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