Question

A 2-pole, 50 Hz, 3-phase induction motor supplies power to a certain load at 2970 rpm. The torque-speed curve of this machine follows a linear relationship between synchronous speed and 95% of synchronous speed. Assume mechanical and stray losses to be zero. If the load torque of the motor is doubled, the new operating speed of the motor, in rpm, is:

• A. 2940
• B. 2812
• C. 2970
• D. 2850

Hint:

In the linear region of a torque-speed curve, torque is directly proportional to slip. So, doubling the torque means doubling the slip, which helps compute the new speed.

Answer 1

The Correct Option is A

Step 1: Compute the synchronous speed. 
Given:
Number of poles \( P = 2 \)
Frequency \( f = 50 \, {Hz} \)
The synchronous speed \( N_s \) is given by: \[ N_s = \frac{120f}{P} = \frac{120 \times 50}{2} = 3000 \, {rpm} \] Step 2: Determine slip. 
The motor runs at 2970 rpm. So, slip \( s \) is: \[ s = \frac{N_s - N}{N_s} = \frac{3000 - 2970}{3000} = 0.01 \] Step 3: Torque-slip relationship. 
Given the torque-speed curve is linear between 3000 rpm and 95% of 3000 rpm: \[ 0.95 \times 3000 = 2850 \, {rpm} \] This implies linear torque-slip relation between \( s = 0 \) (at 3000 rpm) and \( s = 0.05 \) (at 2850 rpm). In this region, torque \( T \propto s \). 
Step 4: When load torque doubles. 
Since \( T \propto s \), to double the torque, slip also doubles: \[ s_{{new}} = 2 \times 0.01 = 0.02 \] \[ N_{{new}} = N_s(1 - s_{{new}}) = 3000 \times (1 - 0.02) = 2940 \, {rpm} \]