PSP EXERCISE 02 172 PSP EXERCISE 05 172 bài tap rolePSP EXERCISE 05 172 bài tap rolePSP EXERCISE 05 172 bài tap rolePSP EXERCISE 05 172 bài tap rolePSP EXERCISE 05 172 bài tap rolePSP EXERCISE 05 172 bài tap rolePSP EXERCISE 05 172 bài tap rolePSP EXERCIS

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EXERCISE 02 -172
1. Consider the power system as Figure 1. The source is wye connection with solid grounding.

Figure 1: The typical network 12 buses.
Parameters of the power system is given in table 1 as below:
Table 1: Thevenin Impedance at buses.
IMPEDANCE (pu)
Nagative
Bus Positive Sequence
Zero Sequence
Sequence
R1
X1
R2
X2
R0
X0
1
0
0.0400
0
0.0400
0
0.0450
2

0.0079 0.3024 0.0079 0.3024 0.0079 0.3074
3
0.1973 0.7551 0.1973 0.7551 0.7657 2.1181
4
0.4139 1.2724 0.4139 1.2724 1.6318 4.1875
5
0.4951 1.4664 0.4951 1.4664 1.9566 4.9636
6
0.5253 3.1325 0.5253 3.1325 1.0937 4.4956
7
0.4680 1.4018 0.4680 1.4018 1.8484 4.7049
8
1.3335 5.3776 1.3335 5.3776 2.7950 8.8747
9
0.5221 1.5311 0.5221 1.5311 2.0649 5.2223
10
2.0590 9.3174 2.0590 9.3174 3.6018 13.0085
11
0.6575 1.8544 0.6575 1.8544 2.6062 6.5156
12 21.9908 65.0156 21.9908 65.0156 23.9396 69.6768

Faults
N(3)
N(2)
N(1)
N(1.1)

NM
N(3)

Table 2: Formula of sequence currents.
I1
I2
I0
U
0
0
Z1
U
U
0
Z1  Z 2
Z1  Z 2
U
U
U
Z 0  Z1  Z 2
Z 0  Z1  Z 2
Z 0  Z1  Z 2
U
 Z 0U
 Z 2U
Z0 Z2
Z1 
 Z 0  Z 2  Z 0 Z1  Z0 Z 2  Z1Z 2 Z 0 Z1  Z 0 Z 2  Z1Z 2
Table 3: Formula of phase currents.
I F a
I F b
I F c
I1

I1

I1
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N(2)

0

N(1)

3I1

(1.1)

N

0

Date:

3I1

3I1

0

0

�
� 3  3Z 0 Z 2
2
�
Z

Z


0
2
�

�
�I1
�
�

�
� 3  3Z 0 Z 2
2
�
Z

Z



0
2
�

�
�I1
�
�

Determine fault currents (at faulted bus) and fill in table 4.

BUS

N(3)
IF-a=IF-b=IF-c

Table 4: Results of the fault calculation
FAULT CURRENTS (A)
N(2)
N(1)
IF-b=IF-c
IF-a
3I0

N(1.1)
IF-b =IF-c

1

2
3
4
5
6
7
8
9
10
11
12
2. Consider the power system as Figure 2.

Figure 2: The typical network 5 buses.
Parameters of the power system is given in table 5 as below:
Table 5: Per-unit reactances of components.
IMPEDANCE (pu)
Positive
Nagative
Components
Zero Sequence
Sequence
Sequence
R1
X1
R2
X2
R0
X0
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3I0


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G1
0
0.15
0
0.15
G2
0
0.15
0
0.15
T1
0
0.10
0
0.10
T2
0
0.10
0
0.10

TL12
0
0.125
0
0.125
TL13
0
0.15
0
0.15
TL23
0
0.25
0
0.25
Determine fault currents (at faulted bus) and fill in table 6.
FAULT CURRENTS (A)
N(3)
N(2)
N(1)
BUS
IF-a=IF-b=IF-c
IF-b=IF-c
IF-a
3I0
1
2
3
4
5

Determine faults currents in branches (TL12, TL13, and TL23)
FAULT CURRENTS (A)
N(3)
N(2)
N(1)
TL
IF-a=IF-b=IF-c
IF-b=IF-c
IF-a
3I0
12
13
23

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0
0
0
0
0
0
0

0.05
0.05
0.10
0.10
0.3
0.35

0.7125

N(1.1)
IF-b =IF-c

3I0

N(1.1)
IF-b =IF-c

3I0


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3. Consider the power system as Figure 3

Figure 3: The practical power system.

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