Desander design, two basins (16 m³/s)

Given: $Q=16$ m³/s, particle $\ge0.2$ mm with $w=0.022$ m/s, horizontal velocity $v=0.2$ m/s, two basins, factor of safety 1.5, $L/B=6$, $C=1$ kg/m³, $\rho_s=2650$ kg/m³, settling (flushing) time 1 day, packing factor 0.5, extra discharge for flushing 10%.

Step 1: Design flow

$$Q'=1.10\times16=17.6\ \text{m}^3/\text{s}$$

Step 2: Surface area

$$A_{total}=\frac{FS\,Q'}{w}=\frac{1.5\times17.6}{0.022}=1200.0\ \text{m}^2\quad\Rightarrow\quad A_{basin}=\frac{1200.0}{2}=600.0\ \text{m}^2$$

Step 3: Plan size of one basin

$$B=\sqrt{\frac{A_{basin}}{6}}=\sqrt{\frac{600.0}{6}}=10.00\ \text{m},\qquad L=6B=60.0\ \text{m}$$

Adopt $B=10$ m, $L=60$ m.

Step 4: Depth of flow

Flow in one basin $=Q'/2=8.8$ m³/s; with $v=0.2$ m/s:

$$H=\frac{Q'/2}{B\,v}=\frac{8.8}{10\times0.2}=4.40\ \text{m}\ \Rightarrow\ \text{adopt }4.5\ \text{m}$$

Step 5: Sediment storage (1 day)

$$\begin{aligned} \text{Mass}&=C\,Q'\,t=1\times17.6\times86400=1520640\ \text{kg}\\ \text{Volume}&=\frac{1520640}{2650\times0.5}=1148\ \text{m}^3\ \text{(both basins)}\\ \text{Depth per basin}&=\frac{1148/2}{600.0}=0.96\ \text{m} \end{aligned}$$

Provide about 1.0 m of sediment storage in the hopper below the flow depth (or flush more often, as in practice).

Answer: 2 basins, each 10 m x 60 m in plan with a flow depth of 4.5 m and 1.0 m storage depth.