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

• A. 1.07

Answer 1

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${}_{g1}$ = $\frac{E_a}{X_1+\left(X_2||X_0\right)}$

where, X${}_0$ = X${}_0$ + 3(Z${}_n$ + Z${}_f$)

X${}_0$ = 3Z${}_n$ + 0.35

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

By solving the equation,

Z${}_n$ = 1.07 p.u.