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.Bus
IMPEDANCE (pu)Positive Sequence Nagative
Sequence Zero Sequence
2 0.0079 0.3024 0.0079 0.3024 0.0079 0.30743 0.1973 0.7551 0.1973 0.7551 0.7657 2.11814 0.4139 1.2724 0.4139 1.2724 1.6318 4.18755 0.4951 1.4664 0.4951 1.4664 1.9566 4.96366 0.5253 3.1325 0.5253 3.1325 1.0937 4.49567 0.4680 1.4018 0.4680 1.4018 1.8484 4.70498 1.3335 5.3776 1.3335 5.3776 2.7950 8.87479 0.5221 1.5311 0.5221 1.5311 2.0649 5.222310 2.0590 9.3174 2.0590 9.3174 3.6018 13.008511 0.6575 1.8544 0.6575 1.8544 2.6062 6.515612 21.9908 65.0156 21.9908 65.0156 23.9396 69.6768
Table 2: Formula of sequence currents.
UZ ZZ
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Determine fault currents (at faulted bus) and fill in table 4.
Table 4: Results of the fault calculation
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
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Trang 53 Consider the power system as Figure 3
Figure 3: The practical power system.
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