Question

The full-load efficiency of a 400 V, 3-phase, 6 pole induction motor is 85%. It draws a line current of 80 A at 0.8 power factor lagging when it delivers full-load at 5% slip. Calculate the shaft output and shaft torque.

Answer 1

The power of the motor is

$$P=VI\cos\phi.$$

The mechanical power output is

$$P_m=\eta P=\eta VI\cos\phi.$$

The synchronous frequency of a 6-pole motor is 1000 min^-1, which corresponds to angular

velocity of

$$\omega=\frac{2\pi n}{60}.$$

The angular velocity of the shaft is

$$\omega_s=w(1-s)=\frac{\pi n}{30}(1-s).$$

The shaft torque:

$$T=\frac{P}{\omega_s}=\frac{30\eta VI\cos\phi}{\pi n(1-s)}=122.4\text{ N}·\text{m}.$$