Fundamentals of Switching Phenomenon

7/29/2019 Fundamentals of Switching Phenomenon

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Module(02)FundamentalsofSwitching Page2of39

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)


16/39


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LV,MV&HVSwitchgear


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


39/39

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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.

Documents

Fundamentals of Switching Phenomenon