Plain sedimentation tank for 3.75 L/s

Data and assumption

$Q=3.75$ l/s; effective depth $H=2.1$ m; detention time $t=4$ h. Assume $L:B=3:1$.

Step 1: Flow and volume

$$Q=3.75\times86\,400=324\,000\ \text{l/day}=324\ \text{m}^3/\text{day}$$ $$V=Qt=3.75\times10^{-3}\times4\times3600=54\ \text{m}^3$$

Step 2: Plan area and dimensions

$$A=\frac{V}{H}=\frac{54}{2.1}=25.71\ \text{m}^2,\qquad 3B^2=25.71\Rightarrow B=2.93\ \text{m},\ L=3B=8.78\ \text{m}$$

Adopt $B=3.0$ m, $L=9.0$ m ($A=27.0$ m²).

Step 3: Checks

$$t=27.0\times2.1/324\times24=4.20\ \text{h},\quad \text{SOR}=\frac{324}{27.0}=12.0\ \text{m}^3/\text{m}^2/\text{day},\quad v=\frac{324}{3.0\times2.1\times1440}=0.036\ \text{m/min}$$

The velocity is below 0.3 m/min (no scouring). The SOR is below the usual 15 to 30, which means a conservative design for such a small flow. Add 0.3 m free board and 0.3 m sludge zone: overall depth 2.7 m.

Answer: tank 9.0 m long $\times$ 3.0 m wide $\times$ 2.1 m effective depth (2.7 m overall).