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 each equals the iron loss. The mechanical losses are one third of no-load losses. The slip is
1. 0.01
2. 0.02
3. 0.03
4. 0.04

Answer 1

Correct Answer - Option 3 : 0.03

Output power = 37 kW

Efficiency (η) = 0.9

\(\Rightarrow η= \frac{output}{input}=0.9\)

\(\Rightarrow input=\frac{37}{0.9}=41.111~kW\)

Losses = input – output

= 41.111 – 37 = 4111 W

Stator copper loss = rotor copper losses = iron loss = P

Mechanical losses = P_m

At no load, iron loss + mechanical losses = no load losses

Given that mechanical losses are one third of no-load losses.

\(\Rightarrow {{P}_{m}}=\frac{1}{3}\left( P+{{P}_{m}} \right)\)     → 1)

Total losses = 4111

⇒ P + P + P + P_m = 4111

⇒ 3P + P_m = 4111     → 2)

By solving both the equations, we get

P = 1174.6 W and P_m = 587.3 W

Rotor input = shaft output + rotor copper losses + mechanical losses

= (37 × 10³) + 1174.6 + 587.3

= 38761.9 W

\(Slip\left( s \right)=\frac{Rotor~copper~loss}{Rotor~input}\)

\(=\frac{1174.6}{38761.9}=0.03\)