NivaarExam PrepOfficial exam papers ↗

18-Env-A5 Air Quality and Pollution Control Engineering · December 2019

Question 5 of 7: Particulate Measurement, SO₂ Combustion Stoichiometry and Settling Velocity

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, December 2019. 3 hours, closed book (candidate-prepared double-sided aid sheet allowed). The paper's notes state that any five (5) of the seven Problems, as they appear in the workbook, constitute a complete paper; all seven Problems are answered in full below.

Reference texts

Problem 5: Particulate Measurement, SO₂ Combustion Stoichiometry and Settling Velocity (20 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) — gravimetric/filter-soiling analysis. A known volume of ambient air is drawn through a pre-weighed filter over a fixed sampling period (typically 24 hours for a hi-vol PM sampler); the filter is conditioned (equilibrated to a controlled temperature/humidity) and re-weighed, and the mass gain divided by the sampled air volume gives the particulate mass concentration ($\mu\text{g/m}^3$). Example: the U.S. EPA Federal Reference Method (FRM) high-volume sampler used at regulatory PM₁₀/PM₂.₅ monitoring stations. Calibration: the sampler's flow rate is periodically verified against a certified orifice or venturi transfer standard traceable to a national metrology standard, and the analytical balance is checked with certified reference weights. This is important because it ensures the reported concentration is accurate and directly comparable across monitoring stations and over time — the basis for regulatory compliance determinations and long-term air-quality trend analysis.

Part (ii) — PM₂.₅ vs. PM₁₀ (four differences). Health — PM₂.₅: fine particles penetrate deep into the alveolar region and can cross into the bloodstream, producing systemic cardiovascular and respiratory effects. Health — PM₁₀: the coarse fraction (2.5–10 µm) deposits mainly in the upper respiratory tract and bronchi, causing irritation and aggravating asthma, but does not reach the deep lung as efficiently. Aesthetics — PM₂.₅: particle diameters close to the wavelength of visible light scatter/absorb light very efficiently, producing regional haze and visibility reduction that can persist and travel long distances. Aesthetics — PM₁₀: coarse particles settle out of the atmosphere much faster, so their nuisance is localized soiling of surfaces and short-range visible dust rather than regional haze.

Part (iii) — SO₂ concentration in flue gas.

Given. 10 mol of fuel with the printed formula $C_7H_{13}$, containing 2% sulphur by mass of the hydrocarbon fuel, burnt with the stoichiometric amount of O₂ supplied as air (21% O₂ / 79% N₂ by volume).

QuantityValue
Moles of fuel, $n_{fuel}$10 mol (as $C_7H_{13}$ — see the check note)
Sulphur content2% by mass of the hydrocarbon fuel
Oxidantstoichiometric O₂, supplied as air (21% O₂/79% N₂ by volume)
Molar massesC = 12.011, H = 1.008, S = 32.07 g/mol

Find. The SO₂ concentration in the flue gas (ppmv).

Approach. Compute the fuel and sulphur mass/mole flows, balance the hydrocarbon combustion equation for the stoichiometric O₂ demand (plus the O₂ consumed oxidizing S to SO₂), scale to air via the 21% O₂ mole fraction (carrying the inert N₂ through unreacted), then divide SO₂ moles by the total flue-gas moles.

  1. Fuel molar mass and sulphur moles. $M_{fuel}=7(12.011)+13(1.008)=97.18\ \text{g/mol}$, so $m_{fuel}=10\times97.18=971.8\ \text{g}$ and $m_S=0.02\times971.8=19.44\ \text{g}$, giving $n_S=19.44/32.07=0.606\ \text{mol S}\rightarrow0.606\ \text{mol SO}_2$.
  2. Hydrocarbon combustion stoichiometry. $C_7H_{13}+10.25\,O_2\rightarrow 7CO_2+6.5H_2O$ (balancing O: $7\times2+6.5=20.5\Rightarrow10.25\ \text{mol O}_2$ per mol fuel), so for 10 mol fuel: $n_{O_2,fuel}=10\times10.25=102.5\ \text{mol}$.
  3. Total O₂ demand and air supplied. Adding the sulphur-oxidation O₂ ($S+O_2\rightarrow SO_2$): $n_{O_2,total}=102.5+0.606=103.1\ \text{mol}$. As air (21% O₂): $n_{N_2}=n_{O_2,total}\times(79/21)=387.9\ \text{mol}$ rides through unreacted.
  4. Flue-gas composition and SO₂ concentration. $n_{CO_2}=7\times10=70\ \text{mol}$, $n_{H_2O}=6.5\times10=65\ \text{mol}$, $n_{SO_2}=0.606\ \text{mol}$; total flue gas $=70+65+0.606+387.9=523.5\ \text{mol}$, so $$C_{SO_2}=\frac{0.606}{523.5}\times10^{6}=\boxed{1{,}158\ \text{ppmv}\ (\approx0.116\%\ \text{by volume})}$$
ResultValue
Sulphur oxidized to SO₂0.606 mol
Total stoichiometric O₂ (as air)103.1 mol O₂ + 387.9 mol N₂
Total flue gas523.5 mol
SO₂ concentration1,158 ppmv (≈0.116% v/v)

Secondary air pollutant formation. Primary combustion products such as SO₂ and NOₓ are not chemically inert once released — SO₂ slowly oxidizes in the atmosphere to SO₃, which reacts with water vapour to form H₂SO₄ aerosol (a key acid-rain precursor), while NOₓ and volatile hydrocarbons react photochemically in sunlight (via the OH-radical oxidation cycle) to form ground-level ozone, peroxyacetyl nitrate (PAN) and secondary organic aerosol. These secondary pollutants form downwind of the source, often hours to days after emission, and are a principal driver of regional smog and haze.

Check: (1) the source prints the fuel formula as $C_7H_{13}$, which has an odd hydrogen count (chemically impossible for a stable neutral hydrocarbon) — the boxed answer uses $C_7H_{13}$ exactly as printed; substituting the nearest valid formula $C_7H_{14}$ changes the SO₂ ppmv by less than 2%, so the conclusion is insensitive to this reading. (2) "2% sulphur" is read as 2% of the hydrocarbon fuel's own mass (not of a sulphur-inclusive total), and the "stoichiometric amount of oxygen" is read as being supplied as ordinary air (21% O₂/79% N₂), so the inert N₂ dilutes the flue gas roughly 6-fold — both are assumptions adopted here.

Part (iv) — terminal settling velocity.

Given. Particle diameter $d_p=15\ \mu\text{m}$, particle density $\rho_p=1500\ \text{kg/m}^3$, air at 25 °C, formula $v_t=g\rho_p d_p^2/(18\mu_g)$.

Find. The terminal settling velocity $v_t$.

Approach. Assume standard dynamic viscosity of air at 25 °C ($\mu_g=1.81\times10^{-5}\ \text{Pa}\cdot\text{s}$), substitute into the supplied Stokes'-law formula, then check the particle Reynolds number confirms the Stokes (laminar, creeping-flow) regime the formula assumes.

  1. Substitute into the terminal-velocity formula. With $g=9.81\ \text{m/s}^2$, $d_p=15\times10^{-6}\ \text{m}$: $$v_t=\frac{(9.81)(1500)(15\times10^{-6})^2}{18(1.81\times10^{-5})}=\boxed{1.02\times10^{-2}\ \text{m/s}\ (1.02\ \text{cm/s})}$$
  2. Check the Stokes regime (particle Reynolds number). With $\rho_{air}=1.18\ \text{kg/m}^3$ at 25 °C: $$Re_p=\frac{\rho_{air}v_t d_p}{\mu_g}=\frac{(1.18)(0.01016)(15\times10^{-6})}{1.81\times10^{-5}}=\boxed{9.94\times10^{-3}\ll1}$$ confirming Stokes' law (laminar creeping flow, no inertial correction needed) applies.
ResultValue
Terminal settling velocity, $v_t$1.02×10⁻² m/s (1.02 cm/s)
Particle Reynolds number, $Re_p$9.94×10⁻³ (Stokes regime confirmed)

Why gravitational settling is only a pre-cleaner. Stokes' law shows $v_t\propto d_p^2$: settling velocity collapses rapidly as particle size decreases, so even the 15 µm particle above — a relatively large particle for control-device design — settles at barely 1 cm/s, and a 1–2 µm particle would settle roughly 50–200× slower still. Achieving useful collection efficiency on such particles under gravity alone would require an impractically long settling chamber (large footprint, low gas velocity to keep residence time high, and re-entrainment risk from any turbulence). Gravity settling chambers are therefore used only as coarse pre-cleaners — removing the large, fast-settling particle fraction and protecting downstream fans and equipment from abrasive wear — while finer particulate is left to devices that apply a much larger effective separating force (centrifugal force in a cyclone, or an electrostatic force in an ESP) to achieve practical collection efficiency in a compact unit.

Check: air viscosity at 25 °C is not supplied in the source text; $\mu_g=1.81\times10^{-5}\ \text{Pa}\cdot\text{s}$ and $\rho_{air}=1.18\ \text{kg/m}^3$ are standard tabulated values (as instructed, "make any appropriate assumptions about the air viscosity at 25 °C").