So this example we will be looking at a Monod Growth Kinetics and the problem statement
is; That if we wanted to produce a recombinant protein in E. Coli once the cell density has
reached 400 million Cells/Liter. How long would we have to wait for us to reach this cell density
if we started an initial inoculation density of 10,000 cells/L, and a glucose concentration
of 100 g/L. And we know that for E. Coli growth on glucose Umax is 1.3 inverse hours. Ks, the Ks constant
is 2.2E-5 g/L. The death constant is 0.02 inverse hours, and we are going to assume
that 1 cells is approximately 1 ng of cellular mass.
So the first thing that we are going
to do is write mole balances for the concentration of cells and substrate. So for our cells
we are going to have a change on our concentration of cells over time. So since we are in a batch
reactor it is going to be equal to the rate of growth of our cells minus the rate of death
of our cells, and so for our substrate the change in the concentration per time will be equal
to the negative rate of growth of our cells, and so now second we are going to define our
rate law expressions. For monod growth kinetics the rate of growth is equal to mu max times
the concentration of substrate times the concentration of cells over the constant ks plus the concentration
of our substrate. And the rate of death is equal to the death constant times the concentration
of cells. So third we need to calculate the concentrations of our cells initially and
the desired concentration based off of the mass we assumed per cell. So our initial cell
concentration in g/L is going to be the 10,000 cells/L that we started with times 1 ng per
cell, and to get it in to grams per liter we are going to have that 1 gram is 10 to the 9th nanograms. So we
are going to have 10^-6 g/L of cells initially, and then the final concentration of our desired
cells also in g/L will be the 400 million cells per liter that we want to reach. Times the same 1 ng/cell,
and 1 gram per 10^9 ng. So we want to get to a point of 0.4 g/L of cells. So now we
can plug in these mole balances for the cells and substrates, and then the rate laws for
the growth and our death of our cells. As well as the initial concentrations for our
cells and substrate, and then vary time to try to figure out how long it would take us to get to this
0.4 g/L of cell. So here is the output from plugging in all of those expressions into
polymath, and after varying the time. You can see that the top line of Cc. The concentration
of our cells our final value is 0.40 and then the time it took use to get there is the last
variable t, which is 8.28 hours.