A maximum demand controller in a factory acts by
Answer: A
Non-critical loads such as an air compressor, a water heater or a battery charger are dropped for a few minutes so the recorded demand does not cross the threshold. It saves the demand charge without affecting production.
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A plant consuming 12 000 MWh a year to produce 30 000 tonnes has a specific consumption of
Answer: B
Correct answer: 400 kWh/tonne
\[ \frac{12\,000\ \mathrm{MWh}}{30\,000\ \mathrm{t}} = \frac{12\,000\,000\ \mathrm{kWh}}{30\,000\ \mathrm{t}} = 400\ \mathrm{kWh/t}. \] The conversion from megawatt hours to kilowatt hours is where the arithmetic usually goes wrong, and the answers offered differ by exactly that factor of a thousand.
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A 3 per cent reduction target on a specific consumption of 400 kWh/tonne requires it to reach
Answer: C
Correct answer: 388 kWh/tonne
A three per cent reduction means retaining 97 per cent: \[ 400 \times 0.97 = 388\ \mathrm{kWh/t}. \] Note that the target is on the specific figure rather than the absolute one, so a plant can meet it while consuming more energy in total, provided it produces proportionately more.
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Saving 360 MWh of electricity is equivalent to about
Answer: D
Correct answer: 31 tonnes of oil equivalent
One tonne of oil equivalent is 11 630 kWh, so \[ \frac{360\,000}{11\,630} = 31\ \mathrm{toe}. \] The conversion matters because the scheme measures targets and issues certificates in tonnes of oil equivalent, so every fuel and every kilowatt hour must be brought to that common unit before anything can be added up.
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A plant that exceeds its target by 500 tonnes of oil equivalent earns
Answer: A
Correct answer: 500 certificates
One energy saving certificate represents one tonne of oil equivalent saved beyond the target, so a 500 toe over-achievement earns 500 certificates. The one-to-one relation is what makes the certificates fungible across sectors: a tonne saved in a cement works is worth the same as a tonne saved in a smelter.
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A cement of clinker factor 0.70 requires, for one million tonnes of cement,
Answer: B
Seventy per cent of the cement's mass is clinker and the balance is fly ash, slag and gypsum. Lowering the factor to 0.65 would avoid fifty thousand tonnes of clinker and all the kiln fuel and process carbon dioxide that goes with it.
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Reducing a blast furnace's coke rate from 500 to 450 kg per tonne of hot metal saves, on 2 million tonnes,
Answer: C
Correct answer: 100 000 tonnes of coke
The saving is the reduction in rate times the tonnage: \[ (500 - 450)\ \mathrm{kg/t} \times 2 \times 10^6\ \mathrm{t} \] \[ = 50 \times 2 \times 10^6\ \mathrm{kg} = 100\,000\ \mathrm{t}. \] Fifty kilograms per tonne looks a small change and is a very large quantity of coke, which is why the coke rate is watched so closely.
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A smelter using 13 500 kWh per tonne of aluminium consumes, for 100 000 tonnes a year,
Answer: D
Correct answer: 1350 GWh
\[ 13\,500\ \mathrm{kWh/t} \times 100\,000\ \mathrm{t} = 1.35 \times 10^9\ \mathrm{kWh} \] \[ = 1350\ \mathrm{GWh}. \] That is the output of a 150 MW station running continuously, for a single smelter - which is why smelters are built beside hydro schemes and why their power contracts are negotiated over decades.
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Evaporating 1 tonne of water requires about 2260 MJ, which is
Answer: A
Correct answer: 628 kWh
One kilowatt hour is 3.6 MJ, so \[ \frac{2260}{3.6} = 628\ \mathrm{kWh}. \] The figure explains why drying is so costly: removing a tonne of water takes more energy than most processes put into a tonne of product, and it is why mechanical dewatering is always preferred to thermal drying where it is possible.
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A 20 kW machine left idling four hours a day for 300 days consumes
Answer: B
Correct answer: 24 000 kWh
\[ 20\ \mathrm{kW} \times 4\ \mathrm{h} \times 300 = 24\,000\ \mathrm{kWh}. \] This is the archetypal housekeeping saving: nothing is produced during those hours and the remedy costs nothing but discipline. It is also why such savings need continuing attention - the machine will be left running again next month unless someone keeps checking.
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