Calculate the kVA output and phase currents for a 400 kW motor with a 3300 V line voltage, 94% efficiency, and 0.8 lagging power factor.

The alternator output power is the motor input power.

$$P_{in\_motor} = \frac{P_{out\_motor}}{\eta_{motor}}$$ $$P_{in\_motor} = \frac{400 \times 10^3 \text{ W}}{0.94}$$ $$P_{in\_motor} \approx 425531.91 \text{ W}$$

The kVA output of the alternator is the apparent power input to the motor.

$$S_{alternator} = \frac{P_{in\_motor}}{\cos\theta}$$ $$S_{alternator} = \frac{425531.91 \text{ W}}{0.8}$$ $$S_{alternator} = 531914.89 \text{ VA}$$ $$S_{alternator} = \boxed{\text{531.91 kVA}}$$

For a three-phase system, the apparent power is $S = \sqrt{3} V_L I_L$.

$$I_L = \frac{S_{alternator}}{\sqrt{3} V_L}$$ $$I_L = \frac{531914.89 \text{ VA}}{\sqrt{3} \times 3300 \text{ V}}$$ $$I_L = \frac{531914.89}{5715.76} \text{ A}$$ $$I_L \approx 93.06 \text{ A}$$

The alternator is star-connected. For a star connection, the line current is equal to the phase current.

$$I_{ph\_alternator} = I_L$$ $$I_{ph\_alternator} = \boxed{\text{93.06 A}}$$

The motor is delta-connected. For a delta connection, the phase current is the line current divided by $\sqrt{3}$.

$$I_{ph\_motor} = \frac{I_L}{\sqrt{3}}$$ $$I_{ph\_motor} = \frac{93.06 \text{ A}}{\sqrt{3}}$$ $$I_{ph\_motor} \approx \frac{93.06}{1.732} \text{ A}$$ $$I_{ph\_motor} = \boxed{\text{53.73 A}}$$