16-Civ-A3 Elementary Environmental Engineering · December 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, December 2014 — 98-Civ-A3 Environmental Engineering. Three hours, closed book with one candidate-prepared double-sided aid sheet. Seven problems, each worth 20 marks; any five constitute a complete paper (maximum 100 marks), and only the first five answers in the work book are marked. All seven problems are solved below, because the set is a study resource rather than an exam attempt.

Reference texts. Davis & Cornwell, Introduction to Environmental Engineering; Mihelcic & Zimmerman, Environmental Engineering: Fundamentals, Sustainability, Design; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery; Crittenden et al. (MWH), Water Treatment: Principles and Design; CCME Canadian Environmental Quality Guidelines; Impact Assessment Agency of Canada, Impact Assessment Act guidance.

Check: two source inconsistencies are carried through deliberately. (1) Problem 1(i) prints the dipropylene glycol formula as C₆H₁₄O₂ (118.2 g/mol); the actual compound is C₆H₁₄O₃ (134.2 g/mol). (2) The same sentence states the dose as “76 kg (1000 mol)”, which implies a molar mass of 76 g/mol and matches neither formula — 1000 mol of the real compound is 134 kg. The mole quantity is the load-bearing datum for a closed-system balance, so 1000 mol is adopted and both molar masses are reported where a mass concentration is asked for. NOTE 1 on page 1 expressly invites this kind of stated assumption.

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.

1 (i) — Closed-tank Henry's law equilibrium (6 marks)

Given. A sealed rigid tank is charged with a fixed quantity of a soluble organic and then allowed to reach equilibrium between its water and its head space; no material leaves the vessel.

Given data — Problem 1(i)
Quantity	Symbol	Value
Total tank volume	V	30 m³
Water volume (half full)	V_w	15 m³ = 15 000 L
Air (head-space) volume	V_a	15 m³ = 15 000 L
Temperature	T	21 °C = 294.15 K
Dipropylene glycol added	n_T	1000 mol
Henry's law constant	K_H	10⁶ mol/(L·atm)
Universal gas constant	R	0.08206 L·atm/(mol·K)

Find. The equilibrium aqueous concentration C of dipropylene glycol in the water and its equilibrium partial pressure p in the sealed air space.

Figure 1(i). Closed two-phase storage tank. The 1000 mol dose partitions between the water and the head space until Henry's law is satisfied at the interface.

Approach. Write a closed-system mole balance on the glycol, express the aqueous inventory through Henry's law and the gaseous inventory through the ideal gas law, and solve the single resulting linear equation for the partial pressure.

State the closed-system mole balance. Nothing enters or leaves the sealed tank, so the dose is distributed between exactly two inventories: $$n_T = n_w + n_a = C\,V_w + \frac{p\,V_a}{RT}$$ where $C$ is the aqueous molar concentration and $p$ the head-space partial pressure.
Introduce Henry's law to eliminate one unknown. In the concentration form appropriate to the stated units, $C = K_H\,p$, so the balance collapses to a single unknown: $$n_T = p\left(K_H V_w + \frac{V_a}{RT}\right)$$ The two grouped terms are the phase capacities, each in mol/atm, and their ratio decides the split before any arithmetic is done.
Evaluate the two capacities. Converting the volumes to litres, as $K_H$ is expressed per litre, $$K_H V_w = (10^{6})(15{,}000) = 1.50\times10^{10}\ \text{mol/atm}$$ $$\frac{V_a}{RT} = \frac{15{,}000}{(0.08206)(294.15)} = 621.6\ \text{mol/atm}$$ The water can hold roughly twenty-four million times as much glycol per unit of driving pressure as the air can. Dipropylene glycol is a fully miscible, very low volatility polyol, so this enormous asymmetry is physically expected rather than an arithmetic slip.
Solve for the equilibrium partial pressure. $$p = \frac{n_T}{K_H V_w + V_a/(RT)} = \frac{1000}{1.50\times10^{10} + 621.6}$$ $$\boxed{p = 6.67\times10^{-8}\ \text{atm} \approx 6.8\times10^{-3}\ \text{Pa}}$$
Back out the aqueous concentration. Substituting into Henry's law, $$C = K_H\,p = (10^{6})(6.667\times10^{-8})$$ $$\boxed{C = 6.67\times10^{-2}\ \text{mol/L}}$$ Using the formula printed on the paper, C₆H₁₄O₂ at 118.2 g/mol, this is $7.88\ \text{g/L} \approx 7{,}880$ mg/L; with the true formula C₆H₁₄O₃ at 134.2 g/mol it is 8.94 g/L.
Check the balance closes. The water holds $C V_w = (0.06667)(15{,}000) = 1000.0$ mol and the air holds $pV_a/RT = 4.1\times10^{-5}$ mol. The two sum to the 1000 mol charged, and 99.99999 % of the dose stays dissolved — effectively the entire spill remains in the water phase.

The engineering message is more useful than the numbers themselves. Because the partition is so lopsided, head-space monitoring would be a hopeless way to detect this contaminant: the vapour above the liquid is eight orders of magnitude below any practical detection limit, while the water is grossly contaminated at nearly 8 g/L. Detection and any subsequent treatment must be done on the aqueous phase.

Final results — Problem 1(i)
Quantity	Result
Equilibrium partial pressure in the air space	6.67 × 10⁻⁸ atm
Equilibrium aqueous concentration	6.67 × 10⁻² mol/L
as mass, using C₆H₁₄O₂ (118.2 g/mol)	7 880 mg/L
as mass, using C₆H₁₄O₃ (134.2 g/mol)	8 945 mg/L
Fraction of the dose remaining dissolved	> 99.999 %

1 (ii) — Total nitrogen (7 marks)

Given. A nitrogen speciation report in which three of the four fractions are expressed as the ion rather than as elemental nitrogen.

Given data — Problem 1(ii)
Species	Reported concentration	Basis
Ammonia	40 mg/L	as NH₃
Nitrite	2 mg/L	as NO₂⁻
Nitrate	10 mg/L	as NO₃⁻
Organic nitrogen	20 mg/L	as N
Atomic weights	H = 1, N = 14, O = 16

Find. The total nitrogen concentration of the sample on a common “as N” basis.

Approach. Convert each ion-basis result to an elemental-nitrogen basis by the mass fraction of nitrogen in that ion, then sum; report Total Kjeldahl Nitrogen alongside total nitrogen because that is the split the laboratory actually measures.

Build the molar masses from the given atomic weights. $$M_{\mathrm{NH_3}} = 14 + 3(1) = 17,\qquad M_{\mathrm{NO_2^-}} = 14 + 2(16) = 46,\qquad M_{\mathrm{NO_3^-}} = 14 + 3(16) = 62$$ Each ion contains exactly one nitrogen atom, so the conversion factor is simply $A_N/M_{\text{ion}}$.
Convert each ion-basis result to nitrogen. Applying $C_{\text{as N}} = C_{\text{ion}} \times A_N / M_{\text{ion}}$: $$\mathrm{NH_3\text{-}N} = 40\left(\tfrac{14}{17}\right) = 32.94\ \text{mg/L}$$ $$\mathrm{NO_2^-\text{-}N} = 2\left(\tfrac{14}{46}\right) = 0.61\ \text{mg/L}, \qquad \mathrm{NO_3^-\text{-}N} = 10\left(\tfrac{14}{62}\right) = 2.26\ \text{mg/L}$$ The organic fraction is already reported as N and is carried over unchanged at 20 mg/L.
Assemble Total Kjeldahl Nitrogen. TKN is the reduced nitrogen the Kjeldahl digestion recovers — organic plus ammonia, and nothing oxidised: $$\mathrm{TKN} = 20 + 32.94 = \boxed{52.9\ \text{mg/L as N}}$$
Add the oxidised species to obtain total nitrogen. $$\mathrm{TN} = \mathrm{TKN} + \mathrm{NO_2^-\text{-}N} + \mathrm{NO_3^-\text{-}N} = 52.94 + 0.61 + 2.26$$ $$\boxed{\mathrm{TN} = 55.8\ \text{mg/L as N}}$$

The speciation itself is diagnostic. Ammonia dominates and the oxidised forms are trivial, which is the fingerprint of a fresh, essentially raw or very lightly treated wastewater in which nitrification has not yet proceeded. A well-nitrified secondary effluent would show the mirror image, with nitrate carrying almost all of the nitrogen and ammonia near the detection limit. At 55.8 mg/L as N this water is far above any receiving-water objective, and the ammonia fraction alone is acutely toxic to fish at the pH and temperature typical of a Canadian receiving stream.

Final results — Problem 1(ii)
Fraction	mg/L as N
Ammonia nitrogen (NH₃-N)	32.9
Nitrite nitrogen (NO₂⁻-N)	0.61
Nitrate nitrogen (NO₃⁻-N)	2.26
Organic nitrogen	20.0
Total Kjeldahl Nitrogen (TKN)	52.9
Total Nitrogen (TN)	55.8

1 (iii) — MPN versus CFU enumeration (7 marks)

Both methods estimate the density of fecal indicator bacteria in a water sample, but they answer subtly different questions and their numbers are not interchangeable.

The colony-forming unit (CFU) method is a direct count. A measured volume of sample is drawn through a 0.45 µm membrane filter, the filter is laid on a selective and differential agar — mFC for thermotolerant coliforms, mEI or mE for enterococci, as in the plate photographed on the exam page — and incubated at a diagnostic temperature. Every viable cell or clump of cells capable of growth develops into one visible colony, and the analyst counts the colonies in the statistically valid range (typically 20 to 80 per filter), reporting the result as CFU per 100 mL. The method is quick, cheap, gives a physical object to look at, and can be run in the field.

The most probable number (MPN) method is a statistical inference. The sample is dispensed into a series of replicate tubes or into a sealed multi-well tray at several dilutions, each containing a defined growth medium; after incubation each vessel is scored simply as positive or negative for a diagnostic reaction — gas production in lauryl tryptose broth, or the fluorogenic and chromogenic response of a defined-substrate medium such as Colilert. The pattern of positives and negatives is then compared against a Poisson-based probability table, which returns the bacterial density most likely to have produced that pattern, reported as MPN per 100 mL with 95 % confidence limits.

Difference 1 — the nature and precision of the estimate. A CFU result is an enumeration with comparatively tight, count-limited precision, whereas an MPN result is a maximum-likelihood estimate whose confidence interval is inherently wide, often spanning a factor of three or more for the common 15-tube configurations. Two laboratories reporting the same MPN may differ materially in true density, so an MPN slightly over a regulatory limit is much weaker evidence of an exceedance than a CFU slightly over the same limit. Conversely MPN has no upper counting ceiling issue and returns a defensible number where a plate would be overgrown.

Difference 2 — what is actually being counted, and the effect of sample character. A colony arises from a single viable unit, which may be one cell or an aggregate of many attached to a particle; a turbid, high-solids or chlorinated sample therefore biases the CFU count low, because clumped cells count once and injured cells that are viable but non-culturable on a selective agar fail to grow at all. The MPN method disperses the sample in liquid medium and typically uses a less inhibitory defined substrate, so it recovers stressed organisms more reliably and generally returns numbers higher than the parallel CFU result on the same sample — commonly by a factor of about 1.5 to 2 on secondary effluent. Membrane filtration also becomes impractical on turbid water, where the filter blinds before an adequate volume passes.

Interpretation rule for compliance work. Because of these differences, a permit must specify the analytical method, and trend data must never mix the two. When a laboratory changes methods, the two should be run in parallel for a period so that the historical record can be interpreted correctly.

Topic. Equilibrium partitioning, nitrogen speciation and microbial enumeration — the three quantitative languages of environmental engineering practice.

Key relations. Henry's law in the concentration form $C = K_H p$ and its pressure form $p = H C$, with $H = 1/K_H$; the closed-system mole balance $n_T = C V_w + pV_a/RT$; species-to-element conversion $C_{\text{as N}} = C_{\text{ion}} A_N / M_{\text{ion}}$; $\mathrm{TN} = \mathrm{TKN} + \mathrm{NO_2^-\text{-}N} + \mathrm{NO_3^-\text{-}N}$; the Poisson likelihood underlying the MPN tables.

Why this works. A sealed vessel conserves mass, so the only freedom the system has is how it distributes a fixed inventory between phases; Henry's law supplies the single constitutive relation needed to close the balance. Because both phase inventories are linear in the partial pressure, the balance factors into a pressure times a total capacity, and the ratio of the two capacity terms is by itself the answer to “where does the contaminant go?”. The nitrogen conversions work because each of the three ions contains exactly one nitrogen atom, so the mass fraction of nitrogen is a pure stoichiometric ratio, independent of concentration.

Common pitfall. Two dominate. First, using $K_H$ with cubic-metre volumes: the constant here is per litre, and forgetting the factor of 1000 shifts the answer by three orders of magnitude. Second, summing nitrogen species on their reported bases — adding 40 + 2 + 10 + 20 = 72 mg/L is the classic wrong answer, roughly 29 % high, because three of the four numbers include the mass of hydrogen or oxygen that is not nitrogen. A third, subtler trap is quoting MPN and CFU results as though they were the same measurement.

Source. Davis & Cornwell, Introduction to Environmental Engineering, Ch. 2 (materials and energy balances) and Ch. 4 (water quality, nitrogen series); Metcalf & Eddy, Wastewater Engineering, Ch. 2 (constituents, TKN and TN) and the microbiological enumeration discussion in Ch. 2; APHA Standard Methods 9221/9222 for MPN and membrane filtration.

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