Question

estimate the time required for sand (.005-1 mm), silt (.002-.05 mm), and clay (<.002 mm) particles to settle to the bottom of a 2-m deep settling basin. Use strokes law, assuming spherical particles with the density of quartz (2.65 g/cm)

Answer 1

We know

From Stone's law, V_settling=

Diameter of the particles as given

D_sand=1mm

D_silt=0.05mm

D_clay=0.002mm

And for solid particle

Density of water

A) Velocity of settling for sand,

V= (2.65-1)×981×1×10^-2/(18×8.90×10^-3)=101.039 cm/sec

And time =200cm/101.039=1.97 sec

B) Velocity of settling for silt,

V=(2.65-1)×981×(5×10^-3)^2/(18×8.90×10^-3)=0.25 cm/sec

So, time =200/0.25=791.77 sec

C) velocity of settling for clay,

V=(2.65-1)×981×(2×10^-4)^2/(18×8.90×10^-3)=4.04×10^-4 cm/sec

Time =200/(4.04×10^-4)=494856.95 sec