A sedimentation basin in a water treatment plant has a length = 48 m, width = 12 m, and depth = 3 m. The flow rate = 4 m 3 / s; particle specific gravity = 1.1; water density = 10 3 kg/m 3; and dynamic viscosity = 1.307 10 - 3 N. sec/m 2. What is the minimum particle diameter that is removed at 85%?

The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.

Solution

Given:

Length = 48 m

Width = 12 m

Depth = 3m

Flow rate = 4 m 3 / s

Water density = 10 3 kg/m 3

Dynamic viscosity = 1.307 10 - 3 N. sec/m

Now,

At the minimum particular diameter it is stated as follows:

The Reynolds number = 0.1

Thus,

0.1 = ρVTD/μ

VT = Dp² (ρp - ρ) g / 10μ²

Where

gn = The case/issue of sedimentation

VT = Terminal velocity

So,

0.1 = Dp³ (ρp - ρ) g / 10μ²

This becomes,

0.1 = 1000 * dp³ (1100-1000) g 0.1 / 10 * (1.307 * 10 ^⁻3) ²

= 3.074 * 10 ^⁻6 = dp³ (. g01 * 10^6)

dp³=3.1343 * 10 ^⁻12

Dp minimum = 1.474 * 10 ^⁻4 meters.