WEBVTT
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On stability of bioreactors and we will continue
from where we left, so what we did was we
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looked at this model of the chemostat for
any generalised form of the specific growth
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rate mu. If you go back to your notes you
will see and then we looked at that and we
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looked at the steady state for that and we
derived the criteria using Herberts criteria,
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we derived the criteria for the stability
of the states.
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And then in the last 5 minutes or 10 minutes
of the lecture what we did was, we put in
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our form of the specific growth rates being
mu here as you see on the screen now and we
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derived the matrix because we had. So, using
the Herberts criteria we had to derive the
01:03.820 --> 01:10.320
matrix A and then we have to look at the terms
of the matrix and there some couple of criteria
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one is that two of the terms are greater and
then the product of the other two less than
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0.
So, that was the criteria. So using that if
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you look here, we did this and I am just going
to quickly go through this, this is the last
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thing I remember having done. So, f 1, f 2
lets go where I can show you what f 1, f 2
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are my two functions.
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So, f 1, f 2 are my two functions out here
which represent the balance equation for the
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cell and the substrate respectively, so f
1 represents the balance equation for the
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cell and f 2 represents the balance equation
for the substrate.
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So, what we did is we took the matrix A which
was del f 1, del X, del f 1, del S, del f
01:57.579 --> 02:04.579
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
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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,
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now next we go and apply our criteria.
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So, this is my matrix and so this is how it
looks. Now for it to be stable so, this is
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A and A minus lambda gives the characteristic
equation for the Eigen values which is this.
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So, this is the characteristic equation. Now,
when I arrange them in terms of increasing
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decreasing lambda power, this is what I get
and my B 1 is this and B 2 is this. So, my
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criteria was that B 1 has to be greater than
0. I believe greater than 0 and B 2 has to
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be as greater than 0 too.
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So, these are my criteria. So, if you go back,
this is my characteristic equation and the
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necessary criteria for the real parts of the
roots to be negative is that both, the coefficients
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have to be greater than 0. So what I need
to satisfy is B 1 and B 2. So, actually you
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know strictly speaking, it is just B 1, B
2 minus B 3, but B 3 is 0 out here because
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it is a 2 by 2 matrix.
So, for a 2 by 2 matrix and what I told you
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is that one of the things is that most of
the equations that I am going to give you
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or you are going to do in the chemostat are
2 by 2 matrices. So, it is kind of simplifies
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your life all you need to do is look at the
characteristic equation for lambda and then
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say that the 2 coefficients of the characteristic
equation one for lambda to the power 1 and
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the other for lambda to the power 0, they
both have to be positive.
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So, if both the coefficients are positive
then it is straight forward and ensure that
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the roots have real negative, but then intuitively
also, you can realise that if the coefficients
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are positive then obviously the roots are
going to be negative. Similarly, if both the
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coefficients are negative then the roots are
going to be positive.
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So, if both the coefficients are positive
then the roots are negative that is all you
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need to remember out here. So, any time we
do this, the easiest way is to create the
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characteristic equation for lambda, write
the characteristic equation, I mean you can
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go do the Harveys way also you know, But I
find it lot easier to actually be able to
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write the characters, you feel lot more confident
about it to be able to write the characteristic
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equation for lambda and then equate or in
equate rather the two coefficients to be greater
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than 0.
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So, that is it. So, next what we do is, we
now use a Monod Growth Model. Till now we
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had been using a generalised mu meneralised
value of mu now, we use the Monod Growth Model,
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because unless we use the Monod Growth Model
or any other growth model we cannot come up
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with the specific criteria. So in the Monod
Growth Model everything else remains the same,
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remember so all the analysis that we did till
now is valid for all kinds of growth models
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provided you know it is just a two equation
system.
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So, here mu in Monod Growth Model is given
by mu max S over K s plus S and so del mu
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del S is given as mu max K s over K s plus
S square. So, as you can see straight away
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over here del mu del S is greater than 0.
These are things that we have to quickly figure
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out so then del mu del S is greater than 0
for all values of S because its K s is positive
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mu max is positive and it is a positive number
in the denominator.
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So, it is greater than 0 for all values of
x S. So, in the previous analysis we have
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shown already when we did the steady state
analysis, that here what you see on the screen
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S naught is less than D K s over mu max plus
D. If that is the case then the reaction does
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not happen at all. Why does the reaction does
not happen at all? Because that is not enough
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substrate for the reaction to happen physically
speaking and mathematically. What you found
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was that you got a unfeasible or else trivial
solution. So, if you remember go back and
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remember that S naught greater than D K s
over mu max minus D and mu max had to be greater
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than D. Why is mu max had to be greater than
D? Because one of the criteria for steady
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state was mu max equals D right the straightway,
steady state criteria was mu max equals D.
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Now, if mu equals D, now mu equals mu max
time some number which is less than 1. So,
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obviously then mu max has to be greater than
D. So, then the denominator is positive out
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here, but the reaction can only happen if
S naught is greater than D K s over mu max
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plus D K s. So, we will consider that case
only here, because there is no point considering
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the other cases. And if you need I will go
back and show you the chart we had, you know
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we have a nice chart which showed here, this
one see,
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So, look at this side S naught less than D
K s over mu max plus D K s, these are all
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trivial solutions and this one you get trivial
or non trivial solutions, depending on the
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D greater than mu max or D less than mu max.
Now, D greater than mu max is not possible,
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just as I explained because D equals mu. So,
mu equals mu max time some number which is
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less than one. So, obviously D has to be less
than mu max so the in that region alone you
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can get a non trivial solution fine.
So, let us work through this all we need to
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do is actually find the criteria. So, we understand
this, that S naught has to be greater than
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D K s over mu max plus D K s.
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Now, let us consider the case of sterile feed.
Why do we consider that? Just to make our
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calculations a little simpler sterile feed
means X naught equals 0. Now, so again we
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are looking in the region of D less than mu
max and S s s, S s s naught less than D K
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s over mu max plus K s, so in that region
S s s that you get is given by D K s over
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mu max plus D K s that is what we got last
time. And if you go back and put these values
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into your the values that we got into your
A matrix, so A matrix is here.
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So, if you go and put these values, this becomes
0. Why because, mu equals D, so this becomes
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0 and the rest of them are all non 0 numbers.
So, that is the only term that becomes 0,
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but by becoming 0 it makes a calculation quite
simple because as you say this a B 2 will
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be simply minus a 1 2 times a 2 1 and B 1
will be simply minus a 2 2, is it clear because
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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
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1, so this is my A matrix over here this term
being 0.
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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
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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
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greater than 0 or in other words, a 1 2 times
a 2 1 less than 0. So, my two criteria B 1
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greater than 0 and B 2 greater than 0 that
is, the two coefficients greater than 0 translate
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too because a 1 1 is 0. Just as I explained
just now, translate to a 2 2 less than 0 and
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a 1 2 a 2 1 less than 0 that is all.
So, that would be that is criteria for stable
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steady state, now a 2 2 is this number over
here as you can see, so minus a 2 2 is minus
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of this number so this number D is positive
number so del mu del S is.
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What is it?
Positive?
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Positive, I just explained why so, this is
a positive number right? This whole thing
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is positive, D is positive, del mu del S is
positive, obviously X is positive, then Y
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is positive, so this whole thing put together
is a negative number. So, this is automatically
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satisfied. Is it clear? a 2 2 less than 0
is automatically satisfied. Now the other
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criteria is the only one that we have to take
care of.
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So, this is a product of these two minus a
1 2 times a 2 1 so, D is what you will get
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is D times X s s over del mu, del S over Y
and minus of that and there is a minus out
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here so this whole thing comes out to be positive
so D X s s del mu, del S over Y has to be
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greater than 0, so that is my criteria. For
greater than 0, I think I will come back to
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that, I think I do not have written it here
clearly.
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So, basically my criteria for D less than
mu max is simply, del mu del S times D times
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X s s times Y s. Now, you are going to tell
me that is it greater than 0, so you are going
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to tell me is that correct or not? As I told
is it stable or not? And the answer is Yes,
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It is stable, but why? We are not going to
satisfy anything, it is automatically satisfied
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everything is automatically satisfied for
this criteria, why because, just from this,
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let me here just because of this del mu del
S is this number mu max K s over K s plus
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S square which is positive for all values
of S.
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So automatically is that clear or do I need
to explain? So, I need to explain this quickly
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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
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X Y fine now D is positive, X is positive,
Y is positive and del mu del S is positive,
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so obviously this whole thing is positive
right?
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So, this criteria satisfies, look over here
you have a 2 2 this number has to be less
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than 0. So, this is positive, so obviously
the negative of the whole thing is negative,
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so this is also satisfied automatically, clear?
So, both criteria automatically satisfied,
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as a result this condition is steady state
is stable. The next one is the trivial state
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as I explained, that you know.
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So, D greater than mu max obviously you do
not have anything, but the trivial solution,
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but we are still looking at the stability
of it and the reason we are still looking
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at it the stability of it is because whether
the trivial stage itself is stable, you know
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a trivial state means what? That nothing really
happens, you let the thing go in and it just
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goes out, but even is that stable or does
that degenerate some other unstable steady
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state.
So, we are trying to look at that and the
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A matrix is slightly different because this
D is no longer equals mu. So this term is
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not 0, on the other hand this term turns out
to be 0 when you go and do the algebra yourself.
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But I am not going to, you know the details
of this one because this is a trivial state,
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but still so you get that a 1 1 plus a 2 2,
when you add these two up equals this number
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which is less than 0. Why is that less than
0? Because D is greater than mu max.
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So, if D is greater than mu max obviously
2 D is greater than mu max, 2 D is greater
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than mu therefore, this number is less than
0. So, this criteria is satisfied and the
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second one is a 1 1 times a 2 2 minus this
is 0. So, the product of these two have to
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be greater than 0. So, this criteria again
is satisfied because D is greater than mu
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D is greater than mu max therefore, it is
greater than mu so this is positive. So, product
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is positive so what we figure out is that,
the Trivial steady state is also stable.
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So, both criteria satisfied and both lambdas
are negative, so trivial steady state is also
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stable. Is that clear to all of you? Or do
you want me to repeat any parts of it? I am
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going a little fast because I want to finish
something. So, then next thing that we will
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be going to do today is that apply the same
calculation, then the same process and the
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same mechanism that we did to multiple steady
states and the case of substrate inhibition.
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I do not know if you have forgotten, let me
remind you.
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So, this was what the case of substrate inhibition
was, so till now we did the Monod Growth Model
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with the simple thing. Now you know, as I
said that we did the case of multiple substrates
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and similar kind of multiple substrates.
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What we get is mu, so we can do a similar
kind of analysis using multiple substrate.
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So, the one that I am doing is the case of
substrate inhibition.
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So, the reaction for that substrate inhibition
was X plus S giving X S and X S reacts with
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S again to give X S 2 which does not produce
new cell.
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So if you remember, these were the all the
derivations. Let me not go through them, you
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know if this X t was the sum of the 3 X plus
X S plus x s 2.
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And then we went through this and this is
what we found, so why are we interested in
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this? Because this is the only thing that
is different from the previous case, the value
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of mu itself, because everything the balance
equations the rest of the analysis everything
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is same. So if you remember the reason, we
did it in the previous classes, we went up
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and did all the analysis without putting in
any value of mu at all. The reason we did
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that is because the basic equations irrespective
of the form of mu as a same, right?
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So, whatever form of mu we take, be it substrate
inhibition, be it multiple substrates, you
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know other kinds of inhibitors does not be
a problem at all. We can just put in that
16:16.800 --> 16:22.379
form, everything into that form. So, let us
go back here so you see it is a, this is my
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form mu which is mu max S, 1 plus K s plus
this term you know apart from Monod. You have
16:28.149 --> 16:35.149
this extra term of S square over K i, K s,
so what do you think you know? So this is
16:35.249 --> 16:41.240
not important for us. So, what do you think
the innovate is because you see that del mu
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del S could be 0, but we will come to that
later.
16:44.170 --> 16:51.110
So, what do you think? Even if I ask you that
so these are basic equations, what would be
16:51.110 --> 16:58.110
the effect of substrate inhibition on the
system? What do you think would be the effect?
17:06.010 --> 17:13.010
What do you think would be the effect of substrate
inhibition? These equations are written for
17:17.040 --> 17:24.040
sterile feed, I have written it down there.
What we got last time, let us review what
17:31.100 --> 17:36.850
we got last time without the substrate inhibition
is that there is one trivial state and one
17:36.850 --> 17:43.230
non trivial state and both are stable, that
is what we found, if I do a quick review.
17:43.230 --> 17:48.539
So, first question is two questions. What
do you think in terms of, what you will get
17:48.539 --> 17:54.779
in terms of trivial states and non trivial
states? a and b is that? What do you think
17:54.779 --> 17:58.520
about their stability just an intuition, we
will work out the whole thing you know whole
17:58.520 --> 18:05.520
maths in the rest of the class, but just some
intuition.
18:06.390 --> 18:10.500
Multiple
Yes, multiple states in a way. Last time also
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you had multiple states, because you had a
trivial state and a non trivial state. So,
18:15.120 --> 18:18.840
two states that was still multiple, but what
do you mean by multiple states here? How many
18:18.840 --> 18:25.840
states? I mean
that is, it should be a easy answer because
what happens to the order of the equation?
18:31.970 --> 18:34.539
Increases?
Increases by?.
18:34.539 --> 18:41.539
By 1 order of magnitude in S, so from 2 to
3 so how many.
18:44.049 --> 18:48.500
3 yes, So that is the straight answer for
the second order. Last time you had a second
18:48.500 --> 18:53.289
order equation, you got two steady states.
This time you have a cubic equations so obviously,
18:53.289 --> 18:58.700
you are going to have three steady state so
that is one part of the answer and the other
18:58.700 --> 19:04.460
part obviously, you cannot tell without doing
the analysis. But what do you, what can you
19:04.460 --> 19:11.460
tell you now?
What are the intuition telling you, what happens
19:12.580 --> 19:15.250
when there are three states?
19:15.250 --> 19:19.500
Yes, so let me show you this, you know, I
think I have drawn this diagram several times
19:19.500 --> 19:25.230
before, so let us say this is X conversion
just like here something like that and this
19:25.230 --> 19:32.230
is the reactor Damkohler number which depends
on the residence time of the chemo stat, exactly
19:32.480 --> 19:38.799
the same thing, so this is how it is going
to be, so these are.
19:38.799 --> 19:42.850
What are these called? I have told you this
before.
19:42.850 --> 19:47.779
Still, what does the name mathematically?
19:47.779 --> 19:54.779
So, limit points they may or may not be ignition
or extension depends on the catalyst and stuffs
19:56.559 --> 20:01.340
like that, you know, if you are doing a chemical
reaction, but in mathematically the generate
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term for these is limit points. So, why this
was important is because between these limit
20:08.690 --> 20:15.690
points in this region of this limit points,
you have three solutions and beyond that one
20:17.440 --> 20:24.440
solution and one solution.
So, between this only you have the cubic solution.
20:26.549 --> 20:30.390
What it means is that, when we solve it, you
will see that outside the certain range in
20:30.390 --> 20:36.120
the of the parameter, there is a single solution
because you know the determinant goes off,
20:36.120 --> 20:39.370
determinant goes to 0 or something like that
typically the determinant goes to 0 and the
20:39.370 --> 20:44.820
outside range and you end up with a single
solution and within that you have just three
20:44.820 --> 20:51.820
solutions, if I look at the stuff again.
So, limit points these two and this is the
20:56.870 --> 21:03.870
ignited branch, this is the stable solution,a
middle one and this is the stable solution.
21:17.000 --> 21:24.000
So, the bottom branch is the is the stable
solution, the top branch is a stable solution
21:26.820 --> 21:30.760
and the middle branch is an unstable solution.
Why is it an unstable? Is because, if you
21:30.760 --> 21:36.899
for some reason going to the unstable branch
and allow the system to stay there it will
21:36.899 --> 21:42.590
either go to the top branch of the ignited
branch or go to the bottom branch or the extinguish
21:42.590 --> 21:47.730
branch depending on whichever is closer to,
it does not stay there for very long.
21:47.730 --> 21:54.730
So, your aim is never to operate close to
the unstable branch and how do you ensure
21:56.240 --> 22:03.240
that? As an engineer for example, you have
to operate your plant or your reactor in a
22:05.850 --> 22:11.669
certain parameter range, parameter range means
say, residence time or liquid volume or the
22:11.669 --> 22:18.669
substrate concentration, whatever it is, how
will you ensure that it does not go to the
22:19.779 --> 22:26.779
unstable branch? You choose your parameters
such that, it does not go to the unstable
22:27.149 --> 22:34.149
branch, right? How do you choose?
How do you choose your parameters such that
22:36.840 --> 22:42.279
it does not go with the unstable branch?
22:42.279 --> 22:46.570
Yes, but there is a procedure for that and
that is called, let me see if you have heard
22:46.570 --> 22:53.570
of this Neutral Stability and that is exactly
what we are doing now. What it does is, basically
23:03.039 --> 23:07.140
it draws you, essentially draw this in the
parametric plain. For example, see here, it
23:07.140 --> 23:12.779
could be in the plain of Damkohler number
verus Pekkle number, something like that or
23:12.779 --> 23:16.500
Pekkle number here.
Let us say, Pekkle number here and Damkohler
23:16.500 --> 23:23.500
number here, then what it gives you? It gives
you a curve like this because from here this
23:24.480 --> 23:28.600
diagram that I drew now how do you know? You
never know for otherwise, what you have to
23:28.600 --> 23:32.860
do if you choose every Damkohler number, go
and make some simulations, find out whether
23:32.860 --> 23:36.010
your solution is in the stable branch or unstable
branch.
23:36.010 --> 23:39.090
That is a complete waste of your time, you
cannot keep doing it. So what you need to
23:39.090 --> 23:45.820
do is, you need to isolate or figure out in
the parametric plain of Pekkle versus Damkohler.
23:45.820 --> 23:52.100
Whether the system is going to be stable or
unstable so this is the part where it can
23:52.100 --> 23:59.100
be unstable and the system is stable here,
which means that you choose parameter such
24:02.010 --> 24:07.830
that it stays in the middle of this range,
it stays within this stable path. It could
24:07.830 --> 24:11.840
be the other way around also, it could be
that inside is stable or outside this is unstable
24:11.840 --> 24:16.299
very unlikely or typically you know because
of Pekkle increases it becomes unstable. So,
24:16.299 --> 24:20.460
but there is a possibility that it can happen,
stable can be inside and unstable can be outside.
24:20.460 --> 24:24.600
So, our aim to do this, you know our engineering
aim to do all this analysis is to be able
24:24.600 --> 24:29.029
to come up with this Neutral Stability Diagram.
So that we do not have to do the calculations
24:29.029 --> 24:34.360
every time we can figure out in the parametric
space. What part is stable and what part is
24:34.360 --> 24:40.940
unstable as a result as soon as you know in
the reactor, as soon as you choose your parameters,
24:40.940 --> 24:45.370
you know your flow rate. Damkohler and Pekkle
number between them contain all the parameters,
24:45.370 --> 24:51.210
essentially you know flow rate volume of the
reaction reactor, reaction time, the size
24:51.210 --> 24:56.039
of the, you know the diffusion length scale
in the reactor and the diffusion coefficient
24:56.039 --> 25:01.279
of the species in the reactor.
So, all these parameters are contained between
25:01.279 --> 25:05.039
the Pekkle and Damkohler number. So, if you
look here. So, essentially that you choose
25:05.039 --> 25:11.730
your Pekkle and Damkohler such that your system
stays here or here or here and does not go
25:11.730 --> 25:16.149
anywhere here so, that is your range. So,
you can directly choose the parameters you
25:16.149 --> 25:21.059
do not have to do calculations all the time,
once you get the Neutral Stability Curve,
25:21.059 --> 25:25.659
this curve once and for all you can directly
choose your parameter and make sure that is
25:25.659 --> 25:30.710
in the stable value. So, that is our aim,
not to just solve one problem, not to just
25:30.710 --> 25:34.950
solve for the steady state and analysis stability
of the steady state, but to get a Neutral
25:34.950 --> 25:37.769
Stability Curve, so that you can use that
at all times.
25:37.769 --> 25:43.220
So, that is what we are trying to do over
here. So, as I discussed that this being a
25:43.220 --> 25:48.309
Cubic equation this will have three solutions.
And out of these three solutions, it turns
25:48.309 --> 25:52.840
out one will be a trivial solution and the
two would be non trivial solutions, so then
25:52.840 --> 25:55.399
we have to look at the stability of these
solutions.
25:55.399 --> 25:59.649
So, the trivial solution as we have looked
before, it is simply going to be X s s equals
25:59.649 --> 26:06.460
0 and S s s equals S naught and the non trivial
solutions, if you solve these you know so
26:06.460 --> 26:13.409
how do I solve this, remember what we did
was we did a invariance right? We did an invariance.
26:13.409 --> 26:20.149
So, you multiply this by the this second equation
by Y and add these two up and as a result
26:20.149 --> 26:24.990
these term will go, then you can express X
as a function of S and then you put it back
26:24.990 --> 26:29.830
over here and then solve it. So, you solve
it by the method of invariance and this is
26:29.830 --> 26:36.830
what you get, this is to only remind you,
Valid for mu max greater than D, this is mu
26:41.080 --> 26:48.080
max not equal to D rather this is valid for
mu max not equals to D, why is this valid
26:48.610 --> 26:51.970
for mu max not equals D? Because mu max equals
D what will happen is this squadratic form,
26:51.970 --> 26:56.950
Remember the you taken out the Trivial solution,
you have a Cubic equation to start with you,
26:56.950 --> 27:02.510
take out one of the solutions which is the
trivial solution then you have a quadratic
27:02.510 --> 27:07.880
form. So, this is the quadratic form you are
solving. So, if you know this thing is like
27:07.880 --> 27:14.549
the Cubic form, it will be some S times a
1 square plus a 2 S plus a 3 equals 0.
27:14.549 --> 27:21.330
So, this is the form once you put that, so
this will give you the sorry this is not S,
27:21.330 --> 27:28.330
S minus S naught so this part will be the
trivial solution this is trivial which will
27:29.419 --> 27:34.110
give you S equals S naught and this is the
non trivial part. So, this is a quadratic
27:34.110 --> 27:41.110
you can solve, but what I am trying to say
is that a 1 equals mu max minus D. So, what
27:46.070 --> 27:52.610
you need to do is when mu max equals D then
you need to drop this completely from the
27:52.610 --> 27:56.500
equation itself. Do not solve the quadratic
and then try to settle around because it will
27:56.500 --> 28:00.200
come in the denominator or something then
the whole thing will blow up.
28:00.200 --> 28:04.590
So, if mu max equals D then just drop this
term completely and then you have a linear
28:04.590 --> 28:11.590
and you have just one single solution which
is you know, given here. So, this solution
28:13.809 --> 28:20.809
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
28:23.340 --> 28:24.960
back to the screen now.
28:24.960 --> 28:31.960
So, the Non Trivial solution is given by S
S equals K I over 2 K D this term mu max minus
28:34.299 --> 28:40.220
D plus minus this determinant and X s s. So,
this is let my S s s and X s s is obtained
28:40.220 --> 28:45.539
as S naught minus S s s times Y, because there
is no material balance right? So, if S naught
28:45.539 --> 28:50.159
minus S has been consumed then Y times that
would be give you the amount of cells that
28:50.159 --> 28:53.330
has been formed.
So, for mu as I said for a 1 equals 0 the
28:53.330 --> 29:00.330
first coefficient equals 0 or for D equals
mu max S s s is square root of K s K s over
29:01.159 --> 29:08.159
K i. So, the two solutions that you get over
here these two so non trivial solutions are
29:09.919 --> 29:16.919
provided that mu max minus D is greater than
this or in other words the discriminant is
29:19.230 --> 29:26.230
greater than 0 this discriminant is positive,
so you get this so mu max minus D is 2D is
29:27.049 --> 29:28.190
it correct.
29:28.190 --> 29:35.190
Now, what we have are essentially my three
states for the special case where mu max equals
29:35.289 --> 29:39.159
d I have this state, but apart from that essentially
for all other cases these are my three states.
29:39.159 --> 29:43.600
Now, what I have to do is I have to look at
the stability of three states, predictably
29:43.600 --> 29:49.389
your trivial state is going to be steady stable
predictably, but we will still have to, we
29:49.389 --> 29:53.500
will still check that, so what is my, you
know, difference that I need to do? Everything
29:53.500 --> 29:58.110
is same or again as before, only my mu is
different so, I need to calculate my del mu
29:58.110 --> 30:04.539
del s. Now what we have looked at the three
states, what do you suspect in terms of the
30:04.539 --> 30:11.000
stability of the states? Something Yes, I
think already said that one of the solutions
30:11.000 --> 30:15.149
is going to be unstable and the other two
are going to be stable. Yes, That is what
30:15.149 --> 30:22.149
I also suspect, but can you create a reason
for that. What could be the reason for that?
30:32.250 --> 30:36.750
What was the driving force the last time we
did for the normal case? What was the driving
30:36.750 --> 30:42.250
force behind making these solutions stable
both the solutions it turned out were stable,
30:42.250 --> 30:49.250
but what was the driving force? What is the
reason for that?
30:50.659 --> 30:53.230
D was greater than mu max? No, D has to be
always greater than mu max for feasible solution
30:53.230 --> 31:00.230
to exist. Is D is now greater than mu it is
not going to work.
31:02.799 --> 31:07.559
Yes, correct the reason behind that was del
mu del S was always positive, but if you remember
31:07.559 --> 31:12.620
from the last time we did in the growth model
del mu del S was not always positive, it goes
31:12.620 --> 31:19.620
to positive and then goes to 0 and then decreases,
right? If you remember, so it turned out that
31:24.880 --> 31:30.200
if del mu del S were not positive last time
for example, then the system would be unstable.
31:30.200 --> 31:34.980
So, what we can predict or intuit I mean,
it is not mandatory that exactly mathematically
31:34.980 --> 31:39.549
we get, but you know we have to always couple
some kind of intuition with the mathematics.
31:39.549 --> 31:43.909
So, what we intuit is that for the region
where the specific growth rate decreases with
31:43.909 --> 31:49.440
substrate the system is going to be unstable.
And physically if you think of it makes sense
31:49.440 --> 31:55.279
that if you increasing the amount of substrate
by the specific growth rate decreasing that
31:55.279 --> 32:01.009
is the really worrying case, right? Physically
speaking and that probably is the reason where
32:01.009 --> 32:05.889
there are sources of instability. So, let
us try and look at it.
32:05.889 --> 32:12.889
So, this is the first thing you will say is
mu over here with the inhibition term in there
32:13.350 --> 32:20.130
and this is my del mu del s or d mu d s over
here and which could be negative or positive
32:20.130 --> 32:25.090
depending on the numerator, because the denominator
is always positive. It is the numerator which
32:25.090 --> 32:30.220
can go negative or positive. So, depending
on whether S is greater than square root of
32:30.220 --> 32:36.950
k i k s, all less than square root of k i
k s, then you have trouble, if f equals square
32:36.950 --> 32:42.590
root of k i k s then you have then what do
you have?
32:42.590 --> 32:44.340
Trivial solutions.
Not the trivial solution, Trivial solution
32:44.340 --> 32:49.960
is 0 this is that solution d equals mu max
and S equals s square root of k s and k s
32:49.960 --> 32:53.840
k s. Trivial solution is this one, let us
not confuse trivial solution with this, this
32:53.840 --> 32:57.919
is trivial solution, these are non trivial
solutions are special case of the non trivial
32:57.919 --> 33:03.710
solution, is this one where d equals mu max.
Is it clear Lisa? Is there any problem understand?
33:03.710 --> 33:08.190
This is probably the hardest thing we have
done in this till now, but if there is a problem
33:08.190 --> 33:15.190
just stop me and so the special case of the
non trivial solution is S square root equals
33:15.500 --> 33:19.559
square root of k i k s, but in general we
have the trivial solution and two non trivial
33:19.559 --> 33:26.559
solutions.
So out of the non trivial solution there is
33:29.100 --> 33:35.279
one possibility of S being equal to k i k
s in which case del mu del S would equal 0,
33:35.279 --> 33:40.159
right? That is one possibility, but that is
a exceptional possibility a special case,
33:40.159 --> 33:45.380
but in general k i k s would have to be greater
than 0 or less than 0. So, depending on whether
33:45.380 --> 33:52.380
it is greater than 0 so square root of k i
k s greater than S or square root of K I K
33:56.009 --> 34:01.179
s less than S, so square root of k i k s is
greater than S and we probably think that
34:01.179 --> 34:05.610
the system is going to be stable fine.
34:05.610 --> 34:09.830
So, we consider the non trivial states over
here and for the non trivial states the mu
34:09.830 --> 34:16.060
equals D for this trivial state is not necessary,
but for the non trivial state mu equals D.
34:16.060 --> 34:23.060
Do you remember we had X times mu minus D
equals 0 so, for trivial state this is X equals
34:24.730 --> 34:31.730
0 for non trivial state mu equals D, right?
Because these are things if you remember,
34:33.480 --> 34:39.109
So, this matrix A that we have, this is the
same matrix that we have remember? Exactly
34:39.109 --> 34:43.879
same, as what we did last time, because we
still have a generalised form of mu, we have
34:43.879 --> 34:49.349
not put in our mu into it. Now we are going
to put in our mu into it so stability criteria
34:49.349 --> 34:56.349
was a 1 1 plus a 2 2 less than 0 or in other
words, this term to be negative and the other
34:58.130 --> 35:04.390
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
35:04.390 --> 35:11.000
these two terms to be positive.
Now, to remind you again the last time we
35:11.000 --> 35:16.400
had a advantage because del mu del S was greater
than 0. So, we could conclude directly that
35:16.400 --> 35:21.440
yes if del mu del S is 0 than greater than
positive then of course, this whole term is
35:21.440 --> 35:25.599
going to be negative. Similarly, del mu del
S is positive then of course, the product
35:25.599 --> 35:32.430
of these two terms is going to be greater
than 0, right? Is it clear to everybody?
35:32.430 --> 35:37.000
So, this we could conclude that directly,
now we cannot conclude that directly and we
35:37.000 --> 35:42.640
have to depend on the, on a case by case basis.
So, this is my criteria now so del mu del
35:42.640 --> 35:48.680
S times X over Y plus D should be greater
than 0 and this one will lead to del mu del
35:48.680 --> 35:55.680
S times D X over Y should be greater than
0 fine now for del mu del S greater for del
35:56.510 --> 36:03.510
mu del S to be greater than 0 my S square.
As I just showed s square has to be less than
36:05.710 --> 36:12.710
k i k s square has to be less than k i k s
so that is my stability criteria, essentially
36:14.650 --> 36:20.730
that s has to be lesser than this is what
we have intuited just by looking at the equation
36:20.730 --> 36:24.849
itself we had intuited this.
So, my stability criteria is that s has to
36:24.849 --> 36:29.500
be less than square root of k i k s and you
can go back and check all rest of it you know,
36:29.500 --> 36:34.829
you can put over here, you can do the rest,
I have skipped that calculation, but the differentiation
36:34.829 --> 36:41.660
is already there you can put go and put back
over here, but you know, coming back to this
36:41.660 --> 36:48.660
model one of the things you need to, though
I do not kind of ask you, I mean I am not
36:48.839 --> 36:54.130
suggesting that you do that in the exam I
want you to put the stability criteria a 1
36:54.130 --> 37:01.130
1 plus a 2 2 greater than less than 0 and
the other two terms a 1 1 minus a 2 2 minus
37:01.980 --> 37:06.670
a 1 2 a 2 1 greater than 0 and then come through
it.
37:06.670 --> 37:11.200
But at the end of the day, if you think, come
to think of it what is your stability criteria
37:11.200 --> 37:17.490
is essentially coming down to del mu del S
should be greater than 0, right? Is it correct?
37:17.490 --> 37:20.750
You do all the calculations and I do not suggest
that you do not do the calculations in the
37:20.750 --> 37:24.079
test or something, you do the calculations
to show that you know everything and this
37:24.079 --> 37:29.400
is the proper way, but check on your calculation
essentially would be what to find out, what
37:29.400 --> 37:34.250
is my criteria such that del mu del S is positive.
So, at the end of the day that is what matters,
37:34.250 --> 37:38.359
but then again do not go and write it straight
away, derive it.
37:38.359 --> 37:44.089
So, this what you see on the screen is now
my Neutral Stability Curve. The thing that
37:44.089 --> 37:51.089
I had been trying to get and look what I have
over here is s naught over d so this is a
37:55.579 --> 37:59.770
major difference between what I showed you,
this multiple solution curve that is called
37:59.770 --> 38:05.660
known as Difurcation diagram. The one that
I just showed you here, this one if I can
38:05.660 --> 38:07.780
go to the, yes,
38:07.780 --> 38:14.359
So, this is my Bifurcation diagram and my
Neutral Stability Curve, would be this now
38:14.359 --> 38:21.359
I need to draw a fresh diagram and over the
Bifurcation Curve and the Neutral Stability
38:22.950 --> 38:29.950
Curve together and there is something I want
you to help me here.
38:32.040 --> 38:36.589
So, as you can see that the extinguish branch
essentially saturates, you know over time
38:36.589 --> 38:43.290
it kind of saturates are over Damkohler number,
let us say this is Damkohler and this is the
38:43.290 --> 38:50.290
conversion X. Now I want you to tell me, help
me with how the Mutual Stability Curve is
38:51.020 --> 38:58.020
going to look like. So this is my Bifurcation
diagram, I want you to help me with how my
39:07.540 --> 39:14.540
neutral stability curve would look like, how
it look like it would have a, would look similar
39:21.880 --> 39:28.880
to what I told you like this, what I drew
over here, but where would this, how would
39:29.819 --> 39:33.520
I place that?
So, let us say my Neutral Stability Curve
39:33.520 --> 39:39.140
in the again X axis is still D a, but y axis
is Pekkle. Do you understand? Do you discern
39:39.140 --> 39:42.960
the difference in the terms of the axis for
the bifurcation and the Neutral Stability
39:42.960 --> 39:49.650
Curve? The major difference is that Bifurcation
Curve is drawn in the Variable Space. Neutral
39:49.650 --> 39:56.650
Stability Curve is drawn in the parameter
space. So, this is the variable space and
40:02.859 --> 40:06.670
this is drawn in the parameter space and there
is, so, Variable space means one variable
40:06.670 --> 40:09.890
versus a parameter, there is, this is, in
the parameter space which means two parameters
40:09.890 --> 40:15.490
and then there is another kind of plot, what
is it called? Where two variables you would
40:15.490 --> 40:21.309
haveand you draw one variable versus another
for let us say, in this case if I want to
40:21.309 --> 40:22.720
draw X versus s space.
40:22.720 --> 40:28.869
What is this called? What is it?
I do not remember.
40:28.869 --> 40:35.869
You do not remember? Who remembers? It is
called the phase space, see if the variable
40:39.309 --> 40:43.900
space here, the parameter space here and the
phase space is where both the variables are
40:43.900 --> 40:50.359
drawn against each other. And so, this is
something that is important and you have to
40:50.359 --> 40:55.240
understand this. Although I explained it before
that you draw you necessarily draw the Neutral
40:55.240 --> 41:00.780
Stability Curve in the parameter space and
not in the variable space or the phase space.
41:00.780 --> 41:04.690
The reason you do that is because you are
not interested in what the value of that,
41:04.690 --> 41:08.109
do you understand? What I am trying to say,
you not interested in what the value of the
41:08.109 --> 41:12.809
parameter variable is going to be. You are
interested in what parameter should you operate
41:12.809 --> 41:18.900
your reactor in. You are not interested whether
X is going to be 0.9 or 0.8 or 0.7 or 0.6,
41:18.900 --> 41:22.500
that is not your concern. You are not even
interested whether it is going to be ignited
41:22.500 --> 41:26.099
or extinguished? That is an important question
by the way, that you know if you look at this
41:26.099 --> 41:30.400
bifurcation diagram over here this is the
ignited branch in this extinguish branch.
41:30.400 --> 41:35.869
It is important actually to figure out whether
the system is going to operate in the ignited
41:35.869 --> 41:41.329
branch or the extinguished branch. Why because,
you want it to operate in the Ignited branch.
41:41.329 --> 41:45.530
Ignited branch, Yes, because you want product,
you know you would not run a reactor without
41:45.530 --> 41:49.250
generating product, so if you are in the extinguish
branch you hardly generate any product. So
41:49.250 --> 41:54.030
that is important, but more important than
that is, whether the system is stable or unstable.
41:54.030 --> 41:59.020
So, that is why the more important thing is
the Neutral Stability go. First come stability,
41:59.020 --> 42:02.630
once the system is stable then you worry about
whether you are generating enough product
42:02.630 --> 42:06.280
or not enough product. You understand what
I am to trying to say, so first is stability,
42:06.280 --> 42:10.500
So, once the system is stable there are two
possibilities that the system could be in
42:10.500 --> 42:14.640
the ignited branch or in the extinguish branch,
if you look over here as we discussed that
42:14.640 --> 42:21.640
this length the bottom length is stable the
top leg is stable, but the middle leg is unstable.
42:21.990 --> 42:25.710
So, even when the system is stable there is
still a possibility of it being the ignited
42:25.710 --> 42:28.510
branch or in the extinguish branch and you
want it in the ignited branch, right? So,
42:28.510 --> 42:34.569
I will come to that. Also how to figure that
out? But your first criteria is, stability.
42:34.569 --> 42:38.930
So, now you tell me that I have explained
so much now, you tell me that how do I draw
42:38.930 --> 42:43.460
my neutral stability curve between the limit
point?
42:43.460 --> 42:50.460
Between the limit point, so this is how we
draw line over here and line over here and
42:53.910 --> 42:57.740
numerical stability curve will be like this
will be like this and then like this. So,
42:57.740 --> 43:04.740
this is the system is unstable over here and
stable over here, so this corresponds to this
43:13.170 --> 43:20.170
branch, corresponds to this branch over here
corresponds to this branch corresponds to
43:20.480 --> 43:26.980
this and the middle branch corresponds to
this, so the two limit points correspond to
43:26.980 --> 43:33.980
the two horizontal. So, two limit points correspond
to the two vertical limits. Please remember
43:38.050 --> 43:45.050
these all things, so the two limit points
correspond to the two vertical limits of the
43:45.550 --> 43:51.130
neutral stability curve.
So, now my problem is solved. So, I now know
43:51.130 --> 43:55.940
that what part is stable and what part is
unstable. I can choose my criteria such that
43:55.940 --> 44:00.940
it is stable, now how do I choose whether
it goes to the steady state or to the ignited
44:00.940 --> 44:06.530
state, extinguish state? That also I can choose
from the Neutral Stability curve. If you look
44:06.530 --> 44:11.440
here, so if I choose something around this
point then it will be in the ignited state,
44:11.440 --> 44:18.440
just say after this vertical second vertical
length. so it will go to the ignited state.
44:19.730 --> 44:25.000
So, if I choose something out here, in the
towards my left hand corner, complete left
44:25.000 --> 44:27.609
hand corner then it will go to the extinguish
state.
44:27.609 --> 44:34.609
So, I want my two things, I want my system
to be stable, I want it to have attained maximum
44:34.710 --> 44:40.420
you know, to give maximum product. So, which
means that first it has to be either on the
44:40.420 --> 44:42.920
ignited and the extinguish branch and the
second would it should be in be the ignited
44:42.920 --> 44:45.890
branch.
So, this Neutral Stability Curve is the very
44:45.890 --> 44:51.309
useful thing in a way because, let me choose
my parameters my flow rate, my reactor size,
44:51.309 --> 44:56.000
my reaction rate, my diffusion coefficient
everything such that say, I choose something
44:56.000 --> 45:00.660
like this, this would be probably the optimum
choice, because it makes sure that it is stable
45:00.660 --> 45:06.540
or you can go here it make sure it is stable
and make sure it is in the ignited branch,
45:06.540 --> 45:13.540
is that clear to all of you?
Now, let us go back to the screen. So, this
45:13.780 --> 45:18.790
was a generic thing I showed and this is the
actual thing. So, this is the wash out case,
45:18.790 --> 45:24.910
so in other words the Trivial solution. Trivial
solution means that no product is being formed.
45:24.910 --> 45:31.420
So, not really worried about this and this
is my Neutral Stability Curve. Looks like
45:31.420 --> 45:36.880
and you know if there is a essentially vertical
limit. Also it goes up a little bit so many
45:36.880 --> 45:41.089
a times, it is little sharp. Sometimes it
is kind of flattened, all these things happen,
45:41.089 --> 45:48.040
but you go a little up. And then a kind of
sharp and so, this is my unstable solution
45:48.040 --> 45:54.520
is middle in the inside here above and the
rest of the system is stable solution. What
45:54.520 --> 45:59.650
is this line over here? This line is not properly
drawn, actually it should touch this due line
45:59.650 --> 46:03.569
over here.
So, any how this line over here corresponds
46:03.569 --> 46:10.569
to, it is written here the steady state value
of the substrate, so this is not a part of
46:10.790 --> 46:16.109
the, let me not confuse you, any which way
this is, this blue line over here, is the
46:16.109 --> 46:23.109
bottom line. This one is not a part of the
Neutral stability curve. It is just concentration
46:25.510 --> 46:29.940
plot or substrate plot relating to this value,
so do not worry so much about this line, just
46:29.940 --> 46:35.300
worry about this one. So, this is my neutral
stability curve. And as we concluded that
46:35.300 --> 46:40.569
the system goes unstable not just with multiple
inhibition. In general the system goes unstable
46:40.569 --> 46:46.200
well when del mu del S is negative and then
it we end up having positive real roots.
46:46.200 --> 46:50.910
So, for del mu del S negative, this is the
part the unstable solution and for del mu
46:50.910 --> 46:55.339
del S positive, everything here below is stable.
So, this is your first encounter with neutral
46:55.339 --> 46:56.240
stability curves, right?
46:56.240 --> 47:02.309
So, just want to make sure that everybody
understand. If there is any question on Neutral
47:02.309 --> 47:09.309
Stability Curve, you can ask me now or may
be a minute later also, this is the overall
47:19.089 --> 47:19.450
stability curve.
47:19.450 --> 47:26.000
And, this is the details of the calculation
that I skipped, you know all those calculations
47:26.000 --> 47:32.069
I skipped you can go through these calculations,
you want me to go through or just not very
47:32.069 --> 47:36.180
kind of Trivial.
So, essentially we have skipped the calculation
47:36.180 --> 47:42.640
because I just showed that del mu del s greater
than 0 is stable. So, but this is the hard
47:42.640 --> 47:48.809
way going through all the details and in everything.
So, this is the better picture. Now, I have
47:48.809 --> 47:55.809
of the this unstable and this is a so. So,
these are rates in gram per litre and d so
47:58.280 --> 48:02.299
what essentially it tells me if you look at
this Neutral Stability Curve. If I want to
48:02.299 --> 48:09.299
be in the stable region then it gives me numbers
on two of the parameters, one is the S naught
48:09.890 --> 48:13.440
the initial substrate concentration I use
and the second one is the D or D is the inverse
48:13.440 --> 48:17.530
of the residence time or in other words residence
time itself.
48:17.530 --> 48:22.559
So, what it tells me is that what should be
the size of the reactor or you know size cum
48:22.559 --> 48:28.839
flow rate? So, residence time is ratio of
these two a and b would be what would be my
48:28.839 --> 48:33.670
initial substrate concentration, this is what
it tells me. So, that I can pick so if you
48:33.670 --> 48:38.599
look at the curve here so I can pick any point
over here for example, I pick a point over
48:38.599 --> 48:44.680
here so that the system is stable and I know
that for my S naught for this system to be
48:44.680 --> 48:50.170
stable . If my S naught is this my dilution
rate has to be this or in other case if my
48:50.170 --> 48:56.089
reactor is given to me, you cannot make changes
to explore rate and its size then I pick my
48:56.089 --> 49:00.349
s naught initial substrate concentration such
that it is in the stable zone.
49:00.349 --> 49:07.349
So, if you need you can go through the calculations
on your own so if you need to feel comfortable
49:09.109 --> 49:15.650
with it you can go, as I told you know there
is this state, one s is outside the quadratic.
49:15.650 --> 49:22.230
So, these are the values in mu max is hour
inverse so these values are in a way important.
49:22.230 --> 49:29.230
If you actually want to plot it so as you
can see that the reaction mu max is sort of
49:29.250 --> 49:36.250
you know gives you the reaction rate and that
is of the order of hours and so rate is 0.6125
49:39.940 --> 49:44.579
the dilution rate is also of the order of
hour inverse and you ensure that these less
49:44.579 --> 49:50.380
than mu max over here.
So, these less than mu max that is ensured
49:50.380 --> 49:57.000
number one and number two they are of the
same order, they follow the stability criteria.
49:57.000 --> 50:04.000
So, that is in a way concludes our discussion
on the stability of these states. Now you
50:09.010 --> 50:15.599
know some assignments you might want to do
on your own is that these stability criteria
50:15.599 --> 50:22.599
that I show over here is for the inhibition,
but let us go back. So, this stability criteria
50:34.680 --> 50:38.970
that you obtained over here is for all possible
values of mu.
50:38.970 --> 50:45.910
So, some of the things that you might want
to do is look at how you change your mu and
50:45.910 --> 50:52.680
that how it changes the stability of the system.
Now it may be little boring to do, that to
50:52.680 --> 50:57.000
start with because this is with the maths,
but once you go through the maths and you
50:57.000 --> 51:00.299
essentially will come out so one of the things
you might want to tell me, is that whether
51:00.299 --> 51:05.930
our intuition that for all del mu del S greater
than less than 0 is the system unstable, that
51:05.930 --> 51:12.930
is the question I posed to you? For different
if I go through this one let us say here,
51:16.730 --> 51:23.730
so system like this for all kinds of del mu
del s less than 0 is the system unstable this
51:24.630 --> 51:30.270
system or any other system, any other kind
of growth model we have. I mean a lot of different
51:30.270 --> 51:37.260
growth models we have studied if I remember.
51:37.260 --> 51:43.940
So, and the Monod, Malthusian logistics all
these different growth models are here so
51:43.940 --> 51:50.220
one question a that I posed to you is that,
are for all these growth models if the system
51:50.220 --> 51:56.710
unstable for del mu del S less than 0, a and
b is that even if is unstable. What kind of
51:56.710 --> 52:01.450
solutions face you get? What kind of Neutral
Stability Curves and in the parametric space
52:01.450 --> 52:08.450
that you get? And this is something that you
might want to look at in at home or something
52:10.329 --> 52:16.630
like that. And you know it is important, to
we just did these two particular cases the
52:16.630 --> 52:22.410
Monod growth model one of the easy things
you can do is simply do the Modified Monod
52:22.410 --> 52:26.309
growth model which is very straight forward
same as Monod growth model almost, because
52:26.309 --> 52:30.740
there is not much effective except for that
k s naught here, but if you take this model
52:30.740 --> 52:34.240
for example, how does it change your whole
system?
52:34.240 --> 52:37.549
And how does it change the Neutral Stability
Curve in the Neutral Stability Space? The
52:37.549 --> 52:42.770
reason I am talking about this is see, till
the point we write the basic equation for
52:42.770 --> 52:47.910
the system and keep any mu out there and do
our analysis it is all valid, but you have
52:47.910 --> 52:52.559
to realise that the growth model when we put
in is an artificial thing. Do you understand
52:52.559 --> 52:57.470
what I am trying to say the growth model itself
is an artificial thing the balance equations
52:57.470 --> 53:02.230
with any mu in there is the real thing because
things are happening that way, but the growth
53:02.230 --> 53:05.780
model is an artificial thing.
So, it could be that I can choose growth model
53:05.780 --> 53:09.650
number one and you can choose growth model
number two for the system and we should not
53:09.650 --> 53:13.549
come up with wholly different criteria for
neutral stability. Do you understand what
53:13.549 --> 53:18.799
I am saying because at the end of the day,
irrespective of what I engineer number one
53:18.799 --> 53:24.059
chooses growth model and engineer number two
chooses different growth model the system
53:24.059 --> 53:27.890
should still run, so you might want to check
with different kind of growth models and try
53:27.890 --> 53:32.069
and see that whether the Neutral Stability
Curve that you get is similar or different.
53:32.069 --> 53:36.839
So, we will stop here and tomorrows lecture
I think if there are any questions on Neutral
53:36.839 --> 53:40.730
Stability. And if there is what you might
want to do is think about it and ask me in
53:40.730 --> 53:44.290
the beginning of the class next class that
is one possibility that is I think, because
53:44.290 --> 53:48.720
this is probably one of the hardest things.
We did this calculations and everything in
53:48.720 --> 53:54.480
this course, so is there anything you want
to ask at this point unintuitive stability
53:54.480 --> 54:01.480
or so we will save it for tomorrow, if there
is any question I will stop you know may be
54:01.630 --> 54:06.490
towards the end of next lecture we will spare
a few minutes and go through and we will start
54:06.490 --> 54:13.490
a new chapter in the next class. Thanks.