Short Circuit 1

8/7/2019 Short Circuit 1

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PS 0523 POWER SYSTEM

MODELLING and ANALYSIS I

SYMMETRICAL FAULT ANALYSIS USING

BUS IMPEDANCE MATRIX


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1 INTRODUCTION

Short circuit study is one of the basic power system analysis problems. It is

also known as fault analysis. Whenever a fault occurs, the bus voltages and

flow of current in the network elements get affected. Faults occur in power

system i) due to insulation failure in the equipments

ii) due to flashover of lines initiated by lightning stroke

iii) due to mechanical damage to conductors and towersiv) due to accidental faulty operation.

Faults can be classified as Shunt faults and Series faults. Series faults are not

so severe as compared to shunt faults. Shunt faults are further classified as

1. Three phase fault

2. Single line to ground fault3. Line to line fault

4. Double line to ground.


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Three phase fault is a symmetrical fault and hence it can be analyzed on perphase basis. All other faults mentioned above are unsymmetrical faults an

they can be analyzed using symmetrical components.

The relative frequency of occurrence of various faults in the power systems, in

the order of severity is as follows;

Three phase fault 5 %

Double line to ground fault 10 %

Line to line fault 15 %

Single line to ground fault 70 %


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When f lt o r , l r e f lt rrent flow thro h the f lt ing l rge

rrent flow in the lines nd reduced us volt ges. Heavycurrent flows may

cause damage to the equi ments. Hence faulty section should e isolated from

the rest of the networ quickly on the occurrence of a fault. his can eachieved y roviding relays and circuit reakers. he rotective relays sense

the occurrence of the fault and send signals to circuit reakers to o en the

circuit under faulty condition. he circuit reakers should able to interrupt

the current flowing after a few cycles.

The calculation of currents in network elements for different types of faultsoccurring at different locations is called SH T T STUDY. The

results obtained from the short circuit study are used to find the relay

settings and thecircuit breaker ratings which are essential for power system

protection.

Short circuit study is carried out using Bus mpedance Matrix. Symmetricalshort circuit study using Bus mpedance Matrix is presented first. ollowed

by this unsymmetrical fault analysis using symmetrical components is

discussed.


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Bus impedance building algorithm

5


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The i m ered d i i en reference . The

. . im edance diagram i hown in ig. .

I admi ance matri can eobtainedas

B s =

2010-10-

10-33.333310-

10-10-26.6667

j

j0.15

Fig.

j0.075j 0.1 j 0.1

j 0.1

3

1 2

0 0


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One ay of finding its bus i pedance at i , Z us, is to in e t the abo e bus

ad ittance at i . If the nu be of buses is o e, it is difficult to get di ect

in e se. lte nati ely Z us at i can be obtained byconst ucting it adding the

element one by one.

The net o g aph of this po e system issho n in Fig. 3.

4

32

3

Fig. 3


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The network cons sting of e ements 1 2 and 3 is a partia network with

buses0 1and2 To this if e ement 4 is added the network will be as

shown in Fig 4.

Now a new bus 3 is created. The added element is a BRANCH. For the

next step network with elements 1 2 3 and 4 will be taken as partial

network. Thiscontains buses 0 1 2 and 3

4

3

21

Fig. 4

2

0

3

1


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When element 5 i ed t thi , the net k ill e hown in Fi . .

In thi e, no new u i eated and the added element link u es 2

and and hence it is called a LI K.

onsiderelements 1 and 2 alonewith impedances j0.15 and j0.075. Its Y us is

iven

Y us =

13.333006.6667j andhence its Z us is iven

Z us = j

0.0750

00.15

1

2

21

5

3

21

Fi .

2

0

3

1

j 0.075

- 1 .

j 0.15

- j 6.6667

1 2


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Note that these element values are the values of the shunt impedances at buses 1

and 2. This result can be extended to more number of buses also. The above ZBus

corresponds to the partial network shown in Fig. 5.

Now add element 4. This it from bus 1 to 3 with an impedance j 0.1. The added

element is a branch. It is from the existing bus 1 and it creates a new bus 3 as

shown in Fig. 6. The modified ZBus matrix is

j 0.1

j 0.075j 0.15

21

Fig. 5

2

4

21

Fig. 6

2


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ZB s = j

0.2500.15

00.0750

0.1500.15

N w a eleme 3. I is r m s 1 2 wi a im e a ce j0.1. Si ce i is

li ki w e exis i ses 1 a 2 as s w i i .7 i is a Li k e wee

ses 1 a 2.

3

321

3

2

1

4

j 0.1

21

i . 7

2


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Th odifi d Z u trix ith th u i

Z u = j

0 3250 150 0750 15

0 150 2500 15

0 07500 0750

0 150 1500 15

Eli in ting u

Z u = j

0 18080 0340 0808

0 0340 05770 0340 08080 0340 0808

N

N


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Nowa ele e . I is rom s2 o3wi a impe a ceof j 0.1. Si ce i is

li ki woof eexis i ses2a 3ass own in i . 8, i isaLink.

T emodifiedZBusma rixwi us is

ZBus j

0. 90.-0.02310.0462-

0.1462-0.18080.03460.0808

0.02310.03460.05770.0346

0.0462-0.08080.03460.0808

5

j 0.14

3

21

i . 8

2

N

3

2

1

N

321


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N

N

ZBus = j

0.26920.1462-0.02310.0462-

0.1462-0.18080.03460.0808

0.02310.03460.05770.0346

0.0462-0.08080.03460.0808

Eliminating bus the final bus impedance matrix is obtained as

ZBus = j

0.10140.04710.0557

0.04710.05570.0386

0.05570.03860.0729

The correctness of the result can be verified by multiplying the values of the tw

matrices ZBus and YBus.

3

2

1

3

2

1

321

1 2 3


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2 S I PE A E A IX - I I G A GO I

Bus pedance atrix usZ of a po er net ork can e obtained by inverting

the bus ad ittance atrix busY , hich is easy to construct. o ever, hen

the order of atrix is arge, direct inversion requires ore core storage and

enor ous co puter ti e. herefore inversion of busY is prohibited for arge

size net ork.

Bus i pedance atrix can be constructed by adding the net ork ele ents

one after the other. sing i pedance para eters, perfor ance equations in

bus fra e of reference can be ritten as

busbusbus Z! (1


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NpNqqp2p21p1p IZIZIZIZ ! .... (3

ith ! oth r bus currents s t to zero, Z! . hus Z can b obtain d by measuring hen 1 .u. current is injected at bus and

leaving the other bus currents as zero. and can be varied from 1 to to

cover all the elementsof busZ .

While making measurements all the buses except one, are open circuited.

Hence, the bus impedance parameters are called open circuit impedances. he

diagonal elements in busZ are kno n as driving point impedances, hile the

off-diagonal elements are called transfer impedances.

While constructing busZ using building algorithm, elements are added one by

one. At any stage, the added element may be a branch or a link as

explained belo .


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The performance equation of the network shown in Fig. 4 is

/

/

/

Fig. 4

b sb sb s IZE ! (4)

where E = an m x 1 vector of voltage mea red with reference to

reference 0

I = an m x 1 vector of c rrent

hen an element p-q i added to the partial network, it may e a ranch or a

link.

Partial

network

1

2

m

0


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Addition of a branch

An element having an impedance of z is added from bus p, creating a ne

bus q as sho n in Fig 5.

/

/

/

Fig. 5

The performance equation of the ne net ork will be

q

m

2

1

E

E

E

E

E

/

/

=

qqqmqpq2q1

mqmmmpm2m1

pqpmppp2p1

2q2m2p2221

1q1m1p1211

..

..

/////

..

/////

..

..

q

m

p

2

1

/

/

(5)

Partial

network

1

2

m

0

p q


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22

/

/

/

/

o th bus urr nts ar

0

0

1

0

0

I ibus

/

/

(6)

Fig. 6

1 p.u.

Partial

n t ork

1

2

i

0

p q

pE

qE


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Corr sponding to this bus urr nt, fro qn. (5) g t

iqq

i

ipp

2i2

1i1

!

!

!

!

!

/

/

(7)

Th r is no urr nt flo in th add d l nt. It is also assum d that th r is

no mutual oupling b t n th add d l m nt and th l m nts in th partian t ork. Th n

qE = pE (8)


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z

//

/

/

/

//

Now th bus urr nts ar

!

q

bus

1

0

0

I /

/

(11)

Using th abov in th p rformanc qn. (5), w g t

Partial

n twor

1

2

0

p q

Fig. 7

1p.u.


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qqq

qmm

qpp

2q2

1q1

ZE

ZE

ZE

ZE

ZE

!

!

!

!

!

/

/

(12)

Th add d l m nt p-q has an imp danc of and th r is no mutual coupling

b tw n th add d l m nt and th l m nts in th partial n twor . Th n fromFig. 7 it is cl ar that

pq EE ! i. . pq EE !

Using qn. (12) in th abov quation, w g t

pqqq zZZ ! (13)

If p happ ns to b th r f r nc bus, th n 0Ep ! and h nc Z 0. Thus

Z ! (14)


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Add n f nk

If h dd d n p-q nk h p du f u n h n

f buZ nn n w h h dd d n u Ne as

sh wn in Fig. 8.

1

/

/ i

/ p

N

q

m

0

Ne

Fig. 8

Partial

n tw rk


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Thi i titio hi h ill li i t l t . The olt e o e

Ne i elected ch that the current through the added link i ero. No the

performance equation i

N

/

/

e

E

E

E

E

m

p

2

1

=

NNNNNN

N

N

N

N

..

..

/////

..

/////

..

..

ZZZZZ

ZZZZZ

ZZZZZ

ZZZZZ

ZZZZZ

mp21

mmmmpm2m1

ppmppp2p1

22m2p2221

11m1p1211

N

/

/

I

I

I

I

I

m

p

2

1

(15)

To determine iN

The elementN

can e determined calculating the oltage at the thN us

.r.t. to us q hen p.u. current is injected into us iand other uses ar

open circuited as sho n in Fig. 9. For this condition, the us current ector is


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i

i

ipp

i22

i11

Ze

ZE

ZE

ZE

ZE

NN

/

/

!

!

!

!

!

(17)

As there is n current in p N, pq EE !N i.e.

qpEEe !

N (18)

Using eqn. (17) in the b e equ ti n

,1,2,......iiqipi !!N (19)

If p happens t be reference bus 0Ep ! and hence 0ip ! . Thus

m,1,2,......iZZ iqi !!N (20)


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To d t rmin NN

Z

Th l m ntNN

Z can b calculat d by m asuring th voltag at th thN bus

w.r.t. busq aft r inj cting 1 p.u. curr nt at th thN bus and ping oth r bus s

op n circuit d as shown in Fig. 10. Now th bus curr nt is

1

/

i

p

N

(21)

q

m

0

Fig. 10

1 p.u.

Partial

n twor

Ne

!

N

/

/

/

1

0

0

0

Ibus

z


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Substituting qn. (21) in qn. (15), p rformanc quation r duc s to

NNN

N

N

N

N

N

/

/

/

Ze

Z

Z

Z

Z

Z

qq

pp

22

11

!!

!

!

!

!

(22)

From th Fig. 10 it is cl ar that !N

i. .

qp !N (23)

Substituting qn. (22) in th abov quation g t

qp ! NNNN (24)

If p happ ns to b r f r nc bus, p ! and h nc Zp !N . Thus

! NNN (25)


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bus

_

j

_

i

busbusbus IZ

ZZ

IZNN

NN

!

= bus

_

j

_

i

bus I]Z

ZZZ[

NN

NN

Thus ! )onmodifi tifore(us)modified(usNN

NN

Z

ZZ

_

j

_

i (29)

ot that_

iZ N

_

jZN ill b a mxm matrix. l m nts of modifi d bus imp danc

matrix can b obtain d as

! )nificatibef re(ji)ifie(ji ZZNN

NN

ZZZ ji (30)i = 1,2,.,

j = 1,2,.,


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Summary of formulas

p is not reference bus p isthereference bus

Added

element

p q ipiq ZZ ! Z iq !

is aqi

i !

qi

i !

Branchqpqq zZZ ! qq zZ !

Added iqipi ZZZ !N iqi ZZ !N

elementN{

!

i

m.,1,2,......i

N{

!

i

m.,1,2,......i

p q qp zZZZ ! NNNN q zZZ !NNN

is a

Link !

nmodiica

i

or

i

modii

d(ji ZZ

NN

NN

Z

ZZ ji

m....,1,2,......jandm....,1,2,......i !!


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Exampl 1

Consid r th po r syst m sho n in Fig. 11. Th valu s mark d ar p.u.

imp danc s. Th p.u. r actanc s of th g n rator 1 and 2 ar 0.15 and 0.07

r sp ctiv ly. Comput th bus imp danc matrix of th g n ratortransmission

n t ork.

j0.1

Fig. 11

1

G

2G

Gj0.1 j0.1

3

1 2


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Sol tio

The gro s is m ered as a d it is taken as reference s. The p. .

impedance diagram is sho n in ig. 12.

ig. 12

j . 5j .1 j .1

j .11 2

j .15

0


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Add ele ent 1 3. It is a branch fr bus 1 and it creates bus 3.

bus !

0.180770.0346150.080770.0346150.057690.034615

0.080770.0346150.08077

Finall add ele ent 2 3. It is a lin between buses 2 and 3. ith bus N

jZbus !

0.269230.1461550.0230750.046155

0.1461550.180770.0346150.08077

0.0230750.0346150.057690.034615

0.0461550.080770.0346150.08077

Eli inating the thN bus, final bus i pedance atri is btained as

jZbus !

0.101430.047140.05571

0.047140.055710.03857

0.055710.038570.07286

1

1 2

2

3

3

1

1

2

2

3

3

N

N

1

1

2

2

3

3


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Consider the circuit shown.

J 0.1

J 0.15 J 0.12

31 2

Its bus impedance is jZbus !

0.370.250.1

0.250.250.1

0.10.10.1

For the circuit

J 0.1

J 0.15 J 0.12

31 2

J 0.8

!busV j

.3..

...

...

.j

!

.

.

.


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For this circuit

J 0.1

J 0.15 J 0.12

31 2

-

+

1.0

!

1.0

1.0

1.0

V

V

V

3

2

1

J 0.1

J 0.15 J 0.12

31 2

J 0.8

For this combined circuit

-1.0 +

!

1.0

1.0

1.0

V

V

V

3

2

1

!

0.2

0.2

0.08

0.8

0.8

0.92


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3 YMMETRICAL AULT A ALYSIS USI G MATRIX

In classical method of symmetrical fa lt analysis, Thevenins impedance is to b

calculated. This impedance is the equivalent impedance bet een the fault poin

and the reference bus. This ill call for net ork reduction if the net ork i

large. ractical systems are very large and the net ork reduction ill be very

tedious. In such cases, the bus impedance matri busZ can be readily used for

the fault analysis.

In this method, once the bus impedance matri is constructed, ith a very few

calculations, bus voltages and currents in various elements can be computed

quickly. When faults are to be simulated at different buses, this method prove

to be good.

Symmetrical short circuit analysis essentially consists of determining the stead

state solution of linear network with balanced sources.


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Normally, in the short circuit study the following assumptions are made.

1. all the shunt parameters like loads, line charging admittances etc. areneglected.

2. all the transformer taps are at nominal position.

. prior to the fault, all the generators are assumed to operate at rated

voltage of 1.0 p.u. with their emfs in phase.

With these assumptions, in the prefault condition, there will not be anycurrent flow in the network and all the bus voltages will be equal to 1.0 p.u.

The linear network that has to be solved comprises of

i) Transmission network

ii) Generation system and

iii) Fault

By properly combining the representations of the above three components, we

can solve the short circuit problem


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Fig. 13

Consider the trans ission networ shown in Fig. 13. Ta ing the ground as t he

reference bus, the bus ad ittance atrix is obtained as

busY =

3232

3311

2121

yyyy

yyyy

yyyy

If we add all the colu ns (or rows ) we get a colu n (or row ) of all ero

ele ents. Hence this bus atrix is singular and hencecorresponding busZ atrixof this trans ission networ does not exist. Thus, when all the shunt para eters

are neglected, busZ atrix will not exist for the trans ission networ .

1y 3y

2

1 32y

1 3

1

2

3

2


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However, connection to ground is established at the generator buses, representing

the generator as a constant voltage source behind appropriate reactance as shown

in Fig. 14.

2

1 3

Fig. 14


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As stated earlier, there is no current flo in the ne t ork in the pre-fault

condition. uring short circuit, injection of bus current arises due to fault.

hen such injection is there, then there ill be current flo in different lines

and bus voltages ill no longer remain at the pre-fault values.

Consider the net ork sho n in Fig. 14. Symmetrical fault occurring at bus 2 can

be simulated by closing the s itch sho n in Fig. 15. ereF

Z is the fault

impedance.

Fig. 15

2

1 3

FZ


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Any general p er system ith a number f generat rs and number f buses

subjected t symmetrical fault at thp bus ill be represented as sh n in Fig. 16.

/

/

/

Fig. 16

In the faulted system there are t types f s urces

1. Current injecti n at the faulted bus

2. Generated v ltage s urces.

The bus v ltages in the faulted system can be btained using Superp siti n

theorem.

f

V

f

V

f

V

Transmission

et or

FZ

p


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Bus voltages due to current injection:

Make all the generator voltages to zero. Then e have enerator-Transmission

system ithout voltage sources. Such net ork has transmission parameters and

generator r

eactanc

es b

etee

n generator bus

es and th

eground. L

et busZ b

ethe

bus impedance matrix of such enerator-Transmission net ork. Then the bus

voltages due to the current injection ill be given by

)F(busbusbus IZV ! (31)

here )(busI is the bus current vector having only one non-zero element. Thus

hen the fault is at the thp bus

!

0

0

I

0

0

I )F(p)F(bus

/

/

ere is the faulted bus current. 32


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Thus, bus voltages due to current injection ill be

!

NNpNN2N1

Npppp2p1

2N2p2221

1N1p1211

bus

ZZZZ

ZZZZ

ZZZZ

ZZZZ

V

..

////

..

////

..

..

0

0

0

(p

/

/=

pN

pp

2p

1p

Z

Z

Z

Z

/

/ pI (33)

Bus voltages due to generator voltages

Make the fault current to be zero. Since there is no shunt element, there ill beno current flo and all the bus voltages are equal to V , the pre-fault voltage

hich ill be normally equal to 1. p.u. Thus, bus voltages due to generator

voltages ill be


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!

1

1

1

1

Vbus

/

/

0V (34)

Thus for the faulted system, wherein both the current injection and generator

sources are simultaneously present, the bus voltages can be obtained by addin

the voltages given by eqns. (33) and (34). Therefore, for the faulted system th

bus voltages are

!)(busV

)(

)(

)(2

)(1

V

V

V

V

/

/=

pN

pp

2p

1p

Z

Z

Z

Z

/

/ )F(pI

1

1

1

1

/

/ 0V (35)

To calculate )(busV we need the faulted bus current )(pI which can be

determined as discussed below.


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The fault can be described as sho n in Fig. 17.

Here pV and pI are the faulted bus voltage and current respectively.

urther V , I and are the fault voltage, current and impedance respectively.

It is clear that ,IZV ! pV = V and pI = - I 36

Therefore

pF)Fp IZV ! 37)

FI

FZ

p

)F(pI

)F(pV

FV

Fig. 17


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Theth

equation extracted from eqn. (35) gives

0)F(ppp)F(p VV (38)

Substituting eqn. (37) in the above, we get

)(Z )(Z

Thus the faulted bus current )( is given by

)F(pI =Fpp

0

ZZ

V

(39)

Substituting the above in eqn. (37), the faulted bus voltage )F(pV is

)F(pV = 0Fpp

FV

ZZ

Z

(40)

Finally voltages at other buses at faulted condition are to be obtained. Theth

i

equation extracted from eqn. (35) gives


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0)F(ppi)F(i VIZV !

Substituting eqn. (39) in the above, e get

0V

ZZ

ZVV

Fpp

pi

0)F(i

! (41)

Kno ing all the bus voltages, current flo ing through the various net ork

elements can be computed as

mk)(m)(k)(mk y)(i ! here mky is the admittanceofelement mk . (42)

When the fault is direct, 0Z F ! and hence

= 0

Z

V , V = 0 and 0

i

0)F(i V

Z

ZVV !

i

N....,1,2,......i

{

! (43)

It is to be noted that hen the fault occurs at the th bus, only the th column

of busZ matrix ( and not the entire busZ matrix ) is required for further

calculations.

i

N....,1,2,......i

{

!


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54

The following are thevarious steps for conducting symmetrical short circuit

analysis.

Step 1 ead i) Transmission line data

ii) enerator reactances dataiii) Faulted bus number p and

iv) Fault impedance .ZF

Step 2 Construct the bus impedance matrix of the transmission networ

including the generator reactances.

Step 3 Compute )F(pI from )F(pI =Fpp

0

ZZ

V

Step 4 Compute p from p = 0

pp

ZZ

Z

Step 5 Compute from 0pp

p

0)( VZZ

ZVV

! p

....,1,2,......!

Step 6 Calculate the element currents from m)F(m)F()F(m )VV(i !


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Example2

Consider the po er system discussed in Example 1. The p.u. impedances are on

a base of 50 MVA and 12 kV. Symmetrical short circuit occurs at bus 3 ith

zero fault impedance. Using busZ matrix determine the fault current, bus voltages

and also the currents contributed by the generators.

Solution

As seen in example1, busZ matrix of the transmission-generator net ork is

jZbus !

0.101430.047140.05571

0.047140.055710.03857

0.055710.038570.07286

Faulted system is sho n in Fig.18.

+ +

_ Fig.18 _ _

3

1

1

2

2

3

3

1 2

j 0.15 j 0.075

1.0 p.u. 1.0 p.u.


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57

Example 3

Fort etransmission-generators stems own in Fig.1 , t e us impedancematrix isobtainedas

jZbus!

0.1014290.0471430.055714

0.0471430.0557140.038571

0.0557140.0385710.072857

Symmetrical t reephasefault wit fault impedance j 0.052143 p.u. occurs

at bus1. Find t ep.u. currents inall t eelementsandmark t em on t e

single linediagram.

1

1

2

2

3

3

Fig. 1

j0.075j 0.1 j 0.1

j 0.1

3

1 2

j0.15

0 0


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58

Solution

Faultoccursat us1andweneedt efirstcolu nofbu , which is

0.055714

0.038571

0.072857

and ZF = j 0.0521

Faulted uscurrent I1(F) = u.p.8j0.125j

10.0521j0.0728 7j

1 !!

Faultcurrent IF = - I1(F) = - j p.u.

F = ZF IF = ( j 0.0521 )(- j ) = 0. 171 p.u.

1(F) = F= 0. 171 p.u.

2(F) = p..u.0.6 10.125

0.0 5711 !

(F) = p..u.0. 20.125

0.055711 !

2

11


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60

SHORT CIRCUIT MVA

Th Shor Cir ui MVA i l o nown FAULTL VELorFAULT MVA.Th

ir ui br rbr ing p i y u b qu l oorgr r h n h hr

ph faul MVA.Th i ationof ir uitbr aker apacity i adeonthe

basis that it ustcleara three phase faultas that is generally the wors

case. ysi ulating threephase faultatapoint,shortcircuit level at that

pointcanbeco putedas

Short Circuit MVA =prefaultvoltage inp.u.xfaultcurrent inp.u.x ase MVA

Unless it isgivenotherwiseprefaultvoltageshall betakenas1.0p.u.


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S l ti


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62

Solution

ju !

0.1014290.0471430.0557140.0471430.0557140.038571

0.0557140.0385710.072857

When three phase fault occursat bus1:

Fault current.

p.3..!

Fault le el 3. MVA

When three phase fault occursat bus 3:

Fault current .p.u. 5910.101 29

1!

Fault le el 9. 9 493 MVA

1

23

21 3

Documents

Short Circuit 1