Max-torque speed and efficiency of 400 V slip ring motor

Given: $V_L = 400$ V (star), 4 poles, 50 Hz, slip-ring voltage at standstill $= 190$ V (star), $Z_1 = 0.5 + j2.5\ \Omega$, $R_2 = 0.06\ \Omega$, $X_2 = 0.3\ \Omega$, no-load input $= 600$ W, $N = 1380$ rpm.

Assumption: the stated 1:1 turns ratio does not agree with 400 V / 190 V, so the measured voltages are used for the effective ratio $K = 190/400$. Rotor values are referred to the stator, the approximate equivalent circuit is used, and the 600 W no-load loss is taken as constant (core + mechanical) loss.

Rotor values referred to stator

$$\begin{aligned} K &= \frac{190}{400} = 0.475, \qquad K^2 = 0.225625 \\ R_2' &= \frac{0.06}{0.225625} = 0.2659\ \Omega, \qquad X_2' = \frac{0.3}{0.225625} = 1.3296\ \Omega \\ X &= X_1 + X_2' = 2.5 + 1.3296 = 3.8296\ \Omega \end{aligned}$$

(i) Speed at maximum torque

$$\begin{aligned} s_m &= \frac{R_2'}{\sqrt{R_1^2 + X^2}} = \frac{0.2659}{\sqrt{0.25 + 14.666}} = \frac{0.2659}{3.8621} = 0.06886 \\ N_m &= (1 - 0.06886) \times 1500 = 1396.7\ \text{rpm} \end{aligned}$$

(ii) Efficiency at 1380 rpm

$$s = \frac{1500 - 1380}{1500} = 0.08, \qquad \frac{R_2'}{s} = \frac{0.2659}{0.08} = 3.324\ \Omega$$ $$\begin{aligned} V_{ph} &= \frac{400}{\sqrt{3}} = 230.94\ \text{V} \\ Z &= (0.5 + 3.324) + j3.8296 = 3.824 + j3.830 = 5.412\ \Omega \\ I_1 &= \frac{230.94}{5.412} = 42.67\ \text{A} \end{aligned}$$ $$\begin{aligned} P_{in} &= 3I_1^2(R_1 + R_2'/s) = 3 \times 42.67^2 \times 3.824 = 20\,890\ \text{W} \\ P_{cu1} &= 3 \times 42.67^2 \times 0.5 = 2731\ \text{W} \\ P_{ag} &= 3 \times 42.67^2 \times 3.324 = 18\,158\ \text{W} \\ P_{cu2} &= sP_{ag} = 0.08 \times 18\,158 = 1453\ \text{W} \\ P_m &= 18\,158 - 1453 = 16\,706\ \text{W} \\ P_{out} &= P_m - 600 = 16\,106\ \text{W} \\ \eta &= \frac{16\,106}{20\,890} = 0.771 = 77.1\% \end{aligned}$$

Loss check: $2731 + 1453 + 600 = 4784$ W $= 20\,890 - 16\,106$.

Answer: (i) maximum torque at about 1397 rpm ($s_m = 0.0689$); (ii) efficiency at 1380 rpm $\approx 77.1\%$.