How to Brew Low Alcohol Beer at Home -- Part 2

In Part I, I described my attempt to make a low alcohol beer by starting with a low gravity beer and then removing the alcohol by freezing the beer and pouring off the alcohol. In this part, I will set out my estimates of how efficient that alcohol removal process was.

THE NUMBERS

My Disclaimer: I am not a chemist, physicist, mathemetician, professional brewer, alchemist, botanist, or floral arranger. However, I am a lawyer and know how to write a disclaimer. Accordingly, none of the following theories, theorems, equations, hypotheses, statements, opinions and conclusions should be relied on. If you need a truly alcohol-free brew for health or safety reasons, please do not rely on my (probably flawed) methodology and results. Instead, buy a tried and tested commercial, non-alcohol beer. The purpose of this post is the amusement of homebrewers.

The Alcohol Content of the Original Beer

I started by figuring the alcohol content of the beer before I did anything unnatural to it. I used the following formulas (from the Summer 1995 Zymurgy):

O.G. 1.031 F.G. 1.012

A%(by weight)=76.08(OG-FG)/(1.775-OG)

=76.08(1.031-1.012)/(1.775-1.031)

=76.08(.019)/1.744

=1.996574585

A%(by volume)=A%(weight)[FG/.794]

=1.996574585[1.012/.794]

=2.544752491

Thus the beer had a very low alcohol content of approximately 2.5% even before doing anything to it.

I next computed the volume of actual alcohol in the beer. The total beer volume was 244 ozs. (7,222.4 ml), and so I multiplied that number by the percent alcohol by volume to determine the number of milliliters of alcohol in the beer: 7,222.4 ml beer x .02544752491 = 203.6779104 ml alcohol

Composition of the Extract

Out of necessity, I made the assumption that the freezing method removed all of the alcohol in the beer with the extract, and that no alcohol remained in the frozen beer. Therefore, all 203.6779104 ml of alcohol were contained in the extract.

I saved 950 ml of the extract and, based on the beginning and ending volumes of beer, estimate that I lost approximately 770 ml of the extract down the sink, on the walls, etc. Therefore, the total volume of the extract was 950 + 770, or 1,720 ml.

Again, out of necessity I assumed that the percentage of alcohol in the extract I lost was the same as the percentage of alcohol in the extract I saved. This does not seem to be an outrageous assumption to make. Then I computed the volume of alcohol that was lost with the lost extract:

[770 ml lost extract/1,720 ml total extract volume] x

203.6779104 ml alcohol = 91.1813901199 ml alcohol lost

203.6779104 ml alcohol - 91.1813901199 ml alcohol lost =

112.496520281 ml alcohol in 950 ml saved extract

Specific Gravity of the Extract Without Alcohol

Next you must compute the specific gravity of the extract as if there were no alcohol in it. I used the number .796 as the S.G. of alcohol in this equation:

(extract gravity)(extract volume)=

(SG alcohol)(alcohol volume)+(SG of extract without

alcohol)(alcohol-free extract volume)

(1.039)(950 ml)=(.796)(112.496520281)+(SG)(837.503479719)

987.05=89.5472301436+(SG)837.503479719

897.502769857=(SG)837.503479719

1.07164064578=SG of 950 ml extract without alcohol

Adjusting the Specific Gravity for Volume

The next calculation adjusts the specific gravity of the extract without alcohol (1.07164064578) for the reduction in volume of the extract that occurred when it was heated. The extract volume was reduced from 950 ml to 600 ml. From the above, though, it was determined that only 837.503479719 ml of the original extract was not alcohol. Assuming there was a proportional decrease in the volume of extract that was alcohol and the volume of extract that was not alcohol, the non-alcohol portion of the extract would have been reduced from 837.503479719 ml to 528.949566144. [The reality is that the non-alcohol part of the extract actually evaporated slightly quicker than the alcohol part. This can be demonstrated through several equations which I have not included here.]

The specific gravity of the non-alcohol part of the extract is adjusted for the reduction in volume as follows:

837.503479719 ml (begin volume)/528.949566144 (end volume)

x 1.07164064578 (SG non-alcohol part)

= 1.11343102248 (SG of non-alcohol part of 600 ml extract)

Calculating the Alcohol Content of the Processed Beer

Finally, taking the results from the foregoing, the new alcohol content of the beer, after the freezing and heating processes, is calculated.

= Efficiency Factor

Efficiency Factor x Original Alcohol Vol.=Actual Alcohol Volume

1.069/1.11343102248 = .60829919797 [efficiency factor]

.60829919797 x 112.496520281 ml alcohol = 68.4315430613 ml alc.

The remaining amount of alcohol in the extract, which is added back to the beer, is 68.4315430613 ml. The total volume of the beer is 7,222.4 ml. Thus, it is simple to do the final calculation and figure out the alcohol content of the beer by volume:

68.4315430613 ml/7,222.4 ml = .0094749035, or approximately 1%

The beer therefore still has about 1% alcohol by volume because the extraction method was only about 61% efficient.

COMPARISON OF BEERS

The untreated beer was only slightly darker (maybe 2-3 SRM) than the beer which had been frozen and had alcohol removed. The untreated beer also had slightly more hoppiness and a little more body. However, on the whole I can not say that the alcohol removal process dramatically changed the beer. While I would certainly prefer a regular pale ale, I found both the treated and untreated versions of this beer to be very drinkable, and certainly recognizable as beer.

METHODS OF IMPROVEMENT

If I were to attempt a no- or low-alcohol beer again, which, by the way, is a lot of work, I would use this same method. Freezing the beer, removing the extract, heating only the extract, and returning the extract to the frozen beer gives the benefits of alcohol removal without losing all of the body, hoppiness and color found in the extract.

Improvement is needed primarily on the efficiency of the heating process to remove alcohol. A longer heating process is necessary. Through a series of other calculations, I estimated that to get a beer of this OG/FG to less than .5% alcohol v/v (the legal definition for a no-alcohol beer), approximately 90% of the extract would have to be evaporated by heating. One-half of the volume that evaporates could be replaced by water without harming the color or flavor of the extract.

Good luck and happy brewing!