In the last post on this topic I discussed the concept of brewhouse efficiency and how to calculate it. Now I want to talk specifically about the mash, how to calculate mash efficiency and what mash efficiency you can expect.
As mentioned in the last post, brewhouse efficiency is a combination of two separate efficiencies: mashing efficiency and lautering efficiency. Mashing efficiency is a measure of how much of the starch in the grain is converted to sugar. When the grains are mashed in, the enzymes in the grain begin the breakdown process. In theory all of the starch can be converted, resulting in a mashing efficiency equal to the FGDB yield. A number of brewing parameters will impact the mash yield: the moisture content of the grain, the crush quality, the mash temperature, the mashing time, and the mash pH (which is a function of the grain bill and the water characteristics). But my goal is not to discuss these factors, but how to measure your mash efficiency so you can work on predicting it.
Measuring the mashing efficiency starts with rearranging the equation for degrees Plato:
ºP = ( 100 * Ms ) / ( Ms + Mw )
Where Ms is the mass of the sugar in the wort and Mw is the mass of the water. What never occurred to me, and I thank Kai for pointing this out, is that we can rearrange the equation to define Ms in terms of Mw:
Ms = -( ºP * Mw ) / ( ºP – 100 )
So if we know the degrees Plato and the amount of water in the mash, we can calculate the sugar and know the mash efficiency. I like to use examples to illustrate the calculations, so assume 5 kilograms of malt with a FGDB yield of 80% is mashed at 3 kilograms of water per kilogram of grain. After one hour a sample of the mash has a gravity of 19 °P. What is the mash efficiency?
The maximum theoretical yield is 80% x 5 = 4 kilograms of sugar. To calculate mash efficiency use °P = 19 and Mw = 5 kg grain x 3 kg water/kg grain = 15 kg water, so that:
Ms = -( ºP * Mw ) / ( ºP – 100 ) = -( 19 * 15 ) / ( 19 – 100 ) = 3.5 kilograms of sugar
Mash Efficiency = 3.5 / 4.0 = 87.5%
So what mash efficiency should the home brewer expect? First of all remember that malt absorbs moisture in the range of 4% by weight, so you can probably start by assuming a maximum mash efficiency of 96%. Beyond that the efficiency depends on a number of factors that have already been mentioned. I am still learning what system really does, but I have found that if I mash for one hour a mash efficiency of 90% works pretty well.
Next time, I will discuss lautering efficiency and how much of the wort sugar we can reasonably expect to extract from the wort.