In this problem what we are going to consider is a system of multiple reactions, and this
is a two part problem. The first is we are going to set up the problem and talk
about what variables we will need to know to deal with this system of multiple reactions
that are exothermic reactions. That exothermicity needs to be explicitly taken into account,
and in the second part of the problem will show the set up of a numerical simulation in order
to try to optimize the performance of the reactor, and we will see that optimization is
actually a very difficult process. But first let's just set up the problem, and that means setting
up balances. In this case we are going to ignore pressure drop for this problem, but we
will not assume isothermal operation, so we will need to set up both an energy balance
and material balances for all our reactants and products. So the reaction network is given
here. It is actually based on a real example of ethylene epoxidation, were A is ethylene,
B is oxygen, C is ethylene oxide, and D, and E are carbon dioxide and water, and so D and E
are not very valuable products, but C is. So you'd like to optimize the amount of C that you
form. You would like to form it at a high rate, but still achieve high selectivity to
C, and what makes this especially difficult as we will see is that the desired reaction
is slightly exothermic, but the complete combustion to form CO2 and water is highly exothermic. It
will tend to release heat and the reaction will end up feeding on itself. The reaction
is going to be conducted an a packed bed reactor with an assumption of minimal pressure drop
in this case. As in the true case of ethylene epoxidation, the reactor is actually a shell
and tube kind of configuration, where we have a tubular reactor with a packed bed of catalyst,
and we will have flow through this reactor and we are going to have a cooling fluid entering,
cooling fluid temperature, that will be water we will say. That will be flowing on the outside
of the tubes to try to cool off the reaction to remove the heat that is generated
from the exotherm of the reactions. So we will need to write an energy balance
that incorporates this. First we will start off by writing the material balances, and
again we are going to do that for each component. We are going to write these rather than in
terms of conversion we will write it in terms of molar flow rates because this is a system
of multiple reactions, and as we have seen that usually makes it advantageous to work in terms of moles. Alright,
so lets write it first for component A. So we can write that dFA/dW is equal to the,
net rate of formation is equal to the net rate of formation of A for reaction 1, plus
the net rate of formation of A through reaction 2, and we see of course these are going to
be negative quantities and so FA is going to decrease with respect to time. Then we
are going to need to know the initial condition for this. It is not specified in this problem. We will
specify at a later point, but I will just not that in order to solve the problem we are
going to need to know what our inlet flow rate of A is. Okay, we can likewise write a balance
for B for example. B is consumed at half the rate of A in reaction 1, B is consumed at 3 times the reaction
of A in reaction 2, and B is consumed at 5/2 times the rate at which C is consumed, so we are tracking
the rate of reaction of C here in this experimental expression. There is actually a typo here.
I should correct this. This depends on the concentration of C, not the concentration of
A, because A is not involved in that reaction. We will need a 5/2 times the rate of net accumulation
of C. Alright, again these are all negative numbers. Once we take into account the negative
signs here, and so oxygen as we know or component B is being consumed in this reaction. So we
expect there to be a negative number, and again I wont bother to write this every time, but
we will also need the initial flow rate of oxygen in order to solve this material balance.
Alright we will go ahead and write a couple more balances. I will go ahead and try to do them in the
margins here. So C we can write a balance on. It is formed through the reaction of A.
In a 1 to 1 stoichiometry, so 1 mole of A makes 1 mole of C, and then it is consumed
in reaction 3, and this is given in terms of the rate of consumption of C. Okay, and
it is not formed in reaction 2. Then finally we can write the balances for D and E, which
are equivalent. Every time you make a mole of D you also make a mole of E in each of
these equations, and they are formed in reaction 2 and 3. And so I will go ahead and write
that here, and in fact FD is always equal to FE, and this is equal to the rate, 2 times the
rate of consumption of A through reaction 1. I want a negative sign here, so that the
number increases. Then 5/2 times the rate of reaction of C through reaction 3. I am
sorry this should be reaction A2. Okay, and both of those will be positive numbers and
will tend to accumulate D and E with respect going down the reactor. These are all our
material balances. The next thing we will need to know are the rates of reaction expressed
in this case in terms of concentration. Those are given here. We know the reaction order and independence.
They are all. half order in one of the reaction, and first order in the other. They have different
rate constants associated with them. We are going to need to know the values of those
rate constants of course, and then more importantly we are actually going to need how those rate constants
vary with temperature. So we are going to need to know k as a function of temperature,
since this is not an isothermal, not necessarily an isothermal reaction, because there is a large exotherm
in this case. So we are going to have to take into account the fact that the temperature
can vary down the length of the reactor, and that can have a corresponding with the fact
that the rate of reaction as we will see in the next part of this problem. So we
will need those values of k, I need their corresponding activation energies and pre-exponential factors,
and then the other thing that we are going to need to do is express concentrations in
terms of molar flow rates, and I won't write this out for each one because it is the same
in every case, but this means for any component , we can write the concentration as being
the molar flow rate of that component i so that we can connect it to our material balances,
divided by the initial volumetric flow rate, but of course the initial volumetric
flow rate can potentially change as we go down the reactor, because we can have a change
in temperature and a change in the number of moles, and those terms allows us to put
just from the ideal gas law we have this kind of relationship, this kind of variation of
the volumetric flow rate which can change as a function of the molar flow rates and the
temperature, and again the pressure is assumed to be constant in this case, so we do not
have a P over P_0 term. So we just write one of these expressions for each of our concentration
variables A through E, and you will see that we need to know things like the inlet volumetric
flow rate. We will need to know the initial molar flow rates of each of our components.
The initial temperature, and of course if we know the initial flow rates of each of
our individual species we can sum them up to get the total flow rate. So I will write
this expression up here, but F_T it is just the sum of F_Ti. Okay, so those are all the
material balances that we need, and if this was in isothermal reaction and we had all the
variables that we talked about, then we would go ahead and solve the material balance for that
isothermal case, but now we need an energy balance because this is not an isothermal
reaction. I have just copied in the correct form of the energy balance relation. I just
wanted to talk about the individual terms in this kind of relationship. So the air that
we are writing the relationship for a packed bed reactor, so we want to keep track of how the temperature
changes as reactants see an increasing mass of catalyst, so dT/dW, and if the temperature
is going up, of course we are going to expect that quantity to be positive. If it is
going down we are going to expect it would be negative, and it would be near 0 for an approximately
isothermal reaction. Lets look at some of the components of this expression.
One of the components have to do with the exotherm of the reaction, and
that is the term that is in parentheses here, and this term we simply multiply the rate
of reaction of a 1 component in each reaction. So you see that we have 3 expressions, here.
3 rates of reaction. This corresponds to the consumption of component 1 in the desired
reaction, and so we need to use the heat of reaction here. That is a variable that
we might need to, for example compute the from the enthapies of formation of the products minus
the reactants. We need to do that in such a way that this is given per mole of A. So we'll
be in units in this problem we will see eventually of the energy per moles of the component, and in
this case it's moles of A because we are defining the reaction rate in terms of moles
of A. Same thing here, we are defining the reaction in terms of moles of A. So we could
look up the heat of reaction in terms of moles of A or compute that per mole of A, and not
for example per mole of oxygen, when we compute the heat of reaction, and the same thing for
component C. That way our units will cancel out correctly, because what essentially this
tells us is the mole of A that are consumed per mass of catalyst per time. So when we multiply
those together what we end up with is an energy per time term. Units ultimately cancel out
in the next example. One useful thing to do here, the signs can get very confusing, so it
is useful to do a sanity check. So these are all exothermic reactions. That means that
all of the heats of reaction should be negative. All of these rates that we looked at the previews
page correspond to consumption of that component, and so as defined they are also going to be
negative numbers, and so this is going to be a positive term that makes dT/dW increase.
Alright, and that makes sense. We expect that the temperature will tend to go up as the
exothermicity of the reaction becomes more important. Usually to avoid getting confused
about the signs it is useful to do that sanity check and make sure each term is bringing
the temperature in the direction that you would expect. Okay so that is the term
that is associated with the reaction. Lets next move to this term in the denominator.
This term has to do with the ability of stream that is flowing through the reactor to absorb
the heat that is being generated in the exothermic reaction, and as we'll see, being removed from heat transfer
in the other term that is in the numerator. Okay, and here is a fairly simple term. We
just need to know the heat capacity of each component in our mixture, and we multiply that
by the molar flow rate, and that gives us the convective heat transfer term. It tells
use how effectively heat can be absorbed or released by the fluid that is flowing
through the reactor, and so we are going to need in addition to the heats of reaction that
we are going to need to have available for us, we are going to need to have heat capacities
for each of our individual reactants. The final term is this heat transfer term that
comes about from the fact that we have a cooling fluid. We have cooling water on the shell
side of our tube and shell reactor geometry. We are going to expect this to have a net
decreasing effect on the temperature. Alright, and so in this case I actually need to change
a sign here, so it's good that I caught this. This should be a negative sign. Alright,
so all these quantities as we will see are positive, they are constants associated with the
heat exchanger itself. The T, the T is the temperature inside the reactor. It is greater
then the temperature of the cooling water. So this is a positive number, but we know that
the effect of the cooling is to bring the temperature down, so we expect an overall
negative term. So we know that the sign here has to be negative. Okay, so we say this should
be a negative quantity. What are these different terms? So the U is the overall heat transfer
coefficient. In many of the problems that you will see in your textbook, this will probably
a term that is just given, but it is a term that you learn how to calculate in heat transfer class
from summing the individual resistance to heat transfer on the tube, wall, and shell
sides. Density is just the catalyst density, and that needs to be in there because of this
term basically. We are not going to go through the detailed derivation of this term. We will
just talk about what it is. This term is basically the amount of surface area you have per volume
of catalyst and then when you divide that by the density that gives you kilograms per volume
of catalyst. So that cancels and you end up getting the surface area that you have available
for heat transfer divided by the kilograms of catalyst, which is what you need in order to determine
dT/dW. So this surface area per volume, typically we would use a tubular geometry. So for the surface
are for a tubular type geometry, the surface area would be the height times the circumference
of the cylinder, so it would be pi*D, or the circumference times the length, and the volume of the reactor
would be again the length,and in this case times the area of the circle. So if you know the
geometry of the reactor, and that is actually something else that you can optimize, then you
can determine the value of A, and in many cases you will just see that it is given for a particular
problem. Again, the temperature is something that we are solving for, the temperature
inside the reactor. The T_A is actually also something that we can potentially solve
for. We can assume that T_A is uniform throughout the cooling fluid or if you really have
a true cocurrent flow, you actually need to write a material balance for that component,
and to do that you get a simpler form. We will go ahead and do that. T_A can vary with catalyst mass, and you expect
to have a similar term here. In this case we wont have a negative sign, because this
heat transfer tends to increase the temperature of the cooling fluid. We do not have a reaction
term. There is no reaction occurring in our cooling fluid, but we do have a convection
term, and in this case it would just be the flow rate of the cooling fluid times the heat
capacity for the cooling fluid. So the balance is very similar, but ends up being simpler
in form. So you see that we are going to need, and this is maybe more of a real world situation,
that if we are going to design this reactor we are going to need to know a lot of parameters
that are not just provided for us necessarily in a problem statement. We are going to
need to figure out for a reactor of this type of geometry, what are reasonable heat transfer
coefficients, what is a reasonable geometry, so this A factor can potentially help us out.
We need to know the density of the catalyst. We need to know all the heat capacities. We
are going to need the kinetics so that we have rate constants for the different rate expressions.
We need to know all the enthalpy changes associated with the reactions, and so in the next part
of this problem we are going to use a numerical solver package, which is POLYMATH in order
to do the solution. So that the problem does not take hours, I am going to assume certain
values for those parameters, but I think it is important to recognize that the values for those parameters
sometimes are not just right at your fingertips they are things that you need to go and determine, and
perhaps do an experiment to determine. But we have now set up the equations that we are
going to need to plug into POLYMATH along with plugging in some of the values of those
those constants in order to be able to simulate the performance of the reactor. So we will
continue that in the next part of this problem.