In this example we will look at cell growth

in a batch reactor and we are using in this case a bacteria to produce a product P. The

rate expression in this case depends on the concentration of cells Cc and the concentration

of the substrate the nutrients that are used for the growth of the cell. This is a rate

constant, and this is another constant. So this notice there is a negative sign here.

So the rate of the substrate, we’re actually see the amount of substrate decrease with

time. We put a minus sign in front of it so the term is positive, and then we also know

for the system. That for every gram of substrate that is consumed a half of a gram of cells form,

and here are the initial concentrations. So our concentrations of the cells are relatively

low, and the concentration of substrate is relatively high. The substrate of course is

going to decrease, so the substrate concentration is going to decease, and the cell concentration is going to increase. Now the problem asked

us to plot the concentration of the substrate, and the concentration of cells as a function

of time. We want to do this until we essentially use up all the substrate. It also asked us

to plot the rates as a function of time. So the solution here is a standard batch reactor.

A couple of things that make it simple is that it is a biological system in a liquid phase so

it is constant volume. So the rate expression, the general form for a batch reactor is the change in the number

of moles of C with respect to time is the rate of reaction of C times the volume, and

actually factor out the volume and cancel. So we can write the rate, and then we can

also write the rate of the concentration of substrate with time as the rate of the substrate

reaction, which is minus mu max, concentration of substrate, concentration of cells, of this

constant and the concentration of cells. So this is one differential equation. Now if

we remember that it says for every gram of substrate consumed, half a gram of cells

form. So that means we can write the rate of concentration of cell change. This Rc is

0.5 times Rs, but with a minus sign. So that we end up with rate expression 0.50, mu max,

k plus Cs. So notice the concentration of C is on the right side, which says as we have

more cells the rate becomes faster, as an auto catalytic behavior that we see with cell

growth. So we have 2 different differential equations, and we have initial conditions,

namely at time equals 0 the concentration of C is 0.40 g/L and the concentration of

S is 15 g/L. So we have 2 ordinary differential equations, and this is an initial value problem.

So we can integrate these relatively straight forward. We know the values of the constants,

k, 0.10 and that is g/L and mu max is 0.50 hours to the minus 1. So we are going to solve

this using POLYMATH. It is relatively easy software that will allow us to calculate these

concentrations and rates as a function of time. So here is the POLYMATH program. We

have the differential equation. We are writing in terms of Rs for the cells, an initial condition,

differential equation for the substrate concentration, and then remember a rate with respect to s has

a minus sign, and the values of the constants. So we will start at t=0, and from trial and

error we find that if we integrate up to 13 hours we have used up all of the substrate. So here

is the change in the substrate concentration with time, and it goes to 0 after about 12

hours and the cell concentration you notice that the rate is actually increasing until

we run out of substrate. So in addition to the concentrations we can ask polymath to

plot the rate. This is Rs which is negative. So the absolute value of the rate if you’d like is increasing.

We are using up the substrate faster until we get to the highest rate of consumption,

and then it starts dropping off rather quickly because we are using up the substrate. We

do not have enough left, and that eventually goes to 0. So the rate, the cell growth is exact same

shape, half the amplitude, and opposite signs. So by solving these differential equations

you can see how the rate is changing as a function of time before we use up all the substrate.