Receiving-Water Assimilative Capacity and the Revised Phosphorus Limit: 16-Civ-B5 Water Supply and Wastewater Treatment, May 2017, Q3

Question 3: Receiving-Water Assimilative Capacity and the Revised Phosphorus Limit (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.

Given.

Quantity	Symbol	Value
WWTP rated capacity (effluent flow)	$Q_e$	5 000 m³/d
River flow upstream of the outfall	$Q_r$	500 000 m³/d
Background (upstream) total phosphorus	$C_{up}$	0.02 mg/L
Existing effluent limits	—	cBOD₅ 15, TSS 15, TP 1.0 mg/L
Permitted deterioration downstream	—	not more than 10 per cent of the upstream concentration

Find. The revised effluent total-phosphorus limit $C_e$ that holds the fully mixed downstream concentration at or below 1.10 times the background, and an assessment of the plant upgrades needed to achieve it.

Figure 3.1 — Mass-balance control volume drawn around the mixing zone. Total phosphorus is treated as conservative over the mixing length, so the balance contains no reaction term and inverts directly for the permissible effluent concentration.

Approach. Draw a steady-state, completely mixed control volume across the river immediately downstream of the outfall, treat total phosphorus as conservative over that short reach, evaluate the balance forwards to show that the existing 1 mg/L permit cannot meet the objective, then invert it for the effluent concentration that just satisfies the objective.

Translate the water-quality objective into a downstream concentration. "Not to deteriorate more than 10 per cent of the upstream concentration" fixes a ceiling on the fully mixed downstream value: $$C_{ds,\max} = 1.10\,C_{up} = 1.10 \times 0.02 = 0.0220\ \text{mg/L}$$ The entire assimilative capacity available to the discharger is therefore the 0.0020 mg/L of headroom between background and this ceiling — a very small allowance, which is what makes this a stringent limit.
Write the conservative mass balance on the control volume. At steady state, mass in equals mass out with no reaction: $$Q_r C_{up} + Q_e C_e = (Q_r + Q_e)\,C_{ds}$$ Because 1 mg/L is numerically 1 g/m³, every term evaluates directly in g/d with no conversion factor, provided both flows are in m³/d. The combined flow is $Q_r + Q_e = 500\,000 + 5\,000 = 505\,000$ m³/d.
Evaluate forwards to condemn the existing permit. Operating at the current 1.0 mg/L limit, the upstream load is $500\,000 \times 0.02 = 10\,000$ g/d and the effluent load is $5\,000 \times 1.0 = 5\,000$ g/d, so $$C_{ds} = \frac{10\,000 + 5\,000}{505\,000} = 0.0297\ \text{mg/L}$$ That is an increase of $(0.0297-0.02)/0.02 = 48.5$ per cent over background — almost five times the permitted 10 per cent. The existing permit is therefore inconsistent with the study's objective and must be tightened.
Invert the balance for the protective effluent limit. Solving the same equation for $C_e$ with $C_{ds} = C_{ds,\max}$: $$C_e = \frac{C_{ds,\max}(Q_r + Q_e) - Q_r C_{up}}{Q_e}$$ $$C_e = \frac{0.0220 \times 505\,000 - 500\,000 \times 0.02}{5\,000} = \frac{11\,110 - 10\,000}{5\,000}$$ $$\boxed{C_e = 0.222\ \text{mg/L total phosphorus}}$$
Restate the limit as a mass load, which is the enforceable form. $$L_e = C_e Q_e = 0.222 \times 5\,000 = 1\,110\ \text{g/d} = \boxed{1.11\ \text{kg/d as P}}$$ against the 5.0 kg/d permitted today. Expressing the requirement as a load rather than a concentration matters: a load cap cannot be met by dilution, and it automatically prevents the plant from buying compliance by increasing its flow. Any future capacity expansion must hold total phosphorus discharge at 1.11 kg/d, which means the concentration limit tightens in proportion to the new flow.
Express the reduction required of the plant. The limit falls by a factor of $1.0/0.222 = 4.5$, that is a further 77.8 per cent removal beyond what the current permit demands. A conventional secondary plant with no deliberate phosphorus removal discharges 4–6 mg/L TP (biomass synthesis alone takes up only about 1 per cent of the volatile solids as phosphorus), so relative to actual current performance the reduction is closer to 95 per cent.

Comment: upgrades required at the WWTP

An effluent limit of 0.222 mg/L TP is achievable but sits well below what secondary treatment delivers on its own, and below what chemical addition alone reliably achieves. Meeting it requires a combination of chemical precipitation and solids removal, and the following options are the realistic ones.

1. Chemical precipitation with a metal salt (the first and cheapest step). Alum, ferric chloride or ferric sulfate is dosed to precipitate phosphate as AlPO₄ or FePO₄ together with a bulk metal hydroxide floc that sweeps colloidal phosphorus. The dose point may be to the primary tank (pre-precipitation), the aeration basin or its outlet (co-precipitation, the usual retrofit), or after secondary clarification (post-precipitation). Stoichiometry is 1 mol Al per mol P, but reaching a low residual demands a substantial excess because the reaction competes with hydroxide precipitation; a molar ratio near 2:1 is typical for a target below 0.5 mg/L. Removing from a secondary effluent of about 4 mg/L TP down to 0.222 mg/L requires $\Delta P = 3.78$ mg/L, that is 0.122 mmol P/L, hence 0.244 mmol Al/L, which as alum $\mathrm{Al_2(SO_4)_3 \cdot 14H_2O}$ (594.4 g/mol, 2 mol Al each) is

$$\begin{aligned} D_{\text{alum}} &= \frac{0.244}{2} \times 594.4 = 72.4\ \text{mg/L} \\ M_{\text{alum}} &= 72.4 \times 5\,000 / 1\,000 = 362\ \text{kg/d} \end{aligned}$$

Consequences the plant must absorb: roughly 122 kg/d of additional chemical sludge (AlPO₄ plus Al(OH)₃), a corresponding increase in thickening, digestion and dewatering duty, alkalinity consumption of about 0.5 mg/L as CaCO₃ per mg/L of alum with the associated pH depression, and a chemical storage, containment and metering installation.

2. Enhanced biological phosphorus removal (EBPR). Adding an anaerobic selector ahead of the aeration basin — an A/O, A²/O, UCT or Bardenpho configuration — cultivates polyphosphate-accumulating organisms that release phosphorus anaerobically while taking up volatile fatty acids, then take up phosphorus in excess aerobically. EBPR routinely delivers 0.5–1.0 mg/L TP without chemical cost or chemical sludge, and it reduces the metal-salt dose needed for the final polish by 60–80 per cent. It requires a readily biodegradable COD to phosphorus ratio of at least about 15–20 to 1 in the influent, which a Canadian municipal sewage with significant infiltration may not have; primary sludge fermentation is the standard remedy. EBPR alone will not reach 0.222 mg/L reliably, and it is sensitive to secondary release in the clarifier and to phosphorus recycled from anaerobic digestion supernatant, which itself may need struvite recovery or supernatant treatment.

3. Tertiary filtration (effectively mandatory at this limit). Below about 0.5 mg/L, most of the residual phosphorus leaves the plant attached to escaping suspended solids rather than in solution — at 1 per cent P in the biomass, a 15 mg/L TSS effluent carries roughly 0.15 mg/L of particulate P by itself, which is two-thirds of the entire new limit. The existing 15 mg/L TSS permit is therefore no longer compatible with the phosphorus objective and must be tightened in step, to about 5 mg/L. This drives a tertiary stage: cloth-media disc filters or continuous-backwash deep-bed sand filters with in-line coagulant addition, or a membrane bioreactor if the plant is being rebuilt. Filtration plus co-precipitation is the standard Canadian route to 0.10–0.20 mg/L TP.

4. Supporting works. Sludge handling must be re-rated for the extra chemical and biological solids; a phosphorus-rich digester supernatant recycle must be intercepted rather than returned blindly to the head of the works; the effluent monitoring programme must move to low-level total-phosphorus methodology, since 0.222 mg/L is near the reporting limit of routine colorimetry and requires persulfate digestion with careful blank control; and the outfall diffuser should be reviewed, since the calculation above assumes complete mixing at the point of compliance.

Recommended package. Co-precipitation with ferric or alum dosed to the mixed liquor, a tightened TSS limit of 5 mg/L achieved by cloth-media tertiary filtration with a small in-line coagulant polish, and — if the influent volatile fatty acid supply permits — an anaerobic selector to reduce the chemical dose and the chemical sludge. That combination reaches 0.1–0.2 mg/L TP with margin against the 0.222 mg/L requirement.

Check: the river flow used is decisive, and 500 000 m³/d appears to be a mean flow. The required limit is very nearly proportional to river flow, since $C_e = (0.002\,Q_r + 0.022\,Q_e)/Q_e$. Canadian provincial assimilative-capacity assessments are normally carried out at a low-flow design condition — the 7Q10, the lowest seven-day average flow with a ten-year return period. If the 7Q10 is 30 per cent of the mean, that is 150 000 m³/d, the same calculation gives $C_e = 0.082$ mg/L; at 100 000 m³/d it gives 0.062 mg/L. The 0.222 mg/L boxed above is therefore the least stringent defensible limit, valid only if the objective is genuinely to be met at mean flow. A permit written for 7Q10 conditions would be three to four times tighter and would push the plant to membrane filtration or a reactive-media polishing stage. This assumption should be confirmed with the regulator before design proceeds. Two further assumptions are noted: total phosphorus is taken as conservative (justified over a short mixing reach, conservative over a long one, since settling and uptake would reduce the downstream value), and complete cross-sectional mixing is assumed at the point of compliance.

Quantity	Value
Permitted downstream concentration, $1.10\,C_{up}$	0.0220 mg/L TP
Downstream concentration under the existing 1 mg/L permit	0.0297 mg/L (48.5 per cent above background — fails)
Revised effluent TP limit	0.222 mg/L
Equivalent mass load limit	1.11 kg P/d (from 5.0 kg P/d today)
Further reduction required relative to the permit	77.8 per cent (factor of 4.5)
Illustrative alum dose at a 2:1 Al:P molar ratio	72.4 mg/L, 362 kg/d, plus about 122 kg/d chemical sludge
Limit if assessed at a 7Q10 of 150 000 m³/d	0.082 mg/L
Upgrade package recommended	Chemical co-precipitation + tertiary filtration (TSS limit to 5 mg/L), optionally with an EBPR anaerobic selector

Topic. Assimilative capacity of a receiving water, and the inversion of a conservative mixing-zone mass balance to set an effluent limit.

Key relations. $Q_r C_{up} + Q_e C_e = (Q_r+Q_e)C_{ds}$; inverted, $C_e = [C_{ds}(Q_r+Q_e) - Q_r C_{up}]/Q_e$; mass load $L = C_e Q_e$ with 1 mg/L equal to 1 g/m³; alum stoichiometry at 1 mol Al per mol P with a practical excess of 1.5–2.5.

Why this works. Over a mixing zone the residence time is minutes to hours, far shorter than the timescale of phosphorus uptake or settling, so no reaction term is needed and the balance is exact rather than approximate. Because it is linear in $C_e$ it inverts in one line, and because the objective is stated as a permitted increment over background, all of the difficulty concentrates in one number: the assimilative headroom $C_{ds,\max} - C_{up}$, here only 0.002 mg/L. Multiply that headroom by the total downstream flow and it becomes the entire mass allowance available to the discharger, 1.11 kg/d, from which the concentration limit follows by dividing by the plant flow. The structure makes the sensitivity obvious: since the headroom is proportional to $C_{up}$ and the mass allowance is proportional to $Q_r$, a pristine river or a low-flow condition collapses the allowance immediately.

Common pitfall. Two recur. First, applying the 10 per cent to the effluent concentration rather than to the downstream river concentration, which produces a nonsensical 0.9 mg/L "limit". Second, mixing units: if the river flow had been quoted in m³/s it must be multiplied by 86 400 before it can be added to a plant flow in m³/d, and forgetting to do so gives an answer wrong by nearly five orders of magnitude. A third, subtler error is quoting the answer only as a concentration; the regulator writes the condition as a load, and a concentration limit alone would permit the load to grow with any future expansion.

Source. Mihelcic & Zimmerman, Environmental Engineering: Fundamentals, Sustainability, Design, 3rd ed., Ch. 4 (steady-state mass balances, mixing zones); Metcalf & Eddy, Wastewater Engineering, 5th ed., §8-6 and §6-4 (chemical and biological phosphorus removal); CCME, Canadian Environmental Quality Guidelines (nutrient guidance for the protection of aquatic life).

Question 3 of 5: Receiving-Water Assimilative Capacity and the Revised Phosphorus Limit

Question 3: Receiving-Water Assimilative Capacity and the Revised Phosphorus Limit (25 marks)

Comment: upgrades required at the WWTP