Question 4.1: A 220 V, 50 kW dc shunt generator was run as a motor on no l...

Input power at no load = VI

= 220 × 8
= 1760 W

Iron, friction and windage, and field copper losses = No-load input – I_{ao} \ ^{2}  R_{a}

= 1760 − 6² × 0.1
= 1760 − 3.6
= 1756.4 W

These are constant losses.
Note that I_{ao} \ ^{2}  R_{a} loss in the armature at no load is very small.
The generator is rated at 50 kW and 220 V.
The generator output current at full load.

=\frac{50  ×  1000}{220}= 227.2  A

Full-load armature Copper loss = I_{a} \ ^{2}  R_{a}

= (229.2)²  ×  0.1
= 5253 W

This is the variable loss.

Efficiency of the generator in percentage = Output/(Output + Constant losses + Variable loss)

=\frac{50  ×  1000  ×  100}{50  ×  1000  +  1756.4  +  5253}

= 89 per cent