Three-Phase Induction Motor on Load

In this chapter, we will explain the behavior of a three-phase induction motor on load.

When we attach a mechanical load to the rotor shaft of the three-phase induction motor, it will begin to slow down and thus the rotating magnetic field (RMF) will cut the rotor conductors at a higher rate. Because of this, the induced EMF and resulting current in the rotor conductors will increase gradually, and producing a higher torque. This torque accelerates the rotor, and the rotor and mechanical load will soon reach a state of equilibrium when the rotor torque and load torque become equal. Once this state is reached, the speed of the motor stops to decrease further, and hence the motor will run at the new speed at a constant rate.

However, the decrease in speed of a three-phase induction motor with increased load is small. It is because, impedance of its rotor circuit is low, and a small drop in the speed produces a large rotor current. This increased rotor current produces a higher torque to meet the increased load demand on the motor shaft. This is why the 3-phase induction motors are considered to be constant speed motors. However, these motors never run at synchronous speed, thus they are also called asynchronous motors.

Technically, the change in load on the three-phase induction motor is met through the adjustment of slip (difference of synchronous speed and rotor speed). Which means, the slip increases slightly with the increase in load on the motor shaft. Due to this, the relative speed between the rotating magnetic field and rotor conductors is increased. Consequently, the rotor current is increased, producing a higher motor torque to meet the increased load demand.

Also, with increasing mechanical load, the increased rotor current is in such a direction so as to decrease the rotating magnetic flux of the stator (according to Lenzs law), thus decreasing the back EMF in the stator windings. The decreased back EMF allows the stator current to increase, and hence increasing the power input to the induction motor.

Concept of Slip in Induction Motor

In a three-phase induction motor, the rotor can never reach the speed of stators rotating magnetic field (called synchronous speed). If it did, there would be no relative motion between the rotating magnetic field and rotor conductors, no induced EMF in the rotor conductors, and hence no torque to rotate the rotor. Therefore, in practice, the speed of rotor of an induction motor is always less than the synchronous speed. This difference is known as slip speed, i.e.,

$$\mathrm{\mathrm{Slip\:speed}\:=\:\mathit{N_{s}-N_{r}}}$$

Where, $\mathit{N_{s}}$ is the synchronous speed and $\mathit{N_{r}}$ is the rotor speed.

$$\mathrm{\mathrm{Synchronous\:speed,}\mathit{N_{s}}\:=\:\frac{120\mathit{f}}{\mathit{P}}}$$

Where,f is the supply frequency and P is the number of poles in induction motor.

The ratio of the slip speed to the synchronous speed is called the slip of the induction motor, i.e.,

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

Also,

$$\mathrm{\mathrm{Percentage\:Slip,}\mathit{s}\:=\:\frac{\mathit{N_{s}-N_{r}}}{\mathit{N_{s}}}\times 100\%}$$

In a practical three-phase induction motor, the change in the slip from no-load to full-load is around 0.1% to 3%.

Numerical Example

An 8-pole 3-phase induction motor is connected to a 60 Hz supply. If it is running at 880 RPM. Calculate the slip.

Solution

Given data,

• Poles,P = 8

• Frequency,f = 60 Hz

• Rotor speed,N_r= 880 RPM

$$\mathrm{\therefore \mathrm{Synchronous\:speed,}\mathit{N_{s}}\:=\:\frac{120\times 60}{8}\:=\:900}$$

Therefore, the slip will be,

$$\mathrm{\mathrm{Slip,}\mathit{s}\:=\:\frac{900-880}{900}\times 100\:=\:2.22\%}$$