Question

The positive , negative and zero sequence impedance of $3-\varphi$ generator are 0.4 pu, 0.3 pu and 0.2 pu respectively. For a line to line fault with fault impedance $Z_f$ and fault current is $I_{f1}=0.2I_f$, where $I_f$ is the fault current with zero fault impedance, then value of $Z_f$ will be ___ pu.

• A. 2.8

Answer 1

The correct option is A 2.8
For given system, fault impedance $Z_f$ is given.

$\therefore L-Lfaultcurrentwillflowthrough{Z}_f$.

$\therefore Including{Z}_{f}inourimpedancewecandraw$,


$Faultcurrent{I}_{f1}=\sqrt{3}{I}_{a1}=\sqrt{3}.\frac{V_{th}}{Z_1+Z_2+Z_f}$

Also, zero sequence impedance diagram,



Also, we know, fault current,

$I_f=\sqrt{3}{I}_{a1}^{\prime }=\sqrt{3}.\frac{V_{th}}{Z_1+Z_2}$

$Alsogiven,I_{f1}=0.2I_f$

$Substitutingvalue,weget:\frac{\sqrt{3}{V}_{th}}{Z_1+Z_2+Z_f}=\mathrm{0.2.}\frac{\sqrt{3}{V}_{th}}{Z_1+Z_2}$

$Z_1+Z_2=0.2(Z_1+Z_2+Z_f)$

$\frac{0.8(Z_1+Z_2)}{0.2}=Z_f$

$Z_f=\frac{0.8(0.4+0.3)}{0.2}=4(0.4+0.3)=2.8pu$