Chapter 10: Symmetrical Components and
Unbalanced Faults, Part II
10.4 Sequence Networks of a Loaded Generator
In the figure to the right is a generator supplying a
threephase load with neutral connected through
impedance
n
Z
to ground. The generator generates
threephase balanced voltages defined as:
2
1
abc
a
E a E
a
The terminal voltage of the generator is found
using Kirchoff's voltage law:
a a s a n n
b b s b n n
c c s c n n
V E Z I Z I
V E Z I Z I
V E Z I Z I
And since
n a b c
I I I I
a a s n n n a
b b n s n n b
c c n n s n c
V E Z Z Z Z I
V E Z Z Z Z I
V E Z Z Z Z I
Or in compact form:
abc abc abc abc
V E Z I
. This equation can be transformed to the
"012" or symmetrical component form thus:
012 012 012
abc
a a a
AV AE Z AI
Multiplying by
1
A
012 012 1 012
012 012 012
abc
a a a
a a
V E A Z AI
E Z I
Where
012 1
2 2
2 2
1 1 1 1 1 1
1
1 1
3
1 1
abc
s n n n
n s n n
n n s n
Z A Z A
Z Z Z Z
a a Z Z Z Z a a
a a Z Z Z Z a a
Performing the above matrix multiplication we have:
0
012 1
2
3 0 0 0 0
0 0 0 0
0 0 0 0
s n
s
s
Z Z Z
Z Z Z
Z Z
E
a
E
b
E
c
Z
s
Z
s
Z
s
Z
n
I
a
I
b
I
c
I
n
+
+
+

 
V
a
V
c
V
b
2
It is important to note that the matrices above are diagonal, and the equations are
completely decoupled. Expanding into three voltage equations we have:
0 0 0
1 1 1
2 2 2
0
( )
0
a a
a a a
a a
V Z I
V E Z I
V Z I
These equations
represent three separate
circuits as shown to the
right, one for the
positivesequence, one
for the negative
sequence, and one for
the zerosequence
networks.
The following important observations are made:
The three sequences are independent.
The positivesequence network is the same one used in the oneline diagram for
studying balanced threephase currents and voltages.
Only the positive sequencenetwork has a voltage source. Thus the positive
sequence voltage will generate only positivesequence currents.
There is no voltage source in the negative and zerosequence networks.
Negative and zerosequence currents cause only negative and zerosequence
voltages.
The neutral of the system is the reference point for the positive and negative
sequence networks. The ground is the reference point for the zerosequence;
hence the zerosequence current cannot flow unless there is a connection to
ground.
The grounding impedance is reflected in the zero sequence network as
3
n
Z
.
Each of the three sequence networks can be solved separately on a per phase
basis. The phase currents and voltages can then be found by superposition, by
adding their symmetrical components of current and voltage respectively.
10.5 Single LineToGround Fault
This is the most common fault on a threephase system. It is illustrated using the
following simple network:
E
a
+

V
1
a
+

V
2
a
+

V
0
a
+

I
1
a
Z
1
I
2
a
Z
2
I
0
a
Z
0
Positivesequence Negativesequence Zerosequence
3
Here the fault occurs between line "a" and ground
through a fault impedance
f
Z
. Assuming the
generator was initially at no load, the boundary
conditions at the fault are:
0
a f a
b c
V Z I
I I
Using the second equation in the symmetrical
components of currents we have:
0
1 2
2 2
1 1 1
1
1 0
3
1 0
a a
a
a
I I
I a a
I a a
Therefore
0 1 2
1
3
aaa a
I I I I
. Thus the three symmetrical components are equal and
each is equal to onethird phase "a" current.
Phase "a" voltage in terms of symmetrical components is:
0 1 2
a a a a
V V V V
0 1 2
1
3
aaa a
I I I I
we have:
0 1 2 0
a a a
V E Z Z Z I
Where
0
3
s n
Z Z Z
and
1 2
s
Z Z Z
. Recall that
a f a
V Z I
and that
0
3
a a
I I
, the
above equation becomes:
0 1 2 0 0
3
f a a a
Z I E Z Z Z I
Or:
0
1 2 0
( )
3
a
a
f
E
I
Z Z Z Z
And the fault current is:
0
1 2 0
3
3
3
a
a a
f
E
I I
Z Z Z Z
Using the symmetrical components of current in equation ++ the symmetrical
components of voltage are found, hence by transformation, the voltages during the fault.
Observing the equations
+ and ++ we notice that
the sequence networks
derived earlier may be
connected as shown here
which satisfies these two
equations.
E
a
E
b
E
c
Z
s
Z
s
Z
s
Z
n
I
a
I
b
=0
I
c
=0
I
n
+
+
+

 
V
a
V
c
V
b
Z
f
E
a
+

V
1
a
+

V
2
a
+

V
0
a
+

I
1
a
Z
1
I
2
a
Z
2
I
0
a
Z
0
3Z
f
4
If the generator neutral is grounded, then
0
n
Z
and
0
s
Z Z
, and for a bolted fault
0
f
Z
.
10.6 LineToLine Fault
This fault is shown in the figure to the right. Note the
neutral is not grounded. The fault is shown between
phases b and c. It is assumed the generator is initially on
noload. The boundary conditions at the fault point are:
0
0
b c f b
b c
a
V V Z I
I I
I
Using
0
a
I
and
c b
I I
0
1 2
2 2
1 1 1 0
1
1
3
1
a
a b
a b
I
I a a I
I a a I
Expanding the matrix equation we have:
0
1 2
2 2
0
1
3
1
3
a
a b
a b
I
I a a I x
I a a I
Note that
1 2
a a
I I
0 1 2
0 2 1 2
0 1 2 2
a a a a
b a a a
c a a a
V V V V
V V a V aV
V V aV a V
Hence
2 1 2
b c a a
f b
V V a a V V
Z I
Using the values of
1
a
V
and
2
a
V
from ++, and note that
2 1
a a
I I
2 1 2 1
a a f b
a a E Z Z I Z I
And substituting for
b
I
from (x) above we have:
E
a
E
b
E
c
Z
s
Z
s
Z
s
I
a
=0
I
b
I
c
+
+
+



V
a
V
c
V
b
Z
f
5
1
1 2 1
2 2
3
a
a a f
I
E Z Z I Z
a a a a
Since
2 2
3
a a a a
, solving for
1
a
I
we have:
1
1 2
a
a
f
E
I
Z Z Z
The phase currents are:
2 1
2 1
1 1 1 0
1
1
a
b a
c a
I
I a a I
I a a I
The fault current is:
2 1
1
3
b c a
a
I I a a I
j I
Using the symmetrical components of currents in equation ++, the symmetrical
components of voltage (hence phase voltages by transformation,) at the fault point are
obtained.
Note that
0
0
a
I
and
2 1
a a
I I
10.7 Double LineToGround
Fault
This fault is not as common as the previous two,
but more likely than the symmetrical fault (all
lines bolted, or connected through a fault to the
neutral.) The double linetoground fault is
illustrated on the right. The fault is a short
between lines b and c, which are grounded
through fault impedance
f
Z
. Again, the
generator is assumed initially at noload. The
boundary conditions at the fault point are:
(0.1)
0 1 2
0
b c f b c
a a a a
V V Z I I
I I I I
We know that the voltages
b
V
and
c
V
are given by:
E
a
+

V
1
a
+

V
2
a
+

I
1
a
Z
1
I
2
a
Z
2
Z
f
E
a
E
b
E
c
Z
s
Z
s
Z
s
I
a
=0
I
b
I
c
+
+
+



V
a
V
c
V
b
Z
f
Z
n
6
(0.2)
0 2 1 2
0 1 2 2
b a a a
c a a a
V V a V aV
V V aV a V
And since
b c
V V
1 2
a a
V V
0 2 1 2 0 1 2 2
0 1 2
0
2
3
b f a a a a a a
f a a a
f a
V Z I a I aI I aI a I
Z I I I
Z I
Using (0.4) and (0.3) in the top equation of (0.2) we have:
(0.5)
0 0 2 1
0 1
3
f a a a
a a
Z I V a a V
V V
Using (++) for the symmetrical components of voltage in the equation above, and solving
for
0
a
I
we have:
(0.6)
1 1
0
0
3
a a
a
f
E Z I
I
Z Z
Using the equation (++) in (0.3) we have:
(0.7)
1 1
2
2
a a
a
E Z I
I
Z
Now using the above two equations in the second equation of (0.1) we have (after some
algebra:)
(0.8)
1
2 0
1
2 0
3
3
a
a
f
f
E
I
Z Z Z
Z
Z Z Z
Note that
a
E
drives the current
1
a
I
, hence the impedance seen by
a
E
is:
2 0
1
2 0
3
3
f
f
Z Z Z
Z
Z Z Z
. Clearly,
this is the impendence
1
Z
in
series with the parallel
combination of
2
Z
and
0
3
f
Z Z
. Thus the sequence
networks can be connected as
shown in the circuit on the
right. Once the sequence
currents are found from (0.8),
(0.7) and (0.6) [in that order,] the phase currents are found using the transformation
matrix
A
. Finally the fault current is found from:
(0.9)
0
3
f b c a
I I I I
E
a
+

V
1
a
+

V
2
a
+

V
0
a
+

I
1
a
Z
1
I
2
a
Z
2
I
0
aZ
0
3Z
f
7
[This is seen from the equation
0
1
3
a a b c
I I I I
and knowing in this case
0
a
I
.]
Example 10.5
The oneline diagram of a simple power
system is shown to the right. The neutral
of each generator is grounded through a
currentlimiting reactor of 0.25/3 per unit n
a 100MVA base. The system data
expressed in per unit on a common 100
MVA base is tabulated below. The
generators are running on noload at their
rated voltage and rated frequency with their
emfs in phase.
Determine the fault current for the
following faults:
(a) A balanced threephase fault at bus
3 through fault impedance
0.1
f
Z j
per unit.
(b) A single linetoground fault at bus 3 through fault impedance
0.1
f
Z j
per unit.
(c) A linetoline fault at bus 3 through fault impedance
0.1
f
Z j
per unit.
(d) A double linetoground fault at bus 3 through fault impedance
0.1
f
Z j
per
unit.
Item Base MVA Voltage rating
1
X
2
X
0
X
1
G
100 20 kV 0.15 0.15 0.05
2
G
100 20 kV 0.15 0.15 0.05
1
T
100 20/220 kV 0.1 0.1 0.1
2
T
100 20/220 kV 0.1 0.1 0.1
12
L
100 220 kV 0.125 0.125 0.3
13
L
100 220 kV 0.15 0.15 0.35
23
L
100 220 kV 0.25 0.25 0.7125
Preliminary calculations:
First we find the Thevenin impedance viewed from bus 3, the faulted bus. First the delta
is changed to a Y as can be seen in the figure below on the right. Now the Yimpedances
are calculated thus:
1
2
3
G
1
G
2
T
1
T
2
8
1
2
3
0.125 0.15
0.0357143
0.525
0.125 0.25
0.0595238
0.525
0.15 0.25
0.0714286
0.525
S
S
S
j j
Z j
j
j j
Z j
j
j j
Z j
j
Combining parallel branches, and
adding the series branch, the positive
sequence Thevenin impedance is:
1
33
0.2857143 0.3095238
0.0714286
0.5952381
0.22
j j
Z j
j
j
The positive and negativesequence networks
are shown to the right. The only difference
between them in this case is that the source is
missing in the negativesequence network.
Hence we have:
2 1
33 33
0.22
Z Z j
Now, the zerosequence
network is constructed
based on the transformer
connections and is shown
in the figures to the right.
Note that the delta is
reduced to its equivalent Y
first, and then the
impedances are combined
to find the equivalent zero
sequence circuit. The
result is the simple circuit
shown below.
Zerosequence
j0.35
j0.25
j0.15
j0.25
j0.125
1
2
3
j0.25
1
2
3
j0.25
j0.25
j0.035714
j0.059524
j0.071428
S
j0.25
j0.35
j0.7125
j0.3
1
2
3
j0.25
j0.05
j0.05
j0.1
j0.1
j0.25
j0.25
j0.05
j0.05
j0.1
j0.1
1
2
3
S
j0.183026
j0.156881
j0.077064
E
a
+

j0.22
Positivesequence Negativesequence
j0.22
9
Now we are ready to solve the problem!
(a) Balanced threephase fault at bus 3.
We assume the generators at noload have a voltage of 1.0 per unit, hence the fault
current is:
3
3
1
33
0
1.0
3.125 pu
0.22 0.1
=820.1 90 V
a
a
f
V
I j
Z Z j j
(b) Single linetogroud fault at bus 3.
We know the sequence currents are the same and given by:
0 1 2 3
3 3 3
1 2 0
33 33 33
0
3
1.0
0.22 0.22 0.35 3 0.1
0.9174 pu
a
f
V
I I I
Z Z Z Z
j j j j
j
And using the matrix A we have:
0 0
3 3 3
2 0
3 3
2 0
3 3
1 1 1 3 2.7523
1 0 0 pu
1 0 0
a
b
c
I I I j
I a a I
I a a I
(c) Linetoline fault at bus 3.
The zerosequence component of current is zero, i.e.
0
3
0
I
. Also from the analysis of
the linetoline done previously we have:
1 2 3
3 3
1 2
33 33
0
1
1.8519 pu
0.22 0.22 0.1
a
f
V
I I j
Z Z Z j j j
The fault current is:
3
2
3
2
3
1 1 1 0 0
1 1.8519 3.2075 pu
1 1.8519 3.2075
a
b
c
I
I a a j
I a a j
(d) Double linetoground fault at bus 3.
From the equations developed earlier for the sequence currents of this fault we have:
10
1 3
3
2 0
33 33
1
33
2 0
33 33
0
1
2.6017 pu
0.22 0.35 0.3
3
0.22
0.22 0.35 0.3
3
a
f
f
V
I j
j j j
Z Z Z
j
Z
j j j
Z Z Z
1 1
2 3 33 3
3
2
33
0 1 0.22 2.6017
1.9438
0.22
a
V Z I j j
I j
Z j
1 1
0 3 33 3
3
0
33
0 1 0.22 2.6017
0.6579
3 0.35 0.3
a
f
V Z I j j
I j
Z Z j j
And using the A matrix we have the phase currents:
3
2
3
2
3
1 1 1 0.6579 0
1 2.6017 4.058 165.93 pu
1 1.9438 4.058 14.07
a
b
c
I j
I a a j
I a a j
The fault current is:
3 3 3
1.9732 90
b c
I F I I
.
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