A 150 kW, 1200 rpm, 250 V, three phase delta connected synchronous motor has synchronous reactance of 1 Ω per phase. if the excitation voltage is fixed at 400 V per phase, power output at a torque angle of 30 degrees is:

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This question was previously asked in

DSSSB JE EE 2019 Official Paper (Held on 5 Nov 2019)

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Answer 1

Concept:

The output power of a three phase synchronous motor is given by

\(P = \frac{{EV}}{{{X_s}}}\sinδ \)

P = output power of the synchronous motor

E = generated e.m.f per phase

V = terminal voltage per phase

X_S = synchronous reactance per phase

δ = load angle

Calculation:

Given:

E = 400, V = 250 V, X_S = 1 Ω, δ = 30°

The output power of the synchronous motor per phase

\(P = \frac{{EV}}{{{X_S}}}\sinδ = \frac{{250\; \times 400\;}}{{1\;}} \times \sin30^\circ \)

P = 50 kW