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
Trang 1EXERCISE 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
Bus
IMPEDANCE (pu) Positive Sequence Nagative
Sequence Zero Sequence
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
Table 2: Formula of sequence currents
N(3)
1
U
N(2)
U
U
N(1)
U
U
U
N(1.1)
0 2 1
U
Z Z Z
0
Z U
2
Z U
Table 3: Formula of phase currents
N(3)
1
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Trang 2N(2) 0 3I1 3I1
N(1)
1
0 2
1 2
3 3
Z Z
I
0 2
1 2
3 3
Z Z
I
Determine fault currents (at faulted bus) and fill in table 4
Table 4: Results of the fault calculation
BUS
FAULT CURRENTS (kA)
1 13.1216 11.3636 12.5967 12.5967 12.8767 12.1122
6 45.4429 39.3547 39.4393 39.4393 43.0372 34.8370
8 26.0515 22.5613 21.2414 21.2414 24.2773 17.9306
10 15.1262 13.0997 13.3447 13.3447 14.3957 11.9385
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
Components
IMPEDANCE (pu) Positive
Sequence
Nagative Sequence Zero Sequence
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Trang 3G1 0 0.15 0 0.15 0 0.05
Determine fault currents (at faulted bus) and fill in table 6
BUS
FAULT CURRENTS (A)
1 25728.7487 22281.6789 26219.4964 26219.4964 25982.4942 26729.2967
2 1809.8881 1567.4010 1832.6147 1832.6147 1821.5139 1855.9449
3 1809.8881 1567.4010 2112.9708 2112.9708 2016.7107 2538.0318
4 1192.8844 1033.0633 1029.2843 1029.2843 1127.8538 905.1540
5 25728.7487 22281.6789 23765.7578 23765.7578 24867.0534 22081.3384
Determine faults currents in branches (TL12, TL13, and TL23)
*Bus 1
TL
FAULT CURRENTS (A)
12 449.0814 388.9159 457.7157 457.7157 453.5473 462.6375
13 140.3379 121.5362 143.0362 143.0362 141.7335 130.5617
23 140.3379 121.5362 143.0362 143.0362 141.7335 130.6592
*Bus 2
TL
FAULT CURRENTS (A)
12 579.1601 501.5674 586.3828 586.3828 582.8515 604.9215
13 180.9875 156.7398 183.2446 183.2446 182.1411 170.7140
23 180.9875 156.7398 183.2446 183.2446 182.1411 170.8443
*Bus 3
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Trang 4FAULT CURRENTS (A)
12 95.4298 82.6446 82.3385 82.3385 90.2263 157.4823
13 715.7235 619.8347 617.5385 617.5385 676.6970 562.4368
23 477.1490 413.2231 411.6924 411.6924 451.1313 342.5931
*Bus 4
TL
FAULT CURRENTS (A)
12 579.1601 501.5674 676.1954 676.1954 645.3865 345.2234
13 180.9875 156.7398 211.3111 211.3111 201.6833 97.1847
23 180.9875 156.7398 211.3111 211.3111 201.6833 97.1847
*Bus 5
TL
FAULT CURRENTS (A)
12 898.1628 777.8318 829.6421 829.6421 868.0836 0.0000
13 280.6759 243.0724 259.2632 259.2632 271.2761 0.0000
23 280.6759 243.0724 259.2632 259.2632 271.2761 0.0000
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Trang 53 Consider the power system as Figure 3
Figure 3: The practical power system
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