WEBVTT
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Welcome to the second lecture of week 3 which
is on examples of laws of comminution. This
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lecture 2 where I am discussing different
example, it will have two different part,
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in part 1, I will consider two examples and
in part 3 I will consider three examples.
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So total using five different example I will
demonstrate how to calculate the power consumption
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in comminution process using three laws of
comminution.
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Example 1, in this example we have considered
a problem where a material is crushed in a
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jaw crusher in which feed of average size
of 60 mm is reduced to 10 mm while consuming
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15 kW/(kg/sec) energy. So we have feed of
60 mm and product of 10 mm and while crushing
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we are using 15 kW/(kg/sec) energy.
Now what we have to compute the energy consumption
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to crush the same material of 80 mm average
size to 20 mm. Further we have to calculate
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energy consumption for crushing same material
and the energy consumption will be computed
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using Kick’s as well as Rittinger’s law.
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Let us start the computation first using Kick’s
law. So as you are very well aware with the
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expression of Kick’s law and it is basically
E/M that is energy consumption for crushing
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M unit of material. It is equal to KK that
is Kick’s constant ln DF/DP. So here we
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have, if you see the first part of this question,
here we have already given the energy consumption
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that is value of E/M is already known to us.
And it is equal to KK ln DF/DP. So in first
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case DF would be 60 mm and DP would be 10
mm.
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So once we use these value in the expression
so E/M is replaced with 15 = KK ln 60/10.
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And while after resolving it we can calculate
the value of Kick’s constant which comes
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out as 8.3717kW/(kg/sec). So here you see
the unit of Kick’s constant is equal to
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that of energy in the present case because
that diameter is in the ratio form, so unit
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will be equal to as of that of energy.
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Once I am having the value of Kick’s constant
I can use this value of Kick’s constant
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to calculate energy consumption while crushing
material from 80 mm average size to 20 mm
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size. Now why we are using same Kick’s constant,
because material for which I am calculating
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the energy consumption is same.
However, its feed as well as product size
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will change and therefore, we can utilize
the KK which we have calculated here as 8.3717
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that we can use.
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So finally we have E/M which we have to compute
equal to Kick’s constant value that is 8.3717
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ln 80/20.
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So using these value we can calculate energy
consumption per unit of feed handled. So the
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value comes as 11.606kW, and here if you see
the expression and if you see the value it
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varies from 60 mm to 10 mm and 80 mm to 20
mm. So it comes under coarse crushing where
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Kick’s law is applicable. However, in the
present example the same problem we have to
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solve using the Rittinger’s law.
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Now as far as Rittinger’s law is concerned
this is the expression of Rittinger’s law
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where E/M=KR'[1/dvsp-dvsf]. So here you can
understand that when we have discussed Rittinger’s
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law in lecture 1 of week 3 there we have represented
this in terms of specific surface of product
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as well as feed.
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And in terms of this dvs that is sorter diameter
which is the volume surface mean diameter.
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And in this particular example as we are given
the feed we are not given the distribution.
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Therefore we cannot use the Rittinger’s
law in terms of specific surface, we have
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to use the Rittinger’s law in terms of Sauter
diameter or volume surface mean diameter.
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So whatever value we are given for feed as
well as product.
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That we can assume as the volume surface,
mean diameter. So for feed volume surface
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mean diameter is equal to 60 mm and volume
surface mean diameter of product is 10 mm,
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so that we have assumed to be used in this
expression.
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So once I use the value of dvsp and dvsf in
this E/M I already know, so here I will write
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the value of E/M which is given as 15 KR'
that we have to compute, 1/10 that is the
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value of volume surface mean diameter of product,
-1/60 60 is the value of volume surface mean
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diameter of feed. So considering all these
value that is energy consumption, mean diameter
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of feed as well as product we can calculate
the value of KR' constant.
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We can say revise Rittinger’s constant and
is equal to 180 kW mm/kg/sec. So this KR'
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value we can use to calculate the power consumption
using Rittinger’s law when we have to crush
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the material from 80 mm size to 20 mm size
which is the second part of the problem, here
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we can calculate E/M value.
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Which is equal to 180 the value of KR´ we
can KR´ as it is over here because we are
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using the same material to be further crushed
from 80 to 20, so here we will again assume
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that volume surface mean diameter of feed
is 80mm and volume surface mean diameter of
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product is 20mm.
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So considering the value of KR´ DVSP and
DVSF we can calculate energy consumption per
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unit feed rate using Rittinger’s law which
comes out as 6.7 kW / (kg/s), so here you
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see we have solved a very simple example in
which a feed size is given, product size is
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given and we have computed the energy consumption
using Kick’s law as well as Rittinger’s
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law. Here Rittinger’s law we have used slightly
differently, instead of surface area we have
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used volume surface mean diameter of feed
and product, so this a simple example how
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to calculate the energy consumption using
Kick’s as well as Rittinger’s law.
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Now here I have considered a complicated problem
as example 2, now in this example what happens.
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We have a ball mill and this ball mill is
fed with fresh feed that is F as 30x 103 kg/
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h or 30t/ h, so feed to the ball mill that
we call fresh feed or we can call it maker
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feed also which is 30t/h and product P is
a screen, using a screen of 295 µm aperture
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size, now where this F and P in ball mill
that we can see from this.
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Figure now if you see this figure this the
ball mill and in this ball mill F1 feed is
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entering, now what is this F then? F is nothing
but the maker feed to ball mill so as far
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as maker feed is concerned we have denoted
this with F and the feed to ball mill is concerned
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that we have denoted with F1. So product of
ball mill that is P is coming from ball mill
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after crushing and this product is passed
through a screen which is having aperture
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size of 295 µm, then this P is passed through
this screen, some of the material is collected
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as undersize and this we have donated as P1,
the oversize of this screen is undesirable
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so that will be recycled back to the crusher.
So in this case F is the maker feed and P1
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is the final product R we have denoted as
recycle, so the problem goes as oversize are
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recycled at a rate of 60 x103 kg/h so this
R value is having the rate of 60 ton/ h, now
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what we have to calculate the energy consumption
of ball mill for open circuit as well as closed
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circuit grinding and further we have to calculate
the effectiveness of 295 µm screen considering
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oversize as the desired product.
For this problem work index is given to us
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that is 12.74 kWh/ ton, now as we know that
work index is related with the Bond’s law
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so we have to calculate energy consumption
in this example suing Bond’s law. Example
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1 I have shown the computation using Kick’s
law as well as Rittinger’s law, here I have
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considered Bond’s law to demonstrate how
the energy consumption will be calculated
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using this.
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Now apart from this we have the huge data
for this problem which is shown in this table.
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Now the first column of this table shows the
mesh size where the mesh size in terms of
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minus and plus are shown, now you understand
very well that is what minus shows and what
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plus sign shows. Minus shows that material
is passed through the screen and plus shows
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that material is retained on a particular
screen, so in minus and plus format we have
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used different mesh size screen and the size
vary from 13.33mm to 0.074mm.
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So the number of screens we have used are
very significant, now what is purpose to use
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this screen is to understand the particle
size distribution of makeup screen that is
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F to the ball mill, the recycle screen which
we have denoted with R and here I am having
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the data of P1 which is the particle size
distribution of final product which is passed
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through 295 µm screen. So along with this
flow rate we have to use this data to calculate
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energy consumption. Now if you understand
this problem what we have to compute is the
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energy consumption for open cycle as well
as closed cycle and this is the diagram we
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are having which is basically.
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Denoted the closed circuit if you understand
the 5th lecture of week 2 where we have discussed
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different mode of operation for size reduction,
so the complete diagram shows the closed assembly
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of.
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Crushing, now in this diagram what is the
open cycle and what is the closed cycle or
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what is open circuit what is closed circuit?
If you understand open circuit it says that
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material passed from one end or to the crusher
and will leave from other end of the crusher
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so considering this if we understand.
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The ball mill and only and we enclose the
ball mill where if I consider input and output
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to this enclosure F1 would be the input and
capital P would be the output, so as far as
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open circuit definition is concerned material
which is passed into the crusher and material
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which is leaving from the crusher, so F1 and
P if I am considering it is the part of open
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circuit and if I consider the recycle part
inside this then we can call this as a closed
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circuit because close circuit has some material
which is undesirable which is continuously
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fed to the ball mill, so if I consider close
circuit.
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Then in it to the closed circuit is capital
F and exit to the closed circuit is P1, so
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if I have to calculate power consumption for
open circuit I have to focus on F1 and P where
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F1 would be the feed and P would be the product,
if I consider closed circuit then f is the
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feed and P1 is the product so I hope you are
understanding what is open circuit and closed
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circuit in a crusher problem. So here what
we have to do, first of fall we have to complete
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the data for computation purpose, now what
is left in the data table.
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Which is given to us, in this data table mass
fraction of F, R and P1 are shown. For closed
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circuit if I consider this closed circuit
F is used as feed and P1 is used as a product,
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so if you consider this table here I am having
the fraction of F as well as P1 so here you
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can see the fraction of F as well as P1, so
for closed cycle size distribution of feed
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as well as product both are given to us however
if I consider the open circuit where feed
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is F1 and product is capital P, so you can
see in this table no data of F1 as well as
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P is known to us.
And therefore we have to compute the size
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distribution of F1 and P to calculate the
power consumption of open cycle, so how I
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can calculate the size distribution of F1
by making material as well as component balance?
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If you consider this F1 that would be nothing
but ? of F + R so here we have F1 = F + R.
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F is given as 30 tons and R is given as 60
ton per hour so this is the mass balance before
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entering to the ball mill and component balance
is F1, X1 = FX + RXR, and from here as we
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have to calculate the mass fraction for F1
stream so we can calculate X1 from this expression.
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X1 would be (FX + RXR/ F + R) divided by basically
F1 and that F1 we have replaced with F and
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R because values of F and R are given to us.
So if I consider the mass fraction for this
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particular interval which is basically – 13.33mm
+ 9.423mm so for this size we can calculate
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X1 as 30 x F that is 0.051 + 60 x 0/ 90, so
using this expression for each size interval
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we can calculate value of X1 so corresponding
to first size that is – 13.33mm + 9.423mm
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the value of X1 has come as 0.017. In similar
line we can calculate the fraction for other
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sizes also and once I will calculate this
we can represent.
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All F1 value in terms of X1 so if you see
initially we have computed X1 as 0.017 that
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I have put over here and further if I want
to calculate this so this into 30 + this into
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60/ 90, so following the same expression we
can calculate mass fraction of F1 which is
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the feed to open circuit, so in this table
all mass fractions of F1 for open circuit
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are shown. In the similar line I can calculate
data for P because F1 we have calculated now
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I have to calculate data for P.
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Now if you make the balance over P what it
will give P = R+P1, so that is the balance
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P=R+P1, if we make the component balance P
xP=R xR + P1xP1. So we have to calculate xP
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that is the distribution of this with respect
to each size so xP would be equal to (R xR+P1
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xP1)/(R+P1). Now how we can calculate P1 value,
because R I know as 60 tons per hour.
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But how I can calculate P1, if you consider
this closed circuit what is the inlet to this
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is F and what is the outlet of this is P1
so if we make the balance over this envelope
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then F would be equal to P1. As in the problem
F is given as 30 tons per hour so the same
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value we can use for P1, so F=P1 for 30 ton
per hour. Now if I consider for this particular
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size here both R and P1 are 0 so no value
we can obtain for P also, further if I consider
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second size that is -9423 +6680 for this we
can calculate (60x0.06+30x0)/90, so xP over
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here we can get 0.04.
So following this method we can calculate
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all fractions of P corresponding to different
sizes, so in this table all mass fraction
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of P for open circuit are shown.
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So this is the complete data for open as well
as closed circuit. In open circuit what are
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the parameter, F1 we should know and P we
will know, so for open circuit F1 and P will
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be utilized to calculate power consumption,
however for closed circuit F and P1 is utilized
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to calculate power consumption, so this is
the complete data for computation of power
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consumption using open as well as closed circuit.
Let us start the computation for open circuit
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where we have to consider F1 is feed and P
as product, for closed cycle F is feed and
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P1 is product.
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So here we have to start calculation for open
circuit considering F1 as well as P and this
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expression we have to use to calculate power
consumption this is nothing but the Bond’s
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law. Why we are using Bond’s law, because
in the beginning we are given the work index
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of the material, so which automatically comes
for Bond’s law. So in this expression you
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see what is dPb and dFb, dPb is corresponding
to the size where 80% of product is passed
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through and dFb is the size where 80% of feed
is passed through a particular screen. So
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if I consider the open cycle.
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There F1 and P are required to estimate dPb
and dFb so what we have done over here we
22:58.630 --> 23:07.110
have done the cumulative fraction of F1 as
well as P. This is the individual fraction
23:07.110 --> 23:12.960
and here I have the cumulative fraction and
you understand that if we do the cumulative
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from bottom we have to represent the screen
or mesh corresponding to negative signs. So
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here we have cumulative of F1 and here we
have cumulative of P, now this two value we
23:29.230 --> 23:35.110
will use for computation of power consumption,
if you understand this graph.
23:35.110 --> 23:43.610
This is nothing but the cumulative plot of
F1 as well as P. Now dPb and dFb I have to
23:43.610 --> 23:52.360
compute according to the 80%, so I will draw
a line of 80% and then draw the line where
23:52.360 --> 24:02.140
I am having the cut at F1 as well as on P.
So dPb basically we can see from this graph
24:02.140 --> 24:13.300
is this value which is 2500 µm and dFb from
this graph we can take as 5100 µm, which
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we have to show in terms of mm so 2.5mm and
5.1mm because that is the requirement to be
24:20.870 --> 24:29.080
used in Bond’s law. So E/M is given here
which is 0.3162 work index and all these value
24:29.080 --> 24:34.540
are known to us, when we put the value over
here we can calculate E/M.
24:34.540 --> 24:40.931
Now if you consider this open circuit here
we have material in total as 90 ton per hour,
24:40.931 --> 24:46.670
because 30 ton from makeup and 60 ton from
recycle, so this value.
24:46.670 --> 24:55.120
We have multiplied with 90 so 68.76 kW would
be the energy consumption for open cycle.
24:55.120 --> 25:00.300
In the similar line for closed cycle we will
use F and P1.
25:00.300 --> 25:10.740
And here we will make the cumulative for f
as well as p1 to know were 80% mass will lie.
25:10.740 --> 25:18.860
We make the plot over here for f as well is
p1, this is the cumulative plot of p1, this
25:18.860 --> 25:26.980
is the cumulative plot of f correspond to
80% we can draw the line and then we can calculate
25:26.980 --> 25:37.600
dpb and dfb as 3.5 mm and 7.15 mm. Considering
this value along with work index we can calculate
25:37.600 --> 25:44.480
the power consumption, we can calculate energy
consumption per unit mass. If I consider closed
25:44.480 --> 25:45.480
cycle
25:45.480 --> 25:52.590
The mass which is entering to this is 30 ton
per hour so that m we can take as 31 so this
25:52.590 --> 26:01.870
is, this we can multiply with 0.6467 total
19.4 kW is the power consumption for this
26:01.870 --> 26:06.430
cycle which is closed. Further we can consider.
26:06.430 --> 26:12.550
Open circuit and we can calculate power consumption
using interpolation, here for open circuit
26:12.550 --> 26:20.580
f1 and p will be considered and you see the
80 % were it will lie from this table we can
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calculate from interpolation also. How I can
use the interpolation over here because between
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2 point straight line will lie so I can definitely
use interpolation, so 80 % will lie over here
26:34.240 --> 26:45.890
so df would be lie in between these two however
as well as product line is concerned 80 % will
26:45.890 --> 26:52.810
lie in between these two so dp will lie in
between these two, so using interpolation
26:52.810 --> 27:03.690
we can calculate and dpb and finding as 2.437
which was 2. 5 in the graphical representation
27:03.690 --> 27:13.280
and dfb value 5.246 mm we can find from the
interpolation. Using these two values along
27:13.280 --> 27:20.820
with work index we can calculate power consumption
it will be multiplied with 90 ton so 73.95
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kW is the power consumption
(Refer Slide Time: 27:23 )
27:23.284 --> 27:30.260
For open cycle. And similarly I can calculate
for closed cycle using interpolation so if
27:30.260 --> 27:36.170
I consider feed to close cycle f I have to
consider and 80 % will lie in between these
27:36.170 --> 27:45.830
two, similarly in p1 80% will lie in between
these two so using interpolation I can calculate
27:45.830 --> 27:55.090
dpb and fdb and further using work index as
well as these values I can calculate power
27:55.090 --> 28:01.540
consumption, it would be multiplied by 30
and final power consumption for closed cycle
28:01.540 --> 28:07.350
is 19.92 kW.
So this is the power consumption for closed
28:07.350 --> 28:14.370
cycle as well as open cycle, now finally we
have to calculate the effectiveness of
28:14.370 --> 28:23.680
295 mm screen, if I have to calculate effectiveness
to this what is the feed to this is p, oversize
28:23.680 --> 28:32.330
is R and P1 is the undersize. So what is the
desired product over here is the oversize,
28:32.330 --> 28:37.360
so to calculate effectiveness of this screen
we have to consider these three columns which
28:37.360 --> 28:45.750
are R, p1as well as p, and you understand
how we calculate the effectiveness we have
28:45.750 --> 28:55.330
to calculate ya, yb and yc, ya is the desire
material in feed so in this case all these
28:55.330 --> 29:05.440
fraction combinely will give the ya, yb is
the r value up to here and yc is basically
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the p1 value up to 295 screen. So using this
ya yb and yc when we put in this particular
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expression we can get the effectiveness which
is coming out as 78.09%.
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So here I am completing example 2 and as far
as this part of lecture is concerned here
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I am stopping, I will consider example three,
four and five in next part of lecture two,
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that is all for now, thank you.