This is an example of a membrane reactor in a simplified version to demonstrate the properties
and the idea of a membrane reactor is if you can preferentially remove or add a component
through the walls of reactor, so you have walls that are porous to at least some of
the components, and so the wall is a membrane and there's some selectivity as to, for example
which components you remove. So in this case we're looking at a dehydrogenation reaction
of A going to B plus hydrogen using a plug flow reactor whose walls are permeable only
to hydrogen. For example, this might be a palladium membrane as the walls of the reactor.
We feed in just pure A to the reactor and it's a reversible reaction and so this is
the motivation for trying to remove hydrogen from the products and therefore shift the
equilibrium further to the right so we have a forward reaction and a reverse reaction,
the feed is at 20 atmospheres. So of course as we go to higher pressures the equilibrium
is pushed back towards reactants. As we go to higher pressures we are increasing the rate
of reaction because of higher concentration and we're providing a driving force for hydrogen
to go through the membrane. So we're going to assume there's no pressure drop through
the plug flow reactor and we're going to assume it's isothermal. We can certainly relax that
assumption just by adding in another equation for the energy balance, but just to demonstrate
the behavior. The total reactor volume is 150 liters and then the flow rate of hydrogen
through the reactor walls, so this is hydrogen only, and say we're modeling this similar
to what behavior might be seen for a palladium membrane where the flux is proportional to
the concentration of hydrogen to the 0.5 power. So this flux... So this is moles per volume
per time and so we've basically built into this equation what's the surface area per
volume of the reactor and for a different shape reactor this number would be different
for the same walls. So we've basically included in this term, in this number here, the surface
area per volume and then the question is, what's the molar flow rates look like as a function
of volume? And so we have the rate of reaction and essentially we're just going to write
down mass balances. Change in the flow rate of A respect to volume is minus r. Alright
so r is in this case, minus r means that A is reacting for a reaction and then A is reforming
in the reverse reaction and B of course is fairly similar except it's positive r and
then the flux of hydrogen, again it's positive r, and then it's minus the flow rate per volume
through the membrane. So essentially we're going to solve these equations simultaneously
which means we have to write the rate, k forward, concentration of A, k reverse, the concentration
of B, concentration of hydrogen. So we have to have equations for the concentration. So
the concentration of A would be the mole fraction of A times the total concentration. The total
concentration if we assume ideal gases from ideal gas law is P over RT and then there's
similar terms for concentration of B and likewise for concentration of hydrogen, same form.
Mole fractions are related to the molar flow rates. So mole fraction of A is the molar
flow rate of A over the total flow rate. Mole fraction of B, molar flow rate over total.
Likewise for mole fraction of hydrogen, same behavior, and the total flow rate is just the flow rate
of A plus the flow rate of B plus the flow rate of hydrogen. Now the units, moles per
liter per minute for the rate and so we're going work in those units and use atmospheres
and the gas constant is going to be 0.08206, so this is atmospheres, liters, moles, Kelvin.
So essentially then the problem is almost finished. We have the equations that we can
put into a differential equation solver. In this case we use POLYMATH, so we put in
the equations and the equations that calculate concentrations. We put in of course, the various
constants that we need for the rate constants and for the pressure and the temperature and
then we solve from volume equals 0 to volume equals 150 liters. And so I've copied in the
POLYMATH program where the flow rate of A, notice the stoichiometric coefficient here
of minus 1, initials conditions for A. The volume, initial value of course is 0 at the
beginning, we're going to integrate to 150 and then flow rate of B, its stoichiometric
coefficient is positive and flow rate of hydrogen also has this term because of the membrane
reactor. So the rate written in terms of the rate constants, concentrations, the flux through
the membrane, the concentration of hydrogen to the 0.5 power, here are the rate constants
and the pressure. So total concentration from ideal gas law is just P over RT, temperature.
We need the concentrations for the rate expressions and we get them in terms of mole fractions
and then the total concentration and the mole fractions we just get from the flow rate.
So the flow rate of A over the total flow rate is the mole fraction of A and so if we
now ask it to solve, we can plot the behavior and here's the output from the POLYMATH program.
We're plotting the flow rates on the y-axis, so these are molar flow rates. This is the
reactant A, flow rate of A, this is the product B and this is the hydrogen. So of course we
form B and hydrogen 1 to 1 stoichiometric ratio, but as the hydrogen concentration increases
then we have a bigger driving force for hydrogen to permeate though the membrane and then as
the reaction slows down we're permeating hydrogen through. That allows the reaction of course
to not stop completely but it's certainly slower and so the hydrogen molar flow rate
moving down the plug flow reactor goes from a maximum and now decreases as the rate that
it's being made from the reaction is slower than the rate that can permeate through the
walls and so a membrane reactor then has potential to allows us to shift the equilibrium of the
reaction by continuously removing hydrogen.