--- title: "Oxidation pond and area procedure" subject: "Sanitary Engineering" chapter: 7 key: "semester_6/sanitary_engineering:oxidation-pond-area-procedure" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/oxidation-pond-area-procedure site: BCT Engineering --- # Oxidation pond and area procedure Sanitary Engineering · Chapter 7: Wastewater Treatment - Asked in: 2077 Chaitra, 4+4 marks - Video: [Oxidation Pond | Waste Water Engineering](https://www.youtube.com/watch?v=HqJ4GCWqJjI) — APSEd - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/oxidation-pond-area-procedure - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7 ## Question Briefly illustrate about oxidation pond. With suitable example, state its procedure to calculate its area. ## Short answer Find sewage flow Q (population × per capita) and BOD load (kg/day = Q × BOD / 1000). Divide load by permissible surface loading (200–400 kg/ha/day) to get area; volume = area × depth, detention = V/Q. Example: 5000 people, 100 lpcd, 250 mg/l BOD, 300 kg/ha/day gives 0.42 ha (about 46 m × 91 m) and 10 days. ## Answer #### Oxidation pond An oxidation pond is a shallow (1-1.5 m) earthen pond in which sewage is stabilised by the symbiotic action of bacteria and algae in sunlight. Bacteria oxidise the organics using oxygen produced by algae; algae use the CO₂ and nutrients released by the bacteria. Settled solids are digested anaerobically at the bottom. It needs little energy and little skill but a large area, and removes 70-90% of BOD. ```text sunlight -> algae --O2--> bacteria <--CO2,NH3,PO4-- ~~~~~~~~~ water surface ~~~~~~~~~ anaerobic sludge layer at bottom ``` #### Procedure to calculate the area 1. Find the sewage flow $Q$ (m³/day) = population × per capita sewage. 2. Find the BOD load: $\text{Load}=Q\times BOD\times10^{-3}$ kg/day. 3. Adopt a permissible surface organic loading $\lambda_s$ (kg BOD/ha/day), 200-400 depending on temperature. 4. Area $A=\dfrac{\text{Load}}{\lambda_s}$ (ha). 5. Adopt depth $d$ (1-1.5 m); volume $V=A\,d$; detention time $t=V/Q$ (check 5-30 days). 6. Fix the length-to-width ratio (2:1 to 3:1) and provide embankment, free board and inlet/outlet. #### Example Population = 5000, sewage 100 lpcd, BOD = 250 mg/l, loading 300 kg/ha/day, depth 1.2 m (assumed data). $$ \begin{aligned} Q &= 5000\times100/1000 = 500\ \text{m}^3/\text{day}\\ \text{Load} &= 500\times250/1000 = 125\ \text{kg/day}\\ A &= \frac{125}{300}=0.4167\ \text{ha}=4167\ \text{m}^2\\ V &= 4167\times1.2 = 5000\ \text{m}^3,\quad t=\frac{5000}{500}=10\ \text{days} \end{aligned} $$ For L : W = 2 : 1, $W=\sqrt{4167/2}=45.6$ m and $L=91.3$ m. **Answer: area about 0.42 ha (about 46 m x 91 m), 10 days detention.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the procedure, the pond area is found from A. BOD load divided by the permissible surface organic loading B. flow divided by depth only C. population divided by 100 D. detention time multiplied by the flow Answer: A. A = Load / lambda_s, with load = Q x BOD x 10^-3 kg/day. (Key point: Basis of area) 2. In the example (5000 persons, 100 lpcd, BOD 250 mg/l, loading 300 kg/ha/day), what is the BOD load? A. 500 kg/day B. 250 kg/day C. 125 kg/day D. 62.5 kg/day Answer: C. Q = 500 m3/day and load = 500 x 250 / 1000 = 125 kg/day. (Key point: Example calculation) 3. After finding the area, how is detention time obtained and checked? A. t = A/d and compared with 1-2 days B. t = V/Q, with V = A x d, and checked to lie in 5-30 days C. t = Q/V and compared with 100 days D. t = load/area and compared with 300 Answer: B. With V = 4167 x 1.2 = 5000 m3 and Q = 500 m3/day, t = 10 days, within 5-30 days. (Key point: Detention check) 4. What length-to-width ratio is adopted for the pond? A. 1 : 1 exactly B. 10 : 1 to 20 : 1 C. 2 : 1 to 3 : 1 D. 1 : 5 Answer: C. A rectangular pond of 2:1 to 3:1 is used, giving about 46 m x 91 m in the example. 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