Slip ring motor max-torque slip and full-load output

Given: $V_L = 380$ V (stator star), rotor star with 180 V between slip rings at standstill, $Z_1 = 0.5 + j2.5\ \Omega$, $R_2 = 0.06\ \Omega$, $X_2 = 0.3\ \Omega$, $T_{st} = 60$ N·m, 4 poles, 50 Hz, full-load slip $= 0.05$.

Method: refer the stator impedance to the rotor side, then use torque ratios (the voltage cancels), with $T_{st}$ as the reference.

Step 1: Turns ratio and referred impedance

Both windings are in star, so the line-voltage ratio equals the phase ratio:

$$\begin{aligned} K &= \frac{180}{380} = 0.47368, \qquad K^2 = 0.22438 \\ R_1' &= K^2 R_1 = 0.22438 \times 0.5 = 0.1122\ \Omega \\ X_1' &= K^2 X_1 = 0.22438 \times 2.5 = 0.5609\ \Omega \\ X &= X_1' + X_2 = 0.5609 + 0.3 = 0.8609\ \Omega \end{aligned}$$

(i) Slip at maximum torque

$$\begin{aligned} s_m &= \frac{R_2}{\sqrt{R_1'^2 + X^2}} = \frac{0.06}{\sqrt{0.1122^2 + 0.8609^2}} = \frac{0.06}{0.8682} = 0.0691 \end{aligned}$$

(Corresponding speed $= 1500(1 - 0.0691) \approx 1396$ rpm.)

(ii) Full-load torque and power

Torque at any slip:

$$T = \frac{3}{\omega_s}\cdot\frac{V'^2\,(R_2/s)}{(R_1' + R_2/s)^2 + X^2}$$

At $s = 0.05$: $R_2/s = 0.06/0.05 = 1.2\ \Omega$.

$$\begin{aligned} \frac{T_{fl}}{T_{st}} &= \frac{R_2/s}{R_2}\cdot\frac{(R_1' + R_2)^2 + X^2}{(R_1' + R_2/s)^2 + X^2} \\ &= \frac{1.2}{0.06}\times\frac{(0.1722)^2 + 0.8609^2}{(1.3122)^2 + 0.8609^2} \\ &= 20 \times \frac{0.7709}{2.4631} = 6.259 \end{aligned}$$ $$T_{fl} = 6.259 \times 60 = 375.6\ \text{N·m}$$

Full-load speed $N = (1 - 0.05) \times 1500 = 1425$ rpm:

$$\begin{aligned} \omega &= \frac{2\pi \times 1425}{60} = 149.23\ \text{rad/s} \\ P_{out} &= T_{fl}\,\omega = 375.6 \times 149.23 = 56\,044\ \text{W} \approx 56.04\ \text{kW} \end{aligned}$$

(Gross mechanical power; mechanical losses are not given.)

Answer: (i) $s_m = 0.0691$; (ii) full-load torque $\approx 375.6$ N·m, power output $\approx 56.04$ kW.