In this problem, we want to develop the equations to determine the plug flow reactor size for
50% conversion of reactant A, where the reaction is A plus 2B goes to 2D.
So we're given information about the pressure and the temperature, the rate of reaction,
how it depends on concentrations, mole fractions, and a volumetric flow rate, and so there's
an inert also in the feed, and so what we want to do is write down the mole balances,
and the change in the molar flow rate of A, with respect to cumulative volume, so F_A
is moles per minute in this case, is the rate of reaction, where the rate of reaction is
-2.5 (the minus sign is very important here because A is reacting away ) and it depends
on the concentrations of A and B. So a similar equation, in terms of the molar flow rate
of B, the rate of reaction of B is twice the rate of reaction of A from the stoichiometry,
and then the product, the mole balance for D, and this is -2 times r_A, so the total
term is positive, we're forming D.
We need initial conditions, and that means at V=0, we need the molar flow rate of A,
it's the mole fraction of A entering times the total flow rate entering, and this mole
fraction is 0.20, while the molar flow rate of B initially, again, mole fraction of B
times the total flow rate, and this is 0.50.
The molar flow rate of D is zero initially.
We're solving for the final volume, and we're going to have to calculate the total molar
flow rate entering, so that's the volumetric flow rate entering times the total concentration
entering.
We're given the total concentration of a gas phase system as pressure over the gas constant
divided by absolute temperature.
If you think of the ideal gas law, you could write this pressure times volume per time,
which is v_0, is moles per time, so that's F_T0 times RT, just ideal gas law, and we're
given that v_0 is 50 liters per minute.
Now, to make these equations complete in order to be able to solve them, we need the concentration
of A and the concentration of B. Well the concentration of A, we can write as the molar
flow rate of A over the volumetric flow rate.
So what's important to keep in mind here is that the volumetric flow rate changes for
this system because we have, if we go back and look at our reaction, we have three moles
reacting to two moles, so our volumetric flow rate is going to change.
Well, we need the concentration of B, and that means we need an equation for the volumetric
flow rate, and it changes because the molar flow rate changes, so again this is the ideal
gas law.
Well, now we have another variable, F total, is the total molar flow rate, and the total
molar flow rate is what's changing, and that's the flow rate of A at any point in the reactor,
plus the flow rate of B, a molar flow rate, plus the flow rate of our product D, plus
the flow rate of our inert.
Now, the flow rate of the inert is the mole fraction of the inert initially, times the
total molar flow rate, and that's 0.3 times the total molar flow rate, and this is the
same as the inert flow at the inlet, inert flow rate is not going to change.
The final calculation that we need to do is to calculate the conversion, because we're
going to carry out this integration until the conversion is 50%, and so the molar flow
rate of A at the inlet, minus the molar flow rate at any point in the reactor, divided
by the molar flow rate at the inlet, this would be our fractional conversion and we
want, at the exit, we'd like the final value to be 0.50.
So we need to integrate, and what we're going to do is solve these three ordinary differential
equations, along with these algebraic equations simultaneously, to determine the size of the
reactor to give us a 50% conversion.