16-Civ-B5 Water Supply and Wastewater Treatment · May 2016

Question 5 of 5: The Activated Sludge Process, and the SRT, SOR and SLR of a Given Plant

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 5: The Activated Sludge Process, and the SRT, SOR and SLR of a Given Plant (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.

(a) Working principle and operation of the activated sludge process (10 marks)

The activated sludge process is a suspended-growth aerobic biological treatment in which a mixed culture of heterotrophic bacteria, protozoa and rotifers is kept in suspension in an aerated tank and fed the settled sewage. The organisms oxidise the soluble and colloidal organic matter partly to carbon dioxide and water for energy, and partly into new cell material:

The essential insight of the process, and the reason it displaced every earlier method, is that the pollutant is not destroyed in the water — it is converted into settleable biomass, which is then removed from the water by gravity. That is why the secondary clarifier in Figure 5.1 is an integral part of the process and not an appendage: without it the biomass would leave with the effluent and the treatment would be undone.

The return activated sludge line is the second essential feature. By settling the biomass and pumping it back to the head of the aeration tank, the plant decouples the residence time of the solids from the residence time of the water. The water passes through in four to eight hours, far too short for slow-growing organisms to establish; the solids stay for five to fifteen days. That decoupling is what makes the process controllable, because the sludge age — the solids retention time — selects which organisms can survive. Only the waste activated sludge stream leaves the system permanently, and the rate at which it is drawn is therefore the master control variable.

In operation the plant is managed through four handles. The wasting rate sets the SRT and therefore the microbial population: a short SRT selects fast-growing heterotrophs and gives carbon removal only, while an SRT above the nitrifier washout value — typically 8–15 days at 10 °C with a safety factor — is required before the slow-growing autotrophic nitrifiers can be retained and ammonia removed. The air supply, from fine-bubble diffusers or surface aerators, must both satisfy the oxygen demand and keep the mixed liquor in suspension; dissolved oxygen is normally controlled to 1.5–2.5 mg/L, since less starves the floc and encourages filaments while more wastes a great deal of blower power. The return sludge ratio, usually 0.25–1.0 of the influent flow, maintains the mixed liquor suspended solids concentration in the reactor, typically 1500–3000 mg/L for a conventional plant. And the food-to-microorganism ratio, $F/M = QS_0/(VX)$, is the loading measure that operators track day to day, with 0.2–0.4 kg BOD per kg MLSS per day being conventional practice.

The characteristic operating problems are all failures of the solids separation rather than of the biology. Filamentous bulking — caused by low food-to-microorganism ratios, low dissolved oxygen, nutrient deficiency or a septic feed — produces a sludge that will not compact and raises the sludge volume index above 150 mL/g. Rising sludge is denitrification in the clarifier, lifting settled solids on nitrogen bubbles. Pin floc results from an excessive sludge age, and foaming from Nocardia or Microthrix. Each is diagnosed from the settling behaviour and corrected through the same four handles. Many variants exist for particular duties — extended aeration and oxidation ditches at long SRT for small communities, step feed and contact stabilisation to redistribute the load, sequencing batch reactors to combine reaction and clarification in one basin, and membrane bioreactors to eliminate the clarifier entirely — but every one of them is the same reaction followed by the same solids separation.

(b) SRT, surface overflow rate and solids loading rate (15 marks)

Given. The examiner supplies a complete conventional activated sludge plant with its secondary clarifier.

Find. The solids retention time of the system, and the surface overflow rate and solids loading rate of the secondary clarifier, with a comment on whether the clarifier is adequately sized.

Approach. Solids retention time is the inventory of solids in the system divided by the rate at which solids leave it, so the calculation is a mass balance and nothing more. The surface overflow rate follows directly from the plant flow and the clarifier area. The solids loading rate needs the return flow, which the question withholds; it is recovered from a solids balance around the clarifier.

1. Convert the concentrations to a mass basis. Since 1 mg/L is 1 g/m³, the solids held in the reactor are $$M = \frac{V X}{1000} = \frac{1000\ \mathrm{m^3}\times 3000\ \mathrm{g/m^3}}{1000} = 3000\ \mathrm{kg}$$
2. Find the rate at which solids leave the system. Wasting is from the return line at $X_w = X_R = 8000$ mg/L, and effluent solids are neglected as the question gives none: $$M_{\text{out}} = \frac{Q_w X_w}{1000} = \frac{80\times 8000}{1000} = 640\ \mathrm{kg/d}$$
3. Divide to obtain the solids retention time. By definition the SRT is the inventory divided by the removal rate, $$\theta_c = \frac{V X}{Q_w X_w} = \frac{1000\times 3000}{80\times 8000} = \frac{3000\ \mathrm{kg}}{640\ \mathrm{kg/d}} = \boxed{4.69\ \text{days}}$$ Note that the concentrations enter only as a ratio, so working in mg/L throughout would have given the same answer without any conversion.
4. Compare the sludge age with the hydraulic detention time to see the decoupling. The nominal hydraulic retention time on the influent flow is $$\tau = \frac{V}{Q} = \frac{1000}{4000} = 0.25\ \mathrm{d} = 6.0\ \mathrm{h}$$ so the solids stay about nineteen times longer than the water. Six hours is a conventional detention time, but an SRT of 4.7 days is short: it is sufficient for carbonaceous removal at summer temperatures, and it is comfortably below the sludge age needed to retain nitrifiers in a cold Canadian winter. This plant is designed for BOD removal, not for nitrification, and if an ammonia limit were imposed the wasting rate would have to be cut — which would raise the mixed liquor concentration and, with it, the loading on the clarifier calculated below.
5. Surface overflow rate from the plant flow alone. Only the influent flow crosses the clarifier weirs; the return sludge is withdrawn from the floor and never rises past the launders, so it is excluded: $$\mathrm{SOR} = \frac{Q}{A} = \frac{4000}{250} = \boxed{16.0\ \mathrm{m^3/m^2\cdot d}}$$ This sits at the bottom of the 16–28 m³/m²·d band normally used for average flow at a conventional activated sludge plant, so the clarifier is conservatively sized for clarification.
6. Recover the return sludge flow from a solids balance. Every kilogram of solids that enters the clarifier must leave in the underflow if the effluent is essentially solids-free, so $$(Q + Q_R)X = Q_R X_R \;\Longrightarrow\; Q_R = \frac{QX}{X_R - X} = \frac{4000\times 3000}{8000-3000} = \boxed{2400\ \mathrm{m^3/d}}$$ a recycle ratio of $R = Q_R/Q = 0.60$, squarely in the usual 0.25–1.0 range and therefore a credible result rather than an artefact of the assumption.
7. Solids loading rate on the total flow entering the clarifier. Unlike the overflow rate, the solids loading counts every kilogram of solids that arrives, however it arrived, so the return flow must be included: $$\mathrm{SLR} = \frac{(Q+Q_R)\,X}{1000\,A} = \frac{(4000+2400)\times 3000}{1000\times 250} = \boxed{76.8\ \mathrm{kg/m^2\cdot d}} \;=\; 3.20\ \mathrm{kg/m^2\cdot h}$$ Metcalf & Eddy give 4–6 kg/m²·h at average flow for conventional activated sludge, so at 3.20 the thickening duty is also conservative and the clarifier has genuine reserve for peak flow.
8. Check the trap. Omitting the return flow would give $\mathrm{SLR} = QX/(1000A) = 48.0\ \mathrm{kg/m^2\cdot d}$, understating the true loading by a factor of 1.60. The distinction is the entire content of this part of the question: the overflow rate is a hydraulic criterion and excludes the return; the solids loading rate is a thickening criterion and includes it. Sizing a secondary clarifier on the overflow rate alone is the classic error, because at peak flow it is almost always the thickening criterion that fails first, and the failure mode is a rising sludge blanket that washes solids over the weirs.