## Go Figure – Calculating Mash Efficiency

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

and,

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.

• Chris  On March 3, 2015 at 8:43 am

I’m trying to simulate my brew setup – mostly by dialing in efficiency and loss characteristics of my brewhouse to predict quickly and accurately my OG..
And I’m trying to predict my sugar content at each step (mashout/sparge/preboil/postboil) to see where my efficiencies vary, etc.

That said, as I see it, there are two ways of doing it:
1) Ms = -( ºP * Mw ) / ( ºP – 100 ) ; where Mw is ~Volume of mash water @ 20°C
2) Ms = °P/100 * SGwort * Volume of wort @ 20°C

Yet, they yield difference results. For example, measuring 10°P in 10L of wort yields:
1) Ms = -(10*10)/(10-100) = 1.11 kg
2) Ms = 10/100 * 1.04 * 10 = 1.04kg

What am I missing?
Grateful for any insight!

• littleboybrew  On March 3, 2015 at 1:15 pm

Chris,

The error you have made is is equation 1. Remember that 10 liters of 10 Plato wort weighs 10.4 kg (sg x volume, 1.04 x 10). Equation 2 tells us that the sugar weighs 1.04 kg, which means the water weighs 9.36 kg.

So Equation 1 should read :

Ms = -(10*9.36)/(10-100) = 93.6 / 90 = 1.04 kg sugar.

I have been getting pretty good results in predicting my efficiency and hence the planned OG of my beers. However I have revisited my spreadsheet calculations and I think I have found a few small errors. I need to brew and see of my accuracy is better than before. I will post soon.

• Chris  On March 4, 2015 at 10:07 am

Thanks, that does make sense.
But that raises another point of misunderstanding on my part!

Is the definition of 1°P not “mass of sugar” as % of total “mass of solute”, or mass of sugar (Ms) / mass of wort (Mwort), whereby the mass of wort = mass of sugar disolved therein + mass of water ?
ie. °P= (mass of sugar)/(mass of wort) = (Msugar)/(Msugar+Mwater).
=> the solute is already broken down in its constituent parts, and I take the volume of wort to be equal to the volume of water = mass of water (SG=1)

I am however assuming the following which may be misinformed, and which explains my errant calculation.
– If I measure a volume of 10L wort solution in my kettle, the volume of water is 10L, the volume taken up by the sugar is minimal (in the interstitial spaces…), and the weight of the wort is SG*10litres. ie. I’m assuming that if I dissolve 1kg sugar in 10L of water, the volume is remaining broadly unchanged (empirical assumption only!), such that any volume measurement of the solution is approximately also a volume of the water.

Grateful for your point of view!

• littleboybrew  On March 4, 2015 at 1:33 pm

Actually the volume does change when you add extract to water and make wort.

To illustrate, remember that there is correlation between the Plato measurement and the specific gravity. For now let us consider specific gravity to be the same as density (grams per cubic centimeter or kg per liter). If you added 1 kg dry malt extract to 10 liters of water the resulting wort would weigh 11 kg, and the °P = 100* (1 / 11) = 9.1. We know 9.1 °P = 1.036 g/cc. So the volume of the wort would be 11 kg / 1.036 kg per liters or about 10.6 liters.

Note I mentioned specific gravity and density are not exactly the same thing. At 4° C, 1 liter of water weighs 1 kg (density = 1 kg / liter). If you heat that liter of water to 100 °C (boiling) it will expand to a volume of 1.04 liters (and the density will drop to 0.96 kg / liter). Using the above example if you mixed a 10 liters of boiling water with a kilogram of DME, the resulting gravity would be 1 kg extract /(1 kg extract + 9.6 kg water) = 9.4 °P.

So you can see where you might get slightly different results if you first measure out a volume of water then heat it and add extract, versus heating some water then measuring out a volume and then adding extract. Not a huge difference, but it can be confusing.

Did that help or just confuse you more?

• Chris  On March 10, 2015 at 8:53 pm

Thank you!