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2.1 INTRODUCTION
Thefollowingphenomenacanbeobservedduringswitchingandthefaultclearingprocessinacircuitbreaker:
1. Asthefaultoccurs,thecurrentincreasestoahighvalueduringthefirsthalfcycleofthewaveandthereafterthe
amplitude
of
the
wave
goes
on
reducing
as
the
waveform
passes
through
the
sub
transient,
transient
and
steadystates.Thewaveformofthecurrentisasymmetricalaboutthenormalzeroaxis.
2. Thevoltageacrossthecircuitbreakerpoleafterthefinalarcextinction(calledthetransientrecoveryvoltage(TRV))hasarelativelyhighamplitudeandrateofrise.Thevoltagehasahighfrequencytransientcomponent
superimposedonapowerfrequencycomponent.
3. Overvoltagescanbegeneratedwhileclosingcircuitbreakeroncapacitorbanksorloadedtransmissionlines.Theseareminimizedbypreclosingresistorsandsurgesuppressors.
Inthischaptertheabovementionedphenomenahavebeenstudiedwithreferencetothebehaviorofcircuitbreaker.
Forthepurposeofanalysis,simpleRLCnetworkshavebeenconsidered.Thegeneratorhasbeenrepresentedbyan
e.m.f.source.Theequationsofvoltageandcurrenthavebeensolvedbysimplerulesofdifferentialcalculus.
2.2 NETWORK PARAMETERS: R, L, C
Anelectricalnetworkcomprisesthefollowingnetworkparameters
Inductance(L) Capacitance(C) Resistance(R)
Theresistancecanbeneglectedasafirstapproximation.
i. InductanceInductanceisdefined
henerydi
dL
= (1)
where L =inductanceofcircuit,Henry.
=fluxlinkageduetocurrenti,Weberturns
I =currentinthecircuit,Amp.
Theelectromotiveforce(e.m.f.)inducedinaninductorisgivenby,
voltsdt
diL
dt
di
di
d
dt
de ===
(2)
Energyininductance(LHenry)attheinstantwhenthecurrentinitisi(A)isgivenby,
Joules2
1 2LiWin= (1Joule=1Wasecond) (3)
Inaninductivecircuitcurrentcannotchangeinstantaneously. Hencewhenthee.m.fisappliedatt=0,thecurrentis
zeroatthe
instant
ofclosing
the
switch.
Also
we
know
that
the
current
lags
behind
applied
voltage
by90
inthe
inductance.
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Consideringsinusoidalvoltageappliedtoan inductance,thecurrentlagsby90,therefore,thevoltageofthecircuit
hasmaximuminstantaneousvalueatthecurrentzero.
While
interrupting
the
current
flowing
through
an
inductive
circuit
such
as
a
transformer
on
no
load,
a
transformer
loaded by an inductor, etc. the circuitbreaker should interrupt the arc at natural current zero of the alternating
currentwave.Ifthearcextinctiontakesplaceatthenaturalcurrentzero,theenergyintheinductance(1/2Li2)iszero.
However,ifthearcissuddenlyinterruptedbeforethenaturalcurrentzero,attheinstantaneousvalueofcurrent,sayi
amperes,theenergy1/2Li2issuddenlyinterruptedbythechoppingofcurrenttoanartificialzerovalue.Duetosucha
phenomenontheinterruptingoflowmagnetizingcurrentsoftransformers,reactorsneedaparticularattention.The
circuitbreakershouldbecapableofinterruptingsuchcurrentswithoutgettingdamagedorwithoutgivingrisetoover
voltageabovethepermissiblelimits.
ii. CapacitanceThewellknowndefinitionofthecapacitorisTwoormoreconductorsseparatedbydielectric(insulating)medium.
ThecapacitanceCisgivenby
faradsdv
dqC = (4)
where C =capacitance,Farads
q =charge,Coulombs
v =voltage,Volts.
Fromtheabovedefinition, it isunderstandablethattransmission lines,bushing,circuitbreakersetc.have inherent
capacitancebetweenphaseandbetweenphaseandground.Insomecasesthecapacitancemaybenegligible.Inh.v.
circuit it becomes important and may not be negligible. In circuitbreaking phenomenon, capacitance plays an
importantrole.Thevoltageacrosscapacitorisgivenby
voltsC
dqdv=
== volts11
idtC
dqC
v
Energyinacapacitorisintheformofelectricfieldandisgivenby
Joules2
1 2CvWC= (5)
where CisinFarads,visinVolts,qisthechargeinCoulombs.
There exists a distributed capacitance between conductors and between conductor and ground in case of
transmission lines.Theflowofalternatingcurrent inthetransmission line isassociatedwithalternatechargingand
dischargingofthiscapacitance.Thecurrentstakenbythecapacitanceforchargingarecalledchargingcurrents.The
chargingcurrentflow intransmission line,even ifthereceivingendisopencircuited.Thevoltageacrossacapacitor
cannotchangeinstantaneously.
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Whileclosingacircuitbreakeronapredominantlycapacitivecircuitlikeacapacitorbank,thecurrentflowing inthe
capacitanceisgivenby
dt
dvCi
=
where i =instantaneousvalueofcurrent,Amperes
C =capacitance,Farads
dvldt =rateofchangeofvoltage,Volts/sec.
The current inrushduring the closingof capacitive circuit canoccurduringprearcingbetween the circuitbreaker
contacts.Thefollowingdutiescanproduceseverestressesonthecircuitbreaker:
Parallelingoftwocapacitorbanks Closingandopeningcapacitorbanks Closingandopeningunloadedtransmissionlinesonnoload
2.3 VOLTAGE EQUATION OF AN RLC SERIES CIRCUIT
ThevoltageequationofanRLCseriescircuitisgivenby
volts1++= idt
CRi
dt
diLe (6)
e =impressedvoltage
Ldt
di =voltageacrossinductor
Ri =voltageacrossresistor
idtC
1 =voltagecrosscapacitor.
Foranalternatinge.m.f,theinducedvoltageeisgivenby
( )+= tEemsin
where,Em=2Ermsand=2f.Angle dependsonmagnitudeofeatt=0.Ifeiszeroati=0,then =0andife=Em
ati=0then= /2.
2.4 SUDDEN SHORT CIRCUIT OF RL SERIES CIRCUIT
Letussee,whathappens,whenswitchSinthecircuitshowninFig2.1issuddenlyclosed.
Figure2.1 RLseriescircuitunderstudy
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Writinganequationforcurrentionthebasisdescribedinsecton2.3,
( ) +==+ tEeRidt
diL
msin (8)
Weshallsolvethisequationtoobtainanexpressionforcurrenti.Eq.(8)isanonhomogeneousdifferentialequation
offirstorder.TheEmcompletesolutionisthesumofacomplementarysolution(ic)andaparticularsolution(ip)i.e.
i=ic+ip (9)
ComplementarySolution,(ic)
TheauxiliaryequationisobtainedbypungtherighthandsideofEq.(8)equaltozero,i.e.
0=+ Ridt
diL
Rearrangingtheterms,
0=+ dtL
R
i
di
Integrating,
KtL
Ri
KtL
Ri
+=
=+
ln
ln
whereKisaconstantofintegrationgivenbyK=lnA,whereAissomeotherconstant.Hence
( )( ) Aei tLR lnlnln +=
( )tLReAi = (10)
This iscomplementarysolutionofcurrent i. It isanexponentiallydecayingcomponentcalledD.C.Component.The
magnitudeofconstantAdependsoninitialconditions.Amaybezero,positiveornegativedependinguponmagnitude
ofeatt=0.
ParticularSolution,(ip)
Takeatrialsolution
( ) ( ) +++= tDtCi sincos (11)
SuchatrialsolutonistakenbecausetheR.H.S.ofEq.(8)isoftheform ( )+tEmsin .Obtain
dt
diofEq.(11)
andsubsttuteinEq.(8).Equatethecoefficientsofliketermsfrombothsidestoget
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222 LR
LEC
m+
= (12)
222 LR
RED m += (13)
SubstitutingthesevaluesofCandDinEq.(11)weget
( ) ( )
++
+++
= tELR
RtE
LR
Li
mmsincos
222222 (14)
Let betheangleofimpedancetriangle
R
L 1tan=
222222cos;sin
LR
R
LR
L
+=
+=
Substitutingsin andcos inEq.(14),
( ) ( )
++
+++
= t
LR
Et
LR
Ei mm sincoscossin
222222 (15)
( ) ++= tLR
E
im
sin222 (16)
Eq.(16)isthepartcularsolutonofEq.(8).ItissinusoidcalledA.C.Component.
CompleteSolution,(i)
i=ip+ic
FromEqs.(10)and(16),weget
( ) ( )
++
+= tLR
EAei mtLR sin
222 (17)
ThisisacompletesolutonofEq.(8).LetusputtheinitalconditontoevaluateA.Att=0;i=0;becausethe
currentininductivecircuitdoesnotchangeinstantaneously.
AssumingRtobetoosmallascomparedwithL;
o90tanand 1222 ==+ R
LLLR
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CaseI.Switchclosedate=0.
Hencee=0att=0,therefore, =0.
Also
i
=
0
at
t
=
0,
and
from
Eq.
(17)
( )o90sin0222
+
+=LR
EA m
L
EA m
=
ThisismaximumvalueofA,hencethed.c.componentismaximumwhenswitchisclosedatvoltagezero.Thiscaseis
calledDoublingEffectbecausepeakvalueisthen2Em/L,atthepeakofthefirstcurrentloop.Thereisaslightdropin
the instantaneous valueof the current from t= 0 to t=/2. Therefore, thepeak value canbe considered tobe
approximately1.8Em/Linsteadof2Em/L.
CaseII.Switchclosedate=Em
Hencee=Ematt=0,therefore =/2.
Alsoi=0att=0,andweget
++=
22sin0
222
LR
EA m
0=A
HenceA is zero, if switch is closedwhene=Em, thereby thed.c. component isalso zero. Fromcases Iand IIwe
observethatthemagnitudeof initialvalueofd.c.component( )tLReA dependsuponthemomentofclosureof
switch,orvoltageattheinstantofoccurrenceofshortcircuit.
Letusnowinterprettheresultsofthesolution.WhenanRLseriescircuitisclosedwithanalternatingvoltagesource,
theresultingcurrentconsistsoftwocomponents,ad.c.componentandana.c.component.Thea.c.component is
superimposedonthed.c.component.Themagnitudeofd.c.componentdependsuponthevoltageattheinstantof
closingtheswitch.Whentheswitchisclosedatvoltagezero,thed.c.componentismaximum(Fig.2.2).Iftheswitchis
closedatvoltagemaximum,d.c.component iszeroandthewaveform issymmetricalaboutthenormalzeroaxisas
showninFig.2.3.
Example 1: A.C. Transient RL circuit.A50Hzsinusoidalvoltageofamplitude400Voltsisappliedtoaseriescircuitof
resistance10Ohmand inductanceof0.1H.Findanexpressionfor thevalueof thecurrentatany instantafer the
voltageisapplied.
a) Find thevalueofd.c.componentofcurrentuponclosing theswitch if instantaneousvalueofvoltage is50Voltsatthattime
b)What valueof instantaneousvoltagewillproduceamaximumd.c. componentof currentupon closing theswitch?
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c) What is the instantaneous value of voltagewhichwill result in the absence of any d.c. component uponclosingtheswitch?
d) Iftheswitch isclosedwhen instantaneousvoltage iszero,findthe instantaneouscurrent0.5,1.5,5.5cycleslater
Figure2.2 Changeofcurrentwith tmeinRLseriescircuituponswitchingatvoltagezero
Figure2.3 Changeofcurrentwith tmeinRLseriescircuituponswitchingatmaximumvoltage
Solution:
Given:R=l0ohms,L=0.1henery,f=50Hz,and =2f=314
( ) ( ) =+=+ 331.031410 222222 LR
radians26.133.7210
4.31tanAngle 1 === o
VErms 400= VEE
rmsm7.56540022 ===
( ) ( ) tttLR eee 1001.010 == SubsttutnginEq.(17),
( )
( ) AtAe
AtAei
t
t
26.1314sin14.17
26.1314sin33
7.565
100
100
++=
++=
(i)
a) Switchclosedatt=0,whene=50V,therefore
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( )+== 0sin7.56550e
radian0885.007.57.565
50sin 1 === o
Sincei=0att=0,thereforefromEq.(i)
( )26.10885.0sin14.170 += A
( ) 79.1526.10885.0sin14.17 ==A
D.C.componentatt=0isgivenby( ) AeAe 79.1578.15 00100 ==
b) Themaximum d.c. componentwill be produced if instantaneous value of applied voltage is zero at theinstantofclosingtheswitch.
c) Thed.c.componentwillvanish ife=Em, i.e.2400=565.7V(instantaneous)atthe instantofclosingtheswitch.
d) Iftheswitch isclosedatzero instantaneousvoltage,theangle iszeroandAwillbe16.32.Thecurrentequationwillthenbe
( ) Atei t 26.1314sin14.1732.16 100 += (ii)0.5 cycles=0.50.02=0.01second
1.5 cycles=0.03second
5.5 cycles=0.11second
SubstituteinEq.(ii),
2.5 SUBTRANSIENT, TRANSIENT AND STEADY STATE
Theanalysisofsuddenshortcircuit,ofanRLseriescircuit (secton2.4)willnowbeapplied to threephaseshort
circuitofanalternator.Theprincipleofoperationofasynchronousgenerator isbasedonarotatingmagneticfield
which generates a voltage in an armaturewinding having resistance and reactance. The current flowing when a
generatorisshortcircuitedissimilartothatgiveninFigs2.2and2.3whichisflowingwhenanalternatingvoltageis
suddenlyappliedtoaresistanceandaninductanceinseries.However,thecurrentflowinginasynchronousmachine
immediatelyaftertheoccurrenceofafault,thatflowingafewcycleslater,andthesustained,orsteadystate,valueof
the faultcurrentdiffer considerablybecauseof the effectof the armature currenton the flux thatgenerates the
voltage inthemachine. Inthealternator,thewaveform ismodifiedbyarmaturereaction.Anoscillogramofthree
phasecurrentsisshowninFig.2.4.Sincethevoltagesgeneratedinthephasesofathreephasemachinearedisplaced
120electricaldegreesfromeachother,theshortcircuitoccursatdifferentpointsonthevoltagewaveofeachphase.
Forthisreasontheunidirectional ordctransientcomponentofcurrentisdifferentineachphase.
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Phase B
Time
Figure2.4 Waveformsofshortcircuitcurrentsinthethreephasesofanalternator.
Ifthedccomponentofcurrentiseliminatedfromthecurrentofeachphase,theresultingplotofeachphasecurrent
versus tmeisthatshowninFig.2.5.ComparisonofFigs.2.3and2.5showsthedifferencebetweenapplyingavoltage
totheordinaryRLcircuitandapplyingashortcircuittoasynchronousmachine.Thereisnodccomponentineitherof
thesefigures.Inasynchronousmachinethefluxacrosstheairgapofthemachine ismuch largeratthe instantthe
short circuitoccurs than it isa few cycles later.The reductionof flux is causedby themmfof the current in the
armature. Such effect is called armature reaction.When a short circuit occurs at the terminals of a synchronous
machine, time is required for the reduction in fluxacross theairgap.As the fluxdiminishes, thearmaturecurrent
decreases because the voltage generated by the airgap flux determines the currentwhichwill flow through the
resistanceandleakagereactanceofthearmaturewinding.
Figure2.5 Currentasafunctonoftmeforasynchronousgeneratorshortcircuitedwhilerunningatnoload.Theunidirectional transientcomponentofcurrenthasbeeneliminatedinredrawingtheoscillogram.
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2.6 SHORTCIRCUIT CURRENTS AND THE REACTANCES OF SYNCHRONOUS MACHINES
CertaintermsthatarevaluableinthecalculationofshortcircuitcurrentinapowersystemcanbedefinedfromFig.
2.5.Thereactancesweshalldiscussaredirectaxisreactances.Thedirectaxisreactance isthatusedforcomputing
voltage
drops
caused
by
that
component
of
the
armature
current
which
is
in
quadrature
(90
o
out
of
phase)
with
the
voltagegeneratedatnoload.Sincetheresistanceinafaultedcircuitissmallcomparedwiththeinductivereactance,
currentduringafaultisalwayslaggingbyalargeangle,andthesocalleddirectaxisreactanceisrequired.
Inthediscussiontofollow,itshouldberememberedthatthecurrentshownintheoscillogramofFig.2.5isthatwhich
flowsinanalternatoroperatingatnoloadbeforethefaultoccurs.
InFig.2.5 thedistanceoa is themaximumvalueof thesustainedshortcircuitcurrent.Thisvalueofcurrent times
0.707isthermsvalue I ofthesustained,orsteadystate,shortcircuitcurrent.Thenoloadvoltageofthealternator
gE dividedbythesteadystatecurrent I iscalledthesynchronousreactanceofthegeneratororthedirectaxis
synchronousreactanceXdsincethepowerfactorislowduringtheshortcircuit.Thecomparativelysmallresistanceof
thearmatureisneglected.
If the envelopeof the currentwave is extendedback to zero timeand the first few cycleswhere thedecrement
appearstobeveryrapidareneglected,theinterceptisthedistanceob.Thermsvalueofthecurrentrepresentedby
this intercept,or0.707 tmesob inamperes, isknownasthetransientcurrent'I .Anewmachinereactancemay
nowbedefined.Itiscalledthetransientreactance,orinthisparticularcasethedirectaxistransientreactance'
dX
andisequalto'IE
g foranalternatoroperatingatnoloadbeforethefault.Iftherapiddecrementofthefirst
few cycles is neglected, the point of intersection that the current envelope makes with the zero axis can be
determined more accurately by plotting on semilogarithmic paper the excess of the current envelope over the
sustainedvaluerepresentedbyoa,asshowninFig.2.6.Thestraightlineportionofthiscurveisextendedtothezero
timeaxis,and the intercept isadded to themaximum instantaneousvalueof the sustained current toobtain the
maximuminstantaneousvalueoftransientcurrentcorrespondingtoobinFig.2.5.
Figure2.6
Excess
ofthe
current
envelope
ofFig.
2.5
over
the
sustained
maximum
current,
plotted
on
semilogarithmicscales.
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Thermsvalueofthecurrentdeterminedbythe interceptofthecurrentenvelopewithzerotime iscalledthesub
transient current"I . InFig.2.5 the subtransientcurrent is0.707 tmes theordinateoc.Subtransientcurrent is
oftencalledthe initialsymmetricalrmscurrent,which ismoredescriptivebecause itconveysthe ideaofneglecting
thedccomponentandtakingthermsvalueoftheaccomponentofcurrentimmediatelyaftertheoccurrenceofthe
fault.Directaxissubtransientreactance"
dX foranalternatoroperatingatnoloadbeforetheoccurrenceofathree
phasefaultatitsterminalsis"IE
g.
The current and reactancesdiscussedabovearedefinedby the followingequations,whichapply toanalternator
operatingatnoloadbeforetheoccurrenceofathreephasefaultatitsterminals:
d
g
X
EoaI ==
2 (18)
'
'
2d
X
EobI
g== (19)
"
"
2d
X
EocI
g== (20)
where
I =steadystatecurrent,rmsvalue
'I =transientcurrent,rmsvalueexcludingdccomponent
"I =subtransientcurrent,rmsvalueexcludingdccomponent
dX =directaxissynchronousreactance
'
dX =directaxistransientreactance
"
dX =directaxissubtransientreactance
gE =rmsvoltagefromoneterminaltoneutralatnoload
oa,ob,oc =interceptshowninFig.2.5.
Equatons(18)to(20)indicatethemethodofdeterminingfaultcurrentinageneratorwhenitsreactancesareknown.
Ifthegenerator isunloadedwhenthefaultoccurs,themachine isrepresentedbythenoloadvoltagetoneutral in
serieswiththeproperreactance.
The resistance is taken intoaccount ifgreateraccuracy isdesired. If there is impedanceexternal to thegenerator
betweenitsterminalsandtheshortcircuit,theexternalimpedancemustbeincludedinthecircuit.
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Althoughmachinereactancesarenottrueconstantsofthemachineanddependonthedegreeofsaturationofthe
magneticcircuit,theirvaluesusuallyliewithincertainlimitsandcanbepredictedforvarioustypesofmachines.Table
2.1givestypicalvaluesofmachinereactancesthatareneededinmakingfaultcalculationsandinstabilitystudies.In
general,
sub
transient
reactances
of
generators
and
motors
are
used
to
determine
the
initial
current
flowing
on
the
occurrenceofashortcircuit.Fordeterminingthe interruptingcapacityofcircuitbreakers,exceptthosewhichopen
instantaneously, subtransient reactance is used for generators and transient reactance is used for synchronous
motors. In stability studies where the problem is to determine whether a fault will cause a machine to lose
synchronismwith the restof the system if the fault is removedaftera certain time interval, transient reactances
apply.
Table2.1 Typicalreactancesofthreephasesynchronousmachines
Turbinegenerators
salientpolegenerators2pole 4pole
Conventional
cooled
Conductor
cooled
Conventional
cooled
Conductor
cooled
With
dampers
Without
dampers
dX 1.76
1.71.82
1.95
1.722.17
1.38
1.211.55
1.87
1.62.13
1
0.61.5
1
0.61.5
qX 1.66
1.631.69
1.93
1.712.14
1.35
1.171.52
1.82
1.562.07
0.6
0.40.8
0.6
0.40.8
'
dX 0.21
0.180.23
0.33
0.2640.387
0.26
0.250.27
0.41
0.350.467
0.32
0.250.5
0.32
0.250.5
"dX 0.13
0.110.14
0.28
0.230.323
0.19
0.1840.197
0.29
0.2690.32
0.2
0.130.32
0.3
0.20.5
2X = "
dX = "
dX = "
dX = "
dX 0.2
0.130.32
0.4
0.30.45
0X variesfrom0.1to0.7of "
dX
Valuesareperunit.Foreachreactancearangeofvaluesislistedbelowthetypicalvalue
Inconclusion,astheshortcircuitoccurs,theshortcircuitcurrentattainshighvalue.Thecircuitbreakercontactsstart
separatingaftertheoperationoftheprotectiverelay.Thecontactsofthecircuitbreakerseparateduring 'transient
state.'The r.m.s.valueof thecurrentat the instantof thecontactseparation iscalled thebreakingcurrentof the
circuitbreakerandisexpressedinkA.
Ifacircuitbreakerclosesonexisting fault,thecurrentwould increasetoahighvalueduringthe first,halfcycleas
shownisFigs.2.2and2.3.
Thehighestpeakvalueofthecurrent isreachedduringthepeakofthe firstcurrent loop.Thispeakvalue iscalled
'making current'of thecircuitbreakerand isexpressed inkA.Though the shortcircuitcurrentvariescontinuously
duringthesubtransientandtransientstates,therepresentativevaluescanbecalculated fromequatons18to20.
Thesubtransient,transientandsteadystatereactancescanbedeterminedexperimentallybyconductingshortcircuit
test.
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ItisclearfromEqs.(18)to(20)thatwhilecalculatngsubtransient,transientandsteadystatecurrents;therespective
reactancesshouldbeconsidered.ShortCircuitCalculationsandtheSelectionofCircuitBreakers
2.7 FAULT CALCULATIONS2.7.1 SYSTEM CONFIGURATION
Powersystemsoperateatvoltageswherekilovolt(kV)isthemostconvenientunitforexpressingvoltage.Also,these
systems transmit large amountofpower, so that kilovoltampere (kVA)andmegavoltampere (MVA) areused to
expressthetotal(generalorapparent)threephasepower.Thesequantities,togetherwithkilovars,amperes,ohms,
flux,andsoon,areusuallyexpressedasaperunitorpercentofareferenceorbasevalue.Theperunitandpercent
nomenclatures arewidely used because they simplify specification and computations, especially where different
voltagelevelsandequipmentsizesareinvolved.
Thisdiscussionisforthreephaseelectricsystemswhichareassumedtobebalancedorsymmetricaluptoapointor
area of unbalance. Thismeans that the source voltages are equal inmagnitude and are 120o displaced in phase
relations,andthattheimpedancesofthethreephasecircuitsareofequalmagnitudeandphaseangle.Fromthisasa
beginning, various shunt and series unbalances can be analyzed, principally by the method of symmetrical
components.
2.7.2 Per Unit and Percent Values
Percentis100
tmes
per
unit,
both
are
used
asama
erofconvenience
orpersonal
choice
and
itisimportant
to
designateeitherpercent(%)orperunit(pu).
Theperunitvalueofanyquantityistheratioofthatquantitytoitsvalue,theratioexpressedasanondimensional
decimalnumber.Thusactualquantities,suchasvoltage(V),current(I),power(P),reactivepower(Q),voltamperes
(VA),resistance(R),reactance(X),andimpedance(Z),canbeexpressedinperunitorpercentasfollows:
quantityofvaluebase
quantityactualunitperinQuantity = (21)
Quanttyinpercent=(quanttyinperunit)x100 (22)
where actualquantity is the scalaror complex valuesof aquantity expressed in itsproperunits, such as volts,
amperes,ohms,orwatts.basevalueofquantityreferstoanarbitraryorconvenientreferenceofthesamequantity
chosenanddesignatedasthebase.Thusperunitandpercentaredimensionlessratioswhichmaybeeitherscalaror
complexnumbers.
Asanexampleforachosenbaseof115kV,voltagesof92,115and161kVbecome0.80,1.00,and1.40puor80%,
100%,and140%,respectvely.
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2.7.2.1 Base Quanttes
The base quantities are scalar quantities, so that phasor notation is not required for the base equations. Thus
equationsforthebasevaluescanbeexpressedwiththesubscriptBtoindicateabasequantityasfollows:
BBB IkVkVApowerbaseFor 3: = (23)
B
BB
kV
KVAIcurrentbaseFor
3: = (24)
B
BB
kVA
xKVZimpedancebaseFor
1000:
2
(25)
andsince MVA=1000kVA (26)
thebaseimpedancecanalsobeexpressedas
B
BB
MVAkVZ
2
= (27)
Inthreephaseelectricpowersystemsthecommonpracticeistousethestandardornominalsystemvoltageasthe
voltagebase,andaconvenientMVAorkVAquanttyasthepowerbase.100MVA isawidelyusedpowerbase.The
systemvoltagecommonlyspecifiedisthevoltagebetweenthethreephases(i.e.,thelinetolinevoltage).Thisisthe
voltage used as a base in Eqs. (23) through (27). As a shortcut and for convenience, the linetoline subscript
designation(ll) isomitted.Withthispractice it isalwaysunderstoodthatthevoltage isthe linetolinevalueunless
indicatedotherwise.Themajorexception is inthemethodofsymmetricalcomponents,where linetoneutralphase
voltage isused.Thisshouldalwaysbespecifiedcarefully,but there issometimesa tendencytooverlookthisstep.
Similarly,currentisalwaysthephaseorlinetoneutralcurrentunlessotherwisespecified.
Power is alwaysunderstood tobe threephasepowerunless otherwise indicated.General power, also known as
complexorapparentpower, isdesignatedbyMVAorkVA,as indicatedabove.Threephasepower isdesignatedby
MVorkV.ThreephasereactivepowerisdesignedbyMVArorkVAr.
2.7.2.2 Per Unit and Percent Impedance Relationships
Perunitimpedanceisspecifiedinohms(Z)fromEq.(21)bysubsttutngEq.(27):22 100 B
B
B
B
Bpu
kV
ZkVAor
Vk
ZMVA
Z
ZZ == (28)
or,inpercentnotation,
22 10
100%
B
B
B
B
Vk
ZkVAor
Vk
ZMVAZ = (29)
wheretheohmvaluesaredesiredfromunit,percentvalues,theequationsare
B
puB
B
puB
kVA
ZVk
orMVA
ZVk
Z
22 1000
= (30)
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wheretheturnsareproportionaltothevoltages.
The
per
unit
impedances
are,
from
Eq.
(21),
(33),
and
(34),
yB
y
x
y
y
x
xB
x
xZ
ohmsZ
N
N
N
N
Z
ohmsZpuZ
22
==
puZZ
ohmsZy
yB
y == (36)
Thustheperunitimpedanceisthesameoneithersideofthebank.
TransformerBankExample
Consideratransformerbankrated50MVAwith34.5kVand161kVwindingsconnectedtoa34.5and161kVpower
system.Thebankreactanceis10%.Nowwhenlookingatthebankfromthe34.5kVsystem,itsreactanceis
10%ona50MVA34.5kVbase (37)
Andwhenlookingatthebankfromthe161kVsystemitsreactanceis
10%ona50MVA161kVbase (38)
Thisequalimpedanceinpercentorperunitoneithersideofthebankisindependentofthebankconnections:wye
delta,deltawye,wyewye,ordeltadelta.
Thismeansthattheperunit(percent)impedancevaluesthroughoutanetworkcanbecombinedindependentlyofthe
voltagelevelsaslongasalltheimpedancesareonacommonMVA(kVA)baseandthetransformerwindingsratings
arecompatiblewiththesystemvoltages.Thisisagreatconvenience.
The actual transformer impedances in ohms are quite different on the two sides of a transformerwith different
voltagelevels.Thiscanbeillustratedfortheexample.ApplyingEq.(33),wehave
kVatjX 5.3438.250100
105.34 2=
= (38)
kVat16184.5150100
101612=
= (39)
ThiscanbecheckedbyEq.(35),wherefortheexamplexisthe34.5kVwindingside,andyisthe161kVwindingside.
Then,
38.284.5116
5.3438.2
2
2
== (40)
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2.7.2.4 Changing Per Unit (Percent) Quanttes to Different Bases
Normally,theperunitorpercentimpedancesofequipmentisspecifiedontheequipmentbase,whichgenerallywill
bedifferentfromthepowersystembase..sinceallimpedancesinthesystemmustbeexpressedonthesamebasefor
per
unit
or
percent
calculations,
it
is
necessary
to
convert
all
values
to
the
common
base
selected.
This
conversion
can
bederivedbyexpressingthesameimpedanceinohmsontwodifferentperunitbases.FromEq.(33)foraMVA1,kV1
base andaMVA2,kV2base,
21
11
kV
ZMVAZ ppu
= (41)
Byrationingthesetwoequationsandsolvingforoneperunitvalue,thegeneralequationforchangingbasesis
1
21
2
2
2
1
2
MVA
kV
kV
MVA
Z
Z
pu
pu = (42)
22
21
1
212
kV
kV
MVA
MVAZZ pupu = (43)
Equaton(43)isthegeneralequatonforchangingfromonebasetoanotherbase.Inmostcasestheturnsratoofthe
transformer isequivalent to thedifferent systemvoltages,and theequipmentratedvoltagesare the sameas the
systemvoltages,sothatthevoltage squaredratoisunity.ThenEq.(43)reducesto
1
212
MVAMVAZZ pupu = (44)
ItisveryimportanttoemphasizethatthevoltagesquarefactorofEq.(43)isusedonlyinthesamevoltageleveland
where slightly different voltage bases exist. It is never used where the base voltages are proportional to the
transformerbankturns,suchasgoingfromthehightothelowsideacrossabank.Inotherwords,Eq.(43)hasnothing
todowithtransferringtheohmicimpedancevaluefromonesideofatransformertotheotherside.
SeveralexampleswillillustratetheapplicatonsofEq.(43)and(44)inchangingperunitandpercentimpedancesfrom
onebasetoanother.
2.7.3 Symmetrical Components
Any unbalanced current or voltage can be determined from the sequence components from the following
fundamentalequations:
oaoa VVVVIIII ++=+== 2121 (45)
oboab VaVaVVaVIaIIaI +++=++= 2212
212
(46)
oCoc VVaaVVIIaaII ++=++= 22
122
1 (47)
whereIa,Ib,andIcorVa,VbandVcaregeneralunbalancedlinetoneutralphasors.
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Fromthese,equationsdefiningthesequencequantitiesfromathreephaseunbalancedsetcanbedetermined:
)()( cbaocbao VVVVIIII ++=++=3
1
3
1 (48)
)()( cbacba VaaVVVIaaIII2
12
131
31 ++=++= (49)
)()( cbacba aVVaVVaIIaII ++=++=2
22
23
1
3
1 (50)
Theselastthreefundamentalequationsarethebasisfordeterminingifthesequencequantitiesexistinanygivenset
ofunbalancedthreephasecurrentsorvoltages.Theyareusedforprotectiverelayingoperationfromthesequence
quantites.Forexample,Fig.2.8showsthephysicalapplicatonofcurrenttransformers(CTs)andvoltagetransformers
(VTs)tomeasurezerosequenceasrequiredbyEq.(48),andusedingroundfaultrelaying.
NetworksoperatingfromCTsorVTsareusedtoprovideanoutputproportionaltoI2orV2andarebasedonphysical
solutonsof(Eq.50).
2.7.3.1 SEQUENCE INDEPENDENCE
Forallpracticalpurposeselectricpower systemsarebalancedor symmetrical from thegenerators to thepointof
singlephaseloadingexceptinanareaofafaultorunbalancesuchasanopenconductor.Inthisessentiallybalanced
area,thefollowingconditionsexist:
1. Positivesequence currents flowing in the symmetrical or balanced network produce onlypositivesequencevoltagedrops;nonegativeorzerosequencedrops.
2. Negativesequencecurrentsflowinginthebalancednetworkproduceonlynegativesequencevoltagedrop;nopositiveorzerosequencevoltagedrops.
3. Zerosequencecurrentsflowinginthebalancednetworkproduceonlyzerosequencevoltagedrops;nopositive ornegativesequencevoltagedrops.Thisisnottrueforanyunbalancedor
nonsymmetrical pointorareasuchasanunsymmetricalfault,openphase,andsoon.Inthese:
4. Positivesequencecurrentflowinginanunbalancedsystemproducespositiveandnegativeandpossiblyzerosequencevoltagedrops.
5. Negativesequence currents flowing inanunbalanced systemproducespositive,negative,andpossiblyzerosequencevoltagedrops.
6. Zerosequence current flowing in an unbalanced system produces all three: positive,negative,andzerosequencevoltagedrops.
This important fundamentalpermits settingup three independentnetworks,one foreachof the three sequences,
whichcaninterconnected onlyatthepointorareaofunbalance.
Beforecontinuingwiththesequencenetworks,areviewofthesourcesoffaultcurrentisuseful.
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Figure2.8 Zerosequencecurrentandvoltagenetworksusedforgroundfaultprotection
2.7.3.2 Sequence Networks
Theserepresentoneofthethreephasetoneutralorgroundcircuitsofthebalancedthreephasepowersystem,and
documentshowthatsequencecurrentswillflowiftheycanexist.Thesenetworksarebestexplainedbyanexample,
soconsider
the
section
ofapower
system
ofFig.
2.9.
Reactance valuesonlyhave been shown for the generator and the transformers. Theoretically, impedance values
shouldbeused,buttheresistancesoftheseunitsareverysmallandnegligibleforfaultstudies.
Figure2.9 Singlelinediagramofasectionofapowersystem.
Itisveryimportantthatallvaluesbespecifiedwithabase[voltageifohmsareused,orMVA(kVA)andkVifperunit
orpercentimpedancesareused].Beforeapplyingthesetothesequencenetworks,allvaluesmustbechangedtoone
commonbase.
Inmost
cases
per
unit
(percent)
values
are
used
and
acommon
base
inprac
tceis100
MVA
atthe
particularsystemkV.
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Positivesequencenetwork
Thisistheusuallinetoneutralsystemdiagramforoneofthethreesymmetricalphasesmodifiedforfaultconditions.
Thepositive sequencenetwork for the systemofFig.2.9 is shown in Fig.2.10.VGandVSare the system lineto
neutral
voltages.
VG
is
the
voltage
behind
the
generator
sub
transient
direct
axis
reactance
Xd'',
and
VS
is
the
voltage
behindthesystemequivalentimpedanceZ1S.
(a) (b)
Figure2.10 PositvesequencenetworkforthesysteminFig.2.9
XTG isthe transformer leakage impedance for thebankbusG,andXHM is the leakage impedance forthebankatH
between theH and M windings. The delta winding L of this threewinding bank is not involved in the positive
sequence network unless a generator or synchronousmotor is connected to this delta or unless a fault is to be
consideredinthedeltasystem.
Forthe linebetweenbusesGandH,Z1GH isthe linetoneutral impedanceofthisthreephasecircuit.Foropenwire
transmission line, an approximate estmatng value is 0.8m/ml for bundled conductors. Typical values for shut
capacitanceoftheselinesare0.2m/milforsingleconductorand0.14/milforbundledconductors.Normally,this
capacitance isneglected,as it is veryhigh in relation toallother impedances involved in fault calculations.These
values should be used for estimating or in the absence of specific line constants. The impedances of cables vary
considerably,sospecificdataarenecessaryforthese.
Theimpedanceangleoflinescanvaryquitewidelydependingonthevoltageandwhethercableoropenwireisused
incomputer faultprograms theanglesareconsideredand included,but forhandcalculation it ispractical inmost
cases to simplify calculatonsbyas summing thatall theequipment involved in the faultcalculatonareat90oor
reactancevaluesonly.Sometimesitmaybepreferredtousethelineimpedancevaluesandtreatthemasreactances.
Unlessthenetworkconsistsofalargeproportionoflowanglecircuits,theerrorofusingallvaluesas90owillnotbe
toosignificant.
LoadisshownconnectedatbusesGandH.NormallythiswouldbespecifiedaskVAorMVAandcanbeconvertedinto
impedance:
3
1000
3
1000 kVVand
kV
loadMVA
loadI LN ==
kVatMVAkVI
VZ
loadload
LNload ===
2
(51)
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Equaton(51) isa linetoneutralvalueandwouldbeused forZLGandZLHrepresentingthe loadsatGandH inFig.
2.10a. If load isrepresented,thevoltagesVGandVSwillbedifferent inmagnitudeandangle,varyingasthesystem
loadvaries.
Thevalueofloadimpedanceusuallyisquitelargecomparedtothesystemimpedances,sothatloadhasanegligible
effecton the faultedphase current. Thus itbecomepracticaland simplifies calculations toneglect load for shunt
faults.Withno load,ZLGandZLHare infinite,VGandVSareequaland inphaseand soare replacedbyacommon
voltageVasinFig.2.10b.Normally,Visconsideredas1pu,thesystemratedlinetoneutralvoltages.
ConventionalcurrentflowisassumedtobefromtheneutralbusN1totheareaorpointofunbalance.Withthisthe
voltagedropV1xatanypointinthenetworkisalways,
= 111 ZIVV x (52)
whereVisthesourcevoltage(VgorVsinFig.2.10a)and 11ZI isthesumofthedropsalonganypathfromtheN1neutralbustothepointofmeasurement.
NegativeSequence Network
Thisnetworkdefinestheflowofnegativesequencecurrentswhentheyexist.Thesystemgeneratorsdonotgenerate
negative sequence, but negativesequence current can flow through their windings. Thus these generators and
sourcesarerepresentedbyanimpedancewithoutvoltage,asshowninFig.2.11.Inthetransformers,lines,andsoon,
thephasesequencesofthecurrentdoesnotchangetheimpedanceencountered,sothesamevaluesareusedasin
thepositive
sequence
network.
(a) (b)
Figure
2.11
Negat
ve
sequence
networks
for
the
system
in
Fig.
2.9:
(a)
network
including
loads;
(b)
network
neglectingloads.
Arotatingmachinecanbevisualizedasatransformerwithonestationaryandonerotatingwinding.Thusdc inthe
field produces positive sequence in the stator. Similarly, the dc offset in the stator ac current produces an ac
component inthefield.Inthisrelativemotionmodelwiththeonewindingrotatingatsynchronousspeed,negative
sequence in the stator results in a doublefrequency component in the field. Thus the negativesequence flux
component in the air gap is alternately between and under the poles at this double frequency. One common
expressionforthenegativesequenceimpedanceofasynchronousmachineis
)(2
1 ""
2 qd
XXX += (53)
ontheaverageofthedirectandquadratureaxessubtransientreactances.
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Zerosequenceimpedancefortransformerbanksisequaltothepositiveandnegativesequenceandisthetransformer
leakageimpedance.Theexceptiontothisisforthreephasecoretypetransformers,wheretheconstructiondoesnot
provideanironfluxpathforzerosequence.Forthesethezerosequencefluxmustpassfromthecoretothetankand
return.
Hence
for
these
types
X0
usually
is
0.85
to
0.9
X1,
and
when
known
the
specific
value
should
be
used.
The lowerrighthanddiagramofFig.2.12isforthesystemconnectedtobusH(Fig.2.9).Currentsoutofthethree
windingtransformerwillflowasshowninthelandmwindings.Thethreecurrentscanflowinthemgroundedwye
sincetheequivalentsourceisshowngroundedwithZ0sgiven.Thusthethreewindingequivalentcircuitisconnected
inthezerosequencenetwork(Fig.2.13)asshown.NotethatintherighthandpartofFig.2.12,ifanyofthewyeconnectonswerenotgrounded,theconnectonswould
bedifferent.
Figure2.13 ZerosequencenetworkforthesystemofFig.2.9
Iftheequivalentsystemorthemwindingwereungrounded,thenetworkwouldbeopenbetweenZmandZos,aszero
sequencecurrentscouldnotflowasshown.Load, ifdesired,wouldbeshown inthezerosequencenetworkonly if
theywerewyegrounded.Deltaloadswouldnotpasszerosequence.
Zerosequenceisalwaysdifferent,asitisaloopimpedance;theimpedanceofthelineplusareturnpatheitherinthe
earth,or inaparallel combinationof theearthandgroundwire, cable sheath, and soon. Thepositivesequence
impedanceisaonewayimpedance:fromoneendtootherend.Asaresult,zerosequencevariesfrom2to6 tmesX1
forlines.Forestimatingopenwritelines,avalueofX0=3or3.5X1iscommonlyused.
Thezerosequenceimpedanceofgeneratorsislowandvariable,dependingonthewindingdesign.Exceptforverylow
voltageunits,generatorsareneversolidlygrounded.InFig.2.9,thegeneratorGisshowngroundedthrougharesistor
R. FaultsonbusGand in the system to the rightdonot involve thegeneratoras faras zero sequence since the
transformerdeltablockstheflowofzerosequencecurrentasshown.
ConventionalcurrentflowisassumedtobefromthezeropotentialbusN0totheareaorpointofunbalance.Thusthe
voltagedropVoxatanypointinthenetworkisalways
Vox=0 i0z0 (55)
wherei0z0isthesumofthedropsalonganypathfromtheN0bustothepointofmeasurement.
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2.7.4 SYMMETRICAL & UNSYMMETRICAL FAULT CALCULATIONS
Theprincipaltypesoffaultsare:threephase,phasetophase,twophasetoground,andonephasetoground.
Example: Fault calculat
ons on a typical system shown in Fig. 2.14
ThesystemofFig.2.14issimilartothatshowninFig.2.9butwithtypicalconstantsforthevariousparts.These
are on the bases indicated, so the first step is to transfer them to a common base. The positive and negative
sequencenetwork(negativethesameaspositiveexceptfortheomissionofthevoltage) isshown inFig.2.15.The
conversiontoacommonbaseof100MVAisshownasnecessary.
Figure2.14 Powersystemexampleforfaultcalculatons
Fora faultatbusG,therighthand impedances(j0.18147+j0.03667+j0.03=j0.2481)areparalleledwiththe lef
hand impedances (j0.20+j0.1375=j0.3375).Reactancevaluesratherthan impedancevaluesareused,as istypical
wheretheresistanceisquitesmallrelatively.( ) ( )
pujXX 1430.05856.0
2481.03375.0
4237.05763.0
21
=
== (56)
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Figure2.15 Positve andnegativesequencenetworksandtheirreductiontoasingleimpedanceforafaultatbusG
inthepowersystemofFig.2.14.
Thedivisionof0.3375/0.5856=0.5763and0.2481/0.5856=0.4237as shownprovidesapartal check,as0.5763+
0.4237mustequal1.0andarethedistributonfactorsindicatngtheperunitcurrentflowoneithersideofthefault.
Thesevaluesareaddedtothenetworkdiagram.Thusforfaultsatbusg,X1=X2=j0.1430puon100MVAbase.
ThezerosequencenetworkforFig.2.14isshowninFig.2.16.Againthereactancevaluesareconvertedtoacommon
100MVAbase.
Theconversionstoacommon100MVAbaseareshownexceptforthethreewindingtransformer.Forthisbank,
puX
puX
puX
ML
HL
HM
18667.0150
100280.0
2400.0150
100360.0
03667.0150
100055.0
==
==
==
and,
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( )
( )
( ) puXL
puX
puX
M
H
1950.003667.018667.02400.02
1
00833.0240.018667.003667.02
1
0450.018667.02400.003667.02
1
=+=
=+=
=+=
(57)
TheseareshowninFig.2.16.
Figure2.16 ZerosequencenetworkanditsreductiontoasingleimpedanceforafaultatbusGinthepowersystemofFig.2.14.
ThisnetworkisreducedforafaultatbusGbyfirstparallelingX0S+ZHwithZLandthenaddingZMandX0GH;
( ) ( )
( )
( )
6709.0
620.0
0083.0
0592.0
280.0
0850.01950.0
3036.06964.0
j
Xj
Zj
j
OGH
M
=
=
Thisistherighthandbranch,parallelingwiththelefthandbranch,
( ) ( )
15407.08709.0
2000.06709.0
2296.07704.0
0 jX == puat100MVA (58)
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Thevalues(0.7704)and(0.2296)shownaddto1.0asacheckandprovidethecurrentdistributiononeithersideof
thebusGfault,asshownonthezerosequencenetwork.Thedistributonfactor0.2296fortherightside isfurther
dividedupby0.69640.2296=0.1599pu in the230kVsystemneutral,and0.30360.2296=0.0697pu in the
three
winding
transformer
H
winding
neutral.
These
are
shown
on
the
zero
sequence
network.
ThreePhase Fault at Bus G
Forthisfault,
I1=IAF=(j1.0)/(j0.143)=6.993pu
=6.993[(100.000)/ 3115)]
=3510.8Aat115kV (59)
The
divisions
of
current
from
the
left
(IAG)
and
right
(IAH)
are:
IAG=0.42376.993=2.963pu (60)
IAH=0.57636.993=4.030 (61)
SinglePhaseToGround Fault at Bus G
Forthisfault,
I1=I2=I0=[j1.0/j(0.143+0.143+0.1541)]=2.272pu (62)
IAF=32.272=6.817pu
=6.817[(100.000)/( 3115)]
=3422.5Aat115kV
Normally,the3Iocurrentsaredocumentedinthesystem,astheseareusedtooperatethegroundrelays.Equations
(45)through(47)providethethreephasecurrents.SinceX1=X2,sothatI1=I2,thesereducetoIb=Ic= I1+I0forthe
phasebandccurrents,sincea+a2= 1.ThecurrentsshownaredeterminedbyaddingI1+I2+I0forIa, I1+I0forIb
andIc,and3I0fortheneutralcurrents.
Inthe115kVsystemthesumofthetwoneutralcurrentsisequalandoppositetothecurrentinthefault.Inthe230
kVsystemthecurrentuptheneutralequalsthecurrentdowntheotherneutral.
Thecalculationsassumednoload,soprefault,allcurrentsinthesystemwerezero.Withthefaultinvolvingphasea
only,itwillbeobservedthatcurrentflowsinthebandcphases.Thisisbecausethedistributionfactorsinthezero
sequencenetwork aredifferent from the positive andnegativesequencedistribution factors.On a radial system
wherepositive,negative, and zerosequence currents flowonly fromone source and in the samedirection, the
distributionfactors inallthreenetworkswillbe1.0,althoughthezerosequence impedancesaredifferentfromthe
positivesequence impedances. Then Ib=Ic=I1 + I0 abovebecomes zero,and fault currentonly flows in the faulted
phase.InthistypeIa=3I0throughoutthesystemforasinglephasetogroundfault.
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2.8 SHORTCIRCUIT LEVEL
QuiteoftenshortcircuitMVAdataaresuppliedforthreephaseandsinglephasetogroundfaultsatvariousbusesor
interconnection points in a power system. The derivation of this and conversation into system impedances is a
follows:
ThreePhaseFaults
MVASC=3faultshortcircuitMVA=(3I3kV)/1000 (63)
whereI3isthetotalthreephasefaultcurrentinamperesandkVisthesystemlinetolinevoltageinkilovolts.From
this,
I3=(1000MVASC)/(3kV) (64)
Z=(VlN)/I3=(1000kV)/(3I3)
=(kV2)/(MVASC) (65)
SubstitutingEq.Zpu=(MVAbaseZ)/(kV2),thepositivesequenceimpedancetothefaultlocationis
Z1=[(MVAbase)/(MVAsc)] pu (66)
Z1=Z2forallpracticalcases.Z1canbeassumedtobeX1unlessX/Rdataareprovidetodetermineanangle.
SinglePhaseToGroundFaults
MVAGSC=GfaultshortcircuitMVA=(3IGkV)/1000 (67)
where IG is the total singlelinetoground fault current in amperes and kV is the system linetoline voltage in
kilovolts.
IG=(1000MVAGSC)/( 3kV) (68)
However,
IG=I1+I2+I0=(3VlN)/(Z1+Z2+Z0)=(3VlN)/ZG) (69)
ZG=[(3kV2)/(MVAGSC)] in (70)
ZG=[(3MVAbase)/(MVAGSC)] pu (71)
ThenZ0=ZGZ1Z2,orinmostpracticalcases,X0=XGX1X2,sincetheresistanceisusuallyverysmallin
relationtothereactance.
Example
Ashortcircuitstudyindicatesthatatbusxinthe68kVsystem,
MVAsc=594MVA
MVAGSC=631MVA
ona100MVAbase.
Thusthetotalreactancetothefaultis
X1=X2=100/594=0.1684pu
XG=300/631=0.4754pu
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X0=0.47540.16840.1684=0.1386pu,
allvaluesona100MVA69kVbase.
Effect of Induction Machines on Short Circuit LevelDuringfaultcondition, inductionmotorwillbe inductiongeneratorwithsubtransientortransientreactance
dependingonthespeedofthecircuitbreaker.Therefore,inductionmachinesas loadswill increasetheshortcircuit
level.
2.9 RUPTURE CAPACITY OF CIRCUIT BREAKERS
Asmostfaultprotectivedevices,suchascircuitbreakersandfuses,operatewellbeforesteadystateconditionsare
reached,generatorsynchronousreactance isalmostneverusedincalculatingfaultcurrentsforapplicationofthese
devices.Asdiscussedbefore,inordertodeterminetheinitialsymmetricalrmscurrent,thesubtransientreactances
of the synchronous generators and motors are used. However, the interrupting capacity of circuit breakers is
determinedusing the subtransient reactance forgeneratorsand transient reactance for the synchronousmotors.
Theeffectsofinductionmotorsareignored.Itisappropriatetousesubtransientreactanceforsynchronousmotorsif
fastactingcircuitbreakersareused.Forexample,modernairblastcircuitbreakersusuallyoperatein2.5cyclesof60
Hz.Oldercircuitbreakersandthoseusedonlowervoltagesmaytake8cyclesormoretooperate.Notethatsofarthe
dcoffset(orunidirectona1currentcomponent)hasbeenexcludedintheabovediscussions. Withfastactingcircuit
breakerstheactualcurrenttobe interrupted is increasedbythedccomponentofthefaultcurrent,andthe initial
symmetrical rmscurrentvalue is increasedbyaspecific factordependingon the speedof thecircuitbreaker.For
example,ifthecircuitbreakeropening tmeis8,3,or2cycles,thenthecorrespondingmultplyingfactoris1.0,1.2,or
1.4,respectvely.Therefore,theinterruptngcapacity(orratng)ofacircuitbreakerisexpressedas
( ) 6"int 10)(3 = IVS prefaulterrupting MVA (72)
where Vprefault =prefaultvoltageatpointoffaultinvolts
I'' =initialsymmetricalrmscurrentinamperes
=multiplyingfactor
ThemultiplyingfactorsandthereactancetypesaregiveninTable2.2.Asdiscussedbefore,theasymmetrical
currentwavedecaysgraduallytoasymmetricalcurrent;therateofdecayofthedccomponentbeingdeterminedby
thel/rofthesystemsupplyingthecurrent.Thetimeconstantfordccomponentdecaycanbefoundfrom
s/circuit RLTdc = or
2
/circuit RLT
dc= cycles (53)
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Table2.2 multplyingfactorsandthereactancetypesofcircuitbreakers
ReactanceQuantityforUseinXs
MultiplyingFactor
Synchronous
Generators
and
Condensers
Synchronous
MotorsInduction
Machines
A. CircuitBreakerInterruptingDuty1. GeneralCase
Eightcycle or slower circuit
breakers
1.0 Subtransient Transient Neglect
Fivecyclecircuitbreaker 1.1
Threecyclecircuitbreaker 1.2
Twocyclecircuitbreaker 1.4
2. Specialcaseforcircuitbreakersat generator voltage only. For
shortcircuit calculations of
morethan500,000kVA(before
the application of any
multiplying factor) fed
predominantly direct from
generators,or through current
limitingreactorsonly:
Subtransient Transient Neglect
Eightcycle or slower circuit
breakers
1.1
Fivecyclecircuitbreaker 1.2
Threecyclecircuitbreaker 1.3
Twocyclecircuitbreaker 1.5
3. Aircircuitbreakersrated600Vandless
1.25 Subtransient Subtransient Subtransient
B. Mechanical Stress andMomentary Duty of CircuitBreakers
1. GeneralCase 1.6 Subtransient Subtransient Subtransient2. At 500 V and below, unless
currentisfedpredominantlyby
directly connected
synchronous machines or
throughreactors
1.5 Subtransient Subtransient Subtransient
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The momentary duty (or rating) of a circuit breaker is expressed as
6" 10)6.1)()((3 = IVS prefaultmomentary MVA (74)
Therefore,thermsmomentarycurrentcanbeexpressedas
"6.1 IImomentary = A (75)
and it isused foroilcircuitbreakersof115kVandabove.Thecircuitbreakermustbeable towithstand this rms
currentduringthefirsthalfcycleafterthefaultoccurs.NotethatiftheI''ismeasuredinpeakamperes,thenthepeak
momentarycurrentisexpressedas
"7.2 IImomentary = A (76)
In the united states, the ratings of circuit breakers are given in theANSI Standards based on symmetrical
currentintermsofnominalvoltage,ratedmaximumvoltage,ratedvoltagerangefactork,ratedcontinuouscurrent,
andratedshortcircuitcurrent.Therequiredsymmetricalcurrentinterruptingcapabilityisdefinedas
voltageoperating
voltagemaximumratedcurrentcircuit-shortRated
The standard dictates that for operatng voltages below 1/k times rated maximum voltage, the required
symmetricalcurrentinterruptingcapabilityof thecircuitbreaker isequal tok times theratedshortcircuitcurrent.
Table2.3givesoutdoorcircuitbreakerratngsbasedonsymmetricalcurrent.Notethattheratedvoltagefactork is
defined as the ratioof ratedmaximum voltage to the lower limitof the rangeofoperating voltage inwhich the
requiredsymmetricalandasymmetrical interruptingcapabilitiesvary in inverseproportiontotheoperatingvoltage.
Therefore, general expressions (which take into account the rated voltage range factor k) for the rms and peak
momentarycurrents,respectively,are
"6.1 kIImomentary = A (77)
and
"7.2 kIImomentary = A (78)
NotcethatinTable2.3thefactorkis1forthenominalvoltagesof115kVandabove.Therefore,equatons(77)and
(78)becomethesameasequatons (75)and (76),respectvely.Asanexample,assumethatacircuitbreakerhasa
ratedmaximumrmsvoltageof38kVandisbeingoperatedat34.5kV.FromTable2.3,theratedvoltagerangefactor
kis1.65,theratedcontnuousrmscurrentis1200A,andtheratedshortcircuitrmssymmetricalcurrentattherated
maximum rms voltageof37kV is22,000A.However, since thecircuitbreaker isusedat34.5 kV, its symmetrical
currentinterruptingcapabilityis
A232,24kV34.5
kV38A)000,22( =
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Thehighestsymmetricalcurrentinterruptingcapabilityis
A36,0001.6522,000kA22,000( =)whichispossiblewhentheoperatingvoltageis
kVk
kV23
65.1
3838=
Table 2.3 Outdoor circuit breaker ratngs based on symmetrical current
Nominal
rmsVoltage
Class(kV)
Rated
Maximumrms
Voltage
(kV)
Rated
VoltageRange
Factor,K
Rated
Continuousrms
Current
(kA)
Rated
Short
Circuitrms
Current
(at Rated
Maximum
kV)(kA)
Rated
InterruptingTime
(cycles)
Rated
Maximumrms
Voltage
Divided
byK(kV)
Maximum
rmsSymmetrical
Interrupting
Capability*
(kA)
14.4 15.5 2.67 0.6 8.9 5 5.8 24
14.4 25.5 1.29 1.2 18 5 12 23
23 25.8 2.15 1.2 11 5 12 24
34.5 38 1.65 1.2 22 5 23 36
46 48.3 1.21 1.2 17 5 40 21
69 72.5 1.21 1.2 19 5 60 23
115 121 1 1.2 20 3 121 20
115 121 1 1.6 40 3 121 40
115 121 1 3 63 3 121 63
138 145 1 1.2 20 3 145 20
138 145 1 1.6 40 3 145 40
138 145 1 2 40 3 145 40
138 145 1 3 80 3 145 80
163 169 1 1.2 16 3 169 16
163 169 1 1.6 31.5 3 169 31.5
163 169 1 2 50 3 169 50
230 242 1 1.6 31.5 3 242 31.5
230 242 1 2 31.5 3 242 31.5
230 242 1 3 63 3 242 63
345 362 1 2 40 3 362 40
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345 262 1 3 40 3 362 40
500 550 1 2 40 2 550 40
500 550 1 3 40 2 550 40
700
765
1
2 40 2 765
40
700 765 1 3 40 2 765 40
*ItisequaltoKtimestheratedshortcircuitrmscurrent
Note that at lower operating voltages the highest symmetrical currentinterruptng capability of 36,000 A
cannotbeexceeded.Theassociatedrmsandpeakmomentarycurrentratings,respectively,are
"momentary kI.I 61= =1.6(36,000A)=57,600Arms
and
"momentary kI.I 72= =2.7(36.000A)=97.200Apeak
A simplified procedure for determining the symmetrical fault current is known as the E/Xmethod and is
describedinsecton5.3.1ofANSIC37.010.Thismethodgivesresultsapproximatngthoseobtainedbymorerigorous
methods. In using this method, it is necessary first to make an E/X calculation. The method then corrects this
calculationtotakeintoaccountboththedcandacdecayofcurrent,dependingoncircuitparametersX/R.
Example
ConsiderthesystemshowninFigure2.17andassumethatthegeneratorisunloadedandrunningattherated
voltagewith the circuit breaker open at bus 3. Assume that the reactance values of the generator are given as
14.021" === XXXd puandX0=0.08pubasedonitsratngs.ThetransformerimpedancesareZ1=Z2=Z0 =
0.05pubasedon itsratngs.Thetransmission lineTL23hasZ1=Z2=0.04puandZ0=0.10pu.Assumethatthefault
point is located on bus 1, and determine the subtransient fault current for a threephase fault in per units and
amperes.Select25MVAasthemegavoltampere baseand8.5and138kVasthelowvoltageandhighvoltagevoltage
bases.
Figure2.17 Transmissionsystemforexampleabove
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Solution
pu143.7j0.14
00.1"
" jX
EI
d
gf ===
o
Thecurrentbaseforthelowvoltagesideis
A1.1698kV)5.8(3
kVA000.25
3 )()( ===
LVB
BLVB
V
SI
Therefore,
A52.129.12143.71.1698" ==fI
Example
Usetheresultsofexampleaboveanddeterminethefollowing:(a)Maximumvaluepossibleofdccurrentcomponent.(b)Totalmaximuminstantaneouscurrent.(c)Momentarycurrent.(d)Interruptingcapacityofthreecyclecircuitbreakeriflocatedatbus1.
Solution
(a) pu1.10)143.7(22 "max, === fdc II (b) 2.20222 "max,"max === fdc III pu(c) 43.11)143.7(6.16.1 " === fmomentary II pu(d) 6"fint 10V3 = ferrupting IS
MVA3.21410)2.1)(52.129,12)(8500(3 6 ==
(e) 6"f
10)6.1(V3
= fmomentaryIS
MVA7.28510)6.1)(52.129,12)(8500(3 6 ==
2.10 SELECTION OF CIRCUITBREAKER
Thesubtransientcurrentistheinitialsymmetricalcurrentanddoesnotincludethedccomponent.Aswehaveseen,
inclusionofthedccomponentresults inarmsvalueofcurrentimmediatelyafterthefault,which ishigherthanthe
subtransientcurrent.Foroilcircuitbreakersabove5kVthesubtransientcurrentmultpliedby1.6isconsideredto
bethermsvalueofthecurrentwhosedisruptiveforcesthebreakermustwithstandduringthefirsthalfcycleafterthe
faultoccurs.Thiscurrentiscalledthemomentarycurrent,andformanyyearscircuitbreakerswereratedintermsof
theirmomentarycurrentaswellasothercriteria.
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The interrupting ratingofa circuitbreakerwas specified in kilovoltamperesormegavoltamperes. The interrupting
kilovoltamperesequal 3 tmesthekilovoltsofthebustowhichthebreakerisconnected tmesthecurrentwhichthe
breakermustbecapableof interruptingwhen itscontactspart.Thiscurrent isofcourse lowerthanthemomentary
current
and
depends
on
the
speed
of
the
breaker,
such
as
8,
5,
3,
or
1
cycles,
which
is
a
measure
of
thet
me
from
theoccurrenceofthefaulttotheextinctionofthearc.
The currentwhichabreakermust interrupt isusuallyasymmetrical since it still contains someof thedecayingdc
component.A schedule of preferred ratings for ac highvoltage circuit breakers specifies the interrupting current
ratingsofbreakersintermsofthecomponentoftheasymmetricalcurrentwhichissymmetricalaboutthezeroaxis.
Thiscurrentisproperlycalledtherequiredsymmetricalinterruptingcapabilityorsimplytheratedsymmetricalshort
circuitcurrent.Oftentheadjectivesymmetricalisomitted.Selectionofcircuitbreakersmayalsobemadeonthebasis
oftotalcurrent(dccomponentincluded).Weshalllimitourdiscussiontoabrieftreatmentofthesymmetricalbasisof
breakerselection.
Breakersare identifiedbynominalvoltageclass,suchas69kV.Alongother factorsspecifiedare ratedcontnuous
current,ratedmaximumvoltage,voltagerangefactorK,andratedshortcircuitcurrentatratedmaximumkilovolts.K
determinestherangeofvoltageoverwhichratedshortcircuitcurrent tmesoperatngvoltageisconstant.Fora69kV
breakerhavingamaximumratedvoltageof72.5kV,avoltagerangefactorKof1.21,andacontnuouscurrentratng
of1200A,theratedshortcircuitcurrentatthemaximumvoltage(symmetricalcurrentwhichcanbe interruptedat
72.5kV) is19,000A.Thismeansthattheproduct72.519,000 istheconstantvalueofratedshortcircuitcurrent
tmesoperatngvoltageintherange72.5to60kVsince72.5/1.21equals60.Theratedshortcircuitcurrentcannotbe
exceeded.At69kVtheratedshortcircuitcurrentis
Breakersofthe115kVclassandhigherhaveakof1.0.
A simplified procedure for calculating the symmetrical shortcircuit current, called the E/Xmethod, disregards all
resistance,allstaticloads,andallprefaultcurrent.SubtransientreactanceisusedforgeneratorsintheE/Xmethod,
andforsynchronousmotorstherecommendedreactanceisthe ofthemotor tmes1.5,whichistheapproximate
valueofthetransientreactanceofthemotor. Inductonmotorbelow50hpareneglected,andvariousmultiplying
factorsareappliedtothe oflargerinductionmotordependingontheirsize.Ifnomotorsarepresent,symmetrical
shortcircuitcurrentequalssubtransientcurrent.
TheimpedancebywhichthevoltageVfatthefaultisdividedtofindshortcircuitcurrentmustbeexaminedwhenthe
E/Xmethod isused. In specifyingabreaker forbusk this impedance isZkkof thebus impedancematrixwith the
propermachinereactances. IftheratioofX/Rofthis impedance is15or less,abreakerofthecorrectvoltageand
kilovoltamperesmaybeusedifitsinterruptingcurrentratingisequaltoorexceedsthecalculatedcurrent.IftheX/R
rato isunknown, thecalculatedcurrent shouldbenomore than80%of theallowedvalue for thebreakerat the
existingbusvoltage.TheANSIapplicationguidespecifiesacorrectedmethodtoaccountforacanddctimeconstants
for thedecayof the current amplitude if theX/R rato exceeds 15. The correctedmethod also considersbreaker
speed.
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Thisdiscussionoftheselectionofcircuitbreakersispresentednotasastudyofbreakerapplicationsbuttoindicate
theimportanceofunderstandingfaultcalculations.Thefollowingexampleshouldclarifytheprinciple.
Example:A25,000
kVA
13.8
kVgenerator
with
=15%
isconnected
through
atransformer
toabus
which
supplies
fouridentcalmotors,asshowninFig.2.18.Thesubtransientreactance ofeachmotoris20%onabaseof
5000 kVA, 6.9 kV. The threephase rating of the transformer is 25,000 kVA, 13.8/6.9 kV, with a leakage
reactanceof10%.Thebusvoltageatthemotorsis6.9kVwhenathreephasefaultoccursatthepointP.For
thefaultspecified,determine(a)thesubtransientcurrentinthefault,(b)thesubtransientcurrentinbreaker
A,and(c)thesymmetricalshortcircuitinterruptingcurrent(asdefinedforcircuitbreakerapplications)inthe
faultandinbreakerA.
Figure2.18 OnelinediagramforExample
Solution:
(a) Forabaseof25,000kVA,13.8kVinthegeneratorcircuit,thebaseforthemotorsis25,000kVA,6.9kV.Thesubtransientreactanceofeachmotoris
unitper0.15000
000,252.0" ==
dX
Figure2.19isthediagramwithsubtransientvaluesofreactancemarked.ForafaultatP,
unitper0.1=f
V
unitper125.0jZth =
unitper0.8125.0
0.1" jj
If
==
Thebasecurrentinthe6.9kVcircuitis
A720,1620908
A20909.63
000,25
" ==
=
fI
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Figure2.19 ReactancediagramforExample
(b) ThroughbreakerAcomesthecontributionfromthegeneratorandthreeofthefourmotors.Thegeneratorcontributesacurrentof
unitper0.450.0
25.00.8 jj =
Eachmotorcontributes25%oftheremainingfaultcurrent,orj1.0perunitA.ThroughbreakerA,
I"= j4.0+3(j1.0)= j7.0perunit=72090=14,630A
(c) TocomputethecurrentthroughbreakerAtobeinterrupted,replacethesubtransientreactanceofj1.0bythetransientreactanceofj1.5inthemotorcircuitsofFig.2.19.Then
unitper15.025.0375.0
25.0375.0jjZ
th=
+=
Thegeneratorcontributesacurrentof
unitper0.4625.0
375.0
15.0
0.1j
j=
Eachmotorcontributesacurrentof
unitper67.0625.025.0
15.00.1
41 j
j=
Thesymmetricalshortcircuitcurrenttobeinterruptedis:
(4.0+30.67)2090=12,560A
Theusualprocedureistorateallthebreakersconnectedtoabusonthebasisofthecurrentintoafaultonthebus.In
thatcasetheshortcircuitcurrentinterruptngratngofthebreakersconnectedtothe6.9kVbusmustbeatleast
4+40.67=6.67 perunit
or
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6.672090=13,940A
A14.4kVcircuitbreakerhasaratedmaximumvoltageof15.5kVandakof2.67.At15.5kV itsratedshortcircuit
interrupt
ng
current
is
8900
A.
This
breaker
is
rated
for
a
symmetrical
short
circuit
interrupt
ng
current
of
2.67
8900
=23,760A,atavoltageof15.5/2.67=5.8kV.Thiscurrentisthemaximumthatcanbeinterruptedeventhoughthe
breakermaybeinacircuitoflowervoltage.Theshortcircuitinterruptngcurrentratngat6.9kVis
A000,2089009.6
5.15=
Therequiredcapabilityof13,940Aiswellbelow80%of20,000A,andthebreakerissuitablewithrespecttoshort
circuitcurrent.
Recommended