Oxidation pond and area procedure

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.

   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.