Question 4.11: 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.
The synchronous speed, N_{s} = \frac{120 f}{p}= \frac{120 ×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 Ω
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} {2X_{20} }At starting, S = 1, Starting torque T_{st} is
T_{st}=\frac{K R_{2} } { R_{2}^{2}+X_{20}^{2}}Substituting values, \frac{ T_{st}} {T_{m} }= \frac{ K R_{2}} {R_{2}^{2} + X_{20}^{2} } × \frac{ 2X_{20}} { K}= \frac{ 2X_{20} R_{2}} {R_{2}^{2} + X_{20}^{2} }
\frac{T_{st} } {T_{m} }= \frac{2×25×0.5 } {(.5)^{2} +(25)^{2} }= 0.04That is, the starting torque is only 4 per cent of the maximum torque.