Diffusion, theoretical question

[ptyx]

The 'rolling a tank to make sure the nitrox is mixed' story got me thinking (not necessarily a good thing as you'll see).

Assuming a set of doubles (AL80 for example), isolator closed. Right cylinder got filled with 3000psi O2, left cylinder with 3000psi air.

Now the isolator is open. How long does it take to reach equilibrium (EANx60 I guess): seconds? minutes? hours?

(Is there a thermodynamics forum on SB?)

 

[TSandM]

Rick Murchison did this experiment. It ran weeks, as I recall.

 

[jared w]

That's a much harder question than I thought, no luck here.

 

[Scuba Duck]

Diffusion is a function of two things: the mean free path between the molecules, and their kenetic energy.

The mean free path is a function of the density of the gas. That is the spacing between molecules. Using the ideal gas law, and what you know about temperature and pressure, you can determine the molecular density (# of atoms per unit volume), and subsequently the mean free path. As you might expect, as you put more stuff into the tank, the mean free path decreases, and the rate at which the gas diffuses over a given distance slows, as it has many more interactions with other molecules.

Temperature is the measure of the kenetic energy of the gas. You can then relate this kenetic energy to the velocity of an individual molecule. As the temp goes up, the molecules are more mobile, and they move around faster. As you might expect, as the temp goes up, the rate of diffusion goes up.

You have an additional level of difficulty in this problem as all of the diffusion must occur through the isolation valve. I would probably need some internal dimensions of that valve to make a decent shot at this. As you might expect, as this valve gets larger, the diffusion rate through it would increase.

I was going to derrive this mathematically, but have been busy... Look for a more detailed answer later...

 

[Viscya]

Rick Murchison did this experiment. It ran weeks, as I recall.

As I recall it ran for weeks, and still came nowhere close to homogeneous. I believe it just mixed the gas in the valve itself . The tanks were relatively unaffected.

 

[jared w]

It isn't a simple diffusion argument. Diffusion is generally treated as a result of concentration gradients giving a net direction to the flow via the collisions on the order of a mean free path. Strictly speaking the standard form for the diffusion equation, even when you include location and material dependence on the diffusion coefficient, assumes a gas diffusing into nothing (other than whatever boundary conditions you have.) By itself the diffusion equation simply describes the evolution of one gas due to what we think of as viscosity. You get a directionality because of the gradients - the mean free path is ~ 1/concentration, with a constant density you get stuff random walking around. With a concentration gradient stuff at higher concentration collides more frequently and at shorter distances than it does at a lower concentration.

This is a case of 2 concentrations of gases, presumably in equilibrium with their system, being brought into contact with each other.

At this point you've got a contact interface wherein these two populations interact. The two populations have different maxwellian velocity distributions because of the different molecular/atmoic weights so you've basically got one slightly heavier and slower population colliding with a slightly lighter and faster population at the interface. The ratio of the momentums of the gases at the peak of their distributions would be root(m1/m2) so I'd guess that the o2 gas moves into the air and displaces it but the rate at which it does so isn't obvious to me since it is two dense gases, at the same pressure and temperature, trying to mix.

It might be treated using some form of tensorial, binary diffusion treatment or in some other simpler approximation but I can't think of a way not to treat it as two populations equilibrating their distributions which is much more complicated than a simple 3d diffusion equation. That's the realm of collision integrals and statistical mechanics.

I thought their might be some way to make a time argument out of the change in the Gibbs free energy of the system but I can't think of it. I'm sure this has been done before but I couldn't find the answer and unless I'm over thinking it isn't easy. I've got a friend who does stuff like this in the context of strong shocks and hydrogen nuclei equilibrating with electrons so he might know more of the basics than I do if no one turns up a simple answer.

 

[Scuba Duck]

Then we could treat this a concentration gradient problem...

You could make some assumptions to mathematically simplify the problem. If the results told you that it would take forever under mathematically simplified conditions... well you get the picture. The original question posed was how long would it take? I think an order of magnitude is all that is required an not an exact number.

Assume the mixing occurs instantly inside the two tanks. Essentially there is no concentration gradient inside the tank, only in the piping between the two tanks. This takes the problem and makes it one dimensional, concentration is only a function of time and position inside the piping connecting the two tanks. At one end of the pipe, you have the concentration in the Nitrox tank, at the other end of the pipe, you have the concentration in the air tank. This is probably acceptable as the internal dimensions inside the two tanks are much much larger than the dimensions of the inside of the pipe connecting the two tanks. This problem is governed by the rate of diffusion inside the pipe connecting the two tanks.

Probably easiest to model the problem as two tanks with .4 cuft of internal volume (77.4 cuft/200 bar) connected by a 1/8" diameter tube about 1 1/2 feet long. That diameter might be generous considering the internal dimensions of the valves that are in between the two tanks. Anyone know what the orifice size of the tank valves are?

Remember Fick's laws of diffusion.

Fick's law of diffusion - Wikipedia, the free encyclopedia

The first law states that the rate that rate of diffusion (flux) is a function of the concentration gradient and the diffusivity of the molecules. Would also have to determine the diffusivity... mean free path and temperature (KE) will effect your diffusivity.

Mass diffusivity - Wikipedia, the free encyclopedia

The flux of oxygen in one direction is the same as the flux of Nitrogen in the other... has to be, otherwise one tank would increase in pressure.

Have to set up the appropriate boundary conditions, and perform some ugly integral...

The long story is that to reach equilibrium, assuming that you started with AL 80's at 3000 psi, air in one tank, and 32% (or whatever %) in the other, a net volume of 4.2 cuft (for the 32% case) of oxygen would have to diffuse from the Nitrox tank to the air tank at the same time that 4.2 cuft of nitrogen diffused from the air tank to the nitrox tank, with only the concentration gradient to drive the process, through the tiny little tube connecting the two tanks.

As the concentrations of the two tanks come closer together, the concentration gradient decreases, as does the net flux of oxygen and nitrogen.

More to come next week...

 

[jared w]

Not worth continuing. I'll leave it at, 'The above suggestion does not represent the problem posed.' It's a gross over simplification and wrong on many levels.

 

[ptyx]

That's a much harder question than I thought, no luck here.

I recall doing the math with two perfect gases in perfect containers - years ago. It was a gross approximation, bad enough, and it was when I was still fluent in that sort of things.

With high pressure, gas gradients and a valve in the middle, any modeling I can think of/handle myself will just be far enough from reality that it would be useless. I'm perfectly happy with TSandM/Viscya practical approach there

It does make for a interesting modeling exercise though - and I'm curious to see what Scuba Duck (or anyone bold enough to try) will end up with!

 

[frogman62]

This is a non functional discussion because the tube diameter between the tanks if you have a standard twin tank manifold the diameter is between 1/8 and 3/16".
Both sides are at 3000 psi so there is no flow between the two sides.
Drain the tanks and start over with the correct mix and stop wasting your time trying to determine the physics of the issue which is very complex if my college physics and math are still relevent.
Sorry for being so practical but even a couple of physics PhD's could have many hours of discussion and probably not come up with the answer except " a long time"
I know, take a shot at me for being cynical.
Frogman62