Question 4 of 5: Secondary Clarifier Sizing, Solids Loading and Sludge Age

Question 4: Secondary Clarifier Sizing, Solids Loading and Sludge Age (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 diameter of each of the two circular secondary clarifiers; (b) the solids loading rate imposed on them at average flow; (c) the solids retention time (sludge age) of the activated-sludge system.

Approach. Part (a) is a direct application of the definition $\mathrm{SOR} = Q/A$ followed by the geometry of a circle. Part (b) is a solids mass balance across the clarifier inlet, and turns on recognising that the solids arrive with the mixed liquor, which includes the return flow, whereas the hydraulic overflow rate in part (a) does not. Part (c) is a solids mass balance over the whole system: the inventory of biomass divided by the rate at which biomass leaves.

1. Part (a) — find the total clarifier surface area from the overflow rate. By definition the surface overflow rate is the plant flow divided by the plan area of the clarifiers, so $$A_{\text{total}} \;=\; \frac{Q}{\mathrm{SOR}} \;=\; \frac{15\,000\ \text{m}^3/\text{d}}{15\ \text{m}^3/\text{m}^2\!\cdot\!\text{d}} \;=\; 1\,000\ \text{m}^2$$ Only the plant flow $Q$ appears here. The return sludge is drawn off the clarifier floor and never crosses the effluent weirs, so it does not contribute to the upflow velocity that the overflow rate represents.
2. Divide the area between the two units. The two clarifiers are identical and share the flow equally: $$A_{\text{each}} \;=\; \frac{A_{\text{total}}}{n} \;=\; \frac{1\,000}{2} \;=\; 500\ \text{m}^2$$
3. Convert the area of one clarifier to a diameter. For a circular tank $A = \pi D^2/4$, so $$D \;=\; \sqrt{\frac{4A_{\text{each}}}{\pi}} \;=\; \sqrt{\frac{4 \times 500}{\pi}} \;=\; \sqrt{636.6}$$

   which gives

   $$\boxed{D \;=\; 25.2\ \text{m diameter, for each of the two clarifiers}}$$ Checking backwards, two tanks of 25.2 m give $2 \times \pi(25.2)^2/4 = 997.5$ m² and an overflow rate of $15\,000/997.5 = 15.04$ m³/m²·d, which is the design value to three figures. In construction the tank would be rounded up to the next standard mechanism size, typically 25.5 m, giving 1 021 m² and an overflow rate of 14.7 m³/m²·d — slightly conservative, which is the correct direction to round.
4. Part (b) — establish the flow that actually carries solids to the clarifiers. The clarifier inlet receives the mixed liquor, which is the plant flow together with the returned sludge: $$Q_{\text{ML}} \;=\; Q + Q_R \;=\; 15\,000 + 10\,000 \;=\; 25\,000\ \text{m}^3/\text{d}$$ This corresponds to a return ratio $Q_R/Q = 0.67$, a normal value for a conventional plant.
5. Compute the mass of solids applied per day. Both streams entering the clarifier carry mixed liquor at the aeration-tank concentration of 3 000 mg/L, or equivalently 3.0 kg/m³: $$M_{\text{solids}} \;=\; Q_{\text{ML}}\,X \;=\; 25\,000\ \text{m}^3/\text{d} \times 3.0\ \text{kg/m}^3 \;=\; 75\,000\ \text{kg/d}$$
6. Divide by the clarifier surface area to obtain the solids loading rate. Using the total area of 1 000 m² found in step 1, $$\mathrm{SLR} \;=\; \frac{(Q + Q_R)\,X}{A_{\text{total}}} \;=\; \frac{75\,000}{1\,000} \;=\; \boxed{75\ \text{kg/m}^2\!\cdot\!\text{d}\ \ (3.1\ \text{kg/m}^2\!\cdot\!\text{h})}$$ This sits comfortably inside the usual design envelope for conventional activated sludge, which is about 4 to 6 kg/m²·h at average flow with a peak allowance of 8 kg/m²·h, so the clarifiers are thickening-limited well within their capacity. It is worth noting how differently the two loadings behave: the hydraulic loading including the return flow is $25\,000/1\,000 = 25$ m³/m²·d, well above the 15 m³/m²·d overflow rate, which is exactly why the two parameters must be checked separately.
7. Part (c) — compute the mass of solids held in the system. The solids retention time is the inventory of biomass divided by the rate at which it is removed. Taking the aeration tank as the inventory, $$M_{\text{system}} \;=\; V\,X \;=\; 5\,000\ \text{m}^3 \times 3.0\ \text{kg/m}^3 \;=\; 15\,000\ \text{kg}$$
8. Compute the rate at which solids leave the system. Solids leave deliberately with the waste sludge and unavoidably with the effluent. The question gives no effluent suspended-solids concentration, so the effluent term is taken as negligible — a reasonable assumption for a well-operated plant, where it accounts for only a few per cent of the total: $$M_{\text{out}} \;=\; Q_W\,X_W \;=\; 150\ \text{m}^3/\text{d} \times 6.0\ \text{kg/m}^3 \;=\; 900\ \text{kg/d}$$
9. Divide to obtain the solids retention time. Assembling the two, $$\mathrm{SRT} \;=\; \frac{V X}{Q_W X_W + Q_e X_e} \;\approx\; \frac{15\,000\ \text{kg}}{900\ \text{kg/d}} \;=\; \boxed{16.7\ \text{d}}$$ The result is entirely consistent with the plant description. A sludge age of about 17 days is well above the four to five days a conventional plant needs for carbon removal alone, and above the roughly 10 to 15 days that year-round nitrification requires at Canadian winter mixed-liquor temperatures, so this plant should nitrify reliably. The corresponding hydraulic retention time is only $5\,000/15\,000 \times 24 = 8.0$ hours, and the ratio of the two — sludge held fifty times longer than water — is the whole reason the activated-sludge process works.