Transients inPower SystemsKEMA High-Power Laboratory, Arnhem, The NetherlandsThis book is published on the occasion of the ofcial opening of Station 6, anextension to the KEMA High-Power Laboratory facilities: Station 4, Station 5,the open-air test site and the jetty for transformer tests inside a vessel.KEMA High-Power LaboratoryUtrechtseweg 3106812 AR ArnhemThe NetherlandsTelephone +31 26 3562991Telefax +31 26 3511468E-mail hpl@kema.nlTransients inPower SystemsLou van der SluisDelft University of TechnologyThe NetherlandsJOHN WILEY & SONS, LTDChichester New York Weinheim Brisbane Singapore TorontoCopyright 2001 John Wiley & Sons LtdBafns Lane, ChichesterWest Sussex, PO19 1UD, EnglandNational 01243 779777International (+44) 1243 779777e-mail (for orders and customer service enquiries): cs-books@wiley.co.ukVisit our Home Page on http://www.wiley.co.uk or http://www.wiley.comAll Rights Reserved. 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Readers, however, should contact the appropriate companiesfor more complete information regarding trademarks and registration.Other Wiley Editorial OfcesJohn Wiley & Sons, Inc., 605 Third Avenue,New York, NY 10158-0012, USAWiley-VCH Verlag GmbHPappelallee 3, D-69469 Weinheim, GermanyJohn Wiley, Australia, Ltd, 33 Park Road, Milton,Queensland 4064, AustraliaJohn Wiley & Sons (Canada) Ltd, 22 Worcester RoadRexdale, Ontario M9W 1L1, CanadaJohn Wiley & Sons (Asia) Pte Ltd, 2 Clementi Loop #02-01,Jin Xing Distripark, Singapore 129809Library of Congress Cataloguing-in-Publication DataVan der Sluis, Lou.Transients in power systems / by Lou van der Sluis.p.cm.Includes bibliographical references and index.ISBN 0-471-48639-61. Transients (Electricity). 2. Electric power system stability. 3. Electric network analysis. I. TitleTK3226.V23 2001621.31 dc212001026198British Library Cataloguing in Publication DataA catalogue record for this book is available from the British LibraryISBN 0 471 48639 6Typeset in 11/13.5pt Sabon by Laser Words, Madras, IndiaPrinted and bound in Great Britain by Biddles Ltd, Guildford and Kings LynnThis book is printed on acid-free paper responsibly manufactured from sustainable forestry,in which at least two trees are planted for each one used for paper production.ToLinda, Tia, and MaiContentsPreface xi1 Basic Concepts and Simple Switching Transients 11.1 Switching an LR Circuit 31.2 Switching an LC Circuit 61.3 Switching an RLC Circuit 91.4 References for Further Reading 132 Transient Analysis of Three-Phase PowerSystems 152.1 Symmetrical Components in Three-Phase Systems 162.2 Sequence Components for Unbalanced NetworkImpedances 172.3 The Sequence Networks 202.4 The Analysis of Unsymmetrical Three-Phase Faults 222.4.1 The Single Line-to-Ground Fault 222.4.2 The Three-Phase-to-Ground Fault 242.5 References for Further Reading 303 Travelling Waves 313.1 Velocity of Travelling Waves and CharacteristicImpedance 323.2 Energy Contents of Travelling Waves 343.3 Attenuation and Distortion of ElectromagneticWaves 363.4 The Telegraph Equations 383.4.1 The Lossless Line 403.4.2 The Distortionless Line 423.5 Reection and Refraction of Travelling Waves 42viii CONTENTS3.6 Reection of Travelling Waves againstTransformer- and Generator-Windings 453.7 The Origin of Transient Recovery Voltages 493.8 The Lattice Diagram 523.9 References for Further Reading 564 Circuit Breakers 574.1 The Switching Arc 584.2 Oil Circuit Breakers 634.3 Air-Blast Circuit Breakers 644.4 SF6 Circuit Breakers 644.5 Vacuum Circuit Breakers 664.6 Modelling of the Switching Arc 684.7 ArcCircuit Interaction 744.8 References for Further Reading 805 Switching Transients 835.1 Interrupting Capacitive Currents 845.2 Capacitive Inrush Currents 915.3 Interrupting Small Inductive Currents 935.4 Transformer Inrush Currents 955.5 The Short-Line Fault 975.6 References for Further Reading 1056 Power System Transient Recovery Voltages 1076.1 Characteristics of the Transient Recovery Voltage 1106.1.1 Short-Circuit Test Duties based on IEC60056 (1987) 1116.1.2 Short-Circuit Test Duties based onANSI/IEEE Standards 1156.1.3 The Harmonisation between IEC andANSI/IEEE Standards with Respect toShort-Circuit Test Duties 1156.2 The Transient Recovery Voltage for DifferentTypes of Faults 1166.3 References 1197 Lightning-Induced Transients 1217.1 The Mechanism of Lightning 1227.2 Waveshape of the Lightning Current 124CONTENTS ix7.3 Direct Lightning Stroke to Transmission LineTowers 1257.4 Direct Lightning Stroke to a Line 1277.5 References for Further Reading 1348 Numerical Simulation of Electrical Transients 1358.1 The Electromagnetic Transient Program 1378.2 The MNA Program 1428.3 The Xtrans Program 1458.4 The MATLAB Power System Blockset 1528.5 References for Further Reading 1569 Insulation Coordination, Standardisation Bodies,and Standards 1599.1 The International ElectrotechnicalCommission IEC 1609.2 The American National Standards Institute ANSI 1629.3 The Conf erence Internationale des Grands R eseauxElectriques ` a Haute Tension CIGRE 1629.4 The Short-Circuit Testing Liaison STL 1639.5 Standards Related to High-Voltage ElectricalPower Equipment 1649.6 References for Further Reading 16810 Testing of Circuit Breakers 16910.1 The High-Power Laboratory 17010.2 The Historical Development of Circuit BreakerTesting 17210.3 Direct Test Circuits 17410.4 Synthetic Test Circuits 18010.5 Short-Line Fault Testing 18610.6 Measuring Transient Currents and Voltages 18910.6.1 Transducers for Current Measurements 19110.6.2 Transducers for Voltage Measurements 19610.7 Measurement Setup for Transient Voltage andCurrent Measurements 19910.8 References for Further Reading 203Index 205PrefaceThe power system is one of the most complex systems designed, built,and operated by engineers. In modern society, the power system plays anindispensable role, and a comparable quality of life without a constantand reliable supply of electricity is almost unthinkable. Because electricitycannot be stored in large quantities, the operation of the power system hasthe constraint of balancing the production of electricity in the connectedpower stations and the consumption by the connected loads and ofmaintaining constant frequency and constant voltage with the clients.During normal operation, loads are connected and disconnected. Controlactions are therefore continuously necessary the power system is neverin a steady state. On a timescale of years, planning of new powerplants, the erection of new transmission lines, or the upgrading fromexisting lines to higher voltage levels are important items to consider.When we look ahead into the future, the main topic is the economicaloperation what is the expected load and what is the most economicalfuel to be used to heat the boilers in the power stations. When thereliability of the system is analysed with repetitive load-ow calculations,the timescale is usually hours, yet when the dynamic stability is analysedto verify whether the system remains stable after a major disturbance,the power system is studied with an accuracy of seconds. Switchingactions, either to connect or disconnect loads or to switch off faultedsections after a short-circuit, and disturbances from outside, such asa lightning stroke on or in the vicinity of a high-voltage transmissionline, make it necessary to examine the power system on an even smallertimescale, microseconds to milliseconds. We speak in that case of electricaltransients. The time that the electrical transients are present in the systemis short, but during a transient period, the components in the systemare subjected to high current and high-voltage peaks that can causeconsiderable damage.xii PREFACEThis book deals with electrical transients in the power system. Muchhas been learned about transient phenomena since the early days of powersystem operation. Pioneers in this eld were men like Charles ProteusSteinmetz and Oliver Heaviside who focussed on the understanding ofelectrical transients in a more or less general way. They took the analyticalapproach, which is restricted to linear circuits. When a circuit becomesmore complex, the application of this method becomes very laborious andtime-consuming. After the Second World War, new tools were developedand used in studying circuit transient phenomena that were previouslyavoided because of their complexity. The transient network analyser(TNA) was exceptionally useful in studying the behaviour of a largevariety of complex linear and nonlinear circuits. The TNA was a powerfultool for obtaining solutions to problems involving distributed constantsas well as nonlinear impedances. The use of the analogue TNA resultedin the publication of much technical literature. In 1951, Harold Petersonpublished his book Transients in Power Systems with many examplesof TNA studies. Petersons book is a practical survey of the particularphenomena (faults, sudden loss of load, switching surges, and so forth.)that can cause transients and is based on his practical experience withthe General Electric Company in the USA. A classical book is ReinholdRuedenbergs Transient Performance of Power Systems, published in1950 and based on his earlier work written in German. In addition,switchgear design is closely related to electrical transient phenomena, andbooks from authors such as Biermanns and Slamecka, who wrote fromtheir experience with the switchgear divisions of AEG and Siemens, area historical source for the understanding of transient phenomena andswitchgear development.When I joined KEMA in 1977, as a test engineer in the famousde Zoeten high-power laboratory, I entered the world of short-circuittesting. The testing of power system equipment according to IEC andANSI standards, calculating test circuits, measuring high currents andhigh voltages in an electromagnetically hostile environment, and so forthdeepened my knowledge about electrical engineering and about physics.My rst introduction to the subject was Allan Greenwoods ElectricalTransients in Power Systems. Later, I went through many more classicalbooks and papers, which gave me a good overview of the historicaldevelopment of high-voltage circuit breakers. In the fteen years and morethat I had the pleasure of working at KEMA, I learned a lot frommy formercolleagues at the high-power laboratory. Together we designed new testcircuits, developed new measuring equipment and built a computerisedmeasurement system with transient recorders and computer workstations.PREFACE xiiiKEMAs high-power laboratory has always been a front-runner when itcomes to test circuit development and there has always been a strongparticipation in IEC standardisation work.In 1990, I joined the Delft University of Technology as a part-timeprofessor to teach a course in transients in power systems. Since 1992, Ihave been a full-time professor and head of the Power Systems Laboratory.I am always pleased to know that students are very much interested inswitching phenomena and attracted by the operation of high-voltagecircuit breakers and the physical processes that take place during currentinterruption at current zero. The advanced mathematics, together withphysics and sometimes exploding equipment is probably the right mixture.After a couple of years, lecture notes need an update. In 1996, Adriaande Lange contacted me about a PhD research project, and to refresh hisknowledge about switching transients, he attended my lectures and madenotes of what I explained to the audience that was not written down inthe lecture notes. Adriaan also researched extensively literature on circuitbreaker development, current zero phenomena, and testing techniques toacquire a solid base for his thesis. Without the effort of Adriaan, this bookwould not have been written.Chapter 1, Basic Concepts and Simple Switching Transients, summarisesthe fundamental physical phenomena and the mathematical tools to tackletransient phenomena. In fact, basic network theory and a thorough under-standing of simple LR and RLC networks and the behaviour of thetransient voltages and currents after a switching action is a necessity.When analysing transients, one always tries to reduce complex networksto series or parallel networks for a rst-approximation. The three-phaselayout of the power system is treated in Chapter 2, Transient Analysis ofThree-Phase Power Systems, wherein faults that result in severe systemstresses are analysed with symmetrical component networks. The prop-erties of travelling waves, which play an important role in the subject,are treated in Chapter 3, Travelling Waves. Overvoltages caused by oper-ation of high-voltage circuit breakers can only be predicted when thephysical processes between the breaker contacts and the inuence ofthe different extinguishing media on the current interruption is under-stood. The different high-voltage circuit breakers, the current interruptionprocess, and arccircuit interaction are described in Chapter 4, CircuitBreakers. In Chapter 5, Switching Transients, the current and voltageoscillations that occur most often in practice, such as capacitive currentinterruption, capacitive inrush currents, the interruption of small inductivecurrents, transformer inrush currents, and the short-line fault are treated.Chapter 6, Power System Transient Recovery Voltages describes powerxiv PREFACEfrequency transients that quite often result from switching actions andthat can cause considerable damage to the power system components. Anoverview of how the different short-circuit duties are represented in theIEC and IEEE/ANSI standards is given. In Chapter 7, Lightning-InducedTransients, the mechanism of lightning is explained, and the chapterfocuses on the impact of lightning strokes on or in the vicinity of transmis-sion lines and substations. The calculation of electrical transients withoutthe help of a computer is nowadays hardly unthinkable. The mathemat-ical formulation and the numerical treatment of power system transientsis shown in Chapter 8, Numerical Simulation of Electrical Transients.Special attention is given on how to incorporate nonlinear elements, suchas arc models, in transient computer programs such as EMTP, MNA, andXTrans and the MATLAB Power System Blockset. A demo version ofthe XTrans program can be downloaded from http://eps.et.tudelft.nl. Thebackground of the insulation coordination and the relevant IEC-standardsand IEEE/ANSI-standards together with a brief history of IEC, ANSI,CIGRE and STL are given in Chapter 9, Insulation Coordination, Stan-dardisation Bodies, and Standards. Testing of high-voltage circuit breakers(the proof of the pudding is in the eating) in the high-power laboratoryand the related measurements and measuring equipment are described inChapter 10, The Testing of Circuit Breakers.I am very much obliged to my secretary Tirza Drisi who devotedlyedited the manuscript and to Henk Paling who made the excellent draw-ings. Pieter Schavemaker put many hours in painstakingly reading themanuscript to lter out errors and inconsistencies. In Chapter 8, TheNumerical Simulation of Electrical Transients, I fruitfully used the educa-tional and illustrative examples that Pieter developed for the chapter aboutnumerical transient calculations in his thesis. In writing Chapter 6, PowerSystem Transient Recovery Voltages, I received valuable support aboutthe latest developments in IEC and IEEE/ANSI standards from Henk tePaske, Test Engineer at KEMAs high-power laboratory. Martijn Venemafrom KEMA supplied the photos.Lou van der SluisNootdorp, Spring 20011Basic Concepts and SimpleSwitching TransientsThe purpose of a power system is to transport and distribute the elec-trical energy generated in the power plants to the consumers in a safeand reliable way. Aluminium and copper conductors are used to carrythe current, transformers are used to bring the electrical energy to theappropriate voltage level, and generators are used to take care of theconversion of mechanical energy into electrical energy. When we speak ofelectricity, we think of current owing through the conductors fromgener-ator to load. This approach is valid because the physical dimensions ofthe power system are large compared with the wavelength of the currentsand voltages; for 50-Hz signals, the wavelength is 6000 km. This enablesus to apply Kirchhoffs voltage and current laws and use lumped elementsin our modelling of the power system. In fact, the transportation of theelectrical energy is done by the electromagnetic elds that surround theconductors and the direction of the energy ow is given by the Poyntingvector.For steady-state analysis of the power ow, when the power frequencyis a constant 50 or 60 Hz, we can successfully make use of complexcalculus and phasors to represent voltages and currents. Power systemtransients involve much higher frequencies up to kiloHertz and mega-Hertz. Frequencies change rapidly, and the complex calculus and thephasor representation cannot be applied any longer. Now the differen-tial equations describing the system phenomena have to be solved. Inaddition, the lumped-element modelling of the system components hasto be done with care if we want to make use of Kirchhoffs voltage andcurrent laws. In the case of a power transformer, under normal powerfrequencyoperation conditions, the transformer ratio is given by the ratio2 BASIC CONCEPTS AND SIMPLE SWITCHING TRANSIENTSbetween the number of the windings of the primary coil and the numberof the windings of the secondary coil. However, for a lightning-inducedvoltage wave, the stray capacitance of the windings and the stray capaci-tance between the primary and secondary coil determine the transformerratio. In these two situations, the power transformer has to be modelleddifferently!When we cannot get away with a lumped-element representation,wherein the inductance represents the magnetic eld and the capacitancerepresents the electric eld and the resistance losses, we have to do theanalysis by using travelling waves. The correct translation of the physicalpower system and its components into suitable models for the analysisand calculation of power system transients requires insight into the basicphysical phenomena. Therefore, it requires careful consideration and isnot easy.A transient occurs in the power system when the network changes fromone steady state into another. This can be, for instance, the case whenlightning hits the ground in the vicinity of a high-voltage transmissionline or when lightning strikes a substation directly. The majority of powersystem transients is, however, the result of a switching action. Load-break switches and disconnectors switch off and switch on parts of thenetwork under load and no-load conditions. Fuses and circuit breakersinterrupt higher currents and clear short-circuit currents owing in faultedparts of the system. The time period when transient voltage and currentoscillations occur is in the range of microseconds to milliseconds. Onthis timescale, the presence of a short-circuit current during a systemfault can be regarded as a steady-state situation, wherein the energyis mainly in the magnetic eld, and when the fault current has beeninterrupted, the system is transferred into another steady-state situation,wherein the energy is predominantly in the electric eld. The energyexchange from the magnetic eld to the electric eld is when the systemis visualised by lumped elements, noticed by transient current and voltageoscillations.In this chapter, a few simple switching transients are thoroughly anal-ysed to acquire a good understanding of the physical processes that playa key role in the transient time period of a power system. As switchingdevices, we make use of the ideal switch. The ideal switch in closed posi-tion is an ideal conductor (zero resistance) and in open position is an idealisolator (innite resistance). The ideal switch changes from close to openposition instantaneously, and the sinusoidal current is always interruptedat current zero.SWITCHING AN LR CIRCUIT 31.1 SWITCHING AN LR CIRCUITA sinusoidal voltage is switched on to a series connection of an inductanceand a resistance (Figure 1.1). This is in fact the most simple single-phaserepresentation of a high-voltage circuit breaker closing into a short-circuited transmission line or a short-circuited underground cable. Thevoltage source E represents the electromotive forces from the connectedsynchronous generators. The inductance L comprises the synchronousinductance of these generators, the leakage inductance of the power trans-formers, and the inductance of the bus bars, cables, and transmission lines.The resistive losses of the supply circuit are represented by the resistanceR. Because we have linear network elements only, the current owing inthe circuit after closing the switch can be seen as the superposition of atransient current and a steady-state current.The transient current component is determined by the inductance andthe resistance only and is not inuenced by the sources in the network(in this case by the voltage source E). It forms the general solution of therst-order homogeneous differential equation, whereas the steady-statecurrent component is the particular solution of the nonhomogeneousdifferential equation. In the latter case, the transient oscillations aredamped out because their energy is dissipated in the resistive part of thecircuit. Applying Kirchhoffs voltage law gives us the nonhomogeneousdifferential equation of the circuit in Figure 1.1:Emaxsin(t +) = Ri +Ldidt (1.1)The switch can close the circuit at any time instant and the phase anglecan have a value between 0 and 2 rad. To nd the general solution ofthe differential equation, we have to solve the characteristic equation ofL RE= Emax sin(wt + j)Figure 1.1 A sinusoidal voltage source is switched on to an LR series circuit4 BASIC CONCEPTS AND SIMPLE SWITCHING TRANSIENTSthe homogeneous differential equationRi +Li = 0 (1.2)The scalar is the eigenvalue of the characteristic equation. We nd for = (R/L), and thus the general solution for Equation (1.1) isih(t) = C1e(R/L)t(1.3)The particular solution is found by substituting in Equation (1.1) a generalexpression for the currentip(t) = Asin(t +) +Bcos(t +) (1.4)A and B can be determined:A = REmaxR2+2L2 B = LEmaxR2+2L2 (1.5)This results in the particular solution for the currentip(t) = EmaxR2+2L2 sin_t + tan1_LR__ (1.6)The complete solution, which is the sum of the general and particularsolution, isi(t) = ih(t) +ip(t)= C1e(R/L)t+ EmaxR2+2L2 sin_t + tan1_LR__ (1.7)Before the switch closes (Figure 1.1), the magnetic ux in the inductanceL is equal to zero; this remains so immediately after the instant of closing,owing to the law of the conservation of ux. Therefore, at t = 0, theinstant of closing, we can writeC1+ EmaxR2+2L2 sin_ tan1_LR__= 0 (1.8)SWITCHING AN LR CIRCUIT 5This gives us the value for C1; hence, the complete expression for thecurrent becomesi(t) = e(R/L)t_ EmaxR2+2L2 sin_ tan1_LR___+ EmaxR2+2L2 sin_t + tan1_LR__ (1.9)The rst part of Equation (1.9) contains the term exp[(R/L)t] anddamps out. This is called the DC component. The expression between thebrackets is a constant and its value is determined by the instant of closingof the circuit. For [ tan1(L/R)] = 0 or an integer times , the DCcomponent is zero, and the current is immediately in the steady state. Inother words, there is no transient oscillation. When the switch closes thecircuit 90earlier or later, the transient current will reach a maximumamplitude, as can be seen in Figure 1.2.The current in Figure 1.2 is called an asymmetrical current. In the casewhere no transient oscillation occurs and the current is immediately inthe steady state, we speak of a symmetrical current. The asymmetricalcurrent can reach a peak value of nearly twice that of the symmetricalcurrent, depending on the time constant L/R of the supply circuit. Thisimplies that, for instance, when a circuit breaker closes on a short-circuitedhigh-voltage circuit, strong dynamic forces will act on the connected busbars and lines because of the large current involved.0 0.011.001.02.00.02 0.03 0.04 0.05 0.06Time in secondsDC componentCurrent in per unit valueFigure 1.2 The shape of a transient current in an inductive circuit depends on the instantof switching6 BASIC CONCEPTS AND SIMPLE SWITCHING TRANSIENTSWhen the time constant of the supply circuit is rather high, whichis the case for short-circuit faults close to the generator terminals, thetransient and subtransient reactance of the synchronous generator causean extra-high rst peak of the short-circuit current. After approximately20 milliseconds, when the inuence of the transient and subtransientreactance is not present any longer, the synchronous reactance reducesthe root-mean-square value (rms value) of the short-circuit current. Underthese circumstances, an alternating current ows without current zerosfor several periods in the case of a fault in one of the phases because ofthe large DC component. This current cannot be interrupted because thecurrent zero necessary for current interruption is lacking.1.2 SWITCHING AN LC CIRCUITAnother basic network is the series connection of an inductance and acapacitance; this is in fact the most simple representation of a high-voltagecircuit breaker switching a capacitor bank or a cable network. To makeit simple, we rst analyse the case in which a DC source energises thenetwork by closing the (ideal) switch.As can be seen from Figure 1.3, there are two energy-storage compo-nents the inductance storing the magnetic energy and the capacitancestoring the electric energy. After closing the switch, an oscillation can occurin the network. This is due to the fact that an exchange of energy takesplace between the two energy-storage devices with a certain frequency.Applying Kirchhoffs voltage law results inE = Ldidt + 1C_ i dt (1.10)L CEFigure 1.3 A DC source switched on an LC series networkSWITCHING AN LC CIRCUIT 7To solve this differential equation, it is transformed to the Laplace domain,and we get the following algebraic equation:Ep = pLi(p) Li(0) + i(p)pC + Vc(0)p (1.11)In this equation, p is the complex Laplace variable.This can be written asi(p)_p2+ 1LC_= E Vc(0)L +pi(0) (1.12)When we look for the initial conditions, it is clear that i(0) = 0 as thecurrent in the network is zero before the switch closes and because ofthe physical law of conservation of the ux. This is the case immediatelyafter closing of the switch too. In the case of the capacitor, the situation isnot so easy because the capacitor can have an initial voltage, for instance,because of a trapped charge on a capacitor bank.Let us assume that there is no charge on the capacitor and thereforeVc(0) = 0 and let 20 = 1/LC.Equation (1.12) becomesi(p) = E_CL0p2+20(1.13)and back-transformation from the Laplace domain to the time domaingives the solution of Equation (1.10)i(t) = E_CL sin(0t) (1.14)In Equation (1.14), we can recognise two important properties of the LCseries network. After closing the switch at time t = 0, an oscillating current starts toow with a natural frequency0 =LC The characteristic impedance, Z0 = (L/C)1/2, together with thevalue of the source voltage E, determines the peak value of theoscillating current.8 BASIC CONCEPTS AND SIMPLE SWITCHING TRANSIENTSWhen there is a charge present on the capacitor, the current in the Laplacedomain becomesi(p) = [E Vc(0)]_CL0p2+20(1.15)For the capacitor voltage in the Laplace domainVc(p) = Ep pLi(p) = Ep [E Vc(0)] pp2+20(1.16)and after back-transformation into the time domainVc(t) = E [E Vc(0)] cos(0t) (1.17)Figure 1.4 shows the voltage waveforms for three initial values of thecapacitor voltage. From these voltage waveforms in Figure 1.4, it canbe seen that for Vc(0) = 0 the voltage waveform has what is called a(1-cosine) shape and that it can reach twice the value of the peak of thesource voltage. For a negative charge, the peak voltage exceeds this value,as the electric charge cannot change instantly after closing the switch. Inaddition, when the characteristic impedance of the circuit has a low value,for example, in the case of switching a capacitor bank (a large C) and astrong supply (a small L), the peak of the inrush current after closing theswitch can reach a high value.Vc (0) = 100Vc (0) = +100Vc (0) = 0Time1000100200300Vc (t)Figure 1.4 Voltage across the capacitor for three different initial values of the capacitorvoltage. The DC voltage source has the value E = 100 VSWITCHING AN RLC CIRCUIT 91.3 SWITCHING AN RLC CIRCUITIn practice, there is always damping in the series circuit and that can berepresented by adding a resistance in series. When a sinusoidal voltagesource Emaxsin(t +) is switched on in the RLC series circuit at t = 0(see Figure 1.5), Kirchhoffs voltage law leads toEmaxsin(t +) = Ldidt +Ri + 1C_ i dt (1.18)To nd the transient (or natural) responses of the network, we have tosolve the homogeneous differential equation0 = d2idt2 + RLdidt + 1LCi (1.19)The general solution of the homogeneous differential equation isih(t) = C1e1t+C2e2t(1.20)where 1 and 2 are the roots of the characteristic equation0 = 2+ RL + 1LC (1.21)1,2 = R2L __ R2L_2 1LC (1.22)The values of the inductance, capacitance, and resistance are indeedpositive because they are physical components. The absolute value of theexpression [(R/2L)2(1/LC)]1/2is smaller than R/2L. When (R/2L)2(1/LC) is positive, the roots 1 and 2 are negative. When (R/2L)2(1/LC) is negative, the roots 1 and 2 are complex but the real part isL R CEmax sin(wt + j)Figure 1.5 A sinusoidal voltage source is switched on an RLC series circuit10 BASIC CONCEPTS AND SIMPLE SWITCHING TRANSIENTSnegative. This shows that in the general solutionih(t) = C1e1t+C2e2t(1.23)the exponential functions will become zero for large values of t andthe particular solution will remain. This particular solution can bewritten asip(t) = Asin(t +) +Bcos(t +) (1.24)In this particular solution, the constants A and B have to be determined.Equation (1.24) is substituted in Equation (1.18) and this gives us for theparticular solutionip(t) = Emax_R2+_ 1C L_2sin___t + +tan1___1C LR_____(1.25)The complete solution, which is the sum of the general and particularsolution, isi(t) = ih(t) +ip(t) = (C1e1t+C2e2t) + Emax_R2+_ 1C L_2sin___t + +tan1___1C LR_____ (1.26)Three different situations can be distinguished:1. When (R/2L)2> 1/LC, the transient oscillation is overdamped andthe roots of the characteristic Equation (1.21) are both real and negative.The expression for the current becomesi(t) = et(C1et+C2et)+ Emax_R2+_ 1C L_2sin___t + +tan1___1C LR_____(1.27)with = (R/2L) and = [(R/2L)2(1/LC)]1/2SWITCHING AN RLC CIRCUIT 112. When (R/2L)2= 1/LC, the roots of the characteristic equation areequal and real and the transient oscillation is said to be critically damped.The expression for the critically damped current isi(t) = et(C1+C2) + Emax_R2+_ 1C L_2sin___t + +tan1___1C LR_____ (1.28)with = (R/2L).3. In the case that (R/2L)2< 1/LC, the roots 1 and 2 in the generalsolution (Equation (1.23)) are complex.1 = +j and 2 = j with = (R/2L) and = [(1/LC) (R/2L)2]1/2and Equation (1.23) can be written asih(t) = C1et+jt+C2etjt(1.29)and because C2 = C1ih(t) = C1et+jt+(C1et+jt) (1.30)Making use of the property of complex numbers that Z +Z = 2Re(Z)and using the Euler notation ejt= cos(t) +j sin(t), Equation (1.30)can be written asih(t) = 2etRe(C1ejt) (1.31)withC1 = Re(C1) +j Im(C1)Equation (1.31) can be written asih(t) = 2etRe(C1) cos(t) 2etIm(C1) sin(t) (1.32)When we substitute for Re(C1) = (k1/2) and for Im (C1) = (k2/2), theexpression for the general solution is ih(t) = et(k1cos t +k2sint). The12 BASIC CONCEPTS AND SIMPLE SWITCHING TRANSIENTScomplete solution for the oscillating current isi(t) = et[k1cos(t) +k2sin(t)] + Emax_R2+_ 1C L_2sin___t + +tan1___1C LR_____ (1.33)with = (R/2L) and = [(1/LC) (R/2L)2]1/2.In the three cases mentioned herewith (see Figure 1.6), the particularsolution is the same but the general solution is different. The transientcomponent in the current contains sinusoidal functions with angularfrequency , which usually differs from 50- or 60-Hz power frequencyof the particular solution, and this is the cause for the irregular shapeof the current. When the DC component exp(t) = exp[(R/2L)t] =exp[(t/)] has damped out with the damping time constant and thetransient part of the current has been reduced to zero, the steady-stateOscillatingCriticallydampedOverdamped0.01 0.03 0.04 0.05 0.061.001.02.0Current in per unit value Time in secondsFigure 1.6 Overdamped, critically damped, and oscillating response of an RLC seriescircuit after closing the switch at maximum supply voltage t = 0, = +2REFERENCES FOR FURTHER READING 13current lags or leads the voltage of the source. The absolute value of 1/Cand L in the term tan1{[(1/C) (L)]/R} determine if the currentis lagging (in a dominant inductive circuit) or leading (in a dominantcapacitive circuit). After a time span of three times the damping timeconstant = 2L/R, only 5 percent of the initial amplitude of the transientwaveform is present in the network.It is not necessary that after every change of state, such as after aswitching action, transient oscillations occur in a network. It is very wellpossible that the initial conditions and the instant of switching are suchthat immediately after closing of the switch the steady-state situationis present. A good example is the switching of a lossless reactor whenthe supply voltage is at the maximum. In practice, however, this rarelyhappens, and after switching actions, transient oscillations originate in anelectrical network. Power systems have a high quality factor, that is, a largeL and small R and are designed such that the frequency of the transientoscillations is much larger than the power frequency; this avoids steady-state overvoltages because of resonance. However, for higher harmonicfrequencies generated by power electronic equipment, resonance can occurmore easily.1.4 REFERENCES FOR FURTHER READINGBoyce, W. E. and DiPrima, R. C., Elementary Differential Equations and BoundaryValue Problems, 3rd ed., Chapter 3, Wiley & Sons, New York, 1977.Edminister, J. A., Electric Circuits, 2nd ed., Chapter 5, McGraw-Hill, NewYork,1983.Greenwood, A., Electrical Transients in Power Systems, 2nd ed., Chapters 14, Wiley& Sons, New York, 1991.Happoldt, H. and Oeding, D., Elektrische Kraftwerke und Netze, 5th ed., Chapter 17,Springer-Verlag, Berlin, 1978.Ruedenberg, R., Transient Performance of Electric Power Systems: Phenomena inLumped Networks, Chapter 3, McGraw-Hill, New York, 1950.Ruedenberg, R., in H. Dorsch and P. Jacottet, eds., Elektrische Schaltvorgaenge, 5thed., Vol. I, Chapters 2, 3, Springer-Verlag, Berlin, 1974.Slamecka, E. and Waterschek, W., Schaltvorgaenge in Hoch- und Niederspan-nungsnetzen, Chapters 3, 5, Siemens Aktiengesellschaft, Berlin, 1972.Wylie, C. R. and Barrett, L. C., Advanced Engineering Mathematics, 5th ed.,Chapter 6, McGraw-Hill, New York, 1989.2Transient Analysisof Three-Phase Power SystemsIn normal operating conditions, a three-phase power system can betreated as a single-phase system when the loads, voltages, and currents arebalanced. If we postulate plane-wave propagation along the conductors(it is, however, known from the Maxwell equations that in the presenceof losses this is not strictly true), a network representation with lumpedelements can be made when the physical dimensions of the power system,or a part of it, are small as compared with the wavelength of the voltageand current signals. When this is the case, one can successfully use a single-line lumped-element representation of the three-phase power system forcalculation. A fault brings the system to an abnormal condition. Short-circuit faults are especially of concern because they result in a switchingaction, which often results in transient overvoltages.Line-to-ground faults are faults in which an overhead transmission linetouches the ground because of wind, ice loading, or a falling tree limb.A majority of transmission-line faults are single line-to-ground faults.Line-to-line faults are usually the result of galloping lines because of highwinds or because of a line breaking and falling on a line below. Doubleline-to-ground faults result from causes similar to that of the single line-to-ground faults but are very rare. Three-phase faults, when all three linestouch each other or fall to ground, occur in only a small percentage of thecases but are very severe faults for the system and its components.In the case of a symmetrical three-phase fault in a symmetrical system,one can still use a single-phase representation for the short-circuit andtransient analysis. However, for the majority of the fault situations, thepower system has become unsymmetrical. Symmetrical components and,especially, the sequence networks are an elegant way to analyse faults16 TRANSIENT ANALYSIS OF THREE-PHASE POWER SYSTEMSin unsymmetrical three-phase power systems because in many cases theunbalanced portion of the physical system can be isolated for a study, therest of the system being considered to be in balance. This is, for instance,the case for an unbalanced load or fault. In such cases, we attempt tond the symmetrical components of the voltages and the currents at thepoint of unbalance and connect the sequence networks, which are, in fact,copies of the balanced system at the point of unbalance (the fault point).2.1 SYMMETRICAL COMPONENTSIN THREE-PHASE SYSTEMSIn 1918, C. L. Fortescue published a paper called Method of SymmetricalCoordinates Applied to the Solution of Polyphase Networks in theTransactions of the American Institute of Electrical Engineers. In thispaper, he proposed a method to resolve an unbalanced set of n-phasorsinto n 1 balanced n-phase systems of different phase sequence and onezero-phase system in which all phasors are of equal magnitude and angle.This approach will be illustrated for a three-phase system. Figure 2.1shows three such sets of symmetrical components.Va = Va1+Va2+Va0Vb = Vb1+Vb2+Vb0 (2.1)Vc = Vc1+Vc2+Vc0where Va, Vb, Vc are three phasors that are not in balance and Va1, Vb1,Vc1 and Va2, Vb2, Vc2 are two sets of three balanced phasors with an angleof 120between the components a, b, and c. The components of the phasorset Va0, Vb0, Vc0 are identical in amplitude and angle. Equation (2.1) canbe simplied by making use of the a-operator:a = ej2/3Vb0Vc0Va0 Va2Va1Vc1Vb1Va1Vb2Vc2Va2Va0 = Vb0 = Vc0Vc2Vc1VbVcVaVb2Vb1= + +Figure 2.1 A set of three unbalanced voltage phasors resolved in three sets of symmetricalcomponentsSEQUENCE COMPONENTS FOR UNBALANCED NETWORK IMPEDANCES 17The relation between the set of phasors (Va, Vb, Vc) and the positivephasors, negative phasors, and zero phasors is

VaVbVc

=

1 1 11 a2a1 a a2

Va0Va1Va2

(2.2)orVabc = AV012 (2.3)The a-operator rotates any phasor quantity by 120and the inverserelation of Equation (2.2) can be written as

Va0Va1Va2

= 13

1 1 11 a a21 a2a

VaVbVc

(2.4)orV012 = A1Vabc (2.5)In Equation (2.4), 0 refers to the zero sequence, 1 to the positive sequence,and 2 to the negative sequence. The names zero, positive, and negativerefer to the sequence of rotation of the phasors. The positive-sequenceset of phasors (Va1, Vb1, Vc1) is the same as the voltages produced bya synchronous generator in the power system that has phase sequencea-b-c. The negative sequence (Va2, Vb2, Vc2) has phase sequence a-c-b.The zero sequence phasors (Va0, Vb0, Vc0) have zero-phase displacementand are identical. The symmetrical component transformation is uniqueif the matrix operator A is nonsingular. If A is nonsingular, its inverseA1= A exists. The method of symmetrical components applies to anyset of unbalanced three-phase quantities; similarly, for currents we haverelations identical to Equation (2.4) and Equation (2.5).2.2 SEQUENCE COMPONENTSFOR UNBALANCED NETWORK IMPEDANCESA general three-phase system has unequal self-impedances and mutualimpedances, as depicted in Figure 2.2:Zaa = Zbb = ZccZab = Zbc = Zca(2.6)18 TRANSIENT ANALYSIS OF THREE-PHASE POWER SYSTEMSZaaZbbZccVmnIcIbIaZbcn mZcaZab+Figure 2.2 A general three-phase systemBoth the self-impedances and mutual impedances constitute sets of unbal-anced or unequal complex impedances, and even balanced currentsproduce unequal voltage drops between m and n. The voltage-dropequation from m to n can be written in matrix form asVmn =

VmnaVmnbVmnc

=

Zaa Zab ZacZba Zbb ZbcZca Zcb Zcc

IaIbIc

(2.7)By applying the symmetrical components transform to both sides, we getAVmn012 = ZAI012 (2.8)The symmetrical components of the voltage drop are given byVmn012 = A1ZAI012 = Zmn012I012 (2.9)Z is a transform that takes a current vector Iabc into a voltage-drop vectorVmn, both in the a-b-c system. A is a linear operator that transformscurrents and voltages from the 0-1-2 coordinate system into the a-b-csystem. The new impedance matrix Zmn012 can be found directly (seeEquation (2.9)):Zmn012 =

(Zs0+2Zm0) (Zs2Zm2) (Zs1Zm1)(Zs1Zm1) (Zs0Zm0) (Zs2+2Zm2)(Zs2Zm2) (Zs1+2Zm1) (Zs0Zm0)

(2.10)SEQUENCE COMPONENTS FOR UNBALANCED NETWORK IMPEDANCES 19withZs0 = 13(Zaa+Zbb+Zcc)Zs1 = 13(Zaa+aZbb+a2Zcc)Zs2 = 13(Zaa+a2Zbb+aZcc)(2.11)andZm0 = 13(Zbc+Zca+Zab)Zm1 = 13(Zbc+aZca+a2Zab)Zm2 = 13(Zbc+a2Zca+aZab)(2.12)We made use of the property of the a-operator 1 +a +a2= 0 and a3= 1.In Equation (2.12), we made use of the property that mutual impedancesof passive networks are reciprocal, and in this case it means that Zab = Zba,Zac = Zca and so forth. When the impedance matrix of Equation (2.10)is substituted in Equation (2.9), the equation for the positive-sequencecomponent of the voltage drop Vmn012 isVmn1 = (Zs1Zm1)Ia0+(Zs0Zm0)Ia1+(Zs2+2Zm2)Ia2 (2.13)The positive-sequence voltage drop depends not only on Ia1 but also onIa2, and this means that there is a mutual coupling between the sequences.Further, we can conclude that Zmn012 is not symmetric; therefore, themutual effects are not reciprocal and this is a rather disturbing result.This is the reason that we prefer to work with the special cases of bothself-impedances and mutual impedances in which the matrix Zmn012 issimplied. In many practical cases, the mutual impedances can be neglectedbecause they are small compared with the self-impedances. The matrixZmn012, however, is nonsymmetric with respect to Zs and Zm termsand is therefore not made symmetric by eliminating either the self-termsor the mutual terms, and because elimination of self-impedance termscannot be applied (because of the inherent nature of the power system),a simplication must be sought in the special case of equal impedanceand symmetric impedance. In many practical power system problems, theself-impedances or mutual impedances are equal in all the three phases. Insuch cases, Equation (2.11) and Equation (2.12) becomeZs0 = Zaa, Zs1 = Zs2 = 0 (2.14)andZm0 = Zbc, Zm1 = Zm2 = 0 (2.15)20 TRANSIENT ANALYSIS OF THREE-PHASE POWER SYSTEMSIf we substitute Equation (2.14) and Equation (2.15) in Equation (2.10)and examine the result, we see that the off-diagonal terms of Zmn012 areeliminated, that the impedance matrix Zmn012 has become reciprocal,and that zero-coupling exists between the sequences.A less-restrictive case than that of equal impedances is the one in whichthe self-impedances or mutual impedances are symmetric with respect toone phase, for example, for phase a:Zbb = Zcc and Zab = Zca (2.16)In this case, the self-impedances becomeZs0 = 13(Zaa+2Zbb)Zs1 = Zs2 = 13(ZaaZbb)(2.17)and the mutual impedances areZm0 = (Zbc+2Zab)Zm1 = Zm2 = (ZbcZab)(2.18)When Zmn012 is diagonal, it means that the sequences are uncoupledand currents from one sequence produce voltage drops only in thatsequence this is a very desirable characteristic. A symmetric impedancematrix means that there is a mutual coupling between sequences but that itis the reciprocal; the coupling frompositive to negative sequences is exactlythe same as the coupling from negative to positive. This situation can besimulated by a passive network. Anonsymmetric impedance matrix meansthat the mutual coupling is not the same between two sequences; this situ-ation requires, in general, controlled voltage sources but its mathematicalrepresentation is no more difcult than that in the symmetric case. Itrequires computation of all matrix elements instead of computing only theupper or lower triangular matrix, as in the symmetric case. In most of thepractical power system calculations, the self-impedances are considered tobe equal, and except for the case of nonsymmetric mutual impedances,the problem is one of a diagonal or a symmetric matrix representation.2.3 THE SEQUENCE NETWORKSIn the case of an unbalanced load or a fault supplied from balanced orequal-phase impedances, the unbalanced portion of the physical powersystem can be isolated for study, the rest of the power system beingTHE SEQUENCE NETWORKS 21considered as balanced. In such a case, we determine the symmetricalcomponents of voltage and current at the point of unbalance and transformthem to determine the system a-b-c quantities. Therefore, the majorobjective in problem-solving is to nd the sequence quantities, and for thispurpose, sequence networks are introduced.The fault point of a power system is that point to which the unbalancedconnection is attached in the otherwise balanced system. For example, asingle line-to-ground fault at bus M makes bus M the fault point of thesystem and an unbalanced three-phase load at bus N denes N as thefault point. In general terminology, a fault must be interpreted as anyconnection or situation that causes an unbalance among the three phasesof the power system.A sequence network is a copy of the original balanced power systemto which the fault point is connected and which contains the same per-phase impedances as the physical, balanced power system. The valueof each impedance is a value unique to each sequence and it can bedetermined by applying Thevenins theorem by considering the sequencenetwork to be a two-terminal or a one-port network. Because the positive-and negative-sequence currents are both balanced three-phase currentsets, they see the same impedance in a passive three-phase network.The zero currents, however, generally see an impedance that is differentfrom the positive- and negative-sequence impedance. Care must be takenwhen the machine impedance, of the supplying synchronous generatorsor of the asynchronous motors in certain loads, has to be taken intoaccount, because the sequence impedances for electrical machines areusually different for all the three sequences!Sequence networks are drawn as boxes in which the fault point F, thezero-potential bus N (often the neutral connection), and the Theveninvoltage are shown. Figure 2.3 shows the sequence networks for the zero,positive, and negative sequences.By denition, the direction of the sequence current is away from theF terminal. This is because the unbalanced connection is to be attachedZ 0F0Z1 Z2Ia1 Ia2Va1 Va2F1 F2N1 N2VfN0Ia0Va0++++Figure 2.3 Sequence networks for the zero, positive, and negative sequences22 TRANSIENT ANALYSIS OF THREE-PHASE POWER SYSTEMSat F, external to the sequence networks, and the currents are assumedto ow toward this unbalanced connection. The polarity of the voltageis dened to be a rise from N to F this makes Va1 positive for anormal power system. The Thevenin equivalent voltage Vf in the positive-sequence network is the voltage of phase a at the fault point F beforethe fault occurred. The relation for the voltage drop from F to N is animportant one and from Figure 2.3, we can write this voltage drop in amatrix notation as

Va0Va1Va2

=

0Vf0

Z0 0 00 Z1 00 0 Z2

Ia0Ia1Ia2

(2.19)2.4 THE ANALYSIS OF UNSYMMETRICALTHREE-PHASE FAULTSThe symmetrical components and, in particular, the sequence networks arevery useful tools when dealing with unsymmetrical faults. The analysis ofunsymmetrical faults is rather straightforward. First the three-phase circuitof the fault is drawn and currents, voltages, and impedances are labelled,taking into consideration the directions and polarities. Then the boundaryconditions for the unsymmetrical fault conditions are determined in thea-b-c system, and these current and voltage relations are transformedfrom the a-b-c system to the 0-1-2 coordinate system by using the Aor A1transformation. Next, the F- and N-terminals of the sequencenetworks are connected so that the sequence currents comply with theboundary conditions in the 0-1-2 system. The boundary conditions for thesequence voltages determine the connection of the remaining terminals ofthe sequence networks. These rather straightforward steps are illustratedby the analysis of two power system fault cases that are taken as typicalfor circuit breaker testing and standardisation. The single line-to-groundfault is a very common fault type; much less common is a three-phase-to-ground fault. The three-phase-to-ground fault, however, is a severe faultto be interrupted by high-voltage circuit breakers.2.4.1 The Single Line-to-Ground FaultThe three-phase circuit of the single line-to-ground fault (SLG) is drawnin Figure 2.4.THE ANALYSIS OF UNSYMMETRICAL THREE-PHASE FAULTS 23abc+++ ZfFVcVbVa Ia Ib = 0 Ic = 0Figure 2.4 Three-phase circuit diagram of an SLG fault at fault point F of the powersystemThe boundary conditions in the a-b-c systemcan be derived by inspectionof Figure 2.4:Ib = Ic = 0 (2.20)Va = ZfIa (2.21)These boundary equations are transformed from the a-b-c system to the0-1-2 coordinate system:I012 = A1Iabc (2.22)I012 = 13

1 1 11 a a21 a2a

Ia00

= 13Ia

111

(2.23)This implies that all the sequence currents are equal, and when theboundary equations for the voltage relations of Equation (2.21) are trans-formed, we nd the relationVa0+Va1+Va2 = ZfIa = 3ZfIa1 (2.24)The fact that the sequence currents are equal implies that the sequencenetworks must be connected in series. From Equation (2.24), we notethat the sequence voltages add to 3ZfIa1 this requires the addition of24 TRANSIENT ANALYSIS OF THREE-PHASE POWER SYSTEMS+Vf+Va1Z13 ZfF1N1+Va2Z2F2N2+Va0Z0F0N0Ia1Figure 2.5 Sequence network connection for an SLG faultan external impedance. The connection of the sequence networks for asingle line-to-ground fault is depicted in Figure 2.5. From Figure 2.5, wecan writeIa0 = Ia1 = Ia2 = VfZ0+Z1+Z2+3Zf(2.25)and now that the sequence current relations are known, we can determinethe sequence voltage relations from Equation (2.24).2.4.2 The Three-Phase-To-Ground FaultThe three-phase-to-ground fault is, in fact, a symmetrical fault becausethe power system remains in balance after the fault occurs. It is the mostsevere fault type and other faults, if not cleared promptly, can easilydevelop into it. The three-phase circuit of the three-phase-to-ground fault(TPG) is drawn in Figure 2.6.THE ANALYSIS OF UNSYMMETRICAL THREE-PHASE FAULTS 25abc+++ Va Vb VcZfFZgIaIa + Ib + IcIb IcZf ZfFigure 2.6 Three-phase circuit diagram of a TPG fault at fault point F of the powersystemThe boundary conditions in the a-b-c systemcan be derived by inspectionof Figure 2.6:Va = ZfIa+Zg(Ia+Ib+Ic) (2.26)Vb = ZfIb+Zg(Ia+Ib+Ic) (2.27)Vc = ZfIc+Zg(Ia+Ib+Ic) (2.28)The boundary conditions are again transformed from the a-b-c systemto the 0-1-2 coordinate system, and when we write in terms of thesymmetrical components of phase a, we getVa = (Va0+Va1+Va2) = Zf(Ia0+Ia1+Ia2) +3ZgIa0 (2.29)Vb = (Va0+a2Va1+aVa2) = Zf(Ia0+a2Ia1+aIa2) +3ZgIa0 (2.30)Vc = (Va0+aVa1+a2Va2) = Zf(Ia0+aIa1+a2Ia2) +3ZgIa0 (2.31)It is considered that Ia+Ib+Ic = 3Ia0 = 0 because the fault impedancesZf and also the supply voltages in each phase are in balance. Therefore,the currents are also in balance and we can writeI012 = 13

1 1 11 a a21 a2a

IaIbIc

=

0Ia10

(2.32)26 TRANSIENT ANALYSIS OF THREE-PHASE POWER SYSTEMS++Vf ZfZ1N1F1Va1Ia1+ZfZ2N2F2Va2Ia2+Zf + 3ZgZ0N0F0Va0Ia0Figure 2.7 Connection of the sequence network for a TPG faultThis leads to Ia1 = Ia, Ia0 = Ia2 = 0. The sequence networks are thereforeconnected as shown in Figure 2.7. When the fault impedance Zf is small,or even zero, the TPG fault is in fact a short circuit.After clearing a three-phase fault, the power system changes from thesteady-state situation, in which the three-phase short-circuit current isowing, to the state in which only the power frequencyrecovery voltageis present across the contacts of the circuit breaker. In an inductive circuit,the change from one steady state to another is always accompanied by atransient the transient recovery voltage or TRV. At current zero, the arcvoltage and the arc current extinguish and the TRV oscillates from zeroto the crest of the AC power frequencyrecovery voltage (See Chapter 4,Circuit Breakers). One of the breaker poles clears rst and the two lastclearing poles interrupt 90later they, in fact, forma single-phase circuit.Of interest is the value of the AC power frequencyrecovery voltage ofthe rst clearing phase because the TRV oscillates to this value. Let usconsider the situation depicted in Figure 2.8.The system is grounded by means of a neutral impedance at the starpoint of the three-phase delta/wye transformer. When the rst pole of thecircuit breaker interrupts the short-circuit current, let us assume that thisis phase a, the other two poles are still arcing and therefore in a conductingstate. In fact, these two last clearing poles interrupt a double line-to-groundGeneratorPowertransformerZnTransmission line Circuit breakerTPGFigure 2.8 The interruption of a three-phase line-to-ground-fault in a power systemgrounded via a neutral impedanceTHE ANALYSIS OF UNSYMMETRICAL THREE-PHASE FAULTS 27(DLG) fault. The boundary conditions for this DLG fault areIa = 0Vb = (Zf+Zg)Ib+ZgIc (2.33)Vc = (Zf+Zg)Ic+ZgIbThese boundary conditions are transformed from the a-b-c system to the0-1-2 coordinate system and this results inIa = 0 = Ia0+Ia1+Ia2 (2.34)From Equation (2.2) we can writeVb = Va0+a2Va1+aVa2 (2.35)Vc = Va0+aVa1+a2Va2 (2.36)and for the difference,VbVc = j3(Va1Va2) (2.37)From Equation (2.33) we can also writeVbVc = Zf(IbIc) (2.38)Substituting Equation (2.37) into Equation (2.38) givesj3(Va1Va2) = Zf(IbIc) (2.39)(Va1Va2) = Zf

IbIcj3

= Zf(Ia1Ia2) (2.40)orVa1ZfIa1 = Va2ZfIa2 (2.41)Adding Equation (2.35) and Equation (2.36) results inVb+Vc = 2Va0(Va1+Va2) (2.42)and adding Vb and Vc from Equation (2.33) gives usVb+Vc = Zf(Ib+Ic) +2Zg(Ib+Ic) (2.43)28 TRANSIENT ANALYSIS OF THREE-PHASE POWER SYSTEMSIb = Ia0+a2Ia1+aIa2Ic = Ia0+aIa1+a2Ia2 (2.44)Ib+Ic = 2Ia0+(a +a2)Ia1+(a +a2)Ia2 = 2Ia0Ia1Ia2 (2.45)Substituting Equation (2.45) in Equation (2.43), we getVb+Vc = Zf[2Ia0(Ia1+Ia2)] +Zg[4Ia02(Ia1+Ia2)] (2.46)Because Equation (2.42) and Equation (2.46) are equal, we can collectterms and write2Va02ZfIa04ZgIa0 = Va1+Va2Zf(Ia1+Ia2) 2Zg(Ia1+Ia2)(2.47)By using Equation (2.41) and the property that Ia1+Ia2 = Ia0, we nd,after rearranging, thatVa0ZfIa03ZgIa0 = Va1ZfIa1 (2.48)From Equation (2.34), we see immediately that the N-terminals of thesequence networks must be connected to a common node. Equation (2.41)shows us that the voltages across the positive and negative sequencenetworks are equal if an external impedance Zf is added in series witheach network. These conditions are met when all the three sequencenetworks are connected in parallel, as shown in Figure 2.9.From the inspection of the parallel connection of the three sequencenetworks, it follows thatIa1 = VfZ1+Zf+ (Z2+Zf)

Z0+Zf+3Zg

Z0+Z2+2Zf+3Zg(2.49)++Z1VfZfF1N1Ia1Va1+Z2 ZfF2N2Ia2Va2+Z0 Zf + 3ZgF0N0Ia0Va0Figure 2.9 Connection of the sequence networks for a DLG faultTHE ANALYSIS OF UNSYMMETRICAL THREE-PHASE FAULTS 29andVa1 = VfZ1Ia1 (2.50)In the case of a bolted fault to ground, Zf = 0 and Zg = 0, and thetransformation back to the a-b-c domain by means of the Equation (2.3)and the use of the boundary condition that Va0 = Va1 = Va2 result in therecovery voltage across the contacts of pole a:Va = 3Va1 = 3VfZ2Z0/(Z2+Z0)

Z1+ Z2Z0(Z2+Z0) (2.51)The ratio between the voltage across the rst clearing pole and the phasevoltage of the undistorted power system is called the rst-pole-to-clear(FPTC) factorFPTC = VaVf= 3 Z2Z0Z1 (Z2+Z0) +Z2Z0(2.52)The positive-, negative-, and zero-sequence impedances Z1, Z2, and Z0have a resistive and an inductive component Z = R +jX. Under normalsystem conditions, the load determines the power factor, but during ashort circuit, the load is short-circuited and the system is mainly inductive.Therefore, we can put for Z1, Z2, and Z0 the inductance values X1, X2,and X0, respectively.When the fault is relatively far away from the supplying generators, thepositive and negative impedances are equal and we can put X1 = X2 = X.In addition, the neutral connection of the transformer star point is acomplex impedance Zn = Rn+jXn and the zero-sequence impedancebecomes Z0 = jX0+3Zn. When this is substituted in Equation (2.52), wecan write for the FPTC factorFPTC = 3 3Rn+j(X0+3Xn)jX+2[3Rn+j(X0+3Xn)] (2.53)In the case of an ungrounded neutral, the value of the impedance Zn isinnite and the FPTC factor is 1.5. This is an important result becausea considerable number of power systems have an isolated neutral. In thecase of clearing a three-phase-to-ground fault, the peak of the TRV can,when there is little or no damping in the system, become as high as twotimes the crest value of the power frequencyrecovery voltage; in systems30 TRANSIENT ANALYSIS OF THREE-PHASE POWER SYSTEMSwith an isolated neutral, this is two times the FPTC factor, which is threetimes the phase voltage.For solidly grounded systems, Zn = 0, and the expression for the FPTCfactor becomesFPTC = 3X0X+2X0(2.54)When X has a larger value then X0, the FPTC factor is smaller than one.For cases in which X = X0, the FPTC factor for solidly grounded systemsis one, and in an undamped situation, the peak of the TRV for the rstclearing pole is twice the crest value of the system phase voltage.2.5 REFERENCES FOR FURTHER READINGAnderson, P. M., Analysis of Faulted Power Systems, Power Systems EngineeringSeries, IEEE Press, 1995.Fortescue, C. L., Method of symmetrical coordinates applied to the solution ofpolyphase networks, Trans. AIEE, 37(Part II), 1329 ff. (1918).Greenwood, A., Electrical Transients in Power Systems, Chapter 6, Wiley & Sons,New York, 1991.Grainger, J. J. and Stevenson, Jr., W. D., Power System Analysis, Chapters 11 and12, McGraw-Hill, New York, 1994.Peterson, H. A., Transients in Power Systems, Chapter 4, Section 6, Wiley & Sons,New York, Chapman & Hall, London, 1951.Slamecka, E. and Waterschek, W., Schaltvorgaenge in Hoch- und Niederspan-nungsnetzen, Chapters 1 and 2, Siemens Aktiengesellschaft, Berlin, 1972.Taylor, E. O., ed., Power System Transients, Chapter 1, Section: Faults, GeorgeNewnes, 1954.3Travelling WavesPower systems are large and complex systems, but for steady-state anal-ysis, the wavelength of the sinusoidal currents and voltages is still largecompared with the physical dimensions of the network for 50-Hz powerfrequency, the wavelength is 6000 km. For steady-state analysis, a lumped-element representation is adequate for most cases. However, for transientanalysis, this is no longer the case and the travel time of the electromag-netic waves has to be taken into account. A lumped representation of, forinstance, an overhead transmission line by means of pi-sections does notaccount for the travel time of the electromagnetic waves, as can be easilyseen from Figure 3.1.When we make a representation of an overhead transmission line ora high-voltage cable by means of a number of pi-sections, we take theproperties of the electric eld in a capacitance and the properties of themagnetic eld in an inductance into account and connect these elementswith lossless wires.When the switch S is closed, a current ows through the rst inductanceL1 and charges the rst capacitance C1. The accumulation of charge onC1 creates a voltage that causes a current to ow through L2. This currentcharges C2, a voltage builds up across C2, a current ows through L3and so on. This kind of reasoning shows that a disturbance at one end ofthe pi-section network is immediately noticeable at the other end of thenetwork. We know from experience that this is not what happens when asource is switched on in a transmission line; it takes a certain time beforethe current and voltage waves reach the end of the line.A representation of overhead lines and underground cables by meansof lumped elements is not helpful in making us understand the wavephenomena because electromagnetic waves have a travel time. Only whenthe physical dimensions of a certain part of the power system are smallcompared with the wavelength of the transients, the travel time of the32 TRAVELLING WAVESLoadLoadSSuuL1C1C2CnLnL2Figure 3.1 Lumped-element representation of a two-wire transmission lineelectromagnetic waves can be neglected and a lumped-element represen-tation of that particular part of the power system can be used for in-depthanalysis.If the travel time of the current and voltage waves is taken intoaccount and we want to represent the properties of the electric eldby means of a capacitance and the properties of the magnetic eld byan inductance, we call the capacitance and the inductance distributed.An overhead transmission line, a bus bar, or an underground cablehave certain physical dimensions and therefore their overall inductance,capacitance, and resistance is regarded to be equally distributed over theirsize. When we work with distributed elements, we must realize that thecurrents and voltages in, for instance, a line do not necessarily have thesame value over the entire length of the line at the same instant of time.3.1 VELOCITY OF TRAVELLING WAVESAND CHARACTERISTIC IMPEDANCEIf a voltage source u is switched on in a two-wire transmission line att = 0 (see Figure 3.2), the line will be charged by the voltage source.After a small time span t, only a small segment x of the line will becharged instantaneously with a charge Q = Cxu. This charge, equallydistributed over the line segment x, causes an electric eld E around thisline segment and the current, or the ow of charge creates a magnetic eldH around the line segment x.VELOCITY OF TRAVELLING WAVES AND CHARACTERISTIC IMPEDANCE 33H Hi in n+u+uuE ESx x x0 0L/m, C/mFigure 3.2 Electric and magnetic eld around a line segment x of a two-wire transmis-sion lineIf x is made innitely small, the expression for the current isi = limx0Qt = limx0Cuxt = Cudxdt = Cu (3.1)Because the distance x is covered in the time t, x/t is the velocity atwhich the charge travels along the line. The magnetic ux present aroundthe line segment is = Lxi. If this is substituted in Equation (3.1) andx is made innitely small, the expression for the induced electromotiveforce emf in the loop enclosed by the two wires over the distance x isemf = limx0

t = LCu_dxdt_2= LCu2(3.2)Because there cannot be a discontinuity in voltage, this emf equals thevoltage source u; this gives an expression for the wave velocity: = 1LC (3.3)The wave velocity depends only on the geometry of the line and on thepermittivity and the permeability of the surrounding medium.For a 150-kV overhead transmission line with one conductor per phase,a conductor radius of r = 25 mm and an equivalent distance betweenphases of dm = 5.5 m, the inductance and the capacitance values areL =_o2_ln_ dm0.779r_= 1.13 mH/km (3.4)34 TRAVELLING WAVESC = 20ln_dmr_ = 10.3 nF/km (3.5)This results in a wave velocity of line = 293.117 km/s. When we calculatethe distributed inductance and the capacitance for a single-core cross-linked polyethylene (XLPE)insulated 18/30-kV medium-voltage cablewith a 120-mm2copper conductor and a 16-mm2earth screen (N2XS2Y-1x120/16-18/30 kV), the values areL = 0.430 mH/km C = 0.178 F/kmThis results in a wave velocity of cable = 114.302 km/s. On an overheadtransmission line the electromagnetic waves propagate close to the speedof light, but in an underground cable the velocity is considerably lower.When the wave velocity is substituted in Equation (3.1) we geti = Cuv = CuLC (3.6)We notice that the ratio between the voltage and current wave has a xedvalueui =_LC (3.7)This is called the characteristic impedance of a transmission line. Thecharacteristic impedance depends on the geometry of the transmissionline and its surrounding medium only. For the 150-kV overhead line,the characteristic impedance is Zline = 330 , and for the XLPE medium-voltage cable, the characteristic impedance is Zcable = 49 .3.2 ENERGY CONTENTS OF TRAVELLING WAVESAn electromagnetic wave contains energy. This energy is stored in theelectric eld and the magnetic eld. The small line segment x in Figure 3.3has a capacitance Cx and is charged to a voltage u(x0, t0). For the electricenergy contents of the line segment at the time t0, we can writewu = 12Cxu(x0, t0)2(3.8)ENERGY CONTENTS OF TRAVELLING WAVES 35Hi n+u u ESx xx00L/m, C/mFigure 3.3 A segment x of a two-conductor line for transverse electromagnetic eldmode of propagationThe magnetic energy contents of the loop enclosed by the two wires overthe distance x at the time t0 iswi = 12Lxi(x0, t0)2(3.9)When the currentvoltage relation u(x, t) = (L/C)1/2i(x, t) is substitutedin the expression for the electric energy, we get the expression for themagnetic energy, and when the currentvoltage relation is substituted inthe expression for the magnetic energy, we get the expression for theelectric energy. This means that wu/wi = 1 and that u(x, t)2/i(x, t)2=L/C = Z2.The energy of an electromagnetic wave is equally distributed over theelectric and the magnetic eld. The total energy of line segment x at thetime t0 is the sum of the energy in the electric and the magnetic eld:wt = wu+wi = Cxu(x0, t0)2= Lxi(x0, t0)2(3.10)The energy passing through x per second is the power P of the electro-magnetic wave:P = wtt = wt(x/v)= Cu(x0, t0)2 1LC = u(x0, t0)2Z= Li(x0, t0)2 1LC = i(x0, t0)2Z(3.11)The power of an electromagnetic wave can be enormous (for example,26.7 MW for a voltage wave with a crest of 100 kV on an overheadtransmission line with a characteristic impedance Z = 375 ), but the36 TRAVELLING WAVESduration of switching and lightning transients is usually short, rangingfrom microseconds to milliseconds. Still, the energy contents of switchingand lightning transients can easily reach values up to a few MJ and cancause considerable damage to power system equipment.3.3 ATTENUATION AND DISTORTIONOF ELECTROMAGNETIC WAVESUntil now we have looked at the transmission line as a so-called losslessline, which means that the following items have not been taken intoaccount: The series resistance of the conductors The skin-effect for higher frequencies The losses in the dielectric medium between the conductors in ahigh-voltage cable The leakage currents across string insulators The inuence of the ground resistance The corona losses and so onTo include these losses in our analysis, we assume a series resistance Rand a parallel conductance G, evenly distributed along the wires as theinductance L and the capacitance C. We consider again a line segment x(see Figure 3.4).i (x, t )u(x, t )+G x C xR x L x i (x + x, t )u(x + x, t )x+Figure 3.4 Equivalent circuit of a differential-length line segment x of a two-conductortransmission lineATTENUATION AND DISTORTION OF ELECTROMAGNETIC WAVES 37The losses in line segment x at the time t0 arePLoss = i(x0, t0)2Rx +u(x0, t0)2Gx = i(x0, t0)2(R +Z2G)x (3.12)When x becomes innitely small, we can write it as dx. An electromag-netic wave with a power P = i(x0, t0)2Z entering this innitely small linesegment dx loses (when we differentiate P with respect to i(x, t)) powerP = 2i(x0, t0) diZ, while travelling the distance dx. This loss of powerhas a negative sign because the wave energy decreases and is dissipated inR and G:i(x0, t0)2(R +Z2G) dx = 2i(x0, t0) diZ (3.13)This can be generalised for every instant of timedii(x, t)= 0.5_RZ +ZG_ dx didx = 0.5_RZ +ZG_i(x, t)(3.14)The solution of this differential equation isi(x, t) = i(x0, t0)e0.5[(R/Z)+(ZG)]x(3.15)and because of the relation u(x, t) = Zi(x, t) that exists between thevoltage and the current wave, we can write for the voltage waveu(x, t) = Zi(x0, t0)e0.5[(R/Z)+(ZG)]x= u(x0, t0)e0.5[(R/Z)+(ZG)]x(3.16)When the voltage and current waves travel along a transmission line withlosses, the amplitude of the waves is exponentially decreased. This is calledattenuation and is caused by the properties of the transmission line.For overhead transmission lines, G is a very small number and wecan simplify Equation (3.15) for the current and Equation (3.16) for thevoltage wave toi(x, t) = i(x0, t0)e(Rx/2Z)(3.17)andu(x, t) = u(x0, t0)e(Rx/2Z)(3.18)The attenuation is small for a line with a low resistance and/or a largecharacteristic impedance.38 TRAVELLING WAVESWhen the series resistance R and the parallel conductance G can beneglected, both the wave velocity and the characteristic impedance areconstant and the transmission line is said to be lossless.When R/L = G/C, we call the transmission line distortionless; the shapeof the current and voltage waves is not affected and the wave velocity andthe characteristic impedance are constant, similar to a lossless line. Whenthe transmission line is not distortionless, the steepness of the wave frontwill decrease and the general shape of the waves will be more elongatedwhen they travel along the line.3.4 THE TELEGRAPH EQUATIONSThe transmission line equations that govern general two-conductor uni-form transmission lines, including parallel-plate, two-wire lines, andcoaxial lines, are called the telegraph equations. The general transmis-sion line equations are named the telegraph equations because they wereformulated for the rst time by Oliver Heaviside (18501925) when hewas employed by a telegraph company and used to investigate disturbanceson telephone wires.When we consider again a line segment x with parameters R, G, L, andC, all per unit length, (see Figure 3.4) the line constants for segment xare Rx, Gx, Lx, and Cx. The electric ux and the magnetic ux created by the electromagnetic wave, which causes the instantaneousvoltage u(x, t) and current i(x, t), ared(t) = u(x, t)Cx (3.19)andd(t) = i(x, t)Lx (3.20)Applying Kirchhoffs voltage law on the loop enclosed by the two wiresover the distance x, we obtainu(x, t) u(x +x, t) = u = i(x, t)Rx + td(t)=_R +L t_i(x, t) x(3.21)THE TELEGRAPH EQUATIONS 39In the limit, as x 0, this voltage equation becomesu(x, t)x = Li(x, t)t Ri(x, t) (3.22)Similarly, for the current owing through G and the current charging C,Kirchhoffs current law can be applied:i(x, t) i(x +x, t) = i = u(x, t)Gx + t d(t)=_G+C t_u(x, t)x (3.23)In the limit, as x 0, this current equation becomesi(x, t)x = Cu(x, t)t Gu(x, t) (3.24)The negative sign in these equations is caused by the fact that when thecurrent and voltage waves propagate in the positive x-direction i(x, t)and u(x, t) will decrease in amplitude for increasing x. To solve theequations, they are transformed into the Laplace domain by substitutingthe Heaviside operator p = /t; this leaves us with the partial differentialequationsu(x, p)x = (R +pL)i(x, p) (3.25)i(x, p)x = (G+pC)u(x, p) (3.26)When we substitute Z

= R +pL and Y

= G+pC and differentiate oncemore with respect to x, we get the second-order partial differentialequations2u(x, p)x2 = Z

i(x, p)x = Z

Y

u(x, p) = 2u(x, p) (3.27)2i(x, p)x2 = Y

u(x, p)x = Y

Z

i(x, p) = 2i(x, p) (3.28) =_(RG+(RC +GL)p +LCp2) = 1_(p +)22(3.29)40 TRAVELLING WAVESIn this expression, = (1/LC)1/2the wave velocity; (3.30) = 1/2[(R/L) +(G/C)] the attenuation constant (of inuenceon the amplitude of the travelling waves); (3.31) = 1/2[(R/L) (G/C)] the phase constant (of inuence on thephase shift of the travelling waves); and (3.32)Z = (Z

/Y

)1/2= (L/C)1/2[(p + +)/(p + )]1/2the characteristic impedance. (3.33)The solutions of Equation (3.27) and Equation (3.28) in the timedomain areu(x, t) = e xf1(t) +e xf2(t) (3.34)andi(x, t) = 1Z_e xf1(t) e xf2(t)_ (3.35)In these expressions, f1(t) and f2(t) are arbitrary functions, independentof x.3.4.1 The Lossless LineFor the lossless line, the series resistance R and the parallel conduc-tance G are zero, and the propagation constant and the characteristicimpedance are = pLC = p/ (3.36)andZ = Z0 =_LC (3.37)For the characteristic impedance of a lossless transmission line, the symbolZ0 is often used. The solutions for the voltage and current waves reduce tou(x, t) = epx/vf1(t) +epx/vf2(t) (3.38)i(x, t) = 1Z0_epx/vf1(t) epx/vf2(t)_ (3.39)THE TELEGRAPH EQUATIONS 41When we apply Taylors series to approximate a function by a series,assuming that the function can be differentiated a sufciently number oftimes,f (t +h) = f (t) +hf

(t) +_h22!_f

(t) + (3.40)and substituting the Heaviside operator p = d/dt results inf (t +h) =_1 +hp + h22!p2+ _f (t) = ehpf (t) (3.41)Applying this to Equation (3.38) and Equation (3.39), we nd solutionsfor the voltage and current waves in the time domain:u(x, t) = f1_t + xv_+f2_t xv_ (3.42)i(x, t) = 1Z0_f1_t + xv_f2_t xv__ (3.43)In this expression, f1[t +(x/v)] is a function describing a wave propa-gating in the negative x-direction, usually called the backward wave, andForward wavesu(x, t )u(x, t )u(x, t ) = Z0 i (x, t )i (x, t )i (x, t )i (x, t )i (x, t )n+u(x, t ) = Z0 i (x, t )+xxBackward wavesFigure 3.5 Forward- and backward-travelling waves and their polarity42 TRAVELLING WAVESf2[t (x/v)] is a function describing a wave propagating in the positivex-direction, called the forward wave (see Figure 3.5). The minus sign forthe backward current wave related to the backward voltage wave accountsfor the fact that the backward energy wave, when the Poynting vectorP = E H is calculated, has a negative sign because it propagates in thenegative x-direction.3.4.2 The Distortionless LineTypical of power systems is that the overhead lines and undergroundcables have small losses. This means that R/L and G/C are small and thatthe attenuation constant and the phase constant are small comparedwith the rate of change p = /t of the voltage and current waves. For adistortionless line, R/L = G/C, which means that the losses of the electriceld equal the losses of the magnetic eld. As is the case with the losslessline, the voltage and current waves carry an equal amount of energy toevery point on the line, and the rate at which they lose energy at everypoint on the line is also equal. There is, however, no need for an exchangeof energy between the voltage and current wave; therefore, the voltageand current waves are undistorted, keep their original shape, and areonly attenuated in amplitude. This is the case for power systems, and thevoltage wave equation (Equation (3.34)) reduces tou(x, t) = ex/vf1_t + xv_+ex/vf2_t xv_ (3.44)and the current wave equation (Equation (3.44)) can be written asi(x, t) = 1Z0_ex/vf1_t + xv_ex/vf2_t xv__ (3.45)Thus, except for an attenuation constant, the characteristics of a distor-tionless line, are the same as that of a lossless line: a constant wavevelocity = 1/(LC)1/2, and a constant real characteristic impedanceZ0 = (L/C)1/2.3.5 REFLECTION AND REFRACTIONOF TRAVELLING WAVESWhen an electromagnetic wave propagates along a transmission line witha certain characteristic impedance, there is a xed relation between theREFLECTION AND REFRACTION OF TRAVELLING WAVES 43voltage and current waves. But what happens if the wave arrives at adiscontinuity, such as an open circuit or a short-circuit, or at a pointon the line where the characteristic impedance Z = (L/C)1/2changes, forinstance, when an overhead transmission line is terminated with a cableor a transformer? Because of the mismatch in characteristic impedance,an adjustment of the voltage and current waves must occur. At thediscontinuity, a part of the energy is let through and a part of the energyis reected and travels back. At the discontinuity, the voltage and currentwaves are continuous. In addition, the total amount of energy in theelectromagnetic wave remains constant, if losses are neglected. Figure 3.6shows the case in which an overhead transmission line is terminated withan underground cable.When, for the sake of simplicity, both the overhead line and theunderground cable are assumed to be without losses, R = 0 and G = 0,then the expressions for the characteristic impedances areZL =_LLCLandZC =_LCCC(3.46)The forward wave is called the incident wave and it travels on the overheadline toward the cable. The positive x-direction in Figure 3.6 is from leftOverhead transmission lineLine parameters LL, CLCable parameters LC, CCxDiscontinuityat x0Figure 3.6 Overhead transmission line terminated with an underground cable44 TRAVELLING WAVESto right and the line-cable joint is at x0. The incident wave reaches thediscontinuity x0 at t = t0. In our equations, the incident waves havesubscript 1, the reected waves have subscript 2, and the waves that arelet through have subscript 3. Voltage and current waves are continuous atthe linecable joint:L(x0, t) = C(x0, t)andiL(x0, t) = iC(x0, t) (3.47)This leads to the conclusion that for t > t0, at the discontinuity at x0,the incident waves and the reected waves equal the waves that are letthrough. For the voltage waves, we can write1L(x0, t) +2L(x0, t) = 3C(x0, t) (3.48)The reected current wave has a negative sign:1L(x0, t)ZL 2L(x0, t)ZL= 3C(x0, t)ZC(3.49)From Equation (3.48) and Equation (3.49), for the voltage and currentwaves that are let through3C(x0, t) =_ 2ZCZL+ZC_1L(x0, t) (3.50)i3C(x0, t) =_ 2ZLZL+ZC_i1L(x0, t) (3.51)and for the reected waves2L(x0, t) =_ZCZLZL+ZC_1L(x0, t) (3.52)i2L(x0, t) =_ZLZCZL+ZC_i1L(x0, t) (3.53)For a transition from overhead line to cable in a substation, an incomingvoltage wave is reduced in amplitude. When an incoming overhead lineis connected to a substation by means of a cable of short length, thecable protects the substation against overvoltages. When lightning strikesREFLECTION OF TRAVELLING WAVES 45the substation, the opposite occurs a magnied voltage wave leaves thesubstation and travels over the overhead line.When the incident wave encounters a short-circuit at x0, which means,in our example, that ZC = 0, the voltage wave disappears at x0 andthe current wave doubles in amplitude; the wave energy is stored in themagnetic eld only. When the incident wave encounters an open circuitat x0, the voltage wave is doubled in amplitude and the incident currentwave is nullied by the reected current wave; the wave energy is storedin the electric eld only. The voltage doubling that occurs when a voltagesource is switched on an unloaded cable or an unloaded overhead line iscalled the Ferranti-effect, named after the British scientist and engineerSebastiano Ziani de Ferranti (18641930).When a transmission line is terminated with a load impedance ZL,different from the lines characteristic impedance Z0, the ratio of thecomplex amplitudes of the reected and incident voltage waves at the loadis called the voltage reection coefcientr = ZLZ0ZL+Z0(3.54)The voltage reection coefcient r is, in general, a complex quantity witha magnitude smaller than one.3.6 REFLECTION OF TRAVELLING WAVESAGAINST TRANSFORMER AND GENERATORWINDINGSWhen the travelling waves reach windings of transformers and generators,they encounter windings whose physical dimensions are rather smallcompared with the dimensions of the power system itself. The travellingwaves generate oscillations and resonance in the windings and causestanding waves. Because a standing wave results from the superpositionof two waves travelling in opposite directions, the voltages and currentsin the generator windings and transformer windings should be treated astravelling waves. This implicates that we have to know their characteristicimpedance, if we want to calculate the reection and refraction of voltageand current waves against transformers and generators.To understand the characteristic impedance of a power transformer,we assume the transformer to consist of a coil mounted on an ironcore, as shown in Figure 3.7. The height of the cylindrical winding is46 TRAVELLING WAVESTankHVLVrCorehtdwWindingFigure 3.7 Dimensions of a simplied power transformer h 2r, r w, w tlarge compared with its diameter (h 2r); the radius of the winding islarge compared to the gap w between winding and core (r w); andthe gap w between winding and core on itself is large compared withthe thickness t of the winding(w t). The distance between tank andwinding is d, and both the tank and transformer core are assumed to beideal for conducting. The core and the transformer tank are grounded.The boundary conditions provide constraints on the components of theeld vectors as they transit across the boundary between two media.The tangential components of the electric eld intensity vector E and themagnetic eld intensity vector H must be continuous across the boundarybetween two physical media medium 1 and medium 2Et1 = Et2 and Ht1 = Ht2 (3.55)In addition, the normal components of the electric ux density vector Dand the magnetic ux density vector B must be continuous across theboundary between two physical mediaDn1 = Dn2 and Bn1 = Bn2 (3.56)REFLECTION OF TRAVELLING WAVES 47The core and the tank wall are assumed to be perfectly conductive; this ischaracterised by an innite conductivity = . The innite conductivitymakes all the elds in the perfect conductor to be zero. This requiresthat the tangential component of E1 of medium 1 adjacent to medium2 (the perfect conductor) must be zero at the boundary. In addition, thenormal component of B1 must be zero at the boundary. The tangentialcomponent of H1 and the normal component of D1 cannot be zero;otherwise, the resulting elds in medium 1 would be overspecied. Thetangential component of the magnetic eld in medium 1, H1, equals thesurface current density on the interface, and the normal component ofthe electric ux D1 equals the surface charge density on the interface(Figure 3.8). This simplies the mathematical calculations. When the rightcontour and surface is chosen, the capacitance and inductance of thewinding can be calculated and this result can be used to calculate thecharacteristic impedance.When a constant voltage V is present between the winding terminal andthe ground, there is a linear voltage prole along the coil that results in aninner electric eld Einner between coil and core and an outer electric eldEouter between coil and tank. Gauss law for the electric eld says thatMedium 1, , Medium 2perfectconductor = Et 1 = 0Ht 1Dn1KsE1B1Bn1 = 0Bn2