In this screen-cast we are going to work through deriving the differential equations for mass
transfer in Cartesian coordinates.
For those of you familiar with this approach from momentum and heat transfer, we will be
using a shell balance method to derive these equations.
This should help you see where they come from and therefore use this approach given other
conditions.
So let's start with a control volume.
And we are going to say it has sides dx, dy, and dz.
So some differential length in those three coordinates.
So through this control volume that contains A and lets say some other species, and let's
say it flows through this control volume.
So if we go back and think about the Reynolds transport theorem, we could write out the
following integral form of the conservation of mass through this control volume.
Our first term evaluates the net flux of mass in and out of our control volume, through
the control surfaces, hence why I have cs.
So we evaluate the dot product of our velocity times our density, over our differential areas,
for each of the control surfaces the mass transfers through.
And then we also had the following term for the control volume which is our net rate of
accumulation of mass within the control volume with time.
Now we say this has to be equal to zero.
However, since this could be involved in a reaction where there is consumed or produced
we also need a term to account for this reaction.
So what we are going to do is subtract out the net rate of production of A on the left
side so that everything is still equal to zero.
And I will show you what that looks like.
So this is our overall conservation of mass.
So what we are going to do is expand these terms and write it in terms of our control
volume dimensions that we have up above.
So let's start by looking at the mass flux that we have in both directions.
We know the mass flow rate is just the density of the fluid times the volumetric flow.
And we could re-write this as the density of the fluid times the velocity of our species,
in this case we are working with A, times the area A is passing through.
So we could look in the x direction, we could redefine this mass flow rate as a mass flux.
So I am going to re-write it as mAx to indicate that we are looking in the x direction for
the mass flux of A. So really what I am doing is I am dividing by this area on this side
so we are saying this mass flux is equal to rho*va.
So now you see that we have this up in our term, and for our x direction I could write
our mass flux in the x direction times our differential area, in this case is dydz.
So for entering we would evaluate this at x.
And then we would do the same thing on the other side of the control volume.
So if I put an x here and x plus some change in x, differential change in x, delta x, on
the other side then we would also be evaluating this through the same cross sectional area
at x+delta x.
Now the problem with the way we wrote with is that if we have more mass coming into our
control volume than leaving then this value that we have written is going to be positive.
And if you look back up at the equation up here, a positive value for our net flux means
that our accumulation term has to be negative.
But that is not the case here, assuming that there is no reaction, we would have to have
a positive accumulation because that is what is actually occurring, more is going into
the control volume than that is leaving.
So we actually need to write this balance, this net flux, in the opposite manner.
So now we have our mass flux in the x direction times it's cross sectional area to give us
a mass flow rate and we are evaluating this on the exit side minus what is coming in to
give us our net flux term.
So we do the same for our y and our z.
So now we have written our mass flux terms that are coming in and out of our control
volume.
So now let's work on our second term, the rate of accumulation.
So if we look at this equation here we could just re-write this as drhoA/dt times our differential
volume.
In this case deltaXdeltaYdeltaZ.
The rate of generation of A is written in a very similar manner because we have that
on a per volume basis we are just multiplying this by our differential volume, deltaXdeltaYdeltaZ.
And then the dimensions will just be mass over time which were the dimensions for the
other two terms.
Now we just put all those terms together set them equal to zero like we have before, we
are going to divide everything by this volume term, deltaXdeltaYdeltaZ, and evaluate in
the limit as deltaXdeltaYdeltaZ approach zero for our differential volume, and what happens?
We get the following simplification.
We could write these three terms together as you have seen before as our gradient operator
of our mass flux of A.
We are going to bring over our accumulation term and subtract out our reaction term, set
this all equal to 0.
Now we have our differential equation for mass transfer of A. So you can the dimensions
on all of these terms are mass per volume per time.
Our mass flux is being evaluated in these directions so we get a mass per volume per
time.
So we could do the same thing on a molar basis.
So I want to do that really quick.
So if we look at the moles of A entering our control volume, leaving our control volume,
we will have some change of moles with time inside of our control volume plus whatever
moles we generate or take away based on the reaction.
Now another way to write this is on a per volume basis just like we did on the mass
differential equation above.
So now this becomes moles per volume where concentration of A.
We could re-write our moles generated as a reaction term so I am going to write a large
Ra.
So this becomes a molar production of A per volume where as before we wrote it as a little
ra to indicate it was a mass per volume per time.
And one way we could re-write our moles of A in is if we wrote it as a flux of A like
we did above for the flux of mass.
So the flux of A the molar flux of A we write as Na and maybe in this case I am writing
this as N is x and that is going to be times our area.
So this is the same think as moles of molar flow-rate in.
So our molar flow-rate out will be the flux of x times A on the other side.
But as we said we are doing this whole thing on a per volume basis.
So let me write out what this whole thing looks like.
So we have our flux of A in minus our flux of A out.
And recall we had to rewrite this the other way to have it balance properly and if we
multiply this by our area and divide by our volume then we get the following term.
We then can add our accumulation term and subtract out the rate of production of A.
Set this all equal to zero and it almost looks good, except this doesn't quite look the same
as it did for the mass equation.
However, hopefully you see the similarities, the dimensions will give us a length on the
bottom so when we take the flux we are evaluating it on each control surface then indeed we
get that this becomes the gradient of our molar flux of A. so hopefully you can see
how we start with our integral equations looked at a control balance and came up with our
differential equations of mass transfer.