Effect of Slip on Rotor Parameters
In case of a transformer, frequency of the induced e.m.f. in the secondary is same as the voltage applied to primary. Now in case of induction motor at start N = 0 and slip s = 1. Under this condition as long as s = 1, the frequency of induced e.m.f. in rotor is same as the voltage applied to the stator. But as motor gathers speed, induction motor has some slip corresponding to speed N. In such case, the frequency of induced e.m.f. in rotor is no longer same as that of stator voltage. Slip affects the frequency of rotor induced e.m.f. Due to this some other rotor parameters also get affected. Let us study the effect of slip on the following rotor parameters.
- Rotor frequency
- Magnitude of rotor induced e.m.f.
- Rotor reactance
- Rotor power factor
- Rotor current
1. Effect on rotor frequency
In case of induction motor, the speed of rotating magnetic field is,
Ns = (120 f )/P ……….(1)
Where f = Frequency of supply in Hz
At start when N = 0, s = 1 and stationary rotor has maximum relative motion with respect to R.M.F. Hence maximum e.m.f. gets induced in the rotor at start. The frequency of this induced e.m.f. at start is same as that of supply frequency.
As motor actually rotates with speed N, the relative speed of rotor with respect R.M.F. decreases and becomes equal to slip speed of Ns – N. The induced e.m.f. in rotor depends on rate of cutting flux i.e. relative speed Ns
– N. Hence in running condition magnitude of induced e.m.f. decreases so as to its frequency. The rotor is wound for same number of poles as that of stator i.e. P. If fr is the frequency of rotor induced e.m.f. in running condition at slip speed Ns – N then there exists a fixed relation between (Ns – N), fr and P similar to equation (1). So we can write for rotor in running condition,
(Ns – N) = (120 fr)/P , rotor poles = stator poles = P ……….(2)
Dividing (2) by (1) we get,
(Ns – N)/Ns = (120 fr / P)/(120 f / P) but (Ns – N)/Ns = slip s
s = fr/f
fr = s f
Thus frequency of rotor induced e.m.f. in running condition (fr) is slip times the supply frequency (f).
At start we have s = 1 hence rotor frequency is same as supply frequency. As slip of the induction motor is in the range 0.01 to 0.05, rotor frequency is very small in the running condition.
Example: A 4 pole, 3 phase, 50 Hz induction motor runs at a speed of 1470 r.p.m. speed. Find the frequency of the induced e.m.f in the rotor under this condition.
Solution : The given values are,
P = 4, f = 50 Hz, N = 1470 r.p.m.
Ns = (120 f )/ P = (120 x 50)/4 = 1500 r.p.m.
s = (Ns – N)/Ns = (1500-1470)/1500 = 0.02
fr = s f = 0.02 x 50 = 1 Hz
It can be seen that in running condition, frequency of rotor induced e.m.f. is very small.