Current, pf, power and efficiency of 3.3 kV motor

Assumption: the printed $X_s = 55\ \Omega$ cannot be right: with 55 Ω the largest power the motor could take is about $3VE/X_s \approx 0.26$ MW, less than 750 kW. The standard version of this problem has $Z_s = (0.8 + j5.5)\ \Omega$ per phase, which is used here.

Given: $V = 3300/\sqrt3 = 1905.26$ V, $E_f = 4270/\sqrt3 = 2465.29$ V per phase, input $750$ kW, i.e. $250$ kW per phase.

$$Z_s = 0.8 + j5.5 = 5.5579\angle 81.72^\circ\ \Omega$$

Power angle

Power input per phase of a motor:

$$P = \frac{V^2}{Z_s}\cos\theta - \frac{VE_f}{Z_s}\cos(\theta + \delta)$$ $$\begin{aligned} \frac{V^2}{Z_s}\cos\theta &= \frac{1905.26^2}{5.5579} \times \cos81.72^\circ = 94011\ \text{W} \\ \frac{VE_f}{Z_s} &= \frac{1905.26 \times 2465.29}{5.5579} = 845107\ \text{W} \\ 250000 &= 94011 - 845107\cos(\theta + \delta) \\ \cos(\theta+\delta) &= -0.1846 \Rightarrow \theta + \delta = 100.64^\circ \\ \delta &= 100.64^\circ - 81.72^\circ = 18.91^\circ \end{aligned}$$

Armature current and power factor

$$\begin{aligned} \vec E_r &= \vec V - \vec E_f = 1905.26 - 2465.29\angle -18.91^\circ = -426.94 + j799.06 \\ &= 906.0\angle 118.12^\circ\ \text{V} \\ \vec I_a &= \frac{\vec E_r}{Z_s} = 163.01\angle 36.39^\circ\ \text{A} \end{aligned}$$

Power factor $= \cos36.39^\circ = 0.8050$ leading (over-excited, $E_f > V$).

Check: $\sqrt3 \times 3300 \times 163.01 \times 0.8050 = 750$ kW.

Mechanical power and efficiency

$$\begin{aligned} P_{cu} &= 3I_a^2R_a = 3 \times 163.01^2 \times 0.8 = 63.77\ \text{kW} \\ P_{mech} &= 750 - 63.77 = 686.23\ \text{kW} \\ P_{out} &= 686.23 - 30 = 656.23\ \text{kW} \\ \eta &= \frac{656.23}{750} \times 100 = 87.50\ \% \end{aligned}$$

Answer: $I_a = 163.0$ A, pf $= 0.805$ leading, power angle $= 18.91^\circ$, mechanical power developed $= 686.2$ kW, efficiency $= 87.5\ \%$.