Design of chemostats, but before we do that we will do what I promised in the last class
which is a brief review of the history of reactors, chemical reactors.
So, if you look on the screen here that the history of chemical reactors. You know it
is essentially talked about this chemical reactors are of two in major kind. So the
batch reactor the history of that is when we started is the dawn of human civilization
with process of cooking and so on. The second one is the plug flow reactor which started
in the late 18th century. Then a big conceptual leap so the plug flow reactor was being used
from the late 18th century and so essentially the major difference between a batch and a
plug flow is that you had continuous inflow and the outflow in the plug flow. Whereas,
in the batch you just have to put the things in and wait for how long the reaction took.
So, the reaction could have taken a day or two days and still you have to wait to get
the products. Where as in the plug flow reactor, you have a continuous inflow and outflow.
Now a big conceptual leap came in the form of a CSTR reactor, which sort of combined
the advantages of the batch reactor with that as the plug flow reactor. The advantages of
the batch reactor is it is a easy portable vessel, whereas the plug flow reactor is a
long tube and it is hard to carry from one place to another and stuffs like that. So,
that was a advantage of the batch reactor, the advantage of the plug flow reactor is
it is continuous inflow and outflow. So these are the two advantages. So, the CSTR model
CSTR thing what it did is combined the advantages of the batch the portability and the factor
that it requires less space of the batch reactor with the advantage of the plug flow reactor
which is the continuous outflow and inflow. And this was remembered not you know at some
point of time once discovered things so this was not there before that. And this came in
1908 I said this in the last class. This was discovered by two german scientists bodenstein
and wohlgast in 1908. It turns out so this was the this is a big conceptual leap as far
as the whole idea of chemical reactions, and chemical reactors was concerned. Because as
I said it you know it is portable it is takes less space and as at the same time it. It
has continuous inflow and outflow and that kind of revolutionize the whole chemical reactor
mean the whole chemical industry. And if you now look at chemical industries
all most all chemical reactors are CSTRs or continuous inflow you know continuous stirred
tank reactors, only a few percentage of them you know a small percentage of them would
be batch reactors. Now 1908 it turns out is a very event full year for the history of
chemical engineering or chemical reaction engineering. Because you had the tubular reactor
model, the plug flow reactor model rather. And in 1908 again it was irving langmuir who
founded the flux type boundary condition. The which what is now was known as danckwerts
boundary condition but, it is no longer known because there are difficult in problem issues
there. So, as I said that langmuir was he got the noble prize later, because of his
work with the kinetics langmuir kinetics adsorption kinetics. But this was another of his pad
break breaking you know finings the flux type boundary condition. The reason being you know
for us it might seem very mundane at this point of time to think of a flux type boundary
condition. But think of it when somebody was discovering it 100 years back to come up with
the concept of the fact that you know whatever is coming in through convection at one end
the equals what is going out in, because of convection and diffusion at the other end
is quite a leap. The reason being that the concepts of convection and diffusion were
not as properly underlined at that point of time. It was only underlined around 1937 and
I come to that paper which dealt with that, but so they that is that there is a little
bit of fuzziness around these concepts of convection and diffusion and to come up with
the boundary condition in 1908 which is stayed for more than 100 years.
Now and nobody has proved it incorrect yet is quite a conceptual leap so that was done
by langmuir in 1908. So, and he the so that that is a very important boundary condition
and it came to be later known as the danckwerts boundary condition. Which it should not have
been, but we did some unraveling of the real history of behind this. So, it turns out that
after danckwerts this condition this danckwerts boundary condition not danckwerts let us call
it the flux type boundary condition was rediscovered a few times. So, the first time it was rediscovered
by forster and geib two more german scientists in 1934, and in maybe the you know it is not
really rediscovered. May be because bodenstein and wohlgast wrote their paper in german so
forster and geib would have read those papers most likely 1934.
Then it was again rediscovered gerhard damkohler another of the most famous chemical engineers
of along with Langmuir. I think he is probably the most famous chemical engineer of his times
and our times. So, he rediscovered in 1937 it was not rediscovered actually what he did
was this is the land mark paper the 1937 paper of gerhard damkohler. And it was published
in german. And what he did in this paper was he reviewed the like we are reviewing the
whole history of chemical reaction engineering in 1937 paper he reviewed the entire history
of chemical engineering including langmuirs work and forster, and geib’s works and you
know the whole history of bodenstein and wohlgast discovering the CSTR.
You review the entire work of chemical reaction engineering that was available till that point
of time then danckwerts rediscovered. And he was truly trying rediscover you know trying
to show the world that you know that he discovered it because he did it. So, there is a fundamental
difference between what damkohler did and forster and geib did and danckwerts did because
forstein and geib and damkohler they referred to the earlier papers of Langmuir. And showed
that these are they have discovered these boundary conditions whereas, danckwerts buried
everything, he buried all of the previous work and he pretended that he had discovered
the boundary condition. And that pretension worked really well, because
you know these papers of the forster and geib and damkohler were all kind of buried by the
political events of the time which was the second world war. So, because there was germans
and there was enormous amount of hatred gains germans at that point of time, and you know
the whole world opinion against germans at that point of time. So, their works were lost
nobody was reading and the german science nobody was paying any attention to them so
their works were practically lost. And then danckwerts in 1953. He sort of claimed to
rediscover the boundary condition and now it has been un urged much before it was not
us, but around 20 years back it was been found that danckwerts did not discovered this boundary
condition. And people stopped calling it the danckwerts
boundary condition. They just so and there is a enormous confusion. You know whether
you want to call it the langmuir boundary condition, and forster and geib boundary condition.
So, just people decided we will call it the flux type boundary condition. But if I think
I wrote it in the last class but, i’ll write it one more time.
So, the boundary condition simply goes as so whatever is coming in u times C equals whatever is
so u times C in equals. Whatever is going out minus D times del C del X so you have
the tubular reactor over here, and you are trying to find out this is X equals 0. If
you are trying to find out what is the boundary condition at X equals 0 so this is my boundary
condition so whatever is coming in at this point is because of convection u times C in.
You can call it even u in times in, but u is suppose to be the same on both sides so
u times C incoming in. Because of the convection because equals what goes out because of convection
u times C C minus plus what goes out of because of diffusion and diffusion term is minus del
C del X so this was essentially the flux type boundary condition and we still continue 100
years down the line we still continue to use this boundary condition it is a quite phenomenal
impact breaking boundary condition. So you know we if we are trying to write something
like if you second order you know differential will come to that, but if you are writing
a second order problems and you need a second boundary condition. So, what would be the
boundary condition at the other end? So, X equals 0 this is my boundary condition. What
would be then my boundary condition at X equals L? It would be typically it would be del C
del X equals 0 why if L is large as compared to say the radius of the reactors do if the,
if the length is very large what does it mean what means is that all the reactants has reacted
and there is no reactant left at the end to leave. This means that all reactants have
reacted that there is no reactant left so these are the boundary condition anyhow so
let us go back to the history.
So, then again as I said the 1937 paper of gerhard damkohler that was the one of the
most important papers in the history of chemical reaction engineering and it is a two dimensional
convection diffusion reaction equation so we for the first time wrote the two dimensional
convection diffusion reaction equation and this as paper is written in 1937 in german.
And then it was later translated into English. And what are other things it did and let us
recount, because so in this paper. He introduced the idea of radial and axial diffusion so
just as I said you know the whole ideas of different kinds of diffusion and convection
were not very clearly outlined of way back in those days.
And it was damkohler who was who first introduced the idea of radial diffusion and convection
axial diffusion and how they are different. So, it is not at really understood at that
point of time how the radial and axial diffusions are different, than he introduced a parabolic
velocity profile, till then with a plug flow reactor right plug flow reactor model, p f
r model. What is the p f r model? Intel it means that, the velocity profile is flat within
the within the reactor. And the but the velocity profile is not really flat because when you
have flow inside the tube the velocity profile is parabolic. But if you understand that that
in a plug flow reactor there is no way I mean there is difficulty at least to introduce
a plug flow radial velocity profile.
Let me show you why because what you have is u del C del X equals minus R of C, this
is your plug flow reactor model. Now this is in X so this is my reactor, and this is
the X direction, and this is the r direction. Now if u equals u naught 1 minus r over R
square then it is, how do you solve it ? You know so if I put it back over here 1 minus
r square over R square so this R of C how do you solve it you see what I am trying to
say the difficulty of solving it why, because the differential in the X direction and then
you have a term in the radial direction. So, how do you solve it? There is no way to solve
it so the only way of being able to incorporate a radial velocity profile is to actually is
to actually. .
Actually go for a two dimensional description so which is what damkohler, did you know he
went for a two dimensional description and as a result of which he could incorporate
the parabolic velocity profile. He included radial and axial diffusion and he included
the flux type boundary condition. So it was the first time somebody looked at the chemical
reactor in a very comprehensive way incorporating convection radial, and axial diffusion parabolic
velocity profile, as well as flux type boundary condition which includes both convection and
diffusion effects as the at the entrance.
So this was the 1937 paper and then the whole idea you were aware of the idea of residence
time distribution theory. So, and surface renewal in residence time distribution theory
so they came around like the same time from danckwerts you know paper. But again it turns
out that the forster and geib was actually the first one to discover the RTD theory.
The residence time distribution theory the whole idea that you know that some fluid the
different fluid elements in a reactor will have different residence times. You know so
you can say that the reactor residence time is tau, but that does mean that the entire
fluid in the reactor spend the amount of tau in the reactor.
What does the residence time mean? You know when I say that the residence time what is
the residence time mean the amount of time you spend here so the residence time of the
B tech student here is 4 years or the residence time of a Phd student is 4 years something
like that so which means that the amount of time that you spend from input inlet to outlet.
But when I say that the residence time in a CSTR is tau strictly speaking the not so
this was it was discovered essentially by forster and geib that strictly speaking. It
is not that all every element of fluid in the reactor is actually spending the tau amount
of time.
So, typically the residence time distribution is you now goes something like this so this
is how residence time changes. So, the average of this you know if you take the whole of
it and divide it then the average of it is tau, but some elements as you see her[e]-
over here some elements of fluid the so this is the amount of elements of the fluid, some
elements of the fluids spend very small time again. You know some very small so the bell
shaped curve and only a fraction says spend a very large amount of time in the reactor.
So the when I say that reactor is the something you have to remember, when I say there is
a reactor residence time is tau. It does not mean that everything in the reactors spends
tau amount of time it means that the average amount of time spent within in the reactor
by a fluid element is tau.
So, this whole concept was actually started by forster and geib in 1934, and then it was
again you know danckwerts is a great rediscoverer so he was he rediscovered and this theory
in 1953. But he had some inputs also which is he generalized, that it was in a forster
and geib’s paper in 1934 which I got translated through somebody, and read it and it was a
very localized concept. But danckwertz generalized it for the whole reactor thing and it really
you know the his presentation was good. Then last important things that came in the whole
CSTR business was a concept of micromixing, which is mixing at the molecular scale. Because
reaction takes place at the molecular scale, and if reaction has to take place in the molecular
scale. Then there has to be mixing at the molecular scale right because if a and b two
reactants are there and unless they are mixed with the molecular scale they cannot react.
Is that clear? Because the reaction has to take in the molecular scale. So, the concepts
of micromixing maximum mixedness and complete segregation. Complete segregation when there
is no mixing at all between the two components. So, these were propounded by as zweitering
in 1959. So, I think that concludes our brief review of the history of chemical reactor
models what we will do now is that we will so much of this theory as you see is related
to CSTR because as I said CSTR was a big conceptual leap.
So what I will do now we will start try and look at the CSTR for a bioreactions. So, the
CSTR for bioreactors is called chemostats, and this is let me show you the picture and
try and explain. So, this is the central thing that you have over here is a chemostat. So,
it is a CSTR which means it is a continuous stirred tank reactor, which has two parts
with one is continuous another is stirred and for four parts. Actually continuous stirred
tank reactors. So, the continuous means there is a continuous inflow of feed and a continuous
outflow of product, stirred means it is being stirred by a a stirred over here, tanks so
it is a tank and then it is a reactor so the four elements come together.
So as you see that the way it works for the chemostat is that, it is the feed reservoir
over here from which feed is being continuously pumped into the reactor. So, then you have
a buffer solution which is also being pumped, because these are being bioreactions you have
to maintain the p H at the certain values could be 7 could be 5 could be 6 whatever.
but you have to maintain the p H at a certain value so you have a p H controller and a buffer
solution being sent in. You have you know air being taken out and replenished so those
kind of things we have stopped about this you know you need a certain amount of oxygen
into the system and so on. So, air being taken out and you know replenished
in stuffs like that here and then the product. So, the air taken out here and replenished
from here just a while air comes in here just for the oxygen you know that you need to supply
and you know the nutrient that is the carbon, hydrogen, nitrogen in liquid form are a part
of the feed reservoir in the feed reservoir itself. So, the product is taken out now the
product will include these 3 X S and F what is X? X is the cell are growing. S is the
substrate and feed is you know everything else apart from the substrate and the cell.
So, that part of the feed which is not either substrate nor cells. V is at any point of
time the liquid volume in the reactor, why is V important not the volume of the actual
reactors? Beause the residence time of the reactor the average residence time as we call
it now depends on the volume of the reactor and not the V. In volume of the liquid sorry
in the reactor and not the actual volume of the tank. Because the rest of the tank that
has air in it is not of any use. So, V here and then the product is taken out and pumped
out and take put in a product vessel. So, this is a whole flow chart of this system,
and is there any question or do you want me to stop up for anything here or shall we proceed?
We should proceed. So, the next question that comes in is that
we have to in order to understand the dynamics of this process we have to write balances
for this. So, the balance would be straight forward similar to the CSTR balance that you
have written except that it is for two components here. So, what are the two components you
are going to write the balances for? One there will be let us go slowly. So, how many balances
do you think that you will have to write? The first balance would be the overall balance
of volume which is the volume accumulated or volume held back in the reactor, and volume
coming in and volume going out so that is a overall balance then what else we have.
Component Balances. So, one would be the substrate balance. So,
the two major components as you can see in the picture over here, that X S and F so one
would be the substrate balance S then would be the cell balance X so essentially three
of these.
So, let us go and do it so the chemostat as you can see over here is an ideal CSTR. Because
it is stirred and you know of everything and continuous, and so on. So, the first you need
to do a total volume balance. So, the total volume balance is very straight forward d
V d t equals F in minus F out feed coming in minus feed going out. So, the next balance
we do is the mass balance for the cells and mass balance of the cells is simply if you
take this. For example, it you have to multiply it with the cell concentration so d d t of
X V this and these two plus. They are the generation of cells so that is
the only difference between the other component, and the cell is that there is a generation.
I mean overall volume is not been generated where cells are being generated so d t of
X V equals F n times X naught is the amount of cell that is coming in the feed. So, the
feed itself can have some cells minus F out times X is the amount of cells that is in
the outlet products, and mu times X is a specific. You know mu is a specific growth rate per
unit cell mu times X is the total for the total amount of cells times, the volume you
know so because the volume of the liquid should come in over there.
Sir the communications death of cells. Equation is.
Death of cells. Death of cells death of cells has not been included, but you can add another
term minus beta X that we did last time. So, you can just add minus so you instead of mu
you can, you can replace mu by mu hat where mu hat is mu minus beta it is fine. So, it
is straight forward so the last thing that we need to do is the mass balance for substrates
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
over Y. Because I explained this several times in the last couple of lectures Y is just the
el ratio of X over S. Now typically F you know if you want to typically in CSTR what
you would like is that the inlet is equal equals outlet and if that is the case F in
equals F out. Then d V d t is 0 so as a result of which
the V is a constant fine and you can do a lot of simplifications in these equations.
Because V is a constant you can take V out of this equation you can take V out of this
equation fine, and you can get rid of b in some other places because the one of the over
V will come in here. So, when you do the simplifications you get d X d t equals F over V X naught minus
X plus mu X it looks like a lot simpler equation and just as he said that if you want to include
the death of cells replace mu by mu hat where mu hat is mu minus beta being the specific
death rate of cells. So, and the last thing would be the substrate balance equation for
the substrate d S d t equals F over V S naught minus S 1 over Y mu X fine.
So what would be our step now so once we have done this. So, what would you like we will
like to find out how it would changed? But what would be your step? What would you do
you think we should do now. So, if F is a constant and known then how many variables
do we have just two variables, and two equations so what we would intend to do is essentially
solve this, but how do we solve this? I mean you know what kind of solution do you want
first let me ask you that.
What is that.
Possible for what for every thing.
Well I do not think analytical solution is possible one of the things, you can try to
do is get an invariance. I taught you have to get invariance in the last class, you can
get an invariance, and then you can get an analytical solution for combination of X and
S naught for X or X separately. But what we want to do first is the easy step out which
is up in the steady state solution of it.
So F over V is called the so we do that. So, let us define some quantities first F over
V is called the dilution rate of the cells F over V is a dilution rate of the cells,
and what is the unit of Sover V F over V?
Hm what.
Inverse of time which means it is inverse of the residence time of the system so F over
V is 1 over tau, which is the inverse of the residence time of the system. So, when you
do that? You can put here d t and d t of X equals dilution rate d e you know instead
of 1 over tau. In this chapter or in this course we will use the dilution rate d instead
of the residence time. So, D is 1 over tau X naught minus X plus mu X and d s d t is
a d e S naught minus S minus mu X over Y fine. Now mu is we are going to assume a Monod growth
kinetics as I said it is a most popular and most useful growth kinetics there are plenty
others as. I showed you in someone of the earlier lectures but, this is the one that
is that we are going to try and use for now. Now as I said that you know sometimes I might
ask you in the exams for example, to do the similar kind of thing with other growth kinetics.
I can give you more complicated growth kinetics or I can give you growth kinetics where inhibition
is involved or all other kinds of things possible. But then but you have to be aware of how this
is the process so what we’re going to do from now on todays class and next class is
slightly hard and you should pay attention to the process, what is really happening and
how to do it? Because then you should be able to do it for other kinds of systems so that
is fine.
So we put the monod growth kinetics in and so then we go for the steady state solution.
So, at steady state this is your basic equation. So, d X naught so if you look at this here
so we put a steady state out here so D X naught equals D minus mu times X for the X equation
and then similarly, for the substrate also you will have another one. Now the and the
first approximation that we can, we make assumption that we make is that it is a sterile feed.
It is we are solving it for a sterile feed sterile feed means that there is no cell in
the feed that is coming in. It is an assumption of coarse because typically there are cells,
but we are just trying to solve it for this system.
We will solve it for other systems also without sterile feed but, let us try and solve it
for with sterile feed is it clear the sterile feed means, that X naught there are no cells
in the in the incoming feed so when you do that you simply get D equals mu the dilution
rate equals mu. Now mu is this specific growth rate remember so it goes as mu max times S
over K S plus S which equals the dilution rate. So, if you have a fix dilution rate.
So, what this means is? For a sterile feed for the system to have any sort of meaning
full solution, if you have a dilution rate that is fixed then yours steady state substrate
concentration has to be this. You see what I am saying why is that? Because see if X
naught is 0 then d minus mu times X is 0 what is the solution of this equation.
The two solutions of this equation one is that X naught it is X itself is 0 then the
other one is that D equals mu so if D is not equals mu, then the other solution is X naught
X equals 0 which means that no cells are being produced which is worth less because I am
running the whole reactor to produce cells. So, the only feasible solution in this place
is d equals mu. Now when you put mu max X over F plus s equals D then the steady state
substrate concentration is D K s over mu max minus D fine. Provided that X s s is not 0
I mean of coarse and mu max is greater than D mu max has to be greater than D because
you see why, because mu if mu is greater than equals D then this fraction over here is always
less than one so mu max obviously has to be greater than D.
So, from the second equation that is the equation for the substrate this is this is what we
get. Now the substrate concentration and you know D equals mu for X naught equals 0 so
you can substitute back over here D equals mu in this equation. So, you know over here
and then you can get your X that is the steady state concentration X s s in terms of these.
As this because we wrote our S s s before if you remember in the last slide I will just
wait for a few seconds because some of you are writing so and then I will show you if
you need to so if we wrote the S s s in the last slide.
So you can substitute your S s s from the last equation that we wrote before in to this
and is it clear to all of you or do you want me to go through this steps. So, all you need
to do is let me just quickly see show. So, all you need to do S s s is given here as
D K s over mu max minus D all you need to do is substitute for S s s in from that equation
into this equation. And you put D equals mu over here, and then you will get X s s that
steady state concentration of cells in the system goes as if we forget the Y let us you
know ignore the y then it goes as s naught minus D K s over mu max minus D and.
You know this we already did D max equals mu max D equals mu max S naught over K s plus
S naught D equals mu max S over K s plus X so D max would be the maximum value that the
D can take is mu max S naught over K s plus s naught. Why because the maximum values that
the S can take is s naught for the feed concentration clear to everybody, and if there is any point
you need to stop me please stop me and i’ll explain one more time .
So if that is clear then as in this equation as mu max goes to D you see over here. Then
what will happen this is goes to 0 the denominator here goes to 0 and this goes to infinity as
a result solutions are possible for mu D greater than mu max and D less than mu max is it clear
see. What was what did we get from our initial assumption of sterile feed? We got that D
equals mu fine and one of the things we said is that D is typically if D equals mu then
D would be less than mu max, because mu max is always greater than mu. So we are actually
working in this space of D less than mu max, but just to make sure that you know to make
sure that this a there’s a feasible range. So, D equals mu max there is no solution because
this whole thing blows up the solution is only possible in the D less than mu max, D
greater than mu max. But obviously we are working in the space of D less than mu max
clear no confusion. So, this is what I said that we are working in the space of D, D less
than mu max because it is not possible for D if D equals mu max mu. Then it is not possible
for D to be greater than mu max because mu max is obviously greater than mu .
Now for X s s equals 0 D greater than mu max which is not in the you know which is not
in the it is in the feasible range, but it is not of any interest to us if no cells are
being produced. it is of no interest to us so this is a just another possibility so for
X s naught equals 0 D has to be less than mu max clear. From not necessarily from the
mathematics but, the physics of the problem. V has to be less than mu max because V is
D equals mu and if D greater than mu max and X s s is 0. So, my steady state solution is
that for D less than mu max I have my steady state solution as this and for D greater than
mu max it is 0. So, what does this mean? what it means is that? Why am I why am? I you know
trying to answer this question? Why am I trying to a kind of boil everything down to d and
mu max? Both of these two are parameters of the system d and mu max but, what is the difference
between these parameters.
Absolutely yeah, see the mu max is a parameter that cannot be changed it cannot be manipulated
whereas, D is a parameter which can be manipulated D is one over the residence time of the reactor.
How can I change the residence time of the reactor? Just flow rate yeah just basically
by changing the flow rate or the volume that of the liquid in the system. So, what this
is trying to say is that here so what this is trying to say over here is that these two
cases, that if you want if you want cells to be produced your d has to be less than
your mu max or in other words your dilution rate has to be less than the maximum specific
growth rate of the cell. If your max dilution rate is greater than
the maximum specific growth rate of the cell then there are no cells which were have going
to be produced. What does this physically mean for you what it physically means is?
That if dilution is very large then it will sweep this as the way you know it would. It
would take the things away with it so if D is very large for example, then what it means
what it means is the residence time in the reactor is very low fine because D goes as
one over tau. If the residence time of the reactor is very low what does it mean it means
that the cells are not getting enough time to react for and grow. So, this is the physical
interpretation so in bioreactor system in biochemical engineering, they say that dilution
of the system dilution rate being very large means that the system is being excessively
diluted and as a result. Cells cannot be grown but, if you look at
it from a chemical engineering point of view ,then we convert it into the residence time.
So, D very large means residence times very time very small which means that the cells
are not getting enough time in the reactor to grow.
So, that is why it happens? From mathematically let us look a little bit of the mathematical
space of it. So, if this is the case that is D is less than mu max, then your steady
state X s s I just gave you here. This is the steady state solution for the cell and
the first thing here is the steady state solution for the substrate. Do you need to write the
steady state solution for cell? I think you wrote it already, but if you want to it will
give you a few seconds so the steady state solution for the substrate now is given by
D K S over mu max minus D. So, this has to be satisfied S naught greater than D K S over
mu max minus D so because mu max is greater than D so this denominator is positive so
the second criteria that we get. So there are two criteria that we have to
satisfy one is that the residence time of the system or dilution rate of the system.
One is the residence time is one over dilution rate dilution rate of the system has to be
greater than mu max. And the second criteria we get is given here S naught, which is the
initial concentration of substrate in the feed has to be greater than this quantity
D K S over mu max minus D. So, both of these two criteria have to be satisfied in order
to be able to produce cells so that your X s s is greater than 0 both of these two criteria
to be satisfied fine is it clear. So, both these straight two criteria have
to be satisfied, in order for it be, if they are not satisfied then X s s and S s s are
in the not in the feasible domain. So, what happened was that? If you remember the that
how do we get the first criteria we got the first criteria by showing that if mu max is
less than d, then your X s s is 0 or negative. We got the second criteria by showing that
if your S naught is not greater than this, then your S s s would be negative or 0. Because
see what happens is that both the substrate and the cell that are coming out have to be
positive numbers both of them. Both the substrate so for for the positivity of the cell you
need to ensure that D is greater than D is less than mu max for the positivity of the
substrate you need to ensure this quantity. That is the initial amount of substrate that
you to put into the system is greater than D K S over mu max minus D S and so and the
second regime is D no. I do not know this is a case of X s s equals 0 and S equals at
S naught so what this means is the nothing is happening in the reactor for the same case.
But nothing is happening in the reactor. So which means that you start with cells this
is what kind of solution is this trivial solution. So, this is known as a trivial solution whereas,
X s s is 0 that is you put in cells is or input solution is free of cells when you are
putting it in now no cells. In no cells out and as a result whatever substrate you put
in is whatever substrate that leaves the system so this is a trivial solution.
Why am I going through? This is because you know one of the things you kind of forget
like let us go back, and show you this equation here. For example so a system like this you
know guys, you know very calmly you go and write the one single solution that is there
possible. But when you are doing a mathematical analysis you have to take into account all
kinds of solutions. So, one of the solution is a trivial solution and the trivial solution
as you will see a little later may be in next lecture is important also. So, why we are
doing this is trying to look at the entire space composed of all different kinds of solutions.
So, the first one we discussed is the non trivial solution the second one is the trivial
solution.
And this is summary of the behavior of the chemo stats. So, these are my equations which
I am sure you had already written down. So, there is no point spending any more time on
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
S and what I will next do is summarize the results also the ones we looked at.
So this is a feasible range this is the feasible range of solution here the rest as we discussed
that there are two constraint for this things to happen that feasible range means both positive
results for X s s and S s s that is the substrate and the and the cells going out of the reactor
are both positive numbers. So, the two constraints for this one is that the dilution rate has
to be less than mu max. The second one is the initial concentration of the cell is greater
than this number. Now if any of these two constraints are violated then what you end
up having is a trivial solution as you have over here all these are trivial solutions.
Because X s s is 0 and the steady state concentration of the substrate is same as, what you put
in so what you need to remember? you know the best way to remember this is that these
two constraints have to be satisfied in they are then in this space where both these constraints
are satisfied and intersect You get the solution anything else you get trivial solution. So
this is essentially the steady state analysis of the chemostat. Now what we are going to
do is? Look at so when you in a chemical reactor or any kind of you know continuous reactor,
you want to run these reactors at steady states fine. You would like to why do you want to
run these reactors as steady states because. Control is easy. Control is easier because
whenever you have dynamics of the system coming in control gets harder, but what happens in
a real system? you know we are running it at the pilot scale or even forget plant scale
even at the pilot when you are running and even in the lab scale, what happens? You are
trying to run it at the steady state but, you are probably not able to run it at the
steady state, why does that happen? You can have your control as an everything and you
will see that, when you do even experiments in the control lab you will see that you you’re
trying to manipulate your control variables. So, that the system is at steady state the
level of the fluid in a in a you tube or anything for that matter and you will see that it keeps
oscillating it is very hard to retain it at the study state. Why is that?
What. Disturbance will be there.
What kind of disturbances that is , what kind of disturbances?
fluctuations. The key word here is as he said fluctuations
or what we call in mathematical balance is perturbation there are these natural noises
and natural perturbations that are there in the system. And why do these natural noises
and perturbations coming every angle? You know you think that this table probably has
no vibration but, it is full of vibrations. So , where ever you putting in your reactor
as lots of vibrations in there, and these vibrations lead to perturbation or fluctuations
in the system. And as it turns out that you know these system many of these reaction systems
are pretty sensitive. The reason these are sensitive is that, I talked about I do not
remember. I think I did talk about the case for the autocatalytic reaction in this class
itself at some point of time. So, that one is example of why these systems are very sensitive?
The reason these systems are very sensitive is because they are non-linear so when they
are non-linear. So a small perturbation in one of the variables
can lead to large perturbations in the entire system, because of non-linearity this is a
multiplying effect. So, small perturbation in one you know leads to another perturbation
in other. And these two effects kind of multiply, and then the wholes the whole kind of whole
amount of perturbation that is generated in the system is multiplied and kind of exaggerated.
So, this you know there is a whole analysis for that and that analysis is known as stability
analysis so we will that is that is an extent we are going to look at.
So what we are going to do is stability analysis of bioreactors and you know we will start
today, because it is a complicated thing and we would not finish today and we will continue
into the next lecture. So, stability analysis is that analysis of the system when you apply
perturbation to a study state. So, if a system you have done this partially. I believe in
the control class process dynamics and control and we are going to use some of those things.
That you had learnt over there so what is the idea of a stability analysis the ideas
of a stability analysis. You first obtain the steady state of the system then you perturb
the steady state slightly. And you see how does a system react?
How this how this system react? There are two possibilities two major possibilities
three actually but, let us say two major possibilities, one is that the system will go back. Go back
to the steady state stabilize the other is, system will move away from the steady state
which is destabilize. And the third possibility is the minor possibility is that the system
remains exactly as you put it you know you perturb it. And the perturbation continuous
that little perturbation you gave continuous like that it does not grow, does not decay.
But that s that is a minor possibility because that does not happen. The two major possibilities.
One is the perturbation that you give decays as a result of which remember the perturbation
that you give decays as a result of which the system returns to the steady state.
And the second one is the perturbation that you gave grows as a result of which the system
destabilizes so how do we mathematically analyze. And understand this so this is what we will
go we are going to do over the next lecture. So, the method that we use as a liapunov’s
here the liapunov’s method of linear stability analysis for a chemical reactor, and you probably
used it in your control class. It was found the and amnndson just type over here m u n
d Amundsen this should not be n not n n m u n d amundson 1955 amundson it turns out
that he is known as the father of modern chemical engineering as we know it .
You know the statement and so the chemical engineering department which is the best in
the world he was the chair of that for 20 20 5 5 years. And then the chemical engineering
department at university of minnesota is known as a Amundsen hall. And the major most important
award of chemical reaction engineering is also named after Amundsen. So, he is the father
of modern chemical engineering, why I am calling it modern chemical engineering is that? It
is with amundson he is the one who introduced along with birds to attain light food of coarse,
but he is the one who introduced mathematics or applied mathematics to chemical engineering.
And as a result of which chemical engineering has a form that it has today so birds to attain
light food are responsible for introducing mathematics.
Applied mathematics to transport phenomena. So, the way you we talk about heat transfer
mass transfer and momentum transfer is, because of with the changes that were in you know
a brought in by birds to attain light food in 1950s. And the very famous book that they
have. So, we do not look at heat transfer just is and the terms of a heat transfer equipment
or mass transfer as a distillation column. But we understand the basics of the system,
and we understand that everything that follows be it distillation be it adsorption or absorption
heat exchange in a heat exchanger. Everything is a product of the very basics
of heat mass and momentum transfer and if as you have done in your transport phenomena
lecture. There is a tremendous analogy between heat mass and momentum transfer so if you
describe one set of systems the momentum transfer equations. You can use the same systems a
boundary layer and so on to describe heat transfer and mass transfer so this was the
it is a concept that was introduced by birds to attain light food to chemical engineering
and we owe it you know we owe it to them. Similarly, we owe it to amundson neal amundson
for introducing the whole of this similar kind of approach a very mathematical way of
looking and a very basic way of looking to chemical engineering.
Before that chemical engineering is also just about using chemical reactors and getting
products out. But then he introduces whole approach and one of the major things did he
did was looking at stability. He also did introduce bifurcation and analysis, but so
these two major contribution is stability analysis, and bifurcation analysis. And what
we study in todays lecture in next lecture is essentially a contribution of Amundsen.
The linear stability analysis and 1955. So, what we do is that? We have to write the CSTR
equation in an unsteady state form now so we looked at the steady state we had the unsteady
state equations we wrote it before, but we did not solve it at that point of time .
Now either are we are going to solve the unsteady state equation now but we are going to analyze
the stability of the steady state using the unsteady form. Now I want to give you a generalized
way of looking at it and therefore, what I do is I use a victorial analysis a victorial
form of it. So, if you remember we had two components X and the F the cell and the substrate
now we write it in the victorial form d C d t equals f C and C is a concentration vector
which is composed of S and X fine now similarly, just because C is a is a vector f also has
to be a vector . So, f is vector of functions and P over here
is a vector of parameters. So, P also involves lot of these parameters and let us chat them
out So, it includes the dilution rate it includes the Y the el ratio the k S michaelis the monod
growth kinetic constant mu max the maximum growth rate and the initial substrate concentration
fine. And So the solution of the steady state solution would be if I wanted to write in
mathematical form what would it be here.
If S steady state just f S it is so we will write it like this f C s s so f remains the
same C s s comma P equals 0. So, let me ask you this you know so what we are trying to?
Let me try and explain first what we are trying to do is, not obtain the unsteady state solution
of the problem as I said what we are trying to do is, give a little perturbation to the
system and try to see how the system functions. So, what we are trying to look at is, how
it functions in the vicinity of the steady state, or in the neighborhood of the steady
state. So, as soon as I say that so what rings a you know brings a bell? So I am trying I
have the steady state and I am trying to look at what happens in the vicinity or the neighborhood
of the steady state so what should I do mathematically.
How.
No what is the theory? What is the theory that we are going to use? What is the theory
you use to look at deviations from anything small deviations from anything? So, F over
here if you look at the screen over here f at unsteady states slight for small perturbations
at the unsteady state is slightly deviated from F at the steady state how do we quantify
that deviation what is the theory we use. What is the theory? We use you you know this
you know you should and you I believe you know all this tailor series expansion. So,
simple when I say it so for tailor series expansion we need to figure out what this
deviation is? So, the deviation is a small deviation about
the steady state is that clear about the steady state, that has to be very clear first we
obtain the steady state solution and then we give a small perturbation. And then look
at deviations around the steady state so X is my deviation around the steady state, which
is given as C t minus C s s C s s being the steady state and C t being the current one
in the presence of deviation. So, what we want to study is? We gave a small deviation
to the variables around the steady state we want to figure out that. how is this system
going to behave is it going to blow up is it at the deviations is going to grow with
time or the deviations are not going to grow with time.
And we will do a linear stability analysis which means that linear stability analysis
would mean that this is a non-linear system, what we have to do?.
Linearise.
Linearise yes So, because I wrote my X as like this now C t could be written as C s
s plus X t why am I writing in terms of X t just. Because I want to do a tailor series
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
it, because I want to you know so what would be the tailor series expansion of this about
C s it would be f C s comma P plus the rest of the term. So, if this I am now showing
for any term so if g which is the function of C s s plus X is a scalar then, and I am
going to go to the vector form of that so let us i'm giving you the simpler thing.
Today which is just tailor series expansion in a scalar form so and then tailor series
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
del g del C s times X del two g del C s square this is not remember. You know one thing I
want to tell you in sometimes people make mistake it is not del g del C s s it is del
g del C evaluated at C s s. Similarly, it is del two g del C square evaluated at C s
s and X’s square over two plus higher order terms, now if am to do quickly tell me if
I am to do a linear stability analysis how many terms should I take in the system.
First two of the terms. So, I have to stop at X if I want to do a linear stability as
soon as I do take X’s square. I do a get an non-linear tailor series. So, I do not
want to do that the I the reason. I want to do a tailor series expansion is, because it
is easy to handle you know if I wanted to solve a non-linear equation, I would have
actually gone and solved the full equation why would I do a tailor series expansion about
it. So a linear analysis is easy to handle and
it would give me ,what I want is? I just need to figure out the earlier initial trend like
you know, whether the election people are always interested in the initial trends because
initial trend show. Basically most of the time what is going to happen? So, similarly,
here also I mean really interested at the initial trends and as soon as I get the initial
trend I will have a ballpark idea of whether the system is going to decay or grow.
So, I will stop here today and from the next class next class, what we have to start with
is? How to convert? Tthis remember our system is victorial, and this is a victorial form
out here. And if you look at the screen so if you so my question is that, how to convert
this tailor series expansion in the first thing. We will do is in from the scalar form
to the vector form, and then we will continue from this so let us stop here and we will
continue from here tomorrow morning.