Effluent phosphorus limit from receiving-water assimilative capacity: 16-Civ-B5 Water Supply and Wastewater Treatment, May 2013, Q3

Question 3: Effluent phosphorus limit from receiving-water assimilative capacity (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. The discharge and the receiving water are fully specified by five numbers, collected below.

Given data — Question 3
Quantity	Symbol	Value
Wastewater treatment plant flow	Q_e	10 000 m³/d
River flow upstream of the outfall	Q_r	5.79 m³/s
Current effluent TP limit	C_e	1.0 mg TP/L
Background TP upstream of the outfall	C_up	0.01 mg/L
TP the river can assimilate	C_assim	0.02 mg/L

Find. Whether the existing 1.0 mg/L effluent limit protects the river against the 0.02 mg/L assimilative ceiling, and if it does not, the numerical effluent limit that would, together with the size of the required tightening.

Figure 3.1. Mixing-zone control volume at the outfall. Total phosphorus is conservative over this reach, so the mass entering with the river and with the effluent must equal the mass leaving in the fully mixed downstream flow.

Approach. Treat the mixing zone as a steady-state, conservative, completely mixed control volume around the outfall: write a total-phosphorus mass balance on it, evaluate the downstream concentration produced by the current limit, compare that with the assimilative ceiling, and then invert the same balance to find the effluent concentration that lands exactly on the ceiling.

Put the two flows into the same units. The plant flow is quoted per day and the river flow per second, and nothing can be compared until they agree: $$Q_e = \frac{10\,000\ \text{m}^3/\text{d}}{86\,400\ \text{s}/\text{d}} = 0.1157\ \text{m}^3/\text{s}$$ against a river flow of Q_r = 5.79 m³/s. The dilution available is therefore Q_r/Q_e = 50.0, or almost exactly fifty parts of river to one part of effluent. That single number already tells us what to expect: a 1.0 mg/L effluent diluted fifty-fold contributes about 0.02 mg/L on its own, which is the entire assimilative ceiling before the background is even counted.
Write the mass balance across the mixing zone. Total phosphorus is conservative over the short reach of a mixing zone — it is neither created nor destroyed, only redistributed between dissolved and particulate forms — so at steady state the mass entering equals the mass leaving: $$Q_r C_{up} + Q_e C_e = (Q_r + Q_e)\,C_{mix}$$ which rearranges to give the fully mixed downstream concentration: $$C_{mix} = \frac{Q_r C_{up} + Q_e C_e}{Q_r + Q_e}$$
Evaluate the downstream concentration produced by the current limit. Substituting the given values with C_e = 1.0 mg/L: $$C_{mix} = \frac{(5.79)(0.01) + (0.1157)(1.0)}{5.79 + 0.1157} = \frac{0.05790 + 0.11574}{5.9057}$$ $$\boxed{C_{mix} = 0.0294\ \text{mg TP/L}}$$ The river can assimilate 0.02 mg/L. A plant discharging at exactly its permitted limit therefore drives the receiving water to 0.0294 mg/L — about 1.47 times the concentration the study identifies as the threshold for deterioration of water quality and aquatic life.
Comment on the current limit. The finding condemns it. The existing 1.0 mg TP/L limit is not protective of the receiving environment: a plant in full compliance with its permit would still push the river 47 % past its assimilative capacity, and the exceedance is not marginal or dependent on a modelling refinement. Note also what has been assumed in the plantʹs favour — complete and instantaneous mixing, no upstream sources other than the stated background, and a river flow of 5.79 m³/s that is treated as always available. Relaxing any of those assumptions makes the picture worse, not better. The limit must be tightened.
Invert the balance for the protective effluent limit. Setting C_mix equal to the assimilative ceiling of 0.02 mg/L and solving the same equation for C_e: $$C_{e,\text{new}} = \frac{C_{assim}(Q_r + Q_e) - Q_r C_{up}}{Q_e} = \frac{(0.02)(5.9057) - (5.79)(0.01)}{0.1157}$$ $$C_{e,\text{new}} = \frac{0.11811 - 0.05790}{0.1157} \quad\Longrightarrow\quad \boxed{C_{e,\text{new}} = 0.520\ \text{mg TP/L}}$$ Substituting 0.520 mg/L back into the balance returns C_mix = 0.0200 mg/L exactly, which confirms the inversion.
State the required change in the two forms a regulator uses. As a concentration, the limit must fall from 1.0 to 0.520 mg TP/L — a tightening of 0.480 mg/L, or 48 % of the present limit. As a mass load, which is how a discharge permit is more usefully written because it cannot be met by dilution: $$L = C_e \times Q_e \;\Rightarrow\; L_{\text{now}} = 10.0\ \text{kg TP/d}, \qquad L_{\text{new}} = 5.20\ \text{kg TP/d}$$ so the permitted phosphorus load must be cut by 4.80 kg/d, slightly more than half.
Cross-check against the riverʹs headroom. An independent route to the same answer confirms it. The river downstream carries (Q_r + Q_e) = 5.906 m³/s, or 510 256 m³/d, and may rise from 0.01 to 0.02 mg/L — a headroom of 0.01 g/m³, which is 5.10 kg TP/d. Adding back the phosphorus the effluent flow would carry even at background concentration, 0.01 g/m³ × 10 000 m³/d = 0.10 kg/d, gives 5.20 kg/d — identical to the limit load computed in Step 6.

Final results — Question 3
Quantity	Value
Effluent flow in consistent units	0.1157 m³/s (dilution 50:1)
Downstream TP at the current 1.0 mg/L limit	0.0294 mg/L
Assimilative ceiling	0.020 mg/L — exceeded by 47 %
Verdict on the current limit	Not protective; must be tightened
Protective effluent TP limit	0.520 mg TP/L
Required tightening	0.480 mg/L, i.e. 48 % of the present limit
Permitted TP load, present → required	10.0 → 5.20 kg TP/d (a cut of 4.80 kg/d)

Check — engineering assumptions behind the 0.520 mg/L figure. Three assumptions are implicit in the calculation and would be stated in an answer paper under page-1 Note 2. First, complete and instantaneous mixing at the outfall: a real plume needs a mixing length, and concentrations within it exceed the fully mixed value, so a regulator would normally define a mixing-zone boundary and could impose a stricter limit than 0.520 mg/L. Second, a river flow of 5.79 m³/s that is treated as always available; assimilative-capacity limits in Canadian practice are set at a low-flow design condition such as the 7Q10 (the seven-day average low flow with a ten-year return period), and because the required effluent concentration is very nearly proportional to river flow, a design low flow of half the stated value would roughly halve the allowable limit again. Third, that total phosphorus behaves conservatively over the mixing zone, which is appropriate for a short reach but conservative for a long one. All three point the same way, so 0.520 mg TP/L should be read as the least stringent defensible limit, not as a target. For context, 0.5 mg/L is a routine chemical-phosphorus-removal target for a conventional plant, whereas going below about 0.1 mg/L requires tertiary filtration.

Topic. Steady-state conservative mass balance on a receiving water, used to convert an assimilative-capacity finding into an effluent discharge limit.

Key relations. Q_rC_up + Q_eC_e = (Q_r + Q_e)C_mix; inverted, C_e = [C_mix(Q_r + Q_e) − Q_rC_up]/Q_e; mass load L = C_eQ_e with 1 mg/L = 1 g/m³; 1 m³/d = 1/86 400 m³/s.

Why this works. A mixing zone has no storage at steady state and phosphorus is not destroyed within it, so the phosphorus arriving per unit time must equal the phosphorus leaving per unit time. That single statement contains the whole problem: written forwards it predicts the river concentration from a proposed limit, and written backwards it converts a water-quality objective into a discharge limit. The reason the answer is so much smaller than the current permit is the arithmetic of dilution — the river offers 50 parts to 1, but the effluent is 100 times the background concentration, and 100 divided by 50 is 2, which is exactly the factor by which the river is over its budget. Expressing the answer as a load as well as a concentration matters in practice, because a load-based permit cannot be met by adding dilution water and it survives changes in plant flow.

Common pitfall. The commonest error is arithmetic rather than conceptual: leaving the plant flow in m³/d while the river is in m³/s, which inflates the effluentʹs influence by a factor of 86 400 and produces a nonsensical answer. The second is omitting Q_e from the denominator — using Q_r instead of (Q_r + Q_e) — which is a small error here but grows quickly on a low-dilution stream. The third is conceptual: comparing the effluent concentration directly with the 0.02 mg/L river criterion, as though the criterion applied at the end of the pipe. It applies in the river, after mixing.

Source. Mihelcic & Zimmerman, Environmental Engineering: Fundamentals, Sustainability, Design, 3rd ed., §4 (material balances on completely mixed systems) and §13 (water quality and receiving-water impact); CCME, Canadian Environmental Quality Guidelines — guidance on nutrient objectives and assimilative capacity; Metcalf & Eddy, Wastewater Engineering, 5th ed., §4 (effluent limits and discharge permits).

Question 3 of 5: Effluent phosphorus limit from receiving-water assimilative capacity

Question 3: Effluent phosphorus limit from receiving-water assimilative capacity (25 marks)