Question

A 3-phase, 50 Hz induction motor with load of 48 kW has an efficiency of 0.8. At this load, stator copper loss and rotor copper loss each are equal to twice of iron loss. The mechanical losses are one fifth of no load loss, then the value of rotor copper loss for induction motor will be

• A. 2124.64 W
• B. 1246.42 W
• C. 4571.42 W
• D. 3241.64 W

Answer 1

The correct option is C 4571.42 W

$\text{Given, Efficiency,}\eta =0.8$

$\text{Assuming iron loss = P}$

$\text{stator copper loss = 2 P}$

$\text{Rotor copper loss = 2 P}$

$\text{No load loss}=P_{NL}$

$\text{Mechanical loss}=P_{ML}$

$\text{No load loss can be expressed as,}$

$P_{NL}=\text{Iron loss + Mechanical loss}$

$=P+P_{ML}$

$\text{Also given,}5P_{NL}=P+P_{ML}$

or $P_{ML}=\frac{P}{4}$

$\text{Efficiency,}\eta =\frac{\text{Output power}}{\text{Output power + losses}}$

$=\frac{48}{48+2P+2P+P}+\frac{P}{4}$

$0.8=\frac{48\times 4}{48\times 4+21P}$

$48\times 4+21P=\frac{48\times 4}{0.8}$

$\text{21P = 240 - 192}$

$\text{P = 2.29 kW}$

$\therefore \text{Rotor copper loss = 4.5714 × 1000}$

$\text{= 4571.42 W}$