In this example we'll look at the calculations for a recycle reactor. So we have a plug flow
reactor volume, V, and into the system we have a feed of reactant A, concentration,
C_A0 and a volumetric flow rate v. We also recycle the same volumetric flow rate so that
the volumetric flow rate going into the plug flow reactor is 2 v and we're mixing the outlet
with the inlet so this concentration is the average of the two, C_A1 and the concentration
leaving the reactor here of course is the same as the concentration here, same as the
concentration here. This overall system has a conversion of 2/3. 2 A goes to 2 R as the
reaction is second-order and the question is, suppose we shut off this recycle,
then what happens to the conversion?
So what we're going to do is write a mass balance
for the plug flow reactor first and for plug flow reactor the change in the molar flow
rate of A. So F_A is the volumetric flow rate into the reactor which is 2 v, concentration,
C_A1. That's for entering, so that's into the reactor. The change in this flow rate
with respect to the cumulative volume is the rate of reaction r_A which in the problem
statement is given as minus k C_A squared. So the stoichiometric coefficient in this
case is built into the rate constant. It's a liquid phase system so the volumetric flow
rate's a constant so I can write this as 2 times the volumetric flow rate, the change
in the concentration with respect to volume and so we can rearrange, separate variables,
the change in the concentration of A over C_A squared is equal to minus the rate constant
times dV over 2 times the volumetric flow rate. We'll integrate from the inlet of reactor
where the volume is 0 until the total volume, V. The inlet of the reactor C_A1, the outlet in our
notation of the figure is C_A final. So this integration just gives us minus 1 over C_A
evaluated between the limits, and the right side the rate constant's a constant, it's
isothermal. The volumetric flow rate's a constant, it's liquid phase v minus 0. So we can substitute
in the limits minus 1 over C_A final plus 1 over C_A1 equals minus k times V over 2
times the volumetric flow rate and so now what we want to do is do a balance around
this point. So we can calculate C_A1 as C_A0 plus C_A final divided by 2. Alright we're
mixing two streams with the same volumetric flow rate and so we have C_A1 now in terms
of C_A0. If x is 2/3 and C_A final is 1/3 of C_A0. So C_A0, this is 1/3 C_A0 divided
by 2. So we now have C_A1 in terms of C_A0, we can go back to our mass balance and substitute
in for C_A final and C_A1 and with that substitution we can essentially just rearrange the equation
and k times the reactor volume times C_A0 over the volumetric flow rate is equal to
3. So these are going to be the same numbers now if we turn off the recycle and so we're
going to do a mass balance again but with a different feed. And so what we're going
to do is apply the plug flow reactor equation, second-order again, now the difference, the
one difference, is that the volumetric flow rate is v instead of 2 v. when there was recycle
it was 2 v, it's only v now. We do the exact same integration and we're trying to solve
for C_A2 and so minus 1 over C_A2 is equal to minus 1 over C_A0 and then k V over v is
3 over C_A0. So this is minus 4 over C_A0 or C_A2 is 1/4 of C_A0 which means the conversion
of A now in the new system is 0.75, so this is with no recycle. With recycle the conversion
was 0.667. So we can check if this is what our expectations would be. When we have recycle
then we're diluting our feed to a lower concentration before it enters the reactor so that lowers
the rate. We're also increasing our volumetric flow rate so it's less time in the reactor.
So expect less conversion with the recycle than without the recycle and indeed that's
what we see.