16-Civ-A3 Elementary Environmental Engineering · May 2014

Question 1 of 7: Material Balance, Reaction Kinetics and Microbiology

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

Notes on this paper

Paper format. National Exams 98-Civ-A3 Environmental Engineering, May 2014 — 3 hours, closed book with one candidate-prepared double-sided aid sheet, approved Casio or Sharp calculator only. Seven questions are offered; any five constitute a complete paper (20 marks each, 100 marks maximum), and only the first five answers in the work book are marked. All seven are solved here, because the set is intended as a study resource rather than an examination script. Section marks are shown in brackets at the left margin of each part, and the marking scheme on page 6 confirms the split.

Reference texts. Davis & Cornwell, Introduction to Environmental Engineering (5th ed.); Mihelcic & Zimmerman, Environmental Engineering: Fundamentals, Sustainability, Design (3rd ed.); Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.); Crittenden et al., MWH’s Water Treatment: Principles and Design (3rd ed.). Canadian regulatory frame: the federal Impact Assessment Act (2019) and the Impact Assessment Agency of Canada, the Canadian Environmental Protection Act (CEPA 1999), CCME Canadian Environmental Quality Guidelines, and Health Canada’s Guidelines for Canadian Drinking Water Quality (GCDWQ).

Check: Henry’s law constant units in Question 1(i). The paper writes the constant as “0.30 (mol/atm)”, which is dimensionally incomplete — a Henry’s constant in the concentration/pressure form must carry a volume in the denominator. It is taken here as 0.30 mol/(L·atm), i.e. the aqueous-concentration form $C_{aq}=K_H\,p$. That reading is confirmed by the published value for ethyl acetate, $H \approx 1.3\times10^{-3}\ \text{atm}\cdot\text{m}^3/\text{mol}$, whose reciprocal is $\approx 0.77\ \text{mol}/(\text{L}\cdot\text{atm})$ — the same order of magnitude. Per NOTE 1 on page 1, this assumption is stated with the answer.

Question 1: Material Balance, Reaction Kinetics and Microbiology (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) — Equilibrium partitioning of ethyl acetate (6 marks)

Given. A sealed, rigid vessel holds equal volumes of water and air at constant temperature, and the entire charge of solvent must distribute itself between the two phases.

Given data — sealed storage tank
Quantity	Symbol	Value
Total tank volume	$V$	10 m³
Water volume	$V_w$	5 m³ = 5000 L
Air (headspace) volume	$V_a$	5 m³ = 5000 L
Temperature	$T$	21 $^\circ\text{C}$ = 294.15 K
Ethyl acetate added	$n_T$	88 kg = 1000 mol (M = 88 g/mol)
Henry’s law constant	$K_H$	0.30 mol/(L·atm)
Universal gas constant	$R$	0.08206 L·atm/(mol·K)

Find. The equilibrium aqueous concentration $C_w$ of ethyl acetate and the equilibrium partial pressure $p$ of ethyl acetate in the headspace.

Closed two-phase system: the 1000 mol charge splits between dissolved and vapour inventories, linked by Henry's law.

Approach. Write a total mole balance on the sealed tank (nothing leaves), express the aqueous inventory through Henry’s law and the vapour inventory through the ideal gas law, and solve the single resulting linear equation for the partial pressure.

State the closed-system mole balance. Because the tank is sealed and no reaction occurs, the charge simply redistributes between the two phases: $$n_T = n_{water} + n_{air} = C_w V_w + \frac{p\,V_a}{RT}$$ where $C_w$ is in mol/L and $V_w$, $V_a$ in litres.
Introduce Henry’s law to eliminate one unknown. At equilibrium the dissolved concentration is fixed by the partial pressure above it: $$C_w = K_H\,p = 0.30\,p$$ Substituting leaves $p$ as the only unknown: $$n_T = K_H p V_w + \frac{p V_a}{RT} = p\left(K_H V_w + \frac{V_a}{RT}\right)$$
Evaluate the two capacity terms. The aqueous capacity is $$K_H V_w = 0.30 \times 5000 = 1500\ \text{mol/atm}$$ and the gas-phase capacity is $$\frac{V_a}{RT} = \frac{5000}{0.08206 \times 294.15} = \frac{5000}{24.14} = 207.1\ \text{mol/atm}$$ The water can hold roughly seven times more ethyl acetate per unit of driving pressure than the headspace can, which already tells us where most of the solvent will end up.
Solve for the equilibrium partial pressure. Summing the capacities and inverting, $$p = \frac{n_T}{K_H V_w + V_a/RT} = \frac{1000}{1500 + 207.1} = \frac{1000}{1707.1}$$ $$\boxed{p = 0.586\ \text{atm}}$$
Back-substitute for the aqueous concentration. Applying Henry’s law once more, $$C_w = K_H p = 0.30 \times 0.586 = 0.1757\ \text{mol/L}$$ and converting to mass units with $M = 88$ g/mol, $$\boxed{C_w = 0.176\ \text{mol/L} = 15.5\ \text{g/L} = 15\,470\ \text{mg/L}}$$
Check the balance closes. The dissolved inventory is $n_w = 0.1757 \times 5000 = 879$ mol and the vapour inventory is $n_a = 0.586 \times 5000 / 24.14 = 121$ mol. Their sum is $879 + 121 = 1000$ mol, which recovers the charge exactly, so the partition is internally consistent: about 88 % of the solvent sits in the water and 12 % in the air space.

Final results — Question 1(i)
Quantity	Value
Equilibrium partial pressure of ethyl acetate, $p$	0.586 atm
Equilibrium aqueous concentration, $C_w$	0.176 mol/L (15 470 mg/L)
Moles dissolved in water	879 mol (87.9 %)
Moles in the air space	121 mol (12.1 %)

Check: the idealised answer exceeds the pure-solvent vapour pressure. Ethyl acetate has a saturation vapour pressure of only about 0.10 atm at 21 $^\circ\text{C}$, so a computed partial pressure of 0.586 atm is thermodynamically unreachable — in reality a separate liquid ethyl-acetate phase would form and the headspace would cap out near 0.10 atm. Likewise 15.5 g/L, while below the roughly 80 g/L aqueous solubility, is far outside the dilute range where a linear Henry’s law is trustworthy. The question is posed as a linear-partitioning exercise and the linear answer is what is marked; the physical limitation is noted here per NOTE 1 and is exactly the judgement an examiner rewards.

Topic. Equilibrium partitioning of a volatile organic compound between water and a closed headspace.

Key relations. Closed-system mole balance $n_T = C_wV_w + n_a$; Henry’s law in the concentration/pressure form $C_w = K_Hp$ (equivalently $p = H C_w$ with $H = 1/K_H$); ideal gas law $n_a = pV_a/RT$.

Why this works. A sealed vessel conserves mass, so the only question is how the fixed inventory divides. Henry’s law and the ideal gas law each make one phase’s inventory linear in the same variable, the partial pressure, so the balance collapses to one equation in one unknown. The grouped coefficients $K_HV_w$ and $V_a/RT$ are the two phases’ capacities in mol/atm, and their ratio — here 1500 : 207 — predicts the split before any arithmetic is done. This capacity view is what lets engineers judge instantly whether a spill will volatilise or stay dissolved.

Common pitfall. Two errors dominate. First, mixing volume units: $K_H$ is per litre, so 5 m³ must become 5000 L before it multiplies anything. Second, assuming the compound is “mostly in the air because it is volatile” and never writing the balance — ethyl acetate is highly water-soluble and its low Henry’s constant keeps almost 90 % of it dissolved.

Source. Mihelcic & Zimmerman, Environmental Engineering: Fundamentals, Sustainability, Design, 3rd ed., Ch. 3 (Environmental Chemistry — phase partitioning and Henry’s law); Davis & Cornwell, Introduction to Environmental Engineering, 5th ed., Ch. 2.

Part (ii) — Total nitrogen concentration (7 marks)

Given. A groundwater sample is reported with each nitrogen species expressed as the whole ion, except organic nitrogen, which is already expressed as N.

Given data — nitrogen speciation of the groundwater
Species	Reported concentration	Reporting basis	Molar mass
Ammonium, $\text{NH}_4^+$	25 mg/L	as $\text{NH}_4^+$	14 + 4(1) = 18 g/mol
Nitrite, $\text{NO}_2^-$	5 mg/L	as $\text{NO}_2^-$	14 + 2(16) = 46 g/mol
Nitrate, $\text{NO}_3^-$	15 mg/L	as $\text{NO}_3^-$	14 + 3(16) = 62 g/mol
Organic nitrogen	20 mg/L	as N	—

Find. The total nitrogen concentration of the sample, expressed on a common basis of mg/L as N.

Only ammonium is materially reduced by conversion to an N basis; nitrite and nitrate are small either way, and organic-N is already as N.

Approach. Concentrations reported “as the ion” cannot be added to one another. Convert each to mg/L as N by multiplying by the mass fraction of nitrogen in that ion, then sum.

Establish the conversion rule. For a species containing one nitrogen atom, $$C_{\text{as N}} = C_{\text{as species}} \times \frac{A_N}{M_{species}}$$ with $A_N = 14$ g/mol. The ratio is simply the mass fraction of the ion that is nitrogen.
Convert the ammonium. With $M_{\text{NH}_4} = 18$ g/mol, $$C_{\text{NH}_4\text{-N}} = 25 \times \frac{14}{18} = 25 \times 0.7778 = 19.44\ \text{mg/L as N}$$ Ammonium loses more than a fifth of its reported mass on conversion, because four hydrogen atoms are a significant share of a light ion.
Convert the nitrite. With $M_{\text{NO}_2} = 46$ g/mol, $$C_{\text{NO}_2\text{-N}} = 5 \times \frac{14}{46} = 5 \times 0.3043 = 1.52\ \text{mg/L as N}$$
Convert the nitrate. With $M_{\text{NO}_3} = 62$ g/mol, $$C_{\text{NO}_3\text{-N}} = 15 \times \frac{14}{62} = 15 \times 0.2258 = 3.39\ \text{mg/L as N}$$ The two oxidised species together contribute under 5 mg/L as N, even though they were reported as 20 mg/L of ion — oxygen dominates their mass.
Note the organic fraction needs no conversion. Organic nitrogen is already reported as N, so it enters the sum unchanged at 20.0 mg/L as N.
Sum on the common basis. $$C_{TN} = 19.44 + 1.52 + 3.39 + 20.0$$ $$\boxed{C_{TN} = 44.35\ \text{mg/L as N}}$$
Report the standard sub-totals. Total Kjeldahl nitrogen captures the reduced forms only: $$TKN = C_{\text{NH}_4\text{-N}} + C_{org\text{-N}} = 19.44 + 20.0 = 39.4\ \text{mg/L as N}$$ and the oxidised fraction $\text{NO}_x\text{-N} = 1.52 + 3.39 = 4.91$ mg/L as N. TKN is what a laboratory actually reports from a digestion, so quoting it demonstrates that the analytical route is understood, not just the arithmetic.

Final results — Question 1(ii)
Fraction	mg/L as N
Ammonium nitrogen ($\text{NH}_4$-N)	19.44
Nitrite nitrogen ($\text{NO}_2$-N)	1.52
Nitrate nitrogen ($\text{NO}_3$-N)	3.39
Organic nitrogen	20.00
Total Kjeldahl nitrogen (TKN)	39.44
Total nitrogen	44.35

For context, this water is heavily impacted: the Health Canada maximum acceptable concentration for nitrate in drinking water is 45 mg/L as $\text{NO}_3^-$, i.e. 10 mg/L as $\text{NO}_3$-N, so nitrate alone is not yet at the limit, but the ammonium and organic loads signal a nearby sewage or manure source and would drive substantial oxygen demand and disinfection-byproduct formation downstream.

Topic. Reporting bases for nutrient analyses and conversion to a common elemental basis.

Key relations. $C_{\text{as N}} = C_{\text{as ion}} \times A_N/M_{ion}$; $TKN = \text{NH}_3\text{-N} + \text{organic-N}$; $TN = TKN + \text{NO}_2\text{-N} + \text{NO}_3\text{-N}$.

Why this works. Total nitrogen is a statement about nitrogen atoms, and atoms are conserved through every transformation in the nitrogen cycle — ammonification, nitrification, denitrification. Reporting each species as the ion attaches a different amount of non-nitrogen mass (four hydrogens, two or three oxygens) to the same one nitrogen atom, so ion-basis numbers are simply not commensurable. Converting by mass fraction strips that ballast away and leaves quantities that may legitimately be added, subtracted across a treatment train, and compared with a discharge permit.

Common pitfall. Adding the four reported numbers straight to get 65 mg/L is the classic error and overstates the true load by nearly 50 %. A second pitfall is converting the organic nitrogen a second time; it is already on an N basis and must be left alone.

Source. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery, 5th ed., Ch. 2 (Constituents in Wastewater) and Ch. 7; Davis & Cornwell, Introduction to Environmental Engineering, 5th ed., Ch. 2.

Part (iii) — Mechanism of disinfection (7 marks)

The question permits one disinfectant to be selected. Ozone is chosen here, with brief comparative notes on chlorine and UV so the mechanism is seen in context.

Ozone inactivates microorganisms by direct oxidative attack on cellular structures. Dissolved ozone is a powerful oxidant with a standard reduction potential of about +2.07 V, and in water it acts along two parallel routes. The molecular route is a direct electrophilic and cycloaddition attack on electron-rich sites — carbon–carbon double bonds in membrane lipids, sulphur-containing and aromatic amino-acid residues in enzymes, and the purine and pyrimidine bases of nucleic acids. The radical route arises from ozone decomposition in water, particularly at elevated pH, generating hydroxyl radicals ($\text{OH}\cdot$) that are even stronger and almost completely non-selective oxidants.

For bacteria, the dominant lethal event is disruption of the cytoplasmic membrane. Ozone cleaves unsaturated lipids and oxidises membrane proteins, so permeability control is lost, the cell leaks its contents, and lysis follows within seconds. For viruses, which have no membrane to attack in the same way, inactivation proceeds by oxidation of the protein capsid — destroying the receptor sites needed to attach to and infect a host cell — and by direct damage to the nucleic acid core. For protozoan cysts and oocysts such as Giardia and Cryptosporidium, the robust outer wall must first be breached, after which the oxidant reaches the internal sporozoites; this is why cysts demand much higher exposure than bacteria.

Design practice quantifies exposure through the CT concept, the product of residual disinfectant concentration $C$ (mg/L) and contact time $T$ (minutes), evaluated at the plant’s $t_{10}$ — the time for the first 10 % of the flow to pass, which conservatively represents short-circuiting through the contactor. Required CT values are tabulated by organism, temperature and pH, and increase sharply as temperature falls. Ozone’s great advantage is the magnitude of its potency: the CT needed for a 3-log Giardia inactivation is on the order of 1–2 mg·min/L, roughly two orders of magnitude below free chlorine, and ozone is one of the very few disinfectants effective against Cryptosporidium, which is essentially chlorine-resistant.

The corresponding limitation is that ozone is unstable and leaves no lasting residual in the distribution system, so a Canadian plant using ozone must still add a secondary disinfectant — typically chloramine — to satisfy the distribution-residual expectation in the Guidelines for Canadian Drinking Water Quality. Ozone must also be generated on site from oxygen or dry air, which carries a significant capital and power cost, and in bromide-bearing source waters it forms bromate, a regulated carcinogen with a Canadian MAC of 0.01 mg/L. By contrast, chlorine acts chiefly as hypochlorous acid ($\text{HOCl}$), a smaller uncharged molecule that penetrates the cell wall and oxidises sulphydryl-containing respiratory enzymes; it is cheap and leaves a durable residual, but it is weak against cysts and generates trihalomethanes with natural organic matter. UV inactivates by an entirely non-chemical route — photons near 254 nm are absorbed by adjacent pyrimidine bases in DNA and RNA, forming cyclobutane dimers that block replication — which makes it outstandingly effective against Cryptosporidium at low dose, but it leaves no residual at all, does not oxidise taste-and-odour compounds, and is defeated by high turbidity.

Topic. Mechanisms of chemical and physical disinfection in drinking water treatment.

Key relations. Chick–Watson law $\ln(N/N_0) = -k\,C^n t$; the CT design criterion $CT = C \times t_{10}$; log inactivation $= -\log_{10}(N/N_0)$; UV dose (fluence) $= I \times t$ in mJ/cm².

Why this works. Every disinfectant works by damaging a structure the organism cannot repair fast enough — membrane, capsid protein, respiratory enzyme, or genome. Because the damage accumulates with both the intensity of the agent and the duration of contact, inactivation follows an approximately first-order law in exposure, which is exactly what the CT product captures. Different organisms present different first barriers — a lipid membrane, a protein coat, a cyst wall — so a single plant usually needs a multi-barrier train: an oxidant for bacteria and viruses, UV or ozone for chlorine-resistant protozoa, and a stable residual for the distribution system.

Common pitfall. Treating “dose” as the amount of chemical added rather than the residual actually maintained in contact with the water. Oxidant demand from natural organic matter, iron and manganese consumes much of the applied dose before any disinfection occurs, so CT must always be built on the measured residual and on $t_{10}$, never on the nominal hydraulic retention time.

Source. Crittenden et al., MWH’s Water Treatment: Principles and Design, 3rd ed., Ch. 13 (Disinfection); Health Canada, Guidelines for Canadian Drinking Water Quality; Davis & Cornwell, 5th ed., Ch. 6.

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