Rotor copper loss and efficiency of 35 kW induction motor

Given: $P = 4$, $f = 50$ Hz, $N = 1440$ rpm, input $P_{in} = 35$ kW, stator losses $= 1$ kW, friction (mechanical) losses $= 1.5$ kW.

Power flow: input → (minus stator losses) → air-gap power → (minus rotor Cu loss) → gross mechanical power → (minus friction) → output.

Synchronous speed and slip

$$\begin{aligned} N_s &= \frac{120 f}{P} = \frac{120 \times 50}{4} = 1500\ \text{rpm} \\ s &= \frac{N_s - N}{N_s} = \frac{1500 - 1440}{1500} = 0.04 \end{aligned}$$

(i) Rotor copper loss

$$\begin{aligned} P_{ag} &= P_{in} - \text{stator losses} = 35 - 1 = 34\ \text{kW} \\ P_{cu2} &= s \, P_{ag} = 0.04 \times 34 = 1.36\ \text{kW} \end{aligned}$$

(ii) Efficiency

$$\begin{aligned} P_{m} &= P_{ag} - P_{cu2} = 34 - 1.36 = 32.64\ \text{kW} \\ P_{out} &= P_m - \text{friction} = 32.64 - 1.5 = 31.14\ \text{kW} \\ \eta &= \frac{P_{out}}{P_{in}} = \frac{31.14}{35} = 0.8897 = 88.97\% \end{aligned}$$

Answer: (i) Rotor copper loss = 1.36 kW; (ii) efficiency = 88.97%.