Question

A 3-phase induction motor has an efficiency of 0.9 when the load is 37 kW. At this load the stator copper and rotor copper loss equals the iron loss. The mechanical losses are one-third of the no load loss, then the slip is________.

• A. 0.03

Answer 1

The correct option is A 0.03
$\text{Given,}\eta =0.9$
$\text{Let,}$
$\text{P = Stator copper loss = Rotor copper loss = iron loss}$

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

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

$\text{No load losses = Iron loss + Mechanical losses}$

$P_{NL}=P+P_{mL}$

$3P_{mL}=P+P_{mL}$

$P_{mL}=\frac{P}{2}$
$\text{Efficiency,}\eta =\frac{OutputPower}{Outputpower+losses}$


$=\frac{37}{37+P+P+P+\frac{P}{2}}$

$0.90=\frac{37}{37+3.5P}$

$\text{P = 1.1746 = 1.175 kW}$

$\text{Air gap power}=P_{output}+\text{Mechanical losses + Rotor copper losses}$

$\text{Now slip,}$ $s=\frac{\text{Rotor copper losses}}{\text{Air gap power}}$

$=\frac{1.175}{37+\frac{1.175}{2}+1.175}=0.03$