Starting current and rotor resistance for 150 kW slip ring motor

Given: 150 kW, 3000 V (star), 50 Hz, 4 poles, slip-ring rotor (star), stator/rotor ratio $= 4$, $R_2 = 0.1\ \Omega$, $L_2 = 3.61$ mH per phase. Stator impedance neglected.

Rotor quantities at standstill

$$\begin{aligned} E_1 &= \frac{3000}{\sqrt{3}} = 1732.05\ \text{V/phase} \\ E_2 &= \frac{E_1}{4} = 433.01\ \text{V/phase} \\ X_2 &= 2\pi f L_2 = 2\pi \times 50 \times 3.61\times10^{-3} = 1.1341\ \Omega \\ Z_2 &= \sqrt{0.1^2 + 1.1341^2} = 1.1385\ \Omega \end{aligned}$$

(a) Starting current with slip rings shorted

$$\begin{aligned} I_{2,st} &= \frac{E_2}{Z_2} = \frac{433.01}{1.1385} = 380.3\ \text{A (rotor)} \\ I_{1,st} &= \frac{I_{2,st}}{4} = \frac{380.3}{4} = 95.08\ \text{A (stator/line)} \end{aligned}$$

(b) External resistance to limit starting current to 30 A

Taking 30 A as the stator (line) current, the rotor current must be:

$$I_2 = 30 \times 4 = 120\ \text{A}$$ $$\begin{aligned} Z_{2,new} &= \frac{433.01}{120} = 3.6084\ \Omega \\ R_{total} &= \sqrt{Z_{2,new}^2 - X_2^2} = \sqrt{3.6084^2 - 1.1341^2} = 3.4256\ \Omega \\ R_{ext} &= R_{total} - R_2 = 3.4256 - 0.1 = 3.3256\ \Omega \end{aligned}$$

Answer: (a) starting current $\approx 95.1$ A in the stator (380.3 A in the rotor); (b) external resistance $\approx 3.33\ \Omega$ per phase in the rotor circuit.