WEBVTT
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Design of chemostats, but before we do that
we will do what I promised in the last class
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which is a brief review of the history of
reactors, chemical reactors.
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So, if you look on the screen here that the
history of chemical reactors. You know it
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is essentially talked about this chemical
reactors are of two in major kind. So the
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batch reactor the history of that is when
we started is the dawn of human civilization
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with process of cooking and so on. The second
one is the plug flow reactor which started
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in the late 18th century. Then a big conceptual
leap so the plug flow reactor was being used
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from the late 18th century and so essentially
the major difference between a batch and a
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plug flow is that you had continuous inflow
and the outflow in the plug flow. Whereas,
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in the batch you just have to put the things
in and wait for how long the reaction took.
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So, the reaction could have taken a day or
two days and still you have to wait to get
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the products. Where as in the plug flow reactor,
you have a continuous inflow and outflow.
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Now a big conceptual leap came in the form
of a CSTR reactor, which sort of combined
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the advantages of the batch reactor with that
as the plug flow reactor. The advantages of
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the batch reactor is it is a easy portable
vessel, whereas the plug flow reactor is a
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long tube and it is hard to carry from one
place to another and stuffs like that. So,
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that was a advantage of the batch reactor,
the advantage of the plug flow reactor is
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it is continuous inflow and outflow. So these
are the two advantages. So, the CSTR model
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CSTR thing what it did is combined the advantages
of the batch the portability and the factor
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that it requires less space of the batch reactor
with the advantage of the plug flow reactor
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which is the continuous outflow and inflow.
And this was remembered not you know at some
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point of time once discovered things so this
was not there before that. And this came in
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1908 I said this in the last class. This was
discovered by two german scientists bodenstein
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and wohlgast in 1908. It turns out so this
was the this is a big conceptual leap as far
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as the whole idea of chemical reactions, and
chemical reactors was concerned. Because as
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I said it you know it is portable it is takes
less space and as at the same time it. It
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has continuous inflow and outflow and that
kind of revolutionize the whole chemical reactor
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mean the whole chemical industry.
And if you now look at chemical industries
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all most all chemical reactors are CSTRs or
continuous inflow you know continuous stirred
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tank reactors, only a few percentage of them
you know a small percentage of them would
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be batch reactors. Now 1908 it turns out is
a very event full year for the history of
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chemical engineering or chemical reaction
engineering. Because you had the tubular reactor
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model, the plug flow reactor model rather.
And in 1908 again it was irving langmuir who
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founded the flux type boundary condition.
The which what is now was known as danckwerts
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boundary condition but, it is no longer known
because there are difficult in problem issues
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there. So, as I said that langmuir was he
got the noble prize later, because of his
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work with the kinetics langmuir kinetics adsorption
kinetics. But this was another of his pad
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break breaking you know finings the flux type
boundary condition. The reason being you know
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for us it might seem very mundane at this
point of time to think of a flux type boundary
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condition. But think of it when somebody was
discovering it 100 years back to come up with
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the concept of the fact that you know whatever
is coming in through convection at one end
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the equals what is going out in, because of
convection and diffusion at the other end
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is quite a leap. The reason being that the
concepts of convection and diffusion were
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not as properly underlined at that point of
time. It was only underlined around 1937 and
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I come to that paper which dealt with that,
but so they that is that there is a little
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bit of fuzziness around these concepts of
convection and diffusion and to come up with
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the boundary condition in 1908 which is stayed
for more than 100 years.
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Now and nobody has proved it incorrect yet
is quite a conceptual leap so that was done
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by langmuir in 1908. So, and he the so that
that is a very important boundary condition
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and it came to be later known as the danckwerts
boundary condition. Which it should not have
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been, but we did some unraveling of the real
history of behind this. So, it turns out that
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after danckwerts this condition this danckwerts
boundary condition not danckwerts let us call
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it the flux type boundary condition was rediscovered
a few times. So, the first time it was rediscovered
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by forster and geib two more german scientists
in 1934, and in maybe the you know it is not
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really rediscovered. May be because bodenstein
and wohlgast wrote their paper in german so
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forster and geib would have read those papers
most likely 1934.
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Then it was again rediscovered gerhard damkohler
another of the most famous chemical engineers
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of along with Langmuir. I think he is probably
the most famous chemical engineer of his times
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and our times. So, he rediscovered in 1937
it was not rediscovered actually what he did
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was this is the land mark paper the 1937 paper
of gerhard damkohler. And it was published
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in german. And what he did in this paper was
he reviewed the like we are reviewing the
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whole history of chemical reaction engineering
in 1937 paper he reviewed the entire history
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of chemical engineering including langmuirs
work and forster, and geib’s works and you
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know the whole history of bodenstein and wohlgast
discovering the CSTR.
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You review the entire work of chemical reaction
engineering that was available till that point
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of time then danckwerts rediscovered. And
he was truly trying rediscover you know trying
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to show the world that you know that he discovered
it because he did it. So, there is a fundamental
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difference between what damkohler did and
forster and geib did and danckwerts did because
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forstein and geib and damkohler they referred
to the earlier papers of Langmuir. And showed
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that these are they have discovered these
boundary conditions whereas, danckwerts buried
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everything, he buried all of the previous
work and he pretended that he had discovered
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the boundary condition.
And that pretension worked really well, because
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you know these papers of the forster and geib
and damkohler were all kind of buried by the
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political events of the time which was the
second world war. So, because there was germans
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and there was enormous amount of hatred gains
germans at that point of time, and you know
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the whole world opinion against germans at
that point of time. So, their works were lost
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nobody was reading and the german science
nobody was paying any attention to them so
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their works were practically lost. And then
danckwerts in 1953. He sort of claimed to
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rediscover the boundary condition and now
it has been un urged much before it was not
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us, but around 20 years back it was been found
that danckwerts did not discovered this boundary
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condition.
And people stopped calling it the danckwerts
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boundary condition. They just so and there
is a enormous confusion. You know whether
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you want to call it the langmuir boundary
condition, and forster and geib boundary condition.
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So, just people decided we will call it the
flux type boundary condition. But if I think
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I wrote it in the last class but, i’ll write
it one more time.
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So, the
boundary condition simply goes as so whatever
is coming in u times C equals whatever is
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so u times C in equals. Whatever is going
out minus D times del C del X so you have
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the tubular reactor over here, and you are
trying to find out this is X equals 0. If
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you are trying to find out what is the boundary
condition at X equals 0 so this is my boundary
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condition so whatever is coming in at this
point is because of convection u times C in.
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You can call it even u in times in, but u
is suppose to be the same on both sides so
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u times C incoming in. Because of the convection
because equals what goes out because of convection
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u times C C minus plus what goes out of because
of diffusion and diffusion term is minus del
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C del X so this was essentially the flux type
boundary condition and we still continue 100
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years down the line we still continue to use
this boundary condition it is a quite phenomenal
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impact breaking boundary condition.
So you know we if we are trying to write something
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like if you second order you know differential
will come to that, but if you are writing
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a second order problems and you need a second
boundary condition. So, what would be the
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boundary condition at the other end? So, X
equals 0 this is my boundary condition. What
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would be then my boundary condition at X equals
L? It would be typically it would be del C
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del X equals 0 why if L is large as compared
to say the radius of the reactors do if the,
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if the length is very large what does it mean
what means is that all the reactants has reacted
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and there is no reactant left at the end to
leave. This means that all reactants have
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reacted that there is no reactant left so
these are the boundary condition anyhow so
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let us go back to the history.
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So, then again as I said the 1937 paper of
gerhard damkohler that was the one of the
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most important papers in the history of chemical
reaction engineering and it is a two dimensional
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convection diffusion reaction equation so
we for the first time wrote the two dimensional
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convection diffusion reaction equation and
this as paper is written in 1937 in german.
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And then it was later translated into English.
And what are other things it did and let us
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recount, because so in this paper. He introduced
the idea of radial and axial diffusion so
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just as I said you know the whole ideas of
different kinds of diffusion and convection
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were not very clearly outlined of way back
in those days.
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And it was damkohler who was who first introduced
the idea of radial diffusion and convection
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axial diffusion and how they are different.
So, it is not at really understood at that
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point of time how the radial and axial diffusions
are different, than he introduced a parabolic
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velocity profile, till then with a plug flow
reactor right plug flow reactor model, p f
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r model. What is the p f r model? Intel it
means that, the velocity profile is flat within
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the within the reactor. And the but the velocity
profile is not really flat because when you
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have flow inside the tube the velocity profile
is parabolic. But if you understand that that
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in a plug flow reactor there is no way I mean
there is difficulty at least to introduce
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a plug flow radial velocity profile.
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Let me show you why because what you have
is u del C del X equals minus R of C, this
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is your plug flow reactor model. Now this
is in X so this is my reactor, and this is
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the X direction, and this is the r direction.
Now if u equals u naught 1 minus r over R
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square then it is, how do you solve it ? You
know so if I put it back over here 1 minus
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r square over R square so this R of C how
do you solve it you see what I am trying to
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say the difficulty of solving it why, because
the differential in the X direction and then
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you have a term in the radial direction. So,
how do you solve it? There is no way to solve
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it so the only way of being able to incorporate
a radial velocity profile is to actually is
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to actually.
.
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Actually go for a two dimensional description
so which is what damkohler, did you know he
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went for a two dimensional description and
as a result of which he could incorporate
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the parabolic velocity profile. He included
radial and axial diffusion and he included
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the flux type boundary condition. So it was
the first time somebody looked at the chemical
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reactor in a very comprehensive way incorporating
convection radial, and axial diffusion parabolic
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velocity profile, as well as flux type boundary
condition which includes both convection and
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diffusion effects as the at the entrance.
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So this was the 1937 paper and then the whole
idea you were aware of the idea of residence
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time distribution theory. So, and surface
renewal in residence time distribution theory
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so they came around like the same time from
danckwerts you know paper. But again it turns
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out that the forster and geib was actually
the first one to discover the RTD theory.
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The residence time distribution theory the
whole idea that you know that some fluid the
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different fluid elements in a reactor will
have different residence times. You know so
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you can say that the reactor residence time
is tau, but that does mean that the entire
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fluid in the reactor spend the amount of tau
in the reactor.
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What does the residence time mean? You know
when I say that the residence time what is
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the residence time mean the amount of time
you spend here so the residence time of the
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B tech student here is 4 years or the residence
time of a Phd student is 4 years something
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like that so which means that the amount of
time that you spend from input inlet to outlet.
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But when I say that the residence time in
a CSTR is tau strictly speaking the not so
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this was it was discovered essentially by
forster and geib that strictly speaking. It
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is not that all every element of fluid in
the reactor is actually spending the tau amount
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of time.
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So, typically the residence time distribution
is you now goes something like this so this
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is how residence time changes. So, the average
of this you know if you take the whole of
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it and divide it then the average of it is
tau, but some elements as you see her[e]-
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over here some elements of fluid the so this
is the amount of elements of the fluid, some
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elements of the fluids spend very small time
again. You know some very small so the bell
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shaped curve and only a fraction says spend
a very large amount of time in the reactor.
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So the when I say that reactor is the something
you have to remember, when I say there is
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a reactor residence time is tau. It does not
mean that everything in the reactors spends
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tau amount of time it means that the average
amount of time spent within in the reactor
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by a fluid element is tau.
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So, this whole concept was actually started
by forster and geib in 1934, and then it was
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again you know danckwerts is a great rediscoverer
so he was he rediscovered and this theory
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in 1953. But he had some inputs also which
is he generalized, that it was in a forster
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and geib’s paper in 1934 which I got translated
through somebody, and read it and it was a
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very localized concept. But danckwertz generalized
it for the whole reactor thing and it really
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you know the his presentation was good. Then
last important things that came in the whole
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CSTR business was a concept of micromixing,
which is mixing at the molecular scale. Because
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reaction takes place at the molecular scale,
and if reaction has to take place in the molecular
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scale. Then there has to be mixing at the
molecular scale right because if a and b two
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reactants are there and unless they are mixed
with the molecular scale they cannot react.
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Is that clear? Because the reaction has to
take in the molecular scale. So, the concepts
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of micromixing maximum mixedness and complete
segregation. Complete segregation when there
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is no mixing at all between the two components.
So, these were propounded by as zweitering
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in 1959. So, I think that concludes our brief
review of the history of chemical reactor
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models what we will do now is that we will
so much of this theory as you see is related
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to CSTR because as I said CSTR was a big conceptual
leap.
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So what I will do now we will start try and
look at the CSTR for a bioreactions. So, the
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CSTR for bioreactors is called chemostats,
and this is let me show you the picture and
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try and explain. So, this is the central thing
that you have over here is a chemostat. So,
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it is a CSTR which means it is a continuous
stirred tank reactor, which has two parts
16:32.350 --> 16:38.620
with one is continuous another is stirred
and for four parts. Actually continuous stirred
16:38.620 --> 16:42.680
tank reactors. So, the continuous means there
is a continuous inflow of feed and a continuous
16:42.680 --> 16:47.860
outflow of product, stirred means it is being
stirred by a a stirred over here, tanks so
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it is a tank and then it is a reactor so the
four elements come together.
16:51.620 --> 16:56.490
So as you see that the way it works for the
chemostat is that, it is the feed reservoir
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over here from which feed is being continuously
pumped into the reactor. So, then you have
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a buffer solution which is also being pumped,
because these are being bioreactions you have
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to maintain the p H at the certain values
could be 7 could be 5 could be 6 whatever.
17:11.439 --> 17:16.639
but you have to maintain the p H at a certain
value so you have a p H controller and a buffer
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solution being sent in. You have you know
air being taken out and replenished so those
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kind of things we have stopped about this
you know you need a certain amount of oxygen
17:26.139 --> 17:31.820
into the system and so on.
So, air being taken out and you know replenished
17:31.820 --> 17:38.820
in stuffs like that here and then the product.
So, the air taken out here and replenished
17:39.769 --> 17:44.529
from here just a while air comes in here just
for the oxygen you know that you need to supply
17:44.529 --> 17:50.960
and you know the nutrient that is the carbon,
hydrogen, nitrogen in liquid form are a part
17:50.960 --> 17:57.960
of the feed reservoir in the feed reservoir
itself. So, the product is taken out now the
17:59.019 --> 18:05.499
product will include these 3 X S and F what
is X? X is the cell are growing. S is the
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substrate and feed is you know everything
else apart from the substrate and the cell.
18:09.820 --> 18:16.820
So, that part of the feed which is not either
substrate nor cells. V is at any point of
18:17.499 --> 18:22.440
time the liquid volume in the reactor, why
is V important not the volume of the actual
18:22.440 --> 18:26.190
reactors? Beause the residence time of the
reactor the average residence time as we call
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it now depends on the volume of the reactor
and not the V. In volume of the liquid sorry
18:32.570 --> 18:37.399
in the reactor and not the actual volume of
the tank. Because the rest of the tank that
18:37.399 --> 18:43.229
has air in it is not of any use. So, V here
and then the product is taken out and pumped
18:43.229 --> 18:49.769
out and take put in a product vessel. So,
this is a whole flow chart of this system,
18:49.769 --> 18:56.769
and is there any question or do you want me
to stop up for anything here or shall we proceed?
18:57.529 --> 19:01.809
We should proceed.
So, the next question that comes in is that
19:01.809 --> 19:06.519
we have to in order to understand the dynamics
of this process we have to write balances
19:06.519 --> 19:11.029
for this. So, the balance would be straight
forward similar to the CSTR balance that you
19:11.029 --> 19:17.789
have written except that it is for two components
here. So, what are the two components you
19:17.789 --> 19:22.720
are going to write the balances for? One there
will be let us go slowly. So, how many balances
19:22.720 --> 19:29.509
do you think that you will have to write?
The first balance would be the overall balance
19:29.509 --> 19:35.220
of volume which is the volume accumulated
or volume held back in the reactor, and volume
19:35.220 --> 19:40.710
coming in and volume going out so that is
a overall balance then what else we have.
19:40.710 --> 19:43.009
Component Balances.
So, one would be the substrate balance. So,
19:43.009 --> 19:47.649
the two major components as you can see in
the picture over here, that X S and F so one
19:47.649 --> 19:52.299
would be the substrate balance S then would
be the cell balance X so essentially three
19:52.299 --> 19:52.739
of these.
19:52.739 --> 19:59.340
So, let us go and do it so the chemostat as
you can see over here is an ideal CSTR. Because
19:59.340 --> 20:05.139
it is stirred and you know of everything and
continuous, and so on. So, the first you need
20:05.139 --> 20:09.470
to do a total volume balance. So, the total
volume balance is very straight forward d
20:09.470 --> 20:16.470
V d t equals F in minus F out feed coming
in minus feed going out. So, the next balance
20:28.690 --> 20:33.320
we do is the mass balance for the cells and
mass balance of the cells is simply if you
20:33.320 --> 20:38.889
take this. For example, it you have to multiply
it with the cell concentration so d d t of
20:38.889 --> 20:44.450
X V this and these two plus.
They are the generation of cells so that is
20:44.450 --> 20:50.470
the only difference between the other component,
and the cell is that there is a generation.
20:50.470 --> 20:54.769
I mean overall volume is not been generated
where cells are being generated so d t of
20:54.769 --> 21:00.129
X V equals F n times X naught is the amount
of cell that is coming in the feed. So, the
21:00.129 --> 21:06.309
feed itself can have some cells minus F out
times X is the amount of cells that is in
21:06.309 --> 21:12.159
the outlet products, and mu times X is a specific.
You know mu is a specific growth rate per
21:12.159 --> 21:18.399
unit cell mu times X is the total for the
total amount of cells times, the volume you
21:18.399 --> 21:23.099
know so because the volume of the liquid should
come in over there.
21:23.099 --> 21:28.269
Sir the communications death of cells.
Equation is.
21:28.269 --> 21:33.830
Death of cells. Death of cells death of cells
has not been included, but you can add another
21:33.830 --> 21:40.830
term minus beta X that we did last time. So,
you can just add minus so you instead of mu
21:41.869 --> 21:47.179
you can, you can replace mu by mu hat where
mu hat is mu minus beta it is fine. So, it
21:47.179 --> 21:54.109
is straight forward so the last thing that
we need to do is the mass balance for substrates
21:54.109 --> 22:01.109
fine so which is d V t of S V equals F n times
S naught minus F out times S minus mu X V
22:02.059 --> 22:06.609
over Y. Because I explained this several times
in the last couple of lectures Y is just the
22:06.609 --> 22:13.609
el ratio of X over S. Now typically F you
know if you want to typically in CSTR what
22:37.940 --> 22:42.489
you would like is that the inlet is equal
equals outlet and if that is the case F in
22:42.489 --> 22:47.059
equals F out.
Then d V d t is 0 so as a result of which
22:47.059 --> 22:54.059
the V is a constant fine and you can do a
lot of simplifications in these equations.
22:54.330 --> 22:58.609
Because V is a constant you can take V out
of this equation you can take V out of this
22:58.609 --> 23:05.320
equation fine, and you can get rid of b in
some other places because the one of the over
23:05.320 --> 23:12.320
V will come in here. So, when you do the simplifications
you get d X d t equals F over V X naught minus
23:12.359 --> 23:17.999
X plus mu X it looks like a lot simpler equation
and just as he said that if you want to include
23:17.999 --> 23:24.999
the death of cells replace mu by mu hat where
mu hat is mu minus beta being the specific
23:26.789 --> 23:33.789
death rate of cells. So, and the last thing
would be the substrate balance equation for
23:47.659 --> 23:54.659
the substrate d S d t equals F over V S naught
minus S 1 over Y mu X fine.
24:08.039 --> 24:12.570
So what would be our step now so once we have
done this. So, what would you like we will
24:12.570 --> 24:18.059
like to find out how it would changed? But
what would be your step? What would you do
24:18.059 --> 24:25.059
you think we should do now. So, if F is a
constant and known then how many variables
24:37.249 --> 24:44.249
do we have just two variables, and two equations
so what we would intend to do is essentially
24:45.859 --> 24:52.299
solve this, but how do we solve this? I mean
you know what kind of solution do you want
24:52.299 --> 24:55.739
first let me ask you that.
24:55.739 --> 25:02.609
What is that.
25:02.609 --> 25:05.929
Possible for what for every thing.
25:05.929 --> 25:11.529
Well I do not think analytical solution is
possible one of the things, you can try to
25:11.529 --> 25:16.769
do is get an invariance. I taught you have
to get invariance in the last class, you can
25:16.769 --> 25:20.929
get an invariance, and then you can get an
analytical solution for combination of X and
25:20.929 --> 25:26.830
S naught for X or X separately. But what we
want to do first is the easy step out which
25:26.830 --> 25:29.909
is up in the steady state solution of it.
25:29.909 --> 25:36.739
So F over V is called the so we do that. So,
let us define some quantities first F over
25:36.739 --> 25:43.739
V is called the dilution rate of the cells
F over V is a dilution rate of the cells,
25:43.769 --> 25:47.009
and what is the unit of Sover V F over V?
25:47.009 --> 25:49.499
Hm what.
25:49.499 --> 25:55.219
Inverse of time which means it is inverse
of the residence time of the system so F over
25:55.219 --> 26:02.219
V is 1 over tau, which is the inverse of the
residence time of the system. So, when you
26:07.820 --> 26:13.849
do that? You can put here d t and d t of X
equals dilution rate d e you know instead
26:13.849 --> 26:20.289
of 1 over tau. In this chapter or in this
course we will use the dilution rate d instead
26:20.289 --> 26:25.909
of the residence time. So, D is 1 over tau
X naught minus X plus mu X and d s d t is
26:25.909 --> 26:32.909
a d e S naught minus S minus mu X over Y fine.
Now mu is we are going to assume a Monod growth
26:42.690 --> 26:47.269
kinetics as I said it is a most popular and
most useful growth kinetics there are plenty
26:47.269 --> 26:51.340
others as. I showed you in someone of the
earlier lectures but, this is the one that
26:51.340 --> 26:57.279
is that we are going to try and use for now.
Now as I said that you know sometimes I might
26:57.279 --> 27:04.279
ask you in the exams for example, to do the
similar kind of thing with other growth kinetics.
27:05.700 --> 27:10.190
I can give you more complicated growth kinetics
or I can give you growth kinetics where inhibition
27:10.190 --> 27:15.549
is involved or all other kinds of things possible.
But then but you have to be aware of how this
27:15.549 --> 27:19.549
is the process so what we’re going to do
from now on todays class and next class is
27:19.549 --> 27:24.019
slightly hard and you should pay attention
to the process, what is really happening and
27:24.019 --> 27:28.639
how to do it? Because then you should be able
to do it for other kinds of systems so that
27:28.639 --> 27:29.779
is fine.
27:29.779 --> 27:33.989
So we put the monod growth kinetics in and
so then we go for the steady state solution.
27:33.989 --> 27:39.529
So, at steady state this is your basic equation.
So, d X naught so if you look at this here
27:39.529 --> 27:46.529
so we put a steady state out here so D X naught
equals D minus mu times X for the X equation
27:46.989 --> 27:52.109
and then similarly, for the substrate also
you will have another one. Now the and the
27:52.109 --> 27:59.109
first approximation that we can, we make assumption
that we make is that it is a sterile feed.
27:59.559 --> 28:04.229
It is we are solving it for a sterile feed
sterile feed means that there is no cell in
28:04.229 --> 28:08.440
the feed that is coming in. It is an assumption
of coarse because typically there are cells,
28:08.440 --> 28:10.639
but we are just trying to solve it for this
system.
28:10.639 --> 28:14.799
We will solve it for other systems also without
sterile feed but, let us try and solve it
28:14.799 --> 28:19.039
for with sterile feed is it clear the sterile
feed means, that X naught there are no cells
28:19.039 --> 28:25.200
in the in the incoming feed so when you do
that you simply get D equals mu the dilution
28:25.200 --> 28:32.200
rate equals mu. Now mu is this specific growth
rate remember so it goes as mu max times S
28:34.099 --> 28:40.969
over K S plus S which equals the dilution
rate. So, if you have a fix dilution rate.
28:40.969 --> 28:45.339
So, what this means is? For a sterile feed
for the system to have any sort of meaning
28:45.339 --> 28:52.339
full solution, if you have a dilution rate
that is fixed then yours steady state substrate
28:52.769 --> 28:58.190
concentration has to be this. You see what
I am saying why is that? Because see if X
28:58.190 --> 29:05.070
naught is 0 then d minus mu times X is 0 what
is the solution of this equation.
29:05.070 --> 29:09.669
The two solutions of this equation one is
that X naught it is X itself is 0 then the
29:09.669 --> 29:16.379
other one is that D equals mu so if D is not
equals mu, then the other solution is X naught
29:16.379 --> 29:20.149
X equals 0 which means that no cells are being
produced which is worth less because I am
29:20.149 --> 29:25.809
running the whole reactor to produce cells.
So, the only feasible solution in this place
29:25.809 --> 29:32.809
is d equals mu. Now when you put mu max X
over F plus s equals D then the steady state
29:34.960 --> 29:41.960
substrate concentration is D K s over mu max
minus D fine. Provided that X s s is not 0
29:45.029 --> 29:49.489
I mean of coarse and mu max is greater than
D mu max has to be greater than D because
29:49.489 --> 29:56.489
you see why, because mu if mu is greater than
equals D then this fraction over here is always
29:56.549 --> 30:00.889
less than one so mu max obviously has to be
greater than D.
30:00.889 --> 30:05.799
So, from the second equation that is the equation
for the substrate this is this is what we
30:05.799 --> 30:12.779
get. Now the substrate concentration and you
know D equals mu for X naught equals 0 so
30:12.779 --> 30:19.589
you can substitute back over here D equals
mu in this equation. So, you know over here
30:19.589 --> 30:26.589
and then you can get your X that is the steady
state concentration X s s in terms of these.
30:31.349 --> 30:36.639
As this because we wrote our S s s before
if you remember in the last slide I will just
30:36.639 --> 30:40.629
wait for a few seconds because some of you
are writing so and then I will show you if
30:40.629 --> 30:43.450
you need to so if we wrote the S s s in the
last slide.
30:43.450 --> 30:49.429
So you can substitute your S s s from the
last equation that we wrote before in to this
30:49.429 --> 30:56.429
and is it clear to all of you or do you want
me to go through this steps. So, all you need
31:01.359 --> 31:05.859
to do is let me just quickly see show. So,
all you need to do S s s is given here as
31:05.859 --> 31:11.489
D K s over mu max minus D all you need to
do is substitute for S s s in from that equation
31:11.489 --> 31:16.749
into this equation. And you put D equals mu
over here, and then you will get X s s that
31:16.749 --> 31:21.830
steady state concentration of cells in the
system goes as if we forget the Y let us you
31:21.830 --> 31:28.830
know ignore the y then it goes as s naught
minus D K s over mu max minus D and.
31:34.839 --> 31:41.239
You know this we already did D max equals
mu max D equals mu max S naught over K s plus
31:41.239 --> 31:46.269
S naught D equals mu max S over K s plus X
so D max would be the maximum value that the
31:46.269 --> 31:51.129
D can take is mu max S naught over K s plus
s naught. Why because the maximum values that
31:51.129 --> 31:58.129
the S can take is s naught for the feed concentration
clear to everybody, and if there is any point
31:58.139 --> 32:02.830
you need to stop me please stop me and i’ll
explain one more time .
32:02.830 --> 32:09.830
So if that is clear then as in this equation
as mu max goes to D you see over here. Then
32:11.839 --> 32:17.879
what will happen this is goes to 0 the denominator
here goes to 0 and this goes to infinity as
32:17.879 --> 32:24.879
a result solutions are possible for mu D greater
than mu max and D less than mu max is it clear
32:27.539 --> 32:31.899
see. What was what did we get from our initial
assumption of sterile feed? We got that D
32:31.899 --> 32:38.899
equals mu fine and one of the things we said
is that D is typically if D equals mu then
32:40.239 --> 32:44.820
D would be less than mu max, because mu max
is always greater than mu. So we are actually
32:44.820 --> 32:51.139
working in this space of D less than mu max,
but just to make sure that you know to make
32:51.139 --> 32:55.919
sure that this a there’s a feasible range.
So, D equals mu max there is no solution because
32:55.919 --> 33:00.669
this whole thing blows up the solution is
only possible in the D less than mu max, D
33:00.669 --> 33:04.409
greater than mu max. But obviously we are
working in the space of D less than mu max
33:04.409 --> 33:10.989
clear no confusion. So, this is what I said
that we are working in the space of D, D less
33:10.989 --> 33:16.089
than mu max because it is not possible for
D if D equals mu max mu. Then it is not possible
33:16.089 --> 33:19.859
for D to be greater than mu max because mu
max is obviously greater than mu .
33:19.859 --> 33:26.859
Now for X s s equals 0 D greater than mu max
which is not in the you know which is not
33:29.489 --> 33:33.869
in the it is in the feasible range, but it
is not of any interest to us if no cells are
33:33.869 --> 33:37.729
being produced. it is of no interest to us
so this is a just another possibility so for
33:37.729 --> 33:44.669
X s naught equals 0 D has to be less than
mu max clear. From not necessarily from the
33:44.669 --> 33:50.159
mathematics but, the physics of the problem.
V has to be less than mu max because V is
33:50.159 --> 33:56.209
D equals mu and if D greater than mu max and
X s s is 0. So, my steady state solution is
33:56.209 --> 34:02.080
that for D less than mu max I have my steady
state solution as this and for D greater than
34:02.080 --> 34:09.080
mu max it is 0. So, what does this mean? what
it means is that? Why am I why am? I you know
34:11.359 --> 34:17.510
trying to answer this question? Why am I trying
to a kind of boil everything down to d and
34:17.510 --> 34:24.510
mu max? Both of these two are parameters of
the system d and mu max but, what is the difference
34:27.109 --> 34:29.500
between these parameters.
34:29.500 --> 34:36.500
Absolutely yeah, see the mu max is a parameter
that cannot be changed it cannot be manipulated
34:39.460 --> 34:43.789
whereas, D is a parameter which can be manipulated
D is one over the residence time of the reactor.
34:43.789 --> 34:47.250
How can I change the residence time of the
reactor? Just flow rate yeah just basically
34:47.250 --> 34:52.720
by changing the flow rate or the volume that
of the liquid in the system. So, what this
34:52.720 --> 34:59.720
is trying to say is that here so what this
is trying to say over here is that these two
35:00.170 --> 35:05.930
cases, that if you want if you want cells
to be produced your d has to be less than
35:05.930 --> 35:10.380
your mu max or in other words your dilution
rate has to be less than the maximum specific
35:10.380 --> 35:14.750
growth rate of the cell.
If your max dilution rate is greater than
35:14.750 --> 35:18.730
the maximum specific growth rate of the cell
then there are no cells which were have going
35:18.730 --> 35:22.410
to be produced. What does this physically
mean for you what it physically means is?
35:22.410 --> 35:27.519
That if dilution is very large then it will
sweep this as the way you know it would. It
35:27.519 --> 35:33.569
would take the things away with it so if D
is very large for example, then what it means
35:33.569 --> 35:39.099
what it means is the residence time in the
reactor is very low fine because D goes as
35:39.099 --> 35:43.079
one over tau. If the residence time of the
reactor is very low what does it mean it means
35:43.079 --> 35:49.010
that the cells are not getting enough time
to react for and grow. So, this is the physical
35:49.010 --> 35:54.440
interpretation so in bioreactor system in
biochemical engineering, they say that dilution
35:54.440 --> 35:59.500
of the system dilution rate being very large
means that the system is being excessively
35:59.500 --> 36:02.569
diluted and as a result.
Cells cannot be grown but, if you look at
36:02.569 --> 36:05.920
it from a chemical engineering point of view
,then we convert it into the residence time.
36:05.920 --> 36:10.829
So, D very large means residence times very
time very small which means that the cells
36:10.829 --> 36:14.430
are not getting enough time in the reactor
to grow.
36:14.430 --> 36:18.690
So, that is why it happens? From mathematically
let us look a little bit of the mathematical
36:18.690 --> 36:24.700
space of it. So, if this is the case that
is D is less than mu max, then your steady
36:24.700 --> 36:31.299
state X s s I just gave you here. This is
the steady state solution for the cell and
36:31.299 --> 36:35.829
the first thing here is the steady state solution
for the substrate. Do you need to write the
36:35.829 --> 36:39.490
steady state solution for cell? I think you
wrote it already, but if you want to it will
36:39.490 --> 36:46.490
give you a few seconds so the steady state
solution for the substrate now is given by
36:53.200 --> 37:00.200
D K S over mu max minus D. So, this has to
be satisfied S naught greater than D K S over
37:09.099 --> 37:16.099
mu max minus D so because mu max is greater
than D so this denominator is positive so
37:16.420 --> 37:19.720
the second criteria that we get.
So there are two criteria that we have to
37:19.720 --> 37:24.640
satisfy one is that the residence time of
the system or dilution rate of the system.
37:24.640 --> 37:27.680
One is the residence time is one over dilution
rate dilution rate of the system has to be
37:27.680 --> 37:33.279
greater than mu max. And the second criteria
we get is given here S naught, which is the
37:33.279 --> 37:37.519
initial concentration of substrate in the
feed has to be greater than this quantity
37:37.519 --> 37:43.970
D K S over mu max minus D. So, both of these
two criteria have to be satisfied in order
37:43.970 --> 37:50.970
to be able to produce cells so that your X
s s is greater than 0 both of these two criteria
37:52.460 --> 37:59.460
to be satisfied fine is it clear.
So, both these straight two criteria have
38:05.339 --> 38:09.750
to be satisfied, in order for it be, if they
are not satisfied then X s s and S s s are
38:09.750 --> 38:14.789
in the not in the feasible domain. So, what
happened was that? If you remember the that
38:14.789 --> 38:18.789
how do we get the first criteria we got the
first criteria by showing that if mu max is
38:18.789 --> 38:25.789
less than d, then your X s s is 0 or negative.
We got the second criteria by showing that
38:25.849 --> 38:32.390
if your S naught is not greater than this,
then your S s s would be negative or 0. Because
38:32.390 --> 38:37.269
see what happens is that both the substrate
and the cell that are coming out have to be
38:37.269 --> 38:44.269
positive numbers both of them. Both the substrate
so for for the positivity of the cell you
38:44.880 --> 38:48.819
need to ensure that D is greater than D is
less than mu max for the positivity of the
38:48.819 --> 38:52.329
substrate you need to ensure this quantity.
That is the initial amount of substrate that
38:52.329 --> 38:59.329
you to put into the system is greater than
D K S over mu max minus D S and so and the
38:59.789 --> 39:06.789
second regime is D no. I do not know this
is a case of X s s equals 0 and S equals at
39:17.789 --> 39:24.789
S naught so what this means is the nothing
is happening in the reactor for the same case.
39:25.029 --> 39:29.740
But nothing is happening in the reactor. So
which means that you start with cells this
39:29.740 --> 39:36.740
is what kind of solution is this trivial solution.
So, this is known as a trivial solution whereas,
39:39.869 --> 39:46.869
X s s is 0 that is you put in cells is or
input solution is free of cells when you are
39:50.369 --> 39:55.490
putting it in now no cells. In no cells out
and as a result whatever substrate you put
39:55.490 --> 39:58.410
in is whatever substrate that leaves the system
so this is a trivial solution.
39:58.410 --> 40:02.809
Why am I going through? This is because you
know one of the things you kind of forget
40:02.809 --> 40:09.809
like let us go back, and show you this equation
here. For example so a system like this you
40:11.960 --> 40:16.180
know guys, you know very calmly you go and
write the one single solution that is there
40:16.180 --> 40:19.359
possible. But when you are doing a mathematical
analysis you have to take into account all
40:19.359 --> 40:24.589
kinds of solutions. So, one of the solution
is a trivial solution and the trivial solution
40:24.589 --> 40:30.019
as you will see a little later may be in next
lecture is important also. So, why we are
40:30.019 --> 40:33.519
doing this is trying to look at the entire
space composed of all different kinds of solutions.
40:33.519 --> 40:37.079
So, the first one we discussed is the non
trivial solution the second one is the trivial
40:37.079 --> 40:38.119
solution.
40:38.119 --> 40:44.210
And this is summary of the behavior of the
chemo stats. So, these are my equations which
40:44.210 --> 40:48.259
I am sure you had already written down. So,
there is no point spending any more time on
40:48.259 --> 40:54.829
this so d X d t equals this and d S d t equals
this and mu equals mu max S over k s plus
40:54.829 --> 41:00.750
S and what I will next do is summarize the
results also the ones we looked at.
41:00.750 --> 41:07.750
So this is a feasible range this is the feasible
range of solution here the rest as we discussed
41:09.349 --> 41:14.019
that there are two constraint for this things
to happen that feasible range means both positive
41:14.019 --> 41:19.539
results for X s s and S s s that is the substrate
and the and the cells going out of the reactor
41:19.539 --> 41:26.539
are both positive numbers. So, the two constraints
for this one is that the dilution rate has
41:26.720 --> 41:31.769
to be less than mu max. The second one is
the initial concentration of the cell is greater
41:31.769 --> 41:36.869
than this number. Now if any of these two
constraints are violated then what you end
41:36.869 --> 41:41.109
up having is a trivial solution as you have
over here all these are trivial solutions.
41:41.109 --> 41:46.329
Because X s s is 0 and the steady state concentration
of the substrate is same as, what you put
41:46.329 --> 41:50.670
in so what you need to remember? you know
the best way to remember this is that these
41:50.670 --> 41:55.150
two constraints have to be satisfied in they
are then in this space where both these constraints
41:55.150 --> 42:01.779
are satisfied and intersect You get the solution
anything else you get trivial solution. So
42:01.779 --> 42:08.779
this is essentially the steady state analysis
of the chemostat. Now what we are going to
42:09.309 --> 42:16.309
do is? Look at so when you in a chemical reactor
or any kind of you know continuous reactor,
42:17.599 --> 42:22.200
you want to run these reactors at steady states
fine. You would like to why do you want to
42:22.200 --> 42:26.269
run these reactors as steady states because.
Control is easy. Control is easier because
42:26.269 --> 42:31.869
whenever you have dynamics of the system coming
in control gets harder, but what happens in
42:31.869 --> 42:36.839
a real system? you know we are running it
at the pilot scale or even forget plant scale
42:36.839 --> 42:40.950
even at the pilot when you are running and
even in the lab scale, what happens? You are
42:40.950 --> 42:45.769
trying to run it at the steady state but,
you are probably not able to run it at the
42:45.769 --> 42:51.220
steady state, why does that happen? You can
have your control as an everything and you
42:51.220 --> 42:55.089
will see that, when you do even experiments
in the control lab you will see that you you’re
42:55.089 --> 42:59.190
trying to manipulate your control variables.
So, that the system is at steady state the
42:59.190 --> 43:03.190
level of the fluid in a in a you tube or anything
for that matter and you will see that it keeps
43:03.190 --> 43:08.849
oscillating it is very hard to retain it at
the study state. Why is that?
43:08.849 --> 43:12.579
What.
Disturbance will be there.
43:12.579 --> 43:14.200
What kind of disturbances that is , what kind
of disturbances?
43:14.200 --> 43:20.680
fluctuations.
The key word here is as he said fluctuations
43:20.680 --> 43:26.099
or what we call in mathematical balance is
perturbation there are these natural noises
43:26.099 --> 43:29.190
and natural perturbations that are there in
the system. And why do these natural noises
43:29.190 --> 43:33.230
and perturbations coming every angle? You
know you think that this table probably has
43:33.230 --> 43:37.420
no vibration but, it is full of vibrations.
So , where ever you putting in your reactor
43:37.420 --> 43:44.049
as lots of vibrations in there, and these
vibrations lead to perturbation or fluctuations
43:44.049 --> 43:49.410
in the system. And as it turns out that you
know these system many of these reaction systems
43:49.410 --> 43:54.789
are pretty sensitive. The reason these are
sensitive is that, I talked about I do not
43:54.789 --> 43:59.190
remember. I think I did talk about the case
for the autocatalytic reaction in this class
43:59.190 --> 44:05.950
itself at some point of time. So, that one
is example of why these systems are very sensitive?
44:05.950 --> 44:08.930
The reason these systems are very sensitive
is because they are non-linear so when they
44:08.930 --> 44:12.130
are non-linear.
So a small perturbation in one of the variables
44:12.130 --> 44:17.059
can lead to large perturbations in the entire
system, because of non-linearity this is a
44:17.059 --> 44:21.930
multiplying effect. So, small perturbation
in one you know leads to another perturbation
44:21.930 --> 44:26.390
in other. And these two effects kind of multiply,
and then the wholes the whole kind of whole
44:26.390 --> 44:31.180
amount of perturbation that is generated in
the system is multiplied and kind of exaggerated.
44:31.180 --> 44:37.039
So, this you know there is a whole analysis
for that and that analysis is known as stability
44:37.039 --> 44:41.420
analysis so we will that is that is an extent
we are going to look at.
44:41.420 --> 44:46.490
So what we are going to do is stability analysis
of bioreactors and you know we will start
44:46.490 --> 44:50.059
today, because it is a complicated thing and
we would not finish today and we will continue
44:50.059 --> 44:57.059
into the next lecture. So, stability analysis
is that analysis of the system when you apply
44:57.859 --> 45:02.490
perturbation to a study state. So, if a system
you have done this partially. I believe in
45:02.490 --> 45:07.509
the control class process dynamics and control
and we are going to use some of those things.
45:07.509 --> 45:10.859
That you had learnt over there so what is
the idea of a stability analysis the ideas
45:10.859 --> 45:15.119
of a stability analysis. You first obtain
the steady state of the system then you perturb
45:15.119 --> 45:19.660
the steady state slightly. And you see how
does a system react?
45:19.660 --> 45:24.369
How this how this system react? There are
two possibilities two major possibilities
45:24.369 --> 45:31.369
three actually but, let us say two major possibilities,
one is that the system will go back. Go back
45:31.559 --> 45:37.150
to the steady state stabilize the other is,
system will move away from the steady state
45:37.150 --> 45:41.200
which is destabilize. And the third possibility
is the minor possibility is that the system
45:41.200 --> 45:44.880
remains exactly as you put it you know you
perturb it. And the perturbation continuous
45:44.880 --> 45:49.839
that little perturbation you gave continuous
like that it does not grow, does not decay.
45:49.839 --> 45:54.559
But that s that is a minor possibility because
that does not happen. The two major possibilities.
45:54.559 --> 45:58.589
One is the perturbation that you give decays
as a result of which remember the perturbation
45:58.589 --> 46:03.130
that you give decays as a result of which
the system returns to the steady state.
46:03.130 --> 46:07.359
And the second one is the perturbation that
you gave grows as a result of which the system
46:07.359 --> 46:11.809
destabilizes so how do we mathematically analyze.
And understand this so this is what we will
46:11.809 --> 46:16.680
go we are going to do over the next lecture.
So, the method that we use as a liapunov’s
46:16.680 --> 46:23.000
here the liapunov’s method of linear stability
analysis for a chemical reactor, and you probably
46:23.000 --> 46:30.000
used it in your control class. It was found
the and amnndson just type over here m u n
46:30.529 --> 46:37.529
d Amundsen this should not be n not n n m
u n d amundson 1955 amundson it turns out
46:37.700 --> 46:42.880
that he is known as the father of modern chemical
engineering as we know it .
46:42.880 --> 46:46.390
You know the statement and so the chemical
engineering department which is the best in
46:46.390 --> 46:51.759
the world he was the chair of that for 20
20 5 5 years. And then the chemical engineering
46:51.759 --> 46:58.609
department at university of minnesota is known
as a Amundsen hall. And the major most important
46:58.609 --> 47:02.289
award of chemical reaction engineering is
also named after Amundsen. So, he is the father
47:02.289 --> 47:06.109
of modern chemical engineering, why I am calling
it modern chemical engineering is that? It
47:06.109 --> 47:11.359
is with amundson he is the one who introduced
along with birds to attain light food of coarse,
47:11.359 --> 47:17.380
but he is the one who introduced mathematics
or applied mathematics to chemical engineering.
47:17.380 --> 47:21.960
And as a result of which chemical engineering
has a form that it has today so birds to attain
47:21.960 --> 47:27.509
light food are responsible for introducing
mathematics.
47:27.509 --> 47:31.470
Applied mathematics to transport phenomena.
So, the way you we talk about heat transfer
47:31.470 --> 47:38.470
mass transfer and momentum transfer is, because
of with the changes that were in you know
47:39.369 --> 47:45.490
a brought in by birds to attain light food
in 1950s. And the very famous book that they
47:45.490 --> 47:50.749
have. So, we do not look at heat transfer
just is and the terms of a heat transfer equipment
47:50.749 --> 47:54.470
or mass transfer as a distillation column.
But we understand the basics of the system,
47:54.470 --> 47:59.240
and we understand that everything that follows
be it distillation be it adsorption or absorption
47:59.240 --> 48:05.599
heat exchange in a heat exchanger.
Everything is a product of the very basics
48:05.599 --> 48:10.259
of heat mass and momentum transfer and if
as you have done in your transport phenomena
48:10.259 --> 48:14.400
lecture. There is a tremendous analogy between
heat mass and momentum transfer so if you
48:14.400 --> 48:19.039
describe one set of systems the momentum transfer
equations. You can use the same systems a
48:19.039 --> 48:22.940
boundary layer and so on to describe heat
transfer and mass transfer so this was the
48:22.940 --> 48:26.720
it is a concept that was introduced by birds
to attain light food to chemical engineering
48:26.720 --> 48:32.609
and we owe it you know we owe it to them.
Similarly, we owe it to amundson neal amundson
48:32.609 --> 48:37.319
for introducing the whole of this similar
kind of approach a very mathematical way of
48:37.319 --> 48:41.069
looking and a very basic way of looking to
chemical engineering.
48:41.069 --> 48:45.200
Before that chemical engineering is also just
about using chemical reactors and getting
48:45.200 --> 48:49.230
products out. But then he introduces whole
approach and one of the major things did he
48:49.230 --> 48:55.190
did was looking at stability. He also did
introduce bifurcation and analysis, but so
48:55.190 --> 48:59.170
these two major contribution is stability
analysis, and bifurcation analysis. And what
48:59.170 --> 49:04.279
we study in todays lecture in next lecture
is essentially a contribution of Amundsen.
49:04.279 --> 49:11.279
The linear stability analysis and 1955. So,
what we do is that? We have to write the CSTR
49:12.680 --> 49:16.359
equation in an unsteady state form now so
we looked at the steady state we had the unsteady
49:16.359 --> 49:19.829
state equations we wrote it before, but we
did not solve it at that point of time .
49:19.829 --> 49:24.049
Now either are we are going to solve the unsteady
state equation now but we are going to analyze
49:24.049 --> 49:29.900
the stability of the steady state using the
unsteady form. Now I want to give you a generalized
49:29.900 --> 49:35.430
way of looking at it and therefore, what I
do is I use a victorial analysis a victorial
49:35.430 --> 49:41.400
form of it. So, if you remember we had two
components X and the F the cell and the substrate
49:41.400 --> 49:48.400
now we write it in the victorial form d C
d t equals f C and C is a concentration vector
49:48.410 --> 49:54.839
which is composed of S and X fine now similarly,
just because C is a is a vector f also has
49:54.839 --> 50:01.519
to be a vector .
So, f is vector of functions and P over here
50:01.519 --> 50:06.470
is a vector of parameters. So, P also involves
lot of these parameters and let us chat them
50:06.470 --> 50:13.470
out So, it includes the dilution rate it includes
the Y the el ratio the k S michaelis the monod
50:14.579 --> 50:19.980
growth kinetic constant mu max the maximum
growth rate and the initial substrate concentration
50:19.980 --> 50:26.049
fine. And So the solution of the steady state
solution would be if I wanted to write in
50:26.049 --> 50:30.430
mathematical form what would it be here.
50:30.430 --> 50:35.829
If S steady state just f S it is so we will
write it like this f C s s so f remains the
50:35.829 --> 50:42.829
same C s s comma P equals 0. So, let me ask
you this you know so what we are trying to?
50:43.119 --> 50:48.740
Let me try and explain first what we are trying
to do is, not obtain the unsteady state solution
50:48.740 --> 50:52.769
of the problem as I said what we are trying
to do is, give a little perturbation to the
50:52.769 --> 50:57.619
system and try to see how the system functions.
So, what we are trying to look at is, how
50:57.619 --> 51:02.249
it functions in the vicinity of the steady
state, or in the neighborhood of the steady
51:02.249 --> 51:09.249
state. So, as soon as I say that so what rings
a you know brings a bell? So I am trying I
51:10.319 --> 51:14.869
have the steady state and I am trying to look
at what happens in the vicinity or the neighborhood
51:14.869 --> 51:18.599
of the steady state so what should I do mathematically.
51:18.599 --> 51:20.960
How.
51:20.960 --> 51:23.319
How.
51:23.319 --> 51:30.319
No what is the theory? What is the theory
that we are going to use? What is the theory
51:31.960 --> 51:38.960
you use to look at deviations from anything
small deviations from anything? So, F over
51:41.359 --> 51:47.359
here if you look at the screen over here f
at unsteady states slight for small perturbations
51:47.359 --> 51:53.329
at the unsteady state is slightly deviated
from F at the steady state how do we quantify
51:53.329 --> 52:00.329
that deviation what is the theory we use.
What is the theory? We use you you know this
52:06.559 --> 52:12.339
you know you should and you I believe you
know all this tailor series expansion. So,
52:12.339 --> 52:19.339
simple when I say it so for tailor series
expansion we need to figure out what this
52:20.809 --> 52:23.900
deviation is?
So, the deviation is a small deviation about
52:23.900 --> 52:27.390
the steady state is that clear about the steady
state, that has to be very clear first we
52:27.390 --> 52:31.420
obtain the steady state solution and then
we give a small perturbation. And then look
52:31.420 --> 52:35.819
at deviations around the steady state so X
is my deviation around the steady state, which
52:35.819 --> 52:42.819
is given as C t minus C s s C s s being the
steady state and C t being the current one
52:43.049 --> 52:49.190
in the presence of deviation. So, what we
want to study is? We gave a small deviation
52:49.190 --> 52:55.160
to the variables around the steady state we
want to figure out that. how is this system
52:55.160 --> 52:58.890
going to behave is it going to blow up is
it at the deviations is going to grow with
52:58.890 --> 53:02.059
time or the deviations are not going to grow
with time.
53:02.059 --> 53:05.690
And we will do a linear stability analysis
which means that linear stability analysis
53:05.690 --> 53:08.660
would mean that this is a non-linear system,
what we have to do?.
53:08.660 --> 53:08.990
Linearise.
53:08.990 --> 53:15.430
Linearise yes So, because I wrote my X as
like this now C t could be written as C s
53:15.430 --> 53:19.569
s plus X t why am I writing in terms of X
t just. Because I want to do a tailor series
53:19.569 --> 53:26.569
expansion fine. So, my d X d t would be written
as f s C ss plus X comma P fine why do I write
53:28.519 --> 53:31.670
it, because I want to you know so what would
be the tailor series expansion of this about
53:31.670 --> 53:38.670
C s it would be f C s comma P plus the rest
of the term. So, if this I am now showing
53:39.369 --> 53:46.369
for any term so if g which is the function
of C s s plus X is a scalar then, and I am
53:47.749 --> 53:51.130
going to go to the vector form of that so
let us i'm giving you the simpler thing.
53:51.130 --> 53:56.779
Today which is just tailor series expansion
in a scalar form so and then tailor series
53:56.779 --> 54:03.779
expansion of g s s is given as this. So, g
and g C s s plus X is simply g g at C s plus
54:04.119 --> 54:11.109
del g del C s times X del two g del C s square
this is not remember. You know one thing I
54:11.109 --> 54:17.009
want to tell you in sometimes people make
mistake it is not del g del C s s it is del
54:17.009 --> 54:24.009
g del C evaluated at C s s. Similarly, it
is del two g del C square evaluated at C s
54:24.990 --> 54:31.869
s and X’s square over two plus higher order
terms, now if am to do quickly tell me if
54:31.869 --> 54:36.589
I am to do a linear stability analysis how
many terms should I take in the system.
54:36.589 --> 54:42.329
First two of the terms. So, I have to stop
at X if I want to do a linear stability as
54:42.329 --> 54:48.109
soon as I do take X’s square. I do a get
an non-linear tailor series. So, I do not
54:48.109 --> 54:52.599
want to do that the I the reason. I want to
do a tailor series expansion is, because it
54:52.599 --> 54:56.880
is easy to handle you know if I wanted to
solve a non-linear equation, I would have
54:56.880 --> 54:59.859
actually gone and solved the full equation
why would I do a tailor series expansion about
54:59.859 --> 55:04.619
it.
So a linear analysis is easy to handle and
55:04.619 --> 55:09.970
it would give me ,what I want is? I just need
to figure out the earlier initial trend like
55:09.970 --> 55:13.420
you know, whether the election people are
always interested in the initial trends because
55:13.420 --> 55:17.450
initial trend show. Basically most of the
time what is going to happen? So, similarly,
55:17.450 --> 55:21.150
here also I mean really interested at the
initial trends and as soon as I get the initial
55:21.150 --> 55:25.450
trend I will have a ballpark idea of whether
the system is going to decay or grow.
55:25.450 --> 55:30.239
So, I will stop here today and from the next
class next class, what we have to start with
55:30.239 --> 55:35.779
is? How to convert? Tthis remember our system
is victorial, and this is a victorial form
55:35.779 --> 55:41.920
out here. And if you look at the screen so
if you so my question is that, how to convert
55:41.920 --> 55:45.910
this tailor series expansion in the first
thing. We will do is in from the scalar form
55:45.910 --> 55:51.349
to the vector form, and then we will continue
from this so let us stop here and we will
55:51.349 --> 55:52.209
continue from here tomorrow morning.