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Air Pollutants and Sources Questions and Answers

31.

A solvent tank loses \( 0.6 \) per cent of its \( 40 \) tonne annual throughput by evaporation. The volatile organic compound emission is

Answer: C

Correct answer: 240 kg

Take the stated fraction of the throughput:

\[ 40\,000\ \mathrm{kg} \times 0.006 = 240\ \mathrm{kg\ per\ year} \]

The distractor \( 2400 \) treats 0.6 per cent as 6 per cent. This is a fugitive emission - it escapes by evaporation from the tank rather than through any stack - which is why it is estimated from a loss factor applied to throughput rather than measured. Vapour recovery and floating roofs are the usual controls.

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32.

Methane has a hundred year global warming potential of \( 28 \). An emission of \( 1500 \) tonnes is equivalent to

Answer: D

Correct answer: 42000 tonnes of carbon dioxide

Multiply the emission by the global warming potential:

\[ 1500 \times 28 = 42\,000\ \mathrm{tonnes\ CO_2\ equivalent} \]

The distractor \( 53.6 \) divides instead of multiplying. The hundred-year potential of 28 reflects methane being a far stronger absorber than carbon dioxide but much shorter-lived, which is why its twenty-year potential is around 80 - the choice of time horizon materially changes how a methane reduction is valued against a carbon dioxide one.

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33.

An area source emits \( 3.2 \) tonnes of PM10 a year from \( 4.0 \) square kilometres. The emission density is

Answer: A

Correct answer: 0.8 tonnes per km\(^2\) per year

Emission density is the total emission divided by the area:

\[ \frac{3.2\ \mathrm{t/yr}}{4.0\ \mathrm{km^2}} = 0.8\ \mathrm{t\ per\ km^2\ per\ year} \]

The distractor \( 12.8 \) multiplies instead of dividing. Expressing an area source this way is what allows it to be entered into a dispersion model, which distributes the emission over the grid cells the source covers rather than treating it as a plume from a single point.

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34.

An unvented kitchen reaches \( 1200 \) micrograms per cubic metre of PM2.5 for \( 3 \) hours a day against an outdoor level of \( 60 \). The daily exposure ratio between the two, over those hours, is

Answer: B

Correct answer: 20 to 1

Over the hours in question the ratio is simply the two concentrations:

\[ \frac{1200}{60} = 20 \]

So for those three hours the cook breathes air twenty times more polluted than the outdoor air. This is why household air pollution from solid fuel cooking is among the largest environmental health burdens worldwide: the exposure is intense, it falls disproportionately on women and young children, and it is invisible to the ambient monitoring network entirely.

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35.

Statement I: Ground level ozone is controlled by regulating ozone emissions.

Statement II: Ozone is formed in the atmosphere from oxides of nitrogen and volatile organic compounds in sunlight.

In the light of the above statements, choose the correct answer.

Answer: C

Correct answer: Statement I is incorrect but Statement II is correct

Statement I is incorrect. Ground level ozone cannot be controlled by regulating ozone emissions, because ozone is not emitted - it is a secondary pollutant.

Statement II is correct and is why: ozone is formed in the atmosphere from oxides of nitrogen and volatile organic compounds reacting in sunlight. Control must therefore address those precursors, which is far harder than controlling a primary pollutant, since the chemistry is non-linear - reducing nitrogen oxides can raise local ozone in some urban conditions even while lowering it downwind.

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36.

Statement I: Sulphur dioxide emissions scale with the sulphur content of the fuel.

Statement II: Sulphur in the fuel oxidises almost completely during combustion.

In the light of the above statements, choose the correct answer.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both statements are correct and the second explains the first. Sulphur in a fuel oxidises almost completely to sulphur dioxide during combustion, so essentially all of it appears in the flue gas.

Emissions therefore scale directly with the fuel's sulphur content, which is what makes fuel specification such a powerful control: capping sulphur in marine fuel or in diesel cuts emissions in proportion, with no equipment required. Where high-sulphur fuel must be burned, flue gas desulphurisation removes the product instead, typically at 90 to 95 per cent.

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37.

Statement I: Thermal nitrogen oxide formation is insensitive to flame temperature.

Statement II: The reaction between atmospheric nitrogen and oxygen becomes significant above about 1300 degrees.

In the light of the above statements, choose the correct answer.

Answer: A

Correct answer: Statement I is incorrect but Statement II is correct

Statement I is incorrect, and the reason shows why. Thermal NOx formation is extremely sensitive to flame temperature: the Zeldovich mechanism has a high activation energy, so the rate rises steeply above about 1300 degrees Celsius and roughly doubles for each further increment of a few tens of degrees.

Statement II is correct as stated. That sensitivity is what makes temperature the primary control variable - low-NOx burners, staged combustion, flue gas recirculation and water injection all work by lowering or flattening the peak flame temperature rather than by treating the exhaust.

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38.

Statement I: Aerodynamic diameter is preferred to physical diameter for describing particles.

Statement II: It combines size, shape and density into the property that governs settling and deposition.

In the light of the above statements, choose the correct answer.

Answer: B

Correct answer: Both Statement I and Statement II are correct

Both statements are correct and the second explains the first. Aerodynamic diameter is defined as the diameter of a sphere of unit density having the same settling velocity as the particle, so it folds size, shape and density into a single number.

That is exactly the property governing how a particle settles, is collected in a sampler, and deposits in the respiratory tract. A dense compact particle and a light fluffy one of the same physical size behave quite differently; only the aerodynamic diameter predicts either, which is why PM10 and PM2.5 are defined in those terms.

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39.

Statement I: Ammonia emissions from agriculture affect urban particulate levels.

Statement II: Ammonia reacts with sulphuric and nitric acids to form ammonium sulphate and nitrate aerosol.

In the light of the above statements, choose the correct answer.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both statements are correct and the second explains the first. Ammonia is the principal atmospheric base, and it neutralises sulphuric and nitric acid to form ammonium sulphate and ammonium nitrate - both major constituents of fine particulate.

Because ammonia and the acid precursors are transported over long distances before reacting, agricultural emissions from the countryside contribute measurably to urban PM2.5. This is why particulate control cannot be pursued through combustion sources alone, and why agriculture has entered air quality policy in many countries.

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40.

Statement I: Indoor solid fuel combustion is a minor health concern compared with ambient pollution.

Statement II: Concentrations in an unvented kitchen can exceed outdoor levels by a factor of twenty.

In the light of the above statements, choose the correct answer.

Answer: D

Correct answer: Statement I is incorrect but Statement II is correct

Statement I is incorrect. Household air pollution from solid fuel combustion is among the largest environmental health burdens worldwide, responsible for millions of premature deaths a year - not a minor concern.

Statement II is correct and is why: concentrations in an unvented kitchen can exceed outdoor levels twentyfold, and the exposure is concentrated on the women and young children who spend most time near the stove. The ambient monitoring network cannot see any of it, which is part of why the problem was recognised late.

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