Energy conservation

Question

a shunt generator gives full load out put of  30 kw of a terminal voltage of 200 volt. The armature and shunt fild resistance are 0.05 ohm and 50 ohm respectively. the iron  and friction losses are 1000 watt . calculate the generated emf, Cooper losses,effeciency

Answer 1

Line current I_L = 30*1000/200 = 300A

I_f = 200/50 = 4A

Armature current I_a = I_L - I_f = 300-4 = 296A

Induced EMF = V + I_aR_a

=200 + 296*0.05 = 214.8V

 Copper losses = 0.05*296*296 = 4380 W

Efficiency = Output power / input power +losses

= 30*1000 / (30000+5380) = 85%