Activated-Sludge Aeration Tank, Secondary Clarifier Loadings and Oxygen Demand: 16-Civ-B5 Water Supply and Wastewater Treatment, December 2014, Q4

Question 4: Activated-Sludge Aeration Tank, Secondary Clarifier Loadings 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.

Quantity	Symbol	Value
Average plant flow	$Q$	$50\,000\ \text{m}^3/\text{d}$
Solids retention time	$\theta_c$	$10\ \text{d}$
Aeration-tank MLSS	$X$	$3\,000\ \text{mg/L}$
Waste sludge flow	$Q_w$	$500\ \text{m}^3/\text{d}$
Waste sludge solids	$X_w$	$8\,000\ \text{mg/L}$
Total secondary clarifier volume	$V_{sc}$	$12\,000\ \text{m}^3$
Clarifier side water depth	$H$	$3.5\ \text{m}$
Primary-clarifier effluent BOD₅	$S_0$	$150\ \text{mg/L}$

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

Figure 4.1 — Conventional activated-sludge flowsheet with the data of Question 4. The RAS stream returns solids to the aeration tank and is counted in the clarifier solids loading rate but not in its surface overflow rate.

Approach. Part (a) follows from the definition of solids retention time as the solids inventory divided by the solids wastage rate, which gives the tank volume directly and hence the hydraulic retention time. Part (b) converts the clarifier volume and depth to a plan area, then applies the two loading definitions — the surface overflow rate on the plant flow alone, the solids loading rate on the plant flow plus the return sludge, whose magnitude comes from a solids balance around the clarifier. Part (c) uses the oxygen mass balance around the aeration tank: the ultimate oxygen demand of the BOD removed less the oxygen equivalent of the biomass wasted rather than oxidised.

Part (a) — write the definition of solids retention time. The SRT is the mass of solids held in the system divided by the mass leaving it per day. Neglecting the solids lost over the effluent weir, which is standard for a well-operated plant, all wastage occurs through the sludge draw-off:
$$\theta_c = \frac{V X}{Q_w X_w}$$
where $V$ is the aeration-tank volume. Every concentration appears as a ratio, so mg/L may be used throughout without conversion.
Solve for the aeration-tank volume. Rearranging,
$$V = \frac{\theta_c\,Q_w X_w}{X} = \frac{(10\ \text{d})(500\ \text{m}^3/\text{d})(8\,000\ \text{mg/L})} {3\,000\ \text{mg/L}}$$ $$\boxed{V = 13\,333\ \text{m}^3 \ (\approx 13.3 \times 10^3\ \text{m}^3)}$$
Checking backwards: $VX = 13\,333 \times 3\,000 = 40.0 \times 10^6\ \text{g}$, that is $40\,000\ \text{kg}$ of MLSS in the tank, against a wastage of $Q_w X_w = 500 \times 8\,000 = 4\,000\ \text{kg/d}$; the quotient is $10\ \text{d}$ as required.
Convert the volume to a hydraulic retention time. The HRT is the tank volume divided by the influent flow — the recycle is deliberately excluded, because HRT is defined on the flow through the process, not on the flow through the tank:
$$\tau = \frac{V}{Q} = \frac{13\,333\ \text{m}^3}{50\,000\ \text{m}^3/\text{d}} = 0.267\ \text{d}$$ $$\boxed{\tau = 6.4\ \text{hours}}$$
Both results sit squarely in the conventional plug-flow activated-sludge range ($\theta_c = 5$ to $15\ \text{d}$, $\tau = 4$ to $8\ \text{h}$, $X = 1\,500$ to $4\,000\ \text{mg/L}$), which is the first sanity check to make. The corresponding food-to-microorganism ratio is $F/M = QS_0/(VX) = 7\,500/40\,000 = 0.19\ \text{kg BOD}_5$ per kg MLSS per day, and the volumetric organic loading is $0.56\ \text{kg BOD}_5\,\text{m}^{-3}\text{d}^{-1}$ — again typical of a conventional plant, and both consistent with an SRT that is long enough to nitrify in warm weather.
Part (b) — convert the clarifier volume to a plan area. The loading rates are defined per unit of surface area, so the side water depth is used only to extract that area:
$$A = \frac{V_{sc}}{H} = \frac{12\,000\ \text{m}^3}{3.5\ \text{m}} = 3\,428.6\ \text{m}^2$$
This is the combined area of all secondary clarifiers in service.
Compute the surface overflow rate. Only the flow that leaves over the weirs is counted, so the return activated sludge is excluded:
$$\mathrm{SOR} = \frac{Q}{A} = \frac{50\,000\ \text{m}^3/\text{d}}{3\,428.6\ \text{m}^2}$$ $$\boxed{\mathrm{SOR} = 14.6\ \text{m}^3\,\text{m}^{-2}\,\text{d}^{-1} \ (= 0.61\ \text{m/h})}$$
Design guidance for activated-sludge secondary clarifiers is roughly $16$ to $28\ \text{m}^3\,\text{m}^{-2}\,\text{d}^{-1}$ at average flow, so at $14.6$ the clarifiers are conservatively loaded hydraulically and have useful headroom for peak wet-weather flow.
Find the return sludge flow from a solids balance. The solids loading rate needs the total flow entering the clarifier, so the recycle must be quantified. Taking the underflow concentration as the waste sludge concentration $X_R = X_w = 8\,000\ \text{mg/L}$ and neglecting effluent solids, a solids balance about the clarifier gives $Q_R X_R = (Q + Q_R) X$, hence
$$Q_R = \frac{Q X}{X_R - X} = \frac{(50\,000)(3\,000)}{8\,000 - 3\,000} = 30\,000\ \text{m}^3/\text{d}$$
a recycle ratio of $R = Q_R/Q = 0.60$, comfortably within the usual 0.25 to 1.0 range for a conventional plant.
Compute the solids loading rate. Every kilogram of solids that reaches the clarifier floor counts, however it arrived, so here the recycle is included:
$$\mathrm{SLR} = \frac{(Q + Q_R)\,X}{A} = \frac{(50\,000 + 30\,000)(3\,000)\ \text{g/d}}{3\,428.6\ \text{m}^2} = \frac{240\,000\ \text{kg/d}}{3\,428.6\ \text{m}^2}$$ $$\boxed{\mathrm{SLR} = 70.0\ \text{kg}\,\text{m}^{-2}\,\text{d}^{-1} \ (= 2.9\ \text{kg}\,\text{m}^{-2}\,\text{h}^{-1})}$$
Against a typical average-flow design value of $70$ to $120\ \text{kg}\,\text{m}^{-2}\,\text{d}^{-1}$ and a peak limit near $200$, the clarifiers are adequately but not generously sized for thickening. Had the recycle been omitted, the answer would have been $43.8\ \text{kg}\,\text{m}^{-2}\,\text{d}^{-1}$, understating the thickening duty by a factor of 1.6 and giving the false impression of substantial spare capacity — this is the single most common error in the calculation. The clarifier hydraulic retention time on the combined flow, $12\,000/80\,000 = 0.15\ \text{d}$ or $3.6\ \text{h}$, is also reasonable.
Part (c) — set up the oxygen balance. The oxygen actually required is the ultimate oxygen demand of the organic matter removed, less the oxygen equivalent of the cell tissue that is wasted as sludge instead of being oxidised. In the standard form,
$$R_O = \frac{Q\,(S_0 - S)}{f} - 1.42\,P_{X,\text{bio}}$$
where $f = \mathrm{BOD_5}/\mathrm{BOD_u} = 0.68$ for municipal wastewater, and $1.42\ \text{kg O}_2$ per kg of cell tissue is the oxygen equivalent of $\mathrm{C_5H_7NO_2}$. Nitrogenous demand is excluded, as the question asks only for BOD removal.
Evaluate the BOD removed and its ultimate demand. Taking a well-treated soluble effluent BOD₅ of $S = 10\ \text{mg/L}$,
$$Q(S_0 - S) = (50\,000\ \text{m}^3/\text{d})(150 - 10)\ \text{g/m}^3 = 7\,000\ \text{kg BOD}_5/\text{d}$$
and converting the five-day value to the ultimate demand,
$$\frac{Q(S_0 - S)}{f} = \frac{7\,000}{0.68} = 10\,294\ \text{kg/d as O}_2$$
Deduct the oxygen equivalent of the wasted biomass. The plant wastes $P_X = Q_w X_w = (500)(8\,000) = 4\,000\ \text{kg TSS/d}$; at a volatile fraction of 0.80 the biological solids are $P_{X,\text{bio}} = 3\,200\ \text{kg VSS/d}$. Substituting,
$$R_O = 10\,294 - (1.42)(3\,200) = 10\,294 - 4\,544\ \text{kg/d}$$ $$\boxed{R_O = 5\,750\ \text{kg O}_2/\text{d}}$$
equivalent to $0.82\ \text{kg O}_2$ per kg of BOD₅ removed, or an average uptake rate of about $240\ \text{kg/h}$. Because oxygen transfer efficiency in a diffused system under process conditions is typically only 8 to 12 percent, this actual oxygen requirement corresponds to a standard oxygen requirement roughly 1.7 to 2.0 times larger — the number that would actually size the blowers.

Check: the assumptions behind part (c), and their weight. The question supplies neither the effluent BOD nor the volatile fraction of the waste sludge, so three values are assumed and each is declared here. (1) Effluent soluble BOD₅ $S = 10\ \text{mg/L}$; taking $S = 0$ instead raises the answer to $6\,485\ \text{kg/d}$, about 13 percent higher. (2) Volatile fraction $\mathrm{MLVSS/MLSS} = 0.80$; applying the $1.42$ factor to the whole $4\,000\ \text{kg TSS/d}$ would give $4\,614\ \text{kg/d}$, about 20 percent lower, but that is not defensible because the $1.42$ oxygen equivalent belongs to cell tissue, which is volatile matter. (3) $f = 0.68$, the standard municipal value.

A second cross-check is worth recording. Estimating the biomass from synthesis theory rather than from the reported wastage, with $Y = 0.40\ \text{kg VSS/kg BOD}$ and $k_d = 0.06\ \text{d}^{-1}$, gives $P_{X} = Y Q (S_0 - S) / (1 + k_d \theta_c) = 1\,750\ \text{kg VSS/d}$ and hence $R_O = 7\,809\ \text{kg/d}$, or $1.12\ \text{kg O}_2$ per kg BOD removed — the textbook range. The discrepancy is real and traceable: the data as given imply an observed yield of $4\,000/7\,000 = 0.57\ \text{kg TSS}$ per kg BOD, which is high for a 10-day SRT and suggests either substantial inert solids carried over from the primary clarifier or an over-stated wastage rate. The boxed answer uses the plant’s own reported wastage, as the question intends; a designer sizing blowers for this plant should carry the higher figure as the aeration design case.

Part	Quantity	Value
(a)	Aeration-tank volume, $V$	$13\,333\ \text{m}^3$
(a)	Hydraulic retention time, $\tau$	$0.267\ \text{d} = 6.4\ \text{h}$
(a)	MLSS inventory / $F\!/\!M$	$40\,000\ \text{kg}$ / $0.19\ \text{d}^{-1}$
(b)	Clarifier plan area, $A$	$3\,428.6\ \text{m}^2$
(b)	Surface overflow rate	$14.6\ \text{m}^3\,\text{m}^{-2}\,\text{d}^{-1}$
(b)	Return sludge flow / ratio	$30\,000\ \text{m}^3/\text{d}$ / $R = 0.60$
(b)	Solids loading rate	$70.0\ \text{kg}\,\text{m}^{-2}\,\text{d}^{-1}$
(c)	BOD₅ removed	$7\,000\ \text{kg/d}$
(c)	Oxygen demand, $R_O$	$5\,750\ \text{kg O}_2/\text{d}$

Topic. Suspended-growth process design: solids retention time as the master variable, secondary-clarifier hydraulic and solids loading, and the oxygen balance about an aeration tank.

Key relations. $\theta_c = VX/(Q_w X_w)$; $\tau = V/Q$; $F/M = QS_0/(VX)$; $A = V_{sc}/H$; $\mathrm{SOR} = Q/A$; $Q_R = QX/(X_R - X)$; $\mathrm{SLR} = (Q + Q_R)X/A$; $R_O = Q(S_0 - S)/f - 1.42 P_{X,\text{bio}}$ with $f = 0.68$.

Why this works. Solids retention time is the master variable of activated sludge because it sets the net specific growth rate of the biomass: only organisms that can double within $\theta_c$ are retained, which is why nitrifiers require a long SRT and why $\theta_c$, not HRT, determines effluent quality. Once $\theta_c$ and the wastage are fixed, the required solids inventory follows, and dividing by the operating MLSS gives the tank volume — so the aeration tank is really sized by the biology, with HRT as a derived quantity rather than a design input. The clarifier obeys a different logic: it must simultaneously clarify, which is a velocity comparison and therefore depends on flow per unit area, and thicken, which is a mass-flux comparison and therefore depends on solids per unit area. The two duties fail in different ways — clarification failure shows as turbid effluent, thickening failure as a rising blanket — which is why both loadings must be checked and why the recycle enters one and not the other. The oxygen balance closes the same accounting on electrons: carbon removed from solution is either respired, consuming oxygen, or built into cells, which are wasted; the $1.42$ factor is simply the stoichiometric oxygen demand of $\mathrm{C_5H_7NO_2}$.

Common pitfall. Including the return activated sludge in the surface overflow rate, or omitting it from the solids loading rate — the first oversizes the tank by 60 percent for a typical recycle ratio, the second understates the thickening duty by the same factor. A second frequent error in part (c) is forgetting to divide by $f = 0.68$: oxygen must be supplied for the ultimate demand, not the five-day value, and omitting the conversion understates the requirement by roughly a third. A third is applying the $1.42$ factor to total suspended solids rather than to volatile suspended solids.

Source. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery, 5th ed., §7-6 (SRT and biomass production), §8-2 to §8-4 (activated-sludge process design and oxygen requirements), §8-10 (secondary clarifier design, SOR and SLR); Davis & Cornwell, Introduction to Environmental Engineering, 5th ed., §9-6.

Question 4 of 5: Activated-Sludge Aeration Tank, Secondary Clarifier Loadings and Oxygen Demand

Question 4: Activated-Sludge Aeration Tank, Secondary Clarifier Loadings and Oxygen Demand (25 marks)