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Calling all math geeks: Force carbonation math!

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Sorry for the delay in responding...just falling back onto the brewing wagon after a hiatus :)

Are you asking for the derivation of the general equation form or the one will all the constants?

The general form was whipped up in my head based on what I thought the curve should look like (asymptotic).

The constants derivations was more of a "throw spaghetti at the wall and see what sticks" method - assisted by excel.

I plotted the volumes versus temperature for 5 different pressures (30, 20, 10, 5 and 1 psi). These fit and exponential curve pretty well so I added trendlines and displayed the equation for each of those (C*EXP(k*T).

Then I plotted the C values versus the psi (five points) and fit another trendline (linear) to those to get an equation that was C = 0.17*P +3.04 (where P was the psi).

By repeating this with the k values and psi I got k = 0.0006*LN(P)+ 0.02.

I combined these into the general form equation along with the aforementioned "K = 0.014" (which is multiplied by the time in the general equation) to yield the overall equation.

Wish I could attach the spreadsheet I used...attaching a screen shot instead.

View attachment carb math plots.pdf
 
Sorry for the delay in responding...just falling back onto the brewing wagon after a hiatus :)

Are you asking for the derivation of the general equation form or the one will all the constants?

The general form was whipped up in my head based on what I thought the curve should look like (asymptotic).

The constants derivations was more of a "throw spaghetti at the wall and see what sticks" method - assisted by excel.

I plotted the volumes versus temperature for 5 different pressures (30, 20, 10, 5 and 1 psi). These fit and exponential curve pretty well so I added trendlines and displayed the equation for each of those (C*EXP(k*T).

Then I plotted the C values versus the psi (five points) and fit another trendline (linear) to those to get an equation that was C = 0.17*P +3.04 (where P was the psi).

By repeating this with the k values and psi I got k = 0.0006*LN(P)+ 0.02.

I combined these into the general form equation along with the aforementioned "K = 0.014" (which is multiplied by the time in the general equation) to yield the overall equation.

Wish I could attach the spreadsheet I used...attaching a screen shot instead.
Thanks for finally responding. Glad you didn't die or something :D It'll take me a little time to review/absorb this. I might be back with questions or comments.

Brew on :mug:
 
Thanks!
for anyone wanting to copy and paste a spreadsheet formula, that's:

B2: e, aka, "=exp(1)" or 2.718281828
B3: PSI of regulator
B4: time in Hours
B5: temperature in F

=((0.17*B3*(1-B2^(-0.014*B4)))+3.04)*B2^((0.0006*LN(B3*(1-B2^(-0.014*B4)))-0.02)*B5)


It appears that the formula is presuming 1.15 vols starting. If i were better at math I'd modify the formula for that as a variable -- though it may not really matter significantly given the type of curve.
Bumping this with a easy link to the google sheet I've been using for years.
Make your one copy of it, or download as Excel, to modify values.

force carb calculator spreadsheet:
https://docs.google.com/spreadsheets/d/1y3tNmGr9JBByQ8vH4g8QwKSfGWr6Le4fpZgpSkDcf24/edit?usp=sharing
 
To approach the problem mathematically (which is the original topic), I believe one should should start with Graham's law of diffusion. The equation requires the molar masses of the gas and solvent - not sure how to get that, but perhaps it's easy if you know the sp gravity and also PV=nRT. n = m/M. So, PV=(m/M)RT. M is the molarity, m is mass. specific gravity is (sorta) density, density is V/m, etc. Some dimensional analysis, spreadsheet, I think it can be done this way.

Anyway, start here.

https://chem.libretexts.org/Bookshe.../2.9:_Graham's_Laws_of_Diffusion_and_Effusion
 
To approach the problem mathematically (which is the original topic), I believe one should should start with Graham's law of diffusion. The equation requires the molar masses of the gas and solvent - not sure how to get that, but perhaps it's easy if you know the sp gravity and also PV=nRT. n = m/M. So, PV=(m/M)RT. M is the molarity, m is mass. specific gravity is (sorta) density, density is V/m, etc. Some dimensional analysis, spreadsheet, I think it can be done this way.

Anyway, start here.

https://chem.libretexts.org/Bookshe.../2.9:_Graham's_Laws_of_Diffusion_and_Effusion
Graham's law is for diffusion of gases thru orifices or pores, not for adsorption/desorbtion of gases at a liquid gas interface, so I don't think it plays a role here.

The question is whether the rate controlling step is the net adsorption of CO2 by the beer at the interface, or the diffusion of CO2 thru the beer after being absorbed. If there are convection currents within the beer due to temperature gradients in the keg, then they would likely overwhelm the slower diffusion process, so that adsorption would be the rate controlling step.

Assuming adsorption is the rate controlling step, the rate of adsorption should be a decaying exponential function, since the adsorption rate is driven by the difference between the headspace partial pressure and the pressure that would be in equilibrium with the beer at its instantaneous level of carbonation. The further you are from equilibrium the faster CO2 is absorbed, and the closer to equilibrium the slower the absorption. The instantaneous carb level (V[t} or P[t]) then becomes a constant times (1 minus the exponentially decaying absorption rate)

If diffusion thru the beer is the rate controlling step, then the problem is one of diffusion into a non-infinite sink. The math for this gets pretty complicated (and I can't remember it,) but I believe it ends up following something known as the "Error Function." The Error Function is similar to a exponetial function, but it tails off differently IIRC.

If someone collected enough data, carefully enough, we might be able to determine whether absorption more closely follows an exponential or Error function. This would allow an inference to be made about what is actually the rate controlling step.

An easier approach is just to assume the rate and instantaneous carb levels follow exponential functions, and empirically determine the coefficients that give a best fit for different pressures and temperatures and call it a day.

Brew on :mug:
 
It's been soooo long... Does the density of the solution (1.030 wee heavy, vs 1.002 lager) affect the equilibrium diffusion level? I've had some heavy beers that just never seemed to be carbed enough, no matter how long they were left at a regular setting like 45F/12PSI.
 
It's been soooo long... Does the density of the solution (1.030 wee heavy, vs 1.002 lager) affect the equilibrium diffusion level? I've had some heavy beers that just never seemed to be carbed enough, no matter how long they were left at a regular setting like 45F/12PSI.
It's plausible that both the equilibrium carb level vs. temp and pressure, and the rate of carbonation is affected by the beer's SG. The ASBC table (which is the basis for all of the carb charts and calculators) is for a specific beer SG (1.005 IIRC.)

Brew on :mug:
 
It's been soooo long... Does the density of the solution (1.030 wee heavy, vs 1.002 lager) affect the equilibrium diffusion level? I've had some heavy beers that just never seemed to be carbed enough, no matter how long they were left at a regular setting like 45F/12PSI.
Yes, that is indeed the case as diffusion rate is inversely proportional to the viscosity (not the density) of the beer. From this it follows that higher FG (higher FG = higher viscosity) beers carbonate more slowly and that all alse being equal carbonation rate is directly proportional to temperature (higher temperature = lower viscosity = higher diffusion rate).

@doug293cz Diffusion in the liquid is indeed the rate controlling factor for carbonation as diffusion in the gas phase, which is itself the rate controlling factor for adsorption, is orders of magnitude faster than diffusion in the solution. This results in near-instantaneous saturation of the surface layer and from this point on diffusion of the gas in the liquid is the controlling factor that determines how fast new gas molecules can be adsorbed at the interface.
 
My physics knowledge is very limited and it’s been a long day, so this may be off base, but this seems relevant, especially part 2: https://lib.dr.iastate.edu/cgi/viewcontent.cgi?article=14332&context=rtd

Would be great if someone with a better background in the subject could take a look and see if it’s useful

EDIT: Looks like it’s about CO2 coming OUT of solution over time, so the opposite of what we care about. But surely there’s some research on the subject
 
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That paper is 100 years old so not exactly cutting-edge but still valid nonetheless. In any case gas dissolution and evolution are the same phenomenon just going in opposite directions and therefore governed by the same laws (Fick's law) irrespective of direction.
 
Otoh, I'm not a fan of weight. Don't like picking weight up, don't like carrying it.
I'm on a mission to shed weight :D

11/1/2019: 223
10/29/2020: 198. And still falling. Turns out one can live on one meal a day plus beer...

Cheers! (Anyone remember "The End"? :D)
 
and with my knowledge of weight i can tell you, what ever your basal metabolic rate is, you were 239 calories under it for the last year...on the avg.


figuring 3500 calories a pound of body weight, 365 day or so...so 25lb'sx3500/365 that would tell you how many calories you burn normally, and you don't need to worry that much to cut 239 on avg, i lost 30lbs in 3 years just cutting it buy 50 on avg....

i like weight.....

(and there's a lot more to eating then calories. you got to get vitamins, minerals, amino acids, esential lipids! gotta make that meal count! ;))
 
Phhhht! Way too complicated.
I just ditched the midnite tube of Ritz crackers, peanut butter, strawberry preserve, a butter knife and a paper towel.
Or, alternately, settling down with a quart of Brigham's mocha chip ice cream and a soup spoon.
At least a couple thousand calories with either option - and totally at the wrong time of the daily cycle, of course :D

Anyway...at 6'5" 198 and 70-ish, I'm feeling much more spry than carrying that extra 25 pounds.
Ever pick up 25 pounds? That's like a countertop microwave. Crazy to just drop that without really giving up much, amiright?

Cheers! (And no beers were harmed in the offing! :ban: )
 
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Phhhht! Way too complicated.


LOL, to each their own, my 6'2" 170 something ass think my picnic taps are still just fine after 16 years of use..but you or someone here, did help out with the 3/16" upgrade from 1/4"..i don't get any foamy pours, but with 10' i start getting bored filling a glass!


edit: but in my case i gotta neuter my beer like spaying neutering your dog/cat
 
i would say in my experience it take 7-8 calories to maintain a pound of body weight...so a 239 defecit is unstanable, unless you plan on starving. just trying to be helpful.
 
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