Slip for maximum torque and full-load output of slip ring motor

Given: $V_L = 400$ V, stator in delta (so stator phase voltage $= 400$ V), rotor in star with 190 V between slip rings at standstill, $Z_1 = 0.5 + j2.5\ \Omega$, $R_2 = 0.06\ \Omega$, $X_2 = 0.3\ \Omega$, $T_{max} = 150$ N·m, 4 poles, 50 Hz, full-load slip $= 0.04$.

Step 1: Turns ratio and stator impedance referred to rotor

$$\begin{aligned} E_{2,ph} &= \frac{190}{\sqrt{3}} = 109.70\ \text{V}, \qquad E_{1,ph} = 400\ \text{V} \\ K &= \frac{109.70}{400} = 0.27424, \qquad K^2 = 0.075208 \\ R_1' &= K^2 R_1 = 0.075208 \times 0.5 = 0.0376\ \Omega \\ X_1' &= K^2 X_1 = 0.075208 \times 2.5 = 0.1880\ \Omega \\ X &= X_1' + X_2 = 0.1880 + 0.3 = 0.4880\ \Omega \end{aligned}$$

(i) Slip at maximum torque

$$\begin{aligned} s_m &= \frac{R_2}{\sqrt{R_1'^2 + X^2}} = \frac{0.06}{\sqrt{0.0376^2 + 0.4880^2}} = \frac{0.06}{0.48947} = 0.1226 \end{aligned}$$

(Speed at maximum torque $= 1500(1 - 0.1226) \approx 1316$ rpm.)

(ii) Full-load torque and power

$$T = \frac{3}{\omega_s}\frac{V'^2 (R_2/s)}{(R_1' + R_2/s)^2 + X^2}, \qquad T_{max} = \frac{3}{\omega_s}\frac{V'^2}{2\left(R_1' + \sqrt{R_1'^2 + X^2}\right)}$$

At $s = 0.04$: $R_2/s = 0.06/0.04 = 1.5\ \Omega$.

$$\begin{aligned} \frac{T_{fl}}{T_{max}} &= \frac{(R_2/s)\cdot 2\left(R_1' + \sqrt{R_1'^2 + X^2}\right)}{(R_1' + R_2/s)^2 + X^2} \\ &= \frac{1.5 \times 2(0.0376 + 0.48947)}{(1.5376)^2 + 0.4880^2} \\ &= \frac{1.5 \times 1.05414}{2.6024} = 0.6076 \end{aligned}$$ $$T_{fl} = 0.6076 \times 150 = 91.14\ \text{N·m}$$

Full-load speed $N = (1 - 0.04) \times 1500 = 1440$ rpm:

$$\begin{aligned} \omega &= \frac{2\pi \times 1440}{60} = 150.80\ \text{rad/s} \\ P_{out} &= T_{fl}\,\omega = 91.14 \times 150.80 = 13\,744\ \text{W} \approx 13.74\ \text{kW} \end{aligned}$$

(This is the gross mechanical power developed; mechanical losses are not given.)

Answer: (i) $s_m = 0.1226$; (ii) full-load torque $= 91.14$ N·m, power output $\approx 13.74$ kW.