Current, pf and efficiency of 100 hp motor

Assumptions: 1 hp = 746 W; "full load" means 100 hp shaft output; the 4 kW excitation and friction losses are supplied by the developed power.

$$\begin{aligned} V &= \frac{440}{\sqrt3} = 254.03\ \text{V},\quad E_f = \frac{400}{\sqrt3} = 230.94\ \text{V} \\ Z_s &= 0.1 + j1 = 1.0050\angle 84.29^\circ\ \Omega \\ P_{dev} &= 74.6 + 4 = 78.6\ \text{kW} = 26.2\ \text{kW per phase} \end{aligned}$$

Load angle

Mechanical power developed per phase: $P_{dev} = \dfrac{E_fV}{Z_s}\cos(\theta - \delta) - \dfrac{E_f^2}{Z_s}\cos\theta$

$$\begin{aligned} 26200 &= 58376\cos(84.29^\circ - \delta) - 5281 \\ \cos(84.29^\circ - \delta) &= 0.5393 \Rightarrow 84.29^\circ - \delta = 57.37^\circ \\ \delta &= 26.92^\circ \end{aligned}$$

Current and power factor

$$\begin{aligned} \vec I_a &= \frac{\vec V - \vec E_f}{Z_s} = \frac{254.03 - 230.94\angle -26.92^\circ}{1.0050\angle 84.29^\circ} \\ &= 108.30 - j37.30 = 114.54\angle -19.00^\circ\ \text{A} \end{aligned}$$

Power factor $= \cos19.00^\circ = 0.9455$ lagging (under-excited, 400 V < 440 V).

Efficiency

$$\begin{aligned} P_{in} &= \sqrt3 \times 440 \times 114.54 \times 0.9455 = 82.54\ \text{kW} \\ P_{cu} &= 3 \times 114.54^2 \times 0.1 = 3.94\ \text{kW} \quad (\text{check: } 82.54 - 3.94 = 78.6\ \text{kW}) \\ \eta &= \frac{74.6}{82.54} \times 100 = 90.38\ \% \end{aligned}$$

Answer: $I_a = 114.5$ A, pf $= 0.946$ lagging, efficiency $= 90.4\ \%$.