# Chemical Forums

## Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: fb1234 on October 31, 2018, 04:10:16 AM

Title: Fick's Diffusion for water evaporation INTO air bubble
Post by: fb1234 on October 31, 2018, 04:10:16 AM
Hey! I'm working on a problem that tackles the uptake of water vapour in air bubbles (e.g. in a bubble column) and I'm having a surprisingly tough time getting some normal looking answers for the amount of water that is being diffused/absorbed into the bubble.

I thought that Fick's law of diffusion would be a very simple way to initially approximate this water uptake, but the amount of water that's being evaporated into the bubble greatly exceeds (e.g. bubble volume ~O(10e-9) m3, water vapour flow into bubble ~O(10e-5) kg/s) the mass of the bubble itself, so the numbers are just way off. The formula I'm using is the following:

Flow into bubble = 4*pi*D*(∆c)/(1/r - 1/R)

Where:
Flow into bubble = [kg/s]
D = diffusion coefficient ~O(10e-5) [m2/s]
∆c = change in concentration ~O(10e-2) [kg/m3] across gas/liquid film around bubble
r = inner radius of bubble ~O(10e-3) [m]
R = outer radius of bubble ~O(10e-3) [m] (e.g. r + film thickness)

I derived it from...

Flow into bubble = D*A*(-dc/dr) = D*4*pi*(r^2)*(-dc/dr)

... which is just Fick's second law applied to a spherical object.

I'm thinking that the huge water uptake might be an issue with the thin film, as the radial coordinates are inappropriate for the tiny thickness of the film which is ~O(10e-7).

Is there anything inherently wrong with the way I'm approaching this problem? Any help would be much appreciated!
Title: Re: Fick's Diffusion for water evaporation INTO air bubble
Post by: mjc123 on October 31, 2018, 05:51:07 AM
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Flow into bubble = 4*pi*D*(∆c)/(1/r - 1/R)
This is dimensionally incorrect. How did you get it from Fick's first (not second) law?
Quote
D = diffusion coefficient ~O(10e-5) [m2/s]
Where do you get this from? I find values of ca. 2e-9 m2/s for the self-diffusion coefficient. Have you found something different for diffusion across a surface?
Quote
∆c = change in concentration ~O(10e-2) [kg/m3] across gas/liquid film around bubble
Shouldn't that be something more like 103 kg/m3 from liquid to vapour?
I'm wondering if a diffusion approach is appropriate at all, as evaporation, unlike diffusion, is an energy-absorbing process, and its rate depends on how much energy (if any) you put into the liquid.
Title: Re: Fick's Diffusion for water evaporation INTO air bubble
Post by: fb1234 on October 31, 2018, 06:46:30 AM
Thank you for the reply! Really appreciate it, as I knew I was making some silly mistakes and wasn't even sure where to start.

Quote
This is dimensionally incorrect. How did you get it from Fick's first (not second) law?
I actually got this from some university material, but it seems I may have miscalculated it. The problem they showed was a tank of water exposed to air, therefore making the water evaporate slowly and diffuse as vapour outwards into the air. They demonstrated how to calculate the flowrate of water vapour from the surface of the water out into the air, which they did via Fick's law of diffusion. I figured maybe it would be possible to rearrange that into a spherical coordinate for a bubble, so I got the following:

Φm = D*A*(-dc/dr) = D*(4*pi*r^2)*(-dc/dr)

Φm/(D*4*pi*r^2)*dr = -dc

Integrating both sides with the following boundary conditions: c(r) = c, c(R) = c0 (e.g. water concentration is higher at the outside surface (r = R) of the bubble)

Φm/(D*4*pi)*(1/r - 1/R) =  (c - c0)

Φm = (D*4*pi)*(∆c)/(1/r - 1/R)

I already knew something was off as the radii don't match up to the concentrations in order to make the flowrate positive, but I was hoping maybe something could be done to fix that, but I guess I was being too hopeful.

Quote
Where do you get this from? I find values of ca. 2e-9 m2/s for the self-diffusion coefficient. Have you found something different for diffusion across a surface?

I actually just got this off of wikipedia. The diffusion coefficient of water vapour in air is given as 0.282 cm2/s.

Quote
Shouldn't that be something more like 103 kg/m3 from liquid to vapour?
I'm wondering if a diffusion approach is appropriate at all, as evaporation, unlike diffusion, is an energy-absorbing process, and its rate depends on how much energy (if any) you put into the liquid.

I see what you mean, and I should have actually clarified initially that the concentration c0 is defined as the partial pressure of water vapour times the molar mass of water. So it isn't actually kg/m3 of water in liquid water, but the concentration of water vapour immediately at the gas-liquid interface, so at r = R.

Overall I was also skeptical of the whole diffusion approach, especially since there are so many other factors involved (e.g. convection, eddies, etc), but I don't have the expertise in the slightest to try to solve the model with something like CFD. I found some theses online that included diffusion using a modified Fick's law ( Φm'' = (ka)*Vliquid*(Ca - Ca,eq) ), so I tried to potentially imitate those. Do you possibly have any other suggestions on how to model this kind of situation?

Title: Re: Fick's Diffusion for water evaporation INTO air bubble
Post by: mjc123 on October 31, 2018, 08:24:38 AM
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Φm = (D*4*pi)*(∆c)/(1/r - 1/R)
OK, I hadn't spotted you'd dropped r2 from the area, my mistake.
Quote
The diffusion coefficient of water vapour in air is given as 0.282 cm2/s.
Ah, vapour in air. But you still haven't got the rate of transfer across the surface. If you assume evaporation is faster than diffusion, so C0 is maintained constant at a value determined by the equilibrium vapour pressure of water, the approach should work. But if evaporation is slower, C0 will decrease. I don't know which is the case.
Are you taking into account that pressure inside a bubble is higher? (I don't know offhand if it makes much difference for mm-sized bubbles.)
Quote
the concentration c0 is defined as the partial pressure of water vapour times the molar mass of water
That doesn't give a concentration. You want mol fraction(=partial pressure/total pressure)*molar mass/molar volume
(Are you remembering to use molar mass in kg/mol not g/mol? And molar volume in m3/mol not L/mol?)