics and the total head efficiency of a centrifugal compressor are shown in Figs. 8.35 and 8.36 respectively. Part of the curve which is on the left of the maximum pressure ratio point is inoperable due to surge, and the line joining these points is called the surge line. On the higher mass flow part the range is limited due to the choke point. The peak efficiencies at each speed are quite close to the surge line. When the compressor is used in the gas turbine power plant then the characteristics of compressor and turbine must be matched properly. Otherwise problems will be experienced either due to surging or due to low efficiency. Worked out Examples 8.1 A centrifugal compressor under test gave the following data: Speed Inlet total head temperature Outlet and inlet total head pressure Impeller dia

: : : :

11,500 rev/min 21◦ C 4 bar, 1 bar 75 cm

If the slip factor is 0.92, what is the compressor efficiency?

314

Gas Turbines

Solution

02 T

02’

01 s Fig. 8.37

11500 60

u

=

π × 0.75 ×

WC

=

μu2 =

W

=

Cp (T02 − T01 )

T02

=

187.63 + 294 1.005

p01

=

1 bar p02 p01

=

0.92 × 451.62 1000

=

451.6 m/s =

480.7 K

γ−1 γ

T02

=

T01

=

ηc

=

437 − 294 × 100 480.7 − 294

187.63 kJ/kg

294 × =

4 1

0.286

=

437 K Ans

76.6%

⇐=

8.2 A centrifugal compressor has to deliver 35 kg of air per sec. The impeller is 76 cm diameter revolving at 11,500 rpm with an adiabatic efficiency of 80%. If the pressure ratio is 4.2:1, estimate the probable axial width of the impeller at the impeller tip if the radial velocity is 120 m/s. The inlet conditions are 1 bar and 47◦ C. Solution T02

=

320 × 4.20.286

T02

=

T01 +

T02 − T01 ηc

=

320 +

482.4 − 320 = 523 K 0.8

=

482.4 K

Centrifugal Compressors

315

02 T

02’

01 s Fig. 8.38

Ignoring the effects of velocity of flow ρ2

=

p02 4.2 × 105 3 = 2.8 kg/m = RT02 287 × 523

Atip

=

35 2.8 × 120

Axial width

=

0.1042 = 0.0436 m = 4.36 cm π × 0.76

=

0.1042 m2 Ans

⇐=

8.3 A centrifugal compressor has an inlet eye 15 cm diameter. The impeller revolves at 20,000 rpm and the inlet air has an axial velocity of 107 m/s, inlet stagnation temperature 294 K and inlet pressure 1.03 kg/cm2 . Determine (i) theoretical angle of the blade at this point and (ii) Mach number of the flow at the tip of the eye. Solution Peripheral speed of eye tip,

20000 = 157.08 m/s 60

u1

=

π × 0.15 ×

β1

=

tan−1

axial velocity ca1 = tan−1 peripheral velocity u1

=

tan−1

107 157.08

=

Ans

34.26◦

⇐=

Relative velocity at eye tip, =

u1 cos β1

=

157.08 cos 34.26

=

190 m/s

316

Gas Turbines

w1

ca1 = c1 α1

β

1

u1

Fig. 8.39

Velocity of sound,

-

=

γR T01 −

c2a 2Cp

1072 1.4 × 287 × 294 − 2 × 1005

= =

1 2

340.35 m/s

Mach number at the tip =

190 Relative velocity at eye tip = Velocity of sound 340.35

=

0.558

Ans

⇐=

8.4 A centrifugal compressor takes in gas at 0◦ C and 0.7 bar and delivers at 1.05 bar. The efficiency of the process compared with the adiabatic compression is 83%. The specific heat of the gas at constant-pressure and constant-volume are 1.005 and 0.717 respectively. Calculate the final temperature of the gas and work done per kg of gas. If the gas were further compressed by passing through a second compressor having the same pressure ratio and efficiency and with no cooling between the two compressors, what would be the overall efficiency of the complete process? Solution T02

T02

p02 p01

γ−1 γ

=

T01

=

=

306.6 K

=

T01 +

T02 − T01 ηc

=

273 +

306.6 − 273 0.83

273 ×

=

1.05 0.7

313.5 K

0.286

Ans

⇐=

Centrifugal Compressors

317

03 03’

T

02’ 02 01

s Fig. 8.40

Wc

=

1.005 × (313.5 − 273) = 40.7 kJ/kg

Ans

⇐=

With additional compressor 1.05 0.7

0.286

T03

=

313.5 ×

=

T03

=

T02 +

=

313.5 +

T03

=

273 × (1.5 × 1.5)

ηoverall

=

344.3 − 273 × 100 359.9 − 273

352 K

T03 − T02 ηc 352.0 − 313.5 0.83

=

0.286

= =

359.9 K 344.3 K

82%

Ans

⇐=

8.5 Determine the impeller diameters and the width at the impeller exit and the power required to drive the compressor, from the following given data: Speed (N ) Mass flow rate (m) ˙ Pressure ratio (r) Isentropic efficiency (ηc ) Slip factor (μ) Flow coefficient at impeller exit(φ) Hub diameter of the eye Axial velocity of air at entry to and exit from the impeller Stagnation temperature at inlet Stagnation pressure at inlet

: : : : : : :

12,500 rev/min 15 kg/s 4:1 75% 0.9 0.3 15 cm

: : :

150 m/s 295 K 1.0 bar

Assume equal pressure ratio in the impeller and diffuser.

318

Gas Turbines

Solution u2

=

ca2 φ

150 0.3

=

500 m/s

mμu ˙ 22

=

15 × 0.9 × 5002 1000

=

3375 kW

u2

=

πD2

D2

=

500 × 60 π × 12500

=

0.7639 m

T1

=

T01 −

c2a 2Cp

=

295 −

=

283.8 K

=

p01

T1 T01

=

0.8733 bar

ρ1

=

0.8733 × 105 287 × 283.8

A1

=

m ˙ ρ1 ca1

=

0.0934 m2

=

0.0934

D1

=

0.376 m

p3 p1

=

4

p2 p1

=

p3 p2

p22

=

4p21

p2

=

2p1

T2

=

283.8 × 20.286

T2

=

283.8 +

=

Power input

p1

π 2 D − 0.152 4 1

N 60

γ γ−1

=

1502 2 × 1005

283.8 295

3.5

3

1.07 kg/m

15 1.07 × 150

=

=

=

Ans

⇐=

Ans

⇐=

2 × 0.8733 = 1.7466 bar =

346 − 283.8 0.75

346 K =

366.73 K

Centrifugal Compressors

ρ2

=

1.7466 × 105 284.6 × 366.73

A2

=

πD2 W2

W2

=

1 15 × 1.67 × 150 π × 0.7639

=

0.025 m

=

319

1.67 kg/m3

m ˙ ρ2 ca2

=

=

Ans

⇐=

2.5 cm

8.6 A single sided centrifugal compressor is to deliver 14 kg of air per second when operating at a pressure ratio of 4:1 and a speed of 12,000 rpm. The total head inlet conditions may be taken as 288 K and 1.033 kgf/cm2 . Assuming a slip factor as 0.9, a power input factor of 1.04 and an isentropic efficiency (based on total head) of 80%, estimate the overall diameter of the impeller. If the Mach number is not to exceed unity at the impeller tip and 50% of the losses are assumed to occur in the impeller, find the minimum possible depth of the diffuser. Solution c2

ca2

w2 β2

α2 ct2

cw2 u2

Fig. 8.41

p03 p01

=

ηc (T03 − T01 ) 1+ T01

4

=

1+

T03

=

462.95 K

T03 − T01

=

ΨμU 2 Cp

U

=

πDN 60

=

γ γ−1

0.8 × (T03 − 288) 288

3.5

(462.95 − 288) × 1.005 × 103 1.04 × 0.9 433.4

1 2

= 433.4 m/s

320

Gas Turbines

D

=

60 × U πN

=

69 cm

=

60 × 433.4 π × 12000

=

0.69 m Ans

⇐=

Minimum depth of the diffuser will be at impeller tip so if we find for the impeller, radial velocity at tip and density at tip and knowing W , we can calculate the area of passage and thus the depth of diffuser which is also the depth of impeller tip. For Mach number, M at impeller tip M

=

c2 √ γRT3

c22

=

γRT3

T03

=

T3 + Temperature equivalent of c2

=

T3 +

c22 2 × 1.005 × 103

=

T3 +

γRT3 2 × 1.005 × 103

=

T3 1 +

γR 2 × 1.005 × 103

=

T3 1 +

1.4 × 287 2 × 1.005 × 103

T3

=

T03 1.2

c2

=

γRT3

=

393.71 m/s

(since M = 1)

=

=

1.2 × T3

462.95 = 385.8 K 1.2 √ = 1.4 × 287 × 385.8

Now, we know that c22

=

c2t2 + c2a2

c2a2

=

c22 − c2t2

=

393.712 − (0.9 × 433.4)

ca2

=

53.48 m/s

T02 − T01

=

ηc (T03 − T01 )

T02

=

139.96 + T01 = 139.96 + 288 = 427.96 K

=

393.712 − (μU )2 2

=

0.8 × (462.95 − 288)

Centrifugal Compressors

321

The overall loss =

T03 − T02

=

462.95 − 427.96

=

34.99 K

=

2.37

Since, 50% of the loss occurs in the impeller, so =

T3 − Loss in impeller

=

385.8 −

p2 p01

=

T2 T01

p2

=

1 × 2.37

ρ2

=

p2 2.37 × 105 3 = 2.24 kg/m = RT2 287 × 368.305

A

=

m ˙ ρ2 ca2

T2

34.99 2

=

γ γ−1

=

3.5

368.305 288

= =

368.305 K

2.37 bar

14 2.24 × 53.48

=

0.117 m2

Total area of flow in impeller A

=

2πr × Depth of diffuser

=

A 2πD/2

=

5.4 cm

Depth of diffuser

=

0.117 π × 0.69

=

0.054 m Ans

⇐=

8.7 A centrifugal compressor runs at 10,000 rpm and delivers 600 m3 /min of free air at a pressure ratio of 4:1. The isentropic efficiency of compressor is 82%. The outer radius of impeller (which has radial blades) is twice the inner one and neglect the slip coefficient. Assume that the ambient air conditions are 1 bar and 293 K. The axial velocity of flow is 60 m/s and is constant throughout. Determine (i) power input to the compressor, (ii) impeller diameters at inlet and outlet and width at inlet, and (iii) impeller and diffuser blade angles at inlet. Solution γ−1 γ

T03

=

T01 × (4)

T03

=

T01 +

=

293 + 173.85

= 293 × 40.286 = 435.56 K

T03 − T01 ηc =

=

293 +

466.85 K

435.56 − 293 0.82

322

Gas Turbines

03 03’

T

c1

w1 = ca1

01

α root

1

s

u1

Fig. 8.42

u22

=

Cp ΔT

=

=

174719.25 m2 /s2

u2

=

418 m/s

WC

=

u22

D2

=

418 × 60 π × 10000

D1

=

D2 2

=

T1

=

T01 −

c2a 602 = 291.2 K = 293 − 2Cp 2 × 1005

=

1005 × 173.85

Ans

174.719kW/kg/s

T1 T01

=

⇐=

0.8 m Ans

0.4 m

⇐=

γ γ−1

291.2 293

3.5

p1

=

p01

ρ1

=

p1 RT1

=

1.171 kg/m

Q˙

=

ca Aflow at root

Wroot

=

1 10 × 60 π × 0.4

u1

=

π × 10000 πN ×D = × 0.4 = 209.4 m/s 60 60

αroot

=

tan−1

60 209.4

αtip

=

tan−1

60 418

=

= 0.9787 bar

0.9787 × 105 287 × 291.2

= 3

= =

= =

ca πD1 Wroot 0.133 m

Ans

⇐=

Ans

16◦

⇐=

8.17◦

⇐=

Ans

Centrifugal Compressors

323

8.8 A centrifugal compressor has a pressure ratio of 4:1 with an isentropic efficiency of 80% when running at 15000 rpm and inducing air at 293 K. Curved vanes at inlet give the air a prewhirl of 25◦ to the axial direction at all radii and the mean dia of eye is 250 mm. The absolute air velocity at inlet is 150 m/s. Impeller tip dia is 600 mm. Calculate the slip factor. Solution c1

w1

ο

25

ca1

ο

β1

65

ct1

cw1 u1

Fig. 8.43

T02 T01

=

p02 p01

T02

=

293 ×

γ−1 γ

4 1

0.286

=

435.56 K

Isentropic temperature rise T02 − T01

=

435.56 − 293

=

142.56 K

=

Isentropic temperature rise Isentropic efficiency

=

142.56 0.8

Actual temperature rise, ΔT

=

178.2 K

Power input per unit mass flow rate = Cp × ΔT = 1.005 × 178.2 = 179 KJ/kg

ct1 At exit,

=

c1

=

150 m/s

u1

=

π × Mean dia of eye × 15000 60

=

π × 0.250 × 15000 60

=

150 × sin 25

c1 sin 25

=

=

196.35 m/s

63.4 m/s

324

Gas Turbines

u2

=

π × Impeller tip dia × 15000 60

=

π × 0.6 × 15000 60

Power input unit mass flow rate

=

=

471.24 m/s

u2 ct2 − u1 ct1

3

=

471.24 × ct2 − 196.35 × 63.4

ct2

=

406.27 m/s

μ

=

ct2 u2

179 × 10

=

406.27 471.24

=

0.862

Ans

⇐=

8.9 Determine the number of radial impeller vanes using Stanitz formulae for a centrifugal compressor which requires 180 kJ of power input per unit mass flow rate and is running at 15000 rpm. Guide vanes at inlet give the air a prewhirl of 25◦ to the axial direction at all radii and the mean dia of eye is 250 mm. Impeller tip dia is 600 mm. The absolute air velocity at inlet is 150 m/s. Solution w1

ο

c1

25

ο

ca1

β1

65

ct1

cw1 u1

Fig. 8.44

Power input per unit mass flow rate

u1

u2

ct1

=

u2 ct2 − u1 ct1

=

π × Mean dia of eye × N 60

=

π × 0.25 × 15000 60

=

π × Impeller tip dia × N 60

=

π × 0.6 × 15000 60

=

c1 sin 25 = 150 × sin 25 = 63.39 m/s

=

=

196.35 m/s

471.24 m/s

Centrifugal Compressors

180 × 103

=

471.24ct2 − (196.35 × 63.39)

ct2

=

196.35 × 63.39 + 180 × 103 471.24

=

408.38 m/s

325

Stanitz formulae for radial impellers is given by μ

=

1−

1.98 z

where z is number of impeller vanes. μ

=

ct2 u2

0.866

=

1−

z

≈

15

Therefore,

=

408.38 471.24

=

0.866

0.63 × π z Ans

⇐=

8.10 A centrifugal compressor compresses 30 kg of air per second at a rotational speed of 15000 rpm. The air enters the compressor axially, and the conditions at the exit sections are radius = 0.3 m, relative velocity of air at the tip = 100 m/s at an angle of 80◦ C with respect to plane of rotation. Take p01 = 1 bar and T01 = 300 K. Find the torque and power required to drive the compressor and also the head developed. Solution c2

cr2

w2 β2

ct2

cw2 u2

Fig. 8.45

N 15000 = π × 0.6 × = 471.24 m/s 60 60

u2

=

πD2

ct2

=

u2 − w2 cos β1 = 471.24 − 100 × cos 80

=

453.88 m/s

326

Gas Turbines

Torque

Power

W

W

=

F r = mct2 r = 30 × 453.88 × 0.3

=

4084.92 Nm

=

T ω = 4084.92 × 2 × π ×

=

6.417 × 106 W

=

6.417 × 103 kW

=

u2 ct2

=

213886.41

=

Cp T01

=

Ans

⇐= 15000 60

Ans

⇐=

471.24 × 453.88

p02 p01

213886.41

=

1005 × 300 ×

p02 p01

=

6.531

p02

=

6.531 bar

γ−1 γ

−1 p02 p01

γ−1 γ

−1

Ans

⇐=

Review Questions 8.1 What is a centrifugal compressor and what are its advantages? 8.2 Where do the centrifugal compressors find application and why? 8.3 With a neat sketch explain the essential parts of a centrifugal compressor. 8.4 With a suitable diagram explain the working principle of a centrifugal compressor. 8.5 What are the three types of blade shapes possible and how they are classified? 8.6 With a neat sketch explain the inlet and exit velocity triangles for various types of blades. 8.7 Briefly explain the flow through the following components: (i) the inlet casing, (ii) the inducer,

Centrifugal Compressors

327

(iii) the impeller, and (iv) the impeller channel. 8.8 Briefly explain with suitable diagram how does the blade shape affect the performance of the compressor. 8.9 What is meant by slip? 8.10 Define slip factor and derive an expression for the same. 8.11 Why diffusers are necessary in a centrifugal compressor? 8.12 Explain the details of vaned and vaneless diffusers? 8.13 What is meant by volute? Explain the purpose of volute casing. 8.14 Explain briefly the following performance parameters: (i) power input factor, (ii) pressure coefficients, and (iii) compressor efficiency. 8.15 Briefly explain the phenomena of surge and chocking in centrifugal compressors. Exercise 8.1 Find the actual shaft power required to the compressor of Jumbo turbo-jet engine from the following data: Pressure ratio : 3:1 γ : 1.4 Adiabatic efficiency based on shaft power : 75% Mass flow of the air compressed : 21 kg/s Temperature of inlet air : 45◦ C Neglect any ram effect due to the forward speed of turbojet. Ans: 3303.56 kW 8.2 A centrifugal compressor of 40.6 cm diameter revolving at 18000 rpm delivers air at an isentropic efficiency of 0.78. What would be the approximate pressure ratio expected if the machine was at 6000 m altitude where p0 = 35 cm of Hg and T0 = 248 K. Calculate the actual delivery temperature and the power required to deliver air at the rate of 0.5 kg/s. Neglect effects of inlet and exit velocities. Ans: (i) 5 (ii) 434.8 K (iii) 73.2 kW 8.3 A single stage turbo blower compresses 1000 m3 /min of air at 1 bar and 15◦ C through a pressure ratio of 1.33 with an index of compression as 1.6, assuming the power input to the blower to be 800 kW and Cp = 1.005, calculate the isentropic efficiency of the blower. Assume R = 287 kJ/kg K. Ans: 82.35%

328

Gas Turbines

8.4 A centrifugal air compressor delivers 20 kg/s of air with a total head pressure ratio of 4:1. The speed of the compressor is 12,000 rpm. Inlet total temperature is 15◦ C, slip factor 0.9, power inlet factor 1.04, and the total head isentropic efficiency as 80%. Calculate the overall diameter of the impeller. Ans: 0.69 m 8.5 For a single-sided impeller of a centrifugal compressor, the following conditions are given: Speed Ambient stagnation temperature Ambient stagnation pressure Hub diameter at eye Outer diameter of the eye Impeller tip dia Axial velocity at inlet to and exit from compressor

: : : : : : :

3600 −7◦ C 1 bar 12.5 cm 20 cm 100 cm 130 m/s

Establish the principal dimensions of the compressor. Ans: (i) 0.0305 m (ii) 0.00516 m (iii) 73.8◦ (iv) 34.6◦ 8.6 A single-sided centrifugal compressor has the internal dia of eye 15 cm. The compressor delivers air at the rate of 9 kg/s with a pressure ratio of 4.4 to 1 at 20,000 rpm. The axial velocity is 150 m/s with no pre-whirl. Initial condition of air are pressure 1 bar and temperature 20◦ C. Assuming adiabatic efficiency as 80%, the ratio of whirl speed to tip speed as 0.95 and neglecting all other losses, calculate the rise of total temperature, tip speed, tip dia and external dia of eye. Ans: (i) 200.64 K (ii) 460.71 m/s (iii) 44 cm (iv) 29.35 cm 8.7 A double-sided centrifugal compressor has impeller eye root and tip dia of 18 cm and 32 cm respectively and is required to rotate at 16000 rpm. Find values for the impeller vane angles at root and tip of the eye if the angle is given 70◦ of prewhirl at all radii. The axial component of the inlet velocity is constant over the eye and is about 150 m/s. Find also the maximum Mach number at the eye. Inlet conditions may be assumed as 288 K and pressure 1 bar. Ans: (i) 57.33◦ (ii) 35.11◦ (iii) 0.784 8.8 Air enters axially in a centrifugal compressor fitted with radial blades at head of 1.033 bar and temperature of 288 K. The air leaves the diffuser with negligible velocity. The tip diameter of the impeller is 30 cm and the speed 25000 rpm. Neglect all losses. Calculate the temperature and pressure of the air as it leaves the compressor. Ans: (i) 441.45 K (ii) 4.61 bar 8.9 A centrifugal compressor has an impeller tip speed of 366 m/s. Determine the absolute Mach number of the flow leaving the radial vanes of the impeller when the radial component of velocity at impeller exit is

Centrifugal Compressors

329

30 m/s and the slip factor is 0.9. Given that the flow area at impeller exit is 0.1 m2 and ηi = 90%, calculate the mass flow rate. Assume p01 = 1.033 bar. Ans: (i) 0.878 (ii) 5.65 kg/s 8.10 In a centrifugal compressor, air leaving the guide vanes has a velocity of 90 m/s at 20◦ to the axial direction at the outer radius of the eye. Assuming frictionless flow through the guide vanes, determine the inlet relative Mach number and the impeller efficiency. Other details of the compressor and its operating conditions are Impeller entry tip dia, Dti Impeller exit tip dia, Dte Slip factor μ Radial component of velocity at impeller exit, cr2 Rotational speed of impeller, N Static pressure at impeller exit, p2 Ambient total temperature at inlet Ambient total pressure at inlet Absolute angle

: : : :

0.45 m 0.76 m 0.9 53 m/s

: : : : :

11000 rpm 2.3 bar 15◦ C 1.033 bar 70◦ Ans: (i) 0.721 (ii) 91.6%

8.11 Air enters the impeller of a centrifugal compressor in an axial direction at a rate of 2 kg/s with a pressure of 1.033 bar and a temperature of 288 K. Measured angle of air entry is 55◦ from the axial direction and leaves in a radial direction. The diameters of the eye root and tip at inlet are one-fifth and one-half respectively of the impeller outside dia, if the impeller is driven at 20000 rpm, determine (i) the diameter of the impeller, and (ii) the temperature rise through the impeller. State the assumptions made. Ans: (i) 29.8 cm (ii) 96.9 K 8.12 Determine the slip factor in a single-sided centrifugal compressor fitted in the aircraft flying with a speed of 230 m/s at an altitude where the pressure is 0.25 bar and the static temperature is 220 K. The mean dia of eye is 25.5 cm and the impeller tip dia is 54 cm. Rotational speed of the compressor is 16000 rpm and the inlet duct of the impeller eye contains fixed vanes which give the air prewhirl of 65◦ with respect to prewhirl speed at all radii. Stagnation pressure at the compressor outlet is 1.75 bar. Take the power input factor as 1.04 and isentropic efficiency as 0.8. Ans: 0.91 8.13 The following data refers to a design of a single-sided centrifugal compressor:

330

Gas Turbines

Air mass flow (m) ˙ Eye tip dia (Dc ) Eye root dia (Dr ) Overall dia of the impeller (Do ) Slip factor (μ) Isentropic efficiency (ηc ) Power input factor (Pif ) Rotational speed (N ) Inlet stagnation pressure (p01 ) Inlet stagnation temperature (T01 )

: : : : : : : : : :

9 kg/s 300 mm 150 mm 500 mm 0.9 80% 1.04 18000 1.1 bar 295 K

Compute (i) pressure ratio of the compressor, (ii) inlet angle of the impeller vane at the root and tip radii of the eye, and (iii) the axial depth of the impeller channel at the periphery of the impeller. Assume that half the total loss occurs with impeller. Ans: (i) 4.75 (ii) 45.16◦ (iii) 26.70◦ (iv) 1.65 cm 8.14 A centrifugal compressor operates with prewhirl and is run with a rotor tip speed of 500 m/s. If slip factor is 0.95, isentropic efficiency of compressor is 0.85 and power input factor is 1.035, calculate the following for operation in standard sea level air: (i) pressure ratio, (ii) the work required per kg of air and (iii) the power required for a flow of 30 kg/s. Take Cp = 1.005 kJ/kg K, ambient temperature 17◦ C. Ans: (i) 6.63 (ii) 246.09 kJ/kg (iii) 7382.7 kW 8.15 An aircraft engine is fitted with a single-sided centrifugal compressor. The aircraft flies with a speed of 850 km/h at an altitude where the pressure is 0.23 bar and the temperature 217 K. The inlet duct of the impeller eye contains fixed vanes which give the air pre-whirl of 65◦ at all radii. These inner and out diameters of the eye are 180 and 330 mm respectively, the diameter of the impeller tip is 540 mm and the rotational speed 16000 rpm. Estimate the stagnation pressure at the compressor outlet when the mass flow is 216 kg per minute. Neglect losses in inlet duct and fixed vanes, and assume that the isentropic efficiency of the compressor is 0.80. Take the slip factor as 0.9 and the power input factor as 1.04. Ans: 1.908 bar

Centrifugal Compressors

331

Multiple Choice Questions (choose the most appropriate answer) 1. A single state modern centrifugal compressor have a pressure ratio (a) 2:1 (b) 4:1 (c) 6:1 (d) 8:1 2. A centrifugal compressor is best suited for the (a) simple cycle (b) intercooled cycle (c) reheat cycle (d) heat exchange cycle 3. In the inlet casing of the centrifugal compressor, pressure (a) increases (b) remains constant (c) decreases (d) initially increases and then remains constant 4. In order to have ideal energy transfer in the centrifugal compressor (a) heat transfer should be high (b) the air should enter the compressor with high tangential component of the absolute velocity (c) the air should leave the impeller with a tangential velocity equal to the impeller velocity (d) none of the above 5. The exit blade angle (β2 ) for a forward curved blade should be (a) > 90◦ (b) = 90◦ (c) < 90◦ (d) can be any angle 6. The pressure rise in a centrifugal compressor is due to (a) internal effect (b) external effect (c) both internal and external effect (d) none of the above

332

Gas Turbines

7. For better performance of the centrifugal compressor, the slip factor μ should be (a) close to 1 (b) close to 0.5 (c) close to 0 (d) can be of any value 8. The compressor isentropic efficiency in terms of pressure coefficient (ψp ), slip factor (μ) and power input factor (Pif ) is given by (a) ηc =

Pif ×μ ψp

(b) ηc =

Pif μ×ψp

(c) ηc =

μ×ψp Pif

(d) ηc =

ψp Pif ×μ

9. With respect to mass flow rate the frictional losses are (a) more than incidence losses (b) less than incidence losses (c) equal to incidence losses (d) initially lower and after a certain mass flow rate it is higher 10. It is better to operate a centrifugal compressor with respect to mass flow rate (a) close to the surge line (b) left side of the surge line (c) right side of the surge line (d) both at left and right side of the surge line Ans:

1. – (b) 6. – (c)

2. – (d) 7. – (a)

3. – (c) 8. – (d)

4. – (c) 9. – (d)

5. – (a) 10. – (c)

9 AXIAL FLOW COMPRESSORS INTRODUCTION The principal type of compressor being used nowadays, in majority of the gas turbine power plants and especially in aircraft applications, is the axial flow compressor. Although in olden days, the turboprop engines incorporated the centrifugal compressors, the recent trend, particularly for high-speed and long-range applications, is towards the axial flow type. This dominance is mainly due to the ability of the axial flow compressor to satisfy the basic requirements of the aircraft gas turbine. The basic requirements of compressors for aircraft gas-turbine application are well-known. In general, they include (i) high air-flow capacity per unit frontal area, (ii) high pressure ratio per stage, (iii) high efficiency, and (iv) discharge direction suitable for multistaging. Because of the demand for rapid engine acceleration and for operation over a wide range of flight conditions, a high level of aerodynamic performance must be maintained over a wide range of mass flow rates and speeds. Physically, the compressor should be designed in such a way to have a minimum length and also its weight must be as low as possible. The mechanical design should be simple, so as to reduce manufacturing time and cost. Further, the resulting structure should be mechanically rugged and must have high reliability. 9.1

HISTORICAL BACKGROUND

The basic concepts of multistage axial flow compressor operation have been known for approximately 100 years, being presented to the French Academic

334

Gas Turbines

des Sciences in 1853 by Tournaire. One of the earliest experimental axial flow compressors (1884) was obtained by C. A. Parsons by running a multistage reaction-type turbine in reverse. Efficiencies for this type of unit were quite low. It was mainly because the blading was not designed for the condition of a pressure rise in the direction of flow. Beginning at the turn of twentieth century, a number of axial flow compressors were built, in some cases with the blade design based on propeller theory. However, the efficiency of these units was still low (50 to 60 per cent). Further, development of the axial flow compressor was retarded by the lack of sufficient knowledge of fluid mechanics at that time. The advances in aviation during the period of World War I and the rapidly developing background in fluid mechanics and aerodynamics gave new impetus to research on compressors. The performance of axial flow compressors was considerably improved by the use of isolated-airfoil theory. As long as moderate pressure ratios per stage were desired, isolated-airfoil theory was quite capable of producing compressors with reasonably high efficiency. Compressors of this class were used in machinery as ventilating fans, air conditioning units, and steam generator fans. Beginning in the middle of 1930’s, interest in the axial flow compressor was greatly increased as the result of the quest for air superiority. Efficient superchargers were necessary for reciprocating engines in order to increase engine power output as well as improved high-altitude aircraft performance. With the development of efficient compressor and turbine components, turbojet engines for aircraft also began receiving attention. In 1936 the Royal Aircraft Establishment in England began the development of axial flow compressor for jet propulsion. A series of high performance compressors was developed, culminating in the F.2 engine in 1941. In Germany, concentrated research work ultimately resulted in the use of axial flow compressors in the Jumo 004 and the B.M.W. 003 turbojet engines. In the United States, aerodynamic research results were applied to obtain high-performance axial flow units. In the development of all these units, increased stage pressure ratios were sought by utilizing high blade cambers and closer blade spacings. Under these conditions the flow patterns about the blades began to affect each other, and it became apparent that the isolated-airfoil approach was inadequate. Aerodynamic theory was therefore, developed specifically for the case of a lattice or cascade of airfoils. In addition to theoretical studies, systematic experimental investigations of the performance of airfoils in cascade were conducted to provide the required design information. By 1945, compressors of high efficiency could be developed by incorporating aerodynamic principles in design and development. Since that time, considerable research has been directed at extending aerodynamic limits in an attempt to maximize compressor and gas-turbine performance. One of the major developments in this direction has been the successful extension of allowable relative inlet Mach numbers without sacrificing efficiency. The subject of allowable blade loading, or blade surface diffusion, has also been attacked with a degree of success. Accompanying improvements led to an

Axial Flow Compressors

335

increase in the understanding of the physics of flow through axial flow compressor blading. This resulted in corresponding improvements in techniques of aerodynamic design and high efficiencies and axial flow machines became a reality. 9.2

GEOMETRY AND WORKING PRINCIPLE

As already stated elsewhere, it may be noted that an axial compressor is a pressure producing machine. The energy level of air or gas flowing through it, is increased by the action of the rotor blades which exert a torque on the fluid. This torque is supplied by an external source – an electric motor or a gas-turbine. Besides its applications in the industrial gas turbine units the multistage axial compressor is the principal element of all gas-turbine power plants for land and aeronautical applications. An axial flow compressor [Fig.9.1(a) and (b)] consists of an alternating sequence of fixed and moving sets of blades. The sets of fixed blades are spaced around an outer stationary casing, called the stator. The sets of moving blades are fixed to a spindle called the rotor. The rotor and stator banks must be as close as possible for smooth and efficient flow. The radius of the rotor hub and the length of the blades are designed so that there is only a very small tip clearance at the end of the stator and rotor blades. One set of stator blades and one set of rotor blades constitute a stage.

(a) Drum type rotor

(b) Disc type rotor

336

Gas Turbines

kinetic energy imparted to the working fluid. It also redirects the fluid at an angle suitable for entry into the rotating blades of the following stage. Usually at entry one more stator is provided to guide the air correctly into the first rotor. These blades are sometimes referred to as the Inlet Guide Vanes (IGV). The details are shown in Fig.9.2. In many compressors there are one to three rows of diffuser or straightener blades installed after the last stage to straighten and slow down the air before it enters into the combustion chamber.

Inlet guide vanes (IGV)

Stage

Rotor blades

Stator blades (diffuser)

Fig. 9.2 An axial flow compressor stage

9.3

STAGE VELOCITY TRIANGLES

The flow geometry at the entry and exit of a compressor stage is described by the velocity triangles at these stations. A minimum number of data on velocity vectors and their directions are required to draw a complete set of velocity triangles. Compressors have a finite cross-section at the entry and exit. Therefore, the magnitude of velocity vectors and their directions vary over these sections. On account of this, an infinite number of velocity triangles are required to fully describe the flow. This is obviously not possible. On the other hand, a single pair of velocity triangles will only represent onedimensional flow through the stage. In view of this, mean values of velocity vectors and their directions are defined for blade rows of given geometries and flow conditions. These values make it possible to draw the mean velocity triangles for the stage. The velocity triangles for a compressor stage contain, besides the peripheral velocity (u) of the rotor blades both the absolute (c) and relative (w) fluid velocity vectors. These velocities are related by the following well-known vector equation: c

=

u+w

(9.1)

where c is absolute velocity vector, u is peripheral velocity vector and w is

Axial Flow Compressors

337

relative velocity vector. This simple relation is frequently used and is very useful in drawing the velocity triangles for turbomachines. For instance the velocity triangles shown in Fig.9.3 are for a general stage which receives air or gas with an absolute velocity c1 and angle α1 (from the axial direction) from the previous stage. In the case of the first stage in a multistage machine the axial direction of the approaching flow is changed to the desired direction (α1 ) by providing a row of blades upstream of the rotor which are called inlet guide vanes (IGV) or upstream guide vanes (UGV). Therefore, the first stage experiences additional losses arising from flow through the guide vanes.

Upstream guide vanes Entry velocity triangle w1 1

w2

β2

β1 α1

ca1

wt1

Rotor blades

h1 , p1 h 01,p01 c1 ct1

u

α2

c2

Exit velocity triangle 2

ca2

h 2 , p2 h 02, p02

c t2

wt2 u

Diffuser blades

3

ca3

α

3

c3

h3 , p3 h 03,p03

Fig. 9.3 Velocity triangles for a compressor stage For a general stage, the entry to the rotor, exit from the rotor and the diffuser blade row (stator) are designated as stations 1,2 and 3 respectively. The air angles in the absolute and the relative systems are denoted by α1 , α2 , α3 and β1 , β2 respectively as seen in Fig.9.3. If the flow is repeated in another stage c1

=

c3

and

α1

=

α3

Subscripts a and t (Fig.9.3) denote axial and tangential directions respectively. Thus the absolute swirl or whirl vectors ct1 and ct2 are the tangential components of absolute velocities c1 and c2 respectively. Similarly, wt1 and wt2 are the tangential components of the relative velocities w1 and w2 respectively.

338

Gas Turbines

The following trigonometrical relations obtained from velocity triangles (Fig.9.3) will be used throughout this chapter. From velocity triangles at the entry: ca1

= c1 cos α1

=

w1 cos β1

(9.2)

ct1

= c1 sin α1

=

ca1 tan α1

(9.3)

wt1

= w1 sin β1

=

ca1 tan β1

(9.4)

u

= ct1 + wt1

(9.5)

u

= c1 sin α1 + w1 sin β1

(9.6)

u

= ca1 (tan α1 + tan β1 )

(9.7)

From velocity triangles at the exit: ca2

= c2 cos α2

=

w2 cos β2

(9.8)

ct2

= c2 sin α2

=

ca2 tan α2

(9.9)

wt2

= w2 sin β2

=

ca2 tan β2

(9.10)

u

= ct2 + wt2

(9.11)

u

= c2 sin α2 + w2 sin β2

(9.12)

u

= ca2 (tan α2 + tan β2 )

(9.13)

For constant axial velocity through the stage: ca1 ca

= ca2

=

ca3

=

ca

= c1 cos α1

=

w1 cos β1

= c2 cos α2

=

w2 cos β2

(9.14)

(9.15)

Equations 9.7 and 9.13 give u ca

=

1 φ

=

tan α1 + tan β1

=

tan α2 + tan β2

(9.16)

This relation can also be presented in another form using Eqs. 9.5 and 9.11, ct1 + wt1 = ct2 + wt2 ct2 − ct1

=

ca (tan α2 − tan α1 ) =

wt1 − wt2

(9.17)

ca (tan β1 − tan β2 )

(9.18)

Equations 9.17 and 9.18 give the change in the swirl components across the rotor blade row. For steady flow in an axial machine, this is proportional to the torque exerted on the fluid by the rotor.

Axial Flow Compressors

339

Table 9.1 illustrates what happen to the flow velocity and pressure when air passes through the stage of the compressor. Table 9.1 Variations occurring in an axial flow compressor Absolute velocity, c

Relative velocity, w

Flow width

Static pressure, p

Total pressure, p0

Rotor

Increases

Decreases

Increases

Increases

Increases

Stator

Decreases

–

Increases

Increases

About constant

9.3.1

Work Input to the Compressor

Now let us derive an expression for work input in terms of velocity and blade angles. The derivation is based on the assumption that the axial velocity, ca , remains constant throughout the machine. This assumption simplifies the design calculations as well as expression for work input and pressure rise. To maintain the axial velocity from the first stage to the last stage (Fig. 9.1) constant throughout the machine, the area of flow is made converging as pressure is increasing in every stage. Now, from Eq. 9.16 u =

ca (tan α1 + tan β1 )

=

ca (tan α2 + tan β2 )

(9.19)

also W

=

u(ct2 − ct1 )

=

u(ca tan α2 − ca tan α1 )

=

uca (tan α2 − tan α1 )

(9.20)

(9.21)

In terms of angle, β, it can be written as W

=

uca (tan β1 − tan β2 )

(9.22)

According the Euler’s⎡ energy equation (Eq. 3.13) E

=

⎤

1⎢ 2 2 2 2 2 2 ⎥ ⎣ c1 − c2 + u1 − u2 + w2 − w1 ⎦ 2 I

II

III

For axial flow compressors (u = u1 = u2 ), the above equation will reduce to 1 2 1 c − c21 + w12 − w22 W = (9.23) 2 2 2 To obtain higher efficiencies the work input should be as minimum as possible. To achieve this, proper care in the design of blade and flow geometries are essential.

340 9.4

Gas Turbines

WORK DONE FACTOR

Owing to secondary flows and the growth of boundary layers on the hub and casing of the compressor annulus, the axial velocity along the blade height is far from uniform. This effect is not so prominent in the first stage of a multistage machine but is quite significant in the subsequent stages. Figure 9.4 depicts the typical axial velocity distributions in the first and last stages of a multistage axial compressor. The degree of distortion of the axial velocity distributions in the last stage will depend on the number of stages. On account of this, the axial velocity in the hub and tip regions is Casing

Actual Flow

Annulus height

Mean First

Last

Hub

Axial Flow Compressors

341

The air angles β2 and α1 are fixed by the cascade geometry of the rotor blades and the upstream blade row. Therefore, assuming (tan α1 + tan β2 ) and u as constant, Eq. 9.28 relates work to the axial velocity at various sections along the blade height. The velocity triangles of Fig.9.3 are redrawn in Fig.9.5 for the design value (mean value shown in Fig.9.4), and the reduced (ca − Δca ) and increased (ca + Δca ) values of the axial velocity. Increased incidence w1

α1 c1

u u u Reduced incidence c2

w c - Δca 2 β ca a 2

ca+Δca

Reduced ca Design ca Increased ca

Reduced c a Design ca Increased ca

u u u

Fig. 9.5 Effect of axial velocity on the stage velocity triangles and work It is seen from the velocity triangles that the work absorbing capacity decreases with an increase in the axial velocity and vice versa. Therefore, the work absorbing capacity of the stage is reduced in the central region of the annulus and increased in the hub and tip regions. However, the expected increase in the work at the hub and tip is not obtained in actual practice on account of higher losses. Therefore, the net result is that the stage work absorbing capacity is less than that given by Euler’s equation based on a constant value of the axial velocity along the blade height. This reduction in the work absorbing capacity of the stage is taken into account by a factor known as “workdone factor”. This varies from 0.98 to 0.85 depending on the number of stages. Thus, the work done factor accounts for the effect of boundary layer and tip clearance. It is an empirical factor which reduces the capacity of compressor. It is denoted by Ω. It takes into account the axial velocity distribution also which is otherwise assumed constant. Therefore, the workdone on air becomes W

= Ωuca (tan β1 − tan β2 )

(9.29)

In terms of temperature difference, we have ΔTs Cp ΔTs ΔTs

= T02 − T01

(9.30)

= Ωuca (tan β1 − tan β2 ) =

Ωuca (tan β1 − tan β2 ) Cp

(9.31)

342

Gas Turbines

In fact, the stage temperature rise will be less due to the three dimensional effects in the compressor annulus. Experiments have shown that in order to get the actual energy transfer the result obtained should be multiplied by a factor Ω which is the work done factor (refer Eq. 9.29). Work done factor is really a measure of the ratio of the actual work absorbing capacity of the stage to its ideal value as calculated from the equation. The explanation of this is based on the fact that the axial velocity contribution is not constant across the annulus. The magnitude increases as the flow velocity across the annulus has compensating effects in respect of work capacity. Unfortunately, the influence of the boundary layer and tip clearance has an adverse effect on this compensation and the net result is a loss in total work capacity which is accounted for by the work done factor, Ω. 9.5

ENTHALPY–ENTROPY DIAGRAM

Figure 9.6 shows the enthalpy–entropy diagram for a general axial flow compressor stage. Static and stagnation values of pressures and enthalpies at various stations are as shown in Fig.9.6. 1–2 –3 shows isentropic compression whereas 1–2–3 shows actual compression. The stagnation point 03 corresponds to the final state at the end of isentropic compression.

1/2 c22

02

03

p

3

02

03’

p

Stator

03

p

3’

p

2’

3

h

1/2 c32

2

2

p

01

01’

p 01

1/2 c12

Rotor

1

1 s

Fig. 9.6 Enthalpy–entropy diagram of an axial compressor From Fig.9.3 it can be seen that air enters the rotor blades with lower absolute velocity (c1 ) but with large relative velocity (w1 ) whereas it leaves the rotor with large c2 and lower w2 . However, when it comes out of the

Axial Flow Compressors

343

diffuser blades c3 is reduced which will be close to (very slightly higher) c1 . Hence the stagnation pressure p01 will be slightly higher than the static pressure p1 by 12 c21 which is shown in Fig.9.6. However, the stagnation pressure p02 will be much higher than the static pressure p2 as can be seen from the Fig.9.6. However, the flow occurs at constant enthalpy as can been seen in the figure. That is h02

=

h03

1 h2 + c22 2

=

1 h3 + c23 2

(9.32)

Further, it should be noted that the actual energy transformation process (1 − 2) and (2 − 3) in the rotor and diffuser blade rows occur with stagnation pressure loss and increase in entropy. However, the relative stagnation enthalpy remains constant. h01

9.6

rel

=

h02

1 h1 + w12 2

=

1 h2 + w22 2

rel

(9.33)

COMPRESSOR STAGE EFFICIENCY

The efficiency of the compression process can now be defined based on ideal and actual process as shown in the h–s diagram (Fig.9.6). The ideal work input to the stage Wideal

= h03 − h01

(9.34)

= Cp (T03 − T01 )

(9.35)

This is the minimum stage work input required to obtain a static pressure rise of p3 − p1 . However, the actual process due to various losses and the associated irreversibilities will require a higher magnitude of work input for the same pressure rise. This is given by Wactual

=

h03 − h01

(9.36)

=

Cp (T03 − T01 )

(9.37)

The compressor stage efficiency pertaining to total (stagnation) conditions at entry and exit can be written as ηpc

=

Wideal Wactual

(9.38)

=

T03 − T01 T03 − T01

(9.39)

The magnitude of the stage work can now be written in terms of actual velocities and air angles from the velocity triangles. Using Eq. 9.21, we can

344

Gas Turbines

write Wactual

=

h03 − h01

=

uca (tan α2 − tan α1 )

=

uca (tan β1 − tan β2 )

=

1 2 1 c − c21 + w12 − w22 2 2 2

Knowing the stage pressure ratio, the isentropic enthalpy drop, h03 − h01 , can be calculated and thereby ηpc can be evaluated from the Eq. 9.38. 9.7

PERFORMANCE COEFFICIENTS

In order to evaluate the performance of the compressor same dimensionless performance coefficients are found useful in various analyses. We will discuss them briefly in the following section. 9.7.1

Flow Coefficient

It is defined as the ratio of axial velocity to peripheral speed of the blades φ

=

ca u

Flow coefficient is sometimes called as compressor-velocity ratio. It may be noted that φ is sensitive to changes in angle of incidence, and as such it is a useful parameter for representing the stalling characteristics of the compressor. 9.7.2

Rotor Pressure Loss Coefficient

It is defined as the ratio of the pressure loss in the rotor due to relative motion of air to the pressure equivalent of relative inlet velocity Yrel

9.7.3

=

p01

rel − p02 rel 1 2 2 ρw1

(9.40)

Rotor Enthalpy Loss Coefficient

It is defined as the ratio of the difference between the actual and isentropic enthalpy to the enthalpy equivalent of relative inlet velocity ξrel

=

h2 − h2 1 2 2 w1

=

Cp (T2 − T2 ) 1 2 2 w1

(9.41)

Because of friction and churning, the enthalpy at the outlet will be more and thereby more work input will become necessary.

Axial Flow Compressors

9.7.4

345

Stator or Diffuser Pressure Loss Coefficient

It is defined as the ratio of the pressure loss in the diffuser due to flow velocity to the pressure equivalent of actual inlet velocity of the diffuser. YD 9.7.5

p02 − p03 1 2 2 ρc2

=

(9.42)

Stator or Diffuser Enthalpy Loss Coefficient

It is defined as the ratio of the difference between the actual and isentropic enthalpy the enthalpy equivalent of absolute velocity of flow at diffuser inlet. ξD 9.7.6

=

h3 − h3 1 2 2 c2

=

Cp (T3 − T3 ) 1 2 2 c2

(9.43)

Loading Coefficient

It is defined as the actual stagnation enthalpy rise in the stage to enthalpy equivalent of peripheral speed of the rotor. Ψ =

h03 − h01 u2

=

W u2

Substituting for W from Eq. 9.22, Ψ =

uca (tan β1 − tan β2 ) u2

=

φ(tan β1 − tan β2 )

=

Cp (T03 − T01 ) u2

=

φ(tan α2 − tan α1 )

(9.44) (9.45)

Loading coefficient in terms of ηpc can be written as Ψ =

Cp (T03 − T01 ) u2 ηpc

(9.46)

Some designers define the loading coefficient as the ratio of stage work to the blade kinetic energy W ψ = (9.47) 1 u2 2 But we will use ψ without the factor 2, i.e., Eq.9.46.

9.8

DEGREE OF REACTION

The degree of reaction prescribes the distribution of the stage pressure rise between the rotor and the diffuser blade rows. This in turn determines the

346

Gas Turbines

cascade losses in each of these blade rows. The degree of reaction for axial compressors can also be defined in a number of ways: it can be expressed either in terms of enthalpies, pressures or flow geometry. For an actual compressor stage the degree of reaction is defined as R

=

actual change of enthalpy in the rotor actual change of enthalpy in the stage

=

h2 − h1 h3 − h1

=

T2 − T1 T3 − T1

(9.48)

For c1 = c3 , h3 − h1

= h03 − h01

=

u(ct2 − ct1 )

From Eq. 9.33, we have h2 − h1

=

1 2 w − w22 2 1

Now, degree of reaction, R, can be written as R

=

h2 − h 1 h03 − h01

(9.49)

=

w12 − w22 2u(ct2 − ct1 )

(9.50)

Equation 9.50 can be further expressed in terms of air angles.

But

R

=

c2a tan2 β1 − tan2 β2 2uca (tan β1 − tan β2 )

R

=

1 ca (tan β1 + tan β2 ) 2 u

(9.51)

ca = φ, and u

1 (tan β1 + tan β2 ) 2

= tan βm

Therefore, R

= φ tan βm

(9.52)

Equation 9.51 can be rearranged to give 1 ca [(tan β1 + tan α1 ) − (tan α1 − tan β2 )] R = 2 u From Eq. 9.27 tan β1 + tan α1 Therefore,

=

u ca

Axial Flow Compressors

R

1 1 ca − (tan α1 − tan β2 ) 2 2 u

=

347

(9.53)

This is a useful relation in terms of the geometry of flow and can be used to study the effect of air angles and the required cascade geometry (to provide these air angles) on the degree of reaction of an axial compressor stage. 9.8.1

Low Reaction Stages

A low reaction stage has a lesser pressure rise in its rotor compared to that in the diffuser, i.e., (Δp)rel < (Δp)D . In such a stage the quantity (tan α1 − tan β2 ) is positive or in other words α1 > β2 (Fig.9.3 and Eq. 9.53). The same effect can be explained in another manner. In Eq. 9.53 ca tan α1

= ct1

=

u − wt1

ca tan β2

= wt2

=

u − ct2

Therefore, after substituting these values in Eq. 9.53 R

=

wt2 1 1 ct1 − − 2 2 u u

(9.54)

=

1 1 − (ct2 − wt1 ) 2 2u

(9.55)

This equation relates the degree of reaction to the magnitudes of swirl or the whirl components approaching the rotor and the diffuser. Thus a low degree of reaction is obtained when the rotor blade rows remove less swirl compared to the diffuser blade rows, i.e., wt1 < ct2 Figure 9.7 shows the enthalpy–entropy diagram for such a stage. The swirl removing ability of a blade row is reflected in the static pressure rise across it. In a low-degree reaction stage the diffuser blade rows are burdened by a comparatively larger static pressure rise which is not desirable for obtaining higher efficiencies. 9.8.2

Fifty Per cent Reaction Stages

One of the ways to reduce the burden of a large pressure rise in a blade row is to divide the stage pressure rise equally between the rotor and diffuser. To approach this condition (Fig.9.8). h 2 − h1

= h3 − h2 =

1 (h3 − h1 ) 2

This when substituted in Eq. 9.48 gives R = 12 . Equation 9.50 for R =

1 2

gives

(9.56)

348

Gas Turbines

p

3

3 3’ Enthalpy

Diffuser

h3- h 2 p

2

2

2’

h2 - h 1

Rotor

p

1

1 Entropy

Fig. 9.7 Enthalpy–entropy diagram for a low reaction stage R
(Δp)D . Figure 9.9 shows the enthalpy diagram. For such a stage the quantity (tan α1 − tan β2 ) in Eq. 9.53 is negative, giving R > 12 . Therefore, for such a stage α1 < β2

and

wt1 > ct2

(9.62)

Figure 9.10 shows the velocity triangles for such a stage. It can be observed that the rotor blade row generates a higher static pressure on account of the larger magnitude of the swirl component wt1 at its entry. The swirl component ct2 passed on to the diffuser blade row is relatively smaller, resulting in a lower static pressure rise therein. Since the rotor blade rows have relatively higher efficiencies, it is advantageous to have a slightly greater pressure rise in them compared to the diffuser.

350

Gas Turbines

p

3

3 3’ Enthalpy

Diffuser

h3 - h 2 p

2

2

2’ h2- h 1 Rotor p

1

1 Entropy

Fig. 9.9 Enthalpy–entropy diagram for a high reaction stage R >

β1 α1

w1

u

w2 wt2

β2 α2

u

c2 c t2

u

ct1

wt1

Rotor blades

c1

Diffuser blades

α3

c3

Fig. 9.10 High reaction stage R > 12 , α1 < β2

1 2

Axial Flow Compressors

9.9

351

FLOW THROUGH BLADE ROWS

After studying the geometry and thermodynamics of the flow through a compressor stage, further insight can be obtained by looking at the flow in the individual blade rows. Therefore, the two parts of the h–s diagram in Fig.9.6 (for the stage) are redrawn in Figs.9.11 and 9.12. The similarity between these diagrams must be noted. p

p

01 rel

02 rel

Enthalpy

01 rel

02 rel

h 01 rel = h02 rel

1/2 w22

2 1/2 w2’

p

2

2 2’ 1/2 w12

p

1

1 Entropy

Fig. 9.11 Enthalpy–entropy diagram for flow through rotor blade row The flow over a small pressure rise can be considered incompressible, i.e., density can be assumed to remain constant with little sacrifice in accuracy. 9.9.1

Rotor Blade Row

The flow process as observed by an observer sitting on the rotor is shown in Fig.9.11. The initial and final pressures are p1 and p2 for both isentropic and adiabatic processes. In the isentropic process the flow will diffuse to a velocity w2 giving the stagnation enthalpy and pressure as h01 rel and p01 rel respectively. h2 − h1

=

1 (p2 − p1 ) ρ

=

1 (Δp)rel ρ

(9.63)

The actual process gives the final velocity w2 and stagnation pressure p02 rel . Here the same static pressure rise (Δp)rel occurs with a greater change in the kinetic energy 12 w12 − w22 . In the ideal or isentropic process this is 1 2 1 w − w22 < w12 − w22 2 1 2

352

Gas Turbines

p

02

02

p

h02= h03

1/2 c 32

1/2 c23’ Enthalpy

03

03

p

3

3 3’

1/2 c 22

p

2

2 Entropy

Fig. 9.12 Enthalpy–entropy diagram for flow through diffuser (stator) blade row This difference is due to the losses and the increase in entropy. The efficiency of the rotor blade row can now be defined by ηrel

=

h2 − h1 h2 − h1

= 1−

=

(h2 − h1 ) − (h2 − h2 ) h2 − h1

h 2 − h2 h2 − h1

(9.64) (9.65)

Assuming perfect gas and substituting from Eq. 9.41 ηrel

= 1− = 1−

w12 ξrel 2Cp (T2 − T1 ) ξrel 1−

(9.66) (9.67)

w22 w12

The assumption of incompressible flow is not required in Eqs. 9.65, 9.66 and 9.67 h2 − h2

= (h2 − h1 ) − (h2 − h1 )

For incompressible flow, substituting from Eq. 9.63 1 2 1 w − w22 − (p2 − p1 ) h2 − h2 = 2 1 ρ =

1 ρ

1 p1 + ρw12 2

1 − p2 + ρw22 2

Axial Flow Compressors

1 (p01 ρ

=

rel

− p02

rel )

=

(Δp0 )rel ρ

353

(9.68)

Substituting this in Eq. 9.65 ηrel

= 1−

(Δp0 )rel ρ(h2 − h1 )

(9.69)

Substitution from Eq. 9.40 gives ηrel

9.9.2

= 1−

Yrel 1−

(9.70)

w22 w12

Stator Blade Row

The ideal and actual flow processes occurring in the diffuser blade row are shown in Fig. 9.12. Its efficiency is again defined similar to that of rotor blade row. ηD

=

h3 − h2 h3 − h2

=

1−

=

(h3 − h2 ) − (h3 − h3 ) h3 − h2

h 3 − h3 h3 − h2

(9.71) (9.72)

Substituting in terms of the enthalpy loss coefficient ηD

=

1−

c22 ξD 2Cp (T3 − T2 )

=

1−

ξD 1−

c23 c22

(9.73)

For incompressible flow, h 3 − h3

=

1 (p02 − p03 ) ρ

=

(Δp0 )D ρ

(9.74)

Therefore, the diffuser efficiency can be expressed in terms of the diffuser stagnation pressure loss coefficient

9.10

ηD

=

1−

ηD

=

1−

(Δp0 )D ρ(h3 − h2 ) YD 1−

c23 c22

(9.75) (9.76)

FLOW LOSSES

Principal aerodynamic losses occurring in most of the turbomachines arise due to the growth of the boundary layer and its separation on the blade and passage surfaces. Others occur due to wasteful circulatory flows and the formation of shock waves. Non-uniform velocity profiles at the exit of the

354

Gas Turbines

cascade lead to another type of loss referred to as the mixing or equalization loss. Aerodynamic losses occurring in a turbomachine blade cascade can be grouped in the following categories. 9.10.1

Profile Loss

As the term indicates, this loss is associated with the growth of the boundary layer on the blade profile. Separation of the boundary layer occurs when the adverse pressure gradient on the surface or surfaces becomes too steep; this increases the profile loss. The pattern of the boundary layer growth and its separation depend on the geometries of the blade and the flow. Positive and negative stall losses occur on account of increased positive or negative incidences respectively. Generally, the suction surface of a blade is more prone to boundary layer separation. The separation point depends besides the blade profile on factors like the degree of turbulence, Reynolds number and the incidence. If the flow is initially supersonic or becomes supersonic on the blade surface additional losses occur due to the formation of shock waves resulting from the local deceleration of supersonic flow to subsonic. 9.10.2

Annulus Loss

The majority of blade rows in turbomachines are housed in casings. The axial compressor stage has a pair of fixed and moving blade rows. In stationary blade rows a loss of energy occurs due to the growth of the boundary layer on the end walls. This also occurs in the rotating row of blades but the flow on the end walls in this case is subjected to effects associated with the rotation of the cascade. The boundary layer on the floor (hub) of the blade passages is subjected to centrifugal force, whereas that on the ceiling (outer casing) is scrapped by the moving blades. 9.10.3

Secondary Loss

This loss occurs in the regions of flow near the end walls owing to the presence of unwanted circulatory or cross flows. Such secondary flows develop on account of turning of the flow through the blade channel in the presence of annulus wall boundary layers. Figure 9.13 depicts the pressure gradients across a blade channel and the secondary and trailing vortices. Static pressure gradients from the suction to the pressure side in the regions away from the hub and tip are represented by the curve AB. In the vicinity of the hub and tip or the end walls the pressure gradient curve CD is not so steep on account of much lower velocities due to the wall boundary layers. Thus the static pressures at the four corners of the section of flow under considerations are pB > pD > pC > pA These pressure differentials across the flow near the end walls give rise

Axial Flow Compressors

B D

p C A

y

C A

Tip

355

B

D

C

D

Over deflection

A

B

Under deflectio

Secondary vortices

Blade wakes Trailing vortices

Hub

A

B

C

D

Fig. 9.13 Secondary flow in a cascade of blades to circulatory flows which are superimposed on the main flow through the blade passage. As a result of this, secondary vortices in the streamwise direction are generated in the blade passages. These vortices, besides wasteful expenditure of fluid’s energy, transport (DC) low energy fluid from the pressure to the suction side of the blade passage, thus increasing the possibility of an early separation of the boundary layer on the suction side. The flow nearer the hub and tip is over-deflected while that slightly away from the end walls is under-deflected as shown in Fig.9.13. The secondary vortices in the adjacent blade channels induce vortices in the wake regions (as shown in Fig.9.13). These trailing vortices lead to additional losses. It is worth observing here that the secondary flows in the cascade also affect the profile and annulus losses. The magnitude of the loss due to secondary flow depend on the fraction of the passage height that is affected by this flow. Blade passages of very low height (aspect ratio) or high hub-tip ratio are likely to be fully occupied by secondary vortices as shown in Fig.9.14(a), and experience higher secondary losses. In contrast to this longer blades [Fig.9.14(b)] have a large proportion of the flow free of secondary flows and therefore experience comparatively lower secondary losses. If the total losses in a blade passage are measured along its height, they appear as peaks near the hub and tip on account of secondary losses. The flow in the central region which is outside the influence of secondary flows (particularly in longer blades) can be assumed to suffer only profile loss. Figure 9.15 illustrates this pattern of losses along the blade height. 9.10.4

Tip Clearance Loss

This loss arises due to the clearance between a moving blade and the casing. In a turbine rotor blade ring the suction sides lead and the pressure sides

356

Gas Turbines

Tip

Tip Hub

Hub

(a) High hub-tip ratio

(b) Low hub-tip ratio

Fig. 9.14 Secondary vortices in short and long blades

Y=

Secondary loss

Δ p0 1 2

ρ c22

Hub

Profile loss Tip Blade height

Fig. 9.15 Variation of losses along the blade height trail. On account of the static pressure difference, the flow leaks from the pressure side towards the suction side as shown in Fig.9.16. However, due to the scrapping up of the casing boundary layer by the blade tips, the scrapped up flow opposes the aforementioned tip leakage. The tip clearance and secondary flows are closely related to each other and it is often convenient to estimate them together. 9.11

STAGE LOSSES

Figure 9.17 shows the energy flow diagram for an axial flow compressor stage. Figures in the brackets indicate the order of energy or loss corresponding to 100 units of energy supplied at the shaft. The stage work (h03 − h01 ) is less than the energy supplied to the shaft by the prime mover on account of bearing and disc friction losses. All the stage work does not appear as energy at the stator entry on account of aerodynamic losses in the rotor blade row. After deducting the stator (diffuser) blade row losses from the energy at its entry, the value of the ideal or isentropic work required to obtain the stage pressure rise is obtained. The cascade losses in the rotor and stator would depend on the degree of reaction. The values shown in the energy flow diagram are only to give

Axial Flow Compressors

357

Suction side Pressure side

Tip leakage Scraped flow

Motion of blades (scraping)

Fig. 9.16 Flow through tip clearance Energy from the prime mover or shaft work (100)

Stage work (98) Rotor aerodynamic losses (9)

Shaft loses (2) Disc Bearing friction loss loss

Secondary Profile Annulus loss loss loss Energy at the stator entry (89)

Tip leakage

Isentropic work (82) Stator aerodynamic losses(7)

Secondary loss Profile loss Annulus loss

Fig. 9.17 Energy flow diagram for an axial flow compressor stage an example. The ratio of the isentropic work (82) and the actual stage work (98) gives the stage efficiency, whereas the overall efficiency is directly obtained as 82%. 9.12

PRESSURE RISE CALCULATION IN A BLADE RING

In case of aerofoil blading we already know that lift, L, is given by L =

CL ρw2 A 2

(9.77)

D

CD ρw2 A 2

(9.78)

and the drag, D, is given by =

358

Gas Turbines

Now R

=

F

=

L sin θ + D cos θ

where F = R is the resultant force in kg and θ is the angle as shown in Fig.9.18. R is the force on each blade and if n is the number of blades in each blade ring then, (L sin θ + D cos θ) nd Torque on shaft,τ = (9.79) 2 and so the power required (W τ ) P

=

2πN nd

(L sin θ + D cos θ) 2

(9.80)

where N is in rev/m and d is diameter of wheel P

=

nπdN (L sin θ + D cos θ)

and if the isentropic efficiency is known then we can obtain the pressure ratio as follows : γ−1 T01 P = m ˙ Cp r γ − 1 (9.81) ηc

L

f=R θ D

Fig. 9.18 Aerofoil blade pressure rise where

T01 ηc r From Eqs. 9.80

: inlet temperature of air : isentropic efficiency of compression : pressure ratio and 9.81

r=

ηc nπdN (L sin θ + D cos θ) + 1 T01 m ˙

γ γ−1

(9.82)

Thus, we will get the pressure ratio by substituting the values of other terms. This will be pressure ratio that can be obtained in a stage or in a blade ring.

Axial Flow Compressors

9.13

359

PERFORMANCE CHARACTERISTICS

The performance characteristics of axial compressors or their stages at various speeds can be presented in terms of the plots of the following parameters: (i) Pressure rise vs flow rate, Δp

= f (Q)

=

f (m)

(ii) Pressure ratio vs non-dimensional flow rate (Fig.9.19), √ p2 m T01 = f p1 p01 (iii) Loading coefficient vs flow coefficient (Fig.9.20), ψ

= f (φ)

(9.83)

The actual performance curve based on measured values is always below the ideal curve obtained theoretically on account of losses. This is shown in Fig.9.21. The surge point and stable and unstable flow regimes have been explained in the following sections. 9.13.1

Off-design Operation

A compressor gives its best performance while operating at its design point, i.e., at the pressure ratio and flow rate for which it has been designed. However, like any other machine or system, it is also expected to operate away from the design point. Therefore, a knowledge about its behaviour at off-design operation is equally important. Off-design characteristic curves can be obtained theoretically from Eq. 9.44. ψ

=

φ(tan β1 − tan β2 )

tan α2

=

1 − tan β2 φ

=

1 − φ(tan β2 + tan α1 )

But,

=

φ(tan α2 − tan α1 )

(9.84)

Therefore, ψ

(9.85)

The quantity (tan β2 + tan α1 ) can be assumed constant in a wide range of incidence up to the stalling value is. This is justified in view of small variations in the air angles at the rotor and stator exits. Therefore, writing α1 = α3 , A =

tan β2 + tan α3

(9.86)

If the design values are identified by the superscript∗ , Eq. 9.85 along with Eq. 9.86 can be written as

Gas Turbines

Su

rge

lin

e

360

p

N

p

T 01

02

01

m

= constan

T 01

p

01

Fig. 9.19 Performance curves of a compressor

50% reaction stage Δw/u 2= Δ p / ρ u

2

1 α = β = constant 1

2

0.5

0 0

0.5 φ = u/ca

1.0

Fig. 9.20 Variation of pressure coefficient with flow coefficient for an axial flow compressor stage

Axial Flow Compressors

361

Pressure rise

Surge point Stage losses

Ideal

Stable

l tua

Ac

Unstable

Flow rate

Fig. 9.21 Ideal and actual performance curves for an axialcompressor ψ∗

=

A =

1 − Aφ∗

(9.87)

1 − ψ∗ φ∗

At off-design conditions ψ

=

1 − Aφ

=

1 − (1 − ψ ∗ )

φ φ∗

(9.88)

This equation also gives the off-design characteristic of an axial flow compressor. Figure 9.22 shows theoretical values of A, the curves are falling while for negative values rising characteristics are obtained. The actual curves will be similarly modified but slightly on account of losses. 9.13.2

Surging

Unstable flow in axial compressors can be due to two reasons: (i) separation of flow from the blade surfaces called stalling, and (ii) complete breakdown of the steady through flow called surging. Both these phenomena occur due to off-design conditions of operation and are aerodynamically and mechanically undesirable. Sometimes, it is difficult to differentiate between operating conditions leading to stalling and surging. It may be noted that the flow in some regions stalls without surging taking place. Surging affects the whole machine while stalling is a local phenomenon. Some typical performance characteristic curves at different speeds (N1 , N2 , etc.,) are shown in Fig.9.23. The surge phenomenon is explained with the

362

Gas Turbines

A = negative

Loading coefficient

1.0

A=0

A = positive

0.75 0.50

0.5

0.25 0.0 0.0

0.25

0.50

1.0 1.0

0.75

Flow coefficient

Fig. 9.22 Off-design characteristic curves for axial compressor stage aid of one of the curves in this figure. Let the operation of the compressor at a given instant of time be represented by point A(pA , m ˙ A ) on the characteristic N3 curve. If the flow rate through the machine is reduced to m ˙ B by closing a valve on the delivery pipe, the static pressure upstream of valve is increased. This higher pressure, pB is matched with the increased delivery pressure (at B) developed by the compressor. With further throttling of the flow (to m ˙ C and m ˙ S ), the increased pressures in the delivery pipe are matched by the compressor delivery pressures at C and S on the characteristic curve. Surge cycle D S

Pressure r se

E

C

B

A

N4

S N3

Surge line

S

N2 N1 mE mD mS mB mA Flow rate

Fig. 9.23 Surging in compressors The characteristic curve at flow rates below m ˙ S provides lower pressure as at D and E. However, the pipe pressures due to further closure of the valve (point D) will be higher than these. This mismatching between the

Axial Flow Compressors

363

pipe pressure and compressor delivery pressure can only exist for a very short time. This is because the higher pressure in the pipe will blow the air towards the compressor, thus reversing the flow leading to a complete breakdown of the normal steady flow from the compressor to the pipe. During this very short period the pressure in the pipe falls and the compressor regains its normal stable operation (say at point B) delivering higher flow rate (m ˙ B ). However, the valve position still corresponds to the flow rate m ˙ D . Therefore, the compressor operating conditions return through points C and S to D. Due to the breakdown of the flow through the compressor, the pressure falls further to pE and the entire phenomenon, i.e., the surge cycle EBCSDE is repeated again and again. The frequency and magnitude of this to-and-fro motion of the air (surging) depend on the relative volumes of the compressor and delivery pipe, and the flow rate below m ˙ S. Surging of the compressor leads to vibration of the entire machine which can ultimately lead to mechanical failure. Therefore, the operation of compressors on the left of the peak of the performance curve is injurious to the machine and must be avoided. Surge points (S) on each curve corresponding to different speeds can be located and a surge line drawn as shown in Fig.9.23. The stable range of operation of the compressor is on the right-hand side of this line. There is also a limit of operation on the extreme right of the characteristics when the mass-flow rate cannot be further increased due to choking. This is obviously a function of the Mach number which itself depends on the fluid velocity and its state.

9.13.3

Stalling

As stated earlier, stalling is the separation of flow from the blade surface. At low flow rates (lower axial velocities), the incidence is increased as shown in Fig.9.5. At large values of the incidence, flow separation occurs on the suction side of the blades which is referred to as positive stalling. Negative stall is due to the separation of flow occurring on the pressure side of the blade due to large values of negative incidence. However, in a great majority of cases this is not as significant as the positive stall which is the main subject under consideration in this section. In a high pressure ratio multistage compressor the axial velocity is already relatively small in the higher pressure stages on account of higher densities. In such stages a small deviation from the design point causes the incidence to exceed its stalling value and stall cells first appear near the hub and tip regions. The size and number of these stall cells or patches increase with the decreasing flow rates. At very low flow rates they grow larger and affect the entire blade height. Large-scale stalling of the blades causes a significant drop in the delivery pressure which can lead to the reversal of flow or surge. The stage efficiency also drops considerably on account of higher losses. The axisymmetric nature of the flow is also destroyed in the compressor annulus.

364 9.13.4

Gas Turbines

Rotating Stall

Figure 9.24 shows four blades (1, 2, 3 and 4) in a compressor rotor. Owing to some distortion or non-uniformity of flow one of the blades (say the third) receives the flow at increased incidence. This causes this blade (number three) to stall. On account of this the passage between the third and fourth blades is blocked causing deflection of flow in the neighbouring blades. As a result, the fourth blade again receives flow at increased incidence and the second blade at decreased incidence. Therefore, stalling occurs on the fourth blade also. This progressive deflection of the flow towards the left clears the blade passages on the right on account of the decreasing incidence and the resulting unstalling. Thus the stall cells or patches move towards the left-hand side at a fraction of the blade speed. In the relative system they appear to move in a direction opposite to that of the rotor blades. However, on account of their (stall cell) lower speed as compared to that of the rotor, they move at a certain speed in the direction of the rotation in the absolute frame of coordinates. Rotating stall cells, develop in a Increased incidence Reduced incidence

Air 4

3

2

Propagating

1

u

Unstalling

Fig. 9.24 Stall propagation in a compressor blade row variety of patterns at different off-design conditions as shown in Fig.9.25. The blades are subjected to forced vibrations on account of their passage through the stall cells at a certain frequency. The frequency and amplitude of vibrations depend on the extent of loading and unloading of the blades, and the number of stall cells. Under these conditions the blades can fail due to resonance. This occurs when the frequency of the passage of stall cell through a blade coincides with its natural frequency. Both the efficiency and delivery pressure drop considerably on account of rotating stall.

Stall cells

Axial Flow Compressors

9.14

365

COMPARISON OF AXIAL AND CENTRIFUGAL COMPRESSORS

In this section, we will compare the axial and centrifugal compressors under twelve different headings, such as (i) type of flow (ii) pressure ratio per stage (iii) isentropic efficiency (iv) frontal area (v) flexibility of operation (vi) part load performance (vii) effect of deposits (viii) starting torque (ix) suitability for multistaging (x) delivery pressure (xi) applications (xii) efficiency with respect to speed (i) Type of flow For the axial compressor the flow direction is parallel to the axis of the machine whereas for the centrifugal compressor it is radial. There is no difference in radius for the axial flow compressor between inlet and exit whereas there is large radius difference in the case of centrifugal compressors. (ii) Pressure ratio In a single stage the centrifugal compressor is capable of developing a pressure ratio close to 5, whereas in the case of axial flow compressor it is only about 1.25. The supersonic centrifugal compressors can go upto a pressure ratio of 10. In order to achieve the pressure ratio of a single stage of centrifugal compressor, multistaging is required in axial flow machines. (iii) Isentropic efficiency The isentropic efficiency of axial flow compressors are higher (86 to 88%) compared to centrifugal compressor (80 to 82%). Higher isentropic efficiency of modern axial flow compressor is due to the aerofoil blades. (iv) Frontal area The frontal area of a centrifugal compressor is larger due to flow entering at one radius and leaving at another larger radius. Since there is no radius difference at entry and exit for an axial compressor the frontal area is smaller.

366

Gas Turbines

(v) Flexibility of operation Comparatively higher due to adjustable prewhirl and diffuser vanes in the case of centrifugal compressors. Due to the absence of above two factors, the flexibility of operation is quite limited in the case of axial flow compressors. (vi) Part load performance Since variation in mass flow rate can be tolerated to some extent in centrifugal compressors, part load performance is better whereas it is poor in axial flow compressors. (vii) Effect of deposits Deposits on the surface of the rotor are a major factor in the operation and performance of compressors. However, deposits do not adversely affect the performance in the case of centrifugal compressors whereas it adversely affects the operation and performance of the axial flow machines. (viii) Starting torque The starting torque required is low in the case of centrifugal compressors whereas it is high in the case of axial flow machines. (ix) Multistaging Comparatively difficult in centrifugal compressors. Maximum of only two staging is possible. The axial flow machines are more suitable for multistaging. (x) Delivery pressure Higher delivery pressure (about 40 bar) is possible in centrifugal machines whereas only 20 bar is possible in axial flow compressors. (xi) Applications Centrifugal compressors are used for blowing in steel mills, low pressure refrigeration systems, big central air-conditioning plants, fertilizer industry, supercharging of IC engines and gas pumping over a long distance pipe lines. Axial flow machines are mainly used in jet engines due to higher isentropic efficiency and lower frontal area. It is also employed in stationary industrial gas turbine power plants and steel mills. (xii) Efficiency with respect to speed The characteristics are more flat in centrifugal compressors whereas it is less flat in axial flow compressors. To put it in a nutshell, the comparison of axial and centrifugal machines are given in Table 9.2. Worked out Examples 9.1 A 10 stage axial flow compressor provides an overall pressure ratio of 5:1 with an overall isentropic efficiency of 87%. When the temperature of air at inlet is 15◦ C. The work is equally divided between the stages. A 50% reaction is used with a blade speed of 210 m/s and a constant axial velocity of 170 m/s. Estimate the blade angles. Assume a work done factor of 1.

Axial Flow Compressors

367

Table 9.2 Comparison of Axial Flow Compressors and Centrifugal Compressors Aspect of comparison

Axial flow compressors

Centrifugal compressors

Type of flow

Parallel to the axis

Radial

Pressure ratio/stage

About 1.25

5

Isentropic efficiency

Higher (86 to 88%)

Lower (80 to 82%)

Frontal area

Smaller

Larger

Flexibility of operation

Limited

Higher

Part load performance

Poor

Better

Effect of deposits

Adverse effect

No adverse effect

Starting torque

High

Low

Suitability for multistaging

More suitable

Difficult

Delivery pressure

Lower (20 bar)

Higher (about 40 bar)

Applications

Jet engines, stationary industrial gas turbine power plants and steel mills.

Steel mills, low pressure refrigeration, big central air-conditioning plants, fertilizer industry, IC engines, gas pumping

Efficiency with respect to speed

Less flat

More flat

Solution Temperature increase per stage ΔTstage

γ−1 T1 r γ −1 nηc

=

ΔToverall n

=

288 × 50.286 − 1 10 × 0.87

=

Increase in temperature per stage, ΔTs

=

Ωuca (tan β1 − tan β2 ) Cp

=

19.35 K

368

Gas Turbines

β α1

w1

1

c1

ca1 wt1

ct1 u

T

02 02’

β 2

w2

c2

α2

ca2 c t2

wt2 u

01 s

Fig. 9.26

=

ΔTs Cp 19.35 × 1005 = 0.545 = Ωuca 1 × 210 × 170

tan β1 + tan β2

=

u ca

tan β1

=

0.89

β1

=

41.67◦

β2

=

tan−1 (1.235 − 0.89)

tan β1 − tan β2 For a 50% reaction,

=

210 170

=

1.235

Ans

⇐= =

19◦

Ans

⇐=

9.2 Air at 1.0132 bar and 288 K enters an axial flow compressor stage with an axial velocity 150 m/s. There are no inlet guide vanes. The rotor stage has a tip diameter of 60 cm and a hub diameter of 50 cm and rotates at 100 rps. The air enters the rotor and leaves the stator in the axial direction with no change in velocity or radius. The air is turned through 30.2◦ as it passes through rotor. Assume a stage pressure ratio of 1.2. Assuming the constant specific heats and that the air enters and leaves the blade at the blade angles, (i) construct the velocity diagram at mean dia for this stage, (ii) mass flow rate, (iii) power required, and (iv) degree of reaction Solution To construct velocity diagram

Axial Flow Compressors

β1

w1

α1

c1=ca1

u

c2

β2 α 2

w2

ca2 c t2

w t2 u

Fig. 9.27

u

β1

=

π(dhub + dtip ) ×N 2

=

π × (0.50 + 0.60) × 100 2

=

tan−1

u ca

= tan−1

=

172.78 150

172.76 m/s = 49◦ 2

As air is deflected by 30◦ , β2

=

49◦ 2 − 30◦ = 19◦

In the figure let Wt2 = x x = ca tan β2

=

150 × tan 19◦ = 51.65 m/s

tan α2

=

u−x ca

α2

=

38.92◦

m ˙

=

π 2 d − d2hub ca ρ2 4 tip

T1

=

T01 −

T2

=

T1

=

172.78 − 51.65 = 80.75 150

C12 1502 = 276.8 K = 288 − 2Cp 2 × 1005

p2 p1

γ−1 γ

= 276.8 × 1.20.286 = 291.6 K

369

370

Gas Turbines

p2

=

1.216 bar

ρ2

=

1.216 × 105 287 × 291.6

m ˙

=

π Ans × 0.602 − 0.502 × 150 × 1.453 = 18.83 kg/s ⇐= 4

P

=

˙ β1 − tan β2 ) Ωuca m(tan

=

1 × 172.76 × 150 × 18.83 × (tan 49.2 − tan 19) 1000

=

397.297 kW

3

=

1.453 kg/m

Ans

⇐=

Degree of reaction R

=

ca (tan β1 + tan β2 ) 2u

=

150 × (tan 49◦ 2 + tan 19◦ ) = 0.65 2 × 172.76

Ans

⇐=

9.3 An axial flow air compressor of 50% reaction design has blades with inlet and outlet angles of 45◦ and 10◦ respectively. The compressor is to produce a pressure ratio of 6:1 with an overall isentropic efficiency of 0.85 when inlet static temperature is 37◦ C. The blade speed and axial velocity are constant throughout the compressor. Assuming a value of 200 m/s for blade speed find the number of stages required if the work done factor is (i) unity and (ii) 0.87 for all stages. Solution β α1

w1

1

c1

ca1 wt1

ct1 u

T

02 02’

β 2

w2

c2

α2

ca2 c t2

wt2 u

01 s

Fig. 9.28

Increase in temperature per stage,

Axial Flow Compressors

ΔTs

=

Ωuca (tan β1 − tan β2 ) Cp

ca

=

u tan β1 + tan β2

=

170.02 m/s

=

371

200 tan 45 + tan 10

Number of stages when Ω = 1 1 × 200 × 170.02 × (tan 45 − tan 10) ΔTs = 1005

ΔToverall

ΔToverall ΔTstage

=

27.86 K

=

T1 γ−1 rp γ − 1 ηc

=

244.12 K

=

244.12 27.86

=

=

8.76

310 × 60.286 − 1 0.85

≈

9

Ans

⇐=

Number of stages when Ω = 0.87 0.87 × 200 × 170.02 × (tan 45 − tan 10) = ΔTs 1005

ΔToverall ΔTstage

=

24.24 K

=

244.12 24.24

=

10.07

Ans

⇐=

9.4 Find the polytropic efficiency of an axial flow compressor from the following data: The total head pressure ratio : 4 Overall total head isentropic efficiency : 85% Total head inlet temperature : 290 K The inlet and outlet air angles from the rotor blades of the above compressor are 10◦ and 45◦ respectively. The rotor and stator blades are symmetrical. The mean blade speed and axial velocity remain constant throughout the compressor. Assuming a value of 220 m/s for blade speed and the work done factor as 0.86, find the number of stages required. Also find the inlet Mach number relative to rotor at the mean blade height of the first stage. Assume R = 284.6 kJ/kg K. Solution ηc

r

= r

γ−1 γ

γ−1 γ

−1

1 ηpc

−1

where ηp is the polytropic efficiency.

372

Gas Turbines β = α2 1

α =β 2 1

β α 1 1

w1

c1

ca1 wt1

ct1 u

wt2

β2

α2

c2

ca2 ct2

w2 u

Fig. 9.29

0.85

=

40.286 − 1 40.286 η1p − 1

ηp

=

0.286 ×

log 4 × 100 = 87.65% log 1.572

Ans

⇐=

Since, the stages are symmetrical, degree of reaction is 50% and also α1 = β2 and α2 = β1 . From the figure, u = ca

=

ca (tan 10 + tan 45) 220 0.17633 + 1

=

187 m/s

The stage temperature rise, ΔTs

=

Ωuca (tan α2 − tan α1 ) JCp

=

0.86 × 220 × 187 × (tan 45 − tan 10) = 29 K 1005

T02 T01

=

T02

=

p02 p01

γ−1 γηp

1.572 × 290

= =

0.286

4 0.8764 455.9 K

=

1.572

Axial Flow Compressors

373

Total temperature rise T02 − T01

=

455.9 − 290 = 165.9 K ≈ 166 K

Total number of stages = Mach number at inlet

166 29 =

Total temperature rise Temperature rise in one stage

= =

5.72

≈

Ans

6

⇐=

w1 √ γRT1

w1

=

ca cos 45

=

187 0.7071

=

264.5 m/s

c1

=

ca cos 10

=

187 cos 10

=

190 m/s

T1

=

T01 −

M

=

264.5 √ 1.4 × 284.6 × 272

c21 1902 = 272 K = 290 − 2Cp 2 × 1005 =

0.8

Ans

⇐=

9.5 An axial flow compressor takes in 1000 m3 /min of free air at 0.9 bar and 15◦ C. The blades are of aerofoil type having projected area and blade length as 19.25 cm2 and 6.75 cm respectively. The blade ring mean diameter is 60 cm and speed is 6000 rpm. On each blade ring there are 50 blades and the blades occupy 10% of the axial area of flow. Values of CL and CD are 0.6 and 0.05 respectively at zero angle of incidence. Assuming isentropic compression, calculate the pressure rise per blade ring and the power input per stage. Assume axial inlet.

Solution β 1

w

c = ca u

Fig. 9.30

ρ

=

p1 RT1

=

0.9 × 105 3 = 1.09 kg/m 287 × 288

374

Gas Turbines

Q

=

Aca

and A

=

kπDm h

ca

=

Q 60A

=

145.55 m/s

u

=

πDm N 60

tan β1

=

u 188.5 = 1.295 = ca 145.55

β1

=

52.33◦

w

=

L

=

CL ρw2 Ac 2

=

0.6 × 1.09 × 238.152 × 19.25 × 10−4 = 35.7 N 2

=

CD ρw2 Ac 2

=

0.05 × 1.09 × 238.152 × 19.25 × 10−4 = 2.98 N 2

1000 × 104 60 × (1 − 0.1) × π × 60 × 6.75

=

Blade velocity

D

=

π × 0.60 × 6000 = 188.5 m/s 60

145.552 + 188.52

=

238.15 m/s

Power input/stage P

=

(L cos β1 + D sin β1 )un

=

(35.7 × cos 52.33 + 2.98 × sin 52.33) × 188.5 × 50 × 10−3

r

Ans

=

227.85 kW

m ˙

=

1000 × 1.09 60

P m ˙

=

Cp T1 γ−1 r γ −1 ηc

=

P ηc +1 × m ˙ Cp T1

=

227.85 × 1 +1 18.33 × 1.005 × 288

γ−1 γ

r0.286

⇐= =

18.167 kg/s

Axial Flow Compressors

375

p2 p1

r

=

1.160

=

p2

=

0.98 × 1.160

=

Ans

1.14 bar

⇐=

9.6 Air at a temperature of 290 K enters a ten stage axial flow compressor at the rate of 3 kg/s. The pressure ratio is 6.5 and the isentropic efficiency is 90%, the compression process being adiabatic. The compressor has symmetrical blades. The axial velocity of 110 m/s is uniform across the stage and the mean blade speed of each stage is 180 m/s. Determine the direction of the air at entry to and exit from the rotor and the stator blades and also the power given to the air. Assume Cp = 1.005 kJ/kg K and γ = 1.4. Solution β

w1

1

α

1

c1

ca1 c t1

wt1 u

β

w2

2

c2

α2

ca2 c t2

w t2 u

Fig. 9.31

Assume that the temperature change is constant in each stage, then the power may be obtained by considering the overall conditions. γ−1 γ

T2 T1

=

T2

=

290 × 6.50.286

ηc

=

T2 − T1 T2 − T1

p2 p1

=

495.33 K

376

Gas Turbines

0.90

=

495.33 − 290 T2 − 290

T2

=

518.14 K

Therefore, power given to the air, =

mC ˙ p ΔT

=

687.84 kW

=

3 × 1.005 × (518.14 − 290)

Temperature change per stage ΔTs

=

ΔT 10

518.14 − 290 10

=

=

22.81 K

Work done/kg of air second =

uΔct

=

180 × Δct

Also work done/per kg of air per second

Δct

=

Cp ΔTs

=

1005 × 22.81 180

=

180Δct =

127.36 m/s

For symmetrical stages, Δct

=

ca (tan β1 − tan β2 )

127.36

=

110 × (tan β1 − tan β2 )

R

=

ca (tan β1 + tan β2 ) 2u

u

=

ca (tan β1 + tan β2 )

β1

=

54.41◦

⇐=

β2

=

13.5◦

⇐=

and when R = 0.5

Ans

Ans

9.7 An axial flow compressor has an overall pressure ratio of 4.0 and mass flow of 3 kg/s. If the polytropic efficiency is 88 per cent and the stagnation temperature rise per stage must not exceed 25 K, calculate the number of stages required and the pressure ratio of the first and last stages. Assume equal temperature rise in all stages. If the absolute velocity approaching the last rotor is 165 m/s at an angle of 20◦ from the axial direction, the work done factor is 0.83, the velocity diagram is symmetrical, and the mean diameter of the last stage rotor is 18 cm, calculate the rotational speed and the length of the last stage

Axial Flow Compressors

377

rotor blade at inlet to the stage. Ambient conditions are 1.01 bar and 288 K. Solution β = α2 1

α =β 2 1

β α 1 1

w1

c1

ca1 wt1

ct1 u

β2

wt2

α2

c2

ca2 ct2

w2 u

Fig. 9.32

ΔTstage

=

ηpc

=

γ−1 γ

n−1 n

=

γ−1 1 γ ηpc

=

T01

p02 p01

=

451.92 K

=

T02 − T01

=

163.92 K

Number of stages

=

Temperature rise across Total temperature rise per stage

Ns

=

163.92 25

T02

ΔToverall

25 K n n−1

n−1 n

0.286

= 288 × 4 0.88

=

=

451.92 − 288

6.56

≈

7

Ans

⇐=

378

Gas Turbines

Pressure ratio across first stage r1

ηc

=

r

=

γ−1 γ

−1

γ−1 1 γ ηpc

r

γ γ−1

ηc ΔTs T01

1+

=

0.855

r1

=

1+

ΔTs

=

40.286 − 1 40.325 − 1

=

−1

0.855 × 25 288

3.5

= 1.285

Ans

⇐=

163.92 ΔToverall = = 23.42 K 7 7

Pressure ratio across last stage: Temperature at inlet to last stage, ΔTs

=

T02 − T0

T0

=

T02 − ΔTs

=

451.92 − 23.42

s

s

r

s

=

ηc ΔT0 1+ T01

r

s

=

1+

s

=

428.5 K

γ γ−1

0.855 × 23.42 428.5

3.5

= 1.173

For symmetrical blade, R = 0.5 =

20◦

α1

=

β2

α2

=

β1

u

=

ca (tan β1 + tan β2 )

ca

=

C1 cos α1 = 165 × cos 20 = 155.05 m/s

u

=

155.05 × (tan β1 + tan β2 )

ΔT0s

=

ωuca (tan β1 − tan β2 ) Cp

u(tan β1 − tan β2 )

=

1005 × 23.42 0.83 × 155.05

tan2 β1 − tan2 β2

=

182.90 155.05

=

= 1.1796

182.9

Ans

⇐=

Axial Flow Compressors

tan2 β1

=

1.1796 + tan2 β2

tan1 β1

=

1.1796 + tan2 20

tan β1

=

1.1455

β1

=

48.88◦

u

=

ca (tan β1 + tan β2 )

=

155.05 × (tan 48.88 + tan 20)

=

234.04 m/s

u

=

πDN 60

N

=

60 × u πD

=

413.87 rps

=

1.3121

Ans

⇐=

60 × 234.04 1 × π × 0.18 60

=

Ans

⇐=

Pressure ratio across the last stage = 1.173 Total pressure at inlet to the last stage, =

4 1.173

=

3.41 bar

Total temperature at inlet to the last stage, =

428.5 K

=

T0 −

=

428.5 −

=

414.96 K

Static temperature, Tst

pst

c21 2Cp 1652 2 × 1.005 × 1000

p0

= T0 Tst

379

(γ−1)/γ

=

3.048 bar

ρ

=

3.048 × 105 287 × 414.96

m ˙

=

Aca ρ

=

3.41

=

=

428.5 3.5 414.96

3

2.56 kg/m

ca ρπDm h

380

Gas Turbines

Length of last stage, h

=

m ˙ ca ρπDm

=

3 × 100 2.56 × π × 0.18 × 155.05

=

1.336 cm

Ans

⇐=

9.8 The first stage of an axial compressor is designed on free vortex principles, with no inlet guide vanes. The rotational speed is 6000 rev/min and the stagnation temperature rise is 20 K. The hub-tip ratio is 0.60, the work done factor is 0.93 and the isentropic efficiency of the stage is 0.89. Assuming an inlet velocity of 140 m/s and ambient conditions of 1.01 bar and 288 K, calculate (i) the tip radius and corresponding rotor air angles β1 and β2 , if the Mach number relative to the tip is limited to 0.95, (ii) the mass flow entering the stage, (iii) the stage stagnation pressure ratio and power input, and (iv) the rotor air angles at the root section. Take R = 0.287 kJ/kg K, Cp = 1.005 kJ/kg K and γ = 1.4. Solution

w1

02 T

β 1

ca1= c

1

02’ u

01 s w2

β

2

c2

α2

ca2 c t2

w t2 u

Fig. 9.33

Axial Flow Compressors

T1

w1

C12 2Cp

=

T01 −

=

278.25 K

=

M1

=

=

288 −

381

1402 2 × 1005

γRT1 √ 0.95 × 1.4 × 287 × 278.25 = 317.65 m/s

Assuming axial inlet, the velocity triangle at inlet can be drawn as shown in Fig. cos β1

=

c1 w1

140 317.65

=

=

0.4407

Tip radius corresponding rotor angles β1 = 63.85◦

Ans

⇐=

u

=

w1 sin β1

u

=

285.13 m/s

ΔTs

=

Ω

tan β1 − tan β2

=

Cp ΔTs uΩca

=

1005 × 20 285.13 × 0.93 × 140

=

0.5414

tan β2

=

tan 63.85 − 0.5414

β2

=

56.23◦

T1

=

278.25 K

p1

=

ρ1

=

p1 RT1

=

1.12 kg/m

Rtip

=

60 × 285.13 60u = = 0.454 m 2πN 2 × π × 6000

Rroot

=

0.6 × Rtip = 0.6 × 0.454 = 0.2724 m

=

317.65 × sin 63.85

uca (tan β1 − tan β2 ) Cp

1.4954 Ans

⇐=

p01 T01 T1

=

3.5

=

1.01 3.5 288 278.25

= 0.8953 bar

0.8953 × 105 0.287 × 103 × 278.25

= 3

Ans

⇐=

382

Gas Turbines

Rm

=

1 (0.454 + 0.2724) 2

h

=

Rtip − Rroot

=

0.454 − 0.2724

=

ρ2πrm hca

=

1.12 × 2 × π × 0.3632 × 0.1816 × 140

=

64.98 kg/s

m

=

Input power

3.5

1+

ηs ΔT03 T01

=

1+

0.89 × 20 288

mC ˙ p ΔT0

=

1306 kW

0.1816 m

Ans

=

=

0.3632 m

⇐=

Stagnation pressure ratio, r

=

=

3.5

=

1.234

Ans

⇐=

64.98 × 1.005 × 20 Ans

⇐=

Rotor air angle at root section, =

2πRroot N 60

=

2 × π × 0.2724 × 6000 = 171.15 m/s 60

tan β1

=

171.15 140

β1

=

50◦ 71

tan β2

=

tan β1 −

Cp ΔT03 Ωuroot ca

=

1.2225 −

1005 × 20 = 0.32 0.93 × 171.15 × 140

=

17.74◦

uroot

β2

=

1.2225 Ans

⇐=

Ans

⇐=

9.9 Determine the stage efficiency, ηs and work done factor Ω of an axial flow compressor, if the actual pressure ratio developed was 1.35 and actual temperature rise was 30 K. The blade inlet and outlet angles are 47◦ and 15◦ respectively. The peripheral and axial velocities are 225 m/s and 180 m/s respectively.

Axial Flow Compressors

383

Solution rs

γ γ−1

ΔT0s ηs 1+ T01

=

γ−1

ηs

rs γ − 1 T01

=

=

ΔT0s 1.350.286 − 1 × 300 × 100 30 Ans

⇐=

=

89.6%

ΔT0s

=

Ω uca (tan β1 − tan β2 ) Cp

Ω

=

ΔT0s Cp uca (tan β1 − tan β2 )

=

30 × 1005 225 × 180 × (tan 47 − tan 15)

=

0.925

Ans

⇐=

9.10 A 50% reaction, axial flow compressor runs at a mean blade speed of 250 m/s. The pressure ratio developed by the machine is 1.3. Determine the blade and air angle if the mean flow velocity was 200 m/s. Condition at inlet are 1 bar and 300 K. Solution β α1

w1

1

c1

ca1 wt1

ct1 u

Fig. 9.34

ΔT

=

r

γ−1 γ

− 1 T1

=

1.30.286 − 1 × 300

=

23.38K

Degree of reaction R

=

ca (tan β1 + tan β2 ) 2u

0.5

=

200 (tan β1 + tan β2 ) 2 × 250

Ans

⇐=

384

Gas Turbines

tan β1 + tan β2

=

1.25

ΔT

=

uca (tan β1 − tan β2 ) Cp

tan β1 − tan β2

=

23.38 × 1005 250 × 200

β1

=

40.7◦

⇐=

β2

=

21.3◦

⇐=

α1

=

β2

α1

=

21.3◦

⇐=

α2

=

40.7◦

⇐=

=

0.47 Ans

Ans

For 50% reaction, and α2

=

β1 Ans

Ans

9.11 A helicopter gas turbine requires an overall compressor pressure ratio of 10:1. This is to be obtained using a two-spool layout consisting of a four-stage axial compressor followed by a single-stage centrifugal compressor. The polytropic efficiency of the axial compressor is 92 per cent and that of the centrifugal is 83 per cent. The axial compressor has a stage temperature rise of 30 K, using a 50 per cent reaction design with a stator outlet angle of 20◦ . If the mean diameter of each stage is 25.0 cm and each stage is identical, calculate the required rotational speed. Assume a work done factor of 0.86 and a constant axial velocity of 150 m/s. Assuming an axial velocity at the eye of the impeller, an impeller tip diameter of 33.0 cm, a slip factor of 0.90 and a power input factor of 1.04, calculate the rotational speed required for the centrifugal compressor. Ambient conditions are 1.01 bar and 288 K. Solution β α1

w1

1

c1

ca1 wt1

ct1 u

Fig. 9.35

For symmetrical stages (R = 0.5)

Axial Flow Compressors

u

=

ca (tan β1 + tan β2 )

385

(1)

Temperature rise across the stage is uca ΔT0s = Ω (tan β1 − tan β2 ) Cp =

ΔT0s Cp Ωca (tan β1 − tan β2 )

Cp ΔT0s

=

Ωc2a tan2 β1 − tan2 β2

tan2 β1

=

Cp ΔT0s + tan2 β2 Ωc2a

=

1005 × 30 + tan2 20 0.86 × 1502

β1

=

52.44

u

=

150 × (tan 52.44 + tan 20)

=

249.66 m/s

=

249.66 π × 0.25

u

(2)

Equating (1) and (2)

Nac

= =

=

1.6906

πDNac Ans

317.88 rps

⇐=

Total pressure ratio across the axial compressor is γ η nΔT0s ( pc γ−1 ) = 1+ r1 T1 =

0.92× 1.4 0.4 ) 4 × 30 ( 1+ 288

=

3.07

Centrifugal compressor : It is installed after the axial compressor. Outlet of axial

=

Inlet to centrifugal

Pressure ratio across centrifugal compressor 10 r2 = 3.2573 = 3.07 T02 is outlet temperature of axial compressor T02

1 γ−1 ηpc γ

=

T01 [r1 ]

=

0.4 288 × [3.07]( 1.4×0.92 )

=

Temperature at the outlet of centrifugal compressor,

408.02 K

386

Gas Turbines

T03

1 γ−1 ηpc γ

=

T02 [r2 ]

=

0.4 × 1 408.02 × [3.2573]( 1.4 0.83 )

=

612.67 K

=

T03 − T02

=

612.67 − 408.02

Cp ΔTsc

=

pif μu2

u

=

ΔTsc

Ncc

=

204.65 K

204.65 × 1005 1.04 × 0.9

=

468.76 m/s

=

πDNcc

=

468.76 = 452.15 rps π × 0.33

Ans

⇐=

Review Questions 9.1 What are the basic requirements of compressors for aircraft applications? Do axial flow compressors meet them? Explain. 9.2 With a suitable sketch explain the working principle of an axial flow compressor. 9.3 What is meant by a stage and explain in detail the stage velocity triangles. 9.4 Derive an expression for work input to the compressor and explain. What is meant by work done factor? 9.5 Explain the following performance coefficients: (i) flow coefficient (ii) rotor pressure flow coefficient (iii) rotor enthalpy loss coefficient (iv) diffuser pressure loss coefficient (v) diffuser enthalpy loss coefficient (vi) loading coefficient 9.6 Define degree of reaction and derive an expression for the same. 9.7 What is meant by low degree of reaction and high degree of reaction? How do you differentiate these two? 9.8 Show that when the degree of reaction is 50% the blades are symmetrical.

Axial Flow Compressors

387

9.9 Briefly explain the flow through a compressor and bring out the details of various losses in an axial flow compressor. 9.10 Derive an expression to calculate the pressure ratio in a stage. 9.11 Explain briefly the design and off-design characteristics of an axial flow compressor. 9.12 Explain the phenomena of surging and stalling in an axial flow compressor. Explain also the rotating stall. Exercise 9.1 Air enters an axial flow compressor at 1 bar and 20◦ C at low velocity. It is compresses through a pressure ratio of 11. Find the final temperature and pressure at outlet from the compressor. Take the compressor efficiency as 85%. Ans: (i) 632.65 K (ii) 11 bar 9.2 An axial flow compressor stage has blade root, mean and tip velocities of 150, 200 and 250 m/s. The stage is to be designed for a stagnation temperature rise of 20 K and an axial velocity of 150 m/s, both constant from root to tip. The work done factor is 0.93. Assuming 50% reaction at mean radius calculate the stage air angles. Also calculate stage air angles and the degree of reaction at root and tip for a free vortex design. Ans: Mid : 45.76◦; 17.04◦; Root : 45◦ ; 2.256◦ ; 0.52 Tip : 59.04◦; 47.47◦; 0.83 9.3 Recalculate the stage air angles for the same data as in the previous question for a stage with 50 per cent reaction at all radii. Ans: Root : 44.42◦ ; 1.14◦ Mean : 45.76◦; 17.04◦; Tip : 48.24◦; 28.59◦ 9.4 An axial flow compressor has an overall pressure ratio of 4.0 and mass flow of 160 kg/min. If the polytropic efficiency is 0.88 and the stagnation temperature rise per stage must not exceed 25◦ C, calculate the number of stages required and the pressure ratio of the first and last stages. Assume equal temperature rise in all stages. If the absolute velocity approaching the last rotor is 155 m/s at an angle of 20◦ from the axial direction, the work done factor is 0.83, the velocity diagram is symmetrical, and the mean diameter of the last stage rotor is 180 mm, compute the rotational speed and the length of the last stage rotor blade at inlet to the stage. Ambient conditions are 1.01 bar and 288 K. Also calculate air angles. Ans: (i) 7 (ii) 1.265 (iii) 1.173 (iv) 24355 rpm (v) 1.255 cm (vi) 50.5◦ ; 20◦

388

Gas Turbines

9.5 An axial flow compressor was tested and found that it gave a pressure ratio of 3 atmospheres and a temperature rise of 125◦ C. A 2000 kW motor was used to drive the compressor. Determine the compressor efficiency and the mass flow of air delivered, if the mechanical efficiency to be 95% and pressure and temperature at inlet were 1 atm and 300 K respectively. Ans: (i) 88.6% (ii) 15.124 kg/s 9.6 A pressure ratio of 1.5 is to be achieved in a single-stage of an axial flow compressor. The air approaches the rotor at an absolute velocity of 200 m/s and an angle of 12◦ . The speed of the compressor is 8000 rpm and has a mean radius 30 cm. Determine the blade angles and the blade height if the mass flow is 10 kg/s. Conditions at inlet are stagnation pressure = 1 bar and stagnation temperature = 300 K. Compressor stage efficiency is 95%. Ans: (i) 47◦ (ii) 15.5◦ (iii) 28 mm 9.7 A multistage axial flow compressor, with symmetrical blading is to deliver 10 kg/s with a pressure ratio of 4. The working fluid air enters the stage with a speed of 200 m/s and at an angle of 12◦ . The mean blade speed of the rotor is 250 m/s and rotational speed is 5000 rpm. Determine the blade angles, number of stages required for compressor and blade height, if the stage efficiency is 90% and inlet stagnation temperature and pressure are 300 K and 1 bar. Ans: (i) 46.81◦ (ii) 12◦ (iii) 4 stages (iv) 17.57 mm 9.8 A 50% reaction, axial flow compressor has inlet and exit blade angles of 45◦ and 15◦ respectively. The axial flow velocity is to be maintained at 200 m/s. Determine the mass flow, blade height, tangential force, axial force and the resultant force acting on the blade to produce a power of 300 kW at a speed of 5000 rpm and a blade spacing of 9 mm. Take p01 = 1 bar and T01 = 300 K. Ans: (i) 8.08 kg/s (ii) 13.77 mm (iii) 1183 N (iv) 4.44 N (v) 1183 N 9.9 From the cascade test of a 50% reaction axial flow compressor, the loading coefficient ψ was found to be 0.43 with the velocity of air flow in the axial direction being 200 m/s and the rotor speed 250 m/s. Determine the blades angles and pressure ratio if the maximum compressor blade efficiency were to be 95%. Assume solidity ratio (the ratio of chord length to blade spacing) = 1, p01 = 1 bar and T01 = 300 K. Ans: (i) 41.8◦ (ii) 19.61◦ (iii) 1.32 9.10 For an axial flow cascade, with usual notation derive the drag and lift coefficients as CD

=

Δp0 s cos3 αm 1 2 l cos2 α1 2 ρc1

Axial Flow Compressors

CL

= 2

s

389

(tan α1 − tan α2 ) cos αm − CD tan αm

From the above show that the maximum diffuser (blade) efficiency can be written as D ηDmax = 1 − 2 L State the assumption made. Also show that for a 50% reaction axial compressor, the pressure rise in a rotor blade in a cascade can be derived as Δprotor =

1 2 ρc (CL sin αm − CD cos αm ) 2 m s

A 50% reaction, aspect ratio (the ratio of blade height to chord length) 3 and blade height 10 cm, axial compressor cascade was tested and found to have a blade efficiency of 90% and a lift coefficient of 0.8. The mean axial velocity was 200 m/s and the inlet and exit blade angle were 45◦ and 15◦ respectively. Determine the lift and drag forces exerted on the blade, and pressure rise achieved in the stage with the flow coefficient of 0.5. Density of air at inlet is 0.9 kg/m3 . Take solidity ratio as 1. Ans: (i) 67.3 N (ii) 3.4 N (iii) 0.2 bar 9.11 An axial flow compressor stage was found to have a drag coefficient of 0.04. The blades are symmetric, the inlet and exit blade angles were 50◦ and 15◦ respectively. The mean axial velocity is 200 m/s. Determine the blade efficiency and the actual pressure ratio. Take solidity ratio as 1 and inlet static temperature as 275 K. Ans: (i) 96.86% (ii) 1.833 9.12 A three-stage axial flow compressor developing a pressure ratio of 3 delivers 10 kg/s of air. The fluid (air) enters the rotor with a velocity of 220 m/s and at an angle of 15◦ . The velocity diagram is symmetrical. Determine the speed of the compressor if the blade height is restricted to 25 mm at the inlet. Take inlet conditions, p01 = 1 bar, T01 = 300 K, degree of reaction = 0.5 and polytropic efficiency = 0.9. Take inlet conditions as stagnation state. Ans: 8091 rpm 9.13 An axial compressor is fitted with 50% reaction blading, the blade inlet and outlet angles being 50◦ and 15◦ when measured from the axial direction. The mean diameter of a certain blade pair is 85 cm and the speed is 5500 rpm. Calculate the axial velocity and the isentropic efficiency of the stage if the pressure ratio of compression is to be 1.4 when the air inlet temperature is 25◦ C. Ans: (i) 167.69 m/s (ii) 79.8%

390

Gas Turbines

9.14 An axial flow compressor has a flow coefficient of 0.8 and the loading coefficient is 0.88. If the blades are symmetrical, calculate the blade angles and the speed of the compressor. Take axial velocity as 200 m/s and mean blade diameter as 47.75 cm. Ans: (i) 49.6◦ (ii) 4.289◦ (iii) 10000 rpm 9.15 An axial flow compressor runs at 9000 rpm and the mean blade diameter for the fourth stage rotor is 55 cm. The rotor blades are 90 mm high and the mass flow rate is 45 kg/s. At entry to the pressure are 345 K and 1.7 bar. While the air leaves the previous row of stator blades at an angle of an angle of 28◦ to the axial direction. Calculate the stage temperature rise, work input and the pressure ratio for the compressor stage, given that the rotor blades deflect the air through 18◦ , that the work done factor is 0.88 and isentropic efficiency is 85%. Ans: (i) 18.8 K (ii) 850.23 kW (iii) 1.163 9.16 From the cascade test of a 50% reaction axial flow compressor, the lift CL was found to be 0.8 with the velocity of air flow in the axial direction being 200 m/s and the rotor speed 250 m/s. Determine the temperature rise and blades angles if the maximum compressor blade efficiency were to be 95%. Assume solidity ratio (the ratio of chord length to blade spacing) as 1, p01 = 1 bar and T01 = 300 K. Ans: (i) 23.83 k (ii) 40.84

◦

(iii) 21.08

Multiple Choice Questions (choose the most appropriate answer) 1. For aircraft application a compressor must have (a) low air flow capacity (b) high frontal area (c) high pressure ratio per stage (d) low efficiency 2. Modern multistage axial flow compressors have a pressure ratio (a) 2 (b) 6 (c) 12 (d) 18 3. The three velocities of a compressor c, u and w is related by (a) u = w − c

(b) c = u + w

(c) w = c − u

(d) c/u = w

◦

Axial Flow Compressors

391

4. In the axial flow compressor the absolute velocity in the stator (a) increases (b) decreases (c) initially increases and then decreases (d) remains constant 5. In the axial flow compressor the absolute velocity in the rotor (a) increases (b) decreases (c) initially increases and then decreases (d) remains constant 6. The work absorbing capacity of an axial flow compressor (a) increases with increase in the axial velocity (b) decreases with increase in the axial velocity (c) remain the same with increase in the axial velocity (d) has no relation between them 7. For a axial flow compressor the loading coefficient for the given stage work is (a) directly proportional to u (b) inversely proportional to u (c) inversely proportional to u2 (d) directly proportional to u2 8. The flow coefficient is defined as (a) φ = ca − u

(b) φ = ca + u (c) φ = ca /u (d) φ = ca × u 9. The degree of reaction is defined as actual enthalpy change in rotor actual enthalpy change in stage actual enthalpy change in stage (b) actual enthalpy change in rotor actual enthalpy change in rotor (c) actual enthalpy change in stator actual enthalpy change in stator (d) actual enthalpy change in rotor (a)

392

Gas Turbines

10. In a multistage axial flow compressor, the axial velocity at higher stage is (a) small (b) high (c) remains same (d) none of the above Ans:

1. – (c) 6. – (b)

2. – (d) 7. – (c)

3. – (b) 8. – (c)

4. – (b) 9. – (a)

5. – (a) 10. – (a)

10 COMBUSTION SYSTEMS INTRODUCTION The combustion process is of critical importance in a gas turbine cycle. It is because in this process the chemical energy of the fuel is converted to heat energy, which is later converted into work by the turbine. Therefore losses incurred in the combustion process will have a direct effect on the thermal efficiency of the cycle. The combustion system is not yet amenable to a complete theoretical analysis as the other components of a gas turbine. Although sufficient experience has been gained to enable a new design of a combustion system that would give a qualitative idea of its performance, the design can only be perfected by testing the component independently and modifying it in the light of the test results. This is due to the basic problem inherent in itself. In the gas turbine combustion system initially a mixing of fuel and air under conditions in which the resulting flame is self-sustaining should be accomplished first. Further, the chemical reaction should be complete. Thus, the combustor design involves the formation of turbulent zones, with the complication of both aerodynamics and thermo-chemical effects. The details of aerodynamics is available for the design of flow around stream lined bodies and also the flow through channels. The production of exactly controlled turbulent mixing zones is so complex that one has to resort to empiricism. Thus, in spite of the tremendous research effort on combustion, there is still no exact mechanism for flame stabilization in steady flows which would be generally agreed upon. The formulation of rules for similarity in combustion is still only in a tentative form. However, it may be noted that during the last four or five decades, gas turbine combustor technology has developed gradually and continuously rather than through any dramatic change. Despite this, the challenge to ingenuity in design is greater than ever before. New concepts and tech-

394

Gas Turbines

nology are still needed to satisfy current and projected pollutant emission regulations. Further, there is a growing emphasis on engines that can utilize a much broader range of fuels. With this in mind, in this chapter we will discuss the general principles and broad correlations which have been established over the years for the design of gas turbine combustion system. 10.1

COMBUSTION THEORY APPLIED TO GAS TURBINE COMBUSTOR

In any combustion process obtaining complete reaction between fuel and air has a chemical aspect and a physical aspect. Chemical aspects are concerned with rate of reaction etc., whereas physical aspects are concerned with particle size, injection mixing and evaporation. To understand these aspects completely, one has to refer to the various combustion theories presented from time to time in the combustion literature. There are three recognized postulations as to the combustion mechanism: (i) Carbon preferential burning which states that carbon in the hydrocarbon fuels burns before the hydrogen. (ii) Hydrogen preferential burning which states that hydrogen in the hydrocarbon fuels burns before the carbon. (iii) Hydroxylation which states that there is an initial uniting of oxygen with the hydrocarbon to form a hydroxylated compound. Through chain reactions of molecules, atoms and radicals, hydroxylated compound burns to CO, CO2 and H2 O. The hydroxylation theory, although not complete, seems to have met with more general acceptance than the other two theories. The modern theory is based on the statistics of probability as well as kinetics. It is known from kinetic theory of gases that the individual molecules are in motion at some average velocity but with a wide difference between the velocities of the slowest and fastest molecules. For combustion reaction to take place the process requires the collision of molecules of fuel and oxygen. The collision must have a sufficiently high energy level so that the molecules are broken down into atoms and radicals. Since, the temperature is a function of the molecular activity, raising the temperature increases the probability and intensity of collision of high velocity molecules. Therefore, there will be an increase in the intensity of combustion. Once the initial combustion starts it proceeds by means of chain reactions. Acceleration of the reaction is brought about by chain branching and is retarded by chain breaking. Free atoms of H and O are the free radicals and they may act as the chain carriers. Chain carriers may be destroyed through a surface reaction. In normal combustion, the relationship between chain branching and chain breaking is such that the reaction proceeds more or less at a constant rate.

Combustion Systems

395

The process of combustion in a gas turbine combustion involves the following four important steps: (i) formation of reactive mixture (ii) ignition (iii) flame propagation (iv) cooling of combustion products with air Under ideal situations, each of these steps would be completed separately before proceeding to the next step. However, space restrictions under the normal operating conditions do not permit the achievement of these ideal conditions. The intimate and uniform mixing of fuel and air is an essential prerequisite to complete the combustion process. The closer this mixing approaches molecular dimensions, there is a better possibility for complete combustion. To achieve this, the fuel should be atomized first, thus providing for good distribution with large surface area for evaporation at the same time. Apart from providing for good distribution and vaporization of the fuel, it is necessary to maintain an optimum air-fuel ratio to ensure ignition and sufficiently fast combustion. Rich mixture, i.e., insufficient air will result in cracking of the fuel with the formation of amorphous carbon which is difficult to burn. Although insufficient air is a cause of carbon formation, the problem is intimately associated with improper mixing. On the other hand, lean mixture or poor atomization and poor mixing will lead to failure of combustion. In actual practice it is necessary to use mixtures in order to satisfy this latter requirement, viz., failure of combustion, since the atomization and distribution can never be achieved on a molecular scale. There is a fixed range for the air to travel through the combustor for stable combustion. This limits the range of air-fuel ratios between the rich and lean mixtures. Operation outside this range results in unstable combustion causing vibrations and combustion failure. Figure 10.1 shows a typical stability curve as a function of air-fuel ratio and mass flow rate. The stable region lies between the rich and lean limits. It will be noticed that the stability range and air-fuel ratio range decreases as the air velocity is increased. Normally combustion chambers are designed with an inlet air velocity not exceeding 80 m/s at design load. In order to cool the products of combustion to a temperature acceptable to turbine blades, it is necessary to use a total air-fuel ratio far in excess of those permitting stable combustion. This difficulty is usually avoided by admitting a satisfactory amount of primary air so as to maintain stable combustion. The products of combustion are then cooled by introducing additional air called secondary air. The air-fuel ratio calculated with respect to the sum of primary and secondary air is known as the total air-fuel ratio. From an analytical point of view it is convenient to consider the gas turbine combustor flame tube as comprising three main zones: primary, intermediate or secondary and dilution. Combustion takes place mainly in the

396

Gas Turbines

Air-fuel ratio

200 We ak

150 100

lim it

Stable region

50 Rich

0 0

t

limi

0.25 0.5 0.75 1.0 Air mass flow kg/s

1.25

Fig. 10.1 Typical stability curve

primary zone. In the intermediate zone a small amount of air is injected into the hot gases coming out of the primary zone, to lower their temperature and thereby encourage the recombination of dissociated species. Finally, in the dilution zone, the combustion gases are mixed with the amount of air needed to achieve the required turbine inlet temperature. Aerodynamics plays a key role in all three zones. For different fuels there are correspondingly different ranges of inflammability within which propagation of combustion will take place once the initial reaction starts. For a given fuel this range is affected by liming rich and lean mixture beyond which a self-propagation combustion reaction cannot take place. The range of inflammability is affected mainly by three factors: (i) pressure, (ii) temperature and (iii) the thoroughness of mixing of fuel and air. The above three factors determine the ignition characteristics of the fuel. Thus, the spontaneous combustion will occur in a reactive mixture only if the mixture is heated to its self ignition temperature. At pressures below atmospheric at altitudes, the ignition temperature is much higher than at sea level condition. This will pose problems for aircraft gas turbines. It may be noted that the reaction becomes self propagating only when the reaction rate exceeds the energy transfer to the surrounding unburnt mixture. A certain minimum time is required for the initial reaction to take place. It should be noted that this time, called ignition delay, is dependent on the temperature of the mixture. If the fuel is heavy so that evaporation of the droplets is slow the rate of chemical reaction will be slow due to poor mixing. Initially a small spark is sufficient to initiate combustion. When fuels with high ignition temperatures or where high convective heat-transfer rates are used especially in a high velocity stream, a pilot flame may be necessary. The rate of flame propagation depends primarily on the range of inflammability, the pressure and the temperature of the mixture, and the shape of the combustion chamber. Hence, these factors must be given due

Combustion Systems

397

consideration. To some extent the rate of flame propagation is also affected by the expansion of hot gases behind the flame front. In a quiescent reactive mixture, it is expected that the rate of reaction and the rate of flame propagation would increase. However, in the constant-pressure combustion process used in the gas turbine power plant it is found that after the initial acceleration the rate of flame propagation becomes steady. This steady value is a function of physical conditions enumerated above. Flame speeds in the constant-pressure combustion process are comparatively low (2 to 5 m/s). Therefore, a region of low stream velocity must be provided in which the required rate of flame propagation can be achieved. Turbulence is another important requirement in the process of mixing and has a marked effect on the rate of flame propagation. However, increased turbulence has a corresponding effect on the heat transfer rate and it is reasonable to assume that there is an optimum turbulence level beyond which flame extinction would occur. When the flow velocity exceeds that of the flame, ignition may still take place in a low velocity region provided behind the turbulent wake of a flame stabilizer aiding the propagation of the flame downstream. 10.2

FACTORS AFFECTING COMBUSTION CHAMBER DESIGN

Experience gained from the design of commercial heating plants, such as oil-fired boilers is of little use to the designer of combustion systems for gas turbines. A number of requirements peculiar to the gas turbine make the design problem quite different, and these may be summarized as follows: (i) The temperature level of the gases after combustion must be comparatively low to suit the highly stressed turbine blade materials. (ii) At the exit of the combustion chamber the temperature distribution must be of known form if a high turbine performance is to be realized and the blades are not to suffer from local overheating. It need not be uniform, but may increase with radius over the turbine annulus because the blade stresses decrease from root to tip. (iii) Combustion must be maintained in a stream of air moving with a high velocity in the region of 30–60 m/s, and stable operation is required over a wide range of air-fuel ratio from full load to idling conditions. The air-fuel ratio might vary from about 60:1 to 120:1 for simple gas turbines and from 100:1 to 200:1 if a heat-exchanger is employed. The heat dilution implied by these figures is necessary to satisfy the turbine material considerations. (iv) The formation of carbon deposits (coking) must be avoided, particularly the hard brittle variety. Small particles carried into the turbine along with the high-velocity gas stream can erode the blades. Further, aerodynamically excited vibration in the combustion chamber

398

Gas Turbines

might cause sizable pieces of carbon to break free resulting in the irreparable damage to the turbine. (v) In aircraft gas turbines, combustion must be stable over a wide range of chamber pressure because this parameter changes with altitude and forward speed. Probably the only feature of the gas turbine which eases the combustion engineer’s problem is the peculiar interdependence of compressor delivery air density and mass flow which leads to the velocity of the air at entry to the combustion system being reasonably constant over the operating range. For aircraft applications, there are additional limitations like small space and low weight, which are, however, slightly offset by somewhat shorter endurance requirements. Aircraft engine combustion chambers are normally constructed of light gauge heat-resisting alloy sheet (approximately 0.8 mm thick), but are only expected to have a life of some 10,000 hours. Combustion chambers for industrial gas turbine plant may be constructed on much sturdier lines but, on the other hand, a life of about 100,000 hours is required. Refractory linings may eventually be used in heavy chambers. However, the effects of hard carbon deposits breaking free, apply with even greater force to refractory material. We have already seen that the gas turbine cycle is very sensitive to component inefficiencies, and it is of the utmost important that the aforementioned requirements should be met without sacrificing combustion efficiency. That is, it is essential that over most of the operating range all the fuel injected should be completely burnt and the full calorific value realized. Also, because any pressure drop between inlet and outlet of the combustion chamber leads to both an increase in specific fuel consumption and reduction in specific power output, it is essential to keep the pressure loss to a minimum. It may be noted that the smaller the space available for combustion the shorter will be the time available for the necessary chemical reactions. Hence, it is more difficult to meet all the requirements and still obtain a high combustion efficiency with low pressure loss. Clearly in this respect the designer of combustion systems for industrial gas turbines has much easier task than its counterpart in the aircraft field. 10.3

FACTORS AFFECTING COMBUSTION CHAMBER PERFORMANCE

The main function of a gas turbine combustor is to effect the chemical combination of oxygen of the air supplied by the compressor with carbon and hydrogen components of the fuel in such a manner that a steady stream of gas at a uniform temperature is produced. Even after years of research work we have gained only limited experience in the design of combustion system for gas turbines. A number of factors peculiar to the gas turbine affect the performance of the combustion chamber. We will briefly discuss them in this section.

Combustion Systems

10.3.1

399

Pressure Loss

It is evident that the turbulence is necessary for rapid combustion. However, this will cause some pressure drop in the combustion chamber. This loss is usually be regarded as a parasitic loss and hence should be minimized. The pressure losses are caused by two factors, viz., pressure drop due to friction and that due to the accelerations accompanying heat addition. Thus, the overall pressure loss in a gas turbine combustion system arises due to (i) cold losses, and (ii) hot losses. The combined pressure loss due to both heating and friction is the sum of the pressure losses determined separately as cold losses and hot losses. Thus the overall stagnation pressure loss can be regarded as the sum of the fundamental loss (a small component which is a function of T02 /T01 ) and the frictional loss. Our knowledge of friction in ordinary turbulent pipe flow at high Reynolds number would suggest that when the pressure loss is expressed non-dimensionally in terms of the dynamic head it will not vary much over the range of Reynolds number in which combustion systems operate. Experiments have shown, in fact, that the overall pressure loss can often be expressed adequately by an equation of the form Pressure loss factor, P LF

=

Δp0 2 m ˙ /2ρ1 A2m

The above pressure loss factor can be brought to the form P LF

=

K2

T02 −1 T01

(10.1)

where K2 is the constant arising out of hot losses. However, it may be noted that due to friction and turbulence cold losses also do occur which is constant. Hence, the total pressure loss factor is given by P LF

=

Cold loss + Hot loss

=

K1 + K2

T02 −1 T01

(10.2)

Note that rather than ρ1 c21 /2, a conventional dynamic head is used based on a velocity calculated from the inlet density, air mass flow rate, m, ˙ and maximum cross sectional area Am of the chamber. This velocity – sometimes known as the reference velocity – is more representative of conditions in the chamber, and the convention is useful when comparing results from chambers of different shape. Equation 10.2 is illustrated in Fig.10.2. If K1 and K2 are determined from a combustion chamber on a test rig from a cold run and a hot run, then Eq.10.2 enables the pressure loss to be estimated when the chamber is operating as part of a gas turbine over a wide range of conditions of mass flow, pressure ratio and fuel input.

400

Gas Turbines

Pressure loss factor

40 Fundamental pressure loss

30 20 10

Cold loss K1

0 1

2 Temperature ratio

T02

3

T 01

Fig. 10.2 Variation of pressure loss factor To give an idea of relative orders of magnitude, typical values of P LF at design operating conditions for tubular, turbo-annular and annular combustion chambers are 35, 25 and 18 respectively. There are two points which must be remembered when considering pressure loss data. Firstly, the velocity of the air leaving the last stage of an axial compressor is quite high – say 150 m/s – and some form of diffusing section is introduced between the compressor and combustion chamber to reduce the velocity to about 60 m/s. It is a matter of convention, depending upon the layout of the gas turbine, as to how much of the stagnation pressure loss in this diffuser is included in the P LF of the combustion system. In other words, it depends on where the compressor is deemed to end and the combustion chamber begin. Any pressure loss due to turbulence and skin friction in an unheated gas stream is usually quoted in dimensionless form as a friction of the dynamic head of the flow. It may be shown by dimensional analysis that this dimensionless group, or pressure loss factor is a function of Reynolds number and Mach number if compressibility effects are considered. Over the range of operating conditions met within combustion systems the effects of these two variables are small, so that the pressure loss factor is approximately constant for any given combustion chamber. Once this factor has been determined experimentally for one particular set of flow conditions, i.e., mass flow, inlet pressure and temperature, the frictional pressure loss may be estimated for any other set of inlet conditions. 10.3.2

Combustion Intensity

The size of the combustion chamber is determined primarily by the rate of heat release required. The nominal heat release rate can be found from mf ˙ ΔHc , where m ˙ is the mass flow rate, f the fuel-air ratio and ΔHc the net calorific value of the fuel. Enough has been said for the reader to appreciate that the larger the volume which can be provided the easier it will

Combustion Systems

401

be to achieve a low pressure drop, high efficiency, good outlet temperature distribution and satisfactory stability characteristics. The design problem is also eased by an increase in the pressure and temperature of the air entering the chamber, for two reasons. Firstly, an increase will reduce the time necessary for the ‘preparation’ of the fuel and air mixture (evaporation of droplets etc.,) making more time available for the combustion process itself. Note that since the compressor delivery temperature is a function of the compressor delivery pressure, the pressure (usually expressed in atmospheres) is an adequate measure of both. Secondly, it may be noted that the combustion chamber pressure is important because of its effect on the rate at which the chemical reactions proceed. An indication of the nature of this dependence can be obtained from chemical kinetics,, i.e., kinetic theory applied to reacting gases. By calculating the number of molecular collisions per unit time and unit volume which have an energy exceeding a certain activation value E, it is possible to obtain the following expression for the ratio at which a simple bimolecular gas reaction proceeds. Rate of reaction ∝ mj mk ρ2 σ 2 T 1/2 M −3/2 e−E/R0 T ρ and T have their usual meaning and the other symbols denote: mj , mk : local concentrations of molecules of species, j and k σ : mean molecular diameter M : mean molecular weight E : activation energy R0 : universal gas constant Substituting for ρ in terms of p and T we can for our purpose simplify the expression to Rate of energy ∝ p2 f (T ) Now T is maintained at a high value by having an approximately stoichiometric mixture in the primary zone. However, we are concerned here with the independent variable p. It is not to be expected that the theoretical exponent 2 will apply to the complex set of reactions occurring when a hydrocarbon fuel is burnt in air. Experiments with homogeneous mixtures in stoichiometric proportions suggest that it should be 1.8. At first sight it appears therefore that the design problem should be eased as the pressure is increased according to the law p1.8 . In fact there is reason to believe that under design operating conditions the chemical reaction rate is not a limiting factor. In an actual combustion chamber where physical mixing processes play such an important role, an exponent of unity is more realistic. This is not to say that under extreme conditions – say at high altitude – the performance will not fall off more in accordance with the p1.8 law. A quantity known as the combustion intensity has been introduced to take account of the foregoing effects. One definition used is Combustion intensity =

Heat release rate Combustion volume × pressure

402

Gas Turbines

Another definition employs p1.8 , with the units kW/m3 atm1.8 . However, it is defined, certainly the lower the value of the combustion intensity the easier it is to design a combustion system which will meet all the desired requirements. It is quite inappropriate to compare the performance of different systems on the basis of efficiency, pressure loss etc., if they are operating with widely differing orders of combustion intensity. In aircraft systems, the combustion intensity is in the region 2 − 5 × 104 kW/m3 atm, while in industrial gas turbines the figure may be a tenth of this owing to the larger volume of combustion space available and to the fact that when a heat-exchanger is employed a smaller heat release is required. 10.3.3

Combustion Efficiency

The combustion efficiency is computed from the chemical analysis of the gases. This is not easy, as not only it is difficult to obtain representative samples from the high velocity stream, but also, owing to the high air-fuel ratios used, the constituents to be measured are of a very small proportion of the whole sample. Ordinary apparatus, such as the Orsat, is not adequate for this purpose and special methods have been evolved. If an overall combustion efficiency is all that is required, however, and not an investigation state of the combustion process at any stage, it is possible to use an efficiency defined by the energy balance equation. Since the mean specific heat may be assumed to be same for both the actual and theoretical temperature rise, this efficiency may be written as Combustion efficiency =

Actual total − head temperature rise Theoretical total − head temperature rise

The actual temperature-rise is found by direct measurement at the inlet and outlet of the chamber, while the theoretical temperature rise may be found from the already available curves, using measured values of air-fuel ratio and inlet total-head temperature. Now this expression for efficiency is the result of an energy balance, i.e., a balance of such terms as ma Cp T0 . Since there is always in practice a variation in velocity as well as a variation in temperature across the section of a combustion chamber, it is necessary to use, not the ordinary arithmetic mean of number of readings across a section, but what is known as the weighted mean. 10.4

FORM OF COMBUSTION SYSTEM

In an open-cycle gas turbine, where compressor sucks fresh air continuously from the atmosphere, the direct burning fuel in the working stream is always used. Because the combustion is self-sustaining and continuous an electric spark is required only for initiating the process. Frequently the air leaving the compressor is split into a number of streams, each supplying to a tubular (can type) combustion chamber. These chambers are spaced around the shaft connecting the compressor and turbine. Each of these chambers has its own fuel injection system fed from a

Combustion Systems

403

common supply line. This type of layout is particularly suitable for gas turbines using centrifugal compressors where the air stream is already divided by the diffuser vanes. A single annular combustion chamber surrounding the rotor shaft would seem to be ideal for use with axial compressors. In this system maximum use is made of the space between the compressor and turbine with lower pressure losses and an engine of minimum diameter. However, there are certain disadvantages in the single annular chamber. They are (i) It is difficult to obtain a uniform fuel-air distribution and a uniform outlet temperature distribution, in spite of employing a large number of fuel jets. (ii) The annular chamber is structurally weaker and it is difficult to avoid buckling of the hot flame tube walls, more so in the case of large engines. (iii) The development of an annular chamber will have to be carried out on a single chamber requiring a test facility capable of supplying the full engine air mass flow. This requires a huge layout and involves enormous cost. However, for industrial gas turbines, with no importance for the space occupied, the combustion may be carried out in one or two large cylindrical chambers feeding the turbine via a scroll or volute. Large combustion chambers might enable adjustable valves or dampers to be incorporated for regulating the airflow to different points in the chamber. 10.5

REQUIREMENTS OF THE COMBUSTION CHAMBER

The main function of a gas turbine combustion chamber is to provide for the complete combustion of fuel and air, the air being supplied by the compressor and the products of combustion being delivered to the turbine. In carrying out this function, the combustion chamber must fulfill the following requirements. (i) Complete combustion of the fuel must be achieved. Any unburnt fuel, in the sense of the full heating value per kg not having been realized, is directly reflected in the fuel consumption or thermal efficiency. (ii) The total pressure loss must be minimum. Combustion pressure loss is one of the flow losses which is called the parasitic losses and its effect has already been discussed. (iii) Carbon deposits must not be formed under any expected condition of operation. As all normal fuels contain carbon, there is a tendency for free carbon to be deposited on the walls of the combustor or carried through

404

Gas Turbines

will appear as smoke. Any accumulation of the carbon will upset the designed flow pattern and increase pressure loss due to blockage. This can cause damage to the turbine blades when it comes out of the combustion chamber at a temperature and at high velocity. Smoke is always objectionable, although a light haze is usually allowable, if not desirable. (iv) Ignition must be reliable and accomplished with ease over wide range of atmospheric conditions. As the gas turbine is a steady flow machine, ignition is required only at the time of starting, the combustion being self-sustaining once the fuel is ignited by outside means. This allows a considerable simplification of the ignition system. Nevertheless, ignition of a liquid fuel under all conditions is not always easy, taken in conjunction with the fact that the igniting device is likely to be permanently exposed to the combustion gases. (v) Temperature and velocity distribution at turbine inlet must be controlled. It is very essential that the gas temperature at the turbine inlet be either uniform or have a designed gradient. At the high general level of temperature the strength of the blade material may be critical within small limits. Poor mixing may produce some hot spots at some sections. The ideal turbine inlet temperature distribution should be such that the stress levels in the blade are minimum. A uniform combustor outlet temperature is not necessarily the most desirable. Control of the temperature distribution is difficult to achieve to the desirable limits and sometimes the best that can be obtained without compromising other qualities is to avoid excessive hot spots. Velocity distribution is not so important, but some attention to this may be necessary to avoid particularly bad conditions, for instance where the turbine immediately follows a bend. (vi) The volume and weight of the combustor must be kept within reasonable limits. These criteria are always desirable, of course, but in some applications, the aircraft gas turbine being the prime example, they may be limiting. Even for stationary industrial application, the size and the weight of the combustion chamber must be suitable for the turbine as a whole and this imposes restrictions, on design, as compared with other steady flow combustion devices. (vii) Reliability and endurance should be ascertained. These are again desired criteria, but importance of the former is paramount in gas turbines, because a failure may wreck the turbine. Owing to the high speed of the turbine, even a small solid element can cause considerable damage. Reliability and endurance are differentiated, because in an aircraft engine, for example, a relatively short life

Combustion Systems

405

may be acceptable, while the reliability for that period is absolutely essential. All these performance criteria must be satisfied to some degree. The importance of each depending on the application and the problem is to arrive at a satisfactory compromise. It must be noted that in most cases it is not only the main operating or design point conditions that must be satisfied, but part-load and idling conditions as well. Although both pressure and mass flow rate may vary considerably, they are to a large extent independent. Together with the adiabatic relationship of compressor delivery pressure, the result is that the air velocity at entry to the combustor is relatively constant. This is fortunate, because the velocity is a control parameter in combustion performance. 10.6

THE PROCESS OF COMBUSTION IN A GAS TURBINE

Combustion of a liquid fuel involves the following: (i) the mixing of a fine spray of fuel droplets with air, (ii) vapourization of the droplets, (iii) the breaking down of heavy hydrocarbons into lighter fractions, (iv) the intimate mixing of molecules of these hydrocarbons with oxygen molecules, and finally (v) the chemical reactions themselves. A high temperature is necessary if all these processes are to occur sufficiently rapidly for combustion in a moving air stream to be completed in a small space by the combustion of an approximately stoichiometric mixture. It should be clear that the combustion of a gaseous fuel presents fewer problems, but much of what follows is still applicable. Since the overall air-fuel ratio is in the region of 100:1, while the stoichiometric ratio is approximately 15:1, the first essential is that the air should be introduced in stages. Three such stages can be distinguished. (i) About 15–20 per cent of the air is introduced around the jet of fuel in the primary zone to provide the necessary high temperature for rapid combustion. (ii) Some 30 per cent of the total air is then introduced through holes in the flame-tube in the secondary zone to complete the combustion. For high combustion efficiency, this air must be injected carefully at the right points in the process, to avoid chilling the flame locally and drastically reducing the reaction rate in that neighbourhood. (iii) In the tertiary or dilution zone the remaining air is mixed with the products of combustion to cool them down to the temperature required at inlet to the turbine. Sufficient turbulence must be promoted

406

Gas Turbines

so that the hot and cold streams are thoroughly mixed to give the desired outlet temperature distribution, with no heat streaks which would damage the turbine blades. Having described the way in which the combustion process is accomplished, it is now possible to see how incomplete combustion and pressure losses arise. Due to poor fuel injector design leads to fuel droplets being carried along the flame-tube wall. This causes incomplete combustion due to local chilling of the flame at points of secondary air entry. This can easily reduce the reaction rate. Because of this some of the products into which the fuel has decomposed are left in their partially burnt state. Due to this the temperature at the downstream end of the chamber is normally below that at which the burning of these products can be expected to take place. Light hydrocarbons into which the fuel has decomposed have a higher ignition temperature than the original fuel. It may be seen that it is difficult to prevent some chilling from taking place, particularly if space is limited and the secondary air cannot be introduced gradually enough. Usually devices are used to increase large-scale turbulence and more uniform distribution of the secondary air throughout the burning gases. Because of which the combustion efficiency will be improved but at the expense of increased pressure loss. A satisfactory compromise must somehow be reached. 10.7

COMBUSTION CHAMBER GEOMETRY

Discussion about the combustion process in the previous section serves to indicate the necessary features of practical combustion chamber. First of all, it is necessary for the reaction zone to operate near the stoichiometric mixture strength in order to develop the highest possible temperature for a rapid reaction or rate of flame propagation. The stoichiometric mixture strength for most hydrocarbon is 15:1. With an overall air-fuel ratio of 60:1 in a gas turbine, only about one quarter of the total air must be admitted to the reaction zone. Rest of the air is to be admitted to reduce the gas temperature to the required turbine inlet temperature. This reaction zone is often called the primary zone. The primary zone must have flame stabilizers such as baffles to establish a recirculation zone. But in addition to stabilization, a vigorous mixing action must be provided in order to mix air and fuel and then to mix unburnt mixture with burnt gases. The stability parameter indicates that it is better to provide a small number of large baffles than a large number of small baffles. The necessity of high temperature for significant reaction rate shows that diluting air must either be added only when it is certain that the reaction has gone to completion or that it must be added in slow degree so that the reaction is not immediately quenched. Because of the problem of mixing air and fuel, it is common to introduce some air as secondary air, i..e, it has a function in the combustion process in promoting mixing and ensuring adequate oxygen for the fuel for complete combustion downstream of the nominal reaction zone. This zone in which the secondary air is added

Combustion Systems

407

is called secondary or mixing zone. Finally, to bring down the temperature to the required level acceptable to the turbine blades the remaining air is introduced to mix with the products of combustion. This zone is called the tertiary or dilution zone. The necessary outline structure of many combustion chambers then appears as in Fig.10.3, yielding a typical annular form, with a central linear or flame tube containing the primary zone and baffle, surrounded by the outer casing or air casing. The annular space serves the purpose not only for separating the required primary air from the total air, but of providing a cooling air stream which limits the temperature of the liner, which contains a reaction zone where the gas temperature may reach locally a value of the order of 2000 K, corresponding to the stoichiometric combustion temperature. In actual practice, the stabilizing baffle is often considerably more complex than the simple bluff bodies. Fuel

Primary Air or zone outer casing

Air

Gap for cooling air

Gas Baffle

Liner or flame tube

Diluting and mixing zone

Fig. 10.3 Typical combustion chamber

10.7.1

Primary Zone or Flame Stabilizing Zone

The zonal method of introducing air cannot by itself give a self piloting flame in air stream which is moving in order of magnitude faster than the flame speed in a burning mixture. The essential requirement is therefore to provide a recirculation zone which directs some of the burning mixture in the primary zone back into the incoming fuel and air to give the primary air a considerable tangential components at entry, by means of swirl vanes, which mixes air and fuel over the cross-section and gives a recirculation pattern by means of vortex action. A free vortex has increasing tangential velocity with decreasing radius. By the Bernoulli equation, the higher velocity at the center entails a lower static pressure and thus a radial pressure gradient. The resulting circulation pattern is shown in Fig.10.4. If fresh air is admitted downstream then this is drawn upstream towards the fuel injection region. The use of swirl is a long established practice with oil and gas firing, but if used along, is liable to give rather coarse mixing, with some regions over-stimulated and other lacking.

408

Gas Turbines

Swirl vanes

Fig. 10.4 Flow pattern in the primary zone The basic Lucas combustor (Fig.10.5) combines a small swirler around the fuel injector, with large number of small holes in surrounding cone, the latter admitting air in jets, which provide finer degree of general turbulence. Figure 10.5 also shows another necessary feature of many combustors, in the flared passage preceding the baffle, which acts as a metering device and flow straightener for the primary air. Often, the velocity distribution in the duct from the compressor is highly irregular. If uncorrected, would lead to poor mixing, stabilization and temperature distribution in the combustor. Such poor distribution may occur due to separated flow in the diffuser or to a curved passage necessary from the geometry of the combustors.

Fig. 10.5 A Lucas combustion chamber with inlet swirler Another simple stabilization pattern, basis of G.E. combustors is shown in Fig.10.6. Here, the recirculation zone is formed by eddies from a row of holes just downstream of the fuel injection point. The location of these primary stabilizing jets, together with admission of secondary and diluting air further downstream, is critical. This requires considerable test work.

Fig. 10.6 A typical G.E. combustor

Combustion Systems

409

Many types of primary zone stabilizing arrangements are possible. Those described above are simply among those which are used in considerable numbers. These cases are all for downstream fuel injection, but often equally good or better results can be obtained by upstream injection, as shown schematically in Fig.10.7. The advantage of this, is the possibility of improved mixing, as air may be injected directly into the center of the fuel injection zone. The disadvantage is the possibility of overheating the fuel pipe and injector unless they are sufficiently cooled, as they are located directly in the reaction zone.

Air

Fig. 10.7 Flame stabilization with upstream injection

10.8

MIXING AND DILUTION

The problem of adding approximately three-quarters of the total air to onequarter of reactant gases in the dilution zone is not an easy job, because of the necessity for a good temperature distribution with minimum pressure loss. In the typical cylindrical construction, the air is admitted progressively through holes or slots, with usually a small gap being left between the liner and the casing to provide a film cooling air for the final contraction to the nozzle entry. An excellent method of maintaining the liner walls at the desired temperature is by means of film cooling. This can be achieved by making the liner in overlapping sections so that each has a layer of cool air flowing over it. The size and the number of dilution air holes is a compromise between a large number of small holes to give fine scale mixing and a small number of large holes to give better penetration. It is important that the mixing air must reach the axis of the combustion chamber, as otherwise at the core a high temperature gas will persist. A slot with major axis in the flow direction gives good penetration, but tends to weaken the liner. The liner is not subjected to a large pressure difference, but may distort owing to local high-temperature gradients. Further, the combustion process is accompanied by vibrations of high frequency and the whole combustion chambers is subjected to a vibration fatigue effect too. 10.9

COMBUSTION CHAMBER ARRANGEMENTS

Comparison of various gas turbines and jet engines gives that, at present, three arrangements of combustion chambers are being practiced, viz.,

410

Gas Turbines

(i) a large single chamber, (ii) multiple chambers, and (iii) the annular chamber. A large single chamber, is usually employed in the case of heavy industrial power plants where space and weight are of secondary importance. Since, space is not so important, relatively low gas velocities can be used which tends towards higher combustion efficiency and low pressure loss. Further, the absence of a weight restriction permits the use of heavy sections, which if properly designed gives long life. Moreover, high operating temperatures without buckling can be achieved. The multiple chamber arrangement is common practice for aircraft applications as well as few stationary power plants or marine engines. Its main advantage is the ease with which the desired total combustion chamber volume can be achieved. This is beneficial from the design point of view since if any increase in overall dimensions for that of the compressor and turbine is required this can be achieved with least difficulty. Secondly, individual chambers can be removed and replaced without disturbing the rest of the assembly. The design also fits nicely in with the general diffuser design of centrifugal compressor, and little difficulty is encountered in obtaining a fairly uniform gas temperature at exit from chambers. The third type of combustion chamber, the annular chamber, is illustrated in Fig.10.8. It consists of an annular passage connecting the compressor delivery with the turbine nozzle, permitting the formation of a continuous sheet of hot gas which flows from the former to the latter. Inside this annular chamber is supported the inner flame tube, which in this case is also annular. Fuel is injected into the inner annular chamber by series of nozzles located at the compressor end of the chamber. Burner Basket To jet

Annulus

Combustion Systems

411

With near perfect mixing and combustion of the fuel and air an ideal flow situation can be envisaged in such designs. Further, with less pressure loss and turbine nozzles being fed with a continuous sheet of gas round the periphery of the machine. The main difficulty associated with such a chamber is to supply fuel in such a way as to produce the desired temperature distribution. In general the fuel is fed by means of limited number of nozzles, possibly 15 to 20 arranged around the inlet of the inner flame holder. It is believed that recent advances have overcome this difficulty. A further difficulty associated with the annular type of chamber is that of providing adequate cooling air to the internal surfaces, bearings, etc., which may be located inside the inner annular wall, since this space is completely blocked off from the outside atmosphere by the compressor, combustion chamber and turbine wheel. 10.10

SOME PRACTICAL PROBLEMS

To conclude this chapter we will discuss some of the problems which have so far not been mentioned but which are none the less important. These are concerned with (i) flame tube cooling, (ii) fuel injection, (iii) ignition, (iv) use of cheaper fuels, and (v) pollution. 10.10.1

Flame-tube Cooling

One problem which has assumed greater importance as permissible turbine inlet temperatures have increased is that of cooling the flame-tube. The tube receives energy by convection from the hot gases and by radiation from the flame. It loses energy by convection to the cooler air flowing along the outer surface and by radiation to the outer casing, but this loss is not sufficient to maintain the tube wall at a safe temperature. A common practice is to leave narrow annular gaps between overlapping sections of the flame tube, as shown in Fig.10.9(a), so that a film of cooling air is swept along the inner surface. Corrugated ‘wigglestip’, spot welded to successive lengths of flame-tube, provides a tube of adequate stiffness with annular gaps which do not vary too much with thermal expansion. Another method, illustrated in Fig.10.9(b), is to use a ring of small holes with an internal splash ring to deflect the jets along the inner surface. A film of cool air both insulates the surface from the hot gases, and removes energy received by radiation. Although empirical relations are available from which it is possible to predict convective heat transfer rates when film cooling a plate of known temperature, the emissivities of the flame and flame-tube can vary so widely that prediction of the flame-tube temperature from an energy balance is not possible with any accuracy. Even in this limited aspect of combustion chamber design, final development is still a matter of trial and error on the

412

Gas Turbines

(a)

(b)

Fig. 10.9 Film cooling of flame-tube test rig. The emissivity of the flame varies with the type of fuel, tending to increase with the specific gravity. Carbon dioxide and water vapour are the principle radiating components in non-luminous flames, and soot particles in luminous flames. It is worth noting that vaporizer systems ease the problem, because flames from premixed fuel vapour-air mixtures have a lower luminosity than those from sprays of droplets. Higher turbine inlet temperatures imply the use of lower air-fuel ratios, with consequently less air available for film cooling. Furthermore, the use of a higher cycle temperature is usually accompanied by the use of a higher cycle pressure ratio to obtain the full benefit in terms of cycle efficiency. Thus the temperature of the air leaving the compressor is increased and its cooling potential is reduced. If permissible turbine inlet temperatures increase much beyond 1500 K, some form of porous material may have to be used for the flame-tube: ‘effusion cooling’, as it is called, is very much more economical in cooling air than film cooling. 10.10.2

Fuel Injection

The use of liquid fuel requires that it be injected in metered quantity in a state as finely divided as possible, so that evaporation and mixing are rapid. Therefore, the injection system must fulfill the requirements, viz., (i) meter the fuel flow, and (ii) atomize the fuel. The metering function of a plain hole is a combination of applied pressure and the cross-sectional area of the metering orifice, the flow being determined by an equation of the form m t where

=

m ˙

=

kAρ

2p ρ

(10.3)

m : kg of fuel k : constant ρ : weight per cum of fuel A : nozzle area sq. meter t : time of flow, s p : injection pressure at nozzle, kg/m2 This points out one of the difficulties of fuel injection. It may be noted that the flow varies as the square root of the pressure. For example, if

Combustion Systems

413

1.5 bar is assumed as the lowest value at which a spray can be formed at minimum flow then it can be shown that a top pressure of 375 bar is necessary for full flow. The throttle valve, provided in the fuel system, in general determines the pressure applied to the nozzle side of the system which in conjunction with the hole size will fix the flow in accordance with the Eq.10.3. When more than one nozzle is connected with manifold, as is common, it is essential that they are all calibrated to divide the fuel equally between the various outlets within a rather narrow tolerance. The atomization produced by nozzle is necessary to divide the fuel flow into a large number of small droplets to permit rapid oxidation with a short length and complete combustion without the production of smoke. A simple atomizer is a nozzle with a plain hole, which will, if the fuel is supplied to it under pressure, results in production of a turbulent stream of fuel form into droplets because of surface tension. This spray is characterized by a rather narrow cone angle, and a high penetrative power suitable for diesel-engine injectors. The injectors for oil burners, jet engines and for similar applications, where penetration is not an important consideration, produce a softer spray but one that is widely dispersed over a large area to provide additional surface for oxidation. The type of injector generally used is the centrifugal or swirl type atomizer shown in Fig.10.10. The essential of this injector are means of feeding the oil to the periphery of a vortex or swirl chamber in a roughly tangential direction producing a vortex in which the angular velocity varies inversely as the radius. The discharge orifice is small compared with the vortex chamber and the rotating liquid spills out in a thin sheet, the quantity varying with the pressure of supply. Thus, there are two factors acting on the fuel, (i) the pressure acting axially, and (ii) the centrifugal force acting radially. The result is that when the flow rate is high enough, this spinning sheet of liquid will break up into ligaments forming drops which will travel from the orifice forward and outward giving the hallow cone of fuel spray. This atomization may occur with at little as 0.7 to 1.05 bar in a well designed nozzle when using kerosene as fuel. The requirements of good atomizer can be listed as follows: (i) Atomization must be effected over a wide range of fuel flows and pressures. (ii) No moving parts at the nozzle and of simple design. (iii) It should distribute the fuel uniformly throughout the air flow. (iv) It should be easily reproducible to high degree of accuracy to enable multi-injector systems to operate with uniform distribution. (v) The weight and size should be low to reduce heat absorption to a minimum.

414

Gas Turbines

Two or three types of swirl atomizers have been developed. One such is the simple single-outlet type of Fig.10.8 where all the fuel flow leaves via the one spray orifice. Air core

Cone angle of spray

Axial force Resultant direction Discharge orifice Swirl chamber

Tangential feed ducts

Fig. 10.10 Swirl type atomizer Spray A

A

Inlet

Tangential ducts Air core

Valve

Split outlet

Section AA’

Combustion Systems

415

Another type of the burner, known as the Duplex, has two orifices in parallel feeding into the swirl chamber. One of the ports does not come into operation until the fuel pressure exceeds a certain value. This naturally gives a stepped characteristic which is not entirely satisfactory. 10.10.3

Ignition

The first step in starting a gas turbine is to accelerate the compressor to a speed which gives an air flow capable of sustaining combustion. In some cases this is achieved by feeding compressed air from an external supply directly to the turbine driving the compressor. If is more common, however, to employ an electric motor or small auxiliary turbine connected to the main shaft by a reduction gear and clutch. The auxiliary turbine may operate from a compressed air supply or, as in the case of military aircraft where independence of ground equipment is desirable, from a cartridge. During the period of acceleration the ignition system is switched on and fuel is fed to the burners. An igniter plug is situated near the primary zone in one or two flame-tubes or cans. Once the flame is established, suitably placed interconnecting tubes between the cans permit ‘light round’, i.e., flame propagation from one flame-tube to another. Light round presents few problems in annular chambers. With an aircraft gas turbine, there is the additional requirement that re-ignition must be possible under windmilling conditions if for any reason the flame is extinguished at altitude. The ignition performance can be expressed by an ignition loop which is similar to the stability loop of Fig.10.1 but lying inside it. That is, at any given air mass flow the range of air-fuel ratio within which the mixture can be ignited is smaller than that for which stable combustion is possible once ignition has occurred. The ignition loop is very dependent on combustion chamber pressure, and the lower the pressure the more difficult is the problem of ignition. Relight of an aircraft engine at altitude is thus the most stringent requirement. Although high-tension sparking plugs similar to those used in piston engines are adequate for ground starting, a spark of much greater energy is necessary to ensure ignition under adverse conditions. A surface discharge igniter, yielding a spark having an energy of about three joules at the rate of one spark per second, is probably the most widely used type for aircraft gas turbines in which fuel is injected as a spray of droplets. One example of a surface-discharge igniter is shown in Fig.10.12. It consists of a central and outer electrode separated by a ceramic insulator except near the tip where the separation is by a layer of semiconductor material. When a condenser voltage is applied, current flows through the semiconductor which becomes incandescent and provides an ionized path of low resistance for the energy stored in the capacitor. Once ionization has occurred, the main discharge takes place as an intense flashover. To obtain good performance and long life the location of the igniter is critical; it must protrude through the layer of cooling air on the inside of the flame-tube wall to the outer edge of the fuel spray, but not so far as to be seriously wetted by the fuel.

416

Gas Turbines Semiconductor

Ceramic Earthed electrod

Central electrode

Fig. 10.12 Surface-discharge igniter For vapourizing combustion chambers, some form of torch igniter is necessary. This comprises a spark plug and auxiliary spray burner in a common housing. The location is not critical, but it is bulkier and heavier system than the surface-discharge type. The torch igniter is particularly suitable for industrial gas turbines, and has the advantage that the auxiliary burner can be supplied with distillate fuel from a separate tank when a heavy oil is used as the main fuel. 10.10.4

Use of Cheap Fuels

The gas turbine has found a use in process industries where the power output is desired wholly or in part as compressed air, or steam from a waste-heat boiler. It is particularly suitable when there is a gaseous byproduct of the process which can be used as fuel for the gas turbine. One obvious example is the use of a gas turbine in a steel mill for supplying air to the blast furnaces while using blast furnace gas a fuel. There is little difficulty in designing a combustion system to handle gaseous fuels and it is not this form of cheap fuel with which we are concerned here. That the gas turbine has taken so long to compete seriously with other forms of prime mover, except for aircraft applications where small size and low weight are vital, is largely due to the difficulties encountered when trying to use residual oil. This cheap fuel is the residue from crude oil following the extraction of profitable light fractions. Some of its undesirable characteristics are (i) high viscosity requiring heating before delivery to the atomizers. (ii) tendency to polymerize to form tar or sludge when overheated. (iii) incompatibility with other fuels with which it might come into contact, leading to jelly-like substances which can clog the fuel system. (iv) high carbon content leading to excessive carbon deposits in the combustion chamber.

Combustion Systems

417

(v) presence of vanadium, the vanadium compounds formed during combustion causing corrosion in the turbine. (vi) presence of alkali metals, such as sodium, which combine with sulphur in the fuel to form corrosive sulphates; (vii) relatively large amount of ash, causing build up of deposits on the nozzle blades with consequent reduction in air mass flow and power output. Characteristics (i), (ii), (iii) and (iv) have a nuisance value but the problems which they give rise can be overcome without much difficulty. (v), (vi) and (vii) on the other hand have proved to be serious. The rate of corrosion from (v) and (vi) increases with turbine inlet temperature, and early industrial gas turbines designed for residual oil operated with temperatures not much greater than 900 K to avoid the problem. Such a low cycle temperature inevitably meant a low cycle efficiency. It has now been found that the alkali metals can be removed by washing the fuel oil with water and centrifuging the mixture, and that fuel additives such as magnesium can be used to neutralize the vanadium. Although the latter treatment costs very little, the former is expensive particularly when carried out by the consumer on a relatively small scale: the fuel is no longer ‘cheap’ and the gas turbine becomes barely competitive. With regard to (vii), i.e., build up of deposits, it has been found that applications involving intermittent operation (e.g., locomotive propulsion) are more suitable than those involving continuous running (e.g., base load electric power generation). This is because expansions and contractions associated with thermal cycling cause the deposits to crack off. No satisfactory solution has been found for continuous running plant, and high maintenance costs would be an additional factor making the gas turbine uneconomic. Finally, various experimental coal burning gas turbines have been built but without success. The problems of ash removal after combustion, or treatment of the coal itself to remove the impurities leading to ash, have prevented the gas turbine from competing successfully with coal burning steam power plant. 10.10.5

Pollution

From the related facts that a gas turbine uses very high air-fuel ratios and provides a combustion efficiency of almost 100 per cent over most of the operating range, it should be clear that the gas turbine is never likely to be a major contributor to atmospheric pollution. The main trace pollutants as from any air-breathing engine burning fossil fuel, are (i) unburnt hydrocarbons and carbon monoxides, (ii) oxides of nitrogen, and (iii) oxides of sulphur.

418

Gas Turbines

The concentration is negligible except under idling conditions when the combustion chamber operating temperature is at its lowest. With turbojet engines the order of magnitude is about 20–30 ppm (parts per million) by weight under normal operating conditions. These pollutants arise due to local chilling of the chemical reactions. The concentration is therefore minimal with an annular combustion chamber or single large tubular chamber because such configurations provide a smaller area of cool flame-tube wall per unit volume than do cannular or multi-tubular designs. In any air breathing engine a small fraction of the nitrogen is oxidized, and the nitric oxide so formed in the exhaust continues to oxidize slowly in the atmosphere to form nitrogen dioxide. Although non-toxic, it is undesirable because it plays an important part in the formation of ‘smog’. The oxidation of nitrogen increases with combustion temperature, and in a turbojet exhaust the concentration of nitric oxide is likely to be negligible under idling conditions rising to approximately 100 ppm by weight at full power. Oxides of sulphur are detrimental not only to plant and animal life, but also to the engine itself because of corrosive acids to which they give rise. The kerosene used as fuel for most gas turbine, however, has a very low sulphur content and the sulphur emission is negligible. Furthermore, the latest oil refineries are now able to produce fuel economically with a much lower sulphur content than has been specified by users in the past. Some aircraft gas turbines have exhibited a smoky exhaust indicating the presence of small particles of unburnt carbon. An amount quite negligible from the point of view of atmospheric pollution can have a pronounced visible effect due to the light scattering properties of the particles. A smoky exhaust can be virtually eliminated during development testing by modifications to the primary zone and fuel burners, and it is unlikely to arise at all when the fuel and air are premixed as in vaporizer systems. It may be concluded that the good characteristics of the gas turbine are likely to increase its competitiveness in the twenty first century. Review Questions 10.1 Explain in detail the combustion theory applied to a gas turbine combustion system. 10.2 What are the various factors those affect the combustion chamber performance? Explain. 10.3 Describe briefly the factors affecting the combustion chamber design. 10.4 What are the various forms of a combustion system? 10.5 What are the basic requirements of a combustion chamber? Explain. 10.6 Explain the process of combustion in gas turbine combustion chambers.

Combustion Systems

419

10.7 With a neat sketch explain the combustion chamber geometry bringing out the various zones that play a part in the process of combustion. 10.8 What are the various possibility of combustion chamber arrangements? 10.9 Mention the various practical problems in the operation of a combustion chamber. 10.10 Write (i) (iii) (v)

short notes on flame tube cooling, ignition, pollution.

(ii) (iv)

fuel injection, use of cheaper fuels, and

Multiple Choice Questions (choose the most appropriate answer) 1. Combustion in a gas turbine combustion system is at (a) constant-volume (b) constant-pressure (c) constant temperature (d) constant enthalpy 2. Combustion process in a combustor is (a) isobaric (b) isochoric (c) isentropic (d) isothermal 3. The recognized postulations as to the combustion mechanism in a combustion chamber is due to (a) carbon preferential burning (b) hydrogen preferential burning (c) hydroxylation (d) all of the above 4. Acceleration of the reaction in a gas turbine combustion system is brought about by (a) bombardment of nucleus (b) chain breaking (c) chain branching (d) none of the above

420

Gas Turbines

5. Air-fuel ratio in a gas turbine is in the range of (a) 20 to 30 (b) 30 to 40 (c) 40 to 60 (d) 60 to 100 6. Large amount of combustion takes place in a combustor in the (a) primary zone (b) secondary zone (c) dilution zone (d) in all places 7. The outlet temperature of the combustor is around (a) 1600 K (b) 1400 K (c) 1200 K (d) 1000 K 8. In the primary zone the air-fuel ratio is (a) 13 (b) 15 (c) 17 (d) 20 9. The major pollutant coming out of the combustion chamber is (a) CO (b) SO2 (c) NOx (d) unburnt hydrocarbons 10. The maximum temperature from the combustor is limited because (a) it is difficult to burn the fuel (b) the air-fuel ratio is too lean (c) combustion chamber walls cannot sustain high temperature (d) turbine blades cannot accept very high temperatures Ans:

1. – (b) 6. – (a)

2. – (a) 7. – (a)

3. – (d) 8. – (b)

4. – (c) 9. – (c)

5. – (d) 10. – (d)

11 IMPULSE AND REACTION TURBINES INTRODUCTION Work can be extracted from a gas at a higher inlet pressure to the lower back pressure by allowing it to flow through a turbine. In a turbine as the gas passes through, it expands. The work done by the gas is equivalent to the change of its enthalpy. It is a well known fact that the turbines operate on the momentum principle. Part of the energy of the gas during expansion is converted into kinetic energy in the flow nozzles. The gas leaves these stationary nozzles at a relatively higher velocity. Then it is made to impinge on the blades over the turbine rotor or wheel. Momentum imparted to the blades turns the wheel. Thus, the two primary parts of the turbine are (i) the stator nozzles, and (ii) the turbine rotor blades. Most turbines possess more than one stage with their respective wheels mounted on a common shaft. Figure 11.1 shows a turbine stage. The stage consists of a ring of fixed nozzle blades followed by the rotor blade ring. However, a nozzleless stage with only the rotor is also possible as in the case of an inward flow radial turbine. In certain requirements, the flow in a turbine stage may be partly radial and partly axial which is called a mixed stage. It combines the advantage of both axial and radial types. Normally, a turbine stage is classified as (i) an impulse stage, and (ii) a reaction stage. An impulse stage is characterized by the expansion of the gas which occurs only in the stator nozzles. The rotor blades act as directional vanes

422

Gas Turbines

Inlet Nozzle blade row Stage Rotor blade row Axial clearance

Outlet

Fig. 11.1 A turbine stage to deflect the direction of the flow. Further, they convert the kinetic energy of the gas into work by changing the momentum of the gas more or less at constant-pressure. A reaction stage is one in which expansion of the gas takes place both in the stator and in the rotor. The function of the stator is the same as that in the impulse stage, but the function in the rotor is two fold. (i) the rotor converts the kinetic energy of the gas into work, and (ii) contributes a reaction force on the rotor blades. The reaction force is due to the increase in the velocity of the gas relative to the blades. This results from the expansion of the gas during its passage through the rotor. 11.1

A SINGLE IMPULSE STAGE

Impulse machines are those in which there is no change of static or pressure head of the fluid in the rotor. The rotor blades cause only energy transfer and there is no energy transformation. The energy transformation from pressure or static head to kinetic energy or vice versa takes place in fixed blades only. To give an example, the transfer of kinetic energy to the rotor in an impulse turbine from a high velocity fluid occurs only due to the impulsive action of the fluid on the rotor. Figure 11.2 shows an impulse turbine stage. As can be seen from the figure that in the rotor blade passages of an impulse turbine there is no acceleration of the fluid, i.e., there is no energy transformation. Hence, the chances are greater for separation due to boundary layer growth on the blades surface. Due to this, the rotor blade passages of the impulse machine suffer greater losses giving lower stage efficiencies. The paddle wheel, Pelton wheel and Curtis steam turbine are some of the examples of impulse machines. 11.2

A SINGLE REACTION STAGE

The reaction machines are those, in which, changes in static or pressure head occur both in the rotor and stator blade passages. Here, the energy

Impulse and Reaction Turbines

423

Nozzle Rotor

Energy transfer Energy transformation

Velocity Pressure

Fig. 11.2 An impulse turbine stage transformation occurs both in fixed as well as moving blades. The rotor experiences both energy transfer as well as energy transformation. Therefore, reaction turbines are considered to be more efficient. This is mainly due to continuous acceleration of flow with lower losses. Figure 11.3 shows a single stage reaction turbine along with pressure and velocity changes when the fluid passes through a turbine stage. The degree of reaction of a turbomachine stage may be defined as the ratio of the static or pressure head change occurring in the rotor to the total change across the stage. A 50% or half degree reaction machine has some special characteristics. Axial-flow turbines with fifty per cent reaction have symmetrical blades in their rotors and stators. It may be noted that the velocity triangles at the entry and exit of a 50% reaction stage are also symmetrical. Hero’s turbine, the lawn sprinkler and Parson’s steam turbine are some of the examples of reaction machines. 11.3

MULTISTAGE MACHINES

It may be noted that for a given rotor speed, only a limited change in the energy level of the fluid can occur in single stage machines. This is true for both turbines and compressors. Hence, it may become necessary that when a large change in the energy level is required, more than one stage becomes a necessity. Multistage machines may employ either impulse or reaction stages or a combination of both. Impulse machines may utilize any one of the following:

424

Gas Turbines Nozzle Rotor

Energy transfer Energy transformation

Velocity Pressure

Fig. 11.3 A reaction turbine stage (i) a large pressure drop in a number of pressure stages (ii) a high kinetic energy in a number of velocity stages (iii) a combination of the above two It may be noted that gas turbine plants have a smaller number of stages due to comparatively lower values of the pressure ratio employed, compared to other similar power plants. 11.4

VELOCITY TRIANGLES OF A SINGLE STAGE MACHINE

The flow geometry at the entry and exit of a turbomachine stage is described by the velocity triangles at these stations. As already mentioned earlier, the velocity triangles for a turbomachine contain the following three components (i) the peripheral velocity, (u), of the rotor blades, (ii) the absolute velocity, (c), of the fluid, and (iii) the relative velocity, (w), of the fluid. These velocities are related by the following well-known vector equation: c

=

u+w

(11.1)

This simple relation is frequently used and is very useful in drawing the velocity triangles for turbomachines. The notation used here to draw velocity triangles correspond to the x-y coordinates; the suffix (a) identifies components in the axial direction and the suffix (t) refers to the tangential direction. Air angles in the absolute system are denoted by alpha (α), where as those in the relative system are represented by beta (β).

Impulse and Reaction Turbines

425

The velocity triangles at the entry and exit of a general turbine stage are shown in Fig. 11.4. Since the stage is axial, the change in the mean diameter between its entry and exit can be neglected. Therefore, the peripheral or tangential velocity (u) remains constant in the velocity triangles. Axial and tangential components of both absolute and relative velocities are also shown in Fig. 11.4. Static and stagnation values of pressure and enthalpy in the absolute and relative systems are indicated.

α1 c1 Station

1

c a1

p1, p01 , h 1 ,h 01

c t1

Nozzle blades Station

p2, p02 , p 02 rel

2

ca2 Station

h 2, h 02 ,h 02 rel

β2

2

α2

w t2

u c t2

Rotor blades Station

c2

w2

p3, p03 , p 03 rel

3 w3

β3

h 3, h 03 ,h 03 rel

α3

c 3 c a3 c t3

u w t3

Fig. 11.4 Velocity triangles for a turbine stage The following trigonometrical relations can be deduced from the velocity triangles shown in Fig. 11.4. ca2

= c2 cos α2

=

w2 cos β2

ct2

= c2 sin α2

=

wt2 + u

= w2 sin β2 + u u sin (α2 − β2 )

=

c2 sin (90 + β2 )

u c2

=

sin (α2 − β2 ) cos β2

ca3 ct3

=

c3 sin α3

= w3 cos β3 = wt3 − u

= =

(11.2)

(11.3) =

c2 cos β2 (11.4)

c3 cos α3 w3 sin β3 − u

(11.5) (11.6)

426

Gas Turbines

ct2 + ct3

=

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (w2 sin β2 + u) + (w3 sin β3 − u) ⎪ ⎬

=

wt2 + wt3

=

c2 sin α2 + c3 sin α3

=

(11.7)

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

w2 sin β2 + w3 sin β3

It is often assumed that the axial velocity component remains constant through the stage. For such a condition

ca

ca

=

=

ca1

= ca2

=

ca3

c1 cos α1

= c2 cos α2

=

c3 cos α3

ca

= w2 cos β2

=

w3 cos β3

(11.8)

Equation 11.7 for constant axial velocity yields a useful relation, tan α2 + tan α3 11.5

= tan β2 + tan β3

(11.9)

EXPRESSION FOR WORK OUTPUT

Though force and torque are exerted on both stationary and moving blades alike, work can only be done on the moving rotor blades. Thus the rotor blades transfer energy from the fluid to the shaft. The stage work in an axial turbine (u3 = u2 = u) can be written as W

=

u2 ct2 − u3 ct3

=

u{ct2 − (−ct3 )}

=

u(ct2 + ct3 )

⎫ ⎪ ⎪ ⎪ ⎪ ⎬

(11.10)

⎪ ⎪ ⎪ ⎪ ⎭

This equation can also be expressed in another useful form.‡ W

=

u2

ct3 ct2 + u u

(11.11)

The first term cut2 in the bracket depends on the nozzle or fixed blade angle (α2 ) and the ratio σ = cu2 . The contribution of the second term cut3 to the work is generally small. It is also observed that the kinetic energy of the fluid leaving the stage is greater for larger values of ct3 . The leaving loss from the stage is minimum when ct3 = 0, i.e., when the discharge from the stage is axial (c3 = ca3 ). However, this condition gives lesser stage work as can be seen from Eqs. 11.10 and 11.11. ‡ Usually, Eq. 11.10 will be written with a minus sign between c t2 and ct3 . Whenever this is written with a plus sign it is implied that ct3 is negative.

Impulse and Reaction Turbines

427

If the swirl at the exit of the stage is zero;, i.e., ct3 = 0, Eq. 11.11 yields : ⎫ ct2 2: W = ⎪ u u ct3=0 ⎪ ⎪ ⎪ ⎬ : c2 sin α2 2: = u (11.12) u ct3=0 ⎪ ⎪ ⎪ ⎪ : ⎭ sin α2 = u2 : σ

ct3=0

Thus, when the exit swirl is zero, the work output is a function of speed ratio, fixed blade angle and peripheral speed. 11.6

BLADE LOADING AND FLOW COEFFICIENTS

Performance of turbomachines are characterized by various dimensionless parameters. For example, the loading coefficient (ψ) and the flow coefficient (φ) have been defined as ψ

W u2

=

(11.13)

ca (11.14) u Since the work, (W ), in Eq. 11.13 is frequently referred to as the blade or stage work, the coefficient (ψ) would also be known as the blade or stage loading coefficient. From Eqs. 11.10, 11.13 and 11.14, for constant axial velocity (ca ), it can be shown that φ

ψ

=

φ(tan α2 + tan α3 )

=

=

φ(tan β2 + tan β3 )

(11.15)

The φ − ψ plots are useful in comparing the performances of various stages of different sizes and geometries. 11.7

BLADE AND STAGE EFFICIENCIES

Even though the blade and stage work (outputs) are the same, the blade and stage efficiencies need not be equal. This is because the energy inputs to the rotor blades and the stage (fixed blade ring plus the rotor) are different. The blade efficiency is also known as the utilization factor ( ) which is an index of the energy utilizing capability of the rotor blades. Thus, =

ηb

=

Rotor blade work Energy supplied to the rotor blades

=

W Erb

(11.16)

The blade work can be written as the sum of changes in the various kinetic energies: W

=

u2 ct2 + u3 ct3

=

1 2 1 1 c − c23 + u22 − u23 + w32 − w22 2 2 2 2

(11.17)

428

Gas Turbines

The energy supplied to the rotor blades is the absolute kinetic energy in the jet at the entry plus the kinetic energy change within the rotor blades. Erb =

1 2 1 2 1 c + w − w22 + u22 − u23 2 2 2 3 2

(11.18)

Substituting Eqs. 11.17 and 11.18 into Eq. 11.16, we get =

ηb

=

c22 − c23 + u22 − u23 + w32 − w22 c22 + (w32 − w22 ) + (u22 − u23 )

(11.19)

For axial machines, u = u2 = u3 , =

ηb

=

c22 − c23 + w32 − w22 c22 + (w32 − w22 )

(11.20)

To avoid any confusion between the stage and blade efficiencies, the term utilization factor ( ) will be used in place of blade efficiency in this book. A stage has already been defined. The rows of nozzle and rotor blades between stations 1 and 2, and 2 and 3 respectively have been represented by the blades as shown in Fig. 11.4. As pointed out earlier, the stage output is the same as the rotor blade work which is given by Eqs. 11.10, 11.11 and 11.17, but the energy input to the stage is different from that of the rotor blades. The stage efficiency (ηst ) takes into account the aerodynamic losses in the fixed and moving blade rows of the stage. 11.8

MAXIMUM UTILIZATION FACTOR FOR A SINGLE IMPULSE STAGE

Figure 11.5 shows a single stage impulse turbine. In this, there is no static pressure change in the rotor. The pressure and velocity variations of the fluid passing through the stage is also shown in Fig. 11.5. Because of the pressure drop in the nozzle blade row, the absolute velocity of the fluid increases correspondingly and transformation of energy occurs in the nozzle. However, the transfer of energy occurs across the rotor blade row. Therefore the absolute fluid velocity decreases through the rotor blade row, as shown in the figure. The velocity triangles for a single impulse stage are as shown in Fig. 11.6. For frictionless flow in the absence of any pressure drop through the rotor blades the relative velocities at their entry and exit are very much the same (w3 ≈ w2 ). To obtain this condition the rotor blade angles must be equal. Therefore the utilization factor is given by

=

u(ct2 + ct3 ) 1 2 2 c2

Substituting from Eq. 11.7 and noting w2 = w3 and β2 = β3 , we get

Impulse and Reaction Turbines

N

429

R

N = Nozzle R = Rotor

c p

Velocity Pressure

Fig. 11.5 Variation of pressure and velocity through a single stage impulse turbine =

4uw2 sin β2 c22

From the velocity triangle at the entry (Fig. 11.6), = Substituting σ =

u c2

ηb

4u(c2 sin α2 − u) c22

and rearranging, =

=

4 σ sin α2 − σ 2

(11.21)

This shows that the utilization factor is a function of the blade-to-gas speed ratio and the nozzle angle. =

f (σ, α2 )

Now, the value of σ corresponding to maximum utilization can be determined. For given value of α2 , differentiating Eq. 11.21, d = sin α2 − 2σ = 0 dσ σopt

=

u c2

=

1 sin α2 2

(11.22)

c2 sin α2

=

ct2

=

2u

(11.23)

=

w3 sin β3

Equations 11.3 and 11.23 give w2 sin β2

=

u

(11.24)

Substituting this in Eq. 11.6 yields ct3 = 0. This means that the exit from the stage should be axial, i.e., c3 = ca3 . This result can also be obtained by mere physical interpretation of the mechanics of flow. Combining Eq. 11.22 and 11.12,

430

Gas Turbines

β2

α2

ca2

w t2

c2

w2

u

c t2

α

w3

w2 =k

u

β3

3

c3 ca3 c t3

wt3

Fig. 11.6 Velocity triangles for a single impulse stage with negative swirl at exit Wct3=0

=

Wopt

=

2u2

(11.25)

Velocity triangles to satisfy these conditions are shown in Fig. 11.7. The combination of Eqs. 11.21 and 11.22 gives max

=

sin2 α2

(11.26)

The rotor blade angles can also be determined for these conditions u (11.27) tan β3 = tan β2 = c2 cos α2 Substituting from Eq. 11.22, tan β3

=

tan β2

=

1 tan α2 2

(11.28)

Thus, for maximum utilization factor, the rotor blade angles are fixed by the nozzle air angle at exit. 11.9

VELOCITY-COMPOUNDING OF MULTISTAGE IMPULSE TURBINE

When the pressure drop to be achieved in an impulse turbine is large; it will not be possible to achieve in one stage. If this has to be achieved in a single stage then this would lead to either a larger diameter or a very high rotational speed. Thus, a single stage utilizing a large pressure drop will have very high peripheral speed of its rotor which is not possible to achieve in practice. Therefore, machines with large pressure drops employ more than one stage.

Impulse and Reaction Turbines

431

c1

β2 ca

α2

wt2

w2

c2

u c t2 = 2u

w 3=

w2

β

3

u

ca = c 3 ct3 = 0

u

Fig. 11.7 Velocity triangles for a single impulse stage with maximum utilization factor One of the methods that is followed in multistage expansion in impulse turbines is to generate high velocity of the fluid. This can be accomplished by expanding the gas through a large pressure drop in the nozzle blade row. The high velocity fluid then transfers its energy in a number of stages by incorporating many rotor blade rows separated by rows of fixed guide blades. Figure 11.8 shows a two-stage velocity compounded impulse turbine. The decrease in the absolute velocity of the fluid across the two rotor blade rows (R1 and R2 ) is mainly due to the energy transfer. The slight decrease in the fluid velocity through the fixed guide blades (F ) is attributed to losses. Since, the turbine is of the impulse type, the pressure of the fluid remains constant after its expansion in the nozzle blade row. Such stages are referred to as velocity or Curtis stages. Figure 11.9 shows the velocity triangles for the above mentioned turbine. To achieve maximum utilization factor, the exit from the last stage (in this case the second-stage rotor) must be in the axial direction, i.e., α3 = 0 and c 3 = ca . The following two assumptions are usually made for this type of turbines: 1. Equiangular flow through rotor and guide blades β2

=

β3 ;

α3

=

α2 ;

β2

=

β3

c2 ;

w2

=

w3

2. Frictionless flow over the blades w2

=

w3 ;

c3

=

For maximum utilization factor, ct3 = 0 and u

=

w3 sin β3

=

w3 sin β2 = w2 sin β2

(11.29)

432

Gas Turbines

N

F

R1

R2 F = Fixed blade N = Nozzle R = Rotor

c p Velocity Pressure

Fig. 11.8 Variation of pressure and velocity through a two-stage velocity compounded impulse turbine

β2

ca

c2 w2 3u ct2 = 4u

α2

w3=w 2

c3 ct3 =2u

u β2’

ca

α2’

u α3 β3 ca

c2’= c 3 w2’

u u c t2’ = 2u ’ =0 =’ w 2 β3’ α c3’3’= ca w3 ct3’= 0 u

Fig. 11.9 Velocity triangles for a two-stage velocity compounded impulse turbine with maximum utilization factor

Impulse and Reaction Turbines

ct2

ct2

=

2u =

c2 sin α2

ct3

=

3u =

w3 sin β3

=

c2 sin α2

= =

=

w2 sin β2 + u

c2 sin α2

=

4u

σopt

=

u c2

433

c3 sin α3

(11.30)

w2 sin β3

(11.31)

Substituting from Eq. 11.31,

=

1 sin α2 4

(11.32)

Similarly, for a three-stage velocity compounded turbine σopt

=

1 sin α2 6

=

1 sin α2 2n

Thus for n velocity stages, σopt

(11.33)

Values of work in the first and second stages, from Eq. 11.7 are given by WI

=

u(w2 sin β2 + w3 sin β3 )

WII

=

u(w2 sin β2 + w3 sin β3 )

WI

=

2uw2 sin β2

WI

=

6u2

From assumptions 1 and 2,

From Eq. 11.31, (11.34)

This expression can also be written directly from the velocity triangles shown in Fig. 11.9. Similarly Eq. 11.29 gives WII

=

2u sin β2

=

2u2

(11.35)

The total work of the velocity compounded impulse turbine (with two stages) is WI + WII

=

8u2

(11.36)

The expressions in Eqs. 11.34, 11.35 and 11.36 can be represented by the following general relations. If the number of velocity stages is n with individual stages identified by i = 1, i = 2, . . . , i = n, then the work done by the ith stage is given by

434

Gas Turbines

Table 11.1 Values of stage and turbine work, blade-to-gas speed ratio and maximum utilization factor for velocity compounded impulse turbine Work n

σopt i=1

i=2

i=3

i=4

1

2u2

2

6u2

2u2

3

10u2

6u2

2u2

4

14u2

10u2

6u2

2u2

5

18u2

14u2

10u2

6u2

Wi

=

i=5

2u2

max

WT 2u2

1 2

sin α2

sin2 α2

8u2

1 4

sin α2

sin2 α2

18u2

1 6

sin α2

sin2 α2

32u2

1 8

sin α2

sin2 α2

50u2

1 10

sin α2

sin2 α2

2{2(n − i) + 1}u2

(11.37)

The total turbine work with n velocity stages is given by WT

=

n )

Wi

=

2n2 u2

(11.38)

i=1

Since the turbine is of the impulse type, the input energy is equal to Therefore, the utilization factor from Eq. 11.38 is max

=

WT u 2 1 2 = 4n c c 2 2 2

=

sin2 α2

1 2 2 c2 .

2

= 4n2 σ 2

(11.39)

Equations 11.33 and 11.39 yield max

This shows that the maximum value of the utilization factor for a given value of the nozzle angle remains unaltered with the number of velocity stages under the assumed conditions. Values in Table 11.1 have been computed from Eqs. 11.33, 11.37 and 11.38. It is observed that the turbine work in the velocity compounded machines increases drastically by employing more velocity stages. This is a great advantage because the pressure of the gas is reduced quickly. This gives a much smaller number of stages which will reduce the overall turbine length and only a small part of the turbine casing is subjected to high pressure gas. However, the work in the subsequent velocity stages goes on decreasing, e.g., the third stage in the three-stage turbine only produces one-ninth of the total work. Therefore, velocity stages beyond three have no practical importance. It may be noted that the pressure ratios across gas turbines are much lower as compared to steam turbines. Therefore, the problem of reducing

Impulse and Reaction Turbines

435

the gas pressure quickly in the casings of gas turbines is not that serious. It may be noted that the losses in the velocity stages are higher, compared to the reaction stages. Therefore, velocity compounding has little to offer in the field of gas turbines except in some special applications. 11.10

PRESSURE COMPOUNDING OF MULTISTAGE IMPULSE TURBINE

In the previous sections we have seen the details of velocity compounding of the impulse turbine. There are two major problems in velocity-compounded stages: (i) the nozzles have to be of the convergent-divergent type for generating high velocity. This results in a more expensive, and difficult design of the nozzle blade rows; and (ii) high velocity at the nozzle exit leads to higher cascade losses. Shock waves are generated if the flow is supersonic which further increase the losses. In order to avoid these problems another method of utilizing a high pressure ratio is employed in which the total pressure drop is divided into a number of impulse stages. These are known as pressure-compounded or Rateau stages. On account of the comparatively lower pressure drop, the nozzle blade rows are subsonic (M < 1). Therefore, such a stage does not suffer from the disabilities of the velocity stages. Figure 11.10 shows the variation of pressure and velocity of gas through the two pressure stages of an impulse turbine. This is equivalent to using two stages of the type shown in Fig. 11.5 in series. Velocity triangles for two Rateau stages, each having an axial exit are shown in Fig. 11.11. It may be noted that the nozzle blades in each stage receive flow in the axial direction. Some designers employ pressure stages upto the last stage. This gives a turbine of shorter length as compared to the reaction type, with a penalty on efficiency. 11.11

THE REACTION TURBINE

In a turbine if the degree of reaction is unity, then we call that machine as the pure reaction machine. But in practice it is impossible to have a 100% degree of reaction and therefore a combination of impulse and reaction is employed. In a reaction turbine there will be number of stages and the pressure changes through each row of blades, both moving and stationary. Leakage is a serious problem in reaction blading in the high pressure part where the blades are short and percentage tip clearance area is relatively large compared to the total area. Leakage depends upon the clearance, the velocity ratio and the stage pressure ratio. Hence, the sealing problem at blade tips is quite critical because the pressure gradient exits across the blades in reaction machines.

436

Gas Turbines

N1

R1

N2

R2

N = Nozzle c p

R = Rotor

Velocity Pressure

Fig. 11.10 Variation of pressure and velocity through a two-stage pressure compounded impulse turbine 11.12

MULTISTAGE REACTION TURBINES

A brief introduction of reaction stages and the degree of reaction has been given already in connection with the axial flow compressor. Figure 11.12 shows two reaction stages and the variation of pressure and velocity of the gas in them. The gas pressure decreases continuously over both fixed and moving rows of blades. Since the pressure drop in each stage is smaller as compared to the impulse stages, the gas velocities are relatively low. Besides this the flow is accelerating throughout. These factors make the reaction stages aerodynamically more efficient though the tip leakage loss increases on account of the relatively higher pressure difference across the rotor blades. Multistage reaction turbines employ a large pressure drop by dividing it into smaller values in individual stages. Thus the reaction stages are like the pressure-compounded stages (see Section 11.10) with a new element of ‘reaction’ introduced in them, i.e., of accelerating the flow through blade rows also. 11.12.1

Enthalpy-Entropy Diagram

Figure 11.13 shows the enthalpy-entropy diagram for flow through a general turbine stage with some degree of reaction. This should be studied with the velocity triangles in Fig. 11.4. Both static and stagnation values of the pressures and enthalpies are indicated at various stations. In general, the gas approaches the stage with a finite velocity (c1 ). Therefore, the values of pressures p1 and p01 and enthalpies h1 and h01 have been differentiated. The relation between the static and stagnation

Impulse and Reaction Turbines

437

c2 w2 u =w w3

2

c3=ca

u

Second stage nozzle ring c 2’ w’2 u =’ w

2’

w3

c 3’= ca

u

Fig. 11.11 Velocity triangles for a two-stage pressure-compounded impulse turbine

F1

R1

F2

R2 F = Fixed blade R = Rotor

c p

Velocity Pressure

Fig. 11.12 Variation of pressure and velocity through a two-stage reaction turbine

438

Gas Turbines

p

01

01 1/2c21

02 p02

p

1

Enthalpy

1

p

Ws

1/2 c 22

02 rel

0 2 rel

Wa

h01= h 02

0 3 rel p03 rel

= h03 rel h 02 rel

2

2

1/2 w 2 p

2

2’

2 1/2 w 3 03

Δs p 0 3"

p

03

1/2 c 23 p

03"

3

3 3’

3"

Entropy

Fig. 11.13 Enthalpy-entropy diagram for flow through a turbine stage values has already been explained. The reversible adiabatic (isentropic) expansion of the gas through fixed and moving rows of blades is represented by processes 1–2 and 2 –3 respectively. Stagnation points O1 and O3 are defined by the following relations: h01 h03

1 = h1 + c21 2

(11.40)

1 = h3 + c203 2

(11.41)

The ideal work in the stage is given by Ws

= h01 − h03

=

Cp (T01 − T03 )

(11.42)

The actual expansion process (irreversible adiabatic) is represented by the curve 1-2-3, and the actual states of the gas at the exits of the fixed and moving blades are represented by points 2 and 3 respectively. Since, there is only energy transformation and no energy transfer (work) between states 1 and 2, the stagnation enthalpy remains constant. h01

=

h02

=

1 h2 + c22 2

(11.43)

The loss due to irreversibility (Δs = s2 − s2 ) is given by the enthalpy loss coefficient h2 − h2 2Cp ξN = = (T2 − T2 ) (11.44) 1 2 c22 2 c2

Impulse and Reaction Turbines

439

The stagnation pressure loss coefficient corresponding to Eq. 11.44 is defined by (Δp0 )N p01 − p02 YN = = (11.45) 1 1 2 2 ρc 2 2 2 ρc2 The expansion process (2–3) in the moving blade rows represents both transformation and transfer of energy. Therefore, the difference in the absolute stagnation enthalpies at states 2 and 3 gives the actual value of the stage work. Wr

h03

=

h02 − h03

=

h01 − h03

=

Cp (T01 − T03 )

(11.46)

=

1 h3 + c23 2

(11.47)

However, to an observer moving with the rotor the relative flow appears as the absolute flow in the nozzle to a stationary observer. Therefore, for him (in the relative system) the stagnation enthalpy in the moving frame of coordinates remains constant. h02rel 1 h2 + w22 2

= h03rel 1 = h3 + w32 2

(11.48)

The enthalpy and pressure losses for the moving row of blades are given by the following coefficients: ξR

=

h3 − h3 1 2 2 w3

YR

=

p02rel − p03rel 1 2 2 ρw3

=

2Cp (T3 − T3 ) w32 =

(Δp0 )R 1 2 2 ρw3

(11.49)

(11.50)

It can be proved that for most of the applications (M < 1), there is little difference between the values of the loss coefficients based on enthalpy and stagnation pressure, i.e., ξ ≈ Y . 11.12.2

Degree of Reaction

Degree of reaction of a turbine stage can be defined in a number of ways; it can be expressed in terms of pressures or velocities or enthalpies or the flow geometry in the stage. (a) A definition on the basis of the isentropic change in the stage is given by the following relation: R

=

Isentropic change of enthalpy in the rotor Isentropic change of enthalpy in the stage

440

Gas Turbines

=

;2 ;31 3

Δhr

=

Δhs

h2 − h3 h1 − h3

(11.51)

Assuming ρ = constant, Eq. 11.51 can be transformed in terms of pressures. Using Δh = Δp ρ , ; 2 Δp ;2 Δp p2 − p3 ρ 3 ≈ ;31 (11.52) R = ; 1 Δp = p1 − p3 Δp 3

ρ

3

(b) Definitions in Eqs. 11.51 and 11.52 are not widely used in practice. Since the actual enthalpy changes and work are more important for turbine stages, a definition of the degree of reaction based on these quantities is more logical and useful. It may be noted that the change of enthalpy through the rotor and the total change in the stage are h2 − h3 h01 − h03

=

1 2 1 w3 − w22 + u22 − u23 2 2

= h02 − h03

=

(u2 ct2 − u3 ct3 )

(11.53) (11.54)

The ratio of these quantities gives another definition of the degree of reaction. h2 − h 3 R = (11.55) h01 − h03 If c1 ≈ c3 , R

=

h2 − h3 h1 − h3

(11.56)

Equations 11.53 and 11.54 when substituted in Eq. 11.55 yield R

=

1 2

w32 − w22 + 12 u22 − u23 u2 ct2 − u3 ct3

(11.57)

For axial machines, u3 = u2 = u. Therefore, from Fig. 11.4, R

=

w32 − w22 2u(ct2 + ct3 )

(11.58)

(c) Equation 11.58 can also be expressed in terms of the geometry of flow through the stage. From velocity triangles (Fig. 11.4), w22

2 = c2a2 + wt2

w32

2 = c2a3 + wt3

For constant axial velocity,

Impulse and Reaction Turbines

w32 − w22

2 2 = wt3 − wt2

= c2a tan2 β3 − tan2 β2 w32 − w22

441

(11.59)

= (wt3 + wt2 )(wt3 − wt2 )

Using Eq. 11.7 and the velocity triangles, w32 − w22

= ca (ct2 + ct3 )(tan β3 − tan β2 )

(11.60)

Equation 11.60 when put into Eq. 11.58 yields a widely used definition for the degree of reaction. ca R = (tan β3 − tan β2 ) (11.61) 2u But φ

=

ca u

Now, the mean axial velocity cam can be written as 1 (ca1 + ca2 ) cam = 2 Similarly,

(11.62)

ctm

=

1 (ct2 − ct1 ) 2

(11.63)

tan αm

=

ctm cam

(11.64)

Therefore,

Hence, for constant velocity tan αm

=

1 (tan α2 − tan α1 ) 2

(11.65)

tan βm

=

1 (tan β3 − tan β2 ) 2

(11.66)

R

= φ tan βm

(11.67)

Two more useful relations are obtained assuming ca to be constant. From the velocity triangle at the entry, Eq. 11.3 can be written as ca tan α2 tan β2

= ca tan β2 + u = tan α2 −

u ca

(11.68)

Eliminating tan β2 , between Eqs. 11.61 and 11.68, and rearranging, ca 1 + (tan β3 − tan α2 ) (11.69) R = 2 2u Equation 11.6 can similarly be written as

442

Gas Turbines

p

1

Enthalpy

1

p =p 2

3

h2 = h 3 2,3 Entropy

Fig. 11.14 Zero reaction or impulse stage, enthalpy-entropy diagram for reversible flow ca tan α3 tan β3

= ca tan β3 − u = tan α3 +

u ca

(11.70)

Equation 11.70 when substituted in Eq. 11.69 gives ca (tan α3 − tan α2 ) R = 1+ 2u 11.12.3

Zero Degree Reaction Stage

Though a single-stage impulse turbine has already been discussed in Sec.11.8, it is further discussed here in the light of degree of reaction. (a) Figure 11.14 shows the enthalpy-entropy diagram for reversible adiabatic expansion in an impulse stage. Expansion of the gas occurs only in the nozzle blade row and its thermodynamic state remains unchanged between stations 2 and 3. Therefore the air angles, static pressure and enthalpy at the entry and exit of the rotor in this stage are the same. β 2 = β3

p2 = p3

h2 = h3

These conditions when put into Eqs. 11.51, 11.52, 11.55, 11.56 and 11.61, give the degree of reaction as zero. On account of the reversible nature of flow, such a stage is of no practical importance. (b) Figure 11.15 depicts the enthalpy-entropy diagram for irreversible adiabatic expansion in an impulse stage with equal pressures at the rotor

Impulse and Reaction Turbines

443

p1

Enthalpy

1

h3 h2

2 1/2 w 3

1/2 w 22

= h 03 rel h 02 rel p =p 3

2

3 2 2’ Entropy

Fig. 11.15 Impulse stage (negative degree of reaction), enthalpy-entropy diagram for irreversible flow entry and exit. On account of losses, both in the nozzle and rotor blade rows, the gas experiences an increase in entropy. Therefore, for this stage p2

=

p3

w2

>

w3

h2




β3

These conditions give the degree of reaction as negative (Eqs. 11.55, 11.56, 11.58 and 11.61). However, Eq. 11.52 defines such a stage as impulse. (c) Figure 11.16 shows the enthalpy-entropy diagram of a real (adiabatic) and a truly zero reaction stage. Here a small pressure drop in the rotor compensates for the deceleration of the flow in the rotor blades (Fig. 11.15),. Thus the enthalpies and relative velocities remain constant between the rotor entry and exit. Equation 11.52 for such a stage gives the value of reaction as positive, whereas Eqs. 11.55, 11.56, 11.58 and 11.61 yield R = 0. 11.12.4

Fifty Per Cent Reaction Stage

For fifty per cent degree of reaction, Eq. 11.56, gives R

=

h2 − h3 h1 − h3

=

1 2

444

Gas Turbines

p

1

Enthalpy

1

h02 rel= h03 re p

1/2w32

2

1/2 w 2

2

p

3

2

h2 = h3

3

2’

Entropy

Fig. 11.16 Zero reaction stage, enthalpy-entropy diagram for irreversible flow This gives equal values to the actual enthalpy drops in the fixed and rotor blade rows (Fig. 11.12). h1 − h2

= h2 − h3

=

1 (h1 − h3 ) 2

(11.71)

Figure 11.17 shows the enthalpy-entropy diagram for such a stage. The flow and cascade geometries in the fixed and moving blade rows are such that they provide equal enthalpy drops. This does not imply that the pressure drops in the two rows are also the same. From Eq. 11.69, 1 2 tan β3 β3

=

1 ca + (tan β3 − tan α2 ) 2 2u

= tan α2 = α2

(11.72)

This condition when put into Eq. 11.9 yields tan β2 β2 Equation 11.57 for R =

1 2

= α3

gives

1 2 1 w − w22 + u22 − u23 2 3 2 For axial turbines,

= tan α3

=

1 (u2 ct2 − u3 ct3 ) 2

(11.73)

Impulse and Reaction Turbines

445

p

1

1 h 1 - h2 = 1/2 (h1 - h3 ) Enthalpy

p2 2

h 2 - h3 = 1/2 (h1 - h3 ) p 3

3

Entropy

Fig. 11.17 Enthalpy-entropy diagram for a 50% reaction stage w32 − w22

1 1 2 c − c23 + w32 − w22 2 2 2

=

= c22 − c23

(11.74)

This with Eqs. 11.72 and 11.73 gives w3

= c2

(11.75)

w2

= c3

(11.76)

Thus, it is observed that the velocity triangles at the entry and exit of a rotor in a fifty per cent reaction turbineare similar. These are shown in Fig. 11.18. Utilization Factor Equations 11.75 and 11.76 when put into Eq. 11.20 gives the value of the utilization factor for the fifty per cent reaction stage. =

=

2 c22 − c23 c22 + (c22 − c23 ) 1− 1−

c23 c22 c23 2c22

From the entry velocity triangle in Fig. 11.18,

=

c22 − c23 c22 − 12 c23 (11.77)

446

Gas Turbines

c1 = c3

Fixed blade row

β2

α2

ca

c2 = w 3

w2 = c3 u

Rotor blade row

α3

β3

w3

ca

u

Fig. 11.18 Velocity triangles for a 50% reaction stage w22

=

c22 + u2 − 2uc2 cos (90 − α2 )

c23

=

c22 + u2 − 2uc2 sin α2

c23 c22

=

1+

σ

=

u c2

c23 c22

=

1 + σ 2 − 2σ sin α2

c23 c22

=

2σ sin α2 − σ 2

(11.79)

c23 2c22

=

1 1 + 2σ sin α2 − σ 2 2

(11.80)

1− 1−

u c2

2

−2

u c2

for R =

2

1 2

sin α2

Equations 11.79 and 11.80 when put into Eq. 11.77 give =

(11.78)

2σ sin α2 − σ 2 1 + (2σ sin α2 − σ 2 )

Impulse and Reaction Turbines

=

2 1+

1 2σ sin α2 − σ 2

447

(11.81)

The utilization factor is maximum when the factor 2σ sin α2 − σ 2 is maximum. d 2σ sin α2 − σ 2 = 0 dσ 2 sin α2

=

2σ

σopt

=

u c2

=

sin α2

(11.82)

This condition in Eq. 11.81 gives max

=

2 sin2 α2 1 + sin2 α2

(11.83)

Equation 11.82 gives u =

c2 sin α2

=

ct2

Since the entry and exit velocity triangles are symmetrical, u =

w3 sin β3

=

wt3

These relations give right-angled velocity triangles (Fig. 11.19) at the entry and exit. β2

=

0

α3

=

0

The exit swirl is zero (ct3 = 0), i.e., the exit from the stage is axial. The stage work for maximum utilization from Eq. 11.11 or 11.12, or direct from the velocity triangles (Fig. 11.19) is Wopt

=

u2

(11.84)

This is half of the value for a single-stage impulse and one-eighth of the two-stage velocity compounded impulse type. Equation 11.58 can be written as 1 2 w − w22 = u(ct2 − ct3 )R 2 3 + , 1 2 1 2 2 2 c − c3 + w3 − w2 R = 2 2 2 w32 − w22

=

R c2 − c23 1−R 2

This when put into Eq. 11.20 gives the utilization factor for an axial turbine as

448

Gas Turbines

ca

α2

c2 ct2=u

w3

β3

ca ct3 = 0

u

Fig. 11.19 Velocity triangles for a 50% reaction stage at maximum utilization factor c22 − c23 + 1 +

=

c22 +

R 2 1−R (c2

R 1−R

− c23 )

This on simplification gives a general relation between the utilization factor and the degree of reaction =

c22 − c23 c22 − Rc23

For R = 1, this equation gives indeterminate value. Therefore Eq. 11.71 is used.

11.12.5

Hundred Per Cent Reaction Stage

Hundred per cent degree of reaction implies that the entire change in the static properties occurs in the rotor. Equation 11.56 for this condition gives h2 − h3 h1 − h3

R

=

h1

= h2

=

1

when R = 1

(11.85)

Equation 11.58 gives 1 2 w − w22 2 3

= u(ct2 + ct3 )

which shows that the stage work is wholly due to the change in the kinetic energy occurring in the rotor. Rewriting the quantity on the right-hand side (refer Eq. 11.17 and note that u2 = u3 ),

Impulse and Reaction Turbines

w2 β α 2 2 u w3

c2 β 3 c 3=

u

449

c2 ct2

α3=α2

ct2 = ct3

Fig. 11.20 Velocity triangles for a 100% reaction turbine stage 1 2 w − w22 2 3

1 2 1 c − c23 + w32 − w22 2 2 2

=

(11.86)

c2

= c3

when R = 1

(11.87)

α2

= α3

when R = 1

(11.88)

Further Eq. 11.71 gives

These conditions give the velocity triangles as shown in Fig. 11.20. It is observed that the rotor blades are highly staggered. On account of the large difference of static pressure across the rotor and high relative velocity at its exit, the losses are relatively higher. A stage with a degree of reaction higher than hundred per cent requires h2

>

h1

c2




α2

These conditions show that for such a stage the flow is decelerating in the nozzle or upstream fixed blades. This is obviously undesirable. 11.12.6

Negative Reaction

A negative degree of reaction in a turbine stage, such as R > 100% is also undesirable. For such a stage h3 w3 β2

> < >

h2 w2 β3

(Eq. 11.55) (Eq. 11.58) (Eq. 11.67)

These conditions will require the diffusion of flow in the rotor blade row. Such a condition may be, accepted only as a necessary evil in long blades where it is necessary to limit the high degree of reaction at the blade tips.

450 11.13

Gas Turbines

BLADE-TO-GAS SPEED RATIO

It has been seen in the earlier sections that the stage work and utilization factor are governed by the blade-to-gas speed ratio parameter (velocity ratio) σ = cu2 . Its value for maximum utilization was determined for some stages. Therefore, efficiencies of the turbine stages can also be plotted against this ratio. Such plots for some impulse and reaction stages are shown in Fig. 11.21. The performance of steam turbines is often presented in this form. The curves in Fig. 11.21 also show the optimum values of the velocity ratio and the range of off-design for various types of stages. The fifty per cent reaction stage shows a wider range. Another important aspect that is depicted here is that in applications where high gas velocities (due to high-pressure ratio) are unavoidable, it is advisable to employ impulse stages to achieve practical and convenient values of the size and speed of the machine. Sometimes it is more convenient to use an isentropic velocity ratio. R = 50%

1.0

ε η

0.8

N

Single-stage impulse

ηN 0.6 Two-stage impulse

0.4 0.2

σopt

σopt 0

0.2

0.4 σ = c2

0.6

σopt 0.8

1.0

Fig. 11.21 Variation of utilization factor and stage efficiency with blade-togas speed ratio This is the ratio of the blade velocity and the isentropic gas velocity that would be obtained in its isentropic expansion through the stage pressure ratio. u σs = (11.89) cs 1 2 c 2 s

=

cs

=

2(h01 − h03 )

cs

=

2Cp T01 1 − r

h01 − h03

(Fig. 11.13) (11.90)

For a perfect gas, 1−γ γ

(11.91)

A relation between velocity ratios and the degree of reaction can be derived for isentropic flow. From Eq. 11.51,

Impulse and Reaction Turbines

=

h2 − h3 h1 − h3

=

1−

h1 − h2

≈

1 2 c 2 2

h1 − h3

≈

1 2 c 2 s

R

h1 − h2 h1 − h3

Therefore, R

=

451

c2 2 u cs 2 u

c2 1 − 22 cs

=

1−

σ

=

σs

(11.93)

σ

=

1 σs √ ηnoz

(11.94)

σ

=

√ 2σs

(11.95)

=

1−

σs2 σ2

(11.92)

For R = 0,

With a nozzle efficiency, ηnoz ,

For R = 12 ,

11.14

LOSSES AND EFFICIENCIES

Losses occurring in a turbomachine blade rows (stationary or moving) have been described in Sec.9.10. Besides these losses, additional losses occur in an actual turbine due to disc and bearing friction. Figure 11.22 shows the energy flow diagram for the impulse stage of an axial turbine. Numbers in brackets indicate the order of energy or loss corresponding to 100 units of isentropic work (h01 − h03 ). It is seen that the energy reaching the shaft after accounting for stage cascade losses (nozzle and rotor blade aerodynamic losses) and leaving loss is about 85% of the ideal value; shaft losses are a negligible proportion of this value. If the turbine is of the partial admission type it incurs additional losses as shown in the figure. The flow losses discussed in section 9.10 are also appropriate to turbine cascades.

11.15

PERFORMANCE GRAPHS

Performance of various types of gas turbine stages is often presented in terms of plots of the loading coefficient (ψ) against the values of the flow coefficient (φ).

452

Gas Turbines Ideal or isentropic work (100)

Energy nozzle exit (95)

Leaving loss (3)

Ideal stage work (92)

Actual stage work

Shaft work (84)

Disc friction loss

Nozzle aero losses (5)

Secondary loss

Partial admission losses

Profile loss

Annulus loss

Rotor area losses

Shaft losses (1)

Pumping Bearing (for partial admission) loss

Mixing Loss

Expansion Loss

Secondary loss

Shear flow Loss

Profile loss

Annulus loss

Jet dispersion Loss

Leakage Loss

Tip leakage loss

Fig. 11.22 Energy flow diagram for an impulse turbine 11.15.1

Zero Degree Reaction

Equation 11.16 for zero degree reaction (β2 = β3 ) gives ψ = 2φ tan β2 = 2φ tan β3

(11.96)

At maximum utilization factor (Fig. 11.7), u 1 tan β3 = = ca φ ψ 11.15.2

= 2

(11.97)

Fifty Per Cent Reaction

Equation 11.68 for R = 12 (α2 = β3 ) gives u tan β2 = tan α2 − ca tan β2

= tan β3 −

1 φ

This relation when put in Eq. 11.16 gives ψ

= 2φ tan β3 − 1

=

2φ tan α2 − 1

(11.98)

Impulse and Reaction Turbines

453

At maximum utilization factor (Fig. 11.19), ψ

= 1

(11.99)

Equation 11.61 gives R

ca ca tan β3 − tan β2 2u 2u

=

For axial exit from the stage, ca tan β2 = 1 u Therefore, R

ca 1 − tan β2 2 2u

=

ca tan β2 u

= 1 − 2R

(11.100)

Equation 11.16 gives ψ

=

ca ca tan β2 + tan β3 u u

=

ca tan β2 + 1 u

= φ tan β2 + 1

(11.101)

This is a general equation for a stage with an axial exit. Equation 11.100 when put in Eq. 11.101 yields ψ

= 2(1 − R)

(11.102)

This for R = 0 and R = 12 gives the same results as in Eqs. 11.97 and 11.99 respectively. This is not valid for R = 1 or R < 0 because then the exit from the stage cannot remain axial. Equation 11.102 shows that the stage loading decreases with an increase in the degree of reaction. Figure 11.23 shows the φ-ψ plots for various axial turbine stages. 5.0

β2 = β3

o

o

60

10.0 o

30

2.5

0

0.5

o

70

o

o

60 5.0

o

45

ψ

α = 80

1.0

φ (a) Zero reaction

β = 60 2

o

45

ψ

ψ

5.0

0

0.5

1.0

φ (b) 50% reaction

o

30

2.5

0

0.5

1.0

φ (c) Axial exit

Fig. 11.23 Performance charts for various axial turbine stages

454

Gas Turbines

Worked out Examples 11.1 A multistage gas turbine is to be designed with impulse stages, and is to operate with an inlet pressure and temperature of 6 bar and 900 K and an outlet pressure of 1 bar. The isentropic efficiency of the turbine is 85%. All the stages are to have a nozzle outlet angle of 75◦ and equal outlet and inlet blade angles. Mean blade speed of 250 m/s and equal inlet and outlet gas velocities. Estimate the maximum number of stages required. Assume Cp = 1.15 kJ/kg K, γ = 1.333 and optimum blade speed ratio. Solution Let the suffix f s, 1, 2 and 3 represent final stage, exit from the nozzle, inlet to the moving blade and exit from the moving blade respectively. Let the isentropic temperature at the exit of the final stage Tf s . Now T02 T0f s

=

T0f s

=

p02 p0f s T02 1.56

γ−1 γ

900 1.56

=

0.333 1.333

6 1

=

=

=

1.56

576.9 K

Actual overall temperature drop ΔToverall

=

ηT T02 − T0f s

=

0.85 × (900 − 576.9)

=

274.6 K

Applying energy equation to the nozzle of any stage, c21 2

=

h2 +

h 1 − h2

=

1 2 c − c21 2 2

Cp (T1 − T2 )

=

1 2 c − c21 2 2

h1 +

c22 2

(1)

For optimum blade speed ratio, σopt

=

u c2

c2

=

2u sin α1

=

sin α2 2 =

2 × 250 sin 75

=

517.6 m/s

For the given conditions, the inlet and exit velocity triangles are as shown in the Fig. and also note that c3 = C1 .

Impulse and Reaction Turbines

c1

=

c3

=

c2 cos α2

=

133.96 m/s

=

c22 − c21 2Cp

=

108.68 K

=

455

517.6 × cos 75

Now from Eq.1 T1 − T2

517.62 − 133.962 2 × 1.15 × 103

=

For the impulse stage there is no expansion in rotor. Therefore, T1 − T2 will be the stage temperature drop. Hence, number of stages n is given by n

=

ΔToverall ΔTstage

Stages required

=

3

=

274.6 108.68

=

2.57 Ans

⇐=

11.2 A gas turbine having single stage rotates at 10000 rpm. At entry to the nozzles the total head temperature and pressure of the gas is 700◦ C and 4.5 bar respectively and at outlet from the nozzle the static pressure is 2.6 bar. At the turbine outlet annulus the static pressure is 1.5 bar. Mach number at outlet is limited to 0.5 and gas leaves in an axial direction. The outlet nozzle angle is 70◦ to the axial direction and the nozzle friction loss may be assumed to be 3% of the isentropic temperature drop from total head at entry to static conditions at outlet nozzle pressure. Calculate (i) the gas angles at entry and outlet from the wheel showing them on velocity diagrams for mean blade section, (ii) output power developed by the turbine shaft. Assume the mean blades diameter as 64 cm, gas mass flow rate as 22.5 kg/s, turbine mechanical efficiency = 99%, Cp = 1.147 kJ/kg K and γ = 1.33. Solution Let the suffix 1, 2 and 3 represent the entry to the nozzle blade, exit from the nozzle blade (entry to rotor blade) and exit from the rotor blade respectively. Let T02 be the isentropic temperature after expansion in the nozzle. γ−1 γ

T02 T01

=

T02p rime

=

0.8727 × 973

ηnozzle

=

0.97

p02 p01

2.6 4.5

=

=

=

0.333 1.333

=

0.8727

849.14 K

T01 − T02 T01 − T02

=

973 − T02 973 − 849.14

456

Gas Turbines

α1 c1 ca1

w1

β1

u c t1

w2

c2

β2

u Fig. 11.24

T02

=

c2

=

2Cp ΔT

=

2 × 1147 × (973 − 852.9)

u

=

From Fig.11.24 wt2 =

852.9 K

πDN 60

=

524.9 m/s

3.14 × 0.64 × 10000 = 334.93 m/s 60

=

c2 sin α2 − u = 524.9 × sin 70 − 334.93 = 158.3 m/s

ca

=

c2 cos α2

tan β2

=

c2 sin α2 − u ca

β2

=

41.4◦

=

524.9 × cos 70 =

158.3 179.53

= =

179.53 m/s 0.8815 Ans

⇐=

Assuming rotor losses are negligible T3

T2

=

p2 p3

c3

γ−1 γ

=

M

γRT3

=

265.33 m/s

=

=

852.9 2.6 1.5

0.330 1.333

0.5 ×

=

744 K

√ 1.33 × 284.6 × 744

Impulse and Reaction Turbines

tan β3

=

u ca

334.93 265.33

β3

=

51.6◦

ct2

=

c2 sin α2 = 524.9 × sin 70 = 493.2 m/s

=

=

457

1.2623 Ans

⇐=

Power developed =

ηm mu(c ˙ t2 + ct3 )

=

muc ˙ t2

=

0.99 × 22.5 × 334.93 × 493.2 1000

=

3679.53 kW

Ans

⇐=

11.3 In a single-stage impulse turbine the nozzle discharges the fluid on to the blades at an angle of 65◦ to the axial direction and the fluid leaves the blades with an absolute velocity of 300 m/s at an angle of 30◦ to the axial direction. If the blades have equal inlet and outlet angles and there is no axial thrust, estimate the blade angle, power produced per kg/s of the fluid and the blade efficiency. Solution Since the axial thrust is zero, ca3 =

ca2

ca

c

w1

ca1

=

β1

α1

u ct1

w2 c 2

c a2

β2

α2

u

c t2

Fig. 11.25

w3

=

w2

ca3

=

ca2 = ca = ca3 × cos α3

458

Gas Turbines

=

300 × cos 30

=

c2

=

ca2 cos α2

259.8 = 614.7 m/s cos 65

u

=

ct2 − wt2

c2 sin α2 − ca2 tan β2

=

ca3 tan β3 − c3 sin α3

= =

259.8 m/s

wt3 − ct3

Since ca2 = ca3 = ca and β2 = β3 = β, 2ca tan β

=

c2 sin α2 + c3 sin α3

tan β

=

614.7 × sin 65 + 300 × sin 30 2 × 259.8

=

1.3609

β

=

53.7◦

β2

=

β3

u

=

C2 sin α2 − ca2 tan β2

=

614.7 × sin 65 − 259.8 × tan 53.7

=

203.43

=

c2 sin α2

=

557.1 m/s

=

c3 sin α3

=

150 m/s

=

203.43 × (557.1 + 150) × 10−3

=

144 kJ/kg

σ

=

u c2

Blade efficiency

=

4 σ sin α2 − σ 2

=

4 × 0.33 × sin 65 − 0.332 = 0.761

=

76.1%

ct2

ct3

WT

=

=

Ans

⇐= Ans

53.7◦

⇐=

=

614.7 × sin 65

=

300 × sin 30

203.43 614.7

=

0.33

Ans

⇐=

11.4 Gas at 7 bar and 300◦ C expands to 3 bar in an impulse turbine stage. The nozzle angle is 70◦ with reference to the exit direction. The rotor blades have equal inlet and outlet angles, and the stage operates

Impulse and Reaction Turbines

459

with the optimum blade speed ratio. Assuming that the isentropic efficiency of the nozzles is 0.9, and that the velocity at entry to the stage is negligible, deduce the blade angle used and the mass flow required for this stage to produce 75 kW. Take Cp = 1.15 kJ/kg K. Solution

c

w1

ca1

β1

α1

u ct1

w2 c 2

c a2

β2

α2

u

c t2

Fig. 11.26

Let the suffix 1, 2 and 3 represent the entry to the nozzle blade, exit from the nozzle blade (entry to rotor blade) and exit from the rotor blade respectively. Let T02 be the isentropic temperature after expansion in the nozzle. γ−1 γ

0.333 1.333

T02 T01

=

T02p rime

=

573 × 0.8092

T02

=

T01 − ηn × (T01 − T02 )

=

573 − 0.9 × (573 − 463.67)

c2

p02 p01

3 7

= =

=

463.67 K

=

2Cp ΔT

=

2 × 1150 × (573 − 474.6)

=

474.6 K

=

475.73 m/s

For optimum blade speed ratio u 1 sin α2 = c2 2 u

=

475.75 × sin 70 2

0.8092

=

223.52 m/s

460

Gas Turbines

ct2 − u ca2

c2 sin α2 − u ca

=

wt2 ca2

=

475.73 × sin 70 − 223.52 475.73 × cos 70

β2

=

54◦

ct2

=

c2 sin α2

=

447 m/s

ca2

=

w2 cos β2

=

c2 cos α2

w2

=

c2 cos α2 cos β2

=

475.73 × cos 70 cos 54

=

276.8 m/s

w

=

w2

ct3

=

w3 sin β3 − u = 276.8 × sin 54 − 223.52

=

0.416 m/s

WT

=

mu ˙ (ct2 + ct3 ) 1000

m ˙

=

75 × 1000 = 0.75 kg/s 223.52 × (447 + 0.416)

tan β2

=

=

=

1.3737 Ans

⇐=

=

=

w3

475.73 × sin 70

=

276.8 m/s

Ans

⇐=

11.5 The following particulars of a single stage turbine of free vortex type is given below: Total head inlet pressure : 4.6 bar Total head inlet temperature : 700◦ C Static head pressure at mean radius : 1.6 bar Mean blade diameter/blade height : 10 Nozzle loss coefficient : 0.10 Nozzle outlet angle : 60◦ Determine the gas temperatures, velocities and discharge angle at the blade root and tip radii. Take Cp = 1.147 and γ = 1.33 and mass flow rate = 20 kg/s. Solution Let the suffix 1, 2 and 3 represent the entry to the nozzle, exit of the nozzle (entry to the moving blade) and exit from the moving blade respectively. Let T01 be the isentropic temperature after expansion in the nozzle.

Impulse and Reaction Turbines

c1

w1

ca1

461

α1

β1

u ct1

Fig. 11.27

T2 T01

=

T2

=

γ−1 γ

pe p01

0.77 × 973

0.33 1.33

=

1.6 4.6

=

749.2 K

=

0.77

Nozzle loss coefficient =

T2 − T2

T02

=

T2 +

c22 2Cp

=

973 − T2

0.1

=

T2 − 749.2 973 − T2

T2

=

769.5 K

c2

=

2Cp ΔT

=

2 × 1147 × (973 − 769.5)

(1)

c22 2Cp

c22 2Cp

=

683.25 m/s

=

c2 cos α2

=

341.63 m/s

=

p2 RT2

=

0.73 kg/m3

Aca ρ

=

m ˙

A

=

20 = 0.08 m2 341.63 × 0.73

ca

ρ

=

=

683.25 × cos 60

1.6 × 105 284.6 × 769.5

462

Gas Turbines

A

=

Dm

=

πDm h

=

πDm

0.8 × 10 10

=

Dm 10

2 πDm 10

=

0.5 m

Mean blade diameter = 50 cm Blade height =

D 10

= 5 cm

Mean blade radius = 25 cm At the root Droot

=

Dm − h

rroot

=

22.5 cm

Dtip

=

Dm + h

rtip

=

27.5 cm

=

50 − 5

=

45 cm

=

50 + 5

=

55 cm

At the tip

For a free vortex ct × r must be constant ct

root

× rroot

=

ct

ct

=

c2 sin α2

=

591.71 m/s

mean

mean

× rmean =

683.25 × sin 60

ct2

root

=

591.71 × 0.25 0.225

α2

root

=

tan−1

ct2 root ca

=

tan−1

657.45 341.63

=

1.924

=

=

ct2 root sin α2 root

=

741.2 m/s

=

T02 +

=

733.5 K

c2

T2

root

root

=

657.45 m/s

62.5◦ =

Ans

⇐=

657.45 sin 62.5 Ans

⇐=

c22 root 741.22 = 973 + 2Cp 2 × 1147 Ans

⇐=

Impulse and Reaction Turbines

463

At the tip ct2

α2

c2

T2

tip

tip

tip

tip

=

ct2

× rmean 591.71 × 0.25 = rtip 0.275

mean

=

537.9 m/s

=

tan−1

=

57.6◦

=

ct2 tip sin α2 tip

=

537.9 = 637.1 m/s sin 57.6

=

973 −

ct2 tip 537.9 = tan−1 ca 341.63 Ans

637.12 2 × 1147

⇐=

=

Ans

⇐= 796 K

Ans

⇐=

Review Questions 11.1 Explain the principle of operation of a turbine and what are the primary parts of a turbine? 11.2 How do you differentiate between an impulse and a reaction turbine? With neat sketches explain the working of an impulse and a reaction stage. 11.3 With a neat sketch explain a single stage velocity triangle and derive an expression for the work output. 11.4 What do you understand by blade and stage efficiency? Derive an expression for blade efficiency. 11.5 Explain what is meant by velocity compounding of a multistage impulse turbine. 11.6 Explain what do you understand by pressure compounding of a multistage impulse turbine. 11.7 Explain with a sketch and h-s diagram, the working of a reaction turbine. 11.8 Define degree of reaction and derive an expression for the same. 11.9 Explain the following: (i) zero per cent reaction stage, (ii) fifty per cent reaction stage, and

464

Gas Turbines

(iii) hundred per cent reaction stage. 11.10 Explain briefly the performance graphs of an reaction turbine. Exercise 11.1 In a single-stage impulse turbine the nozzle discharges the hot gas on to the blades at a velocity of 750 m/s. The mass flow rate of gas is 100 kg/s. The turbine rotates at 20000 rpm. The mean diameter of wheel is 31.5 cm. The nozzles are inclined at an angle of 20◦ to the plane of wheel rotation. Calculate (i) power developed by the blades in MW, (ii) energy lost in the blades in MW, and (iii) determine the blade efficiency of the turbine. Assume the blade friction coefficient as 0.92 and outlet blade angle as 25◦ to the plane of rotation. Ans: (i) 24.86 MW (ii) 1.58 MW (iii) 93.6% 11.2 The mean diameter of the blades of an impulse turbine with a single row wheel is 105 cm and the speed is 3000 rpm. The nozzle angle is 72◦ with respect to axial direction, the ratio of blade speed to gas speed is 0.42 and the ratio of the relative velocity at outlet from the blades to that at inlet is 0.84. The outlet angle of the blade is to be made 3◦ less than the inlet angle. The mass flow rate is 8 kg/s. Calculate the following: (i) tangential thrust on blades, (ii) axial thrust on blades, (iii) power produced, and (iv) blade efficiency. Ans: (i) 3024.8 N (ii) 83.12 N (iii) 499 kW (iv) 91% 11.3 Define utilization factor. Show that the utilization factor of a stage of an impulse turbine with single row wheel can be derived as = 2 σ sin α2 − σ 2 (1 + kc) where σ =

u c2 ; k

=

w3 w2

and c =

sin β3 sin β2 .

State the assumptions made.

Calculate the maximum utilization factor for an equiangular blade for a nozzle angle of 70◦ . Take the velocity coefficient as 0.83. If the utilization factor is to be 90% of the maximum, what are the possible ratios of the blade speed to gas speed? Ans: (i) 80.8% (ii) 0.615 or 0.325

Impulse and Reaction Turbines

465

11.4 Gas with a velocity of 240 m/s relative to blades enters an impulse moving row at an angle of 60◦ with respect to axial direction. The tangential velocity of blades is 183 m/s. The work developed in blades is estimated as 75 kJ/kg of gas. Find the blade efficiency and the blade friction coefficient for relative velocities. Assume symmetrical blades. Ans: (i) 89.7% (ii) 0.97 11.5 The nozzles of a two-row velocity compounded stage have outlet angles of 68◦ with reference to the axial direction and utilize an isentropic enthalpy drop of 200 kJ/kg of gas. All moving and guide blades are symmetrical, and mean blade speed is 150 m/s. Assuming an isentropic efficiency of the nozzle of 90%, find all the blade angles and calculate the specific power output produced by the stage. The velocity at inlet to the stage can be neglected. Ans: (i) β1 = β2 = 61◦ ; β3 = 25.3◦ ; α2 = α3 = 48.7◦ (ii) 153.75 kW Multiple Choice Questions (choose the most appropriate answer) 1. An impulse turbine stage is characterized by the expansion of the gases in (a) stator nozzles (b) rotor nozzles (c) both stator nozzles and rotor nozzles (d) none of the above 2. A reaction turbine stage is characterized by the expansion of the gases in (a) stator nozzles (b) rotor nozzles (c) both stator nozzles and rotor nozzles (d) none of the above 3. In an impulse turbine stage, pressure (a) increases in stator and decreases in rotor (b) decreases in stator and increases in rotor (c) decreases in stator and remains constant in rotor (d) remains constant in both stator and rotor 4. In an impulse turbine stage, velocity (a) increases in stator and decreases in rotor (b) decreases in stator and increases in rotor

466

Gas Turbines

(c) increases in stator and remains constant (d) remains constant in both stator and rotor 5. In a reaction stage turbine, the velocity (a) (b) (c) (d)

increases in stator and remains constant in rotor increases in stator and decreases in rotor decreases in stator and remains constant in rotor remains constant throughout

6. In a reaction stage turbine, the pressure (a) (b) (c) (d)

increases in stator and remains constant increases in stator and decreases in rotor decreases in stator and remains constant in rotor decreases in stator and also in rotor

7. The loading coefficient for a reaction turbine for constant axial velocity is given by (a) ψ = φ(tan α2 + tan α3 ) (b) ψ = φ(tan α2 − tan α3 ) (c) ψ = φ(tan α2 × tan α3 ) (d) ψ = φ

tan α2 tan α3

8. Degree of reaction of a reaction turbine is given by (a) R = (b) R = (c) R = (d) R =

ca 2u (tan β3 ca 2u (tan β3 ca 2u (tan β3 ca tan β3 2u tan β2

+ tan β2 ) − tan β2 ) × tan β2 )

9. Multistage reaction turbine are employed to achieve (a) (b) (c) (d)

a large pressure drop a large mass flow rate a large volume flow rate none of the above

10. Multistage reaction turbine are employed to achieve (a) (b) (c) (d)

a high velocity of fluid a high pressure of fluid a high temperature of fluid all of the above Ans:

1. – (a) 6. – (d)

2. – (c) 7. – (a)

3. – (c) 8. – (b)

4. – (a) 9. – (a)

5. – (b) 10. – (a)

12 TRANSONIC AND SUPERSONIC COMPRESSORS AND TURBINES INTRODUCTION The pressure rise in a compressor is due to the conversion of kinetic energy to potential energy with higher static pressure at the exit. Therefore, the compressor outlet pressure depends upon the inlet velocity of the compressor. Nowadays, higher pressure ratios are achieved with reasonably good efficiency. Based on the Mach number, which is the ratio of fluid velocity to the velocity of sound, the compressor can be broadly classified into (i) subsonic compressors (ii) supersonic compressors However, the Mach number between 0.8 and 1.2 is considered as transonic. Mach number between 1.2 and 5 is considered as supersonic and above 5, it is hyposonic. The recent trend in the design of compressors is to achieve higher and higher pressure ratio per stage. The pressure rise or the specific work is proportional to the square of the inlet velocity. The question is why not increase the velocity at inlet in a subsonic compressor? It should be noted (i) (ii) (iii)

subsonic sonic supersonic

(M < 1) (M = 1) (M > 1)

468

Gas Turbines

that the pressure ratio attainable in a single stage in a subsonic compressor is limited by the rapid decrease in efficiency. Further, with increase in inlet velocity shock waves will form which will increase the losses. There is also a possibility of flow separation and the boundary layer development, both of which affect the flow. Hence, a transonic or supersonic compressor should be designed appropriately from the beginning taking into account all the above problems. 12.1

THE SUPERSONIC COMPRESSOR

As stated in the Section 12, the increased losses are largely due to the presence of shock waves and the flow separation. This is due to the interaction of the shock waves with the blade surface boundary layer. On the other hand, with supersonic velocities compression occurs through shock waves. This enables a high pressure ratio to be obtained in a machine with short axial length. Thus the supersonic compressors appear to have an advantage and may be expected to be compact machines. It may be noted that higher pressure ratios are obtained at the expense of efficiency. Figure 12.1 shows the details of subsonic, transonic and supersonic compressors.

Subsonic � (M = 0 to 0.8)

Transonic (M = 0.8 to 1.2)

Supersonic (M = 1.2 to 5.0)

Fig. 12.1 Compressor size for a given pressure ratio and mass flow rate When supersonic flow approaches a blade passage, depending on the blade geometry at inlet, a normal shock will form (Fig. 12.2). The normal shock is considered as a discontinuity. The flow properties will change over a negligibly small distance. The static pressure, temperature and entropy will increase while the velocity and stagnation pressure are reduced. Mach number is reduced from supersonic to subsonic. Higher the inlet Mach number, higher will be these changes and the shock is considered strong. The flow does not undergo any turning through a normal shock. The supersonic stage concept started around nineteen forties. The actual development of a supersonic stage was much delayed because of the non-availability of proper materials to withstand the high stresses due to higher rotational speeds. Since, the supersonic compressor operates in the high subsonic or transonic range during off-design speeds, the distinction between the transonic and the supersonic compressor is rather vague.

Transonic and Supersonic Compressors and Turbines

469

A compressor with supersonic relative velocity everywhere at inlet to rotor blade row is by convention called a supersonic compressor. If the relative velocity is subsonic in some part of the inlet then it is known as a transonic compressor. In a transonic axial flow compressor the Mach number based on the tip peripheral speed may be as high as 1.4. It is possible to set a stage pressure ratio about 1.6 to 1.8. In supersonic compressors, still higher values are possible.

M>1 N M 1

3 M1 1 1

1

2

3

2 M1 1

M 3 1

M1r>1

M 1 1.0

Fig. 12.7 CW profile

12.3

SUPERSONIC RADIAL COMPRESSORS

The axial velocity of the air at the entrance of the compressor is mostly subsonic. However, depending on the rotational speed of the impeller, the relative velocity into the inducer becomes supersonic. In most of the supersonic radial compressors used for gas turbines, the supersonic conditions have been mostly limited to the inlet of the inducer and the exit of the impeller. This could cause a surge condition. Most of the results presented in literature on supersonic radial compressors, report that the peak compressor efficiency is obtained with inlet Mach numbers in the range of 0.6 to 0.85. Further increase in Mach number reduces the efficiency drastically. There has been hardly any mention of shocks being stabilized efficiently in rotors of centrifugal compressors. However, the Mach number leaving the impeller of a 12:1 pressure ratio impeller has been found to be 1.30. At this Mach number high fluctuations and non-uniform inlet conditions are found to develop. It has been reported that a Mach number of unity is almost exceeded in the case of even 4:1 pressure ratio impellers. The traditional method to overcome such high supersonic inlet velocities in the diffuser is to provide a short vaneless diffuser at the entry. This allows sufficient diffusion to a subsonic velocity before entering any vane system. It has been observed that leaving such a vaneless diffusing space is a disadvantage. It is due to the fact that the boundary layer gets skewed relative to the main stream flow. Further, the flow which is subjected to a strong pressure gradient separates and invariably the diffuser stalls.

Transonic and Supersonic Compressors and Turbines

12.4

475

SUPERSONIC AXIAL FLOW TURBINE STAGES

Similar to the supersonic compressors, supersonic axial flow turbines have also been developed to obtain higher specific work. In the following sections the details are discussed. 12.4.1

Turbine stages with Supersonic velocities

Multistage steam turbines are usually provided with a Curtis velocity compounded turbine with large specific work. The maximum pressure drop takes place across the first nozzle blade row. As a result of this high pressure drop, the absolute velocity leaving the nozzle and often the relative velocity entering the rotor blade row will be supersonic. Gas turbines used in rocket engines to drive the liquid fuel pumps also operate under very high pressure ratios resulting in supersonic velocities. In large output steam turbines, the rotor blades are very long in the last stage and they operate with high tip velocities with relative flow Mach number around the range of 1.5 to 1.8 while leaving the rotor blades. Typical blade passages for such velocities are shown in Fig. 12.8. Such turbine stages with supersonic velocities can be classified as follows: c2

w2

Th roa t

u

Convergent divergent passage

w1

c1

M1r > 1.0

u

Fig. 12.8 Typical tip section of a steam turbine last stage

(i) Supersonic stages with supersonic absolute velocity at nozzle exit and supersonic relative velocity in the rotor blade row. For such conditions, the rotor is invariably, of the impulse type and the axial velocity would be subsonic: (ii) Transonic stages with supersonic absolute velocities at nozzle exit and supersonic relative velocities at rotor exit. For such conditions, the rotor is often designed with a high degree of reaction. The essential difference between a subsonic and supersonic turbine can be stated as follows: In subsonic turbines the flow incidence at the rotor

476

Gas Turbines

blade is mainly controlled by the upstream nozzle outlet angle and the rotor blade speed u. In supersonic turbines, the relative velocity is supersonic. Therefore, the incidence is mainly controlled by the rotor blade leading edge. Thus the rotor blade of a supersonic turbine fixes its own incidence. Thereby, the static pressure and the flow angle downstream of the preceding nozzle is controlled. 12.4.2

Effect of Turbine Blade Speed

Let us examine the supersonic flow in such a turbine (Fig. 12.9) under design conditions. Neglecting the nozzle trailing edge thickness and rotor blade leading edge thickness, the incidence on to the rotor would be zero. The flow through the nozzle and rotor would be shock free. The Mach number of the flow leaving the nozzle will be equal to the design value. The flow direction will be the same as that of the nozzle angle with almost zero deviation, Fig. 12.9(a).

c2

w2

u

(a) u - design

Rotor blade

c2 u

c2

w2 u (b) u - increased

w2 (c) u - decreased

Fig. 12.9 Effect of turbine blade speed on the velocity triangle If the blade speed u is now increased the velocity triangle will now change as in Fig. 12.9(b). However, the relative flow direction will be the same as the rotor blade direction at inlet resulting in zero incidence. This is obtained by deflecting the absolute flow leaving the nozzle through an increased angle or decreased α2 . This is caused by the creation of shock and expansion waves at the trailing end of the nozzle blades. The flow from the nozzle first passes through an oblique trailing edge shock wave. Then there is a reflection of the shock wave from the adjacent blade and again through an expansion fan which comes off the trailing edge of the adjacent blade. The mass flow will remain the same as per the design, because, the inlet conditions remain unchanged. The pressure and density downstream

Transonic and Supersonic Compressors and Turbines

477

of the nozzle will be higher than that of design conditions. Therefore, the axial velocity will reduce. The exit Mach number M2 will be less than the design M2 . Similarly, if the blade speed is reduced it would result in flow condition shown in Fig. 12.9(c). The pressure at the exit of the nozzle will be less than its design value and M2 would be greater than the design M2 . The nozzle will operate at a higher pressure ratio (P3 /P2 ) than the design condition. Such conditions of operation are termed ’supercritical’ conditions. The above explanation is based on the assumptions that (i) the trailing edge of the nozzle blades are very thin, and (ii) the rotor blade leading edges are also very thin. Definite wedge angle of the nozzle trailing edge even with sharp edge will give rise to two oblique shock waves at the trailing edge. Definite thickness of the trailing edge will produce more disturbance at the trailing edge. If the rotor blades are not very thin then at the leading edge, the operating incidence is not zero, but, is slightly positive measured relative to the flat region of the suction leading edge. This incidence is a function of the inlet Mach number and the blade leading edge geometry and is called the unique incidence. Shocks originate from the leading edge and expansion waves from the chamfer end. The entering relative flow passes through a wave pattern made up of a shock and expansion waves as shown in Fig. 12.10. As the relative flow

M2oo β 2 oo

M 2r oo S

M2

β =β 2b

β2b

β2

tu

Chamfer wedge angle

2

(S-tu) sin (180 - β2b) 2 inlet to blade row 2 oo head of waves or eixt of nozzle Shock Expansion wave Fig. 12.10 Supersonic impulse rotor incidence

478

Gas Turbines

reaches the wedge end of the blade chamfer a shock waves is produced. This turns the flow parallel to the tangential direction, i.e., parallel to the blade chamfered surface. As the flow reaches the chamfered end, it has, to be turned parallel to the blade suction surface inlet region. This is done by the expansion waves emanating from the chamfer end. If the deflection through the waves is small then the flow can be considered to be isentropic and then it is a problem of simple area change. Review Questions 12.1 Based on Mach number how are compressors classified? 12.2 Explain subsonic and supersonic compressors with figures. 12.3 With sketches explain the different types of supersonic compressors. 12.4 Explain the blade profile requirement of supersonic compressors. 12.5 With a diagram explain the turbine stages with supersonic velocities. 12.6 With a neat sketch explain the effect of turbine blade speed on velocity triangle. Multiple Choice Questions (choose the most appropriate answer) 1. The Mach number between 0.8 and 1.2 is called (a) subsonic (b) supersonic (c) hypersonic (d) transonic 2. In supersonic compressors, high pressure ratio is obtained due to shock which makes the machine size to be (a) long (b) medium (c) short (d) none of the above 3. In supersonic compressor if the relative velocity is subsonic at some part then it is known as (a) subsonic compressor (b) transonic compressor (c) supersonic compressor (d) hypersonic compressor

Transonic and Supersonic Compressors and Turbines

479

4. In a supersonic compressor the pressure ratio that can be obtained per stage is (a) 1.7 (b) 1.5 (c) 1.4 (d) 1.2 5. A supersonic compressor with shock in rotor and shock in stationary blade is called (a) normal shock compressor (b) oblique shock compressor (c) distributed shock compressor (d) none of the above 6. For high subsonic inlet Mach numbers (M = 0.9) (a) circular arc blades are preferred (b) double circular arc blades are preferred (c) straight blades are preferred (d) slightly cambered blades are preferred 7. The peak efficiency of a supersonic compressor is obtained with inlet Mach number in the range of (a) M = 0.6 to 0.8 (b) M = 1.0 to 1.2 (c) M = 1.4 to 1.6 (d) M = 1.8 to 2.0 8. Supersonic stages in turbines with supersonic absolute velocity at nozzle exit and supersonic relative velocity at the rotor blade row must be of (a) reaction type (b) impulse type (c) mixed type (d) none of the above 9. Transonic stages in turbines with supersonic absolute velocities at nozzle exit and supersonic relative velocities at rotor exit must be of (a) reaction type (b) impulse type

480

Gas Turbines

(c) mixed type (d) none of the above 10. If the turbine nozzle operates at a higher pressure ratio than the design value then it is called (a) super critical condition (b) critical condition (c) subcritical condition (d) abnormal condition 11. By increasing the axial component of Mach number, Max , through the stage, the frontal area (a) decreases (b) increases (c) remains the same (d) does not depend on Mach number 12. Peak efficiency of a supersonic compressor is obtained with inlet Mach number in the range of (a) 0.4 to 0.6 (b) 0.6 to 0.8 (c) 0.8 to 1.0 (d) 1.0 to 1.2 Ans:

1. – (a) 6. – (b) 11. – (a)

2. – (c) 7. – (a) 12. – (b)

3. – (b) 8. – (b)

4. – (a) 9. – (a)

5. – (c) 10. – (a)

13 INLETS AND NOZZLES INTRODUCTION Compressors, combustion systems and turbines were discussed in detail in the last four chapters. There are two more additional components, viz., inlets and nozzles to be considered especially in the case of power plants for jet propulsion. The inlet and exhaust nozzle are the two engine components that directly interfere with the internal air flow and the flow about the aircraft. In fact, integration of the engine and the airframe is one of the most complex problems and has a major impact on the performance especially in aircraft system. The purpose of any aircraft gas turbine engine inlet is to provide a sufficient air supply to the compressor with as minimum a pressure loss as possible. This is to be achieved with as small a drag force on the airplane as possible. It also has the purpose of being a diffuser that reduces the velocity of the entering air as efficiently as possible. The purpose of the exhaust nozzle is to increase the velocity of the exhaust gas before discharge from the nozzle. Further, it should collect and straighten the gas flow. For large values of specific thrust, the kinetic energy of the exhaust gas must be high. This requires a high exhaust velocity. The pressure ratio across the nozzle controls the expansion process. The maximum thrust for a given engine is obtained when the exit pressure pe equals the ambient pressure p0 . 13.1

INLETS

The inlet interchanges the organized kinetic and random thermal energies of the gas in an essentially adiabatic process. However, the perfect (no-loss) inlet would correspond to an isentropic process. The primary purpose of the inlet is to bring the air required by the engine from free-stream conditions to the conditions required at the entrance of the compressor with minimum total pressure loss. The best work output from a compressor is obtained when the flow of air is at a Mach number of about 0.5. It is a known fact that the engine performance heavily depends on the installation losses in

482

Gas Turbines

the inlet such as additive drag, forebody or cowl drag, bypass air, boundarylayer bleed air, etc., Hence, the design of the inlet should be done with great care so as to minimize these losses. The performance of an inlet is related to the following characteristics: (i) high total pressure ratio, (ii) controllable flow matching requirements, (iii) good uniformity of flow, (iv) low installation drag, (v) good starting and stability, (vi) low signatures (acoustic, radar, etc.,), and minimum weight, and (vii) low cost while meeting life and reliability goals. An inlet’s overall performance must be determined by simultaneously evaluating all these characteristics. It may be noted that improvement in one, is often achieved at the expense of another. The design and operation of subsonic and supersonic inlets differ considerably due to the characteristics of the flow. For the subsonic inlets, near-isentropic internal diffusion can be easily achieved, and the inlet flow rate adjusts to the demand. For a supersonic inlet, the internal aerodynamic performance is a major design problem. It should be appreciated that achieving efficient and stable supersonic diffusion over a wide range of Mach numbers is very difficult. In addition, the supersonic inlet must be able to capture its required mass flow rate. This requires variable geometry to minimize inlet loss and drag and provide stable operation. Because of these differences, in the following sections we consider the subsonic and supersonic inlets separately, beginning with the subsonic inlet and diffuser. 13.2

SUBSONIC INLETS

Most subsonic aircraft have their engines placed in nacelles. Hence, in this section we may not deal with the inlet alone but include the nacelle at subsonic Mach numbers. The cross section of a typical subsonic inlet and its geometric parameters are shown in Fig.13.1. The inlet area A1 , is based on the flow cross section at the inlet height. It may be noted that the subsonic inlet can draw in airflow whose free-stream area A0 is larger than the inlet area A1 . Variable inlet geometry is not required (except sometimes blow-in doors are used to reduce installation drag during takeoff). The details in this section on subsonic inlets are based on a fixed-geometry inlet. The operating conditions of an inlet depend on the flight velocity and mass flow requirements. Figure 13.2 shows the streamline pattern for three typical subsonic conditions. Figure 13.2(a) shows acceleration of the fluid external to the inlet which will occur when the inlet operates at a velocity lower than the design value or at a mass flow higher than the design value.

Inlets and Nozzles

483

Highlight (hl)

d fan d max

A l d hl d t

2

Al=

π d hl 4

L

Fig. 13.1 Subsonic inlet nomenclature Figure 13.2(b) show the configuration for correct mass flow for the design conditions. Figure 13.2(c) shows deceleration of the fluid external to the inlet which will occur at a velocity higher than design or a mass flow lower than design.

(a) Static operation (b) Low-speed operation (c) High-speed operation (or insufficient air (correct engine air) (more air than the for engine) engine needs) Fig. 13.2 Typical streamline patterns for subsonic inlet A list of the major design variables for the inlet and nacelle includes the following: (i) Inlet total pressure ratio and drag during cruise (ii) Engine location on wing or fuselage (iii) Aircraft attitude envelope (iv) Inlet total pressure ratio and distortion levels required for engine operation (v) Engine-out windmilling airflow and drag (vi) Integration of diffuser and fan flow path contour

484

Gas Turbines

(vii) Integration of external nacelle contour with thrust reverser and accessories (viii) Flow field interaction between nacelle, pylon, and wing (ix) Noise suppression requirements 13.3

DIFFUSER

The flow within the inlet is required to undergo diffusion in a divergent duct. This reduction in flow velocity creates an increase in static pressure that interacts with the boundary layer. If the pressure rise due to diffusion occurs more rapidly than turbulent mixing can re-energize the boundary layer, the boundary layer will assume the configurations shown in Fig.13.3. If the flow in an inlet separates, the total and static pressures are reduced from their corresponding nonseparated flow values.

(a) Turbulent(b) Intermediate (c) Separation (d) Reversed flow Fig. 13.3 Boundary layer development under adverse pressure gradient As shown in Fig.13.4, the rate of area increase in a diffuser has a direct effect on the behavior of flow in the diffuser. If the rate of area increase is greater than that needed to keep the boundary layer energized and attached, the flow may be characterized by unsteady zones of stall. The turbulent mixing is no longer able to overcome the pressure forces at all points in the flow, and local separation occurs at some points. The total pressure decreases markedly due to the irreversible mixing of a fairly large portion of low-velocity fluid with the main flow. If the diffuser walls diverge rapidly, the flow will separate completely and behave much as a jet, as shown in Fig.13.4(d). The rate of area increase without stall for a diffuser depends on the characteristics of the flow at the entrance and on the length of the divergent section. Figure 13.5 shows the results for two-dimensional straight-walled diffusers. The results are for incompressible flow, and they do not give a qualitatively valid indication of the sensitivity of any diffuser to rapid divergence of flow area. For the design of an optimum diffuser, research has shown that the boundary layer profile should maintain a constant shape. The boundary layer thickness will, of course, increase as the flow moves down the diffuser. The stipulation of a constant shape for the boundary layer profile implies

Inlets and Nozzles

(a) Well-behaved flow

485

(b) Large transitory stall

(c) Steady stall

(d) Jet flow

Fig. 13.4 Types of flow in straight-walled diffusers

40

2θ

Flow separation

H

20 2θ

L 10 0

Without flow separation 1

2

3

4

6

8

20

L/H

Fig. 13.5 Flow separation limits in two-dimensional straight-walled diffusers

the assumption that mixing re-energizes the profile at the same rate as the static pressure depletes it. In the presence of an adverse pressure gradient (static pressure increasing in the direction of flow), boundary layers tend to separate when the boundary layer is not re-energized rapidly enough by turbulent mixing. If vortices are generated by vortex generators in pairs, regions of inflow and outflow exist. These carry high-energy air into the boundary layer and low-energy air out. Figure 13.6 shows how vortex generators re-energize a boundary layer. By using vortex generators together with a short, wide-angle diffuser, it may be possible to have a lower total pressure loss than with a long diffuser without vortex generators. Here, the reduced skin friction losses associated with flow separation are traded against vortex losses. The use of shorter diffusers may reduce weight and facilitate engine installation.

486

Gas Turbines

After reenergizing Trailing vortex

Before Vortex generator

Fig. 13.6 The role of vortex generator 13.4

SUPERSONIC INLETS

It is much more difficult to design supersonic inlets than the subsonic inlets. The inlet used on a supersonic aircraft is the design that optimizes performance for the mission for which the aircraft is designed. The supersonic inlet must have the following characteristics: (i) provide adequate subsonic performance, (ii) good pressure distribution at the compressor inlet, (iii) high pressure recovery ratios, and (iv) must be able to operate efficiently at all ambient pressures and temperatures during take-off, subsonic flight, as well as its supersonic design condition. Both axisymmetric and two-dimensional inlets have been designed and used. Variable geometry center bodies, which change the inlet geometry, for better off-design operation, and boundary-layer bleed have been incorporated into supersonic inlets. The inlet performance characteristics that are used to assess the performance of supersonic inlets and those with the largest influence on aircraft performance and range are (i) total pressure recovery, (ii) cowl drag, (iii) boundary layer bleed flow, (iv) capture-area ratio (mass flow ratio), and (v) weight.

Inlets and Nozzles

487

Supersonic inlets are usually classified by their percent of internal compression. Internal compression refers to the amount of supersonic area change that takes place between the cowl lip and the throat. This is illustrated in Figure 13.7 which identifies the center body, cowl lip, capture area and throat for a supersonic inlet. External compression

Internal compression

Cowl

Cowl lip Throat area

ck

Capture area

liq

ho es

u

Ob

Centre body

488

Gas Turbines

Inlets and Nozzles

Nozzle entrance

489

Nozzle throat

Fig. 13.10 Convergent exhaust nozzle 13.5.2

Convergent-Divergent (C-D) Nozzle

The convergent-divergent nozzle has two sections (i) convergent duct and (ii) divergent duct. The section of the nozzle where the cross-sectional area is minimum is called the throat of the nozzle. Most convergent-divergent nozzles used in aircraft are not simple ducts. They incorporate variable geometry and other aerodynamic features. The convergent-divergent nozzle is used if the nozzle pressure ratio pte /p0 is high (greater than about 6). High-performance engines in supersonic aircraft generally have some form of a convergent-divergent nozzle. If the engine incorporates an afterburner, the nozzle throat is made in such a way that the operating conditions of the engine upstream of the afterburner remain unchanged (in other words, vary the exit nozzle area so that the engine performance does not change when the afterburner is in operation. Further, the exit area must be varied to match the flow conditions in order to produce the maximum available thrust. Earlier supersonic aircraft used ejector nozzles in their high performance turbojets. Use of such a nozzle permitted the following: (i) bypassing varying amounts of inlet air around the engine, (ii) providing engine cooling, (iii) good inlet recovery, and (iv) reduced boat-tail drag. Ejector nozzles can also receive air from outside the nacelle directly into the nozzle for better overall nozzle matching – these are called two-stage ejector nozzles. For the modern high performance afterburning turbofan engines, simple convergent-divergent nozzles are used without secondary air. 13.5.3

Nozzle Functions

An exhaust nozzle may be thought of as a device dividing the power available from the main burner exit gas between the requirements of the turbine and the jet power. Thus the nozzle serves as a back-pressure control for the engine and an acceleration device converting gas thermal energy to kinetic

490

Gas Turbines

energy. A secondary function of the nozzle is to provide required thrust reversing and/or thrust vectoring. 13.5.4

Engine Backpressure Control

The throat area of the nozzle is one of the main means available to control the thrust and fuel consumption characteristics of an existing engine. In preliminary engine cycle analysis, the throat area of the nozzle is fixed by selection of specific values for the engine design parameters and the design mass flow rate. Assumption of constant areas can establish the off-design operating characteristics of the engine and the resulting operating lines for each major component. Changing the nozzle throat area from its original design value will change the engine design and the operating characteristics of the engine at both on- and off-design conditions. At times, it is necessary to change the off-design operation of an engine in only a few operating regions, and variation of the throat area of the exhaust nozzle may provide the needed change. At reduced engine corrected mass flow rates (normally corresponding to reduced engine throttle settings), the operating line of a multistage compressor moves closer to the stall or surge line. This has already been discussed in Sections 8.11 and 9.13. Steady-state operation close to the stall or surge line is not desirable since transient operation may cause the compressor to stall or surge. The operating line can be moved away from the stall or surge line by increasing the exhaust nozzle throat area. This increase in nozzle throat area reduces the engine backpressure and increases the corrected mass flow rate through the compressor. Large changes in the exhaust nozzle throat area are required for afterburning engines to compensate for the large changes in total temperature leaving the afterburner. The variable-area nozzle required for an afterburning engine can also be used for back pressure control at its nonafterburning settings. One advantage of the variable-area exhaust nozzle is that it improves the starting of the engine. Opening the nozzle throat area to its maximum value reduces the backpressure on the turbine and increases its expansion ratio. Thus the necessary turbine power for starting operation may be produced at a lower turbine inlet temperature. Also, since the backpressure on the gas generator is reduced, the compressor may be started at a lower engine speed, which reduces the required size of the engine starter. 13.5.5

Exhaust Nozzle Area Ratio

Maximum engine thrust is realized for ideal flow when the exhaust nozzle flow is expanded to ambient pressure (pe = p0 ). When the nozzle pressure ratio is above choking, supersonic expansion occurs between aft-facing surfaces. A small amount of under-expansion is less harmful to aircraft and engine performance than overexpansion. Overexpansion can produce regions of separated flow in the nozzle and on the aft end of the aircraft, reducing aircraft performance.

Inlets and Nozzles

491

The exhaust nozzle pressure ratio (pte = p0 ) is a strong function of flight Mach number. Whereas convergent nozzles are usually used on subsonic aircraft, convergent-divergent nozzles are usually used for supersonic aircraft. When afterburning engine operation is required, complex variablegeometry nozzles must be used. Most of the nozzles shown in Fig.13.11 are convergent-divergent nozzles with variable throat and exit areas. The throat area of the nozzle is controlled to satisfy engine backpressure requirements, and the exit area is scheduled with the throat area.

13.5.6

Thrust Reversing and Thrust Vectoring

The need for thrust reversing and thrust vectoring is normally determined by the required aircraft and engine system performance. Thrust reversers are used on commercial transports to supplement the brakes. In-flight thrust reversal has been shown to enhance combat effectiveness of fighter aircraft. Two basic types of thrust reversers are used: the cascade-blocker type and the clamshell type (Fig.13.12). In the cascade-blocker type, the primary nozzle exit is blocked off, and cascades are opened in the upstream portion of the nozzle duct to reverse the flow. In the clamshell type, the exhaust jet is split and reversed by the clamshell. Since both types usually provide a change in effective throat area during deployment or when deployed, most reversers are designed such that the effective nozzle throat area increases during the brief transitory period, thus preventing compressor stall. The exhaust system of the Concorde, the supersonic passenger aircraft, has two nozzles, a primary nozzle and a secondary nozzle. The secondary nozzle is positioned as a convergent nozzle for takeoff and as a divergent nozzle for supersonic cruise. Attempts to develop thrust vectoring nozzles for combat aircraft has increased in the last decade. Vectoring nozzles have been used on vertical takeoff and landing (VTOL) aircraft, and are proposed for future fighters to improve maneuvering and augment lift in combat. Thrust vectoring at augmented power settings is being developed for use in future fighters. However, cooling of the nozzle walls in contact with the hot turning or stagnating flows is very difficult and will require increased amounts of nozzle-cooling airflow.

13.5.7

Nozzle Coefficients

Nozzle performance is ordinarily evaluated by two dimensionless coefficients; the gross thrust coefficient and the discharge or flow coefficient. Figure 13.13 shows a convergent-divergent exhaust nozzle with the geometric parameters used in the following definitions of nozzle coefficients. Only total losses downstream of station 8 (as marked in the figure) are included in the gross thrust coefficients.

492

Gas Turbines

Short convergent

Con-di iris

Iris

Simple ejector

Fully variable ejector

Blown-in-door ejector

Plug

Isentropic ramp

Fig. 13.11 Typical nozzle concepts for afterburning engines

Inlets and Nozzles

Primary propulsion nozzle

493

Deployed clamshell

* *

Blocker Cascade reverse Clamshell reverser

Fig. 13.12 Thrust reversers 13.5.8

Gross Thrust Coefficient

The gross thrust coefficient Cgtc is the ratio of actual gross thrust Factual to the ideal gross thrust Fideal or Cgtc

≈

Factual Fideal

(13.1)

Empirically derived coefficients are applied to Eq.13.1 to account for the losses and directionality of the actual nozzle flow. Each engine organization uses somewhat different coefficients, but each of the following basic losses are accounted for: (i) Thrust loss due to exhaust velocity vector angularity (ii) Thrust loss due to the reduction in velocity magnitude caused by friction in the boundary layers (iii) Thrust loss due to loss of mass flow between stations 7 and 9 from leakage through the nozzle walls (iv) Thrust loss due to flow nonuniformities 13.5.9

Discharge or Flow Coefficient

The ratio of the actual mass flow m ˙ 8 to the ideal mass flow m ˙ 8i is called the discharge coefficient CD (refer Fig.13.13): CD

≈

m ˙8 m ˙ 8i

(13.2)

This coefficient can be shown to be equal to the ratio of the effective one-dimensional flow area required to pass the total actual nozzle flow A8e , to nozzle physical throat area A8 as follows: CD

=

m ˙8 m ˙ 8i

=

ρ8 c8 A8e ρ8 c8 A8

=

A8e A8

(13.3)

494

Gas Turbines

θ

α A9 Ls

A8 Station

7

8

9

A8

Primary nozzle throat area

A9

Secondary nozzle exit area

α

Secondary nozzle half angle

θ

Primary nozzle half angle

Ls

Secondary nozzle length

Fig. 13.13 Nozzle geometric parameters The variation of the discharge coefficient with nozzle pressure ratio is shown in Fig.13.14(a) for a conical convergent nozzle. When the nozzle is choked, the discharge coefficient reaches a maximum value CDmax . The value of CDmax is a function of the primary nozzle half-angle θ, as shown in Fig.13.14(b). Figure 13.14(c) shows the variation in discharge coefficient for a convergent-divergent nozzle with nozzle pressure ratio. Note the change in behavior of CD between that of the convergent-divergent nozzle and that of the convergent nozzle as the nozzle pressure ratio drops below choking. This is due to the venturi behavior of the convergent-divergent nozzle. 13.5.10

Velocity Coefficient

The velocity coefficient Cv is the ratio of the actual exit velocity c9 to the ideal exit velocity c9i corresponding to pt9 = pt8 , or Cv

≈

c9 c9i

(13.4)

and represents the effect of frictional loss in the boundary layer of the nozzle. It is mainly a function of the nozzle ratio A8 /A9 and the half-angle α, as shown in Fig.13.15. 13.5.11

Angularity Coefficient

The angularity coefficient CA represents the axial friction of the nozzle momentum; thus it is proportional to the thrust loss due to the nonaxial

Inlets and Nozzles

1.00 C D max

0.98

CD

0.96 0.94 pt8 /p 0 (a) Convergent nozzle

0.92

0

10 20 30 40 (b) CD max vs θ

C D max

CD

1

2

pt8 /p

3

4

0

(c) C-D nozzle Fig. 13.14 Nozzle discharge coefficient

α 2 4 6 8

1

10 CA

12 14

0.97 1

A 9 /A 8 Fig. 13.15 C-D nozzle velocity coefficient

2

495

496

Gas Turbines

V9

α αj

Fig. 13.16 Local angularity coefficient exit of the exhaust gas (see Fig.13.16). For a differential element of flow, this coefficient is the cosine of the local exit flow angle αj . The local flow angle αj varies from zero at the centerline to α at the outer wall; thus, the nozzle angularity coefficient is the integral of αj . across the nozzle. 1 CA ≈ cos αj dm ˙ (13.5) m ˙ Figure 13.17 presents the correlation of the angularity coefficient with the area ratio A8 /A9 and half-angle α. The details are based on analytical evaluations of the inviscid flow field in convergent-divergent nozzles for practical nozzle geometries. o

α =o 15 12 o 9 CV

6

o

o

4 o

1

o

2

1

o

3

A 9 /A 8

4

Fig. 13.17 C-D nozzle angularity coefficient

13.5.12

Nozzle Performance

Many nozzle coefficients simplify to algebraic expressions or become unity for the special case of one-dimensional adiabatic flow. This is a useful limit

Inlets and Nozzles

497

for understanding each coefficient and for preliminary analysis of nozzle performance using engine cycle performance data. The gross thrust coefficient for one-dimensional flow of a calorically perfect gas can be shown to be

Cgtc

. / / /1 − = CD Cv / 0 1−

p9i pt8 p0 pt8

(γ−1)/γ

⎡

γ−1 ⎢ 1+ 2γ

(γ−1)/γ ⎣

pt9 p9

1 − pp09 (γ−1)/γ

−1

⎤ ⎥ ⎦

(13.6)

The extent of overexpansion in nozzles is limited by flow separation resulting from the interaction of the nozzle boundary layer and the strong oblique shock waves at the exit of the nozzle. In extreme overexpansion, it is observed that the oblique shock waves moved from the exit lip into the nozzle, the flow downstream of the shock waves was separated in the vicinity of the wall, and as a result, the wall static pressure downstream of the shock waves was nearly equal to the ambient pressure, p0 . A simple estimate for the ratio of the pressure just preceding the shock waves, ps , to the ambient pressure, p0 , is given by ps p0

≈

0.37

(13.7)

This flow separation limit can be included in the one-dimensional gross thrust coefficient of Eq.13.6 for isentropic flow by considering the effective exit pressure (p9 = p9i ) to be the pressure just preceding the shock wave (ps ). The design area ratio A9 /A8 of convergent-divergent nozzles is selected such that the nozzle flow does not separate due to overexpansion for most throttle settings. This is because the increase in gross thrust coefficient associated with flow separation does not normally offset the accompanying increase in installation loss. Nozzle pressure ratios are 3 to 5 in the subsonic cruise speed range of turbofan and turbojet engines. Typically, a subsonic engine uses a convergent exhaust nozzle. This is because, in the nozzle pressure range of 3 to 5, the convergent nozzle gross thrust (interception of lines with vertical axis, A9 /A8 = 1) is 1 to 3 percent below the peak gross thrust (p9 = p0 ). Consequently, there may be insufficient gross thrust increase available in going to a convergent-divergent nozzle on a subsonic cruise turbofan or turbojet engine to pay for the added drag and weight of such a nozzle. In some applications, this loss in gross thrust coefficient of a convergent nozzle is too much, and a C-D nozzle is used. The design pressure ratio across the nozzle increases rapidly with supersonic flight Mach number. At Mach 2, a pressure ratio of about 12 is typical. At this pressure ratio, the convergent nozzle gross thrust penalty will be of the order of 10 percent. Substitution of convergent-divergent nozzles for convergent nozzles provides large thrust gains for supersonic aircraft.

498

Gas Turbines

Review Questions 13.1 What is the purpose of an aircraft gas turbine inlet and nozzle? 13.2 Briefly explain inlets. 13.3 What is the purpose of a diffuser in the inlet? Explain. 13.4 Briefly explain the supersonic inlets. 13.5 What are the various functions of an exhaust nozzle? 13.6 What are the two types of nozzles used in an aircraft engine. Briefly explain them. 13.7 Summarize the functions of an exhaust nozzle.

13.8 Write short notes on (i) engine back pressure control, (ii) exhaust nozzle area ratio, and (iii) thrust reversing. 13.9 Explain the following performance coefficients: (i) nozzle coefficient, (ii) cross thrust coefficient, (iii) flow coefficient, (iv) velocity coefficient, and (v) angularity coefficient. 13.10 Briefly describe the nozzle performance. Multiple Choice Questions (choose the most appropriate answer) 1. The purpose of any aircraft gas turbine inlet is to provide (a) sufficient air supply to the compressor (b) to have minimum drag force (c) reduction in velocity (d) all of the above 2. The best work output from a compressor is obtained when air flows at a Mach number of about (a) 0.1 (b) 0.3

Inlets and Nozzles

499

(c) 0.5 (d) 0.6 3. Because of the diffusion in the inlet casing (a) air velocity decreases and pressure increases (b) air velocity and pressure decreases (c) air velocity and pressure increases (d) air velocity increases and pressure decreases 4. As the flow takes place through an inlet diffuser, the boundary layer thickness (a) remains constant (b) increases (c) decreases (d) initially increases and then remains constant 5. A supersonic inlet must have (a) adequate subsonic performance (b) good pressure distribution at the compressor inlet (c) must be able to operate at all ambient conditions (d) all of the above 6. The function of an exhaust nozzle (a) decelerate the flow (b) allow for cooling of walls (c) increase jet noise (d) none of the above 7. Supersonic aircrafts are provided with (a) convergent nozzle (b) divergent nozzle (c) convergent divergent nozzle (d) none of the above 8. The exhaust nozzle pressure ratio is a strong function of (a) Reynolds number (b) Mach number (c) Prandtl number (d) Schmidt number

500

Gas Turbines

9. Subsonic aircrafts are provided with (a) convergent nozzle (b) divergent nozzle (c) convergent divergent nozzle (d) none of the above 10. Performance of a exhaust nozzle is a function of (a) gross thrust coefficient (b) flow coefficient (c) velocity and angularity coefficient (d) all of the above Ans:

1. – (d) 6. – (b)

2. – (c) 7. – (c)

3. – (a) 8. – (b)

4. – (b) 9. – (a)

5. – (d) 10. – (d)

14 BLADES INTRODUCTION Turbine blades are one of the most important components in a gas turbine power plant. These are components across which flow of high pressure gases takes place to produce work. A blade can be defined as the medium of transfer of energy from the gases to the turbine rotor. The blade as an entity, is subjected to a large number of forces, some are inevitable and some are caused by the rotation of blade. The blade is subjected to forces in the three directions, viz., (i) the rotor driving force along the radial direction, (ii) the axial forces caused by the gas flow, and (iii) the forces acting normal to the turbine shaft due to the centrifugal forces. Further, the blade is subjected to differential thermal stresses, erosioncorrosion and a host of other hostile parameters hampering its smooth functioning. In this chapter we will briefly discuss the details of blade materials, manufacturing, fixing and cooling. 14.1

BLADE MATERIALS

Researchers around the world are trying to develop new materials which have high strength and stability at high temperatures to meet the demands of the turbine designers. Different alloy compositions have been developed which have a good stability to withstand the thermal stresses. These alloys have been modified to give good erosion- corrosion characteristics to the blades. The method of manufacture of the blades plays an important role in determining the strength, uniformity of micro-structure and properties of the blades. Among the material that have been found to be suitable for use in blades are steels, titanium alloys and nickel base alloys. All the three types

502

Gas Turbines

of alloys which are mainly used, have varying proportions of chromium and aluminium to improve the strength and corrosion at high temperatures. Titanium alloys are preferred to steel because of its lower density (nearly 50%). Titanium has superior oxidation resistance and like other materials it also has a decreasing strength with temperature. However, by alloying with other elements its strength can be stabilized. Latest types of titanium alloys have good properties upto about 550◦C (IMI 829). Titanium also has a lower coefficient of thermal expansion which helps to reduce thermal stresses. The main disadvantage with titanium alloys has been the high reactivity at very high temperatures. Titanium dissolves oxygen at high temperatures. It even reduces the refractory lining. Hence, usually titanium is worked upon with vacuum or with an inert gas atmosphere (argon). The above mentioned alloy IMI 829 has composition Ti–5.5, Al–3.5, Sn–3, Zr–1, Nb–0.3 and Mo–0.3. Nickel alloys have also been developed extensively and are currently being used for turbines. These alloys have superior strength and oxidation resistance even though nickel by itself has poor oxidation resistance. This weakness is overcome by alloying with chromium. In alloys, chromium is generally 15–30% and forms chromium oxide (Cr2 O3 ) a protective layer and chromium carbide. Other elements generally added are aluminium, titanium and niobium to improve the strength at high temperatures. Current alloys also use cobalt, hafnium, boron, zirconia, molybdenum etc., Nimonic 118 the most advanced alloy developed has working capacity approximately upto 925◦ C and has composition 9.0% (Ti + Al + Nb), 3.5% Mo, 14.8% Cr, 14.9% cobalt. For working with nickel alloys it is necessary to use vacuum melting techniques to encourage the use of hardened metals and reduce trace elements which affect the ductility. Materials working under high stress and high temperature have a particular rate of creep. It means elongation under stress is not a fixed quantity as it is in case of normal temperature. Elongation continues to increase with time and blades gradually reduce the original gap provided at their tips. Therefore contacts with casing results in failure. The repeated heating and cooling of the material affect the physical properties of the materials. 14.1.1

Factors to be Considered in the Selection of Materials

Since, the turbine blades are the most conditioned members of a gas turbine, they must withstand (i) high operating temperatures; Material should possess good strength at high temperatures (ii) centrifugal tensile stresses due to rotational speeds as high as 30000 rpm (iii) bending stresses due to equivalent impulse load of the gases acting on the cantilever blade at a certain distance from fixing, also, it must have high creep strength.

Blades

503

(iv) hot erosive and corrosive effects due to the high temperature combustion products, for example, CO2 and CO with water vapour (v) varying temperature and should have structural stability when exposed to varying temperatures Further, it should posses the following characteristics: (i) castability or forgeability, (ii) weldability, (iii) machinability, and (iv) no embrittlement during operation. Turbine blade material must therefore be selected from the consideration of the working conditions, taking into account the above factors. 14.2

MANUFACTURING TECHNIQUES

The above discussion brings out the fact that there are many restrictions and requirements which will affect the manufacturing process. The various techniques of manufacturing blades are (i) forging, (ii) casting, (iii) fabrications, and (iv) powder metallurgy. In the following sections we will briefly discuss the various manufacturing techniques employed. 14.2.1

Forging

Forging is essentially a process of impacting a piece of metal to give it a certain shape. In a gas turbine forging was used in the beginning for the manufacture of both compressor and turbine blades. But, presently it is mostly used for compressor and low pressure stages of the turbine blades. This is because, as the need arises for alloys of greater strength becomes difficult to forge blades. Also, the required section of holes in hollow blades could not be achieved by this method. Blades being manufactured by this technique can be categorized as (i) precision forging, and (ii) oversized forging.

504

Gas Turbines

In the first method, direct blade profile is obtained during the process of forging. In the second method, oversized blades are manufactured and then machined to obtain the final shape. The amount of machining is generally 1 to 1.5 mm Either type of forging can be made by any one of the four processes (i) extrusion, (ii) coining, (iii) upsetting, and (iv) final forging. The bar stock which is initially circular, is suitably modified to ensure that it fits into the forging die. The bar stock is usually called use. In the upsetting operation the part which will later play the role of shroud is developed and its shape is also achieved. There are essentially two types of upsetters: (i) hydraulic, and (ii) electric. The section obtained is usually not tapered and hence this process is usually split up into two processes to achieve the taper. Before the upsetting process the use is coined. The head of the blade is produced to the required shape. After the use undergoes the above operations it is checked and corrected for various defects such as removal of flash around the heads, removal of scale and forging lubricants by blasting or chemical means. Controlled etching is also done to remove the surface layers of metals contaminated during the heating operations and application of forging lubricant etc., In the final forging operation the blade shape is developed with a controlled application of force and with minimum flash. To remove the flash a certain amount of dimensional accuracy is attained through preforming. The removal of the flash becomes necessary to avoid wear due to dies. The various materials used for forging compressor blades are the following: (i) Aluminium alloy such as RR59 is used in the Dart and Avon engines. Aluminium is forged at 450◦C although no coating is as such necessary to prevent oxidation but graphite is often used. The blades are then solution treated at about 510◦ C and quenched to avoid precipitation during cooling. Because the blades are distorted, they are corrected for profile and then aged at 150 – 200◦ C. The blades are then anodized to produce a corrosion resistant surface. (ii) Steel blades are produced by coating with a thin layer of nickel and forging at 1100◦C. The nickel is necessary to prevent decarburization by the reducing atmosphere and to act as a lubricant. The nickel

Blades

505

layer is stripped off after forging. The blade is then heated to about 1050◦C to dissolve the hardening metals. The distortions are removed by cold coining operations. In recent times the final forge heating is also done as a heat treatment of the blade. Then the blade is rapidly cooled and tempered to give the required hardness. The blades are then checked and polished by blasting or barrelling. Titanium blades are manufactured by first coating with a thin layer of glass or by electroplating static coating of glass. This is done to prevent sticking of the blades to the dies. During forging operations it is found that oxygen is often dissolved. This is removed by etching with hydrofluoric acid mixed with either nitric acid or ferric sulphate. Then distortion of the blades is corrected between hot dies. Hydrogen absorption is another problem and very careful monitoring is required in the entire manufacturing process and excess hydrogen is removed by heating in vacuum. Some commonly used alloys are IMI 679, IMI 550, Ti 811 and Ti 6242. Blades with nickel based alloys are extensively used in the last stages of compressor and turbine blades. As the temperature requirement increases, the strength requirement of the alloys also goes on increasing. New manufacturing techniques are to be developed. The introduction of vacuum melting helps to reduce the trace elements in the material and improved forgeability. Increased homogenization also helps to improve forgeability. Besides the problem encountered in forging, nickel base alloys have limited ductility in a narrow forgeability range. This makes it necessary to heat the entire forging uniformly to produce a uniform grain structure. Further, there is a need for avoiding differential flow of metal in the forging to avoid shear and cracking. The forging is accomplished in small steps so that there is uniform distribution of energy to the blade. Both nickel plating or glass powder coating is given to ensure a lubricant action and non-sticking quality before oxidation. After the final forging, the lubricant is removed and the blade is age hardened. The oversize blades are subjected to blasting, to remove any oxide layers and then ready for machining. Close finish blades are subjected to finishing operations very carefully, such as etching. With the use of the higher and higher temperatures which are close to the thermal yield strength of the blades – blade cooling becomes necessary. To achieve this, holes are drilled in the blades through which either compressed air or water would flow to cool the blade. The simplest technique of blade manufacture to meet this requirement is to forge the blade and then drill the holes in it. But better techniques have been developed which produce better results. They are (i) tadpole method, and (ii) extrusion method. In the tadpole method a complex form is first formed. The blade thus becomes a solid piece. This is achieved by the regular extrusion and preforming. The blade is checked very closely for dimensional accuracy and holes are drilled (3 in number) along the cross-section of the blade. The blades are then subjected to coining operations after being heated to 1100◦ C. The coining operation is performed in a screw press. The operation

506

Gas Turbines

converts the section of the blade to that of the required aerofoil shape. The nickel coating is then removed and finishing operation is done on the blade. This technique is extensively used for the high pressure turbine blades of Rolls-Royce, Tyne and some Avon engines. The alloys generally used are Nimonic 105. In the extrusion method a certain length of a circular bar is taken and subjected to extrusion process to create a certain shape and cross-section. Holes are drilled and filled with nickel rods and then firmly welded. The bar is then once again subjected to extrusion to create not only the aerofoil section but also the head. The blade is once again upset to create the shroud if found necessary. This technique requires a certain amount of machining. 14.2.2

Casting

The blades manufactured by this technique are intricate and any amount of complexity can be adopted. This technique uses either precision casting or the lost wax method and is mainly used for nickel and cobalt alloys. The technique also employs vacuum casting methods. The main advantage of this process is the high dimensional accuracy that can be achieved (± 0.06 mm). The investment casting technique essentially involves 7 main stages. (i) Producing an injection wax pattern which may be removed by heating (ii) Assembling the wax patterns into a runner system (iii) Formation of refractory castings around the wax pattern (iv) Removing the wax from the refractory castings (v) Hardening the refractory castings by firing in a kiln (vi) Melting and casting the molten metal (vii) Removing the part from the casting material and inspecting the part The process is started with the formation of the wax pattern. This is achieved by solidifying wax in a die. Since the overall process efficiency is dependent on the dimensional accuracy of the die, the dies are machined very very accurately (of the order of 0.005 mm on CNC machines). The essential properties that are to be satisfied by a wax material are (i) low ash content, (ii) dimensional and chemical stability, (iii) good joining and fabrication conditions, (iv) predictable expansion and contracting characteristics, (v) minimum amount of lead and bismuth, and (vi) no residue formation.

Blades

507

Usually the waxes used are a blend of natural and mineral waxes available. This is to achieve suitable characteristics. The natural waxes (carnauba and candellia) are obtained from South American plants while mineral or montan waxes are obtained from European lignite. A recent trend is the use of polystyrene because of its low contraction and excellent die reproduction. The wax is injected in the die under pressure and allowed to partially dry in the die. This is achieved by using an injection machine. The cores used are of three types. (i) Soluble cores These are made from a mixture of mica (silica), sodium carbonate and polyethylene glycol. In this method the core is inserted in a wax pattern die and wax is poured in the die. The core is removed by immersion in agitated water or dilute acid. The cavity left is filled with investment material. This is a very simple technique. (ii) Ceramic cores These are prepared from silicon or zircon and small quantities of other materials such as cristobalite or urea may be added to improve the strength at room temperature operations. The cores are pressed together. (iii) Silica rods These are used where it is difficult to get ceramic shapes such as radial cooling holes. These are produced by using extruded or drawn vitreous silica rods as cores. Silica has advantages due to easiness of its removal by leaching. The wax patterns are formed by injecting wax around the core but with about 3 mm space above to help the core to be supported by the investment material when it is formed around the wax pattern. Varnish or polystyrene are coated over one point to allow the use to float and prevent distortion due to differential core and shell thermal expansion. The next step is to join the runner and riser assembly which is also made of wax into the pattern. This is done carefully to prevent distortion of the wax pattern. The wax pattern is then dipped in a refractory slurry and the slurry covering is allowed to dry. A layer of refractory grit is then stuck on the refractory layer which is still drying. The process is repeated until a certain thickness of the investment is achieved. The number of coatings applied depends on the required shell strength and on the weight of the casting besides depending on the properties. Current investment materials use silica or hydrolyzed ethyl silicate and refractory materials such as alumina, zircon, alumino silicates, silica etc., To improve consistency and good coating properties of the investment material it is necessary to add wetting, defoaming, deaerating and suspension agents to the slurry mix. Where grain refinement of the casting is important, materials such as cobalt oxide or cobalt aluminate are added to the primary face coat of the shell. The removal of the wax from the investment is achieved by subjecting the mould to high pressure steam or by direct heating at temperatures in the range of 1000◦ C (flash dewaxing). The outer layer of the wax is melted and absorbed by the investment material to allow for expansion of the later

508

Gas Turbines

melting wax. Steam dewaxing is preferred as it allows the removal of the wax material. The moulds are then fired at temperatures above 900◦ C to form a ceramic bond and this is done in a slightly oxidizing atmosphere to allow the last traces of wax to be removed. The alloys are prepared by either air melting or vacuum melting. The alloys may be prepared either from virgin materials or from scrap material. In the virgin material method the various elements such as nickel, cobalt, chromium, molybdenum and tungsten and elements as boron in nickelboron and carbon in chromium-carbon are added. Melting techniques are followed with great care to prevent excessive melt temperatures. For this protective stage fluxes are often used. The alloys prepared from the scrap material available from the previous casting are usually adjusted for chemical composition. In the vacuum casting technique virgin alloy is prepared by careful layering of individual materials like nickel chromium, cobalt and refractory metals as carbon. The elements are melted slowly to let out the gas from the furnace by carbon boil (carbon getting oxidized). Reaction elements like aluminium and titanium are then added with good furnace temperature control. The elements like zirconium, hafnium, boron are added near the end in the melting process. Integrally cast turbine wheels are also being produced off late. The casting presents difficult problems for the foundry-man as high casting temperature are required to produce good dimension and decrease the porosity of the hub. For good blades the grain structure should be uniform for creep resistance and the hub should have fine grains, free from porosity. This is achieved using grain refining agents while the microporosity is controlled only by hot isostatic pressing. 14.3

BLADE FIXING

The blades of a gas turbine are to be fitted carefully to the discs. This is served by the blade root which is either welded or fixed to the disc circumference. While considering the root the following aspects are to be kept in mind. (i) The root should be able to transmit the centrifugal force and vibrations to the turbine disc. (ii) It should be easy to mount the blades on the turbine discs. (iii) The manufacturing process should be affordable. (iv) The temperature effects on the blade and the disc should not cause expansion due to high temperature and should not lead to any looseness or deformation of the attachment. Consideration should also be given to the need for cooling the blade. Some degree of freedom is usually given to the blades to counteract vibration.

Blades

509

For turbines the configuration of the blade root disc assembly plays an important role in fixing. Early turbine blades were fitted to the discs by the Delaval root fixing which was later superseded by the fir tree fixing which are used in majority of the gas turbines nowadays. The machining of serrations is very important to ensure the blade load to be shared equally by all the serrations. The root is manufactured by broaching the blade and the disc separately. The blade is free in the serrations when the turbine is stationary and is stiffened by the centrifugal forces when the disc rotates. Sufficient clearance is to be obviously provided at the tip of the blade. Various methods are used to develop a good fit for the blade and prevent it from sliding out of the serration. Some of the blades are having screws between the bulb root and the groove in the turbine disc. Hollow blades are occasionally held by pins introduced from one side and a common plate on the other side some times holes are drilled radially through the shoulder of the disc rim upto the blade seating groove when it is aligned with the rim hole and a steel pin driven through the rim hole into the blade. The following are some of the methods for fixing blade to the wheel: (i) Bulb Root Method A bulb is made in the root of the blade which is fixed to the wheel with the help of steel pins welded and inserted in the holes. This method is employed in Metro Vick gas turbine. Endwise movement of the blade is usually prevented by swaging over the metal of the rotor around the root portions of the blades. Figure 14.1 shows the details.

Fig. 14.1 Bulb root method of blade fixing (ii) Tee and Double Tee Figure 14.2 illustrates the method of fixing the blade using Tee and double Tee technique. A T–slot is made in the wheel and after keeping the blade in position a wedge is placed to keep it in tact. (iii) Anchor Pin As shown in the Fig. 14.3 the blades are fixed by inserting the pin through the hole in the wheel and in the blade root and screwing the nut over the threaded portion.

510

Gas Turbines

(a) Tee

(b) Double Tee

Fig. 14.2 Blade fixing by T

Fig. 14.3 Anchor pin method of blade fixing (iv) Fir Tree Method The details are shown in Fig. 14.4. This method has certain advantages, viz., (a) rotor is not weakened, (b) contact is distributed, (c) raper is provided, and (d) allowance is given for thermal expansion. This type of fixing is used on the W2 B turbine and subsequently on the Rolls-Royce engines. The blade is driven into the rotor teeth and locked by turning over the lip at the apex of the fir cone on the blade. (v) Grub Screw Method A screw is fitted half in the rotor drum and half in the blade root. This method is used on German gas turbines for securing the blades in the rotor (Fig. 14.5). In this case the blade is a dovetail section and is secured by Grub-screws against end movements. (vi) Welding The blade can also be fixed by welding as shown in Fig. 14.6. 14.4

PROBLEMS OF HIGH TEMPERATURE OPERATION

It is always the aim of a gas turbine designer to increase power output, reduce weight and also to reduce the fuel consumption. Improvement in the Carnot efficiency η = 1 − TT21 of a heat engine with an increase in the

Blades

Pitch at tip

Pitch at root

A gap of 100 microns

Fig. 14.4 ’Fir-tree root’ method of blade fixing

Rotor

Grub screw

Fig. 14.5 Grub screw method of blade fixing

Blade Welding material

Wheel

Fig. 14.6 Blade fixing by welding

511

512

Gas Turbines

entry temperature, T1 is well known. Hence, attempts to increase the power to weight ratio have lead to the development of gas turbines operating at higher and higher entry temperatures. The maximum temperature of a gas turbine plant cycle occurs at the entry of the first stage of the turbine. We already know the effect of the maximum cycle temperature at various values of the pressure ratio on the thermal efficiency (refer Chapter 6). Employment of high temperature for a given pressure ratio leads to a higher thermal efficiency, higher power to weight ratio and lower specific fuel consumption. Referring to Chapter 6, Figs.6.7 and 6.8 show the effect of pressure ratio and maximum cycle temperature on the specific power output and efficiency of gas turbine power plants. From these two figures, the advantages of operating the power plant at higher gas temperature is evident. However, employment of high temperature gas, in gas turbines require materials which can withstand the effects of high temperature operation. Further, high temperature operation affects all the components in the engine with varying degree. However, the life-limiting component is usually the high pressure turbine blade, where failure is associated with thermal fatigue, oxidation or corrosion and creep. If a turbine blade is heated rapidly to a high temperature it causes uneven temperature distribution and as a result of this, severe thermal stresses are developed within the material. Beyond certain temperature (650 - 800◦ C) the blade material does not remain elastic and continues to stretch under the applied forces. This is called creep. If this phenomenon exists for a long time fracture can occur. There are two options open to the designer to overcame the problems of high temperature. (i) A new material may be sought which is capable of operating at high temperatures. (ii) By blade cooling which maintains the temperature of the blade at a value low enough to preserve the desired material properties. From the point of view of new materials, many alloys have been specially developed for gas turbine blades operating at high temperatures having manganese, molybdenum, copper, columbium, silicon, tungsten, vanadium and zirconium in them. They are used in various proportions to obtain certain desired properties in the alloys. Ceramic materials and mixtures of ceramics with metals, whilst having some very desirable properties have been shunned because they have equally disagreeable shortcomings such as brittleness and proneness to failure in thermal shock. The use of high gas temperatures at entry is intimately linked with the characteristics of materials that can be used in such applications. We have already discussed the properties required in the high temperature materials employed in gas turbines. However, we will recall them once again. The requirements are: (i) high strength at the maximum possible temperature,

Blades

513

(ii) low creep rate, (iii) resistance to corrosion and oxidation, (iv) resistance to fatigue, and (v) ease in manufacture, i.e., machinability, castability, weldability, etc., 14.5

BLADE COOLING

Blade cooling is the most effective way of maintaining high operating temperatures making use of the available material. Blade cooling may be classified based on the cooling site as external cooling and internal cooling. Another classification based on the cooling medium is liquid cooling and air cooling. 14.5.1

External Cooling

As the name implies, the external surface of the gas turbine blade is cooled by making use of compressed air form the compressor. The quantity of air required for this purpose is from 1 to 3% of the main flow entering the turbine stage by which blade metal temperatures can be reduced by about 200–300◦C. By employing suitable material (nickel-based alloys) an average blade temperature of 800◦ C can be used. This can permit maximum gas temperature of about 1400 K. Still higher temperatures can be employed with nickel, chromium and cobalt based alloys. Other methods of external cooling are film cooling and transpiration or effusion cooling. Film cooling to local areas can be applied by drilling holes on precision cast blades by the process of Electrical Discharge Machining, EDM, which removes material by multiple spark discharge action at the end of a special electrode. By this process very accurate, smooth, low residual stress holes can be produced upto a dia. of 0.15 mm. The cooling air flowing out of these small holes forms a thin film over the blade surfaces. Besides cooling the blade surface it decreases the heat transfer from the hot gases to the blade metal. This is shown in Fig. 14.7.

Cooling air passages

Fig. 14.7 Film cooling In the transpiration cooling (Fig. 14.8) the air is allowed to effuse or sweat form the pores of the porous blade metal. This provides a blanket of cool air, insulating the metal of the turbine blade from the hot gas. Effusion

514

Gas Turbines

of the coolant over the entire blade surface causes uniform cooling of the blade. Sintered wire composites are the most commonly used transpiring materials. Some of the disadvantages of this method are, low strength at high temperature, low modulus of elasticity, very small pore size and poor oxidation resistance. Envelope of film

Porous wall

Fig. 14.8 Transpiration or effusion cooling

14.5.2

Internal Cooling

Internal cooling of blades is achieved by passing air or liquid through internal cooling passages from hub towards the blade tip. The internal passages may be circular or elliptical as shown in Fig. 14.9 and are distributed over the entire surface of the blade. The cooling of the blades is achieved by conduction and convection. Relatively hotter air escapes to the main flow from the blade tips after traversing the entire blade length in the cooling passages. Internal cooling passages

Fig. 14.9 Convection cooling Hollow blades can also be manufactured with a core and internal cooling passage as in Fig. 14.10. The cooling air is first admitted into the inner shell (1) from which it comes out through narrow and long slots (2) and (3) and impinges the inner surface of the leading and trailing edges respectively. The air form (2) after impinging, flows over the inner shell and comes out by the holes (5) on the concave surface near the trailing edge. This type of cooling is called impingement cooling. Based on the cooling medium employed, blade cooling may be classified into liquid cooling and air cooling.

Blades

Convection Cooling air

515

3 4

2 Jet impingement

1 Core

5

Fig. 14.10 Impingement cooling 14.6

LIQUID COOLING

Because of the higher rates of heat transfer the use of water or other liquids as coolants have been tried. Liquid cooling has been tried on long life turbines as the heat transfer rates are higher for these turbines. Even though this is one of the most effective methods of removing heat, it must be noted that it may overcool the blade which is not desirable. When liquid is used as coolant the equipment becomes additionally complex. When water is used, the disadvantage is that either the vapourization must be allowed for or water must be circulated at high pressure above its vapour pressure so as to prevent vapourization. There is another disadvantage in using water as coolant since, it has a high heat transfer coefficient and because of this the heat transfer rate becomes quite high as already mentioned. For this reason organic liquids with low vapour pressure than water for the same temperature have been tried. If organic coolants are used there is a possibility that it could be mixed with the fuel before burning. A liquid cooling system is more advantageous when the liquid is recirculated, so that it can be used over and over again. Because of the complexity this method is not suitable for aircraft applications. The use of water as coolant has been tried on gas turbines for power generation. Figure 14.11 shows a water cooled blade in which water flows through internal passages in the blade. It is almost impossible to eliminate corrosion or the formation of deposits in this system. 14.7

AIR COOLING

Most of the mechanical complexities that are present with liquid cooling virtually disappear when air is used as the cooling medium. So much so, all present day aircraft gas turbine featuring turbine cooling fall into this category. In all these systems air is bled from the high pressure end of the compressor and delivered to the blades and vanes to be cooled. Quantity of the coolant required is about 1 to 3% of engine air flow per turbine blade row. Advantage is taken of the fact that the cooling air is some 600◦C cooler than the gas surrounding the blades. After passing through the

516

Gas Turbines

Top cover

Plug

Plug Inlet

Outlet

Blades

14.7.2

517

Requirements for Efficient Blade Cooling

In a conventional cooled blade, cooling is obtained due to convection by passing cooling air through internal passages within the blade. The success in obtaining the large reduction in metal temperature at the expense of a small quantity of cooling flow is governed by the skill in devising and machining the cooling passages. Because the internal cooling relies on the cooling air scrubbing against the cooling surface, the internal surface area must be large and the velocity of the cooling air must be high. This implies that the cross-sectional flow area of the passages must be small. The design of the blade internal geometry for cooling is more complex because of the various aerodynamic, heat transfer, stress and mechanical design criteria that must be satisfied. The most successful designs have incorporated radial passages through which cooling air passes, escaping at the tip.

14.7.3

Heat Transfer in Cooling Passages

As already mentioned, the cooling passage along the blade span must be devised so as to give a large internal surface area. Also for efficient heat transfer the cooling air passing through the passages must have high velocities which implies that the cross-sectional area of the passages must be small. Taking these two factors into consideration, a factor termed Z factor is introduced which could be used to compare the relative merits of various cooling passage configurations. The coolant geometric parameter Z

=

Sc1.2 C2 × C 1.2 Ac

(14.1)

where Sc is internal surface area per unit length of the blade, Ac is total cross sectional area of the cooling passage in one blade, C is true chord of the blade. The Z factor determines the relative blade temperature. A high value of Z means the ratio of heat transferred per unit temperature difference to the coolant to the total heat transferred per unit temperature difference to the blade from gas is high. Increasing Z would mean a reduction in the mean blade temperature. The blade temperature varies rapidly when Z is small, but is progressively less sensitive to changes in Z for large values. Values of Z above 150 are required for good cooling but there is little point in going beyond 250 for values of Z. One way of achieving a high value of Z is to have a large number of small spanwise holes. By using elliptically shaped holes, rather than round ones, the number required can be kept to a minimum. The influence of Z on the reduction in midspan average temperature for various percentages of cooling air flow is shown in Fig. 14.13.

Gas Turbines

Reduction in midspan average metal temperature (o C)

518

300

200 150 100 Z=50

200 100

0

0

1

2 3 4 5 Percentage cooling air flow

6

Fig. 14.13 The effect of Z factor on midspan cooling 14.8

PRACTICAL AIR COOLED BLADES

It is the problem of manufacturing more than anything else which has practically restricted the development of the air cooled blades. Even the simplest cooled blades are so expensive to manufacture that to justify their use they must be doubly reliable. In fact a cooled blade which will not last a major proportion of the aircraft life, say 10000 hours, almost prices out the engine out of the market on the basis of high overhaul costs. Different cooled blade designs are conveniently classified according to their method of manufacture as follows: (i) fabricated, (ii) forged, and (iii) cast. 14.8.1

Fabricated Designs

Fabricated blades were the fore runners in the field of blade cooling. The blade was nothing more than a plain hollow blade. The amount of cooling obtained was trivial because of the very poor Z factor. 14.8.2

Forged Blades

Instead of one hole as in previous one, this employs three spanwise holes. This cooling configuration was conceived in the early 1950. Tyne, Conway and Spey blades are the most important ones belonging to this category made by Rolls-Royce. The Tyne Blade Cooling air enters at the blade root, passes up the trailing edge hole, down the centre hole and up the leading edge hole, to exhaust at the blade tip (Fig. 14.14). The cooling air flow, expressed as a percentage of the main turbine flow is just under 1% and this cools the mid-span section of the blade by an average of 40◦ C. Blade life is increased from 2,000 hours uncooled to an achieved life of 6,500 hours Z factor is 33. Conway Blade In this design cooling air flow from one side of the blade root to both leading and trailing edges, turns over at the blade tip and flows down the centre passage and then exhausts at the blade root on the

Blades

3 cm

519

Z = 33.0 20 Variation from average temp. 10 0 -10 Span

Take-off temperature Cooling air flow Mid span cooling Material Predicted life Achieved life Life without cooling

1242 K 0.8% Data from 40 oC Canadair CL44 operation N 105 7000 h 6500+ h 2000 h

Fig. 14.14 Tyne triple pass turbine blade opposite side of the blade (Fig. 14.15). Cooling air used is 1.4% of main turbine flow and this reduces the midspan average temperature by 120◦C. From a life of 75 hours at uncooled stage life is increased to 13,000 hours after cooling. Z factor is 31. The low Z value of both Tyne and Conway blade implies that if the air is passed straight through the blade the cooling would be very inefficient. Therefore more than one pass of the cooling air is used. Spey Blade This type of blade has elliptical shaped cooling

7 cm

Z = 31.0 100

Variation from average temp.

50 0 -50 Span Take-off temperature Cooling air flow Mid span cooling Material Predicted life Achieved life Life without cooling

1310 K 1.4% 120 oC Boeing 707-420 and N 105 DCB-40 operation 13000+ h 15000 h 75 h

Fig. 14.15 Conway double pass turbine blade holes formed by a unique manufacturing method. Cooling air enters from both sides of the root and exhausts from the tip of the blade, thus giving the single pass cooling system (Fig. 14.16). The cooling air flow used is 2% of the main turbine flow and this reduces the midspan average blade temperature by 220◦ C. If uncooled this blade would last for 12 minutes, with air cooling the predicted life is 10000 hours.

520

Gas Turbines

300 200

Variation from average temp.

100 0 -100

Cooling air flow Mid span cooling Material Predicted life Life without cooling

2.0% 200 o C N 108 10000 h 0.2 h or 12 min

Span Data from BAC I-II operation

Blades

521

Review Questions 14.1 What are the forces the blades are subjected to? Explain. 14.2 Explain the various materials that can be used for blades. 14.3 Explain the factors to be considered in the selection of blade materials. 14.4 Briefly explain the various manufacturing techniques for blades. 14.5 Give an account of various blade fixing techniques. 14.6 Why the blades are to be cooled? 14.7 Explain the problems encountered in high temperature operation. 14.8 What is meant by internal and external cooling? Explain. 14.9 What are the requirements for blade cooling? 14.10 Give an account of practical air-cooled blades. Multiple Choice Questions (choose the most appropriate answer) 1. A turbine rotor blade is one, which transfers energy (a) from gases to the turbine rotor (b) from turbine to the rotor gases (c) there is no energy transfer (d) to the compressor 2. The most preferred material for blades is (a) steel alloys (b) titanium alloys (c) brass alloys (d) aluminium alloys 3. Turbine blades must withstand (a) high operating pressure (b) high operating temperature (c) high operating volume (d) all of the above 4. Modern blades are manufactured by (a) forging (b) turning

522

Gas Turbines

(c) machining (d) wire cutting 5. Forging can be done by (a) (b) (c) (d)

upsetting extrusion coining any of the above

6. For the casting the cores are usually (a) (b) (c) (d)

soluble cores ceramic cores silica cores any of the above

7. The materials used for gas turbine blades must have (a) (b) (c) (d)

high strength high resistance to corrosion high resistance to fatigue all of the above

8. External cooling of gas turbine blade is achieved using (a) (b) (c) (d)

liquid air exhaust gas none of the above

9. When liquid cooling is employed in internal cooling of blades, the water should be pressurized (a) (b) (c) (d)

below the vapour pressure above the vapour pressure upto the vapour pressure to any pressure

10. The most stressed part in the blade is (a) (b) (c) (d)

tip middle section root side Ans:

1. – (a) 6. – (d)

2. – (b) 7. – (d)

3. – (b) 8. – (a)

4. – (a) 9. – (b)

5. – (d) 10. – (c)

15 COMPONENT MATCHING AND PERFORMANCE EVALUATION INTRODUCTION In the last six chapters, we dealt with in detail the various components of a gas turbine power plant. For example, Chapter 8 and 9 are concerned with compressors whereas Chapter 10 is devoted for combustion systems. In Chapter 11 there is a detailed discussion on turbines and in Chapter 13 we have analyzed the inlets and nozzles. In Chapter 14 we have discussed briefly the blade materials, manufacturing, fixing and cooling. It may be recalled that we have already discussed the various possible cycle arrangements and also their analysis both at ideal and practical conditions. How the efficiency of each component plays a role in determining the overall performance of the system was brought to light. From practical cycle analysis (Chapter 6) it is possible to calculate, for instance, the pressure-ratio, which, for any given maximum cycle temperature will give the highest overall efficiency, and the mass-flow required to give the desired output. From such preliminary calculations it is possible to choose the most suitable design data for any particular application. Then, it becomes easier to design the individual components of a gas turbine so that the complete unit will give the required performance at the design point. However, the problem remains to examine more critically the characteristics of the entire plant as a unit, the methods of control, governing, and the operation at a condition other than design load. To a large extent the desired performance characteristics of the gas turbine power plant can be accomplished by the control of various design features of the components. Therefore, design decisions for the components

524

Gas Turbines

must be governed by what is best for the whole plant rather than what is necessarily the best for the individual component. What is best for the whole plant is dictated by economic considerations, the purpose the plant is to serve, and such matters as weight and space restrictions. This chapter includes how the performance or the characteristic curves for different components such as turbine and compressor are represented graphically and then by means of graphs, how to link these components so as to make the combination run in perfect equilibrium. A matching study is an investigation of the interplay of the engine geometry and engine parameters such as pressure ratio, airflow, rotor speed, component efficiencies, pressure drops, areas and so on. Such a study must be conducted to answer questions about the steady-state and transient operation of a gas turbine engine. Most of the discussion in this chapter deals with the steady-state operation of a single-spool aircraft engines. If one has to effect perfect matching of components, then, equilibrium running diagram should be constructed taking into account all the parameters mentioned in the previous paragraphs. In this connection the following questions are to be answered: (i) When a gas turbine is operated at an off-design condition with a fixed engine geometry, what happens to the component parameters and resulting component match? (ii) How does the magnitude of turbine inlet temperature influence the resulting component match? (iii) When the jet nozzle area of a turbojet engine is changed, does the component match point change? If so, how to arrive at the new match point? (iv) What effect does afterburner operation have on the match point for the gas generator of an afterburning turbojet engine? Is it possible for an afterburning turbojet engine gas generator to operate at the same match point with the afterburner and without afterburner turning on? (v) How does diffuser water injection or operation with a low calorific value fuel influence the engine match point? (vi) How does variation of the turbine nozzle area influence the engine match? (vii) How does engine bleed, power extraction and/or turbine blade cooling, influence the engine match? Answers to these questions involve the interplay of a large number of variables and would require a detailed computer code. Since, it is out of preview of this chapter, our analysis will be based on simplifying assumptions. The discussion regarding engine matching in this chapter is intended to give answers, in a general way, to many of the above questions. The equations

Component Matching and Performance Evaluation

525

developed to answer these questions contain a number of assumptions that may be questionable in an actual engine. They are made here to simplify the problem so that one can see the trends without becoming deeply involved in a number of finer details. Engine matching, because of its complexity is done almost exclusively on high speed digital computers but, since all new computer programs must be checked for accuracy and are constantly being modified, one must understand the principles of matching techniques. The geometry and flow areas of a given power plant are established at the “design point”. At all other operating conditions, the components must be “matched” to determine the pressure ratio, airflow, rotor speed, efficiency and so on. The “match point” is defined as the steady-state operating point for a gas turbine when the compressor and turbine are balanced in rotor speed, power, and flow, the operating points at the various power settings defining the operating line for the given engine configuration. No matter what type of engine is being considered, the conservation of mass, energy, and momentum must be satisfied. The first two lead to the “match point”, the third to the thrust developed by the engines for aircraft application. For satisfying the principle of conservation of mass requires: (i) The flow through the turbine must equal to the flow through the compressor plus the fuel added, minus any air extracted. Care must be taken into account for any extracted air from the system and exactly where it re-enters the system. (ii) The exhaust system (nozzle) flow characteristics must also be satisfied. For satisfying the principle of conservation of energy requires that the power developed by each turbine equal the power required to drive each compressor plus losses and power extracted. When dealing with the general matching trends of a single-spool turbojet engine, several simplifying assumptions are made. These assumptions hold true unless stated otherwise and include the following: (i) No burner or exhaust system pressure losses will be included; that is, it is assumed that pressure remains constant throughout the combustion chamber and also from the exit of the turbine to the inlet of the exhaust nozzle. (ii) The mass rate of flow is assumed to be the same through the compressor, turbine, and exhaust nozzle. This means that the mass of fuel added is being neglected, no air is extracted from or after the compressor, and turbine cooling is not being used on the engine being studied. (iii) No power is extracted and no losses occur between the turbine and the compressor, i.e., bearing losses are neglected.

526

Gas Turbines

(iv) The turbine nozzle and jet nozzle areas are constant values as determined for the design point conditions. (v) It is assumed that air is the working fluid throughout and that the specific heat ratio, γ, has a constant value of 1.4. Using the above assumptions, we can construct the equilibrium running diagram. In order to draw the same, first we should know the compressor and turbine performance characteristics. 15.1

PERFORMANCE CHARACTERISTICS

Performance curves of compressor and turbine may be plotted for delivery pressure and temperature against the mass flow for different speeds. These curves will reveal how the compressor and turbine will respond to the changes in different variables. But these are not the only variables which are responsible for performance variations. There are other parameters such as inlet pressure and temperature and also the physical properties of the working fluid. If full variation of all these variables is allowed over the entire working range of the components, it would involve large number of tedious experiments and thus precise presentation of the results will be next to impossible. To overcome this complication, the technique of dimensional analysis is helpful. This consists of arranging the variable in picture in particular groups of variables which are dimensionless. This way the complete performance characteristics of any compressor or turbine may be represented only by two sets of curves. These details have already been dealt in Chapter 3. 15.2

EQUILIBRIUM RUNNING DIAGRAM

It may be recalled that using dimensional analysis a set of performance characteristics can be obtained and plotted. Typical performance characteristic curves are as shown in Fig. 15.1. When the components are to be linked together in an engine use of these characteristic curves are very helpful. The problem is to find corresponding operating points on the characteristic curves of each component when the unit is running at a uniform speed. The equilibrium running points for a series of speeds may be plotted on the compressor characteristics and joined up to form an equilibrium running diagram. Once the operating conditions have been determined it is simpler to find performance data such as power output or thrust and specific fuel consumption. There are two schemes for which the equilibrium running, points may be obtained. (i) When a separate power-turbine is provided to take the load the compressor turbine runs the compressor only (Fig. 15.2).

Component Matching and Performance Evaluation

(a) t

A

N T 01

Sur ge

Y

line

line

(b) Sur ge

p03 p01

527

T 02 Y T01

X m T01

A

N T 01

X

p01

m T01

p01

Compressor characteristics

p03 p04 N T03

m T03

p03

(d)

Choking mass flow

Choking mass flow

(c)

T 03 T04 N T03

m T03

p03

Turbine characteristics

Fig. 15.1 Compressor and turbine characteristics (ii) Only one turbine is there which runs both the compressors as well as the load. (Fig. 15.3). Torque-speed characteristics for these two schemes are as shown in Fig. 15.4. The characteristic is poor when the turbine is coupled to compressor as well as load. Curve 1 is for the 1st scheme in which turbine is coupled to the compressor only and there is a separate power turbine. Curve 2 is for 2nd scheme in which a single turbine drives the compressor and the load. 15.3

TO FIND THE EQUILIBRIUM POINTS

Now, let us see how to find the equilibrium points from the characteristic curves. First of all for equilibrium running, the following conditions must be satisfied. (i) Speed of the turbine and compressor is same. Therefore, √ N T N √ √ 03 = √ T01 T03 T01

(15.1)

Gas Turbines

Heat exchanger Combustion chamber

Air Fuel

HPT

HPC

Compressor

Turbine Fuel

Combustion chamber Power LPT

Separate power turbine

output

Fig. 15.2 open-cycle gas turbine with separate power turbine Combustion chamber 1 Air

Product of combustion

3

2

4

Fuel

Power output

Compressor

Turbine

Fig. 15.3 Turbine running the compressor and load Sepa

rate p

ower

1

turbi

ne

Torque

528

sor

s pre 2 m co d to loa d e l nd up a

Co

Speed

Fig. 15.4 Torque–speed characteristics

Component Matching and Performance Evaluation

=

N √ √ t T03

529

(15.2)

where t = T03 /T01 .

(ii) Mass flow through the compressor and turbine are assumed to be same. Hence, √ √ √ m ˙ T01 m ˙ T03 p03 T01 √ = × (15.3) p01 p03 p01 T03 or

√ m ˙ T01 p01

=

√ m ˙ T03 1 √ R p03 t

(15.4)

where R is the pressure ratio p03 /p01 . It may be noted that the fuel supply increases the mass of gas in turbine but some amount of the compressed air is also taken out for cooling which is approximately equal to the amount of fuel supply. (iii) Power output of the turbine must be equal to the work input required to run the compressor. Hence Cpa (T02 − T01 )

= Cpg (T03 − T04 )ηmech

or

or

(T02 − T01 )

=

Cpg (T03 − T04 )ηmech Cpa

T02 − T01 T01

=

Cpg T03 − T04 T03 ηmech Cpa T03 T01

T02 − T01 T01

=

Cpg T03 − T04 t ηmech Cpa T03

(15.5)

where ηmech is the mechanical efficiency of the transmission. To satisfy the above conditions, the compressor and turbine characteristics diagrams are to be slightly changed. The changes to be effected are the following: (i) Instead of TT02 and TT03 in y-axis in Fig. 15.1, graphs should be plotted 01 04 T02 −T01 04 for T01 and T03T−T in the y axis. 03 (ii) Instead of pp02 the graphs should be plotted for pp03 where p03 ac01 01 counts for pressure losses in the piping combustion chamber and heat exchanger.

530

Gas Turbines

Because of these changes in the variables in the y-axis, the shape of the curves will not change, but, the values will change. The revised graphs are as shown in Fig. 15.5. (a)

A

Z

T02 - T01 T01

N T 01

Sur ge

Y

t

Sur ge

p03 p01

line

line

(b)

X m T01

A

N T 01

X

p01

m T01

p01

Compressor characteristics

B

N T03

T 03 - T04 T03

B P N T03

p m T03

Choking mass flow

p03 p04

q

(d)

Choking mass flow

(c)

p03

p

m T03

p03

Turbine characteristics

Fig. 15.5 Revised characteristic diagram of compressor and turbine

15.4

PROCEDURE TO FIND EQUILIBRIUM POINT

To find out the equilibrium points and lines, the following procedure is to be followed: (i) Assume some value of t and N . (ii) On the speed curve of particular value of √N choose a point A, on T01 the compressor√characteristics [Fig. 15.5(a)] and find out the values, √ p03 m ˙ T01 m ˙ T01 viz., pp03 and . Let us denote = Y and = X. p01 p01 p01 01 √

01 (iii) On the same speed curve of √N and with X = m˙ p01T01 find T02T−T T01 01 [Fig. 15.5(b)] of the compressor characteristics and let it be Z.

(iv) Using Eq. 15.2, find the value of √N T01

and t. Similarly find

√

m ˙ T03 p03

√N T03

corresponding to the value of

from Eq. 15.4.

Component Matching and Performance Evaluation

(v) On the speed curve and let it be P .

√N T03

and with

√ m ˙ T03 p03

find

T03 −T04 T03

531

in [Fig. 15.5(d)]

(vi) After getting Z and P substitute in the Eq. 15.5 and see whether the equation is satisfied or not. (vii) If this equation is satisfied, point A is the equilibrium running point for that N and t. (viii) If Eq. 15.5 is not satisfied, choose another point on the same speed curve and again find the values of Z and P . Carry out this trial till Eq. 15.5 is satisfied. That point will be the point A which satisfies Eq. 15.5. (ix) Similar points can be tried on other speed curves. (x) The line joining all such points on all speed curves is known as equilibrium running line for that particular value of t. (xi) For some other value, the equilibrium running line will be different and thus many lines can be drawn for many values, of t. This will be known as equilibrium running diagram which is as show in Fig. 15.6.

03

p 01

Su

p

rge

lin

e

t

N T01

m T01 p01 Fig. 15.6 Equilibrium running diagram for compressor-turbine combination

15.5

PERFORMANCE EVALUATION OF SINGLE-SPOOL TURBOJET ENGINE

To understand the details of engine matching a single-spool turbojet engine will be considered for performance evaluation. This study can be used as a building block, to help the understanding of other types.

532

Gas Turbines

Figure 15.7 is a schematic diagram of a single-spool turbojet engine showing various components and station numbers.

CC

Air

Exhaust gases

T

C CC

(0)

(1)

(2)

(3)

(4)

(5)

0-1 Inlet 1-2 Compressor 2-3 Combustion chambe 3-4 Turbine 4-5 Nozzle Fig. 15.7 Schematic diagram of a single-spool turbojet engine The turbine flow characteristics that will be used in this analysis are shown in Fig. 15.8. Note that in Fig. 15.8, the variation of turbine flow parameter as a function of expansion ratio which is shown as a single curve. This has been done without effecting corrections for turbine rotational speed. This type of turbine flow characteristics is considered mainly: (i) to simplify the analysis, and

Turbine flow parameter, m T03

p03

(ii) because of the negligible effect that turbine rotational speed has on the turbine flow parameter and efficiency.

A

B

C

D

Choking expansion ratio

Turbine expansion ratio, p03 / p04

Fig. 15.8 Turbine flow characteristics From Fig. 15.8 it can be seen that the turbine of the single-spool turbojet engine operates at choked condition over a wide region of its operating spectrum. Thus, most of the time, the turbine operating point will be to the right of point A (choking, expansion ratio) as shown in Fig. 15.8.

Component Matching and Performance Evaluation

533

When the turbine flow parameter becomes constant (the turbine is choked), the turbine expansion ratio can still continue to increase. For a fixed turbine inlet temperature and turbine efficiency, the turbine work is dependent on the turbine expansion ratio. Whether the “match point” for the turbine is at A, B, C, D, or another point depends on the flow characteristics of what follows the turbine For the single-spool turbojet engine we should consider the flow characteristics of the exhaust nozzle also. The relationship between the turbine flow characteristics and compressor flow characteristics can be expressed algebraically by the following equality: ⎤ ⎡ √ √ m ˙ g3 T03 ˙ g3 ⎣ 1 m ˙ a1 T01 T03 m ⎦ = (15.6) p03 p02 p03 A3 δ01 θ01 m ˙ a1 p p A p02

p01

01 std

3

01 01 and θ01 = TTstd where δ01 = ppstd For a constant-pressure in the combustion chamber and assuming mass flow rate through the compressor to be same as through the turbine m ˙ g3 = m ˙ a1 , √fixed turbine nozzle area (A3 = constant), and a choked turbine

m ˙ g3 T03 p03 A3

is also constant, Eq. 15.6 becomes p02 p01

=

C1

√ m ˙ a1 θ01 δ01

T03 θ01

(15.7)

where C1 is a constant. For a fixed T03 /θ01 , Eq. 15.7 reduces to p02 p01

=

C2

√ m ˙ a1 θ01 δ01

(15.8)

which is the equation of a straight line passing through the point (0,0). The slope of this straight line, C2 , increases for increasing values of (T03 /T01 ), that is, turbine inlet temperature over compressor inlet temperature. Plotting Eq. 15.8 on a typical compressor map yields the results shown in Fig. 15.9. Equation 15.8 shows that the pressure ratio is zero when the airflow is zero. This, of course, is impossible. What is wrong? Equation 15.8 assumes a choked turbine. At low air flows, the turbine unchokes so that the constant T03 /θ01 lines curve into a pressure ratio of 1.0 at zero airflow. This is shown in Fig. 15.9 by the dashed lines. For a given T03 /θ01 , where is the operating point? Is it at A, B, C, or D as shown in Fig. 15.9. This is fixed by, the compressor work required, which, for an energy balance, is fixed by the turbine work, the turbine work being fixed by the turbine inlet temperature and expansion ratio, the turbine expansion ratio being fixed by the nozzle flow characteristics. These details are discussed in the following paragraphs. For the single-spool turbojet engine illustrated in Fig. 15.7 the ideal work of compression for constant specific heat and unit mass flow is given

e

in

l ge ur

T 03 = θ 01

s con

tan

t

Gas Turbines

Compressor pressure ratio, p02 / p01

534

D B

S

C

A nt nsta

T 03 = co θ 01

1 0

Corrected air flow, m T01

δ 01

0 0

Fig. 15.9 Equation 15.8 plotted on a typical compressor map by wC,i

=

h02 − h01

=

Δh0C,i

(15.9) =

Cpa (T02,i − T01 )

(15.10)

Equation 15.10, when divided by compressor inlet temperature over the standard temperature, becomes, for constant specific heats Δh0C,i θ01

=

p02 p01

Cpa Tstd

(k−1)/k

−1

(15.11)

Equation 15.11 shows that the ideal work of compression divided by θ01 is a function of pressure ratio only. When variable specific heats are considered, the ideal work of compression divided by θ01 is approximately a function of pressure ratio. The actual compressor work depends on the compressor efficiency. Therefore, Δh0Ca θ01

=

Δh0C,i θ01 Cpa Tstd

1 ηC (k−1)/k

p02 p04

=

−1

ηC

(15.12)

The ideal turbine work for turbine of the single-spool turbojet engine illustrated in Fig. 15.7 is, for constant specific heats, wT,i

= Δh0T,i

=

(T03 − T04,i )

(15.13)

Component Matching and Performance Evaluation

= Cpg T03 1 −

(1−k)/k

p03 p04

(15.14)

or Δh0T,i θ03

p03 p04

= Cpg Tstd 1 −

535

(1−k)/k

(15.15)

Equation 15.14 illustrates that the ideal work developed by a turbine is a function of the expansion ratio and the turbine inlet temperature. Equation 15.15 illustrates that the ideal work developed by a turbine divided by θ03 is a function of the expansion ratio only. The actual turbine work depends on the turbine efficiency, or Δh0T,a θ03

=

Δh0T,i ηT θ03

(15.16) p03 p04

= Cpg Tstd ηT 1 −

(1−k)/k

(15.17)

We have already seen that for a choked turbine, the compressor pressure ratio is a function of turbine inlet temperature over theta and the compressor corrected mass rate of flow is given by Eq. 15.7 which is reproduced here. √ p02 T03 m ˙ a1 θ01 = C1 (15.18) p01 δ01 θ01 This results in several straight lines as illustrated in Fig. 15.9, each straight line being for a different turbine inlet temperature over theta. Equation 15.7 is based on conservation of mass and, as illustrated in Fig. 15.9, does not give any due consideration regarding, whether the engine will be operating at point A, B, C, or D. The actual steady-state operating point occurs where the turbine power is equal to the power required to drive the compressor; that is, it is determined by, an energy balance. Based on the assumptions stated earlier (no power extraction, no losses between compressor and turbine, and m ˙ a1 = m ˙ g3 ), Δh0Ca θ01

=

Δh0T,a θ03

T03 T01

(15.19)

Combining Eqs. 15.12, 15.17 and 15.19 yields, for the case where Cpa = cpg , and a = (k − 1)/k. p02 p01

a

−1

= ηC ηT 1 −

p03 p04

−a

T03 T01

(15.20)

This equation illustrates that the compressor pressure ratio is a function of compressor efficiency, turbine efficiency, turbine expansion ratio, and the ratio of turbine inlet temperature to compressor inlet temperature, or

Gas Turbines

Choked

Unchoked

Choking expansion ratio

m 4 T04

Exhaust nozzle flow parameter

536

Exhaust nozzle expansion ratio, p04 / pamb

Fig. 15.10 Typical exhaust nozzle flow characteristics p02 p01

= f ηC , ηT ,

p03 , p04

T03 T01

(15.21)

Equation 15.21 illustrates that for a fixed T03 /T01 and constant values of ηC and ηT , the compressor pressure ratio is a function of the turbine expansion ratio. The turbine expansion ratio for a specified T03 /T01 is fixed by the exhaust nozzle flow characteristics. A typical nozzle flow characteristic is given in Fig. 15.10, in which the exhaust expansion ratio, is specified as p04 /pamb instead of p0 /pstd . The turbine and exhaust nozzle flow parameters, when combined, will yield the following expression: √ m ˙ g4 T04 p04 A4

=

√ m ˙ g3 T03 p03 A3

p03 p04

T04 T03

A3 A4

(15.22)

For a given turbine polytropic efficiency, the turbine actual temperature ratio is related to the turbine expansion ratio according to the following equation: T04 = T03

p04 p03

ηp (k−1)/k

(15.23)

˙ g4 = m ˙ g yields Combining Eqs. 15.22 and 15.23 and assuming that m ˙ g3 = m √ m ˙ g T04 p04 A4

=

√ m ˙ g T03 p03 A3

p03 p04

[1−0.5ηp (k−1)/k]

A3 A4

(15.24)

Equation 15.24 shows that for exhaust and turbine nozzles of fixed area, the expansion ratio has a constant value once the turbine and exhaust nozzles become choked. This holds true only if the turbine polytropic efficiency (and therefore the overall turbine efficiency) is a constant.

Component Matching and Performance Evaluation

Compressor pressure ratio, p02 / p01

6.5

537

Steady-state operating line

6.0 e

5.5

e urg

5.0

lin

4.16 t= 3.82 3.47

S

4.5 η

4.0

% 95 = c

0%

3.5 3.0

ηc

% = 92

ηc

η

=9

12

14

16

18

% 80

η

2.5 10

= c

20

%

c

=

70

22

24

Corrected mass rate of flow, m 1 θ 01 / δ01 (kg/s)

Fig. 15.11 Hypothetical compressor map with steady-state operating line

15.6

OPERATING LINE

It is now important to evaluate what happens to the steady-state operating conditions of an engine as the turbine inlet temperature and/or flight conditions are changed. This will be done by assuming a hypothetical compressor map for a single-spool turbojet engine, fixing the areas at the design point, then determining what happens to the engine operation as the turbine inlet temperature is varied. This can be better illustrated by a numerical example, with the results shown on a hypothetical compressor map in Fig. 15.11. A single-spool turbojet engine, having the compressor performance illustrated in Fig. 15.11, has the following design point conditions at sea level √ m ˙ a1 θ01 static on a standard day. pp02 = 5.0; = 22.7 kg/s; T03 = 1200 K; δ01 01 ηT = 92% (constant). Calculate, neglecting the mass of fuel added, the turbine expansion ratio. Assume no pressure drop in the combustion chamber and constant specific heats with Cp = 1.004 kJ/kg K and k = 1.40. From the compressor map (Fig. 15.11) at the design point conditions, it is determined that the compressor efficiency ηc is 90%. Therefore, from Eq. 15.12, since θ01 = 1.0. Wca

ΔhT,i

= Δhca =

1 × 1.004 × 288.15 × 5(0.4/1.4) − 1 = 187.9 kJ/kg 0.90

=

ΔhT,a ηT

=

ΔhC,a ηT

538

Gas Turbines

θ03

=

187.9 0.92

=

1200 288.15

From Eq. 15.15 p03 = p04

1−

=

204.24

=

≈

204 kJ/kg

4.16

204.24 4.16 × 1.004 × 288.15

−(1.4/0.4)

= 1.91

Note that the expansion ratio across the exhaust nozzle, for the conditions assumed, is p04 5.0 = 2.62 = pamb 1.91 which is well above the pressure ratio required for the exhaust nozzle to be choked. It is next important to determine the operating line for this engine. This is illustrated by means of another example below. Determine, for the engine defined in the above example, the steady-state operating points when the turbine inlet temperature is (i) 1100 K (ii) 1000 K. For T03 = 1100 K; T01 = 288.15 K; Tstd = 288.15 K; θ01 = 1.0; θ03 = 3.82. From Eq. 15.17 ΔhT,a θ01

=

3.82 × 1.004 × 288.15 × 0.92 1 − 1.91(−0.4/1.4) 1.0

= 171.8 kJ/kg At the design point conditions, using Eq. 15.7 one may determine C1 , C1

=

5.0 × 1 √ 22.7 × 1200

=

0.00636

Therefore. the relationship between compressor pressure ratio and corrected mass rate of flow is, √ p02 T03 m ˙ a1 θ01 × = 0.00636 × p01 δ01 θ01 or, for θ03 = 3.82, i.e., T03 = 1100 K, the relationship becomes √ p02 m ˙ a1 θ01 = 0.211 × p01 δ01

(15.25)

Equation 15.25, which is an expression of the conservation of mass, is plotted on Fig. 15.11 and is labelled T03 /T01 = 3.82. Since both the turbine and exhaust nozzle are choked and the turbine efficiency is a constant, the turbine expansion ratio is a constant and is equal to 1.91.

Component Matching and Performance Evaluation

539

The actual turbine work, as determined by Eq. 15.17, is ΔhT,a θ03

=

1.004 × 288.15 × 0.92 × 1 − 1.91(−0.4/1.4)

=

44.97 kJ/kg

Note that this is a constant value for this engine as long as the turbine and exhaust nozzle are choked. For T03 = 1100 K, i.e., θ03 = 3.82 ΔhT,a θ01

=

ΔhC,a θ01

=

44.97 × 3.82 1

=

ΔhT,a × θ03 θ01 =

171.79 kJ/kg

It is now necessary to determine the compressor pressure ratio that satisfies ΔhC,i θ01

=

171.79

The ideal compressor work over θ01 is determined from Eq. 15.11. For a pressure ratio of 4.4. The result is ΔhC,a θ01

=

1.004 × 288.15 × 4.4(0.4/1.4) − 1

=

152.65 kJ/kg

The compressor efficiency, for an ideal work of 152.65 and an actual work of 171.79 is 152.65 = 0.888 ηC = 171.79 The results for several pressure ratios are, for an actual work of 171.79 ΔhC,a are given in Table 15.1. Therefore, a θ01 = 171.79 line may be constructed on the compressor map. This is shown in Fig. 15.12, which is a portion of the compressor map. ΔhC,a The point of intersection between the TT03 = 3.82 line and θ01 = 01 171.79 line is the steady-state operating point at sea level condition on a standard day for a turbine inlet temperature of 1100 K. From the plot, this point is p02 p01 ηC √ m ˙ a1 θ01 δ01 For T03 = 1000 K

= 4.49 = 90.4% = 21.4 kg/s

540

Gas Turbines

Table 15.1 p02 /p01

ΔhC,a /θ01 (kJ/kg)

ηC

155.3 152.5 149.6 146.6 143.6

0.905 0.888 0.872 0.854 0.837

Steady-state operating line

5.0

p /p 02 01

Compressor pressure ratio

4.5 4.4 4.3 4.2 4.1

t=

3.82

2%

4.5 ηc

=9

9 1.7 17 = Δh % 0% 90 θ1 = =8 ηc ηc

4.0

c,a

3.5

%

η

c

=

70

23 20 21.5 (kg/s) / m Corrected mass rate of flow, 1 θ 01 δ01

Fig. 15.12 Compressor map for t = 3.82 θ03

=

1000 288.15

=

3.47

ΔhT,a θ01

=

ΔhC,a θ01

=

44.9 × 3.47

=

155.8 kJ/kg

Once again, plotting on a portion of the compressor map (Fig. 15.13), one determines that the steady-state operating point is p02 = 4.05 p01 ηC √ m ˙ a1 θ01 δ01

= 90.3% = 20.1 kg/s

Three points of the steady-state operating line have now been determined. The steady-state operating line is shown on the hypothetical compressor map (Fig. 15.11). The exhaust nozzle is choked above a compressor pressure ratio of approximately 3.6. Below this compressor pressure ratio the exhaust nozzle

Component Matching and Performance Evaluation

541

Table 15.2 p02 /p01

ΔhC,a /θ01 (kJ/kg)

ηC

143.6 140.6 137.5 134.3

0.921 0.902 0.882 0.862

5.0

Steady-state operating line t=

3.47

%

p /p 02 01

Compressor pressure ratio

4.1 4.0 3.9 3.8

4.5 4.0 3.5

η

= c

92

5.8 15 c,a = h Δ θ 01

%

η

= c

90

ηc

%

%

0 =8

η

= c

70

20 22.5 25 / (kg/s) m θ δ Corrected mass rate of flow, 1 01 01

Fig. 15.13 Compressor map for t = 3.47 will not be choked. This mean that even though the turbine may be operating choked, the turbine expansion ratio will change as the turbine inlet temperature varies if the steady-state operating point is below a compressor pressure ratio of 3.6 for the engine illustrated in Fig. 15.11. It is important to keep in mind the many simplifying assumptions in the preceding analysis. These included the following: (i) The mass rate of flow through each component is the same. (ii) There is no pressure drop in the combustion chamber or between the turbine exit and exhaust nozzle inlet. (iii) No power is extracted and bearing losses are neglected. (iv) The turbine and exhaust nozzle areas have fixed values. (v) There are constant specific heats with k = 1.4. (vi) There is constant turbine efficiency. (vii) The turbine and exhaust nozzle are choked.

542

Gas Turbines

One should next consider what happens when one or more of these simplifying assumptions is changed. How should the match point for a specified T03 /T01 change if the mass of fuel added is taken into account? In Eq. 15.6, m ˙ g3 /m ˙ a1 > 1.0. Therefore, the constant C1 of Eq. 15.7 will increase, since all other quantities in Eq. 15.6 remain constant. This means that the slope of each of the T03 /T01 lines will increase. Since, the power developed by the turbine must be equal to the power required to drive the compressor, m ˙ a1 ΔhC,a

=

m ˙ g3 ΔhT,a

(15.26)

or Eq. 15.19 becomes ΔhC,a θ01

=

ΔhT,a θ03

m ˙ g3 m ˙ a1

T03 T01

(15.27)

This means that the match point for specified T03 /T01 will shift as illustrated in Fig. 15.14, where the original conservation of mass and energy lines are shown as solid lines, the new ones as dashed lines. t

tan

t=

ns co

Δ h c,a θ1

= constant

Fig. 15.14 Effect of fuel addition on T03 /T01 line How would the match point for a specified T03 /T01 change if there is a 5% pressure drop in the combustion chamber? In Eq. 15.6, p03 /p02 < 1.0. Therefore, since all other quantities in Eq. 15.6 remain constant, the constant C1 , in Eq. 15.7 will increase. Equation 15.19 will not change. Therefore, the match point will shift as illustrated in Fig. 15.15. The preceding two examples illustrate how the conservation of mass and conservation of energy “lines” change when one of the assumptions is changed. The reader now should be able to determine how the match points, and therefore the operating line, will shift if the exhaust nozzle area is increased, what must be changed and how for the gas generator to operate at the same “match point” when an afterburner is in operation. and/or how diffuser water injection or operation with a low calorific value fuel will influence the engine match point.

Component Matching and Performance Evaluation

543

tant

t=

s con

Fig. 15.15 Effect of combustion chamber pressure drop on T03 /T01 line 15.7

GENERAL MATCHING PROCEDURE

The preceding section discussed component matching of a single-spool turbojet with many simplifying assumptions. The general matching procedure must take into account variable specific heats, the actual products of combustion, the fact that air may be extracted at an intermediate stage or at the exit from the compressor, turbine cooling may be used, the mass of fuel added, the pressure drop in the combustion chamber, turbine flow characteristics as a function of rotor speed, the fact that turbine efficiency is not a constant, and so on. It is obvious that the general matching problem is very complicated, yet it must be done to select the best engine design for a given application. A general matching procedure is outlined below for a two-spool turbojet engine. It is assumed that the two-spool turbojet engine has the station numbers shown in Fig. 15.16 and that the compressor, turbine, and nozzle performance characteristics as shown in Fig. 15.17 are known and stored in a high end computer where the matching study is being performed.

Air

Exhaust gases

(0)

(1) (1.5) (2)

0-1 Inlet 2-3 Combustion chamber 4-5 Nozzle

(3) (3.5) (4)

(5)

1-2 Compressor (L.P.C. + H.P.C.) 3-4 Turbine (L.P.T. + H.P.T.)

Fig. 15.16 Schematic diagram of a two-spool turbojet engine The following are the step-by-step procedure to obtain matching points. It is assumed that the operating conditions (altitude and speed of the aircraft) and turbine inlet temperature are known.

Gas Turbines

p

p

02

01.5

544

m 1 θ 01 / δ1

m 1.5

(b) High-pressure compressor

η HPT

m 3 T03 / p03 A 3

(a) Low-pressure compressor

θ 01.5 / δ 01.5

p03.5 / p

p03 / p 03.5

04

(d) High-pressure turbine

η LPT

m 3.5 T03.5 / p03.5A 3.5

(c) High-pressure turbine

p03.5 / p

p03.5 / p

04

04

(f) Low-pressure turbine

m 4 T04 / p A 4 04

(e) Low-pressure turbine

p04 / p amb (g) Exhaust nozzle

Fig. 15.17 Compressor, turbine and nozzle characteristics for a two-spool turbojet engine

Component Matching and Performance Evaluation

545

(i) Knowing the flight condition fix T01 and p01 . √

and m˙ a1δ01θ01 . Since (ii) Assume an LPC operating point (assume pp01.5 01 C ηLP C and N√LP are known from the LPC map [Fig. 15.17(a)], NLP C , θ01 m ˙ a1 , ΔhLP C , m ˙ a1,5 , T01.5a and p01.5 can be calculated. m ˙

√ θ

01.5 02 (iii) Assume an LPC pressure ratio, pp01.5 . Calculate a1,5 then read δ01.5 N HP C from the HPC map [Fig. 15.17(b)] ηHP C , √θ . Calculate NHP C , 01.5 ΔhLP C,a , m ˙ g2 , T02,a and p02 .

(iv) Since T03 is known, determine the fuel-air ratio and combustion chamber pressure drop. Calculate m ˙ g3 and p03 . (v) Assume a HPT expansion ratio. Calculate N√HP T . Determine from the θg3

turbine performance characteristics ηHP T [Fig. 15.17(d)] and [Fig. 15.17(c)]. Calculate ΔhHP T,a and m ˙ g3 .

√ m ˙ g3 T03 p03 A3

(vi) Check to determine if ΔhHP T,a and m ˙ g3 are within a preset tolerance. If not repeat steps 3 through 5 until a match is obtained. (vii) Once the high-pressure spool has been matched, m ˙ g3.5 p03.5 T03.5a and NLP C are known. Assume an LPT expansion ratio, pp03.5 . Deter04 mine ηLP C [Fig. 15.17(f)] and ΔhLP T,a and m ˙ g3.5 .

√ m ˙ g3.5 T03.5 p03.5 A3.5

[Fig. 15.17(e)]. Calculate

(viii) Check to determine if ΔhLP T,a and m ˙ g3 are within a preset tolerance. If not repeat steps 2 through 7 until a match is obtained. (ix) Once the low-pressure spool has been matched m ˙ g4 p04 and T04,a are known. Determine from the exhaust nozzle flow characteristics √ m ˙ T [Fig. 15.17(g)] pg404 A404 . Calculate A4 and compare with the known exhaust nozzle A4 . An engine match exists and the thrust, thrustspecific fuel consumption, and other desired values may be calculated with a preset tolerance. If a match does not exist, repeat 2 through 9 until a match has been achieved. It should be obvious, from the matching procedure described above, that the component matching study for a gas turbine is quite a complex problem. The procedure described is for a two-spool turbojet engine. It should be obvious that the problem of complexity increases as one studies a turbofan engine. Keep in mind that three-spool static pressure-balanced engines are currently being built. A matching procedure for this type of engine is left to the reader. 15.8

TRANSIENT OPERATION

The discussion so far in this chapter has dealt with the stead- state operation of a gas turbine. It is next important to determine what happens when the engine undergoes a power change.

546

Gas Turbines

T 03 = constant Steady state operating line

N / θ01 = const. p

02

e rg

B

e lin

Su

A

m 1 θ 01 δ1 Fig. 15.18 Transient operation Consider the case where a single-spool gas turbine engine, with the steady-state operating line shown in Fig. 15.18, is to be accelerated from point A (low power setting) to point B (high power setting). To change the steady-state operating point from A to B, the mass rate of flow of fuel must be increased. Increasing the mass rate of flow of fuel will initially increase the engine to a higher T03 /T01 line than the initial equilibrium √ value without appreciably changing the√ corrected mass flow rate (m ˙ a1 θ01 /δ01 ) or the corrected rotor speed N/ θ01 ). Next, the engine will start to increase in rotor speed (N ), which will increase the corrected mass rate of flow. The exact path that the engine will follow in accelerating from A to B depends on the design characteristics of the engine components and the manner in which the fuel flow to the combustion chamber is changed. One possible acceleration path between steady-state operating points A and B is shown in Fig. 15.18. Acceleration and deceleration paths will take the compressor to opposite sides of the steady-state operating line. One must be careful during transient operations to schedule the fuel flow change so that the compressor does not surge during the change. Review Questions 15.1 What do you understand by the term ‘matching of components’? 15.2 Does inlet temperature influence the component match? Explain. 15.3 Define match point and explain the effect of exit nozzle area on the match point.

Component Matching and Performance Evaluation

547

15.4 Explain the details of component match with and without afterburner operation. 15.5 Mention the various assumptions in the general matching trends. 15.6 What is an equilibrium diagram? 15.7 Explain with sketches the method of finding the equilibrium points. 15.8 Give a brief account of performance evaluation of a single spool turbojet engine. 15.9 What is meant by an operating line and what are the assumptions made? Explain how to determine the operating line. 15.10 Explain the general matching procedure for a two-spool turbojet engine. Multiple Choice Questions (choose the most appropriate answer) 1. At match point the compressor and turbine are balanced for (a) rotor speed (b) power (c) flow rate (d) all of the above 2. In simplified matching technique (a) bearing losses are neglected (b) compressor efficiency is neglected (c) turbine efficiency is neglected (d) varying mass flow consideration is considered 3. The performance map of a compressor is a funtion of (a) mass flow rate and pressure ratio (b) pressure ratio and temperature ratio (c) pressure ratio and specific heat ratio (d) pressure ratio and work ratio 4. The performance map of a turbine is a funtion of (a) pressure ratio and temperature ratio (b) pressure ratio and specific heat ratio (c) pressure ratio and work ratio (d) mass flow rate and pressure ratio

548

Gas Turbines

5. Simplifying assumption used in matching procedure is that (a) constant specific heat (b) constant turbine efficiency (c) chocked exhaust nozzle (d) all of the above Ans:

1. – (d)

2. – (a)

3. – (a)

4. – (d)

5. – (d)

16 ENVIRONMENTAL CONSIDERATIONS AND APPLICATIONS INTRODUCTION The fourteen preceding chapters have covered the various details of the gas turbine, examined possible gas turbine cycles, considered the various components and factors that influence their performance, and then examined the manner in which these components influence one another to determine the engine operating point. One last area will be considered, this being a brief examination of environmental aspects of gas turbines, viz., current gas turbine air and noise regulations and engine modifications that can be effected to reduce the quantity of air pollutants and noise emitted by a gas turbine engine. Further, typical applications of gas turbines will also be discussed. 16.1

AIR POLLUTION

Most of the early air pollution episodes, such as those that occurred in London in 1873, 1952 and 1956, in the Meuse Valley of Belgium in 1930, or in Donora, Pennsylvania in 1948, were associated with sulphur dioxide and particulate emissions. This was due to atmospheric inversions in highly industrialized areas. The first legislation enacted was the Air Pollution Control Act of 1955 by USA. It was quite narrow in scope. It mainly considered prevention and control of air pollution. Further, it placed the primary responsibility on state and local governments. Further, American Congress, because of worsening conditions in urban areas due to air pollution by mobile sources, directed the surgeon general to conduct a thorough study of motor vehicle exhaust effects on human health. As a result, 1955 Act was amended.

550

Gas Turbines

The two acts were the Air Pollution Control Act Amendments of 1960 and Amendments of 1962. The Clean Air Act of 1963 provided for the development of air quality criteria. The Act encouraged state, regional, and local programmes, for the control and abatement of air pollution. Further, for the first time, provided for federal financial aid for research and technical assistance. The Air Quality Act of 1967 initiated a two-year study on the concept of national emission standards from stationary sources which served as the basis for the 1970 legislative act. The 1967 act also required: ”. . . development and issue to the States such criteria of air quality that may be required for the protection of the public health and welfare.” This provision led to a series of documents for common air pollutants such as hydrocarbons, carbon monoxide, sulphur oxides, nitrogen oxides, and so on. Further, a separate document on each pollutant was prepared regarding the air quality criteria which summarized what science at that time was able to measure of the effects of air pollution on humans and the environment. Also, there was the second document summarizing information on techniques for controlling certain emissions. The 1967 act also required that ”The Secretary shall conduct a full and complete investigation and study on the feasibility and practicability of controlling emissions from jet and piston aircraft engines and of establishing national emission standards with and report to Congress the results of such study and investigation within one year from the date of enactment of the Air Quality Act of 1967, together with his recommendations.” The major provisions of the Clean Air Amendments of 1970 included: (i) Primary responsibility for assuring air quality within the entire geographic area comprising of each state lies with such state. (ii) A requirement that national ambient air quality standards (both primary and secondary) are to be established by the Environmental Protection Agency (EPA). (iii) A requirement that standards of performance for new station be established and implemented. (iv) A requirement that industry was to monitor and maintain emission records and make them available to EPA officials. (v) A requirement establishing, aircraft emission standards. The Clean Air Act Amendments of 1977 included many amendments. The one affecting gas turbines was Sec. 225, which stated: ”Any regulations in effect under this section on date of enactment of the Air Act Amendments of 1977 or proposed or promulgated thereafter, or amendments thereto, with respect to aircraft shall not apply if disapproved by the President, after notice and opportunity for public hearing, on the basis of a finding by the Secretary of Transportation that any such

Environmental Considerations and Applications

551

regulation would create a hazard to aircraft safety. Any such finding shall include a reasonably specific statement of the basis upon which the finding, was made.” 16.2

AIRCRAFT EMISSION STANDARDS

The Clean Air Amendments of 1970 required that national and primary ambient air quality standards were to be established by EPA. These standards were published in the Federal Register in 1971 (1) and set maximum concentration limits for carbon monoxide, hydrocarbons, nitrogen dioxide, sulphur dioxide, particulate and oxidant. The Clean Air Amendments of 1970 also required the establishment of aircraft emission standards. Proposed emission standards were published on December 12, 1972 with the final emission standards and test procedures being published on 1973. The standards published in 1973 have done minor modifications over the last twenty years. It is recommended that any reader needing, the latest standards consult the Code of Federal Regulations. The original and current standards are discussed below. Of the pollutants generated in any combustion process, only carbon monoxide (CO), hydrocarbons (HC), nitrogen oxides (NOx ), and smoke have created the most concern in aircraft gas turbine engines and for which emission standards have been issued. The first step in developing aircraft and aircraft engine standards was deciding to classify the engines. They had to be classified in a manner consistent with their design, performance, and construction, giving consideration to their potential for reducing their emissions and the need to do so. Eight classes were originally defined. The details are listed in Table 16.1. A separate class was selected for turboprop engines because the proposed equivalency between shaft and jet thrust was not considered acceptable over the landing-take-off cycle. The Pratt and Whitney JT3D and JT8D engines were given separate categories in order to be able to require separate schedules of smoke retrofits. Supersonic gas turbine engines were included in a separate category because: (i) Some employ afterburners during take-off and acceleration, because of which the combustion pressure and mixing methods, result in higher hydrocarbon and carbon monoxide emissions. (ii) The cycle pressure ratios for engines used in supersonic aircraft usually operate at lower pressure ratios compared to engines designed for subsonic flight. At low flight speed and at low altitude, these engines do not benefit from the ram effect they experience at high flight speed. (iii) They usually do not a employ high-bypass-ratio turbofan engines.

552

Gas Turbines

Table 16.1 Original Engine Classification System for EPA Standard EPA Class Description TI

All aircraft turbofan and turbojet engines of rated power less than 3636 kg thrust, except engines in Class T5.

T2

All aircraft turbofan and turbojet engines of rated power of 3636 kg thrust or greater, except engine in Class T3, T4, or T5.

T3

All aircraft gas turbine engines of the JT3D model family.

T4

All aircraft gas turbine engines of the JT8D model family.

T5

All aircraft gas turbine engines for propulsion of aircraft designed to operate at supersonic flight speeds.

P1

All aircraft piston engines except radial engines.

P2

All aircraft turboprop engines.

APU

Any engine installed in or on an aircraft exclusive of the propulsion engines.

The current engine classes are slightly different from those listed in Table 16.1. The current classifications are listed in Table 16.21. The next major step before aircraft engine emission standards could be set was to determine the engine operating conditions which will enable determining the air pollutants. The current exhaust emission test selected is designed to measure hydrocarbons, carbon monoxide and carbon dioxide emissions for a simulated aircraft landing-takeoff (LTO). This is based on the time, in each mode during a high-activity period at major airports. The details of the standards, regarding the gas turbine engine test in each of the operating modes are listed in Table 16.3 The values given in Table 16.3 are the percentages of rated power settings on a standard day. The standard day is defined as a day with a temperature of 15◦ C, a pressure of 101325 Pa, and a specific humidity of 0.0 kg H2 O/kg dry air. Emission tests are to be conducted on a warmed-up engine, which has achieved a steady operating temperature. The time in each mode is listed in Table 16.4. Gaseous emission standards for hydrocarbons, carbon monoxide, nitric oxide, and smoke have been set in the 1979 standards. The 1994 standards have details regarding gaseous emission standards for hydrocarbons, smoke exhaust emissions, and fuel venting. The 1994 engine fuel venting emission 1 Code of Federal Regulations, Title 14, Part 34 U.S. Government Printing Office, Washington D.C. Jan.1,1994.

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553

Table 16.2 Current Engine Classification System for EPA Standards EPA Class Description TP

All aircraft with turboprop engines

TF

All turbofan or turbojet aircraft engines except engines of Class T3, T8, and TSS

T3

All aircraft gas turbine engines of JT3D model family

T8

All aircraft gas turbine engines of the JT8D model family

TSS

All aircraft gas turbine engines employed for propulsion of aircraft designed to operate at supersonic flight speeds

Table 16.3 Gas Turbine Engine Power Settings for Emission Measurements Aircraft Operating mode Taxi/idle Take off Climb out Descent Approach

TP 7% rated thrust 100% 90% NA 30%

Engine Class TF, T3, T8 TSS 7% rated 7% rated thrust thrust 100% 100% 85% 65% NA 15% 30% 34%

standards apply to all new aircraft gas turbine engines of classes T3, T8, TF and TSS with rated output equal to or greater than 36 kilonewtons manufactured after January 1, 1974. For all in-use aircraft gas turbine engines of classes T3, T8, TF and TSS with rated output equal to or greater than 36 kilonewtons manufactured after February 1, 1974 also these standards apply. The engine fuel venting emission standards also apply to all new aircraft gas turbine engines of class TF with rated outputs less than 36 kilonewtons, all turboprop engines of class TP manufactured on or after January 1, 1975 and all aircraft gas turbine engines of class TF with rated outputs less than 36 kilonewtons and class TP turboprop engines manufactured after January 1, 1975. The engine fuel venting emission standards require that no fuel be discharged into the atmosphere from any new or in-use gas turbine engines after dates listed above. The purpose of this standard is to eliminate the direct discharge into the atmosphere of fuel drained from the engine fuel nozzle manifolds after the engine has been shut down. The details of current smoke exhaust emission standards are listed in Table 16.5. The effective date, maximum smoke number, engine class, and rated output where the standards apply are also listed in this table.

554

Gas Turbines

Table 16.4 Landing-Take-off Cycle for Aircraft Engine Emission Standards Aircraft Operating Mode Taxi/idle Take off Climb out Descent Approach

TP (min) 26.0 0.5 2.5 NA 4.5

Engine Class TF, T3, T8 TSS (min) (min) 26.0 26.0 0.7 1.2 2.2 2.0 NA 1.2 4.0 2.3

Table 16.5 Smoke Exhaust Emission Standard–New Aircraft Gas Turbine Engines Engine class T8 TF T3 T3, T8, TSS, TF TF

Rated output all ≥ 129 kN all ≥ 26.7 kN ≥ 26.7 kN ≥ 26.7 kN

Manufactured on or after Feb.1, 1974 Jan. 1, 1976 Jan. 1, 1978 Jan. 1, 1984 Jan.19, 1984 Aug. 9, 1985

TP

≥ 1000 kW

Jan. 1, 1984

Maximum smoke number 20 83.6 (RO∗ )−0.274 25 83.6 (RO∗ )−0.274 with max of 50 83.6 (RO∗ )−0.274 with max of 50 187 (RO∗∗ )−0.168

* RO is rated output in kilonewtons ** RO is rated output in kilowatts The 1979 standards are listed in Tables 16.6 and 16.7. In the tables “T” standards are pounds pollutants/1000 lb-thrust hours/cycle and “P” standards are pounds pollutant/1000 hp-hours/cycle. Note at that time there were different standards for newly manufactured and newly certified engines and that there were HC, CO and NO standards. The 1994 gaseous exhaust emission hydrocarbon limits are listed in Table 16.8. rPR in Table 16.8 is the engine rated pressure ratio. It should be noted that the hydrocarbon emission standards are based on mass of pollutant emitted to thrust-hours over the landing take-off cycle that is specified in Table 16.4 and at the power settings that are listed in Table 16.3. Other possible ways of expressing gaseous emissions are (i) Pollutant concentration : This has the advantage of being easy to use but does not provide any guide regarding to the mass of pollutant being emitted by the gas turbine engines. (ii) Ratio of mass of pollutants emitted to mass of fuel consumed : This provides a guide as to how “clean” the combustion system is in the gas turbine engine. However, it does not reveal the air pollution

Environmental Considerations and Applications

555

Table 16.6 Gaseous Emission Standards Applicable to Newly Manufactured Aircraft Turbine Engines Engine class T1 T2, T3, T4 P2 T5

HC 1.6 0.8 4.9 3.9

CO 9.4 4.3 26.8 30.1

NOx 3.7 3.0 12.9 9.0

Table 16.7 Gaseous Emission Standards Applicable to Newly Certified Aircraft Gas Turbine Engines Engine class T2, T3, T4 T5

HC 1.0 1.0

CO 7.8 7.8

NOx 5.0 5.0

Effective date Jan.1, 1983 Jan.1, 1984

effects of the complete engine since different engines have different fuel consumption characteristics. (iii) Total mass of pollutants emitted over the LTO cycle : This would be the most useful one if one is interested in estimating total airport emissions. The one selected by EPA, normalizes emissions over the LTO cycle. 16.3

STATIONARY ENGINE EMISSION STANDARDS

The primary federal environmental laws applicable to power generating gas turbine engines are the Clean Air Act Amendments of 1970, 1977, and 1990. The national and primary ambient air quality standards established by the Clean Air Amendments of 1970 which have been discussed in Section 16.2 are very important when considering standards that apply to power generating gas turbine engines. The current standards are published in the Code of Federal Regulations and will be discussed in this section. Subpart CG-Standards of Performance for stationary Gas Turbines of Part 60-Standards of Performance of New Stationary Sources apply to all gas turbine engines whose heat input at peak-load is equal to or greater than 10.7 gigajoules per hour (GJ/h) for gas turbine engine installations, which commenced construction, modification, or reconstruction after October 3, 1977. Maximum emission limits for the oxides of nitrogen, NOx , and sulphur dioxide SO2 , are specified in the standard based on input energy to the gas turbine unit and intended use; that is, utility or industrial application. A unit is considered by EPA as an industrial gas turbine engine if less than one-third of the potential electric output capacity is sold. A unit is

556

Gas Turbines

Table 16.8 Hydrocarbon Gaseous Exhaust Emission Standards – New Commercial Aircraft Gas Turbine Engines Manufactured on or after January 1, 1984 Engine class T3, T8, TF TSS

Rated output ≥ 26.7 kilonewtons all

Maximum emissions grams/kilonewton rated output 19.6 140 (0.92)rP R

considered as a utility gas turbine engine if more than one-third of the potential electric output capacity is sold through a utility system. Current gas turbine engine New Source Performance Standards (NSPS) are listed, in Table 16.9. The emission limits as listed in Table 16.9 are the NSPS limits when on a dry basis and converted to 15% oxygen. The limits are calculated using the following formulae: (i) For electric utility applications when the heat input is greater than 107.2 GJ/h 14.4 NOx,max = 0.0075 +X (16.1) qr (ii) For all uses where the heat input is between 10.7 and 107.2 GJ or for base loaded non-utility applications where the output is less than 30 MW. 14.4 NOx,max = 0.0150 +X (16.2) qr In Eqs. 16.1 and 16.2 qr is the manufacturer’s rated heat release rate at rated load, kilojoules per watt hour, and X is an allowance factor for fuel-bound nitrogen. The value of qr cannot exceed 14.4 kilojoules per watt hour (kJ/wh). The value of 14.4 assumes that the gas turbine engine has a thermal efficiency of 25%. This means that gas turbine engines with thermal efficiencies higher than 25% will be allowed to emit more NOx than is listed in Table 16.9. No fuel bound allowance is made if the percent by weight nitrogen in the fuel is 0.015% or less. It is recommended that the readers consult the latest standards if they are interested in the fuel-bound nitrogen allowance for fuels with higher nitrogen contents. The sulphur dioxide standards are the same for all gas turbine engines. All are limited to 150 ppmv at 15% oxygen and on a dry basis. All operators shall never burn fuel that contains sulphur in excess of 0.8% by weight. There are several exceptions to the emission limits listed in Table 16.9. Two of these are (i) Stationary gas turbine engines which use water or steam injection for control of NOx emissions are exempt when ice fog is deemed a traffic hazard.

Environmental Considerations and Applications

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Table 16.9 Gas Turbine Engine New Source Performance Standards Input energy (based on lower calorific value of fuel) Greater than 107.2 gigajoules/h

Use Electric utility

Emission limit∗ NOx SO2 (ppmv) (ppmv) 75∗∗ 150

Between 10.7 and 107.2 gigajoules/h

All uses

150∗∗∗

150

Base load equal to or less than 30 megawatts

Nonutility

150∗∗∗

150

Less than 10.7 gigajoules/h

All uses

None

150

Greater than 30 megawatts

Nonutility

None

150

Regenerative cycle, input less than 107.2 gigajoules/h

All uses

None

150

All input levels

Emergency, Fire fighting, Military

None

150

* All values corrected to 15% oxygen, dry basis. ** Use Eq. 16.1 *** Use Eq. 16.2 (ii) Stationary gas turbine engines with a peak load heat input equal to or greater than 10.7 GJ/h and less than 107.2 GJ/h that commenced construction prior to October 3, 1982. The emission limits as listed in Table 16.9 are the maximum levels for new engines. Levels can be set at lower levels to prevent significant deterioration in areas that meet National Ambient Air Quality Standards (NAAQS) or in the areas that do not meet the NAAQS air quality levels. 16.4

NOx FORMATION

It might have been noted in the previous section that the oxides of nitrogen, are the predominant emissions from stationary gas turbine engines and the one that is controlled by the standards. The most prevalent NOx emissions are nitric oxide, NO, and nitrogen dioxide, NO2 . Nitric oxide is the one mainly formed in the combustion chamber. Factors that influence the amount of NO formed are (i) peak temperature,

558

Gas Turbines

(ii) percentage of excess air, (iii) pressure, (iv) residence time at peak temperature, and (v) fuel bound nitrogen. The peak temperature is attained when the fuel is burned with the stoichiometric (chemically correct) amount of air. Higher the temperature of the air at the inlet to the combustion chamber, higher the resulting equilibrium adiabatic flame temperature. Burning the fuel with excess air lowers the maximum temperature but increases the availability of oxygen and nitrogen in the products of combustion. It is known that for a fixed air supply temperature and combustion chamber pressure, the amount of NO formed for equilibrium conditions increases form 0% excess air to 30% excess air, then starts to decrease even though the adiabatic equilibrium flame temperature decreases continuously. It is a known fact that increasing the combustion temperature, pressure, increases the equilibrium adiabatic flame temperature but decreases the amount of NO formed. The preceding discussion assumes that equilibrium has been reached. The next important thing is to determine the rate at which the products will reach equilibrium. The basic mechanism presently used to predict the formation of NO had its origin in the work of Zeldovich and coworkers around 1946. The reader is referred to Wark and Warner2 for a development of one mechanism that can be used to determine the rate at which NO is formed. The equation developed by Wark and Warner is c+1

(1 − Y )

c−1

(1 + Y )

e−Mt

=

(16.3)

where Y

=

c =

M In the above [NO] = [NO]e = xN2 = xO2 = T = p =

=

[N O] [N O]e

(16.4) 0.5 −7750/T

2.1 × 104 [xN2 ]

e

0.5

T [xO2 ]

5.4 × 1015 (p)

0.5

0.5 −58330/T

[xN2 ] T

e

(16.5)

(16.6)

equations concentration of NO concentration of NO at equilibrium mole fraction of N2 in the products mole fraction of O2 in the products temperature, K pressure, atm

2 Wark, W. and Warner, C., Air Pollution, Its Origin and Control, EIP A DunDonnelley Publisher, New York, NY, 1976.

Environmental Considerations and Applications

16.5

559

NOx REDUCTION IN STATIONARY ENGINES

It has been shown in Section 16.4 that the higher the temperature and longer the gases are at that temperature, more nitric oxide is formed. As shown by the emission limits in Table 16.9, NOx is the main pollutant from stationary gas turbine engines. The amount of SO2 emitted is limited by the amount of sulphur in the fuel since this is the only source of sulphur. Prior to NOx emission controls, gas turbine engine combustion chambers were designed so that the fuel-air ratio in the primary zone was approximately the stoichiometric value; that is, the percent excess air in the primary zone was 0%. This resulted in maximum temperature. The maximum temperature can be reduced by designing the combustion chamber so that the primary zone either operates fuel rich (insufficient air for complete combustion) or fuel lean (excess air). Both of these conditions can result in increased smoke (fuel rich) or increased carbon monoxide and total hydrocarbon emissions (fuel lean). Several methods can be used to reduce NOx emissions such as water or steam injection or staged combustion or selective catalytic reduction. The most commonly used method of controlling NOx emissions is with water or steam injection into the primary zone of the combustion chamber. The water (or steam) injected acts as a heat sink, resulting in a lower maximum temperature, thereby reducing the amount of NOx formed. The rate at which water is injected is approximately 50% of the fuel flow. Steam rates are usually 100–200% of the fuel flow. An advantage of this method is that it increases the output from the power turbine due to the increased flow through the gas generator and power turbines and the higher specific heat of water. Disadvantages of using water or steam injection are (i) need of a constant supply of pure water (ii) possibility of increase in CO emissions (iii) pressure oscillations in combustion chamber especially with water injection (iv) penalty on heat release rate The following results have been reported in literature when water and steam injection have been tried. (i) An NOx level of 75 ppmvd for an oil-fired simple cycle with water flow which is 50% of the fuel flow. Output was increased 3% when compared with no water injection, and the unit had a heat release rate penalty of 1.8%. (ii) An NOx level of 42 ppmvd for a natural gas fired simple cycle with water flow equal to the fuel flow. Output increased by 5% and the heat release rate penalty was 3%.

560

Gas Turbines

(iii) An NOx level of 42 ppmvd for a natural gas fired combined cycle with steam flow of 1.4 times the fuel flow. Output increased 5%, and the heat release rate penalty was 2%. (iv) An NOx level of 25 ppmvd for a natural gas fired simple cycle with GE’s Quiet combustion chamber. Water injection was used with water flow 1.2 times fuel flow. The output increased by 6%, and the heat release rate penalty was 4%. (v) An NOx level of 25 ppmvd for a natural gas fired combined cycle with steam flow 1.3 times fuel flow. Output increased by 5.5% with a heat release rate penalty of 3%. Staged combustion is currently being tested by a number of manufacturers. It provides a way of achieving NO emission levels of 25 ppmvd or less at 15% oxygen without using water or steam injection. Most of the systems being tested use a two-stage premixed combustor for use with natural gas. The resulting mixture is lean so the amount of NOx is low. Selective catalytic reduction involves injecting ammonia into the gas turbine engine exhaust stream. The exhaust gases then pass over a catalyst where the NOx reacts with the ammonia, (NH3 ), oxygen, (O2 ) and nitrogen, (N2 ) to form water, (H2 O), and nitrogen, (N2 ). When combined with water or steam injection, it is reported that NOx levels of 10 ppm or less can be achieved. One major disadvantage is that the reaction is very much temperature dependent. For a vanadium pentoxide type catalyst, the exhaust gas temperature range for best operation is 600–750 ◦ F. For this reason, the selective catalytic reduction method for reducing NOx emissions is limited to combined cycles only. 16.6

NOISE

Since the introduction of jet-powered commercial airplanes, aircraft noise has been of concern. A continuous effort has been made to develop the technology to design a quiet gas turbine engine. The main concern of the public about aircraft noise is the large number of major airports around the world that have noise restrictions in the form of curfews, night-time limitations, flight restrictions, and/or preferred run ways or routings that take aircraft over water or sparsely populated areas. The major problem with aircraft noise, in terms of number of people exposed and the frequency with which they are exposed, occurs in the vicinity of airports. Unlike emissions, whose sole source is the combustion chamber of the gas turbine engine, noise has many sources. The two major sources of noise are (i) Propulsion system noise (ii) Aircraft noise other than propulsion system noise, including sonic boom, flap noise, etc.,

Environmental Considerations and Applications

561

We will consider only propulsion system noise. Propulsion system noise may be classified into two categories, viz., (i) externally generated noise associated with the exhaust gases from the propulsion system, the propeller of a turboprop powered aircraft, and (ii) internally generated noise associated with the rotating machinery and the combustion process. An important distinction between internally generated and externally generated noise is that internally generated noise can be suppressed, whereas externally generated noise cannot be suppressed. The sources of noise will be divided into three main groups: (i) Jet or exhaust noise (ii) Fan noise (iii) Core noise Each of these groups is discussed below. Jet noise results from the mixing of the high-velocity exhaust stream with the ambient air. A considerable amount of turbulence is generated when these two streams at different velocities mix. With the increase in intensity of the turbulence, the noise starts increasing. Researchers have found that the magnitude of the jet noise increases as the eighth power of the velocity. The dominating noise of the early turbojet engines, the jet roar, was generated behind the jet engine exhaust nozzles where the high exhaust stream mixes with the ambient air. With the introduction of the nonmixed turbofan engine, there were two exhaust streams, therefore two sources of external noise. One source was the turbulent mixing of the fan exhaust steam with the ambient air. The other source was the turbulent mixing of the core exhaust stream with the fan exhaust stream and the ambient air. When a turbojet engine is converted to a turbofan engine with the same core pressure ratio and turbine inlet temperature, the core velocity decreases. The amount of decrease and difference in velocity between the core and fan exhaust streams depends on the fan pressure ratio and bypass ratio. Fan noise is one of the major, if not the predominant, sources of noise in a high-bypass-ratio turbofan engine. Fan noise has different characteristics depending on whether the tip speed of the fan rotor blades is subsonic or supersonic. Fan noise usually is separated into three categories: (i) Broad-band noise, which is essentially the noise generated from the turbulence in the air and by the air load fluctuations as it passes across the blade (rotor and stator) surfaces. (ii) Discrete tone noise, which is noise generated by the fluctuating pressures generated by the interaction between the rotor blades and the

562

Gas Turbines

stationary blades. The frequency of this noise may be predicted by the rotor rotational speed. (iii) Multiple-tone noise, which is associated with the shock waves on the rotor blades caused by supersonic relative flow over the blades. As higher-bypass-ratio engines are built, the exhaust velocity and therefore the jet noise are reduced. Under these conditions, the turbofan core noise becomes more important. Core noise consists of compressor noise, combustion noise, and turbine noise. Compressor and turbine noise are similar to fan noise, resulting mainly from the interaction between the rotating and stationary blades. Combustion noise results from the turbulence generated by the burning of the fuel. 16.7

NOISE STANDARDS

Increased use of jet-powered commercial aircraft, along with a growing concern for the quality of the environment, has resulted in considerable emphasis on the reduction of noise from gas turbine engines since the mid 1960s. The initial national noise regulations were those in Public Law of USA in 1969, which are commonly known as FAR 36. These were modified in 1973. It is suggested that the reader consult the Code of Federal Regulations under Title 14, Part 36, for the latest regulations. Below is an abridged version of these regulations. The noise regulations depend on the number of engines, when application is made, and other engine design conditions. For subsonic aircraft with turbofan engines with a bypass ratio of 2 or more, the current noise limits are as follows: (i) If application was made before January 1, 1967, it had to meet stage 2 noise limits or be reduced to the lowest levels economically and technologically possible. (ii) If application was made on or after January 1, 1967, and before November 5, 1975, it had to meet stage 2 noise limits. (iii) If application was made on or after November 5, 1975, it must meet stage 3 noise levels. For aircraft with turbofan engines with a bypass ratio less than 2, the current noise limits are as follows: (i) If application was made before December 1, 1969, the noise level had to be the lowest level reasonably obtainable through use of procedures and information developed for the flight crew. (ii) If application was made on or after December 1, 1969, and before November 5, 1975, the noise levels has to be no greater than stage 2 noise limits.

Environmental Considerations and Applications

563

(iii) If application is made after November 5, 1975, it must meet stage 3 noise limits. For the Concorde airplane, the noise levels must be reduced to the lowest levels that are economically reasonable, technologically practicable, and appropriate for the Concorde design. Section A36 of the code prescribes in detail the conditions under which the aircraft noise certification tests must be conducted, the measurement procedures that must be used to measure the noise levels, and the corrections that must be applied for variations in atmospheric conditions or flight path. The measurement points are as follows: (i) For take-off, at a point 7000 m from the start of the take-off roll on the extended centerline of the runaway. A typical profile is shown in Figure 16.1. The take-off profile is defined by five parameters, including length of take-off roll, climb angle, and thrust change points.

Microphone

Start of take-off roll

7000 m

Fig. 16.1 Typical take-off profile for noise measurements

(ii) For approach, measurements are taken at a point 2000 m from the touchdown point on the extended centerline of the runaway. A typical profile is shown in Figure 16.2. (iii) For sideline, measurements are made at a point parallel to and 450 meters from the extended centerline of the runaway where the noise level after liftoff is the greatest. The exception is that for Stage 1 and 2 compliance for aircraft powered by more than three turbojet engines, measurements are made at 0.35 nautical miles from the centerline. Stage 2 take-off, sideline, and approach noise limits are shown in Figure 16.3. The Stage 2 noise limits are independent of the number of engines. Stage 3 take-off noise limits are shown in Figure 16.4. Note that for maximum weights of about 390 ton, the limits are constant but depend on the number of engines. In all cases the allowable noise limits are reduced by 4 EPNdB for each halving of the 390 ton maximum weight until a maximum noise limit of 89 EPNdB is reached. This occurs for maximum weights of

Gas Turbines

Microphone

Threshold

2200 m

Effective perceived noise level, EPNdB

Fig. 16.2 Typical approach profile for noise measurements

110 105 100 95 90 10

50 100 500 1000 2000 Maximum aircraft weight, 500 kg

Fig. 16.3 Stage 2 take-off, sideline and approach noise limits

Effective perceived noise level EPNdB

564

110 105 s

100

re mo

95 90

ine

ng

e n3

tha

s ine es gin eng 3 n tha

n 3e

er

few

10

50

500

2000

Maximum aircraft weight, 500 kg

Fig. 16.4 Stage 3 take-off noise limits

Environmental Considerations and Applications

565

Effective perceived noise level, EPNdB

20 ton for aircraft with more than three engines, 28 ton for aircraft with three engines, and 48 ton for aircraft with fewer than three engines. Stage 3 sideline noise limits are shown in Figure 16.5. The allowable sideline noise limits are independent of the number of engines.

110 105 100 95 90

10

50 500 2000 Maximum aircraft weight, 500 kg

Fig. 16.5 Stage 3 sideline noise limits

Effective perceived noise level, EPNdB

Stage 3 approach noise limits are shown in Figure 16.6. These noise limits, like the sideline limits, are independent of the number of engines.

110 105 100 95 90 10

50

500

2000

Maximum aircraft weight, 500 kg

Fig. 16.6 Stage 3 approach noise limits

16.8

NOISE REDUCTION

Noise sources and noise standards were discussed in the last two preceding sections. It is now important to examine what has or can be done to reduce the noise emitted from a gas turbine engine. One must keep in mind that design changes in the engine that reduce noise emissions usually add weight, length and cost. Quite often, features desirable in reducing noise are in conflict with the best aerodynamic design.

566

Gas Turbines

External noise is caused by the mixing of the exhaust stream with the ambient air, with the noise level increasing by approximately the eighth power of velocity. Noise suppressors were used in early commercial airplanes powered by turbojet engines. Shortly thereafter, low-bypass-ratio turbofan (BPR ≡ 1) were installed on commercial aircraft, which slightly lowered the exhaust stream velocity. Most of these aircraft were certified and entered service long before the current noise regulations. For these aircraft, exhaust (external) noise dominates. The only effective way to reduce the noise level of these aircraft is to decrease the exhaust stream velocity. This, of course, can be done by replacing the engine on the aircraft with a turbofan engine with a higher bypass ratio. An effective way to reduce the maximum exhaust velocity, and therefore the exhaust jet noise, on mixed (pressure-balanced) turbofan engines is to use an exhaust gas mixer behind the turbine. An exhaust-gas mixer mixes the high-velocity core exhaust gas stream with the lower-velocity cold stream that has passed through the fan but bypassed the combustion chamber and turbine. It reduces the maximum exhaust stream velocity and eliminates the external turbulent mixing of these two streams. However, it does add weight to the engine. Turbofan engines with bypass ratios of 5 or higher have been installed on many of the commercial aircraft in the last twenty years. With highbypass-ratio turbofan engines, the exhaust velocity, and therefore the exhaust noise, is reduced considerably. With the reduction in the exhaust noise, additional internal noise is heard and becomes a problem. Internal noise, unlike external noise, may either be eliminated or confined. One way to confine the noise is to install an acoustical liner at the inlet to the engine and/or in the exhaust duct. Conventional acoustical treatment is quite effective and is currently used but does add weight to the airplane. Other ways to reduce internal noise and the disadvantages are to: (i) Decrease the fan tip speed. The lowest-noise fans are generally subsonic. Low-speed fans usually require more compressor stages, which leads to a heavier and more costly engine. (ii) Increase the spacing between the rotor and stator. Noise is reduced as the spacing increases, but large spacing tend to increase the length and weight of the engine. (iii) Eliminate the inlet guide vanes, which eliminates one of the rotorstator interactions. (iv) Change the number of rotor and/or stator blades. Changing the ratio changes the frequency of the noise. If the number of blades is decreased, it may affect the turbomachinery performance.

Environmental Considerations and Applications

16.9

567

ASSESSMENT OF THE GAS TURBINE

The most outstanding characteristic of the gas turbine is that of flexibility of arrangement, with respect to choice of cycle, choice of plant arrangement, type of component and type of fuel. This is borne out by actual developments over the past many years, which have shown gas turbines in power output from a few kilowatts to outputs of 30,000 kW in straightforward shaft units and of even greater equivalent thrust horsepower in jet engines. The most outstanding physical characteristics are its small size and low weight per unit power output. The values are much lower than those of other prime movers. Within limits the gas turbines can burn a wide range of fuels with little modification and it is possible to develop a gas turbine which can burn any fuel from a gas to residual oils. It is these factors above all which have led to the use of the gas turbines as it stands at the present time, even though the theoretical analysis has shown that the efficiency, particularly at part load, and the range of satisfactory operation are not very attractive except for more complex arrangements. There are, however, other features which lead to the choice of the gas turbine when an overall balance sheet is made of operation and which in some particular applications are overriding. Below, an attempt is made to summarize the potentialities of the gas turbine, but it should be noted that the relative values of a particular characteristic vary with application. (i) High Specific Output The power output on weight and volume basis is quite high except the most complex cycle. (ii) Relatively Lower Efficiency than Comparable Primemovers Diesel and petrol engines are usually more efficient than gas turbine in the lower output range. This is true both at the design point and, more particularly at part-load. In the higher output range where the diesel engine becomes unacceptable due to size, weight and number of separate unit required, the gas turbine has to compete with the steam turbine. The efficiency of the steam turbine is highly variable depending upon the size and upon the complexity of its cycle (condensing or non-condensing, reheat, etc.,), but it is comparatively high. A closed-cycle gas turbine can compete with all but the largest and most complex steam plants. (iii) High Air Rate This may pose problems in the air intake and disposal of exhaust gas particularly for mobile units (marine and automotive). If advantage can be taken of large volume of exhaust gas at relatively high temperature for use as an auxiliary heating medium, then the high air-rate is useful and the overall efficiency of the complete thermal plant is high. No such problem with respect to air and gas ducts occurs in the closed-cycle plant. (iv) Low Maintenance Cost Here the gas turbines are comparable to the steam turbines in lubricating oil consumption, and is very superior to

568

Gas Turbines

the reciprocating IC engines. With respect to inspection and repair or replacement of components, the gas turbine has advantage, as it is a much simpler plant. It has many fewer accessories than either a steam power plant, a diesel or a petrol engine. It should also be noted that because gas turbine cycle processes are carried out by discrete components, modification is usually easier than in other plants. Thus improved components, such as compressors, combustors or turbines, can be introduced with minimum disturbance to the rest of the plant. (v) Freedom from Auxiliary Services For most of the applications, water supply is not required, which is considered as an advantage with respect to location. Ignition system is simple as it is necessary only for starting. However, for aircraft units, the control and electrical systems are complex, owing to the large variation of ambient conditions. However, this is true for any power plant for an aircraft. The fuel control system for the gas turbine is no more complicated than the carburettor in its developed form for aero-engines. (vi) Low Installation Cost Because of its high output per unit weight and volume, both the building for housing the plant and actual floor foundation required are much smaller and simpler than for other power plants of similar output. This may be an important point from economic considerations. (vii) Starting of a Gas Turbine Starting has both favourable and unfavourable aspects. On the favourable side is that except for very large units, it can be started and put on load in very short time. Usually, the time is much shorter than for the alternative prime mover for the particular application. Turbojets of enormous power are ready for aircraft take-off in a matter of seconds. Gas turbines for large output for electrical generation may require minutes of warm-up time. It is usually much less than the corresponding steam turbine, even assuming steam available at throttle. This characteristic is due to the light construction, which requires the minimum of large metal masses of great thermal capacity. On the unfavourable side is the power required from rest to selfsustaining or idling speed. Not only component efficiencies are very poor at low speed, but the turbine temperature must be limited for the sake of metal properties and to avoid compressor stalling. It is necessary to accelerate the unit after combustion is initiated up to a speed at which it is self-driving at a safe turbine temperature. This speed is usually about one-third of the design speed. The electric motors of some fair size are required which will sustain high current for comparatively larger time. Cartridge or ‘rocket’ starters are used successfully for aircraft turbines when self contained starting means are required otherwise auxiliary electric generators are needed. Again on credit side, starting at low ambient temperatures is usually easier than with reciprocating engines, because of the lack of contact resistance.

Environmental Considerations and Applications

569

(viii) Effect of Ambient Conditions The effect of variation of pressure is not serious and comparable to that on reciprocating engines. The effect of temperature variation is very marked. The plant must be designed so that a certain minimum output is maintained at the highest intake temperature, thus probably having excess capacity for major part of its operation. However, there are many applications when increased output is welcome at low inlet temperatures. (ix) Multifuel Capability A gas turbine can be designed to utilize gaseous fuels or distillate of liquid hydrocarbons with ease. The problem of combustion of heavier and cheaper fuels is not completely solved. If the problem of deposit and corrosion due to ash content can be solved, then there is no inherent reason why any liquid fuel cannot be used. (x) Capital Cost In this case no generalization can be made. Because it is largely number of units which determine the first cost and the industrial production is not yet sufficiently great to have warranted the standardized and mass production methods like other prime movers. For small units, the cost of gas turbines is so much greater than that of reciprocating engines. Therefore, it may seem to be improbable for it ever to compete. 16.10

TYPICAL APPLICATIONS OF GAS TURBINES

Research efforts and investigations have gone into various aspects of gas turbines in the constant endeavour to improve the performance, increase the overall efficiencies, extend the applications and, in general, to overcome the practical problems and difficulties associated with these power units. Majority of these aspects have already been dealt in the last 14 chapters. Finally in the following sections, an account is given about typical applications of gas turbine to locomotive, aircraft, marine and stationary plant. The possible application of gas turbine to automobiles is also discussed. 16.11

THE SMALL GAS TURBINE APPLICATIONS

About sixty years ago it was clear that in the aircraft field the gas turbine has established itself for military purposes, and even on the commercial side the reciprocating engine was fast becoming obsolete. It was, therefore, natural to examine how far the turbine invasion could be expected to penetrate into the automobile and small industrial outputs, say 200 kW and lower. Even at first sight, this area does not appear promising because of the advanced state of development of petrol and diesel engines in this range, together with their relatively very low capital cost due to mass production for automotive use. A gas turbine of small size is likely to be less efficient than that of large ones. It is mainly due to Reynolds number effects and difficulty in maintaining the same relative accuracy of part dimensions and of small clearances. Nevertheless, numerically there are probably many more gas turbines of 200 kW and less in use than those of larger output.

570

Gas Turbines

It is true that their use is predominantly in the military field. However, their application in shaft-power units, a field in which existing reciprocating engines dominate is not yet predominant. 16.12

ELECTRIC POWER GENERATION APPLICATIONS

The efficiency of the gas turbine is not high enough to allow it to compete, with the steam turbine for continuous power generation. However, its other characteristics have led to its use in number of instances. Many gas turbine installations have been built since 1947, for standby and peak-load purposes. These vary from about 4,000 kW to 24,000 kW output, the 10,000 kW to 15,000 kW installations being the most favoured. In January 1948, fifteen Brown Boveri plants, with a total capacity of 92,000 kW, were in operation for power generation purposes in India. Most of the existing gas turbines are operated only on oil or gas. The 10,000 kW gas turbine, working with a normal air inlet at 20◦ C and a cheap fuel oil, is considered superior to the steam turbine, but above this output the latter scores on the ground of efficiency. If the exhaust gases from the gas turbine power station can be used for a waste heat boiler or a water heater, it should prove a little more advantageous in its field, provided that a cheap fuel oil or coal can be burned. Many open-cycle plants are being used for electrical generation, a number of them being only simple cycle plants with efficiencies of 18–22%. The reasons are many. One of the most important reasons being that the location where water is either non-existent or very sparse. Thus several are in use, in desert areas, notably in the Middle East. Many of the installations are in the oil-fields or at oil refineries. In such installations natural or process gas is either ‘free’ or relatively cheap. This provides the best possible fuel for trouble-free combustion and compensating for low efficiency. Some generating units have heat exchangers, thus yielding higher efficiencies. Gas turbines are particularly useful for peak-load or stand-by units. They are self-contained, require no additional steam capacity. They can be put on load with the minimum of delay. They occupy less floor space than other plants for the same capacity. They can also increase the station capacity beyond their own nominal output, because the exhaust gas can be used for air or water preheating, thus raising the existing boiler capacity. 16.13

MARINE APPLICATION

The application of the gas turbine to marine use has been slower than originally anticipated. The slow development in this field is natural on account of number of ships being built is less. The high capital cost of single unit is quite high. The great loss involved in breakdown of equipment above and beyond the direct repair cost involved also inhibit marine applications. The relatively high fuel consumption of simple gas turbines, coupled with the

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difficulties associated with combustion of cheaper fuels, has counteracted expectations. Nevertheless, there are apparent advantages and these are gradually being demonstrated in test vessels. The major advantage once again is low weight and compactness of the gas turbine. Engine rooms of all vessels are always crowded, owing to the importance of each tonne and cubic metre of capacity for pay load. Gas turbine can ‘offer’ either more room or more power in the same space. Other attractive features are the low maintenance costs and the ability to replace individual components with minimum effort and loss of time. There is considerable difference in the requirements of mercantile and naval vessels. The former require a power plant which operates at full speed at most of the time, reduced speed being necessary only during severe weather and for maneuvering in close quarters. A naval vessel, on the other hand, spends a large portion of its life at relatively low speed, requiring top speed only in emergency. Thus part-load efficiency, while highly desirable, is not so important for mercantile marine vessels. It would seem that for the former use, a very high degree of reliability while in latter case, it is high part-load fuel consumption which is more important. The gas turbine also has the advantages of availability and absence of transmitted vibration. The high air-rate poses some problems of intake and exhaust ducts, which must be of considerable size for the open-cycle. On the other hand, closed-cycle seems very well suited to marine propulsion. Another problem lies in the provision of power for astern movement. It is not possible to utilize a condenser vacuum as does the steam turbine installation. A separate reversing gas turbine leads to windage and disc friction losses. Reversing gear mechanism with torque converter is possible, though difficult for high power. Electric propulsion, though readily reversible, adds more cost and maintenance. The reversible-pitch propeller has not yet been extensively used. For naval gas turbines, one solution for part-load efficiency is the use of the closed-cycle, and another is the use of multiple units. For the later, the long-term low output can be supplied by a relatively small unit with good efficiency. Several other simple cycle units can be made to standby which can be coupled immediately to the main shaft when required for high speed operation. For small naval craft, such as motor torpedo-boats or air rescue, boats, the light weight, high-output gas turbine is an excellent proposition. It enables much higher speeds to be attained than would be possible with other types of prime movers. The first marine installation of a gas turbine was successful and has been followed by several others in many countries. 16.14

GAS PUMPING APPLICATIONS

Gas turbines coupled to centrifugal compressors for pumping natural gas have proved to be of one of the most successful uses. The major pumping application is in cross-country pipe-lines, in which booster stations are required at frequent intervals. Such locations are often in isolated regions, in which water is difficult to supply. Furthermore, the gas turbine speed is

572

Gas Turbines

perfectly suited to the centrifugal compressor for the natural gas. The low maintenance and low installation cost make it considerably more attractive than a reciprocating gas or oil engine. Again, the available fuel, that in the pipe line itself, is very suitable. The majority of such units are simple cycle, single-shaft units, although for some uses, a free turbine is advantageous. The addition of a heat exchanger improves efficiency at the expense of first cost. 16.15

LOCOMOTIVES APPLICATIONS

With one outstanding exception, the gas-turbine has not to-date proved popular for locomotive use, although certain apparent advantages led to earlier prediction that this would indeed be so. Almost any power plant would be an advantage over the reciprocating steam locomotive. In almost all countries the universal replacement has been the diesel engine. In India, steam locomotives are being replaced by diesel locomotives and by electric locomotives on all trunk routes. But due to the oil shortage, felt severely after 1973, Government of India is revising its policy of switching over completely to diesel traction. It is quite likely that all the trunk routes may be completely electrified and central power stations based on coal may be built all over the country for meeting the severe oil shortage. The gas turbine has advantages of great power in a small volume and of immediate availability. The major disadvantage is low efficiency, even with a heat exchanger, which cannot be of high effectiveness without having a large space. The gas turbine is also not suited to short journeys, because of its poor part-load efficiency. Thus in India, the gas turbine locomotive does not seem likely to prosper. In USA, on the other hand, loads are higher and distances are greater. Diesel power requires electrical traction and it is possible for the gas turbine to use a direct mechanical drive, thus eliminating some initial cost and complication. Diesel engines require considerably more costly maintenance than gas turbines and the reduction of this item would allow the gas turbine to be considered very favourably if, at the same time, the fuel cost could be made comparable. Here the solution of the problem of heavy fuel combustion would be important or, alternatively, the introduction of coal burning. In the USA, development of a direct coal-burning unit has been developed using pulverized fuel. The major problem is of the erosive effect on the turbine of the inevitable large quantity of ash. In Canada, effort has been directed to the exhaust-heated cycle, eliminating the turbine erosion, but requiring a rather large heat exchanger. 16.16

AUTOMOTIVE APPLICATIONS

Ever since the gas turbine came to public notice, one of the most often asked questions has been its adaptability for automotive use. Its apparent advantages are again weight and volume, simplicity (simple ignition system and absence of water cooling), good starting torque characteristics (with

Environmental Considerations and Applications

573

a power turbine), and absence of vibrations. Against these assets there are disadvantages like poor part-load fuel consumption and necessity for a reduction gear of large ratio. In spite of the ultimate advantages of lightness, simplicity and compactness, absence of necessity for a cooling system and negligible oil consumption, the use of gas turbines for road transport is much less favourable. Even then, most of the major automobile manufacturers in Europe and the USA have engaged in development on suitable gas turbines, from 100 to 250 kW. Considerations of efficiency and load control requirements rule out the simple compressor-turbine coupled arrangement with power take-off from the turbine shaft. Apart from this, the problem of mechanical transmission will be difficult. For automobile application it becomes necessary to use either electrical transmission which is expensive and heavy, or a separate power turbine independent of turbocompressor shaft. This power turbine is geared to the road wheels, while the compressor unit can run upto the speed with the car stationary. The exhaust from the compressor turbine would thus develop a fluid drive effect on the power turbine and exert on it a starting torque. Low thermal efficiency remains the greatest hurdle in the development of a small gas turbine for an automobile. Unless a heat exchanger is used, the specific fuel consumption will probably be twice that of the average petrol engine and three times that of the diesel. While the commercial user will assess lower capital and maintenance costs a set-off against higher fuel consumption and cost, this aspect will appeal less to the private motorist, whose accounts rarely include depreciation and interest. 16.17

AIRCRAFT APPLICATIONS

Nearly all military aircraft are powered by turbojets, from trainers to latest supersonic fighters. The age of the turbojet and turboprop commercial airliner has long been started. The turbojet is satisfactory up to a flight Mach number of 2.5, above which the ram-pressure developed leads to diminishing returns. For military use, it would appear that the area of turbojet is about two decades old, with rockets already taking over. 16.18

PROCESS APPLICATIONS

By process applications, we mean, those in which the major purpose of the gas turbine is not supplying shaft power, although some of the power extraction is possible, but in supplying compressed air or gas for physicochemical processes. The air may be bled off from the compressor for some process use, this representing the excess power of the turbine over the compressor. Alternatively, all the compressed air may be passed, through a ‘reactor’ of some nature, in which it is required to perform a chemical or physical operation, the air then passing to, the turbine, with possibly some

574

Gas Turbines

fuel being required to raise its temperature sufficiently for an energy balance. In other cases, a normal combustor with outside fuel supply is used in the ordinary way, however, the gases are not completely expanded in the turbine, passing to the process at elevated temperature and pressure. This last case is analogous to the turbojet, but with the gases used for chemical process rather than for propulsion. There are many such different processes in the petrochemical field, catalytic cracking, oxygen production and so forth. Included in this category are many different ways of utilizing the gas turbine as an auxiliary component in conjunction with steam cycle. The possible arrangements are many and will undoubtedly be tried out in future. Perhaps one factor militates against their use in new steam power installations is that the latter are now usually of very large output and of high efficiency, utilizing elevated temperatures and pressures. The latter are now quite, often in the supercritical region and represent a considerable advance in technology, so that the inclusion of another new element, the gas turbine is not required at the present time.

16.19

ADDITIONAL FEATURES OF GAS TURBINE ENGINES

Having known a power unit which functionally competes with the petrol and diesel engines, let us look on other features which characterize the gas turbine.

16.19.1

Exhaust

This aspect is of the greatest importance and will undoubtedly play a crucial part in the general acceptance and introduction of gas turbine for road Vehicles. In USA, particularly at present, there is growing concern over the quality of vehicle exhaust gases and newer and newer legalizations are introduced which makes it obligatory for any exhaust system to control such emissions as carbon monoxide, oxides of nitrogen and unburnt hydrocarbons. The turbine exhaust, due to high efficiency possible with its combustion system and lack of flame-chilling, can meet these stringent demands in the near future.

16.19.2

Easy Starting

Due to very low compression ratio during the starting cycle the main resistance to spinning over is the bearing drag. These can be very low and hence the gas turbine is noted for its ability to start reliably at very low temperatures without any additional starting aids. Whilst this is essential for military duties it is also one of the most useful attribute for general operation. Starting at ambient temperatures of −30 ◦ C are quite common and −60 ◦ C has also been achieved in some conventional small gas turbines.

Environmental Considerations and Applications

16.19.3

575

Smoothness

This applies in two ways : (i) in the lack of reciprocating masses which cannot be fully balanced, and (ii) in the two shaft characteristic preventing any torque ‘snatch’ on acceleration or deceleration. The lack of any physical vibration makes the operation of any vehicle less fatiguing and when the comparison is made with diesel trucks this is an important factor in reducing driver fatigue. The excellent torque characteristic permits fewer gear changes, all of which will be very smooth as in the best standard of automobile automatic transmissions. This all adds upto a general driving smoothness which probably has to be experienced to be believed. 16.19.4

Cooling and Heating Aspects

In piston engines it is imperative that cooling is arranged around the combustion areas to avoid excessive metal temperatures, and this plus the general heat to oil by churning and from friction, requires a considerable cooling system. With the gas turbine no cooling system around the combustion chamber or indeed anywhere in the gas paths is required, and it is only necessary to cool the oil which is keeping the bearings and associated gear trains cool and lubricated. Hence, the amount of parasitic weight of the cooling system is greatly reduced. On the other hand, the delay in bringing that engine temperatures upto normal is very short and heaters are able to operate as soon as the engine is upto the speed. This is of considerable advantage in cold climate where devising of screens or equivalent is frequently required before a vehicle can commence operation. 16.19.5

Multifuel Ability

Whilst this matters little for normal automotive use where a suitable fuel will always be available, it is of great benefit in military application. To be able to run on any fuel, ranging from heavy oils to aviation fuels and including paraffin and diesel, is highly desirable as virtually any fuel available in any part of the world will permit the engine to operate satisfactorily. Some fuel systems may require simple adjustments to cover the wide range but others need none and can change fuel while running. 16.19.6

Filtration and Noise

The gas turbine uses large quantity of air than its piston counterpart of similar power and this results in large air inlet and exhaust ducts. Atmospheric pollution has, in time, detrimental effect on the efficiency of the

576

Gas Turbines

compressor by depositing a thin layer of oil, dust rubber, etc., within the diffuser passages. To reduce, this it has been found expedient to filter the air before it enters the engines These filters are, therefore, bulky compared with those on a piston engine but are usually very light as paper elements are used, and also have a considerable intake silencing effect. The exhaust noise level of an engine without heat exchangers is high, but the inclusion of heat exchanger greatly decreases this. If the engine compartment is well designed to avoid gaps and holes and is lined with suitable absorbing material, the general level of noise can be as low as comparable piston engine installations. 16.19.7

Engine Life

There has been no mention yet of an important feature of the gas turbine engine, namely its excellent life between overhauls. The aero-gas turbine has proved conclusively that in that application at least the turbine easily surpasses piston engines. This is principally because, the turbine components work under steady state conditions, i.e., a turbine stress remains constant and in one direction, whilst the piston engine equivalent, say a crankshafts is subjected to alternating stresses continually. There is also the aspect of the sealing by contact, in the piston engine against sealing by close clearances, avoiding friction, in turbine. In addition aero engines work at a high power for most of their lives which is very demanding on lightweight piston engines. For road use the picture is very different. The amount of time the automotive engine is near full power is very small. Over half of the engine’s life is spent below quarter power. This has permitted piston engines to be developed with acceptable lives. In fact, engines can now be said to have sufficient life under normal conditions to outlast the rest of the vehicle. This means that the advantages of long life for gas turbine as applied to automobiles are diminished. However, there are benefits : (i) the complete lack of vibration will reduce failures in other components of the cars, transmissions, etc., and (ii) the likelihood of premature failure will be reduced, which should permit improved warranties to be offered to the public, buying gas turbine cars. The case for the small gas turbine in vehicles has been made by many knowledgeable people-extolling its virtues and showing how its limitations can be overcome. Why then are we not all driving around smoothly in gas turbine cars? The answer lies in two reasons : (i) the very high investment of money, plant and engineers, in existing reciprocating piston engines, and (ii) the economic aspects of manufacturing the gas turbine engine.

Environmental Considerations and Applications

577

The first will be overcome when the second is acceptable and by some public demand such as non-polluting exhaust, or even simply a sales demand. The real problem facing the introduction of the gas turbine engine is its manufacturing cost. True cost must include considerations of the life of the engine in service, and the running costs including any repairs, in addition to the basic manufacturing cost. We already know that the operating costs will be comparable to, or little lower than the piston engine so the real issue lies with the manufacturing cost of the engine. Considered in overall terms, there are less components and a lower overall weight of the engineering hardware, and this should automatically result in a lower overall cost of the unit. The difficulties lie in two fields: (i) the greater proportion of more exotic and hence expensive materials, and (ii) the development of different manufacturing techniques. The gas turbine ‘art’ is at the most interesting stage. The learning curve has overcome the majority of technical problems and is now fairly and squarely on price reduction aspects and tremendous strides are being taken by all the companies involved with small gas turbines. When such very large amount of money is being invested in the development of gas turbines by big American automotive companies, there is bound to be a rapid progress towards mass production of their engine and vehicles using it. As in all competitive manufacturing, components can only be made for their minimum price by mass production, and hence the break-through of gas turbines is likely to be sudden when it occurs. There is no justification for expecting that piston engines will be replaced overnight, there being many fields where the gas turbine will show insufficient benefit to warrant the change. In higher powers, over 250 kW the turbine is already becoming competitive in cost with piston engines but it is unlikely that, below 50–100 kW, will it ever be commercially competitive to replace the established reciprocating engines? The final aspect which must be remembered at all times is the potential of development available in the small gas turbines. Power will increase for the same weight and bulk, fuel consumption will drop further, and technical developments will further improve the torque characteristics and the control aspects. Ceramics will, and are beginning to replace the metals used in the hottest parts of the engine. They have tremendous potential not only because of their ability to withstand very high temperatures but also because of their intrinsic low cost and general availability. With the maximum permitted loading of vehicles likely to become something of the order of 50 tonnes, engines of about 300 kW will be required for commercial vehicles. For this output, the gas turbine is worthy of serious consideration. Another favourable factor is widespread development throughout the world of the motorway Net works on which relatively constant speeds of operation for long periods are practicable. The stage is set

578

Gas Turbines

for the gas turbine to make a big impression on the power units of the world, and when it does it should make the air breathe cleaner and transportation more comfortable. 16.20

TRENDS IN THE FUTURE DEVELOPMENT

(i) There will be higher compressor airflows, pressure ratios and efficiencies with fewer compressor stages, fewer parts and lower cost. Variable-pitch fan or compressor blades will provide reverse thrust for braking. (ii) Engines will be developed to have a lower specific fuel consumption resulting from component design improvements and other changes. (iii) Increased turbine efficiencies with fewer stages to do the necessary work, less weight, lower cost and decreased cooling air requirements will be developed. (iv) Increased turbine temperatures using better materials and improved cooling along with new manufacturing techniques will be used. (v) Less use of magnesium, aluminium and iron alloys but more of nickel and cobalt-base alloys plus increased use of composite materials. (vi) Burning fuel more cleanly with less pollution which will run more quietly. It will make this type of power plant less hostile to the environment. (vii) More airborne and ground engine-condition-monitoring equipment will be used, such as vibration and oil analyzers, and radiometer sensors to measure turbine blade temperature while the engine is operating. Review Questions 16.1 Give a brief account of aircraft emission standards. 16.2 What are the standards for the emission of pollutants from a stationary engine? 16.3 Explain the mechanism of NOx formation and also the methods for its reduction. 16.4 What is noise? What are the sources of noise in a gas turbine engine? 16.5 Explain the noise standards in connection with gas turbine power plants. 16.6 What are the various methods adopted for noise reduction? 16.7 Give a brief account of the assessment of gas turbine power plant.

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16.8 Bring out clearly the role of gas turbine for the following applications: (i) locomotive, (ii) aircraft, (iii) marine, and (iv) stationary power plant. 16.9 What are the additional features of gas turbine engine? 16.10 Explain the trend in the future of gas turbine. Multiple Choice Questions (choose the most appropriate answer) 1. The first legislation enacted to control air pollution was by (a) UK (b) Germany (c) USA (d) Japan 2. The ever first act for pollution control was enacted in the year (a) 1940 (b) 1955 (c) 1960 (d) 1962 3. The APU classification of EP1 class of engine is for (a) all aircraft with piston engine (b) all aircraft with turboprop engine (c) all aircraft for supersonic flights (d) all engines installed in or on an aircraft exclusive of the propulsion engine 4. The TP classification for engine of EPA class for (a) all aircraft with turboprop engine (b) all aircraft with turbojet engine (c) all aircraft with turbofan engine (d) any engine used in aircraft 5. Gaseous emission standards were set in the year (a) 1969 (b) 1975 (c) 1979 (d) 1985

580

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6. The emission standard applicable to newly manufactured aircraft turbine engines of class T1 is (a) HC < 2% (b) CO < 10% (c) NOx < 5% (d) all of the above 7. The dominant factor that influence the amount of NOx in a gas turbine engine is (a) pressure (b) peak temperature and excess air (c) fuel nitrogen (d) residence time 8. Disadvantage of using water injection to reduce NOx in aircraft gas turbine engines is (a) carrying water is very heavy (b) increased heat release which increases the temperature (c) steady increase in pressure in the combustion chamber (d) possibility of increase in CO emission 9. Jet is the result of (a) mixing of fuel and air (b) due to altitude (c) mixing of high velocity exhaust stream with ambient air (d) due to vortex generation behind wings 10. The major application of gas turbine is for (a) aircraft application (b) automotive application (c) locomotive application (d) gas pumping Ans:

1. – (c) 6. – (d)

2. – (b) 7. – (b)

3. – (d) 8. – (d)

4. – (a) 9. – (c)

5. – (c) 10. – (a)

17 ROCKET PROPULSION INTRODUCTION The history of rocket propulsion can be traced back to eighth century when the Greeks learnt the art of flying fire. Though the actual period regarding the use of the principle of rocket is quite uncertain, the Chinese were, perhaps, the first to use this principle in fire arrows as early as 1234. In spite of the fact that a large number of investigators were working on rockets, it was only during the two World Wars that rocket was given its due importance for warfare and other uses. It was in 1919 that Dr. R.H. Goddard from USA published what is called the first report on rockets. After that, with the advent of German V-1 and V-2 rockets, the rockets started establishing. First came the V-1 rocket working on constant-volume explosion type combustion; it was actually a pulsejet. Then came, the V-2 rocket a liquid propulsion device, using liquid oxygen and ethyl alcohol in water as main propellants. During and after the Second World War, Americans studied the captured V-2 rockets and afterwards rockets started establishing themselves for war purposes. The role of the rocket as an aircraft power plant is becoming important day by day. Whereas the primary application of the use of rockets is at present military, they do offer promise for long-range high-speed transport aircraft and also as a power plant for space travel. All the post war efforts culminated into the magnificent moon landing by Neil Armstrong in 1969; and the success of the Russian Lunar vehicle, “Lunokhod” has opened new horizons for the exploration of interstellar systems and a great boost to the insatiable explorative instinct of the man. In many ways the rocket is similar to the other jet power plants that have been discussed in the chapter on Jet Propulsion. Many of the principles that have been developed, apply directly to the rockets. One major difference between the rocket engine power plant and other jet propulsion systems is that it carries its entire propellant (oxidizer and fuel) with it. Other jet propulsion systems depend upon atmospheric air. Thus the basic difference is that the air breathing jet propulsion engine utilizes oxygen from its surrounding medium, whereas the rocket engine utilizes oxidizer,

582

Gas Turbines

which it carries in its own tanks. The fact that it does not depend upon atmospheric oxygen allows it to operate at very high altitudes and even in vacuum. This fundamental distinction between the two types of the power plants produces radically different performance capabilities, which are summarized in Table 17.1. Table 17.1 Performance Differences of the air-breathing engine and the rocket engines Air-breathing engine Rocket engine Altitude limitation. Thrust decreases with

No altitude limitation, travel possible.

space

altitude.

Thrust increases slightly with altitude.

Rate of climb decreases with altitude.

Rate of climb increases with altitude.

Engine ram drag increases with flight speed.

Engine has no ram drag; constant thrust with speed.

Flight speed always less than jet velocity.

Flight speed not limited, can be greater than jet velocity.

Reasonable efficiency and reasonable flight duration.

Low efficiency except at extremely high flight speed for small duration.

It is noted that all the advantages of rocket engine may be offset by its low efficiency. Overall propellant consumption is high because the rocket must carry its oxidizer as well as fuel. For these reasons, the rocket power plant should not be considered as a direct competitor to air-breathing engines. The rocket is really a power plant that can attain performance capabilities beyond the scope of the air-breathing engines. 17.1

CLASSIFICATION OF ROCKETS

The necessary energy and the momentum that can be imparted to a rocket propulsion device by a propellant can be achieved in many ways. For example, chemical, nuclear, or solar energy can be used or an electrostatic or electro magnetic force can also impart the momentum. Alternatively, the propellant can be heated with the help of electric power and expelled in a nozzle. Figure 17.1 gives the classification of rockets on the basis of the methods used for accelerating the propulsion device. Chemical rockets are so far the most widely used rockets. They depend upon the burning of the fuel inside the combustion chamber and expanding it through a nozzle to obtain thrust. The important character of chemical rockets is that the propulsive energy is obtained from within the fuel and the oxidant. The propellant may be solid or liquid. In the free radical propellant system, the propellant has certain free radicals in them, so that

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583

Rockets

Chemical

Solid propellant

Liquid propellant

Nuclear

Free Fusion radical

Fission

Electrodynamic

Ion rocket

Plasma rocket

Photon rocket

Fig. 17.1 Classification of rockets

when they recombine enormous amount of energy is released which can be used for obtaining propulsion. The vast store of the atomic energy can be utilized in case of nuclear propulsion. Fission or fusion both can be used to increase the energy of the propellant. In case of electrodynamics propulsion devices a separate energy source is used and electrostatic or electromagnetic forces accelerate the propellant. The propellant is either discrete charged particles that are accelerated by electrostatic forces or a stream of electrically conducting material which can be accelerated by electromagnetic forces. The first type is called electrostatic or ion rocket, and the second type is called plasma jet. Photon propulsion seems to be the ultimate. In this the propellant being expelled in the form of photons traveling with the velocity of light. As on date photon propulsion does not seem to be technically feasible.

17.2

PRINCIPLE OF ROCKET PROPULSION

The rocket engine in its simplest form consists of a combustion chamber and an expanding nozzle. The fuel and the oxidant, when ignited cause the combustion to proceed at a very fast rate. The exhaust gases produce the required propulsive forces. In other words propellant gases, that are generated in the combustion chamber by burning propellants, are expanded in a nozzle to a supersonic velocity, thereby, requiring a convergent-divergent nozzle. These high velocity gases going out of the nozzle produce the thrust and propel the rocket. Now, the power required to produce an exhaust jet velocity, cj , can be written as P

=

1 2 mc ˙ 2 j

(17.1)

Further, the thrust can be written as F

=

mc ˙ j = mα

(17.2)

584

Gas Turbines

where

m = mass of the rocket m ˙ = mass rate of consumption of propellant cj = exhaust velocity from the nozzle F = thrust P = power required to give an exhaust jet velocity cj α = acceleration of the rocket Combining Eq. 17.1 and 17.2, we get, αcj P = m 2

(17.3)

From the above equation it is clear that the velocity and the thrust that can be obtained from a given type of rocket propulsion system greatly depend upon the power which is available for imparting kinetic energy and the mass of the rocket vehicle. In the case of chemical rockets this energy is extracted from the propellant itself and depends greatly upon the characteristics of the propellant. Such rockets are energy-limited. The exhaust velocity obtained by the use of present day propellants is low which limits their performance. It is same in the case for nuclear rockets also. In the case of ion, plasma, and photon propulsion, this energy is supplied from a separate power source which is usually an electric power obtained from nuclear or solar source. Such rockets are power limited as it is very difficult to produce such a huge amount of power within the small mass of the rocket. In order to obtain high exhaust jet velocity, i.e., high specific impulse for a given thrust, very large amount of power is required. In the case of nuclear rockets the acceptable wall temperature limits the power. In electrodynamic propulsion huge size equipments are needed to convert such large amount of power into kinetic energy. This in turn increases the mass of the vehicle beyond economic and practical limits. If exhaust velocity is low then the mass flow required will be high. Since this mass of propellants has to be carried within the rocket, a compromise has to be made in order to obtain best economy. For a given power, very high velocities may not be of great use as the thrust will be low which will not allow to take any big payload. It must be noted that the useful output of all propulsive devices is thrust, the pull which motivates the craft. Figure 17.2 shows the range of the accelerations and the exhaust velocities obtainable from different types of rocket propulsion systems. For going out of the earth’s gravitational field high acceleration is needed. Either chemical or nuclear systems may be used. Once the gravity limit is crossed the low acceleration available by ion or plasma devices is sufficient for inter-planetary travels. 17.3

ANALYSIS OF AN IDEAL CHEMICAL ROCKET

Figure 17.3 shows the schematic diagram of an ideal chemical rocket and its corresponding T -s diagram.

Rocket Propulsion

585

Chemical Earth surface gravitational acceleration

10

Acceleration m/s

2

Solid core reactor Fusion

reactor

Solar arc

Ion and plasma

Photon

10

0

10

1

10

2

10

3

4

10

10

5

10

6

10

7

Exhaust velocity km/s Fig. 17.2 Range of accelerations and exhaust velocities obtainable by different rocket propulsion systems

p 2

02

p 3

p

T

4

Exhaust Nozzle gases (c j )

3

4

1

4 1

2

3

s

Fig. 17.3 Schematic diagram of a chemical rocket and its corresponding T -s diagram

586

Gas Turbines

The propellant is stored in the combustion chamber at pressure and it is assumed that the burning occurs at constant-pressure. The products of combustion enter the convergence divergent nozzle at point 2 (point 3 corresponds to nozzle throat) and get expanded isentropically in the nozzle from stagnation pressure, p02 to static pressure, p4 , leaving it with an exhaust velocity cj . We also assume that the gases behave as perfect gas. Since the expansion is isentropic, we get . / γ−1 / γ p4 0 cj = (17.4) 2Cp T02 1 − p02 Substituting Cp in terms of universal gas constant R, molecular weight M , and ratio of specific heats γ, we get . / γ−1 / 2γR γ p4 0 cj = T02 1 − (17.5) (γ − 1)M p02 If q is the heat supplied in the form of chemical energy per unit mass of propellant, we get q

= Cp (T02 − T01 )

Then, from Eq.17.5, it follows that . / / 2γR q T01 + cj = 0 (γ − 1)M Cp

1−

p4 p02

γ−1 γ

(17.6)

If At is the area of the throat of the nozzle and pt the pressure at throat it can be shown that the mass flow rate is given by . / γ+1 γ−1 At p02 / 2 0 m ˙ = γ (17.7) γ+1 R T 2 M The thrust produced is given by F

= mc ˙ j + (p4 − pa )A4

(17.8)

where pa is the atmospheric pressure. It should be noted that the term (p4 − pa )A4 has not been neglected because the exhaust velocity cj is supersonic. By substituting Eqs.17.6 and 17.7 into Eq.17.8 the thrust produced by a chemical rocket can be written as . ⎡ ⎤ / γ+1 / γ−1 2 ⎢ At p02 0 ⎥ F = ⎣ γ ⎦ γ+1 R T 02 M

Rocket Propulsion

!

2γR q T01 + × (γ − 1)M Cp

1−

p4 p02

γ−1 γ

+(p4 − pa )A4

587

" (17.9)

As is clear from the above equation, the thrust depends upon the pressure in the combustion chamber, the properties of the propellant and the geometrical shape of the rocket. To obtain high thrust the molecular weight of the propellants must be as low as possible (see Eq.17.9). Most of the propellants, presently in use, produce CO, H2 O, CO2 , N2 and H2 and the average molecular weight of such species is about 25 and seldom less than 20 or so. It means that a limit on the performance of a chemical rocket is imposed by the fact that the molecular weight of the products of combustion cannot be reduced below certain limit. Alternative fuels such as hydrogen and fluorine (molecular weight 8.9) or hydrogen and oxygen (molecular weight 9) or even some light metals like lithium with a molecular weight of 7 can be used. Even with the use of such light materials this limit, cannot be lowered indefinitely because the use of lower molecular weight reactants do not necessarily give lighter products. Another factor, which limits the thrust obtainable, is the maximum allowable temperature as well as the maximum temperature, which can be produced by chemical reactions. At very high temperatures, dissociation does not allow the whole of the heat energy to be converted into the kinetic energy and the maximum obtainable temperatures are limited. The pressurization of combustion chamber, to a certain extent, curbs the dissociation. However, above a pressure of about 20 bar this curbing tendency almost vanishes. Thus, the chemical rocket is energy limited in the sense that in the first instance a large amount of heat release from the propellant is difficult to obtain, and even if obtained, it is still more difficult to utilize it due to dissociation. The effect of variation in the ratio of specific heats is not very much due to the presence of dissociation. The value of γ varies between 1.2 and 1.3 for almost all rocket propulsion fuels. Table 17.2 gives theoretical flame temperatures that can be achieved in chemical rockets and the specific impulse obtainable with different fluids. Apart from the characteristics of the propellant the area ratio of the nozzle and its shape also play an important role in dictating the performance of the chemical rocket. They affect the velocities obtainable as well as the drag on the rocket. 17.4

OPTIMUM EXPANSION RATIO FOR ROCKET

The thrust developed by a rocket is given by F

=

mc ˙ j + (pj − pa )Aj

(17.10)

By increasing the divergent portion of the convergent-divergent nozzle

588

Gas Turbines

Table 17.2 Theoretical Heat Release that can be obtained in Chemical Rockets System

Cyanogen and ozone

Hydrogen and fluorine

Hydrogen peroxide and gasoline

Specific impulse, (s)

≈ 310

≈ 410

≈ 290

Highest temperature

≈ 5000

≈ 4000 K

≈ 3000 K

Lowest molecular weight

between 8 and 9

between 8 and 9

between 8 and 9

Lowest ratio of specific heats

between 1.2 and 1.3

between 1.2 and 1.3

between 1.2 and 1.3

of the rocket the area Aj , velocity cj , and static pressure pj , at nozzle exit can be varied. By differentiating Eq.17.10 with respect to pj and equating to zero; the condition for maximum thrust can be obtained as dF dcj dAj =m ˙ + Aj + (pj − pa ) =0 dpj dpj dpj

(17.11)

Euler’s equation can be written as dpj dcj

=

−ρj cj

(17.12)

By continuity equation m ˙ = ρj cj Aj , then we have dpj dcj

=

−

m ˙ Aj

(17.13)

From Eqs.17.11and 17.13, we have −m ˙

Aj m ˙

+ Aj + (pj − pa )

dAj =0 dpj

(17.14)

or (pj − pa ) Since

dAj dpj

dAj dpj

= 0

(17.15)

= 0, the condition for maximum thrust is given by pj = pa

(17.16)

Rocket Propulsion

589

i.e., the optimum expansion condition is that when the static pressure at the exit of the rocket nozzle is the same as the ambient pressure. Simple physical reasoning can also arrive at this conclusion. Suppose that the length of the diverging nozzle passage is increased over that corresponding to the optimum expansion ratio (pa = pj ); this will result in further expansion in the nozzle and pressure at exit of the nozzle will be less than ambient pressure pa and the second term in the Eq.17.10 will become negative and thrust will decrease. On the other hand, if the length of the nozzle is less than that corresponding to pj = pa will result in lesser expansion and hence in reduced exhaust velocity and thrust will again be reduced. So the condition pj = pa gives the best expansion condition for a rocket nozzle with a given throat diameter. Hence, it is imperative that the design of rocket nozzle is very crucial for producing the required thrust. 17.5

THE CHEMICAL ROCKET

Chemical rockets are classified on the basis of the type of propellant used: (i) Solid propellant rockets (ii) Liquid propellant rockets (iii) Free radical rockets The nature of the propellant used greatly affects the overall shape and size of the rocket. However, major differences in details occur. One typical example is that in the case of solid propellant rockets no fuel-pump and injector is required as in liquid propellant rockets. 17.5.1

Solid Propellant Rocket Engines

Figure 17.4 shows a schematic diagram of a solid propellant rocket. It consists of a combustion chamber, an expansion nozzle, and an igniter. The peculiar characteristic of this type of rocket is that entire propellant is contained in the combustion chamber which has its oxidant within the fuel itself. Therefore, the normal duration of combustion in such a motor is relatively very short (usually less than a minute). Both fuel and oxidizer are homogeneously mixed within the solid propellant grain. The figure indicates that the major components are the propellant grain the case, the nozzle and the igniter (not shown in figure). The igniter, which is usually a pyrotechnic mixture triggered by an electrical signal, initiates combustion over the entire open face of the grain. After ignition the propellant burns more or less evenly and at a steady rate, thus, the name ‘Cigarette burning’ has been used to describe the type of burning for the rocket motor shown in Fig.17.4. This burning produces gases at high pressure and high temperature. The burning causes rapid decomposition and heat release. This heat is conducted back to the propellant and thus the process is self-sustaining. Such a process is called burning by deflagration. The burning rate depends

590

Gas Turbines

Solid propellant

Combustion chamber

Motor wall

Exhaust gases

} Flexible liner

Burning surface

Nozzle

Fig. 17.4 Typical solid propellant motor upon the pressure of the chamber and increases with an increase in pressure. Burning may be controlled or uncontrolled. To obtain maximum thrust whole of the combustion space is filled with the propellant and it is allowed to burn like a cigarette. These gases then flow through the throat area within the grain and are expanded in the exhaust nozzle to a high velocity jet of 1500 to 2400 m/s, depending upon the propellant combination used. The area of the propellant grain over which the combustion takes place determines the magnitude of the exhaust gas generation and hence the thrust of the engine. The thickness of the grain determines the burning time or, duration of thrust production. An active cooling system is not provided since the propellant grain itself frequently insulates the structural case throughout the combustion process. However, insulation is often plated on the inner surface of the case. Therefore, the burning rate is usually controlled by pressure as well as by giving different shapes and sizes to the propellant. This propellant shape is called grain. The various types of grains are shown in Fig.17.5. The pressures developed in a solid propellant rocket are about 80 to 200 bar. The thrust and duration of operation depends upon the shape, size and the material of the propellant. Usually the propellant used is cast or extruded. A solid propellant must release its thermochemical energy without, any additional oxidizer, and must have large energy contents which is to be released rapidly. Hence, a rapid burning substance which contains hydrogen, carbon and oxygen is required. Thus, solid propellants which contains explosives such a nitroglycerin C3 H5 (ONO2 )3 , nitromethane CH3 NO2 , trinitrotoluene C6 H2 (CH3 ) (NO2 )3 are typical. It should be recognized that the solid-propellant rocket motor cannot be shut off once started and that its duration is short. Hence, it is used where the propulsion requirements meet these criteria. Typical areas of solidpropellant rocket motors are rocket-assisted take-off of aircraft (RATO) (units producing 5 to 10 kN of thrust for 10 to 30 seconds), large missile boosters (units with thrust values up to several hundred thousand kN for

Rocket Propulsion

Tubular

Rod and tube

Double anchor

Star

Multifin

Double composition

591

Fig. 17.5 Internal-burning charge designs several seconds), and the main power plant for a multiplicity of missiles including the relatively small projectile like air-to-air missiles (AAM), the large artillery type surface-to-surface missiles (SSM), the surface-to-air missiles (SAM) and some air-to-surface missiles (ASM). Still other uses might include application to one or several stages of long range ballistic missiles and even satellite vehicles. From the foregoing discussion it is apparent that even though the solid-propellant rocket motor has limitations, it is used quite extensively and for varied purposes. The rather widespread use of solid propellant rocket motor is relatively recent, the trend in greater usage is continuing. This trend is primarily due to three factors, viz., (i) Increased use of missiles at the expense of aircraft in the modern air weapons (ii) Improved technology in the solid propellant rocket motor field. Specific impulse during the last fifteen years has increased significantly; new case-bonding techniques have developed to the extent that with proper design, the latter portions of the propellant that burn act as a heat insulator (iii) Inherent advantages of solid-propellant motors over the liquid propellant motor from the military point of view is the storage stability of the solid-propellant and its relatively minor logistics problems Although the use of solid-propellant rocket motors is extensive, the application of liquid-propellant-rocket motor is currently even more extensive in the high thrust units. Also for long duration applications, the liquidpropellant rocket motor has greater potential. A modification of solid-propellant engine is the hybrid engine, which incorporates one solid propellant and one liquid-propellant. Normally, the

592

Gas Turbines

solid propellant is the fuel because of its higher energy potential, while the liquid propellant is the oxidizer. Figure 17.6 schematically illustrates a typical hybrid rocket engine. Liquid oxidizer Injector

Solid propellant gas generator

Solid fuel

Fig. 17.6 Typical hybrid rocket engine The hybrid engine tends to combine the advantages of both solid and liquid propellant engines. It can be operated intermittently and throttled to a degree by varying the oxidizer flow rate. It tends to have simplicity and hence the reliability of the solid-propellant engine. Hybrid engines are still under development. Their advantages would seem to offer good potential for application in the future. 17.5.2

Solid Propellants

In general, the solid propellants can be classified as composite propellants. and double-base propellants. Composite propellants consist of an oxidizer in a ground crystalline shape mixed in the fuel of plastic nature acting as a binder to hold the mixture and to keep a uniform composition. Typical oxidizers are finely ground crystals of perchlorate or nitrate of potassium, lithium or ammonium. The binder may be rubber, asphalt or elastomers. Composite propellants are comparatively difficult to cast. Further, it will have a high oxidizer content, which results in high density of the exhaust. The mechanical properties of the composite propellants depend upon the nature of the binder used. For example, use of polystyrene as a binder results in hard structure of the grain. The burning rate of the propellant can be controlled by controlling the ratio of the mass of the fuel to the oxidizer, by addition of catalyst or by the shape and size of the oxidizer crystals. Double-base propellants or the homogenous propellants are characterized by nitrocellulose containing [C6 H72 (ONO3 )3 ] compounds, which act as a plastic and gives the fuel a colloidal characteristic. These are true monopropellants in that each molecule of the propellant has, in it, the necessary amount of fuel and oxidizer. Double-base propellants, combinations of nitroglycerin [C3 H5 (NO3 )3 ] and nitrocellulose with small quantities of additives are very commonly used. The additives are provided to impart stability of combustion and freedom from ageing which are necessary for achieving high performance. Such propellants are plastic in nature, have a very high viscosity and their appearance is smooth and waxy. These are made in required shape by casting or extrusion. Nowadays this classification of solid propellant has become arbitrary in nature because of the development of such nitrocel-

Rocket Propulsion

593

lulose and nitroglycerin based propellants, which also have oxidizers like potassium perchlorate. In Table 17.3 the typical solid propellants and their chemical formulae along with the important properties are given. Table 17.3 Solid Propellants and their Properties Solid propellant system

Flame temperature

Molecular weight

(K) KClO4 and C2 H2 O NH4 ClO4 and C2 H4 O Nitrocellulose and Nitroglycerine Asphalt and Perchlorate

17.5.3

Specific impulse (s)

1800–3000 1800–2150 2350–3150

25–35 22–25 22–28

165–210 175–240 105–195

2350–2650

–

180–195

Propellants and their Desirable Characteristics

A great variety of propellants have been developed for rocket systems. A few of the more widely used propellants will be discussed in this section along with their various desirable characteristics. (i) Monopropellant A monopropellant, as the name indicates, is one, which is capable of releasing its chemical energy without the addition of an oxidizer. Monopropellants have been used in liquid-propellant rockets. Among the materials that have been used for monopropellants are hydrazine, N2 H4 ; nitromethane, CH3 NO3 ; and hydrogen peroxide, H2 O2 . (ii) Bipropellant fuels Bipropellant systems have been most widely used in liquid propellant rockets. Among the fuels that have been used are ethyl alcohol (ethanol), C2 H5 OH; methyl alcohol (methanol), CH3 OH; gasoline, C8 H18 ; hydrazine, N2 H4 ; furfural alcohol, C4 H3 OCH2 OH; aniline C6 H5 NH2 ; ammonia, NH3 ; and hydrogen, H2 . (iii) Bipropellant-oxidizers Among the liquid oxidizers that have been used for bipropellant systems are oxygen, O2 ; nitric acid, HNO3 ; and hydrogen peroxide, H2 O2 . Table 17.4 gives the several combinations of oxidizers and fuels which have been used for bipropellant systems. The desirable characteristics of propellants are the following: (i) Ability to produce a high chamber temperature. In general, this requirement means a high calorific value per unit mass of propellant.

594

Gas Turbines

Table 17.4 Liquid Bipropellant Combinations Oxidizer

Fuel

Oxygen

gasoline, methane, ethanol, 75% ethanol, 25% water methanol ammonia hydrazine hydrogen

Red-fuming nitric acid

aniline ethanol

White-fuming nitric acid

furfural alcohol

Hydrogen peroxide

hydrazine ethanol methanol

(ii) Low molecular weight of products of combustion. This is one of the most important characteristics in order to obtain a high jet velocity and high specific thrust. (iii) High density to reduce the physical dimensions of the overall missile by allowing a large propellant weight in a given space. In general, requirements (ii) and (iii) conflict and a balance between the two must be reached for any particular application. (iv) Ease of storage and handling. (v) No chemical reaction with the system including tanks, piping, valves, and injection nozzles. (vi) Readily ignitable. (vii) Availability. An ideal propellant which satisfies all of the above requirements is yet to be developed. For example, rocket motors using liquid hydrogen as a fuel are capable of the highest jet velocities but the low density necessitates large fuel tanks; hence greater overall size and increased aerodynamic drag. Further, the extreme low temperatures necessary to maintain hydrogen in the liquid state impose difficult handling and mechanical problems.

Rocket Propulsion

17.5.4

595

Liquid Propellant Rocket Engines

Liquid propellant rocket engines may be divided into several categories. These include cryogenic and storable bipropellant systems, monopropellant systems, and the hybrid systems, which are really a cross between a liquid and a solid propellant systems. Although significantly simpler than bipropellant engines, the performance of monopropellant engines is sufficiently low. It may be noted that these systems are currently used only for relatively small applications. Since the liquid bipropellant system has greater potentialities it will be discussed in details. Figure 17.7 illustrates the principal elements of a liquid bipropellant rocket engine. The rocket engine consists of an oxidizer tank, a fuel tank, various control valve lines and the rocket motor. The rocket motor is the heart of the rocket engine and, therefore, it deserves additional discussion. The basic rocket motor comprises of an injection plate I, a combustion chamber B, and a discharge nozzle N. The function of the injection plate is to receive the liquid oxidizer and fuel and direct them in liquid streams so that they mix up with one another and produce a chemical reaction in the combustion chamber. When the reaction takes place in the combustion chamber, very high pressure and temperature gases are produced which in turn are expanded in nozzle to produce a high supersonic exit velocity (1500 to 4500 m/s). The net thrust is equal to the product of the difference between the exit gas velocities, and zero velocity (because the initial velocity of the fuel and oxidizer with respect to the engine is zero) and mass flow rate. An ignition system is not shown in the engine because the suitable liquids can be chosen which react spontaneously upon mutual impingement. These are called hypergolic combinations1 . A cooling system (not shown in diagram) is also necessary to prevent the walls of the motor from melting because the temperature of reaction often exceeds 2700◦C. Oxidizer tank

Fuel tank

Control valve

B

cj N

I

Control valve

Fig. 17.7 Principal elements of a liquid bipropellant rocket engine In order that the rocket motor might function properly the combustion chamber pressure must be considerably higher than the surrounding pressure (typical values of internal pressures are 20 to 40 bar). The source of 1 Meaning propellant combinations that ignite spontaneously upon contact such as nitric acid and aniline.

596

Gas Turbines

this high combustion chamber pressure must originate upstream of the injector, i.e., a source of pressure is required to force the propellants through the injector nozzles (which requires a pressure drop of about 7 bar or more) and into the combustion chamber. At present there are two general systems of transporting the propellants from their storage tanks to the rocket motors. They are (i) the gas pressurization system, and (ii) pump pressurization system . 17.5.5

Gas Pressurization System

This system is the simplest type of liquid-propellant rocket motor. The details are shown schematically in Fig.17.8. System energizing valve Inert gas supply usually nitrogen of 154 bar 31.5 bar 21 bar Fuel tank Pressure regulators

28 bar

Oxidizer tank Bi-propellant control valve 35 bar

Fig. 17.8 Liquid bipropellant gas pressurization system As illustrated in Fig.17.8 an inert gas (usually nitrogen), a gas which will not react nor be excessively soluble with the fuel or oxidizer, is stored in tanks at, as high a pressure as possible and is supplied through pressureregulator valves to force the liquid propellants through the lines, bipropellant control valves, injector plate and into the combustion chamber. Typical values of pressures are shown at various flow points to show the relative magnitude of pressures required. Since typical mixture ratios (oxidizer flow to fuel flow) are 3 to 5, more pressure and/or larger flow lines are used in the oxidizer flow system. The system is first energized by opening the system-energizing valve, and then the rocket motor is turned on by opening the bipropellant valve. The motor can be stopped and restarted at will by closing or opening the latter valve. The gas pressurization system is a relatively simpler power plant. Its big advantage, however, is the fact that both oxidizer and fuel tanks are pressurized. But the point to note is that when the tanks are of considerable

Rocket Propulsion

597

volume (as they must be for long range applications), the tank walls, which must withstand such high pressures, are excessively thick and heavy. In addition to the increase of propellant tank weight with motor duration, the inert gas volume, and consequently the inert gas tank weight will also increase. Thus, the major limitation of the gas pressurization system is its heavy weight when used for long duration operation. This limitation makes the gas pressurization system applicable only to short duration operations. 17.5.6

Pump-pressurization System

The limitation of gas pressurization system to the application of long-range rocket flight stimulated the development of the pump-pressurization system. A typical liquid bipropellant pump pressurization rocket engine is shown in Fig.17.9. In this system the liquid oxidizer and fuel are stored in tanks at low pressure (thus the tanks can be made light) and forced into the rocket at high pressure by the fuel and oxidizer pumps. The power required for driving the pumps is supplied by a gas turbine which is supplied with steam and oxygen (as shown in this illustration) obtained by decomposing liquid hydrogen peroxide by a catalyst such as calcium or sodium permanganate. The gases used to drive the gas turbine need not be generated from a separate propellant as shown in Fig.17.9, but the use of separate hydrogen peroxide system is popular in rocket motor design, because the gases generated are at relatively low temperature (about 478◦ C), therefore, the problem of high turbine blade temperature is not there. Sometimes the turbine is driven by the gases generated from the main propellant system, which at first examination, may look simpler and more logical. However, when it is remembered that the main propellants are chosen for their high thermochemical activity and consequently react at temperature up to 3100◦C, that portion which is used to operate the turbine must be combined to “run fuel-rich” to keep the turbine temperature at acceptable value, or a third liquid, usually water, is required to dilute the turbine gases down to the temperature at about 875◦C. Because of the third liquid, the gas turbine and the pumps additional lines are necessary. This makes the pump-pressurization system considerably more complex than the gas-pressurization system. The extra complexity is not only due to extra pieces of equipment; but also additional problems arise because of the pumping plant. Perhaps the greatest problem is the design of pumps that will handle the liquids safely and without leaks. Leaks cannot be tolerated because of the possibility of fires or explosions. Since, propellants are selected on the basis of their vigorous reacting ability in the combustion chamber, the risk of fire or explosion due to leaks is very high. To eliminate leaks, special seal design and special seal materials are required. Some of the newer plastics such as teflon have great promise in this application. Liquid oxidizers are generally acids, liquid oxygen, or concentrated hydrogen peroxide; therefore special pump materials are required for the oxidizer pump impeller and casing. The pumping problems have already been satisfactorily solved, and successful applications of

598

Gas Turbines

H2 O2

Oxidizer tank Fuel tank

High pressure gas

Steam & oxygen generated � in the presence of a catalyst

Pump Turbine

Pump

Waste Bi-propellant control valves

Fig. 17.9 Typical pump-pressurization rocket motor pump pressurization system have been used even in piloted aircrafts. 17.5.7

Liquid Propellants

As already discussed there are two basic types of liquid propellants, namely: (i) monopropellant, and (ii) bipropellant. Monopropellants are those, which do not require an oxidizer and can produce thermal energy by decomposition either by a catalyst or by ignition. Typical examples are hydrazine, nitromethane and hydrogen peroxide. Monopropellants are usually hazardous and are not preferred. They are used for limited purposes such as for a source of power to the turbine in a pump feed system, etc., Bipropellant, as the name implies, is a combination of a fuel and an oxidizer, which when combines produces thermal energy. Typical examples of such combinations are hydrogen peroxide and alcohol, liquid oxygen and liquid hydrogen, etc., Similar combinations are given in Table17.5 The above combinations may or may not be hypergolic. Nowadays, a combination of metallic fuels like lithium or beryllium with either oxygen or fluorine and liquid hydrogen, what is called as a tripropellant, is also being used to improve the performance of chemical rockets. Table 17.5 shows the common liquid propellants used along with the approximate obtainable specific impulse from them. Fluorine is the most powerful oxidizer, next is the chlorine trifluoride, which is so active that even the glass-fibre cotton, a fireproof material burns in it. Hydrogen peroxide and ozone are also good oxidizers. Liquid hydrogen has maximum

Rocket Propulsion

599

Table 17.5 Common Bipropellants used in Liquid Propellant Rockets Oxidizer

Fuel

Temperature (K)

Molecular weight

Specific impulse (s)

Hydrogen peroxide(H2 O2 )

Gasoline

2950

21

250

Red fuming nitric acid (HNO3 )

Gasoline

3125

25

240

Alcohol

3350

22

260

Liquid oxygen ∗

Liquid oxygen

Liquid ammonia

—

—

—

Liquid oxygen

Hydrazine (N2 H4 )

3400

18

310

Liquid oxygen

Liquid hydrogen

3000

9

390

Liquid fluorine

Liquid hydrogen

4000

8.9

410

Liquid fluorine

Ammonia

4600

19

360

Nitric acid

Aniline (COH5 NH2 )

—

—

Nitrogen tetroxide

Hydrazine

3260

—

290

Ozone

Cyanogen

520

—

270

Hydrogen peroxide

Ethyl diamene (C2 H4 N2 H5

—

—

—

* Not commonly adopted

600

Gas Turbines

heat value but is highly explosive. Germans used alcohol and liquid oxygen as propellants for V-2 rocket. Kerosene is also used with some powdered metals as rocket fuel. Hydrazine requires comparatively smaller quantities of oxidizer for burning. 17.5.8

Requirements of a Liquid Propellant

Any liquid or a combination of liquids, if capable of producing mainly gaseous products by an exothermic chemical reaction, can be regarded as a liquid propellant. However, to be ideally suitable for liquid propellant rocket it must have certain specific characteristics such as (i) High calorific value. (ii) High density. (iii) Ease of storage and handling. (iv) Low corrosion characteristics. (v) Less toxicity. (vi) Performance independent of ambient temperature. (vii) Smooth ignition and high reliability. (viii) Chemical stability and long shelf life. (ix) Negligible viscosity change with temperature. As far as known none of the presently available propellants meets all these requirements and still lot of development work is needed in this field even after so much successful launchings of rockets and satellites. 17.6

ADVANTAGES OF LIQUID PROPELLANT ROCKETS OVER SOLID PROPELLANT ROCKETS

The following are some of the advantages of liquid-propellant rockets over the solid-propellant rockets: (i) Duration of operation can be increased. (ii) Use of cooling allows the metal walls to keep their strength. (iii) Use of lower pressures in conjunction with cooling makes it possible to use less expensive or non-critical materials. (iv) Size of the combustion chamber is smaller as whole of the fuel need not be stored in it. (v) Comparatively easier to control by controlling the propellant flow.

Rocket Propulsion

601

(vi) Heat losses to the wall are reduced due to preheating of fuel during cooling process. Liquid propellant rockets are extensively used with aircraft, for assisted take-off, ballistic missiles, thrust augmentation, and in space flight. 17.7

FREE RADICAL PROPULSION

If a stable chemical material is supplied with sufficient energy to break the energy bonds some unstable free radicals will be produced. If these free radicals are used in a rocket so that they recombine and release a vast amount of energy, the limitations of a chemical rocket can be partially overcome. Molecular hydrogen with certain concentration of unstable free radicals in it, seems to be the best propellant due to its lower molecular weight. However, it is long time before such a system can be used in practice due to difficulties in generating free radicals, preserving them for sufficient time and controlling the recombination rates. 17.8

NUCLEAR PROPULSION

Work on nuclear rocket engines was begun in 1956 in USA. With the significant increase in emphasis on space flight activities beginning in 1958, National Aeronautics and Space Administration, USA (NASA) took over the development of nuclear technology for nuclear rocket-propulsion systems for manned or unmanned space exploration. Nuclear rocket engine produces very small thrust and thus have quite low thrust-to-weight ratio which are only useful for space operations in gravity-free fields. The maximum exhaust velocity, that can be obtained in a chemical rocket is limited to less than 4.5 km/s because of the lower energy release from the propellant and other limitations such as attainment of a high temperature in the combustion chamber. It is well known that in a nuclear reaction vast amounts of energy, 106 to 108 times that of a chemical reaction can be obtained. This high rate of energy release will result in very low mass of the propelling device and a wider latitude in selection of propellant material. Detailed study of these reactions and related equipment is beyond the scope of this book and only a simple description is given below. There are two types of nuclear reactions which can be used to produce energy, viz., fission and fusion. In fission a heavy mo1ecule is broken into fragments by the bombardment of neutron on its nucleus. When the neutron is absorbed into the nucleus, the heavy molecule splits up into a number of fragments and energy is released in the process. This energy is in the form of heat energy. In addition to this, the reaction also produces about 2.5 neutrons which makes the reaction self-sustaining. Figure 17.10 shows a schematic diagram of a nuclear propulsion unit. It consists of a nuclear core having 92 U235 as the nuclear fuel. The control rods control the reaction. A reflector is provided to avoid leakage of neutrons.

602

Gas Turbines

Hydrogen is used as a propellant, which takes up heat from the reactor and expands in the nozzle. Control rods

Reflector

H 2 from pump

Exhaust

Reactor core

Fig. 17.10 Schematic diagram of solid core nuclear-heated hydrogen rocket In the second type of reaction, i.e., thermonuclear reaction or fusion, charted nuclei of light elements are fused into a heavy nucleus and in the process vast amount of energy, is released. To start fission reaction, coulomb repulsion forces between the nuclei must be overcome, which presents a formidable problem as it requires temperatures of the order of 1060 K. Nevertheless, if nuclear fusion is achieved in near future then the future propulsion devices are very likely to use nuclear energy. Nuclear energy has been successfully used in submarines, ships and some experimental rockets. 17.9

ELECTRO DYNAMIC PROPULSION

The fundamental requirement for space propulsion is the generation of very high exhaust velocities in order to minimize propellant consumption. Electric propulsion engines generate exhaust velocities more than 4 to 10 times those of chemical rocket engines, and thus they have considerable potential for space propulsion. There are three basic types of electric rocket propulsion engines: (i) ion rocket propulsion, (ii) the plasma rocket propulsion, and (iii) the photon propulsion . For each of these engines, the energy is supplied from an energy source separate from propulsion device. 17.9.1

Ion Rocket Propulsion Engines

Figure 17.11 represents a schematic diagram of Electrostatic or ion rocket engine. The major components are (i) propellant supply tank and propellant feed mechanism,

Rocket Propulsion

603

(ii) thrust chamber, and (iii) electric power supply. The thrust chamber incorporates a vapourizing chamber, an ionization chamber, an accelerating grid and an electron emitter. The propellant is first heated and vaporized and then passed on to the ionization chamber, where electrons are stripped to ionize it. The resultant ions are then accelerated electrostatically with a d.c. voltage drop (in excess of 1000 volts) and discharged to produce thrust. The electrons stripped in the ionization chamber are collected and injected into the ion stream aft of the accelerating grid to neutralize the ion beam and prevent the ions from being attracted back to the grid. Removal of electrons is also necessary to maintain a neutral charge on the space vehicle. Electric power supply Electron emitter Ion flow

Propellant� tank

Pump

Vaporizing � chamber Ionization � grid Accelerating � grid

Neutral plasma flow

Fig. 17.11 Schematic diagram of an ion rocket engine

17.9.2

Plasma Rocket Propulsion

Plasma propulsion differs from ion propulsion in that the electrical forces are not directly used to accelerate the propellant; instead the propellant is first heated to a very high temperature and then expanded through a nozzle. It is a hybrid thermal-electrical propulsion system. It relies on expansion of hot plasma for bulk of its thrust but its energy is supplied electrically rather than by chemical combination or by nuclear fusion or fission. A high voltage source, which again may be a nuclear reactor, is used to produce an electric arc to heat up the propellant to a high temperature. The propellant is either hydrogen or helium as both have low molecular weight and high maximum stagnation temperature before they dissociate. Helium is preferable as it has a higher stable temperature. Once the propellant is heated to a high temperature it is expanded in a nozzle to get propulsive force. For starting, electrodes are brought closer than the normal position to produce a spark. The chamber core becomes a good conductor in the presence of positive ions. After the arc is established, the electrodes are taken

604

Gas Turbines

apart to their normal position. Energy leaves the arc in electromagnetic radiation and is consumed because of the increase in the potential energy of ionization and dissociation. The temperature of the material in the region of arc is increased. The propellant is fed in the form of vortex to concentrate the arc into a narrow region. This allows a very high temperature (50,000 K) in the central core-much more than the wall material can withstand since the propellant vortex also cools the wall. This high temperature plasma escapes through the nozzle and a propulsive force is created. Since ionization of hydrogen takes place only at about 10,000 K a large amount of energy supplied to the arc is not available for producing thrust. Unless this energy is used back by proper recombination, the efficiency will be very low. Production of an efficient plasma arc, the difficulty of cooling the rocket wall and the need of very high power for arcing are the major limitations of plasma jet propulsion system. Hot plasma erodes the nozzle, and the life of electrodes is limited to a few minutes. Not much information is available about it but may have a great future potential. The plasma rocket propulsion can be of two types: (i) Arc plasma rocket engine, and (ii) Magneto plasma rocket engine. Arc Plasma Rocket Engine The arc plasma rocket engine is one of the simplest types of electrical propulsion systems being developed. Its schematic diagram is shown in Fig.17.12. The major components are (i) propellant supply tank and propellant feed mechanism, (ii) thrust chamber, (iii) cooling system, and (iv) electric power supply. The thrust chamber contains two electrodes. The nozzle walls serve as the cathode, and an electrode within the chamber is the anode. An arc is formed between these electrodes, and the propellant is heated to an electrically neutral plasma in passing through the arc. Thrust results from the expansion of the heated plasma through the nozzle. The velocity produced is 4000 to 19,500 m/s. The propellant can be used to cool the chamber regeneratively before passing into the chamber. If the rate of propellant consumption is too low to provide enough cooling, a supplementary cooling system with a radiator to reject the excess heat must be provided. The arc plasma or electrothermal rocket is quite similar to the chemical racket, the primary difference being that electrical rather than chemical energy is used for heating the propellant.

Rocket Propulsion

Propellant tank

605

Pump

Electric power supply Cathode Vj Anode

Pump

Radiator for cooling system

Fig. 17.12 Schematic diagram of an arc plasma rocket engine Magneto-Plasma Rocket Engine Another type of electrodynamic propulsion system is the magnetohydrodynamic (MHD) or magneto-plasma rocket engine. Figure 17.13 shows principal elements of such a system. MHD engines produce thrust from the interaction between magnetic and electric fields. The force which results is customarily called the J × B force, where J is the current density and B is the applied magnetic field. The nature of this resulting force, which accelerates the propellant, is analogous to the force which rotates electric motors. Although a great amount of research and development work remains to be done, the ultimate development of an efficient and reliable MHD engine appears assured. The magneto-plasma rocket engine uses an arc chamber Magnetic field Propeller Tank

Ve Pump

Arc chamber

Accelerator Plasma (+,-)

Electric supply

Fig. 17.13 Principal elements of a magneto-plasma rocket engine to produce a conductive plasma, similar to arc plasma engine. The plasma then passes through an accelerator. Two different methods can be used to accelerate the plasma : (i) changing or collapsing of magnetic fields, or (ii) interaction of an electric current and a steady magnetic field.

606

Gas Turbines

The first method is cyclic and can impart extremely high velocities to the plasma (up to 2,40,000 m/s). The second method is continuous but imparts somewhat lower to the plasma (up to 15,000 m/s). 17.10

PHOTON PROPULSION

Photons are electromagnetic quanta of energy which do not carry an intrinsic mass but carry a momentum equal to hf /c where h is Plank’s constant, f the frequency of radiation, and c is the velocity of light. When photons are produced some mass according to Einstein’s equation m = E/c2 , disappears, i.e., in other words if photons are emitted some mass is transferred. Figure 17.14 shows a schematic diagram of a photon propulsion unit in which a hot nuclear power source generates a flux of photons, which is shaped by a photon absorber and then escapes through the nozzle producing a thrust equal to d F

=

hf c

=

dt

1000Pe c

where Pe is the power in jet in kilowatts. As on date such rockets are not feasible, but probably have a great future potential!

Nuclear photon source

Pay load

Photons

Nozzle Biological shield

Shaped absorber

Fig. 17.14 Schematic diagram of a photon propulsion unit

17.11

COMPARISON OF VARIOUS TYPES OF ROCKETS

So far we have discussed various types of rockets. In summary, there are two general measures of the performance of a rocket engine. (i) specific impulse, which will determine the amount of propellant that must be used to accomplish a given task. (ii) fixed weight of the engine, including the necessary tankage, power supply, and structure. The chemical rocket engine is a fairly lightweight device. However, the specific impulse is not high. Solid and liquid propellants in present use

Rocket Propulsion

607

deliver an impulse of around 250 seconds. The best liquid propellants so far conceived and evaluated yield an impulse of about 350 seconds. Certain solid propellants, proposed on the basis of theory alone might yield 300 seconds. The fundamental theory of chemical binding energies precludes the possibility of any substantial gains over these numbers, Even some asyet-undiscovered superfuel is unlikely to raise the specific impulse beyond 400 seconds. Nuclear rocket is not limited by propellant binding energies. However, it is constrained by the temperature limitations of wall materials. Using hydrogen as a propellant, values of specific impulse of perhaps 1,000 seconds or more are feasible. If gaseous containment of the fissioning fuel becomes possible, specific impulses of several thousand seconds might be achieved. This type of rocket engine appears very promising. Research on nuclear rockets and controlled thermonuclear power reactors may yield information useful to the construction of such a device. The primary consideration in obtaining useful thrust from ion or plasma rockets is the construction of lightweight electric power supplies. A gross reduction in electrical generation equipment is required to make the electric rocket useful for flight in the solar system. In any event, the electric rocket is likely to remain a low-thrust device. Therefore, large chemical or nuclear rockets would still be required to boost a space ship from the surface of the earth.

17.12

STAGING

Often, the required velocity (delta-v) for a mission is unattainable by any single rocket. This is because the propellant, tankage, structure, guidance, valves and engines and so on, take a particular minimum percentage of takeoff mass. This becomes too high for the propellant it carries to achieve that velocity. For example the first stage of the Saturn V, carrying the weight of the upper stages, was able to achieve a mass ratio of about 10, and achieved a specific impulse of 263 seconds. This gives a delta-v of around 5.9 km/s whereas around 9.4 km/s delta-v is needed to achieve orbit with all losses allowed for. This problem is frequently solved by staging - the rocket sheds excess weight (usually empty tankage and associated engines) during launch. Staging is either serial where the rockets light after the previous stage has fallen away, or parallel, where rockets are burning together and then detach when they burn out. The maximum speeds that can be achieved with staging is theoretically limited only by the speed of light. However the payload that can be carried goes down geometrically with each extra stage needed, while the additional delta-v for each stage is simply additive.

608

Gas Turbines

17.13

MULTISTAGE ROCKET

A multistage (or multi-stage) rocket is a rocket that uses two or more stages, each of which contains its own engines and propellant. A tandem or serial stage is mounted on top of another stage; a parallel stage is attached alongside another stage. The result is effectively two or more rockets stacked on top of or attached next to each other. Taken together these are sometimes called a launch vehicle. Two stage rockets are quite common, but rockets with as many as five separate stages have been successfully launched. By jettisoning stages when they run out of propellant, the mass of the remaining rocket decreases. This staging allows the thrust of the remaining stages to more easily accelerate the rocket to its final speed and height. In serial or tandem staging schemes, the first stage is at the bottom and is usually the largest, the second stage and subsequent upper stages are above it, usually decreasing in size. In parallel staging schemes solid or liquid rocket boosters are used to assist with lift-off. These are sometimes referred to as ’stage 0’. In the typical case, the first stage and booster engines fire to propel the entire rocket upwards. When the boosters run out of fuel, they are detached from the rest of the rocket (usually with some kind of small explosive charge) and fall away. The first stage then burns to completion and falls off. This leaves a smaller rocket, with the second stage on the bottom, which then fires. Known in rocketry circles as staging, this process is repeated until the final stage’s motor burns to completion. In some cases with serial staging, the upper stage ignites before the separation - the interstage ring is designed with this in mind, and the thrust is used to help positively separate the two vehicles. This is known as ”fire in the hole”. 17.13.1

Upper Stages

The upper stages of space launch vehicles are designed to operate at high altitude, and thus under little or no atmospheric pressure. This allows them to use lower pressure combustion chambers and still obtain nearoptimum nozzle expansion ratios with nozzles of reasonable size. In many low pressure liquid rocket upper stage engines, such as the Aerojet AJ10, propellants are pressure fed without need for complex turbomachinery. Low chamber pressures also generate lower heat transfer rates, which allow ablative cooling of the combustion chambers rather than more elaborate regenerative cooling. 17.13.2

Advantages

The main reason for multi-stage rockets and boosters is that once the fuel is burned, the space and structure which contained it and the motors themselves are useless and only add weight to the vehicle which slows down its future acceleration. By dropping the stages which are no longer useful, the rocket lightens itself. The thrust of future stages is able to provide more acceleration than if the earlier stage were still attached, or a single, large

Rocket Propulsion

609

rocket would be capable of. When a stage drops off, the rest of the rocket is still traveling near the speed that the whole assembly reached at burnout time. This means that it needs less total fuel to reach a given velocity and/or altitude. A further advantage is that each stage can use a different type of rocket motor each tuned for its particular operating conditions. Thus the lower stage motors are designed for use at atmospheric pressure, while the upper stages can use motors suited to near vacuum conditions. Lower stages tend to require more structure than upper as they need to bear their own weight plus that of the stages above them, optimizing the structure of each stage decreases the weight of the total vehicle and provides further advantage. 17.13.3

Disadvantages

Staging requires the vehicle to lift motors which are used only at a later stage. Further, it makes entire system more complex and harder to build. Nevertheless the savings are so great that every rocket ever used to deliver a payload into orbit had staging of some sort. In more recent times the usefulness of the technique has come into question due to developments in technology. In the case of the Space Shuttle the costs of space launches appear to be mostly composed of the operational costs of the people involved, as opposed to fuel or equipment. Reducing these costs appears to be the best way to lower the overall launch costs. New technology that is mainly in the theoretical and developmental stages is being looked at to lower the costs of launch vehicles. 17.14

COMPARISON OF VARIOUS PROPULSION SYSTEMS

We have seen various propulsive devices, which include air breathing engines and rockets. In this section, we will make the comparison of various propulsion systems which respect to propulsive efficiency, overall efficiency and performance details (Figs.17.15, 17.16 and 17.17). The considerations of thrust and propulsive efficiency are by no means the whole story in comparing various propulsion systems. We must consider, for example, the thermodynamic efficiency of the engine cycle, or how efficiently the chemical energy of the fuel is converted to useful work such as driving the propeller and compressor, or accelerating the exhaust jet. The weight and size of the engine, its complexity, and its service life are other factors to be considered. 17.15

PROPULSIVE EFFICIENCY

To go deeper into the subject of propulsive effectiveness, we can consider the propulsive efficiency. Marine engineers call it the Froude efficiency after William Froude (1810-1879) who first used it. Propulsive efficiency is defined as the ratio of useful power output to the rate of energy input. For

610

Ro

ll pe ro

Propulsive efficiency

P

cke t

Tu rbo jet

Gas Turbines

er

Flight velocity, V

0

Fig. 17.15 Propulsive efficiency at different speeds of various system

propulsion

50

Overall efficiency, %

40

Rocket Ramjet

Reciprocating aircraft

30

Turbojet 20

Automobile

10

0 0

800

1600

2400

3200

8000

9000

Speed, kph

Fig. 17.16 Efficiency of various propulsion systems at different speeds

Rocket Propulsion

Power plant

Thrust per unit � engine weight

611

Frontal area� Specific fuel � per unit thrust consumption

Rocket Ramjet

Turbojet

Piston

Fig. 17.17 Performance details of different propulsive systems flight, propulsive efficiency is expressed as: thrust × flight speed It is to be noted that the various propulsion systems operate best in different speed ranges. This is shown qualitatively in Fig.17.15. The propeller is the most efficient propulsive method at low speeds, while the jet engine achieves best efficiency only at relatively high flight speeds. The very high exhaust velocities of the rocket make its propulsive efficiency high only at very high flight speeds. Figure 17.16 shows the overall efficiency of various propulsive devices. Figure 17.17 shows the performance details of different propulsive systems. It is clear that the maximum thrust is obtained from the rockets whereas it is least for piston engines. At the same time it may be noted that the specific fuel consumption is the highest for rockets and lowest for the piston engines. However, from the point of view of frontal area the rocket has the advantage of being the least front area whereas it is maximum for the piston engines. Review Questions 17.1 What is the basic difference between rocket propulsion and jet propulsion? Can rockets work in vacuum? 17.2 How rockets are classified? What is the stage of development of each type? 17.3 Derive a general expression for the thrust produced by a chemical

612

Gas Turbines

rocket and hence discuss the importance of the molecular weight of the propellants. 17.4 What are the factors which limit the thrust obtainable from chemical rockets ? 17.5 What is the condition for the maximum thrust in a chemical rocket? Derive an expression for the same. 17.6 Draw a schematic diagram of a solid propellant rocket and explain its working. What are the applications of this type of rocket? 17.7 Briefly describe the two types of solid propellant rockets. 17.8 Draw a schematic diagram of a liquid propellant rocket. What are the different systems of injecting liquid propellants into the combustion chamber? 17.9 Discuss the mechanism of burning of various types of liquid propellants. 17.10 What is meant by monopropellant and bipropellant fuels for rockets? Give important examples of each type. What is the difference in the application of a monopropellant and a bipropellant rocket engine? 17.11 What are the desirable requirements of a liquid propellant for rockets? 17.12 Compare the advantages and disadvantages of solid and liquid propellants? 17.13 Explain the concept of free radical propulsion. 17.14 Describe how nuclear energy can be used for propulsion of rockets. Has it been used so far? 17.15 Discuss, with suitable sketches, the three systems of electrodynamic rocket propulsion. 17.16 What factors are important in the comparison of propulsive devices? State the optimum operational range, specific fuel consumption and relative weights for various propulsion devices. 17.17 What are the advantages and disadvantages of rocket engines? 17.18 What are the various types of oxidizers in use in rockets? 17.19 Describe the use of hydrogen peroxide as a monopropellant. What is the function of hydrogen peroxide in a bipropellant rocket engine? 17.20 Discuss the relative merits of gas-pressurization rocket power plant and the pump-pressurization system. 17.21 Briefly compare the various rocket propulsion devices. 17.22 What is meant by staging? Explain why multi-staging is required.

Rocket Propulsion

613

17.23 What are the advantages and disadvantages of multistage rockets? 17.24 With a neat graph compare the various propulsive and overall efficiency of various propulsion devices (air breathing and rocket) with respect to flight speed. 17.25 With a neat diagram compare the performance parameters of rocket, Ramjet, turbojet and piton engines. Multiple Choice Questions (choose the most appropriate answer) 1. The first report on rockets was published in the year (a) 1909 (b) 1919 (c) 1929 (d) 1939 2. The magnificent moon landing by Neil Armstrong was in the year (a) 1949 (b) 1959 (c) 1969 (d) 1979 3. For rocket propulsion the nozzle used is (a) convergent type (b) divergent type (c) convergent divergent type (d) any type can be used 4. Burning of propellant occurs at (a) constant-volume (b) constant-pressure (c) constant temperature (d) none of the above 5. Thrust in a rocket depends on (a) pressure in the combustion chamber (b) properties of the propellant (c) geometrical shape of the rocket (d) all of the above

614

Gas Turbines

6. Chemical rockets use propellant which is (a) solid (b) liquid (c) free radical (d) all of the above 7. Burning of a propellant in a rocket is due to (a) subsonic combustion (b) supersonic combustion (c) deflagration (d) chemical decomposition 8. Mono propellant is one which requires (a) fuel (b) fuel and oxidizer (c) oxidizer (d) none of the above 9. Plasma rocket propulsion and ion rocket propulsion comes under the category of (a) electrical propulsion (b) mechanical propulsion (c) chemical propulsion (d) nuclear propulsion 10. If a stable chemical material is supplied with sufficient energy to break the energy bonds and if it is used for rocket propulsion, it is called (a) nuclear propulsion (b) free radical propulsion (c) electrodynamic propulsion (d) plasma rocket propulsion Ans:

1. – (b) 6. – (d)

2. – (c) 7. – (c)

3. – (c) 8. – (c)

4. – (b) 9. – (a)

5. – (d) 10. – (b)

Appendix 1. THE SIMPLE GAS TURBINE CYCLE

2. THE HEAT EXCHANGE CYCLE

616 Gas Turbines

3. THE REHEAT CYCLE

4. THE REHEAT AND HEAT EXCHANGE CYCLE

Appendix

5. THE INTERCOOLED CYCLE

6. THE INTERCOOLED CYCLE WITH HEAT EXCHANGER

617

618

Gas Turbines

Index actual cycle performance, 156 adiabatic flow one-dimensional, 496 adiabatic process, 13 aero-thermodynamic theory, 25 aerodynamic loss annulus, 354 profile, 354 secondary, 354 tip clearance, 355 aerofoil cambered, 56 uncambered, 56 afterburner, 229, 247 air angles, 337, 341 air cooled blades, 518 air pollution, 549 air-fuel ratio, 151 aircraft engine single-spool, 524 airfoil, 54 elementary theory, 54 angle of incidence, 56 angularity coefficient, 494 annulus loss, 354 application aircraft, 573 automotive, 572 gas pumping, 571 gas turbine, 72, 569 locomotive, 572 marine, 570 power generation, 570 process, 573 pulse jet, 224 ramjet, 220 small gas turbine, 569 arc plasma rocket engine, 604

area-velocity relation, 26 arrangement closed-cycle, 63 open-cycle, 63 single-shaft, 63 augmentation thrust, 246, 247 axial flow compressor, 333, 335 geometry, 335 historical background, 333 working principle, 335 backward swept blades, 289 Balje’s formula, 302 basic definitions, 19 binder, 592 bipropellant, 598 bipropellant fuels, 593 bipropellant-oxidizers, 593 blade, 501 air cooled, 518 backward swept, 289 backward-curved, 294 cast, 520 conway, 518 cooling, 513, 514 effect of shape, 294 efficiency, 427 fabricated, 518 fixing, 508 forged, 518 forward-curved, 294 loading, 427 manufacture, 505 materials, 501 radial, 295 radial-curved, 294 radial-tipped, 289 selection of materials, 502 619

620

INDEX

spey, 519 turbine, 501 turbine rotor, 421 tyne, 518 blade cooling requirements, 517 blade fixing, 508 anchor pin , 509 bulb root method, 509 fir tree method, 510 grub screw method, 510 tee and double tee, 509 welding, 510 blade manufacture, 505 extrusion method, 505 tadpole method, 505 blade profiles, 472 blade row rotor, 351 stator, 353 blade shapes, 287 bleed burn cycle, 247, 252 boundary layer, 35, 340, 485 laminar, 36 separation, 36 turbulent, 36 Brayton cycle, 229 camber line, 54 carbon deposits, 403 carbon preferential burning, 394 centrifugal compressor, 279 essential parts of, 280 losses, 310 principle of operation, 282 centrifugal machines, 44 centripetal machines, 44 chemical kinetics, 401 chemical rocket, 589 free radical, 589 liquid propellant, 589 solid propellant, 589 choking, 311 classification rocket, 582 Clausius statement, 12 closed-cycle, 68 advantages, 71

basic requirement of working medium, 71 disadvantages, 71 coefficient angularity, 494 discharge, 493, 494 flow, 493 gross thrust, 493, 497 nozzle, 491 velocity, 494 combustion, 4 efficiency, 402 form of system, 402 intensity, 400, 401 process in gas turbine, 405 spontaneous, 396 combustion chamber, 3 annular, 403 arrangements, 409 geometry, 406 performance, 398 requirement, 403 combustion efficiency, 150 combustion process, 393 combustion systems, 3, 393 combustion theory, 394 complex cycle, 99, 165 component matching, 523 composite propellants, 592 compression, 4 compressor, 3, 47 axial flow, 333 centrifugal, 279, 333 efficiency, 140 flow analysis, 290 low-pressure, 279 medium-pressure, 279 requirement, 333 rotor, 42 stage efficiency, 343 supersonic, 468 turbo-, 279 work input, 339 continuity equation, 26 control surface, 8 cooling air, 515 aspects, 575

INDEX

external, 513 internal, 514 liquid, 515 passage, 517 cycle, 8 actual, 156 aircraft propulsion, 79 arrangement, 63 bleed burn, 247, 252 Brayton, 229 closed, 63, 68 comparison, 101 complex, 99 efficiency, 151 Ericsson, 103 heat exchange, 83, 159 ideal, 79 intercooled, 92, 163 intercooled and reheat, 96 intercooled with heat exchange and reheat, 99, 165 intercooled with heat exchanger, 94 intercooled with reheat, 165 jet propulsion, 213 Joule, 229 open, 63 performance, 79 practical, 3 reheat, 86, 159 reheat and heat exchange, 89, 159 simple, 157 simple gas turbine, 80

blade ring, 303, 305 vaned, 303 vaneless, 302 discharge coefficient, 493, 494 double-base propellants, 592 drag, 56 drag force, 59 dynamic temperature, 138

effect altitude, 246 centrifugal, 44 external, 43 impulse, 43 internal, 44 reaction, 44 suction, 56 efficiency, 451 blade, 427 combustion, 150 compressor, 140, 309 compressor stage, 343 cycle, 151 isentropic, 47 overall, 3, 245 polytropic, 152, 155 propeller, 241 propulsive, 242, 244 ram, 238 small-stage, 152 stage, 427 transmission, 242 turbine, 140, 141 electro dynamic propulsion, 602 plasma rocket, 602, 603 ion rocket, 602 definitions and laws, 7 photon, 602 deflagration, 589 degree of reaction, 45, 345, 423, 439– electrodynamics, 583 emission standards 441 aircraft, 551 fifty per cent, 347 hydrocarbon, 554 high reaction stage, 349 smoke exhaust, 553 hundred per cent, 448 stationary engine, 555 low reaction stage, 347 energy, 9 negative, 449 equation, 15, 42 zero, 452 gravitational potential, 15 degree of turbulence, 35 internal, 10, 15 density, 8 kinetic, 15 diffuser, 280, 294, 302, 484

621

622

INDEX

transfer, 16, 39, 43, 293 transformation, 16, 39, 423 energy equation, 15 engine afterburning, 490 air breathing, 214 backpressure control, 490 gas turbine, 214, 215 hybrid, 591 liquid propellant rocket, 595 NOx reduction, 559 performance, 516 propeller, 214 pulse jet, 215, 220 ramjet, 215, 216 rocket, 214, 581 turbofan, 215 turbojet, 213, 215, 227 turboprop, 213, 215, 224 enthalpy, 10 enthalpy loss coefficient, 438 enthalpy stagnation, 17 enthalpy-entropy diagram, 284, 342, 436, 442 entropy, 12 equilibrium running diagram, 526 Ericsson cycle, 103 Euler’s energy equation, 286 exhaust, 574 exhaust nozzle area ratio, 490 expansion waves, 30

hypersonic, 25 inviscid, 21 isentropic, 34 laminar, 34 losses, 143, 353 mixed, 44 nozzles, 421 quasi-one-dimensional, 26 radial, 44 Rayleigh-type, 34 rotor blade row, 351 secondary, 340, 356 stator blade row, 353 steady, 20 transonic, 25 turbulent, 34 unsteady, 20 flow coefficient, 301, 344, 451 flow process, 14 flow through blade rows, 351 fluid, 19 velocity, 19 fluid dynamic equation, 7 fluid dynamics analysis, 39 forward speed, 246 free radical propulsion, 601 free radical rocket, 589 friction coefficient, 33 friction factor, 36 future development, 578 gas

Fanno line flow, 32 Fanno-type flow, 34 fifty per cent reaction, 452 fifty per cent reaction stage, 347 finite disturbance, 25 flame stabilization, 393 flame stabilizing zone, 407 flow axial, 44 coefficients, 427 compressible, 20 compressor, 290 constant-area duct, 32 Fanno-type, 34 friction, 32 heat transfer, 32

ideal, 11 perfect, 11 real, 11 semi-perfect, 11 gas pressurization system, 596 gas turbine, 2, 4 applications, 72, 569 assessment, 567 combustor, 398 comparison, 73 components, 3, 4 engine, 214, 215 features, 574 parallel flow, 64 practical cycle, 137 process of combustion, 405

INDEX

series flow, 64 simple, 2 technology, 215, 279 gross thrust, 237 gross thrust coefficient, 497 heat exchange cycle, 83 heat exchanger, 63 effectiveness, 147, 148 high reaction stage, 349 high temperature operation, 510 hybrid engine, 591 hydraulic diameter, 33 hydro-turbomachines, 16 hydrogen preferential burning, 394 hydroxylation, 394 hypersonic flow, 25 ideal cycle, 79 assumptions, 80 ideal energy transfer, 286 ideal gas, 11 ignition, 404 ignition delay, 396 impeller, 280, 293 blade shape on performance, 294 channel, 296 components, 280 hub, 281 inducer, 281 radial-vaned, 310 shroud, 281 vanes, 281 impulse machines, 422 turbine, 428 impulse stage, 421 single, 422 impulse turbine pressure compounding of multistage, 435 inducer, 281, 293 inlet casing, 280, 291 inlet diffuser, 238 inlet drag, 237 inlet guide vanes, 287, 336, 337 inlet momentum, 237 inlets, 481

623

design variables, 483 performance of, 482 subsonic, 482 supersonic, 486 intercooled and reheat cycle, 96 intercooled cycle, 92 intercooled cycle with heat exchange and reheat, 99 intercooled cycle with heat exchanger, 94 ion rocket, 583 ion rocket propulsion, 602 irreversible process, 13 isentropic flow, 34 isentropic process, 13 isentropic velocity ratio, 450 jet propulsion, 213 analysis, 213 cycle, 213 principle, 213 Joule cycle, 229 Kelvin-Planck’s statement, 12 laminar flow, 34 leaving losses, 243 lift, 56 lift force, 59 liquid propellant, 598 advantages, 600 bipropellant, 598 monopropellant, 598 requirement, 600 liquid propellant rocket, 589 liquid propellant rocket engine, 595 gas pressurization system, 596 pump pressurization system, 596, 597 loading coefficient, 345, 451 loss, 451 cascade, 356, 451 coefficient, 439 cold, 399 frictional, 310 hot, 399 leaving, 243, 451 mechanical, 150 parasitic, 144

624

INDEX

pressure, 143, 399 profile, 354 stage, 356 tip clearance, 355 total pressure, 403 low reaction stage, 347 Mach angle, 22 Mach cone, 22 Mach lines, 22 Mach number, 17, 22 Mach waves, 22 machine half degree reaction, 423 multistage, 423 machines axial, 44 classification of, 44 impulse, 45 reaction, 45 magneto plasma rocket engine, 604, 605 mass velocity, 32 match point, 525 matching procedure, 543 maximum utilization factor, 430 momentum equation, 7 momentum thrust, 236 monopropellant, 593 multistage rocket, 608 negative energy transfer, 42 negative reaction, 449 net thrust, 236 Newton’s laws second law of, 213 third law of, 213 Newtonian fluid, 21 NOx formation, 557 NOx reduction, 559 noise, 560, 575 broad-band, 561 core, 561, 562 fan, 561 jet, 561 multiple-tone, 562 reduction, 565 sources of, 560

standards, 562 non-flow process, 13 normal shock wave, 28 normal shock waves, 28 nozzle, 435, 481 coefficients, 491 convergent, 488 convergent-divergent, 489 efficiency, 451 ejector, 489 exhaust, 488 flow, 421 flow characteristic, 536 functions, 489 performance, 491, 496 stator, 421 variable-area, 490 nuclear propulsion, 583, 601 nuclear propulsion unit, 601 oblique shock, 30, 31 off-design characteristic, 361 off-design operation, 359 one-dimensional flow area, 493 optimum expansion ratio, 587 overall efficiency, 245 performance characteristics, 310, 523, 526 performance coefficients, 344 performance evaluation, 523 performance graphs, 451 performance parameter, 307 photon propulsion, 602, 606 pi theorem, 50 piloted operation, 215 pilotless operation, 215 plasma jet, 583 plasma rocket propulsion, 602, 603 polytropic efficiency, 152, 155 positive energy transfer, 42 power propulsive, 240, 243 thrust, 244 power input factor, 307 practical cycle, 137 pressure, 8 pressure coefficients, 308

INDEX

pressure drop coefficient, 239 pressure loss, 143 pressure loss factor, 399 pressure thrust, 236 primary zone, 407, 409 process, 8 adiabatic, 13 flow, 14 irreversible, 12, 13 non-flow, 13 reversible, 12, 14 steady flow, 45 profile loss, 354 propellant, 581, 593 characteristics, 593 free radical, 589 grain, 589, 590 liquid, 589 solid, 589, 592, 593 propellant grain, 589, 590 propeller engine, 214 propulsion arc plasma rocket engine, 604 comparison systems, 609 electro dynamic, 602 electrodynamics, 583 free radical, 601 ion rocket, 602 magneto plasma rocket engine, 604, 605 nuclear, 583, 601 photon, 583, 602, 606 plasma rocket, 602, 603 principle, 583 propulsive efficiency, 242, 244, 609 power, 240 propulsive power, 243 pulse jet advantages, 223 applications, 224 disadvantages, 223 engine, 220 pump pressurization system, 596, 597 quasi-one-dimensional flow, 25, 26 radial-tipped blade, 289

625

ram effect, 216, 246 ram efficiency, 238 ram pressure, 216 ramjet advantages, 219 applications, 220 characteristics, 220 disadvantages, 220 engine, 216 performance, 218 thermodynamic cycle, 217 Rayleigh-type flow, 34 reaction machine, 422 pure, 435 reaction stage, 421, 422 fifty per cent, 443 hundred per cent, 448 single, 422 zero degree, 442 reaction turbine, 421, 435 fifty per cent, 445 multistage, 436 reciprocating engine, 214 relative eddy, 296, 297 reversible process, 12 Reynolds number, 21 rocket, 581 advantages, 582 chemical, 582, 584, 589 classification, 582 comparison, 606 engine, 581 free radical, 589 hybrid engine, 591 ion, 583 liquid propellant, 589 optimum expansion ratio, 587 performance capabilities, 582 solid propellant, 589 rocket engine, 214, 581 liquid propellant, 595 rocket propulsion, 581 principle, 583 rotating machines, 39 dimensional analysis, 48 non-dimensional parameters, 51 rotating stall, 364 rotational effects, 41

626

INDEX

rotor blade row, 351 rotor enthalpy loss coefficient, 344 rotor pressure loss coefficient, 344 secondary loss, 354 shock, 470 shock wave, 28, 30 shroud, 281 single-spool turbojet engine, 533 skin friction, 36 slip factor, 297, 299 slip velocity, 297, 299 small-stage efficiency, 152 solid propellant, 592, 593 advantages, 591 composite, 592 double-base, 592 solid propellant rocket, 589 specific heat constant-pressure, 10 constant-volume, 10 specific impulse, 237 specific static thrust, 237 specific thrust, 237 stage efficiency, 427 pressure ratio, 450 stage losses, 356 stage velocity triangles, 336 stage work, 356 staging, 607 stagnation density, 18 enthalpy, 17 pressure, 18, 139 pressure loss coefficient, 439 properties, 138 state, 19 temperature, 17, 138 values, 18 velocity sound, 18 stalling, 361, 363 Stanitz’s formula, 301 state, 8 stator blade row, 353 steady-state operating point, 535 Stodola’s formula, 300 stream tube, 20

streamline, 19 streamtube area-velocity relation, 25 subsonic compressor, 467 suction side, 56 supersonic axial flow compressors, 469 supersonic axial flow turbines, 475 supersonic compressor, 467–469 supersonic radial compressor, 474 surge line, 490 surging, 311, 361 swirl atomizer, 414 system, 7 closed, 8 open, 8 temperature, 8 thermal efficiency, 240 thermal jet, 214 thermodynamic first law of, 7, 9 second law of, 7, 12 throat, 27 thrust, 41, 235, 236, 586 augmentation, 246, 247 basic laws of, 235 equation, 235 gross, 237 net, 236 power, 242, 244 pressure, 236 reversers, 491 specific, 237 tip clearance, 356 tip clearance loss, 355 torque-speed characteristics, 527 transmission efficiency, 242 transonic flow, 25 trigonometrical relations, 425 turbine, 3, 47, 421 blades, 501 efficiency, 140, 141 impulse, 421 reaction, 421 turbine blades, 501 turbine rotor, 42 turbo-compressor, 279 turbojet, 213 advantages, 234

INDEX

627

components, 227 disadvantages, 235 efficiency, 238 engine, 227 performance, 232 thermal efficiency, 239, 240 thermodynamics cycle, 229 turboprop, 213 advantages, 226 applications, 227 components, 224 disadvantages, 227 engine, 224 performance, 226 thermodynamic cycle, 225 turboprop engines, 280 turbulence, 35 turbulent flow, 34

inlet guide, 287, 336, 337 upstream guide, 337 variable specific heat, 149 velocity coefficient, 494 velocity triangle, 287, 297, 336, 341, 349, 424, 428, 431 entry, 287 exit, 289, 295 velocity-compounding, 430 virtual pressure rise, 43 viscosity, 21 dynamic, 21 kinematic, 21 volute casing, 306 volute passage, 307 cross-section, 307 vortex generators, 485

vanes, 283 curved, 283

zone of action, 22 zone of silence, 22

water-alcohol mixture, 250 waves propagation, 23 unique incidence, 477 utilization factor, 427–429, 434, 445, work done factor, 340 work ratio, 143 447