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30 kW shunt generator: emf, copper loss, efficiency (100 W)

Short answer

Load current 150 A, shunt current 4 A, armature current 154 A. Generated emf E = 200 + 154×0.05 = 207.70 V. Copper losses: armature 1185.8 W, shunt 800 W, total 1985.8 W. Efficiency = 30000/(30000+1985.8+100) × 100 = 93.50%.

A dc shunt generator gives full load output of 30 kW at a terminal voltage of 200 V. The armature and shunt field resistances are 0.05 Ω0.05\ \Omega and 50 Ω50\ \Omega respectively. The iron and friction losses are 100 W. Calculate (i) generated emf (ii) copper losses (iii) efficiency.

For a shunt generator Ia=IL+IshI_a = I_L + I_{sh} and E=V+IaRaE = V + I_aR_a.

(i) Generated emf

IL=30000200=150 A,Ish=20050=4 AI_L = \frac{30000}{200} = 150\ \text{A}, \qquad I_{sh} = \frac{200}{50} = 4\ \text{A} Ia=IL+Ish=154 AI_a = I_L + I_{sh} = 154\ \text{A} E=V+IaRa=200+(154)(0.05)=207.70 VE = V + I_aR_a = 200 + (154)(0.05) = 207.70\ \text{V}

(ii) Copper losses

Armature copper loss=Ia2Ra=(154)2(0.05)=1185.8 WShunt field copper loss=Ish2Rsh=(4)2(50)=800.0 WTotal copper loss=1185.8+800.0=1985.8 W\begin{aligned} \text{Armature copper loss} &= I_a^2R_a = (154)^2(0.05) = 1185.8\ \text{W} \\ \text{Shunt field copper loss} &= I_{sh}^2R_{sh} = (4)^2(50) = 800.0\ \text{W} \\ \text{Total copper loss} &= 1185.8 + 800.0 = 1985.8\ \text{W} \end{aligned}

(iii) Efficiency

Total losses == copper loss ++ iron and friction loss =1985.8+100=2085.8 W= 1985.8 + 100 = 2085.8\ \text{W}.

η=outputoutput+losses=3000030000+2085.8×100=93.50%\eta = \frac{\text{output}}{\text{output} + \text{losses}} = \frac{30000}{30000 + 2085.8}\times 100 = 93.50\%

Answer: (i) E=207.70 VE = 207.70\ \text{V}; (ii) copper loss =1985.8 W= 1985.8\ \text{W}; (iii) η=93.50%\eta = 93.50\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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