Question

A 500 kVA, 2.5 kV generator with transient reactance of 8% is connected to a bus through a circuit breaker as shown in below figure. The synchronous motor each rated 250 kVA, 2.5 kV. The sub transient reactance of each of the motor being 20%. If a 3-$\Phi$ short circuit fault occur 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. -j5

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,

$X_g^{\prime }=\frac{j8}{100}=j0.08p.u$

Per unit subtransient reactance of each motor,

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

Per unit reactance diagram is shown below,



Thevenin reactance when viewed from fault terminals,

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

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

Fault current at F,$I_f=\frac{1}{j0.05}=-j20p.u$

The generator contribution is,

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

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

Contribution of motors,

$3I_m=I_f-I_g=-j20-(-j12.5)$

$3I_m=-j7.5$

$I_m=-j2.5p.u.$