Go Figure – What is Efficiency?

After ten years of brewing and ten years of guessing what my mashing and lauterng efficiency might be, I wasn’t so much frustrated as simply convinced there was no better way. I went about guessing what my efficiency might be. Then Kai Troester pointed me in the right direction.  But before we talk about that, let’s review what efficiency really means.

Efficiency is just of measure of how much fermentable and non-fermentable sugar the brewer extracts from the malt. All brewing grains are composed of a variety of materials, such as starches, sugars, proteins, tannins and the like.  Brewers want to extract the starches (and convert them to sugars) and the sugars (for converted malts such as crystal malts).

But how does the brewer know how much sugar to expect?  Maltsters test their brewing grains in special laboratory mashes.  One commonly reported result is the percent yield, fine grind dry basis (FGDB).  In short, the grain is ground fine and mashed in controlled conditions, and the amount of fermentable material is measured.  Most fully modified base malts available to home brewers have a FGDB yield in the range of 75% to 82%.  So one kilogram of malt should yield 0.75 to 0.82 kilograms of sugar.

But of course it is not that simple.  Notice that FGDB specifies dry basis, and brewing malts are not dry.  They absorb moisture, and while I have never personally measured the moisture content of my malt, the moisture level is often given to be 4%.  So that malt with an 80% FGDB yield will actually yield about 77% sugar by weight.

Next, and here where Kai’s insights are particularly valuable, the brewer must consider his mashing efficiency and his lautering efficiency.  If the mash is not conducted efficiently, then the yield will fall further.  Once the mash step is complete, the brewer must then extract the sugars by lautering the grains with water.  And this is another step where sugar can be lost and efficiency suffers.

At the completion of the mashing step, the mass of the sugar in the boil kettle will be less than the mass of the grain or even the amount of sugar predicted by the FGDB method.  I find that usually the brewhouse efficiency is defined as :

Mass of Sugar in Brew Kettle / (FGDB Yield * Mass of the Grain)

I have sometimes seen efficiency expressed as the mass of the sugar in the brew kettle divided by only the mass of the grains.  Obviously this will be a much lower value than the definition above, and there is nothing wrong with calculating in that manner.  However I believe the brewhouse efficiency formula that I have proposed is the most common one, and that it is slightly more indicative of the brewer’s true efficiency.

The last piece of the puzzle is determining how much sugar is in the brew kettle.  Most brewers are probably familiar with the formula for calculating degrees Plato:Where Ms is the mass of the sugar and Mw is the mass of water.

This formula simply expresses the amount of sugar in the wort on a weight percent basis.  As a practical example, consider 25 liters of 10 °P wort.  The first calculation that must be done is covert the volume measurement (25 liters) to a mass.  This is easily accomplished because we can convert the Plato measurement to specific gravity, specifically 10 °P is equivalent to a specific gravity of 1.040.  Therefore the wort has a mass of 25 x 1.040 = 26 kilograms.  And since 10 °P is equivalent to 10% mass by weight, we know the wort contains 2.6 kilograms of sugar.  If the brewer started with 5 kilograms of malt with a FGDB yield of 80%,then  the brewhouse efficiency would be 2.6 / (5 * 0.8) or 65%.

So now we know what efficiency is and how to calculate it.  But how can we control and predict it?  That is a subject for another post.

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