Question 1 of 5: Impact of Wastewater Characteristics on Receiving Waters

Question 1: Impact of Wastewater Characteristics on Receiving Waters (25 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.

A discharge changes a receiving water through five broadly independent pathways: it consumes oxygen, it fertilises, it poisons, it infects, and it heats. The five sub-parts below take those pathways in turn. The unifying idea is that the receiving water, not the pipe, sets the acceptable effluent quality — the same effluent that is harmless in a large, cold, well-aerated river can sterilise a small one.

(i) BOD (5 marks)

Biochemical oxygen demand measures the oxygen that heterotrophic bacteria will consume while they oxidise the organic matter in the discharge. Once mixed into the river, that demand is exerted in the river itself, drawing dissolved oxygen down faster than reaeration across the water surface can replace it. The result is the classic dissolved-oxygen sag: DO falls downstream of the outfall, reaches a minimum at the critical point where deoxygenation and reaeration rates are momentarily equal, and then recovers. The damage is done at that minimum. Salmonids in British Columbia require roughly 5 mg/L for continued rearing and considerably more for incubating eggs, so a sag that dips below that level excludes the fishery from a reach even though the water looks clean. If the sag reaches zero the reach turns anaerobic: sulfate reduction produces hydrogen sulfide, iron and manganese are released from the sediments, the water blackens, and the benthic community collapses to a few pollution-tolerant worms and midge larvae.

Given. A river at 20 °C carrying \(Q_r = 3.0\ \text{m}^3\text{/s}\) with an ultimate BOD of 2.0 mg/L and 9.0 mg/L of dissolved oxygen receives \(Q_e = 0.5\ \text{m}^3\text{/s}\) of poorly treated effluent at \(L_e = 120\) mg/L ultimate BOD and 2.0 mg/L DO. Take \(k_d = 0.23\ \text{d}^{-1}\) and \(k_r = 0.40\ \text{d}^{-1}\).

Find. The minimum dissolved oxygen downstream, and whether the reach still supports salmonids.

Approach. Mix the two streams by conservative mass balance to get the initial BOD and deficit, then apply the Streeter–Phelps deficit equation and locate its maximum.

1. Mix the flows. A conservative mass balance on the mixing zone gives \(C_{mix} = (Q_rC_r + Q_eC_e)/(Q_r + Q_e)\), so \[L_0 = \frac{3.0(2.0) + 0.5(120)}{3.5} = 18.86\ \text{mg/L}, \qquad \text{DO}_{mix} = \frac{3.0(9.0) + 0.5(2.0)}{3.5} = 8.00\ \text{mg/L}.\] The initial deficit is therefore \(D_0 = 9.08 - 8.00 = 1.08\) mg/L.
2. Locate the critical point. Setting \(dD/dt = 0\) in the Streeter–Phelps equation gives \[t_c = \frac{1}{k_r-k_d}\ln\!\left[\frac{k_r}{k_d}\left(1 - \frac{D_0(k_r-k_d)}{k_dL_0}\right)\right] = \frac{1}{0.17}\ln\!\left[1.739\left(1 - \frac{1.08(0.17)}{0.23(18.86)}\right)\right] = 3.00\ \text{d}.\]
3. Evaluate the deficit there. At the critical time the deficit is \(D_c = (k_d/k_r)L_0e^{-k_dt_c}\), so \[D_c = 0.575(18.86)e^{-0.23(3.00)} = 5.44\ \text{mg/L} \;\Longrightarrow\; \boxed{\ \text{DO}_{min} = 9.08 - 5.44 = 3.64\ \text{mg/L}\ }\]

Three days of travel — perhaps 40 km on a slow river — the oxygen has fallen to 3.6 mg/L, well below the 5 mg/L salmonid criterion, and the reach is lost to the fishery even though it never goes anaerobic. That is the practical impact of BOD.

(ii) Phosphorus (5 marks)

Phosphorus is not toxic at the concentrations found in municipal effluent; its impact is entirely through eutrophication. In most Canadian fresh waters phosphorus is the limiting nutrient, so every gramme discharged is converted almost quantitatively into algal and cyanobacterial biomass. The consequences follow in sequence: turbidity and loss of water clarity, which shades out rooted macrophytes and destroys spawning habitat; wide diurnal swings in dissolved oxygen and pH as the bloom photosynthesises by day and respires by night, which is far harder on fish than a steady low DO; a taste-and-odour and cyanotoxin burden on any downstream drinking-water intake; and finally the death and settling of the bloom, whose decay exerts a secondary oxygen demand in the hypolimnion that is very much larger than the BOD the plant discharged in the first place. Anoxic hypolimnetic sediments then release the phosphorus they had stored, so the lake continues to bloom for years after the external load is cut — the internal-loading trap.

Given. A plant discharging \(Q = 5000\ \text{m}^3\text{/d}\) with total phosphorus 4.0 mg P/L and a residual cBOD of 20 mg/L. Algal cells have the Redfield composition \(\text{C}_{106}\text{H}_{263}\text{O}_{110}\text{N}_{16}\text{P}\), molar mass 3553 g/mol, and their aerobic decay consumes 138 mol of oxygen per mole of cell.

Find. The algal biomass the phosphorus can support and the oxygen demand of its decay, compared with the plant’s own residual BOD load.

1. Phosphorus load. \(L_P = CQ/1000 = 4.0(5000)/1000 = 20.0\ \text{kg P/d}\).
2. Convert to biomass. One gramme of phosphorus builds \(3553/30.97 = 114.7\) g of algal dry cell, so the discharge underwrites \(20.0 \times 114.7 = 2294\ \text{kg}\) of algae per day.
3. Convert to oxygen demand. Decay consumes \(138 \times 32 = 4416\) g of oxygen per mole of cell, i.e. \(4416/30.97 = 142.6\) g O₂ per gramme of phosphorus, giving \[\boxed{\ R_{O_2} = 20.0 \times 142.6 = 2851\ \text{kg O}_2\text{/d}\ }\]

The plant’s own residual carbonaceous load is only \(20 \times 5000/1000 = 100\) kg/d. The phosphorus therefore carries roughly 28 times the eventual oxygen demand of the BOD in the same effluent. That single ratio is the whole argument for phosphorus removal: chasing the last few milligrams of BOD is pointless while the nutrient that drives a demand thirty times larger goes untreated.

(iii) Total ammonia, free ammonia and nitrates (5 marks)

These three are stages of the same nitrogen, and each harms the receiving water differently. Total ammonia is the analytical sum of the ionised ammonium ion and the un-ionised ammonia molecule. Ammonium is comparatively benign; free (un-ionised) ammonia is the toxic species, because being uncharged it crosses gill membranes freely and disrupts ion regulation and ammonia excretion in fish. The split between the two is governed by pH and temperature through the equilibrium \(\text{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+\), so an effluent that is safe in a cool, near-neutral river becomes acutely toxic in a warm, algal-bloom river whose afternoon pH climbs above 8.5. Total ammonia also exerts a nitrogenous oxygen demand as nitrifying bacteria oxidise it, adding a second, delayed sag beneath the carbonaceous one. Nitrate, the end product, is essentially non-toxic to fish but is a nutrient: in nitrogen-limited waters, particularly estuaries and coastal receiving waters in British Columbia, it drives the same eutrophication sequence described for phosphorus, and in a drinking-water aquifer it is the cause of methaemoglobinaemia in infants.

Given. A discharge raising the total ammonia to 3.0 mg N/L in a river at 20 °C. The dissociation constant follows Emerson, \(\mathrm{p}K_a = 0.09018 + 2729.92/T\) with \(T\) in kelvin. Compare an afternoon pH of 8.5 with a night-time pH of 7.5.

Find. The free-ammonia concentration at each pH, against the CCME long-term guideline of 0.019 mg/L as N, and the nitrogenous oxygen demand.

1. Dissociation constant. \(\mathrm{p}K_a = 0.09018 + 2729.92/293.15 = 9.40\).
2. Un-ionised fraction. \(f = \left[1 + 10^{\,\mathrm{p}K_a - \mathrm{pH}}\right]^{-1}\), so \(f_{7.5} = (1 + 10^{1.90})^{-1} = 0.0124\) and \(f_{8.5} = (1 + 10^{0.90})^{-1} = 0.111\).
3. Free ammonia. Multiplying by the total, \[\boxed{\ [\text{NH}_3]_{7.5} = 0.037\ \text{mg N/L}, \qquad [\text{NH}_3]_{8.5} = 0.334\ \text{mg N/L}\ }\] Both exceed the 0.019 mg/L guideline; one pH unit multiplies the toxic fraction by almost exactly ten.
4. Nitrogenous oxygen demand. From \(2\,\text{NH}_4^+ + 4\,\text{O}_2 \rightarrow 2\,\text{NO}_3^- + 4\,\text{H}^+ + 2\,\text{H}_2\text{O}\), the ratio is \(4(32)/2(14.01) = 4.57\) g O₂ per g N, so the discharge carries \(4.57(3.0) = 13.7\) mg/L of additional oxygen demand.

The practical lesson is that an ammonia limit written as total ammonia is meaningless without the pH and temperature at which it must be met; Canadian permits therefore state ammonia limits as a table indexed on receiving-water pH and temperature.

(iv) Pathogens, residual chlorine and sodium bisulfite (5 marks)

Pathogens — enteric bacteria such as Salmonella and pathogenic E. coli, viruses such as norovirus and hepatitis A, and protozoan cysts of Giardia and Cryptosporidium — are the historic reason for treating sewage at all. Their impact on a receiving water is not ecological but public-health: they close bathing beaches, force the closure of bivalve harvesting areas, and contaminate downstream drinking-water intakes. Protozoan cysts are the governing organisms because they are resistant to chlorine and survive for weeks in cold water, which is why Canadian practice increasingly favours ultraviolet disinfection over chlorination.

Residual chlorine is the classic case of a treatment step creating a new impact. Chlorine is acutely toxic to fish and invertebrates at concentrations two to three orders of magnitude below the residual carried by a disinfected effluent, and it reacts with effluent organics to form trihalomethanes and haloacetic acids and with ammonia to form chloramines, which are themselves persistent and toxic. Canadian guidelines set the chronic total-residual-chlorine objective at about 0.002 mg/L.

Sodium bisulfite is the reagent added to destroy that residual before discharge, through \(\text{HSO}_3^- + \text{HOCl} \rightarrow \text{SO}_4^{2-} + \text{Cl}^- + 2\,\text{H}^+\). It is effective and fast, but it introduces two impacts of its own: any overdose is itself oxidised in the receiving water, consuming dissolved oxygen, and the reaction releases protons, depressing pH in a poorly buffered stream.

Given. An effluent leaving the contact tank with a 1.5 mg/L chlorine residual, discharged at \(Q_e = 0.5\ \text{m}^3\text{/s}\) into the \(3.0\ \text{m}^3\text{/s}\) river of part (i). Consider a 5 mg/L bisulfite overdose.

Find. Whether dilution alone can meet the chlorine objective, the stoichiometric bisulfite dose, and the oxygen cost of the overdose.

1. Dilution is not enough. Meeting 0.002 mg/L from a 1.5 mg/L residual needs a dilution of \(1.5/0.002 = 750\!:\!1\); the river offers only \(3.5/0.5 = 7\!:\!1\). Dechlorination is therefore mandatory, not optional.
2. Bisulfite dose. One mole of NaHSO₃ destroys one mole of chlorine, so the mass ratio is \(104.06/70.91 = 1.47\) mg per mg Cl₂, giving \[\boxed{\ \text{dose} = 1.47 \times 1.5 = 2.2\ \text{mg/L as NaHSO}_3\ }\]
3. Oxygen cost of an overdose. Excess bisulfite is oxidised by \(\text{HSO}_3^- + \tfrac{1}{2}\text{O}_2 \rightarrow \text{SO}_4^{2-} + \text{H}^+\), i.e. \(16/104.06 = 0.154\) g O₂ per g of reagent, so a 5 mg/L overdose removes \(0.154(5) = 0.77\) mg/L of dissolved oxygen — a fifth of the margin the sag calculation in part (i) had left.

(v) High temperature (5 marks)

A warm discharge acts on the receiving water in four ways at once, and they all push in the same direction. First, oxygen solubility falls as temperature rises, so the ceiling on dissolved oxygen drops just when it is most needed. Second, all biological rates accelerate — the deoxygenation constant, the respiration of fish and invertebrates, and the growth of algae — so the same BOD is exerted faster and the sag becomes deeper and closer to the outfall. Third, warming shifts the ammonia equilibrium toward the toxic un-ionised form, so a temperature rise converts an acceptable ammonia load into a toxic one without any change in the nitrogen discharged. Fourth, the direct thermal effect: cold-water species such as salmonids have narrow thermal tolerances, warming triggers premature emergence and migration cues, and a thermal plume can form a barrier that blocks upstream passage entirely. British Columbia guidance limits the change in ambient temperature in a salmonid stream to roughly 1 °C for this reason.

Given. A receiving water warmed from 20 to 30 °C, with \(k_d = 0.23\ \text{d}^{-1}\) at 20 °C and the usual Arrhenius coefficient \(\theta = 1.047\). Ammonia at pH 8.0.

Find. The loss of oxygen-carrying capacity, the acceleration of the BOD reaction, and the shift in ammonia speciation.

1. Solubility. Saturation falls from 9.08 to 7.54 mg/L, a loss of 1.54 mg/L or 17 per cent of the available oxygen — before a single milligram of BOD is exerted.
2. Reaction rate. \(k_{d,30} = k_{d,20}\,\theta^{\,T-20} = 0.23(1.047)^{10} = 0.364\ \text{d}^{-1}\), an increase of 58 per cent, so the sag deepens and moves upstream toward the outfall.
3. Ammonia speciation. The dissociation constant falls from 9.40 to 9.10, so at pH 8.0 the free fraction rises from 3.8 to 7.4 per cent: \[\boxed{\ \text{warming } 20 \to 30\ ^\circ\text{C at fixed pH nearly doubles the toxic free ammonia}\ }\]