- [Instructor] We're gonna to talk about membrane reactors
in this screencast.
And the reason is when we start looking
at exothermic reactions and we ask what are
the options when the reactions are limited by equilibrium?
So I am showing an energy diagram
for reaction A plus B goes to C plus D.
Exothermic reaction.
The activation energy for the forward reaction is here.
And the activation energy
for the reverse reaction is this value.
As we raise the temperature, the ray
the reverse reaction increases more.
And so our equilibrium conversion decreases.
So increasing the temperature will not help us.
Well, we can lower the temperature
and that will now favor the products.
The problem is it will also lower the reaction rate.
So one's possibility is to lower the temperature
and this will indeed work
but it may then require a very large reactor to
get conversion because the rate will slow down.
The other is to take advantage
of membranes and selectively remove a product.
So for example
if we could selectively remove product C as it forms
then more an A and B would react together to make more C
and D to approach equilibrium.
And so if we can find a membrane that is selective
then we have the potential to shift the equilibrium
to the right and keeping the temperature high.
So keeping the rate of reaction high.
So here is a schematic representation of a membrane reactor.
We're feeding our reactants A and B.
So F is a molar flow rate of A and B into the reactor.
We're assuming for now it's an ideal membrane only C goes
through the membrane, so the membrane is here and
and we have C going through.
And then C is leaving this direction,
leaving reactors A,
B
product C, that didn't go through the membrane
and product D, of course, we have to have some C leaving
or we wouldn't have a driving force
for C to go through the membrane.
And the idea is if we look at,
say the rate
of formation
of product D,
will be a rate constant
for the for reaction,
say concentration
of A and concentration of B, for example,
rate concent reverse reaction concentration C,
concentration D.
And what we're doing is decreasing the concentration
of C by using the membrane reactor.
In this case, I've represented it
as the membrane is a catalyst
since most reactions are catalytic.
So catalytic reaction takes place
in this porous membrane and we preferentially
are exclusively remove C and a real membrane.
Typically, we also are gonna remove some A and B and D
and we need a good enough membrane to selectively
remove C to make it economical.
The other way we could envision a membrane
reactor is to fill a membrane with a catalyst.
And so the catalyst causes reaction.
The membrane is catalytically inert
but is preferential for one of the products.
So here shown the representation
of this where we're flowing.
For example, let's look at a simpler reaction
A reacting to B plus C.
And so flow rate
of a molar flow rate enters
leaves is A and B and C.
And then through the membrane wall
which is here would be for example, C diffusing through.
And the diffusion permeation would take place because
of a pressure difference or concentration difference.
In other words, we need the concentration of C
on the reactor side to be higher than the concentration C
on what is referred to as the permeate side.
We have to have a driving force
for that permeation take place.
Strictly speaking, it's a chemical potential.
We'll look at here in terms of concentration.
So again, we're assuming an ideal reactor.
So now if we write down the mass balances for this reaction
assuming it's plug flow inside the pack bed, so the change
molar flow rate of A, that's FA
cumulative volume, it's gonna be rate reaction
the rate reaction, maybe minus K
CA plus it's okay forward
K, reverse C, B, C, C.
For B, we have the exact same type, except this would be
minus RS of A.
And what's different is when we write the balance for C
we have minus RS of A again, and then we have
an additional term which
represents the flux
through the membrane
times the area for permeation
per volume.
FD is the diameter of the tube.
The inside diameter PI times D times DZ.
So DZ, some differential
distance where Z is distance
down the reactor per differential volume.
And we can of course reduce this minus RS of A.
This differential volume would be PI D squared over 4.
That's cross sectional area times D Z.
So that this just reduces to 4 over D
The only thing we need is the flux of C
which would be the permeance
of C times the concentration of C in the reactor
minus the concentration C on the permeate side.
So this concentration of course
is the same concentration that's
in the reactor that affects the rate of reaction.
This is on the permeated side.
We need a driving force.
Typically we're gonna have of course, a higher pressure
inside reactor to make the membrane reactor work.
And then other thing we have to keep
in mind is our total molar flow rate is going to change.
Now in this case we have one mole going to two.
If this were gas phase reaction
the molar flow rate would increase
which affects the concentrations.
Now we're gonna removing C through the membrane.
So we have to take that
in account and we calculate the total molar flow rate.
Now we can solve these differential equations
same way we would for other plug flow reactors as long
as we have a value for the permeance.
So this is the permeance which is a
property of the membrane.