Activated sludge
The activated sludge process is the most widely used biological method for treating municipal sewage. In an aeration tank, a dense population of bacteria and other microorganisms, the mixed liquor, is kept in suspension with air. The microbes consume dissolved and colloidal organic matter as food, converting it to carbon dioxide, water and new cells. The mixture then flows to a secondary clarifier, where the microbial flocs settle. Most of the settled sludge is returned to the aeration tank to keep the microbe population high; the excess is wasted to sludge treatment.
Control revolves around a few parameters. Mixed liquor suspended solids (MLSS), often 2 000 to 4 000 mg/L in conventional plants, measure the amount of biomass. The food-to-microorganism ratio (F/M) relates the daily BOD load to the biomass. The mean cell residence time, or sludge age, is the average time microbes stay in the system, controlled by how much sludge is wasted. Longer sludge ages favour nitrifying bacteria and give more stable sludge. Dissolved oxygen is kept around 1.5 to 2 mg/L. Sludge settleability is checked with the sludge volume index.
Variants include extended aeration with long sludge age and little excess sludge; sequencing batch reactors, which aerate, settle and decant in one tank in cycles; moving bed biofilm reactors with plastic carriers; and membrane bioreactors that replace the clarifier with membranes for very clear effluent. Aeration energy is the largest running cost, so fine-bubble diffusers and dissolved oxygen control pay back quickly. Bulking sludge, caused by filamentous organisms, is a common operational problem.
Watch the 30-minute settling test daily; it gives early warning of bulking before effluent quality drops.
F/M = Q S₀ / (V X)Sludge age: θc = V X / (Qw Xr + Qe Xe)Hydraulic retention time: HRT = V / QSVI = settled volume in 30 min (mL/L) × 1 000 / MLSS (mg/L)
Problem. An aeration tank of 4 000 m³ treats 10 MLD with settled BOD 170 mg/L at MLSS 3 000 mg/L. Find HRT and F/M.
- HRT = 4 000 / 10 000 × 24 = 9.6 h.
- BOD load = 10 000 × 0.170 = 1 700 kg/day.
- Biomass = 4 000 × 3.0 = 12 000 kg.
- F/M = 1 700 / 12 000 = 0.14 per day, on the low side, typical of extended-type operation.
Answer. HRT ≈ 9.6 h; F/M ≈ 0.14 per day
Quick quiz
1Why is sludge returned to the aeration tank?
To keep a high population of active microorganisms.
2What does sludge age control?
How long microbes stay in the system, affecting nitrification and sludge stability.
3What is sludge bulking?
Poor settling caused by filamentous organisms.
- CPHEEO Manual on Sewerage and Sewage Treatment Systems (2013) (India): Central government manual for sewer networks and sewage treatment plants.
Clauses and documents to read
- CPHEEO Manual on Sewerage and Sewage Treatment Systems, Part A (secondary treatment, cited by chapter)
- Environment (Protection) Rules, 1986 (discharge standards, cited by name)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- Metcalf and Eddy, Wastewater Engineering: Treatment and Resource Recovery, McGraw Hill, 5th edition.
- S. K. Garg, Sewage Disposal and Air Pollution Engineering (Environmental Engineering Vol. II), Khanna Publishers, latest edition.
Free and official sources
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