Question

A 6 pole, 50 Hz, 3-phase induction motor running on full load develops a useful torque of 150 Nm at a rotor frequency of 1.5 Hz. The shaft power output is

• A. 18.2 kW
• B. 11.6 kW
• C. 15.2 kW
• D. 10.4 kW

Answer 1

The correct option is C 15.2 kW

$\text{Synchronous speed,}$
$N_s=\frac{120f_1}{P}=\frac{120\times 50}{6}=1000rpm$

$\text{Slip, s}=\frac{f_2}{f_1}=\frac{1.5}{50}=0.03or\text{3 \%}$

$\text{Rotor speed,}$
$N_r=(1-s)N_s=(1-0.03)\times 1000=970rpm$

${\omega }_r=\frac{2\pi N_r}{60}=\frac{2\pi \times 970}{60}=101.58\text{mech rad/sec}$

$\text{Shaft power output}=P=T{\omega }_r$

$=150\times 101.58=15236W=15.24kW$