When 3-phase stator windings are fed by a 3-phase supply then a rotating magnetic field is setup. This rotating magnetic field while cutting the stationary rotor conductors, induces an e.m.f. in these conductors. As the rotor is a closed circuit, a current flows in the rotor circuit, whose direction is found by Fleming 's right hand rule. Another magnetic field is produced round the rotor conductors due to this rotor current. Now by the interaction of these two magnetic fields a mechanical force is produced on these conductors tending to rotate them.
The rotating speed of the stator field is called synchronous speed. It is given by the formula:
Ns = 120f
Ns = Synchronous speed in r.p.m.
F = Supply frequency in cycles per second.
P = No. of poles of stator.
The actual mechanical speed of the rotor is called rotor speed. It is always less than the synchronous speed of the stator field. It is represented by "N" and is taken in r.p.m.
Slip Speed and Slip
The difference between synchronous speed and the actual speed of the rotor is called slip speed. Slip is represented by "S".
Fractional Slip S = (Ns – N) / Ns
Percentage slip S = (Ns – N) N x 100
Frequency of Rotor Current
At standstill, rotor current frequency is equal to the supply frequency. But when rotor starts rotation then frequency of rotor current depends upon the relative speed or slip speed:
fr = rotor current frequency in cycle/sec.
Slip speed = 120fr / P
Ns — N = 120fr / P --- (1)
S = (Ns — N ) / Ns
Ns - N = S
Ns = S (120f/P) --- (2)
Comparing equation (1) & (2) we get:
S (120f / p) = 120fr / P
fr = Sf
i.e. the rotor current frequency is slip times the supply frequency.
A 4-pole, 3-phase induction motor operates fro a supply whose frequency is 50Hz. Calculate:
- speed at which magnetic field of the stator is rotating;
- speed of rotor when slip is 0.04.
- frequency of rotor current when slip is 0.03
- frequency of rotor current at standstill.
(i) Stator field revolves at synchronous speed:
Ns = 120f/P = (120x50)/4 = 1500 r.p.m.
(ii) Rotor speed:
S = (Ns- N)/Ns or N = Ns (1-S)
N = 1500 (1 - 0.04)
N = 1440 r.p.m.
(iii) frequency of rotor current, fr = Sf
fr = 0.03 x 50 = 1.5 cycles/sec.
(iv) at standstill, S = 1
fr= Sf = 1 x 50
fr= 50 cycles/second.
A 4-pole, 3-phase induction motor is supplied with 50 cycles/sec source of electric power. The frequency of the rotor current is 1.5 cycles/sec. Find the slip and speed of rotor.
fr = Sf or S = fr/f = 1.5/50 = 0.03 or 3%
Ns = 120f/P = (120x 50)4 = 1500 r.p.m.
N = Ns (I - S) = 1500 (1 - 0.03) = 1455 r.p.m.
If the e.m.f. in the rotor of an 8-pole induction motor has a frequency of 1.5Hz and that in the stator is 50Hz, at what speed is the motor running and what is the slip.
fr = Sf
1.5 = S x 50
S = 0.03 or 3%
Ns = 120f/P = (120 x 50)8 = 750 r.p.m.
Percentage S = ((Ns – N)/750) x 100
3 = ((750-N)750) x100
N = 727.5 r.p.m.