Discussion
A plant's specific consumption is 400 kWh/tonne at full output. At 60 per cent output, with fixed loads of 20 per cent of the full-load consumption, it becomes about
Answer: C
Correct answer: 453 kWh/tonne
Split the consumption into a fixed part and a part proportional to output. At full load the fixed part is 20 per cent, so at 60 per cent output \[ E = 0.2 + 0.8(0.6) = 0.68 \text{ of the full-load energy,} \] against 0.6 of the output. The specific figure is therefore \[ \frac{0.68}{0.6} \times 400 = 453\ \mathrm{kWh/t}. \] This is why specific consumption must be normalised for capacity utilisation before it is benchmarked.
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