WEBVTT
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welcome to lecture number ten for the n
p t e l online certification course on bioreactors
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the ten hour course today we will a begin
module three which is on analysis of common
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bioreactor operating modes let me maximizes
a screen here analysis of common bioreactor
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operating modes let's first recall the common
bioreactor operating modes that we saw in
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the previous module we first saw batch as
the operating mode a batch is nothing but
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you add everything in and then at time
start time everything is already there this
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no addition or removal the process takes place
inside the batch reactor with no addition
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or removal and at the end of the batch time
you take the contents out and go for processing
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the batch time begins when everything after
everything has been added and the batch time
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ends before things are removed that is the
batch mode of operation
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we also saw the continuous mode of operation
where there is a continuous stream in and
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a continuous stream out of the bioreactor
and the processes simultaneously take place
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when there is a continuous flow that is
happen in between the two hence the batch
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and the continuous we had the fed batch where
you could you start out as a batch and then
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you could have intermittent feeding or intermittent
removal or even continuous feeding and continuous
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removal or both you know as long as it
is not both continuous in and continuous out
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simultaneously any combination is called any
other combination is called a fed batch these
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are the three common bioreactor operating
modes the batch the continuous and the fed
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batch let us look at the analysis some basic
analysis of a these modes and such an analysis
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would help us answer design kind of a question
question such as how long do we need to
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operate the bioreactor for for us particular
desired goal and so on that that would
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be a very basic design question that one
would want to ask and then based on that they
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could be other complex design questions that
one can answer through such analysis in this
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lecture let us focus on the batch bioreactor
the batch bioreactor we are first taking
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up because it is a very common mode of operation
especially in the industries it is very
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powerful that way simple but powerful it
is easy to operate compare to the other modes
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and that is one of the reasons why it is preferred
in the industry the batch mode of operation
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what is shown here is some of a around data
from our lab a long time ago this is total
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cell concentration remember we talked about
total cell concentration and viable cell concentration
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this is total cell concentration which has
been measured through what we called o d actually
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cell scatter and then calibrated and you
and we have used a calibration curve to convert
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o d to gram per litre i think this is a
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xanthomonas campestris if i remember right
it could i just took one representative batch
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growth the total cell concentration in gram
per litre is on the y axis the time of growth
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is on the x axis as we can see till about
let us say five six hours there is no change
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in the total cell concentration around six
hours the cell concentration starts to increase
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till about twenty eight hours and then it
goes flat the phase in which the initial phase
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in which there is no change in the cell concentration
total cell concentration is called the lag
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phase the region where the cell concentration
changes rapidly is called the log phase we
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look at the reason a little later and then
it reaches what is called
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stationary phase where where there is no
increase in cell concentration
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in slight contrast let us look at the viable
cell concentration profile for the same organism
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the viable cell concentration gram per litre
is given here time in hours this is a total
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cell concentration this scale is pretty much
the same zero to three here a again zero to
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three but this is viable cell concentration
this was most likely obtained by the plating
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method xanthomonas plating method as we can
see there is a good over lap of the viable
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cell concentration with the total cell concentration
here right in in this region here till
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about this region and till about here there
is a good over lap and then at around forty
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hours it's starts going down as expected because
the viable cell concentration would dropped
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here after the after some time the stationary
phase you enter the death phase which a will
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show up in the viable cell concentration profile
it does not show up for a long time in the
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total cell concentration profile let's actual
data so that is a simple batch where probably
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there was one limiting substrate one main
substrate and the cells grow on that substrate
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most likely growths there is another type
of growth that is sometimes seen which is
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called the diauxic growth sometimes i call
it diauxy and so on so fourth which essentially
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arises because of sequential utilization of
substrates one after the other sequential
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utilization for example if there is a mixture
of glucose and lactose
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let us say for a typical organisms such as
a collide then glucose is preferred over lactose
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and consumed first there is well known
mechanism for this to happen in molecular
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term it is known the so on so fourth we
will not get into that the we understand very
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well why this happens in equalize however
for our purposes the growth curve would look
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like this if you plot total cell concentration
x versus time there is an initial lag phase
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and then there is a log phase and then this
seems like stationary phase but it is a actually
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secondary lag phase and then growth occurs
and when you do the analysis of glucose and
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lactose one finds that glucose gets consumed
here first and then lactose gets consumed
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as you can see the
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growth rate which you would get out to
of this which will we see how to get would
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be higher here compare to the growth rate
on lactose this specific growth rate on glucose
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would be higher then the specific rate growth
rate on lactose that's the usual way in which
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happens the preferred substrate gets a taken
up first with a typically higher growth rate
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specific growth rate followed by the other
substrate this is also seened and it's
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good to know this even in a introductory course
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having said this for our analysis let us focus
on simple batch growth that we saw earlier
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one a one organism one substrate and so on
so fourth one limiting substrate batch growth
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as we saw in a with real data cell concentration
versus time a lag phase something called a
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log phase and a stationary phase and if you
plot viable cell concentration you would also
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see a death phase where the viable cell concentration
goes down the solid line is a total cell concentration
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the dotted line or the dash line is the
viable cell concentration now let us do a
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mass balance on cells in the bioreactor broth
the bioreactor broth the liquid in the bioreactor
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is what we are going to take as our system
to do this mass balance on sets this is our
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equation that we have seen right from module
one this is these are mass rates mass rate
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of input minus mass rate of output plus rate
of generation minus rate consumption all on
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a mass bases equals the rate of accumulation
of mass since we are doing the balance on
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cells this would be written for the mass of
cells in the batch bioreactor since it is
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batch and a let us consider only the growth
this phase for a let us say only this stage
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for now in other words there is no death that
we are going to consider there is no rate
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of input there is no rate of output because
it is a batch that's the basic definition
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of a batch and because there is no death there
is no consumption of cells and therefore r
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i r o and r c are zero those three rates as
zero therefore r g which is the growth rate
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on a mass basis equals d m d t here if we
replace m in terms of cell concentration because
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the cell concentrations are the ones that
are easily measured we would like to right
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things in terms of
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measured quantities because those are the
one set are measured and they can be used
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to make decisions so r g on a mass basis is
d d t of m which can be written as x into
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v cell concentration is mass of cells for
unit volume times the volume of the system
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or the broth that would be m and a in the
case of batch growth there is no change in
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the volume of batch right there is no addition
there is no a depletion or taking away things
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from the batch and therefore volume remains
constant if volume is a constant the it's
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not going to change with time and therefore
v can be taken out is a constant here of out
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of out of this derivative and therefore it
becomes v times d x d t now if r x is the
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rate on a volumetric basis ok again those
are the ones that are directly measured so
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we will write it in terms of them then r x
into v would be equal to r g you know this
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is the on a concentration or a volumetric
basis this is the rate in other words the
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rate of change of cell concentration with
time is what r x is therefore rate of change
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of cell concentration times the volume
would be the rate of change of mass concentration
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sorry rate of change of mass not a concentration
rate of change of mass therefore r x times
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v equals r g and that equals v d x d t from
here so if we equate this and this r x is
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nothing but d x d t ok so under this special
case of the batch bioreactor the volumetric
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rate equals the rate of accumulation of the
cell concentration ok this we have already
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seen as a special case of balance and we
also said that this the where your you would
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have been introduced in your school and
that is rather limited to the batch case and
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don't take it forward this aspect will become
clear when we discuss continuous processes
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the a the concept of a rate is a lot more
general where as the rate being equal to an
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accumulation rate is valid only for a batch
system one of the common models for growth
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rate that we have already seen infect the
first module that we saw was r x you know
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this is on a volumetric basis r x equals mu
x or in this case d x d t the accumulation
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rate equals mu x we are replacing d x sorry
we are replacing r x with d x d t d x d t
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equals mu x infect mu can be defined in such
a situation or here as one by x d x d t
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even it becomes easier to see the meaning
of term that's a reason i have given it here
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it is still the meaning of the term is still
valid it's just easier to see it here one
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by x d x d t it is been normalized with respective
cell concentration and therefore it is called
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the specific growth rate this model can be
used to describe batch growth but in parts
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these are the various parts of the batch growth
that we have seen the variation of cell concentration
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with time the lag log and the stationary phase
in the lag phase mu is zero you know d x d
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t is zero there is no change in the cell concentration
in the log phase which is here which is probably
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what we have been exposed to so for mu happens
to be a constant and in the stationary phase
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again there is no change in cell concentration
therefore mu is zero
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there are some phases in between the late
log phase the early log phase or the early
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stationary phase as it is called here these
phases are not very clearly described by this
15:24.410 --> 15:32.199
equation but it may not be very necessary
to know them or to describe them for our ends
15:32.199 --> 15:38.130
to make a design decisions and therefore we
will not get into that there are ways of doing
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them there are ways of including them in the
mathematical representation we will not a
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look into those aspects in this particular
course now let us consider the log phase alone
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that's where we are going to focus on the
reason will become clearer as we go a long
15:58.959 --> 16:06.130
in the log phase mu is a constant therefore
d x d t equals mu x mu can be take in to be
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a constant here if we solve this a straight
forward first order differential equation
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you know the first order differential equation
you need atleast one initial condition to
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solve so the initial condition is as follows
at time t equals t zero which is the beginning
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of the log phase is not the beginning of the
batch it is the beginning of the log phase
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the cell concentration is x zero ok x zero
may not be very different from the concentration
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at the start of the batch but sometimes
there might be a difference it's not change
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much but it has changed a little bit so we
need to keep we need to have clarity on
16:49.620 --> 16:55.279
what x zero actually is because this model
where mu is a constant is applicable only
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to the log phase not any other phase so if
go about solving this equation this few steps
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i have writtened down the steps here d x by
x equals mu d t integrate both sides we get
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a d the integral of d x by x as lon x and
here it is straight forward mu t plus c in
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diffeneted integral we have the initial condition
here which can be used to evaluate c if you
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putten the initial condition at time t
equals t zero x equals x zero therefore lon
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of x equals mu t zero plus c or c equals lon
of x zero minus mu of mu into t zero ok this
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is r c so if you substituted back into the
expression here the solution becomes lon x
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equals mu t plus lon x zero minus mu t not
ok if we combine similar terms on either side
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then lon of x minus lon of x naught which
turns out to be lon x by x naught lon a minus
18:00.990 --> 18:09.460
lon b is lon a by b equals mu into t minus
t naught taking mu common out and therefore
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x equals log x by x naught equals mu into
t minus t naught x equals x naught exponential
18:16.950 --> 18:25.540
of this term mu into t minus t naught this
is the description of the variation in cell
18:25.540 --> 18:38.110
concentration x with time since this is exponential
and this is a log we call the phase as either
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an logarithmic phase which corresponds to
this or an exponential growth phase which
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actually describes the variation of cell concentration
with type that's actually why it is called
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the log phase
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this equation can be used to find the time
needed to reach a desired cell concentration
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that's the straight forward design application
if the parameters mu t not which is the lag
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phase time and x naught which is the concentration
at the beginning of the log phase which
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may not be very different from the initial
cell concentration we can predict the total
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time from the start of the batch t to reach
x as this it is one by mu lon of x by x naught
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this is what we get from our previous thing
here ok one by mu lon of x by x naught if
19:36.260 --> 19:48.030
t naught can be taken as a zero plus our
t naught which is the time that the batch
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spends in the lag phase so lag time it is
here this is the time for logarithmic growth
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starting from x zero to reach x this would
be the total time of the batch if we are interested
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in reaching a certain cell concentration x
beginning with the cell concentration of about
20:07.500 --> 20:16.200
x zero ok i am assuming that a there is no
change in cell concentration in the lag phase
20:16.200 --> 20:23.600
ok having said this let me assign this problem
here we will continue a little bit more
20:23.600 --> 20:29.030
will continue and do a little bit more in
this lecture it'self for completeness but
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i would like you to work this out at the end
of this lecture to get some practice in the
20:35.159 --> 20:42.610
batch bioreactor analysis the problem reach
in a batch bioreactor the concentration after
20:42.610 --> 20:48.340
inoculum was point five gram per litre this
is the initial cell concentration the lag
20:48.340 --> 20:54.630
phase usually last twenty minutes under these
conditions assuming that the cell concentration
20:54.630 --> 21:01.669
at the start of the log phase was not significantly
different from that immediately after inoculation
21:01.669 --> 21:07.400
a estimated the time needed for it to reach
four gram per litre the specific growth rate
21:07.400 --> 21:13.230
for this organism under these conditions is
point five hour inverse b what is the time
21:13.230 --> 21:19.340
needed for the cell concentration to double
in the log phase this is the problem please
21:19.340 --> 21:25.330
work it out it is application of the earlier
derived equation we will of course see the
21:25.330 --> 21:28.809
solution in the next lecture
21:28.809 --> 21:39.010
let us continue now the initial analysis what
we have done so far did not consider the effect
21:39.010 --> 21:45.500
of substrate on the specific growth rate mu
ok we had kind of tacitly assumed that a enough
21:45.500 --> 21:52.070
of substrate was present so that mu equals
mu m you know mu versus s mu changes when
21:52.070 --> 21:58.450
the s changes in the initial stages and then
reaches constant value or a astatic value
21:58.450 --> 22:03.170
mu m when the substrate concentration is above
a certain level we kind of assumed that we
22:03.170 --> 22:09.250
are always had enough substrate so that
we were always at mu m for the specific growth
22:09.250 --> 22:19.440
rate this need not always be the case ok
and we have seen this this is the mathematical
22:19.440 --> 22:25.780
representation of the variation the monod
model of mu's variation with s the specific
22:25.780 --> 22:32.000
growth rate is a mu m times s by k s plus
s ok this variation would effect the growth
22:32.000 --> 22:37.360
rate and therefore that will ultimately effect
the time that it takes to achieve a certain
22:37.360 --> 22:45.419
cell concentration in batch if it is possible
if we incorporate the effect of the substrate
22:45.419 --> 22:54.360
for the log phase the mu needs to be replaced
by mu m s by k s plus s so d x d t equals
22:54.360 --> 23:02.360
mu x is still valued only the mu has been
replaced with the substrate effect to include
23:02.360 --> 23:11.140
the substrate effect mu m s by excuse me k
s plus s now let us consider two cases for
23:11.140 --> 23:19.700
e is of analysis the first case is that the
substrate concentration is much much greater
23:19.700 --> 23:31.900
then the k s value ok recall this k s is the
substrate concentration at which we got half
23:31.900 --> 23:37.270
maximal growth rate ok what we are saying
is that we are somewhere here were the substrate
23:37.270 --> 23:42.640
concentration is much higher than k s that
is the condition that we are looking at first
23:42.640 --> 23:49.789
if that is a case then s is the approximately
equal to k s i will tell you how this happens
23:49.789 --> 23:57.549
mathematically k s and s they are added
in the denominator if s is much much greater
23:57.549 --> 24:00.630
than k s let us say this to illustrate
24:00.630 --> 24:06.150
that k s is one and s is thousand ok whether
it is thousand in the denominator or whether
24:06.150 --> 24:10.520
it is thousand and one in the denominator
it is not going to make of a much difference
24:10.520 --> 24:18.110
to the value of this mu ok so you can very
safely replace k s plus s with just s alone
24:18.110 --> 24:23.659
if s happens to be much much greater than
k s if he replace the denominator by s you
24:23.659 --> 24:35.730
have mu m s by s s s can be canceled and you
will be left with mu m alone and
24:35.730 --> 24:43.090
that's what i shown here which is exactly
the same cases earlier only thing is that
24:43.090 --> 24:49.740
instead of saying d x d t equals to mu x we
could directly say d x d t equals mu m x we
24:49.740 --> 24:55.870
are actually tacitly assumed so earlier even
a in our earlier analysis but here we can
24:55.870 --> 25:04.840
very confidently say that this is mu m times
x it is the maximum specific growth rate now
25:04.840 --> 25:15.520
is the second condition if s is is approximately
equal to k s then we cannot use the above
25:15.520 --> 25:21.340
approximation you cannot say that you can
replace k s plus s with s alone because both
25:21.340 --> 25:31.130
are comparable this cases rather uncommon
in the case of a batch a bioreactor and can
25:31.130 --> 25:37.110
happen when some crucial but unusual substrate
becomes limiting otherswise the culture would
25:37.110 --> 25:41.500
have re stationary phase when this actually
happens that's you in the batch you typically
25:41.500 --> 25:47.039
make sure that you have enough substrate but
there are situations when this can arise and
25:47.039 --> 25:54.360
this also has great relevance in the continuous
bioreactor analysis so let us start looking
25:54.360 --> 25:59.320
at at here where it has some application and
then picking it up for the continuous case
25:59.320 --> 26:07.940
becomes a little
easier also we have two quantities here x
26:07.940 --> 26:14.140
and s that vary with time in the differential
equation and of course you know that it is
26:14.140 --> 26:21.700
preferable to have only one dependent variable
either x or s not both that becomes difficult
26:21.700 --> 26:28.440
here so if you can convert one in terms of
the other s in terms of x preferable because
26:28.440 --> 26:34.510
we are looking at x we have x is all over
if we can express s in terms of x then we
26:34.510 --> 26:41.220
can solve this equation how do we do that
some thoughts on that something related to
26:41.220 --> 26:49.140
some concepts that we have seen earlier in
the in the course in module two yes we could
26:49.140 --> 26:55.480
use what we called what we defined as an yield
coefficient in the earlier lecture or in the
26:55.480 --> 27:00.270
earlier module that relates the cell concentration
and the substrate concentration
27:00.270 --> 27:08.740
we know that y x is x slash s yield of
cells with respect to substrate it's nothing
27:08.740 --> 27:14.270
but the amount of cells produced by the amount
of substrate consumed in terms of the values
27:14.270 --> 27:19.760
that we have defined here x minus x naught
is the amount of cells the concentration of
27:19.760 --> 27:25.300
cells produced by the concentration of the
substrate consumed s naught minus s in a concentration
27:25.300 --> 27:30.809
is amount by unit volume amount by unit volume
since we are dividing one by the other and
27:30.809 --> 27:35.860
the volume happens to be the same you can
cancel out the volumes so the amount of cells
27:35.860 --> 27:41.790
produced by the amount of substrate consumed
can be replaced on both the numerator and
27:41.790 --> 27:46.850
the denominator by the concentrations and
we would have the same yield coefficient ok
27:46.850 --> 27:54.600
that's what we have done here and the assumption
that y x is a is a constant is a good assumption
27:54.600 --> 27:56.230
under many different conditions
27:56.230 --> 28:06.090
that is a case we can use the yield coefficient
to write s in terms of x s naught is a constant
28:06.090 --> 28:11.960
s naught is the total substrate concentration
that is usually known similarly x naught is
28:11.960 --> 28:18.260
the initial cell concentration assuming
this not much of a change in the in the
28:18.260 --> 28:23.640
lag phase that is also known so we are actually
interested in the variables x n relating the
28:23.640 --> 28:30.260
variable s and x so s equals s naught minus
x minus x naught by y x s directly from the
28:30.260 --> 28:39.580
definition here therefore the differential
equation becomes this mu has been written
28:39.580 --> 28:52.799
in terms of um this conversion here mu m s
by k s plus s the s has been replaced with
28:52.799 --> 28:58.559
s naught minus x minus x naught by y x's that
is what happens here similarly k s plus s
28:58.559 --> 29:05.490
the s has been replaced here therefore
this times x and seems a little formatable
29:05.490 --> 29:11.669
it takes time to solve that's all the step
by step solution will take a very long time
29:11.669 --> 29:16.760
to show especially context of this course
this is a total of ten hours this is a probably
29:16.760 --> 29:22.789
given to take forty five minutes to show you
all the algebra so if you are interested and
29:22.789 --> 29:27.029
comfortable in mathematics i would suggest
that you work it out when you have about a
29:27.029 --> 29:31.690
half an hour and so if you good in math it
wont take more than that you might also need
29:31.690 --> 29:36.380
a table of integrals for the part of the
solution you dont want go and derive the well
29:36.380 --> 29:41.279
known results ok so you need to use some well
known results also from the table of integrals
29:41.279 --> 29:48.289
ok here i am just going to present the final
solution we need to take it on phase value
29:48.289 --> 29:53.720
if you are doubt full you go head and solve
it and convince yourself there indeed the
29:53.720 --> 29:59.159
solution the total time for the start of the
batch from the start of the batch to reach
29:59.159 --> 30:08.200
a certain cell concentration x is given as
one by mu m times a certain alpha times log
30:08.200 --> 30:15.600
of x by x zero minus a certain beta times
log of y x's s naught plus x naught minus
30:15.600 --> 30:26.790
x by y x's s naught plus the lag time t zero
where alpha is this combination k s y x's
30:26.790 --> 30:35.350
plus yx's is zero plus x zero divided by y
x's s zero plus x zero and beta is k s y x's
30:35.350 --> 30:41.711
y x's divided by y x's is zero plus x zero
it will take quite a bit of algebra to show
30:41.711 --> 30:47.600
this if you are good at math interested
please go ahead and show it to a convince
30:47.600 --> 30:55.250
yourself that it is the indeed the case ok
i think it's a good time to stop here
30:55.250 --> 31:01.450
for this lecture when we meet next i will
solve the problem that is been assigned but
31:01.450 --> 31:08.010
please make all attempt solve it you need
to pickup the skills of close ended problem
31:08.010 --> 31:15.090
solving atleast and if you follow the method
some time soon you would be able to we able
31:15.090 --> 31:17.980
to solve complex problems see you then