Brew House Efficiency DefinedSaturday, June 27th, 2009
For the all grain brewer, brew house efficiency is of great importance. Brew house efficiency will directly affect your recipe formulation and how many pounds of malt will be required to reach a specific OG at a specific volume in the fermentor. This is a brief explanation of brew house efficiency.
Brew house efficiency is the calculation of the overall efficiency of your brewing system. It takes into consideration the percent of potential grain sugars that are converted in the mash, effectively washed during the latuer and all wort losses in your system. If you do not accurately calculate the brew house efficiency of your brewing system you will find it very difficult, of not impossible, to anticipate the OG of the recipes that you are brewing. Here are some helpful equations to calculate efficiency in your brewing system: The following equations assume 10 pounds of 2-row pale malt in 5 gallons of water, with a mash/lauter efficiency of 75%.
Calculating Extract Potential:
((grain points)*(pounds of grain)) / (volume in gallons) = extract potential
((36)*(10)) / 5 = 72 or 1.072
Calculating recipe OG:
(potential points) * (brew house eff.) = OG
(72) * .(75) 75% = 54 or 1.054
(measured points) / (potential points) = efficiency
(54 or 1.054) / (72 or 1.072) = 75%
Calculating Brew house Efficiency:
(Measured Points * Actual volume) / (Potential Points * Target Volume) = Brew house Efficiency
((54) * (4.5 gallons)) / ((72) * (5.0 gallons)) = 67.5%
Each pound of grain has an extract potential per gallon of water. Typical 2-row pale malt has a potential of 36 points, or 1.036, per gallon of water. Theoretically, if you achieved 100% mash and lauter efficiency (which is impossible) you could use 10 pounds of 2-row pale malt in a 5 gallon batch of beer and achieve an OG of 1.072. Most brewers are accustomed to achieving 75% efficiency during a typical mash and lauter, which would translate into 72 potential points * .75 (75%) = 54 points, or an OG of 1.054 with a volume of 5 gallons. But…..
The above can be considered mash efficiency (mash/lauter efficiency) since it does NOT take into consideration system losses of wort. Typically this is a loss that is created in the boil kettle and is lost to hops, trub or dead space in the boil kettle. In our example, if you end up with 5 gallons of wort in the boil kettle at an OG of 1.072, you have a mash efficiency of 100%, but if you transfer this wort to the fermentor and leave .5 gallons behind due to hops, trub and dead space you have now effectively reduced your system efficiency, your brew house efficiency to 90%. This would be calculated as such.
(Measured Points (72) * Actual volume (4.5 gallons)) / (Potential Points (72) * Target Volume (5.0 gallons) = Brew house Efficiency (90%)
In our same example, if you realized a typical mash/lauter efficiency of 75%, or an OG of 1.054 at a volume of 5.0 gallons in the boil kettle and 4.5 gallons to the fermentor after hop, trub and dead space losses, your brew house efficiency would look like this:
(Measured Points (54) * Actual volume (4.5 gallons)) / (Potential Points (72) * Target Volume (5.0 gallons) = Brew house Efficiency (67.5%)
So as you can see, mash/lauter efficiency is basically your efficiency into the boil kettle, but brew house efficiency is the efficiency throughout your brewing system which takes into consideration exactly what gravity and VOLUME you get into your fermentor. Understanding this calculation and the brew house efficiency of your brewing system will allow you to achieve a precise OG and volume into your fermentor each time, resulting in repeatability.
If all this math just gave you a headache, check out the brewhouse efficiency calculator found at this site.