POWERSYSTEMSTABILITYVolume IElementsof Stability CalculationsIEEEPress445 Hoes Lane, POBox 1331Piscataway, NJ 08855-1331Editorial BoardJohn B. Anderson, Editor in ChiefR. S. BlicqM.EdenD. M. EtterG. F. HoffnagleR. F. HoytJ. D. IrwinS. V. KartalopoulosP. LaplanteA. J. LaubM. LightnerJ. M. F. MouraJ.PedenE. Sanchez-SinencioL. ShawD. J. WellsDudley R. Kay, Director ofBook PublishingCarrie Briggs, Administrative AssistantLisa S. Mizrahi, Review and Publicity CoordinatorValerie Zaborski, Production EditorIEEE Press Power Systems Engineering SeriesDr. Paul M. Anderson, Series Editor. PowerMathAssociates, Inc.Series Editorial Advisory CommitteeDr. Roy BillintonUniversity of SaskatchewanDr. Atif S. DebsGeorgiaInstituteof TechnologyDr. M. El-HawaryTechnicalUniversity ofNovaScotiaMr. RichardG. FarmerArizonaPublicServiceCompanyDr. Charles A. GrossAuburnUniversityDr. G. T. HeydtPurdueUniversityDr. George KaradyArizonaState UniversityDr. DonaldW. NovotnyUniversity of WisconsinDr. A. G. PhadkeVirginiaPolytechnic andState UniversityDr. ChananSinghTexas A&M UniversityDr. E. KeithStanekUniversity of Missouri-RollaDr. J. E. Van NessNorthwestern UniversityPOWERSYSTEMSTABILITYVolume IElements of Stability CalculationsEdward Wilson KimbarkAIEEE'VPRESSroWILEY-\:'l9INTERSCIENCEAJOHNWILEY & SONS, INC., PUBLICAnONIEEE Press Power Systems Engineering SeriesDr. Paul M. Anderson, Series Editor1995 by the Instituteof Electrical and Electronics Engineers, Inc.345 East 47th Street, New York, NY10017-23941948 by EdwardWilsonKimbarkThis is the IEEE reprinting of a book previously published by John Wiley&Sons, Inc. under the title Power SystemStability, VolumeI: Elements ofStabilityCalculations.All rightsreserved.No part ofthis book maybe reproduced in any form,nor may it be stored in a retrieval systemor transmitted in any form,without written permission fromthe publisher.Printed in the UnitedStates of America10 9 8 7 6 5 4 3 2ISBN0-7803-1135-3Library of Congress Cataloging-in-Publication DataKimbark, Edward WilsonPower system stabilityI Edward Wilson Kimbark.p. cm. - (IEEE Press power systems engineeringseries)Originally published: New York : Wiley, 1948-1956.Includes bibliographical references and index.Contents: v.I. Elements of stability calculations - v. 2. Powercircuit breakers and protective relays - v. 3. Synchronousmachines.ISBN 0-7803-1135-3(set)1. Electric power system stability. I. Title. II. Series.TKI010.K56 1995621.319--dc20 94-42999CIPTo my wifeRUTHMERRICKKIMBARKFOREWORDTOTHE1995 REISSUETheIEEEPressEditorial Boardfor thePowerSystemsEngineeringSeries has, for some time, discussedthe possibilityof reprintingclas-sic texts in power system engineering. The objective of this series is torecognizepast works that merit being remembered and to make theseolder worksavailabletoanewgenerationofengineers. Webelievemanyengineers will welcometheopportunityofowningtheir owncopies of these classics.In order to come to an agreement about which text to reprint, a num-ber of candidateswere proposed.After a discussion, the board took avote.TheKimbarkserieswastheoverwhelming choiceforthefirstbooks in the IEEE Power Systems Engineering Classic Reissue Series.The subject of power system stability has been studied and writtenaboutfor decades. It has alwaysbeena challengefor the engineertounderstand the physical descriptionofa systemdescribed' bya hugenumber of differential equations. The system modeling is central to anunderstanding of these large dynamic systems. Modeling is one of thecentral themesofKimbark'sPower SystemStabilitybooks. Hisdis-cussion of the system equations remains as clear and descriptivetodayas it was when first published. Many engineers have seen references tothese works, and may have had difficultyin findingcopies for study.Thisnewprintingpresentsa newchancefortheseengineersto nowhave copies forpersonalstudy and reference.Kimbark presents a method of solving the system equations that wasusedinthedaysofthenetworkanalyzer. Thismethodhasbeenre-placed by digital computer techniques that provide much greaterpower andspeed. However, theolder methodsarestillofhistoricalinterest, moreover, thesestep-by-stepmethodsprovidea convenientway of understanding how a large system of equationscan be solved.EdwardKimbarkwas noted duringhis long careeras an excellentwriterand onewho hadthe uniquecapabilityof explaining complextopicsina clearandinterestingmanner. Thesethreevolumesunderthegeneral titlePower SystemStability, VolumesI, II, andIIIwereoriginallypublishedintheyears 1948, 1950,and1956. Kimbark'sbook, Electrical Transmissionof Power Signals, publishedin1949,viiprovidedageneral treatmentof electricpowernetworksandsignalpropagation.Kimbark studied Electrical Engineering at Northwestern Universityandat theMassachusettsInstitute of Technology, wherehe receivedthe Sc.D. degreein1937. He thenbegana career in teaching andre-searchat the Universityof California, Berkeley, MIT, PolytechnicInstitute, Brooklyn, InstitutoTenologicode Aeronautica(SanJoseCampos Brazil) and, finally, as the Dean of Engineering at Seattle Uni-versity. In 1962 Kimbark joined the Bonneville Power Administrationas head of the systems analysis branch, where he remained until his re-tirement in1976. He continued to work on special tasks at Bonnevilleuntil his death in1982.Kimbark is well-known forhis excellent booksandalsohismanytechnical papers.He was formallyrecognized for his achievements bybeingelectedaFellowintheIEEE, tomembershipintheNationalAcademy of Engineering, and was the recipient of the IEEE HarbishawAward. Hewasawarded a Distinguished ServiceAwardanda GoldMedal for his service to the U.S. Department of the Interior.The IEEE Power Engineering Society is proud to present thisspecial reprintingofallthreevolumes of Power SystemStabilitybyEdward Kimbark.Paul M. AndersonSeries Editor, IEEE PressPower Systems Engineering SeriesviiiPREFACEThis work on power-systemstabilityisintendedfor useby power-systemengineers and by graduatestudents. It grewout of lectures given by the author in agraduate eveningcourse atNorthwesternUniversity during theschool year 1941-2.For the convenienceofthe reader, the workisdividedintothreevolumes. VolumeI coverstheelements ofthe stabilityproblem, the principal factors affectingstability, the ordinarysimplifiedmethods ofmakingstabilitycalculations, andillus-trations of the applicationof these methods instudies whichhavebeen madeon actualpower systems.Volume II covers power circuit breakers and protectiverelays, includingmaterial onrapidreclosing ofcircuitbreakersandontheperformanceof protectiverelaysduringswingsandout-of-step conditions. Such material belongs in a work on sta-bility because themost important meansof improving thetran-sient stabilityofpowersystemsinthe improvement ofcircuitbreakers and of protectiverelaying. It .is expected, however,that the publicationofthis material inaseparatevolumewillmakeit more useful to persons whoare interestedinpower-systemprotection, even thoughtheymaynot beparticularlyconcernedwiththesubject of stability.Justification of thesimplifyingassumptions ordinarily used in'stability calculations andthecarrying out of calculations for theextraordinary cases in which greater accuracy than that affordedby thesimplified methodsisdesiredrequireaknowledge ofthesomewhat complicatedtheoryofsynchronousmachines andoftheir excitationsystems. This material iscoveredinVolumeIII, which is expected to appealto those desiring a deeper under-standing of thesubject than is obtainable fromVolume Ialone.ixx PREFACEIt is myhope thatthistreatisewill proveuseful not onlytoreadersseeking anunderstanding of power-system stabilitybutalso to those desiring information on the followingrelated topics:a-c. calculatingboards, fault studies, circuit breakers, protectiverelaying, synchronous-machine theory, exciters and voltage regu-lators, and the step-by-stepsolution of nonlinear differentialequations.I wish to acknowledgemy indebtedness to the followingpersons:To mywife,RuthMerrickKimbark, for typingthe entiremanuscriptandfor heradvice andinspiration.To Charles A. Imburgia, A. J. Krupy,Harry P. St. Clair, andespeciallyClement A. Streifus for supplying andinterpretinginformationon stability studiesmadeon actualpower systems.To J. E.Hobson, W. A. Lewis, and E.T. B. Gross for review-ingthe manuscript and for making many suggestionsfor itsimprovement.To engineersof the General Electric Companyand of theWestinghouseElectric Corporationfor reviewingcertainpartsof themanuscriptpertainingtoproductsof their companies.To manufacturers, authors, and publishers who suppliedillustrations or gavepermissionfor the useofmaterial previ-ously published elsewhere. Credit for such material is given attheplace where it appears.EnwARDWILSONKIMBARKEvanston, IllinoisJune, 1947CONTENTSCHAPTER PAGEI The StabilityProblem 1II The Swing Equation andIts Solution 15III Solution of Networks 53IVThe Equal-Area Criterion for Stability 122VFurther Consideration of the Two-MachineSystem 149VI Solution of FaultedThree-Phase Networks 193VII Typical StabilityStudies 253INDEX 349CHAPTER ITHE STABILITYPROBLEMGen. t-------f Motorx.IeXo1Definitionsandillustrationsofterms. Power-systemstabilityisatermapplied toalternating-current electric power systems, denoting aconditioninwhichthe varioussynchronousmachinesofthe systemremaininsynchronism, or "instep," witheachother. Conversely,instability denotes acondition involving loss of synchronism, or falling"out of step."Consider the verysimplepowersystemof Fig. 1, consistingofasynchronousgenerator supplyingpower toasynchronous motor overa circuit composedof seriesinductive reactance XL. Each of thesynchronousmachinesmay berep-resented, at least approximately, bya constant-voltagesourceinserieswitha constant reactance. * Thusthegenerator is representedbyEoand Xo; and the motor, by EMandXM. Uponcombining thema-chine reactances and the line re..actanceintoa, singlereactance, wehave an electric circuit consisting of FI 1 S I t hG.. Imp e we-mac me powertwo constant-voltage sources, Eo system.and EM, connected through re-actance Xc= XG +XL +XM. It will be shownthat the powertransmitted from thegenerator tothemotordepends uponthephasedifference0 of thetwo voltages EG andEM. Since these voltages aregeneratedby theflux produced bythefield windings of themachines,theirphasedifferenceis thesameastheelectrical angle betweenthemachine rotors.Thevectordiagramof voltages is shown in Fig. 2. Vectorially,EG= EM+JXI [1](The bold-faceletters here and throughout the bookdenote com-*Eitherequivalent synchronous reactance or transient reactance is used,dependinguponwhether steady-stateortransientconditionsare assumed. Thesetermsaredefined anddiscussed inChapters XII andXV, Vol. III.12 THE STABILITYPROBLEMplex, or vector, quantities). Hence thecurrent isI =Eo- EMjX[2][5][4][3]Thepower output of the generator-andlike-wise thepower inputof themotor, since therejXI is no resistance in theline-is given byP=Re(EoI)R(-E Eo- EM)= e 0---FIG. 2. Vector diagram sxof the systemof Fig. 1. h R " I f -were e means the rea part 0" andEomeansthe conjugateofEo. NowletEM = EM!!andThenEo=EolJ..Eo = Eo/-a[6][7]Substitution of eqs. 5, 6, and7 intoeq. 4 gives(EolJ.. - EMI!J.)P= Re Eo/-a x ~... Re(E;2 /-900_ Ea;M /-900- a)EoEM 0= - X-cos(-90 - 8)EoEM.= -y-sma[8]This equation shows that thepower Ptransmitted from thegeneratortothemotorvaries withthesine of the displacementangle abetweenthetwo rotors, as plotted in Fig. 3. Thecurveis knownasapower-angle curve. The maximumpower that can betransmittedinthesteady state with the given reactance X and the given internal voltagesEo andEM is[9]DEFINITIONS ANDILLUSTRATIONS OF TERMS 3andoccurs at adisplacement angle~ = 90. Thevalueof maximumpower maybe increased by raising eitherof theinternal voltages or bydecreasing thecircuit reactance.The systemis stableonly if thedisplacement angle~ is in therangefrom-90 to+90,in which theslope dP/do is positive; that is, therange in which anincrease in displacement angle resultsin anincreaseintransmittedpower. Suppose that thesystemis operatinginthesteady state at point A, Fig. 3. The mechanical input of the generatorandthe mechanical output ofthe motor, ifcorrectedfor rotationallosses, will beequal tothe electricpowerP. Nowsupposethat apFIG. 3. Power-anglecurve of the system of Fig. 1.small increment of shaft load isadded to the motor. Momentarily theangularposition of the motor with respect to thegenerator,andthere-fore the power input to the motor, is unchanged; but the motor outputhasbeenincreased. Thereis,therefore, anet torque onthe motortendingtoretard it,andits speed decreases temporarily. As aresultof thedecrease inmotor speed, 0 is increased,andconsequentlythepower input is increased, until finally the input and output are again inequilibrium, and steady operation ensues at a newpoint B, higher thanAonthepower-angle curve. (It hasbeentacitlyassumedthat thegeneratorspeed would remainconstant. Actually thegenerator mayhavetoslow downsomewhat inorderforthegovernor ofitsprimemover to operate and increasethe generator input sufficientlytobalance theincreased output.)4 THE STABILITYPROBLEMSuppose thatthemotorinput is increased graduallyuntil thepointC of maximumpower is reached. Ifnow anadditional incrementofloadisput onthe motor, the displacement angle awillincreaseasbefore, butasit does so therewill benoincrease ininput. Insteadtherewill be adecreaseininput, furtherincreasing thedifference be-tween output andinput, andretardingthemotormore rapidly. Themotor will pullout of stepandwill probablystall (unless it iskeptgoing byinduction-motor actionresulting from dampercircuits whichmaybe present) . Pmis thesteady-statestability limitof the system.Itisthemaximum power that canbetransmitted, andsynchronismwill belostif anattempt is madetotransmit morepower thanthislimit.Ifa large increment of load on themotor is added8uddenly,insteadof gradually, the motor may fall out of step even thoughthenew loaddoesnot exceedthe steady-statestabilitylimit. The reasonis asfollows: When the large increment of load is added to the motor shaft,the mechanical power output of the motor greatly exceedsthe electricalpower input, andthe deficiencyofinput issuppliedbydecrease ofkineticenergy. Themotor slows down,andanincrease ofthe dis-placement angle~ andaconsequent increase of inputresult. Inac-cordance withtheassumptionthatthenew loaddoes not exceed thesteady-state stability limit, ~ increases totheproper value for steady-stateoperation, avalue suchthatthemotor input equals theoutputandtheretardingtorquevanishes. Whenthisvalueof 0 is reached,however,the motor is runningtooslowly. Itsangular momentumprevents its speedfromsuddenly increasingto the normal value.Hence itcontinues torun too slowly, andthe displacement anglein-creasesbeyondthe proper value. After the anglehas passedthisvalue, themotorinput exceeds theoutput, andthenet torqueis nowanacceleratingtorque. Thespeedof the motor increasesandap-proachesnormal speed. Beforenormal speedisregained, however,thedisplacement angle mayhaveincreased to such an extent that theoperatingpoint on thepower-angle curve(Fig. 3)not only goes overthehump(point C) but also goes so farover it that themotorinputdecreases toavalueless thanthe output. If this happens, the nettorque changesfroman accelerating torque to a retardingtorque.Thespeed, which is still belownormal,now decreases again, and con-tinues todecrease during all but a small part of each slip cycle. Syn-chronism is definitely lost. Inotherwords, thesystemis unstable.If, however, the sudden increment in load is not too great,the motorwill regain its normal speed before thedisplacement angle becomes toogreat. Thenthenet torqueis still anaccelerating torque andcausesDEFINITIONS ANDILLUSTRATIONS OF TERMS 5themotorspeedtocontinue toincreaseandthustobecomegreaterthannormal. The displacement anglethen decreasesandagainap-proachesitsproper value. Again it overshootsthisvalueon accountof inertia. The rotor of the motor thus oscillates about the newsteady-state angular position. The oscillations finallydieout be..cause of dampingtorques, t whichhave been neglectedinthis ele-mentary analysis. Adamped oscillatory motion characterizes astable system.Withagivensudden increment inload, there isa definiteupperlimit to theload which the motor will carry without pulling out of step.Thisisthe transientstabilitylimit ofthesystemfor thegivencondi-tions. Thetransient stabilitylimit is always below thesteady-statestabilityIimit.t but, unlike the latter, it may have manydifferentvalues, dependinguponthenatureandmagnitudeof thedisturbance.The disturbance maybe a suddenincrease in load, as just discussed, orit maybeasudden increaseinreactanceof the circuit, caused, forexample, bythedisconnection of one of two or more parallel lines as anormal switchingoperation. Themost severetypeof disturbancetowhich a power system is subjected, however, is a short circuit. There-fore,theeffect ofshort circuits (or"faults,"asthey areoftencalled)must be determined in nearlyall stability studies.A three-phase short circuit on theline connectingthegenerator andthemotorentirelycuts off the Bow ofpowerbetweenthemachines.The generator output becomes zero in the pure-reactance circuits underconsideration; themotor input alsobecomeszero. Becauseof theslowness ofactionof the governor of theprimemover driving thegenerator, themechanical powerinputof thegeneratorremainscon-stantforperhaps i sec. Also, sincethepowerandtorqueof the loadon the motor are functions of speed, and since the speed cannot changeinstantly and changes by not more than a few per cent unless and untilsynchronism is lost, the mechanical power output of the motor maybeassumed constant. As theelectrical power ofbothmachines isde-creasedbytheshort circuit, while themechanical powerof bothre-mainsconstant, there is anaccelerating torque on thegenerator andaretardingtorqueonthemotor. Consequently, the generatorspeedsup, themotor slows down, anditis apparent that synchronismwill belost unlessthe short circuit isquicklyremovedsoastorestoresyn..chronizingpowerbetweenthemachinesbeforetheyhavedriftedtootDiscussed in Chapter XIV, Vol. III.tConventional methodsof calculation, however, sometimes indicatethat thetransient stabilitylimitisabovethesteady-statestabilitylimit. Thisparadox isdiscussedin ChapterXV, Vol. III.6 THE STABILITYPROBLEMfarapartin angle andin speed. Iftheshort circuit is on one of twoparallel lines and is not at either end of the line, or if the short circuit isof another typethanthree-phase-that is, one-line-to-ground, line-to-line, ortwo-line-to-ground-thensome synchronizing powercanstillbetransmittedpast thefault, but theamplitudeofthepower-anglecurve is reduced in comparison with that of the pre-fault condition. Insome cases thesystemwill bestableevenwithasustainedshortcir-cuit, whereas in othersthesystemwill be stable only if theshort cir-cuit Is cleared withsufficient rapidity. Whether thesystemis stableduring faults will depend notonly on thesystemitself, but also on thetype offault, locationoffault, rapidityofclearing, andmethodofclearing-that is, whethercleared bythesequential opening of two ormore breakers, orbysimultaneousopening-andwhether ornot thefaultedline is reclosed, Forany constant set of theseconditions,thequestion of whether the system is stable depends upon howmuch powerit wascarryingbeforethe occurrenceof the fault. Thus, for anyspecified disturbance, thereis a value of transmitted power, called thetransient stabilitylimit,below whichthe systemis stableand abovewhich itis unstable.The stability limit is one kind of power limit, but the power limit of asystemis not alwaysdeterminedbythequestionof stability. Evenin a system consisting of a synchronous generator supplying power to aresistance load, thereis a maximum power received bytheload as theresistanceoftheloadisvaried. Clearlythere is8. power limit herewith no question of stability.Multimachinesystems. Few, if any, actualpower systemsconsistofmerelyonegenerator and onesynchronous motor. Most powersystemshavemanygeneratingstations, each withseveralgenerators,and many loads, most of whichare combinations of synchronousmotors, synchronous condensers, induction motors, lamps, heatingdevices, andothers. Thestabilityproblemonsuchapower systemusuallyconcernsthe transmissionofpowerfromonegroupofsyn-chronousmachinestoanother. Asarule, bothgroupsconsist pre-dominantly of generators. Duringdisturbancesthemachines of eachgroup swing more or less together; that is, they retain approximatelytheir relative angular positions, although these vary greatly withrespecttothemachines of theothergroup. Forpurposes of analysisthe machines of each group can be replaced by one equivalent machine.If thisis done, thereis one equivalent generator andone equivalent syn-chronousmotor,even thoughthelatter oftenrepresentsmachines thatare actually generators.Because of uncertainty as towhich machines will swing together, orAMECHANICALANALOGITE OFSYSTEMSTABILITY 7in order toimprove theaccuracy of prediction, itis oftendesirable torepresent thesynchronous machines of apower systembymore thantwoequivalent machines. Nevertheless, qualitativelythebehavior ofthe machinesof anactual system isusually likethat ofatwo-machinesystem. Ifsynchronism is lost, themachines of eachgroupstayto-gether,althoughtheygo outof stepwith theothergroup.Becausethebehaviorof atwo-machine systemrepresents thebe-haviorof amultimachinesystem, at least qualitatively, andbecausethetwo-machine system is very simple in comparison withthemulti-machinesystemwhichit represents, the two-machine systemisex-tremely useful indescribingthe general concepts of power-systemstability and the influence of various factors upon stability. Ac-cordingly, the two-machine system plays a prominent role in this book.A mechanical analogueof system stability.5 A simple mechanicalmodel of the vector diagram of Fig. 2 may be built of two pivoted rigidarms representingthe EGandEMvectors,joinedat their extremitiesbya spring representing the XI vec-tor. (SeeFig. 4.) Lengths rep-resent voltages in the model, justasthey dointhe vector diagram.The lengths of the arms, EG andEM,are fixed in accordancewith theti f t t t 1FIG. 4. A mechanical analogueof theassumpIon 0 cons an In erna systemof Fig. 1.voltages. ThelengthofthespringXI is proportional to the appliedtensileforce (for simplicity, weassumeanideal springwhichreturns to zerolengthif the forceisremoved). Hence thetensile forcecan be considered torepresentthecurrent, and the compliance of the spring(its elongation per unit force),torepresent thereactance.The torque exerted on an arm by the spring is equal to the product ofthe length of the arm, the tensile force of the spring, and the sine of theangle between the arm and the spring. (More torque is exerted by thespring when itis perpendicular to thearm than at anyother angle forthesame tensile force.) Obviously, thetorqueson thetwo arms areequal andopposite. Thetorque, multiplied bythespeed of rotation,givesthemechanicalpower transmittedfromonearmtotheother.Forconvenience of inspection, themechanical model will be regardedas stationary,rather than as rotating at synchronous speed, just as weregard the usual vector diagram as stationary. The formula for torque(or power)inthemodel is analogous tothatfor power inthevectorSuperior numeralsrefer toitems in thelistof References at endof chapter.8 THE STABILITYPROBLEMdiagram, namely: voltageX current X cosine of angle between them.(Since theXIvector is90 aheadof theI vector, thecosine of theangle between EandI is equal to the sine of theangle between E andXI.)Theshaft power of themachines mayberepresentedbyapplyingadditional torques to the arms. Aconvenient methodofapplyingFIG. 5. A mechanical analogue ofthe systemof Fig. 1, suitable forrepresenting transient conditions.FIG. 6. Amechanical analogue of a three-machine system consisting of generator,synchronous condenser, and synchronousmotor.constant equal and opposite torques to the two arms is to attach a drumto each armandtosuspend aweight panfrom apulley hanging on acord, oneendofwhichiswoundoneachdrum, all asindicatedinFig. 5.Asweights are added to the pan in small increments, the two arms ofthe model gradually move fartherapart until the angle 0 between themreaches 90, at which position the spring exerts maximum torque. Iffurtherweights are added,the arms fly apart and continue to rotate inopposite directions until all the cord is unwound from the drums. Thesystemis unstable. The steady-state power limit is reached ato= 90. Althoughfrom 90 to180 thespringforce(current) con-tinues to increase, the angle between arm andspring changes in such away thatthetorquedecreases.Theeffect ofchanging themachinevoltagescanbeshown byat-tachingthespring to clamps which slide along the arms.Theeffect of anintermediatesynchronous-condenser stationinin-BADEFFECTS OF INSTABILITY 9creasingthe steady-state powerlimit canbeshownbyadding athirdpivoted armattachedtoanintermediatepoint ofthespring(Fig. 6).The condenser maintains a fixed internal voltage. Since the condenserhasno shaft input or output,no drum is providedonthe third arminthemodel. Withthe intermediatearm(representingthecondenser)in place, the angle betwe.enthe two outer arms (representing thegenerator andmotor) mayexceed90 without instability, andthepower limit is greater thanbefore.The modelcanbeusedtoillustrate transient stability by providingeach arm with a flywheel such that the combinedmoment of inertia ofthe arm and flywheel is proportional tothat of the corresponding syn-chronous machine together with its prime mover (or load). Thedrumscanbe made toservethispurpose.If not toogreat anincrement of load is sud-denlyadded to thepan, it willbefound thatthe arms oscillatebeforesettlingdown to theirnew steady-statepositions. Theanglebetweenthearmsmayexceed90 duringthese oscilla-tionswithout loss of stability. Ifthe incrementofloadis toolarge, the armswillfly apart andcontinue torotate inopposite directions, indi-cating instability. This may happen eventhoughthetotal loadislessthanthesteady-state stability limit.The effect of switching out one of two parallellines may be simulated byconnecting thearmsbytwosprings inparallel andthensuddenlydisconnectingonespringby burningthepiece FIG. 7. A mechanicalof string by whichthespringis attached. ~ n a l o g u e of theeffect of ahe li be si line fault on the powerThe effect of a fault on t e hne may e SlDlU- t fF" 1BySem0 19lated bysuddenly pushing a point on the springtoward the axle(Fig. 7). The armswill start tomoveapart, andsta-bilitywillbelost unlessthespringisquicklyreleased.Modelsof thiskindhave beenbuilt togive a scale representation ofactual powersystemsofthreeorfour machines, andtheoscillationsofthearmshave beenrecordedby moving-picturecameras.6Therearepractical difficulties, however, in applying the model representationtoacomplicated system. The chief value of the model is toillustratetheelementaryconceptsofstability. Othermethodsofanalysisareused in practice.Badeffectsof instability. When one machine falls out of step withthe others ina system, it nolonger serves itsfunction. If it is a10 THE STABILITYPROBLEMgenerator, it no longer constitutesareliable source of electric power.Ifit is amotor, it no longer delivers mechanical power attheproperspeed,ifat all. If it isacondenser, it nolonger maintains propervoltage at its terminals. An unstable two-machine system, consistingof motor andgenerator, may be compared to a slipping belt or clutchin a mechanical transmission system; instability means thefailure ofthe systemas a power-transmitting link.Moreover, a large synchronous machine outof stepis notonly use-less; 'it is worse thanuseless-it is injurious-because ithas a disturb-ing effect on voltages. Voltages will fluctuateupanddown betweenwide limits. Thusinstabilityhasthesamebadeffect onservice tocustomers' loads as does a fault, except that the effect of instability islikely to lastlonger. Ifinstabilityoccurs as a consequenceof a fault,clearing of thefault itself maynot restore stability. Thedisturbingvoltagefluctuationsthen continueafter the fault has beencleared.The machine, or group of machines,' which is out of step with therestof thesystemmust either bebrought backintosteporelse discon-nected from therest of thesystem. Either operation, if done manu-ally, may takea long time compared with thetime required to clear afault automatically. As arule, thebest waytobringthemachinesbackintostepistodisconnect themandthenre-synchronize them.Protective relayshavebeendevelopedtoopenabreaker at apre-determined location when out-of-step conditions occur. Such relays,however, are not yet in wide use. Preferably the power system shouldbe split up into such parts that each part will have adequate generatingcapacityconnected toitto supply theload of that part. Some over-loadmayhavetobecarriedtemporarilyuntil thesystemisre-syn-chronized.Ordinary protective relays are likely to operate falselyduring out-of-step conditions, thereby tripping the circuit breakers of unfaultedlines. Suchfalse tripping may unnecessarilyinterrupt service totappedloads andmaysplit thesystem apart atsuch pointsthatthegenerating capacity of some'partsis inadequate.]Thetrendinpower-system design hasbeentowardincreasing thereliability of electric power service. Since instability has a badeffectonthe quality ofservice, a powersystemshouldbedesignedandoperated so that instabilityis improbable andwill occur only rarely.Scope of this book. This book will deal with two different phases oftheproblemof power-system stability: (1) methodsof analysisandcalculationtodeterminewhetheragiven systemis stablewhen sub-jectedtoaspecifieddisturbance; (2) anexamination of theeffect ofliThia aspect of relayoperation is discussed fully in Chapter X, Vol. II.HISTORICALREVIEW 11variousfactorsonstability, andaconsiderationof measuresforim-proving stability. Inour discussion these two phases will be related:after a methodof analysis is presented, it will be appliedtoshow theeffectofvaryingdifferentfactors. Amongthesefactors aresystemlayout, circuit impedances, loading of machines andcircuits, typeoffault, fault location,methodof clearing,speed ofclearing, inertia ofmachines, kind of excitation systems used with the machines, machinereactances, neutral groundingimpedance, and damper windingsonmachines.Sincetransient power limits are lower than steady-state powerlimits,andsince anypower systemwill be subject tovariousshocks,themost severeofwhich areshort circuits, thesubject oftransientstabilityismuch more important thansteady-statestability. Ac-cordingly, 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 aproblem associated withtheparallel operation of synchronous machines, it might be suspected thattheproblem appeared when synchronous machines were first operatedinparallel. Thefirstserious problem of paralleloperation, however,was not stability, but hunting. When thenecessity for parallel opera-tionofa-c. generators becamegeneral, most ofthe generators weredriven by direct-connected steam engines. The pulsating torquedelivered bythose engines gave rise tohunting, which was sometimesaggravatedbyresonancebetweenthe' periodofpulsation ofprime-movertorqueandtheelectromechanical period of thepower system.Insome cases improper design or functioningof theengine governorsalsoaggravatedthe hunting. Huntingofsynchronousmotors andconverters wassometimes duetoanothercause, namely, toohigharesistance in thesupply line.The seriousness of hunting was decreased bythe introduction of thedamper winding,inventedbyLelslancinFranceandbyLammeinAmerica. Later, theproblemlargelydisappearedonaccount of thegeneral use of steamturbines, which have no torque pulsations.Nearly all the prime movers in use nowadays, both steam turbines andwaterwheels, give a steady torque. A few generatorsarestill drivenbysteamengines or byinternal combustion engines. These,aswellas synchronous motors driving compressors, haveatendency tohunt,but, on thewhole, hunting is no longer a serious difficulty.In the first ten or twenty years of this century, stability was not yet asignificant problem. Before automatic voltage-controlling devices(generator-voltage regulators, induction feeder-voltage regulators,synchronous condensers, andthelike)hadbeen developed, thepower12 THE STABILITYPROBLEMsystems hadtobe designed tohavegood inherent voltageregulation.This requirement calledforlowreactance incircuits and machines.Asa consequenceof the lowreactances, the stabilitylimits (bothsteady-state andtransient)were well above thenormallytransmittedpower.Thedevelopment of automaticvoltageregulators madeit possibletoincrease generatorreactancesin ordertoobtainamore economicaldesign andtolimitshort-circuitcurrents. Byuseof inductionregu-lators to control feeder voltages, transmission lines of higher impedancebecame practicable. These factors, together withtheincreased use ofgeneratorandbusreactorstodecrease short-circuitcurrents, led toadecrease in theinherentstability of metropolitan power systems.Stability first became animportant problem, however, in connectionwith long-distance transmission, which is usually associated withremote hydroelectric stations feeding intometropolitan load centers. ~Theapplicationof theautomaticgenerator-voltageregulatortosyn-chronous condensers madeitpossible to get good local voltage regula-tionfromahydroelectricstationandatransmissionlineofhighre-actance-and hence of lowsynchronizing power. The high investmentintheselong-distanceprojectsmadeit desirable totransmitasmuchpoweraspossible over agivenline, and there wasatemptationtotransmit normal powerapproaching the steady-statestabilitylimit.Inafew cases instabilityoccurred duringsteady-state operation, andmore frequentlyit occurredbecauseofshort circuits. Thestabilityproblemis still more acuteinconnection withlong-distance transmis-sion from a generating station to a load centerthanitis in connectionwithmetropolitansystems. Itshouldnot be inferred, however, thatmetropolitan systems have no stability problems.Another type of long-distancetransmission whichhas frequentlyinvolved astability problem is theinterconnection betweentwolargepower systems for the purpose of exchanging power to obtain economiesin generationor toprovide reserve capacity. Inmanycases thecon-nectingties were designed totransmit an amount of power 'whichwassmall incomparisonwiththe generating capacityof either system.Consequently, thesynchronizing power which theline could transmitwas not enough toretainstability if asevere fault occurredon eithersystem. Therewas also considerable dangerof steady-statepull-outif thepower on thetieline was not controlledcarefully.Fromabout 1920 the problemofpower-systemstabilitywastheobject of thorough investigation. Tests were made bothon laboratory,Amongsuch hydroelectric stations are BigCreek, Bucks Creek, Pit River,FifteenMile Falls, Conowingo, andBoulderDam.HISTORICAL REVIEW 13set-upsandonactual power systems, methods of analyses werede-velopedandcheckedby tests, andmeasures for improvingstabilityweredeveloped. Someoftheimportantstepsinanalytical develop-ment werethe following:1. Circlediagrams for showing the steady-state performance oftransmission systems. These diagrams consistof a family of circles,eachofwhichisthelocusofthevectorpowerforfixed voltages atboth sending andreceivingendsof the line. The circles aredrawnon rectangular coordinates, the abscissasand ordinates of which are,respectively, activeandreactivepower at either end of theline.Thesediagramsshowclearly themaximumpowerwhichalinewillcarryinthe steady state for giventerminal voltages, aswell astherelationbetweenthepower transmittedandtheangular displace-ment between the voltages at the t\VOends of the line. (Suchdiagrams aredescribed inChapter XV, Vol. III.)2. Improvements in synchronous-machine theory, especially theextension of two-reaction theoryto the transient performance ofbothsalient-poleandnonsalient-polemachines. A number ofnewreactances were definedandused. (SeeChapter XII, Vol. III.)Morerecently, the effect of saturation onthesereactances hasbeeninvestigated. (See Chapters XII and XV, Vol. III.)3. The method of symmetrical components for calculating theeffectof unsymmetrical shortcircuits. (See ChapterVI.) Inthisconnection, methodsofdeterminingthesequenceconstantsofap-paratus by test andbycalculation hadtobedevised.4. Point-by-point methods of solving differentialequations, particu-larly the swing equation (giving angular position of a machineversus time). (See Chapter II.)5. Theequal-area criterionfor stabilityoftwo-machinesystems,obviatingthemore laborious calculationof swingcurvesfor suchsystems. (See Chapter IV.)6. The a-c. calculatingboard ornetwork analyzer for the solutionof complicated a-c. networks. (See Chapter III.)The methods of analysis and calculation now in use are believed to besufficientlyaccuratefordeterminingwhether any givenpowersystemin agivenoperating condition will be stable whensubjected toagivendisturbance. The calculations, however, are rather laborious ,vhenappliedtoalargenumberofdifferent operatingconditionsof acom-plicated powersystem.Methods of analysis will betaken upin the following chapters.Calculatedresultshavebeencheckedinanumberofinstancesby14 THE STABILITY PROBLEMobservationson actual power systemsrecordedwithautomaticoscil-lograph equipment.REFERENCES1. R. D. BOOTHandG. C. DAHL, "Power SystemStability-a Non-mathe-matical Review," Gen. Elec. Rev., vol. 33, pp. 677-81, December, 1930; andvol. 34,pp. 131-41, February, 1931.2. A.I.E.E. Subcommittee on Interconnection and Stability Factors, "FirstReport of Power-System Stability," Elec. Eng., vol. 56, pp. 261-82, February, 1937;3. O.G. C. DAHL, Electric Power Circuits, vol. II, Power-SystemStability, NewYork, McGraw-HillBook Co., 1938.4. ElectricalTransmissionandDistributionReference Book, byCentral StauonEngineersofthe Westinghouse Electric&ManufacturingCompany, East Pitts-burgh, Pa., 1st edition, 1942.a. Chapter8, "Power SystemStability-BasicElementsofTheoryand Ap-plication,"byR. D. EVANS.b.Chapter 9, "SystemStability-Examples of Calculation," by H. N.MULLER, JR.5. S. B. GRISCOM, "A Mechanical Analogy of the Problemof TransmissionStability,"Elec. Jour., vol.23, pp. 230-5, l\1ay, 1926.6. R. C. BERGVALL and P. H. ROBINSON, "QuantitativeMechanical AnalysisofPower SystemTransient Disturbances," A.I.E.E. Trans., vol. 47, pp. 915-25,July, 1928; disc., pp. 925-8. Useofmechanical model withsevenarms for in-vestigating transient stability of Conowingo transmission system.PROBLEMS ON CHAPTER I1. Twosynchronous machinesofequal rating, having internal voltages(voltagesbehindtransient reactance) of1.2 and1.0 perunit, respectively,and transient reactancesof 0.25per unit each, are connected by a line having0.50per unit reactanceand negligibleresistance. Assumethat the angle~ between thetwo machines varies from 0 to 3600by15 steps, andcalculatefor each stepthecurrent, thepower, andthevoltage at each of threepoints:ateachendof thelineandat itsmidpoint. Drawloci of thecurrent andvoltagevectors, markingthevaluesof 0 thereon. Also plot inrectangularcoordinatescurrent, power, andvoltage, all as functionsof o.2. Drawthepower-angle curveanddiscuss theconditionfor stabilityoftwomachinesconnectedthroughseriescapacitivereactancewhich exceedstheinternal inductivereactanceof bothmachines.CHAPTERIITHE SWINGEQUATIONANDITS SOLUTIONReview of thelaws of mechanics; translation. Since a synchronousmachine is a rotating body, thelaws of mechanics applyingtorotatingbodies applyto it. Review of these laws may be advantageous at thispoint. The laws of rotating bodies will be clearer if we first review thelaws which applyto linear motion, or translation.TABLE1FUNDAMENTAL AND DERIVEDQUANTITIES OF MECHANICS, ApPLYINGPARTICULARLYTO Quantity Symbol DefiningEquationUnit andDimensionsIts AbbreviationLength x ... meter (m.) LMass m ... kilogram(kg.) MTime t ... second (sec.) Tdx[1]ur:' Velocity v v =- meterper second (m.dtper sec.)dvAcceleration a a =- [2] meter per second per Lr-2dtsecond(m. per sec.")Force F F = ma [3] *newton(newt.) MLT-2Momentum M' M' = mv [4] *newton-second (newt-MLT-lsec.)Work W W=fFdx[5] joule (j.)ML2T- 2Powerp.P = dW[6J watt (w.) ML2r-3dt* Unofficial name.Thefundamental quantitiesof mechanicsarelength, mass, andtime.The fundamental units (in the m.k,s. system, which is nowthe recognizedsystem of unitsfor electrical work) are themeter, the kilogram, andthesecond. From these fundamental quantities and their units are derivedother quantities,' such as velocity, acceleration, force, momentum,work, andpower, and their units. InTable 1 arelistedthefunda-mental quantitiesandcertainderivedquantitieswiththeirsymbols,definingequations, units and theabbreviations thereof, anddimen-1516 THESWINGEQUATIONANDITSSOLUTIONsions interms ofthe fundamental quantitieslength (L), mass (M),andtime(T).Besides thedefining equations (numbers1 to6 of Table1), certainotherequationsgiving relationsbetween thesequantities which areofinterest arederivedbelow. Substitutionof eq. 1 intoeq. 2 gives fortheacceleration[7][8]Substitution of eq. 2 andeq. 7 in turn intoeq. 3 givesdv d2xF=m-=m-dt dt2Comparisonof eq. 8 withthe timederivative of eq. 4 gives theaddi-tional relationdM'F=-dtDifferentiationof eq. 5 withrespecttox givesF=dWdxSubstitution for dW from eq. 10 intoeq. 6 givesFdxp =.-=FvdtIntegration of eq. 9 withrespect tot givesM' = JFdlFromeqs. 11, 3, and4 we obtainP=Fv = mav = aM'whenceAf' =~a[9][10][11][12].[13][14]The kinetic energy of a moving body may be obtained by finding thework required to set it in motion from rest, as follows:W=JFdx=mJdV dx=mJdV vdtd ~ d ~= mJ: v dv = !mv2= !M'v [15]ROTATION 17Rotation. In treating rotation we must first introduce the concept ofangle. Anangleisdefined, with referencetoacircular arc withitscenter at the vertex of theangle,as the ratio of arc8toradius r, thus:8(}=-r[16]Theunit of anglesodefinedistheradian. Thedimensionsof anglefromthe definition are lengthdivided by length. For the presentpurpose, however, itis well torecognize lengthsthat areperpendiculartooneanother ashavingdifferent dimensions, andtorepresent tan-gentiallengthsbythesymbol Lasbeforeandradial lengthsbythemodifiedsymbol LB. Fromthis viewpoint the dimensions ofangleareLLR-1Thedefinitionsof angular velocity and angular acceleration follow byanalogytothecorrespondingdefinitions oflinear velocityandlinearacceleration. Angular velocityisand angular acceleration isdBe,.,=-d ~dw d28a=-=-dt dt2(17][18]The relations betweenangular displacement, velocity, and accelera-tionand thecorrespondingtangential componentsoflinear displace-ment, velocity, and acceleration, respectively, of a particle of a rotatingbody at distance r fromthe axis of rotation aregivenby:8 = rOv = rwa= ra[19}[20][21]The torque on a bodydue to a tangential force F at a distance r fromthe axis of rotation isT= rF [22]or, considering the total torque as due to the summation of infinitesimalforces, we may writeT = frdF [23]The unit of torque could be called a newton-meter, but thisnameis notentirely satisfactory as it might imply a unit of work. Both work and18 THESWINGEQUATIONANDITSSOLUTIONtorque arethe products of force and distance; but in the case of workthe component of force parallel tothe distance is used, and in the caseof torque the component of force perpendicular tothe distance is used.The two quantities may be distinguished by their dimensional formulas,which are as follows: work, ML2T-2; torque, MLLRT-2 For reasonsthat will appearlater, theunit oftorquemaybecalledthejoule perradian.Whentorque is applied toabody, the body experiencesangular ac-celeration a ~ Each particle experiences a tangential accelerationa =ra, whereris the distance of the particle from the axis of rotation.If themass oftheparticleisdm, thetangentialforcerequiredtoac-celerate it isdF = a dm = r admSince this force acts with lever arm r, the torque required for the particleisdT=rdF =~ a d mand that required forthe whole body isT = aJr2dm =[aHere[= !TJdm[24][25]is knownasthe moment of inertia of the body, The m.k.s. unit"is thekilogram-meter2(kg-m.").Note the analogybetween T=I for rotation and F =rna fortranslation.Thework donein rotating a body through anangledO byexerting atorqueT may befoundas follows: If the torque is assumed tobetheresult ofa number oftangential forcesFacting at different pointsofthe body,T= ErFEach force acts through adistanceds= rdOThe workdoneisdW= LF ds= E F rdO=dO E F r = dO TW= f TdO [26]ROTATIONThisis analogous to eq. 5 fortranslation. AlsoT=dWdO19[27]whichis analogous toeq, 10, and whichexplains why the m.k.s. unit oftorque may becalled the joule per radian.Substitutionfor dWfromeq. 27intoeq. 6 (Table1) gives anex-pression for power inrotary motion,dOP= T-= Twdt[28]whichis analogous toeq. 11 fortranslation.Byanalogywiththe definition of momentum, M' = mv,angularmomentummay bedefinedasM= Iw [29]and by derivations analogous tothose of eqs. 12 and14, we obtain alsoM= JTd ~ = ~ [30]Them.k.s. unitof angularmomentummaybevariouslycalled(fromeq. 29) kilogram-meters2-radiansper second, or (fromeq. 30) watts perTABLE 2QUANTITIES OF MECHANICS ApPLYINGTO ROTATIONQuantity SymbolDefiningUnit andAbbreviation DimensionsEquationAngle8radian(rad.) LLR-t8 0=-rAngulardBradian persecond(rad. LLR-1r-1t w =-velocitydtpersec.)Angulardwradian per second per second LLn-1rr-2a Ol =-accelerationdt(rad. persec,")Torque T T=rFjoule per radian(j. per rad.) MLLRT- 2or newton-meter (newt-m.)Moment of I I =fr2dmkilogram-meter-(kg-m.P) MLR2inertiaAngular M M= Iw joule-second perradian(j- MLLRT-lmomentum sec. perrad.)(radianper secondper second), or joule-seconds per radian. Thelastnameseems best fromthe standpoint of brevity.20 THESWINGEQUATIONANDITSSOLUTIONThe kinetic energy of arotating body may bewrittenW= !Iw2= !Mw[31]which is analogous toeq. 15.Table2summarizesthequantitiesofmechanicsapplyingtorota-tion, withtheir symbols, definingequations, units, anddimensions.Table 3 indicates some analogies and relations between the quantitiesand laws of rotation and thoseof translation.TABLE 3ANALOGIES AND RELATIONS BETWEENTHE QUANTITIESANDLAWS OF TRANSLATION ANDOF ROTATIONTranslation Rotation Relation8 8 8 = r8f}w V = rwa Of. a=ram 11=Jr2dmF T T=rFM' M M=JrdM'F = ma T=IW= JFd8 W=fTdJJP= F =M'a P= Tw =MOt.M' =m M=IwW= !mv2W= !Iw2=tM'v =!MwdM'T=dMF=-dt dtThe swingequation. The laws of rotation, as developedin theforegoingsection, apply to the motion of a synchronous machine.Equation24 states thatthetorqueis equaltotheproduct of angularacceleration and moment of inertia:or101. = T [32][33]HereTis the net torque oralgebraic sumof allthe torques acting onthemachine, includingshaft torque (due to the primemover of ageneratorortotheloadonamotor), torqueduetorotational losses(friction, windage, andcoreloss), andelectromagnetictorque. Elec-THESWINGEQUATION 21tromagnetictorque may he subdivided into torques due to synchronousand toasynchronous(induction)action.Let Ti =shaft torque, corrected for torque due to rotationallossesandlet Tu = electromagnetic torque.Bothof thesearetakenaspositiveforgenerator action, thatis, withmechanical input andelectricaloutput. They arenegative for motoraction, that is, with mechanical output and electrical input. Thenet torque, whichproducesacceleration,' is thealgebraicdifference ofthe accelerating shaft torque andtheretarding electromagnetic torque:[34]In thesteady state thisdifference is zero, andthere is no acceleration.During disturbances of the kinds considered in transient-stabilitystudies, however, the differenceexists, and there isacceleration orretardation, depending onwhether thenet torque Taispositive ornegative.Our problem isto solve eq. 33 so astofind theangular positionfJ ofthemachine rotor as a functionof time t. It is more convenient, how-ever, to measure the angular position and angular velocity withrespectto a synchronously rotating reference axis than with respect to astationary axis. Hence leta =(J - Wtt [35]whereWI istheratednormal synchronous speed. Then, ifwetaketimederivatives, we getda dO- = - - WI [36]dt dtandd2a d2fJdt2= dt2Withthissubstitution eq.33 becomes/d2a= Tdt2[37][38]which is unchanged in form. Writing thetorque as in eq. 34, we haved2aI dt2= Ta = Ti - Tu [38a]22 THESWINGEQUATIONANDITSSOLUTIONIf we multiply this equation by the speed w, we obtaind25Mdt2 = Pa = Pi - Pu[39]where M= I w is the angular momentum.Pi = Tiwis the shaft power input, corrected for rotationallosses.Pu = TuWis the electrical power output, correctedfor electricallosses.Pa = Pi- Pu, is the acceleratingpower, ordifference betweeninput and output,each corrected for losses.Equation 39 is more convenient to use than eq. 38abecause it involvesthe electrical poweroutput ofthemachine, rather thanthetorquecorrespondingtothis output. Equation39will bereferredtohere-after as theswingequation. An equation of thisform maybe writtenfor each machineof thesystem.The angular momentumMis not strictly constant because thespeed w varies somewhat during the swings which followa disturbance.Inpractical cases, however, thechange in speedw before synchronismis lost is so small in comparison tothenormal speedWt that verylittleerror is introducedby theassumption that Mis constant. Hence itiscustomary in solving the swing equation toregardMasconstant andequal to1Wt, thevalueof angular momentumat normal speed. Thisvalueof Mis known as theinertiaconstant of the machine.Theinertiaconstant. Intheswing equation(eq. 39)various con-sistent setsof unitsmaybe used. Inthem.k.s. systemPa will be inwatts, 0 inradians, t in seconds, andMinwattsper (radianpersec-ondper second) or joule-secondsper radian. Inpractical stabilitystudies Pa usually will be expressed either in megawatts or in per unit, *oin electrical degrees, andt in seconds. Hence, if Pais in megawatts,Mmustbe in megawatts per (electrical degree per second per second),or megajoule-seconds per electrical degree (abbreviatedMi-sec, perelec. deg.). If Paisinper unit, Mmust beinunit powersecondssquaredperelectricaldegree. Forbrevitythelatter valueof Mwillbe called a per-unit value.Sometimes theavailableinformationregardingthe angular momen-tum of a machine takes the form of the value of its stored kinetic energyat ratedspeed. Moreoften, however, the designerormanufacturergives thevalues of themoment of inertiaof themachineexpressed .inPer-unit power is power expressed as a decimal fractionof anarbitrarily chosenbase power. See Chapter III for further discussion of per-unit quantities.THE INERTIACONSTANT 23pound-feet/andthespeedinrevolutionsperminute. Ineithercase,beforeone can proceedtoasolutionoftheswingequation, hemustcalculatethe value of the inertia constant Mfromthe data. Theformulasneeded for thispurposewillnowbederived, beginningwiththe onegiving the kinetic energy in terms of the moment of inertia andspeed, andproceedingtovariousformulasforM.Let WR2= moment of inertia inpound-feet'',I = moment of inertia inslug-feet/,n= speedinrevolutionsper minute.w =speed inradiansper second.We =speed inelectrical degrees per second.W= kineticenergy infoot-pounds.N= kinetic energyinmegajoules,M= inertiaconstant inmegajoule-seconds per electrical de-gree.J = frequencyincyclesper second.p= numberof poles.G =ratingof machine inmegavolt-amperes.Byeq.31,usingEnglish units, the kineticenergy isW=!Iw2But21rnw=-60and[40][41][42][43]By substituting eqs.41 and 42 into eq.40 and the result into eq. 43, weobtain746 -6 1 WR2(211"n)2N=550X10 X"2 X32.2 X 60= 2.31X10-10WR2n2By eq. 31 the kinetic energy may bewritten alsoas[44][45]24 THESWINGEQUATION. ANDITSSOLUTIONSolving forM,ButHenceM=2NWeWe = 360/[46][47][49][48]M=2N = ~360/ 180/Bysubstitution of eq. 44 into eq. 48 we obtainWR2n2M= 1.28X10-12-1-'-Sincethe relationamongspeed, frequency, andnumber of poles isnp=120/ [50]eq. 49 may bewritten alsoasM.:::: 1.54X10-10 WR2npor asM=1.84X10-8W ~ 2 jP[51][52][53]The inertia constant inmegajoule-seconds per electricaldegreemaybecalculatedfromeqs. 48, 49, 51, or 52. If per-unit power istobeused instead of megawatts, the value of Mso obtained is merelydividedbythebase power inmegavolt-amperes, givingwhatwe maycall aper-unit valueof M.Another constant whichhasproved very useful is denoted by Handis equal to the kinetic energyat ratedspeed dividedbythe ratedapparent powerof the machine.H = stored energyin joulesratinginvolt-amperesstored energy inkilojoulesratinginkilovolt-amperesstored energy inmegajoules, Nrating inmegavolt-amperes, GInterms of H the inertia constant is, byeq. 48,GHM=180j[54]THE INERTIACONSTANT 25... --............ lY1800 r.am, condensing......., .......

.....

r---.

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

............ ./3,600r.p.m. noncondensin--.

f-.+.J.:..[TThequantityHhasthedesirable propertythat its value, unlikethoseofMor WR2, doesnot varygreatlywith the ratedkilovolt-amperesandspeedofthemachine,but insteadhasacharacteristic101002o 20 40 60 80Generator ratingG (megavolt.amperes)FIG. 1. Storedenergyof large steamturbogenerators, turbine included (fromRef. 1, bypermission).5450514 r.p.m.-

"

200. 400 r.arn..... ...........,,-

....... i\.

",

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

.......lI' ,.

...... --

,....-,/ ",

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

"7-e

......

r.p.m,

1o 20 40 60 80 100Generator rating G (megavolt- amperes)FIG. 2. Storedenergy of large vertical-type water-wheel generators, includingallowanceof 15%for water wheels (fromRef. 1, bypermission).value or setof values for each class of machine. InthisrespectHissimilar to the per-unit reactance of machines. In the absence of moredefinite information, acharacteristicvalueof Hmay beused. Suchvalues aregiven in thecurves of Figs.1 and2 for large steamturbo-26 THESWINGEQUATIONANDITSSOLUTIONType of Machinegeneratorsandfor large vertical-shaft water-wheel generators, respec-tively. Inbothcases theinertiaof theprimemoversis included, asit always shouldbe. For' othermachines the values of Hmaybetaken from Table4.Itwill beobservedthat the valueof Hisconsiderablyhigher forsteamturbogeneratorsthanforwater-wheel generators, rangingfrom3 to10 Mj. per Mva,for theformer andfrom 2 to 4 Mj. per Mva. forthelatter. Average values areabout 6 and3, respectively.TABLE 4AVERAGE VALUES OF STOREDENERGYIN ROTATING MACHINESH, Stored Energy atRated Speed(megajoules per megavolt-ampere)Synchronous motorsSynchronous condenserstLargeSmallRotary convertersInduction motors Principally fromRef.1.t Hydrogencooled, 25%less.2.01.251.02.00.5From 30 to 60% of the total inertia of a steam turbogenerator unit isthat oftheprimemover, whereasonly4to15%ofthe inertiaof ahydroelectric generating unit isthat of the water wheel, includingwater. tInertia constant H,like theper-unit reactanceof machines or trans-formers, may be expressed on eitherof two volt-ampere bases: (a) themachineratingor (b) asystembasearbitrarily selected forapower-system study. The value of Hfor a given machine varies inversely asthebase, whereas per-unit reactancevaries directly as thebase.tEXAMPLE 1Giventhefollowing informationon asteamturbogeneratorunit:Rated outputRated voltageRated speedMoment of inertiaNumberof polesRated frequency85,000kw. at 85% power factor13,200 volts1,800r.p.m,859,000 Ib-ft.2460 cycles per sec.tSome dataonmoments ofinertiaof generators and their primemoversaregiven inTable1, Chapter VII.iPer-unit quantities arediscussed in Chapter III.SOLUTIONOFTHESWINGEQUATION 27compute thefollowingquantities:Kineticenergyinmegajoules atratedspeed.Inertia constant H.Inertia constant Minmegajoule-secondsperelectrical degree.Minper uniton50-Mva. base.Comparethe computedvalueof Hwiththe typical value readfromthecurvesofFig. 1.Solution. Rating = 85,000 =100,000kva. =100Mva.0.85Fromeq. 44kineticenergy atratedspeedisN=2.31 X10-10WR2n2= 2.31 X10-10X859,000 X(1,800)2= 642 Mj.From eq. 53H=642 =6.42Mj. per Mva.100From eq. 48N 642 .M = - = =0.0595Mj-sec. per elec. deg.180/ 180 X60If power isto beexpressedinper unit insteadofinmegawatts, thisresultmust bedividedbythebasepower. ThusM =0.0595=0.00119 'per uniton50-Mva. base50FromFig. 1, 1800-r.p.m. curve, at 100Mva., H=6.5, which agrees wellwiththevalueof Hcomputedabove.Point-by-point solution of theswing equation. The swing equation,a differential equationgoverning the motion of each machine of asystem, is[55]where 8= displacement angle of rotor withrespecttoareference axisrotating at normalspeed.M= inertia constant of machine.Pa= acceleratingpower, or difference betweenmechanicalinputand electrical output after each has been corrected forlosses.t =time.28 THESWINGEQUATIONANDITSSOLUTIONThe solution of this equation gives 0 as a functionof t. A graphof thesolutionis known as aswingcurve. Inspectionof theswing curves ofallthemachinesof asystemwill showwhether themachineswill re-maininsynchronismaftera' disturbance. Manyexamplesofswingcurvesobtainedinstabilitystudies onmultimachinesystemsarein-cludedinChapter VII. Bothstable andunstableconditionsareil-lustrated there.Inamultirnachinesystem, the output and hencethe acceleratingpower of each machine dependupontheangular positions-and, tobemore rigorous, also upon the angular speeds-of all the machines of thesystem. Thus, for a three-machine systemthereare three simulta-neous differential equations like eq. 55:[56a][56b](56c]Formal solutionof such a setof equations is notfeasible. Eventhesimplest case, which was considered in Chapter I,of one finite machineconnected through reactance to an infinite bus, with damping neglected,leads toanequation[57]the formal solutionof which,withPi = 0, involves ellipticintegrals."Equation57, withPi 0,hasbeensolvedbyuseof acalculatingmachinecalled anintegraph ordifferentialanalyzer,3andit is possiblethat machine methods of solution could be applied(although they havenotbeen yet)to solving theswing equations of multimachine systems.Point-by-point solutionis themost feasible andwidely usedwayofsolvingthe swingequations. Suchsolutions, whichare alsocalledstep-by-stepsolutions, areapplicabletothe numerical solutionofallsorts ofdifferential equations. Goodaccuracycanbeattained, andthecomputations aresimple.4-7In a point-by-point solution one or more of the variables are assumedeitherto be constant or to vary according to assumed laws throughouta short interval of time ~ t , so that as a resultof theassumptions madeSOLUTIONOFTHESWINGEQUATION 29the equations canbesolvedfor the changesinthe other variablesduringthesametimeinterval. Then, fromthe values ofthe othervariables attheendof theinterval, new values canbecalculatedforthe variables whichwere assumedconstant. Thesenewvaluesarethenused in thenext time interval.In applying the point-by-point method to the solutionof swingequations, itis customaryto assume that the accelerating power(andhence the acceleration) is constant during each time interval,althoughit hasdifferent valuesindifferent intervals. Whenthisassumptionis made, aformal solutionofeq. 55canbeobtainedwhich is validthroughout aparticular timeinterval; andfromthe formal solutionand thevalues of~ andw at the beginning of theintervalthe values ofaandw at theend of theinterval can be computed for each machine.Before a similar computation can be made for thenextinterval, how-ever, it is necessary toknow thenew valueof accelerating power, ordifference between input andoutput, of each machine. Themechan-icalinputs areusuallyassumedconstant, because of theslownessofgovernor action, but theelectrical outputs arefunctionsoftherela-tiveangular positionsof all themachinesofthesystemandcanbefoundbysolving thenetworktowhich themachinesareconnected.Ifdampingpower is takeninto account, theoutput, including damp-ing, will depend also on therelative angular speeds of all the machines.It therefore becomes clear that thepoint-by-point solution of swingcurves consistsof twoprocesses whichare carried out alternately.Thefirst process is thecomputation of theangularpositions, andper-hapsalso of theangularspeeds, attheendof atimeinterval from aknowledge of thepositions and speeds at thebeginning of theintervaland the acceleratingpower assumedfor theinterval, The secondprocess is thecomputation of theacceleratingpower of each machinefrom the angularpositions(and perhaps speeds) of all machines of thesystem. The secondprocess requires a knowledgeof network solution,a topic which is discussed in Chapter III. Inthepresentchapter theemphasisisonthe first process,namely, the solutionof the swingequation proper. Two different point-by-point methods will bedescribed.Method1 is themore obvious, althoughtheless accurate, of thetwomethods. Inmethod 1 it is assumedthat theacceleratingpower isconstant throughout a time interval lit and has the value computed forthebeginning of theinterval. No further assumptions aremade. Ifeq. 55 is divided by M andintegrated twice with respect to t, Pa beingSee discussion of Assumption1 onp.43.30 THESWINGEQUATIONANDITSSOLUTIONtreated as aconstant,we obtainsuccessivelydo ~ a t- =w=wo+-dt Mand[58][59]These equations give, respectively, w, the excessof speed of the machineovernormal speed, and0, the angular displacement ofthemachinewithrespecttoareference axisrotating atnormal speed. 00and Woare the values of 0 andw, respectively, at the beginning of theinterval.Theseequationsholdforanyinstantof timet duringtheinterval inwhich Pa is constant. We areparticularly interested, however, in thevaluesof 0 andw attheend oftheinterval. Let subscript ndenotequantities attheendof thenthinterval. Likewise letn- 1 denotequantities at theend of the (n- l)th interval, which is thebeginningof thenthinterval. At is the lengthof theinterval. Putting At inplace of t in eqs.58 and59 andusing theappropriatesubscripts, weobtainfor thespeed andangle at theend of thenth intervalAtWn = Wn-l +MPa(n-l) [60](At)2On = On-l + At Wn-l +2MPa(n-l) [61]The increments of speed andangle duringthenth interval areAtaWn = Wn - Wn-l= MPa(n-l) [62](At)2aOn== On - On-l=at Wn-l + 2M Pa(n-l) [63]Equations 60 and61, or eqs. 62 and63, are suitablefor point-by-pointcalculation. However, if one is interestedonly in the angular position(for plotting aswing curve) andnot inthespeed, Wn-l maybe elim-inatedfrom eqs. 61 and63, as follows: Writeanequationlike eq. 61but forthepreceding interval. ( ~ t ) 2On-l = On-2 +~ t Wn- 2 +2MPa(n-2) [64]andsubtract itfrom eq. 61, obtaining(on - On-I) = (On-l - On-2) +At(Wn-l - Wn-2)(At)2+2M (Pa(n-l) - Pa(n-2) [65JSOLUTIONOFTHESWINGEQUATIONBut31and fromeq. 62Making these substitutions intoeq.65, we getdt ( ~ t ) 2a6n= a61l-l + at M Pa(1l-2) +2M(Pa(1l-l) - Pa(1l-2)(dt)2== a8n_l+2M (Pa(n-l) + pa(n-2 [66]Thisequation, which gives the increment in angleduring any intervalintermsoftheincrementfortheprevious interval, maybeusedforpoint-by-point calculations inplaceof eqs.62 and 63. The last termis the second difference of 5, which maybe symbolized by t!A25.The time interval ~ t shouldbeshort enoughtogivetherequiredaccuracy, but not so short astounduly increasethenumber of pointstobecomputedonagivenswingcurve. Example2 will throw somelight ontheeffectofthelengthof intervalupontheaccuracyofthesolution.EXAMPLE2If asynchronousmachine performs oscillations ofsmall amplitudewithrespect toaninfinitebus,itspoweroutput maybeassumedtobedirectlyproportional to itsangular displacement from theinfinite bus. Because thiscase is known to result in sinusoidal oscillations, as maybe verified readily byformal solution ofthe swingequation, it will servewell asacheckontheaccuracy of various point-by-point solutions,Consider a 60-cycle machine for which H =2.7 Mj. per Mva. andwhich isinitially operating in thesteady "tate withinput andoutput of 1.00 unit andan angular displacement of 45elec. deg. with respect to an infinite bus.Upon occurrence of a fault, assume that theinput remainsconstant andthattheoutput is givenby(a)even though theamplitude of oscillationmaybe great. Calculate one cycleof theswingcurvebymeansof (a) theformalsolutionand (b)apoint-by-32 THESWINGEQUATION ANDITSSOLUTIONpoint solution, method1, usingvarious valuesof timeinterval ~ t , 0.05 sec.beingtriedfirst.Solution. (a) Formal solution. Thedifferential equation,with 0 expressedinelectrical radians, isd ~ 2Mdt2 =P; =Pi - P =1 - ;;: 0 (b)andtheinitialconditions areo=~4when t = O.Thesolution isand ~ = odt(c)~-7r 7r 20=2-"4cOS 7rM tTABLE5COMPUTATION OF Swrso CURVE FROMFORMAL SOLUTIONOF SWINGEQUATION (EXAMPLE 2)(d)t(sec.) 382t (deg.) cos 382t45 cos 382t8(deg.)(deg.)0 0.0 1.000 45.0 45.00.05 19.1 0.945 42.5 47.50.10 38.2 0.786 35.4 54.60.15 57.3 0.540 24.3 65.70.20 76.4 0.235 10.6 79.40.25 95.5 -0.096 - 4.3 94.30.30 114.6 -0.416 -18.7 108.70.35 133.7 -0.691 -31.1 121.10.40 152.8 -0.889 -40.0 130.00.45 171.9 -0.990 -44.5 134.50.50 191.0 -0.982 -44.2 134.20.55 210.1 -0.865 -38.9 128.90.60 229.2 -0.653 -29.4 119.40.65 248.3 -0.370 -16.7 106.70.70 267.4 -0.045 - 2.0 92.00.75 286.5 0.284 12.8 77.20.80 305.6 0.582 26.2 63.80.85 324.7 0.816 36.7 53.30.90 343.8 0.960 43.2 46.80.95 362.9 0.999 45.0 45.01.00 382.0 0.927 41.7 48.3whichmaybeverifiedbydifferentiatingeq. dtwicewithrespect tot andthen substituting theresultandeq. d itselfintoeq. b; alsobysubstitutingl= 0 intoeq. d anditsfirst derivative andcomparing theresultswith eqs. c.SOLUTIONOFTHESWINGEQUATIONTheperiodof oscillation is givenbyT = 21r = 1r V21rMV2/1rMThe inertiaconstant is33(e)H 2.7 2 1 dM = - = ---=0.01432unitpowersec. per e ec.ra 7rf 1r X60= -.!!- =2.5X10-4unit power sec."per elec. deg. (or, simply, per unit).180!Ifthe first valueof Mis usedineq. e, theperiodis foundtobeT =1r V21rX0.01432 =0.943sec.If5 isexpressedinelectrical degrees insteadof inradians, thesolutionbecomes5 =90 - 45cos(382t)0(I)The amplitude of oscillation is45.Valuesof 0 at0.05-sec. intervals of t arecomputedfromeq. f inTable5.(b) Point-by-point solution, method 1 (ilt=0.05 sec.) Substitution ofthevaluesof ilt andMintoeqs..62 and63 givesilt 0.05~ W n = M Pa(n-l)= ---Pa(n-l) =200 Pa(n-l) (g)0.00025(ilt)2~ O n =ilt Wn-l +2MPa(n-l) =O.05wn - 1 +5Pa(n.-l) (h)Computation of the swing curve fromeqs.g and h is carried out in Table 6.The swingcurve is plotted as curve 3 in Fig. 3, where it is compared with thecorrect curve (curve1) calculatedfromtheformal solution, part aofthisexample.Comparisonofcurves1 and3 of Fig. 3 showsthatcurve 3,computedbypoint-by-point method 1, althoughhavingverynearlythecorrect period,increases in amplitude approximately29%in each half-cycle, or 67%ineachcycleof oscillation. II Theaccuracyispoor.It seemsreasonable thattheaccuracywouldbeimprovedbytheuseof ashorter interval. Consequently, let us try ilt= 0.0167 sec. (one-third of theformervalue). Theninplaceof eqs. g andh we have111.67 =(1.29)2.dWn= 66.7Pa(n - l )L18n =O.OI67wn- l +0.555Pa(n-l)(i)(j)34 THESWINGEQUATIONANDITSSOLUTIONTABLE 6POINT-By-POINT COMPUTATION OF SWINGCURVE(METHOD1, tlt = 0.05sEc.)(ExAMPLE2)t PuPaliw w 0.05w 5Patl8 8(sec. ) (p.u.) (p.u.) (deg.Zsec.) (deg.Zsec.) (deg.) (deg.) (deg.) (deg.)--------0+0.500 0.500 100 0 0.0 2.5 2.5 45.00.05 0.528 0.472 94 100 5.0 2.4 7.4 47.50.10 0.610 0.390 78 194 9.7 2.0 11.7 54.90.15 0.740 0.260 52 272 13.6 1.3 14.9 66.60.20 0.906 0.094 19 324 16.2 0.5 16.7 81.50.25 1.091 -0.091 -18 343 17.2 -0.5 16.7 98.20.30 1.277 -0.277 -55 325 16.2 -1.4 14.8 114.90.35 1.441 -0.441 -88 270 13.5 -2.2 11.3 129.70.40 1.567 -0.567 -113 182 9.1 -2.8 6.3 141.00.45 1.637 -0.637 -127 69 3.4 -3.2 0.2 147.30.50 1.639 -0.639 -128 -58 -2.9 -3.2 -6.1 147.50.55 1.571 -0.571 -114 -186 -9.3 -2.9 -12.2 141.40.60 1.436 -0.436 -87 -300 -15.0 -2.2 -17.2 129.20.65 1.244 -0.244 -49 -387 -19.4 -1.2 -20.6 112.00.70 1.016 -0.016 -3 -436 -21.8 -0.1 -21.9 91.40.75 0.772 0.228 46 -439 -22.0 1.1 -20.9 69.50.80 0.540 0.460 92 -393 -19.6 2.3 -17.3 48.60.85 0.348 0.652 130 -301 -15.0 3.3 -11.7 31.30.90 0.218 0.792 158 -171 -8.6 3.8 -4.8 19.60.95 0.164 0.836 167 -13 -0.6 4.2 3.6 14.81.00 .. . . .. . .. 154 . .. ... . .. 18.4The detailedcalculations are not shown here, but theyare similar to those ofTable6. Theswing curve is plotted ascurve2 inFig.3. It is foundthatthe amplitudeerror has beenreducedto9%inahalf cycleor 21%inacycle-about one-third of the corresponding errors with dt =0.05 sec.Theseerrorsarestill prettylarge, andthelabor ofcalculationis excessivebecause so manypointsmust be calculated..Two additional curves (4and 5), calculated for larger values of dt(0.10 and0.15 sec., respectively), havebeen plotted in Fig. 3. As mightbeexpected, the amplitudesincreasefaster thanbefore. Besides, the periodis noticeablylengthened.Example 2 has demonstrated that swingcurves calculated by method1 are subject to aconsiderable cumulativeerror which manifestsitselfbyincrease in boththeamplitude and theperiod of oscillation, espe-cially in theamplitude. Thereasonfor theerrors is not hardto find.Consider a time, such as the first half cycle of theharmonic oscillationof. Example2,duringwhich theaccelerationis diminishing. Suchacurve of acceleration versus time is shown in Fig. 4. Inmethod1 theSOLUTIONOFTHESWINGEQUATION 351.0 0.8 0.9 0.7 0.3 0.2 0.15

r ".... ,]'00,[\/,\.ill\ 3 \/,/V'1t\; ')I.' \ \l(i;jVT

\ -.-l\\

'fN' b'"f\i\j

rNbI

--"If,

1 Formalsolution "

2 I1t =0.0167sec.\\3 0--0--0/1t =0.05sec.4 6--------6!:At =0.10sec.))5 0-- - -041t = 0.15sec.""\\\\\r\\ 4\,1

tso3015195180150165135120-45o-30-150.4 0.5 0.6Time t (seconds)FIG. 3. Swing curvescalculated from a formal solutionof theswing equation andbypoint-by-point calculation, method1, withvarious values of /1t(Example 2).a36 THESWINGEQUATIONANDITSSOLUTIONacceleration during eachinterval ~ t is assumed constant at its value forthe beginning of theinterval, asshownby the step functioninFig. 4,andduringthetimeconsideredit is alwaystoogreat. Consequently,the calculated speedbecomesprogressively higher than the true speed,and the calculated advance ofangular position is likewisegreater than the true advance.The secondhalf cycle of os-cillation thus begins with acalculated amplitude greaterthan the truevalue; and, ifno further errors occurred, theoscillation would continuewith thisamplitude. DuringtFIG. 4. True and assumed curves of ac- the second half cycle, how-celeration versus time,point-by-point calou- ever, the acceleration is in-lation, method1. creasing, and during this timethe assumed acceleration isalways toosmall, that is, too negative. The calculated negative speedis thereforetoogreat in absolute value, andthecalculated retardationof angularpositionislikewisetoogreat. Thus thecalculatedampli-tudeincreases with eachswing.Method 2. Most of the error caused by assuming the acceleration tobe constant during a time interval can be eliminated by using the valueofacceleration atthemiddleinsteadof thebeginningoftheinterval.ReferringtoFig. 5a, we see that .theacceleration at themiddleof theinterval isverynearlyequal to theaverageaccelerationduringtheinterval, as isshown by the near equality of the triangular areas aboveandbelowthetrue curve. Inmethod 2this assumption is made:namely, that the acceleration during aninterval is constant at its valuecalculated forthe middleof theinterval. Or, if weconsider that theintervalsused incalculationbegin andendat thepoints of timeatwhich acceleration is calculated, then the assumption must be re-statedsomewhat as follows: Theacceleration, as calculatedat thebeginning of aparticular time interval, is assumed toremain constantfromthe middle of the preceding interval to the middle of the intervalbeingconsidered. Let us, for example, consider calculationsfor thenth interval, which begins at t =(n- l ) ~ t . (See Fig. 5.) Theangular positionat this instant is an-I. Theacceleration an-I, ascalculated at thisinstant, is assumed tobeconstant fromt =(n- -!) Ilt to t =(n- !) IltSOLUTIONOFTHESWINGEQUATION 37Over this perioda changeinspeedoccurs, whichiscalculated asat = = MPa{n.-l) [67]giving thespeed at theend of thistime aswn-l = Wn-l + [68]Asa logical outcomeof the assumption regardingacceleration, thechangeinspeedwouldoccur linearlywith time. Tosimplify thea In-t:n-tU I II IOn --.----1------,-----Ao,,: :8,,-1 __ .1.I0,,-2 In-2 n-l n(0)(b)(e).1. FIG. 5. True and assumedcurvesofacceleration, speed, andangular positionversus time, point-by-point calculation, method2.ensuing calculations, however, thechange in speed is assumed to occuras a step at the middle of the period, that is, at t =(n-which isthesame instant for which theacceleration was calculated. Betweensteps thespeed is assumed to be constant, as shown in Fig. 5b. Fromt= (n- l)L\t tot =nL\t,orthroughout thenthinterval, thespeedwill beconstantat thevaluewn-i. Thechangeinangular positionduring thenth interval is, therefore,L\on= [69]andtheposition at theend of theintervalisOn = 011.-1 + [70]as showninFig. 5c.38 THESWINGEQUATIONANDITSSOLUTIONEquations 67to70maybeusedfor computation; but, ifweareinterestedonlyintheangularpositionandnotinthespeed, we mayusea formula forfromwhich whas been eliminated. Suchaformula will nowbedeveloped. Substitutionofeq. 67into eq. 68,andof the resultintoeq. 69, gives

=+M Pa(n-l) [71]Byanalogy witheq. 69Substituting eq.72 intoeq.71, ,ve obtainthedesiredformula

= + M Pa(n-l)[72][73]which, like eq. 66 of method 1, gives theincrement in angle during anytimeinterval in termsof theincrement for thepreviousinterval. Thelast termagainmaybesymbolizedby Thisformulamakesthecalculationof aswingcurveverysimple. If thespeediswanted, itcan beobtained fromtherelation[74]Method2issimpler tousethanmethod1andismuchmoreac-curate,aswill be shownin Example 3.Beforeproceedingwith this example, it is necessarytogivesomeattentiontotheeffects of discontinuities intheacceleratingpower Pawhich occur, for example,,vhen a faultis appliedorremovedor whenanyswitchingoperationtakes place. If suchadiscontinuityoccursatthe beginningof aninterval, thentheaverage ofthevalues ofP(Jbefore andafter the discontinuity must be used. Thus, in computingtheincrement of angle occurring duringthe first interval after a fault isappliedat t =0, eq.73 becomes[75]wherePaO+is theacceleratingpowerimmediately afteroccurrenceofthefault. Immediatelybeforethefault the systemis inthesteadystate; hencetheaccelerating power,Pao-, andthepreviousincrementof angle, are both equal to zero. If the fault is cleared atthe beginning of themth interval, incalculations for this intervalone should use for Pa(m-l) thevalue!(Pa(m-l)- +Pa(m-l)+), whereSOLUTION OFTHESWINGEQUATION 39Pa(m-l)- istheacceleratingpowerimmediately beforeclearing andPa(m-l)+ isthat immediatelyafter clearingthe fault. If the dis-continuityoccurs at themiddleof aninterval, nospecial procedure isneeded. The increment in angle during such an interval is calculated,as usual, from thevalue of Paat thebeginning of theinterval. Thereasons for thesetwo rules should beclear after studyof Fig. 6. Ifthediscontinuity occurs at some time otherthanthe beginning or the

IIIIIIIIIII Pa(m-l)-II II II II II II II II II ,tm-l m n-l n FIG. 6. Thehandling of discontinuities of accelerating power Pain point-by-pointcalculation, method 2. Ais adiscontinuityoccurringatthebeginning of the mthinterval. PaCm-l) = ! (PaCm-l>+ +PaCm-1)- ) is used in calculating Bisa discontinuity at the middle of the nth interval. Here PaCn-1) is usedincalculatingLl8nmiddle of aninterval, aweighted average of thevalues of Pabeforeandafterthediscontinuityshouldbeused,but theneed forsuch arefinement seldomappears because thetimeintervals used in calcula-tionaresoshort that it issufficientlyaccurate toassumethe dis-continuitytooccureither at the beginning orat the middleofaninterval.EXAMPLE3Work the problem of Example 2 by method2 of point-by-point calculation,using variousvaluesof timeintervalComparetheswing curvesthusobtained withthose obtainedbymethod1 inExample2.Solution. Thepoint-by-point calculations will bebasedoneqs. 70 and73. The value of M from Example 2 is 2.5 X10-4per unit. Theaceelerat-40 THESWINGEQUATIONANDITSSOLUTIONing power is given,as before,by

P = Pc>P = 1--a ,u 900Fort1t = 0.05 sec., which will be tried first, we have(t1t)2= (0.05 )2 = 10M 2.5X10-4andeq. 73 becomes(a)(b)The detailed calculations are carried out in Table 7. Note that, fort =0,theinstant of occurrenceof thefault, thevaluesof Pu andof Pa areentered both for thefault off (at t = 0-)andfor the fault on(at t =0+).Theaveragevalueof Pais used.The order in which items (except t)are entered in thetableis: from left torightacross aline until thecolumnheaded"lOPa"is reached; thendiago-nally downward and to the rightto the"value of 0in the line belowthe startingpoint; thencetothe"Pu"columninthesameline; andso on. Eachnewvalueof flois found by addition of theprevious valueof flo andthevalueoflOPa between the old andnew values of dO. Each new value of 0is found insimilarfashionbyadditionof theprecedingvalueof aandthe valueoff16betweentheold andnew valuesof o.Comparisonof thevaluesof 0 in Table 7 withthecorresponding values inTable5, computed from theformalsolution, shows a maximumdiscrepancyof 1.00in thefirst cycle of oscillation. Theagreement of thepoint-by-pointsolutionwiththeformalsolutionis verysatisfactory.Similarcalculationswere madefor at =0.10sec., 0.15see., 0.20sec., and0.25sec. Thedetails ofcalculationarenot shownhere, but the resultingswing curvesareplottedinFig. 7. Thecurvesnumbered3,4, and5, andthe symbols usedfor them, correspondwiththe curves ofFig. 3for likevaluesof dt. Theformal solutionisnot plottedinFig. 7 butagreeswithcurve3 totheaccuracyof plotting.ItmaybenotedinFig. 7that, ast1t isincreased, the calculatedswingcurves remain and correct inamplitude, but that their periodsdecrease. Evenwhen thepointsare so far apart that they arenotadequateto determine theshapeof thecurve(as for= 0.25 sec., giving fewer than4 pointspercycle), asine-wave canbe drawnthrough them.The curve for at = 0.1sec. wouldbe considered accurate enough forengineeringuse inastability study. Inactual studiestheusual periodsofoscillationarefrom0.5to2 sec., andatiscommonlytakenas0.05or0.1sec. The value0.1sec. usuallysuffices and, inadditiontorequiring onlyhalf as manysteps as0.05sec., has the advantagethat multiplicationordivisionby canbeperformedmerelybyshifting thedecimalpoint.In Table 8 a comparison is made of the errors in the amplitudes and periodsof the curves calculated by methods 1 and 2 in Examples 2 and 3, respectively.TABLE7POINT-ByPOINTCOMPUTATIONOP SWINGCURVE(METHOD2, at = 0.05sEc.)(ExAMPLE3)t (sec.) P(p.u.) Po, (p.u.) 10Po, (deg.) f:.8 (deg.) 8(deg.)0- 1.000 0 ... "...0+0.500 0.500 ........oavg. ... 0.250 2.5..45.02.50.05 0.528 0.472 4.7 47.57.20.10 0.609 0.391 3.9 54.711.10.15 0.731 0.269 2.7 65.813.80.20 0.885 0.115 1.2 79.615.00.25 1.052 -0.052 -0.5 94.614.50.30 1.214 -0.214 -2.1 109.112.40.35 1.350 -0.350 -3.5 121.58.90.40 1.450 -0.450 -4.5 130.44.40.45 1.497 -0.497 -5.0 134.8-0.60.50 1.492 -0.492 -4.9 134.2-5.50.55 1.430 -0.430 -4.3 128.7-9.80.60 1.320 -0.320 -3.2 118.9-13.00.65 1.177 -0.177 -1.8 105.9-14.80.70 1.011 -0.011 -0.1 91.1-14.90.75 0.847 0.153 1.5 76.2-13.40.80 0.697 0.303 3.0 62.8-10.40.85 0.582 0.418 4.2 52.4-6.20.90 0.514 0.486 4.9 46.2-1.30.95 0.499 0.501 5.0 44.93.71.00 0.540 0.460 4.6 48.68.31.05 0.632 0.386 3.7 56.912.01.10 ... .. . ... 68:9411.0 1.1THESWINGEQUATIONANDITSSOLUTION0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Time t(seconds). ~ f I~ . ~~\/V\.\'~ lit:\\II\~ . \ ~ \~h\. \'I ~\ \\'~.'\ \ /,f\.\ ~ \7 :54'I6 p).-v1\ '\t ~/V3/\'l' .. ~ ~. "3Cl0oL\t =0.05 sec.46---------6L\t=0.10 sec.50-----0L\t =0.15 sec.6 ---L\t=0.20 sec.7 ---- L\t =0.25 sec.I I I I Ioo301542150135120-;;;..~ 105..'01390:s:rl~'075~c...1:60..E..u45...1i.Uli;iFIG. 7. Swing curves calculatedbypoint-by-point method2 withvarious valuesof At (Example 3).TABLE8PERCENTAGE ERRORS IN AMPLITUDE ANDPERIODOFSINUSOIDAL OSCILLATION C OMPUTEDBYPOlNT-By-POlNTMETHODS 1AND2WITH VARIOUS V ALUES OF tPer cent Errorin AmplitudePointsPer cent ErrorAtperTrueAt Endof One CycleAtHalfin Period(sec.)CycleCycleMethod 1 Method 2 Method1 Method1 Method e0.0167 56 +21 ...+9 +1 ...0.05 19+68 0 +29 +2 -0.40.10 9.4 + 191 0 + 67 +5 -20. 15 6.2 +300 0 +100 +11 -50.20 4.7 . .. 0 ... . .. -90.25 3.7 . .. 0 . .. . .. -15ASSUMPTIONS MADEINSTABILITYSTUDIES 43Thesuperiorityofmethod2over method1isshownunmistakablybyTable 8and bythe curves. Method 2withat = 0.10sec. yieldsmoreaccurate resultsthanmethod1with at = 0.0167 sec., andtheresultsareobtainedwith less thanone-sixth theamount of calculation.Assumptions commonly made instability studies. In the foregoingdiscussion of the solution of the swing equation, itwas tacitly assumedthat theacceleratingpower Pawas knownat thebeginningofeachinterval, and, indeed, the equationcouldnot be solvedunlessPa wereknown. SincePa= Pi - Pu, boththeinput Pi andtheoutput Pumust be known. In the determinationof Pi andPuthefollowingassumptions areusually made:1. The input remainsconstant during the entire periodof a swingcurve.2. Damping or asynchronouspoweris negligible.3. Synchronous power may be calculated froma steady-statesolutionof thenetwork towhichthemachines areconnected.4. Each machine may be represented in thenetwork by a constantreactance (direct-axistransientreactance) inserieswithaconstantelectromotive force(voltage behind transient reactance).5. The mechanical angle of eachmachine rotor coincides withtheelectrical phaseof the voltage behind transient reactance.Theseassumptions will now bediscussed.1. Theinput is initially equaltotheoutput. Whenadisturbanceoccurs,the output usually undergoesanabrupt change, but the inputisunchanged. Theinput to a generatingunit iscontrolledby thegovernor of its primemover. The governor will not act until thespeedchangeexceeds acertain amount (usually 1%of normal speed),depending on theadjustment of thE{governor, and eventhen there is atime lag before thegovernor changes the input. During swings of thesynchronousmachines thepercentagechange inspeedisverysmalluntil after synchronismis actually lost. Therefore governor action 'isusuallynotafactorindeterminingwhether synchronismwill belost,and, accordingly, it is neglected.2. Theoutput (electricpower) ofasynchronousmachineconsistsof asynchronouspart, dependingontherelativeangular positions ofallthemachinesof thesystem, andanasynchronouspart, dependingontherelativeangular speedsof all the machines. Theasychronouspart may betakenintoaccount if desired, but, asit is usuallyunim-portant incomparisontothe synchronouspart, it is commonlyneg-lected intheinterest ofsimplicity. The calculationof dampingorasynchronous power is discussed in Chapter XIV, Vol. III.44 THESWINGEQUATIONANDITSSOLUTION3. The networkconnecting the machines is not strictly in thesteadystate duringswinging of themachines,both because of suddencircuitchanges, such as application or removalof a fault, andbecauseof themoregradual changeof phaseoftheelectromotiveforces duetotheswinging. However,astheperiodsof oscillation of themachinesarerelatively long (of the order of 1sec.) in comparisonto the timeconstants of the network, the network may be assumed, withoutserious error, to be in the steadystateat all times. Steady-statenetworksolution is presented in Chapter III.4 and5. The assumption that each machinecan be representedby aconstant reactancein series withaconstant voltage, andtheassump-tionthat themechanical positionof therotor coincides withthephaseangle of the constant voltage, are not entirelycorrect. Asa rule,however, theydonot leadto seriouserror inthedeterminationofwhether a givensystemisstable. Sinceexaminationand justifica-tion of these assumptions require a considerable knowledgeof syn-chronous-machine theory, they will notbe attempted at thispoint butwill be postponed toChapter XII, Vol. III.EXAMPLE4A 25-Mva.6o-cycle water-wheel generator delivers 20 Mw. over a double-circuit transmissionline to alargemetropolitansystemwhichmaybe re-gardedasaninfinite bus. The generating unit (includingthewaterwhe 1)hasakineticenergyof2.76Mj. perMva. atratedspeed. Thedirect-axistransient reactanceofthegeneratoris0.30per unit. Thetransmissioncir-cuitshavenegligibleresistances, andeachhasareactanceof O ~ 2 0 perunitona 25-Mva. base. The voltage behind transient reactance of the generatoris1.03per unit, andthe voltage of themetropolitan system is1.00per unit.Athree-phaseshort circuitoccurs atthemiddleofonetransmissioncircuitand is cleared in0.4sec. by the simultaneous opening of the circuit breakersatbothendsoftheline.Calculateandplot theswingcurveofthegeneratorfor 1 sec.Solution. Theswingcurvewill be calculatedbypoint-by-point method2, using atime intervalof 0.05 sec. Beforecommencing thepoint-by-pointcalculations, wemust knowtheinertiaconstant of thegenerator andthepower-angle equations forthree different conditions of the network, namely:(1)before the fault occurs; (2)whilethe faultis on; and(3) after thefaulthasbeencleared. Thepower-angleequationdependsonthereactancebe-tweenthe generator andtheinfinitebus.Network reduction. Figure 8ais a reactance diagramof the system.Before occurrence of the fault the reactance between points AandBisfoundby series andparallelcombinationstobe+0.20 0 Xl= 0.30 - = .40perunit2ASSUMPTIONSMADEINSTABILITYSTUDIES 45Whenthe fault is cleared, one of theparallel circuits is disconnected, makingthe reactanceX3 = 0.30 +0.20= 0.50perunitTheequivalentseriesreactancebetweenthe generatorand theinfinite buswhile the faultis onmaybe foundmost readilybyconverting the Y circuitGABFtoa ~ , eliminatingjunctionG. The resultingcircuit isshown inFig. 8b. Thereactance of thebranch of the ~ between AandBisX = 0.30 +0.20 +0.30X0.202 0.10= 0.50 +0.60= 1.10 perunitThe values ofreactanceoftheothertwobranchesare not neededbecausethesebranches, beingconnecteddirectlyacrosstheconstant-voltagepower0.20(a)1.10--......-rmrn'-.......- .......----o(b)FIG. 8. (a)Reactancediagramof asystemconsisting of ageneratorAsupplyingpowerover a double-circuit transmissionlinetoa largemetropolitansystemB(Example 4). (b) Equivalent circuit of the systemwith athree-phase shortcircuitat the middle of one transmission circuit,pointF of a. Thecircuit of b is obtainedfrom that of a by ay - ~ conversion to eliminate pointG.sources, have noeffectonthepoweroutputsofthesources, althoughtheyincreasethereactivepoweroutputs. Thesameistrueof theO.IO-per-unitreactance atB. The power-angleequationforthe circuitofFig. 8b isthesameasit wouldbewiththesethree shuntbranchesomitted.Power-angle equations. The power-angle equation, giving the outputPuA of generator Aas a function of theangle 0 between voltages EAandEB,isPEAEB ~ C ~uA= --SInu = sIn uXwhereC hasthefollowing values:Before fault, C1= EAEB = 1.03 X1.00 = 2.58Xl 0.4046 THESWINGEQUATIONANDITSSOLUTIONDuring fault, O2 = EAEB=1.03 X1.00 =0.936X21.10After clearing, 03 = EAEB= 1.03 X1.00 = 2.06x, 0.50Inertia constant. By eq. 54M = GH =1.00 X2.76 = 2.56 X10-4 er unit180/ 180 X60 PI nitial conditions. The power output of generator Abeforethe faultwasgivenas 20Mw., whichona 25-Mva. base is0.80per unit. Theinitialangular positionof Awith respect toBisfound by thepre-fault power-angleequation:PuAl =2.58 sin~ = 0.80sin ~ =0.80= 0.3102.58~ = 18.10Immediately after occurrence of the faultthe angular position is unchanged,but the power output changes to that given by the fault power-angle equationPuA2 =0.936 sin~= 0.936 sin18.10= 0.936 X'O.310=0.290 p.u.Point-by-point calculations. Take the time interval as at = 0.05 sec.Thesteps of calculation for eachpoint areas follows:Pa(n-l) =Pi - Pu(n-l) =0.800 - Pu(n-l) per-unitpower(at)2 (0.05)2--Pa(n-l) = 4Pa(n-l) =9.76 Pa(n-l) elec. deg.M 2.56 X10-~ ~ n =.1on - 1 + 9.76Pa (n -l ) elec. deg.an = 8n-l +.1onelec. deg,Pun =C sinOn per-unit powerwhere C =C2 =0.936whilethefaultis on(0 < t < 0.4 sec.).C =C = 2.06 after the faulthasbeencleared (0.4 095

-

_""-......-

,.....---t--..r---.

.......4020oo 2 3 4 5 6 7 8 9 roModified timeTFIG. 10. sin00'=0.90.FIGS. 9 and10. Pre-calculatedswingcurves (copiedfromRef. 2bypermission).PRE-CALCULATEDSWINGCURVES1573. Compute the equivalent inertia constant M by eq. 2.4. Find the familyof curves for the proper value of sin00' and theindividual curve for the proper value of p. Enter thiscurve with theordinate 0' = Oe' and read the corresponding abscissa T =TeInterpolationbetweencurvesorbetweenfamiliesofcurvesmaybenecessary.5. Byeq. 11computethecritical clearingtimetecorrespondingtoTe.Todeterminetheclearingangle correspondingto a given clearingtime, the order of the steps of procedure is altered in a way thatshould beobvious.Theproceduredescribedabovebreaksdownifthefault isofsuchnature thatthere is nosynchronizing powerwhilethe fault is on. InsuchacasePM=0,fromwhichit follows thatp= 00 andT = 0 forall values of t. The pre-calculated curve for this condition is thevertical coordinateaxis, andtherelationbetween0'" andtcannotbedeterminedfromit. However, this relationcan befound byeq. 41ofChapter IV, namely: ~ ( )2M0 - 00t = [15]PaFurthermore, thepre-calculatedcurvescannot beusedtorepresentconditions after clearing afaultbecause the curve forthe proper valueof angle and speed(at the instant of clearing) does not have the propervalueofacceleratingpower or accelerationafter clearing.EXAMPLE1In Example 1of Chapter IV, which deals with a machine connectedthrough reactance toan infinite bus, the critical clearing anglefor the condi-tions of Example 4 of Chapter II was foundbythe equal-area criterion tobe1380 The corresponding critical clearing time, as determined fromtheswing curve, is 0.61 sec. Check this value byuse of the pre-calculated swingcurves.Solution. FromExample 4 of Chapter II, the power-angle equationvalidforthefault condition ispu = 0.936sin 0 per unit,whencePo = 0, PM= 0.936, 'Y = 0;thepower input isPi = 0.80perunit;the inertia constantof thefinite machine isMl =2.56 X10-4perunit;