Question

A 500 kVA, 2.5 kV generator with transient reactance of 8% is connected to bus through a circuit breaker as shown in below. The synchronous motors each rated 250 kVA, 2.5 kV. The sub transient reactance of each of the motors being 20%. If a 3-$\varphi$ short circuit fault occurs at point A. Then the current contribution of each motor in fault current (in p.u.) is (Assume prefault voltage is rated voltage)

• A. -j7.5
• B. -j2.5
• C. -j12.5
• D. -j20

Answer 1

The correct option is B -j2.5

Let the base kVA be 500 kVA and base voltage be 2.5 kV,
Per unit transient reactance of generator,

$X_2{}^{\prime }=\frac{8}{100}=j0.08$ p.u.
Per unit subtransient reactance of each motor,

$X_m{}^{\prime \prime }=j0.2\times \frac{500}{250}=j0.4$ p.u.

Per unit reactance diagram is shown below,


Thevenin reactance when viewed from fault terminals,

$X_{th}=\frac{\frac{j0.4}{3}\times j0.08}{\frac{j0.4}{3}+j0.08}=j0.05$ p.u.

At fault location $V_{th}=$ rated voltage,

Fault current at $F,I_f=\frac{1}{j0.05}=-j20$ p.u.

The generator contribution is,

$I_g=-j20\times \frac{j\frac{0.4}{3}}{j\frac{0.4}{3}+j0.08}$

$I_g=-j12.5$ p.u.

Contribution of motors,

$3I_m=I_f-I_g=-j20-(-j12.5)$
$3I_m=-j7.5$
$I_m=-j2.5$ p.u.