Rectangular sedimentation tank for 3000 persons

Data and assumptions

Population 3000; 120 lpcd; $t=6$ h. Assumed: effective depth $H=3$ m, $L:B=3:1$ (no peak factor is given).

Step 1: Flow and volume

$$Q=3000\times120=360\,000\ \text{l/day}=360\ \text{m}^3/\text{day}$$ $$V=\frac{360}{24}\times6=90.0\ \text{m}^3$$

Step 2: Plan area and size

$$A=\frac{V}{H}=\frac{90.0}{3}=30.0\ \text{m}^2,\qquad 3B^2=30.0\Rightarrow B=3.16\ \text{m},\ L=3B=9.49\ \text{m}$$

Adopt $B=3.2$ m and $L=9.5$ m.

Step 3: Checks

$$t=\frac{3.2\times9.5\times3}{360}\times24=6.08\ \text{h},\quad \text{SOR}=\frac{360}{30.4}=11.8\ \text{m}^3/\text{m}^2/\text{day},\quad v=\frac{360}{3.2\times3\times1440}=0.026\ \text{m/min}$$

The velocity is far below 0.3 m/min. The SOR is low (safe).

Answer: provide a tank 9.5 m $\times$ 3.2 m in plan with 3 m effective depth (add 0.3 m free board and 0.5 m sludge zone, so 3.8 m total).