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Electrical Power Systems C L WADHWA NEW ACADEMIC SCIENCE New Academic Science www.EngineeringBooksPdf.com www.EngineeringBooksPdf.com  This page intentionally left blank www.EngineeringBooksPdf.com  Electrical Power Systems...

Electrical Power Systems C L WADHWA NEW ACADEMIC SCIENCE New Academic Science www.EngineeringBooksPdf.com www.EngineeringBooksPdf.com  This page intentionally left blank www.EngineeringBooksPdf.com  Electrical Power Systems C L WADHWA Former Professor & Head Electrical Engineering Department Delhi College of Engineering Delhi, India NEW ACADEMIC SCIENCE New Academic Science Limited The Control Centre, 11 A Little Mount Sion Tunbridge Wells, Kent TN1 1YS, UK www.newacademicscience.co.uk e-mail: [email protected] www.EngineeringBooksPdf.com Copyright © 2012 by New Academic Science Limited The Control Centre, 11 A Little Mount Sion, Tunbridge Wells, Kent TN1 1YS, UK www.newacademicscience.co.uk e-mail: [email protected] ISBN : 978 1 906574 39 0 All rights reserved. No part of this book may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the copyright owner. British Library Cataloguing in Publication Data A Catalogue record for this book is available from the British Library Every effort has been made to make the book error free. However, the author and publisher have no warranty of any kind, expressed or implied, with regard to the documentation contained in this book. www.EngineeringBooksPdf.com  To My Parents Wife and Children www.EngineeringBooksPdf.com  This page intentionally left blank www.EngineeringBooksPdf.com  Preface “Electrical Power System” has been written primarily for the undergraduate students in Electrical Engineering. The book covers conventional topics like the basics of power systems, line constant calculations, performance of lines, corona, mechanical design of overhead lines etc., and the more advanced topics like load flows studies, economic load dispatch, optimal power flows, state estimation in power systems etc. The book covers a very wide spectrum of electrical power system studies which is normally not available in one single book. The book is so comprehensibly written that at least five to six courses on power systems can be designed. It has been the constant endeavour of the author to understand the difficulties of his students in the classroom and accordingly prepare the lecture notes after consulting various journals and books on electrical power systems. The present book is an outcome of these notes and some research work the author carried out. Chapters 1 to 9 deal into the electrical and mechanical design of overhead and underground transmission networks. The analysis and performance of the system in terms of line constant calculations, efficiency and regulations, corona loss and interference of power lines with communication networks have been studied. A chapter on mechanical design of lines gives in a nutshell all the important aspects of erection of overhead lines. Chapter 5 on HVDC transmission discusses combined characteristics of rectifiers and inverters. Various controls like constant ignition angle, constant extinction angle, and constant currents have been discussed. Advantages and disadvantages of HVDC vs HVAC have been explained. Also, the role of HVDC link in improving system stability has been discussed. Chapter 10 is devoted to the study of voltage-reactive power problems on transmission lines. Chapter 11 defines an effectively grounded system, discusses and compares various systems of neutral grounding. Chapter 12 describes transients in power systems. Travelling waves on transmission lines, capacitance switching and lightning phenomenon have been discussed. Chapter 13 discusses calculation of symmetrical and asymmetrical fault conditions on the system, concept of infinite bus and short circuit capacity of a bus. Electric Power System is the most capital intensive and the most complex system ever developed by man. Not only that the system should be operated most effectively and efficiently, any abnormality in the operation of the system must be detected fast and reliable operation of the protective system must be ascertained. Protective relays is the subject of Chapter 14 of the book wherein various types of relays from conventional electromechanical relays to digital protective relays have been discussed. vii www.EngineeringBooksPdf.com  viii PREFACE Chapter 15 presents material on the conventional circuit breakers like air break C.B., oil C.B., airblast C.B. etc., and the more advanced ones like the vacuum C.B. and SF6 circuit breakers. With the higher and higher operating voltages the impulse insulation levels of the system are increasing. Chapter 16 is devoted to the insulation problems of the system and the solutions in terms of coordinating the insulation levels economically of various equipments on the system have been discussed. Chapter 17 deals into power system synchronous stability for a single machine connected to an infinite bus and multi-machine systems. Various techniques have been explained using algorithms and flow charts. With the advent of digital computers and modern methods of network solution, it has now been possible to analyse the present day large interconnected systems with greater accuracy and short computational effort. Various techniques of load flow solutions of large networks have been discussed and explained using flow charts in Chapter 18. Various techniques have been compared in terms of their complexities and computational efforts. With the advancement in technology of generation and load dispatching it has been possible to maintain the cost of electrical energy almost same even though the cost of fuel and other components have multiplied over the years. Chapter 19 on economic load dispatching discusses some of the classical techniques which even today are being used by the electric utilities. The techniques have been explained with the help of flow charts, algorithms and suitable examples. Chapter 20 deals into the load frequency control or automatic generation control problems of the system. The economics of a.c. power transmission has always forced the planning engineers to transmit as much power as possible through existing transmission lines. The need for higher index of reliability, the availability of hydro-power over long distances from the load centres, the difficulty of acquiring right-of-way for new transmission lines (the so-called corridor crisis) and the increased pressure to maximise the utilisation of both new and existing lines has helped to motivate the development and application of compensation system. Chapter 21 on compensation in power system discusses elaborately both the series and shunt compensation of overhead lines. The concept of FACTS (Flexible A.C. Transmission Systems) has also been introduced. The voltage stability also known as load stability is now a major concern in planning and operation of electric power system. Chapter 22 on power system voltage stability discusses various factors which lead to this problem and methods to improve the voltage stability of the system. State estimation is the process of determining a set of values by making use of the measurements made from the system and since the measurements may not be precise due to inherent errors associated with measuring devices, statistical methods have been discussed in Chapter 23, using the line power flows and maximum likelihood criterion have been discussed in detail with a number of solved problems. Techniques to detect and identify bad data during measurements have also been discussed. www.EngineeringBooksPdf.com  ix PREFACE Unit commitment is a way out to suggest just sufficient number of generating units with sufficient generating capacity to meet a given load demand economically with sufficient reserve capacity to meet any abnormal operating conditions. This aspect has been nicely dealt with suitable examples in Chapter 24. Chapter 25 deals into economic scheduling of hydro-thermal plants and optimal power flows including the multi-objective optimal power flows. Appendix A on formulation of bus impedance matrix is given which is very useful for the analysis of the system, especially for short circuit studies. Power transmission and synchronous machines as power systems elements have been discussed in Appendices B and C respectively. A suitable number of problems have been solved to help understand the relevant theory. At the end of each chapter unsolved problems with their answers have been suggested for further practice. At the end, a large number of multiple choice questions have been added to help the reader to test himself. An extensive bibliography will help the reader to locate detailed information on various topics of his interest. Any constructive suggestions for the improvement of the book will be gratefully acknowledged. Last but not the least, I wish to express my gratitude to my wife Usha, daughter Meenu and son Sandeep for their patience and encouragement during the preparation of the book. C.L. WADHWA www.EngineeringBooksPdf.com  This page intentionally left blank www.EngineeringBooksPdf.com  Contents Preface vii 1 FUNDAMENTALS OF POWER SYSTEMS 1–13 1.1 Single-phase Transmission 2 1.2 The 3-phase Transmission 6 1.3 Complex Power 7 1.4 Load Characteristics 7 1.5 The Per Unit System 9 2 LINE CONSTANT CALCULATIONS 15–38 2.1 Magnetic Flux Density 16 2.2 Inductors and Inductance 18 2.3 Magnetic Field Intensity due to a Long Current Carrying Conductor 18 2.4 Inductance of Two-Wire Transmission Line 19 2.5 Flux Linkages of One Conductor in a Group of Conductors 22 2.6 Inductance of 3-φ Unsymmetrically Spaced Transmission Line 24 2.7 Transposition of Power Lines 25 2.8 Composite Conductors 26 2.9 Inductance of Composite Conductors 27 2.10 Inductance of Double Circuit 3-φ Line 31 2.11 Concept of Geometric Mean Distance 34 2.12 Bundled Conductors 35 2.13 Skin and Proximity Effect 36 3 CAPACITANCE OF TRANSMISSION LINES 39–57 3.1 Electric Field of an Infinite Line of Charge 40 3.2 Potential Difference between Two Points due to a Line Charge 42 3.3 Two Infinite Lines of Charge 42 3.4 Capacitance of a 1-φ Transmission Line 44 3.5 Capacitance of a 3-phase, Unsymmetrically Spaced Transmission Line 47 3.6 Capacitance of a Double Circuit Line 49 3.7 Effect of Earth on the Capacitance of Conductors 53 4 PERFORMANCE OF LINES 59–97 4.1 Representation of Lines 60 4.2 Short Transmission Lines 62 4.3 Medium Length Lines 68 4.4 Long Transmission Lines 75 4.5 ABCD Constants 86 4.6 Ferranti-effect 94 xi www.EngineeringBooksPdf.com  xii CONTENTS 5 HIGH VOLTAGE D.C. TRANSMISSION 99–136 5.1 Rectification 101 5.2 The 3-phase Bridge Rectifier or Graetz Circuit 105 5.3 Inversion 109 5.4 Kinds of d.c. Links 113 5.5 Parallel and Series Connection of Thyristors 114 5.6 Power Flow in HVDC Transmission System 114 5.7 Constant Ignition Angle β Control 117 5.8 Constant Extinction Angle δ Control 118 5.9 Constant Current Control 119 5.10 Actual Control Characteristics 120 5.11 Frequency Control 124 5.12 Reactive VAr Requirements of HVDC Converters 125 5.13 Parallel Operation of d.c. Link with an a.c. Network 128 5.14 Ground Return 129 5.15 Circuit Breaking 130 5.16 Advantages of d.c. Transmission 131 5.17 Disadvantages 134 5.18 Cables 134 5.19 Economic Distances for d.c. Transmission 134 6 CORONA 137–151 6.1 Critical Disruptive Voltage 139 6.2 Corona Loss 143 6.3 Line Design Based on Corona 147 6.4 Disadvantages of Corona 148 6.5 Radio Interference 148 6.6 Inductive Interference between Power and Communication Lines 148 7 MECHANICAL DESIGN OF TRANSMISSION LINES 153–171 7.1 The Catenary Curve 154 7.2 Sag Tension Calculations 156 7.3 Supports at Different Levels 163 7.4 Stringing Chart 166 7.5 Sag Template 166 7.6 Equivalent Span 167 7.7 Stringing of Conductors 168 7.8 Vibration and Vibration Dampers 168 8 OVERHEAD LINE INSULATORS 173–187 8.1 Types of Insulators 175 8.2 Potential Distribution Over a String of Suspension Insulators 177 8.3 Methods of Equalising the Potential 181 9 INSULATED CABLES 189–223 9.1 The Insulation 190 9.2 Extra High Voltage Cables 193 9.3 Grading of Cables 196 www.EngineeringBooksPdf.com  CONTENTS xiii 9.4 Insulation Resistance of a Cable 203 9.5 Capacitance of a Single Core Cable 204 9.6 Heating of Cables 207 9.7 Current Rating of a Cable 214 9.8 Overhead Lines Versus Underground Cables 218 9.9 Types of Cables 218 10 VOLTAGE CONTROL 225–246 10.1 Methods of Voltage Control 228 10.2 Determination of Synchronous Phase Modifier Capacity 237 10.3 Sending End Power Circle Diagram 243 11 NEUTRAL GROUNDING 247–256 11.1 Effectively Grounded System 248 11.2 Ungrounded System 249 11.3 Resonant Grounding 249 11.4 Methods of Neutral Grounding 252 11.5 Generator Neutral Breaker 255 11.6 Grounding Practice 256 12 TRANSIENTS IN POWER SYSTEMS 257–295 12.1 Transients in Simple Circuits 259 12.2 3-phase Sudden Short Circuit of an Alternator 265 12.3 The Restriking Voltage after Removal of Short Circuit 267 12.4 Travelling Waves on Transmission Lines 269 12.5 Attenuation of Travelling Waves 284 12.6 Capacitance Switching 286 12.7 Overvoltage due to Arcing Ground 288 12.8 Lightning Phenomenon 289 12.9 Line Design Based on Lightning 293 13 SYMMETRICAL COMPONENTS AND FAULT CALCULATIONS 297–356 13.1 3-phase Systems 298 13.2 Significance of Positive, Negative and Zero Sequence Components 299 13.3 Average 3-phase Power in Terms of Symmetrical Components 303 13.4 Sequence Impedances 305 13.5 Fault Calculations 308 13.6 Sequence Network Equations 310 13.7 Single Line-to-Ground Fault 312 13.8 Line-to-Ground Fault with Zf 325 13.9 Sequence Networks 329 13.10 Faults on Power Systems 332 13.11 Phase Shift ∆-Y Transformers 333 13.12 Reactors 338 13.13 Concept of Short-circuit Capacity of a Bus 340 14 PROTECTIVE RELAYS 357–476 14.1 Some Definitions 359 14.2 Functional Characteristics of a Protective Relay 360 www.EngineeringBooksPdf.com  xiv CONTENTS 14.3 Operating Principles of Relays 361 14.4 Torque Production in an Induction Relay 362 14.5 Over-current Relays 364 14.6 Directional Overcurrent Relays 366 14.7 The Universal Relay Torque Equation 370 14.8 Differential Relays 377 14.9 Feeder Protection 381 14.10 Distance Protection 387 14.11 Generator Protection 392 14.12 Protection of Transformers 401 14.13 Translay Relay 408 14.14 Carrier Current Protection 410 14.15 Comparators 413 14.16 Static Relays 428 14.17 Digital Protection 453 14.18 Fuses and HRC Fuses 461 14.19 Linear Couplers 465 14.19.1 Current Transformers 465 14.19.2 Potential Transformers 470 15 CIRCUIT BREAKERS 477–512 15.1 Arc in Oil 479 15.2 Arc-interruption Theories 481 15.3 Current Chopping 485 15.4 Oil Circuit Breaker 488 15.5 Air Circuit Breakers 492 15.6 Air Blast Circuit Breakers 494 15.7 Vacuum Circuit Breakers 496 15.8 Sulphur Hexafluoride (SF6) Circuit Breakers 500 15.9 Rating of Circuit Breakers 502 15.10 Testing of Circuit Breakers 505 15.11 Autoreclosing 510 16 INSULATION COORDINATION AND OVERVOLTAGE PROTECTION 513–532 16.1 Volt-time Curve 514 16.2 Overvoltage Protection 518 16.3 Ground Wires 526 16.4 Surge Protection of Rotating Machine 531 17 POWER SYSTEM SYNCHRONOUS STABILITY 533–589 17.1 The Power Flow 536 17.2 The Swing Equation 538 17.3 Steady State Stability 543 17.4 Equal Area Criterion 546 17.5 Critical Clearing Angle 550 17.6 Two Finite Machines 554 17.7 Point-by-point Method 556 17.8 Factors Affecting Transient Stability 562 www.EngineeringBooksPdf.com  CONTENTS xv 17.9 The Role of Automatic Voltage Regulator (AVr) in Improving Stability 563 17.10 The Excitation System 566 17.11 Effect of Grounding on Stability 568 17.12 Prevention of Steady Pull Out 569 17.13 Multi-Machine Stability—Classical Model 573 17.14 Limitations of the Classical Model 587 18 LOAD FLOWS 591–625 18.1 Bus Classification 593 18.2 Nodal Admittance Matrix 594 18.3 Development of Load Flow Equations 598 18.4 Iterative Methods 599 18.5 Newton-Raphson Method 608 18.6 Comparison of Solution Methods 618 18.7 Approximation to Newton-Raphson Method 619 18.8 Line Flow Equations 619 18.9 Fast-decoupled Load Flow 620 19 ECONOMIC LOAD DISPATCH 627–661 19.1 System Constraints 629 19.2 Economic Dispatch Neglecting Losses 632 19.3 Optimum Load Dispatch Including Transmission Losses 635 19.4 Exact Transmission Loss Formula 649 19.5 Modified Coordination Equations 651 19.6 Automatic Load Dispatching 654 19.7 Power Line Carrier Communication (PLCC) 656 20 LOAD FREQUENCY CONTROL 663–678 20.1 Load Frequency Problem 664 20.2 Speed Governing System 667 20.3 Reasons for Limits on Frequency 672 21 COMPENSATION IN POWER SYSTEM 679–717 21.1 Load Compensation 680 21.2 Loadability Characteristic of O/H Lines 685 21.3 Uncompensated Transmission Line 687 21.4 Symmetrical Line 690 21.5 Radial Line with Asynchronous Load 692 21.6 Compensation of Lines 694 21.7 Subsynchronous Resonance 700 21.8 Active Shunt Compensator 702 21.9 Static Compensators 705 21.10 Flexible A.C. Transmission System (FACTS) 713 22 POWER SYSTEM VOLTAGE STABILITY 719–762 22.1 Reactive Power Flow 720 22.2 Difficulties with Reactive Power Transmission 724 22.3 Voltage Stability: Definition and Concept 729 22.4 Power System Loads 734 www.EngineeringBooksPdf.com  xvi CONTENTS 22.5 Generation Characteristics 743 22.6 HVDC Operation 747 22.7 Voltage Stability Analysis: P-V Curves 750 22.8 Methods of Improving Voltage Stability 756 23 STATE ESTIMATION IN POWER SYSTEMS 763–796 23.1 Introduction 764 23.2 State Estimation for Line Power Flow 765 23.3 Maximum Likelihood Criterion 776 23.4 Detection and Identification of Bad Data 786 23.5 State Estimator Linear Model 789 23.6 The Role of State Estimation in Power System Operations 794 24 UNIT COMMITMENT 797–811 24.1 Introduction 798 24.2 Spinning Reserve 800 24.3 Thermal Unit Constraints 801 24.4 Unit Commitment Solution Methods 802 25 ECONOMIC SCHEDULING OF HYDROTHERMAL PLANTS AND OPTIMAL POWER FLOWS 813–831 25.1 Introduction 814 25.2 Problem Formulation 815 25.3 Optimal Power Flow 820 25.4 Problem Formulation 822 25.5 Multi-Objective Optimal Power Flow 826 25.6 Problem Formulation 827 APPENDICES 833–862 Appendix A: Algorithm for Formation of Bus Impedance Matrix 834 Appendix B: The Power Transformer 845 Appendix C: Synchronous Machine 850 OBJECTIVE QUESTIONS 863 ANSWERS TO OBJECTIVE QUESTIONS 944 ANSWERS TO PROBLEMS 948 INDEX 953 www.EngineeringBooksPdf.com  1 FUNDAMENTALS OF POWER SYSTEMS www.EngineeringBooksPdf.com  1 Fundamentals of Power Systems INTRODUCTION The three basic elements of electrical engineering are resistor, inductor and capacitor. The resistor consumes ohmic or dissipative energy whereas the inductor and capacitor store in the positive half cycle and give away in the negative half cycle of supply the magnetic field and electric field energies respectively. The ohmic form of energy is dissipated into heat whenever a current flows in a resistive medium. If I is the current flowing for a period of t seconds through a resistance of R ohms, the heat dissipated will be I 2Rt watt sec. In case of an inductor the energy is stored in the form of magnetic field. For a coil of L henries and a current of I 1 amperes flowing, the energy stored is given by 2 LI 2. The energy is stored between the metallic 1 plates of the capacitor in the form of electric field and is given by 2 CV 2, where C is the capacitance and V is the voltage across the plates. We shall start with power transmission using 1-φ circuits and assume in all our analysis that the source is a perfect sinusoidal with fundamental frequency component only. 1.1 SINGLE-PHASE TRANSMISSION Let us consider an inductive circuit and let the instantaneous voltage be v = Vm sin ωt (1.1) Then the current will be i = Im sin (ωt – φ), where φ is the angle by which the current lags the voltage (Fig. 1.1). The instantaneous power is given by p = vi = Vm sin ωt. Im sin (ωt – φ) = VmIm sin ωt sin (ωt – φ) (1.2) Vm I m = [cos φ – cos (2ωt – φ)] 2 2 www.EngineeringBooksPdf.com  FUNDAMENTALS OF POWER SYSTEMS 3 The value of p is positive when both v and i are either positive or negative and represents the rate at which the energy is being consumed by the load. In this case the current flows in the direction of voltage drop. On the other hand power is negative when the current flows in the direction of voltage rise which means that the energy is being transferred from the load into the network to which it is connected. If the circuit is purely reactive the voltage and current will be 90° out of phase and hence the power will have equal positive and negative half cycles and the average value will be zero. From equation (1.2) the power pulsates around the average power at double the supply frequency. p v i IVIIII cos f f Fig. 1.1 Voltage, current and power in single phase circuit. Equation (1.2) can be rewritten as p = VI cos φ (1 – cos 2ωt) – VI sin φ sin 2ωt (1.3) I II We have decomposed the instantaneous power into two components (Fig. 1.2). p I II p = VI cos f VI sin f Fig. 1.2 Active, reactive and total power in a single phase circuit. (i) The component P marked I pulsates around the same average power VI cos φ but never goes negative as the factor (1 – cos 2ωt) can at the most become zero but it will never go negative. We define this average power as the real power P which physically means the useful power being transmitted. www.EngineeringBooksPdf.com  4 ELECTRICAL POWER SYSTEMS (ii) The component marked II contains the term sin φ which is negative for capacitive circuit and is positive for inductive circuit. This component pulsates and has zero as its aver- age value. This component is known as reactive power as it travels back and forth on the line without doing any useful work. Equation (1.3) is rewritten as p = P(1 – cos 2ωt) – Q sin 2ωt (1.4) Both P and Q have the same dimensions of watts but to emphasise the fact that Q represents a nonactive power, it is measured in terms of voltamperes reactive i.e., V Ar. The term Q requires more attention because of the interesting property of sin φ which is – ve for capacitive circuits and is +ve for inductive circuits. This means a capacitor is a generator of positive reactive V Ar, a concept which is usually adopted by power system engineers. So it is better to consider a capacitor supplying a lagging current rather than taking a leading current (Fig. 1.3). + + V C V C – – I leads V by 90° I lags V by 90° Fig. 1.3 V-I relations in a capacitor. Consider a circuit in which an inductive load is shunted by a capacitor. If Q is the total reactive power requirement of the load and Q′ is the reactive power that the capacitor can generate, the net reactive power to be transmitted over the line will be (Q – Q′). This is the basic concept of synchronous phase modifiers for controlling the voltage of the system. The phase modifier controls the flow of reactive power by suitable excitation and hence the voltage is controlled. The phase modifier is basically a synchronous machine working as a capacitor when overexcited and as an inductor when underexcited. It is interesting to consider the case when a capacitor and an inductor of the same reactive power requirement are connected in parallel (Fig. 1.4). IC IC IL V V IL Fig. 1.4 Power flow in L-C circuit. www.EngineeringBooksPdf.com  FUNDAMENTALS OF POWER SYSTEMS 5 The currents IL and IC are equal in magnitude and, therefore, the power requirement is same. The line power will, therefore, be zero. Physically this means that the energy travels back and forth between the capacitor and the inductor. In C I R one half cycle at a particular moment the capacitor is fully charged and the coil has no energy stored. Half a voltage cycle later the coil stores maximum energy and the Vm sin wt capacitor is fully discharged. The following example illustrates the relationship Fig. 1.5 Relationship between electric field energy and reactive power. between the reactive power and the electric field energy stored by the capacitor. Consider an RC circuit (Fig. 1.5). From Fig. 1.5 V Vω C I= = (1.5) 2 2 R + (1/ωC) R ω 2C 2 + 1 2 and if voltage is taken as reference i.e., v = Vm sin ωt, the current i = Im sin (ωt + φ) Vm ωC ∴ i=. sin (ωt + φ) (1.6) R ω 2C 2 + 1 2 I /ωC 1 where sin φ = = (1.7) 2 2 2 I R + ( I /ωC) R ω C2 + 1 2 2 Now reactive power Q = VI sin φ (1.8) Substituting for I and sin φ, we have VωC 1 V 2 ωC Q=V.. = (1.9) R 2ω 2C 2 + 1 R2ω 2C 2 + 1 R2ω 2C 2 + 1 V 2 ωC ∴ Reactive power = R 2ω 2 C 2 + 1 Now this can be related with the electric energy stored by the capacitor. The energy stored by the capacitor 1 W= 2 Cv2 (1.10) Now v= 1 C z i dt = 1 C VmωC 2 2 2 R ω C +1 cos (ωt + φ) Vm cos (ωt + φ) ω =. R2ω 2C 2 + 1 (1.11) 1C Vm2 cos2 (ωt + φ) V 2 cos2 (ωt + φ) ∴ W= 2. = (1.12) R2ω 2C 2 + 1 R2ω 2C 2 + 1 dW V2 = 2 2 2. 2 cos (ωt + φ). sin (ωt + φ). ωC dt R ω C +1 V 2 ωC =. sin 2(ωt + φ) R ω 2C 2 + 1 2 = Q sin 2(ωt + φ) (1.13) www.EngineeringBooksPdf.com  6 ELECTRICAL POWER SYSTEMS From this it is clear that the rate of change of electric field energy is a harmonically varying quantity with a frequency double the supply frequency and has a peak value equal to Q. In an R-L circuit the magnetic field energy and reactive power in a coil are similarly related. 1.2 THE 3-PHASE TRANSMISSION Assuming that the system is balanced which means that the 3-phase voltages and currents are balanced. These quantities can be expressed mathematically as follows: Va = Vm sin ωt Vb = Vm sin (ωt – 120°) Vc = Vm sin (ωt + 120°) (1.14) ia = Im sin (ωt – φ) ib = Im sin (ωt – φ – 120°) ic = Im sin (ωt – φ + 120°) The total power transmitted equals the sum of the individual powers in each phase. p = Vaia + Vbib + Vcic = Vm sin ωtIm sin (ωt – φ) + Vm sin (ωt – 120°) Im sin (ωt – 120° – φ) + Vm sin (ωt + 120°)Im sin (ωt + 120° – φ) = VI[2 sin ωt sin (ωt – φ) + 2 sin (ωt – 120°) sin (ωt – 120° – φ) + 2 sin (ωt + 120°) sin (ωt + 120° – φ)] = VI[cos φ – cos (2ωt – φ) + cos φ – cos (2ωt – 240° – φ) + cos φ – cos (2ωt + 240° – φ)] = 3VI cos φ (1.15) This shows that the total instantaneous 3-phase power is constant and is equal to three times the real power per phase i.e., p = 3P, where P is the power per phase. In case of single phase transmission we noted that the instantaneous power expression contained both the real and reactive power expression but here in case of 3-phase we find that the instantaneous power is constant. This does not mean that the reactive power is of no importance in a 3-phase system. For a 3-phase system the sum of three currents at any instant is zero, this does not mean that the current in each phase is zero. Similarly, even though the sum of reactive power instantaneously in 3-phase system is zero but in each phase it does exist and is equal to VI sin φ and, therefore, for 3-φ the reactive power is equal to Q3φ = 3VI sin φ = 3Q, where Q is the reactive power in each phase. It is to be noted here that the term Q3φ makes as little physical sense as would the concept of 3-phase currents I3φ = 3I. Nevertheless the reactive power in a 3-phase system is expressed as Q3φ. This is done to maintain symmetry between the active and reactive powers. www.EngineeringBooksPdf.com  FUNDAMENTALS OF POWER SYSTEMS 7 1.3 COMPLEX POWER Consider a single phase network and let V = |V|e jα and I = |I|e jβ (1.16) where α and β are the angles that V and I subtend with respect to some reference axis. We calculate the real and reactive power by finding the product of V with the conjugate of I i.e., S = VI* = |V|e jα |I|e–jβ = |V| |I|e j(α – β) = |V| |I| cos (α – β) + j|V| |I| sin (α – β) (1.17) Here the angle (α – β) is the phase difference between the phasor V and I and is normally denoted by φ. ∴ S = |V| |I| cos φ + j|V| |I| sin φ = P + jQ (1.18) The quantity S is called the complex power. The magnitude of S = P 2 + Q 2 is termed as the apparent power and its units are volt-amperes and the larger units are kVA or MVA. The practical significance of apparent power is as a rating unit of generators and transformers, as the apparent power rating is a direct indication of heating of machine which determines the rating of the machines. It is to be noted that Q is positive when (α – β) is positive i.e., when V leads I i.e., the load is inductive and Q is –ve when V lags I i.e., the load is capacitive. This agrees with the normal convention adopted in power system i.e., taking Q due to an inductive load as +ve and Q due to a capacitive load as negative. Therefore, to obtain proper sign for reactive power it is necessary to find out VI* rather than V*I which would reverse the sign for Q as V*I = |V|e–jα |I|e jβ = |V| |I|e–j(α – β) = |V| |I| cos (α – β) – j|V| |I| sin (α – β) = |V| |I| cos φ – j|V| |I| sin φ = P – jQ (1.19) 1.4 LOAD CHARACTERISTICS In an electric power system it is difficult to predict the load variation accurately. The load devices may vary from a few watt night lamps to multi-megawatt induction motors. The following category of loads are present in a system: (i) Motor devices 70% (ii) Heating and lighting equipment 25% (iii) Electronic devices 5% The heating load maintains constant resistance with voltage change and hence the power varies with (voltage)2 whereas lighting load is independent of frequency and power consumed varies as V 1.6 rather than V 2. www.EngineeringBooksPdf.com  8 ELECTRICAL POWER SYSTEMS For an impedance load i.e., lumped load V2 P=.R R + (2πfL) 2 2 V2 and Q=. (2πfL) (1.20) R 2 + (2πfL) 2 From this it is clear that both P and Q increase as the square of voltage magnitude. Also with increasing frequency the active power P decreases whereas Q increases. The above equations are of the form P = P [f, |V|] (1.21) Q = Q [f, |V|] Composite loads which form a major part of the system load are also function of voltage and frequency and can, in general, be written as in equation (1.21). For this type of load, however, no direct relationship is available as for impedance loads. For a particular composite load an empirical relation between the load, and voltage and frequency can be obtained. Normally we are concerned with incremental changes in P and Q as a function of incremental changes in | V | and f. From equation (1.21). ∂P ∂P ∆P ~ −.| ∆V |+. ∆f ∂|V | ∂f ∂Q ∂Q and ∆Q ~ −.|∆V |+. ∆f (1.22) ∂|V | ∂f The four partial derivatives can be obtained empirically. However, it is to be remembered that whereas an impedance load P decreases with increasing frequency, a composite load will increase. This is because a composite load mostly consists of induction motors which always will experience increased load, as frequency or speed increases. The need for ensuring a high degree of service reliability in the operation of modern electric systems can hardly be over-emphasized. The supply should not only be reliable but should be of good quality i.e., the voltage and frequency should vary within certain limits, otherwise operation of the system at subnormal frequency and lower voltage will result in serious problems, especially in case of fractional horse-power motors. In case of refrigerators reduced frequency results into reduced efficiency and high consumption as the motor draws larger current at reduced power factor. The system operation at subnormal frequency and voltage leads to the loss of revenue to the suppliers due to accompanying reduction in load demand. The most serious effect of subnormal frequency and voltage is on the operation of the thermal power station auxiliaries. The output of the auxiliaries goes down as a result of which the generation is also decreased. This may result in complete shut-down of the plant if corrective measures like load shedding is not resorted to. Load shedding is done with the help of under- frequency relays which automatically disconnect blocks of loads or sectionalise the transmission system depending upon the system requirements. www.EngineeringBooksPdf.com  FUNDAMENTALS OF POWER SYSTEMS 9 1.5 THE PER UNIT SYSTEM In a large interconnected power system with various voltage levels and various capacity equipments it has been found quite convenient to work with per unit (p.u.) system of quantities for analysis purposes rather than in absolute values of quantities. Sometimes per cent values are used instead of p.u. but it is always convenient to use p.u. values. The p.u. value of any quantity is defined as The actual value of the quantity (in any unit) The base or reference value in the same unit In electrical engineering the three basic quantities are voltage, current and impedance. If we choose any two of them as the base or reference quantity, the third one automatically will have a base or reference value depending upon the other two e.g., if V and I are the base voltage and current in a system, the base impedance of the system is fixed and is given by V Z= I The ratings of the equipments in a power system are given in terms of operating voltage and the capacity in kVA. Therefore, it is found convenient and useful to select voltage and kVA as the base quantities. Let Vb be the base voltage and kVAb be the base kilovoltamperes, then Vactual Vp.u. = Vb kVAb × 1000 The base current = Vb Actual current Actual current ∴ p.u. current = = × Vb Base current kVAb × 1000 Base voltage Base impedance = Base current Vb 2 = kVAb × 1000 Actual impedance ∴ p.u. impedance = Base impedance Z. kVAb × 1000 Z. MVAb = 2 = Vb (kVb ) 2 This means that the p.u. impedance is directly proportional to the base kVA and inversely proportional to square of base voltage. Normally the p.u. impedance of various equipments corresponding to its own rating voltage and kVA are given and since we choose one common base kVA and voltage for the whole system, therefore, it is desired to find out the p.u. impedance of the various equipments corresponding to the common base voltage and kVA. If the individual quantities are Zp.u. old, kVAold and Vold and the common base quantities are Zp.u. new, kVAnew and Vnew, then making use of the relation above, Zp.u. new = Zp.u. old. kVAnew FG Vold IJ 2 (1.23) kVAold. VnewH K www.EngineeringBooksPdf.com  10 ELECTRICAL POWER SYSTEMS This is a very important relation used in power system analysis. The p.u. impedance of an equipment corresponding to its own rating is given by IZ Zp.u. = V where Z is the absolute value of the impedance of the equipment. It is seen that the p.u. representation of the impedance of an equipment is more meaningful than its absolute value e.g., saying that the impedance of a machine is 10 ohms does not give any idea regarding the size of the machine. For a large size machine 10 ohms appears to be quite large, whereas for small machines 10 ohms is very small. Whereas for equipments of the same general type the p.u. volt drops and losses are in the same order regardless of size. With p.u. system there is less chance of making mistake in phase and line voltages, single phase or three phase quantities. Also the p.u. impedance of the transformer is same whether referred on to primary or secondary side of the transformer which is not the case when considering absolute value of these impedances. This is illustrated below: Let the impedance of the transformer referred to primary side be Zp and that on the secondary side be Zs, then FG V IJ p 2 Zp = Zs HV K s where Vp and Vs are the primary and secondary voltages of the transformer. ZpI p FG V IJ p 2 Ip Now Zp p.u. = Vp = Zs HV K s. Vp Vp I p Vs I s Zs Is = Zs. = Zs. = Vs 2 Vs 2 Vs = Zs p.u. From this it is clear that the p.u. impedance of the transformer referred to primary side Zp p.u. is equal to the p.u. impedance of the transformer referred to the secondary side Zs p.u.. This is a great advantage of p.u. system of calculation. The base values in a system are selected in such a way that the p.u. voltages and currents in system are approximately unity. Sometimes the base kVA is chosen equal to the sum of the ratings of the various equipments on the system or equal to the capacity of the largest unit. The different voltage levels in a power system are due to the presence of transformers. Therefore, the procedure for selecting base voltage is as follows: A voltage corresponding to any part of the system could be taken as a base and the base voltages in other parts of the circuit, separated from the original part by transformers is related through the turns ratio of the transformers. This is very important. Say, if the base voltage on primary side is Vpb then on the secondary side of the transformer the base voltage will be Vsb = Vpb(Ns/Np), where Ns and Np are the turns of the transformer on secondary and primary side respectively. www.EngineeringBooksPdf.com  FUNDAMENTALS OF POWER SYSTEMS 11 The following example illustrates the procedure for selecting the base quantities in various parts of the system and their effect on the p.u. values of the impedances of the various equipments. Example 1.1: A 100 MVA, 33 kV 3-phase generator has a subtransient reactance of 15%. The generator is connected to the motors through a transmission line and transformers as shown in Fig. E1.1a. The motors have rated inputs of 30 MVA, 20 MVA and 50 MVA at 30 kV with 20% subtransient reactance. The 3-phase transformers are rated at 110 MVA, 32 kV, ∆/110 kV Y with leakage reactance 8%. The line has a reactance of 50 ohms. Selecting the generator rating as the base quantities in the generator circuit, determine the base quantities in other parts of the system and evaluate the corresponding p.u. values. j 50 W 100 MVA, 33 kV 15% Fig. E1.1= Solution: Assuming base values as 100 MVA and 33 kV in the generator circuit, the p.u. reactance of generator will be 15%. The base value of voltage in the line will be 110 33 × = 113.43 kV 32 In the motor circuit, 32 113.43 × = 33 kV 110 The reactance of the transformer given is 8% corresponding to 110 MVA, 32 kV. Therefore, corresponding to 100 MVA and 33 kV the p.u. reactance will be (using Eq. 1.23). 100 FG IJ 32 2 0.08 × 110 × H K 33 = 0.06838 p.u. 50 × 100 The p.u. impedance of line = = 0.3886 p.u. (113.43) 2 FG IJ 100 30 2 The p.u. reactance of motor 1 = 0.2 × H K 30 × 33 = 0.5509 p.u. 100 F 30 I 2 ×G J 20 H 33 K motor 2 = 0.2 × = 0.826 p.u. 100 F 30 I 2 ×G J 50 H 33 K motor 3 = 0.2 × = 0.3305 p.u. www.EngineeringBooksPdf.com 12 ELECTRICAL POWER SYSTEMS The reactance diagram for the system is shown in Fig. E1.1b. j 0.06838 W j 0.3886 W j 0.06838 W j 0.826 W j 0.15 W j 0.5509 W j 0.3305 W 1 2 3 Fig. E1.1> Reactance diagram for Example 1.1. PROBLEMS 1.1. Two generators rated at 10 MVA, 13.2 kV and 15 MVA, 13.2 kV are connected in parallel to a busbar. They feed supply to two motors of inputs 8 MVA and 12 MVA respectively. The operating voltage of motors is 12.5 kV. Assuming base quantities as 50 MVA and 13.8 kV draw the reac- tance diagram. The per cent reactance for generators is 15% and that for motors is 20%. 1.2. Three generators are rated as follows: Generator 1–100 MVA, 33 kV, reactance 10%; Generator 2–150 MVA, 32 kV, reactance 8%; Generator 3–110 MVA, 30 kV, reactance 12%. Determine the reactance of the generator corresponding to base values of 200 MVA, 35 kV. 1.3. A 3-bus system is given in Fig. P1.3. The ratings of the various components are listed below: Generator 1 = 50 MVA, 13.8 kV, X″ = 0.15 p.u. Generator 2 = 40 MVA, 13.2 kV, X″ = 0.20 Generator 3 = 30 MVA, 11 kV, X″ = 0.25 Transformer 1 = 45 MVA, 11 kV, ∆/110 kV Y, X = 0.1 p.u. Transformer 2 = 25 MVA, 12.5 kV, ∆/115 kV Y, X = 0.15 p.u. Transformer 3 = 40 MVA, 12.5 kV, ∆/115 kV Y, X = 0.1 p.u. The line impedances are shown in Fig. P1.3. Determine the reactance diagram based on 50 MVA and 13.8 kV as base quantities in Generator 1. j 50 W G1 G2 j 25 W j 25 W G3 Fig. P1.3 www.EngineeringBooksPdf.com FUNDAMENTALS OF POWER SYSTEMS 13 1.4. Explain clearly the concept of reactive power in single phase and three phase circuits. 1.5. Explain clearly how the magnetic field energy and the reactive power in an inductive circuit are related. 1.6. Explain clearly what you mean by good quality supply and discuss the effect of bad supply on the performance of the system. 1.7. Explain the p.u. system of analysing power system problems. Discuss the advantages of this method over the absolute method of analysis. REFERENCES 1. Electric Energy System Theory—An Introduction, O.I. Elgord, McGraw-Hill, 1971. 2. Elements of Power System Analysis, W.D. Stevenson Jr., McGraw-Hill, 1962. 3. Electric Power Systems, B.M. Weedy, John Wiley & Sons, 1974. www.EngineeringBooksPdf.com  This page intentionally left blank www.EngineeringBooksPdf.com  2 LINE CONSTANT CALCULATIONS www.EngineeringBooksPdf.com  2 Line Constant Calculations INTRODUCTION An electric transmission line can be represented by a series combination of resistance, inductance and shunt combination of conductance and capacitance. These parameters are symbolized as R, L, G and C respectively. Of these R and G are least important in the sense that they do not affect much the total equivalent impedance of the line and hence the transmission capacity. They are of course very much important when transmission efficiency and economy are to be evaluated as they completely determine the real transmission line losses. The resistance of a conductor is given by Power loss in conductor R= ohms (2.1) I2 where R is the effective resistance of the conductor and I the current flowing through the conductor. The effective resistance is equal to the d.c. resistance of the conductor only if the current is uniformly distributed throughout the section of the conductor. The difference in the d.c. resistance and effective resistance to frequencies less than 50 Hz is less than 1% for copper conductors of section less than 350,000 circular mils. The loss on the overhead lines is due to (i) ohmic loss in the power conductors, (ii) corona loss and (iii) leakage at the insulators which support the lines at the towers. This leakage loss is different from the leakage in cables as in cables the leakage is uniformly distributed along its length, whereas the leakage on overhead lines is limited only to the insulators. This could be represented as conductance uniformly distributed along the line. Since the corona loss and the leakage over the insulators is negligibly small under normal operating conditions, the conductance between the conductors of an overhead line is assumed to be zero. 2.1 MAGNETIC FLUX DENSITY A current carrying conductor produces a magnetic field which is in the form of closed circular loops around the conductor. The relation of the magnetic field direction to the current direction 16 www.EngineeringBooksPdf.com  LINE CONSTANT CALCULATIONS 17 can be easily remembered by means of the right hand rule. With the thumb pointing in the direction of the current, the fingers of the right hand encircling the wire point in the direction of the magnetic field. According to Biot-Savart’s law, the magnetic flux density at any point P as produced by a current carrying element shown in Fig. 2.1 is given by µ Idl sin θ dB = (2.2) 4π r2 where dB = infinitesimal flux density at point P, P I = current in the element, dB dl = length of element, r θ = angle between current direction and radius q vector to P, and dI I r = radius vector. In order to determine the magnetic flux density B due to a long, straight or curved conductor, we assume that the conductor is made up of infinitesimal lengths dl and B is given by Fig. 2.1 Flux density to a current B= 4π z µI sin θ r 2 dl (2.3) carrying element. The integration is carried out over the length of the conductor. If relation (2.3) is made use of in evaluating the magnetic flux density B at any point due to an infinite conductor, it is given by µI I B= (2.4) 2πR where R = radial distance of the point from the conductor. The direction of the flux density is normal to the plane R containing the conductor and the radius vector R. If B is now integrated around a path of radius R enclosing the wire once (Fig. 2.2), we have z Bdl = 2µπIR z dl Fig. 2.2 Ampere’s law: Line integral µI =. 2πR = µI of H over a closed path. 2πR or z Hdl = I as H = Bµ (2.5) In words it states that the line integral of H around a single closed path is equal to the current enclosed. This is known as Ampere’s law. If the path of integration encloses N number of turns of wire, each with a current I in the same direction, then z Hdl = NI (2.6) www.EngineeringBooksPdf.com  18 ELECTRICAL POWER SYSTEMS These relations are very much useful in evaluating the flux linkages and hence the inductance of a given system of conductors. Variation of the current in the conductors causes a change in the number of flux linkages. According to Faraday’s laws of electromagnetic induction, this change in flux linkages induces a voltage in the conductors which is proportional to the rate of change of flux linkages. 2.2 INDUCTORS AND INDUCTANCE An inductor is a device which stores energy in a magnetic field. By definition, the inductance L of an inductor is the ratio of its total magnetic flux linkages to the current I through the inductor or Nψ m λ L= = (2.7) I I This definition is satisfactory for a medium for which the permeability is constant. However, the permeability of ferrous media is not constant and for such cases the inductance is defined as the ratio of the infinitesimal change in flux linkage to the infinitesimal change in current producing it, i.e., dλ L= (2.8) dI The unit of inductance is the henry. Mutual inductance between two circuits is defined as the flux linkages of one circuit due to the current in the second circuit per ampere of current in the second circuit. If the current I2 produces λ12 flux linkages with circuit 1, the mutual inductance is λ 12 M12 = henries (2.9) I2 The phasor voltage drop in circuit 1 caused by the flux linkages of circuit 2 is V1 = jωM12I2 = jωλ12 volts. (2.10) 2.3 MAGNETIC FIELD INTENSITY DUE TO A LONG CURRENT CARRYING CONDUCTOR Let us consider a long current carrying conductor with radius R as shown in Fig. 2.3. We will consider here that the current is uniformly distributed across the section of the conductor. The flux linkages here will be both due to internal flux and external flux. The magnetic field intensity due to the current distribution inside the conductor is calculated as follows: Consider a cylinder with radius r < R. The current enclosed by the cylinder will be FG r IJ. 2 I′ = I H RK (2.11) where I is the current through the conductor. www.EngineeringBooksPdf.com  LINE CONSTANT CALCULATIONS 19 I R r H r Fig. 2.3 Variation of H due to current in the conductor for r ≤ R and r > R. Therefore, the magnetic field intensity at a distance r due to this current, using Ampere’s Law, I′ r 1 FG IJ Ir 2 Hr = 2πr =I = R 2πr 2πR2H K (2.12) which means that the magnetic field intensity inside the conductor is directly proportional to the distance from the centre of the conductor. Now consider a cylinder with radius r > R. Applying Ampere’s Law, I H= 2πr which means H is inversely proportional to r outside the conductor. The variation of H as a function of r is shown in Fig. 2.3. It can be shown that the magnetic field density (energy volume density) 1 µH 2 We = 2 From this and the distribution of magnetic field intensity as shown in Fig. 2.3, the following observations are made: (i) Although the volume of the conductor is comparatively small, the field densities are of high magnitude, and the magnetic field energy stored in the conductor is not small. (ii) The presence of the earth will affect the magnetic field geometry insignificantly. 2.4 φ ) TRANSMISSION LINE INDUCTANCE OF TWO-WIRE (1-φ By definition inductance is the flux linkages per ampere (Fig. 2.4). So the objective is to find out the flux linkages to this system of conductors. Now there are two flux linkages: (i) due to internal flux, and (ii) due to external flux. Internal flux linkages: In order to determine the internal flux linkages, we start with the magnetic field intensity H at any distance r < R. www.EngineeringBooksPdf.com  20 ELECTRICAL POWER SYSTEMS R I I D Fig. 2.4 Magnetic field due to one conductor of a 1-φ transmission line. Ir H= (2.13) 2πR 2 µ0I ∴ B = µH = µ0H =. r (as µr = 1) for conductors. 2πR2 This flux density as we see is varying with r. We can assume this to be constant over an infinitesimal distance dr. The flux lines are in the form of circles concentric to the conductor. Therefore, the flux lines passing through the concentric cylindrical shells of radii r and r + dr, dφ = B. Area normal to flux density B = Bdrl where l is the length of wire. In case the inductance per unit is desired, l = 1 metre. ∴ dφ = Bdr µ0 I = rdr 2πR 2 Now flux linkages = Flux × No. of turns. Here since only a part of the conductor (r < R) is being enclosed by the flux lines dφ, Fr I 2 ∴ dλ = dφ GH R JK 2 µ0 I r2 = rdr 2 πR 2 R2 ∴ Total internal flux linkages λ = z 0 R dλ = µ0 I 2πR4 z R 0 r 3 dr µ0 I = (2.14) 8π From this it is clear that the flux linkage due to internal flux is independent of the size of the conductor. www.EngineeringBooksPdf.com  LINE CONSTANT CALCULATIONS 21 External flux linkages: These flux linkages are due to the flux lines outside the conductor. There will be no flux line that encloses both the conductors. This is because for any distance r > D the total current enclosed is zero (single phase line i.e., one conductor is a ‘go’ conductor and the other ‘return’). The magnetic field intensity H due to one conductor at any distance R ≤ r < D, I H= 2πr µ0 I B = µ0H = (µr = 1 as the medium is air) 2 πr The flux density B can be considered uniform over a distance dr. Therefore, as in case of internal flux, the flux lines passing through the concentric cylindrical shells with radii r and (r + dr) will be (per unit length) dφ = B.dr.1 Since this flux encloses only one conductor, therefore, the number of turns enclosed by this flux is one. ∴ dλ = dφ.1 = B.dr.1.1 µ0 I = dr 2 πr Therefore, the total external flux linkages due to current flow in one conductor, λ= zR D− R dλ The lower limit is because we measure the distances from the centre of the conductor and external flux begins from the surface of the conductor and this extends up to the surface of the other conductor and, therefore, the upper limit (D – R) λ= µ0 I 2π z R D− R dr r µ0 I D−R = ln 2π R Since R is small as compared to D i.e. R > 1, r h h K~ + (3.16) r r www.EngineeringBooksPdf.com  46 ELECTRICAL POWER SYSTEMS Since K ≠ 0 only positive sign is taken into account. Therefore πε 0 C= F/metre (3.17) h h2 + −1 r r2 The expression for capacitance obtained above is very accurate. However, it could be assumed that the charge is uniformaly distributed which is not very far from the actual condition h for power system problems where >> 1. The derivation is much more simplified and is as r follows (refer to Fig. 3.4): Since the charge is assumed to be uniformly distributed over the surface of the conductor, this could be considered as concentrated along the axis on conductor. The electric field intensity at point P due to ρL is ρL E+ = (3.18) 2πε 0 x and is directed along a x. Similarly electric field intensity at P due to – ρL ρL E– = (3.19) 2πε 0 (h − x) along a x again, as this time the force experienced by a unit positive charge at P will be towards the negative charge (force of attraction). Total electric field intensity at P E= ρL LM 1 + 1 OP (3.20) 2πε 0 Nx h − xQ The potential difference between the conductors V=– zh−r r E dx = ρL 2πε 0 z r h− r FG 1 + 1 IJ dx H x h − xK ρL h− r = ln x − ln (h − x) r 2πε 0 ρL = [ln (h – r) – ln r – ln {h – (h – r)} + ln (h – r)] 2πε 0 ρL h−r ρ h−r = 2 ln = L ln (3.21) 2πε 0 r πε 0 r Since h >> r, h – r ~ h. ρL h ∴ V= ln πε 0 r www.EngineeringBooksPdf.com  CAPACITANCE OF TRANSMISSION LINES 47 ρL πε 0 or C= = F/metre (3.22) V ln h / r Equation (3.22) corresponds to the expression for capacitance of a single phase transmission line. Compare this expression with the expression for inductance equation (2.17) of a single phase transmission line. Equation for inductance contains a constant term corresponding to the internal flux linkages whereas since charges reside on the surface of the conductor, similar term is absent in the capacitance expression. As a result of this, the radius in the expression for capacitance is the actual outside radius of the conductor whereas for inductance equation (2.18) the radius is the self GMD of the conductor. The concept of self GMD is applicable for inductance calculation and not for the capacitance. Sometimes it is required to know the capacitance between one conductor and a neutral point between them which will be defined as the charge on one of the conductors per unit of voltage difference between the neutral and the conductor. This means the capacitance of one conductor with respect to the neutral plane is two times the capacitance of the single-phase line (Fig. 3.5). 2πε 0 Can = 2Cab = h ln r a b a N b º Cab Can Cbn Fig. 3.5 3.5 CAPACITANCE OF A 3-PHASE UNSYMMETRICALLY SPACED TRANS- MISSION LINE For an untransposed line the capacitances between conductor to neutral of the three conductors are unequal. In transposed lines the average capacitance of each conductor to neutral is the same as the capacitance to neutral of any other phase. The dissymmetry of the untransposed line is slight for the usual transmission lines and, therefore, the calculations for capacitance are carried out as though the lines were completely transposed. The three positions of the conductors are shown in Fig. 3.6. P a Da c b b D b c Dc c b a b a a c Fig. 3.6 Unsymmetrically spaced transposed 3-phase transmission line. www.EngineeringBooksPdf.com  48 ELECTRICAL POWER SYSTEMS Since the potential due to a linear charge is a linear function of the charge it follows that the potentials of more than one charges are linearly superposable. Considering Fig. 3.6, let a point P be at a large distance D from the system such that Da, Db and Dc are approximately same. It is required to find out the potential of conductor a due to charges ρa, ρb and ρc per unit length of the conductors. Since it is a 3-phase balanced system, taking ρa as the reference charge, ρb = ρa ∠– 120 and ρc = ρa ∠120 The potential of conductor ‘a’ with respect to point P due to the charge on the conductor itself, ρa D Vaa = ln a (3.23) 2πε 0 r Similarly, the potential of conductor ‘a’ due to the charges ρb and ρc respectively are ρb D ρc D ln b and ln c 2πε 0 c 2πε 0 b 1 DLM D D ρ a ln a + ρb ln b + ρ c ln c OP ∴ Va′ = 2πε 0 rN c b Q (3.24) Similarly, the potential of conductor a in the other two positions is given by 1 DLM D D ρ a ln b + ρb ln c + ρ c ln a OP Va″ = 2πε 0 rN a c Q (3.25) 1 LMρ Dc D D O a PQ ln + ρb ln a + ρ c ln b and Va′″ = 2πε 0 N a r b (3.26) The average voltage of phase a with respect to point P Va′ + Va″ + Va′ ″ Va = 3 1 LM D D D D D D D D D ρ a ln a 3b c + ρb ln a b c + ρ c ln a b c OP = 6πε 0 N r abc abc Q (3.27) Now ρa + ρb + ρc = 0 ∴ ρb + ρ c = – ρ a Substituting this in the expression (3.27), 1 LM D D D D D D ρ a ln a 3b c + (ρb + ρ c ) ln a b c OP Va = 6πε 0 N r abc Q 1 LMρ ln Da Db Dc − ρ a ln Da Db Dc OP = 6πε 0 N a r 3 abc Q ρa abc = ln 3 6πε 0 r www.EngineeringBooksPdf.com  CAPACITANCE OF TRANSMISSION LINES 49 3 ρa abc = ln 2πε 0 r ρa GMD = ln 2πε 0 r ρa 2πε 0 or C= = F/metre (3.28) Va ln GMD r Since the conductors b and c also occupy the same three positions as occupied by conductor a, the average voltage of the conductors is same and, therefore, the capacitance is also the same. For a symmetrical spacing of the conductors, a=b=c=h 2πε 0 ∴ C= (3.29) h ln r Example 3.1: Determine the capacitance and the charging current per km when the transmission line of example 2.2 is operating at 132 kV. Solution: The radius of conductor = 0.4 cm. The mutual GMD of conductors, Dm = 2.015 metres. 2πε 0 ∴ Capacitance per phase per metre = F/metre 2.015 ln × 10 2 0.4 10 −9 = = 8.928 pF/metre 201.5 18 × ln 0.4 = 8.928 × 10–12 × 103 F/km = 8.928 × 10–9 F/km 132 × 1000 The charging current = × 8.928 × 10–9 × 314 3 = 0.2136 amp/km. Ans. 3.6 CAPACITANCE OF A DOUBLE CIRCUIT LINE Normally two configurations of conductors are used: (i) hexagonal spacing, and (ii) flat vertical spacing. First of all an expression of capacitance for hexagonal spacing is derived. Hexagonal Spacing Since the conductors of the same phase are connected in parallel the charge per unit length is the same (Fig. 3.7). Also, because of the symmetrical arrangement the phases are balanced www.EngineeringBooksPdf.com  50 ELECTRICAL POWER SYSTEMS and the conductors of each phase are also balanced if the effect of ground is neglected. Therefore, the transposition of conductors is not required. D a c' D b 3D 2D b' D c a' Fig. 3.7 Double circuit line—Hexagonal spacing. Assume a point P very far from the system of conductors such that the distances of the conductors from P are almost same. It is to be noted here that point P corresponds to almost zero potential. The potential of conductor a with respect to point P due to the charge on the conductor itself and the charges on conductors b, c, a′, b′ and c′ is given by ρa D ρb D ρc Dc ρ D ρ Db′ Va = ln a + ln b + ln + a′ ln a ′ + b′ ln 2πε 0 r 2πε 0 D 2πε 0 3 D 2πε 0 2 D 2πε 0 3D ρ c′ D + ln c′ (3.30) 2πε 0 D Since ρa = ρa′, ρb = ρb′ and ρc = ρc′ 1 LM FG D D D IJ D FG IJ + ρ FG ln Dc Dc′ IJ OP N H K K H KQ ρ a ln a + ln a ′ + ρb ln b + ln b′ Va = 2πε 0 r 2D D 3D H c 3D + ln D 1 LMρ ln (D D = 2πε 0 Na a a′ ) + ρ b ln ( Db Db′ ) + ρ c ln ( Dc Dc′ ) + ρ a ln 1 + ρb ln 1 + ρ c ln 1 OP 2 Dr 3 D2 3D 2 PQ 1 = (ρ + ρb + ρc) ln (DaDa′) 2πε 0 a + 1 LM ρ a ln 1 + (ρb + ρ c ) ln 1 OP (3.31) 2πε 0 MN 2 Dr 3D 2 PQ Since DaDa′ = DbDb′ = DcDc′. Also since ρa + ρb + ρc = 0, 1 F 1 1 I Va = 2πε 0 ρ a lnGH 2 Dr − ln 3 D2 JK www.EngineeringBooksPdf.com  CAPACITANCE OF TRANSMISSION LINES 51 ρa 3D Va = ln 2πε 0 2r ρa 2πε 0 or C= = F/metre/conductor. (3.32) 3D Va ln 2r Equation (3.32) represents an expression for the capacitance of conductor a alone, whereas there are two conductors per phase a and a′. Therefore, the capacitance of the system per phase will be twice the capacitance of one conductor to neutral, i.e., 4πε 0 C= F/metre/phase (3.33) 3D ln 2r Here expression for capacitance for phase a has been derived. Since the conductors of different phases are symmetrically placed, the expression for capacitance for other phases will also be the same. Flat Vertical Spacing Refer to the system of conductors in Fig. 3.8. The conductors of different phases are not symmetrically placed; therefore, the derivation of capacitance expression will require the transposition of conductors as shown in Fig. 3.8. a h c' c b' b a' d g b f b' a a' c c' c a¢¢ b c¢¢ a b¢¢ 1 2 3 Fig. 3.8 Double circuit flat vertical spacing, transposed line. It is required to find out average voltage of conductor a in the three different positions due to the charge on conductor a and the conductors b, c, a′, b′ and c′. For this we again assume a point very far from the system of conductors such that Da ~ − Db ~− Dc ~ − Da′ ~ − Db′ ~ − Dc′. Since point P is at a very large distance from the system of conductors, the potential of point P is approximately zero. The potential of conductor a in position 1. 1 LM D D D ρ a ln a + ρb ln b + ρ c ln c Va′ = 2πε 0 N r d 2d + ρa ′ ln Da ′ D D + ρb′ ln b′ + ρc ′ ln c′ OP (3.34) f g h Q Using the relations Da ~ − Db ~ − Dc ~ − Da′ ~ − Db′ ~ − Dc′, ρa + ρb + ρc = 0, and ρa = ρa′, ρb = ρb′, ρc = ρc′. www.EngineeringBooksPdf.com  52 ELECTRICAL POWER SYSTEMS Va′ = 1 ρ a ln LM 1 + ρ b ln 1 + ρ c ln 1 OP