Được sự phân công của bộ môn Hệ thống điện, khoa Điện – Điện tử trường đại học Bách Khoa và sự đồng ý hướng dẫn của thầy ThS. Trần Thế Tùng, em đã thực hiện đề tài luận văn tốt nghiệp: “Tìm hiểu phần mềm PSSADEPT và tính toán cho tuyến dây 473TN Cần Thơ

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2)

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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
1
2
3
4
5
6
7
8
9
10
11
12

N(3)
IF-a=IF-b=IF-c
13.1216

8.7670
3.3626
1.9613
1.6956
45.4429
1.7758
26.0515
1.6223
15.1262
1.3338
2.1030

Table 4: Results of the fault calculation
FAULT CURRENTS (kA)
N(2)
N(1)
IF-b=IF-c
IF-a
3I0
11.3636
12.5967
12.5967
7.5131
8.6278
8.6728
2.9121
2.0647
2.0647
1.6986
1.0980

1.0980
1.4684
0.9338
0.9338
39.3547
39.4393
39.4393
1.5379
0.9828
0.9828
22.5613
21.2414
21.2414
1.6223
0.8895
0.8895
13.0997
13.3447
13.3447
1.1551
0.7188
0.7188
1.8213
2.0528
2.0528

N(1.1)
IF-b =IF-c
12.8767
8.6518

3.0058
1.7408
1.5034
43.0372
1.5749
24.2773
1.4380
14.3957
1.1810
2.0789

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
2/4

3I0
12.1122
8.5808
1.4897
0.7624
0.6444
34.8370
0.6794
17.9306
0.6127
11.9385
0.4920
2.0048


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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
25728.7487
22281.6789 26219.4964 26219.4964
2
1809.8881
1567.4010 1832.6147 1832.6147
3
1809.8881
1567.4010 2112.9708 2112.9708
4
1192.8844
1033.0633 1029.2843 1029.2843
5
25728.7487
22281.6789 23765.7578 23765.7578

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
25982.4942 26729.2967
1821.5139 1855.9449
2016.7107 2538.0318
1127.8538
905.1540
24867.0534 22081.3384

Determine faults currents in branches (TL12, TL13, and TL23)
*Bus 1

TL
12
13
23

N(3)
IF-a=IF-b=IF-c
449.0814

140.3379
140.3379

FAULT CURRENTS (A)
N(2)
N(1)
IF-b=IF-c
IF-a
3I0
388.9159
457.7157
457.7157
121.5362
143.0362
143.0362
121.5362
143.0362
143.0362

N(3)
IF-a=IF-b=IF-c
579.1601
180.9875
180.9875

FAULT CURRENTS (A)
N(2)
N(1)
IF-b=IF-c
IF-a

3I0
501.5674
586.3828
586.3828
156.7398
183.2446
183.2446
156.7398
183.2446
183.2446

N(1.1)
IF-b =IF-c
453.5473
141.7335
141.7335

3I0
462.6375
130.5617
130.6592

*Bus 2

TL
12
13
23
*Bus 3

3/4

N(1.1)
IF-b =IF-c
582.8515
182.1411
182.1411

3I0
604.9215
170.7140
170.8443


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TL
12
13
23

2)

Date:

N(3)
IF-a=IF-b=IF-c
95.4298
715.7235
477.1490

FAULT CURRENTS (A)
N(2)
N(1)
IF-b=IF-c
IF-a
3I0
82.6446
82.3385
82.3385
619.8347
617.5385
617.5385
413.2231
411.6924
411.6924

N(3)
IF-a=IF-b=IF-c
579.1601
180.9875
180.9875

FAULT CURRENTS (A)
N(2)
N(1)
IF-b=IF-c
IF-a
3I0
501.5674

676.1954
676.1954
156.7398
211.3111
211.3111
156.7398
211.3111
211.3111

N(3)
IF-a=IF-b=IF-c
898.1628
280.6759
280.6759

FAULT CURRENTS (A)
N(2)
N(1)
IF-b=IF-c
IF-a
3I0
777.8318
829.6421
829.6421
243.0724
259.2632
259.2632
243.0724
259.2632
259.2632

N(1.1)
IF-b =IF-c
90.2263
676.6970
451.1313

3I0
157.4823
562.4368
342.5931

*Bus 4

TL
12
13
23

N(1.1)
IF-b =IF-c
645.3865
201.6833
201.6833

3I0
345.2234
97.1847
97.1847

*Bus 5

TL
12
13
23

4/4

N(1.1)
IF-b =IF-c
868.0836
271.2761
271.2761

3I0
0.0000
0.0000
0.0000


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3. Consider the power system as Figure 3

Figure 3: The practical power system.

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