Settling basin design for 0.25 mm sediment

Given: $Q = 10$ m³/s, particle size $d = 0.25$ mm, trap efficiency $\eta = 95\%$.

Assumptions (not given): basin depth $h = 4.0$ m; water at about 20 °C; fall velocity of 0.25 mm quartz sand from the standard fall-velocity table, $w = 2.70$ cm/s $= 0.027$ m/s; Velikanov coefficient for 95% efficiency $\lambda = 1.65$ (Velikanov table: 90% → 1.28, 95% → 1.65, 98% → 2.05).

1. Horizontal velocity (Camp)

$$v = a\sqrt{d} = 44\sqrt{0.25} = 22\ \text{cm/s} = 0.22\ \text{m/s}$$

2. Width

$$B = \frac{Q}{v\,h} = \frac{10}{0.22 \times 4.0} = 11.36\ \text{m}$$

3. Ideal settling length (no turbulence)

$$\begin{aligned} t &= \frac{h}{w} = \frac{4.0}{0.027} = 148.1\ \text{s} \\ L_0 &= \frac{h\,v}{w} = \frac{4.0 \times 0.22}{0.027} = 32.6\ \text{m} \end{aligned}$$

4. Length considering turbulence (Velikanov)

Turbulent eddies keep particles in suspension longer, so the length is increased:

$$\begin{aligned} L &= \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2} \\ &= \frac{1.65^2 \times 0.22^2 \times (2.0 - 0.2)^2}{7.51 \times 0.027^2} \\ &= \frac{2.7225 \times 0.0484 \times 3.24}{0.005475} = 77.98\ \text{m} \end{aligned}$$

5. Adopted dimensions

Answer: a two-chamber settling basin about 78 m long, 12 m wide (2 × 6 m) and 4 m deep (plus storage depth, inlet and outlet transitions). The turbulence effect more than doubles the length compared with the ideal settling length of 32.6 m. Each chamber has $L/B = 78/6 = 13$, which gives uniform flow.