WEBVTT
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Welcome to the second part of lecture 4 which
is on classification as well as jigging.
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So concept of classification and the equipment
related to classification that we have already
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covered in part 1 of this lecture.
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Now here we will solve some of the examples
on classification and then we will discuss
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concept of jigging.
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Therefore, let us start with the example 1,
and in this example bauxite ore consists of
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bauxite and silica is cleaned through water
flowing in laminar zone in a classifier.
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So you can see that zone is already given
and we have bauxite ore which is the mixture
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of bauxite as well as silica.
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After cleaning three part should be appeared,
first is pure bauxite, second is pure silica
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and then third is the mixture of these.
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The size range of particle is 20-600 µm.
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So this size range is applicable for three
part where bauxite, silica both are having
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same size range from 20-600 µm.
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Densities of bauxite and silica are given
as 2200 kg/m3 and 2800 kg/m3 respectively.
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So what is the point over here that we have
to compute the size range of three part, what
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is the size range of pure bauxite, what is
the size range of pure silica, and what is
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the size range of mixture?
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So let us start the solution of this problem
and before solving it we should understand
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the concept of it.
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If you compare the densities of bauxite as
well as silica, bauxite is having lesser density
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or it is lighter in comparison to silica.
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Therefore, bauxite will settle as overflow
and silica will settle as underflow.
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However, we can have the mixture also, because
what happens the terminal velocity of largest
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particle of bauxite would be equal or would
be larger than the terminal velocity of smallest
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particle of silica.
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So therefore, we have the mixture of the silica
as well as bauxite along with the pure collection
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of these.
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Therefore, we have to decide the size range
of all three sections.
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So these size ranges can be defined, can be
decided based on the settling ratio.
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For example, we are having ore and the overflow
is basically bauxite which is lighter in comparison
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to silica and underflow we have the silica
and third section is we have the mixture of
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bauxite as well as silica.
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So we have to decide the range based on settling
ratio.
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So we are already given the laminar flow where
the settling ratio [dA/dB] = [?B- ?F/ ?A-
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?F]0.5 this equation we already have derived
while studying the classification in part
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1 of this lecture.
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So A we are referring to silica and B to bauxite,
where ?A is 2800 and ?B is 2200.
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So given size range is 20-600 µm which is
applicable for silica also and bauxite also.
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So size of silica that will settle along with
the largest particle of bauxite that is 600
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µm.
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If you see this settling ratio this ratio
speaks about equally falling particle, therefore
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when we are considering that bauxite will
be collected as oversize, so it is very possible
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that largest particle of bauxite will come
into the mixture.
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So what is the size of silica which will settle
along with the largest particle of bauxite?
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So if you see the largest particle of bauxite
we have taken 600 µm and dA we have to calculate.
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When we calculate dA from this it comes out
as 489.9 µm.
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What is the meaning of this, that when largest
particle of bauxite will settle along with
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this silica particle will also settle which
is having diameter 489.9 µm.
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So what is the range of silica which will
go to the underflow, the range will vary from
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489.9 µm and greater than this.
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So you can understand up to 489.9 µm silica
will found in the mixture and the larger particle
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then this will come in the underflow.
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Therefore, the particles of silica greater
than 489.9 µm will be collected as underflow.
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In a similar line we can calculate the size
of bauxite that will settle along with the
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smallest particle of silica, because largest
particle will definitely settle in overflow,
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but smallest particle will remain in the mixture.
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So what is the size of bauxite particle which
will settle with the smallest particle of
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silica, and size of the smallest particle
of silica is 20 µm.
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So following same equation we can calculate
dB because dA we have fixed to 20 µm.
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So dB from here we can calculate which comes
out as 24.5 µm.
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Now what is the meaning of this, this is the
size of bauxite that will settle with the
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smallest particle in silica, and this will
fall in the mixture.
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So the bauxite particles having size lesser
than 24.5 will go to the overflow.
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Yes, therefore the particles of bauxite is
smaller than 24.5 µm will be collected as
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overflow.
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The selected size range therefore, this is
the ore and we have bauxite, the size range
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is 20 to lesser than 24.5 µm, 20 is the minimum
size.
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Silica that is +489.9 µm to 600 µm and mixture
we are having for bauxite as well as silica
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where bauxite size vary 24.5 to 600 µm and
silica particle size vary from 20 to 489.9
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µm.
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So these are the size ranges of overflow,
underflow and the mixture.
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Now here we have second example of classification.
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In this example again I am having the mixture
of two material that is ore and gangue, ore
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is having a density of 2000 kg/ m3
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And gangue is having density 7000 kg/m3 which
we have to separate through hydraulic elutriator,
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so elutriator working, I guess you understand
that here we have the flow of the water and
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the particle which are coming from the top
it if the particle is having larger settling
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velocity then the upward velocity of the water
so particle will settle down otherwise it
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will be collected, it will carried out by
the water and collected as overflow.
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So that is the concept of elutriator, so the
mixture has followed the size distribution
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which is valid for ore as well as gangue,
so if you see the associated table of this
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problem here we have particle size, in this
the size vary from 0.36 to 0.58 mm and this
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range which is 0.36 to .58 it is applicable
for ore also as well as for gangue, so what
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we have to predict over here is the upward
velocity of water in elutriator so that entire
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ore is collected in overflow.
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It means we have to make sure that sharp separation
is possible for ore as well as gangue, ore
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should be collected as a overflow and gangue
will be collected as underflow, so once we
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will decide that velocity we have to ensure
that no gangue should be present in the overflow,
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and here we will use this relationship of
fD which is 20 Re0.5 so you see here we do
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not know the region in which particle is falling
to, for computation purpose we are given the
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fD relationship with respect to Reynolds number.
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So the expression of terminal settling velocity.
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This is the generalized expression where fD
relationship is given to us so here we have
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rewritten the expression and fD relation is
given to us which we will put over here, so
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we have to calculate the terminal settling
velocities so fD relation we will put over
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here as a function of terminal settling velocity
and therefore fD we equated to 24/, this is
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Reynolds number particle where terminal velocity
term is also coming so we have fD in this
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form 20 µf 0.5 / dp0.5 ?f 0.5 and Vt0.5 so
this fD relationship we will use over here.
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So expression of terminal falling velocity
after keeping, after using the expression
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of fD in terms of vt so here we have further
resolved this and finally when we see.
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The expression this is the expression of terminal
settling velocity of particle.
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Now what we have to do after this, we have
to calculate the velocity of water which is
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moving upward in the sense that ore should
be collected as overflow, no ore particle
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should come into the underflow.
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Now what is the meaning of this the, if we
have the range of particle size and if we
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calculate the terminal settling velocity of
largest ore particle, please try to understand
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this.
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If we have to calculate the water velocity
we have to calculate terminal settling velocity
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of ore, now how we will compute this, what
is the size of this?
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From the size range we can take the largest
size because when the largest size terminal
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settling velocity is computed and water velocity
is taken just higher to this it means it will
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make sure that all particle will be collected
as overflow.
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Therefore we have to calculate terminal settling
velocity of ore considering largest size of
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ore and from this table we can see the largest
size that is 0.58 which is 0.0058m and the
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value of parameters are given over here.
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Considering this we can calculate the terminal
settling velocity which is coming out as 0.0437
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m/s therefore water velocity should be higher
than this velocity, upward velocity of water
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in elutriator should be more than 0.0437 m/s.
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Now second part of the question is we have
to make sure that no gangue should be available
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in overflow, so this same expression we will
follow and how we will compute this, we have
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the size range over here, now if we.
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Do not want any gangue particle to be available
in overflow then we have to make sure that
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terminal settling velocity of smallest particle
of gangue should be larger than the water
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velocity which we have said which is just
greater than 0.043m/s, so here we will calculate
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the terminal settling velocity of smallest
particle of gangue and the smallest size is
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0.00036m and here we have different parameters
so we can calculate terminal settling velocity
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of particle which is coming out 0.0896m/s.
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So you see this velocity is greater than the
water velocity which were just larger than
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0.0437 which we have just computed in the
previous slide therefore all gangue will be
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collected as underflow, no gangue particle
will go with the water in overflow, so in
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this way we compute the particle size distribution
when we handle the separation through fluids,
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and here we have another example, in this
problem the mixture of.
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Galena and limestone having the ratio of 3:7
is treated in elutriator using 7mm/s of water
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flowing upward, so water velocity is given
here, the size distribution of each material
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is shown in the table so you see different
size and corresponding cumulative mass percentage
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is given to us so if Stokes’ law will be
applicable what we have to find is the percentage
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of galena in overflow and that is in underflow,
so densities of galena and limestone are given
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over here which shows that galena is heavier
than the limestone.
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So particles of galena having settling velocity
lesser than 7mm per second will be.
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Collected as overflow, now what we have to
calculate over here that how much percentage
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of galena is available in overflow as well
as in underflow?
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For that purpose we should know what are the
fraction of galena which is available in overflow
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so that can be decided by settling velocity
of galena with respect to 7mm per second because
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that is the upward velocity, whatever velocity.
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Of galena particle larger than this that will
come in underflow and similarly we can have
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the case with the limestone, so expression
of.
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Terminal settling velocity here we have to
follow the Stokes’ law, this is the expression
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of Stokes’ law and diameter of particle
with known settling velocity we can calculate
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from this expression.
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So particle size of galena and limestone having
settling velocity = 7mm/s so the same expression
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we will use but we will use different densities,
here we have used density of galena so this
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is the particle size of galena which has settling
velocity 44.5 µm so you can see when we have
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galena 44.5 µm the particles of galena having
size larger than this will settle in underflow.
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And smaller than this will settle in overflow,
in the similar line we can calculate.
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Diameter for limestone which is coming as
86.9 µm.
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So these are the particle size of galena as
well as limestone which has settling velocity
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= 7mm/s therefore fraction of particles of
galena and limestone having diameters more
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than 44.5 µm and 6.9 µm will fall in underflow,
so this is the particle size and respective
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cumulative.
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Mass percentage is given over here so here
you see we have the table of particle size
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as well as cumulative mass percentage, we
can draw this in this graph which shows that
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for corresponding 244.5 µm the mass percentage
is 48.4 for galena and fraction corresponding
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to 86.9 µm for limestone the fraction is
81.5%.
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Therefore the percentage of mass of particles
of galena in underflow which is 48.4% which
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we have computed from the previous graph and
similar value similar mass fraction for particle
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of limestone in an underflow is 81.5%.
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Therefore galena in overflow is basically
100 – 48.4 which is for 51.6% and limestone
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in underflow is 18.5%, this is the fraction
and the ratio of galena as well as limestone
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is given as 3:7 if 100kg feed is available
in this 30kgs galena + 70kg limestone is there.
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So galena in underflow, underflow percentage
is 48.4 so 30 x 0.484 that is 14.52kg and
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galena in overflow is 30 – 14.52 that is
15.48kg, and similarly we can calculate the
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mass of limestone in underflow as well as
in overflow so that is 70 into 815 57.05and
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70- 57.05 which is 12.95 kg limestone is available
in overflow, so the percentage of galena in
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underflow.
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You see the underflow is having 14.52kg/ 14.52+57.05
this limestone is also available in underflow
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so percent of galena in underflow is 20.3%
and in overflow it is coming out as 54.45%,
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so in this way you can calculate the percentage
of a particular material in overflow.
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As well as in underflow, so these three examples
we have considered to illustrate the computation
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how the classification is used to solve.
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Different problems so let us start with the
concept of jigging so classification which
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we have just.
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Seen is possible only if the particles have
different terminal settling velocity in a
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given fluid, equally falling particle cannot
be separated by classification however even
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if the terminal settling velocities of two
particle in a fluid are same their initial
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settling velocity could be different.
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In such cases these particles can be separated
by making use of the difference in their initial
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settling velocity or initial acceleration,
so the process related to this which considers
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the concept of initial settling velocity.
21:19.070 --> 21:26.930
We call it as jigging, so in jigging process
what happens, we separate the particle based
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on their initial settling velocity.
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Now if I consider this concept initial settling
velocity so what happen, this is the force
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balance equation and if you remember when
we were discussing the terminal settling velocity
21:42.470 --> 21:51.640
we have discussed that when particle start
settling at initial stage its velocity is
21:51.640 --> 22:00.930
not significant and therefore frictional force,
therefore kinematic force offer to the particle
22:00.930 --> 22:03.970
is not significant, so in that case.
22:03.970 --> 22:11.210
This parameter will be equal to 0, so when
I consider the initial settling velocity when
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FD0 it means while jigging process particle
is allowed to settle for.
22:17.220 --> 22:25.260
A very short period of time, so when we have
equated FD to 0, so final expression we are
22:25.260 --> 22:34.680
having as a=(1-?f/?p)g, so that a is the initial
acceleration.
22:34.680 --> 22:41.180
And you see the value of a, of the particle
when we refer this particular expression it
22:41.180 --> 22:48.270
does not depend on the size of particle it
only depends on the densities of the particle.
22:48.270 --> 22:54.730
So here the size and shape of the particle
will not come into the picture, initial acceleration
22:54.730 --> 23:04.191
or initial velocity is only dependent on the
densities of material not their size, therefore
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separation of two materials according to densities
may be possible regardless of size, if settling
23:12.640 --> 23:19.270
period are therefore separation of two materials
according to density maybe possible regardless
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of the size if settling periods are extremely
short.
23:23.780 --> 23:29.580
And relative acceleration or relative initial
velocity of two materials we can calculate
23:29.580 --> 23:35.730
by this where aA that is acceleration initial
acceleration of particle A and this is the
23:35.730 --> 23:40.450
initial acceleration of particle B, and this
is the expression.
23:40.450 --> 23:48.620
So the concept of jigging is we have to let
the particle settle for very short time, and
23:48.620 --> 23:52.800
in that time separation is mainly occurred
due to density.
23:52.800 --> 23:58.860
Now the difference between separation on the
basis of initial velocity and terminal settling
23:58.860 --> 24:05.030
velocity is shown here, here if you see this
graph here we have five different graphs.
24:05.030 --> 24:10.890
Now what is the difference between the two,
difference between all these graph that we
24:10.890 --> 24:11.890
will discuss.
24:11.890 --> 24:19.730
Now if we consider this graph it is the velocity
versus time plot, and curve 1 if I consider
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this is for slate particle of size d1.
24:23.510 --> 24:32.350
Curve 2 is coal particle of size d1, now if
you see the graph of curve 1 and curve 2 in
24:32.350 --> 24:39.760
this plot here we have curve 1 is upper than
curve 2, what is the meaning of this that
24:39.760 --> 24:48.450
slate particle will settle with largest velocity
in comparison to coal though both particles
24:48.450 --> 24:50.460
are having same size.
24:50.460 --> 24:55.220
And why it is so, because slate is having
more density in comparison to coal.
24:55.220 --> 24:58.750
In the similar line if we see the.
24:58.750 --> 25:06.690
Curve 3 it has slate particle of size d2 and
d2 is lesser than d1, so here you see curve
25:06.690 --> 25:14.420
2 is having coal with size d1 and curve 3
is having slate of size lesser than that of
25:14.420 --> 25:22.620
coal, so what is the meaning of this, that
smaller particle of slate will settle at the
25:22.620 --> 25:29.500
same time when the coal particle of larger
size will settle, okay.
25:29.500 --> 25:37.140
And here we have curve 4 as well as curve
5, curve 4 is the slate particle of size d3
25:37.140 --> 25:44.290
which is lesser than d2, and coal particle
is having size equal to d3, so these both
25:44.290 --> 25:50.860
curve 4 and 5 is of equal size particle but
as slate is.
25:50.860 --> 25:56.320
Heavier than coal it will settle with large
velocity than coal.
25:56.320 --> 26:03.390
Therefore, if you consider all these curves
here up to here when the particle will settle
26:03.390 --> 26:07.260
here the terminal settling velocity will achieve.
26:07.260 --> 26:13.570
However, if we consider the initial part of
the settling here you see all these dotted
26:13.570 --> 26:19.500
lines are basically for coal and solid lines
are basically for slate.
26:19.500 --> 26:24.880
So what is the meaning of this particular
section that whatever would be the size of
26:24.880 --> 26:34.460
particle all slate particle will settle faster
as far as this particular section is concerned,
26:34.460 --> 26:41.490
all slate particle will have more velocity
in comparison to coal irrespective of size.
26:41.490 --> 26:44.880
So therefore we can say the initial velocity.
26:44.880 --> 26:51.240
Or the initial acceleration of particle will
dependent on the density which is having larger
26:51.240 --> 26:56.390
than density they will start settling faster.
26:56.390 --> 27:02.280
In comparison to the particle having lesser
density, and here we have the industrial equipment
27:02.280 --> 27:07.690
related to the jigging and this we call the
hydraulic jig.
27:07.690 --> 27:13.940
This is the hydraulic jig, so the hydraulic
jig operates by allowing material to settle
27:13.940 --> 27:19.600
for brief period so that particle do not attain
their terminal falling velocity.
27:19.600 --> 27:25.830
Now here what will happen, if you see these
are basically the rectangular section tank
27:25.830 --> 27:34.700
in which the bottom section is tapered and
the upper section is divided in two part by
27:34.700 --> 27:40.380
putting baffle in between, so this is the
baffle and it is divided in two part.
27:40.380 --> 27:46.160
In one section the plunger operates in a vertical
direction so if you see this figure in one
27:46.160 --> 27:55.410
section what happens here we have this diaphragm
and which has the continuous stroke.
27:55.410 --> 28:01.080
And due to this stroke the separation takes
place, how it is that I will explain, and
28:01.080 --> 28:06.700
if you consider the second part of this here
we have the screen and slurry will allow to
28:06.700 --> 28:08.600
enter over here.
28:08.600 --> 28:13.340
So how the jigging takes place, I do not know
whether you understand the meaning of jigging
28:13.340 --> 28:18.410
in normal in dictionary or not it is related
with dancing.
28:18.410 --> 28:25.750
So here the particle used to dance when we
allow them to settle for very short time.
28:25.750 --> 28:30.840
So as far as functioning is concerned what
happens here if you see, here we have.
28:30.840 --> 28:38.310
The valve, so a, the first figure shows the
downward stroke and upper b figure shows the
28:38.310 --> 28:39.310
upward stroke.
28:39.310 --> 28:44.710
When we have the downward stroke it means
when this section is closed it is completely
28:44.710 --> 28:50.670
filled with the liquid so when we put the
stroke over here, this diaphragm gets the
28:50.670 --> 28:59.280
stroke and water which is available over here
it used to moves to this place.
28:59.280 --> 29:05.860
And when water moves or fluid is moved in
this side it opens the bed, bed of particle
29:05.860 --> 29:07.870
which are available over here.
29:07.870 --> 29:14.030
So it opens the bed and for very short time
the particles are allowed to settle.
29:14.030 --> 29:20.970
And when the upper stroke is concerned then
this valve is open.
29:20.970 --> 29:27.470
And water is allowed to enter into this, and
once water is allowed to enter it comes at
29:27.470 --> 29:34.211
this position and then these valve closed
we have the stroke over here, so water move
29:34.211 --> 29:36.700
from this side to this side.
29:36.700 --> 29:39.980
So the material to be separated is fed dry.
29:39.980 --> 29:46.660
Or in suspension over the screen, and is subjected
to pulsating action by liquid which is set
29:46.660 --> 29:53.650
in the oscillation by means of reciprocating
plunger or diaphragm, that just I have explained.
29:53.650 --> 29:58.790
So as water passes upward the belt opens up
and thus tends to rise.
29:58.790 --> 30:00.220
That just we have discussed.
30:00.220 --> 30:06.901
During this upward stroke the input of water
is adjusted so that there is virtually no
30:06.901 --> 30:13.240
flow through the bed, it means liquid will
come over here but then it come downwards
30:13.240 --> 30:16.970
so there is no flow of liquid through this
bed.
30:16.970 --> 30:22.650
So during this period differential settling
takes place and the denser particle tends
30:22.650 --> 30:27.320
to collect near the screen and lighter material
above it.
30:27.320 --> 30:32.920
After a short time the material becomes divided
in four different stages, so what happens
30:32.920 --> 30:41.000
when liquid comes to this and it opens the
bed then for very short time it will start
30:41.000 --> 30:43.210
settling.
30:43.210 --> 30:48.720
So how the settling will occur, the bottom
layer, bottom layer means which is near to
30:48.720 --> 30:55.360
this screen which is just above to the screen,
so bottom layer consist of large particles
30:55.360 --> 31:02.130
of heavy material and next of the large particle
of lighter material so you see bottom most
31:02.130 --> 31:10.340
we have heaviest as well as lightest material,
above this we have the lighter and large particle
31:10.340 --> 31:17.080
and small particle you see is small particle
of higher density that can pass through the
31:17.080 --> 31:20.830
bed of large particle.
31:20.830 --> 31:27.990
Because when large particle are collected
at one place there is space in between are
31:27.990 --> 31:34.200
available for movement of fine particle of
high density, so fine particle of high density
31:34.200 --> 31:41.130
moves from that bed of large particle and
they are collected at the bottom.
31:41.130 --> 31:46.550
So if we consider this particular section
that is heaviest and smallest particle, and
31:46.550 --> 31:53.170
further the finest, and finest as well as
lightest particles which are available that
31:53.170 --> 31:59.280
remains suspended in the slurry and that can
be separated.
31:59.280 --> 32:06.280
So the four section, first is heaviest and
largest second is lightest and largest third
32:06.280 --> 32:12.020
is smallest and heaviest and fourth is smallest
and lightest.
32:12.020 --> 32:18.460
So four sections we are having and these was
about the jigging, now in this particular
32:18.460 --> 32:24.410
lecture we have considered, we have discussed
the concepts of classification and jigging
32:24.410 --> 32:29.870
And then we have discussed the industrial
equipment related to the classification and
32:29.870 --> 32:36.230
jigging along with their working and here
in this the lecture four we have also discussed
32:36.230 --> 32:45.190
three work example to illustrate the use of
classification concept in example in different
32:45.190 --> 32:46.190
problems.
32:46.190 --> 32:54.590
So here we have the references for lecture
five and it was all about lecture five and
32:54.590 --> 32:58.230
here we are ending the course, it was a four
week course.
32:58.230 --> 33:04.580
I hope you will be benefited by this course,
I expect your enthusiastic participation in
33:04.580 --> 33:07.610
this course, wish you all the best, thank
you.