Rectangular sedimentation tank, 20000 population, 0.4 mm/s

Data and assumptions

Population 20 000; per-capita demand is not given, so 120 lpcd is assumed. Settling velocity $v_s=0.4$ mm/s; $L=2B$; detention period $t=2$ h.

Step 1: Flow

$$Q=20\,000\times120=2\,400\,000\ \text{l/day}=2400\ \text{m}^3/\text{day}$$

Step 2: Surface area

For all particles with $v_s=0.4$ mm/s to be removed, the overflow rate must equal $v_s$:

$$v_0=v_s=0.4\times10^{-3}\times86400=34.56\ \text{m}^3/\text{m}^2/\text{day}$$ $$A=\frac{Q}{v_0}=\frac{2400}{34.56}=69.44\ \text{m}^2$$

Step 3: Width and length

$$2B^2=69.44\Rightarrow B=5.89\ \text{m},\qquad L=2B=11.79\ \text{m}$$

Adopt $B=6$ m, $L=12$ m ($A=72$ m²).

Step 4: Depth

$$V=Q\,t=\frac{2400}{24}\times2=200\ \text{m}^3,\qquad H=\frac{V}{A}=\frac{200}{69.44}=2.88\ \text{m}$$

(The same as $H=v_st=0.4\times10^{-3}\times7200=2.88$ m.) Adopt effective depth 3.0 m, so the detention time $=72\times3/2400\times24=2.16$ h.

Add free board 0.3 m and a sludge zone 0.5 m: total depth 3.8 m. Horizontal velocity $=2400/(6\times3\times1440)=0.093$ m/min (OK).

Answer: tank 12 m long $\times$ 6 m wide $\times$ 3.0 m effective depth (3.8 m overall).