Oh geez, this is fairly excellent. I think about this problem fairly often while waiting for my coffee to brew, but I never really thought I'd be able to hash it out anywhere. Rockmastermike, you have made some great points - individual inquiry stagnates so often...
You can't really treat the compressed mass of coffee as a point diffuser. There are two different rates of diffusion at work. Diffusion within the "puck" of grounds (slow) and diffusion at the surface of the puck to the mass of water (fast).
Ah, definitely! The "grounds"-region and "coffee"-region need to be considered distinctly even in our simple approximation. I know also realize that my terminology was off and this has confused you. I said point - I should have just said 1D diffusion problem, since....
Diffusion of oils from within a flattened disk of coffee at the bottom of the container with only the top in contact with water would be much more analogous to a case of diffusion of heat from within a body where only one side of the mass is losing heat (the rest is infinitely insulated) AND the thermal conductivity of the mass is low. In that case the top would get cooler than the bottom of the mass with a temperature gradient given by the conductivity. The bottom cools more slowly.
..I believe that under your above assumptions Fourier's law will be analogous to Fick's, right? So the solution to the above which you suggested should be identical to the "classic" 1D diffusion solution for a finite source? Gradient here will be given by diffusivity, assume diffusivity is low so we are always at equilibrium, etc, just as you said for conduction. (We are of course taking the sane choice and throwing convection out the window! How safe is this, do you think?)
In this case the top of the "puck" donates oils and such, but the coffee inside the compressed mass donates oils and such at a much much slower rate. The rate of diffusion would greatly depend on the permeability of the mass.
Here you've really got it, I think. Once the plunger is down, the dimensionality of the problem is effectively reduced, by a very large amount. We go from jillions of high-surface area particles dispersed throughout a liquid medium - meaning a huge effective area of exchange - to just the surface area of the puck. From that, I bet you could make a strong case based just on dimensional analysis that grounds --> water exchange is practically shut down, at least in comparison to the brewing step. I'm not sure what you mean about permeability though - surely the puck is fully permeated? Any water in the interstices must come close to equilibrium concentrations fairly quickly - errr.. but then I can only hazard that guess because I'm assuming no convection!
With no fluid flow between grains, the rate of diffusion will be very low. Of course once a particular molecule of something works its way up to the water then it goes pretty quick. But in the case of our puck the top of the puck loses soluble material but the concentration remaining of soluble material would be greater the deeper you measured into the puck (just as the temperature rises as you dig into the earth that is losing heat to space). I suppose you could really just treat the puck as one large grain with only the top surface available to the water.
Again, I kind of automatically ignored convection / fluid flow within the puck. Really the more I have to say that, the less I think it's a valid assumption - but it's the only way I have to get a grip on the process. Lack of fluid flow doesn't effect chemical diffusion - "fluid flow" / convection being bulk mechanical processes, while diffusion is atomistic (or er, mollecular in this case!). In any case, your last sentence in the above quote makes the dimensional analysis point even more succinctly!
I can't resist some back of the envelope math any longer...so the simple math myself and rockmastermike (I'm familliar with Fick's Law while rockmastermike has referenced heat conduction, described by Fourier's Law) have been employing boils down to a certain rate - a flux. Flux is a rate which has units of [stuff] over [area]*[time] - the amount of stuff which passes through a small piece of area in a given time. So total flux in our coffee press will be directly proportional to, or at least strongly a function of, total surface area.
For round number's sake, let us just say that we have 20,000** uniformly spherical coffee grounds, each 2mm in diameter, equally dispersed around the coffee (but somehow there isn't any convection or fluid flow...) - total surface area diffusion is then ~ 1 m^2.
Now we press down the plunger: some very brief googling finds me that a Bodum 8-cup press has a 3 and 3/4 inch diameter, so ~ 0.1 meters. This means the coffee grounds puck has a surface area of ~0.05 m^2.
This is all a bit dodgy - I got the 20,000 number by back extrapolating from 1 tablespoon per 6 oz of water, assuming we're making 8 cups and that the grounds are uniformly 1mm in radius. The puck will have a rough surface - not a flat circle, and thus will have greater than 0.05 m^2 surface area, and we also have convection etc. But still I think I made the point - the surface area drops by several orders of magnitude. As rockmastermike put with much fewer words - we go from mixed particles, completely filling the water, to just the surface area available. Extraction will definitely continue, but the rate will probably be 10 or 100 times slower than during the brewing time.
Uh, then, the devil is in the details, however...and despite appearances, I've neglected an enormous number of them.
(Thanks for the welcome, Zach.)
** I had to edit this a few times for algebra mistakes and also forgetting the packing fraction of random spheres. Note also that it is *strongly* dependent on the diameter of the grounds, so I may be guilty of fudging. Input?
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