In this example, I'm going to look at an autothermal reactor, which means that the heat of reaction
from an exothermic reaction is used to preheat the reactor feed. Now in the figure is an
annular region around the packed bed reactor, and so the feed enters this annular region.
As it moves through, heat of reaction is exchanged with the feed gas, so the gas at this point
is being heated from the catalytic reaction, and then this gas enters the reactor much
hotter than the initial feed temperature, and of course the rate of reaction is faster
at the higher temperature. What we want to do is determine the conditions in the reactor
and the exit temperature given our feed temperatures. We're given a temperature T1, and then we
want to determine our final reactor temperature. The problem is that to solve the differential
equations as initial value problems, we need the temperature that's right here, and this
temperature we're going to call T0 in order to make this initial value problem and calculate
and solve the mass and energy balances in the reactor. So what we're going to do is
guess T0, and then calculate T final and T1. If we guess the correct value, we will then
end up with the T1 value that we already knew. If we guess the incorrect value, then we're
going to iterate until we get back our initial starting values. So we want to go through
what the equations are and what the procedure is. There's a couple things we have to be
careful about, and we're going to look at the case for an A goes to B reaction that's
first order and gas phase, and the procedure then is described here. What I'm going to
do is first start with the material balance for the reactor, a material balance on A.
I'm going to write this as the change in the molar flow rate of A, moles over time, with
respect to the cumulative volume, or the cross section times the distance down the reactor.
This is equal to the rate of reaction per volume, which is this, so the differential
equation that we're going to substitute into a numerical integration package where the
material balance is this. Now I'm going to write down mass balances, and so, important:
what we have to be careful about is Z. So at this point, that corresponds to Z equals
0, and this is our Z final. What we want to do is integrate in this direction, so we'll
likewise write down now the energy balance. This is the energy balance for the packed
bed. I'm going to write it in terms of the density of the material, the volumetric flow
rate, so that's the mass flow rate, and this is a mass heat capacity, the change in temperature
with distance down the reactor, is equal to the heat of reaction per mole times the number
of moles reacting, we have the cross sectional area just like we did in the mass balance.
This minus sign is here because overall for an exothermic reaction, this term should be
positive, and then we have the term for heat transfer which is in this form, and it's described
elsewhere why that's the form, radius of the tube containing the catalyst, the cross sectional
area, and an overall heat transfer coefficient. This is the temperature in the annular region,
and this is in the packed bed, so we're referring to in this region, TA, and inside the packed
bed we'll use the single T. And so we need as our final balance, a balance on the annular
region. And this is of a similar form, except we don't have any reaction in that region,
so we have a density, a volumetric flow rate, the same mass heat capacity, change in temperature
A, but I'm going to call this ZA, so this is where we have to be careful. ZA means moving
in this direction, so ZA equals zero here and then it increases as we move to the left,
because that's the direction of flow. Then the term on the right side is the heat transfer
term. That's the only term that we have for heat exchange, there's no reaction, and it
has the opposite sign because we're removing heat from the reactor part and preheating
our feed. We have to be careful with signs. What we're going to end up doing is saying,
as you can see from my drawing, dZ is minus dZA. So therefore, we could take this equation,
remove the A and put a minus sign here, so we can integrate with respect to just one
variable and indeed we can of course multiply both sides by a minus sign and make this plus,
so that we have now three equations that we're going to solve simultaneously as initial value
problems. So at Z=0, we know the flow rate going into the reactor, and at Z=0 what we
don't know is the temperature going into the reactor. So at Z=0, temperature is T0, and
indeed, Z=0, the temperature here is also T0. So T0 remember is this temperature at
Z=0. They have to match since it's the same stream, and that's where we make our guess,
so we guess T0, and then we solve and the energy balance on the annular region at Z=Z
final, temperature is T1, this is what we know. So we know T1, we're going to do this
integration, we're guessing this, and we're going to iterate if we don't get the correct
value for T1 and we guess a different T0. So what I'm going to show you is a POLYMATH
program that goes through this calculation. So here's the POLYMATH program and then the
results, so let's look first at the POLYMATH program. We're given T1 is 316, the feed to
the overall system, and then we adjusted T0, so this is the final result where I've run
the program multiple times, adjusting T0 until at 545 Kelvin, I get as my final temperature
for the annular region the 316 Kelvin that I'm already given. And so what I have is a
mass balance, the energy balance, the annular region energy balance, the heat transfer term,
length is 7.5, this is in meters, and heat of reaction of the various terms. I've accounted
for the fact that the volumetric flow rate of the gas changes with temperature using
an ideal gas law, where volumetric flow rate is related to molar flow rate, which is a
constant, the gas constant times temperature divided by pressure, given the overall heat
transfer coefficient, the radius. Here are the units that are used in this POLYMATH calculation,
and here are the results, so let's look at the flow rate as a function of distance through
the packed bed. Of course, the flow rate of A decreases as the reaction takes place, and
it looks like we have about 70% conversion. This is temperature, again as a function of
distance. Our inlet temperature comes in at 316, this is 545 Kelvin, so the temperature
increases because of the temperature difference between the packed bed and the annular region,
and then this is the temperature in the packed bed, because of the reaction it increases,
and then we get to a point where this temperature difference becomes large and the rate of reaction
slows down because the concentration of reactant is decreased, and now the temperature starts
decreasing the packed bed region. The point to keep in mind is the signs for the energy
balance in the annular region, and using this iteration which is actually very quick because
the solution to a differential equation is very quick. We can determine this using what's
referred to as a shooting problem, making a guess and then solving differential equations.