Low-Alcohol Beer

by Nicholas A. Franke (nafrank@ibm.net)


In the July 6, 1995 issue of the Digest (HBD #1774), I requested information on making alcohol-free beer from the collective. My motivation for making alcohol-free beer was that a friend of ours had recently become pregnant but still wanted to have an occasional beer. I thought I would provide a summary of the responses I received and my experience with making low alcohol beer.


I want to thank everyone who responded to my request. All of the responses were very helpful. I received twenty-four responses to my post, which provided the following input: Coincidentally, the Fall 1995 issue of Zymurgy contained a small blurb regarding an article written by Dr. Siegfried Gunther and Stefan Vetter for Brewing and Beverage Industry International (April 1994 edition) regarding alcohol-free brewing methods. That article divided methods into "biologic" and "physical". The three biologic methods noted were:

1. Interrupting fermentation with a one minute pasteurization at 140 F at the desired alcohol level;
2. Brewing a low-gravity beer and using yeast of the strain Saccharomyces iudwiggii, which will ferment only simple sugars;
3. Mixing the yeast and wort at 32 F and filtering the yeast out after a rest period (duration not specified).

The three physical methods noted were:

1. Heating the finished beer at atmospheric pressure. The authors noted damage to the beer that made this method undesirable;
2. Reverse osmosis. Removing alcohol by passage through diaphragms by use of pressure gradients;
3. Dialysis. Passing the beer through fibers that are bathed in a counterflow dialysate that produces a concentration gradient.


Armed with this information, I decided to merge two of the most popular concepts together. I decided to brew a fairly low gravity beer, and then to freeze it after fermentation so that the alcohol could be poured off. For reasons that I will go into later, and although not planned, I also ended up using a third method of removal--heat--but in a different way that was suggested.

The recipe I used was for a pale ale-type beer, and consisted of the following:

   3#     Domestic Malted Barley
   1.75#  Munich Malt
   1.25#  Cara-pils
  12 ozs. Canadian Wheat
   6 ozs. English Crystal (80L)
   2 ozs. Domestic Crystal (120L)
 .88 ozs. Northern Brewers (8.2% A)--60 mins.
2.69 ozs. Liberty (2.7% A)         --10 mins.
          Wyeast American Ale yeast (#1056)

The mash was a standard infusion mash, except that the Cara-pils was not added until the mash was brought to 158 F. The boil was for 90 mins. A 300 ml yeast starter and 3 tsp. yeast food were pitched into 6 gals. The O.G. was 1.031.

Primary fermentation lasted for four days at 68 F. Secondary fermentation was for another ten days at 68 F. F.G. was 1.012.

270.4 fl. ozs. was separated from the main batch after fermentation was complete for use in the "no alcohol experiment".


The 270.4 "experimental" ozs. were racked into 4, 2-liter plastic soda bottles. Those bottles were placed upside-down (on their caps) in a freezer at between 0 and 10 F for 36 hours. The fluid in the bottles was solid at the end of that period.

By inverting the bottles, the fluid that did not freeze (including the alcohol) was at the cap-end of the bottle, making it easier to pour it off. The biggest surprise of this whole process was that when I opened the cap on the bottles, they virtually exploded. After the first bottle, and cleaning the sink, cabinets, walls and ceilings of alcohol sludge, I opened the remaining three bottles underneath a plastic bucket. Each one of them exploded. I estimate that 770 ml of the alcohol sludge (I will call the "extract") was lost on the walls and down the sink. I did manage to save 950 ml of the extract.

There was an unexpected problem when I removed the 950 ml extract. I had been forewarned that this freezing method removed a lot of body and hop bitterness (thus the 1# Cara-pils and 35 IBUs), but no one had mentioned the loss of color. This was supposed to be a pale ale, and had been the appropriate color when I froze the beer. But when the extract was removed the beer remaining frozen in the 2 liter bottles was almost as clear as ice--no color at all. The extract, on the other hand, was as dark as a stout. The extract also had a S.G. of 1.039 and smelled like alcohol and HOPS (capitalized on purpose and for effect).

So, I was left with the problem of a colorless, no alcohol beer. This is when I decided to employ the third method of de-alcoholization--heat. Numerous people had warned me about the bad effects of heat on the beer for alcohol removal, and so I decided to heat only the extract. After all, that's where the alcohol was. After I removed the alcohol from the extract, I planned to put the extract back in the beer to give it color again.

Those suggesting heat as an alcohol removal method instructed that alcohol evaporates at 78-82 C (I did not check this fact). Therefore, I heated the 950 ml of extract at 174-178 F for 13 minutes, which resulted in 600 ml of extract at 1.069. After cooling, I returned this extract to the melting beer.


After going to all this trouble, I did not want to add any alcohol back into the beer through priming. I was also concerned about the viability of the yeast after the freezing/thawing/heating process. So, I put the thawed beer into a 3 gallon keg and force carbonated the beer for approximately 2.6 volumes of CO(2). The end result was 244 ozs. (7,222.4 ml) of low alcohol beer.


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.


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 the homebrewing collective. If anyone sees an error in any of the following (and I'm sure there are a few), please point it out. I don't mind admitting that I struggled greatly with these equations and the theory underlying them.

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 (taken from the Summer 1995 Zymurgy):

   O.G. 1.031     F.G. 1.012
   A%(by weight)=76.08(OG-FG)/(1.775-OG)
   A%(by volume)=A%(weight)[FG/.794]

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)
  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.
[Actual SG of extract]/[SG of extract without alcohol]
   = 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.


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.


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.

I would be very interested in receiving any comments or suggestions anyone might have.