Question

A 50 kw synchronous motor is tested by driving it by another motor. When the excitation is not switched on, the driving motor takes 800w. When the armature is short-circuited and the rated armature current of 10 a is passed through it, the driving motor requires 2500 w. On open-circuiting the armature with rated excitation, the driving motor takes 1800w. Calculate the efficiency of the synchronous motor at 50% load. Neglect the losses in the driving motor.

Answer 1

Answer:

87.72 percentage

Explanation:

The losses corresponding to 'without excitatio condition are frictional and windage losses

Wf= 800 W

When armature is short circuited, losses will be corresponding to copper losses

W=2500 - 800= 1700 W

When armature is open circuited with excitation losses are corresponding to iron losses

W = 1800 - 800= 1000 W

efficiency=

(25,000/ 25,000+800+1700+ 1000)

=87.72%