Speed, input current and output of 2/3 HP motor

Solve using the double revolving field equivalent circuit, with the magnetising and rotor branches split into forward and backward halves.

Assumption: $X_0 = 105.6\ \Omega$ is taken as the magnetising reactance. Core, friction and windage losses are taken together as the given no-load loss of 122 W, so $R_0$ is not used again (using it would count core loss twice).

Given: $V = 230$ V, $f = 50$ Hz, $P = 6$, $s = 0.04$, $R_1 = 3.04\ \Omega$, $X_1 = 6.2\ \Omega$, $R_2' = 6.26\ \Omega$, $X_2' = 2.12\ \Omega$.

(i) Motor speed

$$\begin{aligned} N_s &= \frac{120f}{P} = \frac{120 \times 50}{6} = 1000\ \text{rpm} \\ N &= N_s(1-s) = 1000 \times 0.96 = 960\ \text{rpm} \end{aligned}$$

(ii) Input current and power factor

Half values: $0.5X_0 = 52.8\ \Omega$, $0.5X_2' = 1.06\ \Omega$, $\dfrac{0.5R_2'}{s} = \dfrac{3.13}{0.04} = 78.25\ \Omega$, $\dfrac{0.5R_2'}{2-s} = \dfrac{3.13}{1.96} = 1.597\ \Omega$.

$$\begin{aligned} Z_F &= \frac{(j52.8)(78.25 + j1.06)}{78.25 + j53.86} = 24.17 + j36.16\ \Omega \\ Z_B &= \frac{(j52.8)(1.597 + j1.06)}{1.597 + j53.86} = 1.533 + j1.085\ \Omega \\ Z &= (3.04 + j6.2) + Z_F + Z_B \\ &= 28.75 + j43.45 = 52.10\angle 56.51^\circ\ \Omega \\ I_1 &= \frac{230}{52.10\angle 56.51^\circ} = 4.415\angle -56.51^\circ\ \text{A} \\ \text{pf} &= \cos 56.51^\circ = 0.552\ \text{lagging} \end{aligned}$$

Input power: $P_{in} = 230 \times 4.415 \times 0.552 = 560.4$ W.

(iii) Output power

$$\begin{aligned} P_{gF} &= I_1^2 R_F = 4.415^2 \times 24.17 = 471.2\ \text{W} \\ P_{gB} &= I_1^2 R_B = 4.415^2 \times 1.533 = 29.9\ \text{W} \\ P_m &= (1-s)(P_{gF} - P_{gB}) = 0.96 \times 441.3 = 423.7\ \text{W} \\ P_{out} &= P_m - P_{NL} = 423.7 - 122 = 301.7\ \text{W} \end{aligned}$$

Efficiency $= 301.7/560.4 = 53.8\%$ (low, as is common for small single-phase motors with large no-load loss).

Answer: (i) 960 rpm; (ii) $I_1 \approx 4.42$ A at 0.552 lagging pf; (iii) $P_{out} \approx 302$ W.