Math is confusing to many folks, but it is very helpful to understanding how variables affect process outcomes.
In general efficiency is how much you got compared to the maximum available. Efficiency is usually expressed as a percentage defined by: 100% * Actual Obtained / Max Available.
In brewing, efficiencies are about how much of the avialable
extract (sugar) you collect vs. how much could be obtained from the grain (if you could get it all.) Volume only comes into play since we can’t directly measure the amount (mass or weight) of extract in wort - we can only measure the volume of wort, and its SG, and use that to calculate the amount of extract.
There are several efficiency measurements of interest in brewing:
- Mash Efficiency = Amount of Extract in BK / Max Potential Extract in Grain
- Mash Efficiency = Conversion Efficiency * Lauter Efficiency
- Conversion Efficiency = Amount of Extract Created in Mash / Max Potential Extract in Grain
- Lauter Efficiency = Amount of Extract in BK / Amount of Extract Created in Mash
- Transfer Efficiency = Amount of Extract in Fermenter / Amount of Extract in BK
- Brewhouse Efficiency = Mash Efficiency * Transfer Efficiency
- Brewhouse Efficiency = Amount of Extract in Fermenter/ Max Potential Extract in Grain
Note that none of the efficiency definitions involve volume, only the amount (mass or weight) of extract. It is possible to calculate the amount of extract if we know the volume of wort and its SG, however we can use a simpler approximation of amount of extract known as
Points. One point will raise the SG of one gallon of wort by 0.001. Thus the number of points in one gallon of wort is:
Points / Gallon = 1000 * (SG - 1)
Rather than doing the math, pts/gal is just taken as the three digits after the decimal point in the SG, and dropping any leading zeros. Thus wort with an SG of 1.037 has 37 pts/gal. You calaculate the total number of points in a volume of wort as: gal of wort * pts/gal, so 5 gal of wort with an SG of 1.037 has 5 gal * 37 pts/gal = 185 points.
We can also express the potential extract in grain in terms of points. Base malts usually contain about 37 points/lb of grain (commonly called points per pound per gallon or “ppg.”) This means that if 1 gal of wort is made with all of the possible extract from one pound of grain, then the wort would have an SG of 1.037. In grain databases, the extract potential is often given in SG rather than ppg, so the potential of a grain with 37 ppg would be listed as 1.037. Note that if you put all of the extract from 1 lb grain with a potential of 1.037 into 1 gal of water, you will have more than one gal of wort (since the extract has volume that adds to the total volume of wort), and the SG of the wort will be less than 1.037. This is a common point of confusion, which often leads to errors in calculating conversion efficiency.
We can use points to calculate efficiency. First we have to determine the number of potential points we start with in our grain bill. To do this we multiply the weight of each grain by the ppg of that grain, and add up the points for each grain. So, if we have 5 lb of grain with a potential of 1.037, and 5 lb with a potential of 1.035, then the total potetial points are 5 lb * 37 ppg + 5 lb * 35 ppg = 185 pts + 175 pts = 360 pts total extract potential. If we mash and sparge that 10 lb of grain with a total of 7.7 gal of water, and collect 6.5 gal of pre-boil wort at an SG of 1.045 then the number of points in our BK will be: 6.5 gal * 45 pts/gal = 292.5 pts. Using the equation for mash efficiency above, our mash efficiency will be: 292.5 pts / 360 pts = .8125 => ~81%.
If we boil that 6.5 gal of wort down to 5.5 gal, we will still have 292.5 pts of extract, since we only evaporate water during the boil, but no extract. The post boil wort will then have: 292.5 / 5.5 gal = 53.18 pts/gal => SG of ~1.053.
If we transfer 5.0 gal of the post-boil wort to the fermenter, then the fermenter will contain: 5.0 gal * 53 pts/gal = 265 points. Brewhouse efficiency then equals: 265 pts / 360 pts = 0.736 => 73.6%.
Now if we had used more brewing water (strike + sparge = ~8.51 gal), and collected 7.3125 gal of wort at 1.040 SG, our BK points would still be: 7.3125 gal * 40 pts/gal = 292.5 pts, and our mash efficiency still would have been ~81%. However, using more brewing water generally improves lauter efficiency, and therefore mash efficiency, so the collected wort would more likely have have an SG of about 1.041, giving BK points of 7.3125 * 41 = ~300 pts, for a mash efficiency of 300 pts / 360 pts => ~83%.
Brew on