Draw the per-unit impedance diagram of the network shown in Fig. 1, calculated to a common base of 50 MVA. Determine: (i) (ii) The rms fault current (in kA) and the fault level (in MVA) for a three-phase symmetrical short- circuit fault at bus no. 4. The rms fault current (in kA) and the fault level (in MVA) for a three-phase symmetrical short- circuit fault at bus no. 5. T₁ 132kV/33kV 50 MVA X=0.15 pu bus 1 1600 MVA Fault Level bus 2 40 km OHL X=0.4 2/km T3 T₂ 132kV/33kV 50 MVA X=0.15 pu 30 km OHL X=0.42/km bus 3 Cable j2.20 bus 4 33kV/11KV 15 MVA X=0.1 pu bus 5 Cable j2.5 Ω Figure 1 T4 33kV/11KV 12 MVA X=0.12 pu bus 6

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Draw the per-unit impedance diagram of the network shown in Fig. 1, calculated to a common base of
50 MVA.
Determine:
(i)
(ii)
The rms fault current (in kA) and the fault level (in MVA) for a three-phase symmetrical short-
circuit fault at bus no. 4.
The rms fault current (in kA) and the fault level (in MVA) for a three-phase symmetrical short-
circuit fault at bus no. 5.
T₁
132kV/33kV
50 MVA
X=0.15 pu
bus 1
1600 MVA Fault Level
bus 2
40 km OHL
X=0.4 2/km
T3
T₂
132kV/33kV
50 MVA
X=0.15 pu
30 km OHL
X=0.42/km
bus 3
Cable
j2.20
bus 4
33kV/11KV
15 MVA
X=0.1 pu
bus 5
Cable
j2.5 Ω
Figure 1
T4
33kV/11KV
12 MVA
X=0.12 pu
bus 6
Transcribed Image Text:Draw the per-unit impedance diagram of the network shown in Fig. 1, calculated to a common base of 50 MVA. Determine: (i) (ii) The rms fault current (in kA) and the fault level (in MVA) for a three-phase symmetrical short- circuit fault at bus no. 4. The rms fault current (in kA) and the fault level (in MVA) for a three-phase symmetrical short- circuit fault at bus no. 5. T₁ 132kV/33kV 50 MVA X=0.15 pu bus 1 1600 MVA Fault Level bus 2 40 km OHL X=0.4 2/km T3 T₂ 132kV/33kV 50 MVA X=0.15 pu 30 km OHL X=0.42/km bus 3 Cable j2.20 bus 4 33kV/11KV 15 MVA X=0.1 pu bus 5 Cable j2.5 Ω Figure 1 T4 33kV/11KV 12 MVA X=0.12 pu bus 6
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