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Oxidation pond theory; 2500 persons

Short answer

Q=450 m³/d (80% of supply), BOD load=112.5 kg/d. Loading gives 2045 m² but detention (12 d × 450 / 1.25 m) gives 4320 m², so use 4320 m². Adopt about 47 m × 93 m at 1.25 m operational depth, with 0.5 m free board; actual loading 260 kg/ha/day is safe.

Describe the theory of oxidation pond. Design an oxidation pond for treating domestic sewage of 2500 persons supplied with 225 lpcd of water. The BOD5 of the wastewater is 250 mg/l. Permissible organic loading for the pond is 550 kg/ha/day and the detention time is 12 days. Assume the width to length ratio of the pond as 1:2 and the operational depth as 1.25 m.

Theory of oxidation pond

An oxidation pond is a shallow earthen basin (1-1.5 m deep) in which sewage is stabilised naturally by bacteria and algae in sunlight.

  • Bacteria in the upper aerobic zone oxidise organic matter with oxygen produced by algae and wind aeration.
  • Algae use sunlight and the CO₂, ammonia and phosphates released by the bacteria, producing oxygen (symbiosis).
  • Settled solids are digested anaerobically at the bottom.
  • Pathogens die by sunlight, high pH and long detention. BOD removal is 70-90%.

Design

Assumption: 80% of the water supplied reaches the pond as sewage.

Q=2500×225×0.8/1000=450 m3/dayBOD load=450×250/1000=112.5 kg/day\begin{aligned} Q &= 2500\times225\times0.8/1000=450\ \text{m}^3/\text{day}\\ \text{BOD load} &= 450\times250/1000=112.5\ \text{kg/day} \end{aligned}

Area from the permissible loading: A1=112.5550=0.205A_1=\dfrac{112.5}{550}=0.205 ha =2045=2045 m².

Area from the detention time (12 days, depth 1.25 m):

V=Q t=450×12=5400 m3,A2=Vd=54001.25=4320 m2V=Q\,t=450\times12=5400\ \text{m}^3,\qquad A_2=\frac{V}{d}=\frac{5400}{1.25}=4320\ \text{m}^2

The larger area governs: A=4320A=4320 m². The loading actually applied is 112.50.432=260\dfrac{112.5}{0.432}=260 kg/ha/day, which is below the permissible 550, so it is safe.

Dimensions (width : length =1:2=1:2):

W=43202=46.5 m (adopt 47 m),L=2W=93.0 m (adopt 93 m)W=\sqrt{\frac{4320}{2}}=46.5\ \text{m (adopt 47 m)},\qquad L=2W=93.0\ \text{m (adopt 93 m)}

Provide a pond of 47 m×93 m×1.2547\ \text{m}\times93\ \text{m}\times1.25 m operational depth, plus 0.5 m free board, with embankment slopes 1 : 2 and an inlet and outlet at opposite ends.

Answer: pond area 43204320 m² (≈47 m×93 m\approx 47\ \text{m}\times93\ \text{m}), depth 1.25 m (1.75 m including free board).

✓ Answer checked by Claude Opus 5.5 (high effort)

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