ieee press - power system stability- edward wilson kimbark

POWER SYSTEM STABILITY

Volume IElements of Stability Calculations

IEEE Press445 Hoes Lane, PO Box 1331Piscataway, NJ 08855-1331

EditorialBoardJohn B. Anderson, Editor in Chief

R. S. BlicqM.EdenD. M. EtterG. F. HoffnagleR. F. Hoyt

J. D. IrwinS. V. KartalopoulosP. LaplanteA. J. LaubM. Lightner

J. M. F. MouraJ.PedenE. Sanchez-SinencioL. ShawD. J. Wells

Dudley R. Kay, Director ofBook PublishingCarrie Briggs,Administrative Assistant

Lisa S. Mizrahi, Review and Publicity Coordinator

Valerie Zaborski, Production Editor

IEEE Press Power Systems Engineering Series

Dr. Paul M. Anderson, Series Editor. PowerMathAssociates, Inc.

Series Editorial Advisory Committee

Dr. Roy BillintonUniversity of Saskatchewan

Dr. Atif S. DebsGeorgiaInstituteof Technology

Dr.M. El-HawaryTechnicalUniversity of

Nova Scotia

Mr. RichardG. FarmerArizonaPublic

ServiceCompany

Dr. Charles A. GrossAuburnUniversity

Dr. G. T. HeydtPurdueUniversity

Dr. George KaradyArizonaState University

Dr. DonaldW. NovotnyUniversity of Wisconsin

Dr. A. G. PhadkeVirginiaPolytechnic and

State University

Dr. Chanan SinghTexas A & M University

Dr. E. Keith StanekUniversity of Missouri-Rolla

Dr. J. E. Van NessNorthwestern University

POWERSYSTEM

STABILITYVolume I

Elements of Stability Calculations

Edward Wilson Kimbark

AIEEE'V PRESS

roWILEY-\:'l9INTERSCIENCEAJOHN WILEY & SONS, INC., PUBLICAnON

IEEE Press Power Systems Engineering SeriesDr. Paul M. Anderson, Series Editor

©1995 by the Institute of Electrical and Electronics Engineers, Inc.345 East 47th Street, New York, NY 10017-2394

©1948 by Edward Wilson Kimbark

This is the IEEE reprinting of a book previously published by John Wiley &Sons, Inc. under the title Power System Stability, Volume I: Elements ofStability Calculations.

All rights reserved. No part ofthis book may be reproduced in any form,nor may it be stored in a retrieval system or transmitted in any form,without written permission from the publisher.

Printed in the United States of America

10 9 8 7 6 5 4 3 2

ISBN 0-7803-1135-3

Library of Congress Cataloging-in-Publication Data

Kimbark, Edward WilsonPower system stability I Edward Wilson Kimbark.

p. cm. - (IEEE Press power systems engineering series)Originally published: New York : Wiley, 1948-1956.Includes bibliographical references and index.Contents: v.I. Elements of stability calculations - v. 2. Power

circuit breakers and protective relays - v. 3. Synchronousmachines.

ISBN 0-7803-1135-3 (set)1. Electric power system stability. I. Title. II. Series.

TKI010.K56 1995621.319--dc20 94-42999

CIP

To my wife

RUTH MERRICK KIMBARK

FOREWORD TO THE 1995 REISSUE

The IEEE Press Editorial Board for the Power Systems EngineeringSeries has, for some time, discussed the possibility of reprinting clas-sic texts in power system engineering. The objective of this series is torecognize past works that merit being remembered and to make theseolder works available to a new generation of engineers. We believemany engineers will welcome the opportunity of owning their owncopies of these classics.

In order to come to an agreement about which text to reprint, a num-ber of candidates were proposed. After a discussion, the board took avote. The Kimbark series was the overwhelming choice for the firstbooks in the IEEE Power Systems Engineering Classic Reissue Series.

The subject of power system stability has been studied and writtenabout for decades. It has always been a challenge for the engineer tounderstand the physical description of a system described' by a hugenumber of differential equations. The system modeling is central to anunderstanding of these large dynamic systems. Modeling is one of thecentral themes of Kimbark's Power System Stability books. His dis-cussion of the system equations remains as clear and descriptive todayas it was when first published. Many engineers have seen references tothese works, and may have had difficulty in finding copies for study.This new printing presents a new chance for these engineers to nowhave copies for personal study and reference.

Kimbark presents a method of solving the system equations that wasused in the days of the network analyzer. This method has been re-placed by digital computer techniques that provide much greaterpower and speed. However, the older methods are still of historicalinterest, moreover, these step-by-step methods provide a convenientway of understanding how a large system of equations can be solved.

Edward Kimbark was noted during his long career as an excellentwriter and one who had the unique capability of explaining complextopics in a clear and interesting manner. These three volumes underthe general title Power System Stability, Volumes I, II, and III wereoriginally published in the years 1948, 1950, and 1956. Kimbark'sbook, Electrical Transmission of Power Signals, published in 1949,

vii

provided a general treatment of electric power networks and signalpropagation.

Kimbark studied Electrical Engineering at Northwestern Universityand at the Massachusetts Institute of Technology, where he receivedthe Sc.D. degree in 1937. He then began a career in teaching and re-search at the University of California, Berkeley, MIT, PolytechnicInstitute, Brooklyn, Instituto Tenologico de Aeronautica (San JoseCampos Brazil) and, finally, as the Dean of Engineering at Seattle Uni-versity. In 1962 Kimbark joined the Bonneville Power Administrationas head of the systems analysis branch, where he remained until his re-tirement in 1976. He continued to work on special tasks at Bonnevilleuntil his death in 1982.

Kimbark is well-known for his excellent books and also his manytechnical papers. He was formally recognized for his achievements bybeing elected a Fellow in the IEEE, to membership in the NationalAcademy of Engineering, and was the recipient of the IEEE HarbishawAward. He was awarded a Distinguished Service Award and a GoldMedal for his service to the U.S. Department of the Interior.

The IEEE Power Engineering Society is proud to present thisspecial reprinting of all three volumes of Power System Stability byEdward Kimbark.

Paul M. AndersonSeries Editor, IEEE Press

Power Systems Engineering Series

viii

PREFACE

This work on power-system stability isintended for use by power-system engineers and by graduatestudents. It grew out of lectures given by the author in agraduate evening course at Northwestern University during theschool year 1941-2.

For the convenience of the reader, the work is divided intothree volumes. Volume I covers the elements of the stabilityproblem, the principal factors affecting stability, the ordinarysimplified methods of making stability calculations, and illus-trations of the application of these methods in studies whichhave been made on actual power systems.

Volume II covers power circuit breakers and protectiverelays, including material on rapid reclosing of circuit breakersand on the performance of protective relays during swings andout-of-step conditions. Such material belongs in a work on sta-bility because the most important means of improving the tran-sient stability of power systems in the improvement of circuitbreakers and of protective relaying. It .is expected, however,that the publication of this material in a separate volume willmake it more useful to persons who are interested in power-system protection, even though they may not be particularlyconcerned with the subject of stability.

Justification of the simplifying assumptions ordinarily used in'stability calculations and the carrying out of calculations for theextraordinary cases in which greater accuracy than that affordedby the simplified methods is desired require a knowledge of thesomewhat complicated theory of synchronous machines and oftheir excitation systems. This material is covered in VolumeIII, which is expected to appeal to those desiring a deeper under-standing of the subject than is obtainable from Volume I alone.

ix

x PREFACE

It is my hope that this treatise will prove useful not only toreaders seeking an understanding of power-system stability butalso to those desiring information on the following related topics:a-c. calculating boards, fault studies, circuit breakers, protectiverelaying, synchronous-machine theory, exciters and voltage regu-lators, and the step-by-step solution of nonlinear differentialequations.

I wish to acknowledge my indebtedness to the followingpersons:

To my wife, Ruth Merrick Kimbark, for typing the entiremanuscript and for her advice and inspiration.

To Charles A. Imburgia, A. J. Krupy, Harry P. St. Clair, andespecially Clement A. Streifus for supplying and interpretinginformation on stability studies made on actual power systems.

To J. E. Hobson, W. A. Lewis, and E. T. B. Gross for review-ing the manuscript and for making many suggestions for itsimprovement.

To engineers of the General Electric Company and of theWestinghouse Electric Corporation for reviewing certain partsof the manuscript pertaining to products of their companies.

To manufacturers, authors, and publishers who suppliedillustrations or gave permission for the use of material previ-ously published elsewhere. Credit for such material is given atthe place where it appears.

EnwARD WILSON KIMBARK

Evanston, IllinoisJune, 1947

CONTENTS

CHAPTER PAGE

I The Stability Problem 1

II The Swing Equation and Its Solution 15

III Solution of Networks 53

IV The Equal-Area Criterion for Stability 122

V Further Consideration of the Two-MachineSystem 149

VI Solution of Faulted Three-Phase Networks 193

VII Typical Stability Studies 253

INDEX 349

CHAPTER I

THE STABILITY PROBLEM

Gen. t-------f Motor

x.Ie

Xo1

Definitions and illustrations of terms. Power-system stability is aterm applied to alternating-current electric power systems, denoting acondition in which the various synchronous machines of the systemremain in synchronism, or "in step," with each other. Conversely,instability denotes a condition involving loss of synchronism, or falling"out of step."

Consider the very simple power system of Fig. 1, consisting of asynchronous generator supplying power to a synchronous motor overa circuit composed of series inductive reactance XL. Each of thesynchronous machines may be rep-resented, at least approximately, bya constant-voltage source in serieswith a constant reactance.* Thusthe generator is represented by Eoand Xo; and the motor, by EMand XM. Upon combining the machine reactances and the line re..actance into a, single reactance, wehave an electric circuit consisting of FI 1 S· I t h·G.. Imp e we-mac me powertwo constant-voltage sources, Eo system.and EM, connected through re-actance X c= XG + XL + XM. It will be shown that the powertransmitted from the generator to the motor depends upon the phasedifference 0 of the two voltages EG and EM. Since these voltages aregenerated by the flux produced by the field windings of the machines,their phase difference is the same as the electrical angle between themachine rotors.

The vector diagram of voltages is shown in Fig. 2. Vectorially,

EG = EM +JXI [1]

(The bold-face letters here and throughout the book denote com-

*Either equivalent synchronous reactance or transient reactance is used, dependingupon whether steady-state or transient conditions are assumed. These terms aredefined and discussed in Chapters XII and XV, Vol. III.

1

2 THE STABILITY PROBLEM

plex, or vector, quantities). Hence the current is

I = Eo - EMjX

[2]

[5]

[4]

[3]

The power output of the generator-and likewise the power input of the motor, since there

jXI is no resistance in the line-is given by

P = Re(EoI)

R (-E Eo - EM)= e 0---

FIG. 2. Vector diagram sxof the system of Fig. 1. h R " I f -were e means the rea part 0" and Eomeans the conjugate of Eo. Nowlet

EM = EM!!and

ThenEo = EolJ..

Eo = Eo/-a

[6]

[7]

Substitution of eqs. 5, 6, and 7 into eq. 4 gives

(EolJ.. - EMI!J.)

P = Re Eo/-a x~

... Re(E;2 /-900_ Ea;M /-900

- a)EoEM 0= - X-cos(-90 - 8)

EoEM.= -y-sma [8]

This equation shows that the power P transmitted from the generatorto the motor varies with the sine of the displacement angle abetweenthe two rotors, as plotted in Fig. 3. The curve is known as a powerangle curve. The maximum power that can be transmitted in thesteady state with the given reactance X and the given internal voltagesEo and EM is

[9]

DEFINITIONS AND ILLUSTRATIONS OF TERMS 3

and occurs at a displacement angle ~ = 90°. The value of maximumpower may be increased by raising either of the internal voltages or bydecreasing the circuit reactance.

The system is stable only if the displacement angle ~ is in the rangefrom -90° to +90°, in which the slope dP/do is positive; that is, therange in which an increase in displacement angle results in an increasein transmitted power. Suppose that the system is operating in thesteady state at point A, Fig. 3. The mechanical input of the generatorand the mechanical output of the motor, if corrected for rotationallosses, will be equal to the electric power P. Now suppose that a

p

FIG. 3. Power-anglecurve of the system of Fig. 1.

small increment of shaft load is added to the motor. Momentarily theangular position of the motor with respect to the generator, and therefore the power input to the motor, is unchanged; but the motor outputhas been increased. There is, therefore, a net torque on the motortending to retard it, and its speed decreases temporarily. As a resultof the decrease in motor speed, 0 is increased, and consequently thepower input is increased, until finally the input and output are again inequilibrium, and steady operation ensues at a new point B, higher thanA on the power-angle curve. (It has been tacitly assumed that thegenerator speed would remain constant. Actually the generator mayhave to slow down somewhat in order for the governor of its primemover to operate and increase the generator input sufficiently tobalance the increased output.)


Suppose that the motor input is increased gradually until the pointC of maximum power is reached. If now an additional increment ofload is put on the motor, the displacement angle a will increase asbefore, but as it does so there will be no increase in input. Insteadthere will be a decrease in input, further increasing the difference between output and input, and retarding the motor more rapidly. Themotor will pull out of step and will probably stall (unless it is keptgoing by induction-motor action resulting from damper circuits whichmay be present) . Pm is the steady-state stability limit of the system.It is the maximum power that can be transmitted, and synchronismwill be lost if an attempt is made to transmit more power than thislimit.

If a large increment of load on the motor is added 8uddenly, insteadof gradually, the motor may fall out of step even though the new loaddoes not exceed the steady-state stability limit. The reason is asfollows: When the large increment of load is added to the motor shaft,the mechanical power output of the motor greatly exceedsthe electricalpower input, and the deficiency of input is supplied by decrease ofkinetic energy. The motor slows down, and an increase of the displacement angle ~ and a consequent increase of input result. In accordance with the assumption that the new load does not exceed thesteady-state stability limit, ~ increases to the proper value for steadystate operation, a value such that the motor input equals the outputand the retarding torque vanishes. When this value of 0 is reached,however, the motor is running too slowly. Its angular momentumprevents its speed from suddenly increasing to the normal value.Hence it continues to run too slowly, and the displacement angle increases beyond the proper value. After the angle has passed thisvalue, the motor input exceeds the output, and the net torque is nowan accelerating torque. The speed of the motor increases and approaches normal speed. Before normal speed is regained, however,the displacement angle may have increased to such an extent that theoperating point on the power-angle curve (Fig. 3) not only goes overthe hump (point C) but also goes so far over it that the motor inputdecreases to a value less than the output. If this happens, the nettorque changes from an accelerating torque to a retarding torque.The speed, which is still below normal, now decreases again, and continues to decrease during all but a small part of each slip cycle. Synchronism is definitely lost. In other words, the system is unstable.

If, however, the sudden increment in load is not too great, the motorwill regain its normal speed before the displacement angle becomes toogreat. Then the net torque is still an accelerating torque and causes

DEFINITIONS AND ILLUSTRATIONS OF TERMS 5

the motor speed to continue to increase and thus to become greaterthan normal. The displacement angle then decreases and again approaches its proper value. Again it overshoots this value on accountof inertia. The rotor of the motor thus oscillates about the newsteady-state angular position. The oscillations finally die out be..cause of damping torques, t which have been neglected in this elementary analysis. A damped oscillatory motion characterizes astable system.

With a given sudden increment in load, there is a definite upperlimit to the load which the motor will carry without pulling out of step.This is the transient stability limit of the system for the given conditions. The transient stability limit is always below the steady-statestability Iimit.t but, unlike the latter, it may have many differentvalues, depending upon the nature and magnitude of the disturbance.The disturbance may be a sudden increase in load, as just discussed, orit may be a sudden increase in reactance of the circuit, caused, forexample, by the disconnection of one of two or more parallel lines as anormal switching operation. The most severe type of disturbance towhich a power system is subjected, however, is a short circuit. Therefore, the effect of short circuits (or "faults," as they are often called)must be determined in nearly all stability studies.

A three-phase short circuit on the line connecting the generator andthe motor entirely cuts off the Bow of power between the machines.The generator output becomes zero in the pure-reactance circuits underconsideration; the motor input also becomes zero. Because of theslowness of action of the governor of the prime mover driving thegenerator, the mechanical power input of the generator remains constant for perhaps i sec. Also, since the power and torque of the loadon the motor are functions of speed, and since the speed cannot changeinstantly and changes by not more than a few per cent unless and untilsynchronism is lost, the mechanical power output of the motor may beassumed constant. As the electrical power of both machines is decreased by the short circuit, while the mechanical power of both remains constant, there is an accelerating torque on the generator and aretarding torque on the motor. Consequently, the generator speedsup, the motor slows down, and it is apparent that synchronism will belost unless the short circuit is quickly removed so as to restore syn..chronizing power between the machines before they have drifted too

tDiscussed in Chapter XIV, Vol. III.tConventional methods of calculation, however, sometimes indicate that the

transient stability limit is above the steady-state stability limit. This paradox isdiscussed in Chapter XV, Vol. III.


far apart in angle and in speed. If the short circuit is on one of twoparallel lines and is not at either end of the line, or if the short circuit isof another type than three-phase-that is, one-line-to-ground, line-toline, or two-line-to-ground-then some synchronizing power can stillbe transmitted past the fault, but the amplitude of the power-anglecurve is reduced in comparison with that of the pre-fault condition. Insome cases the system will be stable even with a sustained short circuit, whereas in others the system will be stable only if the short circuit Is cleared with sufficient rapidity. Whether the system is stableduring faults will depend not only on the system itself, but also on thetype of fault, location of fault, rapidity of clearing, and method ofclearing-that is, whether cleared by the sequential opening of two ormore breakers, or by simultaneous opening-and whether or not thefaulted line is reclosed, For any constant set of these conditions, thequestion of whether the system is stable depends upon how much powerit was carrying before the occurrence of the fault. Thus, for anyspecified disturbance, there is a value of transmitted power, called thetransient stability limit, below which the system is stable and abovewhich it is unstable.

The stability limit is one kind of power limit, but the power limit of asystem is not always determined by the question of stability. Evenin a system consisting of a synchronous generator supplying power to aresistance load, there is a maximum power received by the load as theresistance of the load is varied. Clearly there is 8. power limit herewith no question of stability.

Multimachine systems. Few, if any, actual power systems consistof merely one generator and one synchronous motor. Most powersystems have many generating stations, each with several generators,and many loads, most of which are combinations of synchronousmotors, synchronous condensers, induction motors, lamps, heatingdevices, and others. The stability problem on such a power systemusually concerns the transmission of power from one group of synchronous machines to another. As a rule, both groups consist predominantly of generators. During disturbances the machines of eachgroup swing more or less together; that is, they retain approximatelytheir relative angular positions, although these vary greatly withrespect to the machines of the other group. For purposes of analysisthe machines of each group can be replaced by one equivalent machine.If this is done, there is one equivalent generator and one equivalent synchronous motor, even though the latter often represents machines thatare actually generators.

Because of uncertainty as to which machines will swing together, or

A MECHANICAL ANALOGITE OF SYSTEM STABILITY 7

in order to improve the accuracy of prediction, it is often desirable torepresent the synchronous machines of a power system by more thantwo equivalent machines. Nevertheless, qualitatively the behavior ofthe machines of an actual system is usually like that of a two-machinesystem. If synchronism is lost, the machines of each group stay together, although they go out of step with the other group.

Because the behavior of a two-machine system represents the behavior of a multimachine system, at least qualitatively, and becausethe two-machine system is very simple in comparison with the multimachine system which it represents, the two-machine system is extremely useful in describing the general concepts of power-systemstability and the influence of various factors upon stability. Accordingly, the two-machine system plays a prominent role in this book.

A mechanical analogue of system stability.5§ A simple mechanicalmodel of the vector diagram of Fig. 2 may be built of two pivoted rigidarms representing the EG and EMvectors, joined at their extremitiesby a spring representing the X I vector. (See Fig. 4.) Lengths represent voltages in the model, justas they do in the vector diagram.The lengths of the arms, EG andEM, are fixed in accordancewith the

ti f t t · t 1 FIG. 4. A mechanical analogue of theassump Ion 0 cons an In erna system of Fig. 1.voltages. The length of the springXI is proportional to the applied tensile force (for simplicity, weassume an ideal spring which returns to zero length if the force isremoved). Hence the tensile force can be considered to represent thecurrent, and the compliance of the spring (its elongation per unit force),to represent the reactance.

The torque exerted on an arm by the spring is equal to the product ofthe length of the arm, the tensile force of the spring, and the sine of theangle between the arm and the spring. (More torque is exerted by thespring when it is perpendicular to the arm than at any other angle forthe same tensile force.) Obviously, the torques on the two arms areequal and opposite. The torque, multiplied by the speed of rotation,gives the mechanical power transmitted from one arm to the other.For convenience of inspection, the mechanical model will be regardedas stationary, rather than as rotating at synchronous speed, just as weregard the usual vector diagram as stationary. The formula for torque(or power) in the model is analogous to that for power in the vector

§Superior numerals refer to items in the list of References at end of chapter.


diagram, namely: voltage X current X cosine of angle between them.(Since the XI vector is 90° ahead of the I vector, the cosine of theangle between E and I is equal to the sine of the angle between E andXI.)

The shaft power of the machines may be represented by applyingadditional torques to the arms. A convenient method of applying

FIG. 5. A mechanical analogue ofthe system of Fig. 1, suitable forrepresenting transient conditions.

FIG. 6. A mechanical analogue of a threemachine system consisting of generator,synchronous condenser, and synchronous

motor.

constant equal and opposite torques to the two arms is to attach a drumto each arm and to suspend a weight pan from a pulley hanging on acord, one end of which is wound on each drum, all as indicated inFig. 5.

Asweights are added to the pan in small increments, the two arms ofthe model gradually move farther apart until the angle 0 between themreaches 90°, at which position the spring exerts maximum torque. Iffurther weights are added, the arms fly apart and continue to rotate inopposite directions until all the cord is unwound from the drums. Thesystem is unstable. The steady-state power limit is reached ato= 90°. Although from 90° to 180° the spring force (current) continues to increase, the angle between arm and spring changes in such away that the torque decreases.

The effect of changing the machine voltages can be shown by attaching the spring to clamps which slide along the arms.

The effect of an intermediate synchronous-condenser station in in-

BAD EFFECTS OF INSTABILITY 9

creasing the steady-state power limit can be shown by adding a thirdpivoted arm attached to an intermediate point of the spring (Fig. 6).The condenser maintains a fixed internal voltage. Since the condenserhas no shaft input or output, no drum is provided on the third arm inthe model. With the intermediate arm (representing the condenser)in place, the angle betwe.en the two outer arms (representing thegenerator and motor) may exceed 90° without instability, and thepower limit is greater than before.

The model can be used to illustrate transient stability by providingeach arm with a flywheel such that the combined moment of inertia ofthe arm and flywheel is proportional to that of the corresponding synchronous machine together with its prime mover (or load). Thedrums can be made to serve this purpose.

If not too great an increment of load is suddenly added to the pan, it will be found thatthe arms oscillate before settling down to theirnew steady-state positions. The angle betweenthe arms may exceed 90° during these oscillations without loss of stability. If the incrementof load is too large, the arms willfly apart andcontinue to rotate in opposite directions, indicating instability. This may happen eventhough the total load is less than the steadystate stability limit.

The effect of switching out one of two parallellines may be simulated by connecting the armsby two springs in parallel and then suddenlydisconnecting one spring by burning the piece FIG. 7. A mechanicalof string by which the spring is attached. ~nalogue of the effect of a

he li be si line fault on the powerThe effect of a fault on t e hne may e SlDlU- t f F" 1ByS em 0 19••

lated by suddenly pushing a point on the springtoward the axle (Fig. 7). The arms will start to move apart, and stability will be lost unless the spring is quickly released.

Models of this kind have been built to give a scale representation ofactual power systems of three or four machines, and the oscillationsof the arms have been recorded by moving-picture cameras. 6 Thereare practical difficulties, however, in applying the model representationto a complicated system. The chief value of the model is to illustratethe elementary concepts of stability. Other methods of analysis areused in practice.

Bad effects of instability. When one machine falls out of step withthe others in a system, it no longer serves its function. If it is a


generator, it no longer constitutes a reliable source of electric power.If it is a motor, it no longer delivers mechanical power at the properspeed, if at all. If it is a condenser, it no longer maintains propervoltage at its terminals. An unstable two-machine system, consistingof motor and generator, may be compared to a slipping belt or clutchin a mechanical transmission system; instability means the failure ofthe system as a power-transmitting link.

Moreover, a large synchronous machine out of step is not only useless; 'it is worse than useless-it is injurious-because it has a disturbing effect on voltages. Voltages will fluctuate up and down betweenwide limits. Thus instability has the same bad effect on service tocustomers' loads as does a fault, except that the effect of instability islikely to last longer. If instability occurs as a consequenceof a fault,clearing of the fault itself may not restore stability. The disturbingvoltage fluctuations then continue after the fault has been cleared.The machine, or group of machines,' which is out of step with the restof the system must either be brought back into step or else disconnected from the rest of the system. Either operation, if done manually, may take a long time compared with the time required to clear afault automatically. As a rule, the best way to bring the machinesback into step is to disconnect them and then re-synchronize them.Protective relays have been developed to open a breaker at a predetermined location when out-of-step conditions occur. Such relays,however, are not yet in wide use. Preferably the power system shouldbe split up into such parts that each part will have adequate generatingcapacity connected to it to supply the load of that part. Some overload may have to be carried temporarily until the system is re-synchronized.

Ordinary protective relays are likely to operate falsely during out-ofstep conditions, thereby tripping the circuit breakers of unfaultedlines. Such false tripping may unnecessarily interrupt service totapped loads and may split the system apart at such points that thegenerating capacity of some'parts is inadequate.]

The trend in power-system design has been toward increasing thereliability of electric power service. Since instability has a bad effecton the quality of service, a power system should be designed andoperated so that instability is improbable and will occur only rarely.

Scope of this book. This book will deal with two different phases ofthe problem of power-system stability: (1) methods of analysis andcalculation to determine whether a given system is stable when subjected to a specified disturbance; (2) an examination of the effect of

liThia aspect of relay operation is discussed fully in Chapter X, Vol. II.

HISTORICAL REVIEW 11

various factors on stability, and a consideration of measures for improving stability. In our discussion these two phases will be related:after a method of analysis is presented, it will be applied to show theeffect of varying different factors. Among these factors are systemlayout, circuit impedances, loading of machines and circuits, type offault, fault location, method of clearing, speed of clearing, inertia ofmachines, kind of excitation systems used with the machines, machinereactances, neutral grounding impedance, and damper windings onmachines.

Since transient power limits are lower than steady-state powerlimits, and since any power system will be subject to various shocks,the most severe of which are short circuits, the subject of transientstability is much more important than steady-state stability. Accordingly, the greater part of this book is devoted to transient stability.Chapter XV, Vol. III, deals with steady-state stability.

Historical review. Since stability is a problem associated with theparallel operation of synchronous machines, it might be suspected thatthe problem appeared when synchronous machines were first operatedin parallel. The first serious problem of parallel operation, however,was not stability, but hunting. When the necessity for parallel operation of a-c. generators became general, most of the generators weredriven by direct-connected steam engines. The pulsating torquedelivered by those engines gave rise to hunting, which was sometimesaggravated by resonance between the' period of pulsation of primemover torque and the electromechanical period of the power system.In some cases improper design or functioning of the engine governorsalso aggravated the hunting. Hunting of synchronous motors andconverters was sometimes due to another cause, namely, too high aresistance in the supply line.

The seriousness of hunting was decreased by the introduction of thedamper winding, invented by Lelslanc in France and by Lamme inAmerica. Later, the problem largely disappeared on account of thegeneral use of steam turbines, which have no torque pulsations.Nearly all the prime movers in use nowadays, both steam turbines andwater wheels, give a steady torque. A few generators are still drivenby steam engines or by internal combustion engines. These, as wellas synchronous motors driving compressors, have a tendency to hunt,but, on the whole, hunting is no longer a serious difficulty.

In the first ten or twenty years of this century, stability was not yet asignificant problem. Before automatic voltage-controlling devices(generator-voltage regulators, induction feeder-voltage regulators,synchronous condensers, and the like) had been developed, the power


systems had to be designed to have good inherent voltage regulation.This requirement called for low reactance in circuits and machines.As a consequence of the low reactances, the stability limits (bothsteady-state and transient) were well above the normally transmittedpower.

The development of automatic voltage regulators made it possibleto increase generator reactances in order to obtain a more economicaldesign and to limit short-circuit currents. By use of induction regulators to control feeder voltages, transmission lines of higher impedancebecame practicable. These factors, together with the increased use ofgenerator and bus reactors to decrease short-circuit currents, led to adecrease in the inherent stability of metropolitan power systems.

Stability first became an important problem, however, in connectionwith long-distance transmission, which is usually associated withremote hydroelectric stations feeding into metropolitan load centers.~The application of the automatic generator-voltage regulator to synchronous condensers made it possible to get good local voltage regulation from a hydroelectric station and a transmission line of high reactance-and hence of low synchronizing power. The high investmentin these long-distance projects made it desirable to transmit as muchpower as possible over a given line, and there was a temptation totransmit normal power approaching the steady-state stability limit.In a few cases instability occurred during steady-state operation, andmore frequently it occurred because of short circuits. The stabilityproblem is still more acute in connection with long-distance transmission from a generating station to a load center than it is in connectionwith metropolitan systems. It should not be inferred, however, thatmetropolitan systems have no stability problems.

Another type of long-distance transmission which has frequentlyinvolved a stability problem is the interconnection between two largepower systems for the purpose of exchanging power to obtain economiesin generation or to provide reserve capacity. In many cases the connecting ties were designed to transmit an amount of power 'which wassmall in comparison with the generating capacity of either system.Consequently, the synchronizing power which the line could transmitwas not enough to retain stability if a severe fault occurred on eithersystem. There was also considerable danger of steady-state pull-outif the power on the tie line was not controlled carefully.

From about 1920 the problem of power-system stability was theobject of thorough investigation. Tests were made both on laboratory

,Among such hydroelectric stations are Big Creek, Bucks Creek, Pit River,Fifteen Mile Falls, Conowingo, and Boulder Dam.

HISTORICAL REVIEW 13

set-ups and on actual power systems, methods of analyses were developed and checked by tests, and measures for improving stabilitywere developed. Some of the important steps in analytical development were the following:

1. Circle diagrams for showing the steady-state performance oftransmission systems. These diagrams consist of a family of circles,each of which is the locus of the vector power for fixed voltages atboth sending and receiving ends of the line. The circles are drawnon rectangular coordinates, the abscissas and ordinates of which are,respectively, active and reactive power at either end of the line.These diagrams show clearly the maximum power which a line willcarry in the steady state for given terminal voltages, as well as therelation between the power transmitted and the angular displacement between the voltages at the t\VO ends of the line. (Suchdiagrams are described in Chapter XV, Vol. III.)

2. Improvements in synchronous-machine theory, especially theextension of two-reaction theory to the transient performance ofboth salient-pole and nonsalient-pole machines. A number of newreactances were defined and used. (See Chapter XII, Vol. III.)More recently, the effect of saturation on these reactances has beeninvestigated. (See Chapters XII and XV, Vol. III.)

3. The method of symmetrical components for calculating theeffect of unsymmetrical short circuits. (See Chapter VI.) In thisconnection, methods of determining the sequence constants of apparatus by test and by calculation had to be devised.

4. Point-by-point methods of solving differential equations, particularly the swing equation (giving angular position of a machineversus time). (See Chapter II.)

5. The equal-area criterion for stability of two-machine systems,obviating the more laborious calculation of swing curves for suchsystems. (See Chapter IV.)

6. The a-c. calculating board or network analyzer for the solutionof complicated a-c. networks. (See Chapter III.)

The methods of analysis and calculation now in use are believed to besufficiently accurate for determining whether any given power systemin a given operating condition will be stable when subjected to a givendisturbance. The calculations, however, are rather laborious ,vhenapplied to a large number of different operating conditions of a complicated power system.

Methods of analysis will be taken up in the following chapters.Calculated results have been checked in a number of instances by


observations on actual power systems recorded with automatic oscillograph equipment.

REFERENCES

1. R. D. BOOTH and G. C. DAHL, "Power System Stability-a Non-mathematical Review," Gen. Elec. Rev., vol. 33, pp. 677-81, December, 1930; and vol. 34,pp. 131-41, February, 1931.

2. A.I.E.E. Subcommittee on Interconnection and Stability Factors, "FirstReport of Power-System Stability," Elec. Eng., vol. 56, pp. 261-82, February, 1937;

3. O. G. C. DAHL, Electric Power Circuits, vol. II, Power-System Stability, NewYork, McGraw-Hill Book Co., 1938.

4. Electrical Transmission and Distribution Reference Book, by Central StauonEngineers of the Westinghouse Electric & Manufacturing Company, East Pittsburgh, Pa., 1st edition, 1942.

a. Chapter 8, "Power System Stability-Basic Elements of Theory and Application," by R. D. EVANS.

b. Chapter 9, "System Stability-Examples of Calculation," by H. N.MULLER, JR.5. S. B. GRISCOM, "A Mechanical Analogy of the Problem of Transmission

Stability," Elec. Jour., vol. 23, pp. 230-5, l\1ay, 1926.6. R. C. BERGVALL and P. H. ROBINSON, "Quantitative Mechanical Analysis of

Power System Transient Disturbances," A.I.E.E. Trans., vol. 47, pp. 915-25,July, 1928; disc., pp. 925-8. Use of mechanical model with seven arms for investigating transient stability of Conowingo transmission system.

PROBLEMS ON CHAPTER I

1. Two synchronous machines of equal rating, having internal voltages(voltages behind transient reactance) of 1.2 and 1.0 per unit, respectively,and transient reactances of 0.25 per unit each, are connected by a line having0.50 per unit reactance and negligible resistance. Assume that the angle~ between the two machines varies from 0 to 3600 by 15° steps, and calculatefor each step the current, the power, and the voltage at each of three points:at each end of the line and at its midpoint. Draw loci of the current andvoltage vectors, marking the values of 0 thereon. Also plot in rectangularcoordinates current, power, and voltage, all as functions of o.

2. Draw the power-angle curve and discuss the condition for stability oftwo machines connected through series capacitive reactance which exceedsthe internal inductive reactance of both machines.

CHAPTER II

THE SWING EQUATION AND ITS SOLUTION

Review of the laws of mechanics; translation. Since a synchronousmachine is a rotating body, the laws of mechanics applying to rotatingbodies apply to it. Review of these laws may be advantageous at thispoint. The laws of rotating bodies will be clearer if we first review thelaws which apply to linear motion, or translation.

TABLE 1

FUNDAMENTAL AND DERIVED QUANTITIES OF MECHANICS, ApPLYING

PARTICULARLY TO 1"'RAN~LATION

Quantity Symbol Defining EquationUnit and

DimensionsIts Abbreviation

Length x ... meter (m.) LMass m ... kilogram (kg.) MTime t ... second (sec.) T

dx[1] ur:'Velocity v v =- meter per second (m.

dt per sec.)dv

Acceleration a a =- [2] meter per second per Lr-2dt second (m. per sec.")

Force F F = ma [3] *newton (newt.) MLT-2Momentum M' M' = mv [4] *newton-second (newt- MLT-l

sec.)

Work W W =fFdx [5] joule (j.) ML2T- 2

Power p. P = dW [6J watt (w.) ML 2r- 3

dt

* Unofficial name.

The fundamental quantities of mechanics are length, mass, and time.The fundamental units (in the m.k,s. system, which is now the recognizedsystem of units for electrical work) are the meter, the kilogram, and thesecond. From these fundamental quantities and their units are derivedother quantities,' such as velocity, acceleration, force, momentum,work, and power, and their units. In Table 1 are listed the fundamental quantities and certain derived quantities with their symbols,defining equations, units and the abbreviations thereof, and dimen-

15

16 THE SWING EQUATION AND ITS SOLUTION

sions in terms of the fundamental quantities length (L), mass (M),and time (T).

Besides the defining equations (numbers 1 to 6 of Table 1), certainother equations giving relations between these quantities which are ofinterest are derived below. Substitution of eq. 1 into eq. 2 gives forthe acceleration

[7]

[8]

Substitution of eq. 2 and eq. 7 in turn into eq. 3 gives

dv d2x

F=m-=m-dt dt2

Comparison of eq. 8 with the time derivative of eq. 4 gives the additional relation

dM'F=

dt

Differentiation of eq. 5 with respect to x gives

F= dWdx

Substitution for dW from eq. 10 into eq. 6 gives

Fdxp =.- = Fv

dt

Integration of eq. 9 with respect to t gives

M' = JFdl

From eqs. 11, 3, and 4 we obtain

P = Fv = mav = aM'whence

Af' =~a

[9]

[10]

[11]

[12].

[13]

[14]

The kinetic energy of a moving body may be obtained by finding thework required to set it in motion from rest, as follows:

W = JFdx = mJdV dx = mJdV vdtd~ d~

= mJ: v dv = !mv2 = !M'v [15]

ROTATION 17

Rotation. In treating rotation we must first introduce the concept ofangle. An angle is defined, with reference to a circular arc with itscenter at the vertex of the angle, as the ratio of arc 8 to radius r, thus:

8(}=

r[16]

The unit of angle so defined is the radian. The dimensions of anglefrom the definition are length divided by length. For the presentpurpose, however, it is well to recognize lengths that are perpendicularto one another as having different dimensions, and to represent tangentiallengths by the symbol L as before and radial lengths by themodified symbol LB. From this viewpoint the dimensions of angleare LLR- 1•

The definitions of angular velocity and angular acceleration follow byanalogy to the corresponding definitions of linear velocity and linearacceleration. Angular velocity is

and angular acceleration is

dBe,.,=-

d~

dw d28

a=-=-dt dt2

(17]

[18]

The relations between angular displacement, velocity, and acceleration and the corresponding tangential components of linear displacement, velocity, and acceleration, respectively, of a particle of a rotatingbody at distance r from the axis of rotation are given by:

8 = rO

v = rw

a = ra

[19}

[20]

[21]

The torque on a body due to a tangential force F at a distance r fromthe axis of rotation is

T = rF [22]

or, considering the total torque as due to the summation of infinitesimalforces, we may write

T = frdF [23]

The unit of torque could be called a newton-meter, but this name is notentirely satisfactory as it might imply a unit of work. Both work and


torque are the products of force and distance; but in the case of workthe component of force parallel to the distance is used, and in the caseof torque the component of force perpendicular to the distance is used.The two quantities may be distinguished by their dimensional formulas,which are as follows: work, M L2T-2

; torque, MLLR T-2• For reasons

that will appear later, the unit of torque may be called the joule perradian.

When torque is applied to a body, the body experiences angular acceleration a~ Each particle experiences a tangential accelerationa = ra, where r is the distance of the particle from the axis of rotation.If the mass of the particle is dm, the tangential force required to accelerate it is

dF = a dm = r a dm

Since this force acts with lever arm r, the torque required for the particleis

dT = rdF = ~adm

and that required for the whole body is

T = aJr2 dm = [a

Here

[= !TJdm

[24]

[25]

is known as the moment of inertia of the body, The m.k.s. unit" is thekilogram-meter2 (kg-m.").

Note the analogy between T = I« for rotation and F = rna fortranslation.

The work done in rotating a body through an angle dO by exerting atorque T may be found as follows: If the torque is assumed to be theresult of a number of tangential forces F acting at different points ofthe body,

T = ErF

Each force acts through a distance

ds = rdOThe work done is

dW = L F ds = E F r dO = dO E F r = dO· T

W = f TdO [26]

ROTATION

This is analogous to eq. 5 for translation. Also

T= dWdO

19

[27]

which is analogous to eq, 10, and which explains why the m.k.s. unit oftorque may be called the joule per radian.

Substitution for dW from eq. 27 into eq. 6 (Table 1) gives an expression for power in rotary motion,

dOP = T- = Tw

dt[28]

which is analogous to eq. 11 for translation.By analogy with the definition of momentum, M' = mv, angular

momentum may be defined asM = Iw [29]

and by derivations analogous to those of eqs. 12 and 14, we obtain also

M =JT d~ = ~ [30]

The m.k.s. unit of angular momentum may be variously called (fromeq. 29) kilogram-meters2-radians per second, or (from eq. 30) watts per

TABLE 2

QUANTITIES OF MECHANICS ApPLYING TO ROTATION

Quantity SymbolDefining

Unit and Abbreviation DimensionsEquation

Angle8

radian (rad.) LLR-t8 0=-r

AngulardB

radian per second (rad. LLR-1r-1t» w =-velocity dt per sec.)

Angulardw

radian per second per second LLn- 1rr-2a Ol =-acceleration dt (rad. per sec,")

Torque T T = rF joule per radian (j. per rad.) MLLRT- 2or newton-meter (newt-m.)

Moment of I I = fr 2 dm kilogram-meter- (kg-m.P) MLR2

inertia

Angular M M = Iw joule-second per radian (j- MLLRT-lmomentum sec. per rad.)

(radian per second per second), or joule-seconds per radian. The lastname seems best from the standpoint of brevity.


The kinetic energy of a rotating body may be written

W = !Iw2 = !Mw [31]

which is analogous to eq. 15.Table 2 summarizes the quantities of mechanics applying to rota

tion, with their symbols, defining equations, units, and dimensions.Table 3 indicates some analogies and relations between the quantities

and laws of rotation and those of translation.

TABLE 3

ANALOGIES AND RELATIONS BETWEEN THE QUANTITIES

AND LAWS OF TRANSLATION AND OF ROTATION

Translation Rotation Relation

8 8 8 = r8f} w V = rwa Of. a = ra

m 1 1 = Jr2 dmF T T = rFM' M M = JrdM'

F = ma T = I«

W = JFd8 W=fTdJJP = F» = M'a P = Tw = MOt.

M' = m» M =IwW = !mv2 W = !Iw2

=tM'v =!MwdM' T =dMF=-dt dt

The swing equation. The laws of rotation, as developed in theforegoing section, apply to the motion of a synchronous machine.Equation 24 states that the torque is equal to the product of angularacceleration and moment of inertia:

or101. = T [32]

[33]

Here T is the net torque or algebraic sum of all the torques acting onthe machine, including shaft torque (due to the prime mover of agenerator or to the load on a motor), torque due to rotational losses(friction, windage, and core loss), and electromagnetic torque. Elec-

THE SWING EQUATION 21

tromagnetic torque may he subdivided into torques due to synchronousand to asynchronous (induction) action.

Let Ti = shaft torque, corrected for torque due to rotationallosses

and let Tu = electromagnetic torque.

Both of these are taken as positive for generator action, that is, withmechanical input and electrical output. They are negative for motoraction, that is, with mechanical output and electrical input. Thenet torque, which produces acceleration,' is the algebraic difference ofthe accelerating shaft torque and the retarding electromagnetic torque:

[34]

In the steady state this difference is zero, and there is no acceleration.During disturbances of the kinds considered in transient-stabilitystudies, however, the difference exists, and there is acceleration orretardation, depending on whether the net torque Ta is positive ornegative.

Our problem is to solve eq. 33 so as to find the angular position fJ ofthe machine rotor as a function of time t. It is more convenient, however, to measure the angular position and angular velocity with respectto a synchronously rotating reference axis than with respect to astationary axis. Hence let

a = (J - Wtt [35]

where WI is the rated normal synchronous speed. Then, if we taketime derivatives, we get

da dO- = - - WI [36]dt dt

andd2a d2fJ

dt2 = dt2

With this substitution eq. 33 becomes

/d2a= T

dt2

[37]

[38]

which is unchanged in form. Writing the torque as in eq. 34, we have

d2aI dt2 = Ta = Ti - Tu [38a]


If we multiply this equation by the speed w, we obtain

d25M dt2 = Pa = Pi - Pu [39]

where M = Iw is the angular momentum.Pi = Tiw is the shaft power input, corrected for rotational

losses.Pu = TuW is the electrical power output, corrected for electrical

losses.Pa = Pi - Pu, is the accelerating power, or difference between

input and output, each corrected for losses.

Equation 39 is more convenient to use than eq. 38abecause it involvesthe electrical power output of the machine, rather than the torquecorresponding to this output. Equation 39 will be referred to hereafter as the swing equation. An equation of this form may be writtenfor each machine of the system.

The angular momentum M is not strictly constant because thespeed w varies somewhat during the swings which follow a disturbance.In practical cases, however, the change in speed w before synchronismis lost is so small in comparison to the normal speed Wt that very littleerror is introduced by the assumption that M is constant. Hence it iscustomary in solving the swing equation to regard M as constant andequal to 1Wt, the value of angular momentum at normal speed. Thisvalue of M is known as the inertia constant of the machine.

The inertia constant. In the swing equation (eq. 39) various consistent sets of units may be used. In the m.k.s. system Pa will be inwatts, 0 in radians, t in seconds, and M in watts per (radian per second per second) or joule-seconds per radian. In practical stabilitystudies Pa usually will be expressed either in megawatts or in per unit, *oin electrical degrees, and t in seconds. Hence, if Pais in megawatts,M must be in megawatts per (electrical degree per second per second),or megajoule-seconds per electrical degree (abbreviated Mi-sec, perelec. deg.). If Pa is in per unit, M must be in unit power secondssquared per electrical degree. For brevity the latter value of M willbe called a per-unit value.

Sometimes the available information regarding the angular momentum of a machine takes the form of the value of its stored kinetic energyat rated speed. More often, however, the designer or manufacturergives the values of the moment of inertia of the machine expressed .in

·Per-unit power is power expressed as a decimal fraction of an arbitrarily chosenbase power. See Chapter III for further discussion of per-unit quantities.

THE INERTIA CONSTANT 23

pound-feet/ and the speed in revolutions per minute. In either case,before one can proceed to a solution of the swing equation, he mustcalculate the value of the inertia constant M from the data. Theformulas needed for this purpose will now be derived, beginning withthe one giving the kinetic energy in terms of the moment of inertia andspeed, and proceeding to various formulas for M.

Let WR 2 = moment of inertia in pound-feet'',I = moment of inertia in slug-feet/,n = speed in revolutions per minute.w = speed in radians per second.

We = speed in electrical degrees per second.W = kinetic energy in foot-pounds.N = kinetic energy in megajoules,M = inertia constant in megajoule-seconds per electrical de-

gree.J = frequency in cycles per second.p = number of poles.G = rating of machine in megavolt-amperes.

Byeq. 31, using English units, the kinetic energy is

W = !Iw2

But

21rnw=-

60and

[40]

[41]

[42]

[43]

By substituting eqs. 41 and 42 into eq. 40 and the result into eq. 43, weobtain

746 -6 1 W R2 (211"n)2N = 550 X 10 X "2 X 32.2 X 60

= 2.31 X 10-10WR2n2

By eq. 31 the kinetic energy may be written also as

[44]

[45]

24 THE SWING EQUATION. AND ITS SOLUTION

Solving for M,

But

Hence

M=2NWe

We = 360/

[46]

[47]

[49]

[48]M= 2N =~360/ 180/

By substitution of eq. 44 into eq. 48 we obtain

WR2n2

M = 1.28 X 10-12 -1-'-

Since the relation among speed, frequency, and number of poles is

np = 120/ [50]

eq. 49 may be written also as

M .:::: 1.54 X 10-10 WR2n

por as

M = 1.84 X 10-8 W~2jP

[51]

[52]

[53]

The inertia constant in megajoule-seconds per electrical degree maybe calculated from eqs. 48, 49, 51, or 52. If per-unit power is to beused instead of megawatts, the value of M so obtained is merelydivided by the base power in megavolt-amperes, giving what we maycall a per-unit value of M.

Another constant which has proved very useful is denoted by Handis equal to the kinetic energy at rated speed divided by the ratedapparent power of the machine.

H = stored energy in joulesrating in volt-amperes

stored energy in kilojoules

rating in kilovolt-amperes

stored energy in megajoules, Nrating in megavolt-amperes, G

In terms of H the inertia constant is, by eq. 48,

GHM = 180j [54]

THE INERTIA CONSTANT 25

...~~

--.... ........ lY 1800 r.am, condensing......., .......

~~ .....

~r---.~~

"- ...... ....... .......-"'-

.......,~, y-3,600 r.p.m.'condensing~~

...... ...... ./ 3,600 r.p.m. noncondensin--.~ f-.+.J.:..[

T

The quantity H has the desirable property that its value, unlikethose of M or WR2, does not vary greatly with the rated kilovoltamperes and speed of the machine, but instead has a characteristic

10

1002o 20 40 60 80

Generator ratingG (megavolt.amperes)

FIG. 1. Stored energy of large steam turbogenerators, turbine included (fromRef. 1, by permission).

5

450·514 r.p.m.-~ "~200.400 r.arn..... ~ ..... ......,,-~

~ ....... i\. ~",

~

~ l..)....... "",.",,.....~.......

lI' ~,.~ ......~ --~~ ,....-,/ ~!;I'",~ ...- --- ',", 138· 180r.p.m.~

j~~"7 -e~......

,.~~~ ~80·120 r.p.m,~

1o 20 40 60 80 100

Generator rating G (megavolt- amperes)

FIG. 2. Stored energy of large vertical-type water-wheel generators, includingallowance of 15% for water wheels (from Ref. 1, by permission).

value or set of values for each class of machine. In this respect H issimilar to the per-unit reactance of machines. In the absence of moredefinite information, a characteristic value of H may be used. Suchvalues are given in the curves of Figs. 1 and 2 for large steam turbo-


Type of Machine

generators and for large vertical-shaft water-wheel generators, respectively. In both cases the inertia of the prime movers is included, asit always should be. For' other machines the values of H may betaken from Table 4.

It will be observed that the value of H is considerably higher forsteam turbogenerators than for water-wheel generators, ranging from3 to 10 Mj. per Mva, for the former and from 2 to 4 Mj. per Mva. forthe latter. Average values are about 6 and 3, respectively.

TABLE 4

AVERAGE VALUES OF STORED ENERGY IN ROTATING MACHINES·

H, Stored Energy at Rated Speed(megajoules per megavolt-ampere)

Synchronous motorsSynchronous condenserst

LargeSmall

Rotary convertersInduction motors

• Principally from Ref. 1.t Hydrogen cooled, 25% less.

2.0

1.251.02.00.5

From 30 to 60% of the total inertia of a steam turbogenerator unit isthat of the prime mover, whereas only 4 to 15% of the inertia of ahydroelectric generating unit is that of the water wheel, includingwater. t

Inertia constant H, like the per-unit reactance of machines or transformers, may be expressed on either of two volt-ampere bases: (a) themachine rating or (b) a system base arbitrarily selected for a powersystem study. The value of H for a given machine varies inversely asthe base, whereas per-unit reactance varies directly as the base.tEXAMPLE 1

Given the following information on a steam turbogenerator unit:

Rated outputRated voltageRated speedMoment of inertiaNumber of polesRated frequency

85,000 kw. at 85% power factor13,200 volts1,800 r.p.m,859,000 Ib-ft.2

460 cycles per sec.

tSome data on moments of inertia of generators and their prime movers aregiven in Table 1, Chapter VII.

iPer-unit quantities are discussed in Chapter III.

SOLUTION OF THE SWING EQUATION 27

compute the following quantities:

Kinetic energy in megajoules at rated speed.Inertia constant H.Inertia constant M in megajoule-seconds per electrical degree.M in per unit on 50-Mva. base.

Compare the computed value of H with the typical value read from thecurves of Fig. 1.

Solution. Rating = 85,000 = 100,000 kva. = 100 Mva.0.85

From eq. 44 kinetic energy at rated speed is

N = 2.31 X 10-10 WR2n2

= 2.31 X 10-10 X 859,000 X (1,800)2

= 642 Mj.From eq. 53

H = 642 = 6.42 Mj. per Mva.100

From eq. 48N 642 .

M = - = = 0.0595 Mj-sec. per elec. deg.180/ 180 X 60

If power is to be expressed in per unit instead of in megawatts, this resultmust be divided by the base power. Thus

M = 0.0595 = 0.00119 'per unit on 50-Mva. base50

From Fig. 1, 1800-r.p.m. curve, at 100 Mva., H = 6.5, which agrees wellwith the value of H computed above.

Point-by-point solution of the swing equation. The swing equation,a differential equation governing the motion of each machine of asystem, is

[55]

where 8 = displacement angle of rotor with respect to a reference axisrotating at normal speed.

M = inertia constant of machine.Pa = accelerating power, or difference between mechanical input

and electrical output after each has been corrected forlosses.

t = time.


The solution of this equation gives 0 as a function of t. A graph of thesolution is known as a swing curve. Inspection of the swing curves ofall the machines of a system will show whether the machines will remain in synchronism after a' disturbance. Many examples of swingcurves obtained in stability studies on multimachine systems are included in Chapter VII. Both stable and unstable conditions are illustrated there.

In a multirnachine system, the output and hence the acceleratingpower of each machine depend upon the angular positions-and, to bemore rigorous, also upon the angular speeds-of all the machines of thesystem. Thus, for a three-machine system there are three simultaneous differential equations like eq. 55:

[56a]

[56b]

(56c]

Formal solution of such a set of equations is not feasible. Even thesimplest case, which was considered in Chapter I, of one finite machineconnected through reactance to an infinite bus, with damping neglected,leads to an equation

[57]

the formal solution of which, with Pi = 0, involves elliptic integrals."Equation 57, with Pi ¢ 0, has been solved by use of a calculating

machine called an integraph or differential analyzer,3 and it is possiblethat machine methods of solution could be applied (although they havenot been yet) to solving the swing equations of multimachine systems.

Point-by-point solution is the most feasible and widely used way ofsolving the swing equations. Such solutions, which are also calledstep-by-step solutions, are applicable to the numerical solution of allsorts of differential equations. Good accuracy can be attained, andthe computations are simple.4- 7

In a point-by-point solution one or more of the variables are assumedeither to be constant or to vary according to assumed laws throughouta short interval of time ~t, so that as a result of the assumptions made


the equations can be solved for the changes in the other variablesduring the same time interval. Then, from the values of the othervariables at the end of the interval, new values can be calculated forthe variables which were assumed constant. These new values arethen used in the next time interval.

In applying the point-by-point method to the solution of swingequations, it is customary to assume that the accelerating power (andhence the acceleration) is constant during each time interval, althoughit has different values in different intervals. When this assumptionis made, a formal solution of eq. 55 can be obtained which is validthroughout a particular time interval; and from the formal solutionand the values of ~ and w at the beginning of the interval the values ofaand w at the end of the interval can be computed for each machine.Before a similar computation can be made for the next interval, however, it is necessary to know the new value of accelerating power, ordifference between input and output, of each machine. The mechanical inputs are usually assumed constant, because of the slowness ofgovernor action,§ but the electrical outputs are functions of the relative angular positions of all the machines of the system and can befound by solving the network to which the machines are connected.If damping power is taken into account, the output, including damping, will depend also on the relative angular speeds of all the machines.

It therefore becomes clear that the point-by-point solution of swingcurves consists of two processes which are carried out alternately.The first process is the computation of the angular positions, and perhaps also of the angular speeds, at the end of a time interval from aknowledge of the positions and speeds at the beginning of the intervaland the accelerating power assumed for theinterval, The secondprocess is the computation of the accelerating power of each machinefrom the angular positions (and perhaps speeds) of all machines of thesystem. The second process requires a knowledgeof network solution,a topic which is discussed in Chapter III. In the present chapter theemphasis is on the first process, namely, the solution of the swingequation proper. Two different point-by-point methods will bedescribed.

Method 1 is the more obvious, although the less accurate, of the twomethods. In method 1 it is assumed that the accelerating power isconstant throughout a time interval lit and has the value computed forthe beginning of the interval. No further assumptions are made. Ifeq. 55 is divided by M and integrated twice with respect to t, P a being

§See discussion of Assumption 1 on p. 43.


treated as a constant, we obtain successively

do ~at- =w=wo+-dt M

and

[58]

[59]

These equations give, respectively, w, the excessof speed of the machineover normal speed, and 0, the angular displacement of the machinewith respect to a reference axis rotating at normal speed. 00 and Woare the values of 0 and w, respectively, at the beginning of the interval.These equations hold for any instant of time t during the interval inwhich Pa is constant. We are particularly interested, however, in thevalues of 0 and w at the end of the interval. Let subscript n denotequantities at the end of the nth interval. Likewise let n - 1 denotequantities at the end of the (n - l)th interval, which is the beginningof the nth interval. At is the length of the interval. Putting At inplace of t in eqs. 58 and 59 and using the appropriate subscripts, weobtain for the speed and angle at the end of the nth interval

AtWn = Wn-l + M Pa(n-l) [60]

(At)2On = On-l + At Wn-l + 2M Pa(n-l) [61]

The increments of speed and angle during the nth interval are

AtaWn = Wn - Wn-l = M Pa(n-l) [62]

(At)2aOn == On - On-l = at Wn-l + 2M Pa(n-l) [63]

Equations 60 and 61, or eqs. 62 and 63, are suitable for point-by-pointcalculation. However, if one is interested only in the angular position(for plotting a swing curve) and not in the speed, Wn-l may be eliminated from eqs. 61 and 63, as follows: Write an equation like eq. 61but for the preceding interval

. (~t)2On-l = On-2 + ~t Wn- 2 + 2M Pa(n-2) [64]

and subtract it from eq. 61, obtaining

(on - On-I) = (On-l - On-2) + At(Wn-l - Wn-2)(At)2

+ 2M (Pa(n-l) - P a(n-2») [65J

SOLUTION OF THE SWING EQUATION

But

31

and from eq. 62

Making these substitutions into eq. 65, we get

dt (~t)2a6n = a61l-l + at M Pa(1l-2) + 2M (Pa(1l-l) - Pa(1l-2»)

(dt)2== a8n_ l + 2M (Pa(n-l) + p a(n-2» [66]

This equation, which gives the increment in angle during any intervalin terms of the increment for the previous interval, may be used forpoint-by-point calculations in place of eqs. 62 and 63. The last termis the second difference of 5, which may be symbolized by t!A25.

The time interval ~t should be short enough to give the requiredaccuracy, but not so short as to unduly increase the number of pointsto be computed on a given swing curve. Example 2 will throw somelight on the effect of the length of interval upon the accuracy of thesolution.

EXAMPLE 2If a synchronous machine performs oscillations of small amplitude with

respect to an infinite bus, its power output may be assumed to be directlyproportional to its angular displacement from the infinite bus. Because thiscase is known to result in sinusoidal oscillations, as may be verified readily byformal solution of the swing equation, it will serve well as a check on theaccuracy of various point-by-point solu tions,

Consider a 60-cycle machine for which H = 2.7 Mj. per Mva. and which isinitially operating in the steady "tate with input and output of 1.00 unit andan angular displacement of 45 elec. deg. with respect to an infinite bus.Upon occurrence of a fault, assume that the input remains constant and thatthe output is given by

(a)

even though the amplitude of oscillation may be great. Calculate one cycleof the swing curve by means of (a) the formal solution and (b) a point-by-


point solution, method 1, using various values of time interval ~t, 0.05 sec.being tried first.

Solution. (a) Formal solution. The differential equation, with 0 expressedin electrical radians, is

d~ 2M dt2 = P; = Pi - P« = 1 - ;;: 0 (b)

and the initial conditions are

o= ~4

when t = O.The solution is

and ~=odt

(c)

~-

7r 7r 20=2-"4cOS 7rM t

TABLE 5

COMPUTATION OF Swrso CURVE FROM FORMAL SOLUTION

OF SWING EQUATION (EXAMPLE 2)

(d)

t (sec.) 382t (deg.) cos 382t45 cos 382t

8 (deg.)(deg.)

0 0.0 1.000 45.0 45.00.05 19.1 0.945 42.5 47.50.10 38.2 0.786 35.4 54.60.15 57.3 0.540 24.3 65.70.20 76.4 0.235 10.6 79.40.25 95.5 -0.096 - 4.3 94.30.30 114.6 -0.416 -18.7 108.70.35 133.7 -0.691 -31.1 121.10.40 152.8 -0.889 -40.0 130.00.45 171.9 -0.990 -44.5 134.50.50 191.0 -0.982 -44.2 134.20.55 210.1 -0.865 -38.9 128.90.60 229.2 -0.653 -29.4 119.40.65 248.3 -0.370 -16.7 106.70.70 267.4 -0.045 - 2.0 92.00.75 286.5 0.284 12.8 77.20.80 305.6 0.582 26.2 63.80.85 324.7 0.816 36.7 53.30.90 343.8 0.960 43.2 46.80.95 362.9 0.999 45.0 45.01.00 382.0 0.927 41.7 48.3

which may be verified by differentiating eq. d twice with respect to t andthen substituting the result and eq. d itself into eq. b; also by substitutingl = 0 into eq. d and its first derivative and comparing the results with eqs. c.

SOLUTION OF THE SWING EQUATION

The period of oscillation is given by

T = 21r = 1r V21rMV2/1rM

The inertia constant is

33

(e)

H 2.7 · 2 1 dM = - = ---= 0.01432 unit power sec. per e ec. ra •7rf 1r X 60

= -.!!- = 2.5 X 10-4 unit power sec."per elec. deg. (or, simply, per unit).180!

If the first value of M is used in eq. e, the period is found to be

T = 1r V21r X 0.01432 = 0.943 sec.

If 5 is expressed in electrical degrees instead of in radians, the solutionbecomes

5 = 90° - 45° cos (382t)0 (I)

The amplitude of oscillation is 45°.Values of 0 at 0.05-sec. intervals of t are computed from eq. f in Table 5.(b) Point-by-point solution, method 1 (ilt = 0.05 sec.) Substitution of

the values of ilt and Minto eqs ..62 and 63 gives

ilt 0.05~Wn = M Pa(n-l) = --- Pa(n-l) = 200 Pa(n-l) (g)

0.00025

(ilt)2~On = ilt Wn-l + 2M P a(n-l) = O.05wn - 1 + 5P a (n.-l) (h)

Computation of the swing curve from eqs. g and h is carried out in Table 6.The swing curve is plotted as curve 3 in Fig. 3, where it is compared with thecorrect curve (curve 1) calculated from the formal solution, part a of thisexample.

Comparison of curves 1 and 3 of Fig. 3 shows that curve 3, computed bypoint-by-point method 1, although having very nearly the correct period,increases in amplitude approximately 29% in each half-cycle, or 67% ineach cycle of oscillation. II The accuracy is poor.

It seems reasonable that the accuracy would be improved by the use of ashorter interval. Consequently, let us try ilt = 0.0167 sec. (one-third of theformer value). Then in place of eqs. g and h we have

111.67 = (1.29)2.

dWn = 66.7Pa(n - l )

L18n = O.OI67wn - l + 0.555Pa(n-l)

(i)

(j)


TABLE 6

POINT-By-POINT COMPUTATION OF SWING CURVE

(METHOD 1, tlt = 0.05 sEc.)(ExAMPLE 2)

t Pu Pa liw w 0.05w 5Pa tl8 8(sec. ) (p.u.) (p.u.) (deg.Zsec.) (deg.Zsec.) (deg.) (deg.) (deg.) (deg.)-- ------

0+ 0.500 0.500 100 0 0.0 2.5 2.5 45.00.05 0.528 0.472 94 100 5.0 2.4 7.4 47.50.10 0.610 0.390 78 194 9.7 2.0 11.7 54.90.15 0.740 0.260 52 272 13.6 1.3 14.9 66.60.20 0.906 0.094 19 324 16.2 0.5 16.7 81.50.25 1.091 -0.091 -18 343 17.2 -0.5 16.7 98.20.30 1.277 -0.277 -55 325 16.2 -1.4 14.8 114.90.35 1.441 -0.441 -88 270 13.5 -2.2 11.3 129.70.40 1.567 -0.567 -113 182 9.1 -2.8 6.3 141.00.45 1.637 -0.637 -127 69 3.4 -3.2 0.2 147.30.50 1.639 -0.639 -128 -58 -2.9 -3.2 -6.1 147.50.55 1.571 -0.571 -114 -186 -9.3 -2.9 -12.2 141.40.60 1.436 -0.436 -87 -300 -15.0 -2.2 -17.2 129.20.65 1.244 -0.244 -49 -387 -19.4 -1.2 -20.6 112.00.70 1.016 -0.016 -3 -436 -21.8 -0.1 -21.9 91.40.75 0.772 0.228 46 -439 -22.0 1.1 -20.9 69.50.80 0.540 0.460 92 -393 -19.6 2.3 -17.3 48.60.85 0.348 0.652 130 -301 -15.0 3.3 -11.7 31.30.90 0.218 0.792 158 -171 -8.6 3.8 -4.8 19.60.95 0.164 0.836 167 -13 -0.6 4.2 3.6 14.81.00 .. . . .. . .. 154 . .. ... . .. 18.4

The detailed calculations are not shown here, but they are similar to those ofTable 6. The swing curve is plotted as curve 2 in Fig. 3. It is found thatthe amplitude error has been reduced to 9% in a half cycle or 21% in acycle-about one-third of the corresponding errors with dt = 0.05 sec.These errors are still pretty large, and the labor of calculation is excessivebecause so many points must be calculated .

.Two additional curves (4 and 5), calculated for larger values of dt(0.10 and 0.15 sec., respectively), have been plotted in Fig. 3. As might beexpected, the amplitudes increase faster than before. Besides, the periodis noticeably lengthened.

Example 2 has demonstrated that swing curves calculated by method1 are subject to a considerable cumulative error which manifests itselfby increase in both the amplitude and the period of oscillation, especially in the amplitude. The reason for the errors is not hard to find.Consider a time, such as the first half cycle of the harmonic oscillationof. Example 2, during which the acceleration is diminishing. Such acurve of acceleration versus time is shown in Fig. 4. In method 1 the


1.00.8 0.90.70.30.20.1

5

I~~,

)~

~r " .... , ]'00, [\

/ ,\.ill \

~ 3 \/,/V'1t\;~,')I.' .~ \ \

l(i;jVT~

\ -.-

l\\~ 'f N'I~ \~

b' "f\

i\j

~r NbI ~~~

--"If ,~~~

1 Formal solution 1~\"~2 •••• • • I1t=0.0167 sec.

\\3 0--0--0 /1t=0.05sec.4 6--------6 !:At =0.10sec. ~3 ))5 0-- - -0 41t = 0.15 sec.

" "\\\\\r\

\ 4\ ,

1 '''i~

ts

o

30

15

195

180

150

165

135

120

-45o

-30

-15

0.4 0.5 0.6Time t (seconds)

FIG. 3. Swing curves calculated from a formal solution of the swing equation andby point-by-point calculation, method 1, with various values of /1t (Example 2).

a


acceleration during each interval ~t is assumed constant at its value forthe beginning of the interval, as shown by the step function in Fig. 4,and during the time considered it is always too great. Consequently,the calculated speed becomes progressively higher than the true speed,

and the calculated advance ofangular position is likewisegreater than the true advance.The second half cycle of oscillation thus begins with acalculated amplitude greaterthan the true value; and, ifno further errors occurred, theoscillation would continuewith this amplitude. During

tFIG. 4. True and assumed curves of ac- the second half cycle, how-celeration versus time, point-by-point calou- ever, the acceleration is in-

lation, method 1. creasing, and during this timethe assumed acceleration is

always too small, that is, too negative. The calculated negative speedis therefore too great in absolute value, and the calculated retardationof angular position is likewise too great. Thus the calculated amplitude increases with each swing.

Method 2. Most of the error caused by assuming the acceleration tobe constant during a time interval can be eliminated by using the valueof acceleration at the middle instead of the beginning of the interval.Referring to Fig. 5a, we see that .the acceleration at the middle of theinterval is very nearly equal to the average acceleration during theinterval, as is shown by the near equality of the triangular areas aboveand below the true curve. In method 2 this assumption is made:namely, that the acceleration during an interval is constant at its valuecalculated for the middle of the interval. Or, if we consider that theintervals used in calculation begin and end at the points of time atwhich acceleration is calculated, then the assumption must be restated somewhat as follows: The acceleration, as calculated at thebeginning of a particular time interval, is assumed to remain constantfrom the middle of the preceding interval to the middle of the intervalbeing considered. Let us, for example, consider calculations for thenth interval, which begins at t = (n - l)~t. (See Fig. 5.) Theangular position at this instant is an-I. The acceleration an-I, ascalculated at this instant, is assumed to be constant from

t = (n - -!) Ilt to t = (n - !) Ilt


Over this period a change in speed occurs, which is calculated as

at~wn-l = ~t·an-l = M Pa{n.-l) [67]

giving the speed at the end of this time as

wn-l = Wn-l + ~wn-l [68]

As a logical outcome of the assumption regarding acceleration, thechange in speed would occur linearly with time. To simplify the

a

~ In-t:n-tU I I

I IOn --.----1------,-----Ao,,: :

8,,-1 __ .1. ~------ I

0,,-2 I

n-2 n-l n

(0)

(b)

(e)

.1.~t

FIG. 5. True and assumed curves of acceleration, speed, and angular positionversus time, point-by-point calculation, method 2.

ensuing calculations, however, the change in speed is assumed to occuras a step at the middle of the period, that is, at t = (n - l)~t, which isthe same instant for which the acceleration was calculated. Betweensteps the speed is assumed to be constant, as shown in Fig. 5b. Fromt = (n - l)L\t to t = nL\t, or throughout the nth interval, the speedwill be constant at the value wn-i. The change in angular positionduring the nth interval is, therefore,

L\on = ~t·wn.-l [69]

and the position at the end of the interval is

On = 011.-1 + ~on [70]as shown in Fig. 5c.


Equations 67 to 70 may be used for computation; but, if we areinterested only in the angular position and not in the speed, we mayuse a formula for ~on from which w has been eliminated. Such aformula will now be developed. Substitution of eq. 67 into eq. 68,and of the result into eq. 69, gives

(~t)2~on = ~t·wn_1 + M Pa(n-l) [71]

By analogy with eq. 69

Substituting eq. 72 into eq. 71, ,ve obtain the desired formula

(~t)2~on = ~On-l + M Pa(n-l)

[72]

[73]

which, like eq. 66 of method 1, gives the increment in angle during anytime interval in terms of the increment for the previous interval. Thelast term again may be symbolized by ~20. This formula makes thecalculation of a swing curve very simple. If the speed is wanted, itcan be obtained from the relation

[74]

Method 2 is simpler to use than method 1 and is much more accurate, as will be shown in Example 3.

Before proceeding with this example, it is necessary to give someattention to the effects of discontinuities in the accelerating power Pa

which occur, for example, ,vhen a fault is applied or removed or whenany switching operation takes place. If such a discontinuity occursat the beginning of an interval, then the average of the values of P (J

before and after the discontinuity must be used. Thus, in computingthe increment of angle occurring during the first interval after a fault isapplied at t = 0, eq. 73 becomes

[75]

where PaO+ is the accelerating power immediately after occurrence ofthe fault. Immediately before the fault the system is in the steadystate; hence the accelerating power, Pao-, and the previous incrementof angle, ~oo, are both equal to zero. If the fault is cleared atthe beginning of the mth interval, in calculations for this intervalone should use for Pa(m-l) the value !(Pa(m-l)- + Pa(m-l)+), where


Pa(m-l)- is the accelerating power immediately before clearing andPa(m-l)+ is that immediately after clearing the fault. If the discontinuity occurs at the middle of an interval, no special procedure isneeded. The increment in angle during such an interval is calculated,as usual, from the value of Pa at the beginning of the interval. Thereasons for these two rules should be clear after study of Fig. 6. Ifthe discontinuity occurs at some time other than the beginning or the

~(m-1)+

IIIIIIIIII

I Pa(m-l)-I

I II II II II II II II II ,

tm-l m n-l n ~

FIG. 6. The handling of discontinuities of accelerating power Pain point-by-pointcalculation, method 2. A is a discontinuity occurring at the beginning of the mthinterval. PaCm-l) = ! (PaCm-l>+ + PaCm-1)- ) is used in calculating ~8m. B isa discontinuity at the middle of the nth interval. Here PaCn-1) is used in

calculating Ll8n•

middle of an interval, a weighted average of the values of P a beforeand after the discontinuity should be used, but the need for such arefinement seldom appears because the time intervals used in calculation are so short that it is sufficiently accurate to assume the discontinuity to occur either at the beginning or at the middle of aninterval.

EXAMPLE 3Work the problem of Example 2 by method 2 of point-by-point calculation,

using various values of time interval (~t). Compare the swing curves thusobtained with those obtained by method 1 in Example 2.

Solution. The point-by-point calculations will be based on eqs. 70 and73. The value of M from Example 2 is 2.5 X 10-4 per unit. The aceelerat-


ing power is given, as before, by~

P = Pc> P = 1--a ,u 900

For t1t = 0.05 sec., which will be tried first, we have

(t1t)2 = (0.05 )2 = 10M 2.5 X 10-4

and eq. 73 becomes

(a)

(b)

The detailed calculations are carried out in Table 7. Note that, fort = 0, the instant of occurrence of the fault, the values of Pu and of Pa areentered both for the fault off (at t = 0-) and for the fault on (at t = 0+).The average value of Pais used.

The order in which items (except t) are entered in the table is: from left toright across a line until the column headed "lOPa" is reached; then diagonally downward and to the right to the"value of 0in the line below the startingpoint; thence to the "Pu" column in the same line; and so on. Each newvalue of flo is found by addition of the previous value of flo and the value oflOPa between the old and new values of dO. Each new value of 0 is found insimilar fashion by addition of the preceding value of aand the value of f16between the old and new values of o.

Comparison of the values of 0 in Table 7 with the corresponding values inTable 5, computed from the formal solution, shows a maximum discrepancyof 1.00 in the first cycle of oscillation. The agreement of the point-by-pointsolution with the formal solution is very satisfactory.

Similar calculations were made for at= 0.10 sec., 0.15 see., 0.20 sec., and0.25 sec. The details of calculation are not shown here, but the resultingswing curves are plotted in Fig. 7. The curves numbered 3, 4, and 5, andthe symbols used for them, correspond with the curves of Fig. 3 for likevalues of dt. The formal solution is not plotted in Fig. 7 but agrees withcurve 3 to the accuracy of plotting.

It may be noted in Fig. 7 that, as t1t is increased, the calculated swingcurves remain sinu~oidal and correct in amplitude, but that their periodsdecrease. Even when the points are so far apart that they are not adequateto determine the shape of the curve (as for ~t = 0.25 sec., giving fewer than4 points per cycle), a sine-wave can be drawn through them.

The curve for at = 0.1 sec. would be considered accurate enough forengineering use in a stability study. In actual studies the usual periods ofoscillation are from 0.5 to 2 sec., and at is commonly taken as 0.05 or 0.1sec. The value 0.1 sec. usually suffices and, in addition to requiring onlyhalf as many steps as 0.05 sec., has the advantage that multiplication ordivision by ~t can be performed merely by shifting the decimal point.

In Table 8 a comparison is made of the errors in the amplitudes and periodsof the curves calculated by methods 1 and 2 in Examples 2 and 3, respectively.

TABLE 7POINT-By·POINT COMPUTATION OP SWING CURVE

(METHOD 2, at = 0.05 sEc.)(ExAMPLE 3)

t (sec.) P« (p.u.) Po, (p.u.) 10Po, (deg.) f:.8 (deg.) 8 (deg.)

0- 1.000 0 ... · "...

0+ 0.500 0.500 ... ·.. ...oavg. ... 0.250 2.5 ·.. 45.0

2.50.05 0.528 0.472 4.7 47.5

7.20.10 0.609 0.391 3.9 54.7

11.10.15 0.731 0.269 2.7 65.8

13.80.20 0.885 0.115 1.2 79.6

15.00.25 1.052 -0.052 -0.5 94.6

14.50.30 1.214 -0.214 -2.1 109.1

12.40.35 1.350 -0.350 -3.5 121.5

8.90.40 1.450 -0.450 -4.5 130.4

4.40.45 1.497 -0.497 -5.0 134.8

-0.60.50 1.492 -0.492 -4.9 134.2

-5.50.55 1.430 -0.430 -4.3 128.7

-9.80.60 1.320 -0.320 -3.2 118.9

-13.00.65 1.177 -0.177 -1.8 105.9

-14.80.70 1.011 -0.011 -0.1 91.1

-14.90.75 0.847 0.153 1.5 76.2

-13.40.80 0.697 0.303 3.0 62.8

-10.40.85 0.582 0.418 4.2 52.4

-6.20.90 0.514 0.486 4.9 46.2

-1.30.95 0.499 0.501 5.0 44.9

3.71.00 0.540 0.460 4.6 48.6

8.31.05 0.632 0.386 3.7 56.9

12.01.10 ... .. . ... 68:9

41

1.0 1.1

THE SWING EQUATION AND ITS SOLUTION

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Time t (seconds)

.~fI~.~

~\/ V \.\'

~lit: \\

II \ ~.\~\~h \. \'

I~\ \\'~

.'\ \ /

,f \. \~\7 : 54'I 6 p).-

v 1\ '\t~/ V 3

/ \'l' .

.~~. "

3Cl 0 o L\t =0.05 sec.46---------6 L\t=0.10 sec.50-----0 L\t =0.15 sec.6--- L\t=0.20 sec.7---- L\t =0.25 sec.

I I I I Ioo

30

15

42

150

135

120-;;;..~ 105..'0

1390:s

:rl~'0 75~c...1: 60..E..u

45...1i.Ul

i;i

FIG. 7. Swing curves calculated by point-by-point method 2 with various valuesof At (Example 3).

TABLE 8

PERCENTAGE ERRORS IN AMPLITU DE AND PERIOD OF

SINUSOIDAL O SCILLATIO N C OMPUTED BY POlNT-B y-POlNT

METHODS 1 AND 2 WITH VA RIOUS V ALUES OF t

Per cent Error in AmplitudePoints Per cent Error

At per True At End of One Cycle At Half in Period(sec.) Cycle Cycle

Method 1 Method 2 Method 1 Method 1 Method e0.0167 56 +21 ... +9 +1 ...0.05 19 +68 0 +29 +2 -0.40.10 9.4 + 191 0 + 67 +5 -20.15 6.2 +300 0 +100 +11 -50.20 4.7 . .. 0 ... . .. -90.25 3.7 . .. 0 . .. . .. -15

ASSUMPTIONS MADE IN STABILITY STUDIES 43

The superiority of method 2 over method 1 is shown unmistakably byTable 8 and by the curves. Method 2 with at = 0.10 sec. yields moreaccurate results than method 1 with at = 0.0167 sec., and the results areobtained with less than one-sixth the amount of calculation.

Assumptions commonly made in stability studies. In the foregoingdiscussion of the solution of the swing equation, it was tacitly assumedthat the accelerating power Pa was known at the beginning of eachinterval, and, indeed, the equation could not be solved unless Pa wereknown. Since Pa = Pi - Pu , both the input Pi and the output Pu

must be known. In the determination of Pi and Pu the followingassumptions are usually made:

1. The input remains constant during the entire period of a swingcurve.

2. Damping or asynchronous power is negligible.3. Synchronous power may be calculated from a steady-state

solution of the network to which the machines are connected.4. Each machine may be represented in the network by a constant

reactance (direct-axis transient reactance) in series with a constantelectromotive force (voltage behind transient reactance).

5. The mechanical angle of each machine rotor coincides with theelectrical phase of the voltage behind transient reactance.

These assumptions will now be discussed.1. The input is initially equal to the output. When a disturbance

occurs, the output usually undergoes an abrupt change, but the inputis unchanged. The input to a generating unit is controlled by thegovernor of its prime mover. The governor will not act until thespeed change exceeds a certain amount (usually 1% of normal speed),depending on the adjustment of thE{governor, and even then there is atime lag before the governor changes the input. During swings of thesynchronous machines the percentage change in speed is very smalluntil after synchronism is actually lost. Therefore governor action 'isusually not a factor in determining whether synchronism will be lost,and, accordingly, it is neglected.

2. The output (electric power) of a synchronous machine consistsof a synchronous part, depending on the relative angular positions ofall the machines of the system, and an asynchronous part, dependingon the relative angular speeds of all the machines. The asychronouspart may be taken into account if desired, but, as it is usually unimportant in comparison to the synchronous part, it is commonly neglected in the interest of simplicity. The calculation of damping orasynchronous power is discussed in Chapter XIV, Vol. III.


3. The network connecting the machines is not strictly in the steadystate during swinging of the machines, both because of sudden circuitchanges, such as application or removal of a fault, and because of themore gradual change of phase of the electromotive forces due to theswinging. However, as the periods of oscillation of the machines arerelatively long (of the order of 1 sec.) in comparison to the timeconstants of the network, the network may be assumed, withoutserious error, to be in the steady state at all times. Steady-statenetwork solution is presented in Chapter III.

4 and 5. The assumption that each machine can be represented by aconstant reactance in series with a constant voltage, and the assumption that the mechanical position of the rotor coincides with the phaseangle of the constant voltage, are not entirely correct. As a rule,however, they do not lead to serious error in the determination ofwhether a given system is stable. Since examination and justification of these assumptions require a considerable knowledge of synchronous-machine theory, they will not be attempted at this point butwill be postponed to Chapter XII, Vol. III.

EXAMPLE 4A 25-Mva. 6o-cycle water-wheel generator delivers 20 Mw. over a double

circuit transmission line to a large metropolitan system which may be regarded as an infinite bus. The generating unit (including the water whe 1)has a kinetic energy of 2.76 Mj. per Mva. at rated speed. The direct-axistransient reactance of the generator is 0.30 per unit. The transmission circuits have negligible resistances, and each has a reactance of O~20 per uniton a 25-Mva. base. The voltage behind transient reactance of the generatoris 1.03 per unit, and the voltage of the metropolitan system is 1.00 per unit.A three-phase short circuit occurs at the middle of one transmission circuitand is cleared in 0.4 sec. by the simultaneous opening of the circuit breakersat both ends of the line.

Calculate and plot the swing curve of the generator for 1 sec.Solution. The swing curve will be calculated by point-by-point method

2, using a time interval of 0.05 sec. Before commencing the point-by-pointcalculations, we must know the inertia constant of the generator and thepower-angle equations for three different conditions of the network, namely:(1) before the fault occurs; (2) while the fault is on; and (3) after the faulthas been cleared. The power-angle equation depends on the reactance between the generator and the infinite bus.

Network reduction. Figure 8a is a reactance diagram of the system.Before occurrence of the fault the reactance between points A and B isfound by series and parallel combinations to be

+ 0.20 0 ·Xl = 0.30 - = .40 per unit2

ASSUMPTIONS MADE IN STABILITY STUDIES 45

When the fault is cleared, one of the parallel circuits is disconnected, makingthe reactance

X 3 = 0.30 + 0.20 = 0.50 per unit

The equivalent series reactance between the generator and the infinite buswhile the fault is on may be found most readily by converting the Y circuitGABF to a ~, eliminating junction G. The resulting circuit is shown inFig. 8b. The reactance of the branch of the ~ between A and B is

X = 0.30 + 0.20 +0.30 X 0.202 0.10

= 0.50 + 0.60 = 1.10 per unit

The values of reactance of the other two branches are not needed becausethese branches, being connected directly across the constant-voltage power

0.20

(a)1.10

--......-rmrn'-.......- .......----o

(b)

FIG. 8. (a) Reactance diagram of a system consisting of a generator A supplyingpower over a double-circuit transmission line to a large metropolitan system B(Example 4). (b) Equivalent circuit of the system with a three-phase short circuitat the middle of one transmission circuit, point F of a. The circuit of b is obtained

from that of a by a y-~ conversion to eliminate point G.

sources, have no effect on the power outputs of the sources, although theyincrease the reactive power outputs. The same is true of the O.IO-per-unitreactance at B. The power-angle equation for the circuit of Fig. 8b is thesame as it would be with these three shunt branches omitted.

Power-angle equations. The power-angle equation, giving the outputPuA of generator A as a function of the angle 0 between voltages EA and EB,is

P EAEB • ~ C· ~uA = --SInu = sIn u

X

where C has the following values:

Before fault, C1 = EA EB = 1.03 X 1.00 = 2.58Xl 0.40


During fault, O2 = EAEB = 1.03 X 1.00 = 0.936X2 1.10

After clearing, 0 3 = EAEB = 1.03 X 1.00 = 2.06x, 0.50

Inertia constant. By eq. 54

M = GH = 1.00 X 2.76 = 2.56 X 10-4 er unit180/ 180 X 60 P

Initial conditions. The power output of generator A before the fault wasgiven as 20 Mw., which on a 25-Mva. base is 0.80 per unit. The initialangular position of A with respect to B is found by the pre-fault powerangle equation:

P uAl = 2.58 sin ~ = 0.80

sin ~ = 0.80 = 0.3102.58

~ = 18.10

Immediately after occurrence of the fault the angular position is unchanged,but the power output changes to that given by the fault power-angle equation

P uA2 = 0.936 sin ~

= 0.936 sin 18.10

= 0.936 X'O.310

= 0.290 p.u.

Point-by-point calculations. Take the time interval as at = 0.05 sec.The steps of calculation for each point are as follows:

Pa(n-l) = Pi - Pu(n-l) = 0.800 - Pu(n-l) per-unit power

(at)2 (0.05)2-- P a(n-l) = 4 P a(n-l) = 9.76 Pa(n-l) elec. deg.

M 2.56 X 10-

~~n = .1on - 1 + 9.76Pa (n - l ) elec. deg.

an = 8n- l + .1on elec. deg,

Pun = C sin On per-unit power

where C = C2 = 0.936 while the fault is on (0 < t < 0.4 sec.).C = C« = 2.06 after the fault has been cleared (0.4 < t).

At t = 0 and t = 0.4 sec. there are discontinuities in P; and hence in Ps,and the average value should be used in calculating .10.

The calculations are carried out in Table 9. The swing curve is plotted in

TABLE 9POINT-By-POINT COMPUTATION OF SWING CURVE (EXAMPLE 4)

t C sin 8Pu Pa 9.76P a A8 8

(sec. ) (p.u.) (p.u.) (p.u.) (elec. deg.) (elec, deg.) (elec. deg.)

0- 2.58 0.310 0.800 0.000 18.10+ 0.936 0.310 0.290 0.510 18.1

o RVg. 0.255 2.52.5

0.05 H 0.352 0.330 0.470 4.6 20.67.1

0.10 H 0.465 0.435 0.365 3.6 27.710.7

0.15 " 0.621 0.581 0.219 2.1 38.412.8

0.20 " 0.779 0.730 0.070 0.7 51.213.5

0.25 " 0.904 0.846 -0.046 -0.4 64.713.1

0.30 II 0.977 0.915 -0.115 -1.1 77.812.0

0.35 II 1.000 0.936 -0.136 -1.3 89.810.7

0.40- " 0.983 0.920 100.50.40+ 2.06 " 2.0240.40 avg, 1.472 -0.672 -6.6

4.10.45 II 0.968 1.995 -1.195 -11.6 104.6

-7.50.50 H 0.992 2.045 -1.245 -12.1 97.1

-19.60.55 II 0.976 2.010 -1.210 -11.8 77.5

-31.40.60 H 0.721 1.486 -0.686 - 6.7 46.1

-38.10.65 H 0.139 0.286 0.514 5.0 8.0

-33.10.70 II -0.424 -0.874 1.674 16.3 -25.1

-16.80.75 u -0.668 -1.376 2.176 21.2 -41.9

4.40.80 " -0.609 -1.255 2.055 20.0 -37.5

24.40.85 II -0.227 -0.468 1.268 12.4 -13.1

36.80.90 II 0.402 0.828 -0.028 -0.3 23.7

36.50.95 " 0.868 1.788 -0.988 -9.6 60.2

24.91.00 " 0.996 2.052 -1.252 -12.2 85.1

12.71.05 97.8

47


Fig. 9, together with curves for several other clearing times. The system isstable with 0.4-sec. clearing. This fact becomes evident at 0.5 sec., and theremainder of the swing computation is unnecessary if we merely wish toknow whether the system is stable with the given clearing time.

EXAMPLE 5Determine the critical clearing time of a three-phase short circuit at the

middle of one transmission line of the system of Example 4, assuming thatthe circuit breakers at both ends of the line open simultaneously.

Critical clearing pointI

Sustained fault

...cu

t Ot----+-----I---t----+-~r+_-___+_#_~...f--~-__I__+_-~

250r---...,.----r---r--..-----r----,..--,.----,----nl~-.....

- 50 "-_.-i.-_-a__--'-_---.__~_---.l. _'" _'

o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0Time (seconds)

FIG. 9. Swingcurvesfor the system of Fig. 8 (Example 4).

i 200t----t----I---I----+---+---4---+--.-,;~~"--_+_-_f

i"0

~ 150I-r--I-~~t:_~_r_:_~-r;a=::~__=:r_--r--....

I~ 100t---+----;f----+---bC~~lId_-__+---::I~-I---____.--_+_~__I...ac.9150

Solution. First, the swing curve for a sustained short circuit is computedand plotted. The computations of Example 4 may be used up to t = 0.4sec.; but for t > 0.4 sec. new computations must be made by use of thepower-angleequation for the faulted condition. The swing curve is plottedin Fig. 9. Obviously the system is unstable for a sustained fault. By in..spection of the swing curve an estimate is made of the critical clearing time.Say that we estimate 0.5 sec. The estimate is checked by computing theswing curve for O.5-sec. clearing, starting from the O.5-sec. point of thecomputation for a sustained fault. Computation of only two points or sosuffices to show that the system is stable with 0.5-8ec. clearing. A longerclearing time is then tried. Several more swing curves may need to becalculated, until two clearing times, differing slightly, are found, for one ofwhich the system is stable and for the other of which it is unstable. From

REFERENCES 49

the curves of Fig. 9 it may be concluded that the critical clearing time liesbetween 0.6 and 0.65 sec. The critical clearing angle lies between 136° and147°.

Detailed computations are not given here because they are similar to thoseof Example 4. It should be stated again that usually only a few points needbe calculated on each swing curve, departing from the curve for a sustainedfault at the assumed clearing time.

EXAMPLE 6In Example 4 find the maximum percentage deviation of the speed from

its normal value, both before and after the time when it is first certain thatthe system is stable.

Solution. As was mentioned in Example 4, about 0.5 sec. after occurrenceof the fault it is certain that the system is stable. The relative speed (averaged over a time interval ~t) is ~8/lit, which, since ~t is constant, is proportional to lia. The maximum value of ~o before 0.5 sec. is 13.5 elect deg.;the maximum value after 0.5 sec. is - 38.1 elec. deg. The correspondingrelative speeds are 13.5/0.05 = 270 elect deg, per sec. and -38.1/0.05= -762 elect deg. per sec. The normal speed, corresponding to a frequencyof 60 c.p.s., is 60 X 360 = 21,600 elec. deg. per sec. The percentage deviations from normal speed are (270/21,600)100 = 1.2% and (762/21,600)100= 3.5%.

It appears that a governor set to be sensitive to a 1% change of speedwould have a negligible effect in determining whether the system was stable,although it might have some effect on the ensuing oscillations.

REFERENCES1. A.I.E.E. Subcommittee on Interconnection and Stability Factors, "First

Report of Power System Stability," Elec. Eng., vol. 56, pp. 261-82, February, 1937.2. FREDERICK S. WOODS, Advanced Calculus, New York, Ginn & Co., 1926.

Application of elliptic integrals to the motion of a pendulum, pp. 369-71.3. V. BUSH, "The Differential Analyzer: A New Machine for Solving Differential

Equations," Franklin Insi. Jour., vol. 212, pp. 447-88, October, 1931.4. R. H. PARK and E. H. BANCKER, "System Stability as a Design Problem,"

A.I.E.E. Trans., vol. 48, pp. 170-94, January, 1929. Includes description ofpoint-by-point calculation of swing curves.

5. I. H. SUMMERS and J. B. MCCLURE, "Progress in the Study of System Stability," A.I.E.E. Trans., vol. 49, pp. 132-58, January, 1930. Appendix IV,Sample Swing Curve Calculation, pp. 145-6.

6. F. R. LoNGLEY, "The Calculation of Alternator Swing Curves," A.I.E.E.Trans., vol. 49, pp. 1129-50, July, 1930; disc., pp. 1150-1.

7. O. G. C. DAHL, Electric Power Circuits, vol. II, Power System Stability,New York, McGraw-Hill Book Co., 1938. Four methods of point-by-point calculation of swing curves, pp. 391-401.


PROBLEMS ON CHAPTER II

1. Calculate the inertia constant M of a 25-cycle 12,500-kva. Ll-kv,187.5-r.p.m. water-wheel-generator unit by taking a typical value of H fromFig. 2.

2. In Example 4 plot on one sheet curves of angular position, relativespeed, and acceleration versus time.

3. Work the problem of Example 4, using ~t = 0.1 sec. instead of 0.05sec., and state whether you regard the accuracy of the swing curve so calculated as satisfactory.

4. Work the problem of Example 4 for a clearing time of 0.625 sec. instead of 0.4 sec. Use ~t = 0.05 sec. Is the system stable?

5. What is the critical clearing time of a three-phase short circuit close tothe sending end of one transmission line of the system of Example 41Assume simultaneous opening of both breakers. Which fault location ismore severe, the sending end or the middle of the line? Why?

6. In Probe 5 what is the critical opening time of the first (nearby)breaker if the second (distant) breaker opens 0.5 sec. later than the first one?

7. Compute and plot the swing resulting from the opening of one of thetwo parallel transmission circuits of Example 4 as a normal switchingoperation.

8. The system of Example 4 is modified by the addition of a radial feederto the bus of station A. If a three-phase fault occurs on this feeder at apoint separated from the bus by an impedance of jO.10 per unit and at atime when the feeder is carrying a very light load and when the main transmission lines are loaded as in Example 4, and if the fault is cleared in 0.4sec., is the shock to the system greater or less than that caused by the faultat the middle of .one transmission line cleared in an equal time? Why?

9. Find the steady-state power limit of the system of Fig. 8a with onetransmission line switched out. Make the following assumptions: For eachload the terminal voltage of generator A is adjusted to 1.00 per unit, and thepower factor is unity. Then assume a very small disturbance to occur duringwhich the excitation voltage behind synchronous reactance of 0.90 per unitremains constant.

10. Find the transient power limit of the system of Example 4 with athree-phase fault at the middle of one transmission line cleared in 0.4 sec. bythe simultaneous opening of the circuit breakers at both ends. Assume thepre-fault conditions of generator terminal voltage and power factor to be asdescribed in Probe 9.

11. The system of Example 4 is modified by making the impedance of eachtransmission line 0.10 + jO.20 per unit instead of 0 + jO.20 per unit. Findthe initial angular position of generator A with respect to the infinite bus ifthe power output of A is 0.800 per unit at a voltage of 1.00 per unit and apower factor of 1.00. Also find the power output and accelerating powerimmediately after occurrence of a three-phase short circuit at the middle of

PROBLEMS 51

one line. Does neglecting resistance give an optimistic or a pessimistic result for the stability of a system having one generator and an infinite bus?

12. Show that point-by-point method 2 is equivalent to making thesubstitution

in the swing equation.13. It is known that the differential equation

(a)

has a solution(b)

the real part of which represents a simple harmonic variation of awith t witha period T = 21r/k. Show that the corresponding difference equation

(c)

in which an denotes the value of 8 at a time t = n~t, has a similar solution

where2. k~t

'-' = - sln-1-~t 2

(d)

(e)

14. By using eq. e of Probe 13 calculate the percentage error in period of asimple harmonic oscillation as a function of the number of points per cyclewhen the oscillation is calculated by point-by-point method 2.

25rv

Fault

FIG. 10. Power system-with frequency changer B-e (Probs. 15 and 16).

15. Describe the procedure for calculating swing curves of the machines ofFig. 10, where A is a 60-cycle generator; B-G, a synchronous-synchronous60-to-25-cycle motor-generator set (frequency changer); and D, a 25-cyclesynchronous motor (or generator with such local load that it is equivalent toa motor for stability analysis). State specifically all respects in which theprocedure differs from that for a system in which all machines are electricallyin parallel.


16. Compute swing curves of the four machines of the system of Fig. 10for a sustained three-phase fault at the end of the branch line near B. Dataon the system are as follows:

Reactances in per unit on a system bas6

LinesA-B 0.40C-D 0.20Branch from B to fault 0 .20

MachinesA 0.10B 0.20C 0.15D 0.15

Inertiaconstants (H) on the system baseA 8.0B 1.0C 1.0D 6.0

Initial conditionsLoad of 1.00 per unit transmitted from A to D.Terminal voltages of Band CJ 1.00 per unit.Power factor at terminals of Band C, 1.00.

17. What are the dimensions of inertia constant H? Give a physicalinterpretation.

CHAPTER III

SOLUTION OF NETWORKS

Determination of the swing curves of the several synchronousmachines of a power system, as has been mentioned, consists of twoprocesseswhich must be carried on alternately: (1) the solution of theswing equation of each machine, giving the change in angular positionduring a short interval of time due to a known accelerating power;and (2) the solution of the network to which the machines are connected, giving the output of each machine when the angular positionsof all machines are known. In Chapter II attention was focussed onthe solution of the swing equation, and the network solution requiredfor Example 4 was purposely made as simple as possible by using atwo-machine reactance system. The solution of the swing equationwas found to be very simple. It becomes no more complicated if thenumber of machines is increased except that a similar calculation hasto be made for each machine. The solution of the network, on theother hand, rapidly becomesmore laborious as the number of machinesis increased. The methods of network solution required for stabilitystudies will be outlined in this chapter.

The impedance diagram (positive-sequence* network). Before thenetwork can be solved, however, it must first be established. Thestarting point is~ usually a one-line diagram of the power system to bestudied, showing generators, synchronous condensers and other largesynchronous machines, reactors, transformers, transmission lines, andloads. The diagram is usually limited to the major transmission system. As a rule, distribution circuits and small loads are not shown indetail but are taken into account merely as lumped loads on substationbusses. From the one-line diagram there is prepared a diagram inwhich all significant 'electrical elements of the power system are represented on a single-phase (line-to-neutral) basis by their positivesequence equivalent circuits with proper values of impedance. Thevalues of impedance of individual apparatus are commonly given eitherin actual ohms or in per unit (or per cent) based on the rating of theindividual apparatus. For use in the system impedance diagram the

*This term is defined in Chapter VI.53

54 SOLUTION OF NETWORKS

values so given must be converted either to ohms referred to a commonvoltage or to per-unit values on a common system base. The relationships among these several systems of units will now be discussed.

Per-unit quantities. In the per-unit system the various physicalquantities, such as current, voltage, power, and impedance, are expressed as decimal fractions or multiples of base quantities. When theapparatus base is used, the base quantities are the rated, or full-load,values or are derived from them. Thus, for a 1,000-kva., 66,OOO-to11,OOO-volt single-phase transformer, base power is taken as 1,000 kva.(or kw.); base voltage on the high-voltage side, as 66,000 volts; andbase voltage on the low-voltage side, as 11,000volts. Base current onthe high-voltage side is 1,000/66 = 15.15 amp.; on the low-voltageside it is 1,000/11 = 90.9 amp.; both these values are full-load currents. The base impedance of the high-voltage side is the ratio ofbase voltage of that side to base current of that side, namely,66,000/15.15 = 4,360 ohms. The base impedance of the low-voltageside is 11,000/90.9 = 121.5 ohms, which is (11,000/66,000)2 = 1/36as great as the high-voltage value. These values are the load impedances required on the high- and the low-voltage sides of the transformerto load it fully at rated voltage.

If the transformer is carrying a load of 500 kva. at 0.8 power factorand at rated voltage, the apparent power is 500/1,000 = 0.50 per unit,the active power is 400/1,000 = 0.40 per unit, the voltage on bothsides is 1.00 per unit, and the current on both sides is 0.50 per unit.The per-unit voltages of both windings are nearly equal, differing byonly a small impedance drop. The per-unit currents of both windingsare nearly equal, differing by only a small exciting current.

The short-circuit, or equivalent, impedance of a transformer referred to one side is the same in per unit as when referred to the otherside, although different in ohms. For example, if the short-circuitimpedance of the above-mentioned transformer is 0.070 per unit, it is0.070 X 4,360 = 305 ohms referred to the high-voltage side, or0.070 X 121.5 = 8.5 ohms referred to the low-voltage side.

Per-unit impedances based on the apparatus rating are nearly thesame for all apparatus of the same general design though of differentvoltage and kilovolt-ampere ratings, whereas the impedances in ohmsvary greatly with the rating. Hence typical values of impedance areeasier to tabulate, remember, and compare if expressed in per unit thanif expressed in ohms.

For conversion of self-impedances from ohms to per unit and reverse,the following formulas are used:

PER-UNIT QUANTITIES 55

[2]

[3]

[4]

B· d · h base voltage in volts

ase impe ance In 0 ms = .base current In amperes

base voltage in volts= -----------

(baSe po\ver in vOlt-amperes)

base voltage in volts

(base voltage in volts)2=----------base power in volt-amperes

(base voltage in kilovoltsj''= [1]

base power in megavolt-amperes

. . impedance in ohmsPer-unit Impedance = b · d · h

ase impe ance In 0 ms

impedance in ohms X base power in megavolt-amperes

(base voltage in kilovoltsj''

Impedance in ohms = per-unit impedance X base impedance in ohms

per-unit impedance X (base voltage in kilovolts)2

base power in megavolt-amperes

For mutual impedances between circuits of different base voltages wehave, instead of eq. 1,

B· d base voltage 1 X base voltage 2

ase unpe ance = base power

Although the foregoing relations have been derived for a single-phasecircuit (which may be regarded as one phase of a Y-connected threephase circuit), they apply equally well to a whole three-phase circuit

provided that the base voltage is the line-to-line value (V3 times theline-to-neutral voltage) and the base power is the three-phase value (3times the power per phase). ' These factors cancel in the expression(voltage)2jpower.

For use in a power-system study, all impedances and other quantitiesmust be expressed on a common system base. An arbitrary base poweris selected; for example, 100 Mva. for a large power system, 50 or 20Mva. for a smaller one. The base voltage for each portion of the network is usually the nominal voltage of that portion and, if not stated,is thus understood. For portions of the network connected throughtransformers, however, the ratio of base voltages should equal the turns


ratio of the transformer for the particular tap used, even if the turnsratio differs from the ratio of nominal voltages. As eq. 2 shows, whenper-unit impedances are changed from one base power (megavoltampere base) to another without change of base voltage, they varydirectly as the base power.

Per-cent quantities are 100 times the corresponding per-unit quantities. The decimal point must be watched in the multiplication ordivision of per-cent quantities; for example, the product of a per-centimpedance by a per-cent current is 100 times the proper per-cent voltage drop.

As an alternative to putting all data into per unit on a system base,they are sometimes expressed in ohms, volts, amperes, and so on,referred to a common voltage, usually the nominal voltage of the majorportion of the transmission system. If this is done, impedances of thelines of the major portion are left expressed in actual ohms, whereasthose of lines of other voltages are referred to the selected voltage bymultiplying them by the square of the turns ratio of the interveningtransformer. Impedances given in per unit on a given power base areconverted to ohms at the chosen voltage by eq. 3.

Representation of large synchronous machines. The representationof various circuit elements in the impedance diagram will now be discussed briefly. References will be given to sources of more completeinformation.

Each generator or other large synchronous machine is commonlyrepresented for transient-stability studies by its direct-axis transientreactance Xd' in series with a constant-voltage power source (Fig. 1).The armature resistance of large machines is usually negligible. Thereactances of machines already built Of designed can usually be obtained from the manufacturer. The method of obtaining Xd' from ashort-circuit oscillogram is described in Chapter XII, Vol. III. Formachines of minor importance average values of Xd' may be taken fromTable 2 of Chapter XII. More exact ways of representing synchro-.nous machines are also discussed in that chapter. .

Representation of transformers. Two-circuit transformers are represented most accurately by an equivalent T circuit in which the seriesarms represent the leakage impedances, and the shunt arm, the excitingimpedance. As a rule, the exciting current can be neglected in stability studies; if this is done, the T circuit is reduced to series impedanceZ (Fig. 2), equal in value to the short-circuit, or equivalent, impedanceof the transformer. The value of this impedance is frequently givenon the name plate. It can be measured by means of a short-circuittest. The resistances of large transformers are usually negligible,

REPRESENTATION OF TRANSFORMERS 57

FIG. 2. Representation of a twocircuit transformer, exciting current neglected. V1, and V2 are theprimary and secondary voltages,respectively. Z is the equivalent

impedance.

High Max. Per-cent Max. Per-cent Max.Voltage, Low Reactance Low Reactance Low

Line to Line Voltage~ Voltage~ Voltage(kv.) (kv.) Min. Max. (kv.) Min. Max. (kv.)

ranging from about 0.3 to 1.1%. Typical values of reactance aregiven in Table 1. They depend principally upon the rated voltage:the higher the voltage, the higher the reactance.

x'"

FIG. 1. Representation of a generator or other large synchronousmachine in a transient-stabilitystudy. Xd' is the direct-axis transient reactance, E/ is the internalvoltage "behind" this reactance,

and V is the terminal voltage.

TABLE 1

TYPICAL REACTANCES OF TWO-WINDING POWER TRANSFORMERS

(500 KVA. PER PHASE AND OVER, 1- OR 3-PHASE, 25 OR 60 c.e.s., 55°0. RISE)

(From Ref. la by permission of Westinghouse Electric Corporation)

Per-centReactance~

Min. Max.

0- 15 15 4.5 7.016- 25 15 5.5 8.026- 37 15 6.0 8.038- 50 25 6.5 9.051- 73 25 7.0 10.074- 92 34.5 7.5 10.593-115 34.5 8.0 12.0

116-138 37 8.5 13.0139-161 50 9.0 14.0162-:96 50 10.0 15.0197-230 50 11.0 16.0

253746697373929292

6.57.58.08.59.09.5

10.511.512.5

9.010.511.012.514.015.016.017.018.0

7392

115138161161

9.0 14.010.0 15.510.5 17.011.5 18.012.5 19.014.0 20.0

Three-circuit transformers are represented, with exciting currentneglected, by Y circuits (Fig. 3) such that the resistance of eachbranch is the resistance of the corresponding winding, and the sum ofthe reactances of each pair of branches equals the short-circuit reactance between the corresponding pair of windings with the remainingwinding open." 5 Thus

Xl +X2 = X l 2

Xl +X3 = X13

X 2 + X 3 = X 23

[5a][5b][5cJ


whenceXl c= !(X12 + X 13 - X23 )

X2 = !(XI 2 + X 23 - X13 )

X 3 = !(XI3 + X23 - X 12)

[6a]

[6b]

l6c]

Frequently the reactance of one arm of the Y, as determined byeqs. 6, is found to be negative.t

For estimating purposes the reactance X l 2. between the two mainwindings of a three-winding transformer may be considered equal to

the reactance of a two-windingtransformer having the same kilo-

R1

+ jX 1 h tV2

volt-ampere and voltage ratings-a: (Table 1). The reactances X 13

I 3 1-J~ ~ and X23 between the main wind-v} ings and the tertiary winding

cannot be estimated accuratelybecauseof the wide range in thesereactances in transformers of different designs.!"

FIG. 3. Representation of a three ...circuit transformer, exciting current Transformers of 4. or 5 circuits.

neglected. See Refs. 6 and 7, respectively.Autotransformers are repre

sented in the same manner as transformers with separate windings.The impedances used in the diagram are those between circuit terminals, not those between parts of the winding. The reactance of anautotransformer in per cent (based on the rated kilovolt-amperes delivered) can 'be estimated by multiplying the reactance of a two-winding transformer from Table 1 by the ratio, Ia

rated high voltage - rated low voltagerated high voltage

Autotransformers can have very small reactances if the voltage ratiois close to unity.

[This circumstance causes no difficulty in an algebraic solution, but on an a-c.calculating board (discussed later in this chapter) an inordinately large capacitanceis required for representing a small negative reactance. This difficulty may beavoided either by combining the negative reactance with whatever positive reactance may be in series with it or by representing it by capacitive reactance inseries with inductive reactance that is smaller than the capacitive reactance by theamount of the required negative reactance.

TRANSMISSION LINES AND CABLES 59

Symmetrical banks of three single-phase transformers, that is, Y or Aconnections of identical transformers, are represented in the singlephase impedance diagram as one single-phase transformer, the per-unitimpedance of which, based on the kilovolt-ampere rating of the bank,is equal to the per-unit impedance of one transformer on its ownkilovolt-ampere rating. Three-phase transformers ~re represented insimilar fashion. The no-load phase displacement between primaryand secondary circuits (0 or 1800 for a ~-~ or Y- Y connection, ±30°for a ~-y or y-~ connection) ordinarily can be and is disregarded.

Regulating transformers "for control of ratio, phase shift, or both. SeeRefs. Ib, 8, and 9.

Transmission lines and cables are represented by their nominal orequivalent 7r circuits.l!

In the nominal 7r circuit (Fig. 4a) the series branch has an impedanceequal to the total series impedance Z per phase of the line, and the

z

II ~y tanh(...{ZY'22" VZf/2

~ ~ ~

(a) Nominalr (b) Equivalent 1r

FIG. 4. Circuits for representing transmission lines and cables.

shunt branch at each end has an admittance equal to half the shuntadmittance Y of the line to neutral. Here Z = zl and Y = st. where lis the length of the line and z and yare the series impedance and shuntadmittance per unit length. The series impedance consists of resistance and inductive reactance; the shunt admittance consists practically of capacitive susceptance only, as the shunt conductance ofpower lines is negligible.

In the equivalent 1r circuit (Fig. 4b), the impedance of the seriesbranch is that of the nominal 1T' multiplied by a correction factor,

sinh vZ¥ Eiv'ZY - E-iy'Zf

VZY 2VZY

1 + Zy (Zy)2 (Zy)3= 6+120+ 5040+··· [7a],

and the admittance of each shunt branch is that of the nominal 1r


multiplied by another correction factor,

EiVZ'i/2 _ E-iVZ'i/2

(Ei Y ZY/ 2 + E-iVZ'i/2) v'ZY/2

tanh (V"ZYj2) = -----------v'ZY/2

The correction factors can be found easily by use of Woodruff'scharts. 12

The equivalent 1r is an exact representation of a line at a particularfrequency, whereas the nominal 1r is an approximation, the use ofwhich is justified only if the correction factors are nearly 1, as is true ifZy = zyl2 « 1, hence for short lines. As a general rule, the nominal eis accurate enough for 60-c.p.s. open-wire lines not more than 100mileslong. Longer lines may be represented by an equivalent 1r or by twoor more nominal 1r'S in tandem. Very short lines have negligibleshunt admittance and can be represented by their series impedanceonly.

Each part of a composite line, such as an underground cable in serieswith an aerial line, may be represented by a 1r. By means of one ormore y-~ conversions, the circuit can be reduced to one 1r, usuallyan unsymmetrical one.

When several linesare connected tothesame bus,the shunt capacitancesat that end of all the 1r circuits of these lines are in parallel and may bereplaced by one capacitance from the bus to neutral. If one or morelines are disconnected in the course of a study, the value of this capacitance should be reduced accordingly.

Although T circuits could be used instead of 1r circuits, the 1r circuitsare preferable because they require only one series impedance per lineand one shunt capacitance per bus, whereas the T circuits require twoseries impedances and one shunt capacitance per line.

The constants of lines may be found by measurement 13 or by calculation. In calculation the length of the line, found from a map orother record, is multiplied by the constants per unit length. Theconstants of cables are best found from the manufacturer, but they'may be estimated from published tables.1d•

14 The constants of aeriallines can be found accurately from tables. They depend upon (1)the frequency, (2) the size and kind of conductor, and (3) the spacingbetween conductors. The series resistance depends principally uponsize and kind of conductor and, to a lesser degree, upon frequency.The series inductive reactance depends upon all the foregoing factors.

MISCELLANEOUS EQUIPMENT 61

It can be found most simply as the sum of two terms, one of which(equal to the reactance ascribable to the flux inside a l-ft, radius)depends upon frequency and size and kind of conductor; and theother, upon frequency and spacing. t The shunt capacitive reactancecan be found in like manner. Reference Ic has suitable tables forfinding series and shunt reactances by this method. For estimatesthe series inductive reactance of a 60-c.p.s. aerial line may be taken as0.8 ohm per mile, and the' shunt capacitive susceptance, as 5 micromhosper mile. For other frequencies the values are proportional.

Mutual impedance and admittance between parallel lines are negligible for positive sequence.

Representation of loads. As the loads on a power system vary withthe time of day and of year and from one year to another, one or moreparticular load conditions must be selected for study; for example, theannual peak load and the minimum load may be taken. Estimatedfuture loads are often used. The connected generator and transformer capacity and other features of system operation will dependupon the loads assumed. Loads are assumed to be lumped on thebusses of major stations and substations. They should be expressedas vector power P + jQ, where P is the active power, and Q the reactive power. Each load is then represented by a shunt admittance,

y - P + jQ [8]- V2

where V is the voltage across the load. Ordinarily, the voltages usedin calculating the load admittances must be estimated. Later, theadmittances of the important loads can be revised, if so desired, byusing more accurate values of voltage.

Small tapped loads on transmission lines may be removed and apportioned between the two ends of the line in inverse proportion to theline impedances between the tap and the ends.

Representation of faults. A three-phase short circuit is representedby connecting the point of fault to the neutral bus. The representation of other types of fault is discussed in Chapter VI.

Miscellaneous equipment. Closed circuit breakers and switches,current transformers, and busses have negligible impedance on highvoltage systems and, therefore, are disregarded. Similarly, such shuntelements as potential transformers, lightning arresters, and coupling

tThis method of tabulating reactance of lines was originated by W. A. Lewis andwas published first in Ref. 2d.

If the three distances between conductors are unequal, the geometric mean of thedistances or the arithmetic mean of the corresponding reactance terms is used.


capacitors have impedances so high that they are considered opencircuits.

Representation of remote portions of the system. In studies of partof a large interconnected system it is neither necessary nor feasible torepresent in equal detail all stations and lines of the entire system.Outlying portions can be represented by equivalent circuits. Aremote portion connected at only one point to the portion being studiedcan be replaced, according to Thevenin's theorem, by an impedance inseries with a constant-voltage power source. The impedance is foundby network reduction (discussed later in this chapter) or by a knowledge of the short-circuit kilovolt-amperes at the point of connection.The impedance may be assumed to be pure reactance, in which casethe equivalent circuit is like that of a generator (Fig. 1). The voltageof the source may be assumed as 1 per unit. A remote portion of thenetwork connected at two points to the' portion being studied canbe represented by a power source and a Y circuit.

The inertia constant assigned to the generator of each of theseequivalent circuits should equal the total of the inertia constants of allthe machines therein (for reasons discussed later in this chapter) and,in the absence of more definite information, may be calculated from anaverage value of H (Figs. 1 and 2, Chapter II) and the known aggregate generating capacity. If the total inertia is large and looselycoupled to the portion of the network being studied, it may be con..sidered infinite with little error. The assumption of infiriite inertia fora remote portion of the system renders the computation of swingcurves unnecessary for the equivalent machines representing suchportions.

Check list of data required for transient-stability study.

Generators

Name-plate ratingDirect-axis transient reactanceKinetic energy at rated speed (or moment of inertia and speed),

including prime mover (see Chapter II)Voltage limitsGrounding§Negative-sequence resistance and reactance]Zero-sequence reactance§

§Required only for studies including ground faults. See Chapter VI.[Required only for studies involving unbalanced faults.

DATA FOR TRANSIENT-STABILITY STUDY 63

Transformers, two-circuit

Name-plate ratingEquivalent resistance and reactanceTurn ratio normally usedAvailable taps and tap-changing equipmentConnections§Grounding§

Transformers, multicircuit

Items listed for two-circuit transformersResistance of each winding r a common kilovolt-Reactances between each pair of windings ampere baseKilovolt-ampere rating of each winding

Reactors

ResistanceReactance

Transmission Zines and coble«

Series resistance and inductive reactance of each line sectionShunt capacitance or capacitive susceptance of each line section,

except those for which its effect is obviously negligibleZero-sequence resistance, reactance, and capacitance§

Synchronous condensers and large synchronous motors

Same data as for generators; however, kilovolt-ampere rating ofcondensers may be different for lagging current than for leadingcurrent

Loads

Active and reactive power for each condition to be studiedVoltage limits

Circuit breakers (see Chapters VIII and XI, Vol. II)

Interrupting timeReclosing time }-f t t- I· · dS- I I th I t- 1 au oma IC rec osmg IS usemg e-po e or ree-po e opera Ion

Protective relays (see Chapter IX, Vol. II)

Types and settings

§Required only for studies including ground faults, See Chapter VI.


The alternating-current calculating board. Mter the required datahave been compiled and the impedance diagram has been drawn, thenetwork must be solved in order to find the magnitudes and initialphase positions of the internal voltages of all the synchronous machines. Subsequently, during the determination of swing curves, thenetwork must be solved repeatedly to obtain the power output of allthe machines when their angular positions and internal voltages areknown.

Networks may be solved by two principal methods, the calculatingboard method and the algebraic method. Calculating-board solutionsare to be preferred on account of their rapidity and ease when the number of machines considered is three or more.

The calculating board is essentially a means of representing an electric power network to scale. A three-phase network is represented onthe board on a single-phase basis. One hundred volts on the boardmay represent, for example, 100 kv. line to line (57.7 kv. line to neutral) on the power system, and 100 watts on the board may represent50 Mw. three-phase (16.7 Mw. per phase) on the system. The scalesfor voltage and power are then 1 : 1,000 and 1 : 500,000, respectively.The scales of current and of impedance follow automatically from thescales chosen for voltage and for power, In this particular example,1 amp. on the board would represent 16.7 X 106/ 57.7 X 103 = 289amp. on the power system, and 100 ohms on the board would represent57,700/289 = 200 ohms per phase (Y basis) on the system.

The alternating-current calculating board, also known as a networkanalyzer or network calculator, consists of an assemblage of adjustableresistors, reactors, and capacitors; a number .of sources of a-c. voltageadjustable both.in phase and in magnitude; provisions for connectingthe foregoing units together in any desired manner to form a network;and instruments for measuring scalar and vector values of voltage,current, and power anywhere in the network. The instruments mustbe so designed that insertion of them does not appreciably changeconditions in the network. A modern a-c. calculating board is shownin Fig. 5.

In using the a-c. board in a stability study, the network is first setup to scale, the generators being represented ordinarily by theirdirect-axis transient reactances in series with power sources. Nextthe phase and magnitude of the voltages of the several power sourcesand the impedances of the loads are adjusted to represent the desirednormal or pre-fault operating condition, with respect to machine outputs, terminal voltages, and similar factors. The fault is then appliedat the desired point. If it is a three-phase fault, it is represented

THE ALTERNATING-CURRENT CALCULATING BOARD 65

simply by a short circuit on the network; if it is an unsymmetricalfault, it is represented either by a shunt impedance or by a connectionbetween the sequence networks (to be discussed in Chapter VI). Thevoltages of the power sources (representing voltages behind transientreactance, which are ordinarily assumed constant) are readjusted, ifnecessary, to restore them to their pre-fault values. It is desirable,however, that the power sources have such good voltage regulationthat readjustment is unnecessary, or at least small. Then the power

FIG. 5. A modern alternating-current calculating board (by courtesyofthe GeneralElectric Company).

output of each machine is read by means of a wattmeter. From thesereadings the accelerating power is computed, and from it the angularchange of each machine during the first time interval is found bymethods already discussed. The angular changes so found are thenreproduced on the calculating board by turning the angle-adjustingdials of the powersources. After this has been done, a new set of powerreadings is taken, and so on. The clearing of a fault or any otherswitching operation is represented by an appropriate change of connections on the board at the proper time in the run. The run is continued, the swing curves being plotted as it progresses, until it is apparent whether all the machines will remain in synchronism.

From the foregoing description of the use of an a-c. calculating


board in a stability study, it should be obvious that a direct-currentcalculating board could not be used for the same purpose. The correctrepresentation of phase relations is all-important.

Alternating-current calculating boards are owned and operated by anumber of leading electrical manufacturers, consulting engineeringfirms, engineering colleges, and, both publicly and privately ownedelectric-power utilities.22 Some of these boards are available for hire.The two most widely used types will now be described.

Description of General Electric A-C. Network Analyzer.17, 21

The a-c. calculating board shown in Fig. 5 operates at a frequency of480 c.p.s, ~ and has a nominal, or base, voltage of 50 volts and a nominal, or base, current of 50 rna.** Consequently, the base power is2.5 watts, and the base impedance is 1,000 ohms. All adjusting dialsand instrument scales are marked in percentages of the base quantities.

Generator units. There are a number of power sources for supplyingsingle-phase voltages of adjustable phase and magnitude, which areused principally for representing generators. Each generator unitconsists of two machines, a phase shifter and a single-phase inductionvoltage regulator. The phase shifter has a three-phase stator winding, which receives 220-volt, 480-c.p.s. power from a motor-generatorset, and a single-phase rotor winding, which delivers power at constantvoltage and adjustable phase to the voltage regulator. The phaseshifter provides adjustment of phase through an unlimited angle, andthe voltage regulator affords adjustment of voltage magnitude fromoto 250% of the base voltage. The phase and magnitude adjustmentsdo not affect each other, nor are the adjustments greatly affected byload. The series inductive reactance of the generator unit is compensated by series capacitive reactance. As a result, the voltage regulation at full load is only 2%, and the phase shift produced by ratedcurrent at zero power factor is only 10

• This small voltage regulationis very convenient in stability studies where the voltages of the powersources represent known constant voltages behind transient reactancesof synchronous machines because the angular positions of the severalmachines can be varied without the necessity of readjusting the voltage magnitudes. Furthermore, the phase and magnitude of eachvoltage agree closely with the respective dial settings.

In studies of normal power-system operation the voltage-magnitude

,The board frequency need not be the same as the frequency of the powersystem to be studied.

**The nominal voltage is approximately the average voltage (to neutral) of theboard circuits, and the nominal current is approximately the average current in theprincipal board circuits.

GENERAL ELECTRIC A-C. BOARD 67

adjustment simulates adjustment of the excitation of the representedmachine, and the phase adjustment corresponds to adjustment of thegovernor of the prime mover. Advancing the phase increases thepower output, and retarding the phase decreases it, unless, of course,the angle of maximum power on the power-angle curve is exceeded.

Line-impedance units. There are a large number of line-impedanceunits, each consisting of resistors and three tapped reactors permanently connected in series. These units are used for representing theimpedances of transmission lines, transformers, reactors, and machines, and other low impedances. Information on the range ofresistance and reactance, size of steps, current capacity, and so onof the line-impedance units and other circuit elements is given inTable 2.

As the impedance units constitute a large portion of the bulk andweight of any a-c. calculating board, it is important that they be madesmall. On the other hand, it is desirable that the reactors have areasonably low ratio of resistance to reactance so that they can represent sufficiently well transformers, reactors, and machines having alow ratio of resistance to reactance. Furthermore, the reactanceshould be almost independent of the magnitude of current in thereactor. These requirements for reactors tend to make them large.The problem of conflicting requirements has been solved by the use oflow base power, a higher frequency than 60 c.p.s., and high-grademagnetic material for the cores of the reactors.

The resistance of the line reactors ranges from 3% of the reactancewhen the reactance is set at 70% of the base value to 8% at a settingof 0.2% of the base. The resistance of the reactors is not negligible;therefore an allowance for it (taken as 5% of the reactance setting)should be made in setting the associated resistors.

Load-impedance units. The load-impedance units differ from theline-impedance units chiefly in that they have a higher impedance.They are used to represent loads or other circuits of high impedance.The resistor and reactor can be connected either in series or in parallelwith one another. The parallel connection affords easier adjustmentof the loads because thus the adjustments of active power and of reactive power are nearly independent of each other. The series connection is used for small loads, for loads of unity or zero power factor,and for representing high impedances other than loads. For representing a load of leading power factor, one of the capacitor units(mentioned in the next paragraph) is connected in parallel with a loadimpedance unit.

Capacitors are used chiefly for representing capacitance of cables

Ele

men

t

TA

BL

E2

CIR

CU

ITE

LE

ME

NT

SO

FG

EN

ER

AL

EL

EC

TR

ICA

-C.

NE

TW

OR

KA

NA

LY

'ZE

B

Inst

alle

dat

Sch

enec

tady

in19

37

Nu

mb

er"

Ran

get

Size

ofS

teps

]T

oler

ance

Max

imum

Safe

Val

uest

c 00

Gen

erat

ors.

....

....

..12

....

....

....

....

...

....

....

...

Vol

tage

regu

lati

onab

ou

t2

%..

1,00

0%cu

rren

tV

olta

ge..

....

....

....

....

..0

-25

0%

....

..S

moo

thP

hase

....

....

....

....

....

..U

nlim

ited

....

Sm

ooth

Lin

eim

peda

nces

....

..15

0S

th{T

appe

dre

sist

or±

1%

....

.·.

.}50

0%1

%1

Res

isto

r...

....

....

...

....

..0-

-51%

....

...

moo

....

Sth

.t

3~

0cu

rren

tor

250

vota

gem

oore

SIS

or±

0..

....

.

Rea

ctor

....

....

....

....

....

0-8

1%

....

...

{±

1%

atst

anda

rdcu

rren

t...

.}

0.2

%..

...

.±3~

tt

50

0%

curr

ent

or12

5%vo

ltag

eo

aan

ycu

rren

....

..

C.

{OO

0-1

10

%}

lot

10

1ap

aclt

ors

....

....

....

41

\c.

600 0

110

·±

/0-

---_

._.·

125%

volt

age

.••...~,

/0

····

Af

{8

85

--1

15

%..

...

1%}

1 01

dut

otra

nso

rmers

....

..12

69

.5-1

30

.5%

.0

.5%

....

..±

10

····

····

····

····

····

··5O

O%

curr

enta

n12

5%vo

ltag

e

Mut

ual

reac

tors

....

...

15

...

...

....

....

....

.·

·..

500%

curr

ent

and

125%

volt

age

*The

num

ber

ofci

rcui

tel

emen

tsva

ries

ondi

ffer

ent

anal

yzer

sbu

ilt

by

the

sam

em

anuf

actu

rer.

22

tIn

perc

enta

geof

base

quan

titi

es(5

0vo

lts,

0.05

0am

p.,

1,00

0oh

ms,

0.00

1m

ho).

Loa

dim

peda

nces

.R

esis

tor

.

Rea

ctor

.

500

-1,6

10

%..

..

....

....

0-1,

605%

....

{T

appe

dre

sist

or±

1%

}50

0%cu

rren

tor

125%

volt

age

Sm

ooth

.. .

.S

moo

thre

sist

or±

3%

.

{±

1 %at

stan

dar

dcu

rren

t}

500%

curr

ento

r12

5%vo

ltage

5%

....

....

±3

%at

any

curr

ent

.

00 o e ~ .-3 ...... o Z ~ Z t?-j~ ~ ~ ~ 00


and transmission lines. They are used also for representing capacitorsand, in steady-state studies, for synchronous condensers.

Autotransformers of two designs are provided, one of which will buckor boost the voltage by 1% steps up to a maximum of 15% of theprimary voltage; the other, by 0.5% steps up to 30.5%. They areused for representing transformers whose actual turns ratios differfrom the nominal ratios on which the circuit impedances are calculated.Thus they are valuable for representing transformers of adjustableratio, and one of them must be included in any loop around which theproduct of all transformer ratios is not unity; for example, if two systems of different voltage are connected through two transformers ofunequal turns ratio. The position of zero buck or boost representsthe nominal voltage ratio.

Mutual transformers of ratio 1:1 are used chiefly for representingmutual reactance between parallel transmission lines that are not connected together at either end. They are used also in various equivalent circuits. These transformers are not adjustable, but adjustablemutual impedance is obtained by shunting one winding of the transformer with an adjustable impedance (line-impedance unit).

Connections. A scheme is necessary whereby the foregoing circuitunits can be readily connected together in any desired way to form anetwork representing a particular power system. On the board shownin Fig. 5 each circuit unit terminates in a pair of flexible cords, eachbearing a single-pole plug. Insertion of these plugs into horizontallyadjacent jacks of the vertical jack panels forms an electrical connectionbetween them. The "busses" formed by groups of adjacent plugscan be placed in the same relative positions on the jack panel as on theconnection diagram, thus greatly facilitating the checking of connections. The lowest row of jacks is used as a ground, or neutral, bus,and the plugs inserted in this row are connected even though notadjacent. As an indication of polarity, one cord of each circuit unitis colored green, the other yellow.tt A positive wattmeter readingdenotes power flow from the yellow to the green cord; a negativereading, from the green to the yellow. Like rules hold for the sign ofthe varmeter reading and the direction of lagging reactive power.

Jumper circuits are used to form zero-impedance ties which can bemetered. They are useful in .representing bus-tie breakers, terminalsof 7r circuits representing transmission lines, connections between thesequence networks for representing faults (see Chapter VI), and so on.

ttThe mutual transformers have two pairs of cords, one for the primary winding,the other for the secondary. Like-colored cords have like polarity. The autotransformers have three cords, one of which, colored black, is usually pluggedto the ground bus. The green cord connects to the movable tap.


Instruments. The system of making measurements is one of themost crucial features of an a-c. calculating board. It must satisfythe following requirements:

1. It must be possible to insert the instruments anywhere in thenetwork without appreciably altering conditions in the network.The volt-ampere burden of the instruments determines the minimumpermissiblebase power and thus fixes the size and weight of the entireboard.

2. Because of the large number of readings required in mostpower-system studies, the instruments should be fast in response andeasily readable so that an operator can take readings all day withouteyestrain.

3. The instruments must be suitable for a variety of measurements: scalar values of voltage, current, active power, and reactivepower; and vector values of voltage and current in both rectangularand polar forms.

It has been found desirable to have two sets of instruments: (1)the master instruments, meeting the requirements enumerated above,and (2) instruments permanently connected to each power source.This second set of instruments expedites the adjustment of the powersources. The burden of these instruments is not a matter of suchconcern as that of the master instruments.

The master instruments of the board shown in Fig. 5 consist of a voltmeter, an ammeter, and a wattmeter-varmeter, In order to make theburden of these instruments on the network negligibly small despitethe low power level used there, they are supplied with current andvoltage through two stabilized-feed-back vacuum-tube amplifiers.PFor good legibility each instrument has an 8-inch translucent scale onwhich a spot of light serves as a pointer. The response of these instruments is rapid. Their over-all, accuracy, including the voltagedivider, the current shunts, and the amplifiers, is within ±O.5% of fullscale.

A phase shifter is provided for use with the wattmeter in the measurement of vector current and voltage, as will be described presently.

The following switches, located within reach of the operator, areused in association with the instruments:

A key switch is provided for each circuit unit on the board. Whenany of these switches is thrown, all three instruments are connectedto the selected circuit unit and indicate simultaneously.

The voltage-selector switch determines whether the voltagefurnished to the voltmeter and wattmeter comes from (a) across the


circuit unit, as for reading line drops and losses, (b) from one side ofthe circuit unit to the ground bus, (c) from the other side of the circuit unit to ground, (d) from either side of the circuit unit to a cordthat can be plugged to any desired point of the network, or (e) fromthe instrument phase shifter.

The current-selector switch makes it possible to select currentfrom the circuit unit, total ground current, or current from the instrument phase shifter.

The reversing switch is thrown so as to make the wattmeter readupscale. The reading is recorded with + or - sign, according to,the position of this switch.

The watt-var switch is used to make the wattmeter-varmeterread either active or reactive power. This switch is also used inreading vector current and voltage.

A seven-position range-selector switch provides for the most commonly used combinations of voltage and current ranges: voltage,100% and 200% of base; current, 20%, 100%,500%, and 2,000%.There is an auxiliary switch for selecting the less-used low-voltageranges: 20% and 40%.

Vector measurements. To measure current in rectangular form, thewattmeter-varmeter is furnished with the current to be measured andwith 100% voltage from the instrument phase shifter, which is set atthe desired reference angle (usually 0°). The inphase and quadraturecomponents of current are equal to the watt and var readings, respectively. To read current in polar form, the phase shifter is turned untilthe var reading is zero; then the angle is read on the phase-shifter dial.The watt reading is the current magnitude. Vector values of voltageare read in the same way as currents except that the wattmeter issupplied with the voltage to be measured and with 100% current fromthe phase shifter.

Instruments on the generator units. In addition to the centrallylocated master instruments, there are a wattmeter-varmeter and avoltmeter at the terminals of each power source. The wattmetervarmeter has voltage ranges of 150% and 300% and current ranges of100% and 400%. It is changed from a wattmeter to a varmeter bysubstituting a capacitor for the usual resistor in the potential circuit.The voltmeter can be connected to any point in the network where thegenerator is expected to regulate voltage.

Power supply. Three-phase, 480-c.p.s., 440-volt power is suppliedto the board from a remote-controlled motor-generator set consistingof a motor, a 3-kva. alternator, an exciter, and a phase balancer. Thephase balancer is used for obtaining balanced three-phase voltages


Primary

FIG. 6. Voltage vector diagram of onephase of the three-phase voltage regulatorsof a generator unit of the Westinghouse A-C.Network Calculator. O-L is the locus of

the output voltage.

(required so that the magnitude of the voltage of each power sourceshall be independent of its phase) in spite of the essentially single-phaseloading.

Between the generator and the power sources on the board is a bankof autotransformers with taps for normal voltage and for 50% and 25%of normal voltage. A tap-selector switch on the instrument deskenables the operator to reduce the voltage of all power sources simultaneously if this is necessary to avoid overloading any of the boardcircuits, for example, during the period that a fault is represented onthe power system.

Arrangement. The instrument desk, at which the operator is seatedin Fig. 5, is at the center, flanked on both sides by the jack panels andthe tables holding the flexible cords. Above the instrument panel andjack panels are the generator units. The wings on each end of theboard (only one of which shows in Fig. 5) contain the line- and loadimpedance units, capacitors, and autotransformers, each in a removabledrawer. The board is so arranged that the units requiring the mostfrequent adjustment are closest to the operator.

Description of Westinghouse A-C. Network Calculator.19,20 TheWestinghouse calculating board differs from the General Electricboard, which has been described, chiefly in the following features:

The frequency is 440 c.p.s.The nominal voltage and current are 100 volts and 1 amp., respec

tively.Each generator unit consists of two three-phase induction voltage

regulators and a phase shifter. The three-phase secondary windingsof the two induction regulatorsare in series, and the rotors aremechanically coupled so as togive them equal but oppositeangular displacements when thevoltage handle is turned. Consequently, the three-phase output voltages of the pair ofregulators are constant in phaseas the magnitude is varied.

(See Fig. 6.)tt The output of the regulators goes to the phase shifter.The output voltage of the phase shifter has three ranges: 0 to 100, 0to 200, and 0 to 400 volts.

ttEarly models had only one three-phase voltage regulator ahead of the phaseshifter. Consequently, the voltage adjustment changed the phase also, requiring acompensating movement of the phase shifter.

WESTINGHOUSE A-C. BOARD 73

For representing the internal reactance of each generator there is aspecial reactor having a lower resistance than the line and load reactors.

Each generator is provided with a voltmeter, an ammeter, and awattmeter. These instruments have auxiliary pointers that can bepre-set manually to show the desired operating conditions. The voltmeter can be switched to read either the internal voltage or theterminal voltage of the generator.

The impedance units are marked in ohms at 440 c.p.s. (equal to percent impedance when 100ohms is used as base); the autotransformers,in percentage; and the capacitors, in microfarads. Additional information on the various circuit elements is given in Table 3.

Each load unit is equipped with a tapped autotransformer, called aload adjuster, by means of which the active and reactive power of eachload can be adjusted simultaneously without changing the powerfactor of the load. Usually the settings of the load resistor andreactor are calculated for an estimated bus voltage (say, 100%). Ifthe actual load voltage differs from the estimated value, the load unitdoes not consume the correct value of power, because power in aconstant impedance varies as the square of the voltage. By means ofthe autotransformer, however, the voltage on the resistor and reactorcan be adjusted to 100% even though the bus voltage has anothervalue, It is merely necessary to set the autotransformer tap to correspond to the measured bus voltage. The taps are in 1% steps. Thefunction of the load adjusters on the board is analogous to that offeeder-voltage regulators on an actual power system.

The masterinstruments include a set for measuring scalar values anda set for measuring vector values. The scalar instruments consist of avoltmeter, an ammeter, and a wattmeter-varmeter. They are supplied with voltage and current through two negative-feed-backvacuum-tube amplifiers. The voltage ranges are 125 and 250 volts;the current ranges are 0.06, 0.2, 0.6, 2, and 6 amp.

The vector instruments are an ammeter (ranges 1 and 3 amp.) and avoltmeter (ranges 50 and 500 volts). Each is a two-coil dynamometer instrument, one coil of which is inserted in the network, and theother, consuming the greater part of the power for actuating the instrument, is fed from one or the other phase of the two-phase secondarywinding of an instrument phase shifter. For reading current or voltage in rectangular form, the phase shifter is set on the desired referenceposition (usually 0°), the field coils of the ammeter and voltmeter areconnected to one secondary winding of the phase shifter, and the "inphase" component is read; then the field coils are connected to theother winding, and the "quadrature" component is read. For reading

Ele

men

t

TA

BL

E3

Cm

cUIT

EL

EM

EN

TS

OF

WE

ST

ING

HO

US

EA

-C.

NE

TW

OlU

tC

ALC

ULA

TOR

Inst

alle

dat

Eas

tP

itts

burg

hin

1942

*

Nu

mb

erf

Ran

geSi

zeof

Ste

psT

oler

ance

Max

imum

Safe

Val

ues

~

Gen

erat

ors.

....

....

..1

8..

....

....

....

....

....

....

...

Vol

tage

regu

lati

onV

olta

ge.

....

....

...

....

....

....

0-40

0vo

lts.

..S

moo

thun

der

2%.

Ph

ase.

....

....

....

..

Un

lim

ited

....

Sm

ooth

Gen

.rea

ctor

sO

OW

R/X

)18

()-4

()()

ohm

s{~='i

~~::

::::

}:H

!%of

sett

ing..

Rea

ctor

...

....

....

...

....

....

..0-

2,40

0oh

ms.

Sm

oo

thj

.

Cap

acit

ors.

....

....

...

48

...

....

...

Q-4

.10

p.f.

....

O.O

I,."

f.

Au

totr

ansf

orm

ers.

....

.36

80

-12

4.5

%..

.0

.5%

.

Mut

ualre

acto

rs..

....

.3

6..

....

....

..

Lin

eim

peda

nces

...

...

Res

isto

r.

Rea

ctor

.L

oad

impe

danc

es.

Res

isto

r.

152

.

0-39

9h

{( o

-IOO

)0

.2oh

m..

..}

1 101

ftt·

oID

S..

.(1

00-3

99)

1o

hm

...

.±

2~/0

0B

em

g..

0-30

0o

hm

s...

Sm

oo

tht.

....

....

..±

1!%

ofse

ttin

g..

48.

O3

990

hm{

(0--

1,00

0)

2oh

m..

..}

1 101

ftt

·..

....

...

...-,o

s.

(1,0

00-3

,990

)10

ohm

±2

/00

seln

g..

±1!

%of

sett

ing

..

400

volt

s,3

amp.

3.0

amp.

or~

volts

220

volt

s

{

(0-9

.8oh

ms)

3.0

amp.

(10

-99

.8oh

ms)

2.3

amp.

(100

-399

ohm

s)1

.2am

p.

3.0

amp.

or22

0vo

lts

220

volt

s

1(0-8

ohm

s)3

.0am

p.(1

0-98

ohm

s)2

.3am

p.(1

00-2

98oh

ms)

1.2

amp.

l(300

-998

ohm

s)0

.48

amp.

(1,0

00-3

,990

ohm

s)0

.23

amp.

3.0

amp.

or22

0vo

lts

220

volt

s22

0vo

lts,

3.0

amp.

3.0

amp.

00 ~ ~ ~ ~ o Z ~ Z trj~ ~ o ;.0 ~ 00

*Rep

laci

ngca

lcul

ator

bu

ilt

in19

29.

Has

two

inst

rum

ent

desk

s.tT

he

num

ber

ofci

rcui

tel

emen

tsva

ries

ondi

ffer

ent

calc

ulat

ors

buil

tby

the

sam

em

anuf

actu

rer.

22

~Reactance

valu

esar

eob

tain

edby

aco

mbi

nati

onof

fixe

dai

r-ga

pre

acto

rsin

seri

esw

ith

ava

riab

leai

r-ga

pre

acto

r.

PROCEDURE FOR USING CALCULATING BOARD 75

current or voltage in polar form, the field coils are connected to the"inphase" winding, the phase shifter is turned until the instrumentreads zero, and the angle is read on the phase-shifter dial; the fieldcoils are then connected to the "quadrature" winding, and the magnitude of current is read on the ammeter, or the voltage on the voltmeter.§§

The master instruments are connected to the desired circuit elementby means of a selector consisting of a compact group of twenty-fivepushbuttons. When the operator pushes one button in each of threecolumns, relays are actuated that connect the selected circuit unit tothe metering bus.

The instruments, the circuit selector, and the controls for the motorgenerator set are mounted on a desk which is separate from the rest ofthe board.

Procedure for using calculating board. After the data on the powersystem to be studied have been assembled (see check list on p. 62 ofthis chapter), the following steps must be taken:

1. Choose a suitable base or scale for representing the powersystem on the calculating board.

2. Convert the data to this base or scale.3. Assign board units to the various circuits.4. Connect the board units.5. Set the resistors, reactors, and capacitors.6. Adjust the operating conditions.7. Take readings.8. Convert the readings to system values.

Some of these steps deserve further discussion.Choice of base. The voltage scale ordinarily is such that the nominal

voltage of the power system is represented by the nominal voltage ofthe calculating board. The current scale should be nearly as large asit can be without subjecting any of the board units to overcurrent (seemaximum safe currents in Tables 2 and 3). If the current scale is toosmall, current and power cannot be read accurately in lightly loadedcircuits. A satisfactory scale can be obtained by expressing systemquantities on a megavolt-ampere base in accordance with the followingtabler'"

§§Early models of the Calculator had the vector instruments only. Active andreactive power was found by multiplying the voltage magnitude by the inphase andquadrature components of current.


Maximum Megavolt-Amperesin Any Circuitor Generator

oto 3030 to 6060 to 120

120 to 300300 to 600600 and above

Megavolt-Ampere Baseto be Used for

System Quantities

102040

100200400

System quantities expressed in per cent on the selected megavoltampere base are equal to the corresponding calculating-board quanti ...ties in per cent of the board base. The selection of the base is notcritical because of the wide range of impedances and instrument scaleson the board. To check the suitability of the base selected, the lowestline impedance and highest load impedance (lightest load) may becompared with the range of available board impedances, and the currents of the most heavily loaded circuits should be compared with thecurrent-carrying capacities of the assigned units. If necessary, twoor more generators or line- or load-impedance units can be connectedin parallel to carry larger currents, or two load-impedance units can beconnected in series to represent very small loads. If there are a fewunusually heavy loads, it is better to represent them by two or moreload units in parallel than to use a higher impedance scale. Loadssmaller than 2% of the chosen megavolt-ampere base should be combined with other loads or neglected, because their influence on theproblem is negligible.

If, during a transient-stability study, some circuits are overloadedpart of the time, for example, while a fault is on the system, the basevoltage of the board may be reduced temporarily, impedances beingleft as they were.

Conversion of the data to the chosen base has already been discussed.The resistance of each reactor (taken as 4%of the reactance setting onthe Westinghouse board, or 5% on the General Electric board) shouldbe subtracted from the desired resistance to obtain the proper settingof the series resistor.

A circuitdiagram should be prepared on which are entered the impedances to be used and the assignedcircuit units on the calculating board.Each circuit should be assigned a positive.direction, which either isindicated on the diagram by an arrow pointing toward, or a dot at, the"positive" end of each circuit, or is covered by a blanket rule that theupper or left-hand end of each circuit is positive. Negative ends ofshunt branches (generators, line capacitances, loads, and autotransformers) are connected to the ground, or neutral, bus.

PROCEDURE FOR USING CALCULATING BOARD 77

Connections are made with attention to proper polarity by puttingplugs to be connected together into adjacent jacks. On the Westinghouse board the positive end of each circuit element has a red cord, andthe negative end, a black cord; on the General Electric board theyellow cord is positive, and the green cord, negative.

Setting impedances and capacitances to the values shown on thecircuit diagram is a simple operation. Positions of series-parallelswitches and the like should be checked at the same time.

Checking of all the foregoing steps by another person is recommended because much time can be wasted by mistakes that are notdiscovered until after many readings have been taken.

Adjustment of initial operating conditions. Generators, loads, autotransformers, and synchronous condensers must be adjusted to represent the desired condition of the power system before the occurrenceof a fault.

First, all generators should be set at the same phase angle and at zerovoltage. The power may then be turned on, and the generator voltages may be raised gradually to their normal value, whilethe generatorcurrents are watched for abnormal values caused by mistakes in settingup the network. The loads and terminal voltages of the generatorsare then adjusted approximately to the desired values by means of thephase and voltage controls.

Bus voltages at other points are adjusted as desired by autotransformers or by capacitors or generator units representing synchronouscondensers. If a generator unit is used, its power must be adjusted tozero.

Load impedances, it is assumed, were set initially to draw the required active and reactive power at an estimated voltage. The important loads are now readjusted for the actual load voltage. If loadadjusters are available, they are used for this purpose; otherwise theresistor and reactor (or capacitor) must be readjusted with the aid ofthe master wattmeter-varmeter.

Generating units, condensers, and other elements should now begiven a final, accurate adjustment, which is checked with the masterinstruments.

Limitations of system. Sometimes the desired operating conditionscannot be obtained on the board. Unless a mistake has been made inconnecting or setting the board units, such failure indicates that thedesired condition is one that is impossible on the actual power system.After adjusting the operating conditions, one should determine whetherall bus voltages are within suitable limits and whether any equipmentis overloaded either on the board or on the power system that it rep-


resents. If any conditions are unsatisfactory, consideration shouldbe given to changing either the operating conditions or the networkitself, for example, by adding new lines, transformers, generators,condensers, reactors, or phase-shifting transformers.

Readings. A complete set of readings is recommended at the outsetin order to check the network. Voltage and phase angle of each busand active and reactive power at each end of each circuit are usuallyread and recorded. The readings are multiplied by suitable factors toconvert them to system values which are then marked on a one-linecircuit diagram. The following checks may be made: The algebraicsum of the active power of all circuits on the same bus should be zero;likewise the algebraic sum of the reactive power. The difference ofpower at the two ends of a line may be taken to see whether the [2Rloss is reasonable; a similar test may be made for reactive power and12X. Perhaps the most valuable check, however, is for a personfamiliar with the power system being studied to see whether all resultsappear reasonable.

During the progress of a transient-stability study the only readingsneeded are the power outputs of the generators. Sometimes, however,additional readings are taken, for example, to investigate the operationof protective relays.

Changes of connections. The power to the board should be turnedoff before changing any connections. After any important change ofconditions, such as the addition or removal of a short circuit, the phase.and magnitude of the internal voltages of the generators should bechecked and readjusted if necessary.

Fault close to generator. When a short circuit is placed on or near theterminals of a generator, the current of that generator is high, and thepower factor is low. These conditions are not conducive to goodaccuracy of the wattmeter. Furthermore, the resistance of the reactors representing the generator and other low-resistance circuits between generator and fault may not truly represent those resistances.Under these conditions it is preferable, instead of reading the wattmeter, to compute the generator power as [2R, using the measuredcurrent and the best obtainable data on the resistance. Conservative conclusions regarding stability are reached by assuming this resistance to be zero.

Algebraic solution of networks: power-angle equations. Convenient as the calculating board is for stability studies, it is nevertheless desirable to know how to solve networks algebraically. For atwo-machine system algebraic solutions are usually so simple thatnothing is gained by using a calculating board, and for three- or four-

POWER-ANGLE EQUATIONS 79

[9al[9b]

n. . .

3Network

o

2

machine systems algebraic solutions are still feasible and may beresorted to when a board is not available.

Figure 7 is a schematic representation of a general n-machine network as it presents itself in most power-system network problems. Therectangle represents a passive (impedance) linear network of any form,to which there are applied n external voltages, representing theinternal voltages of the synchronous machines. One side ofeach voltage source is connectedto a common or neutral terminal(0); the other sides are connected to various points of thenetwork, numbered from 1 to n.

FIG. 7. An electric power networkThe voltages E1, E2, • • • En com- represented by a passive linear networkmonly represent voltages behind having external e.mJ.'s connected to it.transient reactances, in whichcase the transient reactances are to be regarded as parts of the net...work inside the rectangle. Let the currents flowing into the networkat the terminals 1, 2, 3, · · · n be calledIj, 12, 13, • • • In.

The vector power (P + jQ) supplied to the network by any machinemay be found by 'multiplying the conjugate of the vector voltage bythe vector current. II II In symbols,

PI + jQI = EII I

P2'+ jQ2 = E2I 2

[9cl

where the bar over the E's denotes their conjugates. Furthermore,since the network has been assumed linear, the currents supplied bythe various machines may be written as linear functions of the appliedvoltages, thus:

n

II = YIIEI + YI2E2 • • • + YlnEn = E YlkEk [lOa]k=l

n

" "Then capacitive reactive power (leading current) is positive, and inductivereactive power (lagging current) is negative, in accordance with the A.S.A. convention. The opposite convention, however, is widely used in power-system studiesand will probably be adopted by the A.S.A. In this case the conjugate of thevector current should be multiplied by the vector voltage.


n

In = YnlEl + Yn2E2 · · · + rnnEn = E YnkEk [10c]k=l

The V's are complex numbers known as the terminal self- and mutualadmittances of the network, or as the driving-point and transfer admittances. If the two subscripts are alike, Y is a self-admittance or driving-point admittance; if unlike, a mutual or transfer admittance.The meaning of these admittances is revealed by a consideration ofwhat happens to eqs. 10 if all the applied voltages except one are madeequal to zero. By letting all voltages except E l become zero, we obtain:

11 = YllEl, 12 = Y21E1, ••• In = YnlEl [11]

permitting us to define the Y's as follows: Y1l is the ratio IlfEl whenall voltages except E1 are zero; Y21 is the ratio 12/ E1 when all voltagesexcept El are zero; and so forth. By letting all voltages except E2

become zero, we can similarly define Y12, Y22, • • • Yn2. These definitions point out how the admittances can be determined by measurement. We shall see how they can be determined by calculation.Before doing so, however, let us turn back to eqs. 9, giving the vectorpower output of each voltage source, and substitute into them eqs. 10.We obtain

PI + jQ1 = E1Y11E1 + E1Y12E2 • • • + E1Y1nEn

[12a]

[12b]

and so forth. These are vector equations. For stability calculationsit is preferable to have scalar equations involving the machine displacement angles o. Therefore let us substitute into eqs. 12. thefollowing:

s, = E 1 fJJ , E2 = E 2&, ... En = En& [13]

or, for the conjugates,

E1 = ElL -81, E2 = E2 / -82, ... En =En / -8n [14]Also let

Y11 = Y11 / S11, Y12 = Y12 / Sl 2, Y21 = Y2l / S21, Y22 = Y22/ S22

[15]

DETERMINATION OF TERMINAL ADMITTANCES 81

and so forth. Equations 12 become

PI + jQI = E12YI1/ e l l + EIE2Y12Le12 - 81 + 82· • •

+ E1EnYlnL81n - 81 + 8nn

= E E1Ek Y l k / 81k - 81 + 8kk=1

P2 + jQ2 = E2EIY21/ e 21 - 82 + 81 +E22Y

22/ 8 22 · • •

+ E2EnY2n / e2n - 82 + 8nn

= L E2EkY2k/e2k - 82 + 8kk=l

[16a]

[16b]

and so forth. Using the relation (which defines the notation),

!.:P = cos cP + j sin cP

and taking the real part of each side of the equations, we obtain

PI = E12Yl l cos 811 + E1E2Y12 cos (812 - 81 + 82) • • •+ ElEnYln cos (Oln - 81 + 8n)

n

= L E1EkYIk cos (elk - 81 + 8k) [17a]k=l

P2 = E2EIY2l cos (821 - 82 + 81) + E22Y

22 cos 822 • • •+ E2En Y 2n cos (e2n - 82 + 8n )

n

= E E2EleY2k cos (82k - 82 + 8le) [17b]k=l

and so forth. Because these equations give the electric power outputof each machine (power input to the network) as functions of theangular positions of all the machines, they may be called power-angleequations. Inasmuch as this chapter deals primarily with networks,the symbol P has been used for electric power instead of Pu, which wasused in Chapter II with the same meaning.

Algebraic solution of networks: determination of terminal admittances. Equations 17 suffice for finding the power output of allmachines when the voltage magnitude and phase (angular displacement) of all machines are known. To obtain such equations, however,we must know how to obtain each Y = Y/8 for the given network.

A clue to a method of doing this is found by considering an nterminal passive linear network having no nodes (junctions) exceptthe terminals. Such a network has, in general, an impedance elementconnected between each pair of terminals (including the neutral), asshown in Fig. 8 for a network of four terminals, including neutral (a


three-machine system). In. this network there are six impedanceelements. For a network of m terminals (or n = m - 1 machines~~)

there are, in general, m(m - 1)/2 = (n + 1) n/2 elements. If anyof these elements are lacking, they may be considered to be present aselements of zero self-admittance (that is, infinite impedance or opencircuit). If two or more elements are in parallel, they may be replaced by an equivalent single element having an admittance equalto the sum of the admittances of the several parallel elements. Fromthis viewpoint any network may be converted to a network having oneand only one element connected between each pair of nodes and havingm(m - 1)/2 elements. This form of network is called a mesh circuit.

oFIG. 8. A four-terminal passive network having one and only one element con

nected between each pair of terminals.

The machine currents in the network of Fig. 8 may be written aslinear functions of the machine voltages and the element self-admittances (lower-case y's), by noting that each machine or terminalcurrent is the sum of several element currents, each of which may bewritten as the product of the element admittance and the voltageacross the element. Thus,

11 = Yl2 (E1 - E2) + Y1S (E1 - Es) + YOlE1 [I8a]

12 = Y12(E2 - E1) + Y2s(E2 - Es) + Y02E2 [18b]. .

13 = Yls(Es - E1) + Y23(Ea - E2) + YosEs lI8e]

Regrouping the terms by E's,

11 = (YOI +Y12 +Yls)E1+ (-YI2)E2 + (-Yls)Ea [I9a]

12 = (-YI2)E1 + (Y02 + Y12 + Y23)E2 + (-Y2s)Ea [19b]

13 = (-Yls)E1 + (-Y2a)E2 + (Yoa + Y13 + Y23)Es [Igel

~'If there are no loads or other shunt branches in the network, then m = n.

NETWORK REDUCTION 83

[20a]

[20b]

[20c]

By comparing these with the type equations (eqs. 10, rewritten forn == 3 as eqs. 20, which follow)

11 = Y I IE1 +Y 12E2 + Y l aEa

12 = Y21El + Y 22E2 + Y 2sEa

l a = Y 3l E I + Y32E2 + YasEa

the terminal admittances are determined as

(Y1l = YOI + Y12 + Y13 [21a]

self-admittances Jl Y 22

= Y02 + Y12 + Y23 [21bJ

Y 33 = Y03 + Y13 + Y2a [21c]

(Y12 = Y2l = -Y12 [22a]

mutual admittances tY 13 = Y3l = -Y13 [22b]

Y23 = Y32 = -Y23 [22c]

That is, each terminal self-admittance is the sum of the element admittances of all the elements connected to that terminal. Each terminalmutual admittance (between two given terminals) is the negative ofthe element admittance of the element connected between the giventerminals,

Algebraic solution of networks: network reduction. Wenow havethe terminal self- and mutual admittances in terms of the elementself-admittances of a network having no nodes except the terminals.In general, the network will have other nodes. These extra nodes,however, may be eliminated by a process known as network reduction.To establish the process, suppose that the extra nodes are initiallyregarded as extra terminals. Suppose, for example, that the networkhas five nodes, of which one is to be eliminated. The terminal voltampere equations for five terminals (neutral and four others) are:

11 = YllEl +Y12E 2 + Y13E 3 + Y14E 4: [23a]

12 = Y 21E1 +Y22E 2 + Y 23E3 + Y 24E4: [23b]

13 = YalEt +Ya2E 2 + YaaEa + Ya4E4 [23c]

14 = Y4lEI +Y42E2+Y 43E3 + Y 44E4: [23d]

Now suppose that node 4 is to be eliminated. It can be eliminatedonly if it has no external connections; that is, if, when considered asa terminal, it is open-circuited. Hence


Solving for E4 and substituting the expression in place of E4 in eqs.23 give

E4

= _ Y41E1 + Y42E2 + Y43E3

Y44

which may be written in the standard form:

11 = Y11'E1+ Y12'E2+ Yls'Es

12 = Y21'E1+ Y22'E2 + Y2s'Es

Is = Ys1'El + YS2'E2+ Ys3'Es

[25]

[26a]

[26b]

[26c]

[27a]

[27b]

[27c]

in which the Y"s are the new terminal admittances. These newterminal admittances are related to the old ones by

y -, Y Y14Y41[28]11 = 11 ---

Y44

Y12, = Y12 - Y14Y42

[29]Y44

Y13, = Y13 - Y14Y43

[30]Y44

Y22, = Y

22- Y24Y42 [31]

Y 44

Y23, = Y

28_ Y24Y43 [32]

Y 44

Y33, = Y

33_ Y34Y43 [33]

Y44

or, in general, by

Yj k' = Yj k - Yj 4Y4k [34]Y44

wherej = 1,2,3; k = 1,2,3.


[35a]

[35b]

[35e]

[35d]

[36a]

[36b]

[36e]

[36d]

[36e]

[36f]

[37a]

[37b]

[37e]

[38a]

[38b]

[3Sc]

Now let us find the relations between the elements of the old andnew networks. Assume that both old and new networks are of thestandard type having one and only one element between each pair ofterminals. Use relations like those of eqs, 21 and 22, namely:

Yll = YOI + Y12 + Y13 + Y14

Y22 = Y02 + Y12 + Y23 + Y24

Y 33 = Y03 + Y13 + Y23 + Y34

Y44 = Y04 + Y14 + Y24 +Y34

Y12 = Y 21 = -Y12

Y 13 = Y31 = -Y13

Y 14 = Y41 = -Y14

Y23 = Y32 = -Y23

Y 24 = Y42 = -Y24

Y34 = Y43 = -Y34

Yl t ' = YOl' + Y12' + Y13'

Y 22' = Y02' + Y12' + Y23'

Y , '+ '+ '33 = Y03 Y13 Y23

Yt 2' = Y21' = -YI2'

Y ' Y' ,13 = 31 = -Y13

Y23' = Y32' = -Y23'

Substitute these relations (eqs. 35 to 38 inclusive) into eqs. 28 to 34inclusive. This substitution works out most easily for the mutualterminal admittances, eqs. 29, 30, and 32. For example, substitutionin eq. 29 gives:

, (-YI4) (-Y24)-Y12 = -Y12 -

Y04 + Y14 + Y24 + Y34or, changing signs,

, + Y14Y24Y12 = Y12

Y04 + Y14 + Y24 + Y34

In similar manner, we obtain

, + Y14Y34Y13 = Y13

Y04 + Y14 + Y24 + Y34

, + Y~Y34Y23 = Y23

Y04 + Y14 + Y24 + Y34

[39]

[40]

[41)


[42]

The equations for terminal self-admittances require a little moremanipulation but. give similar results. For example, substitution ineq. 28 gives

, , , ( -YI4) ( -YI4)YOI + Y12 + Y13 = YOI + Y12 + Y13 + Y14 - + + +

Y04 Y14 Y24 Y34

Subtraction of eqs. 39 and 40 from this gives

, + Y14(Y14 + Y24 + Y34)YOI = YOI Y14 -

Y04 + Y14 + Y24 + Y34

+ Y14(Y04 + Y14 + Y24 + Y34 - Y14 - Y24 - Y34)= YOI

Y04 + Y14 + Y24 + Y34

+ Y04Y14= YOIY04 + Y14 + Y24 + Y34

Similarly

In general

, + Y04Y24Y02 = Y02

Y04 + Y14 + Y24 + Y34

, + Y04Y34Y03 = Y03

Y04 + Y14 + Y24 + Y34

, Yj4Yk4Yjk = Yik +-4--

L Yi4i=O

[43]

[44]

[45]

Equation 45 may be interpreted as follows: Every element of the newor reduced networkis the result of paralleling the corresponding element ofthe old network with an element arising from a star-mesh conversion.The admittance of the mesh element arising from the star-mesh conversion is given by the last term of eq. 45. The star-mesh conversionwith somewhat simplified notation is shown in Fig. 9. Figure 9a showsa four-point star which is to be converted into a four-terminal mesh(Fig. 9b), thereby eliminating the node S. The mesh-element admit-tances in terms of the star-element admittances are:

YIY2 [46a] Y2Y3 [46d]Y12 =- Y23 =-L:y L:yYIY3 [46b] Y2Y4 [46e]Y13 =- Y24=-L:y L:yYIY4 [46c] Y3Y4 [46f]Y14 =- Y34 =-L:y L:y

whereLY ~ Yl + Y2 + Y3 + Y4 [47]


If the star has n points, similar formulas hold, but the expressionfor ~y has n terms. If n = 3, we have a Ydelta conversion. Ifn = 2, we have a simple series combination.

Only if n = 3 is there always a unique inverse conversion, that is,from delta to Y. In general, the conversion from mesh to star forn > 3 isnot possible. If n = 2, the conversion can be made but is notunique; that is, a single element can be split into two elements in seriesin an infinite number of ways.

By one star-mesh conversion, one node is eliminated. By successivestar-mesh conversions, as many nodes as desired may be eliminated.For present purposes all nodes except the terminals are to be eliminated.

2 2

4 4(0) Star circuit (b) Mesh circuit

FIG. 9. Star-mesh conversion.

The formula for star-mesh conversion is simpler when expressed interms of admittances than in terms of impedances. Furthermore, theparallel combinations of the elements resulting from such a conversionwith other elements are made more easily by working with admittancesthan with impedances. For these reasons, in addition to the fact thatwe want terminal admittances in our final expressions for power, itis preferable to work with admittances rather than impedances whilereducing a network, even though the use of impedances is, perhaps,more customary.

The process of network reduction may be summarized as follows,assuming the impedances of the elements to be given:

1. Make all the possible series combinations; also make parallelcombinations of elements having equal impedances.

2. Convert impedances to admittances.3. Eliminate a node by star-mesh conversion, giving preference to

a node with the least number of elements connected to it.4. Make parallel combinations of the new elements resulting from

the conversion and the old elements.


5. Repeat steps 3 and 4 until all nodes except the terminals havebeen eliminated.***To prepare for the power calculations required in a stability study,

perform the following additional step, which is not strictly a part ofthe process of network reduction:

6. Compute the terminal admittances from the element admittances by using equations like 21 and 22.

All these steps must be repeated for a number of different networkconditions. Thus, if the disturbance causing the machines to swing isa fault, the network must be solved for the pre-fault condition, thefault condition, and, unless the fault is sustained, for the post-fault orcleared condition. If fault clearing is accomplished by sequentialopening of two or more breakers, the network must be solved for twoor more fault conditions. Also, if different runs are to be made fordifferent fault locations, different lines connected or disconnected, orother different conditions, network solutions must be made for eachcondition. On a calculating board the required changes in the network for these different conditions are made very simply and quickly.It is clear that the use of a calculating board is desirable in multimachine stability studies.

Determination of initial operating conditions. Even after determining the network terminal admittances for the pre-fault, fault, andpost-fault conditions, we are still not ready to begin calculation of theswing curves, for the initial operating conditions must first be determined. Specifically, we must find the values of magnitude and phaseof the internal voltages of all the synchronous machines. If thesevalues were given, it would be simple to substitute them into the powerequations (like eqs. 17) for the pre-fault condition and thus. to calculate the pre-fault power output of each machine, which is also thepower input to the machine.

However, the initial conditions are not usually known in such terms.Usually the power outputs of the machines are known or assumed; theremaining information may consist of the reactive power outputs atthe terminals (not behind transient reactance), the power factor,terminal voltage, current, or some mixture of such quantities. Occasionally conditions somewhere in the network other than at themachine terminals are given: for instance, voltage at a certain substation bus, or power or current in a certain line. From such data,sometimes insufficient and sometimes conflicting, the initial values of

***Although any network can be reduced by the exclusive use of star-mesh conversions and parallel combinations, in many networks the reduction can be hastened by the use of 6-Y conversions and series combinations.

INITIAL OPERATING CONDITIONS 89

voltage behind transient reactance and of angular positions must bedetermined as well as is possible. As a rule, this can be done only bycut-and-try methods, which are likely to be very laborious if the network is solved algebraically. Even with a calculating board cut-andtry procedures must be adopted, but the work goes much faster whena board is used than when algebraic solutions are relied upon.

lunt Fisk Deering Patten Murphy0.30 0.10 0.08 0.12

A 0.30 0.10 0.08 0.12 B

0.100.16

0.25 0.10 0.05 Dyche

Thorne Ward 0.05

FIG. 10. Three-machine system of Example 1. Line reactances are given in perunit on lOO-Mva. base.

After the initial power outputs, angles, and internal voltages havebeen found, the calculation of the swing curves may be begun.

EXAMPLE 1

The layout of a 60-cycle three-phase three-machine system is given inFig. 10. The reactances of the lines, expressed in per unit on a lOO-Mva.base, are marked on the figure. Line resistances may be neglected. Dataon the generators and on the initial generating-station outputs and bus volt-

TABLE 4

DATA FOR EXAMPLE 1

Steam Number Rating of H Initial InitialXd' Station Bus

Station or of Each Unit (p.u.) (Mj. perOutput Voltage

Hydro Units (Mva.) Mva.)(Mw.) (p.u.)

Lunt Hydro 3 35 0.35 3.00 80 1.05Murphy Steam 4 75 0.21 7.00 230 ,1.00Wieboldt Steam 2 50 0.18 8.00 90 1.00

ages are given in Table 4. All loads may be neglected except those atMurphy and Wieboldt, which are each 200 Mw. at unity power factor.


Find the initial phase and magnitude of the voltage behind transientreactance (Xd') of each generator.

Solution.

Outline.

1. The network between the busses at Lunt, Murphy, and Wieboldtwill be reduced to its simplest form (a delta).

2. The known magnitudes and assumed phase angles of the voltages ofthese three busses will be substituted into the power equations, and theangles will be varied until the calculated values of power input to the network agree with the given generating-station outputs less local load.

3. The current supplied to the network at each terminal will be foundfrom the known terminal 'voltages and admittances.

4. The generators at each station will be combined by paralleling theirimpedances to give one equivalent machine at each station. The internalZl drop of each equivalent machine will be added to the terminal voltageto find the phase and magnitude of the internal voltage.

Node to be retained.

Network element withadmittance.

Element entering intostar-mesh conversion.Element resulting fromstar-mesh conversion.

•@

1.21

I , I I

1. Network reduction. The symbols which will be used are given in Fig.11. First the network of Fig. 10 is simplified by making the obvious series

and parallel combinations and by omittingNode to beeliminated, the open branch from Ward to Thorne.

The result is shown in Fig. 12a, in whichthe names of the stations are abbreviatedto their initial letters. The values of impedance in per unit are encircled. Thesevalues of impedance are converted to admittance values, which are not encircled.The impedance and admittance angles,

FIG. 11. Symbols used in net- which are 900

and -900

respectively, arework reduction. omitted for brevity.

By means of alternate star-mesh conver-sions and parallel combinations of elements, all nodes are eliminated exceptthe three (L, M, and Wi) which are to be retained. The order in whichthey are eliminated is not particularly important; Wa is eliminated first.The star elements radiating from Wa are cross-hatched in Fig. 12a. Theresulting mesh elements are shown by broken lines in Fig. 12b. The starmesh calculations are as follows:

E Y= 10.0+ 5.0+ 10.0 = 25.010.0 X 5.0

YFD = YWiD = 25.0 = 2.0

. = 10.0 X 10.0 = 4 0YFW~ 25.0 ·

One parallel combination is made between Wi and D, with the result shownin Fig. 12c.


We~~+-+-+of+a+++-+~5.00

(e)

L (Q][) F

• 6.66

~

8.08

L 2.00 M@------e, ~ ,

~, '},.~,'J'$', . ~~.

• Wi

40 X 42 = 16.15104

20 X 42

104

2 X 42---= 0.81

104

(h)

L 6.66 F 6.35 M@lllllllq(g) ~o~ te'~

• Wi

D

pF6.66

(b)

(d)

L.>----~------...-(

( i ) e,~.

• WiFIG. 12. Reduction of the network of Fig. 10 (Example I).

Next, node D is eliminated, with results as shown in Fig. 12d. Thecalculations are as follows:

1: Y = 40.0 + 20.0 + 2.0 + 42.0 = 104.0

40 X 20 = 7.70104

40 X 2 = 0.77104

20 X 2 = 0.385104

Four parallel combinations are made, with results as shown in Fig. 12e.


Node P is now eliminated, with results as shown in Fig. 12/. The calculations are as follows:

E y = 11.5 + 24.4 + 14.3 = 50.2

11.5 X 24.4 = 5.58 11.5 X 14.3 = 3.2850.2 50.2

24.4 X 14.3 = 6.9450.2

Three parallel combinations are made, with results as shown in Fig. 12g.Next, node F is eliminated, with the result shown in Fig. 12k. One

parallel combination is made, resulting finally in the Ll network of Fig. 12i.2. Determination of bus voltage angles. The terminal admittances are

calculated from the element admittances as follows:

YLL = 2.00/- 90°+ 2.55/- 90° = 4.55/- 90° p.u.

YMM = 2.0 /-90° + 25.5/-90° = 27.5/-90° p.u,

Yww = 2.6 /-90° + 25.5/-90° = 28.1(-90° p.u,

YLM = YML = -2.00/-90° = 2.00/90° p.u,

YLw = YWL = -2.55/-90° = 2.55/90° p.u,

YMw = YWM = -25.5/-90° = 25.5/90° p.u.

The power inputs to the network are equal to the generator outputs lesslocal load.

P L = 0.80 - 0 0.80 p.u.

PM = 2.30 - 2.00 = 0.30 p.u.

Pw = 0.90 - 2.00 = -1.10 p.u.

The bus voltages are:

EL = 1.05&

EM = 1.00/5M

Ew = 1.00/0 (taken as reference phase)

The power equations are:

PL = EL2y LL cos eLL + ELEMYLM cos (eLM - OL + OM)

+ ELEwYLw cos (8LW - OL+ ow) (a)

PM = EM2YMM cos OMM + EMELYZ.fL cos (eML - OM + OL)+EMEwYMwCOS(aMW-OM+OW) (b)

Pw = Ew2Yww cos Oww+ EwELYWL cos (aWL - Ow + OL)

+ EwE-tllYWM cos (SWAt - Ow + OM) (c)


Because there are no losses in the network, only the first two equationsneed be used. Because eLL = eMM = 8ww = -90°, the self-admittanceterms vanish. eLM = eLW = 8ML = eMW = 90°, and we may use therelation, cos (900

- x) = sin z, Also noting that Ow = 0, we have:

PL ~ ELEMYLM sin (OL - OM) + ELEwYLW sin OL (d)

PM = ELEMYLM sin (OM - 01,) + EMEwY MW sin OM (e)

Substituting numerical values, we obtain:

0.80 = 2.10 sin (OL - OM) + 2.68 sin OL

0.30 = 2.10 sin (OM - OL) + 25.5 sin OM

(J)

(g)

These equations must be solved by trial for OL and OM. Because of the largecoefficient of sin OM we may be sure that OM is small. As a first trial, letOM = o. Equation J then becomes

0.80 = (2.10 + 2.68) sin OL = 4.78 sin OL

sin OL = 0.80 = 0.1675· OL = 9.60

4.78 '

Substitution of this value of OL in eq. g gives

0.30 ~ 2.10 sin (-9.6°) + 25.5 sin OM

25.5 sin OM = 0.30 + 2.10 X 0.1675 = 0.30 + 0.352 = 0.652

sin OM = 0.0256; OM = 1.50

Substitution of OL = 9.60 and OM = 1.5° into eqs. Jand g gives for the righthand members:

PL = 2.10 sin 8.10 + 2.68 sin 9.60 = 2.10 X 0.141 + 2.68 X 0.167

= 0.296 + 0.448 = 0.744 (Should be 0.80.)

PM = 2.10 sin (-8.1°) + 25.5 sin 1.5°

= - 0.296 + 0.652 = 0.356 (Should be 0.30.)

PL will be increased by increasing OL and PM by increasing OM. Since PLis too small and PM too large, OL should be increased and OM decreased.Try OL = 10.3°, OM = 1.4°.

PL = 2.10 sin 8.90 + 2.68 sin 10.3°

= 2.10 X 0.1547 + 2.68 X 0.1788

= 0.324 + 0.479 = 0.803 (Should be 0.800.)

PM = -0.324 + 25.5 sin 1.40

= - 0.324 + 25.5 X 0.0244

= -0.324 + 0.622 = 0.298 (Should be 0.300.)


These values are close enough to the correct values. The bus voltages aretherefore

EL = 1.05/10.3° = 1.032+ jO.188 p.u.

EM = 1.00 /1.4° = 1.000+ jO.024 p.u,

Ew = 1.00 /0.00 = 1.000+ jO.OOO p.u,

3. Determination of currents.

ILW = YLW (EL - Ew) = - j2.55 (0.032+ jO.188)= 0.479 - jO.082 p.u,

IMW = YMW (EM - Ew) = -j25.5 X jO.024 = 0.612+ jO.OOO p.u,

ILM = YLM (EL - EM) = -j2.00 (0.032+ jO.162)= 0.324 - jO.064 p.u,

The terminal currents of the network are

IL = ILw + I LM = 0.803 - jO.146 p.u.

1M = IMW - ILM = 0.288+ jO.064 p.u,

Iw = -ILW - IMW = -1.091 + jO.082 p.U.

The load currents, calculated by the formula

I = p -t. jQE

are2.00LO

1M ' = L - 0 = 2.00L1.4° = 2.000+ jO.049p.u,1.00 -1.4

2.00i!2I w' = -----;n = 2.00i!2 = 2.000+ jO.OOO

1.00~

The generator currents are equal to the currents supplied to the networkplus local load:

At L, IA = I L = 0.803 - jO.146 p.u,

At W, IB = 1M +1M ' = 2.288+ jO.113 p.u,

At W, Ie = Iw + Iw' = 0.909 + jO.082 p.u,

4. Determination of generator internal voltages. The generator reactances,which were given in per unit on the basis of the generator ratings, must beconverted to per unit 011 lOO-Mva. base.


100XA = 0.353 X 35 = 0.333 p.u,

100XB = 0.21 4 X 75 = 0.070 p,u,

100Xc = 0.18--= 0.180 p.u.

2X 50

ZA = iO.a3a p.u,

ZB = jO.070 p.u,

Zc = iO.180 p.u.

Each internal voltage is the vector sum of the terminal (bus) voltage and theinternal ZI drop.

EA = EL + ZAIA = 1.032+iO.I88+ jO.333 (0.803 - jO.146)= 1.032+ jO.188 + 0.049+ jO.268 = 1.081 + jO.456= 1.17/23.0 0 p.u, Ans.

EB = EM+ ZBIB = 1.000+ jO.024 + jO.070 (2.29+ jO.11)= 1.000 +jO.024 - 0.008+ iO.160 = 0.992+ jO.184= 1.01/10.4° p.u, Ans,

Ec = Ew+ZeIc = 1.000+ jO.OOO + jO.180 (0.909 + jO.082)= 1.000 - 0.015+ jO.164 = 0.985+ jO.164= 1.00 /9.5° p.u, Ans.

Check.

PA = EA·IA= 1.081 X 0.803+ 0.456 (-0.146) = 0.868 - 0.067= 0.801 p.u, (Should be 0.80.)

PB = EB·IB = O. 92 X 2.29 + 0.184 X 0.113 = 2.27+ 0.021= 2.291 p.u, (Should be 2.30.)

Pa = Ec·Ie = O. 85 X 0.909+ 0.164 X 0.082 = 0.895+ 0.013= 0.908 p.u, (Should be 0.90.)

EXAMPLE 2

Find the terminal admittances of the network of the three-machine systemof'Example 1, including the loads and the direct-axis transient reactances ofthe machines, (a) when a three-phase short circuit is present at point X,Fig. 10, and (b) after the short circuit has been cleared by opening both endsof the line.

Solution. (a) The short circuit at X is electrically equivalent to a shortcircuit on the Patten bus. In the network .reduction of Example 1, thePatten bus (node P) was preserved until the stage shown in Fig. 12e. Theapplication of the short circuit can be represented by connecting node P tonode 0 (the neutral), a node not shown in Fig. 12 because the network reduced in Example 1 had no shunt elements. Hence P may be relabeled


as O. The admittances of the three equivalent machines and of the two loadsmust be added to the network. All admittances should now be expressed ascomplex numbers, because the load admittances do not have the same angleas do the line and machine admittances. A lOO-Mva. base will be used, asin Example 1.

The machine impedances in per unit on a 100-Mva. base, as found inExample 1, and the corresponding admittances are as follows:

Machine Station Impedance AdmittanceA Lunt 0.33 /900 3.00/-90°B Murphy 0.07 /900 14.3/-900

C Wieboldt 0.18 /900 5.55(-90°

The admittances of the loads at Murphy and at Wieboldt are each2.00~ p.u.

After the machine and load admittances are added, the network of Fig. 12ebecomes that of Fig. 13a.

A series combination is then made between A and F; eliminating node L,and parallel combinations are made between M and 0 and between Wiand O. The computation is as follows:

3.00/ - 90° X 6.66/ - 90°A-F: 9.66~ = 2.07/-90°

M-o: 2.0 - j24.4 = 24.5/-85.3°

Wi-O: 2.0 - j14.33 = 14.5/-82.0°

The resulting network is shown in Fig. 13b. Next, node F is eliminated by astar-mesh conversion, as follows:

E y = (2.07+ 0.77+ 11.5+ 4.81) /-90° = 19.15/-90°

A-M: 2.07 X 0.77 L-900 = 0.083L-90019.15

A-D: 2.07 X 11.5 /-900 = 1.24/-90019.15

A-Wi: 2.07 X 4.81 /-900 = 0.520L-90019·15

M-O: 0.77 X 11.5 L-900 = 0.46/-900

19.15 -

M-Wi: 0.77 X 4.81 L-900 = 0.193L-900

19.15

o-Wi: 11.5 X 4.81 L-900 = 2.88/-90°19.15


B--~~@

(c)

(h)

·CFIG. 13. Reduction of the network of Fig. 10 with a short circuit at X (Example 2).


O-B:

The resulting network is shown in Fig. 13c. Parallel combinations are made,as follows:

M-Wi: (16.15+ 0.19) /-90° = 16.34/-90°

M-o: (2.0 - j24.4) - jO.5 = 2.0 - j24.9 = 25.0/-85.4°

Wi-o: (2.0 - j14.3) - j2.9 = 2.0 - j17.3 = 17.3/-83.3°

The resulting network is shown in Fig. 13d. Node M is eliminated, asfollows:

I: y = (2.0 - j24.9) - j16.3 - j14.3 - jO.1 = 2.0 - j55.6= 55.6/-88.0°

25.0/-85.4° X 14.3~

/

0 = 6.42/-87.4°55.6_-88.0

25.0/-85.4° X 16.34~O-Wi: / ° = 7.34/-87.4°

55.6_-88.0

14.3/-90° X 16.34/-90°B-Wi: L ° = 4.20L-92.0°

55.6_-88.0

A-o:

A-B:

0.083L- 90° X 25.0/- 85.4° /_ -- --- = 0.037,--_-_87_.4_°

, 5.5.6/- 88.0°

0.083/-90° X 14.3/-90°

/

0 = 0.021/-92.0°55.6 _-88.0

0.083~X16.34~ /. A-Wi: / ° =·0.024_-92.0°

55.6_- 88.0 -----

The resulting network is shown in Fig. 13e. Parallel combinations aremade, as follows:

O-Wi: 17.3/-83.3° + 7.34/-87.4° == 2.0 - j17.2 + 0.3 - j7.3

= 2.3 - j24.5 = 24.6/ - 84.6°

A-o: -j1.24 - jO.04 = -jl.28 = 1.28/-90°

A-Wi: -jO.52 - jO.02 = -jO.54 = 0.54/-90°


The resulting network is shown in Fig. 13/. Node Wi is eliminated, asfollows:

L: y = 24.6/-84.6°+ 5.55/- 90°+ 0.54/- 90°+ 4.20/ - 92.0°

= (2.3 - j24.5) - j5.6 - jO.5+ (-0.2 - j4.2)= 2.1 - j34.8 = 34.9/-86.7°

0.54/- 90° X 24.6/ - 84.6°A-O: = 0.38/-87.9°

34.9/ - 86.7°

0.54L- 90° X 4.20/-92.0°A-B: = 0.065/-95.3°

34.9/-86.7°

A-C.. 0.54/-90° X 5.55/-90° / 0

= 0.086 - 93.334.9/-86.7°

420/-92.0° X 24.6/-84.6° _ / °B-D: / ° - 2.96_-89.9

34.9_ -86.7

4.20/-92.0° X 5.55/ - 90°B-C: = 0.668/- 95.3°

34.9/-86.7°

O- C.. 24.6L-84.6° X 5.55/-90°

= 3.91/-87.9°34.9/-86.7°

The resulting network is shown in Fig. 13g. Parallel combinations aremade as follows:

A-B: 0.065L-95.3° + 0.021/-92.0°

= -0.005 - jO.065 - 0.001 - jO.021 = -0.006 - jO.086= 0.086/-94°

A-O: 1.28/-90° + 0.38/-87.9° = -jl.28 + 0.02 - jO.38

= 0.02 - jl.66 = 1.66/-89.2°

B-O: 6.42/-87.4° + 2.96/-89.9° = 0.29 - j6.42 + 0.01 - j2.96

= 0.30 - j9.38 = 9.38/88.2°

The result is the final network shown in Fig. 13k.Calculation oj terminal admittances.

YAA = 1.66/-89.2° + 0.086/-94.0° + 0.086/-93.3°

= 0.02 - jl.66 - 0.01 - jO.09 - 0.01 - jO.09= 0.00 - jl.84 = 1.84/-90.0° p.u. Ans.


YBB = 9.38/-88.2° + 0.086/-94.0° +0.668/-95.3°

= 0.30 - j9.38 - 0.01 - jO.09 - 0.05 - iO.66= 0.24 - j10.13 = 10.14/-88.7° p.u, Ans.

Yee = 3.91/-87.9° + 0.086/-93.3° + 0.668/-95.3°

= 0.15 - j3.91 - 0.01 - jO.09 - 0.05 - iO.66= 0.09 - j4.66 = 4.66/ -88.9° p.u, Ans.

YAB = - 0.086/ - 94.0° = 0.086/86.0° p.u. Am.

YAc = -0.086/-93.3° = 0.086/86.7° p.u. Ans.

YBc = -0.668/-95.3° = 0.668/84.7° p.u. Ans,

(b) To clear the fault, the line from Dyche (D) to Patten (P) is disconnected from the network. In the network reduction of Example 1 this line

A B

• •

FIG. 14. Network resulting from the reduction of the network of Fig. 10 withthe line from Patten to Dyche disconnected (Example 2).

was preserved until the stage shown in Fig. 12c. We may start, therefore,with that diagram and disconnect the line, then further reduce the network,preserving terminals L, M, and Wi. At this point the load and machineadmittances may be attached, as in part a of this solution, and the networkfurther reduced, preserving terminals A, B, C, and O. The details of calculation will not be given here. The resulting network is shown in Fig. 14.The element admittances are:

YAB = 1.12/-100.5° = -0.205 - il.IO p.u.

YBC = 3.06/ -102.6° = -0.66 - j2.98 p.u.

YeA = 0.502/ -100.8° = -0.093 - jO.493p.u.

YAO = 0.335/-10.8° = +0.330 - jO.063 p.u,

YBO = 2.48/-10.7° = +2.44 - jO.46 p.u.

yeo = 1.11/-10.9° = +1.08-jO.21 p.u,


The terminal admittances are:

YAA = YAB + YCA + YAO = +0.03 - jl.66 = 1.66/-89.0° p.u, Ans.

YBB = JAB + YBC + YBO = +1.58 - j4.54 = 4.81/-70.7° p.u, Ans.

Yeo = YBC + YCA + soo = +0.33 - j3.68 = 3.69 /-84.9° p.u, Ana.

YAB = -YAB = +0.205 + jI.IO = 1.12/79.5° p.u. Ans.

YBC = -YBC = +0.66 + j2.98 = 3.06 /77.4° p.u. Ans.

YCA = -YCA = +0.093 + jO.493 = 0.502/79.2° p.u, Ans.

EXAMPLE 3

Calculate and plot swing curves for the three-machine system of Fig. 10with initial conditions as in Example 1, assuming a three-phase short circuitto occur at point X, Fig. 10, and to be cleared (a) in 0.40 sec. and (b) in 0.35sec. Carry the curves far enough to determine whether the system isstable.

Solution. The initial conditions, as calculated in Example 1, are:

EA = 1.17

EB = 1.01

Eo = 1.00

~A = 23.0°

8B = 10.4°

8c = 9.5°

PuA = PiA = 0.80

PUB = PiB = 2.30

Puc = PiC = 0.90

The terminal admittances with the fault on, as calculated in Example 2,part a, are:

Y AA cos 8AA = 0

Y BB cos eBB = 0.24

Y cc cos ecc = 0.09

YAB = 0.086 aAB = 86.0°

Y BC = 0.668 eBC = 84.7°

Y CA = 0.086 eCA = 86.7°

The terminal admittances with the fault cleared, as calculated in Example2, part b, are:

Y AA cos 8AA = 0.03

Y BB cos eBB = 1.58

Yoo cos ecc = 0.33

YAB = 1.12

YBC = 3.06

YCA = 0.502

eAB = 79.5°

aBC = 77.4°

eCA = 79.2°


The power-angle equations have the form of eqs. 17. ·Substituting theforegoing numerical values, we obtain the following expressions, whichapply while the fault is on:

PuA = (1.17)2 X 0 + 1.17 X 1.01 X 0.086 cos (86.0° - ~A + ~B)

+ 1.17 X 1.00 X 0.086 cos (86.7° - ~A + ac)

= 0.10 cos (86.0° - OA + ~B) + 0.10 cos (86.7° - OA + Oc)

PuB = (1.01)2 X 0.24 + 1.01 X 1.17 X 0.086 cos (86.0° - ~B + aA)+ 1.01 X 1.00 X 0.668 cos (84.7° - OB + ~c)

= 0.24 + 0.10 cos (86.0° - OB + OA) + 0.68 cos (84.7° - OB + oc)

Puc = (1.00)2 X 0.09 + 1.00 X 1.17 X 0.086 cos (86.7° - ~c + ~A)

+ 1.00 X 1.01 X 0.668 cos (84.7° - Oc + ~B)

= 0.09 +0.10 cos (86.70- ~c + ~A) + 0.68 cos (84.70

- ac + OB)

and the following expressions, which apply-after the fault has been cleared:

PuA = (1.17)2 X 0.03 + 1.17 X 1.01 X 1.12 cos (79.5° - ~A + OB)

+ 1.17 X 1.00 X 0.502 cos (79.2° - ~A + ~c)

= 0.04 + 1.30 cos (79.5° - OA +DB) + 0.59 cos (79.2° - DA + DC)

PuB = (1.01)2 X 1.58 + 1.01 X 1.17 X 1.12 cos (79.5p- OB +DA)

+ 1.01 X 1.00 X 3.06 cos (77.4° - DB + ~c)

= 1.61 + 1.30 cos (79.5° - OB + DA) + 3.09 cos (77.4° - OB + oc)

Puc = (1.00)2 X 0.33 + 1.00 X 1.17 X 0.502 cos (79.2° - Dc + OA)

+ 1.00 X 1.01 X 3.06 cos (77.4° - ~c + OB)

= 0.33 + 0.59 cos (79.2° - Oc + OA) + 3.09 cos (77.4° - Oc + OB)

The inertia constants of the machines may be calculated from the datagiven in Example 1 by the formula (eq. 54, Chapter II)

M= GH = GH180!' 180 X 60

where M = inertia constant in per unit.G = station rating in per-unit apparent power.H = kinetic energy at rated speed in megajoules per megavolt

ampere of rating.I = frequency in cycles per second.

M = 3 X 35 X 3.00 = 2.92 X 10-4A 100 X 10,800

M = 4 X 75 X 7.00 = 19 45 X 10....B 100 X 10,800 ·

M = 2 X 50 X 8.00 = 7.41 X 10-4c 100 X 10,800


The time interval ilt for point-by-point calculations will be taken as0.10sec. Then

_il_t2

= 0.010 = 34.3MA 2.92 X 10-4

_1l_t2 = 0.010 = 5.14

MB 19.45 X 10--4

_1l_t2

= 0.010 = 13.5Me 7.41 X 10-4

The computations of power are carried out in Tables 5, 6, and 7, and thecomputations of swing curves in Tables 8, 9, and 10. The results of thecomputations of swing curves, namely, the angular positions of the three

- Cleared 0.40sec. ))

--- Cleared 0.3) sec. tis;:;!/~.

~,~

;} /'

"b\J,

/' '9;i'f-. .4 ~~

/ jf

J' jr

) V Vr

~~~-

400

oo 0.5 0.8

lime (seconds)

FIG. 15. Swing curves of the three-machine system of Fig. 10 with a three-phaseshort circuit at point X near Patten bus cleared in 0.35 sec. and in 0.40 sec. Machines Band C stay within about 20 of each other. See also Fig. 16. (Example 3.)

machines as functions of time, and also the angular differences between eachpair of machines, are listed in Table 11. The swing curves are plotted inFigs. 15 and 16. Figure 15 shows the angular position of each machine withrespect to a reference axis rotating at normal speed. All three machinesincrease their speeds on account of the drop in output during the short circuit and the assumption of constant input. Machines Band C swing practically together and are therefore represented by the same curve. Figure 16shows the angular position of machine A with respect to machine B.

If the short circuit is cleared in 0.40 sec., the system is unstable: machine

TA

BL

E5.

CO

MP

UT

AT

ION

OF

Pu

A(E

XA

MP

LE

3)

eA

B8

AB

eA

B-

lJA

Bco

sP

AB

mP

AB

aA

ClJ

AC

SA

C-

aAC

cos

PA

Cm

PA

CP

AA

Pu

A

86.0

12.6

73.4

0.28

60

.10

0.03

86.7

13.5

73.2

0.28

90.

100.

030

0.06

"2

0.2

65.8

0.41

0It

0.04

"2

0.9

65.8

0.41

0It

0.0

4"

0.08

u42

.143

.90.

721

It0.

07"

42.3

44.4

0.71

4"

0.0

7"

0.14

"76

.19

.90.

985

"0.

10"

75.4

11.3

0.98

1"

0.1

0"

0.20

"11

9.9

-33

.90.

830

II

0.08

"11

7.8

-31

.10.

856

It0.

09"

0.17

79.5

119.

9-4

0.4

0.76

21

.30

0.99

79.2

117.

8-3

8.6

0.78

20.

590

.46

0.0

41.

49"

153.

8-7

4.3

0.27

1"

0.35

"15

1.4

-72

.20.

306

"0

.18

"0.

57"

133.

1-5

3.6

0.59

3"

0.77

"13

1.9

-52

.70.

606

"0

.36

"1.

17It

127.

6-4

8.1

0.66

8It

0.87

"12

8.8

-49

.60.

648

"0

.38

It1

.29

"99

.6-2

0.1

0.93

9"

1.22

"10

2.4

-23

.20.

919

"0

.54

"1.

80

TA

BL

E6.

CO

MP

UT

AT

ION

OF

PU

B(E

XA

MP

LE

3)

eA

BO

AB

eA

B+

OA

Bco

sP

AB

mP

BA

aB

Ca

BC

eB

C-

aBC

cos

PB

Cm

PB

CP

BB

Pu

B

86

.012

.698

.6-0

.15

00

.10

-0.0

284

.70.

9'83

.80.

108

0.6

80

.07

0.2

40.

29"

20.2

106.

2-0

.27

9"

-0.0

3"

0.7

84.0

0.10

4"

0.0

7"

0.28

"42

.112

8.1

-0.6

17

"-0

.06

"0

.284

.50.

096

"0

.07

"0.

25"

76.1

162.

1-0

.95

2"

-0.1

0"

-0.7

85

.40.

080

"0

.05

"0.

19"

119.

920

5.9

-0.9

00

"-0

.09

"-2

.186

.80.

056

"0

.04

"0.

1979

.511

9.9

199.

4-0

.94

31.

301-

1.23

77.4

-2.1

79.5

0.18

23.

090

.56

1.61

0.94

"15

3.8

233.

3-0

.59

8"

-0.7

8"

-2.4

79.8

0.17

7u

0.5

5II

1.38

,e13

3.1

212.

6-0

.84

2It

-1.0

9"

-1.2

78.6

0.19

8"

0.61

"1.

13,e

127.

620

7.1

-0.8

90

"-1

.16

"1

.276

.20.

238

"0

.74

II1.

19"

99.6

179.

1-1

.00

0"

-1.3

0"

2.8

74.6

0.26

6II

0.8

2II

1.13

~ 00 o ~ ~ ~ ~ o z o ~ ~ ~ ~ ~ o ~ ~ 00

TA

BL

E7

CO

MP

UT

AT

ION

OF

Pu

C(E

XA

MP

LE

3)

9A

.CO

AC

9A

C+

OA

Cco

sP

AC

mP

CA

eB

CO

BC

eB

C+

OB

Cco

sP

BC

mP

CB

Pce

Pu

C

86.7

13.5

100.

2-0

.17

70.

10-0

.02

84.7

0.9

85

.60.

077

0.6

80.

050.

090

.12

u20

.910

7.6

-0.3

16

"-0

.03

"0

.785

.40.

080

"0.

05"

0.11

"42

.312

9.0

-0.6

29

"-0

.06

"0

.28

4.9

0.08

9"

0.06

"0.

09tc

75

.416

2.1

-0.9

52

"-0

.10

"-0

.784

.00.

104

"0.

07"

0.0

6"

117.

820

4.5

-0.9

10

"-0

.09

"-2

.182

.60.

129

"0.

09"

0.0

979

.211

7.8

197.

0-0

.95

60.

59-0

.56

77.4

-2.1

75.3

0.25

43.

090.

780.

330

.55

"15

1.4

230.

6-0

.63

5"

-0.3

7"

-2.4

75.0

0.25

9"

0.80

"0

.76

"13

1.9

211.

1-0

.85

6u

-0.5

0"

-1.2

76.2

0.23

8"

0.74

"0

.57

u12

8.8

208.

0-0

.88

3"

-0.5

2"

1.2

78.6

0.19

8"

0.61

"0

.42

"10

2.4

181.

6-1

.00

0"

-0.5

9u

2.8

80

.20.

170

"0.

52"

0.2

6

.... !Z ~ ~ ~ ..... Z c o o !Z tj ..... t-3 ..... o ~ .- ~


TABLE 8

COMPUTATION OF 8A (EXAMPLE 3)

t PiA PUA PaA 34.3PaA a8A 8A(sec.) (p.u.) (p.u.) (p.u.) (elec. deg.) (elee. deg.) (elee. deg.)

0+ 0.80 0.06 0.74 23.0Oavg. 0.37 12.7

12.70.1 " 0.08 -0.72 24.7 35.7

37.40.2 " 0.14 0.66 22.6 73.1

60.00.3 " 0.20 0.60 20.6 133.1

SO.60.4- " 0.17 213.70.4+ " 1.490.4 avg. u 0.83 -0.03 -1.0

79.60.5 " 0.57 0.23 7.9 293.3

87.50.6 380.8

80.60.4 0.80 1.49 -0.69 -23.6 213.7

57.00.5 u 1.17 -0.37 -12.7 270.7

44.30.6 " 1.29 -0.49 -16.8 315.0

27.50.7 " 1.80 -1.00 -34.3 342.5

-6.80.8 335.7

INITIAL OPERATING CONDITIONS

TABLE 9

COMPUTATION OF 8B (EXAMPLE 3)

107

t PiB PuB PaB 5.14PaB A8B 8s(see.) (p.u.) (p.u.) (p.u.) (elec. deg.) (elec. deg.) (elee. deg.)

0+ 2.30 0.29 2.01 10.4Oavg. 1.00 5.1

5.10.1 Ie 0.28 2.02 10.4 15.5

15.50.2 u 0.25 2.05 10.5 31.0

26.00.3 u 0.19 2.11 10.8 57.0

36.80.4- u 0.19 93.80.4+ u 0.940.4 avg. ce 0.56 1.74 8.9

45.70.5 u 1.38 0.92 4.7 139.5

50.40.6 189.9

36.80.4 2.30 0.94 1.36 7.0 3.8

43.80.5 It 1.13 1.17 6.0 137.6

49.80.6 II 1.19 1.11 5.7 187.4

55.50.7 " 1.13 1.17 6.0 242.9

61.50.8 304.4


TABLE 10

COMPUTATION OF 8c (EXAMPLE 3)

t Pie PuC PaC 13.5Pac t16c 6c(sec.) (p.u.) (p.u.) (p.u.) (elec. deg.) (elec..deg.) (elec. deg.)

0+ 0.90 0.12 0.78 9.5oavg. 0.39 5.3

5.30.1 H 0.11 0.79 10.'7 14.8

16.00.2 H 0.09 0.81 10.9 30.8

26.90.3 " 0.06 0.84 11.3 57.7

38.20.4- " 0.09 95.90.4+ H 0.550.4 avg. H 0.32 0.58 7.8

46.00.5 " 0.76 0.14 1.9 141.9

47.90.6 189.8

38.20.4 0.90 0.55 0.35 4.7 95.9

42.90.5 " 0.57 0.33 4.5 138.8

47.40.6 H 0.42 0.48 6.5 186.2

53.90.7 H 0.26 0.64 8.6 240.1

62.50.8 302.6

NETWORK REDUCTION BY CALCULATING BOARD 109

A goes out of step with machines Band C, although Band C stay together.If the short circuit is cleared in 0.35 sec., the system is stable. Both conditions are apparent from Fig. 16 at 0.6 sec. The critical clearing time is between 0.35 and 0.40 sec.

Ji 90E8co'is.en:a..oS!Qc< 0

)I I I

VCleared 0.40 see.~

l?~ ...... ...--..~;' "-

I~"

"'..'I \

J Cleared 0.35sec.-~\

V \/

"~V

o 0.5 0.8Time (seconds)

FIG. 16. Swing curve in terms of angular difference between machines A and B.(Example 3.)

TABLE 11

SWING-CURVE DATA (EXAMPLE 3)

t 8,4 6B 80 8AB 8AO 8BO(sec.) (elec. deg.) (elec. deg.) (elec. deg.) (elec. deg.) (elec. deg.) (elec. deg.)

0 23.0 10.4 9.5 12.6 13.5 0.90.1 35.7 15.5 14.8 20.2 20.9 0.70.2 73.1 31.0 30.8 42.1 42.3 0.20.3 133.1 57.0 57.7 76.1 75.4 -0.70.4 213.7 93.8 95.9 119.9 117.8 -2.10.5 293.3 139.5 141.9 153.8 151.4 -2.40.6 380.8 189.9 189.8 190.9 191.0 0.1

0.5 270.7 137.6 138.8 133.1 131.9 -1.20.6 315.0 187.4 186.2 127.6 128.8 1.20.7 342.5 242.9 2'40.1 99.6 102.4 2.80.8 335.7 304.4 302.6 31.3 33.1 1.8

Network reduction by use of calculating board. Sometimes, inorder to set up a network on a calculating board without exceeding thenumber of impedance units available, the network or a portion of it


must be reduced to an equivalent circuit. If the identities of only twoterminals of the network, including neutral or ground if any branchesare connected to it, need be preserved, the network can be reduced toone impedance element; if three terminals are to be preserved, it can bereduced to an equivalent ~ (or Y) circuit; if four terminals, to anequivalent six-element mesh circuit; and so on.

The reduction may be accomplished by calculation, as alreadydescribed, or it may be done very simply with the help of a calculatingboard. If the board method is to be used, the network is set up; one

(a) (b)

FIG. 17. Reduction of four-terminal network, set up on calculating board, to anequivalent four-terminal mesh circuit by voltage and current measurements.

terminal of it is connected to a power source, and the other terminalsare connected through jumpers to the neutral bus as shown in Fig. I7afor a four-terminal network. The applied voltage and the currentsleaving all other terminals are then measured in vector form. If theequivalent circuit (Fig. 17b) were set up, it would yield the same measurements. As there would be no currents in the short-circuited elements (shown by broken lines in Fig. 17b), each terminal currentmeasured would equal the current in one of the remaining elements.Hence the impedances of these elements are:

[48]

[49]

Now, if the voltage is applied between terminal 2 and terminals 1, 3,and 4 joined together, and if similar measurements are made, we have:

E2 E2 E2Z12 = - J Z23 = -, Z24 = -

11 13 14

COMBINING MACHINES 111

If similar measurements are made with the voltage applied to eachterminal in tum, every impedance' is determined twice (as indicatedfor Z12 above), thereby furnishing a check on the work.

If resistance and capacitance are neglected, or if all impedance anglesare assumed equal, a d-c. calculating board can be used in this mannerto determine the admittances in the power-angle equations, which arethen used in calculating the outputs of the machines. .

Combining machines. It is apparent from the foregoing discussionand examples that the amount of work involved in making a stabilitystudy increases tremendously as the number of synchronous machinesincluded becomes greater, especially if the network is solved algebraically. In order to save work, it is important to keep to a minimum thenumber of machines which are separately represented. Even on acalculating board this number must not exceed the number of generatorunits on the board. The number of machines may be reduced bycombining several machines which swing together or almost together toform a single equivalent machine.

If several machines were mechanically coupled (at such speed ratiosthat they ali generated the same frequency), they would be forced toswing together, that is, to have equal velocities and accelerations, eventhough they might not have equal angular positions. Since the inertiaconstant may be defined as the power required to produce unit angularacceleration, and since the power (or torque) required to produce equalacceleration of all the machines is the sum of the powers (or torques)required to accelerate the individual machines, the inertia constant ofthe group is the sum of the inertia constants of the individual machines.If the machines swing together even though not mechanically coupled,conditions in the network are the same as if they were mechanicallycoupled. Therefore, the inertia constant of the equivalent machine istaken as the sum of the inertia constants of the individual machines (referred to a common megavolt-ampere basettt if power is expressed inper unit). It matters not whether the machines are forced to swingtogether by close electrical coupling (low impedance between machines)or whether they merely happen to swing together in spite of being farapart electrically.

If the machines to be combined to form an equivalent machine areconnected in parallel at their terminals, then, by Thevenin's theorem,their effect upon the network is the same as if they were replaced by asingle source of e.m.f., equal to the open-circuit voltage of the group ofmachines, in series with a single impedance, equal to the impedance

tttAs stated in Chapter II, the inertia constant varies inversely as the megavoltampere base.


seen from the terminals when the e.m.f.'s of the machines are zero.Therefore the impedance oj the equivalent machine is a reactance equal tothe parallel combination of the reactances of the individual machines (referred, of course, to a common megavolt-ampere base if expressed inper unit). The equivalent e.m.f, is a sort of average of the e.m.f.'s ofthe individual machines. If the machines swing together, the equivalent e.m.f. is constant in magnitude and has the same frequency as thee.m.f.'s of the machines. The e.m.f. of the equivalent machine is suchthat the equivalent machine initially supplies to the network the sameactive and reactive power as the group of machines that it replaces.

If the machines to be combined are not in parallel at their terminals,their reactances cannot be combined. The fictitious points betweenthe reactance and the source of internal voltage of each machine areconnected together, however, thereby paralleling the several voltagesources and connecting together at one end all the reactances, theother ends of which go to different points of the network. The paralleled voltage sources are replaced by a single source, which, as before,is adjusted to deliver initially to the network the same active andreactive power as the sources which it replaces. The division of thispower among the several points of connection to the network generallywill be different from the original division. The reactances of theindividual machines may be left as part of the network set up on acalculating board; or, if desired, the network may be further reduced.On the calculating board it is easy to combine machines when conditions justify so doing and later to separate them again or to form newcombinations as may seem desirable.fj]

The only conclusive test of whether machines may be combined in agiven case without too much error in the swing curves or in the conclusions regarding system stability is to compute swing curves, firstwith the machines not combined, and then again with them combined,and to compare the results of the two computations. If the swingcurves obtained with the machines not combined show that a group ofmachines swing very nearly together, however, this evidence is sufficient for concluding, without running the combined swing curves, thatthe machines of the group may be combined with negligible effect onthe swing curves of the other machines. Neither of the foregoingcriteria for combining machines saves any work on the cases to which itis applied; thus their only value is to give the computer experiencewhich should develop his judgment on the circumstances under whichcombinations may be made. Therefore some further remarks are inorder.

tttAn example of this practice appears in Study 1, Chapter VII.


The likelihood that machines will swing together is increased by adecrease of impedance of the connections between them, by proximityin their initial angular positions, by similarity of their inertia constants,and by remoteness of the fault or source of the disturbance. It iscustomary to combine all the machines in the same station (unless thestation is operated sectionalized), even though the machines haveunlike ratings, impedances, inertia constants, or initial loadings. Twoor more stations which are connected together by low-impedance tiesmay likewise be combined. Frequently an entire metropolitan systemis represented by a single equivalent machine if the study concerns thestability of the connections between such a system and remote hydroelectric plants or other metropolitan systems.

EXAMPLE 4

The swing curves obtained in Example 3 show that machines Band Cswing very nearly together. Combine them to form a single equivalentmachine called D and compute swing curves of the resulting two-machinesystem for the same conditions as those which hold in Example 3.

Solution. The inertia constant of the equivalent machine is

MD = MB + Me = (19.5 + 7.4) 10-4 = 26.9 X 10-4 per unit

MA = 2.92 X 10-4, as before

The time interval tJ.t for point-by-point calculations will be taken as 0.10sec., as in Example 3.

ilt2- = 34.3, as before1t'IA

0.010 = 3.7226.9 X 10-4

The initial conditions for the three-machine system, as calculated inExample 1 and used in Example 3, are:

EA = 1.17

EB = 1.01

Ee = 1.00

0.04. = 23.0°

OB = 10.4°

Oc = 9.5°

PiA = 0.80

PiB = 2.30

PiC = 0.90

The conditions for machine A still hold. The input and pre-fault output ofmachine D is

PiD = PiB + Pie = 2.30 + 0.90 = 3.20

The values of EB and Ee are nearly equal, so the value of ED may be takenwithout serious error as equal to that of EB = 1.01. The value of OD willlie between the values of OB = 10.4° and oe = 9.5°, probably closer to the


value of OB becauseB is a larger machine than C. A weighted average couldbe used, thus:

~D = ~BPB + ~cPc = lOA X 2.30+ 9.5 X 0.90 = 10.1"PD 3.20

0.172/- 93.6°

YAD = YAB + YAC = 0.086/-94.0° + 0.086/-93.3°

= -0.006 - jO.086- 0.005 - jO.086= -0.011 - jO.172 = 0.172/-93.6°

YDO = YBO + yeo = 9.38/-88.2° + 3.91/-87.9°

= 0.30 - j9.38 + 0.15 - j3.91 = 0.45 - j13.29

= 13.3/-88.0°


YAA = -j1.84 = 1.84/-90.0°, as before

YDD = YAD + YDO = -0.01 - jO.17 + 0.45 - j13.29= 0.44 - j13.46 = 13.5/-88.1°

YAD = -YAD = -0.172/-93.6° = 0.172!86.4°


The power-angle equations are:

PuA = 1.17 X 1.01 X 0.172 cos (86.4° - 8A+ 8D)

= 0.203 cos (86.4° - DAD) = 0.203 sin (3.6° + 8AD)

PuD = (1.01)2 X 0.44 + 1.01 X 1.17 X 0.172 cos (86.4° + 8AD)

= 0.45 + 0.203 sin (3.6° - DAD)

Network reduction and power-angle equations, fault cleared. The reducednetwork of the three-machine system with the fault cleared is shown in Fig.

(0)

1.62/-100.6°

(6)

•oFIG. 19. Reduction of the network of Fig. 14 by joining terminals Band C to

form terminal D. Short circuit cleared. (Example 4.)

14 and is redrawn in Fig. I9a with terminals Band C joined to give terminalD. After making two parallel combinations, as follows, the network ofFig. 19b is obtained.

YAD = YAB + YAe = 1.12/-100.5° + 0.502/-100.8°

= -0.205 - j1.10 - 0.093 - jO.49 = -0.30 - j1.59= 1.62/-100.6°

fDO = YBO + yeo = 2.48/-10.7° + 1.11/-10.9°

= 2.44 - jO.46 + 1.08 - jO.21 = 3.52 - jO.67= 3.58/-10.7°


YA A = 0.03 - j1.66 = 1.66/-89.0°, as before

YDD = YAD + YDO

= -0.30 - jI.59 + 3.52 - jO.67 = 3.22 - j2.26

YAD = -YAD = -1.62/-100.6° = 1.62/79.4°

TA

BL

E12

CO

MP

UT

AT

ION

OF

Pu

AA

ND

Pu

D(E

XA

MP

LE

4)

SA

D,

BA

D9

AD

'+B

AD

sin

PA

Dm

PA

DP

AA

Pu

Ae'

-B

AD

sin

PD

Am

PD

AP

DD

Pu

D

11.0

12.7

23.7

0.40

21.

910.

770.

030.

80-1

.7-0

.03

01.

91-0

.06

3.26

3.20

3.6

12.7

16.3

0.28

10.

203

0.05

70.

000.

057

-9.1

-0.1

58

0.20

3-0

.03

0.45

0.42

"2

0.2

23.8

0.4

04

"0.

082

"0.

082

-16

.6-0

.28

6"

-0.0

6"

0.39

II41

.945

.50.

713

"0.

145

u0.

145

-38

.3-0

.62

0"

-0.1

3"

0.32

"75

.378

.90

.98

r"

0.19

9"

0.19

9-7

1.7

-0.9

49

u-0

.19

"0.

26II

118.

412

2.0

0.84

8"

0.17

2u

0.17

2-1

14

.8-0

.90

8"

-0.1

8"

0.27

10.6

11

8.4

129.

0-0

.777

1.91

1.48

50.

041.

525

-10

7.8

-0.9

52

1.91

-1.8

33.

281.

45u

151.

116

1.7

0.31

4"

0.60

"0.

64-1

40

.5-0

.63

6"

-1.2

2"

2.06

II13

0.2

140.

80.

632

"1.

21"

1.25

-11

9.6

-0.8

70

u-1

.66

"1.

62II

120.

713

1.3

0.75

1"

1.43

"1.

47-1

10

.1-0

.93

9"

-1.7

9"

1.49

cc81

.892

.40.

999

"1.

91"

1.95

-71

.2-0

.94

7"

-1.8

1"

1.47

... S ~ ~ ~ foo-

4 o Z o ~ z ~ ~ o ~ p:: 00

COMBINING MACHINES

The power-angle equations are:

PuA = (1.17)2 X 0.03 + 1.17 X 1.01 X 1.62 cos (79.4° - 6AD)

= 0.04 + 1.91 siu (10.6° +8AD )

PuD = (1.01)2 X 3.22 + 1.01 X 1.17 X 1.62 cos (79.4° + 8AD)

= 3.28 + 1.91 sin (10.6° - 8A D )

117

The computations of power are carried out in Table 12, and the computations of swing curves, in Tables 13 and 14. The angular positions of thetwo machines and the angular difference between them are tabulated asfunctions of time in Table 15. The results agree fairly well with those forthe three-machine system, Table 11, Example 3.

TABLE 13

COMPUTATION OF 8A (EXAMPLE 4)

t PiA PuA PaA 34.3PaA ~8A 8A(sec.) (p.u.) (p.u.) (p.u.) (elec. deg.) (elec, deg.) (elec, deg.)

0- 0.800 0.800 0.000+ " 0.057 0.743Oavg. 0.371 12.7 23.0

12.70.1 " 0.082 0.718 24.6 35.7

37.30.2 " 0.145 0.655 22.4 73.0

59.70.3 " 0.199 0.601 20.6 132.7

SO.30.4- u 0.172 213.00.4+ " 1.520.4 avg. 0.85 -0.05 -1.7

78.60.5 " 0.64 0.1e. 5.5 291.6

84.10.6 375.7

80.30.4 " 1.525 -0.725 -24.8 213.0

55.5.0.5 " 1.25 -0.45 -15.4 268.5

40.10.6 u 1.47 -0.67 -23.0 308.6

17.10.7 " 1.95 -1.15 -39.4 325.7

-22.30.8 303.4


TABLE 14

COMPUTATION OF aD (EXAMPLE 4)·

t PiD PuD PaD 3.72PaD AaD aD(sec.) (p.u.) (p.u.) (p.u.) (elec. deg.) (elec. deg.) (elee. deg.)

0- 3.20 3.20 0.000+ " 0.42 2.78oavg. 1.39 5.2

5.210.3

0.1 " 0.39 2.81 10.4 15.515.6

0.2 . " 0.32 2.88 10.7 31.126.3

0.3 " 0.26 2.94 10.9 57.437.2

0.4- " 0.27 94.6'(>.4+ " 1.450.4 avg. " 0.86 2.34 8.7

45.90.5 2.06 1.14 4.2

50.1140.5

0.6 190.6

0.4 " 1.45 1.75 6.537.2

94.6

0.5 " 1.62 1.58 5.943.7

138.3

" 1.49 1.71 6.449.6

187.90.6

0.7 " 1.47 1.73 6.456.0

243.962.4

0.8 306.3

TABLE 15

SWING-CURVE DATA (EXAMPLE 4)

t aA aD aAD(sec.) (elec. deg.) (elec. deg.) (elec. deg.)

0 23.0 10.3 12.70.1 35.7 15.5 20.20.2 73.0 31.1 41.90.3 132.7 57.4 75.30.4 213.0 94.6 118.40.5 291.6 140.5 151.10.6 375.7 190.6 185.1

0.5 268.5 138.3 130.20.6 308.6 187.9 120.70.7 325.7 243.9 81.80.8 303.4 306.3 -2.9

REFERENCES 119

Treatment of synchronous condensers. In the calculation of swingcurves a synchronous condenser should logically be handled in thesame way as a generator. Its power output is zero initially but notwhile it is swinging. When condensers are treated like generators, it isoften found that, although they swing with large amplitude andshorter period than the generators, because of their smaller inertiaconstants they have little effect on the swing of the generators. Therefore synchronous condensers are sometimes represented in transientstability studies on a calculating board by static capacitors (or byreactors if the condensers operate with lagging current). Each suchcapacitor is in series with a reactor representing the transient reactanceof the condenser. At each step of the swing calculation the capacitanceis readjusted so that its voltage, representing voltage behind transientreactance, is restored to the initial value. Thus the reactive power ofthe synchronous condenser is taken into account, but the active poweris disregarded. This procedure obviates the taking of power readingsand the calculation of the swing curve of the condenser. It affordsanother method, in addition to that of combining machines, of reducingthe number of machines considered in a stability study.

REFERENCES

1. Electrical Transmission and Distribution Reference Book, by Central StationEngineers of the Westinghouse Electric & Manufacturing Company, East Pittsburgh, Pa., 1st edition, 1942.

a. Chapter 16, "Power Transformers and Reactors," by J. E. Hobson and R.L. Witzke.

b. Appendix, Table 7, "Equivalent Circuits of Power and Regulating Transformers."

c. Chapter 3, "Characteristics of Aerial Lines," by Sherwin H. Wright and C.F. Hall.

d. Chapter 6, "Electrical Characteristics of Cables," by H. N. Muller, Jr.2. C. F. WAGNER and R. D. EVANS, Symmetrical Components, New York, Me-

Graw-Hill Book Co., 1933.a. Chapter VI, "Constants of Transformers."b. Chapters VII, VIII, IX, which discuss constants of transmission lines.c. Chapter X, "Constants of Cables."d. Appendix VII, "Characteristics of Conductors."

3. EDITH CLARKE, Circuit Analysis of A-C Power Systems, vol. I, New York,John Wiley & Sons, 1943.

a. Chapter VI, "Transmission Circuits with Distributed Constants."b. Chapter XI, "Impedances of Overhead Transmission Lines."c. Chapter XII, "Capacitances of Overhead Transmission Lines."

4. A. BOYAJIAN, "Theory of Three-Circuit Transformers," A.I.E.E. Trans.,vol. 43, pp. 508-28, February, 1924; disc., p. 529.


5. O. G. C. DAHL, Electric Circuits-Theory and Applications, vol. I,New York, McGraw-Hill Book Co., 1928. Chapter II, "Transformer Impedance and Equivalent Circuits."

6. F. M. STARR, "An Equivalent Circuit for the Four-Winding Transformer,"Gen. Elec. Rev., vol. 36, pp. 150-2, March, 1933.

7. L. C. AICHER, JR., "A Useful Equivalent Circuit for a Five-Winding Transformer," A.I.E.E. Trans., vol. 62, pp. 66-70, February, 1943; disc., p. 385.

8. J. E. CLEM, "Equivalent Circuit Impedance of Regulating Transformers,"A.I.E.E. Trans., vol. 58, pp. 871-3, 1939; disc., pp. 873-4.

9. J. E. HOBSON and W. A. LEWIS, "Regulating Transformers in Power-SystemAnalysis," A.I.E.E. Trans., vol. 58, pp. 874-83, 1939; disc., pp. 883-6.

10. L. F. BLUME, G. CAMILLI, A. BOYAJIAN, and V. M. MONTSINGER, Transformer Engineering, New York, John Wiley & Sons, 1938.

11. L. F. Woodruff, Principles of Electric Power Transmission, New York,John Wiley & Sons, 2nd edition, 1938. Equivalent T and r lines, pp. 112-5.

12. L. F. WOODRUFF, "Complex Hyperbolic Function Charts," Elec. Eng., vol.54, pp. 550-4, May, 1935; disc., p. 1002, Sept., 1935. The charts are also publishedin the book, Ref. 11.

13. C. A. STREIFUS, C. S. ROADHOUSE, and R. B. Gow, "Measured ElectricalConstants of 27Q-Mile 154-Kv. Transmission Line," A.I.E.E. Trans., vol. 63, pp.538-42, July, 1944; disc., pp. 1351-2.

14. DONALD M. SIMMONS, "Calculation of the Electrical Problems of Underground Cables," Elec. Jour., vol. 29, pp. 237-41, 283-7, 336-40, 395-8, 423-6,470-7, 527-30, May to November, 1932.

15. H. L. HAZEN, O. R. SCHURIG, and M. F. GARDNER, "The M.I.T. NetworkAnalyzer: Design and Application to Power System Problems," A.I.E.E. Trans.,vol. 49, pp. 1102-13, July, 1930; disc., pp. 1113-4.

16. H. A. TRAVERS and W. W. PARKER, "An Alternating-Current CalculatingBoard," Elec. Jour., vol. 27, pp. 266-70, May, 1930.

17. H. P. KUEHNI and R. G. LORRAINE, etA New A-C Network Analyzer,"A.I.E.E. Trans., vol. 57, pp. 67-73, February, 1938; disc., pp. 418-22, July, 1938.

18. H. A. THOMPSON, "A Stabilized Amplifier for Measurement Purposes,"A.I.E.E. Trans., vol. 57, pp. 379-84, July, 1938.

19. W. W. PARKER, "The Modern A-C Network Calculator," A.I.E.E. Trans.,vol. 60, pp. 977--82, November, 1941; disc., pp. 1395-8.

20. Westinghouse Electric & Manufacturing Company, Instruction Book onAlternating-Current Network Calculator. These books, furnished to purchasers ofthe calculators, are somewhat different for each model of the calculator.

21. General Electric Company, A-C. Network Analyzer Manual, PublicationGET-1285, Schenectady, 1945.

22. DAN BRAYMER, "Today's Network Calculators Will Plan Tomorrow'sSystems," Elec. Wld., vol. 125, pp. 52-4, January 5, 1946. Table I lists a-c.calculating boards existing or on order 'with ownership, date, frequency, number ofeach type of circuit unit, and name and title of man in charge.

PROBLEMS ON CHAPTER m1. By use of a T-to-r conversion verify the rule stated in the text (p, 61)

for apportioning a small tapped load between the two ends of the line.2. Write an expression for the impedance of a load in terms of its voltage

and vector power.

PROBLEMS 121

3. Verify the rule given (p. 58) for estimating the reactance of an autotransformer.

4. Find the terminal admittances of the network of Fig. 10, considered asa two-machine system, for a short circuit partially cleared by opening thecircuit breaker at the Patten end only.

5. Using the results of Probe 4 and Example 4, compute and plot swingcurves for determining to the nearest 0.05 sec. the critical time of openingthe breaker at Patten if the breaker at Dyche remains closed.

6. Work Example 2 for a new condition (c), short circuit partially clearedby opening the circuit breaker at the Patten end only.

7. Using the results of Probe 6 and Example 3, compute and plot swingcurves for determining to the nearest 0.05 sec. the critical opening time ofthe breaker at Patten to clear partially a three-phase short circuit at X ifthe breaker at Dyche remains closed.

8. Work Example 1 with the following changes: The two lines fromMurphy to Dyche (Fig. 10) are out of service. The initial generator outputs are: Lunt, 105 Mw.; Murphy, 195 Mw.; Wieboldt, 100 Mw.

9. Work Example 2 for the condition described in Probe 8.10. Calculate and plot swing curves for the three-machine system of

Fig. 10 with conditions as described in Probe 8 if a three-phase short circuitoccurs at point X and is cleared in 0.30 sec. by opening breakers at both endsof the line. The results of Probs. 8 and 9 are needed in the solution of thisproblem.

CHAPTER IV

THE EQUAL-AREA CRITERION FOR STABILITY

Applicability of the equal-area criterion. To determine whether apower system is stable after a disturbance, it is necessary, in general, toplot and to inspect the swing curves. If these curves show that theangle between any two machines tends to increase without limit, thesystem, of course, is unstable. If, on the other hand, after all disturbances including switching have occurred, the angles between the twomachines of every possible pair reach maximum values and thereafterdecrease, it is probable, although not certain, that the system is stable.Occasionally in a multimachine system one of the machines may stayin step on the first swing and yet go out of step on the second swingbecause the other machines are in different positions and react differently on the first machine.

In a two-machine system, under the usual assumptions of constantinput, no damping, and constant voltage behind transient reactance,the angle between the machines either increases indefinitely or else,after all disturbances have occurred, oscillates with constant amplitude.In other words, the two machines either fall out of step on the firstswing or never. Under these conditions the observation that themachines come to rest with respect to each other may be taken asproof that the system is stable. There is a simple graphical method,which will be explained in this chapter, of determining whether themachines come to rest with respect to each other. This method isknown as the equal-area criterion for stability. When this criterion isapplicable, its use wholly or partially eliminates the need of computingswing curves and thus saves a considerable amount of work. It isapplicable to any two-machine system for which the assumptionsstated above may be made.

The fact that the assumed conditions are not strictly true does notnecessarily invalidate the criterion. If the input to the machines is notconstant, but is changed by action of governors, the effect of such actiongenerally will not be appreciable until after the first swing and will thenbe in such direction as to aid stability. Whatever damping is presentwill reduce slightly the amplitude of the first swing and will reduce stillfurther the amplitude of subsequent swings.

122

APPLICABILITY OF EQUAL-AREA CRITERION 123

The effect of varying voltage behind transient reactance, Of, what isthe same thing, varying flux linkage of the field winding, deservesconsideration. Upon the occurrence of a fault the field current suddenly increases to the extent required to offset the increased demagnetizing reaction of the armature current and thereby to maintain constantflux linkage of the field circuit. If the machine does not have a voltageregulator, the field current ultimately decays back to its original value,equal to the exciter voltage divided by the field-circuit resistance; and,as it decays, the flux linkage also decays. The time constant of thedecay is of the order of 2 to 5 sec., and during the first swing the fluxlinkages do not decrease much in any machine which does not go out ofstep on that swing, If the fault is sustained for a long time, however,the flux linkages may be so much reduced that the system, althoughsurviving the first swing, will ultimately become unstable. Even if thefault is cleared rapidly, the opening of a line to clear it may decreasethe maximum synchronizing power and therefore increase the angulardisplacement required for a given power transfer and decrease the fluxlinkage for a given field current. Here, as well as for a sustained fault,it is possible to have the machines stay in step during the first swing butgo out of step later. If the system is stable on the first swing and alsostable in the ensuing steady state (under assumption of constant fieldcurrent),* it is reasonable to suppose that it will not be unstable at anyintermediate time; for there will probably be enough damping toreduce the amplitude of swing as fast as the flux decays.

If the machines have voltage regulators, the regulators will tend tomaintain constant terminal voltage, which would require an increaseof field flux linkages. With excitation systems of ordinary speed ofresponse, the regulator and exciter action is too slow to have an appreciable effect during the first swing but is fast enough to prevent lossof synchronism on subsequent swings. By the use of voltage regulatorsit is possible to preserve stability even in some instances when thesystem would be unstable on the basis of constant field current in thesteady state after clearing of the fault. t

From the foregoing discussion it may be seen that, if a two-machinesystem does not lose synchronism during the first swing, it is veryprobably stable, especially if the machines have voltage regulators, andalso that stability or instability on the first swing may be determinedwith good accuracy under the assumptions of constant input, no damping, and constant voltage behind transient reactance. The equal-area

*Steady-state stability is discussed in Chapter XV, Vol. III.[Field decrement and voltage-regulator action arc discussed in Chapters XII

and XIII, Vol. III.

-[2]

[3]

124 THE EQUAL-AREA CRITERION FOR STABILITY

criterion is a useful means of determining whether a two-machinesystem is stable under these assumptions.

The equal-area criterion is applicable to all two-machine systems,whether they actually have only two machines or whether they aresimplified representations of systems with more than two machines.Two-machine systems may be divided into two types, which will beconsidered in turn: (1) those having one finite machine swingingwithrespect to an infinite bus.] and (2) those having two finite machinesswinging with respect to each other.

One machine swinging with respect to an infinite bus. The swingequation of the finite machine is

d20M - = p = p. - P [1]dt2 a , u

where M is the inertia constant of the finite machine, and 8 is theangular displacement of this machine with respect to the infinite bus.

Multiply each member of the equation by 2do/Mdt:

2 d2~ d~ = 2 Pa d~dt2 dt M dt

or

:!. [(do)2] == 2 P a dodt dt M dt

Next multiply each side by dt, obtaining differentials instead of derivatives,

[4]

[5]

[6]

and integrate.

(do) 2 = ~ r Pad~dt M Jao

d~ = w' = '-2-r-aP-dodt \j M J; a

When the machine comes to rest with respect to the infinite bus-a,condition which may be taken to indicate stability-

w' = 0 [7]

tAn infinite bus is a source of voltage constant in phase, magnitude, and frequency and not affected by the amount of current drawn from it. It may be regarded as a bus to which machines having an infinite aggregate rating are connected or, in other words, as a machine having zero impedance and infinite inertia.A large power system often may be regarded as an infinite bus.

ONE MACHINE SWINGING TO INFINITE BUS 125

requiring thatr-J80

Pet do = 0 [8]

This integral may be interpreted graphically (Fig. 1) as the area undera curve of Pa plotted against 0 between limits 00, the initial angle, andOm, the final angle; or, since

[9]

the integral may be interpreted also as the area between the curve ofPi versus 0 and the curve of P u

versus o. The curve of Pi versus Pois a horizontal line, since Pi isassumed constant. The curve ofP u versus 0, known as a powerangle curve, is a sinusoid (derivedin the next section of this chapter) if the; network is linear andif the machine is represented by 8

08m 8

a constant reactance. The area,FIG. 1. The equal-area criterion for

to be equal to zero, must consist stability.of a positive portion AI, forwhich Pi > Pu , and an equal and opposite negative portion A 2, forwhich Pi < Pu. Hence originates the name, equal-area criterion forstability.

The areas A 1 and A?, may be interpreted in terms of kinetic energy.The work done on a rotating body by a torque T acting through anangle 0 - 00 is

[10]

and this work increases the kinetic energy of the body. The accelerating power Pa is proportional to the torque, under the previously madeassumption of nearly constant speed. Hence the work done on themachine to accelerate it, which appears as kinetic energy, is proportional to area At. When the accelerating power becomes negativeand the machine is retarded, this kinetic energy is given up; and, whenit is all given up, the machine has returned to its original speed. Thisoccurs when A 2 = At. The kinetic energies involved in this explanation are fictitious, being calculated in terms of the relative speed ratherthan the actual speed.

126 THE EQUAL-.AREA CRITERION FOR STABILITY

The power-angle equation for the case of one machine and an infinitebus follows directly from the power-angle equation for one machine of amultimachine system Ceq. 17a, Chapter III) if we let subscript 1 denotethe finite machine and subscript 2 denote the infinite bus, and if weput 01 = 0 and 02 = o.

Pu1 = E12 Y l l cos 811 + E 1E2Y12 cos (a12 - 0)

= Pc + PM sin (0 - 'Y) [11]

where Pc = E12Y11 cos all.PM = E lE2Y12.

E1 is the internal voltage of the machine.E2 is the voltage of the infinite bus.Ylt/8l1 and Y12/8 12 are terminal admittances of the net-

work between the machine and the infinite bus, as definedin Chapter III.

'Y = 812 - 90°.

The power-angle curve is, in general, a displaced sinusoid. It issimilar to the simple sinusoid

r, = PM sin 0 [12]

displaced upward by a distance Pc and to the right by a distance'Y. = a12 - 90°, as shown in Fig.2. (NOTE: For a network consisting of resistance and inductive reactance, 8 12 lies between90° and 180°, 'Y lies between 0and 90°, and all lies betweeno and - 90°. For a networkconsisting of inductive react

" ance only, 812 = 90°, 'Y = 0,and 811 = -90°.)

If the network consists ofreactance only, then eq. 11 reduces to eq. 12, and the power

FIG. 2. Power-angle curve of dissipative angle curve is an undisplacednetwork: a displaced sinusoid. The verti- sinusoid.cal displacement is Pa,and the horizontal

displacement is ~. Applications of the criterion.The use of the equal-area

method will be illustrated by applying it to two simple cases:

1. A sustained line fault.

APPLICATIONS OF THE CRITERION 127

2. A line fault cleared after the lapse of a certain time by thesimultaneous opening of the circuit breakers at both ends of the line.

The fault is assumed to occur at point X of the simple system ofFig. 3, which consists of a generator connected through a double-

)(Fault

®-1H:~-~t-eGenerator Infinite

bus

FIG. 3. Power system consisting of a generator connected through a doublecircuit line to an infinite bus. The equal-area criterion for stability of this system

is illustrated in Figs. 4, 5, and 6.

circuit line to an infinite bus. The input to the generator and thevoltage behind transient reactance are assumed constant.

1. Sustained line fault. The power-angle curves, giving the generator output versus displacement angle, are shown in Fig. 4 for two

p

Pml

Output, normalconditions

Pm2

o ~o

FIG. 4. The equal-area criterion applied to a sustained fault on the power systemof Fig. 3. The generator swings from the initial angle 80 to the maximum angle8m determined by equality of areas Al and A2• The system is stable when trans-

mitting power Pi.

conditions: (1) normal, and (2) faulted. The horizontal line at distance Pi above the axis represents the constant input. The initialoperating point is a at the intersection of the input and normal outputcurves. The initial displacement angle is ~o, and the initial relativeangular velocity is zero. When the fault is applied, the operatingpoint drops to b, directly below a on the fault output curve. The dis-


placement angle remains 00 at the instant of fault application. Thereis then an accelerating power, Pa = Pi - Pi; represented by thelength abo As a consequence the generator is accelerated, the displacement angle increases, and the operating point moves along thecurve from b toward c. As it does so, the accelerating power and theacceleration decrease, becoming zero at c. At this point, however, thespeed of the generator is greater than that of the infinite bus, and theangle 0 continues to increase. As it does so, P a becomes negative,representing retarding power. The speed diminishes until at point d,determined by the equality of area Al = abc and area A 2 = cde, itbecomes zero. Here the maximum angular displacement Om is reached.There is still a retarding torque; therefore the speed of the generator

PPm2

Pi t--~~~~II6IW.'-"e""'~"*""I.QaQ..

o

FIG. 5. Application of the equal-area criterion to finding the power limit of thepower system of Fig. 3 with a sustained fault. The input line is raised from itsposition in Fig. 4 until 8m reaches the intersection of the input line with the curve

of output, fault on.

continues to decrease, becoming less than that of the infinite bus. Thedisplacement angle 0 decreases, and the operating point moves from dthrough c toward b. The system is stable. The operating point wouldcontinue to oscillate between band d if there were no damping. Actually the oscillations diminish, and the operating point finally becomesestablished at c.

If the initial load on the generator were increased, as represented byraising the input line, areas Al and A 2 and the maximum angle Omwould increase. The greatest value which Pi could have without themachine going out of step during the existence of the fault would bethat value which makes Om occur at the intersection of the input curveand the fault output curve, as shown in Fig. 5. This is the criticalcondition in which both the speed and the acceleration become zerosimultaneously at angle Om. The value of Pi which makes this condition occur is the transient stability limit.

If the initial load were still larger, then area A2 in Fig. 5 would besmaller than area A 1. The generator would reach point e on the curve,


where the acceleration is zero, with the speed above normal. Consequently s would continue to increase, and, as it did so, the accelerating power would again become positive. The system would be unstable. In this case there is some retardation (between c and e), butnot enough to prevent loss of synchronism.

If input Pi were greater than Pm2, the maximum output with thefault on, there would be no retardation whatever; and, of course, thesystem would be unstable with a sustained fault.

p

Pma -----

Pm2

o 8

FIG. 6. The equal-area criterion applied to the power system of Fig. 3 for a faultcleared at angle oc.

2. Line fault with subsequent clearing. In this case three powerangle curves are needed: (1) for the normal or pre-fault condition withthe system intact, (2) for the fault condition, and (3) for the post-faultor cleared condition with the faulted line disconnected. These curvesare shown in Fig. 6.

As in case 1, the initial angle 00 is determined by the intersection ofthe input line and the pre-fault output curve (point a). Applicationof the fault causes the operating point to drop from a to b on the faultoutput curve, and the accelerating power causes it to move along thecurve from b to c. We may assume that, when point c is reached, thecircuit breakers open, clearing the fault. The operating point thenjumps up to e on the post-fault output curve and travels along thatcurve to j, where area A 2 = defg equals area At = abed.

With a cleared fault, as with a sustained fault, a higher input (andinitial output) would cause point f to move to the right until at the


stability limit f would coincidewith h. A still higher value of Pi wouldlead to instability.

Another factor which would cause f to move to the right is an increase in the time of clearing the fault, resulting in a larger clearingangle 6c• For any given initial load there is a critical clearing angle.If the actual clearing angle is smaller than the critical value, the systemis stable; if larger, the system is unstable.

Ordinarily, the clearing angle 6c is not known directly; instead, theclearing time (sum of relay time and breaker time) is known. Todetermine the clearing angle from a knowledge of the clearing time, theswing curve must be determined up to the time of clearing. A pre-calculated swing curve may be used for this purpose. (See Chapter V.)The use of swing curves is not entirely eliminated but is reduced toa minimum by the equal-area criterion.

3. Other applications of the equal-area criterion are left to the reader.(See Problems at the end of this chapter.)

EXAMPLE 1

By using the equal-area criterion, find the critical clearing angle for theconditions of Example 4, Chapter II. From the swing curve for a sustainedfault find the critical clearing time.

Solution. The network (shown in Fig. 8, Chapter II) has no resistance,and the power-angle curves are therefore undisplaced sinusoids. In Example 4 of Chapter II the amplitudes of the curves were found to be 2.58,0.936, and 2.06 per unit for the pre-fault, fault, and post-fault conditions,respectively. The input was 0.80 per unit. The three output curves andthe input line are plotted in Fig. 7. The initial operating point a lies at theintersection of the input line and the pre-fault output curve, The initialangular displacement 60, as determined from this intersection, is about 180

•

(It was computed as 18.10 in Example 4.) Upon occurrence of the fault theoperating point drops to b on the fault output curve and then moves alongthat curve. As it moves from b to c, the machine is accelerated and acquires kinetic energy proportional to At (the shaded area abc). By countingsmall squares on the graph paper, we find this area to be 19.3 squares. Theoperating point continues to move along the fault output curve, and, as itmoves from c to d, the machine is retarded. Area A2 is found to be -11.4squares, giving a net area out to point d of 19.3 - 11.4 = 7.9 squares.Therefore, when the machine reaches a value of aequal to the abscissa ofpoint d, it still has a positive velocity relative to the infinite bus. As theoperating point moves from d to e, the machine is again accelerated, andobviously it would pull out of step if the fault were not cleared. Indeed, thiswas found to be so in the point-by-point calculation of the swing curve ofFig. 9, Chapter II, for a sustained fault. When the fault is cleared, theoperating point jumps up to the post-fault output curve, or from point e to


180

~ ...............

/ i'<~Pre· fault output

/ ~----- i\V ~ " \

J

/ / <,Post-fault output -, \

1/ ~~\I I Output during faUlt",

I

~~~[\A 4 = - 10.4 sq.I I I I I

all I

~~~\~ Input .-c ~ f~

·v ~ c"\ . I d~~

hr'\~~I"~ ~yj~WK" I A 2 = - 11.4sq. A 3 = 3.3 sq.

I~ %~

~~~~ At = 19.3 sq. I

~~ \I~~b "' I I

I 13Soor I

~ISo 139° 157·I Io

o 60 90 120 s, 150 s;Angular displacement 8 (electrical degrees)

FIG. 7. Determination of critical clearing angle by the equal-area criterion.(Example 1.)

1.0

point g, and then moves along this curve. As it moves from g to h; themachine is retarded. The relative velocity will become zero when the netarea becomes zero. The critical condition is that in which the relativevelocity becomes zero just as the retarding power becomes zero. Geometrically, the condition is that in which the net area out to intersection h iszero. This condition is found by sliding the vertical line efg from left toright or from right to left until the area At + A 2+ As + A 4 = o. This is

3.0

2.0

2.5

0.5

easily done in practice by counting squares in columns from d to the rightand from h to the left, accumulating totals as we go and finding where thetotals are equal and opposite or at least most nearly so. In the presentproblem this equality occurs at 138° or 139°, which is therefore the criticalclearing angle, ac. The shaded area in Fig. 7 is At + A 2 + Aa+ A 4

= 19.3 - 11.4 + 3.3 - 10.4 = 0.8 ~ O.From the swing curve for a sustained fault (Fig. 9, Chapter II) we find

that the time corresponding to 0 = 1380 is t = 0.61 sec. This is the criticalclearing time, t.,

It is interesting to note that the swing curve plotted for a clearing time of0.60 sec. (clearing angle 136.2°) indicates that the system is stable. The

[~3a]


maximum value of 0 on that curve is 147°. For the critical clearing time of0.61 sec. the maximum value of 0 is determined by the abscissa of intersection h, Fig. 7, and is about 157°. The swing curve for a clearing time of0.65 sec. (clearing angle 146.7°) shows that the system is unstable. Thusthere is good agreement between the critical clearing angle determined bythe equal-area criterion and that determined by swing curves calculatedpoint by point.

Two finite machines. A system having two finite machines may bereplaced by an equivalent system having one finite machine and aninfinite bus, so that the swing equations and swing curves of angulardisplacement bet\veen the two machines are the same for both systems.It is necessary to use an equivalent inertia constant, equivalent input,and equivalent output for the equivalent finite machine. The equivalent inertia constant is a function of the inertia constants of the twoactual machines, and the equivalent input and output are functions ofthe inertia constants, inputs, and outputs of the two actual machines.The equivalent system will now be derived.

The swing equations of the t\VO finite machines are as follows:

d201 ~al ~il -- Jlu1dt2 = M 1 = M 1

[ISb]

[14]

The relative angle

will be used, because its value is significant in,showing the stability orinstability of the two-machine system.

d20 d20l d202 Pal ~a2

dt2 = dt2 - dt2 = M1 - M2

Multiply each side of the equation by M1M2/(M1 + M2 ) , obtaining

MlM2 d20 M2Pal -- M1Jla22=

M1 + M2 dt

M2Pui -- M1Pu2

M1 +'M2[15]

which may be written more simply as

d20M 2 = Ps = Pi - Pu,

dt[16]

EQUIVALENT POWER-ANGLE CURVE 133

[17]

Equation 16 is identical in form with eq. 1 for a single finite machineand an infinite bus. Here the equivalent input,

M2P i l - M1Pi 2

Pi = M1+ M

2

and the equivalent output,P

u= M2PU1 - M1Pu 2

Afl + M2[18]

are weighted means of the inputs and outputs, respectively, of the t\VOactual machines, with the signs of P i2 and Pu2 reversed, the weightsbeing inversely as the inertia constants. The equivalent inertiaconstant,

[19]

is smaller than the smaller inertia constant of the t\VO actual machines.The law of combination of the inertia constants is similar to that forthe parallel combination of impedances. This law will appear reasonable if we note that the inertia constant is the accelerating powerdivided by the acceleration and that in a two-machine system theaccelerating power of the generator is nearly§ equal (except in sign) tothat of the motor, while the relative acceleration is the sum of the acceleration of the generator and the retardation of the motor. Thissituation is in contrast to that which prevails in finding the inertiaconstant of a machine equivalent to a group of machines that swingtogether (discussed in Chapter III). There the accelerating power of thegroup is the sum of the accelerating powers of the individual machines,and the accelerations of all machines are equal. Consequently, theinertia constants combine like impedances in series.

Having obtained the equivalent system, we may investigate itsstability either by calculation of its swing curve or by use of the equalarea criterion. The equal-area criterion is the more convenientmethod, but to apply it we must first obtain the power-angle curve ofthe equivalent system.

Equivalent power-angle curve of two finite machines. The powerangle equations of a two-machine system are (by eqs, 17, ChapterIII):

Pul = E12 Y 11 cos 8 11 + E1E2 Y 12 cos (8 12 - 01 + 02) [20]

P'U2 = E2El Y2l cos (821 - 02 + 01) + E22Y 22 cos 822 [21]

§Exactly equal in a purely reactive network, as explained later in the chapter.


Substitute these values of Pul and Pu2 into the expression for equivalent output (eq. 18), and let ~ = ~1 - ~2. The result is

P« = M2E12Y

ll cos ell - MIE22Y22 cos e 22

M1+M2

+ E1E2Yd M2 cos (8 - e12) - M1 cos (8+ e12)] [22]M1+M2

The two cosine terms involving 8 may be combined into a single cosineterm by considering each term as the horizontal projection of a vectorin the position for which the variable ~ is zero. Thus the first term andthe second term are, respectively, the horizontal projections of vectors

FIG. 8. Vector diagram used in the derivation of eqs. 23 to 26.

M2/-812 and - M I!812. (See Fig. 8.) In this position the horizontal projections are M 2 cos (-812) = M 2 cos 812 and -M1 cos812, and the sum is

[23]

The vertical projections are M2 sin (-812) == -M2 sin 812 and- M 1 sin 812, and the sum is

V = - (M1 + M2 ) sin 812 [24]

The magnitude of the sum vector is

M" == VH2 + V2 = V (M2 - M1 )2 cos2 e12 + eMI + M2 )2 sin2 e12

== v' (M 12+M

22) (cos2el2+ sin2012)+2MIM2(sin2812-COS2 812)

== VM 12 + M2

2 - 2M1M2 cos 2812 [25]

and its phase angle is" -1 V -1 - (M! + M 2 ) sin 812

-8 = tan - = tanH (M 2 - M 1) cos 812

= tan-1(:: ~ z: tan e 12) [26]

REACTANCE NETWORK

Hence eq. 22 may be written more simply as

Pu = Pc + PM cos (0 - e/)

= Pc + PM sin (0 - 1')where

Pc = M2E12Y

ll cos ell - MIE22Y22 cos e22

M1+M2

is the vertical displacement (see Fig. 2), and

(M l + M 2 ) 0

'Y = -tan-1 M1

_ M2

tan e12 - 90

135

[27]

[28]

[29]

[30]

is the horizontal displacement, of a sine-wave, the amplitude of which is

PM = E1E2Y12M"

M1+M2

E1E2Y12 VM12 + M2

2 - 2MlM2 cos 2e12==

M1+M2

If we put M2 = 00, eqs. 28, 29, and 30 reduce to the values previouslyderived for one finite machine and an infinite bus (eq. 11).

Reactance network. If the network to which the two machines areconnected contains only reactance, the power-angle equation and theequation for equivalent input are considerably simplified. In thiscase 8 11 = 822 = -90° and 812 = 90°, giving cos 811 = cos 822

= 0, cos 2812 = -1, and tan 812 = 00. Hence Pc = 0, 'Y = 0, andPM = E lE2Y I 2 . The power-angle curve is then an undisplacedsinusoid,

[31]

which is identical to the power-angle curve for one machine connectedto an infinite bus through a reactance network. In other words, if thenetwork contains only reactance, the power-angle equation and curveof a two-machine system are independent of the inertia constants ofthe machines. If, however, the network contains resistance as well asreactance, both the amplitude and displacement of the power-anglecurve depend on the inertia constants.

Since there are no losses in a reactance network, one of the two machines must act as a generator and the other as a synchronous motor,so that, if both are considered as generators, their outputs will be equaland opposite:

Pu2 = -Pul [32]


Initially the inputs are equal to the outputs:

Pit = Put

Pi2 = Pu2

[33]

[34]

giving equal and opposite inputs,

Pi2 = -Pit [35]

Hence the equivalent input, as given by eq. 17, becomes

M 2P it - M t ( - Pit)Pi = M

1+ M

2= P« [36]

which is equal to the actual generator input. In similar fashion theequivalent output, as given by eq. 18, becomes equal to the actualgenerator output.

EXAMPLE 2

By use of the equal-area criterion determine the critical clearing angle ofthe two-machine system of Example 4, Chapter III.

Solution. The following data are obtained from Example 4, ChapterIII:

MA = 2.92 X 10-4 per unit

MD = 26.9 X 10-4 per unit

PiA = 0.80 unit power

PiD = 3.20 unit power

aA O= 23.0°

aDO = 10.3°With the fault on

EA2y AA cos eAA = 0

ED2yDD cos aDD = 0.45 unit power

EAEDYAD = 0.203 unit power

SAD = 86.4°With the fault cleared

EA2YAA cos 8AA = 0.04 unit power

ED2yDD cos aDD = 3.28 unit power

EAEDYAD = 1.91 unit power

SAD = 79.4°

From eq, 19 the equivalent inertia constant is

MAMD 2.92 X 26.9 .M = M,A + MD = 29.8 10-4 = 2.64 X 10-4 per unit

REACTANCE NETWORK

From eq. 17 the equivalent input is

p. _ MDPiA - MAPiD

,- MA+ MD

26.9 X 0.80 - 2.92 X 3.2029.8

21.5 - 9.35 12.15 0 41 it= -- = . unl power29.8 29.8

The output power-angle equation isP; == Po + PM sin (8 - 1)

where by eqs. 28, 29, and 30

Po == MDEA2YAA cos eAA - MAED2y DD cos eDD

MA+MD

P _ E E Y "';MA2 + MD2 - 2MAMD cos 29ADM - A D AD M

A+ MD

1 == -tan-1 (MA + MD tan eAD) - 90°MA- MD

For the faulted condition

P 26.9 X 0 - 2.92 X 0.45 0044.c == = -. unit power

29.8

P = 0.203 ...;(2.92)2+ (26.9)2 - 2 X 2.92 X 26.9 cos (2 X 86.4°)M 29.8

= 0.203 "';8 + 722 - 157 cos 172.8°29.8

= 0.203 "';730 - 157 X (-0.992)29.8

= 0.203 "';730 + 156 = 0.203 v'88629.8 29.8

== 0.203 unit power

== -tan-1 ( 29.8 tan 86.4°) - 90°2.9 - 26.9

== _tan-t 29.8 X 15.9 _ 900-24.0

== -tan-1 (-19.7) - 90° = 87.1° - 90° == -2.9°

137

138 THE EQUAIr-AREA CRITERION FOR STABILITY

For the cleared condition

Po = 26.9 X 0.04 - 2.92 X 3.28 = 1.1 - 12.929.8 29.8

11.8 3 ·= - - = - O. 96 unit power29.8

P = 1.91 V730 - 157cos 158.80

M 29.8

= 1.91 v730 - 157(-0.932)29.8

= 1.91 v730 + 146 = 1.91 V876 = 1.89 unit power29.8 29.8

'Y = -tan-1 ( 29.8 tan 79.4°) - 90°-24.0

= -tan-1 (29.8 X 5.34) _ 900-24.0

= -tan-1(- 6.63) - 90° = 81.4° - 90° = -8.6°

Hence for the fault condition

P« = -0.044 + 0.203sin (8+ 2.9°)

and for the cleared condition

P u = -0.40 + 1.89sin (8+ 8.6°)

The power-angle curves are plotted in Fig. 9. Note that they are displacedsine curves, the displacement being from 0 to 0' and from 0 to' 0". Theinput line is also drawn. The initial angle is

80 = OAO - aDO = 23.0 - 10.3 = 12.7°

This angle is marked on Fig. 9. The point on the input line at this value of 8is the point representing the pre-fault operating condition. It lies on thepre-fault power-angle curve, but there is no need of plotting any more of thatcurve than this one point.

By application of the equal-area criterion, the critical clearing angle isfound to be oc = 100°, and with clearing at this angle the maximum angulardisplacement that is reached is Om = 146°.

Let us see how well these values agree with the results obtained in Examples 3 and 4 of Chapter III. In both these examples the system was foundto be stable if the fault was cleared in 0.35 sec. (clearing angle, approximately96°) and unstable if the fault was cleared in 0.40 sec. (clearing angle, 118°).

SWING CURVE BY GRAPHICAL INTEGRATION 139

Thus, in these previous examples the clearing angle was found to be between96° and 118°, and the value found in the present example lies between theselimits. The equal-area criterion provides an easier method than we hadbefore of determining the critical clearing angle, but it does not enable us tofind directly the critical clearing time. From the swing curve for O.40-sec.clearing (Fig. 16 of Chapter III) we may read the time-O.36 sec.-corresponding to the critical clearing angle of 100°.

The clearing time of 0.35 sec. for which a swing curve was obtained andplotted in Fig. 16 of Chapter III is seen to be only slightly below the critical

I

146-

6".150

13'0· 100·o 00 30 60 90 s, 120

Angular displacement 8 (electrical degrees.)

FIG. 9. Power-angle curves and determination of critical clearing angle by use ofthe equal-area criterion. (Example 2.)

1.00I----I_---L._-+-.--,.~

-~c::s

! 0.50J----+-\rh'---+--4----1I---~--+---t~~~~~r_t-_r____;

value. The maximum angle attained in Example 3 was 135° or 136°, andin Example 4, 131°, whereas the maximum angle under the critical conditionis 146°. The discrepancy between Examples 3 and 4 is not surprising in viewof the proximity to the critical condition, at which a very small difference inaccelerating torque changes a stable system into an unstable one

Determination of swing curve by graphical integration. The type ofdisturbance which is most important in stability studies is a fault applied and subsequently cleared. It is usually desired to determinewhether a system is stable with a given load and given fault-clearingtime, to determine the stability limit for a given clearing time, or todetermine the critical clearing time for a given power. The equal-areacriterion by itself affords information on clearing angle but not onclearing time. The clearing time, however, is of primary importancebecause the circuit breakers and protective relays, by means of whichthe fault is cleared, have definite operating times which are independentof the angular displacements of the machines. Therefore, when the

[37]


equal-area criterion is used, it is necessary to find the clearing anglewhen the clearing time is known, or vice versa. For this procedurea swing curve, carried out as far as the point of clearing, is required.The swing curve for this purpose may be obtained in at least threedifferent ways: (1) by point-by-point calculation, (2) by graphicalintegration, or (3) by selection of a curve from sets of pre-calculatedswing curves. Point-by-point calculation was explained in ChapterII; the method of graphical integration'' will be described in this section; and the use of pre-calculated curves will be discussed in Chapterv.

It has already been pointed out, in deriving the equal-area criterion,that the relative angular speed of a machine is given by:

dB J218- = CJJ' = - Padadt AI ~

Graphically, the integral appearing in eq. 37 is the area under a curve ofPa against 8 or between curves of Pi and Pu against 8; in other words,the same area that was used in the equal-area criterion. In the process of applying the equal-area criterion the area may be determined asa function of 8 by counting squares on the graph paper for successiveincrements of 8. Or, if the power output is a sine function or othersimple function of the angle, the integral may be evaluated formally.In either case w' as a function of Bmay be computed from the area bymeans of eq, 37. By rearranging this equation we obtain:

dBdt = --; [38]

CJJ

and by integrating we get:

[39]

Evaluation of this integral gives t as a function of 8; in other words, itgives a swing curve.

If formal evaluation of this integral were possible, graphical methodswould be unnecessary. As a rule, however, eq. 39 cannot be formallyintegrated even though eq. 37 can be. One simple case in which bothintegrations can be done formally is that in which the acceleratingpower Pa is constant, Then

w' = ~2Pa(~- 80) [40]


and

[41]

In general, however, the integral of eq. 39 must be evaluated graphically. This may be done by plotting a curve of l/w' against 8 andfinding the area under the curve as a function of 8. Some difficultywill be encountered in determining the area under the curve for valuesof a near the initial value 80 and the maximum value 8m because, atthese values of 8, w' is zero and the ordinate of the curve, u«, is accordingly infinite, although the area under the curve is finite. Thisdifficulty can be avoided by assuming P a to be constant over a smallrange of a (until the curve of l/w' comes back on scale) and by usingeq. 41 to compute the time for the machine to swing through this rangeof 8.

EXAMPLE 3

By means of graphical integration obtain the swing curves (a) for a sustained fault on the system of Example 4, Chapter II, and (b) for a faultcleared in 0.4 sec. (These curves were previously computed by the pointby-point method and were plotted in Fig. 9 of Chapter II.)

Solution. (a) The following data are obtained from Example 4 ofChapter II:

Inertia constant, M = 2.56 X 10-'

Initial angle, 80 = 18.10

Input, Pi = 0.80

Output during fault, P 11 = 0.936 sin 8

Therefore the accelerating power during the fault is

Po = P, - P« = 0.80 - 0.936 sin aThe area under the curve of Po against 8 is

Al = r P..da = r (0.80 - 0.936 sin8) daJ 40 JI8.1°

= [0.808+ 0.936 X 57.3cos 8]:8.1'=0.80a+ 53.6 cos 8 - 0.80 X 18.1 - 53.6cos 18.10

= 0.80a+ 53.6 cos 8 - 14.5 - 51.0

= 0.806+ 53.6 cos 6 - 65.5'\1nit power degrees

(a)

(b)


(57.3 is the conversion factor from radians to degrees.)

1 ~ M ~1.28 X 10-4

- = - = seconds per degreeWi 2A1 Al

TABLE 1

(c)

COMPUTATION OF Al AND l/w' FOR SUSTAINED FAULT (EXAMPLE 3, PART a)

a l/w'cos 8 53.6cos 8 0.808 Sum Al (msec,

(deg.)per deg.)

18.1 0.951 51.0 14.5 65.5 0 ClO

20 0.940 50.4 16.0 66.4 0.9 12.025 0.906 48.5 20.0 68.5 3.0 6.530 0.866 46.4 24.0 70.4 4.9 5.145 0.707 37.9 36.0 73.9 8.4 3.960 0.500 26.8 48.0 74.8 9.3 3.775 0.259 13.9 60.0 73.9 8.4 3.990 0 0 72.0 72.0 6.5 4.4

105 -0.259 -13.9 84.0 70.1 4.6 5.3120 -0.500 -26.8 96.0 69.2 3.7 5.9135 -0.707 -37.9 108.0 70.1 4.6 5.3150 -0.866 -46.4 120.0 73.6 8.1 4.0165 -0.966 -51.8 132.0 80.2 14.7 3.0180 -1.000 -53.6 144.0 90.4 24.9 2.3

Al and l/w' are computed in Table 1. A curve of l/w' versus 0 is plotted inFig. 10. This curve is "off scale" for values of 0 between 18.1° and 20°.The time required for the machine to swing through this angle is thereforecomputed by eq. 41, using the value of P« at 0 = 19° as a good approximation to the average value of F; between ~ = 18.1° and ~ = 20°.

Pa (19°) = 0.800 - 0.936 sin 19°

= 0.800 - 0.936 X 0.326

= 0.800 - 0.305 = 0.495 (d)

t = 2M(~ - ~o) = /2 X 2.56 X 10-4(20 - 18.1)r. ~' 0.495

= 0.044sec. (e)

The area A2 under the curve for 10° increments of ~ between 20° and 180° isfound by counting squares on the graph paper and is recorded in Table 2.The cumulative total of the area, the time t, is also tabulated. The value oft computed above for 0 = 20° is included. The total area out to any value


: II II

I ,

\ J

\ J,Before clearing"'" u-~~~~-, ~/] 1/ ~Sustained-- - J ) I"~

. After~Ieari~g a~ IOO~,V~ I~Vl'" '"r-,I ~~tr~l

.....

I Afterclearing at 1350/ I IIo

o 20 40 60 80 100 120 140 160 1806 (electrical degrees)

FIG. 10. Curves for the determination of swing curves by graphical integration.(Example 3.)

12

10

......cu! 8~"'0...8-en

6"'0c:0uQ)

:E'E 4-.-.j3

2

TABLE 2

TABULATION OF AREA A2 UNDER CURVE OF l/w' VERSUS 8 AND COMPUTATION OF

t FOR SUSTAINED FAULT (EXAMPLE 3, PART a)

a A2 t a A2 t(deg.) (sec.) (sec.) (deg.) (sec.) (sec.)

20 0.044 0.044 110 0.053 0.45630 0.071 0.115 120 0.058 0.51440 0.046 0.161 130 0.058 0.57250 0.039 0.200 140 0.052 0.62460 0.037 0.237 150 0.044 0.66870 0.038 0.275 160 0.036 0.70480 0.039 0.314 170 0.029 0.73390 0.042 0.356 180 0.024 0.757

100 0.047 0.. 403

of arepresents the time for the machine to swing to that value of a. Theswing curve, 0 versus t, is plotted in Fig. 11. The points marked by smallcircles were computed by point-by-point method 2 with ~t = 0.05 sec.The two methods give results which agree well.

(b) From Example 4 of Chapter II the output after clearing of the fault is

P u = 2.06 sino (j)


and the accelerating power is therefore

P; = Pi - PUt = 0.80 - 2.06 sino (g)

From part a of this exampIe the clearing angle corresponding to the specifiedclearing time of 0.40 sec. is

(h)

0.90.80.70.60.30.20.1

Sustain~ fatult ......~~

~~L~

.". --,~ C)

~r -,~

Fault cleared at 0.6sec.->-.....

./V ~~ 1\V~ " ,

~l,/' Fault cleared at 0.4sec.-~~

~./ \.........

~~oo

160

200

40

0.4 0.5t (seconds)

FIG. 11. Swing curves determined by graphical integration. (Example 3.) Smallcircles are points on swing curves computed point by point.

The area under the curve of P a against ~ from the clearing angle to anysubsequent angle 0 is

A1' = r pGda = t" (0.80- 2.06sino)daJ8c JlOO°

= [0.800+ 2.06 X 57.3 cos0]~()()'

= 0.800+ 118.0 cos ~ - 0.80 X 100 - 118.0(-0.1736)

= 0.800+ 118.0 cos 0 - 80.0 + 20.5

= 0.80~ + 118.0 cos ~ - 59.5 (i)

The area before clearing, up to the clearing angle, is calculated from eq. b as

0.80 X 100+ 53.6 cos 1000- 65.5 = 80.0 - 9.3 - 65.5 = 5.2

Therefore the total area up to any angle ~ subsequent to clearing is

Al = 5.2 + AI' = 0.80~ + 118.0 cos ~ - 54.3 (j)

Al and l/w' are computed in Table 3 fr~m eqs. j and c, respectively, and1/w' is plotted as a function of 0 in the curve labeled "after clearing at


100°" in Fig. 10. It will be noted that this curve crosses the previous curve(marked "before clearing") at 100°.

TABLE 3

COMPUTATION OF At AND l/w' FOR FAULT CLEARED IN 0.4 SEC.

(EXAMPLE 3, PART b)

Bl/w'

(deg.)cos 8 118.0 cos 8 0.808 Sum At (msec.

per deg.)

0 1.000 118.0 0 118.0 63.7 1.4215 0.966 114.0 12.0 126.0 71.7 1.3430 0.866 102.1 24.0 126.1 71.8 1.3445 0.707 83.4 36.0 119.4 65.1 1.4060 0.500 59.0 48.0 107.0 52.7 1.5675 0.259 30.6 60.0 90.6 36.3 1.8885 0.087 10.3 68.0 78.3 24.0 2.3190 0 0 72.0 72.0 17.7 2.6995 -0.087 -10.3 76.0 65.7 11.4 3.35

100 -0.174 -20.5 80.0 59.5 5.2 4.97102 -0.208 -24.6 81.6 57.0 2.7 6.9104.4 -0.249 -29.4 83.6 54.2 -0.1 00

For a fault cleared at 0.4 sec. (100°) the swing curve is found by taking tequal to the area under the curves of l/w' versus 0, as follows: on the curvemarked "before clearing" to the intersection; thence on the curve marked"after clearing at 100°" to the maximum angle, which is found by settingAl = 0 in eq. j. This angle is about 104.4°. At this point w' changes frompositive to negative, and 1/w' does likewise. The area under the curve ofnegative values of 1/w', taken with negative increments of 0, is positiveand may be found equally well by using the curve plotted for positive valuesof 1/w'• Therefore the curve "after clearing" is followed back past the intersection and as far as one cares to go. Ultimately w' again vanishes, and theordinate l/w' becomes infinite at some negative value of 0; however, thecurve in Fig. 10 is not drawn that far to the left.

The area under the curve in the neighborhood of the maximum angle, sayfrom 104.4° to 102°, may be calculated by eq. 41. The value of P; at 103°will be assumed as the average value of P« throughout this interval of o.At 103°, by eq. g,

Pa(103°) = 0.80 - 2.06 sin 1030

= 0.80 - 2.06 X 0.974

= 0.80 - 2.00

= -1.20 (k)

t = /2 X 2.56 X 10-4(102

- 104.4) = 0.032 sec. (l)"'J -1.20


Equation 41 was derived on the assumption of initial velocity equal to zero,which is true if 104.4° is taken as the initial angle. The time required for themachine to swing from 102° to 104.4°, however, is exactly equal to the timerequired for it to swing back from 104.4° to 102°.

The area A 2 under the curve of l/w' and the cumulative total of this area,the time t, are entered in Table 4. The swing curve is plotted in Fig. 11.

TABLE 4

TABULATION OF AREA A2 UNDER CURVE OF 1I"l VERSUS aAND COMPUTATION OF

t FOR FAULT CLEARED IN 0.4 SEC. (EXAMPLE 3, PART b)

s A 2 t a A2 t(deg.) (sec.) (sec.) (deg.) (sec.) (sec.)

100 ..... 0.400 60. 0.016 0.579102 0.011 0.411 50 0.015 0.594104.4 0.032 0.443 40 0.014 0.608102 0.032 0.475 30 0.014 0.622100 0.011 0.486 20 0.013 0.63590 0.035 0.521 10 0.013 0.64880 0.023 0.544 0 0.014 0.66270 0.019 0.563

This curve agrees well with the points (marked by small circles) computed bypoint-by-point method 2.

The curve of l/w' versus 0 and the swing curve for a fault cleared in 0.6sec. (135°) have also been plotted in Figs. 10 and 11. The swing curve doesnot agree well with the one calculated by the point-by-point method. Theclearing time and angle in this instance are so near the critical values that itis likely that even small discrepancies lead to a considerable divergence ofthe curves.

REFERENCES

1. R. H. PARK and E. H. BANCKER, "System Stability as a Design Problem,"A.I.E.E. Trans., vol. 43, pp. 170-94, 1929.

2. H. H. SKILLING and M. H. YAMAKAWA, "A Graphical Solution of TransientStability," Elec. Eng., vol. 59, pp. 462-5, November, 1940.

3. O. G. C. DAHL, Electric Power Circuits, vol. II, Power System Stability, NewYork, McGraw-Hill Book Co., 1938, pp. 401-12, 443-50.

PROBLEMS ON CHAPTER IV

1. Show by diagrams how the equal-area criterion can be applied toexamine the stability. of a two-machine system subjected to the followingdisturbances:

a. A line fault, cleared by the successive opening of two circuit breakers.b. A fault on a radial feeder, cleared by disconnection of the feeder.

PROBLEMS 147

c. A line fault, cleared by the simultaneous opening of the circuitbreakers at both ends of the line, followed by the subsequent simultaneousreclosing of the same breakers.

d. The opening of one circuit of a double-circuit line as a normal switching operation.

e. A sudden increase of shaft load on 3, synchronous motor.

2. A synchronous motor, supplied with electric power from an infinite busover a circuit of negligible resistance, is operating with an initial shaft loadPo which is suddenly increased by an amount D.P. The power-angle curvehas an amplitude Pm. Derive a formula for the critical load incrementset»; as a function of the initial load PO/Pm. 'Plot the equation. Howmuch error would there be in assuming that the equation could be represented by a straight line between the two points where the true curveintersects the axes of coordinates?

3. Derive a formula for the transient stability limit P L of a two-machinereactance system subjected to a sustained fault, expressing the limit interms of the following quantities:

P« = amplitude of pre-fault power-angle curve

flPm = amplitude of fault power-angle curve

ao = initial angular displacement

Plot a curve of PL/Pm against fl.

4. Derive a formula for the transient stability limit of a two-machinereactance system subjected to a fault on a radial feeder and its subsequentclearing by disconnection of the feeder at a clearing angle oC. Use the notation given in Probe 3.

5. Derive a formula for the transient stability limit of a two-machinereactance system subjected to a fault on a transmission, line and its subsequent clearing by simultaneous opening of the circuit breakers at both endsof the faulty line. The stability limit should be expressed in terms of thequantities listed in Probe 3 and the following additional quantities:

r2Pm = amplitude of post-fault power-angle curve

Oc = clearing angle

6. Find the critical clearing time of a three-phase fault at the middle ofone transmission line of the two-machine system of Example 4, Chapter II,if the initial output of the water-wheel generator is 25 Mw.

7. Find the transient stability limit of the two-machine system of Example 4, Chapter II, for a three-phase fault at the middle of one transmissionline, cleared in 0.2 sec. by the simultaneous opening of both ends of the line.Assume that the power-angle curves obtained in Example 4 are valid for anyvalue of initial power. (It would be more accurate, but more laborious, toassume a different voltage behind transient reactance of the generator for


each different value of initial power, so as to hold the initial terminal voltageconstant.)

8. Using the method of graphical integration, plot the swing curve of thewater-wheel generator of the system of Example 4, Chapter II, for an initialoutput of 1.00 p.u, and a three-phase fault at the middle of one transmissionline cleared in 0.35 sec. by the simultaneous opening of circuit breakers atboth ends of the line.

[1]

CHAPTER V

FURTHER CONSIDERATION OF THE TWO-MACHINE SYSTEM

Pre-calculated swing curves. This chapter continues the consideration of the problem of the stability of two-machine systems. Theequal-area .criterion, which was discussed in Chapter IV, is a veryeffective means of determining whether a given two-machine system isstable when subjected to a given disturbance. The most importanttype of disturbance is the occurrence of a fault and its subsequentclearing by the opening of circuit breakers. If the clearing time isknown, as it usually is, the corresponding clearing angle must be foundbefore the equal-area criterion can be applied; or, conversely, if thecritical clearing angle for a given transmitted power is obtained fromthe equal-area criterion, the corresponding critical clearing time mustbe found in order that the results of the stability study will be in themost useful form.

Perhaps the simplest way to find the clearing angle corresponding toa given clearing time, or the time corresponding to a given angle, is torefer to the appropriate swing curve in a set of swing curves which havealready been calculated and plotted, and which may be appropriatelycalled "pre-calculated swing curves." Such a set of curves, obtainedby solving the swing equation on the M.I.T. integraph, was publishedby Summers and McClure2 and is reproduced here (Figs. 1 to 10) bykind permission of Mr. McClure.

The curves were derived for a sustained fault on a system consistingof a synchronous machine of finite size connected through reactanceto an infinite bus. By the methods presented in Chapter IV, however,the pre-calculated curves can be used with a system of two finite machines connected through any linear network. The usual simplifyingassumptions are made, to wit: constant input, no damping, and constant voltage behind direct-axis transient reactance. To make thecurves generally applicable they are plotted in terms of a dimensionlessvariable, the "modified time" T, defined by eq. 11 below.

The swing equation of a two-machine system is

d2aM dt2 = Pa = Pi - Pu

149

150 THE TWO-MACHINE SYSTEM

[3]

M1M2 equivalent inertia constant in megajoule-where M = M1 + M2 = seconds per electrical degree. [2]

o= 01 - 02 = angular displacement in electrical degrees.

t = time in seconds.

P M2Pil - M 1Pi2 • 1 · •i = M M = equrva ent Input In megawatts

1 + 2

which, either for a reactance network or for M 2 = 00, reduces to

Pi = Pil = input of machine 1 (the generator) [4]

The equivalent power output, dependent on a, is given by the powerangle equation:

which, for a reactance network, reduces to

P« = PM sino

[5]

[6]

Expressions for Pc, 'Y, and PM are given by eqs. 28, 29, and 30 ofChapter IV. For a reactance network the amplitude of the powerangle curve is

[7]

where X 12 is the reactance connecting machines 1 and 2, including thedirect-axis transient reactances of the machines themselves.

Substitution of eq. 5 into eq. 1 gives

d28M dt2 = Pi - Po - PM sin (a.- 7)

or

wherep/ = Pi - Po

0' = 0 - 'Y

[8]

[9]

[10]

[11]

To put eq. 8 into dimensionless form, divide it by PM and then introduce a quantity r, defined by

r;;:P; /7tJPM

T = t\)180 M = t\)' GH

PRE-CALCULATED SWING CURVES 151

[13]

[12]

where f = frequency in cycles per second.GH = kinetic energy, in megajoules, of equivalent generator at

rated speed = 180fM.

The result is:d2~' P ,r u i ., .,

180' dT2 = PM - sin 8 = P - sin 8

if a' is in electrical degrees, or simply

d2~'u • ~,

-=p-SIDud,,2

if a' is in electrical radians. Here

p/ Pi - Pcp = PM = PM [14]

and 0' has been defined in eq. 10.A differential equation has been obtained which is independent of the

inertia constants of the machines and of the constants of the network.The solution of the equation depends on the ratio of the input to theamplitude of the power-angle curve (both input and amplitude beingmeasured from the horizontal axis of symmetry of the sine curve ifthe sine curve is displaced vertically) and on the initial angle 50' andinitial angular speed woo For the present purpose, swing curves fora sustained fault are wanted; hence Pn, PM, and 'Y must be the constants of the power-angle equation for the faulted condition, and theinitial speed will always be zero. The solution then depends only uponp (defined by eq. 14) and 00'.

Each family of curves in Figs. 1 to 10 is for a constant value of sin00" the range covered being from 0 to 0.90 in steps of 0.10. Theindividual curves in each family are for constant values of p, ranging ineach family from a minimum value slightly larger than the value ofsin 00' for that family up to a maximum of 3.00.

The procedure for using the pre-calculated curves to determinecritical clearing time from a given critical clearing angle may be summarized as follows:

1. The power-angle curve (Pc, PM, and 1') for the faulted condition, the power input Pi, and the initial angle 00 are assumed to-beknown, because they are needed for finding the critical clearingangle Dc by the equal-area criterion.

2. Compute p/ = Pi - Pc, P = p//PM , 00' = 00 - 1', sin 50"and oc' = Dc - 'Y.


160

10~ 60c4(

40

20

2 3 4 5 6Modified time T

FIG. 1. sin 00' = O.

10

1098732

160 t----+--__t__

40

20

'0

~ 60c:c(

-~ 120 t---+--t--f-1H1--f-I-I+-"""'i-1I~f---;'+-"""f---J~Q)

"t:J

co 100 r---t---tt-..............,.-++--#-H-"'#--W'~'---E ..----.--1

U

~ 80 t---+--ftl-f1H+-1-""~~~F--:::'''':=;'-4-~~--+-~--+---+_-+---+_-+---+~-t---i

4 5 6Modified time T

FIG. 2. sin 00' = 0.10.

FIGS. 1 and 2. Pre-calculated swing curves (copied from Ref. 2 by permission).


109874 5 6Modified timeT

FIG. 3. sin 80' = 0.20.

321

40

160

....ene120 J--+-~-U-J..I~-I-~~+'~-btc-t---t---ir--t--+-+--;--t--;--t--t---1

l- 1001--4--V-I-I-I-:H+-I-,~~~~~-~~~-+--i-t-;--;----r---t--t.~-a! 80 t--f--+It-#+~-:0CD~ 60<

1098732

160

en~ 120 ....---+--4-1-1--I-I-+++I-~~~~ ~~+---+--t---+---+--f--+-+---+--I---+----4bOQ)

"'0

~ 100 t---+---I'"f-f-"""-N-#--It#---']~~ ~=---+--+--t--+--~~--+--+--I---t--+---t

:soQ)

a; 80 1---+--f.j"""""f+BtI-I-.~~~~__----,--,-+-,-t~~-+-+---f--t---Pli~-+--+--t

~

~ 60c::<

20

40


FIG. 4. sin 80' = 0.30.



109873 4 5 6Modified timeT

FIG. 5. sin 80' = 0.40.

2

160

40

20 t--+--+-...---+---+-t--+---+-t--+--+--+--I---+-+--+--i--+---+--I

-U)

~ 120 I--.......---+++-+I+-It-+-#+-I--i+--#t~~-t--:--~~-i--t---t-+---+-t----t--i---ttoQ)

"0

~ 100 1---+--........#-I-.........+#-,j~~-f#--+::."e-+----+--+---+--+---+--t---i--+--1f---+--+---I

.suQ)

~ 80 1--;--tt-tH'iHfiJ~t-T...,.-t~.............::::--t---t----t~+--t--t---t--t--t--t----1

:0

~ 60c-c

10987321o,-O........-.._......--"'_......._ •..&._oIoo---A.o_oI---I- .....&.- --...._..&.--.._...........-..

o

160 t---+---+-

20 t--+--+-+---+---+--+---+-I---+---+--+---+--I---+--+-+---+---Io-o+----I

40

U)! 120 t---+--..........-+4--#-~~~~++--+-+-- -bIIlI",--+-+--+--t---f---t---t--+---I'---+---f

~"0co 100 1---+--1lI----jHf-,I--It--#-#-.,IJ--,~L.-~~U:suQ):e 80 t---+-~~~

~

~ 60cc(


FIG. 6. sin 80' = 0.50.




FIG. 7. sin 00' = 0.60.

32O'"-...&---'----"--..J"".......,j",----"""-.J.-....r.-....l.---L---I-......,j",_~""'--...-.--L---I...--'-......J___'

o

160 I---+---t--

20 J--+--+--+--+--+---lf--f--+--+--1---+--+-I--+--+--+--+--+-----4---1

-t 120 1-+--~-v-I--W-+-I--#+I---+-J~iL+---+---+-J.~--I:--~4----+---4-~

f-c(ij 100 1---+-~~'-I-+I-~~~+--~~-+---+--+-t----+---+--+---+---f---4---4U

:a~ 80 ~-+-I-I-fI.I-#I:-I--I-I-~~~f==--+--+--1---l-~~I---f---+---I---+-----I----4~

~

~ 60c<

en~ 120t---+--~_f_li-f+_#_t.i~tI---l#__-t-#---f--+::.~--I-+_+__-f---I---+----+~1---I

~-cB 100t--t--~~Jt/_--Ih~~~~+--+--+--+~-+-+--I--I--.J-----I-~~

:a(1)

~ 80 t-~f_IrI_I~~~~r-::.-'f_+--+---:&-~--I-f_1-~~--I-~--l~

°"-........ -..J"".---"'_a...-""'---....I.-......L---J.,;.--I..--L_"'---""""--'----&...--'---L---I-J

o 2 3 4 5 6 1 8 9 10Modified time T

FIG. 8. sin 00' = 0.70.




321O'-J...-,..L--'-...L-...L..--1.-...I.--J.---L.--.L-~---L-..-.I--...a..---L---L.~-.l~.a...--..a

o

FIG. 9. sin 00' = 0.80.

180

160

140

en~ 120~Q,)'0

co 100-g'0Q,)

~ 80"0

~ 60c:c:(

ltJt: ~I ~ ~ol 1/ j8°011) 00 o~o II)

o " 11).'" "' .. ) /': ~ ~;;- ;;-;; ;; n: j- ~ ~~ J j~ ; ~ Q'11 7" Ji7 JI/

1111 III Ii / if 'I :Y r7I '1IJ 'I J '/JIf/ / V /11// "I 'II II r V .~'~.,

/ IJvrII if/7 V VJJflli rJjIfh '/ V ~

V

I ~V/ ~V

~Vf--

~~~""/l;"~ »> 095

~ - ~ _""-......-

~ ~~ ~~

,..... ---t--..r---.~ ~ ~.......

40

20

oo 2 3 4 5 6 7 8 9 ro

Modified time T

FIG. 10. sin 00' = 0.90.



3. Compute the equivalent inertia constant M by eq. 2.4. Find the family of curves for the proper value of sin 00' and the

individual curve for the proper value of p. Enter this curve with theordinate 0' = Oe' and read the corresponding abscissa T = T e•Interpolation between curves or between families of curves may benecessary.

5. By eq. 11 compute the critical clearing time te correspondingto Te.

To determine the clearing angle corresponding to a given clearingtime, the order of the steps of procedure is altered in a way thatshould be obvious.

The procedure described above breaks down if the fault is of suchnature that there is no synchronizing power while the fault is on. Insuch a case PM = 0, from which it follows that p = 00 and T = 0 forall values of t. The pre-calculated curve for this condition is thevertical coordinate axis, and the relation between 0'" and t cannot bedetermined from it. However, this relation can be found by eq. 41

of Chapter IV, namely: ~ ( )2M 0 - 00

t = [15]Pa

Furthermore, the pre-calculated curves cannot be used to representconditions after clearing a fault because the curve for the proper valueof angle and speed (at the instant of clearing) does not have the propervalue of accelerating power or acceleration after clearing.

EXAMPLE 1

In Example 1 of Chapter IV, which deals with a machine connectedthrough reactance to an infinite bus, the critical clearing angle for the conditions of Example 4 of Chapter II was found by the equal-area criterion to be1380

• The corresponding critical clearing time, as determined from theswing curve, is 0.61 sec. Check this value by use of the pre-calculated swingcurves.

Solution. From Example 4 of Chapter II, the power-angle equationvalid for the fault condition is

pu = 0.936sin 0 per unit,whence

Po = 0, PM = 0.936, 'Y = 0;the power input is

Pi = 0.80 per unit;

the inertia constant of the finite machine is

Ml = 2.56 X 10-4 per unit;


and the initial angle is

whencesin ~o = 0.310.

From the problem statement,

Oc = 1380

The following quantities are computed from the data:

Pi' = P, - Pc = Pi = 0.80

p = P/ = 0.80 = 0.854PM 0.936

sin 00' = sin 00 = 0.310

~c' = Dc = 1380

M = M1 = 2.56 X 10-4

~ = J1I"PM = I 11" X 0.936 = 8 0t \} 180M \} 180 X 2.56 X 10-4 ·

The most suitable pre-calculated curve is that for sin 00 = 0.30 andp = 0.85 in Fig. 4. The ordinate 0' = 1380 corresponds to the abscissaIf" = 4.8. Hence T c = 4.8 and

T c 4.8tc = - = - = 0.60 sec.

8.0 8.0

This agrees reasonably well with the previously found value, 0.61 sec.

EXAMPLE 2

Find the clearing angle corresponding to a clearing time of 0.30 sec. on thesystem of Example 4, Chapter III, which consists of two finite machinesconnected through an impedance network. (The equal-area criterion wasapplied to this system in Example 2, Chapter IV.)

Solution. The following data are obtained from Example 2 of ChapterIV:

M = 2.64 X 10-4 per unit

Pi = 0.41 per unit

Pc = -0.044 per unit

PM = 0.203 per unit

'Y = -2.9°

80 = 12.7°

1r X 0.203 = 3.66180 X 2.64 X 10-4

EFFECT OF FAULT-CLEARING TIME

and the following from the statement of this problem:

t = 0.30 sec.

The following quantities are computed from the data:

Pi' = Pi - Pc = 0.41 - (-0.04) = 0.45

P = Pi' = 0.45 = 2.2PM 0.203

00' = 00 - 'Y = 12.7 - (-2.9) = 15.6

sin 00' = 0.269

T~"t= '1180 M=T = 3.66t = 3.66 X 0.30 = 1.1

159

The most suitable pre-calculated curve is that for sin 00 = 0.30 andp = 2.0 in Fig. 4. On this curve 'T = 1.1 corresponds to 0' = 76°.

o= 0' + 'Y = 76° - 3° = 73°

From Table 15, Example 4, Chapter III, at t = 0.3 sec. we find OAD

= 75.3°. This value was obtained by a point-by-point calculation. Theagreement is reasonably good.

Effect of fault-clearing time on transient stability limit. Theamount of power that can be transmitted from one' machine to theother in a two-machine system without loss of synchronism when thesystem is subjected to a fault depends on the duration of the fault.The power limit can be determined as a function of clearing angle bythe equal-area criterion, and the relation between clearing angle andclearing time can be found from the pre-calculated swing curves. It isthen possible to plot a curve of stability limit as a function of clearingtime. Such a curve is shown in Fig. 11. It shows that the transientstability limit of the system can be greatly increased by decreasing thetime of fault clearing from 0.5 sec. or more to 0.2 sec. or less. Thetime of fault clearing is the sum of the time that the protective relaystake to close the circuit-breaker trip circuit and the time required bythe circuit breaker to interrupt the fault current. Frequently a systemwhich is unstable for a particular type of fault and fault location can bemade stable by altering the existing relaying or by modernizing thecircuit breakers so as to decrease the clearing time. *

*Typical values of relay time and breaker time are given in Chapter VIII, Vol.II. Modernization of breakers is discussed in the same chapter. Protectiverelaying is discussed in Chapter IX, Vol. II.


A curve like that of Fig. 11 can be obtained by the following procedure. First, the equal-area criterion isused to determine the stabilitylimit with instantaneous clearing. Truly instantaneous clearing is not

o

~ 1...'2::s...(1)c....~~:cnJ

en !h'----......---_....................---_..............--0.5 1.0 (X)

Fault- clearing time (seconds)

FIG. 11. Curve of stability limit as a function of fault duration.

obtainable in practice, but it may be regarded as the limit approachedas the clearing time is reduced. The stability limit for instantaneousclearing is the same as that for disconnection of the faulted line whenthere is no fault on it. This limit is determined as shown in Fig. 12.

o 00 01 r/2 OmAngle 0

FIG. 12. Determination of stability limit for instantaneous fault clearing by useof the equal-area criterion.

After the power-angle curves of pre-fault and post-fault (cleared) output have been plotted, the horizontal line representing the input,which is equal to the initial output, is shifted up and down until areaA l equals area A2• It should be noted that moving this line changesthe initial angle 80 and thus moves the vertical line bounding area A l ·

EFFECT OF FAULT-CLEARING TIME 161

The value of input determined by equality of the areas is plotted inFig. 11 as point 1 at zero clearing time.

Next the stability limit for a sustained fault is found as shown inFig. 13. The power-angle curve for the fault condition is used in placeof the post-fault curve; in other respects the determination of thestability limit for a sustained fault is like that for instantaneous clearing. The value of stability limit so found is plotted as the asymptote(point 2, Fig. 11) which the curve approaches at large clearing times.

The two extreme values of stability limit have now been found.Any number of convenient values of initial power between these ex-

,....~C:J..CDCo..CD~

2- r1Pm

~t--~~~""~~~~~~J--"'-l-

Angle 0

FIG. 13. Determination of stability limit for sustained fault by use of the equalarea criterion.

tremes may now be,assumed, and the critical clearing angle for each isfound by the equal-area criterion as shown in Fig. 14. Power-anglecurves are drawn for the output before the fault, during the fault, andafter the fault, and the input line is drawn at one of the selected valuesof initial power. The vertical line at the clearing angle ac is shiftedfrom right to left until areas Al and A 2 are equal, thus fixing thecritical clearing angle. The clearing time corresponding to this clearing angle is determined from the appropriate pre-calculated swingcurve. Point 3, Fig. 11, is then plotted, its coordinates being theassumed power and the corresponding critical clearing time. Additional points on the curve are determined in similar fashion.

The procedure described above, in which values of power areassumed and the corresponding clearing times found, is simpler than


the alternative procedure in which clearing times are assumed and thecorresponding power limits found. For in the latter procedure thehorizontal input line is shifted, resulting in a shift also of the verticalline at the initial angle 00, whereas in the former procedure only oneline, the vertical line at the clearing angle oc, is shifted.

The procedure which has been described for finding transient-stability power limit as a function of fault duration is applicable only to twomachine systems. Nevertheless the general conclusion, that decreaseof fault-clearing time improves stability and increases stability limits,

o s,Angle 0

FIG. 14. Determination of stability limit for fault cleared in finite time.

is just as valid for a multimachine system as for a two-machine system. The speeding up of relay and breaker operation is one of themost effective and important means of improving power-systemstability.

EXAMPLE 3

Plot stability limit in per unit as a function of the fault duration for athree-phase short circuit (a) at the middle of one of the parallel transmissionlines of the power system shown in Fig. 15 and (b) at the sending end of oneof the lines. The fault is cleared in both cases by the simultaneous openingof the circuit breakers at both ends of the line. The system consists of ahydroelectric station sending power over two parallel transmission lines atgenerator voltage to a metropolitan system which may be considered aninfinite bus. The following data pertain to the system. The base poweris the aggregate rating of the hydroelectric generators.

Direct-axis transient reactance of hydroelectric generators: 0.35 per unit


Stored energy of hydroelectric generators, H: ~.OO Mj. per Mva. of ratingFrequency: 60 c.p.s,Voltage behind transient reactance of hydroelectric generators: 1.00 per

unitVoltage of infinite bus: 1.00 per unitReactance of each transmission line: 0.40 per unit (neglect resistance)Reactance of transformers at receiving end of lines: 0.10 per unit

Fault Fault(6) 0.20 <e) 0.20 0.10

~r " tJPGHydro station -0-.-40--- -#A Me:OPolitanH = 3.0 system

H=oo

FIG. 15. Two-machine power system. (Example 3.) Reactances are given in perunit on a common base.

~0.35. 0.20 • 0.10~ ~ 0:5 ~

""------<@-.(a) (b)

FIG. 16. Reduction of the network of Fig. 15, pre-fault condition. (Example 3.)

~0.35. 0.40 • 0.10~ ~ 0:5 ~

------(@-.(a) (b)

FIG. 17. Reduction of the network of Fig. 15, post-fault condition. (Example 3.)

~ 0.35 0.40 0.10~

'V o.~o 'V(a) (c)

(d)~0.35 • Q:.!Q-rQ.:.lQ. • 0.10~.

'V I 0.05 '\.i (b)

-----.(@~--_...FIG. 18. Reduction of the network of Fig. 15 with a three-phase short circuit at

the middle of one line. (Example 3.)

Solution. The network joining the two machines, which is considered toconsist of reactance only, is reduced as shown in Figs. 16, 17, and 18 for thepre-fault and post-fault conditions and for a fault at the middle of the line,respectively. For the first two conditions the reduction is accomplished by


~/

'/Pre- fault......V

" - ......) 4 ~~~

:J~~,

"I ~V \ V Fault cieared

1/ J~ 1\, I ,I ) ~

/ V \~/ r\

/ I \IJ ~~~~~~~-.... 1\J~~~

~ Fault at middle of line-~r-, \II ~~ <,

r-, ~If;~ ~Faultat sending end '~

1.081.00

--.~

'c::s~

8----~

0.50

0.32

parallel and series combinations; for the third condition, by two yea conversions and intervening series combinations. With a three-phase fault atthe sending end of one line it is obvious, without reducing the network, thatno power can be transmitted, The resulting reactances between machines,

1.50

oo 30 60 90 120 150 180

o(degrees)

FIG. 19. Determination of stability limit for instantaneous clearing of a fault ateither location and for a sustained fault at the middle of the line. (Example 3.)

and the amplitudes of the power-angle curves (PM = EAEB/XAB = l/XAB),are as follows: .

Condition Reactance PM(per unit) (per unit)

Pre-fault 0.65 1.54Post-fault 0.85 1.18Fault at middle 2.45 0.41Fault at end co 0

The power-angle curves are plotted in Fig. 19. The stability limit forinstantaneous clearing is found by the equal-area criterion in the upper partof this figure. The value thus found, 1.08 per unit, is correct for both faultlocations. The stability limit for a sustained fault at the middle of the lineis found in the lower part of Fig. 19; it is 0.32 per unit. The stability limitfor a sustained three-phase fault at the end of the line is zero.

TA

BL

E1

DE

TE

RM

INA

TIO

NO

FC

LE

AR

ING

TIM

EF

RO

MC

LE

AR

ING

AN

GL

E(E

XA

MP

LE

3)

sin

00Fa

ult

atM

iddl

eof

Lin

eF

ault

atE

nd

ofL

ine

Pi

Pi

00s,

(p.u

.)=

-(d

eg.)

OcT

ct c

=T

c/5.

08Oc

-00

to1.

54p

=P

i/0

.41

(deg

.)fr

ompr

e-ca

lc.

(sec

.)(d

eg

.)(d

eg.)

(see

.)fr

omFi

g.20

curv

esfr

omFi

g.21

1.0

0.65

040

2.5

510

.40.

0846

60

.06

0.9

0.58

536

2.2

600

.70.

1452

160

.10

0.8

0.52

031

2.0

701

.00.

2058

270

.14

0.7

0.45

527

1.7

821

.30.

26..

...

.·.

.0

.60.

390

231.

595

1.6

0.31

7552

0.2

20

.50.

325

191

.211

12

.20.

43..

,..

,·.

.0

.40.

260

151

.013

03

.10.

6195

800

.33

0.35

0.22

813

0.85

144

6.0

1.18

..,

...

·..

0.2

0.13

08

·.,

..,

·.,

·..

120

112

0.5

60

.10.

065

4·.

...

,·.

,·.

'13

813

40

.86

0.05

0.03

21

·..

..'

·.,

·..

150

149

1.28

trj ~ a t-3 ~ ~ d ~ a t'"4 trj >- ~ Z o ~ ......~ tJ

:j

t-l

~


Pre-faulty

J 1/ I r-.......

IJ.

0.32 1-1..1-

Vi!11111I 11j.;

:--.: ~ I-- Faultat middle of line I ! ............. \~ I I I I I I I I I "r-.... 1\

I, I I I I I I I ~

180: 120: : 150111 130 144

i i 604551

30 : : 90:70 82 95

6 (degrees)

FIG. 20. Determination of stability limit as a function of clearing angle by useof the equal-area. criterion, three-phase fault at the middle of the line. Areas for

P = 0.50 are shaded. (Example3.)

Stability limits between 0.32 and 1.08 per unit are assumed with the faultat the middle of the line, and the corresponding critical clearing times arefound in Fig. 20. Stability limits between 0 and 1.08 per unit are assumedwith the fault at the sending end of the line, and the corresponding criticalclearing times are found in Fig. 21. In both figures curves of stability limitas a function of clearing angle are plotted. The values of clearing angle areentered in Table 1.

The values of clearing time corresponding to these values of clearing angleare found from the pre-calculated swing curves if the fault is at the middleof the line and from eq. 15 if the fault is at the end of the line. In the firstcase the numerical value of Tit is needed. Byeq. 11 it is

: = ~7T'!PM = /7T' X 60 X 0,41 = 5.08t GH '\j 3.0 X 1

In the second case we have from eq. 15

2GH (<<5c - «50)

180! r,2 X 3.0 X 1 (<<5 c - «50) =

180 X 60 Pi - P..

Details of the determination of clearing time are given in Table 1.Curves of stability limit as a function of clearing time are plotted in Fig.

22


Pre-fault-V-J

~tf- 17\- .~

~- -- - - 0-'\: 1:___:___~~ ~

~Fault cleared-- -- , I , I~-- ~ I ..-- Stability limit _

, , I~ vs. clearing angle ~~J : II I I

{7lY1I I "-

~I

~~ ~-- -- '- I - 11 - - -'- - - -: I ~

I I I~~ ~~, , I I I I\.. I

if ' , I I I "~ ~~ ~~I I

I I I~-

I I , -- ~- --7:; -t- - 1-,- - ' I -,-~ ~~

I: I ~ ~,

I I I 1'0.., I I

~~~

I I . ' .....~~~ ~~~I I Fault at end of line"

, I

~!"~ ~ ?%~~~v;a~ I I

~V/

~~v/ ij~ r/ /ijj:~ . -~ 1/'// 1'/// /// v/~v/./ ,

1.30

1.081.00

-.-=c:

'"~-~e 0.50

oo 30 : : :60 90:

465258 75 95o(degrees)

120 1~8 150 180

FIG. 21. Determination of stability limit as a function of clearing angle by use ofthe equal-area criterion, three-phase fault at the sending end of the line. Areas

for P = 0.20 and P = 1.00 are shaded. (Example 3.)

1.4 ex>0.5 1.0Fault- clearing time ( seconds)

~'\ 1\

1\\\ f\

'\1\

\ .", <, \r-Fault at middle of line11"--

t\. t'--.... fll---, ~ II ==

'"~L- Fault at sending end of line-

~""""- -

:...-oo

~ 0.5

1.0

FIG. 22. Curve of stability limit as a function of fault-clearing time for the systemof Fig. 15. (Example 3.)


Curves for determining critical clearing time. A more direct way ofdetermining the critical clearing time of a fault on a two-machinesystem than that described in the first part of this chapter has beendeveloped and described by Byrd and Pritchard.f Their methodfacilitates the plotting of curves of stability limit as a function of faultduration. It takes cognizance of the fact that the voltages behindtransient reactances, and hence the amplitudes of the power-anglecurves, usually vary with the power transmitted in a way determinedby operating practice. This fact was disregarded in our previousdiscussion, although allowance could have been made for it.

In Byrd and Pritchard's method two assumptions are utilized inaddition to those which were made in developing the equal-area criterion. They are:

1. That the network is purely reactive.2. That all circuit breakers which open to clear the fault do '80

simultaneously.

Both these assumptions are commonly, although not necessarily, madein using the equal-area criterion in order to simplify the calculations.

The derivation of Byrd and Pritchard's method follows: Let thepower-angle curves of the two-machine reactance system be:

Before the fault: Pu, = Pm sin 0

During the fault: Pu, = T1Pm sin a

After the fault: Pu, = T2Pm sin 0

[16]

[17]

[18]

In other words, Tl and T2 are the ratios of the amplitudes of the powerangle curves during and after the fault, respectively, to that before thefault. Inasmuch as the amplitude of the power-angle curve of areactance network is given by eq. 7,

p _ E1E2m - X

12[19]

[20]

it isapparent that fl and T2 may be expressed in terms of the reactancesbetween machines before, during, and after the fault, thus:

X 12 before faultT1 ==

X 12 during fault

X12 before faultT2 ==

X 12 after fault[21]

CUltVES DETEltMINING CRITICAL CLEAItING TIME 169

It is also apparent that, if the internal voltages E1 and E2 should vary,the amplitudes of all three power-angle curves would be changedproportionately, and the ratios rl and r2 would not be affected.

The three power-angle curves and the input line Pi are drawn inFig. 14 so that the equal-area criterion may be used to find the criticalclearing angle oc. Ordinarily the two shaded areas are equated. It isjust as correct, however, to equate the irregular area under the heavyline to the area of the rectangle below the input line. If this is done,the following equation results, from which Oc can be found:

[22]

Upon setting Pi = Pm sin 00 and evaluating the integrals, eq. 22 becomes:

(Om-OO)Pm sin 80= -rIPm(COSOc-coSOO)-r2Pm(COSom-cosoc) [23]

(om - 00) sin 00 = (r2 - rl) cos Oc + rl cos 00 - T2 cos Om [24]

(om - 00) sin 00 - Tl cos 00 + r2 cos Omcos Oc = [25]

r2 - rlwhere

[26]

Thus, if TI, r2, and 00 are known, the critical clearing angle Oc can befound from eqs, 25 and 26. The corresponding critical modified timeTc can then be found from pre-calculated swing curves (Figs. 1 to 10)which are solutions of the swing equation (13) in which 0' now issimply 0 and p = (sin oO)/rl. The actual clearing time to in secondscan be found from the equation

~180M I GHtc = Tc 1rrlPm = T c '\j~frlPm [27]

which differs from eq. 11 in that PM has been replaced by TIPm, the newsymbol for the amplitude of the power-angle curve during the fault.

The steps which serve to determine T c as a function of Tl, T2, and sin00 were carried out for many values of the independent variables, andthe results were plotted in the form of curves of Tc versus r2 for constantTI and sin 00. See Figs. 23 to 39. Each family of curves is for a constant value of sin 00, the range covered being from 0.10 to 0.90 in stepsof 0.05. The individual curves in each family are for constant valuesof Tl.

170

1.0

0.8

0.6

0.4

0.2

THE TWO-MACHINE SYSTEM

J I

~~'"""~I I Tl I J rr

0 ,"""l- I!! 8 &t)1

8 .....~'""" 1-0 ..... .....

0 ~I-1-0 0 0

" ~1-1- II " II j....... ~

~~ c: ,,"'rJ "...,.....~ II I'rI

I' J J J JII III II r, --If..r "

.,~-~II II r1 'I

(). ~J-'I I.J // ~ ~

~ 'I <""[;;jjl;t~'"""II I~

I~ ~ l...l ~

1...oIl L..I .... L.".1iiJ" I.....l .....~~ looII .... L-I.III~ L-L- -. ...

oo 1.0 2.0 3.0 4.0 5.0

1.0

0.8

0.6

0.4

0.2

FIG. 23. sin 80 = 0.10.

I I I I IIJ I , l II I I I

i-H &t)

~~--~olI II

0 i-&t)

~ 8l --'~IIIt') ..... ", f-1- C\I f- 1('),~r0 ..... 0 .......... i-f- ..... - : IJd ,"""0 _0 _1- 0 1-'-

__ 00

" f- "f-II

~ "1 1- .....I- " " :c: ~ ....... f- ........ J- :: c: f-I- ,-- ........, ........JI

'-I- I- I-i---,J I r 1

j I' " II, , I -,I' j If 'I ~., I II ~ IJ ]

J ,II IJ ~ " ..oil

v II' 1...oIl ~

~ v " " l....lII' L...II I....ll ~ ~~ L-L...o ..... '- 1-1- ....-

oo 1.0 2.0 3.0 4.0 5.0

FIG. 24. sin 80 = 0.15.

FIGS. 23and 24. Curves for determination of clearing time (copied from Ref. 3by permission).

CURVES DETERMINING CRITICAL CLEARING TIME 171

I I I I I I , • • I • I ,

1-+-1--0 i-I----fir) : I ,It)

If? ' raj0 ~i-- ..... ---- ....0 1-1-'-

~Ji-'-----0 0' '0 '

11 ~" ~~- 1-1- - II J II

~ji-i-i-

...0-#i-I--=...-' 4,.'" I 1...'"i-i-i- i-i-

i-i-r- i-I---- : II I, J J J, I I I-I-, J /

'1~~ 2J ~

I , o~~

" v v f\~Ij ,,- ~ " l.",lI'ilI"""

~ ..... ""'.",.- i.ooo'ioooo' ....~III~

., -," 10.-.... ".- --...- ..--...... ,..-~

Tt - 0.25_

Tc

FIG. 25. sin 80 = 0.20.

1.0

0.8

...N 0.6

0.4

0.2o 1.0 2.0 3.0 4.0 5.0

1.0

0.8

a!' 0.6

0.4

I ,~i-~ ~

~~i-i- ~4' II

I III,~ i=--Ill) 01 1:fJt -~K> ~~~.C? -f- ..... t- -i- I-d I' -- --

1° ~ o l- e -l- e:) -....--I- CV i- C\J

" ~ /IT II -i-

" II ~-~- 0J~ I- C:i J

...M~ 0-# ~

a..'" J--~ a.....,1 a....,I-:-- - II - ~ II II

i- ~ i- -i-I ---a.."" - i- J..."",

I' I t-- i-

I/ / 'I IIt)';~~l• J " II

I J J I~ ,,- ~ ~ r-J'I., IJ'(" ;i"~1-J II ~ ~ ~

I J , , ~ ~ ~,~ ~ ~ ~~ [.""II i.lIII .....

~ilI""" ~ilI""" -... ..... ........ 1.1-- ... L-"'~~ -~

".- 1.1 ...._... -... """ --""" --- t

'Tc

FIG. 26. sin 80 = 0.25.

0.2o 1.0 2.0 3.0 4.0 5.0

1.0

0.8

..." 0.6

0.4

1---I--~a.n'tWII I I r I I , t I I I , I I , I ~

ttC:fJ I~ I ::r' /fI1--I- 0 I-S liliI-r- -i- 0 OJ 10·>- °t i- .... f-~c::) l 10. 'O·ti-_ -f- II

" III :1-11 ~

i-- i- II I II 1 II~"""- ... - a.....~

~-~"...,,, c: j J......,~:I-i- - i.: 1 ..."t: i- """Ji- 1-- ,

1 I 'I ~,

II I I I~ IJr II v ~

I~ 'I ~

I I v ~ ~ '", ,~ ~ .,

.~,

~ ~ ~I-' ~

~ ~ ~ ~ - ---,- --,--

p

0.2o 1.0 2.0 3.0 4.0 5.0

FIG. 27. sin 80 = 0.30.

FIGS. 25, 26, 27. Curves for determination of clearing time (copied from Ref. 3by permission).


---~;U) ~;'~'t ~I t~t t~i~II I I .I-

t-~-- - 0 2 2 9i>-fil 1-- ~ :1I rl~

........-- - 0 1-- ~ _i-~

...... - - .... 11 ~ /I II II IIt- II ~-1-- O' J 1 0 , -- ....---- .... - ---- II ~:_" If

" rj'- -----__ a..;:-c: "'~1 .......~ ..... 4.'"'f I-

4.'"'fj

'- _i--- 4.'"'f.,-._ .....,/J .....,

I II 'I 1/ J vII 'I J " If' v

" v I~ I;'" I" L."l

" J v ~ ~ I...... I......II J v l,;Ill IlllI' L....~ 1...- ......

J ~ L."l ~ ~ ..-~iiI"" L-~ ~ ....~- - - ~""111"" '-~""" -~...... _I.- ...

I I

I III I

1.0

0.8

...... 0.6

0.4

0.2o 1.0 2.0 3.0 4.0 5.0

1.0

0.8

....0.6

0.4

FIG. 28. sin 80 = 0.35.

I I I I I II I I ~ I I 1-' 1/ I I I 1/ I II IIT:J:,.It~1-1-

~8r~ [ ....W~ f. .....1-&0 ~~~ ........ril~t~~~:~ II~~~~~~ - t:i tt; ~~ O' ... - O' 1-..... O' II O' - 1-1-

1-1-

~~ ~j~- II t..~ ~ II ~';. .... _~" ~f-""--1- 1/ ~ .... P" 1/ ~~~ ~- .... ....~ II J tt '''''Ij~~~- '''''1} ... -\""if"~I- ...~~ c: 1-- 4."'f ~ f-- ~'"'f '~1~1-I- ~

,I ,

I~ ~,

I I ~ ~ ~,.

~

~ ~ ~ ~ J

J , if" ,J ~ . .

I~ l....IIlI l...oI~ ,-'" ~ .... _..... - ~ ...,,~ ..-~ -- - '---

Tc

FIG. 29. sin 80 = 0.40.

0.2o 1.0 2.0 3.0 4.0 5.0

Tc

FIG. 30. sin 80 = 0.45.


CURVES DETERMINING CLEARING TIME 173

J II II I 1 II J IJI J 'I '(

II J J 'II] v J11

I II I 1/ "JII II 1 ~

5.04.03.02.01.00.5 '-'-__ ---..........,,-'-' ..Io...I~"'-'-'I-Io. ...-.. ...r....I

o

FIG. 31.. sin ao = 0.50.

..."

1.0

0.9

0.8

0.7

I- I I

fS; f1-~gl~, I I II L~~Ij

I I IIt)

~~;:'OJL-L-

t~i1-1- ~ 'I I f I

I- 0~'/- I I ~1-0 0 o Oi'

I-~I- t;;:)' tr ~ II J~ ,...I- " " " l- II : II II i~

I-~ I- III- "... ~... ~ 1-1-~ ti-~- II II c::r ~-I- "... ..............., ......., I-

1-1-(,.....,

J l-I- ~.... " ~,... fo-'I 'I II If 'f

(..., l-I-

J J IJ IJ rt-fo- ....I II I T 1I , i J v

j , , vI I II I~

," ~ '"II J J ,

~

1.11 ~.., l...oillill"'", J ~ ~ ~ I..- "I 'I ~ IlIlII' ..,

--1...-"- _........

I, ~ ~ ,,- L-'-'_....

_L- I.~ ... ""'"

0.6

0.5o 1.0 2.0 3.0 4.0 5.0

Tc

FIG. 32. sin ao = 0.55.

FIGB. 31 and 32. Curves for determination of clearing time (copied from Ref. 3by permission).

174

...e-t 0.8

THE TWO-MACHINE SYSTEM

II , ~ ~ ~ ~ ,~ _~~ __~

~~~~~ ~~~~ -~~ -

FIG. 33. sin 80 = 0.60.

I,O~ I ~ I ~14-~L~_~I+~II I I I Is -~~ ~ ~- t8 /f{ ~ ~ ~~~~ ~>o-F#f8l ~0 -10 0 c;-c; -~" ~.- ()" ~ <:>. f-~ t::)~" "-~ t::)"

" ~ '~ /I I /I ~" zr q I" ... ::: II -~1- /t t.-~_ /t ~II.: ->0-- 4'~-c: I.;.J , .... 1..."" J 1..."" -~ """-1 ~""-~II- "...,-:: '\~2~: ~ ,\'\0

-l- '('\! ~

~ ~ I, I II I I~ ~ ~ lA" I 1:10 _J J I) ~ ~ ~ ~~ ~ ~ ~O" ~~

I I I~ ~ ~ ~ ~1\L..;;;..... 1III!!!~I ,

~ ~ ~ ~f' ,,-L..oo~

If II v ~ ~ ~ ......~---- -~

~ J If' v: ~f' --'" ~"" ..... .- .... ....- I.-~......

'-~ """'10-- ....,,~ iIooolI""~ ......

~ .,

1.0

0.9

..." 0.8

0.7

0.6o 1.0 2.0 3.0 4.0 5.0

1.Q

0.9

0.8

FIG. 34. sin 80 = 0.65.

I

t.~~~&~l$qr!J>-~ ~WI JI , I I I J , II' , .~

, ,- 0

~ Ort' II I O~O 11~<i) ~ I

- - - ~_0 o c::i Ci O' I I ()-:"

-- " "I/Ht Ht: :r-I/O"li- " J

I ~ If' I ~ i~ I I- ...... ""'J "14,....J 4,....tr-- - v- ~r- v- 0(", ~'\ ~ 'D,I-~()~ ~~

t I ,I~ ~ ~ .C'\ ___ ....~-

I II II 1...- ~ ~ ~

I I ~ ~ ~ ~ ,~,~ ~ ~ ~ ~

_.... -", ~ 1...oIlI~ ....- -~ ...~ _....11""-

0.7o 1.0 2.0 3.0 4.0 5.0

FIG. 35. sin 80 = 0.70.



1.0

0.9

0.8

~~ I~t~ ~tfll~~Xi~ ~Wd'lIIi} I I

~~I-~

~ 0'Yo ~ 0:' If' T T 11~ ~~1-1-~O 0 t::i O' O' o 0'//'1 //~:t.1-1- ~ ~ T o ~

~~~ u: :!J ~~I-~,<, ~ ~

~" 'h...--~ ....I ~~ - ~.,

11 } II v " lIi"" ..... I......""

'I' ~ " ~ ."" ~ 1..- .....

J J I/' ~ ~ '" l....olI~ l..o""""~~ ...

'I ~.,.

~ jllIII" L.oo~ '- io"'" _I-'"

0.7o 1.0 2.0 3.0 4.0 5.0

FIG. 36. sin 00 = 0.75.

030r'= .'1 = O. 0 rrl =0.40

oJ1/ ,~r~J~~a ~\ffi1<j-~B:-t\iP .. ~l 17;L

t~~~. ~~ ~ ~o'f.~ a:t a:l,;~ ~ ~o.~~ Iio' \)'b\<)~, 'r .1:I- ;!Ii • Ci

" J-/t ~ ~I- '- 'h~~- t\~,,~ 1.t~.,,~- 1'\I"'~~~- "~~..-""'~ ~o~'z;;..-e

t-,"" 11~ -I~""I-f-<~ ~ tt" ."" ~" "" -- -" ~\-,,"'-~-~11'1 ~ I.' 1....1"'" I..- 1.1I ...... ~ ... 1- ....... L.- .... 11I""

r:r/: ~~ ~ l..o.IlIl-- ~~ ...-....... '--..."'"'" 1--..... 1.-:- ~iI-"'"

1.0

~ 0.9

Tc

FIG. 37. sin 00 = 0.80.

0.8o 1.0 2.0 3.0 4.0 5.0

1.00

0.95

0.90

o 0

~!~~~~O~""I- ~~~~I Wco'>v

- n,F-f- rt> - ~'>~ v

~ ~JI-~' o· f-f- oep co'" Alo,,,

-"I" I- 1/ 1/ rlf- f-I- //~ <~~ <.t;~~~ - ?

I- 4.. ....J"-'r- ~ ....,jf- ..... <)I.~'(,............ ~... ~"'J~ l-I..;

r J I;' I;l'" I.l'"II IJ l' I.... ~ I...- ~'-'

j J " I.l'" i...ll ~ ijll'"

I v ~ ~

--..... jllIII" t-

IJ ~ 1....1 L....o ..... ~,.... II"" ....... "'"'"JI 1....1 l.olII

Tc

FIG. 38. sin 00 = 0.85.

0.85o 1.0 2.0 3.0 4.0 5.0

~ ....~~*~Y/)4:<§l- ~I-lli~'j-t1~$~:1CiL.... II 0 .....

~~D~09..... ~ol- e::; O· O· + 0' ~~I- ~().t!.~~ ~(). O~~~.,.

" - II " 1/ '(~~L.,--f- ~'\....-I~ - '- t~~~ h~ -'" 1\ ~.....~..." -) "...; ~"'1 4..' ..... ~..... _"'" l0- ll""

, I v: ~ - ~""" "'""',- ... ", ~~ IiiiiiiI I.." --" --

--"'"- ~

-~ ...... """'"

IT

1.00

...~ 0.95

0.90o 1.0 2.0 3.0 4.0 5.0

FIG. 39. sin 80 = 0.90.

FIGS. 36, 37, 38, 39. Curves for determination of clearing time (copied from Ref. 3by permission).


1.00.90.80.70.4 0.5 0.6rJ or r2

0.30.20.1

IiJ

v1/

iJ

II'1/

J

VII

/I~

~

vV

WI'"I;"

v:,.,

I~

~l.',.,,.,

~I'

'-'l;l"

I~

I#"oo

0.1

0.3

0.4

0.7

0.9

0.8

0.6

0.2

It is convenient to have an additional curve for determining thestability limits for instantaneous clearing and for sustained faults. Toobtain the equation of such a curve refer to Fig. 12, which shows theapplication of the equal-area criterion to the case of instantaneous

1.0

~c 0.5.~

FIG. 40. Curve for determining the stability limit for a sustained fault or for afault cleared instantaneously (copied from Ref. 3 by permission).

clearing, and equate the area of the rectangle under the Pi line betweenao and 6m to the area under the post-fault power-angle curve betweenthe same limits.

Area of rectangle = Pi(8m - 60) = Pm(6m - 60) sin 60

Area under power-angle curve=T2Pm16msin ~ as =T2Pm (cos ~o - cos 8m)

Equating the two areas, we get

(am - ao) sin ~o = T2(COS ao - cos ~m) [28]


where 8m is as given by eq. 26. Equation 28 expresses implicitly arelation between T2 and sin 80 which is plotted in Fig. 40. To find thestability limit for instantaneous clearing, enter this curve with T2 andread sin 80• Then the stability limit is Pm sin 80• The same curve canbe used to find the stability limit fora sustained fault by entering it with P

the value of TI instead of T2; this becomes evident when Fig. 13 is compared with Fig. 12.

It has already been mentionedthat, as the transmitted power (Pi ==Pm sin 80 ) varies, the internal voltagesE1 and E2 usually vary, and hencePm = E1E2/X12 also varies. For thepresent purpose the most useful way a0------------"'·to show these relations is to plot Pi sin 00

and Pm against sin 80, as shown in F 41 T · 1 f P d10. . ypica curves 0 , anFig. 41. r; versus sin ~o.

The procedure for obtaining a curveof power limit as a function of fault duration by Byrd and Pritchard'smethod will now be summarized.

1. Reduce the reactance network to an equivalent ~ between thetwo machines and neutral for each of the three conditions:

a. Before the fault.b. During the fault.c. After clearing the fault.

Only the reactances X 12 between the two machines are used in whatfollows.

2. Compute the equivalent inertia constant M by eq. 2.3. Calculate and plot curves (like those of Fig. 41) of

a. Pm = EIE2/XI 2 versus sin 80

b. Initial power Pi versus sin 80

maintaining the bus voltages at the values which would be held inactual operation.

4. Compute TI and T2 by eqs. 20 and 21.5. a. Enter the curve of Fig. 40 with Tl and read sin 80• From

Fig. 41 read Pi corresponding to this value of sin 80• This is thestability limit for a sustained fault.

b. Repeat, using T2 instead of TI. The value of Pi thus found isthe stability limit for instantaneous clearing.


6.t Select values of sin 00 which are multiples of 0.05 and whichare between the values found in steps 5a and 5b. For each suchvalue of sin 00 find the proper family of curves from Figs. 23 to 39;find the curve for the value of Tl; enter this curve with T2 and read

Te•7. For each value of sin 00 used in step 6 read the corresponding

values of Pi and Pm from the curves of step 3 (like Fig. 41).8. t For each value of Tc found in step 6 compute the clearing time

te by eq. 27, using the proper value of Pm determined in step 7.9. Plot stability limit Pi as a function of clearing time te• This

curve will look like the one in Fig. 11.

The method which has been described fails if Tl = 0, since for thiscaser = 0 and t is indeterminate from eq. 27. To avoid this difficultya new modified time p is now introduced, related to the actual time t by

t = p J180M = p IGH . [29]7rPm ~7rfPm

and differing from Tin eq. 27 in that Pm, the amplitude of the pre-faultpower-angle curve, is used instead of T1Pm , the amplitude of the faultpower-angle curve. Hence p = T/Vi). The swing equation nowbecomes

p=

with 8 in electrical radians, and it has the solution

2(0 - 00)

.sin 00

[30]

[31]

which will be recognized as eq. 15, transformed by the substitution ofeq. 29 for t, and with 0 - 00 expressed in electrical radians.

Given Tl = 0 and the values of T2 and sin 00, Pc may be found byusing eqs. 25 and 26 to get oc, and then eq. 31 to get Pc. From theresults of such calculations Pc has been plotted in Fig. 42 against sin 00for various values of T2.

If Tl = 0, steps 6 and 8 of the procedure are replaced by steps 6Aand 8A, respectively, which are as follows:

6A. Select values of sin 00 which are between zero and the valuefound in step 5b. Find the curve in Fig. 42 corresponding to the

[If rl = 0, substitute steps 6A and 8A, described on this page and page 180,for steps 6 and 8, respectively.

a ~ ax t:1 ~ t;d ~ a= ~ z ~ z o § :j ~ a &; ~ Z o ~ ~ s: l%j

7.0

6.5

6.0

5.5

5.0

4.5

4.0

3.0

2.5

2.0

1.5

3.5

Pc

Cur

ves

for

dete

rmin

ing

clea

ring

tim

e-i

frl

=0

(cop

ied

from

Ref

.3

bype

rmis

sion

).FI

G.

42.

0.5

1.0

o o

1.0

0.8

0.6

0'0 c: '(

ii

0.4 2

~ ~ co


value of r2. Enter it with the selected values of sin ao and read thecorresponding values of Pc.

8A. For each value of Pc found in step 6A compute the clearingtime tcby eq. 29, using the proper value of Pm determined in step 7.

EXAMPLE 4

Plot a curve of stability limit as a function of fault duration .for a threephase fault at the middle of one of the 132-kv. transmission lines of the powersystem of Fig. 43. The two hydroelectric generators i and k are to be ra.

0.175

~33.0/13.2O.l75~

33.0/13.2

0.30

0-111:0.35

0.35~30)(

6.6/132Fault

.6.'A-0.140

~33.0/13.2

0.140 f----®33.0/13.2

FIG. 43. One-line diagram of power system, with reactances of lines and transformers in per unit on a lOO-Mva. base (Example 4). Data on generators and loads

are given in Table 2. (From Ref. 3 by permisslon.)

garded as one machine of a two-machine system, and the steam turbogenerators, a, b, g, and h, together with loads c, d, e, and J, as the othermachine. Each load is assumed to consist of three equal parts, one ofresistance, one of synchronous motors operating at half load, and one ofinduction motors at 1/2.75 load. The reactances and inertia constants ofthe generators and of the composite loads are given in Table 2, expressed on abase power equal to the rating of the individual generator or load. In Fig.43 the reactances of the lines and transformers are given on a base of 100Mva. and nominal voltage. The system frequency is 60 c.p.s,

Solution. The procedure described on pp. 177-8 will be followed.1. Network reduction. All reactances will be expressed on a system base

of 100 Mva, The reactances in Fig. 43 are already on this base. The


machine reactances in Table 2, given on the machine base, are converted tothe system base by multiplying them by (100 Mva.)/(machine rating inmegavolt-am peres) .

TABLE 2

MACHINE REACTANCES AND INERTIA CONSTANTS (EXAMPLE 4)

Rating,

HXdMachine Kind ~

Mva. P.F. Machine System Machine SystemBase Base Base Base

a Turbogenerator 40 0.8 0.18 0.45 6.0 2.40b " " " " " " "c Composite load 55 0.9 0.56 1.01 1.95 1.07d " " 30 u " 1.86 " 0.58e " " " " " " " u

f u " 55 " " 1.01 " 1.07g Turbogenerator 50 0.8 0.18 0.36 6.0 3.00h " " " " " " "i Hydro generator 29. 4 0.85 0.30 1.02 3.0 0.88k " u " " " " " u

The network is so nearly symmetrical about a horizontal axis that thereduction may be facilitated with very little sacrifice in accuracy by assuming that the two sections of the long receiving bus are tied together and alsothat the load busses d and e are tied together. In addition, machines i andk are assumed to be tied together behind their transient reactances; machines a, b, c, d, e, j, g, and h are assumed to be tied together similarly.Then, after simplifying the hydroelectric-station circuits and the receivingsystem by series and parallel combinations, the circuit of Fig. 44a results.

The reactances of the three parts which are separated from one another bytransformers are given on a common power base of 100 Mva, and on voltagebases equal to the nominal voltages of 6.6, 132.0, and 33.0 kv, It will benoticed that the ratio of the receiving-end transformers, 125.4 kv./33.0 kv.,is not equal to the ratio of base voltages, 132.0 kv./33.0 kv, Therefore thebase voltage for the receiving system should be changed from 33.0 kv. to33.0 X 132.0/125.4 = 34.7 kv.; the reactance of that system on the newbase is

0.13 X (33'0)2 = 0.12 per unit34.7The new value is shown in Fig. 44b. Parallel and series combinations givethe result shown in Fig. 44c for the pre-fault network. Xu = 1.10 per unit.

The faulted network is obtained by grounding the midpoint X of oneline in Fig. 44b. Then a ~-Y conversion, series combinations, and a Y-~

conversion lead to the ~ network of Fig. 44d. X12 = 7.20 per unit.The post-fault network is obtained by omitting one of the parallel lines of


Fig. 44b. Then a series combination leads to Fig. 44e. X 12 = 1.28 perunit.

2. Inertia constant. The inertia constants in Table 2, given on the machine base, are converted to the system base by multiplying them by (rna-

~~~(a)6.6/132 kv. 0.35 125.4/33.0 kv.

051 O.~15 2

~~3~(b)0.35

/>. ~D~ vvv~ (c)

(d)

~vvv~ (e)

FIG. 44. Reduction of the network of Fig. 43 (Example 4). The reduced networksare given by c for the pre-fault condition, by d for the fault condition, and by e for

the post-fault condition.

chine rating)/(100 Mva.). Then

HI = Hi+ Hk = 1.76

H2 = Ha + Hb+ H c + Hd+ He + H/+ H, + Hh = 14.1

H = H1H2 = 1.76 X 14.1 = 1.56HI + H2 15.9

3. Curves of P« and Pi versus sin 00. Assume that 132 kv. (1.00 per unit)is maintained on the high-voltage side of the step-up transformers, and 33.0kv. (0.95 per unit) is maintained on the low-voltage side of the step-downtransformers. The voltages behind transient reactance, E1 and E21 asfunctions of the initial angle 00 between them, can be found from the vectordiagram of Fig. 45 by assigning arbitrary values to the current I and solving,


either graphically or by trigonometric computation, the resulting trianglesfor Ell E2, and 00. Then P« and Pi are calculated from the relations

P _ E1E2 _ E1E2

m - Xl2

- 1.10

andPi = Pm sin 00

0.661 ----~~ 0.121

FIG. 45. Vector diagram for determining E1 and E2 as functions of ~o (Example 4).

TABLE 3

CALCULATION OF Pm AND Pi VERSUS sin ao (EXAMPLE 4)

P, EI E2 ~o sin ~o Pm

0.125 1.106 0.933 7.7 0.134 0.9350.250 1.121 0.934 15.3 0.264 0.9480.375 1.145 0.936 22.7 0.386 0.9710.500 1.177 0.939 30.0 0.499 1.0010.750 1.267 0.950 43.6 0.689 1.0901.000 1.373 0.972 55.9 0.828 1.221

The results of the calculations are given in Table 3. Curves of Pm and Pi asfunctions of sin 00 are plotted in Fig. 46.

4. Computation of fl and f2. By eqs. 20 and 21,

1.10 3fl= -= 0.15

7.20

1.10r2= -= 0.86

1.28

5. Stability limits for sustained fault and [or instantaneous clearing.a. Sustainedfault. Entering the curve of Fig. 40 with fl = 0.153, we read


sin 80 = 0.117. Entering the Pi curve of Fig. 46 with this value of sin 00, weread Pi = 0.110.

b. Instantaneous clearing. Entering the curve of Fig. 40 with '2 = 0.86,

I

/Pm~/'

~~ v

1----- J

/J

~V

V/

/1>;/

1/V

0.5

1.0

oo 05 ID

sin 00

FIG. 46. Curves of Pm and Pi as functions of sin 80 (Example 4).

TABLE 4

CALCULATION OF POWER LIMIT Pi AS A FUNCTION OF CLEARING TIME te(EXAMPLE 4)

1 2 3 4 5 6

sin 80 Pi Pm Tc/tcte

Te (sec.)

0.80 0.933 1.18 4.68 0.13 0.030.75 0.843 1.14 4.57 0.24 0.050.70 0.763 1.10 4.50 0.35 0.080.65 0.690 1.07 4.44 0.45 0.100.60 0.623 1.04 4.37 0.55 0.130.55 0.560 1.02 4.33 0.67 0.150.50 0.500 1.00 4.29 0.78 0.180.45 0.443 0.986 4.26 0.90 0.220.40 0.390 0.973 4.23 1.07 0.250.35 0.336 0.963 4.21 1.26 0.300.30 0.285 0.954 4.19 1.50 0.360.25 0.236 0.946 4.17 1.79 0.430.20 0.188 0.940 4.16 2.29 0.550.15 0.141 0.936 4.15 3.31 0.80

~

SUMMARY OF CALCULATING TRANSIENT STABILITY 185

1\\,~\'i\

",'"r-,

............ ............~ r--.. ::::-

we read sin 00 = 0.825. Entering the Pi curve of Fig. 46 with this value ofsin 00, we read Pi = 0.995.

6, 7, and 8. Stability limii« [orfinite clearing times. These three steps arecarried out in Table 4. In col. 1 are entered values of sin 00 from 0.15 to 0.80at intervals of 0:05. In cola. 2 and 3 are the corresponding values of Pi and

1.0

1.0

1r X 60 X 0.153 v'p = 4.3 vp1 X 1.56 m m

oo 0.5

Clearing timet, (seconds)

FIG. 47. Curve of stability limit as a function of clearing time, three-phase shortcircuit at the middle of one of the 132-kv. lines of the system of Fig. 43 (Example 4).

PM' read from the curves of Fig. 46. In col. 4 is the value of'rcltc, which byeq, 27is

In col. 5 is the value of Te read from the curves. In col. 6 is fe, obtained bydividing Te from col. 5 by Tc/tcfrom col. 4.

9. Curve oj stability limit as a Junction oj clearing time is plotted in Fig. 47.

Summary of methods of calculating transient stability. Byrd andPritchard's method, which has just been illustrated, is probably thebest way of finding the critical switching time corresponding to a giventransmitted power, or of finding the transient stability limit for a givenswitching time, for a two-machine pure-reactance system with only oneinstant of switching after inception of the fault.

If the network is not purely reactive (for example, if line resistanceor shunt loads are to be taken into account), this method is not applicable. However, the equal-area method may be used in conjunctionwith pre-calculated swing curves as described on pp. 149-57 of thischapter.

If the system has three or more machines, their swing curves must be


TABLE 5

RECOMMENDED METHODS OF CALCULATING TRANSIENT STABILITY OF A TwoMACHINE SYSTEM

Number of Recommended Method of Calculating

Instants of A BSwitching Examples Critical Switching Time Stability Limit Corre-or Circuit Corresponding to a sponding to a GivenChange Given Power Switching Time

Sustained fault Use equal-area crite-

1Fault cleared instantly rion or Byrd andSwitching out a sound Pritchard's method

line (steps 1 to 5).

(Pure-reactance network)

Plot part of curve by Byrd and Pritchard'smethod.

Fault cleared in finite (Linear impedance network)time by simultane-

Use equal-area method Cut and try by assum-2 ous opening of all

breakers for finding critical ing power and pro-switching angle, then ceeding as in A.pre-calculated swingcurve for findingswitching time.

Fault cleared by se.,quential opening of2 breakers

3 Fault cleared by si- Dse a combination of swing curves, equal-areamultaneous opening criterion, and successive trials.of all breakers, fol-lowed by simultane-ous reclosing

calculated to a value of time when one can determine whether the system is stable for the assumed power and switching time. The procedure is then repeated for different values of power or of switchingtime.

If there is more than one instant of switching after inception of thefault on a two-machine system (for instance, if a fault is cleared bysuccessive opening of two or more circuit breakers), it is still possibleto use the equal-area criterion as an adjunct to swing curves, makingcontinuation of the swing curves beyond the instant of the last switching operation unnecessary. Suppose, for example, that a fault iscleared by successive opening of two breakers at known times and that

CERTAIN FACTORS AFFECTING STABILITY 187

the power limit is sought. A value of power is assumed, and the swingcurve is calculated till the time of opening of the second breaker.The equal-area criterion for stability is then used. The procedure isrepeated for other values of power until one is found which makes thepositive and negative areas equal. This value is the stability limit.A plot of net area against power enables one to find the stability limitby interpolation after two or three trials.

Table 5 lists the foregoing methods of calculation for two-machinesystems with recommendations concerning which to use in each ofvarious situations.

For estimating purposes two sets of curves which are not reproducedhere will be found useful. One of them" gives, for a two-machinereactance system, curves of critical clearing time tc against sin 00 forconstant rl and r2. Four families of such curves are given, each for avalue of rl corresponding to a different type of short circuit at thesending end of a transmission line and for several values .of r2. Theequivalent inertia constant, H = H1H2/(H1 + H2 ) , is assumed to be1.5. If it has a different .value, the switching time read from thecurves must be multiplied by a correction factor V H/1.5.

The other set of curves" gives the critical clearing time of a fault on ametropolitan power system having generators of H = 8 operating atfull load and 0.85 power factor, based on the assumption that thegenerator or group of generators nearest the fault swings with respectto the remaining generators.

Certain factors affecting stability. With a knowledge of the methods of analyzing the stability of the two-machine system that have beendescribed in the preceding pages, we can proceed to draw a number ofgeneral conclusions regarding the effect on stability of certain featuresof apparatus design, system layout, and operation. The effect of eachfeature must be considered under all three conditions-before thefault, during the fault, and after clearing the fault. Some features oflayout or design promote stability during all three conditions, whereasothers are beneficial during one condition but detrimental duringanother.

In the equal-area criterion for stability, the power-angle curves foreach of the three conditions are used. (Refer to Fig. 14.) The factorsdetermining the relative sizes of the two compared areas Al and A 2 are(1) the clearing angle Oc and (2) the amplitudes of the three power-angle curves, Pm, rIPm, r2Pm» relative to the height of the input(initial power) line Pi. Stability is aided by decreasing area A 1 and byincreasing area A 2 • For a given Pi this may be accomplished, fromthe geometrical viewpoint, chiefly by decreasing the clearing angle Oc


and by increasing the amplitudes of the fault and post-fault curves,T1Pm and T2Pm. It is also helpful to decrease the amplitude of thepre-fault power-angle curve if such reduction is possible without at thesame time decreasing the amplitudes of the other two power-anglecurves, because so doing increases the initial angle 00- If any featurechanges the amplitudes of all three curves proportionally, however, it isbeneficial to increase rather than to decrease the amplitudes.

The clearing angle s, depends on the clearing time and on the equivalent inertia constant (eq. 2). The importance of rapid fault clearing,obtainable by the use of high-speed circuit breakers and fast relaying,has already been stressed. For a given clearing time the clearingangle is decreased by increasing the inertia constant. If the inertiaconstants of the two machines are far from equal, the equivalentinertia constant of the system is very nearly that of the lighter machine; hence it is more effective to increase the inertia of the lightmachine than of the heavy one. Seldom, however, has it provedeconomical to increase the inertia of a generator beyond its normalvalue. t But a consideration of the role of inertia sheds some light onthe relative critical clearing times for hydroelectric generators (averageH = 3) and steam turbogenerators (average H = 6) if the circuitreactances are about the same. Since the time varies as the square rootof the inertia constant Ceq. 27), a fault on a hydroelectric system mustbe cleared in about 70% of the critical clearing time of the corresponding steam system.

The amplitude of the power-angle curves is E1E2/X12, where E1 andE2 are the internal voltages of the two synchronous machines and X 12

is the reactance between these voltages, which is, in general, differentfor each of the three circuit conditions. Increasing the internal voltages increases the amplitude of all three power-angle curves and aidsstability. An increase in internal voltages usually accompanies anincrease of load on the machines, but this does not necessarily mean anincrease of the equivalent input Pi. If both generators have localloads which are increased in the proper ratio, Pi is not affected. (Seeeq.3.)

The amplitude of all the power-angle curves is increased by decreasing the reactance X 12 between the machines. This reactance consistsprincipally of the reactances of the two synchronous machines, transformer reactance, and line reactance. A large part of it is in the machines. The transient reactance of each class of large synchronousmachines (steam turbogenerator, water-wheel generator, condenser,etc.) has a characteristic value and does not vary much in normal

tA notable exception is the Boulder Dam station.

CERTAIN FACTORS AFFECTING STABILITY 189

designs. A lower value of reactance (in per unit) is obtained essentially by building a larger machine and under-rating it. This hasseldom proved an economical way to aid stability.§ The reactances oftransformers also have, for a given size and voltage, normal valuesbelow which it is difficult to go. The reactance of an overhead transmission line is only slightly affected by any change of spacing or conductor size which is practical from other standpoints than stability.A decrease in system frequency, of course, decreases the reactances ofall parts of the system and thereby aids stability. Therefore a frequency of 25 or 50 c.p.s. is preferable to 60 c.p.s. from the standpoint ofstability. Nevertheless, because of other advantages, 60 c.p.s. hasbecome the standard power-system frequency in the United States, anda transmission project operating at a lower frequency would necessarilyinclude frequency-changing equipment at one or both ends. This additional equipment might offset any advantage of the lower frequencyin regard to stability. Up to the present time the stability limitationsof 60-c.p.s. systems have not been serious enough to warrant lowerfrequencies. Series capacitors have been used on some transmissionlines to partially compensate for the inductive reactance of the lines.However, they have not been used on any major project where stabilitywas an important factor, although they have been seriously considered.6,7,g

The most important means of reducing the reactance are (1) toconnect more lines in parallel and (2) to raise the transmission voltages.

The ratio rl of the amplitude of the fault power-angle curve to theamplitude of the pre-fault power-angle curve depends upon the typeand location of the fault. The effect of the type of fault will be discussed in Chapter VI. The effect of location will be consideredbriefly now. A fault on a bus or on a line close to a bus is more severethan a fault of the same type near the middle of a line. Most severeof all is a three-phase short circuit at some point (for example, fault bin Fig. 15) where it entirely blocks the transfer of power from onegenerator to the other; then rl = o. Although a fault near the endof a line and one near the middle of the line are equally probable, themost severe location is usually assumed in a stability study so that theresults will be conservative.

The ratio T2 of the amplitude of the post-fault curve to the amplitudeof the pre-fault curve depends on the number and location of lineswhich are opened to clear the fault; therefore it depends upon thefault location and upon the relaying scheme. The most favorable

§Again the Boulder Dam station is a notable exception.


value of T2 encountered in practice is 1. This value applies to a faulton an unloaded radial feeder cleared by disconnection of the feeder.It is also attained with quick-reclosing schemes which restore thefaulted line to service after a brief time to allow for extinction of thearc. II A fault on a bus, although electrically equivalent before clearingto a fault on a line adjacent to a bus, is more severe in its effects afterclearing than a line fault is because several lines must be opened toclear it. Sometimes a comparative study is made between a smallnumber of high-voltage lines and a large number of low-voltage lineshaving the same parallel reactance in per unit. The large number oflines is preferable to the small number from the standpoint of stability,because to clear a fault requires the opening of one line in either case,

24

24

16

J

16 16

FIG. 48.System for which the effectof high-voltagebussingis considered in Prob, 4.

but one line is a small fraction of the entire number of lines when thisnumber is great. For other reasons, however, a smaller number oflines may be preferable.

When two or more parallel lines are used, the number of intermediatebusses is a factor affecting stability. For example, in the system ofFig. 48 it is worth inquiring whether the addition of the two high-voltage busses shown in broken lines is beneficial or detrimental. If thehigh-voltage busses and circuit breakers are provided, a line fault canbe cleared by switching out one line while leaving all the transformersconnected, whereas, without the high-voltage busses, the transformerswould be switched out with the line. Thus the busses increase thevalue of T2 and are beneficial after clearing of the fault. On the otherhand, during the fault the shock to the system is increased by thebusses; that is, Tt is lessened. Which effect predominates dependsupon the speed of clearing. With fast clearing the busses are beneficial; with slow clearing they are detrimental.

The same principle may determine, although less obviously, theeffectupon stability of changes of layout in more complicated networks.The effect of a given change may be either beneficial or detrimental,depending upon the speed of fault clearing. Needless to say,_ many

USee Chapter XI, Vol. II.

PROBLEMS 191

contemplated changes of connections are conceived from other motivesthan the improvement of stability. They may be intended, forexample, to improve voltage regulation, to increase the reliability ofsupply to certain loads, or to relieve overloading of certain lines.Nevertheless the effect of such changes on stability conditions shouldbe considered.

Most of the conclusionswhich have been drawn from a study of thetwo-machine system apply equally well to a multimachine system.When a muItimachine system is unstable, as a rule, it is split into twogroups of machines which go out of synchronism with each other, whilethe machines within each group stay in synchronism. The groupingmay differ with the fault location. Still, for a given fault location,the general behavior of the multimachine system is similar to that of atwo-machine system. It could be analyzed as a two-machine systemexcept that the machines within a group may swing so far with respectto each other (yet without going out of step) as to have a marked effecton the relations between the two groups. In addition, the groupingfor a given fault location is not always apparent in advance.

REFERENCES

1. R. H. PARK and E. H. BANCKER, "System Stability as a Design Problem,"A.l.E.E. Trans., vol. 48, pp. 170-93, January, 1929.

2. I. H. SUMMERS and J. B. MCCLURE, "Progress in the Study of System Stability," A.I.E.E. Trans., vol. 49, pp. 132-58, January, 1930.

3. H. L. BYRD and S. R. PRITCHARD, JR., "Solution of the Two-MachineStability Problem," Gen. Elec. Rev.,vol. 36, pp. 81-93, February, 1933.

4. R. D. EVANS and W. A. LEWIS, "Selecting Breaker Speeds for Stable Operation," Elec. Wid., vol. 95, pp. 336-40, Feb. 15, 1930.

5. S. B. GRISCOM, W. A. LEWIS, and W. R. ELLIS, "Generalized Stability Solution for Metropolitan Type Systems," A.I.E.E. Trans., vol. 51, pp, 363-72, June,1932; disc., pp. 373-4.

6. E. C. STARR and R. D. EVANS, "Series Capacitors for Transmission Circuits,"A. I. E. E. Trans., vol. 61, pp. 963-73, 1942; disc., pp. 1044-6.

7. R. B. BODINE, C. CONCORDIA, and GABRIEL KRON, "Self-Excited Oscillationsof Capacitor-Compensated Long-Distance Transmission Lines," A.I.E.E. Trans.,vol. 62, pp, 41-4, January, 1943; disc., pp. 371-2.

8. J. W. BUTLER, J. E. PAUL, and T. ·W. SCHROEDER, "Steady-State and Transient-Stability Analysis of Series Capacitors in Long Transmission Lines," A.I.E.E.Trans., vol. 62, pp. 58-65, February, 1943; disc., pp. 377-80.

PROBLEMS ON CHAPTER V

1. Check the results of Example 3, part a, by Byrd and Pritchard'smethod.

2. Work Example 4 for a three-phase fault on a 132-kv. line near thehydroelectric-station high-voltage bus.


3. Work Example 4, assuming that the voltages behind transient reactanceare independent of the initial power transmitted.

4. Find the effect of high-voltage bussing on the stability of the 6o-cyclesystem shown in Fig, 48 (consisting of a generator G1 feeding over a doublecircuit high-voltage line to the infinite bus G2) by plotting stability limit inper unit against fault duration in seconds for a three-phase fault at the sending end of one circuit (a) with no high-voltage bussing and (b) with highvoltage bussing at both ends, as indicated by the broken lines. The reactances of the circuit elements are given in per cent based on the rating ofG1. Assume that the voltage of the infinite bus and the sending-endvoltage of the high-voltage lines are 1.00 per unit for all values of initialpower. Explain why the curves cross.

5. Find the effect of fault location on the system of Fig. 48 with highvoltage bussing, by plotting stability limit in per unit as a function of faultlocation as the three-phase fault moves from the sending end to the receivingend of the line. The clearing time is constant at 0.15 sec.

CHAPTER VI

SOLUTION OF FAULTED THREE-PHASE NETWORKS

Symmetrical components. Most power systems are three-phase,but, as long ,as they are symmetrical, they may be represented on asingle-phase line-to-neutral basis for purpose of calculations. Threephase systems are usually symmetrical or very nearly so during normalconditions and also during three-phase faults. During other types offault, such as line-to-ground or line-to-line, they are unsymmetrical,and the simple single-phase representation no longer suffices. In suchcases the method of symmetrical components! is generally used.

In the method of symmetrical components an unsymmetrical set ofvector currents or voltages is resolved into symmetrical sets of components. Before proving that such resolution is possible, let us firstconsider the inverse process, that of adding symmetrical sets ofvectors to obtain an unsymmetrical set.

Consider a symmetrical set of vectors (Vat, Vbl, Vel of Fig. la)representing the balanced voltages or currents of a three-phase circuit.The vectors are equal in magnitude and 1200 apart in phase. Theirphase order, or phase sequence, is abc, which is called the positivesequence. Consider also another symmetrical set of vectors (Va2, Vb2,

Ve2 of Fig. Ib) of the opposite phase sequence, cba. These vectors aresaid to have negative phase sequence. If now corresponding vectorsof the two sets are added, thus

Va' = Val + Va2 [1]

Vb' = »« + Vb2 [2]

v: = Vel + Ve2 [3]

as shown in Fig. Ic, the resultant vectors, Va', v: Ve', form anunsymmetrical set. They are neither equal in magnitude nor 1200

apart in phase.It has been shown that an unsymmetrical set of vectors results from

the addition of corresponding vectors of two symmetrical sets ofopposite phase sequence, By reversal of the process the resultingunsymmetrical set can be resolved into its symmetrical components.

We may well ask, can any unsymmetrical set of three vectors beresolved into two symmetrical sets, one of positive phase sequence, theother of negative phase sequence? This question is answered if we

193

194 SOLUTION OF FAULTED NETWORKS

notice that, since the sum of the three vectors in each symmetrical set iszero, the sum of the three vectors in the resulting unsymmetrical set isalso zero.

Val + Vbl + Vel = 0 [4]

Va2 + Vb2 + Ve2 = 0 [5]Adding, we get

(Val + Va2) + (Vbl + Vb2 ) + (Vel + Vc2) = 0 [6]or

(a)

[7]

\\\~o VbO VeO

(d)

FIG. 1. Composition of unsymmetrical vectors from symmetrical components.(a) Positive-sequence vectors. (b) Negative-sequence vectors. (c) Addition ofnegative-sequence vectors to positive-sequence vectors. (d) Zero-sequencevectors.(e) Addition of the sum of positive-sequence and negative-sequence vectors to zero-

sequence vectors.

In general, the sum of the three vectors in an unsymmetrical set is notzero but, instead, may have any value. Obviously, an unsymmetricalset whose sum is not zero cannot result from the addition of two sets,each having a sum of zero.

Suppose, however, that we add to the positive-sequence and negativesequence sets a set of equal vectors, Van, VbO, VeO (shown in Fig. Id),known as a zero-sequence set. The resulting set of vectors (shown inFig. Ie) is:

v, = v: + Val = VaO + Val + Va2

Vb = VbO + Vb' = V bO + V bl + Vb2

v, = VeO + V e' = V eO + Vel + Ve2

[Sa]

[8bl

[Sc]

SYMMETRICAL COMPONENTS 195

These vectors are unsymmetrical, and their sum is not zero. Can anyunsymmetrical set of three vectors be resolved into three symmetricalsets-a positive-sequence set, a negative-sequence set, and a zerosequence set? It can if eqs. 8 can be solved for the symmetrical components in terms of the given vectors Va, Vb, Ve• In eqs. 8 there appearto be nine unknown components, but actually only three of them areindependent because the three vectors of each set are determined byanyone of the three, say by the phase a vector, and by the symmetryof the set.

Vbl = Va1!240° = a2Va l

Vel = Val/1200 = aVal

Vb2 = Va2/120° = aVa2

Ve2 = Va2/240° = a2Va 2

VbO= VeO= VaOwhere

a = /1200 = -0.500 + jO.866and hence also

a2 = /2400 = -0.500 - jO.866

a3 = 1

a4 = a

Substituting eqs. 9 into eqs. 8, one obtains

v. = VaO+ Val + Va2

Vb = VaO + a2Va1+ aVa2

v, = VaO+ aVal + a2Va2

Subscript a may now be dropped, giving

Va = Vo+ Vt + V2

v, = v, + a2Vt + aV2

v, = Vo + aVI + a2V2

[9a]

[9b]

[ge]

[9d]

[ge]

[10]

[11]

[12]

[13]

[14a]

[14b]

[14c]

[1Sa]

[I5b]

[15e]

It is desired to solve eqs. 15 for Vo, v; and V2 in terms of Va, Vb, andVe• The simplest method of solution follows. To obtain Vo, add thethree eqs. 15 together, obtaining

Va+ Vb + Vc = 3Vo+ (1 + a + a2) (Vl + V2 )

196

Note that

Hence

SOLUTION OF FAULTED NETWORKS

1 + a +a2 =0 [16]

[17]

To obtain VI, multiply the three eqs. 15 by 1, 8, and a2, respectively,obtaining

v; = Vo + VI + V2

aVb = aVo + VI + a2Y2

a2Ye = a2yo + VI + aV2

Then add, obtaining

Va + aVb + a2y c = (1 + a + a2) (y o+ V2 ) + 3V1whence

VI = i(Va + aVb + a2Ve) [18]

To obtain V2, multiply the three eqs. 15 by 1, a2, and a, respectively,add, and divide by 3, obtaining

V2 = i(Va + a2Vb + aYe) [19]

The symmetrical components of any three given vectors Va, Vb, v, are,therefore

Vo = i(Va + Vb + Ve)

VI = i(Va + aVb + a2Ve)

V2 = l(Ya + a2Vb+ aYe)

[20a]

[20b]

[20c]

Any three given vectors can be resolved into symmetrical componentsof positive, negative, and zero phase sequence by using eqs. 20. Theoperations indicated by eqs. 20 may be performed either graphicallyor arithmetically. If the given vectors comprise a symmetrical set of aparticular phase sequence, the other two components will vanish. Ifthe sum of the given vectors is zero, they have no zero-sequence component. The vectors may represent any alternating quantities associated with a three-phase circuit or machine, such as line-to-neutral,line-to-ground, or line-to-line voltage; current; or flux. In thischapter we are concerned with voltage and current vectors.

The actual or phase voltages are given in terms of their symmetricalcomponents by eqs. 15, and the symmetrical components are given interms of the actual voltages by eqs. 20. The corresponding equationsfor current are:

[21a]

SYMMETRICAL COMPONENTS

Ib = 10 + a211 + aI2

I, = 10 + all + a212

10 = J(la+lb+le}

11 = i n,+ alb + a2I c)

12 = l(la + a2Ib + ale)

197

[21b]

[21c]

[22a]

[22b]

[22c]

An alternative form of eqs. 15 and 20, obtained by using the rectan-gular form of operators a and a2 (see eqs. 10 and 11), is as follows:

Va = v; + (VI + V2 ) [23a]

Vb = Vo - 0.500(V1 + V2) - jO.866(VI - V2) [23b]

Ve = Vo - O.500(V1 + V2) + jO.866 (VI - V2 ) [23c]

Vo = l[Va + (Vb + Vc) ] [24a]

VI = i[Va - O.500(Vb + Vc) + jO.866 (Vb - Ve) ] [24b]

V2 = irva - 0.500(Vb + Vc) - jO.866(Vb - Vc)] [24cJ

This form is convenient if it is desired to compute with complex numbers all in rectangular form. Needless to say, similar equations holdfor currents.

EXAMPLE 1

Calculate the symmetrical components of the following unbalanced lineto-neutral voltages:

Va = 100/90° volts

Vb = 116LQvolts

Vc = 71 /225° volts

Solution 1. The symmetrical components are given by eqs, 20.

v, = 100/90° 0 + jl00

v, = 11612 116+ jO

Vc = 71/225° = -50 - j503Vo = Va+ Vb + Vc = 66+ j50 =,83/37°

Vo = 22 + j17 = 28/37° (Ans.)


s, = 100/90° = 0 + j100

aVb = 116/120° = -58 + j100

a'lYe = 71/105° = -18 + j68

3VI = VI + aVb + a'rYe= -76 + j268 = 279/106°

VI= -25+ j89 = 93/106° (Ans.)

Va = 100/90° = 0+ jl00

a"Vb = 116/240° = -58 - jl00

aVe = 71/345° = +68 - j18

3V2 = Vo + a"Vb + aVe = +10 - j18 = 21/299°

V2 = + 3 - j6 = 7/299° (Ans.)

The symmetrical components are, therefore:

Vo == 22+ j17 = 28/37° volts

VI= -25 + j89 = 93/106° volts

V2 = +3 - j6 = 7/299° volts

Check 1. From the symmetrical components which have just beenfound, the phase voltages will be calculated by eqs. 15 as a check.

Vo= 28/37° = 22+ j17

a"VI = 93/346° = 90 - j23

aV2 = 7/59° = 4 + j6Vb = Vo+ a?:VI + aV2= 116+. jO (Check)

Vo = 28/37° = 22+ j17

aVI = 93/2260 = -65 - j67

aZV2 = 7/179° = - 7 + jOVe = Vo+ aVl + a?:V2 = -50 - j50 (Check)

SYMMETRICAL COMPONENTS

Solution 2. Equations 24 will be used.

Vb= 116+ jO

Vc = -50 - j50

Vb + Ve = 66 - j50

O.500(Vb +Vc) = 33 - j25

Va = 0+ jl00

3Vo = 66 + j50

Va - O.500(Vb + Ve) = -33 + j125

Vb - Ve = 166+ j50

O.866(Vb - Ve) = 143+ j43

jO.866(Vb - Vc) = -43 + j143

3V1 = -76 + j268

3V2 = 10 - jlS

The symmetrical components are

Vo = 22+ j17 volts

VI= -25 + j89 volts

V2 = 3 - j6 volts

agreeing with Solution 1.Check 2. Equations 23 will be used.

VI = -25+ j89

V2 = 3 - j6

VI+ V2 = -22+ j83

O.500(VI +V2) = -11 + j42

Vo = 22+ j17

Va = 0 + jl00 (CMc1c)

Vo - O.500(VI +V2) = 33 - j25

Vi - V2 = -28 + j95

O.866(V1 - V2) = -24 + j82

jO.866(V1 - V2) = -82 - j24

Vb = 115 - jl (Should be 116 - jO)

Vc = -49 - j49 (Should be -50 - j50)

199

200, SOLUTION OF FAULTED NETWORKS

[25a]

[25b]

[25c]

Sequence impedances. The most important property of symmetrical components of current and voltage is that, in a balanced threephase circuit or part of it, the three sequences are independent. Abalanced three-phase circuit is one having similar connections in allthree phases, equal self-impedances in all three phases, and equalmutual impedances between each pair of phases. The statement thatthe three sequences are independent means that currents of each phase

sequence will produce voltage drops of thesame phase sequence only. For example,if positive-sequence currents flow througha balanced circuit, the resulting voltagedrops are exclusively of positive sequence,and their zero-sequence and negativesequence components are zero.

The independence of the sequences in anyFIG. 2. A balanced three- particular form of balanced circuit may bephase circuit element with shown algebraically. This will be done forself- and mutual impedances. the circuit shown in Fig. 2, consisting of

equal series self-impedances Zs in the threephases and equal mutual impedances Zm between each pair of phases.The relation between phase currents la, Ib, I, and phase voltage dropsVa, Vb, Ve is given by the following equations:

·Va = z.r, + Zmlb +Zmle

Vb = Zmla +Zslb + ZmIe

Ve = Zmla + ZmIb + z.r,It is now necessary to find relations between the symmetrical components of voltage and current similar in form to eqs. 25. To obtainthese relations, substitute eqs. 25 into eqs. 20 and then use eqs. 22.

v; = lcva +v, + Ve)

== i CIa + r, + Ie) CZs + 2Zm)

= CZs + 2Zm)Io

VI = leVa + aVb + a2Ve)

= Zs!Cla + alb + a2l e) + Zm!CIb'+ I, +ala + ale +a21a + a2Ib)

= ZsI1 + zmiC-Ia - alb - a2Ie)

= (Z8 - Zm)I1

Similarly',

SEQUENCE IMPEDANCES

These equations may be written as

Vo = Zolo

VI = z.r,V2 = Z2I2

whereZo = Z, + 2Zm

Zl == Z2 = Z, - Zm

201

[26a]

[26b]

[26c]

[27a]

[27b]

Unlike eqs. 25, in which the value of each phase voltage depends uponall three phase currents, in eqs. 26 the voltage of each sequence dependsonly upon the current 'of the same sequence. The ratio of sequencevoltage to sequence current may be called a sequence impedance; thusZo is the zero-sequence impedances Zl, the posiiioe-sequence impedance;and Z2, the negative-sequence impedance. The positive- and negativesequence impedances of this particular circuit are equal; indeed, forany static three-phase circuit they are equal. For rotating machinery,however, the positive-sequence impedance usually differs from thenegative-sequence impedance because of the mutual reactances between rotor and stator windings. The impedances of synchronousmachines are considered further in Chapters XII and XIV, Vol. III.

Equations 27 show that, if Z8 and Zm are of the same sign (or, moreprecisely, if their vectors lie approximately in the same direction, asthey do for a transmission line or cable, where each of the three phasesis considered to consist of a wire with ground return), then the zerosequence impedance is higher than the positive- or negative-sequenceimpedance. If, on the other hand, Zm is of opposite sign from Zs, asit is for the stator windings of a three-phase machine, then the zerosequence impedance is lower than the positive- or negative-sequenceimpedance. Let us show that Zm of the machine is a negative reactance. It is a familiar fact that, if two coils are placed initially withtheir axes parallel and if one coil is then turned with respect to theother, their mutual inductance diminishes, reaching zero at 90°; and,upon further turning of the coil, the mutual inductance becomesnegative and is still so at 120°, which is the angular separation of thecoils of a three-phase machine.

The sequence impedances may be calculated from the phase selfand mutual impedances, as illustrated by eqs. 27 for a particular case;or they may be measured. For example, the positive-sequenceimpedance of an induction motor running at any given speed may bedetermined by impressing known positive-sequence voltages on its


[28a]

[28b]

[28c]Ie + Ie' + Ie" = 0

FIG. 3. A three-phasejunction.

terminals, measuring the resulting positive-sequence currents, andtaking the ratio of voltage to current. The negative-sequence impedance may be measured in similar fashion by impressing negativesequence voltage on the terminals and driving the rotor at the same

speed as before. For measuring the zerosequence impedance a single-phase voltagemay be impressed between the neutralpoint and the three line terminals tied together.

Kirchhoff's laws. In the preceding section it was shown that the symmetricalcomponents of current and voltage obeyOhm's law, V = ZI. It will now be shownthat they obey Kirchhoff's laws as well.Kirchhoff's law of currents states that thesum of all the currents flowing into a junc

tion (node) is zero. At a three-phase junction, as illustrated in Fig.3, this law applies separately to each phase, thus:

la + i: + la" = 0

Ib + Ib' + I b" = 0

If we add the three equations together and divide by 3, we obtain

!(la + Ib + Ie) + !(Ia' + Ib' + Ie') + i(Ia" + Ib" + Ie") = 0

or10 +10' + 10" = 0 [29]

Also, by multiplying eqs. 28a, 28b, and 28c by 1, a, and a2, respectively,

adding the resulting equations together, and dividing by 3, we obtain

11 + 11' + II" = 0 [30]

By a similar procedure, with a and a2 interchanged, we get

[31]

Equations 29, 30, and 31 show that the sequence currents (symmetricalcomponents of current) obey Kirchhoff's current law just as the phasecurrents do.

Kirchhoff's voltage law states that the sum of all the voltage drops(or rises) around a closed path (loop or mesh) equals zero. In a three-

THE SEQUENCE NETWORKS 203

[32a)

[32b]

[32c]

Va + Va' + Va" = 0

Vb + Vb' + Vb" = 0

v, + Ve' + v: == 0

phase circuit the law holds for each phase. Thus for the circuit ofFig. 4 we have

FIG. 4. A three-phase mesh.

These equations are of the same form as eqs. 28; therefore equationscan be derived from them similar to eqs. 29, 30, and 31. They are

v; + Vo' +Vo" = 0 [33a]

VI + VI' + VI" = 0 [33b)

V2 + V2' + V2" = 0 [33c]

Thus the sequence voltages obey Kirchhoff's voltage law.The sequence networks. It has been shown that in a balanced

three-phase network the symmetrical components of voltage and current obey Ohm's law and Kirchhoff's laws, the same laws followed bythe actual voltages and currents. There is thus good justification forthe concept of sequence networks. The actual three-phase network,in which the actual voltages and currents exist, is replaced for purposesof analysis by three fictitious single-phase networks, in which the symmetrical components or sequence voltages and currents exist. Theyare the positive-sequence network, in which the positive-sequence voltage and current exist; the negative-sequence network, in which thenegative-sequence voltage and current exist; and the zero-sequencenetwork, in which the zero-sequence voltage and current exist. The


positive-sequence network is identical to the single-phase impedancediagram considered in Chapter III. The negative-sequence networkis very much like the positive-sequence network but differs from it inthe following respects: (1) ordinarily there are no negative-sequencegenerated electromotive forces; (2) the negative-sequence impedanceof rotating machinery is different from the positive-sequence impedance; and (3) the phase displacement of transformer banks for negative sequence is of opposite sign to that for positive sequence. Thezero-sequence network differs greatly from the other two in that (1)the impedance of transmission lines is higher than for positive sequence;and (2) the equivalent circuits of transformers are different. Consequently, the zero-sequencenetwork is given further consideration laterin this chapter.

In so far as the three-phase network is balanced, the three sequencenetworks are independent; that is, they are not connected or coupledto one another. But wherever the three-phase network is unbalanced,there is a connection or coupling between t\VO or three of the sequencenetworks at the corresponding points of unbalance. The nature of theconnection or coupling between the sequence networks will be determined in the next section for the types of unbalances most importantin stability studies, that is, for short circuits.

Consider the component parts of a three-phase power network tosee whether they are balanced. Rotating three-phase machines,transformer banks consisting of three identical units, cables, andtransposed overhead lines are balanced. Untransposed overheadlines are only slightly unbalanced. Loads, although often unbalancedwhen considered in small portions, are approximately balanced whenthe whole load of a substation is considered. Therefore, all the majorparts of a three-phase power network are either exactly or very nearlybalanced during normal conditions. During fault conditions the samestatement holds for all parts of the network except for the fault itselfif the fault is of some type other than three-phase. Therefore, thesequence networks representing the power system are entirely independent of one another under normal conditions and are connected onlyat the point of fault under unbalanced fault conditions.

The usual three-phase power system has no generated electromotiveforces except symmetrical positive-sequence ones. Therefore thesequence networks representing such a system have no generatedelectromotive forces except those in the positive-sequence network.The negative-sequence and zero-sequence networks, having no electromotive forces within themselves and not being connected to the positive...sequence network under balanced conditions, are "dead." There-

REPRESENTATION OF SHORT CIRCUITS 205

fore only the positive-sequence network need be considered under suchconditions. Under unbalanced fault conditions, however, the negativesequence network and, for some types of faults, also the zero-sequencenetwork are energized by their connection to the positive-sequence network at the point of fault. The nature of these connections willno,v be considered.

Representation of short circuits by connections between thesequence networks. Short circuits on a three-phase system may beclassified into the following types: one-line-to-ground, line-to-line,

One- Llne-to- Ground

;~ Phaseb

g,,:,

Phase c

Line-to. LinePhases band c Phases c anda Phases a andh

!=c :=E =t=

Iwo-Line-to· GroundPhases band c Phases c anda Phases a andb

;=:F =F =Fg~ : ~

Three· PhaseInvolving ground Not involving ground

.~~ :=Eg-:r

FIG. 5. Types of short circuit on a three-phase network.

two-line-to-ground, and three-phase. Three-phase short circuits maybe further classified into those involving ground and those not involving ground. The other types may be further classified by thephases involved. The different types of short circuit are showndiagrammatically in Fig. 5.

The procedure for finding the connections of the sequence networkscorresponding to a fault on a three-phase network consists of four steps:

1. Draw a circuit diagram of the fault on the three-phase network,labelling the pertinent phase voltages and currents by suitable lettersymbols and by arrows denoting their positive directions.

2. Write three equations expressing relations between the phase


(a)

Posmvesequencenetwork

Negativesequencenetwork

Zero·sequencenetwork

a c 14 ....

I ., VaThree· b~

b .....,phase Ie ..... Vb

network C~ "tVe

g~

currents and voltages or particular values thereof imposed by thefault condition.

3. Convert these equations to three corresponding equations insequence currents and voltages. This conversion is accomplishedby using eqs. 15, 20, 21, and 22.

4. Find connections of the sequence networks satisfying theequations obtained in the last step.

The procedure just outlined will now be carried out for each type ofshort circuit, commencing with the line-to-ground short circuit on

phase a. Line-to-ground short circuits on phase b or phase c may betreated by renaming the phases soas always to call the faulted phase,phase a. The relations of the symmetrical components come out mostsimply when the fault is taken onphase a, which was adopted as thereference phase. (That is, the components of the other phase vectorswere expressed in terms of thecomponents of the phase-a vector.)For the same reason line-to-lineand two-line-to-ground faults willbe taken on phases band c, thusmaking the faults symmetrical withrespect to phase a.

The three-phase network, whatever its form, may be symbolizedby a box having four terminals atthe point of fault, namely, phases a,b, and c and the ground, as shownin Fig. 6a. Different types of shortcircuit may be applied to the network by making the appropriateconnections between two, three, orfour of these terminals. Each ofthe three sequence networks may

be represented similarly by a box having two terminals at the pointof fault (line terminal and neutral terminal) as shown in Fig. 6b.The connections between these terminals, corresponding to the connections between terminals of the three-phase network, will be derived.

Let Va, Vb, and Vc denote the line-to-ground voltages at the point of

(b)

FIG. 6. (a) Schematic representation of a three-phase network withterminals at the point of fault. (b)Schematic representation of the cor-

responding sequence networks.


fault, and let la, Ib, and I, denote the phase currents flowing into thefault, as marked on Fig. 6a. Let Vo, Vt, V2 be the symmetrical components of Va' Vb, Vc; and let 10, It, 12 be the symmetrical componentsof la, Ib' r,

Line-to-ground short circuit on phase a. (See Fig. 7a.) The line-to-ground voltage of the grounded phase is zero, and the currents flowinginto the fault from the two unfaulted phases are zero. Hence theequations for phase voltages and currents are:

From eqs. 34 and 15av, + VI + V2 = 0

From eqs. 35, 36, and 22

10 = II = 12 = i1a

[34]

[35]

[36]

[37]

[38]

Hence the relations between symmetrical components of fault voltageand current, corresponding to eqs. 34, 35, and 36 for actual faultvoltages and currents, are

Vo + VI + V2 = 0

10 = II = 12

[39a]

[39b]

A series connection of the three sequence networks (Fig. 7b) satisfiesthese equations.

Line-to-line short circuit on phases band c. (See Fig. 7c.) Theequations for phase voltages and currents are

Vb = Va

I, = 0

Ib+ t, = 0

Substituting eqs. 15 into eq. 40,

v; + a2Vt + aV2 = Vo+ aVI +a2V

2

Subtracting V0 from each side and rearranging, we get

V1 (a2- a) = V2 (a2

- a)whence

[40]

[41]

[42]

[43]

4~t

3q,

bI:

tv:V

aC

+~

bg

-

- (a)

....1 0

.....

~tvo

Zer

o-

o....

1 1

POSe

"tV

Io

o....

1 2'"

Neg

.tV

2

-(b

)

One

-line

-to

·-gro

und

a

3q,

b c g - (c) ::M

oZ

ero

tvo

-....

11.....

7't Vl

POSe

o.....

1 2

Neg

.~t V2

(d)

Line

-to

-line

ao

-+

--f>

b3q

,C

::tm

Jg

v,V

b

(e)

~o

Zer

o-t....

r

'r-Vo 11 ....

Pas.

-t-,

VI 1 2 .....

Neg

.~t-

.,V

2

(I)

Two

•lin

e•t

o-gr

ound

~la

aI C

I~

3q,

6r,

3q,

bI b

co

ct

IeC

tIe

g~

gV

bV

oVe

aV

"v,

Va

(g)

(i)

m1 0

I1f.

1~

~

Zer

otV

o§

Zer

oHv

o0 Z 51

11

~~ ~

POSet+

~_

Pas.

tVI

t3 t;rj

tj Z t;rj

J-31 2

~:d 0

Neg.Jt~

_N

eg.tY

2~

- (h)

(j)

Thr

ee-p

hase

Thr

ee·p

hase

noti

nvol

ving

grou

ndin

volv

ing

grou

nd

FIG

.7.

Con

nect

ions

betw

een

the

sequ

ence

netw

orks

corr

espo

ndin

gto

vari

ous

typ

esof

sho

rtci

rcui

ton

ath

ree-

phas

ene

twor

k.


Substituting eqs. 41 and 42 into the expression for 10 in eq.22a, weobtain

10 = 0 [44]

Substituting eq. 44 into the expressions for Ib and Ie in eqs, 21 andsubstituting the results into eq. 42, we obtain

(a2 + a) (11 +12) = 0or

[45]

Hence the relations between symmetrical components of voltage andcurrent for the line-to-line short circuit are:

VI = V2

11 + 12 = 0

10 = 0

[46a]

[46b]

[46c]

These relations are realized by a parallel connection oj the positive- andnegative-sequence networks, leaving the zero-sequence network opencircuited. (See Fig. 7d.)

Two-line...to-ground short circuit on phases band c. (See Fig. 7e.)The equations for phase voltages and currents are:

[47a]

[47b]

[47c]

These are like eqs. 34 to 36 for a one-line-to-ground short circuit,except that voltage and current are interchanged. Therefore theequations in symmetrical components will be like eqs. 39 except for aninterchange of voltage with current. They are:

10 + 11 + 12 = 0

Vo = VI = V2

[48a]

[48b]

These equations are realized by a parallel connection oj the three sequencenetworks (Fig. 7f).

Three-phase short circuit not involving ground. (See Fig. 7g.) Theequations of phase voltages and currents are:

Va = v, = v,Ia + Ib+ I, = 0

[49]

[50]


Substituting eqs. 49 into eqs. 20, we get

VI = V2 = iVa(l + a + a2) = 0

Substituting eq. 50 into eq. 22a for 10 :

10 = 0

Hence the equations of symmetrical components for this type of faultare:

[5Ia]

[51b]

[5Ic]

[52a]

[52b]

[52c]

[53a]

[53b]

[53c]

They are satisfied by short-circuiting the positive- and negativesequence networks and leaving the zero-sequence network opencircuited (Fig. 7h).

Three-phase short circuit involving ground. (See Fig. 7i.) Theequations of phase quantities are:

v, == 0

Vb == 0

v, = 0

The corresponding equations for symmetrical components are:

Vo = 0

VI = 0

V2 == 0

The equivalent circuit consists of a short circuit on each of the threesequence networks (Fig. 7;).

Ordinarily no distinction need be made between the two types ofthree-phase short circuit, because the zero-sequencenetwork is "dead"whether or not its terminals are short-circuited. The negativesequence network is also dead during a three-phase short circuit, andthe zero-sequence network is dead during a line-to-line short circuit.These dead networks are shown in Fig. 7 for the sake of completeness,but they are disregarded in fault calculations except in the rare instances where (1) the three-phase network contains unsymmetricalgenerated electromotive forces, or (2) there is a second fault or otherunbalance in addition to the one being considered.

Solution of faulted three-phase networks by the method of symmetrical components. Any three-phase network which is symmetrical

SOLUTION OF FAULTED NETWORKS 211

throughout except for one unsymmetrical short circuit can be represented by its three sequence networks connected at the point of faultas shown in Fig. 7. When the appropriate connections have beenmade, the current or voltage anywhere in anyone of the sequence networks is equal to the corresponding component of current or voltageat the corresponding point of the three-phase network. If the valuesof the phase currents and voltages are desired, they can be computedby combining the sequence currents and voltages in accordance witheqs. 15 and 21 or eqs. 23. In stability studies, however, the symmetrical components usually suffice.

This method of solving a faulted three-phase network is well suitedto the calculating board. After the three sequence networks have beenset up on the board, connections can be made readily for representingany type of fault at any location. The sequence currents and voltagescan then be read by plugging in the measuring instruments.

If the network is not too complicated, it is feasible to solve it byalgebraic reduction of the sequence networks.

ExAMPLE 2

A line-to-ground short circuit occurs on bus D of the power system of Fig.8a. Find the symmetrical components of current and voltage and the phasecurrents and voltages throughout the network if the internal voltages ofgenerators A and B are equal and in phase, both having the value jl.05 perunit. The sequence reactances in per unit are as marked on the sequencenetworks of Fig. 8b, and resistances are assumed to be negligible. GeneratorA is ungrounded; B has a solidly grounded neutral.

Solution. The short circuit is assumed to be on phase a. It is represented by connecting the three sequence networks in series as shown bythe broken lines in Fig. Sb. Note that zero-sequence current cannot flowin ungrounded generator A; hence the zero-sequence network is open at thispoint (between Co and 00, Fig. 8b).

The positive-sequence network is reduced as shown in Fig. 9 to one e.m.f,in series with one reactance. During the process of reduction the identityof the e.m.f.'s of generators A and B is retained until step c, so that the results can be used later inExample 3. Then, as the e.m.f.'s are assumed to beequal in the present example, points Al and BI of Fig. 9c are joined to giveFig. 9d. In Figs. 10 and 11 the negative-sequence and zero-sequence networks, respectively, are reduced each to a single reactance. Then in Fig. 12the three reduced networks are connected in series, as the original networkswere in Fig. 8, to represent the line-to-ground short circuit; and the combined circuit is further reduced to an e.m.f, and a reactance. In Fig. 13 thecurrent through the reactance is found by dividing the e.m.I, by the reactance, and the voltages across the sequence networks are found by multiplyingthe current by the respective reactances. The terminal voltages and cur-


(a)

Positive- Sequence Network0 1

Negative· Sequence NetworkO2

0.10

zero-Sequence Network

0 0 0-- - - - - - - ---'(b)

FIG. 8. (a) One-linediagram of a powersystem used to illustrate fault calculations.(b) The sequence networks of the system with values of reactance in per unitmarked thereon. The sequence networks are connected by broken lines so as to

represent a line-to-ground short circuit at point D. (Example 2.)


Al 0.08 D1 0.03 81

0.25 0.20rv _0.13_ rv

(a)°1

(6)

(c)

(e)(d)

0.216

01 01

FIG. 9. Reduction of positive-sequence network (Example 2).

0.293


0.08 D2 0.03

(a)

(6)

D2

0.01 D2

(c) (d) (e)

0.090.193 0.136 0.08

02 02

FIG. 10. Reduction of negative-sequence network (Example 2).


0.14 Do 0.054

(a)

0.054

215

Do 0 00-- - - - - ---..(b)

(d) 0.066

0 0

0.046 0.02::::::::J (c)

FIG. 11. Reduction of zero-sequencenetwork (Example 2).

·105'~1 D2 0.09 O2 Do0.066 00r-VYV~--~--~L ::-=- ..J (0)

jl.05

ctt::J-+-- (b)

FIG. 12. (a) Connection of the reduced sequence networks to represent a lineto-ground fault. (b) Final reduction. (Example 2.)

..---N 3.62

LV +il.OS:JjO.57


il ~ 3.62

..... " ~- 1.54 2.08in

sfrvin

0 ci.~ .~

.....°1

(0) (b) (c)

FIG. 14. Determination of the positive-sequence currents and voltages by reexpansion of the positive-sequence network. See Fig. 9 for values of reactance.

(Example 2.)


rents of the sequence networks thus found are the symmetrical componentsof the voltages at the point of fault and of the current in the fault.

Each sequence network is now re-expanded by going through the steps ofthe network reduction backwards, finding all currents and voltages at each

~D2

~D2

3.62 3.62

3.62 -

~1.50 ~ 2.12

°23.62

(a)

(b) °2 (c)

D2Jo.041 3.621.50 I ( jO.06 jO.03) I 2.12

I 150 I 2.12 IjO.23 t · t jO.29 t jO.26

(d)

1.27 D2 2.35

- jO.OJ ~""'-<f----'.......----+-jO.03- 2.12

0.23 jO.33 ~ /0.26

°2 w)FIG. 15. Determination of the negative-sequence currents and voltages by reexpansion of the negative-sequence network. See Fig. 10 for values of reactance.

(Example 2.)

step. This re-expansion is illustrated in Figs. 14, 15, and 16. Then inFig. 17 the sequence networks are redrawn with the values of the sequencecurrents and voltages marked on them.

The phase currents and voltages are computed from their symmetricalcomponents in Table 1 by use of eqs, 23 and are shown on the circuit diagramof Fig. 18. It should be noted that, since the generator e.m.f.'a were assumed to be imaginary (that is, 90° ahead of the reference phase), and sincethe impedances of the network were assumed to be purely reactive, all thesequence currents are real, and all the sequence voltages are imaginary. Asimilar statement holds for phase a currents and voltages, but not for phasesband c.


(a)

3.62 3.62Do 0----<]0-_--<10--.

! jO.17 ~ I '0.07jO.24 iJ

°00---------'3.08

0.54 Do 3.08

(b)

(e)

jO.08 jO.17jO.24

i jO.l6 jO.09(d)

°0FIG. 16. Determination of the zero-sequence currents and voltages by re-ex-pansion of the zero-sequence network. See Fig. 11 for values of reactance.

(Example 2.)

FIG. 17. Sequence currents and voltages in the power system of Fig. 8(Example 2).

TA

BL

E1

tD 0CA

LCU

LATI

ON

OF

PHA

SEC

UR

REN

TSA

ND

VO

LTA

GES

(EX

AII

PL

E2)

t"'4 d t-3 .....C

urr

ents

Vol

tage

s0 Z

r--

,.,

,,.

"G

en.

Gen

.L

ine

Lin

eL

ine

Lin

eB

us

Bu

sB

us0 ~

AB

CD

CE

ED

tE

DS

CD

E~

111

.54

2.0

81

.29

0.2

51

.16

1.1

6jO

.67

jO.5

7jO

.M>

121.

502

.12

1.2

70

.23

1.1

71

.17

-jO

.23

-jO

.33

-jO

.26

d ~11

-12

0.0

4-0

.04

0.0

20

.02

-0.0

1-0

.01

jO.9

0jO

.90

jO.9

0t.'=

j11

+12

3.0

44

.20

2.5

60

.48

2.3

32

.33

jO.4

4jO

.24

jO.3

8tj

1 00

3.6

20

.54

-0.5

41

.68

1.4

0-j

O.1

6-j

O.2

4-j

O.0

7Z

r,3

.04

7.8

23

.10

-0.0

64

.01

3.7

3jO

.28

0jO

.31

t.'=j

t-3!(

11+

12)

1.5

22

.10

1.2

80

.24

1.16

1.1

6jO

.22

jO.1

2jO

.19

=a1 0

-!(

Il+

12)

-1.5

21

.52

-0.7

4-0

.78

0.5

20

.24

-jO

.38

-jO

.36

-jO

.26

00.

866(

11-

12)

0.0

3-0

.03

0.0

20

.02

-0.0

1-0

.01

jO.7

SjO

.7S

jO.7

8~ ~ ..... co


BusD4.01+jO

0.52+jO.otO.52-jO.Ol

BusE

10.86+jO

BusC3.10+jO

-0.74 - jO.02

- 0.74+jO.02

-1.52 +jO.03 00 ~ ~

~~~....... I I+coco0"":"":00

I

\0\0('t)('t)

000.....................+ I IOcoCO,,-,,-

00I

-O.06+jO- 0.78- j 0.02- 0.78+jO.02

\0\0_NNa ~~....... I I+ coco0"-"-

ddI

FIG. 18. Phase currents and voltages in the network of Fig. 8 with a line-toground short circuit on phase a of bus D. Values of current in arrow directionare marked on the horizontal lines; values of line-to-ground voltage, on the vertical

lines (busses). (Example 2.)

Fault shunts. Since the internal electromotive forces of a threephase synchronous machine are of positive sequence, and since nopower results from the combination of positive-sequence voltages withnegative-sequence or zero-sequence currents, the generated power of asynchronous machine and the synchronizing power between the varioussynchronous machines of a power system are positive-sequence power.Therefore the positive-sequence network is of primary interest in astability study, and the zero- and negative-sequence networks are onlyof secondary interest.

In the positive-sequence network a short circuit can be representedby connecting a shunt impedance Zp at the point of fault. The valueof ZF depends upon the type of fault and upon the impedances Zo andZ2 of the negative- and zero-sequence networks as viewed from thepoint of fault.

The formula for the impedance of the fault shunt may be determinedin either of two ways. The first way is by inspection of the connections between the sequence networks for representing the various typesof short circuits (Fig. 7). The line-to-ground short circuit is represented by connecting in shunt with the positive-sequence network

FAULT SHUNTS 221

{57]

[56]

[541[55J

Two-line-to-ground

Three-phase

at the point of fault the series combination of the zero-sequence andnegative-sequencenetworks; the line-to-line short circuit is representedby thus connecting the negative-sequence network only; and thetwo-line-to-ground short circuit is represented by the parallel combination of the zero- and negative-sequence networks. Hence theimpedances of the fault shunts are as follows:

Type of Impedance ofShort Circuit Fault Shunt, Zl'

Line-to-ground Zo + Z2Line-to-line Z2

ZoZ2z, +Z2

o

The second way of finding the impedance of the fault shunt is totake the two equations

Vo = -ZoIo

V2 = -Z2I2

[58]

[59]

which are independent of the type of fault, together with the threeequations giving relations between, or special values of, the sequencecurrents and voltages at the fault, and to eliminate from the five equations the four quantities Vo, 10, V2, 12, thereby obtaining one equationof the form

[601

[61J

For example, for a line-to-ground short circuit on phase a, eqs. 39 areused. Substitution of.eqs. 58 and 59 into eq. 39a gives

-ZoIo+VI - Z2I2 = 0

VI = ZoIo + Z2I2

By eq. 39b10 and 12 may be replaced by It, giving

VI = (Zo + Z2)II

By comparison of eqs. 60 and 61 the impedance of the fault shunt is

ZF = z, + Z2 [62]

agreeing with eq. 54. The same result would have been obtained ifthe short circuit had been taken on phase b or c instead of on a.

It may be worth pointing out that, although the connections ofFig. 7 are valid even if the zero- or negative-sequence networks containgenerated electromotive forces, the expressions given for the imped-


ances of fault shunts are restricted to the normal condition, in whichsuch e.m.f.'s are lacking.

The positive-sequence network with the fault shunt connected to itmay be reduced in the usual way to the simplest form of network connecting the internal voltages of the synchronous machines. For a twomachine system the reduced network is, of course, a tJ. (or 1r).

EXAMPLE 3

Find the fault shunt for representing a line-to-ground short circuit on busD of the power system of Fig. 8, Example 2, and reduce the positive-sequencenetwork (with shunt attached) to an equivalent 1f' between the two generator

0.293

(a)

(b)

(c)

FIG. 19. Reduction of positive-sequence network with attached fault shunt ofreactance X, representing a line-to-ground short circuit on the power system of

Fig. 8 (Example 3.)

e.m.f.'s and neutral. Repeat for other types of short circuit at the samelocation.

Solution. In Example 2 the zero-sequence and negative-sequence networks were reduced to single impedances as viewed from the point of fault(terminals D and 0). These impedances were Zo = jO.066 and Z2 = jO.09per unit. For a line-to-ground short circuit the impedance of the faultshunt is

Zp = Zo + Z2 = jO.066 + jO.09 = jO.156 per unit

The positive-sequence network was reduced as shown in Fig. 9, and in partc of that figure it has been reduced as far as possible while still retaining the

EFFECT OF TYPE OF FAULT ON STABILITY 223

separate identities of generators A and B. In Fig. 19 the fault shunt isattached, and the network is reduced to an equivalent 1r. Since all thebranches of this 1r are reactive, only the branch between A and B affectsthe power-angle equation. The reactance of that branch, for a line-toground fault, is

X = 0.293 + 0.216 + 0.293 X 0.216 = 0.509 + 0.381AB 0.010+ '0.156

= 0.890 per unit

For any reactance XF of the fault shunt,

X = 0.509 + 0.0633AB 0.010+ X,

Hence for a line-to-line short circuit

X". = X 2 = 0.090

X 0 + 0.0633 + ·AB = 0.5 9 0.100 = 0.509 0.633= 1.14 per unit

For a two-line-to-ground short circuit,

X,. = XoX2 = 0.066 X 0.090 = 0.038x, + X 2 0.156

X + 0.0633 + ·AB = 0.509 0.048 = 0.509 1.32 = 1.83 per unit

For a three-phase short circuit

X F = 0

X + 0.0633 + 33 ·AB = 0.509 0.010 = 0.509 6. := 6.84 per umt

The reactance between generators A and B is lowest for the line-to-groundfault, higher for the line-to-line fault, still higher for the two-line-to-groundfault, and highest of all for the three-phase fault.

Effect of type of fault on stability. The lower the impedance of thefault shunt, the less is the power exchanged between any two synchronous machines for a given angular displacement, and thereforethe lower is the stability limit for a given fault duration. In a twomachine system, such as that of Example 3, it is clear that the lowerthe impedance of the fault shunt (connected in the shunt branch of theT network of Fig. 19a), the higher is the impedance of that branch ofthe reduced network which joins the two generators (branch AlB!)Fig. 19c). The synchronizing power varies inversely as the impedanceof branch AIBI. A similar situation exists on a system of more thantwo machines.


Comparison of the expressions for the impedances of the fault shuntsfor the several types of short circuit (eqs. 54 to 57) shows that thisimpedance is lowest-indeed, is zero-for the three-phase short circuit,higher for the two-line-to-ground short circuit, still higher for the lineto-line short circuit, and highest for the line-to-ground short circuit.It follows that the most severe type of fault, as ·regards power-systemstability, is the three-phase short circuit, followed in order of decreasingseverity by the two-line-to-ground, line-to-line, and one-line-to-ground

~

~ 0.5 J---+---+-++~~~~-+--i---+--f

~e

0.5 0.8 00

Fault duration (seconds)

FIG. 20. Curves of stability limit as a function of fault duration for four types offault at the sending end of a line of the system of Fig. 15, Chapter V (Example 4).

short circuits. The relative severity of the various faults is shown toadvantage by curves of power limit versus fault duration, such as thecurves of Fig. 20. The difference in severity of the several types offault becomes smaller as the fault duration is decreased, but, even withthe fastest feasible clearing times, the difference is usually pronounced.For zero clearing time (which is not attainable in practice) the powerlimit is independent of the type of fault.

In making a stability study some judgment is necessary in choosingthe type of fault which is assumed to occur. The assumption of threephase faults (the most severe type) gives conservative results and thesimplest computation. It is therefore a useful assumption whencomparing the effects of different fault locations, different system layouts, bussing arrangements, load conditions, and similar factors.


However, three-phase faults are of infrequent occurrence, especially onhigh-voltage overhead lines on steel towers, and the power limitsdetermined on this assumption may be unduly pessimistic. For thisreason it has been common practice to assume two-line-to-ground faultsas the most severe condition likely to be encountered. In some locations, such as on busses of isolated-phase construction, it would bereasonable to assume that only line-to-ground faults could occur.

Line-to-ground faults are the type of most frequent occurrence, andthree-phase faults, the least frequent. The relative and absolutefrequency of occurrence of the several types differs on different powersystems, depending largely on the type of line construction.

Sporn and Muller4 give the following figures on the number of faultsof different types occurring on a group of transmission lines most ofwhich are operated at 132 kv.:

Line-to-ground 58Two-line-to-ground 8Three-phase 6

Total 72

In designing a system or a modification of one to improve stability,it appears logical to make estimates both of the value of various degreesof reliability of service and of the cost of achieving them. Then it maybe determined whether the expense of any particular improvement ofreliability is warranted. If a certain system can be made stable duringthree-phase faults at reasonable cost, it may be worth doing; but, if theexpense is excessive, one may have to be satisfied with a system whichis stable during two-line-to-ground faults but not during three-phasefaults. In another system for which the requirements of reliabilityare not so stringent, instability during two-line-to-ground faults maybe tolerated provided that the system is stable during line-to-groundfaults. If still lower standards of reliability prevail, interruptions ofservice can be expected from any type of fault.

EXAMPLE 4

Determine and plot stability limit in per unit as a function of fault durationfor each of the following types of short circuit at the sending end of one lineof the two-machine system of Fig. 15, Example 3, Chapter V: (a) one-lineto-ground, (b) line-to-line, (c) two-line-to-ground.

Positive-sequence reactances are given in Fig. 15 of Chapter V. Negative-and zero-sequence reactances in per unit are as follows:

Generators: X 2 = 0.24, K« = 0.06Each transmission line: X 2 = 0.40, Xo = 0.65Receiving-end transformers: X 2 = Xo = 0.10


The transformers are connected Y-~ with neutral of primary windingssolidly grounded.

Solution. The negative-sequence and zero-sequence networks are shownin Figs. 21a and 22a, respectively. They are reduced to single reactances

(d)

0.24 0.30

(O)~ (c)

i°.l33(b)

FIG. 21. Reduction of the negative-sequence network, fault at sending end ofline (Example 4).

0.650.06 0.425

(0) r::-:=::J (c)

0.06 0.325 0.10

~(b) (d)

FIG. 22. Reduction of the zero-sequence network, fault at sending end of line(Example 4).

in parts b, c, and d of those figures. The values of the reactances areX2 = 0.133 and K« = 0.053 per unit.

The reactances of the fault shunts (XF ) are:

Line-to-ground:Line-to-line:

Two-line-to-ground:

Xo+ X 2 = 0.186 per unitX 2 = 0.133 per unit

XOX2 •X

o+ X ~ = 0.038 per unit

The positive-sequence network is shown in Fig. 16a of Chapter V. Withthe fault shunt attached to it at the proper point it becomes as shown inFig. 23a, and, after a Y-d conversion, it is as shown in Fig. 23b. Thereactance of the branch connecting the two generator e.m.L's is

XAB = 0.35 + 0.30 + 0.35;F 0.30 = 0.65 + 0~~5


Upon substituting the values of X F given above, we get the following valuesof XAB:

Line-to-ground: 1.22 per unitLine-to-line: 1.44 per unitTwo-line-to-ground: 3.41 per unit

0.65+ °i~5

~~~~~(a) (b)

FIG. 23. Reduction of the positive-sequence network with attached fault shunt ofreactance XI' (Example 4).

For the pre-fault condition (Chapter V, Fig. 16) XAB was 0.65 per unit,and for the post-fault condition (Chapter V, Fig. 17) it was 0.85 per unit.Henee n and r2 have the following values:

Line-to-ground.·

Line-fo-line:

Two-line-to-ground.·

All typesojJault.·

0.65 5'1 = -= 0.31.22

rl = 0.65 = 0.451.44

rl = 0.65 = 0.193.41

0.65 7'2=-= 0.60.85

Byrd and Pritchard's curves will now be used. From Fig. 40 of ChapterV values of sin 80 are obtained, corresponding to the values of '1 and r2above. The values of sin 00, multiplied by Pm = EAEB/XAB = 1.00X 1.00/0.65 = 1.54 per unit, give the power limits for sustained fault andfor instantaneous clearing. These values have been entered in Table 2 inthe columns headed "sin 00" and Upi" on the lines for te= 00 and to = o.

Values of sin 00 which are multiples of 0.05 and which lie between thevalues already found for te = 00 and for te = 0 are selected. The curvesof Figs. 23 to 39 of Chapter V are entered with the proper values of sin 00,rl, and '2,and the values of Teare read from the curves and written in Table2. Values of Te/teare calculated from eq. 27 of Chapter V, as follows:

~ == ~1ff1'1P ... == ~1f X 60 X 1'1 X 1.54 ==~to GH 1 X 3.00

and are written in Table 2. The critical clearing time to is then calculatedfrom Te + Tc/te•


TABLE 2

CALCULATION OF POWER LIMIT Pi AS A FUNCTION OF CLEARING TIME tc(EXAMPLE 4)

Type of Fault sin ~o Pi = 1.54sin ~o "cltc t« tc (sec.)0.456 0.70 7.15 00

0.50 0.77 " 3.65 0.51Line-to-ground 0.55 0.85 " 2.45 0.34rl = 0.53 0.60 0.92 It 1.70 0.24r2 = 0.76 0.65 1.00 u 1.15 0.16

0.701 1.08 " 0

0.378 0.58 6.59 00

0.40 0.62 " 4.40 0.67Line-to-line 0.45 0.69 " 2.90 0.44

Tl = 0.45 0.50 0.77 " 2.20 0.33r2 = 0.76 0.55 0.85 " 1.70 0.26

0.60 0.92 " 1.25 0.190.65 1.00 It 0.85 0.13

0.147 0.23 4.28 00

0.20 0.31 It 2.95 0.690.25 0.38 " 2.25 0.530.30 0.46 " 1.75 0.41

Two-line-to-ground 0.35 0.54 " 1.40 0.33Tl = 0.19 0.40 0.62 " 1.20 0.28T2 = 0.76 0.45 0.69 " 1.00 0.23

0.50 0.77 " 0.80 0.190.55 0.85 " 0.65 0.150.60 0.92 u 0.5 0.120.65 1.00 " 0.32 0.07

The results of the calculations are shown by the curves of Fig. 20. Thecurve for a three-phase fault has been copied into this figure from Fig. 22 ofChapter V so that the effects of the four types of fault on stability may becompared.

Effect of fault impedance. In the derivation of the connectionsbetween the sequence networks for representing short circuits,. it wasassumed that the fault itself had no impedance. Actually the faultmay have some impedance, principally the resistance of the arc between conductors. On high-voltage circuits this resistance is usuallynegligible in comparison to the other impedances of the network, andno allowance is made for it in calculation.* In ground faults there maybe additional resistance in the path to ground through the tower foot-

*For valuesof arc resistance seefootnote in sectionon reactance relays in Chapter IX, Vol. II.

EFFECT OF FAULT IMPEDANCE 229

ing or through the object causing the fault. The value of this resistance varies widely, with a median value of around 20 ohms. 5 Thisresistance also is usually neglected in stability studies, thereby simplifying calculation and giving results for the most severe condition.

If, for any reason, it is desirable, fault impedance may readily betaken into account, as shown below.

Line-to-ground fault. Let F be the impedance of the fault betweenconductor a and ground. To facilitate the analysis suppose that at thepoint of fault there are attached to the three-phase network three equalimpedances of value F, as shown in Fig. 24a. This artifice has theadvantage of making the three-phase network symmetrical up topoint M, where a line-to-ground fault of zero impedance may be applied. The added impedance F appears in each sequence networkbetween the point of fault and point M. Terminals 0 and M of thesequence networks are then interconnected as shown in Fig. 24b torepresent a line-to-ground short circuit. Conditions inside the sequence networks are not altered if an impedance 3F is connected inseries with the networks as shown in Fig. 24c instead of three separateimpedances of F each.

Line-to-line fault. Let F be the impedance of the fault betweenconductors band c. Suppose that the fault condition is representedby connecting an impedance F/2 to each of the three conductors at thepoint of fault and then applying the line-to-line short circuit at theother end of these impedances (point M, Fig. 24d). Such a connection is equivalent to connecting impedance F/2 to the line terminal ofeach sequence network and then connecting the positive- and negativesequence networks in parallel as shown in Fig. 24e. Of course, the twoimpedances F/2 which are no,v in series are equivalent to a singleimpedance F.

Two-line-to-ground fault. Let the impedance between conductorsband c be FL , and let the impedance to ground be Fa. For generalityFa might be tapped onto any point of FL, but, as FL is usually muchsmaller than FG, it will be sufficiently accurate for the present purposeto assume that Fa is connected to the center point of FL , giving FL/2

between the junction and each faulted conductor, as shown in Fig. 24f.The impedance FG in the ground connection can be shown to beequivalent to an impedance 3FG in the zero-sequence network. Theconnections between the sequence networks are then as shown inFig. 24g.

Three-phase fault. For symmetrical fault impedances as shown inFig. 24h the connections of the sequence networks are as shown inFig. 24i.

~ ee <:) rn ~ 1-3 t--04 o Z ~ ~ ~ t:1 Z t=rj

t-3 :a o ~ 00

Fa(h

)

a~""-

3cb

b ;~

I~

19

F./

2

(I)

a~~V!

3,1,.

b'i'c~

go.

-.l.--

-J\A

A--I

Zer

o

~J FL

/2

FL

/2

F

a

3c1>

b c go - (d)

F/2

Zer

oo--f-

!V'

0

~'/~

POSe

F/2

3F

~L!:

:J-

ll::J

I~

(b)

(c)

(e)

(g)

(i)

One

•lin

e•t

o-g

roun

dfa

ult

line

-to

-line

faul

tT

wo-

line

-to

-gro

und

faul

tTh

ree

-pha

sefa

ult

FIG

.24.

Con

nect

ions

betw

een

the

sequ

ence

netw

orks

corr

espo

ndin

gto

vari

ous

type

sof

faul

tw

ithfa

ult

impe

danc

e.

°2

Mo

(a)

a3q,b~

c-

g:r:

:vv'

:I

UNSYMMETRICAL OPEN CIRCUITS 231

[66)

Impedance of Fault Shunt, Z,.

z, +Z2 +3F [63]Z2 +F [64]

FL + (Zo +FL/2 +3Fa) (Z2 +FL/2) [65]2 Zo+Z2 +FL + 3Fa

FL2

Line-to-ground

Line-to-line

Three-phase

Two-line-to-ground

Fault shunts. The impedances of the fault shunts, connected acrossthe positive-sequence network at the point of fault to represent theeffect of faults having impedance, may be found from inspection of theconnections in Fig. 24; they are as follows:

Type of Fault(with Fault Impedance)

Type of Fault

Unsymmetrical open circuits and series impedances. Unsymmetrical open circuits may be caused by blown fuses or by single-poleswitching.t The connections between the sequence networks forrepresenting one or two open conductors or an impedance in series withone conductor are given in Fig. 25. The derivations of these connections are similar to those given for.ahort circuits earlier in this chapter.Before these connections are made, the line side of each sequence network is opened at the point of fault, and the line on each side of theopening is considered to be a terminal. (These are the two upperterminals shown on each block in Fig. 25.) The neutral terminal,used for representing short circuits, is not used for representing opencircuits.

The effect of an unsymmetrical open circuit or series impedance onpositive-sequence quantities may be represented by connecting animpedance Zp in series with the positive-sequence network at the pointof fault. The value of .Zp depends upon the impedances Zo of the zerosequence network and Z2 of the negative-sequence network, eachmeasured or calculated from terminals at the point of fault in series,instead of in shunt, with the respective network.

Series Impedance z"Inserted in PositiveSequence Network

One conductor open

Two conductors openThree conductors open

Impedance Z in serieswith one conductor

ZoZ2

ZO + Z2

Zo + Z2

co

1

(l/zo) + (l/z2) + (3fZ)

[67]

[68][69]

[701

fSingle-pole switching is discussed in Chapter XI, Vol. II.

232

abcg


-l!. a a Z ab --!!. b ~c .s. c cg g g g

Z/3

( oZero

0

o oPos.

0

o 0

Neg.0

-.z~o-.P~s.[5JNeg. Neg.

0 0'

'\Nv

() C)

Zero0

() 0

Pos.0

o 0

Neg.0

(a) (b) (c) (d) (e)

Normal One con- Two con- Three con- Impedance Zcondition duetor open ductors open duetors open in one conductor

FIG. 25. Connections between the sequence networks corresponding to varioustypes of open circuit or series impedance.

Simultaneous faults and other double unbalances. Approximately20% of the faults on double-circuit overhead transmission lines on thesame towers involve both circuits. Occasionally two faults occursimultaneously at points which are separated geographically as well aselectrically, particularly on systems which are grounded through highimpedance. A combination of an unsymmetrical short circuit with anunsymmetrical open circuit may occur, as when the short circuit ispartially cleared by the opening of a fuse or single-pole circuit breaker.The conditions described above may be called "double unbalances."In general, a double unbalance cannot be correctly represented byconnecting the sequence networks at each of the two points of fault asthey would be connected for single unbalances at those points, nor canthe effect of a double unbalance on positive-sequence quantities becorrectly represented by connecting into the positive-sequence networkfault shunts or series impedances which have the same values that theywould if the two faults had occurred separately.

For methods of solving three-phase networks having double unbalances see Refs. 2, 3, 6, and 7.

LINES IN THE ZERO-SEQUENCE NETWORK 233

The zero-sequence network. The zero-sequence network plays animportant role in determining the currents and voltages during groundfaults. This network differs from the positive- and negative-sequencenetworks particularly with respect to the representation of lines and oftransformers.

Representation of lines in the zero-sequence network. In anythree-phase circuit positive- and negative-sequence currents are confined to the line conductors. Zero-sequence currents, however, being inphase in all three line conductors, must find a return path elsewhere,usually either through the earth or through both the earth and overhead ground wires or cable sheaths. The spacing between the outgoing and return paths is therefore greater for zero-sequence than forpositive-sequence currents. Accordingly, the zero-sequence seriesinductive reactance is greater; and the shunt capacitive susceptance issmaller, than the corresponding positive-sequence quantities. Thezero-sequence inductive reactance of lines is greater than the positivesequence reactance approximately by the following factors:

Single-circuit aerial lines without ground wires or with steel groundwires. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . .. .. . 3.5

Single-circuit aerial lines with copper or aluminum ground wires. . . . . . . 2Double-circuit aerial lines without ground wires or with steel ground

wires " . . . . .. . . . . . 5.5Double-circuit aerial lines with copper or aluminum ground wires. . . . . . 3Three-phase cables. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3 to 5Single-phase cables. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Although these factors are useful for estimating zero-sequence reactance of lines, more accurate values should be calculated if the conductorsizes and spacings and the earth resistivity are known. Methods ofcalculating zero-sequence resistance, reactance, and capacitance aregiven in Refs. 1, 2, 3, and 21 of Chapter III.

There is a considerable zero-sequence mutual impedance betweenparallel aerial lines on the same towers or even on the same right-ofway. The methods of calculating mutual impedance are also given inthe references. If t\VO lines having self-impedances Za and Zb andmutual impedance Zm are connected together at both ends, they maybe replaced by a single impedance, Fig. 26a. If they are connectedtogether at one end only, they can be represented by the equivalent Ycircuit of Fig. 26b. If they are not connected at either end, they canbe represented on an a-c. board by the equivalent circuit of Fig. 26c,which utilizes an insulating transformer of ratio 1: i.t For ordinary

tWo A. Lewis, in discussing Ref. 8, attributes this circuit to C. F. Wagner.


calculation they can be represented by the mesh circuit of Fig. 26d.8

If a fault is to be represented on one of a pair of coupled lines, the lineson each side of the fault must be considered a separate section. Thus,

Ca)

ZeZb-Z",,'lZa+ Zb - 2Z""

o------JVV\r---

~---6

6'----6

,CJ ----a

(c)

c

(bl I~---a

a'a'----a

(d)6'----b

b'Zb

FIG. ?6. Equivalent circuits for two coupled lines of self-impedances Za and Zband mutual impedance Zm. Lines connected together (a) at both ends, (b) at one

end only, and (c) and (d) at neither end.

I Fault)(

t=ml~

a

b

m(Zb-Zm) (l-m)(Zb-Zm)

FIG. 27. Equivalent circuit of two coupled lines, one of which is faulted.

if there is a fault on one of two lines connected' at both ends, and thefault is distant from one end by a fraction m of the length of the lines,the equivalent circuit is as in Fig. 27. Equivalent circuits for three ormore coupled lines have been devised" but are seldom needed. Twocoupled circuits having appreciable shunt admittance can be represented by the equivalent or nominal circuits given in Ref. 9.

TRANSF9RMERS IN THE SEQUENCE NETWORKS 235

Representation of transformers in the sequence networks. As arule, every important transformer bank on a three-phase power systemconsists either of three identical single-phase transformers having eachset of windings connected in Y or in d or of a three-phase transformerhaving its windings connected likewise. Only 'such transformers having not more than three sets of windings are considered here.

As was mentioned in Chapter III, the positive-sequence equivalentcircuit of a two-circuit transformer bank, with exciting current neglected,consists of a series impedance and an ideal transformer of complexratio. The angle of the complex ratio expresses the phase shift ofpositive-sequence voltage from one side of the transformer bank to theother. For D.-d or Y-Y connections the shift is either 0 or 180°. For!:J,.-Y or Y-!:J,. connections the phase shift, with the standard manner ofnaming the phases, is either +30° or -30°, depending upon the exactconnections. By naming the phases in a different manner, the phaseshift of +300 may become + 1500 or -90°, and the phase shift of -30°may become +90° or -150°. The nomenclature in which the phaseshift is ±90° is the most convenient for computation and is recommended for that purpose.2, 10 If per-unit quantities are used, themagnitude of the complex ratio becomes 1, but the angle remains. Insetting up the positive-sequence network, the phase shift is commonlyneglected. The currents and voltages are first calculated as if therewere no phase shift, and later the phase shift is taken into account if itis of any moment in the use of the results. When phase shift is notrepresented, the positive-sequence equivalent circuit for per-unitquantities consists merely of a series impedance.

The negative-sequence equivalent circuit of a transformer bank is likethe positive-sequence circuit except that the angle of the complexratio is of opposite sign. For example, if the phase shift for positivesequence current and voltage is +90°, the phase shift for negativesequence current and voltage is -90°. Here again the phase shift isusually disregarded in setting up the negative-sequence network, butit may be taken into account later if desired.

The zero-sequence equivalent circuit differs considerably from thepositive- and negative-sequence circuits. For banks of single-phasetransformers the value of impedance is the same, but, instead of thisimpedance being simply in series with the line, each end of the impedance may be either open or grounded, depending upon the connectionsof the bank. The proper connection may be deduced by consideringthe paths of zero-sequence currents. Zero-sequencecurrents are equalin the three phases. It is apparent that such currents cannot flowthrough a set of Y-connected windings if the neutral of the Y is isolated,


but they can flow if the neutral is grounded. Zero-sequence currentscannot flow from a three-phase line into a set of A-connected windings,but they can flow around the delta. If exciting current is neglected,zero-sequence currents can flow in one set of windings of a transformerbank only if zero-sequence currents can flo,v also in the other set ofwindings, The foregoing statements may be summarized as follows:Zero-sequence current cannot flow from a line into a transformer bank

Three-phase circuit

(a)Delta·delta

(b)Ungrounded Yungrounded Y

(c)Grounded Ygrounded Y

(d)Grounded Y

delta

(e)Ungrounded

Y- delta

(I)Ungrounded Y

grounded Y

Zero- sequence circuit

---VV\;---

_l

FIG. 28. Zero-sequence equivalent circuits of two-circuit transformer banks.

unless the windings connected to the line are in Y with groundedneutral and unless the other windings are either in ~ or in Y withgrounded neutral, The zero-sequence equivalent circuits of varioustransformer connections are shown in Fig. 28. Only connections c,grounded Y-grounded Y, and d, grounded Y-~, present a path forzero-sequence current. The other connections are equivalent to anopen circuit in the zero-sequence network,

The zero-sequence impedance of the grounded y-~ bank is equal tothe short-circuit impedance of one transformer. Figure 29 may help

TRANSFORMERS IN THE SEQUENCE NETWORKS 237

to show why this is so. In part a of the figure the actual connectionsare shown. It is assumed that equal currents 10 are impressed on theprimary windings, causing the secondary current 10 ' to circulatearound the delta. Because of the symmetry of the delta, terminalsA, B, and C are at the same potential and may be connected to oneanother, as indicated by broken lines in the figure, without effectivelychanging the circuit for zero sequence. Connecting terminals A, B,and C is equivalent to short-circuiting each secondary winding separately as in part b of the figure. It is apparent that the circuit of Fig.

(a)

1 #0

10 ~

Vo 10

~Iri

Vo1~Io'

0

tvo-

(b)

FIG. 29. Diagram showing that the zero-sequence impedance of a transformerbank connected grounded Y-A is equal to the short-circuit impedance of the trans

formers.

29b presents between each primary line and ground an impedance equalto the short-circuit impedance of one of the transformers. As the linecurrent is 10 and the line-to-ground voltage is Vo, the impedanceVolIo is the zero-sequence impedance. It is equal to the positivesequence impedance because the latter is also the short-circuit impedance.

Grounding impedance. The neutral of the Y-connected windingsmay be grounded either solidly or through a neutral impedance Zn.The current through Zn is 310, causing a voltage drop across it of3ZnIo, which is added to all three line-to-neutral voltages and henceto the zero-sequence component of these voltages. In the zerosequence network the current is 10, and to obtain the correct zero-


sequence voltage drop the neutral impedance must appear in this network as an impedance 3Zn in series with the transformer impedance.

Multicircuit transformers. As mentioned in Chapter III, the positive-sequence equivalent circuit of a three-circuit transformer, withexciting current, phase shift, and ratio neglected, consists of threeimpedances connected in Y between the terminals of the three circuits.

1 2 3

41£;(a)

FIG. 30. Zero-sequence equivalent circuits of three-circuit transformer banks.

As long as phase shift is neglected, the negative-sequence circuit isidentical to the positive-sequence circuit. The zero-sequence circuitconsists of the same set of Y-connected impedances, but the threeterminals of the, Y may not all be connected to the correspondingexternal circuits. Each of the three windings should be consideredseparately, and the corresponding terminal of the Y should be connected according to the same principles already established for twocircuit transformers, namely: (1) if the winding is ~-connected, theline is left open, and the corresponding impedance is grounded; (2) if

TRANSFORMERS IN THE SEQUENCE NETWORKS 239

the winding is Y-connected with grounded neutral, the line is connected to the impedance; and (3) if the winding is Y-connected withisolated neutral, both the line and the impedance are left open. InFig. 30 these principles are applied to several different connections ofthree-circuit transformers. The same principles apply to transformersof four or more circuits.

Three-phase transformers. Three-phase shell-type and three-phasefive-legged core-type transformers resemble banks of three single-phasetransformers in that they have three magnetic circuits which are almost independent of each other. Hence the equivalent circuits ofthese transformers are the same as those for banks of single-phase units.In three-phase three-legged core-type transformers, however, conditions are different. Application of Kirchhoff's law to the fluxesof threelegs shows that there can be no zero-sequence flux in such a transformerexcept that leaking through air paths from top to bottom of the core.Therefore zero-sequence exciting currents produce no zero-sequencevoltage except that induced by this leakage flux, which is small compared with the core flux that would be set up by positive-sequencecurrents of like magnitude but large compared with the leakage fluxbetween primary and secondary windings. In other words, the zerosequence exciting impedance is much smaller than the positivesequence exciting impedance but is about five times as great as theleakage impedance, or equivalent impedance, between windings. Consequently, even a two-circuit transformer of this type should be represented by a T circuit. The shunt branch, representing the zerosequence exciting admittance, cannot be neglected, as it usually is inthe positive-sequence circuit. The effect of the three-legged core issimilar to that produced by adding a high-impedance il-connectedwinding, as is evidenced in the equivalent circuit by the third branch,the terminal of which is grounded. The connections at the otherterminals, corresponding to the actual windings, depend upon theconnections of those windings, as already discussed.

Autotransformers. The commonest type of autotransformer onthree-phase transmission systems has a set of Y-connected tappedwindings with a grounded neutral and a separate A-connected stabilizing winding (formerly called a tertiary winding). It is, therefore, athree-circuit transformer connected Y- Y- il, and its equivalent circuitis the same as that of a transformer with three separate windingssimilarly connected, the impedances being given on the circuit basis.2

The zero-sequence circuit is connected as shown in Fig. 30a without thegrounding impedances. If the neutral of the autotransformer isgrounded through impedance, this impedance is common to both


Y-connected circuits and, when expressed in actual ohms, is both themutual impedance and a component of each self-impedance. Whenexpressed in per unit or in ohms referred to a particular voltage, however, this impedance has three different values, one for the mutualimpedance and two for the two self-impedances. Therefore, the singleimpedance must be represented by a Y circuit, as shown in Fig. 31.If the neutral of an autotransformer is ungrounded, zero-sequencecurrents can pass from one circuit to the other without transformation,the transformer acting merely as a series impedance.

(a)

A B

(b)

FIG. 31. An impedance Z common to two circuits, A and B, must be representedin the per-unit system by a Y circuit if different base voltages are used for the twocircuits. N is the ratio of base voltage A to base voltage B, and Z' is the per-unitvalue of Z on base voltage A. In the zero-sequence network all impedances are

multiplied by 3.

Other symmetrical transformer connections, including regulating transformers. See Ref. 12.

Effect of grounding on stability. Methods of grounding a powersystem modify its zero-sequence impedance. This affects the impedance of the fault shunts for representing ground faults and therebyaffects the severity of such faults.

Most transmission systems are grounded through transformers, thehigh-voltage windings of which are connected in Y with groundedneutral. Such transformers are nearly always used at the points ofsupply and sometimes also at points of load. Occasionally transformers which do not carry load are used solelyfor the purpose of grounding.On high-voltage systems the usual practice is to ground the transformer neutrals solidly; on medium-voltage systems, to ground themeither solidly or through resistors or reactors of low per-unit impedance.Some medium-voltage systems are nominally ungrounded but actuallyare grounded through the relatively high zero-sequence shunt capacitivereactance of the lines. If one or more transformer neutrals aregrounded through reactors, the inductive reactance of which resonateswith the capacitive reactance, the zero-sequence impedance may bemade much higher than that of a nominally ungrounded system.Resonant grounding reactors, known as Petersen coils or ground-fault

EFFECT OF GROUNDING ON STABILITY 241

neutralizers, are used to a small but increasing extent. Transmissionsystems to which generators are connected directly, rather than throughtransformers, are grounded through the generators. The generatorneutrals, like transformer neutrals, may be grounded either solidly orthrough resistors or reactors.

The zero-sequence impedance of a transmission system, viewed fromthe point of fault, depends upon the number of grounding points, uponthe distance of the grounding points from the point of fault, upon thezero-sequence impedance of the grounding transformers (or generators), and upon the impedance of the grounding resistor or reactor,if used.

An increase in the zero-sequence impedance, viewed from the pointof fault, results in a decrease in the severity of a one-line-to-ground ortwo-line-to-ground fault. The effect of a two-line-to-ground faultapproaches that of a line-to-line fault at the same location, and theeffect of a one-line-to-ground fault approaches that of no fault. Therefore, in its effect on system stability during ground faults, groundingthrough impedance is better than solid grounding. Either resistorsor reactors may be used to decrease the severity of ground faults, butreactors are cheaper and commoner than resistors.

At some locations, however, resistors are more effective thanreactors. On a system consisting of several synchronous machinesjoined by low-resistance lines, the occurrence of a fault decreases' theelectrical outputs of the generators and the electrical inputs of theactual or equivalent motors, causing the generators to be acceleratedand the motors to be retarded. The closer the fault is to a particularmachine, the greater is its effect on the electrical output or input of thatmachine, and the greater is the tendency for the machine to be accelerated or retarded, as the case may be. Stability is promoted byanything which increases the output of the generators during the fault,thereby lessening their acceleration, provided that the retardation ofthe motors is not increased correspondingly. A grounding resistor, byconsuming power during a ground fault, exerts on a synchronous machine a braking effect which is greater the closer the fault is to theresistor and the closer the machine is to the fault. A grounding resistor located near a generator is therefore beneficial, as it exerts thegreatest braking effect on the generator for faults closeby, which wouldotherwise result in the greatest decrease of load and, consequently, inthe greatest acceleration. Grounding resistors should not be used,however, near actual or equivalent synchronous motors or near synchronous condensers, for such machines already are retarded by faults.In a two-machine system resistance grounding may be advisable at thesending end and reactance grounding at the receiving end.


Neither grounding resistors nor grounding reactors have any effectduring three-phase faults.

The choice of a grounding system is influenced not only by consideration of stability but also by other factors, among which are:(a) limitation of current to prevent damage to generator windingsfrom excessive mechanical forces; (b) limitation of neutral-to-groundand line-to-ground voltages of transformers to permit the use of gradedinsulation and lightning arresters of lower voltage rating, which givebetter protection; (c) limitation of inductive influenceon communication circuits during ground faults on the powersystem; and (d) obtaining suitable conditions for rapid and selective relay operation to clearground faults. Since the conditions for limiting current conflict withthose for limiting voltage, the choice of grounding method rests oncompromise. The importance of the effect on stability of the methodof grounding has been decreased in recent years by the trend towardhigher speeds of fault clearing.

EXAMPLE 5Determine the effectiveness of grounding reactors or resistors in increasing

the stability limit of the two-machine system of Fig. 32 during two-line-toground faults on the transmission lines. Assume the grounding reactanceor resistance to be 10% on the basis of the kilovolt-ampere rating of.generating station A.

Solution. A fault at either end of one of the transmission lines is moresevere than a fault near the middle of a line; therefore, two-line-to-groundfaults will be assumed, first at point C, then at point D, Fig. 32. Thestability limit depends upon fault duration. For comparing the differentmethods of grounding it will be simplest to find the stability limit for sustained faults. This limit will be found by first obtaining the power-angleequation and then using the equal-area method, or, when the network ispurely reactive, by using Fig. 40 of Chapter V, which was derived by theequal-area method.

The grounding impedances, Zs at the sending end and ZRat the receivingend, appear in the zero-sequence network as 3Zs and 3ZB, respectively.The three sequence networks are connected in parallel at points C and 0(as shown in Fig. 32b) to represent a two-line-to-ground fault at the sendingend; they are connected similarly, but at D and 0, to represent a fault at thereceiving end. The zero-sequence and negative-sequence networks are eachreduced to a single impedance between terminals C and 0 (or D and 0), andthen the positive-sequence network with fault shunt attached is reduced to a7r circuit. Next the power-angle equation is found by the methods of Chapter IV. The details of calculation will not be shown, but the resultingequations are given in Table 3. The corresponding stability limits, foundby the equal-area method, are presented in Table 4.

EFFECT OF GROUNDING ON STABILITY 243

30

30

(a)

1.0

101115

Posltive- sequence network

Zero- sequence network

10

°1III

3Z R I'-- ~~---1

32

3Z s

(b) o

FIG. 32. Two-machine system of Example 5. (a) One-line diagram. (b)Sequence networks interconnected to represent a two-line-to-ground fault at pointC. Reactances are given in per cent on the rating of the generating station.

TABLE 3

POWER-ANGLE EQUATIONS OF THE Two-MACHINE SYSTEM OF FIG. 32 FOR

VABIOUS METHODS OF GROUNDING (EXAMPLE 5)

Power-Angle Equations(power in per cent of rating of station A)

GroundingImpedance(per cent)~~

Z8 Zso 0

jl0 0o jtO

jl0 jtO10 j1010 10

Any Any

Any Any

SLG Fault at SendingEndPu = 34 sin 8Pu = 51 sin 8Pu = 36 sin 8P« = 55 sin 8Pu = 5 +60 sin (8 + 7°)P u = 10 + 56 sin (8 + 10°)

LL Fault at SendingEndPu, = 77 sin 8

Norrnal ConditionsP« = 154 sin 8

SLG Fault at Receiving EndPu, = 40 sin 8Pu, = 42 sin 8Pu, = 55 sin 8Pu, = 59 sin 8Pu, = 4 + 58 sin (8 + 2°)P u, = -5 + 55 sin (8 +8°)LL Fault at Receiving End

Pu, = 77 sin 8

One Line DisconnectedPu, = 129 sin 8


With solid grounding at both ends, the stability limit for a sustained two..line-to-ground fault at the -sending end is 27% of the rated kilovolt-amperesof generating station A; and, for a similar fault at the receiving end, it is32%, as given in the first line of Table 4. The addition of a 10% groundingreactor at either end increases the stability limit considerably for a fault atthat end but only very slightly for a fault at the distant end. The use ofreactors at both ends increases the stability limit for any fault location; theincrease of stability limit over that obtained with solid grounding at both

TABLE 4

Stability Limit for Sustained Fault(in per cent of rating of station A)

A

oo

jtOjtOjtO10

ojto

ojl01010

STABILITY LIMITS OF THE Two-MACHINE SYSTEM OF FIG. 32 FOR

SUSTAINED FAULTS, AS AFFECTED BY GROUNDING REACTORS

AND RESISTORS (EXAMPLE 5)

GroundingImpedance(per cent)~

Zs ZR~

2LG Fault at 2LG Fault atSending End Receiving End

27 3242 3429 4546 4857 5259 41

LL Fault at LL Fault atSending End Receiving End

Any Any 66 66

Stability limit for any fault cleared instantaneously is 123%.

ends is considerable, but the increase over that obtained with a reactor at theend near the fault is slight. The substitution of a 10% resistor for thereactor at the sending end provides an additional increase in the stabilitylimit, although the increase is greater for a fault at the sending end than for afault at the receiving end. The further substitution of a resistor for thereactor at the receiving end results in a small additional increase in stabilitylimit for a fault at the sending end, but the benefit is more than offset by aconsiderable decrease in stability limit for a fault at the receiving end.

The best grounding scheme oj those investigated here is a resistor at the sendingend and a reactor at the receiving end. Even when this scheme is used, thestability limit is lower for a two-line-to-ground fault than for a line-to-linefault.

The improvement in stability limit obtained by the use of groundingimpedance is smaller for a rapidly cleared fault than for a sustained fault,although the stability limit itself is greater.

If grounding resistors are used at one or both ends, the stability limitdepends upon the ratio of the inertia constant of the motor to that of the

TWO-PHASE COORDINATES 245

generator. The dependence of stability limit on the inertia constants maybe understood by recalling that the equivalent power-angle curve of a twomachine system having resistance is a sinusoid displaced upward (accordingto eq. 28, Chapter IV) by an amount

Po = M2E12Y ll cos811 - M1E22 Y 22 cos 8 22

M1+M2

where Ml and M2 are the inertia constants of the generator and of the motor,respectively; E1 and E2 are the internal voltages of the generator and of themotor, respectively; and Fu cos On and Y 22 cos 8 22 are the terminal selfconductances of the network between the internal voltages. If the twoconductances are assumed to be equal, as they would be if the fault wereelectrically midway between the two internal voltages, and if, furthermore,the two internal voltages are assumed to be equal, Po would be zero if theinertia constants were equal. If, however, M2 were greater than MI , Pcwould be positive, indicating a raising of the power-angle curve by theamount of Po and a resultant raising of the stability limit by about the sameamount. If, on the other hand, M2 were smaller than MI , Pc would benegative, indicating a lowering of the curve and of the stability limit. Inthe present example it has been assumed that the generator and motor haveequal values of stored energy per rated kilovolt-ampere but that the ratingof the motor is three times that of the generator, giving M2 = 3Ml • If thetwo-machine system consisted of a hydroelectric station supplying power to ametropolitan system of one or more steam stations, M2 would be considerably larger than MI , both because of the greater capacity of the steam systemand because of the greater stored energy per kilovolt-ampere of the steamturbogenerators; increasing M2/M1from 3 to 00, however, would increasethe stability limit only a small amount. The stability limit is further increased if, as is usually true, the internal voltage of the generator is greaterthan that of the motor (E I > E 2) . In addition to the upward shift of thepower-angle curve when M2 > MI , there is a small shift to the left, which isbeneficial.

A thorough investigation of the methods of grounding would include astudy of the effects of varying the magnitude of the grounding reactance orresistance on power limit and also on the line-to-neutral voltages. Thesematters, however, will not be considered here.

Two-phase coordlnates.P-l" Symmetrical components, which havebeen used in the preceding part of this chapter for analyzing unbalancedthree-phase circuits, are a kind of substituted variables. Another kindof substituted variables, sometimes used for the same purpose, aretwo-phase coordinates. Two-phase components of current and voltage are defined as follows:§

I, = i (2Ia - I b - Ie) [71a]§In Ref. 14 a, {j, and 0 components are defined in a slightly different way.

246 SOLUTION OF FAULTED NET\VORKS

1[7tb]III = - (Ie - Ib)va

Iz = l(la+ Ib + Ie) [71c]

Vz = !(2Va - Vb - Vc) [72a]

1[72b]V1I = va (Vc - Vb)

v, = leVa + Vb + Vc) [72c]

Note that the expression for I, differs from that for Vz by a factor 2.If eqs. 71 and 72are solved for the phase quantities in terms of their

two-phase components, the following expressions are obtained:

la = Ix + !Iz [73a]

1 V3 1 [73b]r, = -"2Ix - 2 r, + '2Iz

1 va 1[73c]I, = -"2Iz + 2 III + 2"Iz

Va = Vx + Vz [74a]

1 va [74b]Vb = -2V - - V + Vx 2 y z

1 va [74c]v, = -7jVx + 2 Vy + v,

Relation to symmetrical components. Two-phase components arerelated to symmetrical components as follows:

r, = 11 + 12

Iy = j(II - 12)

[75a]

[75b]

[75c]

v. = VI + V2 [76a]

Vy = j(V1 - V2 ) [76b]

v, = v; [76c]

Note that the x current and voltage are equal to the sum of the positive- and negative-sequence components, andthe y current and voltageare equal to the difference of the positive- and negative-sequence com-

TWO-PHASE COORDINATES 247

ponents multiplied by j. The z voltage is the zero-sequence voltage,but the z current is twice the zero-sequence current.

Consider a balanced, positive-sequence set of three-phase voltages.Their negative-sequence component is zero. Their two-phase components, byeqs. 76, are Vx = VI and Vy = jV1, which are balancedtwo-phase voltages of phase order y, x. Similarly, the two-phasecomponents of negative-sequence three-phase voltages are two-phasevoltages of phase order x, y.

The substitute networks. The x voltages and currents may be imagined to exist in a single-phase network called the z network; the yvoltages and currents, in another called the y network; and the z voltages and currents, in a third called the z network. These networks areanalogous to the sequence networks used in the method of symmetricalcomponents and may be given the more general name of "substitutenetworks." The substitute networks representing a balanced threephase network whose positive- and negative-sequence impedances areequal are independent of one another. The x and y impedances areequal to each ather and to the positive- and negative-sequence impedances. The z impedances are one-half the corresponding zero-sequenceimpedances.

The positive- and negative-sequence impedances of rotating polyphase machinery are usually unequal. For many purposes, however,they may be assumed equal without serious error, thus making the useof two-phase coordinates practicable.

If the three-phase network contains balanced, positive-sequencegenerated electromotive forces, then the x and y networks containelectromotive forces of equal magnitude, those in the y network leadingthose in the x network by 90°. Ordinarily, the z network contains nogenerated electromotive forces.

The self- and mutual impedances of the substitute networks and theconnections between the networks corresponding to any given impedances and connections of the three-phase network or any portion of itmay be found by the process previously described for symmetricalcomponents and illustrated by the derivation of equivalent circuitsfor representing several types of short circuit.

Connections between the substitute networks for representing shortcircuits are shown in Fig. 33.13 A line-to-ground short circuit isrepresented by connecting the x and z networks in series and leaving they network open. A line-to-line short circuit is represented by shortcircuiting the y network and leaving the x and z networks open. The znetwork is dead and need not be set up. However, the x network inthis case and the y network in the case of a line-to-ground fault, al-


though not connected to anything at the point of fault, contain generated voltages and carry normal load currents.] A two-line-to-groundfault is represented by short-circuiting the y network and parallelingthe x and z networks through a transformer of 2 : 1 ratio. If theimpedances of the z network are given four times their normal value

ao

3<p ~ bgo

o%

-.

[J0--

Z

o

[]

-.[]

[Z]

o 0__.1

1

% I;~p I(a) (6) (c) (d)

One •line•to- ground une-to•line Two . line•to- ground Three- phase

FIG. 33. Connections between the substitute networks (z, y, z) for representingvarious types of short circuit on a three-phase network (34)).

(twice, instead of half, the zero-sequence impedances), the transformer can be omitted.

If the three-phase network is symmetrical except at the point offault, and if the negative-sequence impedances are assumed everywhere equal to the positive-sequence impedances, the x and y networksare not coupled to each other and are identical except for a 900 phasedifference in generated e.m.f.'s. Therefore it is unnecessary to haveboth these networks set up simultaneously on a calculating board.The positive-sequence network is used in turn for both the x and ynetworks. For example, in calculating a line-to-line fault (Fig. 33b),readings are taken in the positive-sequence network, first with nothing

[Under the assumptions made in using a d-e. calculating board, all line-toneutral voltages in this network are equal, and all currents in it are zero.

REFERENCES 249

connected to it at the point of fault, and then again with a short circuitthere. It is not even necessary to shift the generator voltages by 90°because it is a simple calculation to rotate the second set of readingsforward through 90°, that is, to multiply them by j. Readings can betaken for all types of short circuit with only two networks set up on theboard.

Applications oj two-phase coordinates. In general, only the positivesequence network need be set up for a stability study, faults beingrepresented by fault shunts. Sometimes, however, it is desired tostudy the probable operation of protective relays, and for this purposethe currents and voltages while the system is faulted must be found.The method of two-phase coordinates furnishes a convenient way ofcalculating these quantities from readings taken on a calculating boardon which only the z network is set up in addition to the positive-sequence network. .

Two-phase coordinates also find application to the analysis of unbalances which are symmetrical with respect to one phase, for example,single-phase series or shunt impedances, banks of transformers connected in V or T, and untransposed transmission lines.14 ,15,16

REFERENCES

1. C. L. FORTESCUE, "Method of Symmetrical Coordinates Applied to theSolution of Polyphase Networks," A.I.E.E. Trans., vol. 37, pp. 1027-140, 1918.Historical reference.

2. C. F. WAGNER and R. D. Evans, Symmetrical Components as Applied to theAnalysis of Unbalanced Electrical Circuits, New York, McGraw-Hill Book Co.,1933.

3. W. V. LYON, Applications of the Method of Symmetrical Components, NewYork, McGraw-Hill Book ce., 1937.

4. PHILIP SPORN and C. A. MULLER, "Five Years' Experience with UltrahighSpeed Reclosing of High-Voltage Transmission Lines," A.I.E.E. Troms., vol. 60,pp. 241-6, May, 1941.

5. C. L. GILKESON, P. A. JEANNE, and E. F. VAAGE, "Power System Faults toGround: Part II: Fault Resistance," Elec. Eng., vol. 56, pp. 428-33, 474, April,1937.

6. EDITH CLARKE, "Simultaneous Faults on Three-Phase Systems," A.I.E.E.Trans., vol. 50, pp. 919-41, September, 1931.

7. E. W. KIMBARK, "Experimental Analysis of Double Unbalances," Elec. Eng.,vol. 54, pp. 159-65, February, 1935.

8. FRANK M. STARR, "Equivalent Circuits-I," A.I.E.E. Trans., vol. 51, PP.287-98, June, 1932; disc., pp. 321-6.

9. J. C. BALSBAUGH, R. B. Gow, W. P. DOUGLASS, and A. H. LEAL, "EquivalentCircuits-2 Coupled Circuits," Elee. Eng., vol. 55, pp. 366-71, April, 1936; disc.,pp, 1037-9, September, 1936. This paper gives an exactly equivalent five-terminalmesh circuit for two coupled lines not connected to each other at either end. The


discussion gives (1) an alternative circuit using a transformer, and (2) nominal orapproximate values of the impedances of the equivalent circuits with numericalcomparisons of the nominal values with the exact values.

10. BRYCE BRADY, "Symmetrical Notation of Three-Phase Circuits," A.I.E.E.Trans., vol. 60, pp. 955-7, November, 1941.

11. A. N. GARIN, "Zero-Phase-Sequence Characteristics of Transformers,"Gen. Elec. Reo., vol. 43, pp. 131-6, 174-9, March and April, 1940.

12. J. E. HOBSON and W. A. LEWIS, "Equivalent Circuits for Power and Regulating Transformers," Elec. Jour. preprint, January, 1939; also published as Appendix, Table 7, Electrical Transmission and DistributionReference Book by CentralStation Engineers of the Westinghouse Electric and Manufacturing Company, 1stedition, 1942.

13. EDITH L. CLARKE, "Determination of Voltages and Currents during Unbalanced Faults: Use of the Sum and Difference of Positive and Negative SequenceSymmetrical Components," Gen. Elec. Reu., vol. 40, pp. 511-3, November, 1937.

14. EDITH CLARKE, "Problems Solved by Modified Symmetrical Components,"Gen. Elec. Reo., vol. 41, pp, 488-94, 545-9, November and December, 1938.

15. EDWARD W. KIMBARK, "Two-Phase Coordinates of a Three-Phase Circuit,"A.I.E.E. Trans., vol. 58, pp. 894-904, 1939; disc., PP. 904-10.

16. EDITH CLARKE, Circuit Analysis of A-C PowerSystems, vol. I, Symmetricaland RelatedComponents, New York, John Wiley & Sons, 1943. Chapter V, "TwoComponent Networks for Three-Phase Systems"; Chapter X, "Alpha, Beta, andZero Components of Three-Phase Systems."

PROBLEMS ON CHAPTER VI

1 ~ Graphically and algebraically find the symmetrical components of thefollowing sets of phase voltages:

a

Va = 100/0

v, = 100 /240°

v, = 100 /120°

bv, = 100LQ

v, = 100/120°

v, = 100 /240°

C

v, = 100LQ

v, = 100LQ

Vc=100LQ

2. Find the symmetrical components of the following sets of phase currents:

a

I" = 0I b = I

I, = -I

bI a = I

I b = -!I

Ie = -!I

cI a = I

Ib = 0

I, = 0

3. Find the symmetrical components of the following sets of voltages,and check the results by computing the given voltages from them:v, = 100/180°, v. = 100/90°, v, = 100/0.

4. Find the phase currents corresponding to the following symmetricalcomponents: 10 = 0.15/31°, 11 = 0.97/18~, 12 = 0.45/303°.

PROBLEMS 251

5. Work Example 2 with a two-line-to-ground short circuit on phasesband c of bus D instead of a one-line-to-ground short circuit.

6. Work Example 2 with a line-to-line short circuit on phases band c ofbus D instead of a one-line-to-ground short circuit.

7. Work Example 2 with the neutrals of both generators A and B, insteadof generator B only, solidly grounded.

8. Compute and diagram the flow of reactive power of each sequence inthe faulted three-phase network of Example 2.

9. Find the equations similar to eqs. 39 for a line-to-ground short circuiton phase b instead of on phase a. Show that the impedance of the fault shuntis the same for a fault on either phase.

10. Find the expression for the impedance of the fault shunt representing atwo-line-to-ground short circuit, using the second method described in thetext (p. 221).

11. Find an expression for the impedance of a shunt representing a faultconsisting of a line-to-ground short circuit on phase a and a line-to-line shortcircuit on phases band c at the same point of a three-phase circuit.

12. Find the sequence currents in the system of Fig. 8 when the internalvoltages of generators A and B differ in phase by 60°. The magnitude ofeach internal voltage is 1.05 per unit. Use the fault current as referencephase.

13. Find the reactances of shunts for representing each of the four types ofshort circuit at the middle of one of the lines of the system of Fig. 15, Chap..ter V. (The negative- and zero-sequence reactances are given in Figs. 21and 22 of Chapter VI.)

14. Calculate and plot stability limit as a function of fault duration forone-line-to-ground and two-line-to-ground short circuits at the middle of oneof the lines of the two-machine system of Fig. 15, Chapter V. Also plot forcomparison the stability limit for a three-phase short circuit at the samepoint. (See Fig. 22, Chapter V.)

15. Work Example 4 of Chapter V with a two-line-to-ground fault, insteadof a three-phase fault, at the center of one line. Assume that the negativesequence reactance of the system is equal to the positive-sequence reactancewhen viewed from the point of fault, and that the zero-sequence reactance ofthe transmission lines is equal to 3.5 times their positive-sequence reactance,mutual reactance between the two transmission circuits being negligible,

16. Which fault on the system of Example 4, Chapter V, is more severe inits effect on stability, a three-phase short circuit at the middle of one line or atwo-line-to-ground short circuit at the sending end of one line?

17. Which fault on the system of Fig. 8 is more severe in its effect onstability, a one-line-to-ground short circuit on line CD near bus D or a threephase short circuit on one of the parallel lines DE' near bus D? Does theanswer depend upon the clearing time?

18. Prove that the connections of Fig. 24gare correct by writing equationsfor the relations between phase currents and voltages at the fault and then


deriving the corresponding equations for the relations between symmetricalcomponents.

19. If, in Example 4, the line-to-ground fault had a 20-ohm resistance, bywhat percentage would the maximum synchronizing power be increased overthat for a solid fault? System base is 50 Mva., 33 kv.

20. Prove that there is no net power in a three-phase circuit attributableto a voltage of one sequence in conjunction with a current of a differentsequence.

21. Find the stability limit of the two-machine system of Example 5 forsustained one-line-to-ground faults. Consider the following methods ofgrounding: (a) Zs = ZR = 0; (b) Zs = jl0%, ZR = 0; (c) ZB = 0,ZR = jl0%; (d) Zs = ZR = j10%. .

22. In Probe 21 consider an additional grounding method, as follows:(e) Zs = 10%, ZR = jl0%.

23. In Example 5 find the line-to-ground voltages of the sending-endtransformers during a two-line-to-ground fault at the sending end. Consider each of the following grounding methods: (a) Zs = ZR = j10%; (b)Zs = 10%,ZR = j10%.

24. In Example 5 find the power limits for two-line-to-ground faultscleared in 0.25 sec. with the methods of grounding listed in Probe 21. Thefrequency is 60 c.p.s.

25. In Example 5 find the power dissipated in the 10% grounding resistorat the sending end during one-line-to-ground and two-line-to-ground faultsat the sending end. A 10% reactor is used at the receiving end. AssumeEI = 1201!1. and E2 = 1001!1.%.

26. Calculate and plot power dissipated in the grounding resistor of thefollowing simple system during a line-to-ground fault as a function of theresistance in per cent. The system consists of a generator for which thepositive- and negative-sequence reactances are each 20%, a bank of transformers of 10% reactance, connected Ll-Y with neutral grounded through aresistor, and a fault at the secondary terminals.

27. Calculate and plot the higher line-to-neutral voltage (in per cent) ofthe unfaulted phases, during a line-to-ground fault on the system of Probe26, as a function of the grounding resistance in per cent.

28. Verify the equivalent circuits of Figs. 26b and c.29. Derive an equivalent circuit for two coupled lines of different nominal

voltages in which per-unit impedances are used.30. Prove the relations shown in Fig. 31 for representing the ground

impedance of an autotransformer.31. Find the value of series impedance presented to zero-sequence current

by an autotransformer with ungrounded neutral.32. Solve Example 2 by two-phase coordinates. Assume X2 of generators

to equal Xl.

CHAPTER VII

TYPICAL STABILITY STUDIES

In the past twenty years a great number of stability studies havebeen made, usually with the aid of an a-c. calculating board, on existingor projected electric power systems. Through the courtesy of a number of operating and engineering companieswe are enabled to present inthis chapter accounts of several such studies made during the last tenyears. Most of the companies have preferred that their names andthe identity of their stations, substations, and lines be withheld; accordingly, letter symbols have been substituted. Let the reader beassured, however, that neither the stations nor the studies are fictitious.

These typical stability studies have been included in order to illustrate some of the purposes for which stability studies are made, toshow how such studies are carried out, how the types and locations offaults to be studied are chosen, how simplifications of the power system(such as the combining of several stations) are decided upon, and howlimiting or critical conditions are selected for study, and also to give aglimpse at the progress made in the improvement of power-systemreliability.

Some of the stability studies here presented are incidental parts ofmore general studies that included also steady-state operation (loadingof lines, maintenance of voltage, required capacity of synchronouscondensers, and so forth). All these studies involved great masses ofdata and calculations which it was not feasible to reproduce here infull; for example, length, spacing, wire size, zero- and positive-sequence impedances of transmission lines; name-plate data andimpedances of transformers, generators, and other machines; actualand estimated future loads on substations at different times of day andyear; moments of inertia of machines; calculations of per-unitimpedances on a selected system base; calculating-board connectiondiagrams and instrument readings; and calculations of swing curves.The reports on some studies fill one or more volumes. The condensedaccounts of the studies which follow give an inadequate idea of theamount of labor actually spent on the studies. Nevertheless, we hopethat they convey a clear conception of the methods and results.

253

254 TYPICAL STABILITY STUDIES

STUDY 1*

The power systems of company A and company B, when operatingin parallel, experienced difficulties due to transient instability. As anincreased flow of power from company B to company A was contemplated, stable operation would become of greater importance.Study 1 was made to determine what factors entering into the interconnected operation of the two systems were important in causinginstability and how stability might be improved.

Description of systems. A map of the two systems, showinggenerating stations, substations, and transmission lines (with theirvoltages and lengths), appears in Fig. 1. Each station and substationis there identified by a two-letter combination, the first letter of whichdenotes the power company owning or operating it. The combinedsystems extended approximately 620 miles from station BO to stationAV.

Company A had two steam stations (AL and part of AP), severalfairly large hydroelectric stations (AB, AC, and AG), and many smallhydroelectric stations, some of which were neglected or were combinedwith larger nearby stations in the study. The principal load centerswere near AL and AM. The transmission system of this company wascomposedof (1) a backbone of three 132-kv.lines in parallel, about 140miles long, firmly tying together the principal generating stations andload centers; and (2) an extensive 44-kv. network which paralleledthe 132-kv. lines, connecting smaller generating stations and loads.undwhich extended the system south of the terminus of the 132-kv. linesat station AE to station AP and to numerous substations. CompanyA had few instability problems in itself, as the loss of anyone line(either 132-kv. or 44-kv.) would not seriously weaken the ties betweengenerating stations.

Company B had a dozen or so small hydroelectric stations, dividedinto an eastern and a western group. Most of the stations of thewestern group (BB to BF, inclusive) were connected to each other andto station AD of company A by a single 132-kv. line. Stations BGand BH, also in the western group, were connected to the otherstations of this group by two 44-kv. lines. The stations of the easterngroup (RI to RO, inclusive) were joined to one another and to thewestern group chiefly by 66-kv. lines, with some 44-kv. lines. Thus,over most of the length of system B there was only one 132...kv. line, or

*Information concerning this study was obtained through the courtesy ofEbasco Services, Incorporated, New York City, with the concurrence of the operating companies concerned.

STUDY I-DESCRIPTION OF SYSTEMS 255

two or three lower-voltage lines. This system undoubtedly wouldpresent problems in transient instability even if it were not interconnected with the system of company A. In addition to what might

34

BM

38

Company B

@] Steam generating station

@ Hydro generating station

~ Combination generating station

@ Synchronous condenser substation

o Substation

• Fault location

- 132·kv. transmission line

+H++ 66·kv. transmission line

-- 44·kv. transmission line

Figures onlines aretheirlengths in miles~

Two·letter combinations are station designations.

AV

71

Principal lineinterconnecting

companies A and B

o AA

125

AO

FIG. 1. Map of the power systems of companies A and B, Study 1.

be called an inherently unstable arrangement of the transmissionsystem of company B, due principally to the nature of the territoryserved, several hydroelectric stations on this system had generatorswith characteristics which were not well suited to maintainingstability.

TA

BL

E1

LIS

TO

FS

YN

CH

RO

NO

US

MA

CH

INE

S(S

TU

DY

1)

Nu

mb

erof

Un

its

Rat

ing

WR

2(k

ilo-

lb-f

t.")

Kin

etic

Nu

mb

erof

Ru

nn

ing

inK

ind

,S

tati

on

*L

ike

ofX

dE

ner

gy

Lo

adC

on

dit

ion

(%)

ofU

nit

Un

its

Mac

hin

ek

v.

kv

a.r.

p.m

,G

ener

ator

Tur

bine

(Mj.

)1

2

AA

l2

22

Hy

dro

2.4

2,50

018

030

375

443

.0A

A2

11

1It

2.3

625

150

25.5

2.0

AA

32

22

It

2.3

600

150

241

.0

AB

l3

31

Hy

dro

6.6

12,2

2251

413

1,29

514

79.5

AB

22

10

It

6.6

8,20

015

040

2,18

018

511

.3A

B3

11

1It

7.0

8,33

317

1.5

371,

949

170

14.3

AC

33

1H

yd

ro6

.611

,111

180

153,

664

144

28

.4

AE

22

Nt

Con

do6

.67,

500

400

1875

5..

.27

.8

AF

l2

NN

Hy

dro

4.0

1,25

060

015

1A

F2

2N

NIt

1.1

1,00

040

06

3

AG

20

1H

yd

ro6

.918

,750

150

355,

760

402

32.0

AH

22

2H

yd

ro2.

32,

500

600

1320

912

.4

AI

11

1H

yd

ro2

.31,

250

600

91

3.0

AL

l2

20

Ste

am4

.48,

000

1,80

08

.580

3081

.9A

Le

11

1It

4.6

25,0

001,

800

.19

114

9815

8A

U2

11

Hy

dro

0.5

750

330

513

03

3.3

AL

41

11

"2

.31,

000

600

14

.527

9710

.3

AN

1N

NH

yd

ro2.

32,

500

360

13.5

3

l'-'

01 ~ ~ ~ ~ ..... o ;> t""4 00 ~ ;> to ~ t""4 ..... ~ t-< 00 ~ ~ t:1 ..... t?:j

00

TA

BL

E1

(Con

tinu

ed)

Nu

mb

erof

Un

its

Rat

ing

WR

2(k

ilo-

lb-f

t.F

)K

inet

icN

um

ber

ofR

un

nin

gin

Kin

d,

Sta

tion

*L

ike

ofX

dE

ner

gy

Lo

adC

ondi

tion

(%)

ofU

nit

Un

its

Mac

hine

kv

,k

va.

r.p.

m,

Gen

erat

orT

urbi

ne1

2(M

j.)

AP

l3

11

Hy

dro

2.3

2,40

030

010

383

58

.0A

P2

11

1u

2.4

7,86

051

424

254

815

.9A

P3

11

1S

team

13.8

18,7

501,

800

2268

9512

0.8

AP

42

I1

Hy

dro

6.6

875

600

13.5

183

1.7

AP

52

22

"6

.660

030

09

643

1.4

AP

61

11

"2

.32,

400

300

1038

33

8.0

AP

72

01

"6

.659

036

010

343

1.1

AP

82

02

"4

.01,

250

514

131

1.0

AV

2N

NC

ondo

6.6

2,50

01

5.5

37..

.3

.3

BB

11

0H

yd

ro2

.34,

375

164

16.3

1,12

0t

7.6

"4

40

"6

.67,

500

120

652,

670

t9

.7

BC

55

0H

yd

ro2

.31,

400

200

1436

0t

3.6

"1

11

"2

.33,

000

200

3635

0t

3.2

BD

11

IH

yd

ro6

.99,

375

240

321,

160

t16

.9

BE

11

1H

yd

ro2

.260

051

415

24t

1.6

"1

11

u6

.612

,000

240

27.5

2,00

0t

29

.2

BF

22

2H

yd

ro6

.99,

000

138.

530

4,00

0t

19

.4

BG

22

2H

yd

ro2

.31,

250

600

2527

t2

.5"

11

1"

6.9

7,50

030

035

480

t10

.9

00.

.-3 § to< ...... I t; ~ 00 (1 ~ ...

-4 ~ ...-4 o Z o ~ 00.

to< ~ trj s= 00.

~ CJ1 "'""

TA

BL

E1

(Coo

tinue

d)

Nu

mb

erof

Un

its

Rat

ing

WR

2(k

ilo-

Ib-f

t.2

)K

inet

icN

um

ber

ofR

un

nin

gin

Kin

d,

Sta

tion

*L

ike

ofX

dE

ner

gy

Lo

adC

on

dit

ion

Un

its

Mac

hine

(%)

ofU

nit

kv

.k

va.

r.p.

m.

Gen

erat

orT

urbi

ne(M

j.)

11&

BH

l2

22

Hy

dro

6.6

2,00

020

018

574

t5

.8

"1

11

u6

.63,

600

225

1059

3t

7.6

BH

21

11

u2

.35,

500

450

2050

0+

25

.6+

BH

31

10

"2

.33,

125

400

5.7

Bll

11

0C

ondo

4.1

7,50

090

03

1.5

39..

.7

.3B

I2~

30

Hy

dro

0.4

300

257

1631

:0

.5B

I33

30

u2

.362

518

022

140

:1

.2

BK

22

2H

yd

ro2

.285

012

015

286

t1

.0

"2

22

"2

.21,

250

9029

546

t1

.1

"2

22

"2

.21,

562

9036

546

t1

.1

"4

44

"2

.21,

000

109

3627

0t

0.8

BL

11

0C

ondo

2.3

3,00

072

034

22..

.2

.6

BN

l2

21

Hy

dro

6.6

5,00

022

545

1,20

0t

15

.4BN

1&2

11

"0

.550

022

512

126

t1

.6

"1

11

u2

.250

022

58

.510

0t

1.3

BO

11

0H

yd

ro6

.63,

600

225

1059

3t

7.6

*The

num

eral

sap

pen

ded

toth

est

atio

nle

tter

sde

note

sep

arat

est

atio

ns

or

par

tsof

stat

ion

sw

hic

hw

ere

com

bine

din

this

Stu

dy

.tN

=N

egle

cted

.tW

R2

oftu

rbin

esof

com

pan

yB

was

assu

med

tobe

10

%of

gen

erat

or

WR

2 •

~ ~ ~ t-d

H a E: S H t:-t

H ~ ~ ~ ~ tj H ~ m

STUDY I-DESCRIPTION· OF SYSTEMS 259

Interconnected operation of the two systems resulted in additionaltransient stability problems for both systems. As will appear fromthe swing curves obtained in this study, however, faults occurring near

TABLE 2

LIST OF STATIONS HAVING SYNCHRONOUS MACHINES (SroDY 1)

Load Condition 1 Load Condition 2

Kind ofAggregate Kinetic Aggregate Kinetic

StationMachines

Mva. Energy Mva. EnergyRunning (Mj.) Running (Mj.)

AA Hydro 6.8 10.0 6.8 10.0AB It 53.2 264.1 20.5 93.8AC u 33.3 85.2 11.1 28.4AE Condo 15.0 55.6 Off OffAF Hydro 4.5 Neglected 4.5 NeglectedAG It Off Off 18.8 32.0AH It 5.0 24.8 5.0 24.8AI It 1.2 3.0 1.2 3.0AL Steam 41.0 321.8 25cO 158.0

It Hydro 1.8 13.6 1.8 13.6u Combined 42.8 335.4 26.8 171.6

AN Hydro 2.5 Neglected 2.5 NeglectedAP Steam 18.8 120.8 18.8 120.8

u Hydro 14.8 36.4 17.9 39.5u Combined 33.6 157.2 36.7 160.3

AV Cond. 5.0 Neglected Off OffBB Hydro 34.4 46.6 Off OffBe It 10.0 21.4 3.0 3.2BD u 9.4 16.9 9.4 16.9BE II 12.6 30.8 12.6 30.8BF Ie 18.0 38.8 18.0 38.8BG It 10.0 15.9 10.0 15.9BH a 16.1 50.5 13.1 44.8BI Condo 7.5 7.3 Off OffIt Hydro 2.8 5.0 OtT OffIt Combined 10.3 12.3 Off Off

BK Hydro 11.3 9.8 11.3 9.8BL Cond. 3.0 2.6 Off OffBN Hydro 11.0 33.7 6.0 18.3BO It 3.6 7.6 Off Off

station BG or east of it had little relation to the stability of intercon-nected operation, although such faults sometimes presented seriousstability problems for the eastern part of the system of company B.

The 125-mile, 44-kv. line from AA to BA had several small hydro-electric stations (grouped as station AA in the study) and several small

~ 0)

o ~ ~ ~ o > t"'1 rn ~ OJ ~ t"'1 ~ ~ to< ~ § ~ t:rj

00

~l

C"I :+ 2 C7> + s: ~ :; s:

C! 2 +' :it

~ +' ~

o .,p £ AH

AJ

sil

132

ky.

C"I ~ +' ~

C"I ~ + ~ .., ~ + ~

AK

aD~~

:i·...1

--

+.

....

-~

~

AT

~;

e~

~lA

U

~

~; + =:

AR

AS

lJF

~ + ~

o ~ ~G

' :;; + ~

BE

AI'

~4Cky

.

::1::J.

+'~~

gl~£

+'~-

~

~T

BJ

~ ~ ~== ~I

BL 66

ky.

6&ky

.

~T ~T ~TB

L4C

ky.

BM

BK

BL

~ .c eo ; ;::.tC! :e :+ ~ FIG

.2.

Stu

dy

1.P

osit

ive-

sequ

ence

netw

ork

aspr

epar

edfo

ra-

c.ca

lcul

atin

gbo

ard.

Impe

danc

esar

ein

ohm

son

boar

dor

inp

erce

nton

a48

.4-M

w.

base

;sh

un

tca

paci

tanc

esar

ein

mic

rofa

rads

onbo

ard.

STUDY I-LOADING CONDITIONS 261

loads. Although faults on this line would sever it from the interconnected system, they would have little effect on the rest of the system.

Table 1 lists synchronous machines (generators, condensers, andlarge motors) on the systems of both companies with their ratings,transient reactances in per cent based on their own ratings, moments ofinertia (WR2

) in thousands of pound-feet", and amounts of kineticenergy in megajoules. Table 2 lists the synchronous-machine stations(or in some cases groups of neighboring stations) represented in thisstudy, with their aggregate megavolt-ampere ratings and amounts ofkinetic energy. Figure 2 is a diagram of the positive-sequence network,simplified to the extent required for setting it up on the calculatingboard. The zero-sequence diagram is not reproduced here.

The oil circuit breakers and protective relays on these systems wereof slow-speed types at the time this study was made.

Loading conditions. Transient stability was studied for twodifferent loading conditions: (1) estimated August day loads and (2)estimated October night loads. (The loading of the various circuitsfor the two load conditions is shown in Figs. 3 and 4, respectively.)The first condition was one of heavy loads, due largely to pumping forirrigation, and involved a considerable transfer of power from companyB to company A (30 Mw. from station BB to station AD). Thesecond condition was one of light loads but of even greater transfer ofpower (36 Mw.) from company B to company A. In load condition 2,all the station BB generators and five of the six station Be generators were shut down because of shortage of water; hence much of thepower transferred to company A was necessarily transmitted fromsources (chieflyfrom stations BE and BF) more distant than in loadcondition 1.

The calculating board used in this study had ten power sources forrepresenting generators. In order to check connections and settingsof the board and also to help determine which generators could begrouped together for transient stability studies, readings for loadcondition 1 were taken, first with company B generators representedin considerable detail and then with company A generators representedin considerable detail, according to the columns of Table 3 headedRun 1 and Run 2. The readings thus obtained are shown in Fig. 3.The generators were then grouped as shown in the columns of Table 3for runs 3, 4, 5, or 6, according to fault location, to permit their representation by ten power sources on the board. Individual line andgenerator loads were somewhat distorted by the grouping; the readings are not reproduced here. Readings for load condition 2 were takenwith the generators grouped as shown in Table 3 for run 7, and thesereadings are shown in Fig. 4.

STUDY i-LOADING CONDITIONS 263

-.~9·O

-"b~'£) (tn)

--.tn· 8S't

J. ~ ...

~= ~~; 0 -..(&rU ~ "CLt'r)

CQ ....0... ~~'l

~;~..----~...

TABLE 3

GROUPING OF SYNCHRONOUS MACHINES FOR REPRESENTATION BY TEN POWER

SOURCES ON THE CALCULATING BOARD AND VALUE OF KINETIC ENERGY OF EACH

GROUP IN MEGAJOULES (STUDY 1)

Power Source Numbers and Kinetic Energy (Mj.)

Station Runs RunSRun 1 Run 23 and 4-

Run 6 Run 7

BO Static*

BL Static 1 22.5 Off

BI 1 1 66.1 1 66.1

BN 2 2 33.8 1 18.3

BK 3 3 9.8 2 9.8

BHs 4 25.64 Off 2 44.8 3 44.8

BH1 5 19.22 66.4

BG (66kv.) 5 6 10.93 15.9 4 15.9

BG (44 kv.) 6 7 5.0

BE 7 4 30.8 8 30.8 5 30.83 69.6

BF 8 5 38.8 Off6 55.7

BD 9 4 16.9 6 16.9 9 16.9

BC 10 1 5 21.4 7 21.4 7 3.2

BB (6.6 kv.) Load box 26 46.6 Off

BB (44 kv.) 3 8 56.6

AA 4 7 10.0 With AP

AG Off Off Off

AB 58 154.2

AC 6 10 Q()

AF 7(Infinite

8 376.2 9 376.2 bus)

AHOff

AN 8AI

199.49AL 9

9 391.0AE Static

10 548.1AP 10 10 170.2

10 157.1AV Static Off

*uStatic" means represented by capacitor on calculating board.264

STUDY I-SWING CURVES 265

Fault locations. A principal object of the study was to determinemaximum allowable clearing times for faults at locations where theywould affect the stability of the interconnection. It is apparent that afault anywhere on the single-circuit 132-kv. lines from station AD tostation BE would split the system into two parts unless high-speedreclosing were used. Since high-speed reclosing of high-voltage lineswas still in the experimental stage at the time of this study, however,it was not considered. Hence faults on the single-circuit 132-kv. lineswere not studied. Faults on anyone circuit of the three-circuit 132kv. connection between stations AB and AE would not divide thesystem, and the critical clearing time of such faults had to be determined. Since the impedance of two or three 132-kv. lines in parallelis very low, the exact location of a fault on these lines would not makemuch difference. But afault nearstation AD on one of the three 13S-kv.lines to station AC was believed to have a somewhat greater tendencyto produce loss of synchronism between the machines of company Aand those of company B than a fault anywhere else. Consequently,this fault location was chosen for study.

Faults on the 44-kv. transmission system also required consideration.The most severe effect on stability would be produced by a fault nearthe 44-kv. terminals of a large 132-to-44-kv. transformer bank. Twosuch locations were selected for study, one near each end of the 132-kv.system. One was near station AE on the 44-kv. line to station AS, theother near station BE on the 44-kv. line to station BG. The criticalclearing time of faults elsewhereon the 44-kv. lines would be expectedto be longer than for those at the selected locations.

Faults on 44-kv. or 66-kv. lines at some distance from the 132-kv.system would not be expected to have much effect on the stability ofthe interconnection. A fault location nearstation BG on the #-kv. lineto station BH was tried to see whether it would disturb the interconnection.

The four fault locations chosen for study and described above aremarked on the map, Fig. 1.

Two-line-to-ground faults were considered in most cases, and threephase faults in some cases. Nearly always, simultaneous clearing atboth ends of the faulted line was assumed.

Swing curves. Eighteen transient-stability runs were made in thecourse of the study. The swing curves obtained are reproduced inFigs. 5 to 22 inclusive, and the conditions and results of each run aresummarized in Table 4.

In numbering the stability runs, the first number, such as 3 of 3-8-1,refers to the load.run made to obtain proper initial conditions for the

~ C)

C)

TA

BL

E4

SU

IOlA

BY

OJ'

ST

AB

ILIT

YR

UN

S(S

TU

DT

1)

Fau

ltL

ocat

ion

Cle

arin

gT

ime

(sec

.)r

....,

Run

Lo

adF

ault

Lin

eN

ear

On

Nea

rF

ar

Sta

ble

orR

emar

ksO

ondi

tion

"T

ype

Vol

tage

Stat

ion

Lin

eto

En

dE

nd

Uns

tabl

e(k

v.)

t-3

3-8-

11

2LG

44A

EA

S0

.4S

ame

Sta

ble

:;3"

3t/I

II"

u0

.4as

at

Sta

ble

H

3-8-

2a

3-S

-3"

"13

2A

DA

C0

.3ne

arU

nsta

ble

> e-3-

S-4

II"

"II

"0

.2en

dS

tabl

e00

3-8-

5II

2LG

IC"

u0

.4ex

cept

Sta

ble

~

3-S

-6II

"II

""

0.5

whe

reS

tabl

e6; H

3-8-

7"

uII

""

0.6

note

dU

nsta

ble

e- H

3-8-

8a

It

It"

"0

.7U

nsta

ble

Seri

esR

atst

a.B

B~

3-8-

10"

""

""

0.8

Sta

ble

Low

:xti

atst

a,B

Bt< 00

4-8-

1"

3t/I

It"

"0

.3U

nsta

ble

Shu

ntX

atst

a.B

B~

4-8-

2II

2LG

""

"0

.7S

tabl

e"

""

""

d tj

5-8-

1"

"44

BE

BG0

.3U

nsta

ble

t-4

t:J=j

5-8-

2"

It

""

"0

.2S

tabl

e00

5-8-

3"

""

""

0.2

0.4

Sta

ble

6-8-

1a

Itu

BG

BH

0.3

Sta

ble

7-8-

12

a13

2A

DA

C0

.4U

nsta

ble

7-8-

2cc

cc"

""

0.3

Sta

ble

7-8-

3"

"44

BE

BG

0.2

Sta

ble

•Loa

dC

ondi

tion

1:es

tim

ated

Aug

ust

day

load

s;lo

adco

ndit

ion

2:es

tim

ated

Oct

ober

nigh

tloa

ds.

STUDY i-SWING CURVES 267

320 r----oyo---r--.,.---r--r-~-__r_-_..._-....

280 ..--t---t--

2401---t---t--

40

9 (AE, AL)

0.60.2 0.4ne (,seconds)

FIG. 5. Swing curves, Study 1, stability run 3-8-3, load condition 1, three-phasefault near substation AD on one 132-ky. line to station AC, cleared in 0.3 sec.

Unstable.


stability run, and 8-1 refers to the first transient-stability run madewith these initial conditions, 8-2 refers to the second run, and so on.

The usual simplifying assumptions were made (constant voltagebehind transient reactance, constant input, and so forth). A timeinterval ~t = 0.1 sec. was used in point-by-point calculations.

240r---r--r---r----r---r--~-'I'_-,

6(BB)

200 t---t---:--"""~ -+---+----+---+-----f

40 t----I--- ~ -f.---f----+----t~.:(cu·U.....

1-----I.-3+---+----+---+---+---+----i

~r

0 ......- .....-"'"-----.......- ......----......- .....o 0.2 0.4 0.6 0.8Time (seconds)

FIG. 6. Swing curves, Study 1, stability run 3-8-4, load condition 1, three-phasefault near substation AD on one 132-kv. line to station AC, cleared in 0.2 sec.

Stable.

In determining the allowableclearingtime offaults, the usual methodwas first to estimate this time and then to make a run using this estimated time. If conditions proved stable, another run was made usinga clearing time 0.1 sec. longer than before. If the second run showedthe systems to be unstable, then the clearing time of the first run gave

STUDY I-SWING CURVES 269

the answer desired. If the second condition was still stable, then athird run was made with 0.1 sec. longer clearing time.

Runs 3-8-5,3-8-6, and 3-8-7 (Table 4) illustrate the method. For atwo-phase-to-ground fault at station AD on one of the lines to stationAC under load condition 1, run 3-8-5 showed the systems to be stable

200 _-_-or---r---__-.,..--..,...--..,...----...

O--_......_.Io-_""-_.a..-_..I..-_"'--......--'o 0.2 0.4 0.6 0.8

Time (seconds)

FIG. 7. Swing curves, Study 1, stability run 3-8-5, load condition 1, two-line-toground fault near substation AD on one 132-kv. line to station AC, cleared in 0.4

sec. Stable.

for 0.4-sec. clearing. Run 3-8-6 showed the systems to be still stablefor 0.5-sec. clearing. However, run 3-8-7 showed them to be unstablefor 0.6-sec. clearing time. Thus it was shown that 0.5 sec. was themaximum allowable clearing time for these particular conditions.

In some cases it was possible to determine maximum allowableclearing time from a single run because the angular swing of particularmachines was wide enough, and the trend of accelerating and decelerating forces on the calculating sheets was such that the run wasreadily seen to be a borderline case, and additional clearing time would


undoubtedly have caused instability. Sometimes comparison withruns previously made was helpful in such interpretations.

A detailed discussion of the various swing curves follows.Stability during load condition 1; faults on 132-kv. system of com

pany A. The generators were grouped on the power sources of thecalculating board as shown in Table 3 for run 3. Initial conditions for

200 .--or---r---r----r---r--'T"""--"---'

160 t---f----.......~~F-

4Ot-----ir---J---+---+---I--qL-..;::!~-+----I

0.2 0.4 0.6 0.8Time (seconds)


sec. Stable.

the stability runs were those of load condition 1 (Fig. 3). Swingcurves 3-8-3 and 3-8-4 (Figs. 5 and 6) show that the systems wereunstable for O.3-sec. clearing but were stable for O.2-sec. clearing of athree-phase fault at station AD on one of the 132-kv. lines to stationAC. Curves 3-8-5, 3-8-6, and 3-8-7 (Figs. 7,8, and 9) show that, witha two-line-to-ground fault at the same location and with the same loadcondition, the systems were stable for O.5-sec. clearing and unstablefor O.6-sec. clearing.

On all these curves it should be noted that the three main generating

STUDY I-SWING CURVES

280 -----.-----..------------

271

80 1----+----I---+-~~lIr___+-__tf__-_+___t

40 I----+---+---+----+----+-~~-+-~

O.....-~_..r.- ......_....... '___""_____.;;:::I

o 0.2 0.4 0.6 0.8Time (seconds)

FIG. 9. Swing curves, Study 1, stability run 3-8-7, load condition 1, two-line-toground fault near substation AD on one 132-kv. line to station AC. cleared in 0.6

sec. Unstable.


320 --.....--------------.---.....--.....---

280 1---+---0+-

240 t--+--r---

80 I----a.--~~__r_-__+_---+--..._-+_____t

8(AB,AC,AFt

AH,AI,AN)40 I--_...-_-+- --.l~r--__+-~~-+____f

0'---.............-_....._ ......._ ....._ ....._ ......_ ...o 0.2 0.4 0.6 0.8

Time (seconds)

FIG. 10. Swing curves, Study 1, stability run 4-8-1, load condition 1, 15,OOo-kva.shunt reactors added on station BB low-voltage bus, three-phase fault near sub

station AD on one 132-kv. line to station AC, cleared in 0.3 sec. Unstable.

STUDY 1-PROPOSED CHANGES AT STATION BB 273

stations of company A swung almost exactly together, slowing downduring the fault, thus acting like the equivalent motor of a two-machinesystem. Station AA and those stations of company B which wereconnected at 132 kv. (BB to BE, inclusive) swung fairly closely together-although not as closely as the stations of system A-andspeeded up, acting like an equivalent generator. In every casestation BB, the station closest to the fault 'of this group, speeded upmore than the others (although it was closely followed by station AA)and was the first station of this group to pull out of step. StationsBG to BO, which were farther from the fault, were less affected, buttended to stay with the other company B stations.

Study of proposed changes at station BB. As station BB was foundto be the first station of company B to pull out of step during faults onthe 132-kv. transmission system of company A, it was believed thatany measures taken to help this station stay in step might materiallyimprove the stability of the interconnected systems. Three differentchanges were studied.

Use of shunt reactors. The design of the generators at station BBcontemplated normal operation at 80% lagging power factor. Actually, however, these machines were usually operated near unity powerfactor. Such operation caused the machines to have internal voltagelower than normal and tended to make them unstable. Studies weremade to show the effect of connecting shunt reactors to the low-tensionbusses at station BB in order to put about 15-Mvar lagging load on themachines and thereby increase their internal voltages.

The initial load conditions with the reactors added were obtainedin load run 4 and differed only slightly from the conditions of load run3. The same generator grouping was used. Three-phase and twoline-to-ground faults were assumed at the same location as before,namely, near station AD on a 132-kv. line to station AC. Curve4-8-1 (Fig. 10) shows that the system was unstable for 0.3-sec. clearingof a three-phase fault, whereas curve 3-S-4 (Fig. 6) shows that thesystem was stable for O.2-sec. clearing of such a fault without the useof shunt reactors at station BB. Therefore it is apparent that theaddition of the reactors would not increase the maximum allowableclearing time for a three-phase fault by as much as 0.1 sec. Curve4-8-2 (Fig. 11) shows that a clearing time of 0.7 sec. was permissiblefor a two-line-to..ground fault, and curve 3-S-6 (Fig. 8) shows a permissible clearing time of 0.5 sec. if no reactors are used. Therefore,a gain of at least 0.2 sec. in allowable clearing time of a two-line-toground fault would result from the use of the reactors.

Use of series resistors. A method of increasing the stability of hydro-


electric generators by automatically cutting resistors in series withthem during faults and thereby maintaining their load and decreasingtheir acceleration during faults was proposed by R. C. Bergvall.' Asthis method appeared to offer a possibility for stabilizing station BB,it was studied on the board. A resistance of 7.2 ohms per phase, connected in series with each of the four 7,500-kva. 6,600-volt generators,

200...---,.----r---~-....--..,..--.,__-_r_-_r_-_,._-_

0.2 0.4 0.6 1.0Time (seconds)

FIG. 11. Swing curves, Study 1, stability run 4-8-2, load condition 1, 16,OOO-kva.shunt reactors added on station BB low-voltage bus, two-line-to-ground fault near

substation AD on one 132-kv. line to station AC, cleared in 0.7 sec. Stable.

was found to maintain practically fuIlload on the generators during thefault. Resistors of this value were assumed to be cut into the circuit0.1 sec. after occurrence of the fault and short-circuited again 0.2 sec.after clearing of the fault. The result is shown in curve 3-8-8 (Fig. 12),in which a two-line-to-ground fault near station AD on a 132-kv. lineto station AC caused instability when cleared in 0.7 sec. This appeared to be a borderline case, however, and it seemed safe to concludethat 0.6-sec. clearing of the fault would have made the system stable.Comparison of this result with curve 3-8-6 (Fig. 8) for the same condi-

STUDY I-PROPOSED CHANGES AT STATION BB 275

1.00.80.2

~~I~.J----+---+-----+--1~-4--~H-~-+---+-~r------I

.mlS.:...I----+--.J----+---+----+----I-~~--I---_+_---i~I~~IubO ~.5 t \0 -I----+---i---+--__+_-

~ 1.5Q)Cl)4----+----+--__+_---+---i-.51~ I'tn+---+---I----4----+----11--'en~I

o

40

200 ----.....--......--'I"--..--..,----r----r-----r--r-----.

120 1--..;;;;::::::'l..._!:::I----a.~---+--~--+---~--+------"'~~I--~

- 80 1...--.&-_..1.-_..1..._..1-_....1-_.....1-_....1-_-"--......---'- __

o

-40

0.4 0.6Time (seconds)

FIG. 12. Swing curves, Study 1, stability run 3-S-8, load condition 1, two-line-toground fault near substation AD on one 132-kv. line to station AC, cleared in 0.7sec.: 7.2-ohm resistors inserted in station BB generator leads 0.1 sec. after initiation

of fault and removed 0.2 sec. after clearing of fault. Unstable.


tions but without the series resistors showed an increase of permissibleclearing time of only 0.1 sec.

Decreased transient reactance. The four large generators at stationBB had the abnormally high transient reactance of 65% by test. Un-

200.---.,.--or---..,---..,.--...,.---.,--....,---.....----

i 120~ -----f

"'t:J~

CDboc«J~ 80~g(ijc'-~.E 40 t---+--+----t--+---r--~----:~--I---+--~

Ot---+---+---+---+---+---+----+---HI.3IiIl~--=t

- 40 -----......-"""-----_...r.-_....._-'-_....._....I

o 0.2 0.4 0.6 0.8 1.0Time (seconds)

FIG. 13. Swing curves, Study 1, stability run 3-8-10, load condition 1, transientreactance of station BB generators reduced from 65 to 35%, two-line-to-groundfault near substation AD on one 132-kv. line to station AC, cleared in 0.8 sec.

Stable.

doubtedly, this high value of reactance decreased the stability ofstation BB. In order to determine the effect of a lower value oftransient reactance for these machines, curve 3-8-10 (Fig. 13) wastaken with the reactance assumed as 35%. The results showed thatO.S-sec. clearing of a two-line-to-ground fault would make the systemstable. This represented an improvement of 0.3 sec. over the 0.5-sec.

STUDY I-FAULTS BETWEEN STATIONS BE AND BG 277

clearing which was necessary for the same fault conditions but withthe high value of transient reactance.

Faults on the 44-kv. system of company A. Three-phase and twoline-to-ground faults were tried near station AE on the 44-kv. line tostation AS under load condition 1. Curves 3-8-1 and 3-8-2 (Figs. 14

200r---r--.,--w=---_-.---~,.----,...___

0.8

9(AE,AL)

0.2o1:-.- ......---------~--------_........- ......

9 0.4 0.6Time (seconds)

FIG. 14. Swing curves, Study 1, stability run 3-8-1, load condition 1, two-line-toground fault near substation AE on 44-kv. line to substation AS, cleared in 0.4

sec. Stable.

160

40 I--f.--+----I---+-~~~-I---~----t

and 15) show that the systems remained stable for either type of fault'cleared in 0.4 sec. The limiting case for a two-line-to-ground faultwas not obtained, and O.5-sec. clearing was judged to be permissible.A fault on the low-voltage circuits of any of the large stations tappingthe 132-kv. lines of company A would be of approximately the sameseverity in regard to stability.

Faults on the 44-kv. line between stations BE and BG.' The effectof a two-line-to-ground fault near station BE on the 44-kv. line tostation BGunder load condition 1 is shown in curves 5-8-1,5-8-2, and5-8-3 (Figs. 16, 17, and 18). The generators were regrouped as shown


200r---r---__--,.--_---r--~-_-___,

- 80enQ)

f~Cl>

"'C-Cl)

00cco 40Cl>~ 8 (AB, AC, AF,.s'0 AH, AI, AN)>

roE$.s 0

- 40 t----+---+---+---t--+---t----+---"---i---t

- 80 t----t-----+----+--

1---+---+----..-- 3 +---+---+---+---f

~I- 120""'-_-'--_......._--'-_--A._...... ....... _ ....._ ...

o 0.2 0.4 0.6 0.8Time (seconds)

FIG. 15. Swing curves, Study 1, stability run 3-8-2, load condition 1, three-phasefault near substation AE on 44-kv. line to substation AS, cleared in 0.4 sec. Stable.


0.80.6

~--""'_-.I 8 (AA, BB)

7(BC)

0.2O'--_J..-_.L-_..Io.-_..a..___......._....Ioo._....._ ....

o

40 1---I---+---;+.Ji-~---t----t"----1---;

320--,...--..,.----r---rlr---r---r--,---,

2401---I----+-~J+---+--_+-__t-__1-___1

80 l----+----"--~--+--__t_-__+-__1-____1

2801----f---t---f---1---t--t----t---t--,

C1J-soc:co 160 1----#--~:::...-.-+-~.....- .......---+-__1~__1~~~(ijE~ 120~~$!!!!!I!!!!!!!IIiI~~.f--_9 (AB, AC, AF, AH, AI, AN)

I I10(AE, AL, AP, AV)

~ 2001---4---'---#--+--+-~r---+--;--,GJeiP

"C-

0.4Time· (seconds)

FIG. 16. Swing curves, Study 1, stability run 5-8-1, load condition 1, two-line-toground fault near station BE on one 44-kv. line to station BG, cleared in 0.3 sec.

Unstable.


for run 5 in Table 3 to give more detailed representation near the faultand less detailed representation far away from it.

Curve 5-8-1 shows the system to be unstable with O.3-sec. clearing.Station BE, which was closest to the fault, pulled ahead and out of

240.--_r--~r---r--r----r---..---....--..

200I--t---...

lO(AE,AL,AP,AV)

40 t---+--

I---+-- ...., +---+--+--+---+---0+-----1

~Io......- .....-.-.--~-"--~- ......-----'o 0.2 0.4 0.6 0.8

Time (seconds)

FIG. 17. Swing curves, Study 1, stability run 5-8-2, load condition 1, two-line-toground fault near station BE on one 44-kv. line to station BG, cleared in 0.2 sec.

Stable.

step with the rest. Stations BD and BF also speeded up, and stationsBG to BO slowed down. It should be noted that the initial flow ofpower was from the former to the latter group of stations, which maybe regarded as the equivalent generator and motor, respectively, of atwo-machine system. Stations BB and Be were but little affected}and the stations of company A were affected hardly at all.

Curve 5-S-2 shows the system to be stable with O.2-sec. clearing at


./1 (BI,BK,, BL,BN,BO)

4(BE)breakersopened

12CYCles'*140

10 (AE, AL, AP, AV)

240 r--......---,.--.,..--....--..,.---.----.----.

200 1---+---'1---+--..---+---+---+---1

o .....------...--.......-...-.------.......----~o 0.2 0.4 0.6 0.8

Time (seconds)

FIG. 18. Swing curves, Study 1, stability run 5-8-3, load condition 1, two-line-toground fault near station BE on one 44-kv. line to station BG, cleared in 0.2 'sec.

at BE and in 0.4 sec. at BO. Stable.


both ends of the line. It was also stable, as shown by curve 5-8-3,with sequential clearing in 0.2 sec. at station BE and in 0.4 sec. atstation BG.

Faults on 44-kv..line between stations BG and BH. The effect of atwo-line-to-ground fault near station BG on the 44-kv. line to station

240..---r----r--,r--~r---.,r----r.--~r--_

40 J---+---t--

9(BD)I

10 (BB, Be,and company Aas infinite bus)

~ 5(B~1) I«S 120 I---~----l~""-"~-+ 7 (BG 44 kv.)~ I

~ 3 (BJC)~ I

Ci 4(B82)c...~ 80 1---t---t--~L..-....lII~~-fo--.--t----1

-;; 160 .....~~~----+~

I."-

I---t---t--...,-I--+---f--+--+---f:;~I

O'---'-------.....-"--"--------~---.to 0.2 0.4 0.6 0.8Time(seconds)

FIG. 19. Swing curves, Study 1, stability run 6-8-1, load condition 1, two-line-toground fault near station BG on one 44-kv. line to station BH, cleared in 0.3 sec.

Stable.

BH was ascertained for load condition 1. The generators were regrouped, as shown in Table 3 for run 6, to give more detail in theneighborhood of the fault, including the separation of stations BG andBH into two parts each. The machines of company A, together withstations BB and Be, all of which were found to be only slightly affectedby a fault between BE and BG, were represented in this run as an

STUDY I-STABILITY DURING LOAD CONDITION 2 283

infinite bus. Curve 6-8-1 (Fig. 19) shows the system to be stable forO.3-sec. clearing at both ends of the line. The general direction ofpower flow was eastward. During the fault stations west of the fault

360 r----y--r---,.--..---yo----,n-----r----.

280 I---+---+----+-+--I--.......~H---+---+----I

160 1--'--+--~~-.f--7l'o~+--+---+----+---4

120t--+---+---+--~r__~.-+--+---t

0.2 0.4 0.6 0.8Time (seconds)


sec. Unstable.

(equivalent generators) speeded up, and stations east of the fault(equivalent motors) sloweddown.

Stability during load condition 2. In this load condition, as alreadymentioned, station BB and most of the generators at station Be were


shut down, and more power was transmitted from company B tocompany A than in the previous load condition. The generators weregrouped in a manner (shown in Table 3 for run 7) which was judged

360 .---".--....---..----.,.....-.,..--.,.---ro--_

10{AA, AP)

~9(AE,AH,

AI, AL, AN)

~

enCD

~CD~ 240 1---t-----#~~.__ +-I-~lIt---+---t-_4

.!lcoCtV

Wo~

~ 200 1-JIIliIr:---J1----#-+---~-+---J~+---+---t-_oftVC...J!!.E

280 t---t---t----.,..;~__II---._,

1601---+--~~-+----4~--+--+--+-_of

120~-+---+---__-+--+--+--+----t

0.60.2 0.4Time (seconds)

FIG. 21. Swing curves, Study 1, stability run 7-S-2, load condition 2, two-line-toground fault near substation AD on one 132-kv. line to station AC, cleared in 0.3

sec. Stable.

to be suitable for faults at either of two locations-near AD on the132-kv. line to AC or near BE on the 44-kv. line to BG.

Curves 7-8-1 and 7-8-2 (Figs. 20 and 21) show that, for a two-lineto-ground fault at the former location, the systems were stable for

STUDY l-STABILITY DURING LOAD CONDITION 2 285

O.3-sec. clearing and unstable for O.4-sec. clearing. It will be notedthat the permissible clearing time of 0.3 sec. for this load condition wasconsiderably shorter than the permissible clearing time of a similarfault under load condition 1, which was found to be 0.5 sec. Thiscan be attributed to three reasons:

320 r----~-......--_-..,._-_.__-....,._-...._-_..

120 I--+.---"'----t--+---+---t---r---;

80 ,--_",--_.a..-_~_..Io.-_-'- """_'"

o 0.2 0.4 0.6 0.8Time (seconds)

FIG. 22 Swing curves, Study 1, stability run 7-8-3, load condition 2, two-lineto-ground fault near station BE on one 44-kv. line to station BG, cleared in 0.2

sec. Stable.

1. The amount of power transferred from company B to companyA was greater under load condition 2 than under load condition 1.

2. With station BB shut down, much of the power which itformerly supplied had to be transmitted from a greater distance.The result was a greater angular displacement between the generators of the two systems in the steady state and, therefore, less stableoperation in the transient state.


3. When station BB was on the line, regardless of the amount ofpower supplied by it, it served to support the voltage near the middleof the interconnecting line and thus to improve both steady-stateand transient stability. In load condition 2, station BB was shutdown, and this effect was nonexistent.

The permissible clearing time of a three-phase fault was not determined, but, as it was 0.2 sec. under load condition 1, it would be0.2 sec. or less under load condition 2.

Curve 7-8-3 (Fig. 22) for a two-line-to-ground fault near BE on the44-kv. line to BG shows a stable condition for O.2-sec. clearing at bothends. It was thought probable that sequential clearing, 0.2 sec. at BEand 0.4 sec. at BG, would be stable under load condition 2, as it wasunder load condition 1.

Summary of allowable clearing times. The slowest clearing timespermissible with stable operation, as determined from the swing curves,are summarized in Table 5.

TABLE 5

SLOWEST PERMISSIBLE CLEARING TIMES OF FAULTS

(STUDY 1)

Load condition 1 (estimated August day loads)Three phase fault near AD on 132-kv. line to AC:Two-line-to-ground fault near AD on 132-kv. line to AC:Three-phase fault near AE on 44-kv. line to AS:Two-line-to-ground fault near AE on 44-kv. line to AS:Two-line-to-ground fault near BE on 44-kv. line to BG:Two-line-to-ground fault near BG on 44-kv. line to BH:

Load condition 2 (estimated October night loads)Three-phase fault near AD on 132-kv. line to AC:Two-line-to-ground fault near AD on 132-kv. line to AC:Two-line-to-ground fault near BE on 44-kv. line to BG:

*Estimated.

0.2 sec.0.5 sec.0.4 sec.0.5 see."0.2 sec.0.3 sec.

0.2 sec.*0.3 sec.0.2 sec.

Conclusion and recommendations. The study showed that thecharacteristics of the transmission system, especially in regard to thespeed of clearing faults, had a decidedly more important bearing on theproblem of improving stability than did the characteristics of individual machines or stations. It was concluded that no adequate correction of transient instability could be obtained until both relays and oil

STUDY 1-CONCLUSION AND RECOMMENDATIONS 287

circuit breakers on most of the 132-kv. circuits, on many of the 44-kv.circuits, and at major generating stations were modernized to obtainhigh-speed clearing of faults. If relaying .and oil circuit breakerswere modernized, no other changes would be needed in the system toobtain a practicable solution of the instability problems which werestudied.

Three proposals for increasing the stability of the generators atstation BB by changes at that station were studied, but the usefulnessof the proposed changes was limited by the fact that this plant wasusually shut down, because of water conditions, at night during thefall and winter seasons when the greatest amount of power was usuallytransferred from company B to company A and when instability ofthe systems was aggravated by this heavy power flow. Furthermore,changes at station BB would have very little effect in improving stability under most fault conditions which occurred on the system ofcompany B.

Consideration was given to the possibility of making changes affecting generators at plants other than BB. In all cases studied, however,it was found that the machines in a particular area swung closelytogether as a group, almost without regard to characteristics of individual stations. In practically every case the machines nearest thefault showed the greatest tendency to lose synchronism.

From the foregoing observations, it was concluded that any changeswhich might be made at any of the generating stations would belimited in their effect and relatively costly to attain and, therefore,would assume a position of minor importance which did not justifyfurther study.

It was recommended that a definite program of relay and oil circuitbreaker improvement be prepared, taking into consideration theamount of good to be accomplished by various changes and the cost ofthe changes. It was also advised to concentrate on those changeswhich would allow the systems to remain stable after a two-line-toground fault, as three-phase faults are infrequent on properly relayedhigh-voltage systems. Modernization of relay systems and of oilcircuit breakers would not only increase stability but would also decrease damage to equipment and improve service to customers.

It appeared that the most improvement at the least cost could beobtained at many points on the system by relay changes. Modernization of some oil circuit breakers might be accomplished at reasonablecost but would, in general, be more expensive than relay changes toattain the same improvement in stability. Someother breakers shouldbe replaced rather than rebuilt.

288

S Chicago

...anC'f).""

Indiana and SCincinnati


Toledo(Toledo Edison Co.)

SouthPoint 10

lorain(OhioP.S. Co.)

FIG. 23. Central 132-kv. transmission system of American

Akron(OhioEdison Co.)

Massilon(Ohio P.s. Co.)

41.8mf.6.68+j19.0

STUDY 2

Alliance(OhioP.S. Co.)

Newcomerstown

289

..,....,........-5

5'--0West PennPower Co.

Turner(Institute, W. Va.,nearCharleston)

CabinCreekW. Va.

Windsor(Power, W.Va., nearWheelinl)

®- Steam - electric generatintplant

@- Synchronous condenser(s).

®- Equivalent of power system.

~Load.

--- Future 132·kv. line.

-- Existing 132·kv. line.

Impedances aregiven in percenton l00·Mw. 132·kv. base.

en Parallel impedance of double circuitline.

Figures on busses areshuntcapacitive susceptances ofconnected linesin percenton same base.

Gas and Electric Company and interconnections (Study 2).

290 TYPICAL STABILITY. STUDIES

STUDY 2t

In 1940 the Ohio Power Company was adding two new generatingunits (numbers 4 and 5, rated at 94.5 Mva. each) to its steam-electricgenerating station at Philo (on the Muskingum River near Zanesville,Ohio), thereby increasing the aggregate capacity of the station from271.4 Mva. to 460.4 Mva. The generators at Philo were paralleledthrough 132-kv. busses. The 132-kv. oil circuit breakers had a ratedinterrupting capacity of 2,500 Mva, The addition of the new generators would have increased the interrupting duty of these breakers fromapproximately 2,250 Mva. to approximately 3,150 Mva., a value inexcess of their rating, had not steps been taken to limit the shortcircuit currents. For this purpose it was decided to install reactorsbetween two sections of the 132-kv. bus, which will be called sectionsA and B.

In October, 1940, an a-c. calculating-board study was made by theAmerican Gas and Electric Service Corporation of the central systemof the American Gas and Electric Company and interconnected systems, including the expanded Philo station with the bus reactors. Thestudy included short-circuit studies, load studies, and a few stabilityruns.

The system studied is represented in Fig. 23. It extended fromIndiana across Ohio to West Virginia with interconnections to Chicago,Cincinnati, western Pennsylvania,' Vir-ginia, and Tennessee. Steamelectric generating plants of the represented part of the American Gasand Electric Company's system were located at Twin Branch (Mishawaka, Indiana), at Philo, at Cabin Creek, West Virginia, at Logan,West Virginia, and at Windsor (near Wheeling), West Virginia. Thegenerating plants of 'the interconnected systems also were predominantly steam stations. All the transmission lines represented in thestudy operated at 132 kv. Impedances of the lines and of the equivalent generators are given in per cent on a 100-Mva. base in Fig. 23;total shunt capacitive susceptances at each bus, resulting from nominal-e representation of the transmission lines, are also given in per centon the same base. The impedances of the equivalent generators wereobtained by simplifying the circuits of generating or synchronouscondenser stations, using the transient reactances of the machines.For example, Fig. 24 is a simplified diagram of Philo station with themachines and lines grouped on the bus sections in the manner assumed

tData on this study were obtained from the American Gas and Electric ServiceCorporation, New York City, through the courtesy of Philip Sporn, Vice Presidentin Charge of Engineering, Harry P. St. Clair, System Planning Engineer, andCharles A. Imburgia.

STUDY 2 291

CrooksvilleZanesville Howard Torrey North· Newcom· Rutland

No. 1 No. 2 No. 1 No. 2 No. 1 No. 2 east erstown No. 1

Generators

j35.3 j61.7 j25.6 j57.0 j57.0 j33.7 j35.3 j61.1 j25.6

e Circuit breakers assumed to open to clear the fault

FIG. 24. Philo steam-electric generating station. Simplified one-line diagram ofthe connections assumed in Study 2. Impedances are in per cent on 100-Mva.

base. Transient reactances are used for the generators.

(a)

(b)

FIG. 25. Philo plant reduced to two equivalent generators, A and B. Impedancesare in per cent on l00-Mva. base. (a) Normal condition, (b) after clearing fault

on bus section AI.


in the stability study. Data on the generators and transformers atthis station are given in Table 6. Values of impedance on a 100-Mva.base, taken from this table, are marked on Fig. 24. By means of seriesand parallel combinations the circuit was further simplified as shownin Fig. 25. Similar simplifications were performed for the othergenerating stations and interconnected power systems shown in Fig. 23.

TABLE 6

DATA ON GENERATORS AND TRANSFORMERSAT PHILO PLANT (STUDY 2)

GENERATORS

Xd' % on Xd" % onUnit ,..--...... Hon

Number Mva. r.p.m, Rating 100 Mva, Rating 100Mva, 100Mva.

1 42.1 1800 24.0 57.0 15.0 33.6 2.352 42.1 1800 24.0 57.0 15.0 33.6 2.35

3-1 62.4 1800 22.0 35.3 13.0 20.8 3.813-2 62.4 1800 21.0 33.7 14.0 22.4 3.813-3 62.4 1800 22.0 35.3 13.0 20.8 3.81

4HP 50.7 3600 13.0 25.6 10.0 19.7 1.744LP 43.8 1800 27.0 61.7 16.0 36.6 3.225 lIP 50.7 3600 13.0 25.6 10.0 19.7 1.745LP 43.8 1800 27.0 61.7 16.0 36.6 3.22

TRANSFORMERSx%on

~

Unit Mva. Rating 100Mva.

T-l 45 8.9 19.8T-2 45 8.2 18.2T-3-1 63 17.8 28.3T-3-2 63 17.5 27.8T-3-3 63 17.9 28.4T-4 120 13.4: 11.2T-5 120 13.4 11.2

The inertia constants of the equivalent generators are listed inTable 7. Most of these values are based upon detailed information,but a few, particularly those for future additional generating capacity,are estimated at the rate of 6 per 100Mva. for old generators and 5 per100 Mva. for new generators, the new ones being assumed to behydrogen-cooled.

The calculating-board set-up was similar to that of Fig. 23, exceptthat Fort Wayne and the stations and systems west and south of itwere combined and represented by an equivalent generator at Fort

STUDY 2 293

TABLE 7

VALUES OF INERTIA CONSTANT H OF SYNCHRONOUS MACHINES ON l00-MVA.BASE (STUDY 2)

Machines Case1 2 3

Chicago, Indana, and Cincinnati'systems Off 61 76*Twin .Branch generators 1, 2, 3, 172 Mva. 7.8 7.8 7.8Twin Branch generator 4, 94 Mva, Off Off 5.0Fort Wayne condenser, 30 Mva, 1.2 1.2 1.2

Total, Fort .Wayne and West 9 70 90

Fostoria condensers, 40 Mva. 0.7 0.7 Negl.

Toledo Edison Co. system, approx, 100 Mva. 6 6 7.7*

Lorain generating station of Ohio P. S. Co., 36 Mva. 2.1 2.1 2.7*

Canton condensers (Torrey, Sunnyside, Northeast) 2.1 2.1 2.1Akron and Alliance systems, over 500 Mva. 8 20 36.5*

Total, Canton and interconnections 38.6

Windsor generators (Beech Bottom Power Co.), 257 Mva. 15.1 15.1 15.1West Penn Power Co. system, approx. 1,000 Mva. Off 71.2 71.2

Total, Windsor and East 86.3 86.3

Philo A (gens. 1, 2, 3-1, 4) before clearing fault, 241 kyat 13.5 13.5 13.5

Philo A (gens. 1, 2, 3-1, 4) after clearing fault, 199 kva, 11.1 11.1 11.1

Philo B (gens. 3-2, 3-3, 5), 219 kva, 12.6 12.6 12.6

Rutland, future installed capacity of Appalachian ElectricPower Co. and interconnected systems, approx. 1,100Mva. 60

Lancaster, future interconnection to west, approx, 170 Mva. 10

Cabin Creek generators 1 to 8, 232 Mva, t

Logan generators, 102 Mva. tPlants south of Logan and interconnected systems, 375 Mva,

Total, Logan and south

*Estimated future increase.[Combined with Rutland.

60

10

t

t

Off

Off

15

4.422.6

27


Wayne. Furthermore, double-circuit transmission lines were represented on the board by single impedance units set at the impedanceof two lines in parallel; when one circuit was supposed to be switchedout, the setting of the impedance unit was doubled.

The short-circuit studies showed that the reactance between bussections A and B at Philo should be 7.5% on a 100-Mva. 132-kv. basein order to limit the interrupting duty of any circuit breaker to a valueslightly less than 2,500 Mva. The 7.5% reactance would be obtainedfrom two sets of reactors in parallel, as shown in Fig. 24, each having15% reactance.j

Load studies were made to determine the proper normal-load ratingof the reactors and to find the arrangement of transmission lines on thetwo bus sections that would give the best conditions of transmission.In these studies estimated 1942 peak loads were used.

In the course of the load studies, three transient-stability runs weremade to test the stability of the system (modified by the addition ofnew generating units and bus reactors at Philo) with the interconnecting lines to other companies (Chicago, Indiana and Cincinnati, andwestern Pennsylvania) first open, and then closed.

The three transient-stability runs are herein designated cases 1, 2,and 3. In cases 1 and 2 proposed generating stations or interconnections at Lancaster and Rutland were assumed to be in operation, andthe existing generating stations at Logan and Cabin Creek were combined in an equivalent generator at Rutland. The flow of active andreactive power in both cases was as shown in Fig. 26. The flow fromPhilo west toward Fort Wayne was heavy. There were also heavyflows from Rutland to Philo and from Philo to Canton. In both cases1 and 2 a three-phase fault was assumed to occur on section Al ofthe Philo 132-kv. bus and to be cleared in 0.15 sec. (9 cycles) by theaction of I-cycle bus diffffi.ential relays opening the following 8-cyclecircuit breakers (see Fig. 24): Howard line 1, Crooksville line 1,generator 1, and bus-tie reactor 1. The assumptions which have beenstated represent a severe condition regarding stability, because of theheavy power flow over long transmission lines, the severe type of fault,the fault location at the sending end, and the loss of a long, heavilyloaded circuit in clearing the fault. A three-phase fault near Philo onone of the lines to Howard might be expected to produce very nearlythe same effect as a fault at the location assumed.

In case 1 the interconnections to other companies were open; in case2 they were closed. The swing curves for cases 1 and 2 are given in

tThe reactors actually installed were rated 125 Mva., 138 kv., 23% reactance(equivalent to 20% per reactor on 10o-Mva. 132-kv. base).

~ § ~ ~

30.3

-j3

.0

New

com

erst

own

Akr

on

Allia

nce

trs

an ~ ..... + C'f~

47.2

+j1

3.0-o

f-

o :;- N a\ N

Tole

dots

,,

106

Rut

land

and

Sou

thrh

t166

+j3

6.2

FIG

.26

.In

itia

llo

adco

ndit

ions

ofca

ses

1an

d2,

Stu

dy

2.Fl

owof

acti

vean

dre

acti

vepo

wer

isgi

ven

inM

w.+

iM

var.

Bus

volt

ages

are

give

nin

per

cent

of13

2kv

.t..:

>~


Figs. 27 and 28, respectively. In case 1 Fort Wayne and Twin Branchpulled out of step with the rest of the system, but in case 2 these stations, supported by the interconnections to the Chicago and Indiana

0.6

I~,

~ r-,I // I

"\I

V Phil0I:!~.~

VI/V I ---....V ~

Rutland~~__lZ

~I ~.......- ~I

I~Canton condenserI Windsor-"'\ I I I

I

~II .i~I Lanc,aster -

~~~ ~' ZAkron - AllianceI........... l-/-~---...""--. '"

t~~ ......

~,

I~" ~I ~,

I I ~ Lorain -I----, I ~~ -~~ I" I /~~,

., I" '"~~o\~ I I~I

\1 /~Fostoria condenser~~ l,~ ~~ I I ""i'-L -

. I J I I.1"~I Y ~Fort Wayne and West

~I '"-,d r111 <,

r-,;1~I "'I'--·~·t !llo...

~I <,r-,I-120o

0

-en

~cv -20-0-c0:een8-

-40~

«I

ioc<

-60

-80

-100

40

60

20

0.1 0.2 0.3 0.4 0.5Time (seconds) afteroccurrence of fault

FIG. 27. Swing curves, Study 2, case 1. Future set-up. Generating plants atLancaster and Rutland. Interconnections open. Three-phase fault on Philo132-kv. bus section Al cleared in 0.15 sec. by opening Howard line 1, Crooksville

line 1, generator 1, and bus-tie reactor 1. Unstable.

systems, stayed in step. The motion of the other machines was verynearly the same in both cases. Philo A, the machine nearest the fault,lost all its load while the fault was on and speeded up more than anyother machine. Philo B, separated from Philo A by the bus-tiereactor, also speeded up, although less than Philo A did because it lost

STUDY 2 297

j 1 I /v~I T

I ~~PhiloA.... Fault I // "cleared

~ ~~ 0.15 sec. y . ./~Philo B "r\. ___

~~

V~~I I -C><~I'R.utland __V

~

--~ [.,.7' I rr-""""""'" '<r-,~ I r-,

lancaster, I I ......./,-Windsor ~Canton condenser ~. -"""

II " ,- '"-.1-, ~/~ ....... C><c-~ ............

-----""""--- -- I -: / ~~~', ~- ~~ .. ,~ . ~-- c:~

Akro~s, I v\ <-Lorain ~~, ~~~ \ ---Y I~K Alliance II r-. --a--~. I ..

,~ i - \ 1oledo IFostoria i

" "condenser~~

~J'I I I \.I' -...',--....r\r---R= , 'J-/

I ~t--- ............I " FortWayne andWest

---.I I I I I

-60

less load. The other generating stations were affected but .little. Thesynchronous condensers, whose swing curves are shown by broken lines,had very low inertia and swung with a shorter period than did thegenerating stations. Because of their small inertia they had littleeffect on the rest of the system, and they were not represented separately in case 3.

60

co:t:i.~

-40

40

o 0.1 0.2 0.3 0.4 0.5 0.6 0.7Time (seconds) after occurrence of fault

FIG. 28. Swing curves, Study 2, case 2. Future set-up. Generating plants atLancaster and Rutland. Interconnections closed. Three-phase fault on Philo132-kv. bus section Al cleared in 0.15 sec. by opening Howard line 1, Crooksville

line 1, generator 1, and bus-tie reactor 1. . Stable.

Case 3, representing immediate-future conditions, was set up withoutthe proposed generating stations at Lancaster and Rutland, but withthe existing stations south of Philo at Logan and Cabin Creek inoperation, and with interconnections to other companies closed.Loads were as shown in Fig. 29, substantially the same as for cases 1and 2. A three-phase fault was assumed to occur on section A2 of the132-kv. bus at Philo and to be cleared in 0.15 sec. by the opening ofHoward line 2, Crooksville line 2, Rutland line 2, generator 2, and bustie reactor 2. Note that the two groups of lines at Rutland are notbussed. The swing curves (Fig. 30) show that Cabin Creek and


FortWayneand West

40-jl0Crooksville

SouthPoint

lorain11.4 - j3.6

-........-- 100.5

.............. 106

t139 +j6.0

Logan andSouth

FIG. 29. Initial load conditions of case 3, Study 2. Flow of active and

....('t)e",

I('f).......

Po: 40.0 +jl1.0~ ~Io~

98.0Massilon~

24.0 + j19.6

BPhilo

STUDY 2

Akron, Alliance, andCanton condensers

s

40.0 +nos-e--

28.0 - ,56

Newcomerstown

Cabin Creek1040

G t190 - ;14.0

299

North east

....---.....-98.0

t$:sIt'

WindsorandWest ~......- 102

Penn

reactive power in Mw. + j Mvar. Bus voltages are given in per cent of 132 kv.


I<.$1~I ",~I ~01

Cabin Creek~v::?~~I ../gJl /

fI' /~I / / ..-Logan

"31 ~~

~I ~ / ..--........<,~ ~

& ~~III""""""

PhiloB~~~

"......,..- I ~

.......-LOga~-r PhiloAJ~~I

~71 I I I I "'lIiIr-,l.a-- I

Akron - Alliance, Canton condenser -.......-..I J • -,

~ I I J -,A Windsor",I .-'!~,~~

I ~ """.------- ~~ V~---r----.. /Lorain ../

~--- .... ~~ ~---...r----.. .......

~~

I -~ ............... Toled0-:L ~~

I ~ ~

!I -~..........

Fort Wayne and West~----.r---.

I ----r---~

l

60

Logan, which were sending power to Philo, speeded up and pulled outof step with Philo. The pull-out might be attributed partly to the lossof a Philo-Rutland-South Point-Turner circuit and partly to the initialangular displacement between Logan and Cabin Creek, on the one

80

-40

-60

40

-80o 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Time (seconds) afteroccurrence of fault

FIG. 30. Swing curves, Study 2, case 3. Immediate-future set-up. No generation at Lancaster or Rutland. Three-phase fault on Philo 132-kv. bus section A2cleared in 0.15 sec. by opening Howard line 2, Crooksville line 2, Rutland line 2,

generator 2, and bus-tie reactor 2. Unstable.

hand, and Philo, on the other hand, being larger than that betweenRutland and Philo in case 2. Philo A and B swung more nearly together than they did in cases 1 and 2, probably because Philo A wasmore lightly loaded than before (175' instead of 215 Mw.).

Conclusions. The following conclusions were reached:1. The proposed bus reactors at Philo were found to have little

effect on system stability, as evidenced by the fact that the generators

STUDY 3 301

in sections A and B of the plant swung approximately together, evenfor a fault on one bus section.

2. It was also found that, without the benefit of the added inertia ofthe Chicago and Cincinnati interconnections (although these connections are normally closed), the system under the assumed load conditions would lose synchronism between Twin Branch and Philo after athree-phase bus fault at Philo cleared in 9 cycles with the loss of onePhilo-Howard circuit. With the interconnections closed the resultsindicated that synchronism would probably be maintained under thisfault condition.

To increase the margin of stability under such conditions a programof modernizing the Philo line breakers was worked out (and has subsequently been completed), calling for a reduction from 8 to 5 cycles inthe breaker-operating time and from 9 to 6 cycles in over-all clearingtime of faults. Also, the system tie lines between Ohio and Indianahave been strengthened by the construction of a new 132-kv. line fromPortsmouth to Muncie, Indiana, providing a direct line from Turner tothe Indiana area.

3. Likewise, with a three-phase bus fault at Philo followed by loss ofa Philo-Rutland-Turner circuit, the system, under the heavy loadconditions assumed, showed a probable loss in synchronism betweenPhilo and the south. Here again, the subsequent improvementswhich have been mentioned were designed to eliminate this difficulty,and they have undoubtedly provided the necessary margin of stabilityto prevent loss of synchronism under these conditions.

4. The system engineers have also pointed out that the future installation of 3-cycle breakers to replace 5-cycle breakers at the mostimportant locations, together with more extensive application of highspeed reclosing, will provide still greater gains in system stability.

STUDY 3

Figure 31 represents a 132-kv. 60-cycle transmission network connecting a large steam generating 'station A (on the extreme right-handside of the diagram) to a large metropolitan receiving system, whichmay be considered an infinite bus and which is represented by system Jon the extreme left. Connected to the same 132-kv. network are twoother steam generating stations, E and I, and a number of substationswith loads.

The circuit breakers and relays on the network were too slow tomaintain synchronism between A and J through severe faults. Therefore, it was believed best to operate the system sectionalized by openbreakers marked with crosses in Fig. 31. Under this plan each of the

Subs

ta.G

CIo:) ~ t-3 ~ "'d

t-4 o :> t-4 00 ~ :> t:d t-4 t-4 t-4~ to< ~ c: t1 t-

4t:r

j00

Sta.

A

9.9+

J38.

J.O

"-4

5-j6

Mva

Subs

ta.D

18

.2+

j4J.

lO1

2.0

+jS

1.5

05.7

+j2

4.3

n

Subs

ta.F

124

kv

j35

.Hl

--1

4+

j10

Mva

.~

~/220.

12

-j1

01

Mv

a~~

129k

v~ 11

8-

j52

Mva

.

4O

-j2

SM

va.

3.3

+j1

2.S

0..

..2

3+

j24

Mva

Sys

tem

J

oC

ircui

tbre

aker

norm

ally

clos

ed.

1mC

ircui

tbre

aker

open

for

sect

iona

lized

oper

atio

n.

6.F

aultl

ocat

iOn

assu

med

inth

est

udy.

FIG

.31

.O

ne-l

ine

diag

ram

ofth

epo

wer

syst

emof

Stu

dy3,

show

ing

impe

danc

esin

ohm

sat

132

kv.,

init

ial

load

sin

meg

avol

t-am

pere

s,in

itia

lvo

ltage

sin

kilo

volts

,an

das

sum

edfa

ult

loca

tion

s.

STUDY 3 303

three generating units at station A fed toward the metropolitan systemJ over a separate circuit. A fault on anyone circuit and its subsequentclearing disconnected one of the units at A from the rest of the system.

The disadvantage of this mode of operation was that, after clearing afault in or near stations A, B, or C, the loads at A, B, C, or D (depending on the fault location) were fed from stations E, I, and J over suchlong lines that the voltages at the loads became too low. To take themost extreme case, the load at station A could not be supplied overthe long line from system J if a fault occurred in generator At.

The study here described was made to determine which breakerswould have to be speeded up and possibly also equipped with highspeed relaying to make station A stay in synchronism with the rest ofthe system for faults on the I32-kv. network if the network were to beoperated unsectionalized, that is, with the breakers marked by crossesclosed.

The breaker-opening times were about 20 cycles, except for some8-cycle breakers at stations A and E; and the relay times were from 1to 60 cycles, depending on the fault location. It was assumed in thestudy that any breakers requiring faster opening would be rebuilt toopen in 8 cycles. The line relaying would be changed where necessaryto high-speed distance relays having I-cycle operating time; and, ifnecessary for stability, carrier-current pilot equipment would be addedto give simultaneous tripping of the breakers at both ends of the line forfaults at any point of the line. It was further assumed that high-speedbus differential protection would be installed where necessary forstability.

It is common to use two-line-to-ground faults in stability studies,inasmuch as three-phase faults are very infrequent on high-voltagesteel-tower transmission lines. In this study, however, three-phasefaults were assumed, because they are not much more severe than twoline-to-ground faults in their effect on stability and because they aremuch simpler to represent. Only the positive-sequence network needbe considered for three-phase faults, whereas the zero- and negativesequence networks must also be considered for two-line-to-groundfaults.

Three-phase faults were assumed at various locations, the choice ofwhich will be discussed, and for each location swing curves were calculated to determine whether the system would be stable with the existingbreakers and relays and with faster breakers and relays.

To obtain the swing curves, the network of Fig. 31 (with all breakersclosed) was set up on an a-c. calculating board. Each line was represented by a nominal e, although the shunt capacitance at each end of


the line is not shown in Fig. 31. The three generators of station A wererepresented by a single equivalent machine on the assumption that,being connected to the same bus-although sectionalized by reactorsthey would swing nearly alike. The generators of station E were notrepresented, except in one case in which the fault location was near thisstation. For other fault locations it appeared that these generatorswere electricaliy close enough to the metropolitan system J and farenough from the fault so that they would not have much effect. Insuch cases the load of station E was set to represent the true load thereless generation there. At station I two groups of generators on separate busses were represented. All generators were represented bytransient reactance and voltage behind transient reactance. Atstation A the reactances of the three generators were represented byseparate impedance units connected at one end to different bus sections(separated by reactors) and connected at the other end to one powersource, the voltage of which represented voltage behind transientreactance of all three generators. Generator A1 had two windings,the self and mutual reactances of which were represented by a Ycircuit. In accordance with usual practice voltages behind transientreactance were kept constant during the study. Data on the inertia,speed, and kinetic energy of each actual and equivalent generator aregiven in Table 8. An interval of 0.1 sec. was used in the point-bypoint calculations.

The outputs of the various machines, the values of the loads, and thebus voltages were adjusted to typical values (shown on Fig. 31) asnearly as possible to values observed on the system with sectionalizedoperation at time of maximum system load. Loads were representedby constant impedances, which were not changed after the faultwas on.

The most severe fault location is ordinarily close to the sending end ofthe system. Therefore it is logical to try first a fault so located. Theworst fault location on the system of Fig. 31 would be on the low-voltage bus of station A. A three-phase fault here would entirely interruptany exchange of power between the generators of this station and therest of the system. This bus, however, was of isolated-phase construction; therefore the only type of fault which could occur here wasone-line-to-ground, which is not a severe fault from the standpoint ofstability. The closest point to the bus at which a three-phase faultcould occur was on the high-voltage side of one of the transformers(there was no high-voltage bus at A) or on the far side of a set ofreactors feeding a local load (fault location 2). As a matter of fact, amore severe fault location than either of these was on the bus of sub-

STUDY 3 305

TABLE 8

INERTIA OF GENERATORS (STUDY 3)

Rating WR2 SpeedStored

Generator (Mva.) (lb-ft.2) (r.p.m.)energy H 11M(Mj.)

At 100 859,000 1,800 640 6.40AS 50 475,000 1,800 365 7.29AS 50 475,000 1,800 365 7.29

A, total 200 1,809,000 1,370 6.85 7.89

Et 12.5 51,000 1,800 38 3.04E2 12.5 51,000 1,800 38 3.04ES 35 357,000 1,800 266 7.60E4 30 29,100 3,600 87 2.90

E, total 90 429 4.76 25.2

Ila 50 390,000 1,800 291 5.81Ilb 60 391,000 1,800 292 4.86

11, total 110 781,000 583 5.30 18.5

ISa 35 243,000 1,800 181 5.17ISb 100 750,000 1,800 560 5.60

IS, total 135 993,000 741 5.49 14.6

station B, where a three-phase fault would entirely interrupt the flowofpower between generators A and the other generators. It wasimmaterial, while the fault was on, whether it was on one or another bussection of substation B or whether it was just outside the substation onanyone of the seven lines. The worst location as regards conditionsafter the fault was cleared, however, was on the upper bus section(fault location 1), where clearing a fault would require the opening of abus-tie breaker and of three line breakers, whereas a fault in any of theother locations mentioned would require the opening of fewer lines.Consequently, the effect of a three-phase fault at location 1 on the busof substation B was examined first.

A three-phase fault at this location was known to put station A out ofstep if slow breakers were used. Therefore 8-cycle breakers and 1cycle bus differential relays were assumed, giving a clearing time of 9cycles. The swing curves are shown in Fig. 32. Clearly the systemwas stable, although without much margin. Twelve-cycle clearingwould probably be too slow.

7264564824


168

-~r---~

V~ A--

./ r-;1/

~ " r-,

~ V r-,

~V~

V ~ f-- Faultcleared completely 9 f"'v

I -- ~2\ "l~~- ...............

<, /",.

~,It- Faulton -'-..... ~ "".,..-- -"~

I--.t---.~~

~' ~ K----- '" ~I ......... ~

Ir-I""""""

I ¥ 'r-Joo 32 40

Time (cycles)

FIG. 32. Swing curves, Study 3, fault at location 1, cleared completely in 9 cycles.System stable.

306

100

80

~Q)

e~Q) 60-0

jij

EuQ)

40"ii)

Q)

1i".oe<

20

-- A. ~ -1 -:~ r-,I ~

V "'~

~VI

I

~1/ I~ Fault cleared

~~ 15~12 --

I ~~

I ~~~-11 ~

-.................. ~1,,7

k-- Fault on-~J~I I

100

80

20

oo 8 16 24 32

Time (cycles)40 48

FIG. 33. Swing curves, Study 3, fault at location 2, cleared in 15 cycles. Systemstable.

STUDY 3 307

J

24

12

11

o~...._ ........._~--......o 8 16

Time (cycles)FIG. 34. Swing curves, Study 3,fault at location 3, cleared at bothends in 9 cycles. System stable.

It could be inferred, without taking more swing curves, that thesystem would likewisebe stable for a fault on either of the other bussections at substation B cleared in 9 cycles, but would be unstable withthe slow breakers. Therefore all the breakers at B should be speededup, and each bus section should be equipped with high-speed protection.

It could be inferred further that the system would be stable for afault close to substation B on anyone of the seven lines cleared in 9cycles at both ends. Whether sequential clearing would give stabilityis not yet clear, but this point will be 60__-...--r--~~-....

considered later. -Attention was next turned to fault ~

location 2, separated by a reactor ~ 40 I---+---+~from the low-voltage bus of station ~

oA. Fifteen-cycle clearing time was :aassumed, corresponding closely to ! 20~~~..........-t-e~~the existing relay and breaker times ~

of 6 and 8 cycles, respectively. The ~

swing curves are given in Fig. 33.Again, the system was stable.

Fault location 3, just outside station A on one of the lines to substation B, was considered next.Nine-cycle clearing was assumed atboth ends. The swing curves, shown in Fig. 34, indicate that the system was stable by a big margin. Sequential clearing will be considered later.

The next fault was taken at location 4 near station E on one of thelines to substation C. The machine at E was represented by a powersource on the supposition that this machine would be a significantfactor in the stability of the system. Sequential clearing was assumed,9 cycles (1 cycle relay time plus 8 cyclesbreaker time) at E, where thebreaker already was an 8-cycleone, and 80 cycles (60 cycles relay timeplus 20 cycles breaker time) at C. The swing curves are given in Fig.35. The system was stable and would be so even if the line were notopened at C. There seems not to be much margin, however, in theclearing time at E. For this fault location the existing breakers weresatisfactory.

When the fault was taken at the other end of the same line (location5), however, 21-cycle clearing at C (1 cycle relay time plus 20 cyclesbreaker time), followed by 68-cycle clearing at E (60 cycles relay timeplus 8 cycles breaker time) was not fast enough for stability. The

120 r-----r----,...--~-__r__-_,_--.,...__-~--

96842412O'-......I.-I---.l----......-~----- ....- ...........-~o

100

- 80&/)Q)

~aD Faultcleared at CQ)

"tJ

co.g 601JQ)

'iiQ)

boe 40-e

20

J

36 48 60Time (cycles)

FIG. 35. Swing curves, Study 3, fault at location 4, cleared at station E in 9cycles and at substation C in 80 cycles. System stable.

7264564824168

/r I

!/ !

~II

I j~ :I

I V II I

VI

II

J I II I

~~ V-I- Fault cleared Fault cleared --Jat station C at station E--.I

21'"'-' 68'"'-'I

./~ I I

I I~~ I I

---- I I

~ - Fault on-~~ ~;

~ I

I~~

~

~I

I ./ I.........._1/ ~ll ;- ......... ........... I

.......... J II II

J ____Io

o 32 40Time (cycles)

FIG. 36. Swing curves, Study 3, fault at location 5, cleared at substation C in21 cycles and at station E in 68 cycles. System unstable.

308

140

120

100-enQ)

~bOQ) 80"'0

tVu

~"ii 60Q)

co~

40

20

STUDY 3 309

swing curves of Fig. 36 show that station A went out of step with therest of the system. Therefore the breaker at C must be speeded up.Nine-cycle clearing at both ends would certainly give stable operation,because 9-cycle clearing sufficed in the more severe fault locations 1, 2,and 3. Sequential clearing (not tried) probably would be satisfactory.

Attention was now turned to the circuit from B to D, and a threephase fault was taken at location 6 near D. Generator 12 was not

140---_-__-...-----.~...,.___,...._-~__r____r__=__.,....__r____r_r_...__......

1I~~>--FaU" cleared in 27'"

rI!;

II

;

I

T I I I I IFault cleared at B'-.. --L.--

68"" ~II

II

1/.1/

/~- --~~ ~Il-r---'----~~..........~-_ ...... I'--Fault cleared in 68/V

20 '--

120

100-(/)Q,)

ebOQ,) 80"0(iju

'':;:t)Q,)

Q) 60-Q,)

~c<

40

72:

645648

I24168

O'--...I....._~~......~-..~ __..._"___'_....__....._........

o 32 40Time (cycles)

FIG. 37. Swing curves, Study 3, fault at location 6, cleared at substation B in27 cycles, system stable; or in 68 cycles, system unstable.

represented for this case, because, with a fault at 6 either on or cleared,12 could exchange no power with station A and hence would have noeffect on the swinging of A with respect to the infinite bus J. Forsimilar reasons the clearing time at substation D was immaterial to thestability of A. The clearing time at substation B was first assumed as68 cycles (60 cycles existing relay time plus 8 cyclesnew breaker time) ;generator A pulled out of step as shown by the swing curves in brokenlines in Fig. 37. The clearing time was then reduced to 27 cycles,resulting in stability without much margin, as shown by the swingcurves in solid lines in Fig. 37. The required clearing time of 27 cyclesor less at B was too short to permit selective operation between the


breaker at B and the 20-cyclebreaker to the left of D for a fault just tothe left of D. Either the breaker to the left of D must be speeded up,or else carrier-current relaying must be installed on the line from Bto D. With an 8-cycle breaker and J-cycle relay at D, and a margin of20 cycles (required by good practice), the clearing time at B would be1 + 8 + 20 + 8 = 37 cycles, which was still too slow. Hence carriercurrent relaying was needed, and by its use the clearing time at substation B could be reduced to 9 cycles. As already stated, the clearingtime at D was not important; therefore the breaker at D need not bechanged.

Consider next fault location 7 near substation C on the line from Bto C. Before the fault was cleared, conditions were the same as they

E

c

B A

• 8 - cycle breakers needed.- - - Carrier - current relaying needed.

FIG. 38. Location of 8-cycle breakers and carrier-current relaying necessary tomaintain stability with interconnected operation under the conditions of Study 3.

were for the fault at location 5, for which 21-cycleclearing was too slow.Hence the breaker at Cmust be speeded up to 8 cycles,giving a clearingtime at C of 9 cycles. After the breaker at C was opened, conditionswere similar to what they were for the fault at location 6, except thatthe shock during the first 9 cycles was greater for location 7 than for 6,as the reactance between the fault and the bus of substation B wasslightly less for location 7 than for location 6. Therefore the criticalclearing time at B for a fault at 7 must be considerably less than the 27cycles found for location 6. A clearing time of 68 cycles (existing relaytime of 60 cycles plus new breaker time of 8 cycles) was far too slow,and even 37 cycles (second-zone impedance relay time of 29 cycles plusbreaker time of 8 cycles) was too slow, Carrier-current relaying wastherefore needed on each of the two lines between Band C.

For similar reasons carrier-current relaying should be used on each

STUDY 4-DESCRIPTION OF SYSTEMS 311

of the four lines between A and B. Sequential clearing of end-zonefaults appeared to be permissible, however, on the two lines betweenC and E. Eight-cycle bus-tie breakers and high-speed bus relayingshould be used at stations C and E, as well as at B.

Faults were not taken any farther from station A than stations Dand E, because it was known that station A would stay in step, evenwith the existing breakers and relays.

The conclusions of the study are set forth in Fig. 38, which shows' thelocations where 8-cycle breakers were required and the lines to whichcarrier-current relay channels should be added.

STUDY 4§

Three power companies (herein called companies A, B, and C),operating in neighboring states but having no physical connectionswith one another, were considering the construction of interconnectingtransmission lines to permit the interchange of power between them ineither direction. Study 4 was made to determine (a) the transmissioncapacity or power limit of the proposed interconnecting lines, (b) therequired capacity of line terminal equipment, such as transformers andshunt reactors or synchronous condensers, and (c) the modificationsnecessary within the three systems to permit utilization of the fullpower capacity of the interconnecting lines.

Description of systems. The systems involved in the study areportrayed by the impedance diagrams (Figs. 39 and 40), and they arealso described in the following paragraphs. Stations of these systemsare designated by two-letter combinations, the first letter of whichdenotes the company owning the station.

Company A operated a metropolitan power system having a steamelectric generating station of 112-Mva. capacity (station AA) whichfed a number of substations through a l3.8-kv. transmission network(shown in upper right-hand corner of Fig. 40). Station AA was con-nected through two 66-kv. lines to substation AB, from which connections were made to companies D and E and-by one of the proposedlines-to company B.

Company B had a system consisting of two distinct parts. (SeeFig. 39.) One part, shown on the left-hand side of the diagram, hadtwo steam-electric plants (station BF, 25 Mva., and station BG, 32.5Mva.) a few miles apart. These plants served the area in which they

§Information concerning this study was obtained through the courtesy ofEbasco Services, Inc., New York City, with the concurrence of the operating com...panies concerned.

312

DC


-- Usbnllines.- - Proposed lines.

0- Synchronous condenser,

0- Ste.m·· electric I.ner.ti",~st.tlon.

rji\.- Hydroelectnc I.ner.tin,v:..;- 'tatlOn.

0- [qulv.le~t lenerator~ representln,.power

system.

LOid.

:I: :'~r~~~=.:n=...+ Tr.ns'ormer.

lOMv...

FIG. 39. Study 4, part 1. Impedance diagram showing impedances of transmission lines and transformers in per cent on 4o-Mva. base. Values of capacitivesusceptances of lines are in per cent on the same base. Proposed interconnections

are shown in broken lines.

STUDY 4-DESCRIPTION OF SYSTEMS 313

System CequlVlltnt

8j:t

1210

AIC

1.!9AG AD

.... ~,.,; +;-~

~ AC

C? ~~ IE~

AJIt'

2 NN

0;~ ;-

0.;- ~

~ ~cd

:;-s AA::

13.8

+all ~~ CA 1l0kv N

H (I:U~:~)

DA 138 ltv.

~ DA 110 kv7~

~",N,.,;

2 .;-aMva. +' ~

C it N ~'i. fIi 3.28 ~III DB HOb -cd DC 4.15ltv. +

N ~In ltl

"! 3.28 ::~ II':;- IQ

IQDC 110 kv. I ;-

iii~N

SA 601tv

20 Mva. ~.,+In'It...

C?

~ 40 Mva.

DE 110 kv H

RD F69/'Ii

:;- kv.,...~

25G 9.92 DF 110kv

Mvt.

FIG. 46. Study 4, part 2. Impedance diagram with values in per cent on 25-Mva.base. Line capacitance values are in microfarads on calculating board.


were situated and fed a number of smaller towns through a 60-kv.transmission system. The other part, shown on the right-hand side,had one steam-electric plant (station BD, 50 Mva.) feeding a number ofsubstations through 66-kv. radial transmission lines. These two partswere joined through a two-circuit 132-kv. line about 100 miles longfrom substation BC to station BD.

The system of company B was: interconnected at substation BAto the system of company F and at station BD to the system ofcompany G.

Company C, having an extensive system with both steam and hydroelectric plants, was interconnected with several other large systems inadjacent states. It is represented in Figs. 39 and 40 as an equivalentgenerator on the bus of substation CA.

Company D operated asystem (shown in Fig. 40) which includedseveral hydroelectric stations (DA, DE, DF, DG, and DH) of capacities ranging from 8 to 40 Mva. and a I IO-kv. transmission systemconnecting these generating stations to one another, to substations,and to companies A and E. This system extended several hundredmiles from the city served by company A.

Company E had a 28-Mva. steam-electric plant located in a cityabout 60 miles distant from company A.

Company F had a large system; with both steam and hydroelectricplants. Of these only station FA, the plant nearest system C, wasrepresented individually in the study, the remainder of the systembeing represented by a single equivalent generator (shown in lower lefthand corner of Fig. 39).

Company G had steam plants at GB and GD and' a hydroelectricplant at GC. It was interconnected with company B at station BD(lower right-hand corner of Fig. 39).

Company H operated a metropolitan system, situated about half waybetween companies A and B but not connected to them. Its generating capacity was approximately 275 Mva. (See Fig. 40.)

A list of synchronous machines of companies A, B, D, E, and G willbe found in Table 9. As the systems of companies C, F, and H werenot represented in detail, their machines are not listed.

Proposed interconnecting lines. Two lines were proposed, the firstconnecting company A to company B, the second connecting companyB to company C. Both are shown by dashed lines in Fig. 39. For thefirst line two alternative routes were studied, one from substation ABto station BD, the other from substation AB to substation Be. Thelengths of these routes are about 280 and 244 miles, respectively. Thesecond line was to have terminals at BD and CA, and a length of about256 miles.

STUDY 4-PROPOSED INTERCONNECTING LINES 315

TABLE 9

LIST OF SYNCHRONOUS MACHINES (STUDY 4)

Number Rating Total Kinetic,WR2

Station of Like Kind of Xd EnergyUnits Machine (%) (kilo-lb- (Mj.)

kv. kva. r.p.m. ft. 2)

AA 1 Steam 13.2 18,750 1,800 28.0 140 104.8" 1 H " 25,000 H 24.0 207 154.0" 1 u " 25,000 H " 237 176.4u 1 " 13.8 31,250 " 21.0 237 176.4u 1 H u 12,500 3,600 11.0 15.5 46.4u Total 112,500 658.0

AJ 1 Steam 13.8 6,250 3,600 14.0 10.2 29.9

BA 1 Condo 4.16 3,500 720 29.0 38.9 4.7

BC 1 Condo 12.47 3,750 720 29.1 38.9 4.7u 1 H

" 3,750 " 32.0 106.3 12.7Ie Total 7,500 17.4

BD 1 Steam 12.0 18,750 1,800 18.0 102 76.4" 1 " " 31,250 " 22.0 235 176--Ie Total 50,000 252.4

BF 1 Steam 12.5 25,000 3,600 11.0 62* 186

BO 1 Steam 4.16 7,500 3,600 15.8 19* 56.8" 2 Ie " 12,500 1,800 19.0 85.3* 64--" Total 32,500 184.8

DA 3 Hydro 13.8 14,000 150 36.0 7,500 38.8H Total 42,000 116.4

DC 1 Condo 4.15 8,000 900 34.9 40.5 7.6

DE 2 Hydro 6.9 10,000 180 35.0 4,650 34.8a Total

-20,000 69.6

DFl 2 Hydro 6.9 10,000 180 35.0 4,650 34.8DF2 1 " u 20,000 164 42.0 9,870 60.8DF Total 40,000 130.4

DG 3 Hydro 6.9 2,750 112.5 39.0 1,250 3.6

" "-

Total 8,250 10.8

*Estimated.


TABLE 9 (Continued)

Number Rating Total KineticKind of

,WR2

Station of Like Zd EnergyUnits Machine (%) (kilo-Ib- (Mj.)

kv. kva. r.p.m. ft. 2)

DH 2 Hydro 13.8 14,500 257 32.0 1,950 29.6II Total

--29,000 59.2

DI 1 Condo 13.8 15,000 900 35.7 93 17.4

EA 2 Steam 4.15 8,575 3,600 12.0 19.6 58.4II 1 " " 6,250 II 14.0 12.4 37.2II 1 " u 5,000 " 16.0 10.0 29.9

"--

Total 28,400 183.9

EA 1 Condo 4.15 7,500 900 35.1 38.0 7.1

FA SteamII Total 162.4 121

GB 1 Steam 31,250 1,800 19.0 414II 1 Freq.chr. 33,000 300 36.0 6,200II Total 64,250

GO 4 Hydro 5,000 164 32.0 1,010II Total 20,000

GD 1 Steam 15,625 23.0

Both interconnecting lines were to be of 154-kv. single-circuitconstruction, with hollow 250,OOO-circular-mil copper conductors ofO.766-inch diameter, supported by wooden H frames. The equivalentspacing between phase conductors would be 17 ft. Ground wireswould be installed to protect the line against lightning.

The necessary terminal equipment would include transformers, shuntreactors or synchronous condensers, circuit breakers, and protectiverelays.

The transformers at substation AB would have three windings: highvoltage, 154 kv. Y; low voltage, 66 kv. Y; and tertiary, 13.8 kv. ~.

Those at the other end of the A-to-B line (at either BC or BD, depending upon the route selected) would be 154-to-132 kv. Y...connectedautotransformers with 12.5-kv. d-connected tertiary windings. Forthe B-to-C line the transformers at station BD would be autotransformers similar to those on the A-to-B line, and, even if the terminals

STUDY 4-SCOPE OF THE STUDY 317

of both lines were to be at station BD, it was assumed that separatetransformers would be used for each line. At substation CA 154-to110-kv. autotransformers with tertiary windings would be required.The tertiary windings of these transformer banks would be used forconnection of shunt reactors or synchronous condensers to supply thereactive power needed to hold the voltages of the terminal busses atsuitable values. They would also provide a path for zero-sequencecurrent.

To attain reliable transmission of power over the proposed singlecircuit interconnections, without the aid of any parallel ties betweenthe systems, it would be necessary to use high-speed clearing of faultson these lines and high-speed reclosure of terminal breakers. Obtainable circuit breakers had an opening time of 5 cycles and a reclosingtime of 20 cycles with three-pole operation. The assumption ofI-cycle carrier relaying gave an assumed fault-clearing time of 6 cyclesand a reclosing time of 21 cycles, measured from the instant of occurrence of the fault.

Scope of the study. The study included power-flow and transientstability studies. The power-flow studies served to determine: (1)approximate steady-state power limits of the interconnections, (2) thereactive power which had to be supplied by reactors or synchronouscondensers at the terminals of the proposed lines and elsewhere, (3) therequired capacity of the terminal transformers, (4) the need for additional generating, transformer, or transmission-line capacity within theseveral systems, and (5) initial conditions for the transient-stabilitystudies. Transient-stability studies served to determine: (1) thetransient power limits of the interconnections for faults on the interconnecting lines, and (2) the required clearing times of faults on otherlines to prevent impairing the power capacity so determined.

The study was divided into two parts, made at different times andwith different a-c. calculating boards. Part 1 dealt with the powercapacity of the A-to-B and B-to-C interconnections for power flow ineach direction, also with the choice between BC and BD as southernterminal of the A..to-B line; it dealt also with the effect on stability ofinterconnecting systems F and G to system B, and with various internalproblems of systems Band C arising from the proposed interconnections. For brevity those portions of the study pertaining primarily tosystem C and to the B-to-C line are omitted from the present report.Part 2 was concerned with internal problems of system A, including therequired speed of clearing of faults on lines within that system, andalso with the effect on stability of connecting system H to a tap on theA-to-B line.


Loads. Estimated July and December peak loads three years afterdate of study were used. The July loads have practically the sameactive-power component as the December loads but have higherreactive-power components.

Method of determining power limits. Approximate steady-statemomentary power limits of the interconnections were found by increasing the electrical angle between bus voltages at major generatingstations of the different power systems to approximately 45°. Thepractical steady-state power capacity was taken as 5 Mw. below thevalue so found, to allow for momentary fluctuations of about 5 Mw.above the average power carried by the line. Such fluctuations couldbe caused by sudden changes in loads or by faults at distant points.

Transient-stability power limits were found principally by obtainingswing curves by the point-by-point method in order to determinewhether the system would be stable or unstable with various values ofinitial power flow if a two-wire-to-ground fault should occur on theinterconnecting line, followed by 6-cycle clearing and 21-cycle reclosure.The effects of faults on other lines were investigated in similar fashion,except that three-phase faults were assumed on overhead lines withoutground "vires, on double-circuit lines, and on cables, and that variousclearing times were assumed. The number of machines representedin these transient-stability runs varied from four to ten.

In two instances the number of machines was reduced to two,power-angle curves were obtained by use of a calculating board, andcurves of power limit versus switching time were then computed.

The rated power capacity of the interconnecting lines was taken as5 Mw. lower than the transient power limit, to allow for fluctuations.

Simplification of systems. In representing the systems only majorstations, substations, and transmission tie lines were included. Radiallines were represented as loads on the busses from which they radiated.For example, the entire 66-kv. system supplied from station BD wasrepresented as a load on the BD 66-kv. bus.

Lines were represented by nominal sr circuits, the capacitances of alllines on a given bus being represented by one capacitor on that bus.Proposed interconnecting lines were represented by several sections,depending upon location of proposed taps to other systems.

Three-circuit transformers were represented by their equivalent Ycircuits. If one branch of such a circuit had negative reactance (asindicated, for example, in Fig. 39 on the 66-kv. side of the transformerat substation AB), the following change was made in setting up thecircuit on the calculating board: The branch with negative reactancewas set up with zero reactance, and its negative reactance was com-

STUDY 4-SIMPLIFICATION OF SYSTEMS 319

bined with the greater positive reactance of the opposite branch, thuspreserving the correct value of through reactance from primary tosecondary terminals.

TABLE 10

GROUPING OF SYNCHRONOUS MACHINES FOR REPRESENTATION BY POWER SOURCES

ON THE CALCULATING BOARD (STUDY 4)

Power Source Numbers

Station Run Run Run Run Run Run Run Runs Run1 2 3 4 5 6 7 8 to 11 l'----------------

DE, DF, DH 1 1 1 1 1 1------ I 1 1 ----

DA,DG 2 2 2 2 2 2----------------

EA 3 3 3 3 3 3------ ----

AJ 4 42 2 2 ----

AA (north bus) 4 4 4 4 5 5----

AA (south bus) 6 6----------------BD 5 5 5 5 3 3 3 7 7----------------

BF 6 6 Off Off-- 6 ---- 4 4 8 8

BG 7 7 6 4----------------

System C 00* 00 00 00 00 00 00 00 00

--------------I-

GB 8 7 5---- 5 --

GC,GD 9 8 6---- -- --

FA Off 10 9 Off 6 Off 7 Off Off---- -- --

Rest of system F 00 00 00 00

---- -- -- --System H Off Off Off Off 9

* 00 = Infinite bus.

Synchronous condensers were not represented dynamically duringstability studies, but merely by capacitors or reactors.

System C was represented by an equivalent generator on the substation CA bus. System F, exclusiveof station FA, which was shownindividually, was represented by an equivalent generator. In stabilitystudies the equivalent generators at CA and FA were regarded asinfinite. Station DG was combined with station DA. Stations GCand GD were represented by an equivalent generator on the sub-


station GA bus. In part 1 of the study the portion of system D beyondsubstation DD was represented as an equivalent generator on the DDbus, and station AJ was combined with station AA. In part 2 theportion of system B near stations BF and BG (including transformers,lines, and loads, but not generators) was represented by an equivalentcircuit (shown in Fig. 40), the terminals of which were: station BF12.5-kv. bus, station BG 4.16-kv. bus, substation Be 132-kv. bus, andsubstation BA 60-kv. bus. Since generators were not included in thisnetwork, it was possible to change the generator reactances accordingto the number of machines assumed in operation at each station andaccording to whether adjusted synchronous reactances were wanted forsteady-state runs or transient reactances for transient-stability runs.

In many of the transient-stability runs further combinations ofgenerators were made, as shown in Table 10; for example, stations BGand BF were combined in some runs.

Swing curves, part 1. Twelve transient-stability runs were madein the part of the study considered. The swing curves are reproducedin Figs. 41 to 52, inclusive, and the conditions under which they weretaken and the resulting conclusions concerning the stability or instability of each case are summarized in Table 11. The grouping ofgenerators for each stability run is shown in Table 10. Each run willnow be discussed.

Run 1. The company B terminal of the A-to-B line was assumed tobe at substation BC. As a step in finding the power limit of this linefor northward power flow, the power sources on the board were adjusted to give 39.6 Mw. of power received at substation AB from Be.To supply this power, 13 Mw. was transmitted to system B from system C, and a new generating unit was assumed installed and operatingat station BF, resulting in a need also of 20 Mva. of additional transformer capacity at BF. A two-line-to-ground fault was assumed tooccur on the 154-kv. A-to-B line near the sending end. The assumedclearing and reclosing times were 6 and 21 cycles, respectively. Thesending end of any line is ordinarily the worst fault location on the linefrom the standpoint of stability; but" since acceleration of the generators during the 6 cyclesin which the fault is on is small compared withthat during the following 15 cycles In which the line is open and thesystems are completely separated, a fault of given type anywhere onthe line would have substantially the same effect on stability as a faultat the sending end. Inspection of the swing curves (Fig. 41) showsthat the machines of system B swing approximately together and thatthey speed up. The speeding up is due to their dropping load duringthe disturbance. The machines of systems A, D, and E also swing

TA

BL

E11

SU

MM

AR

YO

FST

AB

ILIT

YR

UN

S(S

TU

DY

4)

Pow

erT

rans

fer

Ty

pe

Loc

atio

nof

Fau

ltS

wit

chin

gd/

lo"

Tim

e(c

ycle

s)S

tab

leor

Ru

nA

.of

,"

Rem

arks

r-a

Lin

eN

ear

On

AU

nst

able

Mw

.R

ec'd

at

Fro

mF

ault

r"

Vol

tage

Sta

tio

nL

ine

toC

lear

ing

Rec

losi

ngtr

:1

39

.6A

BB

C2L

G15

4kv

.B

CA

B6

21S

tab

le~ d

13.0

BD

CA

tj

245

.6A

BB

e"

""

u"

"U

nst

able

*Jo<

21.4

BD

CA

~ ,3

40.0

AB

BC

3q,-3

q,13

2kv

.B

DB

e9

30S

tab

le*

in

58

.0B

CB

Db

ut

~ ~4

0.4

BD

CA

crit

ical

Z 04

35

.4B

eA

B2L

G15

4kv

.A

BB

e6

21U

nst

able

a5

""

""

"u

""

IIU

nst

able

•~

630

.0A

BB

DII

uB

DA

B"

"S

tab

le:;:0

74

0.6

AB

BD

""

""

""

Sta

ble

•;3

12.4

BD

CA

..til

826

.0B

DA

B3q

,13

.8kv

.A

A(N

)A

F12

Sta

ble

t-d

·.>-

9"

II"

""

""

18U

nsta

ble

~..

10"

"u

""

AA

(S)

t18

Sta

ble

·......

112

5.5

"u

""

AA

(N)

AF

18U

nsta

ble

AA

bu

ssp

lit

·.12

45.0

u"

2LG

154

kv.

AB

BD

621

Uns

tabl

e~

*Sys

tem

sF

and

Gco

nnec

ted

tosy

stem

Bw

ith

0+

}Opo

wer

inte

rcha

nge.

tFau

lt1,

000

ft.

from

stat

ion

AA

on70

0,00

0ei

r.-m

ilca

ble.

tsy

stem

Hco

nnec

ted

wit

h0

+jO

pow

erin

terc

hang

e.

~ to-


1.00.80.2

- 120 '-----a_.....&._~-....._'__ ~........ ....

o

I"Of~I~

,....40 ~l~en

~I\Ot... ;1.5~Q) L.l.

t"0- IQ)

~ Ie 0taQ)~ta

%!0:>

tic -40....!.5..cQ)

·Ciicl!'

to- -80

together, but they slow down because they take on additional loadduring the disturbance to make up in their systems for the power whichwas received from system B. System C was represented as an infinitebus and, therefore, did not swing. Station BD, which was closer tosystem C than were stations BF and BG, swung less than did stationsBF and BG, as though it were attracted by system C. The curvesclearly show that the system was stable.

80 r-------,----.-----r---r-~-...._-_r_____,r____,..-__T'"-.....,.

1 (DE, DF. DH)


FIG. 41. Swing curves, Study 4, part 1, stability run 1. 39.6 Mw. received atsubstation AB from substation Be, and 13.0 Mw. received at station BD fromsubstation CA. Two-line-to-ground fault-near substation BC on proposed 154-kv.

line to substation AB, cleared in 6 cycles, line reclosed in 21 cycles. Stable.

Run S. The power received at substation AB was increased to 45.6Mw. To supply this increased power, the power sent to system B fromsystem C was increased to 21.4 Mw., while the outputs of stations BD,BF, and BG were left unchanged. Systems F and G were connected tosystem B with zero interchange of active and reactive power. Withthe same fault location and with switching times the same as those inthe preceding run, the system proved unstable. Therefore the powerlimit oj the A-to-B line for northward r'power flow is between J,.O and J,.5Mw. received. The rated power capacity of the line may be taken as35 Mw, Study of the swing curves (Fig. 42) shows that the machines

STUDY 4-SWING CURVES, PART 1 323

of systems Band G and of station FA swung ahead together. Therest of system F was represented as one infinite bus, and system C wasrepresented as another infinite bus. The generators of system B didnot swing as far ahead of system C in this run as they did in the preceding run, in spite of the increase in power transfer. This fact shows

80 ..--.....----r---,...__--r--_,._-_- - __-_-__

40

-enG)G)...r 0"0

G)

00eco System F as infinite busWo.19 -40'0>n;e....!.5't: -80G)'Cijce~

-120

- 160 '"'---..---'--...L-....-"'---..L_--L.._-'-_..L-_I-....--J._...L.........J

o 0.2 0.4 0.6 0.8 1.0 1.2Time (seconds)

FIG. 42. Swing curves, Study 4, part 1, stability run 2. 45.6 Mw. received atsubstation AB from substation BO, and 21.4 Mw. received at station BD fromsubstation CA. Systems F and G connected to system C with 0 + jO powerinterchange. Two-line-to-ground fault near substation Be on proposed 154-kv.line to substation AB, cleared in 6 cycles, line reclosed in 21 cycles. Unstable.

that the connection of systems F and G to system B has a beneficialeffect on stability. Nevertheless, the generators of systems A, D, andE, again swinging together, swung farther behind than previously andpulled out of step with the other generators. Hence the power limit,either with or without the connection of systems F and G to system B, isbetween 40 and 45 Mw., and it may be concluded that the connectionof those systems has only a small effect.

Run 3. One consequenceof selecting substation Be as the company


B terminal of the A-to-B line, while station BD is the terminal of theB-to-C line, is that, when it is desired to transfer power from companyC to company A, this power must flow over the double-circuit 132-kv.

120__......__---t"---.--......--_-__-....--__- __-~-_

~I=80 j.~

~I~;1.5'-

I- 40if-a-i 0...

f System F....coc -4011~5..c.I!encI!to- -80

- 120 t---+-..--+---+---f---of---+----f~lE_FI~~-

- 160 ..........-"'" """"""-_......._ .......-...l"----""_-'"'-_....._.-..-_.a..-.....o 0.2 0.4 0.6 0.8 1.0 1.2

Time (seconds)

FIG. 43. Swing curves, Study 4, part 1, stability run 3. 40.0 Mw. received atsubstation AB from substation BC, 58.0 Mw. receivedat substation BC from stationBD, and 40.4 Mw. received at station BD from substation CA. Systems F andG connected to system C with 0 + jO power interchange. Double-circuit threephase fault near station BD on the 132-kv. lines to substation Be, cleared in 9

cycles, lines reclosedin 30 cycles. Stable but critical.

line from BD to BC, which at the same time may be required to carrypower from station BD to loads in the left-hand portion of system B.To test stability under such conditions, run 3 was made. Generationwas adjusted to send 40.0 Mw. (slightly below the power limit foundabove) to AB from BC and 40.4 Mw. to BD from CA. At the same


time station BD was generating more than was required by the loadin the adjacent part of System B, and the surplus was transmitted tothe distant part, resulting in a total power of 58.0 Mw. received at substation BC over the 132-kv. lines from BD. No new generators wereassumed to be running. Systems F and G were connected to system Bwith zero interchange. A three-phase fault involving both circuits]of the 132-kv. line to Be was assumed to occur near BD and to becleared in 9 cyclesby simultaneous opening of the breakers at both endsof both circuits, followed by their reclosure in 30 cycles. The swingcurves (Fig. 43) show that the system was stable ·but critical. Themachines of systems A, D, and E, again swinging together, sloweddownand almost, but not quite, pulled out of step with the other machines.Stations BD, GB, GC, and GD (which are on the sending side of theopened line) swung ahead and then oscillated; of these stations, BD,which is nearest the fault, swung ahead most rapidly. Stations BFand BG slowed down at first but speeded up after reclosure of thefaulted lines. The power limit wasslightly over 58 Mw. on the BC-BDlines or slightly over 40 Mw, on the AB-BC line. The latter figure isclose to the limit already found for a fault on the AB-BC line and istherefore satisfactory. If the fault should involve only one circuit,the power limit would be higher.

Run 4. It was next decided to investigate the power limit of theA-to-B line when carrying power southward (from A to B). Thepower sources were adjusted to give 35.4 Mw. received at Be from AB.To supply this amount of power, the output of station AA generatorswas increased to 22 Mw. above the rated capacity of the station,requiring that a new generating unit be assumed installed and operating there; and, in addition, 30.5 Mw. was received at substation ABfrom system D. To absorb this amount of power in system B, stationBF was shut down. The B-to-C line was closed with zero active andreactive power interchange as measured at station BD 132-kv. bus.Systems F and G were not connected. Additional transformer capacity was required as follows: 13 Mva. at station AA, 5 Mva, at substation AB, and 12 Mva. at substation Be. A two-line-to-groundfault was assumed to occur near AB on the proposed 154-kv. A-to-Bline, with the usual switching times. The swing curves (Fig. 44) showthe system to be unstable. Systems A, D, E and system B, althoughinitially swinging apart, came back together and remained in synchronism, but both these groups pulled out of step with system C.

[Such a fault might consist either of a three-phase fault on each circuit or of 8,

one-line-to-ground fault on one circuit and a two-line-to-ground fault on the remaining phases of the other circuit.


System C

as infinite bus

en~I

~.f-----+----I\0

.sle';1u...., ...---+---1

~I

80 I----J---+-~~--+--+---+-----:~-J#_---+---r--T----t---t

-;; 240er"'0-CD

J 200 1---1---4---1ftS

fCiE 160 1----I---+-~~-!....-4--~-+--+_-+_-..---+__#_-+---+-----t~CCD'iiic~ 120 1--,..,~~---+---JC.-+---+--4----+--+_-+_-+-i__l_-+---+---t---i

Run 5. Conditions in this run were identical to those in the last ronexcept that systems F and G were connected to system B with zerointerchange. The combined system was again unstable (see Fig. 45),but this time systems Band C stayed in step with each other, whereas

320 ---------....--r------r----r---y---.--.---r--,

40 '--......_.-L._-A--'-...&.-_...-_.l..--.lIo--..Io_--'-_~_.....---'

o 0.2 0.4 0.6 0.8 1.0 1.2Time (seconds)

FIG. 44. Swing curves, Study 4, part 1, stability run 4. 35.4 Mw. received atsubstation BC from substation AB. Two-line-to-ground fault near substation ABon proposed 154-kv. line to substation BC, cleared in 6 cycles, line reclosed in 21

cycles. Unstable,

systems A, D, E pulled ahead and out of step with Band C. Themotion of the A, D, E machines was almost exactly the same in bothruns, but the motion of the system B machines was greatly reduced bythe connection of systems F and G. The power limit of the A-to-Bline for southward power flow was less than 35 Mw. whether systemsF and G were connected or not, and was estimated as being between


I i VII "2' /I Cl)1 1(systemDl')VI -8r~cu_ ....I

~I~ V V~, /~~I

ar.... /~I ~t~ V i/~~I ~ ,,- ./

I- ~I ~'I

V~ Vml I

f- "til I"31 I V 14"(AA. AJ, EA)~, Y ~

I

V~I VI I /I

LVI V

Y; V I

5 (~~stem G~1--' I .~

rr I(3(B~~~~

I ~ 6(FA)I ~""7

System C as infinitebus yo~ ~V"" .

- -.L .L~ V----~ ~~ V _~I I~,,-I V System F as Infinite busI 4 (BG).""",- r---""'"

I I !I

240

280

80

-I"0

~ 200fco8L~

~ 160ca

j1:QI

.~ 120e...

40o 0.2 0.4 0.6 0.8 1.0 1.2

Titne (seconds)

FIG. 45. Swingcurves, Study 4,part 1, stability run 5. Same as run 4 except thatsystems F and G were connected to system C with 0 + jO power interchange.

Again unstable.

30 and 35 Mw. The practical power capacity was taken as 25 Mw.Note that the transient-stability limit for southward flow was less than thatfor northward flow on the same line.

Run 6. The company B terminal of the A-to-B line was now shiftedfrom BC to BD, and northward power transfer was investigated.

300

First the value of 30 Mw. received at AB from BD was tried. StationBF was generating 54 Mw. with the new generator operating, and 23Mva, of additional transformer capacity was required there. Nopower was received from system C, and systems F and G were disconnected. Generation at station AA was reduced to such an extent thatto supply its local load necessitated the receipt of power from systemD as well as from system B and required 11 Mva. of additional trans-


1 (System D)

t----t---I--~IIIIIIIl:-_+__-+_~~2 (AA, AJ, EA)

Q)

"60e«S

~Bg 120

~~-""'- L....coE.!.5

~ 80 t--_--+----+-~_+__-+_~---+--~--+-__t'inccu.=

40 a-----'_--.L.._.....-'--......_""----"_.-.._......._~__

o 0.2 0.4 0.6 0.8 1.0Time (seconds}

FIG. 46. Swing curves, Study 4, part 1, stability run 6. 30.0 Mw. received atsubstation AB from station BD. Two-line-to-ground fault near station BD onproposed 154-kv.line to substation AB, cleared in 6 cycles, line reclosedin 21 cycles.

Stable.

reduced. Power flow conditions were much like those of run 1 exceptthat the generation at AA was smaller, the difference being made upby power received from system D. Also, systems F and G were connected to B. The interconnected systems were stable (Fig. 47), withconsiderably more margin than there was in run 1 (Fig. 41). Theimprovement might be due in part to the change in company B terminal and in part to the connection of systems F and G. The powerlimit of the A-to-B line with terminal-at BD was not exactly determined, but it appeared to be somewhat higher than the correspondinglimit with line terminal at Be, at least for northward power flow.

Swing curves, part 2. In part 2 of the study the required clearingtime of faults on company A's 13.8-kv. system and the effect on stabil-

former capacity at station AA (slightly less than the amount requiredin run 4). A two-line-to-ground fault was applied to the A-to-B lineat the sending end (station BD), cleared in 6 cycles, and reclosed in 21cycles. The system was stable; all the swings were of small amplitude.(See Fig. 46.)

Run 7. The power transfer on the A-to-B line was increased to 40.6Mw. received at substation AB from station BD. To provide the increased power, 12.4 Mw. was received at BD from system C. Toabsorb the increased power, the generation at station AA was further

200 ,.--..,.-_,_--..---y---r---.,.-----,--..,..---.-----r-__


ity of an interconnection with company H were investigated. Thenetwork of Fig. 40 was set up on the calculating board. As comparedwith the set-up of part 1 (Fig. 39), the principal changes were: (a)The base was changed from 40 Mva. to 25 Mva. (b) The l3.8-kv.network of company A was represented in detail, including the separation of station AA bus into two sections by a bus reactor, with thegenerators of each section represented separately, and the separation

200 r----r---,.---r---r---.....---...r----r-_r_-.......--.....---,

- 160 I----I---+---+--.a..--t-+---+----t-- 6 (Ge, GD)

~ I~ 5 (GBl,..,...--+----t'0

- 7·(FA~)

I~1 ~ --t'---t'--.....--+-----t--i---f---t----f

ml"Uul~3 t ~ -+---+--+--.--+----+----t---..+---+--t

~ 1'-

O--......-~- ..........I.-..&.-_"--.......- ......_------.....o 0.2 0.4 0.6 0.8 1.0

Time (seconds)

FIG. 47. Swing curves, Study 4, part 1, stability run 7. 40.6 Mw. received atsubstation AB from station BD and 12.4 Mw. received at station BD from substation CA. Systems F and G connected to system C with 0 + jO power interchange. Two-line-to-ground fault near station BD on proposed 154-kv. line to

substation AR, cleared in 6 cycles, line reclosed in 21 cycles. Stable.

of station AJ from station AA. The 13.8/66-kv. transformers atstation AA were assumed to be of 25-Mva. capacity each, or 50 Mva,total, as found necessary in part 1. (c) The remote part of system Dwas shown in more detail. (d) The part of system B containing stations BF and BG was represented by an equivalent circuit. Thecompany B terminal of the A-to-B line was placed at -station BD.Systems F and G were not connected. Zero active power and about17 Mva, reactive power were supplied to system B from system C.

Run 8. First the required clearing time of faults on the 13.8-kv.


1.21.0

System C as infinite bus

0,40.2- 20 "'- ..-..-_""----'-_........_"--......_ ........_Il.o.- .....-.--'_~

o

140

system of company A was examined. The A-to-B interconnectionwas delivering 26 Mw. to station BD (approximately the practicalpower capacity of the interconnection as found in part 1). The output of station AA generators was 101.5 + j40.7 Mva. Since this output exceeded the capacity of the station, a new generator was assumedon the north bus, and the aggregate capacity was divided equally

160 ...----r---.--r---r-.,...--..,r-----r---r----,r----,--.,.--.,---,

en 120Q)

et)I)Q)

~ 100Q)

1ioc:co~ 80t--+----I-t--++--~~~-+--f-~f---+---+-----f--4.Bgco 60 t---+~_t_-hr--.L-.-f"'"___I__r_~r---+--t-~t:::::;a.....Iiiii:::+---t~......E.s.5C 40 t--:3IIlI'f'"""'---::I~~t---#l--_t_-~--t-__:l~-,t----+---+-----f~r-fQ)

'Vic

~ 20 ~~~_f_--#t----t-~Lt._-~lL-+--+--..--+--f----I---1


FIG. 48. Swing curves, Study 4, part 2, stability run 8. 26.0 Mw. received atstation BD from substation AB. Three-phase fault on 13.8-kv. feeder near station

AA north bus, cleared in 12 cycles. Stable.

between the two busses. The voltages of both bus sections were heldat the same magnitude, but a phase difference between them caused5 Mw. to be transferred from the south to the north bus. Fromsystem D 25 Mw. was delivered to system A. One machine wasassumed to be operating in each system D plant except station DA,where all machines were assumed to be in service. At station EA 15Mva. of additional 110/66-kv. transformer capacity was required.

A three-phase fault was assumed to occur near station AA on a13.8-kv. underground cable feeder supplied from the north bus. With12-cycle clearing the system was stable. (See swing curves, Fig. 48.)


260

240

220

200

180,...

I 160r'0

GJco 140cto

InI 120gCic"- lOaG)..,.5~c:Q)

'iii 80ceto-

60

40

20

0

- 20 .........._-'----a_--'-_...A..---"_........ ....._~..... ....

o 0.2 0.4 0.6 0.8 1.0 1.2Time (seconds)

FIG. 49. Swing curves, Study 4, part 2, stability run 9. Same as run 8, exceptthat clearing time was increased to 18 cycles. Unstable.

Run 9. With I8-cycle clearing of the same fault the systems wereunstable (Fig. 49).

Run 10. The purpose of this run was to show the effect of threephase faults farther away from station AA. The location selectedwas a point about 1,000 ft. from station AA on a 700,OOO-circular-milcable supplied from the south bus. With the additional impedance ofthis cable between station AA and the fault, the voltage at AA re-


mained high enough to prevent the generator output from decreasingas much as before, so that even with I8-cycle clearing the system wasstable (see Fig. 50). The extreme oscillations of the station AJmachine, as shown by its swing curve, should be discounted somewhat,as they did not represent the true motion of the machine but were the

200 r----.---r-.-......-......-r--,---r---r--.----.~__y_o-~__..

180

Xli160 i

~If"04

140 ~

!!.;1

4D ~I"0 120.....~I.!!co Ic

tV100I»

~gto 80c..~.., 60c.~enC

~ 40

20

0

- 20 '--.-.._-L---a._~_"--__'__-'--.._....... -.r.._.....

o 0.2 0.4 0.6 0.8 1.0 1.2Time (seconds)

FIG. 50. Swing curves, Study 4, part 2, stability run 10. Same as run 9 exceptthat fault was 1,000 ft. from station AA south bus on 700,OOO-cir.-mil cable.

Stable.

result of inaccuracies caused by the use in the step-by-step calculationsof a time interval too long in comparison to the period of oscillation ofthe machine.

Run 11. The purpose of this run was to show the effect of removingthe bus-tie reactor at station AA, leaving the north and south bussestied together through only the I3.8/66~kv. transformers and the 13.8kv. network. In every other respect it was identical to run 9. Com-


parison of the swing curves of run 11 (Fig. 51) with those of ron 9(Fig. 49) shows that splitting the bus at AA decreased the rate atwhich systems A and B swung apart, but not sufficiently to allowstability to be regained. Although this run was considered unstable,

220 ------.....------,r-----r---w----r-.....-----,---t--r----,

200

180

160~

enQ,)

~ 1401)0Q,)

"'C-Q,)

ao 120eto

~S

100g-;;E.! 80.5....cQ)

.iii60c:e...40

20

0

- 20 ~---------"'-...._I..---L.-...L-.--J._~_a..._.......,L.._.l....---'o 0.2 0.4 0.6 0.8 1.0 1.2

Time (seconds)

FIG. 51. Swing curves, Study 4, part 2, stability run 11. Same as run 9 exceptthat station AA bus was split. Still unstable.

it represented a borderline condition, indicatingthat the systemswouldbe stable under slightly more favorable conditions, such as appreciablearc impedance in the fault or decreased power transfer on the interconnection.

Run 12. The purpose of this run was to show the effect on stabilityof connecting system H to the A-to-B line. System H was a metro-


politansystem located near the center of the proposed interconnectingline from AR to RD. It was represented as a 275-Mva. generator tiedto the 154-kv. line through a 30-Mva. transformer, with zero interchange of active and reactive power. Power of 45 Mw. was trans-

180

160

140

,.... 120

~if»"" 100.....Ic.,

80&~~Ci 60Efl 40·Iiic:

S 20

0

~20

-400 0.2 0.4 0.6 0.8 1.0 1.2

Time (seconds)

FIG. 52. Swing curves, Study 4, part 2, stability run 12. 45.0 Mw. received atstation BD from substation AB. System H connected to the A-to-B line with0+;0 power interchange. Two-line-to-ground fault near substation AB on theproposed 154-kv. line to station BD, cleared in 6 cycles, line reclosed in 21 cycles.

Unstable.

mitted to BD from AB, necessitating a new generator on AA north busand requiring only one machine to be operated at BD. The powerdelivered to system A from system D was raised to 37.4 Mw. by increasing the output of the system D generators already in operation.With a two-line-to-ground fault at the sending end of the 154-kv. inter-

STUDY 4-POWER-ANGLE CURVES, PART 2 335

-- ~

V ""'"'<~Systems

/ IA,D,E,

I no fault

/" I

'"1I

Q)

/00 -,I; ~

/1 I~ '\l:cv I I~

I I~I I~I I~

I I<.JI I

I,I ~Systems

I I / A,D,E,I ~ --......""- fault onI

V~

I.Y

200

220

i20

100·

80-100 -80 -60 -40 -20 0 20 40 60 80

Rotor electrical angle (desrees)

FIG. 53. Power-anglecurves, Study 4, part 2, run 13. 5 Mw. receivedat stationBD from substation AB. Three-phase fault on 13.8-kv. feeder near station AAnorth bus. Critical clearing angle, determined by equal-area method, is -34°;

critical clearing time, 24 cycles.

connecting line and with the usual clearing and reclosing times" thesystems were unstable. The swing curves (Fig. 52) show that systemsA, D, and E swung ahead together, whereas systems B, C, and Hswung very little. The power limit of the A-to-B line for southwardpower transfer with system H connected to the line was thus shown tobe less than 45 Mw. This limit was found more exactly as describedbelow.

240

Power-angle curves, part 2. Power-angle curves and the equal-areacriterion were used for (1) obtaining a curve showing the critical clearing time of a three-phase fault close to the north bus of station AArequired to maintain stability of systems A, D, E with systems Band Cfor any amount of power transferred from A to B; (2) determining thepower limit of the A-to-B line for southward transfer with system Hconnected.

Critical·clearing time of a three-phase fault near 8tation AA 19.8-kv.


north bus. To use the method mentioned it was necessary to reducethe system to an equivalent two-machine system. Inspection of theswing curves for runs 8 to 11 showed that the machines of systems A,D, and E swung together and, therefore, could be combined with verylittle error to one equivalent machine. The machines of system Blikewise swung together and could be combined, and system C wasregarded as an infinite bus. It was decided to combine the machines

v ",-~ ...-Systems

/ "< A.D,E.I no fault

V I

""/II1

~, "-

v: leu ."/ 1"6"0I

,e:t'O

V I 001.=I ,-5I I'~

I I~I I-I I.~I I=EI

IU

I I

JI-Systems

I I A.D,E,I I fault on.I ------

..... ...........

I ~~ ""'"~

~V80-100 -80 -60 -40 -20 a 20 40 60 80

Rotor electrical angle (degrees)

FIG. 54. Power-anglecurves, Study 4, part 2, run 14. 15 Mw. received at stationBD from substation AB. Three-phase fault on 13.8-kv. feeder near station AA

north bus. Critical clearing angle is -37°; critical clearing time, 18 cycles.

240

100

120

cJ)

t::eu~ 180iE

~ 160...::Jo...Q)

~ 140Q.,0

200

220

of system B with those of system C and to represent them all as aninfinite bus. This procedure could be expected to give somewhatpessimistic results because the speeding up of the machines of system Bto follow those of systems A, D, E was beneficial to stability.

Power-angle curves were obtained on the calculating board by thefollowing procedure: when possible, the initial conditions of each runwere made identical to those of a previous load-flow study. Thepower received at station BD from substation AB was set to 5 Mw.,15 Mw., 25 Mw., and 35 Mw. in runs 13, 14, 15, and 16, respectively;


and the voltages of the 66-kv. bus at AB and of the 132-kv. bus at BDwere adjusted to normal values. The phase angle of the power sourcerepresenting the voltage back of transient reactance of the finiteA-D-E machine was then advanced in steps of about 15°, both thisvoltage and the voltage representing the infinite bus being held constant, and at each step the power output of the finite machine wasrecorded. These results were plotted as the curves labelled "no fault"

- -V' .........

"<~Systems

A,D,E,/ I ~ no fault

1/ I~ -,I

/~ I I'VI! I

"I~l:i

I I.~I.e:I ~I~

IJ 1«1I I~

I 18

I I~Systems

I I

li A, D, E,~ L-.-~ ---... fault on

!~

~~~--

80-100 -80 -60 -40 -20 0 20 40 60 80


FIG. 55. Power-angle curves, Study 4, part 2, run 15. 25 Mw. received at stationBD from substation AB. Three-phase fault on 13.8-kv. feeder near station AA

north bus. Critical clearin.g angle is -33°; critical clearing time, 16 cycles:

240

120

100

200

220

in Figs. 53 to 56, inclusive. The angle scale in these figures is arbitrary and is not the angle between the internal voltage of the finitemachine and the voltage of the infinite bus. For each value of powertransfer a second curve was obtained in similar manner, except withthe fault applied, and with the internal voltages held at the samevalues as before. The critical clearing angle for each value of powertransfer was then obtained by the equal...area method, and the corresponding critical clearing time was found from a swing curve calculated by use of the values of power read from the power-angle curve.


The kinetic energy of the A-D-E machine was taken as 1,361 Mi.The swing curves are given in Fig. 57.

The desired curve of power transfer versus critical clearing time wasthen plotted and is shown in Fig. 58. The results of stability runs 8and 9 are consistent with this curve: for 25-Mw. transfer the systemwas stable with 12-cycle clearing and unstable with 18-cycle clearing.

I~-_~~systems

~ I ~ A,D,E•./ -, no fault

Vi ~ '"220 1----+----4--~+--__+_----t'----_f__:::lIillE:_--t---t---t

240--------.....------.---.-----,---,...----.,

120I----f----+--!-"f-----f---i---+----i---f-----t

100t----t--~........-+--_+_-___t--_+__-__P-----t__-__t

806080'---......- .....--.....--....------......- .....--~-~-100 -80 .. 60 -40 -20 0 20 40

Rotor electrical angle (degree~)

FIG. 56. Power-angle curves, Study 4, part 2, run 16. 35 Mw. received atstation BD from substation AB. Three-phase fault on lS.8-kv. feeder near stationAA north bus. Critical clearing angle is -45°; critical clearing time, 11 cycles,

Power limit oj A-to-B line with system H connected. The swingcurves of run 12 indicate that the machines of systems B, C, and H maybe regarded as an infinite bus and those of systems A, D,and E as asingle finite machine. Power-angle curves with no fault and againwith a two-line-to-ground fault near AB were obtained. It seemedpermissible to use the same power-angle curves for all values of initialpower transfer, because in runs 13 to 16 the voltage behind transientreactance varied only 3% over the range of power transfer that wasconsidered. With the fault cleared, the power transfer over the inter-


connection became zero, but the output of the finite machine did notbecome zero because it was supplying local load. Its output with theline open, however, was independent of angle and was found fromprevious load studies, as follows: The total load on systems A, D, andE, when operating with zero interchange with system B, was determined from one load run, and the total load plus losseswhen delivering

-20 0

..-.- -40 _ -20en en

iQ)

!

-60 i -40"0

Q) Q)

lib libc c:-e -80 -e -60

-100 -800 0.1 0.2 0.3 0.4 0.5 0 0.1 0.2 0.3 0.4

Time (seconds) Time (seconds)(a) (b)

0.1 0.2 0.3 0.4Time (seconds)

(cl)

~

100----::7f~ I

iJ

o

-80o

Q)

bOc:-e -60

- -20~~"0 -40

0.1 0.2 0.3 0.4Time (seconds)

(c)

-80 '---"""---..............-...--...o

o---~--y----,.---.,

_ - 20f---+---+---~~--t~er~ -40 t---+&---~~_+__---1

j-. -60 ~-+----+--+--+--~

FIG. 57. Swing curves used in connection with the power-angle curves of Figs.53 to 56 to determine critical clearing times.

45 Mw. to system B was obtained from another load run. The difference between these figures gave the decrease in losses which wouldoccur on opening the 154-kv. interconnecting line, normal voltageconditions being assumed. This decrease in losses was expressed as apercentage of the power sent into the 154-kv. line. To determine theoutput with the fault cleared, the percentage found as just describedwas added to the input to the 154-kv. line, and the augmented inputwas subtracted from the output of the initial operating conditions.


\\ Unstable,

Stable 1\\,

~

\ ,

oo

~ 30;~Q)

E"0 20~Q,)

.~Q)"0

10 20Clearing time(cycles)

FIG. 58. Curve of power deliveredto station BD from substation ABover proposed 154-kv. interconnecting line versus critical clearing timeof a three-phase fault on a 13.8-kv.feeder near station AA north bus.Based on the results of runs 13 to 16,

Study 4, part 2.

The critical reclosing time was found by use of the equal-area methodand swing curves (Figs. 59 to 61) for transfers of 25, 35, and 45 Mw.,using a clearing time of 6 cycles in every case. The clearing anglecorresponding to the clearing time of 6 cycles was found by computingthe swing curve up to this point. The critical reclosing angle was then

40 found by the equal-area method, andthe corresponding reclosing time wasobtained by extending the swingcurve to this angle.

A curve of critical reclosing timeversus power transfer was plottedand may be seen in Fig. 62. Fromthis curve the power limit for 21cycles reclosing time is read as 42Mw, This is consistent with the result of stability run 12, in which thesystem was found to be unstable with45 Mw. transfer and 21-cycle reclos-

30 ing. The practical power capacitywas taken as 35 Mw.

Conclusions. The following cOI\elusions were among those reachedas a result of Study 4:

Capacity of interconnection betweensystems A and B. The power capacityof the proposed 154-kv. interconnection between systems A andB, based on the maintenance of syn

chronism through a two-line-to-ground fault on the interconnectingline, with 6-cycle clearing and 21-cycle reclosing, and allowing a margin of at least 5 Mw. for momentary fluctuations in power on the line,would be 25 Mw. received in system B or 35 Mw. received in systemA. These values would hold for either of the two routes considered.If system H were connected to the line, the power received in systemB might be increased to 35 Mw, In arriving at these limits it wasassumed that automatic equipment for controlling the interchangepower flow would be installed and used, and that high-speed relayingand switching would be provided on critical circuits within systems Aand B as well as on the interconnecting line itself. The connectionof systems F and G to system B would be beneficial to stability butwould not have enough effect to warrant the assignment of higherpower ratings to the interconnection. The capacity of the intercon-

STUDY 4-CONCLUSIONS 341

I I I

Critical reclosing timeI~~ r- 'I ·1'-

V~Swing curve~ Power output

V ff),Systems A,D,E, ~

1/ V ~ no fault~

V /V

~ '/~~

V / V Critical ~ I~reclosing

/ /V angle",--~ ~

v V ~. "

~- j ,~

K7 Input~\1 ~.~

~-17 7 7 7 ~/ 77 -77 ,

/ f'/f///f'/// '/// '////~ \,'/

-r I

I~~Clearing angle

V,........ [7

.\ -:--10-V ~r.-'//M~ '/// 1//./ 1/// /// I~ Power output7 ~

Systems A,D,E,

'./" Power output

fault on

Systems A, D,E,lineopen

260

0.6~

en 240"'0c0

~ 0.4....,CDE 220i=

0.2~

(I)=co~

~200

E0 ~

jcf 180

'160

140

nection, if a loss of synchronism were allowable during a fault on theinterconnecting line 'or on other critical circuits, would be approximately 35 to 40 Mw. in either direction.

Speed of fault clearing. To avoid instability when the interconnection was delivering its rated load of 25 Mw. to system B, it would be

0.8

12060 80 100 120 140 160 180 200


FIG. 59. Power-angle curves and swing curve, Study 4, part 2, run 17. 25 Mw.received at station BD from substation AB. Two-Jine-to-groundfault near substation AB on proposed 154-kv. interconnecting line to station RD. Clearing

time, 6 cycles. Critical reclosing time, 42 cycles.

necessary to use high-speed relaying and switching on some of thecircuits of company A. Three-phase faults near station AA 13.8-kv.or 66-kv. busses and near substation AB 66-kv. bus must be cleared in12 cycles or less. The time for clearing three-phase faults on the13.8-kv. network might be increased to 18 cycles if the fault wasseparated from the bus by 1,000 feet or more of 700,OOO-circular-milcable or by an equivalent impedance. If the station AA bus wassectionalized, three-phase faults on or near the bus must be cleared in16 cycles or less.


1 I I I I 7Critical reclosing time l/- ,I. ,I. ~1- ~_

I V 11Swing curve:-> I I./:'Power outputI--

/" 1-- biSystems A, D,E, -

/ rnf/ no fault

/ /.V I~ .'/~a..

v V I~ r,,1o.j )~-

V Critical .~ "I /-- IIJV

reclosing

~ ~angle-~'//J.

JV Inputt: J ~[7 I

\~r--- Clearing angle

V,.........

.~ r--r---V r--...~ Power output

~ 4Systems A, D,E,

'/ \ Power outputfault on

Systems A, D, E.line open

260

-.g 0.4c0

~ 2404n-«U

~ 0.2

220

0 -;;;=cv

&200E-...Q)

~e. 180

160

140

High-speed relaying and switching might be necessary also on severalof the transmission lines of company B, but further studies would berequired, after the terminal of the line had been selected, to determinethe extent of such changes.

Terminal transformer capacity. The transformer at substation ABshould have a capacity of 45 Mva. and that at either station BD or

0.6

12060 80 100 120 140 160 180 200


FIG. 60. Power-angle curves and swing curve, Study 4, part 2, run 18. 35 Mw.received at station BD from substation AB. Two-line-to-ground fault near substation AB on proposed 154-kv. interconnecting line to station RD. Clearing time,

6 cycles. Critical reclosing time, 31 cycles.

substation BCshould have a capacity of 50 Mva. Both these transformers should have tertiary windings rated 15 Mva. for connectionto synchronous condensers or static reactors.

The capacity of the 13.8/66-kv. transformers at station AA shouldbe increased from 20 Mva. to 45 Mva,

The line as built. The interconnection between companies A and Bwas built and put into service in 1942. The line was 268.5 miles long

STUDY 4-THE LINE AS BUILT 343

with terminals at substations AB and Be and no intermediate switching stations nor taps. (The route finally selected was longer than theone assumed in the preliminary studies.) It was a single-circuit 154kv. line. The phase conductors were 250,OOO-circular-mil, hollow,hard-drawn copper cables of O.u83-inch diameter, supported by strings

0.4-;;-

260"0e0

! 0.2Q,)

240Ei=

0

220

-~IV

~ 200t]GQ)

E-i~ 180

160

140

~, I I

Critical .........Swing curve_ reclosing time- ~

~-I--I----~-P(. /I- Clearing r I """"'"'"

"",."."

~ Power output~ime~ J I~~/r ;"~ ~/ Systems A,D,E, -~--- 1 'r ~~

~ no fault1/ ~I--~--- I~

~ ~~I

/v InputP,../' -,I I

)/ ~Critical I\.reclosing

....V angle 1\~ Clearing v--: Power output \

ISystems A, D,E, -~angle""",,-- i--=3 fault on--V --~,...

.....~ ~ '///1

.,Vt"" . ",""- Power output

V Systems A, D,E,line open

12060 80 100 120 140 160 180 200


FIG. 61. Power-anglecurves and swing curve, Study 4, part 2, run 19. 45 Mw.received at station BD from substation AB. Two-line-to-ground fault near substation AB on proposed 154-kv. interconnecting line to station BD. Clearing time,

6 cycles. Critical reelosing time, 16 cycles.

of ten 10-inch suspension insulators. They were spaced 14 ft. 6 in.apart in a horizontal plane (18 ft. 3 in. equivalent spacing). Therewere two i-inch high-strength galvanized-steel ground wires 12 ft.above the phase conductors .at structures and 20 ft. above them atmidspan. The structures were wooden H frames. There were threecomplete transposition barrels in the line.

At substation AB there was a 40-Mva. 161/69/13.8-kv. three-winding transformer bank, and at substation Be, a 45-Mva. 161/138/12.5-


~~

'"~ ..

-, Unstable

,,~

Stable "'-.<,-,"'0e~~ 20"'0

J

kv. autotransformer bank. At each terminal there were three 5-Mvar.static reactors connected to the tertiary windings.

Because of the length of the line it was necessary to develop and use anew type of high-speed relay system which could distinguish betweenfaults and power swings even though the magnitude of indicatedimpedance might be nearly the same. 2 ,3 Internal faults in the terminal transformers were cleared by differential relays which tripped

50

40

10

oo 10 20 30 40 50

Reclosing time (cycles)Constant clearing time, 6 cycles

FIG. 62. Curve of power delivered to station BD from substation AB over proposed 154-kv. interconnecting line versus critical reclosing time for a two-line-toground fault on this line near substation AB, cleared in 6 cycles. Based on the

results of runs 17 to 19, Study 4, part 2.

open the low-voltage oil circuit breakers (66 kv. at AB, 132 kv. atBC)and tripped closed a three-pole grounding switch on the 154-kv.side, causing the breakers at the distant end of the line to open.

Staged-fault tests." After the line had been constructed and wasready for service, a number of staged faults were placed on the line tocheck the operation of the new long-line relays and to check the transient-stability power limits found in the calculating-board studies whichhave been described. Some faults were placed on the 154-kv. interconnecting line itself, others on 60-kv., 66-kv., and 132-kv. lines nearits terminals. The first group of fourteen tests was made with systerns A and B synchronized but with almost no power flow on theinterconnection. In these tests a few incorrect relay operations were

TA

BL

E12

STA

GED

Tw

o-L

INE

-To-

GR

OU

ND

FAU

LTT

EST

SO

N15

4-K

v.IN

TER

CO

NN

ECTI

ON

OF

SYST

EMS

AAN

DB

WIT

HPO

WER

INTE

RC

HA

NG

E(S

TUD

Y4)

00 ~

Fau

ltL

oca

tio

nP

ow

erF

low

on

Cle

arin

gR

eclo

sing

d tjT

est

IfA

t,

Inte

rco

nn

ecti

on

Tim

e(c

ycle

s)T

ime

(cyc

les)

Sta

ble

or~

Nu

mb

erL

ine

Vol

tage

Nea

rO

nL

ine

A.A

.A

i

Un

stab

le~

,,

r..

,~

,I

(kv.

)S

tati

on

toM

w.

Rec

'dat

at

AB

atB

CstA

Ba

tBC

U2 t-3 >

115

4A

BB

e15

BC

6.8

6.8

21.2

24

.2S

tab

le0 t:r

j2

154

AB

Be

30B

e6

.57

.521

.524

.5S

tab

lee

366

AB

EA

28B

C10

.5..

....

...

Sta

ble

~3A

132

Be

BD

29B

C10

.0S

tab

le>

...

.....

.~

415

4B

CA

B35

AB

10.0

6.0

26.2

23

.0U

nsta

ble*

E:34A

154

BC

AB

26A

B1

9.0

t5

.534

.522

.0U

nst

able

tt-3

4B15

4B

CA

B31

AB

10.0

6.0

25

.82

2.5

Sta

ble

t:rj

513

2B

eB

D34

AB

8.5

Sta

ble

U2

......

...t-3 00

*Syn

chro

nism

lost

bet

wee

nsy

stem

sB

an

dF

.tR

elay

tro

ub

leat

stat

ion

AB

caus

edsl

owtr

ipp

ing

.

~


M, 1.2<, I I

~ 1Unstable

"IV

~ 4

14"

I~

~B

2'~ 4A

Stable 'I'\..I

~~N

Stable Unstable~ Test ~ A

Calculated 0 •

40

discovered, which led to some changes in the connections and adjustments of the relays and circuit-breaker control. The second group ofeight tests were made with various amounts and directions of powerinterchange. Oscillographic records were made of eighteen currentsor voltages in the main and control circuits, making possible thedetermination of the sequence and time of various operations. Thetests of the second group are summarized in Table 12.

50

oo 10 20 30 40Reclosing time of terminal breakp.rs (cycles)

FIG. 63. Approximate curve of transient-stability power limit as a function ofreclosing time for faults on the 154-kv. interconnection of companies A and B,Study 4. Points from staged-fault tests. and from calculating-board results are

plotted.

With the exception of tests 4 and 4A, in which synchronism was lost,power swings observed throughout the interconnected systems weresmall.

After the application of faults on the 154-kv. line the terminalbreakers successfully cleared the fault and reclosed in every case wherethey were intended to do so. ·The same breakers remained closed inevery test in which a fault was applied to other lines, indicating properfunctioning of the relays during external faults.

Study of the test results revealed that the expected 21- or 22-cycleover-all time from fault to final reclosure of the second breaker was notbeing obtained on account of a delay of from 2 to 4 cycles in the relayaction at the end of the line remote from the fault. This trouble wascorrected subsequently.

REFERENCES 347

In order to correlate the tests with the results of stability runs madeon the calculating board, Fig. 63 was prepared. Here points representing the staged-fault test results and other points representingthe calculating-board results with line terminal at substation Be areplotted with delivered power as ordinate and reclosing time as abscissa.The sloping line MN separates the region of stability from the regionof instability. The test results agree reasonably well with the resultsof the board studies. Exact agreement cannot be expected becausethe conditions existing at the time of the staged-fault tests differredfrom the conditions assumed in the calculating-board studies. Inparticular, the interconnection between systems Band C, assumed inthe calculations, had not yet been built. Even if it' had been, itwould not have been feasible to schedule power flow and generation infive or more power systems in several states in such manner as toduplicate the conditions assumed in calculation.

REFERENCES

1. R. C. BERGVALL, "Series Resistance Method of Increasing Transient-StabilityLimit," A.I.E.E. Trans., vol. 50, pp. 490-4, June, 1931; disc., pp, 494-7.

2. F. C. POAGE, C. A. STREIFUS, D. M. MACGREGOR, and E. E. GEORGE,

"Performance Requirements for Relays on Unusually Long Transmission Lines,"A.l.E.E. Trans., vol. 62, pp, 275-83, June, 1943.

3. A. R. VAN C. WARRINGTON, "Protective Relaying for Long TransmissionLines," A.I.E.E. Trans., vol. 62, pp. 261-8, June, 1943.

4. C. W. MINARD, R. B. Gow, W. A. WOLFE, and E. A. SWANSON, "StagedFault Tests of Relaying and Stability on Kansas-Nebraska 27Q-Mile 154-kv.Interconnection," A.I.E.E. Trans., vol. 62, pp. 358-67, 1943; disc., pp.425-6.

INDEXAcceleration, 15, 16

angular, 17, 19Adjustment of initial operating condi-

tions, 77Admittances, driving-point, 80-6

mutual, 80-6self-,80-6terminal, 80-6, 95transfer, 80-6

Algebraic solution of networks, 78-101Alpha, beta, zero components, 245Alternating-current calculating board,

64American Gas and Electric Company,

288,290Amplifiers for instruments on a-c.

board, 70, 73Analogies between translation and rota-

tion, 20Analogue, mechanical, 7Angle, 17, 19Angular acceleration, 17, 19Angular momentum, 19Angular velocity, 17, 19Apparatus base, 54Assumptions made in stability studies,

43Autotransformers, 58

on a-c. board, 68, 69, 72, 73, 74zero-sequence equivalent circuit of,

239

Bad effects of instability, 9Base impedance, 55Base quantities, for a-c. board, 66,

72choice of, 75

for per-unit system, 54Board, a-c. calculating, 64

349

Braking effect of grounding resistor,241

Breakers, data required on, 63eight-cycle, need for, Study 3, 310

Bus, infinite, 124Bus reactors at Philo, 290, 294, 300Busses, effect of, on stability, 190Byrd, H. L., 168

Cables, data required on, 63equivalent circuits, 59

Calculating board, a-c., 64procedure for using, 75

Capacitance of transmission lines,representation of, 60

Capacitors, on a-c. board, 67, 68, 73, 74series, 189

Carrier-current relaying needed, Study3,310

Check list of data required for transient-stability study, 62

Checks on readings, 78Circuit breakers, data required on, 63

eight-cycle, need for, Study 3, 310Cleared fault, application of equal-area

criterion to, 129Clearing angle, 130

determination of, by pre-calculated. swing curves, 157, 158

Clearing time, 139allowable, Study 1, 286critical, curves for determining, 168-

85determination of, by pre-calculated

swing curves, 157, 1,P8effect of, on transient stability, 159

Combining machines, 111, 264, 319Common voltage, referring quantities

to, 56

350 INDEX

Components, alpha, beta, zero, 245symmetrical, 193

Composite line, 60Condenser, synchronous, data required

on, 63mechanical analogue of, 8Study 1, 296-7treatment of, in transient-stability

study, 119Connections on a-c. board, 69, 77

changes of, 78Conversion, of impedances, 54

star-mesh, 86-7Coordinates, symmetrical, 193

two-phase, 245Correction factors for equivalent 1r cir-

cuit of transmission line, 59Coupled lines, equivalent circuits of,

234Critical clearing angle, 130, 136

determined by pre-calculated swingcurves, 157, 158

Critical clearing time, curves for deter-mining, 168-85

determined by pre-calculated swingcurves, 151, 157

Damper winding, 11D8IDping, 5,43,122,123Data required for transient-stability

study, 62-Dead sequence networks, 210Difference equation (problems), 51Dimensions of quantities of mechanics,

15, 17-9Discontinuities in point-by-point solu-

tion of swing equation, 38Double unbalances, 232Duration of fault, effect of, on stability,

159, 185, 224

Ebasco Services, Inc., 254, 311Energy, kinetic or stored, 16, 20, 23-6

of synchronous machines, 25,256-8, 292, 305, 315-6; seealsoInertia constant

rotational, 20, 23-6Equal-area criterion, 122

Equal-area criterion, Study 4, 335-43Equipment, miscellaneous, 61Equivalent circuits, of coupled lines,

234of remote portions of system, 61of transformer, 56

negative-sequence, 235zero-sequence, 235-40

of transmission lines and cables, 59Equivalent generator, 6Equivalent impedance of transformers,

54,56Equivalent inertia constant of two-

machine system, 133Equivalent input of two-machine

system, 133, 136Equivalent motor, 6Equivalent output of two-machine

system, 133, 136Equivalent pi (71") circuit of transmis-

sion line, 59Equivalent power-angle curve of two

finite machines, 133Equivalent T circuit of transformer,

56

Factors affecting stability, 187Fault, close to generator, 78

effect on stability, 5of duration of, 159, 185, 224of type of, 223shown by mechanical analogue, 9

Fault impedance, 228Fault location, effect of, on stability,

189Fault locations, Study 1, 265

Study 3,302Fault shunts, 220, 231Faulted three-phase networks, solution

of, 193Faults, relative number of various types

of, 225relative severity of various types of,

223representation of, in symmetrical

components, 205, 228-30in two-phase coordinates, 247

simultaneous, 232

INDEX 351

Finite machines, 132Flux linkage of field winding, 123Force, 15, 16Formal solution of swing equation, 28,

32Frequency, effect of, on stability, 189

used on a-c. board, 66, 72Frequency changer, 189

(problem), 51

General Electric A-c. Network Ana-lyzer,66

Generator units of a-c. board, 66, 68,72,74

Generators, data required on, 62representation of, 56

Governor, 5, 122Graphical integration, determination of

swing curve by, 139Grounding, effect of, on stability,

240Grounding impedance, 237

effect of, on stability, 240of autotransformer, 240

H (constant), 24-6High-voltage bussing, effect of, on sta-

bility, 190Historical review, 11Hunting, 11

Impedance, base, 55fault, 228grounding, 238negative-sequence, 201per-unit, 55positive-sequence, 201series, representation of, 231transformer, 56zero-sequence, 201

effect of grounding on, 241mutual, of lines, 233

Impedance diagram, 53Impedance units on a-c. board, 67, 68,

73,74Inertia, moment of, 18, 19Inertia constant, 22

effect of, on stability, 188

Inertia constant, equivalent, of. two-machine system, 133

of combined machines, 111of remote portion of system, 62

Infinite bus, 124Initial operating conditions, 77, 78Input assumed constant, 43Instability, bad effects of, 9

definition of, 1Instantaneous clearing, stability limit

for, 160, 164determination of, by curve, 176

Instruments on a-c. board, 70, 71Integration, graphical, determination of

swing curve by, 139Interconnecting lines, Study 4, 314, 342Interconnection, 12Interconnections, effect of, on stability,

Study 2, 294-301Intermediate busses, effect of, on sta-

bility, 190Internal voltages, effect of, on stabil-

ity, 2,188

Joule (unit), 15Jumper circuits on a-c. board, 69Junction, three-phase, 202

I{inetic energy, 16, 20, 23-6in equal-area criterion, 125of synchronous machines, 25, 256-8,

292, 305, 315-6; see also Inertiaconstant

rotational, 20, 23-6Kirchhoff's laws, 202

Laws of mechanics, 15Length, 15Limitations of system, 77Line-impedance units of a-c. board, 67,

68,73,74Line-to-ground fault, 207, 248

fault shunt, 220, 231with fault impedance, 229

Line-to-line fault, 207, 212, 248fault shunt, 220-231with fault impedance, 229

Lines, transmission, representationof, 59

352 INDEX

Lines, zero-sequence mutual impedanceof, 233

Load adjuster, 73Load-impedance units of a-c. board, 67,

68;73, 74adjustment of, 73, 77

Loads, data required on, 63representation of, 61

Location of fault, effect of, on stability,189

Mass, 15Master instruments of a-c. board, 70,

73,75McClure, J. B., 149Mechanical analogue, 7Mechanics, laws of, 15Mesh, three-phase, 203Mesh circuit, 82, 110Metropolitan systems, 12, 187M.k.s. system, 15Modified time, 149, 169, 178Moment of inertia, 18, 19Momentum, 15, 16

angular, 19Motor-generator set for a-c. board, 71Motors, large synchronous, data re-

quired on, 63Multicircuit transformers, zero-

sequence equivalent circuit of,238

Multimachine system, 6, 122, 191Mutual impedance base, 55Mutual impedance of parallel transmis-

sion lines, 61zero-sequence, 233

Mutual reactors on a-c. board, 68, 69,74

Mutual transformers on a-c. board, 68,69, 74

Negative sequence, 193, 194Negative-sequence equivalent circuit

of transformer, 235Negative-sequence impedance, 201Negative-sequence network, 203, 212Network, negative-sequence, 203, 212

poffitive-sequence, 53,203,212

Network, zero-sequence, 203, 212Network analyzer, 64

General Electric, 66Westinghouse, 72

Network calculator, 64Network reduction, 83, 213-5

algebraic, 83calculating-board method of, 109symbols used in, 90

Network re-expansion, 216-8Network solution, algebraic, 78-101Networks, sequence, 203

solution of, 53solution of faulted three-phase, 193substitute, 247

Newton (unit), 15Nodes, 83Nominal pi (1r) circuit of transmission

line, 59Nominal voltage and current of a-c.

board, 66, 72

Ohio Power Company, 290One-line diagram, 53Open circuits, representation of, by

connections between sequencenetworks, 231

Parallel transmission lines, 61effect of, on stability, 189, 190

Per-cent quantities, 56Per-unit quantities, 54Phase shift in transformers, 59, 235Phase shifters on a-c. board, 66, 72, 73Philo station, 290-2Pi (1r) circuits of transmission line, 59Point-by-point solution of the swing

equation, 27discontinuities, 38errors, 34, 42

Polarity of connections on a-c. board,69,76

Positive direction of circuit units on a-c.board, 69, 76

Positive sequence, 193, 194Positive-sequence impedance, ·201Positive-sequence network, 53, 203

use of, for x and y networks, 248

INDEX 353

Post-fault output curve, 129, 131Power (mechanics), 15, 16

in rotation, 19Power-angle curve, 2, 3, 125, 126

equivalent, of two finite machines,133, 135

Power-angle curves, Study 4, 335-8,341-3

Power-angle equation, 2, 78one machine and infinite bus, 126two finite machines, 133two-machine reactance system, 135

Power limit, 6method of determining, Study 4, 318

Power supply of a-c. board, 71Pre-calculated swing curves, 149-59Pre-fault output curve, 129-31Pritchard, S. R., Jr. 168

Radian, 17, 19Reactance, effect of, on stability, 2, 188

of synchronous machines, 56of transformers, 57of transmission lines, 61

zero-sequence, 233transient, 256-8, 292, 315-6

decrease of, to aid stability, 276Reactive power, sign of, 79Reactors, data required on, 63

grounding, effect of, on stability, 241on a-e. board, 67, 68, 73, 74

correction for resistance of, 76shunt, use of, to aid stability, 273special, on Westinghouse a-c. board,

73Readings on a-c. board, 78Reclosing, high-speed, Study 4, 317Reclosing time, critical, Study 4, 340,

344, 346Reduction, of network, 83

algebraic, 83calculating-board method of, 109symbols used in, 90

of sequence networks, 213-5Re-expansion of networks, 216-8Referring quantities to a common vol-

tage,56Regulating transformers, 59

Regulator, induction voltage, on a-c.board, 66, 72

voltage, 11effect of, 123

Relays, long-line, 344protective, data required on, 63

false operation of, 10Remote portions of system, 62Resistors, grounding, effect of, on sta-

bility, 241on a-c. board, 67, 68, 73, 74series, use of, to aid stability, 273

Rotation, analogies of, to translation,20

laws of, 17, 19

Scalar instruments on a-c. board, 73Sectionalised operation, 301Sequence, negative, 193-4

positive, 193, 194zero, 194

Sequenceirnpedances, 200Sequence networks, 203, 212

connections between, for represen-ting faults, 205

for representing, faults with impe-dance, 228-30

for representing open circuits, 231Series capacitors, 189Series impedances, representation of,

by connections between sequencenetworks, 231

Series resistors, use of, to aid stability,273

Severity of types of faults, 223Short circuit, effect of, on stability, 5

representation of, by connectionsbetween the sequence networks,205, 228-30

by connections between the sub-stitute networks, 247

types of, 205Short-circuit impedance of transform-

ers, 54, 56Shunt reactors used to aid stability, 273Shunts for representing faults, 220, 231Simplification of systems, Study 4,318Simultaneous faults, 232

354 INDEX

Solution, of faulted three-phase net-works, 193

of networks, algebraic, 78-101Stability, certain factors affecting, 187

definition of, 1transient, summary of methods of

calculating, 185Stability limit, effect of fault-clearing

time on, 159, 167effect of type of fault on, 223steady-state, 4transient, 5, 6

Stability studies, assumptions made in,43

typical, 253Staged-fault tests, Study 4, 344Star-mesh conversion, 86-7Steady-state stability limit, 4

Study 4,318Steam turbogenerators, stored energy

of, 25Step-by-step solution of the swing

equation, 27discontinuities, 38errors, 34,42

Stored energy, seeKinetic energyStudies, typical stability, 253-347

Study 1, 254Study 2,290Study 3,301Study 4,311

Substitute networks, 247Summary of methods of calculating

transient stability, 185Summers, I. H., 149Sustained fault, application of equal-

area criterion to, 127stability limit for, 161, 164

determination of, by curve, 176Swing curves, 28, 32, 48, 103, 1.09, 144

determination of, by graphical inte-gration, 139

pre-calculated, 149-59Study 1, 265-85Study 2, 296,.297, 300Study 3, 306-9Study 4, 322-34, 339

Swing equation, 20

Swing equation, formal solution of, 28,32

point-by-point solution of, 27errors, 34, 42

Symmetrical components, 193Synchronous condenser, data required

on, 63mechanical analogue of, 8Study 1, 296-7treatment of, in transient-Btability

study, 119Synchronous machines, kinetic energy

of, 25, 256-8, 292, 305, 315-6reactance of, 256-8, 292, 315-6representation of, 1, 43-4,56

Synchronous motors, data required on,63

System base, 55

T circuit of transmission line, 60Tapped loads, 61

(problem), 120Three-circuit transformer, 318

zero-sequence equivalent circuit of,238

Three-phase fault, 209, 248fault shunt, 220, 231with fault impedance, 229

Three-phase networks, solution offaulted, 193

Three-phase transformers, zero-sequence equivalent circuit of,239

Time, 15modified, 149, 169, 178

Torque, 17, 19Transformer banks, 59Transformers, data. required on, 63

equivalent circuits of, 56impedance of, 56multicircuit, 58

zero-sequence equivalent circuitof, 238

negative-sequence equivalent circuitof, 235

per-unit quantities of, 54phase shift in, 235reactances of (table), 57

INDEX 3S5

Transformers, regulating, 59representation of, 56

in the sequence networks, 235three-circuit, 57, 318

zero-sequenceequivalent circuit of,238

three-phase, 59zero-sequence equivalent circuit

of, 239two-circuit,' 56zero-sequence equivalent circuit of,

235Transient reactance, decrease of, to aid

stability, 276values of, 256-8, 292, 315-6voltage behind, 123

Transient stability,certaiIL factors af-fecting, 187

summary of methods of calculating,185

Transient stability limit, 5, 6determination of, by equal-area ori-

terion,128Study 4, 318, 340, 346

Translation, analogies of, to rotation, 20laws of, 15

Transmission, long-distance, 12Transmission lines, constants of, 61

data required on, 63equivalent circuits, 59

Turbogenerators, stored energy of, 25,256-9, 292, 305, 315-6

Two-line-to-ground fault, 209, 248fault shunt for, 220, 231with fault impedance, 229

Two-machine system, 1, 7, 122, 149equivalent inertia constant of, 133equivalent input of, 133, 136equivalent output of, 133, 136

Two-phase coordinates, 245

Unbalances, double, 232Units of quantities of mechanics, 15,

19

Varmeter on a-c. board, 70, 73Vector instruments, 71, 73Vector measurements on a-c. board, 71,

73Vector power, 79Velocity, 15

angular, 17Voltage, behind transient reactance,

123of transmission, effect of, on stabil-

ity, 189Voltage regulators, 11

effect of, 123

Water-wheel generators, stored energyof,- 25, 256-9, 315-6

Wattmeter-varmeter on a-c. board, 70,73

Westinghouse .A-c. Network Calcula-tor, 72

Work, 15of rotation, 18

WR2 (moment of inertia), 23, 256-9,305, 315-6

Zero sequence, 194Zero-sequence equivalent circuit of

transformers, 235-40Zero-sequenceimpedance,201

effect of grounding OD, 241of transformer, 236

Zero-sequence mutual impedance oflines, 233

Zero-sequence network, 203, 233representation of lines in, 233

Zero-sequence reactance of lines, 233

ieee press - power system stability- edward wilson kimbark

Documents

power system engineering

ieee power systems

system equations

system modeling

title power systemstability

electrical engineering

bonneville power adminis

large system of equationscan