Excitation emf and maximum power of 1.5 MW motor

Reading of data: at 300 V a 1.5 MW motor would take about 2900 A, and $I_aX_q$ would be many times the supply voltage, so "300 V" is a misprint for 3300 V (the standard version of this problem). Star connection, losses neglected.

$$V = \frac{3300}{\sqrt3} = 1905.26\ \text{V},\quad I_a = \frac{1.5\times10^6}{\sqrt3 \times 3300} = 262.43\ \text{A}\ (\text{upf})$$

Excitation emf (two-reaction method)

$$\begin{aligned} \vec E' &= V - jX_qI_a = 1905.26 - j787.30 = 2061.51\angle -22.45^\circ \Rightarrow \delta = 22.45^\circ \\ I_d &= I_a\sin\delta = 100.22\ \text{A} \\ E_f &= E' + (X_d - X_q)I_d = 2061.51 + 1 \times 100.22 = 2161.7\ \text{V per phase} \end{aligned}$$

Line value $= 3744$ V.

Maximum mechanical power

$$P = 3\left[\frac{E_fV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta\right]$$

With $a = E_fV/X_d = 1029665$ W and $b = V^2(1/X_q - 1/X_d) = 302500$ W:

$$\begin{aligned} \cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = 0.2554 \Rightarrow \delta_m = 75.20^\circ \\ P_{max} &= 3a\sin\delta_m + \tfrac{3b}{2}\sin2\delta_m = 2.987 + 0.224 = 3.21\ \text{MW} \end{aligned}$$

Answer: $E_f = 2162$ V per phase (3744 V line); maximum power $\approx 3.21$ MW at $\delta = 75.2^\circ$.