WEBVTT
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There in the last lecture of reaction and
diffusion that occur simultaneously in rectangular
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slab is something that you found little hard.
So, what I will do today before I go into
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the spherical coordinates and or immobilizing
themselves sphere, I will try and recap once
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again what we did with a rectangular slab.
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So, we are looking at this problem let us
go through the to the transparency, and we
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will we are looking at this problem of the
rectangular slab being here and the enzymes
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are immobilized inside that rectangular slab.
So, these are infrared solvents encapsulated,
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and the length is the total length of in the
direction in which the diffusion occur this
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2L, and what we are suppose to do is, this
is a heterogeneous system essentially right,
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but what we did we did a pseudo homogeneous
approximation of it. So, once we do a pseudo
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homogeneous approximation, we assume that
reaction and diffusion occur simultaneously.
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So, did just recapping what we did. So, it
is just a quick question did any of you have
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any questions or doubts about what we did
in the last class about the concept? What
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your problem was with the concept or your
problem was with the calculations?
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Sir concepts.
Concept. So, I will run through the concept
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one more time.
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So, the idea was that you know here let us
go back the idea was that when you have inter
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heterogeneous reactions and see a quick for
example, reaction in a porous catalyst. So,
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reaction itself is heterogeneous, but when
you look at it from microscopically from a
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large scale point of view. Then you can assume
the reaction to be homogeneous. So, what are
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the criteria we discussed two criterions;
one is that the length scale of heterogeneity
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over the microscopic length scale should be
very, very small. So, in this case for example,
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if I am looking at a matrix like this, in
which the enzyme is impregnated or encapsulated
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the length scale of the pore over the entire
thickness of the solid core.
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So, this is a core for example, or the encapsed
total encapsulation if this length scale is
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L say or 2 L and that length scale of the
pore is d the diameter of the pore, then d
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over 2 L should be very small that is my criteria
number one to be very specific I said it will
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conceptual way. So, that is let me write it
down also.
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So, if this is my encapsulation in the whole
thing and this length scale is L and the length
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scale the length scale of a pore for example,
is say d then d over 2 L should be very smaller
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than one for the pseudo, for pseudo homogeneous
model
to be valid and what was the other condition
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we said.
Yeah. So, concentration C s minimum over C
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s, maximum should be of the order one. So,
it there should not be a large difference
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in variation between the maximum and the minimum
concentration. Because see the whole idea
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is that averaging, what is you are doing here
is averaging now? What is the average of any
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population signified? It means that somewhere
you know you are trying to have a sense of
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the total population by looking at the average
now, if that if you have three people over
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there and four people over there and. So,
when you taking average of the four people
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out of which the ages have two of them are
say 5 and 6 years old then, the other two
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ages are 60 and 70 or 65 and 70 then you will
get an average of whatever 20 or whatever
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it is it makes no sense because it gives you
no sense of the population.
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So, that is the concept here is that the maximum
over the minimum over maximum concentration
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should be of the of some order one should
be very small number. So, once you are able
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to do that then we can substitute.
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So, what we do is, we average out this thing.
So, here we are we have the reaction let us
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go back to the reaction. So, the here we have
the reaction. So, this is the R that you see
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over here is the reaction at the small scale
at the scale of the pore. And if there is
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in uniformity model I think uniformity in
the concentration and the minimum to the maximum
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pore scales minimum pore scale size over the
macro scale size of the core scale is much
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smaller than one. Then only can do an averaging
which means that we tend the reaction rate
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and integrated over the volume and divided
by the total volume then you will get a average
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reaction rate which is given over here as
R v bar.
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So, this is the basic concept there should
it should be now. So, then we went over to
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this particular example. So, once we have
been able to do that averaging, what we have
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been able to achieve? We have been able to
convert a heterogeneous model which is much
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difficult to quite difficult to solve to a
homogeneous model. We call it a pseudo homogeneous
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model because it is not really homogeneous
we are just assuming to be homogeneous. So,
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once we have been able to write the pseudo
homogeneous model. So, then we can write a
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reaction rate as time just like a homogeneous
reaction rate which is first order homogeneous
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reaction rate k v times C. And again why you
are doing first order? Because we are looking
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at the two asymptotes of first order and the
0 the order and the Michaels mention kinetics
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some or it going to be in between this. I
am going to come and show you actually how
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the Michaels mention kinetics looks like may
be if not in today’s class the first mandate
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in the next class.
So, once we have been able to write our assume
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on pseudo homogenous model then you simply
assume that. So, when once you assume a pseudo
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homogenous model, what does it come with the
other assumption that diffusion and reaction
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can run parallely. So, in reality. So, what
is a difference? The difference is this model
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that you are writing tries to mimic reality,
but is not in a direct actual representation
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of the reality, it is a it is some solved
form the representation of reality. So, in
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reality diffusion and reaction occur do not
occur parallely they occur in series, but
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when you convert them into a pseudo homogeneous
model, diffusion and reaction can be assumed
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to occur parallely.
So, when you assume them to occur parallely,
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then you can go back and write an equation
of this sort that I have wrote over here.
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The effective length laplacian of C equals
k v time C. So, k v time C being the reaction
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rate and the left hand side being the diffusion
rate. What is that to note here is? The d
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a factor because we are using a pseudo homogeneous
model the effective diffusivity is some total
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of the diffusivity through the solid pore
solid part of the material and through the
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fluid in the pores. So, the because you have
different diffusivities much different diffusivities
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actually through the solid and the and the
fluid and.
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So, if you had a simple fluid that gives you
much higher where this effective diffusivity
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is going to be much lower than that because
of the presence of solids fine. And then the
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solution was I do not know if you have the
solution the solution we got by using the
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two boundary conditions one is that at this
extremity x equals plus minus L actually it
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should not be 1 x equals plus minus L. So,
this boundary and this boundary you assume
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that concentration equals k v k v time C naught
fine. And the other condition was that at
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x equals 0, del C del x equals 0 that comes
to a symmetry. We assume that it is because
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your it is exposed to the same boundary condition
on both sides. We assume symmetry one thing
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I want you to understand at this point is
that you can have this equation, but if I
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had two different boundary conditions on two
sides.
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For example, at x equals plus L I had k a
v times C 1 or C naught 1 and then this x
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equals minus l k v time C naught 2, then you
cannot use the symmetry boundary condition.
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For symmetry, you need symmetry of equation
as well as symmetry of boundary condition
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the other boundary condition should be symmetric.
So, if you have the same slab, and it is exposed
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to one kind of boundary condition, one kind
of concentration on the left hand side; another
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kind of concentration on the right hand side.
So, of course you cannot expect symmetry.
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So, in that case what how do you go and solve
the equation? You do not have boundary condition
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tool. So, what did you do?
If a.
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Condition.
So, the only difference is here you are solving
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for half the slab, but there you are not going
to solve it for half the slab. You are going
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to solve it for the full slab and the boundary
conditions would be on both sides. So, you
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will use k naught C a k a v C naught one on
one side and k a v C naught two on the other
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side. So, this is not going to be symmetry.
So, these are assumption that you have to
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be careful about and should not get too complacent
about these things because you know. So, I
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might give them details. For example, I might
give you the same problem with just two different
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boundary conditions on two sides, and do not
assume symmetry that will be very stupid of
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you to do. So, just you know be careful about
what kind of boundary conditions have been
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used.
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So, then the solution of this is a very standard
solution which is a cos hyperbolic x and b
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cos hyperbolic y, but before we do that what
we did was? We did the equations done the
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equations dimensionless because you know it
is easier for scaling. And when we turned
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the equations dimensionless we got the dimensionless
number Thiele modulus which is k v L square
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over d effective it is a ratio of the transverse
diffusion time over reaction time. Why do
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you call it the transverse diffusion time?
Because it is a diffusion time in the transverse
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direction; so, you have this is the transverse
diffusion time t d and you can have a diffusion
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time in this direction as well which is typically
you know as t d x which is the actual diffusion
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time.
So, anyhow, so because I. So, what would be
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that actual diffusion time it would be the
length scale l over here, if I am to write
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the actual diffusion time the length scale
l over here will be replaced by the actual
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diffusional length.
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So, to repeat again. So, t D r if I call this
is the transverse diffusion time say this
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is L square over D D effective, and if I have
a actual diffusion time, then what would that
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be? So, if this is my 2 L and let us say this
is my B then there would be.
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B square.
B square.
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By.
Not necessarily this is d effective R and
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this is d effective x, because not necessarily
if the diffusion coefficients are going to
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be the same in the radial and the actual coordinates.
So, these are the things I want you to understand.
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because you know the for example, the mid
seems are coming up and one of the things
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I would like you to do when. I give you a
problem in the mid sem is not to just pounce
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under problem and solve it and get some numbers
that is not what I am looking at, what I am
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frankly look at is that, how you approach
a more you know these system that we are modeling
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over here there is some intuition there is
some assumptions there is some understanding
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that come into the modeling. It is not just
about writing pouncing on the on the system
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and writing an equation and solving it because
at the end of the day you might get a solution
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which do not in any way they represent the
system that you are trying to solve.
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So, what is important is to understand the
system? For example, I gave you this example
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of the boundary of the boundary condition
being asymmetric one possibility, the other
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possibility is if you have to consider both
actual and radial diffusions. Then there a
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diffusion times might be very different, and
it is possible there are problems when your
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there could be encapsulations in matrices
where the radial diffusivity could be very
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different from the actual diffusivity, or
in other words the radial diffusion diffusivity
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itself is varying actually. Those kind of
complications can come because these are matrices
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you make and you know you can make them in
certain way.
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So, you see do you see what I am trying to
say? So, the radial you have a certain radial
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diffusivity with the radial diffusivity itself
is varying actually, because how the pore
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structure is varying. The radial pore structures
that you have might vary actually. So, anyhow.
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So, we did that and we used a exact used as
x over l my dimensionless coordinate and theta
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as the dimensional variable C over k v C naught.
And then my dimensionless equation became
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model became del two theta del x s square
equals phi l square theta. And then the solution
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is this is what we wrote and we think we wrote
it slightly differently. We wrote it cos hyperbolic
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and sin hyperbolic, and got rid of the sin
hyperbolic we got just a cos hyperbolic, and
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then what did we do? We calculated eta the
effectiveness factor.
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So, the effectiveness factor is the reaction
rate as it happens inside the matrix itself
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which is a observed reaction rate, total reaction
rate, actual reaction rate. Actually not the
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intrinsic reaction rate, over the reaction
rate , intrinsic reaction rate evaluated the
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surface. That is how I have we have defined
if you remember our eta. So, this is how we
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defined our eta? R v over R v value to that
C s now R v, general R v is much lesser because
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the concentration inside is much lesser now.
If there were this is something that you need
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to understand if there were no diffusional
limitations in the system then your eta is
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going to be one. So, because of diffusional
limitations that occur in the system ,what
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happens your R v? R v is decreased, because
the concentration at any point inside the
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matrix is lower than that at the surface.
So, because the concentration is lower and
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this is being a first order reaction it is
different.
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Now, what would you presume for a 0 the order
reaction? This we had discussed it in the
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different context in the last week’s class
I believe.
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Eta will always be one.
Eta will always be one, why is that? Because
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0 the order reaction is independent of concentration
a and you know, more than that this is the
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whole idea of 0 the order reaction as I told
you is based on the fact that there is plenty
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of substrate. If there is plenty of substrate
then the whole concept of mass transfer limitations
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do not come in at all. This is what we did
in the in the previous scene that we did?
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So, eta is s and as I told you the tan had
hyperbolic phi over phi is what comes out
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now, what happens one interesting thing that
I asked you to do as, remember this formula
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eta s tan hyperbolic phi over phi the reason
being that . This is a generalized formula
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this is something that you can use them and
how do you use this is a and there’s a method
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called shape normalization we heard of this
shape normalization.
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So, shape normalization is essentially that
if you convert if you have any shape be spherical
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be it cylindrical be it is a very weird shape.
You know the pentagon or cone some kind of
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weird shape pyramidal which you cannot capture.
So, you cannot solve the equation in every
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shape all of these shapes. So, what you do
is, you transform these shapes into standard
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shapes for example, a slab. And then you do
all the calculations and when you transform
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them the only thing that is transformed is
eta for phi square, for example, it is go
15:27.379 --> 15:34.379
this k v L square over d effective. So, there
is that is your phi square. Now this L square
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over here, now your shape normalized length
scale, do you see what I am saying. So, this
15:52.720 --> 15:57.490
L square now, so in your definition of phi
square this is. Is it correct right you see?
15:57.490 --> 16:03.100
So, in the definition of L square by a phi
square the L that you have the length scale
16:03.100 --> 16:07.910
that you have is now a shape normalized length
scale. So, which for example, I will give
16:07.910 --> 16:12.139
you in our next class itself I will show you
what it is R over 2 and R over 3 per cylinder
16:12.139 --> 16:16.829
and sphere and. So, you can come up with what
the shape normalized length scale even then
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you can. So, what is the idea is that irrespective
of what from you have you can transform it
16:22.889 --> 16:28.970
into a slab. So, is that a perfect answer?
No, it is not a perfect answer, but the advantage
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is this is very close to the perfect answer.
So, if you say that your eta is, tan hyperbolic
16:35.559 --> 16:42.559
phi over phi and you define your phi this
way in terms of the shape normalized value,
16:44.389 --> 16:47.790
and there is a mechanism for the that you
know R over 2 R over 3 for different shapes
16:47.790 --> 16:50.769
there this how this l is going to vary is
given.
16:50.769 --> 16:55.209
So, what you can do? You can just use that
value shape normalized value of radius and
16:55.209 --> 16:59.339
then use this formula this of this is not
going to be a hundred percent perfect solution
16:59.339 --> 17:04.459
to your question, but you are 95 percent there
you in terms of your eta. So, the advantage
17:04.459 --> 17:08.910
is that if you do not for example, if you
want to use it as a form for a formula and
17:08.910 --> 17:12.809
something and you do not remember it. You
can just remember this single formula and
17:12.809 --> 17:17.640
then we scale your phi based on that fine.
So, this is I think where we stopped and then
17:17.640 --> 17:24.640
I gave you this example of these. Now two
things I asked you to do one who has to be
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able to solve in a cylindrical coordinates.
So, what is the solution of that in the cylindrical
17:29.390 --> 17:36.390
coordinate? I do not have I think I have these
pages over here, I have got that pages.
17:37.290 --> 17:44.290
So, this is the Bessel function solution you
have. So, this is this is one form if I remember
17:47.310 --> 17:53.650
this is one form, or another form you can
do also if you have this r square del 2 C
17:53.650 --> 18:00.650
del r square plus r del C del r minus phi
square r square C equal zero. So, I can call
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my say y hat equals minus phi times R, then
you can convert this to another form let us
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call this not y hat, let us call this xi just
for the sake of. So, then your xi square del
18:30.410 --> 18:37.410
2 C del xi square plus xi into fine plus xi
square C equals 0. So, if I keep this and
18:46.960 --> 18:51.450
this over here next to each other, both is
Bessel functions they are just two different
18:51.450 --> 18:54.770
Bessel functions of two different kinds. Why
is that?.
18:54.770 --> 19:01.180
One is Bessel one is modified Bessel. So,
you can use either form. So, what will be
19:01.180 --> 19:08.180
the solution for the solution of this one?
So, C equals.
19:12.760 --> 19:19.760
a j naught.
Yeah j naught phi plus b.
19:23.220 --> 19:30.220
No k k it is k.
Now.
19:31.360 --> 19:38.090
Y
Use a bold alphabets. So, this is the solution
19:38.090 --> 19:45.090
and then the solution for this one would be
therefore, C equals some a hat, what it be?
19:50.900 --> 19:54.660
I.
I and.
19:54.660 --> 19:55.910
k.
19:55.910 --> 20:00.340
So, these two solutions are possible. So,
either you write it in this form or that form.
20:00.340 --> 20:03.640
So, if it comes with a the negative sign,
then you have the I and k and if it comes
20:03.640 --> 20:06.980
with the positive sign then fine then you
have the j and y. It does not matter it is
20:06.980 --> 20:13.160
the same thing you will get now what after
this you have to evaluate the constants. So,
20:13.160 --> 20:17.570
out of the two these two are Eigen functions
I hope you know that out of these two Eigen
20:17.570 --> 20:22.980
functions these two here and these two here.
What will happen?
20:22.980 --> 20:27.690
Sir one is.
Which one which one.
20:27.690 --> 20:30.700
B second.
The B right in both cases the B and the B
20:30.700 --> 20:34.770
hat would be 0 why is that.
We have finite value.
20:34.770 --> 20:38.260
Finite value at R equals 0 xi xi equal 0.
20:38.260 --> 20:45.260
So, the problem is that you know both these
say the y naught and k naught, they go to
20:45.360 --> 20:52.360
infinity as this is my xi or R xi or y say.
So, as xi or y go to 0 they go to infinity.
20:55.370 --> 20:59.080
So, those two are not allowed. So, you will
you can only have your solution in terms of
20:59.080 --> 21:03.170
J naught and I naught, and then it is simple
because all you need to do. So, if you have
21:03.170 --> 21:09.270
your solution as C equal say a j naught xi.
Then the other solution if I remember you
21:09.270 --> 21:13.960
have at xi equals one or something right xi
equals one is the other concentration that
21:13.960 --> 21:19.480
is given let us look at the screen here.
No not this yeah. So, we are looking here.
21:19.480 --> 21:26.390
So, and. So, this is the same kind of thing.
So, the concentration at xi equals, one say
21:26.390 --> 21:33.390
is some C naught C equal C naught fine. So,
then you are pretty straight forward. So,
21:33.410 --> 21:40.410
a J naught 1 so you can just put it over there.
And a would be equals to C naught over J naught
21:41.280 --> 21:45.430
something like this it may not be xi over
here it may be xi may not be one over here
21:45.430 --> 21:51.900
say even if it is xi is at xi equals, any
value you can put that value over here. So,
21:51.900 --> 21:58.900
if xi equals xi naught then you know you put
xi equals xi naught over here. So, very straight
21:59.660 --> 22:06.660
forward then you essentially your final solution
is C naught J naught xi over J naught xi naught.
22:08.650 --> 22:12.510
So, this is the cylindrical coordinate.
22:12.510 --> 22:18.450
Now, then other question, we have whether
the spherical coordinate what is it? This
22:18.450 --> 22:25.220
is the solution equation is. So, I asked you
to solve it. So, what is the solution?
22:25.220 --> 22:32.220
So, my equation is d effective r square d
d r of r square Del C del r equals k v C,
22:35.870 --> 22:42.870
and my boundary condition are r equals 0 del
C del r equals 0 fine. So, what do you do
23:05.720 --> 23:08.030
over now over here now?
.
23:08.030 --> 23:10.310
C.
r is.
23:10.310 --> 23:17.310
C times.
C by r C by r it is.
23:17.900 --> 23:21.550
C by R equals.
Y
23:21.550 --> 23:26.740
Y it should be yeah fine. So, that. So, then
what.
23:26.740 --> 23:33.540
if changes then it will come.
What is the change two?
23:33.540 --> 23:37.690
You have done it or not done it.
23:37.690 --> 23:41.770
I asked you to do it yesterday and come with.
If you are by k a by d f y is equal to 0.
23:41.770 --> 23:42.020
d 2.
Then.
23:41.980 --> 23:42.630
d 2 y.
d 2 y by dr square.
23:42.630 --> 23:43.560
dr square.
Minus k v by d effective y is equal to 0.
23:43.560 --> 23:47.660
Right.
It can be solvable y is equal to three naught
23:47.660 --> 23:53.370
xi naught.
for some no it is not cos sir it is.
23:53.370 --> 24:00.370
No it is not cos
It is
24:05.190 --> 24:08.130
But at least this form is correct so.
Hyperbolic.
24:08.130 --> 24:14.080
So, you know. So, while you know since none
of you have done it and nobody has done it
24:14.080 --> 24:19.220
right you have done it, but nobody else has
done it. So, I will do it here. So, this is
24:19.220 --> 24:26.220
my equation over here this deffective R square
del R of R square del C del R equals k v.
24:28.690 --> 24:35.690
So, let us call it f equals
C equals f over xi is it what you said something
different is it What you assumed.
24:43.730 --> 24:49.090
I assumed, but you said the other way round
I think.
24:49.090 --> 24:52.070
You can also get it that way.
24:52.070 --> 24:54.370
C by r o equals some other variable.
24:54.370 --> 25:00.720
Anyway you check that now, but this is something
that gives now. So, my del C del R equals
25:00.720 --> 25:07.720
now minus f over xi square plus one over xi
del f del f del R or I mean I you have right
25:15.970 --> 25:22.970
very you will want to say xi over here. So,
if I am correcting this or let us just let
25:23.800 --> 25:25.080
me just use another page.
25:25.080 --> 25:31.730
So, one over my equation that I am now trying
to solve is one over xi square del xi of xi
25:31.730 --> 25:38.730
square del C del xi equals C square times
C, that is what I am trying to say solve?
25:39.960 --> 25:46.960
And my boundary conditions are xi equals 0
del C del xi equals 0 and xi equals one C
25:57.440 --> 26:04.440
equals k v C naught Lisa what are these in
these equations these are in dimensionless
26:04.730 --> 26:09.420
form. So, we started with a dimensional form
and converted it with dimensionless form,
26:09.420 --> 26:14.690
and my phi square in this case is going to
be what k v.
26:14.690 --> 26:21.690
K v times big R square over j affected. So,
now my assumption is C equals f over xi. So,
26:35.430 --> 26:42.430
del C del xi equals minus f over xi square
plus one over xi del C del xi correct then.
26:48.920 --> 26:54.570
So, I will go to the next one.
26:54.570 --> 27:01.570
So, then my xi square del C del xi equals
minus f plus xi into del C del xi, now del
27:04.290 --> 27:11.290
xi of this del xi of this is, this plus and
cancels out. So, one of a xi square del xi
27:38.890 --> 27:45.890
of xi square del C del xi equals one over
xi times del 2 f del xi square. So, my equation
27:58.870 --> 28:05.870
now. So, this equals this now this is equivalent
to phi square of C phi square time C. So,
28:06.210 --> 28:13.210
what I get is? One over xi del 2 f del xi
square equals phi square times C is now f
28:13.340 --> 28:20.340
over xi no.
28:34.120 --> 28:41.120
So, my equation becomes del 2 f del xi square
equals phi square of f. So, this is a equation
28:51.010 --> 28:56.170
that you have solved before, and the solution
you know is cos hyperbolic and sin hyperbolic.
28:56.170 --> 29:00.880
So, only thing that now left is a boundary
condition. So, my boundary condition earlier
29:00.880 --> 29:07.880
was xi equals 0 del C del xi equals 0 now.
So, which means that xi equals 0 del xi of
29:14.530 --> 29:21.530
f over xi equals 0. Which means that this
is minus f over xi square plus 1 over xi del
29:28.070 --> 29:35.070
f del xi equals 0 want to play all through
by xi square, you will get minus f for xi
29:36.040 --> 29:43.040
times del f del xi equals f this is your boundary
condition at xi equals 0 which means that
29:47.850 --> 29:52.700
f equals 0 at xi equals 0 done.
29:52.700 --> 29:59.700
So, this you get your first boundary condition.
The second boundary condition is what was
30:02.190 --> 30:09.190
that the xi equals is one here, xi equals
one C equals k v C naught which means that
30:11.510 --> 30:18.510
f equals again same K v C naught at xi equals
1. Now your solution to your equation is f
30:20.740 --> 30:27.740
is Acos hyperbolic phi xi plus B sin hyperbolic
phi xi. So, first boundary condition was f
30:37.240 --> 30:44.240
equals 0 at xi equals 0 implies that, A is
0, second boundary condition was this one.
30:58.550 --> 31:05.550
So, f equals k v C naught at xi cos 1. So,
which means K v C naught equals b sin hyperbolic
31:09.740 --> 31:16.740
phi, if I remember last time it was my Eigen
function that remained was the cosine hyberpolic.
31:20.900 --> 31:24.980
This time there’s sin hyperbolic remain
because of the boundary conditions. So, which
31:24.980 --> 31:31.980
means that B equals K a v C naught over sin
hyperbolic phi; that means, mine f is K a
31:37.290 --> 31:44.290
v C naught sin hyperbolic. So, this is my
final form for f that we get what is left
31:55.240 --> 32:02.240
now? We have to calculate what?
Convert it into C, converting into C is not
32:03.890 --> 32:04.710
a problem.
32:04.710 --> 32:11.710
So, C is. So, C is f equals f over eta. So,
which is K a v C naught over xi is this straight
32:23.130 --> 32:30.080
forward no, but what is left is we have to
calculate our eta which was C, what is my
32:30.080 --> 32:37.080
definition of eta going to be now C times
C?
32:38.930 --> 32:45.930
C times four pi r square or xi square whatever
phi xi from 0 to 1 divided by xi going from
32:53.660 --> 33:00.660
C naught K a v, and K a v cancels out both
places numerator and denominator the k a v
33:07.810 --> 33:13.690
will cancel out clear. Because eta is reaction
rate devaluated overall divided by reaction
33:13.690 --> 33:17.830
rate devaluated surface. So, the K a v is
in the numerator and cancels out this is,
33:17.830 --> 33:21.260
what remains now this is? So, this is very
straight the denominator is straight forward,
33:21.260 --> 33:24.640
but the numerator is slightly tedious. So,
what I will do is? I will just give you the
33:24.640 --> 33:25.920
solution to this.
33:25.920 --> 33:32.920
So, my eta after I do all these calculation
comes out to be 3 over phi actually it is
33:34.860 --> 33:40.940
here. So, I do not need to write this. So,
this is let us seeing up. So, eta comes out
33:40.940 --> 33:47.940
to be three over phi one over tan hyperbolic
phi minus one over phi. So, what you see is
33:50.330 --> 33:57.220
that? It is a essentially goes a co tan hyperbolic
the reason being that goes A co tan hyperbolic
33:57.220 --> 34:04.220
the reason being that the Eigen function is
changed from cos to sin. So, we now finish
34:04.220 --> 34:08.359
we understand at least all the three, how
you know three different geometries have to
34:08.359 --> 34:13.760
do it, and as I said you know so for you could
be asked to do in any of these geometries
34:13.760 --> 34:18.169
using any of these boundary conditions. So,
you should be prepared for that. So, as is
34:18.169 --> 34:25.169
a slab where the where the Eigen values are.
So, what are the Eigen values for slab let
34:25.169 --> 34:30.990
us chart the Eigen values.
34:30.990 --> 34:37.990
So, slab what are the Eigen values cos hyperbolic
and sin hyperbolic or it could be cos and
34:49.350 --> 34:52.120
sin in which case.
source term.
34:52.120 --> 34:54.610
Called the.
Source term.
34:54.610 --> 35:01.610
Or in other words for the product of not necessarily
for the product we have if it is A, if it
35:01.800 --> 35:08.800
is instead of taking part let us see how can
it be, how can there be a source term for
35:08.820 --> 35:13.610
the substrate may be the substrate is produced
from some other reactions. So, this is the
35:13.610 --> 35:19.250
freak case, but source term let us say, but
I just want to chart that Eigen functions.
35:19.250 --> 35:24.480
So, this is the normal case and this is the
special case if there is a source term I just
35:24.480 --> 35:31.480
want to write all the. So, then next one would
be cylinder would be j naught fine and y naught
35:40.830 --> 35:47.830
or I naught, and k naught. If anyone source
and sink is that clear and then whatever this
35:54.770 --> 36:01.770
sphere this sphere would be cos hyperbolic
one over xi cos hyperbolic, and one over xi
36:06.020 --> 36:13.020
sin hyperbolic or it would be one over xi
cos, and one over xi sin this is for a source
36:18.740 --> 36:22.240
term special case is that clear to all of
you.
36:22.240 --> 36:29.000
So, I am charting out all the three possibilities.
So, rectangular slab is cos hyperbolic sin
36:29.000 --> 36:33.570
hyperbolic typically. If there is a source
term for this substrate it could be instead
36:33.570 --> 36:39.000
of a sink it could be cos or sin, but this
special case again. Cylinder is j naught y
36:39.000 --> 36:44.460
naught typically or you can write it as I
naught k naught also, and sphere is cos hyperbolic
36:44.460 --> 36:50.860
over xi, and sin hyperbolic over xi it could
be cos over xi and sin over xi for special
36:50.860 --> 36:53.630
case when there is a source term.
36:53.630 --> 36:59.100
So, let us now go on to the next. So, is that
clear is that clear to all of you. So, next
36:59.100 --> 37:04.380
now go on to the next issues that we are looking
at and let us look at the screen here. So,
37:04.380 --> 37:08.630
till now what we did was in? And this is a
slightly more complicated form a slightly
37:08.630 --> 37:13.100
advanced form of what we are doing? So, till
now what we did was we looked at encapsulated
37:13.100 --> 37:17.970
matrices where the enzyme was entrapped only
outside. And now it is we’re looking at
37:17.970 --> 37:24.110
a more practical case where the entrapped
enzyme is entrapped outside as well as well
37:24.110 --> 37:29.420
as inside the matrix. So, we looked at enzyme
trapped outside or enzyme kind of attached
37:29.420 --> 37:34.130
outside, then we looked at the case where
enzymes were trapped inside. And now we’re
37:34.130 --> 37:37.520
looking at the case where both these things
happen. So, enzyme is entrapped inside as
37:37.520 --> 37:43.700
well as it is covalently attached to the surface.
So, this is a slightly more complicated case.
37:43.700 --> 37:48.190
So, what you think ? What you are suggestions
on what should be the equation on the boundary
37:48.190 --> 37:55.190
conditions in this case?
We should not.
37:56.120 --> 38:00.840
Right equations will remain the same the reason
being that, if you remember what we did before
38:00.840 --> 38:07.150
was that, this does not affect the you know
then this only effects the boundary condition.
38:07.150 --> 38:10.530
The receptors on the surface only vector boundary
condition it does not affect the equation
38:10.530 --> 38:16.750
right and the equation inside the matrix remains
the same because reaction is occurring only
38:16.750 --> 38:22.640
inside the matrix. So, we are again back to
the slab problem and what you get is d effective
38:22.640 --> 38:28.760
Del 2 C del x square equals k v times C? And
what do you have boundary conditions now?
38:28.760 --> 38:34.190
That boundary is C s.
It will be.
38:34.190 --> 38:41.190
Something C s they have we have to solve them
Bessel equation with the boundary layer solve
38:43.770 --> 38:44.020
it.
So, there is a boundary layer. That is the
38:43.970 --> 38:49.890
thing that you have to what you said is not
correct, but the equation is not correct that
38:49.890 --> 38:53.220
you said, but the concept is correct that
there is a boundary layer. So, we have to
38:53.220 --> 38:57.580
account for that boundary layer and try and
get the concentration across the boundary
38:57.580 --> 39:01.890
layer, and what was a mechanism that we followed
in the last previously? We use the concept
39:01.890 --> 39:08.240
of mass transfer coefficient, we use the concept
of mass transfer coefficient to calculate
39:08.240 --> 39:14.210
what would be my difference across the boundary
layer. So, you do not essentially know what
39:14.210 --> 39:18.990
is your concentration exactly at the surface?
But what you know is a concentration far outside
39:18.990 --> 39:21.130
the boundary layer far from the surface you
know.
39:21.130 --> 39:24.330
And you also know if you know the mass transfer
coefficient you would know how much is the
39:24.330 --> 39:29.080
total driving force, and total transfer that
is be the mass transfer coefficient times
39:29.080 --> 39:35.520
the outside concentrations C naught minus
C at the surface. So, that would be my boundary
39:35.520 --> 39:39.230
condition let me show you and then it will
be clear up to you. So, this step is that
39:39.230 --> 39:45.050
you see over here on the right hand side k
f C naught minus C s forget this term over
39:45.050 --> 39:48.950
here do not do not worry about this term right
now. So, let us look at this term alone C
39:48.950 --> 39:55.420
naught minus C this is you’re the total
driving force. C naught minus C is the total
39:55.420 --> 39:59.380
driving force times the k f which is the mass
transfer coefficient that should be equal
39:59.380 --> 40:05.440
to the amount of flux that is coming in fine
now what is this term doing over here because
40:05.440 --> 40:09.230
your receptors on the surface see if the surface
had no receptors and you just had a boundary
40:09.230 --> 40:13.300
layer then this is what you have, but because
there are receptors on the surface and there
40:13.300 --> 40:18.140
is a first order reaction going on the surface,
you add that as well because that is your
40:18.140 --> 40:23.330
consumption and the surface.
So, whatever is coming in substrate or in
40:23.330 --> 40:30.330
other words whatever is coming in equals what
reacts on the surface plus what goes in. So,
40:30.560 --> 40:34.140
if you bring this to this side for example,
so there is whatever is coming in is? What
40:34.140 --> 40:39.940
reacts plus what goes in fine. Is it clear?
Not clear.
40:39.940 --> 40:46.940
Not clear. So, draw a picture and try and
show. So, this is my matrix now the receptors
40:52.350 --> 40:58.520
are on the surface and the receptors are inside
as well fine. So, if the receptors are inside
40:58.520 --> 41:05.520
then I straight forward write, d effective
del 2 C del x s square equals k v c. So, this
41:06.590 --> 41:13.590
is for the inside. So, governing equation
for inside, the matrix is that clear till
41:13.600 --> 41:18.780
that point now the boundary conditions. So,
first boundary condition is at this point
41:18.780 --> 41:25.780
x equals l let us said. So, what is happening
over here is that if you are. So, d effective
41:28.750 --> 41:35.750
times Del C del x at x equals l should be
equal to k f C minus C naught. So, this is
41:45.090 --> 41:49.080
nothing is happening or in other words you
can just write it like this d effective times
41:49.080 --> 41:56.080
del C del x equals l equals k f C naught minus
C that is how people write it because you
41:56.890 --> 42:03.510
know C naught is typically larger than c.
So, this is what you have of is? There any
42:03.510 --> 42:09.770
problem with this? No, absolutely not know
what happens, because the reaction occurs
42:09.770 --> 42:16.110
on the surface. So, what is coming in does
not all of it go all of it does not go in,
42:16.110 --> 42:21.350
a part of it is actually reacting on the surface,
and how do you in incorporate that? So, take
42:21.350 --> 42:27.750
that out what is reacting on the surface which
is the k s surface reaction rate times C at
42:27.750 --> 42:34.750
x equals l is it clear now. So, this is without
what is happening on the surface. So, this
42:35.030 --> 42:40.820
part let us put it is over here. So, what
is coming in what is coming in here through
42:40.820 --> 42:45.369
mass transfer is, what is going out? But that
is not the reality in this case there are
42:45.369 --> 42:48.730
receptors on the surface.
And a part of it part of what is coming in
42:48.730 --> 42:52.700
is actually reacting with the with what is
there on the surface. So, you have to take
42:52.700 --> 42:56.810
that out. So, what is coming in minus what
reacts on the surface is what is allowed to
42:56.810 --> 42:58.540
go into the matrix.
42:58.540 --> 43:05.540
So, that is my b C one and B C 2 is again
the same the straight forward B C which is
43:08.060 --> 43:14.080
x equals 0 the del C del x is 0 the reason
being that again there is symmetry. So, all
43:14.080 --> 43:17.700
are all around you assume the same boundary
condition. So, all you need to do is go ahead
43:17.700 --> 43:24.700
and solve it. But before we do that we will
be going to make this equation dimensionless.
43:25.000 --> 43:31.350
So, we have to write the equation in dimensionless
forms and we have to define some dimensionless
43:31.350 --> 43:37.600
variables. And can you tell me that, last
time we had one dimensionless number, two
43:37.600 --> 43:42.340
dimensionless variables which were x and theta,
and one dimensionless number which was the
43:42.340 --> 43:46.950
Thiele modulus. In addition do we have another
dimensionless number this time yes we have
43:46.950 --> 43:53.950
and why do we have another dimensionless number.
Because of k L, because see the dimensionless
43:58.960 --> 44:02.920
numbers are always quantitative of the physical
phenomena that is happening; that is the way
44:02.920 --> 44:08.330
you have to look at it not in terms of coefficients
and so on. So, earlier what were the physical
44:08.330 --> 44:15.330
phenomena that were there earlier case. What
were the physical phenomena that were there
44:15.560 --> 44:19.900
the diffusion and reaction?
44:19.900 --> 44:26.900
So, you had phi square, which was a ratio
of diffusion time over reaction time. Now
44:29.450 --> 44:36.450
what you have diffusion, reaction and let
us call this t M external mass transfer. So,
44:45.390 --> 44:51.160
these three phenomena occur simultaneously.
So, you will have therefore, two dimensionless
44:51.160 --> 44:55.260
numbers, one would be say ratio of. So, out
of these three you can get any two independent
44:55.260 --> 45:02.260
dimensionless numbers. Say it could be t D
over t R is 1, and then you can have say t
45:05.000 --> 45:08.390
M over t D, which is you know? I will come
to that.
45:08.390 --> 45:11.650
So, typically these are the two numbers that
are used one is the ratio of the diffusion
45:11.650 --> 45:15.930
time to reaction time phi square and the other
one is the ratio of mass transfers. So, here
45:15.930 --> 45:20.430
for example, what you have is? This is a external
mass transfer, and diffusion is internal mass
45:20.430 --> 45:24.360
transfer. So, what you have is another number
which is the ratio of external to internal
45:24.360 --> 45:29.900
mass transfer, in addition you can form another
number actually. Because you have surface
45:29.900 --> 45:33.780
reaction and the rate for surface reaction
is going to be different for the rates for
45:33.780 --> 45:39.170
from reaction inside the matrix. Why is that?
Because in inside the matrix it is a volume
45:39.170 --> 45:43.119
average rate wherein surface reaction rate
is the intrinsic rate based on the surface
45:43.119 --> 45:47.210
reaction rate. So, you can form another number
there we are not going there, but there’s
45:47.210 --> 45:48.800
a possibility.
45:48.800 --> 45:54.520
So, let us look at what we have over here.
So, we have theta which is C over C naught
45:54.520 --> 46:00.600
k a v x hat which is x over l we have phi
square which is k k v times l square over
46:00.600 --> 46:07.600
the D effective. And then we have this quantity
by biot number which is a ratio of inter phase
46:09.680 --> 46:15.800
mass transfer or in other words internal mass
transfer time ,over inter phase mass transfer
46:15.800 --> 46:21.090
in other words external mass transfer time.
So, this is a new numbers that we have and
46:21.090 --> 46:27.740
you have studied this number elsewhere, but
we just get back to it here. So, phi square
46:27.740 --> 46:34.740
is your ratio of transverse diffusion to reaction
or inters phase mass transfer to reaction,
46:36.160 --> 46:40.380
if you want and biot number is a ratio of
intra phase mass transfer to inter phase mass
46:40.380 --> 46:46.170
transfer in other words internal mass transfer
time over external mass transfer time.
46:46.170 --> 46:53.170
So, if your internal mass transfer is very
slow then as compared to external mass transfer
46:55.210 --> 46:59.440
which is typically the case actually then,
what will happen biot number is much greater
46:59.440 --> 47:03.160
than one ? And if your external mass transfer
is very slow, as compared to internal mass
47:03.160 --> 47:08.230
transfer does not happen typically then your
biot number is much smaller than one and.
47:08.230 --> 47:12.660
So, you have three different competing phenomena
that are occurring over here. And the surface
47:12.660 --> 47:16.310
reaction we are not considering here because
its only few receptors are engaged in the
47:16.310 --> 47:19.530
surface and the major reaction is occurring
at the inside.
47:19.530 --> 47:24.119
So, you have the ratios of these three numbers.
So, you know for let me simplify it for you
47:24.119 --> 47:31.119
phi square equals t D over t R which is internal
mass transfer over reaction. And biot number
47:38.380 --> 47:45.380
is t d over t m which is internal mass transfer
over reaction. So, which means that phi square
47:55.720 --> 48:01.240
over biot number would give you the ratio
of response reaction this is external mass
48:01.240 --> 48:08.240
transfer. I will give you the ratio of external
mass transfer over reaction clear it is clear
48:12.390 --> 48:19.390
to everyone. I am not we are again running
short of time today. So, one. So, I will go
48:23.810 --> 48:26.940
through this quickly a little bit and then
you know one of the things I want you to do
48:26.940 --> 48:30.250
is actually go and look up your notes.
So, if there are problems. So, I will explain
48:30.250 --> 48:34.310
at the beginning of the next class like I
did today, but I do want you to go back to
48:34.310 --> 48:38.830
your notes and have a look at it. So, when
I write in terms of these two dimensionless
48:38.830 --> 48:43.859
variables, and these two dimensionless numbers
my equation that i had, before comes out to
48:43.859 --> 48:48.130
be the same del two theta del x hat square
equals phi square times theta. Boundary condition
48:48.130 --> 48:55.130
is except equal 0 del theta del x 0, and this
boundary condition over here is x equals plus
48:55.520 --> 49:02.520
minus one del theta del x hat equals biot
number one biot number times one minus theta.
49:02.710 --> 49:08.160
Who will tell me?
Krishna pratap you tell me, that is this boundary
49:08.160 --> 49:13.730
condition second boundary condition completely
correct.
49:13.730 --> 49:17.740
Right.
49:17.740 --> 49:24.740
So, this boundary condition had been written
without the reaction into consideration. So,
49:25.450 --> 49:32.450
you get del theta del x hat equals biot number
into one minus theta at x equals one you had
49:32.910 --> 49:36.400
a if you have a reaction determined does not
make much of a difference what happens is
49:36.400 --> 49:41.970
you. Just put a gamma out here because the
reaction term if you remember also has a external
49:41.970 --> 49:48.970
boundary condition which is that x at x equals
one. So, basically what you have is this?
49:49.650 --> 49:56.650
So, you can add your gamma over there and
then gamma will you know have the k s over
49:57.410 --> 50:03.700
k v and so on. So, just a factor over there
is that clear let me let me go back and show
50:03.700 --> 50:09.100
you then it will be clearer again here do
you look at this equation this is k f C naught
50:09.100 --> 50:14.800
minus x at plus minus one minus k s times
x at plus minus one.
50:14.800 --> 50:19.850
So, you can club what I am trying to say is
this and these are two same variables and
50:19.850 --> 50:25.119
you can club these two same variables. So,
what you will have is d effective Del C del
50:25.119 --> 50:32.119
x equals k f C naught minus k f plus k s over
k f C at x equals one right fine. So, this
50:48.359 --> 50:55.359
number is your gamma that is all I am saying.
So, gamma equals k f plus k s over k f fine.
50:58.640 --> 51:04.030
So, in the same you know I have this boundary.
So, so in dimensionless coordinates x equals
51:04.030 --> 51:09.030
plus at x equals plus minus one Del theta
Del x hat is biot number times 1 minus gamma
51:09.030 --> 51:16.030
where gamma it is k f 1 plus k s over k s.
So, does that would not change your solution
51:22.930 --> 51:26.970
or any of these things just the boundary condition
would be slightly altered, as a result your
51:26.970 --> 51:33.970
constant would be slightly altered, but otherwise
it would not change anything else. So, as
51:35.030 --> 51:39.670
we know we do not have a lot of time today.
So, let us quickly summarize this. So, we
51:39.670 --> 51:43.500
did this and there is no reason for us to
repeat. So, theta equals A cos hyperbolic
51:43.500 --> 51:50.500
x plus B cos hyperbolic sin hyperbolic phi
x. So, there is no reason for us to repeat.
51:50.590 --> 51:53.770
So, del theta del x would be 0 at x equals
0.
51:53.770 --> 51:58.880
So, b could turn out to be, and theta would
be this and all you need to do is apply this
51:58.880 --> 52:04.430
boundary condition at this end and you will
get your solution. So, I did this without
52:04.430 --> 52:09.250
the gamma out here, but you can just put the
gamma in there and it would not make any difference
52:09.250 --> 52:10.530
to your equation.
52:10.530 --> 52:15.700
So, this would be your theta that you get
at the final thing the, but again without
52:15.700 --> 52:22.609
the gamma and if you put the gamma in here
if I remember yeah. So, it will come over
52:22.609 --> 52:27.220
here the biot number it would come with a
suffix of gamma. So, it will be biota number
52:27.220 --> 52:31.440
times gamma cos hyperbolic x and this. So,
what I want you to do is just you know go
52:31.440 --> 52:35.500
and redo this and see whether gamma comes
in. I am sure it comes in over here, but just
52:35.500 --> 52:42.500
for you to check now. So, this is once you
write it in this form then you can you can
52:42.859 --> 52:47.140
write it in the tan hyperbolic form as well.
Because sin hyperbolic over cos hyperbolic
52:47.140 --> 52:51.369
and then you get the tan hyperbolic phi over
tan hyperbolic phi and so on.
52:51.369 --> 52:56.869
And these are values at x equals one. So,
this is general theta and this is evaluated
52:56.869 --> 53:02.640
at x equals one the surface concentration.
Now why do I do them? Because I need to evaluate
53:02.640 --> 53:08.000
the eta the effectiveness factor. So, I just
need to evaluate that. So, my R v the average
53:08.000 --> 53:15.000
volumetric average is given by k v over l
times integral of this from 0 to one which
53:15.080 --> 53:22.080
is integral of this is the little complicated
integral and then eta comes out to be tan
53:22.540 --> 53:28.990
hyperbolic phi over phi over biot number tan
hyperbolic phi plus one this is this is what
53:28.990 --> 53:33.340
it comes out to be this is because R v C s
is this number. So, this gets canceled with
53:33.340 --> 53:39.730
this over here and this is what remains.
So, I think you know we are doing a little
53:39.730 --> 53:46.730
probably little fast today. So, I would stop
here, but I think we will probably we might
53:47.480 --> 53:51.940
is it clear to everybody or do we need to
go back to it rather next select next lecture
53:51.940 --> 53:57.640
what I want you to do is. Just go back and
put this gamma over here, that we had. There
53:57.640 --> 54:03.840
is a gamma out here. So, you just and as you
see that my intuition tells me that you know
54:03.840 --> 54:08.130
the gamma comes here with the cos hyperbolic
term. So, wherever there is a cos hyperbolic
54:08.130 --> 54:13.960
term the gamma would come there, and then
you can just then the rest is easy. But again
54:13.960 --> 54:18.190
you know this integrals are not very easy
you know that is that is the problem, when
54:18.190 --> 54:21.560
integrals are actually not very difficult
either, because this is just cos hyperbolic.
54:21.560 --> 54:25.260
So, the denominator is something you do not
need to worry about. So, this is just cos
54:25.260 --> 54:28.830
hyperbolic the integral of that would be sin
hyperbolic, but we you know the thing is that
54:28.830 --> 54:31.950
we can really go through each of these steps
in this class.
54:31.950 --> 54:38.540
So, if you want I still have a couple of minutes
or if you want I can go through this. So,
54:38.540 --> 54:45.540
theta equals biot number, cos hyperbolic phi
x times phi. I guess the gamma would come
54:56.710 --> 55:03.710
here and here that is my guess. So, if you
do that then the integral of theta d x hat
55:06.349 --> 55:12.690
would be simply biot number gamma over this
entire denominator. I am not writing it because
55:12.690 --> 55:19.690
of lack of time integral of cos hyperbolic
phi x d x hat from 0 to 1. So, which would
55:21.510 --> 55:28.510
be this whole thing in square brackets over
one over phi sin hyperbolic phi once you put
55:32.630 --> 55:35.990
the integrals out there.
55:35.990 --> 55:42.990
So, you what you get is your eta comes out
to be tan hyperbolic, because the sin hyperbolic
55:45.840 --> 55:49.980
and the cos hyperbolic you can take this in
denominator, tan hyperbolic phi over because
55:49.980 --> 55:56.980
there is a there is a phi out here. So, that
would come. So, phi times phi over biot number
55:58.670 --> 56:05.670
tan hyperbolic phi plus 1. So, what I want
you to do is, you know as a as homework today
56:06.090 --> 56:10.280
is to go, and add the gamma, and go through
these steps because unless you go through
56:10.280 --> 56:13.330
these steps yourself. You know this is not
something that you will take away from the
56:13.330 --> 56:18.550
class. So, just go through these steps, because
you know as I said in a in that in the test,
56:18.550 --> 56:23.200
for example, you might have to do all these
calculations very quickly within a short period
56:23.200 --> 56:27.270
of time and then there is only one way of
being able to do that is which is do them
56:27.270 --> 56:31.210
yourself. So, I want you to go through these
steps, and if there are problems I will recap
56:31.210 --> 56:38.210
that in the first 5 minutes of the next class,
otherwise I will start with something else.
56:41.119 --> 56:47.420
So, I will stop here. Thanks.