Unbalanced Faults
Faults at the Generator Terminals
1
Unbalanced Faults
• Common unbalanced faults are SLG, LL and LLG faults.
• Unbalanced faults unbalance the network, but only at
the fault location.
• This causes a coupling of the sequence networks.
• How the sequence networks are coupled depends upon
the fault type.
• We’ll derive these relationships for several common
faults.
2
Single Line-to-Ground (SLG) Faults
• With a SLG fault only one phase has non-zero fault current -we’ll assume it is phase A.
 I af
 f
 Ib
 f
 I c
 ? 
  
  0 
 0 
  
3
SLG
Then since
 I 0f 
1  ? 
1 1
 
1
1 f

2 
0


 I f   1    0  I f  I f  I f  I a
 
3
3
 
1  2   0 

 I f 


With the generators normally producing balanced three-phase voltages,
which are positive sequence only, we can write
Va = 0 for a single line to ground fault
V+ + V- +V0 = 0
4
SLG Equivalent Circuit
V   I f ( Z   Z   Z 0 )
I f
Va
Ia
 


0
(Z  Z  Z ) 3
Ia 
3Va
(Z   Z   Z 0 )
Ib  Ic  0
5
Example
Consider a system with sequence impedances given by
Z+ = j0.2577, Z- = j0.2085, and Z0 = j0.14; find the voltages
and currents at the fault point for a single line-to-ground fault.
Solution
The sequence networks are connected in series for a SLG fault.
The sequence currents are given by;
Therefore,
6
Sequence Voltages
7
Phase Voltages
8
Double Line to Ground Fault
• Fault conditions
Ia = 0
Vb= Vc=0
I 
V
Z   Z  // Z 0
Va
V V V 
3


0
V   E  I Z 
I  
I 
o
V
Z
Vo
Zo
9
Double line-to-ground fault
Vb  Vc  0
a
b
Ib
c
Ic
Va 0 
1
V   1 1
 a1  3 
Va 2 
1
Ia  0
1
a
a2
1  Va 
a2  0 
 
a   0 
1
Va1  Va 2  Va 0  Va
3
In
Va 0   0   Z 0
V   V    0
 a1   f  
Va 2   0   0
0
Z1
0
0  I a0 
0   I a1 
Z 2   I a 2 
10
Double line-to-ground fault
 V f  I a1Z1   0   Z 0

   

V

I
Z
a1 1   V f    0
 f
 V f  I a1Z1   0   0


 V f  I a1Z1   0    Z 0

   
 V f  I a1Z1   V f  0
  
 
   
 V f  I a1Z1   0    0
0
Z1
0
0
Z1
0
0  I a0 
0   I a1 
Z 2   I a 2 
0
0 
Z 2 
1
I a0 
  I a1 
 I a 2 
11
Double line-to-ground fault

 V f  I a1Z1  


V

I
Z

V
a1 1
f 
 f
 V f  I a1Z1  



1
Z0
0
0
-
+
Va 1
Vf
Z1
0
-
Z2
Z1
Va 2
-
Z0
Ia2
Vao
+
+
+
I a1
1
Vf
I a1 
Z Z
Z1  2 0
(Z2  Z0 )
I a 0  I a1  I a 2  I a  0
-
0 
 I a0 
0    I a1 

1   I a 2 
Z 2 
0
Ia0
12
Double line-to-ground fault
Example : Find the line currents and the line-to-line voltages when a
double line-to-ground fault occurs at the terminals of the generator
described in single line-to-ground fault example. Assume that the generator
was unloaded and operating at rated voltage when the fault occurred. The
generator has positive sequence reactance of 0.25 , negative sequence
reactance of 0.35 and zero sequence reactance of 0.10 per unit. The neutral
of the generator is solidly grounded. Neglect resistance.
I a1 
Ea
1. 0  j 0
1.0


  j 3.05 p.u.
Z1  Z 2 Z 0 /( Z 2  Z 0 ) j 0.25  ( j 0.35  j 0.10) /( j 0.35  j 0.10) j 0.3278
Va1  Va 2  Va0  Ea  I a1Z1  1  ( j3.05)( j 0.25)
 1.0  0.763  0.237 p.u.
Va 2
0.237

 j 0.68 p.u.
Z2
j 0.35
V
0.237
  a0  
 j 0.2.37 p.u.
Z0
j 0.10
Ia2  
I a0
13
Double line-to-ground fault
I a  I a1  I a 2  I a 0   j 3.05  j 0.68  j 2.37  0
I b  a 2 I a1  aI a 2  I a 0  ( 0.5  j 0.866)(  j 3.05)  ( 0.5  j 0.866)( j 0.68)  j 2.37
 3.230  j 3.555  4.80132.30 p.u.
I c  aI a1  a 2 I a 2  I a 0  ( 0.5  j 0.866)(  j 3.05)  ( 0.5  j 0.866)( j 0.866)  j 2.37
 3.230  j 3.555  4.8047.70 p.u.
I n  3I a 0  3  j 2.37  j 7.11 p.u.
I n  I b  I c  3.230  j 3.555  3.230  j 3.555  j 7.11 p.u.
Va  Va1  Va 2  Va 0  3Va1  3  0.237  0.711 p.u.
Vb  Vc  0
Vab  Va  Vb  0.711 p.u.
Vbc  0
14
Double line-to-ground fault
Vca  Vc  Va  0.711 p.u.
Expressed in amperes and volts,
Ia  0
I b  837  4.80132.30  4017132.30 A
I c  837  4.8047.70  401747.70 A
I n  837  7.11900  5951900 A
Vab  0.711 
13.8
 5.6600 kV
3
Vbc  0
Vca  0.711 
13.8
 5.661800 kV
3
15
Line-to-line fault
Vb  Vc
a
b
Ib
c
Ic
Ia  0
Ib   Ic
 Ia0 
1
 I   1 1
 a1  3 
 I a 2 
1
a
a2
Ia0  0
I a 2   I a1
Va 0 
1
V   1 1
 a1  3 
Va 2 
1
1
1
a
a2
1  0 
a 2   I c 


a   I c 
1  Va 
a 2  Vc 
a  Vc 
16
Line-to-line fault
Va1  Va 2
-
-
+
Va1
Vf
Z1
I a1
+
-
Z2
Va 2
Va 0   0   Z 0
V   V    0
 a1   f  
Va 2   0   0
0
Z1
0
0  I a0 
0   I a1 
Z 2   I a 2 
+
Ia2
0  V f  I a1Z1  I a 2 Z 2
I a1 
Vf
Z1  Z 2
17
Line-to-line fault
Example : Find the subtransient currents and the line-to-line voltages at the fault under
subtransient conditions when a line-to-line fault occurs at the terminals of the generator
described in the sing line-to-ground fault example. Assume that the generator is unloaded
and operating at rated terminal voltage when the fault occurs. Neglect resistance.
I a1 
1.0  j 0
  j1.667 per unit
j 0.25  j 0.35
I a 2   I a1  j1.667 per unit
I a0  0
I a  I a1  I a 2  I a 0   j1.667  j1.667  0
I b  a 2 I a1  aI a 2  I a 0   j1.667(0.5  j 0.866)  j1.667(0.5  j 0.866)
 j 0.833  1.443  j 0.833  1.443  2.886  j 0 per unit
I c   I b  2.886  j 0 per unit
18
Line-to-line fault
in the sing line-to-ground fault example, base current is 837A, and so
Ia  0
I b  2.886  837  2416180 0
I c  2.886  837  241600
Va1  Va 2  1  ( j1.667)( j 0.25)  0.583 per unit
Va 0  0
(Neutral of generator grounded)
Line-to-ground voltages are
Va  Va1  Va 2  Va 0  0.583  0.583  1.16600 p.u.
Vb  a 2Va1  aVa 2  Va 0
Vc  Vb  0.583(0.5  j 0.866)  0.583(0.5  j 0.866)
 0.583 p.u.
19
Line-to-line fault
Line-to-line voltages are
Vab  Va  Vb  1.166  0.583  1.749 per unit
Vbc  Vb  Vc  0.583  0.583  0 per unit
Vca  Vc  Va  0.583  1.166  1.749 per unit
Expressed in volts, the line-to-line voltages are
Vab  1.749 
13.8
 13.9400 kV
3
Vbc  0kV
Vca  1.749 
13.8
 13.94180 0 kV
3
20