Settling basin design for 0.15 mm particles

Given: $Q = 10$ m³/s, $h = 4.0$ m, particle size $d = 0.15$ mm, fall velocity $w = 6$ cm/s $= 0.06$ m/s, $\lambda = 1.50$.

Method: Camp's formula for flow velocity, continuity for width, settling theory with Velikanov's turbulence correction for length.

1. Horizontal velocity (Camp)

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

$$v = 44\sqrt{0.15} = 17.04\ \text{cm/s} = 0.1704\ \text{m/s}$$

2. Width

$$B = \frac{Q}{v\,h} = \frac{10}{0.1704 \times 4.0} = 14.67\ \text{m}$$

3. Ideal settling length

$$\begin{aligned} t &= \frac{h}{w} = \frac{4.0}{0.06} = 66.7\ \text{s} \\ L_0 &= \frac{h\,v}{w} = \frac{4.0 \times 0.1704}{0.06} = 11.36\ \text{m} \end{aligned}$$

4. Length considering turbulence (Velikanov)

$$\begin{aligned} L &= \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2} \\ &= \frac{1.5^2 \times 0.1704^2 \times (2.0 - 0.2)^2}{7.51 \times 0.06^2} \\ &= \frac{2.25 \times 0.02904 \times 3.24}{0.02704} = 7.83\ \text{m} \end{aligned}$$

5. Adopted dimensions

Since the length cannot be less than the ideal settling length, $L_0 = 11.36$ m governs; adopt $L = 12$ m.

Answer: $v = 0.170$ m/s, $B \approx 14.7$ m, Velikanov length $= 7.83$ m; adopt a basin about 12 m long × 15 m wide × 4 m deep, divided into two chambers so that one can be flushed while the other operates.

Note: the given $w = 6$ cm/s is much larger than the real fall velocity of 0.15 mm sand (about 1.5 cm/s), which is why the computed length is short; with the real value the basin would be several times longer.