Settling basin design for 0.2 mm particles

Given: $Q = 12$ m³/s, $h = 3.5$ m, particle size $d = 0.2$ mm, fall velocity $w = 6$ cm/s $= 0.06$ m/s, turbulence coefficient $\lambda = 1.50$.

Method: horizontal velocity by Camp's formula, width from continuity, length by settling theory corrected for turbulence by Velikanov's formula.

1. Horizontal (flow-through) velocity

Camp: $v = a\sqrt{d}$ cm/s with $a = 44$ for $0.1 < d < 1$ mm.

$$\begin{aligned} v &= 44\sqrt{0.2} = 19.68\ \text{cm/s} = 0.197\ \text{m/s} \end{aligned}$$

2. Width of basin

$$\begin{aligned} B &= \frac{Q}{v\,h} = \frac{12}{0.1968 \times 3.5} = 17.42\ \text{m} \end{aligned}$$

3. Length by ideal settling theory (no turbulence)

$$\begin{aligned} t &= \frac{h}{w} = \frac{3.5}{0.06} = 58.3\ \text{s} \\ L_0 &= v\,t = \frac{h\,v}{w} = \frac{3.5 \times 0.1968}{0.06} = 11.48\ \text{m} \end{aligned}$$

4. Length considering turbulence (Velikanov)

$$\begin{aligned} L &= \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2} \\ &= \frac{1.5^2 \times 0.1968^2 \times (\sqrt{3.5} - 0.2)^2}{7.51 \times 0.06^2} \\ &= \frac{2.25 \times 0.03872 \times 2.7916}{0.02704} = 9.00\ \text{m} \end{aligned}$$

5. Adopted dimensions

The length must not be less than the ideal settling length, so the larger value (11.48 m) governs; adopt $L = 12$ m.

Answer: $v = 0.197$ m/s, $B \approx 17.4$ m, turbulent (Velikanov) length $= 9.0$ m; adopted basin about 12 m long × 17.5 m wide × 3.5 m deep (split into two chambers for flushing).

Note: the given $w = 6$ cm/s is high for 0.2 mm sand (real value about 2–2.5 cm/s); with a smaller $w$ the required length becomes much larger, so in practice the basin is made longer, with $L/B$ about 8–10 per chamber.