Go Figure – Lautering Efficiency

The last post on this topic discussed mash efficiency, which was defined as how effectively the brewer converts the grains into wort sugars.  With that step completed, the brewer now needs to extract those sugars from the mash and get them to the boil kettle.  And the effectiveness of this process is what will be called lautering efficiency.

I believe these topics are best illustrated via example, so let us continue with the example presented in the mash efficiency post.   Five  kilograms of malt with a FGDB yield of 80% was mashed at 3 kilograms of water per kilogram of grain or 15 kilograms of water (Note: I giver water measurement by mass, not volume. A beauty of the metric system is that at room temperature one liter of water weighs 0.998 kilograms. (I just round that to 1.000.  There is more to come about water density in another post).  In this example the gravity of the liquid in the wort was measured at 19.0 °P, which corresponded to 3.5 kilograms of sugar and a 87.5% mash efficiency.

So now, just open the spigot on the mash tun and drain the liquor into the brew pot.  The first obvious fact is that  no matter what happens  the wort in the brew pot will be 19.0 °P (we already measured it, remember?).  So, assuming no further rinsing of the grains, our lautering efficiency will be a function of how well the mash/lauter tun drains.  How well will it drain?  It depends on the equipment.  In my brewery I find that I lose about 1.50 liters of water per kilogram of grain.  So back to the example above:

Water Loss = 5.0 kg malt x 1.5 kg water loss/kg malt = 7.5 kg

After starting with 15 kg of water, 7.5 kg are absorbed into the grain and 7.5 kg are now in the boil kettle.  How much sugar is in the boil kettle?  Remember:

Ms = -( ºP * Mw ) / ( ºP – 100 ) = -(19*7.5) / (19 -100) = 1.76 kg sugar

Since the sugar in the mash tun was calculated to be 3.5 kg, we have extracted exactly 50%, which is our lauter efficiency (so far).  The overall brewhouse efficiency is:

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

1.76 / (0.80 * 5.0) = 44%

which is of course pathetic but this is just an example.  Plus, the grains have not been rinsed or sparged in any way.  For arguments sake, assume 7.5 kg of water is now added to mash/lauter tun.  What happens?

Well, first remember that there is again 15.0 kg of water in the mash tun (the 7.5 kg absorbed by the grains plus the 7.5 kg we just added).  There is 1.74 kg of sugar in the mash tun (we started with 3.50 and there is now 1.76 in the boil kettle).  Therefore:

ºP = ( 100 * Ms ) / ( Ms + Mw ) = (100 * 1.74) / (1.74 = 15.0) = 10.4 ºP

If the mash/lauter tun is drained again then again 7.5 kg of water will be lost and 7.5 kg of water will go into the boil kettle.  How much sugar will be carried with this rinse?

Ms = -( ºP * Mw ) / ( ºP – 100 ) = -(10.4 * 7.5) / (10.4 – 100) = 0.87 kg sugar

So now there is 1.76 kg of sugar from the initial draining of the mash/lauter tun and 0.87 kg from the rinsing or 2.63 kg sugar in the boil kettle.  The lauter efficiency is now:

2.63 kg sugar (in boil kettle) / 3.5 kg sugar (in mash) = 75%

and the overall brewhouse efficiency:

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

2.63 / (0.80 * 5.0) = 66%

a much more respectable figure!

I hope that we have now laid the basic groundwork for how I conduct calculations related to mashing and lautering.  It is fine stuff, but how can a brewer use this to plan his brew day and end up making the desired beer?  Well, first we need to establish the targets that need to be hit, then calculate how much water and grain will be needed.  That is where we are headed next.

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  • Tom  On December 24, 2012 at 6:40 pm

    It seems to me that when calculating the lautering efficiency, you would divide by the actual sugar mass calculated for the mash efficiency instead of the FGDB- based sugar mass. Otherwise, wouldn’t you be calculating a combination of mash and lautering efficiency?

    • littleboybrew  On December 26, 2012 at 10:43 am


      I believe I used the correct number. In the example there was 5 kg of malt with a maximum extract of 80% or 4 kg. The actual extract (in the mash) was 3.5 kg, or 87.5% mash efficiency. The lauter efficiency I calculated was 2.63 kg (extract in boil kettle) / 3.5 kg (extract in mash) or 75%.

      The last number I calculated was the overall brewhouse efficiency. I have seen this calculated different ways. I base the efficiency on the FGDB adjusted malt weight, in this case 80% of 5 kg. I have also seen it calculated based on the gross malt weight, which in the example would be 2.63 kg / 5.0 kg or about 53%. If I were running a brewery I would probably care more about this latter value, since effectively converting purchased malt into extract directly affects the profitability. In my case I am just trying to consistently predict my brew strength!

      • Tom  On December 26, 2012 at 12:03 pm

        Now I see! I must have glossed over that important line that said “brewhouse efficiency” after calculating the first runoff. I assumed that was lauter efficiency – teaches me to not read before bed. So, great articles. I really helps having the examples to follow instead of just equations. As a new brewer, I really want to see my efficiencies at each step in the process. Thanks for the reply.

      • littleboybrew  On December 27, 2012 at 3:31 pm

        No problem. Always question authority!

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