Question

If maximum torque of an induction motor is 200 kg-m at a slip of 12% the torque at 6% slip would be

Answer 1

Answer:

The torque at 6% slip will be 160 kg-m

Step-by-step explanation:

We know that

$\frac{T}{Tm} = \frac{2SSm}{Sm^{2} + S^{2} }$

where Tm = maximum torque = 200

Sm = slip at maximum torque = 12% = 0.12

S = slip at which the value of T is to be found out = 6% = 0.06

Putting these values in the above formula we get,

T/200 = (2 x 0.06 x 0.12)/(0.12² + 0.06²)

=> T/200 = 0.0144/0.018

=> T/200 = 0.8

=> T = 200 x 0.8 = 160 Kg-m

Hence the torque at 6% slip will be 160 kg-m