WEBVTT
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So, in todays lecture in biochemical engineering
we are going to look at continue looking at
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the immobilized enzymes. We will start from
where we left in the last class, which was
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the mass transfer co-efficient. So, what we
were looking at in the last class, if you
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remember is intraphase mass transfer with
reaction. There was a flat plate so, if I
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am to remind you. So, we had a flat plate
like this and there is a boundary layer set
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and there these enzymes are receptor. You
know under the receptors that are on bound
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to the surface and, because this is fluid
phase and this is the solid phase in which
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the enzymes are immobilized. So, the enzymes
are enzymes immobilized.
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So, what happens is there is transfer between
the fluid and the solid phase and that mass
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transfer co-efficient was referred to as k
f. Now what we did not or do get the time
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do it in the last class look how this k f
varies. How do you evaluate this k f it turns
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out there lot of empirical relations to evaluate
the k f of the mass transfer co-efficient.
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Now what we looked at in the last class was
flat plate, but if what if it were in the
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flat plate what it was circular tube or sphere
or any other kind of things.
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So, we what I wanted to show you on the screen
today are these relations for mass transfer
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co-efficient these for laminar flows. So,
different kind of geometry, first one is flat
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plate then circular tube then solid spheres
and then finally, falling film. So, it is
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not the first relationship is k, k observed
time that is k observed is mass transfer co-efficient
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time’s z over D m Z being the distance.
So, this is the flat plate, depending on where
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are on the flat plate that goes as Reynolds
to the power half and Schmidt number to the
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power one third.
For A circular pipe you it the k f, which
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is the mass transfer co-efficient is given
by k f D over D m, d being the diameter of
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the pipe over D m. The molecular diffusivity
that again goes as Reynolds to the half and
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Schmidt to the one third, but the co-efficient
is different. So, co-efficient in the last
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one there is like 0.3 and here it is 1.8 so,
it is like 6 times. So, what you see is that
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one important similarity is that that k observed
k Z over D m or k f D over D m. Whatever it
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is Sherwood number actually goes as Reynolds
to the half and Schmidt to the one third.
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So, how does it look when we have forced convection
around solid spheres, this is something probably.
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Have you read in the mass transfer classes
again it goes as the same it again goes as
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Reynolds to the half and Schmidt to the one
third, but there is little attached is that
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there is a constant. Because the shape of
the sphere is the constant with that and if
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you go to the screen, let what you will see
the Sherwood number goes as 2 plus 0.6 Reynolds
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to the half Schmidt to the third. So, that
co-efficient those two co-efficient are there
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Reynolds to the half and Schmidt to the third,
but you have the 2.
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Finally, the scene that we want to look at
is the falling film. So, the falling film
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turns out similar, to the flat plate in the
sense that if the geometry. Similar also is
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the flat kind of thing and it goes again as
Reynolds to the half and Schmidt to the one
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third and the co-efficient then the 0.7, which
is not to different from 0.3.
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So, it is around twice that of the flat plate
so, the difference that you know what would
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be the difference between the falling film
and the flat plate? So, this is for this is
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if I go to this thing this is the flat plate
what would be a falling film line. So, this
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is again a solid surface and this is how this
going this is the falling film. So, what is
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different over here is the effect of...
Gravity
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Effect of gravity here
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So, this is the vertical this is the horizontal.
So, the gravity is going to affect the size
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of the boundary layer and also the velocity
profile in the boundary layer, the thickness
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of the boundary layer and the velocity profile
in the boundary layer.
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So, if you have taken a falling film like
this and compared it with a flat plate, now
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the here what is going to happen here? Is
that what you have at the bottom this is going
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to the liquid is going to drag. In this direction
liquid is going to drag in this direction.
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So, as the liquid goes down the z or in other
words as a length goes down, then the size
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of the boundary layer grows. So, as z goes
high more and more you go down the size of
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the boundary layer is going to grow. Because
it is dragging the fluid is dragging more
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in the in the vertical direction because of
gravity.
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So, as a result which in the two things that
is different over here is the thickness of
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the boundary layer and the velocity profile
and the boundary layer and therefore, the
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Sherwood number is slightly different from
the flat plate. So, what we are looking at
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till now are these cases, if you remember
all the examples. We did in the last couple
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of lectures are examples, where the enzyme
is on the surface. The enzyme is on the surface
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and either it through diffusion or it is through
convection that mass transfer is taking place.
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So, that is the whole idea of today’s lecture
is go the screen is intraphase mass transfer
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and reaction. So, what is different in this
one is that is intraphase. So, let me see
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if I have picture I do not have picture here.
Let see I have picture out here not let me
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go back and show you some so, I can use this
picture.
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So, that is the difference you know. So, if
for example, if you have intraphase mass transfer
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if the enzyme immobilized that is there is
entrapment within the encapsulation within
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the gel of the matrix. Then there is no way
for you to make it react with the substrate
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without the substrate, actually diffusing
into the enzymes. So, that is the different.
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So, intraphase is within the for phase the
mass transfer should happen within the phase,
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and when you have the intraphase mass transfer?
What happens is that the reaction occurs simultaneously.
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So, that is a major difference, this was an
example, we had the only the intraphase mass
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transfer.
So, the solid the enzymes are encapsulate
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or on the surface of the solid and the fluid
comes in and reacts. And then intraphase mass
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transfer then enzymes are on encapsulated
within is this difference clear, the physical
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difference now in addition to this. You can
what can you have you can have enzymes, which
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are on the surface in addition to which being
entrap inside you can have enzymes on the
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surface. Now let us go and look at what the
difference is and how to study this kind of
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models can we make. So, the first important
thing that I want to discuss over here, is
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that the concept of pseudo homogeneous reaction
and we did that before. So, if I can you know
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interact your attention to this.
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So, my enzymes are you know let us say these
are my points of where the enzymes are...
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Now I can go and write homogeneous model for
this heterogeneous model for this is something
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I had discussed, but I still want to go into
details and try to see. If you can understand
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it completely so, if I am to do a heterogeneous
modelling of this. What would I do, I would
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actually let the substrate diffuse to the
surface then from the surface diffuse then
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why do, I separate these two diffusion that
diffusion over here. So, diffusion in fluid
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and then diffusion in this, because porous
solid so, diffusion in the porous of diffusion
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co-efficient for going to be different.
You know it could be diffusion depending on
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if the porous is very small, if it not it
could it would still be different the diffusion
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co-efficient still going to be different.
So, if I am to do a heterogeneous model of
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this my step 1 is diffusion in fluid, step
2 is diffusion in porous solid then step 3
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would be adsorption or you can just go straight
to reaction. If you want reaction and adsorption
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you can say, but what is the challenge? That
we face over there the challenge that we have
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to face over there is that we have to have
a geographical. You know map of where these
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particular enzymes are located within the
matrix, if we are to allow. So, because you
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are doing simulation, if you are doing simulation
of the whole thing you need.
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So, the material reception is defusing into
the whole thing, but for you to figure out
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that for you to be able to quantify. That
you need to know at which point this reacting
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what are these, because see what happens this
within this matrix it become almost like point
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syncs for the substrate. These of points of
reaction become point sync for this substrate
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and you need geographical, you need a map
of where these points syncs out. So, if you
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want to do heterogeneous model is absolutely
not impossible, but it is a little cumbersome
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in terms of the modelling. And then you have
to do the diffusion out of the porous solid
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and out of the fluid that is not the problem
the problem lies here.
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So, what we do is we come up with what is
known as a pseudo homogeneous model. What
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is the assumption of such a model? The assumption
of pseudo homogeneous model, what is the assumption
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of pseudo homogeneous model? The assumption
is that you homogeneous over the entire matrix.
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So, you assume that these receptors are present
uniformly in the entire matrix. So, what you
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take? Is you take you figure out that, what
is the total number of receptors? What is
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the reaction rate for each and multiply that
and get the total amount of reaction rate
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for the receptors and then divided by volume
together volume average reaction rate.
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So, that would be once that is there, then
it is assumed that though these reactions
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are in principal heterogeneous reaction and
they are occurring at certain points. You
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assume that the whole thing occurs uniformly
over the entire matrix so, this is the concept
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of pseudo homogeneous reaction. Now is there
any new concept for you studied this before
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know so, at least I hope, because we do not
want to get over this again and again. Let
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us go back to what we have on the screen?
So, there is a definition exactly what I have
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explained, but just to go through it the heterogeneous
reaction that occurs within a tissue or porous
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matrix containing immobilized enzymes are
called interphase.
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Although the reaction occurs at the interphase
between the fluid and the cell surface or
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at the cell extra cellular matrix the phases
within the tissues is considered to be pseudo
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homogeneous. If the reaction is macroscopically
uniform so, this is the term that we introduced
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here macroscopically uniform. What it means
that as a microscopic level? It is occurring
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at different points inside the matrix, but
at the microscopic level is uniform and when
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at the macroscopic level it is uniform. Then
you can assume it to be pseudo homogeneous
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so, it is a matter of scale, if at the scale
that you are looking your reaction is homogeneous.
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Then you can assume it to be pseudo homogeneous
if, but as a microscopic scale it could be
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heterogeneous is the concept of scale.
So, at the smallest of scale say for example,
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you know if you have surface if you have a
surface like this and you are trying to take
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the derivative at this point. Which point
you take the derivative? You never know or
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if you have surface like this. We can say
that the microscopically at the microscopic
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scale, we can say that this is the singular
point, because you cannot define the derivative
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here and this is a non-singular point. You
can define the derivative here, but that the
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microscopic scale this is not singular. You
see what I am saying, because when you go
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to the level of at the microscopic level this
is no longer as sharp this peak is no longer
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as sharp.
Because when you go to the microscopic level
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this is not singular as a microscopic level.
You can define the derivative at this point
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so, that is the matter of scale the manner
of looking. So, the same thing happens here
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at the microscopic level, it is heterogeneous.
Because these reaction happens at different
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point within the matrix, but at the macroscopic
level you can average them out the macroscopically
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uniform, which means that you can average
them out and the whole concept of averaging
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theory. You know the basic concept of averaging
theory originate from this, where you can
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actually decide that. Whether the system is
microscopically uniform or macroscopically
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uniform, that kind of things.
So, it is microscopically uniform, you know
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there is no question at all, but the whole
concept of averaging comes in these certain
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issues or in these certain cases. Where the
system is microscopically non-uniform, but
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at the macroscopic level, you can decide now
what how do you decide, what is microscopic
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and macroscopic?
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You know that is the question that is important
and what typically is done is that the let
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us call this r. So, is microscopic length
scale versus macroscopic length scale, if
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this is much smaller than 1. Then you assume
this is your criteria, typically you assume
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that it is microscopically uniform so, the
length scale at which you see this heterogeneity.
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This is the heterogeneity length scale and
this is the non-heterogeneity length scale.
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So, that the length scale you see at which
you see heterogeneity over the length scale
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at which you do not see heterogeneity, the
ratio of this two have to be much smaller
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than 1. So, if this is possible then what
you do is you can average it out over the
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macroscopic length scale as soon as the average
it out you gets the pseudo homogeneous model.
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So, pseudo homogeneous model is not a matter
of just intuition, it is a matter it evolves
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the out of you know averaging theory. Whether
in the macroscopic once you have done the
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pseudo homogeneous got the pseudo homogeneous
model. In the pseudo homogeneous model what
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happens? This assumption is that reaction
and diffusion occurs simultaneously. In the
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real model in real life diffusion and reaction
not occurring simultaneously, in the heterogeneous
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model diffuse reaction and not occurring simultaneously
remember. So, first diffusion followed by
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reaction, but in the pseudo homogeneous model
diffusion and reaction is assumed to occur
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parallel.
So, if this is just you know corollary. So,
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to say if the reaction is too fast as compared
to diffusion the concentration at some places
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becomes 0, because it is diffusion control.
This is one of the interesting things that
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can happen in pseudo homogeneous model, I
mean what pseudo homogeneous model cannot
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capture. So, if you have a system like this
and say for example, you have a reaction like
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an autocatalytic reaction that we discussed
in the last class. So, what can happen, because
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the reaction at some places can, if there
is a small perturbation in the system? The
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reaction in some places could be much higher
than the reaction in other places.
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So, the concentration in some places the for
example, the concentration of the substrate
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here could be 0 and next you know too far
from that place the concentration could be
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high. So, what will happen in that case what
would be the consequence of
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that there will be major difference, the large
difference in concentrations, it could be
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next to each other depending on the kinetics,
if you have kinetics of this type A plus 2
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B giving 3 B. Then A could be depleted in
one place and A could be far from the depletion
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from another, it could be presented in large
amount. So, what will happen in that case
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that is a real system I am talking about the
heterogeneous system.
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I am just talking of catalyst pellet here;
you know it cannot be catalyst pellet. So,
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this is just catalyst pellet catalyst pellet
or immobilized enzyme whatever you want to
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call it. So, what can happen what will happen
is that the heterogeneous model is still finding,
19:27.809 --> 19:34.809
but you cannot average it anymore? So, there
is no point in having an average when one
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part of that has a very large concentration
and another part of that very small concentration,
19:39.190 --> 19:44.690
because the average value would not give sense
of the distribution. So, when you want to
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average the criteria number 1 is that the
ratio of the smallest through the largest
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length scale is much lesser than 1.
The criteria number 2 would be that the minimum
19:58.129 --> 20:05.129
over maximum concentration minimum C over
maximum C should be of the order 1. This should
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not be a huge difference it could be 0.1 they
still order 1, but it could be it cannot be
20:11.109 --> 20:18.109
that cannot be 10 to the minus 3 or 10 to
the minus 4 within the matrix. So, which means
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that the variation within the matrix has to
be reasonable they cannot be very large variation.
20:24.729 --> 20:28.669
So, having set these things I think now that
and it does not make sense to go and write
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a model. Unless we understand what are the
constraints? What are the boundaries between
20:32.769 --> 20:34.749
which we writing that model so, having met
that...
20:34.749 --> 20:41.749
Let us we can you know try and write a model
of A. So, what we do is these are for drug
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bio molecules, therapeutic molecules and so
on. If the diffusion is slow then reaction
20:50.859 --> 20:55.889
rate could decrease significantly the reason
is, because this is diffusion and reaction
20:55.889 --> 21:00.499
occur parallelly, but in real life diffusion
decrease reaction. So, if diffusion is very
21:00.499 --> 21:06.239
slow then reaction; obviously, would be hamper
straightforward. So, what we do is that in
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the pseudo homogeneous model what we write
over here is d effective Laplacian C minus
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R of v. Do you all remember by the way how
to write Laplacian in spherical and cylindrical
21:16.460 --> 21:23.460
and all these coordinates, because you know
they will be in. So, why D is effective? Because
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this is not the real diffusion.
21:26.759 --> 21:33.349
Because this is not the real diffusion, the
real diffusion occurs in the spaces between
21:33.349 --> 21:37.340
the porous. So, this is the effective diffusion
co-efficient because you taken an average
21:37.340 --> 21:41.749
out of the whole thing, if the diffusion co-efficient
also an also an average between the diffusion
21:41.749 --> 21:48.749
co-efficient of the solids through the solids
and the spaces between the solids. So, R of
21:49.720 --> 21:55.849
v is let us defining R of R v average as the
microscopic reaction rate. So, what we would
21:55.849 --> 22:00.109
be the microscopic reaction rate as I just
as I said before the summation of the all
22:00.109 --> 22:06.139
the microscopic reaction rates over the volume
divided by the volume. So, R of v average
22:06.139 --> 22:13.139
is the summation of all the microscopic reaction
rates divided by the volume fine. So, this
22:15.830 --> 22:22.830
is given by 1 over minus 1 over v.
So, this is you know you can replace what
22:27.779 --> 22:34.779
you have over here the diffusion reaction
problem from here to here and then what you
22:36.749 --> 22:43.749
do is you simply use the stokes here. So,
just change the Laplacian to the surface into
22:44.629 --> 22:49.349
this straightforward. I think you have done
this before changing from the Laplacian to
22:49.349 --> 22:56.309
the surface integral. So, when you do that
then this is all fine. So, you can write here
22:56.309 --> 23:02.259
R average in this way so, the advantage with
this writing is that you just need to changing
23:02.259 --> 23:06.519
from the surface from the volume to the surface
is that you just need to evaluate. What is
23:06.519 --> 23:10.849
going on to the surface? You really do not
need to care about, what is going on in the
23:10.849 --> 23:11.200
volume?
23:11.200 --> 23:18.200
So, then we define this quantity eta is my
effectiveness factor and I think we did effectiveness
23:23.789 --> 23:30.789
factor in the lecture 1 of the lecture before
also couple of lectures back know. So, this
23:32.700 --> 23:37.710
is something that we are going to do for the
next couple of lectures. So, let us try and
23:37.710 --> 23:44.599
see what the effectiveness factor is essentially
the reaction rate. When the reaction is occurring
23:44.599 --> 23:50.460
over the entire region over the reaction rate.
If it were only occurring at the surface concentration
23:50.460 --> 23:56.029
or only occurring at the surface, if the entire
concentration in other word if I am to explain
23:56.029 --> 23:59.969
little better. If the entire concentration
if the entire amount of material, where is
23:59.969 --> 24:04.580
the surface concentration.
So, that is my denominator entire material
24:04.580 --> 24:11.580
and C s surface concentration over the actual
reaction rate. That happens and this actual
24:18.039 --> 24:23.080
reaction rate you can obtain as I showed here
on the screen, you see actual reaction rate
24:23.080 --> 24:30.080
you can obtain as the volume average fine.
So, why is this important, because see for
24:30.389 --> 24:35.059
a material once you have evaluated. What is
your effective reaction rate? If you want
24:35.059 --> 24:38.570
to know your actual reaction rate all you
need to do is evaluate the actual rate at
24:38.570 --> 24:44.549
the surface and effectiveness and this is
essentially effect on factor. This is very
24:44.549 --> 24:50.259
important thing that you know some of you
are taking the advance reaction engineering
24:50.259 --> 24:52.190
course are you any of you.
24:52.190 --> 24:57.219
Not taken. So, then you will study these things
you know effectiveness factor. The effectiveness
24:57.219 --> 25:04.169
factor essentially measures that if reaction
is occurring at if reaction were occurring
25:04.169 --> 25:09.559
at the surface concentration and if the actual
rate of the reaction. The ratio between these
25:09.559 --> 25:16.559
two I am going to see how to how to incorporate
that in the next few cases so, this is not
25:17.109 --> 25:17.679
important.
25:17.679 --> 25:24.460
So, what we do now is now let us go and actually
does A solve the problem. So, the first problem
25:24.460 --> 25:27.989
we look at the intraphase mass transfer and
the reaction in a rectangular slab. We will
25:27.989 --> 25:32.940
look at the sphere and the rectangle. So,
first thing we will look at today is this
25:32.940 --> 25:39.940
problem in the rectangular slab. So, what
is my what is my basic governing equation
25:41.879 --> 25:48.879
for the rectangular slab what is my governing
equation for the rectangular slab in this
25:55.599 --> 26:00.369
case intraphase mass transfer? And reaction
this you know there in the title if you look
26:00.369 --> 26:06.789
at here intraphase mass transfer and reaction
in the rectangular slab. So, these are interrupt
26:06.789 --> 26:13.789
these receptors are interrupting inside. So,
what would be my governing equation? So, what
26:16.549 --> 26:20.779
the basic phenomena let us talk about it for.
So, essentially the basic phenomenon is heterogeneous
26:20.779 --> 26:24.989
reaction, but we are going to use pseudo homogeneous
model. So, let us make that clear first that
26:24.989 --> 26:28.639
we are essentially it is heterogeneous phenomena
heterogeneous system, but we are going to
26:28.639 --> 26:32.549
use pseudo homogeneous model what is the basic
assumption of pseudo homogeneous model? If
26:32.549 --> 26:37.820
reaction is uniform everywhere and you can
use an average reaction rate A B is that,
26:37.820 --> 26:44.820
because use the average reaction rate diffusion
and reaction occurs parallelly. So, what would
26:47.869 --> 26:52.219
be my model?
26:52.219 --> 26:53.669
So,
26:53.669 --> 27:00.669
So, minus the effective Laplacian of C should
be equal to minus R of v. So, this is my diffusion
27:03.989 --> 27:10.989
reaction because this has been you know C
is a substrate. So, C is a substrate concentration.
27:16.830 --> 27:23.830
So, this reaction is A sync term, because
it being consumed fine. So, this is my basic
27:25.009 --> 27:32.009
equation the reaction rate is here given as
k v times c. So, the Laplacian the effective
27:33.710 --> 27:40.710
transverse Laplacian of C equal k v times
C. Now, what would be my boundary conditions?
27:41.509 --> 27:48.509
Let us talk about that there is several possibilities
with boundary conditions, but let us you know
27:54.759 --> 27:59.339
talk about boundary conditions. So, what are
my two boundaries one would be at the centre,
27:59.339 --> 28:01.749
because we are going to this is the symmetric
we have to assume.
28:01.749 --> 28:05.419
So, depends on how what we assume, if we assume
Laplacian to be A symmetric Laplacian means
28:05.419 --> 28:12.419
in the case what would do you have let us
you know in think you do not really remember
28:15.159 --> 28:22.159
all the Laplacian things. So, this is the
no this is the rectangular slab. So, there
28:22.529 --> 28:29.080
is not you know not too much of question still
there is, but if it were cylindrical then
28:29.080 --> 28:30.029
there is A question of symmetric.
28:30.029 --> 28:35.559
Now what we assume over here, let us put this
assumption very clearly, if this is my rectangular
28:35.559 --> 28:42.559
slab and I am going to consider diffusion
in this region in this direction. Which is
28:46.339 --> 28:53.339
x and diffusion in the y direction is neglected
why, because it is assumed to be too long
28:56.440 --> 29:03.190
fine. So, diffusion is neglected now if I
am doing it, because of symmetry my condition
29:03.190 --> 29:10.190
at this my x equal says this x equal to 0.
So, at I think that is my x equal to 0 my
29:13.239 --> 29:17.649
fine, now what will be the other condition?
29:17.649 --> 29:24.649
Is equal to L what there several possibilities
with this. So, you have to just pick one the
29:27.799 --> 29:33.179
let us talk about the simplest. Now we will
do the little more complicated one in the
29:33.179 --> 29:35.219
next class.
29:35.219 --> 29:40.349
Continuity: is the simplest form. So, there
could have A robin boundary condition, which
29:40.349 --> 29:45.759
is flux type flux with concentration or you
can have simple continuity. So, let us assume
29:45.759 --> 29:52.399
the simple one which is continuity at this
point I think this would be k v C naught.
29:52.399 --> 29:59.399
So, what is k v? So, this is my x equal plus
minus 1 x is dimensionless by the way here.
30:01.070 --> 30:06.190
So, how make dimensionless just by dividing
by l. So, k C equals k A v times C naught
30:06.190 --> 30:12.710
and rather boundary condition is the centre
x equal to 0 Del C Del x equal to 0 fine.
30:12.710 --> 30:19.710
So, can we go ahead and solve this why not
you do that. So, what would be max Laplacian
30:20.139 --> 30:27.139
simply D effective time Del 2 C Del x square
equals k v C. So, again you know we are going
30:30.599 --> 30:35.429
to assume these two different limits of it
1 is the zeroth order and one of the first
30:35.429 --> 30:39.399
order like we had been doing all through.
So, the zeroth order is very easy why, because
30:39.399 --> 30:44.580
it is simply came k v and you can just go
and straight integrate it. And then what the
30:44.580 --> 30:51.580
solution you know you have done the something
time. So, the first order what is the solution
30:52.820 --> 30:54.960
of this?
30:54.960 --> 31:00.469
First order of this what is the solution of
this?
31:00.469 --> 31:02.739
Problem
31:02.739 --> 31:09.739
No, look carefully first
is you have to see what is physically meaning
full A sin x and B cos.
31:17.219 --> 31:20.469
into power minus...
A sin x B cos x what is the problem with that
31:20.469 --> 31:26.959
you know one of them I dropped out. You will
have you will have A sinusoidal wave of concentration
31:26.959 --> 31:31.469
inside the matrix is that physically meaningful.
So, whenever you write solution the thing
31:31.469 --> 31:34.700
it will figure out whether it is physically
meaningful or not and that is not physically
31:34.700 --> 31:39.279
meaningful. So, the physically meaningful
solution would be exponential and you, because
31:39.279 --> 31:44.089
you forgotten how to do these things which
is A strange thing that you know. So, essential
31:44.089 --> 31:49.769
when you have a positive on this side you
will have an exponential. If you have a negative
31:49.769 --> 31:56.269
out there then only you can have that and
that is not possible, because you know that
31:56.269 --> 32:03.269
that is not possible out here.
So, the solution would be simple A cos hyperbolic,
32:03.429 --> 32:10.429
I think what is going to I am going to do
this little differently later that is do dimensionless
32:14.909 --> 32:21.909
and all the stuff, but just to give you A
sense. I think k v over D effective I think
32:22.379 --> 32:29.379
why cosine hyperbolic x and sin hyperbolic
x just from the plus x and minus x I get this.
32:29.639 --> 32:35.159
So, what we are going to do now is use a dimensionless
form of this we can go ahead and solve this,
32:35.159 --> 32:39.529
but the advantage of making the system. Dimensionless
is that you can just scale all the variables
32:39.529 --> 32:43.709
as you want and you will see that it is very
easy to work with the dimensionless system.
32:43.709 --> 32:50.279
So, what we do is make it make the system
non-dimensional and what will happen is, when
32:50.279 --> 32:56.389
we make the system non-dimensional it will
come up with numbers dimensionless numbers.
32:56.389 --> 33:00.839
Which ratios of important time scales in system
and that is one of the added advantages of
33:00.839 --> 33:04.089
making the system dimensionless, it comes
out naturally and these are very important
33:04.089 --> 33:09.029
number. So, the first number we will get is
in this case is A Thiele modulus and I have
33:09.029 --> 33:11.679
already taught discussed A Thiele modulus.
In the last class if I remember which the
33:11.679 --> 33:15.479
ratio of diffusion time is over reaction time
and why is this important here, because these
33:15.479 --> 33:18.229
are the two dominant phenomena in the system
diffusion and reaction.
33:18.229 --> 33:21.989
So, it is very important to have the sense
of the numbers ratios of these numbers why
33:21.989 --> 33:26.429
that important to have a sense, because if
Thiele modulus is given as ratio of diffusion
33:26.429 --> 33:33.429
time to reaction time and Thiele modulus is
very large. That means that diffusion is...
33:34.700 --> 33:38.209
If
Limiting
33:38.209 --> 33:45.209
Limiting. So, that is something that you know
we need to figure out here so, why do not
33:49.089 --> 33:54.669
you do it on your own little bit. So, let
we all help you out now out with this.
33:54.669 --> 34:01.669
So, say x that because we need to do this
unless and otherwise. So, this is all we need
34:06.589 --> 34:13.589
to go here so, D effective times Del 2 C Del
x square equals k v into C. So, x is this
34:19.560 --> 34:26.560
is how define x. So, you get the d effective
L square Del 2 theta Del x hat square equals
34:35.299 --> 34:42.299
k v into theta. So, k v. So, k v L square
over D effective time’s theta, now this
35:01.359 --> 35:08.359
is this could be written as L square over
D effective divided by 1 over k v time’s
35:10.690 --> 35:17.549
theta. So, this is phi square over theta.
So, this is term L square over D effective
35:17.549 --> 35:24.220
is the t D transverse diffusion and this is
the transverse reaction time so, you get theta
35:24.220 --> 35:29.769
phi square. So, this is your basic equation
Del 2 theta Del x square equals phi square
35:29.769 --> 35:35.740
times theta and that phi square phi. Why phi
square is k v L square over D effective and
35:35.740 --> 35:38.670
that is the ratio of the transverse diffusion
time of reaction time.
35:38.670 --> 35:45.670
So, this is my dimensionless model and my
dimensionless boundary conditions are very
35:46.109 --> 35:53.109
straightforward. So, at x equals plus minus
1 you get theta equals 1 and x equals 0 Del
35:53.289 --> 35:59.529
theta Del x equals 0. This is very straightforward
and solution is also something I had discussed.
35:59.529 --> 36:06.529
So, this is the solution, but you should write
it as so, but I know I think it is always
36:10.880 --> 36:17.880
best to write it as A cosine hyperbolic and
A sin hyperbolic, because it is much easier.
36:17.950 --> 36:24.950
So, if you write theta as say A 1 cosine hyperbolic
phi x, because if you look at your reaction
36:27.619 --> 36:34.619
equation over here it is Del 2 theta. So,
equation over here is Del 2 theta Del x square
36:35.549 --> 36:42.549
equals phi square times theta. So, the solution
is theta equals A 1 cosine hyperbolic phi
36:44.559 --> 36:51.559
x plus A 2 sin hyperbolic phi x. So, Del theta
Del x equals phi times A 1 sin hyperbolic
36:56.829 --> 37:03.400
phi x plus A 2 cosine hyperbolic phi x.
37:03.400 --> 37:10.400
So, next step is that at X equal to 0 Del
theta Del x equal to zero. So, Del theta Del
37:17.740 --> 37:24.740
x at x equals to 0 turns out to be phi times
A 2 cosine hyperbolic phi x equals phi times
37:29.910 --> 37:36.910
x equals 0 phi times A 2 which is 0 which
implies that A 2 equals 0. So, what you get
37:40.000 --> 37:47.000
is theta equals A 1 cosine hyperbolic phi
x now your other boundary condition is that
37:48.160 --> 37:55.160
x equals plus minus 1 theta equals one. So,
equals A 1 cosine phi x. So, with my A 1 becomes
38:05.740 --> 38:10.549
1 over cosine hyperbolic phi.
38:10.549 --> 38:17.549
So, my solution is theta equals cosine hyperbolic
phi x divided by cosine hyperbolic phi. So,
38:24.339 --> 38:31.339
this is what I have which is which equals
C over k a v C naught. Now, what we intend
38:40.980 --> 38:47.019
to do now what we intend to do is find my
effectiveness factor eta. So, eta if you remember
38:47.019 --> 38:54.019
was defined as R of the average over R of
the average R of v evaluates with this C s.
38:59.650 --> 39:06.650
So, my C s is x a at x equals 1, C s is defined
as x equals 1 which is given as k v C naught.
39:11.250 --> 39:18.250
So, the denominator is known only you have
to do is evaluate the numerator k a v C naught
39:21.309 --> 39:28.309
clear. So, how do you evaluate the numerator?
39:32.440 --> 39:38.619
I think you can use that, but I think one
of the other ways that is also you can do
39:38.619 --> 39:43.869
you can, because we got the theta. You can
just evaluate the surface and do that and
39:43.869 --> 39:48.880
another option is what I have on the screen
over here just use this straight away calculation
39:48.880 --> 39:55.359
of eta this one. So, because you know what
is the volume of your slab? So, and what you
39:55.359 --> 40:00.039
got over here is this there is A slight typo
which should be R this 1 should be R v average
40:00.039 --> 40:06.000
you know this should be on the top of it.
So, minus R v average is simply going to be
40:06.000 --> 40:13.000
1 over V times R v times d v and that that
equals what we got. So, k C time d v equals
40:20.400 --> 40:27.400
k times k a v times C naught into theta times
d v and d v you can cancel out a v on both
40:35.069 --> 40:40.109
sides.
So, you will get 1 over I think 2 L this goes
40:40.109 --> 40:47.109
from minus L to plus L k a v C naught times
theta into d x is it. So, then you can write
40:50.490 --> 40:57.490
this take this two out and write this as 1
over L going from 0 to L k a v C naught theta
40:57.740 --> 41:04.740
into d x fine . So, what would be the integral
if I have A cos hyperbolic it will simply
41:05.690 --> 41:09.849
be sin hyperbolic integral.
41:09.849 --> 41:16.849
So, know that is not minus R v equal 1 over
L 0 to L k a v C naught theta was what cos
41:23.519 --> 41:30.519
hyperbolic phi x divided by cos hyperbolic
phi I think of d x not this k a v C naught.
41:40.920 --> 41:47.920
Now this integral turns out to be this whole
thing turns out to be k a v C naught over
41:57.869 --> 42:04.869
L slightly complicated integral. This is comes
out to be this tan phi if this cos hyperbolic
42:05.359 --> 42:11.730
phi x 0 to L divided by cos hyperbolic phi
comes out be tan phi over phi. So, you can
42:11.730 --> 42:18.730
do that what you have to do is essentially,
because you know this will be sin. So, you
42:19.490 --> 42:25.490
have A cos, but this 1 over phi will come,
because of the integral. So, this will be
42:25.490 --> 42:32.490
sin hyperbolic phi over phi over cos hyperbolic
phi in the denominator. So, my eta is going
42:34.099 --> 42:41.099
to be what? Does my eta is going to be...
42:44.259 --> 42:51.259
Let us go back here eta is R this R v is this
and this is k actually here and I think the
42:52.640 --> 42:59.640
L should go from here, because there will
be you know L and you know this. If you put
43:02.369 --> 43:07.539
this in the dimensionless coordinate, because
x is in the dimensional coordinate. What do
43:07.539 --> 43:14.539
you used hat over here? So, if I convert this
is a hat and if I convert this with the hat
43:14.710 --> 43:21.710
then you got L over here then the L would
go. So, this is what you have.
43:22.430 --> 43:29.430
Then your eta is going to be R v over R v
C s, now it comes out be k a v C naught over
43:44.029 --> 43:51.029
by k a v C naught tan hyperbolic phi over
phi equal. So, this is what we get and this
44:06.019 --> 44:11.700
is a very important formula to remember for
eta, because I will show you later may be
44:11.700 --> 44:16.319
in the couple of lectures. So, that this is
something that you can use independent of
44:16.319 --> 44:22.289
geometry, because though you change the geometry
and we will do in the little time. We have
44:22.289 --> 44:29.289
another geometry you will see it is difference,
but what happened is that overall be it cylinder
44:30.490 --> 44:37.490
or sphere or A slab the effectiveness factor
is more or less close to tan hyperbolic phi
44:39.490 --> 44:43.400
over phi more or less close to.
So, if you use this number you are not really
44:43.400 --> 44:47.140
very off. So, something this is something
you might want to remember this formula. So,
44:47.140 --> 44:54.140
this is just going through what you have done
over here the same thing you will get the
44:54.480 --> 45:00.269
tan hyperbolic phi over phi. So, what we do
now is that in the next 10, 15 minutes we
45:00.269 --> 45:07.269
have lets quickly do this same thing and I
want you to go take the lead on this is take
45:07.819 --> 45:14.819
exactly the same thing, but this. So, this
is the typo over here this is not A rectangular
45:15.250 --> 45:22.180
slab. So, I want you to do this in A sphere.
So, this is that example, we had before much
45:22.180 --> 45:29.180
before the picture I showed you where the
enzymes are actually trapped here.
45:30.609 --> 45:37.140
So, the enzymes are actually trapped and this
is a very common example, because you know
45:37.140 --> 45:42.380
when you take some of your capsules. For example,
enzymatic capsule they are the spherical like
45:42.380 --> 45:47.960
stuff and the enzymes entrap. So, this is
very physical, very practical example. So,
45:47.960 --> 45:53.079
this is the one that I want you to look at
now. And I showed you how to do it. So, what
45:53.079 --> 46:00.079
I want you to do in the next few minutes are
go and do it for the sphere. So, first write
46:02.430 --> 46:06.460
this is again your basic equation. All you
have to do is write Laplacian in the in the
46:06.460 --> 46:10.859
spherical coordinate and then go and make
it dimensionless and then go and solve it.
46:10.859 --> 46:17.859
So, just star doing it and I will help you.
So, we have written the Laplacian I do not
46:47.509 --> 46:51.539
know some of you might have forgotten, but
if you have not forgotten. Then this is what
46:51.539 --> 46:56.799
you have on the screen is Laplacian and the
spherical coordinate. So, D effective again
46:56.799 --> 47:00.829
over R square d R of R square Del C Del R.
Now what is your boundary conditions going
47:00.829 --> 47:06.609
to be?
47:06.609 --> 47:09.779
Because you have assumed symmetry so, the
basic assumption I want here that this is
47:09.779 --> 47:14.059
a symmetric and once a symmetric if it is
a non-symmetric. For example, let me ask you
47:14.059 --> 47:19.369
this for a spherical substance if this is
non-symmetric. So, what would be the can you
47:19.369 --> 47:26.369
write 2 D model for a non-symmetric here for
a cylinder for example, if it is symmetric?
47:28.630 --> 47:35.630
So, this is my D effective of Laplacian k
C let say k c. So, for a cylinder what is
47:40.170 --> 47:47.170
my Laplacian? 2 D Laplacian is 1 over R d
R of Del C Del R plus Del 2 C Del x square
47:52.519 --> 47:59.519
this is for a cylinder. Now this for a symmetric
cylinder symmetric cylinder, now if I ask
48:07.400 --> 48:14.400
you for a symmetric sphere can you write the
2 D model what is 2 D model for symmetric
48:23.680 --> 48:30.680
sphere. So, let me add that you know let me
close my eyes and add the radial part plus.
48:48.180 --> 48:55.180
What is the other component be there comment?
49:05.749 --> 49:12.749
The theta and phi, but I asked you to write
the 2 D model. So, which one you choose?
49:20.650 --> 49:27.650
Why is that?
49:33.089 --> 49:40.089
Think about it physically the sphere has you
know cylinder, you have this access of symmetry
49:41.420 --> 49:46.420
and you are taking symmetric about the radial
coordinate. But the physically you know how
49:46.420 --> 49:51.529
would you physically denote that about this
access of symmetry you have a symmetry is
49:51.529 --> 49:55.349
it possible to do it. In a sphere it is not
think about that in a cylinder you can do
49:55.349 --> 49:59.180
it, because in cylinder you can fix your access
of symmetry. And you can say about this you
49:59.180 --> 50:04.589
can, but in a sphere theoretically it is possible,
but physically is it possible to denote the
50:04.589 --> 50:09.539
system and say that there is an is an access
of symmetry it is not possible.
50:09.539 --> 50:13.499
Then what would be your answer? The answer
is that it is not possible to write the symmetric
50:13.499 --> 50:20.220
2 D model in sphere. For physical system theoretically
mathematicians can write, but as a engineers
50:20.220 --> 50:24.559
you know there is no point. You know do you
understand what I am trying to say that at
50:24.559 --> 50:28.970
what point how we are going to denote your
access of symmetry and say that only around
50:28.970 --> 50:34.430
this. There is a symmetry, because if you
there is a symmetry in this direction. Then
50:34.430 --> 50:38.240
if there is no physical demarcation of that
and if you turn that sphere around then there
50:38.240 --> 50:44.730
is no symmetry. So, if you write have to write
a symmetric model the way do it. Is actually
50:44.730 --> 50:50.420
gone and write a 3 D model you can have both
in phi and theta.
50:50.420 --> 50:56.140
So, let us go ahead with this I do not think
we will have the time to finish this, but
50:56.140 --> 51:00.400
the boundary condition is such symmetric boundary
conditions, because you have assumed symmetry
51:00.400 --> 51:06.660
r equals 0 Del c Del r equals 0. What will
be the other boundary condition? Straightforward
51:06.660 --> 51:13.589
that at on the surface the concentration is
given by k v C naught. So, how do you solve
51:13.589 --> 51:20.589
this? So, the surface is given by r equals
R m. So, how do you solve this now? So, one
51:31.990 --> 51:36.079
thing that we can do is we can safely close
our eyes and make the system dimensionless
51:36.079 --> 51:42.819
for the sake of simplicity my dimension is
variable would be R m. So, zeta is given by
51:42.819 --> 51:49.819
r over R M then my governing equation is given
by this 1 over xi square Del d xi of xi square
51:50.410 --> 51:55.130
Del c Del xi equal phi square times c.
51:55.130 --> 52:02.130
No, solution is not Bessel function you need
to go and brush up your mathematics. So, let
52:04.220 --> 52:08.380
me go back a little bit we have complete run
out of time, but because we have mentioned
52:08.380 --> 52:13.170
the Bessel function. So, what is the coordinate
is a Bessel function is actually in the cylindrical
52:13.170 --> 52:13.549
coordinate.
52:13.549 --> 52:20.549
So, I did not do the cylinder, but that does
not mean that I have to go back and do the
52:21.529 --> 52:28.529
cylinder now. So, this is my D effective d
r of r Del c Del r equals k v C. So, after
52:30.230 --> 52:34.950
making the dimensionless you gets 1 over r
d r this is the cylindrical coordinate by
52:34.950 --> 52:40.380
the way do not confuse its spherical. So,
let us the spherical where we there, because
52:40.380 --> 52:47.380
the mentioned Bessel function that I am going
back to phi square of C fine. So, what would
52:47.609 --> 52:52.660
be the solution of that the solution of this
is going to Bessel function. So, this you
52:52.660 --> 52:59.660
can write as Del 2 c Del r square plus 1 over
r Del c Del r equals phi square times C multiply
53:03.980 --> 53:06.170
this by r square both sides.
53:06.170 --> 53:13.170
Then we will have r square Del 2 c Del r square
plus r Del c Del r minus phi square r square
53:20.329 --> 53:27.329
C equals zero. So, then your new variable
you have to what is the next step you have
53:27.420 --> 53:34.420
to find a new variable y?.
y equals phi times r. So, then equation becomes;
53:53.539 --> 54:00.539
now you can use the Bessel function solution.
This is you know function solution what I
54:04.980 --> 54:08.819
want you to do I will stop it here today,
but what I want you to do is to complete this
54:08.819 --> 54:11.900
Bessel function solution. Because these are
things that you need to review on your own
54:11.900 --> 54:16.289
and I cannot review. This is not a maths class
complete this Bessel function solution and
54:16.289 --> 54:20.880
come up with some clue and what would be the
solution of the spherical thing. So, the cylinder
54:20.880 --> 54:25.900
the flat plate was very easy, it is cosine
hyperbolic and sin hyperbolic this is the
54:25.900 --> 54:29.289
Bessel function solution. I gave you the solution,
but all you have to do is figure out what
54:29.289 --> 54:32.839
kind of Bessel function first kind, second
kind modified what kind of Bessel function
54:32.839 --> 54:38.690
is going to be and...
So, you figure that out and last thing that
54:38.690 --> 54:43.180
you have to do is, what would be your solution
in the spherical coordinate? Which is the
54:43.180 --> 54:50.170
most pretty interesting thing to do? So, I
will leave at that if there is any question
54:50.170 --> 54:54.650
on the part that we did today. So, what we
essentially did with let me talk conceptually
54:54.650 --> 54:59.930
for 1 minute. About what we did we and the
whole thing whole lecture kind of on this
54:59.930 --> 55:06.140
idea of the pseudo homogeneous model and the
basic idea that we try to discussed is that
55:06.140 --> 55:10.009
for a pseudo homogeneous model. Pseudo homogeneous
model is really actually for a heterogeneous
55:10.009 --> 55:15.630
system, where these different points sources
of sync of reaction and essentially. If the
55:15.630 --> 55:21.599
two conditions on which you can do an averaging
one is that the microscopic scale length.
55:21.599 --> 55:28.599
The microscopic scale over the length is much
smaller than 1 and second is there is a large.
55:28.609 --> 55:33.950
There is no large variation in concentration
over the entire domain; it should not be the
55:33.950 --> 55:37.069
concentration is very smaller at one place
and very large. In another if these two conditions
55:37.069 --> 55:40.859
are satisfied then you can average it over
the volume and come up with the pseudo homogeneous
55:40.859 --> 55:47.859
model. The main characteristic of the pseudo
homogeneous model is that diffusion and reaction
55:48.200 --> 55:52.289
occur parallel in the pseudo homogeneous model,
while it does not occurs parallelly in the
55:52.289 --> 55:56.769
heterogeneous model. And then you can just
write the diffusion equation and solve this
55:56.769 --> 56:00.049
so, what are these are two things? I want
to you to come up with one is the final Bessel
56:00.049 --> 56:04.099
function solution and second is how to solve
it in the spherical coordinates. So, tomorrow
56:04.099 --> 56:07.529
I will start off with that what is that?
56:07.529 --> 56:14.519
No, not a function. So, just a trick you know
you have to do this and it will come. So,
56:14.519 --> 56:15.430
we will stop here thanks.