11 ELC4340 Spring13 Short Circuits

8/13/2019 11 ELC4340 Spring13 Short Circuits

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_11_ELC4340_Spring13_Short_Circuits.doc

Short Circuits

1. Introduction

Voltage sags are due mostly to faults on either transmission systems or distribution feeders.

ransmission faults affect customers o!er a "ide area# possibly do$ens of miles# but distribution faults

usually affect only the customers on the faulted feeder or on ad%acent feeders ser!ed by the samesubstation transformer.

Single&phase faults 'i.e.# line&to&ground( are the most common type of faults# follo"ed by line&to&line#and three&phase. Since single&phase and line&to&line faults are unbalanced# their resulting sag !oltages

are computed using symmetrical components. ransformer connections affect the propagation of

positi!e# negati!e# and $ero se)uence components differently. hus# the characteristics of a !oltage sag

changes as it propagates through a net"or*.

ypically# a transmission !oltage sag passes through t"o le!els of transformers before reaching a 4+0V

load 'e.g.# 13+*V,1-.4*V at the entrance to the facility# and 1-.4*V,4+0V at the load(. 1-0V loadsli*ely ha!e a third transformer 'e.g.# 4+0V,1-0V(. /t is not intuiti!ely ob!ious ho" the sag changes# but

the changes can be computed using symmetrical components and are illustrated in this report.

2. Symmetrical Components

n unbalanced set ofNrelated phasors can be resol!ed intoNsystems of phasors called the symmetricalcomponents of the original phasors. or a three&phase system 'i.e.N2 3(# the three sets are,

1. ositi!e Se)uence & three phasors# e)ual in magnitude# 1-0o

apart# "ith the same se)uence 'a&b&c( asthe original phasors.

-. egati!e Se)uence & three phasors# e)ual in magnitude# 1-0oapart# "ith the opposite se)uence 'a&c&

b( of the original phasors.3. 5ero Se)uence & three identical phasors 'i.e. e)ual in magnitude# "ith no relati!e phase

displacement(.

he original set of phasors is "ritten in terms of the symmetrical components as follo"s,

-106666

aaaa VVVV ++= #

-106666

bbbb VVVV ++= #

-106666

cccc VVVV ++= #

"here 0 indicates $ero se)uence# 1 indicates positi!e se)uence# and - indicates negati!e se)uence.

he relationships among the se)uence components for a&b&c are

ositi!e Se)uence egati!e Se)uence 5ero Se)uence6 6

V Vb a1 1 1= 1-0o 6 6V Vb a- - 1= +1-0

o000

666cba VVV ==

6 6V Vc a1 1 1= +1-0

o 6 6V Vc a- - 1= 1-0o

he symmetrical components of all a&b&c !oltages are usually "ritten in terms of the symmetricalcomponents of phase a by defining

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a= +1 1-0o # so that a- 1 1= + = -40 1-0o o # and a3 1 1= + = 370 0o o .

Substituting into the pre!ious e)uations for6

#6

#6

V V Va b c yields

6 6 6 6V V V V a a a a= + +0 1 - #

-1-

0

6666aaab VaVaVV ++= #

6 6 6 6V V aV a V c a a a= + +0 1

-- .

/n matri8 form# the abo!e e)uations become

=

-

1

0

-

-

6

6

6

1

1

111

6

6

6

a

a

a

c

b

a

V

V

V

aa

aa

V

V

V

#

=

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

6

6

6

1

1

111

3

1

6

6

6

-

-

-

1

0

'1(

or in matri8 form

6 6V T Vabc = 01- # and

6 6V T Vabc01-

1= # '-(

"here transformation matri8 Tis

T a a

a a

=

1 1 1

1

1

-

-

# and T a a

a a

=

1 -

-

1

3

1 1 1

1

1

. '3(

/f6

Vabc represents a balanced set 'i.e.6 6 6

V V a V b a a

= =1 -1-0o #6 6 6

V V aV c a a

= + =1 1-0o (# then

substituting into6 6

V T Vabc01-1= yields

6

6

6

6

6

6

6V

V

V

a a

a a

V

a V

aV

V

a

a

a

a

a

a

a

0

1

-

-

-

-1

3

1 1 1

1

1

0

0

=

=

.

9ence# balanced !oltages or currents ha!e only positi!e se)uence components# and the positi!ese)uence components e)ual the corresponding phase a !oltages or currents.

9o"e!er# balanced !oltages are rare during !oltage sags. :ost often# one phase is affectedsignificantly# and the other t"o less significantly. hus# all three se)uence !oltages -10

6#

6#

6aaa VVV e8ist

during most sags# and these se)uence !oltages are shifted differently by transformers "hen propagating

through a system. ;hen recombined to yield phase !oltages cba VVV 6#

6#

6# it is clear that the form of

phase !oltages must also change as transformers are encountered.

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3. Transformer Phase Shift

he con!entional positi!e&se)uence and negati!e&se)uence model for a three&phase transformer is

sho"n belo". dmittance y is a series e)ui!alent for resistance and lea*age reactance# tap t is the tap'in per unit(# and angle < is the phase shift.

y

Ii ---> Ik --->Bus i

Bus k

t / :1Bus k'

igure 1. ositi!e& and egati!e&Se)uence :odel of hree&hase ransformer

or grounded&"ye,grounded&"ye and delta,delta transformers# < is 0=# and thus positi!e& and negati!e&se)uence !oltages and currents pass through unaltered 'in per unit(. 9o"e!er# for "ye&delta and delta&

"ye transformers# < is se)uence&dependent and is defined as follo"s,

or positi!e se)uence# < is >30= if bus i is the high&!oltage side# or ?30= if bus i is the lo"&

!oltage side

and oppositely

or negati!e se)uence# < is ?30= if bus i is the high&!oltage side# or >30= if bus i is the lo"&

!oltage side

/n other "ords# positi!e se)uence !oltages and currents on the high&!oltage side leadthose on the lo"&

!oltage side by 30=. egati!e se)uence !oltages and currents on the high&!oltage side lagthose on thelo"&!oltage side by 30=.

or $ero&se)uence !oltages and currents# transformers do not introduce a phase shift# but they may bloc*

$ero&se)uence propagation as sho"n in igure -.

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Grounded Wye - Grounded Wye

Grounded Wye - Delta

R + jX

Grounded Wye - Ungrounded WyeR + jX

R + jXUngrounded Wye - Delta

R + jXDelta - Delta

R + jX

igure -. 5ero&Se)uence :odels of a hree&hase ransformer

/t can be seen in the abo!e figure that only the grounded&"ye,grounded&"ye transformer connection

permits the flo" of $ero&se)uence from one side of a transformer to the other.

hus# due to phase shift and the possible bloc*ing of $ero&se)uence# transformers ob!iously play an

important role in unbalanced !oltage sag propagation.

4. System Impedance Matrices

ault currents and !oltage sags computations re)uire elements of the impedance matri8 5 for the study

system. ;hile each of the three se)uences has its o"n impedance matri8# positi!e& and negati!e&se)uence matrices are usually identical. /mpedance elements are usually found by

building the system admittance matri8 @# and then in!erting it to obtain the entire 5#

or by

using Aaussian elimination and bac*"ard substitution to obtain selected columns of 5.

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he admittance matri8 @ is easily built according to the follo"ing rules,

he diagonal terms of @ contain the sum of all branch admittances connected directly to the

corresponding bus.

he off&diagonal elements of Ycontain the negati!e sum of all branch admittances connected

directly bet"een the corresponding busses.

he procedure is illustrated by the three&bus e8ample in igure 3.

1 2 3

ZE

ZA

ZB

ZC

ZDI3

igure 3. hree&Bus dmittance :atri8 E8ample

pplying CL at the three independent nodes yields the follo"ing e)uations for the bus !oltages '"ith

respect to ground(,

t bus 1# 0-11 =

+

AE Z

VV

Z

V#

t bus -# 03-1-- =

+

+

CAB Z

VV

Z

VV

Z

V#

t bus 3# 3-33 I

Z

VV

Z

V

CD

=

+ .

Collecting terms and "riting the e)uations in matri8 form yields

=

+

++

+

33

-

1

0

0

1110

11111

0111

IV

V

V

ZZZ

ZZZZZ

ZZZ

DCC

CCBAA

AAE

#

or in matri8 form#

IYV = #

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Besides being the *ey for fault calculations# the impedance matri8# 1=YZ # is also physically

significant. Consider igure 4.

Power System

All Other Busses

Open Circuited

Applied Current atInduced Voltage

+

-

V

Bus kat Bus j

Ik

Vj

igure 4. hysical Significance of the /mpedance :atri8

'the impedance matri8 includes orton e)ui!alent impedances for the current sources(

/mpedance matri8 element kjz # is defined as

kmNmmIk

jkj

I

Vz

==

=

###-#1#0

#

# '4(

"here kI is a current source attached to bus k# jV is the resulting !oltage at busj# and all busses

e8cept kare open&circuited. he depth of a !oltage sag at bus * is determined directly by multiplying

the phase se)uence components of the fault current at bus * by the matri8 elements kjz # for the

corresponding phase se)uences.

5. Short Circuit Calculations

Short circuit calculations re)uire positi!e# negati!e# and $ero se)uence impedance information#depending on "hether or the fault is balanced or not. or e8ample# the commonly&studied# but relati!elyrare# three&phase fault is balanced. herefore# only positi!e se)uence impedances are re)uired for its

study.

Consider the balanced three&phase fault represented by the one&line diagram in igure D# "here THV

and THZ are the he!enin e)ui!alent circuit parameters for bus *. ault impedance FZ is e8ternal to

the net"or* for "hich THZ is computed.

+

-

Vth

Zth

ZFIF

Bus k

igure D. hree&hase ault at Bus k

'he!enin e)ui!alent THZ at bus * is obtained withoutfault impedance FZ (

he fault current and !oltage are clearly

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FTH

THFk ZZ

VI

+= # and then

+==

FTH

FTH

FkTHTH

Fk ZZ

ZVIZVV .

/n a large po"er system# the he!enin e)ui!alent impedance for a bus is the corresponding diagonal

impedance matri8 element# and the he!enin e)ui!alent !oltage is usually assumed to be 1.0 0 pu.

he type of machine models used "hen building impedance matrices affects the he!enin e)ui!alentimpedances and fault calculations. Fotating machines actually ha!e time&!arying impedances "hensub%ected to disturbances. 9o"e!er# for simplification purposes# their impedances are usually di!ided

into three $ones & subtransient 'first fe" cycles(# transient '7 cycles & 70 cycles(# and steady&state 'longer

than 70 cycles(. ;hen performing fault studies# the time period of interest is usually a fe" cycles# sothat machines are represented by their subtransient impedances "hen forming the impedance matrices.

Ge!eloping the e)uations for fault studies re)uires adept use of both a&b&c and 0&1&- forms of the circuit

e)uations. he use of se)uence components implies that the system impedances 'but not the system!oltages and currents( are symmetric. /n general# there are si8 e)uations and si8 un*no"ns to be sol!ed#

regardless of the type of fault studied. /n e!ery case# it is necessary to first obtain a key fault euation.

9a!ing the *ey fault e)uation# then the resulting bus !oltages and branch currents throughout thenet"or* during the fault can be determined.

/t is common in fault studies to assume that the po"er system is initially unloaded and that all !oltagesare 1.0 per unit. ;hen there are multiple sources# this assumption e!uie"that there are no shunt

elements connected# such as loads# capacitors# etc.# e#ce$t for rotating machines '"hose he!enin

e)ui!alent !oltages are 1.0 pu.(. he he!enin !oltages 'and orton in%ection currents( of rotatingmachines are assumed to be constant during faults. erminal !oltages of machines are not constant

during faults because of he!enin machine impedances e8ist bet"een the internal machine !oltage and

the machine terminals.

Since "ye&delta transformers shift positi!e# negati!e# and $ero se)uence components differently# it isimportant to model transformers according to the rules gi!en earlier. his means that the pre&fault

!oltages all ha!e magnitude 1.0 pu.# but that the pre&fault !oltage angles can be ooo 30or#30#0 + # or

multiples of o30 # depending upon the net transformer phase shift bet"een them and the chosen

reference bus.

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Balanced Three-Phase Fault

Consider the three&phase fault at bus k# as sho"n in igure 7. Bus * can be at an e8isting bus in the 5matri8# or at some fault point along a transmission line 'e.g.# a lightning stri*e(. or the latter# the 5

matri8 must be modified to include the ne" bus *.

igure 7, Balanced hree&hase ault at Bus k

FZ is not included in the impedance matri8 or he!enin e)uation. he 01- he!enin e)uation#

a""umi%g that all othe cue%t i%jectio%" i% the "&"tem ae u%cha%ged# is

Fkkk

ek

Fk

IZVV01-01-#01-

r01-01- = . '3_1(

ote & the minus sign is needed because the fault current has been dra"n as positi!e out"ard.

Fk

V01- consists of the !oltages at bus kduring the fault#

ek

Vr01- consists of the pre&fault !oltages#

Fk

I01-

gi!es the fault currents# and 01-#01- kkZ contains the indi!idual impedance elements e8tracted from theimpedance matri8.

E)uations related to the faulted element# "hich is e8ternal to the 5 matri8# are

F

FkaF

ka Z

VI = #

F

FkbF

kb Z

VI = #

F

FkcF

kc Z

VI = . '3_-(

he relationship bet"een 01- fault currents and abc fault currents is

=

==

=

0

0

1

1111

3

1

-

-1

-

1

0

01-Fka

Fkc

Fkb

F

kaFkabc

Fk

Fk

F

kFk

I

I

II

aa

aaIT

I

II

I . '3_3(

Substituting '3_3( into 01- he!enin e)uation '3_1( yields

age + of -7&

FZ

FkaI

FkbI

FkcI

a

b

c

FZ FZ


9/26


=

=

=

=

0

0

00

00

00

0

0

-#-

1#1

0#0

r-

rr1

r0

-

1

0Fka

kk

kk

kk

ek

eka

ek

ek

Fk

Fk

Fk

I

z

z

z

V

VV

V

V

V

V

.

dding the three ro"s yields

( )1#1r

-10 kkFka

eka

Fka

Fk

Fk

Fk

zIVVVVV ==++ .

SubstitutingF

FkaF

kaZ

VI = from '3_-( into the abo!e e)uation yields

( )1#1r

kkFka

ekaF

Fka

zIVZI = .

Sol!ing for FkaI yields the key fault euation for three!phase "alanced faults

Fkk

ekaF

ka Zz

VI

+=

1#1

r

. '3_4(

9a!ing the *ey fault e)uation# then the resulting bus !oltages and branch currents throughout thenet"or* during the fault can be determined. Fecall from '3_3( that the 01- fault current components

are

00 =

FkI #

Fka

Fk II =1 # 0- =

Fk

I

ll $ero se)uence and negati!e se)uence currents in the net"or* are $ero# so all $ero se)uence andnegati!e se)uence !oltages in the net"or* remain $ero. ositi!e se)uence net"or* !oltages can be

found from the he!enin e)ui!alent circuit e)uation

Fkkj

ej

Fj IZVV 11#1

r11 = .

hen# positi!e se)uence currents in branches can be found using HhmIs La" "ith positi!e se)uencebranch impedances.

age J of -7&


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Single-Phase to Ground Fault

Consider the single&phase fault at bus k# as sho"n in igure .

igure , Single&hase ault at Bus k# hase a



Fkkk

ek

Fk

IZVV01-01-#01-

r01-01- = . '1_1(


F

FkaF

ka Z

VI = # 0=FkbI # 0=

Fkc

I . '1_-(


=

=

=

==

=

3E

3E

3E

0

0

1

1

111

3

1

-

-1

-

1

0

01-Fka

Fka

Fka

Fkc

Fkb

Fka

Fkabc

Fk

Fk

Fk

Fk

I

I

I

I

I

I

aa

aaIT

I

I

I

I . '1_3(

Substituting '1_3( into the 01- he!enin e)uation '1_1( yields

=

==

=

3E

3E3E

00

0000

0

0

-#-

1#1

0#0

r-

rr1

r

0

-

1

0

Fka

Fka

F

ka

kk

kk

kk

ek

eka

ek

e

k

Fk

Fk

F

k

I

II

z

zz

V

VVV

V

VV

.

dding the three ro"s yields

( )-#-1#10#0r

-10 3 kkkkkk

Fkae

kaFka

Fk

Fk

Fk

zzzI

VVVVV ++==++ .

age 10 of -7&

F

kaI

a

b

c

FZ

FkbI

FkcI


11/26


Substituting FFka

Fka

ZIV = from '1_-( into the abo!e e)uation yields

( )-#-1#10#0r

3 kkkkkk

Fkae

kaFFka

zzzI

VZI ++= .

:ultiplying both sides by 3 and sol!ing for FkaI yields the key fault euation for sin#le!phase to

#round faults

Fkkkkkk

ekaF

ka Zzzz

VI

3

3

-#-1#10#0

r

+++= . '1_4(

9a!ing the *ey fault e)uation# then the resulting bus !oltages and branch currents throughout the

net"or* during the fault can be determined. Fecall from '1_3( that the 01- fault current componentsare

3-10

FkaF

k

F

k

F

k

I

III ===.

ll 01- net"or* !oltages can then be found from the 01- he!enin e)uation '1_1(

Fkkj

ej

Fj IZVV 01-01-#01-

r

01-01- = .

hen# 01- fault currents in branches can be found using HhmIs La" and the corresponding positi!e#

negati!e# and $ero se)uence branch impedances. fter"ard# 01- !oltages and currents can be con!ertedto abc. See later sections on ho" to include transformer phase shifts "hen con!erting 01- to abc.

ote that if -#10#0 kkkk zz < # a single&phase fault "ill ha!e a higher !alue than does a three&phase fault.

age 11 of -7&


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Line-to-Line Fault

Consider the line&to&line fault at bus k# as sho"n in igure +.

igure +. Line&to&Line ault Bet"een hases b and c at Bus k



Fkkk

ek

Fk

IZVV01-01-#01-

r01-01- = . 'LL_1(


0=Fka

I # 'LL_-(

F

Fkc

FkbF

kb Z

VVI

= # 'LL_3(

Fkb

Fkc

II = . 'LL_4(


=

=

=

==

=

Fkb

Fkb

Fkb

Fkb

Fkb

Fkb

Fka

Fkabc

Fk

Fk

Fk

Fk

Iaa

Iaa

II

I

I

I

aa

aaIT

I

I

I

I

('

('

0

3

10

1

1

111

3

1

-

-

-

-1

-

1

0

01- . 'LL_D(

Since a 2 11-0K and a-11-0K# then 'a a-( 2 3j . hus# 'LL_D( becomes

=Fkb

Fkb

Fk

I

Ij

I

0

301- . 'LL_7(

Substituting 'LL_7( into 01- he!enin e)uation 'LL_1( yields

age 1- of -7&

FkaI

FkbI

FkcI

a

b

c

FZ


13/26


=

=

=

=

Fkb

Fkb

kk

kk

kk

ek

eka

ek

ek

Fk

Fk

Fk

I

I

z

z

zj

V

VV

V

V

V

V 0

00

00

00

30

0

-#-

1#1

0#0

r-

rr1

r0

-

1

0

.

Simplifying yields

=

Fkbkk

Fkbkk

eka

Fk

Fk

Fk

Izj

Izj

V

V

V

V

-#-

1#1r

-

1

0

3

3

0

. 'LL_(

rom the top ro" it is clear that there "ill be no $ero se)uence !oltages# and thus no $ero se)uencecurrents. ote & this is physically true because there is no ground current.

Feferring bac* to 'LL_3(#

F

Fck

Fk

Fk

Fk

Fk

Fk

F

Fkc

FkbF

kb Z

VaaVVaVVaV

Z

VVI -

-10-1

-0

++++=

= #

F

Fk

Fk

Z

aaVaaV --

-1

+=

F

Fk

Fk

Z

VjVj-1

33 += . 'LL_+(

Substituting the !oltages from 'LL_( into 'LL_+( yields

F

F

kbkk

F

kbkk

e

kaFkb Z

Iz

j

jIz

j

VjI

+

=-#-1#1

r

3333

F

Fkbkk

Fkbkk

eka

Z

IzIzVj -#-1#1r3

= #

and# after combining terms# the key fault euation for line!to!line faults 'LL_J( is obtained.

Fkkkk

ekaF

kb Zzz

VjI

++

=

-#-1#1

r3

'LL_J(

9a!ing the *ey fault e)uation# then the resulting bus !oltages and branch currents throughout thenet"or* during the fault can be determined. Fecall from 'LL_7( that there is no $ero&se)uence fault

current. herefore# there are no $ero&se)uence !oltages in the net"or*. By using the se)uence faultcurrents from 'LL_J( and 'LL_7(# the positi!e and negati!e se)uence components of all net"or*

!oltages can then be found from the 01- he!enin e)uation 'LL_1(.

hen# positi!e and negati!e se)uence fault currents in branches can be found using HhmIs La" and thecorresponding positi!e and negati!e se)uence branch impedances. fter"ard# positi!e and negati!e

se)uence !oltages and currents can be con!erted to abc.

age 13 of -7&


14/26


ote in 'LL_J( that for $ero fault impedance cases# line&to&line fault current magnitudes are slightly

smaller 'i.e.#-

3( than those of three&phase faults.

age 14 of -7&


15/26


Line-to-Line-to-Ground Fault

Consider the line&to&line fault at bus k# as sho"n in igure J.

igure J. Line&to&Line&to&Around ault

he 01- he!enin e)ui!alent circuit e)uation# a""umi%g that all othe cue%t i%jectio%" i% the "&"temae u%cha%ged# is

Fkkk

ek

Fk

IZVV01-01-#01-

r01-01- = . 'LLA_1(

Specifics for the fault# "hich are not included in the system 5 matri8 or he!enin e)uation# follo".

Since phases b and c are tied together#F

kcF

kb VV = . 'LLA_-(

rom HhmsIs la"#

F

FkbF

kc

F

kb Z

VII =+ . 'LLA_3(

Because phase a is not faulted# then

0=Fka

I . 'LLA_4(

Because FkaI is $ero# then

0-10

=++ Fk

Fk

Fk

III . 'LLA_D(

t this point# the ob%ecti!e is to find the key fault euation. inding it re)uires manipulating the abo!e

e)uations until at least one of the 01- components of FkaI can be computed using onlye

kaV

r and

elements of the 5 matri8. his means e8pressing all abc terms# e8cept for eka

Vr # in terms of se)uence

components.

Begin by e8amining FkcFkb II + from 'LLA_3( in terms of se)uence components#

aaIaaIIIaaIIaIIaIII F

kFk

Fk

Fk

Fk

Fk

Fk

Fk

Fk

Fkc

Fkb

++++=+++++=+ --

-10-

-10-1

-0

-

aaIaaII Fk

Fk

Fk

++++= --

-10

-

( ) ( ) ( Fk

Fk

Fk

Fk

Fk

Fk

Fk

Fk

Fk

IIIIIIIII000-10-10

3--11- =+=+=++=

yields

age 1D of -7&

F

kaI

FkbI

FkcI

a

b

c

FZ


16/26


Fk

Fkc

Fkb

III0

3=+ . 'LLA_7(

Substituting 'LLA_7( into 'LLA_3( yields

F

Fk

Fk

Fk

F

FkbF

k Z

aVVaV

Z

VI -1

-0

03

++== . 'LLA_(

E8amining FkV 1 andFk

V- in terms of se)uence components# and recogni$ing 'LLA_-(#

( ) ( )#('3

1

3

1 --1

Fkb

Fka

Fkc

Fkb

Fka

Fk

VaaVVaaVVV ++=++= and

( ) ( )Fkb

Fka

Fkc

Fkb

Fka

Fk

VaaVaVVaVV ('3

1#

3

1 ---

++=++=

yieldsF

kF

k VV

-1 = . 'LLA_+(

Substituting 'LLA_+( into 'LLA_( yields

F

Fk

Fk

F

Fk

FkF

k Z

VV

Z

VaaV

I

101-

0

0

('

3

=

++=

'LLA_J(

rom he!enin e)uation 'LLA_1(#Fkkk

Fk

IzV00#00

= # 'LLA_10(Fkkk

$eka

Fk IzVV 11#11 = # 'LLA_11(

Fkkk

Fk

IzV--#--

= . 'LLA_1-(

ccording to 'LLA_+(# e)uations 'LLA_11( and 'LLA_1-( are e)ual# soFkkk

Fkkk

$eka

IzIzV--#-11#1

= #

"hich yields

-#-

11#1

-kk

$eka

FkkkF

k z

VIzI

= . 'LLA_13(

Substituting 'LLA_10( and 'LLA_11( into 'LLA_J( yields

F

Fkkk

$eka

FkkkF

k Z

IzVIzI 1

1#100#0

03

+=

"hich in turn yields

( )Fkk

F

kkk

$e

kaFk Zz

IzVI30#0

11#10 +

+= 'LLA_14(

Substituting 'LLA_14( and 'LLA_13( into 'LLA_D( yields

( ) 0

3 -#-

11#1

10#0

11#1 =

+++

+

kk

$eka

FkkkF

kFkk

Fkkk

$eka

z

VIzI

Zz

IzV#

age 17 of -7&


17/26


( ) ( ) 0

33 -#--#-

11#1

10#0

11#1

0#0

=

++++

++

kk

$eka

kk

FkkkF

kFkk

Fkkk

Fkk

$eka

z

V

z

IzI

Zz

Iz

Zz

V

( ) ( )

+

+=

++

+ -#-0#0-#-

1#1

0#0

1#1

1

1

3

11

3 kkFkk

$eka

kk

kk

Fkk

kkFk zZz

Vz

z

Zz

zI

( ) ( )

+

+=

++

+ -#-0#01#1-#-1#10#01

1

3

111

3

1

kkFkkkk

$eka

kkkkFkk

Fk zZzz

V

zzZzI 'LLA_1D(

Gefine $z as the parallel combination of Fkk Zz 30#0 + and -#- kkz

( ) -#-0#01

3

11

kkFkk$ zZzz+

+= 'LLA_17(

so that 'LLA_1D( becomes

=

+$kk

$e

kakk$

Fk zz

V

zzI 111

1#11#11 #

=

+

$kk

$eka

kk$

$kkFk

zz

V

zz

zzI

1

1#11#1

1#11 #

$kk

$ekaF

k zz

VI

+=

1#11 'LLA_1(

Substituting 'LLA_17( into 'LLA_1( yields the follo"ing key fault euation for line!to!line!to!

#round faults

( )( )Fkkkk

Fkkkkkk

$ekaF

k

Zzz

Zzzz

VI

3

3

0#0-#-

0#0-#-1#1

1

++

++

=. 'LLA_1+(

Hnce the *ey fault e)uation is e!aluated# then FkI 0 andFk

I- can be found using '-J( and '-+(# and

chec*ed "ith '-0(. fter"ard# the 01- bus !oltages and branch currents at the fault bus and all otherbusses throughout the net"or* can be determined.

More on Transformer Phase Shifts

ransformer phase shifts due to "ye&delta connections ha!e been ignored in 5 matrices and fault

calculations up until this point. hey can be handled after 01- fault calculations are made. Hnce the

se)uence components of all !oltages and currents ha!e been computed in 01-# the phase shifts are

introduced according to the follo"ing section on MCalculation rocedure.N This must "e done "efore

con$ertin# %12 $olta#es and currents to a"c.

age 1 of -7&


18/26


&. Calculation Procedure

Step 1. ic* a system :V base and a VLL base at one point in the net"or*. he system :V base"ill be the same e!ery"here. s you pass through transformers# !ary the system VLL base according to

the line&to&line transformer turns ratio.

Step 2. he system base phase angle changes by 30O each time you pass through a @P 'or P@(

transformer. S/ rules state that transformers must be labeled so that high&side positi!e se)uence

!oltages and currents lead lo"&side positi!e se)uence !oltages and currents by 30O. egati!e se)uencedoes the opposite 'i.e.# &30O shift(. 5ero se)uence gets no shift. he Met 30O N phase shift bet"een a

faulted bus * and a remote bus % is ignored until the last step in this procedure.

Step 3. Begin "ith the positi!e se)uence net"or* and balanced three&phase case. ssume that thesystem is Mat restN "ith no currents flo"ing. his assumption re)uires that the only shunt ties are

machines "hich are represented as he!enin e)ui!alents "ith 1.0 pu !oltage in series "ith subtransient

impedances. Loads 'e8cept large machines(# line capacitance# shunt capacitors# and shunt inductors areignored. Con!ert all linetransformersource impedances to the system base using

-

=

%ewba"e

oldba"e

oldba"e

%ewba"eold

$u%ew$u

V

V

'

'ZZ .

ba"e' is three&phase :V. ba"eV is line&to&line. /f a transformer is comprised of three identical

single&phase units# old$uZ is the impedance of any one transformer on its o"n base# andoldba"e

' is three

times the rated po"er of one transformer. or a delta connection# ba"eV line&to&line is the rated coil

!oltage of one transformer. or a "ye connection# ba"eV line&to&line is the rated coil !oltage multiplied

by 3 .

Step 4. or small net"or*s# you can find the fault current at any bus * Mby handN by turning off all

!oltage sources and computing the positi!e&se)uence he!enin e)ui!alent impedance at the faulted bus#1#kkZ . /gnore the et 30O during this step because actual impedances are not shifted "hen reflected

from one side to the other side of @P transformers. Hnce the he!enin impedance is *no"n# then use

Fkk

$ekF

k ZZ

VI

+=

1#

11 # follo"ed by

Fkkk

$ek

Fk

IZVV11#11

= .

Step 5. he *ey to finding 01- currents in a branch during the fault is to *no" the !oltage on each end

of the branch. or a branch "ith positi!e se)uence impedance 1z bet"een busses % and *# first find

Fkjk

$ej

Fj IZVV 11#11 = .

ositi!e&se)uence current flo" through the branch during the fault is

1

1#1#

1# z

VVI

Fk

FjF

jk

= .

'ote that 1z is the physical positi$e seuence impedance of the "ranch ! it is not an element of the

( matri). n accuracy chec* should be made by ma*ing sure that the sum of all the branch currents

into the faulted bus e)uals FkI 1 .

age 1+ of -7&


19/26


Step &. ;hile continuing to ignore the et 30O# !oltage sag propagation at all other buses % can be

computed "ith

Fkjk

$ej

Fj IZVV 11#11 = .

Step *. or larger net"or*s# MhandN methods are not practical# and the 5 matri8 should be built. ormthe admittance matri8 @# and in!ert @ to obtain 5. /gnore the Met 30O N "hen forming @. :atri8

in!ersion can be a!oided if Aaussian elimination is used to find only the *thcolumn of 5.

Step +. Qnbalanced faults re)uire positi!e and negati!e se)uence impedances. egati!e se)uence

impedances are usually the same as positi!e. aults "ith ground currents re)uire positi!e# negati!e# and$ero se)uence impedances. 5ero se)uence impedances can be larger or smaller and are dramatically

affected by grounding. @P transformers introduce bro*en $ero se)uence paths. refault negati!e

se)uence and $ero se)uence !oltages are al"ays $ero.

Step ,. fter the 01- fault currents are determined# continue to ignore the Met 30O N and use

Fkjk

$ej

Fj IZVV 01-01-#01-01- =

for each se)uence to find 01- bus !oltages.

Step 1%. e8t# compute branch currents 'ignoring the et 30O( bet"een buses % and * for each se)uence

using

0

0#0#

0#z

VVI

Fk

Fj

jk

= #

1

1#1#

1#z

VVI

Fk

Fj

jk

= #-

-#-#

-#z

VVI

Fk

Fj

jk

= .

Step 11. s the last step# include the 'et 3%-bet"een bus % and faulted bus *. Go this by adding the

et 30O to FjV 1 and 1#jkI calculations# and subtracting the et 30O fromFjV - and -#jkI calculations.

This must "e done "efore con$ertin# %12 $olta#es and currents to a"c. hen# use

01-VTVabc = # 01-ITIabc =

to find the abc bus !oltages and branch currents.

age 1J of -7&


20/26


Short Circuit Pro"lem 1

he positi!e&se)uence one&line diagram for a net"or* is sho"n belo". refault !oltages are all 1.0pu.

a. Qse the definitionkmmI

k

j

k

j

jk I

V

I

V

z=

=

=#0

to fill in column 1 of the 5 matri8.

o"# a solidly&grounded three&phase fault occurs at bus 1.

b. Compute the fault current

c. Qse the fault current and 5 matri8 terms to compute the !oltages at busses - and 3.

d. ind the magnitude of the current flo"ing in the line connecting busses - and 3.

age -0 of -7&

1 - 3

1

-

3

%0.-R

%0.3R

%0.1R

%0.0DR

Bus 1

Bus -

Bus 3

%0.1R

>

10

?

>10

?


21/26


Short Circuit Pro"lem 2.

30:V# 1-*V generator is connected to a delta & grounded "ye transformer. he generator and

transformer are isolated and not connected to a Mpo"er grid.N /mpedances are gi!en on e)uipmentbases.

single&phase to ground fault# "ith $ero impedance# suddenly appears on phase a of the 7J*Vtransformer terminal. ind the resulting a&b&c generator currents 'magnitude in amperesand phase(.

Fegarding reference angle# assume that the pre&fault phase a !oltage on the transformerIs 7J*V bus has

angle 2 0.

age -1 of -7&

Aen30:V# 1-*V

Subtransient reactances

1 2 - 2 0.1+pu0 2 0.1-pu

Aenerator is connectedA@ through a %0.D ohm

grounding reactor

2 0.0Dpu

30:V1-*V7J*V

Single phase to

ground faultoccurs on phase a

ransformer

'Gelta&A@(


22/26


Short Circuit Pro"lem 3

one&line diagram for a t"o&machine system is sho"n belo".

he transmission line bet"een busses - and 3 has 1 2 - 2 0.1-pu# 0 2 0.40pu on a 100:V#34D*V base.

Qsing a base of 100:V# 34D*V in the transmission line# dra" one line diagrams in per unit for

positi!e# negati!e# and $ero&se)uences.

hen#

a. Compute the phase a fault current 'in pu( for a three&phase bolted fault at bus -.

b. Compute the phase a fault current 'in pu( for a line&to&ground fault at bus -# phase a.

age -- of -7&

ll !alues in puon e)uipment

base


23/26


age -3 of -7&

/se a 1%% M0 22%k$ "ase in the transmission line.

Bus1 Bus-

Bus3

BusDBus7Bus4

Ste$enson Pro". &.15


24/26


age -4 of -7&

/se 1%% M0 "ase

Bus1 Bus-

Bus3

Bus4 BusD

Bus7 BusBus+ BusJ

1%% M0 13+k0 in the

Ste$enson Pro". &.1&


25/26


Short Circuit Calculations


Balanced hree&hase ault# Ste!enson rob. 7.1D. three&phase balanced fault# "ith

52 0# occurs at Bus 4. Getermine

a. FaI4 'in per unit and in amps(

b. hasor abc line&to&neutral !oltages at the terminals of Aen 1c. hasor abc currents flo"ing out of Aen 1 'in per unit and in amps(


Line to Around ault# Ste!enson rob. 7.1D. Fepeat T4 for phase a&to&ground fault at

Bus 4# again "ith 52 0.

Short Circuit Pro"lem &.

Fepeat T4# Qsing Ste!enson rob. 7.17.

Short Circuit Pro"lem *.

Fepeat TD# Qsing Ste!enson rob. 7.17.

age -D of -7&


26/26


Ste$enson Pro"lem &.15 Phase to round ault at us 4

Enter olar orm 01- Currentsat Aen T1# Compute the BC Currents

___________________________________________________________________________

Enter olar orm 01- 0olta#es at Aen T1# Compute the BC Voltages

6nter

Press

7esults

Documents

11 ELC4340 Spring13 Short Circuits