Calculate brewhouse efficiency

The easiest way to calculate brewhouse efficiency is to go metric! You need four pieces of information to perform this straightforward calculation. The data you need is: post boil wort volume in liters (there are 3.785 liters per gallon), post boil specific gravity (for example, 1.056 kg/L), post boil wort density in ºPlato (to approximate, divide the number behind the decimal of the specific gravity by four — e.g. 56/4 = 14 ºPlato) and the weight of grains used in recipe (in kg).

Once you have this information the calculation is easy. The first thing that is calculated is the weight of extract in the wort. Extract = (volume) x (specific gravity) x (ºPlato — expressed in decimal form). For example, (20 liters) x (1.056 kg wort/liter wort) x (0.14 kg extract/kg wort) (Plato is a weight/weight measure) equals 2.96 kilograms of extract. This is how much stuff you extracted from the grain during mashing and lautering.

The efficiency number is determined by comparing what was extracted to what was used. For example, if 4.5 kilograms (9.9 pounds) of malt was used to produce 20 liters of 1.056 wort, the efficiency is 2.96 kg extract/4.5 kg malt or 0.658. This number can be multiplied by 100 and expressed as a percentage . . . like 66%.

Without going into the nitty gritty details of the “problem” with this number, I do want to point out that this number is pretty crude. The reason is that not all grains used in brewing have the same potential. In technical circles, brewers talk about laboratory or theoretical yields of different ingredients. Some ingredients like pale malt have a laboratory yield of around 78% and most specialty grains have laboratory yields ranging from 55-65%. This means that a pale beer without special malts has a better efficiency than beers made using special malts.

Since brewers, especially commercial brewers, want to get as much out of the grain as possible, it makes it difficult to examine efficiency. A low yield calculated the way I show above may be due to the type of beer being made or a problem in the brewing process. The solution to this dilemma is to compare the yield of a particular mash to its theoretical yield.

Malt specification sheets give the lab yield number and a theoretical yield can be estimated. If your brew contains 8.8 pounds (4 kg) pale malt with a lab yield of 78% and 1.1 pounds (0.5 kg) of crystal malt with a lab yield of 65%, you can estimate the combined lab yield of these grains using a weighted average. Estimated combined yield = (4 kg pale/4.5 kg total malt x 78%) + (0.5 kg crystal/4.5 kg total malt x 65%) = 76.5%.

This number can then be used as something to gauge the performance of your equipment against. If you got a yield of 66% and the lab yield is 76.5%, you can calculate what is know as the brewhouse yield. In this case, it is 66 dived by 76.5 or 86%. Most homebrewers do not calculate brewhouse yield because malt specification sheets are not always available. I hope I haven’t confused matters too much, but that’s how to run the yield calculations. Happy number crunching!

Response by Ashton Lewis.