GATE 2016 EE – Question 51
A 30 MVA, 3-phase, 50 Hz, 13.8 kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%, 15% and 5% respectively. A reactance ($X_n$) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases 'b' and 'c', with a fault impedance of j0.1 p.u. The value of $X_n$ (in p.u.) that will limit the positive sequence generator current to 4270 A is ________.
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Correct answer: 1.05 to 1.15
Explanation
The base current is $\frac{30 \times 10^6}{\sqrt{3} \times 13.8 \times 10^3} = 1255$ A, so the positive sequence current is $\frac{4270}{1255} = 3.40$ pu. For a double line to ground fault, $I_1 = \frac{1}{X_1 + X_2 \parallel X_0'}$, where $X_0' = 0.05 + 3(0.1) + 3X_n$. Then $X_1 + X_2 \parallel X_0' = \frac{1}{3.40} = 0.294$, so $X_2 \parallel X_0' = 0.144$. Solving $\frac{0.15 X_0'}{0.15 + X_0'} = 0.144$ gives $X_0' \approx 3.57$, and $3X_n = 3.57 - 0.35$, so $X_n \approx 1.07$ pu.