Question 4 of 5: Primary clarifier volume, overflow rate, and the choice between a second tank and a deeper tank

Question 4: Primary clarifier volume, overflow rate, and the choice between a second tank and a deeper tank (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 existing clarifier and the plant flow are specified by three numbers, and part (b) then poses two equal-volume expansion options.

Find. (a) the tank volume and the surface overflow rate at average flow; (b) a reasoned recommendation between the two equal-volume expansion options, argued explicitly in terms of what each does to HRT and to SOR.

Approach. Get the volume from the definition of hydraulic retention time, convert it to plan area using the side water depth, and divide the flow by that area to get the overflow rate; then recompute both parameters for each expansion option and decide using the ideal-settling-basin result that discrete removal depends on overflow rate alone.

(a) Volume and surface overflow rate (10 marks)

1. Volume from the retention time. Hydraulic retention time is tank volume divided by volumetric flow, so the volume is simply the flow multiplied by the time the flow is held: $$V = Q \times \mathrm{HRT} = 15\,000\ \frac{\text{m}^3}{\text{d}} \times \frac{2.0\ \text{h}}{24\ \text{h/d}}$$ $$\boxed{V = 1250\ \text{m}^3}$$
2. Plan area from the side water depth. The clarifier is a tank of essentially uniform depth, so its surface area is the volume divided by the side water depth: $$A = \frac{V}{H} = \frac{1250\ \text{m}^3}{3.0\ \text{m}} = 416.7\ \text{m}^2$$ For a circular clarifier that corresponds to a diameter of about 23 m, which is an ordinary size for a plant of this capacity.
3. Surface overflow rate. The overflow rate is the flow divided by the plan area — a velocity, and specifically the settling velocity of the slowest particle that is completely removed in an ideal basin: $$\mathrm{SOR} = \frac{Q}{A} = \frac{15\,000\ \text{m}^3/\text{d}}{416.7\ \text{m}^2}$$ $$\boxed{\mathrm{SOR} = 36.0\ \text{m}^3/(\text{m}^2\cdot\text{d}) = 36.0\ \text{m/d}}$$
4. Check the result against the identity that governs part (b). Because area was itself derived from volume and depth, the overflow rate must collapse to depth divided by retention time: $$\mathrm{SOR} = \frac{Q}{A} = \frac{Q}{V/H} = \frac{H}{V/Q} = \frac{H}{\mathrm{HRT}} = \frac{3.0\ \text{m}}{2.0/24\ \text{d}} = 36.0\ \text{m/d}$$ which agrees. This identity is not an arithmetic curiosity — it is the key to part (b), because it shows that overflow rate and retention time are not independent design choices once depth is fixed. In more familiar units the answer is 1.50 m/h, or 4.17 × 10⁻² cm/s: any particle settling faster than that is fully captured. Both values sit comfortably within the customary design ranges for primary clarification at average flow — roughly 30 to 50 m³/(m²·d) and 1.5 to 2.5 h — so the existing tank is conventionally proportioned.

(b) A second clarifier, or a deeper one? (15 marks)

The question is carefully constructed: both options double the primary clarification volume, so any argument based on volume alone cannot separate them. The decision has to be made on what each option does to the two parameters the question names.

1. Option 1 — a second identical clarifier in parallel. Plan area doubles while depth is unchanged, and the flow divides between two tanks: $$A_1 = 2 \times 416.7 = 833.3\ \text{m}^2, \qquad \mathrm{SOR}_1 = \frac{15\,000}{833.3} = 18.0\ \text{m/d}$$ $$\mathrm{HRT}_1 = \frac{2 \times 1250\ \text{m}^3}{15\,000\ \text{m}^3/\text{d}} \times 24 = 4.0\ \text{h}$$ Each tank receives 7500 m³/d through its original 416.7 m², so the overflow rate is halved and the retention time doubled.
2. Option 2 — double the depth of the existing tank. Volume doubles, but the footprint does not move: $$V_2 = 416.7\ \text{m}^2 \times 6.0\ \text{m} = 2500\ \text{m}^3, \qquad \mathrm{HRT}_2 = \frac{2500}{15\,000}\times 24 = 4.0\ \text{h}$$ $$\mathrm{SOR}_2 = \frac{Q}{A} = \frac{15\,000}{416.7} = 36.0\ \text{m/d} \quad \text{(unchanged)}$$ The two options are therefore identical in volume and identical in retention time, and differ in exactly one respect: only the parallel tank changes the overflow rate.
3. Apply the ideal settling basin result. In an ideal rectangular basin a discrete particle entering at the surface is captured if it reaches the floor within the detention period. Writing that condition out, the tank length divided by the horizontal velocity must exceed the depth divided by the settling velocity, and the depth cancels: $$\frac{L}{v_h} \ge \frac{H}{v_s} \;\Longrightarrow\; v_s \ge \frac{H\,v_h}{L} = \frac{H\,Q}{L\,(WH)} = \frac{Q}{A} = \boxed{v_c = \mathrm{SOR}}$$ The critical settling velocity equals the surface overflow rate and is a function of plan area alone. Depth has vanished, and with it any hope that a deeper tank captures finer particles. Particles slower than v_c are removed only in the fraction v_s/v_c, so lowering the overflow rate improves their capture too.
4. Quantify what each option buys. Option 1 halves the critical settling velocity from 36.0 to 18.0 m/d, that is from 1.50 to 0.75 m/h. Through Stokesʹ law, and taking a typical primary solid at 1400 kg/m³ in water at 20 °C, that moves the smallest fully captured particle from about 44 μm down to about 31 μm — a genuine extension of the captured size range, since the diameter scales with the square root of the settling velocity. Option 2 moves it not at all: the same 44 μm particle remains the smallest fully removed, because the tank is passing the same flow through the same surface.
5. Recommendation. Build the additional identical primary clarifier. It is the only one of the two options that improves settling efficiency, because it is the only one that lowers the surface overflow rate — and in primary sedimentation, which is dominated by discrete and lightly flocculent settling, overflow rate is the controlling design parameter. The doubled retention time that Option 2 also delivers is, on its own, worthless: retention time is a consequence of geometry, not a cause of removal, and the identity SOR = H/HRT proves it. Doubling the depth doubles the retention time solely because it doubles the distance a particle must fall, so the extra time is exactly consumed by the extra fall. Nothing is gained.
6. Note the practical arguments, which point the same way. Beyond the settling theory, a second tank gives redundancy — one unit can be taken out of service for sludge-hopper maintenance or repair without shutting down primary treatment — and it gives turndown flexibility at low night-time flow. A 6 m deep primary tank, by contrast, requires deeper excavation and heavier walls for the increased hydrostatic load, imposes a longer sludge-rake path, and risks the very problem primary clarifiers are prone to: a deep, quiescent sludge blanket going septic, releasing dissolved sulphide and phosphorus back into the liquid stream and undoing part of the removal it achieved. The one legitimate argument for depth is that greater depth gives flocculent particles more opportunity to collide and coalesce, and modern practice does favour side water depths above 3.5 m for that reason — but that is an argument for a moderate depth increase as a refinement, not for doubling depth in place of adding area.