Question

A typical riverborne silt particle has a radius of 20(mu)m and a density of 2xx10^(3) kg//m^(3). The viscosity of water is 1.0 mPI. Find the terminal speed with which such a particle will settle to the bottom of a motionless volume of water.

Answer 1

The terminal speed with which such a particle will settle to the bottom of a motionless volume of water is:

• Radius of the typical river borne slit particle = 20μm = 0.00002 m

• Density of the typical river borne slit particle = 2000 kg/$m^3$

• Viscosity of water, η= 1.0 mP = 0.001 P

• Terminal velocity = (2/9) * ($r^2$ *(ρ-σ)*g)/η

                                     = $\frac{2}{9} * \frac{0.00002^2 (2000-1000)* 9.8}{0.001} = 0.87 mm/sec$

• Therefore, answer = 0.87 mm/sec

Answer 2

Explanation:

The terminal speed with which such a particle will settle to the bottom of a motionless volume of water is:

Radius of the typical river borne slit particle = 20μm = 0.00002 m

Density of the typical river borne slit particle = 2000 kg/\begin{gathered}m^3\\\end{gathered}

m

3

Viscosity of water, η= 1.0 mP = 0.001 P

Terminal velocity = (2/9) * (r^2r

2

*(ρ-σ)*g)/η