Got this from Brewer's Friend, home brewing resources
"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%"
And here's a good explanation also: http://www.beersmith.com/blog/2008/1...-beer-brewing/