Minimum C02 Pressure and Volume to tap a keg of beer 1

[cjschwal]

Please help me settle a bet at work. How many kegs of beer can be pumped with a 20 oz. paintball CO2 tank?

Assume pressure of CO2 tank is 1000 PSI, beer keg is 16 gallons (128 lbs), and lift to spout tap is 36 inches. Volume of paintball tank therefore is ~36 in^3.

Using the potential energy equation PV = mgh my buddy says about 1/8 kegs can be tapped but I say he is "all wet" -- it's more like one (I've seen it)!

He says .128 = 128*36/(1000*36)

Where are we going wrong?

I'll buy you a beer if you can convince me. Thanks!

 

[Zapster]

You can only pressure one keg in your life if you do it wrong.

 

[ivymike]

He says .128 = 128*36/(1000*36)
he's right, but wth does that have to do with the question?

 

[ivymike]

you have 20oz CO2, which is 0.567kg CO2. To dispense, you need 36 inches of water (pressure), or 1.3psi as a minimum.

If the keg was empty, what pressure could be reached using your CO2? Ideal gas law says that at 32F you could pressurize an empty keg to about 64psi if you put 20oz of CO2 into it.

That's enough pressure to dispense, and then some. Even with a large amount of consumed CO2 (as bubbles) you would likely be able to dispense the entire keg.

 

[cjschwal]

Ivy- I've seen the 20 oz bottle lift the keg with my own eyes. These small bottle systems are found pretty commonly online. So the question is, Why is our math off by a factor of eight? (I miss-typed above - the math implies the bottle has enough energy to tap eight kegs! - the problem is I know the bottle is only good for one keg)

 

[ivymike]

I have no idea what your math is supposed to represent physically. You are absolutely correct, however, that 128/1000 is 0.128.

Let me see if my crystal ball has any idea what you're doing:

Assume pressure of CO2 tank is 1000 PSI, beer keg is 16 gallons (128 lbs), and lift to spout tap is 36 inches. Volume of paintball tank therefore is ~36 in^3. Using the potential energy equation PV = mgh my buddy says about 1/8 kegs can be tapped but I say he is "all wet" -- it's more like one (I've seen it)! He says .128 = 128*36/(1000*36)

0.128 kegs = (128 lbs)*(36*in)/(1000psi*36in3)
so the potential energy to lift a keg 36 inches, divided by whatever (1000psi)*(36in3) is supposed to represent, gives a unitless quantity with some special meaning to you.

http://en.wikipedia.org/wiki/Compressed-air_energy_storage#Energy_density_and_efficiency

The maximum energy stored using isothermal compression & expansion is about 100ln(P2/P1)kJ/m3, according to wikipedia. Using your 36in3 and 1000psi, one would calculate 0.25kJ or 2210 lbf*in maximum energy storage. 128lbf * 36in gives 4608 lbf*in. This would suggest that you can lift about half a keg 36 inches. Luckily for you, not all the beer is at the very bottom of the keg.

 

[zdas04]

The missing piece is the Henry's Law constant for CO2 in Beer (I don't have it, but it should be available on the web). The CO2 available for motive force is the amount in the bottle to start with minus the amount that disolves. I wouldn't be surprised if the amount that disolves is a very large number.

David

 

[cjschwal]

Man my simple equation threw everybody off! It was simply the conservation of potential energy, P*Vbottle = m*g*h beer

I chose the conservation formula to debate my buddy with because it's simple with no english/metric conversions (apparently too simple if we need to be working with the isothermal logarithm equation). Anyway to justify my orginal post:

PVbottle = 1000 psi * 36 in^3 * (number of bottles) = mgh beer = 128 lbf x 36 in

The "unitless quantity" is the number of bottles!

So, I appreciate the input thus far but want to learn why the potential energy equation won't hold here. Perhaps it's the compressibility/themodynamic/heat transfer nature of the fluids at play? Perhaps the PV side doesn't represent the potential energy of the CO2?

I do want to thank the group and specifically Ivymike. But we must be missing something conservation-wise since I've seen this product lift (about exactly one 16 gallon) keg:

http://stores.kegconnection.com/Cat...erator+Kits,+20oz+CO2:Fridge+(Shank)+20oz+CO2

Thanks again, cjschwal

 

[ivymike]

read the wikipedia article.

 

[ivymike]

two problems I can see with your calculation:
1) m*g*h, as you've calculated it, does not seem to accurately represent the potential energy to lift the beer out of a keg
2) P*V for the storage container does not represent the available energy for lifting the beer.

Aside from that, the math should have worked great!

 

[zdas04]

I've been thinking about this all night and you are going about it all wrong. First approximation, assume that the beer is irrelevant,the keg starts off full to the overflow, and the CO2 acts as an ideal gas. Then P1V1=P2V2. P1 is the pressure in your cylinder. V1 is the volume of the cylinder. V2 is the volume of the cylinder plus the volume of the keg. Solve for P2 if it is more than your required pressure then you can do another keg.

That will get you in the ball park. Then you can refine it with compressibility factors and Henry's Law constants.

David

 

[ivymike]

hey, i think I've heard that suggestion somewhere before... say, 4th post.

 

[zdas04]

IS THAT WHAT YOU SAID????????????? I just read it again and still can't get what I said from what you said.

I've been thinking about this while I drove today and I think I held the wrong thing constant. You'll use a regulator, so you'll keep pressure in the keg constant and
V1/n1=V2/n2. If V1 is the volume of the CO2 hose downstream of the regulator and N1 is number of moles in that volume at 36 inH2O, and n2 is the number of moles in the keg plus the hose when the beer is gone. The difference between n1 and n2 is the number of moles that left the tank. See if this is less than you had in the tank at 1,000 psig. If it is, then calculate a new pressure with those moles gone and try it again. Repeat until the ending pressure in the paintball bottle is less than 36 inH20.

David

 

[ivymike]

yeah, I said you have a known mass of an ideal gas at a known temperature contained within a vessel of known volume. solve for pressure and compare to the minimum required value to dispense of 1.3psi (36in). I even did all the math and provided the answer.

 

[MortenA]

Anything involing beer will of course attact an engineers mind.

Disregarding energy ballances etc, and CO2 absorbed in the liquid:

Since the pressure i the vessel must remain around 16 psi absolute (i think thats low seems to remember that the vessels here in DK are pressureized to around 2-3 barg but OK.

Sorry im going total metirc on you guys now -:

From mu Moullier diagram i get the density of CO2 at 69 barg, 20 deg C==800 kg/m3.

Assuming that the temperature remains constant (=warm beer unless you cool the beer like we do just priou to tapping) then the desity of co2 at just above atmospheric pressure and 20 deg C is around 2 kg/m3. This means that you can fill 400 times the volume at around atmospheric pressure the volume you have at 69 barg (1000 psig). The co2 bullit is 36 cu.in and the keg is 16 gal so the ratio is around 230. So slightly more than one keg can be emtied. Allowing for absorption of CO2 and since i think the pressure is most like somewhat higher then 1 keg seems right.

Best regards

Morten

 

[MortenA]

I think i must add one note of explanation to my train of thoughts:

When you empty the keg - then you replace the beer with co2. So when the keg is empty you now have a keg full of cos - and therefore the mass ballance must work out (also).

I make some important assumptions - and maybe the reason why your simple energy ballance dosnt work out also lies here?

-I assume that the keg absorbs heat fast enough se that the co2 remains at the same temperature. If the pressure let down was isenthalpic then the co would get cold - seperate into an ice and a gas phase and the average density would be much higher.

-I also assume that co2 dosnt absob in the beer. Im not sure if draugh beer is "pre carbosined" or actually carbonised when the co2 pressure is added. That would make some difference - and i havnttaken the time to look at the Henry's law constant and work out the number.

Best regards

Morten

 

[ivymike]

the beer must be pre-carbonated, because you would commonly use atmospheric air added via a hand pump to dispense, yet it comes out with lots of CO2 bubbles in it.

the CO2 in the paintball cannister is liquid, isn't it? Where'd 1000psi come from anyway?

That's why I didn't use P1V1=P2V2 in my initial calculation, and instead used P2 = N.R.T2/V2 (20oz of CO2 provided). I didn't bother counting the volume of hose&bottle, since they only amount to around 1% of the volume of the keg and it was faster to leave them out (omitting them has no important effect on the accuracy of the result for the problem at hand). I assumed that the beer was to be served very cold. My result was Ideal gas law says that at 32F you could pressurize an empty keg to about 64psi if you put 20oz of CO2 into it.

20oz CO2 = 0.567 kg CO2
32F = 273.16K
R = 8314L-kPa/kmol-K
molecular mass of CO2 is 12+16+16 = 48
# k-moles is 0.567/48 = 0.0118
Volume of the keg is 16gal = 60.57L = 0.06057m3
P = nRT/V = 443kPa = 64psi

if you include the CO2 container in the volume, you get a total volume of 16.156gal or 0.06116m3, so your final pressure is 439kPa or 64psi.

 

[ivymike]

btw, using your figure for the density of CO2 @ 1000psi (800kg/m3), I get only 16.6oz in the 20oz (36ci) bottle. Close... but the actual pressure must give an average density of 961 kg/m3, or we'd need more than 36ci to hold 20oz. The internet seems to confirm that the volume of the 20oz CO2 container should be very close to 36ci.

 

[MortenA]

We dont use hand pumps here in Denmark - but i think you are right.

I tried to skim the previous posts but missed that you have done the same as I - more or less perhaps even more simple!

I use the Moullier diagram - its tested material more than 70 yeras old! But i was a little quick when reading it - and dont know the reference temperature for the 1000 psig/20 OZ. I think its more like 810 kg/m3 at 20 deg. In order to get to 961 it would have to be fairly close to 0 deg C or 32 deg F. This would mean that the pressure could increase up to around 400 barg (==6000 psig) if the temperature increased to 40 deg C! Maybe these little bullits has a bleed/relief valve?

Best regards Morten

One link is here - it covers the operational area and also has data below the critical points (many other free charts dont).

http://www.union.dk/media/CO2-GB.pdf

 

[dcasto]

here, you guys need this chart

http://sdcollins.home.mindspring.com/ForceCarbonation.html

or here

http://www.kegkits.com/carbonation.htm