Mass Transfer - Need help on setting it up

[dweeegs]

Homework Statement

Not looking for solutions, just confused with the problem set up. Need help with the boundary conditions. Here it is:

A porous ceramic sphere of radius R1 is kept saturated with a pure component liquid A. The vapor pressure of A is 50 torr. This sphere is surrounded by a concentric solid spherical surface of radius R2. Species A reacts at the surface r=R2, according to A-->B(s). Species B is deposited as a solid film by this reaction, which is first order.

Assume that the system is in a steady state, and derive an expression for the partial pressure of species A at the reaction surface r=R2. Assume that the space between the two spheres is isothermal and at a uniform pressure of one atmosphere.

AND the professor gave us some extra info:

Because the reaction rate is finite, you must specify a 3^rd type boundary condition on the reactive surface. Assume that the space between the two spheres contains an ideal gas. Solve for the concentration profile and use this expression to obtain the requested result

2. Homework Equations and attempt at a solution

D∇^2*Ca + Ra = 0

Where D = diffusivity. That's the above equation that I have reduced already from its original form.

For starters, I'm assuming no huge velocities, steady state (no time derivative), and from now on I will represent the concentration gradient by dCa/dr, since now I will be assuming mass transfer is in the radial direction only. Spherical coordinates will also be introduced.

For first order kinetics, Ra = -kCa. Rewriting everything consider what I just said,

D*[(1/r^2) d/dr*(r^2 * dCa/dr)] - kCa = 0

I am sure I can solve this... but I don't have the right boundary conditions because I don't quite get the physics of the problem.

Here are my guesses at boundary conditions for the above equation

Boundary Condition #1: at r=R1, Pa = Pao (the partial pressure is as specified; ie 50 torr. This is converted to concentration using the ideal gas law: Ca = (Pao)/RT)

Boundary Condition #2 (the 3rd type BC): Would it be that the flux (Na) at r=R2 would be equal to the diffusivity times the concentration gradient?

I'm not sure what a third type boundary condition in mass transfer is. We haven't gone over that.

Thank you for any help. I don't want assistance in solving the differential equation, just the boundary conditions

 

[Chestermiller]

Your differential equation is incorrect. There is no chemical reaction occurring except at the boundary. The rate of reaction at the boundary is in moles/cm^2/sec, and is proportional to the concentration of A. So the rate constant is in units of cm/sec. The diffusional flux of A at the boundary has to match the rate at which A is consumed by chemical reaction per unit area at the boundary.

 

[dweeegs]

Your differential equation is incorrect. There is no chemical reaction occurring except at the boundary. The rate of reaction at the boundary is in moles/cm^2/sec, and is proportional to the concentration of A. So the rate constant is in units of cm/sec. The diffusional flux of A at the boundary has to match the rate at which A is consumed by chemical reaction per unit area at the boundary.

So let me if I understand you correctly,

The reaction isn't occurring while the gas is diffusing, thus Ra isn't included in the differential equation. However it is introduced when we consider our boundery condition?

Something along the lines of:

-D(dCa/dr) = kCa at r=R2

Is that correct? Thank you for the help. I was looking at another example of diffusion within a film in which there was homogenous reaction.

 

[Chestermiller]

So let me if I understand you correctly,

The reaction isn't occurring while the gas is diffusing, thus Ra isn't included in the differential equation. However it is introduced when we consider our boundery condition?

Something along the lines of:

-D(dCa/dr) = kCa at r=R2

Is that correct? Thank you for the help. I was looking at another example of diffusion within a film in which there was homogenous reaction.

Yes. This is exactly correct.

 

[dweeegs]

Yes. This is exactly correct.

Thank you!