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