Question

A 10 kW, 50 Hz, 6-pole poly phase induction motor has a full-load slip of 0.04. If its friction and windage losses are 4% of the output then full-load electromagnetic torque is_________N-m.

• A. 103.42

Answer 1

The correct option is A 103.42
$\text{Power output,}{P}_0=10kW$

$\text{Frequency,}f=50Hz$

$\text{Poles,}P=6$

$\text{Slip, s = 0.04}$

$\text{Friction and windage losses = 0.4 kW}$

$\text{Mechanical power developed = 10.4 kW}$

$\text{Air gap power,}{P}_g=\frac{\text{Mechanical power developed}}{1-s}$

$=\frac{10.4}{1-0.04}=10.83kW$

$\text{Synchronous speed}=\frac{120\times 50}{60}$

$\text{= 1000 rpm or 104.72 rad/sec}$

$\text{Full load electromagnetic torque,}$

$T_e=\frac{P_g}{{\omega }_s}=\frac{10.83\times {10}^3}{104.72}$

$=103.42N-m$