Rectangular sedimentation tank, 0.2 mm/s, 25000 population

Data

Population 25 000; peak allowance 110 lpcd; $v_s=0.2$ mm/s; $L=2B$; settling period $t=3$ h.

Step 1: Flow

$$Q=25\,000\times110=2\,750\,000\ \text{l/day}=2750\ \text{m}^3/\text{day}$$

Step 2: Surface area

$$v_0=0.2\times10^{-3}\times86400=17.28\ \text{m}^3/\text{m}^2/\text{day},\qquad A=\frac{2750}{17.28}=159.14\ \text{m}^2$$

Step 3: Width and length

$$2B^2=159.14\Rightarrow B=8.92\ \text{m},\qquad L=2B=17.84\ \text{m}$$

Step 4: Depth

$$H=\frac{Q\,t}{A}=\frac{2750\times3/24}{159.14}=2.16\ \text{m}$$

Adopt $B=9.0$ m, $L=18.0$ m, effective depth 2.2 m, free board 0.3 m and sludge zone 0.5 m (total 3.0 m).

Sketch

 PLAN
  +-------+------------------------------+--------+
  | inlet |         L = 18.0 m           | outlet |  B = 9.0 m
  +-------+------------------------------+--------+
 SECTION
  inlet ->  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~  FB 0.3 m
            |      H = 2.2 m           |-> outlet
            |__________________________|
             \__ sludge zone 0.5 m __/ -> drain

Answer: tank 18.0 m $\times$ 9.0 m in plan with 2.2 m effective depth (3.0 m total).