Torque at 1450 rpm from starting torque

Given: 4 poles, 50 Hz, $T_{st} = 150$ N·m, $R_2 = 0.5\ \Omega$, $X_2 = 2\ \Omega$ at standstill, $E_2 = 400$ V/phase, $N = 1450$ rpm. Stator impedance neglected.

Slip

$$\begin{aligned} N_s &= \frac{120 \times 50}{4} = 1500\ \text{rpm} \\ s &= \frac{1500 - 1450}{1500} = 0.03333 \end{aligned}$$

Torque ratio

$$T = \frac{K s E_2^2 R_2}{R_2^2 + (sX_2)^2}, \qquad T_{st} = \frac{K E_2^2 R_2}{R_2^2 + X_2^2}$$

Since $E_2$ is the same in both, it cancels:

$$\begin{aligned} \frac{T}{T_{st}} &= \frac{s(R_2^2 + X_2^2)}{R_2^2 + (sX_2)^2} \\ &= \frac{0.03333 \times (0.25 + 4)}{0.25 + (0.03333 \times 2)^2} \\ &= \frac{0.14167}{0.25 + 0.00444} = \frac{0.14167}{0.25444} = 0.5568 \end{aligned}$$ $$T = 0.5568 \times 150 = 83.52\ \text{N·m}$$

Rotor current at this speed (for information): $I_2 = \dfrac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}} = \dfrac{13.33}{0.5044} = 26.4$ A.

Answer: Torque at 1450 rpm $\approx 83.5$ N·m.