Question

What is the sedimentation velocity (in m/s) of a 250 µm diameter spherical particle with a density of 1.2 g/cm3 in water at 15°C? b. To remove 80% of such particles, what an overflow rate (in m/s) is needed in a sedimentation tank? c. If the tank is an up-flow clarifier (round shape from a top view) with a diameter: depth ratio of 3:1, and the flow to be treated is 100 cfs, what are the diameter (in m), depth (in m) and retention time (in hour) of the tank?

Answer 1

A)Given , diameter of particle = 250 = 0.00025 m.

Density = 1.2 g/cm3 . => G = 1.2 .

As we know at 15°C , viscosity = 1.1375 * 10^-3 Ns/m2.

By stokes law ,

Sedimentation velocity =

=>

=> 0.00000598901 m/s = 5.98 * 10^-6 m/s.

B)

To remove 80% of particles -

Sedimentation velocity = 0.8 * 5.98 * 10^-6 = 4.79 * 10^-6 m/s.

C)

Given , Q= 100 cfs = 100 * 0.305^3 = 2.837 m3/s.

As we know , Q= Area * velocity

=> 2.837 = ( π * D^2)/4 * 5.98 * 10^-6

=> D = 777.23 m .

Hence, Depth = 777.23/3 = 259 m.

Retention Time = Volume/Flow .

Volume = Area * Depth =

122856197.175 m3

Hence, retention Time = 122856197.175/ 2.837 = 43304969.0431 s = 12029 hours

If any doubt kindly comment. Thank you.