Question 4 of 5: Conventional Activated Sludge — Tank Sizing, Clarifier Loading and Oxygen Demand

Question 4: Conventional Activated Sludge — Tank Sizing, Clarifier Loading and Oxygen Demand (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.

Find. (a) the aeration tank volume $V$ and hydraulic retention time $\tau$; (b) the secondary clarifier surface overflow rate and solids loading rate; (c) the oxygen demand in kg/d for carbonaceous BOD removal.

Approach. Part (a) uses the definition of solids retention time — solids in the system divided by solids leaving per day — inverted for the tank volume, then the hydraulic retention time on the influent flow. Part (b) converts the clarifier volume to a plan area, then applies the surface overflow rate on the plant flow alone and the solids loading rate on the plant flow plus the return sludge, the return flow being obtained from a solids balance because the question does not state it. Part (c) uses the standard oxygen-demand balance: ultimate carbonaceous demand removed, less the oxygen equivalent of the cell mass wasted.

Part (a) — Aeration tank volume and HRT (10 marks)

1. Part (a) — write the definition of solids retention time. The sludge age is the mass of solids held in the aeration tank divided by the mass of solids leaving the system each day. Neglecting the solids lost over the effluent weirs (an assumption checked below), everything leaves in the waste stream: $$\theta_c = \frac{V X}{Q_w X_w}$$
2. Invert for the tank volume. This is the only route to $V$, since neither a volumetric loading nor an F/M ratio is given: $$V = \frac{\theta_c\,Q_w X_w}{X} = \frac{10 \times 100 \times 10\,000}{3\,500}$$ Every concentration in this expression appears as a ratio, so mg/L cancels against mg/L and no conversion to kg/m³ is needed. $$\boxed{V = 2\,857\ \text{m}^3}$$
3. Compute the hydraulic retention time on the influent flow. The HRT of the aeration tank is defined on the flow entering the process, not on the mixed-liquor flow $Q + Q_R$; the return sludge is an internal recycle and does not add to the throughput of the plant. $$\tau = \frac{V}{Q} = \frac{2\,857}{10\,000} = 0.2857\ \text{d}$$ $$\boxed{\tau = 6.86\ \text{hours}}$$
4. Check the result against conventional design practice. A conventional (plug-flow) activated-sludge process is normally designed at $\theta_c$ of 3–15 d, $\tau$ of 4–8 h, MLSS of 1 500–4 000 mg/L, an F/M of 0.2–0.4 kg BOD per kg MLVSS per day, and a volumetric loading of 0.3–0.7 kg BOD/m³·d. This design gives an F/M of $$\text{F/M} = \frac{Q S_0}{V X_v} = \frac{10\,000 \times 150}{2\,857 \times 2\,625} = 0.20\ \text{kg BOD/kg MLVSS}\!\cdot\!\text{d}$$ taking MLVSS as 75 per cent of MLSS, and a volumetric loading of $1\,500/2\,857 = 0.53$ kg BOD/m³·d. Both sit inside the conventional envelope, as does the 6.86 h retention time, so the answer is self-consistent.

Part (b) — Secondary clarifier surface overflow rate and solids loading rate (7 marks)

1. Part (b) — convert the clarifier volume to a plan area. For a tank of uniform side water depth, $$A = \frac{V_{sc}}{H} = \frac{3\,500}{3.5} = 1\,000\ \text{m}^2$$ If this is provided as two circular units, each is 500 m², that is 25.2 m diameter.
2. Apply the surface overflow rate on the plant flow only. The return sludge leaves through the underflow and never crosses the effluent weirs, so it is excluded: $$\text{SOR} = \frac{Q}{A} = \frac{10\,000}{1\,000} = \boxed{10.0\ \text{m}^3/\text{m}^2\!\cdot\!\text{d}}$$ This is conservative against the 16–33 m/d normally allowed at average flow for a conventional activated-sludge clarifier, which leaves useful margin for peak wet-weather flow.
3. Obtain the return sludge flow from a clarifier solids balance, since the question does not state it. The solids loading rate cannot be evaluated without $Q_R$. Taking the underflow concentration as the waste sludge concentration, $X_R = X_w = 10\,000$ mg/L, and neglecting effluent solids, a solids balance about the clarifier gives $(Q+Q_R)X = Q_R X_R$, hence $$Q_R = \frac{Q X}{X_R - X} = \frac{10\,000 \times 3\,500}{10\,000 - 3\,500} = 5\,385\ \text{m}^3/\text{d}$$ a recirculation ratio of $R = Q_R/Q = 0.54$, comfortably inside the 0.25–1.0 range typical of conventional plants.
4. Apply the solids loading rate on the total flow entering the tank. Every kilogram of solids must be thickened whether it arrived with the plant flow or the return flow, so the return is included: $$\text{SLR} = \frac{(Q + Q_R)\,X}{A} = \frac{(10\,000 + 5\,385)(3\,500)}{1\,000 \times 1\,000}$$ $$\boxed{\text{SLR} = 53.8\ \text{kg}/\text{m}^2\!\cdot\!\text{d}}$$ Omitting the return flow — the standard error — would have given 35.0 kg/m²·d, low by a factor of 1.54. The computed value is well inside the 70–150 kg/m²·d normally permitted at average flow, so the clarifier is adequate for both clarification and thickening.
5. Note the clarifier detention time as a check. $V_{sc}/Q = 3\,500/10\,000 = 0.35$ d, that is 8.4 h on the plant flow, or 5.5 h on the total flow including the return. Both comfortably exceed the 1.5–2.5 h minimum, confirming that this tank is generously sized.

Part (c) — Oxygen demand for BOD removal (8 marks)

The oxygen actually required is not simply the BOD removed. Part of the organic carbon is converted to new cell mass and leaves the plant as waste sludge rather than being oxidised, so the oxygen requirement is the ultimate carbonaceous demand of the substrate removed, less the oxygen equivalent of the biomass wasted. Since the question asks for the demand "for BOD removal", no nitrogenous demand is included.

1. Part (c) — state the oxygen balance and its two assumptions. $$R_o = \frac{Q\,(S_0 - S)}{f} - 1.42\,P_{x,\text{bio}}$$ where $f = \text{BOD}_5/\text{BOD}_u = 0.68$ converts the 5-day demand to the ultimate demand, and 1.42 is the oxygen equivalent of cell tissue, from the complete oxidation $\mathrm{C_5H_7NO_2} + 5\mathrm{O_2} \rightarrow 5\mathrm{CO_2} + 2\mathrm{H_2O} + \mathrm{NH_3}$, which requires $5 \times 32/113 = 1.42$ g O₂ per g of cells. The question does not state the effluent soluble BOD₅, so a well-performing secondary effluent value of $S = 10$ mg/L is assumed.
2. Compute the BOD₅ removed. $$Q(S_0-S) = 10\,000 \times (150-10) = 1.40\times10^{6}\ \text{g/d} = 1\,400\ \text{kg BOD}_5/\text{d}$$
3. Convert to ultimate carbonaceous demand. The oxygen balance must be written on ultimate demand, because the aeration system supplies oxygen for the whole reaction, not merely the part expressed in five days: $$\frac{Q(S_0-S)}{f} = \frac{1\,400}{0.68} = 2\,058.8\ \text{kg BOD}_u/\text{d}$$
4. Compute the oxygen equivalent of the biomass wasted. The 1.42 factor belongs to volatile solids, since it is the oxygen demand of cell tissue; the question supplies a total-solids wastage, so it must first be converted at the usual 0.80 volatile fraction for waste activated sludge: $$P_{x,\text{bio}} = 0.80\,Q_w X_w = 0.80 \times 100 \times 10\,000 = 8.00\times10^{5}\ \text{g VSS/d} = 800\ \text{kg VSS/d}$$ $$1.42\,P_{x,\text{bio}} = 1.42 \times 800 = 1\,136\ \text{kg O}_2/\text{d}$$
5. Close the balance. $$R_o = 2\,058.8 - 1\,136 = \boxed{923\ \text{kg O}_2/\text{d}}$$ that is an average oxygen uptake of 38.5 kg/h, and 0.66 kg O₂ per kg of BOD₅ removed. Blower sizing would then apply a peaking factor and correct from this actual oxygen requirement to a standard oxygen requirement using the alpha, beta and temperature-correction factors of the diffuser system, typically inflating it by a factor of 1.8–2.5.