Coal of \( 2.4 \) per cent sulphur is burnt at \( 500 \) tonnes a day with complete conversion to sulphur dioxide. The daily emission is
Answer: A
The sulphur is \( 500 \times 0.024 = 12 \) tonnes, and each tonne of sulphur gives two tonnes of sulphur dioxide because the molecular masses are 32 and 64. Emissions therefore scale directly with the fuel's sulphur content.
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A flue gas desulphurisation plant removes \( 92 \) per cent of \( 24 \) tonnes a day of sulphur dioxide. The residual emission is
Answer: B
Correct answer: 1.92 tonnes
With 92 per cent removed, 8 per cent remains:
\[ 24 \times (1 - 0.92) = 24 \times 0.08 = 1.92\ \mathrm{tonnes\ per\ day} \]
The distractor \( 22.1 \) is the quantity removed rather than the residual, and \( 0.92 \) confuses the efficiency with a mass. Note how much a few percentage points of efficiency are worth at this scale: raising removal from 92 to 96 per cent would halve the residual emission again.
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A vehicle emits carbon monoxide at \( 1.8 \) g/km and travels \( 15000 \) km a year. Its annual emission is
Answer: C
Correct answer: 27.0 kg
Multiply the emission factor by the activity level, watching the units:
\[ 1.8\ \mathrm{g/km} \times 15\,000\ \mathrm{km} = 27\,000\ \mathrm{g} = 27.0\ \mathrm{kg\ per\ year} \]
This is the basic operation of an emission inventory - a factor relating emitted mass to an activity that is already recorded, in this case vehicle kilometres. No measurement of the individual vehicle is required, which is exactly why inventories can cover millions of sources.
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A city of \( 800000 \) vehicles each emitting \( 27 \) kg of carbon monoxide a year emits in total
Answer: D
Correct answer: 21600 tonnes
Scale the per-vehicle emission to the fleet:
\[ 800\,000 \times 27\ \mathrm{kg} = 21\,600\,000\ \mathrm{kg} = 21\,600\ \mathrm{tonnes\ per\ year} \]
The step from one vehicle to a city is what makes an area or line source treatment necessary: no inventory tracks 800,000 vehicles individually, so the fleet is represented by activity data and a factor. The arithmetic also shows why a small reduction in the per-vehicle factor is worth a great deal in aggregate.
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Ambient PM2.5 is \( 68 \) micrograms per cubic metre against a guideline of \( 15 \). The exceedance factor is
Answer: A
Correct answer: 4.53
The exceedance factor is the measured concentration divided by the guideline:
\[ \frac{68}{15} = 4.53 \]
The distractor \( 0.22 \) inverts the ratio, and \( 53.0 \) is the arithmetic difference. Expressing the result as a multiple rather than a difference is standard practice because it is comparable across pollutants with quite different guideline values, and it conveys the scale of the problem more directly than an absolute excess.
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A stack emits \( 40 \) g/s of particulate at a gas flow of \( 25 \) m\(^3\)/s. The concentration in the flue gas is
Answer: B
Correct answer: 1600 mg/m\(^3\)
Concentration is the mass emission rate divided by the volumetric gas flow:
\[ C = \frac{40\ \mathrm{g/s}}{25\ \mathrm{m^3/s}} = 1.6\ \mathrm{g/m^3} = 1600\ \mathrm{mg/m^3} \]
The unit conversion is where the marks are lost: a gram per cubic metre is a thousand milligrams per cubic metre. In practice the figure would also have to be corrected to reference conditions - a stated temperature, pressure and oxygen content - before it could be compared with an emission limit.
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A control device must bring \( 1600 \) mg/m\(^3\) down to \( 50 \) mg/m\(^3\). The required collection efficiency is
Answer: C
Correct answer: 96.9 per cent
Collection efficiency is the fraction of the inlet loading removed:
\[ \eta = \frac{C_{in} - C_{out}}{C_{in}} = \frac{1600 - 50}{1600} = 0.969 = 96.9\ \mathrm{per\ cent} \]
The distractor \( 3.1 \) is the fraction penetrating rather than collected. Note how demanding tight emission limits are: reaching \( 50 \) needs 96.9 per cent, but going to \( 20 \) would need 98.75 - a small change in the limit requiring a large change in the device.
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A boiler burns \( 12 \) tonnes of fuel an hour with an emission factor of \( 8.5 \) kg of nitrogen oxides per tonne. The hourly emission is
Answer: D
Correct answer: 102 kg
Multiply the emission factor by the fuel consumption:
\[ 12\ \mathrm{t/h} \times 8.5\ \mathrm{kg/t} = 102\ \mathrm{kg\ per\ hour} \]
The distractor \( 1020 \) misplaces a decimal and \( 70.8 \) divides rather than multiplies. Emission factors of this kind, tabulated by fuel and combustion technology, are what let an inventory estimate a plant's output from records that already exist - here the fuel purchase ledger.
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Particulate of density \( 2000 \) kg/m\(^3\) and physical diameter \( 5 \) micrometres has an aerodynamic diameter of about
Answer: A
Aerodynamic diameter scales as the physical diameter times the square root of the density ratio, so \( 5\sqrt{2} = 7.07 \) micrometres. A dense particle therefore behaves like a larger one in both the lung and a control device.
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A power station emits \( 21 \) tonnes of nitrogen oxides a day and generates \( 4800 \) MWh. The emission rate per unit generated is
Answer: B
Correct answer: 4.375 kg per MWh
Divide the emission by the output, converting tonnes to kilograms:
\[ \frac{21\,000\ \mathrm{kg}}{4800\ \mathrm{MWh}} = 4.375\ \mathrm{kg\ per\ MWh} \]
Normalising emissions to output rather than reporting them as an absolute rate allows plants of different size and utilisation to be compared, and it is the form in which emission performance standards are usually written. It also separates two distinct routes to lower emissions: cleaner combustion, and simply generating more efficiently.
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