CIRCUIT DIAGRAM Figure 4.3 V= supply voltage Motor input = V(I1+I2) Generator output = VI1 ----------------- (a) If we assume both machines have the same efficiency then, ‘η’, η η Output of motor = x input = x V (I1+I2) = input to generator η η ηV η2V(I1+I2)-----(b) Output of generator = x input = x (I1+I2) = Equating (a) and (b), η2V(I1+I2) VI1 = η ~ Therefore, = 11 ………………… &! &!0&# Armature copper loss in motor = (I1 + I2 I4)2ra – Shunt field copper loss in motor = VI4 (8 + 8|) ?: Armature copper loss in generator = Shunt field copper loss in generator = VI3 drawn from supply = VI2 Therefore stray losses = VI2 - + 8 8Z) ?: + (8 + 8|) ?: + = W (say) 12 …… •(8 − €8Z+ €8|• Stray losses/motor = 13 ……………… Therefore for generator Total losses = (8 + 8|) ?: + + = Wg (say) 14 ………………. €8| Output = VI1, therefore ηgenerator = = 15 ……….. %&! %&!0‚ 0'( For motor,
CIRCUIT DIAGRAM Figure 4.3 V= supply voltage Motor input = V(I1+I2) Generator output = VI1 ----------------- (a) If we assume both machines have the same efficiency ‘η’, then, Output of motor = η x input = η x V (I1+I2) = input to generator Output of generator = η x input = η x ηV (I1+I2) = η2V(I1+I2)-----(b) Equating (a) and (b), VI1 = η2V(I1+I2) Therefore, η = ~ &! ………………… 11 &!0&# Armature copper loss in motor = (I1 + I2 – I4)2ra Shunt field copper loss in motor = VI4 Armature copper loss in generator = (8 + 8|) ?: Shunt field copper loss in generator = VI3 drawn from supply = VI2 Therefore stray losses = VI2 - •(8 + 8 − 8Z) ?: + €8Z+ (8 + 8|) ?: + €8|• = W (say) …… 12 Stray losses/motor = ……………… 13 Therefore for generator Total losses = (8 + 8|) ?: + €8| + = Wg (say) ………………. 14 Output = VI1, therefore ηgenerator = %&! = ……….. 15 %&!0‚ 0'( For motor,