# Answer to Question #226500 in Electrical Engineering for Alock kumar

Question #226500

A 25 MVA, 11 kV generators with is connected through a transmission line and a transformer to a bus that supplies three identical motors as shown in Figure. Each motor has on a base of 5 MVA, 6.6 kV. The three-phase rating of the step up transformer is 25 MVA, 11/66 kV with a leakage reactance of 10% and that of the step-down transformer is 25 MVA, 66/6.6 kV with a leakage reactance of 10 %. The bus voltage at the motor is 6.6 kV when a three-phase fault occurs at the point 'P'.

For the specified fault, calculate- (i) The sub-transient current in the fault, (ii) The sub-transient current in the breaker-B

1
2021-08-17T10:11:01-0400

Part 1

For the generator

X_{d(gen)}=(0.2)*[\frac{11}{11}]^2*[\frac{25}{25}]= 0.2 pu

For transformer T1

X_{d(T_1)}=0.1pu

For transformer T2

X_{d(T_2)}=0.1pu

Calculating thevinin equivalent

X_{TH}=[0.2+0.1+0.25+0.1]||[2.25||1.25||1.25]=0.2371 pu

Fault current

I_F= \frac{V_{TH}}{X_{TH}}=\frac{1}{0.2371}=4.2183 pu

Part 2

I_F(amp)=I_F(pu)*I_B= 4.2183*2086.93=9225.14 Amp\\ I_B= \frac{S_B}{\sqrt{3}V_B}= \frac{25*10^6}{\sqrt{3}*6.6*10^3}=2186.93 Amp

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