Separately excited elevator motor with field flux reduced

In a separately excited motor $T = k\phi I_a$ and $E_b = k\phi\omega$. The developed power $E_b I_a = T\omega$.

Given: $T = 300$ N·m (constant), $V = 600$ V, $N_1 = 1500$ rpm, $R_a = 0.5\ \Omega$.

Part 1: Armature current

$$\omega_1 = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s}$$ $$E_{b1} I_{a1} = T \omega_1 = 300 \times 157.08 = 47\,124\ \text{W}$$

With $E_{b1} = 600 - 0.5 I_{a1}$:

$$\begin{aligned} (600 - 0.5 I_{a1}) I_{a1} &= 47\,124 \\ 0.5 I_{a1}^2 - 600 I_{a1} + 47\,124 &= 0 \\ I_{a1} &= \frac{600 - \sqrt{600^2 - 4(0.5)(47\,124)}}{2(0.5)} = 84.49\ \text{A} \end{aligned}$$

(The other root, about 1115 A, is not a practical operating point.)

$$E_{b1} = 600 - 0.5 \times 84.49 = 557.76\ \text{V}$$

Part 2: Flux reduced by 10% ($\phi_2 = 0.9\phi_1$)

Torque is constant, so $\phi_1 I_{a1} = \phi_2 I_{a2}$:

$$I_{a2} = \frac{84.49}{0.9} = 93.88\ \text{A}$$ $$E_{b2} = 600 - 0.5 \times 93.88 = 553.06\ \text{V}$$

Speed:

$$\begin{aligned} N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2} \\ &= 1500 \times \frac{553.06}{557.76} \times \frac{1}{0.9} = 1652.6\ \text{rpm} \end{aligned}$$

Answer: Initially $I_a \approx 84.5$ A. With 10% less flux, $I_a \approx 93.9$ A and speed $\approx 1653$ rpm.