In this screen-cast we are going to work through a diffusion problem where we work in spherical
coordinates.
So looking at our problem statement we have UF6 that is being prepared by exposing pure
uranium pellets to fluorine gas at 1000 K and 1 atm.
The molecular diffusion of fluorine to the pellet gas is thought to be the controlling
step.
The reaction shown here occurs irreversibly and instantaneously on the surface.
And the diffusivity of the fluorine gas through UF6 gas is 0.3 cm^2 per sec at this temperature
and pressure.
Far away form the pellet the gas is basically entirely fluorine.
So for this problem we want to determine the moles per time.
The production of UF6 that is produced when the pellet's diameter is 0.5 cm.
We are given a gas constant value that we may use.
First place we are going to start is by drawing a diagram.
So we have a spherical pellet.
This is uranium.
We are working in spherical coordinates so I am going to label our radial direction little
r.
We have some radius big R. And we know we have fluorine gas that diffuses to the surface
and the UF6 that diffuses away.
So the second thing that we want to do is fill in what we know.
And since we are interested in UF6 I am going to call that species A and refer to that as
we move forward in subscript form.
And last thing I like doing before we really start working on the problem is system behavior
and in this case we look at this problem statement and we know that we are looking at UF6 and
it is not a function of theta or phi so we are just looking in the radial direction and
really we are looking at a change in concentration in that radial direction.
So we are looking at the flux of A away from the surface.
We also know that we are going to assume steady state behavior and we have no reaction in
the gas phase.
So even though there is a reaction on the surface we care about the flux away from the
surface so within the space that we are interested in there is no reaction going on.
So I like to start our mass transfer problems with the continuity equation.
So here is our continuity equation in spherical coordinates.
So we start writing down the things we are going to assume or use in our problem to simplify
this equation.
And so we did say steady state behavior.
We have no accumulation of our UF6 in the gas phase that we are looking at.
So this allows us to get rid of this first term.
We know that the flux is not a function of theta or phi so that allows us to get rid
of any term that we are looking at as a function of theta or a function of phi.
We have a reaction at the surface, not in our gas film so this term is zero.
So as you can see at this point what we are left with is this first term inside the parenthesis.
And so we have i/r^2(d/dr)(r^2NAr)=0.
This allows us to get rid of the r^2 and this is what is left of our continuity equation.
And when we are down to this this is saying that r^2 times our flux in the radial direction
is going to be equal to some constant.
So now we have used the continuity equation and simplified it down so we now need to have
some expression for our flux.
Now if you think about our flux and our flux only in the radial direction you could write
it as the following where we have our concentration of our gas times the diffusivity times the
gradient in terms of mole fraction in the radial direction.
That is our diffusive flux part of the equation and now we are going to add the bulk flow
component.
So that is the mole fraction of A times the flux of A plus the flux of B. So in this case
we are looking at fluorine being the component B. Now what we want to do is re-write the
flux of B in terms of A so that we can combine them together and simplify this equation.
So if we go back to our diagram, you can see that the flux of A, for every one mole of
A that is fluxing away from the surface of the particle we have three moles of fluorine
that has to come to the particle to react.
So our flux of B in this case is going to be 3x that of A and in the opposite equation.
So we say the flux of B in the radial direction is -3*that of A. So now we could plug this
into the parenthesis and if we simplify this, this becomes -2*NAr.
So at this point we are going to simplify this further, bring this to this side and
again get it back into an equation that starts with the flux of A. so now we have our flux
expression for A in the radial direction.
So at this point we want to combine what we have from our flux equation and what we have
from our continuity equation.
So from our continuity equation we said that r^2 times the flux in the radial direction
had to be equal to a constant.
So what we are going to do is multiply both sides by r^2 in our flux sides to get that
to be true.
And what this is going to allow us to do is separate variables and set up the following
integral.
We are going to bring over our r values to the left side.
Since we know this is constant we can bring it out of our integral and set up our integral
on this side and we are going to do the same thing assuming the concentration of our gas
and the diffusivity don't change then we can set up our integral here.
And so now it is just a matter of choosing our appropriate boundary conditions.
So we are going to evaluate the flux of our UF6 from the surface R to a distance infinity.
And we are going to say at R we have a mole fraction ya equal to 1 since the reaction
is going to be instantaneous and we are going to say it is zero at infinity.
When we evaluate these integrals with these boundary conditions you should get the following
equation which will now simplify which you should now be able to see that r^2 times our
flux is equal to a constant.
And this constant is the following.
So we know we have means to calculate the concentration of our gases in this film that
we are looking at.
We have a radius of our particle, we are given a diffusivity, the rest we can calculate,
but unfortunately that doesn't tell us what the rate of production of our UF6 is.
That tells us what the flux times our r^2 values is which is therefore some constant.
Since we are asked in the problem statement to determine a moles per time this is going
to be equal to our flux of A in the radial direction times an area.
Right a flux times our area will give us a moles per time.
So in this case this is very close, but not quite, the cross sectional area that we care
about for our sphere is actually going to be 4pi*r^2 so what we need to do is multiply
both sides by 4pi and this allows us to now calculate, filling in our values with the
rate of production or in this case we could rewrite this as W is equal to 4pi, in this
case lets simplify this and say 2pi* radius, concentration, diffusivity and the natural
log of 3.
So I will go ahead and plug in our values.
After we calculate what the concentration of c is we could calculate c using the ideal
gas law and we will assume that here is just a pressure over a gas constant times our pressure.
This gives us a concentration of 1.218E-5 moles per cm^3.
Let's see if those units work out in the equation we just had.
So if we use a radius given to us as 0.5 cm we now have our concentration, our diffusivity,
0.3 cm^2/sec.
it looks like we should be able to calculate this out.
You should make the appropriate change for the radius, we are using a radius not a diameter,
the pellet diameter was given at 0.5 so the radius is 0.25.
So when you put this all together you should get 6.3E-6 moles per second.
And so that is our production rate of the UF6 away from the surface.
Now one thing to check is it does have a positive sign associated with it because we are moving
in the positive radial direction.
One thing to ask also, based on our assumptions is that even though we do have the uranium
being reacted with the fluorine gas and therefore we would assume that the spherical particle
would decrease in size with time, we are really looking at that diffusion path from the radius
surface to a radius of infinity and we are going to say that that is essentially constant
with time.
And so although we said it was a steady state, really we call this a pseudo steady state
approach for doing this material balance on this particle.