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

Students’ name: 1) 2) Date: EXERCISE 02 -172 Consider the power system as Figure The source is wye connection with solid grounding Figure 1: The typical network 12 buses Parameters of the power system is given in table as below: Table 1: Thevenin Impedance at buses IMPEDANCE (pu) Nagative Bus Positive Sequence Zero Sequence Sequence R1 X1 R2 X2 R0 X0 0.0400 0.0400 0.0450 0.0079 0.3024 0.0079 0.3024 0.0079 0.3074 0.1973 0.7551 0.1973 0.7551 0.7657 2.1181 0.4139 1.2724 0.4139 1.2724 1.6318 4.1875 0.4951 1.4664 0.4951 1.4664 1.9566 4.9636 0.5253 3.1325 0.5253 3.1325 1.0937 4.4956 0.4680 1.4018 0.4680 1.4018 1.8484 4.7049 1.3335 5.3776 1.3335 5.3776 2.7950 8.8747 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 Z1 U U Z1  Z Z1  Z U U U Z  Z1  Z Z  Z1  Z Z  Z1  Z U  Z 0U  Z 2U Z0 Z2 Z1   Z  Z  Z Z1  Z0 Z  Z1Z Z Z1  Z Z  Z1Z Table 3: Formula of phase currents I F a I F b I F c I1 I1 I1 1/4 Students’ name: 1) 2) N(2) N(1) 3I1 (1.1) N Date: 3I1 3I1 0 � �  3Z Z 2 � Z  Z   � � �I1 � � � �  3Z Z 2 � Z  Z   � � �I1 � � Determine fault currents (at faulted bus) and fill in table 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 10 11 12 Consider the power system as Figure Figure 2: The typical network buses Parameters of the power system is given in table 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 Students’ name: 1) 2) Date: G1 0.15 0.15 G2 0.15 0.15 T1 0.10 0.10 T2 0.10 0.10 TL12 0.125 0.125 TL13 0.15 0.15 TL23 0.25 0.25 Determine fault currents (at faulted bus) and fill in table FAULT CURRENTS (A) N(3) N(2) N(1) BUS IF-a=IF-b=IF-c IF-b=IF-c IF-a 3I0 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 3/4 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 Students’ name: 1) 2) Date: Consider the power system as Figure Figure 3: The practical power system 4/4