Lecture 10
Unbalanced Fault Analysis
Dr Mostafa Elshahed
Cairo University
Single Line-to-Ground (SLG) 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.
With a SLG fault only one phase has non-zero fault
current -- we’ll assume it is phase A.
1
1
SLG Faults, cont’d
f
 Ia
 f
 Ib
 f
 I c
 ? 
  
  0
 0
  
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 


2
SLG Faults, cont’d
f
Va
f
Va
 f
Vb
 f
Vc

f
Z f Ia
0


V
 f

1
1 1




 
2
  V f 
  1 



1   2  

 V f 

This means
f
Va
0
 Vf

Vf

Vf
The only way these two constraints can be satisified
is by coupling the sequence networks in series
3
2
SLG Faults, cont’d
With the
sequence
networks in
series we can
solve for the
fault currents
(assume Zf=0)

If
1.050

0

  j1.964  I f  I f
j (0.1389  0.1456  0.25  3Z f )
I  A Is 
f
Ia
  j 5.8 (of course,
f
Ib

f
Ic
 0)
4
Line-to-Line (LL) Faults

The second most common fault is line-to-line,
which occurs when two of the conductors come in
contact with each other. With out loss of generality
we'll assume phases b and c.
Current
f
Relationships: I a
Voltage Relationships:
 0,
f
Ib

f
 Ic
,
0
If
0
Vbg  Vcg
5
3
LL Faults, cont'd
Using the current relationships we get
0
I f 
1  0
1 1
 
1

2 f
 I f   1     I b
3
 
1  2     I f

 I f 

 b
0
If
 0

If
1 f
2
 Ib   
3
Hence


If


If






1 f 2
 Ib   
3



I f

6
LL Faults, con'td
Using the voltage relationships we get
V f0 
1
 
1

V f   1
3
 
1

V f 

Hence
1 f

Vf 
Vag 
3
1 f

V f  Vag 
3
To satisfy

If
1


2
f 

1  Vag
 

2
f

  Vbg  
 

  Vcgf 
 
   
f 
Vbg


f 
Vbg 
2

2


I f

&

Vf



Vf


Vf

Vf
the positive and negative sequence networks must
be connected in parallel
7
4
LL Faults, cont'd
Solving the network for the currents we get

If
1.050

 3.691  90
j 0.1389  j 0.1456
f
 Ia
 f
 Ib
 f
 I c

1 
1 1
0
  0 



2



  3.691  90  6.39
  1 

 


1   2   3.69190   6.39 



8
LL Faults, cont'd
Solving the network for the voltages we get

Vf
 1.050  j 0.1389  3.691  90  0.5370

Vf
  j 0.1452  3.69190  0.5370
Vaf
 f
Vb
 f
Vc

1   0   1.074 
1 1



2



  0.537  0.537
  1 

 


1   2  0.537   0.537 



9
5
Double Line-to-Ground Faults
With a double line-to-ground (DLG) fault two line
conductors come in contact both with each other
and ground. We'll assume these are phases b and c.
f
Ia
f
Vbg
0
f
 Vcg

f
Z f ( Ib

f
Ic
)
10
DLG Faults, cont'd
From the current relationships we get
f
 Ia
 f
 Ib
 f
I
 c
0


I
 f

1
1 1




 
2
  I f 
  1 



2
1    

  I f 

Since
f
Ia
0

0
If


If


If
0
Note, because of the path to ground the zero
sequence current is no longer zero.
11
6
DLG Faults, cont'd
From the voltage relationships we get
f


V
ag

0
V f 
1
1 1
 


1

2
f
V f   1    Vbg  
3
 


2
1 


f

V f 

 Vbg 
Since
f
Vbg
Then
f
Vbg
f
 Vcg

0
Vf

Vf

 (
2

 Vf

  )V f
2
2
But since 1      0 
f
Vbg
0
 Vf
    1

Vf
12
DLG Faults, cont'd
f
Vbg

0
Vf

Vf

f
Z f ( Ib

f
Ic
)
Also, since
f
Ib
f
Ic

0
If
2 
 I f

0
If

I f

I f
2 
 I f
2
Adding these together (with     -1)
f
Vbg
0
Vf

Vf

0
Z f (2 I f

0
3Z f I f


If


If )
with
0
If


I f


If
13
7
DLG Faults, cont'd
The three sequence networks are joined as follows
Assuming Zf=0, then

V
1.050

If  


0
Z  Z // ( Z  3Z f ) j 0.1389  j 0.092
 4.547  0
14
DLG Faults, cont'd

Vf
 1.05  4.547  90  j 0.1389  0.4184

If
 0.4184 / j 0.1456  j 2.874
0
If


I f


If
 j 4.547  j 2.874  j1.673
Converting to phase:
f
Ib
f
Ic
 1.04  j 6.82
 1.04  j 6.82
15
8
Example 9.3
Determine pu and actual value of the current and line voltage at
generator bus, when a SLG fault occurs at bus 2.
16
Fault Impedance (3-ph Fault)
17
9
Fault Impedance (SLG Fault)
18
Fault Impedance (LL Fault)
(LL Fault)
(DLG Fault)
19
10
Fault Variables

Fault Current (If )

Voltage at faulted bus (Vf = If Zf)

Voltage of healthy buses

Currents flow on transmission lines

Generators currents
You must calculate the phase values.
20
Fault Summary




3-ph: Positive sequence only
SLG: Sequence networks are connected in series,
parallel to three times the fault impedance
LL: Positive and negative sequence networks are
connected in parallel; zero sequence network is not
included since there is no path to ground
DLG: Positive, negative and zero sequence
networks are connected in parallel, with the zero
sequence network including three times the fault
impedance
21
11