The torque produced in the induction motor depends on the following factors :
1. The part of rotating magnetic field which reacts with rotor and is responsible to produce induced e.m.f. in rotor.
2. The magnitude of rotor current in running condition.
3. The power factor of the rotor circuit in running condition.
Mathematically the relationship cab be expressed as,
T Φ Φ I2r cos Φ2r ………(1)
where Φ = Flux responsible to produce induced e.m.f.
I2r = Rotor running condition
cos Φ2r = Running p.f. of motor
The flux Φ produced by stator is proportional to i.e. stator voltage.
... Φ Φ E1 ………(2)
while E1 and E2 are related to each other through ratio of stator turns to rotor turns i.e. k.
... E2/E1 = K ………….(3)
Using (3) in (2) we can write, Thus in equation (1), Φ can be replaced by E2.
While I2r = E2r /Z2r = (s E2)/Φ(R22 +(s X2)2) ………….(5)
and cos Φ2r = R2/Z2r = R2/Φ(R22 +(s X2)2) …………(6)
Using (4), (5), (6) in equation (1),
... T = (k s E22 R2)/(R22 +(s X2)2) …………(7)
where k = Constant of proportionality
The constant k is provided to be 3/2 for three phase induction motor.
... k =3/(2 Φ ns) …………(8)
Key Point : ns = synchronous speed in r.p.s. = Ns/60
Using (8) in (7) we get the torque equation as,
So torque developed at any load condition can be obtained if slip at that load is known and all standstill rotor parameters are known.
1.1 Starting Torque
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.
The change in R2 at 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 torque Tst.
Example 1 : A 3 phase,
400 V, 50 Hz, 4 pole induction motor has star connected stator winding. The rotor resistance and reactance are 0.1 Φ and 1 Φ 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 Φ, X2 = 1 Φ, 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 /Φ3 = 400/Φ3 = 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