A 250 kw compund generator is delivering 800 A @230 V. The shunt field current is 12 A.

A 250 kw compund generator is delivering 800 A @230 V. The shunt field current is 12 A. The armature resistance is 0.007 ohm. and series field resistance is 0.002 ohm. the stray loss(mechanical loss) is 5.5 Kw and the generator is connected long shunt. Determine full load Efficiency at rated values.

P_rated = 250kW

Power delivered
P = V * I = 230V * 800A = 184kW

Generator is not at full load. Long shunt so I_se = I_a.

Armature current
I_a = I_sh + I_Load = 12A + 800A = 812A

Generated voltage
E₀ = V_Term + I_a (R_a + R_se) = 230V + 812A (0.007Ω + 0.002Ω) = 237.3V

Electrical power delivered
P_e = P_a = E₀ Ia = 237.3V * 812A = 192.7kW

Total mechanical power input into generator
P_in = P_me + P_e = 5.5kW + 192.7kW = 198.2kW

Efficiency = P_out / P_in * 100% = 184kW / 198.2kW * 100% = 92.8%