Question 9.11: A four-pole 50 Hz, three-phase induction motor has rotor res...
A four-pole 50 Hz, three-phase induction motor has rotor resistance of 0.5 Ω phase. The maximum torque occurs at a speed or 1470 rpm. Calculate the ratio of starting torque to maximum torque.
N_{s}=\frac{120 f}{P}=\frac{120 \times 50}{4}=1500 rpm
Rotor speed, N_{r} = 1200 rpm
Slip, S=\frac{N_{s}-N_{r}}{N_{s}}=\frac{1500-1470}{1500}=0.02
At this slip, torque is maximum. Condition for maximum torque is R_{2}=SX_{20}.
Thus, X_{20}=\frac{R_{2}}{S}=\frac{0.5}{0.02}=25 \Omega
For constant supply voltage, the expression for torque, T is given as
T=\frac{K SR_{2}}{R_{2}^{2}+S^{2} X_{20}^{2}}Value of maximimum torque T_{m} is
T_{m}=\frac{K}{2 X_{20}}At starting, S = 1, Starting torque T_{st} is
T_{st}=\frac{KR_{2}}{R_{2}^{2}+X_{20}^{2}}Substituting values, \frac{T_{st}}{T_{m}}=\frac{K R_{2}}{R_{2}^{2}+X_{20}^{2}} \times \frac{2 X_{20}}{K}=\frac{2 X_{20} R_{2}}{R_{2}^{2}+X_{20}^{2}}
\frac{T_{st}}{T_{m}}=\frac{2 \times 25 \times 0.5}{(.5)^{2}+(25)^{2}}=0.04That is, the starting torque is only 4 per cent of the maximum torque.