On stability of bioreactors and we will continue from where we left, so what we did was we
looked at this model of the chemostat for any generalised form of the specific growth
rate mu. If you go back to your notes you will see and then we looked at that and we
looked at the steady state for that and we derived the criteria using Herberts criteria,
we derived the criteria for the stability of the states.
And then in the last 5 minutes or 10 minutes of the lecture what we did was, we put in
our form of the specific growth rates being mu here as you see on the screen now and we
derived the matrix because we had. So, using the Herberts criteria we had to derive the
matrix A and then we have to look at the terms of the matrix and there some couple of criteria
one is that two of the terms are greater and then the product of the other two less than
0. So, that was the criteria. So using that if
you look here, we did this and I am just going to quickly go through this, this is the last
thing I remember having done. So, f 1, f 2 lets go where I can show you what f 1, f 2
are my two functions.
So, f 1, f 2 are my two functions out here which represent the balance equation for the
cell and the substrate respectively, so f 1 represents the balance equation for the
cell and f 2 represents the balance equation for the substrate.
So, what we did is we took the matrix A which was del f 1, del X, del f 1, del S, del f
2 at del X, del f s 2, del S and we obtained the four terms a 1 ,a 1, a 1 1, a 1 2, a 2
1, a 2 2, so a 1 1 is this, a 1 2 is this one, a 2 1 is this and a 2 2 is this one,
now next we go and apply our criteria.
So, this is my matrix and so this is how it looks. Now for it to be stable so, this is
A and A minus lambda gives the characteristic equation for the Eigen values which is this.
So, this is the characteristic equation. Now, when I arrange them in terms of increasing
decreasing lambda power, this is what I get and my B 1 is this and B 2 is this. So, my
criteria was that B 1 has to be greater than 0. I believe greater than 0 and B 2 has to
be as greater than 0 too.
So, these are my criteria. So, if you go back, this is my characteristic equation and the
necessary criteria for the real parts of the roots to be negative is that both, the coefficients
have to be greater than 0. So what I need to satisfy is B 1 and B 2. So, actually you
know strictly speaking, it is just B 1, B 2 minus B 3, but B 3 is 0 out here because
it is a 2 by 2 matrix. So, for a 2 by 2 matrix and what I told you
is that one of the things is that most of the equations that I am going to give you
or you are going to do in the chemostat are 2 by 2 matrices. So, it is kind of simplifies
your life all you need to do is look at the characteristic equation for lambda and then
say that the 2 coefficients of the characteristic equation one for lambda to the power 1 and
the other for lambda to the power 0, they both have to be positive.
So, if both the coefficients are positive then it is straight forward and ensure that
the roots have real negative, but then intuitively also, you can realise that if the coefficients
are positive then obviously the roots are going to be negative. Similarly, if both the
coefficients are negative then the roots are going to be positive.
So, if both the coefficients are positive then the roots are negative that is all you
need to remember out here. So, any time we do this, the easiest way is to create the
characteristic equation for lambda, write the characteristic equation, I mean you can
go do the Harveys way also you know, But I find it lot easier to actually be able to
write the characters, you feel lot more confident about it to be able to write the characteristic
equation for lambda and then equate or in equate rather the two coefficients to be greater
than 0.
So, that is it. So, next what we do is, we now use a Monod Growth Model. Till now we
had been using a generalised mu meneralised value of mu now, we use the Monod Growth Model,
because unless we use the Monod Growth Model or any other growth model we cannot come up
with the specific criteria. So in the Monod Growth Model everything else remains the same,
remember so all the analysis that we did till now is valid for all kinds of growth models
provided you know it is just a two equation system.
So, here mu in Monod Growth Model is given by mu max S over K s plus S and so del mu
del S is given as mu max K s over K s plus S square. So, as you can see straight away
over here del mu del S is greater than 0. These are things that we have to quickly figure
out so then del mu del S is greater than 0 for all values of S because its K s is positive
mu max is positive and it is a positive number in the denominator.
So, it is greater than 0 for all values of x S. So, in the previous analysis we have
shown already when we did the steady state analysis, that here what you see on the screen
S naught is less than D K s over mu max plus D. If that is the case then the reaction does
not happen at all. Why does the reaction does not happen at all? Because that is not enough
substrate for the reaction to happen physically speaking and mathematically. What you found
was that you got a unfeasible or else trivial solution. So, if you remember go back and
remember that S naught greater than D K s over mu max minus D and mu max had to be greater
than D. Why is mu max had to be greater than D? Because one of the criteria for steady
state was mu max equals D right the straightway, steady state criteria was mu max equals D.
Now, if mu equals D, now mu equals mu max time some number which is less than 1. So,
obviously then mu max has to be greater than D. So, then the denominator is positive out
here, but the reaction can only happen if S naught is greater than D K s over mu max
plus D K s. So, we will consider that case only here, because there is no point considering
the other cases. And if you need I will go back and show you the chart we had, you know
we have a nice chart which showed here, this one see,
So, look at this side S naught less than D K s over mu max plus D K s, these are all
trivial solutions and this one you get trivial or non trivial solutions, depending on the
D greater than mu max or D less than mu max. Now, D greater than mu max is not possible,
just as I explained because D equals mu. So, mu equals mu max time some number which is
less than one. So, obviously D has to be less than mu max so the in that region alone you
can get a non trivial solution fine. So, let us work through this all we need to
do is actually find the criteria. So, we understand this, that S naught has to be greater than
D K s over mu max plus D K s.
Now, let us consider the case of sterile feed. Why do we consider that? Just to make our
calculations a little simpler sterile feed means X naught equals 0. Now, so again we
are looking in the region of D less than mu max and S s s, S s s naught less than D K
s over mu max plus K s, so in that region S s s that you get is given by D K s over
mu max plus D K s that is what we got last time. And if you go back and put these values
into your the values that we got into your A matrix, so A matrix is here.
So, if you go and put these values, this becomes 0. Why because, mu equals D, so this becomes
0 and the rest of them are all non 0 numbers. So, that is the only term that becomes 0,
but by becoming 0 it makes a calculation quite simple because as you say this a B 2 will
be simply minus a 1 2 times a 2 1 and B 1 will be simply minus a 2 2, is it clear because
a 1 1 is 0 so B 1 will simply be minus a 2 2 and B 2 will simply be minus a 1 2, a 2
1, so this is my A matrix over here this term being 0.
So, a 1 1 being 0 so all I need to do is prove that I have my a 2 2 to be less than 0, because
B one is minus of a 2 2 now and what else I need a 1 2 times a 2 1 the negative of that
greater than 0 or in other words, a 1 2 times a 2 1 less than 0. So, my two criteria B 1
greater than 0 and B 2 greater than 0 that is, the two coefficients greater than 0 translate
too because a 1 1 is 0. Just as I explained just now, translate to a 2 2 less than 0 and
a 1 2 a 2 1 less than 0 that is all. So, that would be that is criteria for stable
steady state, now a 2 2 is this number over here as you can see, so minus a 2 2 is minus
of this number so this number D is positive number so del mu del S is.
What is it? Positive?
Positive, I just explained why so, this is a positive number right? This whole thing
is positive, D is positive, del mu del S is positive, obviously X is positive, then Y
is positive, so this whole thing put together is a negative number. So, this is automatically
satisfied. Is it clear? a 2 2 less than 0 is automatically satisfied. Now the other
criteria is the only one that we have to take care of.
So, this is a product of these two minus a 1 2 times a 2 1 so, D is what you will get
is D times X s s over del mu, del S over Y and minus of that and there is a minus out
here so this whole thing comes out to be positive so D X s s del mu, del S over Y has to be
greater than 0, so that is my criteria. For greater than 0, I think I will come back to
that, I think I do not have written it here clearly.
So, basically my criteria for D less than mu max is simply, del mu del S times D times
X s s times Y s. Now, you are going to tell me that is it greater than 0, so you are going
to tell me is that correct or not? As I told is it stable or not? And the answer is Yes,
It is stable, but why? We are not going to satisfy anything, it is automatically satisfied
everything is automatically satisfied for this criteria, why because, just from this,
let me here just because of this del mu del S is this number mu max K s over K s plus
S square which is positive for all values of S.
So automatically is that clear or do I need to explain? So, I need to explain this quickly
through here. So, here a 1 2 times a 2 1 negative of that is X s s, del mu del S times D over
X Y fine now D is positive, X is positive, Y is positive and del mu del S is positive,
so obviously this whole thing is positive right?
So, this criteria satisfies, look over here you have a 2 2 this number has to be less
than 0. So, this is positive, so obviously the negative of the whole thing is negative,
so this is also satisfied automatically, clear? So, both criteria automatically satisfied,
as a result this condition is steady state is stable. The next one is the trivial state
as I explained, that you know.
So, D greater than mu max obviously you do not have anything, but the trivial solution,
but we are still looking at the stability of it and the reason we are still looking
at it the stability of it is because whether the trivial stage itself is stable, you know
a trivial state means what? That nothing really happens, you let the thing go in and it just
goes out, but even is that stable or does that degenerate some other unstable steady
state. So, we are trying to look at that and the
A matrix is slightly different because this D is no longer equals mu. So this term is
not 0, on the other hand this term turns out to be 0 when you go and do the algebra yourself.
But I am not going to, you know the details of this one because this is a trivial state,
but still so you get that a 1 1 plus a 2 2, when you add these two up equals this number
which is less than 0. Why is that less than 0? Because D is greater than mu max.
So, if D is greater than mu max obviously 2 D is greater than mu max, 2 D is greater
than mu therefore, this number is less than 0. So, this criteria is satisfied and the
second one is a 1 1 times a 2 2 minus this is 0. So, the product of these two have to
be greater than 0. So, this criteria again is satisfied because D is greater than mu
D is greater than mu max therefore, it is greater than mu so this is positive. So, product
is positive so what we figure out is that, the Trivial steady state is also stable.
So, both criteria satisfied and both lambdas are negative, so trivial steady state is also
stable. Is that clear to all of you? Or do you want me to repeat any parts of it? I am
going a little fast because I want to finish something. So, then next thing that we will
be going to do today is that apply the same calculation, then the same process and the
same mechanism that we did to multiple steady states and the case of substrate inhibition.
I do not know if you have forgotten, let me remind you.
So, this was what the case of substrate inhibition was, so till now we did the Monod Growth Model
with the simple thing. Now you know, as I said that we did the case of multiple substrates
and similar kind of multiple substrates.
What we get is mu, so we can do a similar kind of analysis using multiple substrate.
So, the one that I am doing is the case of substrate inhibition.
So, the reaction for that substrate inhibition was X plus S giving X S and X S reacts with
S again to give X S 2 which does not produce new cell.
So if you remember, these were the all the derivations. Let me not go through them, you
know if this X t was the sum of the 3 X plus X S plus x s 2.
And then we went through this and this is what we found, so why are we interested in
this? Because this is the only thing that is different from the previous case, the value
of mu itself, because everything the balance equations the rest of the analysis everything
is same. So if you remember the reason, we did it in the previous classes, we went up
and did all the analysis without putting in any value of mu at all. The reason we did
that is because the basic equations irrespective of the form of mu as a same, right?
So, whatever form of mu we take, be it substrate inhibition, be it multiple substrates, you
know other kinds of inhibitors does not be a problem at all. We can just put in that
form, everything into that form. So, let us go back here so you see it is a, this is my
form mu which is mu max S, 1 plus K s plus this term you know apart from Monod. You have
this extra term of S square over K i, K s, so what do you think you know? So this is
not important for us. So, what do you think the innovate is because you see that del mu
del S could be 0, but we will come to that later.
So, what do you think? Even if I ask you that so these are basic equations, what would be
the effect of substrate inhibition on the system? What do you think would be the effect?
What do you think would be the effect of substrate inhibition? These equations are written for
sterile feed, I have written it down there. What we got last time, let us review what
we got last time without the substrate inhibition is that there is one trivial state and one
non trivial state and both are stable, that is what we found, if I do a quick review.
So, first question is two questions. What do you think in terms of, what you will get
in terms of trivial states and non trivial states? a and b is that? What do you think
about their stability just an intuition, we will work out the whole thing you know whole
maths in the rest of the class, but just some intuition.
Multiple Yes, multiple states in a way. Last time also
you had multiple states, because you had a trivial state and a non trivial state. So,
two states that was still multiple, but what do you mean by multiple states here? How many
states? I mean that is, it should be a easy answer because what happens to the order of the equation?
Increases? Increases by?.
By 1 order of magnitude in S, so from 2 to 3 so how many.
3 yes, So that is the straight answer for the second order. Last time you had a second
order equation, you got two steady states. This time you have a cubic equations so obviously,
you are going to have three steady state so that is one part of the answer and the other
part obviously, you cannot tell without doing the analysis. But what do you, what can you
tell you now? What are the intuition telling you, what happens
when there are three states?
Yes, so let me show you this, you know, I think I have drawn this diagram several times
before, so let us say this is X conversion just like here something like that and this
is the reactor Damkohler number which depends on the residence time of the chemo stat, exactly
the same thing, so this is how it is going to be, so these are.
What are these called? I have told you this before.
Still, what does the name mathematically?
So, limit points they may or may not be ignition or extension depends on the catalyst and stuffs
like that, you know, if you are doing a chemical reaction, but in mathematically the generate
term for these is limit points. So, why this was important is because between these limit
points in this region of this limit points, you have three solutions and beyond that one
solution and one solution. So, between this only you have the cubic solution.
What it means is that, when we solve it, you will see that outside the certain range in
the of the parameter, there is a single solution because you know the determinant goes off,
determinant goes to 0 or something like that typically the determinant goes to 0 and the
outside range and you end up with a single solution and within that you have just three
solutions, if I look at the stuff again. So, limit points these two and this is the
ignited branch, this is the stable solution,a middle one and this is the stable solution.
So, the bottom branch is the is the stable solution, the top branch is a stable solution
and the middle branch is an unstable solution. Why is it an unstable? Is because, if you
for some reason going to the unstable branch and allow the system to stay there it will
either go to the top branch of the ignited branch or go to the bottom branch or the extinguish
branch depending on whichever is closer to, it does not stay there for very long.
So, your aim is never to operate close to the unstable branch and how do you ensure
that? As an engineer for example, you have to operate your plant or your reactor in a
certain parameter range, parameter range means say, residence time or liquid volume or the
substrate concentration, whatever it is, how will you ensure that it does not go to the
unstable branch? You choose your parameters such that, it does not go to the unstable
branch, right? How do you choose? How do you choose your parameters such that
it does not go with the unstable branch?
Yes, but there is a procedure for that and that is called, let me see if you have heard
of this Neutral Stability and that is exactly what we are doing now. What it does is, basically
it draws you, essentially draw this in the parametric plain. For example, see here, it
could be in the plain of Damkohler number verus Pekkle number, something like that or
Pekkle number here. Let us say, Pekkle number here and Damkohler
number here, then what it gives you? It gives you a curve like this because from here this
diagram that I drew now how do you know? You never know for otherwise, what you have to
do if you choose every Damkohler number, go and make some simulations, find out whether
your solution is in the stable branch or unstable branch.
That is a complete waste of your time, you cannot keep doing it. So what you need to
do is, you need to isolate or figure out in the parametric plain of Pekkle versus Damkohler.
Whether the system is going to be stable or unstable so this is the part where it can
be unstable and the system is stable here, which means that you choose parameter such
that it stays in the middle of this range, it stays within this stable path. It could
be the other way around also, it could be that inside is stable or outside this is unstable
very unlikely or typically you know because of Pekkle increases it becomes unstable. So,
but there is a possibility that it can happen, stable can be inside and unstable can be outside.
So, our aim to do this, you know our engineering aim to do all this analysis is to be able
to come up with this Neutral Stability Diagram. So that we do not have to do the calculations
every time we can figure out in the parametric space. What part is stable and what part is
unstable as a result as soon as you know in the reactor, as soon as you choose your parameters,
you know your flow rate. Damkohler and Pekkle number between them contain all the parameters,
essentially you know flow rate volume of the reaction reactor, reaction time, the size
of the, you know the diffusion length scale in the reactor and the diffusion coefficient
of the species in the reactor. So, all these parameters are contained between
the Pekkle and Damkohler number. So, if you look here. So, essentially that you choose
your Pekkle and Damkohler such that your system stays here or here or here and does not go
anywhere here so, that is your range. So, you can directly choose the parameters you
do not have to do calculations all the time, once you get the Neutral Stability Curve,
this curve once and for all you can directly choose your parameter and make sure that is
in the stable value. So, that is our aim, not to just solve one problem, not to just
solve for the steady state and analysis stability of the steady state, but to get a Neutral
Stability Curve, so that you can use that at all times.
So, that is what we are trying to do over here. So, as I discussed that this being a
Cubic equation this will have three solutions. And out of these three solutions, it turns
out one will be a trivial solution and the two would be non trivial solutions, so then
we have to look at the stability of these solutions.
So, the trivial solution as we have looked before, it is simply going to be X s s equals
0 and S s s equals S naught and the non trivial solutions, if you solve these you know so
how do I solve this, remember what we did was we did a invariance right? We did an invariance.
So, you multiply this by the this second equation by Y and add these two up and as a result
these term will go, then you can express X as a function of S and then you put it back
over here and then solve it. So, you solve it by the method of invariance and this is
what you get, this is to only remind you, Valid for mu max greater than D, this is mu
max not equal to D rather this is valid for mu max not equals to D, why is this valid
for mu max not equals D? Because mu max equals D what will happen is this squadratic form,
Remember the you taken out the Trivial solution, you have a Cubic equation to start with you,
take out one of the solutions which is the trivial solution then you have a quadratic
form. So, this is the quadratic form you are solving. So, if you know this thing is like
the Cubic form, it will be some S times a 1 square plus a 2 S plus a 3 equals 0.
So, this is the form once you put that, so this will give you the sorry this is not S,
S minus S naught so this part will be the trivial solution this is trivial which will
give you S equals S naught and this is the non trivial part. So, this is a quadratic
you can solve, but what I am trying to say is that a 1 equals mu max minus D. So, what
you need to do is when mu max equals D then you need to drop this completely from the
equation itself. Do not solve the quadratic and then try to settle around because it will
come in the denominator or something then the whole thing will blow up.
So, if mu max equals D then just drop this term completely and then you have a linear
and you have just one single solution which is you know, given here. So, this solution
over here is for mu max for a 1 equals 0 then this is given by K 1 K s. So, if I can go
back to the screen now.
So, the Non Trivial solution is given by S S equals K I over 2 K D this term mu max minus
D plus minus this determinant and X s s. So, this is let my S s s and X s s is obtained
as S naught minus S s s times Y, because there is no material balance right? So, if S naught
minus S has been consumed then Y times that would be give you the amount of cells that
has been formed. So, for mu as I said for a 1 equals 0 the
first coefficient equals 0 or for D equals mu max S s s is square root of K s K s over
K i. So, the two solutions that you get over here these two so non trivial solutions are
provided that mu max minus D is greater than this or in other words the discriminant is
greater than 0 this discriminant is positive, so you get this so mu max minus D is 2D is
it correct.
Now, what we have are essentially my three states for the special case where mu max equals
d I have this state, but apart from that essentially for all other cases these are my three states.
Now, what I have to do is I have to look at the stability of three states, predictably
your trivial state is going to be steady stable predictably, but we will still have to, we
will still check that, so what is my, you know, difference that I need to do? Everything
is same or again as before, only my mu is different so, I need to calculate my del mu
del s. Now what we have looked at the three states, what do you suspect in terms of the
stability of the states? Something Yes, I think already said that one of the solutions
is going to be unstable and the other two are going to be stable. Yes, That is what
I also suspect, but can you create a reason for that. What could be the reason for that?
What was the driving force the last time we did for the normal case? What was the driving
force behind making these solutions stable both the solutions it turned out were stable,
but what was the driving force? What is the reason for that?
D was greater than mu max? No, D has to be always greater than mu max for feasible solution
to exist. Is D is now greater than mu it is not going to work.
Yes, correct the reason behind that was del mu del S was always positive, but if you remember
from the last time we did in the growth model del mu del S was not always positive, it goes
to positive and then goes to 0 and then decreases, right? If you remember, so it turned out that
if del mu del S were not positive last time for example, then the system would be unstable.
So, what we can predict or intuit I mean, it is not mandatory that exactly mathematically
we get, but you know we have to always couple some kind of intuition with the mathematics.
So, what we intuit is that for the region where the specific growth rate decreases with
substrate the system is going to be unstable. And physically if you think of it makes sense
that if you increasing the amount of substrate by the specific growth rate decreasing that
is the really worrying case, right? Physically speaking and that probably is the reason where
there are sources of instability. So, let us try and look at it.
So, this is the first thing you will say is mu over here with the inhibition term in there
and this is my del mu del s or d mu d s over here and which could be negative or positive
depending on the numerator, because the denominator is always positive. It is the numerator which
can go negative or positive. So, depending on whether S is greater than square root of
k i k s, all less than square root of k i k s, then you have trouble, if f equals square
root of k i k s then you have then what do you have?
Trivial solutions. Not the trivial solution, Trivial solution
is 0 this is that solution d equals mu max and S equals s square root of k s and k s
k s. Trivial solution is this one, let us not confuse trivial solution with this, this
is trivial solution, these are non trivial solutions are special case of the non trivial
solution, is this one where d equals mu max. Is it clear Lisa? Is there any problem understand?
This is probably the hardest thing we have done in this till now, but if there is a problem
just stop me and so the special case of the non trivial solution is S square root equals
square root of k i k s, but in general we have the trivial solution and two non trivial
solutions. So out of the non trivial solution there is
one possibility of S being equal to k i k s in which case del mu del S would equal 0,
right? That is one possibility, but that is a exceptional possibility a special case,
but in general k i k s would have to be greater than 0 or less than 0. So, depending on whether
it is greater than 0 so square root of k i k s greater than S or square root of K I K
s less than S, so square root of k i k s is greater than S and we probably think that
the system is going to be stable fine.
So, we consider the non trivial states over here and for the non trivial states the mu
equals D for this trivial state is not necessary, but for the non trivial state mu equals D.
Do you remember we had X times mu minus D equals 0 so, for trivial state this is X equals
0 for non trivial state mu equals D, right? Because these are things if you remember,
So, this matrix A that we have, this is the same matrix that we have remember? Exactly
same, as what we did last time, because we still have a generalised form of mu, we have
not put in our mu into it. Now we are going to put in our mu into it so stability criteria
was a 1 1 plus a 2 2 less than 0 or in other words, this term to be negative and the other
one is a 1 1 a 2 2 minus a 1 2 a 2 1 greater than 0 and in other words, the product of
these two terms to be positive. Now, to remind you again the last time we
had a advantage because del mu del S was greater than 0. So, we could conclude directly that
yes if del mu del S is 0 than greater than positive then of course, this whole term is
going to be negative. Similarly, del mu del S is positive then of course, the product
of these two terms is going to be greater than 0, right? Is it clear to everybody?
So, this we could conclude that directly, now we cannot conclude that directly and we
have to depend on the, on a case by case basis. So, this is my criteria now so del mu del
S times X over Y plus D should be greater than 0 and this one will lead to del mu del
S times D X over Y should be greater than 0 fine now for del mu del S greater for del
mu del S to be greater than 0 my S square. As I just showed s square has to be less than
k i k s square has to be less than k i k s so that is my stability criteria, essentially
that s has to be lesser than this is what we have intuited just by looking at the equation
itself we had intuited this. So, my stability criteria is that s has to
be less than square root of k i k s and you can go back and check all rest of it you know,
you can put over here, you can do the rest, I have skipped that calculation, but the differentiation
is already there you can put go and put back over here, but you know, coming back to this
model one of the things you need to, though I do not kind of ask you, I mean I am not
suggesting that you do that in the exam I want you to put the stability criteria a 1
1 plus a 2 2 greater than less than 0 and the other two terms a 1 1 minus a 2 2 minus
a 1 2 a 2 1 greater than 0 and then come through it.
But at the end of the day, if you think, come to think of it what is your stability criteria
is essentially coming down to del mu del S should be greater than 0, right? Is it correct?
You do all the calculations and I do not suggest that you do not do the calculations in the
test or something, you do the calculations to show that you know everything and this
is the proper way, but check on your calculation essentially would be what to find out, what
is my criteria such that del mu del S is positive. So, at the end of the day that is what matters,
but then again do not go and write it straight away, derive it.
So, this what you see on the screen is now my Neutral Stability Curve. The thing that
I had been trying to get and look what I have over here is s naught over d so this is a
major difference between what I showed you, this multiple solution curve that is called
known as Difurcation diagram. The one that I just showed you here, this one if I can
go to the, yes,
So, this is my Bifurcation diagram and my Neutral Stability Curve, would be this now
I need to draw a fresh diagram and over the Bifurcation Curve and the Neutral Stability
Curve together and there is something I want you to help me here.
So, as you can see that the extinguish branch essentially saturates, you know over time
it kind of saturates are over Damkohler number, let us say this is Damkohler and this is the
conversion X. Now I want you to tell me, help me with how the Mutual Stability Curve is
going to look like. So this is my Bifurcation diagram, I want you to help me with how my
neutral stability curve would look like, how it look like it would have a, would look similar
to what I told you like this, what I drew over here, but where would this, how would
I place that? So, let us say my Neutral Stability Curve
in the again X axis is still D a, but y axis is Pekkle. Do you understand? Do you discern
the difference in the terms of the axis for the bifurcation and the Neutral Stability
Curve? The major difference is that Bifurcation Curve is drawn in the Variable Space. Neutral
Stability Curve is drawn in the parameter space. So, this is the variable space and
this is drawn in the parameter space and there is, so, Variable space means one variable
versus a parameter, there is, this is, in the parameter space which means two parameters
and then there is another kind of plot, what is it called? Where two variables you would
haveand you draw one variable versus another for let us say, in this case if I want to
draw X versus s space.
What is this called? What is it? I do not remember.
You do not remember? Who remembers? It is called the phase space, see if the variable
space here, the parameter space here and the phase space is where both the variables are
drawn against each other. And so, this is something that is important and you have to
understand this. Although I explained it before that you draw you necessarily draw the Neutral
Stability Curve in the parameter space and not in the variable space or the phase space.
The reason you do that is because you are not interested in what the value of that,
do you understand? What I am trying to say, you not interested in what the value of the
parameter variable is going to be. You are interested in what parameter should you operate
your reactor in. You are not interested whether X is going to be 0.9 or 0.8 or 0.7 or 0.6,
that is not your concern. You are not even interested whether it is going to be ignited
or extinguished? That is an important question by the way, that you know if you look at this
bifurcation diagram over here this is the ignited branch in this extinguish branch.
It is important actually to figure out whether the system is going to operate in the ignited
branch or the extinguished branch. Why because, you want it to operate in the Ignited branch.
Ignited branch, Yes, because you want product, you know you would not run a reactor without
generating product, so if you are in the extinguish branch you hardly generate any product. So
that is important, but more important than that is, whether the system is stable or unstable.
So, that is why the more important thing is the Neutral Stability go. First come stability,
once the system is stable then you worry about whether you are generating enough product
or not enough product. You understand what I am to trying to say, so first is stability,
So, once the system is stable there are two possibilities that the system could be in
the ignited branch or in the extinguish branch, if you look over here as we discussed that
this length the bottom length is stable the top leg is stable, but the middle leg is unstable.
So, even when the system is stable there is still a possibility of it being the ignited
branch or in the extinguish branch and you want it in the ignited branch, right? So,
I will come to that. Also how to figure that out? But your first criteria is, stability.
So, now you tell me that I have explained so much now, you tell me that how do I draw
my neutral stability curve between the limit point?
Between the limit point, so this is how we draw line over here and line over here and
numerical stability curve will be like this will be like this and then like this. So,
this is the system is unstable over here and stable over here, so this corresponds to this
branch, corresponds to this branch over here corresponds to this branch corresponds to
this and the middle branch corresponds to this, so the two limit points correspond to
the two horizontal. So, two limit points correspond to the two vertical limits. Please remember
these all things, so the two limit points correspond to the two vertical limits of the
neutral stability curve. So, now my problem is solved. So, I now know
that what part is stable and what part is unstable. I can choose my criteria such that
it is stable, now how do I choose whether it goes to the steady state or to the ignited
state, extinguish state? That also I can choose from the Neutral Stability curve. If you look
here, so if I choose something around this point then it will be in the ignited state,
just say after this vertical second vertical length. so it will go to the ignited state.
So, if I choose something out here, in the towards my left hand corner, complete left
hand corner then it will go to the extinguish state.
So, I want my two things, I want my system to be stable, I want it to have attained maximum
you know, to give maximum product. So, which means that first it has to be either on the
ignited and the extinguish branch and the second would it should be in be the ignited
branch. So, this Neutral Stability Curve is the very
useful thing in a way because, let me choose my parameters my flow rate, my reactor size,
my reaction rate, my diffusion coefficient everything such that say, I choose something
like this, this would be probably the optimum choice, because it makes sure that it is stable
or you can go here it make sure it is stable and make sure it is in the ignited branch,
is that clear to all of you? Now, let us go back to the screen. So, this
was a generic thing I showed and this is the actual thing. So, this is the wash out case,
so in other words the Trivial solution. Trivial solution means that no product is being formed.
So, not really worried about this and this is my Neutral Stability Curve. Looks like
and you know if there is a essentially vertical limit. Also it goes up a little bit so many
a times, it is little sharp. Sometimes it is kind of flattened, all these things happen,
but you go a little up. And then a kind of sharp and so, this is my unstable solution
is middle in the inside here above and the rest of the system is stable solution. What
is this line over here? This line is not properly drawn, actually it should touch this due line
over here. So, any how this line over here corresponds
to, it is written here the steady state value of the substrate, so this is not a part of
the, let me not confuse you, any which way this is, this blue line over here, is the
bottom line. This one is not a part of the Neutral stability curve. It is just concentration
plot or substrate plot relating to this value, so do not worry so much about this line, just
worry about this one. So, this is my neutral stability curve. And as we concluded that
the system goes unstable not just with multiple inhibition. In general the system goes unstable
well when del mu del S is negative and then it we end up having positive real roots.
So, for del mu del S negative, this is the part the unstable solution and for del mu
del S positive, everything here below is stable. So, this is your first encounter with neutral
stability curves, right?
So, just want to make sure that everybody understand. If there is any question on Neutral
Stability Curve, you can ask me now or may be a minute later also, this is the overall
stability curve.
And, this is the details of the calculation that I skipped, you know all those calculations
I skipped you can go through these calculations, you want me to go through or just not very
kind of Trivial. So, essentially we have skipped the calculation
because I just showed that del mu del s greater than 0 is stable. So, but this is the hard
way going through all the details and in everything. So, this is the better picture. Now, I have
of the this unstable and this is a so. So, these are rates in gram per litre and d so
what essentially it tells me if you look at this Neutral Stability Curve. If I want to
be in the stable region then it gives me numbers on two of the parameters, one is the S naught
the initial substrate concentration I use and the second one is the D or D is the inverse
of the residence time or in other words residence time itself.
So, what it tells me is that what should be the size of the reactor or you know size cum
flow rate? So, residence time is ratio of these two a and b would be what would be my
initial substrate concentration, this is what it tells me. So, that I can pick so if you
look at the curve here so I can pick any point over here for example, I pick a point over
here so that the system is stable and I know that for my S naught for this system to be
stable . If my S naught is this my dilution rate has to be this or in other case if my
reactor is given to me, you cannot make changes to explore rate and its size then I pick my
s naught initial substrate concentration such that it is in the stable zone.
So, if you need you can go through the calculations on your own so if you need to feel comfortable
with it you can go, as I told you know there is this state, one s is outside the quadratic.
So, these are the values in mu max is hour inverse so these values are in a way important.
If you actually want to plot it so as you can see that the reaction mu max is sort of
you know gives you the reaction rate and that is of the order of hours and so rate is 0.6125
the dilution rate is also of the order of hour inverse and you ensure that these less
than mu max over here. So, these less than mu max that is ensured
number one and number two they are of the same order, they follow the stability criteria.
So, that is in a way concludes our discussion on the stability of these states. Now you
know some assignments you might want to do on your own is that these stability criteria
that I show over here is for the inhibition, but let us go back. So, this stability criteria
that you obtained over here is for all possible values of mu.
So, some of the things that you might want to do is look at how you change your mu and
that how it changes the stability of the system. Now it may be little boring to do, that to
start with because this is with the maths, but once you go through the maths and you
essentially will come out so one of the things you might want to tell me, is that whether
our intuition that for all del mu del S greater than less than 0 is the system unstable, that
is the question I posed to you? For different if I go through this one let us say here,
so system like this for all kinds of del mu del s less than 0 is the system unstable this
system or any other system, any other kind of growth model we have. I mean a lot of different
growth models we have studied if I remember.
So, and the Monod, Malthusian logistics all these different growth models are here so
one question a that I posed to you is that, are for all these growth models if the system
unstable for del mu del S less than 0, a and b is that even if is unstable. What kind of
solutions face you get? What kind of Neutral Stability Curves and in the parametric space
that you get? And this is something that you might want to look at in at home or something
like that. And you know it is important, to we just did these two particular cases the
Monod growth model one of the easy things you can do is simply do the Modified Monod
growth model which is very straight forward same as Monod growth model almost, because
there is not much effective except for that k s naught here, but if you take this model
for example, how does it change your whole system?
And how does it change the Neutral Stability Curve in the Neutral Stability Space? The
reason I am talking about this is see, till the point we write the basic equation for
the system and keep any mu out there and do our analysis it is all valid, but you have
to realise that the growth model when we put in is an artificial thing. Do you understand
what I am trying to say the growth model itself is an artificial thing the balance equations
with any mu in there is the real thing because things are happening that way, but the growth
model is an artificial thing. So, it could be that I can choose growth model
number one and you can choose growth model number two for the system and we should not
come up with wholly different criteria for neutral stability. Do you understand what
I am saying because at the end of the day, irrespective of what I engineer number one
chooses growth model and engineer number two chooses different growth model the system
should still run, so you might want to check with different kind of growth models and try
and see that whether the Neutral Stability Curve that you get is similar or different.
So, we will stop here and tomorrows lecture I think if there are any questions on Neutral
Stability. And if there is what you might want to do is think about it and ask me in
the beginning of the class next class that is one possibility that is I think, because
this is probably one of the hardest things. We did this calculations and everything in
this course, so is there anything you want to ask at this point unintuitive stability
or so we will save it for tomorrow, if there is any question I will stop you know may be
towards the end of next lecture we will spare a few minutes and go through and we will start
a new chapter in the next class. Thanks.