Question 3 of 5: Solids retention time and solids loading rate; choice of sludge digestion process

Question 3: Solids retention time and solids loading rate; choice of sludge digestion process (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.

Note on the labelling: the printed paper labels the two sub-parts of Question 3 "c." and "d." rather than "a." and "b.". The labels are reproduced above exactly as printed; the answers below follow the same lettering.

(c) Solids retention time and solids loading rate (15 marks)

Solids retention time (SRT, sludge age, mean cell residence time) is the average time a particle of biological solids spends in the system, defined as the mass of solids held in the system divided by the mass leaving it per day:

where $V$ is the aeration-tank volume, $X$ the mixed-liquor suspended solids, $Q_w$ and $X_w$ the flow and concentration of the waste sludge, and $X_e$ the effluent suspended solids. Its significance is that SRT, not hydraulic retention time, is the master control variable of the activated-sludge process. An organism can only persist in a continuous culture if it can double faster than it is removed, so the reciprocal of the SRT is the imposed growth rate: any species whose maximum specific growth rate falls below $1/\theta_c$ is washed out. Selecting an SRT therefore selects the microbial community. Short SRTs of 2 to 4 days give carbon removal only; 8 to 15 days at Canadian wastewater temperatures are needed to retain the slow-growing autotrophic nitrifiers; 20 days and beyond gives extended aeration, in which the sludge substantially oxidises itself. SRT also fixes the observed yield through $Y_{\text{obs}} = Y/(1 + k_d \theta_c)$, so a long SRT means less sludge to dispose of but more oxygen to supply, and it fixes the F/M ratio and the settleability of the floc.

Solids loading rate (SLR) is the mass flux of suspended solids applied to unit plan area of the secondary clarifier per day:

Its significance is that a secondary clarifier has two independent duties and each has its own governing rate. It must clarify — produce a low-solids overflow — which is controlled by the surface overflow rate, and it must thicken — concentrate the returning sludge enough to sustain the mixed-liquor inventory — which is controlled by the solids loading rate. The tank must be sized for whichever duty governs, and on a conventional plant the thickening duty usually does. The single most important distinction to state explicitly is that SOR excludes the return activated sludge while SLR includes it: only the plant flow crosses the effluent weirs, so only $Q$ enters the overflow rate, but every kilogram of solids arriving at the tank must be thickened and removed however it got there, so the recycle flow belongs in the loading rate.

Given. A conventional plant treats $Q = 20\,000\ \text{m}^3/\text{d}$ with $V = 5\,000\ \text{m}^3$ of aeration volume at $X = 3\,500\ \text{mg/L}$ MLSS, wastes $Q_w = 150\ \text{m}^3/\text{d}$ from the return line at $X_w = 9\,000\ \text{mg/L}$, carries $X_e = 12\ \text{mg/L}$ of effluent solids, and has $A = 700\ \text{m}^2$ of clarifier surface. Find. The SRT, the SLR and the overflow rate, and the size of the two errors this calculation most often attracts.

1. Solids retention time, counting both exit routes. $$\theta_c = \frac{V X}{Q_w X_w + (Q-Q_w)X_e} = \frac{5\,000 \times 3\,500}{150 \times 9\,000 + 19\,850 \times 12} = \frac{17.5 \times 10^{6}}{1.588 \times 10^{6}} = \boxed{11.0\ \text{d}}$$ Ignoring the effluent solids gives 13.0 d, an 18 per cent overstatement: a clarifier that is passing solids is quietly wasting sludge, and the operator who trims $Q_w$ to hold "13 days" is actually running at 11.
2. Hydraulic retention time, for contrast. $\tau = V/Q = 5\,000/20\,000 = 0.25\ \text{d} = 6\ \text{h}$, some forty-four times shorter than the SRT. That separation of the two residence times — achieved entirely by recycling the settled solids — is what makes activated sludge work in a tank of buildable size.
3. Return sludge flow from the clarifier solids balance. The paper does not state it, so recover it from a balance on the clarifier with $X_R = X_w$: $$Q_R = \frac{Q X}{X_R - X} = \frac{20\,000 \times 3\,500}{9\,000 - 3\,500} = 12\,727\ \text{m}^3\!/\text{d}, \qquad R = \frac{Q_R}{Q} = 0.64$$
4. Solids loading rate and overflow rate. With $X = 3\,500\ \text{mg/L} = 3.5\ \text{kg/m}^3$, $$\text{SLR} = \frac{(20\,000 + 12\,727)(3.5)}{700} = \boxed{163.6\ \text{kg/m}^2\!\cdot\!\text{d}} = 6.8\ \text{kg/m}^2\!\cdot\!\text{h}, \qquad \text{SOR} = \frac{20\,000}{700} = 28.6\ \text{m}^3\!/\text{m}^2\!\cdot\!\text{d}$$ Both sit inside the usual design windows (SLR 4 to 6 $\text{kg/m}^2\!\cdot\!\text{h}$ average with 8 permitted at peak; SOR 16 to 32 $\text{m}^3\!/\text{m}^2\!\cdot\!\text{d}$), so this clarifier is adequate for both duties.
5. Quantify the classic error. Omitting the recycle from the loading rate gives $20\,000 \times 3.5/700 = 100\ \text{kg/m}^2\!\cdot\!\text{d}$, understating the thickening duty by a factor of 1.64. A clarifier sized on that number will build a deep sludge blanket, lose solids over the weirs at peak flow, and return a dilute sludge that cannot hold the design MLSS — a failure that looks like bulking but is purely a loading error.

(d) Aerobic or anaerobic digestion at a 30-day SRT (10 marks)

Recommendation: aerobic digestion. The decisive point is that an extended-aeration plant operating at a 30-day SRT has already digested its sludge in the aeration basin. Over thirty days the readily biodegradable volatile fraction has largely been oxidised by endogenous respiration, so the waste solids arrive at the digester with a low volatile fraction and very little remaining substrate. There is consequently almost no feedstock for a methanogenic population to work on, and the case for anaerobic digestion — which is fundamentally an energy-recovery case — collapses.

Given. An extended-aeration plant treating $4\,000\ \text{m}^3/\text{d}$ wastes 400 kg TSS/d at 70 per cent volatile, thickened to 1.5 per cent solids; digester feed at $12\,{}^{\circ}\text{C}$ must be raised to a mesophilic $35\,{}^{\circ}\text{C}$. Find. Whether the biogas from anaerobic digestion can even heat its own digester.

1. Volatile solids available and destroyed. The feed carries $400 \times 0.70 = 280\ \text{kg VS/d}$. A 30-day-SRT waste activated sludge achieves only about 25 per cent volatile-solids destruction in a mesophilic digester, against 50 to 60 per cent for a primary sludge, so $280 \times 0.25 = 70\ \text{kg VS/d}$ is destroyed.
2. Biogas and its energy. At a typical $0.90\ \text{m}^3$ of biogas per kg of VS destroyed and a lower heating value of $22\ \text{MJ/m}^3$ at 65 per cent methane, $$E_{\text{gas}} = 70 \times 0.90 \times 22 = 1\,386\ \text{MJ/d}$$
3. Heating duty of the same digester. Thickened to 1.5 per cent, the feed is $400/(0.015 \times 1\,000) = 26.7\ \text{m}^3/\text{d}$, and raising it through $\Delta T = 23\ \text{K}$ requires $$E_{\text{heat}} = 26.7 \times 1\,000 \times 4.186 \times 23 \times 10^{-3} = 2\,568\ \text{MJ/d}$$
4. Compare. $$\boxed{E_{\text{heat}}/E_{\text{gas}} = 2\,568/1\,386 = 1.85}$$ The digester needs 85 per cent more heat than its own gas can supply, before any allowance for boiler efficiency or tank heat loss. Anaerobic digestion here is a net energy consumer, which removes its only real advantage.
5. Size the aerobic alternative. Destroying the same 70 kg VS/d at $2.3\ \text{kg O}_2$ per kg VS destroyed needs $161\ \text{kg O}_2/\text{d}$; at a field transfer of $1.2\ \text{kg O}_2$ per kWh that is $\boxed{134\ \text{kWh/d}}$ of blower power. The digester itself, at a 40-day aerobic SRT, is $26.7 \times 40 = 1\,067\ \text{m}^3$ — an open concrete tank with a blower, and nothing else.

Factors considered in the evaluation. The comparison was made on seven criteria. Feed characteristics: a 30-day-SRT waste sludge is already largely stabilised, with low biodegradable VS, which is what caps the anaerobic gas yield at an uneconomic level. Energy balance: shown above to be negative for the anaerobic option at this scale and feed quality. Plant scale: at $4\,000\ \text{m}^3/\text{d}$ the plant is far below the roughly $20\,000\ \text{m}^3/\text{d}$ threshold at which anaerobic digestion with combined heat and power is normally justified. Capital cost and complexity: anaerobic digestion needs heated, sealed, mixed tanks, gas holders, flares, hydrogen-sulphide scrubbing and Class 1 hazardous-area electrical work; aerobic digestion needs a tank and a blower that the plant's existing operators already understand. Operability and process risk: methanogens are sensitive to temperature, pH and ammonia and a soured digester takes weeks to recover, whereas an aerobic digester is essentially unfailable and tolerates intermittent feeding — a real consideration at a small plant with part-time staff. Product quality: aerobic digestion gives an odourless, well-nitrified, easily handled biosolid, though with poorer dewaterability and a higher polymer demand; anaerobic digestion gives better dewatering and higher solids capture. Sidestream and regulatory effects: anaerobic digestion returns a strong ammonia-rich centrate that can add 15 to 25 per cent to the plant's nitrogen load, while aerobic digestion nitrifies in the digester and returns a much weaker liquor; both routes can meet the vector-attraction and pathogen requirements applied to biosolids in Canadian provincial regulations, aerobic digestion typically by the 38 per cent volatile-solids-reduction or the specific-oxygen-uptake-rate route.

Reasons for the choice. Aerobic digestion wins because at this feed quality the anaerobic route buys nothing: it cannot pay for its own heat, its capital and operating complexity are disproportionate to a small plant, and the sludge it would receive has already had its energy extracted in the aeration basin. The cost of the aerobic route is a continuous 134 kWh/d of aeration power and a poorer dewatering characteristic, which at this scale is a smaller penalty than the gas system it avoids. It is worth adding for the client that a further legitimate option is no further digestion at all: at a 30-day SRT the volatile fraction is often low enough that the sludge already satisfies the stabilisation criteria, in which case thickening and direct dewatering is the lowest-cost route and the aerobic digester serves only as storage and as a compliance buffer.