Question 1 of 7

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) — Downstream phosphorus concentration (6 marks)

Given. An outfall discharges into a flowing river; both streams are characterised by a flow rate and a phosphorus concentration.

Find. The phosphorus concentration $C_d$ in the river immediately downstream of the discharge, once the effluent is fully mixed across the channel, in mg P/L.

Approach. Draw a control volume around the mixing zone and write a steady-state mass balance on phosphorus, which is treated as a conservative substance over the short mixing reach.

1. State the general material balance. For any control volume, $$\text{accumulation} = \text{in} - \text{out} + \text{generation}$$ At steady state the accumulation term vanishes, and phosphorus is neither created nor destroyed over the few hundred metres of the mixing zone, so the generation term vanishes as well. The balance collapses to mass in equals mass out.
2. Write the flow balance. Water is likewise conserved, so the downstream flow is the sum of the two inflows: $$Q_d = Q_u + Q_e = 20 + 1.0 = 21\ \text{m}^3\text{/s}$$
3. Write the mass balance on phosphorus. Mass flux is the product of volumetric flow and concentration, so $$Q_u C_u + Q_e C_e = Q_d C_d$$ Substituting the given data, with the convenient identity that $1\ \text{m}^3\text{/s} \times 1\ \text{mg/L} = 1\ \text{g/s}$: $$(20)(0.05) + (1.0)(5.0) = 1.0 + 5.0 = 6.0\ \text{g/s}$$ The effluent, though it carries only one twenty-first of the water, contributes five-sixths of the phosphorus load.
4. Solve for the downstream concentration. Dividing the total load by the total flow, $$C_d = \frac{Q_u C_u + Q_e C_e}{Q_u + Q_e} = \frac{6.0\ \text{g/s}}{21\ \text{m}^3\text{/s}}$$ $$\boxed{C_d = 0.286\ \text{mg P/L}}$$
5. Check the result for physical sense. The answer must lie between the two feed concentrations, 0.05 and 5.0 mg P/L, and closer to the more abundant stream — it does. Re-multiplying, $21 \times 0.286 = 6.0$ g/s, which recovers the load computed in Step 3.

The number is small in absolute terms but large in environmental terms. Guidance for Canadian rivers places the trigger for accelerated eutrophication near 0.03 mg P/L of total phosphorus, so the mixed river water sits roughly nine to ten times above the level at which nuisance algal growth becomes likely. Phosphorus is almost always the limiting nutrient in fresh water, which is why a single municipal outfall of modest flow can drive a receiving stream into a eutrophic state, and why chemical phosphorus removal at the plant is normally the cheapest available control.

Part (ii) — CSTR outlet concentration (7 marks)

Given. A single completely stirred tank reactor operating at steady state, with first-order destruction of the waste.

Find. The steady-state outlet waste concentration $C$, in mg/L.

Approach. Apply the same material balance as in part (i), but retain the reaction term; complete mixing lets the exit concentration be identified with the tank concentration, which makes the balance algebraic rather than differential.

1. Write the unsteady balance on the waste. Over the reactor control volume, $$V\frac{dC}{dt} = Q C_0 - Q C - kCV$$ where the three terms on the right are, in order, the mass entering, the mass leaving, and the mass destroyed by reaction. The reaction term carries the tank concentration $C$ because complete mixing means the reaction proceeds everywhere at the same rate.
2. Impose steady state. Setting $dC/dt = 0$ and rearranging, $$Q C_0 = QC + kCV = C\,(Q + kV)$$ so that $$C = \frac{Q C_0}{Q + kV}$$
3. Form the hydraulic residence time and the Damköhler number. Dividing numerator and denominator by $Q$ and writing $\tau = V/Q$, $$\tau = \frac{100\ \text{m}^3}{30\ \text{m}^3\text{/d}} = 3.33\ \text{d}, \qquad \mathrm{Da} = k\tau = (0.25)(3.33) = 0.833$$ The Damköhler number compares the time available for reaction against the time the reaction needs; at $\mathrm{Da} = 0.833$ the two are comparable, so partial — not near-complete — destruction is expected before any arithmetic is done.
4. Evaluate the outlet concentration. In terms of $\mathrm{Da}$, $$C = \frac{C_0}{1 + k\tau} = \frac{80}{1 + 0.833} = \frac{80}{1.833}$$ $$\boxed{C = 43.6\ \text{mg/L}}$$ which corresponds to a removal efficiency of $(80 - 43.6)/80 = 45.5\ \%$.
5. Close the mass balance as a check. Mass in is $QC_0 = 30 \times 80 = 2400$ g/d. Mass out is $QC = 30 \times 43.6 = 1309$ g/d, and mass destroyed is $kCV = 0.25 \times 43.6 \times 100 = 1091$ g/d. The two outgoing terms sum to 2400 g/d, matching the input exactly.

It is worth contrasting this result with the ideal plug-flow reactor of identical volume, for which $C = C_0 e^{-k\tau} = 80 e^{-0.833} = 34.8$ mg/L, a removal of 56.5 %. The plug-flow vessel does appreciably better on the same footprint because in a CSTR the entire tank operates at the low exit concentration, and a first-order rate is proportional to concentration. Back-mixing therefore costs removal, which is precisely why full-scale reactors are commonly staged as several smaller tanks in series rather than built as one large one.

Part (iii) — Chlorine disinfection and total residual chlorine (7 marks)

Chlorine added to water as chlorine gas or as sodium hypochlorite hydrolyses immediately to hypochlorous acid, $\mathrm{HOCl}$, which dissociates to the hypochlorite ion, $\mathrm{OCl^-}$, with a $pK_a$ near 7.5. Together these two species constitute the free available chlorine. Hypochlorous acid is by far the more potent of the pair — roughly two orders of magnitude more biocidal than hypochlorite — because it is a small, electrically neutral molecule that diffuses readily through the lipid cell membrane, whereas the negatively charged hypochlorite ion is largely repelled by the similarly charged cell surface. This is the reason disinfection efficiency falls sharply as pH rises above about 7.5, and why pH control is part of disinfection control.

Once inside the cell, chlorine acts as a powerful oxidant on several targets simultaneously rather than through a single mechanism. It oxidises the sulfhydryl groups of enzymes essential to glucose metabolism, destroying respiratory and transport activity; it damages membrane proteins and lipids so that the cell can no longer maintain the ionic gradients it needs and leaks its cytoplasmic contents; and at higher doses it attacks nucleic acids and disrupts DNA replication. The cumulative effect is that the organism is inactivated — rendered incapable of reproducing and causing infection — rather than merely removed. Because the mechanism is a chemical reaction with the cell, the extent of inactivation depends on the product of disinfectant concentration and contact time, the familiar $C\!\cdot\!t$ criterion, and on temperature and pH. Resistance also varies enormously by organism: vegetative bacteria such as E. coli are inactivated in seconds, enteric viruses require considerably more $C\!\cdot\!t$, and the protozoan cysts and oocysts of Giardia and especially Cryptosporidium are so chlorine-resistant that they must be addressed by filtration or ultraviolet light instead.

The reason total residual chlorine is a good indirect indicator of disinfection performance follows from that chemistry. Chlorine added to a wastewater effluent is first consumed by the reduced material present — ammonia, organic nitrogen, sulfide, ferrous iron, nitrite and dissolved organic matter. Only once that chlorine demand has been satisfied does a measurable residual persist. The presence of a residual at the end of the contact tank is therefore direct evidence that the demand was exceeded, that oxidant was still available throughout the contact period, and by implication that the pathogens in the effluent were exposed to an oxidising environment for the full detention time. Total residual chlorine counts both the free forms and the combined forms (the chloramines produced by reaction with ammonia), which matters in wastewater because ammonia is normally abundant and most of the residual is combined rather than free.

The practical merit of the measurement is that it can be made continuously, on-line, in seconds and at negligible cost by amperometric or colorimetric analysers, whereas the quantity actually of interest — the surviving pathogen or indicator-organism density — requires a laboratory culture taking 18 to 24 hours for faecal coliforms and far longer for viruses or protozoa. By the time a microbiological result is available the effluent has long since been discharged. A continuous residual signal, interlocked to the chlorinator and to an alarm, allows the operator to detect and correct a dosing failure within minutes. This is why Canadian approvals for chlorinated effluents almost invariably specify a minimum residual at the end of the contact chamber as an operational surrogate, alongside a periodic bacteriological limit as the verification.

The indicator must nevertheless be used with an understanding of its limits. A residual confirms that oxidant was present but not that it was well distributed: short-circuiting in a poorly baffled contact tank can leave a satisfactory residual at the outlet while a fraction of the flow passed through in a fraction of the design contact time. Particles are the second limitation, since organisms embedded in or shielded by suspended solids are protected from the oxidant, which is why a residual is only meaningful in a well-clarified effluent and why turbidity and chlorine residual are read together. Finally, in Canada the residual is itself a regulated pollutant — chlorine is acutely toxic to fish at concentrations well below 0.1 mg/L, so the Wastewater Systems Effluent Regulations made under the Fisheries Act cap total residual chlorine in the discharge at 0.02 mg/L, and effluents are routinely dechlorinated with sulfur dioxide or sodium bisulfite after the contact tank. The residual is thus maximised where it does the disinfecting work and then deliberately destroyed before discharge.