Question

A two pole, 50 Hz, induction motor supplies 15 kW to a load at a speed of 2950 rpm. The operating speed of the motor if its torque is doubled in _____rpm.

• A. 2900

Answer 1

The correct option is A 2900
$N_s=\frac{120f}{P}=\frac{120\times 50}{2}=3000rpm$

$\text{Hence the motor slip is}$
$s=\frac{N_s-N_m}{N_s}=\frac{3000-2950}{3000}$

$=0.0167=1.67\%$

In the low slip region, the torque speed curve is linear, and the induced torque is directly proportional to slip. Therefore if the torque doubles, then the new slip will be 3.33%. The operating speed of motor is thus,

$N_m=(1-s)N_s$

$=(1-0.0333)\left(3000\right)$

$=2900rpm$