chapter 2 - particle technology (motion of particles through fluids)
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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
(1)
The external force can be expressed as a product of the mass and the acceleration ae of the particle
from this force,
(2)
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
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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
(5)
Motion from gravitational force:
In this case, ae = g
(6)
Motion in a centrifugal field:
ae = rw 2
(7)
In this equation, u is the velocity of the particle relative to the fluid and is directed outwardly along aradius.
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 isthe maximum attainable under the circumstances. This maximum settling velocity is called terminalvelocity.
(8)
(9)
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
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(10)
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:
(11)
2.3 Drag coefficient
Drag coefficient is a function of Reynolds number. The drag curve applies only under restrictedconditions:
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:
(12)
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)
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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 ut
2r (17)
(18)
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
(19)
If Stokes’ law is to apply, NRe,p <1.0. Let us introduce a convenient criterion K
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(20)
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 ofnearby 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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Go to Chapter 1
(Characterisation of solid particles)
Go to Chapter 3
(Size Reduction)
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