Oxidation pond design, 250 m³/d; theory

Theory of oxidation pond

An oxidation pond is a shallow earthen pond in which sewage is treated by the combined action of bacteria and algae in sunlight.

1. Aerobic bacteria in the upper layer oxidise organic matter: $\text{organics}+O_2\rightarrow CO_2+H_2O+\text{new cells}+NH_3+PO_4$.
2. Algae take the CO₂, ammonia and phosphates and, with sunlight, produce oxygen: $CO_2+H_2O+\text{light}\rightarrow\text{algae}+O_2$; this oxygen is used by the bacteria (symbiosis), with additional aeration by the wind.
3. Anaerobic bacteria at the bottom digest the settled sludge to CH₄, CO₂ and H₂S.
4. Sunlight, high pH and long detention destroy pathogens.

  sunlight -> algae --O2--> bacteria
              <--CO2,NH3,PO4--
  ~~~~~~~ surface ~~~~~~~
  anaerobic sludge at bottom

Design

Assumption: operating depth $d=1.2$ m; length : width $=2:1$.

$$\begin{aligned} \text{BOD load} &= 250\times250/1000=62.5\ \text{kg/day}\\ \text{Area} &= \frac{62.5}{450}=0.139\ \text{ha}=1389\ \text{m}^2\\ \text{Volume} &= 1389\times1.2=1667\ \text{m}^3\\ \text{Detention time} &= \frac{1667}{250}=6.67\ \text{days}\\ W &= \sqrt{1389/2}=26.4\ \text{m},\quad L=2W=52.7\ \text{m}\\ \text{Effluent BOD} &= 250\times(1-0.9)=25\ \text{mg/l} \end{aligned}$$

Answer: pond area $1389$ m² (about 27 m x 53 m), depth 1.2 m (plus 0.5 m free board), detention 6.67 days, effluent BOD about 25 mg/l.