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Circular sedimentation tank, population 5500

Short answer

One circular tank of 8.0 m diameter, 3.0 m side water depth (total 4.0 m), detention 2.0 h, surface loading 35.5 m³/m²/day.

What will be the suitable dimensions of a circular sewage sedimentation tank for an industrial area having population of 5500? The average water demand is 180 lpcd. Assume that 75% water reaches the treatment plant and maximum demand is 2.4 times average demand. Dimension of the suspended silica particles available in influent water are larger than 0.14 mm.

Assumptions: sewage reaching the plant =0.75×180=135=0.75\times 180 = 135 lpcd, designed for the maximum rate (2.4×2.4\times average): Qavg=5500×135/1000=742.5Q_{avg}=5500\times 135/1000 = 742.5 m³/day. Detention 2 h and depth 3 m are adopted (usual design values). Silica grains of d=0.14d=0.14 mm (G=2.65G=2.65) at 20°C.

Step 1: Design flow

Q=5500×180×0.75×2.4/1000=1782 m3/dayQ = 5500\times 180\times 0.75\times 2.4/1000 = 1782\ \text{m}^3/\text{day}

Step 2: Capacity

Adopt detention period t=2.0t = 2.0 h and side water depth H=3.0H = 3.0 m (typical for plain sedimentation: 2-3 h, 2.5-4 m).

V=Q t=178224×2.0=148.5 m3,A=VH=148.53.0=49.5 m2V = Q\,t = \frac{1782}{24}\times 2.0 = 148.5\ \text{m}^3,\qquad A = \frac{V}{H} = \frac{148.5}{3.0} = 49.5\ \text{m}^2

Step 3: Diameter

Provide one circular tank, so A1=AA_1=A.

D=4A1π=4×49.5π=7.94 m ⇒ adopt D=8.0 mD = \sqrt{\frac{4A_1}{\pi}} = \sqrt{\frac{4\times 49.5}{\pi}} = 7.94\ \text{m}\ \Rightarrow\ \text{adopt } D = 8.0\ \text{m}

Adopt a total depth of H+0.5 (free board)+0.5 (sludge zone)=4.0H + 0.5\ (\text{free board}) + 0.5\ (\text{sludge zone}) = 4.0 m, with a central inlet well and a peripheral weir.

Step 4: Checks

Surface loading=Q/NπD2/4=178250.3=35.5 m3/m2/day (usual 25-40 for primary tanks)Weir loading=Q/NπD=1782π×8.0=70.9 m3/m/day (should be below about 300)Actual detention=(πD2/4)HQ/N×24=2.03 h\begin{aligned} \text{Surface loading} &= \frac{Q/N}{\pi D^2/4} = \frac{1782}{50.3} = 35.5\ \text{m}^3/\text{m}^2\text{/day}\ (\text{usual 25-40 for primary tanks})\\ \text{Weir loading} &= \frac{Q/N}{\pi D} = \frac{1782}{\pi\times 8.0} = 70.9\ \text{m}^3/\text{m/day}\ (\text{should be below about 300})\\ \text{Actual detention} &= \frac{(\pi D^2/4)H}{Q/N}\times 24 = 2.03\ \text{h} \end{aligned}

Check of removal of the 0.14 mm silica: for 0.1<d<10.1<d<1 mm, Hazen's transition formula applies: vs=60.6(G−1)d3T+70100=60.6×1.65×0.14×1.3=18.2v_s=60.6(G-1)d\dfrac{3T+70}{100}=60.6\times1.65\times0.14\times1.3 = 18.2 mm/s ≈1572\approx 1572 m/day (dd in mm, T=20T=20°C). The tank overflow rate is only 35.5 m³/m²/day, much less than vsv_s, so all silica particles larger than 0.14 mm settle easily; the tank size is governed by the detention time, not by the settling velocity.

   Section (circular, radial flow)
      inlet well        weir/launder
   ->  |__|~~~~~~~~~~~~~~~~~~~~~|_
       |  \                  /  | H = 3.0 m
       |   \______    ______/   | + sludge zone
       +----------\__/----------+
                  sludge hopper -> sludge pipe
        <---------- D = 8.0 m ---------->

Answer: 1 circular sedimentation tank of diameter 8.0 m, side water depth 3.0 m (total depth about 4.0 m), detention 2.0 h, surface loading 35.5 m³/m²/day.

✓ Answer checked by Claude Opus 5.5 (high effort)

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