Frequency of flushing of settling basin

Given: sediment storage $=10\%$ of the basin volume $V_b$; overall trap efficiency $\eta=40\%$; sediment concentration $C=2000$ ppm $=2.0$ kg/m³. Assumption: the settled sediment has a bulk density of $\rho_b=2650\times0.5=1325$ kg/m³ (particle density 2650 kg/m³, packing factor 0.5).

Step 1: Volume of sediment trapped per m³ of water

$$v_s=\frac{\eta\,C}{\rho_b}=\frac{0.40\times2.0}{1325}=6.038 \times 10^{-4}\ \text{m}^3\ \text{of deposit per m}^3\ \text{of water}$$

Step 2: Storage volume

$$V_{st}=0.10\,V_b$$

Step 3: Volume of water that can pass before the storage is full

$$V_w=\frac{V_{st}}{v_s}=\frac{0.10\,V_b}{6.038 \times 10^{-4}}=165.6\,V_b$$

Step 4: Frequency of flushing

The water passes through the basin at the rate $Q$, so the time between flushes is

$$T=\frac{V_w}{Q}=165.6\,\frac{V_b}{Q}=165.6\,t_d$$

where $t_d=V_b/Q$ is the detention time of the basin.

So the basin should be flushed once after about 165.6 detention periods (about every 166 times the time that water takes to pass through it). For example, if the detention time is 10 minutes, then $T=1656$ minutes $\approx27.6$ hours (about 0.9 flushes per day); if $t_d=5$ minutes, $T\approx13.8$ hours.

Answer: flush every $\approx165.6\,V_b/Q$ (about 166 detention times; e.g. about 27.6 h for a 10-minute detention time).