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Sedimentation tank and oxidation pond design

Short answer

Sedimentation tank: 7.5 m diameter, 2.5 m side water depth (total 3.5 m). Oxidation pond: 7000 m² area, 1.2 m depth, dimensions ~59.2 m × 118.3 m, detention 8.4 days.

Design a sedimentation tank and oxidation pond for a town with the following data: Population = 10,000; Sewage flow = 100 lpcd; BOD of incoming sewage = 250 mg/l; BOD in the effluent of oxidation pond should be less than 30 mg/l.

Data

Q=10000×100/1000=1000Q = 10000\times 100/1000 = 1000 m³/day; raw BOD =250=250 mg/l; required effluent BOD <30<30 mg/l.

Part A: Sedimentation tank (circular)

Adopt detention t=2.5t=2.5 h and side water depth H=2.5H=2.5 m.

V=100024×2.5=104.2 m3A=VH=104.22.5=41.7 m2D=4Aπ=7.28 m ⇒ adopt D=7.5 mSurface loading=100044.2=22.6 m3/m2/day\begin{aligned} V &= \frac{1000}{24}\times 2.5 = 104.2\ \text{m}^3\\ A &= \frac{V}{H} = \frac{104.2}{2.5} = 41.7\ \text{m}^2\\ D &= \sqrt{\frac{4A}{\pi}} = 7.28\ \text{m}\ \Rightarrow\ \text{adopt } D=7.5\ \text{m}\\ \text{Surface loading} &= \frac{1000}{44.2} = 22.6\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}

The surface loading is a little below the usual 25-40 m³/m²/day. A lower overflow rate only means better settling, so the tank is on the safe side and acceptable. (If a loading within the range is wanted, adopt t=2t = 2 h, giving D≈6.5D \approx 6.5 m and about 30 m³/m²/day.)

Provide total depth =2.5+0.5+0.5=3.5= 2.5+0.5+0.5=3.5 m. Primary sedimentation removes about 30% of the BOD, so the BOD leaving it is 250×0.7=175250\times0.7=175 mg/l.

Part B: Oxidation pond (facultative)

Adopt a surface organic loading λs=250\lambda_s = 250 kg BOD/ha/day (suitable for a warm climate; usual range 200-400) and depth 1.2 m.

BOD load=Q×BOD=1000×175×10−3=175.0 kg/dayArea=175.0250=0.70 ha=7000 m2Volume=7000×1.2=8400 m3Detention time=VQ=84001000=8.4 days (usual 5-10 days)\begin{aligned} \text{BOD load} &= Q\times BOD = 1000\times 175\times10^{-3} = 175.0\ \text{kg/day}\\ \text{Area} &= \frac{175.0}{250} = 0.70\ \text{ha} = 7000\ \text{m}^2\\ \text{Volume} &= 7000\times 1.2 = 8400\ \text{m}^3\\ \text{Detention time} &= \frac{V}{Q} = \frac{8400}{1000} = 8.4\ \text{days (usual 5-10 days)} \end{aligned}

Take length : width =2:1=2:1: W=7000/2=59.2W=\sqrt{7000/2}=59.2 m, L=2W=118.3L=2W=118.3 m. Provide 2 ponds in parallel (each half this area) so that one can be desludged. Embankment side slope 1 : 2 (inside) to 1 : 1.5 (outside), free board 0.5 m, top width 2.5-3 m.

Efficiency check

Required BOD removal in the pond =175−30175×100=82.9%= \dfrac{175-30}{175}\times100 = 82.9\%, which is within the 80-90% removal normally achieved at this loading and detention time, so the effluent will be below 30 mg/l.

Answer: sedimentation tank D = 7.5 m, depth 2.5 m (+1 m); oxidation pond area 7000 m² (about 0.70 ha), depth 1.2 m, size about 59.2 m x 118.3 m, detention 8.4 days.

✓ Answer checked by Claude Opus 5.5 (high effort)

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