Energy Balance and Benchmarking Questions and Answers

Answer: A

The gross figure is the larger denominator, so the same useful output divided by it gives the smaller number. Comparing two boilers quoted on different bases can reverse the apparent ranking, which is why the basis must always be stated.

Answer: B

Correct answer: 3.6

A kilowatt hour is a thousand watts for 3,600 seconds: \[ 1000 \times 3600 = 3.6\times10^6\ \mathrm{J} = 3.6\ \mathrm{MJ}. \] The conversion is worth knowing by heart, since energy accounts routinely mix electrical units with thermal ones and the two must be brought to a common basis before anything can be added.

Answer: C

Correct answer: 4.19

A calorie was defined as the heat raising a gram of water by one degree, and the mechanical equivalent is 4.187 joules. So a kilocalorie is about 4.19 kilojoules. The figure appears constantly in fuel and steam calculations, where calorific values are still often quoted in kilocalories.

Answer: D

Correct answer: 400 kWh/t

\[ \frac{4{,}800{,}000\ \mathrm{kWh}}{12{,}000\ \mathrm{t}} = 400\ \mathrm{kWh/t}. \] The megawatt hours must be converted to kilowatt hours first; working in the given units and forgetting the factor of a thousand gives 0.4, and misplacing it gives the 40 offered here.

Answer: A

Correct answer: risen to 440 kWh/t despite the lower total

\[ \frac{4{,}400{,}000}{10{,}000} = 440\ \mathrm{kWh/t}. \] Consumption has fallen by 400 MWh and performance has worsened, because output fell further. This is exactly why specific consumption is tracked rather than the total - a bill that falls in a bad year says nothing good about the plant.

Answer: B

Correct answer: 40 kWh/t

The marginal figure is the slope of the line between the two points: \[ \frac{6000 - 2000}{100} = 40\ \mathrm{kWh/t}. \] Dividing the full-output consumption by the output instead gives 60, which is the average specific consumption at full load and includes a share of the fixed load.

Answer: C

Correct answer: one third

\[ \frac{2000}{6000} = \frac{1}{3}. \] A third of the energy is drawn whether or not anything is produced, which is why running at part load is so expensive per tonne and why shutting down properly rather than idling is often the largest single saving available.

Answer: D

Correct answer: 12 000 kWh, some 13 per cent of the total

\[ 90{,}000 - 78{,}000 = 12{,}000\ \mathrm{kWh}, \qquad \frac{12{,}000}{90{,}000} = 13.3\ \mathrm{per\ cent}. \] The proportion is taken against the main meter, which is the true total; taking it against the sub-metered sum gives 15 per cent, which is offered here. Unaccounted energy of this size points to unmetered loads rather than to meter error.

Answer: A

Five hundred kilograms an hour is 0.139 kilograms a second, and multiplying by 42 megajoules a kilogram gives 5.83 megawatts. Converting the hourly figure to a per-second one before multiplying is the step most often fumbled.

Answer: B

Correct answer: 81 per cent

The input is \( 500 \times 42 = 21{,}000 \) MJ an hour, which is \( 21{,}000/3600 = 5.83 \) MW. The efficiency is then \[ \frac{4.7}{5.83} = 0.81. \] Note the units must be reconciled first - megajoules per hour against megawatts - and the answer above 100 per cent offered here is what results from dividing the wrong way round.