Question

A three-phase 415 V, 50 Hz, 6-pole, 960 RPM, 4 HP squirrel cage induction motor drives a constant torque load at rated speed operating from rated supply and delivering rated output. If the supply voltage and frequency are reduced by 20%, the resultant speed of the motor in RPM (neglecting the stator leakage impedance and rotational losses) is ..................... (Round off to the nearest integer).

Hint:

In constant torque loads, slip approximately remains constant when both supply voltage and frequency are reduced proportionally.

Answer 1

Step 1: Rated synchronous speed.
\[ N_s = \frac{120 f}{P} = \frac{120 \times 50}{6} = 1000 \, \text{RPM} \] Given actual speed at rated load: \[ N = 960 \, \text{RPM} \] So slip: \[ s = \frac{N_s - N}{N_s} = \frac{1000 - 960}{1000} = 0.04 \]

Step 2: New frequency and synchronous speed.
Supply frequency reduced by 20%: \[ f_{new} = 0.8 \times 50 = 40 \, \text{Hz} \] So new synchronous speed: \[ N_{s,new} = \frac{120 \times 40}{6} = 800 \, \text{RPM} \]

Step 3: Slip remains the same (constant torque load).
\[ N_{new} = (1-s) N_{s,new} = (1 - 0.04)(800) = 0.96 \times 800 = 768 \, \text{RPM} \]

Final Answer:
\[ \boxed{768 \, \text{RPM}} \]