Three-phase Induction Motor Synchronous Speed and Slip

Synchronous Speed

In an induction motor, the speed at which the rotating magnetic field (RMF) rotates is known as synchronous speed (NS).

The value of the synchronous speed depends upon the number of stator poles (P) in the motor and the supply frequency (f). Therefore, for a given motor of P-poles, the synchronous speed is,

$$\mathrm{\text{Synchronous Speed, } \: N_{S} \: = \: \frac{120f}{P} \: RPM}$$

Slip in Induction Motor

An induction motor cannot run at synchronous speed. If it runs at synchronous speed, there would be no cutting of the flux by the rotor conductors and there would be no induced EMF, no current and no torque. Therefore, the rotor of the induction motor rotates at a speed slightly less than the synchronous speed. For this reason, an induction motor is also known as asynchronous motor.

The difference between the synchronous speed and the actual rotor speed is known as slip speed, i.e.,

$$\mathrm{\text{Slip Speed } \: = \: N_{S} \: - \: N_{r}}$$

Where, Nr is the actual rotor speed.

Generally, the slip speed is expressed as a fraction of the synchronous speed is called the per-unit slip. The per-unit slip is usually called the slip and denoted by "s". Thus,

$$\mathrm{Slip,\: s \: = \: \frac{N_{S} \: - \: N_{r}}{N_{S}}}$$

$$\mathrm{\text{Percentage Slip } \: = \: \frac{N_{S} \: - \: N_{r}}{N_{S}} \: \times \: 100}$$

Following points are important about the slip −

• When the rotor is stationary, i.e., N_r = 0, then the slip, s = 1 or 100 %.
• The slip at full-load varies from about 5 % for small motors to about 2 % for large motors.

Numerical Example 1

A 3-phase induction motor has 26 poles and is connected to 50 Hz AC supply. Find the synchronous speed of the RMF in the motor.

Solution

$$\mathrm{\text{Synchronous speed, } \: N_{S} \: = \: \frac{120f}{P} \: = \: \frac{120 \: \times \: 50}{26} \: = \: 231 \: RPM}$$

Numerical Example 2

A 3-phase induction motor is wound for 8 poles and is supplied from 50 Hz source. Calculate (1) synchronous speed (2) slip of the motor when speed is 720 RPM.

Solution

(1) Synchronous Speed −

$$\mathrm{N_{S} \: = \: \frac{120f}{P} \: = \: \frac{120 \: \times \: 50}{8} \: = \: 750 \: RPM}$$

(2) Slip of the motor, when Nr = 720 RPM −

$$\mathrm{s \: = \: \frac{N_{S} \: - \: N_{r}}{N_{S}} \: = \: \frac{750 \: - \: 720}{750} \: = \: 0.04 \: per \:unit}$$

$$\mathrm{\% s \: = \: \frac{750 \: - \: 720}{750} \: \times \: 100 \: = \: 4 \%}$$