So, in todays lecture in biochemical engineering we are going to look at continue looking at
the immobilized enzymes. We will start from where we left in the last class, which was
the mass transfer co-efficient. So, what we were looking at in the last class, if you
remember is intraphase mass transfer with reaction. There was a flat plate so, if I
am to remind you. So, we had a flat plate like this and there is a boundary layer set
and there these enzymes are receptor. You know under the receptors that are on bound
to the surface and, because this is fluid phase and this is the solid phase in which
the enzymes are immobilized. So, the enzymes are enzymes immobilized.
So, what happens is there is transfer between the fluid and the solid phase and that mass
transfer co-efficient was referred to as k f. Now what we did not or do get the time
do it in the last class look how this k f varies. How do you evaluate this k f it turns
out there lot of empirical relations to evaluate the k f of the mass transfer co-efficient.
Now what we looked at in the last class was flat plate, but if what if it were in the
flat plate what it was circular tube or sphere or any other kind of things.
So, we what I wanted to show you on the screen today are these relations for mass transfer
co-efficient these for laminar flows. So, different kind of geometry, first one is flat
plate then circular tube then solid spheres and then finally, falling film. So, it is
not the first relationship is k, k observed time that is k observed is mass transfer co-efficient
time’s z over D m Z being the distance. So, this is the flat plate, depending on where
are on the flat plate that goes as Reynolds to the power half and Schmidt number to the
power one third. For A circular pipe you it the k f, which
is the mass transfer co-efficient is given by k f D over D m, d being the diameter of
the pipe over D m. The molecular diffusivity that again goes as Reynolds to the half and
Schmidt to the one third, but the co-efficient is different. So, co-efficient in the last
one there is like 0.3 and here it is 1.8 so, it is like 6 times. So, what you see is that
one important similarity is that that k observed k Z over D m or k f D over D m. Whatever it
is Sherwood number actually goes as Reynolds to the half and Schmidt to the one third.
So, how does it look when we have forced convection around solid spheres, this is something probably.
Have you read in the mass transfer classes again it goes as the same it again goes as
Reynolds to the half and Schmidt to the one third, but there is little attached is that
there is a constant. Because the shape of the sphere is the constant with that and if
you go to the screen, let what you will see the Sherwood number goes as 2 plus 0.6 Reynolds
to the half Schmidt to the third. So, that co-efficient those two co-efficient are there
Reynolds to the half and Schmidt to the third, but you have the 2.
Finally, the scene that we want to look at is the falling film. So, the falling film
turns out similar, to the flat plate in the sense that if the geometry. Similar also is
the flat kind of thing and it goes again as Reynolds to the half and Schmidt to the one
third and the co-efficient then the 0.7, which is not to different from 0.3.
So, it is around twice that of the flat plate so, the difference that you know what would
be the difference between the falling film and the flat plate? So, this is for this is
if I go to this thing this is the flat plate what would be a falling film line. So, this
is again a solid surface and this is how this going this is the falling film. So, what is
different over here is the effect of... Gravity
Effect of gravity here
So, this is the vertical this is the horizontal. So, the gravity is going to affect the size
of the boundary layer and also the velocity profile in the boundary layer, the thickness
of the boundary layer and the velocity profile in the boundary layer.
So, if you have taken a falling film like this and compared it with a flat plate, now
the here what is going to happen here? Is that what you have at the bottom this is going
to the liquid is going to drag. In this direction liquid is going to drag in this direction.
So, as the liquid goes down the z or in other words as a length goes down, then the size
of the boundary layer grows. So, as z goes high more and more you go down the size of
the boundary layer is going to grow. Because it is dragging the fluid is dragging more
in the in the vertical direction because of gravity.
So, as a result which in the two things that is different over here is the thickness of
the boundary layer and the velocity profile and the boundary layer and therefore, the
Sherwood number is slightly different from the flat plate. So, what we are looking at
till now are these cases, if you remember all the examples. We did in the last couple
of lectures are examples, where the enzyme is on the surface. The enzyme is on the surface
and either it through diffusion or it is through convection that mass transfer is taking place.
So, that is the whole idea of today’s lecture is go the screen is intraphase mass transfer
and reaction. So, what is different in this one is that is intraphase. So, let me see
if I have picture I do not have picture here. Let see I have picture out here not let me
go back and show you some so, I can use this picture.
So, that is the difference you know. So, if for example, if you have intraphase mass transfer
if the enzyme immobilized that is there is entrapment within the encapsulation within
the gel of the matrix. Then there is no way for you to make it react with the substrate
without the substrate, actually diffusing into the enzymes. So, that is the different.
So, intraphase is within the for phase the mass transfer should happen within the phase,
and when you have the intraphase mass transfer? What happens is that the reaction occurs simultaneously.
So, that is a major difference, this was an example, we had the only the intraphase mass
transfer. So, the solid the enzymes are encapsulate
or on the surface of the solid and the fluid comes in and reacts. And then intraphase mass
transfer then enzymes are on encapsulated within is this difference clear, the physical
difference now in addition to this. You can what can you have you can have enzymes, which
are on the surface in addition to which being entrap inside you can have enzymes on the
surface. Now let us go and look at what the difference is and how to study this kind of
models can we make. So, the first important thing that I want to discuss over here, is
that the concept of pseudo homogeneous reaction and we did that before. So, if I can you know
interact your attention to this.
So, my enzymes are you know let us say these are my points of where the enzymes are...
Now I can go and write homogeneous model for this heterogeneous model for this is something
I had discussed, but I still want to go into details and try to see. If you can understand
it completely so, if I am to do a heterogeneous modelling of this. What would I do, I would
actually let the substrate diffuse to the surface then from the surface diffuse then
why do, I separate these two diffusion that diffusion over here. So, diffusion in fluid
and then diffusion in this, because porous solid so, diffusion in the porous of diffusion
co-efficient for going to be different. You know it could be diffusion depending on
if the porous is very small, if it not it could it would still be different the diffusion
co-efficient still going to be different. So, if I am to do a heterogeneous model of
this my step 1 is diffusion in fluid, step 2 is diffusion in porous solid then step 3
would be adsorption or you can just go straight to reaction. If you want reaction and adsorption
you can say, but what is the challenge? That we face over there the challenge that we have
to face over there is that we have to have a geographical. You know map of where these
particular enzymes are located within the matrix, if we are to allow. So, because you
are doing simulation, if you are doing simulation of the whole thing you need.
So, the material reception is defusing into the whole thing, but for you to figure out
that for you to be able to quantify. That you need to know at which point this reacting
what are these, because see what happens this within this matrix it become almost like point
syncs for the substrate. These of points of reaction become point sync for this substrate
and you need geographical, you need a map of where these points syncs out. So, if you
want to do heterogeneous model is absolutely not impossible, but it is a little cumbersome
in terms of the modelling. And then you have to do the diffusion out of the porous solid
and out of the fluid that is not the problem the problem lies here.
So, what we do is we come up with what is known as a pseudo homogeneous model. What
is the assumption of such a model? The assumption of pseudo homogeneous model, what is the assumption
of pseudo homogeneous model? The assumption is that you homogeneous over the entire matrix.
So, you assume that these receptors are present uniformly in the entire matrix. So, what you
take? Is you take you figure out that, what is the total number of receptors? What is
the reaction rate for each and multiply that and get the total amount of reaction rate
for the receptors and then divided by volume together volume average reaction rate.
So, that would be once that is there, then it is assumed that though these reactions
are in principal heterogeneous reaction and they are occurring at certain points. You
assume that the whole thing occurs uniformly over the entire matrix so, this is the concept
of pseudo homogeneous reaction. Now is there any new concept for you studied this before
know so, at least I hope, because we do not want to get over this again and again. Let
us go back to what we have on the screen? So, there is a definition exactly what I have
explained, but just to go through it the heterogeneous reaction that occurs within a tissue or porous
matrix containing immobilized enzymes are called interphase.
Although the reaction occurs at the interphase between the fluid and the cell surface or
at the cell extra cellular matrix the phases within the tissues is considered to be pseudo
homogeneous. If the reaction is macroscopically uniform so, this is the term that we introduced
here macroscopically uniform. What it means that as a microscopic level? It is occurring
at different points inside the matrix, but at the microscopic level is uniform and when
at the macroscopic level it is uniform. Then you can assume it to be pseudo homogeneous
so, it is a matter of scale, if at the scale that you are looking your reaction is homogeneous.
Then you can assume it to be pseudo homogeneous if, but as a microscopic scale it could be
heterogeneous is the concept of scale. So, at the smallest of scale say for example,
you know if you have surface if you have a surface like this and you are trying to take
the derivative at this point. Which point you take the derivative? You never know or
if you have surface like this. We can say that the microscopically at the microscopic
scale, we can say that this is the singular point, because you cannot define the derivative
here and this is a non-singular point. You can define the derivative here, but that the
microscopic scale this is not singular. You see what I am saying, because when you go
to the level of at the microscopic level this is no longer as sharp this peak is no longer
as sharp. Because when you go to the microscopic level
this is not singular as a microscopic level. You can define the derivative at this point
so, that is the matter of scale the manner of looking. So, the same thing happens here
at the microscopic level, it is heterogeneous. Because these reaction happens at different
point within the matrix, but at the macroscopic level you can average them out the macroscopically
uniform, which means that you can average them out and the whole concept of averaging
theory. You know the basic concept of averaging theory originate from this, where you can
actually decide that. Whether the system is microscopically uniform or macroscopically
uniform, that kind of things. So, it is microscopically uniform, you know
there is no question at all, but the whole concept of averaging comes in these certain
issues or in these certain cases. Where the system is microscopically non-uniform, but
at the macroscopic level, you can decide now what how do you decide, what is microscopic
and macroscopic?
You know that is the question that is important and what typically is done is that the let
us call this r. So, is microscopic length scale versus macroscopic length scale, if
this is much smaller than 1. Then you assume this is your criteria, typically you assume
that it is microscopically uniform so, the length scale at which you see this heterogeneity.
This is the heterogeneity length scale and this is the non-heterogeneity length scale.
So, that the length scale you see at which you see heterogeneity over the length scale
at which you do not see heterogeneity, the ratio of this two have to be much smaller
than 1. So, if this is possible then what you do is you can average it out over the
macroscopic length scale as soon as the average it out you gets the pseudo homogeneous model.
So, pseudo homogeneous model is not a matter of just intuition, it is a matter it evolves
the out of you know averaging theory. Whether in the macroscopic once you have done the
pseudo homogeneous got the pseudo homogeneous model. In the pseudo homogeneous model what
happens? This assumption is that reaction and diffusion occurs simultaneously. In the
real model in real life diffusion and reaction not occurring simultaneously, in the heterogeneous
model diffuse reaction and not occurring simultaneously remember. So, first diffusion followed by
reaction, but in the pseudo homogeneous model diffusion and reaction is assumed to occur
parallel. So, if this is just you know corollary. So,
to say if the reaction is too fast as compared to diffusion the concentration at some places
becomes 0, because it is diffusion control. This is one of the interesting things that
can happen in pseudo homogeneous model, I mean what pseudo homogeneous model cannot
capture. So, if you have a system like this and say for example, you have a reaction like
an autocatalytic reaction that we discussed in the last class. So, what can happen, because
the reaction at some places can, if there is a small perturbation in the system? The
reaction in some places could be much higher than the reaction in other places.
So, the concentration in some places the for example, the concentration of the substrate
here could be 0 and next you know too far from that place the concentration could be
high. So, what will happen in that case what would be the consequence of
that there will be major difference, the large difference in concentrations, it could be
next to each other depending on the kinetics, if you have kinetics of this type A plus 2
B giving 3 B. Then A could be depleted in one place and A could be far from the depletion
from another, it could be presented in large amount. So, what will happen in that case
that is a real system I am talking about the heterogeneous system.
I am just talking of catalyst pellet here; you know it cannot be catalyst pellet. So,
this is just catalyst pellet catalyst pellet or immobilized enzyme whatever you want to
call it. So, what can happen what will happen is that the heterogeneous model is still finding,
but you cannot average it anymore? So, there is no point in having an average when one
part of that has a very large concentration and another part of that very small concentration,
because the average value would not give sense of the distribution. So, when you want to
average the criteria number 1 is that the ratio of the smallest through the largest
length scale is much lesser than 1. The criteria number 2 would be that the minimum
over maximum concentration minimum C over maximum C should be of the order 1. This should
not be a huge difference it could be 0.1 they still order 1, but it could be it cannot be
that cannot be 10 to the minus 3 or 10 to the minus 4 within the matrix. So, which means
that the variation within the matrix has to be reasonable they cannot be very large variation.
So, having set these things I think now that and it does not make sense to go and write
a model. Unless we understand what are the constraints? What are the boundaries between
which we writing that model so, having met that...
Let us we can you know try and write a model of A. So, what we do is these are for drug
bio molecules, therapeutic molecules and so on. If the diffusion is slow then reaction
rate could decrease significantly the reason is, because this is diffusion and reaction
occur parallelly, but in real life diffusion decrease reaction. So, if diffusion is very
slow then reaction; obviously, would be hamper straightforward. So, what we do is that in
the pseudo homogeneous model what we write over here is d effective Laplacian C minus
R of v. Do you all remember by the way how to write Laplacian in spherical and cylindrical
and all these coordinates, because you know they will be in. So, why D is effective? Because
this is not the real diffusion.
Because this is not the real diffusion, the real diffusion occurs in the spaces between
the porous. So, this is the effective diffusion co-efficient because you taken an average
out of the whole thing, if the diffusion co-efficient also an also an average between the diffusion
co-efficient of the solids through the solids and the spaces between the solids. So, R of
v is let us defining R of R v average as the microscopic reaction rate. So, what we would
be the microscopic reaction rate as I just as I said before the summation of the all
the microscopic reaction rates over the volume divided by the volume. So, R of v average
is the summation of all the microscopic reaction rates divided by the volume fine. So, this
is given by 1 over minus 1 over v. So, this is you know you can replace what
you have over here the diffusion reaction problem from here to here and then what you
do is you simply use the stokes here. So, just change the Laplacian to the surface into
this straightforward. I think you have done this before changing from the Laplacian to
the surface integral. So, when you do that then this is all fine. So, you can write here
R average in this way so, the advantage with this writing is that you just need to changing
from the surface from the volume to the surface is that you just need to evaluate. What is
going on to the surface? You really do not need to care about, what is going on in the
volume?
So, then we define this quantity eta is my effectiveness factor and I think we did effectiveness
factor in the lecture 1 of the lecture before also couple of lectures back know. So, this
is something that we are going to do for the next couple of lectures. So, let us try and
see what the effectiveness factor is essentially the reaction rate. When the reaction is occurring
over the entire region over the reaction rate. If it were only occurring at the surface concentration
or only occurring at the surface, if the entire concentration in other word if I am to explain
little better. If the entire concentration if the entire amount of material, where is
the surface concentration. So, that is my denominator entire material
and C s surface concentration over the actual reaction rate. That happens and this actual
reaction rate you can obtain as I showed here on the screen, you see actual reaction rate
you can obtain as the volume average fine. So, why is this important, because see for
a material once you have evaluated. What is your effective reaction rate? If you want
to know your actual reaction rate all you need to do is evaluate the actual rate at
the surface and effectiveness and this is essentially effect on factor. This is very
important thing that you know some of you are taking the advance reaction engineering
course are you any of you.
Not taken. So, then you will study these things you know effectiveness factor. The effectiveness
factor essentially measures that if reaction is occurring at if reaction were occurring
at the surface concentration and if the actual rate of the reaction. The ratio between these
two I am going to see how to how to incorporate that in the next few cases so, this is not
important.
So, what we do now is now let us go and actually does A solve the problem. So, the first problem
we look at the intraphase mass transfer and the reaction in a rectangular slab. We will
look at the sphere and the rectangle. So, first thing we will look at today is this
problem in the rectangular slab. So, what is my what is my basic governing equation
for the rectangular slab what is my governing equation for the rectangular slab in this
case intraphase mass transfer? And reaction this you know there in the title if you look
at here intraphase mass transfer and reaction in the rectangular slab. So, these are interrupt
these receptors are interrupting inside. So, what would be my governing equation? So, what
the basic phenomena let us talk about it for. So, essentially the basic phenomenon is heterogeneous
reaction, but we are going to use pseudo homogeneous model. So, let us make that clear first that
we are essentially it is heterogeneous phenomena heterogeneous system, but we are going to
use pseudo homogeneous model what is the basic assumption of pseudo homogeneous model? If
reaction is uniform everywhere and you can use an average reaction rate A B is that,
because use the average reaction rate diffusion and reaction occurs parallelly. So, what would
be my model?
So,
So, minus the effective Laplacian of C should be equal to minus R of v. So, this is my diffusion
reaction because this has been you know C is a substrate. So, C is a substrate concentration.
So, this reaction is A sync term, because it being consumed fine. So, this is my basic
equation the reaction rate is here given as k v times c. So, the Laplacian the effective
transverse Laplacian of C equal k v times C. Now, what would be my boundary conditions?
Let us talk about that there is several possibilities with boundary conditions, but let us you know
talk about boundary conditions. So, what are my two boundaries one would be at the centre,
because we are going to this is the symmetric we have to assume.
So, depends on how what we assume, if we assume Laplacian to be A symmetric Laplacian means
in the case what would do you have let us you know in think you do not really remember
all the Laplacian things. So, this is the no this is the rectangular slab. So, there
is not you know not too much of question still there is, but if it were cylindrical then
there is A question of symmetric.
Now what we assume over here, let us put this assumption very clearly, if this is my rectangular
slab and I am going to consider diffusion in this region in this direction. Which is
x and diffusion in the y direction is neglected why, because it is assumed to be too long
fine. So, diffusion is neglected now if I am doing it, because of symmetry my condition
at this my x equal says this x equal to 0. So, at I think that is my x equal to 0 my
fine, now what will be the other condition?
Is equal to L what there several possibilities with this. So, you have to just pick one the
let us talk about the simplest. Now we will do the little more complicated one in the
next class.
Continuity: is the simplest form. So, there could have A robin boundary condition, which
is flux type flux with concentration or you can have simple continuity. So, let us assume
the simple one which is continuity at this point I think this would be k v C naught.
So, what is k v? So, this is my x equal plus minus 1 x is dimensionless by the way here.
So, how make dimensionless just by dividing by l. So, k C equals k A v times C naught
and rather boundary condition is the centre x equal to 0 Del C Del x equal to 0 fine.
So, can we go ahead and solve this why not you do that. So, what would be max Laplacian
simply D effective time Del 2 C Del x square equals k v C. So, again you know we are going
to assume these two different limits of it 1 is the zeroth order and one of the first
order like we had been doing all through. So, the zeroth order is very easy why, because
it is simply came k v and you can just go and straight integrate it. And then what the
solution you know you have done the something time. So, the first order what is the solution
of this?
First order of this what is the solution of this?
Problem
No, look carefully first is you have to see what is physically meaning full A sin x and B cos.
into power minus... A sin x B cos x what is the problem with that
you know one of them I dropped out. You will have you will have A sinusoidal wave of concentration
inside the matrix is that physically meaningful. So, whenever you write solution the thing
it will figure out whether it is physically meaningful or not and that is not physically
meaningful. So, the physically meaningful solution would be exponential and you, because
you forgotten how to do these things which is A strange thing that you know. So, essential
when you have a positive on this side you will have an exponential. If you have a negative
out there then only you can have that and that is not possible, because you know that
that is not possible out here. So, the solution would be simple A cos hyperbolic,
I think what is going to I am going to do this little differently later that is do dimensionless
and all the stuff, but just to give you A sense. I think k v over D effective I think
why cosine hyperbolic x and sin hyperbolic x just from the plus x and minus x I get this.
So, what we are going to do now is use a dimensionless form of this we can go ahead and solve this,
but the advantage of making the system. Dimensionless is that you can just scale all the variables
as you want and you will see that it is very easy to work with the dimensionless system.
So, what we do is make it make the system non-dimensional and what will happen is, when
we make the system non-dimensional it will come up with numbers dimensionless numbers.
Which ratios of important time scales in system and that is one of the added advantages of
making the system dimensionless, it comes out naturally and these are very important
number. So, the first number we will get is in this case is A Thiele modulus and I have
already taught discussed A Thiele modulus. In the last class if I remember which the
ratio of diffusion time is over reaction time and why is this important here, because these
are the two dominant phenomena in the system diffusion and reaction.
So, it is very important to have the sense of the numbers ratios of these numbers why
that important to have a sense, because if Thiele modulus is given as ratio of diffusion
time to reaction time and Thiele modulus is very large. That means that diffusion is...
If Limiting
Limiting. So, that is something that you know we need to figure out here so, why do not
you do it on your own little bit. So, let we all help you out now out with this.
So, say x that because we need to do this unless and otherwise. So, this is all we need
to go here so, D effective times Del 2 C Del x square equals k v into C. So, x is this
is how define x. So, you get the d effective L square Del 2 theta Del x hat square equals
k v into theta. So, k v. So, k v L square over D effective time’s theta, now this
is this could be written as L square over D effective divided by 1 over k v time’s
theta. So, this is phi square over theta. So, this is term L square over D effective
is the t D transverse diffusion and this is the transverse reaction time so, you get theta
phi square. So, this is your basic equation Del 2 theta Del x square equals phi square
times theta and that phi square phi. Why phi square is k v L square over D effective and
that is the ratio of the transverse diffusion time of reaction time.
So, this is my dimensionless model and my dimensionless boundary conditions are very
straightforward. So, at x equals plus minus 1 you get theta equals 1 and x equals 0 Del
theta Del x equals 0. This is very straightforward and solution is also something I had discussed.
So, this is the solution, but you should write it as so, but I know I think it is always
best to write it as A cosine hyperbolic and A sin hyperbolic, because it is much easier.
So, if you write theta as say A 1 cosine hyperbolic phi x, because if you look at your reaction
equation over here it is Del 2 theta. So, equation over here is Del 2 theta Del x square
equals phi square times theta. So, the solution is theta equals A 1 cosine hyperbolic phi
x plus A 2 sin hyperbolic phi x. So, Del theta Del x equals phi times A 1 sin hyperbolic
phi x plus A 2 cosine hyperbolic phi x.
So, next step is that at X equal to 0 Del theta Del x equal to zero. So, Del theta Del
x at x equals to 0 turns out to be phi times A 2 cosine hyperbolic phi x equals phi times
x equals 0 phi times A 2 which is 0 which implies that A 2 equals 0. So, what you get
is theta equals A 1 cosine hyperbolic phi x now your other boundary condition is that
x equals plus minus 1 theta equals one. So, equals A 1 cosine phi x. So, with my A 1 becomes
1 over cosine hyperbolic phi.
So, my solution is theta equals cosine hyperbolic phi x divided by cosine hyperbolic phi. So,
this is what I have which is which equals C over k a v C naught. Now, what we intend
to do now what we intend to do is find my effectiveness factor eta. So, eta if you remember
was defined as R of the average over R of the average R of v evaluates with this C s.
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.
So, the denominator is known only you have to do is evaluate the numerator k a v C naught
clear. So, how do you evaluate the numerator?
I think you can use that, but I think one of the other ways that is also you can do
you can, because we got the theta. You can just evaluate the surface and do that and
another option is what I have on the screen over here just use this straight away calculation
of eta this one. So, because you know what is the volume of your slab? So, and what you
got over here is this there is A slight typo which should be R this 1 should be R v average
you know this should be on the top of it. So, minus R v average is simply going to be
1 over V times R v times d v and that that equals what we got. So, k C time d v equals
k times k a v times C naught into theta times d v and d v you can cancel out a v on both
sides. So, you will get 1 over I think 2 L this goes
from minus L to plus L k a v C naught times theta into d x is it. So, then you can write
this take this two out and write this as 1 over L going from 0 to L k a v C naught theta
into d x fine . So, what would be the integral if I have A cos hyperbolic it will simply
be sin hyperbolic integral.
So, know that is not minus R v equal 1 over L 0 to L k a v C naught theta was what cos
hyperbolic phi x divided by cos hyperbolic phi I think of d x not this k a v C naught.
Now this integral turns out to be this whole thing turns out to be k a v C naught over
L slightly complicated integral. This is comes out to be this tan phi if this cos hyperbolic
phi x 0 to L divided by cos hyperbolic phi comes out be tan phi over phi. So, you can
do that what you have to do is essentially, because you know this will be sin. So, you
have A cos, but this 1 over phi will come, because of the integral. So, this will be
sin hyperbolic phi over phi over cos hyperbolic phi in the denominator. So, my eta is going
to be what? Does my eta is going to be...
Let us go back here eta is R this R v is this and this is k actually here and I think the
L should go from here, because there will be you know L and you know this. If you put
this in the dimensionless coordinate, because x is in the dimensional coordinate. What do
you used hat over here? So, if I convert this is a hat and if I convert this with the hat
then you got L over here then the L would go. So, this is what you have.
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
by k a v C naught tan hyperbolic phi over phi equal. So, this is what we get and this
is a very important formula to remember for eta, because I will show you later may be
in the couple of lectures. So, that this is something that you can use independent of
geometry, because though you change the geometry and we will do in the little time. We have
another geometry you will see it is difference, but what happened is that overall be it cylinder
or sphere or A slab the effectiveness factor is more or less close to tan hyperbolic phi
over phi more or less close to. So, if you use this number you are not really
very off. So, something this is something you might want to remember this formula. So,
this is just going through what you have done over here the same thing you will get the
tan hyperbolic phi over phi. So, what we do now is that in the next 10, 15 minutes we
have lets quickly do this same thing and I want you to go take the lead on this is take
exactly the same thing, but this. So, this is the typo over here this is not A rectangular
slab. So, I want you to do this in A sphere. So, this is that example, we had before much
before the picture I showed you where the enzymes are actually trapped here.
So, the enzymes are actually trapped and this is a very common example, because you know
when you take some of your capsules. For example, enzymatic capsule they are the spherical like
stuff and the enzymes entrap. So, this is very physical, very practical example. So,
this is the one that I want you to look at now. And I showed you how to do it. So, what
I want you to do in the next few minutes are go and do it for the sphere. So, first write
this is again your basic equation. All you have to do is write Laplacian in the in the
spherical coordinate and then go and make it dimensionless and then go and solve it.
So, just star doing it and I will help you. So, we have written the Laplacian I do not
know some of you might have forgotten, but if you have not forgotten. Then this is what
you have on the screen is Laplacian and the spherical coordinate. So, D effective again
over R square d R of R square Del C Del R. Now what is your boundary conditions going
to be?
Because you have assumed symmetry so, the basic assumption I want here that this is
a symmetric and once a symmetric if it is a non-symmetric. For example, let me ask you
this for a spherical substance if this is non-symmetric. So, what would be the can you
write 2 D model for a non-symmetric here for a cylinder for example, if it is symmetric?
So, this is my D effective of Laplacian k C let say k c. So, for a cylinder what is
my Laplacian? 2 D Laplacian is 1 over R d R of Del C Del R plus Del 2 C Del x square
this is for a cylinder. Now this for a symmetric cylinder symmetric cylinder, now if I ask
you for a symmetric sphere can you write the 2 D model what is 2 D model for symmetric
sphere. So, let me add that you know let me close my eyes and add the radial part plus.
What is the other component be there comment?
The theta and phi, but I asked you to write the 2 D model. So, which one you choose?
Why is that?
Think about it physically the sphere has you know cylinder, you have this access of symmetry
and you are taking symmetric about the radial coordinate. But the physically you know how
would you physically denote that about this access of symmetry you have a symmetry is
it possible to do it. In a sphere it is not think about that in a cylinder you can do
it, because in cylinder you can fix your access of symmetry. And you can say about this you
can, but in a sphere theoretically it is possible, but physically is it possible to denote the
system and say that there is an is an access of symmetry it is not possible.
Then what would be your answer? The answer is that it is not possible to write the symmetric
2 D model in sphere. For physical system theoretically mathematicians can write, but as a engineers
you know there is no point. You know do you understand what I am trying to say that at
what point how we are going to denote your access of symmetry and say that only around
this. There is a symmetry, because if you there is a symmetry in this direction. Then
if there is no physical demarcation of that and if you turn that sphere around then there
is no symmetry. So, if you write have to write a symmetric model the way do it. Is actually
gone and write a 3 D model you can have both in phi and theta.
So, let us go ahead with this I do not think we will have the time to finish this, but
the boundary condition is such symmetric boundary conditions, because you have assumed symmetry
r equals 0 Del c Del r equals 0. What will be the other boundary condition? Straightforward
that at on the surface the concentration is given by k v C naught. So, how do you solve
this? So, the surface is given by r equals R m. So, how do you solve this now? So, one
thing that we can do is we can safely close our eyes and make the system dimensionless
for the sake of simplicity my dimension is variable would be R m. So, zeta is given by
r over R M then my governing equation is given by this 1 over xi square Del d xi of xi square
Del c Del xi equal phi square times c.
No, solution is not Bessel function you need to go and brush up your mathematics. So, let
me go back a little bit we have complete run out of time, but because we have mentioned
the Bessel function. So, what is the coordinate is a Bessel function is actually in the cylindrical
coordinate.
So, I did not do the cylinder, but that does not mean that I have to go back and do the
cylinder now. So, this is my D effective d r of r Del c Del r equals k v C. So, after
making the dimensionless you gets 1 over r d r this is the cylindrical coordinate by
the way do not confuse its spherical. So, let us the spherical where we there, because
the mentioned Bessel function that I am going back to phi square of C fine. So, what would
be the solution of that the solution of this is going to Bessel function. So, this you
can write as Del 2 c Del r square plus 1 over r Del c Del r equals phi square times C multiply
this by r square both sides.
Then we will have r square Del 2 c Del r square plus r Del c Del r minus phi square r square
C equals zero. So, then your new variable you have to what is the next step you have
to find a new variable y?. y equals phi times r. So, then equation becomes;
now you can use the Bessel function solution. This is you know function solution what I
want you to do I will stop it here today, but what I want you to do is to complete this
Bessel function solution. Because these are things that you need to review on your own
and I cannot review. This is not a maths class complete this Bessel function solution and
come up with some clue and what would be the solution of the spherical thing. So, the cylinder
the flat plate was very easy, it is cosine hyperbolic and sin hyperbolic this is the
Bessel function solution. I gave you the solution, but all you have to do is figure out what
kind of Bessel function first kind, second kind modified what kind of Bessel function
is going to be and... So, you figure that out and last thing that
you have to do is, what would be your solution in the spherical coordinate? Which is the
most pretty interesting thing to do? So, I will leave at that if there is any question
on the part that we did today. So, what we essentially did with let me talk conceptually
for 1 minute. About what we did we and the whole thing whole lecture kind of on this
idea of the pseudo homogeneous model and the basic idea that we try to discussed is that
for a pseudo homogeneous model. Pseudo homogeneous model is really actually for a heterogeneous
system, where these different points sources of sync of reaction and essentially. If the
two conditions on which you can do an averaging one is that the microscopic scale length.
The microscopic scale over the length is much smaller than 1 and second is there is a large.
There is no large variation in concentration over the entire domain; it should not be the
concentration is very smaller at one place and very large. In another if these two conditions
are satisfied then you can average it over the volume and come up with the pseudo homogeneous
model. The main characteristic of the pseudo homogeneous model is that diffusion and reaction
occur parallel in the pseudo homogeneous model, while it does not occurs parallelly in the
heterogeneous model. And then you can just write the diffusion equation and solve this
so, what are these are two things? I want to you to come up with one is the final Bessel
function solution and second is how to solve it in the spherical coordinates. So, tomorrow
I will start off with that what is that?
No, not a function. So, just a trick you know you have to do this and it will come. So,
we will stop here thanks.