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18-Env-A5 Air Quality and Pollution Control Engineering · May 2014

Question 6 of 7: Combustion Stoichiometry, Flue Gas Desulfurisation and Photochemical Smog

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

Notes on this paper

National Exams — May 2014 — 04-Env-A5 / Air Quality and Pollution Control Engineering. 3 hours duration; closed book with a candidate-prepared 8.5×11 in double-sided aid sheet; Casio or Sharp approved calculator only. Any five (5) questions constitute a complete paper (the first five answers as they appear are marked); all seven are solved below for completeness. Each question is worth 20 marks with section marks shown in brackets.

Reference texts. Cooper & Alley, Air Pollution Control: A Design Approach (4th ed.); Wark, Warner & Davis, Air Pollution: Its Origin and Control (3rd ed.); Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Canadian Environmental Protection Act, 1999 (CEPA) and the Canadian Ambient Air Quality Standards (CAAQS) administered by Environment and Climate Change Canada.

Question 6: Combustion Stoichiometry, Flue Gas Desulfurisation and Photochemical Smog (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.

(i) Stoichiometric Fuel/Air Mass Ratio and Product Gas Composition

Given. Complete combustion of octane with a sulphur impurity, at a non-standard air composition:

Given data
QuantitySymbolValue
Fuel—Octane, C8H18
Basis$n_{oct}$1 mol
Sulphur content of total fuel (by mass)$w_S$2%
Air composition (molar)N2:O24:1 (stated, not the usual 3.76:1)

Find. The stoichiometric fuel/air mass ratio, and the percent composition of the product (flue) gas.

Check: consistent with Question 1(i), "octane containing 2% sulphur" is read as the octane being 98% of the total fuel mass, with elemental sulphur the remaining 2% by mass. All carbon and hydrogen are assumed to combust completely to CO2/H2O and all sulphur to SO2, with air supplied at exactly the stoichiometric O2 requirement (so no leftover O2 reports in the products) and at the exam-specified 4:1 N2:O2 molar ratio, which is deliberately different from real air's 3.76:1 and is used as given rather than "corrected" to the textbook value.

Approach. Work on a 1 mol octane basis: back out the moles of sulphur from the 2%-by-mass fuel specification, balance the combustion of both octane and sulphur, size the stoichiometric air at the stated 4:1 N2:O2 ratio, then form the fuel/air mass ratio and the product-gas mole fractions.

  1. Molar mass and sulphur content. $M_{oct} = 8(12.01)+18(1.008) = 114.22\ \text{g/mol}$; for 1 mol octane ($m_{oct}=114.22\ \text{g}$), $m_{fuel} = 114.22/0.98 = 116.55\ \text{g}$, so $m_S = 0.02(116.55) = 2.331\ \text{g}$ and $n_S = 2.331/32.07 = 0.0727\ \text{mol}$.
  2. Balance the combustion. $$\text{C}_8\text{H}_{18} + 12.5\,\text{O}_2 \rightarrow 8\,\text{CO}_2 + 9\,\text{H}_2\text{O}, \qquad \text{S} + \text{O}_2 \rightarrow \text{SO}_2.$$ Required O2: $12.5 + 0.0727 = 12.573\ \text{mol}$.
  3. Air at the stated 4:1 N2:O2 ratio. $n_{N_2} = 4(12.573) = 50.29\ \text{mol}$, so $n_{air} = 12.573+50.29 = 62.86\ \text{mol}$, with molar mass $M_{air} = \dfrac{1(32.00)+4(28.02)}{5} = 28.82\ \text{g/mol}$ and mass $m_{air} = 62.86(28.82) = 1811.5\ \text{g}$.
  4. Fuel/air mass ratio. $$\frac{m_{fuel}}{m_{air}} = \frac{116.55}{1811.5} = \boxed{0.0643\ \ (\approx 1:15.5\ \text{by mass, i.e. AFR}\approx15.5)}.$$
  5. Product gas composition (wet basis). Products are $n_{CO_2}=8$, $n_{H_2O}=9$, $n_{SO_2}=0.0727$, and the unreacted excess $n_{N_2}=50.29$ (no leftover O2, since air was supplied exactly stoichiometric); total $n_{prod}=67.36\ \text{mol}$: $$\boxed{y_{CO_2}=11.9\%,\ \ y_{H_2O}=13.4\%,\ \ y_{SO_2}=0.11\%,\ \ y_{N_2}=74.7\%.}$$
QuantityValue
Sulphur combusted, $n_S$0.0727 mol per mol octane
Stoichiometric air required62.86 mol (1811.5 g) per mol octane
Fuel/air mass ratio0.0643 (AFR ≈ 15.5:1)
Product gas, wet basisCO2 11.9%, H2O 13.4%, SO2 0.11%, N2 74.7%
Product gas, dry basis (informational)CO2 13.7%, SO2 0.12%, N2 86.2%

(ii) Flue Gas Desulfurisation (FGD) Schematic and Process Description

The most common FGD technology is wet limestone scrubbing: raw flue gas from the boiler is contacted with a recirculated limestone (CaCO3) slurry sprayed into an absorber tower. SO2 dissolves into the slurry droplets and reacts to form calcium sulfite, which is oxidized (often with forced air injection) to calcium sulfate dihydrate (gypsum):

$$\text{SO}_2 + \text{CaCO}_3 + \tfrac12\text{O}_2 + 2\text{H}_2\text{O} \rightarrow \text{CaSO}_4\!\cdot\!2\text{H}_2\text{O} + \text{CO}_2.$$

SprayAbsorberTowerMistEliminatorReactionTank /OxidationDewatering(gypsum)Flue gas in(SO2, NOx, PM)Scrubbed gasClean gasto stackLimestoneslurry spraySpent slurryAir(oxidation)CaSO4.2H2OslurryGypsumby-product
Fig. Q6(ii) — Wet limestone FGD: flue gas is scrubbed in the spray absorber tower, passes through a mist eliminator to the stack; spent slurry is oxidized and dewatered to a saleable gypsum by-product.

Scrubbed gas passes through a mist eliminator (to strip entrained slurry droplets) before release, while the spent slurry drains to a reaction tank where forced-air oxidation converts calcium sulfite to the more stable, more easily dewatered gypsum, which is thickened and dewatered to a saleable by-product (e.g., wallboard-grade gypsum) rather than a waste requiring disposal. The two design levers controlling SO2 removal efficiency are the liquid-to-gas (L/G) ratio in the absorber (more slurry contact area per unit gas) and the pH maintained in the recirculating slurry (higher pH drives faster SO2 absorption but risks scaling/blinding of the limestone if pushed too high).

(iii) Roles of Nitrogen and Hydrocarbon Compounds in Photochemical Smog

Nitrogen compounds (NOx). (1) Photolytic ozone-forming cycle — NO2 absorbs UV/visible sunlight and photolyzes, $\text{NO}_2 + h\nu \rightarrow \text{NO}+\text{O}$, and the released oxygen atom combines with atmospheric O2 to form ozone, $\text{O}+\text{O}_2\rightarrow\text{O}_3$ — this is the sole significant tropospheric source of ozone. (2) Ozone titration ("NOx cycle") — the NO produced in step (1) reacts back with O3 to regenerate NO2, $\text{NO}+\text{O}_3\rightarrow\text{NO}_2+\text{O}_2$, which on its own would leave a steady-state O3 level with no net accumulation; NOx therefore both creates the pathway to ozone and, without hydrocarbons, would also cap it.

Hydrocarbon (VOC) compounds. (1) NO-to-NO2 conversion without consuming ozone — VOCs react with hydroxyl radicals to form peroxy radicals (RO2•, HO2•), which oxidize NO to NO2 directly, bypassing the ozone-titration reaction above; this breaks the steady-state cap on ozone from step (2) and allows net ozone accumulation as more NO2 becomes available for photolysis without any O3 being consumed to make it. (2) Formation of secondary toxic/phytotoxic products — certain VOCs (aldehydes and related oxidation products) react further with NOx-derived radicals to form peroxyacetyl nitrate (PAN) and related compounds, which are potent eye irritants and phytotoxins and are themselves a defining component of photochemical smog beyond ozone alone.