Question 1 of 7: Problem 1 — Microbiology, reaction kinetics and material balance

Question 1: Problem 1 — Microbiology, reaction kinetics and material balance (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) The MPN and CFU methods for enumerating fecal indicator bacteria (7 marks)

Both methods answer the same regulatory question — how many fecal indicator organisms are in 100 mL of this water? — but they answer it by fundamentally different logic, and the two answers are not interchangeable numbers.

The most probable number (MPN) method is a statistical estimate obtained by dilution to extinction. A measured sample is diluted in a decimal series, and a set of replicate tubes (classically five tubes at each of three dilutions, now more commonly a sealed 51- or 97-well tray) is inoculated at each dilution with a selective growth medium. After incubation at the defined temperature — 44.5 °C for thermotolerant coliforms, 35 °C for total coliforms — each tube is scored simply as positive or negative for the diagnostic reaction: gas production, acid production, or fluorescence from the cleavage of a chromogenic or fluorogenic substrate such as MUG by the β-glucuronidase enzyme specific to E. coli. The pattern of positives across the dilution series is then converted, through a Poisson maximum-likelihood calculation embodied in standard MPN tables, into the density of organisms most likely to have produced that pattern. The reported result is an estimate expressed as MPN/100 mL, and it always carries a formal 95 % confidence interval.

The colony-forming unit (CFU) method is a direct count. A measured volume of sample — typically 100 mL for a low-turbidity surface water — is drawn under vacuum through a 0.45 µm membrane filter, which retains the bacteria on its surface. The membrane is transferred to a selective differential medium (m-FC agar for thermotolerant coliforms, mTEC or m-ColiBlue24 for E. coli) and incubated. Each retained viable cell or clump of cells that is capable of multiplying under those conditions grows into a visible, differentially coloured colony, and the colonies are counted by eye, ideally within the statistically defensible window of 20–80 target colonies per membrane. The result is reported as CFU/100 mL. Spread-plate and pour-plate variants follow the same logic when the sample is too turbid to filter a useful volume.

First important difference — the nature and precision of the number. A CFU result is a count with a Poisson counting error that is comparatively tight and symmetric: 50 colonies carries a relative standard error of about 14 %. An MPN result is a maximum-likelihood estimate whose confidence interval is wide and strongly asymmetric — for the classical five-tube, three-dilution design, the 95 % limits typically span a factor of three to ten around the point estimate. Two MPN results must therefore differ by roughly a factor of three before the difference is real, whereas a much smaller difference between two CFU counts is defensible. When an MPN value is compared against a numerical compliance limit, the limit is only meaningfully exceeded if the lower confidence bound also exceeds it; treating the tabulated MPN point value as an exact count is the single most common misreading of a microbiology report.

Second important difference — what a “unit” physically is, and the resulting bias with particles. The CFU method counts clumps, not cells: a floc particle carrying two hundred bacteria produces one colony and is recorded as one CFU. Surface waters after a rainfall event carry heavily particle-associated bacteria, so membrane filtration systematically under-reports them, and high turbidity additionally blinds the count by clogging the membrane (limiting the volume that can be filtered) and by allowing background heterotrophs to overgrow the plate. The MPN method scores growth in a liquid medium, where turbulence and the medium itself disperse aggregates, so particle-associated organisms are detected and MPN values on the same water are usually higher than CFU values, sometimes by a factor of two or more. Neither number is “wrong”; they measure slightly different things. The practical consequence is that data must be compared only on a like-for-like basis — against the same method and against a guideline expressed on the same basis — and that a change in laboratory method mid-programme creates a step change in the record that has nothing to do with the water.

In the Canadian regulatory setting both bases appear explicitly. The Guidelines for Canadian Drinking Water Quality set a maximum acceptable concentration for E. coli of none detectable per 100 mL and accept either basis, since the practical question is simply presence or absence. The CCME Guidelines for Canadian Recreational Water Quality set a geometric-mean limit of 200 E. coli per 100 mL over at least five samples, and it is precisely because of the confidence-interval width discussed above that the guideline is written as a geometric mean over a sample set rather than as a single-sample ceiling.

(ii) Total nitrogen concentration of the wastewater sample (7 marks)

Given. A nitrogen speciation on a single wastewater sample, with the atomic weights the question directs be used:

Find. The total nitrogen concentration of the sample, expressed on a common basis of milligrams of elemental nitrogen per litre (mg/L as N).

Approach. Total nitrogen is the sum of the nitrogen atoms contributed by every species, so each species reported as the whole ion or molecule must first be converted to an elemental-N basis by multiplying by the ratio of the atomic weight of nitrogen to the molecular weight of that species; organic nitrogen is already on an N basis and is added unchanged.

1. Establish the molecular weights of the three reported species. Using the atomic weights the question supplies, $$M_{\text{NH}_3} = 14 + 3(1) = 17 \qquad M_{\text{NO}_2} = 14 + 2(16) = 46 \qquad M_{\text{NO}_3} = 14 + 3(16) = 62$$ all in g/mol. The question explicitly directs that all the ammonia be taken as the un-ionised molecule NH₃, so the divisor is 17 and not the 18 that would apply to the ammonium ion NH₄⁺. This is a real distinction: at the near-neutral pH of a municipal wastewater the ammonium ion in fact dominates, and using 18 would change the answer by about 6 %.
2. Convert each reported concentration to an elemental-nitrogen basis. The general conversion is $$C_{\text{as N}} = C_{\text{as species}} \times \frac{A_{\text{N}}}{M_{\text{species}}}$$ Applying it to each species in turn gives, for ammonia, $$C_{\text{NH}_3\text{-N}} = 30 \times \frac{14}{17} = 24.71\ \text{mg/L as N}$$ then for nitrite, $5 \times 14/46 = 1.52$ mg/L as N, and for nitrate, $15 \times 14/62 = 3.39$ mg/L as N. Organic nitrogen is already reported as N and enters at its face value of 10.00 mg/L.
3. Sum the contributions to obtain the total nitrogen. With every term now on the same basis the sum is simply additive: $$\text{TN} = 24.71 + 1.52 + 3.39 + 10.00$$ $$\boxed{\text{TN} = 39.6\ \text{mg/L as N}}$$
4. Report the Kjeldahl fraction as well, because that is what the laboratory actually measures. Total Kjeldahl nitrogen is defined as organic nitrogen plus ammonia nitrogen — the reduced forms that the Kjeldahl digestion recovers — and the oxidised forms are determined separately by colorimetry or ion chromatography: $$\text{TKN} = 24.71 + 10.00 = 34.7\ \text{mg/L as N} \qquad \text{NO}_x\text{-N} = 1.52 + 3.39 = 4.91\ \text{mg/L as N}$$ Quoting both demonstrates that the analytical route behind the numbers is understood, and TN = TKN + NO_x-N is the arithmetic check on the whole calculation.

The speciation itself is diagnostic. Ammonia carries 62 % of the total nitrogen while the oxidised forms together carry only 12 %, which is the signature of a raw or primary-treated municipal wastewater in which nitrification has not occurred. A well-nitrified secondary effluent shows the reverse pattern, with nitrate dominant and ammonia below 1 mg/L as N. At 39.6 mg/L as N this sample is a typical medium-strength domestic wastewater, and it sits an order of magnitude above the 1–10 mg/L as N total-nitrogen limits that receiving-water sensitivity increasingly imposes on Canadian discharges, so nitrogen removal — nitrification followed by anoxic denitrification — would be required.

(iii) Steady-state mass balance on the continuous-stirred tank reactor (6 marks)

Given. A single completely mixed flow reactor with a first-order depletion reaction:

Find. The mass balance equation for the reactor, and the steady-state concentration of S leaving in the discharge.

Approach. Write the general material balance for the conservative statement “accumulation = in − out + generation” over the control volume drawn above, invoke the completely mixed assumption to set the discharge concentration equal to the tank concentration, then set the accumulation term to zero for steady state and solve the resulting linear equation.

1. Write the general material balance over the control volume. For the substance S over the reactor contents of volume V, $$\underbrace{V\frac{dS}{dt}}_{\text{accumulation}} = \underbrace{QS_0}_{\text{in}} - \underbrace{QS_{\text{out}}}_{\text{out}} + \underbrace{r_S V}_{\text{reaction}}$$ Each term carries units of mass per unit time, here kg/s. The reaction term is written as a generation term by convention, and the negative sign inside $r_S$ makes it a sink.
2. Apply the completely mixed assumption. The defining idealisation of a CSTR is that mixing is instantaneous and complete, so there is no concentration gradient anywhere inside the tank and the fluid leaving the outlet is a fair sample of the contents: $$S_{\text{out}} = S$$ This single assumption is what makes the reaction term evaluable, because the rate can now be computed at one concentration for the entire volume rather than integrated along a gradient.
3. Substitute the first-order rate law. With $r_S = -kS$ and $k = 0.1$ s^-1, the balance becomes $$V\frac{dS}{dt} = QS_0 - QS - kSV$$ This is the mass balance equation the question asks for.
4. Impose steady state and solve for S. At steady state nothing accumulates, so $dS/dt = 0$ and the left-hand side vanishes: $$0 = QS_0 - QS - kSV$$ Collecting the terms in S and factoring gives $QS_0 = S(Q + kV)$, so $$S = \frac{QS_0}{Q + kV}$$ Substituting the given values, $$S = \frac{(1)(1)}{1 + (0.1)(10)} = \frac{1}{2}$$ $$\boxed{S = 0.5\ \text{kg/m}^3}$$
5. Interpret the result through the residence time. Dividing the numerator and denominator of the result by Q introduces the hydraulic residence time $\tau = V/Q$ and gives the standard dimensionless form $$\frac{S}{S_0} = \frac{1}{1 + k\tau}$$ Here $\tau = 10/1 = 10$ s, so the Damköhler number is $k\tau = (0.1)(10) = 1.0$. A Damköhler number of exactly unity means the reaction time scale and the residence time scale are equal, which is why the concentration falls to exactly half. The removal efficiency is therefore 50 %.
6. Check the mass flows close. Mass entering is $QS_0 = 1.0$ kg/s; mass leaving in the discharge is $QS = 0.5$ kg/s; mass destroyed by reaction is $kSV = (0.1)(0.5)(10) = 0.5$ kg/s. Since $1.0 = 0.5 + 0.5$, the balance closes exactly, which confirms the algebra.

It is worth noting how much the reactor configuration costs. An ideal plug-flow reactor of the same volume and the same rate constant would give $S = S_0 e^{-k\tau} = e^{-1} = 0.368$ kg/m³, a removal of 63 % rather than 50 %. The complete back-mixing that makes the CSTR easy to analyse also dilutes the incoming substrate immediately to the low outlet concentration, and a first-order reaction proceeds more slowly at a low concentration. This is the reason real activated-sludge and chlorine-contact designs favour plug-flow basins or a series of CSTRs in cascade over a single large mixed tank.