16-Civ-A3 Elementary Environmental Engineering · May 2016

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

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

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.

(i) Blending two methanol–water mixtures (6 marks)

Given. Two batches are combined in a vessel in which nothing reacts and nothing is left behind.

Find. The total mass of the blended product and its composition, expressed as weight percentages of methanol and water.

Approach. Write a steady-state mass balance on the mixer — once for total mass and once for the methanol component — and divide the component result by the total.

1. State the general balance and delete the terms that vanish. For any control volume, $$\text{accumulation} = \text{in} - \text{out} + \text{generation}$$ Blending is a physical operation, so there is no generation, and the mixer operates with no hold-up, so accumulation is zero. The balance collapses to what goes in comes out, applied to the total and to each species independently.
2. Total mass balance. The two feeds are the only inputs: $$m_T = m_1 + m_2 = 100 + 200 = \boxed{300\ \text{kg}}$$ where $m_1$ and $m_2$ are the masses drawn from tanks 1 and 2.
3. Methanol component balance. Each feed carries its own mass fraction $x_i$ of methanol: $$m_{\text{MeOH}} = m_1 x_1 + m_2 x_2 = (100)(0.30) + (200)(0.70)$$ Evaluating the two terms gives $30 + 140 = 170\ \text{kg}$ of methanol in the product.
4. Water by difference — and as an independent check. Subtracting, $m_{\text{H}_2\text{O}} = 300 - 170 = 130\ \text{kg}$. Because the check must not reuse the subtraction, recompute water from its own feed fractions: $(100)(0.70) + (200)(0.30) = 70 + 60 = 130\ \text{kg}$. The two routes agree, so the component balances close.
5. Convert to composition. Dividing each component by the total, $$x_{\text{MeOH}} = \frac{170}{300} = 0.5667 \qquad x_{\text{H}_2\text{O}} = \frac{130}{300} = 0.4333$$ so the product is $\boxed{56.67\ \%\ \text{methanol},\ 43.33\ \%\ \text{water by weight}}$.

The result is a weighted average, and it sits nearer the richer stream because that stream is twice as large: the blend fraction $(1)(0.30)/3 + (2)(0.70)/3$ is pulled two-thirds of the way toward 70 %. This is worth noting because the same lever-arm reasoning is how an operator decides which tank to draw from when trimming a blend on-line, without solving anything.

(ii) Pressure rise on complete decomposition of N₂O₅ (7 marks)

Given. A rigid closed vessel of volume $V$ holds a gas mixture that is 40 % N₂O₅ and 60 % N₂ (both by mole, as is conventional for gas compositions), at some initial pressure $P_1$ and temperature $T$. The dinitrogen pentoxide decomposes completely to N₂O₄ and O₂; the nitrogen is inert.

Find. The rise in pressure inside the vessel when the reaction has gone to completion, expressed relative to the initial pressure.

Approach. At constant volume and temperature the ideal-gas law makes pressure directly proportional to total moles, so the whole question reduces to counting moles before and after on a convenient basis.

1. Balance the reaction. The only stoichiometric route from N₂O₅ to the two named products is $$2\,\text{N}_2\text{O}_5 \longrightarrow 2\,\text{N}_2\text{O}_4 + \text{O}_2$$ so each mole of N₂O₅ consumed yields one mole of N₂O₄ plus half a mole of O₂ — that is, 1.5 mol of product gas from 1 mol of reactant. The mole count increases even though mass, of course, does not.
2. Choose a basis and tabulate the initial charge. Take 1.00 mol of initial gas, which is legitimate because the answer is asked as a ratio. Then $n_{\text{N}_2\text{O}_5,0} = 0.40$ mol and $n_{\text{N}_2,0} = 0.60$ mol, giving $n_1 = 1.00$ mol.
3. Advance the reaction to completion. All 0.40 mol of N₂O₅ reacts, producing 0.40 mol of N₂O₄ and $0.40/2 = 0.20$ mol of O₂. Nitrogen is untouched at 0.60 mol. Summing, $$n_2 = 0.60 + 0.40 + 0.20 = 1.20\ \text{mol}$$
4. Relate moles to pressure. With $V$ and $T$ fixed, $PV = nRT$ gives $$\frac{P_2}{P_1} = \frac{n_2}{n_1} = \frac{1.20}{1.00} = 1.20$$ so the final pressure is 1.20 times the initial pressure and $$\Delta P = P_2 - P_1 = \boxed{0.20\,P_1 \quad (\text{a } 20\ \% \text{ rise})}$$
5. Report the product composition and check mass conservation. The product gas is $0.40/1.20 = 33.33\ \%$ N₂O₄, $0.20/1.20 = 16.67\ \%$ O₂ and $0.60/1.20 = 50.00\ \%$ N₂. As a check on the stoichiometry, the mass that disappeared must reappear: $0.40 \times 108 = 43.2\ \text{g}$ of N₂O₅ against $0.40 \times 92 + 0.20 \times 32 = 36.8 + 6.4 = 43.2\ \text{g}$ of products. Mass closes exactly while moles do not, which is precisely why the pressure moves.

Two features of this answer deserve emphasis. First, the inert nitrogen dilutes the effect: had the vessel been charged with pure N₂O₅, the rise would have been the full 50 % implied by the stoichiometry, and the 60 % of inert gas is what cuts it to 20 %. Second, no rate constant or reaction time appears anywhere, because “proceeds to completion” removes kinetics from the problem entirely — kinetics would govern how long the vessel takes to reach 1.20 $P_1$, never the endpoint itself. Practically, a 20 % overpressure is the number a designer would carry into the relief-device sizing for a vessel holding this mixture.

(iii) Three disinfection terms (7 marks)

Indicator organism. It is neither possible nor affordable to assay drinking water for every pathogen that might be present: the pathogens of concern are numerous, episodic, and often present at densities far below routine detection limits. An indicator organism is a surrogate that is monitored instead. To serve, it must be present whenever faecal contamination is present and absent otherwise, occur in far greater numbers than the pathogens, be at least as persistent in water and at least as resistant to disinfection as they are, not multiply in the distribution system, and be cheap and rapid to enumerate. Escherichia coli and the broader total-coliform group are the classical choices, with enterococci and the spore-forming Clostridium perfringens used where a more chlorine-resistant surrogate is needed. In the Canadian regulatory frame, the Guidelines for Canadian Drinking Water Quality set a maximum acceptable concentration of none detectable per 100 mL for both E. coli and total coliforms, and a positive result triggers boil-water action rather than a search for the specific pathogen. The significance for disinfection is therefore twofold: the indicator both defines the treatment target and is the compliance test that proves the target was met.

Contact time. Disinfection is a rate process, not an event. Inactivation of a microbial population by a chemical disinfectant follows, to first order, the Chick–Watson relationship, in which the surviving fraction depends on the product of disinfectant concentration and exposure time. That product, the CT value, is the operational currency of disinfection: regulators publish required CT values for a given pathogen, log-removal, temperature and pH, and the operator demonstrates compliance by showing that the residual concentration multiplied by the effective contact time meets or exceeds it. Contact time is measured not as the nominal hydraulic residence time of the contact basin but as the time in which 90 % of the water has been retained, denoted $t_{10}$, because short-circuiting means some parcels of water leave far sooner than the average. Baffling a clearwell to raise the ratio $t_{10}/\tau$ is one of the cheapest ways to buy disinfection credit, since it increases the effective time without adding a milligram of chemical. Where contact time is short — a small pressure system feeding directly into distribution, for instance — the same CT must be bought with a much higher dose, with consequent taste, odour and disinfection-by-product penalties.

Log reduction. Because microbial densities span many orders of magnitude, disinfection performance is expressed logarithmically. A log reduction of $n$ means $$n = \log_{10}\!\left(\frac{N_0}{N}\right)$$ where $N_0$ and $N$ are the pathogen densities before and after treatment, so 1-log is 90 % removal, 2-log is 99 %, 3-log is 99.9 % and 4-log is 99.99 %. The logarithmic form matters because the marginal cost of each additional log is roughly constant while the marginal benefit in absolute organisms removed falls steeply; it also makes the credits additive, so that a treatment train earning 2.5-log through filtration and 1.5-log through chlorination delivers 4-log overall. Canadian and international practice sets minimum targets of 3-log for Giardia and Cryptosporidium and 4-log for enteric viruses in surface-water supplies, allocated between physical removal and chemical or UV inactivation. Notably Cryptosporidium oocysts are highly resistant to free chlorine, so their log credit is normally earned by filtration and ultraviolet light rather than by CT — a reminder that the three terms are linked: the indicator tells you whether a problem exists, contact time is the lever you pull, and log reduction is the unit in which the result is banked.