Question 4.2: A four-pole, three-phase induction motor is supplied with 40...
A four-pole, three-phase induction motor is supplied with 400 V, 50 Hz supply. The rotor-circuit resistance is 2 Ω and standstill rotor-circuit reactance is 8 Ω. Calculate the speed at which maximum torque will be developed.
Given, f = 50 Hz, P = 4. R_{2} = 2 Ω, X_{20} = 8 Ω
N_{s}=\frac{120 f}{p}=\frac{120×50}{4}=1500 rpm
Condition for maximum torque is given by
R_{2} = SX_{20}substituting values S=\frac{R_{2}}{X_{20}}=\frac{2}{8}=0.25
at this slip, rotor speed, N_{r} = (1 – S)N_{s}
= (1 – 0.25)1500
= 1125 rpm
The value of maximum torque
We will put the condition for maximum torque, i.e., R_{2} = S X_{20} in the torque equation as
T=K \frac{S R_{2}}{R_{2}^{2}+ S^{2} X_{20}^{2}}putting R_{2}=S X_{20} , Maximum torque T_{m} is
T_{m}=\frac{K .S. S X_{20}}{S^{2} X_{20}^{2}+S^{2} X_{20}^{2}}or, T_{m}=\frac{K S^{2} X_{20}}{2 S^{2} X_{20}^{2}}
or, T_{m}=\frac{K}{2 X_{20}} (4.9)
Thus, we see that the value of maximum torque, T_{m} is independent of rotor resistance R_{2} but the slip at which maximum torque is developed changes with value of R_{2}.