Rectangular sedimentation tank, 0.2 mm/s, 3 hours

Data

Population 20 000; peak allowance 112 lpcd; settling velocity $v_s=0.2$ mm/s; $L=2B$; settling (detention) period $t=3$ h.

Step 1: Flow

$$Q=20\,000\times112=2\,240\,000\ \text{l/day}=2240\ \text{m}^3/\text{day}$$

Step 2: Surface area (overflow rate = settling velocity)

$$v_0=0.2\times10^{-3}\times86400=17.28\ \text{m}^3/\text{m}^2/\text{day},\qquad A=\frac{Q}{v_0}=\frac{2240}{17.28}=129.63\ \text{m}^2$$

Step 3: Width and length

$$2B^2=129.63\Rightarrow B=8.05\ \text{m},\qquad L=2B=16.10\ \text{m}$$

Step 4: Depth

$$H=\frac{V}{A}=\frac{Q\,t}{A}=\frac{2240\times3/24}{129.63}=2.16\ \text{m}$$

(Also $H=v_s\,t=0.2\times10^{-3}\times10\,800=2.16$ m.)

Adopt $B=8.1$ m, $L=16.2$ m and an effective depth of 2.2 m, plus 0.3 m free board and a 0.5 m sludge zone, giving a total depth of 3.0 m.

Answer: required size 16.10 m $\times$ 8.05 m $\times$ 2.16 m effective depth; adopt a tank 16.2 m long $\times$ 8.1 m wide $\times$ 2.2 m effective depth (3.0 m overall).