Settling tank for 41,600 population

Design data and assumptions

Population = 41 600; rate = 120 lpcd; detention period $t = 4$ h; $L:B=4:1$; effective depth $H=3.5$ m. The design flow is taken as the average daily demand (no peak factor is given).

Step 1: Flow

$$Q=41\,600\times120=4\,992\,000\ \text{l/day}=4992\ \text{m}^3/\text{day}$$

Step 2: Volume and plan area

$$V=Q\,t=\frac{4992}{24}\times4=832.0\ \text{m}^3$$ $$A=\frac{V}{H}=\frac{832.0}{3.5}=237.71\ \text{m}^2$$

Step 3: Length and width

$L=4B$, so $4B^2=237.71$:

$$B=\sqrt{\frac{237.71}{4}}=7.71\ \text{m},\qquad L=4B=30.84\ \text{m}$$

Adopt $B=7.8$ m, $L=31.2$ m (area 243.4 m²).

Step 4: Checks

$$t_{actual}=\frac{243.4\times3.5}{4992}\times24=4.09\ \text{h}\ (\approx4\ \text{h, OK})$$ $$\text{SOR}=\frac{Q}{A}=\frac{4992}{243.4}=20.5\ \text{m}^3/\text{m}^2/\text{day}\ (15\text{ to }30,\ \text{OK})$$ $$\text{Velocity}=\frac{Q}{B\,H}=\frac{4992}{7.8\times3.5\times1440}=0.127\ \text{m/min}\ (<0.3,\ \text{OK})$$

Add a free board of 0.3 m and a sludge zone of 0.5 m: total depth $=3.5+0.3+0.5=4.3$ m.

Sketch

 PLAN (not to scale)
  +-------+---------------------------------+--------+
  | inlet |<-------- L = 31.2 m ----------->| outlet |
  | zone  |         settling zone           |  weir  |  B = 7.8 m
  +-------+---------------------------------+--------+
 SECTION
  inlet --> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ FB 0.3 m
  baffle    |           H = 3.5 m           |--> outlet
            |_______________________________|
             \_______ sludge 0.5 m ________/ -> drain

Answer: tank 31.2 m long $\times$ 7.8 m wide $\times$ 3.5 m effective depth (4.3 m total); SOR = 20.5 m³/m²/day and velocity = 0.127 m/min, both within limits.