We want to determine the residence time distribution when we have a number of CSTR's in series
and we do that by solving mass balances for a tracer. And this will be a pulse tracer
that we inject to the first CSTR in the series. And so at time equals zero, we inject a pulse of tracer
into the first CSTR and that means we're not adding any more tracer after that so that
c0 concentration coming into the first CSTR is zero. We'll call this C1 and of course
the concentration leaving the first CSTR is the same as the concentration in the first
CSTR. C1 now is the concentration entering the second one. So C2 is leaving, C3, and
Cn for the nth CSTR. So we're going to set up the mass balance for the first CSTR. I
have the volume of the CSTR where each CSTR has the same volume. The change in the concentration
in the reactor. C1 of course is the same as that leaving. The volumetric flow rate, this
is liters per second, for example. The volumetric flow rate times the concentration entering
minus the volumetric flow rate times the concentration leaving. There's no reaction for the tracer.
So this is the complete mass balance. And then we know that at t=0, that we've already
injected the tracers, so if you like, at t=0.0001, we have the concentration inside the CSTR
is some initial concentration which is a function of how much tracer we inject. So to solve
this mass balance, this is zero, and so we can separate variables to solve this. And
then once we've separated variables, we're going to integrate from time equals zero to some later
time. Integrate from the initial concentration to the concentration at some later time. And
so the left side, if we integrate, we get a log(c1/c0), the right side, first, I'm going
to write this as the integral of dt/tau. So tau is the space time. Volume over the volumetric
flow rate has units of time. Tau is the same for each reactor, the residence time in the
reactor. Now I do the integration on the right side. It's t/tau. Which means the concentration
leaving the reactor is the initial concentration in this exponential decay. We inject tracer
in the first reactor, then the concentration exponentially decays as a function of time.
So now I want to do the same thing for the second reactor. And the difference is there
is tracer flowing into the second reactor. So the change in the concentration in the
reactor, C2, is the volumetric flow rate times what's flowing in. C1, the volumetric flow
rate, what's leaving. Of course, we just calculated C1. Now I can rearrange this equation and
also substitute for volume over volumetric flow rate in terms of tau. And now we need
initial conditions for reactor 2. And that's at t=0, C2=0. We've injected tracer into the
first reactor only so there's been no time for tracer to get into the second reactor.
Now we can solve this differential equation using an integrating factor. I won't go into
explanation. You can find an explanation on integrating factors elsewhere. We're going
to use e^t/tau as our integrating factor. And then we know from differential equations
solutions, we can write the concentration C2 is this integrating factor.times now the
right side of the equation. Integrating this with respect to dt, and then there's going
to be an integration constant and it's going to be divided by our integrating factor. We
can solve for C2, going to bring the term in the denominator. And I do the integration,
you notice I have e to the plus tau over 2, e to the minus, so these cancel. I have C initial as a constant. So I just
end up with t/tau plus my integrating factor and now I apply my initial conditions when t is zero,
the first term disappears. C2 is zero. This means my integrating constant for this solution
is also zero. So I have C2, I can now do the exact same balance for the third reactor where
there's no concentration of tracer in the third reactor at time equals zero, so the initial condition
t=0, C3=0. The feed to the third reactor is concentration of C2. I won't go through the
details. I'm just going to write them down so you can go through them at your own speed.
I've worked through the exact same type of solution, notice the difference now, because
I have t here inside the integral sign, I end up with t squared over 2 and so now if I write the
balance for the fourth reactor, I'm going to end up with t cubed over three and so it looks like
this. So here is the concentration of the fourth reactor and then the general Cn is
this initial concentration*t^(n-1)*e^(-t/tau) and you can see for the fifth reactor, we
have 4*3*2, so this is going to be N minus one factorial, and we have tau to the N minus one also. So we're not
quite finished. We now have the concentration from the nth reactor. We want the residence time
distribution and so we have to normalize and we can look at the first reactor to do this
normalization. The normalization that we're going to do is that the integral of the residence
time distribution, or sometimes referred to as the exit age distribution, from 0 to infinity
should equal 1 because essentially this is a probability of leaving a reactor and the
probability should add to 1. So that means the integral of C1 which is this initial concentration
tau, if we want to normalize this and make it equal to 1, we would essentially say C1
divided by some constant alpha now is going to equal 1. And then what we're saying is
that C1 over alpha is our residence time distributions. So we're going to do this integration, Cinitial/alpha,
integral of e^(-t/tau) divided by -1/tau, evaluated between our limits is equal to 1.
Then we can substitute in the limits. We have -tau*cinitial/alpha, e^(-infinity) is 0, e^0
is 1, so it's -1. This is equal to 1. This says alpha is just tau*cinitial. Remember
what I did is divided C1 by alpha to get E(t). So E(t) was just C1/alpha. C1 was the initial
concentration*e^(-t/tau), alpha was cinitial*tau. So the residence time distribution for the
first reactor, e^(-t/tau)/tau. And then for each of the others, I would also divide by
alpha so my general residence time distribution, E for the nth reactor, can divide this by
alpha so the cinitial is going to cancel, have my exponential term, factorial, and I
have an additional tau, so now I have tau^n. So this is the residence time distribution
for any CSTR in this series of CSTR's where we have tau as the space time, the residence
time in a given CSTR where all the CSTR's are identical.