Short Circuit Studies
Indonesia Clean Energy Development (ICED) project
Indonesia Wind Sector Impact Assessment
Presented 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.
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
I a
I
b
 I c




=
 I a1 I a 2
I
 b1 I b 2
 I c1 I c 2
I a0 
I b 0 
I c 0 
Ib2
Ic1
Ic
120
Ia
Ib
Unbalanced
current phasor
120
120
Ib1
Positive
sequence current
phasors
Ia1
120
Ia0
120
120
Ia2
Ib0
Ic0
Ic2
Negative
sequence current
phasors
Zero sequence
current phasor
Similarly for voltages
Va Va1  Va2  Va0
Vb Vb1  Vb2  Vb0
Vc Vc1  Vc2  Vc0
 Va  1 1 1  Va1 

   2

V

a
a
1
V
  a2 
 b 
 Vc  a a 2 1  Va0 


1 1 1


Tsp  a 2 a 1
a a 2 1


a  1120, a 2  1240  1  120
and sp  Sequenceto phase
1 a a 2 

1
T ps  1 a 2 a 
3
11 1 


ps  Phasetosequence
Va1 
V  1
 a2  3

Va 0 

Va 0 
1 a a 2  Va 



2
1
a
a
V

 b 
1 1 1  V 

 c 
1
Va  Vb  Vc 
3
Vab  Vbc  Vca  0 always
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.
Sequence Currents
Delta Connection
1 a a 2   I a 
 I a1 
 I   1 1 a 2 a   I 
 b
 a2  3 
1 1
 I 
 I a 0 
1

 c
1
I a 0   I a  Ib  I c 
3
a
Ia
I a+ I b + I c = 0
b
Ib
c
Ic
Therefore no zero
sequence current flows
into delta connection.
I0
a
Ia
a
Ia
3I0
I0
Ib
b
Ib
I0
Ic
c
Star connection
Ia+Ib+Ic = 0, There fore no
zero sequence current
flows into star connection
Ic
b
c
Star grounded
Ia+Ib+Ic may not be zero.
Hence path always exists for
zero sequence currents.
Transformer sequence impedance diagram
Star - Delta
+ve and -ve
Zero
Transformer sequence impedance diagram
Star grounded-Delta
+ve and -ve
Zero
Three winding Transformer
Sequence Impedance
a
Zaa
a’
Zbb
b’
b
Zcc
c’
c
 V '  Z aa
 aa '  
 Vbb Zba
 V '  Z
 cc   ca
Z ab
Zbb
Z cb
Z ac Ia 



Zbc Ib 
Z cc Ic 
Let Zaa = Zbb = Zcc = Zs and Zab = Zac = Zba = Zbc =
Zca = Zcb = ZmTsp Vs = [Z] . Tsp.Is
 12 1 1 Va1  Zs Zm Zm   12 1 1 Ia1 
a a 1 Va2   Zm Zs Zm  a a 1 Ia2 

  

 
2
2
 a a 1 Va0  Zm Zm Zs   a a 1 Ia0 
Va1  Zs  Zm
0
0  Ia1 
Va2    0
 Ia2 
s  Zm
Z
0
  
 
Va0   0
0
Zs 2Zm  Ia0 
Zs  Zm
0
0 
Z1,2,0  Zs   0
Zs  Zm
0 
 0
0
Zs  2Zm 
Rotating Machines
• Za,b,c is not symmetric. Even then, the Z1,2.0 is
diagonalized
Va   Zs
V   Z
 b   m2
Vc   Zm1
Zm1
Zs
Zm2
Zm2   I a 
Zm1   Ib 
Zs   I c 
Exercise:
• Find the expression for Z1,2,0
• and Prove that it is a diagonal matrix:
Steady State
Ia
+
In
Zn
a
Ea
Ec
Eb
+
+
Ib
b
Ic
c
Positive Sequence Network
Ia1
a
Reference Bus
Z1
-
Ea
Ec
Ea
Z1
Eb
Z1
Z1
Ib1
b
Ic1
c
+
Va1
Ia1
Va1 = Ea - Ia1Z1
a
Negative Sequence Network
Ia2
a
Reference Bus
Z2
Va2
Z2
Z2
Z2
Ib2
a
b
V a2 = - Ia2 Z2
Ic2
c
Zero Sequence Network
Ia0
a
Reference Bus
Zg0
3Ia0
Zn
Zg0
3Zn
Z0
Va0
Zg0
Zg0
a
Ib0
b
V a3 = - Ia0 Z0
Ic0
c
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
Fault Representation
Vf
Va1
Va2
Z1
Z2
Ia1
Z0
Ia2
Va1 = Vf - Z1.Ia1
Va2 = -Z2 . Ia2
 Va1 


V
 a2 

 Va0 

Va0
Ia0
Va0 = - Z0 . Ia0
 Vf   Z100  Ia1 
0 0Z 0  I 
2
  
  a2 

0 
 
00 Z 0 

Ia0 

Three phase fault representation
a
a
b
b
c
c
Vab = 0
Vbc = 0
Vca = 0
Va = 0
Vb = 0
Vc = 0
Va1 = 0
Va2 = 0
Va0 = 0
0  Vf   Z100 Ia1 
00 0 Z 0I 
2   a2 
    
0 0  00Z 0 Ia0 
Ia1 = Vf  Z1 , Ia2 = 0 , Ia0 = 0
Ia  1 1 1  Ia1 
 
   2
Ib a a 1 Ia2 
Ic  a a 2 1 Ia0 


Ia = Ia1
Ib = a2Ia1
Ic = aIa1
 Ia  =
 Ib  =  Ic 
Fault through impedance :
a
Zf
a
Zf
a
Zf
a
Zf
Z0
Vf
Ia1
, Ia2  0 , Ia0  0
Z1  Z f
a
Zf
a
Zf
Single line to ground fault representation
a
b
Va = 0
c
Ib = 0
Ic
Ib
Ia
Ic = 0
2
a
a

1
a
1
I
  1  2 
I a 2   1 a a 
  3
I a 0  1 1 1 
I a 
I b 
 
I c 
Va1  V f
Va 2 0
  
Va 0  0
 Z1 0 0 I a1 
0 Z 2 0 I a 2 
 
 0 0 Z 0 I a 0 
I a1  I a 2  I a 0
Va1 V f  Z 1 I a1
V a 2  Z 2 I a 2
Va 0  Z 0 I a 0
Va 1  Va 2  Va 0
QVa  Va1  Va 2
I a1  I a 2  I a 0
3V
 V f  Z 1 I a1  Z 2 I a 2  Z 0 I a 0
 Va 0 0V
f
 Z1  Z 2  Z 0
I a  Z1  Z 2  Z 0
f
Fault through Impedance
a
b
Ib
c
Ic
Z
Ia1  Ia2  Ia0 
f
Ia
Vf
Z1  Z2  Z0  3z f
Line to line fault representation
a
Vb = V c
Ia
b
Ia = 0
c
I b = -I c
Ib
Ic
2

 Va 
1
a
a
 Va1 


 1

2
V

1
a
a
V
 b 
 a2  3 
1 1 1  V 


V
 a0 



 c 
Va1 = 1/3 [Va+aVb+a2Vc] = 1/3 [Va+aVb +a2Vb]
Va2 = 1/3 [Va+a2Vb+aVc] = 1/3 [Va+a2Vb+aVb]
Va1 = Va2
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
Va1 = Vf - Z1Ia1
Va2 = Z2Ia2 = Va1 = Vf - Z1Ia1
1
I a 1 
I   1 1
 a2  3 

I a 0 
1
a
a2
1
a 2  I a 
 
a  I b 

1  I c 

Va1  V f   Z1 00  I a1 
    
 I 
V

0

0
Z
0
 a 2     2  a1 
Va 0  0  00Z 0  0 
 
Z2Ia1 = Vf - Z1Ia1
Vf
 I a1 
and
I
a 2   I a1 , I a 0  0, Va 0  0
Z1  Z2 
Fault through impedance
a
Ia
b
c
Ib
Ic
Zf
Ia1   Ia 2 
Z
1
Vf
 Z2  Z f

Double line to ground fault representation
a
Ia
b
c
Vb = 0
Ib
Vc = 0
Ic
Va1  1 1
Va2   1
  3
Va0  1

a a2  Va 
a2 a  0 
1 1  0 
Ia = 0
1
1
1
Va1  Va , Va2  Va , Va0  Va
3
3
3
Va1  Va2  Va0
Ia1 + Ia2 + Ia0 = IA =0
Ia1 = 1/3 (aIb + a2Ic)
Ia2 = 1/3 (a2Ib + aIc)
Ia0 = 1/3 (Ib + Ic)
1
I a 1 
I   1 1
 a2  3 

I a 0 
1
a
a2
1
a2 

a 

1 

0 
I 
 b
I c 
1/3(aIb + a2Ic) + 1/3(a2Ib + aIc) + 1/3(Ib +Ic) = 0
aIb + a2Ic + a2Ib + aIc + Ib + Ic = 0
Va1 = Vf - Z1Ia1
Va2 = - Z2Ia2
Va0 = - Z0Ia0
Va1 = Va2 = Va0
Vf - Z1Ia1 = - Z2Ia2 - Z0Ia0

Vf
I a1  
 Z2Z0  

 Z 1  

Z 2 Z 0  

Va1  Vf  Z1
V   0   0
 a2    
Va 0  0  0
0 0  I a1 
Z2 0  I a 2 
0 Z0  I a 0 
V
V
f
V
a1
Z1
Z2
I a2
I a1
Fault through impedance:
a
b
c
Zf
Zf
Zg
V
a2
a0
Z0
I ao
Ia1 
Vf
 Z2' Z0' 
'
Z1   '
' 

 Z2 Z0 
Z1'  Z1  Z f
Z2'  Z2  Z f
Z0'  Z0  Z f  3Zg
Open in phase B & C
Ib = 0 and Ic = 0
a
Vi
a’
b’
c’
a
b
c
'
'
 Vka  Zaa

I a Zaa I f
1
I a1 
 I   1 1
 a2  3 
 I a 0 
1
a
a2
1
a2 

a 
1 
I a 
0 
 
 0 
1
1
I a1  I a 2  I a 0  I a  I f
3
3
Va1  Vf  Z1I a1
Va 2   Z2 I a 2
Va 0   Z0I a 0
Va1  Vf  Z1
V   0   0
 a2    
Va 0  0  0
0 0  I a1 
Z2 0  I a 2 
0 Z0  I a 0 
1
'
Vka  Vla  Vkl
 Vkl2  Vkl0  Z 'zzI f  3Zaa
I a1
Vf  Vkl
'
3Zaa
I a1  Vkl  Z1I a1  Z2 I a 2  Z0I a 0
I a1  I a 2  I ao 
Vkl
'
Z1  Z2  Z0  3Zaa
• Vkl is the Thevenin’s equivalent voltage, once the
line is removed between buses k & l
Z1  Z1kk  Z1ll  Z1lk  Z1kl
2
2
2
Z 2  Z kk
 Z ll2  Z lk
 Z kl
0
0
0
Z 0  Z kk
 Z ll0  Z lk
 Z kl
3Z 'aa  Z f  Z1f  Z 2f  Z 0f
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.
I lf
Vk  Vl

Z1' 
I 2f

Z '2 . Z '0
Z '2  Z '0
I1f .
Z '0
Z '0  Z '2
Z1l  Z1kk  Z1ll  Z1lk  Z1kl  Z1f
2  2  2
Z l2  Z 2kk  Z ll2  Z lk
Z kl Z f
0  0  0
Z l0  Z 0kk  Z ll0  Z lk
Z kl Z f
I 0f
I1f . Z '2

Z '0  Z '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 required
Consider, Yx = z
x=Y-1z
x= Z.z
Let z have 1 in pth location and 0, at all other location
Solution methodology
Z11 Z12 L Z1p Z1n  0 Z1p 

   
Zp1 Zp2 L Zpp Zpn 1 Zpp
 
x  



M
 M  M 
Zn1 Zn2 L Znp Znn 0 Znp
Yx z
•
•
•
•
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 I1f, I2f and I0f.
Determine the post fault bus voltage
[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 faults other than open fault,
 I f p
 0

 0
 :
   I f
 :
 0












p
For open fault, between k & l,

I
f
p
 0
 M0
 M
 k   I
 M
 M I
 0
f
f







Example:
Xd”=24%
High resistance
Grounding
50 MVA ,13.8/138 kV
11%
G
F
40 Mile, 0.77 Ohm/mile X.
X0 = 3X
E
D
100 MVA ,138/13.8 kV
11%
C
B
Xd”= 9%
A
20% on 100 MVA
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 restriking 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
communication lines is more.
• Ground current can’t be limited
system,
influence
on
Ungrounded system
a
b
c
1.73
3.0
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.
Resistance grounded system :
• The resistance to neutral limits ground current.
Selection of resistance value:
• Amount of ground fault current
Power loss in resistor during ground fault
• Power loss: usually expressed as a percentage of
system rating
R
Xg = 8%
Xg = 16 %
Z1 = 24% , Z2=24% , Z0=3R+j8
I
f

3
Z Z
1
2
Z
Power Loss  I
2
f

0
3  100
j24  j24  3 R  j8


300

R 
 3 R  j56 
2
.R

300
3 R  j56
If : in pu, R : in pu
For three phase system, power loss

I
2
f
R
percentage of 3 phasesystem kVA rating
3
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 ?
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 :
X 0 system
X 1 system
Ro
 1.0
X 1 system
 3.0
69 kV
X1 =
X2 =25%
X1 =7%
=X2=X0
A
69 kV
25 MVA
X1 =34%
X2 =120%
B
For fault at A
X0
For fault at B
X0

X1
X1
7
32

127
127

32  34
66
Re ac tan ce Grounded System :
X0
X1
 3.0 ; but no resonance
resonance
grounding

If Xg = 50  at A , find the nature of grounding at A & B.
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
I
0
0
I
I2
0
I
I
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.