A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW. The magnitude of the per phase excitation emf of the motor, in volts, is ________. (round off to nearest integer).
GATE 2022 EE Q.64: solution and answer
Answer and solution
ANSWER
The magnitude of the per-phase excitation emf is 245 V.
Official answer: 240 to 248Source: Official answer key, GATE 2022 EE (IIT Kharagpur)
Phase (star) voltage: , taken as reference, .
Total electrical input power drawn by the motor = shaft output + losses = (three-phase total).
Line current from (for a star connection, line current = phase/armature current):
For a motor, a leading power factor means the armature current leads the applied voltage: , so
With negligible armature resistance, the motor's voltage equation is , so
Substituting:
Magnitude:
Rounded to the nearest integer: .
KEY IDEA
For a synchronous motor with negligible armature resistance, the phasor equation is V = E + jXsI (current taken as flowing into the motor), so E = V - jXsI. Input electrical power equals shaft output plus internal losses, and for a leading power factor the current phasor leads the reference voltage.
WATCH OUT FOR
Forgetting to add the 2 kW loss to the 10 kW shaft output before computing the line current (using 10 kW instead of 12 kW input power), or using the wrong sign convention for a leading power factor (taking the current phase angle as -36.87° instead of +36.87°, since for a motor a leading power factor means the current phasor leads the voltage phasor).
From the GATE 2022 EE paper. Every question of this paper · Download the full paper · All 11 Synchronous machines questions, with notes
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