Question

Three 6.6 kV, $3-\Phi$, 10 MVA alternators are connected to a common bus. Each alternator has a positive, negative and zero sequence reactance as 0.15 p.u., 0.1 p.u., 0.05 p.u. respectively. A single line ground fault occurs at point F, as shown in figure below.



$I_f$ is the fault current when all the alternator neutrals solidly grounded. $I_f^{\prime }$ is the fault current when only one neutral is solidly grounded and other two neutrals are isolated. Then the value of $|I_f-I_f^{\prime }|=$________ kA.

• A. 6.512

Answer 1

The correct option is A 6.512
Let base MVA be 10 MVA and base kv be 6.6 kV

$\therefore$ Base current = $\frac{10\times {10}^3}{\sqrt{3}\times 6.6}=874.77Amp$

when all the 3 alternators are solidly grounded,



Since all the alternators are operating in parallel, the resultant reactance will be one-third i.e.,

$Z_1=\frac{j0.15}{3}=j0.05p.u.$

$Z_2=\frac{j0.1}{3}=j0.03333p.u.$

$Z_0=\frac{j0.05}{3}=j0.0166p.u.$

Fault current, $I_f=\frac{3}{Z_1+Z_2+Z_0}$

$=\frac{3}{j0.05+j0.0333+j0.0166}=j30p.u.$

$I_f=-j30\times 874.77$

$=-j262243.1Amp$

when only one alternator netral is solidly grounded and the others are isolated.


$Z_1=\frac{j0.15}{3}p.u.$

$Z_2=\frac{j0.1}{3}p.u.$

$Z_0=j0.05p.u.$

$\therefore$ Fault current, $I_f^{\prime }=\frac{3}{Z_1+Z_2+Z_0}$

$=\frac{3}{j0.05+j0.033+j0.05}=-j22.55p.u.$

= -j22.55 $\times$ 874.77

$I_f^{\prime }=-j19731.65Amp$

$\therefore |I_f-I_f^{\prime }|=|-j26.243+j19.731|$

= 6.512 kA