Solution Manual For Protective Relaying: Principles and Applications, 4th Edition

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Protective RelayingPrinciplesandApplications4thEditionJ.LewisBlackburn(Deceased)ThomasDomainSolutionsManualPreparedby the followingcontributors:SeanW.Carr,P.E.Sr. Engineer-Relay & Protection ServicesCommonwealth Edison CompanySean.W.Carr@ComEd.comJamesK.Niemira, P.E.,ME-EPEPrincipal Engineer-Power Systems SolutionsS&C Electric CompanyJim.Niemira@SandC.comJohnR. Bettler, P.E.,MSEEPrincipal Engineer-Relay & Protection ServicesCommonwealth Edison CompanyJohn.Bettler@ComEd.comWilliamJ.Niemira,MSEEAssociate-Power Delivery ServicesSargent & Lundy LLCwniemira@SargentLundy.comAnthony D. Locatelli, P.E.,MSEESr. Engineer-Relay & Protection ServicesCommonwealth Edison Company

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CHAPTER 22.1A wye-connectedgenerator ha s a namep late ratingof200MVA, 20 kV, andits subtransientreactance(X;')is1.2pu. Detennineits reactance in ohms.2.2The generatorofProblem2.1is connected in a power system where the baseisspecifiedas100MYA ,13.8kV.What is the generator reactance(X;')in per unit on this system ba se?2.3Converttheper-unitanswercalculatedinProb lem2.2toohms.Doesthismatchtheva luedetermined in Problem 2.1?2.4Three 5 MVAsingle-phase transformers, each rated 8: 1.39 kY, havea leakage impedanceof6%.These can beconnected in a numberofdifferent ways to supply three identical 5nresisti ve loads.Varioustransformer and loadconnections are outlinedin Table P2.4.Complete the table columns.Use a three-phase base of 15 MVA.TABLEP2.4TotalZas \'icwcdCaseTransformerconnectionLoadconnce.tion toLine-lo-IinebasekVLoadRinfromt he high sideNo.PrimarySecondal)'secondaryHVLVperun itPerunitQWyeWyeWye2WyeWyeDelta3WyeDeltaWye4WyeDeltaWye5DeltaWyeWye6DeltaWyeDelta7DeltaDeltaWye8DeltaDeltaDelta2.5A three-phasegenerator feeds three large synchronous motors over a 16 km,115 kV transmissionline, through a trans former bank, as showninFigureP2.S.Draw an e qui valent single-line reactancediagram with all reactances indicatedinper unitofa100 MVA , 13.8 orl i SkV base.Transformer #1<ldoGenerator~50MVA50 MVA13.8:115 kV13.2 kVX = 12%Xd= 20%Transformer #2~<lM,115kVTransmissionline--16km~X=0.5 Ohm/kmMotors20 MVA115:13.8kVX =10%M, :10 MVA. 13.8 kVXd=20%M2 :5 MVA, 13.8 kV X;;=20%M3:5 MVA, 13.2kVX;;=17%

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2.6Inthe systemofProblem2.5,itisdesired to maintain thevoltage atthe motor bus of1.L O·perunit.The three motors are operating at full rating and 90%pfa. Detennine the voltage requiredatthe generator temlinals assuming that there is novoltageregulating taps or similar equipmentinthis system.b.Whatisthevoltage required behindthe subtransient reactance?2.7The percentimpedanceofatransformer is typically detennined by a short circuit test.Insuch a test,the secondaryofthe transformer is shorted and the voltage on the primary isincreased until ratedcurrentflowsinthe transformer windings.The applied voltage that produces rated currentdividedbythe ratedvoltage of the transformer is equaltothe per-unit impedance ofthe transformer.A short circuit test on a150KVA,7200-240 V transformerprovidesthe following results:Primaryvoltageat20.8 primary amperes=208.8 Va. Determinethe %Z ofthe transformer.b.Calculate the ohmic impedanceofthe tran sformer in primaryand secondarytenns.c.Howmuch currentwouldflowinthe transformer ifits secondarywould become shorted duringnormal operating conditions? (Consider source impeda ncetobe zero.)

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•2.1Findimpedanceinohmsfromper-unit.MVAn=200,kVn=20.X-d= ZP\J=1.2PerEq.(2.17)Zn=2.4ImpedanceX"d=2.4 ohmskVll2X ZruMVAB200

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•2.2 - Method 1Convert per-unitimpedancefromonebaseto another.MVA.=200.M VA2 =100.kV.= 20.kV~=13.8.X"J= Z IPU=1.2PcrEq . (2.33)ZMVA,kV,'=lPU x1.1VAx--,I~IkV2-loa2021.2)(-.--20013 . 821.26024Per- unit impedanceinnew baseZ2PU=1.2602~•2.2 - Method 2Convon impedance from ohms (Result 2. 1) to per- unit.MVAH=100.kVB=13.8.x~,,=Zn=2.4PcrEq.(2.t5)1.26024Per- unit impedanceZpu=1.2602~ZPU=Zn=MVA .x ZnZskvi100)(2 . 413. 8 2

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•2.3Convertimpedance fromper· unit (Result 2.2) bac k to ohms.MVA.=100.kVH=t3.8.X"d= ZP\J=1.26024PcrEq.(2.17)2.4Zn=13. 82Ie1 .26024100ImpedanceX"d=2.~ohms.This matches Res ult 2.1.

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J-~'V.1iIf '.....HV'-v~t:£J~Ve.~~IA.'"P4.bfVI-iv,I..V'IfUIIJ!...~R--Y~y'{y'(I~.'alt2·<I'cf'~-'eJl.-.1't)."bS-~~:'J&I'"r~yyAi3.8S1~.'1ofjo.IJ"t.l.(,1•'J•;,y.6.Y/'$.iS1/,Hod0.01.r~.t{yAAn.es;,I,~c;ocjO",.'67S-~Y~I8I;,<ff)f.JP.Ob~.C-D.YA8',2.<fOg.JP,04/.1.171A6.Yg-1,39()JO.D1-St'I>A~&'I.YiDVf).02-/."67.•J~?;kV¥f3~15.j56 Jd:e-&;VU;-&Wp-::/2V~r+-vJIS'MVIf""'"'-/J~il4SdZe_~lr£",~d-r~.

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Problem2.4ColAColBColCColDColEColFColGColHColIColJColKColLColMPUleakagereactanceequivalentY‐connectedresistiveload(ohmsonLVside)PUequivalentwye‐connectedloadresistanceCaseHVLVLoadconnectionHVLVHVLVequivalentinline(ColJ/ColH)perunit(ColK+j*Coli)ohms(ColL*ColG)1wyewyewye13.8562.40712.7990.3860.0605.00012.94012.940+j0.060165.623+j0.7682wyewyedelta13.8562.40712.7990.3860.0601.6674.3134.313+j0.06055.208+j0.7683wyedeltawye13.8561.39012.7990.1290.0605.00038.81838.818+j0.060496.840+j0.7684wyedeltadelta13.8561.39012.7990.1290.0601.66712.93912.939+j0.060165.613+j0.7685deltawyewye8.0002.4074.2670.3860.0605.00012.94012.940+j0.06055.211+j0.2566deltawyedelta8.0002.4074.2670.3860.0601.6674.3134.313+j0.06018.404+j0.2567deltadeltawye8.0001.3904.2670.1290.0205.00038.81838.818+j0.020165.623+j0.0858deltadeltadelta8.0001.3904.2670.1290.0201.66712.93912.939+j0.02055.208+j0.085Basepower(3‐phaseMVA)HVwindingvoltage(kV)HVdeltaLLvoltage(kV)HVwyeLLvoltage(kV)ZbaseforHVdeltaZbaseforHVwye15.0008.0008.00013.8564.26712.799Basepower(3‐phaseMVA)LVwindingvoltage(kV)LVdeltaLLvoltage(kV)LVwyeLLvoltage(kV)ZbaseforLVdeltaZbaseforLVwye15.0001.3901.3902.4070.1290.386WindingconnectionLLVoltageBase(kV)BaseImpedance(ohms)IntermediatecalculationsTotalZviewedfromHVside

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c.k;yj13~:/Mv'lttJE..t)(kVoL-PY~fu·tJW;:::~O'-VV-w~OLDk\[~~J~thr"1/':;2D1.(~)(~~:;tc~6."1.Io.~{,£/"<1-1X::I;;<'/;C~)=-Zrn.~O.~Yf"'-<-r-~)c=Iv%'~;:)=o,j10I0.>0f'''-;t(/Xd""Jo1,(p.)~;Wll%/~d)r-IA1Z-y./'.2tJ~C~)'"foz>7.I~artA13'f.J"-;;/11,t~)(',~.}y~3111.)'3.IIf>lAUV\12.;"1r.(l~: .~f<'--'"~~-z.-v"r5"lc.v)"V'b......ItS"k."-/ez,,.,.jit-=-/:)z..'2-'s-...t::'l..-,,fble"ro.t/W:-(e'~~.)(II.'M~'"0to05'Ir$Z,l-r-IV•f)~

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JKN2~1~146MVAE10 ·volt·ampProblem 2.6PUE1Prob2.6.aVoltage atgeneratorbusto keep 1PUvoltageatmotorbusYmotor:=1.0·PULoadcurrent ·mot orsatrated powerand0.9PF.Problem does not say explicitly, but weassume that the PF is laggingasthis is the most probable.Total motor bus powerinper unitSntOlor:=IO·MVA + 5·MVA + 5·MVA ./acos(O.9)= (0.18 + 0.087")PU•lOO.MVAJISmotorl=0.2PUarg(Smotor) =25.842 dcgCurrent is found from the relation that complex power is equal to voltage times complexconjugate of current.(Smolo, )Imotor:==(0.18-O.087~PUVmotorp motorl=0.2 PUarg( lmotor) =-25.842dcgUne reactance inPU(from prob. 2.5)ZLinc :=j·0.0605PUT1reactance in PU (fromprob. 2.5)Zn:=j·O.24PUT2reactance in PU (fromprob. 2.5)ZT2 := j·O.50 PUGeneratorbusvoltage is motor bus voltage plus vollage dropinT1,line ,andT2.IVOcu.busl=1.079PUarg(VOcn.bus)=7.671dcgProb2.6 b - Generator internalvoltagebehind subtransient reactance: add intemalvoltagedropto the generator tenninalvoltageGenerator subtransient reactanceinPU(from prob. 2.5)ZGcn:=j·0.366VOen.inlema l:=VOen.bus +Imotor·ZOcn=(I102+ 0.21] PUIVOell.illtem all=I.122PUarg(VGen.inlemal)=10.79Ideg

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JKN 2-1-14Problem 2.7Transformerratedpower and rated voltagesJKN2-1-14page1of 2kVA=:lOOO·voh·amp6MVA=:10·volt·ampPU .: ISrlltcd := 150 -kVAVruled.primary:=72()()·voltVrutcd.sccondary := 240· voltTransformer rated current[150. kVA = 20.833Arated.primary :=7.2.kVIISOOOO·volt·amp=625 Arated.sccondarv :=' J240· voltThis confirms that the shortcircuit test wasinfact performed atrated current . Theimpedancevoltage is therefore equal to the voltage measured duringthetest , or 208.8 volts.U&..The transformer per cent impedance voltage(%IZor%Z) is equal to the measuredimpedancevoltage expressed as per cent of the rated voltage or in thiscase:,,__V~"~S~I__,=O.029I'UVruted.primary208.8 ·volt = 0 .029PU72()()·voltx,.=2.9 %UJLTheequivalent ohmic impedance referredto theprimary or secondaryis equal to thecorresponding base impedance times theperunit impedance.,Vruled.primary-Zbasc.primary :-,(7.2·kV)-=345MlO. IS·MVA345MlXT.ohms.primary:=XrZbasc_primary =10.022nXT.ohms.second,IIY := XrZbasc_sccondmy = 0.0111flV2raled.sccondary= 0 .384nSratcd2(O.24-k V)=0.384nO. ISO·MVA

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2.7.cThere are a few ways to arriveatthe short circuit current.JKN 2-1-14page2of2One way is to divide the rated voltagebythe equivalent ohmic impedanceoneither primaryorsecondary.Alternatively, 1.0 PU voltage canbedivided by the PU impedance to determinethe short circuitcurrent in PU of rated current.The PU short circuit current can thenbemultiplied by the basecurrent (i.e., rated current ) on either primary or secondary as desired.This method can be furthersimplifiedto "divide rated current by per unit impedance."Both methods are illustrated here:vd.rate .pnmary7 18.391 AXT.ohms.primaryvrated.secondary-====~=2 155L724AXT.ohms.sccondarynoo·voltIO.022·ohm240·volt.Olll·ohm718.419 A21621.622A(NOTE:this result can also be used to illustrate the concept of significant digits. The result is21.6 kA, athough the computer intemally carries out calculations to greater precision than issignificant. )Another methodtoarrive at the same result:IISc.pU := -=34.483 PUXl'SC.Primary:= ' Sc.PU· 'ratcd.primary=718.391A34.5·20.8·amp = 717.6 A'SC.secondary := ' Sc.PU· ' ratcd.secondary = 21.552kAAnd finally, another method, often the simplest:lrated.primaryXl718.39 ' A'rated.secondary= 2 1551.724AXT34.5·625·amp=2 1562.5 A20.8·amp = 7 17.241A029625·amp=21551.n4A0.029

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CHAPTER33.1FOllrboxes represent an AC generator, reactor, resistor, and capacitor and are connected to a sourcebus XY as shown in Figure P3.1. From the circuit and phasor diagrams,identifyeach box.yI,II,!13tIJx1234I,13I,~------1r---~---~VXY3.2Two transformer banks are connected to a common bus as shown in Figure P3.2.What are the phase relations betweenthe voltagesV ANandV A'N';VllNand V13'N',VCNandVC'N'?Tsformer'""Baok2~No.1,*~.-----Arr*-----il-----B&_+-_______C~~'N-Tran sformerNo.2Baok23r-::~A'~.B'~'C'~'N'_L~

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