Question

A 3φ, 2-pole, 60 Hz, induction motor operates at 3546 rpm while delivering 20 kW to a load. Neglect all losses. Determine (a) The slip of the motor. (b) The developed torque.

Answer 1

The slip of the motor is 4% and the developed torque is 176.6 Nm.

Step-by-step explanation:

(a) Slip of the motor can be determined using the formula:

Slip = (Ns-N)/(Ns) x 100%

where Ns is the synchronous speed of the motor and N is the actual speed of the motor.

In this case, the synchronous speed can be calculated using the formula:

Ns = (120 x f)/P

where f is the frequency (60 Hz) and P is the number of poles (2).

Given that the motor operates at 3546 rpm, we can convert it to revolutions per second:

N = 3546/60 = 59.1 rps

Substituting the values into the formulas:

Ns = (120 x 60)/2 = 3600 rpm

Slip = (3600 - 3546)/3600 x 100% = 4%

(b) The developed torque of the motor can be calculated using the formula:

Torque = (P x 1000) / (Ns x 2 x π x Slip)

where P is the power (20 kW), Ns is the synchronous speed, and Slip is the slip ratio (4% or 0.04).

Substituting the values into the formula:

Torque = (20 x 1000) / (3600 x 2 x π x 0.04) = 176.6 Nm