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CPE 124 Particle Technology - Study Notes

Dr. Jie Zhang

Chapter 2. Motion of Particles through Fluids

2.1 Motion of particles through fluids

2.1.1 Mechanics of particle motion

Three forces acting on a particle moving through a fluid:

1). The external force, gravitational or centrifugal;

2). The buoyant force, which acts parallel with the external force but in the opposite direction;

3). The drag force, which appears whenever there is relative motion between the particle and the fluid

Drag: the force in the direction of flow exerted by the fluid on the solid is called drag.

2.1.2 Equations for one-dimensional motion of particle through fluid

Consider a particle of mass m moving through a fluid under the action of an external force Fe. Let the velocity of the particle relative to the fluid be u, let the buoyant force on the particle be Fb and let the drag be FD, then


The external force can be expressed as a product of the mass and the acceleration ae of the particle from this force,


The buoyant force is, be Archimedes’ law, the product of the mass of the fluid displaced by the particle and the acceleration from the external force. The volume of the particle is m/r p, the mass of fluid displaced is (m/r p)r , where r is the density of the fluid. The buoyant force is then

Fb = mr ae/r p (3)

The drag force is

FD = CDu2r Ap/2 (4)

where CD is the drag coefficient, Ap is the projected area of the particle in the plane perpendicular to the flow direction.

By substituting the forces into Eq(1), we have


Motion from gravitational force:

In this case, ae = g


Motion in a centrifugal field:

ae = rw 2


In this equation, u is the velocity of the particle relative to the fluid and is directed outwardly along a radius.

2.2 Terminal velocity

In gravitational settling, g is constant. Also, the drag always increases with velocity. The acceleration decreases with time and approaches zero. The particle quickly reaches a constant velocity which is the maximum attainable under the circumstances. This maximum settling velocity is called terminal velocity.



In motion from a centrifugal force, the velocity depends on the radius and the acceleration is not constant if the particle is in motion with respect to the fluid. In many practical use of centrifugal force, du/dt is small. If du/dt is neglected, then


Motion of spherical particles:

If the particles are spheres of diameter Dp, then

m = p Dp3r p/6

Ap = p Dp2/4

Substitution of m and Ap into the equation for ut gives the equation for gravity settling of spheres:


2.3 Drag coefficient

Drag coefficient is a function of Reynolds number. The drag curve applies only under restricted conditions:

i). The particle must be a solid sphere;

ii). The particle must be far from other particles and the vessel wall so that the flow pattern around the particle is not distorted;

iii). It must be moving at its terminal velocity with respect to the fluid.

Particle Reynolds number:


u: velocity of approaching stream

Dp: diameter of the particle

r : density of fluid

m : viscosity of fluid

Stokes’ law applies for particle Reynolds number less than 1.0

CD = 24/NRe,p (13)

From Eq(4)

FD = 3p m ut Dp (14)

From Eq(11)

ut = g Dp2(r p - r )/(18m ) (15)

At NRe,p =1, CD =26.5 instead of 24 from the above equation.

Centrifugal: rw 2 g.

For 1000 < NRe,p <200,000, use Newton’s law

CD = 0.44 (16)

FD= 0.055p Dp2 ut2r (17)


Newton’s law applies to fairly large particles falling in gases or low viscosity fluids.

Terminal velocity can be found by trial and error after guessing NRe,p to get an initial estimate of CD.

2.4 Criterion for settling regime

To identify the range in which the motion of the particle lies, the velocity term is eliminated from the Reynolds number by substituting ut from Stokes’ law


If Stokes’ law is to apply, NRe,p <1.0. Let us introduce a convenient criterion K


Then NRe,p = K3/18. Setting NRe,p = 1 and solving for K gives K=2.6. If K is less than 2.6 then Stokes’ law applies.

Substitution for ut using Newton’s law

NRe,p = 1.75K1.5

Setting NRe,p = 1000 and solving for K gives K = 68.9. Setting NRe,p = 200,000 and solving for K gives K = 2,360.

Stokes’ law range: K < 2.6

Newton’s law range: 68.9 < K < 2,360

when K > 2,360 or 2.6 < K < 68.9, ut is found from using a value of CD found by trial from the curve.

2.5 Hindered settling

In hindered settling, the velocity gradients around each particle are affected by the presence of nearby particles. So the normal drag correlations do not apply. Also, the particles in settling displace liquid, which flows upward and make the particle velocity relative to the fluid greater than the absolute settling velocity. For uniform suspension, the settling velocity us can be estimated from the terminal velocity for an isolated particle using the empirical equation of Maude and Whitmore

us = ut(e )n

Exponent n changes from about 4.6 in the Stokes’ law range to about 2.5 in the Newton’s law region. For very small particles, the calculated ratio us/ut is 0.62 for e =0.9 and 0.095 for e =0.6. With large particles, the corresponding ratios are us/ut = 0.77 and 0.28; the hindered settling effect is not as profound because the boundary layer thickness is a smaller fraction of the particle size.

If particles of a given size are falling through a suspension of much finer solids, the terminal velocity of the larger particles should be calculated using the density and viscosity of the fine suspension. The Maude-Whitmore equation may then be used to estimate the settling velocity with e taken as the volume fraction of the fine suspension, not the total void fraction.

Suspensions of very fine sand in water is used in separating coal from heavy minerals and the density of the suspension is adjusted to a value slightly greater than that of coal to make the coal particles rise to the surface, while the mineral particles sink to the bottom.

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