modern-power-systems-analysis-d-p-kothari-i-j-nagrath-

fl$4.:rl MooernpowLrSystemnnarysis symmetricCatomponents t- From Fig' 10.13bthe negativesequencevoltage of terminal a with respecr Zs= 3Zra Zos FjSiffi to referencebus is (10.53) Voz= - Zzloz (10.s1) in orderfor it to havethesamevoltagefrom a to referencebus.Thereference bushereis, of coursea, t groundpotential. From Fig. 10.14b zero sequencevoltage of point c with respect to the Voo= - Z{oo (10.54). Order of Values of Sequence Impedances of a Synchronous Generator laz a Typical valuesof sequenceimpedancesof a turbo-generatorrated5 MVA, 6.6 kV, 3;000rpm are: (b) Single-phase model Zr = lZ%o(subtransient) Flg. 10.13 Negativesequencenetworkof a synchronoums achine Zr = 20Vo(transient) Zero Sequence Impedance and Network Zr = 7l0Vo(synchronous) we state once again that no zero sequencevoltages are induced in a synchronousmachine. The flow of zero sequencecurrents creates three mmfs Zz= I2Vo which are in time phase but are distributed in space phase by 120\". The tesultant air gap field caused by zero sequencecurrents is therefore zero. Zo= 5Vo Hence,the rotor windings presentleakagereactanceonly to the flow of zero For typical values of positive, negative and zero sequencereactancesof a sequencecurrents (Zos< Zz < Z). synchronousmachinerefer to Tablu 9.1. /66 IO.7 SEOUENCE IMPEDANCES OF TRANSMISSIO\\ LINES Reference bus -a A fully transposedthree-phaseline is completely symmetrical and therefore the Iao per phaseimpedanceoffered by it is independentof the phase sequenceof a balancedsetof currentsI.n otherwords,the impedanceosffired by ii to positive (b)Single-phase model and negative sequenceculrents are identical. The expressionfoi its per phase inductive reactanceaccounting for both self and mutual linkages ias been derived in Chapter2. (a)Three-phasemodel When only zero sequencecurrentsflow in a transmissionline, the currents in each phaseare identical in both magnitudeand phaseangle. part of these 1o.14 Zerosequencenetworkof a synchronoums achine currents return via the ground, while the rest return throu h the overhead ground wires. The ground wires being groundedat severaltowers, the return Zero sequencenetwork modelson a three-and single-phasebasisare shown currentsin the ground wires arenot necessarilyuniform along the entire length. in Figs. 10.14aand b. In Fig. r0.l4a, the cu'ent flowing in the inpedancezn The flow of zerosequencecurrentsthroughthe transrnissioniln\"r, ground wires betweenneutral andground is In = 3lno.The zero sequencevoltage of t\"r-inul and ground createsa magneticfield pattern which is very different from that a with respectto ground, the referencebus. is therefore causedby the flow of positive or negativesequencecurrentswherethe currents Vr1= - 3znioo- Zorlno=,r, (32, + Z0)loo e0.52) rhr, rq. rYov (^ r ^L^^^ u- Jr:rf fr^s- r^ t- ^r I^l u t r U^ f I It L^Vn o an(l tne fetufn Cuffent lS ZefO. T\\e ZefO Prrd'Ds' wh11eZo, is the zerosequenceimpedanceper phaseof the machine. sequenceimpedance of a transmission line also accountsfor the ground Sincethe single-phasezerosequencenetworkof Fig. 10.14bcarriesonlyper impedance(zo = z.to+ 3zri.since the ground impedanceheavily dependson soil conditions,it is esseniialto make somesimplifying assumptionsio obtain phasezero sequencecurrent, its total zero sequenceimpedancemust be analytical results.The zero sequenceimpedanceof transmissionlines usuallv

3Sd I ModernPowerSystemAnatysis Symmetri a rangesfromZ to 3.5 timesthe positivesequenceimpedance*T. his ratiois on ter\" -\"*\"- -\"-t--t transformers thehighersidefor doublecircuit lines without groundwires. \" Beforeconsideringthe zero'sequencenetworksof varioustypesof transformer 10.8 SEOUENCEIMPEDANCES AND NETWORKS OF connectionst,hree importantobservationsare made: It is well known that ahnosl all presentday installationshave three-phase current only if thereis cunent florv on the secondaryside. transformerssince they entail lower initial cost, have smaller spacerequire- mentsandhigherefficiency. (ii) Zero sequencecurrentscan flow in the legsof a star connectiononly if the star point is groundeclwhich proviciesthe necessaryreturn path for The positive sequenceseriesimpedanceof a transformerequalsits leakage zero sequenceculrents.This fact is illustratedby Figs. 10.15aand b. impedance.Sincea transformeris a static device, the leakageimpedancedoes not changewith alterationof phasesequenceof balancedappliedvoltages.The /ao= 0 ':o __----, a transformernegativesequenceimpedanceis also thereforeequal to its leakage - reactance.Thus. for a transformer .{-> Zt= Zz= Zr\"ukug\" (r0.55) /co= 0 r'l --f-' Assuming suchtransformerconnectionsthat zero sequencecunentscanflow on both sides,a transformeroffers a zerosequenceimpedancewhich may differ / o o =o (a) Groundedstar slightly from the correspondingpositive and negativesequencevalues.It is, ( a ) U n g r o u n d esdt a r however,normal practiceto assumethat the seriesimpedancesof all sequences are equal regardlessof the type of transformer. Fig. 10.15 Flowof zerosequencecurrentsin a starconnection The zero sequencemagnetizingcurrentis somewhathigher in a core type (iii) No zero sequencecurrents can flow in the lines connecte.dto a delta than in a shell type transformer. This difference does not matter as the connectionasno returnpathis availablefor theseculrents.Zero sequence magnetizing curent of a transformeris always neglectedin short circuit currents can, howevet, flow in the legs of a delta-such currents are analysis. causedby the presenceof zerosequencevoltagesin the deltaconnection. Above a certain rating (1,000 kvA) the reactanceand impedanceof a This f act is illt r st r at ebdY Fig. 10. 16' transformerare almostequaland are thereforenot distinguished. /ao= 0 a *We can easily comparethe forward path positive and zero sequenceimpedancesof Fig. 10.16 Flowof zeroSequencceurrentsii',a deltaconnection a transmissionline with groundreturnpathinfinitelyaway.Assumethateachline has a self inductance,L and mutual inductance M between any two lines (completely Let us now consider varioustypes of transformerconnections. symmetrical case).The voltage drop in line a causedby positive sequencecurrentsis Case I; Y-Y transformerbank with any one neutrttlgrounded. VAnt=uLlnr+ uMI,,r+ aMI,.1 If any one of the two neutralsof a Y-Ytransformeris ungrounded,zerosequence - [uL+ (& + a) aMllo, = a(L-tu|)Ior currentscannotflow in theungroundedstarandconsequentlyt,hesecannotflow Positive sequencereactance= a(L- fuI) in the groundedstar.Hence,an opencircuit existsin the zerosequencenetwork The voltage drop in line a causedby zero sequencecurrentsis betweenH and L, i.e. betweenthe two parts of the systemconnectedby the transformeras shownin Fig. 10.17. VAno=aLloo+ utMluo+ uMlrs = a(L + 2M)Ino Zero sequencereactance= w(L + 2W Obviously,zerosequencereactanceis much more than positivesequencereactance. This resulthasalreadybeenderivedin Eq. (10.45).

Zg ffi ---tll d--- (seeFig. 10.19).If the star neutral is groundedthrough 2,, an impedance3Zn L appearsin series with Z, in the sequencenetwork- Casb Y-A tran rmer bank with ungroundedstar Y-Ytransformebrankwithone neutragl roundedanditszero This is the special case of Case 3 where the neutral is grounded through sequencenetwork Zn = oo. Thereiore no zero sequence curent can flow in the transformer windings.The zero sequencenetwork thenmodifiesto that shownin Fig. 10.20. Case2: Y-Y transforrnerbank both neutrals grounded {L When both the neutralsof a Y-Y transformerare grounded,a path through the transformer exists for zero sequencecurrentsin both windings via the two Referencebus groundedneutrals.Hence,in the zero sequencenetwork 11andL areconnected by the zero sequenceirnpedanceof the transformeras shown in Fig. 10.1g. H\"ifr6L2 ; ----------o Case3: Y-A transformer bank with grounded y neutral L Fig. 10.20 Y-l transformebrankwithungroundesdtarand its zerosequencenetwork Referencebus Case5: A-A transformer bank o---4TdL=-o Since a delta circuit provides no return path, the zero sequencecurrents cannot HZs flow in or out of A-A transformer;however, it can circulate in the delta windings*. Theretore,there is an open circuit between H anE L and Zo is Fig. 10.18 Y-Ytransformebrankwithneutralsgroundedand itszero connectedto the referencebus on both endsto accountfor any circulating zero s e q u e n c en e t w o r k sequencecurrent in the two deltas (seeFrg. 10.2I). L d-E^a--r--;-1-I--_. Reference bus Fig. 10.19 Y-1 transformebrankwithgroundedyneutralanditszero Fig. 10.21 A-Atranstormebrankanditszerosequencenetwork sequencenetwork 10.9 CONSTRUCTION OF SEOUENCENETWORKS OF A If the neutral of starsideis grounded,zerosequencecurrentscan flow in star POWER SYSTEM becausea path is availableto ground and the balancing zerosequencecurrents can flow in delta.Of courseno zero sequencecurrentscan flow in the line on In the previous sections the sequencenetworks for various power system the delta side. The zero sequencenetwork must therefbrehave a path from the elements-synchronous machines,transformersand lines-have been given. line 1{ on the star side through the zero sequenceimpedance of the transformer Using thesq, complete sequencenetworks of a power system can be easily constructed.To start with, the positive sequencenetwork is constructedby *Such circulating currentswould exist only if zero sequencevoltages are somehow induced in either delta winding.

ModernPowerSystemAnalysis ffi examinationof theone-linediagrarnof the system.It is to be notedthat positive MVA basein all othercircuits and the following voltagebases. sequencveoltagesarepresenitn synchronoums achines(gcneratorasndmotors) Transmissionline voltasebase= 11 x 121 - 123.2kY only. The transitionli'orr-rpositil'e sequencenetwork to negative sequence 10.8 networkis straightforwardS, ince the positiveand negative sequenceirnped- Motor voltagebase= 123.2x +1+21 = 11kV ,) \\rrrrvD qrtu Lt otrJltrt ttlvt Jrr, rrrv LrltlJ vrrarr Thereactanceosf transformersli,ne andmotorsareconvertedto pu values on appropriatbeasesasfollows: neclessariyn positivesequencenetwork to obtainnegatit'esequencenetwork in respect of synchronousmachines. Each machine is representedby Transformreeractanc=e0.1x =3.0 \\f+1+1) / = 0.0s05pu neg;ttivesequenceimpedancet,he negativesequencevoltage being zero. Line reac:l a n c elo=o x 25 = o . 1 6 4 p u The referencebus for positive and negativesequencenetworks is the system ffi neutral.Any impedanceconnectedbetweena neutraland ground is not included in thesesequencenetworks as neither of thesesequencecurrents can flow in Reactanceof motor I = 0.25 x -2xs t -/ 1 0 \\|2 = 0.345pu suchan impedance. 15 \\lll Zero sequencesubnetworksfor various parts of a system can be easily Reactanceof motor 2 - 0.25 x 7?1. 5, rf \\ -lLl o)/ t = 0.69 pu combinedto form completezero sequencenetwork. No voltage sourcesare The requiredpositive sequencenetwork is presentedin Fig. 10.23. presentin the zero sequencenetwork. Any impedanceincluded in generatoror transformerneutralbecomesthree tirnesits valuein a zero sequencenetwork. R e f e r e n c eb u s I Specialcare needsto be taken of transforners in respect of zero sequence network.Zero sequencenetworks of all possibletransformerconnectionshave I Enrt( ) i ) Eara been dealt with in the preceding section. l+ Es( ) +l The procedure for drawing sequence networks is illustrated through the +l I --,1 following examples. ioz:), . r :t ! iio6e - -:----' f-\"-- d ,-66X-L-g io345:-r i oo8o5 trE_xa_m_pl_e10_.2_1 : _ - l A 25 MVA, 1l kV, three-phasegeneratorhasa subtransienrteactanceof 20Vo. T lr cgcrrc ri ttosru p p l i c stw o rrto to rso v 0r' r,tr ansnri ssi ol innc w i Lhtral tsl ol l rrcl ' s atboth endsas shownin the one-linediagramof Fig. I0.22. The motorshave . P -- - - -r--'-- - 9 ratedinptrtsof 15 and7.5 MVA, both 10kV with 25Vosubtransienrteactance. ,.I-6XU -, r- - In j0'164 i 0\"0xf8i 05 Thethree-phasetransformersare both rated30 MVA, I0.8/L2I kV, connection Fig. 10.23 Positivesequencenetworkfor Example10.3 A-Y with leakagereactanceof 70Voeach.The seriesreactanceof the line is 100ohms. Draw the positive and negativesequencenetworks of the system with reactancesmarkedin per unit. I t s' r t - o 'p u(-9t)1, 0'r'' Referencebus e t IYA . (2) r, \\,,1 q' \\-/ Motor I i 0.0805 j4164 loo8o5 Fig. 10.22 Fig. 10.24 Negativesequencenetworkfor Example10.3 Assume that the negative sequence reactance of each machine is equalto its Since all the negative sequence reactances of the system are equal to the subtransientreactance.Omit resistances.Select generator rating as basein the positive sequencereactances,the negative sequencenetwork is identical to the generatorcircuit.

' rti..;;:1 . ::i.l l-4odernPov\"'eSi' vstemAnalvsis symmetriccaotmpblents-------_-1 ffi 3,42'l positive sequencenetwork but for the omissionof voltage sources.The negative Zerosequencreactancoef motor2 - 0.06\" +. (i+)' sequencenetwork is drawn in Fig. 10.24. = 0.164pu Reactancoef currentlimiting reactors= ''.1.i3t = 0.516pu ( 11 ) ' , For the power systemwhose one-line diagramis shown in Fig. 10.25,sketch Reactancoef currentlimiting reactorincludedin zerosequencneetwork' the zero sequencenetwork. = 3 x 0 . 5 1 6- - 1 . 5 4 8p u T1 T2 Zerosequencreeactanceof transmissiolnine = r(yr2:231.D2 = 0.494pu Thezerosequencneetworkis shownin Fig. 10.27. Fig.10.25 Referencebus Solution The zero sequencenetwork is drawn in Fig. 10.26. j1-548 j1.il8 Reference bus 32n io06 jo164 Zost Fig. 1O.27 Zero sequencenetworkof Example 10.5 Zrt Z2 6 ( l i n ee)) 272 Fig. 10.26 Zero sequence network of the system presented in Fig. 10.25 PROIBEIvIS i--i---- - _- *-; 1E xa mp l-e*1-0\". 4| 10.1 Computethe following in polar form 1 (i) o2-t 1ii; I- a- (iii) 3d + 4cy+ 2 (iv) ja \"? Draw the zero sequencenetrvork for the systemdescribedin Example 10.2. 10.2 Thlee identicalresistorsare star connectedand rated2,500 V, 750 kVA. Assumezero sequencereactancesfor the generatorandmotors of 0.06per unit. Currentlimiting reactorsof 2.5 ohms eachare connectedin the neutralof the This thre.e-phaseunit of resistorsis connectedto the I side of a A-Y generatorand motor No. 2. The zero sequencereactanceof the transmission line is 300 ohms. transformer.The following are the voltagesat the resistorload Solution The zero sequencereactance of the transfonner is equal to its lVo6=l 2,000 Y; lVu,l= 2,900 V; lV,ol = 2,500 V positive sequencereactance.Hence Choosebaseas 2,500 V, 750 kVA and determinethe line voltagesand Transformerzero sequencereactance: 0.0805 pu currentsin per unit on the delta sideof the transforrner.It may be assumed that the load neutral is not connectedto the neutral of the transformer Generatorzero seqllencereactances: 0.06 pu secondary. Zero sequencereactanceof motor 1 : 0.06 x 10.3 Determinethe symmelrical componentsof three voltages Vo= 2001ff, Vt = 2001245\" and V, = 2001105' y : 0.082

PowerSystemAnalysis ' \"G-\"e*n' erato2r: \"25 MVA, 11kV, Xtt = ZOVo Three-phasetransformer(each):20 MVA, ll Y1220Y kV, X = I5Vo 10.4 A single-phasreesistiveload of 100kVA is connectedacrossiines bc of The negative sequencereactanceof each synchronousmachine is equal a balancedsupplyof 3 kV. Computethe symmetricacl omponentsof the line currents. machine is 8Vo.Assume that the zero sequencereactancesof lines are 25OVoof their positive sequencereactances. 10.5 A delta connectedresistive load is connectedacrossa balanced three- pnasesupply 250f) X = 5o/o at machine2 rating B at machine,l rating Fig.P-10.5 PhasesequenceABC Fig.P-10.8 of 400 V as shown in Fig. P-10.5. Find the symmetricalcomponentsof r0.9For the power systemof Fig. P-10.9draw the positive,negativeand zero line cunents and delta currents. 10.6 Three resistancesof 10, 15 and 2O ohms are connectedin star acrossa sequencenetworks. The generatorsand transformersareratedas follows: three-phasesupply of 200 V per phase as shown in Fig. P-10.6. The G ener at olr: 25 M VA, 11 kV, Xt t =0. 2, X2 = 0. 15,Xo = 0. 03 pu supply neutral is earthedwhile the load neutralis isolated.Find the G ener at o2r: 15 M VA, 11 kV, Xt =0. 2, Xz= 0. 15,X0 = 0. 05 pu currentsin eachload branch and the voltage of load neutral aboveearth. SynchronousMotor 3: 25 MVA, 11 kV, Xt = 0-2, Xz= 0-2,Xo = 0.1 pu Use themethodof symmetricalcomponents' Transformerl: 25 MVA, I 1 Lll20 Y kV, X = IIVo la 2: 12.5MVA, 11 LlI20 Y kV, X = l07o 3: 10 MVA, I2O Ylll Y kV, X - IjVo Ir Choosea baseof 50 MVA, I I kV in the circuit of generatorl. '0.:-\"t T1 tB te c1 tb I- 1_L--__ _l ls Fig.P-10.6 10.7 The voltagesat the terminalsof a balancedload consistingof three20 \\7 ohm X-connecterdesistorsarc 2OO4O\"1, 00 1255.5'and 200 llsf V. Find the line currents from the symmetrical components of the line Fig.P-10.9 voltagesif the neutralof the load is isolated.What relationexistsbetween the syrnrnetricalcomponentsof the line and phase voltages'/ Find the Note: Zero sequencereactanceof each line is 250Voof its positive power expanded in three 20 ohm resistors from the symmetrical sequencereactance. componentsof currentsand voitages. 10.8. Draw the positive, negativeand zero sequenceimpedancenetworks for the power systemof Fig. P-10.8. Choosea baseof 50 MVA, 220 kV in the 50 0 transmissionlines, and mark all reactancesin pu. The ratings of the generatorsand transformers are:

j,,W voo\"inpo*\",,syrt\"r Rn\"tu\"i, 10.10Considerthe circuit shownin Fig. P-10.10.Suppose Vo, = L00 l0 Xr=12 Q vbn = 60 160 Xob=Xbr=X*=5d) Fig.P-10.10 II.I INTRODUCTION (a) Calculate Io, 16,and1, without using symmetricalcomponent. Chapter 9 was devoted to the treatment of symmetrical (three-phase)faults in (b) Calculate Io, 16,and1\" using symmetrical component. a power system.Sincethe systemremainsbalancedduringsuchfaults,analysis could convenientlyproceed on a single-phasebasis. In this chapter,we shall REFERNECES deal with unsymmetricalfaults. Various types of unsymmetrical faults that occur in power systemsare: Books Shunt Wpe Faults l. wagncr,c.F. and R.D. Evans,symmetrit:acl omponenfsM, cGraw-Hill,Ncw york , 1933. (i) Single line-to-ground(LG) fault 2. Clarke,E., Circuit Analysisof Alternating Current Power SystemsV, ol. 1. Wiley, (ii) Line-to-line (LL) fault (iii) Double line-to-ground(LLG) fault Ncw York, 1943. 3. Austin Stigant,5., MasterEquationsand Tablesfor SymmetricalComponentFault Series Type Faults S t u d i e s ,M a c d o n a l d ,L o n d o n , 1 9 6 4 . (i) Open conductor(one or two conductorsopen) fault. 4. stcvensonw, .D., ElcmentsoJP' ower Sy.stemAnalysis,4thedn,McGraw-Hill, New It was statedin Chapter9, that a three-phase(3L) fault being the most severe must be usedto calculatethe rupturing capacityof circuit breakers,even though York, 1982. this type of fault has a low frequency of occurrence,when compared to the 5. Nagrath,I.J. and D.P. Kothai, Electric Machines,2nd edn.,Tata McGraw-Hill, unsymmetricalfaults listed above.There are,however,situationswhen an LG fault can causegreater fault current than a three-phasefault (this may be so New Delhi, 1997. when the fault location is close to large generating units). Apart fiom tliS, unsymmetricalfault analysis is important for relay setting, single-phase Paper switching and systemstability studies(Chapter12). The probability of two or more simultaneousfaults (cross-countryfaults) on 6. Fortescue,C.L., \"Method of SymmetricalCoordinatesAonlies to the Solution of a power system is remote and is therefore ignored in system design for PolyphaseNetworks',AIEE, 1918,37: 1O27. abnormalconditions.

*g9,S.I Modernpower SystemAnatysis UnsymmetricaFl ault Analysis The method of symmetricarcclmponentspresentedin chapter 10, is a Vat (a) powerful tool for study of unsymmetricalfaults and witl be fully exploited in this chapter. (b) (b) UNSYMMETRICAL FAULTS Consider a general power network shown in Fig. 11.1. It is assumedthat a shundtype fault occursat point F in the system,is a resultof which currenm Io, 16/,, flow out of the system,andvo, v6, v\"are voltagesof lines a, b, c with respectto ground. Vao (c) (c) _l lao Fig.11.1 A generapl owernetwork Fig. 11.2 Sequencenetworkas s seen Fig. 11.3 Theveninequivalentosf the fromthe faultpointF sequencenetWorksas seen f r o mt h e f a u l tp o i n t F Let us also assume that the system is operating at no load before the occurrence of a fault. Therefore, the positive sequence voltages of all Dependingupon the type of fault, the sequencecunentsand voltages are synchronousmachineswill bc identicaland will equaltheprefaultvoltageat F. constrained, leading to a particular connection of sequencenetworks. The Let this voltage be labelledas Eo. sequencecurents and voltages and fault currentsand voltages can then be easily computed.We shall now considerthe varioustypesof faults enumerated As seenfrom F, the power systemwill presentpositive, negative and zero earlier. sequencenetworks, which are schematicallyrepresentedby Figs. Il.2a, b and c. The referencebus is indicatedby a thick line and the point F is identified on 11.3 STNGIE LINE-TO-GROUND (tG) FAULT eachsequencenetwork. Sequencevoltagesat F and sequencecurrentsflowing Figure 11.4showsa line-to-groundfault at F in a powersystemthrougha fault out of the networks at F are also shown on the sequencenetworks. Figures impedanceZI. The phasesare so labelled that the fault occurs on phase a. 11.3a,b, andc respectively,give theThevenin equivalentsof the threesequence networks. Recognizingthat voltageEo is presentonly in the positive sequencenetwork andthat thereis no couplingbetweensequencenetworks,the sequencevoltages at F can be expressedin terms of sequencecurrents and Thevenin sequence impedancesas 1r\" ,,1l l=t\"l1ollto,- l 0 ooll|l-r, \"*,l'l (11.1) lr\" 22 Lv\"o)Lol Io o zo)lr\"o) Fig. 11.4 Single line-to-ground(LG) fault at F

At the fault point F, the currentsout of thepower systemand the line to ground UnsvmmetriFcalultAnatvsis fffiH voltagesare constrainedas follows: . (11.7) 'rat = E o Iu=o (rr.2) (zr* zz* z) +3zr Fault current /o is then given by Ir= 0 11 . 3 ) vo = zllo (11.4) (zr* zz* zo)+32 The symmetrical componentsof the fault currents are The above resultscan also be obtaineddirectly from Eqs. (11.5) and (11.6) by using Vop Vo2and Voefrom Eq. (11.1). Thus (Eo- IolZ) + (- IorZr) + (- I\"&d = 3Zf lot I(4+ 4+ Zo)+ 3zfllor= Eo (11.5) Ior = (zr* Eo +3zr ExpressiEnqe ,r':l; ,';',.*1;,ij;.*.ar componenwrseh, ave zz-t Z) Vot * Voz * Voo= ZfIo = 3zflor (11.6) The voltage of line b to ground under fault condition is co1of1nsA.n5eseqacaputeneiorcnnEicboeq.vfsos.let(a1qg1ue.e5sne)caqenunadelt(sw13o1zr.fk6sl)oatthlrlToshueegqiheufaeonnreicm,etcphueerdsreaeneniqtesua3arzteifoenuqstusraurilgoag\"n*endsttiahnesFesigruisem.s Vu= &Vor+ aVo2* Voo =o(,n2-'^+)*4-',+)(.-\"+) Substitutingfor /o from Eq. (11.8) and reorganizing,we ger Va= Eo 3rlz2r + Z2(& - a)+ zo(& - D (11.e) (Zr*22+Zo)+3Zt The expression for V, can be similarly obtained. Fault Occurring Under Loaded Conditions la,t=laz= t\"o=t h When a fault occursunderbalancedload conditions,positivesequencecurrents alone flow in power system before the occurrence of the fault. Therefore, l\"s=I,,1t=t, negativeand zerosequencenetworksarethesanleaswithoutloacl.The positive /a\\ sequencenetwork must of course carry the load current.To account for load current, the synchronousmachines in the positive sequencenetwork are (b) replacedby subtransientt,ransientor synchronousreactances(dependingupon the time after the occulTenceof fault, when currents are to be determined) and Fig' 11'5 Connectioonf sequencenetworkfor a singleline-to-groun(dLG)fault voltages behind appropriate reactances.This changedoesnot disturb the flow of prefault positive sequencecurrents (see Chapter9). This positive sequence In termsof the Theveninequivalentof sequencenetworks, we can write from network would then be usedin the sequencenetwork conhectionof Fig. 11.5a Fig.11.5b. for computing sequencecurrentsunder fault In casethe positive sequencenetwork is replacedby its Thevenin equivalent asin Fig. 11.5b,the Theveninvoltage equalsthe prefaultvoltage Vi atthefault point F (under loadedconditions).The Thevenin impedanceis th6 impedance betweenF and thereferencebus of the passivepositivesequencenetwork (with voltage generatorsshort circuited).

i'4ot',,1 M o d e r nP o @ i s fni, i, illustratecbl y a two machinesystemin Fig. 11.6.It is seenfiom this from whlch we get (11.11) figure that while the prefault currents flow in the actual positive sequence Io2= - Iol network<lfFig. 11.6a,the samcdo not existin its Theveninequivalentnetwork 1.,0= 0 (11.12) of Fig' 11.6b.Therefore,when the Theveninequivalentof positive sequence network is used fot calculatlng fault eurre+ts, the positive sequeneeeurrents within the network are those due to fault alone and we must superimposeon these the prefault currents. Of course, the positive sequencecurrent into the fault is directly the correctanswer,the prefaultcurrentinto the fault being zero. The symmetrical components of voltages at F under fault are Iv,,f ,[t a o'fl'' lr\",l =i at12 l \"I l Ll r\" ,,- I (11.13) LV\"o) Lr l z r ra) Writing the first two equationsw, e have (b) (c) 3vot= vo + (a+ &) vu - &zftu Fig. 11.6 PositivesequencenetworkanditsTheveninequivalenbtefore 3voz= v, + (tr + ,l) vo- ull Iu occurrencoef a fault from which we get ( 11 . 1 4 ) The aboveremarksarevalid for the positive sequenienetwork, independent .r of the type of fault. 3(va- voz)- (a - &1zrtu- iJi y' to t7.4 LrNE-TO-LrNE (LL) FAULT Now 1 6 = ( & - a ) I o t ( : I o z = - I o t , 1 o o= 0 ) Figure I 1.7 showsa line-to-linefault at F in a power systemon phasesb and c through a fault impedanceZf. The phasescan always be relabelled,suchthat = - jJt I\"r (11.15) the fault is on phasesb md c. Substituting16from Eq. (11.15)in Eq. (11.14),we get Vot- Voz=Zf Iot (11.16) b Equations(11.11)and (11.16)suggestparallelconnectionof positive and negativesequencenetworksthrougha seriesintpedanceV as shown in Figs. 11.8aand b. Since loo= 0 as per Eq.(11.12),the zero sequencenetwork is unconnected. l6 Fig. 11.7 Line-to-line(Lt) faultthrough impedanceZ/ Vaz The currents and voltagesat the fault can be expressedas F _1 [1\":o I lat laz l ,I o = I l II,u: t' V o- Vr= IoZf (11.10) Fig. 11.8 Connectionof sequencenetworksfor a line-to-line(LL) fault l In terms of the Thvenin equivalents, we get from Fig. 11.8b -Ir _i (11 . 1 7 ) The symmetricalcomponentsof the fault currentsare

t UnsymmetricaFl ault Analysis--_1 I SQt* F ro mEq . (1 1 .1 5 )w, e g e t or I u' - - ! L - - jJJ' -i uno , Voo= Vot * 3Zf Ino (l 1.23) Z4t,+*+2222\" +2+z+Zf l (\\1. 1 . 1 8 ) From Eqs. (11.19),(ll.22a) and (11.23),we can draw the connectionof sequencenetworksas shownin Figs. 11.10aand b. The readermay verify this Knowingl,-trwecarlcalsqld{ev,,, andv, orfromwhichvsltagesatthefault, by writing mesh and nodal equationsfor thesefigures. be found. b e If the fault occursfrom loadedconditions,the positive sequencenetworkcan modified on the lines of the laterportion of Sec. 1r.3. 11.5 DOUBLE LrNE-TO-cRouND (Lrc) FAULT Figure 11.9 showsa doubreline-to-groundfault at F in a power system.The fault may in generalhave an impedanceZf as shown. a - atr I IEa I Y/u=9 Vaz Vss b- -,..\\. : 1 +Z1 I laz _.__ | r\"|l ____t,'_o_zr!:]ttr'\"\"o Fig. 11.9 Double line-to-ground(.LLG) fault through impedanceZl (b) The current and voltage (to ground) conditions at the fault are expressed as Fig. 11.10 Connectionof sequencenetworksfor a double line-to-ground (LLG) fault Io=o l ln terms of the Thevenin equivalents, we can write from Fig. 11.10b ot i\", + Io, + I\"o =o| ( 11 . 1 9 ) I ,. =l - @ Eo V,,= V, = Zf (lt, + Ir) = 37rf1,,r, (r1.20) The symmetricalcomponentsof voltagesare given by (rr.21) En 1ll !!\"::v''l,1=, ,*,)l[l_'rl a zt + z2(zo*3zI ) I (22+ zo+ 3zt) (11.24) a2 \"or'l]ll l,lvur .Irul) The aboveresult can be obtainedanalyticallyas follows: S ubst ir ut infgor Vut , Vuz and V, * in t er m sof E, , in Eq. ( 11. 1) and I premultiplyingboth sidesby Z-t (inverseof sequenceimpedancematrix), we fiom which it follows rhat get v,,t= V.z= Llv,, + (a + r11V,,1 (Il.22a) ol[r,,-ztrut 3\"' '=17'Ilto;' I z;' o I v,,o=1\"\" + 2vu) (rr.22b\\ Io ll llEn ,,-,-22,,rr,\"+, ,,3 z fI o z;' ,,nl From Eqs. (11.22a)and (t I.Z2b) ;, ilt?l'l;\"1(rr.2s) voo- vot= - ,r- &1 vu= vt = 3zfloo tr, I o o z o t ) L o )L r , o _ J

406t'l Modern powe1_syglgll 4!g!yg!s (11-I..g2lt0ii)t,iUwyeinggerboth sidesby row marrix tl 1 1l and using Eqs. (11.19)and be more than the three-phasefault current. From Eq. (11.22a)w, e have \\ Eo- Ztlot=- Zzloz O< Substituting loz= - (Ior + Io) fseeEq. (11.19)] E o - Z rl o t= Z z (Io r+ Io s) lx, or ? i l. Negative Fig. 11.12 Sequencenetworksof synchronougseneratogr roundedthrough n e u t r a li m p e d a n c e 'I u^ =o -E o - ( zt+z'\\' (d) Write expressionfor neutral groundingreactance,such that the LG fault ,-l current is less than the three-phasefault current. ,\" 1\"' Solution (a) Figure 11.12 gives the se- Substitutingthis value of Iooin Eq. (rr.26) and simplifying, we finally get quencenetworks of the generator.As stated f= earlier voltage source is included in the \"l positive sequencenetwork only. If the fault takesplacefrom loadedconditions,thepositive sequencenetwork (b) Connectionof sequencenetworksfor a will be modified as discussedin Sec. 11.3. solid LG fault (ZI = 0) is shown in Fig. 11.13,from which we can write the fault ire--+- t-f-lP- -.;-f;-!u-trl l,,r| current as ll)rc - 3 lE , l (i) zxt+&,*3X,, Figure 11.11 showsa synchronousgeneratorwhoseneutralis groundedthrough (c) If the neutralis solidly grounded a reactancaXn.The generatorhas balancedemfs and sequencereactancesX1, X., and Xu such that X, = Xz > Xo. l l o lL G= 3 lE \" l (ii) 2= \\ + x c For a solid three-phasefault (seeFig. 11 . 1 4 ) l I ) r r =I;E:o:l t_ 3 l E \" l (iii) xn Ea Comparing(ii) and (iii), it is easyto seethat l q s =1 1 3 1 . : |----l'F:J\"f[\"\\-\"- / . -)- \\ \" / llol Lc> llol3L Fig.11.13 LG fautt +)- ) An important observationis made here that, E \"n( ) I b when the generator neutral is solidly c grounded,LG fault is more severethan a 3I_ ft l fault. It is so because, Xo * Xr = X, in Fig' i i.i i Synehronousgeneratogr roundedthroughneutralreactance generator.However,for a line Xo D Xt = Xz, \\l so that for a fault on a line sufficiently away (a) Draw the sequencenetworks of the generatoras seenfrom the terminals. frorn generator,3L fault will be ntore severe ql (b) Derive expressionfor fault current for a solid line-to-groundfault on than an LG fault. Xt'4 I phasea. 7i L _ _l Fig. 11.14 Three-phase fauft

-a.1.-.4s^&g...-'r}.}.*'*.3a---}l!-}-}.}'Qtpt| 408 usqelrl@ UnsymmetricaFl aultAnalysis lt'iltrffi (d) with generatorneu,tralgroundedthrough reactance,comparing Eqs.(i) 3x1 f and (iii), we have for LG faurt current to be less than 3L fault ( a ) II fr =- j- 0 . 0' A A A e+r j 0 . 0 7s: A A 4 E , + j 0. n .r1+ 0 . 9 : J J e 0.99+ j0.26s 3 lE , l = 2.827- j0.756 24 + Xo+3X, (b) Cunent in the groundingresisror= If = 2.g27_ j0.756 llrl = 2.926x -:L- = 3.07 kA \" J3 xll 2X, + Xo + 3Xn> 3Xl *^, I(xr - xo) (iv) (c) Voltageacrossgroundingresisror= *1 2 r e.g2i - j0.756) J - 0.932- j0.249 i;;;ili\",i:rl = 0.965 = 6 . 1 3k V EvAgeaToesaswlstnsctoiiamehhngrb1tgaaer1eltwteeaoiknnlc(Vriigisnrin),soe,r2Fsti-ehs0stiosoeggMp-lr'afgoeaVltruleuco'An1dtru,itd5avncitn,endhudfgtlrrayhterrjhu,0eeceal.nsp-t1tpitosoo,ghts(cf,oaicijitrs)t0hu.ie.evrcrs,seo5utaan,tr,ht7rreecteohgonr.eiant1sinttn0eeievparcgcmruatrteo.ihnndTuedgandh-eltsdzehnioerensofrrgtsouaoaregtneroqphesruoasiosoei2ftnpon.ett0rchr,oroeeaafehtngeaomdeinncnr(tepeeiaioairsn)afircsatttothehloolreesefsr.l. Forthe systemof Example 10.3the one-linediagramis redrawnin Fig. I 1.16. Ona baseof 25 MVA and I 1 kV in generatorcircuit, the positive,nega\"tiueand zerosequencenetworksof the systemhave beendrawn alreadyin Figs. 10.23, 10.24and 10.27-Before the occurrenceof a solid LG at bus g, the motors are loadedto draw 15 and 7.5 MW at 10 kV, 0.8 leading power factor. Ifiprefault currentis neglected,calculate the fault partsof the system. current and subtransientcurrent in all What voltage behind subtransient reactancesmust be used in a positive sequencenetwork il prefault current is to be accounted fbr'/ F i g .1 1 . 1 5 (H+----Xo= 0.06pu Tz e Xo = 300 fl 2.5A- d )( | Solution (Note:All valuesare given in per unit.) Fig.11. 16 one- r inediagr amof t hesyst emof Exam pr1e1. 3 Since the two identicar generatorsoperatein paralrer, solution The sequencenetworks given in Figs. 10.23, 10.24and r0.2i are Xr.o= j 0 . 1 8 = i0.0g,Xr\"q=t-'# = j0.075 connectedin Fig. 11.17to simulatea solid LG fault at bus g (seeFig. 11.16). rSeianccteatnhcedesoteasrnpooticnot cmfetihnteospeicctounredg.Tehneerreaftoorires,isolatedi,ts zerosequence Z o \" q = 7 O .+130R n = j 0 . 1 0 + 3 x4 : + [f nreforrlt nv ur rrfrfvarnrfl Do qarrva rnravnSllov^v+L^(./i\\ I (ll)', = 0 . 9 9 +ri-0 . 1 E'l= E',!,t= E',1,=2 vj (prefault voltage at g) For an LG fault, using Eq. (11.1g), we ger =1+=o.eopeu Iy (fault currentfor LG fault) - Io = 3lo, = 3E X t\" o+ X z\" ql Zo\" ,

410 I ModernPowerSystemAnalysis . UnsymmetricFaal uttAnatysir Tm-bl Z z = Z r = j 0 . 1 6p u I I Fiomthescquencneetworkconnection .l''. E\"o( ) 'f o r- _ Vf f J l 02 : l 737;4 ( io'szs = q'901 = - jo.Mlpu dt rb-6T--___ --'> j2.032 0 . 9 9+ i 0 . 6 0 7 Ioz= Ioo= Ior = - j0.447 pu l- io.t+z Faultcurrent= 3loo= 3 x (- j0.447) = - jL341 pu The componentof Io, flowing towards g from the generatorside is ------+-i0.136 *_-- - j0.311 j0.447x j!0:.7?s:=s = - 70.136pu -i0.136 .447 i 0.608 11.712 and its componentflowing towardsg from the motors side is - jo.Ml* ij,0?.s7?s-ss==- j0.311pu I -p.oo, Similarly, the componentof Io2from the generatorside is - j0.136 pu and its componentfrom the motorssideis -70.311.All of Iosflowstowardsg from Fig. 11.17 Connectionof the sequencenetworksof Example 11.3. tnotor2. Subtransiencturrentsare shownon thediagramin pu fora solid line-to-grounfadultat g Fault currentsfrom the generatortowards g are The positive sequencenetwork can now be easily replaced by its Thevenin I r,1 [ r I rl [-ro.r30ll-i0.2721 e q u i v a l e nats s h o w ni n F i g . 1 1 . 1 8 . lll G. lr li rl l l l - o . rl =6 7 0 . 1l3p6u Reference bus Irul=laz l Lr.J lo u2 rJL o j L jo.r36j T and to g fiom motors are r [t..] [t I llf-ro.l tt1 [-.r l,06e l I tol-|a2 a r l l- r o . s ul : l-io .r36l p l\" l l e^\" ll\"ll'l' u L/,J la I)L-j0.447JL-j0.136_l The positive and negative sequencecomponentsof the transmissionline cunents are shifted -90\" and +90o respectively, from the corresponding componentson the generatorside of Tr, i.e. Positivesequencecurrent= - j(-jO.136) - - 0.136 pu Negative sequencecurrent - j(- j0.I36) = 0.136 pu Zero sequencecurrent = 0 ('.' there are no zero secluencecurrents on the transmissionline, seeFrg. ll.17) Now Line a current on the transmission line =-Q.136+0.136+0=0 Iu and I, can be similarly calculated.

4l:iN Modernpower SystemAnatvsis UnsymmetricFaat ultAnglysis Let us now calculatethe voltagesbehind subtransienrteactancesto be used lll4#hF if the load currentsare accountedfor. The per unit motor currents are: F = _ j6.06 i0.09+ i0.075 t - u a tal x*q* Xr\", 25x0.909x0.8 136.86 = 0.66+ j0.495pu UsingEq. (11.15),we have = 0.4125136.86'=0.33+ j0.248pu 1y(fault current) = Io = -i 3X-j6.06):-10.496 25x0.909x0.8 Now Total cuffent drawn by both motors = 0.99 + j0.743 pu Vot = Voz = Eo - Iorxleq = 1'0 - (- j6.06) u0.0e) Mo1tot:rr'^'The voltages behind subtransientreactancesare calculatedbelow: = 0.455 =lllt-r:';:=: \"ri;Yiliir.\"il Voo=- I^/'o-g (\"' /oo = 0) Mo2to:Er!\"2 Voltage of the healthy phase, =l.l:t-l,';:::iir:tfi il r\" V o = V o t * V n z * V o o= 0 . 9 1 Generator,E'{ = 0.909+ j0.525 x I.2375136.86\" = 0.52+ j0.52 = 0.735145.pu It may be noted that with thesevoltages behind subtransienrteactailces,the 3orExample 11.2,assumethat the groundedgeneratoris solidly grounded.Find Thevcnin equivalentcircuit will still be the same as that of Fig. 11.19. Therefore, in calculating fault currents taking into account prefault loading .hefault currentin each phaseand voltage of the healthy phasefor a double condition, we need not calculate EIy E/ft and E(. Using the Thevenin ine-to-groundfault on terminals of the generator.Assume solid fault (Zf - 01. eqtrivalentapproach,we canfirst calculatecurrentscauiedby fault to which the )olution Using Eq. (1I.24) and substitutingthe valuesof Zp* Zr\"rand Z*, load currentscan thenbe added. rom E xam ple1l. 2, we get ( not eZf =0, Z0\"q=j0. 1) Thus, the actual value of positive sequencecurrent from the generator rIa l _- t' '+r 7- 0 Jji00:..l:i10:0 - - ; 7 < ? r u w : l r. ^ - - . ^ - l ^ u s rul -l-e I a u lf ^ - - l r t l! s- i- o . o e{j.10:0.Y0. 07j75:+x^5 0.99+ -j0.743 -j0.136= 0.99+ j0.607 Vot= Vo2= Vo1= Eo- Ior Zr.q= 1 - (- j7.53) (/O.09) andtheactualvalueof positivesequencceurrenftromthemotorsto thefault = 0.323 ls' rI a 2 -_ _ V o z - 0.323 - j4.306 -0.99 -j0.743 -j0.311= - 0.99 -jr.054 j0.07s Zr., In this problem, becauseof large zero sequencereactance,load currentis comparablewith (in fact, more than) the fault current. In a large practical tta 0 -- Voo 0.323 .^A^ system,however, the reversewill be the case,so that it is normal practiceto Zo., = t1 /1 neglectload current without causing an appreciableerror. j0.10 For Example I1.2, assumethat the grounded generatoris solidly grounded.Find Iu= rllot + alo, + Ioo fr}rrrov fr^c\"r1l+r r L tn-rtr u* ar*s#r r l a^ -uJ L r ,v, ^ul rr^L- ^i l $ t r u^ fl et tLl^9 ll-r^€^ at r rr ^t l-l-y -prl-l a-s-e - r a r. r a. = (-0.5 -/0.866) (-j7.53)+ (-O.5+ _i0.866U) 4.306+) j3.23 = - 10.248+ j4.842 - 11.3341154.74' IOf lfne-to-llne laUlt On Ir= el,r, + ozlor* Ino terminalsof the generatorsA. ssume solid fault (Zf - 0). = (-0.s + 70.866)(-j7.s3) + (-0.5 -j0.866) (j4.306)+ j3.23 Solution For the LL fault, using Eq. (11.I7) and substitutingthe valuesof X,\"u and Xr\"u from Example 11.2,we get

4!4\" 1 ModernPowerSystemAnalysis UnsymmetricaFl ault Analysis ,415 - 10.248+ j4.842 - 71.334125.28\" | I P o s i t i v es e q u e n c e network Voltage of the healthy phase, V o = 3 Va t = 3 x 0 .3 2 3= 0.969 Negativesequence network 11.6 OPEN CONDUCTOR FAULTS An open conductor fault is in serieswith the line. Line currents and series voltagesbetweenbrokenendsof the conductorsarerequiredto be determined, lc c i c ' Zerosequence network Fig. 11.19 Currentas ndvoltagesin openconductofrault Fig. 11.20 Sequencenetworks for open conductorlaull at FF/ Figure 11.19 shows currentsandvoltagesin an openconductorfault. The ends [n terms of symtnctrical cotnponents, we can wrlte of the system on the sides of the fault are identified as F, F', while the conductorendsare identifiedas ua /, bb/ andcc'. The set of seriescurrentsand Vonl * Voo,2* Vno,g = O voltagesat the fault are Iot= Io2= Ino-+1\" l-l I f v .l l'' l llt',, lI : lv '.o\". o\", 'r p - v: l l l-, 1' 'P I L / I, L V ,), , The symmetrical componentsof currentsand voltagesare [/.'I 1v,,,,,,1 t\". = | I^ .l : v \":l v\"^\"-',I\"l | l \"' l L / . , ,J l V o u ' o) The sequencenetworks can be drawn for the power systemas seenfrom FF/ and areschematicallyshownin Fig. ll.2O. Theseareto be suitably connected dependingon the type of fault (one or two conductorsopen). Two Conductors Open F IF' Figure lI.2l representsthe fault at FF /with conductors b and c open. The 'tD currentsand voltages due to this fault are expressedas I I Voo'= 0 (rr.27) I 16=Ir-Q (11.28) c\"c' I Fig.11.22 Connectioonf sequence networksfor two conductors F i g 1 1 . 2 1 T w o c o n d u c t o r so p e n open

tt'6 il ModernPowerSystemAnatysis t prefault voltage Vf_. of bus r in series with the passivepositive sequence network(all voltagesourcesshortcircuited).Sincenegativeand zerosequence Equations(11.29) and (11.30) suggesta seriesconnectionof sequence prefault voltages are zero,both these are passivenetworksonly. networksas shown in Fig. II.22. Sequencecurrelrtsand voltagescan now be Reference bus for passive computed. positive sequence Onc Con-luctor Open network For one conductoropen as in Fig. 11.23,the circuit conditionsrequire Vbb,=Vrr,=O (11.31) Io = O (11.32) In termsof symmetricalcomponentstheseconditionscan be expressedas 1 (11.33) Vool= Voor2= Voory= *Voo, (11.34) J Iot* In + Ioo=0 Equations(11.33) and (1I.34) suggesta parallel connection of sequence networksas shown in Fie. 1I.24. [--l Fig. 11.25 Connectionof sequencenetworksfor LG fault on the r th bus (positivesequencenetwork tl representedby itsTheveninequivalent) I V\"\"'z I It may be notedthat subscripta has beendroppedin sequencecurrentsand L-'i-- ' ryl voltages, while integer subscriptis introduced for bus identification. Super- scriptso and/respectively, indicate prefault and postfaultvalues. For the passivepositivesequencenetwork F Vr-\"us= Zr-nusJr-\"ut (11.35) la a where V t - u u s= positivesequencbeus voltagevector (11.36) ,t c + - c', c/ lao i Fig. 11.24 Connectionof sequence networks for one conductor Fig. 11.23 One conductoropen open Z-trl Zr-nus : l- positive sequencebus impedancematrix II.7 BUS IMPEDANCE MATRIX METHOD FOR ANALYSIS OF Zt-nn ) /1 1 2?\\ UNSYMMETRICAL SHUNT FAULTS and \\L r.J t ) Bus impedancemethod of fault analysis, given for symmetrical faults in [/'-I' Chapter9, canbe easilyextendedto thecaseof unsymmetricaflaults.Consider fbr examplean LG fault on the rth bus of a n-bus system.The connectionof Jr-sus= l t t . ' | = p o s i t i vsee q u e n cbeuscunentinjectionvector (l1.38) sequencenetworksto simulatethe fault is shown in Fig. I1.25. The positive l:l sequencenetwork has been replacedhere by its Thevenin equivalent,i,e. rl

t4l8 | todern power SvstemAnalvsis I From the Sequencenetworkconnectionof Fig. 11.25,we can now write As per the sequencenetwork connection,current- IJr_, is injectedonly at the faulted rth bus of the positive sequencenetwork, we have therefore vro-, (rr.47) 2r-,, * zz-,,I Zo-r,+3zf -- tr2f- r ( 11 . 3 e ) ue other types of faults can be simila*Seomputed using Zl-rr, Zr-r, and Zo-n i n p l a c eo f Z r , Z r a n d Z o i n E q s . ( 1 I . 7 ) , ( 1 1 . 1 7 )a n d s u b s t i t u t i n gE q . ( 1 1 . 3 9 )i n E q . ( 1 1 . 3 5 ) ,w e c a n w r i r e t h e p o s i t i v esequence (11.24w) ith E, - Vi-,. voltage at the rth bus of the passivepositive sequencenetwork as Postfaulst equencveoltagesat anybuscannowbecomputedby superposing V,-,'= - Zr-rrlfr-, on prefault bus voltage, the voltage developedowing to the injection of appropriatesequencecurrentat bus r'. Foi passivepositive^sequencne twork,the voltagedevelopedat bus i owing to the injection of - IIr-, at bus r is ( 11 . 4 0 ) Vt-r=- Zr-,Jfr-, (11.48) Thus the passivepositive seguencenetworkpresentsan impedanceZr_ r, to the Hencepostfaultpositivesequencveoltageat bus I is givenby (l r.49) p o s i t i v es e q u e n cceu r r e n tI { _ r . Vl-,= Vi-,- Zr-,,fr-,;i = l' 7' \"'' tt For the negativesequencenetwork Vz-uus= Zz_susJz_nus (11.41) where The negativesequencenetwork is injectedwith current lfr_, at the rth bus prefaultpositivesequencevoltageat bus i only. Therefore, Zr-,, = irth componenot f Zt-\"ut 0 Since the prefault negativesequencebus voltagesare zero, the postfault 0 negativesequencebus voltagesare given by Jz-sus= -,i{, (1r.42) Vf'-'=0+ Vz-r -- - zr-,rl'fr-, (l r.s0) 0 where l r - , , = i r t l t c o l l l p o l l c l t to l ' Z t - t , , t The negative sequencevoltage at the rth bus is then given by Similarly, the postfault zero sequence bus voltages are given by Yr,=- zr rrlf, I (l 1.43) (il.s1) Vd-' = - Zu-''lfu-''; j = l' 2' \"'' tI Thus,thenegativesequencenetworkoffersan impedanceZr_rrto thenegative sequencecurrentltr_, where Sirnilarly,fbr the zero scqucncenetwork = irth component tlf Z9-sLr. Vu-uu,= Zo-susJ,,-\",r, ( 1 1. 4 4 ) With postfaultsequencevoltagesknown at the buses,sequencecurrentsin 0 linescan be comPr'rteads: 0 For line uv, having sequenceadrnittarlcesyl -ur, Jz',u,and yo-r, Jo-sus= - r^{u-r (l l.4s) f , - r r = l t - u , ( V f t - u- v [ - r ) (rr.s2) I f r - r r = ! 2 - , , r (, V t - , ,- V 5 - r , ) I i, r - ur = Jo- u,( Vio- , - Vf o- r l 0 Knowing sequencevoltagesandcurrents,phasevoltagesandcurrentscanbe and Vo_, = - Zs_,.rlfs_,., easilycomputedby the useof the symmetricalcomponent ransformation (r1.46) Vr, = AV\" That is, the zero s.equencenetwork off'ers zrn intpeclance Zrr_,.,t.o the zero Ir,= AI, sequence cuffent l'-*r.

'i[di;,l Modern powSrJy$ern_4nglysis It appearsat first, as if this methodis more laboriousthancomputingfault UnsymmetFriacaulnt narvsis liffiffi Yr-aa= l_+ _+^- =_ jtj.422 currentsfrom Thevenin impedancesof the sequencenetworks, as it requires computationof bus impedancematricesof all the three sequencenetworks.It io.z io.o'os must,however,be pointedout herethatoncethe bus impedancematriceqhave Y r - r e -Y r - a \" = : jr2-4zz beenassembledf,ault analysiscanbe convenientlycarriedout for all the buses, to;;t which, in fact, is the aim of a fault study. Moreover, bus impedancematrices Y r - n Y= t_ , ,= .# - -i78.s1e can be easilymodifiedto accountfor changesin power networkconfiguration. #* Yr-\"-f- i6.0s1 *h: For Example 10.3,positive, negativeand zero sequencenetworkshave been drawn in Figs. 70.23, 10.24 and 10:27.Using the bus impedancemethod of -Yt.- 8 8 * I + I =-i16.769 fault analysis,find fault currentsfor a solid LG fault at (i) bus e and (ii) bus j0.085 j0.345 j0.69 I Also find bus voltages and line currentsin case (i). Assume the prefault currentsto be zero and the prefault voltagesto be 1 pu. d e fI Solution Figure 11.26 shows the connection of the sequencenetworks of Figs. 10.23, 10:24 and 10.27for a solid LG fault at bus e. -t7.422 12.422 0 0 E\"* rv I - B U S _ Iv2 _ B U S _ t2.422 - 18.519 6.097 0 I Positivseequence a 0 6.097 - 18.519 12.422 network 10.345 0 0 12.422 -16.769 @@ rv o - d d - I = _ j0.62I j1.608 'vo - e e -- 'Yt t,- t t -- - - 1 =-i14.446 ' j0.494 j0.0805 Negativesequence 1 Yr-rr=jth=-io'584 network iO'345 i 0.6e Y o - a r =o ' o @@ Yo-,r=jo-f.1u = j2.024 i0.0805 Yo-fs= 0.0 Zero sequence d e f 6o network 0 orl-0.621 -r4.M6 0 0 {@ 2.024 0 @ JU.4v+ /U.UUUs v/ \" r r lIv O_BUS_ 0 2.024 -14.446 0 0 0 -0.584 Fig. 11.26 Connectioonf the sequencenetworksof Example11.6 for an LG 0 faultat bus e 0 Refer to Fig. 11.26to find the elementsof the bus admittancematricesof the lnverting the threematricesaboverendersthe fbllowing threebus impedance three sequencenefworks.as follows: matrices

Analysis 00 0 - - 0.417pu 0.07061 0.00989 0 VL\" = - Zu,,Ifo-, = - j0.0706x (- j2.362) 0.00989 0.07061 0 - - 0.167pu 0 0 r.71233 Vfz-r=- 4-r\"llr1 =-j0.08558 x(- j2.362) The fault currentwith LG fault on busc is = - 0.202pl il= j 0 J 76 3 6+ j 3x1 j = - j 7 . 0 8 6p u (i) e 0 . r 76 3 6+ Vfus=-ZurJfu\"=O Using Eq. (11.52), the currents in various parts of Fig. 1I.26 can be 0 . 0 70 6 1 cornputedas follows: The fault curent with LG faulton bus/'is II+= Yr-pUI+- vI+,) Irr __3x1 )a - - j6.097 (0.728- 0.s84) = _ 70.88 .i0.t82gg+ iOlAZgg+/O\"O?Gi j0.436s9 If ,-0\"= Yr4\" (Vf ,-a - Vf ,-r) = - 7 6 . 8 7 1p u (ii) = - j12.422(0.703- 0.584=) - jL482 Bus voltagesand line currentsin c as e(i) can easily be c om p u t c du s i n g Eqr. (1 1.49)-(tI.sZ). Given below is a samplecalculation for computing I,,t= If t-f\" * If t-,t,= - 70.88 + (- ,tl .482) voltage - - i2.362 at busf andcurrent in line ef which is the sameas obtainedearlier [seeEq. (i)l where If, = 3lut. FrorrrEq. (11.49) IJrsf = YFcf (vf '-, - vf '-) - j12.422 (-0.798- 0.728)= - i0.88 VI-a= Vi_a- Zr_0\"- Ifr_\" Notice that as per Fig. 11.26,it was requiredto be the sameas llr-s\". = t.0- j 0 . t 2 s 7 s ( -' 7'0ttb = 0 . 7 0 3p u Iz-f\" = Yr-r, (Vfr-r - Vf,-r) \\\" 3 )) = - j6. 097 ( - 0. 212+ 0. 417)= - 7O . 884 IL- tr.= nt-1a(Vtt-1- Vfrr,) Vft-t=Vi-t - Z, tu- If,., = - j2.024 (- 0.023+ 0.167)= - jO.29Ipu = r.o- j o . n s 4 j( - r , orru = o.z z sp , Ittn (a) = IJr-f\" * It)-r, + {* ) = - j0.88 + (- 70.88)+ (- j0.291) V J r - \"= V \" r - \" - Z r _ \" o - l Jr _ \" - - jz.os = 1.0- i0.1763(8- j2.363)= 0.584pu Sirnilarly,othercurrentscan be computed. vI-r= vi-s- z,-r\"-I{-, A single line to ground fault (on phasea) occurson the bus I of the systemof = 1.0- j0.08558(- j2.363)= 0.798pu Fig. 11.27.Find vfz-f= - zr-.fJfr-\" = - ./0.1154x7(- j2.362)= - 0.272pu Vfo-r=- Zo-f\"Ifo=-,- j0.00989x (- j2.362) = _ 0.023pu

I qU;l ModernPowerSystemAnatysis UnsymmetrlcFaal ultAnalysis-T f,r'4l# zr-ws=irLt-o0..1o0+5s0o..014o515.ZJ:z-.sus (i) Tero sequencenetwork of the system is drawn in Fig. II.29 and its bus matnx ls ted below. Referencebus Ftg. 11.27 0.15 0.15 (a) Current in the fault. 0.05 0.05 (b) sc current on the transmissionline in all the three phases. : (c) SC current in phasea of the generator. 0.05 0.4 0.05 (d) Voltage of the healrhy phasesof the bus 1. Given: Rating of each machine 1200 kvA, 600 v with x, = x, = rTvo, Fig.11.29 xo = 5vo.Each three-phasetransformeris rated rz0o kvA, 600 v - nlgroo V-Y with leakagereactanceof SVoT, he reactancesof thetransmissionline are Bus 1 to ref'erencebus xr = Xz = 20voandXo = 40voon a baseof 1200kVA, 3300 V. The reactances Zo-sus= i [0.05] of the neutral grounding reactors are5Voon the kVA and voltage base of the machine. Bus 2 to bus I Note: Use Z\"u, method. Zo- n=us/Ll-0o..0o5s 0.051 Solution Figure 11.28 shows the passivepositive sequencenetwork of the 0.451 systemof Fig.l1.27. This alsorepresentsthe negativesequencenetwork for the system.Bus impedancematricesare computedbelow: Bus 2 to referencebus Reference bus z7O-BUS - 'Ll-0o.0.os s00..04_5sl110 . 4 ; l-0.051 . o so . 4 s l 5+ 0.0sl10._4l5-t1o or Z o - s u=sJ.Z1 I[O0 ..0O4Os 500 .. 00405511 (ii) 0.05 0.2 0.05 As perEq.(11.47) II-t = VO Flg. 11.28 Z ; t r * Z z _ t r* Z o _ r +, 3 Z l Bus I to referencebus But V\", = I Pu (sYstemunloadedbefore fault) Zt_svs= j[0.15] Then Bus 2 to Bus I fr_, -j1.0 = - j3.92pu ; ; ; lZr-8,tr,-.=r,lfOo'.1tS5 o l 5 t ,0.105+0.105+0.045 -^2Pu Ifr-t=ltr;=f Bus2 to referencbeus (a) Fault current, I\\ = 3If ,-, = z L Brrs, .-=\", /[ oL'o1 5. r 0.1s1 r Ll-Oo..trs[tto.l'ls o'3s] O) Vfvr = Vor-, = Zt-rr lfr-t so35J*-* \"\" = 1. 0- 70. 105x - j3. 92 = 0. 588;Vot ; = |

Vft-r= Vor-r- Zr-rJfr-ri Vora = 1.0(systemunloadedbeforefault) ( c) r c = l (1 - o's88)t-33\" = 1.0- j0.M5 x - j3.92 = 0.g24 4- ,o; vtr;=- zr-trfr-t ---1.37-i2.38 = -70.105x - j3.92 = 0.412 V{-z=- Zr-r, Ifr-, rLc= j.0:=.1s to - (- 0.412))t3o\" - - j0.045x - j3.92 = _ 0.176 r.37- i2.38 vfo-t= - zurrlL, IIo-c= 0 (seeFig' 1I'29) Ifo_c= et37 - j2.38)+ (1.37- i2.38) - - j0.045x - j3.92 = _ 0.176 = _ j4.76 vt-z=- Zuy IL, Currentin phasesb andt:cof the generatorcan be similarlycalculated. = - 7O.005x - j3.92 = - 0.02 (d) Vfo-r= ZVfvr + Vf,-, + Vfur I{-rz= yvrz (VI_r- Vfr_r) = 0.588 1240\"- 0.4L2 1120\"- 0.176 = - 0.264- j0.866= 0.905l- 107\" = - 1- ( 0 . 5 8 8- 0 . 8 2 4 ) =j l . r 8 j0.2 VIr-t= Vf,-t + Vfr-,+ Vfu' = 0.588ll20' - 0.412 1240\"- 0.176 Ifr-rr= !z-n U{-t - Vfr-r) - - 0.264+ i0.866= 0.905 1107\" = *j .0r.- 2 ' 0 . 4 1 2+ 0 . 1 7 6=) i 1 . 1 8 If o_r,= lo_tz(Vfu, - Vfur) = (- 0'176+ o.o2o=) io.3s PROLBEIVIS ;, 11.1 A 25 MVA, 11 kv geaeratorhas a x\"o= 0'2 pu' Its negativeand zero sequencereactancesarerespectively0.3 and 0.1 pu. The neutralof the generatoiis solidly grounded.Determinethe subtransienct urrent in the generatorand the line-to-line voltagesfor subtransienct onditions when an LG f'ault occurs at the generatorterminals. Assumethat before the occuffence of the fault, the generatoris operatingat no load at rated voltage.Ignore resistances. 11.2RepeatProblem 11.1 for (a) an LL fault; and (b) an LLG fault. Iro _ tz =j l .1 8 + 7 1 .1 8+ j 0.39_ j 2.75 11.3A synchronousgeneratoris rated 25 MVA, 11 kV. It is star-connected with the neutralpoint solidly grounded.The generatoris operatingat no -7 f D. _,t-. z-_ Ji lr . trau - roo load at rated voltage. Its reactancesare Xt' = Xz = 0.20 and Xo = 0'08 r 4.-/ a1vA I r o _rr . 'r1. z /_7t ^Lfvr . .n ^^ rr^r^i-r^+^ +Lll^l s^ -D. ,Jmr l,u-uov+ler ri nv 4orl oo ur rvl rufer or an c i c n t l i n e e t r r r e . n f s f o r ( i ) -S-i-n- oS-l -e J + Jv.Jy pu. \\-aruulalLE = _ j07g line-to-ground fault; (ii) double line fault; (iii) double line-to-ground fault; and (iv) symmetrical three-phasefault. Comparethese currents If,-rz= jl.18 lI20 + 71.1glZ4V + il.3g =j0.79 andcomment. ll.4 For the generatorof ProblemI 1.3,calculatethe valueof reactanceto be includedin the generatorneutralandground,so thatline-to-groundfault

lW:il Mqdernpower Syst€mAnatysis current equalsthe three-phasefault current. What will be the value of t*t. sub-^ the groundingresistancteo achicvcthc samcconclition,l prelaultcurrentignored. with the reactance value (as calculated above) included between neutral and ground, calculate the double line fault current ancl rlso 7 . 5M V A double line-to-ground faul 3 . 3 / 0 , 6k v X = 10o/o 11'5 Two 25 MVA, 11 kv synchronousgeneratorsare connected to a common bus bar which suppliesa feeder. The star point of generatorsis groundedthrougha resistanceof 1.0ohm, while one of the that of the other generatoris isolated.A line-to-ground fault occursat the far end fauft X'= Xz=2oo/o of the feeder.Determine:(a) the fault current; (b) the voltage to ground of the sound phasesof the feeder at the Xo = 5o/o the star point of the groundedgenerator fault ipeosipnet;cattnodgrr.ol u\"nodri.lg\" bi Xn = 2'5o/o with Yr6i* The impedancesto sequencecurrentsof eachgeneratorand feederare given below: F i g .P - l 1 . 8 Generator Feeder n.g A doubleline-to-groundfault occurson lines b and c at point F in the system of Fig. i-tf.q. Find the subffansientcurrent in phase c of (per unit) (ohms/phase) machine 1, assumingprefault currentsto be zero. Both machines are rated 7,200kvA,600 v with reactanceosf x//= xz=lUvo and xo= Positive sequence jo.2 j0.4 5%. llac5 thrcc-phirstcrtnslorutcris rutcd 1.200kVA. 600 V-A/3.300 V-ts with leakagereactanceof 5Vo.The reactancesof the transmission Negative sequence i0.15 j0.4 line areX,=X,-=20vo andXo= 4oToon a baseof 1,200kVA, 3,300V' The reactancesof the neutral grounding reactors are 5Voon the kVA Zero sequence j0.08 j0.8 baseof the machines. rI'6 Determine the fault currentsin each phasefollowing a double line-to- ground short circuit at the terminals of a star-connectedsynchronous generatoroperatinginitially positive, negative and zero on an open circuit voltage or i.o pu. The sequencereactanceof the generator are, respectively,70.35,j0.25 andj0.20, and its star point is isolated from ground. 11'7 A three-phasesynchronousgeneratorhas positive,negativeand zero vn6l sequencereactancepser phaserespectively,of 1.0,0.g and0.4 ohm. The winding resistancesare negligible.The phasesequenceof the generator _r_rr\"*1_ i^ -: is RYB with a no loadvoltageof I I kV betweenlines.A shortcircuit occurs betweenlines I and B and earth at the generatorterminals. ,_L F l g .P ' 1 1 . 9 Calculatesequencecurrentsin phaseR and currentin the earthreturn 11.10 A synchronousmachine1 generating1 pu voltageis connectedthrough circuit, (a) if the gencratorneutralis solidly earthecla; ncl(b) il a Y/Y transformerof reactance0.1 pu to two transmissionlines in generatorneutral is isolated. the parallel. The other ends of the lines are connected through a YN Use R phasevoltageas reference. transformerof reactance0.1 pu to a machine2 generating1 pu voltage' 11.8 A generatorsuppliesa group of identical motors as shown in Fig. For both transformers X, = Xz = Xo' P-11'8. The motors are rated 600 V, 9O%oefficiencyat full load unity Calculatethe currentfed into a fdeodufbrolemlinme-atcoh-ginreou'2n. dTfhaeulsttoanr ptohienltinoef power factor with sum of their output ratings being i rrrrW.The motors 'side terminals of the transformer a f e s h a r i n o e n r r q l l r r so rl ^vos. ul ^\\ ^€ '.r/ t ll \\vt rr rvrv i^ltr l-a^ .L^ €r u -v-'-r' .' a g e , u . 6 p o w e r t a c t o r machine I and of the two transiormers.are soiiriiy grourrded. The reactancesof the machines and lines referred to a common base are lagging and90voefficiency whea an LG fault occurson the low voltage xl Xz xo side of the transformer. ttoreSianptceeludcadifsyeacthoseminpeglfelfeeteecltqyoutfhiveparsleeefqanumut elotnctcouenrrr.eetnwto. Trkhsetogrsoimupuloaftemthoetofarsulctasnobaes Machine 1 0.35 0.25 0.05 Machine 2 0.30 0.20 0.04 Line (each) 0.40 0.40 0.80

Modernpower ,:ill'll iltr\"::,i\"il\"ti .1T:-Toow.enr erworwk ithtwogeneratocrsonnecred ' I lncrrrnmatrinal Farrlt Anahraio lL.'A+\\l i:i,Tl\":\"litill:'l\"ji1..\"\":5.:'\":1.iT1::',:1{:::\":u,1,inil.fn.rffr:::ini\"9,ti1e,fr,u,;s.i;*iJli;l;l*;.'':;;^f;f:\"i#:.liHl::lii.f;fH,i-ffifl\"f#f\",ifi;ffi*\"#,J'#::,irfrfvli:ltfr{'o\"m,,\"v:vt''rTv\\h,reerpr,o-\"tin+-p:Erqfi\"itge*:E\" yuenozeroseque;r1:#;:\"# \"\"*\"t-;;;il#, Generator: Xr = Xz = 0.1 pui Xo = 0.05 pu Positive Negative X, (Broundingreactance)= 0.02 pu Zero G e n e r a t o rI Transformer: Xr: Xr:Xt = 0J?u 0.15 0.15 0.0g X, (grounding reactance)= 0.04 pu Generaror2 0.25 0.25 oo (i.e. neutralisolated) Form the positive, negative and zero sequencebus impedancematrices. Each rransformer 0.15 0.15 0.15 For a solid LG fault at bus 1, calculate the fault current and its contributionsfrom the generatorand ffansformer. Infinite bus 0.15 0.15 0.05 1,L2 Line 0.20 0.20 0.40 _f6l-Y €fff (l.b]) fsnPwloto_aiwt-rhr.-io.nc.abgo.*don,t,(rnhaih)eesgrcieinentqenetudehw-etreioannf-rcdagoeiunrrnosleagi;utnwoandfoncfinratrdhkfuis(enriiotit)torfeatcthnchbrsueoufrouspsrgooamhpwt e\"r.ernhrurAeet;.i;tno_rcaffan\"tiachsuferol.atoremterpmeutrhiAnve.oaclrtusaorgrfeetnohtnes rTft-YA 6L-_' ,q F i g .P - 1 1 . 1 3 Hint: Notice that the line reactancesare not given. Therefore it is convenientto obtain Zt, svs directly rather than by inverting Ir, sus. Also ro, it singular and zs, BUScannotbe obtainedfrom it. In such - _ t\\--7x| )r IIl situation\"sutshe method of unit current injection outlined below can be lI_ J]L | used. ,- I t z 2 , H /\"\\ . .Y' - For a two-buscase F i g .p - 1 1 . 1 1 li;,1=17,',7,',)l';,1 rl'r2 srAcclscteooy'aiqnnstscnutntnteaseeaemlrnc0nccct.tcteo'eBeetrdtndeusltAncg:.solleleic'rlcbttnbs*taeusnetrasrccdarrseIbpitycsooaonrsfrrrscwwnv2hcti'irtrstohaicohonarnuennaeosesuqboctcsurgetloaieae,drndr\"rnrqsltypepoumcoroiaouisnbstnuieuostntinisrevoefteensnbcaetltairilendnyerdsc'egrt2oerb.ivpugTtnerhsrhcenorbegsouba.e'nepgtrnidrohvotr.wecTwdaoehb:sfreycsnapt(aareeprrtr./wosruct0wonraerrierctkr_rctas Injectingunit currentat bus 1 (i.e. Ir = tr, !2= 0), we get pol;itive Nc,,gutivt: Zt:ro Zn= Vt Generator 0.20 0.l5 0.05 Zzt = Vz Transfbrmer OJZ 0.12 0.12 S ir nilar lyinjcct ingr r uitcur r cutut bus 2 1i. c./ r = 0, lz = l) , we get Ztz= Vl T r a n s m i s s i o nL i n e 0.30 0.30 0.50 7:tz= Vz PowerNerwork X X Zou5could thus bc dircctly obtainedby this technique. Under no load condition with 1.0 go v 0.10 ll.l4 Considerthe 2-bussystemof Example11.3.Assumethat a solid LL of 4'0 pu is fed to a three-phase short oltage at each bus bar, a current fault occurson busf Determinethefault currentand voltage(to ground) the positive sequence reactance X of circ-uit on bus bar Z.Deitrmhe of the healtlryphase. power network. the equivarent ,, . generator of the 11.15 Write a computer programmeto be employedfor studying a solid LG fault on bus 2 of the sy'stemshown in Fig. 9.17. our aim is to find the For the same initial conditions, find the faurt current for single line_ fault current and all bus voltagesand the line currents following the to-ground fault on bus bar l. fault. use the impedance data given in Example 9.5. Assume all transformersto be YlA type with their neutrals (on HV side) solidly grounded.

Assume that the positive and negative sequencer,eactancesof the il generatorsare equal,while their zero sequencereactanceis one-fourth of their positive sequencereactance.The zerosequencereactancesof the 12 lines are to be takenas 2.5 times their positive sequencereactancesS. et all prefault voltages = 1 pu. T2.T INTRODUCTION REFERNECES The stabilityof an interconnectepdower systemis its ability to returnto normal Books or stable operation after having been subjectedto some form of disturbance. 1. StevensonW, .D., Elementsof Power SystemAnalysis,4th edn., McGraw-Hill, Conversely, instability means a condition denoting loss of synchronism or New York, 1982. falling out of step.Stabilityconsiderationshavebeenrecognizedas an essential 2. Elgerd, O.I., Electric Energy Systems Theory: An Introduction, 2nd edn., McGraw-Hill, New York, 1982. part of power system planning for a long time. With interconnectedsystems 3. Gross,C.A., Power SystemAnalysis, Wiley, New York, 1979. continually growing in size and extendingover vast geographicalregions,it is 4. NcuenswandeJr,.R., Modern Power SystemsI,nternationaTl extbookCo., Ncw becomingincreasinglymore difficult to maintainsynchrortismbdtweenvarious York, 1971. '. 5. Bergan,A.R. andV. Vittal, Power SystemAnalysis,2nd edn.,PearsonEducation parts of a power system. Asia, Delhi, 2000. The dynamicsof a power systemarecharacterisedby its basic featuresgiven 6. Soman, S.A, S.A. Khapardeand Shubha Pandit, ComputationalMethodsfor tjElt w. Large SparsePower SystemsAnalysis, KAP, Boston, 2002. 1. Synchronoustie exhibits the typical behaviourthat as power transfer is Papers gradually increaseda maximum limit is reachedbeyond which the system 7. Brown, H.E. and C.E. Person,\"Short Circuit Studiesof Large Systemsby the cannotstay in synchronism,i.e., it falls out of step. ImpedanceMatrix Method\",Proc. PICA, 1967,p. 335. 2. The systemis basicallya spring-inertiaoscillatorysystemwith inertia on 8. Smith, D.R., \"Digital Simulation of SimultaneousUrrbalancesInvolving Open the mechanicalside and spring action providedby the synchronoustie wherein and FaultcdConductors\"IE, EE Trans.PnS, 1970,1826. power transfer is proportional to sin d or d (for small E, 6 being the relative internal angleof machines). 3. Because of power transfer being proportional to sin d, the equation determining system dynamics is nonlinear for disturbancescausing large variationsin angle d, Stability phenomenonpeculiar to non-linear systemsas distinguishedfrom linear systems is thereforeexhibited by power systems (stable up to a certain magnitude of disturbanceand unstable for larger disturbances). Accordi^rglypower systemstability problemsare classifiedinto three basic types*-steady state,dynamic and transient. *There are no universally accepted precise definitions of this terminology. For a definition of some important tenns related to power system stability, refer to IEEE StandardDictionary of Electrical and Electronic Terms, IEEE, New York, 19i2.

'l Modernpower SystemAnalysis Dnrrrn' Qrra+^- Or^L:l:r-. I -^- 434 .'l Th\"t study of steady state stability is basically concerned with the machineswhich are not separatedby lines of high reactance are lumped determinationof the upperlimit of machineloadingsbeiorelosing synchronism, togetherand consideredas one equivalent machine.Thus a multimachine providedthe loadingis increasedgradually. system can often be reduced to aq equlyalg4t fery lq4qhrue system- If Synchronlsmii lost, ttrern'achineiof eaitr gioup staytogetheralthough they go Dynamic instability is more probable than steady state instability. Small out of step with other groups.Qualitative behaviourof machinesin an actual i\" systemis usually that of a two machinesystem.Becauseof its simplicity, the disturbances are esntinuaHy oeeurring irr a po*.. system r\"#\"ti\"* two machine systemis extremelyuseful in describingthe generalconceptsof loadings,changesin turbine speeds,etc.) which are small enoughnot to cause powersystemstabilityand the influenceof varioust'actorson stability. It will be seenin this chaptertbata two machine systemcan be regardedas a single thesystemto lose synchronismbut do excitethe systeminto the\"itateof natural machinesystemconnected.toinfinite system. oscillations.The systemis said to be dynamically stableif the oscillationsdo not acquiremore thancertainamplitudeanddie out quickly (i.e.,the systemis Stabilitystudy of a multimachinesystemmustnecessarilybe carriedout on well-damped).In a dynamically a digital computer. large and thesepersistfor a long unstablesystem,the oscillation amplitude is time (i.e.,the systemis underda-p\"a;. rni, kind of instability behaviourconstitutesa seriousthreat to systemsecurityand createsvery difficult operating conditions. Dynamic stability can be signifi- cantly improved through the use of power systemstabilizers.Dynamic system studyhas to be carried out for 5-10 s and sometimesup to 30 s. computer simulationis the only effective meansof studyingdynamic stability problems. I2.2 DYNAMICS OF A SYNCHRONOUS MACHINE The samesimulationprogrammesare,of course,appiicableto transientstability The kinetic energyof the rotor at synchronousmachineis studies. Following a sudden disturbance on a power system rotor speeds,rotor K E = 1 JJ,^ x 10-6MJ angulardifferencesand power transferundergofast changeswhosemagnitudes 2 aredependenut ponthe severityof disturbanceF. or a largedisturbanc\", where -/ = rotor momentof inertia in kg-m2 aro, = synchronousspeedin rad (mech)/s in angulardifferencesmay be so large as to .:ausethe machinesto fal\"lhouuntgo..f step'This type of instability is known as transient instability and is a fast phenomenonusuallyoccurringwithin I s fbr a generat orcloseto the causeof But u.r,n= rotor speed in rad (elect)/s disturbanceT. hereis a large rangeof disturbanceiwhich may occur on a power systemb, ut a fault on a heavilyloacleclilne which requiresopeningthc lipc t<l clearthefault is usually of greatestconcern.The tripping of a loadJdgenerator whorc P = nuurbelo1'rnaclrinepoles or the abrupt dropping of a large load may also causeinstability. KE=+(t(?)'cx.r,.o-.)* whTichheaepffoewcteorfssyhstoermtcisircsuuibtsje(cfateudlt,ms)u,tshtebme doesttesremvienreedtyinpeneoaf rdlyisatullrsbtaanbcileittoy studies' During a fault, electrical power from nearby generatorsis reduced drastically,while power from remotegeneratorsis scarcelyaf1'ecte4ln. some -L M, cases'the systemmay be stable even with a sustainedfault, whereasother 2', systemswill be stable only if the fault is cleared with sufficient rapidity. M = J ( ?\\' u , x' ,1 0 - 6 sWyhsteetmheitsrtehlef,sbyusttaelmsoisosntathbeletoypneoocfcfuarureltn, lcoecoaf taiofnaoufltfadueipt,ernadpsindoittyoonflcyleoanrtihnge \\P/ and method of clearing,i.e., whetherclearedby the sequentialopeningof two = moment of inertia in MJ-s/electrad or more breakersor by simultaneousopeningand whether or not the faulted line We shall detine the inertia constantH such that is reclosed.The transientstability limit is atmostalways lower than the steady statelimit, but unlike the latter,it may exhibit different G H = K E =!2u % M J nature,location and magnitudeof disturbance. valuesdependingon the G = machinerating (base) in MVA (3-phase) Modern power systemshavemany interconnectedgeneratingstations,each H = inertictconstantin MJ/I4VA or MW-s/MVA with severalgeneratorsand many loads.The machineJlocateda-tany onepoint ln a systemnormally act in unison.It is, therefore,commonpracticein stability

{'.3: 6-: PowerSystemStability II 437 il MociernPowerSysiemnnaiysis e Swing Equation i I ;ure 12.1 shows the torque,speedand flow of mechanicaland electrical wersin a synchronousmachine.It is assumedthat the windage,triction and It immediatelyfollows that n-loss torque is negligible. The differential equation governing the rotor namicscan then be written as M = 2GH = GH MJ-s/electrad (r2.r) (ts lt f J- -d:;'to ^ =T^- r\" Nt (r2.3) = ffi 14J-s/electdegree 180-f ot- M is alsocalledthe inertiaconstant. 'here Taking G as base,the inertia constantin pu is 0* = angle in rad (mech) M (pu) = +n f s2lelectrad (r2.2) T* = turbine torque in Nm; it acquiresa negative value for a motoring = H s)'l,e,lecdt egree machine ffi T, = electromagnetictorque developedin Nm; it acquiresnegativevalue The inertia constantH has a characteristicvalue or a range of values for for a motoring machine each classof machines.Table 12.1 lists sometypical inertiaconstants. , Table 12.1 Typicalinertiaconstantsof synchronousmachines* to m_ > Type oJ Mat:hine Intertia Con.slunt H Stored Energy in MW Sec per MVA** Fig.12.1 Flowof mechanicaalndelectricapl owersin a synchronoums achine Turbine Generator Condensing 1 , 8 0 0r p m 9-6 While the rotor undergoesdynamics as per Eq. (12'3), the rotor' speed 7-4 changesby insignificant magnitudefor the time period of interest (1s) [Sec. 3,000rpm 4-3 Non-Condensing 3,000rpni L-J 0.i. Equation (12.3) can thereforebe converted into its more convenient Water wheelGencrzttor 2-4 . - L-. .,..-,=*:** r!:a *nr^r ('nerr.l tn lrvcl lnr qrr at l i n !cv nr r nu rsq rtr:rt n t at thg SVnChfOnOUS lnrltar Tnrm vn Jv 2 \\ \\ l l l l l l l l y lllc l t r l t r l- JUUUU Lv t.25 ,vvrvr tvrrlt I Slow-speed(< 200 rpm) l.00 peed(ur,.). Multiplying both sidesof Eq. ( 12.3)by u),,^'we can write High-speed(> 200 rpm) 2.00 S y n c h l o r r o uCso r r d c r r s c+r4' + t':t);\" x lo-(' - P^ P,.Mw (t2.4) J6\"n, Large dtt Small SynclrrcnouMs otor with load varying li'ortr where 1.0to 5.0andhighcr lor hcavyl'lywhccls p - mechanical ptlwer inPut ln MW electrical power outPttt t n MW; statorcopPelrossis assumed 'ttt negligible. It is observedfrom Table 12.1 th'atthe value of H is considerablyhigher for D steamturbogeneratothr zrntbr watc:rwheelgeneratorT. hirty to sixty per centof the total inertia of a steam turbogeneratour nit is that of the priine mover, t-,- whereasonly 4 -I5Vo of the inertia of a hydroelectricgeneratingunit is that of RewritingEq. (12.4) the waterwheel,including water. ul/;))\\ 2 s2a - P^ P,Mw u.rx, 1o-6ff) where 0, = anglein rad (elect) * Rcprinted with permission of the Westinghous Electric Corporation from ol Mn oi;d-' \\ , ==,P,,--pP\" (12.5) Electrical Transmissionand Distribution ReferenceBook. +* Where range is given, the first figure applies to the smaller MVA sizes. *tc+ Hydrogen-Cooled,25per cent less.

43S I M o d e r np o w e r S v s t e mA n a l v- -s i s t _ PowerSystemStabitity _ I +Sl - (t:.r2) It is moreconveniento measurethe angularpositionof the rotor with respect to a synchronouslyrotating frame of reference.Let Grnu\"h= machine rating (base) Gryrt\",o= system base o = (a;,?;[' ;\"':?: f# ul ar disP1acement from synchronously Equation (12.11) can then be written as (called torque angle/powerangle) (r2.6) +-\" (Yry*l =(p- p.)g'\"\"Gsystem\\- / From Eq. (12.6) \\ \"t c' Grrr,.* dt' ) drg, _ d26 ,Jt2 dt2 (r2.7) or {\"\"\"' o'! = 'P*- PP\"PUin sYstebmase (12.8) \"f dl Hence Eq. (12.5)can be written in termsof d as M, , - d: =2 6P ^ - P e ' w where Hryr,\"*= H*-*u\" \"h\\t.'.+Gt*\"\"l'n / (r2.r3) ot- M With M as definedin Eq. (12.1),we can write = machineinertia constantin systembase GH d2d (r2.9) Machines Swinging Coherently (12.r0) fV-P*-P\"Mw Considerthe swing equationsof two machinesor a commonsystembase. Dividing throughoutby G, the MVA rating of the machine, Hr d261 = P*r- P\"tPu (r2.14) a; (12.rs) a \"f M(pu+)dt' = P*- P,; H2 d262 = P*2- P\"zPu . A; in pu of machinerating as base \"f where M ( P =u )+l r j Sincethe machinerotors swing together(coherentlyor in unison) 6, = [r= [ A ddi ngEqs ( 12. 14)and ( 12. 15) or H dzb = P * - P, pu (IZ.1I) H\"q d26 = P*- P\" (r2.16\" n f ^-d t \" trf dtz This equation(Eq. (12.1D)l}q.(12.11)),is called thestvingecluatioyandit whcrc describestherotordynamicsfbr a synchronousmachine(generating/motoring). Prr=P*r+ Pn (r2.r7) It is a second-ordedr ifferential equationwherethe darnpingterm (proportional Pr=P\"l * Prz to d6ldt)is absenbt ecauseof the assumptionof a losslessmachineanclthe fact H\"q=Hr+ H, that the torqueof damper winding has beenignored. This assumptionleadsto pessimisticresultsin transientstability analysis-dampinghelps to stabilizethe The two machinesswinging coherentlyarethusreducedto a singlemachine as system.Dampingmust of coursebe consideredin a dynamicstability study. in Eq. (12.16).The equivalentinertia in Eq. (12.17) can be wrirten as Sincetheelectricapl owerP, dependsuponthe sineof angled(seeEq.(12.29)), the swing equationis a non-linear second-orderdifferential equation. H\"q, = Hl ,nu.h Gl ^ach/Grystem* Hz ^u\"h G2 -u\"h/G.yr,.. (12.18) Multimachine System The aboveresultsareeasilyextendableto any numberof machinesswinging coherently. In a multimachinesystema commonsystembasemust be chosen. ffimel\"L2rl Let 1 A 50 Hz, four pole turbogeneratorated 100 MVA, 11 kv has an inertia g6nsf anf nf R O MI/I\\/IVA

t4/s' torJern Power Srrctarn Anarrraio teryer-Svmel-eqtsqllv-- -- I 3. Effect of voltage regulatingloop during the transientis ignored, as a consequenctheegeneratedmachineemf remainsconstant.This assumptionalso (a) Find the storedenergy in the rotor at synchronousspeed. leadsto pessimisticresults-voltage regulatorhelpsto stabilizethe system. (b) If the mechanicailnput is suclcJenrlayisedto 80 MW fbr an electrical oad Before the swing equationcan be solved,it is necessaryto determinethe of 50 MW, tind rotor acceleration,neglecting rrlechanicaland electrical dependencoef the electricalpowel otttput(P,,)upon the rotor angle. losses. Simplified Machine Model (c) If the accelerationcalculatedin part (b) is maintainedfor 10 cycles,find the changein torqueangleandrotor speedin revolutionsper minute at the For a nonsalientpole machine,the per phaseinduced emf-terminalvoltage endof this period. equationundersteadyconditionsis )olution E =V + jXolu+ jXol,,;X,r) X,r (r2.Ie) (a) Storedenergy= GH = 100 x g = g00 MJ ( b )P , = 8 0 - 5 0 = 3 0 M W = M4 where I=Ia+ Is (r2.20) dt' andusual symbolsare used. Undertransientcondition r, CH 800 4 MJ-r/elect deg .1 -- 1 8 0x 5 0 Xa-X'a1Xa 180/ 4 d26 = 3o but +'j 6,2 or X'o = Xn sincethe main fleld is on the d-axis Xtd< Xo ; but the differenceis less than in Eq' (I2.I9) r) c Equation(12.19)during the transientmodifiesto (12.2r) ' (12.22) - 337'5 electdeg/s2 Et =V + jxtlo+ jXnln =V + jXq(I - I) + jXotlo (c) 10cycre=s = (Y + jxp + j(X'a - Xq)Id Changein d\"=lT!{ZZI.S) x (0.2)2= 6.75 elect degrees The phasordiagram colrespondingto Eql (12.21) and (12.22)is drawn in = 60 x 337'5 = z. o6 ' 112^rE ! Fig. 12.2. 2x360J {PnVs Since undertransientcondition, X'a 1X, but Xn remainsalmostunaffected, .'. Rotor speedat the end of l0 cvcles it is fairly valid to assttmethat x'a = xq (r2.23) - r 2 o x 5 o+ ' z 8 . t z 5x 0 . 2 4 = 1505.625rpm I2.3 POWER ANGLE EQUATION Fig. 12.2 Phasordiagram-salient pole machine In solvingthe s'uvinegquation(Eq (12.10)),certainsimplifyingassumpticnsare usually made.Theseare: 1. Mechanicalpower input to the machine(P*) remainsconstantduring the period of electromechanicatlransientof interest.In other words,it meansthat the effect of the turbine governingloop is ignored being much slower thanthe speedof the transient.This assumptionleadsto pessimisticresult-governing loo'2p.hReloptsotrosspteaebdiclhizaenthgeessayrserienmsig. nificant-these have alreadybeenignored in formulating the swing equation.

,firZlll r\"orrr r atysi_s Power SvstemStabilitv T*[ECAC,+' \\t2 equation(12.22)now becomes E,=V+jXnI o@ = V + jXotl (r2.24) Ofr The machine mod --- - -D' also applies to a cylindrical rotor machine where synchronousreactance) X,l = X*/(transient Fig. 12.5 Two-busstabilitystudynetwork For the 2-bussystemof Fig. 12.5 Y n u s= fY,, Y\"1 (r2.2s) I-j' Y- :r',[l ; YI, z =Y z r (r2.26) LY^ Fig. 12.3 Simplifiedmachine model Complex power into bus is given by The simplified machineof Fig. 12.3will be used in all stabilitv studies. Pi+ jQi-Elf P.ower Angle'Cunre ncnbsrtrrvFheleeaeytatoteagmtllrlrrwrccueieioaanttollqrhwayaibngrnuFen'tkalfsbiiecnttvpbeats'ceagTru(urtnotllXhmaeorshncpebuneoeislnor))itsptrle.aaslsa,iibhstfletnnoyiauentsasdrttooleibttocvtfruhs,hti,htesaroieltstetydmrhlarbs(mene(tabiehonofgicideettldeihtihanrtaeneiiycnnnlstenorshedc(trctmbaheeeaueucrpstsodborls(eyldnuircleeldsnoesiotsusedw)len)a.t'eolgmenETsnfsrcelnhsstmtt1cvuhhei;,staiyocetentwrnwlirdesisyaaocitdn,tnnbahrsthikenslboreesraoiiremb.lhsndiisgnetucieoyern.ctnleetsuiraaeftmmtisf1luriubevtm2fsadierot.eeiiyott4lnnlwtinofan,rectwe.agefgdosredwheteo)nbwniiinmvlsysleotoletrtarabltarahtkhandacnelpeottegwrsmoea.pemvicrradmtesotphiacvsfnloiltihiasoadanntiirtowecganoyorneeen,tt l At bus I Pr + jQr - Er' (YyE1)* + E, (YpEil* ,/ But Systemnetwork E ' t = l E i l l 6 ; E / = l E ' z l1 4 . Fig. 12.4 Y, u= G rt + jBt i Yr z= lYr zl l0 p Sincein solutionof the swing equationonly real power is involved,we have from Eq. (12.26) I r t = l E l i 2 G t r + l E r tI i E i i i y , r l c o s ( , t i- , h - 0 , 2 ) u 2 . 2 7 ) A similar equationwill hold at bus 2. Let lEllzG, = P, l E t' l lEztI lY7l = Pn, * 4- 6=6 (IZ.ZB) and Qn = x/2 -1 Then Eq. (12.27)can be written as Pr = P, * P,n\"*sin (6 - 1); Power Angle Equation For a purely reactivenetwork Gtt = 0 (.'. P. = 0); losslessnetwork } t z = n 1 2 ,: . J = 0 P,=P^*sin 6 (12.29a)

j0.5 I i0.5 I where P^^', = lE-'' I lxE'' | '' 1lo\" simplified power angle equation (r2.29b) where X = transferreactancebetween nodes(i.e., betweenE{ andEi) 'Ihe graphical plot of power angle equation (Eq.(12.29)) is shownin Fig. 12.6. p\"l I ID max - Generator (Ps6+APe).-....t-----* Peo lE'lt6 (b) F|g.12.7Asimp|esystemwithitsreactancediagram Fig.12.6 Poweranglecurve X r z= 0 ' 2 5+ 0 ' 1 + -0.5 The swing equation(Eq. (12.10)) can now be written as Yl aZt - P^ Pn'us* in dpu (12.30) = 0.6 += Considenr owa morecomplicatedcasewhereina 3-phasefaultoccursat the 7rI dt- midpointof oneof thelinesin whichcasethereactancdeiagrambecomesthat of Fig. 12.8(a). which, as alreadystated,is a non-linearsecond-orderdifferential equationwith no damping. T2.4 NODE ELIMINATION TECHNIOUE Star-Delta Conversion In stability studies,it has been indicated that the busesto be consideredare Convertingthe starat the bus 3 to delta,the networktransformsto that of thosewhich are excitedby the internalmachinevoltages(transientemf's) and Fig. 12.8(b)wherein not the load buseswhich areexcitedby the terminalvoltagesof the generators. Therefore, in Y\"u, formulation for the stability study, the load buses must be o (, eliminated. Three methods are available for bus elimination. These are j0.25 illustrated by the simple systemof Fig. 12.7(a)whose reactancediagram is - aUt( drawnin Fig. I2.7(b).In this simplesituation,bus 3 getseasilyeliminatedby lE/lt6 1ttr parallel combinationof the lines. Thus (a)

446 L ModernpowerSystemAnatysis Power SystemStability l u- t _ , This method obviously is cumbersometo apply for a network of even small complexity andcannotbe computenzed. Node Elimination Technique lE/lt6 Formulate the bus admittancesfor the 3-bus system of Fig. 12.8(a).This CD network is redrawn in Fig. 12.9 wherein instead of reactance branch. admittancesare shown. For this network. @ o (r ra (c) Fig.12.9 Fig. 12.8 v _ 0.25x 0.35+ 0.35x 0.5+ 0.5x0.25 ruus = 1.55 The bus 3 is to be eliminated. This method for a complex network, however, cannot be mechainzedfor In generalfor a 3-bus system preparing a computerprogramme. lfh' '|I=lfYr '^ Y\"t\"lIu'I Y, rvrurlJlvL,%I l Thevenin's Equivalent Llrl Lv,, (r2.31) Yn With ref'crcncclo Fig. l2.ti(a), tho Thcvcnirr'sccluivalcntlbr thc network portion to the left of terminalsa b as drawn in Fig. 12.8(c)wherein bus t has Sinceno sourceis connectedat the bus3 beenmodified to 1/. It =o ,vrr h = 0.25 l E t l l 5 025+0-,5 or YrrVr+YrrVr+YrrVr=O = 0.417|Et I 16 or vz=-?rr- ?Y.r, , (r2.32) Yrt xrh = 0.35x0.25 o'146 Substituting this value of V3 in the remaining two equations of Eq. (12.31), - therebyeliminating Vy 035+025 It =Y,Vt * YrzVz+YttVt Itlow Xt2 =0.146+ 0.5= 0.646* -=( .(^\"t,t - Y , r Y) ru,+' [(' ty' , \"- ! + ) v \" Yr, Yr, )'' *This value is different from that obtained by stardelta transformationas V* is no longerlEtl I { in fact it is 0.417 lEtl 16.

In compactform to be 1.0pu. Solution Y n u s( r e d u c e=dl t), i , ' Yl ,'lr,,')1 (12.33) lY'r, Q23aa) (r2.34b) Y'tl=Y\"-ry (r2.34c) =lfEfr lil=E ^ml = o ' 4I9. ZPx I u '33 Y'12= Y'21= Ytz - YrtYt, (r2.3s) (2) Equivalent circuit with capacitive reactoris shown in Fig. 12.71 (a). Ytt j1.o jo.l i0.25 io.1 i1.o p'e65 tv 2t ,2=- y'rzrz- Y r tYt yT lEnl= 1.2 -i1.0 = lEml= 1.0 In general, in eliminating node n (a) (b) Yo^(old)Y,,(old) Flg.12.11 Yo,(new) = Yry(old) Applying Eq. ( 12.34)to the example in hand Ysu5(reduce=d); l-t.gzt Converting star to delta,the network of Fig. 12.11(a)is reduced to that of 1L 0.646 Fig. 12.11(b)where It then follows that 7 X ( t r a n s f e r ) - / 1 . 3 5 X / 1 . 1 + / 1 . 1-Xj1(._0J 1 . 0 ) + ( - / 1 . 0 ) X J 1 . 3 5 . . X,t=:- = 1 . 5 4 8 (= 1 . 5 5 ) = j0.965 o.646 Steadystatepowerlim.it = 1 . 2 x 1= 1-244pu ffi In the systenrshown in Fig. 12,10,a three-phassetaticcapacitivereactorof (3) With capacitivereactancereplacedby inductivereactancew, e get the reactance 1 pu per phaseis connectedthrough a switch at motor bus bar. equivalenctircuitof Fig. 12.12C. onvertinsgterto delta,we havethe Calculatethe limit of steadystatepower with and without reactorswitch closed. trasferreactanceof Recalculatethe power limit with capacitivereactor replacedby an inductive reactor of the samevalue. i1.35 i1.1 Fig.12.12 _ j 1 . 3 5x . r 1 . l+ \"11.1llx.0 + 1 1 . 0x . t 1 . 3 5 i 1.0 7X(transfer) F i g .1 2 . 1 0 - j3.e35

-!5o' l t\"-t, PowerSystemStability hiffi \"\"attt As alreadycalculatedin this section, Steadystatepower timit -':'!] = 0.304pu Xn = l'55 3.935 = !\".!;!! = 0.6e4pu E x a m p l e1 2 . 3 gt=Treah1nne'es0grfeeapnrtuore)er,redawmutnoiftdrhboeeftrhhFteihingegd. feto1rnal2leno.r7swa(itieanon)igrsttreecdraomecnl iitdvnaieant irlocivnenogF.sl t1i:an.gd0ept huoefpmovwa, rxei=rmt ou1-.t0hiepoui*n\".f.ci naitltcheuabltauctsaen( ltVbheel or P, = 0.694 sin d (ii) (a) Systemhealthy (b) One line shorted(3_phasei)n rhe middle (c) One line open: (c) One line open. It easily follows from Fig. 12.7(b) that .Plot all the threepower anglecurves. Xrz=0.25 + 0.1 + 0.5 = 0.85 Solution P*.*=tit^o]t =r.265 0.85 or P\" = I.265 sin 6 (iii) L,et Vt= lV,l la = | la The plot of the three power angle curves (Eqs. (i), (ii) and (iii)) is drawn in From power angleequation Fig. 12.13.Under healthycondition, the systemis operatedwith P,, = P, = 1.0 pu and 6o= 33.9\",i.e., at the point P on the poweranglecurve 1.79 sin d As _t v- -t-l l v I s t n o = p \" one line is shortedin the middle, Po,remainsfixed at 1.0pu (governing system X act instantaneously)and is further assumedto remain fixed throughout the ( t\"t ) < t= I transient(governingactionis slow), while the operatingpoint instantly shiftsto [025+oJJsrn Q on thc curvc 0.694sin dat d= 33.9\".Noticcthutbccuuscof machineinertiu, the rotor anglecan not changesuddenly. or rr = 20.5\" 1.79 Currentinto infinite bus. 1.265 lV,lla-lVll0\" Pn=1'O 0.694 I _ jx | 1 2 0 . 5 -I l 0 0.694sin 6 i0.3s 0 I A = l + j 0 . 1 8 = 1 . 0 1 61 1 0 . 3 \" 33.90 900 V o l t a g e b e h i n dt r a n s i e n tr e a c t a n c e , Fig.12.13 Poweranglecurue,s- E t = t l t r + . j 0 . 6x ( l + 7 0 . 1 g ) = 0 .8 9 2+ j 0 .6 - 1.075133.9\" (a) Systemhealrhy p^u* = lv )-lEt | -- 1x 1.075=- .t,' tF9,. \\, - , I2.5 SIMPLE SYSTEMS x,, c,5 PU Machine Connected to Infinite Bus P, = I'79 sind (i) Figure 12.14is the circuit modelof a singlemachineconnectedto infinitebus (b) One line shorredin the middle: througha line of reacthnceXr.In this simplecase

PowerSystemStability t- I I lt5' Xtransf.er=X'a* X, (r--n-) From Eq. (12.29b) and d2 6 \"= \".f( P - r-P \"r\\ \"tlH') (r2.39b) l2 (r2.40) it ?: )= ,, ='4U sind= p.u*sind (r2.36) SubtractingEq. (12.39b)from Eq. (12.39a) Xt urrrf.. d2@,;6)=^.Jr(':jr!,] ,.- - P,) The dynamicsof this sysremare describeciln Eq. (12.11) as dtz \\ HrH, ) -#ft= P^-P\"Pu -H*\"q d26 (r2.4r) 7rI (.lt,- (r2.42) or =Pn- P, where 6=4- 6. lE/lt6 rr - HtH, (r2.43) \"eq Hl + Hz Fi1.12.14 Machineconnectetdo infinitebus The electrical power interchange is given by expression Two Machine System The caseof two finite machinesconnectedthrough a line (X\") is illustrated in p'\"= X',0E,+!x4,!+- xd2 ,in6 (12.44) Fig. 12.15whereone of the machinesmustbe generating und th\" othermust be motoring.Understeadycondition, beforethe systemgoesinto dynamicsand The swing equationEq. (12.41) and the power angle equationF;q-(12.aa) have the same form as for a single machine connectedto infinite bus. Thus a two-machinesystemis equivalentto a singlemachineconnectedto infinite bus. Becauseof this, the single-machine(connectedto intinite bus) qYstemwould be studiedextensivelyin this chapter. Fig. 12.15 Two-machinesvstem In the system of Example 12.3,the generatorhas an inertia constant of 4 MJ/ MVA, write the swingequationupon occurrenceof the fault. What is the initial angularacceleration?If this accelerationcan be assumedto remain constantfor Lt = 0.05s, find the rotor angle at the end of this time interval and the new acceleration. P*t=-P*z=P. (12.38a) Solution Swing equation upon occurrenceof fault the mechanicalinput/outputof the two machinesis assumedto remain constant at these values throughout the dynamics (governor action assumedslow). H d'6 -_ or m - 'De During steadystateor in dynamic condition, the electrical power output of the 1 g 0 fd v generatormust be absorbedby the motor (network being lossless).Thus at all time 4 d,26 , s i n6 Prl=-Prz= P\" (12.38b) *\"t#=l-0'694 The swing equationsfor the two machinescan now be written as t4 = z2so(1- 0.6e4sin d;. t L -_\"_' ir ( P ^ r _P \" r_) _ \"( p . - + \\ dt\" dtz ff, ):ut t rr, l (12.39a)

,.4#jirf vooern powersysremAnatysis ffi L ltc..:' I Linearizing about the operatingpoint Qo (P\"0, 4) *\" can write ]i;E!!\";1&t Initial rotor angle do= 33.9\" (calculatedin Example 12.3) LP,=(*). o, (r2.47) a 2s l = 2250(l - 0.694sin 33.9\") The excursionsof A d are then describedby ;dl t ' l ,- n+ no9i+' - P^- (P,o+ aP,'=) - L,P, #Cll ll r : o*+ = 0; rotorspeedcannotchangesuddenly A, (in A,t = 0.05s)= I x 1379x (0.05)2 d,r 2 Mdd'r+'* t .7\" 6 t = 6 + 4,6= 33.9+ 1.7\"= 35.6\" a26l = 2250(l - 0.694sin 35.6\") d. r, l| lr= 0.05s Pd where dt - l34I elect deg/s2 The system stability to small changesis determined from the characteristic Observethat as the rotor angle increases,the electrical power output of the equation generatorincreasesand so the accelerationof the rotor reduces. 12.6 STEADY STATE STABILITY Mp+, [#], =o The steady statestability limit of a particularcircuit of a power systemis whosc two roots are defineclas the maximutnpower that can be transmittedtri fhe receivv'irn6oevrnr;u withoutloss of synchronism. f /t\\tr ,.1 (\\ f{ Considerthe sirnplesystemof Fig. 12.14whoseclynamicsis describecbt y P = + l- \\ u 1 t0 o ) olI- equations L M As long as (0P/0 0o it positive,the roots are purely imaginaryand conjugate and the systembehaviour is oscillatory about do.Line resistanceand damper M* = P^ P\" MW; Eq. (12.8) windings of machine,which havebeenignoredin the abovemodelling, cause dl the system oscillations to decay.The system is thereforestablefor a small M = H ln Pu sYstem incrementin power so long as 7 (12.4s) (aP,/aa|> o (12.48) and p, = !4)! ) ,in 6'=p^u*sin d (12.46' When (0 P/AD, is negative,the roots are real, one positive and the other negativebut of equal magnitudeT. he torqueanglethereforeincreaseswithout x,t bound upon occurrcnccol a small powcr incretrten(t disturbancca) nd the synchronismis soon lost. The systemis thereforeunstablefor For determinationof steadystate stability,the direct axis reactance(X.r)ant, voltagebehind X4 areusedin the aboveequahons. @ Pe/aDo < 0 The plot of Eq. (12.46)is given in Fig. 12.6.Let the systembe operaring @p/A[ois known as synchronizingcofficienr. This is alsocalled stffiess with steadypower transferof P^ = P^with torque angle d as indicatedin the (electrical)of synchronousmachine. figure.Assumea smallincrementAP in theelectricpower with the input from the prime moverremainingfixed at p*(governor r.rforrr\" is slow comparedto Assuming lEl and lVl to remainconstant,the systemis unstable,if

po*e,systmst\"uititv br-{dffi l E l l v l c o sd ^ < o Iae1 _ _L_2_ _x r c o s 3 0 \" X l--e | or 4>90\" L06Jro\" 1.8 (12.4e) Themaximumpowerthatcanbe transmittewd ithoutlossof stabili = 0.577MW (pu)/elecrtad andis given by ( r2 . s 0 ) M(pu)= H = 4 s?/ cr ccrtat r (12.sr) o*ro trx5o lEnvl From characteristicequation P^u* If the systemis operatingbelow the limit of steadystability condition (Eq. P=tr[(*),,\"1*)' 12'48), it may continue to oscillate for a long time if the iamping is low. Persistentoscillations are a threat to system security. The study oi system -+i(WiY\")u ==i4.76 dampingis rhe studyof dynamicalstability. FrequencYof oscillations= 4.76 railsec The above procedure is also applicable for complex systems wherein 4'76 - 0. 758Hz governor action and excitation control are also accountedfor. The describing 2r dif f er e n ti a le q u a ti o ni s l i n e a ri z e calbout the opcrati ngpoi nt. C oncl i ti epfbr steadystate stability is then determinedfrom the correspondingcharacteristic (ii) For 807oloading equation (which now is of order higher than two) s i nd o= + = 0 . 8 o r6 = 5 3 . 1 \" It was assumedin the above account that the internal rnachinc voltage lEl remains constant(i.e., excitationis held constant).The result is that as P-u* loading increases,the terminal voltage lv,l dips heavily which cannot be toleratcdin practice.Thereforc,we mustconsicletrhe steadystatestabilitylimit rqa) - r'Zxrcos53.1\" by assuming that excitation is adjusted for every load increase to keep lv,l constanr.This is how the systemwill be operaiedpractically.It may be \\ 05)rr, 1.8 understocdthat we are still not consideringthe effect of automaticexcitation = 0'4 MW (Pu)/elecrtad control. p=!, (q+k)* =* i3s6 steady statestabilitylimit with lv,l anrt lvl constantis consicleredin Example12.6. Frequencyof oscillations = 3.96 radlsec A synchronousg eneratorof reactance1.20pu is ?q6 bar (l Vl = 1.0pr) throughtransformersanda line ')* connectedto an inf inite bus of totalreactanceof 0.60pu. The generatorno load voltageis I .20 pu andits inertiaconstantis H = 4 MW- silvIVA. The resistanceand machinedampingmay be assumednegligible.The systemfrequencyis 50 Hz. Find the steady state power limit of a system consisting of a generator Calculatethe frequencyof natural oscillationsif the generatoris loaded to equivalent reactance0.50 pu connectedto an infinite bus through a series (1)50Voand (ii) 80Voof its maximum power limit. rcactancgef 1.0pu.The terminalvoltageof thegeneratoirs heldat 1.20pu and the voltage of the infinite bus is 1.0 pu. Solution Solution (i) For 50Voloading The systemis shown in Fig. 12.16.Let the voltage of the infinite bus be taken as reference. s m d o- i t P = 0 . 5 o r4 = 3 0 o ' max

m Analysis Power SrrstemStahilitu EsE V = 7 . 0/ - f f , Vt= !.2 l0 Then A knowledgeof steadystatestability limit is important for variousreasons.A Now I _ 1.210-7.0 system can be operated above its transient stability limit but not above its jI steady-tatelimit. Nowrwith increased fault el,earing speedsjt is possible to make the transient limit closely approachthe steady state limit. lE4t6 Xa= O'5 V = 1.0100 As is clearfrom Eq. (12.51),the methodsof improving steadystatestability I limit of a systemare to reduceX and increaseeither or both lEl and I Vl. If the transmissionlines are of sufficiently high reactance,the stability limit can be Vt=1.219 raisedby usingtwo parallel lines which incidently alsoincreasesthe reliability of the system.Seriescapacitorsare sometimesemployedin lines to get better Flg.12.16 voltage regulation and to raise the stability limit by decreasing the line reactance.Higher excitation voltages and quick excitation system are also r IE = Vt + jXdI = 1.2 l0 + j0.5 I r . 2l g - 1 . 0 employed to improve the stability limit. LJ I2.7 TRANSIENT STABITITY .E= l.g l0 _ 0.5= (t.g cos e_ 0.5)+ 71.gsin 0 It has been shown in Sec. L2.4 that the dynamics of a single synchrono,rs Steadystateporver imit machineconnectedto infinite bus bars is governed by the nonlinear differential part is zero.Thus, is reachedwhenE hasanangleof 6= 90o,i.e.,its real equation 1 . 8 c o0s - 0 . 5 = 0 M,, d'6 = P ^ - P \" iF or 0 = 73.87\" where P, = P-* sin d Now Vt = 1.2/.73.87\"= 0.332+ j t . 1 5 2 or M- _ -d-+2 6- P*- P** sind ot- r _ 0.332+.ir.rs 2_r (r2.s2) = t.t52 + j0.669 E -0.332+ jr.r52+ 70.5(1.152+ j0.668) As said earlier,this equationis known as the swing equation.No closed form - 0.002+ j1.728= 1.728I90. solution exists for swing equationexcept for the simple case P- = 0 (not a practicalcase)which involves elliptical integrals.For small disturbance(say, Steadystatepowerlimit is givenby gradualloading), the equationcan be linearized(seeSec.12.6) leading to the concept of steady state stability where a unique criterion of stability p ^ u* =- l EllVl = -1-i.57 2 8 x=L l'152Pu (APrlAd>0) could be established.No generalizedcriteria are available* for V;-+V determining system stability with large disturbances(called transient stability). The practical approach to the transient stability problem is therefore to list all If insteadt,he generatoer mf is held fixed at a valueof r.2pu, the steady important severedisturbancesalong with their possible locations to which the statepowerlimit wouldbe systernis likely to be subjectedaccording to the experienceand judgement of the power system analyst. Numerical solution of the swing equation (or P*\"=*i# =o'8Pu equationsfor a multimachine case) is then obtainedin the presenceof such disturbancesgiving a plot of d vs. r called the swing curve. If d starts to t',h'oeItvltioaslgtoaaebgtserer.er2vgpeuudtlharaatitinrseglegostuohlpaehtpienolgtpwhseiengrlepin.oiewt rfearrotsoreyrnsm0t.efgtmopshrt'oarlobd'triht.yre5.te2rpmui;ntahlgiseinsehroawtor decreaseafter reaching a maximum value, it is normally assumed that the systemis stableand the oscillationof daroundthe equilibriumpoint will decay tRecent literature gives methods of determining transient stability through Liapunov and Popov's stability criteria, b:rt thesehave not beenof partical use so far.

ffiffif ModerpnowerSystemAnatysis PoweSr ','stemStabilitY ffiffi I andfinally die out. As alreadypointedout in the introduction,important severe p\"rrnon.ntly till clearedmanually. Since in the majority of faults the first distulbancesare a short circuit or a sudden loss of load. ieclosure will be successful,the chances of system stability are greatly enhancedby using autoreclosebreakers. For easeof analysiscertainassumptionsand simplificationsare always made (someof thesehave alreadybeenmade in arriving at the swing equation (Eq. /1 /l <.t\\\\ a rr 11- consequenceuspon accuracyof results. Fig.12.17 1. Transmission line as well as synchronous machine resistance are In the caseof a perrnanentfault, this systemcompletely falls apart. This will ignored.This leads to pessimisticresult as resistanceintroducesdamping term not be the casein a multimachine system.The stepslisted, in fact, apply to a in the swing equationwhich helpsstability.In Example I2.11, line iesistance has been takeninto account. systemof any size. 2. Damping term contributedby synchronousmachinedamperwindings is 1. From prefault loading, determinethe voltage behind transientreactance ignored. This also leads to pessimisticresults for the transientstability limit. and the torque angle 16of the machine with referenceto the infinite bus. 3. Rotor speed is assumedto be synchronous.In fact it varies insignifi- 2. For the specifiedfault, determinethe power transferequation Pr(A during cantly during the courseof the stability transient. ' ^ault. In this system P\" = 0 for a three-phasefault' ' 4. Mechanical input to machineis assumedto remain constantdurins the transient,i.e., regulating action of the generatorloop is ignored. This leais to pessimisticresults. 5. Voltage behind transientreactanceis assumedto remain constant, i.e., action of voltageregulatingloop is ignored.It alsoleadsto pessimisticresults. 6. Shunt capacitancesare not difficult to accountfor in a stability study. Where ignored, no greatly significant error is caused. 7. Loads are modelled as constant admittances.This is a reasonablv accuraterepresentation. Note: Sincerotor speedandhencefrequencyvary insignificantly, the network 5 . From the swing equation starting with fi as obtained in step 1, calculate parametersremain fixed during a stability study. das a function of time using a numerical techniqueof solving thetnon- A digital computer programmeto compute the transient following sudden linear differential equation. disturbanceaanbe suitably modified to include the effect of governlr action and excitation control. 4. After clearanceof the fault, once again determineP, (A and solve further for d (r). In this case,P\"(A = 0 as when the fault is cleared,the system getsCisconnected. 5 . After the transmissionline is switchedon, againfind P\" (0 and continue to calculate d (r). 6 . If 6 (t) goes through a maximum value and startsto reduce,the system is regardedas stable.It is unstableif d(r) continuesto increase.Calculation Upon occulTenceof a severedisturbance,say a short circuit, the power is ceasedafter a suitable length of time. transfer between machinesis greatly reduced, causing the machine torque An importantnumericalmethodof calculating d(t) from the swing equation angles to swing relatively. The circuit breakersnear the fault disconnect the will be giurn in Section12.9.For the single machineinfinite bus bar system, stability can be conveniently determinedby the equal areacriterion presented in the following section. I2.8 EOUAL AREA CRITERION In a systemwhere one machine is swinging with respectto an infinite bus, it is possible to study transient stability by means of a simple criterion, without resorting to the numerical solution of a swing equation.

Considerthe swing equation The stability criterion for power systemsstatedabovecan be convertedintc a simple and easily applicableform for a single machineinfinite bus system. d26 = I P ' 1= &rt ,\"= AF *@^- a c c e l e r a t i npgo w e r Multiplying both sidesof the swing equation* we get [t#), M=! ln pu system (r2.s3) rf 2P\"d6 Mdt Ifrtegrating,we have (r2.s6) where dois the initial rotor anglebefore it beginsto swing due to disturbance. From Eqs. (12.55) and (12.56),the condition for stability can be written as Fig. 12.1g protof 6 vs tforstabreand unstabresystems bffta6fntipohpmlaerwfn!areres'saaldostttrhafujhectfis\"siagorehstosweieturiictnesmtanviarouenydeedebpfsabsga'rillofteitioIesowftitetwoecyssssamhcanreeycoreeausirsssiefsnlrnsttalssseaadteeueyrypoftbidmenassrniofoiwotstslc,sneiudnimtithinhlsaymlrasylrdboebrotbgc1(gianwhlncnreee6onpainaoidtnv0scittetnnicieoimleinsnrgsoori.Haudti.oiannosennaioonmltcnriullingyw,fminasaztf.osutehatosexehttevlaii(d1alddeumnesstn2bianttornooiuo.hi.lnit1l)twqmeeihwFiitgnutnxyeia(hsh.gcceptwsrnyeohl.reiuahsdiensitlrrsdntalewacwizeeehdnes.ombinardadleigelcmttsb,iiih,nnseJgirdampfedoitielntsulvseahaitsptdtrheofadeuyyteiitenbncbcdssoeneaibaltwestxdereeteytyehieiiedmnnfnlsdyiosgesgbgtiiaubeowhcstieatctmsrrauiqeitnietseipenehiuaoo.rTsosdstvasmnahtehntiaest_asmeaiiettstoslbhsariiiretnonlnfhieeibilame)enioasl.otnceraTapwyne)aft,rchodtcin,ttnh6tenutahtuhia(hssrnneataeeteee)l 6 (r2.s7) or [r\"ad -o 6,, The condition of stability can therefore be statedas: the system is stableif the areaunderPo(acceleratingpower) -dcurve reducesto zero at somevalue of d In other words, the positive (accelerating)areaunder Po- 6curvemust equal the negative (decelerating)areaand hence the name 'equal area' criterion of lstability. To illustrate the equal areacriterion of stability, we now considerseveral types of disturbancesthat may occur in a single machine infinite bus bar system. d6 =o Sudden Change in Mechanical Input dt (r2.s4) Figure 12.t9 showsthe transientmodel of a single machinetied to infinite bus bar. The electrical power transmittedis given by and is unstable,if -d-6 > 0 (12.ss) --> Infinite <lt busbar for a sufticiently long time (more than 1 s will genera'y do). Pm lvlr0o Fig. 12.19

ffifftl| ModerPn owesr vstemnnarys,s P, = lEtllvl sin d- P** sind , nud, T#^ e Under steadyoperating condition 4n Ar=)(Pn-P\")d6 P.o = Pro= P** sin do Az=i<r,-P^)d6 6l be possible to find angle d2such that rg condition is finally reached when 41 own in Fig.l2.2L Under this condition, hat 6 . = 6 ^ o = T - 6 t = n ' - s i n+- l: (12.58) 6bfi 6(' s2 % ?'i Q <(4 Flg. 12.20 P\"- 6diagramfor suddenincreasein mechanicainl putto g e n e ra toorf F i g .1 2 .19 This is indicatedby the point ainthe Pr- 6 diagramof Fig. 12.20. Fig. 12.21 Limitingcaseof transientstabilitywith mechanicailnput Let the mechanical input to the rotor be suddenly increased to Pn (by s u d d e n l Yi n c r e a s e d opening the steamvalve). The acceleratingpower 1o = P*t - P, causes'the AdscaooernencythaestcaeluertqirrtttuaehhetreeiinnoretgnilxnyptcbchoeerawesctsaoetkhsemrienicirnneehgtP-aiciu^nsiae,gsnntemaeusberpoglapevyne.ecsrrItatlhimuthaosaittetsahsttocedhcasuteoursledebiandreaaceevtrnineainnasigclhsapreooebbwwaleensfyeoeorbir,nnyAdwm,upisteiohsecilnhoettafhsncesthitcahesanayelidnsnqAtptuehu1amet'l rotor speed to increase(u> a,,r)and so does the rotor angle. At angle 6r,, (tc'hProi'^tiueir-gr'i,ho,unPtyho'iues,sls)rmo,ofoetbotre.rtTnhmoheate\"syiycoosfrstnoecdmmiiltliaiFontinegq.bouef1ey2sod.t2i-on=1ndtth9-o{a0^r\"t=teh.ims9e0asm\"iy'nesasstenotamtblfowolneirlg'l ursaeesmitanhienssteetqaaubdalyelesavtraeetnae Po= P*r- Pr(= P-* sin 4) = O (statepoint atb)but therotor anglecontinues stability only and doesnot apply to the transientstability case' to increaseas t.,) ur.Po now becomesnegative(decelerating),the rotor speed begins to reducebut the anglecontinuesto increasetill at angle 6., a= ur once again (statepoint at c. At c), the-deceleratingareaA, equalsthe accelerating bc areaA, (areasare shaded)j,.e., J ,, Od = 0. Since the rotor is decelerating, 6o the speedreducesbelow ur andthe rotor anglebeginsto reduce.The statepoint now traversesthe P, - 6 curvein the oppositedirection as indicatedby arrows in Fig. 12.20.It is easilyseenthat the systemoscillatesaboutthe new steady statepoint b (6= 4) with angleexcursionup to 6 *d 4.on the two sides. Theseoscillationsaresimilar to the simple harmonic motion of an inertia-spring systemexceptthat thesearenot sinusoidal. As the oscillations decay out becauseof inherent system damping (not modelled),.thesystemsettlesto the new steadystate where P^t = P, = Prn.*sin dl

-t{ri.s.f#ufuf i.xd r rru._oro_e.r.n._FAower uvslem Anarvsrs Power a l':fiffn#;i; Effect of Clearing Time on Stability t rresponding to a clearing angle can be establishedonly by numericalintegrationexceptin this simple case.The equal Let the systemof Fig. 12.22be operatingwith mechanicalinput P^ at a steady areacriterion therefore gives only qualitative answerto systemstability as the angleof d0(Pn,= P\") as shown by the point a on the Pr- 6 cliagramof Fig. time whgn the breakershould be openedis hard to establish. 12.23.If a 3-phasefault occurs at the point P of the outgoing radial line, the electricaloutput of the generatorinstantlyreducesto zero,i.e,, p, = 0 and the Pe statepoint drops to b. The accelerationareaA, begins to increaseand so does therotor angle while the statepoint movesalongbc. At time /. corresponding 'D max to angle 6, the faulted line is clearedby the opening of the line circuit breaker. Pm The values of /, attd 4 are respectively known as clearing time and,clearing angle.The systemonce again becomeshealthyand transmits p, = p,ou,s. in d i.e. the state point shifts to d on the original P, - d curve. The rotor now deceleratesand the deceleratingareaA, begins while the state point moves along de. | /'-\\ -ll c\\o ) - l \\ - /|-t___j I d\", 6r\"* + C r i t i c a lc l e a r i n g angle Fig. 12.24 Criticalclearingangle F19.12.22 As the clearing of the faulty line is delayed,A, increasesand so\\oes d, to find A2 = Ar till 6r = 6^ as shownin Fig. 12.24.For a clearingtime (or angle) If anangle fi canbe found such that A2= Ap the systemis found to be stable. larger than this value,the systemwould be unstableasA, < Ar The maximum The systernfinally settlesdown to the steadyoperatingpoint a rn an oscillatory aiiowabie vaiue of the clearing time and angle for the systemto remain stabie mannerbecauseof inherentdamping. are known respectively as critical clearing time and angle. Pe For this simple case (P, = 0 during fault), explicit relationshipsfor 6, (critical) and t\" (critical) are establishedbelow. All anglesare in radians. ,D max It is easily seenfrom Fig. 12.24 that 4nu*=T- d; (12.59) and P*= Pr* sin 6o (r2.60) Now uct Pm At= (P^ --0) d6= P^ (4, - 6) J h and 6elf.\" 61 6^^ P\"=-o.--/ i JA z = (P** sin d- P^) d6 6,, (3-phasefault) Clearing angle = P.u* (cos d, - cos d-*) - P* (6^o - 6\"i) Fig. 12.23

ffil uoo\"rnpo*\"r systemAnatvsis W For the systemto be stable, A2= A1,which yields t7,, 4/__\\ l_Lr__l - lI Infinite cos{. = P!^rn* (\\ 5-tn,^iu^-(\" d) + cos 4o\"* , . 7t \\--l ) | I bus (r2.61) P m Lrl IVVO\" where 4, = critical clearing angle (r2.62) (a) (12.63) SubstitutingEqs. (1259) and (12.60)in Eq. (12.61), we get 4r = cos-t [(r, _ Z6l sin do_ cos 6o] During the period the fault is persisting,the swing equationis d,2d = rf P^: P, = o IVVO\" 1r: d,r, (b) Integrating twice Fi1.12.25 Singlemachinetiedto infinitebusthroughtwo par:allelilnes 6 = -,rf- P*tz + $ Both thesecurvesare plotted in Fig. 12.26,wherein P-u*n ( P_u*ras (Yo * Xr) 2H > (Ya + Xr ll X).The systemis operatinginitially with a steadypower transfer Pr= P^ at a torqueangle 4 on curve I. 0 \" , = # P ; 2 \" ,1 6 0 (12.64) Immediately on switching off line 2, the electrical operatingpoint shifts to where curve II (point b). Acceleratingenergycorrespondingto areaA, is put into rotor followed by decelerating energy for 6 > q. Assuming that an area A2 /cr= critical clearing time correspondingfo deceleratingenergy (energy out of rotor) can be found such 4, = critical clearing angle that At = Az, the system will be stable and will finally operate at c From Eq. (12.6a) correspondingto a new,rotor angle 6, > 60\"This is so becausea single line offers larger reactanceand larger rotor angle is neededto transferthe same 2 H ( 6 , -, 4 ) (r2.6s) s t e a d yp o w e r . TrfP* ,\" (bothlinesin) / where d, is given by the expressionof Eq, (12.62) An explicitrelationshi the faulted condition p\" = pI'rrrclctenninirtrg., i, porisiblcin thisc a s ea s t l u r i n g closedform. This will not o and so trre ,wing equation can be integratedin be the casein mosi other situations. Sudden Loss of One of parallel Lines ssicnwuodnLFitdsecigeithd.nuien1lsyrg2nsso.owt2fufw5i,dtacpayh.costeihwirndeceguotrrlifateafmnnmwgsaoliicetedhhcneuitnltsrhoetveatfeibetsihidslyeittgsoystiveiyonemsffrnitoenhbpmieyteie,srybagiurti\"sivnrtenghnarwotinuhageFhnsigttwoe. naoredp2yo.a2rfro5atahbllde.e.llBiinneeefsosairses P\"r= lE'llvl sin d= Pm*l sind xa i xt llx2 Immediatelyon switching off line 2, powerangrecurve is given by P\"n= g:+ sin d= pmaxsrin d Fig.12.26 Equalareacriterionappliedto the openingof oneof thetwo linesin parallel ,\\d -T rt7

ffiffi-4l Mod\"rnpo*rr.surt* nrryr', 4=4o*_T_6, Power Sy-t-- St\"blllry which is the samecondition as in the previous example. - Sudden Short Circuit on One of parallel Lines via the healthy line (through higher line Case a: Short circuit at one end of line reactanceX2 in place of Xl ll Xz)7w; ith power angle curve Let us now assumethe disturbanceto be a short circuit at the generatorend of sln sin d line 2 of a double circuit line as shown in Fig. 12.27a.We shall assumethe fault to be a three-phaseone. obviously, P-o[ ( P-\"*r. The rotor now starts to decelerate as shown in Fig. 12.28. The systemwill be stableif a deceleratingareaA, can be found equal to accelerating area A, before d reachesthe maximum allowable value 4o*.At areaA, dependsupon clearingtime /. (correspondingto clearingangle {), clearing time must be less than a certain value (critical clearing time) for the systemto be stable.It is to be observedthat the equal areacriterion helps to determine critical clearing angle and not critical clearing time. Critical clearing time can be obtained by numerical solution of the swing equation (discussedin Section 12.8). P\"y,prefault (2 lines) P6n1,postfault (1 line) X2 , (b) ?t6 F19.12.27Shoftcircuiat t oneendof the line Ffg. 12.28 Equalareacriterionappliedto the systemof Fig. 12.24a, I systemnormal,ll faultapplied,lll faultedlineisolated. Before the occurrenceof a fault, the power angle curve is given by It also easilyfollows that larger initial loading (P.) increasesA, for a given '-- clearing angle(and time) and thereforequicker fault clearingwould be needed to maintain stableoperation. p\"'t= x,)i4+,'xrllrt,x,2,as, ind= p_*, sind Case b: Short circuit away from line ends which is plottedin Fig. 12.25. Upon occulrenceof a three-phasefault at the generatorend of line 2 (see When the fault occursaway from line ends(say in the middle of a line), there is somepower flow during the fault thoughconsiderablyreduced,as different Fig. I2.24a),thegeneratorgetsisolatedfrom the power systemfor purposesof from case a where Pen= 0. Circuit model of the system during fault is now power flow as shown by Fig. 12.27b.Thus during the period the fauit lasts, shown in Fig. 12.29a.This circuit reducesto that of Fig. 12.29cthrough one delta-star and one star-delta conversion.Instead,node elimination technique of The rotor thereforeaccelerate,.i;t:;i\"s dincreases.synchronism will be Section12.3couldbe employedprofitably.The power anglecurveduring fault lost unlessthe fault is clearedin time. is thereforegiven by The circuit breakers at the two ends of the faulted line open at time tc P \" t = | E l l v l s i n d = P m a xsrirn d (correspondingto angle 4), the clearing time, disconnectingthe faulted line. '1r'II

ffiil Modern Power svstem Analvsis ls#z#i '!'vev\" ' ' - '- v' v, -'-\" '' \" '-', -'- +t*F;4-l systemoperationis shown in Fig. 12.30,wherein it is possibleto find an area A, equal-to A, for q. < 4nu*.At the clearing angle d. is increased, area I ai increat\"t und to nna Az = Ar, 4. increasestill it has a value 4n*' t6\" -ooi*,,- ollnrvohlefnr stahilitv This caseof critical clearineangleis shown x,t X6 in Fig. 12.3L G xc @ (b) Pe Pr1,prefault(2 lines) Xr P6111p,ostfault (1 line) Fig. 12.31 Faulton middleof one lineof the systemof Fig. t2.l4a, case of criticalclearingangle (c) Annlvins eoualareaeriterion to the caseof critical clearingangleof Fig. 12.31' Fig. 12.29 we can wnte P\"rand P,u as in Fig. 12.28 and Per as obtained above are all plotted in Fig. I2.3O. Accelerating area A, corresponding to a given clearing angle d is less 4, dntn' Pe j (P^- 4n*u sinfldd= J {r^*r sind- P^)d6 P\"'Prefault(2 lines) 60 6,, ,.a P\"11p;,ostfaul(t 1 line) where P \" 1 1d,u r i n gf a u l t 4,,* =T - sin-r (:t_) (r2.66) Fig. 12.30 Faulton middleof one lineof the systemof Fig. 12.24a V maxIII ./ with d\"< {, Integrating,we get l6* =Q (P^a+ Pmaxrcr os d) | * (P'*,,1 cos d + 16o or P^ (6\", - 6) * P.u*u (cos '[. - cos do) I P* (6** - 6\"r)* P-om (cos fi* - cos 4J = 0

t cos {r = 4naxtn - PmaxII :otd\",* (12.67) l:cnltr:io:ttiec1ga|roel cnelse. arainrgeainngrlaedciaannbseT.chaelceuqluaatetifdoronmmoEoqiii.e(1sa2s.6b7e)alobwoivf et.TheheaannggJeleassirne Give the systemof Fig. 12,33 where a three-phascfault is applied at rhe point P as shown. cos {. - ft r.(6 ^i* - do) - Pmaxcror sdo * prnu*rcuosd,ou* Case c: Reclosure i0.s Pmaxltr - Prnaxn Infinite bus vFlloo Itfratnhseiecnirtcounitebarnedaktehresroeffolirneeva2naisrheeredcolnosceldeasuricncgetshsefufalluy(rit.ye.l,inthee), fault was a transferonce again becomes the power P\"N = P\"r= p*u*I sin d Flg.12.33 Since reclosure restores power transfer, the chances of stable operati'n Find the critical clearing angle for clearing the fault with simultaneousopening improve. A caseof stableoperationis indicated by Fig. 12.32. of the breakers I and 2. T\\e reactancevalues of'various components are For critical clearing angle indicated on the diagram.The generatoris delivering 1.0 pu power at the instanr precedingthe fault. 4 = 4r* = 1T- sin-l 1p_/p*.*r; Solution ut c r With referenceto Fig. 12.31,three separatepower angle curvesare involved. 6rc f. Normal operation (prefault) J @r,- Pmaxsrrin 0 dd = J (p.*m sin d_ pm)d,6 60 6r, dru, + t (P,*r sin d_ p^) d6 Xr=0.2s+ffi+0.05 J 6.- = 0.522pu p,t=rysind:ffirino = 2.3 sin d (il Prefault operatingpower angle is given by 1.0= 2.3 sin 6 or 6o=25.8\" = 0.45 radians IL During fault (Clearingangle) \\ It is clear from Fig. 12.31 that no power is ffansferredduring fault, i.e., =o (Angleof reclosure) ,0.:\"o Fig- 12-32 Faurtin middreof a rineof the systemof Fig. 12.27a where trrj tr, + r; T = time betw,eenclearing anclreclosure. trin 1n 1A

iffi ModernpowerSystemAnatysis U rrr. Post fault operation (fault cleared by openingr the faulted Iiae) E Pn errr=l.2xl.0 sin d= 1.5sin 6 (iii) = - 1.5(cos2.41- cos 6,) - (2.41- 6\") ff = 1.5cos 6\",+ 6r,- I.293 Setting A = Az and solving Pe 6r,- 0.45= 1.5cos 6r,+ 6r,- 1.293 or cos {,. = 0.84311.-5 0.562 or 4, = 55.8\" Thecorrespondinpgoweranglediagramsareshownin Fig. 12.35. Pn=1'O Find the critical clearinganglefor the systemshownin Fig. 12.36for a three- phasefault at the point P. The generatoris delivering 1.0 pu power under prefault conditions. i0.15 i 0.1s lnfinite bus 66=0.4r5ad 6^rr=2.41rdd lF,l=1.2Pu lvF1.otoo Fig. 12.35 .10.15 jo.15 TrrTrwt o -ooi*\"* -^*:^^:Ll^ ^--l^ C f^- ----- ^ . 1 2 . 3 5i)s Flg. 12.36 urour.rLurr psllluDDlulc alilBrtr Al = A2 (Sge Omax l()f afea flg. given by Solution 4ou*=r-sin-l I = 2.4Lradians f, Prefault operation Transfer reactancebetween generator and infinite 1.5 bus is Applying equal area criterion for critical clearing angle { & = 0.25+ 0.17+ 0 . 1 5 + 0 . 2 8 + 0 . 1=50 . 7 1 Ar = P^ (6\",- 6) 2 = 1.0 (6\", - 0.45) = 6c,- 0.45 Pc- ,' = r ' Zxl sin d= 1 . 6 9s i n 6 (i) 0.7 dr* 1 A z =! { r , n - p ^ ) d , 6 The operating power angle is given by 6,, 1.0= 1.69sin ,fr 2.41 or do= 0.633 rad = I ( 1 . 5s i n 6 - 1 ) d d IL Durtng fault The positive sequencereactancediagram during fault is presentedin Fig. 12.37a. J| 6., r2.41 = - 1 . 5 c ods _ d l | 6\

,"J 000 L / 0 0 0 L - - - - - - - - -0-0r 0 \\ - PowerSystemStabilit-v r*Mi#ffi j0.25 Per' rr=U!l sin d = r'2 sin 6 (iii) With referenceto Fig. 12.30and Eq. (12.66),we have 0' j0.14 j0.'15 jo.17 + j0.15 j0.14 V=1.0 + ) E1=t.z To find the cqitical clearingangle,areasA1 andA, arc to be equated. (a) Positivesequencereactancediagramduringfault 6\", j0.25 j0.145 j0.145 ',j0.17 At = l.o (6,,- 0.633-) ,J o.+e5sind dd 60 and dmax lE'l=1.2 A-z = f 1 . 2s i n d d d - 1 . 0( 2 . 1 5 5- 4) V=1.OlOo J| 6-cr Now At =Az ( b ) N e t w o r ka f t e rd d l t a - s t acr o n v e r s i o n or J6r, = 0.633--- 0.495sin d dd o. 6 3J l9l=1'z V=1.0100 2.155 = J[ t.Z rin 6d6- 2.t55+ 6,, ' 6cr (c)Networakfterstar-delctaonversion or - 0.633+ 0.495cosolo' = - 1.2cos ol\"tt -2.155 . Ftg. 12.32 lo.orr la., Converting delta to star*, the reactancenetwork is changed to that of Fig. or - 0.633+ 0.495cos 6,,- 0.399= 0.661+ 1.2cos 6\",- 2.155 12.37(b). Further, upon converting star to delta, we obtain the reactance network of Fig..r2.37(c). The transferreactanceis given 6y or cos 6r, = 0.655 (0.2+s 0.1450).072+s(0.14+50.170) .072+5 (0.25+ 0.145) U 6r,= 49.I\" Xu= ( 0 . 1 4+s 0 . 1 7 ) 0.075 _ 2.424 A generatoroperatingat 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistanceis ignored. A fault takes place rpe-I =- lal! usuinr dv =- 0v..a4/,95sin 6 (ii) reducingthe maximum powertransferableto 0.5 pu whereasbeforethe fault, i.+Z+ this power was 2.0 pu and afterthe clearanceof the fault, it is 1.5 pu. By the use of equal areacriterion, determinethe critical clearingangle. fir. Postfault operation (faulty line switched off) Solution Xrl =0.25 + 0.15+ 0.28+ 0.15+ 0.17= 1.0 *Node elirnination techniquewould be'used for complex network. All the three power angle curvesare shown in Fig. 12.30.

,'ffi| Mod\"rnPo*.. sEl!\"-nAn\"lytit 2.The angular rotor velocity u= d6ldt (over and above synchronousvelocity Ilere P-\"*r =2.0 pu, Pmaxl= 0.5 pu and Pmaxr=rr 1.5 pu Initial loading P^ = 1.0 pu 6r,roz=rsinI (p \\ tffiJ E7-sinl 1 :2.4!rad r>2 n-1 t 1.5 Discretesolution n Af Continuousolution Applying Eq. (r2.67) U un-|/2 t cos {, - 1.0(2.4- 10.523-)0.5cos0.52+3 1.5cos2.41= o ??? u13/2 1.5--0.5 6r,= 70'3\" un-I/T-+tsn4l2 T2.9 NUMERICAT SOTUTION OF SWING EOUATION -t Af n-2 p3l2 n-'l r>112 n In most practical systems,after machine lumping has beendone, there are still more than two machines to be consideredfrom the point of view of system stability. Therefore, there is no choice but to solve thp swing equation of each machine by a numerical techniqueon the digital computer.Even in the caseof a single machine tied to infinite bus bar, the critical clearing time cannot be obtained from equal area criterion and we have to make this calculation ! . .rr-- mr r -r- o- -u- -g n- l - s- -w- -u: -l g- e- -q- -u- aL lu^ ,ulr. zFL^-^ aIU s s v t r I i l r| J- ^u- LPl ^u+riD^ ^t+r^ul ilL('(l *ll^l+ELL^ ll-luLlD $n-i numerlca[y t rttrIc now available for the solution of the swing equation including the powerful 6n-z Runge-Kutta method. He.rewe shall treat the point-by-point method of solution which is a conventional,approximatemethod like all numerical methods but a n-2 n-1 n J- well tried andproven one.We shall illustrate the point-by-point method for one Af machine tied to infinite busbar. The procedureis, however,generaland can be Fig. 12.38 Point-by-point solutionof swing equation applied-to every machine of a multimachine system. Consider the swing equation In Fig. L2.38,the numbering on tl\\t axis pertainsto the end of intervals. At the end of the (n -l)th interval, the accelerationpower is d26 = 1 -- ;T ;e*-P^*sind): PolM; Pa (n_r)--Pm- P-* sh 4-r Q2.68) (* - 9H orinpusystemM = +iT)f) where d_1 has been previously calculated.The changein velocit! (a= d6ldt), causedby the Pa@-r),assumedconstant over At from (n-312) to (n-ll2) is \\ 7t w n- '2- wn- 3t 2=( Lt / M ) Pa@- r ) (12.6e) The solution c(r) is obtainedat discreteintervals of time with interval spread The changein d during the (n-l)th interval is (12.70a) of At uniform throughout. Acceleratingpower and changein speedwhich are L6r-t= 6r-1- 6n-2= A'tun4'2 continuous functions of time are discretrzedas below: and during the nth interval (12.70b) 1. The acceleratingpower Po computed at the beginning of an interval is L6r- 6n- 6n-t= /\\tun-112 assumedto remain constantfrom the middle of the precedinginterval to the middle of the interval being consideredas shown in Fig. t2.38.

,ir Yirtvt r t Ar lr.rl b r ) power SystemStabilitv SubtractingEq. (12.70a\\ from Eq. (12.70b) and using Eq. (12.69),we get [i{8il;r \\Qr ^vtt,u, 1.;,^v^, t nE s^l uf ^l-s^ w s L a l l a^ -P* 1P, ,t . y +uLl^t t s^ +t^E- P - ULy. .- s l^t+t P^ - -l^I Ir cl . ^ull U U , - ^ -t ^- - -r ---r-a- Wtr lltrC(l t() Calculate L'6,= A6,-t + ( A r ) 2 rDa(.n-I) (12.7r) the inertia constant M and the power angle equations under prefault and M postfaultconditions. Usingthis,we canwrite BaseMVA = 20 6n= 6n-t+ L,6n G2.72) IneRia coflstant, Mepu\\ = 1.0x L52 1 8 0/ 1 8 0x 5 0 The processof computation is now repeatedto obtain Pa61, L6r*tand d*t. The time solution in discreteform is thus carried out over the desiredlength of time, = 2.8 x 10+ s2le\\ecdt egree normally 0.5 s. Continuous form of solution is obtained by drawing a ;mooth curve through discrete values as shown in Fig. 12.38. Greateraccuracyof I I Prefault solution can be achi.evedby reducing the time duration of intervals. &'= 0 . 3 5 + 0' 2 = 0 . 4 5 The occurrenceor removal of a fault or initiation of any switching event 2 causesa discontinuityin acceleratingpower Po.lf such a discontinuityoccurs at the beginning of an interval, then the averageof the valuesof Po before and Pd= Pr.*r sin d (i) after thediscontinuitymustbe used.Thus,in computing the incrementof angle = -!'.,-.;l;sxitn . dr = 2.M sin 5 occurring durirrg the first interval after a fault is applied at t = 0, Eq. (I2.7I) becomes 7 , ,,6=, (Ar)t * P a2o + Prefault power transf'er= +20 = 0.9 pu M Initial power angle is given by where Pos* is the acceleratingpower immediately after occurrenceof fault. 2.44sin4=0.9 \\ Immediately before the fault the system is in steady state, so that Poo-= 0 and or 6o= 21.64\" dsis a known value. If the fault is clearedat the beginning of the nth interval, II During fault A positive sequence reactancediagram is shown in Fig. in calculation for this interval one should use for Pa@-r)the value llP\"6-r>- 12.39a.Converting starto delta,we obtain the network of Fig. 12.39b,in which + Po6_9*),where Pa@_r)i-s the acceleratingpower immediately before clearing and Po6_r)+is that immediately after clearing the fault. If the discontinuity t,r,tr= -0. 35x 0. i + A. 2x0. i + 0. 35x0-. 2= I1..AZ-) pu occurs at ihe miciciie of an intervai, no speciai proceciure is neecled.The 0l increment of angle during such an interval is calculated, as usual, from the value of Po at the beginning of the interval. P.u = Pmaxtt sin d The procedureof calculating solution of swing equationis illustrated in the - 1'1x1 r;n d = 0.88sin 6 (ii) following example. 1.25 A 20 MVA, 50 Hz generatordelivers 18 MW over a double circuit line to an Fig. 12.39 infinite bus. The generatorhas kinetic energy of 2.52 MJA4VA at rated speed. The generatortransient reactanceis X/o = 0.35 pu. Each transmissioncircuit has R = 0 and a reactanceof 0.2 pu on a 20 MVA bgq-e.lE/l = 1.1 pu and infinite bus voltage V = 7.0 10\". A three-phaseshort circuit occurs at the mid point of one of the transmissionlines. Plot swing curves with fault clearedby simrrltaneousopening of breakersat both endsof the line at2.5 cycles and6.25 cycles after the occuffence of fault. Also plot the swing curve over the period of 0.5 s if the fault is sustained.