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 'to' 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).

The rms value of the ac component of fault current (I${}_x$) will be

• A. 3.59 kA
• B. 5.07 kA
• C. 7.18 kA
• D. 10.15 kA

Answer 1

The correct option is A 3.59 kA

Fault occurs at mid point 'F'



Thevenins equivalent,

V${}_s$ = V${}_{s1}$ = V${}_{s1}$ = 1.0 $\angle 0$pu

Z${}_1$ = $(Z_{s_1}+\frac{Z_L}{2})(Z_{s_2}+\frac{Z_L}{2})$

as Z${}_{s_1}$ = Z${}_{s_2}$ = 0.001 + j0.01 pu

and Z${}_L$ = 0.006 + j0.06

Z${}_1$ = $\frac{1}{2}(Z_{s_1}+\frac{Z_L}{2})$

= $\frac{1}{2}(0.001+j0.01+\frac{0.006+j0.06}{2})$

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



Fault current = I${}_f$ = $\frac{V_s}{Z_1}=\frac{1\angle 0^0}{0.002+j0.02}$

= 49.75 $\angle$-84.29 p.u.

$\left|I_1\right|$p.u. = 49.75

Base current = $\frac{BaseMVA}{\sqrt{3}\times VoltageBase}$ kA

I${}_B$ = $\frac{100}{\sqrt{3}\times 400}$ kA

$\left|I_f\right|={\left|I_f\right|}_{p.u.}$ x I${}_B$

= 49.75 x $\frac{100}{\sqrt{3}\times 400}$ kA

= 7.18 kA

$I_x^f=\frac{\left|I_f\right|}{2}=\frac{7.18}{2}$kA = 3.59 kA