Transmission line fault analysis using bus impedance matrix method

International Journal of Engineering Research & Technology (IJERT)
ISSN: 2278-0181
Vol. 4 Issue 03, March-2015
Transmission Line Fault Analysis using Bus
Impedance Matrix Method
Prajakta V. Dhole1
Farina S. Khan1
First Year Engineering Department, JSPM's Imperial
College of Engg. & Research, Wagholi, Pune
Savitribai Phule Pune University,
Pune, India
First Year Engineering Department, JSPM's Imperial
College of Engg. & Research, Wagholi, Pune
Savitribai Phule Pune University,
Pune, India
Abstract -- The fault analysis is done for the three phase
symmetrical fault and the unsymmetrical faults. The
unsymmetrical faults include single line to ground, line to line
and double line to ground fault. The method employed is bus
impedance matrix which has certain advantages over
thevenin’s equivalent method. The advantage of this approach
over conventional method is to make the analysis of three
typical un-symmetrical faults, namely single-line-to-ground
fault, line-to-line fault and double-line-to-ground fault more
unified. So it is unnecessary to cumbersomely connect three
sequence networks when calculating the fault voltages at each
bus and fault currents flowing from one bus to its neighboring
bus.
Keywords- Bus impedance matrix; fault analysis; fault
impedance; thevenin’s equivalent.
3.
Line-to-line fault (LL)
4.
Double line-to-ground fault (DLG)
The most common type of faults by far is the SLG fault,
followed in frequency of occurrence by the LL fault, DLG
fault, and three-phase fault.
Out of the above four faults, two are of the line-toground faults. Most of these occur as a result of insulator
flashover during electrical storms. The balanced threephase fault is the rarest in occurrence and the least complex
in so far as the fault current calculations are concerned. The
other three unsymmetrical faults will require the
knowledge and use of symmetrical components.
Unsymmetrical faults cause unbalanced currents to flow in
the system. The method of symmetrical components is a
very powerful tool which makes the calculations of
unsymmetrical faults almost as easy as the calculations of a
three-phase fault.
To analyze un-symmetrical faults, one needs to develop
positive-, negative-, and zero-sequence networks of the
power system under study, based on which one further
need to work out the impedance of three thevenin
equivalent circuits as viewed from faulty point. Then the
positive-, negative- and zero-sequence components of
phase-a faulty-point-to-ground current can be calculated.
To calculate three-phase currents flowing from one bus to
its neighboring bus and three-phase voltages at each bus,
one needs to connect three sequence networks uniquely for
each type of fault. This may make circuit drawing very
cumbersome. Furthermore by using the network with three
sequence networks connected, it is impossible to appreciate
the impedance matrix approach to calculate the sequence
voltage at each bus when fault occurs.
To overcome these two drawbacks, paper [1] introduces
a new approach to unify the analysis of three typical
unsymmetrical faults, namely single-line-to-ground fault,
line-to-line fault and double-line-to-ground fault. This new
method allows the analysis of three typical un-symmetrical
faults to share all steps except one. The only different step
is how to calculate the positive-, negative-, and zerosequence components of phase-a-to-ground fault current at
faulty point. It also makes impedance matrix approach
more understandable when used to calculate the sequence
voltages at each bus.
I.
INTRODUCTION
The steady state operating mode of a power system is
balanced 3-phase ac. However due to sudden external or
internal changes in the system, this condition is disrupted.
When the insulation of the system fails at one or more
points or a conducting object comes in contact with a live
point, a short circuit or fault occurs. A fault involving all
the three phases is known as symmetrical (balanced) fault
while one involving only one or two phases is known as
unsymmetrical fault. Majority of the faults are
unsymmetrical. Fault calculations involve finding the
voltage and current distribution throughout the system
during the fault. It is important to determine the values of
system voltages and currents during fault conditions so that
the protective devices may be
set to detect the fault and isolate the faulty portion of the
system.
II. FAULTS IN A THREE PHASE SYSTEM
1.
2.
Symmetrical three-phase fault
Single line-to-ground fault (SLG)
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948
International Journal of Engineering Research & Technology (IJERT)
ISSN: 2278-0181
Vol. 4 Issue 03, March-2015
All the above four faults (1, 2, 3, 4) are being solved
using the bus impedance matrix.
Fig. 3.
Fig. 1.
Double line to ground fault
Single line to ground fault
III. BUS IMPEDANCE MATRIX METHOD
We can work out a universal representation of all three
typical un-symmetrical faults. This representation is valid
with the imposition of different fault conditions for each
typical un-symmetrical fault, such as for the single-line-toground fault, such as for the single-line-to-ground
fault, the fault conditions being Vka=ZfIfa, IfbIfc0.
Fig. 2.
Line to line fault
In the following formulation, per-unit system is
adopted. Zero-sequence voltage at each bus
contributed by equivalent current source is
determined by
 Y11 0 Y12 0 .. Y1k 0 .. Y1n 0   V1f 0   0 
  0
   0  

 0
 0
 0
 Y21 Y22 .. Y2k .. Y2n   V2f   0 
 :
:
..
:
..
:   :   : 

=

 Y 0 Y  0 .. Y  0 .. Y  0   V  0  -Ifa0 
k1
k2
kk
kn
kf



:
:
:
:
:   :   : 
 :
  0
 0
 0
 0    0   0 

 Yn1 Yn2 .. Ynk .. Ynn   Vnf  
where
 Y11 0
  0
 Y21
 0
Y =  :
 Y  0
 k1
 :
  0
 Yn1
IJERTV4IS030919
Y12 0
..
Y1k 0
..
 0
Y22
:
..
..
 0
Y2k
:
..
..
 0
Yk2
:
..
:
0
Y kk
:
..
:
 0
Yn2
..
 0
Ynk
..
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Y1n 0 

 0
Y2n

: 
 0 
Ykn

: 
 0 
Ynn

949
International Journal of Engineering Research & Technology (IJERT)
ISSN: 2278-0181
Vol. 4 Issue 03, March-2015
So
is the admittance matrix for the sub-transient or transient
zero-sequence network.
 0  0 
 V1f 0    -Z1k
I fa
  0     0  0 
 V2f   -Z 2k I fa 

 :  
:


= 
 V  0    -Z  0  I 0  
 kf   kk fa 
:

 :  
  0     0  0 
 Vnf   -Z nk I fa 
Then
 V1f 0    Y  0
  0    11
 V2f   Y21 0
 :  

= :
 V  0    Y  0
 kf   k1
 :   :
  0     0
 Vnf   Yn1
Y12 0
.. Y1k 0
 0
 0
Y22
:
.. Y2k
..
:
Yk2 0
:
.. Y kk0
:
:
Yn2 0
.. Ynk 0
.. Y1n 0 

.. Y2n 0 
..
: 
.. Ykn 0 

:
: 

.. Ynn 0 
 0 
 0 
 
 : 
0
= Z    0  
 -I fa 
 : 
 
 0 
 0
 Z11
  0
 Z21

= :
 Z 0 
 k1
 :
  0
 Zn1
In a similar way, positive-sequence voltage at each bus
contributed by equivalent current source as is determined
by
 Y111
 1
 Y21
 :

 Y 1
 k1
 :
 1
 Yn1
Where,
 0
 Z11
  0
 Z21

 0  :
Z = 0
 Z 
 k1
 :
  0
 Zn1
 0 
 0 
 
 : 
  0 
-Ifa 
 : 
 
 0 
 0
 0
Z12
.. Z1k
Z220 .. Z2k0
: .. :
Zk20 .. Z kk0
:
:
:
Zn20 .. Znk0
 0 
.. Z1n

.. Z2n0 
.. : 
.. Zkn0 

:
: 

.. Znn0 
 0
Z12
..
 0
Z1k
..
Z220
:
..
..
Z2k0 
:
..
..
Zk20
:
..
:
0
Z kk
:
..
:
Zn20
..
Znk0 
..
 0 
Z1n

Z2n0 
: 
Zkn0  

: 

Znn0 
Y121 .. Y1k1 .. Y1n1   V1f1   0 

  
Y221 .. Y2k1 .. Y2n1   V2f1   0 
: .. : .. :   :   : 
=  1 
Yk21 .. Y kk1 .. Ykn1   Vkf1  -Ifa 


:
:
:
:
:   :   : 


Yn21 .. Ynk1 .. Ynn1   Vnf1   0 
This gives
1
 V1f1   Z11

  1
1
 V2f   Z21
 :   :

 =
 V 1   Z1
kf

  k1
 :   :

1   1
 Vnf   Zn1
1
Z12
..
1
Z1k
Z221
:
..
..
Z2k1
:
Zk21
:
.. Z kk1
:
:
Zn21
..
Znk1
1 
.. Z1n

.. Z2n1 
..
: 
.. Zkn1 

:
: 

.. Znn1 
1 1
 0   -Z1k I fa 
 0   1 1 
   -Z 2k I fa 

 :  
:
 1  =  1 1 
 -I fa   -Z kk I fa 

 :  
:


 

1
1




 0  -Z I 
 nk fa 
If pre-fault current is ignored, then the pre-fault voltage
at each bus is the same and equal to that at fault bus k
before fault occurs, which is assumed to be Vf. So the
positive sequence voltage at each bus when fault occurs
can be written as follows.
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International Journal of Engineering Research & Technology (IJERT)
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 V1f1   V11   V1f1   V   -Z1k1 Ifa1 
 1   1   1   f   1 1 
 V2f   V2   V2f   Vf   -Z2k Ifa 
 :   :   :   :   : 

  =  +
=  +
 V 1   V 1   V 1   Vf  -Z 1 I1 
 kf   k   kf     kk fa 
 :   :   :   :   : 
 1   1   1   Vf   1 1 
 Vnf   Vn   Vnf 
 -Znk Ifa 
Item
Base MVA
Voltage Rating
X1
X2
X0
G1
G2
T1
T2
TL1
TL2
TL3
100
100
100
100
100
100
100
20 kV
20 kV
20/220 kV
20/220 kV
220 kV
220 kV
220 kV
0.15
0.15
0.10
0.10
0.125
0.15
0.25
0.15
0.15
0.10
0.10
0.125
0.15
0.25
0.05
0.05
0.10
0.10
0.30
0.35
0.7125
 Vf -Z1k1 Ifa1 


1 1
 Vf -Z2k Ifa 


:

=
 V -Z 1 I1 
 f kk fa 
:



1 1 
 Vf -Znk Ifa 
In a similar way, the negative-sequence voltage at each
bus can be computed by
 2  2 
 V1f 2    -Z1k
I fa
  2    2  2 
 V2f   -Z 2k I fa 

 :  
:


=
 V  2    -Z  2  I 2  
kk fa

 kf  
:

 :  
  2    2  2 
 Vnf   -Z nk I fa 
 2
 Z11
  2
 Z21
 :
Z 2 =  2
 Z 
 k1
 :
  2
 Zn1
TABLEI.
Fig. 4.
Single line diagram 1
IV. MATHEMATICAL ANALYSIS
A. Sequence impedance networks
Firstly let us obtain the sequence impedance
networks. From the data given in table 4.1 the
following positive, negative and zero sequence
impedance networks are obtained in fig.4.6, 4.7 and
4.8 respectively.
 2
 2
 2 
Z12
.. Z1k
.. Z1n

Z222 .. Z2k2 .. Z2n2 
: .. : .. : 
Zk22 .. Z kk2 .. Zkn2 

:
:
:
:
: 

Zn22 .. Znk2 .. Znn2 
POWER SYSTEM NETWORK PARAMETERS
Fig. 5.
IJERTV4IS030919
Positive Sequence impedance network
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951
International Journal of Engineering Research & Technology (IJERT)
ISSN: 2278-0181
Vol. 4 Issue 03, March-2015
Fig. 9.
Fig. 6.
2
1
Ybus
 Ybus
Fig. 7.
Positive Sequence admittance network
Negative Sequence impedance network
j6.667 
-j18.667 j8
  j8
-j16
j4 
 j6.667
j4 -j10.667 
Zero Sequence impedance network
Fig. 10.
Zero Sequence admittance network
j3.3333 j2.8571 
 -j8.690

Y =  j3.3333 -j14.7368 j1.4035 
 j2.8571 j1.4035 -j4.2606
0
bus
B. IMPEDANCE MATRICES
The impedance matrices are obtained from the admittance
matrices.
Z1bus
Fig. 8.
IJERTV4IS030919
Positive Sequence admittance network
=
 j0.1450

= j0.1050

 j0.1300
j0.1050
j0.1450
j0.1200
j0.1300
j0.1200
j0.2200
 j0.1820
Z0bus =  j0.0545
 j0.1400
j0.0545
j0.0864
j0.0650
j0.1400
j0.0650
j0.3500
Z2bus
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International Journal of Engineering Research & Technology (IJERT)
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Vol. 4 Issue 03, March-2015
C. Single Line To Ground Fault At Bus 3 Through A
Fault Impedance Of J0.1
When single line to ground fault occurs, the sequence
components of fault current at bus three are given by
I1f3  I2f3  I0f3 =
Vk (0)
Z +Z2kk +Z0kk +3Zf
1
kk
V3 (0)
= 1
2
0
Z33 +Z33
+Z33
+3Zf
=
=
100
j0.22+j0.22+j0.35+3(j0.1)
1 0 0
j1.09
At bus 1
0 0
 Vf10   0-Z13
I3  0-j0.14(-j0.9174)   0.1284 
 1  1
1 1 =
 =  0.8807 
V
=
V
(0)-Z
13I 3 
 f1   1

1-j0.13(-j0.9174)  
2 2
 Vf12   0-Z13
I3  0-j0.13(-j0.9174)   0.1193
  
At bus2
 Vf20   0-Z023I30 
 1  1
1 1=
 Vf2  =  V2 (0)-Z23I3 
 Vf22   0-Z223I32 
  

0-j0.065(-j0.9174)
1-j0.120(-j0.9174)  =


0-j0.120(-j0.9174)
 0.0596
 0.8899 


 0.1101
 Vf30   0-Z023I30  0-j0.35(-j0.9174) 
 1  1
=
1 1 =
 Vf3  =  V3 (0)-Z33I3  1-j0.22(-j0.9174) 
2 2
 Vf32   0-Z33
I3  0-j0.22(-j0.9174) 
  
 0.3211
 0.7982 


 0.2018
At bus3
The voltages during fault are
= -j0.9174 p.u.
At bus 1
The fault current is
 Iaf3 
1 1
 b

2
 If3  = 1 a
 Icf3 
1 a
 
0

1   If3


a   I 0f3 
a 2   I0f3 
 Vf1a  1 1
 b 
2
 Vf1  = 1 a
 Vf1c  1 a
 
1   Vf10  1 1
2
  
a   Vf11  = 1 a
a 2   Vf12  1 a
1
a 
a 2 
 0.1284 
 0.8807 


 0.1193
1
a 
a 2 
 0.0596 
 0.8899 


 0.1101
1
a 
a 2 
 0.3211
 0.7982 


 0.2018


0.63300
0
1.0046


120.45


 1.0046120.450 


=
3I0f3 
 
=  0 
 0 
 
At bus 2
 Vf2a  1 1
 b 
2
 Vf2  = 1 a
 Vf2c  1 a
 
3(-j0.9174) 

= 
0



0
1   Vf20  1 1
2
  
a   Vf21  = 1 a
a 2   Vf22  1 a


0.720700
= 0.9757  117.430 


0.9757  117.430 


-j2.7522 
=  0 
 0 
At bus 3
 2.7523  900 


= 
0



0


 Vf3a  1 1
 b 
2
 Vf3  = 1 a
 Vf3c  1 a
 
1   Vf30  1 1
  
2
a   Vf31  = 1 a
a 2   Vf32  1 a
The symmetrical components of voltages during fault at
buses 1, 2 and 3
IJERTV4IS030919
www.ijert.org
(This work is licensed under a Creative Commons Attribution 4.0 International License.)
953
International Journal of Engineering Research & Technology (IJERT)
ISSN: 2278-0181
Vol. 4 Issue 03, March-2015


0.275200

0
= 1.0647  125.56 


1.0647  125.560 


V.
VI.
[1]
CONCLUSION
[2]
This paper presents a method to tackle typical unsymmetrical faults. It is found that the bus impedance
matrix method involves comparatively less computations
than the thevenin’s equivalent method. The proposed
approach has another advantage over traditional method
that it is more intuitive when matrix approach is adopted to
tackle a fault problem.
IJERTV4IS030919
REFERENCES
Daming Zhang,”An alternative approach to analyze unsymmetrical faults in power system”. TENCON 2009, from
ieeexplore.
B. R. Gupta, Power System Analysis and Design, published
by S. Chand and Company Ltd., p.265.
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(This work is licensed under a Creative Commons Attribution 4.0 International License.)
954