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Sedimentation tank 6 m x 18 m x 3 m, 4 ML/day

Short answer

Detention period is 1.94 h and surface overflow rate is 37.04 m³/m²/day. The tank is acceptable with coagulation, but for plain sedimentation the detention time is slightly low and the overflow rate is high.

In a sedimentation tank (dimensions: 6 m wide, 18 m long and 3 m depth), 4 million litres of water passes per day. Calculate (a) detention period and (b) surface overflow rate. Check the values with standard range.

Data

Width B=6B=6 m, length L=18L=18 m, depth H=3H=3 m, flow Q=4Q=4 ML/day =4000=4000 m³/day.

(a) Detention period

Volume=18×6×3=324 m3t=VQ=3244000=0.0810 day=1.94 h\begin{aligned} \text{Volume} &= 18\times6\times3 = 324\ \text{m}^3 \\ t &= \frac{V}{Q} = \frac{324}{4000} = 0.0810\ \text{day} = 1.94\ \text{h} \end{aligned}

(b) Surface overflow rate

SOR=QA=400018×6=37.04 m3/m2/day\text{SOR} = \frac{Q}{A} = \frac{4000}{18\times6} = 37.04\ \text{m}^3/\text{m}^2/\text{day}

Check with the standard range

Horizontal velocity =QB H=40006×3=222.2=\dfrac{Q}{B\,H}=\dfrac{4000}{6\times3}=222.2 m/day =0.154=0.154 m/min (the limit is 0.3 m/min; OK).

ItemComputedUsual range (plain sedimentation)Remark
Detention period1.94 h2 to 4 h (up to 6 h for plain tanks)Just below 2 h, slightly low
SOR37.0 m³/m²/day15 to 30 m³/m²/day (plain); 30 to 40 (with coagulation)Above the plain range, within that for coagulated water
Velocity0.154 m/minbelow 0.3 m/minSatisfactory
Depth3 m2.5 to 5 mSatisfactory
L : B3 : 13:1 to 5:1Satisfactory

Answer: detention period = 1.94 h; SOR = 37.04 m³/m²/day. The tank is satisfactory only if coagulation is used; for plain sedimentation the tank is slightly small (detention a little low and SOR high), so the area should be increased.

✓ Answer checked by Claude Opus 5.5 (high effort)

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