Question

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

$I_f$ is the fault current when all the alternator neutrals solidy grouned $I_f^{\prime }$ 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.,

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

$Z_2=\frac{j0.1}{3}=j0.0333$ p.u.

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

Fault current, $I_f=\frac{3}{Z_1+Z_2+Z_0}$
$=\frac{3}{j0.05+j0.0333+j0.0166}=-j30$ p.u.

$I_f=-j30\times 874.77$
$=-j26243.1$ Amp

When only one alternator neutral is solidly grounded and the others are isolated.


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

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

$Z_0=j0.05$ p.u.

$\therefore$ Fault current $I_f^{\prime }=\frac{3}{z_1+z_2+z_0}$
$=\frac{3}{j0.05+j0.033+j0.05}=-j22.55$ p.u.
$=-j22.55\times 874.77$
$I_f^{\prime }=-j19731.65$ Amp

$\therefore$ $|I_f-I_f^{\prime }|=|-j26.243+j19.731|$
= 6.512 kA