3 size reduction

SIZE REDUCTION
Department of Chemical Engineering
University of Engineering & Technology Peshawar

Introduction
• The size reduction of solids is an energy intensive and
highly inefficient process
• 5% of all electricity generated is used in size reduction
• Based on energy required for the creation of new
surfaces, the industrial scale processes is generally less
than 1% efficient

Introduction
Raw materials often occur in sizes that are too large to be used
and, therefore, they must be reduced in size
The term size reduction is applied to all the ways in which
particles of solids are cut or broken into smaller pieces
The reason that size reduction or comminution is usually carried
out is to increase the surface area of the material
comminution is a generic term for size reduction
crushers and grinders are type of comminuting equipment

Benefits of Size Reduction
maximize the area of solid in contact with the
liquid or gas phase around it
which enhances:
• Reactivity
• Dissolution
• Catalytic effects
• Permits separation of unwanted ingredients by
mechanical methods
• Easy handling

Mechanisms of Size Reduction
There are four mechanisms by which size
reduction may be achieved:
• Impact
• Attrition (rubbing)
• Compression
• Cutting

Impact
(Mechanisms of Size Reduction)
In crushing terminology, impact refers to the
sharp, instantaneous collision of one moving
object against another
Gravity Impact
Dynamic Impact
Coarse, Medium or Fine Products

Attrition
Attrition is a term applied to the reduction of materials by
scrubbing it between two hard surfaces
consumes more power and exacts heavier wear on hammers and
screen bars
practical for crushing the less abrasive materials such as pure
limestone and coal
most useful in the following circumstances:
• when material is friable or not too abrasive
• when a closed-circuit system is not desirable to control top size

Compression
crushing by compression is done between two
surfaces, with the work being done by one or
both surfaces
• if the material is hard and tough
• if the material is abrasive
• if the material is not sticky
• where the finished product is to be relatively
coarse, i.e., 11/2" (38mm) or larger top size
• when the material will break cubically

Cutting (Shear)
Shear/cutting consists of a trimming or cleaving (chop, slice, cut etc)
action rather than the rubbing action associated with attrition
Cutting gives a definite particle size and some time a definite shape,
with few or no fines
Shear is usually combined with other methods. For example, single-roll
crushers employ shear together with impact and compression
• when material is somewhat friable and has a relatively low silica
content
• for primary crushing with a reduction ratio of 6 to 1
• when a relatively coarse product is desired, usually larger than
11/2" (38mm) top size

Ideal Crusher/Grinder
• Have large Capacity
• Require a small power input per unit of
product
• Yield a product of single size or the size
distribution required

Crushing Efficiency, ƞc
The ratio of the surface energy created by
crushing to the energy absorbed by the solid
n
wawbs
c
W
AAe )( 

waA
nW
se : Surface energy per unit area
wbA : Product area per unit mass
: Feed area per unit mass
: Energy absorbed by a unit mass of material
c
wawbs
n
AAe
W

)( 
=

Mechanical Efficiency, ƞm
The ratio of the energy absorbed to the energy input
W
Wn
m 
c
wawbs
n
AAe
W

)( 

c
wawbs
m
W
AAe


)( 

cm
wawbs
m
n AAeW
W

)( 


Power Requirements
If is the feed rate , the power required by
the machine is:
m
mWP 
As volume-surface mean diameter is defined as:
PWS
S
A
D


6
SD










SaaSbbPmC
S
DD
em
P
116



Empirical Relationships for Crushing
• it is impossible to predict from any theory the energy
consumed in size reduction
• there are a number of empirical rules which allow
data from one process to be extrapolated to another
• All are based on the premise that the energy dE
required to affect a small change in size; dL for unit
mass of solids is a simple power function of the size
i.e.:
P
CL
dL
dE


Rittinger’s law (1867)
The energy required for size reduction is
directly proportional to the change in surface
area (new surface created)
This leads to a value of –2 for P
where E is the energy required per unit mass of solid
Crushing efficiency is constant







2
1
1
1
LL
CE

Rittinger’s law
Continue…..
Writing C = KRfc
Where
fc is the crushing strength of the material in N/m2
And
KR is Rittinger's constant for the material gives
KR is not dimensionless







21
11
LL
fKE CR

Kick’s law(1885)
the energy required to reduce a material in size is
directly proportional to the size reduction ratio dL/L,
that is the ratio of the initial particle size to the finial
particle size
This implies that P equal to -1.
If P = -1 then
2
1
ln
L
L
CE 
putting C = KKfc gives:
2
1
ln
L
L
fKE cK
KK is not dimensionless

Applications
Kick's law is more appropriate to coarse crushing
for the grinding of coarse particles in which the
increase in surface area per unit mass is relatively
small
Rittinger's law is more appropriate to fine grinding
the size reduction of fine powders, in which large
areas of new surface are being created,
Rittinger's Law fits the experimental data better

Bond crushing law(1952) and work index
(intermediate relationship with P = -3/2 = -1.5)
The work required to form particles of size Dp
from very large feed is proportional to the
square root of the surface-to-volume ratio of
the product, sp/vp.









12
11
2
LL
CE









qL
EE i
1
1
100
2
where q = L1/L2
C = 5Ei
and

Ei the work index
The work index, Ei, is defined as the gross energy
required in KWH per ton of feed to reduce a very
large feed to such a size that 80% of the product
passes a 100 µm screen
C = 5Ei









qL
EE i
1
1
100
2
In all these equations the particle sizes are defined as the
size of square hole through which 80% of the material will
pass.

Factors Influencing Choice of Size
Reduction Equipment
1. Feed and Product Size
Feed Size Product Size
Coarse Crushers 1500 - 40mm 50 - 5mm
Intermediate Crushers 50 - 5mm 5 - 0.1mm
Fine Crushers. (Grinders) 5 - 2mm <0.1mm
Fine Milling <0.2mm down to 0.01m

Factors Influencing Choice of Size
Reduction Equipment
2. Nature of Material
• Hardness - very hard materials are better in low speed or
low contact machines
• Structure - fibrous materials need tearing or cutting
action
• Moisture content - materials with 5 - 50% moisture do
not flow easily and can be difficult to process
• Stickiness - sticky materials need easily cleaned machines
• Soapiness - if coefficient of friction is low crushing may
be difficult
• Explosives - need inert atmosphere
• Hazardous to health - need good confinement

CRUSHER SELECTION CRITERIA
When selecting a crusher, the following criteria must be
considered:
• Will it produce desired output size and shape at the required
capacity?
• Will it accept the largest input size expected?
• What is its capacity?
• Will it choke or plug?
• Can it pass uncrushable debris without damage to the crusher?
• How much supervision of the unit is necessary?
• Will it meet product specifications without additional crushing
stages and auxiliary equipment?
• What is the crusher’s power demand per ton per hour of
finished product?
• How does it resist abrasive wear?

Crusher Selection Criteria
• Does it operate economically with minimum
maintenance?
• Does it offer dependable and prolonged service
life?
• Is there ready availability of replacement parts?
• Does it have acceptable parts replacement cost?
• Does it have easy access to internal parts?
• Is the crusher versatile?
• How does the initial cost of the machine compare
with its long term operating costs?
• Is experienced factory service readily available?

Types of Size Reduction Equipment
• A. Crushers (coarse and fine)
• B. Grinders (intermediate and fine)
• C. Ultrafine grinders
• D. Cutting machines

Crushers (coarse and fine)
Crushers are slow speed machines for coarse
reduction of large quantities of solids
• Crusher do the heavy work of breaking large pieces
of solid material into small lumps
• A primary crusher operates on run-of -mine material
accepting anything that comes from mine face and
breaking it into 150 to 250 mm lumps
• A secondary crusher reduces these lumps into
particles perhaps 6mm in size

– Jaw crushers
– Gyratory crushers
– Crushing rolls
• Single Roll Crushers
• Double Roll Crusher
Crushers

B. Grinders (intermediate and fine)
reduce crushed feed to powder
• The product from an intermediate grinder
might pass a 40-mesh screen
• most of the product from a fine grinder would
pass a 200-mesh screen with a 74 µm opening

–Hammer mills; impactors
–Rolling-compression mills
–Attrition mills
–Tumbling mills
Grinders (intermediate and fine)

An ultrafine grinder accepts feed particles no
larger than 6 mm and the product size is typically
1 to 5 µm
– Hammer mills with internal classification
– Fluid-energy mills
– Agitated mills
Ultrafine grinders

Cutters give particles of definite size and
shape, 2 to 10mm in length
–Knife cutters
• Dicers
• slitters
Cutting machines

Jaw Crusher
• The swinging jaw makes an angle of 200 to 300 with the fixed jaw
• Feed opening may be up to 2.5m x 2.0m
• processing up to 1200 t/h
• Product size is adjusted by adjusting the gap size
• crushing is done by compression
• 250 to 400 strokes per minute
• accept feed sizes up to 48" (1200mm)
• product size as small as 3/4" (19mm)

Gyratory Crusher
• the head is carried on a heavy shaft pivoted at the
top of the machine
• An eccentric drives the bottom end of the shaft
• Size is controlled by raising and lowering the cone
• The speed of the crushing head is typically 125 to
425 gyrations per minute
• the discharge from a gyratory crusher is
continuous

Crushing Rolls
(Single Roll Crushers)
• typically used as primary crushers
• impact, shear and compression
• there are no screen bars
• the clearance between the breaker plate and the roll
determines the product size
• Applications include
petroleum coke,
coal with rock, coal,
aggregate, limestone,
chemicals, phosphate rock,
shale and many other materials

Crushing Rolls
(Double Roll Crushers)

Smooth roll crusher
(Roll Design)
The limiting size Dp,max of particles that can be
nipped by the rolls, can be estimated from the
simple relation:
Dp,max = 0.08R + d
Where:
R = roll radius
d = half the width of the gap between the rolls
The maximum size of the product is approx. 2d

Smooth roll crusher
(Roll Design)
Fµ Fµ cos θ
F sin θ
r
R F
2d
For a particular operation where the ore/feed size is known, it is
necessary to estimate the diameter of rolls required for a specific
degree of size reduction. To estimate the roll diameter it is
convenient to assume that the particle to be crushed is spherical
and roll surfaces are smooth.
The nip angle is defined as the
angle that is tangent to the roll
surface at the points of contact
between the rolls and the particle.
It depends on the surface
characteristics of the rolls.
Usually the nip angle is between 20° and 30° but in some large
roll crushers it is up to 40°
The nip angle

Smooth roll crusher
(Roll Design)
From simple geometry it can be seen that for a particle of size r,
nipped between two rolls of radius R:
Fµ Fµ cos θ
F sin θ
r
R F
2d
rR
dR
Cos



Figure shows a spherical particle
about to enter the crushing zone of
a roll crusher and is about to be
nipped. For rolls that have equal
radius and length, tangents drawn
at the point of contact of the
particle and the two rolls meet to
form the nip angle
1cos
cos




rd
R

Smooth roll crusher
(Roll Design)
Figure indicates that to estimate
the radius R of the roll, the nip
angle is required. The nip angle
on its part will depend on the
coefficient of friction’µ’, between
the roll surface and the particle
surface
Fµ Fµ cos θ
F sin θ
r
R F
2d
To estimate µ consider a compressive force, F, exerted by the rolls
on the particle just prior to crushing, operating normal to the roll
surface, at the point of contact, and the frictional force between
the roll and particle acting along a tangent to the roll surface at the
point of contact. The frictional force is a function of the
compressive force F and is given by the expression, Fµ

Smooth roll crusher
(Roll Design)
If the vertical components of
these forces considered, and
neglect the force due to gravity,
then it can be seen that at the
point of contact for the particle to
be just nipped by the rolls, the
equilibrium conditions apply
where:
Fµ Fµ cos θ
F sin θ
r
R F
2d cossin FF 
 tan
 1
tan

The friction coefficient is roughly between 0.20 and 0.30 the nip
angle has a value of about 11°-17°

Tumbling Mills
A cylindrical shell slowly turning about a horizontal axis
and filled to about half its volume with a solid grinding
medium forms a tumbling mill
The shell is usually steel, lined with high-carbon steel plate,
porcelain, silica rock, or rubber
The grinding medium is metal rods in a rod mill, lengths of
chain or balls of metal, rubber, or wood in a ball mill, flint
pebbles or porcelain or zircon spheres in a pebble mill
For intermediate and fine reduction of abrasive materials
tumbling mills are unequaled

Tumbling mills
Ball Mill
In its simplest form, the ball mill
consists of a rotating hollow cylinder,
partially filled with balls, with its axis
either horizontal or at a small angle to
the horizontal.
The inner surface of the cylinder is
usually lined with an abrasion
resistant material such as manganese
steel, stoneware or rubber
the ratio of length to the diameter
is usually 1 or 1.5 : 1

Tumbling mills
(Tube mill)
The characteristics of the two mills are
similar although the material remains
longer in the tube mill because of its
greater length, and a finer product is
therefore obtained.
The tube mill is similar to the ball mill in
construction and operation, although
the ratio of length to the diameter is
usually 3 or 4 : 1, as compared with 1 or
1.5 : 1 for the ball mill.

Tumbling mills
(Rod mill)
 In a rod mill, much of the reduction is done by rolling
compression and by attrition as the rods slide downward and
roll over one another.
 The grinding rods are usually steel, 25 to 125 mm (1 to 5 in.)
in diameter, with several sizes present at all times in any
given mill.
 The rods extend the full length of the mill.
 Rod mills are intermediate grinders, reducing a 20-mm (3/4
in.) feed to perhaps I0-mesh, often preparing the product
from a crusher for final reduction in a ball mill.

The load of balls in a ball or tube mill is normally such that
when the mill is stopped, the balls occupy about one-half the
volume of the mill.
The void fraction in the mass of balls, when at rest, is typically
0.40.
When the mill is rotated, the balls are picked up by the mill
wall and carried nearly to the top, where they break contact
with the wall and fall to the bottom to be picked up again.
Centrifugal force keeps the balls in contact with the wall and
with each other during the upward movement.
While in contact with the wall, the balls do some grinding by
slipping and roIling over each other, but most of the grinding
occurs at the zone of impact, where the free-faIling balls strike
the bottom of the mill.
Action in tumbling mills

Forces on ball in ball mill
If the speed is too high, however, the balls
are carried over and the mill is said to be
centrifuging.
The speed at which centrifuging occurs is
called the critical speed.
Little or no grinding is done when a mill is
centrifuging, and operating speeds must be
less than the critical.
The speed at which the outermost balls
lose contact with the wall of the mill
depends on the balance between
gravitational and centrifugal forces
Consider the ball at point A on the periphery of the mill.
Let the radii of the mill and of the ball be R and r, respectively.
The center of the ball is, then R - r meters (or feet) from the axis of the mill.

Let the radius AO form the angle α with the
vertical. Two forces act on the ball.
The first is the force of gravity:
mg/gC
where m is the mass of the ball.
The second is the centrifugal force:
(R - r) ω2/gC
where ω = 2πn and n is the rotational speed.
The centripetal component of the force of
gravity is:
(mg/gC) cos α
this force opposes the centrifugal force. As long as the centrifugal force exceeds
the centripetal force, the particle will not break contact with the wall.

As the angle α decreases, however, the centripetal force increases, and unless
the speed exceeds the critical, a point is reached where the opposing forces are
equal and the particle is ready to fall away.
The angle at which this occurs is found by equating the two forces, giving
At the critical speed, α = 0, cos α = 1, and n becomes the critical speed n,. Then
Tumbling mills run at 65 to 80 percent of the critical speed, with the lower values
for wet grinding in viscous suspensions

OPEN-CIRCUIT AND CLOSED-CIRCUIT
OPERATION
In open circuit grinding, material
makes only one pass through the
mill and is conveyed to storage or
subsequent processing.
In closed circuit grinding each
pass through the mill is followed
by classification. The coarse
material is returned to the mill for
additional grinding while the fine
material is conveyed to storage or
subsequent processing

PROBLEM 2.1
A material is crushed in a Blake jaw crusher such that the average size of particle
is reduced from 50 mm to 10 mm, with the consumption of energy of 13.0
kW/(kg/s). What will be the consumption of energy needed to crush the same
material of average size 75 mm to average size of 25 mm:
(a) assuming Rittinger’s Law applies,
(b) assuming Kick’s Law applies?
Which of these results would be regarded as being more reliable and why?
Solution
Rittinger’s law

The size range involved by be considered as that for coarse
crushing and, because Kick’s law more closely relates the energy
required to effect elastic deformation before fracture occurs, this
would be taken as given the more reliable result
Kick’s law:

PROBLEM 2.2
A crusher was used to crush a material with a compressive strength of
22.5MN/m2. The size of the feed was minus 50 mm, plus 40 mm and the power
required was 13.0 kW/(kg/s). The screen analysis of the product was:
What power would be required to crush 1 kg/s of a material of compressive
strength 45 MN/m2 from a feed of minus 45 mm, plus 40 mm to a product of 0.50
mm average size?

Solution
A dimension representing the mean size of the product is required. Using
Bond’s method of taking the size of opening through which 80 per cent of the
material will pass, a value of just over 4.00 mm is indicated by the data.
Alternatively, calculations may be made as follows:

For the purposes of calculation a mean value of 4.0 mm will be used, which
agrees with the value obtained by Bond’s method
For coarse crushing, Kick’s law may be used as follows:

PROBLEM 2.3
A crusher reducing limestone of crushing strength 70 MN/m2 from 6 mm
diameter average size to 0.1 mm diameter average size, requires 9 kW. The same
machine is used to crush dolomite at the same output from 6 mm diameter
average size to a product consisting of 20 per cent with an average diameter of
0.25 mm, 60 per cent with an average diameter of 0.125 mm and a balance
having an average diameter of 0.085 mm. Estimate the power required, assuming
that the crushing strength of the dolomite is 100 MN/m2 and that crushing
follows Rittinger’s Law.
PROBLEM 2.4
If crushing rolls 1 m diameter are set so that the crushing surfaces are 12.5 mm
apart and the angle of nip is 31◦ , what is the maximum size of particle which
should be fed to the rolls?
If the actual capacity of the machine is 12 per cent of the theoretical, calculate
the throughput in kg/s when running at 2.0 Hz if the working face of the rolls is
0.4 m long and the feed density is 2500 kg/m3.

Solution of Problem 2.4
1 hertz is equal to 1 RPS, 2 hertz is equal to 2 RPS
Each revolution advances one circumference and the circumference is πD
AS the Diameter is 1 m, therefore , 3.14 meter/rev
2 rev/sec X 3.14 meter/rev = 6.28 m/sec
The volumetric flow rate is: m/s X m2 = m3 /s = 6.28 X 0.005 = 0.0314 m3 /s
The actual throughput = 0.0314 X 0.12 = 0.003768 m3/s
= 0.003768 m3/s X 2500 kg/m3 = 9.42 kg/sec

PROBLEM 2.6
A ball-mill 1.2 m in diameter is run at 0.8 Hz and it is found that the mill is
not working satisfactorily. Should any modification in the condition of
operation be suggested?
SOLUTION
The critical speed of the ball mill is:
The actual speed of the ball mill is: 0.8 Hz + 0.8 rps = 0.8 X 60 = 48 rpm
6.0
8.9
14.3*2
1
cn
The critical speed of the ball mill nc= 0.64 rps = 38.4 rpm
Actual speed is 48 rpm which is higher than the critical speed.
The optimum speed of the ball mill lies between 0.5 to 0. 75 of the critical
speed, say 0.60.
The optimum speed should be 38.4 X 0.60 = 23.04 = 23 rpm
The speed of rotation should be halved

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