Good morning everyone, so we will be starting
with our next separation process, it is the Centrifugal Separation Process. And these
will be used whenever you will be having to, you will require to separate the slurries
from solid particles from liquid, so they are not completely soluble, they are called
slurries, when the suspended materials or the solid materials are there in the liquid,
so it is called a slurry. And in order to, of sizable size for example,
they are of the size of the microns and so on, so forth, and whenever you will be applying
a centrifugal force, these particles will be subjected to a body force which is nothing
but, the centrifugal force omega square g sort of thing. And they will be pushed towards
the wall of the of the container, and it will be entirely depending on the density of this
particle. So, heavier particles will be pushed more towards the wall so, if you have slurry
of heavier and lighter materials, the heavier particles will be pushed towards the walls,
and they can be separated out from the wall. So, in centrifugal separation of processes,
the main driving force is centrifugal force, the driving force is centrifugal force
that induces separation, and it will be something like this process can be described by the
schematic. So, the this is the slurry feed will be going into the system, and this the
middle shaft will be rotating and the whole system will be rotating, and solid particles
experience a force directed towards the wall. Because, of this rotation solid particles
in the slurry, they experience a force there is a centrifugal force
towards the wall. Now, there are this centrifugal separation can be operated on two systems,
one is the solid liquid system, there is a slurry system; second one is liquid-liquid
system, where the two liquids will be having a difference in density. So, therefore, it
is not necessary, that the centrifugal separation process can be operating can be operated on
solid liquid system; only it can operate on a liquid-liquid system as well if there is
a density difference of the two liquids that, we are talking about. So, for slurry feed, you will be having, and
the central the bowl will be rotating about the central axis with a speed omega, and the
slurry will be coming and the solid particles will be experiencing a force. So, centrifugal
force towards the wall so therefore, they will be depositing over the walls. So, you
will be getting a layer of solid particles there, and this will be the liquid and you
will be having solid particles there. Now, these solid particles can be separated
out, in case of liquid-liquid system and we are talking of two liquids, one is heavier,
another is lighter. Now, the heavier liquid that will be having higher density, it will
be pushed towards the wall, and they will be existing close to the wall, and you will
be having another liquid to the other side. So, you will be having a heavier liquid towards
the wall so, this is nothing but, a heavier liquid higher density, and this will be the
lighter liquid. So, the heavier liquid can be taken out from
the wall, and the centrifugal separation process will be affected in this system. So, we can
have either a solid liquid system or we can have a liquid-liquid system of different densities
as well. Now, let us look into the principle of centrifugal
separation process it acts, let us centrifugal force acting on a particle of mass m located
at a radial distance r from the center, the force is represented by F c is equal to m
omega square r. The angular velocity is related to the linear velocity by the relation, where
omega is the, m is the mass, omega is the angular velocity, r is the radial distance.
Angular velocity, omega is related to linear velocity as, omega is equal to v by r.
Now, if you have a rotating speed in revolution per minute RPM, RPM is nothing but, the Revolution
Per Minute, rotating speed N this is (r, p, m) revolution per minute, then omega the angular
velocity is given as 2 pi N divided by 60. So, it is rotation per minute so 60 is it
confers into per second, and one full rotation or full revolution means 2 pi radian.
So, these angular velocity will be radian per second, then the centrifugal force in
Newton can be calculated at as F c will be nothing but, m into 4 pi square N square divided
by 3600 times r and these turns out to be 0.01097 m times r N square. So, these the
unit is Newton, the unit of the centrifugal force. So, on the other hand, if you compare this
centrifugal force with the gravitational force the, then we can see how much more the centrifugal
force is compared to the gravitational force. Now, gravitational force working on a mass
m is given as F of g is nothing but, m times g, the ratio of centrifugal to gravitational
force, then can be computed ratio of centrifugal to gravitational force, can thus we computed
F c by F g is equal to r omega square divided by g. So, it will be r over g replace the
expression of omega that is 2 pi N over 60 square of that so, is turns out to be 0.001118
r N square. Now, in this expression we put g is equal
to about 10 it is 9.8, so you put it about 10 meter per second square, thus the force
developed in the centrifuge is r omega square by g times that of the gravity, so this is
expressed now, so centrifugal by this ratio what you can say that, centrifugal force is
this many times more than g or the gravitational force. So, how it is expressed now, it is
expressed we give an example, if we talk about a radius of centrifuge bowl radius of centrifuge
bowl, if that is 0.1016 meter, and if you talk about a rotational speed about 1000 rpm,
revolution per minute. Then F c by F g becomes, I will put the value
F c by F g becomes 0.001118 r N square put the value of r, put the value of N square
of that, so it turns out to be 113.6, so that means, we term that centrifugal force force
is 113.6 times gravitational force. And it is expressed as that, as this and this whole
statement is equivalent to 113.6 g that is the nomenclature.
So, if generally sometimes, we you will see that the centrifuge is operating at 5 g at
10 g that means, it is 10 times the gravitational, the centrifugal force that is generated is
10 times the gravitational force. If your centrifuge is operating at 20 g means, the
centrifugal force that will be generating generated, it is 20 times more than the gravitational
force. Though, if your r becomes 0.2032, then F c by F g will be 227.2, and we will call
that centrifugal force generated is 227.2 g.
That means, centrifugal force generated is 227.2 times more than the two times the gravitational
force. So, that is the nomenclature, sometimes whenever you will be looking into the specification,
that is written on the wall of the centrifuge you will see all these informations, so that
means the centrifugal force generated x times the gravitational force. Now, let us look into the settling rates in
centrifuge, now we are talking about let us say, talking about a slurry system, and always
slurry system it is basically, solid liquid system. In that, so let us first draw a typical
centrifuge, so you draw a schematic of a centrifuge, then all our nomenclature will be fixed; so
we are having the liquid surface, and the whole system is rotating about its central
axis with omega. So, this height of the centrifuge is B, and
let us this distance is r 2, so r 2 is nothing but, the radius of cylindrical centrifuge,
let us say this distance is r 1, so r 1 is the radius, where the liquid surface is located,
so this is nothing but, the liquid surface is located. And r b let us any location
of a solid particle r B is location of a solid particle at any point of time let us say,
and the feed is going like this, so you are you are, you are, you are feeding the slurry
from the bottom a whole thing is rotating. And typical trajectory of a particle will
be will be something like this, this is a typical trajectory of a particle, it is a
parabolic in nature, trajectory of a particle, of a solid particle. Now, at the end of the
residence time particle is at a distance r B from the axis of rotation, r B is location
of the solid particle at the end of residence time. What is the residence time, residence
time is total flow rate, the feed is being fed into the centrifuge at a particular flow
rate, let us say that is q meter cube per second, and you have the you have a certain
volume of the centrifuge, so let us say that is v v meter cube, so the residence time is
nothing but, v by q. So, that is the average residence time of
a of a of a particle that will be staying within the centrifuge, now if the if the if
r B is less than the r 2, the radius of the centrifuge, the particle will be going away
it will not be settling on the wall, it will be going away in the in the product stream,
so there will be no separation. So, our our objective should be the flow rate should be
maintained such that, that r B that is the location of the particle at the end of the
residence time must be less than r 2, so that it will be hitting the wall and staying there,
then only the separation will be possible. That means, if r B is less than r 2 then particle
leaves the bowl leaves the centrifuge with liquid, if r B is equal to r 2 or greater
than r it will be equal to r 2, it cannot be greater than r 2, because there is a physical
limit, if it is equal to r 2, then it deposits it gets deposited on the wall, once it gets
deposited on the wall it will be effectively separated from the liquid. Now, in a stokes
law regime in stokes law regime, what is that Stokes law regime, when the Reynolds number
is less than 0.1 that is called the Stokes law regime. And what this Reynolds number
will be defined on the diameter of the particle and the diameter of the particle will be extremely
small in the order of micron and so, so it will be really small. The terminal velocity
can be expressed a, terminal velocity under in gravitational field
field becomes V t is nothing but, g times D p square D p is the particle diameter 18
mu into due into rho p minus rho. So, g is acceleration due to gravity 10 meter
per second square, D p is diameter of the particle particle diameter, mu is water viscosity
or feed viscosity. Rho p is particle density, and rho is water or the fluid fluid density,
whatever we are dealing with; this is the expression of terminal velocity under the
gravitational field. But, in the centrifugal fields the terminal
velocity becomes, under centrifugal force the expression of terminal velocity is different;
the terminal velocity the g will be replaced by the omega square r, at g will be replaced
by the omega square r. And this terminal velocity becomes omega square r D p square (rho p minus
rho) divided by 18 mu. So, that is the terminal velocity under centrifugal force and compared
to the gravitational force, the g will be replaced by the omega square r, so acceleration
due to gravity will be replaced by the centrifugal acceleration.
Now, this terminal velocity is nothing but, d r d t if you if you look into the definition
of velocity is d r d t and acceleration is d v d t. So therefore, this expression can
be written as d r d t that is the location at any point of time omega square r divided
by 18 mu D p square times rho p minus rho. Now, you can integrate it out, if you integrate
it out see, it becomes just separate the variables d t equal to 18 mu divided by omega square
rho p minus rho D p square integral d r by r, and this will be let us say from r 1 to
r 2 from 0 to T terminal. Now, these will give you an expression of
the time t T or the terminal the the time becomes 18 mu divided by omega square (rho
p minus rho) D p square l n r 2 by r 1. Now, if the volumetric flow rate becomes q and
v is the volume of the liquid column in the centrifuge. If v is volume of centrifuge volume of say
of of liquid column liquid column, in centrifuge, so in fact, there will be two phases in the
centrifuge one, the liquid phase another be the solid phase. And the solid phase volume
will be the solid particle that will be residing on the wall, but that volume will be extremely
small compared to the volume of the centrifuge. So, so volume of the liquid so volume of the
liquid be roughly equal to the volume of the centrifuge, and q is the flow rate, steady
state flow rate of the liquid of liquid, so residence time is nothing but,
as I told earlier it will be v by q, so it and this becomes so now now, you can you can
combine these two and q becomes v by q. So, you just replace t by t in the as v by q in
the above expression. The earlier, expression this become 18 mu divided by omega square
rho p minus rho D p square l n r 2 by r1. And what is the value of a volume volume becomes
nothing but, pi times b (r 2 square minus r 1 square), so q the expression of q is nothing
but, omega square (rho p minus rho) D p square 18 mu l n (r 2 by r 1) multiplied by pi b
r 2 square minus r 1 square. What I saying that this r 1 will be very small, so it becomes
the radius of the centrifuge itself, so the condition for this is nothing the flow rate
should be matched by you should be given by this expression. Now, particles with diameter
smaller then D p calculated from this equation will not reach the wall. So, the conclusion is the interpretation is
particle of diameter less than smaller than D p calculated from above equation, so what
is the expression of D p, D p is nothing but, [q times 18 mu l n (r 2 by r 1) divided by
omega square (rho p minus rho) times pi b (r 2 square minus r 1 square) raised to the
power 1 upon 2]. So, if particle of diameter less than D p calculated from this equation,
then they will not they will not reach the wall of the bowl or they will go they will
go with the effluent stream; they will be going out with the outgoing stream along with
them. And they will go out with exit liquid stream,
now larger particles larger particles larger particle means particle diameter larger than
D p, they hit the wall and get separated from the liquid. Now, therefore, a cut point
or critical diameter of the particle is defined point or critical diameter of particle is
defined and that is defined this way that Dpc. So, it is denoted by nomenclature Dpc that
is the Critical Diameter of the Particle particle, and this is defined as the diameter of particle,
that reaches half the distance between r 1 and r 2 within residence time
t t T. That means, at t is equal to 0, r equal to r 1 plus r 2 by 2 and at t is equal to
t r, r is equal to r 2 therefore, the critical flow rate can be calculated with this with
this cut point. Now, cut point diameter, what is the critical
flow rate, the critical flow rate becomes in that case q c is omega square (rho p minus
rho) D p c square multiplied by V divided by 18 mu l n 2 r 2 divided by r 1 plus r 2.
So, this this turns out to be omega square (rho p minus rho) D p c square and volume
can be replaced as pi b into (r 2 square minus r 1 square) and divided by 18 mu l n (2 r
2 divided by r 1 plus r 2). So, at this flow rate that means, at the flow rate of q c particles
with diameter greater than D p c will settle with the wall, and most of the smaller particle
will remain in the liquid, and they will go out with the liquid. So, at flow rate q c particles, with diameter
greater than D p c, they settle to the wall the wall and other small particle and smaller
particles and what do you mean by these smaller particles, whose diameter is less than D p
c, they remain in liquid and exit with liquid. So, we can find out that the value of the
particle diameter given a flow rate, what will be the typical value of the particle
diameter, which will be settled on the wall or separated from the liquid and diameter
range that, will be going out with the liquid stream.
Now, let us talk about the sigma values and scaling up issues, scaling up of centrifuge
system that is very important; because, you will be conducting all the experiments in
a laboratory scale centrifuge. Now, if you would like to scale it up, you have for the
industrial level centrifuge which are higher which are larger in size, so the scale up
issues are very important. And for scaling up of centrifuge system centrifuge system
a concept of sigma value is used concept of sigma values are used, then what is the concept
of sigma sigma values. Let us let us look into the volumetric flow
rate, volumetric flow rate is given as q c is equal to omega square (rho p minus rho)
D p c square V divided by 18 mu l n (2 r 2 divided by r 1 plus r 2). Now you so, just
separate these two out, so I will be I will be taking the geometric and the operating
conditions on the right hand side, (rho p minus rho) D p c square divided 18 mu multiplied
by omega square times V l n (2 r 2 divided by r 1 plus r 2).
Now, r 1 and r 2 will be defined for a larger centrifuge, volume will be defined for a larger
centrifuge and of course, the rpm will be may be separate or may be equal. So, I will
just take this, now multiply denominator and the numerator by g, so 2 g, so it becomes
2 g here, becomes 2 g here and what is this if you remember, this value is nothing but,
two times terminal velocity due to gravity. So, this will be nothing but, 2 V t g multiplied
by and this is known as a sigma. So, V t g, what is V t g, V t g is nothing
but, the terminal velocity under gravity gravitational force and what
is this sigma sigma is nothing but, omega square times V divided by 2 g l n (2 r 2 divided
by r 1 plus r 2). So, sigma value contains all the operating conditions those are required
for a larger centrifuge, because the geometric parameters volume and r 1 r 2 will be different
for two different centrifuge, the rotational speed may be different for two different centrifuge.
And g is of course, a constant, so one can take up a ratio of sigma values and and look
into the you know, go for the scaling up options, so we will just look into that. So, if you if you put the dimension of sigma,
the expression is already there omega square times pi b (r 2 squares minus r 1 square),
so this whole term is basically nothing but, volume divided by 2 g l n (2 r 2 divided by
r 1 plus r 2). This will be having a unit of meter square, so sigma is a physical characteristic
of centrifuge, but not a feed particle system. So, this sigma is a physical characteristic
of a centrifuge, it is not a characteristic for fluid particle system, it does not contain
any property of the fluid, it does not contain mu it does not contain rho.
So, it is it is entirely, and it is not and it does not contain the particle, also it
does not contain the radius of the diameter of the particle, all this all this fluid particle
characterization that means, diameter and the density viscosity, they will be appearing
in the V t g; that the terminal velocity under gravity. And this sigma value is nothing but,
the characteristic of the centrifuge, so that is the difference between the two. So, physical
interpretation of sigma is very important, what is the physical interpretation sigma,
if you remember sigma has unit of area it means, it is area in meter square of a gravitational
settler, what is the gravitational settler gravitational settler is nothing but, a tank
where you just put the material slurry, and the solid particles will deposit or settle
by under gravity and the terminal velocity is given by V t g.
So, sigma is it is it is area in meter square of a gravitational settler that will leave
that will leave same sedimentation characteristic, as the centrifuge at the same feed rate. So,
basically, it gives an idea how this is compared with a gravity settler because, gravity settler
is very common. So, it gives the area under which the sedimentation characteristic is
almost equal or easily understandable, if you talk about a gravity settler because,
it is it is easier to understand gravity settling method, compare to centrifugal separation
process. So, therefore, in order to scaling up, for
scaling up what we will do will try to maintain the same terminal velocity under gravity same
terminal velocity, so we would like to maintain same terminal velocity of a gravity settler
t g V t g. So, V t g for system one should be equal to V t g for system two, that simply
means q 1 by sigma 1 is equal to q 2 by sigma 2; q 1 is the flow rate of smaller centrifuge
sigma one is the sigma value for the smaller centrifuge, and q 2 is the flow rate of the
higher centrifuge, so it may be 10 times of q 1. And sigma 2 is the characteristic sigma
values for the larger centrifuge, so so that is it, this this is dependable if centrifuge
for centrifugal forces between the two or within a factor of two from each other that
means, one can use use this rule for scaling up, if centrifugal force forces between two
centrifuges are within a factor of two this is a thumb
rule, this is a design thumb rule. That means, if the largest centrifuge, I will be have
will be having centrifugal force, which is slightly less than 2 or less than or equal
to 2, then you can use this rule, I just scaling up option. And if not what happens, if not one has to
use q 1 E 1 sigma 1 is equal to q 2 E 2 sigma 2 and what are E 1 and E 2, E 1 and E 2 these
are basically, the efficiency factors of the centrifuge of centrifuges, these factors are
determined experimentally. So, using this concept of sigma value, one can scale up a
centrifuge from lower version to higher version, which can accumulate more flow rate, and more
separation will be will be effected. You will be in fact, we are we are going to
we will be solving some other problems, example problems on this. Now, next we move on to
the separation of liquid still now, whatever we have discussed we have talked about the
separation of a slurry process that means, solid liquid system. Now, let us move on to the separation of liquids
that means, we are having two liquid stream of different density, so the schematic of
a tubular centrifuge, let us draw a schematic of a tubular centrifuge. I am just writing
a, I am drawing a symmetric part, so these have symmetry the axis about, which the centrifuge
is rotating with a angular speed omega, and let us say this is the portion where you are
you will be having a liquid, these a now will be having a header here, fine, this is the
lighter, lighter fluid, this is the heavier liquid.
So, the heavier liquid will push towards the wall, so it will be heavy liquid let us say
rho h this will be a lighter one liquid rho l, and you are putting the feed
from the bottom, and there is a separator here, that separator separates basically form
the lighter to the heavier liquid other there will be mixed up. Now, let us put the dimensions,
let us say the from the from the center the distance the level of the heavier liquid will
be, let us say r 4 and this distance is r 3, this separation is called, let us say r
2, and this level is r 1 or it is the a feed is going there.
And let us say r 1 is the core, they it is basically air core, so in they in this figure
the geometry as a details are given like this, r 1 is basically the location of lighter liquid
liquid r 2 is nothing but, the liquid-liquid interface; the interface that the that separates
the two liquids location of liquid-liquid interface. And this part is
called a weir w e i r, so this weir is basically separating out, physically the heavier liquid
and a lighter liquid. r 3 is location of weir, and r 4 is a location
of heavy liquid stream or the product, so location of interface can be calculated by
a balance of pressure in the two layers. So, let us and this location of interface
can be calculated by doing a pressure balance between two layers, now the force on the fluid
located at a radial location r, fluid at r is given as F c is equal to m r omega square,
m omega square r that is the mass into acceleration. Now, the differential force over a differential
thickness can be expressed as differential force on differential element D r is expressed
as d F c time d m omega square r the mass containing will be d m.
So, differential mass can be expressed d m is a differential mass, and differential mass
can be given as d m is equal to 2 pi b r d r times rho, so this is the volume within
the differential element and rho is the density of the fluid. So, what is b is nothing but,
the height of bowl, height of the tubular centrifuge. Now, you will be in a position to calculate
the pressure force, d p is differential pressure force what is that force per unit area d F
c divided by area, so it will be omega square rho r d r, and what is A, A is nothing but,
2 pi r b. Now, you can integrate this equation between r 1 and r 2 the pressure difference
between two points So, integrate between two points, 1 and 2 we will be getting
p 2 minus p 1 is nothing but, rho omega square by 2 (r 2 square minus r 1 square).
So, equating now now you can equate pressure on two sides, equate pressure over r 2 to
r 4 and r 2 to r 1, if you equate this pressures then you can get the your location of the
interface rho l omega square by 2 (r 2 square minus r 1 square) between r 1 and r 2 it was
a lighter liquid. And rho a is equal to rho h omega square by 2 r 2 square minus r 4 square
between r 2 and r 4, it is only the heavier liquid.
So, the location of the interface, now you can you will be you will be able to calculate,
location of interface, now becomes r 2 becomes r 2 square becomes rho h r 4 square minus
rho l r 1 square divided by rho h minus rho l. Now, the interface r 2 must be located
such that, r 2 is less than r 3 otherwise no separation occurs between the heavy and
the lighter fraction. So, this this gives you an expression of location of the interface,
that has to be maintained. And r 2 must be less than r 3 otherwise, there
will be mix up of heavier and lighter liquids and there is no separation possible, so that
is basic principles of any centrifuge separation process, and we have discussed the principles
for the slurry that is the solid liquid system, and the liquid-liquid systems, and various
scales up issues as well. So, the idea is if you if you know or may in the laboratory
scale that, given a centrifuge there its dimension etcetera, operating connections given, this
is the extent of separation you are going to do, then if you would like to scale up
you can you can do appropriately. So, once after doing this let us look into
some of the examples, so that the calculation procedures etcetera will be clear to us. Now,
I have probably around three problems for you, so let us look in to the first one, first
example of discuss solution containing particles with density, so let us talk about a slurry
system, slurry system means is it is a solid plus liquid system. And it has containing
it as it as containing particle of density 1461 Kg per meter cube, and it has to be and
this particle has to be clarified removed by the centrifuge.
Solution density is given rho is 801 Kg per meter cube, and viscosity of the solution
solution viscosity is also given viscosity is given as mu is equal to 100 centipoises.
So, it is 100 time viscous than a water, and centrifuge bowl has the following dimensions
r 2 is given as 0.02225 meter, and r 1 is given as 0.00716 meter your height is given
as 0.197 meter. Now, what you have to do you have to calculate
the critical particle diameter, calculate D p c such that, D p c is the exist exist
stream, if n is equal to 23000 rpm and q is given q is equal to 0.002832 meter cube per
hour, so you have to calculate the critical particle diameter of the largest particle
that in the exit stream. So, let us look in to the solution, the solution is straight
forward you have two first you have to convert the rpm into radian per second.
So, it will be 2 pi n divided by 60, so it will be 2 pi into 23000 divided by 60 and
it turns out to be 2410 radian per second, and bowl volume can be calculated v is equal
to pi b r 2 square minus r 1 square, and the if you put the values and these turns out
to be 2.747 into 10 to the power minus 4 meter cube. The viscosity is given as 100 centipoise
that means 100 into 10 to the power minus 3, so it will be 0.1 pascal second.
So, q c you can calculate 0.002832 divided by 3600 this is this is given in meter cube
per hour, so it you can convert in to meter cube per second 7.887 into 10 to the power
minus seven meter cube per second. So, now, if you put all these values in your governing
equation, in the equation of D p c the D p c turns out to be 0.746 micron, so put put
all the values in the expression, so it turns out to be around 0.746 micro that means, the
particles of diameter 0.746 and less, they will be going out with the liquid stream,
and the particles with higher diameter, larger diameter they will be settled on the wall
and they can be separated by this centrifuge. And you can check terminal velocity, and terminal
velocity and then check and you can check the Reynolds number, the Reynolds number will
be less than 1 that means, you are within the stocks regime. So, I have go go move on move on for the next
example, example number 2, again I were talking about a viscous solution it contains a particle
of density, 1200 Kg per meter cube, it has to be clarified by the centrifuge, the solution
density is given as 850 Kg per meter cube, the viscosity of the solution is 80 c p, and
centrifugal bowl will be having dimension r 2 is 0.02 meter, and r 1 is 0.02 meter.
The height is given as 0.25 meter, you have to you have to again find out the particle
you know, critical particle diameter given n is equal to 1500 rpm, and flow rate q is
equal to 0.002 meter cube per hour. So, again you have the steps are straight
forward, you have to calculate the angular velocity 2 pi n over 60, and this turns out
to be 1570 radian per second and bowl volume becomes pi b in to (r 2 square minus r 1 square);
if you put the value it turns out to be 2.355 into 10 to the power minus 4 meter cube. And
the flow rate is given as 0.002 meter cube by hour, so it turns out to be 5.56 into 10
to the power minus 7 meter cube per second. Now, if you put all these values in your governing
equation the equation of D p c, this turns out to be 5.56 into 10 to the power minus
7 is equal to 1570 square (1200 minus 850) into D p c square divided by 80 into 18 into
80 in to 10 to the power minus 3, that is mu l n 2 into 0.02 divided by 0.01 plus 0.02
multiplied by 2.355 in to 10 to the power minus 4. And if you calculate D p c, D p c
turns out to be 2.57 micron, so just check the previous one it was 0.746 micron by changing
the operating conditions and the dimensions slightly, you will be landing up with a by
critical particle diameter of 2.57 micron. So, there is a huge difference that is 0.746
micron, so that is almost 1 micron and and here here you are you getting around 2.5 times
of that by changing, the flow rate and the other operating conditions. So, this gives
an idea that over given the typical operating condition of a centrifuge, what is the typical
diameter of the particles, those cane be separated by using the centrifuge. And you know how
to do the calculations, in the next class first, I will go just solve one more example
for scaling up of a centrifuge, and then I will move on to the next separation process
that is the chromatography separation process, thank you.
