Characterization of solid particles

In general, particles are characterized by the following parameters:

Size

Shape

Composition

Density

Particle shape

- Expressed in terms of the sphericity. This is a measure of how close the

shape is to that of the sphere. Thus, the sphericity of a sphere is unity. For a

non spherical particles, the sphericity is defined by:

particleofareaSurface

particleasvolumesameofsphereofareaSurfaceψ

Mathematically,

pp

p

sD

v6ψ

Where DP = equivalent diameter of particles, sometime defined as the

diameter of sphere of equal volume.

VP = volume of particle.

SP = surface area of one particle.

Quiz 1

Calculate the sphericity of a cube of side 5 cm. Repeat the calculations for

cubes of side 8 cm, 12 cm and 20 cm. What can you deduce from your result?

Solution

Volume of cube of side 5 cm = 125 cm3 Now, calculate diameter of a sphere of

same volume, i.e 125 cm

1256

πD3

cm6.2035π

1256D 3

Surface of sphere of same volume as particle = πD2 = π(6.2035) 2

= 120.9 cm2

Surface area of cube = 6(5) 2 = 150 cm2

0.806150

120.9ψ

(ii)

For a cube of side 8cm. Diameter of sphere of same volume = 9.93 cm.

Surface area of sphere = 309.5 cm2 surface area of cube = 6(18) 2 = 384

0.806384

5.309ψ

The results of the other cube size are summarized below:

Side of cube

Diameter of sphere

Surface area of sphere

Surface area of cube

ψ

12 cm 14.89 696.4 864 0.806

20 cm 24.81 1934.4 2400 0.806

Comment: the results show that the sphericity of a cube is always 0.806,

irrespective of the size.

Note that the sphericity can be obtained directly by using the mathematical

relation directly

pp

p

sD

v6ψ

For a cube size 5 cm;

Vp = 53 = 125 cm3

Dp = 6.2035 cm

Sp = 150 cm2

0.806

1502035.6

1256ψ

Quiz 2

Calculate the sphericty of a rectangular block 5 x 10 x 12 cm

Solution

Volume = 5 x 10 x 12 = 600 cm3

Equivalent Diameter of sphere of same volume:

6006

πD3

cm10.4645π

600x6D 3

p

Surface area of sphere = πD2

= π (10.4645)2

= 344. 021

Surface area of rectangular block

= 2 (5X10) + 2(5X12) + 2(10X12)

= 100 + 120 + 240 = 460 cm

750.460

021.344ψ

Note: for a rectangular slab 10 cm by 20 cm by 24 cm, following the same

procedure, the sphericity is calculated to be 0.75 as well. However a slab 8 cm

by 10 cm by 18 cm, ψ = 0.76, thus, the sphericity is not necessarily the same

when the dimensions of a rectangular slab change except when all dimensions

change by the same factor.

Characteristics continued

Voidage: the fraction of the total volume which is made up of free space

between the particles and is filled with fluid. i.e fraction of the volume of bed

not occupied by solid material sometimes called porosity or fractional voidage.

Why is it important?

Determines the bulk density of the material and hence the volume taken

up by a given mass.

Affects the tendency for agglomeration of the particles.

Influences the resistance which the material offers to the percolation of

fluid through it eg in filtration.

Therefore fractional volume of the bed occupied by solid material is (i-e)

Specific surface area of a particle

Defined as the surface area of the particle divided by its volume

Example: A cube of side a

Surface area = 6a2

Volume = a3

So specific surface = aa

a 663

2

For a sphere,

d

6

6

πd

πdS

3

2

For a rectangular solid of sides a, b, c:

Surface area = 2ab + 2ac + 2bc = 2(ab + ac + bc)

Volume = abc

Specific surface area = 2(ab + ac + bc)/abc.

Specific surface area of a bed .

-Surface area presented to the fluid per unit volume of the bed when the

particles are packed in a bed.

If point contact occurs between particles such that only a small fraction of a

surface area is lost by overlapping,

SB = S (1-ε)

Importance of porosity continued

Influences

Pressure drop for flow through the phase.

Electrical resistivity of the phase.

Effective thermal conductivity.

Reactive surface area.

Porosity of a static bed depends on the following

Particle size and size distribution.

Particle shape.

Surface roughness.

Method of packing.

Size of container relative to the particle diameter.

Usually particulate – solid phase must be dealt with rather than the particles

that exist in it. The phase may be:

Stationary bed (static bed).

Fluidized bed.

Fog.

Example

Determine the voidage of a close packing of 8 spheres of diameter a

Volume of each sphere 6

πa 3

Volume of the 8 sphere 6

πa8 3

Volume of cube = (2a)3 = 8a3

Fraction of solid is 3

3

a8

6πa8

= 5238.06

π

Void fraction = 1 - 0.5238

= 0.476 = 0.48

Particle size distribution

Particulate systems usually consist of particles of a wide range of sizes.

Therefore a quantitative indication of the mean size and the spread of sizes are

required.

Cumulative mass fraction curve

Proportion of particles x smaller than a certain size (d) is plotted against that

size d.

1

dx

dd

O d

Size frequency curve

Slope of cumulative curve is plotted against particle size (d)

dy

dx

d

Maximum of the curve shows the most frequency of occurring size.

Mean size based on volume

3

11

4

11

dn

dndv

Alternatively

33

11 dx

1dv

Mean diameter based on length

11

2

11

ldn

dnd

Example

The size distribution of crushed rock is given below. Convert these data to

obtain the distribution on mass basis and calculate

(a) The mean size based on volume.

(b) The mean diameter based on length.

Size range (mm) No. of particle in range

0 – 2 2000

2 – 4 600

4 – 8 140

8 – 12 40

12 – 16 15

16 – 20 5

20 – 24 2

From the data given, the following table is readily generated.

Di Ni Nidi nidi2 nidi3 nidi4

1 2000 2000 2000 2000 2000

3 600 1800 5400 16200 48600

6 140 840 5040 30240 181440

10 40 400 4000 40000 400000

14 15 210 2940 41160 576240

18 5 90 1620 29160 524880

22 2 44 968 21296 468512

5384 21968 180056 2201672

(a) Mean size based on volume mm12.23180056

2201672

dn

dndv

3

11

4

11

(b) mean size based on length is

mm4.085384

21968

dn

dnd

ii

2

ii

1

Projected Area

Depends on

Size of the solid.

Geometric form of sphere, cylinder.

Orientation with respect to the direction of flow of the fluid.

Example

(1) Cylinder so oriented that its axis is perpendicular to flow direction

AP = LD

L = Length of cylinder

D = Diameter

(2) Cylinder with axis parallel to direction of flow

2

P D4

πA

(3) For a sphere, orientation does not play a part. In all case:

2

P D4

πA

Where D = diameter of the sphere.

Just as the skin friction coefficient is defined as the ratio between shear stress

and the kinetic energy of the flow:

2

212

21 u

AF

uf

A drag coefficient is defined as

P

DP

D

DA

F

u

AF

C2

0212

021

Rearranging,

PDD AuCF2

021

Thus, the force acting on a body can be determined if the drag coefficient is

known.

In general, CD = f (Reρ)

From dimensional analysis,

The Reynolds number is defined as

μ

DGRe Po

Go = Uo ρ

Dp = characteristics length

CD versus Reynolds no charts are available in the literature for various shapes

sphere

Drag coeff

cylinder

μ

ρuDRe oP

Note: variation of CD vs Re is more complicated than variation of f vs Re due to

the complex nature of drag.

Sphere

For low Reynolds numbers, the drag force can be predicted using stokes law

which is

FD = 3 π µ uo DP

Substituting equation (1), the drag coefficient predicted by stokes law is:

pRe

24CD

FILTRATION

Definition: A separation process in which solids from a suspension in a liquid

are removed using a porous medium or screen which retains the solids and

allows the liquid to pass through. Note that the challenges of filtration on an

industrial scale are quite different from Laboratory filtration.

Selection of equipment and operating conditions depend on the following

factors:

Fluid properties, for example, its density, viscosity as well as corrosive

properties.

Particle characteristics such as size, shape, size distribution

Concentration of solids in the suspension

Quantity of material to be handled and its value

The value of the solid or liquid or both

If a slight contamination by contact of the suspension or filtrate with the

various components of the equipment can be tolerated, e.g.

pharmaceutical products.

If any form of pretreatment may be required.

Operating Principles of a typical Filtration system:

1. Slurry (liquid containing the solids in suspension) is passed through the

porous filter medium.

2. A filter cake gradually builds up so that the resistance to flow

progressively increases

3. The initial layers form the effective filter medium in one operational

procedure referred to as cake filtration

4. In the other type, called deep-bed filtration, particles penetrate into the

pores of the medium. Impacts between the particles and the surface of

the medium are for the most part responsible for their removal and

retention.

In general, the rate of filtration depends on the following factors:

The pressure drop between the feed side and the other side of the

filter medium

The area of the filtering surface

The viscosity of the filtrate

The resistance of the filter cake

The choice of the filter medium (septum) for a given filtration system is

critical to ensure an efficient operation. The septum must meet the

following requirements:

It must retain the solids to be filtered, giving a clear filtrate

It must not plug or blind

It must be resistant chemically and strong enough physically to

withstand the process conditions

It must permit the cake formed to discharge cleanly and completely

It must not be prohibitively expensive.

FILTRATION THEORY

Filtration is essentially a special case of flow through porous media. The

difference is that the resistance to flow increases with time as filtration

progresses. Thus, equations such as the Kozeny-Carman equation for flow

through packed beds need to be modified in order to be applicable here. With

time, either the flow rate decreases or the pressure drop will have to increase

if the flow rate is to remain constant. Two types of operation are applicable:

1. Constant-pressure filtration: The pressure drop is held constant and the

flow rate is allowed to decrease.

2. Constant-rate filtration: The pressure drop is progressively increased to

maintain a constant flow rate.

The overall pressure drop at any time is the sum of the pressure drops over the

filter medium and cake. i.e.

∆p = ∆pc + ∆pm

Where ∆p = overall pressure drop

∆pc = pressure drop over cake

∆pm = pressure drop over filter medium

Recall the Kozeny-Carman equation:

32

21150

ps D

u

dL

dP

Here, u is the superficial velocity of the filtrate and dP/dL is used in place of

∆P/L. Expressing the sphericity φs in terms of the surface-volume ratio, the

equation becomes:

3

22/117.4

pp vsu

dL

dP [1]

Now, the linear velocity u can be expressed as:

A

dtdVu [2]

Where V is the volume of filtrate collected from the start of the filtration to

time t. Also, the mass dm of solids in the layer is

dm = ρp(1-ε)AdL [3]

Elimination of dL between equations (1) and (3) gives:

dmA

vsukdP

p

pp

3

2

1 /1

Integration gives

3

2

1 1

A

mvsukP

p

cpp

c

[4]

where ∆Pc = pressure drop over cake

mc = total mass of solids in the cake.

Defining a specific cake resistance as

3

2

1 1

p

pp vsk [5]

Then, A

umP c

c

[6]

By analogy, a filter medium resistance can also be defined as

u

PR m

m

[7]

The total pressure drop is the sum of the pressure drops across the filter cake

and filter medium. So, combining equations (6) and (7),

m

c

mc RA

muPPP

[8]

mc = Vc, V being the total volume of filtrate collected to time t and c the mass

of the particles deposited in the filter per unit volume of filtrate. The linear

velocity is related to V (see equation ). Thus, u and mc can be replaced by

quantities that are easily measurable. That is,

mR

A

cV

PAdV

dt

Constant Pressure Filtration

At constant pressure, the only variables are V and t, ∆P is constant and at the

onset of filtration, i.e. at t = 0, V = 0 and ∆P is equal to ∆Pm.

00

1

qdV

dt

PA

Rm

[9]

At time t, equation (9) becomes

0

11

qVK

qdV

dtc

where PA

cKc

2

Integration between the limits (0, 0) and (t, V) gives

0

1

2 qV

K

V

t c

[10]

A plot of t/V versus V gives a straight line whose slope is Kc/2 and intercept of

1/q0, from which the parameters α and Rm can be determined.

FLOCCULATION

The tendency of the particulate phase of colloidal dispersions to aggregate

(come together) is an important physical property since it influences rates in

separation processes such as sedimentation and filtration. The aggregation of

colloids is known as flocculation (or coagulation).

Sometimes, the particles that make up a suspension are primarily individual

particles and the particle sizes are so low that for all practical purposes, gravity

settling is not possible. That is, the gravity settling rates are too low for the

separation process. Agglomeration or clustering of the particles will improve

the settling rates. Flocculation is the addition of certain chemicals or materials

that facilitate or induce the agglomeration of fine particles in a suspension by

reducing the repulsive forces between the charged particles to achieve the said

objective. Flocculation agents may be cationic, anionic, or non-ionic in

character. Other flocculants include lime, alumina or sodium silicate. These

materials form loose clusters that carry the fines down with them. The size,

shape, and effective density of the flocs are not well defined. Thus, settling

rates cannot be predicted easily.

Commonly occurring attractive and repulsive forces between colloidal particles

are

Van der Waals forces

Electrostatic forces (e.g. Repulsive interaction between similarly charged

double-layers)

Forces due to adsorbed macromolecules.

FLOW OF FLUIDS THROUGH PACKED BEDS

Liquids or gases flow through beds of solid particles. Applications are found in:

Filtration

Reaction through Catalyst Beds (either fixed or fluidized)

In absorption we have a two-phase counter current flow of liquid and

gas through packed towers.

Ion exchange

An Important parameter is the pressure drop across the bed. The starting point

for the estimation is the Hagen-Poiseulle equation for flow of fluids through a

cylindrical tube. For a packed bed, it can be assumed that the bed has straight

cylindrical tubes imbedded in it to account for the void space. In reality, we

know that the void space is actually a network of highly complicated,

interwoven, tortuous paths. The assumption is still in order to make the

analysis easier. At the end of the analysis, a correction factor is included in the

correlation with the aid of actual experimental data.

Pressure drop through the bed is given by the Kozeny-carman equation,

obtained by following the procedure outlined above:

2........................................................ε

ε1

DΦ

μV150

L

ΔP3

2

2

P

2

s

o

Where oV = superficial velocity or empty tower velocity given by:

oV = ε V , V being the average velocity in the channels. This equation is

applicable up to Re = 1.0, I.e low flow rates and the flow is laminar.

µ = fluid viscosity

ε = porosity

φs = sphericity of particle

Dp = Particulate diameter

At high Reynolds number, Re > 1,000 an empirical equation is applicable:

3.................................................ε

ε1

DΦ

Vρ1.75

L

ΔP3

PP

2o

Equation (3) is called the Burker – Plummer equation:

For the entire range of flow rates, a single empirical correlation can be

obtained by combining equations (2) and (3):

)4......(..............................ε

ε1

DΦ

Vρ1.75

ε

ε1

DΦ

μV150

L

ΔP3

Ps

2O

3

2

2

P

2

s

O

Equation (4) is called the Ergun equation.

At low Reynolds number ≤ 1.0 i.e laminar flow, the first part of equation (4)

dominates and ∆P/L is inversely proportional to DP2. For turbulent flow, ∆P/L is

inversely proportional to Dp.

FLUIDIZATION

When a fluid is passed upwards through a bed and the velocity is progressively

increased, the frictional drag on the particles becomes equal to the apparent

weight (i.e. the actual weight of the particles minus the buoyancy) at some

point. The particles offer less resistance to flow at this point as they become

re-arranged and expand. Consequently, the bed voidage increases. With

further increase in velocity, the individual particles separate from one another

and become freely suspended in the fluid. The bed is then said to be fluidized.

Applications of Fluidization

Industrial applications of fluidization include the following:

Catalytic processes such as cracking in petroleum refining, synthesis of

acrylonitrile, including catalyst regeneration

Solid-gas reactions in general

Fluidized bed combustion of coal

Roasting ores

Drying of fine solids

Gas adsorption

Pneumatic transport

Advantages of using fluidized beds are

Good mixing ensures absence of temperature gradients in reactors even

with highly exothermic or endothermic reactions.

High heat transfer rates to either the walls of the reactor or cooling

tubes immersed in the bed are guaranteed due to the high agitation of

the solids.

Ease of transportation of solids from one vessel to another, for example

for catalyst regeneration and return, due to the fluidity of the solids.

Disadvantages include:

Uneven contacting of gas and solid

Erosion of vessel’s internal parts

Attrition (Break-up) of the solids leading to loss of fines

Minimum Fluidization Velocity

Carrying out a force balance, the pressure drop across the bed is equal to the

weight of the bed per unit cross-sectional area, less the buoyant force of the

displaced fluid. That is,

∆P = g(1 – ε)(ρP – ρ)L or,

PMgL

P1 [1]

Recall the Ergun equation for pressure drop in packed beds:

3

2

0

3

2

22

0 175.11150

PSPSD

V

D

V

L

P

[2]

At the point of incipient fluidization, ε = εM, MVV 00

and Equation (1)

combined with Equation (2), after re-arrangement becomes

P

MPS

OM

M

M

PS

OMg

D

V

D

V3

2_

322

75.11150 [3]

For very small particles, the laminar contribution of the Ergun equation is

predominant and the equation for minimum fluidization velocity reduces to

22

3_

1150PS

M

MPOM D

gV

[4]

This correlation is applicable at low Reynolds number (ReP < 1.0). At Re > 1000,

the turbulent contribution in the Ergun equation becomes predominant and

the equation becomes

2

13_

75.1

MPPSOM

gDV [5]

Types of Fluidization

1. Particulate fluidization: characterized by a large but uniform expansion

of the bed at high velocities e.g. when sand is fluidized by water at high

velocities

2. Aggregative or Bubbling fluidization: usually occurs when the fluidizing

fluid is air. Two distinct phases are identifiable when this phenomenon

occurs, a bubble phase and a dense bed of suspended particles. Most of

the gas passes through the bed as bubbles while a small fraction of the

gas flows in the channels between the particles.

3. Turbulent fluidization: With further increase in gas velocity, a point is

reached when the bed expands so much that there is no longer a

dispersed bubble phase. Also called fast fluidization.

MOTION OF PARTICLES THROUGH FLUIDS

1. Forces acting on a particle moving vertically down in a fluid are depicted

below:

FD FB FE

FE = external force due to force of gravity or a centrifugal force field

FE = mae , where ae is the acceleration, and m is the mass of the

particle.

Volume of particle = Pρ

m

Buoyant Force = mass of fluid displaced by the particle X acceleration

from the external force [Archimedes Principle]. Since volume of fluid

displaced is same as volume of mass.

Mass of fluid displaced = Pρ

mρ

Therefore eaρ

mρF

P

B

Drag force FD is given by

2

ρACF P

2

oD

D

u

Carrying out a force balance on the particle:

Resultant force on Particle = FE – FB – FD

Acceleration of Particle dt

du .

So,

.2

ρACa

ρ

mρma

dt

dum P

2

oD

e

P

e

u

m2

ρAC

ρ

ρaa

dt

du P

2

D

P

e

e

u

m2

ρAC

ρ

ρρa P

2

D

P

Pe

u

For a gravitational force field, ae = g. Therefore

)1.....(..............................m2

ρAC

ρ

ρρg

dt

du P

2

D

P

P u

Terminal velocity

At the point du/dt (acceleration) is zero, the particle assumes a constant

velocity called the terminal velocity. Rearranging equ (1), after setting

du/dt = 0,

)2....(..................................................ρCρA

mρρg2

DPP

Pt

u

For a centrifugal force field:

)3..(..................................................ρCρA

mρρ2w

DPP

Pt

ru

Special case of spherical particles:

For a spherical particles of diameter DP:

P

3

P ρDπ6

1m

2

PP Dπ4

1A

Substituting

)4........(..................................................ρC3

Dρρg4

D

PPt

u

At low Re,

Re

24CD

FD = 3 π µ ut DP

μ18

ρρDg P

2

Pt

u

For 1000 < Re < 200,000

CD ≈ 0.44

FD = 0.055 π DP 2 ut

2 ρ

and

ρ

ρg75.1 P

t

PDu ………………………………………………(5)

Equation (5) is called Newton’s law and is applicable only for fairly large

particles falling in gasses or low viscosity fluids.

Note: In the stokes law regime, the terminal velocity is directly proportional to

diameter squared.

ut α DP2

In the Newton’s law regime, the terminal velocity is directly proportional to the

square root of diameter.

Pt Du

SIZE REDUCTION

Size reduction is required in industrial operations whenever the size of a

material is greater than the size for the most efficient use of the material. A

raw material, for example may be obtained from a quarry in large chunks. For

further processing, there may be a need to crush it to smaller particles or even

to a fine powder. The size reduction may be carried out in stages using

completely different machines. The choice of machine for each stage depends

on:

Size of the feed

Size of the product

Compressive strength of the material

Brittleness and stickiness of the material

Reasons for size reduction include

1. Increase surface area since in most reactions involving solid particles,

rate of reation is directly proportional to area of contact with a second

phase.

2. In leaching, weight of extraction is increased because of increased area

of contact as well as reduced distance solvent has to penetrate into the

particle.

3. In drying of porous solids, size reduction also causes both an increase in

area and a reduction in the distance the moisture must travel within the

particle in order to reach the surface.

4. It may be necessary to break a material into very small particles in order

to separate two constituents.

5. Chemical reactivity of fine particles is greater than that of coarse

particles.

6. Colour and covering power of pigments is greatly affected by the size of

the particles.

7. More intimate mixing of solids can be achieved if the particle size is

small.

There are situations when the reverse, that is, size enlargement is required.

Drugs are usually produced in tablet form from powder. For ease of

transportation and elimination of excessive dust, some powders are

pelleted which entails an increase in size. In this course, however, we will

only concern ourselves with size reduction.

ENERGY REQUIREMENTS FOR SIZE REDUCTION

Size reduction is a very inefficient process energy wise in the sense that a large

amount of energy is expended in overcoming friction in moving parts of

machinery and other operational requirements. In addition to frictional losses

in the plant, energy is utilized in the following ways:

Producing elastic deformation of the particles before fracture occurs

Producing inelastic deformation which results in size reduction

Causing elastic distortion of the equipment

Friction between particles and between particles and the machine

Noise, heat and vibration in the plant

It is estimated that only about 10 percent of the total power is usefully

employed.

Some empirical relations are in use for estimation of the energy requirements.

The starting point is the basic energy relationship being the energy dE required

to effect a small change dL in the size of unit mass of material is a simple

power function of the size. i.e.

1............................................................cL

dL

dE P

Riltinger assumed p = - 2. Integration gives

2....................................................

L

1

L

1CE

12

Putting C = KRfC where fC is the crushing strength of the material,

3............................................L

1

L

1fKE

12

cR

Equation (3) is Riltinger’s law

Kick assumed p = -1 and following the same procedure obtained

2

1cK

L

LlnfKE

Bond assumed an intermediate value of P = -3/2 and obtained

21

21

21

q

11

L

1C2

L

1

L

1C2E

2

12

Where q = 2

1

LL

EXAMPLE

A material is crushed in a Blake Jaw crusher such that the average size of

particles is reduced from 80 mm to 20 mm with the consumption of energy of

15.0 kW/(kg/s). What would be the consumption of energy needed to crush

the same material of average size 100 mm to an average size of 30 mm:

(a) Assuming Rittinger’s law applies

(b) Assuming Kick’s law applies?

Which of these results would be more accurate and why?

Data given:

L1 = 80 mm

L2 = 20 mm

E = 15.0 kW/(kgs)

To determine energy consumption needed to crush the same material of

average size 100 mm to an average size of 30 mm.

(a) Applying Riltinger’s law:

12

cRL

1

L

1fKE

80

1

20

1fK15.0 cR

KR FC = (15.0)(80/3) = 400 KW/(Kg.mm)

So, energy required to crush same material from 100 mm to 25 is:

100

1

40

1400E

(b) Applying Kick’s Law:

2

1cK

L

LlnfKE

Substituting values:

20

80lnfK0.15 cK

kg/sKW/10.821.386

15.0fK cK

Energy required to crush same material from 100 mm to 25 mm is

KJ/kg9.9140

100ln10.82E

Because the size range in this problem can be considered coarse grinding,

kick’s law is likely to give a more accurate result since experimental data has

shown that Kick’s law is more applicable for coarse grinding.