# Deriving the Differential Equations of Mass Transfer 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.

### 4 thoughts on “Deriving the Differential Equations of Mass Transfer”

• March 11, 2014 at 5:13 pm

When you decide to change the sign of mass in – mass out, you justify this by noting that you cannot have () accumulation with greater in than out. Could one also justify this when we take the definition of a derivative for each flux direction by saying that final – initial then take z>0. This gives the same effect yes?

• September 16, 2016 at 9:22 pm

Hey thank you so much it was really helpful! 😀

• December 20, 2016 at 2:21 pm
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