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EMF, copper loss and efficiency of 30 kW shunt generator

Short answer

Load current is 150 A, shunt current 4 A, armature current 154 A. Generated emf = 200 + 154×0.05 = 207.7 V. Copper losses are 1185.8 W armature plus 800 W shunt = 1985.8 W; efficiency = 30000/(30000+1985.8+1000) = 90.95%.

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 ohm and 50 ohm respectively. The iron and friction losses are 1000 W. Calculate: (i) generated emf, (ii) copper losses; (iii) efficiency

Given: Pout=30P_{out} = 30 kW, V=200V = 200 V, Ra=0.05 ΩR_a = 0.05\ \Omega, Rsh=50 ΩR_{sh} = 50\ \Omega, iron + friction loss = 1000 W.

Currents

IL=30000200=150 AIsh=20050=4 AIa=IL+Ish=154 A\begin{aligned} I_L &= \frac{30000}{200} = 150\ \text{A} \\ I_{sh} &= \frac{200}{50} = 4\ \text{A} \\ I_a &= I_L + I_{sh} = 154\ \text{A} \end{aligned}

(i) Generated emf

Eg=V+IaRa=200+154×0.05=207.7 V\begin{aligned} E_g &= V + I_a R_a \\ &= 200 + 154 \times 0.05 = 207.7\ \text{V} \end{aligned}

(ii) Copper losses

Pcu,a=Ia2Ra=1542×0.05=1185.8 WPcu,sh=VIsh=200×4=800 WPcu=1185.8+800=1985.8 W\begin{aligned} P_{cu,a} &= I_a^2 R_a = 154^2 \times 0.05 = 1185.8\ \text{W} \\ P_{cu,sh} &= V I_{sh} = 200 \times 4 = 800\ \text{W} \\ P_{cu} &= 1185.8 + 800 = 1985.8\ \text{W} \end{aligned}

(iii) Efficiency

Ploss=1985.8+1000=2985.8 WPin=30000+2985.8=32985.8 Wη=3000032985.8×100=90.95%\begin{aligned} P_{loss} &= 1985.8 + 1000 = 2985.8\ \text{W} \\ P_{in} &= 30000 + 2985.8 = 32985.8\ \text{W} \\ \eta &= \frac{30000}{32985.8} \times 100 = 90.95\% \end{aligned}
QuantityValue
Generated emf207.7 V
Armature copper loss1185.8 W
Shunt copper loss800 W
Total copper loss1985.8 W
Efficiency90.95 %

Answer: (i) Eg=207.7E_g = 207.7 V, (ii) total copper loss =1985.8= 1985.8 W, (iii) η≈90.95%\eta \approx 90.95\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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