Rectangular sedimentation tank for 10 MLD

Data

$Q=10$ MLD $=10\,000$ m³/day; trap all particles larger than 0.04 mm ($d=0.004$ cm); $S_s=2.65$; $T=20$ °C ($\nu=0.01007$ cm²/s); width : length $=1:3$; depth $=3.5$ m.

Step 1: Settling velocity

$$v_s=\frac{981\times1.65\times0.004^2}{18\times0.01007}=0.1429\ \text{cm/s}=123.45\ \text{m/day}\quad(Re=0.057<1)$$

To trap all particles of this size, the overflow rate must equal the settling velocity: $v_0=Q/A=v_s$.

Step 2: Surface area

$$A=\frac{Q}{v_s}=\frac{10\,000}{123.45}=81.01\ \text{m}^2$$

Step 3: Length and width

$L=3B$, so $3B^2=81.01$:

$$B=5.20\ \text{m},\qquad L=3B=15.59\ \text{m}$$

Step 4: Detention time for depth 3.5 m

$$t=\frac{V}{Q}=\frac{81.01\times3.5}{10\,000}=0.0284\ \text{day}=0.68\ \text{h}=40.8\ \text{min}$$

(This also equals $H/v_s=350/0.1429$ s $=40.8$ min.)

Answer: tank 15.59 m long $\times$ 5.20 m wide ($A=81.0$ m²); detention time for H = 3.5 m is 0.68 h (41 min). A longer detention (2 to 4 h) is normally adopted in practice by providing a larger tank to allow for inefficiencies such as short-circuiting.