Input current, pf and torque of single phase IM

The slip, poles and frequency are not given. Assume a 4-pole, 50 Hz motor running at slip $s = 0.05$, and neglect core loss. Use the double revolving field equivalent circuit.

Given: $V = 240$ V, $R_1 = 2\ \Omega$, $X_1 = X_2' = 3.1\ \Omega$, $R_2' = 1.98\ \Omega$, $X_{mag} = 40.17\ \Omega$.

Half values

$$0.5X_m = 20.085\ \Omega,\quad 0.5X_2' = 1.55\ \Omega,\quad \frac{0.5R_2'}{s} = \frac{0.99}{0.05} = 19.8\ \Omega,\quad \frac{0.5R_2'}{2-s} = \frac{0.99}{1.95} = 0.5077\ \Omega$$

Forward and backward impedances

$$\begin{aligned} Z_F &= \frac{(j20.085)(19.8 + j1.55)}{19.8 + j21.635} = 9.287 + j9.938\ \Omega \\ Z_B &= \frac{(j20.085)(0.5077 + j1.55)}{0.5077 + j21.635} = 0.437 + j1.449\ \Omega \end{aligned}$$

Input current and power factor

$$\begin{aligned} Z &= (2 + j3.1) + Z_F + Z_B = 11.724 + j14.487 \\ &= 18.64\angle 51.02^\circ\ \Omega \\ I_1 &= \frac{240\angle 0^\circ}{18.64\angle 51.02^\circ} = 12.88\angle -51.02^\circ\ \text{A} \\ \text{pf} &= \cos 51.02^\circ = 0.629\ \text{lagging} \end{aligned}$$

Input power $= 240 \times 12.88 \times 0.629 = 1944$ W.

Torque developed

$$\begin{aligned} P_{gF} &= I_1^2 R_F = 12.88^2 \times 9.287 = 1540.1\ \text{W} \\ P_{gB} &= I_1^2 R_B = 12.88^2 \times 0.437 = 72.5\ \text{W} \\ P_g &= P_{gF} - P_{gB} = 1467.6\ \text{W} \\ \omega_s &= \frac{2\pi N_s}{60} = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} \\ T &= \frac{P_g}{\omega_s} = \frac{1467.6}{157.08} = 9.34\ \text{N·m} \end{aligned}$$

(Mechanical power developed $= (1-s)P_g = 1394$ W.)

Answer (for s = 0.05, 4-pole, 50 Hz): input current $\approx 12.88$ A, power factor $\approx 0.629$ lagging, torque developed $\approx 9.34$ N·m.