What's the % ABV?

I made a very strong beer with 12.2% ABV. I decided to tame it down so I made another beer with 4.2% ABV and blended the two in a 50/50 blend by volume. So equal parts of each beer. I want to determine the % ABV of the blended result - short of a lab analysis. I know it's not a simple linear dilution (meaning the average of the two) and everything I've come across for dilution assumes diluting with water. Does anyone know how to calculate the resulting ABV of the blend?

http://winemaking.jackkeller.net/blending.asp

a couple of calculators I think the easier one is the second one down, basically ABV of each and then put in the quantity of your 12.2% and change the amount of the Fortifier (4%) until you reach the ABV desired. From what I ran with 5 gal of 12%, if you add 2 gal of 4% you'll be at around 9.7%.

If the volumes were equal it should be an average of the two.

+1, I was thinking the same thing. Why wouldn't it be a straight linear dilution?

5 gallons * 0.122 = 0.61 gallons of ethanol
5 gallons * 0.042 = 0.21 gallons of ethanol

Upon blending you have 0.82 gallons of ethanol in 10 gallons total, or 8.2%. Unless I'm somehow under thinking this.

Why wouldn't it be a straight linear dilution?

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Because volume isn't conserved but mass is so the first thing to do is convert the ABVs of each beer to ABW's by

ABW = ABV*0.791/SG

where SG is the specific gravity of the beer. Now calculate the weight of each volume of beer by W = Vol*SG*.998203. Multiply by the ABWs to find the weight of alcohol in each. Add the weights of alcohol and divide by the sum of the weights of the 2 beers to find the ABW of the mix. Now convert back to ABV. To do that you need the SG of the mixture so measure it or estimated it by a weighted average of the "points" i.e. 40% 1.030 beer with 60% 1.070 beer will probably have an SG of around 1 + .4*30 + .6*70. Now compute the new ABV from ABV = SG*ABW/0.791

I'm trying to calculate a similar thing for a big Belgian I made, wondering how I would know the final ABV with this scenario:

9 gals in fermenter of 1.110

at high krauesen I added 3 gallons of 1.065 sugars

FG = 1.012

I guessed this was around 15%.

Same technique - conservation of mass. Find density of each liquid by multiplying SG of each by 0.998203 (the density of water). Find the mass of each liquid. Convert the SG's to °P and calculate the mass of extract in each from the °P (mass_extract = mass_liquid*°P/100). Calculate the mass of the water in each by subtracting the extract mass from the total mass of each. Sum the masses of the water and of the extract. Divide the sum of the masses of the extracts by the total mass of water plus extract and multiply by 100. This is the effective OG in °P for the mix. Apply the expected RDF or ADF to calculate the apparent or true extract. Subtract from the OG and multiply by the proper Balling factor for the OG. This is the ABW. Divide by 0.791 and multiply by the specific gravity of the finished beer (as obtained from the AE estimate if you estimating before the beer finishes). There will be some error from the water lost to evaporation during the fermentation.

Whew, more complex than expected, thanks, I'll have to print that out and try to figure it. Any water loss was fairly minimal because I got 11 gallons out of the 12, which is typical with my fermenter.

If you go and plug it into a spreadsheet you will probably find it's not as bad as it looks and once it's in the spreadsheet then you don't have to worry about the complexity any more.

Also, if you know that you are losing a gallon to evaporation you can account for that by subtracting 3.78 kg from the total water mass.

ajdelange said:

Same technique - conservation of mass. Find density of each liquid by multiplying SG of each by 0.998203 (the density of water). Find the mass of each liquid. Convert the SG's to °P and calculate the mass of extract in each from the °P (mass_extract = mass_liquid*°P/100). Calculate the mass of the water in each by subtracting the extract mass from the total mass of each. Sum the masses of the water and of the extract. Divide the sum of the masses of the extracts by the total mass of water plus extract and multiply by 100. This is the effective OG in °P for the mix. Apply the expected RDF or ADF to calculate the apparent or true extract. Subtract from the OG and multiply by the proper Balling factor for the OG. This is the ABW. Divide by 0.791 and multiply by the specific gravity of the finished beer (as obtained from the AE estimate if you estimating before the beer finishes). There will be some error from the water lost to evaporation during the fermentation.

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I can not imagine a scenario where this info would be that important to me.
Good luck though.

One can certainly make decent, even good, beer with out understanding any of the underlying principles but limiting one's imagination, in brewing as in anything else, generally limits one's horizons.

I dont really see the point in converting to ABW for calculating dilution(sinse the final result is the same anyway)

The point was explained in #5: you can't do these problems by volume because if you add 10 cc of alcohol to 100 cc of water you don't get 110 cc of solution. You get 109.27

If, for example, you have a liter of a 10% ABV water solution and add a liter of pure water to it the blend winds up at 5.204% ABV. If you tried to do the calculation based on volume you would come up with 5.0%. Doesn't look the same to me.

Numbers in these examples are based on OIML polynomial.

I don't plan on doing that math for this but I assume it's pretty close to the value you'd get if you made the determination in ABV. I think the OP was looking for a ballpark in order to blend two beers.

Depends on your definition of "pretty close". If 5% is close enough to 5.2% for you then Bob's your uncle!

And, of course, the whole concept of "points per pound per gallon" is based on an approximation which will usually get you even closer to the truth than making the conservation of volume assumption with alcohol solutions.

actually most of that loss is due to the fermenter shape and the trub.

ajdelange said:

Also, if you know that you are losing a gallon to evaporation you can account for that by subtracting 3.78 kg from the total water mass.

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ajdelange said:

The point was explained in #5: you can't do these problems by volume because if you add 10 cc of alcohol to 100 cc of water you don't get 110 cc of solution. You get 109.27

If, for example, you have a liter of a 10% ABV water solution and add a liter of pure water to it the blend winds up at 5.204% ABV. If you tried to do the calculation based on volume you would come up with 5.0%. Doesn't look the same to me.

Numbers in these examples are based on OIML polynomial.

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Question on this...

Back calculating from your liter of 10% abv, which would have 100 mL ethanol, and the final blend ABV of 5.204, you're using 1.92 as the final volume of the blend, right?

I know 1.92 L is the volume of a blend of 1 L of water and 1 L of ethanol, but you're not going to have nearly that much volume loss when blending 1 L of 10% ethanol with 1 L of water, are you?

Easy guys, relax, have a home brew.

Any ways, since the ABV of the original beers was not determined by actual analysis, who is to say if the 12.2% and 4.2% is correct?

Dr Malt

discnjh said:

.. but you're not going to have nearly that much volume loss when blending 1 L of 10% ethanol with 1 L of water, are you?

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No. I made a stupid mistake in the math. For 10% diluted 1:1 the numbers are:

For 1 L of solution of ABV 10.00 percent diluted with 1.000 liters of water; 20.00 °C
Desnsity of solution: 0.984710 gram/cc
Mass of 1 L of solution: 984.7104 grams
Volume of EtOH: 100.0000 cc
Mass of EtOH: 78.9239 grams
Mass of H2O: 905.7865 grams
Volume of H2O 907.4187 cc
Volume of EtOH + volume of H2O: 1007.4187 cc; Volume 'loss' 7.419 cc
Volume of dilution water 1000.00
Mass of dilution water 998.2012
Total mass of water: 1903.9877
Total volume of water: 1907.4187
Total mass: 1982.9117
ABW (mass of EtOH/total_mass) 3.9802 percent
ABV 4.9980 percent
Density of diluted solution: 0.9911
Volume of diluted solution 2000.8031
Volume of EtOH + Volume of H2O: 2007.4187; Volume 'loss' 6.616 cc

Thus the ABV is very close to what the linear assumption would have shown and for all intents and purposes you can just use the linear approximation.

Now on the other hand if you are doing this for the TTB from 160 proof spirits:

For 1 L of solution of ABV 80.00 percent diluted with 1.000 liters of water; 20.00 °C
Desnsity of solution: 0.859270 gram/cc
Mass of 1 L of solution: 859.2696 grams
Volume of EtOH: 800.0000 cc
Mass of EtOH: 631.3915 grams
Mass of H2O: 227.8781 grams
Volume of H2O 228.2887 cc
Volume of EtOH + volume of H2O: 1028.2887 cc; Volume 'loss' 28.289 cc
Volume of dilution water 1000.00
Mass of dilution water 998.2012
Total mass of water: 1226.0793
Total volume of water: 1228.2887
Total mass: 1857.4708
ABW (mass of EtOH/total_mass) 33.9920 percent
ABV 40.7774 percent
Density of diluted solution: 0.9468
Volume of diluted solution 1961.8705
Volume of EtOH + Volume of H2O: 2028.2887; Volume 'loss' 66.418 cc

The difference here is appreciable - more than they allow.