Question

Given that V$s1$ = V$s2$ = 1 + j0 p.u. +ve sequence impendance are Z$s1$ = Z$s2$ = 0.001 + j0.01 p.u and Z${}_L$ = 0.006 + j0.06 p.u, 3-$\varphi$. Base MVA = 100, voltage base = 400 kV(L-L).

Nominal system frequency = 50 Hz. The reference voltage for phase 'a' is deifned as V(t) = V${}_m$ cos($\omega$t). the symmetrical 3$\varphi$ fault occurs at centre of the line i.e, and point 'F' at the time '$t_o$' the +ve sequence impendance from source S${}_1$ to point 'F' equals (0.004 + j0.04) p.u. The wave from corresponding to phase 'a' fault current from bus X reveals that decaying d.c. offset current is -ve and in magnitude at its maximum initial vale. Assuming that the negative sequence are equal to +ve sequence impendance and the zero sequence (Z) are 3 times +ve sequence (Z).

Instead of the three phase fault, if a single line ground fault occurs on phase 'a' at point 'F' with zero fault impendance, then the rms of the component of fault current (I${}_x$) for phase 'a' will be

• A. 4.97 pu
• B. 7.0 pu
• C. 14.93 pu
• D. 29.85 pu

Answer 1

The correct option is C 14.93 pu

Z${}_1$ = 0.002 + j0.02 pu

Z${}_2$ = Z${}_1$ = 0.002 +j0.02 pu

Z${}_0$ = 3Z${}_1$ = 0.006 + j0.06 pu



I${}_{a0}$ = I${}_{a1}$ = I${}_{a2}$ = $\frac{V_s}{Z_1+Z_2+Z_0}$

= $\frac{V_s}{Z_1+Z_2+3Z_1}$

= $\frac{V_s}{5Z_1}=\frac{1\angle 0^0}{5\times (0.002+j0.02)}$

= 9.95 $\angle$ - 84.29 p.u.

Fault current = ${\left|I_f\right|}_{p.u.}=3\left|I_{a0}\right|$

= 3 x 9.95 = 29.85 p.u.

I${}_x^f$ = $\frac{{\left|I_f\right|}_{p.u.}}{2}=\frac{29.85}{2}$ = 14.93 p.u.