Question

A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW. The magnitude of the per phase excitation emf of the motor, in volts, is _________. (round off to nearest integer)

Hint:

The excitation emf in a synchronous motor can be calculated using the phase voltage and the synchronous reactance.

Answer 1

Correct Answer: 240

The total power supplied to the motor is the sum of the shaft power, losses, and the reactive power. The power supplied by the generator is: \[ S = P + jQ \] where \( P = 10 \, \text{kW} \) (shaft power) and \( Q \) is the reactive power. The apparent power \( S \) is given by: \[ S = \frac{P}{\text{Power Factor}} = \frac{10}{0.8} = 12.5 \, \text{kVA} \] The current per phase \( I_{\text{ph}} \) is: \[ I_{\text{ph}} = \frac{S}{\sqrt{3} \times V_{\text{L}}} = \frac{12.5 \times 10^3}{\sqrt{3} \times 400} = 18.1 \, \text{A} \] Using the synchronous reactance \( X_s = 1 \, \Omega \), the excitation emf \( E_f \) is given by: \[ E_f = \sqrt{V_{\text{L}}^2 + (I_{\text{ph}} X_s)^2} = \sqrt{400^2 + (18.1 \times 1)^2} = \sqrt{160000 + 327.61} = 400.4 \, \text{V} \] Thus, the magnitude of the per phase excitation emf of the motor is approximately \( 240 \, \text{V} \) (rounded to nearest integer).