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Oxidation pond theory; colony of 15000

Short answer

Assume 80% of 200 lpcd reaches the pond: Q=2400 m³/d, BOD load=816 kg/d. At 300 kg/ha/day, area=2.72 ha; with depth 1.2 m, volume=32640 m³, detention=13.6 days, adopt about 117 m × 234 m. For 9-hour flow at 0.9 m/s, pipe area=0.0823 m², so diameter ≈ 325 mm.

Describe the theory of oxidation pond. A new colony of 15000 populations is supplied with water at 200 lpcd. The sewage from colony is to be treated in an oxidation pond. The BOD and suspended solids are each 340 mg/l. Assuming the organic loading of 300 kg/ha/day, calculate the dimension of the pond. If the daily flow is for 9 hours and the average velocity of sewage is 0.9 m/s, what is the diameter of the inlet pipe required for the oxidation pond?

An oxidation pond (waste stabilisation pond, facultative pond) is a shallow earthen basin, 1-1.5 m deep, in which sewage is purified naturally by the mutual action of bacteria and algae in the presence of sunlight.

  • Aerobic zone (top): aerobic bacteria oxidise organic matter using oxygen produced by algae and by wind aeration: organics+O2→CO2+H2O+new cells+NH3+PO4\text{organics}+O_2\rightarrow CO_2+H_2O+\text{new cells}+NH_3+PO_4.
  • Algae use sunlight, the CO₂, nitrogen and phosphate released by the bacteria to grow and release oxygen: CO2+H2O+light→algae+O2CO_2+H_2O+\text{light}\rightarrow\text{algae}+O_2.
  • Anaerobic zone (bottom): settled solids are digested to CH₄, CO₂ and H₂S.
  • Pathogens die by sunlight, high pH and long detention. Typical detention is 5-30 days, loading 200-400 kg BOD/ha/day, and BOD removal 70-90%.
   sunlight
   \ | /
  ~~~~~~~~~~~~~~~~~~~~~~~~~~ surface (wind aeration)
   Algae --O2--> Bacteria (aerobic zone)
   Algae <--CO2,NH3,PO4-- Bacteria
  - - - - - - - - - - - - - - - - -
   Anaerobic zone: sludge -> CH4, CO2
  ==================== bottom

Design

Assumptions: 80% of the water supplied becomes sewage; depth of pond =1.2=1.2 m; length : width =2:1=2:1.

Q=15000×200×0.8/1000=2400 m3/dayBOD load=2400×340/1000=816 kg/dayArea=816300=2.72 ha=27200 m2W=27200/2=116.6 m,L=2W=233.2 mVolume=27200×1.2=32640 m3,t=326402400=13.6 days\begin{aligned} Q &= 15000\times200\times0.8/1000=2400\ \text{m}^3/\text{day}\\ \text{BOD load} &= 2400\times340/1000=816\ \text{kg/day}\\ \text{Area} &= \frac{816}{300}=2.72\ \text{ha}=27200\ \text{m}^2\\ W &= \sqrt{27200/2}=116.6\ \text{m},\quad L=2W=233.2\ \text{m}\\ \text{Volume} &= 27200\times1.2=32640\ \text{m}^3,\quad t=\frac{32640}{2400}=13.6\ \text{days} \end{aligned}

Adopt a pond of about 117 m×234 m×1.2117\ \text{m}\times234\ \text{m}\times1.2 m (preferably as two ponds in parallel).

Inlet pipe diameter

The daily flow arrives in 9 hours, so

q=24009×3600=0.0741 m3/sApipe=qv=0.07410.9=0.0823 m2d=4Apipeπ=0.324 m=324 mm\begin{aligned} q &= \frac{2400}{9\times3600}=0.0741\ \text{m}^3/\text{s}\\ A_{pipe} &= \frac{q}{v}=\frac{0.0741}{0.9}=0.0823\ \text{m}^2\\ d &= \sqrt{\frac{4A_{pipe}}{\pi}}=0.324\ \text{m}=324\ \text{mm} \end{aligned}

Answer: pond area 2720027200 m² (≈117 m×234 m\approx 117\ \text{m}\times234\ \text{m}, depth 1.2 m, detention 13.6 days); inlet pipe diameter about 325 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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