In this example, we're going to look at a gas-phase reaction in an adiabatic plug flow
reactor. It's a first order reaction, the rate constant depends exponentially on temperature,
and we're given feed conditions, heat of reaction and heat capacity, and the problem asks us
to plot the molar flow rates of A and B as a function of cumulative volume as we move
down the reactor, effectively the same as saying as a function of distance as we move
down the reactor, and to also plot the temperature versus volume as we move down the reactor.
What we're going to write is mass balances and an energy balance for this system, and
then we're going to account for the volumetric flow rate change due to the mole change and
the temperature change using the ideal gas law. So the first part is the mass balance,
and what we have here is the change in the molar flow rate of A, so FA represents the
molar flow rate, is equal to this rate expression, which I'm calling r1, which is just k times
CA. The minus sign is because A is reacting. Now, we could use stoichiometry to account
for the products, or we can just write differential equations. Because we're going to solve this
numerically, we can just write differential equations to solve simultaneously, where now
we have a plus sign here because we're forming products B and C. And when we have differential
equations we need an initial condition, namely at the inlet of the reactor, where volume
is equal to zero. We need a molar flow rate of A, which we're going to have to calculate.
The molar flow rate of B is zero, and the molar flow rate of C is zero, because the
problem says we have pure A coming into the reactor. And we're going to use the ideal
gas law. What we can certainly write is PV=NRT, but for a flow system, were going to write
a pressure times a volumetric flow rate is the molar flow rate, that's the total molar
flow rate. In general, as we move through the reactor, the volumetric flow rate is the
flow rate of the three components, RT over P. For the inlet of the reactor, we're given
the volumetric flow rate. There's only A, there's no B and C, so we can calculate the
flow rate of A coming into the reactor, that's the pressure times that inlet volumetric flow
rate, so substitute subscript 0 to indicate this, over RT. We'll substitute some numbers
in here to get our inlet molar flow rate of A because we need that as an initial condition
for our differential equation. We have our inlet molar flow rate of A, which we used
as our initial condition, and then we're going to use this equation, which remember is essentially
the ideal gas law, in solving to account for the volumetric flow rate change as you move
down the reactor. So we have the mass balances, we now need the energy balance, and essentially
if we look at the heat capacities, we can see the heat of reaction doesn't change with
temperature because the heat capacities with the stoichiometric coefficients sum to zero.
In other words, CpA, where A is a reactant so we put a minus sign, CpB plus CpC is zero,
which means delta H is independent of temperature. And so we can end up simplifying our energy
balance. First we're going to write it in the general form, heat capacity of A, heat
capacity of B, heat capacity of C, so at any point in the reactor, derivative of the temperature
of the reactor with respect to that cumulative volume, that's going to change because of
the heat of reaction and the rate of reaction. this is negative, its negative, so this is
positive, so the right side is positive as we would expect, the temperature is increasing
for an exothermic reaction. Well, we can simplify because of the fact that the heat capacities
of B and C are equal to the heat capacity of A, and the flow rate of B is the flow rate
of C because of stoichiometry, so we can actually write this left side as just the heat capacity
of A times the molar flow rate, and so this then becomes our differential equation for
the energy balance that we're going to substitute into a numerical program to obtain the solution.
We're going to need a few other things. One is minus k CA. We're going to have to substitute
in the value for k, which is given in our problem statement. We're going to have to
use this volumetric flow rate equation, and we also need the concentration of A. We're
going to calculate the flow rate of A, and the flow rate of A divided by the volumetric
flow rate gives us the concentration, and then we need to enter the values of heat of
reaction, heat capacity of A, the gas constant, and the pressure. And we're going to solve
these equations all simultaneously, and we're going to in this case use POLYMATH. It's a
relatively simple method to solve differential equations. So what I'm showing now is the
POLYMATH program, where I've written the mass balances, with the three differential equations
and their initial conditions. We have to substitute in our starting volume, we cal this volume
equals zero, and our final volume, 800 liters. The energy balance that I've just written
with the flow rate and the heat capacity terms in the denominator on the right side, to satisfy
the format for POLYMATH, with the inlet temperature, and then the various parameters that we need,
and then the rate expression, the concentration, the calculation, and the volumetric flow rate
calculation. Here are the results from POLYMATH. First the molar flow rate, so as we move through
the reactor, the molar flow rate of A decreases to a relatively low value, and correspondingly,
the molar flow rate of B increases, and of course that's what also corresponds to molar
flow rate of C, and you'll notice that the rate at FA is decreasing, it actually becomes
faster as we go further down the reactor because the temperature is increasing. So this is
a rather dramatic temperature increase, and it demonstrates what can happen in an adiabatic
reactor. We have to be careful as to whether we could realistically run a reactor this
way, and with this big a temperature increase, we'd have to worry about additional reactions
that take place at the higher temperatures as to whether this would be a good model.
But the temperature increase, and the molar flow rate increase, means that the volumetric
flow rate actually increases significantly. Anther way of looking at it is, this decreases
what the residence time in the reactor would be compared to an isothermal case with no
mole change. So the volumetric flow rate starts close to 2 and ends up above 6 because temperature
expands the gas and we're doubling the number of moles as the reaction takes place. And
so, this is a template to solve adiabatic reactors. If we have multiple reactions, more
complicated kinetics, we can just change one or two terms, and this POLYMATH program can
calculate the profiles of temperature and molar flow rate as a function of distance
or cumulative volume down the reactor.