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 × 10³ W,R_a = 0.05 W, R_sh = 50 W

                             P_i+f = 1000 W

                                  I = P/V = 30 × 10³/200 = 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_a² 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 × 10³/30 × 10³ + 1985.8 + 1000 = 90.95%