Short Circuit Studies

Short Circuit Studies

Indonesia Clean Energy Development (ICED) project

Indonesia Wind Sector Impact AssessmentPresented by:

Dr. K. Balaraman,

Makassar, February 17 to 21, 2014

Short circuit study requirements:

•  Determining the fault level at buses

• Selection of breaker ratings

• Protective device co-ordination

Symmetrical components

• Unbalanced system of ‘n’ related phasors can be resolved to ‘n’ system of balanced phasors

• In each balanced phasor, angle between two phasors and magnitude of each phasor are equal

• Phase Quantities: Ia , Ib & Ic and• Sequence components are:

– Ia1 stands for Positive sequence current– Ia2 stands for Negative sequence current– Ia0 stands for Zero sequence current.

021

021

021

ccc

bbb

aaa

IIIIIIIII

c

b

a

III

=

Sequence components

• Ia1, Ib1 & Ic1 : Same phase sequence as Ia, Ib & Ic

• Ia2, Ib2, & Ic2: Opposite phase sequence as Ia, Ib and Ic

• Ia0, Ib0, & Ic0 : All in-phase

Ia

Ic

Ib

Ia1

Ic1

Ib1

120

120 120 Ia2

Ib2

Ic2

120

120 120

Ia0 Ib0 Ic0

Unbalanced current phasor

Positive sequence current

phasors

Negative sequence current

phasors

Zero sequence current phasor

c0c2c1c

b0b2b1b

a0a2a1a

VVVV

VVVV

VVVV

a0

a2

a1

2

2

c

b

a

VVV

1aa

1aa

111

VVV

Similarly for voltages

sequencetoPhaseps

aa

aa

T

phasetoSequencespand

aa

aa

aaT

ps

sp

1111

1

31

12012401,1201

1

1

2

2

2

2

2

111

alwaysVVV

VVVV

VVV

aa

aa

VVV

cabcab

cbaa

c

b

a

a

a

a

0

31

1111

1

31

0

2

2

0

2

1

Therefore no zero sequence exists in line voltage phasor. Since phase voltage sum is not always zero, in the phase voltage phasor, the zero sequence voltage exists.

III

a aa a

III

I I I I

a

a

a

a

b

c

a a b c

1

2

0

2

2

0

13

111 1 1

13

Delta Connection

Ia

a

b

Ib

Ic

c

Ia+ Ib + Ic= 0

Therefore no zero sequence current flows into delta connection.

Sequence Currents

Ia a

Ib

Ic

b

c

Star connection Star grounded

Ia

I0a

3I0

Ib

Ic

I0

I0

b

c

Ia+Ib+Ic = 0, There fore no zero sequence current flows into star connection

Ia+Ib+Ic may not be zero. Hence path always exists for zero sequence currents.

Star - Delta +ve and -ve Zero

Transformer sequence impedance diagram

Star grounded-Delta +ve and -ve Zero

Transformer sequence impedance diagram

Three winding Transformer

a

b

c

a’

b’

c’

Zaa

Zbb

Zcc

c

b

a

cccbca

bcbbba

acabaa

'cc

'bb

'aa

III

ZZZZZZZZZ

VVV

Let Zaa = Zbb = Zcc = Zs and Zab = Zac = Zba = Zbc = Zca

= Zcb = ZmTsp Vs = [Z] . Tsp.Is

Sequence Impedance

1 1 111

1 1 111

0 00 00 0 2

2

2

1

2

0

2

2

1

2

0

1

2

0

1

2

0

1

a aa a

VVV

Z Z ZZ Z ZZ Z Z

a aa a

III

VVV

Z ZZ Z

Zs Z

III

Z

a

a

a

s m m

m s m

m m s

a

a

a

a

a

a

s m

s m

m

a

a

a

, ,2 00 0

0 00 0 2

ZZ Z

Z ZZ Z

s

s m

s m

s m

Rotating Machines

• Za,b,c is not symmetric. Even then, the Z1,2.0 is diagonalized

Exercise:• Find the expression for Z1,2,0 • and Prove that it is a diagonal matrix:

VVV

Z Z ZZ Z ZZ Z Z

III

a

b

c

s m m

m s m

m m s

a

b

c

1 2

2 1

1 2

a

b

c

Ia

Ib

Ic

Ea

Ec Eb

ZnIn

+

+ +

Steady State

c

a

b

Ia1

Ib1

Ic1

Ea

Ec Eb

Z1

Z1

Z1

Z1

Ea Va1

Ia1

Reference Bus

Va1 = Ea - Ia1 Z1

a

+

-

Positive Sequence Network

c

Ia2

Ib2

Ic2

Z2

Z2

Z2

Z2 Va2

Reference Bus

Va2 = - Ia2 Z2

a b

a

Negative Sequence Network

c

a

b

Ia0

Ib0

Ic0

Zg0

3Zn Va0

Reference Bus

Va3 = - Ia0 Z0

a Zg0

Zg0

Zn 3Ia0 Zg0 Z0

Zero Sequence Network

Generator impedance for fault study :

• Transient (xd’) or sub-transient ( xd”) is considered for positive sequence

• X2 i.e. Negative sequence which is close to xd”. (Approximately)

• X0 is small 0.1 to 0.7 times xd”

Typical values on own rating: Xd 100 to 200 %Xq 60 to 200 %

Xd’ 21 to 41 %

Xd” 13 to 30 %

X2 Xd”

Transmission line:

• Positive sequence impedance = Negative sequence impedance• Zero sequence impedance depends upon: Return path, Ground

wires and Earth resistively• Zero sequence reactance is approximately 2 - 2.5 times

positive sequence impedanceR0 is usually large. May be 5 to 10 times also

• B0 is 65 to 80 % of B positive sequence

Transformers :• All are equal i.e. Zpt = Znt = Zzt for transformer

Va1 Vf

Z1

Ia1

Va2

Z2

Ia2

Va0 Z0

Ia0

Va1 = Vf - Z1.Ia1 Va2 = -Z2 . Ia2 Va0 = - Z0 . Ia0

a0

a2

a1

0

2

1f

a0

a2

a1

III

Z000Z000Z

00V

VVV

Fault Representation

a

b

c

a

b

c

Vab = 0 Va = 0 Va1 = 0Vbc = 0 Vb = 0 Va2 = 0Vca = 0 Vc = 0 Va0 = 0

a0

a2

a1

0

2

1f

III

Z000Z000Z

00V

000

Three phase fault representation

Ia1 = Vf Z1 , Ia2 = 0 , Ia0 = 0

a0

a2

a1

2

2

c

b

a

III

1aa

1aa

111

III

Ia = Ia1

Ib = a2Ia1

Ic = aIa1

  Ia = Ib = Ic

Zf

a

Zf

a

Zf

a

Z0

Zf

a

Zf

a

Zf

a

0I0IZZ

VI a0a2f1

fa1

, ,

Fault through impedance :

ab

c

IcIb Ia

Va = 0

Ib = 0

Ic = 0

Single line to ground fault representation

021

3

ZZZV

If

a 021

021 ZZZV

IIIf

aaa

IZIZIZVVVV 002211021 aaafaaa 000 IZV aa 222 IZV aa

111 IZVV afa

021

02

1

021

000000

00

III

ZZ

ZV

VVV

aaaf

aaa

021 III aaa

22

11111

31 aa

aa

III

cba

021

III

aaa

021 0VVVV aaaa Q

a

b

c

Ia

Ic

Ib

Zf

f021

fa0a2a1 3zZZZ

VIII

Fault through Impedance

a

b

c

Ic

Ib

Ia

V b = V c

Ia = 0

Ib = -I c

c

b

a2

2

a0

a2

a1

VVV

111aa1

aa1

31

VVV

Va1 = 1/3 [Va+aVb+a2Vc] = 1/3 [Va+aVb +a2Vb]Va2 = 1/3 [Va+a2Vb+aVc] = 1/3 [Va+a2Vb+aVb]Va1 = Va2

Line to line fault representation

III

a a

a aIII

a

a

a

a

b

c

1

2

0

2

213

1

11 1 1

Ia0 = 1/3 (Ia +Ib +Ic) = 1/3 (Ib - Ib) = 0

Ia1 = 1/3 (aIb+a2Ic) = 1/3(aIb-a2Ib)

Ia2 = 1/3 (a2Ib + aIc) = 1/3 (a2Ib-aIb) Ia1 = -Ia2

0

000000

00 1

1

0

2

1

0

2

1

a

af

a

a

a

II

ZZZV

VVVVa1 = Vf - Z1Ia1

Va2 = Z2Ia2 = Va1 = Vf - Z1Ia1

Z2Ia1 = Vf - Z1Ia1 I VZ Z and I I I Va

fa a a a1

1 22 1 0 00 0, ,

a

b

cIcIb

Ia

Zf

I IV

Z Z Za af

f1 2

1 2

Fault through impedance

a

b

cIcIb

IaVb = 0

Vc = 0

Ia = 0

VVV

a aa a

Vaaa

a120

2

213

111 1 1

00

V V Va a a1 2 0

1 1 1V V V V V Va a a a a a1 2 03 3 3

, ,

Double line to ground fault representation

III

a a

a a II

a

a

a

b

c

1

2

0

2

213

1

11 1 1

0

Ia1 + Ia2 + Ia0 = IA =0Ia1 = 1/3 (aIb + a2Ic)Ia2 = 1/3 (a2Ib + aIc)Ia0 = 1/3 (Ib + Ic)

1/3(aIb + a2Ic) + 1/3(a2Ib + aIc) + 1/3(Ib +Ic) = 0aIb + a2Ic + a2Ib + aIc + Ib + Ic = 0

VVV

V ZZ

Z

III

a

a

a

f a

a

a

1

2

0

1

2

0

1

2

0

00

0 00 00 0

Va1 = Vf - Z1Ia1

Va2 = - Z2Ia2

Va0 = - Z0Ia0

Va1 = Va2 = Va0

\ Vf - Z1Ia1 = - Z2Ia2 - Z0Ia0

\ I V

Z Z ZZ Z

af

1

12 0

2 0

æèç

öø÷

V a0

Z 1 Z 2 Z 0

V f

I ao

V a2V a1

I a2I a1

Fault through impedance:a

b

c

ZfZf

Zg

I V

Z Z ZZ Z

Z Z ZZ Z ZZ Z Z Z

af

f

f

f g

1

12 0

2 0

1 1

2 2

0 0 3

æèç

öø÷

'' '

' '

'

'

'

bc

b’c’

a a’Ib = 0 and Ic = 0

V V Z I Z Iia

ka

aa a aa f ' '

III

a aa a

I

I I I I I

a

a

a

a

a a a a f

1

2

0

2

2

1 2 0

13

111 1 1

00

13

13

Open in phase B & C

VVV

V ZZ

Z

III

a

a

a

f a

a

a

1

2

0

1

2

0

1

2

0

00

0 00 00 0

V V Z IV Z IV Z I

V V V V V Z I Z IV V

Z I V Z I Z I Z I

I I IV

Z Z Z Z

a f a

a a

a a

ka

la

kl kl kl zz f aa a

f kl

aa a kl a a a

a a aokl

aa

1 1 1

2 2 2

0 0 0

1 2 01

1 1 1 2 2 0 0

1 21 2 0

3

3

3

' '

'

'

f0

f2

f1f

'aa

0kl

0lk

0ll

0kk0

2kl

2lk

2ll

2kk2

1kl

1lk

1ll

1kk1

ZZZZ3Z

ZZZZZ

ZZZZZ

ZZZZZ

• Vkl is the Thevenin’s equivalent voltage, once the line is removed between buses k & l

Solution Methodology:• 1.    Do the load flow with the line isolated• 2.    Then insert the line with two phase open and

perform short circuit study

Open in phase A• Ia = 0• Similar type of analysis like double line fault• Voltage is between two nodes, rather than between

node and ground• Impedance is the Thevenin’s equivalent impedance

between nodes.

IV V

ZZ Z

Z Z

fl k l

1

2 0

2 0

'' '

' '.

Z Z Z Z Z Z

Z Z Z Z Z Z

Z Z Z Z Z Z

lkk ll lk kl f

lkk ll lk kl f

lkk ll lk kl f

11 1 1 1 1

22 2 2 2 2

00 0 0 0 0

I IZ

Z Zf f2 1 0

0 2

.'

' ' II Z

Z Zff01

2

0 2

. '

' '

Solution methodology

• Form the Y bus for +ve, -ve and zero sequence• Do the LU factorisation for +ve, -ve and zero

sequence• To find the driving point impedance in fault study,

particular row or column of Z bus is requiredConsider, Yx = zx=Y-1zx= Z.z

Let z have 1 in pth location and 0, at all other location

Solution methodology

• Pass in z, 1.0 for the desired bus and 0 else where.• Call LU solution - x will be the desired column of Z bus.• Find the sequence fault current I1

f, I2f and I0

f.• Determine the post fault bus voltage

x

Z Z Z ZZ Z Z Z

Z Z Z Z

ZZ

Z

Yx z

p n

p p pp pn

n n np nn

p

pp

np

11 12 1 1

1 2

1 2

101

0

LL

ML

M M

[Vf] = [Z] [If1]

= [Z] {[I0] - [If]}

= [Z] [I0] + [Z][-If]

= [V0] + [V]

V = [Z] [-If]

Y[V] = [-If]

Solve V using LU solution.

Above can be written as :

[Y]p[V]p = [-If]p , p: positive sequence and similarly for other sequences.

For open fault, between k & l,

I k II

f p f

f

00

0

MMMM

For faults other than open fault,

I If p f

p

00

0

:

:

A

BC

DE

F

G

20% on 100 MVA

Xd”= 9%

100 MVA ,138/13.8 kV11%

40 Mile, 0.77 Ohm/mile X.X0 = 3X

50 MVA ,13.8/138 kV11%

Xd”=24% High resistanceGrounding

Example:

Grounding practice in Power System

Advantages of ungrounded system :     • Ground connection normally doesn’t carry current.

Hence elimination saves the cost.• Current can be carried in other phases, with fewer

interruptions.

Limitations of ungrounded system

• With the increase in voltage and line length current has increased and self clearing nature advantage couldn’t be seen

• Arcing ground : Phenomena of alternate clearing and re-striking of the arc, which cause high voltage (surge and transient)

• If grounded, the insulation can be graded in transformer from line to neutral, there by reducing the cost

• In case of ungrounded system, influence on communication lines is more.

• Ground current can’t be limited

a

b

c

3.01.73 1.73

0.0

Perfectly transposed line : Neutral of transformer is at zero potential.Un-transposed line : Neutral is shifted.Capacitance grounded : Perfectly transposed line.

Ungrounded system

Resistance grounded system :

• The resistance to neutral limits ground current. Selection of resistance value:• Amount of ground fault currentPower loss in resistor during ground fault• Power loss: usually expressed as a percentage of

system rating

R

Xg = 16 %Xg = 8%

Z1 = 24% , Z2=24% , Z0=3R+j8

IZ Z Z j j R j R j

Power Loss I RR j

R

f

f

´

æ

èç

ö

ø÷

3 3 100

24 24 3 8

300

3 56

300

3 56

1 2 0

2

2

.

If : in pu, R : in pu For three phase system, power loss

I R

percentage of phasesystem kVA ratingf

2

33

R is in percent p.u.Maximum power loss = ? Selection : How much be the value of ground fault current ?What should be the percent power loss in the ground resistance ?

X system

X systemR

X systemo0

1 130 1 0 . .

X1 =X2 =25%

X1 =7% =X2=X0

X1 =34%X2 =120%

69 kV25 MVA

69 kV

B

A

Effectively grounded system

• Solidly grounded : No impedance between neutral and earth

• Effectively grounded : As per AIEE standard No.32, Section 32 - 1.05, May 1947 :

For fault at AXX

For fault at BXX

ac ce Grounded System

XX

but no resonance resonance grounding

0

1

0

1

0

1

732

12732 34

12766

3 0

³

Re tan :

. ;

If Xg = 50 at A , find the nature of grounding at A & B.

0

0

I

I

I

0

I

I 2

Resonant - Grounded System :

• Capacitance current is tuned or neutralised by a neutral reactor or similar device

• Fault current is made zero by selecting the suitable tap on the ground reactor

Selection of Breaker

• Determine the symmetrical current (maximum) for any type of fault

• Multiply the current by factor obtained from standard tables depending on the duty cycle (2 cycles clearing, 5 cycles clearing etc.)

• Use the above current to specify the interrupting current

References

1. Stagg and A.H. El-Abiad, "Computer Methods in Power System Analysis", McGraw-Hill, 13th print, New Delhi,1988.

2. William D. Stevenson "Elements Of Power System Analysis", McGraw-Hill, Fourth edition, New Delhi, 1982.

3. George L. Kusic, "Computer Aided Power System Analysis", Prentice-Hall, International, N.J.,New Delhi, 1989.

4. J. Arrillaga, C.P. Arnold and B.J.Harker ,"Computer Modeling of Electrical Power Systems", John Wiley and Sons,1983.