Rotor copper loss, frequency, current and efficiency of 11 kW motor

Given: 4 poles, 400 V, 50 Hz, $N = 1450$ rpm, pf $= 0.8$ lag, mechanical power developed $P_m = 11$ kW, stator losses $= 1100$ W, mechanical losses $= 400$ W.

Assumption: "developing 11 kW" is the gross mechanical power developed by the rotor (before mechanical losses).

$$N_s = \frac{120 \times 50}{4} = 1500\ \text{rpm}, \qquad s = \frac{1500 - 1450}{1500} = 0.03333$$

(i) Rotor copper loss

$$P_{ag} : P_{cu2} : P_m = 1 : s : (1 - s)$$ $$\begin{aligned} P_{ag} &= \frac{P_m}{1 - s} = \frac{11\,000}{0.96667} = 11\,379.3\ \text{W} \\ P_{cu2} &= sP_{ag} = 0.03333 \times 11\,379.3 = 379.3\ \text{W} \end{aligned}$$

(ii) Rotor frequency

$$f_2 = sf = 0.03333 \times 50 = 1.667\ \text{Hz}$$

(iii) Line current

$$\begin{aligned} P_{in} &= P_{ag} + \text{stator losses} = 11\,379.3 + 1100 = 12\,479.3\ \text{W} \\ I_L &= \frac{P_{in}}{\sqrt{3}\,V_L\cos\phi} = \frac{12\,479.3}{\sqrt{3}\times 400 \times 0.8} = 22.52\ \text{A} \end{aligned}$$

(iv) Efficiency

$$\begin{aligned} P_{out} &= P_m - \text{mech. losses} = 11\,000 - 400 = 10\,600\ \text{W} \\ \eta &= \frac{10\,600}{12\,479.3} = 0.8494 = 84.94\% \end{aligned}$$

Answer: (i) 379.3 W; (ii) 1.667 Hz; (iii) 22.52 A; (iv) 84.94%.