Power SystemAnalysis and Control (EET 408)
Laboratory Module
EXPERIMENT 6
ANALYSIS THE UNSYMMETRICAL FAULT
1. OBJECTIVE:
To analysis the Unsymmetrical fault by means of MiPower soft ware
2. EQUIPMENT:
MiPower software
3. INTRODUCTION:
For unsymmetrical fault, the solution be used the symmetrical components, by applying
Thevenin’s Theorem, which allows us to find the current in the fault by replacing the
entire system with a single generator and series impedance, and the bus impedance
matrix is applied to the analysis of unsymmetrical faults.
Actually in a symmetrical fault we used the positive-sequence network to solve for the
fault current, and voltages and currents at various other busses during the fault. It is also
possible to create bus impedance matrices for negative- and zero- sequence networks,
and they can be strung together in different ways to determine voltages and currents for
unsymmetrical faults in a power system.
And we have seen that when the network is balanced, the symmetrical components
impedances are diagonal, so, that is possible to calculate Zbus separately for zero-,
positive-, and negative- sequence networks. And also, we have observed that for a fault
at bus k, the diagonal element in the k axis of the bus impedance matrix Zbus is the
Thevenin impedance to the point of fault. In order to obtain a solution for the
unsymmetrical faults, the bus impedance matrix for each sequence network is obtained
separately. The fault formulas for various unsymmetrical faults are summarized as below.
1. Assumed the fault occurs at phase a for single line-to ground, phase b and c for
double line-to-ground and double line-to-ground fault.
2. The fault occurs at bus k through fault impedance Zf.
3. The diagonal impedance are Zkk,0, Zkk,1 and Zkk,2 for zero-, positive- and negative
sequence networks.
4. Voltage for prefault condition in bus k = Vf
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
So, we can give for each unsymmetrical fault using [Z] as below
1. SINGLE LINE-TO-GROUND FAULT
Consider a fault between phase a and ground through an impedance Zf at bus k as
shown in Fig1. The line to-ground fault requires that positive-, negative- and zerosequence networks. Thus, in general for a fault at bus k, the symmetrical components of
fault current is
I k ,a 0  I k ,a1  I k ,a 2 
V fk ,a
Z kk ,a 0  Z kk ,a1  Z kk ,a 2  3Z f
1
The fault phase current is
I k ,abc  CI k ,012
2
Bus k of network
b
a
c
Zf
Fig 1 Single line-to-ground fault at bus k
2 DOUBLE LINE FAULT
For this fault, the configure sequence network as Fig. 2.
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
Bus k of network
a
b
c
Zf
Fig 2 Double line fault at bus k
The symmetrical components of the fault current as given below
I k ,0  0
I k ,1  I k , 2 
Vk , f
3
Z kk ,1  Z kk , 2  Z f
And the fault phase current is
I k ,abc  CI k ,012
4
3. DOUBLE LINE-TO-GROUND FAULT USING [z]
The diagram sequence network for double line-to-ground fault through an impedance Zf
as Fig 3. follow
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
Bus k of network
b
a
c
Zf
Fig 3. A double line-to-ground fault through an impedance Zf at bus k
The symmetrical components of the fault current as given
I ka1 
Z kk ,a1 
I k ,a 2 
Vf k ,a1
Z kk ,a 2 ( Z kk ,a 0  3Z f )
5
Z kk ,a 2  Z kk ,a 0
I k ,a1 Z kk ,a1  Vfk, a1
Z kk ,a 2
6
and
I k ,a 0 
I k ,a1 Z kk ,a1  V fk , a1
Z kk ,a 0
7
The phase fault current at bus k
I k ,abc  CI k ,012
8
Using the sequence components of the fault current the symmetrical components of the
ith bus voltages during fault are obtained
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
Vif ,a 0  0  Z ik ,a 0 I k ,a 0
Vif ,a1  Vif ,a1  Z ik ,a1 I fk ,a1
9
Vif ,a 2  0  Z ik ,a 2 I fk ,a 2
where
Vif,a1 = phase voltage at phase a when prefault at bus i.
The phase voltages during fault are
Vi ,abc  CVi , 012
10
The symmetrical components of fault current in line i to j is given by
I ij ,a 0 
I ij ,a1 
I ij ,a 2 
Vif ,a 0  V jf ,a 0
z ij ,a 0
Vif ,a1  V jf ,a1
z ij ,a1
11
Vif ,a 2  V jf ,a 2
z ij ,a 2
where
zij,a0, zij,a1 and zij,a2 are the zero-, positive- and negative- sequence components of the
actual line impedance between buses i and j. having obtained the symmetrical
components of current, the phase fault current in line i to j is
I ij,abc  CI ij, 012
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
PROCEDURE
Figure 6.1
Line
No
Structure
Ref. No.
Bus
Code
p–q
+ve Seq
Impedance
+ve Seq.
Zero Seq
Susceptance Impedance
Zero Seq.
Susceptance
1
1
1–4
0.035 +0.225
0.0065
0.175+j0.45
0.0039
2
2
1-6
0.04 + 0.215
0.0055
0.2+j0.430
0.0033
3
3
1-5
0.025+j0.105
0.0045
0.125 +j0.21
0.0027
4
4
4–6
0.028 +j0.125
0.0035
0.14 +j0.25
0.0021
5
5
5–6
0.026 +j0.175
0.030
0.13 +j0.35
0.018
* De-Rated MVA for all Transmission Line is 100 MVA
Table 6.1
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
Bus Data
Bus Number
Bus Name
1
Bus 1
2
Bus 2
3
Bus 3
4
Bus 4
5
Bu 5
6
Bus 6
11
3.3
3.3
11
11
11
Nominal Voltage
Table 6.2
Transformer data
Transformer Number
1
2
Transformer Name
2T1
2T2
From Bus-to Bus-
2–4
3-5
Control Bus
2
3
Manufacturer Ref No.
1
2
De-rated MVA
100
100
Nominal tap setting
5
5
+ve /zero Sequence Impedance
0.035
0.042
+ve/zero Sequence R to X Ratio
20
20
Table 6.3
Generator Data
Name
Bus Number
GEN-1
1
GEN 2
2
GEN 3
3
Manufacturer Ref. No.
20
21
22
200
3.3
150
0
140
0.15
100
3.3
100
0
90
0.25
De-rated MVA (MVA)
100
Specified Voltage (kV)
11
Schedule Power (MW)
0
Reactive Power Minimum (Mvar)
0
Reactive Power Maximum (Mvar)
100
Direct axis Transient Reactance
0.2
Table 6.4
Figure 6.1 shows a single line diagram of 6 bus power system. Find the fault currents
and voltages for the following type of Faults at Bus 5.
1. Single Line to Ground Fault.
2. Line-to- line fault.
3. Double Line to Ground Fault.
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
1. Double click the MiPower icon on your pc screen.
2. Follow the procedure in first experiment (Introduction to MiPower) to draw and
input all the data and elements.
3. For transmission line, load and generation database, use the data from Table 6.1
6.2 and 6.3.
4. To solve unsymmetrical fault, on network editor screen, click Solve  Short
circuit Analysis, Short circuit analysis screen analysis will appear.
5. Click Study Info button. In the short circuit studies screen, specify a type of
fault.
For first case, select Single Line to Ground fault. Select bus to simulate on fault.
6. On short Circuit Output Options screen, select details data as shown below.
7. After click Execute. Draw a single line drawing together with the short circuit data.
Print the result.
8. From the result obtain, fill in and complete Table6.4 in the answer sheet.
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
EXPERIMENT 5
ANALYSIS OF UNSYMMETRICAL FAULT
Name: ___________________________________ Date:_________________________
Matrix No ________________________________ Experiment No: ________________
Fault at Bus 5
Single Line to
Line to Line
Double Line to
Ground fault
Fault
Ground Fault
Sequence
+ve
Fault
Current
-ve
zero
Phase
Fault
Current
Phase A
Phase B
Phase C
FAULT MVA
TOTAL FAULT
CURRENT
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
QUESTIONS
1. Repeat the above procedure for Fault at Bus 4. Obtain line current for every
transmission line and voltage magnitude for each bus.
2. A synchronous machine A generating I pu voltage is connected through a starstar transformer, reactance 0.12pu, to two lines in parallel. The other ends of the
line are connected through a star-star transformer of reactance 0.1 pu to a
second machine B, also generating 1 pu voltage. for both transformer X1=X2=X0.
X1
X2
X0
Machine A
0.3
0.2
0.05
machine B
0.25
0.15
0.03
For each line X1=X2=X0= 0.70
Figure 1
a) Draw an equivalent circuit for positive, negative sequence and zero sequence for
above system
b) Calculate the fault current.
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Power SystemAnalysis and Control (EET 408)
Laboratory Module
DISCUSSION
CONCLUSION
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