welcome to lecture number eleven the nptel online certification course on bioreactors
in the last lecture we had a assigned problem a practice problem three point
one let us solve that in this particular lecture the problem read in a in a batch bioreactor
the concentration after inoculation was point five gram per liter the lag phase usually
last twenty minutes under these conditions assuming that the cell concentration at the
start of the lag phase the start of the log phase was not significantly different from
that immediately after inoculation a estimate the time needed for it to reach four gram
per liter the the specific growth rate for this organism under these conditions is point
five hour inverse and what is the time needed for the cell concentration to double
in the log phase is part b so let us look at the solution let me maximize this first
and start looking at the solution we will ask her same questions for a close ended problems
solving what is needed what is needed here is the in part a is the time required to reach
four gram per liter in batch growth what is known or giving the cell concentration just
after inoculation is point five gram per liter therefore that can be taken to be x naught
we are assuming that we are assuming that there is the not much of a change in cell
concentration during the lag phase therefore x naught which is the cell concentration
at the beginning of the log phase is also taken to be point five gram per liter the
specific growth rate in the log phase is given as point five hour inverse which we are taking
into the constant the lag phase is a twenty minutes which is t zero and those are the
things that are known the so we have seen what is needed and what are known are given
now how to connect what is needed with what is known we have already derived the equation
relating the cell concentration and time in batch growth in the previous lecture the equation
was this one by mu lon of x by x naught plus t zero equals equals the t time so we need
time here if we substitute the known values we get one by point five mu mu is constant
point five hour inverse lon four by point five plus t not is twenty minutes and hour
specific growth rate was in hour inverse so we need to work with the consistence system
of unit's in other words the unit's of each of these terms must be the same ok if we check
that we will need to convert the twenty minutes into hours and therefore we are dividing that
by sixty twenty minutes by sixty minutes per hour is one third hour that that equals
the time that is required we have checked for the consistency in unit each term having
the same system of unit's and the need for twenty by sixty we saw it's always good
to check for check the unit's of each term in an equation and ensure that it is the same
if we calculate this left hand we will get t equals four point four nine hours or two
sixty nine point three minutes so this is part a the time that is required to reach
a concentration of four gram per liter starting with the concentration of point five gram
per liter when the specific growth rate was point five hour inverse
for part b what is needed the time for the cell concentration to double in the log phase
what is known or given this written in terms of hour terminology would be x equals two
times x naught the specific growth rate is point five hour inverse and what is how to
connect what is needed to what is given since we are concentrating only on the log phase
the lag time is really irrelevant here we don't have to worry about it therefore one
by mu lon of x by x naught equals t we don't have to worry about plus t naught with this
actual the time form the beginning of the batch so we are concentrating only on the
log phase here if we substitute the known variables one by mu is point five x of
requirement here is two x naught by x naught x naught x naught can get canceled this is
lon two therefore lon two by point five equals t lon two is point six nine three one by point
five lon two point six this is a nice number to remember
point six nine three point six nine three by mu is a time that it usually takes to double
so that would turn out to be t equals one point three nine hours so this was a straightforward
solution i am sure many of you would have a gotten this and with practice we will become
much better i think we will stop this lecture this is the short lecture which kind of a
looks at the solution of the problem when we begin the next lecture we will take module
three forward see you then