Effect of five-times rotor resistance on 40 kW slip ring motor

With negligible stator impedance, torque depends on $R_2/s$:

$$T = \frac{K s E_2^2 R_2}{R_2^2 + (sX_2)^2} = \frac{K E_2^2 (R_2/s)}{(R_2/s)^2 + X_2^2}$$

So for the same torque, $R_2/s$ stays constant, i.e. $s \propto R_2$. Also $s_m = R_2/X_2 \propto R_2$.

Given: rated output 40 kW at $N = 0.96 N_s$, so $s_1 = 0.04$. $s_{m1} = 4 \times 0.04 = 0.16$. New rotor resistance $R_2' = 5R_2$.

Assumption: 40 kW is the mechanical power developed at rated torque (mechanical losses neglected).

Original condition

$$\begin{aligned} P_{ag} &= \frac{P_m}{1 - s_1} = \frac{40}{0.96} = 41.667\ \text{kW} \\ P_{cu2} &= s_1 P_{ag} = 0.04 \times 41.667 = 1.667\ \text{kW} \end{aligned}$$

(i) Rated torque with $5R_2$

$$\begin{aligned} s_2 &= 5 s_1 = 5 \times 0.04 = 0.20 \\ N_2 &= (1 - 0.20) N_s = 0.80\, N_s \end{aligned}$$

Torque and synchronous speed are unchanged, so air-gap power $P_{ag} = T\omega_s$ is unchanged $= 41.667$ kW.

$$\begin{aligned} P_{out} &= (1 - s_2) P_{ag} = 0.8 \times 41.667 = 33.33\ \text{kW} \\ P_{cu2} &= s_2 P_{ag} = 0.2 \times 41.667 = 8.333\ \text{kW} \end{aligned}$$

(ii) Speed at maximum torque

$$\begin{aligned} s_{m2} &= 5 s_{m1} = 5 \times 0.16 = 0.80 \\ N_{m2} &= (1 - 0.80) N_s = 0.20\, N_s \end{aligned}$$

(The maximum torque value itself is unchanged.)

Answer: (i) speed $= 0.8N_s$, output $= 33.33$ kW, rotor Cu loss $= 8.333$ kW; (ii) speed at maximum torque $= 0.2N_s$ (slip 0.8). For example, for a 4-pole 50 Hz machine ($N_s = 1500$ rpm) these speeds would be 1200 rpm and 300 rpm.