

02252010, 03:38 PM

#1

Feedback Score: 0 reviews
Join Date: Mar 2009
Posts: 5

Calculating total gas produced in fermentation


Hi,
I'm trying to answer 2 questions:
 how much gas will be produced during fermentation
 what would the pressure be in a 5g corny if used as primary fermenter will NO VENTING
I tried to calculate this, but the number is absurdly low. What did I do wrong?
CALCULATION:
 Assume 5.5 ABV% in 5 gallons yields 1.041L of ethanol (EtOH).
 1.041L EtOH * (789 g / L) = 821 g EtOH
 821 g EtOH / (46.07 g / mole) = 17.82 moles EtOH
 Yeast metabolism fermentation RXN: C6H12O6 → 2 CH3CH2OH + 2 CO2.
 Therefore Ethanol to CO2 ratio is 1:1 = 17.82 moles CO2
 17.82 mole CO2 * (44.010 g/ mol) = 784 g CO2
 784 g CO2 / (1.977 g/L) = 396 L CO2 (at 0 deg C) (not used)
 "Confined to 1L headspace: P = 17.82 **8.314472 * 273.15 / 1L = 40470 Pa
 40470 Pa = 5 PSI



02252010, 06:58 PM

#2

Feedback Score: 0 reviews
Join Date: Feb 2010
Location: Louisiana
Posts: 831
Liked 28 Times on 21 Posts Likes Given: 3

I can see a use for such a calculation! Maybe the government will cap how much beer we can make (at least lower than it is now) so that we don't harm the environment by pumping CO2 into the atmosphere. I mean what if all the beer makers of the world are the cause for global warming!



02252010, 07:06 PM

#3

Feedback Score: 0 reviews
Join Date: Aug 2007
Location: Southern Maine
Posts: 3,523
Liked 345 Times on 244 Posts Likes Given: 261

The problem is that the rate of fermentation is going to slow as the pressure inside of the vessel increases.



02252010, 07:49 PM

#4

Feedback Score: 0 reviews
Join Date: Jan 2009
Location: Chicago, Il
Posts: 1,327
Liked 6 Times on 6 Posts Likes Given: 1

Quote:
Originally Posted by mooface
Hi,
I'm trying to answer 2 questions:
 how much gas will be produced during fermentation
 what would the pressure be in a 5g corny if used as primary fermenter will NO VENTING
I tried to calculate this, but the number is absurdly low. What did I do wrong?
CALCULATION:
 Assume 5.5 ABV% in 5 gallons yields 1.041L of ethanol (EtOH).
 1.041L EtOH * (789 g / L) = 821 g EtOH
 821 g EtOH / (46.07 g / mole) = 17.82 moles EtOH
 Yeast metabolism fermentation RXN: C6H12O6 → 2 CH3CH2OH + 2 CO2.
 Therefore Ethanol to CO2 ratio is 1:1 = 17.82 moles CO2
 17.82 mole CO2 * (44.010 g/ mol) = 784 g CO2
 784 g CO2 / (1.977 g/L) = 396 L CO2 (at 0 deg C) (not used)
 "Confined to 1L headspace: P = 17.82 **8.314472 * 273.15 / 1L = 40470 Pa
 40470 Pa = 5 PSI

1) Why are we doing this calc at 0 Celsius?
You're going to be fermenting at say 20C (68f?), so we're looking at ~44K Pa. which would be 6.4 or so PSI.
I still think something is wrong here. The guys fermenting in closed vessels (check the fermentation in cornies thread) are using the last couple points to carb beer. One guy kept a sealed corny and IIRC< it was at like 60 psi.
Maybe we're missing CO2 being released when the yeast cleave larger sugars into smaller ones?



02252010, 07:53 PM

#5

Feedback Score: 0 reviews
Join Date: Feb 2008
Location: Mechanicsburg, PA
Posts: 580
Liked 5 Times on 5 Posts

Volume should be cubic meters, not liters.
Multiply your answer by 1000.
http://www.chemicool.com/idealgas.html



02272010, 12:01 AM

#6

Feedback Score: 0 reviews
Join Date: Dec 2006
Posts: 474
Liked 18 Times on 14 Posts

You are right, 5 psi is a bit low. Here’s how I’d estimate it:
Assume an empty corny as this will make it easier, corny volume is 19 liters (5gal X 3.78 l/gal).
Assume about 800 grams CO2 produced, your pretty close, I recall 49% of fermentables are turned into CO2.
So, 800 g CO2 equals 18.2 moles CO2 (44 g per mole CO2)
At standard temperature and pressure 1 mole of any gas is 22.4 liters (STP is 1 atmosphere, 14.7 psi, at 0 degrees C.
So, 18.2 moles CO2 is 408 liters at STP (18.2 X 22.4).
To convert to 19 liters, use the old P1V1 = P2V2 equation, where
P1 = 14.7 psi, V1 = 408 liters, P2 is the unknown, and V2 is 19 liters
Solve the equation for P2, P1V1/V2 = 14.7 X 408 / 19 = 315 psi, at 0 C, will be higher at room temp.
So that’s your answer for putting all that CO2 into an empty corny, with it filled with fermented beer, I have no idea. I would assume it will be higher, but I don’t have any idea what the maximum solubility of CO2 is in beer, so I can’t really say. Of course, the pressure relief valve on a corny is supposed to blow at 130 psi, so maybe that’s the real answer.



03122010, 10:07 PM

#7

Feedback Score: 0 reviews
Join Date: Feb 2010
Location: Cincinnati, Ohio
Posts: 404
Liked 10 Times on 10 Posts Likes Given: 8

You can't do this calculation very easily without some fancy differential equations. The problem is that CO2 will become more soluble as the pressure increases (Henry's Law) and as the solubility increases, the marginal pressure contribution of each additional molecule of CO2 becomes less. At some point it will turn into carbonic acid and the pH will drop and that will affect the solubility. But then some of the carbonic acid will turn into bicarbonate and the solution will buffer itself. All in all you are talking about a very involved problem here.
If you use the combined gas law instead of Boyle's law and assume no solubility of CO2 and 1L of headspace:
P1*V1/T1=P2*V2/T2
or:
P2=P1*V1*T2/(V2*T1)=14.7 PSI*408 L*298.15 K/(1 L * 273.15 K)=~6500 PSI
so HairyDog was about right.



08242012, 12:32 PM

#8

Feedback Score: 1 reviews
Join Date: Jul 2012
Location: Alexandria, VA
Posts: 1,027
Liked 81 Times on 62 Posts Likes Given: 84

I wonder if you are still out there?


Quote:
Originally Posted by AiredAle
You are right, 5 psi is a bit low. Here’s how I’d estimate it:
Assume an empty corny as this will make it easier, corny volume is 19 liters (5gal X 3.78 l/gal).
Assume about 800 grams CO2 produced, your pretty close, I recall 49% of fermentables are turned into CO2.
So, 800 g CO2 equals 18.2 moles CO2 (44 g per mole CO2)
At standard temperature and pressure 1 mole of any gas is 22.4 liters (STP is 1 atmosphere, 14.7 psi, at 0 degrees C.
So, 18.2 moles CO2 is 408 liters at STP (18.2 X 22.4).
To convert to 19 liters, use the old P1V1 = P2V2 equation, where
P1 = 14.7 psi, V1 = 408 liters, P2 is the unknown, and V2 is 19 liters
Solve the equation for P2, P1V1/V2 = 14.7 X 408 / 19 = 315 psi, at 0 C, will be higher at room temp.
So that’s your answer for putting all that CO2 into an empty corny, with it filled with fermented beer, I have no idea. I would assume it will be higher, but I don’t have any idea what the maximum solubility of CO2 is in beer, so I can’t really say. Of course, the pressure relief valve on a corny is supposed to blow at 130 psi, so maybe that’s the real answer.

Wow... so cool.... I really need a math and science refresher....
I would love to see a reference on this "I recall 49% of fermentables are turned into CO2"...



08242012, 02:34 PM

#9

Feedback Score: 0 reviews
Join Date: Aug 2010
Location: McLean/Ogden, Virginia/Quebec
Posts: 7,732
Liked 1028 Times on 813 Posts Likes Given: 31

Balling's findings were that 2.0665 grams of extract were consumed for each gram of ethanol produced along with 0.11 grams of yeast biomass and 0.9565 grams CO2. This, of course, represents an average as some yeast strains will put more of the extract into biomass than others. That aside, if you know OG and FG you can calculate the number of grams of CO2 produced. From there you can fiddle with your PV=nRT and Henry stuff to come up with a volume of CO2. Don't forget to include the vapor pressure of water.



08242012, 03:02 PM

#10

Feedback Score: 0 reviews
Join Date: Jun 2012
Location: Birmingham, AL
Posts: 244
Liked 25 Times on 21 Posts Likes Given: 8

I don't think you would ever be able to realistically be able to calculate a theoretical gas production with any acceptable range of accuracy. Mainly, at well below 5000 PSI the yeast would have ceased fermentation for too many reasons. If you were concerned with the total volume of gas produced, it would be possible to do a normal fermentation with a mass air flow sensor capable of low volume flow rates attached to a carboy. After the fermentation has ceased, you should be able to integrate the flow rate over the elapsed time for a measure of volume of gas produced. If this could be done somewhere around at least 50 times, with the same basic conditions, then you might be able get a good idea on the volume of gas produced.





