A 500-kVa, 3.3-kV, 3-phase star-connected alternator is found to give a short-circuit current of 110√3 A at normal field current. Estimate the magnitude of synchronous reactance if the effective winding resistance per phase is 1 ohm.

A 500-kVa, 3.3-kV, 3-phase star-connected alternator is found to give a short-circuit current of 110√3 A at normal field current. Estimate the magnitude of synchronous reactance if the effective winding resistance per phase is 1 ohm.

Right Answer is:

Xs = √99Ω

SOLUTION

Per phase Synchronous impedance is given by

Zs = Voc/Isc

Where,

Zs = Per phase Synchronous impedance

Voc = Per phase Open circuit voltage of the alternator

Isc = Per phase Short circuit current of the alternator

Given:

Rating of alternator = 500 KVA

Terminal voltage of alternator Vt L-L = 3.3 KV = 3300 V

Short circuit current Isc = 110√3 A

Winding Resistance = 1 Ω

Voc = 3300/√3 V

Isc = 110√3 A ( assume star connected)

Therefore,

${Z_s} = \dfrac{{\dfrac{{3300}}{{\surd 3}}}}{{110\surd 3}}$ = 10 Ω

$\begin{array}{l} {Z_s} = \sqrt {{R^2} + X_s^2} \\ \\ 10 = \sqrt {{1^2} + X_s^2} \end{array}$

100 = 1 + Xs2

Xs = √99Ω

 

Important Reluctance Motor MCQ
Scroll to Top