Question

A sand particle has an average diameter of 1 mm and a shape factor of 0.90 and a specific gravity of 2.1, determine the terminal velocity of the particle settling in water at 20^oC (kinematic viscosity of water = 1.003 x 10^-6 m²/s and specific gravity = 1) Drag coefficient can be computed using the formula:

Cd= 24/Re +( 3/sqrtRe ) + 0.34

  Answer: 0.1419 m/s

Answer 1

Given: Average diameter of particle, D=1 mm=1*10^-3m

Shape factor, =0.90

Specific gravity of water=1 Density of water, _f = 1000 kg/m³

Specific gravity of sand particle=2.1

Density of sand particle, _p= 2.1*Density of water=2.1*1000 kg/m³=2100 kg/m³

Kinematic viscosity of water, =1.003*10^-6 m²/s

where is the Dynamic viscosity of water

Dynamic Viscosity of water, _f = 1.003*10^-3 kg/ms

Settling velocity, u_t is calculated using Stoke's law as follows

where u_t is the Settling Velocity

g is Acceleration due to gravity, g=9.81 m/s²

Reynolds Number, Re is calculated as

Reynolds Number, Re is greater than 1, hence the terminal velocity should be calculated using Newton's law for flow in transition region.

Newtons's law to calculate terminal velocity is

where C_D is the Drag Coefficient and is calculated as

Thus terminal velocity is

Using an iterative method, calculate the terminal velocity until the assumed u_t value is approximately equal to u_t calculated using Newton's law.

The initial terminal velocity value will be u_t =0.5977 m/s.

Assumed Terminal Velocity (m/s)	Reynolds Number, Re	Drag Coefficient, CD	Terminal Velocity using Newton's Law (m/s)	Difference
0.5977	536.32	0.5142	0.1763	0.4214
0.1763	158.19	0.7302	0.1480	0.0283
0.1480	132.8	0.7811	0.1431	4.9*10-3
0.1431	128.40	0.7917	0.1421	1*10-3
0.1421	127.50	0.7939	0.1419	2*10-4
0.1419	127.328	0.7943	0.1419	0

Iteration 2: u_t= 0.1763 m/s

Difference=0.1763-0.148=0.0283

Iteration 3: u_t= 0.148 m/s

Difference=0.148-0.1431=4.9*10^-3

Iteration 4: u_t= 0.1431 m/s

Difference=0.1431-0.1421=1*10^-3

Iteration 5: u_t= 0.1421 m/s

Difference=0.1421-0.1419=2*10^-4

Iteration 6: u_t= 0.1419 m/s

Difference=0.1419-0.1419=0

In this iteration, the terminal velocity calculated using Newton's law is equal to the assumed terminal velocity value.

Therefore, the terminal velocity is u_t = 0.1419 m/s