Question

A 4-pole, 50 Hz, 8 kW, 3-phase induction motor develops rated torque at 1440 rpm. In case the load torque is reduced to $1/4^{th}$ of original load torque (keeping all other parameters constant and assuming linear torque - slip characteristics)
The power output will be

• A. 3.02 kW
• B. 2.06 kW
• C. 1.02 kW
• D. 4.07 kW

Answer 1

The correct option is B 2.06 kW

$N_s=\frac{120\times 50}{4}=1500rpm$

$s_1=\frac{1500-1440}{1500}=0.04$

$\text{Rated torque}=\frac{8000}{2\pi \times 1500}\times 60$

$=50.95N-m$

$T_{e1}\propto slip\left(s_1\right)$

$T_{e1}\propto slip0.04$

$T_{e2}=\frac{T_{e1}}{4}$

$\frac{T_{e2}}{T_{e1}}=\frac{s_2}{s_1}\Rightarrow s_2=0.01$

$N_2=N_s(1-s_2)=1500(1-0.01)=1485rpm$

$\frac{P_2}{P_1}=\frac{T_{e2}\times {\omega }_2}{T_{e1}\times {\omega }_1}=\frac{1}{4}\times \frac{{\omega }_2}{{\omega }_1}$

$P_2=\frac{P_1}{4}\times \frac{{\omega }_2}{{\omega }_1}=\frac{8000}{4}\times \frac{1485}{1440}=2.0625kW$