# Chapter 10: Symmetrical Components and Unbalanced Faults, Part II

Electronics - Devices

Oct 13, 2013 (5 years and 1 month ago)

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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
three-phase load with neutral connected through
impedance
n
Z
to ground. The generator generates
three-phase 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
positive-sequence, one
for the negative-
sequence, and one for
the zero-sequence
networks.

The following important observations are made:

 The three sequences are independent.
 The positive-sequence network is the same one used in the one-line diagram for
studying balanced three-phase currents and voltages.
 Only the positive sequence-network has a voltage source. Thus the positive-
sequence voltage will generate only positive-sequence currents.
 There is no voltage source in the negative- and zero-sequence networks.
 Negative- and zero-sequence currents cause only negative- and zero-sequence
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 zero-sequence;
hence the zero-sequence 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 Line-To-Ground Fault

This is the most common fault on a three-phase 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
Positive-sequence Negative-sequence Zero-sequence
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 t-he 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 one-third 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 Line-To-Line 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
no-load. 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 Line-To-Ground
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 line-to-ground 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 no-load. 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 one-line diagram of a simple power
system is shown to the right. The neutral
of each generator is grounded through a
current-limiting reactor of 0.25/3 per unit n
a 100-MVA base. The system data
expressed in per unit on a common 100-
MVA base is tabulated below. The
generators are running on no-load at their
rated voltage and rated frequency with their
emfs in phase.

Determine the fault current for the
following faults:

(a) A balanced three-phase fault at bus
3 through fault impedance
0.1
f
Z j
 per unit.
(b) A single line-to-ground fault at bus 3 through fault impedance
0.1
f
Z j
 per unit.
(c) A line-to-line fault at bus 3 through fault impedance
0.1
f
Z j
 per unit.
(d) A double line-to-ground 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 Y-impedances
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 negative-sequence networks
are shown to the right. The only difference
between them in this case is that the source is
missing in the negative-sequence network.
Hence we have:
2 1
33 33
0.22
Z Z j 
Now, the zero-sequence
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.
Zero-sequence
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
Positive-sequence Negative-sequence
j0.22
9
Now we are ready to solve the problem!
(a) Balanced three-phase fault at bus 3.

We assume the generators at no-load 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 line-to-groud 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) Line-to-line fault at bus 3.

The zero-sequence component of current is zero, i.e.
0
3
0
I

. Also from the analysis of
the line-to-line 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 line-to-ground 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   

.