Question 5 of 5: Trickling Filter Operation; Activated Sludge SRT and Clarifier Sizing

Question 5: Trickling Filter Operation; 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.

Part (a) — Working principle and operation of a trickling filter (10 marks). A trickling filter is an attached-growth aerobic reactor. Despite the name it does not filter: settled wastewater is distributed over a bed of highly permeable media on which a microbial biofilm grows, and treatment occurs as the liquid trickles down through the void space in a thin film over that biofilm while air moves through the same voids in the opposite direction.

The working principle is diffusion across a fixed film. Primary effluent is applied at the top by a rotary distributor, whose arms sweep the bed and dose each area intermittently; the dosing rate, expressed as the Spuelkraft or flushing intensity in millimetres of liquid per pass, is set by the rotational speed and is a genuine operating variable, since heavy intermittent dosing flushes excess growth and controls filter flies and ponding. The applied liquid spreads as a film over the biofilm. Organic matter and dissolved oxygen diffuse into the film from the liquid, metabolic end products diffuse out, and only the outer 0.1 to 0.2 mm of the film is aerobic; beneath that the film goes anoxic and then anaerobic. Because oxygen must be resupplied continuously from the air rather than from dissolved oxygen in the liquid, ventilation is the single most important operating requirement. Air moves by natural draught, driven by the temperature difference between the ambient air and the wastewater; the draught reverses direction when that difference reverses, and it can stall entirely when the two temperatures are equal. Design practice provides open underdrains, peripheral vents sized at not less than about 1 m² of vent area per 250 m² of filter area, and forced ventilation on deep plastic-media towers. A filter that loses ventilation goes anaerobic, produces hydrogen sulphide and odour, and loses treatment capacity.

The film grows as it metabolises, and when it becomes thick enough that the inner layer is starved and loses its grip — and under the shear of the applied flow — it sloughs, releasing solids into the underdrain flow. Sloughing is normal and necessary, but it means that the filter effluent always carries suspended biomass. The secondary clarifier is therefore an integral part of the process, not a polishing step: without it the sloughed film would appear directly in the effluent as suspended solids and BOD, and the process would fail to meet its limits no matter how well the biology performed. The humus sludge collected there is sent to digestion; there is no return of sludge to the filter, which is one of the fundamental differences from activated sludge.

Recirculation returns a fraction of filter or clarifier effluent to the filter inlet at a ratio $R = Q_R/Q$ typically between 0.5 and 2. It serves four purposes: it maintains a minimum wetting rate so the whole bed stays active at low night-time flows, it dilutes and dampens shock loads of strong or toxic influent, it re-seeds the upper bed with active organisms, and it increases the flushing intensity so that the bed self-cleans. Its effect on performance is captured by the NRC formulation, in which the recirculation factor

$$F = \frac{1 + R}{(1 + 0.1R)^2}$$

enters the efficiency expression $E = 100/\bigl(1 + 0.44\sqrt{W/(VF)}\bigr)$; the diminishing return in the denominator is why recirculation ratios above about 2 buy very little.

Media and loading define the classification. Rock media 1.5 to 2.5 m deep with a specific surface of 40 to 60 m²/m³ support low-rate filters at 1 to 4 m³/m²·d, giving 80 to 90 percent BOD removal and often nitrification; modern cross-flow plastic media, 4 to 12 m deep with 90 to 150 m²/m³ and much greater void fraction, permit high-rate and roughing towers at 10 to 75 m³/m²·d. The characteristic operating problems are ponding from excessive growth or media breakdown, filter flies (Psychoda) in under-dosed low-rate rock filters, odour from inadequate ventilation, and cold-weather performance loss, which in Canadian installations is met with covers and wind screens. Set against these, the trickling filter's virtues are exactly the ones activated sludge lacks: low energy use because ventilation is passive, high resistance to shock and toxic loads because the biomass is attached and cannot wash out, simple operation, and no sludge-bulking mechanism.

Part (b) — SRT and secondary-clarifier area (15 marks).

Given.

Find. The solids retention time of the activated-sludge system, and the plan area of the secondary clarifier such that the surface overflow rate at peak flow does not exceed 40 m³/m²·d.

Assumptions (the question invites them). (1) Sludge is wasted from the return line, because the question pairs a waste volume with the return solids concentration; this is stated explicitly below because it changes the answer by a factor of 2.7. (2) Effluent suspended solids are negligible against the wasted solids, so $X_e \approx 0$. (3) A peaking factor of 2.5 is applied to the average flow to obtain the peak design flow, consistent with a plant of this size (roughly 10 000 people at 400 L/capita·d); Harmon's formula at that population gives 2.96 and Babbitt's 3.15, so 2.5 is at the lower end and the sensitivity is quantified below. (4) Steady state, complete mixing, and a clarifier solids balance with no solids accumulation in the blanket.

Approach. Take the aeration tank as the control volume for solids: SRT is the inventory of solids under aeration divided by the rate at which solids leave the system, which under assumption (2) is the wastage alone. Then take the clarifier: size its plan area from the peak flow at the stated overflow ceiling, and check the resulting tank against solids loading and weir loading before accepting it.

1. Establish where the sludge is withdrawn, because that fixes the SRT. The question gives a waste sludge flow of 80 m³/d and, in the same sentence, a return solids concentration of 8000 mg/L. The two belong together: wasting is from the clarifier underflow, at $X_w = X_R = 8000$ mg/L. Had the same 80 m³/d been drawn from the aeration tank it would carry only $X = 3000$ mg/L, and the answer would be $V/Q_w = 12.5$ d instead — a factor of 2.7. State the withdrawal point before computing anything.

2. Compute the solids inventory under aeration. Since 1 mg/L $=$ 1 g/m³,

   $$M_{\text{tank}} = \frac{V X}{1000} = \frac{1000 \times 3000}{1000} = 3000\ \text{kg of MLSS}$$

3. Compute the solids leaving the system each day. With effluent solids neglected, all solids leave in the wastage:

   $$M_{\text{waste}} = \frac{Q_w X_R}{1000} = \frac{80 \times 8000}{1000} = 640\ \text{kg/d}$$

4. Divide to obtain the solids retention time. SRT is the inventory divided by the daily removal rate,

   $$\theta_c = \frac{V X}{Q_w X_R} = \frac{1000 \times 3000}{80 \times 8000} = \frac{3000\ \text{kg}}{640\ \text{kg/d}}$$

   $$\boxed{\theta_c = 4.69\ \text{days}}$$

   This is a conventional carbonaceous-removal sludge age. It is well above the 3 to 4 days needed to retain heterotrophs, but below the 8 to 10 days that reliable nitrification would require at Canadian winter temperatures — consistent with a plant designed for BOD removal only.

5. Check the hydraulic retention time for consistency. On the influent flow, not on $Q + Q_R$,

   $$\tau = \frac{V}{Q} = \frac{1000}{4000} = 0.25\ \text{d} = 6.0\ \text{hours}$$

   which is the normal range for a conventional plug-flow or completely-mixed activated-sludge basin and confirms the data are internally sensible.

6. Derive the return sludge flow, which the question does not state, from the clarifier solids balance. At steady state the solids entering the clarifier equal those leaving in the underflow, so $(Q + Q_R)X = Q_R X_R$ with $X_e \approx 0$, giving

   $$Q_R = \frac{Q X}{X_R - X} = \frac{4000 \times 3000}{8000 - 3000} = 2400\ \text{m}^3\text{/d}, \qquad R = \frac{Q_R}{Q} = 0.60$$

   a normal recycle ratio for conventional activated sludge. This value is needed for the solids-loading check in step 9.

7. Establish the peak design flow. Applying the assumed peaking factor,

   $$Q_{\text{peak}} = 2.5 \times 4000 = 10\,000\ \text{m}^3\text{/d}$$

8. Size the clarifier at the overflow ceiling. Surface overflow rate is the plant flow crossing the weirs divided by the plan area — the return sludge is not included, because it leaves through the underflow and never crosses the weirs. Setting $\mathrm{SOR} = \mathrm{SOR}_{\max}$ at peak flow,

   $$A_s = \frac{Q_{\text{peak}}}{\mathrm{SOR}_{\max}} = \frac{10\,000}{40}$$

   $$\boxed{A_s = 250\ \text{m}^2}$$

   For a single circular tank, $D = \sqrt{4A_s/\pi} = 17.8$ m. Two units of 125 m² each, $D = 12.6$ m, are preferable in practice so that one can be taken out of service for maintenance without losing secondary treatment.

9. Check the design against the other clarifier criteria. At average flow the overflow rate is $\mathrm{SOR}_{\text{avg}} = 4000/250 = 16.0$ m³/m²·d, within the usual 16 to 32 band. The solids loading rate does include the return flow, since every kilogram of solids must be thickened however it arrived:

   $$\mathrm{SLR}_{\text{peak}} = \frac{(Q_{\text{peak}} + Q_R)X}{1000\,A_s} = \frac{(10\,000 + 2400) \times 3000}{1000 \times 250} = 148.8\ \frac{\text{kg}}{\text{m}^2\cdot\text{d}} = 6.2\ \frac{\text{kg}}{\text{m}^2\cdot\text{h}}$$

   which is below the 8 kg/m²·h peak ceiling; at average flow it is 76.8 kg/m²·d, or 3.2 kg/m²·h, comfortably inside the 4 to 6 design band. Omitting the return flow would have given 120 kg/m²·d at peak and understated the thickening duty by a factor of 1.24. Weir loading for a single peripheral-weir tank is $10\,000/(\pi \times 17.84) = 179$ m³/m·d, below the 250 m³/m·d peak guide. The design is therefore governed by the stated overflow rate, as the question intends, and is not violated by any secondary criterion.

Final Results