The torque produced in the induction motor depends on the following factors :
So torque developed at any load condition can be obtained if slip at
that load is known and all standstill rotor parameters are known.
Starting torque is nothing but the torque produced by an induction
motor as start. At start, N= 0 and slip s = 1. So putting s = 1 in the
torque equation we can write expression for the starting torque Tst as,
Key Point : From the equation (10), it is clear that by changing the starting torque can be controlled.
start is possible in case of slip ring induction motor only. This is the
principle used in case of slip induction motor to control the starting
400 V, 50 Hz, 4 pole induction motor has star connected stator winding.
The rotor resistance and reactance are 0.1 ? and 1 ? respectively. The
full load speed is 1440 r.p.m. Calculate the torque developed on full
load by the motor.
Assume stator to rotor ratio as 2 :1.
Solution : The given values are,
P = 4, f = 50 Hz, R2 = 0.1 ?, X2 = 1 ?, N = 1440 r.p.m.
Stator turns/Rotor turns = 2/1
... K = E2 /E1 = Rotor turns/Stator turns = 1/2 = 0.5
Ns=120f/P = 120×50 / 4 = 1500 r.p.m.
E1line = 400 V …………..Stator line voltage given
... E1ph = E1line /?3 = 400/?3 = 230.94 V
But E2ph /E1ph = 0.5 = K
... E2ph = 0.5 x 230.94 = 115.47 V
Full load slip, s = (Ns-N)/Ns = (1500-1400)/1500 = 0.04
ns = Synchronous speed in r.p.s.
= Ns/60 = 1500/60 = 25 r.p.s.
= 87.81 N-m