Q. A shunt generator gives full load output of 30 kW at a terminal voltage of 200V. The armature and field resistance are 0.05? and 50? resp.. The iron and friction losses are 1000 W. Calculate
(1) generated emf (2) copper losses (3) efficiency
Ans.
Given
V = 200 V, P = 30 × 10^{3} W,R_{a} = 0.05 W, R_{sh} = 50 W
P_{i+f} = 1000 W
I = P/V = 30 × 10^{3}/200 = 150 Amp
I_{sh} = V/R_{sh} = 200/50 = 4A
I_{a} = I +I_{sh} = 150 + 4 = 154 Amp
(a) E = V + I_{a} R_{a} = 200 + 154 × 0.05 = 207.7 volt
(b) Copper losses = I_{a}^{2} R_{a} + I_{a}^{sh} R_{sh}
= (154)^{2} × 0.05 + (4)^{2} × 50 = 1985.8w
(c) Efficiency ? = Output/ Output + Copper losses + iron losses + friction loss
= 30 × 10^{3}/30 × 10^{3} + 1985.8 + 1000 = 90.95%