I am going to show how the residence time distribution which we could determine from
a tracer experiment that is described in other screencasts. So combining a residence time
distribution with the assumption that we have segregated flow in the reactor, we can use
that to calculate conversion for a non-ideal reactor. Whether that is going to be an accurate
determination of the real conversion of the reactor depends on the how good of an approximation
the segregated flow is. So an important thing to keep in mind when we are talking
about modeling a reactor with residence time distribution is that the residence time distribution
does not give us information on mixing, meaning the exchange of matter between different fluid
elements, and so we can make an approximation of complete mixing or we can make an approximation
of segregated flow, and they are going to give us the extremes of what the actual conversion
is going to be. The mixing can be quite important when determining for example
in polymerization reactors the molecular weight distribution, and for polymerization reactors
where we have high viscosities, mixing indeed can be quite important. We have a number of
reaction steps taking place. The equation that we would use if we know the residence
time distribution, and we want to calculate the concentration leaving the reactor,
so I will put a bar to indicate this is the average, so this is the exit concentration
from the reactor, that is an average. What we are going to do is integrate over all the
different times in the residence time distribution, so what we will assume is we have essentially
a number of batch reactors and the residence time distribution will determine how long they stay in the
reactor and how many of them there are. So we are going to calculate, as a function of time, the concentration
and this is for a batch reactor. So we are assuming that if we had a batch reactor how
this concentration depend on time, and then we are going to use the residence time distribution,
which we obtained from a tracer measurement, and essentially we're adding up all of those from zero to long
enough time to account for all of the reaction. So we are envisioning that a large number of
batch reactors are moving through the reactor independently of each other, and how much
time they spend in the reactor we get from the residence time distribution. So this E(t)
in the residence time distribution, in words it's a fraction, and it's the fraction of the packets,
Large number of molecules, but small packets, that do not mix with each other, spend a time
between t, so the t that we are using in the integration at a given time t, and t plus
dt. That is what this fraction is. If we had first order reaction then the residence time
distribution is sufficient to give us the exit conversion, but if we had another order of
reaction, less than first, or greater than first, it is not sufficient. So what we are going
to do to demonstrate this is take an ideal CSTR. An ideal CSTR we could use our normal
way of calculating, just a mole balance, assume complete mixing, and so all reactions is at that
concentration, the final concentration, and we are going to compare that to an ideal CSTR
where we have segregated flow. Here are the parameters we are going to use. So if we are
looking at the complete mixing case. Steady-state, it is the molar flow rate
in to the reactor minus the molar rate out, the rate of reaction times the volume, where
this is the volumetric flow rate, the concentration in, the volumetric flow rate the final concentration
leaving the reactor and then second order reaction minus k, A is our reactant, the concentration
of A squared times volume, and it is important to keep on mind this concentration is the same
as this concentration. We are assuming well mixed, so we can substitute the numbers in
to this and calculate the conversion. So I have shown substitution of the numbers with
the units, reducing to this quadratic equation. The solution to the quadratic equation is
given here, and then the conversion is the inlet concentration, the outlet concentration over the inlet
concentration. Since it's liquid phase we can use concentrations, and the conversion is 0.79.
So this is for complete mixing which is what we are assuming when we use the CSTR and assume
that everything mixes so it's at that final concentration. Now let's contrast that to where we're going
to assume segregated flow. So I've just copied the equation down again for segregated
flow. Second order reaction. If we solve the mass balance and that is something that we
have done in another screencast, the equation relates the concentration as a function
of time, so the rate constant, initial concentration, ideal CSTR the residence time distribution
which is also derived in other screencasts is given here where tau is the volume of the
reactor divided by the volumetric flow rate, so in this case 150 liters over 100 liters per hour, so
tau is 1.5 hours, the base time, in this case, residence time of the reactor. So we can substitute
the numbers in to this equation, the integral. So I've substituted in the numbers without the
units just so this won't look too confusing, but you should certainly double check that the units
are correct, and then if I do this integration numerically I end up with a concentration
of 0.68 mol/L to 2 significant figures, and then I can calculate the conversion in the
same way and so when I do the calculation of conversion, I get 86% conversion compared
to 79%. So the conversion for segregation flow is greater then for complete mixing.
So the important point here is that there is a difference in conversion for an ideal CSTR depending
whether we assume complete mixing or we assume segregated flow. So the residence time distribution
is not sufficient for us to determine what the conversion is going to be in the reactor,
we need some information about the mixing. So I have added at the end here the Mathematica
commands for those who are interested in how I got the numerical values. Certainly could
have used the quadratic equation, but it is certainly very easy particularly when we
have more complicated equations to use the FindRoot command, where basically I have
written down the equation here. Then I indicate the variable that I'm solving for, Ca
and I indicate my guess, my starting value or approximation to the solution. Then over here I have
written the integral, which is a little more difficult to do analytically, very easy to do numerically. We're
integrating from 0 to infinity. So in Mathematica I integrated from 0 to 10, and then 0 to 100, got the same
answer, and so it is just the integral command on Mathematica and it very directly outputs
the solution.