Question

460 V, 25 hp, 60 Hz, wye-connected, four-pole induction motor has the following impedances in ohms per phase referred to the stator circuit:

R1=0.641W, R2=0.332W, X1=1.106W, X2=0.464W, Xm=26.3W
The total rotational losses are 1100 W and are assumed to be constant. Core losses are grouped with rotational losses. For a rotor slip of 2.2% at rated voltage and frequency, find the following motor quantities:
a. Speed

Answer 1

To find the speed of the motor, calculate the synchronous speed and then account for the slip. The motor's synchronous speed is 1800 RPM, and with a slip of 2.2%, the actual rotor speed is 1760.4 RPM.

Step-by-step explanation:

To find the speed of the given 460 V, 25 hp, 60 Hz, wye-connected, four-pole induction motor at a rotor slip of 2.2%, one must first calculate the synchronous speed, which is the speed at which the motor's magnetic field rotates. Given the frequency (f) is 60 Hz and it is a four-pole motor, the synchronous speed (N_s) can be calculated using the formula

N_s = (120 × f) / P = (120 × 60 Hz) / 4 = 1800 RPM.

The actual rotor speed can be found by accounting for the slip (s), which is given as 2.2%. To get the rotor speed (N_r), we use the formula

N_r = N_s × (1 - s) = 1800 RPM × (1 - 0.022) = 1800 RPM × 0.978 = 1760.4 RPM.

The rotor actually turns at 1760.4 RPM under the given conditions.