Plain sedimentation tank for 4 MLD

Assumed data

Flow $Q=4$ MLD $=4000$ m³/day. Assume detention period $t=4$ h (3 to 4 h for plain sedimentation), effective depth $H=3$ m, $L:B=3:1$ (usual 3:1 to 5:1).

Step 1: Volume and area

$$V=\frac{4000}{24}\times4=666.7\ \text{m}^3,\qquad A=\frac{V}{H}=\frac{666.7}{3}=222.2\ \text{m}^2$$

Step 2: Dimensions

$$3B^2=222.2\Rightarrow B=8.61\ \text{m},\quad L=3B=25.82\ \text{m}$$

Adopt $B=8.7$ m, $L=26.0$ m.

Step 3: Checks

$$t=8.7\times26.0\times3/4000\times24=4.07\ \text{h},\quad \text{SOR}=\frac{4000}{226.2}=17.7\ \text{m}^3/\text{m}^2/\text{day}\ (15\text{ to }30),\quad v=\frac{4000}{8.7\times3\times1440}=0.106\ \text{m/min}\ (<0.3)$$

All values are within the usual limits. Provide a free board of 0.5 m and a sludge zone of 0.5 m with the floor sloping 1 in 10 towards the sludge pit, so the overall depth is 4.0 m.

 PLAN
  +--------+--------------------------------+--------+
  | inlet  |          L = 26.0 m            | outlet |  B = 8.7 m
  | baffle |                                |  weir  |
  +--------+--------------------------------+--------+
 SECTION
  inlet -> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ FB 0.5 m
  baffle   |          H = 3.0 m               |-> outlet
           |__________________________________|
            \__ sludge zone 0.5 m (1:10) __/ -> drain

Answer: rectangular plain sedimentation tank 26.0 m $\times$ 8.7 m $\times$ 3.0 m effective depth (4.0 m overall).