Slip Of Induction Motor

When have seen that rotor rotates in the same direction as that of R.M.F. but in steady state attains a speed less than the synchronous speed. The difference between the two speeds i.e. synchronous speed of R.M.F. ( Ns ) and rotor speed (N) is called slip speed. This slip speed is generally expressed as the percentage of the synchronous speed.

So slip of the induction motor is defined as the difference between the synchronous speed ( Ns) and actual speed of rotor i.e. motor (N) expressed as a friction of the synchronous speed ( Ns ). This is also called absolute slip or fractional slip and is denoted as ‘s’.

Thus

The percentage slip is expressed as,

In terms of slip, the actual speed of motor (N) can be expressed as,

At start, motor is at rest and hence its speed N is zero.

This is maximum value of slip s possible for induction motor which occurs at start. While s = 0 given us N = Ns which is not possible for an induction motor. So slip of induction motor can not be zero under any circumstances.

Practically motor operates in the slip range of 0.01 to 0.05 i.e. 1 % to 5 %. The slip corresponding to full load speed of th motor is called full load slip.

Example 1 :
A 4 pole, 3 phase induction motor is supplied from Hz supply. Determine its synchronous speed. On full load, its speed is observed to be 1410 r.p.m. calculate its full load slip.

Solution : Given values are,

P = 4,            f = 50 Hz ,               N = 1410 r.p.m.
Ns = 120f / P = 120 x 50 / 4 = 1500 r.p.m.
Full load absolute slip is given by,
s = ( Ns – N)/ V₂ = (1500-1410 )/ 1500 = 0.06
.^..                            %s = 0.06 x 100 = 6 %

Example 2 : A 4 pole, 3 phase, 50 Hz, star connected induction motor has a full load slip of 4 %. Calculate full load speed of the motor.

Solution : Given values are,

P = 4,           f = 50 Hz,                % s_fl = 4%
s_fl = Full load absolute slip = 0.04
Ns = 120f  / P = 120 x 50 / 4 = 1500 r.p.m.
s_fl = (Ns – N_fl ) / Ns =  where   =  full load speed of motor
.^..                                 0.04 = (1500 – N_fl )/ 1500
.^..                                  N_fl = 1440 r.p.m.