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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-Φ 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
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B
-j2.5
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C
-j12.5
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D
-j5
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Solution

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,

Xg=j8100=j0.08p.u

Per unit subtransient reactance of each motor,

X′′m=j0.2×500250=j0.4p.u.

Per unit reactance diagram is shown below,



Thevenin reactance when viewed from fault terminals,

Xth=j0.43×j0.08j0.43+j0.08=j0.05 p.u.

At fault location Vth= rated voltage

Fault current at F,If=1j0.05=j20p.u

The generator contribution is,

Ig=j20×j0.43j0.43+j0.08

Ig=j12.5p.u.

Contribution of motors,

3Im=IfIg=j20(j12.5)

3Im=j7.5

Im=j2.5p.u.

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