Mass transfer through Desert Fridge... Fick's law?

Tried to solve the problem of mass transfer through the Desert Fridge problem by assuming steady diffusion (there will actually also be convection but lets keep it simple for now) and using the approach suggested by Taylor on page 131 of Multicomponent Mass Transfer (he actually looks at mass transfer from gas to liquid – I have rewritten his approach in reverse) as follow:

The transfer of a liquid to a gas may be regarded as taken place in three phases:

1) Diffusion from the bulk liquid (water contained in the porous sand and clay) to the liquid surface (at the surface of the outer pot).

2) Dissolving of the liquid into the gas (to give a water vapour – air mixture in our case).

3) Diffusion from the liquid surface to the gas (warm air of the surroundings).

Now according to section 3.2.3 Concentrations at interfaces in Transfer Processes by D.K Edwards the mass fractions of water can determine across the air-water interface as follows:

Define imaginary surfaces infinitely close to, and at either side of the interface as shown in the diagram below:
Where:
x= mole fraction
u = liquid phase
s = gas phase (vapour-air mixture)
e= bulk gas phase?

• Mass fraction of water at u (m_H2O_u) = 1
• For (m_H2O_s):

1. Look up saturation pressure of steam at the temperature of the water (p_sat_H20_s) in thermodynamics steam tables

2. Then mole fraction is given by:

So in our case we have:

3. Then mass fraction is then given by:

Where:
m= mass fractions
M = molecular weights (M_air= 29g/mol , M_H2O=18)
x = mole fractions
subscript i is water in our case and j is water and air.

So we have:

Using the definition of mass fractions (i.e sum of mass fractions = 1) we can further simplify giving:


(Nb. this is where the first equation in this sweat sweat cooling example must come from)

• And finally for mass fraction of water in the bulk air (m_h2O_e) we can directly deduce this as the mass fraction of water is the humidity of the air (kg water vapour/kg air) (I think) which can be found using a psychometric chart for a known dry bulb temperature and relative humidity (can this be done analytically or can data from psychometric chart be used in matlab?)

And it turns out we have exactly the same mass concentration profile through a porous sweat cooling wall as shown in diagram below (this was copied from some book which I cant find now) :

Now to solve for the mass transfer rate iv tried using:

a) Fick's law:
where:
j = diffusive mass flux
p = fluid density
D = diffusivity coefficient
dm/dz = mass concentration gradient (m =mass fraction, z= diffusion length)

The density of water vapor will be a function of temperature . Now do we use the temperature of the water at the surface of outer pot or the surrounding air here?

The diffusivity coefficient can be taken as approximately 22 m2/s (or does it vary significantly with temperature?).

Finally, although we have can derive the mass fractions at the various phases (see above) I have no idea how to determine the mass concentration gradient as this will require some knowledge of the diffusion length.
For diffusion through the porous layers, it may be their thickness, but what would it be for diffusion from the surface of the outer pot to the surrounding air?

Stuck here so try another method:

b) Mass transfer driving force…