Rectangular sedimentation tank, 9 million L peak demand

Data

Peak demand $Q=9\times10^6$ l/day $=9000$ m³/day; velocity of flow $v=20$ cm/min $=0.2$ m/min; detention period $t=4$ h $=240$ min.

Step 1: Length of the tank

$$L=v\,t=0.2\times240=48\ \text{m}$$

Step 2: Cross-sectional area

$$Q=\frac{9000}{24\times60}=6.250\ \text{m}^3/\text{min},\qquad A_x=\frac{Q}{v}=\frac{6.250}{0.2}=31.25\ \text{m}^2$$

Step 3: Depth and width

Assume effective depth $H=3.5$ m (usual 3 to 4 m):

$$B=\frac{A_x}{H}=\frac{31.25}{3.5}=8.93\ \text{m}\approx9\ \text{m}$$

Step 4: Checks

$$V=Q\,t=\frac{9000}{24}\times4=1500\ \text{m}^3\ (\text{tank volume}=48\times9\times3.5=1512\ \text{m}^3)$$ $$\text{SOR}=\frac{9000}{48\times9}=20.8\ \text{m}^3/\text{m}^2/\text{day}\ (\text{within }15\text{ to }30)$$

$L:B=48/9=5.3:1$, slightly above the usual 3:1 to 5:1 range. This is acceptable for a long horizontal-flow tank; if a ratio within 5:1 is wanted, take $H=3.2$ m, giving $B=31.25/3.2=9.8$ m and $L:B=4.9:1$. Add 0.5 m free board and a 0.5 m sludge zone: total depth 4.5 m.

 PLAN                                       ----> flow
  +-------+------------------------------+--------+
  | inlet |           L = 48 m           | outlet |  B = 9 m
  +-------+------------------------------+--------+
 SECTION: H = 3.5 m effective + 0.5 m free board + 0.5 m sludge zone = 4.5 m

Answer: tank 48 m long $\times$ 9 m wide $\times$ 3.5 m effective depth (4.5 m total). SOR = 20.8 m³/m²/day.