welcome to this second lecture on bioreactors this is a mock on bioreactors in the last
lecture previous lecture we looked at module one which is introduction after mentioning
the formalities of the course we looked at some day to day examples such as cancer
concerns day to day concerns cancer we looked at what cancer was and how cancer
is treated allopathically and then we said that monoclonal antibodies are used to target
the drugs that kill cancerous cells more directly to the cancer cells these monoclonal antibodies
when required in large amounts are produced through by reactors or bioprocess and that's
how we started then we looked at other examples such as alternative liquid fuels bio ethanol
and bio diesel and saw that they were also produced through bio processes had given you
links to some information on ethanol making ethanol from corn algae to fuels and so on
i hope your seen them they are interesting videos then we looked at insulin which is
a classic drug infact one of the first ones to be produce large scale one of the first
new ones to be produced large scale using bio processes and finally we looked at
curd making and said curd making is also a bio process the vessel in which the curd is
made is nothing but a bioreactor of course it's a simpler form of a bioreactor a bioreactor
is nothing but a vessel highly instrumented and controlled vessel in which bio reactions
take place and bio products are made normally with the multiplication of sets then we looked
at the over view of the bio process we said that it has two major aspects the bioreactor
aspect or the upstream aspect and the downstream processing aspects we said that the raw material
goes into the bioreactor which contains the micro organism and nutrients for it to grow
and the produce the product of interest or it could be an enzyme which is doing an enzymatic
conversion and what is harvested from the bioreactor these liquids stream that is
harvested from the bioreactor contains the product of interest along with the micro organism
and the other material that arose from these two since we are interested only in the product
we need to remove that from the other material and that is what is done through downstream
processing steps to achieve the final purified product then we looked at common bioreactor
types some common bioreactor types we said that the stirred tank bioreactor is a one
of the common types it's nothing but a stirred tank and then an air lift bioreactor
a packed bed bioreactor fluidized bed bioreactor which is nothing but a packed bed with
gigues around there are some advantages there there was a video which lead you to some
detail on the fluidized bed bioreactor and then solid state bioreactors single use bioreactors
that are helpful in meeting the stringent demands or stringent regulatory procedures
for the production of any biological material then also a photo bioreactor that's typically
used with micro algae and other photosynthetic organisms we said there are many other kinds
of reactors these are the typical ones there could be other kinds too then
we said that any bioreactor or many bioreactors where each bioreactor can be operated in three
basic modes either a batch operation where we dump everything in wait for the process
to complete and dump everything out and proceed with further processing or a continuous operation
where there is a continuous input a continuous output the growth product formation flow of
nutrients all happens all happen simultaneously then a fed batch operation which is in between
batch and a continuous operation where there could be intermittent input or output
or both both intermittent output and the intermittent input at the same time some reactors we said
nicely fit into all three modes of the possibilities of operating them in all three
modes of operation some of them don't this is think where we finished up the last lecture
we said that whenever a bio processes involved we typically need a clean slate we want only
the organisms of our interest to grow in the bioreactor and therefore the bioreactor should
be a clean slate when we begin the ways of achieving a clean slate is either through
high temperature or a chemicals vapors or through radiation such as u v and gamma which
kills all the organisms that are present in the bioreactor and then provides us with the
clean slate and then we introduce the organisms of our interest which grows a multiplies produces
the product of interest so let us move further in this lecture from this point we will look
at high temperature in some details of achieve a a clean slate or sterilization as it is
called using high temperature this is specifically thermal sterilization the one that uses high
temperature if we plot the percent viable cell concentration versus time as is shown
here this percent viable cell concentration
we are plotting on a log scale hundred ten one point one at a temperature of fifty which
is typically a higher temperature then many organisms grow at thirty degree c thirty
seven degree c we sum at twenty twenty seven oh sorry twenty five twenty seven and so on
so fifty degree c is some what high for most organisms there are some thermophiles at grow
well at high temperatures but let us not consider them for the time being if it is at fifty
degree c then the viable cell concentration decreases linearly with time at fifty degree
c at sixty degree c again the response is linear but it is steeper it gets killed faster
and similarly at seventy degree c the response is linear but the slope of this curve is larger
compare to the one side fifty degree c or sixty degree c this is the typical response
that is absorbed when viable cells are exposed to high temperature temperature that
is higher than their normal operating temperature this kind of a behavior is applicable for
a suspension of single cells cells separately not clump together and so on and so for and
what this graph tells us is that the logarithm of the viable cell concentration know this
is on a log scale so the logarithm of the viable cell concentration percentage or a
fraction is you know that fraction times under is percentage so fraction is directly proportion
to the time of exposure at the high temperature so this is the kind of behavior that we see
when we expose single cell suspensions to high temperature so how we going to use
this before we look at how to use this lets us understand this a little better so that
it becomes easier to use this at any required situation in any required situation this
kind of a relationship that we saw earlier results when the rate of decrease in viable
cell concentration is directly proportion to the viable cell concentration present at
any time now let me repeat this the rate of decrease of viable cell concentration is directly
proportional to the viable cell concentration present at any time for example this is what
is what we mentioned here but written in mathematical terms the rate of decrease
in concentration of viable cells is proportional to the viable cell concentration x subscript
v
if we call the rate r d the rate of decrease in concentration as r d since it is proportional
we can replace the proportionality by equal to sign and here we need to multiply x v with
the constant let us call this constant k d you might be familiar with this kind of a
first order expression in your earlier courses so we are using a first order kind of a representation
here note that this r d is rate of decrease in concentration which is the concentration
decrease per unit time mass or number of cells per unit volume per unit time that is the
unit here therefore the bases is the concentration bases we need to remember this now lets us
light let us write a balance on cells taking the bioreactor as the system you all gone
through a course on material balances so we are going to write a material balance on cells
and you all know that a balance can be written over a region of focus and that is called
the system region of focus is called the system in this case the bioreactor broth the liquid
part in the bioreactor in which the cells grow that is what we are going to take as
the system and we are going to write a balance on cells on that system you might recall this
equation here the rate of input of the cells into the system which is the bioreactor
broth minus the rate of output of cells from the bioreactor broth plus the rate of generation
of cells in the bioreactor broth minus the rate of consumption of cells in the bioreactor
broth equals the rate of accumulation of cells in the bioreactor broth ok this if you
consider all that can happen to this species cells this is all that can happen you can
either comment to the system go out of the system get generated in the system or get
consumed in the system and the net result of all these things must equal the accumulated
rate or accumulated amount expresses the rate here because all these are rates this
also a rate also important to note that this is on a mass bases this comes from the principle
of mass conservation and that principle is what allows us to write this equal to sign
mass is conserved mass can be either be created nor destroyed we are of course not considering
either nuclear reactions or travelling at the speed of light and in such situations
the mass of a species is conserved the mass of the species before a process equals the
mass of species after the process so that is the bases for this if it's a
not clear when you students take a while to understand this so let us a review the basis
for this equation you may have done some of this in your first course and material balances
so you could take this as a review and try to understand this to the extent needed to
be able to manipulate or write balances according to our need to do that let us consider
a situation that we are all in chennai please very familiar with in summers now this
is the use of water tankers what are unreserved then water tankers are used to get water
for daily needs let and the typical size of a water tanker could be around twelve thousand
liters know this is my representation of a water tanker here i would like you to use
your imagination to see how this is water tanker this is the back of the lorry on which
sea water tanker rest and let us say that this is the filling of the water tanker
with water at it's source
so this typically is either eight thousand liters ten thousand liters twelve thousand
liters or sixteen thousand liters let us consider a tank of twelve thousand liter capacity this
is the case what is the mass of the water in the tank all of you know that if we know
the density we can find out the mass of the water in the tank and the mass happens to
be twelve thousand kg because the density of water is one gram per cc or thousand kilogram
per meter cubed one kilogram per liter and therefore twelve thousand liters comes out
to be twelve thousand kg now the question is how long would it take to fill a tank to
fill this tank twelve thousand liter tank that's the question so people who remember
their material balance scores or who are comfortable with this will immediately ask
the question what is the input rate of water that fills the tank if you know the input
rate of water then you can figure out the time taken to fill the tank quite easily if
the input rate happens to be ten kilogram per second the time would be twelve thousand
sorry thousand two hundred seconds or a twenty minutes if the input rate is twenty the time
would be six hundred seconds or ten minutes if the input rate is fifty kilogram per second
the time would be two forty seconds or four minutes how did we get at this if we know
the rate of water
input the time by the mass is nothing but mass divided by the rate and that is what
this turns out to be for example twelve thousand kilogram by ten kilogram per second
gives you twelve by thousand two hundred seconds which is twenty minutes and so on
a typical time in which the tank gets filled is about ten minutes and therefore let us
choose this as the input rate for further discussion note that we are talking of rates
now let us complicate this process a little bit suppose there is a hole in the tanker
which oozes water at the rate of five kilogram per second how long would it take to fill
the tank the if you recall your course you would say if i know the net rate i can very
easily find the tank so that is the advantage here if you work in terms of rate all these
very standard questions become easier to calculate if you work in terms of volumes
in it becomes rather difficult to see what is happening and so on so for therefore we
work in terms of rate and that's why rate is considered some sort of a standard parameter
in a while dealing with the anvex systems engineering systems it's so on will look that
in a little while in a little while the net rate in this case there is a ten kilograms
coming in five kilograms going out per second ten kilograms per second coming in sorry
i think is twenty kilograms per second coming in five kilograms per second going out so
the net rate happens to be fifteen kilogram per second and once you know the rate you
can find out the time as the volume divided by the net rate or the in this case the
mass divided by the net rate this has to be m twelve thousand kg is divided by fifteen
kilogram per second or eight hundred seconds or about thirteen point three minutes is what
we get
now let us complicate this in a creative fashion suppose that in addition to the leak there
is some mechanism inside the tank it'self that is generating water at one kilogram per
second ok and some other reaction in which the water is used up inside the tank at point
two five kilogram per second all of which simultaneously occur ok this is the crucial
line here they all simultaneously occur if all these simultaneously occur how long would
it take to fill the tank here comfortable with rates by now so you ask me what is the
net rate when if i know the net rate i can find it out how do you find the net rate the
net rate is rate of input minus rate of output plus rate of generation minus rate of consumption
rate of input is twenty kilogram per second rate of output is five kilogram per second
that we already seen now the rate of generation here water is being generated at one kilogram
per second and it is being consumed by a reaction at point two five kilogram per second therefore
the net rate turns out to be fifteen point seven five kilogram per second so this is
the rate at which water gets accumulated inside the system or inside the tank the rate of
change of water mass with time in the tank so that will turn out to be twelve thousand
kgs divided by the net rate which is fifteen point seven five kilogram per second which
turns out to be seven sixty one point nine seconds or twelve point seven minutes
now i gave you this background two are equation rate of input minus rate of output plus rate
of generation minus rate of consumption equals the rate of accumulation and the rate of accumulation
we write as the derivative so that the form is easy to directly use during heat sterilization
of cells you know you are sterilizing let say a stirred tank bioreactor containing medium
it which probably has some cells growing in it already it's not sterilized and we are
going to heat sterilize and kill all the cells that are present there during that process
the cells are being killed in the bioreactor broth it is a system nothing as is happening
ok there is no input there is no output and so on so for therefore the rate of input of
cells mass of cells into the system is zero there is no output of cells from the system
the rate of output of cells again mass of cells or number of cells is zero there is
no generation of cells in the system or let assume that therefore the rate of generation
of cells is zero and therefore the only term that remains from r i minus r o plus r g minus
r c is this minus r c and minus r c the rate of consumption of cells r c is equal to d
d t the rate at which mass of cells gets accumulated in the system this is what we get by considering
it from first principles
the consumption of cells is through the death of cells as they are getting expose to the
high temperature or this is the heat sterilization that we are taking about therefore r c which
is the rate of consumption of cells on a mass basis is r d which was on a volume basis remember
that was a sorry that was on concentration basis a concentration is nothing but mass
divided by the volume therefore to get mass from concentration you need to multiply concentration
by volume therefore to get a rate that is mass based from a rate that is concentration
ra[te]- based you need to multiply the concentration based rate by volume therefore this r d into
v which is the rate of a consumption of cells or the rate of death of cells on a mass
basis this minus comes from our balance equation equals the accumulation rate of the
mass of cells inside the bioreactor when it is being sterilized when is r d into v equals
d d t of m x now we have everything on a mass basis which is nice to have now from the definition
of concentration mass is nothing but concentration times volume or in another words concentration
is mass by volume that's the definition and therefore m x which we have here is x v the
concentration of viable cells into the volume if we use this to replace this m x then we
get this minus r d into v equals m x as x v into v therefore d d t of x v into v and
in this case since a volume is a constant volume of the broth volume of the system that
is the constant we can take volume out of the derivative v into d d t of x v thus if
we now we have a volume here we have a volume here we can cancel them out there is
same volumes therefore minus r d equals minus k d x v which is what we had seen earlier
we had represented r d as first order expression k d into x v the rate of death
of cells equals constant times the viable cell concentration that's the first order
concentration that equals d x v d t since we have we have cancel the volumes here
ya this is the final expression here d d t of x v equals minus k d times x v this is
the first order differential equation which you are all familiar with if we solve this
first order differential equation we get lon or natural log x v naught by x v equals k
d times t time where x v naught is the concentration when we began the process of sterilization
heat sterilization and x v is the viable cell concentration at any time t or the way to
look at this from perspective by design perspective is the time needed for the viable cell concentration
to go down to x v starting from x v naught when the process started is nothing but i
am converting the natural log to log base ten you know that the conversation factor
is two point three naught three therefore natural log x v naught x v is two point three
naught three log to the base ten x v naught x v and k d comes to the denominator here
therefore the time required for this amount of a decrease in viable cell concentration
from x v naught by x v is this expression two point three naught three k d divided by
k d log to the base ten of x v naught by x v the time taken for a ten fold reduction
in the viable cell concentration now when if it was a initially a certain value it is
gone down to one tenth of water towards initially that is the ten fold reduction in viable cell
concentration at any given temperature this is an important parameter for the design of
thermal sterilization ok we are figuring out means by which we collect the information
so that it can be used else where without much information to design the sterilization
process to design the sterilization equipment and so on this time taken for ten fold reduction
in viable cell concentration is called the decimal reduction time typically expressed
as capital d in terms of whatever we have derived at ten fold reduction in viable cell
concentration essentially means that the initials viable cell concentration x v naught divided
by the viable cell concentration at that time t x v must be ten straight forward and if
we substitute this in the expression for time that we had derived earlier this expression
the time taken for the concentration to go from x v naught to x v then we would get an
expression for the decimal reduction time in terms of the variables that we would like
two point three naught three divided by k d log ten to the base ten you all know log
a to the base a is one and therefore this reduces to two point three naught three by
k d
recall that the above analysis is applicable only to a suspension of single cells for which
the rate of kill is directly proportional to the viable cell concentration at any given
time remember this has been observed and then this observation we look at representing
mathematically and we said that the first order decrease would be good way to represent
this that is the rate of decrease of cell concentration at the high temperature is directly
proportional to the viable cell concentration at any time and that is what resulted in the
earlier mathematical expression that we had mathematical expression always generalizes
things it makes it a lot more useful in situations that has not been seen earlier relevant situation
that have not been seen earlier and thereby it leads to design of a sterilization processes
in this case
that was for a suspension of single cells there are other death behaviors that are non
linear let us see whether you are able to guess this first this is a plot of again percent
viable cell concentration hundred ten one point one therefore on a log scale versus
time and can you guess what would give a curve like this earlier it was linear a linear decrease
in this case the decreases something like this ok it doesnt decrease much for quite
a while and then it starts decreasing can you guess what kind of content or viable
cells viable cells at what form would give raise to this the answer is clumped cells
now cells clump with each other then it takes a while for the killing of the internal
parts of the clump compared to the outer parts of the clump so that results in this kind
of a behavior for a plot of percent viable cell concentration versus time let us look
one other non linear behavior this for the variety this is an another non linear vari[ety]-
behavior as you can see it is represented as two lines from hundred to lets say point
two or something like that this is one line and then there is another line here with
a difference slope compare to this when do you think such a behavior would be seen
in a wave already seen that this happens a suspension of single cells only thing is that
we have two lines with different slopes ok so this could arise when you have mixed population
of two different cell types the type differing the the difference in types arising from their
response to a high temperature this decreases these type of cells decrease fast these types
of cells decreases slow the viable cell concentration versus time and this is what this is a
kind of behavior that arises when you have a mixed population of two different cell types
ok now i am going to present the first practice problem this is let me call at one point
one because this is module one that is what the first number indicates and this is problem
one of module one therefore one point one i am going to assign this problem to you and
you can solve it you need not submit this problem what we will this will not count
toward evaluation for this mock however i believe that you need to pick up these skills
of problem solving therefore what i am going to do is when we begin the next lecture i
am going to solve this first for you and then go forward ok the practice problem here reads
a bioreactor needs to be sterilized before use the solution in the bioreactor consist
of single cells with similar thermal response characteristics ok so it is going to be a
single line that's dropping at seventy degree c it takes five minutes for the viable cell
concentration to reduce to twenty percent of it's original value a determine the decimal
reduction time remember we said it was an important parameter so determine that first
determine the decimal reduction time and be how long would it take for the viable cell
concentration to reduce to point one percent of it's original value under the same conditions
ok this is a problem for you you need to know how to solve problems we will probably
touch up on that when we solve the pro[blem]- or atleast i will show you some thinking behind
the solution when we begin the next lecture therefore this is what we did in the introduction
i guess we went through the earlier parts of the introduction when we began this lecture
essentially some examples of a bioprocesses and what exactly is a bioprocess and that
the bioreactor is at the heart of the bioprocess
different types of bioreactors are commonly used some commonly used bioreactors we saw
such as the stirred tank bioreactor the air lift bioreactor the solid state bioreactor
single used bioreactors photo bioreactors packed bed bioreactors fluidized spread bioreactors
and so on there are many different types bioreactor is nothing but an instrumented
controlled vessel in which bioreactions take place typically mediated by cells or it could
be an enzyme that mediates the reaction that's exactly what a bioreactor is then we said
that any bioreactor can be operated in different modes some bioreactors lend them cells to
operation all the three basic modes and some don't for example a stirred tank bioreactor
can be operated in a batch mode in a continuous mode as well as in a fed batch mode whereas
may be a packed bed it's a little difficult to operate only in a batch mode
you can but it takes a lot more effort rather it it it is normally not operated in
a batch mode it is operated in a continuous mode then we looked at the need for
a clean slate that is we want only our organisms to grow and produce the product of interest
and therefore we need to get of all other organisms with that are normally present we
said that organisms are present in the air around us all the time and nothing prevents
them from getting into the bioreactor broth that is open to the atmosphere and therefore
we need to kill all the cells that are initially present there the main modes of killing
include thermal killing a high temperature can be used to kill cells or could use a appropriate
vapours or liquids seventy percent ethanol solution is used to kill cells on your hands
when you when you handle these organisms cell cultures and so on or a formalin vapour
formaldehyde vapours from formalin solution is used to kill the organisms in a space such
as a lab and then you could also use gamma rays and so u v rays gamma rays gamma rays
especially if you are sterilizing lets say syringes and things like that that cannot
be expose to high temperature there might of plastics they cannot be expose to high
temperature so those three modes of killing then we said we will look at the thermal
sterilization and some detail we reaffirmed the need to be clear about the concept of
rate as a basic parameter in any engineering system analysis dynamics system analysis the
rate by it'self is
important the way you measure rate could be very different for example you have speed
which is equivalent of rate and you could measure an average speed by the distance travelled
by time taken but speed is rather central to the kind of to answer the kind of questions
that we would be interested in when we deal with design operation of engineering systems
in this case biological engineering systems and then using rates we said that the rate
of killing of cells by high temperature could be directly proportional to the viable
cell concentration at a particular time of first order kind of expression then we
saw that lead to the definition of a decimal reduction rate is standard parameter and
then also a more general expression before that that gives the time of time that it takes
to go from let us say an initial concentration of x v naught initial viable cell concentrate
of x v naught to certain viable cell concentration of x v at a particular time t then we looked
at this practice problem here ya ya before that we did see that this linear killing
is only one kind of a behavior you could have other kinds of killing such as a curve when
there are clump cells and combination of two straight lines with different slopes
if you have a mix population of thermally different organisms then we looked at this
practice problem you might want to spend some time and solving this it will be helpful
these are the kind of things that engineers are expected to do and it will be good to
pick up the skills when we meet next we will solve this problem and start module two
see you in the next lecture