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18-Env-A5 Air Quality and Pollution Control Engineering · Undated paper

Question 1 of 5: Sources, Classification and Combustion Air Demand of Atmospheric Pollutants

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

18-Env-A5, Air Quality and Pollution Control Engineering — National Exam, May 2019. 3 hours, closed book. The paper's notes state that Question 1 and 2 are compulsory and two (2) others complete a four-question paper; all five Problems are answered in full below.

Reference texts

Problem 1: Sources, Classification and Combustion Air Demand of Atmospheric Pollutants (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (i) — five outdoor air pollutants: source, health impact, one engineering control each.

PollutantSource of originPotential health impactEngineering control
Particulate matter (PM₂.₅/PM₁₀)Incomplete combustion (diesel exhaust, wood smoke), industrial process emissions (cement kilns, smelters), re-entrained road/construction dustFine fraction penetrates deep into the alveoli, aggravating asthma and cardiovascular disease; classified a Group 1 carcinogenFabric-filter baghouse (or an electrostatic precipitator for sub-micron fume) upstream of the stack
Sulphur dioxide (SO₂)Oxidation of fuel-bound sulphur in coal/heavy-fuel-oil combustion; non-ferrous metal smeltingBronchoconstriction and aggravated asthma on acute exposure; precursor to acid depositionWet limestone flue-gas desulfurization (FGD), absorbing SO₂ into a CaCO₃ slurry
Nitrogen oxides (NOₓ)High-temperature combustion (vehicle engines, utility boilers) — thermal NOₓ (Zeldovich mechanism above ∼1,300 °C) and fuel NOₓAirway inflammation; ozone/smog precursor causing indirect respiratory harm across a wider populationSelective catalytic reduction (SCR): NH₃ injected over a catalyst reduces NOₓ to N₂+H₂O
Carbon monoxide (CO)Incomplete combustion under fuel-rich conditions — cold-start vehicle engines, poorly-maintained heating appliancesBinds haemoglobin ∼200× more strongly than O₂, causing hypoxia and, at high concentration, deathOxidation catalytic converter completes CO→CO₂ combustion in the exhaust stream
Ground-level ozone (O₃)Secondary pollutant — formed photochemically from NOₓ and VOCs in sunlight, not directly emittedAirway irritation, reduced exertional lung function, crop/vegetation yield lossVOC vapour-recovery systems at fuel storage/dispensing facilities cut the hydrocarbon precursor supply

Part (ii) — combustion air demand for the coal-fired plant.

Given. The plant burns 60,000 US tons of coal per day; the supplied reaction is the stoichiometric oxidation of carbon, $C+O_2\rightarrow CO_2$.

QuantityValue
Coal consumption rate60,000 US tons/day
Reaction$C+O_2\rightarrow CO_2$ (coal treated as carbon — see the check note)
Molar mass, C12.011 kg/kmol
Molar mass, O₂32.00 kg/kmol
Air composition21% O₂ by mole (molar mass of air ≈ 28.97 kg/kmol)
Plant rating5,000 MW (context only — see the check note)

Find. The mass and volumetric flow rate of air required to stoichiometrically combust the daily coal feed.

Approach. Convert the coal feed to a molar carbon flow, apply the 1:1 stoichiometry of $C+O_2\rightarrow CO_2$ to get the theoretical O₂ demand, then scale to air using the 21% O₂ mole fraction of standard dry air.

  1. Coal feed rate, SI units. $\dot m_{coal}=60{,}000\ \text{ton}\times 907.185\ \tfrac{\text{kg}}{\text{ton}}=5.443\times10^{7}\ \text{kg/day}$.
  2. Molar carbon flow (coal ≈ pure C). $\dot n_C=\dfrac{\dot m_{coal}}{M_C}=\dfrac{5.443\times10^{7}}{12.011}=4.532\times10^{6}\ \text{kmol/day}$.
  3. Stoichiometric oxygen demand. From $C+O_2\rightarrow CO_2$, $\dot n_{O_2}=\dot n_C$, so $$\dot m_{O_2}=\dot n_{O_2}\,M_{O_2}=4.532\times10^{6}\times32.00=\boxed{1.450\times10^{8}\ \text{kg/day}}$$
  4. Scale oxygen to air (21% O₂ by mole). $\dot n_{air}=\dot n_{O_2}/0.21=2.158\times10^{7}\ \text{kmol/day}$, so $$\dot m_{air}=\dot n_{air}\,M_{air}=2.158\times10^{7}\times28.97=\boxed{6.252\times10^{8}\ \text{kg/day}\ (\approx 7{,}236\ \text{kg/s})}$$
  5. Express as a volumetric flow (STP, 0 °C, 101.325 kPa, 22.414 m³/kmol). $$\dot V_{air}=\dot n_{air}\times22.414=\boxed{4.837\times10^{8}\ \text{m}^3/\text{day}\ (\approx 5{,}598\ \text{m}^3/\text{s})}$$
ResultValue
Stoichiometric O₂ demand1.450×10⁸⁸ kg/day (145,020 t/day)
Theoretical air demand (mass)6.252×10⁸⁸ kg/day (≈7,236 kg/s)
Theoretical air demand (volume, STP)4.837×10⁸⁸ m³/day (≈5,598 m³/s)
Check: (1) the plant's 5,000 MW rating is not needed for this calculation — the coal feed rate (60,000 ton/day) is given directly and no heat-rate/thermal-efficiency figure is supplied to connect capacity to fuel rate, so the MW value is contextual only. (2) With no proximate/ultimate coal analysis supplied and only the elemental reaction $C+O_2\rightarrow CO_2$ given, coal is treated as pure carbon — a standard simplifying assumption for a screening-level air-demand estimate; real bituminous coal is typically 65–80% C by mass, so this is an upper-bound estimate. (3) This is the theoretical (stoichiometric) air requirement; real boilers run 15–20% excess air for complete combustion, which is not asked for here.
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