Ieee Press - Power System Stability- Edward Wilson Kimbark

POWER SYSTEM STABILITY Volume I Elements of Stability Calculations

IEEE Press 445 Hoes Lane, PO Box 1331 Piscataway, NJ 08855-1331 Editorial Board John B. Anderson, Editor in Chief R. S. Blicq M.Eden D. M. Etter G. F. Hoffnagle R. F. Hoyt

J. D. Irwin

J. M. F. Moura

S. V. Kartalopoulos P. Laplante A. J. Laub M. Lightner

J.Peden E. Sanchez-Sinencio L. Shaw D. J. Wells

Dudley R. Kay, Director ofBook Publishing Carrie 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

. PowerMath Associates, Inc. Series Editorial Advisory Committee Dr. Roy Billinton

Dr. George Karady

University of Saskatchewan

ArizonaState University

Dr. Atif S. Debs

Dr. Donald W. Novotny

GeorgiaInstituteof Technology

University of Wisconsin

Dr. M. El-Hawary

Dr. A. G. Phadke

TechnicalUniversity of Nova Scotia

VirginiaPolytechnic and State University

Mr. Richard G. Farmer

Dr. Chanan Singh

ArizonaPublic ServiceCompany

Texas A & M University

Dr. Charles A. Gross

Dr. E. Keith Stanek

Auburn University

University of Missouri-Rolla

Dr. G. T. Heydt

Dr. J. E. Van Ness Northwestern University

Purdue University

POWER

SYSTEM STABILITY Volume I Elements of Stability Calculations

Edward Wilson Kimbark

AIEEE PRESS

'V

roWILEY\:'l9INTERSCIENCE

A JOHN WILEY & SONS, INC., PUBLICAnON

IEEE Press Power Systems Engineering Series Dr. 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 of Stability 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 Wilson Power 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. Synchronous machines. ISBN 0-7803-1135-3 (set) 1. Electric power system stability. I. Title. II. Series. TKI010.K56 1995 621.319--dc20 94-42999

CIP

To my wife

RUTH MERRICK KIMBARK

FOREWORD TO THE 1995 REISSUE

The IEEE Press Editorial Board for the Power Systems Engineering Series has, for some time, discussed the possibility of reprinting classic texts in power system engineering. The objective of this series is to recognize past works that merit being remembered and to make these older works available to a new generation of engineers. We believe many engineers will welcome the opportunity of owning their own copies of these classics. In order to come to an agreement about which text to reprint, a number of candidates were proposed. After a discussion, the board took a vote. The Kimbark series was the overwhelming choice for the first books in the IEEE Power Systems Engineering Classic Reissue Series. The subject of power system stability has been studied and written about for decades. It has always been a challenge for the engineer to understand the physical description of a system described' by a huge number of differential equations. The system modeling is central to an understanding of these large dynamic systems. Modeling is one of the central themes of Kimbark's Power System Stability books. His discussion of the system equations remains as clear and descriptive today as it was when first published. Many engineers have seen references to these works, and may have had difficulty in finding copies for study. This new printing presents a new chance for these engineers to now have copies for personal study and reference. Kimbark presents a method of solving the system equations that was used in the days of the network analyzer. This method has been replaced by digital computer techniques that provide much greater power and speed. However, the older methods are still of historical interest, moreover, these step-by-step methods provide a convenient way of understanding how a large system of equations can be solved. Edward Kimbark was noted during his long career as an excellent writer and one who had the unique capability of explaining complex topics in a clear and interesting manner. These three volumes under the general title Power System Stability, Volumes I, II, and III were originally published in the years 1948, 1950, and 1956. Kimbark's book, Electrical Transmission of Power Signals, published in 1949, vii

provided a general treatment of electric power networks and signal propagation. Kimbark studied Electrical Engineering at Northwestern University and at the Massachusetts Institute of Technology, where he received the Sc.D. degree in 1937. He then began a career in teaching and research at the University of California, Berkeley, MIT, Polytechnic Institute, Brooklyn, Instituto Tenologico de Aeronautica (San Jose Campos Brazil) and, finally, as the Dean of Engineering at Seattle University. In 1962 Kimbark joined the Bonneville Power Administration as head of the systems analysis branch, where he remained until his retirement in 1976. He continued to work on special tasks at Bonneville until his death in 1982. Kimbark is well-known for his excellent books and also his many technical papers. He was formally recognized for his achievements by being elected a Fellow in the IEEE, to membership in the National Academy of Engineering, and was the recipient of the IEEE Harbishaw Award. He was awarded a Distinguished Service Award and a Gold Medal for his service to the U.S. Department of the Interior. The IEEE Power Engineering Society is proud to present this special reprinting of all three volumes of Power System Stability by Edward Kimbark. Paul M. Anderson Series Editor, IEEE Press Power Systems Engineering Series

viii

PREFACE

This work on power-system stability is intended for use by power-system engineers and by graduate students. It grew out of lectures given by the author in a graduate evening course at Northwestern University during the school year 1941-2. For the convenience of the reader, the work is divided into three volumes. Volume I covers the elements of the stability problem, the principal factors affecting stability, the ordinary simplified methods of making stability calculations, and illustrations of the application of these methods in studies which have been made on actual power systems. Volume II covers power circuit breakers and protective relays, including material on rapid reclosing of circuit breakers and on the performance of protective relays during swings and out-of-step conditions. Such material belongs in a work on stability because the most important means of improving the transient stability of power systems in the improvement of circuit breakers and of protective relaying. It .is expected, however, that the publication of this material in a separate volume will make it more useful to persons who are interested in powersystem protection, even though they may not be particularly concerned with the subject of stability. Justification of the simplifying assumptions ordinarily used in 'stability calculations and the carrying out of calculations for the extraordinary cases in which greater accuracy than that afforded by the simplified methods is desired require a knowledge of the somewhat complicated theory of synchronous machines and of their excitation systems. This material is covered in Volume III, which is expected to appeal to those desiring a deeper understanding 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 to readers seeking an understanding of power-system stability but also to those desiring information on the following related topics: a-c. calculating boards, fault studies, circuit breakers, protective relaying, synchronous-machine theory, exciters and voltage regulators, and the step-by-step solution of nonlinear differential equations. I wish to acknowledge my indebtedness to the following persons: To my wife, Ruth Merrick Kimb ark, for typing the entire manuscript and for her advice and inspiration. To Charles A. Imburgia, A. J. Krupy, Harry P. St. Clair, and especially Clement A. Streifus for supplying and interpreting information on stability studies made on actual power systems. To J. E. Hobson, W. A. Lewis, and E. T. B. Gross for reviewing the manuscript and for making many suggestions for its improvement. To engineers of the General Electric Company and of the Westinghouse Electric Corporation for reviewing certain parts of the manuscript pertaining to products of their companies. To manufacturers, authors, and publishers who supplied illustrations or gave permission for the use of material previously published elsewhere. Credit for such material is given at the place where it appears.

Enw ARD Evanston, Illinois

June, 1947

WILSON KIMBARK

CONTENTS

CHAPTER

I The Stability Problem II The Swing Equation and Its Solution III Solution of Networks IV The Equal-Area Criterion for Stability

PAGE

1 15 53 122

V Further Consideration of the Two-Machine 149 System VI Solution of Faulted Three-Phase Networks

193

VII Typical Stability Studies

253

INDEX

349

CHAPTER I THE STABILITY PROBLEM Definitions and illustrations of terms. Power-system stability is a term applied to alternating-current electric power systems, denoting a condition in which the various synchronous machines of the system remain 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 a synchronous generator supplying power to a synchronous motor over a circuit composed of series inductive reactance XL. Each of the synchronous machines may be represented, at least approximately, by Gen. t-------f Motor a constant-voltage source in series with a constant reactance. * Thus x the generator is represented by Eo 1 X o and Xo; and the motor, by EM and XM. Upon combining the machine reactances and the line re.. actance into a, single reactance, we have an electric circuit consisting of FIG.. 1 S·ImpIe t we-mach·me power two constant-voltage sources, Eo system. and EM, connected through reactance X c= XG + XL + XM. It will be shown that the power transmitted from the generator to the motor depends upon the phase difference 0 of the two voltages E G and EM. Since these voltages are generated by the flux produced by the field windings of the machines, their phase difference is the same as the electrical angle between the machine rotors. The vector diagram of voltages is shown in Fig. 2. Vectorially, .Ie

E G = EM

+ JXI

[1]

(The bold-face letters here and throughout the book denote com*Either equivalent synchronous reactance or transient reactance is used, depending upon whether steady-state or transient conditions are assumed. These terms are defined and discussed in Chapters XII and XV, Vol. III. 1

2

THE STABILITY PROBLEM

plex, or vector, quantities). Hence the current is I = Eo - EM jX

[2]

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)

[3]

Eo-- -EM) = Re ( -E0 -

sx

2. Vector diagram of the system of Fig. 1. FIG.

[4]

h R e means " the reaI part 0" f and -Eo were means the conjugate of Eo. Now let

= EM!!

[5]

Eo = EolJ..

[6]

Eo = Eo/-a

[7]

EM

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

= Re ( Eo/-a

EolJ.. - EM I!J.) x~

... Re(E;2 /-90 Ea;M /-90 a) 0

= - EoEM X-cos(-90 EoEM.

= -y-sma

0

_

0

-

- 8)

[8]

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

[9]

DEFINITIONS AND ILLUSTRATIONS OF TERMS

3

and occurs at a displacement angle ~ = 90°. The value of maximum power may be increased by raising either of the internal voltages or by decreasing the circuit reactance. The system is stable only if the displacement angle ~ is in the range from -90° to +90°, in which the slope dP/do is positive; that is, the range in which an increase in displacement angle results in an increase in transmitted power. Suppose that the system is operating in the steady state at point A, Fig. 3. The mechanical input of the generator and the mechanical output of the motor, if corrected for rotational losses, will be equal to the electric power P. Now suppose that a p

FIG.

3. Power-angle curve of the system of Fig. 1.

small increment of shaft load is added to the motor. Momentarily the angular position of the motor with respect to the generator, and therefore the power input to the motor, is unchanged; but the motor output has been increased. There is, therefore, a net torque on the motor tending to retard it, and its speed decreases temporarily. As a result of the decrease in motor speed, 0 is increased, and consequently the power input is increased, until finally the input and output are again in equilibrium, and steady operation ensues at a new point B, higher than A on the power-angle curve. (It has been tacitly assumed that the generator speed would remain constant. Actually the generator may have to slow down somewhat in order for the governor of its prime mover to operate and increase the generator input sufficiently to balance the increased output.)

4

THE STABILITY PROBLEM

Suppose that the motor input is increased gradually until the point C of maximum power is reached. If now an additional increment of load is put on the motor, the displacement angle a will increase as before, but as it does so there will be no increase in input. Instead there will be a decrease in input, further increasing the difference between output and input, and retarding the motor more rapidly. The motor will pull out of step and will probably stall (unless it is kept going by induction-motor action resulting from damper circuits which may be present) . Pm is the steady-state stability limit of the system. It is the maximum power that can be transmitted, and synchronism will be lost if an attempt is made to transmit more power than this limit. If a large increment of load on the motor is added 8uddenly, instead of gradually, the motor may fall out of step even though the new load does not exceed the steady-state stability limit. The reason is as follows: When the large increment of load is added to the motor shaft, the mechanical power output of the motor greatly exceedsthe electrical power input, and the deficiency of input is supplied by decrease of kinetic 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 the steady-state stability limit, ~ increases to the proper value for steadystate operation, a value such that the motor input equals the output and the retarding torque vanishes. When this value of 0 is reached, however, the motor is running too slowly. Its angular momentum prevents 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 this value, the motor input exceeds the output, and the net torque is now an 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 the operating point on the power-angle curve (Fig. 3) not only goes over the hump (point C) but also goes so far over it that the motor input decreases to a value less than the output. If this happens, the net torque 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 motor will regain its normal speed before the displacement angle becomes too great. 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 greater than normal. The displacement angle then decreases and again approaches its proper value. Again it overshoots this value on account of inertia. The rotor of the motor thus oscillates about the new steady-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 a stable system. With a given sudden increment in load, there is a definite upper limit 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-state stability Iimit.t but, unlike the latter, it may have many different values, depending upon the nature and magnitude of the disturbance. The disturbance may be a sudden increase in load, as just discussed, or it may be a sudden increase in reactance of the circuit, caused, for example, by the disconnection of one of two or more parallel lines as a normal switching operation. The most severe type of disturbance to which 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 and the motor entirely cuts off the Bow of power between the machines. The generator output becomes zero in the pure-reactance circuits under consideration; the motor input also becomes zero. Because of the slowness of action of the governor of the prime mover driving the generator, the mechanical power input of the generator remains constant for perhaps i sec. Also, since the power and torque of the load on the motor are functions of speed, and since the speed cannot change instantly and changes by not more than a few per cent unless and until synchronism is lost, the mechanical power output of the motor may be assumed 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 a retarding torque on the motor. Consequently, the generator speeds up, the motor slows down, and it is apparent that synchronism will be lost unless the short circuit is quickly removed so as to restore syn.. chronizing power between the machines before they have drifted too

t Discussed 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 is discussed in Chapter XV, Vol. III.

6

THE STABILITY PROBLEM

far apart in angle and in speed. If the short circuit is on one of two parallel lines and is not at either end of the line, or if the short circuit is of another type than three-phase-that is, one-line-to-ground, line-toline, or two-line-to-ground-then some synchronizing power can still be transmitted past the fault, but the amplitude of the power-angle curve is reduced in comparison with that of the pre-fault condition. In some 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 stable during faults will depend not only on the system itself, but also on the type of fault, location of fault, rapidity of clearing, and method of clearing-that is, whether cleared by the sequential opening of two or more breakers, or by simultaneous opening-and whether or not the faulted line is reclosed, For any constant set of these conditions, the question of whether the system is stable depends upon how much power it was carrying before the occurrence of the fault. Thus, for any specified disturbance, there is a value of transmitted power, called the transient stability limit, below which the system is stable and above which it is unstable. The stability limit is one kind of power limit, but the power limit of a system is not always determined by the question of stability. Even in a system consisting of a synchronous generator supplying power to a resistance load, there is a maximum power received by the load as the resistance of the load is varied. Clearly there is 8. power limit here with no question of stability. Multimachine systems. Few, if any, actual power systems consist of merely one generator and one synchronous motor. Most power systems have many generating stations, each with several generators, and many loads, most of which are combinations of synchronous motors, synchronous condensers, induction motors, lamps, heating devices, and others. The stability problem on such a power system usually 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 each group swing more or less together; that is, they retain approximately their relative angular positions, although these vary greatly with respect to the machines of the other group. For purposes of analysis the 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 that are 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 to represent the synchronous machines of a power system by more than two equivalent machines. Nevertheless, qualitatively the behavior of the machines of an actual system is usually like that of a two-machine system. 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 because the 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-system stability 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 mechanical model of the vector diagram of Fig. 2 may be built of two pivoted rigid arms representing the E G and EM vectors, joined at their extremities by a spring representing the X I vector. (See Fig. 4.) Lengths represent voltages in the model, just as they do in the vector diagram. The lengths of the arms, E G and EM, are fixed in accordancewith the ti f t t · t 1 FIG. 4. A mechanical analogue of the assump Ion 0 cons an In erna system of Fig. 1. voltages. The length of the spring XI is proportional to the applied tensile force (for simplicity, we assume an ideal spring which returns to zero length if the force is removed). Hence the tensile force can be considered to represent the current, 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 of the length of the arm, the tensile force of the spring, and the sine of the angle between the arm and the spring. (More torque is exerted by the spring when it is perpendicular to the arm than at any other angle for the same tensile force.) Obviously, the torques on the two arms are equal 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 regarded as stationary, rather than as rotating at synchronous speed, just as we regard 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.

8

THE STABILITY PROBLEM

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 the angle between E and I is equal to the sine of the angle between E and XI.) The shaft power of the machines may be represented by applying additional torques to the arms. A convenient method of applying

FIG. 5. A mechanical analogue of the system of Fig. 1, suitable for representing 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 drum to each arm and to suspend a weight pan from a pulley hanging on a cord, one end of which is wound on each drum, all as indicated in Fig. 5. As weights are added to the pan in small increments, the two arms of the model gradually move farther apart until the angle 0 between them reaches 90°, at which position the spring exerts maximum torque. If further weights are added, the arms fly apart and continue to rotate in opposite directions until all the cord is unwound from the drums. The system is unstable. The steady-state power limit is reached at o = 90°. Although from 90° to 180° the spring force (current) continues to increase, the angle between arm and spring changes in such a way 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 third pivoted arm attached to an intermediate point of the spring (Fig. 6). The condenser maintains a fixed internal voltage. Since the condenser has no shaft input or output, no drum is provided on the third arm in the model. With the intermediate arm (representing the condenser) in place, the angle betwe.en the two outer arms (representing the generator and motor) may exceed 90° without instability, and the power limit is greater than before. The model can be used to illustrate transient stability by providing each arm with a flywheel such that the combined moment of inertia of the arm and flywheel is proportional to that of the corresponding synchronous machine together with its prime mover (or load). The drums 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 that the arms oscillate before settling down to their new steady-state positions. The angle between the arms may exceed 90° during these oscillations without loss of stability. If the increment of load is too large, the arms will fly apart and continue to rotate in opposite directions, indicating instability. This may happen even though the total load is less than the steadystate stability limit. The effect of switching out one of two parallel lines may be simulated by connecting the arms by two springs in parallel and then suddenly disconnecting one spring by burning the piece FIG. 7. A mechanical of string by which the spring is attached. ~nalogue of the effect of a line fault onf F" the 1power The effect of a fault on t hee li hne may bee si SlDlUt ByS em 0 19• • lated by suddenly pushing a point on the spring toward 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 of actual power systems of three or four machines, and the oscillations of the arms have been recorded by moving-picture cameras. 6 There are practical difficulties, however, in applying the model representation to a complicated system. The chief value of the model is to illustrate the elementary concepts of stability. Other methods of analysis are used in practice. Bad effects of instability. When one machine falls out of step with the others in a system, it no longer serves its function. If it is a

10

THE STABILITY PROBLEM

generator, it no longer constitutes a reliable source of electric power. If it is a motor, it no longer delivers mechanical power at the proper speed, if at all. If it is a condenser, it no longer maintains proper voltage at its terminals. An unstable two-machine system, consisting of motor and generator, may be compared to a slipping belt or clutch in a mechanical transmission system; instability means the failure of the 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 between wide limits. Thus instability has the same bad effect on service to customers' loads as does a fault, except that the effect of instability is likely to last longer. If instability occurs as a consequence of a fault, clearing of the fault itself may not restore stability. The disturbing voltage fluctuations then continue after the fault has been cleared. The machine, or group of machines,' which is out of step with the rest of 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 a fault automatically. As a rule, the best way to bring the machines back 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 should be split up into such parts that each part will have adequate generating capacity 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 unfaulted lines. Such false tripping may unnecessarily interrupt service to tapped loads and may split the system apart at such points that the generating capacity of some'parts is inadequate.] The trend in power-system design has been toward increasing the reliability of electric power service. Since instability has a bad effect on the quality of service, a power system should be designed and operated so that instability is improbable and will occur only rarely. Scope of this book. This book will deal with two different phases of the problem of power-system stability: (1) methods of analysis and calculation 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 the effect of varying different factors. Among these factors are system layout, circuit impedances, loading of machines and circuits, type of fault, fault location, method of clearing, speed of clearing, inertia of machines, kind of excitation systems used with the machines, machine reactances, neutral grounding impedance, and damper windings on machines. Since transient power limits are lower than steady-state power limits, and since any power system will be subject to various shocks, the most severe of which are short circuits, the subject of transient stability 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 the parallel operation of synchronous machines, it might be suspected that the problem appeared when synchronous machines were first operated in 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 were driven by direct-connected steam engines. The pulsating torque delivered by those engines gave rise to hunting, which was sometimes aggravated 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 governors also aggravated the hunting. Hunting of synchronous motors and converters was sometimes due to another cause, namely, too high a resistance in the supply line. The seriousness of hunting was decreased by the introduction of the damper winding, invented by Lelslanc in France and by Lamme in America. Later, the problem largely disappeared on account of the general use of steam turbines, which have no torque pulsations. Nearly all the prime movers in use nowadays, both steam turbines and water wheels, give a steady torque. A few generators are still driven by steam engines or by internal combustion engines. These, as well as 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 a significant problem. Before automatic voltage-controlling devices (generator-voltage regulators, induction feeder-voltage regulators, synchronous condensers, and the like) had been developed, the power

12

THE STABILITY PROBLEM

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 (both steady-state and transient) were well above the normally transmitted power. The development of automatic voltage regulators made it possible to increase generator reactances in order to obtain a more economical design and to limit short-circuit currents. By use of induction regulators to control feeder voltages, transmission lines of higher impedance became practicable. These factors, together with the increased use of generator and bus reactors to decrease short-circuit currents, led to a decrease in the inherent stability of metropolitan power systems. Stability first became an important problem, however, in connection with long-distance transmission, which is usually associated with remote 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 investment in these long-distance projects made it desirable to transmit as much power as possible over a given line, and there was a temptation to transmit normal power approaching the steady-state stability limit. In a few cases instability occurred during steady-state operation, and more frequently it occurred because of short circuits. The stability problem is still more acute in connection with long-distance transmission from a generating station to a load center than it is in connection with metropolitan systems. It should not be inferred, however, that metropolitan systems have no stability problems. Another type of long-distance transmission which has frequently involved a stability problem is the interconnection between two large power systems for the purpose of exchanging power to obtain economies in generation or to provide reserve capacity. In many cases the connecting ties were designed to transmit an amount of power 'which was small in comparison with the generating capacity of either system. Consequently, the synchronizing power which the line could transmit was not enough to retain stability if a severe fault occurred on either system. There was also considerable danger of steady-state pull-out if the power on the tie line was not controlled carefully. From about 1920 the problem of power-system stability was the object 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 stability were developed. Some of the important steps in analytical development were the following: 1. Circle diagrams for showing the steady-state performance of transmission systems. These diagrams consist of a family of circles, each of which is the locus of the vector power for fixed voltages at both sending and receiving ends of the line. The circles are drawn on 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 will carry in the steady state for given terminal voltages, as well as the relation between the power transmitted and the angular displacement between the voltages at the t\VO ends of the line. (Such diagrams are described in Chapter XV, Vol. III.) 2. Improvements in synchronous-machine theory, especially the extension of two-reaction theory to the transient performance of both salient-pole and nonsalient-pole machines. A number of new reactances were defined and used. (See Chapter XII, Vol. III.) More recently, the effect of saturation on these reactances has been investigated. (See Chapters XII and XV, Vol. III.) 3. The method of symmetrical components for calculating the effect of unsymmetrical short circuits. (See Chapter VI.) In this connection, 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 machine versus time). (See Chapter II.) 5. The equal-area criterion for stability of two-machine systems, obviating the more laborious calculation of swing curves for such systems. (See Chapter IV.) 6. The a-c. calculating board or network analyzer for the solution of complicated a-c. networks. (See Chapter III.)

The methods of analysis and calculation now in use are believed to be sufficiently accurate for determining whether any given power system in a given operating condition will be stable when subjected to a given disturbance. The calculations, however, are rather laborious ,vhen applied 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

14

THE STABILITY PROBLEM

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, "First Report 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, New York, McGraw-Hill Book Co., 1938. 4. Electrical Transmission and Distribution Reference Book, by Central Stauon Engineers 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 having 0.50 per unit reactance and negligible resistance. Assume that the angle ~ between the two machines varies from 0 to 360 by 15° steps, and calculate for 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 and voltage vectors, marking the values of 0 thereon. Also plot in rectangular coordinates current, power, and voltage, all as functions of o. 2. Draw the power-angle curve and discuss the condition for stability of two machines connected through series capacitive reactance which exceeds the internal inductive reactance of both machines. 0

CHAPTER II THE SWING EQUATION AND ITS SOLUTION Review of the laws of mechanics; translation. Since a synchronous machine is a rotating body, the laws of mechanics applying to rotating bodies apply to it. Review of these laws may be advantageous at this point. The laws of rotating bodies will be clearer if we first review the laws which apply to linear motion, or translation. TABLE 1 FUNDAMENTAL AND DERIVED QUANTITIES OF MECHANICS, ApPLYING PARTICULARLY TO 1"'RAN~LATION

Symbol

Defining Equation

Length Mass Time

x m

...

t

...

Velocity

v

v =-

[1]

Acceleration

a

a

=-

[2]

Force Momentum

M'

F = ma M' = mv

[3] [4]

Work

W

W =fFdx

Power

p.

P

Quantity

F

...

dx dt dv dt

= dW

dt

Unit and Its Abbreviation

Dimensions

meter (m.) kilogram (kg.) second (sec.)

L

ur:'

[5]

meter per second (m. per sec.) meter per second per second (m. per sec.") *newton (newt.) *newton-second (newtsec.) joule (j.)

[6J

watt (w.)

M T

Lr-2 MLT-2 MLT-l ML 2T- 2 ML 2 r- 3

* 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 recognized system of units for electrical work) are the meter, the kilogram, and the second. From these fundamental quantities and their units are derived other 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 dimen15

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), certain other equations giving relations between these quantities which are of interest are derived below. Substitution of eq. 1 into eq. 2 gives for the acceleration

[7] Substitution of eq. 2 and eq. 7 in turn into eq. 3 gives dv d2x F=m-=m-2 dt

[8]

dt

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

dM'

F=-

[9]

dt

Differentiation of eq. 5 with respect to x gives F= dW dx

[10]

Substitution for dW from eq. 10 into eq. 6 gives Fdx p = . - = Fv dt

[11]

Integration of eq. 9 with respect to t gives

M'

= JFdl

[12].

From eqs. 11, 3, and 4 we obtain

P = Fv

= mav = aM'

[13]

whence Af'

= ~a

[14]

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

W = JFdx = mJdV dx = mJdV vdt

=m

J:

d~

v dv

= !mv2 = !M'v

d~

[15]

ROTATION

17

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

8

(}=r

[16]

The unit of angle so defined is the radian. The dimensions of angle from the definition are length divided by length. For the present purpose, however, it is well to recognize lengths that are perpendicular to one another as having different dimensions, and to represent tangentiallengths by the symbol L as before and radial lengths by the modified symbol LB. From this viewpoint the dimensions of angle are LL R- 1• The definitions of angular velocity and angular acceleration follow by analogy to the corresponding definitions of linear velocity and linear acceleration. Angular velocity is

dB

e,.,=-

(17]

d~

and angular acceleration is

dw

d28

a = - = -2 dt dt

[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 rotating body at distance r from the axis of rotation are given by: = rO

[19}

v = rw

[20]

a = ra

[21]

8

The torque on a body due to a tangential force F at a distance r from the axis of rotation is T = rF [22] or, considering the total torque as due to the summation of infinitesimal forces, we may write

T

= frdF

[23]

The unit of torque could be called a newton-meter, but this name is not entirely satisfactory as it might imply a unit of work. Both work and

18

THE SWING EQUATION AND ITS SOLUTION

torque are the products of force and distance; but in the case of work the component of force parallel to the distance is used, and in the case of 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 L 2T-2 ; torque, M LLR T- 2 • For reasons that will appear later, the unit of torque may be called the joule per

radian. When torque is applied to a body, the body experiences angular acceleration a~ Each particle experiences a tangential acceleration a = 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 particle is dT = rdF = ~adm and that required for the whole body is

T

=

aJr

2

dm

= [a

[24]

Here [=

!TJdm

[25]

is known as the moment of inertia of the body, The m.k.s. unit" is the kilogram-meter2 (kg-m."). Note the analogy between T = I« for rotation and F = rna for translation. The work done in rotating a body through an angle dO by exerting a torque T may be found as follows: If the torque is assumed to be the result of a number of tangential forces F acting at different points of the body, T = ErF Each force acts through a distance

ds = rdO The 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.

19

Also

T= dW

[27]

dO

which is analogous to eq, 10, and which explains why the m.k.s. unit of torque 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,

dO P = Tdt

=

Tw

[28]

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

M = 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 (from eq. 29) kilogram-meters2-radians per second, or (from eq. 30) watts per TABLE 2 QUANTITIES OF MECHANICS ApPLYING TO ROTATION

Quantity Angle Angular velocity Angular acceleration Torque Moment of inertia Angular momentum

Defining Equation

Symbol

8

8

0=-

t»

w =-

a

Ol

=-

T

T

= rF

I

I = fr 2 dm

M

r

dB dt

dw

M

dt

=

Iw

Unit and Abbreviation

Dimensions

radian (rad.)

LLR-t

radian per second (rad. per sec.)

LLR-1r- 1

radian per second per second (rad. per sec,") joule per radian (j. per rad.) or newton-meter (newt-m.) kilogram-meter- (kg-m.P)

LL n- 1rr- 2

joule-second per radian (jsec. per rad.)

MLLRT-l

MLL R T- 2 MLR 2

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

20

THE SWING EQUATION AND ITS SOLUTION

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 rotation, 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

Relation

Translation

Rotation

8

8

8

f}

w

V

a

Of.

a = ra

m F

M' F = ma = JFd8 P = F» = M'a

W M' W

= m» = !mv 2

=tM'v dM' F=dt

1 T M

= r8 = rw

1 = Jr 2 dm T = rF M = JrdM'

T = I« W=fTdJJ P = Tw = MOt. M =Iw W = !Iw 2 =!Mw T =dM dt

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

101. = T

[32]

or

[33] Here T is the net torque or algebraic sum of all the torques acting on the machine, including shaft torque (due to the prime mover of a generator 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 synchronous and to asynchronous (induction) action.

Let

Ti

= shaft

torque, corrected for torque due to rotational

losses

and let

Tu

= electromagnetic torque.

Both of these are taken as positive for generator action, that is, with mechanical input and electrical output. They are negative for motor action, that is, with mechanical output and electrical input. The net torque, which produces acceleration,' is the algebraic difference of the 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-stability studies, however, the difference exists, and there is acceleration or retardation, depending on whether the net torque T a is positive or negative. Our problem is to solve eq. 33 so as to find the angular position fJ of the machine rotor as a function of time t. It is more convenient, however, to measure the angular position and angular velocity with respect to a synchronously rotating reference axis than with respect to a stationary axis. Hence let a = (J - Wtt [35] where WI is the rated normal synchronous speed. time derivatives, we get da dO -dt = dt - - WI and d 2a

d 2fJ

Then, if we take [36]

[37]

dt2 = dt2

With this substitution eq. 33 becomes 2

/d a = T dt2

[38]

which is unchanged in form. Writing the torque as in eq. 34, we have d2a I dt2 = T a

= Ti

-

Tu

[38a]

22

THE SWING EQUATION AND ITS SOLUTION

If we multiply this equation by the speed w, we obtain d25

M dt2

= P a = Pi

- Pu

[39]

where M = I w is the angular momentum. Pi = Tiw is the shaft power input, corrected for rotational losses. P u = T uW is the electrical power output, corrected for electrical losses. P a = 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. 38a because it involves the electrical power output of the machine, rather than the torque corresponding to this output. Equation 39 will be referred to hereafter as the swing equation. An equation of this form may be written for each machine of the system. The angular momentum M is not strictly constant because the speed w varies somewhat during the swings which follow a disturbance. In practical cases, however, the change in speed w before synchronism is lost is so small in comparison to the normal speed Wt that very little error is introduced by the assumption that M is constant. Hence it is customary in solving the swing equation to regard M as constant and equal to 1Wt, the value of angular momentum at normal speed. This value 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 P a will be in watts, 0 in radians, t in seconds, and M in watts per (radian per second per second) or joule-seconds per radian. In practical stability studies P a 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, per elec. deg.). If P a is in per unit, M must be in unit power seconds squared per electrical degree. For brevity the latter value of M will be 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 energy at rated speed. More often, however, the designer or manufacturer gives 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 chosen base 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 must calculate the value of the inertia constant M from the data. The formulas needed for this purpose will now be derived, beginning with the one giving the kinetic energy in terms of the moment of inertia and speed, 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 We

= speed in radians per second. = 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 degree. 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

[40]

But [41] 21rn w=-

60

[42]

and [43] By substituting eqs. 41 and 42 into eq. 40 and the result into eq. 43, we obtain 2 (211"n)2 746 -6 1 WR N = 550 X 10 X "2 X 32.2 X 60

= 2.31

X 10-10 WR 2n 2

[44]

By eq. 31 the kinetic energy may be written also as [45]

24

THE SWING EQUATION. AND ITS SOLUTION

Solving for M,

M=2N

[46]

We

But We

= 360/

[47]

Hence

M= 2N =~ 360/

[48]

180/

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

M = 1.28 X 10-12

WR2n 2

-1-'-

[49]

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

= 120/

[50]

eq. 49 may be written also as 2

M .:::: 1.54 X 10-10 WR n p

[51]

or as M

= 1.84 X 10-8 W~2j P

[52]

The inertia constant in megajoule-seconds per electrical degree may be calculated from eqs. 48, 49, 51, or 52. If per-unit power is to be used instead of megawatts, the value of M so obtained is merely divided by the base power in megavolt-amperes, giving what we may call a per-unit value of M. Another constant which has proved very useful is denoted by Hand is equal to the kinetic energy at rated speed divided by the rated apparent power of the machine.

H = stored energy in joules rating in volt-amperes stored energy in kilojoules rating in kilovolt-amperes stored energy in megajoules, N rating in megavolt-amperes, G

[53]

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

GH M

= 180j

[54]

THE INERTIA CONSTANT

25

The quantity H has the desirable property that its value, unlike those of M or WR 2, does not vary greatly with the rated kilovoltamperes and speed of the machine, but instead has a characteristic 10

...

~~

,

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

....... ~

~

"- "'-

.......

~ .....

r---. ~

~

...... .......

,

.......

-

~,

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

T 2

FIG.

o

20 40 60 80 Generator rating G (megavolt.amperes)

100

1. Stored energy of large steam turbogenerators, turbine included (from Ref. 1, by permission).

5

450·514 r.p.m.-~ ~ .....

200. 400 r. arn.....

~ ....... i\.

,/

lI'

~

"7

~

",

"

......,,- ~ ,..... .......

~

~

,. ...... ~ ,.... ...- --- ',", 138· 180 r.p.m.

~!;I' ", ~

j~ ~ ...... ~~

,.~

l..).......

~

~

~ ~

~

-e

~80·120

"",.",

---

~~

r.p.m,

~

1

FIG.

o

20 40 60 80 Generator rating G (megavolt- amperes)

100

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

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

THE SWING EQUATION AND ITS SOLUTION

26

generators and for large vertical-shaft water-wheel generators, respectively. In both cases the inertia of the prime movers is included, as it always should be. For' other machines the values of H may be taken from Table 4. It will be observed that the value of H is considerably higher for steam turbogenerators than for water-wheel generators, ranging from 3 to 10 Mj. per Mva, for the former and from 2 to 4 Mj. per Mva. for the latter. Average values are about 6 and 3, respectively. TABLE 4 AVERAGE VALUES OF STORED ENERGY IN ROTATING MACHINES·

Type of Machine

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

Synchronous motors Synchronous condensers t Large Small Rotary converters Induction motors

2.0 1.25 1.0 2.0

0.5

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

From 30 to 60% of the total inertia of a steam turbogenerator unit is that of the prime mover, whereas only 4 to 15% of the inertia of a hydroelectric generating unit is that of the water wheel, including water. t Inertia constant H, like the per-unit reactance of machines or transformers, may be expressed on either of two volt-ampere bases: (a) the machine rating or (b) a system base arbitrarily selected for a powersystem study. The value of H for a given machine varies inversely as the base, whereas per-unit reactance varies directly as the base.

t

EXAMPLE

1

Given the following information on a steam turbogenerator unit: Rated output Rated voltage Rated speed Moment of inertia Number of poles Rated frequency

85,000 kw. at 85% power factor 13,200 volts 1,800 r.p.m, 859,000 Ib-ft. 2 4

60 cycles per sec.

tSome data on moments of inertia of generators and their prime movers are given 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-M va. base. Compare the computed value of H with the typical value read from the curves 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 = 642 100

=

6.42 Mj. per Mva.

From eq. 48

M=-

N

180/

=

642 180 X 60

=

. 0.0595 Mj-sec. per elec. deg.

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

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

From Fig. 1, 1800-r.p.m. curve, at 100 Mva., H with the value of H computed above.

=

Point-by-point solution of the swing equation.

6.5, which agrees well

The swing equation,

a differential equation governing the motion of each machine of a

system, is [55]

where 8 = displacement angle of rotor with respect to a reference axis rotating at normal speed. M = inertia constant of machine. P a = accelerating power, or difference between mechanical input and electrical output after each has been corrected for losses. t = time.

28

THE SWING EQUATION AND ITS SOLUTION

The solution of this equation gives 0 as a function of t. A graph of the solution is known as a swing curve. Inspection of the swing curves of all the machines of a system will show whether the machines will remain in synchronism after a' disturbance. Many examples of swing curves 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 accelerating power of each machine depend upon the angular positions-and, to be more rigorous, also upon the angular speeds-of all the machines of the system. 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 the simplest case, which was considered in Chapter I, of one finite machine connected 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 possible that machine methods of solution could be applied (although they have not been yet) to solving the swing equations of multimachine systems. Point-by-point solution is the most feasible and widely used way of solving the swing equations. Such solutions, which are also called step-by-step solutions, are applicable to the numerical solution of all sorts of differential equations. Good accuracy can be attained, and the computations are simple.4- 7 In a point-by-point solution one or more of the variables are assumed either to be constant or to vary according to assumed laws throughout a short interval of time ~t, so that as a result of the assumptions made

SOLUTION OF THE SWING EQUATION

29

the equations can be solved for the changes in the other variables during the same time interval. Then, from the values of the other variables at the end of the interval, new values can be calculated for the variables which were assumed constant. These new values are then used in the next time interval. In applying the point-by-point method to the solution of swing equations, it is customary to assume that the accelerating power (and hence the acceleration) is constant during each time interval, although it has different values in different intervals. When this assumption is made, a formal solution of eq. 55 can be obtained which is valid throughout a particular time interval; and from the formal solution and the values of ~ and w at the beginning of the interval the values of aand 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, or difference between input and output, of each machine. The mechanical inputs are usually assumed constant, because of the slowness of governor action,§ but the electrical outputs are functions of the relative angular positions of all the machines of the system and can be found 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 swing curves 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 a knowledge of the positions and speeds at the beginning of the interval and the accelerating power assumed for theinterval, The second process is the computation of the accelerating power of each machine from the angular positions (and perhaps speeds) of all machines of the system. The second process requires a knowledge of network solution, a topic which is discussed in Chapter III. In the present chapter the emphasis is on the first process, namely, the solution of the swing equation proper. Two different point-by-point methods will be described. Method 1 is the more obvious, although the less accurate, of the two methods. In method 1 it is assumed that the accelerating power is constant throughout a time interval lit and has the value computed for the beginning of the interval. No further assumptions are made. If eq. 55 is divided by M and integrated twice with respect to t, P a being § See discussion of Assumption 1 on p. 43.

30

THE SWING EQUATION AND ITS SOLUTION

treated as a constant, we obtain successively

do

~at

-dt =w=wo+M

[58]

and [59]

These equations give, respectively, w, the excessof speed of the machine over normal speed, and 0, the angular displacement of the machine with respect to a reference axis rotating at normal speed. 00 and Wo are 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 in which P a is constant. We are particularly interested, however, in the values of 0 and w at the end of the interval. Let subscript n denote quantities at the end of the nth interval. Likewise let n - 1 denote quantities at the end of the (n - l)th interval, which is the beginning of the nth interval. At is the length of the interval. Putting At in place of t in eqs. 58 and 59 and using the appropriate subscripts, we obtain for the speed and angle at the end of the nth interval Wn = Wn-l On =

On-l

At

+M

P a(n-l)

+ At Wn-l +

[60] (At)2

2M P a(n-l)

[61]

The increments of speed and angle during the nth interval are aWn = Wn - Wn-l aOn == On -

On-l

At

=M =

[62]

Pa(n-l)

at Wn-l

+

(At)2

2M

Pa(n-l)

[63]

Equations 60 and 61, or eqs. 62 and 63, are suitable for point-by-point calculation. 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. 61 but for the preceding interval . (~t)2 On-l = On-2 + ~t W n- 2 + 2M P a(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 (P a(n-l)

-

P a(n-2»)

[65J

SOLUTION OF THE SWING EQUATION

31

But

and from eq. 62

Making these substitutions into eq. 65, we get dt

a6n =

a61l-l

== a8n _ l

+ at M Pa(1l-2) + +

(dt)2

2M (P a(n-l)

(~t)2

2M

(Pa(1l-l) -

+ p a(n-2»

Pa(1l-2») [66]

This equation, which gives the increment in angle during any interval in terms of the increment for the previous interval, may be used for point-by-point calculations in place of eqs. 62 and 63. The last term is the second difference of 5, which may be symbolized by t!A 25. The time interval ~t should be short enough to give the required accuracy, but not so short as to unduly increase the number of points to be computed on a given swing curve. Example 2 will throw some light on the effect of the length of interval upon the accuracy of the solution. EXAMPLE

2

If a synchronous machine performs oscillations of small amplitude with respect to an infinite bus, its power output may be assumed to be directly proportional to its angular displacement from the infinite bus. Because this case is known to result in sinusoidal oscillations, as may be verified readily by formal solution of the swing equation, it will serve well as a check on the accuracy of various point-by-point solu tions, Consider a 60-cycle machine for which H = 2.7 Mj. per Mva. and which is initially operating in the steady "tate with input and output of 1.00 unit and an angular displacement of 45 elec. deg. with respect to an infinite bus. Upon occurrence of a fault, assume that the input remains constant and that the output is given by (a)

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

32

THE SWING EQUATION AND ITS SOLUTION

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 expressed in electrical radians, is d~

M dt2 = P;

=

Pi - P«

2

= 1 - ;;: 0

(b)

and the initial conditions are

o= ~

~=o

and

4

(c)

dt

when t = O. The solution is 7r

7r

0=2-"4cOS

~

-

2

7rM t

(d)

TABLE 5 COMPUTATION OF

Swrso

CURVE FROM FORMAL SOLUTION

OF SWING EQUATION (EXAMPLE

t (sec.)

382t (deg.)

cos 382t

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

0.0 19.1 38.2 57.3 76.4 95.5 114.6 133.7 152.8 171.9 191.0 210.1 229.2 248.3 267.4 286.5 305.6 324.7 343.8 362.9 382.0

1.000 0.945 0.786 0.540 0.235 -0.096 -0.416 -0.691 -0.889 -0.990 -0.982 -0.865 -0.653 -0.370 -0.045 0.284 0.582 0.816 0.960 0.999 0.927

2)

45 cos 382t (deg.)

45.0 42.5 35.4 24.3 10.6 - 4.3 -18.7 -31.1 -40.0 -44.5 -44.2 -38.9 -29.4 -16.7 - 2.0 12.8 26.2 36.7 43.2 45.0 41.7

8 (deg.)

45.0 47.5 54.6 65.7 79.4 94.3 108.7 121.1 130.0 134.5 134.2 128.9 119.4 106.7 92.0 77.2 63.8 53.3 46.8 45.0 48.3

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

SOLUTION OF THE SWING EQUATION

33

The period of oscillation is given by

T=

21r

=

V2/1rM

1r

V21rM

(e)

The inertia constant is

M

= -H = -2.7 - - = 0.01432 unit· power sec. 2 per e1ec. ra d• 7rf

=

1r

X 60

-.!!- = 2.5 X 180!

10-4 unit power sec." per elec. deg. (or, simply, per unit).

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 solution becomes 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 ~Wn =

~On

ilt M P a(n-l)

= ilt Wn-l +

0.05 0.00025

= - - - P a(n-l) = (ilt)2

2M P a(n-l)

200 P a(n-l)

= O.05wn - 1 + 5P a (n.-l)

(g)

(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 the correct curve (curve 1) calculated from the formal solution, part a of this example. Comparison of curves 1 and 3 of Fig. 3 shows that curve 3, computed by point-by-point method 1, although having very nearly the correct period, increases in amplitude approximately 29% in each half-cycle, or 67% in each cycle of oscillation. II The accuracy is poor. It seems reasonable that the accuracy would be improved by the use of a shorter interval. Consequently, let us try ilt = 0.0167 sec. (one-third of the former value). Then in place of eqs. g and h we have

111.67

= (1.29)2.

dWn

= 66.7Pa(n -

L18n

= O.OI67w n - l

(i)

l)

+ 0.555Pa(n-l)

(j)

34

THE SWING EQUATION AND ITS SOLUTION TABLE 6 POINT-By-POINT COMPUTATION OF SWING CURVE

(METHOD

t

Pu

(sec. ) (p.u.)

- - 0.500 0+ 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

0.528 0.610 0.740 0.906 1.091 1.277 1.441 1.567 1.637 1.639 1.571 1.436 1.244 1.016 0.772 0.540 0.348 0.218 0.164

.. .

Pa (p.u.)

0.500 0.472 0.390 0.260 0.094 -0.091 -0.277 -0.441 -0.567 -0.637 -0.639 -0.571 -0.436 -0.244 -0.016 0.228 0.460 0.652 0.792 0.836

. ..

1,

tlt

liw

= 0.05 sEc.)(ExAMPLE 2) w

0.05w

(deg.Zsec.) (deg.Zsec.) (deg.)

100 94 78 52 19 -18 -55 -88 -113 -127 -128 -114 -87 -49 -3 46 92 130 158 167

. ..

0 100 194 272 324 343 325 270 182 69 -58 -186 -300 -387 -436 -439 -393 -301 -171 -13 154

0.0 5.0 9.7 13.6 16.2 17.2 16.2 13.5 9.1 3.4 -2.9 -9.3 -15.0 -19.4 -21.8 -22.0 -19.6 -15.0 -8.6 -0.6

. ..

5P a (deg.)

tl8 (deg.)

8 (deg.)

2.5 2.4 2.0 1.3 0.5 -0.5 -1.4 -2.2 -2.8 -3.2 -3.2 -2.9 -2.2 -1.2 -0.1 1.1 2.3 3.3 3.8 4.2

2.5 7.4 11.7 14.9 16.7 16.7 14.8 11.3 6.3 0.2 -6.1 -12.2 -17.2 -20.6 -21.9 -20.9 -17.3 -11.7 -4.8 3.6

45.0 47.5 54.9 66.6 81.5 98.2 114.9 129.7 141.0 147.3 147.5 141.4 129.2 112.0 91.4 69.5 48.6 31.3 19.6 14.8 18.4

------

...

. ..

The detailed calculations are not shown here, but they are similar to those of Table 6. The swing curve is plotted as curve 2 in Fig. 3. It is found that the amplitude error has been reduced to 9% in a half cycle or 21% in a cycle-about one-third of the corresponding errors with dt = 0.05 sec. These errors are still pretty large, and the labor of calculation is excessive because 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 be expected, the amplitudes increase faster than before. Besides, the period is noticeably lengthened.

Example 2 has demonstrated that swing curves calculated by method 1 are subject to a considerable cumulative error which manifests itself by 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 oscillation of. Example 2, during which the acceleration is diminishing. Such a curve of acceleration versus time is shown in Fig. 4. In method 1 the

SOLUTION OF THE SWING EQUATION

35

195 5

180 165

~

150

I~ /

ill ~

\

3

--"

I If

~

15

\

\.

~,

\ -.-

N'

rb'

f\

" j

,

~~~ ~ ~~

1 2 •••• • • 3 0--0--0 4 6--------6 5 0-- -0

30

[\

V'1 t\; \ \ l(i;jVT ~l\\ ~ \~ I~ 'f i\ Nb I.'') .~

120

)~

,

r " .... , ]'00,

/,/

135

~,

1~\"~

Formal solution I1t 0.0167 sec. /1t 0.05sec. !:At 0.10sec. 41t = 0.15 sec.

= = =

\\

\\ \

-15 -30

1

o

0.1

0.2

0.3

0.4 0.5 0.6 Time t (seconds)

"

"\\

o

-45

~3 ))

0.7

0.8

r\

\ 4 \

t

,

'''i~

s

0.9

1.0

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

36

THE SWING EQUATION AND ITS SOLUTION

acceleration during each interval ~t is assumed constant at its value for the 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 of angular position is likewise greater than the true advance. The second half cycle of oscillation thus begins with a a calculated amplitude greater than the true value; and, if no further errors occurred, the oscillation would continue with this amplitude. During t 4. True and assumed curves of ac- the second half cycle, howceleration versus time, point-by-point calou- ever, the acceleration is inlation, method 1. creasing, and during this time

FIG.

the assumed acceleration is always too small, that is, too negative. The calculated negative speed is therefore too great in absolute value, and the calculated retardation of 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 to be constant during a time interval can be eliminated by using the value of acceleration at the middle instead of the beginning of the interval. Referring to Fig. 5a, we see that .the acceleration at the middle of the interval is very nearly equal to the average acceleration during the interval, as is shown by the near equality of the triangular areas above and below the true curve. In method 2 this assumption is made: namely, that the acceleration during an interval is constant at its value calculated for the middle of the interval. Or, if we consider that the intervals used in calculation begin and end at the points of time at which acceleration is calculated, then the assumption must be restated somewhat as follows: The acceleration, as calculated at the beginning of a particular time interval, is assumed to remain constant from the middle of the preceding interval to the middle of the interval being considered. Let us, for example, consider calculations for the nth interval, which begins at t = (n - l)~t. (See Fig. 5.) The angular position at this instant is an-I. The acceleration an-I, as calculated at this instant, is assumed to be constant from t

= (n -

-!) Ilt

to t = (n -

!) Ilt

SOLUTION OF THE SWING EQUATION

37

Over this period a change in speed occurs, which is calculated as ~wn-l

at

= ~t·an-l = M

[67]

Pa{n.-l)

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, the change in speed would occur linearly with time. To simplify the a

(0)

(b)

~

In-t:n-t I I

U

On

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

8,,-1 __ .1.

~------

0,,-2

I

I

n-2 FIG.

(e)

.1.

n-l

n

~t

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

ensuing calculations, however, the change in speed is assumed to occur as a step at the middle of the period, that is, at t = (n - l)~t, which is the same instant for which the acceleration was calculated. Between steps the speed is assumed to be constant, as shown in Fig. 5b. From t = (n - l)L\t to t = nL\t, or throughout the nth interval, the speed will be constant at the value wn-i. The change in angular position during the nth interval is, therefore, L\on = ~t·wn.-l

[69]

and the position at the end of the interval is On =

as shown in Fig. 5c.

011.-1

+

~on

[70]

38

THE SWING EQUATION AND ITS SOLUTION

Equations 67 to 70 may be used for computation; but, if we are interested only in the angular position and not in the speed, we may use a formula for ~on from which w has been eliminated. Such a formula will now be developed. Substitution of eq. 67 into eq. 68, and of the result into eq. 69, gives ~on = ~t·wn_1

(~t)2

+M

Pa(n-l)

[71]

By analogy with eq. 69 [72] Substituting eq. 72 into eq. 71, ,ve obtain the desired formula ~on

=

~On-l

(~t)2

+M

Pa(n-l)

[73]

which, like eq. 66 of method 1, gives the increment in angle during any time interval in terms of the increment for the previous interval. The last term again may be symbolized by ~20. This formula makes the calculation of a swing curve very simple. If the speed is wanted, it can 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 some attention to the effects of discontinuities in the accelerating power P a which occur, for example, ,vhen a fault is applied or removed or when any switching operation takes place. If such a discontinuity occurs at the beginning of an interval, then the average of the values of P before and after the discontinuity must be used. Thus, in computing the increment of angle occurring during the first interval after a fault is applied at t = 0, eq. 73 becomes (J

[75]

where P aO+ is the accelerating power immediately after occurrence of the fault. Immediately before the fault the system is in the steady state; hence the accelerating power, Pao-, and the previous increment of angle, ~oo, are both equal to zero. If the fault is cleared at the beginning of the mth interval, in calculations for this interval one should use for Pa(m-l) the value !(Pa(m-l)- + Pa(m-l)+), where

SOLUTION OF THE SWING EQUATION

39

is the accelerating power immediately before clearing and Pa(m-l)+ is that immediately after clearing the fault. If the discontinuity occurs at the middle of an interval, no special procedure is

Pa(m-l)-

needed. The increment in angle during such an interval is calculated, as usual, from the value of P a at the beginning of the interval. The reasons for these two rules should be clear after study of Fig. 6. If the discontinuity occurs at some time other than the beginning or the

~(m-1)+ I I I

I I I

I I

I I I Pa(m-l)-I I I I I I I I I I I I I I I I I

I

,

m-l

m

n-l

n

t ~

FIG. 6. The handling of discontinuities of accelerating power Pain point-by-point calculation, method 2. A is a discontinuity occurring at the beginning of the mth interval. PaCm-l) = ! (PaCm-l>+ P aCm-1)- ) is used in calculating ~8m. B is a 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 before and after the discontinuity should be used, but the need for such a refinement 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 an interval. EXAMPLE

3

Work the problem of Example 2 by method 2 of point-by-point calculation, using various values of time interval (~t). Compare the swing curves thus obtained with those obtained by method 1 in Example 2. Solution. The point-by-point calculations will be based on eqs. 70 and 73. The value of M from Example 2 is 2.5 X 10-4 per unit. The aceelerat-

40

THE SWING EQUATION AND ITS SOLUTION

ing power is given, as before, by

P a = Pc> P ,u

~ = 1 -900 -

(a)

For t1t = 0.05 sec., which will be tried first, we have (t1t)2 =

M

(0.05 )2 = 10 2.5 X 10-4

and eq. 73 becomes (b)

The detailed calculations are carried out in Table 7. Note that, for t

= 0, the instant of occurrence of the fault, the values of P u and of P a are

entered 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 to right across a line until the column headed "lOPa" is reached; then diagonally downward and to the right to the"value of 0 in the line below the starting point; thence to the "Pu" column in the same line; and so on. Each new value of flo is found by addition of the previous value of flo and the value of lOP a between the old and new values of dO. Each new value of 0 is found in similar fashion by addition of the preceding value of a and the value of f16 between the old and new values of o. Comparison of the values of 0 in Table 7 with the corresponding values in Table 5, computed from the formal solution, shows a maximum discrepancy of 1.00 in the first cycle of oscillation. The agreement of the point-by-point solution with the formal solution is very satisfactory. Similar calculations were made for at = 0.10 sec., 0.15 see., 0.20 sec., and 0.25 sec. The details of calculation are not shown here, but the resulting swing curves are plotted in Fig. 7. The curves numbered 3, 4, and 5, and the symbols used for them, correspond with the curves of Fig. 3 for like values of dt. The formal solution is not plotted in Fig. 7 but agrees with curve 3 to the accuracy of plotting. It may be noted in Fig. 7 that, as t1t is increased, the calculated swing curves remain sinu~oidal and correct in amplitude, but that their periods decrease. Even when the points are so far apart that they are not adequate to determine the shape of the curve (as for ~t = 0.25 sec., giving fewer than 4 points per cycle), a sine-wave can be drawn through them. The curve for at = 0.1 sec. would be considered accurate enough for engineering use in a stability study. In actual studies the usual periods of oscillation are from 0.5 to 2 sec., and at is commonly taken as 0.05 or 0.1 sec. The value 0.1 sec. usually suffices and, in addition to requiring only half as many steps as 0.05 sec., has the advantage that multiplication or division 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 periods of the curves calculated by methods 1 and 2 in Exam ples 2 and 3, respectively.

TABLE 7 POINT-By·POINT COMPUTATION OP SWING CURVE

(METHOD

t (sec.)

00+

o avg.

P«

(p.u.)

1.000 0.500

...

2, at = 0.05

Po, (p.u.)

sEc.)(ExAMPLE

10Po, (deg.)

...

0 0.500 0.250

...

2.5

3)

f:.8 (deg.)

·

"

· .. · ..

8 (deg.)

...

. ..

45.0

2.5

0.05

0.528

47.5

4.7

0.472

7.2 0.10

0.609

54.7

3.9

0.391

11.1 0.15

0.731

0.269

2.7

65.8 13.8

0.20

0.885

0.115

1.2

79.6 15.0

0.25

1.052

-0.052

94.6

-0.5 14.5

0.30

1.214

-0.214

109.1

-2.1 12.4

0.35

1.350

-0.350

121.5

-3.5 8.9

0.40

1.450

-4.5

-0.450

130.4

4.4 0.45

1.497

-0.497

134.8

-5.0 -0.6

0.50

1.492

-4.9

-0.492

134.2 -5.5

0.55

1.430

-4.3

-0.430

128.7 -9.8

0.60

1.320

118.9

-3.2

-0.320

-13.0 0.65

1.177

-1.8

-0.177

105.9 -14.8

0.70

1.011

-0.011

91.1

-0.1 -14.9

0.75

0.847

0.153

76.2

1.5 -13.4

0.80

0.697

0.303

62.8

3.0 -10.4

0.85

0.582

0.418

4.2

52.4 -6.2

0.90

0.514

4.9

0.486

46.2 -1.3

0.95

0.499

0.501

5.0

44.9

3.7 1.00

0.540

48.6

4.6

0.460

8.3 1.05 1.10

0.632

...

56.9

3.7

0.386

...

.. .

41

12.0 68:9

42

THE SWING EQUATION AND ITS SOLUTION

150 135 120

.

105

:s '0

~ c

75

...

u

1i. Ul

45

\ ~.\~\ \. \'

\ \\' \ \~

\. \~\

,f

v

1: 60 E

\\ ~

\.\'

I~

...

...

~

II

90

:rl

~

\

h~

'0

13

~.~

lit: V /

-;;;

..~

.~fI

.-'I

t~

.' 54

p)

6

/ 1\ ' \ ' . \'l . " .~~

/

V

3

0 3Cl o L\t =0.05 sec. 46---------6 L\t=0.10 sec. 5 0 - - - - - 0 L\t =0.15 sec. 6 - - - L\t =0.20 sec. 7 - - - - L\t =0.25 sec.

i;i 30 15

o o

/ 7:

0.1

I I

I

I

I

0.2

0.3

0.4

0.5 0.6 0.7 Time t (seconds)

0.8

0.9

1.0

1.1

FIG. 7. Swing curves calculate d by point-by-point method 2 with various values of At (Example 3).

TABLE 8 PERCENTAGE ERRORS IN AMPLITU DE AND PERIOD OF SINUSOIDAL O SCILLATIO N C OMPUTE D BY POlNT-B y-POlNT METHODS

1

AND

2

WITH VA RIOUS V ALUES OF

t

Per cent Error in Amplitude At

(sec.)

0.0167 0.05 0.10 0 .15 0.20 0.25

Points per True Cycle

56 19 9.4 6.2 4.7 3.7

Per cent Error in Period

At End of One Cycle

At Half Cycle

Method 1

Method 2

Method 1

Method 1

+21 +68 + 191 +300

...

+9 +29 + 67 +100

+1 +2 +5 +11

. .. . ..

0 0 0 0

0

... ...

. .. ...

Method

e

... -0.4 -2 -5 -9 -15

ASSUMPTIONS MADE IN STABILITY STUDIES

43

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

Assumptions commonly made in stability studies. In the foregoing discussion of the solution of the swing equation, it was tacitly assumed that the accelerating power P a was known at the beginning of each interval, and, indeed, the equation could not be solved unless P a were known. Since P a = Pi - P u , both the input Pi and the output P u must be known. In the determination of Pi and P u the following assumptions are usually made: 1. The input remains constant during the entire period of a swing curve. 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 constant electromotive force (voltage behind transient reactance). 5. The mechanical angle of each machine rotor coincides with the electrical 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 input is unchanged. The input to a generating unit is controlled by the governor of its prime mover. The governor will not act until the speed change exceeds a certain amount (usually 1% of normal speed), depending on the adjustment of thE{governor, and even then there is a time lag before the governor changes the input. During swings of the synchronous machines the percentage change in speed is very small until after synchronism is actually lost. Therefore governor action 'is usually not a factor in determining whether synchronism will be lost, and, accordingly, it is neglected. 2. The output (electric power) of a synchronous machine consists of a synchronous part, depending on the relative angular positions of all the machines of the system, and an asynchronous part, depending on the relative angular speeds of all the machines. The asychronous part 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 or asynchronous power is discussed in Chapter XIV, Vol. III.

44

THE SWING EQUATION AND ITS SOLUTION

3. The network connecting the machines is not strictly in the steady state during swinging of the machines, both because of sudden circuit changes, such as application or removal of a fault, and because of the more gradual change of phase of the electromotive forces due to the swinging. However, as the periods of oscillation of the machines are relatively long (of the order of 1 sec.) in comparison to the time constants of the network, the network may be assumed, without serious error, to be in the steady state at all times. Steady-state network solution is presented in Chapter III. 4 and 5. The assumption that each machine can be represented by a constant reactance in series with a constant voltage, and the assumption that the mechanical position of the rotor coincides with the phase angle of the constant voltage, are not entirely correct. As a rule, however, they do not lead to serious error in the determination of whether 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 but will be postponed to Chapter XII, Vol. III. EXAMPLE

4

A 25-Mva. 6o-cycle water-wheel generator delivers 20 Mw. over a doublecircuit 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-axis transient reactance of the generator is 0.30 per unit. The transmission circuits have negligible resistances, and each has a reactance of O~20 per unit on a 25-Mva. base. The voltage behind transient reactance of the generator is 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 circuit and is cleared in 0.4 sec. by the simultaneous opening of the circuit breakers at 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-point calculations, we must know the inertia constant of the generator and the power-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 fault has 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 is found by series and parallel combinations to be

Xl = 0.30

· = 0.40 per unit + 0.20 2

ASSUMPTIONS MADE IN STABILITY STUDIES

45

When the fault is cleared, one of the parallel circuits is disconnected, making the reactance 0.20 = 0.50 per unit X 3 = 0.30

+

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

= 0.30 + 0.20 + 0.30 X

X

0.20 0.10

2

= 0.50 + 0.60 =

1.10 per unit

The values of reactance of the other two branches are not needed because these branches, being connected directly across the constant-voltage power 0.20

(a)

1.10

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

(b)

8. (a) Reactance diagram of a system consisting of a generator A supplying power 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 circuit at 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. FIG.

sources, have no effect on the power outputs of the sources, although they increase the reactive power outputs. The same is true of the O.IO-per-unit reactance at B. The power-angle equation for the circuit of Fig. 8b is the same as it would be with these three shunt branches omitted. Power-angle equations. The power-angle equation, giving the output PuA of generator A as a function of the angle 0 between voltages EA and EB, is C· • ~ P uA = EAE - -B SIn u = sIn u~ X where C has the following values:

Before fault,

C1 = E A E B = 1.03 X 1.00 = 2.58 Xl 0.40

THE SWING EQUATION AND ITS SOLUTION

46

= 1.03 X

= 0.936

During fault,

O2 = EAEB X2

After clearing,

0 3 = EAEB = 1.03 X 1.00 = 2.06

1.00 1.10

x,

0.50

Inertia constant. By eq. 54

M = GH = 1.00 X 2.76 180/ 180 X 60

= 2.56 X 10-4

er unit P

I nitial conditions. The power output of generator A before the fault was given as 20 Mw., which on a 25-Mva. base is 0.80 per unit. The initial angular 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

= 0.310

2.58

~ =

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)

(at)2 - - P a(n-l)

M

~~n

= Pi =

=

0.800 -

(0.05)2

2.56 X 10-

4 P a(n-l)

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

an = 8n - l Pun

Pu(n-l)

=

+ .1on

C sin On

Pu(n-l)

= 9.76 P a(n-l)

per-unit power elec. deg.

elec. deg.

elec. deg, 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 9 POINT-By-POINT COMPUTATION OF SWING CURVE (EXAMPLE

C

t (sec. )

(p.u.)

00+

2.58 0.936

o RVg. 0.05

sin 8

0.310 0.310

Pu

Pa

(p.u.)

(p.u.)

0.800 0.290

4)

A8 9.76P a 8 (elec. deg.) (elec, deg.) (elec. deg.)

0.000 0.510 0.255

18.1 18.1 2.5

0.470

4.6

2.5 H

0.352

0.330

20.6 7.1

0.10

H

0.465

0.435

0.365

3.6

27.7 10.7

0.15 0.20 0.25 0.30

"

0.621

"

0.779

"

0.904

II

0.977

0.581

0.219

2.1

38.4 12.8

0.730

0.070

0.7

51.2 13.5

0.846

-0.046

-0.4

64.7 13.1

0.915

-0.115

-1.1

77.8 12.0

0.35

II

1.000

0.936

-0.136

-1.3

89.8 10.7

0.400.40+ 0.40 avg,

" 2.06

0.983

"

100.5

0.920 2.024 1.472

-0.672

-6.6

1.995

-1.195

-11.6

4.1 0.45

II

0.968

104.6 -7.5

0.50

H

0.992

2.045

-1.245

-12.1

97.1 -19.6

0.55

II

0.976

2.010

-1.210

-11.8

77.5 -31.4

0.60

H

0.721

1.486

-0.686

- 6.7

46.1 -38.1

0.65

H

0.139

0.286

0.514

5.0

8.0 -33.1

0.70

II

-0.424

-0.874

1.674

16.3

-25.1 -16.8

0.75

u

-0.668

-1.376

2.176

21.2

-41.9 4.4

0.80 0.85

"

-0.609

II

-0.227

-1.255

2.055

20.0

-37.5 24.4

-0.468

1.268

12.4

-13.1 36.8

0.90

II

0.402

0.828

-0.028

-0.3

23.7 36.5

0.95 1.00

"

0.868

"

0.996

1.788

-0.988

-9.6

60.2 24.9

2.052

-1.252

-12.2

85.1 12.7

1.05

97.8 47

THE SWING EQUATION AND ITS SOLUTION

48

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

5

Determine the critical clearing time of a three-phase short circuit at the middle of one transmission line of the system of Example 4, assuming that the circuit breakers at both ends of the line open simultaneously. 250 r---...,.----r---r--..-----r----,..--,.----,----nl~-.....

i

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

i

"0

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

I

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

...

ac

1... t Ot----+-----I---t----+-~r+_-___+_#_~...f--~-__I__+_-~ .9

50

cu

- 50 " - _ . - i . - _ - a_ _- - ' - _ - - - ._ _ o 0.1 0.2 0.3 0.4 0.5

~_---.l.

0.6

_'"

0.7

0.8

_'

0.9

1.0

Time (seconds) FIG.

9. Swingcurves for the system of Fig. 8 (Example 4).

Solution. First, the swing curve for a sustained short circuit is computed and plotted. The computations of Example 4 may be used up to t = 0.4 sec.; but for t > 0.4 sec. new computations must be made by use of the power-angle equation for the faulted condition. The swing curve is plotted in 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 the swing curve for O.5-sec. clearing, starting from the O.5-sec. point of the computation for a sustained fault. Computation of only two points or so suffices to show that the system is stable with 0.5-8ec. clearing. A longer clearing time is then tried. Several more swing curves may need to be calculated, until two clearing times, differing slightly, are found, for one of which 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 lies between 0.6 and 0.65 sec. The critical clearing angle lies between 136° and 147°. Detailed computations are not given here because they are similar to those of Example 4. It should be stated again that usually only a few points need be calculated on each swing curve, departing from the curve for a sustained fault at the assumed clearing time.

6 In 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 that the system is stable. Solution. As was mentioned in Example 4, about 0.5 sec. after occurrence of 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 corresponding relative 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 frequency of 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 speed would have a negligible effect in determining whether the system was stable, although it might have some effect on the ensuing oscillations.

EXAMPLE

REFERENCES 1. 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 of point-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.

50

THE SWING EQUATION AND ITS SOLUTION

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 from Fig. 2. 2. In Example 4 plot on one sheet curves of angular position, relative speed, and acceleration versus time. 3. Work the problem of Example 4, using ~t = 0.1 sec. instead of 0.05 sec., 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 to the sending end of one transmission line of the system of Example 41 Assume simultaneous opening of both breakers. Which fault location is more 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 the two parallel transmission circuits of Example 4 as a normal switching operation. 8. The system of Example 4 is modified by the addition of a radial feeder to the bus of station A. If a three-phase fault occurs on this feeder at a point separated from the bus by an impedance of jO.10 per unit and at a time 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 0.4 sec., is the shock to the system greater or less than that caused by the fault at 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 one transmission line switched out. Make the following assumptions: For each load the terminal voltage of generator A is adjusted to 1.00 per unit, and the power factor is unity. Then assume a very small disturbance to occur during which the excitation voltage behind synchronous reactance of 0.90 per unit remains constant. 10. Find the transient power limit of the system of Example 4 with a three-phase fault at the middle of one transmission line cleared in 0.4 sec. by the simultaneous opening of the circuit breakers at both ends. Assume the pre-fault conditions of generator terminal voltage and power factor to be as described in Probe 9. 11. The system of Example 4 is modified by making the impedance of each jO.20 per unit instead of 0 jO.20 per unit. Find transmission line 0.10 the initial angular position of generator A with respect to the infinite bus if the power output of A is 0.800 per unit at a voltage of 1.00 per unit and a power factor of 1.00. Also find the power output and accelerating power immediately 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 the substitution

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 with = 21r/k. Show that the corresponding difference equation

a period T

(c)

in which an denotes the value of 8 at a time t

= n~t, has a similar solution (d)

where 2. k~t '-' = - sln-1 ~t

(e)

2

14. By using eq. e of Probe 13 calculate the percentage error in period of a simple harmonic oscillation as a function of the number of points per cycle when 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 of Fig. 10, where A is a 60-cycle generator; B-G, a synchronous-synchronous 60-to-25-cycle motor-generator set (frequency changer); and D, a 25-cycle synchronous motor (or generator with such local load that it is equivalent to a motor for stability analysis). State specifically all respects in which the procedure differs from that for a system in which all machines are electrically in parallel.

52

THE SWING EQUATION AND ITS SOLUTION

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

Reactances in per unit on a system bas6 Lines A-B C-D

Branch from B to fault

Machines A B C D

0.40 0.20 0 .20

0.10 0.20 0.15 0.15

Inertia constants (H) on the system base A B

8.0 1.0

C

1.0

D

6.0

Initial conditions Load 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 physical interpretation.

CHAPTER III SOLUTION OF NETWORKS

Determination of the swing curves of the several synchronous machines of a power system, as has been mentioned, consists of two processeswhich must be carried on alternately: (1) the solution of the swing equation of each machine, giving the change in angular position during 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 positions of all machines are known. In Chapter II attention was focussed on the solution of the swing equation, and the network solution required for Example 4 was purposely made as simple as possible by using a two-machine reactance system. The solution of the swing equation was found to be very simple. It becomes no more complicated if the number of machines is increased except that a similar calculation has to be made for each machine. The solution of the network, on the other hand, rapidly becomesmore laborious as the number of machines is increased. The methods of network solution required for stability studies will be outlined in this chapter. The impedance diagram (positive-sequence* network). Before the network can be solved, however, it must first be established. The starting point is~ usually a one-line diagram of the power system to be studied, showing generators, synchronous condensers and other large synchronous machines, reactors, transformers, transmission lines, and loads. The diagram is usually limited to the major transmission system. As a rule, distribution circuits and small loads are not shown in detail but are taken into account merely as lumped loads on substation busses. From the one-line diagram there is prepared a diagram in which 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. The values of impedance of individual apparatus are commonly given either in actual ohms or in per unit (or per cent) based on the rating of the individual 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 common voltage 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 physical quantities, such as current, voltage, power, and impedance, are expressed as decimal fractions or multiples of base quantities. When the apparatus 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; and base voltage on the low-voltage side, as 11,000volts. Base current on the high-voltage side is 1,000/66 = 15.15 amp.; on the low-voltage side 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 of base voltage of that side to base current of that side, namely, 66,000/15.15 = 4,360 ohms. The base impedance of the low-voltage side is 11,000/90.9 = 121.5 ohms, which is (11,000/66,000)2 = 1/36 as great as the high-voltage value. These values are the load impedances required on the high- and the low-voltage sides of the transformer to load it fully at rated voltage. If the transformer is carrying a load of 500 kva. at 0.8 power factor and 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 both sides 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 by only a small impedance drop. The per-unit currents of both windings are 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 other side, although different in ohms. For example, if the short-circuit impedance of the above-mentioned transformer is 0.070 per unit, it is 0.070 X 4,360 = 305 ohms referred to the high-voltage side, or 0.070 X 121.5 = 8.5 ohms referred to the low-voltage side. Per-unit impedances based on the apparatus rating are nearly the same for all apparatus of the same general design though of different voltage and kilovolt-ampere ratings, whereas the impedances in ohms vary greatly with the rating. Hence typical values of impedance are easier to tabulate, remember, and compare if expressed in per unit than if expressed in ohms. For conversion of self-impedances from ohms to per unit and reverse, the following formulas are used:

PER-UNIT QUANTITIES

· d · h B ase impe ance In 0 ms

=

55

base voltage in volts . 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'' base power in megavolt-amperes

[1]

. . impedance in ohms · d · h Per-unit Impedance = b 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

[2]

X base impedance in ohms

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

[3]

For mutual impedances between circuits of different base voltages we have, instead of eq. 1, · d base voltage 1 X base voltage 2 [4] B ase unpe ance = b ase power Although the foregoing relations have been derived for a single-phase circuit (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 the line-to-neutral voltage) and the base power is the three-phase value (3 times the power per phase). ' These factors cancel in the expression (voltage) 2 jpower. For use in a power-system study, all impedances and other quantities must be expressed on a common system base. An arbitrary base power is selected; for example, 100 Mva. for a large power system, 50 or 20 Mva. 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 through transformers, however, the ratio of base voltages should equal the turns

56

SOLUTION OF NETWORKS

ratio of the transformer for the particular tap used, even if the turns ratio differs from the ratio of nominal voltages. As eq. 2 shows, when per-unit impedances are changed from one base power (megavoltampere base) to another without change of base voltage, they vary directly as the base power. Per-cent quantities are 100 times the corresponding per-unit quantities. The decimal point must be watched in the multiplication or division of per-cent quantities; for example, the product of a per-cent impedance 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 major portion of the transmission system. If this is done, impedances of the lines of the major portion are left expressed in actual ohms, whereas those of lines of other voltages are referred to the selected voltage by multiplying them by the square of the turns ratio of the intervening transformer. Impedances given in per unit on a given power base are converted to ohms at the chosen voltage by eq. 3. Representation of large synchronous machines. The representation of various circuit elements in the impedance diagram will now be discussed briefly. References will be given to sources of more complete information. Each generator or other large synchronous machine is commonly represented for transient-stability studies by its direct-axis transient reactance Xd' in series with a constant-voltage power source (Fig. 1). The armature resistance of large machines is usually negligible. The reactances of machines already built Of designed can usually be obtained from the manufacturer. The method of obtaining Xd' from a short-circuit oscillogram is described in Chapter XII, Vol. III. For machines of minor importance average values of Xd' may be taken from Table 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 series arms represent the leakage impedances, and the shunt arm, the exciting impedance. As a rule, the exciting current can be neglected in stability studies; if this is done, the T circuit is reduced to series impedance Z (Fig. 2), equal in value to the short-circuit, or equivalent, impedance of the transformer. The value of this impedance is frequently given on the name plate. It can be measured by means of a short-circuit test. The resistances of large transformers are usually negligible,

REPRESENTATION OF TRANSFORMERS

57

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

"

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

FIG. 1. Representation of a generator or other large synchronous machine in a transient-stability study. Xd' is the direct-axis transient reactance, E/ is the internal voltage "behind" this reactance, and V is the terminal voltage.

TABLE 1 TYPICAL REACTANCES OF TWO-WINDING POWER TRANSFORMERS

(500

1- OR 3-PHASE, 25 OR 60 c.e.s., 55°0. (From Ref. la by permission of Westinghouse Electric Corporation)

KVA. PER PHASE AND OVER,

High Max. Per-cent Voltage, Low Reactance Line to Line Voltage ~ (kv.) (kv.) Min. Max.

0- 15 16- 25 26- 37

15 15 15

38- 50 51- 73

25 25

74- 92 93-115 116-138 139-161

162-:96 197-230

34.5 34.5 37 50 50 50

4.5 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 10.0 11.0

RISE)

Max. Per-cent Max. Per-cent Reactance Low Reactance Low Voltage ~ Voltage ~ (kv.) Min. Max. (kv.) Min. Max.

7.0 8.0 8.0 9.0 10.0 10.5 12.0 13.0 14.0 15.0 16.0

25 37

46 69 73 73

92 92 92

6.5 9.0 7.5 10.5 8.0 11.0 8.5 12.5 9.0 14.0 9.5 15.0 10.5 16.0 11.5 17.0 12.5 18.0

73 92 115 138 161 161

9.0 10.0 10.5 11.5 12.5

14.0 15.5 17.0 18.0 19.0

14.0 20.0

Three-circuit transformers are represented, with exciting current neglected, by Y circuits (Fig. 3) such that the resistance of each branch is the resistance of the corresponding winding, and the sum of the reactances of each pair of branches equals the short-circuit reactance between the corresponding pair of windings with the remaining winding open." 5 Thus [5a] Xl X2 = X l 2 [5b] Xl X3 = X 13 X 2 + X 3 = X 23 [5cJ

+ +

SOLUTION OF NETWORKS

58

whence

Xl

X2 X3

c=

!(X12 + X 13

-

= !(X I 2 + X 23 = !(X I 3 + X 23 -

X 23 )

[6a]

X 13 )

[6b]

X 12 )

l6c]

Frequently the reactance of one arm of the Y, as determined by eqs. 6, is found to be negative. t For estimating purposes the reactance X l 2. between the two main windings of a three-winding transformer may be considered equal to the reactance of a two-winding transformer having the same kiloR 1 + jX 1 h V volt-ampere and voltage ratings 2 -a: (Table 1). The reactances X 13 3 1-J~ ~ and X 23 between the main windv} ings and the tertiary winding cannot be estimated accurately because of the wide range in these reactances 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 represented 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 an autotransformer 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

t

I

rated high voltage - rated low voltage rated high voltage Autotransformers can have very small reactances if the voltage ratio is 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 capacitance is required for representing a small negative reactance. This difficulty may be avoided either by combining the negative reactance with whatever positive reactance may be in series with it or by representing it by capacitive reactance in series with inductive reactance that is smaller than the capacitive reactance by the amount of the required negative reactance.

TRANSMISSION LINES AND CABLES

59

Symmetrical banks of three single-phase transformers, that is, Y or A connections of identical transformers, are represented in the singlephase impedance diagram as one single-phase transformer, the per-unit impedance of which, based on the kilovolt-ampere rating of the bank, is equal to the per-unit impedance of one transformer on its own kilovolt-ampere rating. Three-phase transformers ~re represented in similar fashion. The no-load phase displacement between primary and 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. See Refs. Ib, 8, and 9. Transmission lines and cables are represented by their nominal or equivalent 7r circuits.l! In the nominal 7r circuit (Fig. 4a) the series branch has an impedance equal to the total series impedance Z per phase of the line, and the

z

II

y tanh(...{ZY'2 ~

2"

~

FIG.

VZf/2

~

(a) Nominalr

~

(b) Equivalent 1r

4. Circuits for representing transmission lines and cables.

shunt branch at each end has an admittance equal to half the shunt admittance Y of the line to neutral. Here Z = zl and Y = st. where l is the length of the line and z and yare the series impedance and shunt admittance 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 of power lines is negligible. In the equivalent 1r circuit (Fig. 4b), the impedance of the series branch is that of the nominal 1T' multiplied by a correction factor, sinh

vZ¥

Eiv'ZY -

VZY

E-iy'Zf

2VZY

1 + Zy =

(Zy)2

(Zy)3

6+120+ 5040+··· ,

[7a]

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

1r

SOLUTION OF NETWORKS

60

multiplied by another correction factor, tanh (V"ZYj2) v'ZY/2

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

= -i Y-ZY-2- - - - - - - (E

/

+ E-iVZ'i/2) v'ZY/2

The correction factors can be found easily by use of Woodruff's charts. 12 The equivalent 1r is an exact representation of a line at a particular frequency, whereas the nominal 1r is an approximation, the use of which is justified only if the correction factors are nearly 1, as is true if Zy = zyl2 « 1, hence for short lines. As a general rule, the nominal e is accurate enough for 60-c.p.s. open-wire lines not more than 100miles long. Longer lines may be represented by an equivalent 1r or by two or more nominal 1r'S in tandem. Very short lines have negligible shunt admittance and can be represented by their series impedance only. Each part of a composite line, such as an underground cable in series with an aerial line, may be represented by a 1r. By means of one or more y-~ conversions, the circuit can be reduced to one 1r, usually an unsymmetrical one. When several lines are connected tothesame bus, the shunt capacitances at that end of all the 1r circuits of these lines are in parallel and may be replaced by one capacitance from the bus to neutral. If one or more lines 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 circuits are preferable because they require only one series impedance per line and one shunt capacitance per bus, whereas the T circuits require two series 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 or other record, is multiplied by the constants per unit length. The constants of cables are best found from the manufacturer, but they' may be estimated from published tables. 1d • 14 The constants of aerial lines can be found accurately from tables. They depend upon (1) the frequency, (2) the size and kind of conductor, and (3) the spacing between conductors. The series resistance depends principally upon size 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 the other, upon frequency and spacing. t The shunt capacitive reactance can be found in like manner. Reference Ic has suitable tables for finding series and shunt reactances by this method. For estimates the series inductive reactance of a 60-c.p.s. aerial line may be taken as 0.8 ohm per mile, and the' shunt capacitive susceptance, as 5 micromhos per 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 with the time of day and of year and from one year to another, one or more particular load conditions must be selected for study; for example, the annual peak load and the minimum load may be taken. Estimated future loads are often used. The connected generator and transformer capacity and other features of system operation will depend upon the loads assumed. Loads are assumed to be lumped on the busses of major stations and substations. They should be expressed as 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 V2

[8]

where V is the voltage across the load. Ordinarily, the voltages used in calculating the load admittances must be estimated. Later, the admittances of the important loads can be revised, if so desired, by using 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 the line impedances between the tap and the ends. Representation of faults. A three-phase short circuit is represented by 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 shunt elements as potential transformers, lightning arresters, and coupling tThis method of tabulating reactance of lines was originated by W. A. Lewis and was published first in Ref. 2d. If the three distances between conductors are unequal, the geometric mean of the distances or the arithmetic mean of the corresponding reactance terms is used.

62

SOLUTION OF NETWORKS

capacitors have impedances so high that they are considered open circuits. Representation of remote portions of the system. In studies of part of a large interconnected system it is neither necessary nor feasible to represent in equal detail all stations and lines of the entire system. Outlying portions can be represented by equivalent circuits. A remote portion connected at only one point to the portion being studied can be replaced, according to Thevenin's theorem, by an impedance in series with a constant-voltage power source. The impedance is found by 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 case the equivalent circuit is like that of a generator (Fig. 1). The voltage of the source may be assumed as 1 per unit. A remote portion of the network connected at two points to the' portion being studied can be represented by a power source and a Y circuit. The inertia constant assigned to the generator of each of these equivalent circuits should equal the total of the inertia constants of all the machines therein (for reasons discussed later in this chapter) and, in the absence of more definite information, may be calculated from an average value of H (Figs. 1 and 2, Chapter II) and the known aggregate generating capacity. If the total inertia is large and loosely coupled to the portion of the network being studied, it may be con.. sidered infinite with little error. The assumption of infiriite inertia for a remote portion of the system renders the computation of swing curves unnecessary for the equivalent machines representing such portions.

Check list of data required for transient-stability study.

Generators Name-plate rating Direct-axis transient reactance Kinetic energy at rated speed (or moment of inertia and speed), including prime mover (see Chapter II) Voltage limits Grounding§ 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 rating Equivalent resistance and reactance Turn ratio normally used Available taps and tap-changing equipment Connections§ Grounding§

Transformers, multicircuit Items listed for two-circuit transformers Resistance of each winding Reactances between each pair of windings Kilovolt-ampere rating of each winding

r

a common kilovoltampere base

Reactors Resistance Reactance

Transmission Zines and coble« Series resistance and inductive reactance of each line section Shunt capacitance or capacitive susceptance of each line section, except those for which its effect is obviously negligible Zero-sequence resistance, reactance, and capacitance§

Synchronous condensers and large synchronous motors Same data as for generators; however, kilovolt-ampere rating of condensers may be different for lagging current than for leading current

Loads Active and reactive power for each condition to be studied Voltage limits

Circuit breakers (see Chapters VIII and XI, Vol. II) Interrupting time I· · d Reclosing }-f t t- Ie-poIetime or three-poIe opera t-Ion 1 au oma IC rec osmg IS use Smg

Protective relays (see Chapter IX, Vol. II) Types and settings §Required only for studies including ground faults, See Chapter VI.

64

SOLUTION OF NETWORKS

The alternating-current calculating board. Mter the required data have been compiled and the impedance diagram has been drawn, the network must be solved in order to find the magnitudes and initial phase positions of the internal voltages of all the synchronous machines. Subsequently, during the determination of swing curves, the network must be solved repeatedly to obtain the power output of all the machines when their angular positions and internal voltages are known. Networks may be solved by two principal methods, the calculatingboard method and the algebraic method. Calculating-board solutions are 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 on the board on a single-phase basis. One hundred volts on the board may 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 represent 50 Mw. three-phase (16.7 Mw. per phase) on the system. The scales for voltage and power are then 1 : 1,000 and 1 : 500,000, respectively. The scales of current and of impedance follow automatically from the scales 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 = 289 amp. on the power system, and 100 ohms on the board would represent 57,700/289 = 200 ohms per phase (Y basis) on the system. The alternating-current calculating board, also known as a network analyzer or network calculator, consists of an assemblage of adjustable resistors, reactors, and capacitors; a number .of sources of a-c. voltage adjustable both.in phase and in magnitude; provisions for connecting the 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 must be so designed that insertion of them does not appreciably change conditions in the network. A modern a-c. calculating board is shown in Fig. 5. In using the a-c. board in a stability study, the network is first set up to scale, the generators being represented ordinarily by their direct-axis transient reactances in series with power sources. Next the phase and magnitude of the voltages of the several power sources and the impedances of the loads are adjusted to represent the desired normal or pre-fault operating condition, with respect to machine outputs, terminal voltages, and similar factors. The fault is then applied at 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 unsymmetrical fault, it is represented either by a shunt impedance or by a connection between the sequence networks (to be discussed in Chapter VI). The voltages of the power sources (representing voltages behind transient reactance, which are ordinarily assumed constant) are readjusted, if necessary, to restore them to their pre-fault values. It is desirable, however, that the power sources have such good voltage regulation that readjustment is unnecessary, or at least small. Then the power

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

output of each machine is read by means of a wattmeter. From these readings the accelerating power is computed, and from it the angular change of each machine during the first time interval is found by methods already discussed. The angular changes so found are then reproduced on the calculating board by turning the angle-adjusting dials of the powersources. After this has been done, a new set of power readings is taken, and so on. The clearing of a fault or any other switching 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

66

SOLUTION OF NETWORKS

board in a stability study, it should be obvious that a direct-current calculating board could not be used for the same purpose. The correct representation of phase relations is all-important. Alternating-current calculating boards are owned and operated by a number of leading electrical manufacturers, consulting engineering firms, engineering colleges, and, both publicly and privately owned electric-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 of 480 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 is 2.5 watts, and the base impedance is 1,000 ohms. All adjusting dials and instrument scales are marked in percentages of the base quantities. Generator units. There are a number of power sources for supplying single-phase voltages of adjustable phase and magnitude, which are used principally for representing generators. Each generator unit consists of two machines, a phase shifter and a single-phase induction voltage regulator. The phase shifter has a three-phase stator winding, which receives 220-volt, 480-c.p.s. power from a motor-generator set, and a single-phase rotor winding, which delivers power at constant voltage and adjustable phase to the voltage regulator. The phase shifter provides adjustment of phase through an unlimited angle, and the voltage regulator affords adjustment of voltage magnitude from oto 250% of the base voltage. The phase and magnitude adjustments do not affect each other, nor are the adjustments greatly affected by load. 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 rated current at zero power factor is only 10 • This small voltage regulation is very convenient in stability studies where the voltages of the power sources represent known constant voltages behind transient reactances of synchronous machines because the angular positions of the several machines can be varied without the necessity of readjusting the voltage magnitudes. Furthermore, the phase and magnitude of each voltage 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 power system to be studied. **The nominal voltage is approximately the average voltage (to neutral) of the board circuits, and the nominal current is approximately the average current in the principal board circuits.

GENERAL ELECTRIC A-C. BOARD

67

adjustment simulates adjustment of the excitation of the represented machine, and the phase adjustment corresponds to adjustment of the governor of the prime mover. Advancing the phase increases the power 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-impedance units, each consisting of resistors and three tapped reactors permanently connected in series. These units are used for representing the impedances of transmission lines, transformers, reactors, and machines, and other low impedances. Information on the range of resistance and reactance, size of steps, current capacity, and so on of the line-impedance units and other circuit elements is given in Table 2. As the impedance units constitute a large portion of the bulk and weight of any a-c. calculating board, it is important that they be made small. On the other hand, it is desirable that the reactors have a reasonably low ratio of resistance to reactance so that they can represent sufficiently well transformers, reactors, and machines having a low ratio of resistance to reactance. Furthermore, the reactance should be almost independent of the magnitude of current in the reactor. These requirements for reactors tend to make them large. The problem of conflicting requirements has been solved by the use of low base power, a higher frequency than 60 c.p.s., and high-grade magnetic material for the cores of the reactors. The resistance of the line reactors ranges from 3% of the reactance when the reactance is set at 70% of the base value to 8% at a setting of 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 the line-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 parallel with one another. The parallel connection affords easier adjustment of 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

Ranget

Size of Steps]

.

. .

.••...

85--115%..... 69.5-130.5%.

/0····

0-110% c. 60001

~,

1\

............

1% } 0.5%......

10·

125% voltage

} 500% current or 125% voltage

} 500% current or 125% voltage

.

·

· .. 500% current and 125% voltage

101 d ± 10······················ 5OO%currentan 125%voltage

± 101 /0- - - - _. _. ·

.

± 1% at standard current 5% . . . . . . . . { ±3% at any current

} lot

.

T apped resistor ±1 % Smooth. . .. { Smooth resistor ±3 %

*The number of circuit elements varies on different analyzers built by the same manufacturer. 22 tIn percentage of base quantities (50 volts, 0.050 amp., 1,000 ohms, 0.001 mho).

15... . ..

8 A utotransf ormers...... { 12

Mutual reactors.. . . . . .

0-1,610% ....

........ 0-1,605% ....

50

. {OO 4 C apacltors.... . . . . . . . .

Reactor

Load impedances Resistor

{ ±1% at standard current .... } ±3~ t o a any current .. . ... 500% current or 125% voltage

0.2%... . . .

........ 0-81% .......

Reactor ............

1,000% current

Maximum Safe Valuest

Smooth .... {Tapped ..... · .. } 500%0 current or 125%0 vo1tage S mooth resistor . t or ±1% reSIS ± 3~0.... ...

Voltage regulation about 2% ..

Tolerance

........ 0--51% .......

150

12 ................... .......... . ........ 0-250% ...... Smooth . ....... Unlimited .... Smooth

Number"

ANALY'ZEB

Line impedances ...... Resistor............

Generators. . . . ....... Voltage ............ Phase..............

Element

CIRCUIT ELEMENTS OF GENERAL ELECTRIC

A-C. NETWORK Installed at Schenectady in 1937

TABLE 2

~ 00

~

~

~

t?-j

Z

~

Z

o

.-3 ......

~

00

oe-

00

c

GENERAL ELECTRIC A-C. BOARD

69

and transmission lines. They are used also for representing capacitors and, in steady-state studies, for synchronous condensers. Autotransformers of two designs are provided, one of which will buck or boost the voltage by 1% steps up to a maximum of 15% of the primary voltage; the other, by 0.5% steps up to 30.5%. They are used for representing transformers whose actual turns ratios differ from the nominal ratios on which the circuit impedances are calculated. Thus they are valuable for representing transformers of adjustable ratio, and one of them must be included in any loop around which the product of all transformer ratios is not unity; for example, if two systems of different voltage are connected through two transformers of unequal turns ratio. The position of zero buck or boost represents the nominal voltage ratio. Mutual transformers of ratio 1:1 are used chiefly for representing mutual 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 adjustable mutual 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 circuit units can be readily connected together in any desired way to form a network representing a particular power system. On the board shown in Fig. 5 each circuit unit terminates in a pair of flexible cords, each bearing a single-pole plug. Insertion of these plugs into horizontally adjacent jacks of the vertical jack panels forms an electrical connection between them. The "busses" formed by groups of adjacent plugs can be placed in the same relative positions on the jack panel as on the connection 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 not adjacent. As an indication of polarity, one cord of each circuit unit is colored green, the other yellow. tt A positive wattmeter reading denotes power flow from the yellow to the green cord; a negative reading, from the green to the yellow. Like rules hold for the sign of the varmeter reading and the direction of lagging reactive power. Jumper circuits are used to form zero-impedance ties which can be metered. They are useful in .representing bus-tie breakers, terminals of 7r circuits representing transmission lines, connections between the sequence 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 plugged to the ground bus. The green cord connects to the movable tap.

70

SOLUTION OF NETWORKS

Instruments. The system of making measurements is one of the most crucial features of an a-c. calculating board. It must satisfy the following requirements: 1. It must be possible to insert the instruments anywhere in the network without appreciably altering conditions in the network. The volt-ampere burden of the instruments determines the minimum permissible base power and thus fixes the size and weight of the entire board. 2. Because of the large number of readings required in most power-system studies, the instruments should be fast in response and easily readable so that an operator can take readings all day without eyestrain. 3. The instruments must be suitable for a variety of measurements: scalar values of voltage, current, active power, and reactive power; and vector values of voltage and current in both rectangular and 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 power sources. The burden of these instruments is not a matter of such concern 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 the burden of these instruments on the network negligibly small despite the low power level used there, they are supplied with current and voltage through two stabilized-feed-back vacuum-tube amplifiers.P For good legibility each instrument has an 8-inch translucent scale on which a spot of light serves as a pointer. The response of these instruments is rapid. Their over-all, accuracy, including the voltage divider, the current shunts, and the amplifiers, is within ±O.5% of full scale. 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, are used in association with the instruments: A key switch is provided for each circuit unit on the board. When any of these switches is thrown, all three instruments are connected to the selected circuit unit and indicate simultaneously. The voltage-selector switch determines whether the voltage furnished to the voltmeter and wattmeter comes from (a) across the

GENERAL ELECTRIC A-C. BOARD

71

circuit unit, as for reading line drops and losses, (b) from one side of the 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 cord that can be plugged to any desired point of the network, or (e) from the instrument phase shifter. The current-selector switch makes it possible to select current from the circuit unit, total ground current, or current from the instrument phase shifter. The reversing switch is thrown so as to make the wattmeter read upscale. The reading is recorded with + or - sign, according to ,the position of this switch. The watt-var switch is used to make the wattmeter-varmeter read either active or reactive power. This switch is also used in reading 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-voltage ranges: 20% and 40%. Vector measurements. To measure current in rectangular form, the wattmeter-varmeter is furnished with the current to be measured and with 100% voltage from the instrument phase shifter, which is set at the desired reference angle (usually 0°). The inphase and quadrature components of current are equal to the watt and var readings, respectively. To read current in polar form, the phase shifter is turned until the var reading is zero; then the angle is read on the phase-shifter dial. The watt reading is the current magnitude. Vector values of voltage are read in the same way as currents except that the wattmeter is supplied with the voltage to be measured and with 100% current from the phase shifter. Instruments on the generator units. In addition to the centrally located master instruments, there are a wattmeter-varmeter and a voltmeter at the terminals of each power source. The wattmetervarmeter has voltage ranges of 150% and 300% and current ranges of 100% and 400%. It is changed from a wattmeter to a varmeter by substituting a capacitor for the usual resistor in the potential circuit. The voltmeter can be connected to any point in the network where the generator is expected to regulate voltage. Power supply. Three-phase, 480-c.p.s., 440-volt power is supplied to the board from a remote-controlled motor-generator set consisting of a motor, a 3-kva. alternator, an exciter, and a phase balancer. The phase balancer is used for obtaining balanced three-phase voltages

72

SOLUTION OF NETWORKS

(required so that the magnitude of the voltage of each power source shall be independent of its phase) in spite of the essentially single-phase loading. Between the generator and the power sources on the board is a bank of autotransformers with taps for normal voltage and for 50% and 25% of normal voltage. A tap-selector switch on the instrument desk enables the operator to reduce the voltage of all power sources simultaneously if this is necessary to avoid overloading any of the board circuits, for example, during the period that a fault is represented on the power system. Arrangement. The instrument desk, at which the operator is seated in Fig. 5, is at the center, flanked on both sides by the jack panels and the tables holding the flexible cords. Above the instrument panel and jack panels are the generator units. The wings on each end of the board (only one of which shows in Fig. 5) contain the line- and loadimpedance units, capacitors, and autotransformers, each in a removable drawer. The board is so arranged that the units requiring the most frequent adjustment are closest to the operator. Description of Westinghouse A-C. Network Calculator.19,20 The Westinghouse calculating board differs from the General Electric board, 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., respectively. Each generator unit consists of two three-phase induction voltage regulators and a phase shifter. The three-phase secondary windings of the two induction regulators are in series, and the rotors are mechanically coupled so as to Primary give them equal but opposite angular displacements when the ConFIG. 6. Voltage vector diagram of one voltage handle is turned. phase of the three-phase voltage regulators sequently, the three-phase outof a generator unit of the Westinghouse A-C. put voltages of the pair of Network Calculator. O-L is the locus of regulators are constant in phase the output voltage. as 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, 0 to 200, and 0 to 400 volts. ttEarly models had only one three-phase voltage regulator ahead of the phase shifter. Consequently, the voltage adjustment changed the phase also, requiring a compensating movement of the phase shifter.

WESTINGHOUSE A-C. BOARD

73

For representing the internal reactance of each generator there is a

special reactor having a lower resistance than the line and load reactors. Each generator is provided with a voltmeter, an ammeter, and a wattmeter. These instruments have auxiliary pointers that can be pre-set manually to show the desired operating conditions. The voltmeter can be switched to read either the internal voltage or the terminal voltage of the generator. The impedance units are marked in ohms at 440 c.p.s. (equal to per cent impedance when 100 ohms 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 a load adjuster, by means of which the active and reactive power of each load can be adjusted simultaneously without changing the power factor of the load. Usually the settings of the load resistor and reactor are calculated for an estimated bus voltage (say, 100%). If the actual load voltage differs from the estimated value, the load unit does not consume the correct value of power, because power in a constant impedance varies as the square of the voltage. By means of the autotransformer, however, the voltage on the resistor and reactor can be adjusted to 100% even though the bus voltage has another value, It is merely necessary to set the autotransformer tap to correspond to the measured bus voltage. The taps are in 1% steps. The function of the load adjusters on the board is analogous to that of feeder-voltage regulators on an actual power system. The masterinstruments include a set for measuring scalar values and a set for measuring vector values. The scalar instruments consist of a voltmeter, an ammeter, and a wattmeter-varmeter. They are supplied with voltage and current through two negative-feed-back vacuum-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 a voltmeter (ranges 50 and 500 volts). Each is a two-coil dynamometer instrument, one coil of which is inserted in the network, and the other, consuming the greater part of the power for actuating the instrument, is fed from one or the other phase of the two-phase secondary winding of an instrument phase shifter. For reading current or voltage in rectangular form, the phase shifter is set on the desired reference position (usually 0°), the field coils of the ammeter and voltmeter are connected to one secondary winding of the phase shifter, and the "inphase" component is read; then the field coils are connected to the other winding, and the "quadrature" component is read. For reading

OOW

.

. .

.

.

.

0-300 ohms...

0-399 oh IDS...

ohms

Smootht...........

mg ..

±1!% of setting ..

Be

tt·

:H!% of setting ..

Voltage regulation under 2% .

Tolerance

(o-IOO) 0.2 ohm.... } 11 01 f { (100-399) 1 ohm... . ± 2~/0 0

{~='i ~~::::::}

. . . . . . . . . . . . . . . . . .. Smooth Smooth

Size of Steps

. . . .

±1!% of setting ..

~

volts

3.0 amp.

l

1

220 volts (0-9 .8 ohms) 3.0 amp. (10-99.8 ohms) 2.3 amp. { (100-399 ohms) 1.2 amp. 3.0 amp. or 220 volts 220 volts (0-8 ohms) 3.0 amp. (10-98 ohms) 2.3 amp. (100-298 ohms) 1.2 amp. (300-998 ohms) 0.48 amp. (1,000-3,990 ohms) 0.23 amp. 3.0 amp. or 220 volts 220 volts 220 volts, 3.0 amp.

3.0 amp. or

400 volts, 3 amp.

Maximum Safe Values

*Replacing calculator built in 1929. Has two instrument desks. tThe number of circuit elements varies on different calculators built by the same manufacturer. 22 ~Reactance values are obtained by a combination of fixed air-gap reactors in series with a variable air-gap reactor.

Smoothj O.OI,."f 0.5% .

(0--1,000 ) 2 ohm .... } 11 01 f tt· 990 ......... . .. O-3 , o hm s . { (1,000-3,990) 10 ohm ± 2/0 0 se lng ..

48

152

()-4()()

Reactor. . . . . . . . . . . . . . . . . . . . . . .. 0-2,400 ohms. Capacitors. . . . . . . . . . .. 48. . . . . . . . .. Q-4. 10 p.f. . . .. Autotransformers...... 36 80-124.5%... Mutual reactors....... 36..........

Resistor

Reactor Load impedances

Resistor

Range

A-C. NETWOlUt CALCULATOR Installed at East Pittsburgh in 1942*

ELEMENTS OF WESTINGHOUSE

18.... . . . . .. . . . . . . . . . . .. 0-400 volts. .. . Unlimited....

Numberf

R/X) 18

Line impedances. . . . ..

Gen. reactors

Generators. . . . . . . . . .. Voltage. . . . . . . . . . . . Phase... .. . . . . .. . ..

Element

CmcUIT

TABLE 3

~

~ 00

;.0

o

~

~

trj

Z

~

Z

o

~

~ ~

00

~

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 instrument reads zero, and the angle is read on the phase-shifter dial; the field coils 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 element by means of a selector consisting of a compact group of twenty-five pushbuttons. When the operator pushes one button in each of three columns, relays are actuated that connect the selected circuit unit to the 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 of the board. Procedure for using calculating board. After the data on the power system to be studied have been assembled (see check list on p. 62 of this chapter), the following steps must be taken: 1. Choose a suitable base or scale for representing the power system 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 of the calculating board. The current scale should be nearly as large as it can be without subjecting any of the board units to overcurrent (see maximum safe currents in Tables 2 and 3). If the current scale is too small, current and power cannot be read accurately in lightly loaded circuits. A satisfactory scale can be obtained by expressing system quantities on a megavolt-ampere base in accordance with the following tabler'" §§Early models of the Calculator had the vector instruments only. Active and reactive power was found by multiplying the voltage magnitude by the inphase and quadrature components of current.

76

SOLUTION OF NETWORKS Maximum Megavolt-Amperes in Any Circuit or Generator o to 30 30 to 60 60 to 120 120 to 300 300 to 600 600 and above

Megavolt-Ampere Base to be Used for System Quantities 10 20 40 100 200 400

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 not critical because of the wide range of impedances and instrument scales on the board. To check the suitability of the base selected, the lowest line impedance and highest load impedance (lightest load) may be compared with the range of available board impedances, and the currents of the most heavily loaded circuits should be compared with the current-carrying capacities of the assigned units. If necessary, two or more generators or line- or load-impedance units can be connected in parallel to carry larger currents, or two load-impedance units can be connected in series to represent very small loads. If there are a few unusually heavy loads, it is better to represent them by two or more load units in parallel than to use a higher impedance scale. Loads smaller than 2% of the chosen megavolt-ampere base should be combined with other loads or neglected, because their influence on the problem is negligible. If, during a transient-stability study, some circuits are overloaded part of the time, for example, while a fault is on the system, the base voltage of the board may be reduced temporarily, impedances being left 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 on the Westinghouse board, or 5% on the General Electric board) should be subtracted from the desired resistance to obtain the proper setting of the series resistor. A circuitdiagram should be prepared on which are entered the impedances to be used and the assigned circuit units on the calculating board. Each circuit should be assigned a positive. direction, which either is indicated 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 the upper or left-hand end of each circuit is positive. Negative ends of shunt 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 putting plugs to be connected together into adjacent jacks. On the Westinghouse board the positive end of each circuit element has a red cord, and the negative end, a black cord; on the General Electric board the yellow cord is positive, and the green cord, negative. Setting impedances and capacitances to the values shown on the circuit diagram is a simple operation. Positions of series-parallel switches 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 not discovered 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 occurrence of a fault. First, all generators should be set at the same phase angle and at zero voltage. The power may then be turned on, and the generator voltages may be raised gradually to their normal value, whilethe generator currents are watched for abnormal values caused by mistakes in setting up the network. The loads and terminal voltages of the generators are then adjusted approximately to the desired values by means of the phase and voltage controls. Bus voltages at other points are adjusted as desired by autotransformers or by capacitors or generator units representing synchronous condensers. If a generator unit is used, its power must be adjusted to zero. 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 load adjusters are available, they are used for this purpose; otherwise the resistor and reactor (or capacitor) must be readjusted with the aid of the master wattmeter-varmeter. Generating units, condensers, and other elements should now be given a final, accurate adjustment, which is checked with the master instruments. Limitations of system. Sometimes the desired operating conditions cannot be obtained on the board. Unless a mistake has been made in connecting or setting the board units, such failure indicates that the desired condition is one that is impossible on the actual power system. After adjusting the operating conditions, one should determine whether all bus voltages are within suitable limits and whether any equipment is overloaded either on the board or on the power system that it rep-

78

SOLUTION OF NETWORKS

resents. If any conditions are unsatisfactory, consideration should be given to changing either the operating conditions or the network itself, for example, by adding new lines, transformers, generators, condensers, reactors, or phase-shifting transformers. Readings. A complete set of readings is recommended at the outset in order to check the network. Voltage and phase angle of each bus and active and reactive power at each end of each circuit are usually read and recorded. The readings are multiplied by suitable factors to convert them to system values which are then marked on a one-line circuit diagram. The following checks may be made: The algebraic sum of the active power of all circuits on the same bus should be zero; likewise the algebraic sum of the reactive power. The difference of power at the two ends of a line may be taken to see whether the [2 R loss is reasonable; a similar test may be made for reactive power and 12X. Perhaps the most valuable check, however, is for a person familiar with the power system being studied to see whether all results appear reasonable. During the progress of a transient-stability study the only readings needed are the power outputs of the generators. Sometimes, however, additional readings are taken, for example, to investigate the operation of protective relays. Changes of connections. The power to the board should be turned off before changing any connections. After any important change of conditions, such as the addition or removal of a short circuit, the phase .and magnitude of the internal voltages of the generators should be checked and readjusted if necessary. Fault close to generator. When a short circuit is placed on or near the terminals of a generator, the current of that generator is high, and the power factor is low. These conditions are not conducive to good accuracy 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 measured current 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 a two-machine system algebraic solutions are usually so simple that nothing is gained by using a calculating board, and for three- or four-

79

POWER-ANGLE EQUATIONS

machine systems algebraic solutions are still feasible and may be resorted 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. The rectangle represents a passive (impedance) linear network of any form, to which there are applied n external voltages, representing the internal voltages of the synchronous machines. One side of each voltage source is connected 2 3 n to a common or neutral terminal Network (0); the other sides are cono nected to various points of the network, numbered from 1 to n. FIG. 7. An electric power network The voltages E 1, E 2 , • • • En com- represented by a passive linear network monly represent voltages behind having external e.mJ.'s connected to it. transient reactances, in which case the transient reactances are to be regarded as parts of the net.. work inside the rectangle. Let the currents flowing into the network at the terminals 1, 2, 3, · · · n be calledIj, 12 , 13 , • • • In. The vector power (P + jQ) supplied to the network by any machine may be found by 'multiplying the conjugate of the vector voltage by the vector current. II II In symbols,

...

PI

+ jQI = EII I

P 2' + jQ2

[9al

= E2I 2

[9b]

[9cl where the bar over the E's denotes their conjugates. Furthermore, since the network has been assumed linear, the currents supplied by the various machines may be written as linear functions of the applied voltages, thus: n

II

= YIIEI + YI2E2 • • • + YlnEn = E YlkEk

[lOa]

k=l n

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

SOLUTION OF NETWORKS

80

n

In

= YnlEl + Yn2E2 · · · + rnnEn = k=l E

YnkEk

[10c]

The V's are complex numbers known as the terminal self- and mutual admittances 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 of what happens to eqs. 10 if all the applied voltages except one are made equal to zero. By letting all voltages except E l become zero, we obtain:

= Y21E 1,

11 = YllE l, 12

•••

In = YnlEl

[11]

permitting us to define the Y's as follows: Y1l is the ratio IlfE l when all voltages except E 1 are zero; Y21 is the ratio 12 / E 1 when all voltages except E l are zero; and so forth. By letting all voltages except E 2 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 vector power output of each voltage source, and substitute into them eqs. 10. We obtain PI + jQ1 = E1Y11E 1 + E1Y 12E2 • • • + E1Y1nEn [12a]

[12b]

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

s, = E 1 fJJ,

E2

= E 2 &,

...

En

= En&

[13]

or, for the conjugates,

E1

= ElL -81,

E2

= E 2 / -82,

...

En =En /

-8n

[14]

Also let Y 11 = Y 11 / S11 ,

Y 12

= Y12 / Sl 2 ,

Y 21 = Y 2l / S21 ,

Y 22

= Y 22 / S22 [15]

DETERMINATION OF TERMINAL ADMITTANCES

and so forth. PI

Equations 12 become

+ jQI = E12YI1/ e l l + EIE2Y12Le12 -

+ E1EnYlnL81n n

=

E

E 1Ek Y l k / 8 1k

-

81

k=1

P 2 + jQ2 = E 2EIY21/ n

=

L

+ 82· • • 81 + 8n

81

+ 8k

[16a]

+ 81 + E 22Y22 / 8 22 · • • + E 2En Y 2n / e2n - 82 + 8n 82

e 21 -

E 2EkY2k/ e2k

-

k=l

and so forth.

81

82

+ 8k

[16b]

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

= E 12Yl l cos 811 + E 1E2Y12 cos (8 12 -

+ ElEnYln cos (Oln -

n

=

L

k=l

E1EkYI k

cos

(elk -

81

+

81 82) • • • 81 + 8n)

+ 8k)

[17a]

P 2 = E 2EIY2l cos (821 - 82 + 81) + E 22Y22 cos 822 • • • E 2En Y 2n cos (e2n - 82 + 8n ) n

= E

k=l

+

E 2Ele Y 2k cos (8 2k

-

82

+ 8le)

[17b]

and so forth. Because these equations give the electric power output of each machine (power input to the network) as functions of the angular positions of all the machines, they may be called power-angle equations. Inasmuch as this chapter deals primarily with networks, the symbol P has been used for electric power instead of P u, which was used in Chapter II with the same meaning. Algebraic solution of networks: determination of terminal admittances. Equations 17 suffice for finding the power output of all machines 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) except the terminals. Such a network has, in general, an impedance element connected between each pair of terminals (including the neutral), as shown in Fig. 8 for a network of four terminals, including neutral (a

SOLUTION OF NETWORKS

82

three-machine system). In. this network there are six impedance elements. 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 any of these elements are lacking, they may be considered to be present as elements of zero self-admittance (that is, infinite impedance or open circuit). If two or more elements are in parallel, they may be replaced by an equivalent single element having an admittance equal to the sum of the admittances of the several parallel elements. From this viewpoint any network may be converted to a network having one and only one element connected between each pair of nodes and having m(m - 1)/2 elements. This form of network is called a mesh circuit.

o FIG.

8. A four-terminal passive network having one and only one element connected between each pair of terminals.

The machine currents in the network of Fig. 8 may be written as linear functions of the machine voltages and the element self-admittances (lower-case y's), by noting that each machine or terminal current is the sum of several element currents, each of which may be written as the product of the element admittance and the voltage across the element. Thus, 11 = Yl2 (E1 - E 2) + Y1S (E1

-

12 = Y12(E 2 - E1 ) .

+ Y2s(E2 -

13 = Yls(Es - E1)

+ Y23(Ea -

+ YOlE1 Es) + .Y02E2 E2) + YosEs Es )

[I8a] [18b] lI8e]

Regrouping the terms by E's,

+ Y12 + Yls)E1 + (-YI2)E2 + (-Yls)Ea

[I9a]

= (-YI2)E 1 + (Y02 + Y12 + Y23)E2 + (-Y2s)Ea

[19b]

+ (-Y2a)E2 + (Yoa + Y13 + Y23)Es

[Igel

11 = (YOI 12

13 = (-Yls)E 1 ~'If

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

NETWORK REDUCTION

83

By comparing these with the type equations (eqs. 10, rewritten for

n == 3 as eqs. 20, which follow) Y I I E1

+ Y 12E2 + Y l aEa

[20a]

= Y21 E l l a = Y 3l E I

+ Y 22E2 + Y 2s Ea + Y32E2 + YasEa

[20b]

11

=

12

[20c]

the terminal admittances are determined as

+ Y12 + Y13 Y02 + Y12 + Y23 Y03 + Y13 + Y2a

[21a]

Y2l =

-Y12

[22a]

= -Y13

[22b]

(Y1l = YOI

self-admittances Jl Y 22

=

Y 33 = (Y 12 =

mutual admittances

t

Y 13 =

Y23

Y3l

= Y32 =

-Y23

[21bJ [21c]

[22c]

That is, each terminal self-admittance is the sum of the element admittances of all the elements connected to that terminal. Each terminal mutual admittance (between two given terminals) is the negative of the element admittance of the element connected between the given terminals, Algebraic solution of networks: network reduction.

We now have the terminal self- and mutual admittances in terms of the element self-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 initially regarded as extra terminals. Suppose, for example, that the network has five nodes, of which one is to be eliminated. The terminal voltampere equations for five terminals (neutral and four others) are:

+ Y12E 2 + Y13E 3 + Y14E 4: 12 = Y 21E1 + Y22E 2 + Y 23E3 + Y 24E4: 13 = YalE + Ya2E 2 + YaaEa + Ya4E4 14 = Y4l EI + Y42E2 + Y 43E3 + Y 44E4: 11

= YllE l

t

[23a] [23b] [23c] [23d]

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

SOLUTION OF NETWORKS

84

Solving for E 4 and substituting the expression in place of E4 in eqs. 23 give E = _ Y 41E1 + Y42E2 + Y43E3 [25] 4 Y44

[26a] [26b] [26c]

which may be written in the standard form:

11

12

= Y11'E1 + Y12'E2 + Yls'E s = Y21'E1 + Y22'E2 + Y2s'Es

[27a]

+ YS2'E2 + Ys3'Es

[27c]

Is = Ys1'El

[27b]

in which the Y"s are the new terminal admittances. These new terminal admittances are related to the old ones by y -, 11

=

Y14Y41 11 - - Y44

[28]

Y

Y12

,=Y

-

Y 14Y42

[29]

Y13

, = Y13 -

Y 14Y43

[30]

12

Y44

Y44

Y22

Y23

Y33

,= Y

22

,= Y ,= Y

28

33

-

Y 24Y42

[31]

Y 44 _

Y 24Y43

[32]

Y 44 _

Y34Y43

[33]

Y44

or, in general, by Yj k '

= Yj k -

Y j 4Y4 k Y44

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

[34]

NETWORK REDUCTION

85

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

= YOI + Y12 + Y13 + Y14 Y22 = Y02 + Y12 + Y23 + Y24 Y 33 = Y03 + Y13 + Y23 + Y34 Y44 = Y04 + Y14 + Y24 + Y34 Yll

Y12

=

Y 13

= Y31

Y 21

= = =

[35a]

[35b] [35e]

[35d]

-Y12

[36a]

-Y13

[36b]

-Y14

[36e]

Y23

= Y41 = Y32 =

-Y23

[36d]

Y 24

= Y42 = -Y24

[36e]

Y34

=

[36f]

Y 14

Y43

=

-Y34

+ Y12' + Y13' Y02' + Y12' + Y23'

[37a]

= Y03'+ Y13'+ Y23' Yt 2' = Y21' = -YI2' Y13' = Y' 31 = -Y13 , Y23' = Y32' = -Y23'

[37e]

Yl t ' = Y 22'

=

YOl'

Y33 ,

[37b]

[38a] [38b] [3Sc]

Substitute these relations (eqs. 35 to 38 inclusive) into eqs. 28 to 34 inclusive. This substitution works out most easily for the mutual terminal admittances, eqs. 29, 30, and 32. For example, substitution in eq. 29 gives: , -Y12

=

(-YI4) (-Y24) -Y12 Y04

or, changing signs,

,

Y12 = Y12

+

Y14Y24 Y04

In similar manner, we obtain Y13

Y23

,=

Y13

+

, = Y23 +

+ Y14 + Y24 + Y34

+ Y14 + Y24 + Y34 Y14Y34

Y04

+ Y14 + Y24 + Y34

Y04

+ Y14 + Y24 + Y34

Y~Y34

[39]

[40] [41)

86

SOLUTION OF NETWORKS

The equations for terminal self-admittances require a little more manipulation but. give similar results. For example, substitution in eq. 28 gives , YOI

,

+ Y12 + Y13

,

=

YOI

+ Y12 + Y13 + Y14

( -YI4) ( -YI4)

-

+ Y14 + Y24 + Y34 Y04

Subtraction of eqs. 39 and 40 from this gives

,=

YOI

Similarly

+ Y14 -

+ Y24 + Y34) Y04 + Y14 + Y24 + Y34 Y14(Y04 + Y14 + Y24 + Y34 - Y14 = YOI + Y04 + Y14 + Y24 + Y34 Y04Y14 = YOI + Y04 + Y14 + Y24 + Y34 YOI

Y02

Y03

,=

,=

Y14(Y14

Y02

Y03

+ Y04

+ Y14 + Y24 + Y34

Y04

+ Y14 + Y24 + Y34

+ ,

[43]

Y04Y34

[44]

= Yik + -Yj4Yk4 4-L

i=O

Y34)

[42]

Y04Y24

In general Yjk

Y24 -

[45]

Yi4

Equation 45 may be interpreted as follows: Every element of the new or reduced networkis the result of paralleling the corresponding element of

the 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 conversion with somewhat simplified notation is shown in Fig. 9. Figure 9a shows a four-point star which is to be converted into a four-terminal mesh (Fig. 9b), thereby eliminating the node S. The mesh-element admittances in terms of the star-element admittances are: YIY2 Y12 = -

[46a]

Y23

=-

L:y

[46d]

YIY3 Y13 = -

L:y

[46b]

Y2Y4 Y24=-

L:y

[46e]

YIY4 Y14 = -

[46c]

Y34

Y3Y4

[46f]

L:y

L:y

Y2Y3

=-

L:y

where LY

~

Yl

+ Y2 + Y3 + Y4

[47]

NETWORK REDUCTION

87

If the star has n points, similar formulas hold, but the expression for ~y has n terms. If n = 3, we have a Ydelta conversion. If n = 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 for n > 3 is not possible. If n = 2, the conversion can be made but is not unique; that is, a single element can be split into two elements in series in an infinite number of ways. By one star-mesh conversion, one node is eliminated. By successive star-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 in terms of admittances than in terms of impedances. Furthermore, the parallel combinations of the elements resulting from such a conversion with other elements are made more easily by working with admittances than with impedances. For these reasons, in addition to the fact that we want terminal admittances in our final expressions for power, it is preferable to work with admittances rather than impedances while reducing 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 parallel combinations 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.

88

SOLUTION OF NETWORKS

5. Repeat steps 3 and 4 until all nodes except the terminals have been eliminated. *** To prepare for the power calculations required in a stability study, perform the following additional step, which is not strictly a part of the 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 network conditions. Thus, if the disturbance causing the machines to swing is a fault, the network must be solved for the pre-fault condition, the fault condition, and, unless the fault is sustained, for the post-fault or cleared condition. If fault clearing is accomplished by sequential opening of two or more breakers, the network must be solved for two or more fault conditions. Also, if different runs are to be made for different fault locations, different lines connected or disconnected, or other different conditions, network solutions must be made for each condition. 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, and post-fault conditions, we are still not ready to begin calculation of the swing curves, for the initial operating conditions must first be determined. Specifically, we must find the values of magnitude and phase of the internal voltages of all the synchronous machines. If these values were given, it would be simple to substitute them into the power equations (like eqs. 17) for the pre-fault condition and thus. to calculate the pre-fault power output of each machine, which is also the power 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; the remaining information may consist of the reactive power outputs at the 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 the machine 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 be determined as well as is possible. As a rule, this can be done only by cut-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 when a board is used than when algebraic solutions are relied upon. lunt

Fisk

A

Murphy

Deering Patten

0.30

0.10

0.08

0.12

0.30

0.10

0.08

0.12

B

0.10 0.16

0.25 Thorne

FIG.

0.10

0.05

Dyche

0.05

Ward

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

After the initial power outputs, angles, and internal voltages have been 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 in Fig. 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. Data on the generators and on the initial generating-station outputs and bus voltTABLE 4 DATA FOR EXAMPLE

Station

Steam or Hydro

Lunt Murphy Wieboldt

Hydro Steam Steam

Number Rating of of Each Unit Units (Mva.)

Xd'

1

H

(p.u.)

(Mj. per Mva.)

Initial Initial Station Bus Output Voltage (Mw.) (p.u.)

3 4

35 75

0.35 0.21

3.00 7.00

80 230

1.05 ,1.00

2

50

0.18

8.00

90

1.00

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

90

SOLUTION OF NETWORKS

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

Solution.

Outline. 1. The network between the busses at Lunt, Murphy, and Wieboldt

will be reduced to its simplest form (a delta). 2. The known magnitudes and assumed phase angles of the voltages of these three busses will be substituted into the power equations, and the angles 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 found from the known terminal 'voltages and admittances. 4. The generators at each station will be combined by paralleling their impedances to give one equivalent machine at each station. The internal Zl drop of each equivalent machine will be added to the terminal voltage to find the phase and magnitude of the internal voltage.

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 omitting the open branch from Ward to Thorne. The result is shown in Fig. 12a, in which @ Node to be retained. the names of the stations are abbreviated 1.21 Network element with to their initial letters. The values of imadmittance. pedance in per unit are encircled. These I , I I Element entering into values of impedance are converted to adstar- mesh conversion. mittance values, which are not encircled. Element resulting from The impedance and admittance angles, star- mesh conversion. 0 0 FIG. 11. Symbols used in net- which are 90 and -90 respectively, are work reduction. omitted for brevity. By means of alternate star-mesh conversions and parallel combinations of elements, all nodes are eliminated except the three (L, M, and Wi) which are to be retained. The order in which they are eliminated is not particularly important; Wa is eliminated first. The star elements radiating from Wa are cross-hatched in Fig. 12a. The resulting mesh elements are shown by broken lines in Fig. 12b. The starmesh calculations are as follows:

•

Node to be eliminated,

E Y = 10.0 + 5.0 + 10.0 = 25.0 YFD

= YWiD =

10.0 X 5.0 25.0 = 2.0

. = 10.0 X 10.0

YFW~

25.0

=

40 ·

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

INITIAL OPERATING CONDITIONS L

91

(Q][) F

•

6.66

(e)

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

5.00

L

6.66

p

F

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

(b)

D

@lllllllq L

F

6.66

~o~

(g)

M

6.35

te'~

• Wi L

,

M

,

2.00

@------e ~ (h)

~,

'J'$', .

'},.~,

~~.

• Wi (d) e,

(i )

FIG.

~.

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

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

1: Y = 40.0 + 20.0 + 2.0 + 42.0 = 104.0 40 X 20 = 7.70 104 40 X 2 104

=

20 X 2 104

= 0.385

0.77

40 X 42 = 16.15 104 20 X 42 104

8.08

2 X 42 - - - = 0.81 104

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

SOLUTION OF NETWORKS

92

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

Ey=

11.5

+ 24.4 + 14.3 =

The calcu-

50.2

11.5 X 14.3 = 3.28

11.5 X 24.4 = 5.58

50.2

50.2 24.4 X 14.3 50.2

=

6.94

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:

+ 2.55/- 90° = 4.55/- 90° p.u. /-90° + 25.5/-90° = 27.5/-90° p.u,

Y LL = 2.00/- 90°

= 2.0 Yww = 2.6

YMM

/-90°

+ 25.5/-90° = 28.1(-90° p.u,

= 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. YLM = YML = -2.00/-90°

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

PL

=

0.80 - 0

PM = 2.30 - 2.00 = Pw

=

0.80 p.u. 0.30 p.u.

0.90 - 2.00 = -1.10 p.u.

The bus voltages are: EL = 1.05& EM

= 1.00/5M

E w = 1.00/0 (taken as reference phase)

The power equations are: PL

=

PM

=

Pw

= Ew

+ ELEMY LM cos (eLM - OL + OM) + ELEwYLw cos (8LW - OL+ ow) cos OMM + EMELYZ.fL cos (eML - OM + OL)

EL 2 y LL cos eLL 2YMM

EM

+EMEwYMwCOS(aMW-OM+OW) 2Yww

cos Oww

+ EwELYWL cos (aWL - Ow + OL) + EwE-tllYWM cos (SWAt - Ow + OM)

(a)

(b) (c)

INITIAL OPERATING CONDITIONS

93

Because there are no losses in the network, only the first two equations need be used. Because eLL = eMM = 8 ww = -90°, the self-admittance terms vanish. eLM = eLW = 8ML = eMW = 90°, and we may use the relation, cos (900 - x) = sin z, Also noting that Ow = 0, we have: PL ~ ELEMY LM sin (OL - OM) PM

= ELEM Y LM sin

+ ELEwYLW sin OL

(d)

+ EMEwY MW sin OM

(e)

(OM - 01,)

Substituting numerical values, we obtain:

+ 2.68 sin OL OL) + 25.5 sin OM

0.80 = 2.10 sin (OL - OM)

(J)

0.30 = 2.10 sin (OM -

(g)

These equations must be solved by trial for OL and OM. Because of the large coefficient of sin OM we may be sure that OM is small. As a first trial, let OM = o. Equation J then becomes 0.80 sin OL

=

(2.10

+ 2.68)

sin OL

= 0.80 = 0.1675· 4.78

OL

'

= 4.78 sin OL

= 9.6

0

Substitution of this value of OL in eq. g gives 0.30 ~ 2.10 sin (-9.6°) 25.5 sin OM

=

0.30

+ 2.10 X 0.1675 = 0.30 + 0.352 = 0.652 OM

sin OM = 0.0256; Substitution of OL hand members:

PL

= 9.6

0

+ 25.5 sin OM = 1.5

and OM = 1.5° into eqs.

0

J and g gives for the right-

+

2.10 sin 8.10 2.68 sin 9.6 0 = 2.10 X 0.141 = 0.296 0.448 = 0.744 (Should be 0.80.)

=

+

+ 2.68 X 0.167

+

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 PL is too small and PM too large, OL should be increased and OM decreased. Try OL = 10.3°, OM = 1.4°.

+

PL = 2.10 sin 8.9 0 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

==

0.324 -0.324

+

+ 25.5 sin 1.4

0

+ 25.5 X 0.0244 + 0.622 = 0.298

(Should be 0.300.)

94

SOLUTION OF NETWORKS

These values are close enough to the correct values. The bus voltages are therefore 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 /0.0 = 1.000 + jO.OOO p.u, 0

3. Determination of currents. ILW

IMW ILM

= YLW (E L - Ew) = = 0.479 - jO.082 p.u, =

YMW

j2.55 (0.032 + jO.188)

(EM - Ew) = -j25.5 X jO.024 = 0.612

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

=

YLM

+ jO.OOO p.u,

+ jO.162)

The terminal currents of the network are IL

= ILw + I LM = 0.803 -

1M =

IMW -

Iw =

-ILW -

ILM

jO.146 p.u.

+ jO.064 p.u, -1.091 + jO.082 p.U.

= 0.288

=

IMW

The load currents, calculated by the formula

I = p

-t. jQ E

are

2.00LO

1M '

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

I w'

= -----;n =

0

2.00

i!2

1.00~

2.00i!2

= 2.000 + jO.OOO

The generator currents are equal to the currents supplied to the network plus local load:

At L,

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

At W, IB = 1M

+1

M'

At W, Ie = I w + I w'

= 2.288 + jO.113 p.u, = 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 be converted to per unit 011 lOO-Mva. base.

INITIAL OPERATING CONDITIONS

XA

100

= 0.353 X 35 = 0.333 p.u,

ZA = iO.a3a p.u,

100

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

95

ZB = jO.070 p.u,

100 = 0.18- = 0.180 p.u. 2X 50

Zc

= iO.180 p.u.

Each internal voltage is the vector sum of the terminal (bus) voltage and the internal 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 = 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) = 0.801 p.u, (Should be 0.80.)

+

+

+ jO.082)

= 0.868 -

PB = EB·IB = O. 92 X 2.29 0.184 X 0.113 = 2.27 = 2.291 p.u, (Should be 2.30.)

+

+ 0.021

Pa = Ec·Ie = O. 85 X 0.909 0.164 X 0.082 = 0.895 = 0.908 p.u, (Should be 0.90.) EXAMPLE

0.067

+ 0.013

2

Find the terminal admittances of the network of the three-machine system

of' Example 1, including the loads and the direct-axis transient reactances of the 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 ends of the line. Solution. (a) The short circuit at X is electrically equivalent to a short circuit on the Patten bus. In the network .reduction of Example 1, the Patten bus (node P) was preserved until the stage shown in Fig. 12e. The application of the short circuit can be represented by connecting node P to node 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

96

SOLUTION OF NETWORKS

as O. The admittances of the three equivalent machines and of the two loads must be added to the network. All admittances should now be expressed as complex numbers, because the load admittances do not have the same angle as do the line and machine admittances. A lOO-Mva. base will be used, as in Example 1. The machine impedances in per unit on a 100-Mva. base, as found in Example 1, and the corresponding admittances are as follows: Station Lunt Murphy

Machine A B C

Wieboldt

Impedance 0.33 /90 0 0.07 /90 0 0.18 /90 0

Admittance 3.00/-90° 14.3/-900 5.55(-90°

The admittances of the loads at Murphy and at Wieboldt are each 2.00~ p.u.

After the machine and load admittances are added, the network of Fig. 12e becomes 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 Wi and O. The computation is as follows:

3.00/ - 90° X 6.66/ - 90° A-F:

9.66~

M-o:

2.0 - j24.4 = 24.5/-85.3°

Wi-O:

2.0 - j14.33 = 14.5/-82.0°

= 2.07/-90°

The resulting network is shown in Fig. 13b. Next, node F is eliminated by a star-mesh conversion, as follows:

Ey=

(2.07 + 0.77

+ 11.5 + 4.81) /-90° =

19.15/-90°

A-M:

2.07 X 0.77 L-900 19.15

= 0.083L-900

A-D:

2.07 X 11.5 /-900

= 1.24/-900

19.15

A-Wi:

2.07 X 4.81 /-900 19·15

M-O:

= 0.520L-900

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

97

INITIAL OPERATING CONDITIONS

B

--~~@

(c)

(h)

·C FIG. 13. Reduction of the network of Fig. 10 with a short circuit at X (Example 2).

98

SOLUTION OF NETWORKS

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

Wi-o:

(2.0 - j14.3) - j2.9 = 2.0 - j17.3 = 17.3/-83.3°

= 25.0/-85.4°

The resulting network is shown in Fig. 13d. Node M is eliminated, as follows:

I: y = (2.0 -

j24.9) - j16.3 - j14.3 - jO.1 = 55.6/- 88.0°

O-B:

O-Wi:

B-Wi:

A-o:

A-B:

= 2.0 -

j55.6

25.0/-85.4° X 14.3~ = 6.42/-87.4° 0 55.6_-88.0 / 25.0/-85.4° X 16.34~

°

/

55.6_-88.0

14.3/-90° X 16.34/-90° ° 55.6_ -88.0

L

0.083 _

= 7.34/-87.4°

= 4.20L-92.0°

L90° X --

25.0/- 85.4° / - - - = 0.037,--_-_87_.4_° , 5.5.6/- 88.0°

0.083/-90° X 14.3/-90° 55.6 _ -88.0

= 0.021/-92.0°

0.083~X 16.34~ / ° 55.6_- 88.0

= ·0.024_-92.0° -----

0

/

. A-Wi:

/

The resulting network is shown in Fig. 13e. Parallel combinations are made, as follows: O-Wi:

17.3/-83.3°

+ 7.34/-87.4° == 2.0 -

= 2.3 - j24.5

= 24.6/ -

j17.2

84.6°

A-o:

-j1.24 - jO.04 = -jl.28 = 1.28/-90°

A-Wi:

-jO.52 - jO.02 = -jO.54

= 0.54/-90°

+ 0.3 -

j7.3

99

INITIAL OPERATING CONDITIONS

The resulting network is shown in Fig. 13/. Node Wi is eliminated, as follows:

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°

A-O:

A-B:

0.54/- 90° X 24.6/ - 84.6° 34.9/ - 86.7° 0.54L- 90° X 4.20/- 92.0°

34.9/- 86.7°

. 0.54/-90° X 5.55/-90° A-C. 34.9/-86.7°

= 0.38/-87.9° = 0.065/-95.3°

/ 0 = 0.086 - 93.3

B-D:

420/-92.0° X 24.6/-84.6° _ / ° / ° - 2.96_-89.9 34.9_ -86.7

B-C:

4.20/- 92.0° X 5.55/ - 90° 34.9/-86.7°

= 0.668/- 95.3°

24.6L-84.6° X 5.55/-90° 34.9/-86.7°

= 3.91/-87.9°

- C..

O

The resulting network is shown in Fig. 13g. Parallel combinations are made 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 = 1.84/-90.0° p.u. Ans.

= 0.00 - jl.84

SOLUTION OF NETWORKS

100 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,

Yee = 3.91 /-87.9°

= 0.15 = 0.09 -

+ 0.086/-93.3° + 0.668/-95.3°

j3.91 - 0.01 - jO.09 - 0.05 - iO.66 j4.66 = 4.66 / -88.9° p.u, Ans.

= - 0.086/ - 94.0° = YAc = -0.086/-93.3° = YBc = -0.668/-95.3° = YAB

Ans.

Am.

0.086/86.0° p.u.

0.086/86.7° p.u. Ans.

Ans,

0.668/84.7° p.u.

(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

•

FIG.

B

•

14. Network resulting from the reduction of the network of Fig. 10 with the 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 machine admittances may be attached, as in part a of this solution, and the network further 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 YBC

= 1.12/-100.5° = = 3.06/ -102.6° =

-0.205 - il.IO p.u. -0.66 - j2.98 p.u.

YeA = 0.502 / -100.8° = -0.093 - jO.493 p.u.

= 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, YAO

INITIAL OPERATING CONDITIONS

101

The terminal admittances are:

= 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. Y A B = -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. YAA

EXAMPLE

3

Calculate and plot swing curves for the three-machine system of Fig. 10 with initial conditions as in Example 1, assuming a three-phase short circuit to occur at point X, Fig. 10, and to be cleared (a) in 0.40 sec. and (b) in 0.35 sec. Carry the curves far enough to determine whether the system is stable. Solution. The initial conditions, as calculated in Example 1, are:

EA = 1.17

~A =

EB = 1.01

8B = 10.4°

Eo = 1.00

8c

=

PuA = PiA = 0.80

23.0°

PUB

=

PiB

Puc = PiC

9.5°

= 2.30 = 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

Y AB = 0.086

aAB

= 86.0°

Y BC = 0.668

eBC

= 84.7°

= 0.086

eCA

= 86.7°

Y CA

The terminal admittances with the fault cleared, as calculated in Example 2, part b, are: Y AA cos 8AA = 0.03 Y BB cos eBB

=

1.58

Yoo cos ecc

=

0.33

YAB

= 1.12

eAB

=

YBC

= 3.06

aBC

= 77.4°

YCA

= 0.502

eCA

= 79.2°

79.5°

SOLUTION OF NETWORKS

102

The power-angle equations have the form of eqs. 17. ·Substituting the foregoing numerical values, we obtain the following expressions, which apply 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)

+ ~B) + 0.10 cos (86.7° -

+ 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 (86.7 ~c + ~A) + 0.68 cos (84.7 ac + OB) = 0.10 cos (86.0° - OA

OA

0

0

-

-

and the following expressions, which apply-after the fault has been cleared:

+ 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) (1.01)2 X 1.58 + 1.01 X 1.17 X 1.12 cos (79.5 OB + DA) + 1.01 X 1.00 X 3.06 cos (77.4° - DB + ~c)

PuA = (1.17)2 X 0.03 = PuB =

p

= 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 data given in Example 1 by the formula (eq. 54, Chapter II)

= GH 180!' 180 X 60

M= GH where M

= inertia constant in per unit.

G = station rating in per-unit apparent power. H

I

= kinetic

energy at rated speed in megajoules per megavoltampere of rating. = frequency in cycles per second.

=3X

M A

35 X 3.00 100 X 10,800

M = 4 X 75 X 7.00 B

M c

100 X 10,800

=2X

50 X 8.00 100 X 10,800

= 2.92 X 10-4

= 19 45 X 10.... ·

=

7.41 X 10-4

103

INITIAL OPERATING CONDITIONS

The time interval ilt for point-by-point calculations will be taken as 0.10 sec. Then 2

_il_t =

0.010

MA 2

_1l_t

= 34.3

2.92 X 10-4

=

MB

= 5.14

0.010 19.45 X 10--4

2

_1l_t = 0.010 = 13.5 Me 7.41 X 10-4 The computations of power are carried out in Tables 5, 6, and 7, and the computations of swing curves in Tables 8, 9, and 10. The results of the computations of swing curves, namely, the angular positions of the three

400 Cleared 0.40sec. - - - Cleared 0.3) sec.

;} .

)

!/~.

)

tis;:; ~ ,~

/'

"

b\J,

'9;i'f-

/'

~~

.4

jf

/

J' jr

)

-~

~

o

o

V

~

V

r

lime (seconds)

0.5

0.8

FIG. 15. Swing curves of the three-machine system of Fig. 10 with a three-phase short 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 each pair of machines, are listed in Table 11. The swing curves are plotted in Figs. 15 and 16. Figure 15 shows the angular position of each machine with respect to a reference axis rotating at normal speed. All three machines increase 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 16 shows 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

"u

12.6 20.2 42.1 76.1 119.9 119.9 153.8 133.1 127.6 99.6

86.0

"

,e" ,e

" 79.5

" " "

OAB

eAB

"

It

" "

79.5

" "

12.6

20.2

86.0

42.1 76.1 119.9 119.9 153.8 133.1 127.6 99.6

8AB

eAB

lJAB

+ OAB

98.6 106.2 128.1 162.1 205.9 199.4 233.3 212.6 207.1 179.1

eAB

73.4 65.8 43.9 9.9 -33.9 -40.4 -74.3 -53.6 -48.1 -20.1

eAB -

-1.000

-0.150 -0.279 -0.617 -0.952 -0.900 -0.943 -0.598 -0.842 -0.890

cos

0.286 0.410 0.721 0.985 0.830 0.762 0.271 0.593 0.668 0.939

cos

0.03 0.04 0.07 0.10 0.08 0.99 0.35 0.77 0.87 1.22

PAB

" "

It

"

1.30

" " "

"

0.10

PABm

-0.02 -0.03 -0.06 -0.10 -0.09 1-1.23 -0.78 -1.09 -1.16 -1.30

PBA

TABLE 6.

"

It

" "

1.30

II

"

It

It

0.10

PABm

TABLE 5.

42.3 75.4 117.8 117.8 151.4 131.9 128.8 102.4

20.9

13.5

lJAC aAC

73.2 65.8 44.4 11.3 -31.1 -38.6 -72.2 -52.7 -49.6 -23.2

SAC -

"

"

" "

77.4

" " " "

84.7

aBC

0.9' 0.7 0.2 -0.7 -2.1 -2.1 -2.4 -1.2 1.2 2.8

aBC aBC

83.8 84.0 84.5 85.4 86.8 79.5 79.8 78.6 76.2 74.6

eBC -

3)

0.108 0.104 0.096 0.080 0.056 0.182 0.177 0.198 0.238 0.266

cos

3)

0.289 0.410 0.714 0.981 0.856 0.782 0.306 0.606 0.648 0.919

cos

COMPUTATION OF PUB (EXAMPLE

" " " "

79.2

" " " "

86.7

aAC

COMPUTATION OF PuA (EXAMPLE

II

" "

u

3.09

" " " "

0.68

PBCm

" " " "

0.59

It

" "

It

0.10

PACm

0.82

0.07 0.07 0.07 0.05 0.04 0.56 0.55 0.61 0.74

PBC

0.03 0.04 0.07 0.10 0.09 0.46 0.18 0.36 0.38 0.54

PAC

II

II

"

II

1.61

" "

" "

0.24

PBB

"

It

" "

0.04

" " "

"

0

PAA

0.29 0.28 0.25 0.19 0.19 0.94 1.38 1.13 1.19 1.13

PuB

1.80

1.29

0.06 0.08 0.14 0.20 0.17 1.49 0.57 1.17

PuA

00

~ ~

o

~

~ ~

~

o~

z

o

~

~ ~

o~

00

~

13.5 20.9 42.3 75.4 117.8 117.8 151.4 131.9 128.8 102.4

86.7

"

" "u

79.2

"

tc

"

u

OAC

9A.C

+ OAC

100.2 107.6 129.0 162.1 204.5 197.0 230.6 211.1 208.0 181.6

9AC

-0.177 -0.316 -0.629 -0.952 -0.910 -0.956 -0.635 -0.856 -0.883 -1.000

cos

" "

"u

0.59

" " " "

0.10

PACm

-0.02 -0.03 -0.06 -0.10 -0.09 -0.56 -0.37 -0.50 -0.52 -0.59

PCA

" " " u

77.4

" " " "

84.7

eBC

0.9 0.7 0.2 -0.7 -2.1 -2.1 -2.4 -1.2 1.2 2.8

OBC

+ OBC

3)

85.6 85.4 84.9 84.0 82.6 75.3 75.0 76.2 78.6 80.2

eBC

COMPUTATION OF PuC (EXAMPLE

TABLE 7

0.077 0.080 0.089 0.104 0.129 0.254 0.259 0.238 0.198 0.170

cos

" " " "

3.09

" " "

"

0.68

PBCm

0.05 0.05 0.06 0.07 0.09 0.78 0.80 0.74 0.61 0.52

PCB

0.12 0.11 0.09 0.06 0.09 0.55 0.76 0.57 0.42 0.26

0.09

" " " "

0.33

"

"

" "

PuC

Pce

....

~

.-

~

tj ..... t-3 ..... o

!Z

o o

c

Z

~ .~....

~

!Z

SOLUTION OF NETWORKS

106

TABLE 8 COMPUTATION OF 8A (EXAMPLE

t

PiA

PUA

Pa A

(sec.)

(p.u.)

(p.u.)

(p.u.)

0.80

0.06

0.74 0.37

12.7

-0.72

24.7

0+

Oavg.

3)

34.3P aA a8A (elec. deg.) (elee. deg.)

8A

(elee. deg.)

23.0 12.7

"

0.08

"

0.14

"

0.20

0.40.4+ 0.4 avg.

" " u

0.17 1.49 0.83

-0.03

-1.0

0.5

"

0.57

0.23

7.9

0.1 0.2 0.3

35.7 37.4

0.66

22.6

73.1 60.0

0.60

20.6

133.1 SO.6 213.7 79.6 293.3

87.5

0.6

380.8

80.6 0.4

0.80

1.49

-0.69

-23.6

213.7

57.0 0.5

u

1.17

-0.37

270.7

-12.7 44.3

0.6

0.7 0.8

"

1.29

"

1.80

-0.49

315.0

-16.8

27.5

-1.00

342.5

-34.3 -6.8

335.7

INITIAL OPERATING CONDITIONS

107

TABLE 9 COMPUTATION OF 8B (EXAMPLE

t

PiB

PuB

(see.)

(p.u.)

(p.u.)

2.30

0.29

0+ Oavg.

PaB (p.u.)

5.14PaB (elec. deg.)

2.01 1.00

5.1

2.02

10.4

3) A8B (elec. deg.)

8s (elee. deg.)

10.4 5.1

0.1 0.2

Ie

u

0.28

15.5 15.5

0.25

2.05

10.5

31.0 26.0

0.3

u

0.19

2.11

10.8

57.0 36.8

0.40.4+ 0.4 avg.

u

0.5

u

u

ce

0.19 0.94 0.56

1.74

8.9

1.38

0.92

4.7

93.8 45.7

139.5 50.4

0.6

189.9 36.8

0.4

2.30

0.94

1.36

7.0

3.8 43.8

0.5

It

1.13

1.17

6.0

137.6 49.8

0.6

II

1.19

1.11

5.7

187.4 55.5

0.7 0.8

"

1.13

1.17

6.0

242.9 61.5 304.4

108

SOLUTION OF NETWORKS

TABLE 10 COMPUTATION OF 8c (EXAMPLE 3)

t

Pie

PuC

PaC

(sec.)

(p.u.)

(p.u.)

(p.u.)

0.90

0.12

0.78 0.39

5.3

0.79

10.'7

0+

o avg.

13.5Pac (elec. deg.)

t16c

6c

(elec..deg.)

(elec. deg.)

9.5 5.3

0.1

H

0.11

14.8 16.0

0.2

H

0.09

0.81

10.9

30.8 26.9

0.3 0.40.4+ 0.4 avg.

"

0.06

"

0.09 0.55 0.32

0.58

7.8

0.76

0.14

1.9

H H

0.84

11.3

57.7 38.2 95.9 46.0

0.5

"

141.9 47.9

0.6

189.8 38.2

0.4

0.90

0.55

0.35

4.7

95.9 42.9

0.5

"

0.57

0.33

4.5

138.8 47.4

0.6

H

0.42

0.48

186.2

6.5

53.9 0.7

H

0.26

0.64

240.1

8.6 62.5

0.8

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.

) I

I

I

Cleared 0.40 see.~

V

l?~ ...... ...--. ".~;'

I~

J i

E

90

8 co

'is. en

.

:a oS!

Q c

<

~

0

V

/

V

J

'I

"

"'..

\

Cleared 0.35 sec.-~\

\

"

o

0.5

0.8

Time (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

(sec.)

(elec. deg.)

0 0.1 0.2 0.3 0.4 0.5 0.6

23.0 35.7 73.1 133.1 213.7 293.3 380.8

10.4 15.5 31.0 57.0 93.8 139.5 189.9

9.5 14.8 30.8 57.7 95.9 141.9 189.8

12.6 20.2 42.1 76.1 119.9 153.8 190.9

13.5 20.9 42.3 75.4 117.8 151.4 191.0

0.9 0.7 0.2 -0.7 -2.1 -2.4 0.1

0.5 0.6 0.7 0.8

270.7 315.0 342.5 335.7

137.6 187.4 242.9 304.4

138.8 186.2 2'40.1 302.6

133.1 127.6 99.6 31.3

131.9 128.8 102.4 33.1

-1.2 1.2 2.8 1.8

6B

80

8AB

8AO

8BO

(elec. deg.) (elec. deg.) (elec. deg.) (elec. deg.) (elec. deg.)

Network reduction by use of calculating board. Sometimes, in order to set up a network on a calculating board without exceeding the number of impedance units available, the network or a portion of it

SOLUTION OF NETWORKS

110

must be reduced to an equivalent circuit. If the identities of only two terminals of the network, including neutral or ground if any branches are connected to it, need be preserved, the network can be reduced to one impedance element; if three terminals are to be preserved, it can be reduced to an equivalent ~ (or Y) circuit; if four terminals, to an equivalent six-element mesh circuit; and so on. The reduction may be accomplished by calculation, as already described, or it may be done very simply with the help of a calculating board. If the board method is to be used, the network is set up; one

(a)

(b)

17. Reduction of four-terminal network, set up on calculating board, to an equivalent four-terminal mesh circuit by voltage and current measurements. FIG.

terminal of it is connected to a power source, and the other terminals are connected through jumpers to the neutral bus as shown in Fig. I7a for a four-terminal network. The applied voltage and the currents leaving all other terminals are then measured in vector form. If the equivalent 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 current measured would equal the current in one of the remaining elements. Hence the impedances of these elements are:

[48] 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: Z12

E2

= -

11

J

Z23

E2 13

= -,

Z24

E2 14

=-

[49]

COMBINING MACHINES

111

If similar measurements are made with the voltage applied to each terminal in tum, every impedance' is determined twice (as indicated for Z12 above), thereby furnishing a check on the work. If resistance and capacitance are neglected, or if all impedance angles are assumed equal, a d-c. calculating board can be used in this manner to determine the admittances in the power-angle equations, which are then used in calculating the outputs of the machines. . Combining machines. It is apparent from the foregoing discussion and examples that the amount of work involved in making a stability study increases tremendously as the number of synchronous machines included becomes greater, especially if the network is solved algebraically. In order to save work, it is important to keep to a minimum the number of machines which are separately represented. Even on a calculating board this number must not exceed the number of generator units on the board. The number of machines may be reduced by combining several machines which swing together or almost together to form a single equivalent machine. If several machines were mechanically coupled (at such speed ratios that they ali generated the same frequency), they would be forced to swing together, that is, to have equal velocities and accelerations, even though they might not have equal angular positions. Since the inertia constant may be defined as the power required to produce unit angular acceleration, and since the power (or torque) required to produce equal acceleration of all the machines is the sum of the powers (or torques) required to accelerate the individual machines, the inertia constant of the 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 mechanically coupled. Therefore, the inertia constant of the equivalent machine is taken as the sum of the inertia constants of the individual machines (referred to a common megavolt-ampere basettt if power is expressed in per unit). It matters not whether the machines are forced to swing together by close electrical coupling (low impedance between machines) or whether they merely happen to swing together in spite of being far apart electrically. If the machines to be combined to form an equivalent machine are connected in parallel at their terminals, then, by Thevenin's theorem, their effect upon the network is the same as if they were replaced by a single source of e.m.f., equal to the open-circuit voltage of the group of machines, in series with a single impedance, equal to the impedance tttAs stated in Chapter II, the inertia constant varies inversely as the megavoltampere base.

112

SOLUTION OF NETWORKS

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 to the parallel combination of the reactances of the individual machines (referred, of course, to a common megavolt-ampere base if expressed in per unit). The equivalent e.m.f, is a sort of average of the e.m.f.'s of the individual machines. If the machines swing together, the equivalent e.m.f. is constant in magnitude and has the same frequency as the e.m.f.'s of the machines. The e.m.f. of the equivalent machine is such

that the equivalent machine initially supplies to the network the same active 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 between the reactance and the source of internal voltage of each machine are connected together, however, thereby paralleling the several voltage sources and connecting together at one end all the reactances, the other 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 and reactive power as the sources which it replaces. The division of this power among the several points of connection to the network generally will be different from the original division. The reactances of the individual machines may be left as part of the network set up on a calculating 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 new combinations as may seem desirable.fj] The only conclusive test of whether machines may be combined in a given case without too much error in the swing curves or in the conclusions regarding system stability is to compute swing curves, first with the machines not combined, and then again with them combined, and to compare the results of the two computations. If the swing curves obtained with the machines not combined show that a group of machines swing very nearly together, however, this evidence is sufficient for concluding, without running the combined swing curves, that the machines of the group may be combined with negligible effect on the swing curves of the other machines. Neither of the foregoing criteria for combining machines saves any work on the cases to which it is applied; thus their only value is to give the computer experience which should develop his judgment on the circumstances under which combinations may be made. Therefore some further remarks are in order. tttAn example of this practice appears in Study 1, Chapter VII.

COMBINING MACHINES

113

The likelihood that machines will swing together is increased by a decrease of impedance of the connections between them, by proximity in their initial angular positions, by similarity of their inertia constants, and by remoteness of the fault or source of the disturbance. It is customary to combine all the machines in the same station (unless the station is operated sectionalized), even though the machines have unlike ratings, impedances, inertia constants, or initial loadings. Two or more stations which are connected together by low-impedance ties may likewise be combined. Frequently an entire metropolitan system is represented by a single equivalent machine if the study concerns the stability 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 C swing very nearly together. Combine them to form a single equivalent machine called D and compute swing curves of the resulting two-machine system 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 0.10 sec., as in Example 3. ilt2 = 34.3, as before 1t'IA 0.010 = 3.72 26.9 X 10-4 The initial conditions for the three-machine system, as calculated in Example 1 and used in Example 3, are:

= 1.17 E B = 1.01 Ee = 1.00

= 23.0° OB = 10.4°

EA

0.04.

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 of machine D is PiD = PiB Pie = 2.30 + 0.90 = 3.20

+

The values of E B and Ee are nearly equal, so the value of ED may be taken without serious error as equal to that of EB = 1.01. The value of OD will lie between the values of OB = 10.4° and oe = 9.5°, probably closer to the

SOLUTION OF NETWORKS

114

value of OB because B is a larger machine than C. A weighted average could be 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

YDO

=

YBO

+ yeo =

= 0.172/-93.6°

9.38/-88.2°

+ 0.15 -

= 0.30 - j9.38 = 13.3/- 88.0°

+ 3.91/-87.9°

j3.91

= 0.45 -

j13.29

The terminal admittances are: YAA YDD

= -j1.84 = 1.84/-90.0°, as before

= YAD + YDO =

YAD

=

-YAD =

-0.01 - jO.17

+ 0.45 -

j13.29

= 13.5/-88.1° -0.172/-93.6° = 0.172!86.4°

= 0.44 - j13.46

COMBINING MACHINES

115

The power-angle equations are:

+

= = PuD = 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) (1.01)2 X 0.44 + 1.01 X 1.17 X 0.172 cos (86.4° = 0.45 0.203 sin (3.6° - DAD)

+ 8AD)

+

Network reduction and power-angle equations, fault cleared. The reduced network of the three-machine system with the fault cleared is shown in Fig.

(0)

1.62/-100.6°

(6)

•

o FIG.

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 terminal D. After making two parallel combinations, as follows, the network of Fig. 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°

The terminal admittances are:

= 0.03 - j1.66 = 1.66/-89.0°, as before YDD = YAD + YDO = -0.30 - jI.59 + 3.52 - jO.67 = 3.22 YAD = -YAD = -1.62/-100.6° = 1.62/79.4° YA A

j2.26

118.4

151.1 130.2 120.7 81.8

10.6

u

cc

II

II

II

"

41.9 75.3 118.4

II

20.2

"

BAD

12.7 12.7

,

11.0 3.6

SAD

23.7 16.3 23.8 45.5 78.9 122.0 129.0 161.7 140.8 131.3 92.4

9AD'+ BAD

0.713 0.98r 0.848 -0.777 0.314 0.632 0.751 0.999

" " " "

1.91

"

" " "

1.91 0.203

0.402 0.281

0.404

PADm

sin

0.77 0.057 0.082 0.145 0.199 0.172 1.485 0.60 1.21 1.43 1.91

PAD

0.80 0.057 0.082 0.145 0.199 0.172 1.525 0.64 1.25 1.47 1.95

0.03 0.00

" " " "

0.04

"u

"u

PuA

PuD

PAA

COMPUTATION OF PuA AND

TABLE 12

BAD

4)

sin PDAm

-1.7 -0.030 1.91 -9.1 -0.158 0.203 -16.6 -0.286 " -38.3 -0.620 "u -71.7 -0.949 -114.8 -0.908 " -107.8 -0.952 1.91 -140.5 -0.636 "u -119.6 -0.870 -110.1 -0.939 " -71.2 -0.947 "

e' -

(EXAMPLE

-1.79 -1.81

-1.83 -1.22 -1.66

-0.06 -0.03 -0.06 -0.13 -0.19 -0.18

PDA

"

" "

"

3.28

" " "

"

3.26 0.45

PDD

3.20 0.42 0.39 0.32 0.26 0.27 1.45 2.06 1.62 1.49 1.47

PuD

...

~ ~

00

p::

~

o

~

z ~

~

o

Z

o

foo-4

~

S

COMBINING MACHINES

117

The power-angle equations are: PuA

= (1.17)2 X 0.03 + 1.17 X

1.01 X 1.62 cos (79.4° - 6AD)

+ 1.91 siu (10.6° + 8A D ) (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 )

= 0.04 PuD =

=

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 the two machines and the angular difference between them are tabulated as functions of time in Table 15. The results agree fairly well with those for the three-machine system, Table 11, Example 3.

TABLE 13 COMPUTATION OF 8A (EXAMPLE

4)

t

PiA

PuA

P aA

34.3PaA

~8A

(sec.)

(p.u.)

(p.u.)

(p.u.)

(elec. deg.)

(elec, deg.)

0.800

0.800 0.057

00+ Oavg.

"

0.00 0.743 0.371

12.7

8A

(elec, deg.)

23.0 12.7

"

0.082

"

0.145

"

0.199

0.40.4+ 0.4 avg.

u

"

0.172 1.52 0.85

0.5

"

0.64

0.1 0.2 0.3

0.718

35.7

24.6

37.3 0.655

22.4

73.0 59.7

0.601

132.7

20.6 SO.3

213.0 -0.05

-1.7

0.1e.

5.5

78.6 291.6 84.1 375.7

0.6 80.3 0.4 . 0.5 0.6

"

1.525

"

1.25

u

1.47

-0.725

213.0

-24.8 55.5

-0.45

268.5

-15.4 40.1

-0.67

308.6

-23.0 17.1

0.7 0.8

"

1.95

-1.15

-39.4

325.7 -22.3 303.4

118

SOLUTION OF NETWORKS 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.)

3.20

3.20 0.42

0.00 2.78 1.39

5.2

00+

"

o avg. 0.1

"

0.39

2.81

10.4

0.2 .

"

0.32

2.88

10.7

0.3

"

0.26

2.94

10.9

0.4'(>.4+ 0.4 avg.

" " "

0.27 1.45 0.86

2.34

8.7

2.06

1.14

4.2

0.5

5.2 15.6 26.3 37.2

10.3 15.5 31.1 57.4 94.6

45.9 50.1

140.5 190.6

0.6 0.4 0.5

"

1.45

"

1.62

" "

0.6 0.7

1.75 1.58

6.5 5.9

1.49

1.71

6.4

1.47

1.73

6.4

37.2 43.7 49.6 56.0 62.4

0.8

94.6 138.3 187.9 243.9 306.3

TABLE 15 SWING-CURVE DATA (EXAMPLE

t

4)

(sec.)

aA (elec. deg.)

aD (elec. deg.)

aAD (elec. deg.)

0 0.1 0.2 0.3 0.4 0.5 0.6

23.0 35.7 73.0 132.7 213.0 291.6 375.7

10.3 15.5 31.1 57.4 94.6 140.5 190.6

12.7 20.2 41.9 75.3 118.4 151.1 185.1

0.5 0.6 0.7 0.8

268.5 308.6 325.7 303.4

138.3 187.9 243.9 306.3

130.2 120.7 81.8 -2.9

REFERENCES

119

Treatment of synchronous condensers. In the calculation of swing curves a synchronous condenser should logically be handled in the same way as a generator. Its power output is zero initially but not while it is swinging. When condensers are treated like generators, it is often found that, although they swing with large amplitude and shorter period than the generators, because of their smaller inertia constants they have little effect on the swing of the generators. Therefore synchronous condensers are sometimes represented in transient stability studies on a calculating board by static capacitors (or by reactors if the condensers operate with lagging current). Each such capacitor is in series with a reactor representing the transient reactance of the condenser. At each step of the swing calculation the capacitance is readjusted so that its voltage, representing voltage behind transient reactance, is restored to the initial value. Thus the reactive power of the synchronous condenser is taken into account, but the active power is disregarded. This procedure obviates the taking of power readings and the calculation of the swing curve of the condenser. It affords another method, in addition to that of combining machines, of reducing the number of machines considered in a stability study.

REFERENCES 1. Electrical Transmission and Distribution Reference Book, by Central Station Engineers 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, MeGraw-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.

120

SOLUTION OF NETWORKS

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-System Analysis," 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 published in the book, Ref. 11. 13. C. A. STREIFUS, C. S. ROADHOUSE, and R. B. Gow, "Measured Electrical Constants 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. Network Analyzer: 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 Calculating Board," 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 on Alternating-Current Network Calculator. These books, furnished to purchasers of the calculators, are somewhat different for each model of the calculator. 21. General Electric Company, A-C. Network Analyzer Manual, Publication GET-1285, Schenectady, 1945. 22. DAN BRAYMER, "Today's Network Calculators Will Plan Tomorrow's Systems," 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 of each type of circuit unit, and name and title of man in charge. PROBLEMS ON CHAPTER

m

1. 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 as a two-machine system, for a short circuit partially cleared by opening the circuit breaker at the Patten end only. 5. Using the results of Probe 4 and Example 4, compute and plot swing curves for determining to the nearest 0.05 sec. the critical time of opening the breaker at Patten if the breaker at Dyche remains closed. 6. Work Example 2 for a new condition (c), short circuit partially cleared by opening the circuit breaker at the Patten end only. 7. Using the results of Probe 6 and Example 3, compute and plot swing curves for determining to the nearest 0.05 sec. the critical opening time of the breaker at Patten to clear partially a three-phase short circuit at X if the breaker at Dyche remains closed. 8. Work Example 1 with the following changes: The two lines from Murphy 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 circuit occurs at point X and is cleared in 0.30 sec. by opening breakers at both ends of the line. The results of Probs. 8 and 9 are needed in the solution of this problem.

CHAPTER IV

THE EQUAL-AREA CRITERION FOR STABILITY Applicability of the equal-area criterion. To determine whether a power system is stable after a disturbance, it is necessary, in general, to plot and to inspect the swing curves. If these curves show that the angle between any two machines tends to increase without limit, the system, of course, is unstable. If, on the other hand, after all disturbances including switching have occurred, the angles between the two machines of every possible pair reach maximum values and thereafter decrease, it is probable, although not certain, that the system is stable. Occasionally in a multimachine system one of the machines may stay in step on the first swing and yet go out of step on the second swing because the other machines are in different positions and react differently on the first machine. In a two-machine system, under the usual assumptions of constant input, 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 first swing or never. Under these conditions the observation that the machines come to rest with respect to each other may be taken as proof that the system is stable. There is a simple graphical method, which will be explained in this chapter, of determining whether the machines come to rest with respect to each other. This method is known as the equal-area criterion for stability. When this criterion is applicable, its use wholly or partially eliminates the need of computing swing curves and thus saves a considerable amount of work. It is applicable to any two-machine system for which the assumptions stated above may be made. The fact that the assumed conditions are not strictly true does not necessarily invalidate the criterion. If the input to the machines is not constant, but is changed by action of governors, the effect of such action generally will not be appreciable until after the first swing and will then be in such direction as to aid stability. Whatever damping is present will reduce slightly the amplitude of the first swing and will reduce still further the amplitude of subsequent swings. 122

APPLICABILITY OF EQUAL-AREA CRITERION

123

The effect of varying voltage behind transient reactance, Of, what is the same thing, varying flux linkage of the field winding, deserves consideration. 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 constant flux linkage of the field circuit. If the machine does not have a voltage regulator, 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 the decay is of the order of 2 to 5 sec., and during the first swing the flux linkages do not decrease much in any machine which does not go out of step on that swing, If the fault is sustained for a long time, however, the flux linkages may be so much reduced that the system, although surviving the first swing, will ultimately become unstable. Even if the fault is cleared rapidly, the opening of a line to clear it may decrease the maximum synchronizing power and therefore increase the angular displacement required for a given power transfer and decrease the flux linkage 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 but go out of step later. If the system is stable on the first swing and also stable in the ensuing steady state (under assumption of constant field current), * it is reasonable to suppose that it will not be unstable at any intermediate time; for there will probably be enough damping to reduce the amplitude of swing as fast as the flux decays. If the machines have voltage regulators, the regulators will tend to maintain constant terminal voltage, which would require an increase of field flux linkages. With excitation systems of ordinary speed of response, the regulator and exciter action is too slow to have an appreciable effect during the first swing but is fast enough to prevent loss of synchronism on subsequent swings. By the use of voltage regulators it is possible to preserve stability even in some instances when the system would be unstable on the basis of constant field current in the steady state after clearing of the fault. t From the foregoing discussion it may be seen that, if a two-machine system does not lose synchronism during the first swing, it is very probably stable, especially if the machines have voltage regulators, and also that stability or instability on the first swing may be determined with 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.

124

THE EQUAL-AREA CRITERION FOR STABILITY

criterion is a useful means of determining whether a two-machine system 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 are simplified representations of systems with more than two machines. Two-machine systems may be divided into two types, which will be considered in turn: (1) those having one finite machine swinging with respect to an infinite bus.] and (2) those having two finite machines swinging with respect to each other. One machine swinging with respect to an infinite bus. The swing equation of the finite machine is d20 M dt - 2

= p a = p., -

Pu

[1]

where M is the inertia constant of the finite machine, and 8 is the angular displacement of this machine with respect to the infinite bus. Multiply each member of the equation by 2do/Mdt: 2 d2~ d~ = 2 P a d~

dt2 dt or

:!. [(do)2] dt

-[2]

M dt ==

dt

2 P a do M dt

[3]

Next multiply each side by dt, obtaining differentials instead of derivatives,

[4] and integrate.

r Pad~ a '-2 -rP-do \j M J;

do) 2 = ~ ( dt M Jao

d~ = w' dt

=

a

[5] [6]

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

requiring that

JrPet do = 0 80

125

[8]

This integral may be interpreted graphically (Fig. 1) as the area under a curve of P a plotted against 0 between limits 00, the initial angle, and Om, the final angle; or, since

[9] the integral may be interpreted also as the area between the curve of Pi versus 0 and the curve of P u versus o. The curve of Pi versus P o is a horizontal line, since Pi is assumed constant. The curve of P u versus 0, known as a powerangle curve, is a sinusoid (derived in the next section of this chapter) if the; network is linear and if the machine is represented by 80 8m 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, for which Pi > P u , and an equal and opposite negative portion A 2 , for which Pi < P u. Hence originates the name, equal-area criterion for stability. 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 an angle 0 - 00 is [10]

and this work increases the kinetic energy of the body. The accelerating power P a is proportional to the torque, under the previously made assumption of nearly constant speed. Hence the work done on the machine to accelerate it, which appears as kinetic energy, is proportional to area At. When the accelerating power becomes negative and the machine is retarded, this kinetic energy is given up; and, when it is all given up, the machine has returned to its original speed. This occurs when A 2 = At. The kinetic energies involved in this explanation are fictitious, being calculated in terms of the relative speed rather than the actual speed.

126

THE EQUAL-.AREA CRITERION FOR STABILITY

The power-angle equation for the case of one machine and an infinite bus follows directly from the power-angle equation for one machine of a multimachine system Ceq. 17a, Chapter III) if we let subscript 1 denote the finite machine and subscript 2 denote the infinite bus, and if we put 01 = 0 and 02 = o. P u 1 = E 12 Y

ll

cos 811

+ E 1E2 Y 12 cos (a12

-

0)

= Pc + PM sin (0 - 'Y)

[11]

where Pc = E 12Y 11 cos all. PM = E lE2 Y 12. E 1 is the internal voltage of the machine. E 2 is the voltage of the infinite bus. Ylt/8l1 and Y12/ 8 12 are terminal admittances of the network between the machine and the infinite bus, as defined in Chapter III. 'Y = 812 - 90°. The power-angle curve is, in general, a displaced sinusoid. similar to the simple sinusoid

r, = PM sin 0

It is [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 between 90° and 180°, 'Y lies between 0 and 90°, and all lies between o and - 90°. For a network consisting of inductive react" ance only, 812 = 90°, 'Y = 0, and 811 = -90°.) If the network consists of reactance only, then eq. 11 reduces to eq. 12, and the powerFIG. 2. Power-angle curve of dissipative angle curve is an undisplaced network: 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 the simultaneous opening of the circuit breakers at both ends of the line. The fault is assumed to occur at point X of the simple system of Fig. 3, which consists of a generator connected through a double-

®-1H:~-~t-e )(

Generator

Fault

Infinite bus

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.

FIG.

circuit line to an infinite bus. The input to the generator and the voltage 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, normal conditions

Pm2

o

~o

FIG. 4. The equal-area criterion applied to a sustained fault on the power system of Fig. 3. The generator swings from the initial angle 80 to the maximum angle 8m determined by equality of areas Al and A 2• The system is stable when transmitting power Pi.

conditions: (1) normal, and (2) faulted. The horizontal line at distance Pi above the axis represents the constant input. The initial operating point is a at the intersection of the input and normal output curves. The initial displacement angle is ~o, and the initial relative angular velocity is zero. When the fault is applied, the operating point drops to b, directly below a on the fault output curve. The dis-

128

THE EQUAL-AREA CRITERION FOR STABILITY

placement angle remains 00 at the instant of fault application. There is then an accelerating power, P a = Pi - Pi; represented by the length abo As a consequence the generator is accelerated, the displacement angle increases, and the operating point moves along the curve from b toward c. As it does so, the accelerating power and the acceleration decrease, becoming zero at c. At this point, however, the speed of the generator is greater than that of the infinite bus, and the angle 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, it becomes zero. Here the maximum angular displacement Om is reached. There is still a retarding torque; therefore the speed of the generator P

Pm2 Pi

t--~~~~II6IW.'-"e""'~"*""I.QaQ..

o FIG. 5. Application of the equal-area criterion to finding the power limit of the power system of Fig. 3 with a sustained fault. The input line is raised from its position 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. The displacement angle 0 decreases, and the operating point moves from d through c toward b. The system is stable. The operating point would continue to oscillate between band d if there were no damping. Actually the oscillations diminish, and the operating point finally becomes established at c. If the initial load on the generator were increased, as represented by raising the input line, areas Al and A 2 and the maximum angle Om would increase. The greatest value which Pi could have without the machine going out of step during the existence of the fault would be that value which makes Om occur at the intersection of the input curve and the fault output curve, as shown in Fig. 5. This is the critical condition in which both the speed and the acceleration become zero simultaneously 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 A 2 in Fig. 5 would be smaller than area A 1. The generator would reach point e on the curve,

129

APPLICATIONS OF THE CRITERION

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), but not enough to prevent loss of synchronism. If input Pi were greater than Pm2, the maximum output with the fault on, there would be no retardation whatever; and, of course, the system would be unstable with a sustained fault. p

Pma

-----

Pm2

o FIG.

8

6. The equal-area criterion applied to the power system of Fig. 3 for a fault cleared 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 with the system intact, (2) for the fault condition, and (3) for the post-fault or cleared condition with the faulted line disconnected. These curves are shown in Fig. 6. As in case 1, the initial angle 00 is determined by the intersection of the input line and the pre-fault output curve (point a). Application of the fault causes the operating point to drop from a to b on the fault output curve, and the accelerating power causes it to move along the curve from b to c. We may assume that, when point c is reached, the circuit breakers open, clearing the fault. The operating point then jumps up to e on the post-fault output curve and travels along that curve to j, where area A 2 = defg equals area At = abed. With a cleared fault, as with a sustained fault, a higher input (and initial output) would cause point f to move to the right until at the

130

THE EQUAL-AREA CRITERION FOR STABILITY

stability limit f would coincide with h. A still higher value of Pi would lead 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 clearing angle 6c• For any given initial load there is a critical clearing angle. If the actual clearing angle is smaller than the critical value, the system is stable; if larger, the system is unstable. Ordinarily, the clearing angle 6c is not known directly; instead, the clearing time (sum of relay time and breaker time) is known. To determine the clearing angle from a knowledge of the clearing time, the swing 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 to a 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 the conditions of Example 4, Chapter II. From the swing curve for a sustained fault 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 and the input line are plotted in Fig. 7. The initial operating point a lies at the intersection of the input line and the pre-fault output curve, The initial angular displacement 60, as determined from this intersection, is about 180 • (It was computed as 18.10 in Example 4.) Upon occurrence of the fault the operating point drops to b on the fault output curve and then moves along that curve. As it moves from b to c, the machine is accelerated and acquires kinetic energy proportional to At (the shaded area abc). By counting small squares on the graph paper, we find this area to be 19.3 squares. The operating point continues to move along the fault output curve, and, as it moves from c to d, the machine is retarded. Area A 2 is found to be -11.4 squares, giving a net area out to point d of 19.3 - 11.4 = 7.9 squares. Therefore, when the machine reaches a value of a equal to the abscissa of point d, it still has a positive velocity relative to the infinite bus. As the operating point moves from d to e, the machine is again accelerated, and obviously it would pull out of step if the fault were not cleared. Indeed, this was found to be so in the point-by-point calculation of the swing curve of Fig. 9, Chapter II, for a sustained fault. When the fault is cleared, the operating point jumps up to the post-fault output curve, or from point e to

APPLICATIONS OF THE CRITERION

131

point g, and then moves along this curve. As it moves from g to h; the machine is retarded. The relative velocity will become zero when the net area becomes zero. The critical condition is that in which the relative velocity becomes zero just as the retarding power becomes zero. Geometrically, the condition is that in which the net area out to intersection h is zero. This condition is found by sliding the vertical line efg from left to right or from right to left until the area At A 2 As A 4 = o. This is

+ + +

3.0

~

2.5

/

/

2.0

i'< i\ -----

~Pre·

~

V ~ -.....

r

~

~

~~L

.".

--,-,

C)

Fault cleared at 0.6 sec.-

~~

"

1\,

Fault cleared at 0.4 sec.- ~~

\

0.4

0.3

0.5

~~

0.6

0.7

0.8

0.9

t (seconds) FIG.

11. Swing curves determined by graphical integration. (Example 3.) Small circles are points on swing curves computed point by point.

The area under the curve of P a against subsequent angle 0 is

A1'

~

from the clearing angle to any

r p Gda = JlOO° t" (0.80- 2.06sino)da

= J8

c

= [0.800 + 2.06 X 57.3 cos 0]~()()'

+ 118.0 cos ~ = 0.800 + 118.0 cos 0 = 0.80~ + 118.0 cos ~ -

= 0.800

0.80 X 100 - 118.0(-0.1736) 80.0

+ 20.5

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 100

0 -

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, and 1/w' is plotted as a function of 0 in the curve labeled "after clearing at

SWING CURVE BY GRAPHICAL INTEGRATION

145

100°" in Fig. 10. It will be noted that this curve crosses the previous curve (marked "before clearing") at 100°.

TABLE 3 At

COMPUTATION OF

B

(deg.)

0 15 30 45 60

75 85 90

95 100 102 104.4

AND l/w' FOR FAULT CLEARED IN (EXAMPLE 3, PART b)

0.4

SEC.

l/w' cos 8

118.0 cos 8

0.808

Sum

At

(msec. per deg.)

1.000 0.966 0.866 0.707 0.500 0.259 0.087 0 -0.087 -0.174 -0.208 -0.249

118.0 114.0 102.1 83.4 59.0 30.6 10.3 0 -10.3 -20.5 -24.6 -29.4

0 12.0 24.0 36.0 48.0 60.0 68.0 72.0 76.0 80.0 81.6 83.6

118.0 126.0 126.1 119.4 107.0 90.6 78.3 72.0 65.7 59.5 57.0 54.2

63.7 71.7 71.8 65.1 52.7 36.3 24.0 17.7 11.4 5.2 2.7 -0.1

1.42 1.34 1.34 1.40 1.56 1.88 2.31 2.69 3.35 4.97 6.9 00

For a fault cleared at 0.4 sec. (100°) the swing curve is found by taking t equal to the area under the curves of l/w ' versus 0, as follows: on the curve marked "before clearing" to the intersection; thence on the curve marked "after clearing at 100°" to the maximum angle, which is found by setting Al = 0 in eq. j. This angle is about 104.4°. At this point w' changes from positive to negative, and 1/ w' does likewise. The area under the curve of negative values of 1/w', taken with negative increments of 0, is positive and may be found equally well by using the curve plotted for positive values of 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 the ordinate l/w' becomes infinite at some negative value of 0; however, the curve in Fig. 10 is not drawn that far to the left. The area under the curve in the neighborhood of the maximum angle, say from 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 At 103°, by eq. g,

o.

P a(103°) = 0.80 - 2.06 sin 1030

= 0.80 -

2.06 X 0.974 = 0.80 - 2.00 = -1.20

t=

4(102

/2 X 2.56 X 10-1.20

"'J

(k) - 104.4)

= 0.032 sec.

(l)

146

THE EQUAL-AREA CRITERION FOR STABILITY

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 the machine to swing from 102° to 104.4°, however, is exactly equal to the time required 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

t

FOR FAULT CLEARED IN

1 I"l

0.4 SEC.

VERSUS

a AND COMPUTATION 3, PART b)

OF

(EXAMPLE

s

A2

t

a

A2

t

(deg.)

(sec.)

(sec.)

(deg.)

(sec.)

(sec.)

100 102 104.4 102 100 90 80 70

.....

0.400 0.411 0.443 0.475 0.486 0.521 0.544 0.563

60.

0.011 0.032 0.032 0.011 0.035 0.023 0.019

0.016 0.015 0.014 0.014 0.013 0.013 0.014

0.579 0.594 0.608 0.622 0.635 0.648 0.662

50

40 30 20 10 0

This curve agrees well with the points (marked by small circles) computed by point-by-point method 2. The curve of l/w' versus 0 and the swing curve for a fault cleared in 0.6 sec. (135°) have also been plotted in Figs. 10 and 11. The swing curve does not agree well with the one calculated by the point-by-point method. The clearing time and angle in this instance are so near the critical values that it is likely that even small discrepancies lead to a considerable divergence of the 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 Transient Stability," Elec. Eng., vol. 59, pp. 462-5, November, 1940. 3. O. G. C. DAHL, Electric Power Circuits, vol. II, Power System Stability, New York, 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 to examine the stability. of a two-machine system subjected to the following disturbances:

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 circuit breakers at both ends of the line, followed by the subsequent simultaneous reclosing 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 bus over a circuit of negligible resistance, is operating with an initial shaft load Po which is suddenly increased by an amount D.P. The power-angle curve has an amplitude Pm. Derive a formula for the critical load increment set»; as a function of the initial load PO/Pm. 'Plot the equation. How much error would there be in assuming that the equation could be represented by a straight line between the two points where the true curve intersects the axes of coordinates? 3. Derive a formula for the transient stability limit P L of a two-machine reactance system subjected to a sustained fault, expressing the limit in terms of the following quantities: P« = amplitude of pre-fault power-angle curve flP m

= amplitude of fault power-angle curve

ao =

initial angular displacement

Plot a curve of P L/Pm against fl. 4. Derive a formula for the transient stability limit of a two-machine reactance system subjected to a fault on a radial feeder and its subsequent clearing 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-machine reactance system subjected to a fault on a transmission, line and its subsequent clearing by simultaneous opening of the circuit breakers at both ends of the faulty line. The stability limit should be expressed in terms of the quantities 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 of one 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 transmission line, 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 any value of initial power. (It would be more accurate, but more laborious, to assume a different voltage behind transient reactance of the generator for

148

THE EQUAL-AREA CRITERION FOR STABILITY

each different value of initial power, so as to hold the initial terminal voltage constant.) 8. Using the method of graphical integration, plot the swing curve of the water-wheel generator of the system of Example 4, Chapter II, for an initial output of 1.00 p.u, and a three-phase fault at the middle of one transmission line cleared in 0.35 sec. by the simultaneous opening of circuit breakers at both ends of the line.

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. The equal-area .criterion, which was discussed in Chapter IV, is a very effective means of determining whether a given two-machine system is stable when subjected to a given disturbance. The most important type of disturbance is the occurrence of a fault and its subsequent clearing by the opening of circuit breakers. If the clearing time is known, as it usually is, the corresponding clearing angle must be found before the equal-area criterion can be applied; or, conversely, if the critical clearing angle for a given transmitted power is obtained from the equal-area criterion, the corresponding critical clearing time must be found in order that the results of the stability study will be in the most useful form. Perhaps the simplest way to find the clearing angle corresponding to a given clearing time, or the time corresponding to a given angle, is to refer to the appropriate swing curve in a set of swing curves which have already been calculated and plotted, and which may be appropriately called "pre-calculated swing curves." Such a set of curves, obtained by solving the swing equation on the M.I.T. integraph, was published by Summers and McClure 2 and is reproduced here (Figs. 1 to 10) by kind permission of Mr. McClure. The curves were derived for a sustained fault on a system consisting of a synchronous machine of finite size connected through reactance to 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 simplifying assumptions are made, to wit: constant input, no damping, and constant voltage behind direct-axis transient reactance. To make the curves generally applicable they are plotted in terms of a dimensionless variable, the "modified time" T, defined by eq. 11 below. The swing equation of a two-machine system is d2a M dt2 = P a

= Pi -

149

Pu

[1]

THE TWO-MACHINE SYSTEM

150

M 1M2 equivalent inertia constant in megajoule+ M 2 = seconds per electrical degree. [2]

where M = M 1

o = 01 t

- 02 = angular displacement in electrical degrees.

= time in seconds.

Pi =

M2Pil -

M

1

+

1P M • 1 · • M i 2 = equrva ent Input In megawatts

which, either for a reactance network or for M 2 = Pi

= Pil

[3]

2 00,

reduces to

= input of machine 1 (the generator)

[4]

The equivalent power output, dependent on a, is given by the powerangle equation:

[5] which, for a reactance network, reduces to

P« = PM sino

[6]

Expressions for Pc, 'Y, and PM are given by eqs. 28, 29, and 30 of Chapter 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 the direct-axis transient reactances of the machines themselves. Substitution of eq. 5 into eq. 1 gives d2 8

M dt2 = Pi - Po - PM sin (a.- 7) or [8] where

p/

= Pi - Po

0' = 0 - 'Y

[9] [10]

To put eq. 8 into dimensionless form, divide it by PM and then introduce a quantity r, defined by T

=

r;;:P; /7tJPM t\)180 M = t\)' GH

[11]

PRE-CALCULATED SWING CURVES

where f

151

= frequency in cycles per second.

GH = kinetic energy, in megajoules, of equivalent generator at rated speed = 180fM. The result is:

Pi, d2u~ ' r ., 180' dT2 = PM - sin 8

=P-

.,

sin 8

[12]

if a' is in electrical degrees, or simply d2~' u • ~, -=p-SIDu

d,,2

[13]

if a' is in electrical radians. Here

p/

p

= PM =

Pi - Pc 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 the amplitude of the power-angle curve (both input and amplitude being measured from the horizontal axis of symmetry of the sine curve if the sine curve is displaced vertically) and on the initial angle 50' and initial angular speed woo For the present purpose, swing curves for a sustained fault are wanted; hence Pn, PM, and 'Y must be the constants of the power-angle equation for the faulted condition, and the initial speed will always be zero. The solution then depends only upon p (defined by eq. 14) and 00'. Each family of curves in Figs. 1 to 10 is for a constant value of sin 00" the range covered being from 0 to 0.90 in steps of 0.10. The individual curves in each family are for constant values of p, ranging in each family from a minimum value slightly larger than the value of sin 00' for that family up to a maximum of 3.00. The procedure for using the pre-calculated curves to determine critical 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-be known, because they are needed for finding the critical clearing angle Dc by the equal-area criterion. 2. Compute p/ = Pi - Pc, P = p//PM , 00' = 00 - 1', sin 50" and oc' = Dc - 'Y.

152

THE TWO-MACHINE SYSTEM

160

10 ~ 60 c

4(

40 20

2

3

4

~

~

10

1. sin 00' = O.

FIG.

-

5 6 Modified time T

160

t----+--__t__

120

t---+--t--f-1H1--f-I-I+-"""'i-1I~f---;'+-"""f---J

Q)

"t:J

co 100 E

r---t---tt-..............,.-++--#-H-"'#--W'~'---

..----.--1

U

~ 80 t---+--ftl-f1H+-1-""~~~F--:::'''':=;'-4-~~--+-~--+---+_-+---+_-+---+~-t---i '0

~ 60 c:

c(

40

20

2

3

4

5

6

Modified time T FIG. 2.

FIGS.

sin 00'

7

8

9

10

= 0.10.

1 and 2. Pre-calculated swing curves (copied from Ref. 2 by permission).

153

PRE-CALCULATED SWING CURVES

160

....-

e120 J--+-~-U-J..I~-I-~~+'~-btc-t---t---ir--t--+-+--;--t--;--t--t---1 en

l

- 100 1--4--V-I-I-I-:H+-I-,~~~~~-~~~-+--i-t-;--;----r---t--t .~

-a

!

80 t--f--+It-#+~-

:0 CD

~ 60

<

40

1

2

3

4

5

6

7

8

9

10

Modified time T FIG. 3.

sin 80' = 0.20.

160

en ~

120 ....---+--4-1-1--I-I-+++I-~~~~ ~~+---+--t---+---+--f--+-+---+--I---+----4

bO Q) "'0

~

:so

100 t---+---I'"f-f-"""-N-#--It#---']~~

~=---+--+--t--+--~~--+--+--I---t--+---t

Q)

a;

80

1---+--f.j"""""f+BtI-I-.~~~~ _ _----,--,-+-,-t~~-+-+---f--t---Pli~-+--+--t

~

~ c:: 60 < 40 20 2

3

4

5

6

7

8

9

10

Modified time T FIG. FIGS.

4.

sin 80'

= 0.30.

3 and 4. Pre-calculated swing curves (copied from Ref. 2 by permission).

154

THE TWO-MACHINE SYSTEM

160

U)

~

to

120 I--.......---+++-+I+-It-+-#+-I--i+--#t~~-t--:--~~-i--t---t-+---+-t----t--i---t

Q)

"0

~

.s u

100 1---+--........# -I-.........+#-,j~~-f#--+::."e-+----+--+---+--+---+--t---i--+--1f---+--+---I

Q)

~

80 1--;--tt-tH'iHfiJ~t-T...,.-t~.............::::--t---t----t~+--t--t---t--t--t--t----1

:0

~ 60 c

-c

40 20

t--+--+-...---+---+-t--+---+-t--+--+--+--I---+-+--+--i--+---+--I

2

3

4

5

6

7

8

9

10

Modified time T FIG. 5.

sin 80' = 0.40.

160 t---+---+-

-! U)

120 t---+--..........-+4--#-~~~~++--+-+--

-bIIlI",--+-+--+--t---f---t---t--+---I'---+---f

~

"0

co 100 1---+--1lI----jHf-,I--It--#-#-.,IJ--,~L.-~~ U

:su

:e Q)

80 t---+-~~~

~

~ 60 c

c(

40 20 t--+--+-+---+---+--+---+-I---+---+--+---+--I---+--+-+---+---Io-o+----I

o

, - O....... . - . . _......- - " ' _......._ • ..&._oIoo---A.o_oI---I-

o

1

2

3

4

--...._..&.--.._...........-..

.....&.-

5

6

7

8

9

10

Modified time T FIG. 6. FIGS.

sin 80'

= 0.50.

5 and 6. Pre-calculated swing curves (copied from Ref. 2 by permission).

155

PRE-CALCULATED SWING CURVES

160 I---+---t--

-t

120 1-+--~-v-I--W-+-I--#+I---+-J~iL+---+---+-J.~--I:--~4----+---4-~

(ij

100 1---+-~~'-I-+I-~~~+--~~-+---+--+-t----+---+--+---+---f---4---4

f -c U

:a

~ 80 ~-+-I-I-fI.I-#I:-I--I-I-~~~f==--+--+--1---l-~~I---f---+---I---+-----I----4~ ~

~ 60 c <

20 J--+--+--+--+--+---lf--f--+--+--1---+--+-I--+--+--+--+--+-----4---1 O'"-...&---'----"--..J"".......,j",----"""-.J.-....r.-....l.---L---I-......,j",_~""'--...-.--L---I...--'-......J___'

o

2

3

4

5

6

7

8

9

10

Modified time T FIG. 7.

en

~

sin 00'

= 0.60.

120 t---+--~_f_li-f+_#_t.i~tI---l#__-t-#---f--+::.~--I-+_+__-f---I---+----+~1---I

~ -c

B 100 t--t--~~Jt/_--Ih~~~~+--+--+--+~-+-+--I--I--.J-----I-~~

:a (1)

~

80

t-~f_IrI_I~~~~r-::.-'f_+--+---:&-~--I-f_1-~~--I-~--l~

°"-. .... o

-..J"".---"'_a...-""'---....I.-......L---J.,;.--I..--L_"'---""""--'----&...--'---L---I-J

2

3

4

5

6

1

8

9

10

Modified time T FIG. FIGS.

8. sin 00'

= 0.70.

7 and 8. Pre-calculated swing curves (copied from Ref. 2 by permission).

THE TWO-MACHINE SYSTEM

156

O'-J...-,..L--'-...L-...L..--1.-...I.--J.---L.--.L-~---L-..-.I--...a..---L---L.~-.l~.a...--..a

o

1

4

3

2

5

7

6

8

9

10

Modified time T FIG. 9.

180

ltJt: ;~I Q'11~j J

~ol ~

-

~

1111 III Ii /

'I

140

~

00

j~

I

'1IJ 'I

J

~ Q,) '0

100

f--

~ 80

~

"0

~ 60 c:

c:(

~

~~

2

3

.......

)

V

~

-

j

/ JI/

:Y

/

V ~

1/

Ji7

V

r

~~

= 0.80.

7"

if

JJ flli rJj Ifh '/ V ~ V/ ~ V ~ V ~~ ~ ""/ l;"~ ~ ,.....

I

'0 Q,)

II)

'/JIf/ /

'II II

~ 120

..

~

11// "I / IJ vr II if/ 7

en

co -g

o~o

o " 11).'" ': ~ ~;;;;-;; ;; n:"' 8°011)

160

sin 00'

r7

.~'

~.,

V

V

»> 095 ~

_""-......-

---

t--.. r---.

~

40

20

o

o

4

FIG. 10. FIGS.

5 6 Modified time T

7

8

9

ro

sin 00' = 0.90.

9 and 10. Pre-calculated swing curves (copied from Ref. 2 by permission).

PRE-CALCULATED SWING CURVES

157

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 the ordinate 0' = Oe' and read the corresponding abscissa T = T e• Interpolation between curves or between families of curves may be necessary. 5. By eq. 11 compute the critical clearing time te corresponding to Te. To determine the clearing angle corresponding to a given clearing time, the order of the steps of procedure is altered in a way that should be obvious. The procedure described above breaks down if the fault is of such nature that there is no synchronizing power while the fault is on. In such a case PM = 0, from which it follows that p = 00 and T = 0 for all values of t. The pre-calculated curve for this condition is the vertical coordinate axis, and the relation between 0'" and t cannot be determined 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 represent conditions after clearing a fault because the curve for the proper value of angle and speed (at the instant of clearing) does not have the proper value of accelerating power or acceleration after clearing. EXAMPLE

1

In Example 1 of Chapter IV, which deals with a machine connected through 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 be 1380 • The corresponding critical clearing time, as determined from the swing curve, is 0.61 sec. Check this value by use of the pre-calculated swing curves. Solution. From Example 4 of Chapter II, the power-angle equation valid for the fault condition is

pu

= 0.936 sin 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;

THE TWO-MACHINE SYSTEM

158

and the initial angle is whence sin ~o = 0.310. From the problem statement, Oc = 1380

The following quantities are computed from the data: Pi' p

= P, = P/ = PM

= Pi = 0.80

Pc

0.80 = 0.854 0.936

sin 00' = sin 00 = 0.310 ~c' =

Dc = 1380

M = M1 = 2.56 X 10-4

~= t

J 1I"PM \} 180M

=

I

11" X 0.936 = 80 \} 180 X 2.56 X 10-4 ·

The most suitable pre-calculated curve is that for sin 00 = 0.30 and p = 0.85 in Fig. 4. The ordinate 0' = 1380 corresponds to the abscissa If" = 4.8. Hence T c = 4.8 and T

4.8

tc = - c = 8.0

8.0

= 0.60 sec.

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 the system of Example 4, Chapter III, which consists of two finite machines connected through an impedance network. (The equal-area criterion was applied to this system in Example 2, Chapter IV.) Solution. The following data are obtained from Example 2 of Chapter

IV: 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°

EFFECT OF FAULT-CLEARING TIME

159

and the following from the statement of this problem: t

= 0.30 sec.

The following quantities are computed from the data:

Pi'

=

P=

Pi - Pc

Pi' = PM

= 0.41 -

(-0.04)

0.45 = 2.2 0.203

00' = 00 - 'Y = 12.7 - (-2.9) sin 00'

T~

p

= 15.6

= 0.269

"t= '1180 M= T

= 0.45

1r X 0.203 180 X 2.64 X 10-4

= 3.66t = 3.66 X 0.30

=

3.66

= 1.1

The most suitable pre-calculated curve is that for sin 00 = 0.30 and On this curve 'T = 1.1 corresponds to 0' = 76°.

= 2.0 in Fig. 4.

o= 0' + 'Y = 76° -

3° = 73°

From Table 15, Example 4, Chapter III, at t = 0.3 sec. we find = 75.3°. This value was obtained by a point-by-point calculation.

OAD

The

agreement is reasonably good.

Effect of fault-clearing time on transient stability limit. The amount of power that can be transmitted from one' machine to the other in a two-machine system without loss of synchronism when the system is subjected to a fault depends on the duration of the fault. The power limit can be determined as a function of clearing angle by the equal-area criterion, and the relation between clearing angle and clearing time can be found from the pre-calculated swing curves. It is then possible to plot a curve of stability limit as a function of clearing time. Such a curve is shown in Fig. 11. It shows that the transient stability limit of the system can be greatly increased by decreasing the time of fault clearing from 0.5 sec. or more to 0.2 sec. or less. The time of fault clearing is the sum of the time that the protective relays take to close the circuit-breaker trip circuit and the time required by the circuit breaker to interrupt the fault current. Frequently a system which is unstable for a particular type of fault and fault location can be made stable by altering the existing relaying or by modernizing the circuit 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. Protective relaying is discussed in Chapter IX, Vol. II.

THE TWO-MACHINE SYSTEM

160

A curve like that of Fig. 11 can be obtained by the following procedure. First, the equal-area criterion is used to determine the stability limit with instantaneous clearing. Truly instantaneous clearing is not

...

~

'2 ::s

1

...

(1)

...c.

~

~

:c nJ en '----......

o

FIG.

---_....................---_..............-- !h 0.5

1.0

(X)

Fault- clearing time (seconds)

11. Curve of stability limit as a function of fault duration.

obtainable in practice, but it may be regarded as the limit approached as the clearing time is reduced. The stability limit for instantaneous clearing is the same as that for disconnection of the faulted line when there is no fault on it. This limit is determined as shown in Fig. 12.

o

00

01

r/2

Om

Angle 0 FIG.

12. Determination of stability limit for instantaneous fault clearing by use of 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 area A l equals area A 2 • It should be noted that moving this line changes the 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 in Fig. 11 as point 1 at zero clearing time. Next the stability limit for a sustained fault is found as shown in Fig. 13. The power-angle curve for the fault condition is used in place of the post-fault curve; in other respects the determination of the stability 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

CD Co

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 is found by the equal-area criterion as shown in Fig. 14. Power-angle curves are drawn for the output before the fault, during the fault, and after the fault, and the input line is drawn at one of the selected values of initial power. The vertical line at the clearing angle ac is shifted from right to left until areas Al and A 2 are equal, thus fixing the critical clearing angle. The clearing time corresponding to this clearing angle is determined from the appropriate pre-calculated swing curve. Point 3, Fig. 11, is then plotted, its coordinates being the assumed 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 are assumed and the corresponding clearing times found, is simpler than

THE TWO-MACHINE SYSTEM

162

the alternative procedure in which clearing times are assumed and the corresponding power limits found. For in the latter procedure the horizontal input line is shifted, resulting in a shift also of the vertical line at the initial angle 00, whereas in the former procedure only one line, 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 decrease of 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 the most effective and important means of improving power-system stability. EXAMPLE

3

Plot stability limit in per unit as a function of the fault duration for a three-phase short circuit (a) at the middle of one of the parallel transmission lines of the power system shown in Fig. 15 and (b) at the sending end of one of the lines. The fault is cleared in both cases by the simultaneous opening of the circuit breakers at both ends of the line. The system consists of a hydroelectric station sending power over two parallel transmission lines at generator voltage to a metropolitan system which may be considered an infinite bus. The following data pertain to the system. The base power is the aggregate rating of the hydroelectric generators. Direct-axis transient reactance of hydroelectric generators: 0.35 per unit

EFFECT OF FAULT-CLEARING TIME

163

Stored energy of hydroelectric generators, H: ~.OO Mj. per M va. of rating Frequency: 60 c.p.s, Voltage behind transient reactance of hydroelectric generators: 1.00 per unit Voltage of infinite bus: 1.00 per unit Reactance of each transmission line: 0.40 per unit (neglect resistance) Reactance of transformers at receiving end of lines: 0.10 per unit Fault

Fault

"

-i= -1:fJ t-- -I~~~ C:i

a.....,1

J- - i -

I I/

J

= 0.20.

I-d I'

-~

a..'"

80

~4'01 II

J

~ilI"""

-~

~

~

i-

J

,

0.4

-i-

II

I'

I

I

l-

~

,•

a!' 0.6

0.4

J

J

~

~

1.0

I

I J

FIG. 25. sin

0.8

1...'"

II

J

-,"

~

11

4,.'" I

I

,,-

.,

'0 '

II

i-I--

,

raj

If? '

--0 - II J

i-i-

i-i-r-

I • I ,

• •

I

J

0.6

0.2

,

,

--fir) -.... :

~J -=..-:' ,,

0

0.8

I I

I

1-+- 1--0

171

~

'I

-

~

~

~ ~

---,-

.,

~

~I-'

--

,--

~

'" p

0.2

FIGS.

o

1.0

2.0

3.0

4.0

5.0

FIG. 27. sin 80 = 0.30. 25, 26, 27. Curves for determination of clearing time (copied from Ref. 3 by permission).

THE TWO-MACHINE SYSTEM

172 1.0

-- -~;U) -0 -- 0 ...... -

2 2 9i>-fil ~;'~'t ~It- -t~t~-0.8 ---- .... ~ .... ... -- :-c: - .... 11

/I

__ a..;

II "'~1 'I

II

I II

II

'I

4.'"'f

J

J

" "v v

...... 0.6

~

v

J

II ~

J

0.4

-- L."l

I~

~

l,;Ill

"

'-~"""

O'

~: _ "

- II

. .v..,/

If'

I"

I;'"

I......

......

..-~iiI""

~

~""111""

-~

I I I

~

I1

J If

.I-

t-~ ........

rl~ , 0 -- ....

_i-~

."

...,

rj'-

---

v

L."l

I......

L....~

IlllI'

~

:1 t~i~I~

4.'"'f '- _ i - - - 4.'"'f.,-._ j J

I-

~

J

--

II

II

1/

1-1-1--

...

L-~ _I.-

~

.... ~

1...- ......

I I I I

0.2

I I I

o

2.0

1.0

3.0

FIG. 28. sin 80

1.0 ~~

1-1-

0.8

I II

I I II 1-1-

~~ ~I-

I-

~8r~

~~ ~j ...~~ c: I

.... 0.6

J ,,~

,

I I

[ .... W~ -

1--

4."'f I

I

--

l...oI~

I I

1/

IIT:J:,. It~

I II

~

,-'"

1/

I

O' II ~

1-..... ...O' .... _~" ~f'''''Ij~~~-

O'

II

~ f - - ~'"'f

,

if" l....IIlI

..-~

1-'

= 0.35.

II~~ f. . . 1-&0 ~~~ . . . . ril~t ~~~:~ tt; ~~ P" 1/ ""--1- 1/ .... J ~ ~';. tt .... ' ' '1} ... '~1~1,, , ,.

t:i ~- II t..~ .... ~ II ~-

I

I~

I

~

5.0

4.0

O' ~

~~

1-1-

-\""if"

~

~

~

~

I~

- '--~

_..... . - ... .

....

~

~

~

~ J

~

~

J

~

0.4

0.2

o

1.0

2.0

FIG. 29. sin

Tc

80

3.0

4.0

5.0

= 0.40.

Tc

FIG. 30. sin FIGS.

28, 29, 30.

80 = 0.45.

Curves for determination of clearing time (copied from Ref. 3 by permission).

173

CURVES DETERMINING CLEARING TIME

I II

II

II

J

I II II

1 'I

I J

I II

1

"J

---..........,,-'-'

o

FIG.

I-

I I It)

0.9

III-

0

""... ~... ~

"

"...

"

l-

II

I I

..."

I

,

,,

J

I

I,

'I

~

~

IlIlII'

~

,,-

..,

J If

ti- ~l- I-

I I

~

" ..,

L-'-'

_.... --

~

I..-

1...-"_L- I.~

~

J~ ,...

II

c::r ~-

II II

" ~,... fort-fo- ....

~ ....

(...,

'f

IJ

l-I-

v

v

~ ~

~

I I I I f I

1

I~

J

(,.....,

1-1-

1.11 J

II

II

T

5.0

~~Ij 1-1~ 'I

t~itr

I-

IJ

J

II

II

,

0.7

I-~

L

t;;:)'

II i~ 1-1- ~ ......., I-

,

i

I

I

~'/-

II

I-~ I-

J

, ,

j

I

'OJ L-L-

'I

J

0.8

I

...............

'I

4.0

ao = 0.50.

31.. sin

o Oi' II : II

...r....I

...-..

3.0

f1- ~gl~, ~~;:

fS;

0

1-0

11

J

..Io...I~"'-'-'I-Io.

2.0

1.0

IJ

J

'(

v

~

0.5 '-'-__

1.0 I-

II

'II]

J 1/

J

~ ~

_........

l...oillill"'"

'"

"

... ""'"

0.6

0.5

o

2.0

1.0 FIG.

FIGB. 31 and 32.

3.0

4.0

5.0

Tc

32.

sin

ao

= 0.55.

Curves for determination of clearing time (copied from Ref. 3 by permission).

THE TWO-MACHINE SYSTEM

174

...e-t

0.8 II

,

~

~~~~~

~

~

~~~~

FIG.

1.0

-

~

,~

-~~

__

_~~

~

= 0.60.

33. sin 80

s I,O~ -~~ ~~ ~14-~L~_ ~- t8 /f{ ~I+~I ~ ~ .~f- ~~~~ ~>o-F#f8l ~ c;-c; ()" ~ t::)~" I I I

I

I

0

-10

0

~ I.;.J '~

",

0.9 c:

/I

, ....

J

I

I

If II ~ J

/I

1..."" J

I I

..." 0.8

I

If' v:

-~"

I

~.-

II -~1- /t t.- ~_ /t ~II.: ~"1..."" zr q I" ...::: -~ """-1 ~""-~II- "...,-:: '\~2~: ,\'\0 ~

~

II

, v

I

I~

I

J

I)

~

~

~

~f'

--'"

~

~

~

~~

~

~f'

~

iIooolI""~

,,~

~

~

I~ ~

~

~"" .....

.-

.... ....

,,-

- ...... ...... ......

I.-~

~

t::)"

"-~

->0--

-l-

4'~-

~

I lA" I

~

~

~

1:1 0 _

~ ~O"

~~

1\L..;;;..... 1III!!!~

L..oo~

~---'-~ """'10-- ....

~

'('\!

-~

0.7 0.6

1.0

2.0 FIG. 34.

1.Q

-t.~~~&~ l$qr!J "I/Ht Ht: I

-

0

-_0

o

c::i

O'

Ci

- " "14,..,.

0.9 -- ...... ""'J I I I

0.8

",

,

II

I~

v-"

J

I I O~O I ~

I I ()-:" I ~

0(",

~'\

~

....

-

~

-~

...

~

~

2.0

1.0 FIG.

35. sin 80

~

_....

3.0

~ 11""-

_....

.~ ~

i~

~ ~

~

~

, II' , 11~~ co'> v o· Alo,,, oep " W ~!~~ ~~O~""I~~ ~~I ~............~J -"I" ~ ,jf- ..... E 2 ) . In addition to the upward shift of the power-angle curve when M 2 > MI , there is a small shift to the left, which is beneficial. A thorough investigation of the methods of grounding would include a study of the effects of varying the magnitude of the grounding reactance or resistance on power limit and also on the line-to-neutral voltages. These matters, however, will not be considered here.

Two-phase coordlnates.P-l" Symmetrical components, which have been used in the preceding part of this chapter for analyzing unbalanced three-phase circuits, are a kind of substituted variables. Another kind of substituted variables, sometimes used for the same purpose, are two-phase coordinates. Two-phase components of current and voltage are defined as follows:§

I, =

i (2Ia -

Ib

-

Ie)

§In Ref. 14 a, {j, and 0 components are defined in a slightly different way.

[71a]

SOLUTION OF FAULTED NET\VORKS

246

III = -

1

va (Ie -

[7tb]

Ib)

I z = l(la+ Ib + Ie)

[71c]

Vz = !(2Va

[72a]

V1I =

Vb - Vc )

-

va1 (V

c -

v, = leVa +

Vb

Vb)

[72b]

+ Vc )

[72c]

Note that the expression for I, differs from that for V z by a factor 2. If eqs. 71 and 72 are solved for the phase quantities in terms of their two-phase components, the following expressions are obtained: la = Ix + !Iz 1

r, =

-"2 I x

I, =

-"2I z

1

-

[73a]

2V3 r, + '21 I z

[73b]

va III + 2"Iz

[73c]

+2

1

Va = Vx + Vz 1

[74a]

va

+

Vz

[74b]

v, = -7jVx + 2 Vy + v,

[74c]

Vb = -2 Vx 1

- - 2 Vy

va

Relation to symmetrical components. Two-phase components are related to symmetrical components as follows:

r, = 11 + 12

[75a]

v. =

Iy

[75b]

Vy = j(V1

[75c]

v, = v;

=

j(II - 12)

VI + V2 -

V2 )

[76a] [76b] [76c]

Note that the x current and voltage are equal to the sum of the positive- and negative-sequence components, andthe y current and voltage are 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 V x = VI and V y = jV1, which are balanced two-phase voltages of phase order y, x. Similarly, the two-phase components of negative-sequence three-phase voltages are two-phase voltages 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 y voltages and currents, in another called the y network; and the z voltages and currents, in a third called the z network. These networks are analogous to the sequence networks used in the method of symmetrical components and may be given the more general name of "substitute networks." The substitute networks representing a balanced threephase network whose positive- and negative-sequence impedances are equal are independent of one another. The x and y impedances are equal to each ather and to the positive- and negative-sequence impedances. The z impedances are one-half the corresponding zero-sequence impedances. 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 use of two-phase coordinates practicable. If the three-phase network contains balanced, positive-sequence generated electromotive forces, then the x and y networks contain electromotive forces of equal magnitude, those in the y network leading those in the x network by 90°. Ordinarily, the z network contains no generated electromotive forces. The self- and mutual impedances of the substitute networks and the connections between the networks corresponding to any given impedances and connections of the three-phase network or any portion of it may be found by the process previously described for symmetrical components and illustrated by the derivation of equivalent circuits for representing several types of short circuit. Connections between the substitute networks for representing short circuits are shown in Fig. 33.13 A line-to-ground short circuit is represented by connecting the x and z networks in series and leaving the y network open. A line-to-line short circuit is represented by shortcircuiting the y network and leaving the x and z networks open. The z network is dead and need not be set up. However, the x network in this case and the y network in the case of a line-to-ground fault, al-

248

SOLUTION OF FAULTED NETWORKS

though not connected to anything at the point of fault, contain generated voltages and carry normal load currents.] A two-line-to-ground fault is represented by short-circuiting the y network and paralleling the x and z networks through a transformer of 2 : 1 ratio. If the impedances of the z network are given four times their normal value ao

3

2

:+

C"I

~l

Study 1. Positive-sequence network as prepared for a-c. calculating board. Impedances are in ohms on board or in per cent on a 48.4-Mw. base; shunt capacitances are in microfarads on board.

66 ky.

~T

BL

BL

~T

~T

FIG. 2.

;::

;-

eo

.c

.t

~

~

:+

:e

C!

BE

00

t:rj

~

~

§

to<

~

~

t"'1

~

~ OJ

rn

o

~

~ > t"'1

~

~ 0)

o

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, and large motors) on the systems of both companies with their ratings, transient reactances in per cent based on their own ratings, moments of inertia (WR2 ) in thousands of pound-feet", and amounts of kinetic energy in megajoules. Table 2 lists the synchronous-machine stations (or in some cases groups of neighboring stations) represented in this study, with their aggregate megavolt-ampere ratings and amounts of kinetic energy. Figure 2 is a diagram of the positive-sequence network, simplified to the extent required for setting it up on the calculating board. The zero-sequence diagram is not reproduced here. The oil circuit breakers and protective relays on these systems were of slow-speed types at the time this study was made. Loading conditions. Transient stability was studied for two different loading conditions: (1) estimated August day loads and (2) estimated October night loads. (The loading of the various circuits for the two load conditions is shown in Figs. 3 and 4, respectively.) The first condition was one of heavy loads, due largely to pumping for irrigation, and involved a considerable transfer of power from company B to company A (30 Mw. from station BB to station AD). The second condition was one of light loads but of even greater transfer of power (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 the power transferred to company A was necessarily transmitted from sources (chieflyfrom stations BE and BF) more distant than in load condition 1. The calculating board used in this study had ten power sources for representing generators. In order to check connections and settings of the board and also to help determine which generators could be grouped together for transient stability studies, readings for load condition 1 were taken, first with company B generators represented in considerable detail and then with company A generators represented in considerable detail, according to the columns of Table 3 headed Run 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 3 for runs 3, 4, 5, or 6, according to fault location, to permit their representation by ten power sources on the board. Individual line and generator loads were somewhat distorted by the grouping; the readings are not reproduced here. Readings for load condition 2 were taken with the generators grouped as shown in Table 3 for run 7, and these readings are shown in Fig. 4.

262

TYPICAL STABILITY STUDIES

STUDY i-LOADING CONDITIONS

~

...

-.~9·O

-"b~'£)

;~..----~ --.tn· 8S't J

.

~=

~~

; 0 -..(&rU ~ CQ .... 0...

~

(tn)

...

"CLt'r) ~~'l

263

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

Run 1

Run 2

Runs 3 and 4-

RunS

Run 6

Run 7

1 22.5

Off

BO BL BI

Static*

BN

2

2 33.8

1

18.3

BK

3

3

2

9.8

3

44.8

4

15.9

5

30.8

6

55.7

7

3.2

BHs

Static

1

4

BH1 BG (66kv.) BG (44 kv.)

5

BE

7

BF BD BC

8

1

Off

2

3

9 10

1 2

BB (44 kv.)

3

AA AG

4 Off

AB

5

AC

6

AF

7

AN

Off

1

66.1

69.6

4 25.6

44.8

3

15.9

4

30.8

8 30.8

5

38.8

Off

9 16.9

4

16.9

6

16.9

5

21.4

7

21.4

6

46.6

8

56.6

7

10.0 Off

9.8

2 66.4

6

BB (6.6 kv.) Load box

AH

66.1

5 19.2 6 10.9 7

5.0

Off With AP

Off

10 8 376.2

9 376.2

8 154.2 Q()

(Infinite bus)

8

AI

9 199.4

AL

9

AE

Static

AP

10

AV

Static

9

391.0 10 548.1

10 157.1

10 170.2 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 determine maximum allowable clearing times for faults at locations where they would affect the stability of the interconnection. It is apparent that a fault anywhere on the single-circuit 132-kv. lines from station AD to station BE would split the system into two parts unless high-speed reclosing were used. Since high-speed reclosing of high-voltage lines was still in the experimental stage at the time of this study, however, it was not considered. Hence faults on the single-circuit 132-kv. lines were not studied. Faults on anyone circuit of the three-circuit 132kv. connection between stations AB and AE would not divide the system, and the critical clearing time of such faults had to be determined. Since the impedance of two or three 132-kv. lines in parallel is very low, the exact location of a fault on these lines would not make much difference. But a fault nearstation AD on one of the three 13S-kv. lines to station AC was believed to have a somewhat greater tendency to produce loss of synchronism between the machines of company A and 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 near the 44-kv. terminals of a large 132-to-44-kv. transformer bank. Two such 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, the other near station BE on the 44-kv. line to station BG. The critical clearing time of faults elsewhere on the 44-kv. lines would be expected to 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 of the interconnection. A fault location nearstation BG on the #-kv. line to station BH was tried to see whether it would disturb the interconnection. The four fault locations chosen for study and described above are marked 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 at both ends of the faulted line was assumed. Swing curves. Eighteen transient-stability runs were made in the course of the study. The swing curves obtained are reproduced in Figs. 5 to 22 inclusive, and the conditions and results of each run are summarized 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

2LG

1

"

3-8-1

3-8-2

"u

3-S-6 3-8-7 3-8-8

"

"

cc

"

"

BE

"

AD

BG

"

AC

BH

" "

BG

" " " " " "

u

r

0.2

0.3

0.4

0.3

0.7 0.8 0.3 0.7 0.3 0.2 0.2

0.6

0.4

Same as at near end except where noted

0.4 0.4 0.3 0.2 0.4 0.5

Far

End

End

Near

....

Clearing Time (sec.)

,

•Load Condition 1: estimated August day loads; load condition 2: estimated October night loads.

44

132

a

cc

2

BG

" "

"

u

"

a

It

"

It

"

BE

44

"

" "

5-8-1

5-8-2 5-8-3 6-8-1 7-8-1 7-8-2 7-8-3

"

2LG

" "

It

"

II

3t/I

It

It

3-8-10 4-8-1 4-8-2

a

"

II

II

II

" " " " " " "

"

IC

2LG

II

II

3-8-5

AC

AD

"

u

132

"

II

"

AS

AE

II

On Line to

Near

Station

Line

Voltage (kv.) 44

3-S-3 3-S-4

" "

3t/I

Fault Type

Load Oondition"

Run

Fault Location

SUIOlABY OJ' STABILITY RUNS (STUDT 1)

TABLE 4

Stable Stable Unstable Stable Stable Stable Unstable Unstable Stable Unstable Stable Unstable Stable Stable Stable Unstable Stable Stable

Stable or Unstable

"

" " "

"

Series R at sta. BB Low:xti at sta, BB Shunt X at sta. BB

Remarks

t-4

00

t:J=j

tj

d

~

00

t<

~

eH

H

6;

~

00

a > e-

H

t-3

:;3

~ C) C)

STUDY i-SWING CURVES

267

320 r----oyo---r--.,.---r--r-~-__r_-_..._-....

280 ..--t---t--

2401---t---t--

40

9 (AE, AL)

0.2

0.4

0.6

ne (,seconds) FIG. 5. Swing curves, Study 1, stability run 3-8-3, load condition 1, three-phase fault near substation AD on one 132-ky. line to station AC, cleared in 0.3 sec. Unstable.

TYPICAL STABILITY STUDIES

268

stability run, and 8-1 refers to the first transient-stability run made with these initial conditions, 8-2 refers to the second run, and so on. The usual simplifying assumptions were made (constant voltage behind transient reactance, constant input, and so forth). A time interval ~t = 0.1 sec. was used in point-by-point calculations. 240 r---r--r---r----r---r--~-'I'_-,

6(BB)

200

40

t---t---:--"""~

t----I---

-+---+----+---+-----f

~ -f.---f----+----t ~.:( cu·

U

.....

1-----I.-3+---+----+---+---+---+----i

~r

0 ......- .....-"'"-----.......- ......- - - -......- .....

o

0.2

0.4 Time (seconds)

0.6

0.8

6. Swing curves, Study 1, stability run 3-8-4, load condition 1, three-phase fault near substation AD on one 132-kv. line to station AC, cleared in 0.2 sec. Stable.

FIG.

In determining the allowableclearingtime of faults, the usual method was first to estimate this time and then to make a run using this estimated time. If conditions proved stable, another run was made using a clearing time 0.1 sec. longer than before. If the second run showed the systems to be unstable, then the clearing time of the first run gave

269

STUDY I-SWING CURVES

the answer desired. If the second condition was still stable, then a third 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 a two-phase-to-ground fault at station AD on one of the lines to station AC 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 Time (seconds)

0.6

0.8

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 stable for 0.5-sec. clearing. However, run 3-8-7 showed them to be unstable for 0.6-sec. clearing time. Thus it was shown that 0.5 sec. was the maximum allowable clearing time for these particular conditions. In some cases it was possible to determine maximum allowable clearing time from a single run because the angular swing of particular machines was wide enough, and the trend of accelerating and decelerating forces on the calculating sheets was such that the run was readily seen to be a borderline case, and additional clearing time would

270

TYPICAL STABILITY STUDIES

undoubtedly have caused instability. Sometimes comparison with runs 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 company A. The generators were grouped on the power sources of the calculating board as shown in Table 3 for run 3. Initial conditions for 200 . - - o r - - - r - - - r - - - - r - - - r - - ' T " " " - - " - - - '

160 t---f----.......~~F-

4Ot-----ir---J---+---+---I--qL-..;::!~-+----I

0.2

0.4 Time (seconds)

0.6

0.8

FIG. 8. Swing curves, Study 1, stability run 3-8-6, load condition 1, two-line-toground fault near substation AD on one 132-kv. line to station AC, cleared in 0.5 sec. Stable.

the stability runs were those of load condition 1 (Fig. 3). Swing curves 3-8-3 and 3-8-4 (Figs. 5 and 6) show that the systems were unstable for O.3-sec. clearing but were stable for O.2-sec. clearing of a three-phase fault at station AD on one of the 132-kv. lines to station AC. Curves 3-8-5, 3-8-6, and 3-8-7 (Figs. 7,8, and 9) show that, with a two-line-to-ground fault at the same location and with the same load condition, the systems were stable for O.5-sec. clearing and unstable for O.6-sec. clearing. On all these curves it should be noted that the three main generating

STUDY I-SWING CURVES

271

280 - - - - - . - - - - - . . - - - - - - - - - - - -

80

1----+----I---+-~~lIr___+-__tf__-_+___t

40 I----+---+---+----+----+-~~-+-~

O.....

-~_..r.-

o

0.2

. . . _. . . . 0.4 Time (seconds)

'___""_____.;;:::I

0.6

0.8

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.

TYPICAL STABILITY STUDIES

272

320 --.....--------------.---.....--.....---

280 1 - - - + - - - 0 + -

240 t - - + - - r - - -

80

I----a.--~~__r_-__+_---+--..._-+_____t

8(AB,AC,AF AH,AI,AN)

t

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 substation 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 down during the fault, thus acting like the equivalent motor of a two-machine system. Station AA and those stations of company B which were connected at 132 kv. (BB to BE, inclusive) swung fairly closely together-although not as closely as the stations of system A-and speeded up, acting like an equivalent generator. In every case station BB, the station closest to the fault 'of this group, speeded up more than the others (although it was closely followed by station AA) and was the first station of this group to pull out of step. Stations BG to BO, which were farther from the fault, were less affected, but tended to stay with the other company B stations. Study of proposed changes at station BB. As station BB was found to be the first station of company B to pull out of step during faults on the 132-kv. transmission system of company A, it was believed that any measures taken to help this station stay in step might materially improve the stability of the interconnected systems. Three different changes were studied. Use of shunt reactors. The design of the generators at station BB contemplated normal operation at 80% lagging power factor. Actually, however, these machines were usually operated near unity power factor. Such operation caused the machines to have internal voltage lower than normal and tended to make them unstable. Studies were made to show the effect of connecting shunt reactors to the low-tension busses at station BB in order to put about 15-Mvar lagging load on the machines and thereby increase their internal voltages. The initial load conditions with the reactors added were obtained in load run 4 and differed only slightly from the conditions of load run 3. 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. Curve 4-8-1 (Fig. 10) shows that the system was unstable for 0.3-sec. clearing of a three-phase fault, whereas curve 3-S-4 (Fig. 6) shows that the system was stable for O.2-sec. clearing of such a fault without the use of shunt reactors at station BB. Therefore it is apparent that the addition of the reactors would not increase the maximum allowable clearing time for a three-phase fault by as much as 0.1 sec. Curve 4-8-2 (Fig. 11) shows that a clearing time of 0.7 sec. was permissible for 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-

274

TYPICAL STABILITY STUDIES

electric generators by automatically cutting resistors in series with them during faults and thereby maintaining their load and decreasing their acceleration during faults was proposed by R. C. Bergvall.' As this 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

1.0

Time (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 the fault. Resistors of this value were assumed to be cut into the circuit 0.1 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. line to station AC caused instability when cleared in 0.7 sec. This appeared to be a borderline case, however, and it seemed safe to conclude that 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

200 - - - - . . . . . - -......--'I"--..--..,----r----r-----r--r-----.

120 1--..;;;;::::::'l. . ._!:::I----a.~---+--~--+---~--+------"'~~I--~

40

~

~I

~.J----+---+-----+--1~-4--~H-~-+---+-~r------I

o

.ml S .:...I----+--.J----+---+----+----I-~~--I---_+_---i ~I~

~Iu bO ~ .5 \0 -I----+---i---+--__+_5

t

~

-40

1.

Q)

Cl)4----+----+--__+_---+---i-

.51 ~I

''tn+---+---I----4----+----11--en

~I

- 80 1...--.&-_..1.-_..1..._..1-_....1-_.....1-_....1-_-"--......- - - ' 0.2 o 0.8 0.4 0.6 1.0 Time (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 0.7 sec.: 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.

TYPICAL STABILITY STUDIES

276

tions but without the series resistors showed an increase of permissible clearing time of only 0.1 sec. Decreased transient reactance. The four large generators at station BB had the abnormally high transient reactance of 65% by test. Un200 .---.,.--or---..,---..,.--...,.---.,--....,---.....----

i

120

~

f

-----

"'t:J ~

CD

bo c «J

~

80

~

g

(ij

c

'-

~

.E

40

t---+--+----t--+---r--~----:~--I---+--~

Ot---+---+---+---+---+---+----+---HI.3IiIl~--=t

- 40

o

- - - - -......-"""-----_...r.-_....._-'-_....._....I

0.2

0.4

0.6

0.8

1.0

Time (seconds) FIG. 13. Swing curves, Study 1, stability run 3-8-10, load condition 1, transient reactance of station BB generators reduced from 65 to 35%, two-line-to-ground fault 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 of station BB. In order to determine the effect of a lower value of transient reactance for these machines, curve 3-8-10 (Fig. 13) was taken with the reactance assumed as 35%. The results showed that O.S-sec. clearing of a two-line-to-ground fault would make the system stable. 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 with the 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 to station AS under load condition 1. Curves 3-8-1 and 3-8-2 (Figs. 14 200 r---r--.,--w=---_-.---~,.----,...___

160

9(AE,AL)

40 I--f.--+----I---+-~~~-I---~----t

o

1:-.-

9

---------~--------_ . .......-

......

0.2

0.4

0.6

......

0.8

Time (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.

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 fault was 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 tapping the 132-kv. lines of company A would be of approximately the same severity in regard to stability. Faults on the 44-kv. line between stations BE and BG.' The effect of a two-line-to-ground fault near station BE on the 44-kv. line to station BG under load condition 1 is shown in curves 5-8-1,5-8-2, and 5-8-3 (Figs. 16, 17, and 18). The generators were regrouped as shown

TYPICAL STABILITY STUDIES

278

200 r - - - r - - -_ _--,.--_---r--~-_-___,

en Q)

80

f

~

Cl>

"'C Cl)

00 c co

Cl>

40

~

.s '0

8 (AB, AC, AF,

AH, AI, AN)

>

ro

E

$

.s

0

- 40

t----+---+---+---t--+---t----+---"---i---t

- 80 t----t-----+----+-1---+---+----..--

3 +---+---+---+---f ~I

- 120""'-_-'--_......._ - - ' - _ - - A . _...... o 0.2 0.4 Time (seconds)

....... _

0.6

....._

...

0.8

FIG. 15. Swing curves, Study 1, stability run 3-8-2, load condition 1, three-phase fault near substation AE on 44-kv. line to substation AS, cleared in 0.4 sec. Stable.

STUDY I-FAULTS BETWEEN STATIONS BE AND BG

279

320 --,...--..,.----r---rlr---r---r--,---,

280 1----f---t---f---1---t--t----t---t--,

240 1---I----+-~J+---+--_+-__t-__1-___1

~ GJ

200 1---4---'---#--+--+-~r---+--;--,

e

iP

-so "C

C1J

c: co

~ ~ ~

~--""'_-.I 8 (AA, BB) ....... ---+-__1~__1

160 1----#--~:::...-.-+-~.....-

7(BC)

(ij

E

~

120 ~~$!!!!!I!!!!!!!IIiI~~.f--_9 (AB, AC, AF, AH, AI, AN)

I I

10 (AE, AL, AP, AV)

80 l----+----"--~--+--__t_-__+-__1-____1

40 1---I---+---;+.Ji-~---t----t"----1---;

_....Ioo._....._ ....

O'--_J..-_.L-_..Io.-_..a..___.......

o

0.2

0.4

0.6

0.8

Time· ( seconds)

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.

FIG.

280

TYPICAL STABILITY STUDIES

for run 5 in Table 3 to give more detailed representation near the fault and 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---..---....--..

200 I - - t - - -...

lO(AE,AL,

AP,AV)

40 t - - - + - I - - - + - - ...., +---+--+--+---+---0+-----1

~I

o............-----'0.8 o .....-.-.--~-"--~0.2 0.4 0.6 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 stations BG to BO slowed down. It should be noted that the initial flow of power was from the former to the latter group of stations, which may be regarded as the equivalent generator and motor, respectively, of a two-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

STUDY I-FAULTS BETWEEN STATIONS BE AND BG

281

240 r--......---,.--.,..--....--..,.---.----.----.

200 1 - - - + - - - ' 1 - - - + - - . . - - - + - - - + - - - + - - - 1

10 (AE, AL, AP, AV)

,

./1 (BI,BK, BL,BN,BO)

4(BE) breakers opened 12 CYCles'*1

40

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.

TYPICAL STABILITY STUDIES

282

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. at station BG. Faults on 44-kv..line between stations BG and BH. The effect of a two-line-to-ground fault near station BG on the 44-kv. line to station 240 ..---r----r--,r--~r---.,r----r.--~r--_

-;; 160 .....~~~----+~

I

-~

."

«S

~ ~

120 I---~----l~""-"~-+

5(B~1)

I

7 (BG 44 kv.) I

3 (BJC)

~

I

4(B82)

Ci ...c

~

9(BD) I 10 (BB, Be, and company A as infinite bus)

80

1---t---t--~L..-....lII~~-fo--.--t----1

40

J---+---t--

I---t---t--...,-I--+---f--+--+---f

:; ~I

O'---'-------.....-"--"--------~---.t o 0.2 0.4 0.6 0.8 Time(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 the neighborhood of the fault, including the separation of stations BG and BH into two parts each. The machines of company A, together with stations BB and Be, all of which were found to be only slightly affected by 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 for O.3-sec. clearing at both ends of the line. The general direction of power flow was eastward. During the fault stations west of the fault 360 r - - - - y - - r - - - , . - - . . - - - y o - - - - , n - - - - - r - - - - .

280 I---+---+----+-+--I--.......~H---+---+----I

160 1--'--+--~~-.f--7l'o~+--+---+----+---4

120 t--+---+---+--~r__~.-+--+---t

0.2

0.4

0.6

0.8

Time (seconds) FIG. 20. Swing curves, Study 1, stability run 7-8-1, load condition 2, two-line-toground fault near substation AD on one 132-kv. line to station AC, cleared in 0.4 sec. Unstable.

(equivalent generators) speeded up, and stations east of the fault (equivalent motors) slowed down. Stability during load condition 2. In this load condition, as already mentioned, station BB and most of the generators at station Be were

TYPICAL STABILITY STUDIES

284

shut down, and more power was transmitted from company B to company A than in the previous load condition. The generators were grouped in a manner (shown in Table 3 for run 7) which was judged 360 .---".--....---..----.,.....-.,..--.,.---ro--_

280 t---t---t----.,..;~__II---._, ~

en CD

~

CD ~

240 1---t-----#~~.__

+-I-~lIt---+---t-_4

.!l co C

tV

Wo ~

~ 200 1-JIIliIr:---J1----#-+---~-+---J~+---+---t-_of tV

... C

10{AA, AP)

J!! .E

~9(AE,AH,

AI, AL, AN)

160 1---+--~~-+----4~--+--+--+-_of

120 ~-+---+---__- + - - + - - + - - + - - - - t

0.2

0.4

0.6

Time (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 the 132-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 noted that the permissible clearing time of 0.3 sec. for this load condition was considerably shorter than the permissible clearing time of a similar fault under load condition 1, which was found to be 0.5 sec. This can be attributed to three reasons: 320 r----~-......--_-..,._-_.__-....,._-...._-_..

120 I - - + . - - - " ' - - - - t - - + - - - + - - - t - - - r - - - ;

80 ,--_",--_.a..-_~_..Io.-_-'-

o

0.2

0.4

"""_'"

0.6

0.8

Time (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 company A was greater under load condition 2 than under load condition 1. 2. With station BB shut down, much of the power which it formerly 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 stable operation in the transient state.

286

TYPICAL STABILITY STUDIES

3. When station BB was on the line, regardless of the amount of power supplied by it, it served to support the voltage near the middle of the interconnecting line and thus to improve both steady-state and transient stability. In load condition 2, station BB was shut down, 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 be 0.2 sec. or less under load condition 2. Curve 7-8-3 (Fig. 22) for a two-line-to-ground fault near BE on the 44-kv. line to BG shows a stable condition for O.2-sec. clearing at both ends. It was thought probable that sequential clearing, 0.2 sec. at BE and 0.4 sec. at BG, would be stable under load condition 2, as it was under load condition 1. Summary of allowable clearing times. The slowest clearing times permissible 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:

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.

*Estimated.

Conclusion and recommendations. The study showed that the characteristics of the transmission system, especially in regard to the speed of clearing faults, had a decidedly more important bearing on the problem 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 obtain high-speed clearing of faults. If relaying .and oil circuit breakers were modernized, no other changes would be needed in the system to obtain a practicable solution of the instability problems which were studied. Three proposals for increasing the stability of the generators at station BB by changes at that station were studied, but the usefulness of the proposed changes was limited by the fact that this plant was usually shut down, because of water conditions, at night during the fall and winter seasons when the greatest amount of power was usually transferred from company B to company A and when instability of the 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 of company 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 closely together as a group, almost without regard to characteristics of individual stations. In practically every case the machines nearest the fault showed the greatest tendency to lose synchronism. From the foregoing observations, it was concluded that any changes which might be made at any of the generating stations would be limited in their effect and relatively costly to attain and, therefore, would assume a position of minor importance which did not justify further study. It was recommended that a definite program of relay and oil circuitbreaker improvement be prepared, taking into consideration the amount of good to be accomplished by various changes and the cost of the changes. It was also advised to concentrate on those changes which would allow the systems to remain stable after a two-line-toground fault, as three-phase faults are infrequent on properly relayed high-voltage systems. Modernization of relay systems and of oil circuit 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 be obtained at many points on the system by relay changes. Modernization of some oil circuit breakers might be accomplished at reasonable cost but would, in general, be more expensive than relay changes to attain the same improvement in stability. Some other breakers should be replaced rather than rebuilt.

TYPICAL STABILITY STUDIES

288

lorain Toledo

S Chicago

(OhioP.S. Co.)

(Toledo Edison Co.)

...an C'f)

.""

Indiana and

Cincinnati

S

South Point 10

FIG.

23. Central 132-kv. transmission system of American

289

STUDY 2

Akron

Alliance

(Ohio Edison Co.)

(OhioP.S. Co.)

Massilon

(Ohio P.s. Co.)

..,....,........-5

41.8mf.

6.68+j19.0

Newcomerstown

5'--0

West Penn Power Co.

Windsor

(Power, W. Va., nearWheelinl)

®-

Steam - electric generatintplant

@- Synchronous condenser(s).

®-

Equivalent of power system.

~Load.

-- -

Future 132·kv. line.

- - Existing 132·kv. line.

Turner

(Institute, W. Va., nearCharleston)

Cabin Creek W. Va.

Impedances aregiven in per cent on l00·Mw. 132·kv. base.

en Parallel impedance of double circuitline. Figures on busses areshunt capacitive susceptances of connected linesin percent on 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 generating units (numbers 4 and 5, rated at 94.5 Mva. each) to its steam-electric generating station at Philo (on the Muskingum River near Zanesville, Ohio), thereby increasing the aggregate capacity of the station from 271.4 Mva. to 460.4 Mva. The generators at Philo were paralleled through 132-kv. busses. The 132-kv. oil circuit breakers had a rated interrupting capacity of 2,500 Mva, The addition of the new generators would have increased the interrupting duty of these breakers from approximately 2,250 Mva. to approximately 3,150 Mva., a value in excess of their rating, had not steps been taken to limit the shortcircuit currents. For this purpose it was decided to install reactors between two sections of the 132-kv. bus, which will be called sections A and B. In October, 1940, an a-c. calculating-board study was made by the American Gas and Electric Service Corporation of the central system of the American Gas and Electric Company and interconnected systems, including the expanded Philo station with the bus reactors. The study included short-circuit studies, load studies, and a few stability runs. The system studied is represented in Fig. 23. It extended from Indiana 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 Gas and 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. The generating plants of 'the interconnected systems also were predominantly steam stations. All the transmission lines represented in the study 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 cent on the same base. The impedances of the equivalent generators were obtained 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 the machines and lines grouped on the bus sections in the manner assumed tData on this study were obtained from the American Gas and Electric Service Corporation, New York City, through the courtesy of Philip Sporn, Vice President in Charge of Engineering, Harry P. St. Clair, System Planning Engineer, and Charles A. Imburgia.

STUDY 2 Crooksville Zanesville Howard No. 1 No. 2 No. 1 No. 2

291 Torrey North· Newcom· Rutland No. 1 No. 2 east erstown No. 1

Generators

j35.3

e

j61.7 j25.6 j57.0 j57.0

j33.7

j35.3

j61.1 j25.6

Circuit breakers assumed to open to clear the fault

FIG. 24. Philo steam-electric generating station. Simplified one-line diagram of the 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. Impedances are in per cent on l00-Mva. base. (a) Normal condition, (b) after clearing fault on bus section AI.

292

TYPICAL STABILITY STUDIES

in the stability study. Data on the generators and transformers at this 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 series and parallel combinations the circuit was further simplified as shown in Fig. 25. Similar simplifications were performed for the other generating stations and interconnected power systems shown in Fig. 23. TABLE 6 DATA

ON GENERATORS AND TRANSFORMERS AT PHILO PLANT (STUDY 2) GENERATORS Xd'

% on

Xd" , . . - -......

Unit

Number Mva. 42.1 1 42.1 2 62.4 3-1 62.4 3-2 62.4 3-3 4HP 50.7 43.8 4LP 50.7 5 lIP 43.8 5LP

r.p.m,

1800 1800 1800 1800 1800 3600 1800 3600 1800

% on

Rating 100 Mva, Rating 100 Mva,

24.0 24.0 22.0 21.0 22.0 13.0 27.0 13.0 27.0

57.0 57.0 35.3 33.7 35.3 25.6 61.7 25.6 61.7

15.0 15.0 13.0 14.0 13.0 10.0 16.0 10.0 16.0

33.6 33.6 20.8 22.4 20.8 19.7 36.6 19.7 36.6

Hon 100 Mva. 2.35 2.35 3.81 3.81 3.81 1.74 3.22 1.74 3.22

TRANSFORMERS x%on ~

Unit

Mva.

T-l T-2 T-3-1 T-3-2

45 45 63

T-3-3 T-4

T-5

63 63

120 120

Rating 100 Mva.

8.9 8.2 17.8 17.5 17.9 13.4: 13.4

19.8 18.2 28.3 27.8 28.4 11.2 11.2

The inertia constants of the equivalent generators are listed in Table 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 100 Mva. for old generators and 5 per 100 Mva. for new generators, the new ones being assumed to be hydrogen-cooled. The calculating-board set-up was similar to that of Fig. 23, except that Fort Wayne and the stations and systems west and south of it were 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 Chicago, Indana, and Cincinnati'systems Twin .Branch generators 1, 2, 3, 172 Mva. Twin Branch generator 4, 94 Mva, Fort Wayne condenser, 30 Mva, Total, Fort .Wayne and West

Case 2 1 3 Off 61 76* 7.8 7.8 7.8 Off Off 5.0 1.2 1.2 1.2

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) Akron and Alliance systems, over 500 Mva.

2.1 8

2.1 20

2.1 36.5*

Total, Canton and interconnections Windsor generators (Beech Bottom Power Co.), 257 Mva. West Penn Power Co. system, approx. 1,000 Mva.

38.6 15.1 15.1 15.1 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

Philo B (gens. 3-2, 3-3, 5), 219 kva,

12.6 12.6 12.6

11.1

11.1

Rutland, future installed capacity of Appalachian Electric Power Co. and interconnected systems, approx. 1,100 Mva. 60

60

Off

Lancaster, future interconnection to west, approx, 170 Mva.

10

10

Off

Cabin Creek generators 1 to 8, 232 M va,

t

t

Logan generators, 102 Mva. Plants south of Logan and interconnected systems, 375 Mva,

t

t

Total, Logan and south *Estimated future increase. [Combined with Rutland.

15 4.4

22.6 27

294

TYPICAL STABILITY STUDIES

Wayne. Furthermore, double-circuit transmission lines were represented on the board by single impedance units set at the impedance of two lines in parallel; when one circuit was supposed to be switched out, the setting of the impedance unit was doubled. The short-circuit studies showed that the reactance between bus sections A and B at Philo should be 7.5% on a 100-Mva. 132-kv. base in order to limit the interrupting duty of any circuit breaker to a value slightly less than 2,500 Mva. The 7.5% reactance would be obtained from two sets of reactors in parallel, as shown in Fig. 24, each having 15% reactance.j Load studies were made to determine the proper normal-load rating of the reactors and to find the arrangement of transmission lines on the two 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 were made to test the stability of the system (modified by the addition of new generating units and bus reactors at Philo) with the interconnecting lines to other companies (Chicago, Indiana and Cincinnati, and western 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, and the existing generating stations at Logan and Cabin Creek were combined in an equivalent generator at Rutland. The flow of active and reactive power in both cases was as shown in Fig. 26. The flow from Philo west toward Fort Wayne was heavy. There were also heavy flows from Rutland to Philo and from Philo to Canton. In both cases 1 and 2 a three-phase fault was assumed to occur on section Al of the Philo 132-kv. bus and to be cleared in 0.15 sec. (9 cycles) by the action of I-cycle bus diffffi.ential relays opening the following 8-cycle circuit breakers (see Fig. 24): Howard line 1, Crooksville line 1, generator 1, and bus-tie reactor 1. The assumptions which have been stated represent a severe condition regarding stability, because of the heavy power flow over long transmission lines, the severe type of fault, the fault location at the sending end, and the loss of a long, heavily loaded circuit in clearing the fault. A three-phase fault near Philo on one of the lines to Howard might be expected to produce very nearly the same effect as a fault at the location assumed. In case 1 the interconnections to other companies were open; in case 2 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).

FIG.

26.

-of-

Rutland andSouth

47.2 + j13.0 106

166 + j36.2

, ,

rh t

C'f ~

+

an

~ .....

30.3 -j3.0

Newcomerstown

trs

Akron Alliance

Initial load conditions of cases 1 and 2, Study 2. Flow of active and reactive power is given in Mw. voltages are given in per cent of 132 kv.

N

a\

N

:;-

o

Toledo t s

+i

Mvar.

Bus

t..:>

~

~

~

~

§

TYPICAL STABILITY STUDIES

296

Figs. 27 and 28, respectively. In case 1 Fort Wayne and Twin Branch pulled out of step with the rest of the system, but in case 2 these stations, supported by the interconnections to the Chicago and Indiana

,

60

I I

I

40

~

//

VV __lZVI/ -

20

~r-,

"\

I

Phil0I:!~.

---....

I

Rutland~

~I

~

.......

~~

~

~

V

I

I

-

0

en

~cv

:e -0

~Canton

I Windsor-"'\ I I ~ ~~ ~' -~---...

I

-20

0

en

,

" .,

8- -40 «I

~

ioc

l,~

"i'-L ~I

.

Y

111

Lorain

,

I

;1

I

~Fort

-,

I

I

J

I

"

I

Wayne and West

'"



240

«U

~ 0.2

220 0

~_

V

j

~-

/ v

= -- I V &200 I

-;;;

/ V

I

cv

-...

I

I~

V

e. 180 [7

.'/~a..

r,,1o.

~

" '//J.

Inputt: J

~

~

160

~

~

V

V

,.........

.~

r--r--- r--...

Power output Systems A, D,E,

4

'/

~

\~

I

r--- Clearing angle

Q)

fault on

\ Power output Systems A, D, E. line open

140

120 60

Systems A, D,E, - I - no fault

.~

Critical reclosing

JV

E

- bi

I./:' Power output

I~

V

angle-~

J

7

11

/ rnf/

/. )

l/ 1-

/"

0

~

I

~1I

,I.

.cg 0.4 4n

I

Critical reclosing time

80

100

120

140

160

180

200

Rotor electrical angle (degrees) 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 connection to synchronous condensers or static reactors. The capacity of the 13.8/66-kv. transformers at station AA should be increased from 20 Mva. to 45 Mva, The line as built. The interconnection between companies A and B was 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 the one 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

,

"0

e

I

I

~ .........Swing curve ~

Critical 260 _ reclosing time-

-;;-

P(

~-I--I----~-

0

! 0.2

.

I-

Clearing

J

~ime~ 240 ~---

Q,)

E

i=

0

r/ I

r

1 ' 1 / I~

I--~---

I

-

220

~

180

~~

I

~ Clearing angle""",,--

i--=3

..... ~

.,V

~

~

Power output Systems A, D,E, no fault

~

~~

I

-, I\.

reclosing angle

160

~/

~Critical

..V

E

;"~

InputP,../'

)/

Q)

"",."."

~

/v

~ IV ~ 200 t]G

-i

""""'"'"

I~ ~/r

t"" .

~

V

",

I

--

1\ \ v--:Systems Power output A, D, E, -

~

--

fault on ~,...

'///1

V

""- Power output Systems A, D,E, line open

140

120 60

80

100

120

140

160

180

200

Rotor electrical angle (degrees)

61. Power-angle curves 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. FIG.

of ten 10-inch suspension insulators. They were spaced 14 ft. 6 in. apart in a horizontal plane (18 ft. 3 in. equivalent spacing). There were two i-inch high-strength galvanized-steel ground wires 12 ft. above the phase conductors .at structures and 20 ft. above them at midspan. The structures were wooden H frames. There were three complete 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-

TYPICAL STABILITY STUDIES

344

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 a new type of high-speed relay system which could distinguish between faults and power swings even though the magnitude of indicated impedance might be nearly the same. 2 ,3 Internal faults in the terminal transformers were cleared by differential relays which tripped 50

~~

'"~

40

-,

.. Unstable

,,~

"'-.

Stable "'0

e

~ ~

"'0