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