Calculate weight of beer per gallon

[doug293cz]

I don't think anyone was claiming truly profound changes to mass here

[edit] Although...found this on the Brewing Reddit. I can't vouch for it but there were other posts in the same thread that confirmed/amplified it...

"Turns out for five gallons/18.9 L of 1.060 wort at 75% apparent attenuation, 449.1 L/ 15.86 cubic feet/ 118.64 gal of CO2 is produced (standard temperature and pressure. This amounts to 0.88 kg/ 1.94 lb of CO2!"

For sure that's a lot more mass shed than would be gained if it's then brought up to 2.5 volumes of CO2...

Cheers!

These numbers look about right to me (where I did some similar calcs.)

Brew on

 

[doug293cz]

Wort will lose mass through fermentation, but will gain mass back through conditioning. It would take someone like our resident physics dude @doug293cz to calculate the net change but I suspect fermentation loss exceeds carbonation gain...especially as the OG increases...

Cheers!

Sounds like an interesting question to look at Sunday, when I will have some time.

Brew on

 

[day_trippr]

Pretty sure that Reddit thing answered the basic comparison - I can definitely carbonate much more beer from one 5 pound CO2 cylinder than 2.5 kegs
But the math would be something we can refer to forever, so if you have the time...

Cheers!

 

[whoaru99]

Read somewhere the conversion from volumes CO2 to g/l weight CO2 is 1.96.

If so, in the example of 2.5 volumes:

2.5 vol x 1.96 = 4.9 g/l
4.9 x 18.9 = 92.6g CO2 in 5gal

But not all of that would come from the tank due to fermentation residual CO2, right?

If the beer had 1 vol residual you'd take only 1.5 vol from tank, or ~56g for 5 gal.

5lb CO2 = 2268g
2268 / 56 = ~40

Forty 5gal batches carbed (in this example) from 5lb CO2 (presuming carb only, no losses, no serving, no purging, etc.)?

Seems like a lot to me so probably missed something along the way. Be interested to see what Doug comes up with.

 

[day_trippr]

I'd say there must be something missing because I estimate closer to 20 kegs or less from a full 5 pounder...
[edit]...although to be honest the cylinder I typically use for conditioning is the one I also use for closed transfers so I could be way off and maybe it is possible to condition 40 kegs on a five pound charge...

Cheers!

 

[Peebee]

Whah! ... Got to have a crack at this ...

Stash big bottle of water with your beer and leave it a couple of hours so they're both same temperature (but you don't care what the temperature actually is). Fill reasonable size glass with the beer (a pint at least, but you don't care about exact size; weigh glass carefully before you get it wet). Don't bother with a head, you only want beer. Fill a small jug with the beer while about it. Carefully move glass onto kitchen scales that are good to a gram, or whatever units you use. Make sure the glass is completely full, meniscus too, with that from the jug. Record the weight. Now the important bit ...

Drink the beer ...


... quicker! The water is warming up!


Rinse glass with water and dry it. Repeat with the water you stashed (but don't drink the water, that's what you've got beer for).

You've got weight of glass of water, weight of glass of beer, and weight of glass. Delete weight of glass from that of glass of beer and water. Divide the weight of beer from weight of water. Answer: The Specific Gravity of your beer! You haven't cared what the temperature is, and you haven't cared about the quantities you had ... okay, if the glass was very large you might be past caring anyway? You didn't care about the gas either (unless it stopped you filling the glass properly).


Now fill a keg (similar to the one with the beer) that you've previously weighed with water to the level you consider "full" (or exactly a gallon in a bucket if that's what you want). Weigh it and delete the weight of the keg. Assume the water has a Gravity, or Density, (not Specific Gravity) of exactly one gram-per-millilitre ... if you're really finicky you can lookup the Gravity of water at the temperature it's at (at 20°C it'll be 0.9982 grams per millilitre, gawd I'm sad ain't I ... that's ... ah, forget it, just assume one). Now multiply the actual weight of water by your calculated SG. ... Done!


Don't forget the beer drinking bit.


[EDIT: "grams per litre"? Oops, "grams per millilitre", well it was getting late!]

 

[doug293cz]

Wort will lose mass through fermentation, but will gain mass back through conditioning. It would take someone like our resident physics dude @doug293cz to calculate the net change but I suspect fermentation loss exceeds carbonation gain...especially as the OG increases...​

Cheers!

Ok, let's give this a go. For simplicity, I will use a sucrose/water solution (after all that's what Plato and Brix used, and we pretend that beer behaves the same way.)

Water has a density of 0.9982 g/cm^3 at 68°F (20°C), and sucrose has a density of 1.587 g/cm^3. Let's assume we start with 20 L of a 15°P solution (15% by weight of sucrose.) 15°P corresponds to an SG of 1.06111 (from NIST tables), so 20 L weighs 20 L* 998.2 g/L * 1.06111 = 21,184 g (21.184 kg.) The weight of the sucrose is 21,184 g * 0.15 = 3177.6 g, and the water weighs 21,184 g * 0.85 = 18,006.4 g.

Each 342.30 g of sucrose that is fermented creates 4 * 44.01 g = 176.04 g of CO2, 4 * 46.07 g = 184.28 g of ethanol (EtOH), and consumes 18.01 g of water. So, if we ferment all of the sucrose we create:

3177.6 * 176.04 / 342.30 = 1634.2 g of CO2​

3177.6 * 184.28 / 342.30 = 1710.7 g of EtOH​

3177.6 * -18.01 / 342.30 = -167.2 g of water lost​

​

Thus after fermentation we have a solution containing 1710.7 g of EtOH and 18,006.4 - 167.2 = 17,839.2 g of water (ignoring any water evaporation during fermentation.) The total weight of the solution will be 1710.7 + 17,839.2 = 19,549.9 g. This solution will be 100% * 1710.7 g / 19,549.9 g = 8.75% EtOH by weight (~10.9% ABV.) The solution has an SG of 0.9854, so the volume will be 19,549.9 g / (998.2 g/L * 0.9854) = 19.875 L.

So, we lost 20 - 19.875 = 0.125 L (125 ml or ~ 1/2 cup) of volume during fermentation (again ignoring any water or EtOH evaporation during fermentation.)

What about the CO2? A fermentation done at 68°F will have a residual carbonation level of 0.84 volumes. 1 volume is equal to 1.977 g/L of CO2, so we will have:

19.875 L * 0.84 * 1.977 g/L = 33.0 g of CO2 in solution​

​

1601.2 g of CO2 will have bubbled out of the solution during fermentation. The volume effect of 33 g of CO2 in solution is unknown at this point, but should be small compared to the total volume.

Brew on

 

[khannon]

Ok, let's give this a go. For simplicity, I will use a sucrose/water solution (after all that's what Plato and Brix used, and we pretend that beer behaves the same way.)

Water has a density of 0.9982 g/cm^3 at 68°F (20°C), and sucrose has a density of 1.587 g/cm^3. Let's assume we start with 20 L of a 15°P solution (15% by weight of sucrose.) 15°P corresponds to an SG of 1.06111 (from NIST tables), so 20 L weighs 20 L* 998.2 g/L * 1.06111 = 21,184 g (21.184 kg.) The weight of the sucrose is 21,184 g * 0.15 = 3177.6 g, and the water weighs 21,184 g * 0.85 = 18,006.4 g.

Each 342.30 g of sucrose that is fermented creates 4 * 44.01 g = 176.04 g of CO2, 4 * 46.07 g = 184.28 g of ethanol (EtOH), and consumes 18.01 g of water. So, if we ferment all of the sucrose we create:

3177.6 * 176.04 / 342.30 = 1634.2 g of CO2​

3177.6 * 184.28 / 342.30 = 1710.7 g of EtOH​

3177.6 * -18.01 / 342.30 = -167.2 g of water lost​

​

Thus after fermentation we have a solution containing 1710.7 g of EtOH and 18,006.4 - 167.2 = 17,839.2 g of water (ignoring any water evaporation during fermentation.) The total weight of the solution will be 1710.7 + 17,839.2 = 19,549.9 g. This solution will be 100% * 1710.7 g / 19,549.9 g = 8.75% EtOH by weight (~10.9% ABV.) The solution has an SG of 0.9854, so the volume will be 19,549.9 g / (998.2 g/L * 0.9854) = 19.875 L.

So, we lost 20 - 19.875 = 0.125 L (125 ml or ~ 1/2 cup) of volume during fermentation (again ignoring any water or EtOH evaporation during fermentation.)

What about the CO2? A fermentation done at 68°F will have a residual carbonation level of 0.84 volumes. 1 volume is equal to 1.977 g/L of CO2, so we will have:

19.875 L * 0.84 * 1.977 g/L = 33.0 g of CO2 in solution​

​

1601.2 g of CO2 will have bubbled out of the solution during fermentation. The volume effect of 33 g of CO2 in solution is unknown at this point, but should be small compared to the total volume.

Brew on

Give or take... Wow, there's a giraffe.. Thank you for doing the math.. clearly more sober than I, so I'm going to trust it.

 

[Peebee]

Thanks to @doug293cz for quietly "correcting" my blunder of missing out the "milli" in my earlier post. Thanks also for following my post with something so black/white contrasting it significantly enhances the point I was making!

Yes! My "tongue-in-cheek" post does have a serious point! Apart from the blunder with the missing "milli". Which didn't affect the result ... it was only being waved as a flag to ignore something, that the Density of water isn't exactly one but very nearly is (depending on temperature). And if that isn't enough to confuse some brewers ... the Specific Gravity of water is always exactly one ... whatever the temperature!

The rest of my post is entirely mathematically correct.

We've got our hydrometers stuffed so far up our a****, we don't realise how simple this "Specific Gravity" caper is. Based on mass (weight-ish) divided by volume, that's all. Hydrometers can't weigh anything, they can't measure the volume of anything, but give them a temperature and they can! And do it relatively simply. To show how, using maths, is excruciatingly complicated ... I think that's been demonstrated?

It didn't take me a lifetime working with maths to figure this out: I did use hydrometers. An accident created a vision defect that makes it extremely difficult to fix my vision on closely packed parallel lines ... I can't read hydrometers! The glasses to fix the error are in the same category as X-ray glasses; err, quite expensive? So, I figured out alternative ways.

Archemedes was a good starting point!