Mo ch 1_properties of particulate solid_complete_10.12.2020

Properties of
Particulate Solids
Presented by:-
Dhaval N. Yadav
Lecturer (GES CL II),
Chemical Engineering Department,
Shri K. J. Polytechnic, Bharuch
Email : dhaval.nyadav@gmail.com

Topics to be Covered
• Unit operation and Unit process
• Properties of solids and its Calculation :
• Particle density , Bulk Density,
• Sphericity, Equivalent Diameter,
• Specific surface area , Volume surface mean diameter
• Mass mean diameter , Shape Factor
12-10-2020 D. N. Yadav (dhaval.nyadav@gmail.com) 2

Role of Engineer
• Create finished products from naturally/ synthesized raw
materials which can earn money and benefit the society at
large .
• Create the best from the waste.
ex. Extraction of metal from ores (Natural products),
Production of Petrol and Diesel from Crude Oil,
Use of Synthesized Zeolite for water purification,

Chemical Engineering
• The American Institute of Chemical Engineering defines
“Chemical Engineering is the profession in which the knowledge of
mathematics, chemistry and other natural sciences, gained by
study, experience and practise is applied with judgement to
develop economic ways of using materials and energy for the
benefit of mankind.”
• It is defined as branch of science and engineering which is
concerned with the design and operation of industrial chemical
plants .

Role of Chemical Engineer
• Design, develop and engineer the process and equipment
used.
• Selection of raw materials, type of reactor/equipment etc
• Ensure safety of the chemical plant and workers
• Ensure the product quality and maintain standards

Chemical
Industry
Raw Materials
Finished
Products
Physical Changes
Chemical
Changes
Physical and
Chemical Change
Upstream
Processing :
Pre processing of
raw materials
Reaction :
Converting
processed raw
materials into
products
Downstream
Processing :
Post processing
of products

Upstream Processing
• First and primary step in any chemical industry.
• It is concerned with the pre processing of raw materials. It
includes all physical process like:
Separation of raw material according to their size, shape,
quality etc
Preparing (size reduction, washing, drying, mixing filtration
etc.) the raw material to be fed to the chemical plant

Chemical Reaction
• Comes after the upstream processing and before
downstream processing.
• It is this phase in which the processed raw materials are
converted into desired products.
• It includes both physical as well as a chemical change in the
raw material.
Ex. Suplphonation, oxidation, nitration, polymerization etc

Downstream Processing
• Last but very much essential step in any chemical industry.
• It is concerned with the post processing of finished products.
It includes physical process change like :
Separation of product obtained from the by product, un
reacted/excess chemicals, Purification, Drying of obtained
product etc using various operations like distillation,
evaporation, extraction, settling, etc

Unit Operations
• Major physical changes occur which are useful to chemical
industries are known as unit operation.
• Unit operations are to be done to set up the condition to
carry out chemical changes.
Fluid Flow Operations Fluids transportation, filtration, solids
fluidization
Heat Transfer Process Evaporation, condensation
Mass Transfer Process Gas absorption, distillation, extraction,
adsorption, drying

Thermodynamic
processes
Gas liquefaction, refrigeration
Mechanical
processes
Solid transportation, Crushing,
Pulverization, Screening
Combination Mixing
Separation Distillation, Extraction

Unit Process
• Useful chemical transformations with or without physical
changes occurs in the chemical industries are called as unit
process.
Ex. halogenations, oxidation, reduction, alkylation and
acylation etc.

Mechanical Operations
•It is defined as branch of science and engineering
dealing with the study of particles (solids), its
characterization, handling, processing, conversion
and use of solid both in wet and dry form and size
ranging from microns to centimetres.

TERA T 1012 1 00 000 000 000
GIGA G 109 1 000 000 000
MEGA M 106 1000 000
KILO k 103 1 000
HECTA h 102 100
DECA da 101 10
- - 10 1
DECI d 10-1 0.1
CENTI c 10-2 0.01
MILI m 10-3 0.001
MICRO μ 10-6 0.000,001
NANO n 10-9 0.000,000,001
PICO p 10-12 0.000,000,000,001
FAMTO f 10-15 0.000,000,000,000,001
AATO a 10-18 0.000,000,000,000,000,001
+ve
-ve

Mechanical Unit
Operation
Solid-Solid
Operations
Size Reduction,
Screening, Storage,
Transportation etc
Solid-Fluid
Operation
Filtration,
Sedimentation,
Flotation, Cyclone
Separation etc12-10-2020 D. N. Yadav (dhaval.nyadav@gmail.com) 16

Unit Operation
• Most of the unit operations aim at :
Separation of solids from fluid/suspension : Flirtation
Separation of solids from solids on basis of size : Screening
Separation of solids from wide size distribution of solids based on size
by Gravity Settling
Separation of solid from fluid using Centrifugal Force, Solubility etc
Separation of immiscible liquids using Centrifugal Force or Decantation

Properties of Solids
• Size, Shape, Hardness, Fragile
• Particle Density, Bulk Density
• Surface Characteristics (smooth, rough, porous, non porous
etc)
• Sphericity, Equivalent Diameter, Specific surface area ,
Volume surface mean diameter , Mass mean diameter ,
Shape Factor

• Density
It is defined as mass per unit volume.
Homogenous solid mixture have the same density as bulk
material. Particles obtained by size reduction of a composite
material, may have different density from the bulk material.
Ex. Metal extracted from Ore

• Hardness
It is defined as the resistance of solid from breakage/wear
and tear.
• Fragile
It determines how easily a substance can be broken when
exposed to force of impact. Ex. Coal is both soft as well as
fragile, but Gypsum is soft and not fragile.

Characterization of Solid
Particle
Size Shape Density

Particle Size, Shape, Area??
FEED PRODUCT

Particle Size
REGULAR SIZED PARTICLE
IRREGULAR SIZED PARTICLE
?

Particle Size
1. Maximum Linear Dimension
2. Minimum Linear Dimension
3. Bisecting Line Dimension
4. Equivalent Diameter
Best method is representing the particle
in terms of Equivalent Diameter

Particle Size
• Solids when subjected to size reduction yields a wide range
of particle size distribution .
• The term “size” of crushed material or powder or particulate
material is very relative.
FEED PRODUCT

Particle Size
• It is used to classify the crushed material or powder or
particulate material
Particulate
Material
Fine
(screen/mesh no)
Medium
(in between)
Coarse
(cms)

Taken from Lecture 2 Particle Size, Mechanical Unit Operations, Prof
Nanda Kishore ,
IIT, Guwhati

Diameter Definition
Surface Diameter
(ds)
Diameter of sphere having the same surface
area as of a sphere.
Surface Volume
Diameter (dsv)
Diameter of sphere having the same surface
to volume ratio as the particle.
Volume Surface
Diameter (dvs)
Diameter of sphere having the same volume
to surface ratio as the particle.
Volume Diameter
(dv)
Diameter of sphere having the same volume
as the particle.
Mass Diameter
(dm)
Diameter of sphere having the same mass
as the particle.

Steps to calculate the Volume
Equivalent Diameter
STEPS :
1. Write down the formulae for volume of a given particle shape.
2. Write down the formulae for volume of a sphere.
3. Equate both the volumes and determine the volume
equivalent diameter of given particle.

• Calculate the Volume Equivalent Diameter of short
cylinder such that its length is equal to its diameter.
• Solution:
Particle
Shape
Volume of
Particle
Volume of
Sphere
Short
Cylinder
( L =D)
πr2L =
[πD3]/4
[πDp
3]/6
Equivalent
Diameter
(Dp)
1.1447 D

• Calculate the Volume Equivalent Diameter of short cube
such that its length is equal to its diameter.
• Solution:
Particle
Shape
Volume of
Particle
Volume of
Sphere
Equivalent
Diameter
(Dp)
Short cube
( L =D)
D3 [πDp
3]/6 1.2407 D

• Calculate the Volume Equivalent Diameter of
hemisphere.
• Solution:
Particle
Shape
Volume of
Particle
Volume of
Sphere
Equivalent
Diameter
(Dp)
Hemisphere [πD3]/12 [πDp
3]/6 0.7937 D

Calculate the Volume Equivalent Diameter of Octahedron.
• Solution:
Particle
Shape
Volume of
Particle
Volume of
Sphere
Equivalent
Diameter
(Dp)
Octahedron [20.5 x S]/3 [πDp
3]/6 0.9656 S

Calculate the Volume Equivalent Diameter of Rectangular
Prism of sides (a, b, c) such that (b=a), (c=2a).
• Solution:
Particle
Shape
Volume of
Particle
Volume of
Sphere
Equivalent
Diameter
(Dp)
Prism
= (a x b x c)
= 2a3 [πDp
3]/6 1.5631 a

Calculate the Volume Equivalent Diameter of Rasching
Rings such that (L = Do, Di = 0.5Do)
• Solution:
Particle Shape Volume of
Particle
Volume of
Sphere
Equivalent
Diameter (Dp)
Rasching
Rings
[πDp
3]/6 1.0400 Do

Diameters
TakenfromLecture3ParticleSize,MechanicalUnitOperations,Prof NandaKishore,
IIT,Guwhati

Particle Shape
• The shape of individual (each) particle obtained after size
reduction is conveniently expressed in terms of Sphericity
Φs.
• It is defined as ratio of surface area of sphere of same
volume as particle dived by surface area of particle.
• The surface area of sphere is given as πDp
2
• The volume of sphere is given as πDp
3/6

vp is the volume of one particle
Dp is the equivalent diameter of one particle
sp is the surface area of one particle
Diameter of Sphere
of Same Volume
𝝋 =
൙
𝝅𝑫 𝒑
𝟐
𝝅𝑫 𝒑
𝟑
𝟔
ൗ
𝒔 𝒑
𝒗 𝒑

Material Sphericity Material Sphericity
Spheres, cubes,
short cylinders
1.0 Ottawa Sand 0.95
Rasching Rings
L=Do, Di= 0.5Do
L=Do, Di= 0.75Do
0.60
0.33
Rounded Sand 0.83
Berl Saddles 0.3 Coal Dust 0.73
Crushed Glass 0.65 Flint Sand 0.65
Mica Flakes 0.28

Mixed Particle Size and Analysis
Let,
m Total Mass of Sample
N Number of particles in sample size
ρp Density of Particle
Dp Uniform Particle Size Diameter
vp Volume of Individual Particle
N Sample Size

• For a sample size (N) consisting of equal sized particles of
Diameter (Dp):
• The Total Volume of particle is given as :
• Since volume of one particle is vp, The Total Number Of
Particles is calculated as
• If the surface area of one particle is sp, the Total Surface
Area Of Particles in sample size N is calculated as

• Equation (4) represents the total surface area (A) of particles
in a particular sample size (N).
• The Specific surface area (Aw) of all particles in different
sample size i.e. N1, N2, N3...Nn is sum of individual total
surface area(s) i.e. A1, A2, A3...An
: average particle diameter taken as arithmetic average
of smallest and largest particle diameters in increment

Volume Surface Mean Diameter
• The most preferred method for expressing average particle
size. It is given as
• Substituting value of Aw from Equation (6) in (7) we get

Other Mean Diameters
• The Arithmetic Mean Diameter is calculated as
• The Mass Mean Diameter is calculated as
• The Volume Mean Diameter is calculated as

Number of Particles in a Mixture
• For a given sample size, the volume of any particle in a mixture
is directly proportional to the cube of its volume.
• Then using equation (3) the total number of particles in a
mixture is calculated as
a is the Shape Factor

Steps to Calculate the Sphericity of
Particle
STEPS :
1. Write down the formulae for volume of a given particle shape. (vp)
2. Write down the formulae for volume of a sphere.
3. Equate both the volumes and determine the volume equivalent
diameter of given particle (Dp).
4. Write down formulae for surface area of particle (sp).
5. Calculate the sphericity using the following equation

Calculate the sphericity of a given sphere of Diameter (D)
• Solution:
Volume of
Particle (vp)
[πD3]/6 [πD3]/6
Surface Area of
Particle (sp)
πD2 πD2
Volume of Sphere [πDp
3]/6
Equivalent Diameter (Dp) Dp = D
Sphericity (Φs) 1.0

• Calculate the sphericity of a given sphere of short
cylinder such that its length is equal to its diameter.
• Solution:
Volume of
Particle (vp) = [πr2L] = [πD3/4]
Surface Area of
Particle (sp) = 2πr(L+r) = [3πD2/2]
3]/6
Equivalent Diameter (Dp) 1.1447D

• Calculate the sphericity of a given of short cube such that
its length is equal to its diameter.
• Solution:
Volume of
Particle (vp)
[πD3]/4 [πD3]/4
Surface Area of
Particle (sp)
6D2 6D2
3]/6
Equivalent Diameter (Dp) 1.2407 D

• Calculate the sphericity of a given of hemisphere.
Volume of
Particle (vp)
[πDp
3]/12 [πDp
3]/12
Surface Area of
Particle (sp)
3πD2/4 3πD2/4
3]/6
Equivalent Diameter (Dp) 0.7937 D

• Calculate the sphericity of Rectangular Prism of sides (a,
b, c) such that (b=a), (c=2a).
• Solution:
Volume of
Particle (vp)
= (a x b x c) = 2a3
Surface Area of
Particle (sp)
2ab + 2bc + 2ca = 2a3
3]/6
Equivalent Diameter (Dp) 1.5631 a

Calculate the sphericity of Rasching Rings such that (L = Do,
Di = 0.5Do)
Volume of
Particle (vp)
= [πDo
2L/4]-[πDi
2L/4] = 0.1875πDo
3
Surface Area of
Particle (sp)
=[πDiL] + [πDoL] +
2[(πDo
2/4)-(πDi
2/4)]
= 1.875πDo
2
3]/6
Equivalent Diameter (Dp) 1.400 Do

Taken from Lecture 3 Particle Size, Mechanical Unit Operations, Prof
Nanda Kishore ,
IIT, Guwhati12-10-2020 D. N. Yadav (dhaval.nyadav@gmail.com) 53

References
• McCabe, Warren L., and Julian C. Smith. 1967, Unit Operations of
Chemical Engineering, New York: McGraw-Hill.
• Gavhane K. A. 2015, Unit Operations -1, Nirali Prakashan, Pune India.
• Dr. Shabina Khanam , “Mechanical Operations”, NPTEL Chemical
Engineering, 25 Sept 2016, https://nptel.ac.in/courses/103107123/
• Prof Nanda Kishore, “Mechanical Unit Operations”, NPTEL Chemical
Engineering, 27 May 2019,
https://nptel.ac.in/courses/103/103/103103155/

Mo ch 1_properties of particulate solid_complete_10.12.2020

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