Sequence Networks
• To analyze unbalanced-faults it is necessary to setup the sequence network which includes the
positive, negative and zero sequence impedances. These impedances are given for all elements of a
power system which includes generators, lines, transformers, etc. Also each unbalanced fault is a
combination of these three networks
Z 0 : Zero _ Sequence
Z 1 : Positive _ Sequence
Z 2 : Negative _ Sequence
The Positive Sequence Network
• The positive sequence network represent the total or equivalent impedance obtained from the reduction
and the system equivalent source
Ia
1
+
Z1
+
Va
1
E1
-
-
The Negative Sequence Network
• The negative sequence network is usually (but not always) the same as the positive sequence network
except that does not include the source.
Ia
2
+
Z2
Va
2
The Zero Sequence Network
• The zero sequence network depends on the transformer and generator connection and similar
to the negative sequence it does include a source
Ia
0
+
Z0
Va
0
-
Transformer Zero Sequence Connection
• The transformers zero sequence impedance is based on the type of connection; such as wye grounded
and ungrounded and delta connections. The are several combinations but the following are the most
commonly used
The wye grounded-wye grounded connection
A
A
B
B
C
C
Equivalent Circuit



Z0

The wye grounded- delta connection
A
A
B
B
C
C
Equivalent Circuit



Wye side grounded
Z0

Delta side open
The wye grounded-wye ungrounded connection
A
A
B
B
C
C
Equivalent Circuit




Single Line to Ground Fault
• In a single line-fault the has the following boundary conditions (assuming phase A is faulted)
Ib = 0
Va = Z f  I a
Based on the initial conditions we set I b and Ic equal to zero
Ic = 0
 I a0 
1 1
 1 1 
 I a  = 3 1 a
2
 I a2 

1
a

 
A
B
1  I a 
a 2    0 
a   0 
We could write the sequence current equations as follow:
C
Ia
Zf
Ib = 0
Ic = 0
I a0 = I a1 = I a2 =
1
Ia
3
Single Line to Ground Sequence Network Connection (ALL IN SERIES)
The positive sequence current is:
Ia
1
I a1 =
+
Z
1
Va
We know that all three-sequence current are equal
1
I a0 = I a1 = I a2
E1
Then, the phase a, b and c currents can be defined as
Ia
I a = 3I a1
2
+
Z
2
Va
2
Ia
0
+
0
Va
3Z f
Ib = 0
Ic = 0
-
Z
E1
Z 0 + Z1 + Z 2 + 3Z f
0
-
EXAMPLE: Give the system find the Single Line to Ground find the line to ground fault at Bus B
Bus A
Bus B
Delta-Ygrounded
X 1Trans = j 0.09
Positive Sequence Network
Z 1 = X 1Gen + X 1Trans = j 0.09 + j 0.05
Z 1 = j 0.14
X 1Gen = j 0.09
E1 = 1.00
X 2Trans = j 0.05
Negative Sequence Network
Z 2 = X 2 Gen + X 2Trans = j 0.08 + j 0.05
Z 2 = j 0.13
X 2 Gen = j 0.08
System Network Impedance for a Single Line to Ground Fault
For the zero sequence network we must consider the transformer connection delta connected
on the primary side (13.6 kV) and Wye grounded on the secondary side (220 kV)
X 0 Gen = j 0.04
X 0Trans = j 0.05
Z 0 = X 0Trans = j 0.05
The generator zero sequence impedance is isolated
since the transformer is isolated because of its delta
wye connection so the generator impedance is not
included on the network
System Sequence Impedance Network
Ia
E1
Z 0 + Z1 + Z 2 + 3Z f
I a1 =
1.00
j 0.05 + j 0.13 + j 0.14 + j 0.0
1
+
Z 1 = j 0.14
I a1 =
Va
1
I a1 = 3.125 − 90
E1 = 1.00
Ia
+
Z = j 0.13
2
Va
Ia
Z 0 = j 0.05
I a0 = I a1 = I a2
2
2
Zf =0
I a = I a0 + I a1 + I a2 = 3I a0
-
I a = 3(3.125 − 90)
+
Therefore, The Single Line to Ground Fault is:
0
Va
0
I a = 9.375 − 90
-