ELECTRIC POWER SYSTEM

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Electric Power Systems FiFth Edition B.M. Weedy | B.J. Cory N. JeNkiNs | J.B. ekaNayake | G. strBaC

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Electric Power Systems

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Electric Power SystemsFifth EditionB.M. Weedy, University of Southampton, UKB.J. Cory, Imperial College London, UKN. Jenkins, Cardiff University, UKJ.B. Ekanayake, Cardiff University, UKG. Strbac, Imperial College London, UK

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This edition ﬁrst published 2012# 2012, John Wiley & Sons LtdRegistered ofﬁceJohn Wiley & Sons Ltd, The Atrium, Southern Gate, Chichester, West Sussex, PO19 8SQ, United KingdomFor details of our global editorial ofﬁces, for customer services and for information about how to apply forpermission to reuse the copyright material in this book please see our website at www.wiley.com.The right of the author to be identiﬁed as the author of this work has been asserted in accordance with theCopyright, Designs and Patents Act 1988.All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or trans-mitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise,except as permitted by the UK Copyright, Designs and Patents Act 1988, without the prior permission ofthe publisher.Wiley also publishes its books in a variety of electronic formats. Some content that appears in print maynot be available in electronic books.Designations used by companies to distinguish their products are often claimed as trademarks. All brandnames and product names used in this book are trade names, service marks, trademarks or registeredtrademarks of their respective owners. The publisher is not associated with any product or vendor men-tioned in this book. This publication is designed to provide accurate and authoritative information inregard to the subject matter covered. It is sold on the understanding that the publisher is not engaged inrendering professional services. If professional advice or other expert assistance is required, the servicesof a competent professional should be sought.Library of Congress Cataloging-in-Publication DataElectric power systems / Brian M. Weedy [...et al.]. – 5th ed. p. cm. Includes bibliographical references and index. ISBN 978-0-470-68268-5 (cloth) 1. Electric power systems–Textbooks. 2. Electric powertransmission–Textbooks. I. Weedy, Brian M. TK1001.E4235 2012 621.319’1–dc23 2012010322A catalogue record for this book is available from the British Library.Print ISBN: 9780470682685Set in 10/12.5pt, Palatino-Roman by Thomson Digital, Noida, India

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Contents ixPreface to First Edition xiPreface to Fourth Edition xiiiPreface to Fifth Edition xvSymbols 1 1 1 Introduction 2 1.1 History 4 1.2 Characteristics Inﬂuencing Generation and Transmission 5 1.3 Operation of Generators 12 1.4 Energy Conversion 17 1.5 Renewable Energy Sources 23 1.6 Energy Storage 27 1.7 Environmental Aspects of Electrical Energy 40 1.8 Transmission and Distribution Systems 43 1.9 Utilization Problems 45 45 2 Basic Concepts 55 2.1 Three-Phase Systems 57 2.2 Three-Phase Transformers 61 2.3 Active and Reactive Power 68 2.4 The Per-Unit System 74 2.5 Power Transfer and Reactive Power 75 2.6 Harmonics in Three-Phase Systems 78 2.7 Useful Network Theory Problems

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vi Contents3 Components of a Power System 83 3.1 Introduction 83 3.2 Synchronous Machines 83 3.3 Equivalent Circuit Under Balanced Short-Circuit Conditions 90 3.4 Synchronous Generators in Parallel 94 3.5 The Operation of a Generator on an Inﬁnite Busbar 95 3.6 Automatic Voltage Regulators (AVRs) 100 3.7 Lines, Cables and Transformers 103 3.8 Transformers 124 3.9 Voltage Characteristics of Loads 131 Problems 1344 Control of Power and Frequency 139 4.1 Introduction 139 4.2 The Turbine Governor 142 4.3 Control Loops 146 4.4 Division of Load between Generators 147 4.5 The Power-Frequency Characteristic of an Interconnected System 151 4.6 System Connected by Lines of Relatively Small Capacity 152 Problems 1595 Control of Voltage and Reactive Power 161 5.1 Introduction 161 5.2 The Generation and Absorption of Reactive Power 163 5.3 Relation between Voltage, Power, and Reactive Power at a Node 165 5.4 Methods of Voltage Control: (a) Injection of Reactive Power 170 5.5 Methods of Voltage Control: (b) Tap-Changing Transformers 176 5.6 Combined Use of Tap-Changing Transformers and Reactive-Power Injection 183 5.7 Phase-Shift Transformer 188 5.8 Voltage Collapse 191 5.9 Voltage Control in Distribution Networks 195 5.10 Long Lines 197 5.11 General System Considerations 198 Problems 2006 Load Flows 205 6.1 Introduction 205 6.2 Circuit Analysis Versus Load Flow Analysis 206 6.3 Gauss-Seidel Method 212 6.4 Load Flows in Radial and Simple Loop Networks 216 6.5 Load Flows in Large Systems 219 6.6 Computer Simulations 231 Problems 234

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Contents vii 7 Fault Analysis 239 7.1 Introduction 239 7.2 Calculation of Three-Phase Balanced Fault Currents 241 7.3 Method of Symmetrical Components 247 7.4 Representation of Plant in the Phase-Sequence Networks 251 7.5 Types of Fault 252 7.6 Fault Levels in a Typical System 259 7.7 Power in Symmetrical Components 265 7.8 Systematic Methods for Fault Analysis in Large Networks 265 7.9 Neutral Grounding 270 7.10 Interference with Communication Circuits–Electromagnetic Compatibility (EMC) 274 Problems 275 8 System Stability 281 8.1 Introduction 281 8.2 Equation of Motion of a Rotating Machine 283 8.3 Steady-State Stability 284 8.4 Transient Stability 287 8.5 Transient Stability–Consideration of Time 293 8.6 Transient Stability Calculations by Computer 298 8.7 Dynamic or Small-Signal Stability 301 8.8 Stability of Loads Leading to Voltage Collapse 305 8.9 Further Aspects 309 8.10 Multi-Machine Systems 311 8.11 Transient Energy Functions (TEF) 312 8.12 Improvement of System Stability 314 Problems 315 9 Direct-Current Transmission 319 9.1 Introduction 319 9.2 Current Source and Voltage Source Converters 320 9.3 Semiconductor Valves for High-Voltage Direct-Current Converters 322 9.4 Current Source Converter h.v.d.c. 325 9.5 Voltage Source Converter h.v.d.c. 346 Problems 35210 Overvoltages and Insulation Requirements 355 10.1 Introduction 355 10.2 Generation of Overvoltages 356 10.3 Protection Against Overvoltages 365 10.4 Insulation Coordination 369 10.5 Propagation of Surges 373 10.6 Determination of System Voltages Produced by Travelling Surges 382

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viii Contents 10.7 Electromagnetic Transient Program (EMTP) 391 Problems 39911 Substations and Protection 403 11.1 Introduction 403 11.2 Switchgear 404 11.3 Qualities Required of Protection 415 11.4 Components of Protective Schemes 416 11.5 Protection Systems 424 11.6 Distance Protection 427 11.7 Unit Protection Schemes 429 11.8 Generator Protection 430 11.9 Transformer Protection 432 11.10 Feeder Protection 435 Problems 43912 Fundamentals of the Economics of Operation and Planning 443 of Electricity Systems 444 12.1 Economic Operation of Generation Systems 451 12.2 Fundamental Principles of Generation System Planning 457 12.3 Economic Operation of Transmission Systems 460 12.4 Fundamental Principles of Transmission System Planning 463 12.5 Distribution and Transmission Network Security Considerations 466 12.6 Drivers for Change 467 Problems 473Appendix A Synchronous Machine Reactances 477Appendix B Typical Transformer Impedances 481Appendix C Typical Overhead Line Parameters 487Further Reading 491Index

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Preface to First EditionIn writing this book the author has been primarily concerned with the presentationof the basic essentials of power-system operation and analysis to students in theﬁnal year of ﬁrst degree courses at universities and colleges of technology. Theemphasis is on the consideration of the system as a whole rather than on the engi-neering details of its constituents, and the treatment presented is aimed at practicalconditions and situations rather than theoretical nicety. In recent years the contents of many undergraduate courses in electrical engineer-ing have become more fundamental in nature with greater emphasis on electromag-netism, network analysis, and control theory. Students with this background will befamiliar with much of the work on network theory and the inductance, capacitance,and resistance of lines and cables, which has in the past occupied large parts of text-books on power supply. In this book these matters have been largely omitted, result-ing in what is hoped is a concise account of the operation and analysis of electricpower systems. It is the author’s intention to present the power system as a systemof interconnected elements which may be represented by models, either mathemati-cally or by equivalent electrical circuits. The simplest models will be used consis-tently with acceptable accuracy and it is hoped that this will result in the woodbeing seen as well as the trees. In an introductory text such as this no apology ismade for the absence of sophisticated models of plant (synchronous machines inparticular) and involved mathematical treatments as these are well catered for inmore advanced texts to which reference is made. The book is divided into four main parts, as follows:a. Introduction, including the establishment of equivalent circuits of the compo- nents of the system, the performance of which, when interconnected, forms the main theme.b. Operation, the manner in which the system is operated and controlled to give secure and economic power supplies.c. Analysis, the calculation of voltage, power, and reactive power in the system under normal and abnormal conditions. The use of computers is emphasised when dealing with large networks.

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x Preface to First Editiond. Limitations of transmittable power owing to the stability of the synchronous machine, voltage stability of loads, and the temperature rises of plant. It is hoped that the ﬁnal chapter will form a useful introduction to direct currenttransmission which promises to play a more and more important role in electricitysupply. The author would like to express his thanks to colleagues and friends for theirhelpful criticism and advice. To Mr J.P. Perkins for reading the complete draft, toMr B.A. Carre on digital methods for load ﬂow analysis, and to Mr A.M. Parkeron direct current transmission. Finally, thanks are due to past students who for overseveral years have freely expressed their difﬁculties in this subject. Birron M. Weedy Southampton, 1967

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Preface to Fourth EditionAs a university teacher for 40 years, I have always admired the way that Dr BirronWeedy’s book has stood out from the numerous texts on the analysis and modellingof power systems, with its emphasis on practical systems rather than extensive the-ory or mathematics. Over the three previous editions and one revision, the text hasbeen continually updated and honed to provide the essentials of electrical powersystems sufﬁcient not only for the ﬁnal year of a ﬁrst degree course, but also as aﬁrm foundation for further study. As with all technology, progress produces newdevices and understanding requiring revision and updating if a book is to be of con-tinuing value to budding engineers. With power systems, there is another dimen-sion in that changes in social climate and political thinking alter the way they aredesigned and operated, requiring consideration and understanding of new forms ofinfrastructure, pricing principles and service provision. Hence the need for an intro-duction to basic economics and market structures for electricity supply, which isgiven in a completely new Chapter 12. In this edition, 10 years on from the last, a rewrite of Chapter 1 has brought in fullconsideration of CCGT plant, some new possibilities for energy storage, the latestthinking on electromagnetic ﬁelds and human health, and loss factor calculations.The major addition to system components and operation has been Flexiblea.c. Transmission (FACT) devices using the latest semiconductor power switchesand leading to better control of power and var ﬂows. The use of optimisation tech-niques has been brought into Chapter 6 with powerﬂow calculations but the increas-ing availability and use of commercial packages has meant that detailed codewriting is no longer quite so important. For stability (Chapter 8), it has been neces-sary to consider voltage collapse as a separate phenomenon requiring furtherresearch into modelling of loads at voltages below 95% or so of nominal. Increas-ingly, large systems require fast stability assessment through energy-like functionsas explained in additions made to this chapter. Static-shunt variable compensatorshave been included in Chapter 9 with a revised look at h.v.d.c. transmission. Manyd.c. schemes now exist around the world and are continually being added to so thedescription of an example scheme has been omitted. Chapter 11 now includes manynew sections with updates on switchgear, and comprehensive introductions to

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xii Preface to Fourth Editiondigital (numerical) protection principles, monitoring and control with SCADA, stateestimation, and the concept of Energy Management Systems (EMS) for systemoperation. Readers who have been brought up on previous editions of this work will realisethat detailed design of overhead and underground systems and components hasbeen omitted from this edition. Fortunately, adequate textbooks on these topics areavailable, including an excellent book by Dr Weedy, and reference to these texts isrecommended for detailed study if the principles given in Chapter 3 herein areinsufﬁcient. Many other texts (including some ‘advanced’ ones) are listed in a neworganisation of the bibliography, together with a chapter-referencing key which Ihope will enable the reader to quickly determine the appropriate texts to look up. Inaddition, mainly for historical purposes, a list of signiﬁcant or ‘milestone’ papersand articles is provided for the interested student. Finally, it has been an honour to be asked to update such a well-known book and Ihope that it still retains much of the practical ﬂavour pioneered by Dr Weedy. I amparticularly indebted to my colleagues, Dr Donald Macdonald (for much help with arewrite of the material about electrical generators) and Dr Alun Coonick for hisprompting regarding the inclusion of new concepts. My thanks also go to the vari-ous reviewers of the previous editions for their helpful suggestions and commentswhich I have tried to include in this new edition. Any errors and omissions areentirely my responsibility and I look forward to receiving feedback from studentsand lecturers alike. Brian J. Cory Imperial College, London, 1998Publisher’s NoteDr B. M. Weedy died in December 1997 during the production of this fourth edition.

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Preface to Fifth EditionWe were delighted to be asked to revise this classic textbook. From the earlier edi-tions we had gained much, both as undergraduate students and throughout ourcareers. Both Dr Weedy and Dr Cory can only be described as giants of powersystem education and the breadth of their vision and clarity of thought is evidentthroughout the text. Reading it carefully, for the purposes of revision, was a mostrewarding experience and even after many years studying and teaching powersystems we found new insights on almost every page. We have attempted to stay true to the style and structure of the book while addingup-to-date material and including examples of computer based simulation. We wereconscious that this book is intended to support a 3rd or 4th year undergraduatecourse and it is too easy when revising a book to continue to add material and soobscure rather than illuminate the fundamental principles. This we have attemptednot to do. Chapter 1 has been brought up to date as many countries de-carbonisetheir power sector. Chapter 6 (load ﬂow) has been substantially rewritten and volt-age source converter HVDC added to Chapter 9. Chapter 10 has been revised toinclude modern switchgear and protection while recognising that the young engi-neer is likely to encounter much equipment that may be 30–40 years old. Chapter 12has been comprehensively revised and now contains material suitable for teachingthe fundamentals of the economics of operation and development of power systems.All chapters have been carefully revised and where we considered it would aidclarity the material rearranged. We have paid particular attention to the Examplesand Problems and have created Solutions to the Problems that can be found on theWiley website. We are particularly indebted to Dave Thompson who created all the illustrationsfor this edition, Lewis Dale for his assistance with Chapter 12, and to IPSA Power forgenerously allowing us a license for their power system analysis software. Also wewould like to thank: Chandima Ekanayake, Prabath Binduhewa, Predrag Djapic andJelena Rebic for their assistance with the Solutions to the Problems. Bethany

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xiv Preface to Fifth EditionCorcoran provided the data for Figure 1.1 while Alstom Grid, through Rose King,kindly made available information for some of the drawings of Chapter 11.Although, of course, responsibility for errors and omissions lies with us, we hopewe have stayed true to the spirit of this important textbook. For instructors and teachers, solutions to the problems set out in the book can befound on the companion website www.wiley.com/go/weedy_electric. Nick Jenkins, Janaka Ekanayake, Goran Strbac June 2012

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SymbolsThroughout the text, symbols in bold type represent complex (phasor) quantitiesrequiring complex arithmetic. Italic type is used for magnitude (scalar) quantities.A,B,C,D Generalised circuit constantsa–b–c Phase rotation (alternatively R–Y–B)a Operator 1ﬀ120C Capacitance (farad)D DiameterE e.m.f. generatedF Cost function (units of money per hour)f Frequency (Hz)G Rating of machineg Thermal resistivity (C m/W)H Inertia constant (seconds)h Heat transfer coefﬁcient (W/m2 per C)I Current (A)IÃ Conjugate of IId In-phase currentIq Quadrature currentj 1 ﬀ 90 operatorK Stiffness coefﬁcient of a system (MW/Hz)L Inductance (H)ln Natural logarithmM Angular momentum (J-s per rad or MJ-s per electrical degree)N Rotational speed (rev/min, rev/s, rad/s)P Propagation constant (a þ jb)P Power (W)dP Synchronising power coefﬁcientddp.f. Power factorp Iteration numberQ Reactive power (VAr)q Loss dissipated as heat (W)R Resistance (V); also thermal resistance (C/W)R–Y–B Phase rotation (British practice)S Complex power ¼ P Æ jQS Siemenss Laplace operators Slip

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xvi SymbolsSCR Short-circuit ratioT Absolute temperature (K)t Timet Off-nominal transformer tap ratioDt Interval of timeVÀ1 SiemensU VelocityV Voltage; DV scalar voltage differenceV Voltage magnitudeW Volumetric ﬂow of coolant (m3/s)X0 Transient reactance of a synchronous machineX00 Subtransient reactance of a synchronous machineXd Direct axis synchronous reactance of a synchronous machineXq Quadrature axis reactance of a synchronous machineXs Synchronous reactance of a synchronous machineY Admittance (p.u. or V)Z Impedance (p.u. or V)Z0 Characteristic or surge impedance (V)a Delay angle in rectiﬁers and inverters–d.c. transmissiona Attenuation constant of linea Reﬂection coefﬁcientb Phase-shift constant of lineb (180 À a) used in invertersb Refraction coefﬁcient (1 þ a)g Commutation angle used in convertersd Load angle of synchronous machine or transmission angle across a system (electrical degrees)d0 Recovery angle of semiconductor valvee Permittivityh Viscosity (g/(cm-s))u Temperature rise (C) above reference or ambientl Lagrange multiplierr Electrical resistivity (V-m)r Density (kg/m3)t Time constantf Angle between voltage and current phasors (power factor angle)v Angular frequency (rad/s)Subscripts 1, 2, and 0 refer to positive, negative, and zero symmetrical components,respectively.

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1Introduction1.1 HistoryIn 1882 Edison inaugurated the ﬁrst central generating station in the USA. This fed aload of 400 lamps, each consuming 83 W. At about the same time the Holborn Via-duct Generating Station in London was the ﬁrst in Britain to cater for consumersgenerally, as opposed to specialized loads. This scheme used a 60 kW generatordriven by a horizontal steam engine; the voltage of generation was 100 V directcurrent. The ﬁrst major alternating current station in Great Britain was at Deptford, wherepower was generated by machines of 10 000 h.p. and transmitted at 10 kV to con-sumers in London. During this period the battle between the advocates of alternat-ing current and direct current was at its most intense with a similar controversyraging in the USA and elsewhere. Owing mainly to the invention of the transformerthe supporters of alternating current prevailed and a steady development of localelectricity generating stations commenced with each large town or load centre oper-ating its own station. In 1926, in Britain, an Act of Parliament set up the Central Electricity Board withthe object of interconnecting the best of the 500 generating stations then in operationwith a high-voltage network known as the Grid. In 1948 the British supply industrywas nationalized and two organizations were set up: (1) the Area Boards, which weremainly concerned with distribution and consumer service; and (2) the GeneratingBoards, which were responsible for generation and the operation of the high-voltagetransmission network or grid. All of this changed radically in 1990 when the British Electricity Supply Industrywas privatized. Separate companies were formed to provide competition in the sup-ply of electrical energy (sometimes known as electricity retail businesses) and inpower generation. The transmission and distribution networks are natural monopo-lies, owned and operated by a Transmission System Operator and DistributionNetwork Operators. The Ofﬁce of Gas and Electricity Markets (OFGEM) wasElectric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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2 Electric Power Systems, Fifth Editionestablished as the Regulator to ensure the market in electricity generation andenergy supply worked effectively and to ﬁx the returns that the Transmission andDistribution Companies should earn on their monopoly businesses. For the ﬁrst 80 years of electricity supply, growth of the load was rapid at around7% per year, implying a doubling of electricity use every 10 years and this type ofincrease continues today in rapidly industrializing countries. However in the USAand in other industrialized countries there has been a tendency, since the oil shockof 1973, for the rate of increase to slow with economic growth no longer coupledclosely to the use of energy. In the UK, growth in electricity consumption has beenunder 1% per year for a number of years. A traditional objective of energy policy has been to provide secure, reliable andaffordable supplies of electrical energy to customers. This is now supplemented bythe requirement to limit greenhouse gas emissions, particularly of CO2, and so miti-gate climate change. Hence there is increasing emphasis on the generation of elec-tricity from low-carbon sources that include renewable, nuclear and fossil fuelplants ﬁtted with carbon capture and storage equipment. The obvious way to con-trol the environmental impact of electricity generation is to reduce the electricaldemand and increase the efﬁciency with which electrical energy is used. Thereforeconservation of energy and demand reduction measures are important aspects ofany contemporary energy policy.1.2 Characteristics Inﬂuencing Generation and TransmissionThere are three main characteristics of electricity supply that, however obvious,have a profound effect on the manner in which the system is engineered. They areas follows:Electricity, unlike gas and water, cannot be stored and the system operator traditionally has had limited control over the load. The control engineers endeavour to keep the output from the generators equal to the connected load at the speciﬁed voltage and frequency; the difﬁculty of this task will be apparent from a study of the load curves in Figure 1.1. It will be seen that the load consists of a steady component known as the base load, plus peaks that depend on the time of day and days of the week as well as factors such as popular television programmes.The electricity sector creates major environmental impacts that increasingly determine how plant is installed and operated. Coal burnt in steam plant produces sulphur dioxide that causes acid rain. Thus, in Europe, it is now mandatory to ﬁt ﬂue gas desulphur- isation plant to coal ﬁred generation. All fossil fuel (coal, oil and gas) produce CO2 (see Table 1.1) which leads to climate change and so its use will be discouraged increasingly with preference given to generation by low-carbon energy sources.The generating stations are often located away from the load resulting in transmission over considerable distances. Large hydro stations are usually remote from urban centres and it has often been cost-effective to burn coal close to where it is mined and transport the electricity rather than move the coal. In many coun- tries, good sites for wind energy are remote from centres of population and,

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Introduction 3Load (GW) 150 2006 PJM One Week Summer Electric LoadLoad (GW) 140 S 130 S 120 110 24 100 90 80 70 60 M TW T F S (a) 110 2006 PJM One Week Winter Electric Load 100 90 80 70 60 M TW T F S (b) 42 2010 GB One Week Summer Electric LoadLoad (GW) 37 32 27 22 TW TFSS M M (c) 1800 1600 2010 Sri Lanka One Day Summer Generation 1400Generation (MW) 1200 Thermal 1000 Hydro 800 6 12 18 600 400 200 0 (d)Figure 1.1 Load curves. (a) PJM (Pennsylvania, Jersey, Maryland) control area in theeast of the USA over a summer week. The base load is 70 GW with a peak of 140 GW.This is a very large interconnected power system. (b) PJM control area over a winterweek. Note the morning and evening peaks in the winter with the maximum demandin the summer. (c) Great Britain over a summer week. The base load is around 25 GWwith a daily increase/decrease of 15 GW. GB is effectively an isolated power system.(d) Sri Lanka over 1 day. Note the base load thermal generation with hydro used toaccommodate the rapid increase of 500 MW at dusk

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4 Electric Power Systems, Fifth Edition although it is possible to transport gas in pipelines, it is often difﬁcult to obtain permission to construct generating stations close to cities. Moreover, the con- struction of new electrical transmission is subject to delays in many developed countries caused by objections from the public and the difﬁculty in obtaining permission for the construction of new overhead line circuits.1.3 Operation of GeneratorsThe national electrical load consists of a base plus a variable element, depending onthe time of day and other factors. In thermal power systems, the base load should besupplied by the most efﬁcient (lowest operating cost) plant which then runs24 hours per day, with the remaining load met by the less efﬁcient (but lower capitalcost) stations. In hydro systems water may have to be conserved and so some gener-ators are only operated during times of peak load. In addition to the generating units supplying the load, a certain proportion ofavailable plant is held in reserve to meet sudden contingencies such as a generatorunit tripping or a sudden unexpected increase in load. A proportion of this reservemust be capable of being brought into operation immediately and hence somemachines must be run at, say, 75% of their full output to allow for this spare generat-ing capacity, called spinning reserve. Reserve margins are allowed in the total generation plant that is constructed tocope with unavailability of plant due to faults, outages for maintenance and errorsin predicting load or the output of renewable energy generators. When traditionalnational electricity systems were centrally planned, it was common practice toallow a margin of generation of about 20% over the annual peak demand. A highproportion of intermittent renewable energy generation leads to a requirement fora higher reserve margin. In a power system there is a mix of plants, that is, hydro,coal, oil, renewable, nuclear, and gas turbine. The optimum mix gives the mosteconomic operation, but this is highly dependent on fuel prices which can ﬂuctu-ate with time and from region to region. Table 1.2 shows typical plant andTable 1.1 Estimated carbon dioxide emissions from electricitygeneration in Great BritainFuel Tonnes of CO2/GWh of Electrical OutputCoal 915Oil 633Gas 405Great Britain generation portfolio 452 (including nuclear and renewables)Data from the Digest of UK Energy Statistics, 2010, published by the Depart-ment of Energy and Climate Change.

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Introduction 5Table 1.2 Example of costs of electricity generationGenerating Technology Capital Cost of Cost of electricity Plant £/MW £/MWhCombined Cycle Gas Turbine 720 80Coal 1800 105Onshore wind 1520Nuclear 2910 94 99Data from UK Electricity Generating Costs Update, 2010, Mott MacDonald, reproduced with permissiongenerating costs for the UK. It is clear some technologies have a high capital cost(for example, nuclear and wind) but low fuel costs.1.4 Energy Conversion1.4.1 Energy Conversion Using SteamThe combustion of coal, gas or oil in boilers produces steam, at high tempera-tures and pressures, which is passed through steam turbines. Nuclear ﬁssion canalso provide energy to produce steam for turbines. Axial-ﬂow turbines are gener-ally used with several cylinders, containing steam of reducing pressure, on thesame shaft. A steam power-station operates on the Rankine cycle, modiﬁed to include super-heating, feed-water heating, and steam reheating. High efﬁciency is achieved by theuse of steam at the maximum possible pressure and temperature. Also, for turbinesto be constructed economically, the larger the size the less the capital cost per unit ofpower output. As a result, turbo-generator sets of 500 MW and more have beenused. With steam turbines above 100 MW, the efﬁciency is increased by reheatingthe steam, using an external heater, after it has been partially expanded. Thereheated steam is then returned to the turbine where it is expanded through the ﬁnalstages of blading. A schematic diagram of a coal ﬁred station is shown in Figure 1.2. In Figure 1.3 theﬂow of energy in a modern steam station is shown. In coal-ﬁred stations, coal is conveyed to a mill and crushed into ﬁne powder, thatis pulverized. The pulverized fuel is blown into the boiler where it mixes with asupply of air for combustion. The exhaust steam from the low pressure (L.P.) tur-bine is cooled to form condensate by the passage through the condenser of largequantities of sea- or river-water. Cooling towers are used where the station islocated inland or if there is concern over the environmental effects of raising thetemperature of the sea- or river-water. Despite continual advances in the design of boilers and in the development ofimproved materials, the nature of the steam cycle is such that vast quantities ofheat are lost in the condensate cooling system and to the atmosphere. Advancesin design and materials in the last few years have increased the thermal

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6 Electric Power Systems, Fifth Edition Stack Boiler Cooling Steam tower Turbine Gener- Transmission ator systemCoal Burner Condenser Precipitator Boiler feed High-voltage (dust pump transformerPulverising Pre-heated Forced collector) mill air draft fanFigure 1.2 Schematic view of coal ﬁred generating stationFrom reheater IP 245oC 360 kPa LP Generator566oC 3919kPa 3.6 x 106J/kg LP LP 500 MWFrom superheater566oC 16 MPa 3.5 x106J/kg To reheater HP 366oC 4231 kPa LP Condenser Extraction 3.1 x 106J/kg heater pump 1st stage Main feed Dearator pump turbine Main boiler 31oC GlandTo boiler feed pump 1.3 X 105J/kg steam252oC vent1.1 x 106J/kg 141oC Drain condenser 6.0 X 10 5J/kg cooler HP 110 C LP Extraction pump Generator feed heaters feed heaters 2nd and 3rd coolers stageFigure 1.3 Energy ﬂow diagram for a 500 MW turbine generator (Figure adapted fromElectrical Review)

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Introduction 7efﬁciencies of new coal stations to approaching 40%. If a use can be found for theremaining 60% of energy rejected as heat, fairly close to the power station,forming a Combined Heat and Power (or Co-generation) system then this isclearly desirable.1.4.2 Energy Conversion Using WaterPerhaps the oldest form of energy conversion is by the use of water power. In ahydroelectric station the energy is obtained free of cost. This attractive feature hasalways been somewhat offset by the very high capital cost of construction, especiallyof the civil engineering works. Unfortunately, the geographical conditions necessaryfor hydro-generation are not commonly found, especially in Britain. In most devel-oped countries, all the suitable hydroelectric sites are already fully utilized. Therestill exists great hydroelectric potential in many developing countries but largehydro schemes, particularly those with large reservoirs, have a signiﬁcant impacton the environment and the local population. The difference in height between the upper reservoir and the level of the turbinesor outﬂow is known as the head. The water falling through this head gains energywhich it then imparts to the turbine blades. Impulse turbines use a jet of water atatmospheric pressure while in reaction turbines the pressure drops across the run-ner imparts signiﬁcant energy. A schematic diagram of a hydro generation scheme is shown in Figure 1.4.Normal max. 120t intakereservoir gantry cranewater levelEl. 451m Transmission linesIntake gate Power station Two,190t Intake 80 MW travelling cranes penstock Generator Normal max. tailwater Draft El. 411m tube 84 MW turbine TailraceFigure 1.4 Schematic view of a hydro generator (Figure adapted from Engineering)

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8 Electric Power Systems, Fifth Edition 1.0Efficiency p.u. 0.9 Pelton Kaplan Francis 0.8 0.7 0.6 0.2 0.4 0.6 0.8 1.0 1.2 0 p.u. of full loadFigure 1.5 Typical efﬁciency curves of hydraulic turbines (1 per unit (p.u.). ¼ 100%) Particular types of turbine are associated with the various heights or heads ofwater level above the turbines. These are:1. Pelton: This is used for heads of 150–1500 m and consists of a bucket wheel rotor with water jets from adjustable ﬂow nozzles.2. Francis: This is used for heads of 50–500 m with the water ﬂow within the turbine following a spiral path.3. Kaplan: This is used for run-of-river stations with heads of up to 60 m. This type has an axial-ﬂow rotor with variable-pitch blades. Typical efﬁciency curves for each type of turbine are shown in Figure 1.5.Hydroelectric plant has the ability to start up quickly and the advantage that noenergy losses are incurred when at a standstill. It has great advantages, therefore,for power generation because of this ability to meet peak loads at minimum operat-ing cost, working in conjunction with thermal stations – see Figure 1.1(d). By usingremote control of the hydro sets, the time from the instruction to start up to theactual connection to the power network can be as short as 3 minutes. The power available from a hydro scheme is given by P ¼ rgQH ½WwhereQ ¼ ﬂow rate (m3/s) through the turbine;r ¼ density of water (1000 kg/m3);

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Introduction 9g ¼ acceleration due to gravity (9.81 m/s2);H ¼ head, that is height of upper water level above the lower (m).Substituting, P ¼ 9:81QH ½kW1.4.3 Gas TurbinesWith the increasing availability of natural gas (methane) and its low emissionsand competitive price, prime movers based on the gas turbine cycle are being usedincreasingly. This thermodynamic cycle involves burning the fuel in the compressedworking ﬂuid (air) and is used in aircraft with kerosene as the fuel and for electricitygeneration with natural gas (methane). Because of the high temperatures obtained,the efﬁciency of a gas turbine is comparable to that of a steam turbine, with the addi-tional advantage that there is still sufﬁcient heat in the gas-turbine exhaust to raisesteam in a conventional boiler to drive a steam turbine coupled to another electricitygenerator. This is known as a combined-cycle gas-turbine (CCGT) plant, a schematiclayout of which is shown in Figure 1.6. Combined efﬁciencies of new CCGT genera-tors now approach 60%. The advantages of CCGT plant are the high efﬁciency possible with large unitsand, for smaller units, the fast start up and shut down (2–3 min for the gas turbine,20 min for the steam turbine), the ﬂexibility possible for load following, the compar-ative speed of installation because of its modular nature and factory-supplied units, Grid To the transformer gridGenerator Heat Natural exchangergas feed StackGenerator Gas turbine Hot gases Steam Cooling Tower Steam turbine Water Condenser Water Pump Water spray steam Pond PumpFigure 1.6 Schematic diagram of a combined-cycle gas-turbine power station

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10 Electric Power Systems, Fifth Editionand its ability to run on light oil (from local storage tanks) if the gas supply is inter-rupted. Modern installations are fully automated and require only a few operatorsto maintain 24 hour running or to supply peak load, if needed.1.4.4 Nuclear PowerEnergy is obtained from the ﬁssion reaction which involves the splitting of thenuclei of uranium atoms. Compared with chemical reactions, very large amounts ofenergy are released per atomic event. Uranium metal extracted from the base oreconsists mainly of two isotopes, 238U (99.3% by weight) and 235U (0.7%). Only 235Uis ﬁssile, that is when struck by slow-moving neutrons its nucleus splits into twosubstantial fragments plus several neutrons and 3 Â 10À11 J of kinetic energy. Thefast moving fragments hit surrounding atoms producing heat before coming to rest.The neutrons travel further, hitting atoms and producing further ﬁssions. Hence thenumber of neutrons increases, causing, under the correct conditions, a chainreaction. In conventional reactors the core or moderator slows down the movingneutrons to achieve more effective splitting of the nuclei. Fuels used in reactors have some component of 235U. Natural uranium is some-times used although the energy density is considerably less than for enriched ura-nium. The basic reactor consists of the fuel in the form of rods or pellets situated inan environment (moderator) which will slow down the neutrons and ﬁssion prod-ucts and in which the heat is evolved. The moderator can be light or heavy water orgraphite. Also situated in the moderator are movable rods which absorb neutronsand hence exert control over the ﬁssion process. In some reactors the cooling ﬂuid ispumped through channels to absorb the heat, which is then transferred to a second-ary loop in which steam is produced for the turbine. In water reactors the moderatoritself forms the heat-exchange ﬂuid. A number of versions of the reactor have been used with different coolantsand types of ﬁssile fuel. In Britain the ﬁrst generation of nuclear power stationsused Magnox reactors in which natural uranium in the form of metal rods wasenclosed in magnesium-alloy cans. The fuel cans were placed in a structure orcore of pure graphite made up of bricks (called the moderator). This graphite coreslowed down the neutrons to the correct range of velocities in order to provide themaximum number of collisions. The ﬁssion process was controlled by the insertionof control rods made of neutron-absorbing material; the number and position ofthese rods controlled the heat output of the reactor. Heat was removed from thegraphite via carbon dioxide gas pumped through vertical ducts in the core. Thisheat was then transferred to water to form steam via a heat exchanger. Once thesteam had passed through the high-pressure turbine it was returned to the heatexchanger for reheating, as in a coal- or oil-ﬁred boiler. A reactor similar to the Magnox is the advanced gas-cooled reactor (AGR) whichis still in use in Britain but now coming towards the end of its service life. Areinforced-concrete, steel-lined pressure vessel contains the reactor and heatexchanger. Enriched uranium dioxide fuel in pellet form, encased in stainless steelcans, is used; a number of cans are ﬁtted into steel ﬁtments within a graphite tube to

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Introduction 11 Control Pressurizer Steam rods generatorREACTOR TURBINE GENERATOR Condenser Primary controlled Feedwater Condensate leakage pump heater pump Figure 1.7 Schematic diagram of a pressurized-water reactor (PWR)form a cylindrical fuel element which is placed in a vertical channel in the core.Depending on reactor station up to eight fuel elements are held in place one abovethe other by a tie bar. Carbon dioxide gas, at a higher pressure than in the Magnoxtype, removes the heat. The control rods are made of boron steel. Spent fuel ele-ments when removed from the core are stored in a special chamber and loweredinto a pond of water where they remain until the level of radioactivity has decreasedsufﬁciently for them to be removed from the station and disassembled. In the USA and many other countries pressurized-water and boiling-waterreactors are used. In the pressurized-water type the water is pumped through thereactor and acts as a coolant and moderator, the water being heated to 315 C ataround 150 bar pressure. At this temperature and pressure the water leaves thereactor at below boiling point to a heat exchanger where a second hydraulic circuitfeeds steam to the turbine. The fuel is in the form of pellets of uranium dioxide inbundles of zirconium alloy. The boiling-water reactor was developed later than the pressurized-watertype. Inside the reactor, heat is transferred to boiling water at a pressure of 75 bar(1100 p.s.i.). Schematic diagrams of these reactors are shown in Figures 1.7 and 1.8.The ratio of pressurized-water reactors to boiling-water reactors throughout theworld is around 60/40%. Both pressurized- and boiling-water reactors use light water.1 The practicalpressure limit for the pressurized-water reactor is about 160 bar (2300 p.s.i.), whichlimits its efﬁciency to about 30%. However, the design is relatively straightforwardand experience has shown this type of reactor to be stable and dependable. In theboiling-water reactor the efﬁciency of heat removal is improved by use of thelatent heat of evaporation. The steam produced ﬂows directly to the turbine, caus-ing possible problems of radioactivity in the turbine. The fuel for both light-waterreactors is uranium enriched to 3–4% 235U. Boiling-water reactors are probably thecheapest to construct; however, they have a more complicated fuel make up withdifferent enrichment levels within each pin. The steam produced is saturated andrequires wet-steam turbines. A further type of water reactor is the heavy-water1 Light water refers to conventional H2O while heavy water describes deuterium oxide (D2O).

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12 Electric Power Systems, Fifth EditionREACTORCore Feed pump TURBINE GENERATOR CondenserControl Feedwater heater Demineralizerrods Figure 1.8 Schematic diagram of a boiling-water reactor (BWR)CANDU type developed by Canada. Its operation and construction are similar tothe light-water variety but this design uses naturally occurring, un-enriched orslightly enriched uranium. Concerns over the availability of future supplies of uranium led to the construc-tion of a number of prototype breeder reactors. In addition to heat, these reactorsproduce signiﬁcant new ﬁssile material. However, their cost, together with the tech-nical and environmental challenges of breeder reactors, led to most of these pro-grammes being abandoned and it is now generally considered that supplies ofuranium are adequate for the foreseeable future. Over the past years there has been considerable controversy regarding the safetyof reactors and the management of nuclear waste. Experience is still relatively smalland human error is always a possibility, such as happened at Three Mile Island in1979 and Chernobyl in 1986 or a natural event such as the earthquake and tsunamiin Fukishima in 2011. However, neglecting these incidents, the safety record ofpower reactors has been good and now a number of countries (including Britain)are starting to construct new nuclear generating stations using Light Water Reactors.The decommissioning of nuclear power stations and the long term disposal of spentfuel remains controversial.1.5 Renewable Energy SourcesThere is considerable international effort put into the development of renewableenergy sources. Many of these energy sources come from the sun, for examplewind, waves, tides and, of course, solar energy itself. The average peak solar energyreceived on the earth’s surface is about 600 W/m2, but the actual value, of course,varies considerably with time of day and cloud conditions.1.5.1 Solar Energy–Thermal ConversionThere is increasing interest in the use of solar energy for generating electricitythrough thermal energy conversion. In large-scale (central station) installations thesun’s rays are concentrated by lenses or mirrors. Both require accurately curved sur-faces and steering mechanisms to follow the motion of the sun. Concentrators may

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Introduction 13 gridparabolic hot tankcollectors turbine storage generator condenser Cooling tower cold tank feedwater pump Figure 1.9 Solar thermal generatorbe designed to follow the sun’s seasonal movement, or additionally to track the sunthroughout the day. The former is less expensive and concentration of the sun up to30 times has been obtained. However, in the French solar furnace in the Pyrenees,two-axis mirrors were used and a concentration of 16 000 was achieved. The reﬂec-tors concentrated the rays on to a single receiver (boiler), hence raising steam. An alternative to this scheme (with lower temperatures) is the use of many indi-vidual parabolic trough absorbers tracking the sun in one direction only (Figure 1.9),the thermal energy being transferred by a ﬂuid to a central boiler. In the arid regionsof the world where direct solar radiation is strong and hence solar thermal genera-tion effective the limited supply of water for the steam cycle and for cooling canbe an important consideration. In solar thermal schemes, heat energy storage can beused to mitigate the ﬂuctuating nature of the sun’s energy.1.5.2 Solar Energy-Photovoltaic ConversionPhotovoltaic conversion occurs in a thin layer of suitable material, typically silicon,when hole-electron pairs are created by incident solar photons and the separation ofthese holes and electrons at a discontinuity in electrochemical potential creates apotential difference. Whereas theoretical efﬁciencies are about 25%, practical valuesare lower. Single-crystal silicon solar cells have been constructed with efﬁciencies ofthe complete module approaching 20%. The cost of fabricating and interconnectingcells is high. Polycrystalline silicon ﬁlms having large-area grains with efﬁciencies ofover 16% have been made. Although photovoltaic devices do not pollute theyoccupy large areas if MWs of output are required. It has been estimated that to pro-duce 1012 kWh per year (about 65% of the 1970 US generation output) the necessarycells would occupy about 0.1% of the US land area (highways occupied 1.5% in

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14 Electric Power Systems, Fifth Edition1975), assuming an efﬁciency of 10% and a daily insolation of 4 kWh/m2. Auto-mated cell production can now produce cells at less than US $3 per peak watt.1.5.3 Wind GeneratorsHorizontal axis wind turbine generators each rated at up to 5 MW mounted on90–100 m high towers are now commercially available. The power in the wind is given by Pw ¼ 1 rAU3 ½W 2While the power developed by the aerodynamic rotor is P ¼ CpPw ¼ Cp 1 rAU3 ½W 2wherer ¼ density of air (1.25 kg/m3);U ¼ wind velocity (m/s);A ¼ swept area of rotor (m2).Cp ¼ power coefﬁcient of the rotor The operation of a wind turbine depends upon the wind speed and is shown inFigure 1.10. P PrPower Uc Ur Uf U (Cut in) (Rated speed) (Max speed) Wind speed Figure 1.10 Wind turbine power curve

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Introduction 15 At low wind speeds, there is insufﬁcient energy to operate the turbine and nopower is produced. At the cut-in Uc speed, between 3 and 5 m/s, power startsto be generated until rated power Pr is produced at rated wind speed Ur. Athigher wind speeds, the turbine is controlled, usually by altering the bladepitch angle, to give rated output up to a maximum wind speed Uf. After thisthe blades are ‘furled’ and the unit is shut down to avoid excessive wind load-ing. Typically, wind turbines with rotors of 80 m diameter, rotate at 15–20 rpm,and are geared up to a generator speed of around 1000 r.p.m. All modern largewind turbines operate at variable speed using power electronic converters toconnect the generator to the 50/60 Hz electrical network. This is in order toreduce mechanical loads and to allow the aerodynamic rotor to run at its mosteffective speed.Example 1.1Calculate the number of wind generators required to produce the equivalent energyof a 600 MW CCGT operating at 80% load factor. Assume the average wind speed is8 m/s, rotor diameter is 80 m, and conversion efﬁciency (coefﬁcient of performance, Cp)is 0.45.CalculationPower in the wind: Pwind ¼ 1 rAU3 2 ¼ 1 Â 1:25 Â p Â 402 Â 83 ¼ 1:6 MW 2Power in generator: Pgenerator ¼ 1:6 Â 0:45 ¼ 724 kW Number of turbines required ¼ 600 Â 0.8/0.724 ¼ 663. A regular spacing of turbines at 5 times rotor diameter (400 m), gives 6.25 turbines/km2.Thus a total area of 106 km2 is required. From this calculation, it is apparent that wind generators spread over a wide area(e.g. 11 km Â 10 km) would be required although the ground beneath them could beused for grazing. The saving in CO2 emissions would be approximately:Daily electrical energy generated ¼ 600 Â 0:8 Â 24 ¼ 11:5 GWh.CO2 emissions saved (see Table 1.1) ¼ 11:5 Â 405 ¼ 4657 tonnes=day1.5.4 BiofuelsBiofuels are derived from vegetable matter produced by agriculture or forestryoperations or from waste materials collected from industry, commerce and residen-tial households. As an energy resource, biomass used as a source of heat by burningwood, dung, and so on, in developing countries is very important and contributes

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16 Electric Power Systems, Fifth Editionabout 14% of the world’s energy requirements. Biomass can be used to produce elec-tricity in two ways:1. by burning in a furnace to produce steam to drive turbines; or2. by fermentation in landﬁll sites or in special anaerobic tanks, both of which pro- duce a methane-rich gas which can fuel a spark ignition engine or gas turbine. It can also be co-ﬁred with coal in large steam power stations. If crops are cultivated for combustion, either as a primary source of heat or as aby-product of some other operation; they can be considered as CO2 neutral, in thattheir growing cycle absorbs as much CO2 as is produced by their combustion. Inindustrialized countries, biomass has the potential to produce up to 5% of electricityrequirements if all possible forms are exploited, including household and industrialwaste, sewage sludge (for digestion), agricultural waste (chicken litter, straw, sugarcane, and so on). The use of good farmland to grow energy crops is controversial asit obviously reduces the area of land available to grow food.1.5.5 Geothermal EnergyIn most parts of the world the vast amount of heat in the earth’s interior is too deepto be tapped. In some areas, however, hot springs or geysers and molten lavastreams are close enough to the surface to be used. Thermal energy from hot springshas been used for many years for producing electricity, starting in 1904 in Italy. Inthe USA the major geothermal power plants are located in northern California on anatural steam ﬁeld called the Geysers. Steam from a number of wells is passedthrough turbines. The present utilization is about 900 MW and the total estimatedcapacity is about 2000 MW. Because of the lower pressure and temperatures the efﬁ-ciency is less than with fossil-fuelled plants, but the capital costs are less and, ofcourse, the fuel is free. New Zealand and Iceland also exploit their geothermalenergy resources.1.5.6 Other Renewable Resources1.5.6.1 TidesAn effective method of utilizing the tides is to allow the incoming tide to ﬂow into abasin, thus operating a set of turbines, and then at low tide to release the storedwater, again operating the turbines. If the tidal range from high to low water is h(m) and the area of water enclosed in the basin is A (m2), then the energy in the fullbasin with the tide outside at its lowest level is: ZhE ¼ rgA xdx 0¼ 1 rgh2A ½J 2

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Introduction 17Table 1.3 Sites that have been studied for tidal range generationSite Tidal Range(m) Area (km2) Generators (MW)Passamaquoddy Bay (N. America) 5.5 262 1800Minas-Cohequid (N. America) 10.7 777 19 900San Jose (S. America) 5.9 750Severn (U.K.) 9.8 700 5870 8000 The maximum total energy for both ﬂows is therefore twice this value, and themaximum average power is pgAh2/T, where T is the period of tidal cycle, normally12 h 44 min. In practice not all this energy can be utilized. The number of siteswith good potential for tidal range generation is small. Typical examples of thosewhich have been studied are listed in Table 1.3 together with the size of generat-ing plant considered. A 200 MW installation using tidal ﬂow has been constructed on the La RanceEstuary in northern France, where the tidal height range is 9.2 m (30 ft) andthe tidal ﬂow is estimated at 18 000 m3/s. Proposals for a 8000 MW tidal barragein the Severn Estuary (UK) were ﬁrst discussed in the nineteenth century and arestill awaiting funding. The utilization of the energy in tidal ﬂows has long been the subject of atten-tion and now a number of prototype devices are undergoing trials. In someaspects, these resemble underwater wind turbines, Figure 1.11. The technical andeconomic difﬁculties are considerable and there are only a limited number oflocations where such schemes are feasible.1.5.6.2 Wave PowerThe energy content of sea waves is very high. The Atlantic waves along the north-west coast of Britain have an average energy value of 80 kW/m of wave crest length.The energy is obviously very variable, ranging from greater than 1 MW/m for 1% ofthe year to near zero for a further 1%. Over several hundreds of kilometres a vastsource of energy is available. The sea motion can be converted into mechanical energy in several ways with anumber of innovative solutions being trialled, Figure 1.12. An essential attribute ofany wave power device is its survivability against the extreme loads encounteredduring storms.1.6 Energy StorageThe tremendous difﬁculty in storing electricity in any large quantity has shaped thearchitecture of power systems as they stand today. Various options exist for thelarge-scale storage of energy to ease operation and affect overall economies. How-ever, energy storage of any kind is expensive and incurs signiﬁcant power losses.Care must be taken in its economic evaluation.

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18 Electric Power Systems, Fifth Edition The options available are as follows: pumped storage, compressed air, heat,hydrogen gas, secondary batteries, ﬂywheels and superconducting coils.1.6.1 Pumped StorageVery rapid changes in load may occur (for example 1300 MW/min at the end ofsome programmes on British TV) or the outage of lines or generators. An instanta-neous loss of 1320 MW of generation (two 660 MW generating units) is consideredwhen planning the operation of the Great Britain system. Hence a considerableamount of conventional steam plant must operate partially loaded to respond tothese events. This is very expensive because there is a ﬁxed heat loss for a steamturbogenerator regardless of output, and the efﬁciency of a thermal generating unitis reduced at part load. Therefore a signiﬁcant amount of energy storage capable ofinstantaneous use would be an effective method of meeting such loadings, and byfar the most important method to date is that of pumped storage. Tidal Turbine Sea level Current Figure 1.11 Tidal stream energy (Figure adapted from Marine Current Turbines)

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Introduction 19 A pumped storage scheme consists of an upper and a lower reservoir and turbine-generators which can be used as both turbines and pumps. The upper reservoir typ-ically has sufﬁcient storage for 4–6 hours of full-load generation. The sequence of operation is as follows. During times of peak load on the powersystem the turbines are driven by water from the upper reservoir and the electricalmachines generate in the normal manner. During the night, when only base loadstations are in operation and electricity is being produced at its cheapest, the waterin the lower reservoir is pumped back into the higher one ready for the next day’speak load. At these times of low network load, each generator changes to synchro-nous motor action and, being supplied from the general power network, drives itsturbine which now acts as a pump. Typical operating efﬁciencies attained are: Motor and generator 96% Pump and turbine 77% Pipeline and tunnel 97% Transmission 95%giving an overall efﬁciency of 68%. A further advantage is that the synchronousmachines can be easily used as synchronous compensators to control reactive powerif required. A large pumped hydro scheme in Britain uses six 330 MVA pump-turbine(Francis-type reversible) generator-motor units generating at 18 kV. The ﬂow ofwater and hence power output is controlled by guide vanes associated with the tur-bine. The maximum pumping power is 1830 MW. The machines are 92.5% efﬁcientas turbines and 91.7% efﬁcient as pumps giving an exceptionally high round tripefﬁcient of 85%. The operating speed of the 12-pole electrical machines is 500 r.p.m.Such a plant can be used to provide ﬁne frequency control for the whole British sys-tem. The machines will be expected to start and stop about 40 times a day as well asFigure 1.12 Wave power generation (Figure adapted from Pelamis)

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20 Electric Power Systems, Fifth Edition Comp. To electricity networkAir Turbine Fuel Air storeFigure 1.13 Storage using compressed air in conjunction with a gas turbine generatorprovide frequency response in the event of a sudden load pick up or tripping ofother generators.1.6.2 Compressed-Air StorageAir is pumped into large receptacles (e.g. underground caverns or old mines) atnight and used to drive gas turbines for peak, day loads. The energy stored isequal to the product of the air pressure and volume. The compressed air allowsfuel to be burnt in the gas turbines at twice the normal efﬁciency. The generalscheme is illustrated in Figure 1.13. A German utility has installed a 290 MWscheme. In one discharge/charge cycle it generated 580 MWh of on-peak electricityand consumed 930 MWh of fuel plus 480 MWh of off-peak electricity. A similarplant has been installed in the USA. One disadvantage of these schemes is thatmuch of the input energy to the compressed air manifests itself as heat and iswasted. Heat could be retained after compression, but there would be possiblecomplications with the store walls rising to a temperature of 450 C at 20 bar pres-sure. A solution would be to have a separate heat store that could comprise stacksof stones or pebbles which store heat cheaply and effectively. This would enablemore air to be stored because it would now be cool. At 100 bar pressure, approxi-mately 30 m3 of air is stored per MWh output.1.6.3 Secondary BatteriesAlthough demonstrated in a number of pilot projects (for example, a 3 MW bat-tery storage plant was installed in Berlin for frequency control in emergenciesand a 35 MW battery system is used to smooth the output of a wind farm inJapan) the large-scale use of battery storage remains expensive and the key areawhere the use of secondary batteries is likely to have impact is in electric vehi-cles. The popular lead-acid cell, although reasonable in price, has a low energydensity (15 Wh/kg). Nickel-cadmium cells are better (40 Wh/kg) but more

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Introduction 21 Water + Hydrogen Oxygen Electrodes (porous) Figure 1.14 Hydrogen-oxygen fuel cellexpensive. Still under intensive development and demonstration is the sodium-sulphur battery (200 Wh/kg), which has a solid electrolyte and liquid electrodesand operates at a temperature of 300 C. Modern electric vehicles use Lithium ionbatteries (100–200 Wh/kg) but these remain expensive. Other combinations ofmaterials are under active development in attempts to increase output and stor-age per unit weight and cost.1.6.4 Fuel CellsA fuel cell converts chemical energy to electrical energy by electrochemicalreactions. Fuel is continuously supplied to one electrode and an oxidant (usuallyoxygen) to the other electrode. Figure 1.14 shows a simple hydrogen-oxygen fuelcell, in which hydrogen gas diffuses through a porous metal electrode (nickel). Acatalyst in the electrode allows the absorption of H2 on the electrode surface ashydrogen ions which react with the hydroxyl ions in the electrolyte to form water(2H2 þ O2 ! 2H2O). A theoretical e.m.f. of 1.2 V at 25 C is obtained. Other fuels foruse with oxygen are carbon monoxide (1.33 V at 25 C), methanol (1.21 V at 25 C),and methane (1.05 V at 25 C). In practical cells, conversion efﬁciencies of 80% havebeen attained. A major use of the fuel cell could be in conjunction with a futurehydrogen energy system. Intensive research and development is still proceeding on various types of fuel cell– the most successful to date for power generation being the phosphoric fuel cell. Ademonstration unit used methane as the input fuel and operated at about 200–300 Cto produce 200 kW of electrical power plus 200 kW of heat energy, with overall efﬁ-ciency of around 80%. Compared with other forms of energy conversion, fuel cellshave the potential of being up to 20% more efﬁcient. Much attention is now beinggiven to the high-temperature molten carbonate cell which has a high efﬁciency.1.6.5 Hydrogen Energy SystemsThe transmission capacity of a pipe carrying natural gas (methane) is high com-pared with electrical links, the installed cost being about one tenth of an equivalent

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22 Electric Power Systems, Fifth Editioncapacity H.V. overhead line. For long transmission distances the pressure drop iscompensated by booster compressor stations. A typical gas system uses a pipe ofinternal diameter 0.9 m and, with natural gas, a power transfer of 12 GW is possibleat a pressure of 68 bar and a velocity of 7 m/s. A 1 m diameter pipe carryinghydrogen gas can transmit 8 GW of power, equivalent to four 400 kV, three phasetransmission lines. The major advantage of hydrogen is, of course, that it can be stored; the majordisadvantage is that it must be produced for example, from water by electrolysis.Very large electrolysers can attain efﬁciencies of about 60%. This, coupled with theefﬁciency of electricity production from a nuclear plant, gives an overall efﬁciency ofhydrogen production of about 21%. Alternative methods of production are underlaboratory development, for example, use of heat from nuclear stations to ‘crack’water and so release hydrogen; however, temperatures of 3000 C are required.1.6.6 Superconducting Magnetic Energy Stores (SMES)Continuing development of high-temperature superconductors, where the transi-tion temperature can be around 60–80 K (K is degrees Kelvin where 0 K is absolutezero and 273 K is 0 C) has led to the possibility of storing energy in the magneticﬁeld produced by circulating a large current (over 100 kA) in an inductance. For acoil of inductance L in air, the stored energy is given byE ¼ 1 LI2 ½ J 2 A big advantage of the high-temperature superconductor is that cooling by liquidnitrogen can be used, which is far cheaper than using helium to reach temperaturescloser to absolute zero. Initially, it is expected that commercial units will be used toprovide an uninterruptible supply for sensitive loads to guard against voltage sagsor to provide continuity whilst emergency generators are started. Another use intransmission networks would be to provide fast response for enhanced transient sta-bility and improved power quality.1.6.7 FlywheelsThe most compact energy store known is that of utilizing high-speed ﬂywheels.Such devices coupled to an electrical generator/motor have been employed inbuses on an experimental basis and also in special industrial applications. Forpower systems, very large ﬂywheels constructed of composite high-tensile resist-ing materials have been proposed, but their cost and maintenance problems haveso far ruled them out of economic contention compared with alternative forms ofenergy supply.1.6.8 SupercapacitorsThe interface between an anode and cathode immersed in an electrolyte has a veryhigh permittivity. This property can be exploited in a capacitor to produce a 25 V

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Introduction 23capsule with a capacitance of 0.1 F. Many units in series and parallel would havethe capability of storing many MWh of energy, which can be quickly released fortransient control purposes. To date, higher voltage forms of a commercially use-ful device have not got beyond their employment for pulse or actuatorapplications.1.7 Environmental Aspects of Electrical EnergyIncreasingly, environmental considerations inﬂuence the development of energyresources, especially those involving electricity production, transmission, and distri-bution. Conversion of one form of energy to another produces unwanted side effectsand, often, pollutants which need to be controlled and disposed of. In addition,safety and health are subject to increasing legislation by national and internationalbodies, thereby requiring all engineers to be aware of the laws and regulations gov-erning the practice of their profession. It must be appreciated that the extraction of fossil fuels from the earth is notonly a hazardous business but also, nowadays, one controlled through licensingby governments and state authorities. Hydro plants require careful study andinvestigation through modelling, widespread surveys, and environmental impactstatements to gain acceptance. Large installations of all kinds require, oftenlengthy, planning enquiries which are both time consuming and expensive,thereby delaying the start up of energy extraction and production. As a conse-quence, methods of producing electrical energy which avoid or reduce theenquiry process are to be favoured over those needing considerable consultationbefore receiving the go ahead. This is likely to favour small-scale projects or theredevelopment of existing sites where industry or production facilities arealready operating. In recent years, considerable emphasis has been placed on ‘sustainable devel-opment’, by which is meant the use of technologies that do not harm the environ-ment, particularly in the long term. It also implies that anything we do now toaffect the environment should be recoverable by future generations. Irreversibledamage, for example, damage to the ozone layer or increase in CO2 in the atmo-sphere, should be avoided.1.7.1 Global Emissions from Fossil Fuelled Power StationsIt is generally accepted that the burning of fossil fuels and the subsequent emis-sion of greenhouse gases, particularly CO2, is leading to climate change andpotentially catastrophic increase in the earth’s temperature. Hence concern overthe emission of greenhouse gases is a key element of energy policy and, inEurope, is recognized through the EU Emissions Trading Scheme which requiresmajor emitters of CO2, such as power stations, to purchase permits to emit CO2.This has the effect of making high carbon generation (particularly from coal)increasingly expensive.

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24 Electric Power Systems, Fifth Edition1.7.2 Regional and Local Emissions from Fossil Fuelled Power StationsFossil fuelled power plants produce sulphur oxides, particulate matter, andnitrogen oxides. Of the former, sulphur dioxide accounts for about 95% and is aby-product of the combustion of coal or oil. The sulphur content of coal varies from0.3 to 5%. Coal can only be used for generation in some US states if it is below acertain percentage sulphur. In the eastern USA this has led to the widespread use of coal from western statesbecause of its lower sulphur content or the use of gas as an alternative fuel. Sulphurdioxide forms sulphuric acid (H2SO4) in the air which causes damage to buildingsand vegetation. Sulphate concentrations of 9–10 mg/m3 of air aggravate asthma andlung and heart disease. This level has been frequently exceeded in the past, a notori-ous episode being the London fog of 1952 (caused by domestic coal burning). Itshould be noted that although sulphur does not accumulate in the air it does so inthe soil. Sulphur oxide emission can be controlled by: the use of fuel with less than, say, 1% sulphur; the use of chemical reactions to remove the sulphur, in the form of sulphuric acid, from the combustion products, for example limestone scrubbers, or ﬂuidized bed combustion; removing the sulphur from the coal by gasiﬁcation or ﬂotation processes. European legislation limits the amount of SO2, NOx, and particulate emission, asin the USA. This has led to the retroﬁtting of ﬂue gas desulphurization (FGD) scrub-bers to coal burning plants. Without such equipment coal ﬁred power stations mustbe retired. Emissions of NOx can be controlled by ﬁtting advanced technology burn-ers which can ensure a more complete combustion process, thereby reducing theoxides going up the stack (chimney). Particulate matter, particles in the air, is injurious to the respiratory system, insufﬁcient concentration, and by weakening resistance to infection may well affectthe whole body. Apart from settling on the ground or buildings to produce dirt, afurther effect is the reduction of the solar radiation entering the polluted area.Reported densities (particulate mass in 1 m3 of air) are 10 mg/m3 in rural areas risingto 2000 mg/m3 in polluted areas. The average value in US cities is about 100 mg/m3. About one-half of the oxides of nitrogen in the air in populated areas are due topower plants and originate in high-temperature combustion processes. At levels of25–100 parts per million they can cause acute bronchitis and pneumonia. Increas-ingly, city pollutants are due to cars and lorries and not power plants. A 1000 MW(e) coal plant burns approximately 9000t of coal per day. If this has asulphur content of 3% the amount of SO2 emitted per year is 2 Â 105t. Such a plantproduces the following pollutants per hour (in kg): CO2 8.5 Â 105, CO 0.12 Â 105, sul-phur oxides 0.15 Â 105, nitrogen oxides 3.4 Â 103, and ash. Both SO2 and NOx are reduced considerably by the use of FGD, but at considera-ble cost and reduction in the efﬁciency of the generating unit caused by the power

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Introduction 25used by the scrubber. Gas-ﬁred CCGT plants produce very little NOx or SO2 andtheir CO2 output is about 55% of an equivalent size coal-ﬁred generator. The concentration of pollutants can be reduced by dispersal over a wider area bythe use of high stacks. If, in the stack, a vertical wire is held at a high negative poten-tial relative to the wall, the expelled electrons from the wire are captured by the gasmolecules moving up the stack. Negative ions are formed which accelerate to thewall, collecting particles on the way. When a particle hits the wall the charge is neu-tralized and the particle drops down the stack and is collected. Precipitators haveparticle-removing (by weight) efﬁciencies of up to 99%, but this is misleading asperformance is poor for small particles; of, say, less than 0.1 mm in diameter. Theefﬁciency based on number of particles removed is therefore less. Disposal of theresulting ﬂy-ash is expensive, but the ash can be used for industrial purposes, forexample, building blocks. Unfortunately, the efﬁciency of precipitators is enhancedby reasonable sulphur content in the gases. For a given collecting area the efﬁciencydecreases from 99% with 3% sulphur to 83% with 0.5% sulphur at 150 C. Thisresults in much larger and more expensive precipitator units with low-sulphur coalor the use of fabric ﬁlters in ‘bag houses’ situated before the ﬂue gas enters the stack.1.7.3 Thermal Pollution from Power StationsSteam from the low-pressure turbine is liqueﬁed in the condenser at the lowest pos-sible temperatures to maximize the steam-cycle efﬁciency. Where copious suppliesof water exist the condenser is cooled by ‘once-through’ circulation of sea- or river-water. Where water is more restricted in availability, for example, away from thecoasts, the condensate is circulated in cooling towers in which it is sprayed in noz-zles into a rising volume of air. Some of the water is evaporated, providing cooling.The latent heat of water is 2 Â 106 J/kg compared with a sensible heat of 4200 J/kgper degree C. A disadvantage of such towers is the increase in humidity producedin the local atmosphere. Dry cooling towers in which the water ﬂows through enclosed channels (similarto a car radiator), past which air is blown, avoid local humidity problems, but at amuch higher cost than ‘wet towers’. Cooling towers emit evaporated water to theatmosphere in the order of 75 000 litres/min for a 1000 MW(e) plant. A crucial aspect of once-through cooling in which the water ﬂows directly into thesea or river is the increased temperature of the natural environment due to the largevolume per minute (typically 360 m3/s for a coolant rise of 2.4 C for a 2.4 GWnuclear station) of heated coolant. Because of their lower thermal efﬁciency, nuclearpower stations require more cooling water than fossil-fuelled plants. Extreme caremust be taken to safeguard marine life, although the higher temperatures can beused effectively for marine farming if conditions can be controlled.1.7.4 Electromagnetic Radiation from Overhead Lines, Cables and EquipmentThe biological effects of electromagnetic radiation have been a cause of considerableconcern amongst the general public as to the possible hazards in the home and the

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26 Electric Power Systems, Fifth EditionTable 1.4 Typical electric and magnetic ﬁeld strengths directly under overheadlinesa. Note that the magnetic ﬁeld depends upon the current carriedLine Voltage (kV) Electric Field Magnetic Flux Strength (V/m) Density (mT)400 and 275 tower lines 3000–5000 5–10132 tower line 1000–2000 0.5–233 and 11 wood pole 200 0.2–0.5UK recommendations on exposure limits 9000 360 (to the public) — 40–50Earth’s magnetic ﬁeldData from http://www.emfs.info/. This site is maintained by the National Grid Company, reproducedwith permission.workplace. Proximity of dwellings to overhead lines and even buried cables hasalso led to concerns of possible cancer-inducing effects, with the consequencethat research effort has been undertaken to allay such fears. In general it is consid-ered that the power frequencies used (50 or 60 Hz) are not harmful. The electricﬁeld and magnetic ﬁeld strengths below typical HV transmission lines are given inTable 1.4. Considerable international research and cooperative investigation has now beenproceeding for over 40 years into low-frequency electric and magnetic ﬁeld expo-sure produced by household appliances, video display terminals, and local powerlines. To date there is no ﬁrm evidence that they pose any demonstrable health haz-ards. Epidemiologic ﬁndings of an association between electric and magnetic ﬁeldsand childhood leukaemia or other childhood or adult cancers are inconsistent andinconclusive; the same is likely to be true of birth defects or other reproductiveproblems.1.7.5 Visual and Audible Noise ImpactsThe presence of overhead lines constitutes an environmental problem (perhaps themost obvious one within a power system) on several counts.1. Space is used which could be used for other purposes. The land allocated for the line is known as the right of way (or wayleave in Britain). The area used for this purpose is already very appreciable.2. Lines are considered by many to mar the landscape. This is, of course, a subjective matter, but it cannot be denied that several tower lines converging on a substation or power plant, especially from different directions, is offensive to the eye.3. Radio interference, audible noise, and safety considerations must also be considered. Although most of the above objections could be overcome by the use of under-ground cables, these are not free of drawbacks. The limitation to cable transmittingcurrent because of temperature-rise considerations coupled with high manufactureand installation costs results in the ratio of the cost of transmission underground to

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Introduction 27that for overhead transmission being between 10 and 20 at very high operating volt-ages. With novel cables, such as superconducting, still under development it ishoped to reduce this disadvantage. However, it may be expected that in future alarger proportion of circuits will be placed underground, especially in suburbanareas. In large urban areas circuits are invariably underground, thereby posingincreasing problems as load densities increase. Reduction of radio interference can be achieved in the same way that electromag-netic effects are reduced, but audible noise is a function of the line design. Carefulattention to the tightness of joints, the avoidance of sharp or rough edges, and theuse of earth screen shielding can reduce audible noise to acceptable levels at a dis-tance dependent upon voltage. Safety clearances dependent upon International Standards are, of course,extremely important and must be maintained in adverse weather conditions. Themost important of these is the increased sag due to ice and snow build up in winterand heavy loading of the circuits in summer.1.8 Transmission and Distribution Systems1.8.1 RepresentationModern electricity supply systems are invariably three-phase. The design of trans-mission and distribution networks is such that normal operation is reasonably closeto balanced three-phase working, and often a study of the electrical conditions inone phase is sufﬁcient to give a complete analysis. Equal loading on all three phasesof a network is ensured by allotting, as far as possible, equal domestic loads to eachphase of the low-voltage distribution feeders; industrial loads usually take three-phase supplies. A very useful and simple way of graphically representing a network is the sche-matic or line diagram in which three-phase circuits are represented by single lines.Certain conventions for representing items of plant are used and these are shown inFigure 1.15. A typical line or schematic diagram of a part of a power system is shown inFigure 1.16. In this, the generator is star connected, with the star point connected toearth through a resistance. The nature of the connection of the star point of rotatingmachines and transformers to earth is of vital importance when considering faultswhich produce electrical imbalance in the three phases. The generator feeds twothree-phase circuits (overhead or underground). The line voltage is increased fromthat at the generator terminals by transformers connected as shown. At the end ofthe lines the voltage is reduced for the secondary distribution of power. Two linesare provided to improve the security of the supply that is, if one line develops a faultand has to be switched out the remaining one still delivers power to the receivingend. It is not necessary in straightforward current and voltage calculations to indi-cate the presence of switches on the diagrams, but in some cases, such as stabilitycalculations, marking the location of switches, current transformers, and protectionis very useful.

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28 Electric Power Systems, Fifth Edition Line, cable or busbar (three-phase) M Rotating machine-general Synchronous machine Two-winding transformer Auto transformer Two-winding transformer (alternative symbol) Three-winding transformer Circuit breaker – alternative Isolator Three-phase wye or star connected with the star point solidly connected to an earth or ground electrode Delta connected Current transformerFigure 1.15 Symbols for representing the components of a three-phase power system A short list of terms used to describe power systems follows, with explanations.System: This is used to describe the complete electrical network: generators, circuits, loads, and prime movers.Load: This may be used in a number of ways: to indicate a device or collection of devices which consume electricity; to indicate the power required from a given supply circuit; to indicate the power or current being passed through a line or machine.Busbar: This is an electrical connection of zero impedance (or node) joining several items, such as lines, generators or loads. Often this takes the form of actual bus- bars of copper or aluminium.Earthing (Grounding): This is connection of a conductor or frame of a device to the main body of the earth. This must be done in such a manner that the resistance

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Introduction 29 Lines Busbars Lines Load Figure 1.16 Line diagram of a simple system between the item and the earth is below prescribed limits. This often entails the burying of the large assemblies of conducting rods in the earth and the use of con- nectors of a large cross sectional area. It is usual to earth the neutral point of 3-phase circuits at least once at each voltage level.Fault: This is a malfunctioning of the network, usually due to the short-circuiting of conductors phase-phase or phase-earth.Outage: Removal of a circuit either deliberately or inadvertently.Security of Supply: Provision must be made to ensure continuity of supply to con- sumers, even with certain items of plant out of action. Usually, two circuits in par- allel are used and a system is said to be secure when continuity is assured. This is obviously the item of ﬁrst priority in design and operation.1.8.2 TransmissionTransmission refers to the bulk transfer of power by high-voltage links between cen-tral generation and load centres. Distribution, on the other hand, describes the con-veyance of this power to consumers by means of lower voltage networks. Generators usually produce voltages in the range 11–25 kV, which is increasedby transformers to the main transmission voltage. At substations the connectionsbetween the various components of the system, such as lines and transformers,are made and the switching of these components is carried out. Large amounts ofpower are transmitted from the generating stations to the load-centre substationsat 400 kV and 275 kV in Britain, and at 765, 500 and 345 kV in the USA. The net-work formed by these very high-voltage lines is sometimes referred to as theSupergrid. Most of the large and efﬁcient generating stations feed through trans-formers directly into this network. This grid, in turn, feeds a sub-transmission net-work operating at 132 kV in Britain and 115 kV in the USA. In Britain the lowervoltage networks operate at 33, 11, or 6.6 kV and supply the ﬁnal consumer feedersat 400 V three-phase, giving 230 V between phase and neutral. Other voltages exist

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30 Electric Power Systems, Fifth Edition Large high-efficiency generating station Generator transformers 275 or 400kV (345, 500 or 765kV in USA) (busbars sectionalised) Load To rest Load of system Substation (Grid supply point GSP) LoadSubstationTo loads vialower voltage network Distribution voltage Loads Distributed generation (local generator) Figure 1.17 Part of a typical power systemin isolation in various places, for example the 66 and 22 kV London cable systems.A typical part of a supply network is shown schematically in Figure 1.17. Thepower system is thus made up of networks at various voltages. There exist, ineffect, voltage tiers as represented in Figure 1.18.

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Introduction 31 Large generators – gas, coal, nuclear, hydro 400/275 kV Tie line to other (765 kV, 500/345 kV) systems 132 kV CHP generators or (230/115kV) small local independents Distribution Small distributed generators, e.g. wind, landfill gasFigure 1.18 Schematic diagram of the constituent networks of a supply system. USAvoltages in parentheses Summarizing, transmission networks deliver to wholesale outlets at 132 kV andabove; sub-transmission networks deliver to retail outlets at voltages from 115 or132 kV, and distribution networks deliver to ﬁnal step-down transformers at volt-ages below 132 kV, usually operated as radial systems.1.8.2.1 Reasons for InterconnectionMany generating sets are large and power stations of more than 2000 MW are usedto provide base load power. With CCGT units, their high efﬁciency and cheap long-term gas contracts mean that it is often more economic to use these efﬁcient stationsto full capacity 24 h a day and transmit energy considerable distances than to useless efﬁcient more local stations. The main base load therefore is met by these high-efﬁciency stations which must be interconnected so that they feed into the generalsystem and not into a particular load. To meet sudden increases in load a certain amount of generating capacity, knownas the spinning reserve, is required. This consists of part-loaded generators synchro-nized with the system and ready to supply power instantaneously. If the machinesare stationary a reasonable time is required (especially for steam turbo-alternators)to run up to speed; this can approach 6 h, although small gas turbines can be started

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32 Electric Power Systems, Fifth Editionand loaded in 3 minutes or less. Hydro generators can be even quicker. It is moreeconomic to have certain stations serving only this function than to have each sta-tion carrying its own spinning reserve. The electricity supplies over the entire country are synchronized and a commonfrequency exists: 50 Hz in Europe, 60 Hz in N. America. Interconnection also allows for alternative paths to exist between generators andbulk supply points supplying the distribution systems. This provides security ofsupply should any one path fail.1.8.3 Distribution SystemsDistribution networks differ from transmission networks in several ways, quiteapart from their voltage levels. The number of branches and sources is much higherin distribution networks and the general structure or topology is different. A typicalsystem consists of a step-down (e.g. 132/11 kV) on-load tap-changing transformer ata bulk supply point feeding a number of circuits which can vary in length from afew hundred metres to several kilometres. A series of step-down three-phase trans-formers, for example, 11 kV/433 V in Britain or 4.16 kV/220 V in the USA, arespaced along the route and from these are supplied the consumer three-phase, four-wire networks which give 240 V, or, in the USA, 110 V, single-phase supplies tohouses and similar loads.1.8.3.1 Rural SystemsIn rural systems, loads are relatively small and widely dispersed (5–50 kVA per con-sumer group is usual). In Great Britain a predominantly overhead line system at11 kV, three-phase, with no neutral or single phase for spurs from the main systemis used. Pole-mounted transformers (5–200 kVA) are installed, protected by fuseswhich require manual replacement after operation; hence rapid access is desirableby being situated as close to roads as possible. Essentially, a radial system is sup-plied from one step-down point; distances up to 10–15 miles (16–24 km) are feasiblewith total loads of 500 kVA or so (see Figure 1.19), although in sparsely populatedareas, distances of 50 miles may be fed by 11 kV. Single-phase earth-return systemsoperating at 20 kV are used in some developing countries. Over 80% of faults on overhead distribution systems are transitory due toﬂashover following some natural or man-made cause. This produces unnecessaryfuse-blowing unless auto-reclosers are employed on the main supply; these havebeen used with great success in either single- or three-phase form. The principle,as shown in Figure 1.20, is to open on fault before the fuse has time to operateand to reclose after 1–2 s. If the fault still persists, a second attempt is made toclear, followed by another reclose. Should the fault still not be cleared, therecloser remains closed for a longer period to blow the appropriate protectivefuse (for example, on a spur line or a transformer). If the fault is still not cleared,then the recloser opens and locks out to await manual isolation of the faulty sec-tion. This process requires the careful coordination of recloser operation and

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Introduction 33 Primary transmission line Primary substationsAdditional Reclosers Reclosers Additional feeders feeders Pole - mounted transformers Section Rural point recloser Section point Section point Normally open switch single phase three phase Figure 1.19 Typical rural distribution systemfuse-blowing characteristics where time grading is important. Section switchesoperated by radio or tele-command enable quick resupply routes to be estab-lished following a faulty section isolation. Good earthing at transformer star points is required to prevent overvoltages atconsumers’ premises. Surge protection using diverters or arcing horns is essential inlightning-prone areas.

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34 Electric Power Systems, Fifth EditionOpen Lock outClose 0 2 2.1 5.1 10 12 12.1Timescalesecondsapprox. Figure 1.20 Sequence of operations for a recloser1.8.3.2 Suburban SystemsThese are a development of the rural system into ring mains, with much of thenetwork underground for amenity reasons. The rings are sectionalized so thatsimple protection can be provided. Loads range between 2 and 10 MW/mile2(0.8–4.0 MW/km2). In high-density housing areas, the practice is to run the L.V. mains on either sideof each road, interconnected at junctions by links for sectionalizing (see Figure 1.21).At appropriate points this network is fed by step-down transformers of200–500 kVA rating connected to an H.V. cable or overhead line network.Reinforcement is provided by installing further step-down transformers tappedfrom the H.V. network. It is rare to up-rate L.V. cables as the load grows. Short-circuit levels are fairly low due to the long H.V. feeders from the bulk supplypoints. With the increasing cost of cables and undergrounding works, but withimproved transformer efﬁciency and lower costs due to rationalization and standard-ization, an economic case can be made for reducing the L.V. network and extendingthe H.V. network so that fewer consumers are supplied from each transformer.1.8.3.3 Urban (Town or City) SystemsVery heavy loadings (up to 100 MW/mile2 or 40 MW/km2) are usual, especiallywhere high-rise buildings predominate. Extensive heating and air-conditioningloads as well as many small motors predominate. Fluorescent lighting reduces thepower factor and leads to some waveform distortion, but computer and TV loadsand power electronic motor drives now cause considerable harmonics on all typesof network. Again, a basic L.V. grid, reinforced by extensions to the H.V. network, as required,produces minimum costs overall. The H.V. network is usually in the form of aring main fed from two separate sections of a double busbar substation where10–60 MVA transformers provide main supply from the transmission system(see Figure 1.22). The H.V. network is sectionalized to contain short-circuit levelsand to ease protection grading. A high security of supply is possible by

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Introduction 35 H. V. supply cables (3 ph.) Link boxes at street junctions Step-down transformer L. V. 3 ph. & N network H.V. 3 ph. network Figure 1.21 Principle of suburban distribution systemoverlapping H.V. rings so that the same L.V. grid is fed from several transform-ers supplied over different routes. Failure of one portion of the H.V. system doesnot affect consumers, who are then supported by the L.V. network from adjacentH.V. supplies. In the UK transformers of 500 or 1000 kVA rating are now standard, with one H.V.circuit breaker or high rupturing capacity (HRC) fuse-switch and two isolatorseither side to enable the associated H.V. cable to be isolated manually in the event offailure. The average H.V. feeder length is less than 1 mile and restoration of H.V.supplies is usually obtainable in under 1 h. Problems may arise due to back-feedingof faults on the H.V. system by the L.V. system, and in some instances reverse powerrelay protection is necessary. In new urban developments it is essential to acquire space for transformer cham-bers and cable access before plans are ﬁnalized. High-rise buildings may requiresubstations situated on convenient ﬂoors as well as in the basement. Apart from the supply of new industrial and housing estates or the electriﬁ-cation of towns and villages, a large part of the work of a planning engineer

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36 Electric Power Systems, Fifth Edition From transmission system 4 x 15 MVA transformersCapital 4 x 45 MVA 11 kV duplicatesubstation transformers busbar switchgear To main 33 kV duplicate substation busbar switchgear To main substationMain 4 x 15 MVAsubstation transformers11 kV feeders 11 kV duplicateto open rings busbar switchgear Transformer 500 kVA Key chamber transformer Busbar supplying Transformer distribution 415/240 V network Oil switch To L.V. Service to network Automatic building ocb Cable-end box Link Fuse Circuit breaker Figure 1.22 Typical arrangement of supply to an urban network–British practiceis involved with the up-rating of existing supplies. This requires good loadforecasting over a period of 2–3 years to enable equipment to be ordered andaccess to sites to be established. One method of forecasting is to survey thedemand on transformers by means of a maximum-demand indicator and totreat any transformer which has an average load factor of 70% or above asrequiring up-rating over the next planning period. Another method is to

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Introduction 37analyze consumers’ bills, sectionalized into areas and distributors and, bysurveys and computer analysis, to relate energy consumed to maximumdemand. This method can be particularly useful and quite economic wherecomputerized billing is used. The development of the Smart Grid is leading tomuch greater monitoring being installed on the 400 V network and control onthe 11 kV system. In practice, good planning requires sufﬁcient data on load demands, energygrowth, equipment characteristics, and protection settings. All this information canbe stored and updated from computer ﬁles at periodic intervals and provides thebasis for the installation of adequate equipment to meet credible future demandswithout unnecessary load shedding or dangerous overloading.1.8.4 Typical Power SystemsThroughout the world the general form of power systems follows the same pattern.Voltage levels vary from country to country, the differences originating mainly fromgeographical and historical reasons. Several frequencies have existed, although now only two values – 50 and 60 Hz –remain: 60 Hz is used on the American continent, whilst most of the rest of theworld uses 50 Hz, although Japan still has 50 Hz for the main island and 60 Hz forthe northern islands. The value of frequency is a compromise between higher gen-erator speeds (and hence higher output per unit of machine volume) and the dis-advantage of high system reactance at higher frequencies. Historically, the lowerlimit was set by the need to avoid visual discomfort caused by ﬂicker from incan-descent electric lamps. The distance that a.c. transmission lines can transfer power is limited by the maxi-mum permissible peak voltage between conductor and ground. As voltagesincrease, more and more clearance must be allowed in air to prevent the possibilityof ﬂashover or danger to people or animals on the ground. Unfortunately, the criti-cal ﬂashover voltage increases non-linearly with clearance, such that, with longclearances, proportionally lower peak voltages can be used safely. Figure 1.23 illus-trates this effect. The surge impedance of a line, describedpfoﬃﬃrﬃﬃﬃaﬃﬃﬃ lossless line where the resistanceand conductance are zero, is given by Z0 ¼ L=C, where L is the series inductanceand C is the shunt capacitance per unit length of the line. When a line is terminatedin a resistive load equal to its surge impedance then the reactive power I2XL drawnby the line is balanced by the reactive power generated by the line capacitanceV2=XC. The surge impedance loading is V2=Z0, typical values are shown in Table 1.5. The curve given in Figure 1.24 shows that, in practice, short lines are usuallyloaded above their surge impedance loading (SIL) but, to ensure stability is main-tained, long lines are generally loaded below their SIL. A useful rule-of-thumb toensure stability is to restrict the phase difference between the sending and receivingend voltages of an EHV transmission circuit to 30. Lower voltage circuits will havemuch smaller phase angle differences.

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38 Electric Power Systems, Fifth EditionCritical flashover voltage (kV) 2000 1800 1600 1400 1200 1000 800 600 400 200 246 8 Clearance (m)Figure 1.23 Critical ﬂashover voltage for V-string insulators in a window tower (Figureadapted from Edison Electric Institute) The highest a.c. voltage used for transmission is around 750 kV, although experi-mental lines above 1000 kV have been built. The topology of the power system and the voltage magnitudes used are greatlyinﬂuenced by geography. Very long lines are to be found in North and SouthAmerican nations and Russia. This has resulted in higher transmission voltages,for example, 765 kV, with the possibility of voltages in the range 1000–1500 kV.In some South American countries, for example, Brazil, large hydroelectricresources have been developed, resulting in very long transmission links. Inhighly developed countries the available hydro resources have been utilized anda considerable proportion of new generation is from wind. In geographicallysmaller countries, as exist in Europe, the degree of interconnection is muchhigher, with shorter transmission distances, the upper voltage being about420 kV. Systems are universally a.c. with the use of high-voltage d.c. links for specialistpurposes, for example, very long circuits, submarine cable connections, and back-to-back converters to connect different a.c. areas. The use of d.c. has been limited bythe high cost of the conversion equipment. This requires overhead line lengths of afew hundred kilometres or cables of 30–100 km to enable the reduced circuit costs tooffset the conversion costs.

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Introduction 39 3.5Line load in SIL 3.0 2.5 2.0 1.5 1.0 0.5 0 0 100 200 300 400 500 600 Line length in miles with no series compensationFigure 1.24 Practical transmission-line capability in terms of surge-impedance load-ing (SIL) (Figure adapted from Edison Electric Institute)1.8.4.1 USAThe power system in the USA is based on a comparatively few investor ownedgeneration/transmission utilities responsible for bulk transmission as well as oper-ating and constructing new generation facilities when needed. Delivery is to smallermunicipal or cooperative rural distribution companies or to the distribution arm ofthe transmission utility. The industry is heavily regulated by Federal and StateAgencies and the investor proﬁts are carefully controlled. The loads differ seasonally from one part of the country to another and loaddiversity occurs because of the time zones. Generally, the summer load is thehighest due to extensive use of air-conditioning (Figure 1.1(a) and (b)). Consum-ers are supplied at either 110 V, 60 Hz, single phase for lighting and small-consumption appliances but at 220 V for loads above about 3 kW (usuallycookers, water-heaters and air-conditioners). This type of connection normallyuses a 220 V centre-tapped transformer secondary to provide the 110 V supply.

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40 Electric Power Systems, Fifth Edition1.8.4.2 UKHere, the frequency (as in the whole of Europe) is 50 Hz and the residential supply is230–240 V single phase. The European standard is now 230 V, but with Æ10% toler-ance this voltage can vary between a maximum of 253 V and a minimum of 207 V;equipment and appliance designers must take this allowed variation into account orbeware! Most commercial and industrial loads are supplied at 400–415 V or highervoltages, three phase. The winter load produces the highest peak because of the pre-ponderance of heating appliances, although it is noted that the summer load isgrowing due to increasing use of air-conditioning in commercial and industrialpremises.1.8.4.3 Continental EuropeMany continental countries still mainly have combined generator/transmission util-ities, which cover the whole country and which are overseen by government control.Increasingly these systems are being ‘unbundled’, that is they are being separatedinto different functions (generation, transmission, distribution), each individuallyaccountable such that private investors can enter into the electricity market (seeChapter 12). Interconnection across national boundaries enables electrical energy tobe traded under agreed tariffs, and limited system support is available under dis-turbed or stressed operating conditions. The daily load variation tends to be much ﬂatter than in the UK because of thedominance of industrial loads with the ability to vary demand and because there isless reliance on electricity for heating in private households. Many German andScandinavian cities have CHP plants with hot water distribution mains for heatingpurposes. Transmission voltages are 380–400, 220, and 110 kV with household sup-plies at 220–230 V, often with a three-phase supply taken into the house.1.8.4.4 China and the Paciﬁc RimThe fastest-growing systems are generally found in the Far East, particularly China,Indonesia, the Philippines and Malaysia. Voltages up to 500/750 kV are used fortransmission and 220–240 V are employed for households. Hydroelectric potential isstill quite large (particularly in China) but with gas and oil still being discovered andexploited, CCGT plant developments are underway. Increasing interconnection,including by direct current, is being developed.1.9 Utilization1.9.1 LoadsThe major consumption groups are industrial, residential (domestic) and commer-cial. Industrial consumption accounts for up to 40% of the total in many industrial-ized countries and a signiﬁcant item is the induction motor. The percentage ofelectricity in the total industrial use of energy is expected to continue to increase

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Introduction 41due to greater mechanization and the growth of energy intensive industries, such aschemicals and aluminium. In the USA the following six industries account for over70% of the industrial electricity consumption: metals (25%), chemicals (20%), paperand products (10%), foods (6%), petroleum products (5%) and transportation equip-ment (5%). Over the past 25 years the amount of electricity per unit of industrialoutput has increased annually by 1.5%, but this is dependent on the economic cycle.Increase in consumption of electricity in industrialized countries since about 1980has been no more than 2% per year due largely to the contraction of energy-inten-sive industries (e.g. steel manufacturing) combined with efforts to load manage andto make better use of electricity. Residential loads are largely made up of refrigera-tors, fridge-freezers, freezers, cookers (including microwave ovens), space heating,water heating, lighting, and (increasingly in Europe) air-conditioning. Togetherthese loads in the UK amount to around 40% of total load. The commercial sector comprises ofﬁces, shops, schools, and so on. The consump-tion here is related to personal consumption for services, traditionally a relativelyhigh-growth quantity. In this area, however, conservation of energy measures areparticularly effective and so modify the growth rate. Quantities used in measurement of loads are deﬁned as follows:Maximum Load: The average load over the half hour of maximum output.Load Factor: The units of electricity exported by the generators in a given period divided by the product of the maximum load in this period and the length of the period in hours. The load factor should be high; if it is unity, all the plant is being used over all of the period. It varies with the type of load, being poor for lighting (about 12%) and high for industrial loads (e.g. 100% for pumping stations).Diversity Factor: This is deﬁned as the sum of individual maximum demands of the consumers, divided by the maximum load on the system. This factor mea- sures the diversiﬁcation of the load and is concerned with the installation of sufﬁcient generating and transmission plant. If all the demands occurred simultaneously, that is, unity diversity factor, many more generators would have to be installed. Fortunately, the factor is much higher than unity, espe- cially for domestic loads.Table 1.5 Surge impedance loading and charging MVA for EHV overhead lines. Therange of values at each line voltage is due to variations in line constructionLine Voltage (kV) Surge Impedance Loading (MW) Charging MVAr per 100 miles230 132–138 27–28345 320–390 65–81500 830–910 170–190700 2150 445750 2165 450Data from Edison Electric Institute.

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42 Electric Power Systems, Fifth Edition Case ILoad Load factor = Lmax x 24 = 1 Lmax x 24 Diversity factor = 4 L max A BCD 6 12 18 24 24 h L max D Case II CLoad B Load factor = Lmax x 6 = 0.25 Lmax x 24 Diversity factor = 1 A 6 12 18 24 24 hFigure 1.25 Two extremes of load factor and diversity factor in a system with fourconsumers A high diversity factor could be obtained with four consumers by compellingthem to take load as shown in Case I of Figure 1.25. Although compulsion obviouslycannot be used, encouragement can be provided in the form of tariffs. An example isthe two-part tariff in which the consumer has to pay an amount dependent on themaximum demand required, plus a charge for each unit of energy consumed. Some-times the charge is based on kilovoltamperes instead of power to penalize loads oflow power factor.1.9.1.1 Load ManagementAttempts to modify the shape of the load curve to produce economy of operationhave already been mentioned. These have included tariffs, pumped storage, and theuse of seasonal or daily diversity between interconnected systems. A more directmethod is the control of the load either through tariff structure or direct electricalcontrol of appliances, the latter, say, in the form of remote on/off control of electricwater-heaters where inconvenience to the consumer is least. For many years this hasbeen achieved with domestic time switches, but some schemes use switches radio-controlled from the utility to give greater ﬂexibility. This permits load reductionsalmost instantaneously and defers hot-water and air-conditioning load until aftersystem peaks.

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Introduction 431.9.1.2 Load ForecastingIt is evident that load forecasting is a crucial activity in electricity supply. Forecastsare based on the previous year’s loading for the period in question, updated by fac-tors such as general load increases, major new loads, and weather trends. Bothpower demand (kW) and energy (kWh) forecasts are used, the latter often being themore readily obtained. Demand values may be determined from energy forecasts.Energy trends tend to be less erratic than peak power demands and are consideredbetter growth indicators; however, load factors are also erratic in nature. As weather has a much greater inﬂuence on residential than on industrialdemands it may be preferable to assemble the load forecast in constituent partsto obtain the total. In many cases the seasonal variations in peak demand arecaused by weather-sensitive domestic appliances, for example, heaters and air-conditioning. A knowledge of the increasing use of such appliances is thereforeessential. Several techniques are available for forecasting. These range from simplecurve ﬁtting and extrapolation to stochastic modelling. The many physical factorsaffecting loads, for example, weather, national economic heath, popular TVprogrammes, public holidays, and so on, make forecasting a complex processdemanding experience and high analytical ability using probabilistic techniques.Problems1.1 In the U.S.A. in 1971 the total area of right of ways for H.V. overhead lines was 16 000 km2. Assuming a growth rate for the supply of electricity of 7% per annum calculate what year the whole of the USA will be covered with trans- mission systems (assume area to approximate 4800 Â 1600 km). Justify any assumptions made and discuss critically why the result is meaningless. (Answer: 91.25 years)1.2 The caloriﬁc value of natural gas at atmospheric pressure and temperature is 40 MJ/m3. Calculate the power transfer in a pipe of lm diameter with gas at 60 atm (gauge) ﬂowing at 5 m/s. If hydrogen is transferred at the same velocity and pressure, calculate the power transfer. The caloriﬁc value of hydrogen is 13 MJ/m3 at atmospheric temperature and pressure. (Answer: 9.4 GW, 3.1 GW)1.3 a. An electric car has a steady output of 10 kW over its range of 100 km when running at a steady 40 km/h. The efﬁciency of the car (including batteries) is 65%. At the end of the car’s range the batteries are recharged over a period of 10 h. Calculate the average charging power if the efﬁciency of the battery charger is 90%. b. The caloriﬁc value of gasoline (petrol) is roughly 16 500 kJ/gallon. By assum- ing an average ﬁlling rate at a pump of 10 gallon/minute, estimate the rate of energy transfer on ﬁlling a gasoline-driven car. What range and what cost/km would the same car as (a) above produce if driven by gasoline with

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44 Electric Power Systems, Fifth Edition a 7 gallon tank? (Assume internal combustion engine efﬁciency is 60% and gasoline costs £3 per gallon) (Answer: (a) 4.3 kW; (b) 2.75 MW, 77 km, 27p/km)1.4 The variation of load (P) with time (t) in a power supply system is given by the expression, PðkWÞ ¼ 4000 þ 8t À 0:00091t2 where t is in hours over a total period of one year. This load is supplied by three 10 MW generators and it is advantageous to fully load a machine before connecting the others. Determine: a. the load factor on the system as a whole; b. the total magnitude of installed load if the diversity factor is equal to 3; c. the minimum number of hours each machine is in operation; d. the approximate peak magnitude of installed load capacity to be cut off to enable only two generators to be used. (Answer: (a) 0.73 (b) 65 MW (c) 8760, 7209, 2637 h (d) 4.2 MW)1.5 a. Explain why economic storage of electrical energy would be of great beneﬁt to power systems. b. List the technologies for the storage of electrical energy which are available now and discuss, brieﬂy, their disadvantages. c. Why is hydro power a very useful component in a power system? d. Explain the action of pumped storage and describe its limitations. e. A pumped storage unit has an efﬁciency of 78% when pumping and 82% when generating. If pumping can be scheduled using energy costing 2.0 p/kWh, plot the gross loss/proﬁt in p/kWh when it generates into the system with a marginal cost between 2p and 6p/kWh. f. Explain why out-of-merit generation is sometimes scheduled. (Answer: (e) Overall efﬁciency 64%; Proﬁt max. 2.87p/kWh; Loss max. 1.25 p/kWh) (From Engineering Council Examination, 1996)

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2Basic Concepts12.1 Three-Phase SystemsThe rotor ﬂux of an alternating current generator induces sinusoidal e.m.f.s in theconductors forming the stator winding. In a single-phase machine these stator con-ductors occupy slots over most of the circumference of the stator core. The e.m.f.sthat are induced in the conductors are not in phase and the net winding voltage isless than the arithmetic sum of the individual conductor voltages. If this winding isreplaced by three separate identical windings, as shown in Figure 2.1(a), each occu-pying one-third of the available slots, then the effective contribution of all the con-ductors is greatly increased, yielding a greatly enhanced power capability for agiven machine size. Additional reasons why three phases are invariably used inlarge A.C. power systems are that the use of three phases gives similarly greatereffectiveness in transmission circuits and the three phases ensure that motorsalways run in the same direction, provided the sequence of connection of the phasesis maintained. The three windings of Figure 2.1(a) give voltages displaced in time or phase by120, as indicated in Figure 2.1(b). Because the voltage in the (a) phase reaches itspeak 120 before the (b) phase and 240 before the (c) phase, the order of phase volt-ages reaching their maxima or phase sequence is a-b-c. Most countries use a, b, and cto denote the phases; however, R (Red), Y (Yellow), and B (Blue) has often beenused. It is seen that the algebraic sum of the winding or phase voltages (and currentsif the winding currents are equal) at every instant in time is zero. Hence, if one endof each winding is connected, then the electrical situation is unchanged and thethree return lines can be dispensed with, yielding a three-phase, three-wire system,as shown in Figure 2.2(a). If the currents from the windings are not equal, then it isusual to connect a fourth wire (neutral) to the common connection or neutral point,as shown in Figure 2.2(b).1 Throughout the book, symbols in bold type represent complex (phasor) quantities requiring complexarithmetic. Italic type is used for magnitude (scalar) quantities within the text.Electric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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46 Electric Power Systems, Fifth Editionbω b a a a b c Return wires c c (a) a bc Time (b)Figure 2.1 (a) Synchronous machine with three separate stator windings a, b andc displaced physically by 120. (b) Variation of e.m.f.s developed in the windingswith time (a) (a) Vcn Van Vcn Van Vab Vca(c) (b) (c) (b) Vbc Vbn Vbn Vcn (a) (b)Figure 2.2 (a) Wye or star connection of windings, (b) Wye connection withneutral line

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Basic Concepts 47Vca Vcn Vab c Van Vbc Vca b n Vbn Vbc = (Vbn – Vcn) a Vab (a) (b)Figure 2.3 (a) Phasor diagram for wye connection, (b) Alternative arrangement ofline-to-line voltages. Neutral voltage is at n, geometric centre of equilateral triangle This type of winding connection is called ‘wye’ or ‘star’ and two sets of voltagesexist:1. the winding, phase, or line-to-neutral voltage, that is, Van, Vbn, Vcn; and2. the line-to-line voltages, Vab, Vbc, Vca. The subscripts here are important, Vab, means the voltage of line or terminal a with respect to b and Vba ¼ ÀVab. The corresponding phasor diagram is shown in Figure 2.3(a) and it can be shownthat2 pﬃﬃ pﬃﬃ pﬃﬃ jVabj ¼ jVbcj ¼ jVcaj ¼ 3jVanj ¼ 3jVbnj ¼ 3jVcnj: The phase rotation of a system is very important. Consider the connectionthrough a switch of two voltage sources of equal magnitude and both of rotationa-b-c. When the switch is closed no current ﬂows. If, however, one source is ofreversed rotation (easily obtainpeﬃﬃd by reversing two wires), as shown in Figure 2.4,that is, a-c-b, a large voltage ( 3 Â phase voltage) exists across the switch contactscb0 and bc0, resulting in very large currents if the switch is closed. Also, withreversed phase rotation the rotating magnetic ﬁeld set up by a three-phase windingis reversed in direction and a motor will rotate in the opposite direction, often withdisastrous results to its mechanical load, for example, a pump. A three-phase load is connected in the same way as the machine windings. Theload is balanced when each phase takes equal currents, that is, has equal impedance.With the wye connection the phase currents are equal to the current in the lines. Thefour-wire system is of particular use for low-voltage distribution networks in which2 Elsewhere, throughout the book, magnitude (scalar) quantities are represented by simple italics.

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48 Electric Power Systems, Fifth Edition Vcn Vb n a a Vcb Va n b Vann c c n (a) b Vbn Vc n (b)Figure 2.4 (a) Two generators connected by switch; phase voltages equal for bothsets of windings, (b) Phasor diagrams of voltages. Vcb0 ¼ voltage across switch; Va0a ¼ 0consumers are supplied with a single-phase supply taken between a line and neu-tral. This supply is often 230 V and the line-to-line voltage is 400 V. Distributionpractice in the USA is rather different and the 220 V supply often comes into a housefrom a centre-tapped transformer, as shown in Figure 2.5, which in effect gives achoice of 220 V (for large domestic appliances) or 110 V (for lights, etc.). The system planner will endeavour to connect the single-phase loads such as toprovide balanced (or equal) currents in the three-phase supply lines. At any instantin time it is highly unlikely that consumers will take equal loads, and at the lowerdistribution voltages considerable unbalance occurs, resulting in currents in theneutral line. If the neutral line has zero impedance, this unbalance does not affectthe load voltages. Lower currents ﬂow in the neutral than in the phases and it isusual to install a neutral conductor of smaller cross-sectional area than the main lineconductors. The combined or statistical effect of the large number of loads onthe low-voltage network is such that when the next higher distribution voltage isconsidered, say 11 kV (line to line), which supplies the lower voltage network,the degree of unbalance is small. This and the fact that at this higher voltage, largethree-phase, balanced motor loads are supplied, allows the three-wire system to be 220 V 110 V 110 VFigure 2.5 Tapped single-phase supply to give 220/110 V (centre-tap grounded),US practice

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Basic Concepts 49 Ia a Ia a Iab Vab Ib Ib bIca b Ic Ibc Vbc c Ic c (a) (b)Figure 2.6 (a) Mesh or delta-connected load-current relationships, (b) Practicalconnectionsused. The three-wire system is used exclusively at the higher distribution and trans-mission voltages, resulting in much reduced line costs and environmental impact. In a balanced three-wire system a hypothetical neutral line may be consideredand the conditions in only one phase determined. This is illustrated by the phasordiagram of line-to-line voltages shown in Figure 2.3(b). As the system is balancedthe magnitudes so derived will apply to the other two phases but the relative phaseangles must be adjusted by 120 and 240. This single-phase approach is very conve-nient and widely used in power system analysis. An alternative method of connection is shown in Figure 2.6. The individual phasesare connected (taking due cognizance of winding polarity in machines and trans-formers) to form a closed loop. This is known as the mesh or delta connection. Herethe line-to-line voltages are identical to the phase voltages, that is. Vline ¼ Vphase The line currents are as follows: Ia ¼ Iab À Ica Ib ¼ Ibc À Iab Ic ¼ Ica À Ibc For balanpcﬃeﬃ d currents in each phase it is readily shown from a phasor diagramthat Iline ¼ 3Iphase. Obviously a fourth or neutral line is not possible with the meshconnection. The mesh or delta connection is seldom used for rotating-machine statorwindings, but is frequently used for the windpinﬃﬃgs of one side of transformers.A line-to-line voltage transformation ratio of 1: 3 is obtained when going from aprimary mesh to a secondary wye connection with the same number of turnsper phase. Under balanced conditions the idea of the hypothetical neutral andsingle-phase solution may still be used (the mesh can be converted to a wye usingthe D ! Y transformation). It should be noted that three-phase systems are usually described by their line-to-line voltage (e.g. 11, 132, 400 kV etc.).

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50 Electric Power Systems, Fifth Edition2.1.1 Analysis of Simple Three-Phase Circuits2.1.1.1 Four-Wire SystemsIf the impedance voltage drops in the lines are negligible, then the voltage acrosseach load is the source phase voltage. Example 2.1 For the network of Figure 2.7(a), draw the phasor diagram showing voltages and currents and write down expressions for total line currents Ia, Ib, Ic, and the neutral current In. Solution Note that the power factor of the three-phase load is expressed with respect to the phase current and voltage.a Ia I Za Ima Induction Ib IZb Imb IZc Imc motor -b Ic Zc Zb Za Ia power factor Van angle - φc Vcnn In (a) VcnI Zc Imc φ IZa φφ Ima ω Imb IZb Vbn (b)Figure 2.7 (a) Four-wire system with single-phase unbalanced loads and threephase balanced motor load, (b) Phasor diagram: only the phasors IZa and Imahave been added to show Ia. Ib, and Ic can be found in a similar way. Note: Vanis the reference direction. Rotation is anticlockwise

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Basic Concepts 51The phasor diagram is shown in Figure 2.7(b). The line currents are as follows: Ia ¼ pVﬃﬃ Á 1 þ Imðcos f À j sin fÞ 3 Za ! pVﬃﬃ ðÀ0:5 1 2p 2p Ib ¼ 3 À j0:866Þ Á Zb þ Im cos 3 À f À j sin 3 À f pVﬃﬃ ðÀ0:5 ! 3 1 4p 4p Ic ¼ þ j0:866Þ Á Zc þ Im cos 3 À f À j sin 3 À fThe neutral current (as the induction motor currents do not contribute to In) In ¼ Ia þ Ib þ Ic ¼ pVﬃﬃ ½Ya þ YbðÀ0:5 À j0:866Þ þ YcðÀ0:5 þ j0:866Þ 3where Ya ¼ 1 , Yb ¼ 1 and Yc ¼ 1 . Za Zb Zc2.1.1.2 Three-Wire Balanced SystemsThe system may be treated as a single-phase system using phase voltages. It must beremembered, however, that the total three-phase active power and reactive powerare three times the values delivered by a single phase.2.1.1.3 Three-Wire Unbalanced SystemsIn more complex networks the method of symmetrical components is used (seeChapter 7), but in simple situations conventional network theory can be applied.Consider the source load arrangement shown in Figure 2.8, in which Za ¼6 Zb ¼6 Zc.From the mesh method of analysis, the following equation is obtained. Za þ Zb ÀZb !! Vab ! V ! ÀZb Zb þ Zc i1 Vbc VðÀ0:5 À j0:866Þ Á i2 ¼ ¼ c a Ia Za Vcn Vca Ib i1 Vbc Vab Ic n b Zb n (a) Vbc i2 Vbn n a c Zc Van b Vab (b)Figure 2.8 (a) Three-wire system with unbalanced load, i1 and i2 are loop currents,(b) Phasor diagram-voltage difference of neutral connection n from ground n0 ! n;n0 ¼ neutral voltage when load is balanced

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52 Electric Power Systems, Fifth Edition Hence i1 and i2 are determined and from these Ia, Ib, and Ic are found togetherwith the voltage of the neutral point (n) from ground (see Figure 2.8(b)). Dependingon the severity of the imbalance, the voltage difference from the neutral point n toground can attain values exceeding the phase voltage, that is, the point n can lieoutside the triangle of line voltages. Such conditions produce damage to connectedequipment and show the importance of earthing (grounding) the neutral. The power and reactive power consumed by a balanced three-phase load for bothwye or mesh connections are given by pﬃﬃ pﬃﬃ P ¼ 3VI cos f and Q ¼ 3VI sin fwhereV ¼ line-to-line voltage;I ¼ line current;f ¼ angle between load phase current and load phase voltage. Alternatively P ¼ 3VphIph cos f and Q ¼ 3VphIph sin f As the powers in each phase of a balanced load are equal, a single wattmeter maybe used to measure total power. For unbalanced loads, two wattmeters are sufﬁcientwith three-wire circuits while for unbalanced four wire circuits, three wattmetersare needed. Example 2.2 A three-phase wye-connected load is shown in Figure 2.9. It is supplied from a 200 V, three-phase, four-wire supply of phase sequence a-b-c. The neutral line has a resistance of 5 V. Two wattmeters are connected as shown. Calculate the power recorded on each wattmeter when: a. Ra ¼ 10 V, Rb ¼ 10 V, Rc ¼ 10 V b. Ra ¼ 10 V, Rb ¼ 10 V, Rc ¼ 2 V Solution The mesh method will be used to determine the currents in the lines and hence in the wattmeter current coils. a. As the phase currents and voltages are balanced, no current ﬂows in the neutral line. The line voltage Vab is used as a reference phasor. Hence, Vab ¼ 200 Vbc ¼ ðÀ0:5 À j0:866Þ Â 200

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Basic Concepts 53 a WM Ra CC 200 V i1 b VC Rb 200 V VC i2 CC Rc c WM i3 200 V 5Ω √3 nFigure 2.9 Use of two wattmeters to measure three-phase power, Example 2.2.cc – current coil, vc – voltage coilFrom Figure 2.9, with i3 ¼ 0 20 À10 !! 200 ! À10 20 i1 À100 À j173:2 Á i2 ¼From which i1 ¼ 10 À j5:75 i2 ¼ Àj11:5Power in wattmeter (1) ¼ real part of Vab Â IÃa3 ¼ 200(10 þ j 5.75) ¼ 2 kWPower in wattmeter (2) ¼ real part of Vcb Â IcÃ ¼ ÀVbc Â IÃc ¼ ð100 þ jl73:2Þði2ÞÃ ¼ ð100 þ jl73:2Þðj11:5Þ ¼ 2 kWThe total measured power is therefore 4 kW.3 See Section 2.3 for use of the complex conjugate IÃ.

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54 Electric Power Systems, Fifth EditionThis can be checked by considering an equivalent balanced Y system. Vphase ¼ 2p0ﬃ0ﬃ ¼ 115:5 V 3Actual power consumed ¼ 3 Â 115:52 ¼ 4 kW 10As expected the two wattmeter method gives the correct answer for a three-phaseconnection with no neutral current.b. The line voltages remain unchanged but with the unbalanced load it is now neces-sary to calculate i3 Vab ¼ 200 Vbc ¼ ðÀ0:5 À j0:866Þ Â 200 Vca ¼ ðÀ0:5 þ j0:866Þ Â 200and Vcn ¼ j 2p0ﬃ0ﬃ 3 2 À10 0 3 Á 2 i1 3 ¼ 2 200 3 20 12 À2 57 64 i2 57 6664 À100 À j173:2 7757 À2 7 i3 46 À10 j 2p0ﬃ0ﬃ 0 3Eliminate i1, 7 À2 ! ! 2 Àj173:2 3 À2 7 i2 Á i3 ¼4 j 2p0ﬃ0ﬃ 5 3From which i3 ¼ j10:26 A i2 ¼ Àj21:8 ATherefore, i1 ¼ 10 À j10:9 ANow, Ia ¼ i1 ¼ 10 À j10:9 A Ib ¼ i2 À i1 ¼ Àj21:8 À 10 þ j10:9 ¼ À10 À j10:9 A Ic ¼ i3 À i2 ¼ j10:26 þ j21:8 ¼ j32:06Power in wattmeter (1) ¼ real part of Vab Â IÃa ¼ Re½200ð10 þ j10:9Þ ¼ 2 kW

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Basic Concepts 55 Power in wattmeter (2) ¼ real part of Vcb Â IÃc ¼ ÀVbc Â IÃc ¼ Re½ð100 þ j173:2Þ Â j32:06 ¼ 5:55 kW Actual power consumed ¼ Ia2 Â 10 þ I2b Â 10 þ I2c Â 2 þ I2n Â 5 ¼ 6:96 kW As the loads are not balanced and there is a return current in the neutral, the actualpower consumed by the load is not the sum of the two wattmeter readings.2.2 Three-Phase TransformersThe usual form of the three-phase transformer, that is, the core type, is shown inFigure 2.10. If the magnetic reluctances of the three limbs are equal, then the sum ofthe ﬂuxes set up by the three-phase magnetizing currents is zero. In fact, the core isthe magnetic equivalent of the wye-connected winding. It is apparent from theshape of Figure 2.10 that the magnetic reluctances are not exactly equal, but in anintroductory treatment may be so assumed. An alternative to the three-limbed coreis the use of three separate single-phase transformers. Although more expensive(about 20% extra), this has the advantage of lower weights for transportation, and ab c a bc Primary Secondary ab cFigure 2.10 Three-phase core-type transformer. Primary connected in wye (star), sec-ondary connected in mesh (delta)

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56 Electric Power Systems, Fifth Editiona a′ Va 1:λ Va′b′ b′ n c′ Va′b Va′b′c Vb′ Va n Vc′ Vc Vb Figure 2.11 Windings and phasor diagram of a YD transformerthis aspect is crucial for large sizes. Also, with the installation of four single-phaseunits, a spare is available at reasonable cost. The wound core, as shown in Figure 2.10, is placed in a steel tank ﬁlled with insu-lating oil or synthetic liquid. The oil acts both as electrical insulation and as a coolingagent to remove the heat of losses from the windings and core. The low voltagewindings are situated over the core limbs and the high-voltage windings are woundover the low-voltage ones. The core comprises steel laminations insulated on oneside (to reduce eddy losses) and clamped together. The phasor diagram of the transformer is shown in Figure 2.11. The voltagesacross the two windings are related by the turns ratio (l). In this Y ! D woundtransformer the secondary equivalent phase voltages are À30 out of phase with theprimary phase voltages. With this winding arrangement the transformer is referredto as type Yd1 in British terminology. Y: star HV winding, d: delta LV winding:1: one o’clock or À30 phase shift.2.2.1 AutotransformersThere are several places in a power system where connections from one voltagelevel to another do not entail large transformer ratios, for example, 400/275,500/345, 725/500 kV, and then the autotransformer is used (Figure 2.12). In the autotransformer only one winding is used per phase, the secondary voltagebeing tapped off the primary winding. There is obviously a saving in size, weightand cost over a two-windings per phase transformer. It may be shown that the ratioof the weight of conductor in an autotransformer to that in a double-wound one isgiven by (1 À N2/N1). Hence, maximum advantage is obtained with a relativelysmall difference between the voltages on the two sides. The effective reactance isreduced compared with the equivalent two winding transformer and this can give

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Basic Concepts 57 V1 N1 N 2 V2Figure 2.12 One phase of an autotransformer, V2 ¼ N2 V1 N1rise to high short-circuit currents. The general constructional features of the core andtank are similar to those of double-wound transformers, but the primary and sec-ondary voltages are now in-phase.2.3 Active and Reactive PowerIn the circuit shown in Figure 2.13, let the instantaneous values of voltage andcurrent be pﬃﬃ pﬃﬃ n ¼ 2E sin ðvtÞ and i ¼ 2I sin ðvt þ fÞ The instantaneous power p ¼ vi ¼ EI cos f À EI cos ð2vt þ fÞ also ÀEI cos ð2vt þ fÞ ¼ ÀEIðcos 2vt cos f À sin 2vt sin fÞ ;p ¼ ni ¼ ½EI cos f À EI cos ð2vtÞ cos f þ ½EI sin ð2vtÞ sin f ¼ ½instantaneous real power þ ½instantaneous reactive power Figure 2:14 shows the instantaneous real and reactive power: I E Z = R + jωL Figure 2.13 Voltage source and load

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58 Electric Power Systems, Fifth Edition The mean active power ¼ EI cos f. The mean value of EI sin ð2vtÞ sin f ¼ 0, but its maximum value ¼ EI sin f. The voltage source supplies energy to the load in one direction only. At the sametime an interchange of energy is taking place between the source and the load ofaverage value zero, but of peak value EI sin f. This latter quantity is known as thereactive power (Q) and the unit is the VAr (taken from the alternative name, Volt-Ampere reactive). The interchange of energy between the source and the inductiveand capacitive elements (that is, the magnetic and electric ﬁelds) takes place at twicethe supply frequency. Therefore, it is possible to think of an active or real powercomponent P (watts) of magnitude EI cos f and a reactive power componentQ (VAr) equal to EI sin f where f is the power factor angle, that is the angle betweenE and I. It should be stressed, however, that the two quantities P and Q are physi-cally quite different and only P can do real work.Vm v i p = Active current component i q = Reactive current componentIm i ip iq ωtφ Instantaneous Power = vi Active PowerQ Component = viPP ωt Reactive Power Component = viqFigure 2.14 Real and reactive power (single phase)

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Basic Concepts 59 E I φ1 φ2 Reference axis Figure 2.15 Phasor diagram The quantity S (volt-amperes), known as the complex (or apparent) power, maybe found by multiplying E by the conjugate of I or vice versa. Consider the casewhen I lags E, and assume S ¼ EÃI. Referring to Figure 2.15, S ¼ EeÀjf1 Â Ie jf2 ¼ EIeÀjðf1Àf2Þ ¼ EIeÀjf ¼ P À jQNext assume S ¼ EIÃ ¼ EIe jðf1Àf2Þ ¼ EIejf ¼ P þ jQ Obviously both the above methods give the correct magnitudes for P and Q butthe sign of Q is different. The method used is arbitrarily decided and the usual con-vention to be adopted is that the volt-amperes reactive absorbed by an inductiveload shall be considered positive, and by a capacitive load negative; hence S ¼ EIÃ.This convention is recommended by the International Electrotechnical Commission. In a network the net stored energy is the sum of the various inductive and capaci-tive stored energies present. The net value of reactive power is the sum of the VArsabsorbed by the various components present, taking due account of the sign. VArscan be considered as being either produced or absorbed in a circuit; a capacitiveload can be thought of as generating VArs. Assuming that an inductive load is rep-resented by R þ jX and that the S ¼ EIÃ convention is used, then an inductive loadabsorbs positive VArs and a capacitive load produces VArs. The various elements in a network are characterized by their ability to generate orabsorb reactive power. Consider a synchronous generator which can be representedby the simple equivalent circuit shown in Figure 2.16. When the generator is over-excited, that is, its generated e.m.f. is high, it produces VArs and the complex powerﬂow from the generator is P þ jQ. When the machine is under-excited the generatedcurrent leads the busbar voltage and the apparent power from the generator is

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60 Electric Power Systems, Fifth Edition(a) V (b) E E Xs I Iq I XS V Iq E I XS I (c) Ip V Ref. Ip ω I Ref. ωFigure 2.16 (a) Line diagram of system, (b) Overexcited generator-phasor diagram,(c) Underexcited generator-phasor diagramP À jQ. It can also be thought of as absorbing VArs. The reactive power character-istics of various power-system components are summarized as follows: reactivepower is generated by over-excited synchronous machines, capacitors, cables andlightly loaded overhead lines; and absorbed by under-excited synchronousmachines, induction motors, inductors, transformers, and heavily loaded overheadlines, see Figure 2.17. The reactive power absorbed by a reactance of XL [V] ¼ I2XL [VAr] where I is thecurrent. A capacitance with a voltage V applied produces V2B [VAr], where B ¼ 1/Xc ¼ vC [S] and Xc ¼ 1/vC [V]. At a node, SP ¼ 0 and SQ ¼ 0 in accordance withKirchhoff’s law. +P Under-excited Over-excited Generators Generators –Q Inductive Capacative+Q Loads Loads –P Generator S = VI* P + jQ V or Load I Figure 2.17 Power quadrant

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Basic Concepts 612.4 The Per-Unit SystemIn the analysis of power networks, instead of using actual values of quantities it isusual to express them as fractions of reference quantities, such as rated or full-loadvalues. These fractions are called per unit (denoted by p.u.) and the p.u. value of anyquantity is deﬁned as ÀÁ ð2:1Þ actual value in any unit base or reference value in the same unit Some authorities express the p.u. value as a percentage. Although the use of p.u.values may at ﬁrst sight seem a rather indirect method of expression there are, infact, great advantages; they are as follows:1. The apparatus considered may vary widely in size; losses and volt drops will also vary considerably. For apparatus of the same general type the p.u. volt drops and losses are inpthe same order, regardless of size.2. The use of 3’s in three-phase calculations is reduced.3. By the choice of appropriate voltage bases the solution of networks containing several transformers is facilitated.4. Per-unit values lend themselves more readily to digital computation.2.4.1 Resistance and ImpedanceGiven that base value of voltage is Vb and base value of current is Ib, the base valueof resistance is given by: Rb ¼ Vb ð2:2Þ IbNow from (2.1) and (2.2), the p.u. value of the resistance is given by:Rp:u: ¼ RðVÞ ¼ RðVÞ ¼ RðVÞ Á Ib ¼ voltage drop across R at base current RbðVÞ Vb=Ib Vb base voltageAlso Rp:u: ¼ RðVÞI2b VbIb ¼ power loss at base current base power or volt-amperes; the power loss (p.u.) at base or rated current ¼ Rp.u. 2Power loss (p.u.) at Ip.u. current ¼ Rp:u:I p:u:

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62 Electric Power Systems, Fifth Edition Similarly p:u: impedance ¼ impedance in ohms ¼ ZðVÞIb base voltage Vb base currentExample 2.3A d.c. series machine rated at 200 V, 100 A has an armature resistance of 0.1 V and ﬁeldresistance of 0.15 V. The friction and windage loss is 1500 W. Calculate the efﬁciencywhen operating as a generator.SolutionTotal series resistance in p.u. is given by Rp:u: ¼ 0:25 ¼ 0:125 p:u: 200=100Where Vbase ¼ 200 V and Ibase ¼ 100 AFriction and windage loss ¼ 1500 ¼ 0:075 p:u: 200 Â 100At the rated load, the series-resistance loss ¼ 12 Â 0:125and the total loss ¼ 0.125 þ 0.075 ¼ 0.2 p.u.As the output ¼ 1 p.u., the efﬁciency ¼ 1 1 ¼ 0:83 p:u: þ 0:22.4.2 Three-Phase CircuitsIn three-phase circuits, a p.u. phase voltage has the same numerical value as thecorresponding p.u. line voltage. With a measured line voltage of 100 kV and arated linpe voltage of 132 pkV, the p.u. value is 0.76. The equivalent phase voltagesare 100/ 3 kV and 132/ 3 kV and hence the p.u. value is again 0.76. The actualvalues of R, XL, and XC for lines, cables, and other apparatus are phase values.When working with ohmic values it is less confusing to use the equivalent phasevalues of all quantities. In the p.u. system, three-phase values of voltage, currentpan3dinpcoowrreerctc.an be used without undue anxiety about the result being a factor of

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Basic Concepts 63It is convenient in a.c. circuit calculations to work in terms of base volt-amperes,ðSbaseÞ. Thus, pﬃﬃ Sbase ¼ Vbase Â Ibase Â 3when Vbase is the line voltage and Ibase is the line current in a three-phase system. Hence Ibase ¼ pSﬃﬃ base ð2:3Þ 3Vbase Expression (2.3) shows that if Sbase and Vbase are speciﬁed, then Ibase is determined.Only two base quantities can be chosen from which all other quantities in a three-phase system are calculated. Thus, Vbase .pﬃﬃ .pﬃﬃ 3 Vbase 3 Zbase ¼ ¼ .pﬃﬃ ¼ Vb2ase ð2:4Þ Ibase Sbase Sbase 3VbaseHence Zp:u: ¼ ZðVÞ ¼ ZðVÞ Â Sbase ð2:5Þ Zbase V2base Expression (2.5) shows that Zp.u. is directly proportional to the base VA andinversely proportional to the base voltage squared. If we wish to calculate Zp.u. to a new base VA, then Zp:u:ðnew baseÞ ¼ Zp:u:ðold baseÞ Â Snew base ð2:6Þ Sold baseIf we wish to calculate Zp.u. to a new voltage base, then Zp:u:ðnew baseÞ ¼ Zp:u:ðold baseÞ Â ðVold baseÞ2 ð2:7Þ ðVnew baseÞ22.4.3 TransformersConsider a single-phase transformer in which the total series impedance of the twowindings referred to the primary is Z1 (Figure 2.18).Then the p.u. impedance Zp:u: ¼ Z1 , where I1, and V1, are the base values of the V1=I1primary circuit.

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64 Electric Power Systems, Fifth Edition Z1( Ω) V2 V1 1:N (a) Z (p.u.)V1(p.u.) V2 (p.u.) (b)Figure 2.18 Equivalent circuit of single-phase transformerThe ohmic impedance referred to the secondary is Z2 ¼ Z1N2 ½Vand this in p.u. notation is 0 V2 I2 Zp:u: ¼ Z1N2V2 and I2 are base voltage and current of the secondary circuit. If they are related tothe base voltage and current of the primary by the turns ratio of the transformer then. Zp:u: ¼ Z1 N 2 I1 Á 1 ¼ Z1I1 N V1N V1 Hence provided the base voltages on each side of a transformer are related by theturns ratio, the p.u. impedance of a transformer is the same whether consideredfrom the primary or the secondary side. The winding does not appear in the equiv-alent circuit (Figure 2.18b), the transformer impedance in per unit is only calculatedonce and Equation (2.7) is not required.Example 2.4In the network of Figure 2.19, two single-phase transformers supply a 10 kVA resistanceload at 200 V. Show that the p.u. load is the same for each part of the circuit and calcu-late the voltage at point D.

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Basic Concepts 65 D A Vs B C 200 V 100:400 V 400:200 V X = 0.1 p.u. X = 0.15 p.u. 10 kVA 10 kVAFigure 2.19 Network with two transformers–p.u. approachSolutionThe load resistance is (2002/10 Â 103), that is, 4 V. In each of the circuits A, B, and C a different voltage exists, so that each circuit willhave its own base voltage, that is, 100 V in A, 400 V in B, and 200 V in C. Although it is not essential for rated voltages to be used as bases, it is essential thatthe voltage bases used be related by the turns ratios of the transformers. If this is not sothe simple p.u. framework breaks down. The same volt-ampere base is used for all thecircuits as V1I1 ¼ V2I2 on each side of a transformer and is taken in this case as 10 kVA.The per unit impedances of the transformers are already on their individual equipmentbases of 10 kVA and so remain unchanged. The base impedance in C Zbase ¼ Vb2ase ¼ 2002 ¼ 4V base VA 10000The load resistance (p.u.) in C ¼ 4 ¼ 1 p:u: 4In B the base impedance is Zbase ¼ V2base ¼ 4002 ¼ 16V base VA 10000and the load resistance (in ohms) referred to B is ¼ 4 Â N2 ¼ 4 Â 22 ¼ 16VHence the p.u. load referred to B 16 ¼ 1 p:u: 16 Similarly, the p.u. load resistance referred to A is also 1 p.u. Hence, if the voltagebases are related by the turns ratios the load p.u. value is the same for all circuits.

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66 Electric Power Systems, Fifth Edition j 0.1 p.u. D j 0.15 p.u. VR = I p.u. 1 p.u. VsFigure 2.20 Equivalent circuit with p.u. values, of network in Figure 2.19 An equivalent circuit may be used as shown in Figure 2.20. Let the volt-amperebase be 10 kVA; the voltage across the load (VR) is 1 p.u. (as the base voltage in Cis 200 V). The base current at voltage level C of this single phase circuit Ibase ¼ base VA ¼ 10000 ¼ 50 A Vbase 200The corresponding base currents in the other circuits are 25 A in B, and 100 A in A.The actual load current is 200V/4 V ¼ 50 A ¼ 1 p.u.Hence the supply voltage VS Vs ¼ 1ðj0:1 þ j0:15Þ þ 1 p:u: pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ VS ¼ 12 þ 0:252 ¼ 1:03 p:u: ¼ 1:03 Â 100 ¼ 103 VThe voltage at point D in Figure 2.17 VD ¼ 1ðj0:15Þ þ 1 p:u: pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ VD ¼ 12 þ 0:152 ¼ 1:012 p:u: ¼ 1:012 Â 400 ¼ 404:8 V It is a useful exercise to repeat this example using ohms, volts and amperes. A summary table of the transformation of the circuit of Figure 2.19 into per unit isshown.Section Sbase Vbase (chosen as Ibase (calculated Zbase (calculatedof (common for transformer from Sbase/Vbase) from V2base=Sbase)network network) turns ratio) 100 A 1VA 10 kVA 100 V 25 A 16 VB 10 kVA 400 V 50 AC 10 kVA 200 V 4V

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Basic Concepts 67Example 2.5Figure 2.21 shows the schematic diagram of a radial transmission system. The ratingsand reactances of the various components are shown. A load of 50 MW at 0.8 p.f.lagging is taken from the 33 kV substation which is to be maintained at 30 kV. It isrequired to calculate the terminal voltage of the synchronous machine. The line andtransformers may be represented by series reactances. The system is three-phase. Vs Line 132 kV 33 kV 50 MW 11 kV 132 kV j100Ω 0.8 p.f. 50 MVA lagging 50 MVA X = 12% X = 10% 30 kV Figure 2.21 Line diagram of system for Example 2.5SolutionIt will be noted that the line reactance is given in ohms; this is usual practice. The volt-age bases of the various circuits are decided by the nominal transformer voltages, thatis, 11, 132, and 33 kV. A base of 100 MVA will be used for all circuits. The reactances(resistance is neglected) are expressed on the appropriate voltage and MVA bases. Base impedance for the line ¼ V2base Sbase ¼ ð132 Â 103Þ2 100 Â 106 ¼ 174VHence the p.u. reactance ¼ j100 ¼ j0:575 p:u: 174Per unit reactance of the sending-end transformer j0:1 Â 100 ¼ j0:2 p:u: 50Per unit reactance of the receiving-end transformer j0:12 Â 100 ¼ j0:24 p:u: 50 Load current ¼ pﬃﬃ Â 50 Â 106 Â 0:8 ¼ 1203 A 3 30 Â 103

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68 Electric Power Systems, Fifth EditionBase current for 33 kV, 100 MVA ¼ pﬃ1ﬃ 00 Â 106 ¼ 1750 A 3 Â 33 Â 103Hence the p.u. load current ¼ 1203 ¼ 0:687 p:u: 1750Voltage of the load busbar ¼ 30 ¼ 0:91 p:u: 33The equivalent circuit is shown in Figure 2.22.Also,VS ¼ 0:687 Â ð0:8 À j0:6Þðj0:2 þ j0:575 þ j0:24Þ þ ð0:91 þ j0Þ ¼ 1:328 þ j0:558 p:u:VS ¼ 1:44 p:u: ¼ 1:44 Â 11 kV ¼ 15:84 kVj1p.u. 0.687 p.u. 0.8 p.f. laggingLoad Es j0.2 p.u. j0.575 p.u. j0.24 p.u. Vs VR = 0.91 Figure 2.22 Equivalent circuit for Example 2.52.5 Power Transfer and Reactive PowerThe circuit shown in Figure 2.23 represents the simplest electrical model for a source(with voltage VG) feeding into a power system represented by a load of P þ jQ. Itcan also represent the power ﬂows in a line connecting two busbars in an intercon-nected power system. The voltage at the source end and load end are related by: VG ¼ VL þ ðR þ jXÞI ð2:8Þ

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Basic Concepts 69 RI jX P + jQ SL VL SG VG Figure 2.23 Power transfer between sources pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ If R þ jX ¼ Zﬀu where Z ¼ R2 þ X2 and u ¼ tanÀ1ðX=RÞ, then the current can beobtained as: I ¼ VG À VL ZﬀuTherefore the apparent power at the source end is given by: SG ¼ VGIÃ ¼ VG À VLÃ ð2:9Þ VG Zﬀu If the load end voltage is chosen as the reference and the phase angle between theload end and source end is d: VL ¼ VLﬀ0 ¼ VL and VG ¼ VGﬀd ð2:10ÞSubstituting for VL and VG from equation (2.10) into (2.9) yields: SG ¼ VGejd eÀjd À VG ZeÀju VL ¼ VG2 eju À VGVL ejðuþdÞ Z ZTherefore PG ¼ VG2 cos u À VGVL cos ðu þ dÞ Z Z ð2:11Þ VG2 QG ¼ Z sin u À VGVL sin ðu þ dÞ Z

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70 Electric Power Systems, Fifth EditionSimilarly, SL ¼ À VGeÀjd ¼ VL2 e ju À VLVG e jðuÀdÞ VL VL ZeÀju Z Z PL ¼ V2L cos u À VLVG cos ðu À dÞ Z Z ð2:12Þ VL2 QL ¼ Z sin u À VLVG sin ðu À dÞ ZThe power output to the load is a maximum when cosðu À dÞ ¼ 1, that is, u ¼ d2.5.1 Calculation of Sending and Received Voltages in Terms of Power and Reactive PowerThe determination of the voltages and currents in a network can obviously beachieved by means of complex notation, but in power systems usually power (P)and reactive power (Q) are speciﬁed and often the resistance of lines is negligiblecompared with reactance. For example, if R ¼ 0:1X, the error in neglecting R is0.49%, and even if R ¼ 0:4X the error is 7.7%. From the transmission link shown in Figure 2.23: For the load: VLIÃ ¼ P þ jQ I ¼ P À jQ VLÃThe voltage at the source and load are related by: VG ¼ VL þ ðR þ jXÞI ! ¼ VL þ ðR þ jXÞ P À jQ ð2:13Þ VLÃ ð2:14ÞAs VL ¼ VÃL ¼ VL ðin this caseÞ: P À jQ ! !VL VG ¼ VL þ ðR þ jXÞ RP þ XQ XP À RQ ! VL VL ¼ VL þ þjEquation (2.14) can be represented by the phasor diagram shown in Figure 2.24. DVp ¼ RP þ XQ ð2:15Þ V

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Basic Concepts 71 VG ΔVq IX R + jX P + jQ VL IRVG VL L δ φ Ip Iq ΔVp I (a) (b) Figure 2.24 Phasor diagram for transmission of power through a lineand DVq ¼ XP À RQ ð2:16Þ VIf d is small (as is usually the case in Distribution circuits) then DVq ( VL þ DVpthen VG ¼ VL þ RP þ XQ VLand VG À VL ¼ RP þ XQ VLHence the arithmetic difference between the voltages is given approximately by RP þ XQ VLIn a transmission circuit, R ’ 0 then VG À VL ¼ XQ ð2:17Þ VLthat is, the voltage magnitude depends only on Q. sinÀ1ÀDVq =VG Á and depends The angle of transmission d is obtained from ,only on P.

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72 Electric Power Systems, Fifth Edition Equations (2.15) and (2.16) will be used wherever possible because of their greatsimplicity. Example 2.6 Consider a 275 kV line of length 160 km (R ¼ 0.034 V/km and X ¼ 0.32 V/km). Obvi- ously, R ( X. Compare the sending end voltage VG in Figure 2.25 when calculated with the accu- rate and the approximate formulae. Assume a load of 600 MW, 300 MVAr and take a system base of 100 MVA. (Note that 600 MVA is nearly the maximum rating of a 2 Â 258 mm2 line at 275 kV.) Ip VG Iq IX PX I VL VG QX VG Figure 2.25 Phasor diagram when VG Is speciﬁedSolution À Â 103Á2 ¼ 756 VThe base impedance is 275 For line of length 160 km, 100 Â 106 XðVÞ ¼ 160 Â 0:32 ¼ 51:2 V X ¼ 51:2 ¼ 0:0677 p:u: 756and let the received voltage be VL ¼ 275 kV ¼ 1 p:u:then XP ¼ 0:0677 Â 6 ¼ 0:406 p:u:

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Basic Concepts 73and VL þ QX ¼ 1 þ 3 Â 0:0677 ¼ 1:203 p:u: 0:406Hence d ¼ tanÀ1 1:203 ¼ 18 This is a signiﬁcant angle between the voltages across a circuit and so the use of theapproximate equations for this heavily loaded, long transmission circuit will involvesome inaccuracy. Using equation (2.14) with R neglected. VG2 ¼ ð1 þ 0:0677 Â 3Þ2 þ ð0:0677 Â 6Þ2 VG ¼ 1:27 p:u:An approximate value of VG can be found from equation (2.17) as VL À VG ¼ DV ¼ 0:0677 Â 3 ¼ 0:203 1 VG ¼ 1:203 p:u:that is, an error of 5.6%. With a shorter, 80 km, length of this line at the same load (still neglecting R) gives XP ¼ 0:0338 Â 6 ¼ 0:203 p:u:and VL þ QX ¼ 1 þ 3 Â 0:0338 ¼ 1:101 p:u: :203 d ¼ tanÀ1 1:101 ¼ 10 The accurate formula gives VG2 ¼ 1 þ 0:101Þ2 þ ð0:203Þ2 VG ¼ 1:120 p:u:and the approximate one gives VG ¼ 1 þ :101 ¼ 1:101 p:u:that is, an error of 1.7%. In the solution above, it should be noted that p.u. values for power system calcula-tions are conveniently expressed to three decimal places, implying a measured value to0.1% error. In practice, most measurements will have an error of at least 0.2% and possi-bly 0.5%. If VG is speciﬁed and VL is required, the phasor diagram in Figure 2.25 is used.

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74 Electric Power Systems, Fifth EditionFrom this, VL ¼ tuvuﬃ\"ﬃﬃﬃﬃﬃﬃVﬃﬃﬃGﬃﬃﬃﬃÀﬃﬃﬃﬃﬃQVﬃﬃﬃﬃXGﬃﬃﬃﬃﬃﬃﬃ2ﬃﬃﬃþﬃﬃﬃﬃﬃﬃﬃﬃPVﬃﬃﬃXﬃGﬃﬃﬃﬃﬃﬃ2ﬃﬃ#ﬃﬃ ð2:18ÞIf ð2:19Þ PX ( VG2 À QX then VG VG VL ¼ VG À QX VG2.6 Harmonics in Three-Phase SystemsThe sine waves of the currents and voltages in a power system may not be perfectand so may contain harmonics. Harmonics are created by domestic non-linear loads,particularly computer and TV power supplies but also by large industrial rectiﬁers(e.g. for aluminium smelters). Consider the instantaneous values of phase voltagesin a balanced system containing harmonics up to the third: va ¼ V1 sin vt þ V2 sin 2vt þ V3 sin 3vt ð2:20Þ ð2:21Þ 2p 2p 2p ð2:22Þvb ¼ V1 sin vt À 3 þ V2 sin 2 vt À 3 þ V3 sin 3 vt À 3 2p 4p¼ V1 sin vt À 3 þ V2 sin 2vt À 3 þ V3 sin ð3vt À 2pÞ 2p 4p¼ V1 sin vt À 3 þ V2 sin 2vt À 3 þ V3 sin 3vt 4p 4p 4pvc ¼ V1 sin vt À 3 þ V2 sin 2 vt À 3 þ V3 sin 3 vt À 3 4p 2p¼ V1 sin vt À 3 þ V2 sin 2vt À 2p À 3 þ V3 sin ð3vt À 2pÞ 4p 2p¼ V1 sin vt À 3 þ V2 sin 2vt À 3 þ V3 sin 3vt Here V1, V2 and V3 are the magnitudes of the harmonic voltages. From equations (2.20)–(2.22) the phasor diagram for fundamental, second andthird harmonic components of va, vb and vc shown in Figure 2.26 can be obtained.As shown in Figure 2.26, the fundamental terms have the normal phase rotation (asdo the fourth, seventh and tenth, etc., harmonics). However, the second harmonicterms possess a reversed phase rotation (also the ﬁfth, eighth, eleventh, etc.), andthe third harmonic terms are all in phase (also all harmonics of multiples three).

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Basic Concepts 75 Va1 Va2 Va3 Vb3 Vc3 ω 2ω 3ω Vb2 Vc2 Vc1 Vb1Figure 2.26 Phase rotation of harmonic voltage (fundamental, 2nd and 3rdharmonics) pﬃﬃ When substantial harmonics are present the 3 relation between line and phasequantities no longer holds. As can be seen from Figure 2.26, harmonic voltages otherthan that of triple harmonics (n ¼ 3, 6, 9, etc.) in successive phases are 2p=3 out ofphase with each other. In the resulting line voltages (as Vab is given by (Va À Vb))no triple harmonics exist. The mesh connection forms a complete path for the tripleharmonic currents which ﬂow in phase around the loop. When analysing the penetration of harmonics into the power network an initialapproximation is to assume that the effective reactance for the nth harmonic is ntimes the fundamental value.2.7 Useful Network Theory2.7.1 Four-Terminal NetworksA lumped-constant circuit, provided that it is passive, linear and bilateral, can berepresented by the four-terminal network shown in the diagram in Figure 2.27. Thecomplex parameters A, B, C and D describe the network in terms of the sending-and receiving-end voltages and currents as follows: VS ¼ AVR þ BIR IS ¼ CVR þ DIRand it can be readily shown that ADÀBC ¼ 1. Is AB IR Vs CD VRFigure 2.27 Representation of a four-terminal (two-port) network

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76 Electric Power Systems, Fifth Edition A, B, C, and D may be obtained by measurement and certain physical interpreta-tions can be made, as follows:1. Receiving-end short-circuited: VR ¼ 0; IS ¼ DIR and D ¼ IS IRAlso, VS ¼ BIR and B ¼ VS ¼ short circuit impedance IR2. Receiving-end open-circuited: Here IR ¼ 0; IS ¼ CVR and C ¼ IS VR VS VS ¼ AVR and A ¼ VR Expressions for the constants can be found by complex (that is, magnitudeand angle) measurements carried out solely at the sending end with the receiv-ing end open and short-circuited. Often it is useful to have a single four-terminal network for two or more itemsin series or parallel, for example, a line and two transformers in series. It is shown in most texts on circuit theory that the generalized constants forthe combined network, A0, B0, C0, and D0 for the two networks (1) and (2) incascade are as follows: A0 ¼ A1A2 þ B1C2 B0 ¼ A1B2 þ B1D2 C0 ¼ A2C1 þ C2D1 D0 ¼ B2C1 þ D1D2 For two four-terminal networks in parallel it can be shown that the parame-ters of the equivalent single four-terminal network are: A0 ¼ A1B2 þ A2B1 B0 ¼ B1B2 D0 ¼ B2D1 þ B1D2 B1 þ B2 B1 þ B2 B1 þ D2 C0 can be found from A0D0 À B0C0 ¼ 12.7.2 Delta-Star TransformationThe delta network connected between the three terminals A, B and C of Figure 2.28can be replaced by a star network such that the impedance measured between theterminals is unchanged. The equivalent impedances can be found as follows:

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Basic Concepts 77Impedance between terminals AB with C open-circuited is given by ZOA þ ZOB ¼ ZAB==ðZCA þ ZBCÞSimilarly ZOB þ ZOC ¼ ZBC==ðZAB þ ZCAÞand ZOC þ ZOA ¼ ZCA==ðZBC þ ZABÞ From these three equations ZOA, ZOB and ZOC, the three unknowns, can be deter-mined as ZOA ¼ ZABZCA ZAB þ ZBC þ ZCA ZOB ¼ ZABZBC ZCA ð2:15Þ ZAB þ ZBC þ ZOC ¼ ZBCZCA ZAB þ ZBC þ ZCA2.7.3 Star-Delta TransformationA star-connected system can be replaced by an equivalent delta connection if theelements of the new network have the following values (Figure 2.28): C Z Z BC CA Z OC O Z OA Z OB ZAB AB Figure 2.28 Star-delta, delta-star transformation

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78 Electric Power Systems, Fifth EditionZAB ¼ ZOAZOB þ ZOBZOC þ ZOCZOA ZOCZBC ¼ ZOAZOB þ ZOBZOC þ ZOCZOA ZOAZCA ¼ ZOAZOB þ ZOBZOC þ ZOCZOA ZOBProblems 2.1 The star-connected secondary winding of a three-phase transformer supplies 415 V (line to line) at a load point through a four-conductor cable. The neutral conductor is connected to the winding star point which is earthed. The load consists of the following components: Between a and b conductors a 1 V resistor Between a and neutral conductors a 1 V resistor Between b and neutral conductors a 2 V resistor Between c and neutral conductors a 2 V resistor Connected to the a, b and c conductors is an induction motor taking a balanced current of 100 A at 0.866 power factor (p.f.) lagging. Calculate the currents in the four conductors and the total power supplied. Take the ‘a’ to neutral voltage as the reference phasor. The phase sequence is a-b-c. (Answer: IA ¼ 686 þ j157 A, IB ¼ À506 þ j361 A, IC ¼ À60 þ j204 A, 350 kW 350 kW) 2.2 The wye-connected load shown in Figure 2.29 is supplied from a transformer whose secondary-winding star point is solidly earthed. The line voltage sup- plied to the load is 400 V. Determine (a) the line currents, and (b) the voltage of a 1Ω 1Ω 2Ω b cFigure 2.29 Circuit for Problem 2.2

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Basic Concepts 79 the load star point with respect to ground. Take the ‘a’ to ‘b’ phase voltage as the reference phasor. The phase sequence is a-b-c. (Answer: (a) Ia ¼ 200 À j69:3 A; (b) Vng ¼ À47 V)2.3 Two capacitors, each of 10 mF, and a resistor R, are connected to a 50 Hz three phase supply, as shown in Figure 2.30. The power drawn from the supply is the same whether the switch S is open or closed. Find the resistance of R. (Answer: 142 V) a b R c NS Figure 2.30 Circuit for Problem 2.32.4 The network of Figure 2.31 is connected to a 400 V three-phase supply, with phase sequence a-b-c. Calculate the reading of the wattmeter W. (Answer: 2.33 kW) aW 50 Ω –j 53 Ω j 40 Ω b c Figure 2.31 Circuit for Problem 2.4

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80 Electric Power Systems, Fifth Edition 2.5 A 400 V three-phase supply feeds a delta-connected load with the following branch impedances: ZRY ¼ 100V ZYB ¼ j100V ZBR ¼ Àj100V Calculate the line currents for phase sequences (a) RYB; (b) RBY. (Answer: (a) 7.73, 7.73, 4 A: (b) 2.07, 2.07, 4 A) 2.6 A synchronous generator, represented by a voltage source in series with an inductive reactance X1, is connected to a load consisting of a ﬁxed inductive reactance X2 and a variable resistance R in parallel. Show that the generator power output is a maximum when 1=R ¼ 1=X1 þ 1=X2 2.7 A single-phase voltage source of 100 kV supplies a load through an imped- ance j100 V. The load may be represented in either of the following ways as far as voltage changes are concerned: a. by a constant impedance representing a consumption of 10 MW, 10 MVAr at 100 kV; or b. by a constant current representing a consumption of 10 MW, 10 MVAr at 100 kV. Calculate the voltage across the load using each of these representations. (Answer: (a) (90 À j8.2) kV; (b) (90 À j10) kV) 2.8 Show that the p.u. impedance (obtained from a short-circuit test) of a star- delta three-phase transformer is the same whether computed from the star- side parameters or from the delta side. Assume a rating of G (volt-amperes), a line-to-line input voltage to the star-winding terminals of V volts, a turns ratio of 1 : N (star to delta), and a short circuit impedance of Z (ohms) per phase referred to the star side. 2.9 An 11/132 kV, 50 MVA, three-phase transformer has an inductive reactance of j0.5 V referred to the primary (11 kV). Calculate the p.u. value of reactance based on the rating. Neglect resistance. (Answer: 0.21 p.u.)2.10 Express in p.u. all the quantities shown in the line diagram of the three-phase transmission system in Figure 2.32. Construct the single-phase equivalent cir- cuit. Use a base of 100 MVA. The line is 80 km in length with resistance and reactance of 0.1 and 0.5 V, respectively, and a capacitive susceptance of 10 mS per km (split equally between the two ends).2.11 A wye-connected load is supplied from three-phase 220 V mains. Each branch of the load is a resistor of 20 V. Using 220 V and 10 kVA bases calculate the p.u. values of the current and power taken by the load. (Answer: 0.24 p.u.; 0.24 p.u.)

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Basic Concepts 81Generator Line Load 2 p.u. Line 500 MVA 0.85 p.f. lag500 22/400 kV 400/11 kVMVA 0.1p.u. 0.15 p.u. reactance reactance Figure 2.32 Circuit for Problem 2.102.12 A 440 V three-phase supply is connected to three star-connected loads in par- allel, through a feeder of impedance (0.1 þ j0.5) V per phase. The loads are as follows: 5 kW, 4 kVAr; 3 kW, 0 kVAr; 10 kW, 2 kVAr. Determine: a. line current; b. power and reactive power losses in the feeder per phase; c. power and reactive power from the supply and the supply power factor. (Answer: (a) 24.9 A; (b) 62 W, 310 VAr; (c) 18.2 kW, 6.93 kVAr, 0.94)2.13 Two transmission circuits are deﬁned by the following ABCD constants: 1, 50, 0, 1, and 0.9 h 2, 150 h 79, 9 Â 10À4 h 91, 0.9 h 2. Determine the ABCD con- stants of the circuit comprising these two circuits in series. (Answer: A ¼ 0.9 h 4.85; B ¼ 165.1 h 63.6)2.14 A 132 kV overhead line has a series resistance and inductive reactance per phase per kilometre of 0.156 and 0.4125 V, respectively. Calculate the magni- tude of the sending-end voltage when transmitting the full line capability of 125 MVA when the power factor is 0.9 lagging and the received voltage is 132 kV, for 16 km and 80 km lengths of line. Use both accurate and approxi- mate methods. (Answer: 136.92 kV (accurate), 136.85 kV; 157.95 kV (accurate), 156.27 kV)2.15 A synchronous generator may be represented by a voltage source of magni- tude 1.7 p.u. in series with an impedance of 2 p.u. It is connected to a zero- impedance voltage source of 1 p.u. The ratio of X/R of the impedance is 10. Calculate the power generated and the power delivered to the voltage source if the angle between the voltage sources is 30. (Answer: 0.49 p.u.; 0.44 p.u.)2.16 A three-phase star-connected 50 Hz generator generates 240 V per phase and supplies three delta-connected load coils each having a resistance of 10 V and an inductance of 47.75 mH.

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82 Electric Power Systems, Fifth Edition Determine: a. the line voltage and current; b. the load current; c. the total real power and reactive power dissipated by the load. Determine also the values of the three capacitors required to correct the over- all power factor to unity when the capacitors are (i) star- and (ii) delta- connected across the load. State an advantage and a disadvantage of using the star connection for power factor correction. (Answer: (a) 416 V, 39.96 A; (b) 23.07 A; (c) 15.9 kW, 24.0 kVAr; (i) 441 mF; (ii) 147 mF) (From Engineering Council Examination, 1996)

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3Components of a PowerSystem3.1 IntroductionInvestigations of large interconnected electric power systems, either through man-ual calculations or computer simulation, use the simplest models of the variouscomponents (for example, lines, cable, generators, transformers, and so on) thatshow the phenomenon being studied without unnecessary detail. Representationsbased on an equivalent circuit not only make the principles clearer but also thesesimple models are used in practice. For more sophisticated treatments, especially ofsynchronous machines and fast transients, the reader is referred to the moreadvanced texts given in the Further Reading section at the end of the book. Loads are considered as components even though their exact composition andcharacteristics are not known with complete certainty and vary over time. Whendesigning a power supply system or extending an existing one a prediction of theloads to be expected is required, statistical methods being used. This is the aspect ofpower supply known with least precision. Most of the equivalent circuits used are single phase and employ phase-to-neutralvalues. This assumes that the loads are balanced three-phase which is reasonable fornormal steady-state operation. When unbalance exists between the phases, fulltreatment of all phases is required, and special techniques for dealing with this aredescribed in Chapter 7.3.2 Synchronous MachinesLarge synchronous generators for power generation have one of two forms of rotorconstruction. In the round or cylindrical rotor the ﬁeld winding is placed in slots cutaxially along the rotor length as illustrated in Figure 3.1(a). The diameter isElectric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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84 Electric Power Systems, Fifth Editionrelatively small (1–1.5 m) and the machine is suitable for operation at high speeds(3000 or 3600 r.p.m.) driven by a steam or gas turbine. Hence it is known as a turbo-generator. In a salient pole rotor the poles project, as shown in Figure 3.1(b), and themachine is driven by a hydro turbine or diesel engine at a lower rotational speed.The frequency of the generated voltage (f) and speed are related by f ¼ np 60 where n is speed in r.p.m. and p is the number of pairs of poles. A hydro turbinerotating at 150 r.p.m. thus needs 20 pole pairs to generate at 50 Hz and a largediameter rotor to accommodate the poles. The three-phase currents in the stator winding or armature generate a magneticﬁeld which rotates with the synchronously rotating rotor to which the d.c. ﬁeld isﬁxed. The effect of the stator winding ﬁeld on the rotor ﬁeld is referred to as‘armature reaction’. Figure 3.1(b) indicates the paths of the main ﬁeld ﬂuxes in asalient pole machine from which it can be seen that these paths are mainly in iron,apart from crossing the air gap between stator and rotor. On the other hand, the Cross-field axis Rotor field conductors in slots Field axisSolidsteelpoleMain field (a) flux Cross-field flux (dotted) Rotor Stator Field windings (b)Figure 3.1 (a) Cylindrical rotor, (b) Salient pole rotor, and stator with ﬂux paths

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Components of a Power System 85cross-ﬁeld ﬂux has a longer path in air and hence the reluctance of its path is greaterthan for the main ﬁeld ﬂux. Consequently, any e.m.f.s produced by the main ﬁeldﬂux in the stator are larger than those produced by any cross-ﬁeld ﬂuxes. Machine designers shape the poles in a salient machine or distribute the windingsin a round-rotor machine to obtain maximum fundamental sinusoidal voltage onno-load in the stator windings. The stator winding is carefully designed and distrib-uted such as to minimize the harmonic voltages induced in it and only the funda-mental e.m.f. is signiﬁcant for power system studies. Whether a synchronous machine has a cylindrical or salient-pole rotor, its actionis the same when generating power and can be best understood in terms of the‘primitive’ representation shown in Figure 3.2(a). The rotor current If produces amagnetomotive force (m.m.f.) Ff rotating with the rotor and induces a sinusoidalvoltage in the stator winding. When the stator windings are closed through an exter-nal circuit, current ﬂows, thereby delivering power to the external load. These stator currents produce an m.m.f., shown in Figure 3.2(a) for phase A as Fa,which interacts with the rotor m.m.f. Ff, thereby producing a retarding force on therotor that requires torque (and hence power) to be provided from the prime mover.The vector m.m.f.s in Figure 3.2(b) are drawn for an instant at which the current I in Cross-field axis Phase AFfN Fa Ff Fa RLω F Main S field δ If axis (b) Phase A E j Xs (a) j Xm j XL jIXm I Va Iω Va jIXL E (e) Vt δ(c) I IRL Vt ω (d)Figure 3.2 (a) Primitive machine at the instant of maximum current in phase Aconductors and ﬁeld m.m.f. Ff. (b) Corresponding space diagram of vector m.m.f.swith resultant F. (c) Terminal phasor diagram, (d) Phasor diagram with voltage repre-sentation of magnetic conditions, (e) Equivalent circuit for round-rotor machine

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86 Electric Power Systems, Fifth Editionphase A is a maximum and the voltage is at an angle f as shown in Figure 3.2(c).Currents in phases B and C of the stator (equal and opposite at this instant) also con-tribute along the axis of Fa. The resultant air-gap m.m.f. is shown as F ¼ Fa þ Ff, usingvectorial addition as in Figure 3.2(b), and this gives rise to a displaced air-gap ﬂuxsuch that the phase of the stator-induced e.m.f. E is delayed from that produced onno-load for the same rotor position by the torque angle shown as d in Figure 3.2(d). Detailed consideration of generator action taking full account of magneticpath saturation requires a simulation of the ﬂuxes, including all winding currentsand eddy currents. However, for power system calculations the m.m.f. summa-tion of Figure 3.2(b) can be transferred to the voltage-current phasor diagram ofFigure 3.2(d) by rotation through 90 clockwise and scaling F to Va. Then, Ffbecomes E and Fa corresponds to IXm where Xm is an equivalent reactance repre-senting the effect of the magnetic conditions within the machine. Effects due tosaturation can be allowed for by changing Xm and relating E to the ﬁeld currentIf by the open circuit (no-load) saturation curve (see later). The machine terminalvoltage Vt is obtained from Va by recognizing that the stator phase windingshave a small resistance RL and a leakage reactance XL (about 10% of Xm) resultingfrom ﬂux produced by the stator but not crossing the air gap. The reactances XLand Xm are usually considered together as the synchronous reactance XS, shownin the equivalent circuit of Figure 3.2(e). The machine phasor equation is then E ¼ Vt þ IðRL þ jXSÞ In most power system calculations RL is neglected and a simple phasor diagramresults.3.2.1 Two-Axis RepresentationIf a machine has saliency, that is, the rotor has salient poles or the rotor has non-uniform slotting in a solid rotor, then the main ﬁeld axis and the cross-ﬁeld axishave different reluctances. The armature reaction m.m.f., Fa in Figure 3.2(b), can beresolved into two components Fd and Fq, as shown in Figure 3.3(b) for the primitivemachine shown in Figure 3.3(a). The m.m.f./ﬂux/voltage transformation will nowbe different on these two component axes (known as the direct and quadratureaxes) as less ﬂux and induced e.m.f. will be produced on the q axis. This differenceis reﬂected in the voltage phasor diagram of Figure 3.3(d) as two different reactancesXad and Xaq, the component of stator current acting on the d axis Id being associatedwith Xad, and Iq with Xaq. The resolution of the stator m.m.f. into two components is represented in Figure 3.3(a)by two coils identiﬁed by the axes on which they produce m.m.f. Repeating theprevious transformation from m.m.f. F to voltages, Figure 3.3(b) transforms to Fig-ure 3.3(c) by considering the two-axis components of current and then results inFigure 3.3(d). Saturation of the two axes is represented by modifying Xad and Xaq.Because I has been resolved into two components it is no longer possible for asalient machine to be represented by an equivalent circuit. However, equationscan be obtained for the Vd and Fq components of the terminal voltage Vt as:

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Components of a Power System 87 q coil q axis Ff Fa d axis d coil Fq F N S q coil Fd If (b) d coil i (a) Eq jIqXaq I Iq jIdXad Id Va (c) Vt jIXL Vq IRL Vd (d)Figure 3.3 (a) Primitive machine – d and q axes, (b) The m.m.f. vectors resolved ind and q axes. Fd and Fq are components of Fa., (c) Current phasors resolved into Idand Iq components, (d) Phasor diagram of voltages, E of Figure 3.2(d) now becomes Eq Vq ¼ Eq À jIdðXad þ XLÞ À IqRL Vd ¼ ÀjIqðXaq þ XLÞ À IdRL It will be seen that the resistive terms (if included) contain the currents from theother axis. The term Eq is the e.m.f. representing the action of the ﬁeld windingand in many power system studies can be taken as constant if no automatic voltageregulator (AVR) action is assumed. A single equation can be written in complex form aswhere Eq ¼ V þ IRL þ jIXL þ jðIdXad þ IqXaqÞand ¼ V þ IRL þ jðIdXd þ IqXqÞ Xd ¼ Xad þ XL Xq ¼ Xaq þ XL In a salient machine, Xad ¼6 Xaq and torque and power can be developed with noﬁeld current, the stator m.m.f. ‘locking’ in with the direct axis as in a reluctance

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88 Electric Power Systems, Fifth Editionmachine. The effect depends upon the saliency but is generally small compared withthe maximum power rating with ﬁeld current. Consequently, for simple multi-machine power system calculations the equivalent circuit of Figure 3.2(e) can be used. The transient behaviour of synchronous machines is often expressed in two axisterms, the basic concept being similar to that shown in Figure 3.3. However, a vari-ety of conventions are employed and care must be taken to establish which directionof current is being taken as positive. In many texts, if a synchronous machine is run-ning as a motor, the current entering the machine is taken as positive. Here we haveassumed that current leaving a generator is positive. To produce accurate voltage and current waveforms under transient conditionsthe two-axis method of representation with individually coupled coils on the d andq axes is necessary. Since these axes are at right-angles (orthogonal), then ﬂuxeslinking coils on the same axis do not inﬂuence the ﬂuxes linking coils on the otheraxis, that is, no cross-coupling is assumed. Differential equations can be set up toaccount for ﬂux changes on both axes separately, from which corresponding volt-ages and currents can be derived through step-by-step integration routines. Thisform of calculation is left for advanced study. The use of ﬁnite elements can extendthe accuracy of calculation still further.3.2.2 Effect of Saturation on XS - the Short-Circuit RatioThe open-circuit characteristic is the graph of generated voltage against ﬁeld currentwith the machine on open circuit and running at synchronous speed. The short-circuit characteristic is the graph of stator current against ﬁeld current with the ter-minals short-circuited. Both curves for a modern machine are shown in Figure 3.4.Here, XS is equal to the open-circuit voltage produced by the same ﬁeld current thatproduces rated current on short circuit, divided by this rated stator current. XS isconstant only over the linear part of the open-circuit characteristic (the air-gap line)when saturation is ignored. The actual value of XS at full-load current will obviouslybe less than this value and several methods exist to allow for the effects of saturation. The short-circuit ratio (SCR) of a generator is deﬁned as the ratio between the ﬁeldcurrent required to give nominal open-circuit voltage and that required to circulatefull-load current in the armature when short-circuited. In Figure 3.4 the short-circuitratio is AH/AK, that is, 0.63. The SCR is also commonly calculated in terms of theair-gap line and the short-circuit curve, this giving an unsaturated value. To allowfor saturation it is common practice to assume that the synchronous reactance is1/SCR, which for this machine is 1.58 p.u. Economy demands the design of machinesof low SCR and a value of 0.55 is common for modern machines. Unfortunately,transient stability margins are reduced as synchronous reactance increases (i.e. SCRreduces) so a conﬂict for design between economy and stability arises.3.2.3 TurbogeneratorsDuring the 1970s, the ratings of steam-driven turbine generators reached more than1000 MW to obtain improvements in efﬁciency made possible by large turbine plantand their lower capital costs per megawatt.

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Components of a Power System 89Stator current p.u.1.5 1.5 Open circuit F Stator voltage p.u.1.0 1.0curve D0.5 0.5 B SaturatioAnilrignaep line C Short circuit curve 0 GH K A 1 p.u. 2 p.u. Field current for armature reaction Field current for leakage Field current flux (XL = 0.1 p.u.) p.u.Figure 3.4 Open- and short-circuit characteristics of a synchronous machine.Unsaturated value of XS ¼ FK=CK. With operation near nominal voltage, the saturationline is used to give a linear characteristic with saturation XS ¼ DK=CK Large generators are cooled by hydrogen circulating around the air-gap andde-ionized water pumped through hollow stator conductors, a typical statisticbeing:500 MW, 588 MVA winding cooling rotor, hydrogen; stator, water; rotor diameter 1.12 m; rotor length 6.2 m; total weight 0.63 kg/kVA; rotor weight 0.1045 kg/kVA.

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90 Electric Power Systems, Fifth Edition 0 5mFigure 3.5 Sectional view of a 1000 MVA, 1800 r.p.m., 60 Hz generator (Figureadapted from IEE/IET) A cross-sectional view of a 1000 MVA steam turbogenerator is shown inFigure 3.5. The most important problems that were encountered in the development and useof large generators arose from (1) mechanical difﬁculties due to the rotation of largemasses, especially stresses in shafts and rotors, critical speeds and torsional oscilla-tions; (2) the large acceleration forces produced on stator bars which must be with-stood by their insulation; (3) the need for more effective cooling. Mica paper andglass bonded with epoxy resin were used as stator insulation, the maximum permis-sible temperature being 135 C. Semiconductor rectiﬁers are employed to produce the d.c. excitation. Earlydesigns of generating sets had d.c. exciters mounted on the main shaft. This typewas followed by an a.c. exciter from which the current is rectiﬁed so that a.c. andd.c. slip-rings are required. In more modern machines, a.c. exciters with integralfused-diode or thyristor rectiﬁers are employed, thus avoiding any brush gear andconsequent maintenance problems, the semiconductors rotating on the main shaft. Reliability of turbogenerators is all-important, and the loss resulting from an out-age of a large conventional generator for one day is roughly equivalent to 1% of theinitial cost of the machine.3.3 Equivalent Circuit Under Balanced Short-Circuit ConditionsA typical set of oscillograms of the currents in three stator phases when a synchro-nous generator is suddenly short-circuited is shown in Figure 3.6(a). In all threetraces a direct-current component is evident and this is to be expected from a knowl-edge of transients in R-L circuits. The magnitude of the initial transient direct cur-rent in any phase present depends upon the instant at which the short circuit isapplied and the speed of its decay depends on the power factor of the circuit. As

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Components of a Power System 91 Time a - phase b - phase c - phase d.c. component Instant of short circuit (a) b I″ I ′ Envelope of 50 Hz waveform aStator current c 1 4 I 0 2 4 6 8 10 12 cycles 0.2(s) 8 Time (s) Instant ofshort circuit (b)Figure 3.6 (a) Oscillograms of the currents in the three phases of a generator when asudden short circuit is applied, (b) Trace of a short-circuit current when direct-currentcomponent is removed

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92 Electric Power Systems, Fifth Editionthere are three voltages mutually at 120 it is possible for only one to have the maxi-mum direct-current component. Often, to clarify the physical conditions, the direct current component is ignoredand a trace of short-circuit current, as shown in Figure 3.6(b), is considered. Immedi-ately after the application of the short circuit the armature current endeavours tocreate an armature reaction m.m.f., as already mentioned, but the main ﬂux cannotchange to a new value immediately as it is linked with low-resistance circuits con-sisting of: (1) the rotor winding which is effectively a closed circuit; (2) the damperbars, that is, a winding which consists of short-circuited turns of copper strip set inthe poles to dampen oscillatory tendencies, and (3) the rotor body, often of forgedsteel. As the ﬂux remains unchanged initially, the stator currents are large and canﬂow only because of the creation of opposing currents in the rotor and damperwindings by what is essentially transformer action. Owing to its higher resistance,the current induced in the damper winding decays rapidly and the armature currentstarts to fall. After this, the currents in the rotor winding and body decay, the arma-ture reaction m.m.f. is gradually established, and the generated e.m.f. and stator cur-rent fall until the steady-state condition on short circuit is reached. Here, the fullarmature reaction effect is operational and the machine is represented by the syn-chronous reactance XS. These effects are shown in Figure 3.6, with the high initialcurrent due to the damper winding and then the gradual reduction until the fullarmature reaction is established. To represent the initial short-circuit conditions twonew models must be introduced. If, in Figure 3.6(b), the envelopes of the 50 Hzwaves are traced, a discontinuity appears. Whereas the natural envelope continuesto point ‘a’ on the stator current axis, the actual trace ﬁnishes at point ‘b’; the reasonsfor this have been mentioned above. To account for both of these conditions, twonew reactances are needed to represent the machine, the very initial conditionsrequiring what is called the subtransient reactance (X00) and in the subsequentperiod the transient reactance (X0). In the following deﬁnitions it is assumed that thegenerator is on no-load prior to the application of the short circuit and is of theround-rotor type. Let the no-load phase voltage of the generator be E volt (r.m.s.),then, from Figure 3.6(b), the subtransient reactance X00 ¼ Epﬃﬃ 0b= 2where pﬃﬃ is the r.m.s. value of the subtransient current (I00); the transient 0b 2reactance X0 ¼ Epﬃﬃ 0a= 2where pﬃﬃ is the r.m.s. value of the transient current (I0 ), and ﬁnally 0a 2 XS ¼ Epﬃﬃ 0c= 2

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Components of a Power System 93Table 3.1 Constants of synchronous machines – 60 Hz (all values expressed as per uniton rating)Type of machine XS (or Xd) Xq X0 X00 X2 X0 RLTurbo-alternator 1.2–2.0 1–1.5 0.2–0.35 0.17–0.25 0.17–0.25 0.04–0.14 0.003–0.008Salient pole 0.16–1.45 0.4–1.0 0.2–0.5 0.13–0.35 0.13–0.35 0.02–0.2 0.003–0.0015 (hydroelectric) 1.5–2.2 0.95–1.4 0.3–0.6 0.18–0.38 0.17–0.37 0.03–0.15 0.004–0.01Synchronous compenstorX2 ¼ negative sequence reactance.X0 ¼ zero sequence reactance.X0 and X00 are the direct axis quantities.RL ¼ a.c. resistance of the stator winding per phase. Typical values of X00, X0 and XS for various types and sizes of machines are givenin Table 3.1. If the machine is previously on load, the voltage applied to the equivalentreactance, previously E, is now modiﬁed due to the initial load volt-drop. ConsiderFigure 3.7. Initially, the load current is IL and the terminal voltage is V. The voltagebehind the transient reactance X0 is E0 ¼ ILðZL þ jX0Þ ¼ V þ jILX0and hence the transient current on short circuit ¼ E0 . jX0 XI IL ZL V EI Figure 3.7 Modiﬁcation of equivalent circuit to allow for initial load current

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94 Electric Power Systems, Fifth Edition AB ERA ERB S ωA EYAR EBB ωB EYBYB EBA (b) (a)Figure 3.8 (a) Generators in parallel, (b) Corresponding phasor diagrams3.4 Synchronous Generators in ParallelConsider two machines A and B (as shown in Figure 3.8(a)), the voltages of whichhave been adjusted to be equal by their ﬁeld regulators, and the speeds of which areslightly different. In Figure 3.8(b) the phase voltages are ERA, and so on, and thespeed of machine A is vA radians per second and of B, vB radians per second. If the voltage phasors of A are considered stationary, those of B rotate at a relativevelocity (vB À vA) and hence there are resultant voltages across the switch S of(ERA À ERB), and so on, which reduce to zero once during each relative revolution.If the switch is closed at an instant of zero voltage, the machines are connected(synchronized) without the ﬂow of large currents due to the resultant voltagesacross the armatures. When the two machines are in synchronism they have acommon terminal-voltage, speed and frequency. Any tendency for one machine toaccelerate relative to the other immediately results in a retarding or synchronizingtorque being set up due to the current circulated. Two machines operating in parallel, with EA ¼ EB and on no external load, arerepresented by the equivalent circuit shown in Figure 3.9(a). If A tries to gain speedthe phasor diagram in Figure 3.9(b) is obtained and I ¼ ER/(ZA þ ZB). The circulat-ing current I lags ER by an angle tanÀ1ðX=RÞ and, as in most machines X ) R, thisangle approaches 90. This current is a generating current for A and a motoring cur-rent for B; hence A is generating power and tending to slow down and B is receivingpower from A and speeding up. Therefore, A and B remain at the same speed, ‘instep’, or in synchronism. Figure 3.9(c) shows the state of affairs when B tries to gainspeed on A. The quality of a machine to return to its original operating state after amomentary disturbance is measured by the synchronizing power and torque. It isinteresting to note that as the impedance of the machines is largely inductive, therestoring powers and torques are large; if the system were largely resistive it wouldbe difﬁcult for synchronism to be maintained. Normally, more than two generators operate in parallel and the operation of onemachine connected in parallel with many others is of great interest. If the remainingmachines in parallel are of such capacity that the presence of the generator under

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Components of a Power System 95A EA EB B I ERZA ZB EA EB (a) ω ER (b) EB EA Iω (c)Figure 3.9 (a) Two generators in parallel–equivalent circuit, (b) Machine A in phaseadvance of machine B. (c) Machine B in phase advance of machine Astudy causes no difference to the voltage and frequency of the other, they are said tocomprise an inﬁnite busbar system, that is, an inﬁnite system of generation. In prac-tice, a perfect inﬁnite busbar is never fully realized but if, for example, a 600 MWgenerator is removed from a 30 000 MW system, the difference in voltage andfrequency caused will be very slight.3.5 The Operation of a Generator on an Inﬁnite BusbarIn this section, in order to simplify the ideas as much as possible, the resistance ofthe generator will be neglected; in practice, this assumption is usually reasonable.Figure 3.10(a) shows the schematic diagram of a machine connected to an inﬁnitebusbar along with the corresponding phasor diagram. If losses are neglected thepower output from the turbine is equal to the power output from the generator.The angle d between the E and V phasors is known as the load angle and dependson the power input from the turbine shaft. With an isolated machine supplyingits own load the load dictates the power required. When connected to an inﬁnitebusbar the load delivered by the machine is no longer directly dependent on theconnected load. By changing the turbine output, and hence d, the generator can bemade to take on any load that the operator desires subject to economic and technicallimits.

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96 Electric Power Systems, Fifth Edition XS V E T I Infinite IX S E busbar δV (a) I (b)Figure 3.10 (a) Synchronous machine connected to an inﬁnite busbar, (b) Corre-sponding phasor diagram From the phasor diagram in Figure 3.10(b), the power delivered to the inﬁnitebusbar ¼ P ¼ VI cos f per phase, but (using the sine rule) E þ fÞ ¼ IXS sinð90 sin dHence I cos f ¼ E sin d XS Power delivered ¼ VE sin d ð3:1Þ XS This expression is of extreme importance as it governs, to a large extent, the oper-ation of a power system. It could have been obtained by neglecting resistance andhence setting u in Equation (2.11) to 90. Equation (3.1) is shown plotted in Figure 3.11. The maximum power is obtained atd ¼ 90. If d becomes larger than 90 due to an attempt to obtain more than Pmax,increase in d results in less power output and the machine becomes unstable andloses synchronism. Loss of synchronism results in the interchange of current surgesbetween the generator and network as the poles of the machine pull into synchro-nism and then out again; that is, the generator ‘pole slips’. If the power output of the generator is increased by small increments with the no-load voltage kept constant, the limit of stability occurs at d ¼ 90 and is known as thesteady-state stability limit. There is another limit of stability due to a sudden largechange in conditions, such as caused by a fault, and this is known as the transientstability limit; it is possible for the rotor to oscillate beyond 90 a number of times. Ifthese oscillations diminish, the machine is stable. The load angle d has a physicalsigniﬁcance; it is the angle between like radial marks on the end of the rotor shaft of

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Components of a Power System 97 Pmax P Po 0 90o 180o δ Electrical degreesFigure 3.11 Power-angle curve of a synchronous machine. Resistance and saliencyare neglectedthe machine and on an imaginary rotor representing the system. The marks are inidentical physical positions when the machine is on no-load. The synchronizingpower coefﬁcient¼ dP ½watts per radian ddand the synchronizing torque coefﬁcient ¼ 1 dP vS dd3.5.1 The Performance Chart of a Synchronous GeneratorIt is convenient to summarize the operation of a synchronous generator connected toan inﬁnite busbar in a single diagram or chart which allows an operator to seeimmediately whether the machine is operating within the limits of stability andrating. Consider Figure 3.12(a), the phasor diagram for a round-rotor machine connectedto a constant voltage busbar, ignoring resistance. The locus of constant IXS, hence,

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98 Electric Power Systems, Fifth Edition Locus of constant E (circle centre 0 ) ps E IXs Locus IXs (circle centre 0) δV 0 I0 q o Locus of (a) Phasor diagram constant ITheoretical MVAr limit 150 Power MW Scale 0.8 p.f. ldg.25 MVAE = 2.5 p.u.E = 2 p.u. 100 E =1.5 p.u. 75 g f h e 50 E = V =1 p.u. MVAa 25bcδ0 d –25 0 25 50 j 75 100 Leading Lagging MVAr(absorbing VArs) (generating VArs) (b) Figure 3.12 Performance chart of a synchronous generatorand constant MVA, is a circle, and the locus of constant excitation E is another circle.Therefore, provided V is constant, then0s is proportional to VI or MVAps is proportional to VI sin f or MVArsq is proportional to VI cos f or MW.

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Components of a Power System 99 To obtain the scaling factor for MVA, MVAr, and MW, the fact that at zero excita-tion, E ¼ 0 and IXS ¼ V, is used, from which I is V/Xs at 90 leading to 000 corre-sponding to VAr/phase. The construction of a chart for a 60 MW machine follows (Figure 3.12(b)).Machine Data60 MW, 0.8 p.f., 75 MVA11.8 kV, SCR 0.63, 3000 r.p.m.Zbase ¼ 11:82 ¼ 1:856V 75XS ¼ 1 p:u: ¼ 1:59pu ¼ 2:94V=phase 0:63 The chart will refer to complete three-phase values of MW and MVAr. When theexcitation and hence E apreﬃﬃreduced to zero, the current leads V by 90 and is equal to(V/XS), that is, 11 800/ð 3 Â 2.94). The leading VArs correspond to this, given by3V2/XS ¼ 47.4 MVAr. With centre 0 a number of semicircles are drawn of radii equal to various MVAloadings, the most important being the 75 MVA circle. Also, arcs with 00 as centreare drawn with various multiples of 000 (or V) as radii to give the loci for constantexcitations. Lines may also be drawn from 0 corresponding to various power factors,but for clarity only 0.8 p.f. lagging is shown, that is, the machine is generating VArs.The operational limits are ﬁxed as follows. The rated turbine output gives a 60 MWlimit which is drawn as shown, that is, a line for example, which meets the 75 MVAline at g. The MVA arc governs the thermal loading of the machine, that is, the statortemperature rise, so that over the portion gh the output is decided by the MVA rat-ing. At point h the rotor heating becomes more decisive and the arc hj is decided bythe maximum excitation current allowable, in this case assumed to be 2.5 p.u. Theremaining limit is that governed by loss of synchronism at leading power factors.The theoretical limit is the line perpendicular to 000 at 00 (i.e. d ¼ 90), but, in practice,a safety margin is introduced to allow a further increase in load of either 10 or 20%before instability. In Figure 3.12(b) a 10% margin is used and is represented by ecd:it is constructed in the following manner. Considering point ‘a’ on the theoreticallimit on the E ¼ 1 p.u. arc, the power 00a is reduced by 10% of the rated power (i.e.by 6 MW) to 00b; the operating point must, however, still be on the same E arc and bis projected to c, which is a point on the new limiting curve. This is repeated forseveral excitations, ﬁnally giving the curve ecd. The complete operating limit is shown shaded and the operator should normallywork within the area bounded by this line, provided the generator is running at ter-minal rated voltage. If the voltage is different from rated, for example, at À5% ofrated, the whole operating chart will shrink by a pro-rata amount, except for theexcitation and turbine maximum power line. Note, therefore, that for the generator

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100 Electric Power Systems, Fifth Editionto produce rated power at reduced voltage, the stator requires to operate on over-current. This may be possible if an overcurrent rating is given.Example 3.1Use the chart of Figure 3.12 to determine the excitation and the load angle at full load(60 MW, 0.8 p.f. lag) and at unity power factor, rated output. Check by calculation.SolutionFrom the diagram, the following values are obtained:0.8 p.f. lag Excitation E Angle du.p.f. 2.3 p.u. 33 1.7 p.u. 50Check Power ¼ VE sin d XS sin d ¼ 60 Â 106 Â 2:94 11800 Â ð11800 Â 2:3Þ d ¼ 0:55rad ¼ 33:4degRepeating check in per unit (on 75 MVA, 11.8 kV base) sin d ¼ 0:8 Â 1:58 2:3 d ¼ 0:55rad3.6 Automatic Voltage Regulators (AVRs)Excitation of a synchronous generator is derived from a d.c. supply with variablevoltage and requires considerable power to produce the required operating ﬂux.Hence an exciter may require 1 MW or more to excite a large generator’s ﬁeld wind-ing on the rotor. Nowadays, such a power ampliﬁer is best arranged through an a.c.generator driven or overhung on the rotor shaft feeding the rotor through diodes,also mounted on the shaft and rotating with it (see Figure 3.13). The diodes can bereplaced by thyristor controllers and the power ampliﬁer can be eliminated. In general the deviation of the terminal voltage from a prescribed value is passedto control circuits and thus the ﬁeld current is varied. A general block diagram of atypical control system is given in Figure 3.14. The most important aspect of the excitation system is its speed of response betweena change in the reference signal Vr and the change in the excitation current If.

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Components of a Power System 101 Rotor If If winding Power amp. Vref same shaftFigure 3.13 Excitation arrangements for a synchronous generator using rotatingdiodesControl system theory using Bode diagrams and phase margins is required to designappropriate responses and stability boundaries of a generator excitation system.3.6.1 Automatic Voltage Regulators and Generator CharacteristicsThe equivalent circuit used to represent the synchronous generator can be modiﬁedto account for the action of a regulator. Basically there are three conditions toconsider:1. Operation with ﬁxed excitation and constant no-load voltage (E), that is, no regu- lator action. This requires the usual equivalent circuit of E in series with XS.Vr + + Amplifier Exciter Generator V _ G1 G3 G2 G4 _ sτ 3 1+sτ2 1+sτ 4 G5s 1+sτ5 Feed backFigure 3.14 Block diagram of a continuously acting closed-loop automatic voltageregulator

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102 Electric Power Systems, Fifth Edition Non-continuously MVA limit acting AVR 1 p.u.. No regulator Power Continuously acting AVR 1 p.u.. Leading VAr (absorbing)Figure 3.15 Performance chart as modiﬁed by the use of automatic voltageregulators2. Operation with a regulator which is not continuously acting, that is, the terminal voltage varies with load changes. This can be simulated by E0 and a reactance smaller than the synchronous value. It has been suggested by experience, in prac- tice, that a reasonable value would be the transient reactance, although some authorities suggest taking half of the synchronous reactance. This mode will apply to most modern regulators.3. Terminal voltage constant. This requires a very-fast-acting regulator and the near- est approach to it exists in the forced-excitation regulators (that is, AVRs capable of reversing their driving voltage to suppress it quickly) used on generators supplying very long lines.Each of the above representations will give signiﬁcantly different values of maxi-mum power output. The degree to which this happens depends on the speed of theAVR, and the effect on the operation chart of the synchronous generator is shown inFigure 3.15, which indicates clearly the increase in operating range obtainable. Itshould be noted, however, that operation in these improved leading power-factorregions may be limited by heating of the stator winding. The actual power-anglecurve may be obtained through a step-by-step process by using the graduallyincreasing values of E in EV sin d. XWhen a generator has passed through the steady-state limiting angle of d ¼ 90with a fast-acting AVR, it is possible for synchronism to be retained. The AVR, inforcing up the voltage, increases the power output of the machine so that instead ofthe power falling after d ¼ 90, it is maintained and dP/dd is still positive. This is

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Components of a Power System 103Eδ X p.u. Vs 0o Vt (a) Infinite 2 Excitation busbar limit 2.5 p.u.P and voltage p.u. E P Vt 1 0 90o 180o δ (b)Figure 3.16 (a) Generator feeding into an inﬁnite busbar (b) variation of outputpower P, generator voltage E, and terminal voltage Vt with load (transmission angle d).Perfect AVRillustrated in Figure 3.16(b), where the P À d relation for the system in Figure 3.16(a)is shown. Without the AVR the terminal voltage of the machine Vt will fall withincreased d, the generated voltage E being constant and the power reaching a maxi-mum at 90. With a perfect AVR, Vt is maintained constant, E being increased withincrease in d. As P ¼ (EVS/X) sin d, it is evident that power will increase beyond 90until the excitation limit is reached, as shown in Figure 3.16(b).3.7 Lines, Cables and Transformers3.7.1 Overhead Lines –Types and ParametersOverhead lines are suspended from insulators which are themselves supportedby towers or poles. The insulation of the conductors is air. The span betweentwo lowers depends on the allowable sag in the line, and for steel towers with very

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104 Electric Power Systems, Fifth Edition47.5 m25.3 m 18.9 m 7.3 mFigure 3.17 400 kV double-circuit overhead-line tower. Two conductors per phase(bundle conductors)high-voltage lines the span is normally 370–460 m (1200–1500 ft). Typical supportingstructures are shown in Figures 3.17 and 3.18. There are two main types of tower:1. Those for straight runs in which the stress due to the weight of the line alone has to be withstood.2. Those for changes in route, called deviation towers; these withstand the resultant forces set up when the line changes direction. When specifying towers and lines, ice and wind loadings are taken into account,as well as extra forces due to a break in the conductors on one side of a tower. Forlower voltages and distribution circuits, wood or reinforced concrete poles are usedwith conductors supported in horizontal formation. The live conductors are insulated from the towers by insulators which take twobasic forms: the pin-type and the suspension type. The pin-type insulator is shownin Figure 3.19 and it is seen that the conductor is supported on the top of the insula-tor. This type is used for lines up to 33 kV. The two or three porcelain ‘sheds’ or‘petticoats’ provide an adequate leakage path from the conductor to earth and are

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Components of a Power System 105Figure 3.18 Typical pole-type structuresConductor tied downwith soft copper wire Sheds Porcelain or glass Steel pinFigure 3.19 Pin-type insulators

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106 Electric Power Systems, Fifth Edition Sheds Pin Figure 3.20 Suspension-type insulatorsshaped to follow the equi-potentials of the electric ﬁeld set up by the conductor-tower system. Suspension insulators (Figure 3.20) consist of a string of interlinkingseparate discs made of glass or porcelain. A string may consist of many discsdepending upon the line voltage; for 400 kV lines, 19 discs of overall length 3.84 m(12 ft 7 in) are used. The conductor is held at the bottom of the string which is sus-pended from the tower. Owing to the capacitances existing between the discs, con-ductor and tower, the distribution of voltage along the insulator string is notuniform, the discs nearer the conductor being the more highly stressed. Methods ofcalculating this voltage distribution are available, but are of limited value owing tothe shunting effect of the leakage resistance (see Figure 3.21). This resistancedepends on the presence of pollution on the insulator surfaces and is considerablymodiﬁed by rain and fog.3.7.1.1 Inductance and CapacitanceThe parameters of interest for circuit analysis are inductance, capacitance and resist-ance. The derivation of formulae for the calculation of these quantities is given inmany reference and text books. It is intended here merely to quote these formulaeand discuss their application. The inductance of a single-phase two-wire line is ! m0 dÀrL ¼ 4p 1 þ 4 ln r H=m

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Components of a Power System 107 Tower C R C1 R C R C2 C1 C2 C R C2 C1 C ConductorFigure 3.21 Equivalent circuit of a string of four suspension insulators. C ¼ self capaci-tance of disc; C1 ¼ capacitance disc to earth; C2 ¼ capacitance disc to line; R ¼ leak-age resistanceSince d ) r ! m0 d L ¼ 4p 1 þ 4 ln r ! ¼ m0 ln e1=4 d m0 d 4p 4 þ 4 ln r ¼ p ln reÀ1=4 dDeﬁning r0 ¼ reÀ1=4 and substituting m0 ¼ 4p Â 10À7; L ¼ 4 ln r0 Â 10À7 where d is the distance between the centres, and r is the radius, of the conductors. When performing load ﬂow and balanced-fault analysis on three-phase systems itis usual to consider one phase only, with the appropriate angular adjustments madefor the other two phases. Therefore, phase voltages are used and the inductancesand capacitances are the equivalent phase or line-to-neutral values. For a three-phase line with equilateral spacing (Figure 3.22) the inductance and capacitancewith respect to the hypothetical neutral conductor are used, and this inductance can

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108 Electric Power Systems, Fifth Edition R dNd Y dB Radius rFigure 3.22 Overhead line with equilateral spacing of the conductorsbe shown to be half the loop inductance of the single-phase line, that is, the induc-tance of one conductor. The line-neutral inductance for equilateral spacing and d ) r ! m0 dL ¼ 8p 1 þ 4 ln r H=mThe capacitance of a single-phase line (again d ) r) C ¼ pe0 F=m ln½d=r With three-phase conductors spaced equilaterally, the capacitance of each line tothe hypothetical neutral is double that for the two-wire circuit, that is. C ¼ 2pe0 F=m ln½d=r Although the conductors are rarely spaced in the equilateral formation, it can beshown that the average value of inductance or capacitance for any formation of con-ductors can be obtained by the representation of the system by one of equivalentequilateral spacing (Figure 3.23). The equivalent spacing deq between conductors isgiven by pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ deq ¼ 3 d12 Â d23 Â d31 The use of bundle conductors, that is, more than one conductor per insulator,reduces the reactance; it also reduces conductor to ground surface voltage gradients

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Components of a Power System 109 a d12 d31 b d23 cFigure 3.23 Geometric spacing of conductorsand hence corona loss and radio interference. When two circuits are situated on thesame towers the magnetic interaction between them should be taken into account. Unsymmetrical conductor spacing results in different inductances for each phasewhich causes an unbalanced voltage drop, even when the load currents are bal-anced. The residual or resultant voltage or current induces unwanted voltages intoneighbouring communication lines. This can be overcome by the interchange of con-ductor positions at regular intervals along the route, a practice known as transposi-tion (see Figure 3.24). In practice, lines are rarely transposed at regular intervals andtransposition is carried out where physically convenient, for example at substations.In short lines the degree of unbalance existing without transposition is small andmay be neglected in calculations.3.7.1.2 ResistanceOverhead-line conductors usually comprise a stranded steel core (for mechanicalstrength) surrounded by aluminium wires which form the conductor. It should benoted that aluminium and ACSR (aluminium conductor steel reinforced) a cb b ac c ba Figure 3.24 Transposition of conductors

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110 Electric Power Systems, Fifth Editionconductors are sometimes described by area of a copper conductor having the samed.c. resistance, that is their copper equivalent. In Table 3.2a, 258 mm2 (0.4 in2) ACSRconductor implies the copper equivalent. The actual area of aluminium is approxi-mately 430 mm2, and including the steel core the overall cross-section of the line isabout 620 mm2, that is a diameter of 28 mm. Electromagnetic losses in the steelincrease the effective a.c. resistance, which increases with current magnitude, givingan increase of up to about 10%. With direct current the steel carries 2–3% of the totalcurrent. The electrical resistivities of some conductor materials are as follows inohm-metres at 20 C: copper 1.72 Â 10À8; aluminium 2.83 Â 10À8 (3.52 Â 10À8 at80 C); aluminium alloy 3.22 Â 10À8. Table 3.2a gives the parameters for various overhead-line circuits for the linevoltages operative in Britain, Table 3.2b gives values for international lines, andTable 3.2c relates to the USA.3.7.2 Representation of LinesThe manner in which lines and cables are represented depends on their length andthe accuracy required. There are three broad classiﬁcations of length: short, mediumand long. The actual line or cable is a distributed constant circuit, that is it has resist-ance, inductance and capacitance distributed evenly along its length, as shown inFigure 3.25. Except for long lines the total resistance, inductance and capacitance of the lineare concentrated to give a lumped-constant circuit. The distances quoted are only arough guide.Table 3.2a Overhead-line constants at 50 Hz (British) (per phase, per km)Voltage: No. and area 132 kV 275 kV 400 kVof conductors (mm)2: l Â 113 l Â 258 2 Â 113 2 Â 258 2 Â 258 4 Â 258Resistance (R) V 0.155 0.068 0.022 0.034 0.034 0.017 0.41 0.404 0.335 0.323 0.323 0.27Reactance (XL) V 7.59 7.59 9.52 9.52 9.52 10.58Susceptance (1/Xc) S Â 10À6 0.22 0.22 0.58 0.58 0.845 0.945Charging current (Ic) A 373 371 302 296 296 258Surge impedance V 47 47 250 255 540 620 MW 2.6 5.9 4.3 9.5 9.5 15.8Natural load —XL/R ratio MVA 125 180 525 760 1100 2200Thermal rating: MVA 100 150 430 620 900 1800Cold weather (below 790 1580 5 C) MVA 80 115 330 480Normal (5–18 C)Hot weather (above 18 C)

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Table 3.2b Typical characteristics of bundled-conductor E.H.V.Number of Country Line voltage and Diameter ofsubconductors number of circuits (in subconductoin bundle parentheses)1 Japan kV mm Canada 275 (2) 27.9Average (one Australia 300 (2) 35.0 conductor): Russia 330 (1) 45.0 Italy 330 (1) 30.22 380 (1) 50.0 JapanAverage (two Canada 275 (2) 25.3 conductors): Australia 360 (1) 28.1 U.S.A. 330 (1) 31.83 Italy 345 (1) 30.44 Russia 380 (1) 31.5 330 (1) 28.0 Sweden Russia 380 (1) 31.68 Germany 525 (1) 30.2 Germany 380 (2) 21.7 Canada 500 (1) 22.86 735 (1) 35.04

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lines Components of a Power Systemors Radius of circle on Resistance Inductive Susceptance which subconductors of bundle reactance at at 50 Hz are arranged 50 Hz V/km mS/km mm 0.0744 V/km 3.01 — 0.0451 0.511 2.33 — 0.0367 0.492 2.78 — 0.065 0.422 2.82 — 0.0294 0.404 2.84 — 0.048 0.398 2.75 0.442 200 0.0444 4.35 299 0.04 0.374 3.57 190.5 0.0451 0.314 3.34 228.6 0.0315 0.341 3.55 200 0.0285 0.325 3.57 200 0.04 0.315 3.43 0.040 0.321 3.58 260 0.323 230 0.018 3.85 202 0.0212 0.29 3.88 323 0.0316 0.294 4.3 323 0.0285 0.260 4.0 0.0121 0.272 3.9 0.279 111

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Table 3.2c Overhead-line parameters – USALine voltage (kV) 345Conductors per phase at 18 in spacing 1 ExpandedConductor code 1.750Conductor diameter (inches) 28 35.3Phase spacing (ft) 0.3336GMD (ft) 0.4325 0.7661 8 < XA 0.0777 0.105760 Hz inductive reactance V=mile: XD 0.1834 XA þ XD 8 X0 < X0D X0A60 Hz capacitive reactance MV-miles: þ X0DZ0(V) 374.8 318Natural loading MVAConductor d.c. resistance at 25 C (V/mile) 0.0644Conductor a.c. resistance (60 Hz) at 50 C (V/mile) 0.0728XA ¼ component of inductive reactance due to ﬂux within a 1 ft radius.XD ¼ component due to other phases.Total reactance per phase ¼ XA þ XD. !XA ¼ 0:2794 log10 1 V=mile. ½NðGMRÞðAÞNÀ1 1=NXD ¼ 0.2794 log10 (GMD) V/mile. 0XA0 and XD0 A are similarly deﬁned for capacitive reactance and X ¼ 0:0683 log10GMR ¼ geometric mean radius in feet;GMD ¼ geometric mean diameter in feet; N ¼ number of conductors per phase; A ¼ S/2 sin (p/N); N > 1 A ¼ 0; N ¼ 1; S ¼ bundle spacing in feet; r ¼ conductor radius in feet.Data adapted from Edison Electric Institute.

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345 500 500 735 735 112 Electric Power Systems, Fifth Edition2 2 4 3 4Curlew Chuker Parakeet Expanded Pheasant 1.246 1.602 0.914 1.750 1.382 28 38 38 56 56 35.3 47.9 47.9 70.6 70.6 0.1677 0.1529 0.0584 0.0784 0.0456 0.4325 0.4694 0.4694 0.5166 0.5166 0.6002 0.6223 0.5278 0.5950 0.5622 0.0379 0.0341 0.0126 0.0179 0.0096 0.1057 0.1147 0.1147 0.1263 0.1263 0.1436 0.1488 0.1273 0.1442 0.1359293.6 304.3 259.2 276.4 1955405 822 965 1844 0.0709 0.0805 0.0871 0.0510 0.162 0.0644 0.0979 0.0599 0.179 0.0728 ! 1 MV-miles; X 0 ¼ 0; 0683 log10 ðGMDÞMV-miles.½NrðAÞNÀ1 1=N D

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Components of a Power System 113Table 3.3 Underground cable constants at 50 Hz (per km)Size mm2 132 kV 275 kV 400 kV 355 645 970 1130 1935Rating (soil g ¼ 120 C cm/W) A 550 870 1100 1100 1600 MVA 125 200 525 525 1100Resistance (R) at 85 C VReactance (XL) V 0.065 0.035 0.025 0.02 0.013Charging current (Ic) A 0.128 0.138 0.22 0.134 0.22XL/R Ratio 7.90 10.69 15.70 17.77 23.86 2.0 4.0 8.8 6.6 16.83.7.2.1 The Short Line (up to 80 km, 50 miles)The equivalent circuit is shown in Figure 3.26 and it will be noticed that both shuntcapacitance and leakage resistance have been neglected. The four-terminal networkconstants are (see Section 2.7): A¼1 B¼Z C¼0 D¼1 The regulation of a circuit is deﬁned as: received voltage on no load ðVSÞ À received voltage on load ðVRÞ received voltage on load ðVRÞ It should be noted that if I leads VR in phase, that is a capacitive load, then VR > VS,as shown in Figure 3.27. rL r L rL r L r L rL C RC RC RC R C RFigure 3.25 Distributed constant representation of a line: L ¼ inductance of lineto neutral per unit length; r ¼ a.c. resistance per unit length; C ¼ capacitance line toneutral per unit length; R ¼ leakage resistance per unit length

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114 Electric Power Systems, Fifth Edition R LIVS VRFigure 3.26 Equivalent circuit of a short line – representation under balanced threephase conditions3.7.2.2 Medium-Length Lines (up to 240 km, 150 miles)Owing to the increased length, the shunt capacitance is now included to form eithera p or a T network. The circuits are shown in Figure 3.28. Of these two versions the prepresentation tends to be in more general use but there is little difference in accu-racy between the two. For the p network:VS ¼ VR þ IZ I ¼ IR þ VR Y 2 YIS ¼ I þ VS 2from which VS and IS are obtained in terms of VR and IR giving the following constants:A ¼ D ¼ 1 þ ZY 2B ¼ Z ZY 4C¼ 1 þ Y VSI IX IZ IR ω VRFigure 3.27 Phasor diagram for short line on leading power factor load

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Components of a Power System 115 IS R L I IR R L R L 2 2 IS 2 2 IR VS Y Y VR VS VC Y VR 2 2 (a) (b)Figure 3.28 (a) Medium-length line – p equivalent circuit, (b) Medium-length line – Tequivalent circuitSimilarly, for the T network: VS ¼ VC þ ISZ 2 VC ¼ VR þ IRZ 2 IS ¼ IR þ VCYgiving A ¼ D1¼þ1Z4þYZ2YY B ¼ C¼Y3.7.2.3 The Long Line (above 240 km, 150 miles)Here the treatment assumes distributed parameters. The changes in voltage and cur-rent over an elemental length Dx of the line, x metres from the sending end, aredetermined, and given below: DVx ¼ zDx Â Ix and DIx ¼ yDx Â Vxwherez ¼ impedance/unit lengthy ¼ shunt admittance/unit lengthIf Dx ! 0, then @Vx ¼ zIx ð3:2Þ @x

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116 Electric Power Systems, Fifth Edition @Ix ¼ yVx ð3:3Þ @xBy differentiating (3.2) and substituting from (3.3) @2Vx À zyVx ¼ 0 ð3:4Þ @x2Similarly, @2Ix À zyIx ¼ 0 ð3:5Þ @x2 Solution to equations (3.4) and (3.5) takes the form Vx ¼ A1 cosh Px þ A2 sinh Pxand Ix ¼ B1 cosh Px þ B2 sinh Px. When x ¼ 0, as Vx ¼ VS and Ix ¼ IS, the voltageand current at x metres from the sending end are given by Vx ¼ VS cosh Px À ISZ0 sinh Px VS Ix ¼ IS cosh Px À Z0 sinh Px ð3:6Þwhere P ¼ propogation constant ¼ ða þ jbÞ ¼ pﬃzﬃﬃyﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ ðR þ jvLÞðG þ jvCand Z0 ¼ crhﬃyzaﬃﬃr¼actserGﬃRiﬃﬃsﬃﬃtþþﬃiﬃﬃcﬃﬃjjﬃvvﬃiﬃmﬃCLﬃﬃﬃpedance ¼ ð3:7ÞwhereR ¼ resistance/unit lengthL ¼ inductance/unit lengthG ¼ leakage/unit lengthC ¼ capacitance/unit length Z0 is the input impedance of an inﬁnite length of the line; hence if any line is ter-minated in Z0 its input impedance is also Z0. The propagation constant P represents the changes occurring in the transmittedwave as it progresses along the line; a measures the attenuation, and b the angular

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Components of a Power System 117 pﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃphase-shift. With a lossless line, where R ¼ G ¼ 0, P ¼ jv LC and b ¼ LC. With avelocity of propagation of 3 Â 105 km/s the wavelength of the transmitted voltageand current at 50 Hz is 6000 km. Thus, lines are much shorter than the wavelengthof the transmitted energy.Usually conditions at the load are required when x ¼ l in equations (3.6). VR ¼ VS cosh Pl À ISZ0 sinh Pl IR ¼ IS cosh Pl À VS sinh Pl Z0Alternatively, VS ¼ VR cosh Pl þ IRZ0 sinh Pl IS ¼ VR cosh Pl þ IR sinh Pl Z0The parameters of the equivalent four-terminal network are thus, pﬃﬃﬃﬃﬃﬃ A ¼ D ¼ cosh ZY rﬃﬃﬃ B ¼ Z pﬃﬃﬃﬃﬃﬃ ð3:8Þ C ¼ sinh ZY rYﬃﬃﬃ Y pﬃﬃﬃﬃﬃﬃ Z sinh ZYwhereZ ¼ total series impedance of lineY ¼ total shunt admittance of lineThe easiest way to handle the hyperbolic functions is to use the appropriate series. A ¼ D ¼ pﬃﬃﬃﬃﬃﬃ ¼ 1 þ YZ þ Y2Z2 þ Y3Z3 þ Á Á Á Á ÁÁ cosh ZY 2 24 720 B ¼ þ YZ þ Y2Z2 þ Y3Z3 Z1 6 120 5040 C ¼ þ YZ þ Y2Z2 þ Y3Z3 Y1 6 120 5040 Usually not more than three terms are required, and for (overhead) lines less than500 km (312 miles) in length the following expressions for the constants hold

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118 Electric Power Systems, Fifth Edition IS IR VS B A-1 VR B A-1 BFigure 3.29 Equivalent circuit to represent accurately the terminal conditions of along lineapproximately: A ¼ D¼ þ1 þZ6YZ2Y B ¼ Z1 C ¼ þ ZY Y1 6 An exact equivalent circuit for the long line can be expressed in the form of the psection shown in Figure 3.29. The application of simple circuit laws will show thatthis circuit yields the correct four-terminal network equations. Figure 3.29 is only forconditions at the ends of the line; if intermediate points are to be investigated, thenthe full equations must be used. If only the ﬁrst term of the expansions are used, then B¼Z A À 1 ¼ Y B 2that is, the medium-length p representation.Example 3.2The conductors of a 1.6 km (1 mile) long, 3.3 kV, overhead line are in horizontal forma-tion with 762 mm (30 in) between centres. The effective diameter of the conductors is3.5 mm. The resistance per kilometre of the conductors is 0.41 V. Calculate the line-to-neutral inductance of the line. If the sending-end voltage is 3.3 kV (50 Hz) and the loadis 1 MW at a lagging p.f. of 0.8, estimate the voltage at the load busbar and the powerloss in the line.

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Components of a Power System 119Solution pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃThe equivalent equilateral spacing is given by de ¼ 3 d12 Â d23 Â d31 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃIn this case de ¼ 3 762 Â 762 Â 1524 pﬃﬃ de ¼ 762: 3 2 ¼ 960 mmThe inductance (line to neutral) ! m0 dÀ r H=m L ¼ 8p 1 þ 4 ln r ln9601À:751:75!H=m ¼ 4p Â 10À7 1 þ4 8p ¼ 1:3105 Â 10À6H=mThe total inductance of 1.6 km Ltotal ¼ 2:11 mHThe inductive reactance XL ¼ 2pfLtotal ¼ 2p Â 50 Â 2:11 Â 10À3 ¼ 0:66 VResistance of line Rtotal ¼ 1:6 Â 0:41 ¼ 0:66 VImpedance of the line Z ¼ 0:66 þ j0:66 V An estimate of the voltage drop is required and so the distribution approximationwill be used. Choosing 1 MVA and 3.3 kV as the bases for the calculation Zbase ¼ Vb2ase ¼ 3:3 Â 103 ¼ 10:89 V Sbase 106 Z ¼ 0:0602 þ j0:0602 per unit P ¼ 1 MW ð1 per unitÞ Q ¼ 0:75 MVAr ð0:75 per unitÞUsing the distribution approximation in per unit DV ¼ PR þ XQ ¼ 1 Â 0:0602 þ 0:75 Â 0:0602 ¼ 0:1053 ¼ 10:53%The voltage drop is therefore 347 V (3.3 kV line-line) and 201 V (1.9 kV line-neutral)

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120 Electric Power Systems, Fifth Edition The apparent power of the load is 1.25 MVA or 1.25 per unit. Assuming the load voltage to be 1 per unit, this is also the current in per unit. The line loss is 1.252 Â 0.0602 ¼ 0.094 per unit or 94 kW for all three phases. As the sending end voltage and receiving end power are speciﬁed, a more precise iterative calculation may be undertaken with Equation (2.13)Example 3.3A 150 km long overhead line with the parameters given in Table 3.2a for 400 kV, quadconductors is to be used to transmit 1800 MW (normal weather loading) to a load with apower factor of 0.9 lagging. Calculate the required sending end voltage using three linerepresentations and compare the results.SolutionFrom Table 3.2 a R ¼ 0:017V=km XL ¼ 0:27V=km 1 ¼ 10:58 Â 10À6S=km XcChoosing a base of 2000 MVA and 400 kV. Zbase ¼ 4002 ¼ 80V 2000For a 150 km line Zpu ¼ 150 ð0:017 þ j0:27Þ ¼ 0:032 þ j0:506 per unit 80 Short-line representation:Load power, S ¼ 1800 þ j870 MVAS ¼ VIÃ and as the receiving end voltage is at 400 kV, I ¼ 1800 À j 870 ¼ 0:9 À j0:435 per unit 2000 2000Hence Vs ¼ VR þ IZ Vs ¼ 1:249 þ j0:442 per unit jVSj ¼ 530 kV

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Components of a Power System 121Medium-line representation: Vs ¼ AVR þ BIR A ¼ 1 þ ZY 2 B ¼Z B ¼ 0:032 þ j0:506 per unit Y ¼ 10:58 Â 10À6 Â 150 Â 80 ¼ 0:127 per unit A ¼ 1 þ ZY ¼ 0:968 þ j0:002 per unit 2Hence Vs ¼ AVR þ BIR ¼ ð0:968 þ j0:002Þ þ ð0:032 þ j0:506Þð0:9 À j0:435Þ ¼ 1:217 þ j0:444 jVsj ¼ 518 kVLong-line representation: A ¼ rcosYZﬃﬃhﬃ psinZﬃﬃﬃhﬃYﬃﬃp¼ﬃZﬃﬃ1ﬃYﬃﬃþ¼Y2ZZð1þþY2Y264ZZ2 þ ÁÁÁ þ Á Á Á Á ÁÁ B ¼ þ Y2Z2 120 YZ ¼ À0:064 þ j4:047 Â 10À3 Y2Z2 ¼ 4:115 Â 10À3 À j5:202 Â 10À4 A ¼ 0:968 þ j2:002 Â 10À3 B ¼ 0:031 þ j0:501 VS ¼ AVR þ BIR ¼ 1:214 þ j0:439 per unit jVSj ¼ 518 kV (Note: This is an extremely high voltage for normal operation and would not betolerated. A 400 kV system would be designed for only about 10% steady-state over-voltage, that is 400 þ 40 ¼ 440 kV. In practice, either the reactance of the line wouldbe reduced by series capacitors and/or the power factor at the receiving end wouldbe raised to at least 0.95 lag by the use of shunt capacitors or synchronouscompensators).3.7.2.4 The Natural LoadThe characteristic impedance Z0 (equation 3.7) is known as the surge impedance ifthe line is considered to be lossless and all resistances are neglected. When a line is

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122 Electric Power Systems, Fifth Editionterminated in its characteristic impedance the power delivered is known as thenatural load. For a loss-free line supplying its natural load the reactive powerabsorbed by the line is equal to the reactive power generated, that is. V2 ¼ I2XL XCand V ¼ Z0 ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ rﬃﬃﬃ I XLXC L C At this load, V and I are in phase all along the line and optimum transmissionconditions are obtained. In practice, however, the load impedances are seldom inthe order of Z0. Values of Z0 for various line voltages are as follows (correspondingnatural loads are shown in parentheses): 132 kV, 150 V (50 MW); 275 kV, 315 V(240 MW); 380 kV, 295 V (490 MW). The angle of the impedance varies between0 and À15. For underground cables Z0 is roughly one-tenth of the overhead-linevalue. Lines are operated above the natural loading, whereas cables operate belowthis loading.3.7.3 Parameters of Underground CablesIn traditional paper insulated cables with three conductors contained within a leador aluminium sheath, the electric ﬁeld set up has components tangential to thelayers of impregnated paper insulation in which direction the dielectric strength ispoor. Therefore, at voltages over 11 kV, each conductor is separately screened toensure only radial stress through the paper. Cables using cross-linked polyethylene(XPLE) insulation are now commonly used and again the electric stresses arearranged to be radial. The capacitance of single-conductor and individuallyscreened three-conductor cables is readily calculated. For three-conductorunscreened cables one must resort to empirical design data. A typical high-voltageXLPE insulated cable is shown in Figure 3.30. The capacitance (C) of single-corecables may be calculated from design data by the use of the formula C ¼ 2pe0er ln R rwhere r and R are the inner and outer radii of the dielectric and er is the relativepermittivity of the dielectric. This expression also holds for three core cables witheach conductor separately screened. Owing to the symmetry of the cable the positive phase-sequence values of C and Lare the same as the negative phase-sequence values (i.e. for reversed phase rotation).The series resistance and inductance are complicated by the magnetic interaction

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Components of a Power System 123 Conductor Conductor screen Insulation Bedding Screen Armour Sheath Figure 3.30 150 kV XLPE insulated 3 core cable with ﬁbre optic communicationsbetween the conductor and sheath. The effective resistance of the conductor is thedirect current resistance modiﬁed by the following factors: the skin effect in the con-ductor; the eddy currents induced by adjacent conductors (the proximity effect); andthe equivalent resistance to account for the I2R losses in the sheath. The determina-tion of these effects is complicated and is left for advanced study. The parameters of the cable having been determined, the same equivalent circuitsare used as for overhead lines, paying due regard to the selection of the correctmodel for the appropriate length of cable. Owing to the high capacitance of cablesthe charging current, especially at high voltages, is an important factor in decidingthe permissible length to be used. Table 3.3 gives the charging currents for self-contained low-pressure oil-ﬁlled(LPOF) cables.

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124 Electric Power Systems, Fifth Edition RP XP ( N1 )2Rs ( N1 )2Xs N2 N2 RO XO Figure 3.31 Equivalent circuit of a two-winding transformer3.8 TransformersThe equivalent circuit of one phase of a transformer referred to the primary windingis shown in Figure 3.31. The resistances and reactances can be found from the well-known open- and short-circuit tests. In the absence of complete information for eachwinding, the two arms of the T network can each be assumed to be half the totaltransformer impedance. Also, little accuracy is lost in transferring the shunt branchto the input terminals. In power transformers the current taken by the shunt branch is usually a verysmall percentage of the load current and is neglected for most power systemcalculations.3.8.1 Phase Shifts in Three-Phase TransformersConsider the transformer shown in Figure 3.32(a). The red phases on both circuitsare taken as reference and the transformation ratio is 1: N. The correspondingphasor diagrams are shown in Figure 3.32(b). Although no neutral point is availablein the delta side, the effective voltages from line to earth are denoted by ER0n, EY0n,and EB0n. Comparing the two phasor diagrams, the following relationships are seen: ER0n ¼ line-earth voltage on the delta-side ¼ NERnﬀ 30that is, the positive sequence or normal balanced voltage of each phase is advancedthrough 30. Similarly, it can be shown that the positive sequence currents areadvanced through 30. By consideration of the negative phase-sequence phasor diagrams (these are pha-sors with reversed rotation, that is R-B-Y) it will readily be seen that the phase cur-rents and voltages are shifted through À30. When using the per unit system thetransformer ratio does not directly appear in calculations and the phase shifts areoften neglected.

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Components of a Power System 125 R IR Primary I I RI R n 1:N Y I Secondary (a) I I B IB Y Y I I BI IY BEBR E Bn ERY E BI YI E BIn 30O E RIn E Rn 30O ER IBI E Yn E YIn E Y IR I EYB (b)Figure 3.32 (a) Star-delta transformer with turns ratio 1: N. (b) Corresponding phasordiagrams (N taken as 1 in diagrams) In star-star and delta-delta connected transformers there are no phase shifts.Hence transformers having these connections and those with star-delta connectionsshould not be connected in parallel. To do so introduces a resultant voltage acting inthe local circuit formed by the usually low transformer impedances. Figure 3.33shows the general practice on the British network with regard to transformers withphase shifts. It is seen that the reference phasor direction is different at differentvoltage levels. The larger than 30 phase shifts are obtained by suitablerearrangement of the winding connections.3.8.2 Three-Winding TransformersMany transformers used in power systems have three windings per phase, the thirdwinding being known as the tertiary. This winding is provided to enable compensa-tion equipment to be connected at an economic voltage (e.g. 13 kV) and to provide a

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126 Electric Power Systems, Fifth Edition 275 kV or 400 kV Red Auto transformer Red 132 kV Red 66 kV Earthing transformer Red 33 kV Red 11 kv or 6.6 kV Red 400 V Figure 3.33 Typical phase shifts in a power system–Britishcirculating current path for third harmonics so that these currents do not appearoutside the transformer. The three-winding transformer can be represented under balanced three phaseconditions by a single-phase equivalent circuit of three impedances star-connectedas shown in Figure 3.34. The values of the equivalent impedances Zp, Zs and Ztmay be obtained by test. It is assumed that the no-load currents are negligible. LetZps ¼ impedance of the primary when the secondary is short-circuited and the ter- tiary openZpt ¼ impedance of the primary when the tertiary is short-circuited and the second- ary open

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Components of a Power System 127 p(a) s t p Zp st ZS s(b) Zt V1 t T12(c) st V2 Zp Zs Zt T13 V3Figure 3.34 (a) Three-winding transformer (b) and (c) equivalent circuitsZst ¼ impedance of the secondary when the tertiary is short-circuited and the primary openThe above impedances are in ohms referred to the same voltage base. Hence, Zps ¼ Zp þ Zs Zpt ¼ Zp þ Zt Zst ¼ Zs þ Zt Zp ¼ 1 À þ Zpt À Á 2 Zps Zst Zs ¼ 1 À þ Zst À Á 2 Zps Zpt Zt ¼ 1 À þ Zst À Á 2 Zpt Zps It should be noted that the star point st in Figure 3.34b is purely ﬁctitious and thatthe diagram is a single-phase equivalent circuit. In most large transformers the

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128 Electric Power Systems, Fifth Editionvalue of Zs is very small and can be negative. All impedances must be referred tocommon volt-ampere and voltage bases. The complete equivalent circuit is shownin Figure 3.34(c) where T12 and T13 provide any off nominal turns ratio.3.8.3 AutotransformersThe symmetrical autotransformer may be treated in the same manner as two andthree-winding transformers. This type of transformer shows to best advantagewhen the transformation ratio is limited and it is widely used for the inter-connection of the transmission networks working at different voltages, for example275 to 132 kV or 400 to 275 kV. The neutral point is solidly grounded, that is con-nected directly to earth without intervening resistance.3.8.4 Earthing (Grounding) TransformersA means of providing an earthed point or neutral in a supply derived from a delta-connected transformer may be obtained by the use of a zigzag transformer, showngrounding the 33 kV system in Figure 3.33. By the interconnection of two windingson each limb, a node of zero potential is obtained.3.8.5 HarmonicsDue to the non-linearity of the magnetizing characteristic of transformers the cur-rent waveform is distorted and hence contains harmonics; these ﬂow through thesystem impedances and set up harmonic voltages. In transformers with delta-connected windings the third and ninth harmonics circulate round the delta and areless evident in the line currents. An important source of harmonics is power elec-tronic devices. On occasions, the harmonic content can prove important due mainly to the possi-bility of resonance occurring in the system, for example, resonance has occurredwith ﬁfth harmonics. Also, the third-harmonic components are in phase in the threeconductors of a three-phase line, and if a return path is present these currents addand cause interference in neighbouring communication circuits and increase theneutral return current. When analysing systems with harmonics it is often sufﬁcient to use the normalvalues for series inductance and shunt capacitance but one must remember to calcu-late reactance and susceptance at the frequency of the harmonic. The effect on resist-ance is more difﬁcult to assess: however, it is usually only required to assess thepresence of harmonics and the possibility of resonance.3.8.6 Tap-Changing TransformersA method of controlling the voltages in a network lies in the use of transformers, theturns ratio of which may be changed. In Figure 3.35(a) a schematic diagram of anoff-load tap changer is shown; this, however, requires the disconnection of the trans-former when the tap setting is to be changed. Many larger transformers have on-load changers, one basic form of which is shown in Figure 3.35(b). In the position

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Components of a Power System 129 (a)Line Winding S1 S2 RR BA CSelector Selector switch switch Taps Taps Winding Neutral point (b)Figure 3.35 (a) Tap-changing transformer, (b) On-load tap-changing transformer(reactor type). S1, S2 transfer switches, R centre-tapped reactorshown, the current divides equally in the two halves of reactor R, resulting in zero-resultant ﬂux and minimum impedance. S1 opens and the total current passesthrough the other half of the reactor. Selector switch B then moves to the next con-tact and S1 closes. A circulating current now ﬂows in the reactor R superimposed onthe load current. Now S2 opens and C moves to the next tapping; S2 then closes andthe operation is complete. Six switch operations are required for one change in tapposition. The voltage change between taps is often 1–1.25% of the nominal voltage.This small change is necessary to avoid large voltage disturbances at consumer bus-bars. Figure 3.35(b) shows reactors used to limit the circulating current during the

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130 Electric Power Systems, Fifth Edition33 kV Current transformer 33/11 kV IL – L VL Load Voltage IC – L ZC α transformer LDC V IC – Lx ZC αMotor Motor V – IC – L x ZC α controlFigure 3.36 Schematic diagram of a control system for an on-load tap-changingtransformer incorporating line drop compensation (LDC)tap-change operation. An alternative technique is to use high speed switching witha resistor to limit the circulating current. A schematic block diagram of the on-load tap systems is shown in Figure 3.36. Theline drop compensator (LDC) is used to allow for the voltage drop along the feeder1 p.u. P consumed Q generatedVoltage p.u. Q P 0.7Figure 3.37 P- V and Q-V curves for a synchronous motor

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Components of a Power System 131to the load point, so that the actual load voltage is seen and corrected by thetransformer. The total range of tapping varies with the transformer usage, atypical ﬁgure for generator transformers is þ2 to À16% in 18 steps each of 1%.3.8.7 Typical Parameters for TransformersThe leakage reactances of two-winding transformers increase slightly with their rat-ing for a given voltage, that is from 3.2% at 20 kVA to 4.3% at 500 kVA at 11 kV. Forlarger sizes, that is 20 MVA upwards, 10–20% is a typical value at all voltages. Forautotransformers the impedances are usually less than for double wound transform-ers. Parameters for large transformers are as follows:1. 400/275 kV autotransformer, 500 MVA, 12% impedance, tap range þ10 to À20%;2. 400/132 kV double wound, 240 MVA, 20% impedance, tap range þ5 to À15%. It should be noted that a 10% reactance implies up to ten times rated current onshort circuit. Winding forces depend upon current squared and so a transformermust be designed to withstand high forces caused by short circuits.3.9 Voltage Characteristics of LoadsThe variation of the power and reactive power taken by a load with various volt-ages is of importance when considering the manner in which such loads are to berepresented in load ﬂow and stability studies. Usually, in such studies, the loadon a substation has to be represented and is a composite load consisting ofindustrial and domestic consumers. A typical composition of a substation load isas follows:Induction motors 50–70% (air-conditioners, freezers, washers, fans, pumps, etc.)Lighting and heating 20–25% (water heating, resistance heaters, etc.)Synchronous motors 10%(Transmission and distributionlosses 10–12%) Increasingly electronic equipment draws a signiﬁcant fraction of the load.3.9.1 LightingIncandescent (ﬁlament) lights are independent of frequency and consume noreactive power. The power consumed does not vary as the (voltage)2, but approxi-mately as (voltage)1.6. However, in many countries the use of incandescent lights isbeing reduced due to their low efﬁciency. Fluorescent (both traditional and com-pact) and sodium/mercury lamps, can take distorted currents and so contribute tonetwork harmonics.

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132 Electric Power Systems, Fifth Edition I X2 X1 Xm X = X1+ X2 r2 V sFigure 3.38 Equivalent circuit of an induction motor: X1 ¼ stator leakage reactance;X2 ¼ rotor leakage reactance referred to the stator; Xm ¼ magnetizing reactance; r2 ¼rotor resistance; s ¼ slip p.u. Magnetizing losses have been ignored and the statorlosses are lumped in with the line losses3.9.2 HeatingThis maintains constant resistance with voltage change and hence the power varieswith (voltage)2. The above loads may be described as static.3.9.3 Synchronous MotorsThe power consumed is approximately constant with the applied voltage. For agiven excitation the VArs change in a leading direction (i.e. reactive power is gener-ated) with network voltage reduction. The P-V, Q-V, generalized characteristics areshown in Figure 3.37.3.9.4 Induction MotorsThe P-V, Q-V characteristics may be determined by the use of the simpliﬁedcircuit shown in Figure 3.38. It is assumed that the mechanical load on the shaftis constant. The electrical power Pelectrical ¼ Pmechanical ¼ P P ¼ 3I2r2 s

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Components of a Power System 133The reactive power consumed ¼ 3V2 þ 3I2ðX1 þ X2Þ XmAlso, from Figure 3.38 X ¼ X1 þ X2 P ¼ 3I2 r2 ¼ rs223þV2X2! r2 s s ð3:9Þ ¼ 3V2r2s r22 þ ðsXÞ2 The well-known power-slip curves for an induction motor are shown in Figure3.39. It is seen that for a given mechanical torque there is a critical voltage and acorresponding critical slip scr. If the voltage is reduced further the motor becomesunstable and stalls. This critical point occurs when dP ¼ 0 ds PmaxPower p.u.1 p.u. V = 1 p.u. 0 Scr V = 0.8 p.u. Slip (s) p.u. V = 0.6 p.u. V = 0.45 p.u. 1Figure 3.39 Power-slip curves for an induction motor. If voltage falls to 0.6 p.u. at fullload P ¼ 1, the condition is critical (slip scr)

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134 Electric Power Systems, Fifth Editionthat is when r22 À ðsXÞ2 ½r22 þ ðsXÞ22 V2r2 ¼ 0so that r2 X s ¼ Pmax ¼ 3V2 2XAlternatively, for a given output power, P rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 Vcritical ¼ 3 PXProblems(Note: All machines are three-phase unless stated otherwise.) 3.1 When two four pole, 50 Hz synchronous generators are paralleled their phase displacement is 2 mechanical. The synchronous reactance of each machine is 10 V/phase and the common busbar voltage is 6.6 kV. Calculate the synchro- nizing torque. (Answer: 968 Nm) 3.2 A synchronous generator has a synchronous impedance of 2 p.u. and a resist- ance of 0.01 p.u. It is connected to an inﬁnite busbar of voltage 1 p.u. through a transformer of reactance j0.1 p.u. If the generated (no-load) e.m.f. is 1.1 p.u. calculate the current and power factor for maximum output. (Answer: 0.708 p.u.; 0.74 leading) 3.3 A 6.6 kV synchronous generator has negligible resistance and synchronous reactance of 4 V/phase. It is connected to an inﬁnite busbar and gives 2000 A at unity power factor. If the excitation is increased by 25% ﬁnd the maximum power output and the corresponding power factor. State any assumptions made. (Answer: 31.6 MW; 0.95 leading) 3.4 A synchronous generator whose characteristic curves are given in Figure 3.4 delivers full load at the following power factors: 0.8 lagging, 1.0, and 0.9 lead- ing. Determine the percentage regulation at these loads. (Answer: 167, 119, 76%) 3.5 A salient-pole, 75 MVA, 11 kV synchronous generator is connected to an inﬁnite busbar through a link of reactance 0.3 p.u. and has Xd ¼ 1.5 p.u. and Xq ¼ 1 p.u., and negligible resistance. Determine the power output for a load

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Components of a Power System 135 angle 30 if the excitation e.m.f. is 1.4 times the rated terminal voltage. Calcu- late the synchronizing coefﬁcient in this operating condition. All p.u. values are on a 75 MVA base. (Answer: P ¼ 0.48 p.u.; dP/dd ¼ 0.78 p.u.) 3.6 A synchronous generator of open-circuit terminal voltage 1 p.u. is on no-load and then suddenly short-circuited; the trace of current against time is shown in Figure 3.6(b). In Figure 3.6(b) the current 0c ¼ 1.8 p.u., 0a ¼ 5.7 p.u., and 0b ¼ 8 p.u. Calculate the values of Xs, X0 and X00. Resistance may be neglected. If the machine is delivering 1 p.u. current at 0.8 power factor lagging at the rated terminal voltage before the short circuit occurs, sketch the new envelope of the 50 Hz current waveform. (Answer: Xs ¼ 0.8 p.u.; X0 ¼ 0.25 p.u.; X00 ¼ 0.18 p.u.) 3.7 Construct a performance chart for a 22 kV, 500 MVA, 0.9 p.f. generator having a short-circuit ratio of 0.55. 3.8 A 275 kV three-phase transmission line of length 96 km is rated at 800 A. The values of resistance, inductance and capacitance per phase per kilometre are 0.078 V, 1.056 mH and 0.029 mF, respectively. The receiving-end voltage is 275 kV when full load is transmitted at 0.9 power factor lagging. Calculate the sending-end voltage and current, and the transmission efﬁciency, and com- pare with the answer obtained by the short-line representation. Use the nomi- nal p and T methods of solution. The frequency is 60 Hz. (Answer: Vs ¼ 179 kV per phase) 3.9 A 220 kV, 60 Hz three-phase transmission line is 320 km long and has the following constants per phase per km: Inductance 0.81 mH Capacitance 12.8 mF Resistance 0.038 V Ignore leakage conductance. If the line delivers a load of 300 A, 0.8 power factor lagging, at a voltage of 220 kV, calculate the sending-end voltage. Determine the p circuit which will represent the line. (Answer: Vs ¼ 241 kV)3.10 Calculate the A B C D constants for a 275 kV overhead line of length 83 km. The parameters per kilometre are as follows: Resistance 0.078 V Reactance 0.33 V Admittance (shunt capacitative) 9.53 Â 10À6 S The shunt conductance is zero. (Answer: [A ¼ 0.98917 þ j 0.00256; B ¼ 6.474 þ j 27.39; C ¼ (À1.0126 Â 10À6 þ j 7.8671 Â 10À4)]

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136 Electric Power Systems, Fifth Edition3.11 A 132 kV, 60 Hz transmission line has the following generalized constants: A ¼ 0.9696 ﬀ 0.49 B ¼ 52.88 ﬀ 74.79 V C ¼ 0.001177 ﬀ 90.15S If the receiving-end voltage is to be 132 kV when supplying a load of 125 MVA 0.9 p.f. lagging, calculate the sending-end voltage and current. (Answer: 165 kV, 498 A)3.12 Two identical transformers each have a nominal or no-load ratio of 33/11 kV and a leakage reactance of 2 V referred to the 11 kV side; resistance may be neglected. The transformers operate in parallel and supply a load of 9 MVA, 0.8 p.f. lagging. Calculate the current taken by each transformer when they operate ﬁve tap steps apart (each step is 1.25% of the nominal voltage). Also calculate the kVAr absorbed by this tap setting. (Answer: 307 A, 194 A for three-phase transformers, 118 kVAr)3.13 An induction motor, the equivalent circuit of which is shown in Figure 3.40 is connected to supply busbars which may be considered as possessing a voltage and frequency which is independent of the load. Determine the reactive power consumed for various busbar voltages and construct the Q-V character- istic. Calculate the critical voltage at which the motor stalls and the critical slip, assuming that the mechanical load is constant. (Answer: scr ¼ 0.2, Vc ¼ 0.63 p.u.)3.14 A 100 MVA round-rotor generator of synchronous reactance 1.5 p.u. supplies a substation (L) through a step-up transformer of reactance 0.1 p.u., two lines each of reactance 0.3 p.u. in parallel and a step-down transformer of reactance 0.1 p.u. The load taken at L is 100 MW at 0.85 lagging. L is connected to a local generating station which is represented by an equivalent generator of 75 MVA and synchronous reactance of 2 p.u. All reactances are expressed on a base of 100 MVA. Draw the equivalent single-phase network. If the voltage at L is toXs = 0.2 p.u. IsV 3 p.u. r2= R2 ; R2 = 0.04 p.u. SFigure 3.40 Equivalent circuit of 500 kW, 6.6 kV induction motor in Problem 3.13. Allp.u. values refer to rated voltage and power (P ¼ 1 p.u. and V ¼ 1 p.u.)

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Components of a Power System 137132 kV (A) 48 km 132 kV 258 mm2 OHL 80 km 258 mm2 OHL 48 km OHL 113 mm2 11 kV 11 kV (C) (D)(B) 66 kV Load 66 kV 66 kV Cable Load Load (2 + j4) Ω Figure 3.41 Line diagram of system in Problem 3.15 be 1 p.u. and the 75 MVA machine is to deliver 50 MW, 20MVAr, calculate the internal voltages of the machines. (Answer: E1 ¼ 2 p.u.; E2 ¼ 1.72 p.u.; d2L ¼ 35.45)3.15 The following data applies to the power system shown in Figure 3.41. Generating station A: Four identical turboalternators, each rated at 16 kV, 125 MVA, and of synchronous reactance 1.5 p.u. Each machine supplies a 125 MVA, 0.1 p.u. transformer connected to a busbar sectioned as shown. Substation B: Two identical 150 MVW, three-winding transformers, each having the following reactances between windings: 132/66 kV windings 10%; 66/11 kV windings 20%; 132/11 kV windings 20%; all on a 150 MVA base. The secondaries supply a common load of 200 MW at 0.9 p.f. lagging. To each tertiary winding is connected a 30 MVA synchronous compensator of synchronous reactance 1.5 p.u. Substation C: Two identical 150 MVA transformers, each of 0.15 p.u. reac- tance, supply a common load of 300 MW at 0.85 p.f. lagging. Generating station D: Three identical 11 kV, 75 MVA generators, each of 1 p.u. synchronous reactance, supply a common busbar which is connected to

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138 Electric Power Systems, Fifth Edition an outgoing 66 kV cable through two identical 100 MVA transformers. Load 50 MW, 0.8 p.f. lagging. Determine the equivalent circuit for balanced operation giving component values on a base of 100 MVA. Treat the loads as impedances.3.16 Distinguish between kW, kVA, and kVAr. Explain why a. generators in large power systems usually run overexcited, ‘generating’ VAr. b. remote hydro-generators need an underexcited rating so that they can absorb VAr. c. loss of an overexcited generator in a power system will normally cause a drop in voltage at its busbar. A load of 0.8 p.u. power and 0.4 p.u. VAr lagging is supplied from a busbar of 1.0 p.u. voltage through an inductive line of reactance 0.15 p.u. Determine the load terminal voltage assuming that p.u. current has the same value as p.u. VA. (Answer: 0.95 p.u.) (From Engineering Council Examination, 1996)3.17 Sketch the performance chart of a synchronous generator indicating the main limits. Consider a generator with the following nameplate data: 500 MVA, 20 kV, 0.8 p.f. (power factor), Xs ¼ 1.5 p.u. a. Calculate the internal voltage and power angle of the generator operating at 400 MW with cosw ¼ 0.8 (lagging) with a 1 p.u. terminal voltage. b. What is the maximum reactive power this generator can absorb from the system? c. What is the maximum reactive power this generator can deliver to the system, assuming a maximum internal voltage of 2.25 p.u. d. Place the numerical values calculated on the performance chart. A graphical solution is acceptable. (Answer: (a) 2.25 p.u., 32; (b) 333 MVAr; (c) 417 MVAr) (From Engineering Council Examination, 1997)

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4Control of Power andFrequency4.1 IntroductionIn a large power system, the control of power and frequency is related only weaklyto the control of reactive power and voltage. For many purposes the operation of thegovernors controlling the power of the prime movers of generating units can be con-sidered independently to the AVRs that control the excitation and hence the reactivepower and voltage of the generators. By dealing with power and frequency sepa-rately from voltage and reactive power control, a better appreciation of the opera-tion of power systems can be obtained. This separation is followed in Chapter 4(Control of Power and Frequency) and Chapter 5 (Control of Voltage and ReactivePower). In a large interconnected system, many synchronous generators, big and small,are directly connected and hence all have the same frequency. In many power sys-tems (e.g. in Great Britain) the control of power is carried out by the decisions andactions of control engineers, as opposed to systems in which the control and alloca-tion of load to machines is effected completely automatically. Fully automatic con-trol systems are sometimes based on a continuous load-ﬂow calculation bycomputers. The allocation of the required power amongst the generators has to be decidedbefore the load appears. Therefore the load must be predicted in advance. An analy-sis is made of the loads experienced over the same period in previous years; accountis also taken of the value of the load immediately previous to the period under studyand of the weather forecast. The probable load to be expected, having been decided,is then allocated to the various turbine-generators. Load cycles of three power systems are shown in Figure 1.1. The PJM control areain the USA has peaks of 140 GW in the summer, a lower demand in the winter and arate of increase of around 15 GW/h. For the isolated Great Britain system the rate ofElectric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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140 Electric Power Systems, Fifth Editionincrease during the week shown was around 4 GW/h. For the smaller Sri Lankapower system the rate of increase was 250 MW/h, provided mainly by hydro gener-ators. The ability of machines to increase their output quickly from zero to full load,and subsequently reduce their output is important. It is extremely unlikely that the output of the machines at any instant will exactlyequal the load on the system. If the output is higher than the demand the machineswill tend to increase in speed and the frequency will rise, and vice versa. Hence thefrequency is not a constant quantity but continuously varies; these variations arenormally small and have no noticeable effect on most consumers. The frequency iscontinuously monitored against standard time-sources and when long-term tenden-cies to rise or fall are noticed, the control engineers take appropriate action by regu-lating the generator outputs. Should the total generation available be insufﬁcient to meet the demand, the fre-quency will fall. If the frequency falls by more than around 1 Hz the reduced speedof power station pumps and fans may limit the output of the power stations and aserious situation arises. When there is insufﬁcient generation, although the lowerfrequency will cause some reduction in power demand, the system voltage must bereduced (which leads to a further reduction in load), and if this is not sufﬁcient thenloads will have to be disconnected and continue to be disconnected until the fre-quency rises to a reasonable level. All utilities have a scheme of planned load shed-ding based on under-frequency relays set to reduce loads in blocks to preventcomplete shut-down of the power system in extreme emergencies. Figure 4.1 shows the frequency of the Great Britain power system when two largegenerating units tripped in rapid succession, for unrelated reasons. The frequencydropped quickly and the fall was only ﬁnally arrested by load shedding whenit reached 48.8 Hz. Figure 4.1 also shows the morning increase in system load of$ 18 GW from around 05:30–10:30 and a similar reduction in load in the evening.Apart from during the incident around 11:35 when two generating units tripped,the frequency was maintained close to 50 Hz by the governors of generators. When an increase in load occurs on the system, the speed and frequency of all theinterconnected generators fall, since the increased energy requirement is met fromthe kinetic energy of the machines. This causes an increase in steam or water admit-ted to the turbines due to the operation of the governors and hence a new load bal-ance is obtained. Initially, the boilers have a thermal reserve of steam, in their boilerdrums, by means of which sudden changes can be supplied until a new ﬁring ratehas been established. Modern gas turbines have an overload capability for a fewminutes which can usefully be exploited in emergency situations. The stored kinetic energy of all the generators (and spinning loads) on the systemis given byKE ¼ 1 Iv2 ½ J 2 ÀÁI ¼ moment of inertia of all generators kgm2v ¼ rotational speed of all generatorsðrad=sÞ

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Control of Power and Frequency 141 44000 50.2 41000 50 38000 49.8 35000 49.6 32000 49.4 Hz MW 49.2 29000 49 26000 48.8 23000 48.6 2000005:00 06:00 07:00 08:00 09:00 10:00 11:00 12:00 13:00 14:00 15:00 16:00 17:00 18:00 19:00 20:00 21:00 22:00 23:00 00:00Figure 4.1 Variation of load (LH axis) and frequency (RH axis) of GB system. Two genera-tors tripped, for unrelated reasons, at around 11:35 (Figure adapted from National Grid)When Pm ¼ Pe (Figure 4.2) the rotational speed of the generators is maintained andthe frequency is constant at 50 (or 60) Hz. When Pm < Pe the rotational speed of thegenerators, and hence system frequency reduces. When Pm > Pe the rotational speedof the generators, and hence system frequency increases.Steam/ Water Control Pm ω Generator Pe input valve Turbine GovernorFigure 4.2 Control of frequency. Angular speed is measured and controls an inletvalve of the turbine ﬂuid

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142 Electric Power Systems, Fifth EditionThe torque balance of any spinning mass determines the rotational speed. Tm À Te ¼ I dv dt In power systems it is conventional to express the inertia in per unit as an HconstantH ¼ 1 Iv2S ½Ws=VA 2 Sratedwhere Srated is the MVA rating of either an individual generator or the entire powersystem, vS is the angular velocity (rad/s) at synchronous speed. Thusdv ¼ vs2 ðTm À TeÞdt 2HSratedwhich in per unit may be written as dv ¼ 1 ðPm À PeÞ dt 2H Thus the rate of change of rotational speed and hence frequency depends on thepower imbalance and the inertia of the spinning masses.4.2 The Turbine GovernorA simpliﬁed schematic diagram of a traditional governor system is shown inFigure 4.3. The sensing device, which is sensitive to change of speed, is the time-honoured Watt centrifugal governor. In this, two weights move radially outwardsas their speed of rotation increases and thus move a sleeve on the central spindle.This sleeve movement is transmitted by a lever mechanism to the pilot-valve pistonand hence the servo-motor is operated. A dead band is present in this mechanism,that is, the speed must change by a certain amount before the valve commencesto operate, because of friction and mechanical backlash. The time taken for the mainsteam valve to move due to delays in the hydraulic pilot-valve and servo-motorsystems is appreciable, 0.2–0.3 s. The governor characteristic for a large steam turbo alternator is shown inFigure 4.4 and it is seen that there is a 4% drop in speed between no load and fullload. Because of the requirement for high response speed, low dead band, and accu-racy in speed and load control, the mechanical governor has been replaced in largemodern turbo generators by electro hydraulic governing. The method normally

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Control of Power and Frequency 143 Spring Governor (driven from turbine shaft)Speeder (Initial position) Turbine Systemmotor Generator Oil at Hydraulic constant servo-motor pressure Displacement Pilot d valve Main inletWorking fluid valve From steam boiler or upper water reservoirFigure 4.3 Governor control system employing the Watt governor as sensing deviceand a hydraulic servo-system to operate main supply valve. Speeder-motor geardetermines the initial setting of the governor positionused to measure the speed is based on a toothed wheel on the generator shaft andmagnetic-probe pickup. The use of electronic controls requires an electro hydraulicconversion stage, using conventional servo-valves. An important feature of the governor system is the mechanism by means of whichthe governor sleeve and hence the main-valve positions can be changed andadjusted quite apart from when actuated by the speed changes. This is accom-plished by the speed changer, or ‘speeder motor’, as it is sometimes termed. The 1.04 Speed p.u. 1.02 1.0 0.5 1.0 0 Power output p.u.Figure 4.4 Idealized governor characteristic of a turboalternator with 4% droop fromzero to full load

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144 Electric Power Systems, Fifth EditionSpeed p.u. 1.04 Nominal 1.02 speed 1.0 0.98 Alternative 0.96 governor settings 0 P1 P2 P3 0.5 1.0 Power output p.u.Figure 4.5 Effect of speeder gear on governor characteristics. P1, P2, and P3 arepower outputs at various settings but at the same speedeffect of this adjustment is the production of a family of parallel characteristics, asshown in Figure 4.5. Hence the power output of the generator at a given speed maybe adjusted independently of system frequency and this is of extreme importancewhen operating at optimum economy. The torque of the turbine may be considered to be approximately proportional tothe displacement d of the main inlet valve. The expression for the change in torquewith speed may be expressed approximately by the equation T ¼ T0ð1 À kNÞ ð4:1Þwhere T0 is the torque at speed N0 and T the torque at speed N; k is a constant forthe governor system. As the torque depends on both the main-valve position andthe speed, T ¼ fðd; NÞ. There is a time delay between the occurrence of a load change and the new oper-ating conditions. This is due not only to the governor mechanism but also to the factthat the new ﬂow rate of steam or water must accelerate or decelerate the rotor inorder to attain the new speed. In Figure 4.6 typical curves are shown for a turbo-generator which has a sudden decrease in the electrical power required, perhapsdue to an external power network fault, and hence the retarding torque on the tur-bine shaft is suddenly much smaller. In the ungoverned case the considerable time-lag between the load change and the attainment of the new steady speed is obvious.With the regulated or governed machine, due to the dead band in the governor

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Control of Power and Frequency 145 Turbine torque Electrical Speed (no governor) torque Speed (governor control)Torque and speed TimeFigure 4.6 Graphs of turbine torque, electrical torque, and speed against time whenthe load on a generator suddenly fallsmechanism, the speed-time curve starts to rise, the valve then operates, and the ﬂuidsupply is adjusted. It is possible for damped oscillations to be set up after the loadchange. An important factor regarding turbines is the possibility of overspeeding, whenthe load on the shaft is lost completely, with possible drastic mechanical breakdown.To avoid this, special valves are incorporated to automatically cut off the energysupply to the turbine. In a turbogenerator normally running at 3000 r.p.m. this over-speed protection operates at about 3300 r.p.m. Example 4.1 An isolated 75 MVA synchronous generator feeds its own load and operates initially at no-load at 3000 r.p.m., 50 Hz. A 20 MW load is suddenly applied and the steam valves to the turbine start to open after 0.5 s due to the time-lag in the governor system. Calcu- late the frequency to which the generated voltage drops before the steam ﬂow meets the new load. The stored energy for the machine is 4 kWs per kVA of generator capacity. Solution For this machine the stored energy at 3000 r.p.m. ¼ 4 Â 75 000 ¼ 300 000 kWs Before the steam valves start to open the machine loses 20 000 Â 0:5 ¼ 10 000 kWs of the stored energy in order to supply the load.

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146 Electric Power Systems, Fifth EditionThe stored energy / ÀspeedÁ2. Therefore the new frequency rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 300 000 À 10 000 ¼ 300 000 Â 50 Hz ¼ 49:2 Hz4.3 Control LoopsThe machine and its associated governor and voltage-regulator control systems maybe represented by the block diagram shown in Figure 4.7. Two factors have a large inﬂuence on the dynamic response of the prime mover:(1) entrained steam between the inlet valves and the ﬁrst stage of the turbine(in large machines this can be sufﬁcient to cause loss of synchronism after thevalves have closed); (2) the energy stored in the reheater which causes the output of Voltage Vr regulatorFrom Exciter Vboiler Steam Turbine Low Infinite valve High pressure Network busbar pressure Reheat Field winding Δω ω δ Controller Δδ ωr Servometer and δr hydraulic relaysFigure 4.7 Block diagram of complete turboalternator control systems. The governorsystem is more complicated than that shown in Figure 4.3 owing to the inclusion of theload-angle d in the control loop. Sufﬁxes r refer to reference quantities and D to theerror quantities. The controller modiﬁes the error signal from the governor by takinginto account the load angle

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Control of Power and Frequency 147the low-pressure turbine to lag behind that of the high-pressure side. The transferfunction prime mover torque valve openingaccounting for both these effects is G1 G2 ð4:2Þ ð1 þ ttsÞð1 þ trsÞwhereG1 ¼ entrained steam constant;G2 ¼ reheater gain constant; tt ¼ entrained steam time constant; tr ¼ reheater time constant. The transfer function relating steam-valve opening d to changes in speed v due tothe governor feedback loop is Dd ðsÞ ¼ À Á G3G4G5 ð4:3Þ Dv 1 þ tgs ð1 þ t1sÞð1 þ t2sÞwhere tg ¼ governor-relay time constant t1 ¼ primary-relay time constant t2 ¼ secondary-valve-relay time constantG3G4G5 ¼ constants relating system-valve lift to speed changeBy a consideration of the transfer function of the synchronous generator with theabove expressions the dynamic response of the overall system may be obtained.4.4 Division of Load between GeneratorsThe use of the speed changer enables the steam input and electrical power output ata given frequency to be changed as required. The effect of this on two machines canbe seen in Figure 4.8. The output of each machine is not therefore determined by thegovernor characteristics but can be varied by the operating personnel to meet eco-nomic and other considerations. The governor characteristics only completelydecide the outputs of the machines when a sudden change in load occurs or whenmachines are allowed to vary their outputs according to speed within a prescribed

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148 Electric Power Systems, Fifth Edition 1.08 Speed or frequency p.u. Machine A 1.04 Machine A - new governor setting Machine B 1.0 0 Load B Load p.u. Load A 0.96Figure 4.8 Two machines connected to an inﬁnite busbar. The speeder gear ofmachine A is adjusted so that the machines share load equallyrange in order to keep the frequency constant. This latter mode of operation isknown as free-governor action. It has been shown in Chapter 2 (Equation (2.15)) that when d is small the voltagedifference between the two ends of an interconnector of total impedance R þ jXis given by DVp ¼ VG À VL ¼ RP þ XQ VL Also the angle between the voltage phasors (that is, the transmission angle) d isgiven by DVq VG sinÀ1where DVq ¼ XP À RQ VLWhen X ) R, that is for most transmission networks, DVq / P DVp / Q

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Control of Power and Frequency 149 P1 System P2 Q1 R + jX Q2 V1 V2Local GA Local GBload loadFigure 4.9 Two generating stations linked by an interconnector of impedance(R þ jX). The rotor of A is in phase advance of B Hence, (1) the ﬂow of power between two nodes is determined largely by thetransmission angle; (2) the ﬂow of reactive power is determined by the scalar volt-age difference between the two nodes. These two facts are of fundamental importance to the understanding of the opera-tion of power systems. Consider the two generator power system of Figure 4.9. The angular advance ofGA is due to a greater relative energy input to turbine A than to B. The provision ofthis extra steam (or water) to A is possible because of the action of the speeder gearwithout which the power outputs of A and B would be determined solely by thenominal governor characteristics. The following simple example illustrates theseprinciples.Example 4.2Two synchronous generators operate in parallel and supply a total load of 200 MW. Thecapacities of the machines are 100 MW and 200 MW and both have governor droopcharacteristics of 4% from no load to full load. Calculate the load taken by eachmachine, assuming free governor action.SolutionLet x megawatts be the power supplied from the 100 MW generator. Referring toFigure 4.10, 4 ¼ a 100 x

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150 Electric Power Systems, Fifth Edition Speed and frequency 1.04 No-load speed p.u. α100 MW x 100 MW Common speed for 1 p.u. load of 200 MW 200 MW m/c A m/c B 200 MWFigure 4.10 Speed-load diagram for Example 4.2For the 200 MW machine, 4 ¼ a x 200 200 À ; 4x ¼ 800 À 4x 100 200and x ¼ 66.6 MW ¼ load on the 100 MW machine. The load on the 200 MWmachine ¼ 133.3 MW. It will be noticed that when the governor droops are the same the machines share thetotal load in proportion to their capacities or ratings. Hence it is often advantageous forthe droops of all turbines to be equal.Example 4.3Two units of generation maintain 66 kV and 60 kV (line) at the ends of an interconnectorof inductive reactance per phase of 40 V and with negligible resistance and shunt capac-itance. A load of 10 MW is to be transferred from the 66 kV unit to the other end.Calculate the necessary conditions between the two ends, including the power factor ofthe current transmitted.Solution(using Equations (2.15) and (2.16))

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Control of Power and Frequency 151As R ’ 0DVq ¼ XP ¼ 40 Â 3:33 Âpﬃ1ﬃ 06 ¼ 3840 V VL 60 000= 3also DVqpﬃﬃ ¼ sin d ¼ 0:10166 000= 3Hence the 66 kV busbars are 5 440 in advance of the 60 kV busbars.DVp ¼ 66 000pÀﬃﬃ60 000 ¼ XQ ¼ 40Q pﬃﬃ 3 VL 000= 3 60henceQ ¼ 3 MVAr per phase ð9 MVAr totalÞThe p.f. angle w ¼ tanÀ1 (Q/P) ¼ 42 and hence the p.f. ¼ 0.74.4.5 The Power-Frequency Characteristic of an Interconnected SystemThe change in power for a given change in the frequency in an interconnected sys-tem is known as the stiffness of the system. The smaller the change in frequency fora given load change the stiffer the system. The power-frequency characteristic maybe approximated by a straight line and DP/Df ¼ K, where K is a constant (MW perHz) depending on the governor and load characteristics. Let DPG be the change in generation with the governors operating ‘free acting’resulting from a sudden increase in load DPL. The resultant out-of balance power inthe system DP ¼ DPL À DPG ð4:4Þand therefore K ¼ DPL À DPG ð4:5Þ Df DfDPL measures the effect of the frequency characteristics of the load and DfDPG / ðPT À PGÞ, where PT is the turbine capacity connected to the network and PGthe output of the associated generators. When steady conditions are again reached,the load PL is equal to the generated power PG (neglecting losses): hence,K ¼ K1PT À K2PL, where K1 and K2 are the power-frequency coefﬁcients relevant tothe turbines and load respectively.K can be determined experimentally by connecting two large separate systemsby a single link, breaking the connection and measuring the frequency change. Forthe British system, tests show that K ¼ 0:8PT À 0:6PL and lies between 2000 and

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152 Electric Power Systems, Fifth Edition 62.5 60.0 57.5Frequency (Hz) 55.0 52.5 50.0 47.5 21.31 21.32 21.33 21.34 21.35 21.29 21.30 Time (h)Figure 4.11 Decline of frequency with time of New York City system when isolatedfrom external supplies (Reproduced with permission from IEEE c 1977)5500 MW per Hz, that is, a change in frequency of 0.1 Hz requires a change in therange 200–550 MW, depending on the amount of plant connected. In smaller sys-tems the change in frequency for a reasonable load change is relatively large andrapid-response electrical governors have been introduced to improve the power-frequency characteristic. In 1977, owing to a series of events triggered off by lightning, New York City wascut off from external supplies and the internal generation available was much lessthan the city load. The resulting fall in frequency with time is shown in Figure 4.11,illustrating the time-frequency characteristics of an isolated power system.4.6 System Connected by Lines of Relatively Small CapacityLet KA and KB be the respective power-frequency constants of two separate powersystems A and B, as shown in Figure 4.12. Let the change in the power transferred

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Control of Power and Frequency 153 System A Δ Pt System B KA (a) KB KA ΔP fA KB fB (b)Figure 4.12 (a) Two interconnected power systems connected by a tie-line, (b) Thetwo systems with the tie-line openfrom A to B when a change resulting in an out-of-balance power DP occurs in sys-tem B, be DPt, where DPt is positive when power is transferred from A to B. Thechange in frequency in system B, due to an extra load DP and an extra input of DPt,from A, is ÀðDP À DPtÞ=KB(the negative sign indicates a fall in frequency). The dropin frequency in A due to the extra load DPt is ÀDPt=KA, but the changes in frequencyin each system must be equal, as they are electrically connected. Hence, ÀðDP À DPtÞ=KB ¼ ÀDPt=KA KA ;DPt ¼ þ KA þ KB DP ð4:6ÞNext, consider the two systems operating at a common frequency f with A exportingDPt to B. The connecting link is now opened and A is relieved of DPt and assumes afreqency fA, and B has DPt more load and assumes fB. Hence fA ¼ f þ DPt and fB ¼ f À DPt KA KBfrom which DPt ¼ KAKB ð4:7Þ fA À fB KA þ KB

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154 Electric Power Systems, Fifth Edition Hence, by opening the link and measuring the resultant change in fA and fB thevalues of KA and KB can be obtained. In practice, when large interconnected systems are linked electrically to others bymeans of tie-lines the power transfers between them are usually decided by mutualagreement and the power is controlled by regulators or Automatic Generation Con-trol (AGC). As the capacity of the tielines is small compared with the systems, caremust be taken to avoid excessive transfers of power and corresponding cascadetripping.4.6.1 Effect of Governor CharacteristicsA fuller treatment of the performance of two interconnected systems in the steadystate requires further consideration of the control aspects of the generators. A more complete block diagram for steam turbine-generators connected to apower system is shown in Figure 4.13.DP0 ¼ change in speed-changer setting;DP ¼ change in power output of prime movers;DL ¼ change load power; Df ¼ change in frequency; R ¼ governor droop, that is drop in speed or frequency when machines of an area range from no load to full load Figure 4.13 can be used to represent a number of coherent generators with similarcharacteristics. For this system the following equation holds: MsDf þ DDf ¼ DP À DL 1/R ΔL Governor Turbine – Power system– 1 ΔPu 1 ΔP + 1 Δf– M s+D τG s+1 τr s+1ΔPFigure 4.13 Block diagram for a turbine generator connected to a power system

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Control of Power and Frequency 155where:M is the angular momentum of the spinning generators and loads (see Chapter 8)D is the damping coefﬁcient of the load that is, change of power drawn by the load with frequencyTherefore, the change from normal speed or frequency, Df ¼ 1 D ðDP À DLÞ Ms þThis analysis holds for steam-turbine generation; for hydro-turbines, the large iner-tia of the water must be accounted for and the analysis is more complicated. The representation of two systems connected by a tie-line is shown in Figure 4.14.The general analysis is as before except for the additional power terms due to thetie-line. The machines in each individual power system are considered to be closelycoupled and to possess one equivalent rotor. 1/R1 Δ L1 – Governor Turbine ΔP1 + – Power system Δf1 – 1 – 1 Δ Pu1 1 Δ P1 τr1s+1 M1 s+D1 τG1 s+1 Area 1 1/s P12 T12 + – Δ P2 1/s Governor Turbine + Area 2 Δf2 – 1 ΔPu2 1 ΔP2 + 1 M2 s+D2 τG2 s+1 τr2 s+1 –– ΔP Δ L2 1/R 2Figure 4.14 Block control diagram of two power systems connected by a tie-line

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156 Electric Power Systems, Fifth EditionFor system (1), M1sDf1 þ D1Df1 þ T12ðd1 À d2Þ ¼ DP1 À DL1 ð4:8Þwhere T12 is the synchronizing torque coefﬁcient of the tie-line. For system (2), M2sDf2 þ D2Df2 þ T12ðd2 À d1Þ ¼ DP2 À DL2 ð4:9ÞThe steady-state analysis of two interconnected systems may be obtained from thetransfer functions given in the block diagram.The response of the Governor is given by À1 tGs þ 1 DPU ¼ 1 R Df þ DP0 ð4:10ÞIn the steady state, from equation (4.10), 1 DP1 ¼ R1 Df ð4:11Þand 1 DP2 ¼ R2 Df ð4:12ÞSimilarly, from equations (4.8) and (4.9), in the steady state, 1 D1 þ R1 Df1 þ T12ðd1 À d2Þ ¼ ÀDL1 ð4:13Þand 1 D2 þ R2 Df2 þ T12ðd2 À d1Þ ¼ ÀDL2 ð4:14Þ ð4:15ÞAdding equations (4.13) and (4.14) gives 1 1 D1 þ R1 Df1 þ D2 þ R2 Df2 ¼ ÀDL1 À DL2In a synchronous system, Df1 ¼ Df2 ¼ Df and equation (4.15) becomes 1 1 D1 þ R1 þ D2 þ R2 Df ¼ ÀDL1 À DL2

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Control of Power and Frequency 157 ð4:16Þand ð4:17Þ Df ¼ ÀDL1 À DL2 D1 1 þ D2 þ 1 þ R2 R1From equations (4.13) and (4.14), ÀÀ D1 þ 1=R1Þ ÀDL1 D2 þ 1=R2Þ þ DL2 T12ðd1 À d2Þ ¼ À À D2 þ 1=R2Þ þ D1 þ 1=R1ÞExample 4.4Two power systems, A and B, each have a regulation R of 0.1 p.u. and a damping factorD of 1(on their respective capacity bases). The capacity of system A is 1500 MW and of B1000 MW. The two systems are interconnected through a tie-line and are initially at60 Hz. If there is a 100 MW load increase in system A, calculate the change in the steadystate values of frequency and power transfer.Solution DA ¼ 1500 MW=Hz DB ¼ 1000 MW=Hz RA ¼ 0:1 Â 60 ¼ 6 Hz=MW 1500 1500 RB ¼ 0:1 Â 60 ¼ 6 Hz=MW 1000 1000 From equation (4.16), Df ¼ ÀDLA À DLB ¼ À100 ¼ À0:034 Hz DA þ DB þ 1 þ 1 1500 þ 1000 þ 1500 þ 1000 RA RB 6 6 À ÁÀ Á ÀDLA DB þ 1=RB þ DLB DA þ 1=RA PAB ¼ TAB ðdA À dBÞ ¼ ðDB þ 1=RBÞ þ ðDA þ 1=RAÞ þ 1000 7000 À100 1000 6 À100 6 1500 þ 1000 þ ¼ 1500 þ 1000 ¼ 17500 ¼ À40 MW 6 6 6

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158 Electric Power Systems, Fifth Edition4.6.2 Frequency-Bias-Tie-Line ControlConsider three interconnected power systems A, B and C, as shown in Figure 4.15,the systems being of similar size. Assume that initially A and B export to C, theirpreviously agreed power transfers. If C has an increase in load the overall frequencytends to decrease and hence the generation in A, B and C increases. This results inincreased power transfers from A and B to C. These transfers, however, are limitedby the tie-line power controller to the previously agreed values and thereforeinstructions are given for A and B to reduce generation and hence C is not helped.This is a severe drawback of what is known as straight tie-line control, which can beovercome if the systems are controlled by using consideration of both load transferand frequency, such that the following equation holds:X ð4:18Þ DP þ BDf ¼ 0 Pwhere DP is the net transfer error and depends on the size of the system and thegovernor characteristic, and Df is the frequency error and is positive for high fre-quency. B is known as the frequency bias factor and is derived from Equation (4.17). In the case above, after the load change in C, the frequency error is negative (i.e.low frequency) for A and B and the sum of DP for the lines AC and BC is positive.For correct control, XX PA þ BADf ¼ PB þ BBDf ¼ 0 SystemPs A and B take no regulating action despite their fall in frequency. In C, DPC is negative as it is importing from A and B and therefore the governorspeeder motors in C operate to increase output and restore frequency. This system isknown as frequency-bias-tie-line control and is often implemented automatically ininterconnected systems.A B PAC PBC CFigure 4.15 Three power systems connected by tie-lines

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Control of Power and Frequency 159Problems4.1 A 500 MVA, 2 pole, turbo-alternator delivers 400 MW to a 50 Hz system. If the generator circuit breaker is suddenly opened and the main steam valves take 400 ms to operate what will be the over-speed of the generator? The stored energy of the machine (generator and turbine) at synchronous speed is 4 kWs/ kVA. (Answer 3117 r.p.m or 52 Hz)4.2 Two identical 60 MW synchronous generators operate in parallel. The governor settings on the machines are such that they have 4 and 3% droops (no-load to full-load percentage speed drop). Determine (a) the load taken by each machine for a total of 100 MW; (b) the percentage adjustment in the no-load speed to be made by the speeder motor if the machines are to share the load equally. (Answer: (a) 42.8 and 57.2 MW; (b) 0.83% increase in no-load speed on the 4% droop machine)4.3 a. Explain how the output power of a turbine-generator operating in a constant frequency system is controlled by adjusting the setting of the governor. Show the effect on the generator power-frequency curve. b. Generator A of rating 200 MW and generator B of rating 350 MW have gov- ernor droops of 5 and 8%, respectively, from no-load to full-load. They are the only supply to an isolated system whose nominal frequency is 50 Hz. The corresponding generator speed is 3000 r.p.m. Initially, generator A is at 0.5 p.u. load and generator B is at 0.65 p.u. load, both running at 50 Hz. Find the no load speed of each generator if it is disconnected from the system. c. Also determine the total output when the ﬁrst generator reaches its rating. (Answer: (b) Generator B 3156 r.p.m; generator A 3075 r.p.m; (c) 537 MW) (From Engineering Council Examination, 1996)4.4 Two power systems A and B are interconnected by a tie-line and have power frequency constants KA and KB MW/Hz. An increase in load of 500 MW on system A causes a power transfer of 300 MW from B to A. When the tie-line is open the frequency of system A is 49 Hz and of system B 50 Hz. Determine the values of KA and KB, deriving any formulae used. (Answer: KA 500 MW/Hz; KB 750 MW/Hz)4.5 Two power systems, A and B, having capacities of 3000 and 2000 MW, respec- tively, are interconnected through a tie-line and both operate with frequency- bias-tieline control. The frequency bias for each area is 1% of the system capac- ity per 0.1 Hz frequency deviation. If the tie-line interchange for A is set at 100 MW and for B is set (incorrectly) at 200 MW, calculate the steady-state change in frequency. (Answer: 1 Hz; use DPA þ BAf ¼ DPB þ BBf)

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160 Electric Power Systems, Fifth Edition 1000 MW BA Figure 4.16 Interconnected systems of Problem 4.6 (b)4.6 a. i. Why do power systems operate in an interconnected arrangement? ii. How is the frequency controlled in a power system? iii. What is meant by the stiffness of a power system? b. Two 50 Hz power systems are interconnected by a tie-line, which carries 1000 MW from system A to system B, as shown in Figure 4.16. After the out- age of the line shown in the ﬁgure, the frequency in system A increases to 50.5 Hz, while the frequency in system B decreases to 49 Hz. i. Calculate the stiffness of each system. ii. If the systems operate interconnected with 1000 MW being transferred from A to B, calculate the ﬂow in the line after outage of a 600 MW gener- ator in system B. (Answer: (b) (i) KA 2000 MW/Hz, KB 1000 MW/Hz; (ii) 1400 MW) (From Engineering Council Examination, 1997)

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5Control of Voltage andReactive Power5.1 IntroductionThe approximate relationship between the magnitude of the voltage difference oftwo nodes in a network and the ﬂow of power was shown in Chapter 2 to be DV % DVp ¼ RP þ XQ ðfrom 2:15Þ VAlso it was shown that the transmission angle d is proportional to d / DVq ¼ XP À RQ ðfrom 2:16Þ VHence it may be seen that for networks where X ) R, that is, most high voltagepower circuits, DV, the voltage difference, is determined mainly by Q while theangle d is controlled by P. Consider the simple system linking two generating stations A and B, as shown inFigure 5.1(a). Initially the system is considered to be only reactive and R is ignored.The machine at A is in phase advance of that at B and V1 is greater than V2; hencethere is a ﬂow of real power from A to B. This can be seen from the phasor diagramshown in Figure 5.1(b). It is seen that Id and hence P is determined by ﬀd and thevalue of Iq and hence Q mainly, by V1 À V2. In this case V1 > V2 and reactive poweris transferred from A to B. By varying the generator excitations such that V2 > V1,the direction of the reactive power is reversed, as shown in Figure 5.1(c). Hence, real power can be sent from A to B or B to A by suitably adjusting theamount of steam (or water) admitted to the turbine, and reactive power can be sentElectric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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162 Electric Power Systems, Fifth Edition P1 System P2 I jX Q1 Q2 V1 V2 V2 IqX IXV1 A B GB IdX V2 IX V1 δ I δ ILoad GA Load Id Iq ω ω (a) (b) (c)Figure 5.1 (a) System of two generators interconnected, (b) Phasor diagram whenV 1 > V 2. Id and Iq are components of I. (c) Phasor diagram when V 2 > V 1in either direction by adjusting the voltage magnitudes. These two operations areapproximately independent of each other if X ) R, and the ﬂow of reactive powercan be studied almost independently of the real power ﬂow. The phasor diagrams show that if a scalar voltage difference exists across a largelyreactive link, the reactive power ﬂows towards the node of lower voltage. Fromanother point of view, if, in a network, there is a deﬁciency of reactive power at apoint, this has to be supplied from the connecting lines and hence the voltage at thatpoint falls. Conversely if there is a surplus of reactive power generated (for example,lightly loaded cables generate positive VArs), then the voltage will rise. This is aconvenient way of expressing the effect of the power factor of the transferred cur-rent, and although it may seem unfamiliar initially, the ability to think in terms ofVAr ﬂows, instead of exclusively with power factors and phasor diagrams, willmake the study of power networks much easier. If it can be arranged that Q2 in the system in Figure 5.1(a) is zero, then there willbe no voltage drop between A and B, a very satisfactory state of affairs. Now assume that the interconnecting system shown in Figure 5.1(a) has someresistance and that V1 is constant. Consider the effect of keeping V2, and hence thevoltage drop DV, constant. From equation (2.15) Q2 ¼ V2DV À RP2 ¼ K À R P2 ð5:1Þ X Xwhere K is a constant and R is the resistance of the system. If this value of Q2 does not exist naturally in the circuit then it will have to beobtained by artiﬁcial means, such as the connection at B of capacitors or inductors.

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Control of Voltage and Reactive Power 163If the value of the power changes from P2 to P20 and if V2 remains constant, then thereactive power at B must change to Q02 such thatQ20 À Q2 ¼ R ÀP02 À Á X P2that is, an increase in real power causes an increase in the reactive power needed tomaintain V2. The change, however, is proportional to (R/X), which is normally small. It is seen that voltage can be controlled by the injection into the network ofreactive power of the correct sign. Other methods of a more obvious kind for con-trolling voltage are the use of tap-changing transformers.5.2 The Generation and Absorption of Reactive Power5.2.1 Synchronous GeneratorsSynchronous generators can be used to generate or absorb reactive power. Anover-excited machine, that is, one with greater than nominal excitation, generatesreactive power whilst an under-excited machine absorbs it. Synchronous generatorsare the main source of supply to the power system of both positive and negativeVArs. The ability to generate or absorb reactive power is shown by the performancechart of a synchronous generator. Reactive power generation (lagging power factoroperation) is limited by the maximum excitation voltage allowable before the rotorcurrents lead to overheating. In Figure 3.12 this is 2.5 p.u. The ability to absorb reactive power is determined by the short-circuit ratio(1/synchronous reactance) as the distance between the power axis and the theoreti-cal stability-limit line in Figure 3.12 is proportional to the short-circuit ratio. In mod-ern machines the value of the short-circuit ratio is made low for economic reasons,and hence the inherent ability to operate at leading power factors (absorbing VArs)is not large. For example, a 200 MW 0.85 p.f. machine with a 10% stability allowancehas a capability of absorbing 45 MVAr at full power output. The VAr absorptioncapacity can, however, be increased by the use of continuously acting voltage regu-lators, as explained in Chapter 3.5.2.2 Overhead Lines and TransformersWhen fully loaded, overhead lines absorb reactive power. With a current I amperesﬂowing in a line of reactance per phase X(V) the VArs absorbed are I2X per phase.On light loads the shunt capacitances of longer lines may become dominant andhigh voltage overhead lines then become VAr generators. Transformers always absorb reactive power. A useful expression for the quantitymay be obtained for a transformer of reactance XT p.u. and a full load rating of3VfIrated

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164 Electric Power Systems, Fifth EditionThe ohmic reactance XðVÞ ¼ XT Vf IratedTherefore the VArs absorbed ¼ 3I2XT Vf Irated ¼ 3I2 Vf2 XT VfIrated À Á2 3IVf ¼ XT 3I rated Vf ¼ ðVA of loadÞ2 Á XT rated VA of transformer5.2.3 CablesCables are generators of reactive power owing to their high shunt capacitance. A 275 kV, 240 MVA cable produces 6.25–7.5 MVAr per km; a 132 kV cable roughly1.9 MVAr per km; and a 33 kV cable, 0.125 MVAr per km.5.2.4 LoadsA load at 0.95 power factor implies a reactive power demand of 0.33 kVAr per kW ofpower, which is more appreciable than the mere quoting of the power factor wouldsuggest. In planning a network it is desirable to assess the reactive power require-ments to ascertain whether the generators are able to operate at the required powerfactors for the extremes of load to be expected. An example of this is shown inFigure 5.2, where the reactive losses are added for each item until the generatorpower factor is obtained. Example 5.1 In the radial transmission system shown in Figure 5.2, all p.u. values are referred to the voltage bases shown and 100 MVA. Determine the power factor at which the generator must operate. Solution Voltage drops in the circuits will be neglected and the nominal voltages assumed. Starting with the consumer load, the VArs for each section of the circuit are added in turn to obtain the total Busbar A, P ¼ 0.5 p.u. Q ¼ 0

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Control of Voltage and Reactive Power 165 160 km 48 km E 0.1 p.u. D 0.04 p.u.C 0.1 p.u. B 0.1 p.u. AG 0.1 p.u. 0.1 p.u. 0.1 p.u. Load 50 MW p.f. = 1 275 kV 0.04 p.u. 132 kV 11 kV 200 MW 0.8 p.f. laggingFigure 5.2 Radial transmission system with intermediate loads. Calculation ofreactive-power requirementI2 Â loss in 132 kV lines and transformers ¼ P2 þ Q2 XCA ¼ 0:52 0:1 V2 12 ¼ 0:025 p:u:Busbar C, P ¼ 2 þ 0:5 p:u: ¼ 2:5 p:u: Q ¼ 1:5 þ 0:025 p:u: ¼ 1:525 p:u:I2X loss in 275 kV lines and transformers ¼ 2:52 þ 1:5252 0:07 12 ¼ 0:6 p:u:If the I2X loss in the large generator-transformer is ignored, the generator must deliverP ¼ 2.5 and Q ¼ 2.125 p.u. and operate at a power factor of 0.76 lagging.5.3 Relation between Voltage, Power, and Reactive Power at a NodeThe voltage V at a node is a function of P and Q at that node, that is. V ¼ fðP; QÞThe voltage also depends on that of adjacent nodes and the present treatmentassumes that these are inﬁnite busbars.

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166 Electric Power Systems, Fifth Edition The total differential of V, dV ¼ @V dP þ @V dQand using @P @Q @P : @V ¼ 1 and @Q : @V ¼ 1 @V @P @V @Q ð5:2Þ dP dQ dV ¼ ð@ P =@ V Þ þ ð@ Q =@ V ÞIt can be seen from equation (5.2) that the change in voltage at a node is deﬁned bythe two quantities @P @Q @V and @VAs an example, consider a line with series impedance (R þ jX) and zero shunt admit-tance as shown in Figure 5.3. From equation (2.15), ðV1 À VÞV À PR À XQ ¼ 0 ð5:3Þwhere V1, the sending-end voltage, is constant, and V, the receiving-end voltage,depends on P and Q. From equation (5.3) @P ¼ V1 À 2V ð5:4Þ @V RAlso, @Q ¼ V1 À 2V ð5:5Þ @V X R + jX P + jQ V1 V LoadFigure 5.3 Single-phase equivalent circuit of a line supplying a load of P þ jQ from aninﬁnite busbar of voltage V1

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Control of Voltage and Reactive Power 167Hence, dV ¼ dP þ dQ ¼ RdP þ XdQ ð5:6Þ @P/@V @Q/@V V1 À 2V For constant V and DV, RdP þ XdQ ¼ 0 and dQ ¼ À(R/X)dP, which is obtainabledirectly from Equation (5.1). Normally, @Q=@V is the quantity of greater interest. It can be found experimen-tally using a load-ﬂow calculation (see Chapter 6) by the injection of a known quan-tity of VArs at the node in question and calculating the difference in voltageproduced. From the results obtained, DQ ¼ Qafter À Qbefore DV Vafter À VbeforeDV should be small for this test, a few per cent of the normal voltage, thereby givingthe sensitivity of the node to the VAr change. From the expression, @Q ¼ V1 À 2V @V Xproved for a single line, it is evident that the smaller the reactance associated with anode, the larger the value of @Q=@V for a given voltage drop, that is, the voltagedrop is inherently small. The greater the number of lines meeting at a node, thesmaller the resultant reactance and the larger the value of @Q=@V. Obviously,@Q=@V depends on the network conﬁguration, but a high value would lie in therange 10–15 MVAr/kV. If the natural voltage drop at a point without the artiﬁcialinjection of VArs is, say, 5 kV, and the value of @Q=@V at this point is 10 MVAr/kV,then to maintain the voltage at its no-load level would require 50 MVAr. Obviously,the greater the value of @Q=@V, the more expensive it becomes to maintain voltagelevels by injection of reactive power.5.3.1 @Q=@V and the Short-Circuit Current at a NodeIt has been shown that for a connecting circuit of reactance X with a sending-endvoltage V1 and a received voltage V @Q ¼ V1 À 2V @V XIf the three-phases of the connector are now short-circuited at the receiving end(i.e. a three-phase symmetrical short circuit applied), the current ﬂowing in the lines I ¼ V1 assuming R(X X

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168 Electric Power Systems, Fifth EditionWith the system on no-load V ¼ V1 and @@QV ¼ VX1Hence the magnitude of @Q=@V is equal to the short-circuit current. With normaloperation, V is within a few per cent of V1 and hence the value of @Q=@V at V ¼ V1gives useful information regarding reactive power/voltage characteristics for smallexcursions from the nominal voltage. This relationship is especially useful as theshort-circuit current will normally be known at all substations. Example 5.2 Three supply points A, B, and C are connected to a common busbar M. Supply point A is maintained at a nominal 275 kV and is connected to M through a 275/132 kV trans- former (0.1 p.u. reactance) and a 132 kV line of reactance 50 V. Supply point B is nomi- nally at 132 kV and is connected to M through a 132 kV line of 50 V reactance. Supply point C is nominally at 275 kV and is connected to M by a 275/132 kV transformer (0.1 p.u. reactance) and a 132 kV line of 50 V reactance. If, at a particular system load, the line voltage of M falls below its nominal value by 5 kV, calculate the magnitude of the reactive volt-ampere injection required at M to re-establish the original voltage. The p.u. values are expressed on a 500 MVA base and resistance may be neglected throughout. Solution The line diagram and equivalent single-phase circuit are shown in Figures 5.4 and 5.5. It is necessary to determine the value of dQ/dV at the node or busbar M; hence the current ﬂowing into a three-phase short-circuit at M is required.275 kV j 50Ω M 132 kV A 0.1 p.u. j 50Ω j 50Ω B 0.1 p.u. C 275 kVFigure 5.4 Schematic diagram of the system for Example 5.2

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Control of Voltage and Reactive Power 169 M j 0.1 j1.43 j1.43 j1.43 j 0.1 NFigure 5.5 Equivalent single-phase network with the node M short-circuited toneutral)The base value of reactance in the 132 kV circuit assuming a 500 MVA system base is Zbase ¼ 1322 ¼ 35 V 500Therefore the line reactances XL ¼ j50 ¼ j1:43 p:u: 35The equivalent reactance from M to N ¼ j0.5 p.u. Hence the fault MVA at M ¼ 500 ¼ 1000 MVA 0:5and the fault current ¼ pﬃ1ﬃ 000 Â 106 ¼ 4380 A 3 Â 132 Â 103 pﬃﬃIt has been shown that @QM/ 3@VM ¼ three-phase short-circuit current when QM andVM are three-phase and line values @QM ¼ 4380 Â pﬃﬃ ¼ 7:6 MVAr=kV @VM 3Assuming the natural voltage drop at M ¼ 5 kV.

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170 Electric Power Systems, Fifth Edition Therefore the value of the injected VArs required to offset this drop ¼ 7:6 Â 5 ¼ 38 MVAr An alternative approach is to consider: The source impedance is 0.5 p.u. on a 132 kV, 500 MVA base A 5 kV voltage drop is 5/132 ¼ 0.038 p.u. The current ﬂow for this volt drop is 0:038/0:5 ¼ 0:076 p:u: At close to 1 p.u. voltage this is also the reactive power ﬂow. Q ¼ 0:076 p:u: ¼ 0:076 Â 500 ¼ 38 MVAr5.4 Methods of Voltage Control: (a) Injection of Reactive PowerIn transmission systems with X ) R, busbar voltages can be controlled by the injec-tion or absorption of reactive power. However, controlling network voltage throughreactive power ﬂow is less effective in distribution networks where the higher cir-cuit resistances lead to the reactive power ﬂows having less effect on voltage andcausing an increase in real power losses. Although reactive power does no real work, it does lead to an increase in themagnitude of current in the networks and hence real power losses. Electricitysuppliers often penalize loads with a poor power factor by applying chargesbased on kVAh (or even kVArh) in addition to kWh or even basing part of thecharge on peak kVA drawn. The provision of static capacitors to improve thepower factors of factory loads has been long established. The capacitancerequired for the power-factor improvement of loads for optimum economy isdetermined as follows. Let the tariff of a consumer be based on both kVA and kWhcharge ¼ $A Â kVA þ $B Â kWhA load of P kilowatts at power factor w1, lagging has a kVA of P/cosw1. If this powerfactor is improved to cos w2, the new kVA is P/cos w2. The saving is therefore 1À1saving ¼ $PA cos w1 cos w2

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Control of Voltage and Reactive Power 171The reactive power required from the correcting capacitors Pðtan w1 À tan w2Þ kVArLet the cost per annum in interest and depreciation on the capacitor installation be$C per kVAr or $CPðtan w1 À tan w2ÞThe net saving ! 1 1 ¼ P $A cos w1 À cos w2 À $Cðtan w1 À tan w2ÞThis saving is a maximum when ÀÁ ! d saving sin w2 1 dw2 ¼ P $A À cos2 w2 þ $C cos2 w2 ¼0that is when sin w2 ¼ $C/$A. It is interesting to note that the optimum power factor is independent of the origi-nal one. The improvement of load power factors in such a manner will help to allevi-ate the whole problem of VAr ﬂow in the distribution system. The main effect of transmitting power at non-unity power factors is to increaselosses and reduce the ability of the circuits to transport active power. Thus bothoperating and capital costs are increased by low power factor. It is evident fromequation (2.15) that, for circuits with a signiﬁcant X/R ratio, the voltage drop islargely determined by the reactive power Q. At non-unity power factors the linecurrents are larger, giving increased I2R losses and hence reduced thermal capa-bility. One of the obvious places for the artiﬁcial injection of reactive power is atthe loads themselves. In general, four methods of injecting reactive power are available, involving theuse of:1. static shunt capacitors;2. static series capacitors;3. synchronous compensators;4. static VAr compensators and STATCOMs.5.4.1 Shunt Capacitors and ReactorsShunt capacitors are used to compensate lagging power factor loads, whereasreactors are used on circuits that generate VArs such as lightly loaded cables. Theeffect of these shunt devices is to supply or absorb the requisite reactive power tomaintain the magnitude of the voltage. Capacitors are connected either directly to abusbar or to the tertiary winding of a main transformer. In the USA they are often

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172 Electric Power Systems, Fifth Editioninstalled along the routes of distribution circuits to minimize the losses and voltagedrops. Unfortunately, as the voltage reduces, the VArs produced by a shunt capaci-tor or absorbed by a reactor fall as the square of the voltage; thus, when neededmost, their effectiveness drops. Also, with light network load when the voltage ishigh, the capacitor output is large and the voltage tends to rise to excessive levels,requiring some capacitors or cable circuits to be switched out by local overvoltagerelays.5.4.2 Series CapacitorsCapacitors can be connected in series with overhead lines and are then used toreduce the inductive reactance between the supply point and the load. One majordrawback is the high overvoltage produced across the capacitor when a short-circuitcurrent ﬂows through the circuit, and special protective devices need to be incorpo-rated (e.g. spark gaps) and non-linear resistors. The phasor diagram for a line with aseries capacitor is shown in Figure 5.6(b). The relative merits between shunt and series capacitors may be summarized asfollows:1. If the load VAr requirement is small, series capacitors are of little use.2. With series capacitors the reduction in line current is small; hence if thermal con- siderations limit the current, little advantage is obtained and shunt compensation should be used.3. If voltage drop is the limiting factor, series capacitors are effective; also, voltage ﬂuctuations due to arc furnaces, and so on, are evened out. XL XC I VRVS I _Φ IXL IXC VS(without capacitor) (a) IXCXL= ω L VS (with capacitor) I(XL - XC)XC= 1 VR ωC ω ΦIVC (b)Figure 5.6 (a) Line with series capacitor, (b) Phasor diagram for ﬁxed VR

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Control of Voltage and Reactive Power 1734. If the total line reactance is high, series capacitors are very effective in reducing voltage drops and stability is improved. Both shunt and series capacitors need to be applied with care as they can bothlead to resonance with the inductive reactance of the power system. Shuntcapacitors are benign as long as their network is connected to the main powersystem and the voltage is controlled. However, if a section of network containingboth shunt capacitors and induction generators is isolated then self-excitation ofthe induction generators can lead to very high resonant voltages. The use of seriescapacitors, although very effective in reducing voltage drop on heavily loadedcircuits, can lead to sub-synchronous resonance with rotating machines. Capacitorsare not commonly used in distribution systems in the UK, partly because ofconcerns over resonance.5.4.3 Synchronous CompensatorsA synchronous compensator is a synchronous motor running without a mechanicalload and, depending on the value of excitation, it can absorb or generate reactivepower. As the losses are considerable compared with static capacitors, the powerfactor is not zero. When used with a voltage regulator the compensator can automat-ically run overexcited at times of high load and underexcited at light load. A typical275 kV To AVR 415 V supply (via VT) CB Earthing 11 kV transformer To AVR66 kV CT Synchronous compensatorFigure 5.7 Typical installation with synchronous compensator connected to tertiary(delta) winding of main transformer. A neutral point is provided by the earthing trans-former shown. The Automatic Voltage Regulator (AVR) on the compensator is con-trolled by a combination of the voltage on the 275 kV system and the current output;this gives a droop to the voltage-VAr output curve which may be varied as required

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174 Electric Power Systems, Fifth EditionVoltage p.u. 1.07 5% Droop 1.05 1.0 0.95 +40 Generate -20 0Absorb Q MVArFigure 5.8 Voltage-reactive power output of a typical 40 MVAr synchronouscompensatorconnection of a synchronous compensator is shown in Figure 5.7 and the associatedVolt-VAr output characteristic in Figure 5.8. The compensator is run up as an induc-tion motor in 2.5 min and then synchronized. A great advantage is the ﬂexibility of operation for all load conditions. Althoughthe cost of such installations is high, in some circumstances it is justiﬁed, for exam-ple at the receiving-end busbar of a long high-voltage line where transmission atpower factors less than unity cannot be tolerated. Being a rotating machine, itsstored energy is useful for increasing the inertia of the power system and for ridingthrough transient disturbances, including voltage sags.5.4.4 Static VAr Compensators (SVCs) and STATCOMsSynchronous compensators are rotating machines and so are expensive and havemechanical losses. Hence they are being superseded increasingly by power elec-tronic compensators: SVCs and STATCOMs. SVCs use shunt connected reactors and capacitors controlled by thyristors. Thereactive power is provided by the shunt elements (capacitors and inductors), as dis-cussed in Section 5.4.1 but these are controlled by thyristors. The output of the

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Control of Voltage and Reactive Power 175TCR / MSR MSC / TSC TCR / TSC TCR / MSCFigure 5.9 Possible combinations of controlled reactors and capacitors forming anSVC. TCR: Thyristor Controlled Reactor, MSR: Mechanically Switched Reactor, MSC:Mechanically Switched Capacitor, TSC: Thyristor Switched CapacitorThyristor Controlled Reactor (TCR) is controlled by delaying the switching on of thethyristor within the 50/60 Hz cycle. The thyristor switches off when the currentdrops to zero. The ﬁring angle of the thyristor can be varied within each cycle andhence the VAr absorption by the TCR controlled. As shown in Figure 5.9, TCRs maybe used with Mechanically or Thyristor Switched Capacitors to create an SVC toexport and import VArs. When a capacitor is connected to a strong voltage source, very large currents canﬂow. Hence Thyristor Switched Capacitors are only operated in integral cycles andthe operation of the thyristors is timed so that they switch when there is no instanta-neous voltage across the capacitor. A STATCOM (Static Compensator) is also a power electronic device to providereactive power but it operates on a different principle (Figure 5.10). A STATCOMconsists of a Voltage Source Converter (VSC) connected to the power systemthrough a coupling reactance (L). The VSC uses very large transistors that can beturned on and off to synthesize a voltage sine wave of any magnitude and phase.VSTATCOM is a 50/60 Hz sine wave kept in phase with Vterminal (Figure 5.10). Ifthe magnitude of VSTATCOM is greater than that of Vterminal then reactive power isgenerated by the STATCOM while if the magnitude of VSTATCOM is less than that ofVterminal then reactive power is absorbed by the STATCOM. A very small phaseangle is introduced between VSTATCOM and Vterminal so that a small amount of realpower ﬂows into the STATCOM to charge the DC capacitor and provide for thelosses of the converter. However, the principle of operation is that the reactivepower is provided by the interaction of the two voltage magnitudes across thereactor. The DC capacitor is only used to operate the power electronics and control

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176 Electric Power Systems, Fifth EditionVterminal ISTATCOM (Generating)Q VSTATCOM_1 VSTATCOM_2 L Vterminal ISTATCOM VSTATCOMVSC ISTATCOM VDC (Absorbing) Q (a) (b) Figure 5.10 Operation of a STATCOMthe ripple current. STATCOMs can be controlled very fast and have a smallerphysical equipment footprint than SVCs.5.5 Methods of Voltage Control: (b) Tap-Changing TransformersThe basic operation of the tap-changing transformer has been discussed in Chapter3. By changing the transformation ratio, the voltage in the secondary circuit isvaried. Hence voltage and reactive power control is obtained. In distribution circuits, tap-changing transformers are the primary method ofvoltage control. In a distribution transformer, the tap-changer compensates for thevoltage drop across the reactance of the transformer but also for the variations inthe voltage applied to the primary winding caused by changes of load within thehigh voltage network. In transmission circuits reactive power is dispatched by alter-ing the taps of transformers and this, in turn, controls the network voltages.5.5.1 Use of Tap-Changing Transformers to Control Voltage in a Distribution SystemConsider the 40 MVA 132/11 kV transformer with a reactance of 13% on its ratingshown in Figure 5.11. It is equipped with an on-load tap-changer that is used tomaintain constant voltage at an 11 kV busbar by compensating for variations in thevoltage of the 132 kV network and for the voltage drop across the transformer. Thevariation of network voltage at the 132 kV transformer busbar for heavy and lightloading conditions, and the loads of the transformer are given in Table 5.1. Activepower losses in the transformer are ignored and it is assumed that the value of thereactance of the transformer is not inﬂuenced by the change in the turns ratio.

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Control of Voltage and Reactive Power 177 t:1 X Vr VsFigure 5.11 Tap changing transformer in a distribution circuit t is the fraction of the nominal transformation ratios, that is the tap ratio/nominalratio. For example, a transformer of nominal ratio 132 to 11 kV when tapped to give144 to 11 kV has a t of 144/132 ¼ 1.09. Choosing SBASE of 40 MVA and VBASES of 132 kV and 11 kV. Under heavy loading conditions,Vr ¼ 11 kV, 1 p.u.Vs ¼ 120 kV, 0.909 p.u.Q ¼ 13.94 MVAr, 0.3485 p.u. DV ¼ Q X ¼ 0:3485 Â 0:13 ¼ 0:05 p:u: VS 0:909 Vrt ¼ ðVs À DVÞ t ¼ ðVs À DVÞ ¼ 0:909 À 0:05 ¼ 0:86 Vr 1Under light loading conditions,Vr ¼ 11 kV, 1 p.u.Vs ¼ 145 kV, 1.1 p.u.Q ¼ 2.11 MVAr, 0.053 p.u.Table 5.1 Loading of transformer Power Voltage Desired Voltage Load Factor at 132 kV at 11 kV of Load Busbar Vs Busbar VrHeavy loading conditions 32 MVALight loading conditions 4 MVA 0.90 120 kV 11 kV 0.85 145 kV 11 kV

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178 Electric Power Systems, Fifth EditionDV ¼ Q X ¼ 0:053 Â 0:13 ¼ 0:006 p:u: VS 1:1Vrt ¼ ðVs À DVÞt ¼ ðVs À DVÞ ¼ 1:1 À 0:006 ¼ 1:094 Vr 1 A radial distribution circuit with two tap-changing transformers, is shown in theequivalent single-phase circuit of Figure 5.12. V1 and V2 are the nominal voltages; at R +jX IIs Line IR –Φ or (P + jQ)V1 Vs Vr V2 Load1 : ts tr : 1 (a) X VrVs t :1 (b) X 2 P t Vs t Q Vr R + jX t 2 rV1( ts ) V2 (c) trFigure 5.12 (a) Coordination of two tap-changing transformers in a radial transmis-sion link (b) and (c) Equivalent circuits for dealing with off-nominal tap ratio, (b) Singletransformer, (c) Two transformers

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Control of Voltage and Reactive Power 179the ends of the circuit the actual voltages are tsV1 and trV2. It is required to deter-mine the tap-changing ratios needed to compensate completely for the voltage dropin the line. The product tstr will be made unity; this ensures that the overall voltagelevel remains in the same order and that the minimum range of taps on both trans-formers is used. (Note that all values are in per unit; t is the off-nominal tap ratio.) Transfer all quantities to the load circuit. The line impedance becomes ðR þ jXÞ=tr2; Vs ¼ V1ts and, as the impedance hasbeen transferred Vr ¼ V1ts. The input voltage to the load circuit becomes V1ts=trand the equivalent circuit is as shown in Figure 5.10(c). The arithmetic voltage drop ¼ V1 ts À V2 % RP þ XQ tr tr2V2 When tr ¼ 1/ts t2s V1V2 À V22 ¼ ðRP þ XQÞt2sAnd Á1! XQÞ 2 V2 ¼ 1/2 t2s V1 Æ tsÀts2V12 À 4ðRP þ ð5:7Þ If ts is speciﬁed then tr is deﬁned. There are then two values of V2 for a given V1,one low current, high voltage and one high current and low voltage. Only the highvoltage, low current solution is useful in a power system.Example 5.3A 132 kV line is fed through an 11/132 kV transformer from a constant 11 kV supply. Atthe load end of the line the voltage is reduced by another transformer of nominal ratio132/11 kV. The total impedance of the line and transformers at 132 kV is (25 þ j66) V.Both transformers are equipped with tap-changing facilities which are arranged so thatthe product of the two off-nominal settings is unity. If the load on the system is 100 MWat 0.9 p.f. lagging, calculate the settings of the tap-changers required to maintain thevoltage of the load busbar at 11 kV. Use a base of 100 MVA.SolutionThe line diagram is shown in Figure 5.13. As the line voltage drop is to be completelycompensated, V1 ¼ V2 ¼ 132 kV ¼ 1 p.u. Also, ts Â tr ¼ 1. The load is 100 MW,48.3 MVAr., that is, 1 þ j0.483 p.u. Using equation (5.7) 1 ¼ 1/2 tS2 1 Æ tS Àt2S1 À 4ð0:14 Â 1 þ 0:38 Â Á1! 0:48Þ 2 ;2 ¼ t2S Æ tSÀt2S À Á1 1:28 2 À À tS2 Á2 ¼ tS2 ÀtS2 À Á ;2 1:28

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180 Electric Power Systems, Fifth Edition V1 = 132kV V2 = 132kV VS Z Vr 1 :tS tr : 1 P, Q Figure 5.13 Schematic diagram of system for Example 5.3Hence, ts ¼ 1:21 and tr ¼ 1/1:21 ¼ 0:83These settings are large for the normal range of tap-changing transformers (usually notmore than Æ20% tap range). It would be necessary, in this system, to inject VArs at theload end of the line to maintain the voltage at the required value. A transformer at the receiving end of a line does not improve the VAr ﬂow in thecircuit and the current in the supplying line is increased if the ratio is reduced. Incountries with long and inadequate distribution circuits, it is often the practice toboost the received voltage by a variable ratio transformer so as to maintain ratedvoltage as the power required increases. Unfortunately, this has the effect of increas-ing the primary supply circuit current by the transformer ratio, thereby decreasingthe primary voltage still further until voltage collapse occurs.5.5.2 Use of Tap-Changing Transformers to Despatch VArs in a Transmission SystemIn transmission networks VArs may be dispatched by the adjustment of tap settingson transformers connecting busbars. Consider the situation in Figure 5.14(a), inwhich Vs and Vr are constant voltages representing the two connected systems. Thecircuit may be rearranged as shown in Figure 5.14(b), where t is the off-nominal (perunit) tap setting; resistance is zero. The voltage drop between busbars X QT Vs t2 Vr DV ¼ t À Vr ¼ ÁHence, À À Vr2t2Á1/X ¼ QT VsVrt

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Control of Voltage and Reactive Power 181 Vs X t : 1 Vr Infinite busbars (a) X / t2 Vs QT Vr Vs / t (b)Figure 5.14 (a) Two power systems connected via a tap-change transformer,(b) Equivalent circuit with impedance transferred to receiver sideand tð1 À tÞV2/X ¼ QT when Vs ¼ Vr ¼ V ð5:8ÞWhent < 1, QT is positive, that is a ﬂow of VArs into Vrt > 1, QT is negative, a ﬂow of VArs out of VrAlso, QT ¼ t(1 À t)S, where S ¼ short-circuit level, that is V2/X. Thus, by suitable adjustment of the tap setting, an appropriate injection of reactivepower is obtained. The idea can be extended to two transformers in parallel between networks. If onetransformer is set to an off-nominal ratio of, say, 1.1 : 1 and the other to 0.8 : 1 (i.e. inopposite directions), then a circulation of reactive power occurs round the loop,resulting in a net absorption of VArs. This is known as ‘tap stagger’ and is a compar-atively inexpensive method of VAr absorption.Example 5.4A synchronous generator (75 MVA, 0.8 p.f., 11.8 kV and XS ¼ 1.1 p.u.) is connectedthrough an 11/275 kV tap changing transformer (75 MVA, XT ¼ 0.15 p.u., tap range ¼þ/À20%) to a very large 275 kV power system, as shown in Figure 5.15.a. What is the value of the internal emf and power angle of the generator when it exports 60 MW of active and zero MVAr of reactive power to the system? With the

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182 Electric Power Systems, Fifth EditionEV t:1 XS XT 275kVFigure 5.15 Tap changer in a transmission circuit – Example 5.4 transformer tap in the neutral position, what is the value of reactive power output at the generator terminals?b. What is the value of the transformer tap at which 20 MVAr is imported from the 275 kV system, if the terminal voltage of the generator V is maintained at 1 p.u. when the generator does not export any active power.SolutionChoosing a common base SBASE ¼ 75 MVA VBASES ¼ 11 kV; 275 kVConverting to the common base (Equation (2.7)) XS ¼ 1:1 Â 11:82 ¼ 1:26 p:u: 11 XT ¼ 0:15 p:u: X ¼ XS þ XT ¼ 1:41 p:u:At 60 MW and 0 MVAr exported into the 275 kV system, I ¼ 0:8 þ j0 p:u:and the generator internal voltage and power angle areE ¼ 1 þ IX ¼ 1 þ 0:8 Â j 1:41 ¼ 1 þ j1:13E ¼ 1:5ﬀ48 With the transformer tap in the neutral position the reactive power output at thegenerator terminals isV ¼ 1 þ IX ¼ 1 þ 0:8 Â j0:15 ¼ 1 þ j0:12S ¼ VIÃ ¼ ð1 þ j0:12Þ0:8 ¼ 0:8 þ j0:096 p:u:Q ¼ 0:096 Â 75 ¼ 7:2 MVAr

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Control of Voltage and Reactive Power 183 If 20 MVAr is absorbed from the 275 kV system and V and the voltage of the 275 kVsystem are at 1 p.u. then using Equation (5.8) tð1 À tÞV2/XT ¼ QT tð1 À tÞ1/0:15 ¼ À20 75 tð1 À tÞ ¼ À0:04 t2 À t À 0:04 ¼ 0 t ¼ 1:04The solution for t near 1 is chosenThus the transformer taps are set to þ4%5.6 Combined Use of Tap-Changing Transformers and Reactive-Power InjectionA common practical arrangement is shown in Figure 5.16, where the tertiary wind-ing of a three-winding transformer is connected to a VAr compensator. For givenload conditions it is proposed to determine the necessary transformation ratios withcertain outputs of the compensator. The transformer is represented by the equivalent star connection and any lineimpedance from V1 or V2 to the transformer can be lumped together with the trans-former branch impedances. Here, VN is the phase voltage at the star point of theequivalent circuit. The secondary impedance (XS) is usually approaching zero andhence is neglected. Resistance and losses are ignored. P2Q2V1 P S V2 V1 XP VL XS T12 V2T XT V3Compensator C P3= 0 T23 (a) Q3 V3 C (b)Figure 5.16 (a) Schematic diagram with combined tap-changing and synchronouscompensation, (b) Equivalent network

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184 Electric Power Systems, Fifth Edition The allowable ranges of voltage for V1 and V2 are speciﬁed and the values of thethree-phase real and reactive power; P2, Q2, P3, and Q3 are given. P3 is usually takenas zero. The volt drop V1 to VL is given by DVp % XpQV2/N3or DVp % Xp Qp2 ﬃﬃ VL 3 pﬃﬃwhere VL is the line voltage ¼ 3VN and Q2 is the total VArs. Also, DVq % Xp Pp2 ﬃﬃ VL 3 À þ DVpÁ2 þ À Á2 ¼ V12 ; VN DVq(see phasor diagram of Figure 2.24; phase values used)and Qp2 ﬃﬃ2 2 pVLﬃﬃ VL 3 3 þ Xp þ Xp2 pPﬃﬃ 2 ¼ V21 3VL ;ÀV2L þ XpQ2Á2 þ Xp2P22 ¼ V12LV2L pﬃﬃ where V1L is the line voltage ¼ 3V1 ;V2L ¼ V21L À 2XpQ2 Æ 1 rhﬃﬃﬃVﬃﬃﬃ21ﬃﬃLﬃﬃÀﬃﬃVﬃﬃﬃﬃ12ﬃﬃLﬃﬃﬃÀﬃﬃﬃﬃﬃ4ﬃﬃXﬃﬃﬃpﬃﬃQﬃﬃﬃﬃ2ﬃﬃÁﬃﬃﬃÀﬃﬃﬃﬃﬃ4ﬃﬃXﬃﬃﬃﬃp2ﬃﬃPﬃﬃﬃ22ﬃﬃiﬃ 2 2 Once VL is obtained, the transformation ratio is easily found. The procedure isbest illustrated by an example. Example 5.5 A three-winding grid transformer has windings rated as follows: 132 kV (line), 75 MVA, star connected; 33 kV (line), 60 MVA, star connected; 11 kV (line), 45 MVA, delta con- nected. A VAr compensator is available for connection to the 11 kV winding. The equivalent circuit of the transformer may be expressed in the form of three wind- ings, star connected, with an equivalent 132 kV primary reactance of 0.12 p.u., negligi- ble secondary reactance, and an 11 kV tertiary reactance of 0.08 p.u. (both values expressed on a 75 MVA base).

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Control of Voltage and Reactive Power 185 60 MW T12 V2 V1 0 N T12 V2 V1 N 120 kV 34 kV 143 kV 30 kV 30 MVAr 0.8 p.u. T23 T23 C C (a) (b)Figure 5.17 Systems for Example 5.4. (a) System with loading condition 1.(b) System with loading condition 2In operation, the transformer must deal with the following extremes of loading:1. Load of 60 MW, 30 MVAr with primary and secondary voltages governed by the lim- its 120 kV and 34 kV; compensator disconnected.2. No load, primary and secondary voltage limits 143 kV and 30 kV; compensator in operation and absorbing 20 MVAr.Calculate the range of tap-changing required. Ignore all losses.SolutionThe value of XP, the primary reactance (in ohms) ¼ 0:12 Â 1322 ¼ 27:8 V 75 Similarly, the effective reactance of the tertiary winding is 18.5 V. The equivalent starcircuit is shown in Figure 5.17. The ﬁrst operating condition is as follows: P1 ¼ 60 MW Q1 ¼ 30 MVAr V1L ¼ 120 kVHence,V 2 ¼ 1 À 0002 À 2 Â 27:8 Â 30 Â 106Á L 2 120 Æ 1 qÂﬃﬃ1ﬃﬃﬃ2ﬃﬃ0ﬃﬃﬃ0ﬃﬃﬃ0ﬃﬃ0ﬃﬃﬃ2ﬃﬃÀﬃﬃ1ﬃﬃ2ﬃﬃﬃ0ﬃﬃﬃ0ﬃﬃ0ﬃﬃﬃ0ﬃﬃ2ﬃﬃﬃÀﬃﬃﬃﬃﬃ4ﬃﬃﬃÂﬃﬃﬃﬃﬃ2ﬃﬃ7ﬃﬃﬃ:ﬃ8ﬃﬃﬃﬃÂﬃﬃﬃﬃ3ﬃﬃﬃ0ﬃﬃﬃÂﬃﬃﬃﬃﬃ1ﬃﬃ0ﬃﬃﬃ6ﬃﬃÁﬃﬃﬃÀﬃﬃﬃﬃﬃ4ﬃﬃﬃÂﬃﬃﬃﬃﬃ2ﬃﬃ7ﬃﬃﬃ:ﬃ8ﬃﬃ2ﬃﬃﬃÂﬃﬃﬃﬃﬃ6ﬃﬃﬃ0ﬃﬃ2ﬃﬃﬃÂﬃﬃﬃﬃﬃ1ﬃﬃ0ﬃﬃﬃ1ﬃ2ﬃﬃÃﬃﬃ 2 122 ¼ 63:6 Æ 2 108 À 124:4 Â 108 ;VL ¼ 111 kV

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186 Electric Power Systems, Fifth Edition The second set of conditions are: V1L ¼ 143 kV P2 ¼ 0 Q2 ¼ 20 MVAr Again, using the formula for VL, VL ¼ 138:5 kV The transformation ratio under the ﬁrst condition ¼ 111/34 ¼ 3:27 and, for the second condition ¼ 138:5/30 ¼ 4:64 The actual ratio will be taken as the mean of these extremes, that is, 3.94, varying by Æ0.67 or 3.94 Æ 17%. Hence the range of tap-changing required is Æ17%.Example 5.6In the system shown by the line diagram in Figure 5.18, each of transformers TAand TB have tap ranges of Æ 10% in 10 steps of 1.0%. It is necessary to ﬁnd thevoltage boost needed on transformer TA to share the power ﬂow equally betweenthe two lines. The system data is as follows (on a common base):All transformers: XT ¼ 0.1 p.u.Transmission lines: R ¼ R0 ¼ 0X ¼ 0.20 p.u.X0 ¼ 0.15 p.uVA ¼ 1:1ﬀ5VB ¼ 1:0ﬀ0SolutionWe must ﬁrst calculate the current sharing in the two parallel lines:I1 ¼ 1:1ﬀ5 À 1:0ﬀ0 ¼ 0:2397 À j0:2397 j0:4I2 ¼ 1:1ﬀ5 À 1:0ﬀ0 ¼ 0:2740 À j0:2740 j0:35Any boost by transformer TA will cause a current to circulate between the two busbarsbecause the voltages VA and VB are assumed to be held constant by the voltage regula-tors on the generators.

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Control of Voltage and Reactive Power 187VA 33 : 132kV 132 : 33kV VB TA I1 TB R + jXEA EB I2 TC R + jX TD (a) i0.1 j0.2 i0.1Vboost Icirc i0.1 j 0.15 i0.1 (b)Figure 5.18 (a) Line diagram of system for Example 5.6. (b) Equivalent networkwith voltage boost Vboost acting To equalize the currents, a circulating current is required, as in Figure 5.18(b),givingIcirc ¼ I2 À I1 2Icirc ¼ 0:0343 À j0:0343 ¼ 0:0241ﬀ À 45 2;VBoost ¼ 0:0241ﬀ À 45 Â j0:75 ¼ 0:0180ﬀ45V To achieve this boost, ideally TA should be equipped with a phase changer of 45 andtaps to give 1.8% boost. In practice, a tap of 2% would be used in either an in-phaseboost (such as obtainable from a normal tapped transformer) or a quadrature boost(obtainable from a phase-shift transformer. In transmission networks it should be notedthat because of the generally high X/R ratio, an in-phase boost gives rise to a quadra-tive current whereas a quadrature boost produces an in-phase circulating current,thereby adding to or subtracting from the real power ﬂow. Circulating reactive current by adjusting the taps in transformers in parallel circuitshas been used to de-ice lines in winter by producing extra I2R losses for heating. Twotransformers in parallel can be tap-staggered to produce I2X absorption under light-load, high-voltage conditions.

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188 Electric Power Systems, Fifth EditionVYB VY1B R Y (a) VB B VYB VR θ VY1B (b) VR1 VYFigure 5.19 (a) Connections for one phase of a phase shift transformer. Similar con-nections to other two phases. (b) Corresponding phasor diagram5.7 Phase-Shift TransformerA quadrature phase shift can be achieved by the connections shown in Figure 5.19(a).The booster arrangement shows the injection of voltage into one phase only; it isrepeated for the other two phases. In Figure 5.19(b), the corresponding phasordiagram is shown and the nature of the angular shift of the voltage boost VYBindicated. By the use of tappings on the energizing transformer, several values ofphase shift may be obtained.Example 5.7In the system shown in Figure 5.20, it is required to keep the 11 kV busbar at constantvoltage. The range of taps is not sufﬁcient and it is proposed to use shunt capacitorsconnected to the tertiary winding. All impedances are referred to 33 kV. The impedance of the overhead line, ZLreferred to 33 kV ¼ (2.2 þ j5.22) V. For the three-winding transformer the measuredimpedances between the windings and the resulting equivalent star impedances Z1, Z2and Z3 are given in Table 5.2.SolutionThe equivalent circuit referred to 33 kV is shown in Figure 5.20(b). The voltage at point C (referred to 33 kV) is 33p0ﬃ0ﬃ 0 À DVp 3

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Table 5.2 Data for three-winding transformerWinding MVA Voltage p.u. ZMVA kV referred to nameplate VAP-S 15 33/11 0.008 þ j0.1P-T 5 33/1.5 0.0035 þ j0.0595S-T 5 11/1.5 0.0042 þ j0.0175

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p.u. Z Z(V) referred Equivalent Control of Voltage and Reactive Powerreferred to 33 kV side Z(V) referredto 15 MVA (ZBASE ¼ 72.6 V) to 33 kV side. (Equation (3.14))0.008 þ j0.1 0.58 þ j7.26 Z1 ¼ 0.212 þ j8.210.0105 þ j.1785 0.76 þ j12.96 1 ðZPS þ ZPT À ZST Þ0.0126 þ j0.0525 0.915 þ j3.81 2 Z2 ¼ 0.368 À j0.945 1 ðZPS þ ZST À ZPT Þ 2 Z3 ¼ 0.547 þ j4.76 1 ðZPT þ ZST À ZPS Þ 2 189

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190 Electric Power Systems, Fifth Edition 33 kV (constant) Tap range +- 10% 11 kV Load 10 MVA 0.8 p.f. lagging A BC (a) 0.368 - j0.945Ω 2.2 + j5.22Ω 0.212 + j8.21Ω Z2 C AN 10 MVA ZL Z1 0.547 + j4.76Ω L 0.8 p.f. lagging B load 33 kV Z3 Tertiary L load (b)Figure 5.20 (a) Line diagram for Example 5.7. (b) Equivalent network – referredto 33 kVwhere DVp % RP þ XQ V ;DVp % 2:78 Â 8=3 Â 106 þ 12p:4ﬃ8ﬃ 5 Â 6=3 Â 106 33 000= 3 DVp ¼ 7:4 þ 24:95 ¼ 1:703 kV and VC ¼ 17:3 kV 19Vc referred to 11 kV ¼ 17.3/3 ¼ 5.77 kV (phase) or 10 kV (line). In order to maintain11 kV at C, the voltage is raised by tapping down on the transformer. Using the fullrange of 10%, that is t ¼ 0.9, the voltage at C is 17:3 pﬃﬃ ðð33 Â Â3 0:9Þ=11Þ ¼ 11 kVThe true voltage will be less than this as the primary current will have increased by (1/0.9) because of the change in transformer ratio. The tap-changing transformer is notable to maintain 11 kV at C and the use of a static capacitor connected to the tertiarywill be investigated. Consider a shunt capacitor of capacity 5 MVAr (the capacity of the tertiary). Assume the transformer to be at its nominal ratio 33/11 kV. The voltage drop topoint N

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Control of Voltage and Reactive Power 191 ¼ 2:412 Â 8=3 Â 106 þ 13:42 Â 1=3 Â 106 VN % 19 kV ¼ 0:574 kV VN ¼ 19 À 0:574 ¼ 18:43 kVTherefore the volt drop N to CDVC ¼ 0:368 Â 8=3 Â 106 À 0:945 Â 6=3 Â 106 ¼ À0:049 kV 18:43 Â 103 ;VC ¼ 18:43 À 0:049 ¼ 18:381 kVReferred to 11 kV, VC ¼ 10.65 kV (line). Hence, to have 11 kV the transformer will tapsuch that t ¼ (1 À 0.35/11) ¼ 0.97, that is, a 3% tap change, which is well within therange and leaves room for load increases. On no-load DVp ¼ 2:959 Â 0 þ 18:19 Â ðÀ5=3Þ kV 19 ¼ À30:3 ¼ À1:594 kV 19 VC ¼ 19 þ 1:6 ¼ 20:6 kV ðphaseÞOn the 11 kV side VC ¼ 11:9 kV ðlineÞtherefore the tap change will have to be at least 8.1%, which is well within the range.5.8 Voltage CollapseVoltage collapse is an important aspect of system stability. Consider the circuit shown in Figure 5.21(a). If VS is ﬁxed (i.e. an inﬁnite busbar),the graph of VR against P for given power factors is as shown in Figure 5.21(b). InFigure 5.21(b), Z represents the series impedance of a 160 km long, double-circuit,400 kV, 260 mm2 conductor overhead line. The fact that two values of voltage existfor each value of power is easily understood by considering the analytical solution ofthis circuit. At the lower voltage a very high current is taken to produce the power. The seasonal thermal ratings of the line are also shown, and it is apparent that forloads of power factor less than unity (lagging) the possibility exists that, before thethermal rating is reached, the operating power may be on that part of the character-istic where small changes in load cause large voltage changes and voltage instabilitywill have occurred. In this condition the action of tap-changing transformers is

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192 Electric Power Systems, Fifth Edition Z P, Q VS VR (a) Hot weather Normal Cold weather rating rating rating 1.2 p.f. = 0.90 lead 1.1 =p1.p.f0.p.f=..f.=0=0.90.95.99l7elealeadadd 1.0 p.f. = p.f. =p.0f..9=70lo.9g9 p.f. 0.9 0.90 log logVR (p.u.) 0.8 p.f. = 0.95 log 0.7 0.6 0.5 0.4 0.3 0.2 0.1 500 1000 1500 2000 Power (MW) (b)Figure 5.21 (a) Equivalent circuit of a line supplying a load P þ jQ. (b) Relationbetween load voltages and received power at constant power factor for a 400 kV,2 Â 260 mm2 conductor line, 160 km in length. Thermal ratings of the line areindicatedinteresting. If the receiving-end transformers ‘tap up’ to maintain the load voltage,the line current increases, thereby causing further increase in the voltage drop. Itwould, in fact, be more proﬁtable to ‘tap down’, thereby reducing the current andvoltage drop. It is feasible therefore for a ‘tapping-down’ operation to result inincreased secondary voltage, and vice versa. The possibility of an actual voltage collapse depends upon the nature of the load. Ifthis is stiff (constant power), for example induction motors, the collapse is aggravated.If the load is soft, for example heating, the power falls off rapidly with voltage and thesituation is alleviated. Referring to Figure 5.21 it is evident that a critical quantity is thepower factor; at full load a change in lagging power factor from 0.99 to 0.90 will pre-cipitate voltage collapse. On long lines, therefore, for reasonable power transfers it isnecessary to keep the power factor of transmission approaching unity, certainly above

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Control of Voltage and Reactive Power 193 Infinite busbar C CC 2 22 X XX C CC 2 22Load Zg GLFigure 5.22 Line diagram of three long lines in parallel – effect of the loss of one line.GL ¼ local generators0.97 lagging, and it is economically justiﬁable to employ VAr injection by static capaci-tors, synchronous compensators or Static VAr Compensators (SVCs) close by the load. A problem arises with the operation of two or more lines in parallel, for examplethe system shown in Figure 5.22, in which the shunt capacitance has been repre-sented as in a p section. If one of the three lines is removed from the circuit becauseof a fault, the system series reactance will increase from XL/3 to XL/2, and thecapacitance, which normally improves the power factor, decreases to 2C from 3C.Thus the overall voltage drop is greatly increased and, owing to the increased I2XLloss of the lines and the decreased generation of VArs by the shunt capacitances, thepower factor decreases; hence the possibility of voltage instability. The same argu-ment will, of course, apply to two lines in parallel. Example 5.8 Figure 5.23 shows three parallel 400 kV transmission circuits each 250 km long. The parameters of the circuits are Resistance : 0:02 ohm=km Inductance : 1:06 mH=km Capacitance : 0:011 mF=km

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194 Electric Power Systems, Fifth Edition G D L1 Demand L2 L3 Figure 5.23 Diagram of Example 5.8 The load demand varies betweenDemand PD (MW) QD (MVAr)Peak 1100 532.8Off-peak 220 106.6 A power ﬂow calculation for the two loading conditions shows the voltage at thedemand busbar to be 0.894 during peak demand and 1.015 for off-peak demandconditions.VG (p.u.) PG (MW) QG (MVAr) VD (p.u.) PD (MW) QD (MVAr)1 1117.5 451.7 0.894 1100 532.81 220.6 À304.4 1.015 220 106.6 To maintain the demand busbar voltage at 1 p.u. reactive compensation is needed 499:5 MVar ðcapacitiveÞ during peak demand condition 83:3 MVAr ðinductiveÞ under off-peak conditionVG (p.u.) PG (MW) QG (MVAr) VD (p.u.) PD (MW) QD (MVAr) Compensation (MVAr)1 1112.9 À166.2 1 1100 532.8 499.51 220.5 À216.3 1 220 106.6 À83.3 If one line is lost during peak demand (N-1 security requires that no load is shed ifone circuit trips) the voltage at the receiving end would reduce to 0.745 p.u. and reactivecompensation of 625.5 MVAr (capacitive) would be required to bring the voltage to 1 p.u.VG (p.u.) PG (MW) QG (MVAr) VD (p.u.) PD (MW) QD (MVAr) Compensation (MVAr)1 1139.9 983 0.745 1100 532.8 01 1119.7 40.6 1 1100 532.8 625.5

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Control of Voltage and Reactive Power 195 Usually, there will be local generation or compensation feeding the receiving-endbusbars at the end of long lines. If this generation is electrically close to the loadbusbars, that is low connecting impedance Zg, a fall in voltage will automaticallyincrease the local VAr generation, and this may be sufﬁcient to keep the reactivepower transmitted low enough to avoid large voltage drops in the long lines. Often,however, the local generators supply lower voltage networks and are electricallyremote from the high-voltage busbar of Figure 5.20, and Zg is high. The fall in volt-age now causes little change in the local generator VAr output and the use of static-controlled capacitors at the load may be required. As Zg is inversely proportional tothe three-phase short-circuit level at the load busbar because of the local generation,the reactive-power contribution of the local machines is proportional to this faultlevel. When a static or synchronous compensator reaches its rated limit, voltage canno longer be controlled and rapid collapse of voltage can follow because any VArsdemanded by the load must now be supplied from sources further away electricallyover the high-voltage system. In the UK and some other countries, many generators are some distance from theload-centres. Consequently, the transmission system operator is required to installlocal ﬂexible VAr controllers or compensators to maintain a satisfactory voltage atthe delivery substations supplying the local distribution systems. Such ﬂexible con-trollers, based on semiconductor devices which can vary the VAr absorption in areactor or generation in a capacitor, are called FACTS (Flexible a.c. TransmissionSystem). Typical values of compensation required for a 400 kV or 500 kV network are: Peak load ¼ 0:3 kVAr=kW generating VArs Light load ¼ 0:25 kVAr=kW absorbing VArs5.9 Voltage Control in Distribution NetworksSingle-phase supplies to houses and other small consumers are tapped off fromthree-phase feeders connected between one phase and the neutral. Althoughefforts are made to allocate equal loads to each phase the loads are not appliedat the same time and some unbalance occurs. In the distribution network (Britishpractice) shown in Figure 5.24 an 11 kV distributor supplies a number of lateralfeeders in which the voltage is approximately 400 V and then each phase, loadedseparately. The object of design is to keep the consumers’ nominal 230 V supply withinÀ6/ þ 10% of the declared voltage. The main 33/11.5 kV transformer is controlledwith an on-load tap changer to maintain the 11 kV busbar at a voltage approxi-mately 5% above 11 kV. The distribution transformers have a secondary phasevoltage of 433/250 V which is some 8.5% higher than the nominal value of 230 V.These transformers have taps of þ/À5% that are only adjustable when the trans-former is off-circuit (isolated).

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196 Electric Power Systems, Fifth Edition 33/11.5 kV Main distributor B ASupply 11/0.43 kV 400 V C RY B off-circuit 11/0.43 kV Laterals tap change RY B Consumers Figure 5.24 Line diagrams of typical radial distribution schemes A typical distribution of voltage drops would be as follows: main 11 kV feederdistributor, 6%; 11/0.433 kV transformer, 3%; 400 V circuit, 7%; consumer circuit,1.5%; giving a total of drop at full load of 17.5%. On very light load (10% of fullload) the corresponding drop may be 1.5%. To offset these drops, various voltage boosts are employed as follows: main trans-former, þ5%; distribution transformer, inherent boost of þ4% (i.e. 433/250 V sec-ondary) plus a 2.5% boost through off-circuit taps. These add to give a total boost of11.5%. Hence the consumers’ voltage varies between (þ 11.5 À 17.5), that is À6% and(þ11.5 À 1.5) that is þ10%, which is permissible. There will be a difference in con-sumer voltage depending upon the position of the lateral feeder on the main distrib-utor; obviously, a consumer supplied from C will have a higher voltage than onesupplied from B. In some circuits the voltage control of the 33/11.5 kV transformeris compounded with a measurement of current through the transformer. This isknown as Line Drop Compensation and allows the voltage to be controlled at aremote point of the 11 kV feeder. L (R + jX) Ω per unit length x i dx Figure 5.25 Uniformly loaded distribution line

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Control of Voltage and Reactive Power 1975.9.1 Uniformly Loaded Feeder from One EndIn areas with high load densities a large number of tappings are made fromfeeders and a uniform load along the length of a feeder may be considered toexist. Consider the voltage drop over a length dx of the feeder distant x metresfrom the supply end. Let i A be the current tapped per metre and R and X be theresistance and reactance per phase per metre, respectively. The length of thefeeder is L (m) (see Figure 5.25). The voltage across dx ¼ Rixdx cos f þ Xixdx sin f, where cos f is the power factor(assumed constant) of the uniformly distributed load. The total voltage drop ZL ZL¼ R ixdx cos f þ X ixdx sin f 00¼ Ri L2 cos f þ Xi L2 sin f 2 2¼ LR I cos f þ LX I sin f 2 2where I ¼ Li, the total current load. Hence the uniformly distributed load may berepresented by the total load tapped at the centre of the feeder (half its length).5.10 Long LinesOn light loads the capacitive charging VArs of a line exceeds the inductive VArsconsumed and the voltage rises, causing problems for generators. Shunt reactorsare switched in circuit at times of light load to absorb the generated VArs. A 500 km line can operate within Æ10% voltage variation without shunt reactors.However, with, say, an 800 km line, shunt reactors are essential and the effects ofthese are shown in Figure 5.26. For long lines in general, it is usual to divide thesystem into sections with compensation at the ends of each section. This controls thevoltage proﬁle, helps switching, and reduces short circuit currents. Shunt compen-sation can be varied by switching discrete amounts of inductance. A typical 500 kV,1000 km scheme uses compensation totalling 1200 MVAr. Improvement in voltage proﬁle may be obtained by compensation, using FACTSdevices, at intermediate points, as well as at the ends of the line, as shown in Figure5.26. If the natural load is transmitted there is, of course, constant voltage along theline with no compensation. If the various busbars of a sectioned line can be main-tained at constant voltage regardless of load, each section has a theoretical maxi-mum transmission angle of 90. Thus, for a three section line a total angle of muchgreater than 90 would be possible. This is illustrated in Figure 5.27 for a three-section, 1500 km line with a unity power factor load.

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198 Electric Power Systems, Fifth Edition (a) 1.4V (p.u.) 1.2 (b) 1.0 (c) 0.8 (d) 0 200 400 600 800 Distance from sending end (km)Figure 5.26 Voltage variation along a long line; (a) on no load with no compensa-tion; (b) on no load with compensation at ends; (c) on no load with compensation atends and at centre, (d) transmitting natural load, compensation at ends and centre5.10.1 Sub-Synchronous ResonanceWith very long lines the voltage drop from the series inductance can be very large.Series capacitors are then installed to improve the power transfer capacity and theseeffectively shorten the line electrically. The combination of series capacitors and the natural inductance of the line(plus that of the connected systems) creates a resonant circuit of sub-synchronous resonant frequency. This resonance can interact with the generator-shaft critical torsional frequency, and a mechanical oscillation is superimposed onthe rotating generator shaft that may have sufﬁcient magnitude to cause mechanicalfailure. Sub-synchronous resonance has been reported, caused by line-switching in a situ-ation where trouble-free switching was normally carried out with all capacitors inservice, but trouble occurred when one capacitor bank was out of service. Althoughthis phenomena may be a rare occurrence, the damage resulting is such that, at thedesign stage, an analysis of possible resonance effects is required.5.11 General System ConsiderationsBecause of increasing voltages and line lengths, and also the wider use of under-ground circuits, the light-load reactive-power problem for an interconnected systembecomes substantial, particularly with modern generators of limited VAr absorption

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Control of Voltage and Reactive Power 199 One Two Three V 0o SeriesSource V δo Load V 0o ShuntConstant Voltage Constant Voltage 2 Series compensation 60% 40% 60% 20% 40% 0 20% Power (p.u.) 1 Zero 20% C 40% Shunt compensation L 0 90o 180o δFigure 5.27 Power-angle curves for 1500 km line in three sections. Voltages at section-busbars maintained constant by variable compensation. Percentage of series andshunt compensation indicated (Figure adapted from IET)capability. At peak load, transmission systems need to increase their VAr generation,and as the load reduces to a minimum (usually during the night) they need to reducethe generated VArs by the following methods, given in order of economic viability:1. switch out shunt capacitors;2. switch in shunt inductors;3. run hydro plant on maximum VAr absorption;4. switch out one cable in a double-circuit link;5. tap-stagger transformers;6. run base-load thermal generators at maximum VAr absorption.

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200 Electric Power Systems, Fifth EditionProblems 5.1 An 11 kV supply busbar is connected to an 11/132 kV, 100 MVA, 10% reac- tance transformer. The transformer feeds a 132 kV transmission link consist- ing of an overhead line of impedance (0.014 þ j0.04) p.u. and a cable of impedance (0.03 þ j0.01) p.u. in parallel. If the receiving end is to be main- tained at 132 kV when delivering 80 MW, 0.9 p.f. lagging, calculate the power and reactive power carried by the cable and the line. All p.u. values relate to 100 MVA and 132 kV bases. (Answer: Line (23 þ j35) MVA; cable (57 þ j3.8) MVA) 5.2 A three-phase induction motor delivers 500 hp 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. (Answer: 365 kVA, 0.274) 5.3 The load at the receiving end of a three-phase overhead line is 25 MW, power factor 0.8 lagging, at a line voltage of 33 kV. A synchronous compensator is situated at the receiving end and the voltage at both ends of the line is main- tained at 33 kV. Calculate the MVAr of the compensator. The line has resist- ance of 5 V per phase and inductive reactance of 20 V per phase. (Answer: 25 MVAr) 5.4 A transformer connects two inﬁnite busbars of equal voltage. The transformer is rated at 500 MVA and has a reactance of 0.15 p.u. Calculate the VAr ﬂow for a tap setting of (a) 0.85:1; (b) 1.1:1. (Answer: (a) 425 MVAr; (b) À367 MVAr) 5.5 A three-phase transmission line has resistance and inductive reactance of 25 V and 90 V, respectively. With no load at the receiving end, a synchronous com- pensator there takes a current lagging by 90; the voltage is 145 kV at the send- ing end and 132 kV at the receiving end. Calculate the value of the current taken by the compensator. When the load at the receiving end is 50 MW, it is found that the line can oper- ate with unchanged voltages at the sending and receiving ends, provided that the compensator takes the same current as before, but now leading by 90. Calculate the reactive power of the load. (Answer: 83.5 A; QL ¼ 24.2 MVAr) 5.6 Repeat Problem 5.3 making use of @Q/@V at the receiving end. 5.7 In Example 5.3, determine the tap ratios if the receiving-end voltage is to be maintained at 0.9 p.u. of the sending-end voltage. (Answer: ts ¼ 1.19; tr ¼ 0.84)

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Control of Voltage and Reactive Power 201 D TA A TB B C t = 0.93 6 MVA t = 0.95 1.2 MVA 0.15 MVA t:1 0.8 lag 0.8 lag 0.8 lag 415 V t:1 Figure 5.28 Line diagram for system in Problem 5.85.8 In the system shown in Figure 5.28, determine the supply voltage necessary at A to maintain a phase voltage of 240 V at the consumer’s terminals at C. The data in Table 5.3 apply.Table 5.3 Data for Problem 5.8Line or Rated Voltage Rating Nominal Tap Impedance (V)Transformer kV MVA Ratio 0.0217 þ j0.00909BC 0.415 10 30.69/11 1.475 þ j2.75AB 11 2.5 10.45/0.415 1.475 þ j2.75DA 33 1.09 þ j9.8TA 33/11 Referred to 33 kV 0.24 þ j1.95TB 11/0.415 Referred to 11 kV(Answer: 33 kV)5.9 A load is supplied through a 275 kV link of total reactance 50 V from an inﬁnite busbar at 275 kV. Plot the receiving-end voltage against power graph for a con- stant load power factor of 0.95 lagging. The system resistance may be neglected.5.10 Describe two methods of controlling voltage in a power system.a. Show how the scalar voltage difference between two nodes in a network is given approximately by: DV ¼ RP þ XQ VEach phase of a 50 km, 132 km overhead line has a series resistance of 0.156V/km and an inductive reactance of 0.4125 V/km. At the receiving end thevoltage is 132 kV with a load of 100 MVA at a power factor of 0.9 lagging.Calculate the magnitude of the sending-end voltage.b. Calculate also the approximate angular difference between the sending- end and receiving-end voltages.(Answer: (c) 144.1 kV; (d) 4.55)(From Engineering Council Examination, 1997)

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202 Electric Power Systems, Fifth Edition5.11 Explain the limitations of tap-changing transformers. A transmission link (Figure 5.29(a) connects an inﬁnite busbar supply of 400 kV to a load bus- bar supplying 1000 MW, 400 MVAr. The link consists of lines of effective impedance (7 þ j70)V feeding the load busbar via a transformer with a maximum tap ratio of 0.9:1. Connected to the load busbar is a compensa- tor. If the maximum overall voltage drop is to be 10% with the trans- former taps fully utilized, calculate the reactive power requirement from the compensator. (Answer: 148 MVAr)Note: Refer the voltage and line Z to the load side of transformer in Fig-ure 5.29(b). VR ¼ VS À 0RP VþRXt2QC1A t @B t25.12 A generating station consists of four 500 MW, 20 kV, 0.95 p.f. (generating VArs) generators, each feeding through a 525 MVA, 0.1 p.u. reactance trans- former onto a common busbar. It is necessary to transmit 2000 MW at 0.95 p.f. lagging to a substation maintained at 500 kV in a power system at a distance of 500 km from the generating station. Design a suitable VS (7 + j70)Ω t :1 VR 1000 MWSupply (a) 400 MVAr 400 kV 0.91 (p.u.) Compensator VR VS ( VS ) (Z t 2) t (b) Equivalent circuitFigure 5.29 Circuits for Problem 5.11

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Control of Voltage and Reactive Power 203 transmission link of nominal voltage 500 kV to achieve this, allowing for a reasonable margin of stability and a maximum voltage drop of 10%. Each generator has synchronous and transient reactances of 2 p.u. and 0.3 p.u. respectively, and incorporates a fast-acting automatic voltage regulator. The 500 kV transmission lines have an inductive reactance per phase of 0.4 V/km and a shunt capacitive susceptance per phase of 3.3 Â 10À6 S/km. Both series and shunt capacitors may be used if desired and the number of three-phase lines used should be not more than three – fewer if feasible. Use approximate methods of calculation, ignore resistance, and state clearly any assumption made. Assume shunt capacitance to be lumped at the receiving end only. (Answer: Use two 500 kV lines with series capacitors compensating to 70% of series inductance)5.13 It is necessary to transmit power from a hydroelectric station to a load centre 480 km away using two lines in parallel for security reasons. Assume sufﬁcient bundle conductors are used such that there are no thermal limitations, and the effective reactance per phase per km is 0.44 V and that the resistance is negligible. The shunt capacitive susceptance of each line is 2.27 Â 10À6 S, per phase per km, and each line may be repre- sented by the nominal p-circuit with half the capacitance at each end. The load is 2000 MW at 0.95 lagging and is independent of voltage over the permissible range. Investigate, from the point of view of stability and voltage drop, the feasibil- ity and performance of the link if the sending-end voltage is 345, 500, and 765 kV assuming the transmission angle is not to exceed 30. The lines may be compensated up to 70% by series capacitors and at the load-end compensators of 120 MVAr capacity are available. The maximum permissible voltage drop is 10%. As two lines are provided for security reasons, your studies should include the worst-operating case of only one line in use.5.14 Explain the action of a variable-tap transformer, showing, with a phasor dia- gram, how reactive power may be despatched from a generator down a mainly reactive line by use of the taps. How is the level of real power despatch controlled? Power ﬂows down an H.V. line of impedance 0 þ j0.15 p.u. from a generator whose output passes through a variable-ratio transformer to a large power system. The voltage of the generator and the distant large system are both kept at 1.0 p.u. Determine the tap setting if 0.8 p.u. power and 0.3 p.u. VAr are delivered to a lagging load at the power system busbar. Assume the reactance of the transformer is negligible. (Answer: t ¼ 1.052) (From Engineering Council Examination, 1995)

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204 Electric Power Systems, Fifth Edition5.15 Two substations are connected by two lines in parallel, of negligible imped- ance, each containing a transformer of reactance 0.18 p.u. and rated at 120 MVA. Calculate the net absorption of reactive power when the trans- former taps are set to 1:1.15 and 1:0.85, respectively (i.e. tap-stagger is used). The p.u. voltages are equal at the two ends and are constant in magnitude. (Answer: 32 MVAr)

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6Load Flows6.1 IntroductionA load ﬂow (sometimes known as a power ﬂow) is power system jargon for thesteady-state solution of an electrical power network. It does not essentially differfrom the solution of any other type of network except that certain constraints arepeculiar to power systems and, in particular, the formulation is non-linear leadingto the need for an iterative solution. In previous chapters the manner in which the various components of a powersystem may be represented by equivalent circuits has been demonstrated. Itshould be stressed that the simplest representation of items of plant should alwaysbe used, consistent with the accuracy of the information available. There is nomerit in using very complicated machine and line models when the load and otherdata are known only to a limited accuracy, for example, the long-line representa-tion should only be used where absolutely necessary. Similarly, synchronous-machine models of more sophistication than those given in this text are neededonly for very specialized purposes, for example in some stability studies. Usually,the size and complexity of the network itself provides more than sufﬁcient intellec-tual stimulus without undue reﬁnement of the components. Often, in high voltagenetworks, resistance may be neglected with little loss of accuracy and an immensesaving in computation. Load ﬂow studies are performed to investigate the following features of a powersystem network:1. Flow of MW and MVAr in the branches of the network.2. Busbar (node) voltages.3. Effect of rearranging circuits and incorporating new circuits on system loading.4. Effect of temporary loss of generation and transmission circuits on system load- ing (mainly for security studies).5. Effect of injecting in-phase and quadrature boost voltages on system loading.Electric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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206 Electric Power Systems, Fifth Edition6. Optimum system running conditions and load distribution.7. Minimizing system losses.8. Optimum rating and tap-range of transformers.9. Improvements from change of conductor size and system voltage. Planning studies will normally be performed for minimum-load conditions(examining the possibility of high voltages) and maximum-load conditions (investi-gating the possibility of low voltages and instability). Having ascertained that a net-work behaves reasonably under these conditions, further load ﬂows will beperformed to optimize voltages, reactive power ﬂows and real power losses. The design and operation of a power network to obtain optimum economy is ofparamount importance and the furtherance of this ideal is achieved by the use ofcentralized automatic control of generating stations through system control centres.These control systems often undertake repeated load ﬂow calculations in close toreal time. Although the same approach can be used to solve all load ﬂow problems, forexample the nodal voltage method, the object should be to use the quickest andmost efﬁcient method for the particular type of problem. Radial networks willrequire less sophisticated methods than closed loops. In very large networks theproblem of organizing the data is almost as important as the method of solution,and the calculation must be carried out on a systematic basis and here the nodal-voltage method is often the most convenient. Methods such as network reductioncombined with the Thevenin or superposition theorems are at their best withsmaller networks. In the nodal method, great numerical accuracy is required in thecomputation as the currents in the branches are derived from the voltage differ-ences between the ends. These differences are small in well designed networks sothe method is ideally suited for computation using digital computers and the perunit system.6.2 Circuit Analysis Versus Load Flow AnalysisThe task of load ﬂow analysis is conceptually similar to that of traditional circuitanalysis, but there is a key difference, which is critical for understanding the special-ized methods used for load ﬂow calculations. In circuit analysis, given all the values of impedances in the circuit and given theparameters of all voltage or current generators in the circuit, all nodal voltages andbranch currents can be calculated directly. The key feature of this analysis is that therelationship between nodal voltages and branch current is linear (i.e. in the formV ¼ Z Â I). In Load Flow analysis, loads and sources are deﬁned in terms of powers notimpedances or ideal voltage or current generators. All power network branches,transformers or overhead and underground circuits, are deﬁned as impedances.The relationship between voltage, power and impedances is non-linear and appro-priate methods for solving non-linear circuits need to be used.

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Load Flows 207 Generator Line Load Figure 6.1 Simple two-busbar system6.2.1 Power Flow in a Two-Busbar SystemConsider a two-busbar (node) system of a generator that supplies a load over atransmission line (Figure 6.1). Figure 6.2 shows the equivalent circuit of the system in Figure 6.1. The line resist-ance is R and reactance is X. The shunt susceptance of the line is neglected. InFigure 6.2 known variables are shown by solid lines (with an arrow showing thedirection) and unknown variables are shown by dotted lines. Initially it is assumed that the power the generator injects and the generator volt-age are both speciﬁed. The equation that links the complex power of the generatorwith the current and the voltage at the generator busbar is: SG ¼ PG þ jQG ¼ VGIÃ ð6:1Þ The power and voltage are known at the same busbar, then from equation (6.1),the current through the line can be calculated directly: I ¼ PG À jQG ð6:2Þ VGÃOnce the current is known, the voltage at the load busbar is given by: VL ¼ VG À ðR þ jXÞI ð6:3Þ R jX I PG+ jQG VG VLFigure 6.2 Power ﬂow in a two-busbar system. The generator power and voltage aregiven

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208 Electric Power Systems, Fifth Editionor by substituting from equation (6.2) for the current into equation (6.3): PG À jQG ! VGÃ VL ¼ VG À ðR þ jXÞ ð6:4ÞBy deﬁning VG ¼ VGﬀd: VÃG ¼ VGﬀ À dTherefore: !! RPG þ XQG XPG À RQG VL ¼ VGﬀd À VGﬀ À d Àj VGﬀ À d ð6:5Þ As the current is known from the voltage and power at the generator busbar, theload voltage can be calculated directly from equation (6.5). Now consider the situation shown in Figure 6.3. In this case, the voltage is speci-ﬁed at the generator busbar, while the power is known at the load busbar (and thevoltage at the load is unknown). The load power is SL ¼ PL þ jQL ¼ VLIÃ ð6:6Þ Although the complex power of the load is known, the load voltage is unknownand hence the current cannot be calculated directly. The generator power is: SG ¼ PG þ jQG ¼ VGIÃ ð6:7Þ Again the current cannot be calculated, as the power at the generator is unknown.To calculate the power at the generator the losses in the line need to be known; forwhich the current is required. R jX I PG+ jQG VG VLFigure 6.3 Power ﬂow in a two-busbar system. The load power and generator voltageare given

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Load Flows 209 In summary from equations (6.6) and (6.7), the current cannot be calculateddirectly, as power at the load, SL, and the voltage at the generation, VG (at oppositeends of the circuit) are given. The expression that links the voltages at the generator and load is obtained bycombining equations (6.3) and (6.6) as: PL À jQL ! VLÃ VL ¼ VG À ðR þ jXÞ ð6:8Þ Equation (6.8) does not have a closed form solution. This equation, in relation toVL is non-linear as it contains the product of the voltages at the load as shownbelow: VLVÃL ¼ VGVLÃ À ðR þ jXÞðPL À jQLÞ ð6:9ÞSolving Equation (6.8) requires an iterative method. The solution can start with an value of VLð0Þ, ﬁnd VÃLð0Þ. This can theninitial new value of then calculated. The process be substituted into equation (6.8)and a VLð1Þ is repeated for several iterationsuntil the voltage of one iteration converges to the next iteration.Once the value of voltage VL at the load end is calculated, the current can be cal-culated from equation (6.6) and the losses can be calculated as:Active losses ¼ I2R,The active power output of the generator is: PG ¼ PL þ I2RSimilarly,Reactive losses1 ¼ I2X,The reactive power production of the generator is: QG ¼ QL þ I2X Example 6.1 The system shown in Figure 6.4 feeds a load of 20 MVA, 0.95 p.f. lagging (absorbing VArs). Draw a two busbar equivalent circuit for the network and carry out two itera- tions of the iterative solution.1 Active (real power) losses are dissipated as heat. Reactive power ‘losses’ are simply the difference inreactive power at the ends of the circuit. No energy is lost as heat.

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210 Electric Power Systems, Fifth Edition j4Ω 0.8 + j1Ω Load132 kV 20 MVA 240 mm2 cable 0.95 p.f. lag 10 km 132/33 kV 33/6.6 kV 50 MVA 25 MVA 10% 6% Figure 6.4 Line diagram of the system of Example 6.1SolutionBy choosing Sbase ¼ 50 MVA, the different impedances in Figure 6.4 can be convertedinto p.u. values as: Base quantity p.u. valueSource impedance 1322=50 ¼ 348:5 V j14/348.5 ¼ j0.04132/33 kV transformer 332=50 ¼ 21:78 V j0.1Cable 0.037 þ j0.04633/6.6 kV transformer 0:06 Â 50=25 ¼ j0:12 For the load: P ¼ 20 Â 0:95 ¼ 19 MW ¼ 0.38 p.u. and Q ¼ 20 Â sinðcosÀ1ð0:95ÞÞ ¼6:25 MVAr ¼ 0.125 p.u. The two busbar equivalent circuit is given in Figure 6.5. In the equivalent circuit theresistance is the resistance of the cable and equivalent reactance is the addition ofsource, transformer and cable reactances. If the load busbar voltage is VL then from equation (6.8): 0:38 À j0:125! VLÃ VL ¼ 1 À ð0:037 þ j0:306Þ ð6:10Þ Initially assume that VðL0Þ ¼ 1ﬀ0 p:u: By substituting the ﬁrst guess of VL into theright hand side of equation (6.10): VðL1Þ ¼ 1 À ð0:037 þ j0:306Þð0:38 À j0:125Þ ¼ 0:95 À j0:11 ¼ 0:956ﬀ À 6:6 0.037 p.u. j 0.306 p.u.1.0 p.u. SL= 0.38 + j 0.125 p.u.Figure 6.5 Two bus equivalent circuit of the system in Example 6.1

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Load Flows 211The complex conjugate of VL is taken VLÃð1Þ ¼ 0:95 þ j0:11 ¼ 0:956ﬀ6:6 The new value of VLÃ is substituted into the right hand side of equation (6.10) to ﬁndthe second iteration of VL: 0:38 À j0:125! 0:956ﬀ6:6 VLð2Þ ¼ 1 À ð0:037 þ j0:306Þ ¼ 1 À ð0:037 þ j0:306Þð0:418ﬀ À 24:8Þ ¼ 1 À ð0:037 þ j0:306Þð0:379 À j0:175Þ ¼ 0:933 À j0:109 ¼ 0:94ﬀ À 6:7 This iterative procedure is repeated until the difference between the nth iterativevalue of voltage and the (n þ 1)th iterative value of voltage becomes very small.6.2.2 Relationship Between Power Flows and Busbar VoltagesIn Chapter 2, the active and reactive power ﬂow in a circuit was discussed. For atransmission circuit as R ( X, Z % X and u % 90. Therefore from Equation (2.11): PG ¼ VGVL sin d X QG ¼ V2G À VGVL cos d ¼ VGðVG À VLÞ cos d X X X The active power at the generator busbar is equal to the active power at the loadas there are no active power losses in the circuit ðR ¼ 0Þ. These expressions illustrate some key aspects of transporting active and reactivepower across the network. Active power ﬂow (P) requires a difference in phaseangle between the busbar voltages while reactive power ﬂow (Q) requires a differ-ence in voltage magnitude between generator and load busbars. Power systems are operated with relatively constant voltages and the differencesin voltage magnitudes between various nodes are not allowed to be large. There areno such strict constraints on differences in phase angles across a line, but these areusually less than 30. With d 30, P is sensitive to changes in sin d but cos d remainsclose to 1. It is possible to transmit a signiﬁcant amount of active power over high voltageoverhead AC transmission lines of several hundred kilometres. Reactive power can-not be transmitted over long distances, as this would require signiﬁcant voltagedrops that are unacceptable. We also observe that since X ) R, the reactive lossesare much larger than the active losses. Hence reactive power is supplied near theneed usually by over-excited generators or shunt capacitive compensators.

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212 Electric Power Systems, Fifth Edition6.3 Gauss-Seidel MethodThe Gauss-Siedel method is a simple iterative technique for load ﬂow calculations.The Gauss-Seidel method ﬁrst approximates the load and generation by ideal cur-rent sources (converting powers into current injections using assumed values ofvoltages). The iteration process is then carried out using injected complex poweruntil the voltages converge. A three-busbar system is shown in Figure 6.6. The admittances of circuits rather than their impedances are used. The complexadmittance of the branch i À j between nodes i and j is deﬁned as: yij ¼ Rij 1 jXij þ The current balance equations for each of the nodes are ﬁrst derived. The injectedcurrent to Bus 1 is equal to the sum of the currents leaving this busbar throughbranches 1–2 and 1–3. I1 ¼ I12 þ I13Then the branch currents are related to the busbar voltages using Ohm’s law: I1 ¼ y12 Â ðV1 À V2Þ þ y13 Â ðV1 À V3ÞRearranging this equation, the following is obtained: I1 ¼ ðy12 þ y13ÞV1 À y12V2 À y13V3 ð6:11Þ 1 2 I12 X12 = j0.1 p.u. 200 MW I13 y12 = -j10 p.u. 50 MVAr50 MW20 MVAr X13 = j0.05 p.u. X23 = j0.125 p.u. y13 = -j20 p.u. I23 y23 = -j8 p.u. 3 100 MW 50 MVArFigure 6.6 Three-busbar network for Gauss-Siedel load ﬂow

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Load Flows 213The same procedure is repeated for Bus 2 and Bus 3 I2 ¼ Ày12V1 þ ðy12 þ y23ÞV2 À y23V3 ð6:12Þ I3 ¼ Ày13V1 À y23V2 þ ðy13 þ y23ÞV3 ð6:13ÞEquations (6.11)–(6.13) can be written in matrix form: 2 I1 3 2 ðy12 þ y13Þ À Ày12 Á Ày13 32 V1 3 46 I2 75 ¼ 46 Ày12 y12 þ y23 À Ày23 Á 5746 V2 75 ð6:14Þ I3 Ày13 Ày23 y13 þ y23 V3where: 2 I1 3I ¼ 64 I2 75 is the vector of nodal current injections I3 2 V1 3V ¼ 64 V2 75 is the vector of nodal voltagesV3 2 ðy12 þ y13Þ À Ày12 Á Ày13 3 YBUS ¼ 46 Ày12 y12 þ y23 À Ày23 75 Á Ày13 Ày23 y13 þ y23is the admittance (or YBUS) matrix, with diagonal elements and off-diagonal ele-ments deﬁned as:Diagonal elements XN ðN is the number of nodesÞ Yii ¼ yij j¼1 j¼6 iOff diagonal elements Yij ¼ Àyjiwhere Yij is the element at the ith row and jth column of the YBUS matrix. Substituting the values of admittances, the YBUS matrix of the example becomes: 2 Àj30 j10 j20 3 YBUS ¼ 64 j10 Àj18 j8 75 j20 j8 Àj28

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214 Electric Power Systems, Fifth Edition The three equations (6.11) to (6.13) are linearly dependent, that is by adding equa-tions (6.12) and (6.13), equation (6.11) can be derived. This means that one of theequations can be eliminated and only two used to obtain the solution. If the current injections were available, then the nodal voltages could be calcu-lated simply from the matrix equation by ﬁnding the inverse of the reduced admit-tance matrix and then multiplying it by the current injection vector. However, in theformulation of the load ﬂow the nodal currents are not available but real andreactive power injections are speciﬁed. Thus the Gauss-Siedel technique proceeds asfollows. A reference for the voltage magnitude and phase is set, and in this case it isassumed that the voltage at Bus 1 is known: V1 ¼ 1ﬀ0The current injection at Bus 2 is: I2 ¼ Y12V1 þ Y22V2 þ Y23V3 ð6:15ÞThe injected current as a function of complex power and voltage at Bus 2 is:S2 Ã ¼ P2 À jQ2 ¼ Y12V1 þ Y22V2 þ Y23V3 ð6:16Þ V2 V2ÃFrom equation (6.16) the voltage of Bus 2 is: 1 P2 À jQ2 ! Y22 V2ÃV2 ¼ À ðY12V1 þ Y23V3Þ ð6:17Þwhere P2 and Q2 are the active and reactive power ﬂows into Bus 2. The unknown complex voltage V2 appears on both sides of (6.17). The Gauss-Seidel method is to update the value of the voltage, in this case V2 on the left-handside of (6.17), using the expression on the right-hand side, with values of the volt-ages already evaluated, in the present or previous iteration. \"#V2ðpÞ ¼ 1 P2 À jQ2 À ðY12V1 þ Y23V3ðpÀ1ÞÞ ð6:18Þ Y22 V2ðpÀ1ÞÃwhere p is the iteration number Similarly, for Bus 3: \"#Vð3pÞ ¼ 1 P3 À jQ3 À ðY13V1 þ Y23Vð2pÞÞ ð6:19Þ Y33 V3ðpÀ1ÞÃ

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Load Flows 215 In general: 23 ViðpÞ ¼ 1 4Pi À jQi À XiÀ1 Yij Vðj pÞ À XN YijVjðpÀ1Þ5 ð6:20Þ Yii ViðpÀ1ÞÃ j¼1 j¼iþ1 The process can be started by initializing the value of all voltages to 1ﬀ0. Hencefor the iteration p ¼ 1, the voltage at Bus 2 is given by: 1 À2 þ j0:5 ! Àj18 1 Vð21Þ ¼ À ðj10 Â 1 þ j8 Â 1Þ ¼ 0:9722 À j0:1111 ¼ 0:9785ﬀ À 6:5198 Similarly, for Bus 3 1 À1 þ j0:5 ! Àj28 1 V3ð1Þ ¼ À ðj20 Â 1 þ j8 Â ð0:9722 À j0:1111ÞÞ ¼ 0:9742 À j0:0675 ¼ 0:9765ﬀ À 3:9612 Table 6.1 shows how values of voltages change in this process. The convergence of the iteration is shown in Figure 6.7. Table 6.1 and Figure 6.7 show that, in this example, the iteration process con-verges after ﬁve or six iterations. The changes in voltage magnitudes can be used as a criterion for convergence. Inthis case normally a maximum change in voltage from the previous iteration, forexample 0.001 p.u is set. Alternatively, the power mismatches at each node, at eachiteration can be calculated. The power mismatch is the difference between the com-plex power injected into a node and the power leaving a node through the networkTable 6.1 Changes of voltage magnitudes and phase angles in the Gauss-SiedelexampleIteration Voltage Magnitude (pu) Voltage Angle (Degrees) Bus 1 Bus 2 Bus 3 Bus 1 Bus 2 Bus 31 1 0.978 551 0.976 539 249 0 À6.51 980 À3.96 1212 1 0.957 666 0.967 134 203 0 À8.38 021 À4.45 3583 1 0.949 463 0.96 419 005 0 À8.71 939 À4.55 1104 1 0.947 275 0.963 412 677 0 À8.81 964 À4.58 1335 1 0.946 675 0.963 197 267 0 À8.84 810 À4.58 9746 1 0.946 508 0.963 137 161 0 À8.85 594 À4.59 206

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216 Electric Power Systems, Fifth Edition 0.98 Bus 2 0.97 Bus 3 Voltage magnitude (p.u.) 0.96 0.95 0.94 1 234 5 6 Iteration Figure 6.7 Convergence of Gauss-Siedel iterationbranches. When the power mismatch is less then a threshold, the iteration process isstopped.6.4 Load Flows in Radial and Simple Loop NetworksDistribution systems are normally operated as radial networks and a simple itera-tive procedure can be used to solve load ﬂows in them. One of the common methodsused is the forward and backward method. The procedure used in this method is:Step 1 Assume an initial nodal voltage magnitude and angle for each busbar (1ﬀ0 p.u.Step 2 voltage is often used)Step 3Step 4 Start from the source and move forward towards the feeder and lateral ends and calculate the current using equation (6.6) Start from the feeder and lateral ends and move towards the source while calculating the voltage at each busbar using equation (6.3) Repeat Steps 2 and 3 until the termination criterion ðVðipÞ À ViðpÀ1Þ < eÞ is met at all busbarsExample 6.2For the system shown in Figure 6.8 carry out two iterations of the forward and back-ward method.

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Load Flows 217 L1 = 30 + j14.5 MVA j14Ω 0.8 + j1Ω L2 = 20 + j 6.5 MVA132 kV 132/33 kV 33/6.6 kV –50 MVA 25 MVA 10% 6% Figure 6.8 Line diagram of the system in Example 6.2SolutionThe equivalent circuit of the radial network shown in Figure 6.8 (in per unit on a 50MVA base) is given in Figure 6.9. Neglecting losses in line 2–3, the current I1 is given by: I1 ¼ ð0:6 À j0:29Þ þ ð0:38 À j0:125Þ V2Ã ð6:21Þ ¼ 0:98 À j0:415 V2Ã ð6:22Þ ð6:23ÞSimilarly, I2 is given by: ð6:24Þ I2 ¼ 0:38 À j0:125 V3ÃFrom equation (6.3), two voltage equations can be written as: V2 ¼ 1 À j0:14I1 V3 ¼ V2 À ð0:037 þ j0:166ÞI2Step 1: Assume V2 ¼ V3 ¼ 1ﬀ0Step 2: From equations (6.21) and (6.22): Ið11Þ ¼ 0:98 À j0:415 Ið21Þ ¼ 0:38 À j0:125Step 3: Substituting the values of I1 and I2 in equations (6.23) and (6.24), voltages can be found as: V2ð1Þ ¼ 1 À j0:14ð0:98 À j0:415Þ ¼ 0:942 À j0:137 ¼ 0:952ﬀ À 8:29

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218 Electric Power Systems, Fifth Edition Bus 1 Bus 2 j 0.166 p.u. Bus 3 j 0.14 p.u. 0.037 p.u. I2 L2 = 0.38 + j0.125 p.u. V31.0 kV V1 I1 L1 = 0.6 + j0.29 p.u. V2 Figure 6.9 Equivalent circuit of the system in Example 6.2 V3ð1Þ ¼ 0:942 À j0:137 À ð0:037 þ j0:166Þð0:38 À j0:125Þ ¼ 0:907 À j0:196 ¼ 0:928ﬀ À 12:15Repeating Step 2: Ið12Þ ¼ 0:98 À j0:415 ¼ 0:956 À j0:58 0:952ﬀ8:29 Ið22Þ ¼ 0:38 À j0:125 ¼ 0:372 À j0:218 0:928ﬀ12:15Repeating Step 3: Vð22Þ ¼ 1 À j0:14ð0:956 À j0:58Þ ¼ 0:92 À j0:133 ¼ 0:93ﬀ À 8:23 V3ð2Þ ¼ 0:96 À j0:152 À ð0:037 þ j0:166Þ0:372 À j0:218 ¼ 0:87 À j0:188 ¼ 0:89ﬀ À 12:2Step 4: Repeat Steps 2 and 3 until the termination criterion (Vðinþ1Þ À VðinÞ < e) is metExample 6.3Repeat Example 6.2 using the Gauss-Seidel method.Solution:The admittances of the branches are: y12 ¼ 1 ¼ Àj7:143 j0:14 y23 ¼ 0:037 1 j0:166 ¼ 1:279 À j5:739 þ

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Load Flows 219Therefore the YBUS matrix is: 2 j7:143 3 Àj7:143 1:279 À j12:881 0 À1:279 þ j5:739 À1:279 þ j5:739 577 YBUS ¼ 466 j7:143 1:279 À j5:739 0From equation (6.18): \"#Vð2pÞ ¼ 1:279 1 À0:6 þ j0:29 À ðj7:143 Â 1 þ ðÀ1:279 þ j5:739ÞV3ðpÀ1Þ Þ À j12:881 Vð2pÀ1ÞÃFrom equation (6.19): \"# Vð3pÞ ¼ 1:279 1 j5:739 À0:38 þ j0:125 À ðÀ1:279 þ j5:739Þ Â V2ðpÞ À Vð3pÀ1ÞÃThese two equations are solved iteratively.6.5 Load Flows in Large SystemsIn large practical power systems, it is necessary to carry out many load ﬂows forboth planning and operation. These take into account the complex impedances ofthe circuits, the limits caused by circuit capacities, and the voltages that can be satis-factorily provided at all nodes in the system. Systems consisting of up to 3000 bus-bars, 6000 circuits, and 500 generators may have to be solved in reasonable timescales (e.g. 1–2 min) with accuracies requiring 32 or 64 bits for numerical stability.The many system states that are possible in a day’s operation may have to be consid-ered. Even though the Gauss-Seidel method could be employed for large networks,its convergence is slow (even with acceleration factors2) and sometimes it fails toconverge. For large power systems the Newton-Raphson or fast decoupled methodsare more commonly used. The speed of convergence of these methods is of extremeimportance as the use of these methods in schemes for the automatic control ofpower systems requires very fast load-ﬂow solutions. For large meshed networks, the following types of busbar are speciﬁed for load-ﬂow studies:1. Slack, swing, or ﬂoating bus: A single busbar in the system is speciﬁed by a volt- age, constant in magnitude and phase (the usual practice is to set the phase angle to 0). The generator at this node supplies the losses to the network; this is neces- sary because the magnitude of the losses will not be known until the calculation of currents is complete, and this cannot be achieved unless one busbar has no2 The correction in voltage from VðpÞ to Vðpþ1Þ is multiplied by a factor so that the number of iterationsrequired to reach the speciﬁed convergence can be greatly reduced.

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220 Electric Power Systems, Fifth Edition power constraint and can feed the required losses into the system. The location of the slack node can inﬂuence the complexity of the calculations; the node most closely approaching an inﬁnite busbar should be used.2. PQ bus: At a PQ bus the complex power S ¼ P Æ jQ is speciﬁed whereas the volt- age magnitude and angle are unknown. All load buses are PQ buses.3. PV bus: At a PV bus P and jVj are speciﬁed; voltage angle and Q are unknown. Q is adjusted to keep jVj constant. Most generator busbars are PV busbars. In Section 6.3, the Gauss-Seidel method was illustrated on a network with slackand PQ busbars. To include PV Buses in the Gauss-Seidel method, Q needs to beguessed. From equation (6.16): XN SiÃ ¼ ViÃ YijVj ¼ Pi À jQi j¼1 \"# Hence, QðipÞ ¼ ÀIm VÃi PN YijVj j¼1During the iterative solution it is assumed Sði pÞÃ ¼ Pi À jQði pÞTentatively solve for Vðipþ1Þ 23Viðpþ1Þ ¼ 1 6664VSiððippÞÞÃÃ À XN Yij Vðj pÞ 5777 Yii j¼1 j¼6 iBut since jVij is speciﬁed, replace Viðpþ1Þ by jVij.6.5.1 YBUS Matrix with Tap-Changing TransformersThe formation of the Y matrix for a network having simple passive components wasshown in Section 6.3. When the taps of a transformer are set to its nominal ratiothe transformer is represented by a single series impedance. When the taps are off-nominal, adjustments have to be made as follows. Consider a transformer of ratio t:1 connected between two nodes i and j; the seriesadmittance of the transformer is Yt. By deﬁning an artiﬁcial node x between the volt-age transforming element and the transformer admittance (Figure 6.10(a)), the fol-lowing equations can be written: Si ¼ ViIÃi ð6:25Þ ð6:26Þ Sx ¼ Vi IjÃ t

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Load Flows 221From Figure 6.10(a), Vi ! t Ij ¼ Vj À Yt ð6:27ÞAs the t : 1 transformer is ideal, Si ¼ ÀSx. Therefore from (6.25)–(6.27): Ij 1 Vi ! t t t Ii ¼ À ¼ À Vj À Yt Yt Yt ð6:28Þ t2 t ¼ Vi À VjFrom equations (6.27) and (6.28), the YBUS matrix is given by: \" # 2 Yt À Yt 3 # Ii 466 t \" Vi Ij ¼ t2 Vj ð6:29Þ 577 Yt À t Yt From equation (6.29) it is seen that for the node on the off-tap side of the trans-former (that is, for node j), the following conditions apply: i t :1 Yt j Ii x Ij Si Vi Sx Vj (a) i Yt j t Yt ( 1- t ) Yt ( t -1 ) t2 t (b)Figure 6.10 (a) Equivalent circuit of transformer with off-nominal tap ratio. Trans-former series admittance on non-tap side. (b) The p section to represent transformerwith off-nominal tap ratio

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222 Electric Power Systems, Fifth Editionwhen forming Yjj use Yt, for the transformer, and when forming Yij, use ÀYt=t forthe transformer.For the tap-side node (for node i) that the following conditions apply:when forming Yii use Yt=t2, and when forming Yij use ÀYt=t. These conditions may be represented by the p section shown in Figure 6.10(b),although it is probably easier to modify the mutual and self-admittances directly.6.5.2 The Newton-Raphson MethodAlthough the Gauss-Seidel was the ﬁrst popular method for load ﬂow calculations,the Newton-Raphson method is now commonly used. The Newton-Raphsonmethod has better convergence characteristics and for many systems is faster thanthe Gauss-Seidel method; the former has a much larger time per iteration butrequires very few iterations (four is general), whereas the Gauss-Siedel requires upto 30 iterations, the number increasing with the size of system. Consider an n-node power system. For the kth busbar, having links connectingnode k to nodes j (depending on the number of links this could be from 1 to n À 1) ofadmittance Ykj, Pk þ jQk ¼ VkIÃk ¼ Vk XnÀ1 À ÁÃ ð6:30Þ YkjVj j¼1Let Vk ¼ ak þ jbk and Ykj ¼ Gkj þ jBkjThen, Pk þ jQk ¼ ðak þ jbkÞ XnÀ1 ÂÀ þ ÁÀ þ jbjÁÃÃ ð6:31Þ Gkj jBkj aj j¼1from which, XnÀ1 Â À ÁÀ ÁÃ Pk ¼ ak ajGkj À bjBkj þ bk ajBkj þ bjGkj ð6:32Þ ð6:33Þ j¼1 XnÀ1 Â À ÁÀ ÁÃ Qk ¼ bk ajGkj À bjBkj À ak ajBkj þ bjGkj j¼1 Hence, there are two non-linear simultaneous equations for each node. Note that(n À 1) nodes are considered because the slack node n is completely speciﬁed.

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Load Flows 223 In order to explain the basic iterative procedure to solve these two non-linearsimultaneous equations, equation (6.32) is deﬁned as: Pk ¼ g1ða1; . . . . . . ; anÀ1; b1; . . . . . . ; bnÀ1Þ ð6:34Þ Note that Gkj and Bkj are constants. If the initial guess of variables are aÃ1; . . . . . . ; anÃÀ1; b1Ã; . . . . . . ; bÃnÀ1 and errors areDa1; . . . . . . ; DanÀ1; Db1; . . . . . . ; DbnÀ1, then equation (6.34) can be written as:PÃk þ DPk ¼ g1 Àa1Ã þ Da1; . . . . . . ; aÃnÀ1 þ DanÀ1; b1Ã þ Db1; . . . . . . ; bnÃÀ1 þ Á ð6:35Þ DbnÀ1 By expanding equation (6.35) using Taylor’s series and neglecting higher orderterms: DPk ¼ @Pk Da1 þ @Pk Da2 þ Á ÁÁ @Pk DanÀ1 @a1 @a2 @anÀ1 þ @Pk Db1 þ @Pk Db2 þ Á Á Á @Pk DbnÀ1 @b1 @b2 @bnÀ1Similar equations hold for DQ in terms of Da and Db.These equations may be expressed generally using the Jacobian Matrix: DP1 @P1 ÁÁÁ @P1 @P1 ÁÁÁ @P1 Da1 @a1 @anÀ1 @b1 ÁÁÁ @bnÀ1 ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ @PnÀ1 ÁÁÁ @PnÀ1 @PnÀ1 ÁÁÁ @PnÀ1 DPnÀ1 ¼ @a1 @anÀ1 @b1 @bnÀ1 DanÀ1 DQ1 @Q1 ÁÁÁ @Q1 @Q1 ÁÁÁ @Q1 Db1 ð6:36Þ @a1 @anÀ1 @b1 @bnÀ1 ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ DQnÀ1 @QnÀ1 ÁÁÁ @QnÀ1 @QnÀ1 ÁÁÁ @QnÀ1 DbnÀ1 @a1 @anÀ1 @b1 @bnÀ1 Jacobian matrixFor convenience, denote the Jacobian matrix by JA JB JC JD

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224 Electric Power Systems, Fifth Edition The elements of the matrix are evaluated for the values of P, Q and V at each itera-tion as follows. For the submatrix JA and from equation (6.32), diagonal elements are given by: @Pk ¼ 2akGkk À bk Bkk þ bkBkk þ XnÀ1 À À Á ð6:37Þ @ak ajGkj bjBkj j¼1 j6¼k This element may be more readily obtained by expressing some of the quantitiesin terms of the current at node k, Ik, which can be determined separately at eachiteration. Let XnÀ1 À ÁÀ Á Ik ¼ ck þ jdk ¼ ðGkk þ jBkkÞðak þ jbkÞ þ Gkj þ jBkj aj þ jbj j¼1 j¼6 kfrom which, ck ¼ akGkk À bkBkk þ XnÀ1 À ajGkj À bjBkj Á j¼1 j¼6 kand dk ¼ akBkk þ bkGkk þ XnÀ1 À ajBkj þ bjGkj Á So that, j¼1 j¼6 k @Pk ¼ akGkk þ bkBkk þ ck @akOff-diagonal elements, where k ¼6 j, are given by: @Pk ¼ akGkj þ bkBkj ð6:38Þ @ajFor JB, @Pk ¼ ÀakBkk þ bkGkk þ dk @bk

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Load Flows 225and @Pk ¼ ÀakBkj þ bkGkj ðk ¼6 jÞ @bj For JC, @Qk ¼ ÀakBkk þ bkGkk À dkand @ak @Qk ¼ ÀakBkj þ bk Gkj ðk ¼6 jÞ @aj For JD, @Qk ¼ ÀakGkk À bk Bkk þ ckand @bk @Qk ¼ ÀakGkj À bkBkj ðk ¼6 jÞ @bj The process commences with the iteration counter ‘p’ set to zero and all the nodesexcept the slack-bus being assigned voltages, usually 1 p.u. From these voltages, P and Q are calculated from equations (6.32) and (6.33). Thechanges are then calculated: DPpk ¼ PkðspecifiedÞ À Ppk and DQpk ¼ QkðspecifiedÞ À Qkpwhere p is the iteration number.Next the node currents are computed as Ipk ¼ Ppk þ jQkpÃ ¼ ckp þ jdpk VPThe elements of the Jacobian matrix are than formed, and from equation (6.36), Da ¼ JA JB À1 DP Db JC JD DQ Hence, a and b are determined and the new values, apkþ1 ¼ apk þ Dapk andbpkþ1 ¼ bpk þ Dbpk, are obtained. The process is repeated ðp ¼ p þ 1Þ until DP and DQare less than a prescribed tolerance.

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226 Electric Power Systems, Fifth Edition In the previous development of the Newton Raphson method the admittancesand voltages were assumed to be in rectangular form. Sometimes the quantities areexpressed in polar form. The polar form of the equations is: Pk ¼ PðV; dÞ ð6:39Þ Qk ¼ QðV; dÞ For a link connecting nodes k and j of admittance, if Vk ¼ Vkﬀuk, Vj ¼ Vjﬀuj andYkj ¼ Gkj þ jBkj, from equation (6.30) the power at a busbar is Sk ¼ Pk þ jQk ¼ VkIÃk XnÀ1 À ÁÀ Á ¼ Vkﬀdk Gkj À jBkj Vjﬀ À dj j¼1 XnÀ1 À ÁÀ Á ¼ VkVj Gkj À jBkj cos dkj þ jsin dkj j¼1 where dkj ¼ dk À dj.Therefore, XnÀ1 À Á Pk ¼ VkVj Gkj cos dkj þ Bkj sin dkj j¼1 ¼ Vk2Gkk þ XnÀ1 À cos dkj þ Bkj sin Á ð6:40Þ VkVj Gkj dkj ð6:41Þ j¼1 j¼6 k XnÀ1 À Á Qk ¼ VkVj Gkjsin dkj À Bkjcos dkj j¼1 ¼ ÀV2k Bkk þ XnÀ1 À dkj À Bkjcos Á VkVj Gkjsin dkj j¼1 j¼6 kFor a load busbar, DPk ¼ XnÀ1 @Pk Ddj þ XnÀ1 @Pk DVj ð6:42Þ @dj @Vj ð6:43Þ j¼1 j¼1 DQk ¼ XnÀ1 @Qk Ddj þ XnÀ1 @Qk DVj @dj j¼1 @Vj j¼1

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Load Flows 227 For a generator PV busbar, only the DPk equation is used as Qk is not speciﬁed.The mismatch equation is: DPpÀ1 ¼ JA JB Ddp ð6:44Þ DQpÀ1 JC DVp JD VpÀ1Ddp is the correction to PQ and PV buses and DVp=VpÀ1 is the correction to PQbuses. For buses k and j, off-diagonal elements are given by: JA : @Pk ¼ À sin dkj À Bkj cos Á @dj VkVj Gkj dkj JB : Vj @Pk ¼ À cos dkj þ Bkj sin Á @Vj VkVj Gkj dkj JC : @Qk ¼ À cos dkj þ Bkj sin Á @dj ÀVkVj Gkj dkj JD : Vj @Qk ¼ À sin dkj À Bkj cos Á @Vj VkVj Gkj dkjAlso, diagonal elements are given by: JA : @Pk ¼ XnÀ1 À sin dkj þ Bkj cos Á @dk VkVj ÀGkj dkj j¼1 j6¼k ¼ ÀQk À BkkV2kSimilarly, JB : Vk @Pk ¼ Pk þ Gkk V2k @Vk JC : @Qk ¼ Pk À Gkk V k2 @dk JD : Vk @Qk ¼ Qk À Bkk V k2 @Vk The computational process can be enhanced by pre-ordering and dynamic order-ing, deﬁned as Pre-ordering, in which nodes are numbered in sequence of increasing number of connections.

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228 Electric Power Systems, Fifth Edition Dynamic ordering, in which at each step in the elimination the next row to be operated on has the fewest non-zero terms.6.5.3 Fast Decoupled Load FlowIn a power system, the changes in voltage angle mainly control ﬂows of P and thevoltage magnitudes of busbars mainly change Q. Therefore the coupling betweenthe Pk À Vj and Qk À dj components is weak. Hence the submatrices JB and JC can beneglected. Equation (6.44) can then be reduced to two equations: ÂDPpÀ1Ã ¼ ½JA½Ddp ð6:45Þ ð6:46Þ ÂDQpÀ1Ã ¼ ½JD DVp ! VpÀ1 The decoupled technique is further simpliﬁed by assuming that: under normalloading conditions, the angle difference across transmission lines is negligible, thusdkj % 0 and for lines and cables it is reasonable to assume Bkj ) Gkj, thus Gkj can beneglected. Therefore:Off-diagonal element of JA : @Pk ¼ ÀVkVjBkj ¼ À VkVj @dj XkjDiagonal element of JA : @Pk ¼ XnÀ1 VkVjBkj ¼ XnÀ1 VkVj @dk Xkj j¼1 j¼1 j6¼k j¼6 kOff-diagonal element of JD : Vj @Qk ¼ ÀVkVjBkj ¼ À VkVj @Vj XkjDiagonal element of JD : Vk @Qk ¼ XnÀ1 ¼ XnÀ1 VkVj @Vk VkVjBkj Xkj j¼1 j¼1 j6¼k j6¼k where Xkj is the branch reactance.This method is called the fast decoupled load ﬂowExample 6.4Using the fast decoupled method calculate the angles and voltages after the ﬁrst itera-tion for the three-node network of Figure 6.11. Initially V1 is 230 kV V2 is 220 kV and V3is 228 kV all at u ¼ 0. Busbar 2 is a load consuming 200 MW, 120 MVAr; busbar 3 is a

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Load Flows 229 1 3 2Figure 6.11 Three busbar network for fast decoupled load ﬂowgenerator node set at 70 MW and 228 kV. V1 is an inﬁnite busbar. The YBUS matrix isgiven by (all quantities are in Siemens): 2 0:00819 À j0:049099 À0:003196 þ j0:019272 À0:004994 þ j0:030112 3YBUS ¼ 46 À0:003196 þ j0:019272 0:007191 À j0:043099 À0:003995 þ j0:02409 57 À0:003995 þ j0:02409 À0:004994 þ j0:030112 0:008989 À j0:053952From equation (6.45) ! !! DP2 JA22 JA23 Dd2 DP3 ¼ JA32 JA33 Dd3 ð6:47ÞJA22 is given by @P2 ¼ X3 ¼ V 2 ðV1 B12 þ V 3 B23 Þ @d2 Vk Vj Bkj j¼1 j6¼2 ¼ 220 Â ð230 Â 0:019272 þ 228 Â 0:02409Þ Â 106 ¼ 2:1835 Â 109JA23 is given by @P2 ¼ ÀV2V3B23 @d3 ¼ À220 Â 228 Â 0:02409 Â 106 ¼ À1:20835 Â 109

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230 Electric Power Systems, Fifth EditionJA32 is given by @P3 ¼ ÀV3V2B32 @d2 ¼ À1:20835 Â 109JA33 is given by @P3 ¼ X2 ¼ V3 ðV1 B13 þ V2 B32 Þ @d3 V k Vj Bkj j¼1 j6¼3 ¼ 228 Â ð230 Â 0:030112 þ 220 Â 0:02409Þ Â 106 ¼ 2:7874 Â 109From (6.40):P2 ¼ V2V1G21 þ V22G22 þ V2V3G23 ¼ 220 Â ðÀ230 Â 0:003196 þ 220 Â 0:007191 À 228 Â 0:003995Þ Â 106 ¼ À14:0624 MWP3 ¼ V3V1G31 þ V3V2G32 þ V23G33 ¼ 228 Â ðÀ230 Â 0:004994 À 220 Â 0:003995 þ 228 Â 0:008989Þ Â 106 ¼ 5:0096 MWTherefore, DP2 ¼ À200 þ 14:0624 ¼ À185:9376 MW DP3 ¼ 70 À 5:0096 ¼ 64:99 MWNow from equation (6.47):\" #\" 2:1835 Â 109 À1:20835 Â 109 #\" #À185:9376 Â 106 À1:20835 Â 109 2:7874 Â 109 Dd2 64:99 Â 106 ¼ Dd3\" Dd2 # 2 À2:1835 Â 109 À1:20835 Â 109 3À12 À185:9376 Â 106 3 4 54 5 ¼Dd3 À1:20835 Â 109 À2:7874 Â 109 64:99 Â 106 \" À0:0951 # \" À5:45 # ¼ rad ¼ À1:026 À0:0179 d2 ! À5:45 ! d3 À1:026 ; ¼¼

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Load Flows 231From equation (6.41)Q2 ¼ ÀV2V1½G21 sinðd2 Àd1ÞÀB21 cosðd2 Àd1ÞÀV22B22 ÀV2V3½G23 sinðd2 Àd3ÞÀB23 cosðd2 Àd3Þ ¼ À220Â230½À0:003196ÂsinðÀ5:45ÞÀ0:019272ÂcosðÀ5:45ÞÀ2202Â0:043099 À220Â228½À0:003995ÂsinðÀ5:45 þ1:026ÞÀ0:02409ÂcosðÀ5:45 þ1:026ÞQ2 ¼ 58:68 MVAr ½DQ2¼À120þ58:68¼À61:32 MVArFrom equation (6.46): ÂÃ ! ½DQ2 ¼ JD2 DV2 V2JD2 is given by Vk @Qk ¼ X3 ¼ ðV2V1B21 þ V2V3B23Þ @Vk Vk Vj Bkj j¼1 j6¼2 ¼ 220 Â ð230 Â 0:019272 þ 228 Â 0:02409Þ Â 106 ¼ 2:1835 Â 109 DV2 ¼ DQ2 Â V2 JD2 ¼ À61:32 Â 106 Â 220 Â 103 ¼ À6:178 kV 2:1835 Â 109Hence V1 ¼ 230 kV V2 ¼ 220 À 6:178 ¼ 213:82 kV V3 ¼ 228 kVThe process is repeated for the next iteration, and so on, until convergence is reached.6.6 Computer SimulationsThe line diagram of a part of a transmission system is shown in Figure 6.12, withdetails of the overhead lines given in Table 6.2.

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232 Electric Power Systems, Fifth Edition 120 MVA F E 150% T3 G3 120 MVA LYNX, 83 km 12.5/132 kV ZEBRA, 17 km 5%2 x 100 MVA D 120% A1 T2 B CG2 ZEBRA, 70 km P = 60 MW A2 T1 Q = 30 MVArG1 2 x 100 MVA P = 130 MW 13/132 kV Q = 60 MVAr 6% Figure 6.12 System used for simulation studiesTable 6.2 Overhead line dataOverhead line Cross-section Resistance Reactance Capacitance mm2 mV/km mV/km nF/km 387.0 9.50ZEBRA (on a double circuit 400 76.0 132 kV tower) 175 401.0 8.90 178.0LYNX (on a single circuit 132 kV tower) The problem has been solved using a load ﬂow program, IPSA3; a brief descrip-tion of the arrangement of essential data is given. Most commercially available loadﬂow programs require data to be entered in per unit. Therefore a base value of S waschosen as 100 MVA and voltage bases selected as the rated primary and secondaryvoltages of the transformers. Previously the capacitance of the lines has been neglec-ted. However for HV cable circuits and EHV overhead lines, when the capacitance isknown, it can be converted into susceptance for load ﬂow programs. Tables 6.3-6.5show the input data for the lines, busbars and transformers. For double circuit linesthe per unit data is given for a single circuit. Transformer taps were changed tomaintain the 132 kV busbar voltage at a reference.3 The IPSA load ﬂow program uses a variant of the fast-decoupled load ﬂow technique.

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Load Flows 233Table 6.3 Input line data (On 100 MVA, 132 kV base, the impedance base is:Zbase ¼ 1322=100 ¼ 174:24 V)Line Resistance (p.u.) ¼ Reactance (p.u.) ¼ Susceptance (p.u.) ¼ R=Zbase X=Zbase 2pfC Â ZbaseB-C 76 Â 10À3 Â 70=174:24 387 Â 10À3 Â 70=174:24 100p Â 9:5 Â 10À9 Â 70 Â 174:24 ¼ 0:0305 ¼ 0:1555 ¼ 0:0364C-D 76 Â 10À3 Â 17=174:24 387 Â 10À3 Â 17=174:24 100p Â 9:5 Â 10À9 Â 17 Â 174:24E-D ¼ 0:0074 ¼ 0:0378 ¼ 0:0088 178 Â 10À3 Â 83=174:24 401 Â 10À3 Â 83=174:24 100p Â 8:9 Â 10À9 Â 83 Â 174:24 ¼ 0:0848 ¼ 0:1910 ¼ 0:0404Table 6.4 Busbar dataBusbar Type V d PG QG PL QL Xs(p.u.) (MW) (MVAr) 0.6 (MW) (MVAr) — —A1 and A2 Slack 1.0 0 — — 130 (combined) — 60 — —B PQ (load) — —— þ100 to 60 — 30 —C PQ (load) — —— À60 — 1.5 Â 100/F PV (generator) 1.05 — 120 120 ¼ 1.25Table 6.5 Transformer dataTransformer Resistance (p.u.) with X/R ¼ 10 Reactance (p.u.)T1 0.006 0.06T2 0.006 0.06T3 0.0042 0.05 Â 100/120 ¼ 0.042 The system shown in Figure 6.12 was implemented in the IPSA load ﬂow softwarepackage. The two generators at busbars A1 and A2 were combined to create the slackbusbar. Figure 6.13 shows the load ﬂow solution. P (upper value) and Q (lower value)ﬂows are shown for each line and transformer. Table 6.6 shows the line ﬂows andlosses.

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234 Electric Power Systems, Fifth Edition 120.000 39.403 120.000 E 119.406 23.756 60.000 39.406 1.050 33.567 1.688 30.000 -3.3º F 42.123 23.756 1.050 B 29.127 1.688 -0.8º 1.000 -1.3º 42.123 84.560 29.127 61.401 84.560 130.000 D 61.393 0.943 A1& A2 C 60.000 - 4.2º 1.000 0.941 0.0º - 4.7º Figure 6.13 Load ﬂow resultsTable 6.6 Line and transformer ﬂowsFrom bus To bus Flow Losses MW MVAr MW MVArA1 and A2 B 84.560 61.401 0.315 3.147B A1 and A2 84.247 58.259 0.315 3.147 C 42.123 (in each line) 29.127 (in each line) 1.666 (total) 1.632 (total)C B 41.290 (in each line) 28.314 (in each line) 1.666 (total) 1.632 (total)D D À23.756 (in each line) À1.688 (in each line) 0.094 (total) 1.08 (total)F C À23.757 (in each line) À1.146 (in each line) 0.094 (total) 1.08 (total)E E 107.514 32.292 11.901 1.271 E 120 39.403 0.584 5.837 D 119.406 33.567 11.901 1.271Problems6.1 Consider a simple power system composed of a generator, a 400 kV overhead transmission line and a load, as shown in Figure 6.14. If the desired voltage at the consumer busbar is 400 kV, calculate the: i. Active and reactive losses in the line. ii. Voltage magnitude and phase angle at busbar 1. iii. Active and reactive output of the generator at busbar 1.

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Load Flows 235 1 2 (2.7+ j30.4)Ω (600 + j210) MVA. Figure 6.14 400 kV system for Problem 6.1 (Answer (i) 6.8 MW, 76.7 MVAr; (ii) 422.35 kV (L-L), 6.0; (iii) 606.8 MW, 286.7 MVAr)6.2 A load of 1 þ j0.5 p.u. is supplied through a transmission line by a generator G1 that maintains its terminal voltage at 1 p.u. (Figure 6.15).G1 1 2 (1+ j 0.5) p.u. j 0.4 p.u. Figure 6.15 System for Problem 6.2i. Form the YBUS matrix for this system.ii. Perform two iterations of the Gauss-Seidel load ﬂow. À2:5j 2:5j ! 2:5j À2:5j(Answer: (i) Ybus ¼ ; (ii) V1 ¼ 1:0 p.u., V2 ¼ 0:6 À j0:3 p.u.)6.3 List the information which may be obtained from a load-ﬂow study. Part of apower system is shown in Figure 6.16. The circuit reactances and values of realand reactive power (in the form P Æ jQ) at the various busbars are expressed asper unit values on a common MVA base. Resistance may be neglected. By theuse of an iterative method, calculate the voltages at the stations after the ﬁrstiteration.(Answer: V2 ¼ 1:0333 À j0:0333 p.u., V3 ¼ 1:1167 þ j0:2333 p.u., V4 ¼ 1:0556Àj0:0556 p.u.)6.4 A 400 kV interconnected system is supplied from busbar A, which may be con- sidered to be an inﬁnite busbar. The loads and line reactances are as indicated in Figure 6.17. Determine the ﬂow of power in line AC using two iterations of voltages at each bus. (Answer: PAC ¼ 1.16 GW)

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236 Electric Power Systems, Fifth Edition V = 1.05 Slack 0.1 p.u. 1 2 j0.5 p.u. j 0.5 p.u. j1.0 p.u. j 1.0 p.u. 3 4 (0.5 + j 0.2) p.u. Load Generating (0.4 + j 0.05) p.u. station Figure 6.16 System for Problem 6.3 400 kV j4 Ω B A j 10 Ω 1 GW j4 Ω j6 Ω j5 Ω DC 2 GW 3 GW Figure 6.17 System for Problem 6.46.5 Determine the voltage at busbar (2), voltage at busbar (3) and the reactive power at bus (3) as shown in Figure 6.18, after the ﬁrst iteration of a Gauss- Seidel load ﬂow method. Assume the initial voltage to be 1ﬀ0 p.u. All the quantities are in per unit on a common base. (Answer: 0.99 À j0.0133, 1.0 þ j0.0015 and Q3 ¼ 0.4 p.u.)6.6 Active power demand of the 132 kV system shown in Figure 6.19 is supplied by two generators G1 and G2. System voltage is supported by generator G2 and a

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Load Flows 237 V = 1 p.u. Slack j 0.2 p.u. V = 1 p.u. 1 P = 0.6 p.u. j 0.05 p.u. 2 3 j 0.025 p.u. (0.8 + j 0.6) p.u. Figure 6.18 System for Problem 6.5large synchronous compensator SC (see Figure 6.19), which both maintain thevoltage at 1 p.u. at their respective nodes. Generator G1, connected at node 1,has no reactive power capacity available for voltage control. i. Form the Ybus matrix for this system.ii. Perform two iterations of the Gauss-Seidel load ﬂow. 23 À7:5j 2:5j 5j(Answer: (i) Ybus ¼ 4 2:5j À6:5j 4j 5; (ii) V1 ¼ 0:9825 þ j0:0520 p.u., 5j 4j À9jV2 ¼ 1:0 p.u., V3 ¼ 0:9903 À j0:1386) G1 1 j 0.4 p.u. 2 G2P= 0.9 p.u. V2 = 1.0 p.u. P = 0.6 p.u. j 0.2 p.u. j 0.25 p.u. S =1.5 + j 0.3 p.u. 3 V3 = 1.0 p.u. SC Figure 6.19 132 kV system for Problem 6.6

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238 Electric Power Systems, Fifth Edition V = 1.01 p.u. 1 j 0.1 p.u. j 0.2 p.u. 4 j 0.4 p.u. 2P = 0.2 p.u. P = 0.5 p.u.Q = 0.0 p.u. j 0.1 p.u. Q = 0.2 p.u. j 0.2 p.u. 3 P = 0.4 p.u. Q = 0.1 p.u. Figure 6.20 System for Problem 6.86.7 In Figure 6.20 the branch reactances and busbar loads are given in per unit on a common base. Branch resistance is neglected. Explain brieﬂy why an iterative method is required to determine the busbar voltages of this network. Form the YBUS admittance matrix for this network. Using busbar 1 as the slack (reference) busbar, carry out the ﬁrst iteration of a Gauss-Seidel load-ﬂow algorithm to determine the voltage at all busbars. Assume the initial voltages of all busbars to be 1.01 p.u. (Answer: V2 ¼ 0:997 À j0:03, V3 ¼ 0:9968 À j0:0415, V4 ¼ 1:0056 À j0:026) (From Engineering Council Examination, 1995)

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7Fault Analysis7.1 IntroductionCalculation of the currents which ﬂow when faults of various types occur is anessential part of the design of a power supply network. Typically, fault currents areobtained using computer packages by applying faults at various points in the net-work. The magnitudes of the fault currents give the engineer the current settings forthe protection to be used and the ratings of the circuit breakers. The types of fault commonly occurring in practice are illustrated in Figure 7.1,and the most common of these is the short circuit of a single conductor to groundor earth. Often, the path to earth contains resistance Rf in the form of an arc, asshown in Figure 7.1(f). Although the single-line-to-earth fault is the most common,calculations are frequently performed with the three-line, balanced short circuit(Figure 7.1(d) and (e)). This is usually the most severe fault and also the most ame-nable to calculation. The same currents ﬂow for the faults shown in 7.1(d) and 7.1(e). The causes of faults are summarized in Table 7.1, which gives the distribution offaults, due to various causes, on the England and Wales Transmission System in atypical year. Table 7.2 shows the components affected. In tropical countries the inci-dence of lightning is much greater than in the UK, resulting in larger numbers offaults. As well as fault current If (inpkﬃAﬃ ), fault MVA is frequently used as a rating; this isobtained from the expression 3VLIf , where VL is the nominal line voltage of thefaulted part in kV. The fault MVA is often referred to as the fault level. The calcula-tion of fault currents can be divided into the following two main types:1. Faults short-circuiting all three phases when the network remains balanced elec- trically. For these calculations, normal single-phase equivalent circuits may be used as in ordinary load-ﬂow calculations.Electric Power Systems, Fifth Edition. B.M. Weedy, J.B. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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240 Electric Power Systems, Fifth Edition L-E (a) L - L (b) L-L-E (c) L - L - L - E (d) (e) (f) Three-phase s.c. Rf Figure 7.1 Common types of fault or short circuit (s.c.) L ¼ line, E ¼ earthTable 7.1 Causes of overhead-line faults, England and Wales system 66 kV andabove Faults/160 km of line/yearLightning 1.59Dew, fog, frost 0.15Snow, ice 0.01Gales 0.24Salt spray 0.01Total 2 faults per 160 km, giving a total of 232 faults on system/yearTable 7.2 Distribution of faults, England and Wales Transmission systemType Number of faults/yearOverhead lines 289Cables 67Switchgear 56Transformers 59Total 471

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Fault Analysis 2412. Faults other than three-phase short circuits when the network is electrically unbalanced. To facilitate these calculations a special method for dealing with unbalanced networks is used, known as the method of symmetrical components. The main objects of fault analysis are:1. to determine maximum and minimum three-phase short-circuit currents;2. to determine the unsymmetrical fault current for single and double line-to earth faults, line-to-line faults, and sometimes for open-circuit faults;3. investigation of the operation of protective relays;4. determination of rated interrupting capacity of breakers;5. to determine fault-current distribution and busbar-voltage levels during fault conditions.7.2 Calculation of Three-Phase Balanced Fault CurrentsThe action of synchronous generators on three-phase short circuits has beendescribed in Chapter 3. There it was seen that, depending on the time from the inci-dence of the fault, either the transient or the subtransient reactance should be usedto represent the generator. For specifying switchgear, the value of the current ﬂow-ing at the instant at which the circuit contacts open is required. It has been seen,however, that the initially high fault current, associated with the subtransient reac-tance, decays with the passage of time. Modern air-blast circuit breakers usuallyoperate in 2 1/2 cycles and SF6 breakers within 1 1/2 cycles of 60 or 50 Hz alternat-ing current, and are operated through extremely fast protection. Older circuit break-ers and those on lower voltage networks usually associated with relatively cruderprotection can take in the order of eight cycles or more to operate. In calculations offault currents on a transmission system it is usual to use the subtransient reactanceof the generators and to ignore the effects of induction motors, which are connectedto lower voltage networks. The simple calculation of fault currents ignores the direct-current component, themagnitude of which depends on the instant in the cycle that the short circuit occurs.If the circuit breaker opens a reasonable time after the incidence of the fault, thedirect-current component will have decayed considerably. With fast acting circuitbreakers the actual current to be interrupted is increased by the direct-current com-ponent and it must be taken into account. In simple, manual calculations the sym-metrical r.m.s. value may be modiﬁed by the use of multiplying factors to takeaccount of the direct-current component of the fault current. Although these multi-pliers are determined by Standards and depend on the X/R ratio of the circuit to thefault, they are of the form: 8-cycle circuit breaker opening time, multiply by 1; 3-cycle circuit breaker opening time, multiply by 1.2; 2-cycle circuit breaker opening time, multiply by 1.4.

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242 Electric Power Systems, Fifth Edition X’’ IfLine voltage =VLSubtransientreactance =X’’Figure 7.2 Voltage source with short circuit and single-phase equivalent circuit Consider a generator with a short circuit across the three terminals, as shownin Figure 7.2. Deﬁning the voltage base as VL and the MVA base of Sb, the baseimpedance is, Zb ¼ VL2 Sb X00ðp:uÞ ¼ X00ðVÞ ZbThe short-circuit current Is:c: ¼ X00 1 ð7:1Þ ðp:uÞThe three-phase short-circuit level in p.u.¼ Vnominal Â Is:c: ¼ 1 Â Is:c: ¼ X00 1 ðp:uÞThe three-phase short-circuit level in volt-amperes ¼ X00 Sb ð7:2Þ ðp:uÞ Hence the short-circuit level contribution is immediately obtained if the subtran-sient reactance X00ðp:u:Þ of the generator is known. Similarly if the impedanceZðp:u:Þ from an inﬁnite busbar to the point of the fault on a network is known thenthe short-circuit level may be determined immediately.Example 7.1An 11.8 kV busbar is fed from three synchronous generators having the following rat-ings and subtransient reactances, 20 MVA; X000:08 p:u:; 60 MVA; X000:1 p:u:; 20 MVA; X000:09 p:u:

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Fault Analysis 243 11.8 kV F j0.24 j0.1 j0.27 (a) F Short (b) circuit 0.24 0.1 0.27 Xeq (d) F (c)Figure 7.3 Line diagram and equivalent circuits for Example 7.1 Calculate the fault current and MVA if a three-phase symmetrical fault occurs on thebusbars. Resistance may be neglected. The voltage base will be taken as 11.8 kV and theVA base as 60 MVA.SolutionThe subtransient reactance of the 20 MVA machine on the above base is(60/20) Â 0.08, that is, 0.24 p.u., and of the 20 MVA machine (60/20) Â 0.09, that is,0.27 p.u. These values are shown in the equivalent circuit in Figure 7.3. As the gen-erator e.m.f.s are assumed to be equal, one source may be used (Figure 7.3(c)). Theequivalent reactance is Xeq ¼ 1=0:24 þ 1 þ 1=0:1 ¼ 0:056 p:u: 1=0:27Therefore the fault MVAand the fault current ¼ 60=0:056 ¼ 1071 MVA p10ﬃﬃ71 Â 106 ¼ 52 402 A 3 Â 11800

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244 Electric Power Systems, Fifth EditionExample 7.2Figure 7.4 shows two 11 kV generators feeding a 132 kV double circuit line through gen-erator transformers. Each 132 kV line is 40 km long and with an inductive reactance of0.4 V/km. Calculate the fault level for a three-phase fault at: Gen.1 40km Gen.2 11 kV 11 kV125 MVA 75 MVAX = 28% X =18% 11/132 kV B 132/11 kV 125 MVA (a) 90 MVA X = 18% X = 12% A j 0.133 j 0.240j 0.224 j 0.144 j 0.092 j 0.046 j 0.046 B A (b) j 0.224 j 0.144 j 0.133 j 0.240 j 0.046 A (c) j 0.133 j 0.240 j 0.224 j 0.144 j 0023 B j 0.023 (d) j 0.012Figure 7.4 Line diagram and equivalent circuits for Example 7.2

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Fault Analysis 245a. Point A on the busbar close to generator 1.b. Point B at the mid-point of one of the lines. Use a 100 MVA base for the calculation, express your answer in Amperes and ignorethe effect of any system loads.SolutionOn 100 MVA, 132 kV base Zb ¼ 1322 ¼ 174:24 V 100On 100 MVA base, the reactances of the generators, transformers and the line are:Generator X j0:28 Â 100=125 Reactance in p.u. on 100 MVAGenerator Y j0:18 Â 100=75Generator transformer of X j0:18 Â 100=125 j0.224Generator transformer of Y j0:12 Â 100=90 j0.240Line j0:4 Â 40=174:24 j0.144 j0.133 j0.092 The equivalent single-phase network of generator and line reactances is shown inFigure 7.4(b). For a fault at A, the equivalent circuit is reduced to the network shown inFigure 7.4(c) (note that the double circuit line is replaced by a single equivalentreactance). The network in Figure 7.4(c) was reduced to give the ﬁnal single equivalent reac-tance, Xeq ¼ 0:196 p:u: The fault level at point A ¼ 100 ¼ 510:4 MVA 0:196Therefore fault current ¼ 510:4 Â 106pﬃ3ﬃ Â 11 Â 103 ¼ 26 789 AIn order to obtain the fault level at B, the equivalent circuit shown in Figure 7.4(b)was replaced by the network shown in Figure 7.4(d) by the use of the delta-star trans-formation. A further transformation is carried out on the network in Figure 7.4(d) togive the ﬁnal single equivalent reactance, Xeq ¼ 0:209 p:u:The fault level at point B ¼ 100 ¼ 479:1 MVA 0:209Therefore the fault current ¼ p4ﬃ3ﬃ7Â9:111ÂÂ101603 ¼ 25 146 A

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246 Electric Power Systems, Fifth Edition G p.u. G p.u. X p.u. X p.u. X p.u. X p.u. X p.u. (a) (b)Figure 7.5 Connection of artiﬁcial reactors: (a) feeder connection; (b) ring system.Transient reactance of machines, G p.u.; reactance of artiﬁcial reactors X p.u7.2.1 Current Limiting ReactorsThe impedances presented to fault currents by transformers and machines whenfaults occur on substation or generating station busbars are low. To reduce thehigh fault current which would do considerable damage mechanically and ther-mally, artiﬁcial reactances are sometimes connected between bus sections. Thesecurrent-limiting reactors usually consist of insulated copper strip embedded inconcrete formers; this is necessary to withstand the high mechanical forces pro-duced by the current in neighbouring conductors. The position in the circuitoccupied by the reactor is a matter peculiar to individual designs and installa-tions (see Figure 7.5(a) and (b)).Example 7.3A 400 kV power system contains three substations A, B, and C having fault levels(GVA) of 20, 20, and 30, respectively. The system is to be reinforced by three lines each of reactance j5 V connectingtogether the three substations as shown in Figure 7.6(a). Calculate the new fault level atC (three-phase symmetrical fault). Neglect resistance.SolutionThe equivalent circuit for the network is shown in Figure 7.6(b). The substations arerepresented by a voltage source (400 kV) in series with the value of reactance to givethe speciﬁed initial fault level. Select Sb ¼ 60 GVA. Hence the effective reactance ‘behind’ A when subject to a three-phase short circuit 60=20 ¼ 3 p.u. Similarly, the effective reactance at B ¼ 3 p.u. and atC ¼ 2 p.u.With Vb ¼ 400 kV and Sb ¼ 60 GVA, Zb ¼ À Â 103Á2 ¼ 2:67 V 400 Â 109 60

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Fault Analysis 247 N A j5Ω B 33 j5Ω j5Ω C F (a) 0.622 0.622 2 If 0.622 F (b)Figure 7.6 (a) Networks for Example 7.3 (b) Reduced network The 5 V mesh is transformed into a star with arms of value 1:66 V and then convertedinto p.u. value by dividing by Zb. The equivalent circuit with the fault on C afterreinforcement is shown in Figure 7.6(b). Fault current at C with equivalent reactance1.1 p.u. gives, fault level at C, 60=1:1 GVA ¼ 54:66 GVA. Note that this value is very high and that the result of reinforcement is always toincrease the fault levels. The maximum rating of a 400 kV circuit breaker is about50 GVA, hence either a busbar must be run split between two incoming feeders or acurrent limiting reactor must be installed.7.3 Method of Symmetrical ComponentsThis method formulates a system of three separate phasor systems which, whensuperposed, give the unbalanced conditions in the circuit. It should be stressed thatthe systems to be discussed are essentially artiﬁcial and used merely as an aid tocalculation. The various sequence-component voltages and currents do not exist asphysical entities in the network, although they could be monitored by special ﬁlters. The method postulates that any three-phase unbalanced system of voltages andcurrents may be presented by the following three separate systems of phasors:1. a balanced three-phase system in the normal a-b-c (red-yellow-blue) sequence, called the positive phase-sequence system;2. a balanced three-phase system of reversed sequence, that is a-c-b (red- blue-yellow), called the negative phase-sequence system;

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248 Electric Power Systems, Fifth Edition ω ω ω ω IB I1B + I2R I2B + I0R = I1R I0Y I1Y I2Y I0B IR Positive sequence Negative sequence Zero sequence IY Real system Figure 7.7 Real system and corresponding symmetrical components3. three phasors equal in magnitude and phase revolving in the positive phase rota- tion, called the zero phase-sequence system. In Figure 7.7 an unbalanced system of currents is shown with the correspondingsystem of symmetrical components. If each of the red-phase phasors are added, thatis I1R þ I2R þ I0R, the resultant phasor will be IR in magnitude and direction; similarreasoning holds for the other two phases. To express the phasors algebraically, use is made of the complex operator ‘a’(sometimes denoted by l or h) which denotes a phase-shift operation of þ120 anda multiplication of unit magnitude; that is. Vﬀf Â a ¼ Vﬀf Â 1ﬀ120 ¼ Vﬀðf þ 120Þ a ¼ e j2p=3 and a3 ¼ e j3Â2p=3 ¼ 1Also, a2 þ a ¼ ðÀ0:5 À j0:866Þ þ ðÀ0:5 þ j0:866Þ ¼ À1 ;a3 þ a2 þ a ¼ 0and ;1 þ a þ a2 ¼ 0 For positive-sequence phasors, taking the red phasor as reference, I1R ¼ I1Rej0 ¼ reference phase (Figure 7.8) I1Y ¼ I1RðÀ0:5 À j0:866Þ ¼ a2I1Rand I1B ¼ I1RðÀ0:5 þ j0:866Þ ¼ aI1R

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Fault Analysis 249 ω I1B Ref. I1R I1Y Figure 7.8 Positive-sequence phasorsFor negative-sequence quantities, I2R ¼ I2Rð1 þ j0Þ ¼ reference phase (Figure 7.9) I2Y ¼ I2RðÀ0:5 þ j0:866Þ ¼ aI2R I2B ¼ I2RðÀ0:5 À j0:866Þ ¼ a2I2RReturning to the original unbalanced system of currents, IR, IY, and IB, IR ¼ I1R þ I2R þ I0R IY ¼ I1Y þ I2Y þ I0Y ¼ a2I1R þ aI2R þ I0R IB ¼ I1B þ I2B þ I0B ¼ aI1R þ a2I2R þ I0RHence in matrix form, 23 2 32 3 IR 1 1 I0R 1 a 577664 I1R 775 ð7:3Þ 646 IY 577 ¼ 466 1 a2 IB 1 a a2 I2R ω I2Y Ref I2R I 2B Figure 7.9 Negative-sequence phasors

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250 Electric Power Systems, Fifth EditionInverting the matrix, 2 3 2 1 32 IR 3 4 5 4 a2 54 IY 5 I0R ¼ 1 1 1 ð7:4Þ I1R 3 1 a IB I2R 1 a a2The above also holds for voltages, that is. ÂÃ ½Eactual ¼ ½Ts E1;2;0where ½Ts is the symmetrical component transformation matrix, 21 1 1 3 4 1 a2 a 5 1 a a2 In a three-wire system the instantaneous voltages and currents add to zero. There isno neutral connection and hence no single phase currents. A fourth wire or connectionto earth must be provided for single-phase currents to ﬂow. In a three-wire system thezero phase-sequence components are replaced by zero in equations (7.3) and (7.4). Where zero-phase currents ﬂow, I0R ¼ IR þ IY þ IB ¼ IN 3 3where IN is the neutral current. ; IN ¼ 3I0R ¼ 3I0Y ¼ 3I0B In the application of this method it is necessary to calculate the symmetrical com-ponents of the current in each line of the network and then to combine them toobtain the actual values. The various phase-sequence values are obtained by consid-ering a network derived from the actual network in which only a particularsequence current ﬂows; for example, in a zero-sequence network, only zero-sequence currents and voltages exist. The positive-sequence network is identical tothe real, balanced equivalent network, that is it is the same as used for three-phasesymmetrical short-circuit studies. The negative sequence network is almost thesame as the real one except that the values of impedance used for rotating machinesare different and there are no generated voltages from an ideal machine. The zero-sequence network is considerably different from the real one. The above treatment assumes that the respective sequence impedances in each ofthe phases are equal, that is Z1R ¼ Z1Y ¼ Z1B, and so on. Although this covers mostcases met in practice, unequal values may occur in certain circumstances, for exam-ple an open circuit on one phase. The following equation applies for the voltagedrops across the phase impedances: 23 23 2 0 03 23 VR V0R ZR I0R 46 VY 57 ¼ ½TS46 V1R 57 ¼ 46 0 ZY 0 75½TS46 I1R 57 VB V2R 0 0 ZB I2R

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Fault Analysis 251From which, 2 32 3Therefore VR ZRðI0R þ I1R þ I2RÞ 64 VY 57 ¼ 46 ZYðI0R þ a2I1R þ aI2RÞ 75 VB ZYðI0R þ aI1R þ a2I2RÞV0R ¼ 1 ðVR þ VY þ VBÞ 3 ¼ 1 À þ a2ZY þ Á þ 1 À þ aZY þ a2ZBÁ þ 1 I0RðZR þ ZY þ ZBÞ 3 I1R ZR aZB 3 I2R ZR 3and, similarly, expressions V1R and V2R may be obtained. It is seen that the voltage drop in each sequence is inﬂuenced by the impedancesin all three phases. If, as previously assumed, ZR ¼ ZY ¼ ZB ¼ Z, then the voltagedrops become V1R ¼ I1RZ, and so on, as before.7.4 Representation of Plant in the Phase-Sequence Networks7.4.1 The Synchronous Machine (see Table 3.1)The positive-sequence impedance Z1 is the normal transient or subtransient value. Neg-ative-sequence currents set up a magnetic ﬁeld rotating in the opposite direction to thatof the positive-sequence currents and which rotates round the rotor surface at twice thesynchronous speed; hence the effective impedance (Z2) is different from Z1. The zero-sequence impedance Z0 depends upon the nature of the connection between the starpoint of the windings and the earth and the single-phase impedance of the stator wind-ings in parallel. Resistors or reactors are frequently connected between the star pointand earth for reasons usually connected with protective gear and the limitation of over-voltages. Normally, the only voltage sources appearing in the networks are in the posi-tive-sequence one, as the generators only generate positive-sequence e.m.f.s.7.4.2 Lines and CablesThe positive- and negative-sequence impedances are the normal balanced values.The zero-sequence impedance depends upon the nature of the return path throughthe earth if no fourth wire is provided. It is also modiﬁed by the presence of an earthwire on the towers that protect overhead lines against lightning strikes. In theabsence of detailed information the following rough guide to the value of Z0 may beused. For a single-circuit line Z0=Z1 ¼ 3:5 with no earth wire and Z0=Z1 ¼ 2 withone earth wire. For a double-circuit line, Z0=Z1 ¼ 5:5. For underground cables,Z0=Z1 ¼ 1 À 1:25 for single core and 3–5 for three-core cables.7.4.3 TransformersThe positive- and negative-sequence impedances are the normal balanced ones. Thezero-sequence connection of transformers is, however, complicated, and depends on

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252 Electric Power Systems, Fifth EditionTable 7.3 Zero sequence representation of transformersConnections of Representation CommentsWindings per PhasePrimary Secondary Z0 Zero-sequence currents free to ﬂow in both primary and secondary circuits Z0 No path for zero-sequence currents in primary circuits Z0 Single-phase currents can circulate in the delta but not outside it Z0 No ﬂow of zero-sequence currents possible Z0 No ﬂow of zero-sequence currents possibleT Tertiary winding provides path for zero- sequence currentsthe nature of the connection of the windings. Table 7.3 lists the zero-sequence repre-sentation of transformers for various winding arrangements. Zero-sequence cur-rents in the windings on one side of a transformer must produce the correspondingampere-turns in the other, but three in-phase currents cannot ﬂow in a star connec-tion without a connection to earth. They can circulate round a delta winding, but notin the lines outside it. Owing to the mutual impedance between the phases, Z0 ¼6 Z1.For a three-limb transformer, Z0 < Z1; for a ﬁve-limb transformer, Z0 > Z1. Anexample showing the nature of the three sequence networks for a small transmissionlink is shown in Figure 7.10.7.5 Types of FaultIn the following, a single voltage source in series with an impedance is used to rep-resent the power network as seen from the point of the fault. This is an extension ofThevenin’s theorem to three-phase systems. It represents the general method usedfor manual calculation, that is the successive reduction of the network to a single

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Fault Analysis 253 A ZL ZT C R ZT ZL B ZT Line diagram N NEE Positive E Negative Z1B sequenceZ1A Z1B sequence Z1C Z2A ZT Z2C ZT ZTZT ZT ZT ZL ZL ZL N ZL 3R Z B0 Zero Z C0 Z A0 Z T0 sequence ZT0 Z L0 ZT0 Z L0Figure 7.10 Typical transmission link and form of associated sequence networksimpedance and voltage or current source. The network is assumed to be initially onno-load before the occurrence of the fault, and linear, so that superposition applies.7.5.1 Single-Line-To-Earth FaultThe three-phase circuit diagram is shown in Figure 7.11, where the three phases areon open-circuit at their ends. Let I1, I2 and I0 be the symmetrical components of IR and let V1, V2 and V0 be thecomponents of VR. For this condition, VR ¼ 0, IB ¼ 0, and IY ¼ 0. Also, ZR includescomponents Z1, Z2, and Z0.From equation (7.4) 1 3 I0 ¼ ðIR þ IY þ IBÞ I1 ¼ 1 À þ aIY þ a2 Á 3 IR IB I2 ¼ 1 À þ a2IY þ Á 3 IR aIB

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254 Electric Power Systems, Fifth Edition ZR VR R IR E If n a2 E ZY VY aE Y If VB B IY ZB IBFigure 7.11 Single line-to-earth fault-Thevenin equivalent of system at point of faultHence, I0 ¼ IR ¼ I1 ¼ I2 ðas IY þ IB ¼ 0ÞAlso, 3 VR ¼ E À I1Z1 À I2Z2 À I0Z0 ¼ 0Eliminating I0 and I2, we obtain E À I1ðZ1 þ Z2 þ Z0Þ ¼ 0hence I1 ¼ Z1 þ E þ Z0 ð7:5Þ The fault current, Z2 ð7:6Þ So If ¼ IR ¼ 3I1 If ¼ Z1 þ 3E þ Z0 Z2

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Fault Analysis 255 VRY VRY VRn n n VBR VBR VYB VYB Pre-fault Post-faultFigure 7.12 Pre- and post-fault phasor diagrams-single-line-to-earth faultThe e.m.f. of the Y phase ¼ a2E, and (from equation (7.4)) IY ¼ I0 þ a2I1 þ aI2 VY ¼ a2E À I0Z0 À a2I1Z1 À aI2Z2 The pre-fault and post-fault phasor diagrams are shown in Figure 7.12, where itshould be noted that only VYB remains at its pre-fault value. It is usual to form an equivalent circuit to represent equation (7.5) and this can beobtained from an inspection of the equations. The circuit is shown in Figure 7.13 andit will be seen that I1 ¼ I2 ¼ I0, and I1 ¼ Z1 þ E þ Z0 Z27.5.2 Line-To-Line FaultIn Figure 7.14, E ¼ e.m.f. per phase and the R phase is again taken as the referencephasor. In this case, IR ¼ 0, IY ¼ ÀIB, and VY ¼ VB. I1 I2 I0 Z1 Z2 Z0 EFigure 7.13 Interconnection of positive-, negative-, and zero-sequence networks forsingle-line-to-earth faults

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256 Electric Power Systems, Fifth Edition Z R VR R IR IR E n ZY IY VY Y If aE a 2E Z B I B VB B Figure 7.14 Line-to-line fault on phases Y and B From equation (7.4), I0 ¼ 0and I1 ¼ 1 À À a2Á 3 IY a I2 ¼ 1 IYÀa2 À Á 3 a ; I1 ¼ ÀI2As VY ¼ VB (but not equal to zero) a2E À a2I1Z1 À aI2Z2 ¼ aE À aI1Z1 À a2I2Z2 ; Eða2 À aÞ ¼ I1ÂZ1ða2 À aÞ þ Z2ða2 À Ã aÞ because I1 ¼ ÀI2hence I1 ¼ Z1 E Z2 ð7:7Þ þ This can be represented by the equivalent circuit in Figure 7.15, in which, ofcourse, there is no zero-sequence network. If the connection between the two lineshas an impedance Zf (the fault impedance), this is connected in series in the equiv-alent circuit.

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Fault Analysis 257 Z1 E I2 I1 Z2 ZfFigure 7.15 Interconnection of sequence networks for a line-to-line fault (includingfault impedance Zf , if present)7.5.3 Line-To-Line-To-Earth Fault (Figure 7.16) IR ¼ 0 VY ¼ VB ¼ 0and IR ¼ I1 þ I2 þ I0 ¼ 0 a2E À a2I1Z1 À aI2Z2 À I0Z0 ¼ VY ¼ 0and aE À aI1Z1 À a2I2Z2 À I0Z0 ¼ VB ¼ 0 E n a2E Z R IR VR aE R ZY IY VY Y Z B I B VB B Figure 7.16 Line-to-line-to-earth fault

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258 Electric Power Systems, Fifth Edition Z1 I1 E Z2 Z0 I2 I0 Figure 7.17 Interconnection of sequence networks–double line-to-earth faultHence, I1 ¼ Z1 þ E þ Z0Þ ð7:8Þ ½Z2Z0=ðZ2 I2 ¼ ÀI1 Z2 Z0 Z0 ð7:9Þ þ ð7:10Þand I0 ¼ ÀI1 Z2 Z2 Z0 þ These can be represented by the equivalent circuit as shown in Figure 7.17. The inclusion of impedances in the earth path, such as the star-point-to earth con-nection in a generator or transformer, modiﬁes the sequence diagrams. For a line-to-earth fault an impedance Zg in the earth path is represented by an impedance of 3Zgin the zero-sequence network. Zg can include the impedance of the fault itself, usu-ally the resistance of the arc. As I1 ¼ I2 ¼ I0 and 3I1 ﬂows through Zg in the physicalsystem, it is necessary to use 3Zg to obtain the required effect. Hence, If ¼ Z1 þ Z2 3E þ 3Zg þ Z0 Again, for a double-line-to-earth fault an impedance 3Zg is connected as shown inFigure 7.18. Zg includes both machine neutral impedances and fault impedances. The phase shift introduced by star-delta transformers has no effect on the magni-tude of the fault currents, although it will affect the voltages at various points. If thepositive-sequence voltages and currents are advanced by a certain angle then thenegative-sequence quantities are retarded by the same angle for a given connection.

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Fault Analysis 259 Z1 Z2 Z0 E 3ZgFigure 7.18 Modiﬁcation of network in Figure 7.17 to account for neutralimpedance Zg7.6 Fault Levels in a Typical SystemIn Figure 7.19, a section of a typical system is shown. At each voltage level the faultlevel can be ascertained from the reactances given. It should be noted that the short-circuit level will change with network conditions, and there will normally be twoextreme values: that with all plant connected and that with the minimum plant nor-mally connected. The short-circuit MVA at 275 kV busbars in Britain is normally X = 0.115 p.u. Infinite busbar 2 x 150 MVA Max. plant 0.0107 p.u. Min. plant 0.0143 p.u. Transformers each 240 MVA X = 0.062 p.u. 132 kV Transformers each 90 MVA X = 0.244 p.u. 33 kV Transformers each 15 MVA X = 0.666 p.u. 11 kV 1MVA X = 4.75 p.u. 415 kVFigure 7.19 Typical transmission system. All reactances on a 100 MVA base

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260 Electric Power Systems, Fifth Edition10 000 MVA, but drops to 7000 MVA with minimum plant connected. Maximumshort-circuit (three-phase) levels normally experienced in the British system are asfollows: 275 kV, 15 000 MVA; 132 kV, 3500 MVA; 33 kV, 750/1000 MVA; 11 kV, 150/250 MVA; 415 V, 30 MVA. As the transmission voltages increase, the short-circuit currents also increase, andfor the 400 kV system, circuit breakers of 35 000 MVA breaking capacity arerequired. In order to reduce the fault level the number of parallel paths is reducedby sectionalizing. This is usually achieved by opening the circuit breaker connectingtwo sections of a substation or generating station busbar. One great advantage ofdirect-current transmission links in parallel with the alternating-current system isthat no increase in the short-circuit currents results.7.6.1 Circuit Parameters with Faults7.6.1.1 Fault ResistanceThe resistance of the fault is normally that of the arc and may be approximated by RaðVÞ ¼ 44 V for V < 110 kV and V in kV and If in A Ifand RaðVÞ ¼ 22 V for V > 110 kV If For example, a 735 kV line would have an arc resistance of 4 V with a fault currentof 4 kA, assuming no resistance in the ground return path. The overall grounding resistance depends on the footing resistance (resistance oftower metalwork to ground) of the towers (RT) and also on the resistance per sectionof the ground wires (RS), where present. The situation is summarized in Figure 7.20,where usually the effective grounding resistance is smaller than the individual Point of fault RS RS RS RT RT RTFigure 7.20 Equivalent network of towers with footing resistance (RT) and ground wiresections (resistance RS)

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Fault Analysis 261tower footing resistance. There is normally a spread in the values of RT but normallyRT should not exceed 10 V (ground wires present). Typical values of fault resistance with fault location are as follows: at source, 0 V;on line with ground wires, 15 V; on line without ground wires, 50 V.7.6.1.2 X/R RatioThe range of X/R values for typical voltage class (Canadian) are as follows: 735 kV,18.9–20.4; 500 kV, 13.6–16.5; 220 kV, 2–25; 110 kV, 3–26. The X/R value decreases withseparation of the fault point from the source and can be substantially decreased bythe fault resistance. For distribution circuits, X/R is lower, and, although data is limited, typical valuesare 10 (at source point) and 2–4 on a medium voltage overhead line.Example 7.4A synchronous machine A generating 1 p.u. voltage is connected through a star-startransformer, reactance 0.12 p.u., to two lines in parallel. The other ends of the lines areconnected through a star-star transformer of reactance 0.1 p.u. to a second machine B,also generating 1 p.u. voltage. For both transformers, X1 ¼ X2 ¼ X0. Calculate the current fed into a double-line-to-earth fault on the line-side terminals ofthe transformer fed from A. The relevant per unit reactances of the plant, all referred to the same base, are asfollows: For each line: X1 ¼ X2 ¼ 0:30, X0 ¼ 0:70. For generators: X1 X2 X0Machine A 0.30 0.20 0.05Machine B 0.25 0.15 0.03 The star points of machine A and of the two transformers are solidly earthed.SolutionThe positive-, negative-, and zero-sequence networks are shown in Figure 7.21. All perunit reactances are on the same base. From these diagrams the following equivalentreactances up to the point of the fault are obtained: Z1 ¼ j0:23 p:u: Z2 ¼ j0:18 andZ0 ¼ j0:17 p:u: The red phase is taken as reference phasor and the blue and yellow phases areassumed to be shorted at the fault point. From the equivalent circuit for a line-to-linefault, I1 ¼ Z1 þ E þ Z0Þ ½Z2Z0=ðZ2 ¼ j½0:23 þ ½0:17 Â 1 þ 0:18Þ ¼ Àj3:15 p:u: 0:18=ð0:17

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262 Electric Power Systems, Fifth EditionA TA TB B EE FEA EB E Positive sequence network 0.42 0.500.30 0.30 0.25 F1 F10.12 0.10 0.30Negative sequence network0.20 0.30 0.15 0.32 0.40 F2 0.30 F2 0.10 0.12 O.C. O.C. Zero sequence network 0.17 0.480.05 0.70 0.03 F0 F0 0.70 0.10 0.12Figure 7.21 Line diagram and sequence networks for Example 7.4I2 ¼ ÀI1 Z0 Z2 þ Z0 ¼ Àj3:15 Â 0:17 ¼ Àj1:53 p:u: 0:17 þ 0:18I0 ¼ ÀI1 Z2 Z2 þ Z0 0:18 ¼ Àj3:15 Â 0:17 þ 0:18 ¼ Àj1:62 p:u:IY ¼ I0 þ a2I1 þ aI2 ¼ j1:62 þ ðÀ0:5 À j0:866ÞðÀj3:15Þ þ ðÀ0:5 þ j0:866Þðj1:53Þ¼ À4:05 þ j2:43 p:u:

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Fault Analysis 263 IB ¼ I0 þ aI1 þ a2I2 ¼ j1:62 þ ðÀ0:5 þ j0:866ÞðÀj3:15Þ þ ðÀ0:5 À j0:866Þðj1:53Þ ¼ 4:05 þ j2:43 p:u: IY ¼ IB ¼ 4:72 p:u: The correctness of the ﬁrst part of the solution can be checked as IR ¼ I0 þ I1 þ I2 ¼ j1:62 À j3:15 þ j1:53 ¼ 0Example 7.5An 11 kV synchronous generator is connected to a 11/66 kV transformer which feeds a66/11/3.3 kV three-winding transformer through a short feeder of negligible imped-ance. Calculate the fault current when a single-phase-to-earth fault occurs on a termi-nal of the 11 kV winding of the three-winding transformer. The relevant data for thesystem are as follows: Generator: X1 ¼ j0:15 p:u:, X2 ¼ j0:10 p:u:, X0 ¼ j0:03 p:u:, all on a 10 MVA base; starpoint of winding earthed through a 3 V resistor. 11/66kV Transformer: X1 ¼ X2 ¼ X0 ¼ j0:1 p:u: on a 10 MVA base; 11 kV windingdelta connected and the 66 kV winding star connected with the star point solidly earthed. Three-winding transformer: A 66 kV winding, star connected, star point solidlyearthed; 11 kV winding, star connected, star-point earthed through a 3 V resistor; 3.3 kVwinding, delta connected; the three windings of an equivalent star connection to repre-sent the transformer have sequence impedances,66 kV winding X1 ¼ X2 ¼ X0 ¼ j0:04 p:u:;11 kV winding X1 ¼ X2 ¼ X0 ¼ j0:03 p:u:;3.3 kV winding X1 ¼ X2 ¼ X0 ¼ j0:05 p:u:;all on a 10 MVA base. Resistance may be neglected throughout.Solution:The line diagram and the corresponding positive -, negative -, and zero-sequencenetworks are shown in Figure 7.22. A 10 MVA base will be used. The 3 V earthingresistor has the following p.u. value:3 Â 10 Â 106 or 0:25 p:u:ð11Þ2 Â 106 Much care is needed with the zero-sequence network owing to the transformer con-nections. For a line-to-earth fault, the equivalent circuit shown in Figure 7.13 is used,from whichI1 ¼ I2 ¼ I0 and If ¼ I1 þ I2 þ I0

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264 Electric Power Systems, Fifth Edition Hence 3Â1 If ¼ Z1 þ Z2 þ Z0 þ 3Zg ¼ j0:32 þ j0:27 3 j0:075 þ 0:75 þ ¼ 3 ¼ 3 p:u: 0:75 þ j0:66 1ﬀ41 If ¼ 3pÂﬃﬃ 10 Â 106 ¼ 1575 A 3 Â 11000 11 kV 66 kV t p s o/c 11 kV 3Ω F IF 3Ω 1 p.u. j0.05 1 p.u. j0.32 p.u. j0.15 j0.1 j0.04 F1 F1 j0.03 Positive sequence diagram j0.05 j0.1 j0.1 j0.04 F2 j0.27 F2 j0.03 Negative sequence diagram 3 x 0.25 j0.05 j0.075 j0.03 j0.1 j0.4 F0 3 x 0.25 j0.03 3 x 0.25 F0 Zero sequence diagramFigure 7.22 Line diagram and sequence networks for Example 7.5

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Fault Analysis 2657.7 Power in Symmetrical ComponentsThe total power in a three-phase network VaIaÃ þ VbIÃb þ VcIcÃwhere Va, Vb, and Vc are phase voltages and Ia, Ib, and Ic are line currents. Inphase (a), Pa þ jQa ¼ ðVa0 þ Va1 þ Va2ÞðIaÃo þ IaÃ1 þ IÃa2Þ ¼ ðVa0IÃa0 þ Va1IÃa1 þ Va2IÃa2Þ þ ðVa0IaÃ1 þ Va1IÃa2 þ Va2IaÃ0Þ þ ðVa0IaÃ2 þ Va1IaÃ0 þ Va2IÃa0Þwith similar expressions for phases (b) and (c). In extending this to cover the total three-phase power it should be noted that IÃb1 ¼ ða2Ia1ÞÃ ¼ ða2ÞÃIaÃ1 ¼ aIaÃ1 Similarly, IÃb2, IÃc1, and IcÃ2 may be replaced. The total power ¼ 3ðVa0IaÃ0 þ Va1IÃa1 þ Va2IaÃ2Þ that is 3 Â (the sum of the individualsequence powers in any phase).7.8 Systematic Methods for Fault Analysis in Large NetworksThe methods described so far become unwieldy when applied to large networksand a systematic approach utilizing digital computers is used. The generators arerepresented by their voltages behind the transient reactance, and normally the sys-tem is assumed to be on no-load before the occurrence of the three-phase balancedfault. The voltage sources and transient reactances are converted into current sour-ces and the admittance matrix is formed (including the transient reactance admit-tances). The basic equation ½Y½V ¼ ½I is formed and solved with the constraint thatthe voltage at the fault node is zero. It is preferable, on grounds of storage and time,not to invert the matrix but to use Gaussian elimination methods. The computationefﬁciency may also be improved by utilizing the sparsity of the Y matrix. The meshor loop (impedance matrix) method may be used, although the matrix is not so eas-ily formed. The following example illustrates a method suitable for determination of balancedthree-phase fault currents in a large system by means of a digital computer.

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266 Electric Power Systems, Fifth Edition Example 7.6 Determine the fault current in the system shown in Figure 7.23(a) for the balanced fault shown.X = j0.1 X = j0.1 10 10 p.u. 10 10 j0.5 B p.u. A j0.3 j0.3 2 10 3 10 3 10 p.u. 10 p.u. C j0.5 D 2 10 F 10X = j0.1 X = j0.1 (a) F (b)Figure 7.23 (a) Circuit for Example 7.6 and (b) equivalent circuit. All values areadmittances (i.e. Àj Y). The generators, that is 1 p.u. voltage behind Àj10 p.u. admit-tance, transform to Àj10 p.u. current sources in parallel with Àj10 p.u. admittanceSolutionThe system in Figure 7.23(a) is replaced by the equivalent circuit shown in Figure 7.23(b)by converting a voltage source in series with transient reactance to a current source inparallel with the same reactance. The nodal admittance matrix is then formed. Finally,equation ½Y½V ¼ ½I is formed with VD ¼ 0.0 p.u. 2 15:33 À2:00 0 À3:33 32 VA 3 2 j10 3 j6666646 À2:00 15:33 À3:33 0 77777576666466 VB 7775777 ¼ 6646666 j10 7577777 0 À3:33 15:33 À2:00 VC j10 À3:33 0 À2:00 15:33 0 j10 À IFIf the bottom row is eliminated and every element is multiplied by 1.5, we get: 2 32 3 2 3 23 À3 0 VA 15 4666 À3 23 À5 75776466 VB 5777 ¼ 6664 15 7577 0 À5 23 VC 15from which by Gaussian elimination (3 Â 1st row þ 23 Â 2nd row):

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Fault Analysis 267 \" #\" # \" # 520 À115 VB 390 ¼ À5 23 VC 15Thus, VB ¼ 0:9394 p:u: VC ¼ 0:8564 p:u: VA ¼ 0:7747 p:u:And IF ¼ jð10 þ 3:33 Â 0:7747 þ 2 Â 0:8564Þ ¼ j14:2926 p:u:Fault MVA ¼ 14.2926 Â 100 ¼ 1429.26 MVA Instead of the nodal admittance method, the bus impedance method may be usedfor computer fault analysis and has the following advantages:1. Matrix inversion is avoided, resulting in savings in computer storage and time.2. The matrices for the sequences quantities are determined only once and retained for later use; they are readily modiﬁed for system changes.3. Mutual impedances between lines are readily handled.4. Subdivisions of the main system may be incorporated. The system is represented by the usual symmetrical component sequence net-works and, frequently, the positive and negative impedances are assumed to beidentical. Balanced phase impedances for all items of plant are assumed as are equalvoltages for all generators. In the bus impedance method the network loop matrix, that is ½E ¼ ½Z½I, is set upin terms of the various loop currents. First, the buses of interest are short-circuited tothe neutral. Consider a fault on one of the buses (k) only, currents in all the other bus-ses short circuited to the neutral will be zero and, from equation ½E ¼ ½Z½I,1:0 ¼ ZkkIk, where Ik ¼ fault current with three-phase symmetrical fault on k. Similarly,the currents with balanced faults on each of the other buses may be easily determined.7.8.1 Computer SimulationsPart of the transmission system shown in Figure 6.12 was used. For fault calcula-tions, normally transient and subtransient reactances and time constants of the gen-erators are required. Table 7.4 gives the data for each generator on the machine base.Table 7.4 Machine dataGenerator Xs R X0 X00 T0 T00G1 and G2 1.2 0.012 0.35 0.25 1.1 0.035G3 1.5 0.015 0.27 0.20 1.8 0.035

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268 Electric Power Systems, Fifth EditionF E2494.378 2188.998A1 & A2 D3067.517 2272.960 B C 2607.749 2306.461 (a)F E2625.190 2309.776A1 & A2 D3081.343 2168.482 B C 2623.861 2196.066 (b)Figure 7.24 (a) Fault level in MVA at each busbar assuming all the busbar voltagesare 1 p.u. (b) fault level calculated using actual pre-fault busbar voltages In fault calculations so far it has been assumed that the busbar voltages are 1 p.u.Figure 7.24(a) shows the fault level of each busbar in MVA of the system shown inFigure 6.12 with 1 p.u. voltage on all the busbars. However pre-fault operating volt-ages of a network depend on the load ﬂow and the resulting fault currents are

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Fault Analysis 269 150 Fault current (kA) 20 40 60 80 100 100 50 0 -50 -100 -150 Time Red Blue Yellow (a)Fault current (kA) 8 20 40 60 80 100 6 4 2 0 0 -2 -4 -6 Time Yellow Red Blue (b)Figure 7.25 Three phase fault currents (a) for a fault at busbar A1 (b) for a fault atbusbar D. Red phase has maximum DC offsetdifferent when actual pre-fault operating voltages are used. Figure 7.24(b) shows thefault levels calculated using the actual busbar voltages.

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270 Electric Power Systems, Fifth Edition Figure 7.25(a) shows the fault current from the synchronous generator A1 for ashort circuit at its terminals. As described in Section 3.3, the subtransient and tran-sient current with DC offset can clearly be seen in this ﬁgure. However when thefault is away from the generator the DC offset currents decay within a few cycles asshown in Figure 7.25(b).7.9 Neutral Grounding7.9.1 IntroductionFrom the analysis of unbalanced fault conditions it has been seen that the connectionof the transformer and generator neutrals greatly inﬂuences the fault currents andvoltages. In most high-voltage systems the neutrals are solidly grounded, that isconnected directly to the ground, with the exception of generators which aregrounded through a resistance to limit stator fault currents. The advantages of solidgrounding are as follows:1. Voltages to ground are limited to the phase voltage.2. Intermittent ground faults and high voltages due to arcing faults are eliminated.3. Sensitive protective relays operated by earth fault currents clear these faults at an early stage. The main advantage in operating with neutrals isolated is the possibility of main-taining a supply with a ground fault on one line which places the remaining conduc-tors at line voltage above ground. Also, interference with telephone circuits isreduced because of the absence of zero-sequence currents. With normal balancedoperation the neutrals of an ungrounded or isolated system are held at groundpotential because of the presence of the system capacitance to earth. For the generalcase shown in Figure 7.26, the following analysis applies:N a Z ag b c Z cg Z bg GroundFigure 7.26 Line-to-ground capacitances in an ungrounded system

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Fault Analysis 271 Vag þ Vbg þ Vcg ¼ 0 ð7:11Þ Zag Zbg ZcgAlso,where Vag ¼ Van þ VngVan ¼ voltage of line (a) to neutralVng ¼ voltage of neutral to groundSimilarly, Vbg ¼ Vbn þ Vngand Vcg ¼ Vcn þ VngSubstituting in equation (7.11) and separating terms, Van Vbn Vcn 1 1 1 Zag þ Zbg þ Zcg þ Vng Zag þ Zbg þ Zcg ¼0 ð7:12ÞThe equation 1þ1þ1 Zag Zbg Zcg ¼ Yggives the ground capacitance admittance of the system.7.9.2 Arcing FaultsConsider the single-phase system in Figure 7.27 at the instant when the instanta-neous voltages are v on line (a) and -v on line (b), where v is the maximum instanta-neous voltage. The sudden occurrence of a fault to ground causes line (b) to assumea potential of À2v and line (a) to become zero. Because of the presence of both L andC in the circuit, the sudden change in voltages by v produces a high-frequency oscil-lation of peak magnitude 2v superimposed on the power frequency voltages (seeChapter 10) and line (a) reaches -v and line (b) À3v, as shown in Figure 7.28. Theseoscillatory voltages attenuate quickly due to the resistance present. The current inthe arc to earth on line (a) is approximately 90 ahead of the fundamental voltage,and when it is zero the voltage will be at a maximum. Hence, if the arc extinguishesat the ﬁrst current zero, the lines remain charged at Àv for (a) and À3v for (b). Theline potentials now change at power frequency until (a) reaches À3v, when thearc could restrike causing a voltage change of À3v to 0, resulting in a transient

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272 Electric Power Systems, Fifth Edition L b υ aN υ L CC EFigure 7.27 Single-phase system with arcing fault to groundovervoltage of þ3v in line (a) and þ5v in line (b). This process could continue andthe voltage build up further, but the resistance present usually limits the peak volt-age to under 4v. A similar analysis may be made for a three-phase circuit, againshowing that serious overvoltages may occur with arcing faults because of theinductance and shunt capacitance of the system. This condition may be overcome in an isolated neutral system by means of an arcsuppression or Petersen coil. The reactance of this coil, which is connected betweenthe neutral and ground, is made in the range 90–110% of the value required to neu-tralize the capacitance current. The phasor diagram for the network of Figure 7.29(a)is shown in Figure 7.29(b) if the voltage drop across the arc is neglected. pﬃﬃ Ia ¼ Ib ¼ 3VvC 5v Transient Time 3v v 3v vVoltageFigure 7.28 Voltage on line (a) of Figure 7.27

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Fault Analysis 273 Va a Vb b N c Vc Petersen IL C Ib Ia coil L Ia+ I b C C IL (a) a N b Vac VNC Vbc Ib Ia 30o c 30o Ground potential Ia+ I b (b) ILFigure 7.29 (a) System with arc suppression coil, (b) Phasor diagram of voltages andcurrents in part (a)and pﬃﬃ pﬃﬃalso, Ia þ Ib ¼ 3 Â 3VvC ¼ 3VvC IL ¼ V vL

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274 Electric Power Systems, Fifth EditionFor compensation of the arc current, V ¼ 3VvC vL ;L ¼ 1 3v2CAnd XL ¼ 1 3vC This result may also be obtained by analysis of the ground fault by means of sym-metrical components. Generally, isolated neutral systems give rise to serious arcing-fault voltages if thearc current exceeds the region of 5–10 A, which covers most systems operatingabove 33 kV. If such systems are to be operated with isolated neutrals, arc-suppres-sion coils should be used. Most systems at normal transmission voltages havegrounded neutrals.7.10 Interference with Communication Circuits– Electromagnetic Compatibility (EMC)When power and telephone lines run in parallel, under certain conditions voltagessufﬁcient to cause high noise levels may be induced into the communication circuits.This may be caused by electromagnetic and electrostatic unbalance in the power lines,especially if harmonics are present. The major problem, however, is due to faults toground producing large zero-sequence currents in the power line, which inductivelyinduce voltage into the neighbouring circuits. The value of induced voltage dependson the spacing, resistivity of the earth immediately below, and the frequency. If thetelephone wires are a twisted pair or are situated close together and transposed, novoltage is induced between the communication conductors. However, a voltage canexist between the pair of wires and ground. These induced longitudinal voltages canbe controlled by connecting the communication circuits to ground through an induc-tance which produces little attenuation at communication frequencies of 400–3500 Hz. Capacitive coupling can occur if open communication circuits are run along thesame route as power lines. Interference from underground cable circuits is muchless (10%) than that from overhead lines. Because of right-of-way constraints, telephone and power distribution lines runparallel along the same street in many urban areas. However, the interference inrural areas is often greater because communication lines and plant may beunshielded or have higher shield resistances, and unlike urban areas there is noextensive network of water and gas pipes to share the ground return currents. Resistive coupling between power and communication circuits can exist:

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Fault Analysis 275 because of physical contact between them; via paths through the soil between telephone and power grounds. Various formulae exist to calculate the value of mutual inductance-in H/mbetween circuits with earth return. These assume that er (soil) is unity, displacementcurrents are much less than conduction currents, and the length of conductor isinﬁnite. During line-to-ground faults, induced voltages into communication circuits maybe sufﬁcient to be a shock hazard to personnel. Although in transmission circuits,equal current loading may be assumed in the phases, this is not the case in the lowervoltage distribution circuits where signiﬁcant residual currents may ﬂow. Most tele-phone circuit standards now require up to 15 kV isolation if communication circuitsare to be connected into substations for monitoring, control and communicationpurposes. Particular care must be taken with bonding the sheath of communicationcircuits brought into buildings where the power distribution system is also bondedto earth and to the building structure.Problems 7.1 Four 11 kV generators designated A, B, C, and D each have a subtransient reactance of 0.1 p.u. and a rating of 50 MVA. They are connected in parallel by means of three 100 MVA reactors which join A to B, B to C, and C to D; these reactors have per unit reactances of 0.2, 0.4, and 0.2, respectively. Calculate the volt-amperes and the current ﬂowing into a three-phase symmetrical fault on the terminals of machine B. Use a 50 MVA base. (Answer: 937.5 MVA; 49 206 A) 7.2 Two 100 MVA, 20 kV turbogenerators (each of transient reactance 0.2 p.u.) are connected, each through its own 100 MVA, 0.1 p.u. reactance transformer, to a common 132 kV busbar. From this busbar, a 132 kV feeder, 40 km in length, supplies an 11 kV load through a 132/11 kV transformer of 200 MVA rating and reactance 0.1 p.u. If a balanced three-phase short circuit occurs on the low-voltage terminals of the load transformer, determine, using a 100 MVA base, the fault current in the feeder and the rating of a suitable circuit breaker at the load end of the feeder. The feeder impedance per phase is (0.035 þ j0.14) V/km. (Answer: 431 MVA) 7.3 Two 60 MVA generators of transient reactance 0.15 p.u. are connected to a busbar designated A. Two identical machines are connected to another busbar B. Busbar A is connected to busbar B through a reactor, X. A feeder is sup- plied from A through a step-up transformer rated at 30 MVA with 10% reactance.

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276 Electric Power Systems, Fifth Edition Calculate the reactance X, if the fault level due to a three-phase fault on thefeeder side of the 30 MVA transformer is to be limited to 240 MVA. Calculatealso the voltage on A under this condition if the generator voltage is 13 kV(line).(Answer: X ¼ 0.075 p.u.; VA ¼ 10.4 kV)7.4 A single line-to-earth fault occurs on the red phase at the load end of a 66 kV transmission line. The line is fed via a transformer by 11 kV generators con- nected to a common busbar. The line side of the transformer is connected in star with the star point earthed and the generator side is in delta. The posi- tive-sequence reactances of the transformer and line are j10.9 V and j44 V, respectively, and the equivalent positive and negative-sequence reactances of the generators, referred to the line voltage, are j18 V and j14.5 V, respectively. Measured up to the fault the total effective zero sequence reactance is j150 V. Calculate the fault current in the lines if resistance may be neglected. If a two- line-to-earth fault occurs between the blue and yellow lines, calculate the cur- rent in the yellow phase. (Answer: 391 A; 1486 A)7.5 A single-line-to-earth fault occurs in a radial transmission system. The follow- ing sequence impedances exist between the source of supply (an inﬁnite bus- bar) of voltage 1 p.u. to the point of the fault: Z1 ¼ 0.3 þ j0.6 p.u., Z2 ¼ 0.3 þ j0.55 p.u., Z0 ¼ 1 þ j0.78 p.u. The fault path to earth has a resistance of 0.66 p.u. Determine the fault current and the voltage at the point of the fault.(Answer: If ¼ 0.74 p.u.; Vf ¼ (0.43 À j0.23) p.u.)7.6 Develop an expression, in terms of the generated e.m.f. and the sequenceimpedances, for the fault current when an earth fault occurs on phase (A) of athree-phase generator, with an earthed star point. Show also that the voltageto earth of the sound phase (B) at the point of fault is given by pﬃﬃ Àj 3EA½Z2 À aZ0 VB ¼ Z1 þ Z2 þ Z0 Two 30 MVA, 6.6 kV synchronous generators are connected in parallel andsupply a 6.6 kV feeder. One generator has its star point earthed through aresistor of 0.4 V and the other has its star point isolated. Determine: (a) thefault current and the power dissipated in the earthing resistor when an earthfault occurs at the far end of the feeder on phase (A); and (b) the voltage toearth of phase (B). The generator phase sequence is ABC and the impedancesare as follows:To positive-sequence currents Generator p.u. Feeder V/p.h. j0.2 j0.6

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Fault Analysis 277 30 MVA xM 15% x x xx xM Grid x xx 11 kV x infeed xM 166.6 MVA 30 MVA 15% 132 kV xM x 5 MVA A Figure 7.30 System for Problem 7.7To negative-sequence currents j0.16 J0.6To zero-sequence currents j0.06 j0.4 Use a base of 30 MVA. (Answer: (a) 5000ﬀ À 58:45 A; 10 MW; (b) À2627 À j1351 V7.7 An industrial distribution system is shown schematically in Figure 7.30. Each line has a reactance of j0.4 p.u. calculated on a 100 MVA base; other system parameters are given in the diagram. Choose suitable short-circuit ratings for the oil circuit breakers, situated at substation A, from those commercially available, which are given in the table below.Short circuit (MVA) 75 150 250 350Rated current (A) 500 800 1500 2000 The industrial load consists of a static component of 5 MVA and four largeinduction motors each rated at 6 MVA. Show that only three motors can be

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278 Electric Power Systems, Fifth Edition A BG1 T1 T2 G2 LineFigure 7.31 Circuit for Problem 7.9 started simultaneously given that, at starting, each motor takes ﬁve times full- load current at rated voltage, but at 0.3 p.f.7.8 Explain how the Method of Symmetrical Components may be used to repre- sent any 3 p.h. current phasors by an equivalent set of balanced phasors. A chemical plant is fed from a 132 kV system which has a 3 p.h. symmetri- cal fault level of 4000 MVA. Three 15 MVA transformers, connected in paral- lel, are used to step down to an 11 kV busbar from which six 5 MVA, 11 kV motors are supplied. The transformers are delta-star connected with the star point of each 11 kV winding, solidly earthed. The transformers each have a reactance of 10% on rating and it may be assumed that X1 ¼ X2 ¼ X0. The ini- tial fault contribution of the motors is equal to ﬁve times rated current with 1.0 p.u. terminal voltage. Using a base of 100 MVA, a. calculate the fault current (in A) for a line-to-earth short circuit on the 11 kV busbar with no motors connected; b. calculate the 3 p.h. symmetrical fault level (in MVA) at the 11 kV busbar if all the motors are operating and the 11 kV busbar voltage is 1.0 p.u. (Answer: (a) 22 kA; (b) 555 MVA) (From Engineering Council Examination, 1996)7.9 Describe the effect on the output current of a synchronous generator following a solid three-phase fault on its terminals. For the system shown in Figure 7.31 calculate (using symmetrical components): a. the current ﬂowing in the fault for a three-phase fault at busbar A; b. the current ﬂowing in the fault for a one-phase-to-earth fault at busbar B; c. the current ﬂowing in the faulted phase of the overhead line for a one-phase- to earth fault at busbar B. Generators G1 and G2: X100 ¼ X200 ¼ j0:1 p.u.; 11 kV Transformers T1 and T2: X1 ¼ X2 ¼ X0 ¼ j0:1 p.u.; 11/275 kV

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Fault Analysis 279 (Earthed star-delta) Line: Z1 ¼ Z2 ¼ j0:05 p.u., Z0 ¼ j0:1 p.u.; 275 kV (All p.u. values are quoted on a base of 100 MVA) Assume the pre-fault voltage of each generator is 1 p.u. and calculate the symmetrical fault currents (in amps) immediately after each fault occurs. (Answer: (a) 1.89 kA; (b) 2.18 kA; (c) 0.89 kA) (From Engineering Council Examination, 1997)7.10 Why is it necessary to calculate short-circuit currents in large electrical systems? A generator rated at 400 MW, 0.8 power factor, 20 kV has a star-connected stator winding which is earthed at its star point through a resistor of 1 V. The generator reactances, in per unit on rating, are: X1 ¼ 0:2 X2 ¼ 0:16 X0 ¼ 0:14 The generator feeds a delta-star-connected generator transformer rated at 550 MVA which steps the voltage up to a 275 kV busbar. The transformer star- point is solidly earthed and the transformer reactance is 0.15 p.u. on its rating. The 275 kV busbar is connected only to the transformer. Assume that for the transformer X1 ¼ X2 ¼ X0. Using a base of 500 MVA calculate the base current and impedance of each voltage level. Calculate the fault current in amperes for: a. a 275 kV busbar three-line fault; b. a 275 kV single-line-to earth fault on the busbar; c. a 20 kV three-line fault on the generator terminals; d. a 20 kV single-line-to-earth fault on the generator terminals. (Answer: (a) 3.125 kA; (b) 4.1 kA; (c) 72.15 kA; (d) 11.44 kA) (From Engineering Council Examination, 1995)

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8System Stability8.1 IntroductionThe stability of a power system is its ability to remain in an operating equilibriumwhen subjected to the disturbances that are inevitable in any network made up ofgenerating plant supplying loads. The disturbance may be small (e.g. caused bychanges in load) or large (e.g. due to fault). After the disturbance a stable systemreturns to a condition of acceptable voltages and power ﬂows throughout thesystem. Figure 8.1 shows how power system stability may be divided into several aspectsin order to make the problem easier to address. These include the loss of synchro-nism between synchronous generators (angle stability) either due to faults and largedisturbances (transient stability) or oscillations caused by changes in load and lackof damping (dynamic or small-signal stability). Voltage instability can be caused bylarge induction motor loads drawing excessive amounts of reactive power duringnetwork faults when the network voltage is depressed or by a lack of reactive powerwhen excessive real power ﬂows in a network. When the rotor of a synchronous generator advances beyond a certain criticalangle, the magnetic coupling between the rotor and the stator fails. The rotor, nolonger held in synchronism with the rotating ﬁeld of the stator currents, rotates rela-tive to the ﬁeld and pole slipping occurs. Each time the poles traverse the angularregion within which it could be stable, synchronizing forces attempt to pull the rotorinto synchronism. It is usual practice to disconnect the generator from the system ifit commences to slip poles, as pole slipping causes large current to ﬂow and hightransient torques. Synchronous, or angle, stability may be divided into: dynamic and transient sta-bility. Dynamic stability is the ability of synchronous generators, when operatingunder given load conditions, to retain synchronism (without excessive angular oscil-lations) when subject to small disturbances, such as the continual changes in load orgeneration and the switching out of lines. It is most likely to result from the changesElectric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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282 Electric Power Systems, Fifth Edition Power system stabilityAngle stability Voltage stabilityTransient stability Dynamic stability Induction motor Network voltage stability collapseFigure 8.1 Main forms of power system stabilityin source-to-load impedance resulting from changes in the network conﬁguration orsystem state and is a consequence of lack of damping in the power system. Often,this is referred to as small-signal stability. Transient stability is concerned with sudden and large changes in the networkcondition, such as those brought about by short-circuit faults. The maximum powerthat can be transmitted, the stability limit, taking into account fault conditions isusually less than the maximum stable steady-state load. The stability of an asynchronous motor load is controlled by the voltage acrossit; if this becomes lower than a critical value, induction motors may becomeunstable and stall. This is, in effect, a voltage instability problem. When the volt-age at the terminals of an asynchronous (induction) motor drops, perhaps due toa fault on the power system, its ability to develop torque is reduced and themotor slows down. If the fault on the network persists the motor stalls anddraws very large amounts of reactive power. These reactive power ﬂows depressthe voltage at the motor terminals further and the section of network has to beisolated. This form of voltage instability is a well known hazard in oil reﬁneriesand chemical plants that have large induction motor loads. Voltage instabilitycan also occur in large national power systems when the loading of transmissionlines exceeds the stable (approximately horizontal) section of the P-V curve, seeFigure 5.21. Once the load of the transmission line approaches the ‘nose’ of theP-V curve instability can occur. In a power system it is possible for either angle or voltage instability tooccur, and in practice one form of instability can inﬂuence the other. Angle orsynchronous stability has traditionally been considered more onerous andhence has been given more attention in the past. Recently, with the increasinguse of static VAr compensators and the experience of large national black-outscaused by a deﬁcit of reactive power, the study of voltage collapse has becomeimportant.

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System Stability 2838.2 Equation of Motion of a Rotating MachineThe kinetic energy of a rotating mass, such as a large synchronous generator, is KE ¼ 1 Iv2 ½J 2and the angular momentum is M ¼ Iv ½Js=radwhere v is the synchronous speed of the rotor (rad/s) and I is the moment of inertia(kgm2). The inertia constant (H) is deﬁned as the stored energy at synchronous speeddivided by the rating of the machine G expressed in volt-amperes (VA). Hence H ¼ 1 Iv2 ½Ws=VA ð8:1Þ 2 Gand the stored kinetic energy is KE ¼ GH ¼ 1 Iv2 ¼ 1 Mv ½J 2 2The angular velocity can be expressed in electrical degrees per second, ve, as ve ¼ 360f ½electrical degrees per secondwhere f is the system frequency in Hz. Then GH ¼ 1 Mð360f Þ ½J 2with M ¼ GH ÂÃ 180f Js=electrical degree The inertia constants (H) of generating sets tend to similar values. The inertia con-stant for steam or gas turbo-generator units decreases from 10 Ws/VA (generatorand turbine together) for machines up to 30 MVA to values in the order of 4 Ws/VAfor large machines, the value decreasing as the rating increases. For salient-polehydro-electric units, H depends on the number of poles; for machines operating inthe speed range 200–400 r.p.m. the value increases from about 2 Ws/VA at 10 MVA

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284 Electric Power Systems, Fifth Editionrating to 3.5 Ws/VA at 60 MVA. A mean value for synchronous motors is 2 Ws/VA.Large wind turbines have an inertia constant in the range 2–5 Ws/VA. The net accelerating torque on the rotor of a machine is DT ¼ mechanical torque input - electrical torque output ¼ I d2d dt2 ; d2d ¼ DT ¼ ðDTvÞv ¼ DPv dt2 I 2KE 2 Iv2 2where DP ¼ net power corresponding to DT, DP ¼ Pmech À Pelec d2d ¼ DP ð8:2Þ dt2 M In equation (8.2) a reduction in electrical power output results in an increase in d. Sometimes, the mechanical power input is assumed to be constant and the equa-tion of motion becomes d2d ¼ À DP dt2 Mor M d2d þ DP ¼ 0 dt28.3 Steady-State StabilityThe steady-state stability limit is the maximum power that can be transmitted in anetwork between sources and loads when the system is subject to small distur-bances. The power system is, of course, constantly subjected to small changes asload variations occur. To calculate the limiting value of power, small increments ofload are added to a model of the system; after each increment the generator excita-tions are adjusted to maintain constant terminal voltages and a load ﬂow is carriedout. Eventually, a condition of instability is reached. The steady-state stability limits of synchronous machines have been discussed inChapter 3 (Section 3.5). It was seen that provided the generator is connected to abusbar with a high fault level and operates within the ‘safe area’ of the performancechart (Figure 3.12), stability is assured; usually, a 20% margin of safety on the lead-ing power factor, absorbing VArs side is allowed and the limit is extended by theuse of automatic voltage regulators. Often, the performance charts are not useddirectly and the generator-equivalent circuit employing the synchronous impedanceis used. The normal maximum operating load angle for modern machines is in the

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System Stability 285order of 60 electrical degrees, and for the limiting value of 90 this leaves 30 tocover the transmission network. In a complex system a reference point must betaken from which the load angles are measured; this is usually a point where thedirection of power ﬂow reverses.The simplest criterion for steady-state synchronous stability is @P > 0, that is the @dsynchronizing coefﬁcient must be positive. The use of this criterion involves thefollowing assumptions:1. generators are represented by constant impedances in series with their internal voltages;2. the input torques from the turbines are constant;3. changes in speed (frequency) are ignored;4. electromagnetic damping in the generators is ignored;5. the changes in load angle d are small.The degree of complexity to which the analysis is taken has to be decided, forexample the effects of machine inertia, governor action and automatic voltage regu-lators can be included; these items, however, greatly increase the complexity of thecalculations. The use of the criterion @P ¼ 0 alone gives a pessimistic or low result @dand hence an inbuilt factor of safety.In a system with several generators and loads, the question as to where the incre-ment of load is to be applied is important. A conservative method is to assume thatthe increment applies to one machine only, determine the stability, and then repeatfor each of the other machines in turn. Alternatively, the power outputs from all butthe two generators having the largest load angles are kept constant.For calculations made without the aid of computers it is usual to reduce the net-work to the simplest form that will keep intact the generator nodes. The values ofload angle, power and voltage are then calculated for the given conditions, @P deter- @dmined for each machine, and, if positive, the loading is increased and the processrepeated.In a system consisting of a generator supplying a load through a network of linesand transformers of effective reactance, XT @P ¼ EV cos d @d XTwhere E and V are the supply- and receiving-end voltages and d the total anglebetween the generator rotor and the phasor of V. The power transmitted is obviously increased with higher system voltages andlower reactances, and it may be readily shown that adding series capacitance in theline increases the steady-state stability limit, although with an increased risk ofinstability due to resonance (a phenomenon known as sub-synchronous resonance).

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286 Electric Power Systems, Fifth EditionThe determination of @P is not very difﬁcult if the voltages at the loads can be @dassumed to be constant or if the loads can be represented by impedances. Use canbe made of the P-V, Q-V characteristics of the load if the voltages change appreci-ably with the redistribution of the power in the network; this process, however, istedious to undertake manually.Example 8.1For the system shown in Figure 8.2, investigate the steady-state stability.SolutionUsing Equation (2.18) (Section 2.5.1) and working from the inﬁnite busbar voltage, thevoltage at Point A is given by VA ¼ vtuu\"ﬃﬃﬃﬃﬃﬃﬃVﬃﬃﬃﬃþﬃﬃﬃﬃﬃQﬃﬃVﬃﬃXﬃﬃﬃﬃﬃﬃﬃ2ﬃﬃﬃþﬃﬃﬃﬃﬃﬃﬃﬃPﬃﬃVﬃXﬃﬃﬃﬃﬃﬃﬃ2ﬃﬃ#ﬃﬃ ¼ tuuvﬃ\"ﬃﬃﬃﬃﬃﬃ1ﬃﬃﬃþﬃﬃﬃﬃﬃ0ﬃﬃ:ﬃﬃ2ﬃﬃﬃÂﬃ1ﬃﬃﬃﬃ0ﬃﬃ:ﬃﬃ5ﬃﬃﬃﬃﬃﬃ2ﬃﬃﬃþﬃﬃﬃﬃﬃﬃﬃﬃ0ﬃﬃ:ﬃ5ﬃﬃﬃﬃÂﬃ1ﬃﬃﬃﬃ0ﬃﬃ:ﬃ5ﬃﬃﬃﬃﬃﬃ2ﬃﬃ#ﬃﬃﬃ ¼ 1:105 p:u:at an angle of 5.19 to the inﬁnite busbar. j 0.25 p.u. A j 0.5 p.u. V constant Large system E1 Lines Lines assumed infinite Xs = j 1.5 p.u. P = 0.5 p.u. busbar Load at 1 p.u. Q = 0.2 p.u. voltage Z 0.5 p.u. MVA at 0.8 p.f. lagging Hence Z = 2 -36.8º p.u. = 1.6 - j 1.2 p.u. 1 j 1.75 A j 0.5 p.u. (0.5 + j 0.2) p.u. E1 IR 2 Z = 2 - 36.8o V = 1 p.u. Figure 8.2 Circuit for Example 8.1

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System Stability 287The reactive power absorbed by the line from point A to point 2 (the inﬁnite busbar) is IR2 X ¼ P2 Vþ2Q2X ¼ 0:52 þ 0:220:5 12 ¼ 0:145 p:uThe actual load taken by A (if represented by an impedance) is given by V2A ¼ 1:1052 ¼ 0:49 þ j0:37 p:u Z 2ﬀ À 36:8The total load supplied by link from generator to A ¼ ð0:5 þ 0:49Þ þ jð0:2 þ 0:145 þ 0:37Þ ¼ 0:99 þ j0:715 p:uInternal voltage of generator E1 ¼ uvtuﬃ\"ﬃﬃﬃﬃﬃﬃ1ﬃﬃ:ﬃ1ﬃﬃﬃ0ﬃﬃ5ﬃﬃﬃﬃþﬃﬃﬃﬃﬃ0ﬃﬃ:ﬃ7ﬃﬃﬃ1ﬃ1ﬃ5ﬃ:ﬃ1ﬃﬃÂﬃ0ﬃﬃ5ﬃ1ﬃﬃﬃ:ﬃ7ﬃﬃ5ﬃﬃﬃﬃﬃﬃ2ﬃﬃﬃþﬃﬃﬃﬃﬃﬃﬃﬃﬃ0ﬃﬃ:ﬃ9ﬃﬃﬃ19ﬃﬃ:ﬃÂ1ﬃﬃﬃ0ﬃﬃ15ﬃﬃ:ﬃﬃ7ﬃﬃ5ﬃﬃﬃﬃﬃﬃ2ﬃﬃ#ﬃﬃ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ 2:73ﬀ35:02 ½5:006 þ 2:458Hence, the angle between E1 and V is 35:02 þ 5:19 ¼ 40:21 Since this angle is much less than 90, the system is stable. (Note that the formula P ¼ (EV/X) sin d is not valid here as the load at A provides aresistance in the equivalent network. If the angle between E, and V by the approximatecalculation shown above approaches 90, a computer based calculation is called for.)8.4 Transient StabilityTransient stability is concerned with the effect of large disturbances. These are usuallydue to faults, the most severe of which is the three-phase short circuit which governsthe transient stability limits in the UK. Elsewhere, limits are based on other types offault, notably the single-line-to-earth fault which is by far the most frequent in practice. When a fault occurs at the terminals of a synchronous generator the real poweroutput of the machine is greatly reduced as the voltage at the point of faultapproaches zero and the only load on the machine is that of the inductive circuits ofthe generator. However, the input power to the generator from the turbine has nottime to change during the short period of the fault and the rotor gains speed to storethe excess energy. If the fault persists long enough the rotor angle will increase con-tinuously and synchronism will be lost. Hence the time of operation of the protec-tion and circuit breakers is all important.

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288 Electric Power Systems, Fifth EditionAn aspect of importance is the use of auto-reclosing circuit breakers. These openwhen the fault is detected and automatically reclose after a prescribed period (usu-ally less than 1 s). If the fault persists the circuit breaker reopens and then recloses asbefore. This is repeated once more, when, if the fault still persists, the breakerremains open. Owing to the transitory nature of most faults, often the circuit breakersuccessfully recloses and the rather lengthy process of investigating the fault andrestoring the line is avoided. The length of the auto-reclose operation must be con-sidered when assessing transient stability limits; in particular, analysis must includethe movement of the rotor over this period and not just the ﬁrst swing. ddIf, in equation (8.2), both sides are multiplied by 2 dt then dd d2d ¼ d dd2 ¼ 2 DP dd 2 dt dt2 dt dt M dt ;dddt2 2 Zd ð8:3Þ M DPdd ¼ d0If the machine remains stable during a system disturbance, the rotor swings until itsangular velocity dd is zero; if dd does not become zero the rotor will continue to move dt dt Rand synchronism is lost. The integral of DPdd in Equation (8.3) represents an area onthe P-d diagram. Hence the criterion for stability is that the area between the P-d curveand the line representing the power input P0 must be zero. This is known as the equal-area criterion. It should be noted that this is based on the assumption that synchronismis retained or lost on the ﬁrst swing or oscillation of the rotor, which may not alwaysbe the case. Physically, the criterion means that the rotor must be able to return to thesystem all the energy gained from the turbine during the acceleration period.A simple example of the equal-area criterion may be seen by an examination ofthe switching out of one of two parallel lines which connect a generator to an inﬁnitebusbar (Figure 8.3). Initially the generator delivers P0 at an angle d0 through bothlines. When one line is switched out, the reactance of the circuit and hence the angled increases. If stability is retained, the two shaded areas (A1 and A2) are equal andthe swinging rotor comes initially to rest at angle d2, after which the damped oscilla-tion converges to d1. In this particular case the initial operating power and anglecould be increased to such values that the shaded area between d0 and d1 (A1) couldbe equal to the area between d1 and d3, where d3 ¼ 180 À d1; this would be the condi-tion for maximum input power. If it swings beyond d3, the rotor continues to accel-erate and instability results.The power-angle curves for the condition of a fault on one of two parallel lines areshown in Figure 8.4. The fault is cleared when the rotor has swung to d1 and theshaded area d0 to d1 (A1) indicates the energy stored. The rotor continues to swinguntil it reaches d2 when the two areas A1 and A2 are equal. In this particular case P0is the maximum operating power for a fault clearance time corresponding to d1 and,conversely, d1 is the critical clearing angle for P0. If the angle d1 is decreased (for

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System Stability 289Power P Po A2 Power - angle curve (2 lines) A1 Power - angle curve (1 line) Infinite busbar 0 δ0 δ1 δ2 δ3 180o δFigure 8.3 Power-angle curves for one line and two lines in parallel. Equal-areacriterion. Resistance neglected Pm Pre - fault (2 lines) Infinite busbar P2 Post - fault (1 line)Power P A2 Po A1 P1 Fault Power - angle curve with fault 0 δ0 δ1 δ2 180o δFigure 8.4 Fault on one line of two lines in parallel. Equal-area criterion. Resistanceneglected. d1 is critical clearing angle for input power P0

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290 Electric Power Systems, Fifth Editionexample, by faster clearance of the fault) it is possible to increase the value of P0without loss of synchronism. The general case where the clearing angle d1 is not critical is shown in Figure 8.5.Here, the rotor swings to d2, where the shaded area from d0 to d1 (A1) is equal to thearea d1 to d2 (A2). The time corresponding to the critical clearing angle is called the critical clearingtime for the particular (normally full-load) value of power input. The time is of greatimportance to protection and switchgear engineers as it is the maximum time allow-able for their equipment to operate without instability occurring. The critical clear-ing angle for a fault on one of two parallel lines may be determined as follows: Applying the equal-area criterion to Figure 8.5: Zd1 Zd2 ðP0 À P1sin dÞdd þ ðP0 À P2sin dÞdd ¼ 0 d0 d1 ;½P0d þ P1cos ddd01 þ ½P0d þ P2cos ddd12 ¼ 0from which cos d1 ¼ P0ðd0 À d2Þ þ P1cos d0 À P2cos d2 ð8:4Þ P1 À P2and the critical clearing angle is obtainedIf d1 is the critical clearing angle then it may be seen that P0 d2 ¼ 180 À sinÀ1 P2 Pm P2 Power - angle curve (2 lines) Post - fault curve (1 line) Power P A2 Po P1 A1 With fault 0 δ0 δ1 δ2 180o δ Figure 8.5 Situation as in Figure 8.4, but d1 not critical

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System Stability 291 It should be noted that a three-phase short circuit on the generator terminals or ona closely connected busbar absorbs zero power and prevents the generator output-ting any real power to the system. Consequently, P1 ¼ 0 in Figures 8.4 and 8.5.Example 8.2A generator operating at 50 Hz delivers 1 p.u. power to an inﬁnite busbar through a net-work in which resistance may be neglected. A fault occurs which reduces the maximumpower transferable to 0.4 p.u., whereas before the fault this power was 1.8 p.u. and afterthe clearance of the fault it is 1.3 p.u. By the use of the equal-area criterion, determinethe critical clearing angle.SolutionThe appropriate load-angle curves are shown in Figure 8.4. P0 ¼ 1 p.u., P1 ¼ 0.4 p.u.,P2 ¼ 1.3 p.u., and Pm ¼ 1.8 p.u. 1 d0 ¼ sinÀ1 1:8 ¼ 33:8 electrical degrees 1 d2 ¼ 180 À sinÀ1 1:3 ¼ 180 À 50:28 ¼ 129:72 electrical degreesApplying equation (8.4) (note that electrical degrees must be expressed in radians) cosd1 ¼ 1ð0:59 À 2:96Þ þ 0:4 Â 0:831 À 1:3 Â ðÀ0:64Þ 0:4 À 1:3 ¼ 0:562 ;d1 ¼ 55:8 In a large system it is usual to divide the generators and spinning loads into asingle equivalent generator connected to an inﬁnite busbar or equivalent motor. Themain criterion is that the rotating machines should be electrically close when form-ing an equivalent generator or motor. If stability with faults in various places isinvestigated, the position of the fault will decide the division of machines betweenthe equivalent generator and motor. A power system (including generation) at thereceiving end of a long line would constitute an equivalent motor if not largeenough to be considered an inﬁnite busbar.8.4.1 Reduction to Simple SystemWith a number of generators connected to the same busbar the inertia constant (H)of the equivalent machine is: He ¼ H1 S1 þ H2 S2 . . . þ Hn Sn Sb Sb Sbwhere S1 . . . Sn ¼ MVA of the machines and Sb ¼ system base MVA. For example, consider six identical machines connected to the same busbar, eachhaving an H of 5 MWs/MVA and rated at 60 MVA. Making the system base MVA

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292 Electric Power Systems, Fifth Editionequal to the combined rating of the machines (360 MVA), the inertia constant of theequivalent coherent machine is: He ¼ 5 Â 60 Â 6 ¼ 5 MWs=MVA 360 It is important to remember that the inertia of the spinning loads must beincluded; normally, this will be the sum of the inertias of the induction motors andtheir mechanical loads. Two synchronous machines connected by a reactance may be reduced to oneequivalent machine feeding through the reactance to an inﬁnite busbar system. Theproperties of the equivalent machine are found as follows. The equation of motion for the two-machine system is: d2d DP1 DP2 1 dt2 M1 M2 1 M2 ¼ À ¼ M1 þ ðP0 À Pm sin dÞwhere d is the relative angle between the machines. Note that DP1 ¼ ÀDP2 ¼ P0 À Pm sin d ð8:5Þwhere P0 is the input power and Pm is the maximum transmittable power. Consider a single generator of Me with the same power transmitted to the inﬁnitebusbar system as that exchanged between the two synchronous machines. Then, Me d2d ¼ P0 À Pm sin d dt2therefore, Me ¼ M1M2 M1 þ M2 This equivalent generator has the same mechanical input as the ﬁrst machine andthe load angle d it has with respect to the busbar is the angle between the rotors ofthe two machines. Often, the maximum powers transferable before, during, and after a fault need tobe calculated from the system conﬁguration reduced to a network between the rele-vant generators. The use of network reduction by nodal elimination is most valuablein this context; it only remains then for the transfer reactances to be calculated, asany shunt impedance at the reduced nodes does not inﬂuence the powertransferred. With unbalanced faults more power is transmitted during the fault period thanwith three-phase short circuits and the stability limits are higher.

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System Stability 2938.4.2 Effect of Automatic Voltage Regulators and GovernorsThese may be represented in the equation of motion as follows, M d2d þ Kd dd ¼ ðPmech À DPmechÞ À Pelec ð8:6Þ dt2 dtwhere: Kd ¼ damping coefﬁcient; Pmech ¼ power input;DPmech ¼ change in input power due to governor action; Pelec ¼ electrical power output modiﬁed by the voltage regulator.Equation (8.6) is best solved by digital computer.8.5 Transient Stability–Consideration of Time8.5.1 The Swing CurveIn the previous section, attention has been mainly directed towards the determinationof the angular position of the rotor; in practice, the corresponding times are moreimportant. The protection engineer requires allowable times rather than angles whenspecifying relay settings. The solution of equation (8.1) with respect to time is per-formed by means of numerical methods and the resulting time-angle curve is knownas the swing curve. A simple step-by-step method will be given in detail and refer-ences will be made to more sophisticated methods used for digital computation. In this method the change in the angular position of the rotor over a short timeinterval is determined. In performing the calculations the following assumptionsare made:1. The accelerating power DP at the commencement of a time interval is considered to be constant from the middle of the previous interval to the middle of the inter- val considered.2. The angular velocity is constant over a complete interval and is computed for the middle of this interval.These assumptions are probably better understood by reference to Figure 8.6.From Figure 8.6, d2d dt2 v 1 À v 3 ¼ Dt ¼ DPnÀ1 Dt 2 2 M nÀ nÀThe change in d over the (n À 1)th interval, that is from times (n À 2) to (n À 1) ¼ dnÀ1 À dnÀ2 ¼ DdnÀ1 ¼ vnÀ 3Dt 2as v 3 is assumed to be constant 2 nÀ

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294 Electric Power Systems, Fifth Edition (a) ΔPn-2 ΔPn-1 ΔPn ΔP n-2 n-1 n Δt Δt Time Angular velocity with respect to (b) synchronous ω ωωnn--1322 n- 3 n-12 2 Time Angle change (c) Δδn Δδn-1 n-2 n-1 n TimeFigure 8.6 (a), (b), and (c) Variation of DP, v and Dd with time. Illustration of step-by-step method to obtain d-time curve

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System Stability 295 Discontinuity ΔP n-2 n-1 n Time Figure 8.7 Discontinuity of DP in middle of a period of timeOver the nth interval, Ddn ¼ Ddn À DdnÀ1 ¼ v 1DtFrom the above, nÀ 2 Ddn À DdnÀ1 ¼ Dt vnÀ 1 À v 3 2 2 nÀ ¼ Dt:Dt: DPnÀ1 M ;Ddn ¼ DdnÀ1 þ DPnÀ1 ðDtÞ2 ð8:7Þ M It should be noted that Ddn and DdnÀ1 are the changes in angle. Equation (8.7) is the basis of the numerical method. The time interval Dt usedshould be as small as possible (the smaller Dt, however, the larger the amount oflabour involved), and a value of 0.05 s is frequently used. Any change in the opera-tional condition causes an abrupt change in the value of DP. For example, at thecommencement of a fault (t ¼ 0), the value of DP is initially zero and then immedi-ately after the occurrence it takes a deﬁnite value. When two values of DP apply, themean is used. The procedure is best illustrated by an example.Example 8.3In the system described in Example 8.2 the inertia constant of the generator plus turbineis 2.7 p.u. Obtain the swing curve for a fault clearance time of 125 ms.Solution H ¼ 2:7 p:u:; f ¼ 50 Hz; G ¼ 1 p:u: ;M ¼ HG ¼ 3 Â 10À4 p:u: 180f

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296 Electric Power Systems, Fifth EditionA time interval Dt ¼ 0.05 s will be used. Hence ðDtÞ2 ¼ 8:33 MThe initial operating angle 1 d0 ¼ sinÀ1 1:8 ¼ 33:8Just before the fault the accelerating power DP ¼ 0. Immediately after the fault, DP ¼ 1 À 0:4 sin d0 ¼ 0:78 p:u: The ﬁrst value is that for the middle of the preceding period and the second is for themiddle of the period under consideration. The value to be taken for DP at the com-mencement of this period is (0.78/2), that is 0.39 p.u. At t ¼ 0, d ¼ 33.8. Dt1 ¼ 0:05 s; DP ¼ 0:39 p:u: Ddn ¼ DdnÀ1 þ ðDtÞ2 DPnÀ1 M;Ddn ¼ 0 þ 8:33 Â 0:39 ¼ 3:25;d0:05 ¼ 33:8 þ 3:25 ¼ 37:05Dt2 DP ¼ 1 À 0:4 sin 37:05 ¼ 0:76 p:u: Ddn ¼ 3:25 þ 8:33 Â 0:76 ¼ 9:58;d0:1 ¼ 37:05 þ 9:58 ¼ 46:63Dt3 DP ¼ 1 À 0:4 sin 46:63 ¼ 0:71 p:u: Ddn ¼ 9:58 þ 8:33 Â 0:71 ¼ 15:49;d0:15 ¼ 46:63 þ 15:49 ¼ 62:12 The fault is cleared after a period of 0.125 s. As this discontinuity occurs in the mid-dle of a period (0.1–0.15 s), no special averaging is required (Figure 8.7). If, on the other hand, the fault is cleared in 0.15 s, an averaging of two values wouldbe required.

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System Stability 297 Angle (electrical degrees) 200 First swing 10 Second swing 150 100 50 Steady state Backswing reached 0 24 6 8 Time (s) Figure 8.8 Typical swing curve for generatorFrom t ¼ 0.15 s onwards, P ¼ 1 À 1.3 sin d (note change to P-d curve of Figure 8.5)Dt4 DP ¼ 1 À 1:3 sin 62:12 ¼ À0:149 p:u: Ddn ¼ 15:49 þ 8:33 Â ðÀ0:149Þ ¼ 14:25 ;d0:2 ¼ 62:12 þ 14:25 ¼ 76:37Dt5 DP ¼ 1 À 1:3 sin 76:4 ¼ À0:26 p:u: Ddn ¼ 14:25 þ 8:33 Â ðÀ0:26Þ ¼ 12:08 ;d0:25 ¼ 76:37 þ 12:08 ¼ 88:39Dt6 DP ¼ 1 À 1:3 sin 88:390 ¼ À0:3 p:u: Ddn ¼ 12:08 þ 8:33 Â ðÀ0:3Þ ¼ 9:59 ;d0:3 ¼ 78:39 þ 9:59 ¼ 97:98 If this process is continued, d commences to decrease and the generator remainsstable. If computed by hand, a tabular calculation is recommended, as shown in Table 8.1 This calculation should be continued for at least the peak of the ﬁrst swing, but ifswitching or auto-reclosing is likely to occur somewhere in the system, the calculationof d needs to be continued until oscillations are seen to be dying away. A typical swingcurve shown in Figure 8.8 illustrates this situation. Different curves will be obtained for other values of clearing time. It is evident fromthe way the calculation proceeds that for a sustained fault, d will continuously increaseand stability will be lost. The critical clearing time should be calculated for conditionswhich allow the least transfer of power from the generator. Circuit breakers and theassociated protection operate in times dependent upon their design; these times can bein the order of a few cycles of alternating voltage. For a given fault position a faster

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298 Electric Power Systems, Fifth EditionTable 8.1 Tabular calculation of dn ÀDt2Át(s) DP M Á DP Ddn dn0À 0.00 — — 33.80þ 0.78 — — 33.8 0.39 3.25 3.25 37.05 0.05 0.76 6.33 9.58 46.63 0.1 0.71 5.91 15.49 62.12 0.15 À0.149 À1.24 14.25 76.37 0.2 À0.26 À2.17 12.08 88.39 0.25 ... Pre - fault power P0 p.u. 2.0 1.6 1.2 0.8 0.4 0.4 0.8 1.2 Critical clearing time (s) Figure 8.9 Typical stability boundaryclearing time implies a greater permissible value of input power P0. A typical relation-ship between the critical clearing time and input power is shown in Figure 8.9 – this isoften referred to as the stability boundary. The critical clearing time increases withincrease in the inertia constant (H) of turbine generators. Often, the ﬁrst swing of themachine is sufﬁcient to indicate stability.8.6 Transient Stability Calculations by ComputerIt is obvious that a digital computer program can be readily written to carry out thesimple studies of Section 8.5. If a load-ﬂow program is readily available, thenimproved accuracy will be obtained if, for each value of dn, the actual power output ofthe generators is calculated. At the same time, the effect of the excitation system andthe governor movement can be included. Such calculations make use of numericalintegration packages based on mathematical concepts. Techniques such as trapezoidalintegration provide fast and sufﬁciently accurate results for many stability studies;more accurate techniques, such as Runge-Kutta (fourth order), predictor-corrector

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System Stability 299routines, and so on, can be employed if the improved accuracy and longer run timescan be economically justiﬁed. Most commercial stability programs offer variousoptions for inclusion of generator controls, system switching and reclosing, compensa-tor modelling, and transformer tap-change operation, according to some input crite-ria. Packages dealing with 1000 generators, 2000 lines, and 1500 nodes are available.Example 8.4Consider the network of Figure 8.10. Data for generators, transformers, lines and loadsare given in Tables 8.2-8.5 T1 BUS1 LN1 BUS2 T2G1 20/220kV LN2 220/20kV LN3 G2 L1 BUS3 T3 L2 20/220kV G3 Figure 8.10 Network for Example 8.4Table 8.2 Generator data G1 G2 G3Rating (MVA) 750 750 250 1.7 1.7 1.6Xs (p.u) 0.05 0.05 0.06Rs (p.u) 0.35 0.35 0.3Xd0(p.u) 0.25 0.25 0.25Xd00 (p.u) 8.0 8.0 8.0Td0 (s) 0.03 0.03 0.03Td00 (s) 6.5 6.5 6.0H (s) Table 8.3 Transformer data T1 T2 T3 Rating (MVA) 750 750 250 Xl (p.u) 0.15 0.15 0.12 R (p.u) 0 0 0

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300 Electric Power Systems, Fifth Edition Table 8.4 Line data LN1 LN2 LN3 LN4 X (V) 115 115 115 9 R (V) 11 11 11 0.9 Y (mS) 1450 1450 1450 115 Table 8.5 Load data L1 L2 P (MW) 600 1000 Q (MVAr) 50 100 The network shown in Figure 8.10 was implemented in the IPSA computer simula- tion package. First a fault applied to Bus 3 at 2 sec and cleared after 80 ms (less than the critical clearance angle). The angle of G1 with respect to that of G2 and power through line LN1 is shown in Figure 8.11(upper trace). Similar results for a clearance time of 600 ms (greater than the critical clearance angle) is shown in Figure 8.11(lower trace).Rotor angle of G1 (deg) 25 120 Power through LN1 (MW) 100 20 5 10 15 5 10 15 20 Time (s) 80 Time (s) 15 60Rotor angle of G1 (deg) 1 2 34 40 Power through LN1 (MW) 10 Time (s) 20 1234 5 5 0 Time (s) 20 0 0 0 150 100 200 150 50 100 0 50 -50 0 -100 50 -50 -100 -150 -200 0Figure 8.11 Angle of generator G1 and power through line LN1 before and aftera fault at BUS 3. For 80 ms clearance time (upper trace). For 400 ms clearancetime (lower trace)

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System Stability 301As can be seen from Figure 8.11(upper trace) the angle between the two generatorsswings but comes back to a stable point after about 25 ms. When the fault clearancetime is longer than the critical clearance time, the machines lose synchronism and poleslipping occurs, as shown in Figure 8.11(lower trace).8.7 Dynamic or Small-Signal StabilityThe power system forms a group of interconnected electromechanical elements, themotion of which may be represented by the appropriate differential equations. Withlarge disturbances in the system the equations are non-linear, but with smallchanges the equations may be linearized with little loss of accuracy. The differentialequations having been determined, the characteristic equation of the system is thenformed, from which information regarding stability is obtained. The solution of thedifferential equation of the motion is of the form d ¼ k1ea1t þ k2ea2t þ . . . . . . kneantwhere k1, k2,.,kn are constants of integration and a1, a2,.,an are the roots of thecharacteristic equation as obtained through the well-known eigenvalue methods.If any of the roots have positive real terms then the quantity d increases continu-ously with time and the original steady condition is not re-established. The crite-rion for stability is therefore that all the real parts of the roots of the characteristicequation, that is eigenvalues, be negative; imaginary parts indicate the presenceof oscillation. Figure 8.12 shows the various types of motion. The determination UnstableMovement Decay envelope Stable with damped oscillation Disturbance, timeFigure 8.12 Types of response to a disturbance on a system

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302 Electric Power Systems, Fifth Edition ΔPPower P P0 ΔP Δδ Δδ 0 δ 0 Angle δFigure 8.13 Small disturbance–initial operation on power-angle curve at P0,d0. Linearmovement is assumed about P0,d0.of the roots is readily obtained through an eigenvalue analysis package, butindirect methods for predicting stability have been established, for example theRouth-Hurwitz criterion in which stability is predicted without the actual solu-tion of the characteristic equation. No information regarding the degree of stabil-ity or instability is obtained, only that the system is, or is not, stable. Oneadvantage of using eigenvalues is that the characteristics of the control loopsassociated with governors and automatic voltage regulators may be incorporatedin the general treatment. For a generator connected to an inﬁnite busbar through a network of zero resist-ance it has been shown in Equation (3.1) that P ¼ VE sin d: XWith operation at P0 and d0 (Figure 8.13), we can write d2Dd dt2 @P M ¼ ÀDP ¼ ÀDd @d 0where the change in output power P causing an increase in d is positive and refers tosmall changes in the load angle d such that linearity may be assumed. @P Ms2Dd þ @d Dd ¼ 0 ð8:8Þ 0Where

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System Stability 303 s ¼ d dtHere, Ms2 þ (dP/dd)0 ¼ 0 is the characteristic equation which has two roots ÆuutvÀﬃﬃﬃﬃﬃﬃﬃ@ﬃﬃﬃPﬃMﬃ.ﬃﬃﬃ@ﬃﬃﬃdﬃﬃﬃﬃﬃ0ﬃﬃWhen (dP/dd)0 is positive, both roots are imaginary and the motion is oscillatoryand undamped; when (dP/dd)0 dis¼n9e0ga, tiv@eP,.b@odth90ro¼ot0s,aarnedrethale, one positive and onenegative and stability is lost. At system is at the limit.If damping is accounted for, the equation becomes @P Ms2Dd þ KsDd þ @d Dd ¼ 0 ð8:9Þ 0and the characteristic equation is Ms2 þ Ks þ @P ¼0 ð8:10Þ @d 0 s1; s2 ¼ ÀK Æ rﬃKﬃﬃﬃ2ﬃﬃﬃÀﬃﬃﬃﬃﬃ4ﬃﬃMﬃﬃﬃﬃﬃﬃﬃﬃ@ﬃﬃPﬃﬃ.ﬃﬃﬃﬃ@ﬃﬃdﬃﬃﬃﬃﬃ 2Mwhere K is the damping coefﬁcient, assumed to be constant, independent of d.Again, if @P is negative, stability is lost. The frequency of the oscillation is given @dby the roots of the characteristic equation.If the excitation of the generator is controlled by a fast-acting automatic voltageregulator without appreciable dead zone, the excitation voltage E is increased asincrements of load are added. Hence the actual power-angle curve is no longerthat for constant E (refer to Chapter 3) and the change of power may be obtained bylinearizing the P-V characteristic at the new operating point (1), when dP DP ¼ dE DE 1The complete equation of motion is now @P @P Ms2Dd þ KsDd þ @d Dd þ @E DE ¼ 0 ð8:11Þ 1 1 Without automatic voltage control the stability limit is reached when d ¼ 90; withexcitation control the criterion is obtained from the characteristic equation of (8.11).Example 8.5A synchronous generator of reactance 1.5 p.u. is connected to an inﬁnite busbar system(V ¼ 1 p.u.) through a line and transformers of total reactance 0.5 p.u. The no-load

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304 Electric Power Systems, Fifth Editionvoltage of the generator is 1.1 p.u. and the inertia constant H ¼ 5 MWs per MVA. All perunit values are expressed on the same base; resistance and damping may be neglected.Calculate the frequency of the oscillations set up when the generator operates at a loadangle of 60 and is subjected to a small disturbance. The system frequency is 50 Hz.SolutionThe nature of the movement (iesqguoavteiornne(d8.b1y0)t)h. eThsiigsnchoafnthgeesqwuahnetnityK2un¼d4eMr t[email protected]@otdsi;ginnin the equation for s1 and s2this example K ¼ 0 and the motion is undamped.The roots of the characteristic equations give the frequency of oscillation; whend0 ¼ 60, @P 1:1 Â 1 cos60 @d ¼ 2 60 ¼ 0:275 p:u: rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ @P 1 s1 and s2 ¼ Æj @d Á M ¼ Æjuutvﬃ5ﬃﬃﬃÂﬃﬃ0ﬃﬃ:ﬃpﬃ2ﬃﬃ7ﬃÂﬃ15ﬃﬃﬃ5ﬃﬃﬃ0ﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ Æj 8:64 ¼ 2:94 rad=sTherefore frequency of oscillation ¼ 2:94 ¼ 0:468 Hz 2pand the periodic time ¼ 1 ¼ 2:14 s 0:4688.7.1 Effects of Governor ActionIn the above analysis the oscillations set up with small changes in load on asystem have been considered and the effects of governor operation ignored.After a certain time has elapsed the governor control characteristics commenceto inﬂuence the powers and oscillations, as explained in Section 4.3. It is nowthe practice to represent both the excitation system and the governor systemwith the dynamic equations of the generator in the state-space form, fromwhich the eigenvalues of the complete system with feedback can be determined.Using well-established control design techniques, appropriate feedback pathsand time constants can be established for a range of generating conditions anddisturbances, thereby assuring adequate dynamic stability margins. Further

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System Stability 305 jX IV E P = f1 (V ) QS Q = f2 (V )Figure 8.14 System with a load dependent on voltage as follows: P ¼ f1(V) and Q ¼f2(V); Qs ¼ supply of VArs from E ¼ Q þ I2X; E ¼ supply voltageinformation on these design processes can be found in advanced controltextbooks.8.8 Stability of Loads Leading to Voltage CollapseIn Chapter 5 the power-voltage characteristics of a line supplying a load were con-sidered. It was seen that for a given load power-factor, a value of transmitted powerwas reached, beyond which further decreases of the load impedance producegreatly reduced voltages, that is voltage instability (Figure 5.21). If the load is purelystatic, for example represented by an impedance, the system will operate stably atthese lower voltages. Sometimes, in load-ﬂow studies, this lower voltage conditionis unknowingly obtained and unexpected load ﬂows result. If the load contains non-static elements, such as induction motors, the nature of the load characteristics issuch that beyond the critical point the motors will run down to a standstill or stall.It is therefore of importance to consider the stability of composite loads that willnormally include a large proportion of induction motors.The process of voltage collapse may be seen from a study of the V-Q character-istics of an induction motor under load, from which it is seen that below a certainvdQol.tage the reactive power consumed increases with decrease in voltage until dV ! 1, when the voltage collapses. In the power system the problem arisesowing to the impedance of the connection between the load and inﬁnite busbar andis obviously aggravated when this impedance is high, that is connection is electri-cally weak. The usual cause of an abnormally high impedance is the loss of one lineof two or more forming the connection. It is proﬁtable, therefore, to study the pro-cess in its basic form – that of a load supplied through a reactance from a voltagesource (Figure 8.14). dP. ddAlready, two criteria for load instability have been given, that is ¼ 0 anddQ. ! 1; from the system viewpoint, voltage collapse takes place whendE.dV ¼ 0 or dV. ! 1. Each value of E yields a corresponding value for V, and dV dEthe plot of E-V is shown in Figure 8.15; also the plot of V-X for various values of E isshown in Figure 8.16. In these graphs the critical operating condition is clearly

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306 Electric Power Systems, Fifth Edition 1.6E p.u. 1.4 dE + dE dV dV Vcr 1.2 1.0 0.7 0.8 0.9 1.0 1.1 1.2 V p.u.Figure 8.15 The E-V relationship per system in Figure 8.14. Vcr ¼ critical voltage afterwhich instability occursshown and the improvement produced by higher values of E is apparent, indicatingthe importance of the system-operating voltage from the load viewpoint. In the circuit shown in Figure 8.14, and from Equation (2.17),E ¼ V þ QX if PX ( V2 þ QX V V V dE dQ 1; dV ¼ 1 þ dV :XV À QX V2which is zero at the stability limit and negative in the unstable region. E1 E2 E3 E4 V 0 Xcr XFigure 8.16 The V-X relationship. Effect of change in supply voltage E. Xcr ¼ criticalreactance of transmission link

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System Stability 307 ð8:12ÞAt the limit, dE ¼ 0 and dQ ¼ QX À 1 V ¼ QÀV dV dV V2 X VXAlso from Equation (2.17), Q ¼ E À V V X X ; dQ ¼ E À 2V dV X XExample 8.6A load is supplied from a 275 kV busbar through a line of reactance 70 V phase-to-neu-tral. The load consists of a constant power demand of 200 MW and a reactive powerdemand Q which is related to the load voltage V by the equation: ðV À 0:8Þ2 ¼ 0:2ðQ À 0:8Þ Q ¼ 5ðV À 0:8Þ2 þ 0:8 This is shown in Figure 8.17, where the base quantities for V and P, Q are 275 kV and200 MVA. Examine the voltage stability of this system, indicating clearly any assumptionsmade in the analysis.Solution E ¼ tuuvﬃ\"ﬃﬃﬃﬃﬃﬃVﬃﬃﬃﬃþﬃﬃﬃﬃﬃQﬃﬃﬃﬃXﬃﬃﬃﬃﬃﬃﬃ2ﬃﬃﬃþﬃﬃﬃﬃﬃﬃﬃﬃPﬃﬃﬃXﬃﬃﬃﬃﬃﬃﬃ2ﬃﬃ#ﬃﬃIt has been shown that VVIf PX ( V þ QX V Vthen QX E¼ V þ Vand dE dQ X QX dV ¼ 1 þ dV V À V2 ¼0

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308 Electric Power Systems, Fifth Edition Q 1.2 p.u. 0.6 p.u. V 1 p.u. 1.2 p.u. 0.8 p.u. 1 p.u. V 0.8 p.u. Figure 8.17 Example 8.6. Reactive power-voltage characteristicIn this problemP ¼ 200 MW, 1 p.u.Q ¼ 200 MVAr, 1 p.u.E ¼ 275 kV, 1 p.u. And X ¼ 70 Â 200 ¼ 0:185 p:u: 2752 Now QX E¼ V þ V À 0:8Þ2 ! 0:2 ;1 ¼ V þ 0:185 ðV þ 0:8 V V2 ¼ V À 0:925 V2 À 0:59 þ 1:48 V À 0:148giving, 1:925 V2 À 2:48 V þ 0:74 ¼ 0 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ;V ¼ þ2:48 Æ 2:482 À 4 Â 1:925 Â 0:74 2 Â 1:925 ¼ 2:48 Æ 0:67 3:85

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System Stability 309Taking the upper value, V ¼ 0.818 p.u.,Q at this voltage is Q ¼ ð0:818 À 0:8Þ2 þ 0:8 ¼ 0:8016 p:u: 0:2 dQ dV ¼ 10ð0:818 À 0:8Þ dE ¼ 1 þ ½10ð0:818 À 0:8Þ Â 0:185 Â 0:818 À 0:8016 Â 0:185 dV 0:8182 0:027 À 0:148 ¼ 1þ 0:8182which is positive, that is the system is stable.(Note: PX/V % 0.16 and (V2 þ QX)/V % 1; therefore the approximation is reasonable.) When the reactance between the source and load is very high, the use of tap-changing transformers is of no assistance. Large voltage drops exist in the supplylines and the ‘tapping-up’ of transformers increases these because of the increasedsupply currents. Hence the peculiar effects from tap-changing noticed when condi-tions close to a voltage collapse have occurred in practice, that is tapping-up reducesthe secondary voltage, and vice versa. One symptom of the approach of critical con-ditions is sluggishness in the response to tap-changing transformers. Studies into the voltage collapse phenomenon in interconnected power systemshave shown how difﬁcult it is to predict its occurrence or, more usefully, estimatethe margin available to collapse at any given operating condition. One approach isto increase all loads at constant power factor until the load ﬂow diverges. Failure ofthe load ﬂow to converge is an indication that voltage collapse may be imminent.Unfortunately, this does not provide a realistic case since (1) any increase in real-power load will result in reserve generation being brought on line at various nodesof the system not previously used for generation input; and (2) as voltage falls, loadsbehave non-linearly (see Figure 8.15) but our knowledge of load behaviour belowabout 0.9 p.u. voltage is extremely sparse. Consequently, only worst-case situationscan be assumed if voltage margins are to be estimated. The voltage collapse phenomenon may be studied using the transient stabilityfacility of many computer programs. In the network shown in Figure 8.10, the loadon BUS 1 was increased in steps of 90 þ j9 MVA. Figure 8.18 shows the voltage col-lapse on that busbar as the load increases.8.9 Further Aspects8.9.1 Faults on the Feeders to Induction MotorsA common cause of the stalling of induction motors (or the low-voltage releasesoperating and removing them from the supply) occurs when the supply voltage is

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Voltage of BUS1 (p.u.)310 Electric Power Systems, Fifth Edition 1.0 0.95 0.9 0.85 0.8 0.75 0.7 0.65 0.6 0 5 10 15 20 25 Time (s) Figure 8.18 Voltage collapse on the network shown in Figure 8.10either zero or very low for a brief period because of a fault on the supply system,commonly known as ‘voltage dip’ or ‘voltage sag’. When the supply voltage isrestored the induction motors accelerate and endeavour to attain their previousoperating condition. In accelerating, however, a large current is taken, and this, plusthe fact that the system impedance has increased due to the loss of a line, results in adepressed voltage at the motor terminals. If this voltage is too low the machines willstall or cut out of circuit.8.9.2 Steady-State Instability Due to Voltage RegulatorsConsider a generator supplying an inﬁnite busbar through two lines, one of which issuddenly removed. The load angle of the generator is instantaneously unchangedand therefore the power output decreases due to the increased system reactance,thus causing the generator voltage to rise. The automatic voltage regulator of thegenerator then weakens its ﬁeld to maintain constant voltage, that is decreases theinternal e.m.f., and pole-slipping may result.8.9.3 Dynamic StabilityThe control circuits associated with generator AVRs, although improving steady-state stability, can introduce problems of poorly damped response and eveninstability. For this reason, dynamic stability studies are performed, that is steady-state stability analysis, including the automatic control features of the machines (seeSection 8.7). The stability is assessed by determining the response to small stepchanges of rotor angle, and hence the machine and control-system equations areoften linearized around the operating point, that is constant machine parametersand linear AVR characteristic. The study usually extends over several seconds ofreal-system time.

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System Stability 3118.10 Multi-Machine SystemsIn networks interconnecting many generators and dynamic loads, it is a complicatedtask, even with a fast digital computer, to solve a set of dynamic equations to estab-lish if a given state of the system is stable or unstable following a credible distur-bance. Obviously, many hundreds of studies may need to be run with differentfaults and the output data assessed in some way. It must be remembered that ifeach dynamic machine (generator or load) is to be represented by its swing trajec-tory, starting from an initial steady-state angle (as in Figure 8.8), then a multitude oftrajectories up to 5 s will be presented by the computer output. Some machines willprobably not be affected by the disturbance and their angles remain within a fewdegrees of their initial angle. Others will show oscillations but their mean anglecould gradually diverge from the more or less unaffected machines – these diverg-ing machines would indicate that the system is unstable and, if continued, wouldsplit up, by action of the interconnecting circuit-protection systems, into two ormore ‘islands’. A common feature of such studies often shows that machine anglesoscillate at different frequencies but that they all gradually change their angles inunison from the initial angles, implying that the system is stable but that, subse-quent to the initiating disturbance, the frequency of the whole system is eitherincreasing or decreasing. To enable a digital study to assess its own results, criteria need to be built into thesoftware. One of these is whether or not the system angles are within a norm.Another is, in the case of instability, which parts of the system are splitting awayfrom other parts. To achieve this assessment, a concept known as ‘Centre of Inertia’(COI) is employed. This is similar to determining the centre of gravity of a mechani-cal system by writing Xi¼n MtotdCOI ¼ Midi i¼1where Mtot is the total area angular momentum and dCOI is the average angle of themachines in the system. Consequently, we now have, for an area of the system, that Mtot d2 dCOI þ DParea ¼ 0 ðsee equation ð8:1ÞÞ dt2where DParea is the combined power being input (negative if output) into that area.By use of conditional statements in the software used for post-analysis of a stabilitystudy, areas of the system which swing together, that is whose angles are withinspeciﬁed limits around the dCOI (known as coherency), can be identiﬁed. Note alsothat the power transfer across the area boundary DParea before the disturbance canalso be established for each area, once the vulnerable areas are known. To ensurestability, further studies need to be undertaken with revised generator schedulingsuch that critical DParea ﬂows are reduced to a sustainable level. System operators

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312 Electric Power Systems, Fifth Editionshould then observe these area ﬂow limits to ensure stability under credible systemcontingencies. In most systems, by proper design of transient machine controllers orreinforcement of the transfer capability of the vulnerable circuits, only a few criticalareas remain where power transfer is limited by stability considerations. For exam-ple, in the UK, the Scotland-England transfer over two double-circuit lines is oftenlimited in this way; similarly under some conditions, circuits from North Wales tothe rest of the National Grid system are ﬂow-restricted.8.11 Transient Energy Functions (TEF)A fast stability study in large systems is essential to establish viable operating condi-tions. The use of Lyapunov-type functions for this purpose is left for advancedstudy. However, the equal-area criterion (EAC) is a form of energy function whichcan be used as a screening tool to enable a fast assessment of multi-machine stabilityto be made. Figure 8.19 shows three curves similar to those of Figure 8.5. From equation (8.5) we have that Me d2d ¼ P0 À P2 sin d dd2We have also that dd ¼ v dt Pm Pre-fault P2 Post-fault P A3 A2 P0 A1 P1 Faulted system δ0 δP δ1 δ2 Angle δFigure 8.19 Equal-area criterion applied to an equivalent machine connected toinﬁnite busbar through a system

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System Stability 313And dv ¼ P0 À P2 sin d dt Menoting that v is the incremental speed from the steady-state speed vs. At anymoment, the energy in this system consists of two components, namely:Mev the incremental kinetic energy stored in the rotating masses of the machines atany instant;(P0 ÀP2 sin d) the potential energy due to the excess electrical power at any instant. The energy balance in the system during the disturbance period can be obtainedby integrating the two energy components over the period. However, we mustremember that prior to the disturbance under steady-state conditions the energyput into the system equalled the energy being taken out. Our development of theEAC only took into account the energy increments or decrements from steady-stateconditions; therefore we need only concern ourselves with the incremental kineticand potential energy functions. We can compute a value for the total transientenergy in the system, TEF, by using Zv1 Zd1 ½J TEF ¼ Mev dv À ðP0 À P2 sin dÞdd 0 dPwhere v1 is the incremental angular speed at angle d1 and dP is the angle at whichthe system settles down following the disturbance; both angles are shown in Fig-ure 8.19. In fact, for any intermediate value of v or d the value of TEF is an indication thatthere is a surplus or deﬁciency of incremental energy and that the system is still in adynamic or oscillation state. Only when d ¼ dP and v ¼ 0, that is no increment on thesteady-state synchronous speed, will the oscillations have died away and the systembe stable. If the system is to regain a steady-state condition, the area A2 must equal A1, andthe maximum angle attained at the limiting conditions is d2, where the incrementalspeed v would again be zero. Under these conditions we see that Zd1 TEFlimit ¼ À ðP0 À P2 sin dÞdd since v ¼ 0 dP TEFlimit represents the maximum energy that the system can gain due to a distur-bance and subsequently dissipate by transfer over the system, if it is to remain sta-ble. Knowing the value of TEFlimit from the above equation enables a quick

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314 Electric Power Systems, Fifth Editiondetermination of stability by calculating TEF as the disturbance proceeds. This canbe done readily as the step-by-step integration proceeds because values of v and dwill be available. Provided that the value of TEF remains less than TEFlimit the sys-tem is stable and will settle down to a new angle dP. It is usual to compute TEF atfault clearance and assume afterwards that no more energy is added (rather, it isdissipated by system damping and losses), thereby checking that TEF < TEFlimit It is worth noting that in Figure 8.19: vR11. The kinetic energy-like term Mev dv is the area A1. 0 Rd12. The potential energy-like term ðP0 À P2 sindÞdd is the area A3.3. TEF is equal to areas A1 þ A3. dP4. TEFlimit is the area A2 þ A3.5. Equating TEF and TEFlimit produces, A1 þ A3 ¼ A2 þ A3 i:e: A1 ¼ A2which is the Equal Area Criterion for the system. This shows that for stability the total energy gained during a disturbance mustequal the capability of the system to transfer that energy away from the systemafterwards. Considerable effort has been put into the determination of dynamic sys-tem boundaries through improved calculations of the TEF.8.12 Improvement of System StabilityApart from the use of fast-acting AVRs the following techniques are in use:1. Reduction of fault clearance times, 80 ms is now the norm with SF6 circuit break- ers and high-speed protection (see Chapter 11).2. Turbine fast-valving by bypass valving-this controls the accelerating power by closing steam valves. Valves which can close or open in 0.2 s are available. CCGTs can reduce power by fuel control within 0.2 to 0.5 s.3. Dynamic braking by the use of shunt resistors across the generator terminals; this limits rotor swings. The switching of such resistors can be achieved by thyristors.4. High-speed reclosure or independent pole tripping in long (point-to-point) lines. In highly interconnected systems the increase in overall clearance times on unsuccessful reclosure makes this technique of dubious value. Delayed auto- reclose (DAR) schemes with delays of 12–15 s are preferable if the voltage sag can be tolerated.5. Increased use of H.V. direct-current links using thyristors and GTOs also allevi- ates stability problems.6. Semiconductor-controlled static compensators enabling oscillations following a disturbance to be damped out.

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System Stability 3157. Energy-storage devices, for example batteries, superconducting magnetic energy stores (SMES) with fast control, providing the equivalent of a UPS.Problems8.1 A round-rotor generator of synchronous reactance 1 p.u. is connected to a transformer of 0.1 p.u. reactance. The transformer feeds a line of reactance 0.2 p.u. which terminates in a transformer (0.1 p.u. reactance) to the LV side of which a synchronous motor is connected. The motor is of the round-rotor type and of 1 p.u. reactance. On the line side of the generator transformer a three-phase static reactor of 1 p.u. reactance per phase is connected via a switch. Calculate the steady-state power limit with and without the reactor connected. All per unit reactances are expressed on a 10 MVA base and resistance may be neglected. The internal voltage of the generator is 1.2 p.u. and of the motor 1 p.u. (Answer: 5 MW and 3.13 MW for shunt reactor)8.2 In the system shown in Figure 8.20, investigate the steady state stability. All per unit values are expressed on the same base and the resistance of the system (apart from the load) may be neglected. Assume that the inﬁnite busbar voltage is 1 p.u.8.3 A generator, which is connected to an inﬁnite busbar through two 132 kV lines in parallel, each having a reactance of 70 V/phase, is delivering 1 p.u. to the inﬁnite busbar. Determine the parameters of an equivalent circuit, consisting of a single machine connected to an inﬁnite busbar through a reactance, which represents the above system a. Pre-fault. b. When a three-phase symmetrical fault occurs halfway along one line. c. After the fault is cleared and one line isolated. Infinite busbar1 X 0.09 p.u. XL 0.5 p.u. X 0.09 p.u. X 0.1 p.u. 2 XL 0.5 p.u. E1 1.5 p.u. P = 0.8 p.u. X 0.28 p.u. Q = 0.3 p.u. L 0.13 + j0.08 p.u. Figure 8.20 Line diagram of system in Problem 8.2

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316 Electric Power Systems, Fifth Edition If the generator internal voltage is 1.05 p.u. and the inﬁnite busbar voltage is 1.0 p.u, what is the maximum power transfer pre-fault, during the fault and post-fault? Determine the swing curve for a fault clearance time of 125 ms. The generator data are as follows: Rating 60 MW at power factor 0.9 lagging. Transient reactance 0.3 p.u. Inertia constant 3 kWs/kVA. (Answer: 1.62 p.u., 0.6 p.u., 1.24 p.u.)8.4 An induction motor and a generator are connected to an inﬁnite busbar. What is the equivalent inertia constant of the machines on 100 MVA base? Also calculate the equivalent angular momentum. Data for the machines are:Induction motor Rating 40 MVA;Generator: Inertia constant 1 kWs/kVA. Rating 30 MVA; Inertia constant l0 kWs/kVA. (Answer 3.4 Ws/VA, 0.00038 p.u.)8.5 The P-V, Q-V characteristics of a substation load are as follows:V 1.05 1.025 1 0.95 0.9 0.85 0.8 0.75P 1.03 1.017 1 0.97 0.94 0.92 0.98 0.87Q 1.09 1.045 1 0.93 0.885 0.86 0.84 0.85 The substation is supplied through a link of total reactance 0.8 p.u. and negli- gible resistance. With nominal load voltage, P ¼ 1 and Q ¼ 1 p.u. By determin- ing the supply voltage-received voltage characteristic, examine the stability of the system by the use of dE/dV. All quantities are per unit.8.6 A large synchronous generator, of synchronous reactance 1.2 p.u., supplies a load through a link comprising a transformer of 0.1 p.u. reactance and an over- head line of initially 0.5 p.u. reactance; resistance is negligible. Initially, the voltage at the load busbar is 1 p.u. and the load P þ jQ is (0.8 þ j0.6) p.u. regard- less of the voltage. Assuming the internal voltage of the generators is to remain unchanged, determine the value of line reactance at which voltage instability occurs. (Answer: unstable when X ¼ 2.15 p.u.)8.7 A load is supplied from an inﬁnite busbar of voltage 1 p.u. through a link of series reactance 1 p.u. and of negligible resistance and shunt admittance. The

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System Stability 317 load consists of a constant power component of 1 p.u. at 1 p.u. voltage and a per unit reactive power component (Q) which varies with the received voltage (V) according to ðV À 0:8Þ2 ¼ 0:2ðQ À 0:8Þ All per unit values are to common voltage and MVA bases. Determine the value of X at which the received voltage has a unique value and the corresponding magnitude of the received voltage. Explain the signiﬁcance of this result in the system described. Use approxi- mate voltage-drop equations. (Answer: X ¼ 0.25 p.u.; V ¼ 0.67 p.u.)8.8 Explain the criterion of stability based on the equal-area diagram. A synchronous generator is connected to an inﬁnite busbar via a generator transformer and a double-circuit overhead line. The transformer has a reac- tance of 0.15 p.u. and the line an impedance of 0 þ j0.4 p.u. per circuit. The gen- erator is supplying 0.8 p.u. power at a terminal voltage of 1 p.u. The generator has a transient reactance of 0.2 p.u. All impedance values are based on the gen- erator rating and the voltage of the inﬁnite busbar is 1 p.u. a. Calculate the internal transient voltage of the generator. b. Determine the critical clearing angle if a three-phase solid fault occurs on the sending (generator) end of one of the transmission line circuits and is cleared by disconnecting the faulted line. (Answer: (a) 1.035 p.u. (b) 64 (From Engineering Council Examination, 1995)8.9 A 500 MVA generator with 0.2 p.u. reactance is connected to a large power sys- tem via a transformer and overhead line which have a combined reactance of 0.3 p.u. All p.u. values are on a base of 500 MVA. The amplitude of the voltage at both the generator terminals and at the large power system is 1.0 p.u. The generator delivers 450 MW to the power system. Calculate a. the reactive power in MVAr supplied by the generator at the transformer input terminals; b. the generator internal voltage; c. the critical clearing angle for a 3 p.h. short circuit at the generator terminals. (Answer: (a) 62 MVAr; (b) 1.04 p.u.; (c) 84) (From Engineering Council Examination, 1996)

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9Direct-Current Transmission9.1 IntroductionThe established method of transmitting large quantities of electrical energy is to usethree-phase alternating-current. However, there is a limit to the distance that bulka.c. can be transmitted unless some form of reactive compensation is employed. Forlong overhead lines either alternating current with reactive compensation (con-nected in shunt or series) or direct current may be used. If undersea crossingsgreater than around 50 km are required, then, because of the capacitive chargingcurrent of a.c. cables, d.c. is the only option. Figure 9.1 shows the distances at which d.c. becomes cheaper than a.c for over-head line and submarine cable transmission. The terminal converter stations of ad.c. scheme are more expensive than a.c. substations but the overhead line ischeaper. The choice of whether to use a.c. or d.c. is usually made on cost. With theincreasing use of high-voltage, high-current semiconductor devices, converter sta-tions and their controls are becoming cheaper and more reliable, so making d.c.more attractive at shorter distances. The main technical reasons for high-voltage direct-current (h.v.d.c.) transmissionare for the:1. interconnection of two large a.c. systems without having to ensure synchronism and be concerned over stability between them (for example, the UK-France cross- channel link of 2000 MW);2. interconnection between systems of different frequency (for example, the connec- tions between north and south islands in Japan, which use 50 and 60 Hz systems);3. long overland transmission of high powers where a.c. transmission towers, insulators, and conductors are more expensive than using h.v.d.c. (for example, the Nelson River scheme in Manitoba – a total of 4000 MW over more than 600 km).Electric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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320 Electric Power Systems, Fifth Edition Scheme Lines & Stations Cost Break even d.c. d.c. Distance Converter Stations a.c. a.c. 800km 50km Stations Overhead Submarine Transmission Line Circuit Distance Figure 9.1 Costs of d.c. and a.c. transmission The main advantages of h.v.d.c. compared with h.v.a.c. are:1. two conductors, positive and negative to ground, are required instead of three, thereby reducing tower or cable costs;2. the direct voltage can be designed equivalent to the peak opf the alternating volt- age for the same amount of insulation to ground (i.e. Vd.c. ¼ 2Va.c.);3. the voltage stress at the conductor surface can be reduced with d.c., thereby reducing corona loss, audible emissions, and radio interference;4. h.v.d.c. infeeds do not increase signiﬁcantly the short-circuit capacity required of switchgear in the a.c. networks;5. fast control of converters can be used to damp out oscillations in the a.c. system to which they are connected. Disadvantages of h.v.d.c. are:1. the higher cost of converter stations compared with an a.c. transformer substation;2. the need to provide ﬁlters and associated equipment to ensure acceptable wave- form and power factor on the a.c. networks;3. limited ability to form multi-terminal d.c. networks because of the need for coordi- nated controls and the present lack of commercially available d.c. circuit breakers.9.2 Current Source and Voltage Source ConvertersTraditionally high power h.v.d.c. schemes have used thyristors with the d.c. currentalways ﬂowing in the same direction (current source converters). These are usedextensively for point-point transmission of bulk power and are available at d.c. voltageratings of up to Æ800 kV and are able to transmit 6500 MW over a single overhead

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Direct-Current Transmission 321Vac L VacCSC Id VSC C Vd (constant) (constant)(a) Current source converter (b) Voltage source converter Figure 9.2 Commonly used converters for h.v.d.c. schemesh.v.d.c. link. These schemes use converters employing thyristors and a large inductoris connected on the d.c. side. As the inductor maintains the d.c. current more or lessconstant (other than a small ripple) these converters are called current source convert-ers (CSCs) (Figure 9.2(a)). In these converters while the thyristors are triggered on by agate pulse, they turn off when the current through them falls to zero. Thus currentsource converters are also called line (or naturally) commutated converters. Current source converter h.v.d.c. has a number of advantages. It can be made upto very high power and d.c. voltage ratings and the thyristors are comparativelyrobust with a signiﬁcant transient overload capability. As the thyristors switch offonly when the current through them has dropped to zero, switching losses are low. However, it suffers from a number of disadvantages. Both the rectiﬁer andinverter always draw reactive power from the a.c. networks and a voltage source,usually synchronous generation, is required at each end of the d.c. link to ensurecommutation of the valves. When the a.c. voltage drops, due to a fault on the a.c.network, the valves may stop operating and experience commutation failure. Thea.c. current contains signiﬁcant harmonics and, although these can be reduced by 12or even 24 pulse connection of the converters, large harmonic ﬁlters are required(which also provide some of the reactive power). Conventionally, the ﬁlters useopen terminal switchgear and air-insulated busbars and so both the valve hall andthe ﬁlters occupy a large area. Although there have been several three terminal CSCh.v.d.c. schemes constructed, this technology is mainly applied for bulk transfer ofpower between two stable a.c. power systems. Recently large transistors called Insulated Gate Bipolar Transistors (IGBTs) havebeen used in h.v.d.c. systems where the d.c. voltage is always the same polarity andthe current reverses to change the direction of power ﬂow. These schemes are usedin underground and submarine cable links of up to 1000 MW. The converters usedfor these schemes have a large capacitor on the d.c. side thus maintaining the d.c.voltage more or less constant. They are referred to as voltage source converters(VSCs) (Figure 9.2(b)). As IGBTs can be turned on and off by their gate voltage, theseconverters are also called forced commutated converters. VSC h.v.d.c. schemes offer the following advantages. They:1. can operate at any combination of active and reactive power;2. have the ability to operate into a weak grid and even black-start an a.c. network;

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322 Electric Power Systems, Fifth Edition3. have fast acting control;4. can use voltage polarized cables;5. produce good sine wave-shapes in the a.c. networks and thus use small ﬁlters; The main disadvantages of VSC h.v.d.c. schemes are that:1. presently their rating is very much lower than CSC h.v.d.c. schemes.2. their power losses are higher.9.3 Semiconductor Valves for High-Voltage Direct-Current ConvertersThe rapid growth in the use of h.v.d.c. since about 1980 has been due to the develop-ment of high-voltage, high-current semiconductor devices. These superseded the pre-viously used complex and expensive mercury arc valves that employed a mercurypool as cathode and a high-voltage graded column of anodes, with the whole enclosedin steel and ceramic to provide a vacuum tight enclosure. Nowadays, the semi-conductor devices are stacked to form a group which is able to withstand the designvoltages and to pass the desired maximum currents – this group is termed a ‘valve’.9.3.1 ThyristorsThyristors are manufactured from silicon wafers and are four-layer versions of thesimple rectiﬁer p-n junction, as shown in Figure 9.3(a). The p layer in the middle isconnected to a gate terminal biased such that the whole unit can be prevented frompassing current, even when a positive voltage exists on the anode. By applying apositive pulse to the gate, conduction can be started, after which the gate controlhas no effect until the main forward current falls below its latching value Anode I I Forward on-stateP Ig V voltage dropN ReverseP Gate (b) leakage current LatchingN Ig1 Ig2 current Cathode Forward V breakdown(a) voltage Forward Reverse leakage breakdown current voltage 1% of rated (c)Figure 9.3 (a) Structure of a four-layer thyristor. (b) Symbol, (c) Thyristor characteristic:Ig gate current to switch thyristor on at forward voltage

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Direct-Current Transmission 323(see Figure 9.3(c)). This current must be kept below the latching value for typically100 ms before the thyristor is able to regain its voltage hold-off properties. (Note thatin forward conduction there is still a small voltage across the p-n junctions, implyingthat power is being dissipated – hence the semiconductor devices must be cooledand their losses accounted for.) In practice, many devices, each of rating 8.5 kV and up to 4000 A, are stacked in avalve to provide a rating of, say, 200 kV, 4000 A. Valves are connected in series towithstand direct voltages up to 800 kV to earth on each ‘pole’. Each thyristor can be15 cm in diameter and 2 cm depth between its anode and cathode terminals. A typi-cal device is shown in Figure 9.4(a) and a valve in Figure 9.4(b).Figure 9.4 (a) High-power thyristor silicon device (Reproduced with permission fromthe Electric Power Research Institute, Inc.). (b) Thyristor valves in converter station(Reproduced with permission from IEEE.)

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324 Electric Power Systems, Fifth Edition R1 R0 C1Light pulses to gate L Figure 9.5 Circuitry associated with each thyristor In order to turn on a thyristor a pulse of current from a gate circuit that is at thesame potential as the cathode is required. Alternatively, a light trigged thyristor istriggered by a light pulse sent through a ﬁbre-optic channel. When many thyristors are connected in series to form a valve, it is necessary to:1. obtain a uniform voltage distribution across each thyristor.2. retain uniform transient voltage distributions with time.3. control the rate of rise of current. These are achieved by the auxiliary electrical circuitry shown in Figure 9.5. Theinductor, L, limits the rate of rise of current during the early stage of conduction.The chain R1, C1 bypasses the thyristor, thus controlling the negative recovery volt-age. The d.c. grading resistor, R0, ensures uniform distribution of voltage acrosseach thyristor in a valve.9.3.2 Insulated Gate Bipolar TransistorsRecently the insulated gate bipolar thyristor (IGBT) has been used in h.v.d.c.schemes. The IGBT is a development of the MOSFET, in which removal of the volt-age from the gate switches off the through current, thereby allowing power to beswitched on or off at any point of an a.c. cycle. IGBTs have ratings up to 4 kV and1000 A, although the on-state voltage drop and switching losses are larger than witha thyristor. When a sufﬁcient voltage is applied to the Gate with respect to the Emitter, itinverts the p region below the Gate (shaded area) thus forming a diode between theEmitter (n region) and Collector (p substrate). As shown in Figure 9.6(c), then when

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Direct-Current Transmission 325 Gate Collector 5000 VGE = 11V Emitter IC 4000nn 3000 VGE = 10 V p Gate VCE IC(A) 2000 VGE = 9 V 1000 VGE = 8 V n p substrate VGE Collector Emitter 4 8 12 16 20 VCE (V) (a) (b) (c)Figure 9.6 (a) structure of an IGBT (b) symbol, (c) characteristicin addition the Emitter to Collector voltage is greater than 0.7 V, conduction betweenEmitter and Collector commences.9.4 Current Source Converter h.v.d.c.A converter is required at each end of a d.c. line and operates as a rectiﬁer (a.c. tod.c.) or an inverter (power transfer from d.c. to a.c.). The valves at the sending endof the link rectify the alternating current, providing direct current which is transmit-ted to the inverter. Here, it is converted back into alternating current which is fedinto the connected a.c. system (Figure 9.7(a)). If a reversal of power ﬂow is required,the inverter and rectiﬁer exchange roles and the direct voltage is reversed (Figure 9.7(b)). This is necessary because the direct current can ﬂow in one direction only(anode to cathode in the valves), so to reverse the direction (or sign) of power thevoltage polarity must be reversed. The alternating-current waveform injected by the inverter into the receiving enda.c. system, and taken by the rectiﬁer, is roughly trapezoidal in shape, and thus pro-duces not only a fundamental sinusoidal wave but also harmonics of an orderdependent on the number of valves. For a six-valve bridge the harmonic order is6n Æ 1, that is 5, 7, 11, 13, and so on. Filters are installed at each converter station totune out harmonics up to the 25th.9.4.1 RectiﬁcationFor an initial analysis of the rectiﬁer, a three-phase arrangement is shown in Fig-ure 9.8(a). Figure 9.8(b) shows the currents and voltages in the three phases of thesupply transformer. With no gate control, conduction will take place betweenthe cathode and the anode of highest potential. Hence the output-voltage wave isthe thick line and the current output is continuous. As the waveform segment from

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326 Electric Power Systems, Fifth Edition a.c. system Id a.c. system RL Vdr Vdi (a) Id RL Vdi Vdr (b)Figure 9.7 (a) Symbolic representation of two alternating current systems connectedby a direct-current link; Vdr ¼ direct voltage across rectiﬁer, Vdi ¼ direct voltage acrossinverter, (b) System as in part (a) but power ﬂow reversedpÀp to pþp repeats, the mean value of the direct-output voltage, V0, can be23 23 p p p p3.obtained by integrating the sine wave from 2 À 3 to 2 þ Therefore: 1 ðp þ p V^ sinðvtÞdðvtÞ 3 V0 ¼ 2 ¼ V2^ps=in3pp3p2 À¼p3 pﬃﬃ ð9:1Þ V^ 33 ¼ 0:83V^ 2p 3where V^ is the peak a.c. voltage.Deﬁning the r.m.s. line-to-line voltage as VL, from equation (9.1) V0 can beobtained as: pﬃﬃ pﬃﬃ pﬃﬃ VL 33 p2ﬃﬃ 33 p3ﬃﬃ p V0 ¼ V^ 2p ¼ VL 3 Â 2p ¼ 2 ¼ 0:675VL ð9:2Þ

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Direct-Current Transmission 327 a Ld Vdr Vdi b Id c (a) 0 V0 Timeπ - π2 3 Phase (a) Time 2π + π 2π 3 3 VoltageCurrent Id Phase (b) Time Phase (c) Time (b)Figure 9.8 (a) Three-phase rectiﬁer; Vdi ¼ voltage of inverter, (b) Waveforms of anodevoltage and rectiﬁed current in each phase

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328 Electric Power Systems, Fifth Edition Time Voltage V Current Id γ Power factor angleFigure 9.9 Waveforms of voltage and current showing effect of the commutationangle g. A lagging power factor is produced9.4.1.1 CommutationOwing to the inductance present in the circuit, the current cannot change instan-taneously from Id to 0 in one anode and from 0 to Id in the next. Hence, twoanodes conduct simultaneously over a period known as the commutation time oroverlap angle (g). When the valve in phase (b) commences to conduct, it short-circuits the (a) and (b) phases, the current eventually becoming zero in the valveof phase (a) and Id in the valve of phase (b). This is shown in Figure 9.9. It can beseen from the diagram that the overlap angle shifts the current peak with respectto the voltage peak by power factor angle (this is zero when g ¼ 0 as shown inFigure 9.8).9.4.1.2 Gate ControlA positive pulse applied to a gate situated between anode and cathode controls theinstant at which conduction commences, and once conduction has occurred the gateexercises no further control. In the voltage waveforms shown in Figure 9.10 the con-duction in the valves has been delayed by an angle a by suitably delaying the appli-cation of positive voltage to the gates. Ignoring the commutation angle g, the newdirect-output voltage with a delay angle of a is,

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Direct-Current Transmission 329 Voltage Time αγFigure 9.10 Waveforms of rectiﬁer with instant of ﬁring delayed by an angle a bymeans of gate controlV00 ¼ 1 ðp þ p þ a V^ sinðvtÞdðvtÞ 2p=3 p2 À p3 À a 2 3 3V^ ðp þa 3V^ sinp3cos 2p 3 2p ¼ p À a cosðvtÞdðvtÞ ¼ 2 a À 3 ¼ 3pﬃ3ﬃV^ cos a pV00 ¼ V0 cos a ð9:3Þwhere V0 is the maximum value of direct-output voltage as deﬁned byequation (9.1).9.4.1.3 Bridge ConnectionThe bridge arrangement shown in Figure 9.11 is the common implementation ofCSC h.v.d.c schemes, mainly because the d.c. output voltage is doubled. There arealways two valves conducting in series. The corresponding voltage waveforms areshown in Figure 9.12 along with the currents (assuming ideal rectiﬁer operation). The sequence of events in the bridge connection is as follows (see Figures 9.11 and9.12). Assume that the transformer voltage VA is most positive at the beginning ofthe sequence, then valve 1 conducts and the current ﬂows through valve 1 and theload then returns through valve 6 as VB is most negative. After this period, VCbecomes the most negative and current ﬂows through valves 1 and 2. Next, valve 3takes over from valve 1, the current still returning through valve 2. The completesequence of valves conducting is therefore: 1 and 6; 1 and 2; 3 and 2; 3 and 4; 5 and4; 5 and 6; 1 and 6. Control may be obtained in exactly the same manner as previ-ously described.

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330 Electric Power Systems, Fifth Edition Id 135 VA I1 I3 I5 Vdr 2 O Vdr Vdi VB AI N VC IB Vdr 2 IC I4 I6 I2 462 Figure 9.11 Bridge arrangement of valves The direct-voltage output with the bridge can be calculated by deﬁning a ﬁctitiousmid-point on the d.c. side (point O of Figure 9.11). The Bridge arrangement maythen be redrawn as shown in Figure 9.13. This is the same as two three-phase rectiﬁ-ers shown in Figure 9.8(a), but having an output voltage of Vdr/2 (here Vdr is the d.c. VA VB VC VA VB Vdr /2 0 Voltage waveforms Time i1 i2 i3 i4 iA = i1 – i4 Figure 9.12 Idealized voltage and current waveforms for bridge arrangement

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Direct-Current Transmission 331 VA 1 Ld VB 3 VC 5 Vdr 2N O VA 4 Vdr 2 VB 6 Ld VC 2Figure 9.13 Equivalent circuit of the bridge arrangementvoltage of the bridge circuit). Hence, from equation (9.2), the mean d.c. voltage forthe bridge rectiﬁer is VL pﬃﬃ p3ﬃﬃ p 3 2VLV0 ¼ 2 Â 2 ¼ p ¼ 1:35VL ð9:4Þ If the analysis used to obtain equation (9.3) is repeated for the bridge, it will beshown that V00 ¼ V0 cos a.9.4.1.4 Current Relationships in the Bridge CircuitThe voltage waveforms with delay and commutation time accounted for are shownin Figure 9.14. Commutation from one valve to another can be explained by follow-ing the dark line in the positive part of Figure 9.14(a). Consider point P, where valve3 is conducting and continues to conduct until point Q. At Q, valve 5 is triggered.Then the positive busbar (common cathode) voltage is increased from the phase (b)voltage to the average of phase (b) and (c) voltages (this is equal to half the inverseof the phase (a) voltage as shown in the dotted lines of Figure 9.14(a)). From R to S,both valves (3 and 5) conduct. At S, the commutation process ﬁnishes and only valve5 conducts, then the positive busbar voltage is equal to the phase (c) voltage. During the commutation process when two valves are conducting simulta-neously, the two corresponding secondary phases of the supply transformer are

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332 Electric Power Systems, Fifth Edition α(a) 1 γ3 (c) 5 (a) 1 R (b) P QS Voltages f624 6 Id Current in valves 1, 3 and 5 Current in phase (a) (a) Supply (anode) voltage Area lostCommon anode voltage f sin( π 3π ) 2 ( π 3π) αγ 2 (b)Figure 9.14 (a) Voltage and current waveforms in the bridge connection, includingcommutation (g) and delay (a). Rectiﬁer action, (b) Expanded waveforms showingvoltage drop due to commutationshort-circuited and if the voltage drop across the valves is neglected the followinganalysis applies. When two phases of the transformer each of leakage inductance L henriesare effectively short-circuited, the short-circuit current (iS) is governed by theequation,

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Direct-Current Transmission 333 2L diS ¼ V^ L sin vt dtV^ L sin vt is the voltage across two phases that are short circuited by two conduct-ing valves ; iS ¼ À V^ L cos vt þ A 2L vwhere A is a constant of integration and V^ L is the peak value of the line-to-linevoltage. Now vt ¼ a when iS ¼ 0 ; A ¼ V^ L cos a 2vLAlso, when vt ¼ a þ g iS ¼ Id ; Id ¼ V^ L ½cos a À cosða þ gÞ 2vL ¼ pVﬃﬃ L ½cos a À cosða þ gÞ 2vL pﬃﬃ 3 2where VL ¼ r.m.s. line-to-line voltage and, as for the bridge circuit, V0 ¼ p VL. Hence Id ¼ ppVﬃ0ﬃ Á p1ﬃﬃ ½cos a À cosða þ gÞ 32 2X ð9:5Þ ¼ pV0 Á ½cos a À cosða þ gÞ 6XThe mean direct-output voltage, with gate delay angle a only considered,has been shown to be V0 cos a. With both a and the commutation angle g, thevoltage with a only will be modiﬁed by the subtraction of a voltage equal tothe mean of the area under the anode voltage curve lost due to commutation.Referring to Figure 9.14(b), the supply b(aynVo^dsein) vp2oÀltap3gecoissðgvitvÞ e(nnobtey V^ sinðvtÞand the common anode voltage is given that withrV^essipnecp2t Àtop3th).e supply voltage this is a cosine function with a peak value Therefore the area between input voltage wave and common

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334 Electric Power Systems, Fifth Editionanode voltage is ðp À p3p3VVV^^^þþhhhÀÀsaaicsnþiopn3sgi½p3Vpc2^oÀÀssiaanp3ðÀÀþvtcgaÞodþsðþvðagtsÞþiÀnþgðÞcp3aaoþÀsgaV^p2sÀÀinp3þp2coÀasp3p3Àscsionisnððavp2þtÞÀdgðÞpv3þtÞsincðoas pþ3gsÞinþasiinp2 p3sin i À a 2 ¼ ¼ p ¼ 2 À Voltage drop (mean value of lost area) ¼ V^ sinðp=3Þ ½cos a À cosða þ gÞ 2ðp=3Þ ¼ V0 ½cos a À cosða þ gÞ 2 The direct-voltage output, Vd ¼ V0 cos a À V0 ½cos a À cosða þ gÞ 2 ð9:6Þ V0 ¼ 2 ½cos a þ cosða þ gÞ Adding equations (9.5) and (9.6), 3X Id þ Vd ¼ V0 cos a p ;Vd ¼ V0 cos a À RcId ð9:7Þwhere Rc ¼ 3X p Equation (9.7) may be represented by the equivalent circuit shown in Figure 9.15,the term RcId represents the voltage drop due to commutation and not a physicalresistance drop. It should be remembered that V0 is the theoretical maximum valueof direct-output voltage and it is evident that Vd can be varied by changing V0 (con-trol of transformer secondary voltage by tap changing) and by changing a.9.4.1.5 Power FactorApplying Fourier analysis to the phase (a) current waveform shown in Figure 9.14(a) and neglecting the effect of the commutation angle g on the current waveform(Figure 9.12), the instantaneous value of phase (a) current is obtained as: pﬃﬃ 23 1 1 1 1 p 5 7 11 13ia ¼ Id cosðvtÞ þ cosð5vtÞ À cosð7vtÞ þ cosð11vtÞ À cosð13vtÞ þ . . . : : ð9:8Þ

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Direct-Current Transmission 335 V0 cos α RC Id VdFigure 9.15 Equivalent circuit representing operation of a bridge rectiﬁer. Reactanceper phase X (V)The r.m.s value of the fundamental line current pﬃﬃ pﬃﬃ 23 p1ﬃﬃ 6 IL ¼ p 2 Id ¼ p Id ð9:9ÞNeglecting losses and by equating a.c. input power to d.c. output power pﬃﬃ 3VLIL cos f ¼ VdIdSubstituting for IL and Vd from (9.6) and (9.9): pﬃﬃ Â pﬃﬃ Id cos f ¼ V0 ½cos a þ cosða þ gÞ Â Id 3VL 6 2 p pﬃﬃSince V0 ¼ 3 2VL, the power factor is given approximately (as this was derived pby neglecting losses) by cos f ¼ 1 ½cos a þ cosða þ gÞ ð9:10Þ 29.4.2 InversionWith rectiﬁer operation the output current Id and output voltage Vd are such thatpower is absorbed by a load. For inverter operation it is required to transfer powerfrom the direct-current to the alternating-current systems, and as current can ﬂowonly from anode to cathode (i.e., in the same direction as with rectiﬁcation), thedirection of the associated voltage must be reversed. An alternating-voltage systemmust exist on the primary side of the transformer, and gate control of the convertersis essential.

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336 Electric Power Systems, Fifth Edition (a) (b) (c) Vd = 0 Time α = 90oFigure 9.16 Waveforms with operation with a ¼ 90, direct voltage zero. Transitionfrom rectiﬁer to inverter action If the bridge rectiﬁer is given progressively greater delay the output voltagedecreases, becoming zero when a is 90, as shown in Figure 9.16. With further delaythe average direct voltage becomes negative and the applied direct voltage (from therectiﬁer) forces current through the valves against this negative or back voltage. Theconverter thus receives power and inverts. The inverter bridge is shown inFigure 9.17(a) and the voltage and current waveforms in Figure 9.17(b). Hence thechange from rectiﬁer to inverter action, and vice versa, is smoothly obtained bycontrol of a. This may be seen by consulting Figures 9.9 (a ¼ 0), 9.16 (a ¼ 90), and9.17(b) (a > 90). Consider Figure 9.17. With valve 3 conducting, valve 5 is triggered at time A andas the cathode is held negative to the anode by the applied direct voltage (Vd), cur-rent ﬂows, limited only by the circuit impedance. When time B is reached, the volt-age applied across anode-to-cathode is zero and the valve endeavours to ceaseconduction. The large d.c.-side inductance Ld, however, which has previously storedenergy, now maintains the current constant (e ¼ ÀLdðdi=dtÞ and if Ld ! 1,di=dt ! 0). Conduction in valve 5 continues until time C, when valve 1 is triggered.As the anode-to-cathode voltage for valve 1 is greater than for valve 5, valve 1 willconduct, but for a time valves 5 and 1 conduct together (commutation time), the cur-rent gradually being transferred from valve 5 to valve 1 until valve 5 is non-conduct-ing at point D. The current changeover must be complete before (F) by a time (d0)equal to the recovery time of the valves. If triggering is delayed to point F, valve 5,which is still conducting, would be subject to a positively rising voltage and wouldcontinue to conduct into the positive half-cycle with breakdown of the inversionprocess. Hence, triggering must allow cessation of current ﬂow before time F. The angle (d) between the extinction of valve 1 and the point F, where the anodevoltages are equal, is called the extinction angle, that is sufﬁcient time must beallowed for the gate to regain control. The minimum value of d is d0. For inverter operation it is usual to replace the delay angle a by b ¼ 180 À a, theangle of advance. Hence b is equal also to ðg þ dÞ.

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Direct-Current Transmission 337 Id Ld a + cb 4 62 4 (a) By-pass 1 35 (if required) (a) – 2 (c) 4 (a) (b) 60 t D -Vd 5 C Voltages BF Current in valves 1, 3 and 5Valve 3 A 5 1 3 5 α γδ Current in valves 4,6,2 A Current in phase (a) 4 β 3 Id 62 (b)Figure 9.17 (a) Bridge connection–inverter operation, (b) Bridge connection–invertervoltage and current waveforms Replacing a by (180 À b) and g by (b À d) the following are obtained from equa-tions (9.5), (9.6) and (9.10): Id ¼ V0 Á ½cosð180 À bÞ À cosð180 À dÞ 2Rc Id ¼ V0 Á ½cos d À cos b ð9:11Þ 2Rc ÀVd ¼ V0 ½cosð180 À bÞ þ cosð180 À dÞ 2

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338 Electric Power Systems, Fifth Edition Id Rc Id – Rc + + Vd V0 cos β Vd V0 cos δ – – (a) (b)Figure 9.18 (a) Equivalent circuit of inverter in terms of angle of advance, b (b) Equiv-alent circuit of inverter in terms of extinction angle, d(note that the direction of Vd is now reversed) Vd ¼ V0 ½cos d þ cos b ð9:12Þ 2 ð9:13Þ cos f ¼ 1 ½cos d þ cos b leading ð9:14Þ 2 ð9:15ÞBy substituting for V0 cos b from equation (9.11) into equation (9.12): 2 Vd ¼ ½V0 cos d À IdRcBy substituting for V0 cos d from equation (9.11) into equation (9.12): 2 Vd ¼ ½V0 cos b þ IdRc Equations (9.14) and (9.15) may be used to describe inverter operation. Thereforetwo equivalent circuits are obtained for the bridge circuit as shown in Figure 9.18(a)for constant b and Figure 9.18(b) for constant d.9.4.3 Complete Direct-Current LinkThe complete equivalent circuit for a d.c. transmission link under steady-state oper-ation is shown in Figure 9.19. If both inverter and rectiﬁer operate at constant delayangles the current transmitted Id ¼ Vdr À Vdi RLFrom equations (9.7) and (9.15) for constant b operation: Id ¼ ½V0r cos a À IdRcr À ½V0i cos b þ IdRci RL ; Id ¼ V0r cos a À V0i cos b ð9:16Þ RL þ Rcr þ Rci

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Direct-Current Transmission 339 R cr Id RL R ci + + Voi cos βVor cos α – Vdr Vdi –Figure 9.19 Equivalent circuit of complete link with operation with given delay anglesfor rectiﬁer and for inverterFrom equations (9.7) and (9.14) for constant d operation: Id ¼ ½V0r cos a À IdRcr À ½V0i cos d À IdRci RL ; Id ¼ V0r cos a À V0i cos d ð9:17Þ RL þ Rcr À Rciwhere RL is the loop resistance of the line or cable, and Rcr and Rci are the effectivecommutation resistances of the rectiﬁer and inverter, respectively. The equations governing the operation of the inverter may be summarized as fol-lows: pﬃﬃ Vd ¼ 3 2VL 3vL p cos b þ p Id constant b ð9:18Þ constant d ð9:19Þ pﬃﬃ 3vL Vd ¼ 3 2VL p p cos d À Id(Note that L is the leakage inductance per phase of the inverter transformer. For the complete direct current link:1. The magnitude of the direct current can be controlled by variation of a, b (or d), V0r and V0i (the last two by tap-changing of the supply transformers).2. Vdr becomes maximum and the power factor becomes closer to unity when a ¼ 0 (see equation (9.10)). However a is maintained at a minimum value, amin, to make sure that the converter valves have a small positive voltage during turn on.3. Inverter control using constant delay angle has the disadvantage that if d and hence b are too large, excessively high reactive-power demand results. It can be seen from Figure 9.17(b) that the inverter currents are then signiﬁcantly out of phase with the anode voltages and hence a large requirement for reactive power

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340 Electric Power Systems, Fifth Edition Converter A Converter B 3 phase Id V Constant Phase Current Phase Constant V extinction locked transducer Tap angle control oscillator Tapcontroller Id locked extinction controller Id oscillator angle control Constant Constant current current control controlα/β α/β Control for converter A Control for converter BFigure 9.20 Schematic diagram of control of an h.v.d.c. system is established. This also can be seen from equation (9.13) where power factor becomes small with large b and d.4. A reduction in the direct voltage (this could happen during grid faults) to the inverter results in an increase in the commutation angle g, and if b is made large to cover this the reactive-power demand will again be excessive. Therefore it is more usual to operate the inverter with a constant d, which is achieved by the use of suitable control systems.9.4.4 Control of h.v.d.c. LinkA schematic diagram of the control systems is shown in Figure 9.20. The rectiﬁerand inverter change roles as the required direction of power ﬂow dictates, and it isnecessary for each device to have dual-control systems. In Figure 9.21 the full char-acteristics of the two converters of a link are shown, with each converter operatingas rectiﬁer and inverter in turn. The thick line PQRST represents the operation ofConverter A. Its operation can be described as follows:1. From P to Q, Converter A acts as a rectiﬁer and the optimum characteristic with a minimum a value, amin, is shown.2. As the current rating of the valves should not be exceeded, constant-current (CC) control is employed from Q to R. With constant-current control, a is increased, thus Vdr is reduced to maintain constant current in the d.c. link.3. The output voltage-current characteristic crosses the Vd ¼ 0 axis at point R, below which Converter A acts as an inverter. Constant current operation can still be maintained by decreasing the angle of advance, b.4. At S, b reaches d0 and Converter A operates in constant excitation angle control. A similar characteristic is shown for Converter B (VWXYZ), which commences asan inverter and with constant-current control (b increasing) eventually changes to

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Direct-Current Transmission 341Vd Vd = V0 cos αmin- Rcr Id Q Constant δ control P YZ CC control Vd = V0 cos α - Rcr Id A rectifier X Is B inverter R Id A inverter B rectifier Vd = V0 cos β + RciId TV W S Vd = V0 cos δ0 – RciId Figure 9.21 Voltage-current characteristics of converters with compoundingrectiﬁer operation. It is seen that the current setting for Converter A is larger thanthat for Converter B. Figure 9.22 shows the voltage current characteristics in the ﬁrst quadrant whereConverter A operates as a rectiﬁer and Converter B operates as an inverter. In therectiﬁer the transmitted current is regulated in CC control by varying the delayangle and hence Vdr. This is described by QXR. The inverter characteristic is chosensuch that Iorder for the inverter is slightly lower than that of the rectiﬁer. Thereforeunder normal operation the inverter operates in the constant extinction angle con-trol and point X gives the operating condition. If, for example, the d.c. link powerneeds to be halved, Iorder of the rectiﬁer is halved and the new operating pointbecomes Z. In case of an a.c. side remote fault where the d.c. voltage is depressed and therectiﬁer side PQ characteristics falls below the inverter constant excitation controlcharacteristic, the inverter changes its role and tries to maintain the d.c. current ﬂowin the same direction. The new operating point is Y and the current is maintained atvalue B thus the power transmitted is smaller than before. The rectiﬁer now operatesat a constant a (closer to amin). In case of an a.c. side fault closer to the rectiﬁer, the d.c. voltage may be depressedto zero and in order to assist the d.c. link to recover from the fault, the original staticcharacteristics of the rectiﬁer shown in Figure 9.21 is modiﬁed using a voltage

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342 Electric Power Systems, Fifth EditionVdP Z Q Inverter (Constant δ A X control) Y Rectifier (CC control) S, Iorder /2 R, Id , BR T I orderO IminFigure 9.22 Inverter and rectiﬁer operation characteristics with constant-currentcompounding. The operating point is where the two characteristics intersectdependent current limit (R0S0T0 characteristic). Under this operation a may increaseto 90. The rectiﬁer tap-changer is used to keep the delay angle close to amin so thatreactive power requirement is reduced. Further, when the current control is trans-ferred to the inverter (point Q), the rectiﬁer tap-changer tries to restore the originalconditions. The value of the voltage margin is chosen to avoid frequent operation inthe inverter control region. To summarize, with normal operation the rectiﬁer operates at constant currentand the inverter at constant d; under emergency conditions the rectiﬁer is at zeroﬁring-delay (that is a ¼ 0) and the inverter is at constant current.9.4.5 Transmission SystemsThe cheapest arrangement is a single conductor with a ground return. This,however, has various disadvantages. The ground-return current results in thecorrosion of buried pipes, cable sheaths, and so on, due to electrolysis. Withsubmarine cables the magnetic ﬁeld set up may cause signiﬁcant errors in ships’compass readings, especially when the cable runs north-south and unacceptableenvironmental impact. This system is shown in Figure 9.23(a). Two variationson two-conductor schemes are shown in Figure 9.23(b) and (c). The latter hasthe advantage that if the ground is used in emergencies, a double-circuit systemis formed to provide some security of operation. Note that in the circuits of Fig-ure 9.23(c), transformer connections provide twelve pulse operation, therebyreducing harmonics generated on the a.c. sides.

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Direct-Current Transmission 343 (a) Vd (b) Vd Vd (c)Figure 9.23 Possible conductor arrangements for d.c. transmissions: (a) groundreturn; (b) two conductors, return earthed at one end; (c) double-bridgearrangement9.4.6 HarmonicsA knowledge of the harmonic components of voltage and current in a power systemis necessary because of the possibility of resonance and also the enhanced interfer-ence with communication circuits. The direct-voltage output of a converter has awaveform containing a harmonic content, which results in current and voltage har-monics along the line. These are normally reduced by a smoothing inductor. The

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344 Electric Power Systems, Fifth Edition 17 n=7 16I 7 as a % of fundamental I(1) 15 α = 80o 14 α = 40o 13 12 α = 30o 11 10 α = 20 o o α = 0o α= 9 10 8 7 10 20 30 40 6 Angle of overlap ϒ (degrees) 5 0Figure 9.24 Variation of seventh harmonic current with delay angle a and commuta-tion angle (Reproduced with permission from the International Journal of ElectricalEngineering Education)currents produced by the converter currents on the a.c. side contain harmonics. Thecurrent waveform in the a.c. system produced by a delta-star transformer bridgeconverter is shown in Figures 9.14 and 9.17. As given in equation (9.8), the order ofthe harmonics produced is 6n Æ 1, where n is the number of valves. Figure 9.24 shows the variation of the seventh harmonic component with bothcommutation (overlap) angle (g) and delay angle (a). Generally, the harmonicsreduce with decreasing in g this being more pronounced at higher harmonics.Changes in a for a given g value do not cause large decreases in the harmonic com-ponents, the largest change being for values between 0 and 10. For normal opera-tion, a is less than 10 and g is perhaps of the order of 20; hence the harmonics aresmall. During a severe fault, as discussed before, a may reach nearly 90, g is small,and the harmonics produced are large. The harmonic voltages and currents produced in the a.c. system by the convertercurrent waveform may be determined by representing the system components bytheir reactances at the particular harmonic frequency. Most of the system compo-nents have resonance frequencies between the ﬁfth and eleventh harmonics. It is usual to provide ﬁlters (L-C shunt resonance circuits) tuned to the harmonicfrequencies. A typical installation is shown in detail in Figure 9.25. At the funda-mental frequency the ﬁlters are capacitive and help to meet the reactive-powerrequirements of the converters.

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230 kV Coal Creek 49 49 Y/Y 32.5 Mvar Y/D 34 Mvar 57 Mvar Y/D 32.5 Mvar Y/Y 34 Mvar 57 MvarFigure 9.25 Single-line diagram of the main circuit of an h.v.d.duced with permission from IEEE)

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99 kV.1250 A Dickinson Direct-Current Transmission 345 kV Y/Y D/Y 31.5 Mvar 33.5 Mvar 68 Mvar 137 Mvar D/Y Y/Y 31.5 Mvar99 kV.1250 A 33.5 Mvar 68 Mvar 137 Mvar 68 Mvar.c. scheme showing ﬁlter banks and shunt compensation. (Repro- 345

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346 Electric Power Systems, Fifth Edition9.4.7 Variable CompensatorsRapid control of the reactive power compensation is achieved using TCR (thyristor-controlled reactor), MSR/MSC (mechanically switched reactor or capacitor) andTSC (thyristor-switched capacitor). Their selection depends upon the applicationand speed of response required for stability and overvoltage control.9.5 Voltage Source Converter h.v.d.c.The converters considered so far use naturally commutated thyristors and are con-nected to a d.c. circuit that has a large inductance thus maintaining the ﬂow of cur-rent. Current always ﬂows in the same direction (from anode to cathode of thethyristors) and power is reversed by changing the voltage polarity of the d.c. circuit.This conventional h.v.d.c. technology has been used since the mid-1950s with mer-cury arc valves and from around 1980 with thyristors. Since about 2000, an alternative technology using voltage source converters(VSC), has become available although at lower power levels than conventional h.v.d.c. The valves of VSC h.v.d.c. use semiconductor devices that can be turned on andoff, that is, they are force commutated. This ability to turn the valves off as well ason, allows the converters to synthesise a voltage wave of any frequency, phase andmagnitude, within the rating of the equipment. Hence both the rectiﬁer and inverterof a VSC h.v.d.c. link can operate at any power factor, exporting as well as importingreactive power from the a.c. systems. VSC h.v.d.c. does not need synchronous gener-ators to be present in the a.c. networks and in fact can supply a passive (dead) load.In contrast to conventional h.v.d.c. the power ﬂow through the link is reversed bychanging the direction of the d.c. current while the d.c. voltage remains of the samepolarity. The voltage of the d.c. link is maintained by capacitors. Using a VSC, amuch closer approximation to a sine wave is created and so the large ﬁlters of con-ventional h.v.d.c. schemes are not needed.9.5.1 Voltage Source ConverterFigure 9.26 shows a one-leg voltage source converter. When switch T1 is on, the volt-age at A becomes Vd with respect to N and Vd/2 with respect to O. When switch T2 ison, the voltage at A becomes zero with respect to N and ÀVd/2 with respect to O. Thetwo switches use IGBTs that can be switched off as well as on rather than thyristorswhich turn off only when the current ﬂowing through it drops to zero. If each IGBT isturned on and off for 10 ms, the resulting output waveform is a square wave at 50 Hz. For three-phase applications, three one-leg converters are placed in each phase asshown in Figure 9.27. This is the same bridge topology of a CSC (shown in Fig-ure 9.11) but with important differences. The d.c. link voltage is held constant witha capacitance rather than the d.c. link current being maintained with an inductance.The connection to the a.c. circuit is through a coupling reactor and so the fundamen-tal operation of the VSC is of a voltage source behind a reactor in a manner similar tothat of the synchronous generator described in Chapter 3. The VSC is normally connected to a transformer to bring the output voltage to beequal to the a.c. system voltage. If the transformer is connected in star, then with

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Direct-Current Transmission 347+ Vd Vd /2 T1 VAN T1 on T1 off O T2 off T2 on A Vd /2 T2 Vd /2 VAO – N -Vd /2 Figure 9.26 One-leg VSCrespect to the neutral point the voltage of the a-phase looks the same as VAO of Fig-ure 9.26. Other phase voltages will be phase shifted by 120. The simple bridge VSC of Figure 9.27 with each device switched once per 50 Hzcycle is not normally used due to the high harmonic content at the output. A squarewave contains all the odd harmonics (excluding the triplen harmonics). As IGBTs canbe switched many times during one a.c. cycle, the two switches of each phase areswitched using a sine-triangular pulse width modulation (PWM) technique. Theswitching instances are determined by comparing a sinusoidal modulating signalwith a triangular carrier signal (see Figure 9.28). A sinusoidal modulating signal rep-resenting the desired output voltage is compared with a high frequency triangularcarrier signal. When the magnitude of the carrier is higher than the modulating sig-nal, the upper switch is turned on. On the other hand, when the magnitude of thecarrier is lower than the modulating signal, the lower switch is turned on.+ A B T1A T1B T1C Vd C T2A T2B T2C – Figure 9.27 Bridge VSC

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348 Electric Power Systems, Fifth Edition Modulating signal Carrier signal Frequency fc Frequency fm Amplitude Ac Amplitude Am 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 PWM waveform Fundamental 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 Figure 9.28 PWM waveform (only half a cycle is shown) In high power applications, such as h.v.d.c., each switch in the bridge is made up ofa large number of IGBTs connected in series to form a valve. It is important to operateevery IGBT in a valve at the same time. This requires complex circuitry. Another dis-advantage of this arrangement is that high frequency switching (around 2 kHz) of highcurrent and voltage results in much higher losses than the equivalent CSC arrange-ment where each thyristor valve is switched on and turns off only once in each cycle. An alternative VSC topology is to connect a large number of one-leg converters inseries to form a multi-level converter. One leg of a three-level converter is shown inFigure 9.29. The ground shown in the ﬁgure is the solidly earthed neutral of the threephase converter transformer.1 Point A provides 0, Vd/2 and À Vd/2 depending on theswitches that are conducting. For example, the right hand side ﬁgure of Figure 9.29shows the switching sequence when the current is at unity power factor. When T23and T24 are on (which connects the positive rail to point A), the voltage at A withrespect to ground becomes the voltage of the positive rail, that is Vd/2. On the otherhand when T21 and T22 are on, the voltage at A with respect to ground becomes thesame as the voltage of the negative rail, that is ÀVd/2. During the positive half cycle of1 For ease of explanation it was assumed that the windings of the converter transformer connected to theVSC is star connected and the neutral is earthed.

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Direct-Current Transmission 349+ Vd /2 VAN Current zero Vd /2 T14 Vd /2 A -Vd /2 VAN T24 T23 on T23 on T23 on T21 on T21 on T21 on T13 T14 on T24 on T14 on T12 on T22 on T12 on Vd /2 One possible switching sequence T23 Vd -Vd /2 T12 Vd /2 T22 T11 Vd /2 T21–Figure 9.29 A three-level modular converter with the switching sequence for unitypower factor operationthe current, if T23 and T14 are on, then the voltage at A with respect to ground becomeszero. During the negative half cycle of the current the voltage at A with respect toground becomes zero, if T21 and T12 are on. However, when the power factor of thecurrent is lagging or leading, the switching sequence becomes more complicated. A ﬁve-level modiﬁed modular converter could be obtained with the same numberof one-leg converters as Figure 9.29, but with two series switches as shown inFigure 9.30. Switch T1 is on during the positive half cycle and upper two one-legconverters constructing the positive half cycle of the output. On the other hand,switch T2 is on during the negative half cycle and lower two one-leg converters con-structing the negative half cycle of the output. Other multi-level conﬁgurations such as diode-clamped topology and thecapacitor-clamped topology exist. However industrial practice is presently converg-ing on modular topologies as this approach makes the implementation easier. A

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350 Electric Power Systems, Fifth Edition+ T14 Vd /4 Vd T24– Vd /2 T13 VAN Vd /4 Vd /2 Vd/4 T23 Series A -Vd /4 T1 switch VAN -Vd /2 T2 T12 T13 on T23 on T23 on T11 on T21 on T21 on Vd/4 -Vd /2 T14 on T14 on T24 on T12 on T12 on T22 on T22 T1 on; T2 off T1 off; T2 on T11 One possible switching sequence Vd/4 T21 Figure 9.30 A ﬁve-level modular convertermulti-level converter used for h.v.d.c. application may have as many as 300 levels tomake a waveform close to a sinusoid. Compared to the bridge shown in Figure 9.27, the modular converter has lowerlosses as the valves in a bridge VSC operating with PWM turn on and off manytimes per cycle. In the modular VSC each valve switches only once per cycle. How-ever modular converters require complex balancing circuits to balance the voltage ineach d.c. capacitor.9.5.2 Control of VSC h.v.d.c.The VSC connected to the a.c. system can be considered as a voltage behind a reac-tance as shown in Figure 9.31. The reactance X represents the leakage reactance of

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Direct-Current Transmission 351Vd VSC X P +jQ I VVSC θ Va.c. 0oFigure 9.31 Equivalent circuit of VSC a.c. sidethe transformer and any additional coupling reactance used between the trans-former and VSC. The resistance is neglected. The VSC output voltage is VVSCﬀu andthe a.c. voltage is Va:c:ﬀ0. Active and reactive power transfer to the grid is:P þ jQ ¼ Va:c: Â IÃ ¼ Va:c:ﬀ0 Â Va:c:ﬀ0 VVSCﬀÀ u À ÀjX Va:c:VVSCﬀÀ u À Va2:c: X ¼ j ; P ¼ Va:c:VVSC sin u ð9:20Þ X ð9:21Þ Q ¼ Va:c:½VVSC cos u À Va:c: X From equations (9.20) and (9.21), it is clear that by controlling both magnitude andangle of VSC output voltage, active and reactive power output of the VSC h.v.d.c.can be controlled. The complete VSC h.v.d.c. scheme is shown in Figure 9.32. One converter controlsactive power ﬂow as well as reactive power or a.c. voltage. The other converter con-trols the d.c. link voltage and reactive power or a.c. voltage. These control actionscan be assigned to either of the two converters. The controllers are normally implemented by transforming the voltage andcurrent into a d-q rotating reference frame as shown in Figure 9.33 (d.c. voltageand reactive power control is shown). The d-q frame is locked, using a phaselocked loop (PLL), on to the phasor of the a.c. system. The controller has twocontrol loops. The outer control loop controls the DC voltage or power ﬂowand reactive power. The inner current control loops regulate the d and q com-ponents of the currents.

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352 Electric Power Systems, Fifth Edition VSC Vd1 Vd2 VSC Va.c.2 Va.c.1 Va.c.2_refVa.c.1_ref +- AC Switching -+ Vd_ref -+ Vd_ref ACvoltage signals for voltagecontrol converter DC Switching control voltage Active and valves control signals for Active and reactive reactive power Inner control DC converter power (current loops) voltage valves control Inner control (current loops) Reactive Active Active Reactive power power power power control control control control Q ref1 Pref1 Pref2 Q ref2 Figure 9.32 Complete control system for VSC based h.v.d.c X P,Q Vd Three-phase voltages PLL Three-phase currents Switching signals abc-dq for converter valves ψ Vd DC voltage d-axis dq-abc Reactive Q abc-dq current q-axis power Iq Id +- control control control - Vd Vq currentVd_ref +- control -+ I q_ref + I d_ref ++ Id Decupling Decupling Iq Q ref terms terms Figure 9.33 d-q controller for VSC h.v.d.cProblems 9.1 A bridge-connected rectiﬁer is fed from a 230 kV/120 kV transformer from the 230 kV supply. Calculate the direct voltage output when the commutation angle is 15 and the delay angle is (a) 0; (b) 30; (c) 60. (Answer: (a) 160 kV; (b) 127 kV; (c) 61.5 kV)

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Direct-Current Transmission 3539.2 It is required to obtain a direct voltage of 100 kV from a bridge-connected rec- tiﬁer operating with a ¼ 30 and g ¼ 15. Calculate the necessary line second- ary voltage of the rectiﬁer transformer, which is nominally rated at 345 kV/ 150 kV; calculate the tap ratio required. (Answer: 94 kV line; 1.6)9.3 If the rectiﬁer in Problem 9.2 delivers 800 A d.c, calculate the effective reac- tance X (V) per phase. (Answer: X ¼ 13.2 V)9.4 A d.c. link comprises a line of loop resistance 5 V and is connected to trans- formers giving secondary voltage of 120 kV at each end. The bridge-connected converters operate as follows:Rectiﬁer: Inverter:a ¼ 10 d0 ¼ 10; Allow 5 margin on d0 for dX ¼ 15 V X ¼ 15 V Calculate the direct current delivered if the inverter operates on constant d control. If all parameters remain constant, except a, calculate the maximum direct current transmittable. (Answer: 612; 1104 A)9.5 The system in Problem 9.4 is operated with a ¼ 15 and on constant b control. Calculate the direct current for g ¼ 15. (Answer: 482 A)9.6 For a bridge arrangement, sketch the current waveforms in the valves and in the transformer windings and relate them, in time, to the anode voltages. Neglect delay and commutation times. Comment on the waveforms from the viewpoint of harmonics.9.7 A direct current transmission link connects two a.c. systems via converters, the line voltages at the transformer-converter junctions being 100 kV and 90 kV. At the 100 kV end the converter operates with a delay angle of 10, and at the 90 kV end the converter operates with a d of 15. The effective reactance per phase of each converter is 15 V and the loop resistance of the link is 10 V. Determine the magnitude and direction of the power delivered if the inverter operates on constant d control. Both converters consist of six valves in bridge connection. Calculate the percentage change required in the voltage of the transformer, which was originally at 90 kV, to produce a transmitted current of 800 A, other controls being unchanged. Comment on the reactive power requirements of the converters. (Answer: 1.55 kA; 207 MW; 6.5%)

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354 Electric Power Systems, Fifth Edition 9.8 Draw a schematic diagram showing the main components of an h.v.d.c. link connecting two a.c. systems. Explain brieﬂy the role of each component and how inversion into the receiving end a.c. system is achieved. Discuss two of the main technical reasons for using h.v.d.c. in preference to a.c. transmission and list any disadvantages. 9.9 A VSC is fed from a 230kV/120 kV transformer from the 230 kV supply. Total series reactance between the VSC and the a.c. supply is 5 V (including the transformer leakage reactance). The output voltage of the VSC is 125 kV and leads the a.c. supply voltage by 5. What is the active and reactive power delivered to the a.c. system? (Answer: 262 MW; 109 MVAr)9.10 A VSC based h.v.d.c. link is connected into a 33 kV a.c. system with a short circuit level of 150 MVA. The VSC can operate between 0.7–1.2 pu voltage and has a 21.8 V coupling reactor. Assume a voltage of 1 pu on the inﬁnite busbar, and using a base of 150 MVA calculate: a. The maximum active power that the d.c. link can inject into the a.c. system. b. The angle of the VSC when operating at 10 MW exporting power. c. The maximum reactive power (in MVAr) that the d.c. link can inject into the a.c. system. d. The maximum reactive power (in MVAr) that the d.c. link can absorb from the a.c. system. (Answer: 45 MW; 12.8, 7.88 MVAr, 10.4 MVAr)

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10Overvoltages andInsulation Requirements10.1 IntroductionThe insulation requirements for lines, cables and substations are of critical importancein the design of power systems. When a transient event such as a lightning strike, afault or a switching operation occurs, the network components can be subjected tovery high stresses from the excessive currents and voltages that result. The internalinsulation of individual items of plant (for example, generators and transformers) isdesigned to withstand the voltage transients that are anticipated at the location of theequipment. In addition, the insulation of the overall system, for example the numberof insulators per string, and clearances of overhead lines, is designed to withstandthese transient voltages. In some cases, additional devices such as over-running earthwires, lightning arrestors or arc gaps are provided to protect items of plant. In earlier years, lightning largely determined the insulation requirements of boththe transmission and distribution system. With the much higher transmission volt-ages now in use and with large cable networks, the voltage transients or surges dueto switching, that is the opening and closing of circuit breakers, have become themajor consideration. A factor of major importance is the contamination of insulator surfaces caused byatmospheric pollution. This modiﬁes the performance of insulation considerably,which becomes difﬁcult to assess precisely. The presence of dirt or salt on the insula-tor discs or bushing surfaces results in these surfaces becoming slightly conducting,and hence ﬂashover occurs. A few terms frequently used in high-voltage technology need deﬁnition. They areas follows:1. Basic impulse insulation level or basic insulation level (BIL): This is the refer- ence level expressed as the impulse crest (peak) voltage with a standard wave ofElectric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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356 Electric Power Systems, Fifth EditionVoltage p.u. 1.0 0.9 0.50 Time Wave tail time Wavefront timeFigure 10.1 Basic impulse waveform; Shape of lightning and switching surges; thelightning impulse has a rise time of 1.2 ms and a fall time to half maximum value of 50 ms(hence 1.2/50 wave). Switching surges are much longer, the duration times varyingwith situation; a typical wave is 175/3000 ms1.2 Â 50 ms wave (see Figure 10.1). The insulation of power system apparatus, asdemonstrated by suitable tests, shall be equal or greater than the BIL. The twostandard tests are the power frequency and 1.2/50 impulse-wave voltage with-stand tests. The withstand voltage is the level that the equipment will withstandfor a given length of time or number of applications without disruptive dischargeoccurring, that is a failure of insulation resulting in a collapse of voltage and pas-sage of current (sometimes termed ‘sparkover’ or ‘ﬂashover’ when the dischargeis on the external surface). Normally, several tests are performed and the numberof ﬂashovers noted. The BIL is usually expressed as a per unit of the peak (crest)value of the normal operating voltage to earth; for example for a maximum oper-ating voltage of 362 kV, 1 p:u: ¼ pﬃﬃ Â 3p6ﬃ2ﬃ ¼ 300 kV 2 3 so that a BIL of 2.7 p.u. ¼ 810 kV.2. Critical ﬂashover voltage (CFO): This is the peak voltage with a 50% probability of ﬂashover or disruptive discharge (sometimes denoted by V50).3. Impulse ratio (for ﬂashover or puncture of insulation): This is the impulse peak voltage to cause ﬂashover or puncture divided by the crest value of power- frequency voltage.10.2 Generation of Overvoltages10.2.1 Lightning SurgesA thundercloud is bipolar, often with positive charges at the top and negative at thebottom, usually separated by several kilometres. When the electric ﬁeld strengthexceeds the breakdown value a lightning discharge is initiated. The ﬁrst discharge

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Overvoltages and Insulation Requirements 357 Cloud 40 ms 40 ms Time time interval Stepped Return Dart Return leader leader stroke Dart leader Return Ground Figure 10.2 Sequence of strokes in a multiple lightning strokeproceeds to the earth in steps (stepped leader stroke). When close to the earth afaster and luminous return stroke travels along the initial channel, and several suchleader and return strokes constitute a ﬂash. The ratio of negative to positive strokesis about 5 : 1 in temperate regions. The magnitude of the return stroke can be as highas 200 kA, although an average value is in the order of 20 kA.Following the initial stroke, after a very short interval, a second stroke to earthoccurs, usually in the ionized path formed by the original. Again, a return strokefollows. Usually, several such subsequent strokes (known as dart leaders) occur, theaverage being between three and four. The complete sequence is known as a multiple-stroke lightning ﬂash and a representation of the strokes at different time intervals isshown in Figure 10.2. Normally, only the heavy current ﬂowing over the ﬁrst 50 msis of importance and the current-time relationship has been shown to be of the formi ¼ ipeakðeÀat À eÀbtÞ. For a 1.2/50 ms wave, a ¼ 1:46 Â 104 and b ¼ 2:56 Â 106. When a stroke arrives on an overhead conductor, equal current surges of theabove waveform are propagated in both directions away from the point of impact.The magnitude of each voltage surge set up is therefore 1 Z0 ipeak ÀeÀat À eÀbt Á where 2 ,Z0 is the conductor surge impedance. For a current peak of 20 kA and a Z0 of 350 Vthe voltage surge will have a peak value of (0.5 Â 350 Â 20 Â 103) that is 3500 kV.When a ground or earth wire is positioned over the overhead line, a stroke arriv-ing on a tower or on the wire itself sets up surges ﬂowing in both directions alongthe wire. On reaching neighbouring towers the surge is partially reﬂected and trans-mitted further. This process continues over the length of the line as towers areencountered. If the towers are 300 m apart the travel time between towers and backto the original tower is (2 Â 300)/(3 Â 108), that is 2 ms, where the speed of propaga-tion is 3 Â 108 m/s (the speed of light). The voltage distribution may be obtained bymeans of the Bewley lattice diagram, to be described in Section 10.6.If an indirect stroke strikes the earth near a line, the induced current, which isnormally of positive polarity, creates a voltage surge of the same waveshape which

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358 Electric Power Systems, Fifth Editionhas an amplitude dependent on the distance from the ground. With a direct strokethe full lightning current ﬂows into the line producing a surge that travels awayfrom the point of impact in all directions. A direct stroke to a tower can cause aback ﬂashover due to the voltage set up across the tower inductance and footingresistance by the rapidly changing lightning current (typically 10 kA/ms); thisappears as an overvoltage between the top of the tower and the conductors (whichare at a lower voltage).10.2.2 Switching Surges–Interruption of Short Circuits and Switching OperationsSwitching surges consist of damped oscillatory waves, the frequency of which isdetermined by the system conﬁguration and parameters; they are normally ofamplitude 2–2.8 p.u., although they can exceed 4 p.u. (per-unit values based on peakline-to-earth operating voltage). A simple single phase network shown in Figure 10.3(a) is used to explainswitching surges. If the transformer leakage inductance is L, the series resistance CB (a) Short circuitRL CBsin ωt C (b) Restriking voltage (expanded time scale) P QArc voltage Time System CB open circuit voltage Fault current (c)Figure 10.3 Restriking voltage set up on fault interruption, (a) System diagram,(b) Equivalent circuit, (c) Current and voltage waveforms

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Overvoltages and Insulation Requirements 359 R Ls s 1/Cs v(s) Figure 10.4 Laplace equivalent circuit with step appliedof the circuit is R and the transformer and connections capacitance to ground isC, then the equivalent circuit after a fault may be drawn as Figure 10.3(b). Whena fault occurs the protection associated with the circuit breaker (CB) sends a con-trol signal to open. At time instant P the contacts of the CB start to open butcurrent continues to ﬂow through the arc across the circuit breaker contacts.When the arc between the circuit-breaker contacts is extinguished (at time instantQ), the full system voltage (recovery voltage) suddenly appears across the R-L-Ccircuit and the open gap of the circuit breaker contacts as shown in Figure 10.3(b). The resultant voltage appearing across the circuit is shown in Figure 10.3(c).It consists of a high-frequency component superimposed on the normal systemvoltage, the total being known as the restriking voltage. During the time forwhich this high-frequency term persists, the change in power frequency term isnegligible. Therefore the equivalent circuit of Figure 10.3(b) may be analyzed bymeans of the Laplace transform assuming that a step input of V^ appears at Q.The Laplace equivalent is shown in Figure 10.4. The voltage across the circuit breaker contacts, that is across C is given by: vðsÞ ¼ V^ =s Â 1=Cs ¼ sðLCs2 V^ þ 1Þ ð10:1Þ R þ Ls þ 1=Cs þ RCsDeﬁning v0 ¼ p1ﬃﬃﬃﬃﬃﬃ and a ¼ R , equation (10.1) can be written as: LC 2L vðsÞ ¼ sðs2 V^ v02 v20Þ þ 2as þ V^ v02 ð10:2Þ aÞ2 þ ðv02 ¼ s½ðs þ À a2Þ

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360 Electric Power Systems, Fifth EditionTo obtain the inverse Laplace transform, equation (10.2) is written in the form: \"# vðsÞ ¼ V^ A À ðs þ Bs þ D À a2Þ ð10:3Þ s aÞ2 þ ðv20Equating the numerators of equations (10.2) and (10.3): Aðs þ aÞ2 þ Aðv02 À a2Þ À Bs2 À Ds ¼ v20 Therefore A ¼ 1; A À B ¼ 0; ;B ¼ 1; 2aA À D ¼ 0; ;D ¼ 2a. Now, equation(10.2) can be rewritten as: \"# vðsÞ ¼ V^ 1 À ðs þ s þ 2a À a2 Þ ð10:4Þ s aÞ2 þ ðv20By taking inverse Laplace transforms:vðtÞ ¼ V^ 1 À eÀat qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ À a eÀat qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ! ð10:5Þ cos v02 À a2 t v0 sin v20 À a2 t pﬃﬃﬃﬃﬃﬃﬃﬃ In practical systems R ( L=C and thus a ( v0. Therefore equation (10.5) can besimpliﬁed to: vðtÞ ¼ V^ Â1 À eÀatcos Ã ð10:6Þ v0t Equation (10.6) deﬁnes the restriking voltage in Figure 10.3(c) where v0 is the nat-ural frequency of oscillation and a is the damping coefﬁcient. The maximum valueof the restriking voltage is 2V^ .Example 10.1Determine the relative attenuation occurring in ﬁve cycles in the overvoltage surge setup on a 66 kV cable fed through an air-blast circuit breaker when the breaker opens on asystem short circuit. The network parameters are as follows: R ¼ 7:8 V L ¼ 6:4 mH C ¼ 0:0495 mFSolutionFrom the network parameters: v0 ¼ p1ﬃﬃﬃﬃﬃﬃ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ1ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ LC 6:4 Â 10À3 Â 0:0495 Â 10À6 ¼ 5:618 Â 104 rad=s

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Overvoltages and Insulation Requirements 361 that is transient frequency ¼ 5:618 Â 104=2p Hz ¼ 8:93 kHz. a ¼ R ¼ 2 Â 7:8 10À3 ¼ 609:4 2L 6:4 Â Hence, time for ﬁve cycles ¼ 8:93 5 103 ¼ 0:56 ms Â The maximum voltage set up with the circuit breaker opening on a short circuit faultwould be V^ ¼ 2 Â p66ﬃﬃ Â pﬃﬃ ¼ 107:78 kV 3 2 From equation (10.6), vðtÞ at t ¼ 0.56 ms is: hi vðtÞ ¼ 107:78 1 À eÀ609:4Â0:56Â10À3 cosð5:618 Â 104 Â 0:56 Â 10À3 ¼ 42:4 kVThe initial rate of rise of the surge is very important as this determines whetherthe contact gap, which is highly polluted with arc products, breaks down agraiﬃﬃnﬃ after 1 Lthe initial open circuit occurs. From (10.5) if a ¼ v0, that is R ¼ 2 C andvðtÞ ¼ V^ ½1 À eÀat. Under this critically damped case, the severity of the transienrt iﬃﬃsﬃ 1 Lreduced. A critically damped surge can be obtained by connecting a resistor of 2 Cacross the contacts of the circuit breaker.10.2.3 Switching Surges–Interruption of Capacitive CircuitsThe interruption of capacitive circuits is shown in Figure 10.5. At the instant of arcinterruption, the capacitor is left charged to value V^ . If the capacitance, C, is largethen the voltage across the capacitor decays slowly. The voltage across the circuitbreaker is the difference between the system voltage and the voltage across thecapacitor. At the negative crest of the system voltage, that is ÀV^ , the gap voltage is2V^ . If the gap breaks down, an oscillatory transient is set up (as previously dis-cussed), which can increase the gap voltage to up to 3V^ , as shown in Figure 10.5.10.2.4 Current ChoppingCurrent chopping arises with vacuum circuit breakers or with air-blast circuit break-ers which operate on the same pressure and velocity for all values of interruptedcurrent. Hence, on low-current interruption the breaker tends to open the circuit

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362 Electric Power Systems, Fifth Edition Capacitive L Vc current CB C (a) Current interruption Vc Vˆ Time System Restriking voltage voltage (expanded time scale) Approaching -3Vˆ (b) Figure 10.5 Voltage waveform when opening a capacitive circuitbefore the current natural-zero, and the electromagnetic energy present is rapidlyconverted to electrostatic energy, that is. rﬃﬃﬃ 1 1 L 2 Li02 ¼ 2 Cv2 and v ¼ i0 C ð10:7ÞThe voltage waveform is shown in Figure 10.6.The restriking voltage, vr, can be obtained from equation (10.7) by includingresistance and time: rﬃﬃﬃ vr ¼ i0 L eÀatsin v0t ð10:8Þ Cwhere v0 ¼ p1ﬃﬃﬃﬃﬃﬃ and i0 is the value of the current at the instant of chopping. High LCtransient voltages may be set up on opening a highly inductive circuit such as atransformer on no-load.10.2.5 FaultsOvervoltages may be produced by certain types of asymmetrical fault, mainly onsystems with ungrounded neutrals. The voltages set up are of normal operating

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Overvoltages and Insulation Requirements 363V Vr t V t VarcifioFigure 10.6 Voltage transient due to current chopping: if ¼ fault current; v ¼ systemvoltage, i0 ¼ current magnitude at chop, vr ¼ restriking voltagefrequency. Consider the circuit with a three-phase earth fault as shown inFigure 10.7. If the circuit is not grounded the voltage across the ﬁrst gap to open is1.5 Vphase. With the system grounded the gap voltage is limited to the phase voltage.This has been discussed more fully in Chapter 7.10.2.6 ResonanceIt is well known that in resonant circuits severe overvoltages occur and, dependingon the resistance present; the voltage at resonance across the capacitance can behigh. Although it is unlikely that resonance in a supply network can occur at normalsupply frequencies, it is possible to have this condition at harmonic frequencies.Resonance is normally associated with the capacitance to earth of items of plant andis often brought about by an opened phase caused by a broken conductor or a fuseoperating. In circuits containing windings with iron cores, for example transformers, a con-dition due to the shape of the magnetization curve, known as ferroresonance, ispossible. This can produce resonance with overvoltages and also sudden changesfrom one condition to another. A summary of important switching operations is given in Table 10.1.

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364 Electric Power Systems, Fifth Edition V V V V VVFigure 10.7 Three-phase system with neutral earthing (grounding)Table 10.1 Summary of the more important switching operations (Reproduced withpermission from Brown Boveri Review, December 1970)Switching Operation System Voltage Across Contacts1. Terminal short circuit 02. Short line fault 03. Two out-of-phase systems voltage depends on 0 grounding conditions in systems 04. Small inductive currents, current chopped (unloaded transformer) See Figure 10.55. Interrupting capacitive currents capacitor banks, lines, and cables on no-load

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Overvoltages and Insulation Requirements 36510.3 Protection Against Overvoltages10.3.1 Modiﬁcation of TransientsWhen considering the protection of a power system against overvoltages, the transi-ents may be modiﬁed or even eliminated before reaching the substations. If this isnot possible, the lines and substation equipment may be protected, by variousmeans, from ﬂashover or insulation damage. By the use of over-running earth (ground) wires, phase conductors may beshielded from direct lightning strokes and the effects of induced surges fromindirect strokes lessened. The shielding is not complete, except perhaps for a phaseconductor immediately below the earth wire. The effective amount of shielding isoften described by an angle f, as shown in Figure 10.8; a value of 35 appears toagree with practical experience. Obviously, two earth wires horizontally separatedprovide much better shielding. It has already been seen that the switching-in of resistance across circuit-breakercontacts reduces the high overvoltages produced on opening, especially on capaci-tive or low-current inductive circuits. An aspect of vital importance, quite apart from the prevention of damage, is themaintenance of supply, especially as most ﬂashovers on overhead lines cause nopermanent damage and therefore a complete and lasting removal of the circuit fromoperation is not required. Transient faults may be removed by the use of autoreclos-ing circuit breakers. In distribution circuits and many transmission systems all threeEarth wire Earth wireshielding angle oPhaseconductorFigure 10.8 Single earth wire protection; shielding angle f normally 35

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366 Electric Power Systems, Fifth Editionphases are operated together. Some countries use single phase auto-reclose toimprove transient stability. Only the faulted phase is opened and re-closed whiletwo phases remain connected. It is uneconomic to attempt to modify or eliminate most overvoltages, andmeans are required to protect the various items of power systems. Surge diver-tors are connected in shunt across the equipment and divert the transient toearth; surge modiﬁers are connected to reduce the steepness of the wavefront-hence reduce its severity. A surge modiﬁer may be produced by the connectionof a shunt capacitor between the line and earth or an inductor in series with theline. Oscillatory effects may be reduced by the use of fast-acting voltage or cur-rent injections.10.3.2 Surge DivertersThe basic requirements for diverters are that they should pass no current at normalvoltage, interrupt the power frequency follow-on current after a ﬂashover, andbreak down as quickly as possible after the abnormal voltage arrives.10.3.2.1 Rod GapThe simplest form of surge diverter is the rod gap connected across a bushing orinsulator as shown in Figure 10.9. This may also take the form of rings (arcingrings) around the top and bottom of an insulator string. The breakdown voltagefor a given gap is polarity dependent to some extent. It does not interrupt thepost-ﬂashover follow-on current (i.e. the power-frequency current which ﬂows inthe path created by the ﬂashover), and hence the circuit protection must operate,but it is by far the cheapest device for plant protection against surges. It is usu-ally recommended that a rod gap be set to a breakdown voltage at least 30%below the voltage-withstand level of the protected equipment. For a given gap,the time for breakdown varies roughly inversely with the applied voltage; thereexists, however, some dispersion of values. The times for positive voltages arelower than those for negative ones. Typical curves relating the critical ﬂashover voltage and time to breakdown forrod gaps of different spacings are shown in Figure 10.10.10.3.2.2 Lightning ArrestorEarly designs of lightning arresters consisted of a porcelain bushing containing anumber of spark gaps in series with silicon carbide discs, the latter possessing lowresistance to high currents and high resistance to low currents. The lightning arres-tor obeys a law of the form V ¼ aIk (for a SiC arrestor k ¼ 0:2) where a depends onthe material and its size. The overvoltage breaks down the gaps and then the power-frequency current is determined by the discs and limited to such a value that thegaps can quickly interrupt it at the ﬁrst current zero. The voltage-time characteristicof a lightning arrester is shown in Figure 10.11; the gaps break down at S and the

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Overvoltages and Insulation Requirements 367 rod gapFigure 10.9 Insulating cross-arm for a double-circuit 420 kV line (Italian)Impulse voltage (1.5/40 μs) kV crest 1500 1000 127 cm 500 76 cm 0 25 cm 5 10 Breakdown time (μs)Figure 10.10 Breakdown characteristics of rod gaps

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368 Electric Power Systems, Fifth Edition S RVoltage Time Figure 10.11 Characteristic of lightning arrestercharacteristic after this is determined by the current and the discs; the maximumvoltage at R should be in the same order as at S. For high voltages a stack of severalsuch units is used. Although with multiple spark gaps, diverters can withstand high rates of rise ofrecovery voltage (RRRV), the non-uniform voltage distribution between the gapspresents a problem. To overcome this, capacitors and non-linear resistors are con-nected in parallel across each gap. With the high-speed surge, the voltage is mainlycontrolled by the gap capacitance and hence capacitive grading is used. At power-frequencies a non-linear resistor provides effective voltage grading. 10 9Voltage (kV) 8 7 6 1 10 102 103 104 Current (A) 5 20oC 4 150oC 10-2 10-1Figure 10.12 Current-voltage characteristic of standard ZnO blocks, 80 mm in diame-ter and 35 mm thick. The reference voltage of the block is 4.25 kV d.c. and its continu-ous rating is 3 kV (effective value); This characteristic is given by V ¼ aI0:03 (Reproducedwith permission from the Electricity Association)

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Overvoltages and Insulation Requirements 369 Developments in materials have produced an arrester which does not requiregaps to reduce the current after operation to a low value. Zinc oxide material isemployed in stacked cylindrical blocks encased in a ceramic weatherproof insulator.A typical current-voltage characteristic is shown in Figure 10.12, where it can beseen that with a designed steady-state voltage of 4.25 kV, the residual current isbelow 10À2 A, depending upon temperature. At an effective rating of 3 kV the cur-rent is negligible. Consequently, no action is required to extinguish the current afteroperation, as may be required with a gapped arrester. Metal oxide arresters up to the highest transmission voltages are now employedalmost to the exclusion of gapped devices.10.4 Insulation CoordinationThe equipment used in a power system comprises items having different break-down or withstand voltages and different voltage-time characteristics. In order thatall items of the system are adequately protected there is a need to consider the sys-tem as a whole and not items of plant in isolation, that is the insulation protectionmust be coordinated. To assist this process, standard insulation levels are recom-mended for power frequency ﬂashovers, lightning impulses and switching impulsesand these are summarized in Tables 10.2 (British substation practice) and 10.3 (U.Spractice). The reduced basic insulation impulse levels for switching impulses areonly speciﬁed for higher voltages. When there are more than one withstand voltagespeciﬁed for a given nominal voltage, the standards relevant to different apparatusshould be referred to in order to ﬁnd the exact insulation level. Coordination is made difﬁcult by the different voltage-time characteristics ofplant and protective devices, for example a gap may have an impulse ratio of 2 for a20 ms front wave and 3 for a 5 ms wave. At the higher frequencies (shorter wave-fronts) corona cannot form in time to relieve the stress concentration on the gap elec-trodes. With a lightning surge a higher voltage can be withstood because a dischargerequires a certain discrete amount of energy as well as a minimum voltage, and theapplied voltage increases until the energy reaches this value.Table 10.2 Recommended BILs at various operating voltages following IEC 60071 (BSEN 60071-1:2006)Nominal Standard lightning Impulse Power frequency Switching impulse withstand (peak kV)voltage (kV) withstand (peak kV) withstand (peak kV)11 60 28 75 90132 550 230 650 275400 1175 675 950 1300 1425

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370 Electric Power Systems, Fifth EditionTable 10.3 Recommended BILs at various operating voltages following IEEE 1313.1-1996Nominal Standard lightning Impulse Power frequency Switching impulse withstand (peak kV) withstand (peak kV)voltage (kV) withstand (peak kV)23 150 50115 350 140 185 450 230 550345 900 650 975 750 1050 825 1175 900 1300 975 1050500 1300 1175 1425 1300 1550 1425 1675 1500 1800 1300765 1800 1425 1925 1550 2050 1675 1800 Figure 10.13 shows the voltage-time characteristics for the system elementscomprising a substation. The difﬁculty of coordination can be explained using thevoltage time characteristics of the rod gap and the transformer. If they are subjectedto a surge having a rate of rise less than the critical slope shown in Figure 10.13 (longrise time), then the rod gap breaks down before the transformer thus protecting it. 1500 Critical slope Peak voltage (kV) 1000 (e) 500 (d) (f) (b) (c) (a) 0 10 μs 20Figure 10.13 Insulation coordination in an H.V. substation. Voltage-time characteris-tics of plant. Characteristics of (a) rod gap (b) lightning arrester, (c) Transformer,(d) Line insulator string, (e) Busbar insulation, (f) Maximum surge applied waveform

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Overvoltages and Insulation Requirements 371 100Flashover probability (per cent) 90 Switching Lightning surge 80 70 60 50 40 30 20 10 0 CFO V Crest voltageFigure 10.14 Flashover probability–peak surge voltage. Lightning ﬂashover is close-grouped near the critical ﬂashover voltage (CFO), whereas switching-surge probabilityis more widely dispersed and follows normal Gaussian distribution (Reproduced withpermissions from Westinghouse Electrical Corporation)On the other hand if the applied surge has a rate of rise greater than the critical slope(short rise time), the transformer breaks down before the rod gap. As can be seen,the arrester protects the transformer for all surges. Up to an operating voltage of 345 kV the insulation level is determined by light-ning, and the standard impulse tests are sufﬁcient along with normal power fre-quency tests. Impulse tests are normally performed with a voltage wave shown inFigure 10.1; this is known as a 1.2/50 ms wave and typiﬁes the lightning surge. Athigher network operating voltages, the overvoltages resulting from switching arehigher in magnitude and therefore they decide the insulation. The characteristics ofair gaps and some solid insulation are different for switching surges than for thestandard impulse waves, and closer coordination of insulation is required becauseof the lower attenuation of switching surges, although their ampliﬁcation by reﬂec-tion is less than with lightning. Research indicates that, for transformers, the switch-ing impulse strength is of the order of 0.95 of the standard (lightning impulse) valuewhile for oil-ﬁlled cables it is 0.7 to 0.8. The design withstand level is selected by specifying the risk of ﬂashover, forexample for 550 kV towers a 0.13% probability has been used. At 345 kV, design iscarried out by accepting a switching impulse level of 2.7 p.u. At 500 kV, however, a2.7 p.u. switching impulse would require 40% more tower insulation than that gov-erned by lightning. The tendency is therefore for the design switching impulse levelto be forced lower with increasing system operating voltage and for control of thesurges to be made by the more widespread use of resistance switching in the circuitbreakers or use of surge diverters. For example, for the 500 kV network the level is

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372 Electric Power Systems, Fifth Edition Front of wave Chopped wave Minimum insulation withstand BIL orVoltage (crest value) full wave Protective 0.83 x BIL Maximum IR margin Protective margin Discharge voltage 60 Hz withstand test Maximum arrester sparkover Arrester reseal ratingVoltage (kV crest) Front of wave Chopped wave BIL 1800 1660 1780 BIL 1550 1800 x 0.83 = 1494 1640 BIL 1425 1500 BIL 1300 1550 x 0.83 = 1286 1425 x 0.83 = 1183 1300 x 0.83 = 1079 Maximum arrester sparkover Impulse 10 200 Switching 2000 60 Hz surge Time μsFigure 10.15 Coordination diagrams relating lightning-arrester characteristics to systemrequirements Top: Explanation of characteristics. Bottom: Typical values for 500 kV system(300 kV to ground) (Reproduced with permissions from Hubbell Power Systems Inc.)2 p.u. and at 765 kV it is reduced to 1.7 p.u.; if further increases in system voltageoccur, it is hoped to decrease the level to 1.5 p.u. The problem of switching surges is illustrated in Figure 10.14, in which ﬂashoverprobability is plotted against peak (crest) voltage; critical ﬂashover voltage (CFO) isthe peak voltage for a particular tower design for which there is a 50% probability ofﬂashover. For lightning, the probability of ﬂashover below the CFO is slight, butwith switching surges having much longer fronts the probability is higher, with thecurve following the normal Gaussian distribution; the tower clearances must bedesigned for a CFO much higher than the maximum transient expected. An example of the application of lightning-arrester characteristics to systemrequirements is illustrated in Figure 10.15. In this ﬁgure, the variation of crest

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Overvoltages and Insulation Requirements 373 Wavefront + +++ + Switch i v Velocity u (a) (b) vi _____ x dx v iu √v= i Z0= LCFigure 10.16 Distribution of charge and current as wave progresses along a previ-ously unenergized line, (a) Physical arrangement, (b) Symbolic representationvoltage for lightning and switching surges with time is shown for an assumed BIL ofthe equipment to be protected. Over the impulse range (up to 10 ms) the fronts ofwave and chopped-wave peaks are indicated. Over the switching surge region theinsulation withstand strength is assumed to be 0.83 of the BIL (based on transformerrequirements). The arrester maximum sparkover voltage is shown along with themaximum voltage set up in the arrester by the follow on current (10 or 20%). A pro-tective margin of 15% between the equipment withstand strength and the maximumarrester sparkover is assumed.10.5 Propagation of SurgesThe basic differential equations for voltage and current in a distributed constant lineare as follows: @2v ¼ LC @2v and @2i ¼ LC @2i @x2 @t2 @x2 @t2and these equations represent travelling waves. The solution for the voltage may beexpressed in the form, pﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃ v ¼ F1 t À x LC þ F2 t þ x LCthat is one wave travels in the positive direction of x and the other in the negative.Also, it may be shown that because @v=@x ¼ ÀLð@i=@tÞ, the solution for current isi ¼ rﬃCLﬃﬃhF1 À pﬃﬃﬃﬃﬃﬃ À þ pﬃﬃﬃﬃﬃﬃi noting that rﬃﬃﬃ ¼ 1 t x LC F2 t x LC C Z0 L

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374 Electric Power Systems, Fifth Edition In more physical terms, if a voltage is injected into a line (Figure 10.16) a corre-sponding current i will ﬂow, and if conditions over a length dx are considered, theﬂux set up between the go and return wires, F ¼ iLdx;where L is the inductance per unit length. As the induced back e.m.f., ÀdF=dt, is equal to the applied voltage vv ¼ À dF ¼ ÀLi dx ¼ ÀiLU ð10:9Þ dt dtwhere U is the wave velocity Also, charge is stored in the capacitance over dx, that is. Q ¼ idt ¼ vCdx and ð10:10Þ i ¼ vCUFrom (10.9) and (10.10), pﬃﬃﬃﬃﬃﬃ vi ¼ viLCU2 and U ¼ 1= LC:Substituting for U in (10.10), rﬃﬃﬃi ¼ vCU ¼ pvﬃCﬃﬃﬃﬃﬃ ¼ v C ¼ v LC L Z0where Z0 is the characteristic or surge impedance. For single-circuit three-phase overhead lines (conductors not bundled) Z0 lies inthe range 400–600 V. For overhead lines, U ¼ 3 Â 108 m/s, that is the speed of light,and for cables U ¼ 3 Â 108 m=s pﬃeﬃrﬃﬃmﬃﬃﬃrﬃwhere er is usually from 3 to 3.5, and mr ¼ 1. From equations (10.9) and (10.10), 1 1 i 2 Li2 ¼ 2 ðiLUÞ U ¼ 1 v Â Cv ¼ 1 Cv2 22 The incident travelling waves of vi and ii, when they arrive at a junction or dis-continuity, produce a reﬂected current ir and a reﬂected voltage vr which travel

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Overvoltages and Insulation Requirements 375 Z08 vi i = v1 Z (= ) z0 vi 2vi 2vi vi (a) (b)Figure 10.17 Application of voltage to unenergized loss-free line on open circuit atfar end. (a) Distribution of voltage, (b) Distribution of current. Voltage source is aneffective short circuitback along the line. The incident and reﬂected components of voltage and currentare governed by the surge impedance Z0, so that vi ¼ Z0ii and vr ¼ ÀZ0ir In the general case of a line of surge impedance Z0 terminated in Z (Figure 10.17),the total voltage at Z is v ¼ vi þ vr and the total current is i ¼ ii þ ir. Also, vi þ vr ¼ Zðii þ irÞ Z0ðii À irÞ ¼ Zðii þ irÞand hence Z0 À Z ir ¼ Z0 þ Z ii ð10:11Þ

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376 Electric Power Systems, Fifth EditionAgain, vi þ vr ¼ Zðii þ irÞ vi À vr ¼Z Z0or, vr ¼ Z À Z0 vi ¼ avi ð10:12Þ Z þ Z0 ð10:13Þ ð10:14Þ Z À Z0where a is the reﬂection coefﬁcient equal to Z þ Z0From (10.12) and (10.11), the total voltage and current is given by: Z À Z0 2Z v ¼ vi þ vr ¼ vi þ Z þ Z0 vi ¼ Z þ Z0 vi Z0 À Z 2Z0 i ¼ ii þ ir ¼ ii þ Z0 þ Z ii ¼ Z0 þ Z ii From equations (10.13) and (10.14), if Z ! 1, v ¼ 2vi and i ¼ 0. This is shown inFigure 10.17. As shown in the ﬁrst plot, a surge of vi travels towards the open circuitend of the line. This creates a current surge of i ¼ vi=Z0. When this surge reaches theopen circuit end, a reﬂected voltage surge of vi and a current surge of Ài are createdand they start travelling back down the line. Therefore the total voltage surge is now2vi and the current is zero as shown in the second plot. At the voltage source, aneffective short circuit, a voltage wave of Àvi is reﬂected down the line. The reﬂectedwaves will travel back and forth along the line, setting up, in turn, further reﬂectedwaves at the ends, and this process will continue indeﬁnitely unless the waves areattenuated because of resistance and corona. From equations (10.13) and (10.14), if Z ¼ Z0 (matched line), a ¼ 0, that is there isno reﬂection. If Z > Z0, then vr is positive and ir is negative, but if Z < Z0, vr is nega-tive and ir is positive.Summarizing, at an open circuit the reﬂected voltage is equal to the incident volt-age and this wave, along with a current (Àii), travels back along the line; note that atthe open circuit the total current is zero. Conversely, at a short circuit the reﬂectedvoltage wave is (Àvi) in magnitude and the current reﬂected is (ii), giving a totalvoltage at the short circuit of zero and a total current of 2ii. For other terminationarrangements, Thevenin’s theorem may be applied to analyze the circuit. The voltageacross the termination when it is open-circuited is seen to be 2vi and the equivalentimpedance looking in from the open-circuited termination is Z0; the termination isthen connected across the terminals of the Thevenin equivalent circuit (Figure 10.18).Consider two lines of different surge impedance in series. It is necessary to deter-mine the voltage across the junction between them (Figure 10.19). From (10.13): 2Z1 2vi vAB ¼ Z1 þ Z0 vi ¼ bvi ¼ Z1 þ Z0 Z1 ð10:15Þ

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Overvoltages and Insulation Requirements 377vi Z0 Z1 Z0 2vi Z1 (b) (a)Figure 10.18 Analysis of travelling waves–use of Thevenin equivalent circuit, (a) Sys-tem, (b) Equivalent circuit Z0 A Z1 A 2vi Z 0vi v vAB B (b) Z1 B (a)Figure 10.19 Analysis of conditions at the junction of two lines or cables of differentsurge impedanceThe wave entering the line Z1 is the refracted wave and b is the refraction coefﬁcient,that is the proportion of the incident voltage proceeding along the second line (Z1).From (10.14), with vi ¼ Z0ii i¼ 2vi ð10:16Þ Z0 þ Z From equations (10.15) and (10.16), the equivalent circuit shown in Figure 10.19(b)can be obtained. When several lines are joined to the line on which the surge originates(Figure 10.20), the treatment is similar, for example if there are three lines havingvi A Z1 Z0 A Z0 Z1 2vi Z1 Z1 Z1 Z1 B B (a) (b)Figure 10.20 Junction of several lines, (a) System, (b) Equivalent circuit

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378 Electric Power Systems, Fifth Edition (a) Z 1 Z2 I (b) V Z1 IV Z2 V (c) IFigure 10.21 Surge set up by fault clearance, (a) Equal and opposite current (I)injected in fault path, (b) Equivalent circuit, (c) Voltage and current waves set up at apoint of fault with direction of travelequal surge impedances (Z1) then 2vi 2vi Z1 i¼ Z1=3 þ Z0 and vAB ¼ Z1=3 þ Z0 3An important practical case is that of the clearance of a fault at the junction of twolines and the surges produced. The equivalent circuits are shown in Figure 10.21; thefault clearance is simulated by the insertion of an equal and opposite current (I) atthe point of the fault. From the equivalent circuit, the magnitude of the resultingvoltage surges (v) Z1Z2 ¼I Z1 þ Z2and the currents entering the lines are Z1 Z2 I Z1 þ Z2 and I Z1 þ Z2The directions are as shown in Figure 10.21(c).10.5.1 Termination in Inductance and Capacitance10.5.1.1 Shunt CapacitanceUsing the cTahpeavceitnoirnCeqisuvivCa¼len2vt iÀc1irÀcueiÀt ta=Zs0CsÁh,owwhnerien Figure 10.22, the voltage riseacross the t is the time commencing withthe arrival of the wave at C. The current through C is given by, i ¼ dvC ¼ 2vi eÀt=Z0 C dt Z0

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Overvoltages and Insulation Requirements 379 Z0 2vi C LFigure 10.22 Termination of line (surge impedance Zo) in a capacitor C orinductance LThe reﬂected wave, vr ¼ vC À vi ¼ 2vi 1 À eÀt=Z0C À vi ¼ vi 1 À 2eÀt=Z0C ð10:17Þ When t ¼ 0, vC ¼ 0 and i ¼ 2vi=Z0 and when t ! 1, vC ¼ 2vi and i ¼ 0. That is,as to be expected, the capacitor acts initially as a short circuit and ﬁnally as anopen circuit.10.5.1.2 Shunt InductanceAgain, from the equivalent circuit shown in Figure 10.22 but with an inductor thevoltage across the inductance is vL ¼ 2vieÀðZ0=LÞtand hi ð10:18Þ vr ¼ vL À vi ¼ vi 2eÀðZ0=LÞt À 1Here, the inductance acts initially as an open circuit and ﬁnally as a short circuit.10.5.1.3 Capacitance and Resistance in ParallelConsider the practical system shown in Figure 10.23(a), where C is used to modifythe surge. vi Z1 A Z1 Z2 2vi Z 2 C C B (a) (b)Figure 10.23 Two lines surge impedances Z1 and Z2 grounded at their junctionthrough a capacitor C. (a) System diagram, (b) Equivalent circuit

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380 Electric Power Systems, Fifth EditionFrom Figure 10.23(b), the open-circuit voltage across AB without the capacitor: Z2 ¼ 2vi Z1 þ Z2Equivalent Thevenin resistance ¼ Z1Z2Voltage across AB is Z1 þ Z2 vAB ¼ 2vi Z2 À ð10:19Þ Z1 þ Z2 1 eÀtðZ1 þZ2 Þ=Z1 Z2 CThe reﬂected wave is given by (vAB À vi).Example 10.2An overhead line of surge impedance 500 V is connected to a cable of surge impedance50 V through a series resistor (Figure 10.24(a)). Determine the magnitude of the resistorsuch that it absorbs maximum energy from a surge originating on the overhead lineand travelling into the cable. Calculate:a. the voltage and current transients reﬂected back into the line, andb. those transmitted into the cable, in terms of the incident surge voltage;c. the energies reﬂected back into the line and absorbed by the resistor. Answer: Let the incident voltage and current be vi and ii, respectively. From the equivalentcircuit (Figure 10.24(b)), vB ¼ 2vi Â Z2 R Z1 þ Z2 þAs vi ¼ Z1ii, iB ¼ Z1 2vi þ R ¼ Z1 2Z1ii R þ Z2 þ Z2 þPower absorbed by the resistance ¼ Z1 2Z1 ii !2 þ Z2 þ RRThis power is a maximum when \"# d R ¼0 dR ðZ1 þ Z2 þ RÞ2 ðZ1 þ Z2 þ RÞ2 À R Â 2ðZ1 þ Z2 þ RÞ ¼ 0 ;R ¼ Z1 þ Z2

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Overvoltages and Insulation Requirements 381 vi A RB iB Z1 A ii R Z 1 Z2 (Cable) 2vi B vrA Z2 i rA (b) (a) vB vrA vi iB i i i rA (c)Figure 10.24 (a) System for Example 10.2. (b) Equivalent circuit, (c) Voltage andcurrent surgesWith this resistance the maximum energy is absorbed from the surge. Hence R shouldbe 500 þ 50 ¼ 550 V.a. the reﬂected voltage at A vrA ¼ 2viðZ2 þ RÞ À vi ¼ Z2 þ R À Z1 vi Z1 þ Z2 þ R Z1 þ Z2 þ R ¼ 50 þ 550 À 500 vi ¼ 0:091 vi 500 þ 50 þ 550And Z1 À Z2 À R ! Z1 þ Z2 þ R irA ¼ iB À ii ¼ ii 500 À 50 À 550 ! 500 þ 50 þ 550 ¼ ii ¼ À0:091 iib. Voltage and current transmitted into the cable iB ¼ 2 Â 500 Â ii ¼ 0:91ii 500 þ 50 þ 550 vB ¼ 0:091vi

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382 Electric Power Systems, Fifth Editionc. The surge energy entering Z2 ¼ vBiB ¼ 0:082viii The energy absorbed by R ¼ ð0:91iiÞ2 Â 550 vi ¼ 455 Z1 ii ¼ 0:91viiiand the energy reﬂected ¼ viiið1 À 0:082 À 0:91Þ ¼ 0:008viii The waveforms are shown in Figure 10.24(c).10.6 Determination of System Voltages Produced by Travelling SurgesIn the previous section the basic laws of surge behaviour were discussed. The calcu-lation of the voltages set up at any node or busbar in a system at a given instant intime is, however, much more complex than the previous section would suggest.When any surge reaches a discontinuity its reﬂected waves travel back and are, inturn, reﬂected so that each generation of waves sets up further waves which coexistwith them in the system. To describe completely the events at any node entails, therefore, an involvedbook-keeping exercise. Although many mathematical techniques are availableand, in fact, used, the graphical method due to Bewley (1961) indicates clearlythe physical changes occurring in time, and this method will be explained insome detail.10.6.1 Bewley Lattice DiagramThis is a graphical method of determining the voltages at any point in a transmis-sion system and is an effective way of illustrating the multiple reﬂections whichtake place. Two axes are established: a horizontal one scaled in distance along thesystem, and a vertical one scaled in time. Lines indicating the passage of surges aredrawn such that their slopes give the times corresponding to distances travelled. Ateach point of change in impedance the reﬂected and transmitted waves are obtainedby multiplying the incidence-wave magnitude by the appropriate reﬂection andrefraction coefﬁcients a and b. The method is best illustrated by Example 10.3.Example 10.3A loss-free system comprising a long overhead line (Z1) in series with a cable (Z2) willbe considered. Typically, Z1 is 500 V and Z2 is 50 V.SolutionReferring to Figure 10.25, the following coefﬁcients apply: Line-to-cable reﬂection coefﬁcient,a1 ¼ 50 À 500 ¼ À0:818 ðnote order of values in numeratorÞ 50 þ 500

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Overvoltages and Insulation Requirements 383 α1 β 1 α =1 β2 α2 Distance α βLong line Z1 Cable Z1 Z2 O/C ( =1 ) vi Z2 =0 P Time vi β 1 vi α 1v i vi βvi β 2β 1vi β 1 vi β 1 α2vi β 1 α2v i β 2 β 1 α v i t β 1 α22v i β 1 α22v i 2 β 1 vi β 1 β 2 α 2v i vi 2 α1v i (b) β2 β1vi β 1 α2v i α1v i P (a)Figure 10.25 Bewley lattice diagram–analysis of long overhead line and cablein series, (a) Position of voltage surges at various instants over ﬁrst completecycle of events, that is up to second reﬂected wave travelling back alongline, (b) Lattice diagramLine-to-cable refraction coefﬁcient, b1 ¼ 2 Â 50 ¼ 0:182 50 þ 500Cable-to-line reﬂection coefﬁcient, a2 ¼ 500 À 50 ¼ 0:818 500 þ 50Cable-to-line refraction coefﬁcient, b1 ¼ 2 Â 500 ¼ 1:818 50 þ 500

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384 Electric Power Systems, Fifth Edition As the line is long, reﬂections at its sending end will be neglected. The remote end of the cable is considered to be open-circuited, giving an a of 1 and a b of zero at that point. When the incident wave vi (see Figure 10.25) originating in the line reaches the junction, a reﬂected component travels back along the line (a1vi), and the refracted or transmitted wave (b1vi) traverses the cable and is reﬂected from the open- circuited end back to the junction (1 Â b1vi). This wave then produces a reﬂected wave back through the cable (1 Â b1a2vi) and a transmitted wave (1 Â b1b2vi) through the line. The process continues and the waves multiply as indicated in Figure 10.25(b). The total voltage at a point P in the cable at a given time (t) will be the sum of the voltages at P up to time t, that is b1við2 þ 2a2 þ 2a22Þ, and the voltage at inﬁnite time will be 2b1við1 þ a2 þ a22 þ a32 þ a42þ . . . .Þ. The voltages at other points are similarly obtained. The time scale may be deter- mined from a knowledge of length and surge velocity; for the line the latter is of the order of 300 m/ms and for the cable 150 m/ms. For a surge 50 ms in duration and a cable 300 m in length there will be 25 cable lengths traversed and the terminal voltage will approach 2vi. If the graph of voltage at the cable open-circuited end is plotted against time, an exponential rise curve will be obtained similar to that obtained for a capacitor. The above treatment applies to a rectangular surge waveform, but may be modi-ﬁed readily to account for a waveform of the type illustrated in Figure 10.1. In thiscase the voltage change with time must also be allowed for and the process is morecomplicated.10.6.2 Short-Line FaultsA particular problem for circuit breakers on h.v. systems is interrupting current if asolid short circuit occurs a few kilometres from the circuit breaker. This situation isillustrated in Figure 10.26.It will be recalled that in Figure 10.3, fault interruption gives rise to a restrikingtransient on the supply side of the breaker. The sudden current interruption sends astep surge down the short line to the fault, which is reﬂected at the short-circuitpoint back to the breaker. In a manner similar to the Bewley lattice calculation ofFigure 10.25, a surge is propagated back and forth in Z0 as shown in Figure 10.26(b).The resulting combination of restriking voltage on the supply side and the continu-ously reﬂected surge on the line side of the breaker produces a steeply rising initialrestriking voltage, greater than the restriking transient of Figure 10.3. Since thisoccurs just following current zero, it provides the most onerous condition of all forthe circuit breaker to deal with, and must be included in the series of proving testson any new design.The ﬁrst peak of the line-side component of the transient recovery voltage may beobtained from a knowledge of the fault current If and the value of Z0. Assumingideal interruption at current zero, the rate of rise of transient recovery voltage isdv ¼ Z0 di ¼ Z0 pﬃﬃ ¼ pﬃﬃ Â 10À6 ðV=msÞdt dt 2If 2vZ0If 1=v

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Overvoltages and Insulation Requirements 385 L Surge SolidEC impedance fault S Z0 Short line Vs Vline (a)Voltage Vs t (μ s) Vline VS0 0 (b) Voltage across Initial restrikingbreaker S voltage rise t (μ s) (c)Figure 10.26 Short-line fault, (a) System diagram, (b) Supply-side and line-side transi-ents, (c) Transient voltage across circuit breaker contactsThe amplitude of the ﬁrst peak of reﬂected wave is V ¼ t dv ðVÞ dtwhere t is twice the transit time of the surge through Z0 and v is the system angularfrequency. To this value should be added any change of vs on the supply side, calcu-lated as given by equation (10.6).10.6.3 Effects of Line LossAttenuation of travelling waves is caused mainly by corona which reduces thesteepness of the wavefronts considerably as the waves travel along the line.

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386 Electric Power Systems, Fifth EditionAttenuation is also caused by series resistance and leakage resistance and thesequantities are considerably larger than the power-frequency values. The determina-tion of attenuation is usually empirical and use is made of the expression vx ¼ vieÀgx,where vx is the magnitude of the surge at a distance x from the point of origination.If a value for g is assumed, then the wave magnitude of the voltage may be modiﬁedto include attenuation for various positions in the lattice diagram. For example, inFigure 10.25(b), if eÀgx is equal to aL for the length of line traversed and aC for thecable, then the magnitude of the ﬁrst reﬂection from the open circuit is aLviaCb1 andthe voltages at subsequent times will be similarly modiﬁed. Considering the power and losses over a length dx of a line shown in Figure 10.27where resistance and shunt conductance per unit length R(V) and G(VÀ1). The power loss ¼ i2Rdx þ ðv À dvÞ2Gdx % i2Rdx þ i2Z02Gdx Power entering the line section ¼ vi ¼ i2Z0 Power leaving the line section ¼ ðv À dvÞði À diÞ ¼ ði À diÞ2Z0 % i2Z0 À 2iZ0di Therefore,vi À ðv À dvÞði À diÞ ¼ i2Rdx þ i2Z02Gdx À Z02GÁdx 2iZ0di ¼ i2 R þ di 1 R i ¼ 2 Z0 þ GZ0 dx ! 1 RThis gives a time constant of the current as 2 Z0 þ GZ0 x and a surge ofamplitude ii entering the line section decaying due to resistance and conductance, ! À1 Ri ¼ ii exp 2 Z0 þ GZ0 x ð10:20ÞAlso, it may be shown that ! 1 Rv ¼ vi exp À 2 Z0 þ GZ0 x ð10:21Þ i Rdx i - di v Gdx v - dv dxFigure 10.27 Lossy line section (L and C are not shown)

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Overvoltages and Insulation Requirements 387and the power at x is ! R vi ¼ viii exp À Z0 þ GZ0 x ð10:22Þ If R and G are realistically assessed (including corona effect), attenuation may beincluded in the travelling-wave analysis.10.6.4 Digital MethodsThe lattice diagram becomes very cumbersome for large systems and digital meth-ods are usually applied. Digital methods may use strictly mathematical methods,that is the solution of the differential equations or the use of Fourier or Laplacetransforms. These methods are capable of high accuracy, but require large amountsof data and long computation times. The general principles of the graphicalapproach described above may also be used to develop a computer program whichis very applicable to large systems. Such a method will now be discussed in moredetail. A rectangular wave of inﬁnite duration is used. The theory developed in the pre-vious sections is applicable. The major role of the program is to scan the nodes of thesystem at each time interval and compute the voltages. In Figure 10.28 a particularsystem (single-phase representation) is shown and will be used to illustrate themethod. The relevant data describing the system are given in tabular form (see Table 10.4).Branches are listed both ways in ascending order of the ﬁrst nodal number and arereferred to under the name BRANCH. The time taken for a wave to travel along abranch is recorded in terms of a positive integer (referred to as PERIOD) which con-verts the basic time unit into actual travel time. Reﬂection coefﬁcients are stored andreferred to as REFLECT and the corresponding refraction coefﬁcients are obtained,that is ð1 þ aijÞ. Elements of time as multiples of the basic integer are also shown inTable 10.4. The method is illustrated by examining the system after the arrival of the rectan-gular wave at node 3 at Time (0). This voltage (magnitude 1 p.u.) is entered in theBRANCH (3, 2), TIME (0) element of the BRANCH-TIME matrix. On arrival at node1 400 Ω,1 2 400 Ω, 2 3 400 Ω ,1 400 Ω ,1 0Ω 4Figure 10.28 Application of digital method. Each line is labelled with surge imped-ance and surge travel time (multiples of basic unit), for example 400 V 1

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388 Electric Power Systems, Fifth EditionTable 10.4 Data for lattice calculations 2 2 3 44 3 4 2 12BRANCH i 1 12 2 1 2 11 À1 0 À1/3 0 À1/3 j2 41 0 1 2/3 1 2/3PERIOD 1 11 1REFLEC (aij) À1/3 0 0 À1/3 2/3ð1 þ aijÞ 2/3 1 1TIME 0 1 2 2/3 30 2/3 42 at time equal to zero plus PERIOD (3, 2), two waves are generated, on BRANCH(2, 1) and BRANCH (2, 4), both of magnitude 1ð1 þ a32Þ, that is 2/3. A reﬂectedwave is also generated on BRANCH (2, 3), of magnitude 1 Â a32, that is À1/3. Thesevoltages are entered in the appropriate BRANCH in the TIME (2) row of Table10.4. On reaching node 1, TIME (3), a refracted wave of magnitude 2/3ð1 þ a12Þ,that is 2/3, is generated on BRANCH (1, 4) and a reﬂected wave 2/3Âa12, that is 0,is generated on BRANCH (1, 2). This process is continued until a speciﬁed time isreached. All transmitted waves for a given node are placed in a separate NODE-TIME array; a transmitted wave is considered only once even though it could beentered into several BRANCH-TIME elements. Current waves are obtained bydividing the voltage by the surge impedance of the particular branch. The ﬂowdiagram for the digital solution is shown in Figure 10.29. It is necessary for the programs to cater for semi-inﬁnite lines (that is lines so longthat waves reﬂected from the remote end may be neglected) and also inductive/capacitive terminations. Semi-inﬁnite lines require the use of artiﬁcial nodes labelled(say) 0. For example, in Figure 10.30, node 2 is open-circuited and this is accountedfor by introducing a line of inﬁnite surge impedance between nodes 2 and 0, andhence, if Z!1 a12 ¼ Z À Z0 ¼ 1 Z þ Z0 Although the scanning will not ﬁnd a refraction through node 2, it is necessary forthe computer to consider there being one to calculate the voltage at node 2. Lineswith short-circuited nodes may be treated in a similar fashion with an artiﬁcial lineof zero impedance. Inductive/capacitive terminations may be simulated by stub lines. An inductanceL (H) is represented by a stub transmission line short-circuited at the far end and ofsurge impedance ZL ¼ L=t, where t is the travel time of the stub. Similarly, a capaci-tance C (farads) is represented by a line open-circuited and of surge impedanceZC ¼ t=C. For the representation to be exact, t must be small and it is found neces-sary for ZL to be of the order of 10 times and ZC to be one-tenth of the combinedsurge impedance of the other lines connected to the node.

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Overvoltages and Insulation Requirements 389 Set all array elements to zero Read and store data Form REFLECT array Form TIME arrayForm array of artificial nodes for treatment of semi-infinite lines Scan BRANCH-TIME array Is there a non-zero voltage No at the time being considered YesCalculate time of arrival of incident wave at new node No Is there reflections YesCalculate change in branch voltage, store in BRANCH-TIME arrayCalculate change in branch voltage, store in NODE-TIME array Increment time by one unit No Maximum time limit being reached Yes Write output, stopFigure 10.29 Flow diagram of digital method for travelling-wave analysisFor example, consider the termination shown in Figure 10.31(a). The equivalentstub-line circuit is shown in Figure 10.31(b). Stub travel times are chosen to be shortcompared with a quarter cycle of the natural frequency, that is. pﬃﬃﬃﬃﬃﬃ ¼ 2p pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ 0:315 Â 10À4 s2p LC 4 0:01 Â 4 Â 10À8 4

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390 Electric Power Systems, Fifth Edition Z0 Z = 18 0 8 2 8Figure 10.30 Treatment of terminations–use of artiﬁcial node 0 and line of inﬁniteimpedance to represent open circuit at node 2 Let t ¼ 5 Â 10À6 s for both L and C stubs (corresponding to the total stub length of1524 m). Hence, ZC ¼ t ¼ 2:5 Â 10À6 ¼ 67:5 V C 4 Â 10À8 400 Ω 0.01 H 0.04 μF 2 3 1 (a) 400 Ω 67.5Ω 4000 Ω (b) 67.5Ω 0 2 4000 Ω 400 Ω 400 Ω (c)0 3 4 0 0Figure 10.31 Representation of line terminated by L-C circuit by means of stublines,(a) Original system, (b) Equivalent stub lines, (c) Use of artiﬁcial nodes to representopen- and short-circuited ends of stub lines

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Overvoltages and Insulation Requirements 391 3 Voltage 2 1 0 4 6 8 10 12 14 02 MicrosecondsFigure 10.32 Voltage-time relationship at nodes 1 and 4 of system in Figure 10.31,—— computer solution; ......., transient analyzer (Reproduced with permission from IEEE)and ZC ¼ L ¼ 0:01 ¼ 4000 V t 2:5 Â 10À6 The conﬁguration in a form acceptable for the computer program is shown inFigure 10.31(c). Reﬁnements to the program to incorporate attenuation, waveshapes, and non-linear resistors may be made without changing its basic form. A typical application is the analysis of the nodal voltages for the system shown inFigure 10.29. This system has previously been analyzed by a similar method byBarthold and Carter (1961) and good agreement found. The print out of nodal volt-ages for the ﬁrst 20 ms is shown in Table 10.5, and in Figure 10.33 the voltage plot fornodes 1 and 4 is compared with a transient analyzer solution obtained by Bartholdand Carter.10.6.5 Three-Phase AnalysisThe single-phase analysis of a system as presented in this chapter neglects themutual effects which exist between the three phases of a line, transformer, and soon. The transient voltages due to energization may be further increased by thismutual coupling and also by the three contacts of a circuit breaker not closing at thesame instant.10.7 Electromagnetic Transient Program (EMTP)The digital method previously described is very limited in scope. A much morepowerful method has been developed by the Bonneville Power Administration and

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392 Electric Power Systems, Fifth EditionTable 10.5 Digital computer printout: node voltages (system of Figure 10.29)Time (ms) Node 1 2340 0.0000E þ 00 0.0000E-01 1.0000E 00 0.0000E þ 001 0.0000E þ 00 0.0000E-01 1.0000E 00 0.0000E þ 002 0.0000E þ 00 6.6670E-01 1.0000E 00 0.0000E þ 003 6.6670E-01 6.6670E-01 1.0000E 00 6.6670E-014 1.3334E þ 00 6.6670E-01 1.0000E 00 1.3334E þ 005 1.3334E 00 1.5557E þ 00 1.0000E 00 1.3334E þ 006 1.5557E þ 00 1.7779E þ 00 1.0000E 00 1.5557E þ 007 2.0002E 00 1.7779E þ 00 1.0000E 00 2.0002E 008 2.2224E 00 2.0743E 00 1.0000E 00 2.2224E 009 2.2965E 00 1.7779E þ 00 1.0000E 00 2.2965E 0010 1.8520E þ 00 1.8520E þ 00 1.0000E 00 1.8520E þ 0011 1.4075E þ 00 1.9508E þ 00 1.0000E 00 1.4075E þ 0012 1.5062E 00 1.0617E þ 00 1.0000E 00 1.5062E þ 0013 1.1604E þ 00 7.6534E-01 1.0000E 00 1.1604E þ 0014 4.1944E-01 8.2297E-01 1.0000E 00 4.1955E-0115 8.2086E-02 2.6307E-01 1.0000E 00 8.2086E-0216 À7.4396E-02 1.6435E-01 1.0000E 00 À7.4396E-0217 7.8720E-03 1.0732E-02 1.0000E 00 7.8720E-0318 9.3000E-02 À2.5556E-01 1.0000E 00 9.3000E-0219 À1.7043E-01 4.1410E-01 1.0000E 00 À1.7043E-0120 1.5067E-01 6.2634E-01 1.0000E 00 1.5067E-01 iR v I (a) (b) Figure 10.33 (a) Lumped inductance, (b) Equivalent circuitis known as EMTP. This is widely used, especially in the USA and now has a num-ber of implementations. It is assumed that the variables of interest are known at the previous time stept À Dt, where Dt is the time step. The value of Dt must be small enough to give rea-sonable accuracy with a ﬁnite difference method.

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Overvoltages and Insulation Requirements 39310.7.1 Lumped Element ModellingConsider an inductance L (Figure 10.33(a)), v ¼ L di ð10:23Þ dtEquation (10.23) can be rearranged to form: ð10:24Þ Zt iðtÞ ¼ iðt À DtÞ þ vdt tÀDtApplying the trapezoidal rule gives: iðtÞ ¼ iðt À DtÞ þ Dt ½vðtÞ þ vðt À DtÞ 2L ¼ vðtÞ Dt þ iðt À DtÞ þ vðt À DtÞ Dt ð10:25Þ 2L 2L ¼ vðtÞ þ I Rwhere R ¼ 2L and I ¼ iðt À DtÞ þ vðt À DtÞ Dt Dt 2L Here, R is constant and I varies with time. The equivalent circuit is shown inFigure 10.33(b). A similar treatment applies to capacitance (C) (see Figure 10.34(a)). Here, i ¼ C dv Z i dt dt ;v ¼ C C R i v I (a) (b) Figure 10.34 Equivalent circuits for transient analysis

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394 Electric Power Systems, Fifth EditionIntegrating and applying the trapezoidal rule gives vðtÞ ¼ vðt À DtÞ þ 1 Zt idt C tÀDt ¼ vðt À DtÞ þ Dt ðiðtÞ þ iðt À DtÞÞ 2CRearranging gives: iðtÞ ¼ 2C vðtÞ À iðt À DtÞ À 2C vðt À DtÞ Dt Dt ð10:26Þ vðtÞ ¼ R þ IWhere R ¼ Dt and I ¼ Àiðt À DtÞ À vðt À DtÞ 2C 2C Dtgiving the equivalent circuit of Figure 10.34(b). Resistance is represented directly by, iðtÞ ¼ vðtÞ ð10:27Þ RThe procedure for solving a network using EMTP is as follows:1. Replace all the lumped elements by models described in equations (10.25), (10.26) and (10.27).2. From initial conditions, determine iðt À DtÞ ¼ ið0Þ and vðt À DtÞ ¼ vð0Þ.3. Solve for iðtÞ and vðtÞ. Increment time step by Dt and calculate new values of i and v, and so on. The analysis is carried out by use of the nodal admittance matrix and Gaussian elimination. If mutual coupling between elements exists then the repre- sentation becomes very complex.Example 10.4The equivalent circuit of a network is shown in Figure 10.35. Determine the networkwhich simulates this network for transients using the EMTP method after the ﬁrst timestep of the transient of 5 ms.SolutionBy replacing all the lumped elements by models described in equations (10.25), (10.26)and (10.27), the equivalent circuit shown in Figure 10.36(a) was obtained.

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Overvoltages and Insulation Requirements 395 RL1 ¼ RL2 ¼ 2 Â 0:0005 ¼ 200 V 0.03 Ω 5 Â01.50mÀH6 0.5mH1 MV RC1 ¼ RC2 ¼ 2 5 Â 10À6 ¼ 41:7 V μF Â 0:060Â.0160μÀF6 0.06 2mH RL3 ¼ 2 Â 0:002 ¼ 800 V 5 Â 10À6 Figure 10.35 Figure for Example 10.4At t ¼ 0, note current sources are zero and equivalent circuit is reduced to Figure10.36(b). By applying the mesh method to Figure 10.36(b): 2 106 3 ¼ 2 241:7 À41:7 32 3 664 0 757 4 À41:7 241:7 0 i1 À41:7 À41:7 54 i2 5 0 0 800 i3Invert 23 2 0:0043 0:0007 32 106 3 i1 0:0043 0:0007 0:0043 775664 0 757 3:8 Â 10À5 0:0002 0 466 i2 775 ¼ 466 3:8 Â 10À5 i3 0:0002 i1 ¼ 4300 A, i2 ¼ À38 A and i3 ¼ À200 From Figure 10.36(a) and (b):IL1 ¼ 4300 A, IL2 ¼ À38 A and IL3 ¼ À200 AIC1 ¼ i1 À i2 ¼ 4338 A and IC2 ¼ i2 À i3 ¼ 238 A The equivalent circuit after 5 ms is shown in Figure 10.36(c). The process is thenrepeated for the next 5 ms step using Figure 10.36(c) as the starting condition.10.7.2 SwitchingThe various representations are shown in Figure 10.37. A switching operationchanges the topology of the network and hence the ½Y matrix. If the ½Y matrix isformed with all the switches open, then the closure of a switch is obtained by theadding together of the two rows and columns of ½Y and the associated rows of ½i.

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396 RL1 Electric Power Systems, Fifth Edition 1 MV IL1 RL2 RC1 IL2 RC2 RL3 IL3 IC1 IC2 200 Ω (a) 0.03 Ω 200 Ω106 i1 41.7 Ω i3 800 Ω 200 Ω i2 (b) 0.03 Ω 200 Ω 4300A 38 A 200 A 41.7Ω10 6 41.7 Ω 4338 A 238 A 800 Ω (c) Figure 10.36 The equivalent circuit of Figure 10.35 Another area where switching is used is to account for non-linear c À i chara-cteristics of transformers, reactors, and so on. The representation is shown inFigure 10.38(a) and (b), in which R ¼ 2bk Dt

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Overvoltages and Insulation Requirements 397 R= , i=0 Open 8 v=0 Closed R = 0, Tdelay Tclose closing opening Time controlled switch Figure 10.37 Representation of switchΨ 3 4 2 1 i (A) (a) R I (b)Figure 10.38 (a) Non-linear characteristic; (b) circuit representation

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398 Electric Power Systems, Fifth Editionand I ¼ vðt À DtÞ=R þ iðt À DtÞ ð10:28Þwhere bk ¼ incremental inductance. If c is outside of the limits of segment K, the operation is switched to either k À 1or k þ 1. This changes the ½Y matrix. Because of the random nature of certain events, for example switching timeor lightning incidence, Monte Carlo (statistical) methods are sometimes used.Further information can be obtained from the references at the end of this book.10.7.3 Travelling-Wave ApproachLines and cables would require a large number of n circuits for accurate representa-tion. An alternative would be the use of the travelling-wave theory. Consider Figure 10.39, iðx; tÞ ¼ f1ðx À UtÞ þ f2ðx þ UtÞ vðx; tÞ ¼ Z0f1ðx À UtÞ þ Z0f2ðx þ UtÞwhere U ¼ speed of propagation and Z0 ¼ characteristic impedance. At node k, ikðtÞ ¼ vkðtÞ þ Ik Z0where Ik ¼ ikðt À tÞ À vmðt À tÞ=Z0 x=0 x=d k m ik im Vk Vm Ground Figure 10.39 Transmission line with earth return

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Overvoltages and Insulation Requirements 399 Z0 ik ( t ) vk ( t ) Ik Ground Figure 10.40 Equivalent circuit for single-phase lineand t ¼ d ¼ travel time U The equivalent circuit is shown in Figure 10.40. An advantage of this method isthat the two ends of the line are decoupled. The value of Ik depends on the currentand voltage at the other end of the line t seconds previously, for example ift ¼ 0:36 ms and Dt ¼ 100 ms, storage of four previous times is required. Detailed models for synchronous machines and h.v.d.c. converter systems aregiven in the references.Problems 10.1 A 345 kV, 60 Hz system has a fault current of 40 kA. The capacitance of a bus- bar to which a circuit breaker is connected is 25 000 pF. Calculate the surge impedance of the busbar and the frequency of the restriking (recovery) volt- age on opening. (Answer: 727 V, 8760 Hz) 10.2 A highly capacitive circuit of capacitance per phase 100 mF is disconnected by circuit breaker, the source inductance being 1 mH. The breaker gap breaks down when the voltage across it reaches twice the system peak line-to-neu- tral voltage of 38 kV. Calculate the current ﬂowing with the breakdown, and its frequency, and compare it with the normal charging current of the circuit. (Answer: 24 kA, 503 Hz; note I ¼ 2Vp=Z0)

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400 Electric Power Systems, Fifth Edition 10.3 A 10 kV, 64.5 mm2 cable has a fault 9.6 km from a circuit breaker on the sup- ply side of it. Calculate the frequency of the restriking voltage and the maxi- mum voltage of the surge after two cycles of the transient. The cable parameters are (per km), capacitance per phase ¼ 1.14 mF, resistance ¼ 5.37 V, inductance per phase ¼ 1.72 mH. The fault resistance is 6 V. (Answer: 374 Hz; 16 kV) 10.4 A 132 kV circuit breaker interrupts the fault current ﬂowing into a symmetri- cal three-phase-to-earth fault at current zero. The fault infeed is 2500 MVA and the shunt capacitance, C, on the source side is 0.03 mF. The system fre- quency is 50 Hz. Calculate the maximum voltage across the circuit breaker and the restriking-voltage frequency. If the fault current is prematurely chopped at 50 A, estimate the maximum voltage across the circuit breaker on the ﬁrst current chop. (Answer: 215.5 kV; 6.17 kHz; 43 kV) 10.5 An overhead line of surge impedance 500 V is connected to a cable of surge impedance 50 V. Determine the energy reﬂected back to the line as a percent- age of incident energy. (Answer: reﬂected energy ¼À 0.67 Â incident surge energy) 10.6 A cable of inductance 0.188 mH per phase and capacitance per phase of 0.4 mF is connected to a line of inductance of 0.94 mH per phase and capaci- tance 0.0075 mF per phase. All quantities are per km. A surge of 1 p.u. magni- tude travels along the cable towards the line. Determine the voltage set up at the junction of the line and cable. (Answer: 1.88 p.u) 10.7 A long overhead line has a surge impedance of 500 V and an effective resist- ance at the frequency of the surge of 7 V/km. If a surge of magnitude 500 kV enters the line at a certain point, calculate the magnitude of this surge after it has traversed 100 km and calculate the resistive power loss of the wave over this distance. The wave velocity is 3 Â 105 km/s. (Answer: 250 kV; 375 MW) 10.8 A rectangular surge of 1 p.u. magnitude strikes an earth (ground) wire at the centre of its span between two towers of effective resistance to ground of 200 and 50 V. The ground wire has a surge impedance of 500 V. Determine the voltages transmitted beyond the towers to the earth wires outside the span. (Answer: 0.44vi from 200 V tower and 0.17vi from 50 V tower) 10.9 A system consists of the following elements in series: a long line of surge impedance 500 ft, a cable (Z0 of 50 V), a short line (Z0 of 500 V), a cable (Z0 of 50 V), a long line (Z0 of 500 V). A surge takes 1 ms to traverse each cable (they are of equal length) and 0.5 ms to traverse the short line connecting the cables. The short line is half the length of each cable. Determine, by means of a

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Overvoltages and Insulation Requirements 401 500 Ω 50 Ω 500 Ω 50 Ω 500 Ω Cable Line Cable E 1.0 0.182 0.331 0 Microseconds 1 0.06 2 3 4 5 6 7 8 Voltage 1.0 0.8 0.6 6 8 10 12 14 16 18 20 22 24 0.4 Microseconds 0.2 0 024 Figure 10.41 Solution of Problem 10.9 lattice diagram, the p.u. voltage of the junction of the cable and the long line if the surge originates in the remote long line. (Answer: see Figure 10.41)10.10 A 3 p.h., 50 Hz, 11 kV star-connected generator, with its star point earthed, is connected via a circuit breaker to a busbar. There is no load connected to the busbar. The capacitance to earth on the generator side terminals of the circuit breaker is 0.007 mF per phase. A three-phase-to-earth short circuit occurs at the busbar with a symmetrical subtransient fault current of 5000 A. The fault is then cleared by the circuit breaker. Assume interruption at current zero. (a) Sketch the voltage across the circuit breaker terminals of the ﬁrst phase to clear. (b) Neglecting damping, calculate the peak value of the transient recovery voltage of this phase. (c) Determine the time to this peak voltage and hence the average rate of rise of recovery voltage. (Answer: (b) 17.96 kV; (c) 16.7 ms, 1.075 kV/ms) (From Engineering Council Examination, 1996)

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402 Electric Power Systems, Fifth Edition10.11 A very long transmission line AB is joined to an underground cable BC of length 5 km. At end C, the cable is connected to a transmission line CD of 15 km length. The transmission line is open-circuit at D. The cable has a surge impedance of 50 V and the velocity of wave propaga- tion in the cable is 150 Â 106 m/s. The transmission lines each have a surge impedance of 500 V. A voltage step of magnitude 500 kV is applied at A and travels along AB to the junction B with the cable. Use a lattice diagram to determine the voltage at:a. D shortly after the surge has reached D;b. D at a time 210 ms after the surge ﬁrst reaches B;c. B at a time 210 ms after the surge ﬁrst reaches B. Sketch the voltage at B over these 210 ms. (Answer: (a) 330 kV, (b) 823 kV and (c) 413 kV) (From Engineering Council Examination, 1995)

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11Substations and Protection11.1 IntroductionChapter 7 described techniques for the analysis of various types of faults whichoccur in a power system. The major uses of the results of fault analysis are for thespeciﬁcation of switchgear and the setting of protective relays, both being housed insubstations. However, the design of all items in an electrical plant is inﬂuenced byknowledge of fault conditions, for example, the bursting forces experienced byunderground cables. Circuit-breaker ratings are determined from the fault currents that may ﬂow attheir particular locations and are expressed in kA (fault current ﬂow) or MVA (shortcircuit level). The maximum circuit-breaker rating is of the order of 50 000–60 000 MVA, and this is achieved by the use of several interrupter heads in seriesper phase. Not only has the circuit breaker to extinguish the fault-current arc, withthe substation connections it has also to withstand the considerable forces set up byshort-circuit currents, which can be very high. Knowledge of the currents resulting from various types of fault at a location isessential for the effective speciﬁcation and setting of system protection. Faults on apower system result in high currents with possible loss of sychronism of generatorsand must be removed very quickly. Automatic means, therefore, are required todetect abnormal currents and voltages and, when detected, to open the appropriatecircuit breakers. It is the object of protection to accomplish this. In a large intercon-nected network, considerable design knowledge and skill is required to remove thefaulty part from the network and leave the healthy remainder working intact. There are many varieties of automatic protective systems, ranging from simpleover-current relays to sophisticated systems transmitting high-frequency signalsalong the power lines. The simplest but extremely effective form of protection is theover-current relay, which closes contacts and hence energizes the circuit-breakeropening mechanisms when currents due to faults pass through the equipment.Electric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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404 Electric Power Systems, Fifth Edition The protection used in a network can be looked upon as a form of insurance inwhich a percentage of the total capital cost (about 5%) is used to safeguard appara-tus and ensure continued operation of the power system when faults occur. In ahighly industrialized society the maintenance of an uninterrupted supply to con-sumers is of paramount importance and the adequate provision of protection sys-tems is essential. It is important the protection system discriminates between thoseitems of plant that are faulted and must be removed from the system and those thatare sound and should remain in service. Summarizing, protection and the automatic tripping (opening) of associated cir-cuit breakers has two main functions: (1) to isolate faulty equipment so that theremainder of the system can continue to operate successfully; and (2) to limit dam-age to equipment caused by overheating and mechanical forces .11.2 SwitchgearFor maintenance to be carried out on high voltage plant, it must be isolated from therest of the network and hence switches must be provided on each side. Differentdesigns of these switches have very different capabilities and costs. Circuit breakersare able to interrupt fault current but some isolators can interrupt load current whileothers can only be operated off-circuit that is with no voltage applied. Some isola-tors are able to close on to faults safely but not interrupt fault current. In additionearth (ground) switches are used to connect to earth the isolated equipment whichis to be worked on. Switchgear must always be operated within its capability as oth-erwise very considerable danger will arise. Owing to the high cost of circuit breakers, much thought is given, in practice, toobtaining the largest degree of ﬂexibility in connecting circuits with the minimumnumber of circuit breakers. Popular arrangements of switches are shown inFigure 11.1 and a typical feeder bay layout of a transmission substation inFigure 11.2. High-voltage circuit breakers, 11 kV and above, take ﬁve basic forms: oil immersed or bulk oil air blast small oil volume sulphur hexaﬂuoride (SF6) vacuum. In addition, air circuit breakers with long contact gaps and large arc-splitter plateshave been used at up to 11 kV, but they are bulky and expensive compared with theﬁve types listed above. All high voltage circuit breakers operate by their contacts separating and creatingan arc that is extinguished at a subsequent current-zero. The voltage across the sepa-rated contacts is then established in the medium of the interrupter (oil, air, SF6 orvacuum).

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Substations and Protection 405 Main Busbars Reserve Busbar isolator switches Circuit breakers Line isolators (a) Isolator switch (series switch) Reactor (b)(c) (d) Line tap arrangements (e)Figure 11.1 Possible switchgear arrangements for transmission and generation sub-stations, (a) Double-busbar selection arrangements, (b) Double-ring busbars with con-necting reactor, busbars can be isolated for maintenance, but circuits cannot betransferred from one side of the reactor to the other, (c) Open-mesh switching stations,transformers not switched, (d) Open mesh switching stations, transformers switched,(e) Closed-mesh switching stations. (Isolators are sometimes called series switches.)Arrangements can be indoors or outdoors

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406 Electric Power Systems, Fifth Edition8.0 mSealing end VT Disconnector CT Breaker Busbar Disconnector Disconnector Busbar 45.8 mFigure 11.2 Typical arrangement of a transmission substation bay. Width is 46 m foroutdoor (air) insulation and 8 m for indoor (SF6) insulation (Figure adapted from Alstom)11.2.1 The Bulk-Oil Circuit BreakerA cross-section of a bulk oil circuit breaker is shown in Figure 11.3(a). There are twosets of contacts per phase. The lower and moving contacts are usually cylindricalcopper rods and make contact with the upper ﬁxed contacts. The ﬁxed contactsconsist of spring-loaded copper segments which exert pressure on the lower contactrod, when closed, to form a good electrical contact. On opening, the lower contacts move rapidly downwards and draw an arc. Whenthe circuit breaker opens under fault conditions many thousands of amperes passthrough the contacts and the extinction of the arc (and hence the effective open-circuiting of the switch) are major engineering problems. Effective opening is only pos-sible because the instantaneous voltage and current per phase reduces to zero duringeach alternating current cycle. The arc heat causes the evolution of a hydrogen bubblein the oil and this high-pressure gas pushes the arc against special vents in a devicesurrounding the contacts, sometimes called a ‘turbulator’ (Figure 11.3(b)). As the lowestcontact moves downwards the arc stretches and is cooled and distorted by the gas andso eventually breaks. The gas also sweeps the arc products from the gap so that the arcdoes not re-ignite when the voltage rises to its full open-circuit value. Bulk oil circuitbreakers have been used widely at voltages from 6.6–150 kV and many are still inservice. For new installations they are being superseded by vacuum and SF6 designs.11.2.2 The Air-Blast Circuit BreakerFor voltages above 120 kV the air-blast breaker has been popular because of the fea-sibility of having several contact gaps in series per phase. Air normally stored at1.38 MN/m2 is released and directed at the arc at high velocities, thus extinguishingit. The air also actuates the mechanism of the movable contact. Figure 11.4 shows a132 kV air-blast breaker with two breaks (interrupters) per phase and its associatedisolator or series switch. Schematic diagrams of two types of air-blast head areshown in Figure 11.5 while the arrangement adopted for very high voltages, withgrading resistors, is shown in Figure 11.6.

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Substations and Protection 407 HV Terminal HV Terminal Operating RodMetal Tank Fixed Contact Moving Contact(a) Gas bubble Fixed contact Arc Path of Oil forced cool oil out Arc splitters (b) Moving contactFigure 11.3 (a) Cross-section of a bulk-oil circuit breaker. Three phases are placed inone tank (Figure adapted from Alstom). (b) Cross-jet explosion pot for arc extinction inbulk-oil circuit breakers (Figure adapted from Alstom)

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408 Electric Power Systems, Fifth Edition Series switch Terminal Interrupter Current Terminal Trip coil Blast pipe transformerTrip valve Change-over Air receiver valve Blast-valve piston Closing coil Closing Series-switch valve pistonFigure 11.4 Schematic arrangement for a 132 kV air-blast circuit breaker with twointerrupters per phase 3 11 3 2 2 1 1 4 4 (a) (b)Figure 11.5 Schematic diagrams of two types of air-blast head, (a) Axial ﬂow with axiallymoving contact, (b) Axial ﬂow with side-moving contact. 1 ¼ Terminal, 2 ¼ moving con-tact, 3 ¼ ﬁxed contact, and 4 ¼ blast pipe (Figure adapted from Siemens)

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Substations and Protection 409 6 interrupter heads per phaseporcelain resistorsupport grading capacitorinsulators main interruptor air receiver resistor interruptor Figure 11.6 275 kV air-blast circuit breaker Although air-blast circuit breakers have been developed and installed up to thehighest voltages they have now been largely superseded for new installations by theSF6 gas circuit breaker11.2.3 Small- or Low-Oil-Volume Circuit BreakersHere, the quenching mechanisms are enclosed in vertical porcelain insulationcompartments and the arc is extinguished by a jet of oil issuing from the moving(lower contact) as it opens (Figure 11.7). The volume of oil is much smaller than inthe bulk-oil type, thereby reducing hazards from explosion and ﬁre. Althoughmany oil circuit breakers are in use, particularly at distribution voltages, modernpractice is to use vacuum or SF6 to avoid the presence of ﬂammable liquids forcircuit interruption purposes.11.2.4 Sulphur Hexaﬂuoride (SF6) GasThe advantages of using sulphur hexaﬂuoride (SF6) as an insulating and inter-rupting medium in circuit breakers arise from its high electric strength and out-standing arc-quenching characteristics. SF6 circuit breakers are much smaller thanair blast circuit breakers of the same rating, the electric strength of SF6 at atmo-spheric pressure being roughly equal to that of air at the pressure of 10 atm. Temper-atures in the order of 30 000 K are likely to be experienced in arcs in SF6 and theseare, of course, well above the dissociation temperature of the gas (about 2000 K);however, nearly all the decomposition products are electronegative so that the elec-tric strength of the gas recovers quickly after the arc has been extinguished. Filtersare provided to render the decomposition products harmless and only a smallamount of ﬂuorine reacts with metallic parts of the breaker. A sectional view of anSF6, switchgear unit is shown in Figure 11.8.

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410 Electric Power Systems, Fifth Edition HV Terminal HV Terminal Interrupter Chamber Fixed (Porcelain enclosure) contact Arc Extinction Moving Device Contact Support Column Operating Operating rod MechanismFigure 11.7 Minimum oil circuit breaker (Figure adapted from Alstom) The switchgear also contains all necessary measuring and other facilities as fol-lows: SF6-insulated toroidal current transformers and voltage transformers, cableterminations, gas storage cylinders, cable isolators, grounding switches, bus isola-tors, and busbar system. Such substations are of immense value in urban areasbecause of their greatly reduced size compared with air blast switchgear. SF6 circuitbreakers rated at 45 GVA are available and designs for 1300 kV have been produced.

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Substations and Protection 411 5 4 34 1. Circuit breaker7 2. Spring - operated 6 1 mechanism 3. Disconnector 4. High - speed earthing switch 5. Low - speed earthing switch 6. Current transformer 7. Cable connection 2Figure 11.8 Sectional view of an SF6 insulated switchgear unit (Figure adapted fromAlstom) SF6 switchgear is now widely used at lower voltages from 6.6 to 132 kV for distri-bution systems. A typical interrupter for outdoor use, shown in Figure 11.9, isenclosed in a sealed porcelain cylinder with SF6 under about 5 atm pressure. Themovement of the contact forces gas into the opening contact by a ‘puffer’ action,thereby forcing extinction. SF6 interrupters are very compact and robust; theyrequire very little maintenance when mounted in metal cabinets to formswitchboards. SF6 is an extremely powerful greenhouse gas and if released into the atmospherehas an effect on global warming some 20 000 times more powerful than an equiva-lent amount of CO2. Hence during switchgear maintenance every effort is made torecover the SF6 and avoid venting it to atmosphere11.2.5 Vacuum InterrupterA pair of contacts opening in vacuum draws an arc which burns in the vaporizedcontact material. Consequently, the contact material and its arcing root shape arecrucial to the design of a commercial interrupter. A typical design is illustrated inFigure 11.10. The main advantages of a vacuum interrupter are: (1) the very small damage nor-mally caused to the contacts on operation, so that a life of 30 years can be expected

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412 Electric Power Systems, Fifth Edition washer nozzle and stream former contact cluster rooting electrode molecular sieveFigure 11.9 An SF6 single-pressure puffer-type interrupter (Figure adapted fromSiemens)without maintenance; (2) the small mechanical energy required for tripping; and(3) the low noise caused on operation. The nature of the vacuum arc depends on thecurrent; at low currents the arc is diffuse and can readily be interrupted, but at highcurrents the arc tends to be constricted. Electrode contour geometries have been pro-duced to give diffuse arcs with current densities of 106–108 A/cm2; electron veloc-ities of 108 cm/s are experienced in the arc and ion velocities of 106 cm/s.

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Substations and Protection 413 11 1 9 10 87 2 63 4 5 1. Fixed-contact stem 7. Moving-contact stem 2. Sputter-shield mounting 8. Bellows 3. Fixed contact 9. Sputter shield for bellows 4. Sputter shield 10. Moving contact 5. Grading-contact guide 11. Glass-ceramic bodyFigure 11.10 Constructional features of an 11 kV vacuum interrupter) (Figureadapted from Siemens) Considerable progress has been made in increasing the current-breaking capacityof vacuum interrupters, but not their operating voltage. Today, they are availablewith ratings of 500 MVA at 30 kV and are extensively used in 11, 20, and 25 kVswitchboards. A three-phase vacuum breaker with horizontally mounted interrupt-ers is shown in Figure 11.11 with SF6 insulation surrounding the interrupters.11.2.6 Summary of Circuit-Breaker RequirementsA circuit breaker must fulﬁl the following conditions:1. Open and close in the shortest possible time under any network condition.2. Conduct rated current without exceeding rated design temperature.

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414 Electric Power Systems, Fifth Edition Voltage Transformer Upper busbarsSecondaryWiringPanelVacuum IsolatableInterrupter Fuse Linkin SF6 CableMechanism Termination PointCircuit CurrentBreaker TransformersTruck Lower Busbars Note: In single busbar unit the lower busbars are replaced by an earthing point, allowing circuit earthing through the breaker in the position shown.Figure 11.11 Double-busbar MV switchgear, horizontal isolation (Figure adaptedfrom Alstom)3. Withstand, thermally and mechanically, any short-circuit currents.4. Maintain its voltage to earth and across the open contacts under both clean and polluted conditions.5. Not create any large overvoltage during opening and closing.6. Be easily maintained.7. Be not too expensive. Although air has now been largely overtaken by SF6 as an interrupting medium athigh voltages, it has given good performance over the years as it was easily able toachieve 2-cycle ($40 ms) interruption after receipt of a tripping signal to the operat-ing coil. The increasing need, at voltages up to 725 kV, to reduce interruption timesto 11/2 cycles to maintain stability and to reduce fault damage has meant that SF6switchgear has been required. A 60 kA rating has been possible with resistiveswitching to avoid unnecessary overvoltages, and a much lower noise level on oper-ation has further conﬁrmed SF6 as the preferred medium. Being factory-sealed andfully enclosed has considerably reduced radio-interference and audible noise due tocorona discharge.

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Substations and Protection 415 Increasing performance and low maintenance has led to the development of on-line monitoring of the health of switchgear and the associated connections and com-ponents (known as ‘condition monitoring’). This is particularly required for SF6enclosed units as even small dust particles can cause breakdown. Fast protectiongear is essential to match these developments in switchgear.11.3 Qualities Required of ProtectionThe effectiveness of protective gear can be described as:1. Selectivity or discrimination – its effectiveness in isolating only the faulty part of the system.2. Stability – the property of remaining inoperative with faults occurring outside the protected zone (called external faults).3. Speed of operation – this property is more obvious. The longer the fault current continues to ﬂow, the greater the damage to equipment. Of great importance is the necessity to open faulty sections of network before the connected synchro- nous generators lose synchronism with the rest of the system. A typical fault clearance time in h.v. systems is 80 ms, and this requires very high-speed relaying as well as very fast operation of the circuit breakers.4. Sensitivity – this is the level of magnitude of fault current at which operation occurs, which may be expressed in current in the actual network (primary cur- rent) or as a percentage of the current-transformer secondary current.5. Economic considerations – in distribution systems the economic aspect almost overrides the technical one, owing to the large number of feeders and transform- ers, provided that basic safety requirements are met. In transmission systems the technical aspects are more important. The protection is relatively expensive, but so is the system or equipment protected, and security of supply is vital. In trans- mission systems two separate protective systems are used, one main (or primary) and one back-up.6. Reliability – this property is self-evident. A major cause of circuit outages is mal- operation of the protection itself. On average, in the British system (not including faults on generators), nearly 10% of outages are due to mal-operation of protec- tion systems.11.3.1 Protective Zones and Back-Up ProtectionEach main protective scheme protects a deﬁned area or zone of the power system.These zones are designed to overlap so that the entire power system is protectedwith fast acting main protection. Overlapping zones of protection are shown inFigure 11.12. Back-up protection, as the name implies, is a completely separate arrangementwhich operates to remove the faulty part should the main protection fail to operate.The back-up system should be as independent of the main protection as possible,possessing its own current transformers and relays. Often, only the circuit-breakertripping and voltage transformers are common.

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416 Electric Power Systems, Fifth Edition 5 Motors 5 3 Station C Station A 2 Station B 3 3 3 3 1 1 4 2 GEN.Gen. 4Gen. 4 4 3 Station D 4Figure 11.12 Line diagram of a typical system and the overlapping zones of protection In distribution the application of back-up protection is not as widespread as intransmission systems; often it is sufﬁcient to apply it at strategic points only or over-lap the main protection further. Remote back-up is slow and usually disconnectsmore of the supply system than is necessary to remove the faulty part.11.4 Components of Protective Schemes11.4.1 Current Transformers (CTs)In order to obtain currents which can be used in control circuits and that are propor-tional to the system (primary) currents, current transformers are used. Often the pri-mary conductor itself, for example a busbar, forms a single primary turn (barprimary). Whereas instrument current transformers have to remain accurate onlyup to slight over-currents, protection current transformers must retain proportional-ity up to at least 20 times normal full load. The nominal secondary current rating ofcurrent transformers is now usually 1 A, but 5 A has been used in the past. A major problem can exist when two current transformers are used which shouldretain identical characteristics up to the highest fault current, for example in pilotwire schemes. Because of saturation in the silicon steel used and the possible exis-tence of a direct component in the fault current, the exact matching of such currenttransformers is difﬁcult.

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Substations and Protection 41711.4.1.1 Linear CouplersThe problems associated with current transformers have resulted in the develop-ment of devices called linear couplers, which serve the same purpose but, havingair cores, remain linear at the highest currents. These are also known as Rogowskicoils and are particularly suited to digital protection schemes.11.4.2 Voltage (or Potential) Transformers (VTs or PTs)These provide a voltage which is much lower than the system voltage, the nominalsecondary voltage being 110 V. There are two basic types: the wound (electromag-netic), virtually a small power transformer, and the capacitor type. In the latter atapping is made on a capacitor bushing (usually of the order of 12 kV) and thevoltage from this tapping is stepped down by a small electromagnetic voltagetransformer. The arrangement is shown in Figure 11.13; the reactor (X) and the V Lightning protectionConductorFoil gaptubes Tappaper/ oil Voltage transformer C To relaysCapacity Xbushing (a) I1 X C1 V C2 C Relay I2 burden (b)Figure 11.13 Capacitor voltage transformer, (a) Circuit arrangement, (b) Equivalentcircuit–burden ¼ impedance of transformer and load referred to primary winding

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418 Electric Power Systems, Fifth Editioncapacitor (C) constitute a tuned circuit which corrects the phase-angle error of thesecondary voltage. In h.v. systems, the capacitor divider is a separate unit mountedwithin an insulator. It can also be used as a line coupler for high frequency signal-ling (power line carrier, see Section 11.10).11.4.3 RelaysA relay is a device which, when supplied with appropriately scaled quantities,indicates an abnormal or fault condition on the power system. When the relay con-tacts close, the associated circuit-breaker trip-circuits are energized and thebreaker opens, isolating the faulty part of the power network. Historically electro-magnetic and semiconductor relays were installed and are still in use on the sys-tem. Modern practice is to install digital (numerical) protection. Although nowalmost all new relays use micro-processors and the measured quantities aremanipulated digitally, the underlying techniques are often those developed forelectro-mechanical relays.11.4.3.1 Induction-Disc RelayThis was used in the classic implementation of an over-current relay and was basedoriginally on the design of electro-mechanical energy meters. An aluminium discrotates between the poles of an electromagnet which produces two alternating mag-netic ﬁelds displaced in phase and space. The eddy currents due to one ﬂux and theremaining ﬂux interact to produce a torque on the disc. In early relays the ﬂux dis-placement was produced by a copper band around part of the magnet pole (shadingring) which displaced the ﬂux contained by it. Later designs of these electro-mechanical relays employed a wattmetric principle with two electromagnets, asshown in Figure 11.14. The current in the lower electromagnet is induced by transformer action from theupper winding and sufﬁcient displacement between the two ﬂuxes results. This,however, may be adjusted by means of a reactor in parallel with the secondarywinding. The basic mode of operation of the induction disc is indicated in the phasor dia-gram of Figure 11.15. The torques produced are proportional to F2i1 sin a andF1i2 sin a, so that the total torque is proportional to F1F2sin a or i1i2sin a as F1 isproportional to i1 and F2 to i2. This type of relay is fed from a current transformer (CT) and the sensitivity maybe varied by the plug arrangement shown in Figure 11.14. The time to the closing ofthe contacts is altered by adjusting the angle through which the disk has to rotate. The operating characteristics are shown in Figure 11.16. To enable a single charac-teristic curve to be used for all the relay sensitivities (plug settings) a quantityknown as the current (or plug) setting multiplier is used as the abscissa instead ofcurrent magnitude, as shown in Figure 11.16. The time multiplier adjusts the anglethrough which the disk rotates and so translates the curve vertically.

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Substations and Protection 419 Input Plug Tapping Bridge Upper magnet Disc carrying contacts Lower magnetFigure 11.14 Induction-disc relay12i1 i2 i 2 (90- α) 1 (a) i1 α 2 i2 i1 (b)Figure 11.15 Operation of disc-type electromagnetic relay, (a) Fluxes, (b) Phasordiagram. i1 and i2 are induced currents in disc

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420 Electric Power Systems, Fifth Edition 30 TMS = 1.0 10Time (secs) 1.0 0.1 2.0 3.0 5.0 10 20 1.0 Current as multiple of the relay setting (plug setting multiplier)Figure 11.16 Time-current characteristics of a typical induction disc in terms of theplug-setting multiplier. TMS ¼ time multiplier setting This relay characteristic is known as Inverse Deﬁnite Minimum Time (IDMT). Theoperating characteristic of a standard IDMT relay is deﬁned as: Standard IDMT curve: t ¼ 0:14 Â TMS ð11:1Þ PSM0:02 À 1TMS: Time Multiplier SettingPSM: Plug Setting Multiplier PSM ¼ I. IS I measured current IS relay setting current To illustrate the use of this curve (usually shown on the relay casing) the follow-ing example is given.

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Substations and Protection 421Example 11.1Determine the time of operation of a 1 A, 3 s overcurrent relay having a Plug Setting of125% and a Time Multiplier of 0.6. The supplying CT is rated 400 : 1 A and the faultcurrent is 4000 A.SolutionThe relay coil current for the fault ¼ (4000/400) Â 1 ¼ 10 A. The nominal relay coil cur-rent is 1 Â (125/100) ¼ 1.25 A. Therefore the relay fault current as a multiple of the PlugSetting ¼ (10/1.25) ¼ 8 (Plug Setting Multiplier). From the relay curve (Figure 11.16), thetime of operation is 3.3 s for a time setting of 1. The time multiplier (TM) controls thetime of operation by changing the angle through which the disc moves to close the con-tacts. The actual operating time ¼ 3.3 Â 0.6 ¼ 2.0 s. This can be obtained directly fromequation (11.1) as: t ¼ 0:14 Â TMS ¼ 0:14 Ã 0:6 ¼ 1:98 s PSM0:02 À 1 80:02 À 1 Induction-disc relays may be made responsive to real power ﬂow by feeding theupper magnet winding in Figure 11.14 from a voltage via a potential transformerand the lower winding from the corresponding current. As the upper coil will consistof a large number of turns, the current in it lags the applied voltage by 90, whereasin the lower (small number of turns) coil they are almost in phase. Hence, F1 is pro-portional to V, and F2 is proportional to I, and torque is proportional to F1F2 sin a,that is to VI sin (90 À a), or VI cos a (where a is the angle between V and I). The direction of the torque depends on the power direction and hence the relay isdirectional. A power relay may be used in conjunction with a current operated relayto provide a directional overcurrent property.11.4.3.2 Balanced BeamThe basic form of this relay is shown in Figure 11.17. The armatures at the ends ofthe beam are attracted by electromagnets which are operated by the appropriateparameters, usually voltage and current. A slight mechanical bias is incorporated tokeep the contacts open, except when operation is required. The pulls on the armatures by the electromagnets are equal to K1V2 and K2I2,where K1 and K2 are constants, and for operation (that is, contacts to close), that is. K1V2 > K2I2then rﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃ V < K2 or Z < K2 I K1 K1

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422 Electric Power Systems, Fifth Edition Beam Restraining Contacts coil Operating coil Figure 11.17 Schematic diagram of balanced-beam relay This shows that the relay operates when the impedance it ‘sees’ is less than a pre-determined value. The characteristic of this impedance relay, when drawn on an Rand jX axes, is a circle as shown in Figure 11.18.11.4.3.3 Distance RelaysThe balanced-beam relay, because it measures the impedance of the protected line,effectively measures distance to the fault. Two other relay forms also may be usedfor this purpose:1. The reactance relay which operates when V sin f < constant, having the straight line characteristic shown in Figure 11.18. I2. The mho or admittance relay, the characteristic of which is also shown in Figure 11.18. X Reactance Admittance (mho) 0 -R +R 0 Impedance -XFigure 11.18 Characteristics of impedance, reactance, and admittance (mho)relays shown on the R-X diagram

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Substations and Protection 423 These relays operate with any impedance phasor lying inside the characteristiccircle or below the reactance line. In practice modern distance relays calculate theoperating characteristics using digital micro-processors and some manufacturersoffer the ability to create more complex shapes (for example, trapezoids).11.4.3.4 Solid-State RelaysElectromechanical relays are vulnerable to corrosion, shock vibration and contactbounce and welding. They require regular maintenance by skilled personnel andover their life are expensive. Hence they were replaced initially with solid state elec-tronic relays that fulﬁlled similar functions using analogue and simple digital circuits. These relays are extremely fast in operation, having no moving parts and are veryreliable. Detection involving phase angles and current and voltage magnitudes aremade with appropriate electronic circuits. Most of the required current-time charac-teristics may be readily obtained. Inverse-characteristic, overcurrent and earth-faultrelays have a minimum time lag and the operating time is inversely related to somepower of the input (for example, current). In practical static relays it is advantageousto choose a circuit which can accommodate a wide range of alternative inverse timecharacteristics, precise minimum operating levels, and deﬁnite minimum times.11.4.3.5 Digital (Numerical) RelayingWith the advent of integrated circuits and the microprocessor, digital (numerical)protection devices are now the norm. Having monitored currents and voltagesthrough primary transducers (CTs and VTs), these analogue quantities are sampledand converted into digital form for numerical manipulation, analysis, display andrecording. This process provides a ﬂexible and very reliable relaying function,thereby enabling the same basic hardware units to be used for almost any kind ofrelaying scheme. With the continuous reduction in digital-circuit costs and increasesin their functionality, considerable cost-beneﬁt improvement ensues. It is now oftenthe case that the cost of the relay housing, connections and EMC protection domi-nates the hardware but, as is usual in digital systems, the software developmentand proving procedure are the most expensive items in the overall scheme. Since digital relays can store data, they are particularly suited to post-fault analysisand can even be used in a self-adaptive mode, which is impossible with traditionalelectro-mechanical or analogue electronic devices. Additionally, numerical relays arecapable of self-monitoring and communication with hierarchical controllers. By thesemeans, not only can fast and selective fault clearance be obtained, but also fault loca-tion can be ﬂagged to mobile repair crews. Minor (non-vital) protection-system faultscan also be indicated for maintenance attention. With the incorporation of a satellite-timing signal receiver using Global Positioning Satellite (GPS) to give a 1 ms synchro-nized signal, faults on overhead circuits can be located to within 300 m. The basic hardware elements of a digital relay are indicated in Figure 11.19. Thesignals coming from CTs and VTs (only a single input is shown even through mea-surements of each phase are connected to the relay) are ﬁrst connected to a surgeprotection circuit so as to prevent any surges entering into the relay. As these signals

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424 Electric Power Systems, Fifth Edition From Surge S/H Analogue CTs protection -to-digital From S/H converter CTs Surge Sample protection and hold ADC CB status Multiplexer Anti- aliasing filter Digital Keyboard CPU input and display CB trip Digital Serial Memory output Comms Figure 11.19 Basic components of a digital relaycontain dc offset, harmonics and noise they are ﬁrst ﬁltered using anti-aliasing ﬁl-ters. Then the ﬁltered analogue signals are converted to digital signals by the Ana-logue to Digital Converter (ADC). Since the ADC takes a ﬁnite conversion time,sample and hold circuits are used to retain the signals coming from CTs and VTswhile the multiplexer passes one signal to the ADC. Digitised signals are then proc-essed by the relay algorithms held in the memory. The memory has a volatile com-ponent (RAM) where some processing data is stored and non-volatile components(EPROM and ROM) where relay algorithms, and event records are kept. In numerical relays (also known as IEDs, Intelligent Electronic Devices, as theyperform more than protection functions) the hardware is similar for different relays.Depending on the application different relay algorithms are stored in the EPROM.For example, if the relay is an overcurrent relay, the relay algorithm reads settingssuch as the type of characteristic (standard inverse, very inverse, and so on), thetime and current settings. Digitized current measurements from the ADC are com-pared with the current setting to check whether the fault current is greater than thesetting. If the fault current exceeds the setting then a trip signal is generated with anappropriate delay. For the advanced student, many good references to numericalrelaying, as well as information from the major equipment manufacturers, are nowavailable (see Further Reading at the end of this book).11.5 Protection SystemsThe application of the various relays and other equipment to form adequateschemes of protection forms a large and complex subject. Also, the various schemesmay depend on the methods of individual manufacturers. The intention here is to

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Substations and Protection 425present a survey of general practice and to outline the principles of the methodsused. Some schemes are discriminative to fault location and involve several parame-ters, for example time, direction, current, distance, current balance, and phase com-parison. Others discriminate according to the type of fault, for example negative-sequence relays for unbalanced faults in generators, and some use a combination oflocation and type of fault. A convenient classiﬁcation is the division of the systems into unit and non-unittypes. Unit protection signiﬁes that an item of equipment or zone is being uniquelyprotected independently of the adjoining parts of the system. Non-unit schemes arethose in which several relays and associated equipment are used to provide protec-tion covering more than one zone. Non-unit schemes represent the most widelyused and cheapest forms of protection and these will be discussed ﬁrst.11.5.1 Over-Current Protection SchemesThis basic method is widely used in distribution networks and as a back-up in trans-mission systems. It is applied to generators, transformers and feeders. The arrange-ment of the components is shown in Figure 11.20. In the past the relay normally Circuit breaker CTs aa I residualbc Overcurrent c b relaysTrip Auxiliary (b)coil switch Trip circuit (a) a b c I ab Ica (c) IbcFigure 11.20 Circuit diagram of simple overcurrent protection scheme, (a) CTs in starconnection, (b) Phasor diagram of relay currents, star connection, (c) CTs in deltaconnection

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426 Electric Power Systems, Fifth Edition A B C DE x x xx xSupply 2s 1.5s 1s 0.5s 0.1stransformer FFigure 11.21 Application of overcurrent relays to feeder protectionemployed was the induction-disc type (Figure 11.14), but numerical over-currentrelays are now used. On an underground cable feeder, fault currents only reduce slightly as the posi-tion of the fault moves further from the in-feed point of supply. The impedance ofthe feeder is small compared with that of the source. Therefore it is necessary to usedifferent operating times to provide discrimination. Grading of relays across trans-formers, which introduce a large impedance in the circuit relies more on the varia-tions in fault current. A simple underground cable feeder is shown in Figure 11.21. Assume that thedistribution network has slow-acting circuit breakers operating in 0.3 s and therelays have true inverse-law characteristics. The relay operating times are graded toensure that only that portion of the feeder remote from the in-feed side of a fault isdisconnected. The operating times of the protection with a fault current equal to200% of full load are shown. Selectivity is obtained with a through-fault of 200% full load, with the faultbetween D and E as illustrated because the time difference between relay operationsis greater than 0.3 s. Relay D operates in 0.5 s and its circuit breaker trips in 0.8 s. Thefault current ceases to ﬂow (normal-load current is ignored for simplicity) and theremaining relays do not close their contacts. Consider, however, the situation whenthe fault current is 800% of full load. The relay operating times are now: A, 0.5 s [i.e.2 Â (200/800)]; B, 0.375 s [i.e. 1.5 Â (200/800)]; C, 0.25 s; D, 0.125 s; and the time forthe breaker at D to open is 0.125 þ 0.3 ¼ 0.425 s. By this time, relays B and C willhave operated and selectivity is not obtained. This problem can only be addressedby extending the settings on relays A to C. This illustrates the fundamental draw-back of this system, that is for correct discrimination to be obtained the times ofoperation close to the supply point become large.11.5.2 Directional Over-Current Protection SchemesTo obtain discrimination in a loop or networked system, relays with an added direc-tional property are required. For the system shown in Figure 11.22, directional andnon-directional over-current relays have time lags for a given fault current asshown. Current feeds into fault at the location indicated from both directions, andthe ﬁrst relay to operate is at B (0.6 s). The fault is now fed along route ACB only,

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Substations and Protection 427 Supply 1.4s A 1.4s0.2s 0.2s C1.0s 1.0s F 0.6s B 0.6sFigure 11.22 Application of directional overcurrent relays to a loop network. $ Relayresponsive to current ﬂow in both directions; ! relay responsive to current ﬂow in direc-tion of arrowand next the relay at C (1 s) operates and completely isolates the fault from the sys-tem. Assuming a circuit-breaker clearance time of 0.3 s, complete selectivity isobtained at any fault position. Note, however, that directional relays require a volt-age input, non-directional over-current relays do not.11.6 Distance ProtectionThe shortcomings of over-current relays graded by current and time led to the wide-spread use of distance protection. The distance between a relay and the fault is pro-portional to the ratio (voltage/current) measured by the distance relay. Relaysresponsive to impedance, admittance (mho), or reactance may be used. Although a variety of time-distance characteristics are available for providing cor-rect selectivity, the most popular one is the stepped characteristic shown inFigure 11.23. Here, A, B, C, and D are distance relays with directional propertiesand A and C only measure distance when the fault current ﬂows in the indicateddirection. Relay A trips its associated breaker instantaneously if a fault occurswithin the ﬁrst 80% of the length of feeder 1. For faults in the remaining 20% offeeder 1 and the initial 30% of feeder 2 (called the stage 2 zone), relay A initiatestripping after a short time delay. A further delay in relay A is introduced for faultsfurther along feeder 2 (stage 3 zone). Relays B and D have similar characteristicswhen the fault current ﬂows in the opposite direction.

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428 Electric Power Systems, Fifth Edition x Feeder - 1 xx Feeder - 2 x A BC FD Stage 3 back-up relayTime Stage 2 - relay Discriminating Stage - 1 margin instantaneous Circuit breaker relay operating time 80% 30%Figure 11.23 Characteristic of three-stage distance protectionThe selective properties of this scheme can be understood by considering a faultsuch as at F in feeder 2, when fault current ﬂows from A to the fault. For this fault,relay A starts to operate, but before the tripping circuit can be completed, relay Ctrips its circuit breaker and the fault is cleared. Relay A then resets and feeder 1remains in service. The time margin of selectivity provided is indicated by the verti-cal intercept between the two characteristics for relays A and C at the position F, lessthe circuit-breaker operating time. It will be noted that the stage 1 zones are arranged to extend over only 80% of afeeder from each end. The main reason for this is because practical distance relaysand their associated equipment have errors, and a margin of safety has to beallowed if incorrect tripping for faults which occur just inside the next feeder is tobe avoided. Similarly, the stage 2 zone is extended well into the next feeder toensure deﬁnite protection for that part of the feeder not covered by stage 1. Theobject of the stage 3 zone is to provide general back-up protection for the rest of theadjacent feeders. The characteristics shown in Figure 11.23 require three basic features: namely,response to direction, response to impedance and timing. These features need notnecessarily be provided by three separate relay elements, but they are fundamentalto all distance protective systems. As far as the directional and measuring relays areconcerned, the number required in any scheme is governed by the considerationthat three-phase, phase-to-phase, phase-to-earth, and two-phase-to-earth faultsmust be catered for. For the relays to measure the same distance for all types offaults, the applied voltages and currents must be different. With electro-mechanicalrelays it was common practice, therefore, to provide two separate sets of relays, oneset for phase faults and the other for earth faults, and either of these caters for three-phase faults and double-earth faults. Each set of relays was, in practice, usually fur-ther divided into three, since phase faults may concern any pair of phases, and, sim-ilarly, any phase can be faulted to earth. With digital relays, a pre-selection ofrelaying quantities allows just one processor to deal with all types of fault.

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Substations and Protection 42911.7 Unit Protection SchemesWith the ever-increasing complexity of modern power systems the methods of pro-tection so far described may not be adequate to afford proper discrimination, espe-cially when the fault current ﬂows in parallel paths. In unit schemes, protection islimited to one distinct part or element of the system that is disconnected if any inter-nal fault occurs. The protected part should be tripped for an internal fault, butremain connected with the passage of current ﬂowing into an external fault.11.7.1 Differential RelayingAt the extremities of the zone to be protected, the currents are continuously com-pared and balanced by suitable relays. Provided that the currents ﬂowing into andout of the zone are equal in magnitude and phase, no relay operation will occur. If,however, an internal fault (inside the protected zone) occurs, this balance will bedisturbed (see Figure 11.24) and the relay will operate. The current transformers atCurrent transformer Current to fault outside protected zone Line protected(a) P Relay R Pilot wires Q Fault (b) P R QFigure 11.24 Circulating current, differential protection (one phase only shown),(a) Current distribution with through-fault–no current in relay, (b) Fault on line, unequalcurrents from current transformers and current ﬂows in relay coil. Relay contacts closeand trip circuit breakers at each end of the line

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430 Electric Power Systems, Fifth Edition Current transformer I2 Operating I2 (I1 - I2) RelayProtected current Restraint operates zone (I1 - I2) Relay does not operate I1 I1 Relay I2 (b) (a)Figure 11.25 (a) Differential protection–circuit connections (one phase only)–relaywith bias, (b) Characteristic of bias relay in differential protection. Operating currentplotted against circulating or restraint currentthe ends of each phase should have identical characteristics to ensure perfect bal-ance on through-faults. Unfortunately, this is difﬁcult to achieve and a restraint, orbias, is applied (see Figure 11.25) that carries a current proportional to the full sys-tem current and restrains the relay operation on large through-fault currents. Thecorresponding characteristic is shown in Figure 11.25b. This principle (circulatingcurrent) may be applied to generators, feeders, transformers, and busbars, and pro-vides excellent selectivity. By suitable connections and current summation, teed ormulti-ended circuits can be protected using the same principles.11.8 Generator ProtectionLarge generators are invariably connected to their own step-up transformer and theprotective scheme usually covers both items. A typical scheme is shown inFigure 11.26, in which separate differential circulating-current protection schemesare used to cover the generator alone and the generator plus transformer. Whendifferential protection is applied to a transformer the current transformer on eachside of a winding must have ratios which give identical secondary currents. In many countries the generator neutral is often grounded through a distributiontransformer. This energizes a relay which operates the generator main and ﬁeldbreakers when a ground fault occurs in the generator or transformer. The groundfault is usually limited to about 10 A by the distribution transformer or a resistor,although an inductor has some advantages. The ﬁeld circuit of the generator mustbe opened when the differential protection operates in order to avoid the machinefeeding the fault.

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Substations and Protection 431Differential c.b. Main power transformer Negative sequence relayrelay R Auxiliary power Transformer transformer c.b. differential R relay Gen. Generator R differential relayGenerator R Overvoltageneutral ground relay Distribution Resistor transformer Figure 11.26 Protection scheme for a generator and unit transformer The relays of the differential protection on the stator windings (see Figure 11.27)are set to operate at about 10–15% of the circulating current produced by full-loadcurrent in order to avoid current-transformer errors. If the phase e.m.f. generated bythe winding is E, the minimum current for a ground fault at the star-point end, andhence with the whole winding in circuit, is E/R, where R is the neutral effectiveresistance. For a fault at a fraction x along the winding from the neutral I1 = I2 I1 I2 I1 = I2 I1 I2 (a) I1 = I2 E I2 (b) (c) I1 xFigure 11.27 Generator winding faults and differential protection, (a) Phase-to phasefault, (b) Intertum fault, (c) Phase-to-earth fault

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432 Electric Power Systems, Fifth Edition(Figure 11.27(c)), the fault current is xE/R and 10–15% of the winding isunprotected. With the neutral grounded via the transformer, R is high and earth faultsare detected by a sensitive relay across the transformer secondary. With an interturnfault (turn-to-turn short circuit) on a phase of the stator winding (Figure 11.27(b)),current balance at the ends is retained and no operation of the differential relay takesplace, the relays operate only with phase-to-phase and ground faults. On unbalanced loads or faults the negative-sequence currents in the generatorproduce excessive heating on the rotor surface and generally (I22t) must be limited toa certain value for a given machine (between 3 and 4 (p.u.)2 s for 500 MW machines),where t is the duration of the fault in seconds. To ensure this happens, a relay isinstalled which detects negative-sequence current and trips the generator mainbreakers when a set threshold is exceeded. When loss of excitation occurs, reactive power (Q) ﬂows into the machine, and ifthe system is able to supply this, the machine will operate as an induction generator,still supplying power to the network. The generator output will oscillate slightly asit attempts to lock into synchronism. Relays are connected to isolate the machinewhen a loss of ﬁeld occurs, which can be readily detected by a reactance relay.11.9 Transformer ProtectionA typical protection scheme is shown in Figure 11.28(a), in which the differentialcirculating-current arrangement is used. The speciﬁcation and arrangement ofthe current transformers is complicated by the main transformer connections andratio. Current-magnitude differences are corrected by adjusting the turns ratio ofthe current transformers to account for the voltage ratio at the transformer termi-nals. In a differential scheme the phase of the secondary currents in the pilot wiresmust also be accounted for with star-delta transformers. In Figure 11.28(a) the pri-mary-side current transformers are connected in delta and the secondary ones instar. The corresponding currents are shown in Figure 11.28(b) and it is seen that theﬁnal (pilot) currents entering the connections between the current transformersare in phase for balanced-load conditions and hence there is no relay operation. Thedelta current-transformer connection on the main transformer star-winding alsoensures stability with through earth-fault conditions which would not be obtainedwith both sets of current transformers in the star connection. The distribution of cur-rents in a Y-D transformer is shown in Figure 11.29. Troubles may arise because of the magnetizing current inrush when energizationoperates the relays, and often restraints sensitive to third-harmonic components ofthe current are incorporated in the relays. As the inrush current has a relatively highthird-harmonic content the relay is restrained from operating. Faults occurring inside the transformer tank due to various causes give rise to thegeneration of gas from the insulating oil or liquid. This may be used as a means offault detection by the installation of a gas/oil-operated relay in the pipe between thetank and conservator. The relay normally comprises hinged ﬂoats and is known asthe Buchholz relay (see Figure 11.30). With a small fault, bubbles rising to ﬂow into

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Substations and Protection 433 Protected transformer iA IA Primary Secondary Ia ia ib iB IB Ib ic iC IC Ic R Y B Relay (a) operating inputs IA R iA- iCPrimary IB B IC IA iC- i B IB IC Y i B- i A R Ia IaSecondary Ib Ic B Ia (b) Ic Ic Ib Y Ib CT Secondary Pilot wire Transformer currents currents currentsFigure 11.28 Differential protection applied to Y-D transformers, (a) Current trans-former connections, (b) Phasor diagrams of currents in current transformersthe conservator are trapped in the relay chamber, disturbing the ﬂoat which closescontacts and operates an alarm. On the other hand, a serious fault causes a violentmovement of oil which moves the ﬂoats, making other contacts which trip the maincircuit breakers.

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434 Electric Power Systems, Fifth Edition Ia 3000/5 11.5 kV 30 MVA 69 kV Ia- Ic 600/5 aA Ia Ia Ia Ib - Ia Ia - Ic b Ib Ib Ic - Ib Ib- Ia c Ib IbB Ic Ic OP A Ic RC Ic Ic - Ib Ia - Ic Ia- Ic Ib- Ia OP B Ib- Ia R Ic - Ib OP C Ic - Ib RFigure 11.29 Currents in Y-D transformer differential protection. OP ¼ operate input;R ¼ restraint input Alarm Relay operation Float to disconnect Float To expansion chamber Transformer (conservator tank Figure 11.30 Schematic diagram of Buchholz relay arrangement

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Substations and Protection 43511.10 Feeder Protection11.10.1 Differential Pilot WireThe differential system already described can be applied to feeder protection. Thecurrent transformers situated at the ends of the feeder are connected by insulatedwires known as pilot wires. In Figure 11.24, P and Q must be at the electrical mid-points of the pilot wires, and often resistors are added to obtain a geographicallyconvenient midpoint. By reversing the current-transformer connections(Figure 11.31) the current-transformer e.m.f.s oppose and no current ﬂows in thewires on normal or through-fault conditions. This is known as the opposed voltagemethod. As, under these conditions, there are no back ampere-turns in the current-transformer secondary windings, on heavy through-faults the ﬂux is high andsaturation occurs. Also, the voltages across the pilot wires may be high under thiscondition and unbalance may occur due to capacitance currents between the pilotwires. To avoid this, sheathed pilots are used. In some differential protection schemes it is necessary to transmit the secondarycurrents of the current transformers considerable distances in order to comparethem with currents elsewhere. To avoid the use of wires from each of the three CTsin a three-phase system, a summation transformer is used which gives a singlephase output, the magnitude of which depends on the nature of the fault. Thearrangement is shown in Figure 11.32, in which the ratios of the turns are indicated.On balanced through-faults there is no current in the winding between c and n. Thephase (a) current energizes the 1 p.u. turns between a and b and the phasor sum of Iaand Ib ﬂows in the 1 p.u. turns between b and c. The arrangement gives a muchgreater sensitivity to earth faults than to phase faults. When used in phase-comparison systems, however, the actual value of output current is not importantand the transformer usually saturates on high-fault currents, so protecting the sec-ondary circuits against high voltages. Pilot wires may be installed underground or strung on towers. In the lattermethod, care must be taken to cater for the induced voltages from the power-lineconductors. Sometimes it is more economical to rent wires from the telephone com-panies, although special precautions to limit pilot voltages are then required.CT CT RFigure 11.31 Pilot-wire differential feeder protection – opposed voltage connections

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436 Electric Power Systems, Fifth Edition CTs One end of To protection system protected zone 7 p.u. turns a 1 b c1 5 n Figure 11.32 Summation transformerA typical scheme using circulating current is shown in Figure 11.33, in which a mix-ing or summation device is used. With an internal fault at F2 the current enteringend A will be in phase with the current entering end B, as in H.V. networks thefeeder will inevitably be part of a loop network and an internal fault will be fedfrom both ends. VA and VB become additive, causing a circulating current to ﬂow,which causes relay operation. Thus, this scheme could be looked on as a phase-com-parison method. If the pilot wires become short-circuited, current will ﬂow and therelays can give a false trip. In view of this, the state of the wires is constantly moni-tored by the passage of a small d.c. current. F1 CT’s F2 CT’s F3 X AX BX Mixing Insulating network VA transf. Mixing VB network Restraint Pilot Operating wires Restraint Sensing units OperatingFigure 11.33 Differential pilot-wire practical scheme using mixing network (or summa-tion transformer) and biased relays. VA ¼ VB for external faults, for example at F1 andF3; V A ¼6 V B for internal faults, for example at F2

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Substations and Protection 43711.10.2 Carrier-Current ProtectionBecause of pilot capacitance, pilot wire relaying is limited to line lengths below40 km (30 miles). Above this, distance protection may be used, although for discrim-ination of the same order as that obtained with pilot wires, carrier current equip-ment may be used. In carrier-current schemes a high-frequency signal in the band80–500 kHz and of low power level (1 or 2 W) is transmitted via the power-line con-ductors from each end of the line to the other. It is not convenient to superimposesignals proportioned to the magnitude of the line primary current, and usually thephases of the currents entering and leaving the protected zone are compared. Alter-natively, directional and distance relays are used to start the transmission of a car-rier signal to prevent the tripping of circuit breakers at the line ends on throughfaults or external faults. On internal faults other directional and distance relays stopthe transmission of the carrier signal, the protection operates, and the breakers trip.A further application, known as transferred tripping, uses the carrier signal to trans-mit tripping commands from one end of the line to the other. The tripping commandsignal may take account of, say, the operation or nonoperation of a relay at the otherend (permissive intertripping) or the signal may give a direct positive instruction totrip alone (intertripping). Carrier-current equipment is complex and expensive. The high-frequency signalis injected on to the power line by coupling capacitors and may be coupled either toone phase conductor (phase-to-earth) or between two conductors (phase-to-phase),the latter being technically better but more expensive. A schematic diagram of aphase-comparison carrier-current system is shown in Figure 11.34. The wave or linetrap is tuned to the carrier frequency and presents a high impedance to it, but a lowX To Line trapX Line lineX trap Coupling Oscillator Coupling capacitors equipmentSummation LS Modulator network Amplifier Fault- LSdetectingnetwork HS Trip Phase Receive Band relay discrimin- Amplifier pass HS ator filterFigure 11.34 Block diagram measuring and control equipment for carrier-currentphase-comparison scheme. LS ¼ low-set relay; HS ¼ high-set relay

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438 Electric Power Systems, Fifth Edition External fault Internal fault End A End B End A End B Primary Primary current current Secondary Secondary current current Transmitted Transmitted signal signal Rectified Rectified received received signal signal Locally Locally derived derived voltage voltage Tripping Tripping output output Figure 11.35 Waveforms of transmitted signals in carrier-type line protectionimpedance to power-frequency currents; it thus conﬁnes the carrier to the protectedline. Information regarding the phase angles of the currents entering and leaving theline is transmitted from the ends by modulation of the carrier by the power current,that is by blocks of carrier signal corresponding to half-cycles of power current(Figure 11.35). With through faults or external faults the currents at the line ends areequal in magnitude but 180 phase-displaced (that is, relative to the busbars, itleaves one bus and enters the other). The blocks of carrier occur on alternate half-cycles of power current and hence add together to form a continuous signal, whichis the condition for no relay operation. With internal faults the blocks occur in thesame half-cycles and the signal comprises non-continuous blocks; this is processedto cause relay operation. The currents from the current transformers are fed into a summation devicewhich produces a single-phase output that is fed into a modulator (Figure 11.34).This combines the power frequency with the carrier to form a chopped 100%modulated carrier signal, which is then ampliﬁed and passed to the line-couplingcapacitors. The carrier signal is received via the coupling equipment, passedthrough a narrow-bandpass ﬁlter to remove any other carrier signals, ampliﬁed,and then fed to the phase discriminator which determines the relative phasebetween the local and remote signals and operate relays accordingly. The equip-ment is controlled by low-set and high-set relays that start the transmission of thecarrier only when a relevant fault occurs. These relays are controlled from a start-ing network. Although expensive, this form of protection is very popular on over-head transmission lines.

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Substations and Protection 43911.10.3 Voice-Frequency SignallingIncreasingly, as the telephone network expands with the use of ﬁbre-optic cablingand high-frequency multiplexing of communication channels, it is no longer possi-ble to obtain a continuous metallic connection between the ends of unit-protectionschemes. Consequently, differential protection, utilizing voice frequencies (600–4000 Hz), has been developed. The pilot wires (after the insulating transformers)feed into a voltage-to-frequency (v.f.) converter with send and receive channels. Thesummated or derived 50 (or 60) Hz relaying signal is frequency-modulated onto thechannel carrier and demodulated at the far end. The signal is then compared withthe local signal and if a discrepancy is detected then relay operation occurs. It isimportant to ensure that both signal magnitude and phase are faithfully reproducedat each end after demodulation, independently of the channel characteristics whichcan change or distort during adverse transmission conditions. The usual methodsbuilt into the v.f. channel relays to ensure reliability utilize an automatic control ofsignal level and a regular measurement (say, every 100 ms) of channel delay so thatphase correction can be applied to the demodulated signal.Problems11.1 A 132 kV supply feeds a line of reactance 13 V which is connected to a 100 MVA 132/33 kV transformer of 0.1 p.u. reactance as shown in Figure 11.36. The transformer feeds a 33 kV line of reactance 6 V which, in turn, is con- nected to an 80 MVA, 33/11 kV transformer of 0.1 p.u. reactance. This trans- former supplies an 11 kV substation from which a local 11 kV feeder of 3 V reactance is supplied. This feeder energizes a protective overcurrent relay through 100/1 A current transformers. The relay has a true inverse-time char- acteristic and operates in 10 s with a coil current of 10 A. If a three-phase fault occurs at the load end of the 11 kV feeder, calculate the fault current and time of operation of the relay. (Answer: 1575 A; 6.35 s)132 kV T1 33kV T2 11 kVFigure 11.36 Circuit for Problem 11.1

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440 Electric Power Systems, Fifth Edition11.2 A ring-main system consists of a number of substations designated A, B, C, D, and E, connected by transmission lines having the following impedances per phase (V): AB (1.5 þ j2); BC (1.5 þ j2); CD (1 þ j1.5); DE (3 þ j4); EA (1 þ j1). The system is fed at A at 33 kV from a source of negligible impedance. At each substation, except A, the circuit breakers are controlled by relays fed from 1500/5 A current transformers. At A, the current transformer ratio is 4000/5. The characteristics of the relays are as follows:Current (A) 7 9 11 15 20Operating times of relays at A, D and C 3.1 1.95 1.37 0.97 0.78Operating times at relays at B and E 4 2.55 1.8 1.27 1.02 Examine the sequence of operation of the protective gear for a three-phase symmetrical fault at the midpoint of line CD. Assume that the primary current of the current transformer at A is the total fault current to the ring and that each circuit breaker opens 0.3 s after the clos- ing of the trip-coil circuit. Comment on the disadvantages of this system.11.3 The following currents were recorded under fault conditions in a three-phase system: IA ¼ 1500 ﬀ 45 A, IB ¼ 2500 ﬀ 150 A, IC ¼ 1000 ﬀ 300 A. If the phase sequence is A-B-C, calculate the values of the positive, negative and zero phase-sequence components for each line. (Answer: I0 ¼À200 þ j480 A, IA1 ¼ 20 À j480, IA2 ¼ 1240 þ j1060)11.4 Determine the time of operation of 1 A standard IDMT over-current relay hav- ing a Plug Setting (PS) of 125% and a Time Multiplier setting (TMS) of 0.6. The CT ratio is 400 : 1 and the fault current 4000 A. (Answer: 2 sec)11.5 The radial circuit shown in Figure 11.37 employs two IDMT relays of 5 A. The plug settings of the relays are 125% and time multiplier of relay A is 0.05 sec. 200/5 100/5 x x B A F = 1400A Figure 11.37 Circuit for Problem 11.5

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Substations and Protection 441 66 kV 11 kV 200/1 600/1 Figure 11.38 Circuit for Problem 11.6 Find the time multiplier setting of relay B to coordinate two relays for a fault of 1400 A. Assume a grading margin of 0.4 sec. (Answer: TMS 0.15)11.6 A 66 kV busbar having a short circuit level of 800 MVA is connected to a 15 MVA 66/11 kV transformer having a leakage reactance of 10% on its rating as shown in Figure 11.38. i. Write down the three-phase short circuit current (in kA) for a fault on the 66 kV terminals of the transformer. ii. Calculate the three-phase short circuit current (in kA) for a fault on the 11 kV feeder. iii. The 11 kV relay has a Current Setting (Plug Setting) of 100% and Time Mul- tiplier of 0.5. Use the IDMT characteristic to calculate the operating time for a three-phase fault on the 11 kV feeder. iv. The 66 kV relay has a Current Setting (Plug Setting) of 125%. Choose a Time Multiplier to give a grading margin of 0.4 seconds for a fault on the 11 kV feeder. v. What is the operating time for a three-phase fault on the 66 kV winding of the transformer? vi. In practice, what form of transformer protection would operate ﬁrst for a fault on the 66 kV winding? (Answers: 7 kA, 6.6 kA, 1.4 sec, 0.5, 1 sec)11.7 A three-phase, 200 kVA, 33/11 kV transformer is connected as delta-star. The CTs on the 11 kV side have turns ratio of 800/5. What should be the CT ratio on the HV side? (Answer: 150/5)11.8 A three zone distance protection relay is located at busbar A as shown in Figure 11.39 (square). The VT ratio is 132 kV/220 V and CT ratio is 1000/1. Find the impedance setting for zone 1 and zone 2 protection of the relay

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442 Electric Power Systems, Fifth Edition B j 0.1 j 1.0A j 0.5 j 0.2 D C P = 0.4 j 0.2 P = 0.2 Q = 0.0 Q = 0.0 P = 0.6 Q = 0.0Figure 11.39 Circuit for Problem 11.8assuming that zone 1 covers 80% of line section AB and zone 2 covers 100% ofline section AB plus 30% of the shortest adjacent line. (All quantities are in p.u. on 100 MVA, 132 kV base)(Answers: j0.135, j0.27 p.u.)

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12Fundamentals of theEconomics of Operationand Planning of ElectricitySystemsFigure 12.1 shows the structure of the traditional electricity system with its fourmain sectors: generation, bulk transmission, distribution and demand. In most of the industrialized countries electricity systems have been designedsince the 1950s to support economic growth and beneﬁt from developments ingeneration technology. The conventional power system is characterized by smallnumbers of very large generators, mainly coal, oil, hydro and nuclear, and morerecently gas fuelled generation. Typical power station ratings would be from afew hundred MWs to several thousand MWs. These stations are connected to avery high voltage transmission network (operating typically at 275/220 kV and400 kV). The role of the transmission system is to provide bulk transport of elec-tricity from these large stations to demand centres, that is cities. The electricity isthen taken by the distribution networks through a number of voltage transforma-tions (typically Extra High Voltage (132/110 kV, 33/35 kV), High Voltage (20/11/10 kV)) and Low Voltage (400/230 V). These networks provide the ﬁnal deliveryof electricity to consumers. The ﬂow is unidirectional from higher to lower volt-age levels. In a typical consumer bill in most European countries, generation costs (invest-ment and operation) make up about 60% while transmission and distribution net-work costs are in the order of 10% and 30% respectively. Annual electricity losses intransmission and distribution networks in the majority of industrialized countriesare around 1–3% and 4–7% respectively.Electric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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444 Electric Power Systems, Fifth Edition Large Central Source of energy Generation and system control Transmission Bulk transport of electricity EHV Distribution Delivery system, HV Distribution radial networks LV DistributionDemandFigure 12.1 Schematic diagram of the conventional power system with four mainsectors: generation, bulk transmission, distribution and demand Key drivers of system operation and planning are economy and security. Morerecently these concerns have been supplemented by sustainability considerationsaimed at limiting the CO2 emissions that accompany the use of fossil fuel in powerstations.12.1 Economic Operation of Generation SystemsThe input-output characteristic of a fossil fuel generating set is of great importancewhen economic operation is considered. Typical characteristics of a fossil fuelledgenerator are shown in Figure 12.2. The graph of the heat rate input (GJ/h) againstoutput (MW) is known as the Willans line. For large turbines with a single valve,and for gas turbines, the heat rate is approximately a no-load offset with a reason-ably straight line over the operating range, Figure 12.3(a). Most steam turbines inBritain are of this type. With multivalve turbines (as used in the USA) the Willansline curves upwards more markedly Figure 12.2(b). The incremental heat rate is deﬁned as the slope of the input output curve at anygiven output. The value taken for the incremental heat rate of a generating set issometimes complicated because if only one or two shifts are being operated (thereare normally three shifts per day) heat has to be expended in banking boilers whenthe generator is not required to produce output. Instead of plotting heat rate or incremental heat rate against power output for theturbine-generator, the incremental fuel cost may be used. This is advantageous

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Fundamentals of the Economics of Operation and Planning 445Heat input (GJ/h) (a) (b) No-load 1 heat loss 0.8 Output (p.u.)Figure 12.2 Input-output characteristic of a turbogenerator set. (a) single valvesteam turbine or gas turbine, (b) steam turbine with multiple valveswhen allocating load to generators for optimum economy as it incorporates differ-ences in the fuel costs of the various generating stations. Usually, the graph of incre-mental fuel cost against power output can be approximated by a straight line(Figure 12.3). Consider two turbine-generator sets having the following differentincremental fuel costs, dC1/dP1 and dC2/dP2, where C1 is the cost of the fuel inputto unit number 1 for a power output of P1 and, similarly, C2 and P2 relate to unitnumber 2. It is required to load the generators to meet a given requirement in themost economic manner. Obviously the load on the machine with the higher dC/dPwill be reduced by increasing the load taken by the machine with the lower dC/dP.This transfer will be beneﬁcial until the values of dC/dP for both sets are equal, afterwhich the machine with the previously higher dC/dP now becomes the one with the ) dc dp ( (1)cost (2) fuel P1 P2 Electrical power output p.u.IncrementalFigure 12.3 Idealized graphs of incremental fuel cost against output for twomachines sharing a load equal to P1 and P2

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446 Electric Power Systems, Fifth Editionlower value, and vice versa. There is no economic advantage in further transfer ofload, and the condition when dC1/dP1 ¼ dC2/dP2 therefore gives optimum econ-omy; this can be seen by considering Figure 12.3. The above argument can beextended to several machines supplying a load. Generally, for optimum economy theincremental fuel cost should be identical for all contributing turbine-generator sets on freegovernor action. In practice, most generators will be loaded to their maximumoutput. These concepts will now be further explained and illustrated on economic des-patch problems, using examples with multiple thermal generation units.Example 12.1Generating units ﬁred with fossil fuels are characterized by their input–output curvesthat deﬁne the amount of fuel required to produce a given and constant electricalpower output for one hour. Consider two coal-ﬁred steam units with minimum stablegeneration of 100 MW and 230 MW (that is, the minimum power that can be producedcontinuously) and maximum outputs of 480 MW and 660 MW. On the basis of measure-ments taken at the plants, the input–output curves of these units are estimated asH1ðP1Þ ¼ 110 þ 8:2P1 þ 0:02P21 ½MJ=h ð12:1ÞH2ðP2Þ ¼ 170 þ 9:6P2 þ 0:01P22 ½MJ=h ð12:2Þ The hourly cost of operating these units is obtained by multiplying the input–outputcurve by the cost of fuel (expressed in £/MJ). Assuming that the cost of coal is 1.3 £/MJ,the cost curve of these generating units is given by the following expressions:C1ðP1Þ ¼ 143 þ 10:66P1 þ 0:026 þ P12 ½£=h ð12:3ÞC2ðP2Þ ¼ 221 þ 12:48P2 þ 0:013 þ P22 ½£=h ð12:4Þ System demand is assumed to be 750 MW. If the minimum and maximum produc-tion constraints are temporarily ignored, the task of allocating production can bedescribed as an optimization problem, with the objective function to be minimized rep-resenting the total production costs, that is C1ðP1Þ þ C2ðP2Þ, while ensuring that thejoint production of the units equals to the system demand, that is P1 þ P2 ¼ 750 MW:min C1ðP1Þ þ C2 ðP2ÞP1 ;P2subject to ð12:5ÞP1 þ P2 ¼ 750 Using the supply-demand balance equation, one of the variables can be eliminated,so that the number of variables reduces to one. At the optimum, the derivative of theobjective function should be equal to zero:

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Fundamentals of the Economics of Operation and Planning 447m)Pi1ndPdC11hðP141Þ3 þ C2ð750 À P1Þ ) d ½C1ðP1Þ þ C2 ð750 À P1Þ ¼ 0 þ dP1 Á ð750 À P1Þ þ 0:013 i 10:66P1 þ 0:026P12 þ 221 þ 12:48 Á ð750 À P1Þ2 ¼ 0 ð12:6Þ Finding the derivative of the cost function and then solving the resulting linear equa-tion gives P1 ¼ 273:33 MW. Given that the joint production of the units must equal sys-tem demand, that is P1 þ P2 ¼ 750 MW, the production of the second unit isP2 ¼ 476:67 MW. We now also observe that both units operate within the constraints on their outputs,that is within minimum and maximum generation, and hence the solution obtainedrepresents the true optimum. A key feature of this solution, is that the marginal cost of each of the units, at theoptimal levels of production are equal: dC1ðP1Þ ¼ 10:66 þ 0:052P1 ¼ 24:87 ½£=MWh dP1 p1 ¼273:33 0:026P2 ð12:7Þ dC2ðP2Þ ¼ 12:48 þ p2 ¼476:67 ¼ 24:87 ½£=MWh dP2 The marginal cost of 24.87 £/MWh is the cost of increasing production from the pres-ent demand of 750 MW by one additional MW. Alternatively, the original problem (12.5) can be expressed more formally using thecorresponding Lagrange function: min C1ðP1Þ þ C2ðP2Þ þ l Á ½750 À ðP1 þ P2Þ ð12:8Þ P1 ;P2 ;l Note that in this formulation the term l Á ½750 À ðP1 þ P2Þ is added to the objectivefunction of Equation (12.5) to form Equation (12.8). However, the value of this termis zero and hence the value of the objective functions Equations (12.5) and (12.8) arethe same, as the generation-load balance constraint will be satisﬁed (that is,P1 þ P2 ¼ 750 MW). Optimality conditions require that the ﬁrst partial derivatives of the objective func-tion with respect to all three variables (P1; P2 and l) are zero: d fC1ðP1Þ þ C2ðP2Þ þ l Á ½750 À ðP1 þ P2Þg ¼ 0 dP1 d fC1ðP1Þ þ C2ðP2Þ þ l Á ½750 À ðP1 þ P2Þg ¼ 0 ð12:9Þ dP2 d fC1ðP1Þ þ C2ðP2Þ þ l Á ½750 À ðP1 þ P2Þg ¼ 0 dl

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448 Electric Power Systems, Fifth EditionThis leads to: 10:66 þ 0:052P1 ¼ l ð12:10Þ 12:48 þ 0:026P2 ¼ l P1 þ P2 ¼ 750 Equations (12.10) are three linear equations with three unknowns, that is P1; P2 and l.Solving these gives: P1 ¼ 273:33 MW ð12:11Þ P2 ¼ 476:67 MW l ¼ 24:87½£=MWh Note that the ﬁrst two expressions in (12.10) represent the marginal cost of the twounits respectively and that the optimality condition requires that these are equal.1 Parameter l is a Lagrange multiplier associated with the production-demand bal-ance equation and represents numerically the change in the total production costs thatwould correspond to an incremental change in demand around the operating point of750 MW. This parameter hence represents system marginal cost at a demand of 750 MW. This solution is also presented in Figure 12.4. Finally, the total cost of supplying this demand over an hour is CT ¼ C1ðP1Þ þ C2ðP2Þ ¼ 14 122:77 ½£ ð12:12ÞMarginal Gen 1cost £/MWh Gen 2 24.87 P1 = 273.33 MW P2 = 476.66Figure 12.4 Marginal cost curves of two generators with output and the systemmarginal cost indicated for demand of 750 MW1 This will be the case when the units operate within their operating constraints but not at the limitsof these constraints.

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Fundamentals of the Economics of Operation and Planning 449Which gives the average production cost aC ¼ CT ¼ 18:83 ½£=MWh ð12:13Þ P1 þ P2 In this particular case the marginal production cost is higher than the average pro-duction cost. If demand is to be charged at the marginal cost (which corresponds to theeconomic optimum), there will be some proﬁt that the generators will make from sup-plying this load.Example 12.2Four generators are available to supply a power system peak load of 472.5 MW. Thecost of power Ci(Pi) from each generator, and maximum output: C1ðP1Þ ¼ 200 þ 15P1 þ 0:20P12 ½£=h Max: output 100 MW ð12:14Þ C2ðP2Þ ¼ 300 þ 17P2 þ 0:10P22 ½£=h Max: output 120 MW C3ðP3Þ ¼ 150 þ 12P3 þ 0:15P23 ½£=h Max: output 160 MW C4ðP4Þ ¼ 500 þ 2P4 þ 0:07P24 ½£=h Max: output 200 MW The spinning reserve is to be 10% of peak load. As in the previous example, the objective is to calculate the optimal loading of eachgenerator, system marginal cost, and the cost of operating the system at peak. We will start with deriving the marginal cost for each of the generators. These aregiven (in £/MWh) by: dC1 ¼ 15 þ 0:40P1 P1 100 MW dP1 P2 120 MW P3 160 MW dC2 ¼ 17 þ 0:20P2 P4 200 MW dP2 ð12:15Þ dC3 dP3 ¼ 12 þ 0:30P3 dC4 ¼ 2 þ 0:14P4 dP4 The marginal cost curves are plotted in Figure 12.5 for each of the generators up totheir maximum output. From the curves, at 30 £/MWh the outputs are: P1 ¼ 23 MW ð12:16Þ P2 ¼ 64 MW P3 ¼ 60 MW P4 ¼ 200 MW Total ¼ 347 MW

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450 Electric Power Systems, Fifth Edition 80Marginal cost (£/MWh) 60 P2 44 £ /MWh P4 40 P1 P3 20 0 40 80 120 160 200 MW Generator output Figure 12.5 Marginal cost curves for four generatorsHence, P4 runs at the full output and the remaining generators must sum to 272.5 MW. At 40£/MWh the outputs sum to 446 MW. At 41 £/MW P2 reaches its maximum of120 MW, and P1 ¼ 40 MW and P3 ¼ 98 MW (total 458 MW). This leaves an additional 14.5 MW to be found from P1, and P3. Adjusting the marginal cost to 44 £/MWh provides 44 MW from P1 and 108 MWfrom P3, giving a total of 472 MW, which is considered to be near enough when usingthis graphical method. The spinning reserve on these generators is 108 MW, more thanenough to cover 10% of 472.5 MW. The cost of operating the system for 1 h at peak is C1ðP1Þ ¼ 200 þ 15:44 þ 0:20 Á 442 ¼ 1247 ½£=h ð12:17Þ ð12:18ÞSimilarly ð12:19Þ C2ðP2Þ ¼ 3780 ½£=h C3ðP3Þ ¼ 3195 ½£=h C4ðP4Þ ¼ 3700 ½£=hThis gives a total of 11 922£/h. This gives an average cost of aC ¼ 11; 922 ¼ 25:2 ½£=MWh 472:5whereas the system marginal cost is 44£/MWh.

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Fundamentals of the Economics of Operation and Planning 45112.2 Fundamental Principles of Generation System PlanningWhen planning investment in generation systems, it is important to consider bothinvestment and operating costs. Generally, generation planning time horizons may bedecades to ensure that load growth is adequately met. We will consider a static generation planning problem in order to highlight basicprinciples that are used to optimise generation technology mixes. In order to com-pare the cost associated with different generation technologies, operating costs areconsidered over a period of one year, while the investment costs are annuitized. Thetotal annual operation and investment costs are then divided by the installed capac-ity expressed in £/kW. Consider three generation technologies i. Base-load generation, characterized by high capital (IBL) and low per unit operat- ing costs (OBL), such as nuclear.ii. Mid-merit generation, characterized by medium capital (IMM) and per unit oper- ating costs (OMM), andiii. Peak-load generation, characterized by low capital (IPL) and high per unit operat- ing costs (OPL). The objective of the generation planning exercise is to determine the capacitiesof individual generation technologies to be installed so that the total investmentand operating costs are minimized. The choice of generation type depends onexpected operating hours, which is illustrated in Figure 12.6. It should be£/kWy DSM Peaking: low Mid merit : mid capital & high op capital & mid op costs costs (eg. OCGT) (eg. CCGT)Generation cost Cheapest way of supplying Base load: high a kW of demand capital & low op Cheaper not to supply costs (eg. nuclear) t1 Hours t0 required t3 t2 Figure 12.6 Determining operating hours of individual plant technologies

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452 Electric Power Systems, Fifth Editionstressed that the expected operating times are a function of ﬁxed and operatingcosts of generation only. Base-load plant will be operating under all demand condition for the entire year(t0 ¼ 8760 h). The operating time for mid-merit plant is t1 and for peaking-plant is t2.For the duration of t3, a portion of demand will be curtailed. For operating time t1 the per unit cost of investment and operation per year(£/kW/y) of mid-merit or base-load plant are equal:IMM þ t1 Á OMM ¼ IBL þ t1 Á OBL ð12:20Þ In other words, we are indifferent if we invest in 1 MW of base load or mid-meritgeneration technology, if this unit will be operating for a period of t1. It is now straightforward to evaluate the operating time t1 that determines theannual duration of operation of mid-merit plant: t1 ¼ IBL À IMM ð12:21Þ OMM À OBL For values of annuitized investment costs of IBL ¼ 170 [£/kW/y] and IMM ¼50 [£/kW/y] and operating costs of OBL ¼ 0.01 [£/kWh] and OMM ¼ 0.04 [£/kWh]associated with base-load and mid-merit generation respectively, the operating timet1 ¼ 4,000 h (per annum). Similarly, the operating time t2, that determines the duration of the operation ofpeak-load generation, can be calculated by comparing annuitized investment costand annual operating costs of mid-merit and peak-load generation: t2 ¼ IMM À IPL ð12:22Þ OPL À OMM For values of investment and operating costs of peak-load plant of IPL ¼30 [£/kW/y] OPL ¼ 0.06 [£/kWh] the operating time t2 ¼ 1,000 h (per annum). Not all demand should be served, that is it is not economically efﬁcient to supplyall demand. The operating time t3, indicates the period of time in which some of thedemand should be curtailed. The duration of interruptions will be determined bythe value that consumers attribute to receiving electricity. This is referred to as theValue of Lost Load (VOLL) expressed in £/kWh which represents the amount thatcustomers receiving electricity would be willing to pay to avoid a disruption in theirelectricity service. The operating time t3, is determined by comparing the annuitizedinvestment cost and annual operating costs of peak-load generation against the costof not supplying the load:t3 ¼ IPL OPL % IPL ð12:23Þ VOLL À VOLL Given the assumed value of investment cost of peaking plant IPL ¼ 30 [£/kW/y]and assuming that the VOLL ¼ 5 £/kWh ()OPL) the duration of efﬁcient interrup-tions would be 6 hours per year.

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Fundamentals of the Economics of Operation and Planning 453MW Demand side measures Peaking capacity Mid merit capacity Base load capacity Hours 8760t3 t2 t1 t0Figure 12.7 Determining the capacities of different generation plant technologies onthe basis of operating hours calculated in Figure 12.6 Once the critical times are determined, we then use the annual load durationcurve2 (LDC) that links generating capacity requirements and capacity utilization,to determine the corresponding capacities of each generation type, as illustrated inFigure 12.7. The plant mix determined by the capacities indicted in Figure 12.7, will minimizethe overall generation investment and operating costs. For reliable power supply sufﬁcient production capacity has to be provided tocover the consumption and additional reserves required for reliable power systemoperation due to the unanticipated events, outages of generation or demandincreases. In order to mitigate the effects of these uncertainties, the installed capacity2 The LDC is derived from a chronological load curve but the demand data is ordered in descending orderof magnitude, rather than chronologically.

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454 Electric Power Systems, Fifth Edition Probability Peak Generation 100% Demand available at Generation forecast Peak MW Risk of insufficient generation Planning Margin (Loss of Load Probability)Figure 12.8 Capacity margins and risk of available generation not being able tomeet demandof generation is greater than expected peak demand. The difference between availa-ble production capacity and peak load is called the capacity margin (see Figure 12.8). Historically, a capacity margin of around 20% was considered to be sufﬁcient toprovide adequate generation security. The two lines in Figure 12.8 represent the probability density function of availablegeneration (on the right) and probability density function of peak demand (line onthe left). Clearly the probability that all installed generators will be available is low,but also the probability that all installed generation will be unavailable is evenlower; hence the bell-shape type probability density function. We also observe someuncertainty around expected peak demand. In the case that the available generationis less than peak demand, then demand will need to be curtailed. Generation adequacy is often assessed by determining the likelihood of therebeing insufﬁcient generation to meet demand, or in other words, by calculatingthe risk that supply shortages will occur. Clearly, the larger the generationcapacity margin the lower the risk of supply shortages. The risk of supply short-age can be quantiﬁed by the probability that peak demand will not be met bythe available generation. This is called Loss of Load Probability (LOLP). TheLOLP has been used as an index to measure generation capacity adequacy andrisks of interruptions. The typical value of LOLP considered to be acceptable isaround or below 10%. This is often interpreted to mean that demand curtail-ment, occurring once in 10 years on average, would be considered acceptable.Similarly, the index LOLE (Loss of Load Expectation) has also been used inmany countries to assess generation capacity adequacy and risks of interrup-tions. The LOLE is the sum of LOLPs calculated for every hour of system opera-tion during a year, multiplied by the corresponding number of hours in a year.Hence, LOLE represents the expected number of hours (in a year) in whichdemand may exceed available generation and hence some demand would becurtailed.

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Fundamentals of the Economics of Operation and Planning 455Example 12.3Calculate the LOLP (at peak demand) and LOLE for the system in Figure 12.9. In this example a simple three-generator system is considered. Their capacities(P1, P2 and P3) and corresponding availabilities (A1, A2 and A3) are indicated onFigure 12.9(a). The ﬁrst step is to evaluate various State Probabilities and CumulativeProbabilities as indicated in Table 12.1.P1 = 60MW P2 = 190MW P3 = 110MW A1= 97% A 2= 98% A 3= 95% MW Load260 MW (a)145 MW Annual Load Duration Curve45 MW Hours 1900 h 2000 h 8760 h (b)Figure 12.9 Evaluating LOLP and LOLE for a simple generation system

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456 Electric Power Systems, Fifth EditionTable 12.1 State and cumulative probabilities for the system of Figure 12.9(a)Available State Cumulative ProbabilityGeneration (MW) Probability Probability of available generation being greater than or equal to generation stated in column 1360 0.90 307 0.90 307300 0.02 793 0.93 100250 0.04 753 0.97 853190 0.00 147 0.98 000170 0.01 843 0.99 843110 0.00 057 0.99 90060 0.00 097 0.99 9970 0.00 003 1.00 000The ﬁrst column in the table indicates all the states in which this system can be found interms of available generation capacity. For example, when all generators are available,the total generation capacity equals 360 MW and the probability of this state is equal tothe product of individual generator availabilities (0.97Ã0.98Ã0.95 ¼ 0.90 307). Similarly,if generator 1 and 2 are available, while generator 3 is not, the total available capacity is250 MW, and the probability of that state is 0.04 753, which is the product of availabili-ties of units 1 and 2 with unavailability of unit 3, that is 0.97Ã0.98Ã(1–0.95) ¼ 0.04 753.The last column represents the cumulative probability density function, indicating theprobability that the available generation capacity is greater than or equal to the genera-tion stated in column 1. From such a table we can calculate the LOLP for the peak load of 260 MW. If thegeneration system is to supply this load, the available capacity must be greater than orequal to 260 MW. This will be the case if either all three units are available (availablegeneration capacity to supply demand is 360 MW), or the two larger units are available(units 2 and 3) and unit 1 is unavailable (the available generation capacity to supplydemand is 300 MW). In all other states of the system, the available generation capacityis less than 260 MW. Therefore, the probability of the generation system being able tosupply load of 260 MW is 0.931 and LOLP is 6.9%. Evaluation of LOLE requires calculation of LOLPs for each demand level (see Figure12.9(b)). For a load demand of 145 MW, the probability that available generation will behigher or equal to this load is 99.843%, as indicated in the third column in the table.Hence the probability that the available generation will not be able to meet demand,that is LOLP, is 0.157%. Finally, for the off-peak demand of 45 MW, the probabilitythat available generation will be higher or equal to this load is 99.997%, and henceLOLP, is 0.003%. Given the duration of the individual demand levels given in Figure12.9(b), the expected number of hours of inadequate supply in which demand wouldexceed available generation in the time horizon of one year (LOLE), can be calculatedas follows: LOLE ¼ 100 Ã 0:069 þ 1900 Ã 0:00157 þ 6760 Ã 0:00003 ¼ 6:9 þ 2:983 þ 0:203 ¼ 10:09 h

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Fundamentals of the Economics of Operation and Planning 45712.3 Economic Operation of Transmission SystemsThree characteristics of the transmission network determine its performance andassociated operating costs: (i) level of congestion (ii) losses and (iii) interruptionscaused by outages of transmission circuits. These factors may impact the optimalgeneration despatch discussed in Section 12.1. The discussion will start with theexamination of the effect of the ﬁrst characteristics. The transmission network is generally characterized by signiﬁcant power ﬂowsfrom exporting to importing areas. This is the case when generation cost in export-ing area is lower than the generation cost in importing area. This can be interpretedas location arbitrage: the transport of power from low cost area to high cost area bytransmission network facilities. However, in order to keep the power ﬂows over the transmission network withinpermissible limits it may be required that occasionally generators in exporting areaswith lower marginal costs are constrained off, and generators in importing areaswith higher marginal cost are constrained-on, which increases the cost of the opera-tion of the generation system. In other words, the presence of transmission networkconstraints will alter the despatch discussed in Section 12.1. The process of optimis-ing the operation of the generation system while maintaining the network ﬂowswithin the prescribed limits is known as the Security Constrained Optimal PowerFlow (SC-OPF) problem. The SC-OPF is distinguished from simple economic des-patch as the solution obtained (generation outputs of individual generators) notonly minimizes the total generation costs but also simultaneously respects the con-straints of the transmission system including limits in both pre- and post-contin-gency states. This could also include optimisation of network control devices suchas phase shifting transformers. In the long term, the costs of network congestion are managed by appropriateinvestment in network capacity3 as will be analyzed in the next section.Example 12.4To examine the impact of limited network capacity, Example 12.1 is considered again,with the two generators G1 and G2 connected to two different busbars (busbar 1 andbusbar 2 respectively) that are connected by two identical transmission lines of 200 MWcapacity each. Note that the total system load of 750 MW is split between the two bus-bars: 670 MW in busbar 1 and 80 MW in busbar 2, as indicated in Figure 12.10. Assum-ing that the economic despatch obtained in Example 12.1 (unconstrained economicdespatch) can be implemented, the corresponding power ﬂows are evaluated and pre-sented in Figure 12.10 (losses are ignored in this exercise). Given that production inbusbar of 2 of 476.7 MW and that local demand is 80 MW, power transfer from busbar 2to busbar 1 is 396.7 MW. In busbar 1, local demand of 670 MW is supplied from localproduction of 273.3 MW and the contribution from busbar 2 of 396.7 MW.3 Through cost beneﬁt based transmission network design, the costs of network congestion are to bebalanced with the cost of network reinforcement.

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458 Electric Power Systems, Fifth Edition Busbar 1 max flow = 200 MW Busbar 2273.3 MW 198.3 MW 476.7 MW 198.3 MWG1 G2 D1 = 670 MW max flow = 200 MW D2 = 80 MWFigure 12.10 Implementing unconstrained economic despatch in Example 12.1The resulting power ﬂow from busbar 2 to busbar 1 of F ¼ 396.7 MW appears to beacceptable given that the ﬂows in individual circuits are within their capacities. How-ever, modern transmission networks operate in accordance with N-1 security criteria,which requires that no transmission circuit should be overloaded (or cause any voltageand stability problems) in case of a single circuit outage. In this particular case, a faulton one of the lines would result in a signiﬁcant overload on the remaining line, as theﬂow would increase from 198.3 MW to 396.7 MW, as illustrated in Figure 12.11. In order to avoid this unacceptable operating state, generation despatch in the intactsystem (with both lines in service) will need to change. This can be formally stated as the following optimisation problem: min C1 ðP1Þ þ C2ðP2Þ P1 ;P2 st P1 þ P2 ¼ 750 ð12:24Þ F 200 F ¼ 670 À P1 ¼ P2 À 80 Busbar 1 max flow = 200 MW Busbar 2273.3 MW 396.7 MW 476.7 MWG1 max flow = 200 MW G2 D1 = 670 MW D2 = 80 MWFigure 12.11 Post fault power ﬂow leading to overload of the remaining circuit

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Fundamentals of the Economics of Operation and Planning 459 Busbar 1 max flow = 200 MW Busbar 2470 MW 100 MW 280 MW 100 MWG1 G2 D1 = 670 MW max flow = 200 MW D2 = 80 MWFigure 12.12 Security constrained economic despatch and secure transferbetween busbars 1 and 2The solution could however be found by inspection, as it is obvious that more efﬁcientgenerator G2 in busbar 2 will be required to reduce its output. In order to re-establishpower balance, the generator in busbar 1 will need to increase its output by exactly thesame amount. The new, security constrained optimal despatch is shown in Figure 12.12. Clearly, outage of one transmission circuit would not now overload the remaining line.The corresponding despatch is therefore called security constrained economic despatch andthe resulting secure power transfer of F ¼ 200 MW satisﬁes the N-1 operation rules. How-ever, this new despatch is characterized by higher generation operating costs. Substitutingthe modiﬁed generation outputs (indicated in Figure 12.12) in expressions 12.3 and 12.4,the total generation costs of the security-constrained despatch can be calculated:CTSC ¼ C1ð470Þ þ C2ð280Þ ¼ 15 631:2 ½£=h ð12:25Þ The increase in costs above the cost of the unconstrained schedule given by Equation(12.12) is clearly driven by the limited capacity of the transmission network that con-nects export and import areas. The cost of constraints, also called cost of security orout-of-merit generation cost, can be calculated as the difference in costs between secu-rity-constrained (SC) and unconstrained (U) generation dispatches:DCS ¼ CTSC À CUT ¼ 15 631:2 À 14 122:77 ¼ 1 508:43 ½£=h ð12:26Þ Another important observation is that because of the network constraints, the twogenerators operate at different marginal costs. This is in contrast to Example 12.1 inwhich both generators operate at the same marginal cost. Hence in a constrained sys-tem, instead of a system marginal cost that would be applied to all network nodes, mar-ginal costs are different in different locations:dC1ðP1Þ ¼ 10:66 þ 0:052P1 ¼ 35:1 ½£=MWh ð12:27Þ dP1 ¼ 12:48 þ 0:026P2p1¼470 ¼ 19:76 ½£=MWhdC2ðP2Þ p2 ¼280 dP2

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460 Electric Power Systems, Fifth Edition The marginal generation cost in the exporting area (busbar 2) is lower than in the importing area (busbar 1). This can be compared with system marginal cost of 24.8 [£/MWh] calculated in Example 12.1. There are many limitations in the above treatment for example not considering losses, reactive power being neglected and the limitations on power ﬂows driven stabil- ity constraints is not taken into account.12.4 Fundamental Principles of Transmission System PlanningReinforcement of existing networks increases the amount of power that can be trans-ported securely from exporting to importing areas. However, investments in newtransmission equipment are costly and should therefore be undertaken only if theycan be justiﬁed economically. In order to deliver maximum economic beneﬁts, theelectricity supply industry should follow the path of least-cost long-term develop-ment. This requires a coordinated approach to the optimization of the generationand transmission operation and development. Optimizing the transmission networkin isolation from the generation resources would almost certainly not meet theabove objective. The cost-beneﬁt approach to optimizing transmission investment is presented inFigure 12.13. Establishing the optimal level of network capacity that should beinvested into should balance (i) the beneﬁts associated with reduction of out-of-merit generation costs (cost of constraints) and costs of losses and (ii) cost of net-work investment that is dependent on the network capacity built. The optimal levelof network capacity that should be built corresponds to the minimum of the totalcosts composed of cost of constraints and network investment. Cost Total Cost (£/y) Cost ofCost of InvestmentConstraints& Losses Optimal capacity Network Capacity (MW)Figure 12.13 Cost-beneﬁt approach to transmission planning

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Fundamentals of the Economics of Operation and Planning 461Example 12.5To illustrate the basic principles for transmission expansion problem, we will considerthe above example and discuss the economic impact of reinforcing the networkbetween busbars 1 and 2, above the existing secure transfer capacity of 200 MW. Thiswill include an analysis of the impact of network reinforcement that will provide addi-tional secure transfer capability of DF on cost of security DCS. The out-of-merit generation cost for generation production of P1 and P2 in busbars 1and 2 respectively, is given by:DCS ¼ CSTC À CUT ¼ À þ 10:66P1 þ 0:026P21 þ 221 þ 12:48P2 þ 0:013P22Á À 14 122:77 143 ð12:28Þ By applying power balance equations to busbar 1 and busbar 2, generation produc-tions can be expressed as a function of the additional power ﬂow DF that can be facili-tated through the network reinforcement: P1 þ DF þ 200 ¼ D1 ¼> P1 ¼ 470 À DF ð12:29Þ P2 ¼ 200 þ DF þ D2 ¼ DF þ 280 By substituting generation productions P1 and P2 given in (12.29) into equation(12.28), we ﬁnally obtain: DCS ¼ 0:039DF2 À 15:34DF þ 1508:4 ð12:30Þ Adding secure transfer capacity of DF (on top of the existing 200 MW capacity),reduces network constraint (out-of-merit generation) costs, as shown in Figure 12.14.Cost (£/h) 1600 1400 1200 20 40 60 80 100 120 140 160 180 200 1000 Additional secure capacity (MW) 800 600 400 200 0 0Figure 12.14 Cost of network constraints as a function of reinforced networkcapacity (DF)

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462 Electric Power Systems, Fifth EditionNote that for the unconstrained economic despatch in Example 12.1, generationP1 ¼ 273.33 MW and P2 ¼ 476.67 MW, constraint costs are zero (DCS ¼ 0). Asshown in Example 12.4, this despatch would result in the maximum power ﬂowfrom busbar 2 to busbar 1 of 396.7 MW, and given that the existing networkcapacity is 200 MW, the network reinforcement should not exceed 196.7 MW(DF 196:7 MW). As indicated in Figure 12.13, the reduction in out-of-merit generation costs need to beconsidered together with the cost of network reinforcement that enables this increase inefﬁciency of system operation (reduction in out-of-merit generation costs). For the sake of simplicity, it is assumed that the annuitized cost of reinforcing thetransmission corridor is a linear function of additional secure capacity delivered DF: ÂÃ ð12:31Þ CDF ¼ 2 Á k Á L Á DF £=y Where L is the length of the line in kilometers, k is the proportionality factor rep-resenting the annuitized marginal cost of reinforcing 1 km of the transmission lineand its dimensions are £/(MWÁkmÁyear). The multiplication factor 2 signiﬁes thattwo circuits would be reinforced. Dividing this annuitized cost by the number ofhours in a year (t0 ¼ 8760 h), the transmission investment costs are expressed perhour and can now be directly compared with hourly savings in generation costs. Ifwe assume, for this example, that the line is 300 km long and that k ¼ 87.6£/(MWÁkmÁyear) CDF ¼ CDF=8760 ¼ 6 Á DF ½£=h ð12:32ÞCost (£/h) 1600 Out-of-merit generation cost (£/h) 1400 Transmission cost (£/h) 1200 Total cost (£/h) 1000 20 40 60 80 100 120 140 160 180 200 800 Additional secure capacity (MW) 600 400 200 0 0Figure 12.15 Finding optimal additional secure capacity by minimizing the sumof out-of-merit generation cost and transmission investment costs

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Fundamentals of the Economics of Operation and Planning 463 Busbar 1 max flow = 320 MW Busbar 2350 MW 160 MW 400 MW 160 MWG1 G2 D1 = 670 MW max flow = 320 MW D2 = 80 MWFigure 12.16 Generation despatch and ﬂows after reinforcing the corridor andproviding 120 MW of additional secure transferIn order to determine the optimal capacity of network reinforcement, the sum of net-work costs and out-of-merit generation costs needs to be minimized: È DFÞ þ À0:039DF2 À 15:34DF þ ÁÉ ) 0:078DF À 9:34 ¼ 0 ð12:33Þmin ð6 1508:4Which gives optimal network reinforcement: DF ¼ 120 MWChanges in out-of-merit generation operating costs, transmission reinforcement costand the total costs, as a function of additional capacity DF are presented in Figure 12.15.Figure 12.16 shows the optimal despatch and secure power transfer over the trans-mission corridor after the circuit has been reinforced to provide an additional 120 MWof secure transfer capacity.The transmission reinforcement, will impact the marginal costs are two locations: dC1ðP1Þ ¼ 28:9 ½£/MWh dP1 P1 ¼360 dC2ðP2Þ ð12:34Þ dP2 P2 ¼400 ¼ 22:9 ½£/MWh We observe that the price difference between the two locations (28.9 [£/MWh] À 22.9[£/MWh] is equal to the incremental investment costs of the line: dCDF ¼ 6 ½£/MWh ð12:35Þ dDF12.5 Distribution and Transmission Network Security ConsiderationsHistorically, the design and structure of electricity distribution and transmissionnetworks was driven by an overall design philosophy developed to support large-scale generation technologies. Modern network design standards specify minimumrequirements for transmission capacity, taking into account both network securityand economics. Generally, the network must be able to continue to function after a

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464 Electric Power Systems, Fifth EditionMW 33/66 kV 132 kV 11 kV100 10 0.4 kV11 3 Time to restore supplies (hours) 24n-2 n-1 Redundancy n-0Figure 12.17 Philosophy of network design standards, level of redundancy increaseswith the amount of demand suppliedloss of a single circuit (or a double circuit on the same tower). After a loss of a circuitdue to fault (e.g. lightning strike) the remaining circuits that take over the load of thefaulty line, must not become overloaded. In the majority of industrialized countries, the level of security in distribution andtransmission networks is deﬁned in terms of the time taken to restore power sup-plies following a predeﬁned set of outages. Security levels in distribution systemsare graded according to the total amount of peak power supplied and potentiallybeing curtailed due to outages of network assets. In general, network redundancyhas been speciﬁed according to a principle that the greater the amount of powerwhich can be lost, the shorter the recommended restoration time. A simpliﬁedillustration of this network design philosophy, as implemented in the UK, is pre-sented in Figure 12.17. For instance small demand groups, less than 1 MW peak, areprovided with the lowest level of security, and have no redundancy (N-0 security).This means that any fault will cause an interruption and the supply will be restoredonly after the fault is repaired. It is expected that this could take up to about 24 hours. n-0 ¼ no redundancy in security (must wait for repair of network) n-1 ¼ one level of network redundancy; and n-2 ¼ two levels of network redundancy For demand groups with peak loads between 1 MW and 100 MW, although a sin-gle fault may lead to an interruption, the bulk of the lost load should be restoredwithin 3 hours. This requires the presence of redundancy, as 3 hours is usually insuf-ﬁcient to implement repairs, but it does allow network reconﬁguration activities.

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Fundamentals of the Economics of Operation and Planning 465 1 0.8Coincidence factor 0.6 0.4 0.2 0 100 10,000 1 10 1,000 100,000 Number of customersFigure 12.18 Load coincidence factor as a function of the number of typicalhouseholdsSuch network designs are often described as providing n-1 security. For demandgroups with peak load larger than 100 MW, the networks should be able to providesupply continuity to customers following a single circuit outage (with no loss of sup-ply) but also provide signiﬁcant redundancy to enable supply restoration following afault on another circuit superimposed on the existing outage, that is n-2 security. This network design practice has effectively determined the characteristicsregarding quality of service as experienced by end customers. The performance of11 and 0.4 kV networks has a dominant effect on the overall quality of service seenby end customers. The vast majority of interruptions (about 90%) have their cause inthese networks. This is primarily driven by the radial design of these networks, asany fault on a circuit leads to an interruption to some customers (unlike transmis-sion and higher voltage level distribution networks that operate interconnected).4 When considering the design of network equipment to deal with peak demandconditions, it is important to consider diversity of demand associated with their useof different appliances. This is fully exploited both in system design and operation.The capacity of an electricity system supplying several thousand households wouldbe only about 10% of the total capacity that would be required if each individualhousehold were to be self sufﬁcient (provide its own generation capacity). Distribu-tion electricity networks are essential for achieving this signiﬁcant beneﬁt of loaddiversity. However, no material gains in the capacity of the electricity supply systemwould be made from increasing further the number of the households. This phe-nomenon is illustrated in Figure 12.18, which shows how the demand coincidencefactor changes with the number of households.4 On average electricity consumers in Europe experience less than one interruption per year lasting aboutan hour. Consumers in urban areas have on the whole fewer interruptions than rural customers.

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466 Electric Power Systems, Fifth EditionThe coincidence factor is the ratio between maximum coincident total demand ofa group of households and the sum of maximum demands of individual consumerscomprising the group. In other words, the coincidence factor represents the ratio ofthe capacity of a system required to supply a certain number of households and thetotal capacity of the supply system that would be required if each household wereself sufﬁcient. For a case with a single house having a peak demand of P1 while the aggregatepeak of n houses being Pn, the coincidence factor is given by: jn ¼ Pn ð12:36Þ n Á P1 As indicated in Figure 12.18 experience suggests that the coincidence factorreduces with the number of households (appliances) following approximately theexpression below: jn ¼ j1 þ 1 pÀﬃjﬃﬃ1 ð12:37Þ nwhere j1 represents the coincidence factor of inﬁnite number of households(appliances). Equation (12.37) shows that for j1 ¼ 0.1, and for the number of households (appli-ances) being 100, the aggregate peak demand (diversiﬁed peak) will be only 20% ofthe sum of the individual peaks (non-diversiﬁed peak). Note that the diversity effectsaturates for the number of households beyond 10 000.12.6 Drivers for ChangeAs discussed, the location of generation relative to demand is the dominant factordriving the design and operation of electricity networks. Furthermore, the type ofgeneration technology used, together with the pattern of usage, will make an impacton the actual network operation and development. Finally, advances in technologymay open up new opportunities for achieving further improvement in efﬁciency ofoperation and investment in transmission and distribution networks. With this in mind, there are four main drivers that may change the conventionalphilosophy system operation and development. First, generation, transmission and distribution systems in most developed coun- tries were expanded considerably in the late 1950s and early 1960s. These assets are now approaching the end of their useful life. It is expected that a signiﬁcant proportion of these assets will need to be replaced in the next two decades. Second, a number of countries are committed to respond to the climate change challenge and the energy sector, and in particular the electricity, will be required to deliver the changes necessary. In the last decade, many countries have sup- ported deployment of distributed generation (DG) of various technologies (partic- ularly renewables and CHP) to reduce carbon emissions and the need to improve system efﬁciency. These generation technologies range from kW size domestic PV

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Fundamentals of the Economics of Operation and Planning 467 and micro CHP systems to several hundred MWs of wind generation connected to EHV (extra high voltage) distribution networks (132 kV). Very large windfarms are connected directly to transmission networks. This trend is likely to accelerate in the next decade and beyond as a key part of future energy policy. Third, there have recently been some major advances made in information and communication technologies (ICT), that in principle, could enable the develop- ment of signiﬁcantly more sophisticated approaches to control and management of the system and hence increase the efﬁciency of operation and utilization of net- work investments. Finally, there has been signiﬁcant research and development effort invested in the development of a number of control devices and concepts, such as FACTS (ﬂexible AC transmission systems), storage and demand-side management which could be used to provide real time control of power ﬂows in the network increasing utiliza- tion of transmission circuits. Similarly, greater automation of distribution network control could facilitate increase in utilization of network circuits. In summary, the need to respond to climate change, improve efﬁciency of the sys-tem and increase fuel diversity and enhance security of supply, coupled with rapidlyageing assets and recent development in ICT, opens up the question of the strategyfor infrastructure replacement in particular the design and investment in future elec-tricity networks. This coincidence of factors presents an opportunity to re-examinethe philosophy of the traditional approaches to system operation and design anddevelop a policy that will provide a secure, efﬁcient and sustainable future energysupply. Although this requirement may not necessarily lead to an immediate radicalchange in the system, like-for-like replacement of assets is unlikely to be optimal.Problems12.1 The input-output curve of a coal-ﬁred generating unit (with a maximum out- put of 550 MW) is given by the following expression: HðPÞ ¼ 126 þ 8:9P þ 0:0029P2½MJ=h If the cost of coal is 1.26 £/MJ, calculate the output of the unit when thesystem marginal cost isa. 13 [£/MWh] andb. 22 [£/MWh].(Answer: (a) P ¼ 244.4 MW, (b) P ¼ 550 MW)12.2 The incremental fuel costs of two units in a generating station are as follows: dF1 ¼ 0:003P1 þ 0:7 dP1 dF2 ¼ 0:004P2 þ 0:5 dP2

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468 Electric Power Systems, Fifth Edition marginal cost are in £/MWh and unit outputs are in MW. Assuming continuous running with a total load of 150 MW calculate the saving per hour obtained by using the most economical division of load between the units as compared with loading each equally. The maximum and minimum operational loadings are the same for each unit and are 125 MW and 20 MW. (Answer: P1 ¼ 57 MW, P2 ¼ 93 MW; saving £1.12 per hour) 12.3 What is the merit order used for when applied to generator scheduling? A power system is supplied by three generators. The functions relating the cost (in £/h) to active power output (in MW) when operating each of these units are: C1ðP1Þ ¼ 0:04P12 þ 2P1 þ 250 C2ðP2Þ ¼ 0:02P22 þ 3P2 þ 450 C3ðP3Þ ¼ 0:01P23 þ 5P3 þ 250 The system load is 525 MW. Assuming that all generators operate at the same marginal cost, calculate: a. the marginal cost; b. optimum output of each generator; c. the total hourly cost of this despatch. (Answer: (a) £10/MWh; (b) P1 ¼ 100 MW, P2 ¼ 175 MW, P3 ¼ 250 MW; (c) £4562.5/h) 12.4 A power system is supplied from three generating units that have the follow- ing cost functions: Unit A : 13 þ 1:3 PA þ 0:037 PA2 ½$=h Unit B : 23 þ 1:7 PB þ 0:061 PB2 ½$=h Unit C : 19 þ 1:87 PC þ 0:01 PC2 ½$=h a. How should these units be dispatched if a load of 380 MW is to be sup- plied at minimum cost? b. If, in addition to supplying a 380-MW load, the system can export (sell) energy to the neighbouring country in which the system marginal costs is 10.65 $/MWh. What is the optimal amount of power that should be exported? c. Repeat problem (b) if the outputs of the generating units are limited as PMA AX ¼ 110 MW follows: PBMAX ¼ 85 MW PMC AX ¼ 266 MW

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Fundamentals of the Economics of Operation and Planning 469 GW Annual Load Duration Curve 30 24 7.5 Hours 900 h 8760 h Figure 12.19 Load Duration Curve of Problem 12.5 (Answer: a. PA ¼ 77:60 MW; PB ¼ 43:79 MW; PC ¼ 258:61 MW b. Pex ¼ 258:71 MW c. Pex ¼ 69:36 MW)12.5 For the system with the load duration curve given in Figure 12.19 below, design a generation system to minimize the total investment and operating costs. The cost characteristics of different generation technologies are given below. Value of lost load is 8000 £/MWh.Technology Investment cost [£/kW/y] Operating costs [£/MWh]Base Load 250 5Mid Merit 80 50Peak Load 30 90 Calculate the amount of energy produced by each technology and the energy not served. (Answer: base load 18.0 GW capacity and 131.3 TWh energy production, mid merit 5.3 GW and 13.3 TWh energy production, peak load generation 6.7 GW and 3.5 TWh energy production, expected energy unserved 47 MWh)12.6 A power system contains ﬁve identical generators of capacity of 120 MW and availability of 96% that supply demand with the Load Duration Curve pre- sented in Figure 12.20. Calculate the LOLP at peak and LOLE. (Answer: LOLP ¼ 1.48%, LOLE ¼ 4.1 h)

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470 Electric Power Systems, Fifth Edition MW470 MW Annual Load Duration Curve315 MW Hours240 MW115 MW 180 h 2500 h 5400 h 8760 hFigure 12.20 Load Duration Curve of Problem 12.612.7 A power system consists of two areas that are currently not connected and the national transmission operator is considering building an interconnector. Generation costs in the two areas are: CAðPAÞ ¼ 6PA þ 0:025PA2 CBðPBÞ ¼ 10PB þ 0:11PB2 with the demand in the two areas being DA ¼ 600 MW and DB ¼ 1000 MW. Calculate the beneﬁts in enhancing the efﬁciency of the overall generation system operation that an interconnector of a. 200 MW and b. 400 MW would create. (Answer: (a) 33 400 £/h, (b) 56 000 £/h)12.8 A power system of a small country is composed of two regions that are not connected. Generators 1 and 2 are located in the Northern Region while gen- erators 3 and 4 are located in the Southern Region. The load in the Northern Region is 100 MW and the load in the Southern Region is 420 MW. Marginal costs of these generators are: Northern Region MC1 ¼ 3 þ 0:02P1 ½£=MWh MC2 ¼ 4 þ 0:04P2 ½£=MWh

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Fundamentals of the Economics of Operation and Planning 471Southern Region MC3 ¼ 3:6 þ 0:025P3½£=MWh MC4 ¼ 4:2 þ 0:025P4½£=MWh a. Calculate the marginal costs in both regions and the corresponding gener- ation dispatches. b. The transmission company is considering building a 450km long trans- mission link between the two regions. Show that the optimal transmis- sion capacity required to connect the two regions should be about 250 MW, when the annuitized investment cost of transmission is 37£/ MW.km.year. (Answer: a. MCnorth ¼ 4:67 £/MWh; P1 ¼ 83:33 MW P2 ¼ 16:67 MW b. MCsouth ¼ 9:15 £/MWh; P3 ¼ 222 MW P4 ¼ 198 MW12.9 Two areas, A and B, of a power system are linked by transmission link, with a secure capacity of 900 MW. System load is concentrated in area B. In winter, the load is 3,500 MW while summer load is 2000 MW. The cost of generation in the areas can be modelled by the following expressions, where P is in MW:CðPAÞ ¼ 70 þ PA þ 0:001 PA2 ½£=h for area A; andCðPBÞ ¼ 50 þ 2PB þ 0:002 P2B ½£=h for area B:a. Determine the optimal levels of generation in areas A and B for each of the winter and summer seasons neglecting the constraints of the transmission system. Calculate the marginal cost of production in each season.b. If necessary, modify the levels of generation computed in (a) to take into consideration the capacity of the existing transmission link. What are the marginal cost prices in areas A and B in each of the seasons?c. Assuming that the duration of the winter period is 2500 h, calculate the generation cost in each season. What are the total annual generation costs?d. The transmission company is considering doubling the transmission capacity between the two areas in order to reduce generation cost. Assum- ing that the annuitized investment cost of the reinforcement is 1 000 000 £/year, determine if this proposed investment is justiﬁed.(Answer:a. Winter: MC ¼ 6 £/MWh; PA ¼ 2500 MW, PB ¼ 1000 MW Summer: MC ¼ 4 £/MWh; PA ¼ 1500 MW, PB ¼ 500 MW

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472 Electric Power Systems, Fifth Edition b. Winter: PA ¼ 900 MW; MCA ¼ 2:8 £/MWh PB ¼ 2600 MW; MCB ¼ 12:4 £/MWh Summer: PA ¼ 900 MW; MCA ¼ 2:8 £/MWh PB ¼ 1100 MW; MCB ¼ 6:4 £/MWh c. Winter: Cw ¼ £35:85 m, Summer: £33.61 m, Annual costs: £69.46 m d. The case for proposed investment is justiﬁed.12.10 A distribution substation will supply 160 households of two types. Large detached houses (45) with expected peak demand of 12 kW and smaller terraced houses (115) with peak demand of 8 kW. Coincidence factors for an inﬁnite number of detached and terraced houses are 0.12 and 0.18 respec- tively. Assuming that peaks of both types of houses coincide, calculate the peak demand of the proposed 11 kV/0.4 kV substation. (Answer: PDS ¼ 230 kW)

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Appendix ASynchronous MachineReactancesElectric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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474 Appendix A: Synchronous Machine ReactancesTable A.1 Typical percentage reactances of synchronous machines at 50 Hz – BritishType and rating of machine Positive sequence Negative Zero Short-circuit X00 X0 Xs sequence sequence ratio11 kV Salient-pole alternator 22.0 33.0 110without dampers 12.5 17.5 201 X2 X0 –11.8 kV, 60 MV, 75 MVA 10.0 14.0 175 2.0 6.0 0.55Turboalternator 14.0 19.0 195 0.6811.8 kV, 56 MW, 70 MVA 20.0 28.0 206 13.5 6.7 0.55Gas-turbine turboalternator 16.0 21.5 260 0.5811.8 kV, 70 MW, 87.5 MVA 19.0 25.5 265 13.0 5.0 0.40Gas-turbine turboalternator 20.5 28.0 255 0.4013.8 kV, 100 MW, 125 MVA 23.0 28.0 207 16.0 7.5 0.40Turboalternator 0.5016.0 kV, 275 MW, 324 MVA 22.4 9.4Turboalternator18.5 kV, 300 MW, 353 MVa 18.0 6.0Turboalternator22 kV, 500 MW, 588 MVA 19.0 11.0Turboalternator23 kV, 600 MW, 776 MVA 20.0 6.0–12.0Turboalternator 26.0 15.0

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Table A.2 Approximate reactance values of three-phase 60-HzApparatus Positive sequence Synchronous Xs Transient X0 Average Range Average Range2-pole turbine generator 1.65 1.22–1.91 0.27 0.20–0.35(45 psig inner-cooled H2) 1.72 1.61–1.86 0.23 0.188–0.3032-pole turbine generator 1.49 1.36–1.67 0.281 0.265–0.30(30 psig H2 cooled) 1.25 0.6–1.5 0.3 0.2–0.54-pole turbine generator 1.25 0.6–1.5 0.3 0.2–0.5(30 psig H2 cooled) 1.85 1.25–2.20 0.4 0.3–0.5Salient-pole generator and 2.2 1.5–2.65 0.48 0.36–0.6motors (with dampers)Salient-pole generator(without dampers)Synchronous condensers(air-cooled)Synchronous condensers(H2 cooled at 1/2 psigrating)Wa ith permission of Westinghouse Corp.

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z generating equipment – US (Values in per unit on rated kVA base) Appendix A: Synchronous Machine Reactances Negative sequence X2 Zero sequence X0Subtransient X00Average Range Average Range Average Range0.21 0.17–0.25 0.21 0.17–0.25 0.093 0.04–0.140.14 0.116–0.17 0.14 0.116–0.17 0.042 0.03–0.0730.19 0.169–0.208 0.19 0.169–0.208 0.106 0.041–0.18250.2 0.13–0.32 0.2 0.13–0.32 0.18 0.03–0.230.3 0.2–0.5 0.48 0.35–0.65 0.19 0.03–0.240.27 0.19–0.3 0.26 0.18–0.4 0.12 0.025–0.150.32 0.23–0.36 0.31 0.22–0.48 0.14 0.03–0.18 475

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Table A.3 Principal data of 200–500 MW turbogenerators – Russi Values of param Units 12Power MW/MVA 200/235 200/2Cooling of winding m Water Hydr m Hydrogenc Hydr (stator) kg/kVA (N/kVA) 1.075 1.075 kg/kVA (N/kVA) 4.35 5.10 (rotor) % 0.93 (9.1) 1.3 (1 % 0.18 (1.76) 0.205Rotor diameter % 188.0 184.0 % 27.5 29.5Rotor length % 19.1 19.0 s 188.0 184.0Total weight 8.5 8.37 2.3 3.09Rotor weight Xsa X0 X00 Xq X0 t1ba For all reactances, unsaturated values are given.b Without the turbine.c Hydrogen pressure for columns 1–4 is equal to 3 atm (304,000N/m2) and for co(Source: Glebov, I, A, C.I.G.R.E., 1968, Paper 11-07.)

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ia 476 Appendix A: Synchronous Machine Reactancesmeters of turbogenerators of various types 34 5 6235 300/353 300/353 500/588 500/588 rogen Water Hydrogen Water Water rogen Hydrogen Hydrogen Hydrogen Water5 1.075 1.120 1.125 1.120 6.1 5.80 6.35 6.2012.7) 0.98 (9.6) 1.05 (10.3) 0.64 (6.26) 0.63 (6.17)5 (2.01) 0.16 (1.57) 0.158 (1.55) 0.11 (1.08) 0.1045 (1.025)0 169.8 219.5 248.8 241.3 25.8 30.0 36.8 37.30 17.3 19.5 24.3 24.3 169.8 219.5 248.8 241.3 8.8 9.63 15.0 14.6 2.1 2.55 1.7 1.63olumn 5, 4 atm (405 000 N/m2).

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Appendix BTypical TransformerImpedancesElectric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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Table B.1 Standard impedance limits for power transformers abHighest Low-voltage At kVA base equal to rating of largest cavoltage winding (BIL kV)winding Self-cooled (OA), self-cooled/forced-a(BIL kV) (OA/FA) self-cooled/forced-air, force (OA, FOA) For intermediate Standard impedance (%) BIL use value for next higher Ungrounded Ground BIL listed neutral operation operatio Min. Max. Min.110 and 110 and below 5.0 6.25 6.0 below 6.75 110 5.0 6.25 7.0150 110 5.5 7.0 6.5200 150 5.75 7.5 7.25250 150 5.75 7.5 7.75350 200 6.25 8.5 7.0450 200 6.25 8.5 7.75 250 6.75 9.5550 200 6.75 9.5 250 7.25 10.75650 350 7.75 11.75 200 7.25 10.75 350 8.25 13.0 450 8.5 13.5 200 7.75 11.75 350 8.5 13.5

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bove 10 000 kVA (60 Hz) 478 Appendix B: Typical Transformer Impedancesapacity winding, 55 Cair cooled Forced-oil cooleded-oil cooled (FOA and FOW) Standard impedance (%)ded neutral Ungrounded Grounded neutral on neutral operation operation Max. Min. Max. Min. Max. 8.75 8.25 10.5 9.5 10.25 8.25 10.5 10.5 14.5 9.75 9.0 12.0 11.25 16.0 10.75 9.75 12.75 12.0 17.25 11.75 9.5 12.75 10.75 16.5 10.75 10.5 14.25 12.0 18.0 12.0 10.25 14.25 12.75 19.5 11.25 15.75 11.75 18.0 11.25 15.75 12.75 19.5 12.0 17.25 12.75 18.0 (continued) 12.0 18.0 13.25 21.0 14.0 22.5 12.75 19.5 14.0 22.5

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Table B.1 (Continued )Highest Low-voltage At kVA base equal to rating of largest cavoltage winding (BIL kV)winding Self-cooled (OA), self-cooled/forced-a(BIL kV) (OA/FA) self-cooled/forced-air, force (OA, FOA) For intermediate Standard impedance (%) BIL use value for next higher Ungrounded Ground BIL listed neutral operation operatio Min. Max. Min.750 450 9.25 14.0 8.5825 250 8.0 12.75 7.5900 450 9.0 13.75 8.251050 650 10.25 15.0 9.251175 250 8.5 13.5 7.751300 450 9.5 14.25 8.75 650 10.75 15.75 9.75 250 8.25 450 9.25 750 10.25 250 8.75 550 10.0 825 11.0 250 9.25 550 10.5 900 12.0 250 9.75 550 11.25 1050 12With permission of Westinghouse Corp.

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apacity winding, 55 C Appendix B: Typical Transformer Impedancesair cooled Forced-oil cooleded-oil cooled (FOA and FOW) Standard impedance (%)ded neutral Ungrounded Grounded neutral on neutral operation operation Max. Min. Max. Min. Max. 13.5 15.25 24.5 14.0 22.5 11.5 13.5 21.25 12.5 19.25 13.0 15.0 24.0 13.75 21.5 14.0 16.5 25.0 15.0 24.0 12.0 14.25 22.5 13.0 20.0 13.5 15.75 24.0 14.5 22.25 15.0 17.25 26.25 15.75 24.0 12.5 13.75 21.0 14.0 15.25 23.5 15.0 16.5 25.5 13.5 14.75 22.0 15.0 16.75 25.0 16.5 18.25 27.5 14.0 15.5 23.0 15.75 17.5 25.5 17.5 19.5 29.0 14.5 16.25 24.0 17.0 18.75 27.0 18.25 20.75 30.5 479

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Appendix CTypical Overhead LineParametersElectric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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482 Appendix C: Typical Overhead Line ParametersTable C.1 Overhead-line parameters–50 Hz (British)Parameter 275 kV 400 kV 2 Â 113 mm2 2 Â 258 mm2 2 Â 258 mm2 4 Â 258 mm2Z1 V/km 0.09 þ j0.317 0.04 þ j0.319 0.04 þ j0.33 0.02 þ j0.28Z0 V/km 0.2 þ j0.87 0.14 þ j0.862 0.146 þ j0.862 0.104 þ j0.793Zmo V/km 0.114 þ j0.487 0.108 þ j0.462 0.108 þ j0.45 0.085 þ j0.425Zp V/km 0.127 þ j0.5 0.072 þ j0.5 0.075 þ j0.507 0.048 þ j0.45Zpp V/km 0.038 þ j0.183 0.033 þ j0.182 0.035 þ j0.177 0.028 þ j0.172B1 mmho/km 3.60 3.65 3.53 4.10B0 mmho/km 2.00 2.00 2.00 2.32Bm0 mtmho/km 5.94 7.00 7.75 8.50Z1 Positive-sequence impedance.Z0 Zero-sequence impedance of a d.c. 1 line.Zm0 Zero-sequence mutual impedance between circuits.Zp Self-impedance of one phase with earth return.Zpp Mutual impedance between phases with earth return.B1 Positive-sequence susceptance.B0 Zero-sequence susceptance of a d.c. 1 line.Bm0 Zero-sequence mutual susceptance between circuits.Note: A d.c. 1 line refers to one circuit of a double-circuit line in which the other circuit is open at bothends.The areas quoted are copper equivalent values based on 0.4 in2 and 0.175 in2.

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Table C.2 Overhead line data – a.c. (60 Hz) and d.c. linesRegion: 500–550 kV–a.c.Utility: Paciﬁc Canad So. California Edison Co. OntarioLine name or number Lugo-Eldorado PinardVoltage (nominal), kV; a.c. or d.c. 500; a.c. 500; a.cLength of line, miles 176 228Originates at Lugo sub PinardTerminates at Eldorado sub HanmeYear of construction 1967–69 1961–6Normal rating/cct, MVA 1000 –Structures S S, A(S)teel, (W)ood, (A)lum 4.21 3.7Average number/mile S (S-56) V(A-51Type: (S)q, (H), guy (V) or (Y) 850 1110Min 60-Hz ﬂashover, kV 1; 1 1; 1Number of circuits: Initial; Ultimate – S, ACrossarms (S)teel, (W)ood, (A)lum – –Bracing (S)teel, (W)ood, (A)Ium 19 515 10 800;Average weight/structure, lb No guys NilInsulation in guys (W)ood, (P)orc, kVConductors ACSR 84/19 ACSR 1.762; 2156 0.9; 583AI, ACSR, ACAR, AAAC, 5005 2.511 0.615Diameter, in.; MCM 2; 18 4; 18Weight 1 cond./ft, lb S RNumber/phase; Bundle spacing, in 2400 2500Spacer (R)igid, (S)ring, (P)reformDesigned for–A/phase

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700–750 kV h.v.d.c. Appendix C: Typical Overhead Line Parametersda Canada Mountain Paciﬁc o Hydro Quebec Hydroelectric U.S. Bureau of Los Angeles Dept.d-Hanmer Co. Reclamation of W. & P. c. Manics.-Levis Oregon-Mead Dalles-Sylmard TS 735; a.c. 750 (Æ375); d.c. 750 (Æ375); d.c. er TS 236 560 560 Manicouagan Oregon border Oregon border 63 Levis sub Mead sub near, Los Angeles 1964–65 1966–70 1969 1, S-51) 1700 1350 MW 1350 MW ; 4730 S S S or A 3.8 4.5 4.5 18/7 -; (S-71) S, T (S-72, 73) S, T (S-72, 73) 3 – – 915 1; 1 1; 1 1; 1 S S S or A S S S or A 67 3000 11 000 – – None – ACSR 42/7 ACSR 96/19 ACSR 84/19 -; 1361 -; 1857 1.82,2300 1.468 2.957 2.68 4; 18 2; 16 2; 18 R S or P – – 1800 3380 (continued) 483

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Table C.2 (Continued ) Paciﬁc 500–550 kV–a.c. H CanadRegion: HPh. conﬁg. (V)ertical, (H)orizontal, 32 40 (T)riangular 1500; 2764 1400; 2 46 at 130 F; 19.14 5.0 at 6Phase separation, ft 40; 12.5 33; 10.5Span: Normal, ft; Maximum, ftFinal sag, ft; at F; Tension, 103 lb Preformed NilMinimum clearance: Ground, ft; Stockbridge Spacer 1105–4896 800–13 Structure, ft 5 –Type armour at clampsType vibration dampers 2080 1800Line altitude range, ftMax corona loss, kW/three-ph. D D mile 2V 1 or 2PInsulation 27; 53=4 Â 10; 30 23; 53=4Basic impulse level, kV D DTangent: (D)isk or (L)ine-postNumber strings in (V) or (P)arallel 2V 61=4 Â 103=8; 40 3P 53=4Number units; Size; Strength, 103 lb 25; 26;Angle: (D)isk or (L)ine-postNumber strings in (V) or (P)arallel None NoneNumber units; Size; Strength; 103 lbInsulators for struts D DTerminations: (D)isk or (L)ine-postNumber strings in (P)arallel 2P 3PNumber units; Size; Strength, 103 lb 25; 61=4 Â 103=8; 40 26; 53=4BIL reduction, kV or steps 3.5 steps –BIL of terminal apparatus, kV 1150–1425 1675–1

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da 700–750 kV Mountain h.v.d.c. 484 Appendix C: Typical Overhead Line Parameters Canada H Paciﬁc 2550 H H 60 F; 3.4 5 50 38–41 39–40 1400; 2800 1150; 1800 1175; – r damper 226 at 120 F; 6.2 40 at 120 F; 12.4 35 at 60 F; 13.0 300 45; 18.3 35; 7.75 35; 7.8P None – None 4 Â 10; 25 – Stockbridge =4 Â 10; 25 Tuned spacer 2000–7000 400–8000 – – =4 Â 10; 25 50–2500 1900 1 74 in. snow/h 2 – – – D D D Single Single –; –; – –; –; – 4V D D Single – 35; 5 Â 10; 15 –; –; – –; –; – – None D D D – 4P 4P 61=410; 36 –; –; – –; –; – 35; – – – 1300 – (continued) – – –; –; – – 2050–2200

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Table C.2 (Continued )Region: 500–550 kV–a.c. Paciﬁc CanadProtection 2; 7 No. 6 AW 2; 5=16Number shield wires; Metal and 24.5; 28 20; 48 Size – 15 C LShield angle, deg; Span clear, ft S; 0 S; –Ground resistance range, V 420; None 480/43Counterpose: (L)inear, (C)rowfoot None NoneNeutral gdg: (S)ol, (T)sf; 0.5 1Arrester rating, kV; Horn gap, in Ph. compar., carr. Dir’l coArc ring diameter, in.: Top; Bottom 2; None 3; NonExpected outages/100 miles/yearType relaying Æ10 —Breaker time, cyc; Reclos, cycles None NoneVoltage Regulation None None None NoneNominal, % –; – –Synch, cond. MVA; Spacing, milesShunt capac. MVA; Spacing, miles None NoneSeries capac. MVA; Compens., % C; – C; 50–9Shunt reactor, MVA; Spacing, miles None –Communication M; 6700 –; –(T)elephone circuits on structures(C)arrier; Frequency, kHz(P)ilot wire(M)icrowave; Frequency, MHz(With permission of Edison Electric Insitute)

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da 700–750 kV Mountain h.v.d.c. Appendix C: Typical Overhead Line Parameters 6 in. galv. Canada 1;– Paciﬁc 32; None ompar. 2; 7=16 in. galv. 30;- 2;–ne – 20; 50 – 20;- 98 – S; – 30 max. L –; – – S; – –; – S;– 636; – – –; – –; – – –; – – –; – 0.2 – – 6; 24 Variable –; – –; – — –; – Grid control –; – –; – None 330; 236 – On a.c. terminals None –; – – None – None None None None C; 42.0 and 50.0 M; – None P None M; 777.35 – 485

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Further ReadingThis bibliography lists some of the many excellent books on power systems. A key isprovided to indicate the most important books that should be looked at ﬁrst if seek-ing more information or historic developments on topics in the denoted chapters.All books will have more references to consult.Book Chapter 3yAdkins, B. and Harley, R. (1978) General Theory of Alternating Current 8y Machines, Chapman and Hall, London. 9y 9yAnderson, P.M. and Fouad, A.A. (1977) Power System Control and 6y Stability, Iowa State University Press, Ames, Iowa. 12Ã 12ÃArrillaga, J. (1983) High Voltage Direct Current Transmission, Peter 10ÃÃ Peregrinus on behalf of the Institution of Electrical, Engineers, UK. 10yArrillaga, J., Liu, Y.H., and Watson, N.R. (2007) Flexible Power 1,11y Transmission: The HVDC Options, John Wiley & Sons.Arrillaga, J. and Arnold, C.P. (1990) Computer Analysis of Power Systems, John Wiley & Sons, Chichester.Berrie, T.W. (1990) Power System Economics, Peter Peregrinus on behalf of the Institution of Electrical, Engineers, UK.Berrie, T.W. (1992) Electricity Economics and Planning, Peter Peregrinus, on behalf of the Institution of Electrical, Engineers, UK.Bewley, L.W. (1961) Travelling Waves on Transmission Systems, Dover Books, New York.Bickford, J.P., Mullineux, N., and Reed, J.R. (1976) Computation of Power System Transients, Peter Peregrinus on behalf of the Institution of Electrical Engineers, UK.British Electricity International: Modern Power Station Practice (1991), Vols L. and K, Pergamon, Oxford.Electric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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488 Further ReadingDebs, A.S. (1988) Modern Power System Control and Operation, Kluwer 4,5y Academic, Boston. 1,10,11y 6,7ÃÃE.H.V. Transmission Line Reference Book (1968), Edison Electric Institute, 1,10,11ÃÃ New York. 11y 3,4,5,6,7yEl Abiad, A.H. and Stagg, G.W. (1968) Computer Methods in Power System 3y Analysis, McGraw-Hill, New York. 10,11yElectrical Transmission and Distribution Reference Book (1964), 8,10y Westinghouse Electric Corporation, East Pittsburgh, Pennsylvannia. 11yElectricity Council (1991) Power System Protection, vols 1–3, Peter 6,7y Peregrinus on behalf of the Institution of Electrical Engineers, UK. 1,3y 3,4,6,7,8yElgerd, O.I. (1983) Electric-Energy Systems Theory—An Introduction, 10ÃÃ 2nd edn, McGraw-Hill. 2,6y 3ÃFitzgerald, A.E., Kingsley, C., and Umans, S. (1990) Electric Machinery, 10y 5th edn, McGraw-Hill. 11yFlurscheim, C.H. (ed.) (1987) Power Circuit Breaker Theory and Design, 12ÃÃ Peter Peregrinus on behalf of the Institution of Electrical, Engineers, 12Ã UK. 8yFouad, A.A. and Vittal, V. (1992) Power System Transient Stability Analysis Using the Transient Energy Function Method, Prentice Hall, Englewood Cliffs, New Jersey.Giles, R.L. (1970) Layout of E.H.V. Substations, Cambridge University Press, Cambridge.Glover, J.D. and Sarma, M. (1994) Power System Analysis and Design, 2nd edn, PWS, Boston, Massachusetts.Gonen, T. (1986) Electric Power Distribution System Engineering, McGraw-Hill.Grainger, J.J. and Stevenson, W.D. Jr (1994) Power System Analysis, McGraw-Hill International.Greenwood, A. (1971) Electrical Transients in Power Systems, Wiley- Interscience.Gross, C.A. (1986) Power System Analysis, 2nd edn, John Wiley & Sons, New York.Hindmarsh, J. (1984) Electric Machines, 4th edn, Pergamon, London.HVDC: Connecting to the Future, (2010), Alstom Grid, Stafford.Johns, A.T. and Salman, S.K. (1996) Digital Protection for Power Systems, Peter Peregrinus on behalf of the Institution of Electrical Engineers, UK.Kirchmayer, L.K. (1958) Economic Operation of Power Systems, John Wiley & Sons, New York.Kirschen, D.S. and Strbac, G. (2004) Fundamentals of Power System Economics, John Wiley and Sons, Chichester, UK.Kundur, P. (1994) Power System Stability and Control, McGraw-Hill, New York.

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Further Reading 489Lakervi, E. and Holmes, E.J. (1995) Electricity Distribution Network Design, 1,5y Peter Peregrinus on behalf of the Institution of Electrical Engineers, UK. 10Ã 11ÃLander, C.W. (1996) Power Electronics, 3rd edn, McGraw-Hill. 8yLooms, J.S.T. (1990) Insulators for High Voltages, Peter Peregrinus on 11y 8y behalf of the Institution of Electrical Engineers, UK. 8yMachowski, J., Bialek, J.W., and Bumby, J.R. (1997) Power System 8y 10y Dynamics and Stability, John Wiley & Sons, Chichester. 3ÃÃNetwork Protection and Automation Guide, (2011), 5th edn, Alstom Grid, 12ÃÃ 7y Stafford UK. 3yPai, M.A. (1981) Power System Stability. Analysis by the Direct Method of 12ÃÃ 3Ã Lyapunov, North Holland Publishing, Amsterdam. 8yPai, M.A. (1989) Energy Function Analysis for Power System Stability, Kluwer Academic, Boston, Massachusetts.Pavella, M. and Murthy, P.J. (1994) Transient Stability of Power Systems, Theory and Practice, John Wiley & Sons, Chichester.Rudenberg, R. (1969) Transient Performance of Electrical Power Systems, MIT Press, Boston.Say, M.G. (1976) Alternating Current Machines, 4th edn, Pitman, London.Schweppe, F.C., Caramanis, M.C., Tabors, R.D., and Bohn, R.E. (1988) Spot Pricing of Electricity, Kluwer Academic Publishers, Boston, MA.Tleis, N.D. (2008) Power Systems Modelling and Fault Analysis: Theory and Practice, Newnes, Oxford.Transmission Line Reference Book, 345 kV and Above, (1975), Electric Power Research Institute, Palo Alto, California.Turvey, R. and Anderson, D. (1977) Electricity Economics: Essays and Case Studies, Johns Hopkins University Press.Weedy, B.M. (1980) Underground Transmission of Electric Power, John Wiley & Sons, Chichester.Wood, A.J. and Wollenberg, B.F. (1996) Power Generation, Operation and Control, 2nd edn, John Wiley & Sons, New York.Key: y consult for advanced study, Ã read for background knowledge, ÃÃ classic book.

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Indexacceleration factor 219 characteristic surge impedance 374amenity 23 circuit breakers (switchgear)arcing faults 270–272, 274arc suppression coil (Petersen coil) 272–274 air blast 406, 408, 409armature reaction 84, 86 arrangement 405audible noise 26, 27, 414 bulk oil 406automatic voltage regulators (AVR) minimum oil 410 requirements 413 effect on generator stability 310 small oil volume 410 speed of response 100 sulphur hexaﬂuoride (SF6) 404, 406, 409 types of 102 vacuum 406, 409auto-reclose 32, 288, 366 combined cycle gas turbine (CCGT) 9auto-transformer 56, 57, 128, 131 commutation angle 328, 333, 334, 340,back-up protection 415, 416, 428 344, 352basic insulation level (BIL) 355 comparison a.c. and d.c. 320Bewley lattice calculation 383 compensators 174, 345boilers (steam) 5boiling water reactor 11, 12 mechanically switched 175British grid system 1,444 thyristor controlled reactor 175Buchholz relay 432,434 thyristor switched 175, 346bundle conductors 104, 108 contamination of insulators 355busbar protection 430 control systems power system 139cables turbo-alternators 141 capacitance 122 converters parameters 113 current source (CSC) 320, 321, 325 polyethylene 123 inverter 325 underground 113, 122 rectiﬁer 322, 325 solid state 423capacitor voltage transformer 417 voltage source (VSC) 320–322, 346–352capacity margin 454 coordination of insulation 370carrier current protection 437 corona 369, 376, 385, 387 costs of energy 5Electric Power Systems, Fifth Edition. B.M. Weedy, B.J. Cory, N. Jenkins, J.B. Ekanayake and G. Strbac.Ó 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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492 Indexcoupling-power and communication lines electromagnetic compatibility (EMC) 274 274–275critical clearing angle 290, 291 electromagnetic radiation 25critical clearing time 290, 297, 298 electromagnetic relay 419critical ﬂashover (CFO) voltage electromagnetic transient program (EMTP) rod gaps 366 391, 392, 394 surge diverters 371, 372 energy functions 312, 313critical voltage 133, 306 energy storage 13, 17–18cross-ﬁeld ﬂux 84, 85 environmental considerationscurrent limiting reactor 246, 247current transformers 432, 433, 435 atmospheric 23, 25 thermal pollution 25damping of generator swings 301–304 transmission lines 22, 26, 37delta-star transformation 76–77 equal area criterion (EAC) 288, 289, 290, 312demand side 453, 467 equation of motion 283, 284, 292, 293design of insulation 369 equivalent circuitsdesign of system 27 long line 115differential relaying 429 medium line 114direct current transmission 319 short line 113 European supply system 31, 32 bridge connection 329, 332 control 340, 350, 351 fault analysis 239–270 equivalent circuits 321 fault level 239, 242 ﬁlters 345 faults harmonics 321 reasons for use 319 arcing 270 semi-conductor valves 322 balanced three-phase 241 thyristors/IGBT 322 critical clearing time 290distance protection 427, 428 double-line to ground 257distribution networks 32, 47 line-to-ground (earth) 253 representation 27 line-to-line 255–257 underground 34 systematic methods 265–270 voltages 48 types of 252–259diversity factor 41, 42 ﬂashoverdivision of load between generators probability 371 critical voltage 38 147–151 ﬂywheels 22dynamic braking 314 four-terminal networks 75dynamic stability 281, 310 four-wire system 47, 50 frequency control 139–159earth (ground) wire 365 fuel cells 21economic dispatch 444 fuel costs 5, 445, 446economic operation 4, 457 fuel mix for generation 4, 5, 445, 448 generating systems 444 gas turbines 9 transmission systems 457–460 Gauss-Seidel method 212–216, 219economics 443 generation operation of system 444–450electric ﬁeld base-load 451, 452 biological effects 25 mid-merit 451, 452 environmental limits 26 peak-load 451, 452

P:522

Index 493generation system planning 451 equivalent circuit 338generator protection 430 extinction angle 336, 338, 340generators isolator 405, 406, 410 inﬁnite busbar 95, 96 Jacobian matrix 223, 225 parameters 106, 473–477 Japan — supply system 37 stored energy 283 sub-transient reactance 92 large systems transient reactance 92, 93 fault analysis 265geothermal energy 16 load ﬂows 219governors characteristics 145, 147, 149 lattice diagram 383, 386, 387 control systems 139, 143 lightning arrester 366, 368, 370, 372 free action 143 lightning surges 356 speeder setting 144 linear coupler 417 time constants 147 line current 47–50grid supply point (GSP) 30 linesharmonics 74, 128 equivalent circuits 110–113hydroelectric generation 7, 8 equivalent spacing 108hydrogen energy systems 21–22 insulators 107 parameters 103, 110, 482–485impulse ratio 356, 369 towers 103incremental costs 445, 448 transposition 109incremental fuel cost 444–446 load coincidence factors 465induction disc relay 418, 419, 421 load duration curve (LDC) 453, 455, 469, 470induction motors 132 load ﬂows complex 212, 214, 215 critical slip 133 decoupled 228 critical voltage 133 direct methods 206inertia constant (H) 283 fast decoupled 219, 228inﬁnite busbar 95 Gauss-Seidel 212input-output characteristic 444, 445 iterative methods 209insulation mismatch 215, 216 coordination 369 Newton-Raphson 219, 222 design 371 loadsinsulators base 2 contamination 355 components 83 cross-arm 367 composite 131 towers 104–107 daily curves 3integrated gate bipolar thyristor (IGBT) diversity 39 forecasting 36, 43 321, 324, 325, 346–348 management 42, 467interconnected systems 154, 156 voltage characteristics 131interference with communication 274 long lines, equations 115interruption loss of load probability (LoLP) 454 capacitive current 362, 364 magnetomotive force (m.m.f.) 85 fault 358 marginal costs 448 inductive current 364 medium length lines 114inverter mesh analysis 51 action 336 control 338

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494 Indexmetal oxide arrester 368 power 11moment of inertia 283 complex 59multimachine systems 94 ﬂows 205multiterminal d.c. system 320 instantaneous 57, 58 mean 58networks sign 59 design standards 463, 464 loop 216 power-factor improvement 170 radial 216 power ﬂow, see Load ﬂows redundancy 464 pressurized water reactor (PWR) privatisation 1neutral grounding (earthing) 270 propagation constant 116nodal voltage analysis 206 protectionnuclear environmental factors 12nuclear ﬁssion 5 back-up 415nuclear reactors 10 classiﬁcation 425 differential 429 boiling water 11, 12 distance 427, 428 breeder 12 feeder 435 CANDU 12 generator 430 magnox 10, 11 qualities 415nuclear safety 12 relays 241, 270, 403 systems 403, 428oil circuit breaker 407 transformer 433optimal power ﬂow (OPF) 457 unit schemes 429optimal system operation 457 pumped storage 18–20ordering of matrix elements 227outages 90 radio interference (RI) 26, 27, 109overcurrent protection 420, 425 reactance, see generators, parameters reactive power directional 421overhead lines 163 control 176, 351 injection 170–171, 183, 214 parallel operation 186, 187 plant characteristics 131 parameters 103, 482–485 sign of 59overvoltages 356 reclosers 32 due to current chopping 363 rectiﬁer 325–334 due to fault currents 362 control 325–326 due to lightning 356 equivalent circuit 335 switching capacitive circuits 361 gate control 325, 328–329 voltage output 326–329parameters reﬂection coefﬁcient 376 of cables 113, 122 refraction coefﬁcient 377 of overhead lines 103, 110, 482–485 regulator 2, 163, 173, 203, 284, 285, 293, 310 of synchronous machines 93, 474–476 relays of transformers 131, 478–479 balanced beam 421–422 digital 423–424penetration of harmonics 75 distance 422, 427–428performance chart (generator) 98 hinged armature 432per unit system 61–68 induction cup 421phase shifts induction disc 418 negative sequence 425 in transformers 124–125pilot wire protection 435–436planning margin 454

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Index 495 numerical 423–424 suburban system 34 solid state 423 sulphur oxides 24renewable energy 4, 12–17 summation transformer 435–436reserve 4, 31, 32, 140, 309, 453 supercapacitors 22resonance 363 surge diverter 366restriking voltage 358–360, 362, 384 surge impedance 374–375, 378–379, 388right of way (wayleave) 26, 274 surge modiﬁer 366rural systems 32 swing curve 293–295 switchgear 404–415safety clearances 27 switching operations 358, 363–364salient pole generators 84–86 switching surges 356, 358–361saturation in machines 86, 88 symmetrical components 247–251security power in sequence networks 251 distribution and transmission 463–467 transformation matrix 250 of supply 29, 32, 34, 415, 467 synchronizing torque 94, 97sequence networks 252–259 synchronous compensators 173shock hazard 275 synchronous generator 83–100, see alsoshort-circuit currents 167–168, 242, 260, generators 403, 414short line fault 364, 384, 385 tap change transformers 128–131, 180–181slack (swing) busbar 219 tap stagger 181solid state (semi-conductors) tariffs 40 termination of lines 378–380 converters for d.c. 321 three-phase relationships 45–78 relays 418, 423 three-wirestability of loads 305 balanced load 51–52 of protection 415 unbalanced load 51 rotor angle 287–291 tidal energy 16–17 steady state 284–286 tie lines 153–154, 158 transient 282, 287–299 frequency-bias control 158star connection 46–47, 125 load-frequency control 153star-delta (wye-mesh) transformation 77 transfer impedance 179steady-state stability transformers effect of governor 304 booster 188 effects of damping 303 connections 55, 251 linear analysis 301, 302 grounding (earthing) 128 practical considerations 284–286 off-nominal tap 221steam generation 5 parameters 131, 478–479stiffness of system 151 phase-shift 188storage protection of 432–434 batteries 18, 20 three-phase 55, 124–125 compressed air 18, 20 three-winding 125–127, 183–184 ﬂywheels 18, 22 transient stability 287–290 heat 18 computation 293–295 pumped 18–20, 43 energy functions 312 superconducting 18, 22, 27 equal area criterion 290, 312stub line 388–389 transmission 29–31substations 403, 410 capability 21, 463subsynchronous resonance 198 network security 463–464

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496 Indextransmission system planning 460 control by reactive power injectiontransposition of conductors 109 170–176travelling waves 373–377, 385–387turbines control by tap-changing 176–179 control in distribution networks 195 gas 9, 84 restriking 358, 360, 362, 384 hydro 7, 84, 155 transformers 410, 415, 417 steam 5, 9turbo-generator 5, 88–90, 283 wave power 17, 19 55,two-axis representation 86 wayleaves (rights of way) 26 Willans line 444unbalanced wind generators 14, 15 three-wire systems 51–52 winding connections for transfor mers three-phase analysis 247 125urban system 34 Wye-mesh transformation 49 Wye (star) connection 47, 55value of lost load (VoLL) 452visual impacts 26 zerovoice frequency signaling 439 phase sequence 248, 250voltage sequence characteristics of loads 131–134 zinc oxide arrester 369 collapse 191–192, 305, 309