Activated Sludge Process & Trickling Filters

1. Activated Sludge Process — Overview

The Activated Sludge Process (ASP) is the most widely used aerobic biological secondary wastewater treatment method for medium and large STPs. It is a suspended growth system where a highly active microbial population (the “activated sludge”) is maintained in an aeration tank and brought into intimate contact with wastewater to decompose organic matter.

1.1 Basic Process Flow

Primary effluent → Aeration tank (aerobic biological treatment) → Secondary clarifier (biomass separation) → Treated effluent (discharge)

Secondary clarifier → Return Activated Sludge (RAS) → back to aeration tank inlet (maintains MLSS)

Secondary clarifier → Waste Activated Sludge (WAS) → sludge treatment (controls sludge age)

1.2 The Activated Sludge — What It Is

The activated sludge is a brown, flocculent mass of microorganisms — mainly bacteria (Zoogloea, Pseudomonas, Bacillus, Nitrosomonas, Nitrobacter), protozoa, fungi, and rotifers — that form settleable aggregates (flocs). These organisms are “activated” because they are in a metabolically active state, continuously feeding on organic matter. The sludge is “recycled” (returned) to maintain a high concentration of active biomass in the aeration tank — this is what distinguishes ASP from simple aeration of wastewater.

2. ASP Design Parameters

3. F/M Ratio

3.1 Relationship Between F/M and Treatment Efficiency

Low F/M (extended aeration) → longer sludge age → more complete oxidation → higher BOD removal, nitrification → less sludge. High F/M (high-loaded) → rapid growth → less complete oxidation → lower BOD removal, bulking risk → more sludge. Conventional ASP balances treatment efficiency with sludge management.

4. Sludge Volume Index (SVI)

5. Sludge Age (SRT / θ_c)

6. Return Sludge Ratio

7. Oxygen Requirements

8. ASP Variants

9. Trickling Filters

A trickling filter (biofilter) is an attached growth aerobic treatment system where wastewater is distributed over a packing media (rock, plastic) on which a biological film grows. As the wastewater trickles down through the media, aerobic microorganisms in the biofilm consume the organic matter. Air moves upward through the media by natural convection (or forced draft in modern plastic media filters) providing oxygen.

9.1 Components

• Rotary distributor: Distributes sewage uniformly over the media surface; rotates by reaction of water jets; 1–3 rpm
• Filter media: Broken stone/gravel (40–75 mm, depth 1.5–3 m) or plastic packing (corrugated sheets, rings, saddles; depth 3–8 m)
• Underdrain system: Collects treated effluent and allows air circulation from below
• Recirculation pump: Returns treated effluent (or secondary effluent) back to the distributor to dilute influent and keep media moist

9.2 Trickling Filter Classification

10. NRC Formula for Trickling Filters

The National Research Council (NRC) formula (1946) is the standard empirical formula for estimating BOD removal efficiency in a trickling filter, widely used in India for design per CPHEEO guidelines.

10.1 NRC Formula for Single-Stage Filter

10.2 NRC Formula for Two-Stage Filters (in Series)

10.3 Recirculation in Trickling Filters

Recirculation returns a portion of the clarifier effluent (or filter effluent) back to the filter inlet to:

• Dilute strong influent (preventing overloading of the biofilm)
• Keep the media continuously wetted (prevents biofilm drying in hot Indian summers)
• Increase hydraulic loading (helps slough off old biofilm and prevents ponding)
• Improve performance at low flows

11. Worked Examples (GATE CE Level)

Example 1 — Aeration Tank Volume from F/M Ratio (GATE CE 2022 type)

Problem: An activated sludge plant treats 5 MLD of primary effluent with BOD₅ = 150 mg/L. Design F/M = 0.30 kg BOD/kg MLSS/day; MLSS = 2500 mg/L. Find (a) the aeration tank volume and (b) the HRT.

Answer: V = 1000 m³; HRT = 4.8 hours

Example 2 — Return Sludge Ratio (GATE CE 2021 type)

Problem: An activated sludge system has MLSS = 3000 mg/L in the aeration tank. The settled sludge volume after 30 minutes in an Imhoff cone test is 350 mL/L. Find (a) the SVI, (b) the RAS concentration X_r, and (c) the return sludge ratio R.

Answer: SVI = 117 mL/g (good settling); X_r = 8571 mg/L; R = 0.54

Example 3 — Sludge Age Calculation (GATE CE type)

Problem: An aeration tank has volume V = 1000 m³, MLSS = 2500 mg/L. Waste sludge (WAS) is withdrawn from the secondary clarifier at Q_w = 50 m³/day with X_w = 8000 mg/L. Secondary effluent SS X_e = 20 mg/L; Q = 5000 m³/day. Find the sludge retention time θ_c.

Answer: Sludge age θ_c = 5 days (conventional ASP)

Example 4 — Trickling Filter Efficiency (NRC Formula, GATE CE 2020 type)

Problem: A single-stage high-rate trickling filter receives primary effluent with BOD = 150 mg/L at flow Q = 2 MLD. Filter volume V = 500 m³. Recirculation ratio R = 1. Find the BOD removal efficiency and the effluent BOD.

Answer: E = 78.9%; Effluent BOD = 31.7 mg/L

Example 5 — Oxygen Requirement in ASP (GATE CE type)

Problem: An activated sludge system removes 200 kg BOD/day. MLVSS in aeration tank = 2000 mg/L; V = 1000 m³. Using a’ = 0.60 kg O₂/kg BOD removed and b’ = 0.08 kg O₂/kg MLVSS/day, find the total daily oxygen requirement.

Answer: O₂ required = 280 kg/day (120 kg synthesis + 160 kg endogenous)

Example 6 — Two-Stage Trickling Filter (GATE CE type)

Problem: A two-stage trickling filter system treats 3 MLD of primary effluent with BOD = 160 mg/L. Stage 1: V₁ = 600 m³, R₁ = 1; Stage 2: V₂ = 400 m³, R₂ = 0. Find the overall BOD removal efficiency.

Answer: E₁ = 76.4%; E₂ = 50%; E_total = 88.2%; Final BOD = 18.9 mg/L (meets IS 2490)

12. Common Mistakes

Mistake 1 — Using MLSS Instead of MLVSS in Oxygen Demand Calculation

Error: Computing endogenous O₂ demand using total MLSS instead of volatile MLVSS (active fraction).
Root Cause: Only the volatile (organic) fraction of MLSS respires and has endogenous oxygen demand. The inorganic (fixed) solids are inert. MLVSS ≈ 0.7–0.8 × MLSS for mixed liquor. Using total MLSS overestimates endogenous O₂ demand by 20–30%.
Fix: If MLVSS is not given, use MLVSS = 0.8 × MLSS (or the given volatile fraction). O₂_endogenous = b’ × MLVSS (kg) × volume (m³) = b’ × MLVSS concentration × V.

Mistake 2 — Substituting R as Percentage in NRC Recirculation Factor

Error: Using R = 100 (for 100% recirculation) instead of R = 1 in the formula F = (1 + R)/(1 + R/10)².
Root Cause: R is defined as a ratio (dimensionless fraction): R = Q_{recirculation}/Q_influent. R = 1 means return flow equals influent flow (100% recirculation ratio). Using R = 100 gives F ≈ 101/110² ≈ 0.0083 — clearly wrong (F should be ≥ 1).
Fix: Express R as a decimal ratio: 50% recirculation → R = 0.5; 100% → R = 1.0; 200% → R = 2.0. F = (1 + R)/(1 + R/10)². For R = 0.5: F = 1.5/1.05² = 1.5/1.1025 = 1.361.

Mistake 3 — Computing F/M with BOD in kg and MLSS in mg/L Without Unit Consistency

Error: Computing F/M = (Q in m³/day × S₀ in mg/L) / (V in m³ × X in mg/L) and getting m³/day instead of day⁻¹.
Root Cause: The F/M formula works if consistent units are used. Using mg/L for both S₀ and X, and m³ for both Q and V: F/M = [m³/day × mg/L] / [m³ × mg/L] = day⁻¹ ✓. The mg/L and m³ cancel correctly. No unit conversion is needed if this form is used throughout.
Fix: F/M = (Q × S₀)/(V × X). Use consistent units throughout: Q in m³/day, S₀ and X in mg/L, V in m³. The mg/L in numerator and denominator cancel; result is in day⁻¹. Verify: for Q = 5000 m³/day, S₀ = 150 mg/L, V = 1000 m³, X = 2500 mg/L: F/M = (5000 × 150)/(1000 × 2500) = 750,000/2,500,000 = 0.30 day⁻¹ ✓.

Mistake 4 — Forgetting the (1–E₁) Factor in the Second-Stage NRC Formula

Error: Using the standard single-stage NRC formula for the second stage without modifying the constant 0.4432 term to 0.4432/(1–E₁).
Root Cause: In a two-stage system, the second stage receives partially treated wastewater (lower BOD concentration). The NRC formula was empirically developed for first-stage filters treating raw/primary effluent. For second-stage filters, the formula is modified to account for the lower-biodegradability of the remaining BOD.
Fix: Second stage: E₂ = 1/[1 + {0.4432/(1–E₁)} × √(W₂/(V₂F₂))]. The term 0.4432/(1–E₁) increases the denominator (reducing E₂), reflecting the harder-to-treat nature of the remaining BOD after first-stage treatment.

Mistake 5 — Confusing SVI with Settleability When SVI is Very Low

Error: Concluding that SVI = 30 mL/g indicates “excellent sludge quality” without noting this may indicate a dense, pin-floc or non-biological sludge that settles very fast but may have poor bioactivity.
Root Cause: SVI < 50 mL/g is technically “excellent settling” but in activated sludge, extremely low SVI often indicates a non-biological or inorganic sludge (excessive inerts, lime sludge contamination) or over-thickened sludge with poor microbial activity. Good, active biological sludge typically has SVI = 80–150 mL/g.
Fix: Interpret SVI in context: 50–150 mL/g = normal healthy ASP sludge. < 50 mL/g may indicate inorganic precipitation or overconditioned sludge. > 150 mL/g indicates bulking. Report SVI with MLSS to give the complete picture.

13. Frequently Asked Questions

Q1. What causes sludge bulking in the activated sludge process, and how is it controlled?

Sludge bulking occurs when filamentous microorganisms (Microthrix parvicella, Sphaerotilus natans, Thiothrix, type 021N, and others) overgrow the floc-forming organisms in the activated sludge. Filamentous organisms form a loose, open floc structure that occupies much more volume than non-filamentous floc — SVI rises above 150–200 mL/g and the secondary clarifier fills with uncompacted sludge that washes over the effluent weir, causing solids carryover and treatment failure. Bulking is triggered by: low dissolved oxygen in the aeration tank (DO < 2 mg/L — filamentous organisms outcompete floc-formers at low O₂); low F/M conditions (starvation promotes filamentous growth); nutrient deficiency (N or P limitation); low pH (< 6.5); high soluble carbohydrate loading; or long HRT without adequate sludge wasting. Control measures include: increasing aeration to maintain DO ≥ 2 mg/L; increasing F/M by reducing HRT or increasing sludge wasting; chlorinating the return sludge (selective toxicity kills filaments); adding nutrients; and in severe cases, adjusting the influent distribution (selector reactors — high-BOD plug flow at the inlet of the aeration tank creates conditions unfavourable for filamentous growth). Selector reactors are now standard in new ASP designs in India to prevent bulking proactively.

Q2. How does the extended aeration process differ from conventional ASP, and why does it produce less sludge?

Extended aeration operates at F/M = 0.05–0.15 day⁻¹ (conventional: 0.2–0.5) and HRT = 18–36 hours (conventional: 4–8 hours). The very long HRT means food is severely limited relative to the microbial population — organisms consume not only the incoming BOD but also their own cell mass (endogenous respiration, or “self-oxidation”). When cells cannot find external carbon for growth, they metabolise their own stored energy and cell constituents, reducing the net biomass production. At very low F/M, the rate of new cell synthesis roughly equals the rate of endogenous decay — net sludge production approaches zero. This is why extended aeration plants (often the oxidation ditch variant) require minimal sludge management — a significant operational advantage for Indian small towns where sludge dewatering and disposal infrastructure is limited. The trade-off is much larger aeration tank volume (4–5× conventional) and higher energy consumption per unit BOD removed. Extended aeration also typically achieves partial nitrification (conversion of ammonia to nitrate) due to the long SRT, which supports nitrifying bacteria — an advantage for effluent quality but increases oxygen demand.

Q3. Why is the NRC formula considered empirical, and what are its limitations?

The NRC formula was developed from operational data on trickling filters at military installations in the USA during World War II (1946 National Research Council study). It is empirical because it was fitted to field observations rather than derived from fundamental mass transfer and microbial kinetics. Its key limitations are: (1) it was calibrated on rock-media filters at 20°C — it is less accurate for plastic media (super-rate) filters which have different surface area to volume ratios and flow patterns; (2) it does not explicitly account for temperature — performance decreases significantly below 15°C (uncommon in most of India but relevant in hill stations and Himalayan regions); (3) it assumes steady-state conditions; (4) the formula was developed for domestic sewage of a specific character — industrial wastewater or high-strength influent may give different results. Despite these limitations, the NRC formula remains the standard design tool in Indian STP design because it is simple, gives conservative estimates, and has been validated against many Indian trickling filter installations. For plastic media filters, modified versions or manufacturer’s specific correlations should be used. The CPHEEO Sewerage Manual recommends NRC for standard rock-media filters and provides supplementary equations for plastic media.

Q4. What is the significance of sludge age (SRT) for nitrification and nutrient removal in ASP?

Nitrification (conversion of NH₄⁺ to NO₃⁻) is carried out by slow-growing autotrophic bacteria — Nitrosomonas (NH₄⁺ → NO₂⁻) and Nitrobacter (NO₂⁻ → NO₃⁻). Their minimum SRT for stable operation is approximately 10–15 days at 20°C (longer at lower temperatures). If SRT falls below this minimum — which happens in high-loaded, short-HRT conventional ASP — nitrifying bacteria are washed out of the system faster than they can grow, and nitrification fails. This is why conventional ASP (SRT = 5–15 days, F/M = 0.2–0.5) provides variable nitrification depending on whether the SRT stays above the minimum, while extended aeration (SRT = 20–30 days) reliably achieves complete nitrification. For biological nitrogen removal (BNR), both nitrification and denitrification are needed: nitrification converts NH₄⁺ to NO₃⁻ (aerobic zone, SRT > 10 days); denitrification converts NO₃⁻ to N₂ gas (anoxic zone using influent BOD as carbon source). The SBR process achieves both in a single tank by alternating aerobic and anoxic phases. For biological phosphorus removal (BPR), an anaerobic zone is added before the aerobic zone — phosphorus-accumulating organisms (PAOs) release phosphate under anaerobic conditions and over-accumulate it under aerobic conditions; excess phosphorus is removed with the WAS. These advanced nutrient removal systems (available in SBR, modified Ludzack-Ettinger (MLE), A₂/O, UCT configurations) are increasingly required for STPs discharging to sensitive water bodies under NMCG and NGRBA (National Ganga River Basin Authority) regulations.