POWER SYSTEM STABILITY Volume I Elements of Stability Calculations
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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
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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 ~ <,
J
...............
"
/ /
Post-fault output
1/
II
1.0
all
~yj
~
0.5
I~
~
o o
%~
~
Output during faUlt", I
I
~~~
Input ~
·v
I
\
-, \
~~\ = - 10.4 sq. ~~ ~[\ I
.-c
I A = - 11.4sq. A = 3.3 sq. .
2
d~~
I
3
hr'\~ ~I"~~ I
I I
I
13Soor I 139° 157· I
I
60
~~ ~\
f~
~
c"\
At = 19.3 sq.
"'
I
ISo
A4
I
I
~WK"
~~b
I
fault output
90
120
s,
150
s;
\
~
180
Angular displacement 8 (electrical degrees) FIG.
7. Determination of critical clearing angle by the equal-area criterion. (Example 1.)
easily done in practice by counting squares in columns from d to the right and from h to the left, accumulating totals as we go and finding where the totals are equal and opposite or at least most nearly so. In the present problem this equality occurs at 138° or 139°, which is therefore the critical clearing angle, ac. The shaded area in Fig. 7 is At + A 2 + A a + A 4 = 19.3 - 11.4 3.3 - 10.4 = 0.8 ~ O. From the swing curve for a sustained fault (Fig. 9, Chapter II) we find that the time corresponding to 0 = 1380 is t = 0.61 sec. This is the critical clearing time, t., It is interesting to note that the swing curve plotted for a clearing time of 0.60 sec. (clearing angle 136.2°) indicates that the system is stable. The
+
132
THE EQUAL-AREA CRITERION FOR STABILITY
maximum value of 0 on that curve is 147°. For the critical clearing time of 0.61 sec. the maximum value of 0 is determined by the abscissa of intersection h, Fig. 7, and is about 157°. The swing curve for a clearing time of 0.65 sec. (clearing angle 146.7°) shows that the system is unstable. Thus there is good agreement between the critical clearing angle determined by the equal-area criterion and that determined by swing curves calculated point by point.
Two finite machines. A system having two finite machines may be replaced by an equivalent system having one finite machine and an infinite bus, so that the swing equations and swing curves of angular displacement bet\veen the two machines are the same for both systems. It is necessary to use an equivalent inertia constant, equivalent input, and equivalent output for the equivalent finite machine. The equivalent inertia constant is a function of the inertia constants of the two actual machines, and the equivalent input and output are functions of the inertia constants, inputs, and outputs of the two actual machines. The equivalent system will now be derived. The swing equations of the t\VO finite machines are as follows: d201 ~al ~il -- Jlu1 dt2 = M 1 = M1
[~3a]
[ISb]
The relative angle will be used, because its value is significant in,showing the stability or instability of the two-machine system. d20
d202
d20l
dt2 = dt2
-
~a2
Pal
dt2 = M 1
-
Multiply each side of the equation by M 1M2/(M1
M lM2 d20 M 2P al 2= M 1 + M 2 dt
--
[14]
M2
+ M 2 ) , obtaining
M 1Jla2 M 2Pui -- M 1Pu 2 M 1 +'M2
[15]
which may be written more simply as d20 dt
M 2 = Ps = Pi - Pu,
[16]
EQUIVALENT POWER-ANGLE CURVE
133
Equation 16 is identical in form with eq. 1 for a single finite machine and an infinite bus. Here the equivalent input,
M 2P i l - M 1Pi 2 Pi = M 1 + M2 and the equivalent output, P = M 2PU1 - M 1Pu 2 u Afl + M 2
[17]
[18]
are weighted means of the inputs and outputs, respectively, of the t\VO actual machines, with the signs of P i2 and P u2 reversed, the weights being inversely as the inertia constants. The equivalent inertia constant,
[19] is smaller than the smaller inertia constant of the t\VO actual machines. The law of combination of the inertia constants is similar to that for the parallel combination of impedances. This law will appear reasonable if we note that the inertia constant is the accelerating power divided by the acceleration and that in a two-machine system the accelerating power of the generator is nearly§ equal (except in sign) to that of the motor, while the relative acceleration is the sum of the acceleration of the generator and the retardation of the motor. This situation is in contrast to that which prevails in finding the inertia constant of a machine equivalent to a group of machines that swing together (discussed in Chapter III). There the accelerating power of the group is the sum of the accelerating powers of the individual machines, and the accelerations of all machines are equal. Consequently, the inertia constants combine like impedances in series. Having obtained the equivalent system, we may investigate its stability either by calculation of its swing curve or by use of the equalarea criterion. The equal-area criterion is the more convenient method, but to apply it we must first obtain the power-angle curve of the equivalent system. Equivalent power-angle curve of two finite machines. The powerangle equations of a two-machine system are (by eqs, 17, Chapter
III):
+ E 1E2 Y 12 cos (8 12 - 01 + 02) = E 2El Y 2l cos (821 - 02 + 01) + E 22 Y 22 cos 8 22
P u l = E 12 Y 11 cos 8 11
[20]
P'U2
[21]
§Exactly equal in a purely reactive network, as explained later in the chapter.
134
THE EQUAL-AREA CRITERION FOR STABILITY
Substitute these values of P ul and P u2 into the expression for equivalent output (eq. 18), and let ~ = ~1 - ~2. The result is P« = M 2E12Yll cos ell - MIE22Y22 cos e 22
M 1+M2 + E 1E2 Yd M 2 cos (8 - e12) - M 1 cos (8 M 1 + M2
+ e12)]
[22]
The two cosine terms involving 8 may be combined into a single cosine term by considering each term as the horizontal projection of a vector in the position for which the variable ~ is zero. Thus the first term and the second term are, respectively, the horizontal projections of vectors
FIG.
8. Vector diagram used in the derivation of eqs. 23 to 26.
M 2/- 812 and - M I!8 12. (See Fig. 8.) In this position the horizontal projections are M 2 cos (-812) = M 2 cos 812 and -M1 cos 812, and the sum is [23]
The vertical projections are M 2 sin (-812) == -M2 sin 812 and - M 1 sin 812, and the sum is
V
= -
(M 1
+ M 2 ) sin 812
[24]
The magnitude of the sum vector is
M" == VH 2 + V 2
= V (M2 - M 1 )2 cos2 e12 + eMI + M 2 )2 sin2 e12 == v' (M 12+M2 2 ) (cos2 el2+ sin2 012)+2MIM2(sin2812-COS2 8 12) == VM 12 + M 2 2 - 2M1M2 cos 2812
and its phase angle is -8
"
= tan-1 -HV = tan-1 1
= tan- ( : :
~
z:
-
[25]
(M! + M 2 ) sin 812 (M 2 - M 1 ) cos 8 12
tan
e 12)
[26]
135
REACTANCE NETWORK
Hence eq. 22 may be written more simply as P u = Pc
= Pc
where
+ PM cos (0 - e/)
+ PM sin (0 -
Pc = M2E12Yll cos ell
[27]
1')
MIE22Y22 cos e22
-
M 1 + M2
[28]
is the vertical displacement (see Fig. 2), and 'Y
l + M2 ) = -tan-1 ( M M _ M tan e12 1
-
0
90
[29]
2
is the horizontal displacement, of a sine-wave, the amplitude of which is
PM = E 1E2Y12M" M 1+M2 ==
E 1E 2 Y 12 VM 12 + M 22 - 2MlM2 cos 2e12 M 1+M2
[30]
If we put M 2 = 00, eqs. 28, 29, and 30 reduce to the values previously derived for one finite machine and an infinite bus (eq. 11). Reactance network. If the network to which the two machines are connected contains only reactance, the power-angle equation and the equation for equivalent input are considerably simplified. In this case 8 11 = 822 = -90° and 812 = 90°, giving cos 811 = cos 822 = 0, cos 2812 = -1, and tan 812 = 00. Hence Pc = 0, 'Y = 0, and PM = E lE2 Y I 2 . The power-angle curve is then an undisplaced sinusoid, [31] which is identical to the power-angle curve for one machine connected to an infinite bus through a reactance network. In other words, if the network contains only reactance, the power-angle equation and curve of a two-machine system are independent of the inertia constants of the machines. If, however, the network contains resistance as well as reactance, both the amplitude and displacement of the power-angle curve depend on the inertia constants. Since there are no losses in a reactance network, one of the two machines must act as a generator and the other as a synchronous motor, so that, if both are considered as generators, their outputs will be equal and opposite: [32] Pu2 = -Pul
136
THE EQUAL-AREA CRITERION FOR STABILITY
Initially the inputs are equal to the outputs: Pit
=
Put
[33]
Pi2
= Pu2
[34]
= -Pit
[35]
giving equal and opposite inputs, Pi2
Hence the equivalent input, as given by eq. 17, becomes Pi
M 2P it - M t ( - Pit) M M
=
1
+
2
= P«
[36]
which is equal to the actual generator input. In similar fashion the equivalent output, as given by eq. 18, becomes equal to the actual generator output. EXAMPLE
2
By use of the equal-area criterion determine the critical clearing angle of the two-machine system of Example 4, Chapter III. Solution. The following data are obtained from Example 4, Chapter
III:
= 2.92 X 10-4 per unit MD = 26.9 X 10-4 per unit MA
PiA
= 0.80 unit power
PiD
= 3.20 unit power
aA O =
23.0°
= 10.3°
aDO
With the fault on E A 2y
AA
cos eAA = 0
2y
DD
cos aDD = 0.45 unit power
ED
EAEDYAD SAD
= 0.203 unit power = 86.4°
With the fault cleared EA
2YAA
ED 2yDD
cos 8AA
= 0.04 unit power
cos aDD = 3.28 unit power
EAEDYAD = 1.91 unit power SAD
= 79.4°
From eq, 19 the equivalent inertia constant is
MAMD
M = M,A + MD
=
2.92 X 26.9
29.8
.
10-4 = 2.64 X 10-4 per unit
REACTANCE NETWORK
137
From eq. 17 the equivalent input is p. _
MDPiA -
,-
MAPiD
MA+ MD 26.9 X 0.80 - 2.92 X 3.20 29.8 21.5 - 9.35
29.8
12.15 0 41 it =- = . unl power
29.8
The output power-angle equation is P;
==
Po
+ PM sin (8 -
1)
where by eqs. 28, 29, and 30
Po == MDEA 2YAA cos eAA - MAED2y DD cos eDD MA+MD P
"';MA2 + MD2
_ E E Y M -
1
==
A
D
-
2MAMD cos 29AD
MA + MD
AD
-tan- 1 (MA + MD tan eA D ) MA- MD
90°
-
For the faulted condition 26.9 X 0 - 2.92 X 0.45 = -0044. . unit power P c == 29.8
= 0.203 ...;(2.92)2 + (26.9)2 -
P
2 X 2.92 X 26.9 cos (2 X 86.4°) 29.8
M
= 0.203 "';8
+ 722 -
157 cos 172.8° 29.8
= 0.203 "';730 -
157 X (-0.992) 29.8
= 0.203 "';730 + 156 = 0.203 v'886 29.8
==
0.203 unit power
==
-tan- 1 (
==
_tan- t 29.8 X 15.9 _ 900
29.8
29.8 tan 86.4°) - 90° 2.9 - 26.9 -24.0
==
-tan- 1 (-19.7) - 90° = 87.1° - 90°
==
-2.9°
138
THE EQUAIr-AREA CRITERION FOR STABILITY
For the cleared condition
Po = 26.9 X 0.04 - 2.92 X 3.28 = 1.1 - 12.9 29.8
= - -11.8 = -
3
·
O. 96 unit power
29.8
= 1.91 V730 -
P
29.8
0
157 cos 158.8 29.8
M
= 1.91 v730 - 157(-0.932) 29.8
+
= 1.91 v730 146 = 1.91 V876 = 1.89 unit power 29.8 29.8 'Y = -tan- 1 ( 29.8 tan 79.4°) - 90°
-24.0
= -tan- 1 (29.8 X
5.34) _ 900 -24.0
= -tan-1 (-
6.63) - 90° = 81.4° - 90°
= -8.6°
Hence for the fault condition
P« = -0.044 + 0.203 sin (8 + 2.9°) and for the cleared condition
P u = -0.40
+ 1.89 sin (8 + 8.6°)
The power-angle curves are plotted in Fig. 9. Note that they are displaced sine curves, the displacement being from 0 to 0' and from 0 to' 0". The input line is also drawn. The initial angle is
80 =
OAO -
aDO
= 23.0 -
10.3 = 12.7°
This angle is marked on Fig. 9. The point on the input line at this value of 8 is the point representing the pre-fault operating condition. It lies on the pre-fault power-angle curve, but there is no need of plotting any more of that curve than this one point. By application of the equal-area criterion, the critical clearing angle is found to be oc = 100°, and with clearing at this angle the maximum angular displacement that is reached is Om = 146°. Let us see how well these values agree with the results obtained in Examples 3 and 4 of Chapter III. In both these examples the system was found to be stable if the fault was cleared in 0.35 sec. (clearing angle, approximately 96°) and unstable if the fault was cleared in 0.40 sec. (clearing angle, 118°).
SWING CURVE BY GRAPHICAL INTEGRATION
139
Thus, in these previous examples the clearing angle was found to be between 96° and 118°, and the value found in the present example lies between these limits. The equal-area criterion provides an easier method than we had before of determining the critical clearing angle, but it does not enable us to find directly the critical clearing time. From the swing curve for O.40-sec. clearing (Fig. 16 of Chapter III) we may read the time-O.36 sec.-corresponding to the critical clearing angle of 100°. The clearing time of 0.35 sec. for which a swing curve was obtained and plotted in Fig. 16 of Chapter III is seen to be only slightly below the critical
-
1.00 I----I_---L._-+-.--,.~
~
c
::s
!
0.50 J----+-\rh'---+--4----1I---~--+---t~~~~~r_t-_r____;
100·
13'0·
o FIG.
00
30
s,
60 90 120 Angular displacement 8 (electrical degrees.)
I
1466".150
9. Power-angle curves and determination of critical clearing angle by use of the equal-area criterion. (Example 2.)
value. The maximum angle attained in Example 3 was 135° or 136°, and in Example 4, 131°, whereas the maximum angle under the critical condition is 146°. The discrepancy between Examples 3 and 4 is not surprising in view of the proximity to the critical condition, at which a very small difference in accelerating torque changes a stable system into an unstable one
Determination of swing curve by graphical integration. The type of disturbance which is most important in stability studies is a fault applied and subsequently cleared. It is usually desired to determine whether a system is stable with a given load and given fault-clearing time, to determine the stability limit for a given clearing time, or to determine the critical clearing time for a given power. The equal-area criterion by itself affords information on clearing angle but not on clearing time. The clearing time, however, is of primary importance because the circuit breakers and protective relays, by means of which the fault is cleared, have definite operating times which are independent of the angular displacements of the machines. Therefore, when the
140
THE EQUAL-AREA CRITERION FOR STABILITY
equal-area criterion is used, it is necessary to find the clearing angle when the clearing time is known, or vice versa. For this procedure a swing curve, carried out as far as the point of clearing, is required. The swing curve for this purpose may be obtained in at least three different ways: (1) by point-by-point calculation, (2) by graphical integration, or (3) by selection of a curve from sets of pre-calculated swing curves. Point-by-point calculation was explained in Chapter II; the method of graphical integration'' will be described in this section; and the use of pre-calculated curves will be discussed in Chapter
v.
It has already been pointed out, in deriving the equal-area criterion, that the relative angular speed of a machine is given by:
J218
-dB = CJJ' = -AI dt
~
Pada
[37]
Graphically, the integral appearing in eq. 37 is the area under a curve of P a against 8 or between curves of Pi and P u against 8; in other words, the same area that was used in the equal-area criterion. In the process of applying the equal-area criterion the area may be determined as a function of 8 by counting squares on the graph paper for successive increments of 8. Or, if the power output is a sine function or other simple function of the angle, the integral may be evaluated formally. In either case w' as a function of B may be computed from the area by means of eq, 37. By rearranging this equation we obtain:
dB
dt = --;
[38]
CJJ
and by integrating we get:
[39]
Evaluation of this integral gives t as a function of 8; in other words, it gives a swing curve. If formal evaluation of this integral were possible, graphical methods would be unnecessary. As a rule, however, eq. 39 cannot be formally integrated even though eq. 37 can be. One simple case in which both integrations can be done formally is that in which the accelerating power P a is constant, Then
w' = ~2Pa(~-
80 )
[40]
SWING CURVE BY GRAPHICAL INTEGRATION
141
and [41] In general, however, the integral of eq. 39 must be evaluated graphically. This may be done by plotting a curve of l/w' against 8 and finding the area under the curve as a function of 8. Some difficulty will be encountered in determining the area under the curve for values of a near the initial value 80 and the maximum value 8m because, at these values of 8, w' is zero and the ordinate of the curve, is accordingly infinite, although the area under the curve is finite. This difficulty can be avoided by assuming P a to be constant over a small range of a (until the curve of l/w' comes back on scale) and by using eq. 41 to compute the time for the machine to swing through this range of 8.
u«,
EXAMPLE
3
By means of graphical integration obtain the swing curves (a) for a sustained fault on the system of Example 4, Chapter II, and (b) for a fault cleared in 0.4 sec. (These curves were previously computed by the pointby-point method and were plotted in Fig. 9 of Chapter II.) Solution. (a) The following data are obtained from Example 4 of Chapter II:
Inertia constant,
M
= 2.56 X
10-'
0
Initial angle,
80 = 18.1
Input,
Pi
= 0.80
Output during fault, P 11 = 0.936 sin 8 Therefore the accelerating power during the fault is
Po = P, - P« = 0.80 - 0.936 sin a
(a)
The area under the curve of Po against 8 is
Al =
r P..da = r
J 40
JI8.1°
(0.80 - 0.936 sin 8) da
= [0.808 + 0.936 X 57.3 cos 8]:8.1' = 0.80a + 53.6 cos 8 - 0.80 X 18.1 -
+ 53.6 cos 8 = 0.806 + 53.6 cos 6 -
= 0.80a
53.6 cos 18.10
14.5 - 51.0 65.5'\1nit power degrees
(b)
142
THE EQUAL-AREA CRITERION FOR STABILITY
(57.3 is the conversion factor from radians to degrees.)
~-M
-1 =
2A1
Wi
=
~1.28 X
Al
4
10- seconds per degree
(c)
TABLE 1 COMPUTATION OF
a (deg.)
18.1 20 25 30
45 60
75 90 105 120 135 150 165 180
Al
cos 8
0.951 0.940 0.906 0.866 0.707 0.500 0.259 0 -0.259 -0.500 -0.707 -0.866 -0.966 -1.000
AND
l/w'
FOR SUSTAINED FAULT (EXAMPLE
53.6 cos 8 51.0 50.4 48.5 46.4 37.9 26.8 13.9 0
-13.9 -26.8 -37.9 -46.4 -51.8 -53.6
0.808
Sum
Al
14.5 16.0 20.0 24.0 36.0 48.0 60.0 72.0 84.0 96.0 108.0 120.0 132.0 144.0
65.5 66.4 68.5 70.4 73.9 74.8 73.9 72.0 70.1 69.2 70.1 73.6 80.2 90.4
0 0.9 3.0 4.9 8.4 9.3 8.4 6.5 4.6 3.7 4.6 8.1 14.7 24.9
3,
PART
a)
l/w' (msec, per deg.) ClO
12.0 6.5 5.1 3.9 3.7 3.9 4.4 5.3 5.9
5.3 4.0 3.0 2.3
Al and l/w' are computed in Table 1. A curve of l/w' versus 0 is plotted in Fig. 10. This curve is "off scale" for values of 0 between 18.1° and 20°. The time required for the machine to swing through this angle is therefore computed by eq. 41, using the value of P« at 0 = 19° as a good approximation to the average value of F; between ~ = 18.1° and ~ = 20°.
Pa (19°)
= 0.800 -
0.936 sin 19°
= 0.800 - 0.936 X 0.326
= 0.800 t=
0.305
=
2M(~ - ~o) =
r.
= 0.044 sec.
0.495
(d)
/2 X 2.56 X 10-4(20 - 18.1) 0.495
~'
(e)
The area A2 under the curve for 10° increments of ~ between 20° and 180° is found by counting squares on the graph paper and is recorded in Table 2. The cumulative total of the area, the time t, is also tabulated. The value of t computed above for 0 = 20° is included. The total area out to any value
SWING CURVE BY GRAPHICAL INTEGRATION
:
12
143
II
II
10
...... cu
! ~
I,
8
\ \
...
"'0
,
8-
en c:
6
"'0
0
u Q)
J
Before clearing"'"
~
-,
:E
-
'E 4
--
.-.j3 2
-
~ /]
u-
J
)
o o
~~
I I
Afterclearing at 135
20
40
60
tr
/
80
~
1/
~Sustained
I"~
'"r-,
~l
0
I
J
I~Vl'"
. After ~Ieari~g a~ IOO~, V~ .....
~~
I
I
100
120
140
160
180
6 (electrical degrees) FIG.
10. Curves for the determination of swing curves by graphical integration. (Example 3.) TABLE 2
TABULATION OF AREA
t
a (deg.)
20 30 40 50 60 70 80 90
100
A2 UNDER
CURVE OF
l/w'
8 AND COMPUTATION OF 3, PART a)
VERSUS
FOR SUSTAINED FAULT (EXAMPLE
A2 (sec.)
t
a
(sec.)
(deg.)
A2 (sec.)
(sec.)
0.044 0.071 0.046 0.039 0.037 0.038 0.039 0.042 0.047
0.044 0.115 0.161 0.200 0.237 0.275 0.314 0.356 0.. 403
110 120 130 140 150 160 170 180
0.053 0.058 0.058 0.052 0.044 0.036 0.029 0.024
0.456 0.514 0.572 0.624 0.668 0.704 0.733 0.757
t
of a represents the time for the machine to swing to that value of a. The swing curve, 0 versus t, is plotted in Fig. 11. The points marked by small circles were computed by point-by-point method 2 with ~t = 0.05 sec. The two methods give results which agree well. (b) From Example 4 of Chapter II the output after clearing of the fault is Pu
= 2.06 sino
(j)
144
THE EQUAL-AREA CRITERION FOR STABILITY
and the accelerating power is therefore
P; = Pi - PUt = 0.80 - 2.06 sino
(g)
From part a of this exampIe the clearing angle corresponding to the specified clearing time of 0.40 sec. is (h)
200
Sustain~ fatult ...... ~~
160
. /V V~
40
.........
o
~
o
./
~
l,/'
0.2
0.1
~ ~ >-.....
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
"
<e)
(6) 0.20
~r
0.20
tJPG
-0-.-40---
Hydro station
H = 3.0
0.10
-# A Me:OPolitan system
H=oo FIG.
15. Two-machine power system. (Example 3.) Reactances are given in per unit on a common base.
~ 0.35.
0.20
~
• 0.10
~
0:5
~
""------<@-. (a) FIG.
(b)
16. Reduction of the network of Fig. 15, pre-fault condition. (Example 3.)
~ 0.35.
0.40
~
• 0.10
~
0:5
~
------(@-. (b)
(a) FIG.
17. Reduction of the network of Fig. 15, post-fault condition. (Example 3.)
~ 0.35 'V
~ 'V
0.40
0.10~
o.~o
'V
(a)
(c)
(b)
(d)
0.35 • Q:.!Q-rQ.:.lQ. • 0.10~. I 0.05
'\.i
-----.(@~--_...
FIG.
18. Reduction of the network of Fig. 15 with a three-phase short circuit at the middle of one line. (Example 3.)
Solution. The network joining the two machines, which is considered to consist of reactance only, is reduced as shown in Figs. 16, 17, and 18 for the pre-fault and post-fault conditions and for a fault at the middle of the line, respectively. For the first two conditions the reduction is accomplished by
164
THE TWO-MACHINE SYSTEM
parallel and series combinations; for the third condition, by two yea conversions and intervening series combinations. With a three-phase fault at the sending end of one line it is obvious, without reducing the network, that no power can be transmitted, The resulting reactances between machines, 1.50
/
~
'/ Pre- fault......
V
)" 1.08
:J~ ~
1.00
,
- ......
4 ~~~
"\ V
I ~V
--.
,1/
~
'c ::s ~
I
8-
---
)
I
J~
Fault cieared
1\,
~
/V
~
\
~/
0.50
r\
\
/I IJ
0.32
~
J~ ~ ~
II
~ ~~ ~ ~ ~-....
~
Fault at middle of line- ~
~~
If;~
o o
~Fault at sending
30
60
90
o(degrees)
r-,
1\ <,
end
120
150
\ ~ '~
r-,
180
FIG. 19. Determination of stability limit for instantaneous clearing of a fault at either location and for a sustained fault at the middle of the line. (Example 3.)
and the amplitudes of the power-angle curves (PM = EAEB/XAB = l/XAB), are as follows: . Reactance PM Condition (per unit) (per unit) Pre-fault 0.65 1.54 Post-fault 0.85 1.18 Fault at middle 2.45 0.41 Fault at end co 0 The power-angle curves are plotted in Fig. 19. The stability limit for instantaneous clearing is found by the equal-area criterion in the upper part of this figure. The value thus found, 1.08 per unit, is correct for both fault locations. The stability limit for a sustained fault at the middle of the line is found in the lower part of Fig. 19; it is 0.32 per unit. The stability limit for a sustained three-phase fault at the end of the line is zero.
0.8 0.7 0.6 0.5 0.4 0.35 0.2 0.1 0.05
0.9
0.650 0.585 0.520 0.455 0.390 0.325 0.260 0.228 0.130 0.065 0.032
1.0
1.54
=-
Pi
sin 00
(p.u.)
Pi
1
4
8
40 36 31 27 23 19 15 13
(deg.)
00
· ..
· ., · ..
2.5 2.2 2.0 1.7 1.5 1.2 1.0 0.85
p =Pi/0.41
. .'
.. ,
.. ,
70 82 95 111 130 144
60
51
from Fig. 20
(deg.)
Oc
· ., · ., · .,
0.4 0.7 1.0 1.3 1.6 2.2 3.1 6.0
from pre-calc. curves
Tc
Fault at Middle of Line
· .. · .' · ..
0.08 0.14 0.20 0.26 0.31 0.43 0.61 1.18
tc = Tc/5.08 (sec.)
,
120 138 150
112 134 149
80
...
95
..
..,
6 16 27 . .. 52
.. ,
75
...
58
52
46
from Fig. 21
Oc - 00 (deg.)
Fault at End of Line
3)
(deg.)
s,
DETERMINATION OF CLEARING TIME FROM CLEARING ANGLE (EXAMPLE
TABLE 1
0.56 0.86 1.28
· ..
~ Z
0.22 · .. 0.33
t-l
~
tJ:j
~
......
~
o
>-
trj
t'"4
~a
d
~ ~
t-3
· ..
0.06 0.10 0.14
(see.)
to
a
~
trj
166
THE TWO-MACHINE SYSTEM Pre-faulty
1/
J
I
r-.......
I J. 1-1..1-
0.32
Vi! 11111
I
11j.;
:--.:
~
I I
30
I
! .............
I
I
I
I
I
I
I
I
I
I
I-- Faultat middle of line
~
I I
I, I i i 60
4551
I
: 70
: 90: 82 95
\
"r-.... 1\
: 150 : 120: 111 130 144
~ 180
6 (degrees)
FIG. 20. Determination of stability limit as a function of clearing angle by use of the equal-area. criterion, three-phase fault at the middle of the line. Areas for P = 0.50 are shaded. (Example3.)
Stability limits between 0.32 and 1.08 per unit are assumed with the fault at the middle of the line, and the corresponding critical clearing times are found in Fig. 20. Stability limits between 0 and 1.08 per unit are assumed with the fault at the sending end of the line, and the corresponding critical clearing times are found in Fig. 21. In both figures curves of stability limit as a function of clearing angle are plotted. The values of clearing angle are entered in Table 1. The values of clearing time corresponding to these values of clearing angle are found from the pre-calculated swing curves if the fault is at the middle of the line and from eq. 15 if the fault is at the end of the line. In the first case the numerical value of Tit is needed. Byeq. 11 it is
: = ~7T'!PM = t
GH
/7T' X 60 X 0,41 = 5.08 3.0 X 1
'\j
In the second case we have from eq. 15
2GH 180!
(<<5 c
-
r,
«50 )
2 X 3.0 X 1 (<<5 c - «50) = 180 X 60 Pi - P..
Details of the determination of clearing time are given in Table 1. Curves of stability limit as a function of clearing time are plotted in Fig. 22
167
EFFECT OF FAULT-CLEARING TIME 1.30
J
1.08 f1.00 ~-
-- - -- --
-.-=
--
c:
'"
V-
Pre- fault-
17\-
~~ ,
I
~ , ,
I I
'- I : I
e 0.50
7:; ~
~ ~~
o o
I I
J : II
-~~ -- -- {7lY1, , ~-
.~
0-'\:
if ' ,I I I
I I
-
I I
~~ ,
I
V/
..-- Stability limit vs. clearing angle
"-
~Fault cleared
~
~~
~~ ~ ~ ~~ ~ ~ -- -,- ~ -- ~ ~ ~ ~ " 1'0....... ~ ~ ~ ~ ~ -
I
~-
I I . ' Fault at end of line" ~
~ ~ ~~ ~
?%~~~v;a~ ~ !" .v/
ij~ / //
1/'// 1'///
30
_
I
-'I
- -
I
I I I I I I
I
I
I
I I I
- '
: I
,
I I\..
I I
,
-t- I
I
I
~
,, 1-,I
11
~t 1:___: ___~~
: : :60 465258
75
r/ /ijj: ~
90: 95
v/~ v/./
120
I
I
~, ~ 1~8 150
~
180
o(degrees)
FIG. 21. Determination of stability limit as a function of clearing angle by use of the equal-area criterion, three-phase fault at the sending end of the line. Areas for P = 0.20 and P = 1.00 are shaded. (Example 3.)
1.0
~
'\ 1\ 1\\
\ f\
'\
, ." 1\
\
~ 0.5
<,
t\.
-,
o
I
~
11"-fll--
I
'" o
\r- Fault at middle of line
t'--.... ~
==
L- Fault at sending end of line
~
-
""""-
:...-
0.5
1.0
1.4
ex>
Fault- clearing time ( seconds)
FIG. 22. Curve of stab ility limit as a function of fault-clearing time for the system of Fig. 15. (Example 3.)
168
THE TWO-MACHINE SYSTEM
Curves for determining critical clearing time. A more direct way of determining the critical clearing time of a fault on a two-machine system than that described in the first part of this chapter has been developed and described by Byrd and Pritchard.f Their method facilitates the plotting of curves of stability limit as a function of fault duration. It takes cognizance of the fact that the voltages behind transient reactances, and hence the amplitudes of the power-angle curves, usually vary with the power transmitted in a way determined by operating practice. This fact was disregarded in our previous discussion, although allowance could have been made for it. In Byrd and Pritchard's method two assumptions are utilized in addition to those which were made in developing the equal-area criterion. They are: 1. That the network is purely reactive. 2. That all circuit breakers which open to clear the fault do simultaneously.
'80
Both these assumptions are commonly, although not necessarily, made in using the equal-area criterion in order to simplify the calculations. The derivation of Byrd and Pritchard's method follows: Let the power-angle curves of the two-machine reactance system be: Before the fault: Pu,
= Pm sin 0 sin a
[17]
= T2Pm sin 0
[18]
During the fault: Pu, = After the fault:
P u,
[16]
T1Pm
In other words, Tl and T2 are the ratios of the amplitudes of the powerangle curves during and after the fault, respectively, to that before the fault. Inasmuch as the amplitude of the power-angle curve of a reactance network is given by eq. 7,
p _ E1E2 m X 12
[19]
it is apparent that fl and T2 may be expressed in terms of the reactances between machines before, during, and after the fault, thus: T1
==
T2
==
X 12 before fault X 12 during fault
X 12 before fault X 12 after fault
[20] [21]
CUltVES DETEltMINING CRITICAL CLEAItING TIME
169
It is also apparent that, if the internal voltages E 1 and E 2 should vary, the amplitudes of all three power-angle curves would be changed proportionately, and the ratios rl and r2 would not be affected. The three power-angle curves and the input line Pi are drawn in Fig. 14 so that the equal-area criterion may be used to find the critical clearing angle oc. Ordinarily the two shaded areas are equated. It is just as correct, however, to equate the irregular area under the heavy line to the area of the rectangle below the input line. If this is done, the following equation results, from which Oc can be found:
[22] Upon setting Pi = Pm sin 00 and evaluating the integrals, eq. 22 becomes:
(Om-OO)Pm sin 80= -rIPm(COSOc-coSOO)-r2Pm(COSom-cosoc)
(om - 00) sin 00
=
cos Oc
=
(r2 -
(om -
+ rl cos 00 - T2 cos Om 00) sin 00 - Tl cos 00 + r2 cos Om
rl) cos Oc
r2 - rl
[23]
[24] [25]
where
[26] Thus, if TI, r2, and 00 are known, the critical clearing angle Oc can be found from eqs, 25 and 26. The corresponding critical modified time Tc can then be found from pre-calculated swing curves (Figs. 1 to 10) which are solutions of the swing equation (13) in which 0' now is simply 0 and p = (sin oO)/rl. The actual clearing time to in seconds can be found from the equation
tc
~180M
I GH
= T c 1rrlPm = T c '\j~frlPm
[27]
which differs from eq. 11 in that PM has been replaced by TIPm, the new symbol for the amplitude of the power-angle curve during the fault. The steps which serve to determine T c as a function of Tl, T2, and sin 00 were carried out for many values of the independent variables, and the results were plotted in the form of curves of T c versus r2 for constant TI and sin 00. See Figs. 23 to 39. Each family of curves is for a constant value of sin 00, the range covered being from 0.10 to 0.90 in steps of 0.05. The individual curves in each family are for constant values of Tl.
THE TWO-MACHINE SYSTEM
170
1.0
~~ '"""~ I!! ~'""" 1-0
0
0
~I-
"
.......
.....
II
~
c:
~
~~
&t)1 ..... .....
0
1-0
~ 1-1-
0
"
II
"... ,
,,"'rJ
I'II II
0.4
J
J
III
II
r
r1
II
.,r,
"
~~
It')
0
~
8l
_0 f-II
f- ........ JI-
.......
'-I-
J
, j
., ,
I'
,
I f-1-
j
l II
5.0
I I I I II
~~ --~ol "1 --,, c: C\I
_1-
~ ::
0
--'~II
~r 0
1('),
f-
i-f- .....
-__ 0:
IJ
1-'1- .....
f-I- I,--I-i-
"
" ........
r
"
1 II
'I
If
IJ
~
:
........JI
-,
I
II
I
I.....l .....
4.0
i-H &t)
I
,I'
-. ...
23. sin 80 = 0.10.
..........
f- "
L-L-
3.0
IIJ
~
",
,"""0
" c:
0.6
.....
..... 0
d
0.8
I I i-&t)
~
L.".1iiJ"
2.0 FIG.
<""[;;jjl;t~'"""
l...l L-I.III~
looII ....
1.0
I I
().
// ~ ~
I~
.... ~
L..I
1...oIl
0
~-~ ~J-
'I
I.J
II
1.0
--If..
'I
~
I~
o o
J
J
'I
0.2
j
I'r
II
I
0.6
rr
I J
8 .....
,"""l-
8
0.8
Tl
I I
J I
]
~
J
0.4
II
IJ
v
~
II' L...o
0.2
o o
L...II
.....
1.0
I....ll '-
v
~
II' ~
"
" ~~
2.0
..oil
~
1...oIl
"
1-1-
....
-
l....l L-
3.0
4.0
5.0
FIG. 24. sin 80 = 0.15. FIGS.
23 and 24. Curves for determination of clearing time (copied from Ref. 3 by permission).
CURVES DETERMINING CRITICAL CLEARING TIME 1.0
I I
I
It)
~i--
0
.....
1-1-'-
"
~j ... N
-i-I-i-'---
~~-
1-1-
i-i-ii-i-i-
i-I--
...0-#
0'
I
,
I
0.4
Ij
"
~
0.2
o
-I-
/
'1~~ 2 ~ o~
v ~
"
..... ""'.",.10.-.... ".-
v
--...-
..-- ...... ,..-
3.0
2.0
f\~ ....~III ~
l.",lI'ilI""" i.ooo'ioooo'
0.25_
Tt -
5.0
4.0
Tc
1.0
I Ill)
,
.C?
~i-~ ~
- f - .....
~~i e
t-
o
" /IT ~
1°
0.8
~
~ i-
... M
0-#
-i-l-i-
o
-- a....,I-: --
"
~
"
,
_... -... -..... ... """ ~
1.0
1---
I-r- - i -
i-_ -f-
~""" I-i-
tI-S
WI I--~a.n' 0
1 - - I-
-
0 II
~
~~
"
10·
a.....~
..." t:
II
I II
I
I~
,
.~ ~
~
~
t
i--
i.lIII .....
L-"'~
I ,
I
I
t
5.0
II I
i-
I
I
,
::r'
1 0.
l
II
c:
~- ~"...,,,
1--
,
I
/fI
~
'O·t
1
II J......,~:
j
,
~
I~
I
;i"~1-
~
4.0
'I
IJ ~
,v ~
~
... ---
I~ I
"""Ji-
v
I
~
~
= 0.25.
i- .... f-~c::)
I
r
,
i-
~
3.0
I ,
>:1-11 ~
II
I
1
..." 0.6
I
80
t)';~~l
~ r-J'I. '("
[.""II
1.1--
--"""
lili ttC:fJ°t
OJ
... i.: 1-
r
t--
J
II
J..."",
ii-
II
~
........
26. sin
I I I
-
II
IJ
'Tc
1.0
II
~
~
2.0 FIG.
C\J
i-
I-
II
,,-
~
~ilI"""
1.1 ....
-
CV
0J~
'I
,
II
~ -~
-a.."" -
I
I~
-....-
e:)
II
/
J J
".-
I I
II,~ -~K> -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~
If'
~
~
0.7
o
I , I I I J
J
~
1...~
1...oIlI~
= 0.65.
~
II
~
i-
v- ~r-
4,....
~
sin 80
5.0
4.0
3.0
~WI >-~ ~ ~ Ort' I :r-I/O"l
tr-- -
J
t I
FIGS.
.,
~
o
4.0
,, I
I I 'D,I-
~()~ ~~
.C'\ ___ ....~,~
5.0
= 0.70.
33, 34, 35. Curves for determination of clearing time (copied from Ref. 3 by permission).
CURVES DETERMINING CRITICAL CLEARING TIME 1.0
IIi}
~Wd'l 0'Yo ~ ~~ I~t~ 0 t::i O'~tfll O' o 0' ~ ~~Xi~
u: :!J ~~I-~
~O
~~~
0.9
11
J J
'I' I/'
.,.
~
'I
~
II
}
~
~ ~
"
~
jllIII"
//'1 //~:t.
.. I
-
~~
v
I
o ~
~
T
~
~~ I-~
'h...-I......""
~ l..o""""
io"'"
'-
I
T T 11~ ~~ 1-1-
If'
.....
.""
L.oo~
0:'
.,
,<,
~" ~
l....olI~
'"
~
~
lIi""
"
~
1-1-
175
~~
_I-'"
...
1..- .....
0.8 0.7
o
1.0 FIG.
r'rr =
4.0
3.0
2.0
5.0
36. sin 00 = 0.75.
030
= O. 0
'1
1.0
~
1/
.
~l 17 ,~r~J \)'b\<)~, 'rCi.1: ~~. ~~ ~~a a:t a:l,;~ ~B:-t\iP ~ ~o'f.~ ~\ffi1<j~ ~o.~~ Iio' I/t ~ ~I- '- 'h~~- t\~,,~ 1.t~.,,~~o~'z;;..-e 1'\I"'~~~- "~~..-""'~
t~ 11 -I~""I-f-<~ ~
" Jt-,"" ~
0.9 0.8
oJ
;L
= . 0.40 l
11'1 ~ r:r/: ~~
tt"
I.' ~
~~
o
- '--
... .......
1.0
... "" 1--..... ~
1.1I ......
I..-
1....1"'"
l..o.IlIl--
~"
.""
..."'"'"
2.0
--
1- .......
;!Ii
•
~\-,,"'-~-~
-"
L.-
.... 11I""
1.-:- ~iI-"'"
3.0
4.0
5.0
Tc
FIG.
1.00 o
rt> - co'~'>~ co'> v o· Alo,,, oep " W ~!~~ ~~O~""I~~ ~~I ~............~J -"I" ~ ,jf- ..... <)I.~ ~ <~~ <.t;~~~ 0 n,
-
F-f-
I-~'
~...
r
~"'J~
1/
J
j
I
0.90 0.85
IJ
JI
I..;
~ ~
1....1
I;l'"
l'
J
v
1....1
I.l'"
"
I.l'"
~
i...ll
~
--
1.0
....... "'"'"
II""
ijll'"
~'-'
t-
..... jllIII"
~,....
L....o .....
I...-
~
I....
l.olII
o
- '(,?
l-
I;'
II IJ
//
rlf- f-I-
4......J"-'r- ....
I-
v
f-f-
1/
I-
0.95
37. sin 00 = 0.80.
2.0
3.0
4.0
5.0
Tc
FIG.
1.00
...~ 0.95
....
~
" -)- "...; II
..."
,
I
I.." I
T
o
"
1/
-- - --"'"
~"'1 4..' v: ~ --"
L.... II 0 ~I- lli~'j~I~(). O~~~.,.~~ D~09..... ~().t!.~~ 1\ ~ '(~~L.,- - f - ~'\....-I~ - '- t~~~ h~ -'" ..... ~
.....
1.0
~""" "'""',~
... ...... """'" ~.....
",
..... ~
l0- ll""
_"'" ~~
Iiiiiii
-~
2.0 FIG.
FIGS.
t1~$~:1Ci
ol-~~*~Y/)4:<§le::; O· O· + 0' ~ I
0.90
38. sin 00 = 0.85.
39. sin
3.0 80
4.0
5.0
= 0.90.
36, 37, 38, 39. Curves for determination of clearing time (copied from Ref. 3 by permission).
THE TWO-MACHINE SYSTEM
176
It is convenient to have an additional curve for determining the stability limits for instantaneous clearing and for sustained faults. To obtain the equation of such a curve refer to Fig. 12, which shows the application of the equal-area criterion to the case of instantaneous 1.0
Ii J
v 1/
0.9
iJ II' 1/
0.8
J
V II
0.7
/ I~ ~
0.6 ~
c 0.5
v
.~
V WI'" I;"
,., v:
0.4
0.3 I~
,., ,.,
0.2
~
l.'
~
I'
0.1 l;l"
'-'
I~
o I#"
o
FIG.
0.1
0.2
0.3
0.4
0.5 r J or r2
0.6
0.7
0.8
0.9
1.0
40. Curve for determining the stability limit for a sustained fault or for a fault cleared instantaneously (copied from Ref. 3 by permission).
clearing, and equate the area of the rectangle under the Pi line between ao and 6m to the area under the post-fault power-angle curve between the same limits. Area of rectangle = Pi(8m - 60) = Pm(6m - 60) sin 60
1
sin ~
T2(COS
ao -
Area under power-angle curve = T2Pm
Equating the two areas, we get (am - ao) sin ~o =
6m
as = T2Pm (cos ~o cos ~m)
cos 8m)
[28]
CURVES DETERMINING CRITICAL CLEARING TIME
177
where 8m is as given by eq. 26. Equation 28 expresses implicitly a relation between T2 and sin 80 which is plotted in Fig. 40. To find the stability limit for instantaneous clearing, enter this curve with T2 and read sin 80 • Then the stability limit is Pm sin 80 • The same curve can be used to find the stability limit for a sustained fault by entering it with P the value of TI instead of T2; this becomes evident when Fig. 13 is compared with Fig. 12. It has already been mentioned that, as the transmitted power (Pi == Pm sin 80 ) varies, the internal voltages E 1 and E 2 usually vary, and hence Pm = E 1E2/X12 also varies. For the present purpose the most useful way a 0------------"'· to show these relations is to plot Pi sin 00 and Pm against sin 80 , as shown in F10. 41. Typica · 1 curves 0 f P, an d Fig. 41. r; versus sin ~o. The procedure for obtaining a curve of power limit as a function of fault duration by Byrd and Pritchard's method will now be summarized. 1. Reduce the reactance network to an equivalent ~ between the two machines and neutral for each of the three conditions:
a. Before the fault. b. During the fault. c. After clearing the fault. Only the reactances X 12 between the two machines are used in what follows. 2. Compute the equivalent inertia constant M by eq. 2. 3. Calculate and plot curves (like those of Fig. 41) of
a. Pm = E IE2/XI 2 versus sin 80 b. Initial power Pi versus sin 80 maintaining the bus voltages at the values which would be held in actual operation. 4. Compute TI and T2 by eqs. 20 and 21. 5. a. Enter the curve of Fig. 40 with Tl and read sin 80 • From Fig. 41 read Pi corresponding to this value of sin 80 • This is the stability limit for a sustained fault.
b. Repeat, using T2 instead of TI. The value of Pi thus found is the stability limit for instantaneous clearing.
178
THE TWO-MACHINE SYSTEM
6. t Select values of sin 00 which are multiples of 0.05 and which are between the values found in steps 5a and 5b. For each such value of sin 00 find the proper family of curves from Figs. 23 to 39; find the curve for the value of Tl; enter this curve with T2 and read T e•
7. For each value of sin 00 used in step 6 read the corresponding values of Pi and Pm from the curves of step 3 (like Fig. 41). 8. t For each value of Tc found in step 6 compute the clearing time te by eq. 27, using the proper value of Pm determined in step 7. 9. Plot stability limit Pi as a function of clearing time te• This curve will look like the one in Fig. 11. The method which has been described fails if Tl = 0, since for this caser = 0 and t is indeterminate from eq. 27. To avoid this difficulty a new modified time p is now introduced, related to the actual time t by
t
= p J180M = p 7rPm
I GH
~7rfPm
. [29]
and differing from Tin eq. 27 in that Pm, the amplitude of the pre-fault power-angle curve, is used instead of T1Pm , the amplitude of the fault power-angle curve. Hence p = T/Vi). The swing equation now becomes [30]
with 8 in electrical radians, and it has the solution p=
2(0 - 00)
.sin 00
[31]
which will be recognized as eq. 15, transformed by the substitution of eq. 29 for t, and with 0 - 00 expressed in electrical radians. Given Tl = 0 and the values of T2 and sin 00, Pc may be found by using eqs. 25 and 26 to get oc, and then eq. 31 to get Pc. From the results of such calculations Pc has been plotted in Fig. 42 against sin 00 for various values of T2. If Tl = 0, steps 6 and 8 of the procedure are replaced by steps 6A and 8A, respectively, which are as follows: 6A. Select values of sin 00 which are between zero and the value found in step 5b. Find the curve in Fig. 42 corresponding to the [If rl = 0, substitute steps 6A and 8A, described on this page and page 180, for steps 6 and 8, respectively.
'(ii
c:
'0
0
o
2
0.4
0.6
0.8
1.0
o
0.5
FIG. 42.
1.0 2.0
2.5 3.0 Pc
3.5 4.0
4.5
5.0
5.5
6.0
Curves for determining clearing time -if rl = 0 (copied from Ref. 3 by permission).
1.5
6.5
7.0
~
co
~
~
s: l%j
~
~
o
~Z
&;
a
~
:j
§
o
z
~
z
a=
~
t;d
~
t:1
ax
~
a
THE TWO-MACHINE SYSTEM
180
value of r2. Enter it with the selected values of sin corresponding values of Pc.
ao and read the
8A. For each value of Pc found in step 6A compute the clearing time tc by eq. 29, using the proper value of Pm determined in step 7. EXAMPLE
4
Plot a curve of stability limit as a function of fault duration .for a threephase fault at the middle of one of the 132-kv. transmission lines of the power system of Fig. 43. The two hydroelectric generators i and k are to be ra. 0.175
~
33.0/13.2 O.l75~
33.0/13.2 0.30
0.35
0-111:
~30
6.6/132
0.35 )(
Fault
.6.'A0.140
~ 33.0/13.2
0.140
f----®
33.0/13.2 43. One-line diagram of power system, with reactances of lines and transformers in per unit on a lOO-Mva. base (Example 4). Data on generators and loads are given in Table 2. (From Ref. 3 by permisslon.)
FIG.
garded as one machine of a two-machine system, and the steam turbogenerators, a, b, g, and h, together with loads c, d, e, and J, as the other machine. Each load is assumed to consist of three equal parts, one of resistance, one of synchronous motors operating at half load, and one of induction motors at 1/2.75 load. The reactances and inertia constants of the generators and of the composite loads are given in Table 2, expressed on a base power equal to the rating of the individual generator or load. In Fig. 43 the reactances of the lines and transformers are given on a base of 100 Mva. and nominal voltage. The system frequency is 60 c.p.s, Solution. The procedure described on pp. 177-8 will be followed. 1. Network reduction. All reactances will be expressed on a system base of 100 Mva, The reactances in Fig. 43 are already on this base. The
CURVES DETERMINING CRITICAL CLEARING TIME
181
machine reactances in Table 2, given on the machine base, are converted to the system base by multiplying them by (100 Mva.)/(machine rating in megavolt-am peres) .
TABLE 2 MACHINE REACTANCES AND INERTIA CONSTANTS (EXAMPLE
,
Rating Kind
Machine
b c
Machine System Machine System Base Base Base Base
0.8
0.18
0.45
6.0
"
"
"
"
"
"
Composite load 55
" " "
" "u
e g h i k
P.F.
Turbogenerator 40
d
f
H
Xd
~
Mva.
a
0.9
30
u
"
" "
55 Turbogenerator 50
"
"
0.8
"
0.56
" " " 0.18
"
1.01 1.86
"
1.01 0.36
"
1.95
" "
"
6.0
"
Hydro generator 29. 4
0.85
0.30
1.02
3.0
u
"
"
"
"
"
4)
"
2.40
"
1.07 0.58 u
1.07 3.00
"
0.88 u
The network is so nearly symmetrical about a horizontal axis that the reduction may be facilitated with very little sacrifice in accuracy by assuming that the two sections of the long receiving bus are tied together and also that the load busses d and e are tied together. In addition, machines i and k are assumed to be tied together behind their transient reactances; machines a, b, c, d, e, j, g, and h are assumed to be tied together similarly. Then, after simplifying the hydroelectric-station circuits and the receiving system by series and parallel combinations, the circuit of Fig. 44a results. The reactances of the three parts which are separated from one another by transformers are given on a common power base of 100 Mva, and on voltage bases equal to the nominal voltages of 6.6, 132.0, and 33.0 kv, It will be noticed that the ratio of the receiving-end transformers, 125.4 kv./33.0 kv., is not equal to the ratio of base voltages, 132.0 kv./33.0 kv, Therefore the base voltage for the receiving system should be changed from 33.0 kv. to 33.0 X 132.0/125.4 = 34.7 kv.; the reactance of that system on the new base is 0.13 X
(33'0)2 34.7 = 0.12 per unit
The new value is shown in Fig. 44b. Parallel and series combinations give the result shown in Fig. 44c for the pre-fault network. Xu = 1.10 per unit. The faulted network is obtained by grounding the midpoint X of one line in Fig. 44b. Then a ~-Y conversion, series combinations, and a Y-~ conversion lead to the ~ network of Fig. 44d. X 12 = 7.20 per unit. The post-fault network is obtained by omitting one of the parallel lines of
THE TWO-MACHINE SYSTEM
182
Fig. 44b. Then a series combination leads to Fig. 44e. X 12 = 1.28 per unit. 2. Inertia constant. The inertia constants in Table 2, given on the machine base, are converted to the system base by multiplying them by (rna-
~~~(a) 6.6/132 kv.
0.35
125.4/33.0 kv.
O.~15
2
~~3~(b) 051
0.35
~D
/>.
~ vvv ~
(c)
(d)
~vvv~
(e)
FIG. 44. Reduction of the network of Fig. 43 (Example 4). The reduced networks are given by c for the pre-fault condition, by d for the fault condition, and by e for the post-fault condition.
chine rating)/(100 Mva.). HI
=
Hi+ H k
=
Then 1.76
H2 = Ha + Hb+ H c + Hd+ He + H/+ H, H
=
+ Hh = 14.1
H 1H2 = 1.76 X 14.1 = 1.56 HI H2 15.9
+
3. Curves of P« and Pi versus sin 00. Assume that 132 kv. (1.00 per unit) is maintained on the high-voltage side of the step-up transformers, and 33.0 kv. (0.95 per unit) is maintained on the low-voltage side of the step-down transformers. The voltages behind transient reactance, E 1 and E 21 as functions of the initial angle 00 between them, can be found from the vector diagram of Fig. 45 by assigning arbitrary values to the current I and solving,
CURVES DETERMINING CRITICAL CLEARING TIME
183
either graphically or by trigonometric computation, the resulting triangles for Ell E 2, and 00. Then P« and Pi are calculated from the relations
P _ E1E2 X l2
m -
_
E 1E2
-
1.10
and
Pi = Pm sin 00 0.661
FIG.
0.121
----~~
45. Vector diagram for determining E 1 and E 2 as functions of ~o (Example 4). TABLE 3 CALCULATION OF
Pm AND Pi
sin
VERSUS
P,
EI
E2
~o
0.125 0.250 0.375 0.500 0.750 1.000
1.106 1.121 1.145 1.177 1.267 1.373
0.933 0.934 0.936 0.939 0.950 0.972
7.7 15.3 22.7 30.0 43.6 55.9
ao
sin
The results of the calculations are given in Table 3. functions of sin 00 are plotted in Fig. 46. 4. Computation of fl and f2.
(EXAMPLE ~o
0.134 0.264 0.386 0.499 0.689 0.828
4)
Pm
0.935 0.948 0.971 1.001
1.090 1.221
Curves of Pm and Pi as
By eqs. 20 and 21, 1.10
fl= - =
7.20
0.15
3
1.10 1.28
r2= - = 0.86 5. Stability limits for sustained fault and [or instantaneous clearing. a. Sustainedfault. Entering the curve of Fig. 40 with fl = 0.153, we read
THE TWO-MACHINE SYSTEM
184
sin 80 = 0.117. Entering the Pi curve of Fig. 46 with this value of sin 00, we read Pi = 0.110. b. Instantaneous clearing. Entering the curve of Fig. 40 with = 0.86,
'2
I
1.0 1--
---
Pm ~ / '
~
~
J
0.5
/ /1>;
oV
v
/ ~V
1/
/
J
V
/
o
05
ID
sin 00 FIG. 46.
Curves of Pm and Pi as functions of sin 80 (Example 4).
TABLE 4 CALCULATION OF POWER LIMIT Pi AS A FUNCTION OF CLEARING TIME t e (EXAMPLE 4)
1
2
3
4
5
6 te
sin 80
Pi
Pm
Tc/tc
Te
(sec.)
0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15
0.933 0.843 0.763 0.690 0.623 0.560 0.500 0.443 0.390 0.336 0.285 0.236 0.188 0.141
1.18 1.14 1.10 1.07 1.04 1.02 1.00 0.986 0.973 0.963 0.954 0.946 0.940 0.936
4.68 4.57 4.50 4.44 4.37 4.33 4.29 4.26 4.23 4.21 4.19 4.17 4.16 4.15
0.13 0.24 0.35 0.45 0.55 0.67 0.78 0.90 1.07 1.26 1.50 1.79 2.29 3.31
0.03 0.05 0.08 0.10 0.13 0.15 0.18 0.22 0.25 0.30 0.36 0.43 0.55 0.80
~
SUMMARY OF CALCULATING TRANSIENT STABILITY
185
we read sin 00 = 0.825. Entering the Pi curve of Fig. 46 with this value of sin 00, we read Pi = 0.995. 6, 7, and 8. Stability limii« [orfinite clearing times. These three steps are carried out in Table 4. In col. 1 are entered values of sin 00 from 0.15 to 0.80 at intervals of 0:05. In cola. 2 and 3 are the corresponding values of Pi and
1.0
,
1\ \
~
\ 'i\
",
'"r-,
o o
............ ............
~
r--.. ::::-
0.5
1.0
Clearing time t, (seconds)
FIG. 47. Curve of stability limit as a function of clearing time, three-phase short circuit at the middle of one of the 132-kv. lines of the system of Fig. 43 (Example 4).
PM' read from the curves of Fig. 46. In col. 4 is the value of'rcltc, which by eq, 27 is 1r
X 60 X 0.153 1 X 1.56
In col. 5 is the value of T e read from the curves. dividing Te from col. 5 by T c/t cfrom col. 4.
v'p = 4.3 vp m
m
In col. 6 is fe, obtained by
9. Curve oj stability limit as a Junction oj clearing time is plotted in Fig. 47.
Summary of methods of calculating transient stability. Byrd and Pritchard's method, which has just been illustrated, is probably the best way of finding the critical switching time corresponding to a given transmitted power, or of finding the transient stability limit for a given switching time, for a two-machine pure-reactance system with only one
instant of switching after inception of the fault. If the network is not purely reactive (for example, if line resistance
or shunt loads are to be taken into account), this method is not applicable. However, the equal-area method may be used in conjunction with pre-calculated swing curves as described on pp. 149-57 of this chapter. If the system has three or more machines, their swing curves must be
THE TWO-MACHINE SYSTEM
186
TABLE 5 RECOMMENDED METHODS OF CALCULATING TRANSIENT STABILITY OF A
Two-
MACHINE SYSTEM
Recommended Method of Calculating
Number of Instants of Switching or Circuit Change
Examples
1
Sustained fault Fault cleared instantly Switching out a sound line
B A Critical Switching Time Stability Limit CorreCorresponding to a sponding to a Given Given Power Switching Time Use equal-area criterion or Byrd and Pritchard's method (steps 1 to 5).
(Pure-reactance network) Plot part of curve by Byrd and Pritchard's method.
2
3
Fault cleared in finite (Linear impedance network) time by simultaneUse equal-area method Cut and try by assumous opening of all for finding critical ing power and probreakers switching angle, then ceeding as in A. pre-calculated swing curve for finding switching time. Fault cleared by se., quential opening of 2 breakers Fault cleared by si- Dse a combination of swing curves, equal-area multaneous opening criterion, and successive trials. of all breakers, followed by simultaneous reclosing
calculated to a value of time when one can determine whether the system is stable for the assumed power and switching time. The procedure is then repeated for different values of power or of switching time. If there is more than one instant of switching after inception of the fault on a two-machine system (for instance, if a fault is cleared by successive opening of two or more circuit breakers), it is still possible to use the equal-area criterion as an adjunct to swing curves, making continuation of the swing curves beyond the instant of the last switching operation unnecessary. Suppose, for example, that a fault is cleared by successive opening of two breakers at known times and that
CERTAIN FACTORS AFFECTING STABILITY
187
the power limit is sought. A value of power is assumed, and the swing curve is calculated till the time of opening of the second breaker. The equal-area criterion for stability is then used. The procedure is repeated for other values of power until one is found which makes the positive and negative areas equal. This value is the stability limit. A plot of net area against power enables one to find the stability limit by interpolation after two or three trials. Table 5 lists the foregoing methods of calculation for two-machine systems with recommendations concerning which to use in each of various situations. For estimating purposes two sets of curves which are not reproduced here will be found useful. One of them" gives, for a two-machine reactance system, curves of critical clearing time tc against sin 00 for constant rl and r2. Four families of such curves are given, each for a value of rl corresponding to a different type of short circuit at the sending end of a transmission line and for several values .of r2. The equivalent inertia constant, H = H1H2/(H1 + H2 ) , is assumed to be 1.5. If it has a different .value, the switching time read from the curves must be multiplied by a correction factor V H /1.5. The other set of curves" gives the critical clearing time of a fault on a metropolitan power system having generators of H = 8 operating at full load and 0.85 power factor, based on the assumption that the generator or group of generators nearest the fault swings with respect to the remaining generators. Certain factors affecting stability. With a knowledge of the methods of analyzing the stability of the two-machine system that have been described in the preceding pages, we can proceed to draw a number of general conclusions regarding the effect on stability of certain features of apparatus design, system layout, and operation. The effect of each feature must be considered under all three conditions-before the fault, during the fault, and after clearing the fault. Some features of layout or design promote stability during all three conditions, whereas others are beneficial during one condition but detrimental during another. In the equal-area criterion for stability, the power-angle curves for each of the three conditions are used. (Refer to Fig. 14.) The factors determining the relative sizes of the two compared areas Al and A 2 are (1) the clearing angle Oc and (2) the amplitudes of the three powerangle curves, Pm, rIPm, r2Pm» relative to the height of the input (initial power) line Pi. Stability is aided by decreasing area A 1 and by increasing area A 2 • For a given Pi this may be accomplished, from the geometrical viewpoint, chiefly by decreasing the clearing angle Oc
188
THE TWO-MACHINE SYSTEM
and by increasing the amplitudes of the fault and post-fault curves, T1Pm and T2Pm. It is also helpful to decrease the amplitude of the pre-fault power-angle curve if such reduction is possible without at the same time decreasing the amplitudes of the other two power-angle curves, because so doing increases the initial angle 00- If any feature changes the amplitudes of all three curves proportionally, however, it is beneficial to increase rather than to decrease the amplitudes. The clearing angle s, depends on the clearing time and on the equivalent inertia constant (eq. 2). The importance of rapid fault clearing, obtainable by the use of high-speed circuit breakers and fast relaying, has already been stressed. For a given clearing time the clearing angle is decreased by increasing the inertia constant. If the inertia constants of the two machines are far from equal, the equivalent inertia constant of the system is very nearly that of the lighter machine; hence it is more effective to increase the inertia of the light machine than of the heavy one. Seldom, however, has it proved economical to increase the inertia of a generator beyond its normal value. t But a consideration of the role of inertia sheds some light on the relative critical clearing times for hydroelectric generators (average H = 3) and steam turbogenerators (average H = 6) if the circuit reactances are about the same. Since the time varies as the square root of the inertia constant Ceq. 27), a fault on a hydroelectric system must be cleared in about 70% of the critical clearing time of the corresponding steam system. The amplitude of the power-angle curves is E 1E2/X12 , where E 1 and E 2 are the internal voltages of the two synchronous machines and X 12 is the reactance between these voltages, which is, in general, different for each of the three circuit conditions. Increasing the internal voltages increases the amplitude of all three power-angle curves and aids stability. An increase in internal voltages usually accompanies an increase of load on the machines, but this does not necessarily mean an increase of the equivalent input Pi. If both generators have local loads which are increased in the proper ratio, Pi is not affected. (See eq.3.) The amplitude of all the power-angle curves is increased by decreasing the reactance X 12 between the machines. This reactance consists principally of the reactances of the two synchronous machines, transformer reactance, and line reactance. A large part of it is in the machines. The transient reactance of each class of large synchronous machines (steam turbogenerator, water-wheel generator, condenser, etc.) has a characteristic value and does not vary much in normal tA notable exception is the Boulder Dam station.
CERTAIN FACTORS AFFECTING STABILITY
189
designs. A lower value of reactance (in per unit) is obtained essentially by building a larger machine and under-rating it. This has seldom proved an economical way to aid stability.§ The reactances of transformers also have, for a given size and voltage, normal values below which it is difficult to go. The reactance of an overhead transmission line is only slightly affected by any change of spacing or conductor size which is practical from other standpoints than stability. A decrease in system frequency, of course, decreases the reactances of all parts of the system and thereby aids stability. Therefore a frequency of 25 or 50 c.p.s. is preferable to 60 c.p.s. from the standpoint of stability. Nevertheless, because of other advantages, 60 c.p.s. has become the standard power-system frequency in the United States, and a transmission project operating at a lower frequency would necessarily include frequency-changing equipment at one or both ends. This additional equipment might offset any advantage of the lower frequency in regard to stability. Up to the present time the stability limitations of 60-c.p.s. systems have not been serious enough to warrant lower frequencies. Series capacitors have been used on some transmission lines to partially compensate for the inductive reactance of the lines. However, they have not been used on any major project where stability was an important factor, although they have been seriously considered. 6 ,7,g The most important means of reducing the reactance are (1) to connect more lines in parallel and (2) to raise the transmission voltages. The ratio rl of the amplitude of the fault power-angle curve to the amplitude of the pre-fault power-angle curve depends upon the type and location of the fault. The effect of the type of fault will be discussed in Chapter VI. The effect of location will be considered briefly now. A fault on a bus or on a line close to a bus is more severe than a fault of the same type near the middle of a line. Most severe of all is a three-phase short circuit at some point (for example, fault b in Fig. 15) where it entirely blocks the transfer of power from one generator to the other; then rl = o. Although a fault near the end of a line and one near the middle of the line are equally probable, the most severe location is usually assumed in a stability study so that the results will be conservative. The ratio T2 of the amplitude of the post-fault curve to the amplitude of the pre-fault curve depends on the number and location of lines which are opened to clear the fault; therefore it depends upon the fault location and upon the relaying scheme. The most favorable §Again the Boulder Dam station is a notable exception.
190
THE TWO-MACHINE SYSTEM
value of T2 encountered in practice is 1. This value applies to a fault on an unloaded radial feeder cleared by disconnection of the feeder. It is also attained with quick-reclosing schemes which restore the faulted line to service after a brief time to allow for extinction of the arc. II A fault on a bus, although electrically equivalent before clearing to a fault on a line adjacent to a bus, is more severe in its effects after clearing than a line fault is because several lines must be opened to clear it. Sometimes a comparative study is made between a small number of high-voltage lines and a large number of low-voltage lines having the same parallel reactance in per unit. The large number of lines is preferable to the small number from the standpoint of stability, because to clear a fault requires the opening of one line in either case, 16 24
J
24 FIG.
16 16 48. System for which the effectof high-voltagebussingis considered in Prob, 4.
but one line is a small fraction of the entire number of lines when this number is great. For other reasons, however, a smaller number of lines may be preferable. When two or more parallel lines are used, the number of intermediate busses is a factor affecting stability. For example, in the system of Fig. 48 it is worth inquiring whether the addition of the two high-voltage busses shown in broken lines is beneficial or detrimental. If the high-voltage busses and circuit breakers are provided, a line fault can be cleared by switching out one line while leaving all the transformers connected, whereas, without the high-voltage busses, the transformers would be switched out with the line. Thus the busses increase the value of T2 and are beneficial after clearing of the fault. On the other hand, during the fault the shock to the system is increased by the busses; that is, Tt is lessened. Which effect predominates depends upon the speed of clearing. With fast clearing the busses are beneficial; with slow clearing they are detrimental. The same principle may determine, although less obviously, the effect upon stability of changes of layout in more complicated networks. The effect of a given change may be either beneficial or detrimental, depending upon the speed of fault clearing. Needless to say,_ many USee Chapter XI, Vol. II.
PROBLEMS
191
contemplated changes of connections are conceived from other motives than the improvement of stability. They may be intended, for example, to improve voltage regulation, to increase the reliability of supply to certain loads, or to relieve overloading of certain lines. Nevertheless the effect of such changes on stability conditions should be considered. Most of the conclusions which have been drawn from a study of the two-machine system apply equally well to a multimachine system. When a muItimachine system is unstable, as a rule, it is split into two groups of machines which go out of synchronism with each other, while the machines within each group stay in synchronism. The grouping may differ with the fault location. Still, for a given fault location, the general behavior of the multimachine system is similar to that of a two-machine system. It could be analyzed as a two-machine system except that the machines within a group may swing so far with respect to each other (yet without going out of step) as to have a marked effect on the relations between the two groups. In addition, the grouping for a given fault location is not always apparent in advance. REFERENCES
1. R. H. PARK and E. H. BANCKER, "System Stability as a Design Problem," A.l.E.E. Trans., vol. 48, pp. 170-93, January, 1929. 2. I. H. SUMMERS and J. B. MCCLURE, "Progress in the Study of System Stability," A.I.E.E. Trans., vol. 49, pp. 132-58, January, 1930. 3. H. L. BYRD and S. R. PRITCHARD, JR., "Solution of the Two-Machine Stability Problem," Gen. Elec. Rev., vol. 36, pp. 81-93, February, 1933. 4. R. D. EVANS and W. A. LEWIS, "Selecting Breaker Speeds for Stable Operation," Elec. Wid., vol. 95, pp. 336-40, Feb. 15, 1930. 5. S. B. GRISCOM, W. A. LEWIS, and W. R. ELLIS, "Generalized Stability Solution for Metropolitan Type Systems," A.I.E.E. Trans., vol. 51, pp, 363-72, June, 1932; disc., pp. 373-4. 6. E. C. STARR and R. D. EVANS, "Series Capacitors for Transmission Circuits," A. I. E. E. Trans., vol. 61, pp. 963-73, 1942; disc., pp. 1044-6. 7. R. B. BODINE, C. CONCORDIA, and GABRIEL KRON, "Self-Excited Oscillations of Capacitor-Compensated Long-Distance Transmission Lines," A.I.E.E. Trans., vol. 62, pp, 41-4, January, 1943; disc., pp. 371-2. 8. J. W. BUTLER, J. E. PAUL, and T. ·W. SCHROEDER, "Steady-State and Transient-Stability Analysis of Series Capacitors in Long Transmission Lines," A.I.E.E. Trans., vol. 62, pp. 58-65, February, 1943; disc., pp. 377-80. PROBLEMS ON CHAPTER V
1. Check the results of Example 3, part a, by Byrd and Pritchard's method. 2. Work Example 4 for a three-phase fault on a 132-kv. line near the hydroelectric-station high-voltage bus.
192
THE TWO-MACHINE SYSTEM
3. Work Example 4, assuming that the voltages behind transient reactance are independent of the initial power transmitted. 4. Find the effect of high-voltage bussing on the stability of the 6o-cycle system shown in Fig, 48 (consisting of a generator G1 feeding over a doublecircuit high-voltage line to the infinite bus G2) by plotting stability limit in per unit against fault duration in seconds for a three-phase fault at the sending end of one circuit (a) with no high-voltage bussing and (b) with highvoltage bussing at both ends, as indicated by the broken lines. The reactances of the circuit elements are given in per cent based on the rating of G1. Assume that the voltage of the infinite bus and the sending-end voltage of the high-voltage lines are 1.00 per unit for all values of initial power. Explain why the curves cross. 5. Find the effect of fault location on the system of Fig. 48 with highvoltage bussing, by plotting stability limit in per unit as a function of fault location as the three-phase fault moves from the sending end to the receiving end of the line. The clearing time is constant at 0.15 sec.
CHAPTER VI SOLUTION OF FAULTED THREE-PHASE NETWORKS
Symmetrical components. Most power systems are three-phase, but, as long ,as they are symmetrical, they may be represented on a single-phase line-to-neutral basis for purpose of calculations. Threephase systems are usually symmetrical or very nearly so during normal conditions and also during three-phase faults. During other types of fault, such as line-to-ground or line-to-line, they are unsymmetrical, and the simple single-phase representation no longer suffices. In such cases the method of symmetrical components! is generally used. In the method of symmetrical components an unsymmetrical set of vector currents or voltages is resolved into symmetrical sets of components. Before proving that such resolution is possible, let us first consider the inverse process, that of adding symmetrical sets of vectors to obtain an unsymmetrical set. Consider a symmetrical set of vectors (Vat, Vbl, Vel of Fig. la) representing the balanced voltages or currents of a three-phase circuit. The vectors are equal in magnitude and 1200 apart in phase. Their phase order, or phase sequence, is abc, which is called the positive sequence. Consider also another symmetrical set of vectors (Va2 , Vb2, Ve2 of Fig. Ib) of the opposite phase sequence, cba. These vectors are said to have negative phase sequence. If now corresponding vectors of the two sets are added, thus Va' =
Vb' =
v:
=
+ V a2 »« + Vb2 Vel + V e2
[1]
Val
[2] [3]
v:
as shown in Fig. Ic, the resultant vectors, Va', Ve', form an unsymmetrical set. They are neither equal in magnitude nor 1200 apart in phase. It has been shown that an unsymmetrical set of vectors results from the addition of corresponding vectors of two symmetrical sets of opposite phase sequence, By reversal of the process the resulting unsymmetrical set can be resolved into its symmetrical components. We may well ask, can any unsymmetrical set of three vectors be resolved into two symmetrical sets, one of positive phase sequence, the other of negative phase sequence? This question is answered if we 193
SOLUTION OF FAULTED NETWORKS
194
notice that, since the sum of the three vectors in each symmetrical set is zero, the sum of the three vectors in the resulting unsymmetrical set is also zero. Val Vbl Vel = 0 [4]
+ + V a2 + Vb2 + Ve2 = 0
[5]
Adding, we get (Val + Va2 )
+
(Vbl
+ V b2 ) + (Vel + V c2 ) = 0
[6]
or
[7]
(a)
\\\
~o VbO VeO (d)
1. Composition of unsymmetrical vectors from symmetrical components. (b) Negative-sequence vectors. (c) Addition of negative-sequence vectors to positive-sequence vectors. (d) Zero-sequence vectors. (e) Addition of the sum of positive-sequence and negative-sequence vectors to zerosequence vectors. FIG.
(a) Positive-sequence vectors.
In general, the sum of the three vectors in an unsymmetrical set is not zero but, instead, may have any value. Obviously, an unsymmetrical set whose sum is not zero cannot result from the addition of two sets, each having a sum of zero. Suppose, however, that we add to the positive-sequence and negativesequence sets a set of equal vectors, Van, VbO, VeO (shown in Fig. Id), known as a zero-sequence set. The resulting set of vectors (shown in
Fig. Ie) is:
v, = v: + Val = VaO + Val + Va2
[Sa]
+ V bl + Vb2
[8bl [Sc]
+ Vb' =
Vb =
VbO
v, =
VeO + V e'
V bO
= V eO + Vel
+ Ve2
SYMMETRICAL COMPONENTS
195
These vectors are unsymmetrical, and their sum is not zero. Can any unsymmetrical set of three vectors be resolved into three symmetrical sets-a positive-sequence set, a negative-sequence set, and a zerosequence set? It can if eqs. 8 can be solved for the symmetrical components in terms of the given vectors Va, Vb, Ve • In eqs. 8 there appear to be nine unknown components, but actually only three of them are independent because the three vectors of each set are determined by anyone of the three, say by the phase a vector, and by the symmetry of the set. [9a] Vbl = Va1!240° = a2Va l
= Val/120 = aVal Vb2 = Va2/120° = aVa2
[9b]
Ve2 = Va2/240° = a2Va 2
[9d]
VbO = VeO= VaO
[ge]
0
Vel
where
[ge]
a = /1200 = -0.500 + jO.866
[10]
a 2 = /240 0 = -0.500 - jO.866
[11]
a3
[12]
and hence also
=1 a4 = a
[13]
Substituting eqs. 9 into eqs. 8, one obtains
v. = VaO+ Val + Va2 Vb = VaO + a2Va1 + aVa2 v, = VaO+ aVal + a a2 2V
[14a] [14b] [14c]
Subscript a may now be dropped, giving
Va = Vo + Vt
[1Sa] + V2 [I5b] v, = v, + a t + aV2 [15e] v, = Vo + aVI + a2V2 It is desired to solve eqs. 15 for Vo, v; and V2 in terms of Va, Vb, and 2V
Ve• The simplest method of solution follows. To obtain Vo, add the three eqs. 15 together, obtaining
Va + Vb + Vc = 3Vo + (1 + a
+ a2 ) (Vl + V2 )
SOLUTION OF FAULTED NETWORKS
196
Note that
1+a
+a = 0 2
[16]
Hence
[17] To obtain VI, multiply the three eqs. 15 by 1, 8, and a2, respectively, obtaining = Vo VI V2
v;
aVb = 2
a Ye =
+ + aVo + VI + a Y a 2y o + VI + aV2 2
2
Then add, obtaining
Va
+ aVb + a2y c =
(1
+ a + a2 ) (y o + V2 ) + 3V1
whence
VI = i(Va
+ aVb + a2Ve )
[18]
To obtain V2 , multiply the three eqs. 15 by 1, a2, and a, respectively, add, and divide by 3, obtaining
V2
= i(Va + a2Vb + aYe)
The symmetrical components of any three given vectors Va, Vb, therefore Vo = i(Va + Vb + Ve )
[19]
v, are, [20a]
VI = i(Va
+ aVb + a Ve )
[20b]
V2 = l(Ya
+ a Vb + aYe)
[20c]
2
2
Any three given vectors can be resolved into symmetrical components of positive, negative, and zero phase sequence by using eqs. 20. The operations indicated by eqs. 20 may be performed either graphically or arithmetically. If the given vectors comprise a symmetrical set of a particular phase sequence, the other two components will vanish. If the sum of the given vectors is zero, they have no zero-sequence component. The vectors may represent any alternating quantities associated with a three-phase circuit or machine, such as line-to-neutral, line-to-ground, or line-to-line voltage; current; or flux. In this chapter we are concerned with voltage and current vectors. The actual or phase voltages are given in terms of their symmetrical components by eqs. 15, and the symmetrical components are given in terms of the actual voltages by eqs. 20. The corresponding equations for current are: [21a]
SYMMETRICAL COMPONENTS
197
Ib = 10
[21b]
I, =
+ a211 + aI2 10 + all + a 212
[21c]
10 = J(la+lb+le}
[22a]
11 =
12 =
i n, + alb + a2I c ) l(la + a2Ib + ale)
[22b] [22c]
An alternative form of eqs. 15 and 20, obtained by using the rectangular form of operators a and a 2 (see eqs. 10 and 11), is as follows:
Va =
v; + (VI + V2 )
[23a]
Vb = Vo - 0.500(V1
+ V2 )
Ve = Vo - O.500(V 1
+ V2 ) + jO.866 (VI
-
jO.866(V I
V2 )
[23b]
- V2 )
[23c]
-
+ (Vb + Vc ) ] i[Va - O.500(V + Vc ) + jO.866 (Vb irva - 0.500(Vb + Vc ) - jO.866(Vb -
Vo = l[V a VI =
V2 =
b
[24a]
Ve ) ]
[24b]
Vc)]
[24cJ
This form is convenient if it is desired to compute with complex numbers all in rectangular form. Needless to say, similar equations hold for currents. EXAMPLE
1
Calculate the symmetrical components of the following unbalanced lineto-neutral voltages: Va = 100/90° volts
Vb
= 116LQ volts
Vc
=
71 /225° volts
Solution 1. The symmetrical components are given by eqs, 20.
v, = 100 /90° v, =
11612
+ jl00 116 + jO 0
Vc = 71/225° = -50 - j50 3Vo = Va + Vb + Vc = 66 + j50 =,83/37° Vo =
22
+
j17
= 28/37°
(Ans.)
SOLUTION OF FAULTED NETWORKS
198
s, =
100/90° =
0
+ j100
aVb = 116/120° = -58 + j100
a'lYe = 71/105° = -18 + j68 3VI
= VI + aVb + a'rYe = -76 + j268 = 279/106° VI = -25+ j89 Va = 100/90° =
0
=
93/106°
(Ans.)
+ jl00
a"Vb = 116/240° = -58 - jl00 aVe = 71/345° = +68 -
j18
+ aVe = +10 -
j18
3V2 = Vo + a"Vb
V2 = + 3 -
= 21/299°
j6 =
7/299°
(Ans.)
The symmetrical components are, therefore: Vo ==
22 + j17 = 28/37° volts
VI = -25 + j89 = 93/106° volts V2 = +3 -
j6
=
7/299° volts
Check 1. From the symmetrical components which have just been found, the phase voltages will be calculated by eqs. 15 as a check.
Vo = 28/37° =
a"VI = 93/346° = aV2 = 7/59° = Vb = Vo + a?:VI aV2 =
+
Vo = 28/37° = aVI = 93/2260
22
+ j17
90 - j23
4 + j6 116 jO (Check)
+.
22 + j17
= -65 -
j67
aZV2 = 7/179° = - 7 + jO Ve = Vo + aVl a?:V2 = -50 - j50 (Check)
+
SYMMETRICAL COMPONENTS
Solution 2. Equations 24 will be used.
Vb= 116+
Vc Vb
O.500(Vb
= -50 -
+ Ve =
+V
c)
=
Va = 3Vo =
Va - O.500(Vb
+V
e)
j50
66 -
j50
33 -
j25
0+ jl00 66 + j50
= -33
Vb - V e =
+ j125
166+ j50
+ j43 = -43 + j143
O.866(Vb - Ve) = jO.866(Vb - Vc )
jO
143
3V1 = -76 + j268 3V2 =
10 -
jlS
The symmetrical components are
Vo =
22 + j17 volts
VI = -25 + j89 volts
V2 =
3-
j6 volts
agreeing with Solution 1. Check 2. Equations 23 will be used.
VI = -25+ j89 V2 =
+V O.500(VI + V VI
2
2)
= -11
Va =
+V
2)
j6
= -22+ j83
Vo =
Vo - O.500(VI
3-
=
+
j42
22+ j17 0
+ jl00
33 -
(CMc1c)
j25
Vi - V2 = -28 + j95
O.866(V1 - V2 ) = -24
+
j82
= -82 Vb = 115 -
j24 jl
(Should be 116 - jO)
Vc = -49 -
j49
(Should be -50 - j50)
jO.866(V1 - V2)
199
200,
SOLUTION OF FAULTED NETWORKS
Sequence impedances. The most important property of symmetrical components of current and voltage is that, in a balanced threephase circuit or part of it, the three sequences are independent. A balanced three-phase circuit is one having similar connections in all three phases, equal self-impedances in all three phases, and equal mutual impedances between each pair of phases. The statement that the three sequences are independent means that currents of each phase sequence will produce voltage drops of the same phase sequence only. For example, if positive-sequence currents flow through a balanced circuit, the resulting voltage drops are exclusively of positive sequence, and their zero-sequence and negativesequence components are zero. The independence of the sequences in any FIG. 2. A balanced three- particular form of balanced circuit may be phase circuit element with shown algebraically. This will be done for self- and mutual impedances. the circuit shown in Fig. 2, consisting of equal series self-impedances Zs in the three phases and equal mutual impedances Zm between each pair of phases. The relation between phase currents la, I b, I, and phase voltage drops Va, Vb, Ve is given by the following equations:
z.r, + Zmlb + Zmle Vb = Zmla + Zslb + ZmIe Ve = Zmla + ZmIb + z.r,
·Va =
[25a] [25b] [25c]
It is now necessary to find relations between the symmetrical components of voltage and current similar in form to eqs. 25. To obtain these relations, substitute eqs. 25 into eqs. 20 and then use eqs. 22.
v; = lcva + v, + Ve ) == i CIa + r, + Ie) CZs + 2Zm ) = CZs
+ 2Zm)Io
VI = leVa + aVb + a2Ve ) = Zs!Cla + alb + a2l e) + Zm!CIb'+ I, 2 = ZsI1 + zmiC -Ia - alb - a Ie)
= (Z8 - Zm)I1 Similarly',
+ ala + ale + a21a + a2Ib)
SEQUENCE IMPEDANCES
201
These equations may be written as Vo = Zolo
[26a]
z.r,
[26b]
V2 = Z2I2
[26c]
Zo = Z, + 2Zm
[27a]
VI =
where
Zl == Z2
= Z, -
Zm
[27b]
Unlike eqs. 25, in which the value of each phase voltage depends upon all three phase currents, in eqs. 26 the voltage of each sequence depends only upon the current 'of the same sequence. The ratio of sequence voltage to sequence current may be called a sequence impedance; thus Zo is the zero-sequence impedances Zl, the posiiioe-sequence impedance; and Z2, the negative-sequence impedance. The positive- and negativesequence impedances of this particular circuit are equal; indeed, for any static three-phase circuit they are equal. For rotating machinery, however, the positive-sequence impedance usually differs from the negative-sequence impedance because of the mutual reactances between rotor and stator windings. The impedances of synchronous machines are considered further in Chapters XII and XIV, Vol. III. Equations 27 show that, if Z8 and Zm are of the same sign (or, more precisely, if their vectors lie approximately in the same direction, as they do for a transmission line or cable, where each of the three phases is considered to consist of a wire with ground return), then the zerosequence impedance is higher than the positive- or negative-sequence impedance. If, on the other hand, Zm is of opposite sign from Zs, as it is for the stator windings of a three-phase machine, then the zerosequence impedance is lower than the positive- or negative-sequence impedance. Let us show that Zm of the machine is a negative reactance. It is a familiar fact that, if two coils are placed initially with their axes parallel and if one coil is then turned with respect to the other, their mutual inductance diminishes, reaching zero at 90°; and, upon further turning of the coil, the mutual inductance becomes negative and is still so at 120°, which is the angular separation of the coils of a three-phase machine. The sequence impedances may be calculated from the phase selfand mutual impedances, as illustrated by eqs. 27 for a particular case; or they may be measured. For example, the positive-sequence impedance of an induction motor running at any given speed may be determined by impressing known positive-sequence voltages on its
202
SOLUTION OF FAULTED NETWORKS
terminals, measuring the resulting positive-sequence currents, and taking the ratio of voltage to current. The negative-sequence impedance may be measured in similar fashion by impressing negativesequence voltage on the terminals and driving the rotor at the same speed as before. For measuring the zerosequence impedance a single-phase voltage may be impressed between the neutral point and the three line terminals tied together. Kirchhoff's laws. In the preceding section it was shown that the symmetrical components of current and voltage obey Ohm's law, V = ZI. It will now be shown that they obey Kirchhoff's laws as well. FIG. 3. A three-phase Kirchhoff's law of currents states that the junction. sum of all the currents flowing into a junction (node) is zero. At a three-phase junction, as illustrated in Fig. 3, this law applies separately to each phase, thus:
+ i: + la" = 0
[28a]
+ Ib' + I b" = 0 Ie + Ie' + Ie" = 0
[28b]
la Ib
[28c]
If we add the three equations together and divide by 3, we obtain
!(la + I b + Ie) + !(Ia'
+ I b' + Ie') + i(Ia" + Ib" + Ie") = 0
or
10
+ 10' + 10"
=
0
[29]
Also, by multiplying eqs. 28a, 28b, and 28c by 1, a, and a 2, respectively, adding the resulting equations together, and dividing by 3, we obtain 11
+ 11' + II" = 0
[30]
By a similar procedure, with a and a2 interchanged, we get [31] Equations 29, 30, and 31 show that the sequence currents (symmetrical components of current) obey Kirchhoff's current law just as the phase currents do. Kirchhoff's voltage law states that the sum of all the voltage drops (or rises) around a closed path (loop or mesh) equals zero. In a three-
THE SEQUENCE NETWORKS
phase circuit the law holds for each phase. Fig. 4 we have
Va + Va' + Va"
203
Thus for the circuit of
=0
[32a)
+ Vb' + Vb" = 0 v, + V + v: == 0
[32b]
Vb
e'
FIG.
[32c]
4. A three-phase mesh.
These equations are of the same form as eqs. 28; therefore equations can be derived from them similar to eqs. 29, 30, and 31. They are
v; + Vo' + Vo" VI + VI' V2
+ V2 '
= 0
[33a]
+ VI" = 0 + V2" = 0
[33b) [33c]
Thus the sequence voltages obey Kirchhoff's voltage law. The sequence networks. It has been shown that in a balanced three-phase network the symmetrical components of voltage and current obey Ohm's law and Kirchhoff's laws, the same laws followed by the actual voltages and currents. There is thus good justification for the concept of sequence networks. The actual three-phase network, in which the actual voltages and currents exist, is replaced for purposes of analysis by three fictitious single-phase networks, in which the symmetrical components or sequence voltages and currents exist. They are the positive-sequence network, in which the positive-sequence voltage and current exist; the negative-sequence network, in which the negative-sequence voltage and current exist; and the zero-sequence network, in which the zero-sequence voltage and current exist. The
204
SOLUTION OF FAULTED NETWORKS
positive-sequence network is identical to the single-phase impedance diagram considered in Chapter III. The negative-sequence network is very much like the positive-sequence network but differs from it in the following respects: (1) ordinarily there are no negative-sequence generated electromotive forces; (2) the negative-sequence impedance of rotating machinery is different from the positive-sequence impedance; and (3) the phase displacement of transformer banks for negative sequence is of opposite sign to that for positive sequence. The zero-sequence network differs greatly from the other two in that (1) the impedance of transmission lines is higher than for positive sequence; and (2) the equivalent circuits of transformers are different. Consequently, the zero-sequence network is given further consideration later in this chapter. In so far as the three-phase network is balanced, the three sequence networks are independent; that is, they are not connected or coupled to one another. But wherever the three-phase network is unbalanced, there is a connection or coupling between t\VO or three of the sequence networks at the corresponding points of unbalance. The nature of the connection or coupling between the sequence networks will be determined in the next section for the types of unbalances most important in stability studies, that is, for short circuits. Consider the component parts of a three-phase power network to see whether they are balanced. Rotating three-phase machines, transformer banks consisting of three identical units, cables, and transposed overhead lines are balanced. Untransposed overhead lines are only slightly unbalanced. Loads, although often unbalanced when considered in small portions, are approximately balanced when the whole load of a substation is considered. Therefore, all the major parts of a three-phase power network are either exactly or very nearly balanced during normal conditions. During fault conditions the same statement holds for all parts of the network except for the fault itself if the fault is of some type other than three-phase. Therefore, the sequence networks representing the power system are entirely independent of one another under normal conditions and are connected only at the point of fault under unbalanced fault conditions. The usual three-phase power system has no generated electromotive forces except symmetrical positive-sequence ones. Therefore the sequence networks representing such a system have no generated electromotive forces except those in the positive-sequence network. The negative-sequence and zero-sequence networks, having no electromotive forces within themselves and not being connected to the positive...sequence network under balanced conditions, are "dead." There-
REPRESENTATION OF SHORT CIRCUITS
205
fore only the positive-sequence network need be considered under such conditions. Under unbalanced fault conditions, however, the negativesequence network and, for some types of faults, also the zero-sequence network are energized by their connection to the positive-sequence network at the point of fault. The nature of these connections will no,v be considered. Representation of short circuits by connections between the sequence networks. Short circuits on a three-phase system may be classified into the following types: one-line-to-ground, line-to-line, One- Llne-to- Ground
;~
Phase c
Phaseb
g,,:,
Phases band c
Line-to . Line Phases c and a
!=c :=E
Phases a and h
=t=
Iwo- Line-to· Ground Phases band c
Phases c and a
Phases a and b
:
~
;=:F =F =F g~
Three· Phase Involving ground Not involving ground
.~~
:=E
g-:r
FIG.
5. Types of short circuit on a three-phase network.
two-line-to-ground, and three-phase. Three-phase short circuits may be further classified into those involving ground and those not involving ground. The other types may be further classified by the phases involved. The different types of short circuit are shown diagrammatically in Fig. 5. The procedure for finding the connections of the sequence networks corresponding to a fault on a three-phase network consists of four steps: 1. Draw a circuit diagram of the fault on the three-phase network, labelling the pertinent phase voltages and currents by suitable letter symbols and by arrows denoting their positive directions. 2. Write three equations expressing relations between the phase
206
SOLUTION OF FAULTED NETWORKS
currents and voltages or particular values thereof imposed by the fault condition. 3. Convert these equations to three corresponding equations in sequence currents and voltages. This conversion is accomplished by using eqs. 15, 20, 21, and 22. 4. Find connections of the sequence networks satisfying the equations obtained in the last step. The procedure just outlined will now be carried out for each type of short circuit, commencing with the line-to-ground short circuit on phase a. Line-to-ground short circuits on phase b or phase c may be a c 1 I ., Va treated by renaming the phases so b .... Three· ., b~ phase as always to call the faulted phase, Vb Ie . . . C~ network " a. The relations of the symphase tVe g~ metrical components come out most simply when the fault is taken on (a) phase a, which was adopted as the reference phase. (That is, the components of the other phase vectors were expressed in terms of the Zero·sequence network components of the phase-a vector.) For the same reason line-to-line and two-line-to-ground faults will be taken on phases band c, thus Posmvesequence making the faults symmetrical with network respect to phase a. The three-phase network, whatever its form, may be symbolized Negativeby a box having four terminals at sequence the point of fault, namely, phases a, network b, and c and the ground, as shown (b) in Fig. 6a. Different types of short FIG. 6. (a) Schematic representacircuit may be applied to the nettion of a three-phase network with work by making the appropriate terminals at the point of fault. (b) connections between two, three, or Schematic representation of the corfour of these terminals. Each of responding sequence networks. the three sequence networks may be represented similarly by a box having two terminals at the point of fault (line terminal and neutral terminal) as shown in Fig. 6b. The connections between these terminals, corresponding to the connections between terminals of the three-phase network, will be derived. Let Va, Vb, and Vc denote the line-to-ground voltages at the point of 4 ....
REPRESENTATION OF SHORT CIRCUITS
207
fault, and let la, Ib, and I, denote the phase currents flowing into the fault, as marked on Fig. 6a. Let Vo, Vt, V2 be the symmetrical components of Va' Vb, Vc; and let 10 , It, 12 be the symmetrical components of la, Ib' r, Line-to-ground short circuit on phase a. (See Fig. 7a.) The line-toground voltage of the grounded phase is zero, and the currents flowing into the fault from the two unfaulted phases are zero. Hence the equations for phase voltages and currents are: [34]
[35] [36] From eqs. 34 and 15a
v, + VI + V2
= 0
[37]
= i1a
[38]
From eqs. 35, 36, and 22
10
= II = 12
Hence the relations between symmetrical components of fault voltage and current, corresponding to eqs. 34, 35, and 36 for actual fault voltages and currents, are
Vo + VI + V2 = 0 10
[39a]
= II = 12
[39b]
A series connection of the three sequence networks (Fig. 7b) satisfies these equations. Line-to-line short circuit on phases band c. (See Fig. 7c.) The equations for phase voltages and currents are
Vb
=
Va
[40]
I, = 0
[41]
Ib + t, = 0
[42]
Substituting eqs. 15 into eq. 40,
v; + a2V t
+ aV2 = Vo + aVI + a2V2
Subtracting V0 from each side and rearranging, we get
V1 (a2
-
a)
= V2 (a2
-
a)
whence
[43]
..... .,
12
VI
12
t+~
Neg.Jt~ -
POSe
11
o
Hv
FIG.
7.
Vo
Y2
Three· phase involvingground
(j)
_ Neg.
VI
t ~
_ Pas.
t ~
Zero
t Vea V"
t I 1f.1
(i)
g
C
3q, b
a
Connections between the sequence networks corresponding to various types of short circuit on a three-phase network.
(h)
Neg. ~t- V2
-t-
Zero
~
10
Three- phase not involvingground
V2
Pas.
,....
11
'r-Vo
(g)
t Ie ~ Vb Vo Va
(I)
~
t
Vl
t
..... 1 2
7'
.... 11
-t. . r
~o
c
g
Two •line•to- ground
o Neg.
POSe
-
.....
(e)
v, Vb
C::tmJ
g
Zero
3q,
(d)
V2
t
.... 12 '"
" tVI
o Zero
::Mtvo
(c)
-
c g
Line-to -line
o
tvo
.... 11
~
Va
~lar,
3q, 6
(b)
Neg.
POSe
o
Zero
.....
.... 10
b
ao-+--f>b
One-line -to ·-ground
-
o
-
(a)
-
gC -
I: +~
4~t tv:
3q, b
3q, b
a
v,
I CI Ib Ie
~
~
0
:d
J-3
t;rj
Z
tj
t;rj
t3
~
~
51
Z
0
§
m
~
co
REPRESENTATION OF SHORT CIRCUITS
209
Substituting eqs. 41 and 42 into the expression for 10 in eq.22a, we obtain 10 = 0 [44] Substituting eq. 44 into the expressions for Ib and Ie in eqs, 21 and substituting the results into eq. 42, we obtain (a2
+ a) (11 + 1
= 0
2)
or
[45] Hence the relations between symmetrical components of voltage and current for the line-to-line short circuit are: [46a]
VI = V2 11
+ 12 = 10
0
[46b]
=0
[46c]
These relations are realized by a parallel connection oj the positive- and negative-sequence networks, leaving the zero-sequence network opencircuited. (See Fig. 7d.) Two-line.. to-ground short circuit on phases band c. (See Fig. 7e.) The equations for phase voltages and currents are: [47a] [47b] [47c] These are like eqs. 34 to 36 for a one-line-to-ground short circuit, except that voltage and current are interchanged. Therefore the equations in symmetrical components will be like eqs. 39 except for an interchange of voltage with current. They are:
10 + 11
+ 12 = 0
Vo = VI = V2
[48a] [48b]
These equations are realized by a parallel connection oj the three sequence networks (Fig. 7f). Three-phase short circuit not involving ground. (See Fig. 7g.) The equations of phase voltages and currents are:
Va Ia
= v, = v,
+ Ib + I,
=
0
[49] [50]
210
SOLUTION OF FAULTED NETWORKS
Substituting eqs. 49 into eqs. 20, we get
VI = V2
= iVa(l + a + a2 ) = 0
Substituting eq. 50 into eq. 22a for 10 :
10
=0
Hence the equations of symmetrical components for this type of fault are: [5Ia] [51b]
[5Ic] They are satisfied by short-circuiting the positive- and negativesequence networks and leaving the zero-sequence network opencircuited (Fig. 7h). Three-phase short circuit involving ground. (See Fig. 7i.) The equations of phase quantities are:
v, == 0
[52a]
Vb == 0
[52b]
v, =
[52c]
0
The corresponding equations for symmetrical components are: Vo = 0
[53a]
VI = 0
[53b]
V2 == 0
[53c]
The equivalent circuit consists of a short circuit on each of the three sequence networks (Fig. 7;). Ordinarily no distinction need be made between the two types of three-phase short circuit, because the zero-sequence network is "dead" whether or not its terminals are short-circuited. The negativesequence network is also dead during a three-phase short circuit, and the zero-sequence network is dead during a line-to-line short circuit. These dead networks are shown in Fig. 7 for the sake of completeness, but they are disregarded in fault calculations except in the rare instances where (1) the three-phase network contains unsymmetrical generated electromotive forces, or (2) there is a second fault or other unbalance in addition to the one being considered. Solution of faulted three-phase networks by the method of symmetrical components. Any three-phase network which is symmetrical
SOLUTION OF FAULTED NETWORKS
211
throughout except for one unsymmetrical short circuit can be represented by its three sequence networks connected at the point of fault as shown in Fig. 7. When the appropriate connections have been made, the current or voltage anywhere in anyone of the sequence networks is equal to the corresponding component of current or voltage at the corresponding point of the three-phase network. If the values of the phase currents and voltages are desired, they can be computed by combining the sequence currents and voltages in accordance with eqs. 15 and 21 or eqs. 23. In stability studies, however, the symmetrical components usually suffice. This method of solving a faulted three-phase network is well suited to the calculating board. After the three sequence networks have been set up on the board, connections can be made readily for representing any type of fault at any location. The sequence currents and voltages can then be read by plugging in the measuring instruments. If the network is not too complicated, it is feasible to solve it by algebraic reduction of the sequence networks. ExAMPLE 2
A line-to-ground short circuit occurs on bus D of the power system of Fig. 8a. Find the symmetrical components of current and voltage and the phase currents and voltages throughout the network if the internal voltages of generators A and B are equal and in phase, both having the value jl.05 per unit. The sequence reactances in per unit are as marked on the sequence networks of Fig. 8b, and resistances are assumed to be negligible. Generator A is ungrounded; B has a solidly grounded neutral. Solution. The short circuit is assumed to be on phase a. It is represented by connecting the three sequence networks in series as shown by the broken lines in Fig. Sb. Note that zero-sequence current cannot flow in ungrounded generator A; hence the zero-sequence network is open at this point (between Co and 0 0, Fig. 8b). The positive-sequence network is reduced as shown in Fig. 9 to one e.m.f, in series with one reactance. During the process of reduction the identity of the e.m.f.'s of generators A and B is retained until step c, so that the results can be used later inExample 3. Then, as the e.m.f.'s are assumed to be equal in the present example, points Al and BI of Fig. 9c are joined to give Fig. 9d. In Figs. 10 and 11 the negative-sequence and zero-sequence networks, respectively, are reduced each to a single reactance. Then in Fig. 12 the three reduced networks are connected in series, as the original networks were in Fig. 8, to represent the line-to-ground short circuit; and the combined circuit is further reduced to an e.m.f, and a reactance. In Fig. 13 the current through the reactance is found by dividing the e.m.I, by the reactance, and the voltages across the sequence networks are found by multiplying the current by the respective reactances. The terminal voltages and cur-
212
SOLUTION OF FAULTED NETWORKS
(a)
Positive- Sequence Network
01
Negative· Sequence Network
O2 0.10
zero-Sequence Network
00
0-- - - - - - - ---' (b)
8. (a) One-linediagram of a powersystem used to illustrate fault calculations. (b) The sequence networks of the system with values of reactance in per unit
FIG.
marked thereon. The sequence networks are connected by broken lines so as to represent a line-to-ground short circuit at point D. (Example 2.)
SOLUTION OF FAULTED NETWORKS
Al
0.08
0.25
rv
D1
0.03
0.20
_0.13_
213
81
rv (a)
°1
(6)
(c)
(d)
0.293
(e)
0.216
01
01
FIG. 9. Reduction of positive-sequence network (Example 2).
214
SOLUTION OF FAULTED NETWORKS
0.08
D2 0.03
(a)
(6)
D2
D2
0.01 (d)
(c)
(e)
0.09
0.193
0.136
02 FIG.
0.08
02
10. Reduction of negative-sequence network (Example 2).
SOLUTION OF FAULTED NETWORKS
215
Do 0.054
0.14
(a)
0.054 (b)
Do (d)
00
0.066
00 FIG.
0-- - - - - ---..
::::::::J 0.046
0.02
(c)
11. Reduction of zero-sequence network (Example 2).
·105'
~1
D2 0.09 O2 Do0.066 00
r-VYV~--~--~
L
::-=-
..J
(0)
jl.05
ctt::J -+--
FIG.
(b)
12. (a) Connection of the reduced sequence networks to represent a lineto-ground fault. (b) Final reduction. (Example 2.)
..---N LV jO.57
3.62
+il.OS:J
SOLUTION OF FAULTED NETWORKS
216
.....
in
0
.~
~
il
in
sfrv .....
(0)
"ci
°1
3.62
~-
1.54
2.08
.~
(b)
(c)
FIG. 14. Determination of the positive-sequence currents and voltages by reexpansion of the positive-sequence network. See Fig. 9 for values of reactance. (Example 2.)
SOLUTION OF FAULTED NETWORKS
217
rents of the sequence networks thus found are the symmetrical components of the voltages at the point of fault and of the current in the fault. Each sequence network is now re-expanded by going through the steps of the network reduction backwards, finding all currents and voltages at each
~ ~
°2
(b)
I
·
(c)
°2
3.62 jO.03) I
I t jO.29
150
2.12
D2
I ( jO.06
jO.23 t
3.62
~
1.50
3.62
Jo.041 1.50
D2
~
3.62
-
3.62
(a)
D2
2.12
I t
2.12
jO.26 (d)
1.27
D2
2.35
- jO.OJ ~""'-
jO.33
°2
~ /0.26 w)
15. Determination of the negative-sequence currents and voltages by reexpansion of the negative-sequence network. See Fig. 10 for values of reactance. (Example 2.) FIG.
step. This re-expansion is illustrated in Figs. 14, 15, and 16. Then in Fig. 17 the sequence networks are redrawn with the values of the sequence currents and voltages marked on them. The phase currents and voltages are computed from their symmetrical components in Table 1 by use of eqs, 23 and are shown on the circuit diagram of Fig. 18. It should be noted that, since the generator e.m.f.'a were assumed to be imaginary (that is, 90° ahead of the reference phase), and since the impedances of the network were assumed to be purely reactive, all the sequence currents are real, and all the sequence voltages are imaginary. A similar statement holds for phase a currents and voltages, but not for phases band c.
218
SOLUTION OF FAULTED NETWORKS 3.62 3.62 Do 0----<]0-_--<10--.
!
jO.17
jO.24
~
I '0.07
(b)
i
J
°00---------' 3.08 (a)
(e)
0.54
i
Do
jO.08 jO.24
3.08
jO.17 (d)
jO.l6
jO.09
°0
FIG. 16. Determination of the zero-sequence currents and voltages by re-expansion of the zero-sequence network. See Fig. 11 for values of reactance. (Example 2.)
FIG. 17. Sequence currents and voltages in the power system of Fig. 8 (Example 2).
1.50
r,
!(11 + 12) 10 - !(Il + 12) 0.866(11 - 12)
0.04 3.04 0 3.04 1.52 -1.52 0.03
12
11 - 12 11 + 12 10
11
Gen.
A 1.54
r--
B 2.08 2.12 -0.04 4.20 3.62 7.82 2.10 1.52 -0.03
Gen. Line CD 1.29 1.27 0.02 2.56 0.54 3.10 1.28 -0.74 0.02
,.
Line CE 0.25 0.23 0.02 0.48 -0.54 -0.06 0.24 -0.78 0.02
Currents
0.52 -0.01
1.16
Line EDt 1.16 1.17 -0.01 2.33 1.68 4.01
,
(EXAIIPLE
Line EDS 1.16 1.17 -0.01 2.33 1.40 3.73 1.16 0.24 -0.01
CALCULATION OF PHASE CURRENTS AND VOLTAGES
TABLE 1
,
jO.7S
-jO.38
Bus C jO.67 -jO.23 jO.90 jO.44 -jO.16 jO.28 jO.22
2) ,.
Bus D jO.57 -jO.33 jO.90 jO.24 -jO.24 0 jO.12 -jO.36 jO.7S
Voltages
jO.38 -jO.07 jO.31 jO.19 -jO.26 jO.78
jO.90
Bus E jO.M -jO.26
"
co
.....
~
~
0
=a
t-3
Z t.'=j
tj
t.'=j
~
> d
~
~
0
Z
0
t-3 .....
tD
0 t"'4 d
SOLUTION OF FAULTED NETWORKS
220
BusD
BusE
4.01 + jO 0.52 +jO.ot O.52-jO.Ol \0\0
('t)('t)
0 00 .....................
+
I
I
OcoCO ,,-,,-
BusC
3.10+ jO -0.74 - jO.02 - 0.74 + jO.02
00 I
10.86+jO
-1.52 +jO.03
00
~ ~
\0\0
_NN
~~~ ....... I I
a ~~
0"":"":
0"-"-
....... I
+coco I
I
+ coco
00
-O.06+jO - 0.78- j 0.02 - 0.78 + jO.02
dd I
FIG. 18. Phase currents and voltages in the network of Fig. 8 with a line-toground short circuit on phase a of bus D. Values of current in arrow direction are marked on the horizontal lines; values of line-to-ground voltage, on the vertical lines (busses). (Example 2.)
Fault shunts. Since the internal electromotive forces of a threephase synchronous machine are of positive sequence, and since no power results from the combination of positive-sequence voltages with negative-sequence or zero-sequence currents, the generated power of a synchronous machine and the synchronizing power between the various synchronous machines of a power system are positive-sequence power. Therefore the positive-sequence network is of primary interest in a stability study, and the zero- and negative-sequence networks are only of secondary interest. In the positive-sequence network a short circuit can be represented by connecting a shunt impedance Zp at the point of fault. The value of ZF depends upon the type of fault and upon the impedances Zo and Z2 of the negative- and zero-sequence networks as viewed from the point of fault. The formula for the impedance of the fault shunt may be determined in either of two ways. The first way is by inspection of the connections between the sequence networks for representing the various types of short circuits (Fig. 7). The line-to-ground short circuit is represented by connecting in shunt with the positive-sequence network
FAULT SHUNTS
221
at the point of fault the series combination of the zero-sequence and negative-sequence networks; the line-to-line short circuit is represented by thus connecting the negative-sequence network only; and the two-line-to-ground short circuit is represented by the parallel combination of the zero- and negative-sequence networks. Hence the impedances of the fault shunts are as follows: Type of Short Circuit Line-to-ground Line-to-line
Impedance of Fault Shunt, Zl' Zo + Z2 Z2 ZoZ2 Two-line-to-ground z, +Z2 Three-phase o
[541 [55J [56] {57]
The second way of finding the impedance of the fault shunt is to take the two equations Vo = -ZoIo [58]
V2
-Z2I 2
=
[59]
which are independent of the type of fault, together with the three equations giving relations between, or special values of, the sequence currents and voltages at the fault, and to eliminate from the five equations the four quantities Vo, 10 , V2 , 12 , thereby obtaining one equation of the form [601 For example, for a line-to-ground short circuit on phase a, eqs. 39 are used. Substitution of.eqs. 58 and 59 into eq. 39a gives
-ZoIo
+ VI -
Z2I2
=0
VI = ZoIo + Z2I2 By eq. 39b 10 and 12 may be replaced by It, giving
VI = (Zo
+ Z2)II
[61J
By comparison of eqs. 60 and 61 the impedance of the fault shunt is ZF =
z, + Z2
[62]
agreeing with eq. 54. The same result would have been obtained if the short circuit had been taken on phase b or c instead of on a. It may be worth pointing out that, although the connections of Fig. 7 are valid even if the zero- or negative-sequence networks contain generated electromotive forces, the expressions given for the imped-
SOLUTION OF FAULTED NETWORKS
222
ances of fault shunts are restricted to the normal condition, in which such e.m.f.'s are lacking. The positive-sequence network with the fault shunt connected to it may be reduced in the usual way to the simplest form of network connecting the internal voltages of the synchronous machines. For a twomachine system the reduced network is, of course, a tJ. (or 1r). EXAMPLE
3
Find the fault shunt for representing a line-to-ground short circuit on bus D of the power system of Fig. 8, Example 2, and reduce the positive-sequence network (with shunt attached) to an equivalent 1f' between the two generator 0.293
(a)
(b)
(c)
FIG. 19. Reduction of positive-sequence network with attached fault shunt of reactance X, representing a line-to-ground short circuit on the power system of Fig. 8 (Example 3.)
e.m.f.'s and neutral. Repeat for other types of short circuit at the same location. Solution. In Example 2 the zero-sequence and negative-sequence networks were reduced to single impedances as viewed from the point of fault (terminals D and 0). These impedances were Zo = jO.066 and Z2 = jO.09 per unit. For a line-to-ground short circuit the impedance of the fault shunt is Zp = Zo Z2 = jO.066 jO.09 = jO.156 per unit
+
+
The positive-sequence network was reduced as shown in Fig. 9, and in part c of that figure it has been reduced as far as possible while still retaining the
EFFECT OF TYPE OF FAULT ON STABILITY
223
separate identities of generators A and B. In Fig. 19 the fault shunt is attached, and the network is reduced to an equivalent 1r. Since all the branches of this 1r are reactive, only the branch between A and B affects the power-angle equation. ground fault, is
X
= 0.293 AB
The reactance of that branch, for a line-to-
+ 0.216 + 0.293 X 0.216 = 0.010 + '0.156
0.509
+ 0.381
= 0.890 per unit
For any reactance X F of the fault shunt,
= 0.509 +
X AB
0.0633 0.010+ X,
Hence for a line-to-line short circuit X". = X 2 = 0.090
X AB = 0.509 + 0.0633 0.100
= 0.509 + 0.633 =
·
1.14 per unit
For a two-line-to-ground short circuit, X,. =
XoX2 = 0.066 X 0.090 = 0.038
x, + X 2
0.156
+ 1.32 = 1.83 per unit· X AB = 0.509 + 0.0633 0.048 = 0.509 For a three-phase short circuit
XF = 0 · X AB = 0.509 + 0.0633 0.010 = 0.509 + 6.33 := 6.84 per umt
The reactance between generators A and B is lowest for the line-to-ground fault, higher for the line-to-line fault, still higher for the two-line-to-ground fault, and highest of all for the three-phase fault.
Effect of type of fault on stability. The lower the impedance of the fault shunt, the less is the power exchanged between any two synchronous machines for a given angular displacement, and therefore the lower is the stability limit for a given fault duration. In a twomachine system, such as that of Example 3, it is clear that the lower the impedance of the fault shunt (connected in the shunt branch of the T network of Fig. 19a), the higher is the impedance of that branch of the reduced network which joins the two generators (branch AlB!) Fig. 19c). The synchronizing power varies inversely as the impedance of branch AIB I. A similar situation exists on a system of more than two machines.
SOLUTION OF FAULTED NETWORKS
224
Comparison of the expressions for the impedances of the fault shunts for the several types of short circuit (eqs. 54 to 57) shows that this impedance is lowest-indeed, is zero-for the three-phase short circuit, higher for the two-line-to-ground short circuit, still higher for the lineto-line short circuit, and highest for the line-to-ground short circuit. It follows that the most severe type of fault, as ·regards power-system stability, is the three-phase short circuit, followed in order of decreasing severity by the two-line-to-ground, line-to-line, and one-line-to-ground
~
~ 0.5 J---+---+-++~~~~-+--i---+--f
~
e
0.5
0.8
00
Fault duration (seconds) FIG. 20. Curves of stability limit as a function of fault duration for four types of fault at the sending end of a line of the system of Fig. 15, Chapter V (Example 4).
short circuits. The relative severity of the various faults is shown to advantage by curves of power limit versus fault duration, such as the curves of Fig. 20. The difference in severity of the several types of fault becomes smaller as the fault duration is decreased, but, even with the fastest feasible clearing times, the difference is usually pronounced. For zero clearing time (which is not attainable in practice) the power limit is independent of the type of fault. In making a stability study some judgment is necessary in choosing the type of fault which is assumed to occur. The assumption of threephase faults (the most severe type) gives conservative results and the simplest computation. It is therefore a useful assumption when comparing the effects of different fault locations, different system layouts, bussing arrangements, load conditions, and similar factors.
EFFECT OF TYPE OF FAULT ON STABILITY
225
However, three-phase faults are of infrequent occurrence, especially on high-voltage overhead lines on steel towers, and the power limits determined on this assumption may be unduly pessimistic. For this reason it has been common practice to assume two-line-to-ground faults as the most severe condition likely to be encountered. In some locations, such as on busses of isolated-phase construction, it would be reasonable to assume that only line-to-ground faults could occur. Line-to-ground faults are the type of most frequent occurrence, and three-phase faults, the least frequent. The relative and absolute frequency of occurrence of the several types differs on different power systems, depending largely on the type of line construction. Sporn and Muller 4 give the following figures on the number of faults of different types occurring on a group of transmission lines most of which are operated at 132 kv.: Line-to-ground Two-line-to-ground Three-phase Total
58
8 6
72
In designing a system or a modification of one to improve stability, it appears logical to make estimates both of the value of various degrees of reliability of service and of the cost of achieving them. Then it may be determined whether the expense of any particular improvement of reliability is warranted. If a certain system can be made stable during three-phase faults at reasonable cost, it may be worth doing; but, if the expense is excessive, one may have to be satisfied with a system which is stable during two-line-to-ground faults but not during three-phase faults. In another system for which the requirements of reliability are not so stringent, instability during two-line-to-ground faults may be tolerated provided that the system is stable during line-to-ground faults. If still lower standards of reliability prevail, interruptions of service can be expected from any type of fault. EXAMPLE
4
Determine and plot stability limit in per unit as a function of fault duration for each of the following types of short circuit at the sending end of one line of the two-machine system of Fig. 15, Example 3, Chapter V: (a) one-lineto-ground, (b) line-to-line, (c) two-line-to-ground. Positive-sequence reactances are given in Fig. 15 of Chapter V. Negativeand zero-sequence reactances in per unit are as follows:
Generators: Each transmission line: Receiving-end transformers:
X 2 = 0.24, K« = 0.06 X 2 = 0.40, X o = 0.65 X 2 = X o = 0.10
SOLUTION OF FAULTED NETWORKS
226
The transformers are connected Y-~ with neutral of primary windings solidly grounded. Solution. The negative-sequence and zero-sequence networks are shown in Figs. 21a and 22a, respectively. They are reduced to single reactances 0.24
0.30
(O)~
i°.l
(b)
FIG.
33
(c)
(d)
21. Reduction of the negative-sequence network, fault at sending end of line (Example 4). 0.65
r::-:=::J 0.06
(0)
0.06
0.325
0.425
0.10
~(b) FIG.
22.
(c)
(d)
Reduction of the zero-sequence network, fault at sending end of line (Example 4).
in parts b, c, and d of those figures. The values of the reactances are X 2 = 0.133 and K« = 0.053 per unit. The reactances of the fault shunts (X F ) are:
Line-to-ground: Line-to-line:
Xo
Two-line-to-ground:
X
+ X2 =
0.186 per unit
X 2 = 0.133 per unit
X OX2
•
+ X ~ = 0.038 per unit o
The positive-sequence network is shown in Fig. 16a of Chapter V. With the fault shunt attached to it at the proper point it becomes as shown in Fig. 23a, and, after a Y-d conversion, it is as shown in Fig. 23b. The reactance of the branch connecting the two generator e.m.L's is XAB
= 0.35
+ 0.30 + 0.35;F 0.30 =
0.65
+ 0~~5
EFFECT OF TYPE OF FAULT ON STABILITY
227
Upon substituting the values of X F given above, we get the following values of XAB: Line-to-ground: 1.22 per unit Line-to-line: 1.44 per unit Two-line-to-ground: 3.41 per unit 0.65 + °i~5
~~~~~ (b)
(a) FIG.
23. Reduction of the positive-sequence network with attached fault shunt of reactance XI' (Example 4).
For the pre-fault condition (Chapter V, Fig. 16) X AB was 0.65 per unit, and for the post-fault condition (Chapter V, Fig. 17) it was 0.85 per unit. Henee n and r2 have the following values:
Line-to-ground.·
0.65 5 '1 = -= 0.3 1.22
Line-fo-line:
rl = 0.65 = 0.45
1.44
Two-line-to-ground.· rl = 0.65 = 0.19 3.41
All types ojJault.·
0.65 7 '2=-= 0.6 0.85
Byrd and Pritchard's curves will now be used. From Fig. 40 of Chapter V values of sin 80 are obtained, corresponding to the values of '1 and r2 above. The values of sin 00, multiplied by Pm = EAEB/XAB = 1.00 X 1.00/0.65 = 1.54 per unit, give the power limits for sustained fault and for instantaneous clearing. These values have been entered in Table 2 in the columns headed "sin 00" and Up i" on the lines for te = 00 and to = o. Values of sin 00 which are multiples of 0.05 and which lie between the values already found for te = 00 and for te = 0 are selected. The curves of Figs. 23 to 39 of Chapter V are entered with the proper values of sin 00, rl, and '2, and the values of Teare read from the curves and written in Table 2. Values of Te/t e are calculated from eq. 27 of Chapter V, as follows:
~ == ~1ff1'1P ... == ~1f X to
GH
60 X 1'1 X 1.54 == 1 X 3.00
~
and are written in Table 2. The critical clearing time to is then calculated from T e + Tc/t e•
228
SOLUTION OF FAULTED NETWORKS TABLE 2
CALCULATION OF POWER LIMIT
Pi
AS A FUNCTION OF CLEARING TIME t c
(EXAMPLE
Type of Fault Line-to-ground rl = 0.53 r2 = 0.76
Line-to-line Tl = 0.45 r2 = 0.76
Two-line-to-ground Tl = 0.19 T2 = 0.76
sin ~o 0.456 0.50 0.55 0.60 0.65 0.701
Pi
4)
= 1.54sin ~o 0.70 0.77 0.85 0.92 1.00 1.08
"cltc 7.15
t«
tc (sec.)
" "
3.65 2.45 1.70 1.15
0.51 0.34 0.24 0.16
It
u
00
"
0.378 0.40 0.45 0.50 0.55 0.60 0.65
0.58 0.62 0.69 0.77 0.85 0.92 1.00
6.59
0.147 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65
0.23 0.31 0.38 0.46 0.54 0.62 0.69 0.77 0.85 0.92 1.00
4.28
"
"
" " " It
It
" " " " "
" "u
"
0 00
4.40 2.90 2.20 1.70 1.25 0.85
0.67 0.44 0.33 0.26 0.19 0.13
2.95 2.25 1.75 1.40 1.20 1.00 0.80 0.65 0.5 0.32
0.69 0.53 0.41 0.33 0.28 0.23 0.19 0.15 0.12 0.07
00
The results of the calculations are shown by the curves of Fig. 20. The curve for a three-phase fault has been copied into this figure from Fig. 22 of Chapter V so that the effects of the four types of fault on stability may be compared.
Effect of fault impedance. In the derivation of the connections between the sequence networks for representing short circuits,. it was assumed that the fault itself had no impedance. Actually the fault may have some impedance, principally the resistance of the arc between conductors. On high-voltage circuits this resistance is usually negligible in comparison to the other impedances of the network, and no allowance is made for it in calculation. * In ground faults there may
be additional resistance in the path to ground through the tower foot*For valuesof arc resistance see footnote in sectionon reactance relays in Chapter IX, Vol. II.
EFFECT OF FAULT IMPEDANCE
229
ing or through the object causing the fault. The value of this resistance varies widely, with a median value of around 20 ohms. 5 This resistance also is usually neglected in stability studies, thereby simplifying calculation and giving results for the most severe condition. If, for any reason, it is desirable, fault impedance may readily be taken into account, as shown below. Line-to-ground fault. Let F be the impedance of the fault between conductor a and ground. To facilitate the analysis suppose that at the point of fault there are attached to the three-phase network three equal impedances of value F, as shown in Fig. 24a. This artifice has the advantage of making the three-phase network symmetrical up to point M, where a line-to-ground fault of zero impedance may be applied. The added impedance F appears in each sequence network between the point of fault and point M. Terminals 0 and M of the sequence networks are then interconnected as shown in Fig. 24b to represent a line-to-ground short circuit. Conditions inside the sequence networks are not altered if an impedance 3F is connected in series with the networks as shown in Fig. 24c instead of three separate impedances of F each. Line-to-line fault. Let F be the impedance of the fault between conductors band c. Suppose that the fault condition is represented by connecting an impedance F /2 to each of the three conductors at the point of fault and then applying the line-to-line short circuit at the other end of these impedances (point M, Fig. 24d). Such a connection is equivalent to connecting impedance F/2 to the line terminal of each sequence network and then connecting the positive- and negativesequence networks in parallel as shown in Fig. 24e. Of course, the two impedances F/2 which are no,v in series are equivalent to a single impedance F. Two-line-to-ground fault. Let the impedance between conductors band c be F L , and let the impedance to ground be Fa. For generality Fa might be tapped onto any point of FL, but, as FL is usually much smaller than F G, it will be sufficiently accurate for the present purpose to assume that Fa is connected to the center point of F L , giving F L/2 between the junction and each faulted conductor, as shown in Fig. 24f. The impedance F G in the ground connection can be shown to be equivalent to an impedance 3FG in the zero-sequence network. The connections between the sequence networks are then as shown in Fig. 24g. Three-phase fault. For symmetrical fault impedances as shown in Fig. 24h the connections of the sequence networks are as shown in Fig. 24i.
a
°2
I
Mo
3F
a
POSe
F/2
~'/~
line -to -line fault
(e)
L!::J
0
F/2
Zeroo--f-!V'
-(d)
go
c
3c1> b
Zero
(I)
with fault impedance.
Fa
Three -phase fault
(i)
~ FL/2
I
I~
FL/2
Two-line -to -ground fault
(g)
(h)
;~
a~""-
3cb b
~J 19
F./2
g o.-.l.---J\ AA--I
'i'c~
- ll::J
F
a~~V!
3,1,. b
FIG. 24. Connections between the sequence networks corresponding to various types of fault
(c)
~
One• line •to -ground fault
(b)
(a)
g:r::vv':
3q,b~ c -
~
ee
t=rj
00
~
:ao
t-3
Z
t:1
~
~ ~
Z
o
t--04
~1-3
rn
<:)
UNSYMMETRICAL OPEN CIRCUITS
231
Fault shunts. The impedances of the fault shunts, connected across the positive-sequence network at the point of fault to represent the effect of faults having impedance, may be found from inspection of the connections in Fig. 24; they are as follows: Type of Fault (with Fault Impedance) Line-to-ground Line-to-line
Two-line-to-ground
Impedance of Fault Shunt, Z,.
z, + Z2 + 3F Z2 +F FL
2
+ (Zo +FL/2 + 3Fa ) (Z2 + FL/2) Zo + Z2 + FL + 3Fa FL
Three-phase
[63] [64] [65] [66)
2
Unsymmetrical open circuits and series impedances. Unsymmetrical open circuits may be caused by blown fuses or by single-pole switching. t The connections between the sequence networks for representing one or two open conductors or an impedance in series with one conductor are given in Fig. 25. The derivations of these connections are similar to those given for.ahort circuits earlier in this chapter. Before these connections are made, the line side of each sequence network is opened at the point of fault, and the line on each side of the opening is considered to be a terminal. (These are the two upper terminals shown on each block in Fig. 25.) The neutral terminal, used for representing short circuits, is not used for representing open circuits. The effect of an unsymmetrical open circuit or series impedance on positive-sequence quantities may be represented by connecting an impedance Zp in series with the positive-sequence network at the point of fault. The value of .Zp depends upon the impedances Zo of the zerosequence network and Z2 of the negative-sequence network, each measured or calculated from terminals at the point of fault in series, instead of in shunt, with the respective network. Type of Fault
Series Impedance z" Inserted in PositiveSequence Network ZoZ2
One conductor open
ZO
Two conductors open Three conductors open Impedance Z in series with one conductor
Zo
+ Z2
[67]
+ Z2
[68] [69]
co
(l/zo)
+
1 (l/z2)
+ (3fZ)
fSingle-pole switching is discussed in Chapter XI, Vol. II.
[701
SOLUTION OF FAULTED NETWORKS
232 a
a
-l!. b c g
b
c g
--!!.
.s.g
Zero
o Pos.
g
P~s.
0
o
g
z~o
0
o
Z a ~
-. -.
o
(
a b c
0
Neg.
Neg.
0
0
[5J
c
Z/3
'\Nv C)
()
Zero 0
()
0
Pos. 0
Neg.
o 0 Neg.
0'
0
(a)
(b)
(c)
(d)
(e)
Normal condition
One conduetor open
Two conductors open
Three conduetors open
Impedance Z in one conductor
FIG.
25. Connections between the sequence networks corresponding to various types of open circuit or series impedance.
Simultaneous faults and other double unbalances. Approximately 20% of the faults on double-circuit overhead transmission lines on the same towers involve both circuits. Occasionally two faults occur simultaneously at points which are separated geographically as well as electrically, particularly on systems which are grounded through high impedance. A combination of an unsymmetrical short circuit with an unsymmetrical open circuit may occur, as when the short circuit is partially cleared by the opening of a fuse or single-pole circuit breaker. The conditions described above may be called "double unbalances." In general, a double unbalance cannot be correctly represented by connecting the sequence networks at each of the two points of fault as they would be connected for single unbalances at those points, nor can the effect of a double unbalance on positive-sequence quantities be correctly represented by connecting into the positive-sequence network fault shunts or series impedances which have the same values that they would if the two faults had occurred separately. For methods of solving three-phase networks having double unbalances see Refs. 2, 3, 6, and 7.
LINES IN THE ZERO-SEQUENCE NETWORK
233
The zero-sequence network. The zero-sequence network plays an important role in determining the currents and voltages during ground faults. This network differs from the positive- and negative-sequence networks particularly with respect to the representation of lines and of transformers. Representation of lines in the zero-sequence network. In any three-phase circuit positive- and negative-sequence currents are confined to the line conductors. Zero-sequence currents, however, being in phase in all three line conductors, must find a return path elsewhere, usually either through the earth or through both the earth and overhead ground wires or cable sheaths. The spacing between the outgoing and return paths is therefore greater for zero-sequence than for positive-sequence currents. Accordingly, the zero-sequence series inductive reactance is greater; and the shunt capacitive susceptance is smaller, than the corresponding positive-sequence quantities. The zero-sequence inductive reactance of lines is greater than the positivesequence reactance approximately by the following factors: Single-circuit aerial lines without ground wires or with steel ground 3.5 wires. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . .. .. . Single-circuit aerial lines with copper or aluminum ground wires. . . . . . . 2 Double-circuit aerial lines without ground wires or with steel ground 5.5 wires " . . . . .. . . . . . Double-circuit aerial lines with copper or aluminum ground wires. . . . . . 3 Three-phase cables. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3 to 5 Single-phase cables. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Although these factors are useful for estimating zero-sequence reactance of lines, more accurate values should be calculated if the conductor sizes and spacings and the earth resistivity are known. Methods of calculating zero-sequence resistance, reactance, and capacitance are given in Refs. 1, 2, 3, and 21 of Chapter III. There is a considerable zero-sequence mutual impedance between parallel aerial lines on the same towers or even on the same right-ofway. The methods of calculating mutual impedance are also given in the references. If t\VO lines having self-impedances Za and Zb and mutual impedance Zm are connected together at both ends, they may be replaced by a single impedance, Fig. 26a. If they are connected together at one end only, they can be represented by the equivalent Y circuit of Fig. 26b. If they are not connected at either end, they can be represented on an a-c. board by the equivalent circuit of Fig. 26c, which utilizes an insulating transformer of ratio 1: i.t For ordinary tWo A. Lewis, in discussing Ref. 8, attributes this circuit to C. F. Wagner.
SOLUTION OF FAULTED NETWORKS
234
calculation they can be represented by the mesh circuit of Fig. 26d. 8 If a fault is to be represented on one of a pair of coupled lines, the lines on each side of the fault must be considered a separate section. Thus, ZeZb-Z",,'l
Za+ Z b - 2Z ""
Ca)
o------JVV\r---
c
I~---a
(bl
~---6
, (c)
CJ
----a
6'----6
a' (d)
a'----a 6'----b b' Zb
FIG. ?6. Equivalent circuits for two coupled lines of self-impedances Za and Zb and mutual impedance Zm. Lines connected together (a) at both ends, (b) at one end only, and (c) and (d) at neither end.
I
a
Fault )(
b
t=ml~
m(Zb-Zm)
FIG.
(l-m)(Zb-Zm)
27. Equivalent circuit of two coupled lines, one of which is faulted.
if there is a fault on one of two lines connected' at both ends, and the fault is distant from one end by a fraction m of the length of the lines, the equivalent circuit is as in Fig. 27. Equivalent circuits for three or more coupled lines have been devised" but are seldom needed. Two coupled circuits having appreciable shunt admittance can be represented by the equivalent or nominal circuits given in Ref. 9.
TRANSF9RMERS IN THE SEQUENCE NETWORKS
235
Representation of transformers in the sequence networks. As a rule, every important transformer bank on a three-phase power system consists either of three identical single-phase transformers having each set of windings connected in Y or in d or of a three-phase transformer having its windings connected likewise. Only 'such transformers having not more than three sets of windings are considered here. As was mentioned in Chapter III, the positive-sequence equivalent circuit of a two-circuit transformer bank, with exciting current neglected, consists of a series impedance and an ideal transformer of complex ratio. The angle of the complex ratio expresses the phase shift of positive-sequence voltage from one side of the transformer bank to the other. For D.-d or Y-Y connections the shift is either 0 or 180°. For !:J,.-Y or Y-!:J,. connections the phase shift, with the standard manner of naming the phases, is either +30° or -30°, depending upon the exact connections. By naming the phases in a different manner, the phase shift of +30 0 may become + 1500 or -90°, and the phase shift of -30° may become +90° or -150°. The nomenclature in which the phase shift is ±90° is the most convenient for computation and is recommended for that purpose.2, 10 If per-unit quantities are used, the magnitude of the complex ratio becomes 1, but the angle remains. In setting up the positive-sequence network, the phase shift is commonly neglected. The currents and voltages are first calculated as if there were no phase shift, and later the phase shift is taken into account if it is of any moment in the use of the results. When phase shift is not represented, the positive-sequence equivalent circuit for per-unit quantities consists merely of a series impedance. The negative-sequence equivalent circuit of a transformer bank is like the positive-sequence circuit except that the angle of the complex ratio is of opposite sign. For example, if the phase shift for positivesequence current and voltage is +90°, the phase shift for negativesequence current and voltage is -90°. Here again the phase shift is usually disregarded in setting up the negative-sequence network, but it may be taken into account later if desired. The zero-sequence equivalent circuit differs considerably from the positive- and negative-sequence circuits. For banks of single-phase transformers the value of impedance is the same, but, instead of this impedance being simply in series with the line, each end of the impedance may be either open or grounded, depending upon the connections of the bank. The proper connection may be deduced by considering the paths of zero-sequence currents. Zero-sequencecurrents are equal in the three phases. It is apparent that such currents cannot flow through a set of Y-connected windings if the neutral of the Y is isolated,
SOLUTION OF FAULTED NETWORKS
236
but they can flow if the neutral is grounded. Zero-sequence currents cannot flow from a three-phase line into a set of A-connected windings, but they can flow around the delta. If exciting current is neglected, zero-sequence currents can flow in one set of windings of a transformer bank only if zero-sequence currents can flo,v also in the other set of windings, The foregoing statements may be summarized as follows: Zero-sequence current cannot flow from a line into a transformer bank Three-phase circuit
Zero- sequence circuit (a)
Delta·delta
(b)
Ungrounded Yungrounded Y
(c) Grounded Ygrounded Y
(d)
Grounded Y delta
---VV\;---
_l
(e)
Ungrounded Y- delta
(I)
Ungrounded Y grounded Y FIG.
28. Zero-sequence equivalent circuits of two-circuit transformer banks.
unless the windings connected to the line are in Y with grounded neutral and unless the other windings are either in ~ or in Y with grounded neutral, The zero-sequence equivalent circuits of various transformer connections are shown in Fig. 28. Only connections c, grounded Y-grounded Y, and d, grounded Y-~, present a path for zero-sequence current. The other connections are equivalent to an open circuit in the zero-sequence network, The zero-sequence impedance of the grounded y-~ bank is equal to the short-circuit impedance of one transformer. Figure 29 may help
TRANSFORMERS IN THE SEQUENCE NETWORKS
237
to show why this is so. In part a of the figure the actual connections are shown. It is assumed that equal currents 10 are impressed on the primary windings, causing the secondary current 10 ' to circulate around the delta. Because of the symmetry of the delta, terminals A, B, and C are at the same potential and may be connected to one another, as indicated by broken lines in the figure, without effectively changing the circuit for zero sequence. Connecting terminals A, B, and C is equivalent to short-circuiting each secondary winding separately as in part b of the figure. It is apparent that the circuit of Fig.
(a)
10
#
~
10
~Iri
Vo 10
(b)
Vo 1
~Io'
0
-
tvo
FIG. 29. Diagram showing that the zero-sequence impedance of a transformer bank connected grounded Y-A is equal to the short-circuit impedance of the transformers.
29b presents between each primary line and ground an impedance equal to the short-circuit impedance of one of the transformers. As the line current is 10 and the line-to-ground voltage is Vo, the impedance VolIo is the zero-sequence impedance. It is equal to the positivesequence impedance because the latter is also the short-circuit impedance. Grounding impedance. The neutral of the Y-connected windings may be grounded either solidly or through a neutral impedance Zn. The current through Zn is 310, causing a voltage drop across it of 3ZnIo, which is added to all three line-to-neutral voltages and hence to the zero-sequence component of these voltages. In the zerosequence network the current is 10, and to obtain the correct zero-
238
SOLUTION OF FAULTED NETWORKS
sequence voltage drop the neutral impedance must appear in this network as an impedance 3Zn in series with the transformer impedance. Multicircuit transformers. As mentioned in Chapter III, the positive-sequence equivalent circuit of a three-circuit transformer, with exciting current, phase shift, and ratio neglected, consists of three impedances connected in Y between the terminals of the three circuits.
1
2
3
41£;
FIG.
(a)
30. Zero-sequence equivalent circuits of three-circuit transformer banks.
As long as phase shift is neglected, the negative-sequence circuit is identical to the positive-sequence circuit. The zero-sequence circuit consists of the same set of Y-connected impedances, but the three terminals of the, Y may not all be connected to the corresponding external circuits. Each of the three windings should be considered separately, and the corresponding terminal of the Y should be connected according to the same principles already established for twocircuit transformers, namely: (1) if the winding is ~-connected, the line is left open, and the corresponding impedance is grounded; (2) if
TRANSFORMERS IN THE SEQUENCE NETWORKS
239
the winding is Y-connected with grounded neutral, the line is connected to the impedance; and (3) if the winding is Y-connected with isolated neutral, both the line and the impedance are left open. In Fig. 30 these principles are applied to several different connections of three-circuit transformers. The same principles apply to transformers of four or more circuits. Three-phase transformers. Three-phase shell-type and three-phase five-legged core-type transformers resemble banks of three single-phase transformers in that they have three magnetic circuits which are almost independent of each other. Hence the equivalent circuits of these transformers are the same as those for banks of single-phase units. In three-phase three-legged core-type transformers, however, conditions are different. Application of Kirchhoff's law to the fluxesof three legs shows that there can be no zero-sequence flux in such a transformer except that leaking through air paths from top to bottom of the core. Therefore zero-sequence exciting currents produce no zero-sequence voltage except that induced by this leakage flux, which is small compared with the core flux that would be set up by positive-sequence currents of like magnitude but large compared with the leakage flux between primary and secondary windings. In other words, the zerosequence exciting impedance is much smaller than the positivesequence exciting impedance but is about five times as great as the leakage impedance, or equivalent impedance, between windings. Consequently, even a two-circuit transformer of this type should be represented by a T circuit. The shunt branch, representing the zerosequence exciting admittance, cannot be neglected, as it usually is in the positive-sequence circuit. The effect of the three-legged core is similar to that produced by adding a high-impedance il-connected winding, as is evidenced in the equivalent circuit by the third branch, the terminal of which is grounded. The connections at the other terminals, corresponding to the actual windings, depend upon the connections of those windings, as already discussed. Autotransformers. The commonest type of autotransformer on three-phase transmission systems has a set of Y-connected tapped windings with a grounded neutral and a separate A-connected stabilizing winding (formerly called a tertiary winding). It is, therefore, a three-circuit transformer connected Y- Y- il, and its equivalent circuit is the same as that of a transformer with three separate windings similarly connected, the impedances being given on the circuit basis. 2 The zero-sequence circuit is connected as shown in Fig. 30a without the grounding impedances. If the neutral of the autotransformer is grounded through impedance, this impedance is common to both
240
SOLUTION OF FAULTED NETWORKS
Y-connected circuits and, when expressed in actual ohms, is both the mutual impedance and a component of each self-impedance. When expressed in per unit or in ohms referred to a particular voltage, however, this impedance has three different values, one for the mutual impedance and two for the two self-impedances. Therefore, the single impedance must be represented by a Y circuit, as shown in Fig. 31. If the neutral of an autotransformer is ungrounded, zero-sequence currents can pass from one circuit to the other without transformation, the transformer acting merely as a series impedance. A
B
(a)
(b) FIG.
31. An impedance Z common to two circuits, A and B, must be represented
in the per-unit system by a Y circuit if different base voltages are used for the two circuits. N is the ratio of base voltage A to base voltage B, and Z' is the per-unit value of Z on base voltage A. In the zero-sequence network all impedances are multiplied by 3.
Other symmetrical transformer connections, including regulating transformers. See Ref. 12. Effect of grounding on stability. Methods of grounding a power system modify its zero-sequence impedance. This affects the impedance of the fault shunts for representing ground faults and thereby affects the severity of such faults. Most transmission systems are grounded through transformers, the high-voltage windings of which are connected in Y with grounded neutral. Such transformers are nearly always used at the points of supply and sometimes also at points of load. Occasionally transformers which do not carry load are used solely for the purpose of grounding. On high-voltage systems the usual practice is to ground the transformer neutrals solidly; on medium-voltage systems, to ground them either solidly or through resistors or reactors of low per-unit impedance. Some medium-voltage systems are nominally ungrounded but actually are grounded through the relatively high zero-sequence shunt capacitive reactance of the lines. If one or more transformer neutrals are grounded through reactors, the inductive reactance of which resonates with the capacitive reactance, the zero-sequence impedance may be made much higher than that of a nominally ungrounded system. Resonant grounding reactors, known as Petersen coils or ground-fault
EFFECT OF GROUNDING ON STABILITY
241
neutralizers, are used to a small but increasing extent. Transmission systems to which generators are connected directly, rather than through transformers, are grounded through the generators. The generator neutrals, like transformer neutrals, may be grounded either solidly or through resistors or reactors. The zero-sequence impedance of a transmission system, viewed from the point of fault, depends upon the number of grounding points, upon the distance of the grounding points from the point of fault, upon the zero-sequence impedance of the grounding transformers (or generators), and upon the impedance of the grounding resistor or reactor, if used. An increase in the zero-sequence impedance, viewed from the point of fault, results in a decrease in the severity of a one-line-to-ground or two-line-to-ground fault. The effect of a two-line-to-ground fault approaches that of a line-to-line fault at the same location, and the effect of a one-line-to-ground fault approaches that of no fault. Therefore, in its effect on system stability during ground faults, grounding through impedance is better than solid grounding. Either resistors or reactors may be used to decrease the severity of ground faults, but reactors are cheaper and commoner than resistors. At some locations, however, resistors are more effective than reactors. On a system consisting of several synchronous machines joined by low-resistance lines, the occurrence of a fault decreases' the electrical outputs of the generators and the electrical inputs of the actual or equivalent motors, causing the generators to be accelerated and the motors to be retarded. The closer the fault is to a particular machine, the greater is its effect on the electrical output or input of that machine, and the greater is the tendency for the machine to be accelerated or retarded, as the case may be. Stability is promoted by anything which increases the output of the generators during the fault, thereby lessening their acceleration, provided that the retardation of the motors is not increased correspondingly. A grounding resistor, by consuming power during a ground fault, exerts on a synchronous machine a braking effect which is greater the closer the fault is to the resistor and the closer the machine is to the fault. A grounding resistor located near a generator is therefore beneficial, as it exerts the greatest braking effect on the generator for faults close by, which would otherwise result in the greatest decrease of load and, consequently, in the greatest acceleration. Grounding resistors should not be used, however, near actual or equivalent synchronous motors or near synchronous condensers, for such machines already are retarded by faults. In a two-machine system resistance grounding may be advisable at the sending end and reactance grounding at the receiving end.
242
SOLUTION OF FAULTED NETWORKS
Neither grounding resistors nor grounding reactors have any effect during three-phase faults. The choice of a grounding system is influenced not only by consideration of stability but also by other factors, among which are: (a) limitation of current to prevent damage to generator windings from excessive mechanical forces; (b) limitation of neutral-to-ground and line-to-ground voltages of transformers to permit the use of graded insulation and lightning arresters of lower voltage rating, which give better protection; (c) limitation of inductive influenceon communication circuits during ground faults on the power system; and (d) obtaining suitable conditions for rapid and selective relay operation to clear ground faults. Since the conditions for limiting current conflict with those for limiting voltage, the choice of grounding method rests on compromise. The importance of the effect on stability of the method of grounding has been decreased in recent years by the trend toward higher speeds of fault clearing. EXAMPLE
5
Determine the effectiveness of grounding reactors or resistors in increasing the stability limit of the two-machine system of Fig. 32 during two-line-toground faults on the transmission lines. Assume the grounding reactance or resistance to be 10% on the basis of the kilovolt-ampere rating of. generating station A. Solution. A fault at either end of one of the transmission lines is more severe than a fault near the middle of a line; therefore, two-line-to-ground faults will be assumed, first at point C, then at point D, Fig. 32. The stability limit depends upon fault duration. For comparing the different methods of grounding it will be simplest to find the stability limit for sustained faults. This limit will be found by first obtaining the power-angle equation and then using the equal-area method, or, when the network is purely reactive, by using Fig. 40 of Chapter V, which was derived by the equal-area method. The grounding impedances, Zs at the sending end and ZR at the receiving end, appear in the zero-sequence network as 3Zs and 3ZB, respectively. The three sequence networks are connected in parallel at points C and 0 (as shown in Fig. 32b) to represent a two-line-to-ground fault at the sending end; they are connected similarly, but at D and 0, to represent a fault at the receiving end. The zero-sequence and negative-sequence networks are each reduced to a single impedance between terminals C and 0 (or D and 0), and then the positive-sequence network with fault shunt attached is reduced to a 7r circuit. Next the power-angle equation is found by the methods of Chapter IV. The details of calculation will not be shown, but the resulting equations are given in Table 3. The corresponding stability limits, found by the equal-area method, are presented in Table 4.
EFFECT OF GROUNDING ON STABILITY
243
30 30
(a)
32
10
15
10
11
1.0
Posltive- sequence network
°1 Zero- sequence network
3Z s
'--
3Z R ~~--
(b)
I I I I
-1
o
32. Two-machine system of Example 5. (a) One-line diagram. (b) Sequence networks interconnected to represent a two-line-to-ground fault at point C. Reactances are given in per cent on the rating of the generating station.
FIG.
TABLE 3 POWER-ANGLE EQUATIONS OF THE Two-MACHINE SYSTEM OF FIG.
VABIOUS
Grounding Impedance (per cent) ~~
Z8
o jl0 o
METHODS OF GROUNDING (EXAMPLE
SLG Fault at Sending End
0
P u = 34 sin 8 P u = 51 sin 8 P u = 36 sin 8 P« = 55 sin 8 P u = 5 60 sin (8 + 7°) P u = 10 + 56 sin (8 + 10°)
jl0 10 10
Any
Any
Any
Any
FOR
Power-Angle Equations (power in per cent of rating of station A)
Zs
0 jtO jtO j10 10
32
5)
+
LL Fault at Sending End Pu, = 77 sin 8 N orrnal Conditions P« = 154 sin 8
SLG Fault at Receiving End Pu, = 40 sin 8 Pu, = 42 sin 8 Pu, = 55 sin 8 Pu, = 59 sin 8 Pu, = 4 + 58 sin (8 + 2°) P u, = -5 + 55 sin (8 + 8°) LL Fault at Receiving End Pu, = 77 sin 8 One Line Disconnected Pu, = 129 sin 8
SOLUTION OF FAULTED NETWORKS
244
With solid grounding at both ends, the stability limit for a sustained two.. line-to-ground fault at the -sending end is 27% of the rated kilovolt-amperes of generating station A; and, for a similar fault at the receiving end, it is 32%, as given in the first line of Table 4. The addition of a 10% grounding reactor at either end increases the stability limit considerably for a fault at that end but only very slightly for a fault at the distant end. The use of reactors at both ends increases the stability limit for any fault location; the increase of stability limit over that obtained with solid grounding at both TABLE 4 STABILITY LIMITS OF THE Two-MACHINE SYSTEM OF FIG.
32
FOR
SUSTAINED FAULTS, AS AFFECTED BY GROUNDING REACTORS AND RESISTORS (EXAMPLE
Grounding Impedance (per cent) ~
Zs
o jto
o
jl0 10 10
Any
5)
Stability Limit for Sustained Fault (in per cent of rating of station A) A
~
2LG Fault at Sending End
2LG Fault at Receiving End
o o
27 42
32 34
jtO jtO jtO
29
45
ZR
10
Any
46 57 59 LL Fault at Sending End 66
48 52 41 LL Fault at Receiving End 66
Stability limit for any fault cleared instantaneously is 123%.
ends is considerable, but the increase over that obtained with a reactor at the end near the fault is slight. The substitution of a 10% resistor for the reactor at the sending end provides an additional increase in the stability limit, although the increase is greater for a fault at the sending end than for a fault at the receiving end. The further substitution of a resistor for the reactor at the receiving end results in a small additional increase in stability limit for a fault at the sending end, but the benefit is more than offset by a considerable decrease in stability limit for a fault at the receiving end. The best grounding scheme oj those investigated here is a resistor at the sending end and a reactor at the receiving end. Even when this scheme is used, the stability limit is lower for a two-line-to-ground fault than for a line-to-line fault. The improvement in stability limit obtained by the use of grounding impedance is smaller for a rapidly cleared fault than for a sustained fault, although the stability limit itself is greater. If grounding resistors are used at one or both ends, the stability limit depends upon the ratio of the inertia constant of the motor to that of the
TWO-PHASE COORDINATES
245
generator. The dependence of stability limit on the inertia constants may be understood by recalling that the equivalent power-angle curve of a twomachine system having resistance is a sinusoid displaced upward (according to eq. 28, Chapter IV) by an amount
Po = M2E12 Y ll cos811 - M1E22 Y 22 cos 8 22 M 1 + M2
where M l and M 2 are the inertia constants of the generator and of the motor, respectively; E 1 and E 2 are the internal voltages of the generator and of the motor, respectively; and Fu cos On and Y 22 cos 8 22 are the terminal selfconductances of the network between the internal voltages. If the two conductances are assumed to be equal, as they would be if the fault were electrically midway between the two internal voltages, and if, furthermore, the two internal voltages are assumed to be equal, Po would be zero if the inertia constants were equal. If, however, M 2 were greater than MI , Pc would be positive, indicating a raising of the power-angle curve by the amount of Po and a resultant raising of the stability limit by about the same amount. If, on the other hand, M 2 were smaller than MI , Pc would be negative, indicating a lowering of the curve and of the stability limit. In the present example it has been assumed that the generator and motor have equal values of stored energy per rated kilovolt-ampere but that the rating of the motor is three times that of the generator, giving M 2 = 3Ml • If the two-machine system consisted of a hydroelectric station supplying power to a metropolitan system of one or more steam stations, M 2 would be considerably larger than MI , both because of the greater capacity of the steam system and because of the greater stored energy per kilovolt-ampere of the steam turbogenerators; increasing M 2/M1 from 3 to 00, however, would increase the stability limit only a small amount. The stability limit is further increased if, as is usually true, the internal voltage of the generator is greater than that of the motor (E I > 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
~b go
[Z]
[]
o % -.
[J
-. o 0 _ _ .11
0Z
-
o
[]
%
I;~p I
(a)
(6)
(c)
(d)
One •line•to- ground
une-to•line
Two . line•to- ground
Three- phase
FIG.
33. Connections between the substitute networks (z, y, z) for representing various types of short circuit on a three-phase network (34)).
(twice, instead of half, the zero-sequence impedances), the transformer can be omitted. If the three-phase network is symmetrical except at the point of fault, and if the negative-sequence impedances are assumed everywhere equal to the positive-sequence impedances, the x and y networks are not coupled to each other and are identical except for a 900 phase difference in generated e.m.f.'s. Therefore it is unnecessary to have both these networks set up simultaneously on a calculating board. The positive-sequence network is used in turn for both the x and y networks. For example, in calculating a line-to-line fault (Fig. 33b), readings are taken in the positive-sequence network, first with nothing [Under the assumptions made in using a d-e. calculating board, all line-toneutral voltages in this network are equal, and all currents in it are zero.
REFERENCES
249
connected to it at the point of fault, and then again with a short circuit there. It is not even necessary to shift the generator voltages by 90° because it is a simple calculation to rotate the second set of readings forward through 90°, that is, to multiply them by j. Readings can be taken for all types of short circuit with only two networks set up on the board. Applications oj two-phase coordinates. In general, only the positivesequence network need be set up for a stability study, faults being represented by fault shunts. Sometimes, however, it is desired to study the probable operation of protective relays, and for this purpose the currents and voltages while the system is faulted must be found. The method of two-phase coordinates furnishes a convenient way of calculating these quantities from readings taken on a calculating board on which only the z network is set up in addition to the positivesequence network. . Two-phase coordinates also find application to the analysis of unbalances which are symmetrical with respect to one phase, for example, single-phase series or shunt impedances, banks of transformers connected in V or T, and untransposed transmission lines.14 ,15,16
REFERENCES 1. C. L. FORTESCUE, "Method of Symmetrical Coordinates Applied to the Solution of Polyphase Networks," A.I.E.E. Trans., vol. 37, pp. 1027-140, 1918. Historical reference. 2. C. F. WAGNER and R. D. Evans, Symmetrical Components as Applied to the Analysis of Unbalanced Electrical Circuits, New York, McGraw-Hill Book Co., 1933. 3. W. V. LYON, Applications of the Method of Symmetrical Components, New York, McGraw-Hill Book ce., 1937. 4. PHILIP SPORN and C. A. MULLER, "Five Years' Experience with UltrahighSpeed Reclosing of High-Voltage Transmission Lines," A.I.E.E. Troms., vol. 60, pp. 241-6, May, 1941. 5. C. L. GILKESON, P. A. JEANNE, and E. F. VAAGE, "Power System Faults to Ground: Part II: Fault Resistance," Elec. Eng., vol. 56, pp. 428-33, 474, April, 1937. 6. EDITH CLARKE, "Simultaneous Faults on Three-Phase Systems," A.I.E.E. Trans., vol. 50, pp. 919-41, September, 1931. 7. E. W. KIMBARK, "Experimental Analysis of Double Unbalances," Elec. Eng., vol. 54, pp. 159-65, February, 1935. 8. FRANK M. STARR, "Equivalent Circuits-I," A.I.E.E. Trans., vol. 51, PP. 287-98, June, 1932; disc., pp. 321-6. 9. J. C. BALSBAUGH, R. B. Gow, W. P. DOUGLASS, and A. H. LEAL, "Equivalent Circuits-2 Coupled Circuits," Elee. Eng., vol. 55, pp. 366-71, April, 1936; disc., pp, 1037-9, September, 1936. This paper gives an exactly equivalent five-terminal mesh circuit for two coupled lines not connected to each other at either end. The
SOLUTION OF FAULTED NETWORKS
250
discussion gives (1) an alternative circuit using a transformer, and (2) nominal or approximate values of the impedances of the equivalent circuits with numerical comparisons of the nominal values with the exact values. 10. BRYCE BRADY, "Symmetrical Notation of Three-Phase Circuits," A.I.E.E. Trans., vol. 60, pp. 955-7, November, 1941. 11. A. N. GARIN, "Zero-Phase-Sequence Characteristics of Transformers," Gen. Elec. Reo., vol. 43, pp. 131-6, 174-9, March and April, 1940. 12. J. E. HOBSON and W. A. LEWIS, "Equivalent Circuits for Power and Regulating Transformers," Elec. Jour. preprint, January, 1939; also published as Appendix, Table 7, Electrical Transmission and DistributionReference Book by Central Station Engineers of the Westinghouse Electric and Manufacturing Company, 1st edition, 1942. 13. EDITH L. CLARKE, "Determination of Voltages and Currents during Unbalanced Faults: Use of the Sum and Difference of Positive and Negative Sequence Symmetrical Components," Gen. Elec. Reu., vol. 40, pp. 511-3, November, 1937. 14. EDITH CLARKE, "Problems Solved by Modified Symmetrical Components," Gen. Elec. Reo., vol. 41, pp, 488-94, 545-9, November and December, 1938. 15. EDWARD W. KIMBARK, "Two-Phase Coordinates of a Three-Phase Circuit," A.I.E.E. Trans., vol. 58, pp. 894-904, 1939; disc., PP. 904-10. 16. EDITH CLARKE, Circuit Analysis of A-C Power Systems, vol. I, Symmetrical and RelatedComponents, New York, John Wiley & Sons, 1943. Chapter V, "Two Component Networks for Three-Phase Systems"; Chapter X, "Alpha, Beta, and Zero Components of Three-Phase Systems." PROBLEMS ON CHAPTER VI 1 ~ Graphically and algebraically find the symmetrical components of the following sets of phase voltages:
Va = 100/0
v, =
b 100LQ
v, = 100LQ
v,
v, =
100 /120°
v, =
a
= 100 /240°
v, =
100 /120°
v,
= 100 /240°
C
100LQ
Vc=100LQ
2. Find the symmetrical components of the following sets of phase currents: b c a Ia = I Ia = I I" = 0
Ib
=I
Ib
=
-!I
Ie = -!I
I, = -I
Ib I,
=0 =0
3. Find the symmetrical components of the following sets of voltages, and check the results by computing the given voltages from them: = 100/180°, = 100/90°, v, = 100/0. 4. Find the phase currents corresponding to the following symmetrical components: 10 = 0.15/31°, 11 = 0.97 /18~, 12 = 0.45/303°.
v,
v.
PROBLEMS
251
5. Work Example 2 with a two-line-to-ground short circuit on phases band c of bus D instead of a one-line-to-ground short circuit. 6. Work Example 2 with a line-to-line short circuit on phases band c of bus D instead of a one-line-to-ground short circuit. 7. Work Example 2 with the neutrals of both generators A and B, instead of generator B only, solidly grounded. 8. Compute and diagram the flow of reactive power of each sequence in the faulted three-phase network of Example 2. 9. Find the equations similar to eqs. 39 for a line-to-ground short circuit on phase b instead of on phase a. Show that the impedance of the fault shunt is the same for a fault on either phase. 10. Find the expression for the impedance of the fault shunt representing a two-line-to-ground short circuit, using the second method described in the text (p. 221). 11. Find an expression for the impedance of a shunt representing a fault consisting of a line-to-ground short circuit on phase a and a line-to-line short circuit on phases band c at the same point of a three-phase circuit. 12. Find the sequence currents in the system of Fig. 8 when the internal voltages of generators A and B differ in phase by 60°. The magnitude of each internal voltage is 1.05 per unit. Use the fault current as reference phase. 13. Find the reactances of shunts for representing each of the four types of short circuit at the middle of one of the lines of the system of Fig. 15, Chap.. ter V. (The negative- and zero-sequence reactances are given in Figs. 21 and 22 of Chapter VI.) 14. Calculate and plot stability limit as a function of fault duration for one-line-to-ground and two-line-to-ground short circuits at the middle of one of the lines of the two-machine system of Fig. 15, Chapter V. Also plot for comparison the stability limit for a three-phase short circuit at the same point. (See Fig. 22, Chapter V.) 15. Work Example 4 of Chapter V with a two-line-to-ground fault, instead of a three-phase fault, at the center of one line. Assume that the negativesequence reactance of the system is equal to the positive-sequence reactance when viewed from the point of fault, and that the zero-sequence reactance of the transmission lines is equal to 3.5 times their positive-sequence reactance, mutual reactance between the two transmission circuits being negligible, 16. Which fault on the system of Example 4, Chapter V, is more severe in its effect on stability, a three-phase short circuit at the middle of one line or a two-line-to-ground short circuit at the sending end of one line? 17. Which fault on the system of Fig. 8 is more severe in its effect on stability, a one-line-to-ground short circuit on line CD near bus D or a threephase short circuit on one of the parallel lines DE' near bus D? Does the answer depend upon the clearing time? 18. Prove that the connections of Fig. 24g are correct by writing equations for the relations between phase currents and voltages at the fault and then
SOLUTION OF FAULTED NETWORKS
252
deriving the corresponding equations for the relations between symmetrical components. 19. If, in Example 4, the line-to-ground fault had a 20-ohm resistance, by what percentage would the maximum synchronizing power be increased over that for a solid fault? System base is 50 Mva., 33 kv. 20. Prove that there is no net power in a three-phase circuit attributable to a voltage of one sequence in conjunction with a current of a different sequence. 21. Find the stability limit of the two-machine system of Example 5 for sustained one-line-to-ground faults. Consider the following methods of grounding: (a) Zs = ZR = 0; (b) Zs = jl0%, ZR = 0; (c) ZB = 0, . ZR = jl0%; (d) Zs = ZR = j10%. 22. In Probe 21 consider an additional grounding method, as follows: (e) Zs = 10%, ZR = jl0%. 23. In Example 5 find the line-to-ground voltages of the sending-end transformers during a two-line-to-ground fault at the sending end. Consider each of the following grounding methods: (a) Zs = ZR = j10%; (b) Zs = 10%,ZR = j10%. 24. In Example 5 find the power limits for two-line-to-ground faults cleared in 0.25 sec. with the methods of grounding listed in Probe 21. The frequency is 60 c.p.s. 25. In Example 5 find the power dissipated in the 10% grounding resistor at the sending end during one-line-to-ground and two-line-to-ground faults at the sending end. A 10% reactor is used at the receiving end. Assume E I = 120 and E 2 = 100 26. Calculate and plot power dissipated in the grounding resistor of the following simple system during a line-to-ground fault as a function of the resistance in per cent. The system consists of a generator for which the positive- and negative-sequence reactances are each 20%, a bank of transformers of 10% reactance, connected Ll- Y with neutral grounded through a resistor, and a fault at the secondary terminals. 27. Calculate and plot the higher line-to-neutral voltage (in per cent) of the unfaulted phases, during a line-to-ground fault on the system of Probe 26, as a function of the grounding resistance in per cent. 28. Verify the equivalent circuits of Figs. 26b and c. 29. Derive an equivalent circuit for two coupled lines of different nominal voltages in which per-unit impedances are used. 30. Prove the relations shown in Fig. 31 for representing the ground impedance of an autotransformer. 31. Find the value of series impedance presented to zero-sequence current by an autotransformer with ungrounded neutral. 32. Solve Example 2 by two-phase coordinates. Assume X2 of generators to equal Xl.
1!1.
1!1.%.
CHAPTER VII
TYPICAL STABILITY STUDIES In the past twenty years a great number of stability studies have been made, usually with the aid of an a-c. calculating board, on existing or projected electric power systems. Through the courtesy of a number of operating and engineering companies we are enabled to present in this chapter accounts of several such studies made during the last ten years. Most of the companies have preferred that their names and the identity of their stations, substations, and lines be withheld; accordingly, letter symbols have been substituted. Let the reader be assured, however, that neither the stations nor the studies are fictitious. These typical stability studies have been included in order to illustrate some of the purposes for which stability studies are made, to show how such studies are carried out, how the types and locations of faults to be studied are chosen, how simplifications of the power system (such as the combining of several stations) are decided upon, and how limiting or critical conditions are selected for study, and also to give a glimpse at the progress made in the improvement of power-system reliability. Some of the stability studies here presented are incidental parts of more general studies that included also steady-state operation (loading of lines, maintenance of voltage, required capacity of synchronous condensers, and so forth). All these studies involved great masses of data and calculations which it was not feasible to reproduce here in full; for example, length, spacing, wire size, zero- and positive-sequence impedances of transmission lines; name-plate data and impedances of transformers, generators, and other machines; actual and estimated future loads on substations at different times of day and year; moments of inertia of machines; calculations of per-unit impedances on a selected system base; calculating-board connection diagrams and instrument readings; and calculations of swing curves. The reports on some studies fill one or more volumes. The condensed accounts of the studies which follow give an inadequate idea of the amount of labor actually spent on the studies. Nevertheless, we hope that they convey a clear conception of the methods and results. 253
254
TYPICAL STABILITY STUDIES STUDY 1*
The power systems of company A and company B, when operating in parallel, experienced difficulties due to transient instability. As an increased flow of power from company B to company A was contemplated, stable operation would become of greater importance. Study 1 was made to determine what factors entering into the interconnected operation of the two systems were important in causing instability and how stability might be improved. Description of systems. A map of the two systems, showing generating stations, substations, and transmission lines (with their voltages and lengths), appears in Fig. 1. Each station and substation is there identified by a two-letter combination, the first letter of which denotes the power company owning or operating it. The combined systems extended approximately 620 miles from station BO to station AV. Company A had two steam stations (AL and part of AP), several fairly large hydroelectric stations (AB, AC, and AG), and many small hydroelectric stations, some of which were neglected or were combined with larger nearby stations in the study. The principal load centers were near AL and AM. The transmission system of this company was composed of (1) a backbone of three 132-kv.lines in parallel, about 140 miles long, firmly tying together the principal generating stations and load centers; and (2) an extensive 44-kv. network which paralleled the 132-kv. lines, connecting smaller generating stations and loads.und which extended the system south of the terminus of the 132-kv. lines at station AE to station AP and to numerous substations. Company A had few instability problems in itself, as the loss of anyone line (either 132-kv. or 44-kv.) would not seriously weaken the ties between generating stations. Company B had a dozen or so small hydroelectric stations, divided into an eastern and a western group. Most of the stations of the western group (BB to BF, inclusive) were connected to each other and to station AD of company A by a single 132-kv. line. Stations BG and BH, also in the western group, were connected to the other stations of this group by two 44-kv. lines. The stations of the eastern group (RI to RO, inclusive) were joined to one another and to the western group chiefly by 66-kv. lines, with some 44-kv. lines. Thus, over most of the length of system B there was only one 132...kv. line, or *Information concerning this study was obtained through the courtesy of Ebasco Services, Incorporated, New York City, with the concurrence of the operating companies concerned.
STUDY I-DESCRIPTION OF SYSTEMS
255
two or three lower-voltage lines. This system undoubtedly would present problems in transient instability even if it were not interconnected with the system of company A. In addition to what might o AA
Company B
125
34 BM
38
Principal line interconnecting companies A and B
@]
Steam generating station
@
Hydro generating station
~
Combination generating station
@
o
Synchronous condenser substation Substation
•
Fault location
+H++
132·kv. transmission line 66·kv. transmission line
- - 44·kv. transmission line Figures onlines aretheirlengths in miles~ Two·letter combinations are station designations.
71
AV AO FIG.
1. Map of the power systems of companies A and B, Study 1.
be called an inherently unstable arrangement of the transmission system of company B, due principally to the nature of the territory served, several hydroelectric stations on this system had generators with characteristics which were not well suited to maintaining stability.
1 2
3 1 1
3
2
N N
0
2
1
2 1 1 1
1
2
3 2 1
3
2
2 2
2
2
1
2 1
2 1
1
ABl AB2 AB3
AC
AE
AFl AF2
AG
AH
AI
ALl ALe AU AL4
AN
N
2
2
1
11,111 7,500 1,250 1,000 18,750 2,500
6.6 6.6 4.0 1.1 6.9 2.3
Hydro
N Hydro
"
Hydro
2.3
2.3
It
1 1
4.4
4.6 0.5
Steam
1
Hydro
Hydro
Hydro
It
Hydro
Condo
It
It
2,500
1,000
8,000 25,000 750
1,250
600
12,222 8,200 8,333
6.6 6.6 7.0
Hydro
It
2.3
2,500 625
2.4 2.3 2.3
Hydro It
kva.
Rating kv.
Kind of Machine
0
1
2
1
N N
Nt
1
1
1 0
2
2 1
2
Number of Units Running in Load Condition
AAl AA2 AA3
Station*
Number of Like Units
TABLE 1
360
600
1,800 1,800 330
600
600
150
400
600
400
180
514 150 171.5
180 150 150
r.p.m,
LIST OF SYNCHRONOUS MACHINES (STUDY
1)
,
13.5
14.5
8.5 .19 5
9
13
35
15 6
18
15
13 40 37
24
30 25.5
(%)
Xd
80 114 130 27
5,760
755
3,664
1,295 2,180 1,949
375
Generator
3
30 98 3 97
1
209
402
81.9 158 3.3 10.3
3.0
12.4
32.0
27.8
... 1 3
28.4
79.5 11.3 14.3
3.0 2.0 1.0
(Mj.)
of Unit
Kinetic Energy
144
14 185 170
44
Turbine
WR 2 (kilo-lb-ft.")
~
t?:j 00
t:1 .....
~
00
t""4 ..... ~ t-<
~
to
;>
~
00
;> t""4
~ ..... o
~ ~
~
l'-' 01
1 4
5 1
1
1 1
2
2 1
BB
BC
BD
BE
BF
BG
"
"
"
"
2
AV
2
2 1 2
2
I
2
1
1
2
2
1 1
1
2 1
2
1
1
I
0 1
0
4
5 1
0
N
1 1 2 1 1 2
1
1
1
N
0 0
1
1 1 1
3 1
APl AP2 AP3 AP4 AP5 AP6 AP7 AP8
Station*
Number of Units Number of Running in Like Load Condition Units 1 2
"
Hydro
Hydro
u
Hydro
Hydro
"
Hydro
"
Hydro
Condo
" " " "
Steam Hydro
u
Hydro
Kind of Machine
2,400 590 1,250 2,500
6.6
2.3 6.9
6.9
2.2 6.6
6.9
2.3 2.3
2.3 6.6
2,400 7,860 18,750 875
2.3 2.4 13.8 6.6 6.6 2.3 6.6 4.0
1,250 7,500
9,000
12,000
600
9,375
1,400 3,000
4,375 7,500
600
kva.
kv,
Rating
TABLE 1 (Continued)
300
600
138.5
514 240
240
200 200
164 120
360 514
600 300 300
514 1,800
300
r.p.m,
,
25 35
30
15 27.5
32
14 36
16.3 65
15.5
10 24 22 13.5 9 10 10 13
(%)
Xd
27 480
4,000
24 2,000
1,160
360 350
1,120 2,670
37
383 254 68 18 64 383 34
Generator
t t
t
t
t
t t t t
2.5 10.9
19.4
1.6 29.2
16.9
3.6 3.2
7.6 9.7
3.3
...
t
8.0 15.9 120.8 1.7 1.4 8.0 1.1 1.0
(Mj.)
Kinetic Energy of Unit
5 8 95 3 3 3 3 1
Turbine
WR 2 (kilo-lb-ft.F)
"'""
~
CJ1
00.
s=
~ trj
to<
00.
o~
Z
...-4
~ o
...-4
~
~ 00 (1
t;
I
......
to<
§
00.
.-3
1
2 2 1
1
BL
BNl BN1&
BO
1
2 1 1
1
2 2 2 4
1 3 3
1
1 1
2
1
0
1 1 1
0
2 2 2 4
0 0 0
2 1 1 0
1&
Number of Units Running in Load Condition
Hydro
u
"
Hydro
Condo
" " "
Hydro
u
Condo Hydro
"
u
u
Hydro
Kind of Machine
6.6
6.6 0.5 2.2
2.3
2.2 2.2 2.2 2.2
4.1 0.4 2.3
6.6 6.6 2.3 2.3
kv.
3,600
5,000 500 500
3,000
850 1,250 1,562 1,000
7,500 300 625
2,000 3,600 5,500 3,125
kva.
Rating
225
225 225 225
720
120 90 90 109
900 257 180
225 450 400
200
r.p.m.
,
10
45 12 8.5
34
15 29 36 36
31.5 16 22
20
10
18
(%)
Xd
593
1,200 126 100
22
286 546 546 270
140
39 31
574 593 500
Generator
15.4 1.6 1.3
...
t t t t
7.6
2.6
t
7.3 0.5 1.2
5.8 7.6 25.6 5.7
Kinetic Energy of Unit (Mj.)
1.0 1.1 1.1 0.8
t t t
: :
...
+ +
t t
Turbine
W R 2 (kilo-Ib-ft. 2 )
*The numerals appended to the station letters denote separate stations or parts of stations which were combined in this Study. tN = Neglected. tWR2 of turbines of company B was assumed to be 10% of generator WR 2•
"
" " "
2 2 2 4
3
~
1
2 1 1 1
BK
Bll BI2 BI3
BH2 BH3
"
BHl
Station*
Number of Like Units
TABLE 1 (Cootinued)
t:-t
H
m
~
H
tj
~
~
~ ~
H
S
E:
a
~ ~
t-d H
~
259
STUDY I-DESCRIPTION· OF SYSTEMS
Interconnected operation of the two systems resulted in additional transient stability problems for both systems. As will appear from the swing curves obtained in this study, however, faults occurring near TABLE 2 LIST OF STATIONS
HAVING
SYNCHRONOUS MACHINES (SroDY
Load Condition 1
Station
Kind of Machines
AA AB AC
Hydro
AE
Condo Hydro
AF AG AH AI AL It
u
AN AP u
u
AV
BB Be BD
BE BF BG BH BI It It
BK BL BN BO
It
u
Kinetic Energy (Mj.)
Aggregate Mva. Running
Kinetic Energy (Mj.)
6.8 53.2 33.3 15.0 4.5
10.0 264.1 85.2 55.6
6.8 20.5 11.1
10.0 93.8 28.4
Off
Neglected Off
4.5 18.8 5.0 1.2 25cO 1.8 26.8 2.5 18.8 17.9 36.7
Off Neglected
It
Off
5.0 1.2 41.0 1.8 42.8 2.5 18.8 14.8 33.6 5.0 34.4 10.0 9.4 12.6 18.0 10.0 16.1 7.5 2.8 10.3 11.3 3.0 11.0 3.6
It
It
u II
Ie It
a
Condo Hydro Combined Hydro Cond. Hydro It
Load Condition 2
Aggregate Mva. Running
It
Steam Hydro Combined Hydro Steam Hydro Combined Cond. Hydro
1)
24.8 3.0 321.8 13.6 335.4 Neglected
120.8 36.4 157.2 Neglected
46.6 21.4 16.9 30.8 38.8 15.9 50.5 7.3 5.0 12.3 9.8 2.6 33.7
7.6
32.0 24.8 3.0 158.0 13.6 171.6 Neglected
120.8 39.5 160.3
Off Off
Off Off
3.0 9.4 12.6 18.0 10.0 13.1
3.2 16.9 30.8 38.8 15.9 44.8
Off
Off Off Off 9.8 Off 18.3
OtT Off 11.3 Off 6.0 Off
Off
station BG or east of it had little relation to the stability of interconnected operation, although such faults sometimes presented serious stability problems for the eastern part of the system of company B. The 125-mile, 44-kv. line from AA to BA had several small hydroelectric stations (grouped as station AA in the study) and several small
BK
BL
BM
6& ky.
4Cky.
~
~
~
== ~I
~T
BJ
~
+' ~-
gl~£
::1+'~ ::J.~
~
4Cky.
AI'
~
+
:;;
G'
~
~
o
+
~
~ AS
=:+
~ ;
AR
lJF
~l
e
~
~
AU
AT
~
;
-
~
~ ~
~
~
+
~
..,
+ -~~ . :i ·...1 -.... aD
AK
~
C"I
.,p
o
AH
AJ
+ +' £
~
C"I
sil
+' ~
~
132 ky.
:it
+'
C!
2
:;-
s:
~
+ s:
C7>
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
'"
<,
~I
r-,
"'I'--
!llo...
<,
I
o
0.1
-I----
'" ~ ~o
·~·t ~I
-120
~
/ ~ Fostoria condenser ~~
dr
-100
I
~,
-~ / ~~ ~
I~I
.1"
-80
,
~
"
I
\1
~I
I
~~
I
\~ ~~
I
I
<
-60
......
-
l-/ ZAkron - Alliance
'"
""--.
t~ ~ I I ~
~"
I
~I I Lanc,aster
.i~
I...........
c
condenser
I
0.2
0.3
0.4
r-,
0.5
0.6
Time (seconds) afteroccurrence of fault FIG. 27. Swing curves, Study 2, case 1. Future set-up. Generating plants at Lancaster and Rutland. Interconnections open. Three-phase fault on Philo 132-kv. bus section Al cleared in 0.15 sec. by opening Howard line 1, Crooksville line 1, generator 1, and bus-tie reactor 1. Unstable.
systems, stayed in step. The motion of the other machines was very nearly the same in both cases. Philo A, the machine nearest the fault, lost all its load while the fault was on and speeded up more than any other machine. Philo B, separated from Philo A by the bus-tie reactor, also speeded up, although less than Philo A did because it lost
STUDY 2
297
less load. The other generating stations were affected but .little. The synchronous condensers, whose swing curves are shown by broken lines, had very low inertia and swung with a shorter period than did the generating stations. Because of their small inertia they had little effect on the rest of the system, and they were not represented separately in case 3. 60
j
.... Fault
I I
1
I
cleared
40
~
/v
0.15 sec. ~
y I
lancaster, c
I
I
.~
~,
~
'
""""""'"
~Canton
,-
.
I
I I
v\
0.2
<-Lorain
\ ---Y
-""" ~
\
~
I I
1oledo
..
I
i
I
r\ R= r---
0.3
............
c: ~
--a--~.
condenser~~
"
I 0.1
"
,--....
c-
-----
r-.
I' -...'
-60
I
condenser
'" C>< ~ ~/~ ....... / ~~~', ,~ . ~--
Akro~ s, ~~~ II ,~ i Fostoria
"'I o
-:
~
__V C>< ~ ~
I'R.utland
rr-
~
I
~ K Alliance
-40
I
/,-Windsor
-.1-, " -""""--- --~~I ..
:t:i
r\. ___
~Philo B
II
o
"- " ~
. ./
--~ [.,.7' I
T
~ ~PhiloA
//
V~ ~
~
I
~
\.
~J
,
'J-/
~
t--- ............
---.
FortWayne and West I
I
0.4
I
I
0.5
0.6
0.7
Time (seconds) after occurrence of fault FIG. 28. Swing curves, Study 2, case 2. Future set-up. Generating plants at Lancaster and Rutland. Interconnections closed. Three-phase fault on Philo 132-kv. bus section Al cleared in 0.15 sec. by opening Howard line 1, Crooksville line 1, generator 1, and bus-tie reactor 1. . Stable.
Case 3, representing immediate-future conditions, was set up without the proposed generating stations at Lancaster and Rutland, but with the existing stations south of Philo at Logan and Cabin Creek in operation, and with interconnections to other companies closed. Loads were as shown in Fig. 29, substantially the same as for cases 1 and 2. A three-phase fault was assumed to occur on section A2 of the 132-kv. bus at Philo and to be cleared in 0.15 sec. by the opening of Howard line 2, Crooksville line 2, Rutland line 2, generator 2, and bustie reactor 2. Note that the two groups of lines at Rutland are not bussed. The swing curves (Fig. 30) show that Cabin Creek and
TYPICAL STABILITY STUDIES
298
lorain 11.4 - j3.6
Fort Wayne and West
-........- - 100.5
40-jl0
Crooksville
South Point
.............. 106
t
139 + j6.0
Logan and South FIG.
29. Initial load conditions of case 3, Study 2. Flow of active and
STUDY 2
299
Akron, Alliance, and Canton condensers
s 40.0 + nos
Po: 40.0 +jl1.0 ~ I
~
-e--
o ~
North east 98.0
....---.....-98.0
Massilon
t$
~
24.0 + j19.6
:s It'
28.0 - ,56
B Philo
Newcomerstown
Windsor andWest
Penn
~...... -
....
102
('t)
e",
I
....... ('f)
Cabin Creek 1040
G
reactive power in Mw.
+j
t
190 - ;14.0
Mvar.
Bus voltages are given in per cent of 132 kv.
TYPICAL STABILITY STUDIES
300
Logan, which were sending power to Philo, speeded up and pulled out of step with Philo. The pull-out might be attributed partly to the loss of a Philo-Rutland-South Point-Turner circuit and partly to the initial angular displacement between Logan and Cabin Creek, on the one 80
I <.$1 ~I
~I
60
01
Cabin
~I
../
gJl
~I
40
"31
~I
~
"......,..-
~
&
.......-LOga~-r
l.a-- ~
-40
~
o
I
I
J J
I
I
"'lIiI
I
-,
-
~
~~
~
~
/Lorain ../V
~
""".--~
~
~ .......
I -~ ...............
~
-,
.-'!
Toled0-:L ~ ~
-
~
~ .......... ----. r---. Fort Wayne and West ~
----r---
l 0.1
~
•
,
~
I I
PhiloB~ ~
Windsor",
!
-60
<,
I
I
r----..
/
..-Logan
/
~~ r-, Akron - Alliance, Canton condenser -.......-..
I
....---...
~
..--........
?~
I
I
r----.. ~
~ III""""""
/
fI'
/
I
I
-80
~
~
/
v::
",
PhiloAJ
I
A
------
~
71
~
Creek~
~
0.2
0.3
0.4
0.5
0.6
~
--~
~
0.7
Time (seconds) afteroccurrence of fault
30. Swing curves, Study 2, case 3. Immediate-future set-up. No generation at Lancaster or Rutland. Three-phase fault on Philo 132-kv. bus section A2 cleared in 0.15 sec. by opening Howard line 2, Crooksville line 2, Rutland line 2, generator 2, and bus-tie reactor 2. Unstable. FIG.
hand, and Philo, on the other hand, being larger than that between Rutland and Philo in case 2. Philo A and B swung more nearly together than they did in cases 1 and 2, probably because Philo A was more lightly loaded than before (175' instead of 215 Mw.). Conclusions. The following conclusions were reached: 1. The proposed bus reactors at Philo were found to have little effect on system stability, as evidenced by the fact that the generators
STUDY 3
301
in sections A and B of the plant swung approximately together, even for a fault on one bus section. 2. It was also found that, without the benefit of the added inertia of the Chicago and Cincinnati interconnections (although these connections are normally closed), the system under the assumed load conditions would lose synchronism between Twin Branch and Philo after a three-phase bus fault at Philo cleared in 9 cycles with the loss of one Philo-Howard circuit. With the interconnections closed the results indicated that synchronism would probably be maintained under this fault condition. To increase the margin of stability under such conditions a program of modernizing the Philo line breakers was worked out (and has subsequently been completed), calling for a reduction from 8 to 5 cycles in the breaker-operating time and from 9 to 6 cycles in over-all clearing time of faults. Also, the system tie lines between Ohio and Indiana have been strengthened by the construction of a new 132-kv. line from Portsmouth to Muncie, Indiana, providing a direct line from Turner to the Indiana area. 3. Likewise, with a three-phase bus fault at Philo followed by loss of a Philo-Rutland-Turner circuit, the system, under the heavy load conditions assumed, showed a probable loss in synchronism between Philo and the south. Here again, the subsequent improvements which have been mentioned were designed to eliminate this difficulty, and they have undoubtedly provided the necessary margin of stability to prevent loss of synchronism under these conditions. 4. The system engineers have also pointed out that the future installation of 3-cycle breakers to replace 5-cycle breakers at the most important locations, together with more extensive application of highspeed reclosing, will provide still greater gains in system stability. STUDY 3
Figure 31 represents a 132-kv. 60-cycle transmission network connecting a large steam generating 'station A (on the extreme right-hand side of the diagram) to a large metropolitan receiving system, which may be considered an infinite bus and which is represented by system J on the extreme left. Connected to the same 132-kv. network are two other steam generating stations, E and I, and a number of substations with loads. The circuit breakers and relays on the network were too slow to maintain synchronism between A and J through severe faults. Therefore, it was believed best to operate the system sectionalized by open breakers marked with crosses in Fig. 31. Under this plan each of the
FIG.
31.
SystemJ
Circuit breaker openfor sectionalized operation.
18.2+j4J.lO
Substa.D 9.9+J38.J.O "-45-j6 Mva
One-line diagram of the power system of Study 3, showing impedances in ohms at 132 kv., initial loads in megavolt-amperes, initial voltages in kilovolts, and assumed fault locations.
6. FaultlocatiOn assumed in the study.
Circuit breakernormallyclosed.
12-j101Mva 129kv
o
118 - j 52 Mva.
~
~~
~/220.
~
12.0+jS1.50 j35.Hl --14 + j 10 Mva.
Substa. F
5.7+j24.3n
1m
3.3+j12.S0 .... 23+j24 Mva
124 kv
Substa. G
4O-j2S Mva.
Sta.A
CIo:)
00
t:rj
t1 t-4
c:
~
to<
~
t-4 t-4
t:d t-4
:>
~
00
t-4
:>
o
t-4
"'d
t-3 ~
~
STUDY 3
303
three generating units at station A fed toward the metropolitan system J over a separate circuit. A fault on anyone circuit and its subsequent clearing disconnected one of the units at A from the rest of the system. The disadvantage of this mode of operation was that, after clearing a fault in or near stations A, B, or C, the loads at A, B, C, or D (depending on the fault location) were fed from stations E, I, and J over such long lines that the voltages at the loads became too low. To take the most extreme case, the load at station A could not be supplied over the long line from system J if a fault occurred in generator At. The study here described was made to determine which breakers would have to be speeded up and possibly also equipped with highspeed relaying to make station A stay in synchronism with the rest of the system for faults on the I32-kv. network if the network were to be operated unsectionalized, that is, with the breakers marked by crosses closed. The breaker-opening times were about 20 cycles, except for some 8-cycle breakers at stations A and E; and the relay times were from 1 to 60 cycles, depending on the fault location. It was assumed in the study that any breakers requiring faster opening would be rebuilt to open in 8 cycles. The line relaying would be changed where necessary to high-speed distance relays having I-cycle operating time; and, if necessary for stability, carrier-current pilot equipment would be added to give simultaneous tripping of the breakers at both ends of the line for faults at any point of the line. It was further assumed that high-speed bus differential protection would be installed where necessary for stability. It is common to use two-line-to-ground faults in stability studies, inasmuch as three-phase faults are very infrequent on high-voltage steel-tower transmission lines. In this study, however, three-phase faults were assumed, because they are not much more severe than twoline-to-ground faults in their effect on stability and because they are much simpler to represent. Only the positive-sequence network need be considered for three-phase faults, whereas the zero- and negativesequence networks must also be considered for two-line-to-ground faults. Three-phase faults were assumed at various locations, the choice of which will be discussed, and for each location swing curves were calculated to determine whether the system would be stable with the existing breakers and relays and with faster breakers and relays. To obtain the swing curves, the network of Fig. 31 (with all breakers closed) was set up on an a-c. calculating board. Each line was represented by a nominal e, although the shunt capacitance at each end of
304
TYPICAL STABILITY STUDIES
the line is not shown in Fig. 31. The three generators of station A were represented by a single equivalent machine on the assumption that, being connected to the same bus-although sectionalized by reactorsthey would swing nearly alike. The generators of station E were not represented, except in one case in which the fault location was near this station. For other fault locations it appeared that these generators were electricaliy close enough to the metropolitan system J and far enough from the fault so that they would not have much effect. In such cases the load of station E was set to represent the true load there less generation there. At station I two groups of generators on separate busses were represented. All generators were represented by transient reactance and voltage behind transient reactance. At station A the reactances of the three generators were represented by separate impedance units connected at one end to different bus sections (separated by reactors) and connected at the other end to one power source, the voltage of which represented voltage behind transient reactance of all three generators. Generator A1 had two windings, the self and mutual reactances of which were represented by a Y circuit. In accordance with usual practice voltages behind transient reactance were kept constant during the study. Data on the inertia, speed, and kinetic energy of each actual and equivalent generator are given in Table 8. An interval of 0.1 sec. was used in the point-bypoint calculations. The outputs of the various machines, the values of the loads, and the bus voltages were adjusted to typical values (shown on Fig. 31) as nearly as possible to values observed on the system with sectionalized operation at time of maximum system load. Loads were represented by constant impedances, which were not changed after the fault was on. The most severe fault location is ordinarily close to the sending end of the system. Therefore it is logical to try first a fault so located. The worst fault location on the system of Fig. 31 would be on the low-voltage bus of station A. A three-phase fault here would entirely interrupt any exchange of power between the generators of this station and the rest of the system. This bus, however, was of isolated-phase construction; therefore the only type of fault which could occur here was one-line-to-ground, which is not a severe fault from the standpoint of stability. The closest point to the bus at which a three-phase fault could occur was on the high-voltage side of one of the transformers (there was no high-voltage bus at A) or on the far side of a set of reactors feeding a local load (fault location 2). As a matter of fact, a more severe fault location than either of these was on the bus of sub-
STUDY
305
3
TABLE 8 INERTIA OF GENERATORS (STUDY
Rating (Mva.)
WR2 (lb-ft.2 )
AS
100 50 50
859,000 475,000 475,000
A, total
200
1,809,000
Generator
At AS
Et E2
ES E4
12.5 12.5 35 30
51,000 51,000 357,000 29,100
E, total
90
Ila Ilb
50 60
390,000 391,000
11, total
110
781,000
ISa ISb
35 100
243,000 750,000
IS, total
135
993,000
3)
Speed (r.p.m.)
Stored energy (Mj.)
1,800 1,800 1,800
640 365 365
6.40 7.29 7.29
1,370
6.85
38 38 266 87
3.04 3.04 7.60 2.90
429
4.76
291 292
5.81 4.86
583
5.30
181 560
5.17 5.60
741
5.49
1,800 1,800 1,800 3,600
1,800 1,800
1,800 1,800
H
11M
7.89
25.2
18.5
14.6
station B, where a three-phase fault would entirely interrupt the flowof power between generators A and the other generators. It wasimmaterial, while the fault was on, whether it was on one or another bus section of substation B or whether it was just outside the substation on anyone of the seven lines. The worst location as regards conditions after the fault was cleared, however, was on the upper bus section (fault location 1), where clearing a fault would require the opening of a bus-tie breaker and of three line breakers, whereas a fault in any of the other locations mentioned would require the opening of fewer lines. Consequently, the effect of a three-phase fault at location 1 on the bus of substation B was examined first. A three-phase fault at this location was known to put station A out of step if slow breakers were used. Therefore 8-cycle breakers and 1cycle bus differential relays were assumed, giving a clearing time of 9 cycles. The swing curves are shown in Fig. 32. Clearly the system was stable, although without much margin. Twelve-cycle clearing would probably be too slow.
TYPICAL STABILITY STUDIES
306
100
V
e
~
~
60
E
u Q) "ii)
40
~
V
Q)
\ 20
~V
V
~
f--
Faultcleared completely 9 f"'v
'-
"l
<,
.....
I--. t---.
",.
~
~
----
I
~~-
/ ~ ~ "".,..--
It- Faulton I
o
" r-, r-,
1/
I
1i".o e
<
A-- r-;
~
~ Q) Q)
r--- ~
-~
./
80
-0 jij
~
~'
-
............... ~2 ~,
"
~
¥ 'r-J
I
o
--
8
32
24
16
48
40
'"
.........
K-~ ~ r-I""""""
56
64
72
Time (cycles) FIG.
32. Swing curves, Study 3, fault at location 1, cleared completely in 9 cycles. System stable.
100
.
80
-:
~V
~
~
~
1/
~
I I
..................
Fault cleared 15~
12
~~
-
~
~1,,7
~~
k-- Fault on- ~ I
-~
-11
J~
I
8
~
"'
I
o o
r-,
VII
~~
20
-- -
A
~
1 I
16
24
32
40
48
Time (cycles) FIG.
33. Swing curves, Study 3, fault at location 2, cleared in 15 cycles. System stable.
STUDY 3
307
It could be inferred, without taking more swing curves, that the system would likewise be stable for a fault on either of the other bus sections at substation B cleared in 9 cycles, but would be unstable with the slow breakers. Therefore all the breakers at B should be speeded up, and each bus section should be equipped with high-speed protection. It could be inferred further that the system would be stable for a fault close to substation B on anyone of the seven lines cleared in 9 cycles at both ends. Whether sequential clearing would give stability 60 _ _-...--r--~~-.... is not yet clear, but this point will be considered later. Attention was next turned to fault ~ location 2, separated by a reactor ~ 40 I---+---+~ from the low-voltage bus of station ~ o A. Fifteen-cycle clearing time was 12 assumed, corresponding closely to 20 ~~~..........-t-e~~ the existing relay and breaker times ~ 11 of 6 and 8 cycles, respectively. The ~ swing curves are given in Fig. 33. J o~...._ ........._~-- ...... Again, the system was stable. 24 8 16 o Time (cycles) Fault location 3, just outside station A on one of the lines to sub- FIG. 34. Swing curves, Study 3, fault at location 3, cleared at both station B, was considered next. ends in 9 cycles. System stable. Nine-cycle clearing was assumed at both ends. The swing curves, shown in Fig. 34, indicate that the system was stable by a big margin. Sequential clearing will be considered later. The next fault was taken at location 4 near station E on one of the lines to substation C. The machine at E was represented by a power source on the supposition that this machine would be a significant factor in the stability of the system. Sequential clearing was assumed, 9 cycles (1 cycle relay time plus 8 cycles breaker time) at E, where the breaker already was an 8-cycle one, and 80 cycles (60 cycles relay time plus 20 cycles breaker time) at C. The swing curves are given in Fig. 35. The system was stable and would be so even if the line were not opened at C. There seems not to be much margin, however, in the clearing time at E. For this fault location the existing breakers were satisfactory. When the fault was taken at the other end of the same line (location 5), however, 21-cycle clearing at C (1 cycle relay time plus 20 cycles breaker time), followed by 68-cycle clearing at E (60 cycles relay time plus 8 cycles breaker time) was not fast enough for stability. The
-
:a
!
120 r-----r----,...--~-__r__-_,_--.,...__-~--
100
&/)
Q)
80
~
aD
Fault cleared at C
Q)
"tJ
co
.g 60 1J Q)
'ii Q)
bo
e
-e 40
20
I.-I---.l----. ... 12 36
O'-......
o
J
-~-----
24
48
60
....- . . . . . . 84
-~
96
Time (cycles)
FIG. 35. Swing curves, Study 3, fault at location 4, cleared at station E in 9 cycles and at substation C in 80 cycles. System stable. 140
r
/
/
120
en
bO
"'0
80
~
~~
60
Q)
./
co ~
40
----
o
..........
~
I
I
I
Fault cleared at station C
21'"'-'
68'"'-'
--J I
~~
~
I I
~~
I
_1/
.......... J
16
I
I
I
......... ........... 8
I
I I
~
./
;
~
I
I I
~
;
~ll
I
J ____
I I
o
Fault cleared at station E--.
I I
Fault on-
-
V-
I-
I
~~
~-
20
I
I
J II
tV u "ii
I
I
V
~
Q)
:
V
I I
Q)
I I
j~
I
100
!
~
I
-
I
!
I
I
24
32
40
48
56
64
72
Time (cycles) FIG.
36. Swing curves, Study 3, fault at location 5, cleared at substation C in 21 cycles and at station E in 68 cycles. System unstable. 308
STUDY 3
309
swing curves of Fig. 36 show that station A went out of step with the rest of the system. Therefore the breaker at C must be speeded up. Nine-cycle clearing at both ends would certainly give stable operation, because 9-cycle clearing sufficed in the more severe fault locations 1, 2, and 3. Sequential clearing (not tried) probably would be satisfactory. Attention was now turned to the circuit from B to D, and a threephase fault was taken at location 6 near D. Generator 12 was not 140 - - - _ -_ _-...-----.~...,.___,...._-~__r____r__=__.,....__r____r_r_...__ ......
I
120
-e
1I~~>--FaU" cleared in 27'"
100
.1/
( /) Q,)
1/
bO
Q,)
"0
80
(ij
I
u t)
'':;:
-
;
I I I I I Fault cleared at B'-.. --L.-68"" ~
T
I
Q,)
Q)
;
60
Q,)
~
c
I
<
20 '--
/~- --~~ ~
......
_ ~ ~
o
r
I
I
Il-r---'----~~..........~- _ ...... I'--Fault cleared in 68/V
O'--...I.....
FIG.
!
I
40
8
16
:
__..._"___'_....__....._........
I
~-..~
24
I
32 40 Time (cycles)
48
56
64
72
37. Swing curves, Study 3, fault at location 6, cleared at substation B in 27 cycles, system stable; or in 68 cycles, system unstable.
represented for this case, because, with a fault at 6 either on or cleared,
12 could exchange no power with station A and hence would have no effect on the swinging of A with respect to the infinite bus J. For similar reasons the clearing time at substation D was immaterial to the stability of A. The clearing time at substation B was first assumed as 68 cycles (60 cycles existing relay time plus 8 cycles new breaker time) ; generator A pulled out of step as shown by the swing curves in broken lines in Fig. 37. The clearing time was then reduced to 27 cycles, resulting in stability without much margin, as shown by the swing curves in solid lines in Fig. 37. The required clearing time of 27 cycles or less at B was too short to permit selective operation between the
310
TYPICAL STABILITY STUDIES
breaker at B and the 20-cycle breaker to the left of D for a fault just to the left of D. Either the breaker to the left of D must be speeded up, or else carrier-current relaying must be installed on the line from B to D. With an 8-cycle breaker and J-cycle relay at D, and a margin of 20 cycles (required by good practice), the clearing time at B would be 1 + 8 + 20 + 8 = 37 cycles, which was still too slow. Hence carriercurrent relaying was needed, and by its use the clearing time at substation B could be reduced to 9 cycles. As already stated, the clearing time at D was not important; therefore the breaker at D need not be changed. Consider next fault location 7 near substation C on the line from B to C. Before the fault was cleared, conditions were the same as they B
E
A
c
• 8 - cycle breakers needed. - - - Carrier - current relaying needed. FIG. 38. Location of 8-cycle breakers and carrier-current relaying necessary to maintain stability with interconnected operation under the conditions of Study 3.
were for the fault at location 5, for which 21-cycle clearing was too slow. Hence the breaker at C must be speeded up to 8 cycles, giving a clearing time at C of 9 cycles. After the breaker at C was opened, conditions were similar to what they were for the fault at location 6, except that the shock during the first 9 cycles was greater for location 7 than for 6, as the reactance between the fault and the bus of substation B was slightly less for location 7 than for location 6. Therefore the critical clearing time at B for a fault at 7 must be considerably less than the 27 cycles found for location 6. A clearing time of 68 cycles (existing relay time of 60 cycles plus new breaker time of 8 cycles) was far too slow, and even 37 cycles (second-zone impedance relay time of 29 cycles plus breaker time of 8 cycles) was too slow, Carrier-current relaying was therefore needed on each of the two lines between Band C. For similar reasons carrier-current relaying should be used on each
STUDY 4-DESCRIPTION OF SYSTEMS
311
of the four lines between A and B. Sequential clearing of end-zone faults appeared to be permissible, however, on the two lines between C and E. Eight-cycle bus-tie breakers and high-speed bus relaying should be used at stations C and E, as well as at B. Faults were not taken any farther from station A than stations D and E, because it was known that station A would stay in step, even with the existing breakers and relays. The conclusions of the study are set forth in Fig. 38, which shows'the locations where 8-cycle breakers were required and the lines to which carrier-current relay channels should be added. STUDY 4§
Three power companies (herein called companies A, B, and C), operating in neighboring states but having no physical connections with one another, were considering the construction of interconnecting transmission lines to permit the interchange of power between them in either direction. Study 4 was made to determine (a) the transmission capacity or power limit of the proposed interconnecting lines, (b) the required capacity of line terminal equipment, such as transformers and shunt reactors or synchronous condensers, and (c) the modifications necessary within the three systems to permit utilization of the full power capacity of the interconnecting lines. Description of systems. The systems involved in the study are portrayed by the impedance diagrams (Figs. 39 and 40), and they are also described in the following paragraphs. Stations of these systems are designated by two-letter combinations, the first letter of which denotes the company owning the station. Company A operated a metropolitan power system having a steamelectric generating station of 112-Mva. capacity (station AA) which fed a number of substations through a l3.8-kv. transmission network (shown in upper right-hand corner of Fig. 40). Station AA was connected through two 66-kv. lines to substation AB, from which connections were made to companies D and E and-by one of the proposed lines-to company B. Company B had a system consisting of two distinct parts. (See Fig. 39.) One part, shown on the left-hand side of the diagram, had two steam-electric plants (station BF, 25 Mva., and station BG, 32.5 Mva.) a few miles apart. These plants served the area in which they §Information concerning this study was obtained through the courtesy of Ebasco Services, Inc., New York City, with the concurrence of the operating com... panies concerned.
TYPICAL STABILITY STUDIES
312
lOMv. . .
DC
- - Usbnllines. - - Proposed lines.
0-
Synchronous condenser,
0Ste.m·· electric I.ner.ti", ~st.tlon. rji\.-
v:..;-
Hydroelectnc I.ner.tin, 'tatlOn.
0- [qulv.le~t lenerator ~
representln,.power system.
LOid.
:I: :'~r~~~=.:n=...
+
Tr.ns'ormer.
39. Study 4, part 1. Impedance diagram showing impedances of transmission lines and transformers in per cent on 4o-Mva. base. Values of capacitive susceptances of lines are in per cent on the same base. Proposed interconnections are shown in broken lines.
FIG.
STUDY 4-DESCRIPTION OF SYSTEMS
313
AIC
H DA
~
(I:U~:~)
DA 110 kv
'i.
~
N
3.28
DB HOb -
DC 4.15 ltv. N
In
::
3.28
~
~
.;-
it
"!
~ C?
~
fIi
cd
;-
,.,;
+'
III
....,.,;
N
2 C
7~
",
~
aMva.
1.!9
:;-
DC 110 kv.
AG
138 ltv.
I
iii
+ ~ ltl
;-
AC
0;
~ 0
~
~ It' N N
2
.;-
~
s
;~
cd
AA
::
~
~
~
AJ
:;-
II' IQ IQ
~+
IE
~
AD
13.8
N
~ .,
20 Mva.
+ In
... 'It
C?
~
40 Mva.
H
DE 110 kv
R
D F69 kv.
/'Ii
:;-
,...
~ 25
Mvt. G
9.92
DF 110kv
SA 601tv
1210
8j:t
System C equlVlltnt
+
all ~
FIG.
CA 1l0kv
~
N
46. Study 4, part 2. Impedance diagram with values in per cent on 25-Mva. base. Line capacitance values are in microfarads on calculating board.
314
TYPICAL STABILITY STUDIES
were situated and fed a number of smaller towns through a 60-kv. transmission system. The other part, shown on the right-hand side, had one steam-electric plant (station BD, 50 Mva.) feeding a number of substations through 66-kv. radial transmission lines. These two parts were joined through a two-circuit 132-kv. line about 100 miles long from substation BC to station BD. The system of company B was: interconnected at substation BA to the system of company F and at station BD to the system of company G. Company C, having an extensive system with both steam and hydroelectric plants, was interconnected with several other large systems in adjacent states. It is represented in Figs. 39 and 40 as an equivalent generator on the bus of substation CA. Company D operated system (shown in Fig. 40) which included several hydroelectric stations (DA, DE, DF, DG, and DH) of capacities ranging from 8 to 40 Mva. and a I IO-kv. transmission system connecting these generating stations to one another, to substations, and to companies A and E. This system extended several hundred miles from the city served by company A. Company E had a 28-Mva. steam-electric plant located in a city about 60 miles distant from company A. Company F had a large system; with both steam and hydroelectric plants. Of these only station FA, the plant nearest system C, was represented individually in the study, the remainder of the system being represented by a single equivalent generator (shown in lower lefthand corner of Fig. 39). Company G had steam plants at GB and GD and' a hydroelectric plant at GC. It was interconnected with company B at station BD (lower right-hand corner of Fig. 39). Company H operated a metropolitan system, situated about half way between companies A and B but not connected to them. Its generating capacity was approximately 275 Mva. (See Fig. 40.) A list of synchronous machines of companies A, B, D, E, and G will be found in Table 9. As the systems of companies C, F, and H were not represented in detail, their machines are not listed. Proposed interconnecting lines. Two lines were proposed, the first connecting company A to company B, the second connecting company B to company C. Both are shown by dashed lines in Fig. 39. For the first line two alternative routes were studied, one from substation AB to station BD, the other from substation AB to substation Be. The lengths of these routes are about 280 and 244 miles, respectively. The second line was to have terminals at BD and CA, and a length of about 256 miles.
a
STUDY 4-PROPOSED INTERCONNECTING LINES
315
TABLE 9 LIST OF SYNCHRONOUS MACHINES (STUDY
Number Station of Like Kind of Machine Units
AA " "u
1
Steam
1
H
1 1
u
u
1
u
" H
,
Rating kv.
13.2
" "
13.8 u
Total
4)
Total
Kinetic Energy
Xd
WR2
(%)
(kilo-lb-
18,750 1,800 H 25,000 H 25,000 31,250 " 12,500 3,600 112,500
28.0 24.0 21.0 11.0
"
140 207 237 237 15.5
104.8 154.0 176.4 176.4 46.4 658.0
6,250 3,600
14.0
10.2
29.9
kva.
r.p.m.
ft. 2 )
(Mj.)
AJ
1
Steam
BA
1
Condo
4.16
3,500
720
29.0
38.9
4.7
1
Condo
12.47
H
"
3,750 3,750 7,500
720
1
29.1 32.0
38.9 106.3
4.7 12.7 17.4
BC u Ie
BD
"
Total
1 1
Ie
12.0
"
"
18,750 1,800 31,250 " 50,000
18.0 22.0
102 235
12.5
25,000 3,600
11.0
62*
186
7,500 3,600 12,500 1,800 32,500
15.8 19.0
19* 85.3*
56.8 64 184.8
14,000 42,000
150
36.0
Total
1
Steam
BO
1 2
Steam
DA
"
Steam
BF
" "
13.8
Ie
4.16
"
Total
13.8
76.4 176 -252.4
38.8 116.4
3
Hydro Total
DC
1
Condo
4.15
8,000
900
34.9
DE
2
Hydro Total
6.9
10,000 20,000
180
35.0 4,650
34.8 69.6
DFl DF2 DF
2 1
Hydro
6.9
10,000
"
u
20,000
180 164
35.0 4,650 42.0 9,870
34.8 60.8 130.4
DG
3
H
a
" *Estimated.
40.5
40,000
Total Hydro Total
7,500
6.9
"
2,750 8,250
112.5 39.0
1,250
7.6
3.6
-10.8
TYPICAL STABILITY STUDIES
316
TABLE 9 (Continued) Number Station of Like Kind of Units Machine
Rating
kv.
kva.
r.p.m.
,
Total
(%)
Kinetic Energy (kilo-Ib- (Mj.) ft. 2 )
Zd
WR2
2
Hydro Total
13.8
14,500 29,000
257
32.0
1,950
29.6 -59.2
DI
1
Condo
13.8
15,000
900
35.7
93
17.4
EA
2
Steam
4.15
II
1 1
"
"u
8,575 3,600 II 6,250 5,000 " 28,400
12.0 14.0 16.0
19.6 12.4 10.0
58.4 37.2
900
35.1
38.0
DH II
II
" EA
Total
1
FA
II
7,500
162.4 31,250 1,800 33,000 300 64,250
19.0 36.0
414 6,200
164
32.0
1,010
Steam Freq.chr. Total
4
Hydro Total
5,000 20,000
1
Steam
15,625
II
GD
4.15
1 1
II
GO
Condo Steam Total
II
GB
"
29.9 -183.9
7.1 121
23.0
Both interconnecting lines were to be of 154-kv. single-circuit construction, with hollow 250,OOO-circular-mil copper conductors of O.766-inch diameter, supported by wooden H frames. The equivalent spacing between phase conductors would be 17 ft. Ground wires would be installed to protect the line against lightning. The necessary terminal equipment would include transformers, shunt reactors or synchronous condensers, circuit breakers, and protective relays. The transformers at substation AB would have three windings: high voltage, 154 kv. Y; low voltage, 66 kv. Y; and tertiary, 13.8 kv. ~. Those at the other end of the A-to-B line (at either BC or BD, depending upon the route selected) would be 154-to-132 kv. Y...connected autotransformers with 12.5-kv. d-connected tertiary windings. For the B-to-C line the transformers at station BD would be autotransformers similar to those on the A-to-B line, and, even if the terminals
STUDY 4-SCOPE OF THE STUDY
317
of both lines were to be at station BD, it was assumed that separate transformers would be used for each line. At substation CA 154-to110-kv. autotransformers with tertiary windings would be required. The tertiary windings of these transformer banks would be used for connection of shunt reactors or synchronous condensers to supply the reactive power needed to hold the voltages of the terminal busses at suitable values. They would also provide a path for zero-sequence current. To attain reliable transmission of power over the proposed singlecircuit interconnections, without the aid of any parallel ties between the systems, it would be necessary to use high-speed clearing of faults on these lines and high-speed reclosure of terminal breakers. Obtainable circuit breakers had an opening time of 5 cycles and a reclosing time of 20 cycles with three-pole operation. The assumption of I-cycle carrier relaying gave an assumed fault-clearing time of 6 cycles and a reclosing time of 21 cycles, measured from the instant of occurrence of the fault. Scope of the study. The study included power-flow and transientstability studies. The power-flow studies served to determine: (1) approximate steady-state power limits of the interconnections, (2) the reactive power which had to be supplied by reactors or synchronous condensers at the terminals of the proposed lines and elsewhere, (3) the required capacity of the terminal transformers, (4) the need for additional generating, transformer, or transmission-line capacity within the several systems, and (5) initial conditions for the transient-stability studies. Transient-stability studies served to determine: (1) the transient power limits of the interconnections for faults on the interconnecting lines, and (2) the required clearing times of faults on other lines to prevent impairing the power capacity so determined. The study was divided into two parts, made at different times and with different a-c. calculating boards. Part 1 dealt with the power capacity of the A-to-B and B-to-C interconnections for power flow in each direction, also with the choice between BC and BD as southern terminal of the A..to-B line; it dealt also with the effect on stability of interconnecting systems F and G to system B, and with various internal problems of systems Band C arising from the proposed interconnections. For brevity those portions of the study pertaining primarily to system C and to the B-to-C line are omitted from the present report. Part 2 was concerned with internal problems of system A, including the required speed of clearing of faults on lines within that system, and also with the effect on stability of connecting system H to a tap on the A-to-B line.
318
TYPICAL STABILITY STUDIES
Loads. Estimated July and December peak loads three years after date of study were used. The July loads have practically the same active-power component as the December loads but have higher reactive-power components. Method of determining power limits. Approximate steady-state momentary power limits of the interconnections were found by increasing the electrical angle between bus voltages at major generating stations of the different power systems to approximately 45°. The practical steady-state power capacity was taken as 5 Mw. below the value so found, to allow for momentary fluctuations of about 5 Mw. above the average power carried by the line. Such fluctuations could be caused by sudden changes in loads or by faults at distant points. Transient-stability power limits were found principally by obtaining swing curves by the point-by-point method in order to determine whether the system would be stable or unstable with various values of initial power flow if a two-wire-to-ground fault should occur on the interconnecting line, followed by 6-cycle clearing and 21-cycle reclosure. The effects of faults on other lines were investigated in similar fashion, except that three-phase faults were assumed on overhead lines without ground "vires, on double-circuit lines, and on cables, and that various clearing times were assumed. The number of machines represented in these transient-stability runs varied from four to ten. In two instances the number of machines was reduced to two, power-angle curves were obtained by use of a calculating board, and curves of power limit versus switching time were then computed. The rated power capacity of the interconnecting lines was taken as 5 Mw. lower than the transient power limit, to allow for fluctuations. Simplification of systems. In representing the systems only major stations, substations, and transmission tie lines were included. Radial lines were represented as loads on the busses from which they radiated. For example, the entire 66-kv. system supplied from station BD was represented as a load on the BD 66-kv. bus. Lines were represented by nominal sr circuits, the capacitances of all lines on a given bus being represented by one capacitor on that bus. Proposed interconnecting lines were represented by several sections, depending upon location of proposed taps to other systems. Three-circuit transformers were represented by their equivalent Y circuits. If one branch of such a circuit had negative reactance (as indicated, for example, in Fig. 39 on the 66-kv. side of the transformer at substation AB), the following change was made in setting up the circuit on the calculating board: The branch with negative reactance was set up with zero reactance, and its negative reactance was com-
STUDY 4-SIMPLIFICATION OF SYSTEMS
319
bined with the greater positive reactance of the opposite branch, thus preserving the correct value of through reactance from primary to secondary terminals. TABLE 10 GROUPING OF SYNCHRONOUS MACHINES FOR REPRESENTATION BY POWER SOURCES ON THE CALCULATING BOARD (STUDY
4)
Power Source Numbers Station
DE, DF, DH
Run
Run
Run
Run
Run
Run
Run
Runs
1
2
3
4
5
6
7
8 to 11
1
1
1
I
1
1
- - --
1
DA,DG
2
EA
3
AJ
- -- - - -- - - - - - - - -l '- -- -- 2
2
2
3
3
3
4
AA (south bus)
BD BF
5
BG
7
System C
6
00*
GB GC,GD FA
Off
Rest of system F System H
1
1
2
2
3
3
4
4
5
5
6
6 --
00
00
Off
Off
- - - - - - - - - - - - - - --
- - --
- -- - - 2
AA (north bus)
Run
4
4
2
2
4
- - --
- - --
- -- -- -- - - -- -- 3 5 5 3 3 7 5 7 - - - - - - - - - - - - - - -Off 6 Off -- 6 ---- 4 8 4 8 7 4 6 - - - - - - - - - - - - - - -- - - - - - - - - - - -- - I 00
00
8
7
00
---9 8 ---9 Off 10
------00
Off
00
00
5
5
-6
--
00
Off
-6 -7 --
00
00
00
Off
-Off
-Off
-9
* 00 = Infinite bus. Synchronous condensers were not represented dynamically during stability studies, but merely by capacitors or reactors. System C was represented by an equivalent generator on the substation CA bus. System F, exclusive of station FA, which was shown individually, was represented by an equivalent generator. In stability studies the equivalent generators at CA and FA were regarded as infinite. Station DG was combined with station DA. Stations GC and GD were represented by an equivalent generator on the sub-
320
TYPICAL STABILITY STUDIES
station GA bus. In part 1 of the study the portion of system D beyond substation DD was represented as an equivalent generator on the DD bus, and station AJ was combined with station AA. In part 2 the portion of system B near stations BF and BG (including transformers, lines, and loads, but not generators) was represented by an equivalent circuit (shown in Fig. 40), the terminals of which were: station BF 12.5-kv. bus, station BG 4.16-kv. bus, substation Be 132-kv. bus, and substation BA 60-kv. bus. Since generators were not included in this network, it was possible to change the generator reactances according to the number of machines assumed in operation at each station and according to whether adjusted synchronous reactances were wanted for steady-state runs or transient reactances for transient-stability runs. In many of the transient-stability runs further combinations of generators were made, as shown in Table 10; for example, stations BG and BF were combined in some runs. Swing curves, part 1. Twelve transient-stability runs were made in the part of the study considered. The swing curves are reproduced in Figs. 41 to 52, inclusive, and the conditions under which they were taken and the resulting conclusions concerning the stability or instability of each case are summarized in Table 11. The grouping of generators for each stability run is shown in Table 10. Each run will now be discussed. Run 1. The company B terminal of the A-to-B line was assumed to be at substation BC. As a step in finding the power limit of this line for northward power flow, the power sources on the board were adjusted to give 39.6 Mw. of power received at substation AB from Be. To supply this power, 13 Mw. was transmitted to system B from system C, and a new generating unit was assumed installed and operating at station BF, resulting in a need also of 20 Mva. of additional transformer capacity at BF. A two-line-to-ground fault was assumed to occur on the 154-kv. A-to-B line near the sending end. The assumed clearing and reclosing times were 6 and 21 cycles, respectively. The sending end of any line is ordinarily the worst fault location on the line from the standpoint of stability; but" since acceleration of the generators during the 6 cyclesin which the fault is on is small compared with that during the following 15 cycles In which the line is open and the systems are completely separated, a fault of given type anywhere on the line would have substantially the same effect on stability as a fault at the sending end. Inspection of the swing curves (Fig. 41) shows that the machines of system B swing approximately together and that they speed up. The speeding up is due to their dropping load during the disturbance. The machines of systems A, D, and E also swing
BC CA
AB BD AB
2LG
"
AB
"
"
35.4
"
" "
45.0
"
u
"u
+
2LG
" " "
3q,
AB
CA
"
BD
BD
154 kv.
AA(S) AA(N) AB
"
AA(N)
13.8 kv.
" " "
"
BD
u
AB
BD
"
"
" u
154 kv.
132 kv.
"
BC
Near Station
d/lo"
Location of Fault
Line Voltage 154 kv.
,
AF BD
" t
AF
"
AB
"
Be
Be
u
AB
"
18 18 6
12 18
" " "
6
9
"
6
A
"
21
.. ·. ·.
·.
" "
II
21
30
"
Reclosing 21
Switching Time (cycles)
4)
Clearing
r
(STUDY
On Line to
TABLE 11 STABILITY RUNS
*Systems F and G connected to system B with 0 + }Opower interchange. t Fault 1,000 ft. from station AA on 700,000 eir .-mil cable. tsystem H connected with 0 jO power interchange.
10 11 12
" " u
II
8
25.5
BD BD
12.4 26.0
9
AB AB
30.0
40.6
6 7
II
CA
3q,-3q,
5
40.4
58.0
AB
4
3
BD
"
2LG
Type of Fault
BC BD Be
Be
From
Rec'd at
-a
CA BC
13.0 45.6 21.4 40.0
39.6
Mw.
A.
Power Transfer
BD
2
1
Run
r
SUMMARY OF
Stable Unstable Stable Unstable Unstable
Stable but critical Unstable Unstable Stable Stable
Unstable
Stable
Stable or Unstable
~
AA bus split
•
•
*
*
Remarks ~
tr:
~ to-
.....
~
>-
t-d
..til
;3
:;:0
~
a
Z 0
~ ~
in
~
,
Jo<
tj
d
322
TYPICAL STABILITY STUDIES
together, but they slow down because they take on additional load during the disturbance to make up in their systems for the power which was received from system B. System C was represented as an infinite bus and, therefore, did not swing. Station BD, which was closer to system C than were stations BF and BG, swung less than did stations BF and BG, as though it were attracted by system C. The curves clearly show that the system was stable. 80
r-------,----.-----r---r-~-...._-_r_____,r____,..-__T'"-.....,.
I
"Of ,.... en
t...
~I~
~l~ ~I\O ;1.5 L.l. t I
40
~ Q)
"0 Q)
~ e ta
I
0
Q) ~
ta %! 0 :>
ti c
...
.!
..
-40
.5 cQ)
·Cii
c
l!'
to-
-80
- 120
1 (DE, DF. DH)
'-----a_.....&._~-....._'__
o
0.2
0.4
0.6
........
~
0.8
....
1.0
Time (seconds) FIG. 41. Swing curves, Study 4, part 1, stability run 1. 39.6 Mw. received at substation AB from substation Be, and 13.0 Mw. received at station BD from substation CA. Two-line-to-ground fault-near substation BC on proposed 154-kv. line to substation AB, cleared in 6 cycles, line reclosed in 21 cycles. Stable.
Run S. The power received at substation AB was increased to 45.6 Mw. To supply this increased power, the power sent to system B from system C was increased to 21.4 Mw., while the outputs of stations BD, BF, and BG were left unchanged. Systems F and G were connected to system B with zero interchange of active and reactive power. With the same fault location and with switching times the same as those in the preceding run, the system proved unstable. Therefore the power limit oj the A-to-B line for northward r'power flow is between J,.O and J,.5 Mw. received. The rated power capacity of the line may be taken as 35 Mw, Study of the swing curves (Fig. 42) shows that the machines
STUDY 4-SWING CURVES, PART 1
323
of systems Band G and of station FA swung ahead together. The rest of system F was represented as one infinite bus, and system C was represented as another infinite bus. The generators of system B did not swing as far ahead of system C in this run as they did in the preceding run, in spite of the increase in power transfer. This fact shows 80 ..--.....----r---,...__--r--_,._-_-
- __- _ -__
40
-... en
G) G)
r
"0
0
G)
00 e co
Wo .19
'0 >
System F as infinite bus
-40
n;
...e
.!
.5
't: -80 G) 'Cij
c
e
~
-120
- 160 '"'---..- --'--...L-....-"'---..L_--L.._-'-_..L-_I-....--J._...L.........J o 0.2 0.4 0.6 0.8 1.0 1.2 Time (seconds)
42. Swing curves, Study 4, part 1, stability run 2. 45.6 Mw. received at substation AB from substation BO, and 21.4 Mw. received at station BD from substation CA. Systems F and G connected to system C with 0 + jO power interchange. Two-line-to-ground fault near substation Be on proposed 154-kv. line to substation AB, cleared in 6 cycles, line reclosed in 21 cycles. Unstable. FIG.
that the connection of systems F and G to system B has a beneficial effect on stability. Nevertheless, the generators of systems A, D, and E, again swinging together, swung farther behind than previously and pulled out of step with the other generators. Hence the power limit, either with or without the connection of systems F and G to system B, is between 40 and 45 Mw., and it may be concluded that the connection of those systems has only a small effect. Run 3. One consequenceof selecting substation Be as the company
TYPICAL STABILITY STUDIES
324
B terminal of the A-to-B line, while station BD is the terminal of the B-to-C line, is that, when it is desired to transfer power from company C to company A, this power must flow over the double-circuit 132-kv. 120__......__---t"---.--......--_-__-....--__-
__
-~-_
~I=
j.~
80
~I~
;1. 5
'-
-i f -i
I
40
-a
...
0
f
System F
....co c 11 -40 ~5
.. c
.I! en
c
I!
to-
-80
- 120
t---+-..--+---+---f---of---+----f~lE_FI~~-
""""""-_......._ ....... -...l"----""_-'"'-_....._.-..-_.a..-..... 0.4 0.6 0.8 1.0 1.2 Time (seconds) FIG. 43. Swing curves, Study 4, part 1, stability run 3. 40.0 Mw. received at substation AB from substation BC, 58.0 Mw. receivedat substation BC from station BD, and 40.4 Mw. received at station BD from substation CA. Systems F and G connected to system C with 0 jO power interchange. Double-circuit threephase fault near station BD on the 132-kv. lines to substation Be, cleared in 9 cycles, lines reclosedin 30 cycles. Stable but critical. - 160 ..........-"'"
o
0.2
+
line from BD to BC, which at the same time may be required to carry power from station BD to loads in the left-hand portion of system B. To test stability under such conditions, run 3 was made. Generation was adjusted to send 40.0 Mw. (slightly below the power limit found above) to AB from BC and 40.4 Mw. to BD from CA. At the same
STUDY 4-SWING CURVES, PART 1
325
time station BD was generating more than was required by the load in the adjacent part of System B, and the surplus was transmitted to the distant part, resulting in a total power of 58.0 Mw. received at substation BC over the 132-kv. lines from BD. No new generators were assumed to be running. Systems F and G were connected to system B with zero interchange. A three-phase fault involving both circuits] of the 132-kv. line to Be was assumed to occur near BD and to be cleared in 9 cyclesby simultaneous opening of the breakers at both ends of both circuits, followed by their reclosure in 30 cycles. The swing curves (Fig. 43) show that the system was stable ·but critical. The machines of systems A, D, and E, again swinging together, sloweddown and almost, but not quite, pulled out of step with the other machines. Stations BD, GB, GC, and GD (which are on the sending side of the opened line) swung ahead and then oscillated; of these stations, BD, which is nearest the fault, swung ahead most rapidly. Stations BF and BG slowed down at first but speeded up after reclosure of the faulted lines. The power limit was slightly over 58 Mw. on the BC-BD lines or slightly over 40 Mw, on the AB-BC line. The latter figure is close to the limit already found for a fault on the AB-BC line and is therefore satisfactory. If the fault should involve only one circuit, the power limit would be higher. Run 4. It was next decided to investigate the power limit of the A-to-B line when carrying power southward (from A to B). The power sources were adjusted to give 35.4 Mw. received at Be from AB. To supply this amount of power, the output of station AA generators was increased to 22 Mw. above the rated capacity of the station, requiring that a new generating unit be assumed installed and operating there; and, in addition, 30.5 Mw. was received at substation AB from system D. To absorb this amount of power in system B, station BF was shut down. The B-to-C line was closed with zero active and reactive power interchange as measured at station BD 132-kv. bus. Systems F and G were not connected. Additional transformer capacity was required as follows: 13 Mva. at station AA, 5 Mva, at substation AB, and 12 Mva. at substation Be. A two-line-to-ground fault was assumed to occur near AB on the proposed 154-kv. A-to-B line, with the usual switching times. The swing curves (Fig. 44) show the system to be unstable. Systems A, D, E and system B, although initially swinging apart, came back together and remained in synchronism, but both these groups pulled out of step with system C. [Such a fault might consist either of a three-phase fault on each circuit or of 8, one-line-to-ground fault on one circuit and a two-line-to-ground fault on the remaining phases of the other circuit.
326
TYPICAL STABILITY STUDIES
Run 5. Conditions in this run were identical to those in the last ron except that systems F and G were connected to system B with zero interchange. The combined system was again unstable (see Fig. 45), but this time systems Band C stayed in step with each other, whereas 320 ---------....--r------r----r---y---.--.---r--,
en
~I
~.f-----+----I \0 .sl -;; 240
e ';1
-
...., ...---+---1
e r
u
"'0
~I
CD
J
ftS
200 1 - - - 1 - - - 4 - - - 1
f
Ci
E 160 1----I---+-~~-!....-4--~-+--+_-+_-..---+__#_-+---+-----t
~
C CD
'iii
c
~ 120 1--,..,~~---+---JC.-+---+--4----+--+_-+_-+-i__l_-+---+---t---i System C 80
as infinite bus
I----J---+-~~--+--+---+-----:~-J#_---+---r--T----t---t
40 '--......_.-L._-A--'-...&.-_...-_.l..--.lIo--..Io_--'-_~_.....- - - '
o
0.2
0.4
0.6
0.8
1.0
1.2
Time (seconds) FIG. 44. Swing curves, Study 4, part 1, stability run 4. 35.4 M w. received at substation BC from substation AB. Two-line-to-ground fault near substation AB on proposed 154-kv. line to substation BC, cleared in 6 cycles, line reclosed in 21 cycles. Unstable,
systems A, D, E pulled ahead and out of step with Band C. The motion of the A, D, E machines was almost exactly the same in both runs, but the motion of the system B machines was greatly reduced by the connection of systems F and G. The power limit of the A-to-B line for southward power flow was less than 35 Mw. whether systems F and G were connected or not, and was estimated as being between
327
STUDY 4-SWING CURVES, PART 1
30 and 35 Mw.
The practical power capacity was taken as 25 Mw.
Note that the transient-stability limit for southward flow was less than that
for northward flow on the same line. Run 6. The company B terminal of the A-to-B line was now shifted from BC to BD, and northward power transfer was investigated. 300
I
280
I I
i
I
"2' Cl)1
I
~I~
~,
-
ar....
240 ~~I ~I
~
I
I-
~I
I
~ 200 f - "til
fco
8L
I
~ca 160
j
1: QI
.~ 120
...e
80
I
LV ;
V V
1--'
----
I
I
I
o
~I
I V
I
YI
V
/
System C as infinitebus
yo ~
. L~
-.L
V
~~
I~ ,,-I 4 (BG).
I
(AA. AJ, EA)
.~
(3(B~ ~~
I
0.2
14"
V 5 (~~stem G~
I
-
~
~
V i/~
I
I
~
I
40
Y
I
rr
V
V
I
~,
~
./
I
ml
"31
,,-
~'I
....
V
V /
/
~t~
~I
"0
V
1 (systemDl')
-8r~ cu_
I
V
/
""",-
r---""'"
_~
V
~ ~ ~ ""7
~V
""
6(FA)
.
V
System F as Infinite bus
!
0.4
0.6 Titne (seconds)
0.8
1.0
1.2
FIG. 45. Swing curves, Study 4,part 1, stability run 5. Same as run 4 except that systems F and G were connected to system C with 0 + jO power interchange. Again unstable.
First the value of 30 Mw. received at AB from BD was tried. Station BF was generating 54 Mw. with the new generator operating, and 23 M va, of additional transformer capacity was required there. No power was received from system C, and systems F and G were disconnected. Generation at station AA was reduced to such an extent that to supply its local load necessitated the receipt of power from system D as well as from system B and required 11 Mva. of additional trans-
TYPICAL STABILITY STUDIES
328
former capacity at station AA (slightly less than the amount required in run 4). A two-line-to-ground fault was applied to the A-to-B line at the sending end (station BD), cleared in 6 cycles, and reclosed in 21 cycles. The system was stable; all the swings were of small amplitude. (See Fig. 46.) Run 7. The power transfer on the A-to-B line was increased to 40.6 Mw. received at substation AB from station BD. To provide the increased power, 12.4 Mw. was received at BD from system C. To absorb the increased power, the generation at station AA was further 200 , . - - . . , . - _ , _ - - . . - - - y - - - r - - - . , . - - - - - , - - . . , . . - - - . - - - - - r -_ _
Q)
"60 e
«S
~
B g
co
E .!
.5
~
'in
120 ~~-""'-
L....
1 (System D) 80 t--_--+----+-~_+__-+_~---+--~--+-__t
c
cu
.=
t----t---I--~IIIIIIIl:-_+__-+_~~ 2 (AA, AJ, EA)
40 a-----'_--.L.._.....-'--......_ " " - - - - " _ . - . . _......._ ~_ _ o 0.2 0.4 0.6 0.8 1.0 Time (seconds}
46. Swing curves, Study 4, part 1, stability run 6. 30.0 Mw. received at substation AB from station BD. Two-line-to-ground fault near station BD on proposed 154-kv.line to substation AB, cleared in 6 cycles, line reclosed in 21 cycles. Stable. FIG.
reduced. Power flow conditions were much like those of run 1 except that the generation at AA was smaller, the difference being made up by power received from system D. Also, systems F and G were connected to B. The interconnected systems were stable (Fig. 47), with considerably more margin than there was in run 1 (Fig. 41). The improvement might be due in part to the change in company B terminal and in part to the connection of systems F and G. The power limit of the A-to-B line with terminal-at BD was not exactly determined, but it appeared to be somewhat higher than the corresponding limit with line terminal at Be, at least for northward power flow. Swing curves, part 2. In part 2 of the study the required clearing time of faults on company A's 13.8-kv. system and the effect on stabil-
STUDY 4-SWING CURVES, PART 2
329
ity of an interconnection with company H were investigated. The network of Fig. 40 was set up on the calculating board. As compared with the set-up of part 1 (Fig. 39), the principal changes were: (a) The base was changed from 40 Mva. to 25 Mva. (b) The l3.8-kv. network of company A was represented in detail, including the separation of station AA bus into two sections by a bus reactor, with the generators of each section represented separately, and the separation 200
r----r---,.---r---r---.....---...r----r-_r_-.......--.....---,
- 160 I----I---+---+--.a..--t-+---+----t-- 6 (Ge, GD)
~
I
~ '0 -
5 (GBl,..,...--+----t 7·(FA~)
I ~ 1 ~ - - t ' - - - t ' - -.....- - + - - - - - t - - i - - - f - - - t - - - - f ml"U ul~
3 t ~ -+---+--+--.--+----+----t---..+---+--t ~
1'-
O--......-~- ..........I.-..&.-_"--.......-
o
0.2
0.4
0.6
......_ - - - - - -.....
0.8
1.0
Time (seconds) FIG. 47. Swing curves, Study 4, part 1, stability run 7. 40.6 Mw. received at substation AB from station BD and 12.4 Mw. received at station BD from substation CA. Systems F and G connected to system C with 0 + jO power interchange. Two-line-to-ground fault near station BD on proposed 154-kv. line to substation AR, cleared in 6 cycles, line reclosed in 21 cycles. Stable.
of station AJ from station AA. The 13.8/66-kv. transformers at station AA were assumed to be of 25-Mva. capacity each, or 50 Mva, total, as found necessary in part 1. (c) The remote part of system D was shown in more detail. (d) The part of system B containing stations BF and BG was represented by an equivalent circuit. The company B terminal of the A-to-B line was placed at -station BD. Systems F and G were not connected. Zero active power and about 17 Mva, reactive power were supplied to system B from system C. Run 8. First the required clearing time of faults on the 13.8-kv.
TYPICAL STABILITY STUDIES
330
system of company A was examined. The A-to-B interconnection was delivering 26 Mw. to station BD (approximately the practical power capacity of the interconnection as found in part 1). The output of station AA generators was 101.5 + j40.7 Mva. Since this output exceeded the capacity of the station, a new generator was assumed on the north bus, and the aggregate capacity was divided equally 160 ...----r---.--r---r-.,...--..,r-----r---r----,r----,--.,.--.,---, 140
120
en Q)
e t)I) Q)
~ 100 Q)
1io c: co
~
80 t--+----I-t--++--~~~-+--f-~f---+---+-----f--4
.B
g co
E
60
t---+~_t_-hr--.L-.-f"'"___I__r_~r---+--t-~t:::::;a.....Iiiii:::+---t~......
40
t--:3IIlI'f'"""'---::I~~t---#l--_t_-~--t-__:l~-,t----+---+-----f~r-f
20
~~~_f_--#t----t-~Lt._-~lL-+--+--..--+--f----I---1
.s
.5 C Q)
'Vi c
~
- 20 "'-
o
System as infinite bus ..-..-_""----'-_........_"--...... _ ........C _Il.o.1.0 0,4 0.6 0.8 0.2
.....-.--'_~
1.2
Time (seconds) FIG. 48. Swing curves, Study 4, part 2, stability run 8. 26.0 Mw. received at station BD from substation AB. Three-phase fault on 13.8-kv. feeder near station AA north bus, cleared in 12 cycles. Stable.
between the two busses. The voltages of both bus sections were held at the same magnitude, but a phase difference between them caused 5 Mw. to be transferred from the south to the north bus. From system D 25 Mw. was delivered to system A. One machine was assumed to be operating in each system D plant except station DA, where all machines were assumed to be in service. At station EA 15 Mva. of additional 110/66-kv. transformer capacity was required. A three-phase fault was assumed to occur near station AA on a 13.8-kv. underground cable feeder supplied from the north bus. With 12-cycle clearing the system was stable. (See swing curves, Fig. 48.)
STUDY 4-SWING CURVES, PART 2
331
260
240 220
200
180
,...
Ir
160
'0 GJ
co c 140 to
InI
g
120
Ci c
"G) .., lOa
.5
~
c: Q)
'iii
c
e
80
to-
60 40
20 0 - 20 .........._-'----a_--'-_...A..---"_........ ..... o 0.2 0.4 0.6 0.8 Time (seconds)
.....
_~
FIG.
1.0
....
1.2
49. Swing curves, Study 4, part 2, stability run 9. Same as run 8, except that clearing time was increased to 18 cycles. Unstable.
Run 9. With I8-cycle clearing of the same fault the systems were unstable (Fig. 49). Run 10. The purpose of this run was to show the effect of threephase faults farther away from station AA. The location selected was a point about 1,000 ft. from station AA on a 700,OOO-circular-mil cable supplied from the south bus. With the additional impedance of this cable between station AA and the fault, the voltage at AA re-
332
TYPICAL STABILITY STUDIES
mained high enough to prevent the generator output from decreasing as much as before, so that even with I8-cycle clearing the system was stable (see Fig. 50). The extreme oscillations of the station AJ machine, as shown by its swing curve, should be discounted somewhat, as they did not represent the true motion of the machine but were the 200 r----.---r-.-......-......-r--,---r---r--.----.~__y_o-~__..
180 Xli
i
160
~I
f"04
! ..... 4D
"0
~
140
!
.;1 ~I
120
~I
.!! co
I
c
tV
I» 100
~
g
.
80
c
60
to c
~ .., .~
en C
~
40 20 0
- 20
.......
'--.-.._-L---a._~_"--__'__-'--.._
o
0.2
0.4
0.6
0.8
- . r . . _.....
1.0
1.2
Time (seconds) FIG.
50. Swing curves, Study 4, part 2, stability run 10. Same as run 9 except
that fault was 1,000 ft. from station AA south bus on 700,OOO-cir.-mil cable.
Stable.
result of inaccuracies caused by the use in the step-by-step calculations of a time interval too long in comparison to the period of oscillation of the machine. Run 11. The purpose of this run was to show the effect of removing the bus-tie reactor at station AA, leaving the north and south busses tied together through only the I3.8/66~kv. transformers and the 13.8kv. network. In every other respect it was identical to run 9. Com-
STUDY 4-SWING CURVES, PART 2
333
parison of the swing curves of run 11 (Fig. 51) with those of ron 9 (Fig. 49) shows that splitting the bus at AA decreased the rate at which systems A and B swung apart, but not sufficiently to allow stability to be regained. Although this run was considered unstable, 220 ------.....------,r-----r---w----r-.....-----,---t--r----, 200 180
160 ~
en
Q,)
~
1)0
-
140
Q,)
"'C
Q,)
ao e 120 to
~
S
g
100
-;; E .!
80
.5
....c Q)
.iii c:
...e
60 40 20
0 - 20 ~---------"'-...._I..---L.-...L-.--J._~_a..._.......,L.._.l....---'
o
FIG.
0.2
0.4
0.6 0.8 Time ( seconds)
1.0
1.2
51. Swing curves, Study 4, part 2, stability run 11. Same as run 9 except that station AA bus was split. Still unstable.
it represented a borderline condition, indicatingthat the systems would be stable under slightly more favorable conditions, such as appreciable arc impedance in the fault or decreased power transfer on the interconnection. Run 12. The purpose of this run was to show the effect on stability of connecting system H to the A-to-B line. System H was a metro-
TYPICAL STABILITY STUDIES
334
politansystem located near the center of the proposed interconnecting line from AR to RD. It was represented as a 275-Mva. generator tied to the 154-kv. line through a 30-Mva. transformer, with zero interchange of active and reactive power. Power of 45 Mw. was trans180 160 140
,.... 120
~
if»
"".....
I., c &
~ ~
Ci
E
f
l S
100
80 60 40
·Iii c:
20 0 ~20
-40 0
0.2
0.4
0.6 0.8 Time (seconds)
1.0
1.2
FIG. 52. Swing curves, Study 4, part 2, stability run 12. 45.0 Mw. received at station BD from substation AB. System H connected to the A-to-B line with 0+;0 power interchange. Two-line-to-ground fault near substation AB on the proposed 154-kv. line to station BD, cleared in 6 cycles, line reclosed in 21 cycles. Unstable.
mitted to BD from AB, necessitating a new generator on AA north bus and requiring only one machine to be operated at BD. The power delivered to system A from system D was raised to 37.4 Mw. by increasing the output of the system D generators already in operation. With a two-line-to-ground fault at the sending end of the 154-kv. inter-
STUDY 4-POWER-ANGLE CURVES, PART 2
335
connecting line and with the usual clearing and reclosing times" the systems were unstable. The swing curves (Fig. 52) show that systems A, D, and E swung ahead together, whereas systems B, C, and H swung very little. The power limit of the A-to-B line for southward power transfer with system H connected to the line was thus shown to be less than 45 Mw. This limit was found more exactly as described below. 240
-V I
220
/
200
I
/"
I I
Q)
00
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I~
/1
l:c
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I
A,D,E, no fault
'"
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, I I
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80 -100
~Systems
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i20
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1
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~
-60
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~Systems
/ ""- A,D,E, --...... fault on
-40 -20 0 20 Rotor electrical angle (desrees)
40
60
80
FIG. 53. Power-anglecurves, Study 4, part 2, run 13. 5 Mw. receivedat station BD from substation AB. Three-phase fault on 13.8-kv. feeder near station AA north bus. Critical clearing angle, determined by equal-area method, is -34°; critical clearing time, 24 cycles.
Power-angle curves, part 2. Power-angle curves and the equal-area criterion were used for (1) obtaining a curve showing the critical clearing time of a three-phase fault close to the north bus of station AA required to maintain stability of systems A, D, E with systems Band C for any amount of power transferred from A to B; (2) determining the power limit of the A-to-B line for southward transfer with system H connected. Critical·clearing time of a three-phase fault near 8tation AA 19.8-kv.
TYPICAL STABILITY STUDIES
336
north bus. To use the method mentioned it was necessary to reduce
the system to an equivalent two-machine system. Inspection of the swing curves for runs 8 to 11 showed that the machines of systems A, D, and E swung together and, therefore, could be combined with very little error to one equivalent machine. The machines of system B likewise swung together and could be combined, and system C was regarded as an infinite bus. It was decided to combine the machines 240 220 200 cJ)
t:: eu ~
iE
,
180
~ 160
...o
::J
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~ 140
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V
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,-5 I'~
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I I
80 -100
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...-Systems A.D,E. no fault
I
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~
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I
V
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~
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~
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...........
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-40 -20 a 20 Rotor electrical angle (degrees)
I-Systems A.D,E, fault on.
""'"
~
40
60
80
FIG. 54. Power-angle curves, Study 4, part 2, run 14. 15 Mw. received at station BD from substation AB. Three-phase fault on 13.8-kv. feeder near station AA north bus. Critical clearing angle is -37°; critical clearing time, 18 cycles.
of system B with those of system C and to represent them all as an infinite bus. This procedure could be expected to give somewhat pessimistic results because the speeding up of the machines of system B to follow those of systems A, D, E was beneficial to stability. Power-angle curves were obtained on the calculating board by the following procedure: when possible, the initial conditions of each run were made identical to those of a previous load-flow study. The power received at station BD from substation AB was set to 5 Mw., 15 Mw., 25 Mw., and 35 Mw. in runs 13, 14, 15, and 16, respectively;
STUDY 4-POWER-ANGLE CURVES, PART 2
337
and the voltages of the 66-kv. bus at AB and of the 132-kv. bus at BD were adjusted to normal values. The phase angle of the power source representing the voltage back of transient reactance of the finite A-D-E machine was then advanced in steps of about 15°, both this voltage and the voltage representing the infinite bus being held constant, and at each step the power output of the finite machine was recorded. These results were plotted as the curves labelled "no fault" 240 220
/
200
V' I
l:i
I.~
I' "
I.e: ~
I~
1«1 I~
I
18
I
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I I
~Systems
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-40
-20
A, D, E, fault on
---...
~
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A,D,E, no fault
-,
I~
IJ
-80
~
I
I I
-100
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~Systems
I
VI!
80
.........
~
I
/~
100
-
I
1/
120
-
0
20
~--
40
60
80
Rotor electrical angle (degrees) FIG. 55. Power-angle curves, Study 4, part 2, run 15. 25 Mw. received at station BD from substation AB. Three-phase fault on 13.8-kv. feeder near station AA north bus. Critical clearin.g angle is -33°; critical clearing time, 16 cycles:
in Figs. 53 to 56, inclusive. The angle scale in these figures is arbitrary and is not the angle between the internal voltage of the finite machine and the voltage of the infinite bus. For each value of power transfer a second curve was obtained in similar manner, except with the fault applied, and with the internal voltages held at the same values as before. The critical clearing angle for each value of power transfer was then obtained by the equal.. area method, and the corresponding critical clearing time was found from a swing curve calculated by use of the values of power read from the power-angle curve.
338
TYPICAL STABILITY STUDIES
The kinetic energy of the A-D-E machine was taken as 1,361 Mi. The swing curves are given in Fig. 57. The desired curve of power transfer versus critical clearing time was then plotted and is shown in Fig. 58. The results of stability runs 8 and 9 are consistent with this curve: for 25-Mw. transfer the system was stable with 12-cycle clearing and unstable with 18-cycle clearing. 240--------.....------.---.-----,---,...----., I
~ -_~~systems
~
I ~ A,D,E• ./ -, no fault 220 1----+----4--~+--__+_----t'----_f__:::lIillE:_--t---t---t
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120 I - - - - f - - - - + - - ! - " f - - - - - f - - - i - - - + - - - - i - - - f - - - - - t 100 t----t--~........- + - - _ + _ - _ _ _ t - - _ + _ _ - _ _ P - - - - - t _ _ - _ _ t
80'---......- .....- -.....--....------......- .....--~-~ 60 80 -100 -80 .. 60 -40 -20 0 20 40
Rotor electrical angle (degree~) FIG. 56. Power-angle curves, Study 4, part 2, run 16. 35 Mw. received at station BD from substation AB. Three-phase fault on lS.8-kv. feeder near station AA north bus. Critical clearing angle is -45°; critical clearing time, 11 cycles,
Power limit oj A-to-B line with system H connected. The swing curves of run 12 indicate that the machines of systems B, C, and H may be regarded as an infinite bus and those of systems A, D,and E as a single finite machine. Power-angle curves with no fault and again with a two-line-to-ground fault near AB were obtained. It seemed permissible to use the same power-angle curves for all values of initial power transfer, because in runs 13 to 16 the voltage behind transient reactance varied only 3% over the range of power transfer that was considered. With the fault cleared, the power transfer over the inter-
STUDY 4-POWER-ANGLE CURVES, PART 2
339
connection became zero, but the output of the finite machine did not become zero because it was supplying local load. Its output with the line open, however, was independent of angle and was found from previous load studies, as follows: The total load on systems A, D, and E, when operating with zero interchange with system B, was determined from one load run, and the total load plus losses when delivering -20
0
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_ -20
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FIG.
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i J
o
Time (seconds)
0.1 0.2 0.3 Time (seconds)
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57. Swing curves used in connection with the power-angle curves of Figs. 53 to 56 to determine critical clearing times.
45 Mw. to system B was obtained from another load run. The difference between these figures gave the decrease in losses which would occur on opening the 154-kv. interconnecting line, normal voltage conditions being assumed. This decrease in losses was expressed as a percentage of the power sent into the 154-kv. line. To determine the output with the fault cleared, the percentage found as just described was added to the input to the 154-kv. line, and the augmented input was subtracted from the output of the initial operating conditions.
TYPICAL STABILITY STUDIES
340
The critical reclosing time was found by use of the equal-area method and swing curves (Figs. 59 to 61) for transfers of 25, 35, and 45 Mw., using a clearing time of 6 cycles in every case. The clearing angle corresponding to the clearing time of 6 cycles was found by computing the swing curve up to this point. The critical reclosing angle was then 40 found by the equal-area method, and the corresponding reclosing time was obtained by extending the swing curve to this angle. ~ 30 ; A curve of critical reclosing time \ Unstable ~ versus power transfer was plotted E Stable and may be seen in Fig. 62. From "0 20 ~ this curve the power limit for 21 .~ cycles reclosing time is read as 42 Q) "0 Mw, This is consistent with the re~ sult of stability run 12, in which the system was found to be unstable with , 45 Mw. transfer and 21-cycle recloso 30 ing. The practical power capacity 10 20 o was taken as 35 Mw. Clearing time (cycles) Conclusions. The following cOI\FIG. 58. Curve of power delivered to station BD from substation AB elusions were among those reached over proposed 154-kv. interconnect- as a result of Study 4: ing line versus critical clearing time Capacity of interconnection between of a three-phase fault on a 13.8-kv. systems A and B. The power cafeeder near station AA north bus. pacityof the proposed 154-kv. interBased on the results of runs 13 to 16, connection between systems A and Study 4, part 2. B, based on the maintenance of synchronism through a two-line-to-ground fault on the interconnecting line, with 6-cycle clearing and 21-cycle reclosing, and allowing a margin of at least 5 Mw. for momentary fluctuations in power on the line, would be 25 Mw. received in system B or 35 Mw. received in system A. These values would hold for either of the two routes considered. If system H were connected to the line, the power received in system B might be increased to 35 Mw, In arriving at these limits it was assumed that automatic equipment for controlling the interchange power flow would be installed and used, and that high-speed relaying and switching would be provided on critical circuits within systems A and B as well as on the interconnecting line itself. The connection of systems F and G to system B would be beneficial to stability but would not have enough effect to warrant the assignment of higher power ratings to the interconnection. The capacity of the intercon-
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,
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STUDY 4-CONCLUSIONS
341
nection, if a loss of synchronism were allowable during a fault on the interconnecting line 'or on other critical circuits, would be approximately 35 to 40 Mw. in either direction. Speed of fault clearing. To avoid instability when the interconnection was delivering its rated load of 25 Mw. to system B, it would be 0.8
I
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140
120 60
80
100 120 140 160 Rotor electrical angle (degrees)
180
200
FIG. 59. Power-angle curves and swing curve, Study 4, part 2, run 17. 25 Mw. received at station BD from substation AB. Two-Jine-to-groundfault near substation AB on proposed 154-kv. interconnecting line to station RD. Clearing time, 6 cycles. Critical reclosing time, 42 cycles.
necessary to use high-speed relaying and switching on some of the circuits of company A. Three-phase faults near station AA 13.8-kv. or 66-kv. busses and near substation AB 66-kv. bus must be cleared in 12 cycles or less. The time for clearing three-phase faults on the 13.8-kv. network might be increased to 18 cycles if the fault was separated from the bus by 1,000 feet or more of 700,OOO-circular-mil cable or by an equivalent impedance. If the station AA bus was sectionalized, three-phase faults on or near the bus must be cleared in 16 cycles or less.
342
TYPICAL STABILITY STUDIES
High-speed relaying and switching might be necessary also on several of the transmission lines of company B, but further studies would be required, after the terminal of the line had been selected, to determine the extent of such changes. Terminal transformer capacity. The transformer at substation AB should have a capacity of 45 Mva. and that at either station BD or 0.6
1
260
-
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4
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80
100
120
140
160
180
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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
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Critical 260 _ reclosing time-
-;;-
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~
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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
~~
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40
-,
.. Unstable
,,~
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Stable "'0
e
~ ~
"'0
<,
20
-,
J
10
o
o
10
20
30
40
50
Reclosing time ( cycles) Constant clearing time, 6 cycles FIG. 62. Curve of power delivered to station BD from substation AB over proposed 154-kv. interconnecting line versus critical reclosing time for a two-line-toground fault on this line near substation AB, cleared in 6 cycles. Based on the results of runs 17 to 19, Study 4, part 2.
open the low-voltage oil circuit breakers (66 kv. at AB, 132 kv. at BC)and tripped closed a three-pole grounding switch on the 154-kv. side, causing the breakers at the distant end of the line to open. Staged-fault tests." After the line had been constructed and was ready for service, a number of staged faults were placed on the line to check the operation of the new long-line relays and to check the transient-stability power limits found in the calculating-board studies which have been described. Some faults were placed on the 154-kv. interconnecting line itself, others on 60-kv., 66-kv., and 132-kv. lines near its terminals. The first group of fourteen tests was made with systerns A and B synchronized but with almost no power flow on the interconnection. In these tests a few incorrect relay operations were
31 34
AB
BD
132
5
*Synchronism lost between systems Band F. tRelay trouble at station AB caused slow tripping.
Be
26
154 154
4
4A 4B
66
132 154
29 35
28
30
15
Be Be EA BD AB AB
Mw.
AB AB AB Be BC BC BC
(kv.)
154 154
, A
AB AB AB
BC Be BC BC AB
Rec'd at
Power Flow on Interconnection
On Line to
,
Near Station
At
Fault Location
Line Voltage
If
3A
2 3
1
Number
Test
TABLE 12
,
...
10.0 19.0t 10.0
.. .
AB
6.8 6.5 10.5
at
r..
.A.
8.5
10.0 6.0 5.5 6.0
.. .
,
6.8 7.5
atBC
Clearing Time (cycles)
...
25.8
26.2 34.5
... ...
21.2 21.5
stAB
~
Ai
,
...
22.5
23.0 22.0
. .. . ..
24.2 24.5
atBC
Reclosing Time (cycles)
STAGED Two-LINE-To-GROUND FAULT TESTS ON 154-Kv. INTERCONNECTION OF SYSTEMS A AND B WITH POWER INTERCHANGE (STUDY 4)
Stable Stable Stable Stable Unstable* Unstablet Stable Stable
Stable or Unstable
I
~
00
t-3
t:rj U2
t-3
E:3
> ~
~
e
t:rj
0
>
t-3
U2
~
~
tj
00 ~
d
TYPICAL STABILITY STUDIES
346
discovered, which led to some changes in the connections and adjustments of the relays and circuit-breaker control. The second group of eight tests were made with various amounts and directions of power interchange. Oscillographic records were made of eighteen currents or voltages in the main and control circuits, making possible the determination of the sequence and time of various operations. The tests of the second group are summarized in Table 12. 50
M,
1.2
<,
I
40
I
Unstable
1
~
IV
"4 " ~
1
4 I~
~B 2'
~ 4A
'I'\..
Stable I
~~
N Stable Unstable ~
o
o
Test
~
Calculated
0
10
20
A
• 30
40
Reclosing time of terminal breakp.rs (cycles)
FIG. 63. Approximate curve of transient-stability power limit as a function of reclosing time for faults on the 154-kv. interconnection of companies A and B, Study 4. Points from staged-fault tests. and from calculating-board results are plotted.
With the exception of tests 4 and 4A, in which synchronism was lost, power swings observed throughout the interconnected systems were small. After the application of faults on the 154-kv. line the terminal breakers successfully cleared the fault and reclosed in every case where they were intended to do so. ·The same breakers remained closed in every test in which a fault was applied to other lines, indicating proper functioning of the relays during external faults. Study of the test results revealed that the expected 21- or 22-cycle over-all time from fault to final reclosure of the second breaker was not being obtained on account of a delay of from 2 to 4 cycles in the relay action at the end of the line remote from the fault. This trouble was corrected subsequently.
REFERENCES
347
In order to correlate the tests with the results of stability runs made on the calculating board, Fig. 63 was prepared. Here points representing the staged-fault test results and other points representing the calculating-board results with line terminal at substation Be are plotted with delivered power as ordinate and reclosing time as abscissa. The sloping line MN separates the region of stability from the region of instability. The test results agree reasonably well with the results of the board studies. Exact agreement cannot be expected because the conditions existing at the time of the staged-fault tests differred from the conditions assumed in the calculating-board studies. In particular, the interconnection between systems Band C, assumed in the calculations, had not yet been built. Even if it' had been, it would not have been feasible to schedule power flow and generation in five or more power systems in several states in such manner as to duplicate the conditions assumed in calculation.
REFERENCES 1. R. C. BERGVALL, "Series Resistance Method of Increasing Transient-Stability Limit," A.I.E.E. Trans., vol. 50, pp. 490-4, June, 1931; disc., pp, 494-7. 2. F. C. POAGE, C. A. STREIFUS, D. M. MACGREGOR, and E. E. GEORGE, "Performance Requirements for Relays on Unusually Long Transmission Lines," A.l.E.E. Trans., vol. 62, pp, 275-83, June, 1943. 3. A. R. VAN C. WARRINGTON, "Protective Relaying for Long Transmission Lines," A.I.E.E. Trans., vol. 62, pp. 261-8, June, 1943. 4. C. W. MINARD, R. B. Gow, W. A. WOLFE, and E. A. SWANSON, "StagedFault Tests of Relaying and Stability on Kansas-Nebraska 27Q-Mile 154-kv. Interconnection," A.I.E.E. Trans., vol. 62, pp. 358-67, 1943; disc., pp.425-6.
INDEX Braking effect of grounding resistor, 241 Breakers, data required on, 63 eight-cycle, need for, Study 3, 310 Bus, infinite, 124 Bus reactors at Philo, 290, 294, 300 Busses, effect of, on stability, 190 Byrd, H. L., 168
Acceleration, 15, 16 angular, 17, 19 Adjustment of initial operating conditions, 77 Admittances, driving-point, 80-6 mutual, 80-6 self-,80-6 terminal, 80-6, 95 transfer, 80-6 Algebraic solution of networks, 78-101 Alpha, beta, zero components, 245 Alternating-current calculating board, 64 American Gas and Electric Company, 288,290 Amplifiers for instruments on a-c. board, 70, 73 Analogies between translation and rotation, 20 Analogue, mechanical, 7 Angle, 17, 19 Angular acceleration, 17, 19 Angular momentum, 19 Angular velocity, 17, 19 Apparatus base, 54 Assumptions made in stability studies, 43 Autotransformers, 58 on a-c. board, 68, 69, 72, 73, 74 zero-sequence equivalent circuit of, 239
Cables, data required on, 63 equivalent circuits, 59 Calculating board, a-c., 64 procedure for using, 75 Capacitance of transmission lines, representation of, 60 Capacitors, on a-c. board, 67, 68, 73, 74 series, 189 Carrier-current relaying needed, Study 3,310 Check list of data required for transientstability study, 62 Checks on readings, 78 Circuit breakers, data required on, 63 eight-cycle, need for, Study 3, 310 Cleared fault, application of equal-area criterion to, 129 Clearing angle, 130 determination of, by pre-calculated . swing curves, 157, 158 Clearing time, 139 allowable, Study 1, 286 critical, curves for determining, 16885 determination of, by pre-calculated swing curves, 157, 1,P8 effect of, on transient stability, 159 Combining machines, 111, 264, 319 Common voltage, referring quantities to, 56
Bad effects of instability, 9 Base impedance, 55 Base quantities, for a-c. board, 66, 72 choice of, 75 for per-unit system, 54 Board, a-c. calculating, 64 349
350
INDEX
Components, alpha, beta, zero, 245 symmetrical, 193 Composite line, 60 Condenser, synchronous, data required on, 63 mechanical analogue of, 8 Study 1, 296-7 treatment of, in transient-stability study, 119 Connections on a-c. board, 69, 77 changes of, 78 Conversion, of impedances, 54 star-mesh, 86-7 Coordinates, symmetrical, 193 two-phase, 245 Correction factors for equivalent 1r circuit of transmission line, 59 Coupled lines, equivalent circuits of, 234 Critical clearing angle, 130, 136 determined by pre-calculated swing curves, 157, 158 Critical clearing time, curves for determining, 168-85 determined by pre-calculated swing curves, 151, 157 Damper winding, 11 D8IDping, 5,43,122,123 Data required for transient-stability study, 62Dead sequence networks, 210 Difference equation (problems), 51 Dimensions of quantities of mechanics, 15, 17-9 Discontinuities in point-by-point solution of swing equation, 38 Double unbalances, 232 Duration of fault, effect of, on stability,
159, 185, 224
Ebasco Services, Inc., 254, 311 Energy, kinetic or stored, 16, 20, 23-6 of synchronous machines, 25, 256-8, 292, 305, 315-6; seealso Inertia constant rotational, 20, 23-6 Equal-area criterion, 122
Equal-area criterion, Study 4, 335-43 Equipment, miscellaneous, 61 Equivalent circuits, of coupled lines, 234 of remote portions of system, 61 of transformer, 56 negative-sequence, 235 zero-sequence, 235-40 of transmission lines and cables, 59 Equivalent generator, 6 Equivalent impedance of transformers, 54,56 Equivalent inertia constant of twomachine system, 133 Equivalent input of two-machine system, 133, 136 Equivalent motor, 6 Equivalent output of two-machine system, 133, 136 Equivalent pi (71") circuit of transmission line, 59 Equivalent power-angle curve of two finite machines, 133 Equivalent T circuit of transformer, 56 Factors affecting stability, 187 Fault, close to generator, 78 effect on stability, 5 of duration of, 159, 185, 224 of type of, 223 shown by mechanical analogue, 9 Fault impedance, 228 Fault location, effect of, on stability, 189 Fault locations, Study 1, 265 Study 3,302 Fault shunts, 220, 231 Faulted three-phase networks, solution of, 193 Faults, relative number of various types of, 225 relative severity of various types of, 223 representation of, in symmetrical components, 205, 228-30 in two-phase coordinates, 247 simultaneous, 232
INDEX Finite machines, 132 Flux linkage of field winding, 123 Force, 15, 16 Formal solution of swing equation, 28, 32 Frequency, effect of, on stability, 189 used on a-c. board, 66, 72 Frequency changer, 189 (problem), 51 General Electric A-c. Network Analyzer,66 Generator units of a-c. board, 66, 68, 72,74 Generators, data required on, 62 representation of, 56 Governor, 5, 122 Graphical integration, determination of swing curve by, 139 Grounding, effect of, on stability, 240 Grounding impedance, 237 effect of, on stability, 240 of autotransformer, 240
H (constant), 24-6 High-voltage bussing, effect of, on stability, 190 Historical review, 11 Hunting, 11 Impedance, base, 55 fault, 228 grounding, 238 negative-sequence, 201 per-unit, 55 positive-sequence, 201 series, representation of, 231 transformer, 56 zero-sequence, 201 effect of grounding on, 241 mutual, of lines, 233 Impedance diagram, 53 Impedance units on a-c. board, 67, 68, 73,74 Inertia, moment of, 18, 19 Inertia constant, 22 effect of, on stability, 188
351
Inertia constant, equivalent, of. twomachine system, 133 of combined machines, 111 of remote portion of system, 62 Infinite bus, 124 Initial operating conditions, 77, 78 Input assumed constant, 43 Instability, bad effects of, 9 definition of, 1 Instantaneous clearing, stability limit for, 160, 164 determination of, by curve, 176 Instruments on a-c. board, 70, 71 Integration, graphical, determination of swing curve by, 139 Interconnecting lines, Study 4, 314, 342 Interconnection, 12 Interconnections, effect of, on stability, Study 2, 294-301 Intermediate busses, effect of, on stability, 190 Internal voltages, effect of, on stability, 2,188 Joule (unit), 15 Jumper circuits on a-c. board, 69 Junction, three-phase, 202 I{inetic energy, 16, 20, 23-6 in equal-area criterion, 125 of synchronous machines, 25, 256-8, 292, 305, 315-6; see also Inertia constant rotational, 20, 23-6 Kirchhoff's laws, 202 Laws of mechanics, 15 Length, 15 Limitations of system, 77 Line-impedance units of a-c. board, 67, 68,73,74 Line-to-ground fault, 207, 248 fault shunt, 220, 231 with fault impedance, 229 Line-to-line fault, 207, 212, 248 fault shunt, 220-231 with fault impedance, 229 Lines, transmission, representation of, 59
INDEX
352
Lines, zero-sequence mutual impedance of, 233 Load adjuster, 73 Load-impedance units of a-c. board, 67,
68;73, 74 adjustment of, 73, 77 Loads, data required on, 63 representation of, 61 Location of fault, effect of, on stability, 189
Mass, 15 Master instruments of a-c. board, 70, 73,75
McClure, J. B., 149 Mechanical analogue, 7 Mechanics, laws of, 15 Mesh, three-phase, 203 Mesh circuit, 82, 110 Metropolitan systems, 12, 187 M.k.s. system, 15 Modified time, 149, 169, 178 Moment of inertia, 18, 19 Momentum, 15, 16 angular, 19 Motor-generator set for a-c. board, 71 Motors, large synchronous, data required on, 63 Multicircuit transformers, zerosequence equivalent circuit of, 238
Multimachine system, 6, 122, 191 Mutual impedance base, 55 Mutual impedance of parallel transmission lines, 61 zero-sequence, 233 Mutual reactors on a-c. board, 68, 69, 74 Mutual transformers on a-c. board, 68, 69, 74 Negative sequence, 193, 194 Negative-sequence equivalent circuit of transformer, 235 Negative-sequence impedance, 201 Negative-sequence network, 203, 212 Network, negative-sequence, 203, 212 poffitive-sequence, 53,203,212
Network, zero-sequence, 203, 212 Network analyzer, 64 General Electric, 66 Westinghouse, 72 Network calculator, 64 Network reduction, 83, 213-5 algebraic, 83 calculating-board method of, 109 symbols used in, 90 Network re-expansion, 216-8 Network solution, algebraic, 78-101 Networks, sequence, 203 solution of, 53 solution of faulted three-phase, 193 substitute, 247 Newton (unit), 15 Nodes, 83 Nominal pi (1r) circuit of transmission line, 59 Nominal voltage and current of a-c. board, 66, 72 Ohio Power Company, 290 One-line diagram, 53 Open circuits, representation of, by connections between sequence networks, 231 Parallel transmission lines, 61 effect of, on stability, 189, 190 Per-cent quantities, 56 Per-unit quantities, 54 Phase shift in transformers, 59, 235 Phase shifters on a-c. board, 66, 72, 73 Philo station, 290-2 Pi (1r) circuits of transmission line, 59 Point-by-point solution of the swing equation, 27 discontinuities, 38 errors, 34, 42 Polarity of connections on a-c. board, 69,76 Positive direction of circuit units on a-c. board, 69, 76 Positive sequence, 193, 194 Positive-sequence impedance, ·201 Positive-sequence network, 53, 203 use of, for x and y networks, 248
INDEX Post-fault output curve, 129, 131 Power (mechanics), 15, 16 in rotation, 19 Power-angle curve, 2, 3, 125, 126 equivalent, of two finite machines, 133, 135 Power-angle curves, Study 4, 335-8, 341-3 Power-angle equation, 2, 78 one machine and infinite bus, 126 two finite machines, 133 two-machine reactance system, 135 Power limit, 6 method of determining, Study 4, 318 Power supply of a-c. board, 71 Pre-calculated swing curves, 149-59 Pre-fault output curve, 129-31 Pritchard, S. R., Jr. 168 Radian, 17, 19 Reactance, effect of, on stability, 2, 188 of synchronous machines, 56 of transformers, 57 of transmission lines, 61 zero-sequence, 233 transient, 256-8, 292, 315-6 decrease of, to aid stability, 276 Reactive power, sign of, 79 Reactors, data required on, 63 grounding, effect of, on stability, 241 on a-e. board, 67, 68, 73, 74 correction for resistance of, 76 shunt, use of, to aid stability, 273 special, on Westinghouse a-c. board, 73 Readings on a-c. board, 78 Reclosing, high-speed, Study 4, 317 Reclosing time, critical, Study 4, 340, 344, 346 Reduction, of network, 83 algebraic, 83 calculating-board method of, 109 symbols used in, 90 of sequence networks, 213-5 Re-expansion of networks, 216-8 Referring quantities to a common voltage,56 Regulating transformers, 59
353
Regulator, induction voltage, on a-c. board, 66, 72 voltage, 11 effect of, 123 Relays, long-line, 344 protective, data required on, 63 false operation of, 10 Remote portions of system, 62 Resistors, grounding, effect of, on stability, 241 on a-c. board, 67, 68, 73, 74 series, use of, to aid stability, 273 Rotation, analogies of, to translation, 20 laws of, 17, 19 Scalar instruments on a-c. board, 73 Sectionalised operation, 301 Sequence, negative, 193-4 positive, 193, 194 zero, 194 Sequenceirnpedances, 200 Sequence networks, 203, 212 connections between, for representing faults, 205 for representing, faults with impedance, 228-30 for representing open circuits, 231 Series capacitors, 189 Series impedances, representation of, by connections between sequence networks, 231 Series resistors, use of, to aid stability, 273 Severity of types of faults, 223 Short circuit, effect of, on stability, 5 representation of, by connections between the sequence networks, 205, 228-30 by connections between the substitute networks, 247 types of, 205 Short-circuit impedance of transformers, 54, 56 Shunt reactors used to aid stability, 273 Shunts for representing faults, 220, 231 Simplification of systems, Study 4,318 Simultaneous faults, 232
354
INDEX
Solution, of faulted three-phase networks, 193 of networks, algebraic, 78-101 Stability, certain factors affecting, 187 definition of, 1 transient, summary of methods of calculating, 185 Stability limit, effect of fault-clearing time on, 159, 167 effect of type of fault on, 223 steady-state, 4 transient, 5, 6 Stability studies, assumptions made in, 43 typical, 253 Staged-fault tests, Study 4, 344 Star-mesh conversion, 86-7 Steady-state stability limit, 4 Study 4,318 Steam turbogenerators, stored energy of, 25 Step-by-step solution of the swing equation, 27 discontinuities, 38 errors, 34,42 Stored energy, see Kinetic energy Studies, typical stability, 253-347 Study 1, 254 Study 2,290 Study 3,301 Study 4,311 Substitute networks, 247 Summary of methods of calculating transient stability, 185 Summers, I. H., 149 Sustained fault, application of equalarea criterion to, 127 stability limit for, 161, 164 determination of, by curve, 176 Swing curves, 28, 32, 48, 103, 1.09, 144 determination of, by graphical integration, 139 pre-calculated, 149-59 Study 1, 265-85 Study 2, 296,.297, 300 Study 3, 306-9 Study 4, 322-34, 339 Swing equation, 20
Swing equation, formal solution of, 28, 32 point-by-point solution of, 27 errors, 34, 42 Symmetrical components, 193 Synchronous condenser, data required on, 63 mechanical analogue of, 8 Study 1, 296-7 treatment of, in transient-Btability study, 119 Synchronous machines, kinetic energy of, 25, 256-8, 292, 305, 315-6 reactance of, 256-8, 292, 315-6 representation of, 1, 43-4,56 Synchronous motors, data required on, 63 System base, 55 T circuit of transmission line, 60 Tapped loads, 61 (problem), 120 Three-circuit transformer, 318 zero-sequence equivalent circuit of, 238 Three-phase fault, 209, 248 fault shunt, 220, 231 with fault impedance, 229 Three-phase networks, solution of faulted, 193 Three-phase transformers, zerosequence equivalent circuit of, 239 Time, 15 modified, 149, 169, 178 Torque, 17, 19 Transformer banks, 59 Transformers, data. required on, 63 equivalent circuits of, 56 impedance of, 56 multicircuit, 58 zero-sequence equivalent circuit of, 238 negative-sequence equivalent circuit of, 235 per-unit quantities of, 54 phase shift in, 235 reactances of (table), 57
3S5
INDEX Transformers, regulating, 59 representation of, 56 in the sequence networks, 235 three-circuit, 57, 318 zero-sequenceequivalent circuit of, 238
three-phase, 59 zero-sequence equivalent circuit of, 239 two-circuit,' 56 zero-sequence equivalent circuit of, 235 Transient reactance, decrease of, to aid stability, 276 values of, 256-8, 292, 315-6 voltage behind, 123 Transient stability,certaiIL factors affecting, 187 summary of methods of calculating,
185
Transient stability limit, 5, 6 determination of, by equal-area oriterion,128 Study 4, 318, 340, 346 Translation, analogies of, to rotation, 20 laws of, 15 Transmission, long-distance, 12 Transmission lines, constants of, 61 data required on, 63 equivalent circuits, 59 Turbogenerators, stored energy of, 25, 256-9, 292, 305, 315-6 Two-line-to-ground fault, 209, 248 fault shunt for, 220, 231 with fault impedance, 229 Two-machine system, 1, 7, 122, 149 equivalent inertia constant of, 133 equivalent input of, 133, 136 equivalent output of, 133, 136 Two-phase coordinates, 245
Unbalances, double, 232 Units of quantities of mechanics, 15, 19 Varmeter on a-c. board, 70, 73 Vector instruments, 71, 73 Vector measurements on a-c. board, 71, 73 Vector power, 79 Velocity, 15 angular, 17 Voltage, behind transient reactance, 123 of transmission, effect of, on stability, 189 Voltage regulators, 11 effect of, 123 Water-wheel generators, stored energy of,- 25, 256-9, 315-6 Wattmeter-varmeter on a-c. board, 70, 73 Westinghouse .A-c. Network Calculator, 72 Work, 15 of rotation, 18 WR2 (moment of inertia), 23, 256-9,
305, 315-6
Zero sequence, 194 Zero-sequence equivalent circuit of transformers, 235-40 Zero-sequenceimpedance,201 effect of grounding OD, 241 of transformer, 236 Zero-sequence mutual impedance of lines, 233 Zero-sequence network, 203, 233 representation of lines in, 233 Zero-sequence reactance of lines, 233
