Question

A two pole 50 Hz, induction motor supplies 15 kW to a load at a speed of 2950 rev/min. If the torque is doubled, then the power supplied by the motor is _______kW. (Assume torque-speed curve as linear at this slip).

• A. 29.48

Answer 1

The correct option is A 29.48
Synchronous speed $N_s=\frac{120f}{P}=\frac{120\times 50}{2}=3000rpm$

$Slip=\frac{3000-2950}{3000}=0.01667$

Since no value was given for mechanical losses,
$T=\frac{P}{\omega }=\frac{15\times {10}^3}{2950\times \frac{2\pi }{60}}=48.55N-m$

Given torque speed curve is linear if torque doubles then the slip will be 0.03334

The operating speed of the motor is
$N_m=(1-s)N_s$
$=(1-0.03334)3000$
$=2900rpm$

$P=T\omega$
$=97.1\times 2900\times \frac{2\pi }{60}=29.48kW$