Another Efficiency Question

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Another Efficiency Question

Postby Shaun » Fri Jan 16, 2009 3:26 am

Ok, so I have been reading through the forum and found several posts concerning how to calculate efficiency. I understand the mouse hover technique and have done that to figure mine. My question is that the figure that BTP is actually giving you is the "Brew House Yield" and not a true "efficiency"...right?
To calculate my efficiency prior to getting BTP I was using the BYO method by first figuring the weight of extract in the wort and then comparing the amount of extract to amount of grain used:

Extract = (Volume) x (SG) x (*Plato<--in decimal form)

So in my Cream Ale I used 11.5 lbs of grain and ended up with 6 gal of OG 1.056
wort.
(22.7L) x (1.056) x (.14) = 3.36 kg of Extract

3.36kg of extract out of 5.22kg of grain (3.36/5.22) = .64

So in this case, my efficiency was 64%. Without actually knowing the potential yield of my grain bill, that was as good as I could get. You can imagine how bummed I was when I figured my efficiency after constantly hearing about everybody getting 80-90% efficiency. So then I used the tastybrew.com efficiency calculator and it gave me an efficiency of 78.7%. This made me feel quite a bit better, but BTP gives me 86% efficiency! While I would love to talk about my 86% efficiency percentage, I'm not sure which figure to use... I guess it only really matters which calculation EVERYBODY ELSE uses so I can better judge my system.

Which one is correct?
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RE: Another Efficiency Question

Postby wottaguy » Fri Jan 16, 2009 9:12 am

Hi Shaun....

There's a decent primer about this subject at Palmer's website.

Here's the direct url: http://www.howtobrew.com/section2/chapter12-4.html

Hope this helps...

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Efficiency

Postby slothrob » Fri Jan 16, 2009 12:09 pm

The equation you've been using is not for calculating efficiency, it's for converting weight of a fermentable into gravity points. It assumes that 1# will produce a solution of ~1.046 in 1 gallon. That is what you get from pure sugar, which grain can never equal.

Pale malt is about 80% sugars, so your equation can be used to calculate that 1# of average malt has the potential to yield about (46*0.8=37) 37 points per pound per gallon. Most of the malts I've used have actually come in at about 34-36 points per pound per gallon potential yield according to the lab analysis.

In your recipe, BTP is using 34 ppppg to arrive at 86%. If the grain you are using has that potential, then that is the correct value for your efficiency. This is the efficiency we are talking about when we discuss mash or extract efficiency. This is also the efficiency that BTP displays in the main window.

Brewhouse efficiency is usually used to discuss the combined effect of mash efficiency and volume losses due to transfers. I prefer to think of these as two separate processes since the steps you take to optimize your mash yield have nothing to do with how much liquid you decide to leave behind during racking. Obviously this would become important in a comercial operation where they are trying to optimize the number of bottles they can fill per pound of grain.
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Thanks

Postby Shaun » Fri Jan 16, 2009 9:55 pm

That makes perfect sense. I appreciate it.
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