Question #130959

30mva 13.8kv 3-phase alternator has subtransient of 15% and negative and zero sequence reactances of 15% and 5%. The alternator supplies two motors over a transmission line having transformers.Motors rated inputs 20mva & 10mva both at 12.5kv with 20% subtransient reactances and negative & zero sequence reactances are 20% & 5% respectively. Current limiting reactors of 2ohm each in neutral of alternator of large motor.3 phase transformers are rated 35mva 21/132 kV delta/star with leakage reactance of 10%, series reactance of the line is 100ohm & zero sequence reactance of the line is 175 ohms. Use a base of 30mva and 22kv and determine the fault current when an L-L-G fault occurs at point P. Assume a prefault voltage of 138 kv

Expert's answer

Solution

The base voltage on the line side of the transformer = 13.8*132/21 =86.742 kV

The base voltage on the motor side of the transformer = 86.742 *21/132 = 13.8 kV

The per cent reactance of transformer = 10 * (21/13.8)^{2} *30/35 = 19.8%

The per cent reactance of motor 20 * (12.5/13.8)^{2} *30/20 = 24.6 %

Zero sequence network

The neutral reactance = 2*3 * 30/13.8^{2} *100 = 94.5%

The zero sequence reactance of line = 200* 30/138^{2} *100 = 31.5%

L-L-G fault

*I*_{a1} = (1+*j0.0*)/(*j*0.19224) = -*j*5.2 p.u

*I*_{a2}* = *(+*j*5.2**j*0.06767)/*j*0.21367 = *j*1.647

*I*_{a0} = *j*3.553

The fault current is

*I*_{b} + *I*_{c} = 3*l*_{a0}* *= 3 * *j*3.553 p.u

Therefore the fault current = 3*3.553*114.3 = 1538 amps

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