Total copper loss is:

1985.8 W (armature 1185.8 W + shunt 800 W)

Efficiency uses total losses of:

copper loss plus iron and friction = 2985.8 W, giving 90.95%

Asked 2/21
2074 Bhadra (old course)

30 kW shunt generator: emf, copper loss, efficiency

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+1000) × 100 = 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.

For a shunt generator $I_a = I_L + I_{sh}$ and $E = V + I_aR_a$ .

(i) Generated emf

I_L = \frac{30000}{200} = 150\ \text{A}, \qquad I_{sh} = \frac{200}{50} = 4\ \text{A}

I_a = I_L + I_{sh} = 154\ \text{A}

E = V + I_aR_a = 200 + (154)(0.05) = 207.70\ \text{V}

(ii) Copper losses

\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 $= 1985.8 + 1000 = 2985.8\ \text{W}$ .

\eta = \frac{\text{output}}{\text{output} + \text{losses}} = \frac{30000}{30000 + 2985.8}\times 100 = 90.95\%

Answer: (i) $E = 207.70\ \text{V}$ ; (ii) copper loss $= 1985.8\ \text{W}$ ; (iii) $\eta = 90.95\%$ .

✓ Answer checked by Claude Opus 5.5 (high effort)

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30 kW shunt generator: emf, copper loss, efficiency

(i) Generated emf

(ii) Copper losses

(iii) Efficiency

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