Power system operation and control 9788131755914, 9789332501157, 8131755916, 9788131793916, 8131793915

Table of contents :
Cover......Page 1
About the Author......Page 7
Contents......Page 8
Preface......Page 10
1.1.1 Unit Commitment Problem......Page 12
1.2.1 Input–Output or Heat Characteristic......Page 13
1.2.2 Cost Curve and Incremental Fuel Cost Characteristic......Page 14
1.3 Objective Function and Constraints for the Economic Dispatch Problem......Page 19
1.4.1 Nonlinear Function Optimization Considering Equality Constraints......Page 20
1.4.2 Nonlinear Function Optimization Considering Equality and Inequality Constraints......Page 22
1.5 Economic Dispatch Problem – Neglecting Transmission Line Losses......Page 23
1.5.1 Solution of Economic Dispatch Problem – Neglecting Losses and Generator Limits......Page 24
1.5.2 Algorithm – Economic Dispatch Without Considering Generator Limits......Page 25
1.5.3 Solution of Economic Dispatch Problem – Neglecting Losses and Including Generator Limits......Page 30
1.5.4 Algorithm – Economic Dispatch Including Generator Power Limits......Page 31
Questions from Previous Question Papers......Page 35
Competitive Examination Questions......Page 37
2.2 Economic Scheduling Problem Considering Losses......Page 40
2.3 Derivation of the Transmission Loss Formula......Page 42
2.4 Solution of the Economic Scheduling Problem – Considering Transmission Line Losses......Page 51
2.4.1 Algorithm......Page 53
2.5 Economic Dispatch Considering Losses – the Classical Method......Page 62
2.5.1 Algorithm......Page 63
Questions from Previous Question Papers......Page 71
Competitive Examination Questions......Page 73
3.1.1 Long Range Hydro Thermal Scheduling......Page 76
3.2 Long Range Hydro Thermal Scheduling......Page 77
3.2.1 Determination of Optimal Contribution of Thermal Plant......Page 79
3.2.2 Short Term Hydro Thermal Scheduling......Page 81
Questions from Previous Question Papers......Page 85
4.1 Introduction......Page 86
Mathematical Model of the Speed Governing System......Page 87
4.4 Generator Load Model......Page 90
4.5 Representation of Loads......Page 91
4.6 Turbine Model......Page 92
4.7 Synchronous Machines......Page 93
4.8 the Swing Equation......Page 95
4.9.1 Exciter System Block Diagram......Page 96
4.9.2 Ieee Type-1 Excitation Model:-......Page 97
Question from Previous Question Papers......Page 102
5.1 Introduction......Page 104
5.2 Control Area Concept......Page 105
5.3 Isolated Block Diagram Representation of Single Area Frequency Control......Page 106
5.4 Steady State Response......Page 107
5.5 Dynamic Response......Page 110
5.6 State Space Model for Single Area......Page 111
5.7 Matrix Representation of All State Equations......Page 112
Questions from Previous Question Papers......Page 122
Competitive Examination Questions......Page 123
6.1.1 Flat Frequency Control of Inter Connected System......Page 126
Assumptions Made in the Analysis of a Two-Area System......Page 127
6.2 Two-Area State Space Model Representation......Page 128
6.3.1 Uncontrolled Case......Page 132
6.3.2 Controlled Case......Page 135
Question from Previous Question Papers......Page 141
Competitive Examination Questions......Page 142
7.2 Proportional Plus Integral Controller......Page 144
7.3 Load Frequency Control and Economic Dispatch Control......Page 147
Questions from Previous Question Papers......Page 150
Competitive Examination Question......Page 151
8.2 Power Factor......Page 152
8.3 Reactive Power......Page 154
8.4 Causes of Low Power Factor......Page 156
8.4.1 Disadvantages of Low Power Factor......Page 157
8.5 Reactive Power Flow in an Uncompensated Transmission Line......Page 160
8.6 Necessity for Reactive Power Compensation in a Transmission Line......Page 161
8.7.1 Synchronous Generator Control......Page 162
8.7.2 Capacitor Control......Page 163
8.8 Power Factor Correction (Loadcompensation)......Page 165
8.8.1 Illustration of Power Factor Correction......Page 166
Question from Previous Question Papers......Page 167
Answers to Selected Competitive Examination Questions......Page 168
Index......Page 170

Citation preview

Power System Operation and Control

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Power System Operation and Control

Professor and Head, Department of Electrical and Electronics Engineering, JNTU College of Engineering, Jagityal, Karimnagar (D), Andhra Pradesh

Chennai • Delhi • Chandigarh

Copyright © 2011 Dorling Kindersley (India) Pvt. Ltd. Licensees of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISBN 9788131755914 eISBN 9789332501157 Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India

In Memory of My Father, Dr N. H. Sitarama Sarma.

About the Author N. V. Ramana is Professor and Head, Department of Electrical and Electronics Engineering in Jawaharlal Nehru Technological University (Hyderabad) College of Engineering, at Jagityal, Karimnagar (D), Andhra Pradesh. His areas of interest include power system operations and control, and dynamics and design of intelligent controllers for power system problems. He is an M.Tech. from S. V. University, Tirupati and a PhD degree holder from JNTU, the university he is associated with since 1992. Professor Ramana has about 18 years of teaching and two years of industrial experience. He has more than 20 research papers to his credit. He has presented papers at international conferences held in Singapore, Vancouver (Canada) and Las Cruces (the USA). An assiduous academician, Professor Ramana has visited a number of universities in the course of his illustrious career including the New Mexico State University, Georgia Tech, Virginia Tech (the USA) and Nan Yang University (Singapore) besides all the IITs in India. He is presently guiding eight PhD students.

Contents About the Author

vi

Preface

ix

1

Economic Operation of Power System—I 1 1.1 Introduction 1 1.2 Characteristics of Thermal Units 2 1.3 Objective Function and Constraints for the Economic Dispatch Problem 8 1.4 Lagrange Multiplier Method – An Overview 9 1.5 Economic Dispatch Problem – Neglecting Transmission Line Losses 12 Questions from Previous Question Papers 24 Competitive Examination Questions 26

2

Economic Operation of Power System—II 2.1 Introduction 2.2 Economic Scheduling Problem Considering Losses 2.3 Derivation of the Transmission Loss Formula 2.4 Solution of the Economic Scheduling Problem – Considering Transmission Line Losses 2.5 Economic Dispatch Considering Losses – The Classical Method Questions from Previous Question Papers Competitive Examination Questions

40 51 60 62

3

Hydro Thermal Scheduling 3.1 Introduction 3.2 Long Range Hydro Thermal Scheduling Questions from Previous Question Papers

65 65 66 74

4

Modelling of Turbine, Generators and Automatic Controllers 4.1 Introduction 4.2 Modelling of Turbine Speed-Governor Controller 4.3 Modelling of Steam Turbine 4.4 Generator Load Model 4.5 Representation of Loads 4.6 Turbine Model 4.7 Synchronous Machines 4.8 The Swing Equation

75 75 76 79 79 80 81 82 84

29 29 29 31

4.9 5

Excitation Question from Previous Question Papers

85 91

Single Area Load Frequency Control 5.1 Introduction 5.2 Control Area Concept 5.3 Isolated Block Diagram Representation of Single Area Frequency Control 5.4 Steady State Response 5.5 Dynamic Response 5.6 State Space Model for Single Area 5.7 Matrix Representation of all State Equations Questions from Previous Question Papers Competitive Examination Questions

93 93 94 95 96 99 100 101 111 112

6

Two-Area Load Frequency Control 6.1 Load Frequency Control of Two-Area System 6.2 Two-Area State Space Model Representation 6.3 Steady State Analysis Question from Previous Question Papers Competitive Examination Questions

115 115 117 121 130 131

7

Load Frequency Controllers 7.1 Introduction 7.2 Proportional Plus Integral Controller 7.3 Load Frequency Control and Economic Dispatch Control Questions from Previous Question Papers Competitive Examination Question

133 133 133 136 139 140

8

Reactive Power Control 8.1 Introduction 8.2 Power Factor 8.3 Reactive Power 8.4 Causes of Low Power Factor 8.5 Reactive Power flow in an Uncompensated Transmission Line 8.6 Necessity for Reactive Power Compensation in a Transmission Line 8.7 Methods to Improve Power Factor 8.8 Power Factor Correction (Load Compensation) Question from Previous Question Papers

141 141 141 143 145 149 150 151 154 156

Answers to Selected Competitive Examination Questions

157

Index

159

Preface It gives me immense pleasure and satisfaction to present this textbook on Power System Operation and Control to cater to the needs of B.Tech. electrical engineering students. The book has been written in a lucid manner for the benefit of beginner students. The basics are covered well and expounded with a view to assist the teachers in helping the student to assimilate the essence of the subject. Novel techniques have been adopted wherever required to drive home the complex aspects without compromising on standard and quality. Questions and problems from previous years’ examinations of JNTU (Hyderabad, Anantapur and Kakinada) have been incorporated to enable students to sail through their term-end tests with ease. Th e book also includes objective questions from IES, GATE and other competitive examinations to empower students preparing for competitive examinations. Solved problems provided in each chapter illustrate and reinforce the concepts explained therein. I welcome constructive criticisms and suggestions for improving the content and presentation of this text. These will be gratefully acknowledged.

ACKNOWLEDGEMENTS Thanks are due to my PhD guide Dr N. Yadaiah, Professor and Head, Department of EEE, JNTU Anantapur, for giving me the idea to write this book. I am also indebted to my colleague Dr S. V. Jayaram Kumar, Professor, Department of EEE, College of Engineering, JNTU Hyderabad, for his expert suggestions. I would not have been able to complete this book if it were not for the cooperation extended by my wife Dr P. V. Naga Prapurna, my mother N. Satyavaty and my daughter N. Divya Sai. I dedicate this book to the memory of my father Dr N. H. Sitarama Sarma. Dr N. V. Ramana

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Economic Operation of Power System—I 1.1

INTRODUCTION

The problem of economic operation of a power system or optimal power flow (OPF) can be stated as: Allocation of the load (MW) among the various units of generating station and among the various generating stations in such a way that, the overall cost of generation for the given load demand is minimum. This is an optimization problem, the objective of which is to minimize the power generation cost function subject to the satisfaction of a given set of equality and inequality constraints. The problem is analyzed, solved and then implemented under online condition of the power system. The input data for the problem comes from conventional power flow study. For a given load demand, power flow study can be used to calculate of active and reactive power generations, line flows and losses. The study also furnishes some control parameters such as the magnitude of voltage and voltage phase differences. The economic scheduling problem can be understood as an outcome of multiple power flow studies, where a particular power flow study result is considered more appropriate in terms of cost of generation. The solution to this problem cannot be optimal unless otherwise all the constraints of the system are satisfied. We discuss the economic scheduling problem in the following sections, but first we consider the constraints that need to be addressed. The problem of economic operation of the power system involves two subproblems, namely, unit commitment (UC) and economic dispatch (ED). While UC is an off-line problem, ED is an area of online concern. The result of UC is an initial solution of the ED problem.

1.1.1 Unit Commitment Problem Generally the load curve of a plant where there is a variation of load vs time in a season, follows a particular trend or pattern. Hence these curves can be predicted with minimum degree of error. Let N be the number of generating units available in a plant. The problem of UC is to forecast how the plant load should be allocated amongst the N units in such a way that the overall cost of generation of the plant is minimum. In other words, the solution of UC for N available units is a schedule of the units that must be kept in ‘on’ (committed) condition or in ‘off’ (decommitted) condition. The problem of UC is complex, since a large number of constraints such as

2

Power System Operation and Control

start-up and shut-down costs and times, fuel cost per KWh, minimum and maximum generating capacities need to be addressed.

1.1.2 Economic Dispatch Problem In contrast to the UC problem, ED assumes that N number of units in the different generating stations are already committed or in ‘on’ condition. The problem here is to forecast how the load should be shared amongst them in such a way that the overall cost of generation is minimum. The solution to UC is the starting point of the ED problem. That is, once a decision on the units that are to be committed is taken, then ED resolves the distribution of load amongst the committed units. If all the units are in ‘on’ condition, UC does not exist and we have to take up only the ED problem. Since UC is beyond the scope of this chapter, let us focus our discussion on the economic dispatch problem for the following case studies.

ED Problem Case Studies

Case 1 Optimal load allocation amongst thermal units neglecting transmission losses.

Case 2 Optimal load allocation amongst thermal units considering transmission losses.

Case 3 Economic dispatch of load amongst the thermal and hydro units – hydro thermal coordination

This chapter deals with Case 1 while the following two chapters present the other two case studies.

1.2 CHARACTERISTICS OF THERMAL UNITS The cost of power generation by any thermal unit depends upon its characteristics. Performance of any unit can be understood by studying the generator’s input–output characteristics and the cost curve. Characteristics of individual machines are drawn on graphs and later converted into suitable mathematical equations using curves fitting methods. These mathematical equations are used to solve the economic scheduling problem. The input of a thermal unit is expressed in kcal/hr in terms of heat supplied or in Rs/hr in terms of cost of fuel, while the output is expressed in of MW.

1.2.1 Input–Output or Heat Characteristic Heat curve is a graph drawn between fuel input in Btu/h or kcal/hr and power output in MW on x- and y-axis respectively, typical curve for a thermal unit takes the form as shown in Fig. l.l. The slope of the curve at different powers can be calculated. The slope is defined as heat rate or incremental fuel rate which is the ratio of fuel input to the corresponding power output and has the units Btu/MWh or kcal/MWh.

Fuel (kcal/hr)

Economic Operation of Power System—I

3

Power MW ) )

Pmin

Pmax

Fig 1.1 Heat curve of a thermal unit

Incremental fuel rate =

dF input = dP output

Heat rate (slope of the heat curve) can be calculated at every point on the heat curve and another characteristic incremental heat curve can be obtained. This graph is drawn between incremental fuel rate in kcal/MWh and output in MW. The graph is shown in Fig. 1.2. Incremental fuel characteristic is very important because it measures the thermal efficiency of a thermal unit and provides an estimate of fuel input needed for every additional MW generation. Different units have different incremental fuel rates for a given MW value. For optimal power generation, it is economical to allocate the load to the unit with minimum incremental fuel rate.

1.2.2 Cost Curve and Incremental Fuel Cost Characteristic

Incremental fuel rate (kcal / MWh)

Knowing the specific heat (in kcal/kg) of the coal used, fuel input (in kcal/hr) and incremental fuel rate (in kcal/MWh) can be converted into kg/hr and kg/MWh respectively. Further, if cost of each kilogram of coal is known, then Kg/hr and Kg/MWh can be converted into Rs/hr and Rs/MWh.

Pmin

Pmax

Power (MW)

Fig 1.2 Incremental fuel rate characteristic of a thermal unit

Power System Operation and Control

Incremental fuel cost (Rs /MWh)

4

Power (MW)

Fig 1.3 Cost curve

Now, the incremental fuel cost curve can be drawn by taking incremental fuel cost in Rs/MWh as an input on the y-axis and power in MW as an output on the x-axis. The cost curve is shown in Fig. 1.3. Similarly, the input–output characteristic given in kcal/hr versus MW can be converted into a cost curve drawn as Rs/hr versus MW as shown in Fig. 1.4. Let the fuel cost curve for the ith generator be in the form as shown in Fig. 1.4. The curve can be fitted in a quadratic equation as: Ci = αi + βi Pi + γi Pi2 Rs/hr

(1.1)

and incremental fuel cost can be written as: dCi = 2γi Pi + βi Rs/MWh dPi

Example 1.1 The fuel input characteristics for two plants are given by F1 = (800 + 22 P1 + 0.04 P12 ) × 106 kcal /hr

Fuel cost (Rs / hr)

F2 = (1000 + 15 P2 + 0.045 P22 ) × 106 kcal /hr

Power (MW)

Pmin

Pmax

Fig 1.4 Cost curve of a thermal unit

(1.2)

Economic Operation of Power System—I

5

Table 1.1 The inputñoutput characteristics P1 (MW) 10 20 30 40 50 60 70 80 90 100

F1 × 106 kcal/hr 1024 1256 1496 1744 2000 2264 2536 2816 3104 3450

P2 (MW) 10 20 30 40 50 60 70 80 90 100

F2 × 106 kcal/hr 1154.5 1318 1490.5 1672 1862.5 2062 2270.5 2488 2714.5 2950

where P1 and P2 are in MW. (i) plot the input–output characteristic for each plant; (ii) plot the heat rate characteristic for each plant; (iii) assuming the cost of fuel as Rs 110/ton calculate the incremental production cost characteristics in Rs/MWh for each plant. Plot the characteristics against power produced in MW. Solution: (i) Input–output characteristic: Consider the power in steps of 10 MW. Fuel input characteristics are given as F1 = (800 + 22 P1 + 0.04 P12 ) × 106 kcal/hr F2 = (1000 + 15 P2 + 0.045 P22 ) × 106 kcal/hr By substituting P1 and P2 in the above equations, F1 and F2 values can be computed. The input–output characteristics are plotted in Fig. 1.5 (see Table 1.1). (ii) Heat rate characteristic: Let P1 = 10 MW at Plant 1 Heat rate at this plant =

1024 × 106 kcal/hr F1 = = 102.4 × 106 kcal/MWh P1 10 MW

At Plant 2 P2 = 10 MW Heat rate at this plant =

F2 1154.5 × 106 kcal/hr = = 115.45×106 kcal/MWh P2 10 MW

Repeat the above calculation for powers in steps of 10 MW till power at both the plants becomes 100 MW. This is shown in Fig. 1.6. (Table 1.2)

6

Power System Operation and Control

Table 1.2 Heat rate calculations Heat rate ×106 for Plant 1 kcal/MWh 102.4 62.8 49.86 43.6 40 37.25 36.22 35 2 34.48 34.5

Output MW 10 20 30 40 50 60 70 80 90 100

Heat rate ×106 or Plant 2 kcal/MWh 115.45 65.9 49.68 41.8 37.25 34.36 32.42 31.1 30.56 29.5

(iii) Incremental production cost characteristic: † Calorific value of fuel in the plants is 5600 kcal/hr Cost of fuel = Rs 110 per ton = Rs 110/1000 per kg 1 kg 110 110 Rs × = Rs/kcal 6 dF1 1000 kg 5600 kcal 6 5.6 × 10 = (0.08 P1 + 22) × 10 kcal/MWh dP1 110 Rs/MWh IC1 = (0.08 P1 + 22) × 106 × 5.6 × 106 = (432.1 + 1.54 P1 ) Rs/MWh −

At 100 MW generation the cost is IC1 = (432.1 + 1.54 × 200) = 740 Rs/MWh † Incremental production cost involes cost of land, cost of maintenance, cost of equipment, cost of installation, cost of wages etc. Since it is cumbersome to obtain these values seperately, these are normally taken as a percentage of incremental fuel cost.

3500 P1

kcal / hr

3000 2500

P2

2000 1500 1000

0

10 20 30 40 50 60 70 80 90 100 Power in MW

Fig 1.5 The inputñoutput characteristic curve

Economic Operation of Power System—I

7

120

kcal/MWh

110 100 90 80 70 60 50 40 30 20 10

P1 P2 20

30

40 50 60 70 power (MW)

80

90 100

Fig 1.6 Heat rate curves The incremental production cost is obtained by adding the maintenance costs of 10% (Table 1.3). Hence, the incremental production cost at Plant 1 = (432.1 + 1.54 P1 ) × 1.1 = 475.31 + 1.69 P1 At P1 = 10 MW IPC1 = (432.1 + 1.69 × 10) = 492.25 Rs/MWh Similarly at Plant 2 dF2 = (0.09 P2 + 15) × 106 kcal/MWh dP2 110 = (0.09 P2 + 15) × 106 5.6 × 106 = 1.76 P2 + 294 Rs/MWh Considering the maintenance cost at 10% The incremental production cost at Plant 2 is = (1.76 P2 + 294) × 1.1 = (1.936 P2 + 323.4) Rs/MWh Table 1.3 Incremental production costs Power(P) 10 20 30 40 50 60 70 80 90 100

IPC at Plant 1 Rs/MWh 492.4 509.12 526 542.9 558.8 576.71 593.6 610.5 627.41 644.3

IPC at Plant 2 342.76 362.11 381.48 400.83 420.2 439.56 458.9 478.28 497.64 517

8

Power System Operation and Control

At P2 = 10 MW IPC2 = (323.4 + 1.936 × 10) = 324.76 Rs/MWh In a similar fashion calculate IPC for all in steps of 10 MW till power is 100 MW. The characteristics are shown in Fig. 1.7.

1.3 OBJECTIVE FUNCTION AND CONSTRAINTS FOR THE ECONOMIC DISPATCH PROBLEM As discussed earlier, the ED problem deals with the optimal allocation of total load demand PD amongst n-number of units. The objective function for the ED problem is presented below. The problem is to minimize the function Eq. (1.3) subject to the satisfaction of constraints. CTotal = f (PG 1 , PG 2 , . . . , PGn )

(1.3)

where CTotal is the total cost of generation and PG 1 , PG 2 , . . . , PGn are individual generations of n-number of units. The ED problem may be treated as parameter (cost) optimization subject to the satisfaction of system constraints. System constraints are of two types: (i) equality constraints, (ii) inequality constraints. Equality constraints These constraints are due to functional dependencies. Let f and g have functional dependencies and the problem is to minimize the function f (x1 , x2 , x3 ), where x1 , x2 , x3 are variables whose values are computed such that the given function f is minimized . The computed values of x1 , x2 , and x3 are said to be optimal when they minimize both the function f and satisfy the following equality constraints which are described by a set of equations: gi (x1 , x2 , x3 ) = 0 for i = 1, 2, 3, . . . , for x set of equations. In the ED problem, the equality constraint functional dependency equations are static power flow equations. For optimized scheduling of generation cost for the given demand, there is a necessity for static power flow equations to be satisfied. The other equality constraint is that sum of power generation by individual units must be equal to the total power demand as: n 

PGi = PD for i = 1, 2, . . . , n.

(1.4)

i=1

or gi (PG 1 , PG 2 , . . . , PGn ) =

n  i=1

PGi − PD = 0.

(1.5)

Economic Operation of Power System—I

9

Inequality constraints The following inequality constraints may be included in the ED problem. (1) Each generating unit of a given n number must operate at a limit equal to, or below and above its maximum and minimum limits respectively. Pimin ≤ Pi ≤ Pimax i = 1, 2, . . . , n.

(1.6)

(2) Reactive power (Q ) generated by individual generating units should be less than the maximum value to avoid rotor over-heating and should be more than minimum value to ensure optimal power transfer capability of the system. (3) Total MVA rating of individual units, to ensure there is no excess heating produced in the unit. (4) Over-loading condition of equipment like transformers and lines. (5) Limits to voltage magnitudes and angles to ensure that the system has better stability margins. Low voltage gives rise to instability and unsatisfactory operation of loads dominated by induction motor type. High voltage problems shall lead to insulation breakdowns. (6) Transformer tap positions: The tap position should be between the minimum and maximum values. (7) Reserve capacities of generating units: To meet the sudden increase of power demand and to maintain good steady-state stability of the system. The above constraints are basically control parameters, which are required to be within their limits for satisfactory operation of the power system. Though the primary interest is minimization of objective functions, owing to operational limitations these constraints are important and hence must be included in the study.

1.4 LAGRANGE MULTIPLIER METHOD – AN OVERVIEW Nonlinear function optimization problems such as ED can be solved by the Lagrange multiplier method.

1.4.1 Nonlinear Function Optimization Considering Equality Constraints Let the problem be to minimize the function f (x1 , x2 , . . . , xn )

(1.7)

Subject to k number of equality constraints gi (x1 , x2 , . . . , xn ) = 0 for i = 1, 2, . . . , k.

(1.8)

The constrained function f can be written as an unconstrained function with the help of the Lagrange method as: £=f +

k  i=1

λi gi .

(1.9)

10

Power System Operation and Control

In Eq. (1.12) £ is the Lagrange function and λ is the Lagrange multiplier. The necessary condition for minimum £ can be obtained from:  ∂gi ∂f ∂£ = + λi = 0 for i = 1, 2, . . . , n. ∂xi ∂xi ∂xi

(1.10)

∂£ = gi = 0 for i = 1, 2, . . . , k. ∂λi

(1.11)

k

i=1

Equations (1.11) represents the original constraints. This method is adopted to solve the ED problem and is explained in detail in the following sections.

Example 1.2 Minimize f(x,y) = Subject to g(x,y) =

5 xy 2 x 2 + y2

– 64 =0

Solution: The Lagrange function is: Ł(x, y, λ) = f (x, y) + λg (x, y) = 5x −1 y −2 + λ(x 2 + y 2 − 64) The necessary conditions for the minimization of f (x, y) is given by dl = −5x −2 y −2 + 2xλ = 0 dx dl = −10x −1 y −3 + 2yλ = 0 dy dl = x 2 + y 2 − 64 = 0 dλ

(1) (2) (3)

From (1), 2λ = 5x −3 y −2

(4)

2λ = 10x −1 y −4

(5)

From (2),

1 From (4) and (5), the relation x = ( √ ) y can be obtained. 2 This relation, along with (3), gives the optimum solution. Replacing y in (3), x 2 + 2x 2 = 64 (or) and

3x 2 = 64 x = 4.618

Economic Operation of Power System—I

11

replacing x in (3),

Simplifying,

1 2 y + y 2 = 64 2  64 × 2 y= = 6.532 3

1.4.2 Nonlinear Function Optimization Considering Equality and Inequality Constraints Majority of optimization problems contain both equality and inequality constraints. Let the problem be to minimize the function: f (x1 , x2 , . . . , xn )

(1.12)

subject to k number of equality constraints gi (x1 , x2 , . . . , xn ) = 0 for i = 1, 2, . . . , k.

(1.13)

and m number of inequality constraints qi (x1 , x2 , . . . , xn ) ≤ 0 for i = 1, 2, . . . , m.

(1.14)

The inequality constraints, as mentioned earlier, are independent control parameters or undetermined quantities. These constraints are bounded to certain limits. By introducing m vector of μ undetermined quantities, the constrained function f can be written as an unconstrained function with the help of Lagrange method as: k m   £=f + λi gi + μ j qj (1.15) i=1

j =1

The necessary condition for minimum £ can be obtained from:  ∂gi  ∂gi ∂f ∂£ = + λi + μj = 0 for i = 1, 2, . . . , n. ∂xi ∂xi ∂xi ∂xi k

m

i=1

j =1

∂£ = gi = 0 for i = 1, 2, . . . , k. ∂λi ∂£ = gj ≤ 0 for j = 1, 2, . . . , m. ∂μj μj gj = 0 and μj > 0 for j = 1, 2, . . . , m.

(1.16)

(1.17) (1.18) (1.19)

Note that Eq. (1.17) is the original equality constraint. The necessary conditions discussed above are known as the Kuhn-Tucker conditions.

12

Power System Operation and Control

Example 1.3 Minimize f(x,y) = x 2 + y 2 Subject to inequality constraint: g(x, y) = (x 2 + y 2 − 64) = 0 Subject to equality constraint: u(x, y) = x + y ≥ 4 Solution: The constrained objective function can be converted into an unconstrained function by using the Lagrange method as: L = x 2 + y 2 + λ(x 2 + y 2 − 64) + μ(x + y − 4) The necessary conditions are: dl = 2x + 2λx + μ = 0 dx dl = 2y + 2λy + μ = 0 dy

(1) (2)

dl = x 2 + y 2 − 64 = 0 dλ dl =x +y−4 =0 dμ

(3) (4)

Solving the above equations for x, y shall yield an optimal solution, which minimizes the objective function. The procedure is similar to that given in Example 1.2.

1.5 ECONOMIC DISPATCH PROBLEM – NEGLECTING TRANSMISSION LINE LOSSES The problem of ED can be easily solved when transmission line losses are neglected. As the losses are neglected, the system model can be understood as shown in Fig. 1.7, where n number of generating units are connected to a common bus bar, collectively meeting the total power demand PD . It should be understood that share of power demand by the units does not involve losses. Since transmission losses are neglected, total demand PD is the sum of all generations of n-number of units. For each unit, a cost functions Ci is assumed and the sum of all costs computed from these cost functions gives the total cost of production CT . CT =

n  i=1

Ci for i = 1, 2, . . . , n

(1.20)

Economic Operation of Power System—I

G1

C1 P1

G2

C2

Cn

P2

13

Gn Pn

PD

Fig 1.7 System with n-generators where the cost function of the ith unit, from Eq. (1.1) is: Ci = αi + βi Pi + γi Pi2 Now, the ED problem is to minimize CT , subject to the satisfaction of the following equality and inequality constraints. Equality constraint The total power generation by all the generating units must be equal to the power demand. n  Pi = PD (1.21) i=1

where Pi = power generated by ith unit PD = total power demand. Inequality constraint Each generator should generate power within the limits imposed. Pimin ≤ Pi ≤ Pimax i = 1, 2, . . . , n

(1.22)

Economic dispatch problem can be carried out by excluding or including generator power limits, i.e., the inequality constraint.

1.5.1 Solution of Economic Dispatch Problem – Neglecting Losses and Generator Limits The constrained total cost function Eq. (1.20) can be converted into an unconstrained function by using the Lagrange multiplier as:   n  Pi (1.23) £ = CT + λ PD − i=1

The conditions for minimization of objective function can be found by equating partial differentials of the unconstrained function to zero as

14

Power System Operation and Control

∂CT ∂£ = + λ (0 − 1) = 0 ∂Pi ∂Pi From Eq. (1.24) λ= Since Ci = C1 + C2 + · · · + Cn

∂CT ∂Pi

(1.24)

(1.25)

∂CT ∂Ci ∂Ci = = =λ ∂Pi ∂Pi ∂Pi

(1.26)

From the above equation the coordinate equations can be written as: ∂C2 ∂C3 ∂Cn ∂C1 = = = ··· = =λ ∂P1 ∂P2 ∂P3 ∂Pn

(1.27)

Using Eq. (l.2),

∂Ci = λ = βi + 2γi Pi ∂Pi The second condition can be obtained from the following equation: n  Pi − PD = 0

(1.28)

(1.29)

i=1

Equations Required for the ED solution For a known value of λ, the power generated by the ith unit from Eq. (1.29) can be written as: λ − βi Pi = (1.30) 2γi Eq. (1.30) is valid, subject to the satisfaction of the equality constraint, which can be written as: n  λ − βi i=1

2γi

n n   1 βi = PD ⇒ λ = = PD 2γi 2γi i=1

From Eq. (1.31) the required value of λ is: n  βi PD + 2γi i=1 λ= n  1 i=1

(1.31)

i=1

(1.32)

2γi

The value of λ can be calculated from Eq. (1.32) and then can be substituted in Eq. (1.31) to compute the values of Pi for i = 1, 2, . . . , n for optimal scheduling of generation.

1.5.2 Algorithm – Economic Dispatch Without Considering Generator Limits Step 1: Read data: Cost coefficients αi ,βi ,γi , and total power demand PD . Step 2: Compute λ value by using Eq. (1.32). Step 3: Obtain economic dispatch solution using the Eq. (1.31) for all the units.

Economic Operation of Power System—I

15

Example 1.4 The incremental costs of the two generating units are as follows: I C1 = (0.1 P1 + 25)Rs/MWh I C2 = (0.2 P2 + 18)Rs/MWh Given that the minimum and maximum powers are 10MW and 100MW for each unit, allocate a total load of 100 MW optimally. Solution: The incremental production cost of the two generating units is given as IC1 = (0.1 P1 + 25)Rs/MWh

(1)

IC2 = (0.2 P2 + 18)Rs/MWh

(2)

For optimal scheduling equate (1) and (2) 0.1 P1 + 25 = 0.2. P2 + 18

(3)

0.1 P1 − 0.2 P2 = −7

(4)

Simplifying the above,

Given total load = 100 MW Hence P1 + P2 = 100MW Solving equations (3) and (4) we get P1 = 43.33 MW and P2 = 56.66MW Both units are operating within their generation limits.

Example 1.5 For the two units in Example 1.4, instead of allocating load optimally, the load is shared equally. Calculate the net loss in Rs/hr for the process. Solution: If loads are shared equally, then the load supplied to the units will be 50 MW each P1 = 50 MW, 50 C1 =

P2 = 50 MW

 (0.1 P1 + 25) dP1 = 0.05 502 − 43.332 + 25 (50 − 43.33)

43.33

= 197.87 Rs/hr (increase in cost – loss) 50 C2 = 56.66

 (0.2 P2 + 18) dP2 = 0.01 502 − 56.662 + 18 (50 − 56.66)  = −190.9 Rs/hr decrease in cost − prof it

16

Power System Operation and Control

Table 1.4 Time 12 Midnight−6 AM 6 AM−12 Noon 12 Noon−3 PM 3 PM−10 PM 10 PM−12 Midnight

Total load 50 MW 120 MW 180 MW 200 MW 80 MW

So. if load is shared equally, Net cost = C1 + C2 = (197.87) + (−190.9) = 6.97 Rs/hr. (net loss)

Example 1.6 The load for the power system consisting of two units in Example 1.3 is varied as shown in Table 1.4. For the shown load variations, allocate the load optimally. Solution: The incremental costs of the two units are IC1 = (0.1 P1 + 25)Rs/MWh

(1)

IC2 = (0.2 P2 + 18)Rs/MWh

(2)

For optimal scheduling equate (1) and (2). 0.1 P1 − 0.2 P2 = −7

(3)

(i) When total load is 50 MW (12 Midnight – 6 A.M) P1 + P2 = 50

(4)

Solving equations (3) and (4) P1 = 10 MW

P2 = 40 MW

(ii) When total load is 120 MW (6 A.M – 12 Noon ) P1 + P2 = 120

(5)

Solving equations (3) and (5) P1 = 56.66 MW

P2 = 63.33 MW

(iii) When total load is 180 MW (12 Noon – 3 P.M) P1 + P2 = 180 Solving equations (3) and (6) P1 = 96.66 MW

P2 = 83.33 MW

(6)

Economic Operation of Power System—I

17

(iv) When total load is 200 MW (3 P.M – 10 P.M) P1 + P2 = 200

(7)

Solving equations (3) and (7) P1 = 110 MW

P2 = 90 MW

(v) When total load is 80 MW (10 P.M – 12 Midnight) P1 + P2 = 80

(8)

Solving equations (3) and (8) P1 = 30 MW P2 = 50 MW

Example 1.7 Incremental fuel cost of two generating units are: IC1 = 30 + 0.2 P1 ; IC2 = 25 + 0.3 P2 Rs/MWh, respectively. Find the savings in fuel cost in rupees annually for optimal scheduling of a total load of 140 MW, as compared to equal distribution of the same load between the two units. Solution: Incremental fuel cost of two generators are given as IC1 = 30 + 0.2 P1

(1)

IC2 = 25 + 0.3 P2

(2)

For optimal scheduling IC1 = IC2 0.2 P1 − 0.3 P2 = −5

(3)

P1 + P2 = 140

(4)

and Solving (3) and (4) we get

P1 = 74MW

P2 = 66MW

For equal sharing of load P1 = P2 = 70MW For saving in fuel cost dF1 = 30 + 0.2 P1 dP1 Integrating the above equation within the limits, we calculate the change in fuel cost F1 70 F1 =

74 (0.2 P1 + 30) dP1 = 0.1 P12 70 + 30 [P1 ]74 70 = −57.6 − 120 = −177.6 (loss)

74

dF2 = 25 + 0.3P2 dP2

18

Power System Operation and Control

Integrating the above equation within the limits we calculate F2 70 F2 =

70 (0.3 P2 + 25) dP2 = 0.15 P22 66 + 25 [P2 ]70 66 = 81.6 + 100 = 181.6 (profit)

66

Net saving = F = F1 = F2 = −177.6 + 181.6 = 4 Rs/hr Net saving in rupees annually = 4 Rs/hr × 8760 hrs in a year = Rs 35,040/-

Example 1.8 The fuel cost functions for three thermal units in Rs/hr are given by: C1 = 400 + 5.2 P1 + 0.002 P12 C2 = 500 + 5.3 P2 + 0.004 P22 C3 = 300 + 5.6 P3 + 0.007 P32 where P1 , P2 and P3 are in MW. For a total load of 1000 MW, find the economic dispatch solution. Solution: Using Eq. (1.33), λ is found to be

λ=

5.2 5.3 5.6 + + 0004 0.008 0.014 1 1 1 + + 0.004 0.008 0.014

1000 +

= 7.532 Rs/MWh Substituting for λ in Eq. (1.31), the optimal dispatch is: P1 =

7.532 − 5.2 = 583 MW 2(0.002)

P2 =

7.532 − 5.3 = 279 MW 2(0.004)

P3 =

7.532 − 5.6 = 138 MW 2(0.007)

Equality constraint PD −

3  i=1

Pi = 0

i.e., 1000 − (583 + 279 + 138) = 0 is satisfied.

Economic Operation of Power System—I

19

1.5.3 Solution of Economic Dispatch Problem – Neglecting Losses and Including Generator Limits For the satisfactory operation of a thermal unit, the power generated by it should not be more than Pmax or less than Pmin This constraint needs to be satisfied owing to thermal lismits imposed on the unit. Now the problem of economic dispatch is, the minimization of the total cost of generation subject to the equality constraint given by Eq. (1.21) and the inequality constraint given by Eq. (1.22). Following the discussion in Section 1.4.2, the necessary Kuhn–Tucker conditions that minimize the Lagrange function by including inequality constraints given by Eq(1.22) are: ⎫ If generations are ⎪ ⎪ ⎪ dCi min ≤ P ≤ P max ⎪ ⎪ = λ for P within limits: i ⎪ i i dPi ⎪ ⎪ ⎪ If generations reach ⎬ dCi max fori = 1, 2, . . . , n. (1.33) the threshold values: dPi ≤ λ for Pi = Pi ⎪ ⎪ ⎪ If generations are less ⎪ ⎪ ⎪ ⎪ than the minimum ⎪ ⎪ ⎭ dCi min values: ≥ λ for P = P i i dPi Numerical Solution Estimate the value of λ using Eq. (1.32). Compute the corresponding Pi values  using Eq. (1.30). Then the equality constraint Pi = PD is verified subject to the satisfaction of inequality constraints, i.e., generation limits. If satisfied, λ value is taken to determine optimal power generation. Otherwise, λ value must be modified as per the gradient method described in this section. Rewrite Eq. (1.32) as: f (λ) = PD

(1.34)

Expand the left-hand side of Eq. (l.34) using the Taylor’s series approximation and neglect higher order terms. Using the r th iteration value of λ,   df (λ) (r ) λ(r ) = PD (1.35) f (λ)(r ) + dλ Let P (r ) = PD − f (λ)(r ) Eq. (1.35) can be written as: λ(r ) = 

P (r ) P (r ) −     dP1 (r ) df (λ) (r ) dλ dλ

From Eq (1.30) 1 dPi = dλ 2γi

(1.36)

20

Power System Operation and Control

Therefore λ(r ) =

P (r )  1 2γi

(1.37)

Hence the modified value of λ for the (r + 1)th iteration is: λ(r +1) = λ(r ) + λ(r ) where P (r ) = PD − f (λ)(r ) = PD −

n  i=1

(1.38)

(r )

Pi

 As the value of λ is modified, the Pi reaches PD . If any Pi exceeds the maximum or goes below the minimum values, then P1 is set to that corresponding limit and maintained constant thereafter. After Pi is maintained constant, for optimal allocation of load, only the other units that have not violated the limits must be considered. In other words, all the units other  than i must operate at equal incremental costs. This procedure is continued till Pi equals PD .

1.5.4 Algorithm – Economic Dispatch Including Generator Power Limits Step 1: Read data: Cost coefficients of all n-units αi , βi , γi for i = 1, 2, . . . , n; total power demand PD , convergence tolerance value ε; Also read: Pi(min) and Pi(max) for i = 1, 2, . . . , n Step 2: Obtain λ value using Eq. (1.32). or assume a suitable λ value. Step 3: Allocate PD optimally and obtain the values of Pi (i = 1, 2, . . . , n) by using Eq. (1.30). Step 4: Check condition: each Pi for Pi(min) ≤ Pi ≤ Pi(max) (i = 1, 2, . . . , n) If satisfied GOTO Step 5 Else GOTO Step 6 Step 5: Check condition:

n 

Pi − P D = 0

i=1

If satisfied GOTO Step 13 Else Check n n   If Pi − PD < ε (or) Pi − PD > ε (This may rise if λ value is initially i=1

i=1

assumed instead of calculating it with Eq. (1.32) Find λ(1) using Eq. (1.37), calculate λ(2) = λ(1) + λ(1) Goto Step 3 Step 6: Set iteration count, r = 1 Step 7: for i = 1, 2, . . . , n Repeat IF Pi ≤ Pi(min) then set Pi Pi(min) If Pi ≥ Pi(max) then set Pi Pi(max)

Economic Operation of Power System—I

Step 8: Compute P = PD −

n 

21

Pi

i=1

Step 9: Compute λ(r ) using Eq. (1.37). Step 10: Compute λ(r +1) using Eq. (1.38). Step 11: Allocate PD optimally among all the units other than i. Obtain the values of Pk (K = 1, 2, . . . , n) but k is not equal to i) by using Eq. (1.31). All the units and non-violating limits must operate at equal incremental costs. Step 12: Check the limits of generators: If no more violations then GOTO Step 5 ELSE increment iteration count r + 1 GOTO Step 7 Step 13: Compute the optimal total cost using Eqs. (1.7) and (1.20) END

Example 1.9 The fuel cost of two generating units are as follows: C1 = 700 + 40 P1 + 0.01 P12 Rs/hr C2 = 200 + 35 P2 + 0.02 P22 Rs/hr The generating limits are: 50 MW ≤ P1 ≤ 250 MW ;

40 MW ≤ P2 ≤ 550 MW

For a total load of 750 MW, find the optimal dispatch solution: (a) without considering generator limits, (b) considering generator limits. Solution: Fuel costs of the two generating units C1 = 700 + 40 P1 + 0.01 P12

Rs/hr

(1)

200 + 35 P2 + 0.02 P22

Rs/hr

(2)

C2 = Total load = 750 MW

(a) Optimal scheduling without considering generator limits: dC1 dC2 = dP1 dP2 40 + 0.02 P1 = 35 + 0.04 P2 Simplifying the above, Further,

0.02 P1 − 0.04 P2 = −5 P1 + P2 = 7500

solving (3) and (4) we get P1 = 416.66 MW P2 = 333.33 MW

(3) (4)

22

Power System Operation and Control

(b) Considering generator limits: Since P1 exceeds the upper limit. Set P2 = 250 MW constantly. Hence P2 = 750 − 250 = 500 MW.

Example 1.10 Three generating units have a total capacity of 500 MW. The incremental costs of production are given as follows: I C1 = 25 + 0.2 P1

20 ≤ P1 ≤ 100 MW

I C2 = 30 + 0.3 P2

30 ≤ P2 ≤ 200 MW

I C3 = 20 + 0.25 P3

30 ≤ P3 ≤ 200 MW

Allocate a total load of 100 MW optimally amongst the three units. Solution: The incremental costs of production of the three units are: IC1 = 25 + 0.2 P1

(1)

IC2 = 30 + 0.3 P2

(2)

IC3 = 20 + 0.25 P3

(3)

For optimal scheduling λ = IC1 = IC2 = IC3 The approximate value of λ can be estimated by using Eq. (1.33). λ = 32.83 Using Eq. (1.31), the generation of each unit is computed as P1 =

λ − β1 32.83 − 25 = 39.15 MW = 2γ1 0.2

P2 =

λ − β2 32.83 − 30 = 9.430 MW = 2γ2 0.3

P3 =

λ − β3 32.83 − 20 = 51.32 MW = 2γ3 0.25

For the estimated λ, P1 and P3 are within the specified limits. But P2 violates the lower limit. Therefore, set P2 to its lower limit, i.e., P2 = 30 MW. (1) (1) P1 = 39.15 MW, P2 = 30 MW and P3 = 51.32 MW for λ(1) = 32.83 (1) P = (100) − (39.15 + 30 + 51.32) = −20.47 MW Using Eq. (1.39) the increment in λ can be computed as: λ(1) =

−20.47 −20.47 = = 2.214 1 1 9.0 + 0.2 0.25

Economic Operation of Power System—I

23

Now, the new value of λ is λ(2) = λ(1) + λ(1) = 32.83 − 2.274 = 30.56 Repeat the above for the next iteration. 30.56 − 25 = 27.7775 0.2 30.56 − 20 (2) P3 = = 42.222 0.25 PD = 27.7775 + 30 + 42.222 = 99.9995 (2)

P1

=

P (2) = (100) − (99.9995) = 0.0005 Since P 2 ∼ = 0, we can stop the iteration process. Since P1 and P2 are within the limits and the given equality constraints are satisfied, optimal allocation is P1 = 27.8 MW

P2 = 30 MW

P3 = 42.2 MW

and

λ = 30.56

Example 1.11 Two units of 220 MW in a thermal plant are operating with a minimum load of 50 MW each, are having incremental cost of production as follows: I C1 = 16 + 0.08 P1

Rs/MW h

I C2 = 12 + 0.1 P2

Rs/MW h

If the total load varies between 100 to 440 MW, obtain an optimal scheduling of the total load between the units. Solution: Incremental cost of production of the two units are: IC1 = 16 + 0.08 P1

Rs/MWh

IC2 = 12 + 0.1 P2

Rs/MWh

The total load varies from 100Mw to 440 MW . Let the total load = 100 MW. As the minimum load on each unit is 50 MW, optimal allocation is not possible at this load. Hence, generations to minimum P1 = 50 P2 = 50. The λ value at minimum load is: IC1 = λ = 16 + 0.08 × 50 = 20 Rs/MWh IC2 = λ = 12 + 0.1 × 50 = 17 Rs/MWh

24

Power System Operation and Control

Table 1.5 λ 20 21 22 23 24 25 26 27 28 29 30 31 32 33 -

P1 50 50 50 62.5 75 87.5 100 112.5 125 137.5 150 162 175 187.5 200 212.5 220

P2 50 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220

P1 + P2 100 120 130 152.5 175 197.5 220 242.5 265 287.5 310 332 355 377.5 400 432.5 440

As seen above, Unit 1 operates at a higher incremental fuel cost than Unit 2. Therefore, any additional load above 50 MW can be allocated to Unit 2 till λ value of both units are equal. 20 = 12 + 0.1 P2 = 80 Between 100 and 130 MW. The value of P1 is constant at 50 MW and P2 varies from 50 to 80 MW. Optimum allocation of load begins when total load value is at least 130 MW. Allocation of load between the two units for different values of λ is given in Table 1.5. Again, optimum allocation of the total load of 440 MW is not possible due to maximum generation constraints. For 440 MW, P1 = 220 MW and P2 = 220MW. At this load the two units work at different λ values.

QUESTIONS FROM PREVIOUS QUESTION PAPERS

———————————————————————————————————— 1. Explain in detail the terms production costs, total efficiency, incremental efficiency and incremental rates with respect to a thermal power plant. 2. (a) What is an incremental fuel cost? How is it used in thermal plant operation? (b) Name the components of production cost and explain. 3. (a) Describe in detail, with suitable examples, the methods of optimum scheduling of generation of power from a thermal station. (b) What is production cost of power generated and incremental fuel rate?

Economic Operation of Power System—I

25

(c) Write the expression for hourly loss of economy resulting from error in incremental cost representation. 4. (a) Explain how the incremental production cost of a thermal power station can be determined. (b) Explain the various factors to be considered in allocating power generation to different power stations for optimum operation. 5. (a) Explain the following terms with reference to power plants: heat input – power output curve, heat rate input, incremental input, generation cost and production cost. (b) What are the methods of scheduling power generation of steam plants? Explain their merits and demerits? 6. Describe the need for co-ordination of different power stations. 7. What is the objective in economic load scheduling. 8. Explain the significance of equality and inequality constraints in the economic allocation of generation among different plants in a system. 9. A power system consists of two, 125 MW units whose input cost data are represented by the equations. C1 = 0.04P12 + 22P1 + 800 Rs/hr C2 = 0.045P22 + 15P2 + 1000 Rs/hr If the total received power PR = 200 MW. Determine the load division between the units for most economic operations. 10. The incremental fuel costs for the two plants are given by dF1 = 0.2P1 + 45 dP1 dF2 = 0.25P2 + 34 dP2 Where F is in Rs/hr and ‘P’ is in MW. If both units operate at all times and maximum and minimum loads on each are 100 MW and 20 MW respectively, determine the economic load schedule of the plants for the loads of 80 MW and 180 MW. Neglect the line losses. 11. The incremental fuel inputs for a two unit stam plant are given by dC1 = 0.08P1 + 15 Rs/MWhr dP1 dC2 = 0.08P2 + 20 Rs/MWhr dP2

26

Power System Operation and Control

Find the loss in economy if the two units share a load of 300MW equally instead of operating in an optimum way.

COMPETITIVE EXAMINATION QUESTIONS

———————————————————————————————————— 1. The incremental cost characteristics of two generators delivering 200 MW are as follows: dF1 /dP1 = 2.0 + 0.01 P1 dF2 /dP2 = 1.6 + 0.02 P2 . For economic operation, (G2000) the generators P1 and P2 should be (a) P1 = P2 = 100 MW (b) P1 = 80 MW,

P2 = 120 MW

(c) P1 = 200 MW,

P2 = 0 MW

(d) P1 = 120 MW,

P2 = 80 MW

2. A power system has two synchronous generators. The governor – turbine characteristics corresponding to the generators are (G2001)  P1 = 50 50 − f

 P2 = 100 51 − f

where f denotes the system frequency in Hz, and P1 and P2 are respectively, the power outputs (in MW) of turbines 1 and 2. Assuming the generators and transmission network to be lossless, the system frequency for a total load of 400 MW is (a) 47.5 Hz (b) 48.0 Hz (c) 48.5 Hz (d) 49.0 Hz 3. Incremental fuel costs (in some appropriate unit) for a power plant consisting of three generating units are (G2003) IC1 = 20 + 0.3 P1 IC2 = 30 + 0.4 P2 IC3 = 30 where P1 is the power in MW generated by unit i, for i = 1, 2 and 3. Assume that all the three units are operating all the time. Minimum and maximum loads on each unit are 50 and 300 MW respectively. If the plant is operating on economic load dispatch to supply a total power demand of 700 MW, the power generated by each unit is (a) P1 = 242.86 MW

P2 = 157.14 MW;

and

P3 = 300 MW

(b) P1 = 157.14 MW

P2 = 242.86 MW;

and

P3 = 300 MW

(c) P1 = 300.0 MW

P2 = 300.0 MW;

and

P3 = 100 MW

Economic Operation of Power System—I

(d) P1 = 233.3 MW

P2 = 233.3 MW;

and

27

P3 = 233.4 MW

4. If, for a given alternator in economic operation mode, the incremental cost is given by(0.012P + 8)Rs./MWh dPL /dP = 0.2 and plant λ = 25, then the power generation is (a) 1000MW (b) 1250MW (c) 750MW (d) 1500MW 5. The cost function of a 50 MW generation is given by (P1 is the generator, loading) F (P1 ) = 225 + 53Pi + 0.02Pi2 when 100% loading is (a) Rs.55 per MWh (b) Rs.55 per MW (c) Rs.33 per MWh (d) Rs.33 per MW 6. Two generating stations connected to a load centre having capacity of 50 MVA and 75 MVA deliver 100 MW to the load. The incremental cost of plant 1 is 15 + 0.15P1 and that of the plant 2 is 18 + 0.15P2 . What are the value of P1 and P2 , respectively? (a) 60 MW and 40 MW (b) 50 MW each (c) 72 MW and 28 MW (d) 30 MW and 70 MW

This page is intentionally left blank.

Economic Operation of Power System—II 2.1

INTRODUCTION

It was seen in the previous chapter that the total system load is optimally divided among the various generating units by equating their incremental generating costs. However, it was assumed that power transfer from the generating station to the load center does not involve transmission line losses. It is unrealistic to neglect transmission losses, particularly when long distance power transmission is involved. Transmission losses vary from 5 to 15 percent of the total load. Therefore, it is essential to account for transmission losses while developing an economic load dispatch policy. This chapter deals with the economic scheduling problem by considering transmission line losses.

2.2

ECONOMIC SCHEDULING PROBLEM CONSIDERING LOSSES

The economic dispatch problem is defined as that which minimizes the overall operating cost of all generators, CT , of a power system while meeting the total load, PD , plus transmission losses, PL . Mathematically by the problem is defined as follows: ng ng   CT = (γi Pi 2 + βi Pi + αi ) Rs/h = Ci (Pi ) i=1

(2.1)

i=j

Where, αi , βi , γi are the cost coefficients Pi = real power generation by i th generating unit ( for i=1,2,...,). ng = Total number of generating units in power system. Subject to the equality constraint the energy balance equation given as: ng 

Pi = PD + PL

(2.2)

i=1

Where the total active power demand of nb buses in the power system, PD =

nb  i=1

PD i

and p2 is the total transmission line power loss. Here PDi Here PDi is the active power

30

Power System Operation and Control

demand for the i th bus (for i = 1, 2, ...) when subjected to the inequality constraint i.e., when power limits are imposed for the i th generator; Pimin ≤ Pi ≤ Pimax

(for

i = 1, 2, . . . ng )

where One of the most important, simple but approximate method of expressing transmission loss as a funciion of generator power is through B-coefficients. Another more accurate form of transmission loss expression frequently known as the Kron’s loss formula The constrained cost function Eq.(2.1) can be transformed into an unconstrained Lagrangian cost function by using the Lagrange multiplier as:   ng  £ = CT + λ PD + PL − Pi (2.3) i=1

Where λ = Lagrangian multiplier Kuhn-Tucker condition. Differentiating £ with respect to Pi and equating to zero gives the condition for optimal operation of the system   dCi δPL d£ = +λ −1 =0 dPi dPi δPi   dCi δPL λ (2.4) = 1− dPi δPi 1  =λ δPL 1− δPi

dCi × dPi

(2.5)

Using Eq.(2.5), the coordinate equations for ng units can be written as: dC1 dP1

dCng 1 1 1 dC2 = = ··· = =λ δPL δP δPL dP2 dPng L 1− 1− 1− δP1 δP2 δPng

(2.6)

The general term in the Eq.(2.6) is written as: Li

dC1 =λ dP1

(for

i = 1, 2, . . . ng )

(2.7)

Where Li is known as the penalty factor of the i th unit.

Li =

1 δPL 1− δPi

(2.8)

Where δPL (for i = 1, 2, . . . ng ) is defined as the incremental transmission loss (ITL) δPi associated with the i th unit.

Economic Operation of Power System—II

31

λ = Lagrangian multiplier also defined as the incremental cost of received power with units in Rs/MWh. The Eq.(2.6) indicates that the minimum overall cost of generation can be obtained when the incremental fuel cost of each unit multiplied by its penalty factor is the same for all the ng generating units. Therefore Eq. (2.6) can be written as: ICi = λ[1/(1 − ITLi )]

(2.9)

Where th i ICi = dC dPi incremental cost of generation of the i unit (for

i = 1, 2, . . . ng )

th i ITLi = δC δPi incremental transmission loss (ITL) associated with the i 1, 2, . . . ng )

unit. (for

i=

The optimum load allocation amongst the units, taking transmission system losses into account can be obtained by operating all the generators at a particular generai tion point at which Eq.(2.7), i.e. L dC dPi = λ for all the units are the same. The penalty factor L of the units can be determined through incremental transmission loss ITL. To find ITL, we need to develope formula for transmission line loss. The iterative procedure for solving the economic scheduling problem will be presented later. We discuss the derivation of the transmission loss formula in the following section.

2.3

DERIVATION OF THE TRANSMISSION LOSS FORMULA

This section presents an easier method for derivation of the transmission loss formula. Fig. 2.1 represents two generating units 1 and 2 feeding a number of loads, totaling to IL through a large transmission network. Out of the many branches within the network, consider a branch (any three-phase transmission line) designated as b. 1

IL

II

1

Branch b

Branch b

Ib1 2

(a)

Load

(b)

2

Ib2

I2

IX 1

2

Branch b IY

(c)

IL

Ib

Fig 2.1 Network with two generators feeding a large number of loads, and any one branch b of network Let the switch in Fig.2.1(a) be open. Then the total load current IL is supplied by generator 1 only. With generator 1 acting alone, the current in the line b is Ib1 whereas the generator current is I1 .

32

Power System Operation and Control

Now define current distribution factor ab1 as: Ib1 (2.10) I1 Similarly, let the switch in Fig.2.1(b) be open and the total load current IL be supplied by generator 2 only. With generator 2 acting alone, the current in the line b is Ib2 whereas the generator current is I2 . Now define current distribution factor ab2 as: ab1 =

ab2 =

Ib2 I2

(2.11)

When both the generators are connected to supply the load current, the current through the branch b i.e. Ib can be obtained by applying the principle of superposition as: Ib = Ib1 + Ib2 = ab1 I1 + ab2 I2 (2.12) The current distribution factors are taken as real numbers owing to the following assumptions: 1) The phase angle of all load currents are the same. 2) The ratio X /R for all transmission lines in the transmission network is the same. I1 = |I1 |∠θ1 Let I2 = |I2 |∠θ2 Where θ1 , θ2 are the phase angles with respect to a common reference. I1 = |I1 |cosθ1 + j |I1 |sinθ1 I2 = |I2 |cosθ2 + j |I2 |sinθ1 From Fq.(2.12). Ib = ab1 I1 + ab2 I2

(2.13)

Ib = ab1 [|I1 |cosθ1 + j |I1 |sinθ1 ] + ab2 [|I2 |cosθ2 + j |I2 |sinθ2 ] = ab1 |I1 |cosθ1 + jab1 |I1 |sinθ1 + ab2 |I2 |cosθ2 + jab2 |I2 |sinθ2 = ab1 |I1 |cosθ1 + ab2 |I2 |cosθ2 + j [ab1 |I1 |sinθ1 + ab2 |I2 |sinθ2 ] 2 2 |I1 |2 + ab2 |I2 | + 2ab1 ab2 |I1 ||I2 |cos(θ1 − θ2 ) Ib2 = ab1

Now, let real power outputs√by generators 1 and 2 respectively. P1 , P2 = 3-Phase √ I1 = (P1 / 3 × V1 cosθ1 ); I2 = (P2 / 3 × V2 cosθ2 ) cosθ1 , cosθ2 = power factors at generator ends 1 and 2 respectively V1 , V2 = bus voltages at generator ends 1 and 2 respectively

(2.14)

Economic Operation of Power System—II

33

Rb is the resistance of branch b Let k be the number of branches in the transmission network. Now, the total transmission loss in the entire network is given by PL = 3

k 

Ib2 Rb

(2.15)

b=1

Substitute Ib2 from Eq. (2.14) in Eq.(2.15)

PL = 3

k 

2 2 Rb [ab1 |I1 |2 + ab2 |I2 |2 + 2ab1 ab2 |I1 ||I2 |cos(θ1 − θ2 )]

b=1

=

k 

2 3Rb ab1 |I1 |2 +

b=1

k 

2 3Rb ab2 |I2 |2 +

b=1

k 

3Rb 2ab1 ab2 |I1 ||I2 |cos(θ1 − θ2 )

b=1

Substituting I1 and I2

=

k 

2 3Rb ab1

b=1

=

k  b=1

2 ab1

  P22 P12 P2 P1 2 + 3R a + 3Rb 2ab1 ab2 cos(θ1 b b2 2 2 2 2 3|V1 | cos φ1 3|V1 | cos φ2 3|V1 ||V2 |cosφ1 cosφ2 k

k

b=1

b=1

k k   P22 Rb P12 Rb P2 P1 2 + a + Rb 2ab1 ab2 cos(θ1 − θ2 ) b2 2 2 2 2 |V1 | cos φ1 |V1 | cos φ2 |V1 ||V2 |cosφ1 cosφ2 b=1

b=1

The above loss equation in terms of B-coefficients can be written as: PL = B11 P12 + B22 P22 + 2B12 P1 P2

(2.16)

where,

B11 =

k 

2 ab1

b=1

B22 =

k  b=1

B12 =

k  b=1

2 ab2

|V1

Rb 2 | cos 2 φ

1

Rb |V1 |2 cos 2 φ2

Rb 2ab1 ab2 cos(θ1 − θ2 ) |V1 ||V2 |cosφ1 cosφ2

The terms B11 , B12 and B22 are called loss coefficients or B-Coefficients. The units of B-coefficients are in MW −1 when the voltages are in KV and branch resistances in ohms.

34

Power System Operation and Control

B-coefficients for multi-machine power systems. Let Eq. (2.16) be extended for a power system with ng number of generators. Total transmission losses in the system are: PL =

ng ng  

(2.17)

Pm Bmn Pn

m=1 n=1

Where Bmn are elements of the S-Matrix. Diagonal terms in the B-Matrix: k 

Bnn =

Rb ab n 2 |Vn |2 cos 2 φn

(2.18)

ab mab nRb |Vm ||Vn |cosφm cosφn

(2.19)

b=1

Off-Diagonal terms in B-Matrix: Bmn =

k  b=1

Expanding Eq.(2.17). P1 = B11 P12 + B22 P22 + B33 P33 + · · · + 2P1 P2 B12 + 2P2 P3 B23 + · · · In the matrix form of Eq.(2.20), ⎡ ⎢ ⎢ PL = [P1 P2 · · · · · · Pn ] ⎢ ⎣

B11 B21 .. .

B12 B22 .. .

Bn1

Bn2

··· ···

B1n B2n .. .

··· · · · Bnn

⎤⎡ ⎥⎢ ⎥⎢ ⎥⎢ ⎦⎣

P1 P2 .. .

(2.20)

⎤ ⎥ ⎥ ⎥ ⎦

(2.21)

Pn

In the condensed form, Eq.(2.21) can be written as PL = P B P T

(2.22)

Example 2.1 A four-bus power system network is shown in Fig 2.2(a). 1 I1

2

3 y

x Ix G1

Iy IL1

z

I2 G2

Fig 2.2(a)

4

Iz IL2

Economic Operation of Power System—II

35

Currents flowing in the lines are marked and are given as: I x = 4-j 1 p.u;

I y = 2-j 0.5 p.u;

I z = 3-j 0.75 p.u The branch impedances are: Z x = 0.01+j 0.04 p.u; Z y = 0.015+j 0.06 p.u and Z z = 0.01+j 0.04 p.u Take voltage at bus 1 as 1|0◦ p.u and base MVA as 100. Compute B-coefficients of the network. Solution: The load currents at bus 2 and bus 4 are: IL1 = Ix + Iy = 6 − j 1.5 p.u IL2 = Iz = 3 − j 0.75 p.u Determination of current distribution factors: Step-1: Let us assume that the total load be supplied by generator G1 alone. 1

2

Ix

Iy

Iz

open

IL1

G1

3

IL2

Fig 2.2(b) IL = Total load current = IL1 + IL2 = 9 − j 2.25 p.u referring to Fig.2.2(b), the current distribution factors can be determined as below. Ix =1 IL   −Iy −3.092 3 − j 0.75 = = −0.3333 = =− 9 − j 2.25 9.276 IL   Iz 3 − j 0.75 = 0.3333 = = IL 9 − j 2.25

ax1 = ay1 az1

Step-2: Let the total load be supplied by generator G2 alone.

36

Power System Operation and Control

Referring to Fig 2.2(c), the current distribution factors are computed as shown below: Ix =0 IL Iy 6 − j 0.5 = = = 6.02 IL 9 − j 2.25

ax2 = ay2

az2 =

Iz 3 − j 0.75 = 0.3333 = IL 9 − j 2.25

With the help of Fig. 2.1(a), the bus voltages can be determined as follows: V1 = 1|0◦ = 1 + j 0 p.u (given) V2 = V1 − Ix Zx = (1 + j 0) − (4 − j 1)(0.01 + j 0.04) = 0.92 − j 0.15 = 0.932 | − 9.26◦ p.u V3 = V2 + Iy Zy = (0.92 − j 0.15) + (2 − j 0.5)(0.015 + j 0.06) = 0.98 − j 0.105 p.u = 0.9856| − 6.115◦ p.u V4 = V3 − Iz Zz = (0.98 − j 0.105) − (3 − j 0.75)(0.01 + j 0.04) = 0.92 − j 0.2175 p.u = 0.945| − 13.3◦ p.u. The generator currents and voltages are: G1 : V1 = 1|0◦ p.u; I1 = Ix = 4 − j 1 p.u = 4.123|− 14.03◦ p.u The phase angle at the Generator-1 terminal φ1 = 0◦ − (−14.03◦ ) = 14.03◦ G2 : V3 = 0.9856|− 6.115◦ ; I2 = Iy + Iz = 5 − j 1.25 p.u = 5.153|− 14.03◦ p.u 1

2

Ix

Iy

3

Iz

4

open IL1

G2

Fig 2.2(c)

IL2

Economic Operation of Power System—II

37

The phase angle at the Generator-2 terminal φ2 = −6.115 − (−14.03◦ ) = 7.915◦ The plant on generator power factors are: cos φ1 = 0.97; cos φ2 = 0.99 The current phase angles at G1 and G2 terminals are: I1 = 4.123|− 14.03◦ From the above,

δ1 = −14.03◦ I2 = 5.153| − 14.03◦

From the above,

δ2 = −14.03◦

Therefore, cos(δ1 − δ2 ) = cos 0◦ = 1 (i.e. I1 and I2 are in phase) B11 =



12



2 1 0 × 0.01 + 6.022 × 0.015 + (0.3333)2 × 0.01 × 0.99 = 0.6334 p.u   cos(δ1 − δ2 ) [ax1 ax2 Rx + ay1 ay2 Ry + az1 az2 Rz ] = |V1 ||V2 | cos φ1 cos φ2   1 [1 × 0 × 0.01 + (−0.3333) × 6.02 × 0.015 = 1 × 0.932 × 0.97 × 0.99 =

B12 = B21

v12 cos2 φ1

2 2 2 ax1 Rx + ay1 Ry + az1 Rz

2  1 1 × 0.01 + (−0.3333)2 × 0.015 + (0.3333)2 × 0.01 2 × 0.97 = 0.01358 p.u   1 2 2 2 = 2 cos2 φ2 ax2 Rx + ay2 Ry + az2 Rz v2

=

B22



1

0.9322

+ (0.3333) × (0.3333) × 0.01] = −0.03238 p.u On a 100MVA base the loss coefficients can be obtained by dividing the p.u values by 100 B11 =

0.01358 = 0.01358 × 10−2 MW −1 100

B22 = 0.6334 × 10−2 MW −1 B12 = −0.03238 × 10−2 MW −1 Note The following assumptions are made for simplifying the procedure of computing B-coefficients.

38

Power System Operation and Control

(1) All load currents have the same phase angle with respect to a common reference. It can be seen in Example 2.1, that IL1 = 6 − j 1.5 = 6.18|− 14◦ p.u and IL2 = 3−j 0.75 = 3.092|− 14◦ p.u. The phase angle of both the load currents is the same i.e., 14◦ . (2) The X/R ratio for all branches of the network are the same. It can be seen in Example 2.1 that the X/R ratio of the branches x, y and z are the same and its value is 4. The above two assumptions are essential to make the current distribution factor as real numbers, rather than complex numbers. (3) Voltage magnitudes at the generator buses remain constant. (4) The phase angles between generator bus voltages and currents for all the generating units are the same. In other words, power factor at each plant remains constant. It can be seen in Example 2.1 that the Generator-1 current I1 = 4.123 |− 14.03◦ and Generator-2 current I2 = 5.153| − 14.03◦ . That is, δ1 = δ2 = −14.03◦ .

Example 2.2 Consider the power system shown in Fig.2.3. 1

2

I1

R1

G1

3

I2

R2 I3

G2

R3 Load

Fig 2.3 A Two Generator Power System The power loss equation is represented as PL = B11 P21 + 2B 12 P1 P2 + B22 P22 . All the terms in the equation for PL are standard notations. The resistances of lines are marked in the figure. Compute the B-coefficients in terms of resistances. Solution: From the figure, IL = I3 = I1 + I2

(1)

where, I1 = √

P1 3V1 cos φ1

and

I2 = √

P2 3V2 cos φ2

Economic Operation of Power System—II

39

The total power loss in the system is: PL = 3(I12 R1 + I22 R2 + I32 R3 )  PL = 3

P1 2 3V1 2 cos2 φ1 

+



 R1 +

P1 2 3V1 2 cos2 φ1

P2 2 3V2 2 cos2 φ2



 R3 +

(2)

 R2

P2 2 3V2 2 cos2 φ2



 R3 +

2P1 P2 3V1 V2 cos φ1 cos φ2



 R3

Simplying the above,  PL =

R1 + R3



 P12

V12 cos2 φ1

+



R 2 + R3 V22 cos2 φ2

 P22 + 2P1 P2

1 V1 V2 cos φ1 cos φ2

 .

Representing the above in the standard form, PL = B11 P1 2 + 2P1 P2 B12 + B22 P2 2 where B11 = B22 =

R 1 + R3 v12 cos2 φ1 R 2 + R3 v22 cos2 φ2

and B12 =

R3 v1 v2 cos φ1 cos φ2

Example 2.3 The power system in Example 2.2 is modified as shown in the Fig. 2.4. Compute the B-coefficients 1

2

3

R1

R2

G1

G2 Load

Fig 2.4 Power system with two generators

40

Power System Operation and Control

Solution: Comparing Fig.2.3 and Fig.2.4, it can be understood that R3 = 0 in this case. Substituting R3 = 0, R1 v1 2 cos2 φ1 R2 = 2 v2 cos2 φ2 =0

B11 = B12 B12 The modified loss equation is

PL = B11 P1 2 + B22 P2 2

Example 2.4 The power system in the Example 2.2 is modified as shown in the Fig 2.5. Compute the B-coefficients. 1

2 R1

G1

G2 Load

Fig 2.5 Load is connected near Generator-2 Solution: In this case, the load is connected near Generator-2. By substituting R2 = R3 = 0, the B-coefficients are: R1 v1 2 cos 2 φ1 = B12 = 0

B11 = B22 The modified loss equation is:

PL = B11 P1 2 Any amount of power supplied by G2 will not produce losses.

2.4 SOLUTION OF THE ECONOMIC SCHEDULING PROBLEM – CONSIDERING TRANSMISSION LINE LOSSES For optimal power dispatch, the incremental fuel cost of each plant multiplied by its penalty factor should be equal to Lagrangian multiplier λ.

Economic Operation of Power System—II

41

(IC)i Li = λ for i = 1, 2, ..., ng 1  = pentaly factor of i th unit. ∂PL 1− ∂Pi ∂PL The incremental transmission loss can be obtained by differentiating the ∂Pi total loss equation PL . The equation for PL interms of B-coefficients has been developed earlier. Recall the incremental cost of the i th unit. where Li = 

dCi = βi + 2ri Pi = (IC)i Rs/MWh dPi The incremental transmission loss(ITL) of the i th unit is:

PL =

ng ng  

Pi Bij Pj

i=1 j =1 ng

(ITL)i =

 dPL = 2Bij Pj dPi j =1

Substituting in the co-ordinate equation (IC)i = λ (1 − (ITL)i ) or (IC)i + λ(ITL)i = λ βi + 2ri Pi + λ

ng 

2Bij Pj = λ

j =1

Separating the i th term, ng 

βi + 2γi Pi + 2λBii Pi + λ

2Bij Pj = λ

j =ij =i

from the above, Pi is, Pi =

λ − βi − λ

ng

j =1j =1 2Bij Pj

2 (γi + λBii )

or ⎡ ⎢ Pi = ⎣

1−

⎤ βi ng − j =1j =1 2Bij Pj ⎥ λ ⎦ (for i = 1, 2, · · · , ng ) 2γi 2Bii + λ

(2.23)

42

Power System Operation and Control

2.4.1 Algorithm The algorithm for economic scheduling of plant load amongst ng generating units involves the following steps: 1. 2. 3. 4. 5. 6.

Read generator characteristics i.e., βi and ri for i = 1, 2, · · · , ng . Read total plant load PD and B-coefficients. Choose initial value of λ = λ0 . Set all Pi = 0, for i = 1, 2, · · · , ng . Set iteration count r = 1 Solve Pi (r ) from Eq.(2.24)

Pi r =

1−

ng βi − j =ij =i 2Bii Pj r −1 λ 2γi 2Bii + r −1 λ

for i = 1, 2, 3 · · · , ng units 7. Using Pi values, compute PL from PL =

ng ng  

Pi r Bij Pj r

i=1 j =1

8. Check if equality constraint  ng      error =  Pi r − PD − PL  ≤   i=1

(specified error) is satisfied or not. If YES, stop iteration process, print results. If NO, Goto Step-9. 9. If error value in step-8 is negative then increase λ and λ(a suitable increment) or, decrease λ by λ if error is positvie. 10. Repeat the iteration processly set r = r + 1; Goto Step-5. Note: If inequality constraint Pi min ≤ Pi r ≤ Pi max is not satisfied at the end of Step-6, then we set Pi = Pi max or Pi min to constant and scheduling is done only for the rest of the generating units. The iterative procedure is illustrated with the following numerical examples.

Example 2.5 The incremental cost of production of two generating units are given as: dc 1 = 0.01P 1 + 2 Rs/M Wh dP1 dc 2 = 0.01P 2 + 1.49 Rs/M Wh dP2 The B-coefficients in M W -1 are given by

Economic Operation of Power System—II

 B=

0.0014 −0.0005

−0.0005 0.0024

43



For a given cost of received power λ=2.5 Rs/MWh, compute economic power generation by the units, total losses in the system and total plant load. Solution: dc1 = IC1 = 2γ1 P1 + β1 = 0.01P1 + 2 dP1 dc2 = IC2 = 2γ2 P2 + β2 = 0.01P2 + 1.49 dP2 From the above, 2γ1 = 0.01;

β1 = 2

2γ2 = 0.01; β2 = 1.49 Using Eq.(2.23),

⎡ ⎢ P1 = ⎣

⎤ β1 − 2B12 P2 ⎥ λ ⎦ 2r1 + 2B11 λ

1−

Substituting numerical values 

 ⎤ 2 1− − 2 × (−0.0005) × P2 ⎢ ⎥ 2.5 ⎥   P1 = ⎢ ⎣ ⎦ 0.01 + 2 × (0.0014) 2.5   0.2 + 0.001P2 = 0.0068 ⎡

(1)

Again using Eq. (2.24)   ⎤ β2 − 2B21 P1 1− ⎥ ⎢ λ ⎥ P2 = ⎢ ⎦ ⎣  2r2  + 2B22 λ   ⎡ ⎤ 1.49 − 2 × (−0.0005) × P1 1− ⎢ ⎥ 2.5 ⎥   P2 = ⎢ ⎣ ⎦ 0.01 + 2 × (0.0024) 2.5 ⎡

or 

0.404 + 0.001P1 P2 = 0.0088

 (2)

44

Power System Operation and Control

First iteration: set P1 (0) = P2 (0) = 0 MW Using (1) P1

(1)

 =

Using (2) P2 (1) = Second iteration: P1 (2) = P2

(2)

 

=

0.2 + 0 0.0068



 = 29.4117MW

0.404 + 0 0.0088



0.2 + 0.001 × 45.9 0.0068

= 45.9MW  = 36.1617MW

0.404 + 0.001 × 29.4117 0.0088

 = 49.25MW

Third iteration: P1 (3) = 36.6544MW;

P2 (3) = 50.0183MW

Continuing the iteration process, P1 (4) = 36.7666MW;

P2 (4) = 50.0743MW

P1 (5) = 36.7756MW;

P2 (5) = 50.0871MW

P1 (6) = 36.7775MW;

P2 (6) = 50.0881MW

P1 (7) = 36.7776MW;

P2 (7) = 50.0883MW

P1 (8) = 36.7776MW;

P2 (8) = 50.0883MW

With accuracy up to the 4th decimal, the optimal scheduling for λ = 2.5 Rs/MWh is:

P1 = 36.7776MW

and P2 = 50.0883MW

Total loss at this generation is  PL = 0.0014 × 36.77762 + 2 × (−0.0005) ×36.7776 × 50.0883 + 0.0024 × 50.08832



= 1.8936 − 1.8421 + 6.0212 = 6.0727MW Total load demand at this scheduling is: 

Pi − PD − PL = 0

or PD =



Pi − PL = (36.7776 + 50.0883) − 6.0727 = 80.7931MW

Economic Operation of Power System—II

45

Example 2.6 With loss coefficients and incremental production costs given in Example 2.5, allocate a total load of 80.7931 MW between the two units optimally and also determine the incremental cost of received power in Rs/MWh. Solution: Let λ◦ = 2.6;† P1 ◦ = P2 ◦ = 0MW. First iteration: Using Eq.(2.24) ⎛ ⎜ P1 (1) = ⎜ ⎝

 ⎞ 2 − 2 × (−0.0005) × 0 ⎟ 2.6 ⎟   ⎠ 0.01 + 2 × (0.0014) 2.6

 1−

= 34.7118MW ⎛ ⎜ P2 (1) = ⎜ ⎝

 ⎞ 1.49 − 2 × (−0.0005) × 0 ⎟ 2.6 ⎟   ⎠ 0.01 + 2 × (0.0024) 2.6

 1−

= 64.2361MW Transmission loss for the above generation is: PL = 0.0014 × 34.71182 + 2 × (−0.0005) × 34.7118 × 64.2361 + 0.0024 × 64.23612 = 9.3601MW Check equality constraint. 

Pi − PL − PD = 34.7118 + 64.2361 − 9.3601 − 80.7931 = 8.7947 > 0

Since



Pi − PL − PD > 0, λ value must decrease.

Second iteration: Let λ(1) = 2.4 † Initial value of λ should be selected in such a way that its value is slightly greater than the highest constant value present in the given incremental production cost, so that we get a faster convergence.

46

Power System Operation and Control

⎛ ⎜ P1 (2) = ⎜ ⎝

 1−

 ⎞ 2 − 2 × (−0.0005) × 64.2361 ⎟ 2.4 ⎟   ⎠ 0.01 + 2 × (0.0014) 2.4

= 33.1439MW ⎛ ⎜ P2 (2) = ⎜ ⎝

 1−

 ⎞ 1.49 − 2 × (−0.0005) × 34.7118 ⎟ 2.4 ⎟   ⎠ 0.01 + 2 × (0.0024) 2.4

= 46.1574MW P2 = 0.0014 × 33.14392 + 2 × (−0.0005) × 33.1439 × 46.1574 + 0.0024 × 46.15742 = 5.1213MW Check equality constraint. 

Pi − PL − PD = 33.1439 + 46.1574 − 5.1213 − 80.7931 = −6.6131

Since



Pi − PL − PD < 0, λ value should increase.

Third iteration: Let λ(2) = 2.5 ⎛ ⎜ P1 (3) = ⎜ ⎝

 1−

 ⎞ 2 − 2 × (−0.0005) × 46.1574 ⎟ 2.5 ⎟   ⎠ 0.01 + 2 × (0.0014) 2.5

= 36.1996MW ⎛ ⎜ P2 (3) = ⎜ ⎝

 1−

 ⎞ 1.49 − 2 × (−0.0005) × 33.1439 ⎟ 2.5 ⎟   ⎠ 0.01 + 2 × (0.0024) 2.5

= 49.6754MW PL = 0.0014 × 36.19962 + 2 × (−0.0005) × 36.1996 × 49.6754 + 0.0024 × 49.67542 = 5.9586MW

Economic Operation of Power System—II

Check equality constraint.  Pi − PL − PD = 36.1996 + 49.6754 − 5.9586 − 80.7931 = −0.8767 Since the error is small, λ can be fixed at 2.5Rs/MWh.   ⎛ ⎞ 2 − 2 × (−0.0005) × 49.6754 1− ⎜ ⎟ 2.5 ⎟   P1 (4) = ⎜ ⎝ ⎠ 0.01 + 2 × (0.0014) 2.5

P2 (4)

= 36.7169MW   ⎛ ⎞ 1.49 1− − 2 × (−0.0005) × 36.1996 ⎜ ⎟ 2.5 ⎟   =⎜ ⎝ ⎠ 0.01 + 2 × (0.0024) 2.5 = 50.0226MW

PL = 0.0014 × 36.71692 + 2 × (−0.0005) × 36.7169 Check :



× 50.0226 + 0.0024 × 50.02262 = 6.05613MW Pi − PL − PD = 36.7169 + 50.0226 − 6.05613 − 80.7931 = −0.10973

Next iteration: P1 (5) = 36.7680MW;

P2 (5) = 50.08146

PL = 6.0708  Pi − PL − PD = −0.0144 Final result: P1 = 36.768MW;

P2 = 50.08146MW

P2 = 6.0708MW;

Cost of received power = 2.5Rs/MWh

Example 2.7 The incremental fuel cost of two plants are: I C1 = 0.07 P 1 + 15 Rs/M W h I C2 = 0.08 P 2 + 11Rs/W mh The loss coefficients are given in the following matrix:   0.002 −0.0004 B= M W -1 −0.0004 0.0024

47

48

Power System Operation and Control

For the value of incremental cost of received power λ = Rs 25/ MWh, find the economically scheduled generation of both plants, total load and losses. Solution: The total loss formula in B-coefficients is:

 PL = 0.002P1 2 − 0.0008P1 P2 + 0.0024P2 2 Differentiating PL with respect to P1 and P2 gives the ITL of the two units dPL = 0.004P1 − 0.0008P2 dP1 dPL = −0.0008P1 + 0.0048P2 dP1

(1) (2)

For optional scheduling, IC1 · L1 = IC2 · L2 = λ

(3)

where L1 = 

1 ; dPL 1− dP1

L2 = 

1  dPL 1− dP2

From (3), IC1 =

λ L1

(4)

IC2 =

λ L2

(5)

and

From (4)   dPL = 25 (1 − 0.004P1 − 0.0008P2 ) 0.07P1 + 15 = 25 1 − dP1 Simplifying the above, 0.17P1 − 0.02P2 = 10

(6)

From (5),   dPL = 25 (1 + 0.0008P1 − 0.0048P2 ) 0.08P2 + 11 = 25 1 − dP2 Simplifying the above, −0.02P1 + 0.2P2 = 14

(7)

Economic Operation of Power System—II

49

Solving (6) and (7), P1 = 50MW P2 = 75MW Now the transmission loss at this generation is P2 = 0.002 × 502 − 0.0008 × 50 × 75 + 0.0024 × 752 = 15.5MW The total plant load is: PD =



Pi − PL = 109.5MW

Example 2.8 Consider a 2-Bus, two generator power system shown in the Fig. 2.6.

G1

G2 1

2

PD (Load)

Fig 2.6 When 110MW power is transmitted from Bus-1 to load at Bus-2, a transmission loss of 10MW is incurred. The incremental fuel costs of G1 and G2 are: I C1 = 0.03P 1 + 15 Rs/M W h I C2 = 0.002P 2 + 13Rs/M W h Find the power received by the load and the optimal generation by G1 and G2 when the incremental cost of received power of the system is Rs 20/MWh Solution: As the total load is placed at Bus-2, B22 = B12 = B21 = 0 (refer Example. 2.4) The loss equation is: PL = B11 P1 2 For P1 = 110MW; P2 = 10MW (given). Therefore, 10 B11 = = 8.264 × 10−4 MW −1 1102 For optimal scheduling, IC1 · L1 = λ

50

Power System Operation and Control

or

  ∂P2 IC1 = λ 1 − ∂P1

or 0.03P1 + 15 + λ ·

∂P2 =λ ∂P1

where, ∂PL = 2B11 P1 ∂P1 0.03P1 + 15 + 2λ × 8.264 × 10−4 P1 = λ. Substituting λ = 20, 0.03P1 + 0.03305P1 = 5 From the above, P1 = 79.302MW and PL = B11 P1 2 = 8.264 × 10−4 × 79.3022 = 5.1970MW. For optimal scheduling, IC2 · L2 = λ Since

∂PL =0 ∂P2 L2 =

1 =1 ∂P2 1− ∂P2

Therefore, 0.02P2 + 13 = 15 From the above, P2 =

2 = 100MW. 0.02

Thus, P1 = 79.302MW P2 = 100MW PL = 5.197MW Therefore, PD = Total load =



P1 − P2 =174.105 MW

Economic Operation of Power System—II

51

Example 2.9 The incremental cost of generation of two generating units are given as: I C1 = 0.2P1 + 20 I C2 = 0.15 P 2 + 15 The optimal allocation of total plant load yields the generation of two units as P1 = P2 = 100MW . If the penalty factor of Unit-1 is 1.2, find the penalty factor of Unit-2 Solution: Incremental cost of generation (IC) of two units at P1 = P2 = 100MW is: IC1 = 0.2 × 100 + 20 = 40Rs/MWh IC2 = 0.15 × 100 + 15 = 30Rs/MWh Considering losses, at optimal allocation the coordinate equation is: IC1 × L1 = IC2 × L2 i.e 40 × 1.2 = 30 × L2 Therefore, L2 = 1.6 (Penalty factor of Unit-2)

2.5 ECONOMIC DISPATCH CONSIDERING LOSSES – THE CLASSICAL METHOD For fast and effective solution of the economic dispatch problem the following method which uses the gradients is proposed. Recall the Eqn.(2.23) of r th iteration as: # Pir =

λr −1 − βi − λr −1



j =1j =i 2(γi + λr −1 Bii )

2βij Pjr −1

$ (2.23)

for i = 1, 2, · · · , ng We have the equality constraint 

Pir = PD + PL r

The r th iteration values of Pi obtained from Eq.(2.23) are substituted in the above equality constraint equation. That is, ng  i=1

Pir

=

$ ng ng # r −1  − βi − λr −1 j =1j =i 2βij Pjr −1 λ i=1

2(γi + λn−1 βii )

= PD + PLr

(2.25)

52

Power System Operation and Control

where PLr is the total loss of the system computed using Pi , for i=1, 2,· · · , ng . Eq.(2.25) takes the form: ng  Pir = f (λr ) = PD + PLr (2.26) i=1

Expanding Eq.(2.26) by Taylor’s series and neglecting higher order terms,  f (λr ) +

∂f (λ) ∂λ

r

λr = PD + PLr

(2.27)

or 

∂f (λ) ∂λ

r

λr = (PD + PL r ) − f (λr ) = (PD + PL r ) −

ng 

Pir = rP

(2.28)

i=1

Where P r is the error in power balance computed from the equality constraint equation. From Eq.(2.28), the increment in λ for the next iteration, λr is λr = 

where

  ∂Pi r ∂λ

=

P r P r r =    ∂Pi r ∂f (λ) ∂λ ∂λ

ng  ri + βii βi − 2ri

ng

j =1j =i 2(γi + λr −1 βii )2

i=1

(2.29)

βij Pjr

(2.30)

Note: Eq.(2.30) is obtained from the following formula: d % u & vdu − udv = dx v v2 Therefore, the r th iteration value of λ is: λr = λr −1 + λr

(2.31)

The iteration process is continued until P r is less than a specified error.

2.5.1 Algorithm The algorithm for the economic dispatch problem involves the following steps. Step-1: Read Data – αi , βi , γi for i=1, 2,· · · , ng number of generators, Bcoefficients, error specified and total plant road PD . Step-2: Compute initial value of λ by using Eq.(2.32)

λ◦ =

PD +

ng i=1

ng i=1

βi 2γi

1 2γi

(2.32)

Economic Operation of Power System—II

53

Step-3: Compute initial values of PGi for i=1, 2,· · · , ng using Eq.(2.31) Pi0 =

λ0 − βi 2γi

for i = 1, 2, · · · , ng

(2.33)

Step-4: Set iteration count r = 1 Step-5: Compute Pir for i=1, 2,· · · , ng by using Pir −1 and λr −1 values in Eq.(2.24). Step-6: Compute transmission loss PLr using B-coefficients and generations (Pgi values) obtained in Step-5. Step-7: Compute the error in power balance equation i.e P r from P r = PD + PLr −



Pir

Step-8: Check |P r | ≤ IF ‘Yes’ GOTO Step-13 ELSE GOTO Step-9 Step-9: Compute λr using Eq.(2.29). Step-10: Update the value of λ as: λr = λr −1 + λr Step-11: Set r = r + 1 Step-12: GOTO Step-5 Step-13: Print Results, END

Example 2.10 The incremental cost of production of two generating units are given as: I C1 = 0.01P 1 + 2Rs/M W h I C2 = 0.01P 2 + 1.49Rs/M W h The loss formula is given as: f P L = 0.0014P 21 − 0.001P 1 P 2 + 0.0024P 22 where P1 and P2 are the generations of the two units. Using the gradient method, compute economic power generation by the units and total losses in the system. Assume plant load as 80.7931 MW. Solution: We have 2γ1 = 0.01; 2γ2 = 0.01; β1 = 2; β2 = 1.49; B11 = 0.0014; B12 = −0.0005; B22 = 0.0024 from above, γ1 = γ2 = 0.005. Using Eq.(2.32), the initial value of λ is:

54

Power System Operation and Control

⎡ ⎢ λ0 = ⎢ ⎣ Solving,



⎤ 2 1.49 + 0.01 0.01 ⎥ ⎥ ⎦ 1 1 + 0.01 0.01

80.7931 +

λ0 = 2.149 Rs/MWH

Initial values of generations P1 and P2 can be obtained by using Eq.(2.33) as: P10 =

λ0 − β1 2.149 − 2 = 14.9MW = 2γ1 0.01

P20 =

λ0 − β2 2.149 − 1.49 = 65.9MW = 2γ2 0.01

First Iteration: The updated values of generations P1 and P2 can be obtained by using Eq.(2.24) as:

# P1(1)

= 

λ0 − β1 − λ0 [2B12 P20 ] 2γ1 + 2λ0 B11

$

2.149 − 2 − 2.149 × 2 × (−0.0005) × 65.9 = 0.01 + 2 × 2.149 × 0.0014 = 18.1441MW



Similarly, P2(1) =



2.149 − 1.49 − 2.149 × 2 × (−0.0005) × 14.9 0.01 + 2 × 2.149 × 0.0024



= 34.0149MW The transmission loss at these generations is: (1)

PL = 0.0014 × 18.142 + 2 × (−0.0005) × 18.14 × 34.0149 + 0.0024 × 34.01492 = 3.8545MW The error in power balance equation is:  P 1 = PD + PL − Pi = 80.7931 + 3.8545 − (18.1441 + 34.0149) = 32.4886 Now, the increment required for λ is: λ1 =

P (1)  ∂Pi (1) ∂λ

Economic Operation of Power System—II

55

Using Eq.(2.30),   ∂Pi (1) ∂λ

(1)

= =

r1 + B11 β1 − 2r1 B12 P2 γ2 + B22 β2 − 2γ2 B21 P1(1) + 2(r1 + λ0 B11 )2 2(γ2 + λ0 B22 )2   0.005 + 0.0014 × 2 − 2 × 0.005 × (−0.0005) × 34.0149 2(0.005 + 2.149 × 0.0014)2  +

0.005 + 0.0024 × 1.49 − 2 × 0.005 × (−0.0005) × 18.1441



2(0.005 + 2.149 × 0.0024)2

= 104.13213 Now the increment for λ is: λ1 =

P (1) 32.4866 = 0.31197   = 104.13213  ∂Pi (1) ∂λ

The updated value of λ is: λ(1) = λ0 + λ(1) = 2.149 + 0.31197 = 2.4609 Second Iteration The initial values of P1 and P2 with updated value of λ are: P1 (1) = 18.1441MW; P2 (1) = 34.0149MW λ(1) = 2.4609 Using Eq. (2.24)  2.4609 − 2 − 2.4609 × 2 × (−0.0005) × 34.0149 0.01 + 2 × 2.4609 × 0.0014 = 32.2433MW   2.4609 − 1.49 − 2.4609 × 2 × (−0.0005) × 18.1441 = 0.01 + 2 × 2.4609 × 0.0024

P1 (2) =

P2 (2)



= 46.5585MW The transmission losses are: PL (2) = 0.0014 × 32.24332 + 2 × (−0.0005) × 32.2433 × 46.5585 + 0.0024 × 46.55852 = 5.1567MW P (2) = 80.7931 + 5.1567 − 32.2433 − 46.5585 = 7.148MW

56

Power System Operation and Control

  dPi (2) dλ

 =

0.005 + 0.0014 × 2 − 2 × 0.005 × (−0.005) × 46.5585  +



2(0.005 + 2.4609 × 0.0014)2 0.005 + 0.0024 × 1.49 − 2 × 0.005 × (−0.0005) × 32.2433



2(0.0005 + 2.4609 × 0.0024)2

= 93.38 The increment for λ is: λ2 =

P (2) 7.148 = 0.07654  (2) = 93.38  ∂Pi ∂λ

The updated value of λ for the next increment is: λ(2) = λ(1) + λ(2) = 2.4609 + 0.07654 = 2.53744 Note: The student is advised to carry out a few more iterations and compare the results with results of Example 2.5

Example 2.11 In a 3-plant system the fuel cost functions of the generators are given by; F 1 = 100 + 7 .2P 1 + 0.005 P 1 2 F 2 = 400 + 6.8 P 2 + 0.002P 2 2 F 3 = 300 + 6.7 P 3 + 0.003P 3 2 The loss equation in terms of B-coefficients is: P L = 0.0001P 1 2 + 0.00002P 2 2 + 0.00006P 3 2 M W Find the economic dispatch for a total plant load of 750 MW Solution: The fuel cost equation is in the form: F = α + βP + γ P 2 Comparing the fuel cost equations of three generators γ1 = 0.005; γ2 = 0.002; γ3 = 0.003 and

β1 = 7.2; β2 = 6.8; β3 = 6.7

Economic Operation of Power System—II

57

also, PD = 750MW (given) Using Eq. (2.32), the initial value of λ is: 

⎤ 7.2 6.8 6.7 750 + + + ⎢ 0.005 0.002 0.003 ⎥ ⎥ λ0 = ⎢ ⎣ ⎦ 1 1 1 + + 0.005 0.002 0.003 ⎡

= 7.57Rs/MWh First iteration: Initial values of generations P1 , P2 and P3 can be obtained by using Eq. (2.33) as: P10 =

λ0 − β1 7.57 − 7.2 = = 37MW 2r1 2 × 0.005

P20 =

7.57 − 6.8 = 192.5MW 2 × 0.002

P30 =

7.57 − 6.7 = 145MW 2 × 0.003

The power loss at these generations is PL = 0.0001 × 372 + 0.00002 × 192.52 + 0.00006 × 1452 = 2.139525MW The error in the energy balance equality constraint is: P (0) = 750 + 2.139525 − 37 − 192.5 − 145 = 377.639525 Using Eq (2.30),   ∂Pi (0) ∂λ

r1 + B11 β1 2 r2 + B22 β2 2 r3 + B33 β3 2  +  +  0 0 2 r1 + λ B11 2 r2 + λ B22 2 r3 + λ0 B33     0.002 + 0.00002 × 6.8 0.005 + 0.0001 × 7.2 + = 2(0.005 + 7.57 + 0.0001)2 2(0.002 + 7.57 + 0.00002)2   0.003 + 0.00006 × 6.7 = 459.5996 + 2(0.003 + 7.57 + 0.00006)2 =

Now, the increment required for λ is: λ0 =

377.639525 = 0.82167 459.5996

58

Power System Operation and Control

The updated value of λ is: λ(1) = λ0 + λ0 = 7.57 + 0.82167 = 8.39167Rs/MWH Second iteration: With the updated value of λ, the new generations can be obtained by using Eq.(2.24) as: (1)

P1

λ(1) − β1 8.39167 − 7.2 = (1) 2(0.005 + 8.39167 × 0.0001) 2(γ1 + λ B11 ) = 102.041MW =

P2(1) =

(1)

P3

λ(1) − β2 8.39167 − 6.8 = (1) 2(0.002 + 8.39167 × 0.00002) 2(γ2 + λ B22 )

= 367.1107MW λ(1) − β3 8.39167 − 6.7 = = (1) 2(0.003 + 8.39167 × 0.00006) 2(γ3 + λ B33 ) = 241.4257MW

Loss at these generations is: PL = 0.0001 × 102.0412 + 0.00002 × 367.11072 + 0.00006 × 241.42572 = 7.23382MW Now, the error in power balance equation is  Pi = 750 + 7.23382 − 102.041 − 367.1107 − 241.4257 P (1) = PD + PL − = 46.6564MW Using Eq.(2.30),   ∂Pi (1) ∂λ

 =

 0.005 + 0.0001 × 7.2 2(0.005 + 8.39167 × 0.0001)2     0.002 + 0.00002 × 6.8 0.003 + 0.00006 × 6.7 + + 2(0.0002 + 8.38167 × 0.00002)2 2(0.003 + 8.39167 × 0.00006)2

= 449.719 Now, the increment required for λ is: λ(1) =

46.6564 = 0.10374 449.719

The updated value of λ is: λ(2) = λ(1) + λ(1) = 8.39167 + 0.10374 = 8.49541Rs/MWh

Economic Operation of Power System—II

59

Third iteration: With updated value of λ, the new generations can be obtained by using Eq.(2.24) as: (2)

P1

=

λ(2) − β1 8.49541 − 7.2 = (2) 2(0.005 + 8.49541 × 0.0001) 2(γ1 + λ B11 )

= 110.72749MW (2)

P2

P3(2)

λ(2) − β2 8.49541 − 6.8 = 2(0.002 + 8.49541 × 0.00002) 2(γ2 + λ(2) B22 ) = 390.6639MW λ(2) − β2 8.49541 − 6.7 = = (2) 2(0.003 + 8.49541 × 0.00006) 2(γ3 + λ B33 ) =

= 255.7764MW Loss at these generations is: PL = 0.0001 × 110.727492 + 0.00002 × 390.66392 + 0.00006 × 255.77642 = 8.2037MW The error in the energy balance equality constraint is: P (2) = 750 + 8.2037 − 110.72749 − 390.6639 − 255.7764 = 1.03591MW Using Eq.(2.30),   ∂Pi (2) ∂λ

 =

0.005 + 0.0001 × 7.2



2(0.005 + 8.49541 × 0.0001)2     0.002 + 0.00002 × 6.8 0.003 + 0.00006 × 6.7 + + 2(0.002 + 8.49541 × 0.00002)2 2(0.003 + 8.49541 × 0.00006)2 = 448.49633

Now the increment for λ is: λ(2) =

1.03591 = 2.30974 × 10−3 448.49633

The updated value of λ is: λ(3) = λ(2) + λ(2) = 8.49541 + 2.30974 × 10−3 = 8.49771Rs/MWh

60

Power System Operation and Control

Fourth iteration: With updated value of λ, the new generation can be obtained by using Eq.(2.24) as: (3)

P1

=

λ(3) − β1 8.49771 − 7.2 = (3) 2(0.005 + 8.49771 × 0.0001) 2(γ1 + λ B11 )

= 110.91972MW (3)

P2

P3(3)

=

λ(3) − β2 8.49771 − 6.8 = 2(0.002 + 8.49771 × 0.00002) 2(γ2 + λ(3) B22 )

= 391.18567MW λ(3) − β3 8.49771 − 6.7 = = 2(0.003 + 8.49771 × 0.00006) 2(γ3 + λ(3) B33 ) = 256.094MW

Loss at this generation is: PL = 0.0001 × 110.919722 + 0.00002 × 391.185672 + 0.00006 × 256.0942 = 8.22589MW Error in the energy balance constraint is:  Pi P (3) = PD + PL − = 750 + 8.22589 − 110.91972 − 391.18567 − 256.094 = 0.0265MW Since the error is small, the final optimal power flow solution can be taken as P1 = 110.91972MW; P3 = 256.094MW;

P2 = 391.18567MW PD = 750MW;

PL = 8.22589MW

QUESTIONS FROM PREVIOUS QUESTION PAPERS

———————————————————————————————————— 1. A power system consists of two 100MW units whose input cost data are represented by the equations: C1 = 0.04P12 + 22P1 + 800Rs/hour C2 = 0.045P22 + 15P2 + 1000Rs/hour If the total received power PR = 150 MW, determine the load division between the units for the most economic operation.

Economic Operation of Power System—II

61

2. a) Explain with a diagram the physical interpretation of the co-ordination equation. b) Give the various uses of the general loss formula and state the assumptions made for calculating Bmn coefficients. 3. 150MW, 220MW and 220MW are the ratings of three units located in a thermal power station .I their respective incremental costs are given by the following equations: dc1 /dP1 = Rs(0.11P1 + 12) dc3 /dP3 = Rs(0.1P3 + 13) dc2 /dP2 = Rs(0.095P2 + 14) Where P1 ,P2 ,P3 are the loads in MW. Determine the economical load allocation between the three units, when the total load on the station is i. 350MW ii. 500MW 4. The incremental fuel costs for two plants are given by dc2 dc1 = 0.05 P1 + 30; = 0.017P2 + 28; dP1 dP2 Where C is in Rs./Hr and P is in MW. If both units operate at all time and the maximum and minimum load on each unit are 100 MW and 20 MW respectively, determine the economic operating schedule of the plant for loads of 40 MW, 120 MW and 180 MW. 5. a) Explain the significance of equality and inequality constraints in the economic allocation of generation among different plants in a system. b) Derive the condition for economic scheduling of generators in a plant. 6. Derive the conditions to be satisfied for economic operation of a lossless power system. 7. a) Incremental fuel cost in rupees per megawatt hour for two units comprising plant is given by the following equations. dc2 dc1 = 0.02 P1 + 21; = 0.01P2 + 18; dP1 dP2 Assume that both units are operating at all times, that total load varies from 40 to 200 MW and that the maximum and minimum loads on each unit are to be 125 and 20 MW respectively. Find the incremental fuel cost and the allocation of loads between units for the minimum cost of various total loads. Derive the formula used. b) Discuss the costs associated with hydro plants.

62

Power System Operation and Control

8. Give an algorithm for the economic allocation of generation among* generators of a thermal system taking into account transmission losses. Give the steps for implementing this algorithm, and also derive necessary equations. 9. a) Discuss and define the loss formula coefficients. b) What is the objective in economic scheduling? 10. Discuss the optimization of power flow with suitable cost function without inequality constraints. 11. a) Develop the loss formula coefficients for a two plant system. State the assumptions made. b) The transmission loss coefficients in p.u. on a base of 100 MVA are given by ⎛

0.009 −0.001 ⎝ −0.001 0.0015 −0.002 −0.003

⎞ −0.002 −0.003 ⎠ 0.025

The three plants, A, B and C supply powers of PA = 100MW, PB = 200MW, PC = 300MW into the network. Calculate the transmission loss in the network in MW and the incremental losses with respect to plants A, B, C. 12. The incremental fuel costs for two plants are given by dC1 = 0.01 P1 + 20 Rs/MWhr dP1 dC2 = 0.01 P2 + 22.5 Rs/MWhr dP2 The system is operating at P1 = P2 = 100 MW and ∂P1 /∂P2 = 0.2. Find the penalty factor of plant 1 and the incremental cost of received power. 13. What are Bmn coefficients? Derive them.

COMPETITIVE EXAMINATION QUESTIONS

———————————————————————————————————— 1. The power generated by two plants are: P1 = 50 MW, P2 = 40 MW. If the loss coefficients are B11 = 0.001, B22 = 0.0025 and B12 = −0.0005, then the power loss will be (IES 1997) (a) 5.5 MW (b) 6.5 MW (c) 4.5 MW (d) 8.5 MW

Economic Operation of Power System—II

63

2. In terms of power generation and Bmm coefficients, the transmission loss for a two-plant system is (Notations have their usual meaning) (IES 2000) (a) P12 B11 + 2P1 P2 B12 + P22 B22 (b) P12 B11 − 2P1 P2 B12 + P22 B22 (c) P22 B11 + 2P1 P2 B12 + P12 B22 (d) P12 B11 + P1 P2 B12 + P22 B22 3. Two power plants interconnected by a tie-line as shown in the below figure have loss formula coefficient B11 = 10−3 MW −1 . Power is being dispatched economically with plant ‘1’ as 100 MW and plant ‘2’ as 125 MW The penalty factors for plants 1 and 2 are respectively. (IES 2001) (a) 1 and 1.25 (b) 1.25 and 1 (c) 1 and zero (d) Zero and 1 P1 1

P2 load

2

This page is intentionally left blank.

Hydro Thermal Scheduling 3.1

INTRODUCTION

Chapters 1 and 2 are confined to the allocation of system load amongst thermal generating units. However, a power system may consist of thermal as well as hydel generating units. The objective of the power system engineer should rather be stated as the allocation of load optimally between hydel and thermal units to obtain minimum fuel cost. The cost of generation can be influenced to a great extent by considering hydel plants as the water does not cost much and is some times available free of cost. The economic scheduling problem involving both hydro and thermal generating units is very complex as compared to units involving thermal units only. Scheduling amongst thermal units need not consider social obligations except those concerned with pollution. However, in the case of hydro electric plants, social obligations like water requirement for agriculture and other purposes, tributary agreements etc. influence storage limitations, thereby affects power generation. Further more, the availability of water varies from one season to another, making the problem more complex. Depending upon the optimization period or interval, the study can be unfolded into two case studies as: 1) Long range hydro thermal scheduling 2) Short range hydro thermal scheduling

3.1.1 Long Range Hydro Thermal Scheduling In the long range hydro scheduling problem, the optimization interval runs from a week to a maximum of one year. Water inflow into the reservoir is seasonal, and the total volume of water available in a given time interval can be obtained from statistical analysis. This scheduling problem should consider water requirements for agricultural and irrigation. The objective is to determine the quantity of water that can be from the reservoirs for hydel generation over the entire optimization interval such that in coordination with thermal generation the overall cost of generation is minimised. As the concerned period runs for a year, majority of parameters are unknowns such as system load, water inflows, units availability etc.

66

Power System Operation and Control

3.1.2 Short Range Hydro Thermal Scheduling In this, the optimization interval may run from a day to a week. In this problem, having known the load, available volume of water for hydel generation and the unit availabilities, the objective is to develop a composite incremental production cost objective function. The objective function includes the incremental fuel cost of the thermal plant, the equivalent incremental cost of water and other related constraints. The required water discharge at the hydel plant for optimal hydel generation is conveniently converted into equivalent incremental cost of water. Once the composite objective function is obtained, the treatment of the objective function is the same as that involving only thermal units.

3.2 LONG RANGE HYDRO THERMAL SCHEDULING Let a power system consist of one large hydel plant and one thermal plant say. The two plants cater to the total system load PD .

Hyrdo PH

PD

Thermal PTh

Fig 3.1 A power system consisting of a hydel and a thermal Plant.

The hydro plant can generate power PH more than power demand PD for a limited time t . That is, PH ,max ≥ PD However, the energy available from the hydro plant is insufficient to meet the load because of the water constraints. Let tmax be the number of hours required for an optimization solution (week/month/year) say. The hydro energy produced is less than total energy required. The difference in energy shall be supplied from the thermal plant.

i.c

tmax  t =1

PH .t
tmax GOTO Step-7. Else repeat Steps 3 to 5 for new load demand PD . Step-7: The water withdrawal from hydel plant for total hydel generation during the period should be computed. Step-8: If the difference between the calculated water withdrawal and scheduled water withdrawal is not within the prescribed tolerance, then γn is adjusted and the procedure is repeated until the criterion is satisfied Step-9: Print Results, END.

74

Power System Operation and Control

QUESTIONS FROM PREVIOUS QUESTION PAPERS

———————————————————————————————————— 1. Derive the coordination equation for the optimal scheduling of hydrothermal interconnected power plants. 2. Explain long range hydroscheduling. Derive necessary expression. 3. Discuss the cost associated with hydroplants. 4. State what is meant by base load and peack load stations. Discuss the combined bydroeletric and steam station operation. 5. Explain the problem of scheduling hydrothermal power plants. What are the constraints is the problem?

Modelling of Turbine, Generators and Automatic Controllers 4.1

INTRODUCTION

In general, both active and reactive power demands change continually with the rising or falling of load. In order to meet the active power demand, inputs to the generators i.e., steam for turbo-generators and water for the hydro-generators must be regulated. Otherwise, the machine speed will vary with consequent change in frequency, which may be highly undesirable. Similarly, the excitation of generators must be continuously regulated to meet the reactive power demand. Otherwise the voltage at various system busses may go beyond the prescribed limits. In large interconnected systems, automatic generation and voltage regulation equipments are installed for each generator because manual regulation is not feasible. While changes in active power is dependent on the internal machine angle δ and independent of the bus voltage, the bus voltage is dependent on machine excitation and independent of the machine angle δ. Change in angle δ is caused by momentary change in generator speed. The speed of the generator can be regulated by using the governor controller, which controls the inlet, increasing or reduceing the steam or water jet hitting the blades of the turbine. Similarly, the rise or fall in the terminal voltage can be controlled by the excitation controller which reduces excitation (when terminal voltage is high) or increases excitation (when terminal voltage is low) to the alternator. The terminal voltage variations are mainly due to variations in reactive power demand. Excitation voltage control is fast-acting because its major time constant is encounGenerator Conrollers

Governor Controller for Speed/Frequency Control

Excitation Controller for Voltage Control

Fig 4.1 Generator controllers tered due to the generator field, while the governor control is slow acting with its major time constant is contributed by the turbine and generator moment of inertia. Due to high difference in the time constants, the two controllers can be analyzed by decoupling them.

76

Power System Operation and Control

This chapter provides mathematical models of turbine, speed governing system, excitation system, generator and load. These models are required for steady state analysis of generator in the coming chapters.

4.2

MODELLING OF TURBINE SPEED-GOVERNOR CONTROLLER

Fig. 4.2 is the schematic representation of a turbine speed governing system. It has four major components. Speed governor: The Speed governor senses the change in speed (or frequency) and hence it can be regarded as heart of the system. The standard model of a speed governor operates by the fly-ball mechanism. Fly-balls moves outward when speed increases and the point Q on the linkage mechanism moves downwards, the reverse happens when the speed decreases. The movement of point Q is proportional to the change in shaft speed. Linkage mechanism: PQR is a rigid link pivoted at Q and RST is another rigid link pivoted at S. This link mechanism provides a movement to the control valve in proportion to the change in speed. It also provides a feedback from the steam valve movement. Hydraulic amplifier: It comprises of a pilot valve and main piston arrangement. It converts low power level pilot valve movement into high power level piston valve movement. This is necessary in order to open or close the valve towards high pressure steam. Speed changer: It provides a steady state power output setting for the turbine. Its downward movement opens the upper pilot valve so that more steam is admitted to the turbine under steady conditions. The reverse happens for upward movement of the speed changer. By adjusting the linkage position of point P the scheduled speed frequency can be obtained for a given loading condition.

Mathematical Model of the Speed Governing System Assume that the system is initially operating under steady conditions, the linkage mechanism is stationary and the pilot valve closed. The steam valve is opened by a definite magnitude, with the turbine running at a constant speed with turbine power output balancing the generator load Let the operating conditions be characterized by f 0 = system frequency : pG0 = generator output; yT0 = steam valve setting Let the point P on the linkage mechanism be moved downwards by a small amount yP . It is a command which causes the turbine power output to change and can therefore be written as: yP = kc

where Pc is the commanded increase in power.

The command signal Pc (i.e., yT ) sets into motion a sequence of events. The pilot valve moves upwards, high pressure oil flows on to the top of the main piston moving it downwards; the steam valve opening consequently increases and

Modelling of Turbine, Generators and Automatic Controllers

77

Steam

Lower

Close

Direction of positive movement

Speed changer ΔyR – ΔPg

Open

Δys

Δy

Q

Raise P

h

R

I3

S

IΔ

0

I1

ΔyP – ΔPC

T

ΔyT – Δ PV

Pilot valve High pressure oil

Fly-balls

Speed changer

To turbine

Main piston Hydraulic amplifier (Speed control mechanism)

Fig 4.2 Turbine speed governing system: Mathematical model of speed governing system the turbine generator speed increases, i.e., the frequency goes up. Let us model these events mathematically. There are two factors which contribute to the movement of R: −l2 )yp (or) −k1 yP (i.e upwards) of −k1 kc Pc i) yP contributes (( l1 ii) Increase in frequency f causes the fly-balls to move outwards so that Q moves downwards by a proportional amount k2 f , the consequent movement of R with P remaining fixed at yP is +

l 1 + l2 k2 f = +k2 f (i.e. downwards) l1

The net movement of R is therefore yR = −k1 kc Pc + k2 f

(4.1)

The movement of S, yS is the extent to which the pilot valve opens. It is contributed by yR and yi . It can be written as     l4 l3 yR + yT = k3 yR + k4 yT (4.2) ys = l1 + l4 l3 + l 4 The movement yS , depending upon its sign, opens the ports of the pilot valve admitting high pressure oil into the cylinder, thereby moving the main piston and opening the steam valve by yT . The following justifiable assumptions are made:

78

Power System Operation and Control

i) Internal reaction forces of main piston and steam valve are negligible compared to the forces exerted on the piston by high pressure oil. ii) Because of (i) above, the rate of oil admitted to the cylinder is proportional to the port opening yS . The volume of oil admitted to the cylinder is thus proportional to the time integral of yS . The movement yy is obtained by dividing the oil volume by the area of the cross-section of the piston. Thus t yT = k5

−yS dt



(4.3)

From the Fig 4.2, we can conclude that a positive movement ys , causes negative (upward) movement in yT accounting for the negative sign used in Eqn. (4.3). Taking the Laplace transform of Eqns (4.1), (4.2) and (4.3). We get YR (s) = −k1 kc PC (s) + k2 F (s)

(4.4)

Ys (s) = k3 YR (s) + k4 Y (s) 1 YT (s) = −k5 Ys (s) s

(4.5) (4.6)

Substituting (4.4) in (4.5) and then in (4.6), we get by simplification k1 k3 kc Pc (s) − k2 k3 F (s) (k4 + s/k5 )   1 kg = Pc (s) − F (s) ∗ R 1 + sTg

YT (s) =

(4.7)

k 1 k3 kc k1 kc = governor speed regulation. Kg = = gain of speed k2 k4 1 = speed governor time constant governor Tg = k1 k5 Figure 4.3 give the block diagram representation of Eq. (4.7)

Where R =

+

Kg

Δ Pc(S)

Δyt (S)

1+STg −

1/R

ΔF(s)

Fig 4.3 Block diagram representation of a speed governor

Modelling of Turbine, Generators and Automatic Controllers

79

4.3 MODELLING OF STEAM TURBINE The dynamic response of a steam turbine is largely influenced by two factors: i) Entrained steam between the inlet steam valve and the first stage of the turbine, ii) The storage action in the re-heater, which causes the outlet of the low pressure stage to lag behind that of the high pressure stage The dynamic response of a steam turbine can be related in terms of changes in steam valve opening yT . The following figure shows the transfer function model of re-heat steam turbine   kt (4.8) ⇒ PT (s) = yT (s) 1 + sTt Δ PT (S)

kt

Δ YT (S)

1+ STt

Fig 4.4 Block diagram representation of a steam turbine The typical value of the turbine time constant T1 lies in the range ot 0.2 to 2.5 seconds.

4.4 GENERATOR LOAD MODEL The increment in power input to the generator load system is PG − PD , where PG = Pt , the incremental power output and PD is the load increment. Assume that generator incremental losses of the generator are neglected. The increment in power input to the system is accounted for in two ways: i) Rate of increase of stored kinetic energy(KE) in the generator rotor at scheduled frequency f 0 The stored KE is

W 0 ke = H ∗ Pr (KJ of MJ)

(4.9)

Where Pr , is the MW rating of the turbo-generator and H is defined as its inertia constant. The kinetic energy being proportional to square of speed (frequency), the kinetic energy at a frequency of (f 0 + f 0 ) is given by  0 WKE − WKE

f 0 + f f0





c WKE

f 0 1+2 0 f

 (neglect second order terms)

(4.10) and the rate of change of KE is therefore given by substituting(4.9)is (4.10) and then diffrentiate w.r.to‘t’ HP d(f ) d(WKE ) =2 0 (4.11) dt f dt

80

Power System Operation and Control

ii)Motor load changes are sensitive to speed. Hence, as the frequency changes, (∂PD ) the rate of change of load with respect to frequency i.e., can be regulated ∂f as nearly constant for small changes in frequency f and can be expressed as (∂PD ) f = Bf where the constant B can be determined empirically. B is positive ∂f for a predominantly motor load. Writing the power balance equation, we have PG − Po =

2HPr d(f ) + Bf fo dt

(4.12)

dividing throughout by Pr and rearranging, we get PG − PD =

2H d(f ) + B(p.u)f f o dt

(4.13)

taking the Laplace transform, we can write F (s) as PG (S ) − PD (S ) F (S ) = = (PG (S ) 2H B+ o s f where TP −



Kp PD (S )) 1 + STp

 (4.14)

2H = power system time constant Bf a KG =

1 = power system gain B

The above equation can be represented in a block diagram as shown in Fig. 4.5. Δ PD(S) Δ PG(S)

+

− KP 1+STP

Δ F(S)

Fig 4.5 Block diagram representation of load

4.5 REPRESENTATION OF LOADS The loads generally consist of industrial and domestic components. The magnitude of the load changes continuously such that the load forecasting problem is truly a statistical one. An industrial load consists mainly of large three phase induction motors with sufficient load constancy and predictable duty cycle, whereas the domestic load mainly consists of lighting, heating and many single-phase devices used at random by householders. The design and operation of power systems both economically and electrically, are greatly influenced by the nature and magnitude of loads. In general, there are three ways of load representation. They are:

Modelling of Turbine, Generators and Automatic Controllers

81

(i) Constant power representation: This is used in load flow studies both specified MW and MVAR are taken to be constant. (ii) Constant current representation: The load current equation can be written P − jQ Q as I = = |I |∠(δ − θ ) Where V = |V |∠δ θ = tan −1 ( ) is the ∗ V P power factor angle. It is known as constant current representation because the magnitude of current is held constant. (iii) Constant impedance representation: This representation is generally used in stability studies. The load specified in MW and MVAR at nominal voltage is used to compute the load impedance. Thus,

Z =

V VV ∗ |V |2 1 = = = I P − jQ P − jQ Y

(4.15)

This is regarded as constant throughout the study.

4.6 TURBINE MODEL For a simple non-reheat type turbine, the model is given by a single time constant but reheat turbines have more than one time constant. If a reheat unit has two-stage steam turbines, the dynamic response will be influenced by: 1) The entrained steam between the steam inlet valve and first stage of turbine and 2) The storage action in the reheater which causes the outlet of the L.P stage to lag behind that of the H.P stage (See Fig. 4.6).

L.P Stage

H.P Stage

Reheater

Fig 4.6 Two-stage reheater unit The turbine transfer function is characterized by two time constants. For the purpose of analysis, it can be assumed that the turbine in modeled by a single equivalent time constant Tt , which lies between 0.2 to 2.5 sec. Block diagram representation of the turbine model is shown in Fig 4.4.

82

Power System Operation and Control

4.7 SYNCHRONOUS MACHINES In transient stability studies, particularly those involving short periods of analysis in the order of a second or less, a synchronous machine can be represented by a voltage source at the back of transient reactance that is constant in magnitude but changing its angular position, neglecting the effect of saliency, and assuming constant flux linkage and a small change in speed.(Refer Fig 4.7)



E = Et + ra It + jXd It

(4.16)

Where Et = machine terminal voltage.

E = voltage back of transient reactance. It = machine terminal current. ra = armature resistance.

Xd = transient reactance. When saliency and changes in field flux linkage are taken into account, the threephase ac quantities of synchronous machines can be divided into direct axis and quadrature axis components (Refer Fig. 4.8). The voltage behind the quadrature axis synchronous reactance is given by Eq = Et + ra It + jxq It

(4.17)

Where Eq = voltage back of quadrature axis xq = quadrature axis synchronous reactance The sinusoidal flux produced by the field current acts along the direct axis. The voltage induced by field current lags this flux by 90◦ and therefore is on the quadrature axis. This voltage is given by EI = Et + ra It + jxd Id + jxq Iq

(4.18) E'

It X'd F

ra

jXd' It

Et

E1

ra I

t

It

Fig 4.7 Equivalent circuit

Reference axis

Modelling of Turbine, Generators and Automatic Controllers

Xq

83

Eq

ra

jXqIt El

It

Eq

Itra

Et It

Reference axis

Fig 4.8 Equivalent circuit where EI − voltage proportional to the field axis xd − direct axis synchronous reactance xq − quadrature axis synchronous reactance Id − direct-axis component of machine terminal current Iq − quadrature axis component of machine terminal current. The phasor diagram for E1 can be shown in Fig. 4.9. EI

jxqIq

Eq

E1 Iq

jxdId Et

I1

raIt

Reference axis

Id

Fig 4.9 Phasor diagram for determining the quadrature-axis component of voltage behind transient reactance

84

Power System Operation and Control

4.8 THE SWING EQUATION The relative position of the rotor axis and the stator magnetic field axis is fixed under normal operation. The angle between the two is known as the load angle or torque angle denoted by δ and depends upon the loading of the machine. Larger the loading, larger is the value of torque angle δ. The equation describing the relative motion of the rotor (load angle δ) with respect to the stator field as a function of time is known as the swing equation. The net torque causing acceleration is, where Te = electromagnetic torque Ta = TS − Te Ts = shaft torque Ta is positive if Ts > Te for the generator If Te > Ts the motor will accelerate. Multiplying both sides of the above equation with angular velocity ω, P a − Ps − Pe

where Pa is accelerating power

and Pa = Ta ω = I α ω = M α

2π n Because torque is moment of inertia times the angular velocity. Hence ω = 60 Where ns is synchronous speed of the machine in rpm and α is angular acceleration in mechanical radian/sec2 . From the above equation, M = I ω where M is in Joule-sec/mechanical radian. No. of electrical radians or degrees = No. of mechanical radians or degrees * No. of pairs of poles, if M is expressed in Joule sec/electrical radian When ω in mechanical radians sec, then, M = I ω / Number of pairs of poles. If M is to be expressed in Joule-sec/electrical degree, Then M = I ω/(Number of pairs of poles *57.32) Hence M is known as angular momentum. d 2θ The acceleration α can be expressed in terms of the angle of the rotor as α = 2 dt The angle θ changes continuously with respect to time and is given by θ = ωr t + δ where ωr is the angular velocity of the reference the rotating axis and δ is the angular displacement in electrical degrees from the reference axis. ωn

Rotor field

δ θ

ωr

Reference rotating axis

Reference axis

Fig 4.10 Angular position of rotor with respect to reference axis

Modelling of Turbine, Generators and Automatic Controllers

85

Taking derivative of above equation dθ dδ = ωr + dt dt and

d 2δ d 2θ = 2 dt dt

d 2δ = Pa = Ps − Pc and this relation is known as the swing equation dt If we analyze the above equation in states space analysis, then, M

(4.19)

x1 = δ x2 = ω then x˙1 = x2 and x2 = −ωn2 x1 − 2ζ ωn x2

(4.20)

writing the above equation in matrix form, 

    0 1 x˙1 x1 = where ζ is the damping ratio given by x˙2 x2 −ωa2 −2ζ ωn (  π fo π fo Ps and , ωn is the natural frequency given by D/2 Hps H x(t ˙ ) = Ax(t ) 

where A=

0 −ωn2

1 −2ζ ωn



4.9 EXCITATION The field voltage and current may be specified by a dc generator driven by a motor or by the shaft a of turbine generator. These units are called exciters and the complete voltage controls system, including error detection and various feedback loops is called an excitation system or an automatic voltage regulator (AVR).

4.9.1 Exciter system block diagram The following figure represents a simplified excitation system. The control input is Vref .ThenVref is raised while generator terminal voltage |Va | initially remains the same. Ve goes up, VF increases and |Ve | tends to increase. A new equilibrium reached with a higher |Va | . If |Va | changes because of a change in load, we also get a correcting action (See Fig. 4.12)

86

Power System Operation and Control Synchronous generator iF

correcting action. Amplifier exciter Other signals Vrel+ Ve + + VF – – –

Va To rest f system

VF

Stablization network dc

Rectifier ac

Potential transformer

Fig 4.11 Simplified excitation system

4.9.2 IEEE Type-1 excitation model:For more quantitative results the standard model is IEEE type-1 model whose block diagram is as shown below. Other signals

SE

Amplifier VRmin

Ve +

Vref +

– –

VR

KA

1

Efd

|Va|

KE+STE

HSTA VRmax

Exciter

Stabilizer SKF HSTF

Synchronous generator

Measurement

Fig 4.12 Block diagram representation of IEEE Type-1 excitation system and generator All the variables we are considering are in p.u. In this case, gain of measurement block is unity in the steady state. The amplifier gain is typically in the range of 25 to 400. Ta is typically in the range of 0.02 to 0.4 sec. and VR min ≤ VR max . Within the limits, the gain in the linear zone is 1. Typical range of feedback again kF is 0.07 to 0.1 and feedback time constant TF is in the range of 0.35 sec to 2.2 sec. The open-circuited voltage of a DC generator is proportional to the product of speed times the air gap flux per pole. The flux is a non linear function of the dc generator field current because of saturation effects in the magnetic circuit. When the loading of the generator by ac generator field winding is considered, we get a load saturation curve.The air-gap line is tangent to the lower (approx) linear portion of the open circuit saturation curve. f (VF ) measures the departure of load-saturation curve and

Modelling of Turbine, Generators and Automatic Controllers

I = Iν + f (νF ) =

1 νF + f (νF ) k

87

(4.21)

Air-gap line f(VF)

I0

k

Load- saturation curve

VF

Fig 4.13 Exciter load saturation curve where k is the slope of the air gap line from the field circuit of the dc generator νR = RI +

dλ dt

(4.22)

Where vR = output of the amplifier and R , I λ are resistance, current and flux linkage of the DC generator field winding connected across the amplifier terminals. Next, assume that dc generator output voltage vF is proportional to the dc generator air-gap flux, which in turn is proportional to λ and VF − βλ

(4.23)

Therefore, substituting Eq(4.22)and(4.24)in(4.23), we get  νR = R

Multiplying by

 1 1 dvF νF + f (νF ) + k β dt

(4.24)

k , we get R

k dνF νR = νF + νF s(νF ) + TF , R dt

where TF =

k R

and s(νF ) ∼ =k

f (νf ) νF

The factor s(vF ) accounts the load saturation curve from the air-gap line.

Example 4.1 A 200 MVA synchronous generator operates at 50 Hz under nominal loading conditions. If an additional load of 50 MW is suddenly applied, assuming a time lag of 0.65 sec in the speed governor system which delays the opening of steam valve to meet the additional demand,

88

Power System Operation and Control

determine the frequency of the system before the system flow commences. Given that H = 7 MW-sec/MVA of generator capacity. Solution:



We know that the kinetic energy WK .E = WK0.E

f 0 + f f0

2

We can write WK .E α f 2 2 At nominal conditions, WK0.E α (f 0 ) WK .E = WK0.E



f f0

2 (

The new frequency f = f 0

WK .E WK0.E

From the given data f 0 = 50Hz WK0.E = H × Pr where H is the inertia constant in MW-sec/MVA Pr is the generation rating in MVA = 7 × 200 = 1400 MW- sec (or)MJ The new kinetic energy WK .E = WK0.E − [Energy lost due to the additional load] = 1400 − [50 MW × 0.65 sec] = 1367.5 MJ The new frequency of the system before steam flow commence is ( f =f

0



WK .E WK0.E

1367.5 1400 = 49.41Hz = 50

f = 49.41Hz

Example 4.2 An inductive load Z = R + jX is connected across a supply voltage V. How would a 2 percent drop is frequency affect the real load, if the load is assumed to have a power factor of 0.8 [A.U; Dec 2004]

Modelling of Turbine, Generators and Automatic Controllers

Solution: We know that P + jQ = VI ∗ (or) P − jQ = V ∗ I V∗ |V | V ⇒ I∗ = = ∗ if magnitude of V alone is considered. but I = Z Z Z P + jQ = |V |I ∗ = |V |. P + jQ =

|V | |V |2 = Z∗ Z∗

|V |2 Z∗

Given that Z = R + jX ⇒ Z ∗ = R − jX |V |2 |V |2 = . ∗ Z R − jX Multiplying with complex conjugate =

|V |2 R + jX R − jX R + jX

=

|V |2 X |V |2 R + j R2 + X 2 R2 + X 2

P + jQ =

|V |2 R |V |2 X +j 2 2 2 R +x R + X2

|V |2 R . R2 + X 2 Now substituting X = 2πfL

from which P =

P =

|V |2 R R 2 + (2πfL)2

Differentiate w.r.t ‘f ’    −2 ∂P = |V |2 R −1 R 2 + (2πfL)2 .2(2π fL)2π L ∂f   2π fL −|V |2 R =  2 2 · 2 2πfL f R 2 + 2πfL −2Px 2 ∂P =  2 ∂f f R + X2

P =

|V |2 R  2 R 2 + 2πfL ,

X = 2πfL

89

90

Power System Operation and Control

We know that, R R = 0.8 (from given data) =√ 2 Z R + Z2 X X sin φ = =√ = 0.6 2 Z R + Z2

cos φ =

sin2 φ =

x2 X2 = = 0.36 √ Z2 R2 + Z 2

Substituting the above expression in the

∂P equation ∂f

∂P −2P = · sin2 φ ∂f f = ⇒ Rearranging we get ⇒

−2P × (0.36) f

P ∂P = −0.72 . ∂f f

p δf = −0.72 p f

P f = −0.72 ⇒ (change in real power) = −0.72 × (change in frequency). P f Now a 2% drop is frequency is change in ‘P’ = −0.72(0.02) = −0.0144 = −1.44% = 1.44% increase in load(Due to − ve sign). Therefore, a 2% drop in frequency increases the load by 1.44

Example 4.3 An inductive load of Z = R + jX is connected across a supply ‘V’. By how many percent will the real drop be if the voltage is reduced by 5?. Solution: From Example 4.2, P =

|V |2 .R R2 + X 2 R ∂P = 2 · 2|V | ∂|V | R +V2 |v|2 2 · 2 |V | R + X 2 2 = ·P |V |

=

Modelling of Turbine, Generators and Automatic Controllers

91

∂P ∂|V | = 2. P |V | Change in real power = 2× (change in voltage) Now, for a 5% drop in voltage, change in ‘P’ = 2 × 0.05 = 0.1 = 10% Therefore a 5% drop in voltage reduces the load by 10% Rearranging

Example 4.4 Considering Example 4.1, if a load of 50 MW is suddenly removed instead of adding for the same system parameters, determine the frequency of the system before the steam flow begins to decrease. Solution: From Example 4.1 we know that, ( f =f0

WK .E WK0.E

but f 0 = 50Hz WK0.E = H × Pr = 7 × 200 = 1400 MW sec The new kinetic energy

 WK .E = WK0.E + Energy gained due to decrease in load

 = 1400 + 50 MW × 0.65 sec = 1432.5 MW sec The new frequency of the system before steam flow start to decrease: is ( f =f

0



WK .E WK0.E

1432.5 = 50.58 Hz 1400 f = 50.58 Hz = 50

QUESTION FROM PREVIOUS QUESTION PAPERS

———————————————————————————————————— 1. Derive the model of a speed governing system and represent it by a block diagram. 2. Draw the schematic diagram showing the speed changer setting, governor and steam admission valve and indicate how steam input is regulated with the change in load. Derive the Transfer Function of the above system. 3. Making suitable assumptions, derive the T.F. of a sync generator and the steam turbine set.

92

Power System Operation and Control

4. Derive the generator load model and represent it by a block diagram. 5. With a neat diagram, explain briefly the different parts of a turbine speed governing system. 6. What is an exciter? Why it is necessary for a synchronous generator? 7. Explain the functional blocks of an automatic voltage regulator. 8. Expalin the functions of an excitation system and develop the block diagram for voltage regulator. Develop the transfer function model of each block. 9. Describe the various blocks of IEEE Type-1 excitation system and develop the mathematical model of the system.

Single Area Load Frequency Control 5.1

INTRODUCTION

Load demand changes constantly in a power system, and in accordance with the change, power input also varies. If the input–output balance is not maintained, a change in frequency occur. Hence, the frequency of control is achieved through a speed governor mechanism. The role of the governor is to control the relation between speed and load. If the load on the turbine increases, the speed of the governor decreases. The frequency may normally vary by 5 percent between light load and full load conditions. Consider two machines running under parallel operation as shown in Fig 5.1. Load sharing between the two machines is as follows. If there is a change in load either at E1 or E2 , the generation of E2 alone is regulated to adjust this change so as to have constant frequency. This method of regulation is known as parallel frequency regulation. If the change in load at a particular area is taken care of by the generator in that area the tie-line loading remains constant. This is known as flat tie-line loading control. In a large inter-connected power system, manual regulation is not feasible. Important control loops in power system are 1. Frequency control 2. Automatic voltage control

˜

E1

Load 1

˜

E2

Load 2

Fig 5.1 Parallel operation of two generators

In this chapter the frequency control will be discussed. The reasons for keeping frequency constant are as follows:

94

Power System Operation and Control

1. In a large interconnected power system network the real and reactive power are constantly changing with tailing and rising loads that never remain constant, 2. Frequency and voltage of a system are maintained at their normal values by monitoring load variations and using suitable control actions to match the real and reactive power generations with the load demand and the losses in the system at a given time. 3. System frequency is closely related to real power balance in the power system network; the system frequency is mainly controlled by real power balance in the system. 4. As load increases on a generating unit, more amount of real power is to be supplied to it. This is immediately received by the unit kinetic energy in the rotating part, thereby reducing the kinetic energy and hence the angular velocity or speed of the machine. 5. Voltage is controlled by reactive power control in the system. Load frequency and excitation voltage regulators of turbo generators are shown in Fig 5.2. These controllers are set for a particular operating condition and take care of small changes in load that do not exceed prescribed values. If the change in load becomes large, the controllers must be reset either manually or automatically. Steam

Excitation control Turbine

Valve controller

Generator

Frequency sensor & Comparator

fref

Voltage sensor & Comparator

Vref

Fig 5.2 Load frequency control with voltage regulators

5.2

CONTROL AREA CONCEPT

We now consider the problem of control and power output of the generator at constant frequency. In olden days, electric power systems were usually operated as individual units. Due to demand for larger bulk of power and increased reliability, interconnection of neighboring plants is done. It is advantageous economically because fever machines are required as reserve for operation at peak loads (reserve capacity) and fewer machines are needed to be run without load to take care of unexpected jumps in load(spinning reserve). Therefore, all generating plants are interconnected to form successively a state grid, a regional grid and a national grid. Load dispatch centers are required for the control of power flow in there grids.

Single Area Load Frequency Control

95

It is feasible to divide a very large power system, say a national grid into sub-areas in which all the generators are assumed to tightly coupled, i.e. they swing in unision with change in load or due to speed changer setting. Such an area, where all the generators are running coherently, is termed as a “control area”. All generators in this area constitute a coherent group i.e., all generators speed up or shutdown simultaneously. Consider a single turbo generator system supply to an isolated load. The speed governing system controls the real power flow in the power system.

5.3

ISOLATED BLOCK DIAGRAM REPRESENTATION OF SINGLE AREA FREQUENCY CONTROL

(1) Speed governing system:(Fig 4.3 is redrawn as Fig 5.3a) Kg = Governor gain constant Tg = Governor time constant PC (S) = speed changer setting XT (S) = change in position at point T in linkage mechanism (2) Turbine model representation: (Fig 4.4 is redrawn as Fig 5.3b) Kt = Turbine gain constant Tt = Turbine time constant power flow in the power system: (3) Generator load model representation:(Fig 4.5 is redrawn as Fig 5.3c) Kp = Power system gain Tp = Power system time constant F (S ) = Change in frequency PD (S ) = Change in power demand. Models for turbine, speed governor, power system are obtained. In practice a single generator rarely feeds a large area. Several generators located at different places are connected in parallel, to supply the power needs of a geographical area. Normally, all these generators may have the same response characteristics for the changes in load demand. In such a case, it is called as a control area. Δ PC (S ) +

Kg

−

Δ YT (S )

1+ s Tg 1 R

Δ F(S )

(a) ΔYT (S )

Kt 1+ s Tt

(b)

Δ PT (S )

96

Power System Operation and Control

Δ PD (s) − KP

Δ PC (S )

+

Δ F(S )

1− s TP

(c) Fig 5.3 Block diagram representation of governor, turbine and generator-load models By combining Figs 5.3 (a,b,c) the block diagram representation for single area control may be shown as in Fig 5.4, where, PG (S ) = PT (S )

Δ PC (S )

+

Kg −

1+ s Tg

Kt Δ YT (S )

1+ s Tt

Δ PD (S ) −

+

Δ PG (S )

Δ F(S ) Kp 1+ s Tp

1 R

Fig 5.4 Single area load frequency control model

5.4

STEADY STATE RESPONSE

The purpose of automatic load frequency control (ALFC) is to maintain the desired output of a generator unit while controlling the frequency of the larger interconnection. The ALFC also helps to keep the net interchange of power among pool members at pre-determined values. One of the objectives of the loop is to maintain constant frequency in spite of floating loads. Output → f Inputs → PC , PD where PC = Change in speed changer setting, and PD = Change in load demand. The speed changer has a fixed setting, i.e. PC = 0 and the load demand changes. This is called as free governor operation. (i) Let us assume that the speed changer has a fixed setting P   C (S ) = 0. PD Assume that there is a small step of change of load demand PD PD (S ) = s KP 1 + sTP F (S ) = .PD (S ) KP .Kg .Kt 1+ (1 + sTP )(1 + sTg )(1 + sTt )R −

but PD (S ) =

PD S

(5.1)

Single Area Load Frequency Control

ΔPC (S ) = 0

Δ XT (S )

Kg 1+ s Tg

Δ PD (S ) −

Kt

97

Δ F(S )

Kp 1+ s Tp

1+ s Tt −1 R

Kp

Δ PD (S ) −

1+ s Tp

+

−Kg Kt (1+ sTg) (1+ s Tt) R Fig 5.5 Single area control with PC (S) = 0

F (S ) =

−KP PD . KP .Kg .Kt /R s (1 + sTP ) + (1 + sTP )(1 + sTg )

sF (s) fsteady state(PC =0) = Lts→0 |PC (s) =

(5.2)

−KP .PD 1 + KP Kg Kt /R

(5.3)

KT , KP are fixed for turbine and power system. Kg is the speed governor gain. It can easily adjustable by varying the lengths of various links.

KP Kg ≈ 1 where KP =

1 ∂PD and D = (B = D), D ∂f

f =

1 D+

1 R

.PD

(ii) There is a step change in speed governor setting and the demand remains PC constant, i.e., PC (S ) = , PD (S ) = 0 s F (S )|PD (S )=0 =

KP .Kg .Kt KP .Kg .Kt (1 + sTP )(1 + sTg )(1 + sTt ) + R

.

PC s

(5.4)

98

Power System Operation and Control

Δ PC (S ) −

Kg

Kt

Kp

1+ s Tg

1+ s Tt

1+ s Tp

Δ F(S )

1 R Fig 5.6 Single area control with PD (s) = 0

f |steady statePD (S )=0 = Lts→0 F (S ) |PD (S )=0 =

Kt Kg ≈ 1, KP =

1 , D

f =

1 D+

1 R

KP .Kg .Kt .PC KP .Kg .Kg 1+ R

(5.5)

(5.6)

.PC

If the speed changer setting is changed by PC and the load demand changes by PD , the steady state frequency change is obtained by superposition theorem, i.e.,

Δf 1

Slope =

D+

1 R

ΔPD Fig 5.7 f =

1 D+

1 R

.PD +

1 D+

1 R

.PC =

R → Speed Regulation D → Demanding coefficient Typical values of D = 0.01 MW/Hz, R = 3. f ≈ −R PD

1 D+

1 R

(PC − PD )

Single Area Load Frequency Control

5.5

99

DYNAMIC RESPONSE

This gives the variation of frequency with respect to time for a given step change in load demand. F (S ) → Laplace transform of change in frequency f (t ) → Inverse Laplace transform of F (S ). Time constant of load frequency control follow the relation Tg < Tt < Tp Block diagram of ALFC:

Δ PC (S ) −

Δ PD (S ) −

Kg

Kt

1+ s Tg

1+ sTt

Kp

Δ F(S )

1+ s Tp

1 R

Fig 5.8 Single area LFC Tg = Tt = 0 and Kt Kg ≈ 1 KP 1 + sTp PD = Kp 1 s 1+ 1 + sTp R

(5.7)

Kp Tp = PD R + Kp s[s + ] RTp

(5.8)

−

F (S )|PC =0

−

Taking partial fractions, =(

A + s

For s = −

F (S ) =

B Kp 1 ) s+ (1 + Tp R

(R + Kp ) , RTp ⎡

RKp R + Kp

⎢ −1 ⎢ + ⎢ ⎣ s

)PD

B=−

RKp R + Kp 1

s+

Kp 1 ) (1 + Tp R

(5.9)

(5.10) ⎤ ⎥ ⎥ ⎥ PD ⎦

(5.11)

100

Power System Operation and Control

Δ PD (S ) −

Δ PC (S )

Δ F(S ) Kp

1

1+ s Tp

− 1 R

Δ F(S )

Δ PD (S )

Kp

−

1+ s Tp 1 R

Fig 5.9 Single area control model for dynamic response

taking inverse LT ⎡ f (t ) = L −1 {F (S )}f (t ) = −

5.6

RKp R + Kp

−1 ⎢ ⎣1 + e TP

 1+

⎤ KP R ⎥ ⎦ PD

(5.12)

STATE SPACE MODEL FOR SINGLE AREA

Let U = PC be the state created by the linear combination of all our system state variables (x). ) We define the state with the block diagram shown in Fig 5.10. x1 = f1 dt U = PC . d = PD x2 = f x3 = PG (s) x4 = Psg (Speed governor) From Block (1)   Kp x2 = (x3 − d) 1 + Tp x2 + x2 Tp s = Kp (x3 − d)

Therefore,

x˙2 = −

Kp Kp 1 x2 + x3 − d Tp Tp Tp

Single Area Load Frequency Control

4

U = ΔPC (S)

1 s

−

B1

1

3

2

1

1

1 + s Tg

1 + Tt s

ΔPG (S)

ΔF(S)

Kp

+

101

1 + Tp

−

ΔPD (S) = d

1 R

Fig 5.10 Single area state space model diagram From Block (2)

 x3 =

1 1 + Tt s

 (x4 )

(5.13)

x4 = x3 + x˙3 Tt x˙3 =

(5.14)

−x3 x4 + Tt Tt

From Block (3) x4 = (U −

1 1 x2 )( ) R 1 + Tg s

(5.15)

1 x2 ) R −1 1 U x˙4 = x2 − x4 + Tg R Tg Tg x4 + x4 Tg s = (U −

From Block (4) x1 =

5.7

x˙1 ⎢ x˙2 ⎢ ⎣ x˙3 x˙4

(5.17)

x˙1 = x2

MATRIX REPRESENTATION OF ALL STATE EQUATIONS ⎡

⎡

1 x2 s

(5.16)

0

1 −1 Tp

⎢ ⎢ 0 ⎢ ⎥ ⎢ ⎥=⎢ ⎦ ⎢ 0 0 ⎢ ⎢ −1 ⎣ 0 RTg ⎤

0 Kp Tp −1 Tt 0

0

⎤

⎥⎡ ⎥ 0 ⎥ ⎥⎢ ⎢ 1 ⎥ ⎥⎣ ⎥ Tt ⎥ −1 ⎦ Tg

⎤

⎡

x1 ⎢ ⎢ x2 ⎥ ⎥+⎢ ⎢ ⎦ x3 ⎣ x4

0 0 0 1 Tsg

⎤

⎡

0

⎥ ⎢ Kps ⎥ ⎢ ⎥ (U ) + ⎢ − Tps ⎥ ⎢ ⎦ ⎣ 0 0

⎤ ⎥ ⎥ ⎥ (d) ⎥ ⎦

(5.18)

102

Power System Operation and Control

It is in the form of x˙ = Ax + BU + Fd U – control vector d – disturbance vector x – state vector Equation (5.18) is converted into standard state space model form. i.e. x˙ = Ax + Bu is written as follows: If x is made up of the sum of transient and steady state value, and U is also made up of transient and steady state value, then 

x = x 1 + xss U = U 1 + Uss

Therefore,

 . then substitute in the equation x˙ = Ax + BU + Fd

x˙ = Ax 1 + Axss + BU 1 + BUss + Fd

(5.19)

If a sudden disturbance occurs and the system reaches the steady state value, then x˙ = 0 Therefore Axss + BUss + BU 1 + Fd = 0 and the transient term is also zero under steady state. Hence x˙ = 0 ⇒ Axss + BUss + Fd = 0

Example 5.1 Consider the block diagram model of LFC (load frequency control) given in Figure 5.11. Make the following approximation (1 + sT g )(1 + sT t ) = 1 + (T g + T t )s = 1 + sT eq . Solve for  f(t) with parameter PD =0.01. Solution: Consider input PC (S ) = 0, then there is only one input i.e., PD (S ). Redrawing the above block diagram as in Fig 5.12, Δ PD (S ) Δ PC (S ) −

Kg

Kt

1+ s Tg

1+ s Tt

−

1 R

Fig 5.11 Single are LFC

Δ F(S ) Kp

1+ s Tp

Single Area Load Frequency Control

Δ PD (S )

103

Δ F(S )

Kp 1+ s Tp −Kg Kt (1+ sTg) (1+ s Tt) R

Fig 5.12 Reduced block diagram from Fig 5.11 F (s) = PD (s)

Kp /(1 + sTp ) Kp Kg Kt 1− 1 + sTp R (1 + sTg )(1 + sTt )

100 0.01 1 + 20s = , but given PD = 100 s 1− 3(1 + 0.9s) A B 300(1 + 0.9s) 0.01 = + sF (s) = (1 + 20s)(2.7s − 97) 1 + 20s 2.7s − 97

(1)

(2)

(3)

By equating the same order coefficients of s 3 × 0.9 = 2.7A + 20B (4) 3 = –97A + 1B (5) Solving Equations (1) and (2). A = −3.075 × 10−2 , B = 1.714 × 10−2 f (t ) = Lt sF (s) = −0.01537e −0.05t + s→0

0.00635e 35.92t . Therefore

f (t ) = −0.01537e −0.05t + 0.00635e 35.92 t

Example 5.2 Determine the automatic load frequency control (ALFC loop parameters for a control area having the following data: Total rated area capacity = 2000 MW Normal operating load = 1000 MW Inertia constant H= 5 Regulation R = 2.4 H/ MW Solution: D=

∂PD 10 16.67 = = 16.67MW/Hz or D = = 8.33 × 10−3 p.uMW/Hz (1) ∂f 0.60 2000

Tp =

2 × 50 2H = = 200.8 sec f0 D 60 × 8.3 × 10−3

(2)

KP =

1 1 = = 120 Hz/p.uMW D 8.3 × 10−3

(3)

104

Power System Operation and Control

Example 5.3 For an isolated single area, consider the following data: Area capacity = 1000 MW Normal operating load = 500 MW Inertia constant = 5 sec Speed Regulation R = 5 % =2.5 Hz/p.u MW Operating frequency f0 =50 Hz The load decreases by 1 percent. For a decrease in frequency by 1 percent, find the gain and time constant of the power system represented by a first order transfer function. If it is an uncontrolled area then find out the change in frequency due to an increase of load by 75 MW. Find in what proportion (the increase in load demand is met by the increase in generation, and the decrease in load due to drop in frequency? Solution: D= TP =

∂PD 0.01 × 500 = = 10MW/Hz ∂f 0.01 × 50

2 × 50 2H = = 20 sec f0 D 0.01 × 50

KP =

1 1 = = 100Hz/p.u MW D 0.01

PD = 75MW = 0.075p.uMW.R = 2.5Hz/p.u MW 1 f = − .PD = −0.1825Hz 1 D+ R Increase in generation APG, 1 . R

KP .PD KP 1+ R = 73.17MW

PG =

KP .PD KP 1+ R = −1.8293MW

Decrease in load = −D.f = −D

Example 5.4 A control area had a total rated of capacity 1000 MW. The speed regulation ‘R ’ for all the units in the area is 2 Hz/p.u MW. A one percent change in frequency causes a one percent change in load. If the system is operating at half of the rated capacity and the load increase by 2 percent, (a) find the static frequency drop. (b) if the speed governor loops were open what would be the frequency drop. Solution: (a) R = 2Hz/p.u MW ∂PD 0.01 × 500 D= = = 10 MW/Hz ∂f 0.01 × 50

Single Area Load Frequency Control

105

1 1 = 0.01 + = 0.51p.u MW/Hz R 2 1 f = − .PD = −0.196 Hz β (b) Speed governor loop is open, then R → infinite 1 β=D+ = D = 0.01p.u MW/Hz ∞ 1 f = − .PD = −1 Hz β β=D+

Example 5.5 An uncontrolled isolated power system has the following parameters: Rated output =350 MW Regulation=0.06 p.u Inertia constant =5 Turbine time constant =0.6 sec Governor time constant =0.3 sec Nominal frequency=50 Hz The load varies by 0.7 percent for a 1 percent change in frequency. Determine the steady state frequency deviation in Hz for a load change of 50 MW. 0.7 × 350 ∂PD 4.9 2.45 100 D= = = 4.9 MW/Hz = = 0.014 p.u MW = 1 ∂f 0.5 350 × 50 100 R = 0.06 p.u (Note: R unit is Hz/p.u MW only, therefore the numerator p.u in the given data is converted to Hz by multiplying with its base.) R = 0.06 × 50 = 3 Hz/p.u MW 1 1 = = 71.428 Hz/p.u MW D 0.014 2H 2×5 Tp = ◦ = = 14.285 sec f D 50 × 0.014

kp =

We know that for the uncontrolled case,

⇒ F (S ) =

F (S ) =

kp /1 + sTp .(−PD (S )) kp kg kt 1 1+ . . . 1 + sTl 1 + sTg 1 + sTp R kp (1 + sTt )(1 + sTg )(−PD (S ))   1 (1 + sTt )(1 + sTg )(1 + sTp ) + kg kt kp R

106

Power System Operation and Control

Δ PD (S ) −

Δ F(S )

Kp 1+ s Tp

+

Kt

Kg

1+ s Tt

1+ s Tg

−1 R

Fig 5.13 Single area LFC for Pc = 0 kg kt = 1, and substituting for kp , Tt , Tg , PD (S ), R in above equation. PD (S ) = PD /s = (50/350)/s = 0.143/s F (S ) = = but

71.428(1 + 0.6s)(1 + 0.3s)(−0.143/s) (1 + 0.6s)(1 + 0.3s)(1 + 14.285s) + 71.428(1/3) 71.428(1 + 0.6s)(1 + 0.3s)(−0.143/s) (1 + 0.6s)(1 + 0.3s)(1 + 14.285s) + 23.81

(71.428)(−0.143) = −0.429 Hz 23.81 Now actual frequency of the system f = f ◦ + f |steady state = 50 + (−0.429) f = 49.571 Hz

f |steady state = lts→0 sF (S ) =

Example 5.6 An uncontrolled single area has two alternators with speed governors. Generating unit ‘1’ is rated 600 MVA, with 5 percent speed regulation is p.u. and generating unit ’2’ is rated at 400MVA, with 4 percent speed regulation in p.u. Under nominal operating conditions with f ◦ =50 Hz, both the units share a load of 700 MW. With an assumption that no frequency dependent load is connected to the system, find the new frequency and load share between each generators for an additional load of 85 MW, if the initial share is 500 MW and 200 MW on alternators ‘1’ and ‘2’ respectively. Assume a base of 800 MW. Solution: R1 = 0.05 p.u = 0.05 × 50 = 2.5 Hz/p.u MW R2 = 0.04 p.u = 0.04 × 50 = 2.0 Hz/p.u MW 85 = 0.10625 800 −PD −(0.10625) = = = −0.1181 Hz 1 1 1 1 + + R1 R2 2.5 2.0

PD = f |steady state

Single Area Load Frequency Control

107

New frequency f = f ◦ + f |steady state = 50 + (−0.1181) = 49.88 Hz −f |steady state = 0.04724 p.u MW = 0.04724 × 800 = 37.792 MW R1 Therefore, PG 1 = PG◦ 1 + PG 1 = 500 + 37.792 = 537.792 MW But, PG 1 =

and PG 2 =

−f |steady state = 0.0591 p.u MW = 0.0591 × 800 = 47.24 MW R1

Therefore, PG 2 = PG◦ 2 + PG 2 = 247.24 MW

Example 5.7 Repeat Example 5.6 assuming frequency dependent load such that load increases by 1 percent for 1 percent increase in frequency. Solution: ∂PD 0.01 × 800 D= = = 16 MW/Hz = 0.02 p.u MW/Hz ∂f 0.01 × 50 −PD −(0.10625) = = −0.1152 Hz f |steady state = 1 1 1 1 D+ + 0.02 + + R1 R2 2.5 2 f = 50 + (−0.1154) = 49.8845 Hz   −f |steady state 0.1154 ◦ × 800 = 536.928 MW = 500 + PG 1 = PG 1 + R1 2.5   −f |steady state 0.1154 ◦ PG 2 = PG 2 + = 200 + × 800 = 246.16 MW R2 2 Note: In this case, because of load frequency dependency the actual load is 783 MW, when compared to the previous case (Example 5.6) where the load is 785 MW.

Example 5.8 Obtain the dynamic response of an uncontrolled isolated power system with the following loop parameters R=2 Hz/p.u MW, kp =75 Hz/ p.u MW, Tp =20 sec, PD =0.02 p.u. Solution: We know that the dynamic response of an uncontrolled isolated power system is ⎡ given by   ⎤ kp 1 1+ t − Rkp ⎢ R ⎥ ⎢1 − e Tp ⎥ PD f (t ) = − ⎦ R + kp ⎣ ⎡

1 − 2 × 75 ⎢ 20 = ⎣1 − e 2 + 75

 ⎤ 75 1+ t 2 ⎥ ⎦ 0.02



108

Power System Operation and Control

  = −1.95 1 − e −1.925t 0.02 f (t ) = −0.039 + 0.039e −1.925t

(1)

Based on Equation (1) the response can be obtained as shown below: t = 0, f (t ) = 0 t = 1, f (t ) = −0.033 Hz t = 2, f (t ) = −0.038 Hz t = 3, f (t ) = −0.038 Hz

0

1

2

3

4

5

t

−0.01 −0.02

Δf/Steady State

−0.03 −0.04

Δf(t) Fig 5.14 Dynamic response of single area LFC

Example 5.9 A single area load frequency control system has the following parameters with free governor action: Total capacity P a = 1800 MW Nominal operating load P D = 900 MW Nominal frequency f ◦ = 50 Hz Inertia constant H = 5 sec Regulation R = 4 percent k p = 90 Hz/p.u MW Given that Tp =20 sec,Tg = Tt =0.4 sec, 1 percent change in frequency causes 1 percent change is load. (a) Check whether the ALFC is stable or not. (b) Determine the location of three eigen values.

Single Area Load Frequency Control

109

Solution: (a) From Fig 5.5

Δ PD (S ) −

Kp

Δ F(S )

1+ s Tp

+

−Kg Kt (1+ s Tt) (1+ s Tg) R Fig 5.15 Single area LFC with Pc (s) = 0 −kp (1 + sTp ) F (S ) = kp kg kt PD (S ) 1+ (1 + sTp )(1 + sTg )(1 + sTt )R =

−kp R (1 + sTg )(1 + sTt ) (1 + sTp )(1 + sTg )(1 + sTt )R + kp kg kt

Therefore the characteristic equation is (1 + sTp )(1 + sTg )(1 + sTt )R + kp kg kt = 0 ⇒ (1 + sTg + sTp + s 2 Tg Tp )(1 + sTt )R + kp kg kt = 0 ⇒ R + sRTg + sRTp + s 2 RTg Tp + sRTt + s 2 RTg Tt + s 2 RTp Tt + s 3 RTt Tg Tp + kp kg kt = 0 ⇒ s 3 (RTt Tg Tp ) + s 2 R (Tg Tp + Tg Tt + Tp Tt ) + sR (Tg + Tp + Tt ) + (kp + R ) = 0 kg kt = 1 ⇒ s 3 Tg Tt Tp + s 2 (Tg Tp + Tg Tt + Tp Tt ) + s(Tg + Tp + Tt ) + (kP /R + 1) = 0 Substituting Tp = 20, Tg = Tt = 0.4, kp = 90, R = 0.04 × 50 = 2 Hz/p.u MW we get ⇒ s 3 (20 × 0.42 ) + s 2 (20 × 0.4 + 0.42 + 20 × 0.4) + s(20 + 0.8) + (90/2 + 1) = 0 ⇒ 3.2s 3 + 16.16s 2 + 20.8s + 46 = 0

110

Power System Operation and Control

Using Routh’s array s3 s2 s1 s0

3.2 16.16 11.69 11.69

20.8 46 0

Since all the elements in the first column are > 0, the system is stable. (b) Finding the root of the equation 3.2s 3 + 16.16s 2 + 20.8s + 46 = 0 x1 = −4.31, x2 = −0.3672 ± j 1.788 The Eigen values are −4.31, −0.3672 + j 1.788, −0.3672 − j 1.788.

Example 5.10 A generator has a primary ALFC loop characterizer by R=4 percent. The secondary loop is open. It is feeding a load consisting of three equal resistors ‘R ’ connected in a star network. The frequency is 50 Hz. The resistances are decreased until the frequency drops to 49 Hz. Find by how many percent the resistance ‘R’ has to be decreased. Assume that the AVR loop keeps the terminal voltage constant. Solution: Per phase equivalent of the above data is shown below: Resistance ‘R ’ is decreased until the frequency drops to 49 Hz.

V G R

Fig 5.16 Single generator connected to a 'R' load

But we know that the power delivered to R is given by P = V 2 /R Watts and the change in frequency is f = 1Hz.

Single Area Load Frequency Control

111

With R = 4 percent, → 0.04 × 50 = 2 Hz/p.u MW It shows that 2 Hz causes a 100 percent change in load Therefore f (1 Hz) causes =

100 × 1 = 50 percent change in load. 2

Let the new load be Rx = xR where ‘R ’ is the given load. Then Px =

V2 V2 = R Rx x

but increase in load =

Px − P = 0.5 P

  V2 1 V2 V2 − 1 − 1−x R x = = 0.5 = xR 2 R = x V V2 R R x(0.5) = 1 − x ⇒ x = 0.667 Decrease in R is (100 − 66.67) = 33.33 percent Resistance R has to be decreased by 33.33 percent for a frequency decrease of 1 Hz

QUESTIONS FROM PREVIOUS QUESTION PAPERS

———————————————————————————————————— 1. a) Briefly explain the control area concept and control area error 2. a) Obtain the dynamic response of load frequency control of an isolated power system for first order approximation. b) Obtain the dynamic response of load frequency controller with and without integral control action. 3.(a) Explain the necessity of maintaining a constant frequency in power system operation. (b) Two generators rated 200 MW and 400 MW are operating in parallel. The drop characteristics of their governors are 4 percent and 5 percent respectively from no load to full load. Assuming that the generators are operating at 50 Hz at no load, how would a load of 600 MW be shared between them? What will be the system frequency at this load? Assume free governor operation. Repeat the problem if both the governors have a drop of 4. 4. A control area has a total rated capacity of 10,000MW. The regulation R for all the units in the area is 2Hz/p.u MW. A 1 percent change in frequency causes a 1 percentage change in load. If the system operates at half of the rated capacity and the load increases by 2 percent,

112

Power System Operation and Control

a) Find the static frequency drop. b) If the speed governor loop were open, what would be the frequency drop? Derive the formula used. 5. Draw a block diagram representation of load frequency control for a single area system. 6.(a) Obtain the dynamic response of a load frequency controller of an isolated power system at first order approximation. (b) Obtain the dynamic response of load frequency controller with and without integral control action. 7.(a) Explain the concept of “control area” in the load frequency control of a power system. (b) Show how the steady state error of frequency in a typical load frequency control of a power system is reduced to zero.

COMPETITIVE EXAMINATION QUESTIONS

———————————————————————————————————— 1. The combined frequency regulation of machines in area of capacity 1500 MW and operating at a nominal frequency of 60 KHz is 0.1 pu on its own base capacity. The regulation in Hz/MW will be (IES 1997) (a) 0.1/1500 (b) 60/1500 (c) 6/1500 (d) 60/150 2. Load frequency control uses

(IES 1999)

(a) proportional controllers alone (b) integral controllers alone (c) both proportional and integral controllers (d) either proportional or integral controllers 3. The speed regulation parameter R of a control area is 0.025 Hz/MW and load frequency constant D is 2 MW/Hz. The area frequency response characteristic (AFRC) is (IES 2000) (a) 42.0 MW/Hz (b) 40.0 MW/Hz (c) 20.0 MW/Hz (d) 2 MW/Hz 4. The electrical stiffness of a synchronous generator connected to a very large rid can be increased by (IES 2002)

Single Area Load Frequency Control

113

(a) increasing the excitation or the power angle of the machine (b) reducing the excitation or the synchronous reactance of the machine. (c) increasing the synchronous reactance of the machine (d) Operating the generator at a much lower MW level compared to the steadystate limit 5. For 800 MJ stored energy in the rotor at synchronous speed, what is the inertia constant H for a 50 Hz, four pole turbo-generator rated 100 MVA, 11kv? (IES 2005) (a) 2.0 MJ/MVA (b) 4.0 MJ/MVA (c) 6.0 MJ/MVA (d) 8.0 MJ/MVA 6. For a synchronous generator connected to an infinite bus through a transmission line, how are the change of volatage ()V and the change of frequency ()V related to the active power (P) and the relative power (Q)? (IES 2006) (a) ()V is proportional to P and ()f to Q (b) ()V is proportional to Q and ()f to P (c) Both V and  f are proportional to P (d) Both V and  f are proportional to Q

This page is intentionally left blank.

Two-Area Load Frequency Control 6.1

LOAD FREQUENCY CONTROL OF TWO-AREA SYSTEM

In single-area frequency control we could represent the frequency deviations by a single variable f . However, in the case of two-area frequency control, we assume each area to be individually strong, and have inter connected through a weak tie-line. This therefore leads us to the assumption that the frequency deviation in the two areas can be represented by two variables f1 , and f2 respectively. An extended power system can be divided into a number of load frequency control areas inter connected by means of tie-lines. Consider a two-area case connected by a single tie-line as shown in Fig 6.1. Power systems are inter connected for economy and continuity of power supply. For an inter connected operation, incremental efficiencies, fuel costs, water availabilities, spinning reserve allocation and area commitments are important considerations in preparing load dispatch schedules.

Tie-line 1

2

Area-1

Area-2

Fig 6.1 Inter connected two-area system

6.1.1 Flat Frequency Control of Inter Connected System Let two generating stations are connected by a tie-line. If there is a load increment at station B, the kinetic energy of the generators reduce to absorb the load, and generations increase at both stations A and B. Further, the frequency will be less than normal at the end of the governor response period as shown in Fig 6.2 (b). The load increment will be shared partly by A and partly by B. The tie-line power flow will change and hence if a frequency controller is placed at B then it will shift the

116

Power System Operation and Control

governor characteristics at B parallel to itself as shown in Fig. 6.2. (c). The frequency will therefore be restored to its normal value, reducing the change in generation in A to zero. If there is a load increment comes at station A, initially the generation in both A and B changes to absorb the additional load. However, the additional load is absorbed by B only. Station A does not absorb the load changes in the steady state. It is possible that in an inter connected operation, a given station can be made to absorb the load changes occurring elsewhere in the system so long as the controlling station has the capability to absorb the change. The same analysis can be applied to a two-area system.

P tie

A

B

(a) f

A

B ΔPB

ΔPA

P PA

PB

0 (b) f B

A Δ PA

Δ PB

P P (c) Fig 6.2 (a) Two inter connected stations. (b) Uncontrolled system with load more on station B and (c) Frequency controller located at station B

Assumptions made in the analysis of a two-area system (1) The overall governing characteristics of the operating units in any area can be represented by linear curves of frequency versus generation.

Two-Area Load Frequency Control

117

(2) The governors in both areas start acting simultaneously to changes in their respective areas. (3) Secondary control devices act after the initial governor response is complete.

6.1.2 Tie-Line Bias Control With tie-line bias controllers located in both areas the control action is complete in all the cases. For the case where the tie-line bias controller in ‘A’ initiates action for local load changes, the response is as shown in Fig. 6.3. The response to controller action is similar for load changes in B when the controller in B initiates regulating action. The governor action is not changed until the load change becomes effective in the area. This avoids unnecessary change in generation, frequency or tie-line power. The controller in the area where load change occurs acts is such a manner that the area absorbs its own load change. A single shift to the governor characteristics is necessary to restore both frequency and tie-line power to normal. A smooth cooperative regulation is thus achieved with a tie-line bias control scheme for the two area system.

f A

fs

B

PA

PB t4 t2

t3 t1 A

0

B

t0..

Fig 6.3

6.2 TWO-AREA STATE SPACE MODEL REPRESENTATION The block diagram representation of a two area system is as shown in Fig. 6.4. State space variables for two-area load frequency control are x1 = t1

x2 = PG1

x3 = PGST

 x4 =

ACE1 dt

U1 = P4

d1 = PD1

118

Power System Operation and Control

x5 = f2 x6 = PG2 x7 = 87PGS 2  x8 = ACE2 dt U2 = Pc2 d2 = PD2 From block (1) x1 =

Kps1 Kps1 Kps1 −1 x1 + x2 − x9 − d1 Tps1 Tps1 Tps1 Tps1

(6.1)

−1 x3 x2 + 2 Tt1 Tt1

(6.2)

−1 1 1 x1 − x3 = U1 R1 Tsg 1 Tsg 1 Tsg 1

(6.3)

From block (2) x2 = From block (3) x3 = From block (4)

x4 = B1 x1 + x9

(6.4)

From block (5) x5 =

Kps2 [x6 − d2 + a12 x9 ] 1 + Tps2 S

x5 (1 + Tps2 S ) = Kps2 [x6 − d2 + a12 x9 ] x5 =

Kps2 Kps2 Kps2 −1 x5 + x6 − a12 x9 − d2 Tps2 Tps2 Tps2 Tps2

(6.5)

−1 1 x6 + x7 X7 Tt 2 T t2

(6.6)

−1 1 1 x5 − x7 + U2 R2 Tsg 2 Tsg 2 Tsg 2

(6.7)

From block (6) x6 = From block (7) x7 =

x8 = B2 x5 − a12 x9 x9 =

(6.8)

2πT12 (x1 − x5 ) S

x9 = 2πT12 (x1 − x2 ) In the matrix, the form above nine equations are written as

(6.9)

B1

ACE1

1 S

4

1 R1

X4

X3

Kg1 1+STg1

3

X2

Kt1 1+TtS

2 Kp1

1

PD1=d

1+Tps1S

Pne1

X1

s

2πT12

X9

Kt2 1+STt

Kps2 1+Tps2S PD2=d

6

∆Ptie2 5

-a12

X6

S

1 R2

1

8 Kg2

X7 1+STg2

7

X8

B2

ACE2

Two-Area Load Frequency Control 119

Fig 6.4 Block diagram representation for state space model for two-area control

120

Power System Operation and Control

[x] = [A][x]−1 [B][U ] + [f ][D] Where [x] = [x1 , x2 , x3 , x4 · · · · · · · · · x9 ]T is state vector, [U ] = [U1 , U2 ]T is control vector D = [d1 , d2 ]T is a disturbance vector. ⎛

−1 ⎜ Tps1 ⎜ ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ ⎜ −1 ⎜ ⎜ RTsg 1 ⎜ ⎜ ⎜ B1 ⎜ ⎜ ⎜ A=⎜ ⎜ 0 ⎜ ⎜ ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ ⎜ 0 ⎜ ⎝ 2π T12

⎛

Kps1 Tps1

0

0

0

0

0

0

−1 Tt1

1 Tt1

0

0

0

0

0

0

−1 Tsg 1

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

−1 Tps2

Kps2 Tps2

0

0

0

0

0

0

−1 Tt2

−1 Tt2

0

0

0

0

−1 R2 Tsg 2

0

−1 Tsg 2

0

0

0

0

0

0

0

0

0

0

0

2π T12

0

0

0

0 0

⎜ ⎜ ⎜ 1 ⎜ ⎜ ⎜ Tsg 1 ⎜ ⎜ ⎜ 0 ⎜ ⎜ 0 B=⎜ ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎝ 0

0 0 0 0 0 0 1 Tsg 2 0 0

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

⎛ −K ps1 ⎜ Tps1 ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎜ ⎜ f =⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎝ 0

Kps1 Tps1

⎟ ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ ⎟ kps2 ⎟ ⎟ a12 T ps2 ⎟ ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ ⎟ −a12 ⎟ ⎟ ⎠ 0

⎞ 0 0 0 0 −Kps2 Tps2 0 0 0 0

⎞

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Two-Area Load Frequency Control

121

6.3 STEADY STATE ANALYSIS 6.3.1 Uncontrolled Case Consider Fig 6.4 with Pc1 (S ) = Pc2 (S ) = 0

(6.10)

In the case when Pc1 = Pc2 = 0, the speed changer setting is fixed irrespective of whether the load changer is in area 1 and area 2. Let a sudden change occur at both the areas 1 and 2. In incremental steps PD1 and PD2 . Due to load change the frequencies drops by f1 and f2 . If load increment in the two areas is equal and if the frequency drop is also equal i.e., if f1 = f2 = f (ss). From the block diagram in Fig. 6.4 the incremental increase of generation is written as Kt 1 Kg 1 −1 f1 (S ) R1 (1 + Tt 1 S )(1 + Tg 1 S )   −1 Kt 1 Ksg 1 ≈ 1 = f1 (S ) Tg 1 = Tt 1 = 0 R1

PG 1 (S ) =

=

−1 fss (S ) R1

(6.11)

We have, PG 1 − PD1 (S ) − Ptie1 (S ) =

2H d fss (S ) + B1 fss (S ) f0 df

(6.12)

Since fss (S ) is a steady state value d fss (S ) ≈ 0 dt For area 1 PG 1 − PD1 (S ) − Ptie1 (S ) = B1 fSS (S )

(6.13)

PG 2 − PD2 (S ) − Ptie1 (S ) = B2 fSS (S )

(6.14)

For area 2

Also, we have a12 =

Area rating 1 Area rating 2

Ptie1 = −a12 Ptie2

122

Power System Operation and Control

using above (6.11) and (6.12)equations(6.13) and (6.14)are written as 1 fss (S ) − PD1 (S ) = B1 f1 ss(S ) + Ptie1 (S ) R1

(6.15)

1 fss (S ) − PD2 (S ) = B2 f2ss (S ) − a12 Ptie1 (S ) R2

(6.16)

−

−

Multiplying (6.15) with a12 and subtracting it from (6.16), we will get  a12

   1 1 f1 (S ) + PD1 (S ) + a12 [B1 f1 (S )] − B2 f2 (S ) + f2 (S ) + PD2 (S ) = 0 R1 R2 f1 (S ) = f2 (S ) = f (S )

f (s.s) =

−(a12 PD1 + PD2 )     1 1 + B2 + a12 B1 + R1 R2

Similarly, from (6.15)and(6.16) Ptie1 is obtained as we have −1 f1 − PD1 = B1 f1 (S ) + Ptie1 R1 −1 f2 − PD2 = B2 f2 (S ) − a12Ptie1 R2 From (6.15) Ptie1

  1 f1 − PD1 = − B1 + R1

(6.17)

From (6.16)

−a12 Ptie1

  1 f2 − PD2 = − B2 + R2









1 Multiplying (6.17) by B2 + R2  Ptie1

1 B2 + R2



1 = − B1 + R1

(6.18)

  1 and (6.18) by B1 + , R1 1 B2 + R2



 f1 − PD1

1 B2 + R2

 (6.19)

Two-Area Load Frequency Control

123

       1 1 1 1 = − B1 + B2 + f2 − PD2 B1 + (6.20) −a12 Ptie1 B1 + R1 R1 R2 R1 Subtracting (6.20) from (6.19)  Ptie1

1 B2 + R2



 + a12

1 B1 + R1

    1 1 = B1 + PD2 − B2 + PD1 f1 = f2 R1 R2



   ⎞ 1 1 PD2 − B2 + PD1 B + ⎜ 1 R1 ⎟ R2 ⎟     =⎜ ⎝ ⎠ 1 1 B1 + a12 + B2 + R1 R2 ⎛

Ptie1

(6.21)

⎞

⎛

⎟ − (PD2 + a12 PD1 )   ⎟ 1 1 ⎠ B1 + a12 + B2 + R1 R2

⎜ f (s.s) = ⎜ ⎝

Let β1 = B1 +

1 R1

and

β2 = B2 +

(6.22)

1 R2

Then, Eqs. (6.21), (6.22) can be modified as Ptie1 = f (s.s) =

β1 PD2 − β2 PD1 β2 + a12 β1

(6.23)

− (PD2 + a12 PD1 ) β2 + a12 β1

(6.24)

If the two areas are identical then, β1 = β2 = β; a12 = 1

(i.e), R1 = R2 = R

and

B1 = B2 = B

Eqs. (6.21), (6.22) can be modified as PD1 + PD2 2β

(6.25)

β (PD2 − PD1 ) 2β

(6.26)

f (s.s) = P(tie1) =

(PD2 − PD1 ) 2 If a step load change occurs in Area 1 only, then =

PD2 = 0 Therefore,

124

Power System Operation and Control

f (s.s) = Ptie1 =

−PD1 2β −PD1 2

(6.27) (6.28)

From Eqn. (6.27), it may be observed that with an interconnection and a step change in load at area 1 only , the steady state change in frequency is 50 percent of the steady state change of frequency in a single area. f (s.s) =

−PD β

for single area ALF

(6.29)

f (s.s) =

−PD 2β

for two-area ALFC

(6.30)

Similarly from Ptie1 equation(Eqn. 6.28) it is observed that if a load change occurs in any area, 50 percent is generated from Area 1.

6.3.2 Controlled Case In the case of an isolated single-area system we used the integral control i.e., we let the speed changer command by a signal obtained by first amplifying, and then integrating the frequency error to bring the steady state deviation in frequency to zero . In the two-area system, our new control starategy should be such that both steady state frequency and tie-line deviations vanish due to change in either area. Area control error is to be defined to take care of frequency deviation and tie-line  power deviation. Under normal operating conditions f (i.e., ) f1 and f2 and Ptie1 must be zero. f =

− (PD2 + a12 PD1 ) β2 + a12 β2

Ptie1 =

β1 PD2 − β2 PD1 β2 + a12 β2

But these two parameters are present under normal operating conditions. Hence these are undesirable and, as in the case of single area control, we use the PI controller to bring these two errors to zero. f = Actual − Normal As with single area control, the speed is changed according to the changes of f and Ptie1 We define the ACE1 (area control error) = B1 f1 + Ptie1

(6.31)

ACE2 = B2 f2 + Ptie2

(6.32)

Where B1 and B2 are frequency bias parameters. Now,

Two-Area Load Frequency Control

125

 Pc1 = −Ki1

ACE1 dt 

= −Ki1

B1 f1 + Ptie1 dt

(6.33)

 Pc2 = −Ki2

ACE2 dt 

= −Ki2

B2 f2 + Ptie2 dt

(6.34)

Ki1 and Ki2 are integrator gains if PD1 and PD2 are applied simultaeously. Under steady state the outputs of all the blocks in two-area ALFC are constant. Under these conditions ACE1 (s.s) = B1 f1 (s.s) + Ptie1 (s.s)

(6.35)

ACE2 (s.s) = B2 f2 (s.s) + Ptie2 (s.s)a

(6.36)

if f1 (s.s) a = f2 (S .S ) = f (s.s) and Ptie2 (s.s) = −a12 Ptie1 (s.s). We know that under normal or steady state, the two errors must be zero. B1 f1 (s.s) + Ptie1 (s.s) = 0 −a12 Ptie1 (s.s) + B2 f (s.s) = 0 

1 −a12

B1 B2

 

Ptie1 (s.s) f (s.s)

 =0

(6.37)

Here under steady state f and Ptie must be zero i.e., each area has to supply its own load by itself and the errors must be zero. If Ptie1 , f is zero and |B2 + a12 |  = 0, the value of determinant will be zero. Similarly, if B1 = 0 and B2  = 0, the determinant  = 0, and vice versa.

Example 6.1 A two-area load frequency control has identical area parameters as shown below: Total area capacity = 1700 MW Nominal frequency = 50 Hz Inertia constant =6 Regulation = 4 percent Nominal operating load = 1000 MW Damping coefficient =0.02 p.u MW/Hz For a load increase of 50 MW in Area 1, find  f (s.s) and P t ie1 .

126

Power System Operation and Control

Solution: We know that, f (s.s) =

Therefore a12 =

− (a12 PD1 + PD2 )     1 1 + B2 + a12 B1 + R1 R2

1700 = 1, B1 = B2 = B = 0.02 p.u MW/Hz, PD2 = 0 1700

R1 = R2 = 0.04 × 50 = 2Hz p.u MW; PD1 =

−((0.029) + 0) = −0.0279 Hz 1 1 1(0.02 + + (0.02 + ) 2 2

Therefore f (s.s) =

Also, Ptie1 =

But β1 = β2 = B1 +

Therefore Ptie,1 =

50 = 0.029 p.u MW 1700

β1 PD2 − β2 PD1 β2 + α12 β1

1 1 1 = B2 + = 0.02 + = 0.52 p.u MW/Hz R1 R2 2

0.52(0) − 0.52(0.029) = −0.0145p.u = −24.65 MW 0.52 + 0.52

Example 6.2 A two-area load frequency control has the following specifications: Area 1: Rated capacity 100MW Regulation 4 Load 60MW Area 2: Rated capacity 30MW Regulation 3 Load 10MW For a change in load of 10MW in Area 2, find the power transmitted in the line and the operating frequency. Solution: f (s.s) =

− (a12 PD1 + PD2 )     1 1 a12 B1 + + B2 + R1 R2

PD1 = 0; PD2 =

10 100 = 0.3 p.u MW; a12 = = 3.33 30 30

Two-Area Load Frequency Control

B1 =

∂PD2 ∂PD1 = 0, B2 = = 0 (not given in the data ) ∂f ∂f R1 = 0.04 × 50 = 2Hz/p.uTh MW, R2 = 0.03 × 50 = 1.5Hz/p.uTh MW 1 = 0.5p.uTh Mw/Hz, R1 1 β2 = B2 + = 0.67p.uTh Mw/Hz R2 −[3.33 × 0 + 0.3 = −0.1285 Hz Therefore f (s. s) = 3.33(0.5) + 0.67 β1 = B1 +

Therefore operating frequency = f deg + f (s. s) = 49.87 Hz

Ptie1 =

β1 PD2 − β2 PD1 β2 + a12 β1

0.5(0.3) − 0.67(0) 0.67 + 3.33(0.5) 0.15 = 2.335 = 0.064p.uMW =

Therefore Ptie1 = 0.064p.uMW = 6.423MW Ptie1 = −a12 Ptie,1 = −21.41MW

Example 6.3 A two-area system interconnected through a tie-line has the following parameters. Maximum capacity Regulation Nominal frequency Gain, Kp

Area 1 2500 MW 2.5 percent 50 Hz 80 Hz/ p.u Th MW

Area 2 1500 MW 3 percent 50 Hz 70 Hz /rm p.u Th MW

Find  f(s. s) and tie-line power for a load change of 200MW in Area 1. Solution:

PD2 = 0, 200 = 0.08p.uTh MW PD1 = 2500

127

128

Power System Operation and Control

R1 =

2.5 × 50 = 1.25Hz/p.uTh MW 100

R2 =

3 × 50 = 1.5 Hz/p.uTh MW 100

B1 =

1 1 = 0.0125p.uMW/Hz = Kp1 80

B2 =

1 1 = = 0.0143p.uTh MW/Hz Kp2 70

β1 = B1 +

1 = 0.8125 R1

β2 = B2 +

1 = 0.6809 R2

a12 = f (s. s) = =

2500 = 1.667 1500 − (PD2 + a12 PD1 ) β2 + a12 β1 −(0 + 1.667 × 0.08) 0.6809 + 1.667 × 0.8125

f (s. s) = −0.066Hz Therefore f = f deg + f (s. s) = 49.93 Hz Ptie1 =

β1 PD2 − β2 PD1 β2 + α12 β1

=

0.8125(0) − 0.6809(0.08) 0.6809 + 1.667(0.8125)

=

−Th 0.0545 = −0.0268 = −Th 66.9 MW 2.0353

Ptie2 = −a12 Ptie1 = 111.52MW

Example 6.4 Two alternators rated 500MW and 800MW are operating is parallel with governor droop characteristics of 4 percent and 5 percent respectively, from no load to full load. If a load of 900MW is connected to the system, find the share of load for each alternator, and also the system operating frequency. Assume the no load frequency to be 50 Hz. Solution: The state of the above system is represented in Fig. 6.5.:

Two-Area Load Frequency Control P

50 Hz

W

U x Q

129

500 MW

V

900 - x

R Δf S

Alternator 1

800 MW

A

T

Alternator 2

Fig 6.5

Let Alternator 1 share ‘x’ MW of load from a total of 900MW. Therefore Alternator 2 shares (900–x) MW of load. From PVU and PRQ , we know that UV PV = QR PR Therefore

50 − f x = 500 PA − RA

(a)

But at full load for Alternator 1, the regulation is 4 percent

i.e.,

no load − full load 50 − RA =R ⇒ = 0.04 full load RA Therefore RA = 48.07 Hz

Similarly, Alternator 2,

50 − SA = 0.05 ⇒ SA = 47.62 Hz SA

(b)

(c)

Substituting (b) in (a), we get x(PA − 48.07) = (50 − f )500 Therefore x(50 − 48.07) = (50 − f )500 ⇒ x = 12953.7 − 259.06f Similarly from PVW and PST VW PV 900 − x 50 − f = =⇒ = ⇒ x = −15906.7 + 336.13f ST PS 800 50 − 47.62

(d)

130

Power System Operation and Control

Solving (d) and (e) x = 391.85 MW , f = 48.49 Hz Therefore Load shared by Alternator 1 = 391.85MW Load shared by Alternator 2=900 – 391.85 = 508.15 MW frequency f = 48.49 Hz

QUESTION FROM PREVIOUS QUESTION PAPERS 1. a) Draw the block diagram of load frequency control of a two-area control system b) What is area control error? What are the control strategies? 2. a) What is load frequency control in a two-area power system? Why is it essential to maintain constant frequency in an inter-connected power system? b) Two power stations A, B are interconnected by a tie-line and an increase in load of 250 MW on system B causes a power transfer of 150 MW from A to B. When the tie-line is open, the frequency of System A is 50 c/s and that of system B is 49.5 c/s. Determine the value of KA and KB , which are the power frenquency constants of the two generators. 3. Explain the power frequency characteristics of an interconnected power system. 4. Derive an expression for steady state change of frequency and tie-line power transfer of a two-area power system. 5. Give typical block diagrams for a two-area system interconnected by a tie-line and explain each block. Also deduce relations to determine the frequency of oscillations of tie-line power and static frequency drop. 6. Two generators rated 200 MW and 400 MW are operating in parallel. The drop characteristic of thin governors are 4% and 6% respectively from no load to full load. Assuming that the generators are operating at 50Hz, how can a load of 500MW be shared between them. 7. Two generators rated 200 MW and 400 MW are operating in parallel.The drop characteristics of their governors are 4% and 5% respectively from no load to full load. The speed changes are so set that the generators operate at 50Hz sharing a load of 600 MW in the ratio of thier ratings. If the load reduces to 400 MW, how will it be shared among the generators and what will the system frequency be? 8. Two generators rated 300 MW and 400 MW are operating in parallel. The drop characteristic of thier governors are 4% and 6% respectively from no load to full load. The speed changes of the governors are set so that a load of 400 MW is shared among the generators at 50Hz in the ratio of their ratings. What are the no load frequencies of the generators

Two-Area Load Frequency Control

131

COMPETITIVE EXAMINATION QUESTIONS

———————————————————————————————————— 1. The following data pertain to two a1tematot working in parallel and supplying a total load of 80 MW: Machine 1:40 MVA with 5% speed regulation. Machine 2:60 MVA with 5% speed regulation. The load sharing between machines 1 and 2 will be: (IES 1997) P1 P2 (a) 48 MW 32 MW P2 P1 (b) 40 MW 40 MW (c) 30 MW (d) 48 MW 2. Two generators rated at 200 MW and 400 MW are operating in parallel. Both the governors have a drop of 4%, when the total load is 300 MW They share the load as (suffix ’1’ is used for generator 200 MW and suffix ’2’ is used for generator 400 MW) (IES 1999) (a) P1 = 100MW and P2 = 200MW (b) P1 = 150MW and P2 = 150MW (c) P1 = 200MW and P2 = 100MW (d) P1 = 200MW and P2 = 400MW 3. Two alternators each having 4% speed regulation are working in parallel. Alternator ’1’ is rated for 12 MW and alternator ’2’ is rated for 8 MW when the total load is 10 MW the Loads shared by alternators 1 and 2 would be respectively. (IES 2000) (a) 4 MW and 6 MW (b) 6 MW and 4 MW (c) 5 MW and 5 MW (d) 10 MW and zero 4. Two generators rated 200 MW and 400 MW having governor droop characteristics of 4% and 5% respectively are operating in parallel. If the generators operate on no load at 50 Hz, the frequency at which they would operate with a total load of 600 MW is (IES 2002) (a) 48.50 Hz (b) 47.69 Hz (c) 46.82 Hz (d) 49.04 Hz 5. Consider a power system with two plants S1 and S2 connected through a tie line as shown above. When the load-frequency control of the system is considered, the ‘flat tie-line control’ system is preferred over the ‘flat frequency regulation system’, because (IES 2003)

132

Power System Operation and Control S2

S1

load

Tie Line

load

(a) It is advantageous to control the frequency from any one particular plant without disturbing the other one during load-swings on their S1 and S2 areas (b) This ensures that only the more efficient plant’s input is controlled for load, variation in any area (c) Only the tie line is required to absorb the load-swing (d) The load-change in a particular area is taken care of by the generator in that area resulting in the tie-line loading to remain constant.

Load Frequency Controllers 7.1

INTRODUCTION

In Chapters 5 and 6 we discussed about steady state and dynamic frequency deviation. For a normal system operation the maximum permissible limit of frequency deviation is ± 0.5 Hz. For a satisfactory operation of the system we need some control strategy that keeps the deviation to a minimum. A commonly used controller is a proportional plus integral controller. Whenever load varies, the presence of an integral controller aids us to bring the steady state frequency deviation to zero with faster dynamic response.However, due to load variation the system loses its economic operation and hence, in addition to an integral controller, an economic dispatch controller should also be present for a satisfactory operation of the system. In this chapter we shall discuss the steady state operation of a single-area system, initially with an integral controller, and later with economic dispatch control.

7.2

PROPORTIONAL PLUS INTEGRAL CONTROLLER

The block diagram of an uncontrolled single-area system with an integral controller, is shown in Fig 7.1 The integral controller gives signal to the speed changer ie Pc . Hence we can write  Pc = − Ki f (t )dt (7.1) ΔPD(s) −Ki S

ΔPC(s) +

Integral Contrller

Kg −

1 + STg

Kt 1 + STt

+

−

Δf(s) Kp

ΔF(s)

1 + STp

1 R

Fig 7.1 Block diagram of a proportional plus integral controller

134

Power System Operation and Control

Taking Laplace transformation Pc (S ) = −

Ki f (S ) S

(7.2)

The negative sign indicates that for a positive frequency error, the speed changer signal should be decreased and vice versa. Steady State Response: For a step load increase PD (S ) = PD /S . Therefore, from Fig 7.1



f (S ) = 1+

KP 1 + STp



(KP /1 + STp )    (−PD /S ) Kg Kt Ki 1 + 1 + STt 1 + STg S R

fsteady state = Lt S . F (s)

(7.4)

S →0

= LtS →0 S

 1+

KP 1 + STp =



(7.3)

(KP /1 + STp )    (−PD /S ) (7.5) Kg Kt Ki 1 + 1 + STt 1 + STg s R

KP   (−PD ) Ki 1 1 + KP Kt Kg + 0 R KP .PD =− =0 1+∞ fsteady state = 0

(7.6) (7.7)

Hence from (7.7) it is clear that the presence of an integral controller makes the steady state deviation in frequency, zero. Dynamic response: The dynamic response for the system shown in Fig 7.1 can be derived assuming Tg = Tt = 0 and the product Kg Kt = 1. Hence, eqn(7.3) reduces to

F (S ) =

=

=

(KP /1 + STp )(−PD /S )   KP Ki 1 1+ + 1 + STp S R −KP (PD )/S   1 Ki (1 + STp ) + KP + S R KP (PD ) S + S 2 TP + KP Ki +

KP S R

(7.8)

Load Frequency Controllers



= Tp

 S2

+

−KP (PD )   1 KP K P Ki + S+ Tp RTp Tp

−KP /Tp (PD )    1 K P Ki 1 K P 2 S + s+ + Tp R KP Tp

F (s) = 

Considering the denominator of Equation(7.10)   KP Ki 1 KP 1 S2 + S+ + Tp R KP Tp

135

(7.9)

(7.10)

(7.11)

From eqn(7.11) it is evident that the nature of the poles depends on the value of Ki . Hence, considering (7.11) as characteristic equation, the poles are

Poles1,2 =

KP − Tp



1 1 + R KP 2



*   2  + + KP 1 KP Ki 1 + − 4 + , Tp R KP Tp ± 4

(7.12)

Case-I: For a critical case, the discriminant is zero.   2   KP 1 1 KP Ki,critical Hence, =0 + −4 Tp R KP Tp   2 KP 1 1 KP ki,critical ⇒ + =4 Tp R KP Tp     2 Tp KP 1 KP 1 1 2 1 2 ki,critical = . 2 = + + 4KP Tp R KP 4Tp R KP but KP =

1 2H and Tp = D Df ◦ Ki,critical =

f◦ 8H



1 +D R

2

Case-II: If Ki > Ki,critical then the roots are complex. Hence, 1 KP − 2 Tp

(7.13)



1 1 + R KP



(   2   1 KP 1 KP Ki 1 − ± = −σ ± j ω + 4 Tp R KP Tp

Thus Poles1,2 ⇒ (S + σ )2 + ω2

(7.14)

136

Power System Operation and Control

f (t ) contains e −σ t sin ωt and e −σ t cos ωt , whose osilations are damped in nature. Case-III If Ki < Ki,critical , then the roots are real, hence Poles1,2 ⇒ (S + α1 )(S + α2 ) f (t ) contains the terms e −α1 t and e −α2 t which indicate exponential decay. In either of the above Case II and Case III, it is evident that f (t ) will approach zero. The dynamic response of an isolated system with integral controller is shown in Fig 7.2

Δf (t) t (s)

− 0.01 − 0.02 − 0.03 Increasing Ki

Without integral controller

Fig 7.2 Dynamic response of an integrated system with integral controller.

7.3

LOAD FREQUENCY CONTROL AND ECONOMIC DISPATCH CONTROL

As mentioned earlier, the presence of an integral controller aids us to attain better dynamic response with zero steady state deviation in frequency whenever there is a variation in load. This variation in load also disturbs the an economic dispatch of load. Hence, in addition to an integral controller, an economic dispatch controller should also be provided for satisfactory operation of the power system. The block diagram of a load frequency control with economic dispatch control is shown in Fig 7.3.

Load Frequency Controllers

Integral Controller

Economic dispatch contorl

+

Generator Control

+ EDC

Generator

137

To Load

DESIRED (from central dispatch control)

ACTUAL

Fig 7.3 Block diagram of LFC with EDC As shown in the block diagram, if there is any variation in load, the load frequency controller, which is fast-acting (within zero seconds) responds and actualises the speed changer (generator control) accordingly. The economic dispatch controller, which is slow acting (one minute), responds to the load change together with the input from the central dispatch control and an error signal is generated. This error signal together with the error signal generated by integral controller actualises the speed changer. Since the load frequency controller is fast acting and economic dispatch controller is slow acting, the power system may operate with EDC error only for few seconds before the EDC generates a command signal.

Example 7.1 Consider an isolated power system as shown is Fig 7.1 with the following parameters Tp =12 sec, KP =80 Hz/p.u MW, R=2 Hz/p.u MW, Ki =0.2, for a load change of 0.2 p.u MW. Obtain the time error in seconds and cycles. Assume Tg = Tt = 0 and system frequency is 50 Hz. Solution: From equation (7.3) KP (1 + STp )   (−PD /S ) F (S ) = Kg Ki 1 KP Kt . . + 1+ . (1 + STp ) (1 + STg ) (1 + STt ) s R From the given data Tg = Tt = 0, Tp = 12, KP = 80, R = 2, Ki = 0.2, PD = 0.2 also Kg Kt = 1.     −0.2 −0.2 80 80 . . (1 + 12S ) S (1 + 12S ) S     =    Hence, F (s) = 80 80 0.2 1 0.4 + S 1+ 1+ . + . 1 + 12S S 2 1 + 12S 2S

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Power System Operation and Control

 0.2 −16 /(1 + 12S ) −80 × 2S 5 S   = = 0.4 + S (2S + 24S )2 + 32 + 80S (1 + 12S ) + 80 2S (1 + 12S ) 

= F (S ) = A=

24S 2

−32 + 82S + 32

−1.33 A B = + S 2 + 3.416S + 1.33 (S + 0.448) (S + 2.968) F (S ) × (S + 0.448) −1.33 = = −0.5278 S = −0.448 −0.448 + 2.968

F (S ) × (S + 2.968) −1.33 = = −0.5278 S = −2.968 −2.968 + 0.448 −0.5278 0.5278 F (S ) = + (S + 0.448) (S + 2.968) B=

Taking inverse Laplace transform f (t ) = 0.5278(−e −0.448t + e −2.968t ) f (t ) = 0.5278(e −2.968t − e −0.448t )Hz  Time error in cycles =

∞



∞

f (t )dt =

0



0.5278(e −2.968t − e −0.448t )dt

0

e −0.448t e −2.968t − −2.968 −0.448   1 1 = 0.5278 − 2.968 0.448



= 0.5278

= −1.00 cycles (−ve sign indicates we loose that many cycles) Time error in seconds =

1.00 = 0.02 sec. 50

Example 7.2 For an isolated, integral controlled power system find the value of Ki,critical . The system parameters are given below Total capacity=1000 MW Load=600 MW Inertia constant =5.0 sec Regulation R=4 %

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139

The nominal operating frequency is 50 Hz and the system load increases by 1 percent if the system frequency increases by 1 percent . Solution: We know from Eqn (7.14) ki,critical

D=

f◦ = 8H



1 +D R

2 (a)

∂PD 600 × 0.01 12 = = 12 MW /Hz = = 0.012 p.u MW/Hz ∂f 50 × 0.01 1000

R = 0.04 × 50 = 2 Hz/p.u MW f ◦ = 50 Hz and H = 5 sec from given data. Hence, Ki,critical =

50 [0.5 + 0.012]2 = 0.3278 8×5 ki,critical = 0.3278

QUESTIONS FROM PREVIOUS QUESTION PAPERS

———————————————————————————————————— 1. Explain the proportional plus integral control for load frequency control for a single area system. 2. With a neat block diagram explain the load frequency control with economic dispatch control. 3. a) With a neat block diagram explain the load frequency control for a single area system. b) Two generators rated 250MW and 500 MW are operating in parallel The droop characteristics are 4 percent and 6 percent respectively . Assuming that the generators are operating at 50 Hz at no load, how would a load of 750MW be shared? What is the system frequency? Assume free governor action. 4. a) Explain the load frequency control problem in a multi-area power system. 5. a. Draw the block diagram representation of load frequency control. b. For two-area load frequency control with integral controller blocks, derive an expression for steady values of change in frequency and tie-line power for simultaneously applied unit step load distrurbance inputs in the two areas. 6. Draw the block diagram of a power system showing the governor, turbine and syngenerator indicating their transfer functions for a step disturbance of small

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Power System Operation and Control

change PD . Obtain the response of increment in frequency, making suitable assumptions (a) without proportional plus integral controller, and (b) with proportional plus integral control.

COMPETITIVE EXAMINATION QUESTION

———————————————————————————————————— 1. Consider the following statements regarding load frequency control: (IES 2004) (1) Time constant of automatic load frequency control is about 15 seconds (2) Integral control eliminates static frequency drop. (3) In tie-line load bias control, the control signal for each area is proportional to change in frequency as well as charge in tie-line power. Which of the statements given above are correct? (a) (b) (c) (d)

1,2 and 3 1 and 2 1 and 3 2 and 3

Reactive Power Control 8.1

INTRODUCTION

The modern electric utility industry began in the 1880s. On September 4, 1882, the first commercial power station, located on Pearl Street in lower Manhattan, known as Pearl Street electricity generating station went into operation providing light and electricity power to customers in a one square mile area. This was a DC power system built by Thomas Alva Edison, which was comprised of a steam driven DC generator connected through an electrical cable system at 110V and 59 customers spread over an approximate area with 1.5 km radius. But now electrical energy is generated, transmitted and distributed exclusively in the form of alternating current (AC). The main drawback of DC generation is that it can neither be generated in bulk nor stepped up to higher levels for efficient transmission. Development of transformers changed the intact scenario of electrical power systems. Invention of induction motors (single phase and poly phase) by Nikola Tesla and the massive increase in load demand called for restructuring of the power systems. A majority of the loads in the present power system are inductive in nature, and whenever they are supplied from an AC source, they draw a current which is lagging by the applied voltage at an angle and immediately the power factor comes into picture.

8.2

POWER FACTOR

The cosine of the phase angle between two electrical quantities such as voltage and current in an AC circuit is known as power factor. For instance, consider a load supplied by a voltage V and drawing a current of magnitude I , which lags the voltage by an angle φ. The phasor diagram of the circuit is given, in Fig. 8.1. The power factor of the circuit is given by, Power Factor = Cos(φ) The load current I can be resolved into two components. 1. Horizontal component, I Cos φ in phase with the applied voltage V 2. Vertical component, I Sin φ, 90° out of phase with the applied voltage V The horizontal component, I Cos φ is known as active or wattful component and the vertical component, I Sin φ is known as the reactive or wattless component. The reactive component is a measure of the power factor. Greater the reactive component, larger the phase angle (φ) between the applied voltage and the current drawn by the load. Hence the power factor Cos φ is low. Therefore, a circuit having small reactive current I Sin φ will have high power factor and vice-versa.

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Power System Operation and Control

o

I Cos φ

A

V

φ

I

I Sin φ B

Fig 8.1 Phase diagram, I lags V Power factor is a characteristic of alternating current (AC) circuits. Its value is always between (0.0) and (1.0), the higher the number, the greater or better the power factor. If a purely resistive load is connected to a power supply, the current and voltage will be in phase with each other, and the angle φ equals to zero. Consequently, the power factor will be unity and the electrical energy flows in a single direction across the network in each cycle. Inductive loads such as transformers and motors (any type of wound coil) generate reactive power with current waveform lagging the voltage. Capacitive loads such as capacitor banks or buried cables generate reactive power with current phase leading the voltage. Both types of loads will absorb energy during part of the AC cycle, and store it in the device’s magnetic or electric field, only to return this energy back to the source during the rest of the cycle. The power factor concept can also be analyzed in terms of power drawn by the circuit, from the power triangle shown in Fig. 8.2

A

V I Cos φ

B

φ

VI

V I Sin φ

C Fig 8.2 Power triangle The right angled triangle ABC shown in Fig. 8.2 is known as the power triangle and is directly drawn from the triangle OAB in the phasor diagram shown in Fig. 8.1 Each side of the current triangle OAB of Fig. 8.1 is multiplied by voltage V , to get the power triangle ABC shown in Fig. 8.2.

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143

To understand the power factor, we will first start with the definition of some basic terms: From the power triangle ABC, AB = V I Cos φ, which represents Active Power (also called Actual Power or Working Power or Real Power ) and is measured in Watts or Kilowatts i.e., W or KW. It is the power that actually enables the equipment and performs useful work. BC = V I Sin φ, which represents Reactive Power and is measured in Volt Ampere Reactive i.e.,VAR’s or KVAR’s. It is the power that a magnetic equipment (transformer, motor etc.,) needs to produce the magnetizing flux. AC = V I , which represents Apparent Power and is measured in Volt Amperes or Kilo Volt Amperes i.e.,VA’s or KVA’s. It is the "vectorial summation" of KVAR and KW. The power factor can also be defined in terms of power by inferring some points from the power triangle. The apparent power V I has two components. One is V I Cos φ i.e., Active Power and the other one is V I Sin φ i.e., Reactive Power at right angles to each other. From the power triangle which is a right angled one, we can write, AC2 = AB2 + BC2 (Apparent power)2 = (Active power)2 + (Reactive power)2 (KVA)2 = (KW)2 + (kVAR)2 AB Activepower kW Power factor, Cos φ = = = AC Apparentpower kVA So, the power factor can also be defined as the ratio of active power to apparent power. It is clear from the power triangle that the lagging reactive power due to the lagging current is responsible for the low power factor. The smaller the reactive power, higher the power factor and vice versa.

8.3

REACTIVE POWER

In a purely resistive AC circuit, voltage and current waveforms are in phase and reverse their polarity at the same instant in each cycle (illustrated in Fig. 8.3). When reactive loads are present (such as capacitors or inductors), energy is stored in the form of electric or magnetic fields respectively in the loads during part of the AC cycle and results in a time difference between the current and voltage waveforms (for a capacitor, current leads voltage; for an inductor, current lags voltage). This stored energy returns to the source during the rest of the cycle and is not available to do work at the load. This energy, continuously flowing back and forth (to and fro), is known as reactive power while the active power flows from one point of the network to another. For a complete cycle, the net reactive power flow is zero as the energy flowing in one direction for half a cycle is equal to the energy flowing in the opposite direction in the next half of the cycle. Let us consider a simple analogy to understand these terms active power, reactive power and apparent power clearly.

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Power System Operation and Control

Voltage

Current Power

Average Power

0

90

180

270

360

Fig 8.3 Waveform Consider that you have ordered a chilled Pepsi (cool drink), feeling thirst on a sunny day. Now observe the content of the glass immediately after Pepsi was poured into it. You observe some bubbles (foam) at the top of the glass. The total content of the glass is the summation of the foam and the cool drink. Now we correlate this setup with our problem. This correlation is illustrated in Fig. 8.4.

Foam

KVAR KVA

Total Content Pepsi

KW

Fig 8.4 A glass of chilled Pepsi

It is obvious that the thirst quenching quantity is the cool drink and not the bubbles (foam) formed at the top of the glass. In other words, we can say that while useful work (thirst quenching) is done by Pepsi, the foam does not quench the thirst. But it is part of the total content and we have to accept it. Let the thirst-quenching portion of the glass (Pepsi) be represented by KW Let the foam formed on the top of the glass be represented by KVAR. The total content of the glass, KVA, is this summation of KW (Pepsi) and KVAR (the foam). Practically KVA is the vectorial summation of KW and kVAR.

Reactive Power Control

145

We know that, the power factor is, P.F = KW / KVA Since KVA is equal to the vectorial summation of KW and KVAR, P.F = KW/ vectorial sum(KW + KVAR) ∼ = Pepsi/Pepsi + Foam Thus, for a given KVA we can deduce that, • The more bubbles (foam) the glass contains (the higher the percentage of KVAR), lower the ratio of Pepsi (KW) to Pepsi plus foam (KVA). Therefore the power factor will be low. • The less (bubbles) foam the glass contains (the lower the percentage of KVAR), the higher the ratio of Pepsi (KW) to Pepsi plus foam (KVA). Therefore the power factor will be more, or approaches unity. We all know that if we poul Pepsi into a glass, it generally has a head of foam on it. Let us imagine a new policy – You pay only for the Pepsi and not for the foam. While the foam is just aerated Pepsi, it is not really a pleasing quantity in that form. If the glass of Pepsi contains half foam, you pay half the price. It is the same principle applied to electricity generation – the consumer only pays for working or real power and not the reactive power.

8.3.1 Sources of Reactive Power The main sources of reactive power are given below. 1. Overhead Lines: When current flows through a line it produces a magnetic field that absorbs reactive power, given by I 2 X where I is the current flowing through the line and X is the reactance of the line in ohms per phase. A lightly loaded overhead line is a net generator of reactive power, whereas a heavily loaded line is a net absorber of reactive power. However, underground cables have a small inductance and relatively large capacitances due to closeness, large size of conductors and high relative permitivity of the dielectric material used. Hence, they generate reactive power. 2. Transformers and Motors: Transformers produce magnetic fields and therefore absorb reactive power. Inductive loads such as motors (any type of wound coil) absorb reactive power with their current waveform lagging the voltage.

8.4 CAUSES OF LOW POWER FACTOR Circuits containing purely resistive loads such as filament lamps, strip heaters, cooking stoves, etc., operate with a power factor of unity. When the power factor is 1, all the energy supplied by the source is consumed by the load. Circuits containing inductive or capacitive elements usually have a power factor below 1. When the power factor is equal to 0, the energy flow is entirely reactive, and stored energy in the load returns to the source at the end of each cycle. Good power factor is considered to be greater than 0.85 or 85%.

146

Power System Operation and Control

1. Most A.C motors (any type of wound coil) are of the induction type i.e., single-phase or three-phase induction motors generate reactive power with the current waveform lagging the voltage, and consequently operating at a poor power factor. 2. Lighting loads such as arc lamps and electric discharge lamps operate at a low lagging power factor. 3. Industrial heating furnaces such as arc furnaces usually operate on the principle of striking an arc, and operate at a low lagging power factor. 4. Normal power factor (NPF) ballasts typically have a value of (0.4) – (0.6). Ballasts with a power factor greater than (0.9) are considered to be high power factor ballasts (HPF).

8.4.1 Disadvantages of Low Power Factor Power factor plays a vital role in A.C circuits since the power consumed depends upon the operating power factor. It is evident that Real power P = V I Cosφ I = P /V Cosφ From the above expression, for a fixed power and voltage the load current is inversely proportional to the power factor. Lower the power factor, higher will be the load current and vice versa. Low power factor results in the following disadvantages: 1. Increase in copper losses When a load has a power factor lower than 1, more current is required to deliver the same amount of useful energy. Copper losses (I 2 R losses) increase with increase in current. This results in poor efficiency with increase in losses. 2. Large conductor size To transmit a given power at a constant voltage, the conductor will have to carry relatively more current at a lower power factor. This obliges greater conductor size i.e., the cross-sectional area of transmission lines, cables and motor conductors has to be designed based on the increased current and entails more cost. This is illustrated in Example 8.1 for better understanding. 3. Large KVA rating Electric machinery like alternators, transformers, motors and switchgear are always rated in KVA. And KVA = KW / Cos φ. From this relation it is clear that KVA rating of the equipment is inversely proportional to the power factor. The lower the power factor, the larger will be the KVA rating. Therefore at low power factor, the KVA rating of the equipment should be made more, making the equipment size larger and expensive. 4. Poor voltage regulation A large current with a low lagging power factor results in greater voltage drops (IZ) in power system components like alternators, transformers, transmission and

Reactive Power Control

147

distribution lines. This increased voltage drop results in poor voltage regulation and decreased voltage at the receiving end, and impairs the performance of consumer loads. In order to maintain receiving end or consumer end voltages within permissible limits, additional equipment like voltage regulators, booster transformers and on-load tap changing transformers are required, which are expensive in nature. 5. Reduced handling capacity Since an increase in the reactive component of the current prevents full utilization of installed capacity, the lagging power factor reduces the handling capacity of all the elements of the system. 6. Increase generation and transmission costs The significance of the power factor lies in the fact that utility companies supply customers with Volt-Amperes, but bill them for Watts. Power factors below 1.0 require a utility to generate more than the minimum Volt-Amperes necessary to supply the working power i.e., the real power (Watts). This increases generation and transmission costs. Utilities may charge additional costs or penalize the customers (not domestic loads but large consumers like industries) who have a power factor below some limit. This case is illustrated in Example 8.2.

Example 8.1 Consider a 10 KW, single phase A.C motor having an efficiency of 90 percent, operating at a terminal voltage of 230 V. Calculate the current drawn by the motor when (i) it is operating at a power factor 1 (unity p.f) and (ii) it is operating at a 0.7 lagging power factor. Solution: Given that Output of the motor = 10000 W Efficiency of the motor = 90 percent = 0.9 output 10000 Inputofthemotor = = = 11111.11 efficiency 0.9 KVA input of the motor at unity power factor = 11111.11/1 = 11111.11 KVA. Input full load current drawn by the motor operating at unity power factor is 11111.11/230 = 48.31 A. KVA input of the motor at 0.7 power factor = 11111.11 / 0.7 = 15873.01 KVA. Input full load current drawn by the motor operating at 0.7 power factor is 15873.01/230 = 69.01A. Therefore, the motor working at unity power factor is drawing a current of 48.31 A while that which works with a lower power factor of 0.7 is drawing a current of 69.01 A, which was approximately 20 A more.

Example 8.2 An alternator is supplying a load of 500 KW at unity power factor. If the same machine is operated at a power factor 0.7 lagging, calculate the KW capacity of the supplying load.

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Power System Operation and Control

Solution: Given that the alternator is supplying 500 KW at a power factor = 1 Therefore its KVA rating at unity power factor is, KVA = KW / Cos φ = 500 /1 => 500 kVA. Now, the same machine is operated at a p.f of 0.7 lag, KVA = KW / Cos φ => KW = 500 × 0.7 = 350 kW. Therefore we can conclude that, A 500 KVA alternator working at unity power factor can supply a load of 500 KW and the same alternator working at a power factor 0.7 can supply a load of 350 KW only. Decrease in KW power supplied by the alternator = 500 − 350 = 150 kW.

Example 8.3 Assume a 4 KV single-phase circuit, which feeds a load of 450 kW and operates at a lagging power factor of 0.75. If it is desired to improve the power factor, determine the following: a) The reactive power consumption. b) The new corrected power factor after installing a shunt capacitor bank with a rating of 400 KVAR Solution: Before installing the shunt capacitors for power factor correction, The current drawn by the load can be calculated by P = V I Cos φ i 450 = 4 × 1 × 0.70 1 = 450/4 × 0.70 I = 160.714A Apparent power = S1 = V ∗ I = 4×160.714 = 642.85 KVA a) Reactive power consumption can be calculated by Q1 = S1∗ Sin φ1

[ Cos φ1 = 0.70, therefore φ1 = 45.570 ]

= 642.85 × Sin(45.57°) ∼ = 459 KVAR After power factor correction by installing a 450 KVAR capacitor bank Q2 = Q1 − Qc = 459 − 400 = 59 KVAR

Reactive Power Control

149

b) Therefore the new corrected power factor can be calculated by 1/2

Cos φ2 = P /[P 2 + (Q1 − QC )2 ] 1/2

= 450/[4502 + 592 ] = 450/453.85

= 0.9915 or 99.15 percent.

8.5 REACTIVE POWER FLOW IN AN UNCOMPENSATED TRANSMISSION LINE Let us consider a transmission line connecting two busses (a simple model) as shown in Fig 8.5. Let the R value of the transmission line be zero. Hence the losses in the line is zero. Let the load be such that it draws | I | amps at a terminal voltage of | I | volts. Now, the reactive power absorbed by the line will be QL =| I 2 | XL

1 G

2 R + jXL Transmission line

L O A D

Fig 8.5 Uncompensated transmission line We know that a transmission line consists of series inductance and shunt capacitance. Case I:

If the reactive power generated by the line given by QC =

the reactive power absorbed, then QL = QC => | I 2 | XL = QC = =>

|V2 | = XL X C | I2 |

|V2 | equals XC

|V2 | XC

(  L |V | 1 = XL XC = 2 π f L × = => |I | 2πf C C  L |V | = be the natural impedance of the line. Let Zn = |I | C

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Power System Operation and Control

Therefore, the natural power transmitted in the line is given by Pn =

|V2 | or | I 2 | . Zn Zn

The above expression indicates that the reactive power generated is equal to the reactive power absorbed, this occurs when the line is terminated by a resistance equal to its loading surge inpedance. This condition introduces a flat voltage profile over the complete distance of the transmission line as shown in Fig . 8.5(a) Case II: If the reactive power generated in the line is more than the reactive power absorbed i.e., QC > QL then, |V2 | > | I 2 | XL XC => | V 2 | > | I 2 | XL XC => | V 2 | > | I 2 | (2 π f L) ×  V Let Z =  I

   > L.  C

1 2πf C

=> | V 2 | > | I 2 | (L/C )

This implies that Z > Zn , where Zn is the natural inpedance of the line. This condition introduces a voltage rise over the complete distance of the transmission line as shown in Fig. 8.6(b) Case III: If the reactive power absored by the line is more than the reactive power generated (i.e) QL > QC then | I 2 | XL >

|V2 | XL

|V2 | < | I 2 | XL XC    V  L Let Z =   < I C =>

Which implies that Z < Zn , where Zn is the natural inpedance of the line. This condition introduces a voltage, say over the complete distance of the transmission line as shown in Fig. 8.6(c)

8.6 NECESSITY FOR REACTIVE POWER COMPENSATION IN A TRANSMISSION LINE As explanined in Section 8.4, the voltage profile varies with distance along the transmission line. Power system engineers will be happy it they are able to maintain

Reactive Power Control

Z > Zn | V|

(b)

Z = Zn

Z < Zn

151

(a) (c)

distance in km Fig 8.6 Voltage variation over a transmission line. a flat voltage profile throughout the distance of the transmission line, which is the ideal case. Power system research engineers have worked for several decades to develop compensation techniques for transmission line, which can maintain the voltage profile close to the ideal case. These compensations are essential because, if the voltage in the system is boosted up due to heavy injection of reactive power, it will lead to an increase in load with drop in frequency causing system collapse. On the other hand, if the voltage in the system decreases due to low reactive power injection in the line, it leads to voltage collapse. Every transmission line has its own limitations, namely, thermal limit and stability limit. The thermal limit of a transmission line depends upon the ambient temperature, nature of wind and the current flowing in the conductor. Beyond a certain limit, it is not possible to operate the transmission line since it will damage the conductor. The stability limit of a transmission line includes steady state stability limit, transient stability limit, dynamic stability limit and voltage stability limit. Hence compensation in a transmission line is very essential in order to maintain these limitations within strict limits for satisfactory operation and overall system performance.

8.7 METHODS TO IMPROVE POWER FACTOR 8.7.1 Synchronous Generator Control This can be treated as the principal source of reactive power control. Synchronous machines can be made to generate or absorb reactive power depending upon their level of excitation. This control is based on the principle that an over-excited synchronous machine generates reactive power and an under-excited synchronous machine absorbs reactive power. The principal advantage of this control is that it offers flexibility for all load conditions, since it supplies reactive power when overexcited (peak load period) and consumes reactive power when it is under excited

152

Power System Operation and Control

(off-peak load period). There is smooth variation in generation of reactive power by this method, when compared to the step-by-step variation by the capacitor control method.

8.7.2 Capacitor Control Capacitors can also be treated as chief sources of reactive power available at the load end or receiving end. They are again classified into two types by the virtue of their mode of connection: (a) series capacitors and (b) shunt capacitors. The objective of using a capacitor either in series or parallel is to compensate reactive power and to consequently improve the power factor and voltage profile of the system. a) Series Capacitors: The series capacitors (capacitors which are connected in series with the lines) directly neutralize the inductive reactance of the system to which it is connected. Since the effect of the series capacitor can be considered as a negative reactance in series, the net impedance will be Z = R + j (XL − Xc ). Therefore, the voltage drop IZ is reduced. Application of a series capacitor to a feeder and its vector diagram representation is shown in Fig. 8.7(a) and Fig. 8.7(b) Voltage drop of the feeder without series capacitor is expressed as VD = I R Cos φ + I XL Sin φ With series capacitor

Without series capacitor R

XL

Vs

XL

R

I

Vr

XC

Vr

Vs

Vs

IXc IXL

Vs

IZ

φ

φ

IR Cosφ

IXL

IZ'

Vr IR

I

IZ

IXL Sinφ

Vr IR

IR Cos φ I(XL– XC) Sinφ

Fig 8.7 (a) Without series capacitor (b) With series capacitor

Reactive Power Control

153

Where, R = Resistance of the feeder XL = Inductive reactance of the feeder φ = Angle between the receiving end voltageVr and current I in the feeder Resultant voltage drop of the feeder with the application of series capacitor is VD = I R Cos φ + I (XL . Xc ) Sin φ Where, XC = Capacitive reactance of the series capacitor After placing a capacitor in series with a feeder, the resultant impedance will be Z = R + j (XL − Xc ). Therefore, IXc is the reduction in voltage drop after series compensation. • If XL = XC then VD = I ∗ R i.e., the line has only resistive drop owing to the resistance of the line and zero inductive drop. • If Cosφ = 1 then Sinφ = 0 and therefore, I (XL .Xc )Sinφ = 0 and VD = IR . In such unity power factor applications, series capacitors have practically no value. • The power factor should be lagging if the voltage drop is to decrease considerably between the sending end and the receiving end by applying a series capacitor. • For long transmission lines where the net reactance is high, series capacitors are effective for enhancing system stability. b) Shunt capacitors: Shunt capacitors (capacitors connected in parallel with the lines) are widely used in distribution systems for compensating reactive power and to improve the power factor. Shunt capacitors draw a leading current which counteracts the lagging component of the inductive load current (some or the entire part) at the point of installation. Thus, it modifies the characteristics of inductive load by drawing leading current. A shunt capacitor has a similar effect as an over-excited synchronous generator or a motor. The application of a shunt capacitor to a feeder and its vector diagram representation is shown in Fig. 8.8(a) and Fig. 8.8(b) Voltage drop of the feeder without shunt capacitor is expressed as VD = IR R + Ix XL where, R = Resistance of the feeder XL = Inductive reactance of the feeder IR = Real power component of the current (in phase) Ix = Reactive component of current (out of phase, lagging the voltage by 90°)

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Power System Operation and Control X1

R

XL

R

I

I Vr

Vs

r

Vs

Vr IC

Vs

Vs IX

IX φ

Ic I

Vr I

φ

IR

Ic

Vr IR

I

Fig 8.8 (a) Without shunt capacitor (b) With shunt capacitor

Resultant voltage drop of the feeder with the application of shunt capacitor is VD = IR R + IX XL − IC XL Where, Ic = reactive component of the current leading the voltage by 90° (out of phase) The voltage rise due to the installation of shunt capacitor can be expressed from the difference between the above two expressions VR = IC XL Volts.

8.8 POWER FACTOR CORRECTION (LOAD COMPENSATION) Loads on electric power systems include two components,namely, real power and reactive power. Real power is exclusively generated at generating stations while reactive power can be generated either by generating plants or at the load site through capacitors. When the reactive power is supplied by power plants, the size of system components like generators, transformers, transmission and distribution lines and protective equipments will be much larger than existing size, and hence it is not economically feasible. The application of shunt capacitors can diminish these conditions by decreasing the reactive power demand on generators. Reduction in like current from capacitor locations leads to released generation, transmission and distribution substation capacities, reduction in losses, and loadings in lines and distribution transformers. Shunt capacitors are the most economical source to meet the reactive power requirements of inductive loads and transmission lines operating with a lagging power factor.

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155

8.8.1 Illustration of Power Factor Correction Assume that a load is supplied with a real power P , lagging reactive power Qi and apparent power Si with a lagging power factor of Cosφ1 = P /S1 1/2

Cosφ1 = P /(P 2 + Q1 2 )

When Qc KVAR is installed across the load and the reactive power of the load is compensated, the net reactive power will be Q2 = Q1 − QC and consequently the power factor is corrected (improved) from Cosφ1 to Cosφ2 . The new corrected power factor is expressed as, Cosφ2 = P /S2 Cosφ2 = P /(P 2 + Q2 2 )1/2 Cosφ2 = P /[P 2 + (Q1 − QC )2 ]1/2 Therefore, we can conclude that by connecting a capacitor across the load and supplying a reactive power of QC , the apparent power is decreased from S1 KVA to S2 KVA, and that the reactive power is decreased from Q1 KVAR to Q2 KVAR. Power factor correction is illustrated in Figs 8.9 and 8.10 P

P

Q 2 = Q 1− Q C

Q1

C

LOAD

Fig 8.9 Circuit representation of P.F correction P φ2 Q2 φ1 Q1

S2 QC

S1

Fig 8.10 Illustration of power factor correction phasor notation

156

Power System Operation and Control

QUESTION FROM PREVIOUS QUESTION PAPERS 1. Explain about the losses that occur due to VAR flow in power systems. 2. Discuss in detail about the generation and absorption of reactive power in power system components. 3. Explain the reason for variations of voltages in a power system and explain any one method to improve its voltage profile. 4. What is load compensation? Discuss its objective in a power system. 5. Write short notes on compensated and uncompensated transmission lines. 6. Explain briefly about shunt and series compensation of transmission lines 7. A three-phase induction motor delivers 500Hp at an efficiency of 0.91, the operating power factor being 0.76 lagging. A loaded synchronous motor with a power consumption of 100 KW is connected in parallel with the induction motor. Calculate the necessary KVA and the operating power factor of the synchronous motor if the overall power factor is to be unity. 8. Find the rating of a synchronous compensator connected to the teritary winding of a 132KV start connected, 33KV star connected, 11KV data connected three-winding transformer to supply a load of 66 MW at 0.8pf lagging at 33KV across the secondary. The equivalent primary and tertiary winding reactances are 32  and 0.16 respectively, while the secondary winding reactance is negligible. Assume that the primary side voltage is essentially constant at 132 KV and maximum of nominal setting between transformer primary and secondary is 1:1.1

Answers to Selected Competitive Examination Questions CHAPTER 1 1. [d];

2. [b];

3. [a];

4. [a];

5. [b];

6. [a]

3. [a];

4. [d];

5. [d];

6. [b]

3. [b];

4. [c];

5. [d]

CHAPTER 2 1. [c];

2. [a]

3. [c]

CHAPTER 5 1. [c];

2. [c];

CHAPTER 6 1. [a];

2. [a];

CHAPTER 7 1. [d]

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Index A accelerating power 84 active power 29, 75, 113, 143 air-gap line 86–87 angular momentum 84 apparent power 143, 148, 155 area control error 124, 130 automatic load frequency control 96, 103, 140 automatic voltage control 93 automatic voltage regulator 85, 92 B B-coefficients 30, 33, 35, 37–42, 48, 53, 56 for multi-machine power systems 34 C capacitor control 152 characteristics of thermal units 2 incremental fuel cost 3, 4, 6, 17, 24–26, 31, 40, 47, 49, 61–62, 66 input-output heat characteristic 2 compensation in a transmission line 150–151 control area 94–95, 103–104, 111–112, 115 coordinate equations 14, 30 cost curve 2, 3, 4, 69 current distribution factor 32, 35–36, 38 D damping ratio 85

E economic dispatch 1–2, 8, 12–14, 18–20, 29, 51–52, 56, 133, 136–137, 139 considering transmission line losses 29, 40 equality constraints 1, 8–9, 11, 13, 19, 23, 25, 61, 62, 71 including generator limits 19 inequality constraints 1, 8, 9, 11, 13, 19, 25, 61, 62, 71 neglecting transmission line losses 12 neglecting generator limits 13, 19 objective function 8–9, 12–13, 66–67, 72 the classical method 51 economic operation 1, 25–27, 60–61, 67, 133 economic scheduling 1, 2, 29, 31, 40, 42, 61–62, 65 electromagnetic torque 84 entrained steam 79, 81 excitation 75–76, 85–86, 92, 94, 113, 151 exciter system 85 block diagram 85–86 IEEE type-1 86, 92 saturation curve 86–87 F feedback gain 76, 85–86 feedback time constant 86 flat tie-line 93 free governor operation 96, 111 frequency constant 93, 112

160

Index

frequency control 75, 93–96, 99, 102–103, 108, 111–112, 115–117, 125–126, 130–131, 133, 136–137, 139–140 frequency deviations 115 frequency error 124, 134 G generator load model 79, 92, 95 block diagram 78–81 H heat rate 2, 3, 5–7, 25 hydraulic amplifier 76–77 hydro thermal scheduling 65–74 block diagram 70 energy contribution 68 long range 65–66 short term 70 short range 65–66 I incremental cost 15–16, 20–23, 25–27, 31, 41–42, 45, 48–49, 51, 53, 61–62, 66, 72–73 incremental fuel rate 2, 3, 24 incremental production cost 5–7, 15, 25, 45, 66 incremental transmission loss 30–31, 41 inertia constant 79, 88, 103–105, 108, 113, 125, 138 integral controller 112, 133–134, 136–137, 139, 140 block diagram 133 dynamic response 79, 81, 99, 100, 107, 108, 111, 112, 133, 134, 136 steady state response 96 integrator gains 125 inter connected stations 116 K kinetic energy 79, 88, 91, 94, 115 Kuhn-Tucker condition 30 Kron’s loss formula 30 L Lagrange multiplier method 9

Lagrangian multiplier 30–31, 40, 67 linkage mechanism 76, 95 load curve 1 load frequency control 93–94, 96, 99, 102–103, 108, 111, 112, 115, 117, 125–126, 130, 136–137, 139–140 loads, 15, 25–26, 31, 61, 80, 94, 96, 131, 141–143, 145–147, 154 constant power 1, 95 constant current 1 contant impedance 81 load-saturation 86 loss coefficients 33, 37, 45, 47, 62 M motors 80, 141–142, 145–146 N natural frequency 85 natural impedance 149 nonlinear function 9, 11 O optimal power flow 1, 60 overhead lines 145 P penalty factor 30–31, 40, 51, 62–63 plant load 1, 42–43, 49, 51, 53, 56 power factor 32, 37, 38, 81, 88, 141, 142–143, 145–149, 151–156 methods to improve 151 power factor correction 148, 154–155 power flow study 1 power system gain 80, 95 power system time constant 80, 95 R reactive power 1, 9, 75, 94, 142–146, 148–156 sources 45, 152 re-heater 79 reserve capacities 9 rotor over-heating 9 S series capacitors 152–153 shaft torque 84

Index

shunt capacitors 148, 152–154 single area frequency control 95 block diagram 95–96 dynamic response 79, 81, 99–100, 107–108, 111–112, 133–134, 136 state space model 100–102, 117 steady state response 96 speed changer 76–77, 91, 95–96, 98, 121, 124, 133–134, 137 speed governor 76, 78, 87, 93, 95, 97, 100, 104–106, 112 mathematical model 76–77, 92 spinning reserve 94, 115 states space analysis 85 steam turbine 79, 81 block diagram 78–81 transfer function model 79, 92 swing equation 84–85 synchronous generator control 151 synchronous machines 82, 151 equivalent circuit 82–83 phasor diagram 83, 141–142 T Taylor’s series 19, 52 thermal efficiency 3

161

three-phase transmission line 31 tie-line loading 93, 132 transformers 9, 141–142, 145–147, 154 transmission losses 2, 12, 29, 34, 55, 62 two-area system 115–116, 124 assumptions 116 block diagram 119 flat frequency control 115 state space model 100–102, 117, 119 steady state 76, 86, 96–98, 102, 105–108, 112, 116, 121, 124–125, 130, 133–134, 136, 151 tie-line bias control 117 two-stage reheater 81 U uncompensated transmission line 149 unit commitment 1, committed 1–2 decommitted 1 shut-down 2 start-up 2 W water discharge 66, 70–72