Deduce Mash efficiency from Brewhouse efficiency

[dude1]

I always focused on brewhouse eff so far, because I found it more relevant.
This said, is there any formula, or at least an approximation, to calculate the mash eff based on the brewhouse eff?

For example, with my BIAB system, I reach approx. 79% brewhouse eff.
How much would that be in mash eff?

Thanks

 

[McKnuckle]

Remove the post-boil component from your brewhouse efficiency, and you'll have your mash efficiency. Mash efficiency is the product of conversion efficiency times lauter efficiency, or the total efficiency reached at the pre-boil stage of brewing.

Divide the wort volume you transferred to your fermenter by the post-boil total volume, after cooling. That will produce a percentage. Example; you ended with 5.5 gallons after chilling, then left 0.25 in the kettle when you transferred to the fermenter, or 5.25 gallons.

5.25 / 5.5 = 95.45% of the wort was collected.

Now take your brewhouse efficiency and divide it by that percentage to get your mash efficiency:

79% / 95.45% = 82.77% mash efficiency

 

[Big Monk]

Remove the post-boil component from your brewhouse efficiency, and you'll have your mash efficiency. Mash efficiency is the product of conversion efficiency times lauter efficiency, or the total efficiency reached at the pre-boil stage of brewing.

Divide the wort volume you transferred to your fermenter by the post-boil total volume, after cooling. That will produce a percentage. Example; you ended with 5.5 gallons after chilling, then left 0.25 in the kettle when you transferred to the fermenter, or 5.25 gallons.

5.25 / 5.5 = 95.45% of the wort was collected.

Now take your brewhouse efficiency and divide it by that percentage to get your mash efficiency:

79% / 95.45% = 82.77% mash efficiency

What strikes me as odd is when programs use Brewhouse Efficiency to estimate extract. Brewhouse Efficiency is a “bean counter” metric that deals with volume, not extract.

More specifically, Brewhouse Efficiency doesn’t describe the wort production phase like Mash Efficiency does and I very rarely see the distinction made between the 2.

I always use them as 2 separate metrics, i.e. I’ll accept much lower values for Brewhouse Efficiency without losing sleep.

 

[CascadesBrewer]

When I am anal about my measurements I find that my mash efficiency is about 10% higher than my brewhouse efficiency for my 5 gal BIAB batches...but it would depend on your process/equipment. I don't really have many places to loose wort (no hoses, pumps, chillers, etc.).

An annoying thing to me about using BeerSmith for this calculation is that it just calculates a "efficiency into the fermenter" value. So if I dump my entire kettle into the fermenter I get more efficiency on paper since I have more volume and the same gravity reading...but then I will end up with about the same amount going into a keg either way.

 

[McKnuckle]

Yeah, I agree; the only thing I care about is my conversion efficiency. That's the number dictating my gravity and IBU.

I mill the same, mash and lauter the same, and boil (mostly) the same every time - and if I choose to leave some crud in my kettle it's a last minute choice, and never enough volume to make a difference in any case.

I prefer to focus on conversion efficiency, since it's the very first thing I can influence in the process.

 

[doug293cz]

Mash Efficiency = Brewhouse Efficiency * Post-boil Volume / Volume in Fermenter

(This is just a more succinct way of saying what @McKnuckle said.)

Brew on