WEBVTT
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welcome to lecture thirteen the n p t e l
online certification course on bioreactors
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we are towards the end of module three we
had sign a problem in the last class we
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will work that out and with with that we can
complete module three the problem statement
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reads as processed fungi are good biosorbents
for toxic trace metals and thus they can be
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used to remove chromium mercury cobalt and
others from industry effluents a suitable
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fungus needs to be produced at five hundred
grams per hour for the above purpose that's
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the production rate the growth limiting substrate
concentration at the inlet of a chemostat
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to produce fungus is fifty grams per liter
the fungus follows monod kinetics with the
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maximum specific growth rate be equal to point
five hour inwards the substrate concentration
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that corresponds to the half maximal growth
rate is one gram per liter it is aerobic growth
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with the yield cell yield coefficient from
substrate of point five there is y x slash
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s is point five find the minimum size of a
chemostat needed for the above for a given
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inlet flow rate
the solution is thus follows we will ask our
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same questions and tried to answer them what
is needed the minimum size of a chemostat
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needed for the above given condition for a
given initial for given inlet volumetric
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flow rate what is know or given the production
rate is five hundred gram per hour that is
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given the maximum specific growth rate
is point five hour inverse the k s is one
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gram per liter y x s is point five all these
are given and also s i the inlet substrate
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concentration in the feet is fifty gram per
liter monod kinetic for growth is given now
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how do we connect what is needed to what is
given
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if the volume needs to be minimized we need
to be operating at the maximum productivity
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r m ok that's what sense only if the productivity
is maximized can the volume be minimized to
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get the best out our inlet flow rate f is
fixed and the dilution rate is defined as
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f by v or the volume of the reactor is f by
d this transposing this and so to minimize
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v we need to maximize d because v equals f
by d right when v is minimum for a constant
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f d has to be maximum only then v will be
minimum so we need to find essentially d m
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which will also maximize the productivity
because d m into mu m and so on a mu m into
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x was are productivity so that will also maximize
productivity we have already derived d
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m as mu m times one minus square root of
k s by k s plus s i that we have already derived
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in our previous lecture
now if we substitute the values that are given
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in the problem we will get d m equals point
five is mu m one minus k s is one gram per
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liter and s i is fifty gram per liter this
is a square root and if we calculate the value
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it will turn out to be point four three hour
inverse you need a calculated to do this if
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we know f we can find v m the minimum size
of the chemostat we don't know f but we know
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that the productivity is five hundred gram
per hour ok from this we will have to somehow
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find f and there by find the minimum value
of the reactor volume and we also know
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that r m is f times x m this is where we get
the relation between f and r m r m is f times
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x m x m is the outlet cell concentration at
maximum productivity right we need v m we
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know r m so we are trying to find f and therefore
v m by using r m
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we also know the yield coefficient we we know
that the cell concentration is y x s times
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s i minus s therefore x m is if you substitute
the conditions for the maximum then we would
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get x m here maximum cell concentration you
get x m here so y x s times s i minus s m
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this is the condition this is the substrate
concentration the outlet which corresponds
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to the maximum cell concentration x m maximum
productivity to find s m let us use the relationship
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between d and s i hope you are able to follow
all the inter related aspects that's an essential
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feature of many problems and that's what takes
time to develop to be able to see the various
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connectivities so mu m equals s m sorry
mu m s m by k s plus s m is d m so mu m s
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m is d m times k s plus s m this transposing
and just collecting all the common terms s
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m into mu m minus d m equals d m times k s
so s m equals d m k s by mu m minus d
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let us do the above step by step through substitution
of the relevant values for our situation s
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m here d m k s by mu m minus d d m what we
found from the earlier case is point four
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three now the calculation point four three
k s is one mu was point five d m is point
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four three this turns out to be six point
one four gram per liter so x m from s m we
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calculating x m and therefore d m and so on
that's the way we going to go about it so
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s m was six point one four gram per liter
x m is y x times s i minus s m the substitution
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of this we get point five into fifty minus
six point one four twenty one point nine three
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gram per liter and r m is f times x m will
come to this a little bit it came a little
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early doesnt matter r m equals f times x m
so r m is five hundred f times twenty one
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point nine three f equals five hundred by
twenty one point nine three that's twenty
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two point eight liters per hour so the corresponding
minimum volume is f by d m right from the
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definition of the dilution rate we set to
minimize v will have to maximize m we have
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found out the maxim d m as point four three
we went through all these steps to get f from
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the productivity is twenty two point eight
so twenty two point eight by point four three
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happens to be fifty three point zero two liters
ok so this is the minimum volume of the chemostat
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for the given set of conditions in the problem
right ok when we meet next we will continue
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the course we will start module four see you