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Line - to - Line - to - Ground Fault (LLG Fault)

A 30 MVA, 3-p...

Question

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 phase 'b' and 'c' with a fault impendance 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
1.07

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Solution

The correct option is A 1.07
Base current,

I $_{B}$ = $\frac{30 \times 10^{3}}{\sqrt{3} \times 13.8}$ = 1255.109 A

I $_{f}$ = 4270 A

I $_{p . u .}$ = $\frac{4270}{1255.109}$ = 3.402 p.u.

I $_{g 1}$ = $\frac{E_{a}}{X_{1} + (X_{2} | | X_{0})}$

where, X $_{0}$ = X $_{0}$ + 3(Z $_{n}$ + Z $_{f}$ )

X $_{0}$ = 3Z $_{n}$ + 0.35

3.402 = $\frac{1.0}{0.15 + \frac{0.15 \times (3 Z_{n} + 0.35)}{0.15 + 3 Z_{n} + 0.35}}$

By solving the equation,

Z $_{n}$ = 1.07 p.u.

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