Question 5 of 5: Facultative lagoon; activated-sludge SRT and clarifier sizing

Question 5: Facultative lagoon; activated-sludge SRT and clarifier sizing (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) Working principle and operation of a facultative lagoon (10 marks)

A facultative lagoon is a shallow earthen basin, typically 1.5 to 2.5 m deep, in which wastewater is treated over weeks to months by a naturally stratified community with no mechanical aeration and no sludge recycle. Its name comes from the middle layer, where organisms switch facultatively between dissolved oxygen and other electron acceptors as conditions change through the day. The defining principle is a self-sustaining algal–bacterial symbiosis: algae in the sunlit upper layer photosynthesise and release oxygen, aerobic heterotrophs use that oxygen to oxidise the incoming organic matter and release carbon dioxide and nutrients, and the algae take those back up. Sunlight, not a blower, supplies the oxygen.

The aerobic surface layer occupies the top 0.3 to 0.6 m, as deep as light penetrates. Oxygen comes from algal photosynthesis, supplemented by wind-driven surface reaeration, and it is here that the soluble BOD is oxidised and pathogen die-off is fastest — sunlight, elevated pH and long retention together give lagoons a better bacterial and viral reduction than many mechanical plants. Dissolved oxygen swings dramatically over 24 hours, from supersaturation of 15 mg/L or more in the afternoon to near zero before dawn when photosynthesis stops and respiration continues; pH swings with it, from about 7.5 at night to 9 or above in the afternoon as the algae strip carbon dioxide out of the bicarbonate buffer. Those swings are not a malfunction — the high afternoon pH is what strips ammonia and precipitates phosphorus — but they do mean a single grab sample characterises nothing.

The facultative middle layer is intermittently aerobic. Organisms there use oxygen when the aerobic layer deepens in the afternoon and nitrate or sulphate when it does not, and this layer is what gives the lagoon its notable resistance to shock loads: there is no fragile inventory of solids to lose. The anaerobic benthic layer is the settled sludge blanket, where acid-forming and methanogenic bacteria digest the accumulated solids in situ, releasing methane, carbon dioxide and hydrogen sulphide. Because this digestion proceeds continuously, the sludge accumulates only 20 to 60 mm per year and a well-designed lagoon needs desludging on a decadal, not annual, cycle — the single biggest operating advantage of the technology.

Given. A small community discharging $1\,500\ \text{m}^3/\text{d}$ at $\text{BOD}_5 = 180\ \text{mg/L}$, designed at a cold-climate areal loading of $25\ \text{kg BOD/ha}\!\cdot\!\text{d}$ and 1.8 m liquid depth. Find. The lagoon area and the retention time. The load is $1\,500 \times 180/1\,000 = 270\ \text{kg BOD/d}$, so the area is $270/25 = \boxed{10.8\ \text{ha}}$, the volume is $10.8 \times 10^4 \times 1.8 = 194\,400\ \text{m}^3$, and the hydraulic retention time is $194\,400/1\,500 = \boxed{130\ \text{d}}$. Areal loading, not detention time, is the governing design parameter, because the oxygen supply is proportional to the sunlit surface and not to the volume — which is also why lagoons are shallow and sprawling rather than deep and compact.

Operation in the Canadian context. Design loadings here are far lower than in warm climates — 10 to 30 against 50 to 90 kg $\text{BOD/ha}\!\cdot\!\text{d}$ — for two reasons. Biological rates roughly halve for every $10\,{}^{\circ}\text{C}$ of cooling, and ice and snow cover eliminate both photosynthesis and surface reaeration for several months, so the lagoon runs anaerobically under the ice and stores rather than treats. Canadian lagoons are consequently sized for winter storage and operated on seasonal or controlled discharge: they are held through the winter and released in spring and autumn windows when the receiving water has the assimilative capacity, which conveniently avoids discharging during low summer flows. Spring turnover releases the accumulated products and produces a characteristic odour and effluent-quality excursion. Multiple cells in series, with the first heavily loaded and later cells polishing, give better performance than one large cell, and the last cell doubles as the discharge control. The persistent weakness is effluent suspended solids: the algae that make the process work are themselves 30 to 100 mg/L of TSS in the effluent, so a lagoon that easily meets a BOD limit may fail a TSS limit, and rock filters, intermittent sand filters or chemical addition in the final cell are the usual remedies.

(b) SRT and secondary clarifier area (15 marks)

Given. A conventional activated-sludge plant with the following data.

Find. The solids retention time of the system, and the secondary clarifier surface area that holds the overflow rate at or below $40\ \text{m}^3/\text{m}^2\!\cdot\!\text{d}$ under peak flow.

Approach. Compute the SRT from a solids balance on the whole system, being explicit about the withdrawal point; recover the unstated return flow from a solids balance on the clarifier; assume and justify a peaking factor to convert the average flow into a peak flow; then size the clarifier on the stated overflow-rate ceiling and confirm the result against the other loading limits a clarifier must satisfy.

1. Establish where the sludge is withdrawn. The question pairs a waste volume, $Q_w = 300\ \text{m}^3/\text{d}$, with the return concentration, $X_R = 8\,000\ \text{mg/L}$. That combination only makes sense if wasting is from the underflow, so the solids leave at $X_R$. This single reading decides the answer and is worth stating before any arithmetic: wasting the same 300 m³/d from the aeration tank at $X = 3\,000\ \text{mg/L}$ would give $V/Q_w = 8.3\ \text{d}$, a factor of 2.7 different.
2. Solids retention time. Neglecting the effluent solids, which the question does not give, $$\theta_c = \frac{V X}{Q_w X_R} = \frac{2\,500 \times 3\,000}{300 \times 8\,000} = \frac{7.50 \times 10^{6}}{2.40 \times 10^{6}} = \boxed{3.13\ \text{d}}$$ For context the hydraulic retention time is $\tau = V/Q = 0.25\ \text{d} = 6\ \text{h}$, so the solids are held about twelve and a half times longer than the water.
3. Recover the return sludge flow. The clarifier receives $Q + Q_R$ at $X$ and discharges $Q_R$ at $X_R$ with a clear overflow, so a solids balance gives $$Q_R = \frac{Q X}{X_R - X} = \frac{10\,000 \times 3\,000}{8\,000 - 3\,000} = 6\,000\ \text{m}^3\!/\text{d}, \qquad R = \frac{Q_R}{Q} = 0.60$$ This is needed for the loading checks in step 6; the question omits it, as these papers usually do.
4. Assume a peaking factor, and justify it. The overflow-rate limit is stated at peak flow but no peak flow is given, so this is the one free variable and the paper's instruction to "make suitable assumptions" applies here. At a Canadian per-capita flow of 350 L/cap·d the plant serves $10\,000/0.350 \approx 28\,600$ people, and Harmon's formula gives $$PF = 1 + \frac{14}{4 + \sqrt{P/1000}} = 1 + \frac{14}{4 + \sqrt{28.6}} = 2.50$$ so $PF = 2.5$ is adopted and $Q_{\text{peak}} = 2.5 \times 10\,000 = 25\,000\ \text{m}^3/\text{d}$.
5. Clarifier area at the overflow-rate ceiling. Since $\text{SOR} = Q/A$ and only the plant flow crosses the weirs, $$A = \frac{Q_{\text{peak}}}{\text{SOR}_{\max}} = \frac{25\,000}{40} = \boxed{625\ \text{m}^2}$$ As a single circular tank that is $D = \sqrt{4A/\pi} = 28.2\ \text{m}$; as two tanks in parallel, $312.5\ \text{m}^2$ each and $D = 19.9\ \text{m}$. The average-flow overflow rate is then $10\,000/625 = 16\ \text{m}^3/\text{m}^2\!\cdot\!\text{d}$, comfortably inside the usual 16 to 32 design range.
6. Check the duties the question did not ask about. A clarifier sized only on overflow rate is not yet a design. The solids loading rate must include the recycle: $$\text{SLR}_{\text{peak}} = \frac{(Q_{\text{peak}} + Q_R)X}{A} = \frac{(25\,000 + 6\,000)(3.0)}{625} = 148.8\ \text{kg/m}^2\!\cdot\!\text{d} = 6.2\ \text{kg/m}^2\!\cdot\!\text{h}$$ which is inside the $8\ \text{kg/m}^2\!\cdot\!\text{h}$ peak ceiling; at average flow it is only $3.2\ \text{kg/m}^2\!\cdot\!\text{h}$. Leaving the recycle out would have given $120\ \text{kg/m}^2\!\cdot\!\text{d}$, understating the thickening duty by a quarter. Weir loading decides the tank count: a single 28.2 m tank gives $25\,000/(\pi \times 28.2) = 282\ \text{m}^3/\text{m}\!\cdot\!\text{d}$, above the customary $250\ \text{m}^3/\text{m}\!\cdot\!\text{d}$ limit, whereas two 19.9 m tanks give $199\ \text{m}^3/\text{m}\!\cdot\!\text{d}$ and pass. Two tanks of 312.5 m² are therefore recommended, which also allows one to be taken out of service without shutting the plant.
7. Interpret the SRT. A 3.1-day sludge age is a high-rate, carbon-removal-only operation. Nitrifiers grow at $\mu = 0.75\ \text{d}^{-1}$ at $20\,{}^{\circ}\text{C}$, falling to $0.44\ \text{d}^{-1}$ at $12\,{}^{\circ}\text{C}$ with $\theta = 1.07$, so the washout SRT is 2.3 d and a design SRT with a safety factor of 2.5 would be 5.7 d. At 3.1 d this plant will not nitrify reliably in winter, and its effluent will carry essentially all of its influent ammonia. That is a legitimate design for a plant with only a BOD and TSS permit, but it should be stated rather than left implicit.