welcome to lecture number three in this n p t e l online certification course on bioreactors
at in our course in the first two lectures we got introduced to what bioreactors where
what kind of products they make what are the common types of bioreactors that i used what
are the modes of operation predominantly batch continuous and semi batch or fed batch and
then we started looking at some details the first aspect that we looked at was how to
have a clean slate because we said we kill all organisms that are there we create a clean
slate and then we introduce the organisms of our interest into the bioreactor which
produces the product of interest to us to achieve the clean slate there are three
methods possible three main methods possible temperature based or chemicals based chemical
weapons and so on or radiation based gamma u b any things like that then we saw thermal
sterilization the temperature based aspect in some detail so that we had enough information
to we able to design a priory this sterilization process thats a whole point about analyzing
and so on it also gives us a lot of insights but the practical aspect is to be able to
design appropriate processes let me open this particular presentation this is what
we saw the thermal sterilization the response of viable cells to high temperature we have
on the y axis the percent viable cell concentration on a long scale versus time on the x axis
and the relationship is linear when this on a long scale also we said that the decrease
the slope of the decease increases has the temperature increases this is the kind
of response that we were trying to analyze we said that we could analyze this situation
if we considered the rate of decrease in concentration to be directly proportional to the concentration
of viable cells a sort of a first order relationship using that and the fact that this operation
is happening in a batch by a reactor we wrote a balance for it and arrived at get there
this was the governing equation d d t of x v is minus k d into x v and if you solve this
we would get lon of x v not by x v equal k d t that gives you the time there the time
needed for the viable cell concentration to go down to x v starting from x v naught is
two point three naught three by k d log to the base ten x v naught by x v as we can realize
this is natural log log to the base e when we convert that to log to the base ten you
have a factor of two point three naught three but otherwise this is essentially this equation
transports the main thing here is that we have the time that is necessary for the viable
cell concentration to go down from x v naught to x v under certain set of conditions which
is characterized by this death constant k d or this sterilization constant k d so this
is what we arrived at and then we also looked at something called a decimal reduction rate
which is the time that is required for a ten fold decrease in viable cell concentration
at a given temperature for ten fold reduction all we needed to do
was replace x v naught by x v have to be ten and then we got the decimal reduction time
d as two point three naught three by k d at this point a problem was assigned the practice
problem one point one it red a bioreactor needs to be sterilize before use the solution
in the bioreactor consists of single cells with similar thermal response characteristics
at seventy degree c it takes five minutes for the viable cell concentration to reduce
to twenty percent of its original value a determine the decimal reduction time and b
how long would it take for the viable cell concentration to reduce to point one percent
of its original value under the same conditions this was the question that was or the problem
that was assigned when we pretty much finished up the previous lecture lecture number two
i hope you would have a chance to work this out i will as promised earlier work
out problems that are assigned in the beginning of the next lecture we will start this lecture
by solving this problem first let me first talk a little bit about problem
solving problem solving is a higher level skill its a same way as somebody being able
to either sing well or dance well or play a game well or come what may you think of
skills and you have a certain association with it similarly problem solving is also
a skill its acts is a higher level skill in the cognitive domain some people would be
naturally good at it as have some people who a who have a natural ability in a music in
dance and so on so forth if you have the natural ability then this initial part or the details
of the problem solving may not be very necessary for you you are already had a good level
i think you should continue with whatever you are doing assign do the problems that
are assigned and you should be fine however in my experience i found that most
people do not have problem solving as that natural skill it is necessary for us to pick
up that skill because it is necessary for this profession and if you dont have a
natural skill but you need to do something then you need to pick it up to pick it up
of course we practice we put an effort and then we get to a certain level of be able
to do problems i am sure with the consistent good effort you would be able to solve problems
we will leave it ourselves to what are called closed ended problems in this course closed
ended problems are problems in which everything that is needed for the solution of the problem
is either given in the problem statement or is known through material in your notes and
so on so forth this is distinct from open ended problems that may not have all the
information that is required state it in the problem or known might need to extend a few
things and then get information from other sources literature and so on and then solve
the problem infact in my regular courses i do assign as
long open ended problem solve a problem to be solved over the entire length of the semester
its a very good learning experience for the students this format does not allow us do
that effectively so let us at least look at close ended problem solving for this course
if you are interested in knowing or in developing the problem solving skill in a
systematic fashion you may want to look at this book this is a the title of the book
is strategies for creative problem solving the authors are scott fogler and steven e
leblanc this is a good book you might want to take a look at that book if you want to
know a little more about problem solving if you really want to get into it and become
an expert with sustained effort with that let me go on to how we are going to go about
closed ended problem solving the first question to ask is what is needed what does the problem
want us to find once that is clear write it down then ask the question what is given or
known through the problem through the material that is relevant to the problem that was covered
in the lectures and so on once both those are clear then we will ask the question how
do we connect the first one with the second one how do we connect the needs of this
particular problem with givens are knowns in this problem and while doing that we will
also ask are there any basic principles that we can rely on to be able to solve this problem
in fact while presenting the solution i am going to use only the first three what
is needed what is given or known how do we connect the needs with givens and knowns and
take this as a part of the third aspect thats how we are going to do it now let us go about
solving the problem that is we already read this we need to find the decimal reduction
time in part a and in part b how long would it take for the viable cell concentration
to reduce to point one percent of its original value some information is given
the solution in the bioreactor consist of single cells with similar thermal response
characteristics therefore you can assume linear dependence of the log of the viable
cell concentration with time as we saw in class so in the lecture and also its says
that at seventy degree c it takes five minutes for the viable cell concentration to reduce
to twenty twenty percent of its original value ok having seen this let us go about the solution
first question as we said was what is needed let us take one part of it after an other
part a the decimal reduction time so that is clear so let us ask the second question
what is known or given we know that it is a single cell suspension we discuss this earlier
too and also we know that at seventy degree c it takes five minutes or three hundred seconds
for the viable cell concentration to go from hundred percent to twenty percent or in other
words a five fold reduction takes a takes the three hundred seconds we see that i have
converted this five minutes into three hundred seconds ok
it is usually good to work in a consistent system of units whatever that might be and
for the purposes of this um course we will mostly use the s i units and thats the reason
why i have converted five minutes to three hundred seconds it is good to do this a priory
and then go about the actual solution or before hand and go about actual solution then
the third main question how to connect what is needed to what is given if you think about
it i am sure quite a lot of you would have solved this you think about it we have derived
the decimal reduction time which we also review a few minutes ago as two point three naught
three by k d therefore we need k d if we have k d we can find out d which is the decimal
reduction time which is what is you required in part a
how do you find k d to do that let us go back and look at the basis for the derivation of
decimal reduction time we i think we went through it during the review process
so let me just say this gives the time for the reduction and viable cell concentration
from x v naught to x v we derived the time to be two point three naught three by k d
log to the base ten x v naught by x v we have the information on the time taken for x v
naught by x v to be five a nother was the five fold reduction form hundred percent to
twenty percent so that if you substitute here in in otherwords this is becomes five right
so we we need five minutes or three hundred seconds for a five fold reduction therefore
three hundred equals two point three naught three by k d log five to the base ten so k
d is the only variable here and k d if we tansforces or essential multiply both sides
of the equation you can do the same things to both side of both sides of the equation
that you know if we multiply this side by k d and this side also by k d we get
and divide this side by three hundred and this side also by three hundred we get k d
equals two point three naught three by three hundred log five to the base ten which turns
out to be five point three seven into ten power minus three seconded percent
i would like you to verify this particular answer if you have this answer fine but please
verify i have deliberately not verified my answers during the solution process alone
i have done that for the exam questions and so on but during the in a demonstration
of the solutions of the problems the practice problems that are assigned i have deliberately
not verified my answers if you find an error please point it out that will be a nice exercise
for you now that we have k d we can find the decimal reduction time as two point three
naught three by k d which turns out to be two point three naught three by five point
three seven written power minus three or four twenty eight point nine seconds because this
was in second inverse we get this in seconds if we divide this by sixty we get seven point
two minutes ok therefore it takes seven point two minutes for ten fold reduction in the
viable cell concentration under these conditions now let us consider part b again will ask
the same questions what is needed the time taken for the viable cell concentration to
reduce to point one percent of its original value under the same conditions what is known
or given single cell suspension the same as earlier at seventy degree c it takes three
hundred seconds for the viable cell concentration to go from hundred percent to twenty percent
and also we found in part a the value of k d and the value of the decimal reduction time
this is what is known how do we connect what is known to what is given the time for reduction
and viable cell concentration from x v naught x v we have already derived and seen earlier
also that equals two point three naught three by k d log to the base ten of x v naught by
x v when the viable cell concentration reduces to point one percent of its original value
this is the condition that is of interest to us in problem b we know that x v naught
by x v sorry x v by x v naught is point one if x v by x v naught is point one percent
this is this is percent if we converted into fraction this will turn out to be point one
by hundred you know percentage is always out of hundred so fraction is when we need
the fraction we need to divided by hundred this is point naught naught one or ten part
minus three so x v naught by x v if we flip it around
it will be one by ten power minus three or ten power three and thus the time required
for the viable cell concentration to reduce to point one percent of its original value
would be this substitution of the various know aspects now into this equation two point
three naught three by five point three seven to ten power minus three log to the base ten
of ten power three that answer to be one two eight six point six seconds or twenty one
point four minutes this is a one way to solve the problem i have shown you all these steps
hopefully you understood this steps and if you be acted if its not natural to you
and if you be acted i am sure you will be able to solve closed ended problems i think
we will finish of lecture three with this these lectures would be of variable times
they will all add up to around ten hours the each lecture would be a variable time we
will meet in the next lecture lecture four see you there