Boil off question... Confused....

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Mpavlik22

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Ok so i brew all grain. My preboil volume is 7 Gal and my post boil is 5.75 Gal. Ive figured out my evap rate to be 12% (or .84Gal/Hr) at those volumes. I always do a 90 Min boil. (7Gal*0.12)*1.5HR=1.26 Gal Boil Off volume over 90 minutes.

My question: I want to brew a barleywine and i am limited on my mush tun size so i am going to do a smaller batch. If i want a 3 Gal post boil, using the 12% my preboil is... 3.66 Gal????????? Thats what the math says... (3.66*.12)*1.5HR=0.66 Gal Boil Off Volume.

But wait... i am using the same pot and and same heat so shouldn't my boil off rate be the same as the prior at .84Gal/Hr? So if its .84 Gal/Hr boil off that would mean 4.26 Gal preboil to give me 3 gal post boil which is 19.75% evap rate??? (4.26Gal*0.1975)*1.5HR=1.26 Gal Boil off.

How can my evap rate change from 12% to 19.75% when i scale my recipe from 5.75 to 3 Gal batches?
 
Realistically, there is no such thing as a boiloff percentage. If you boil off .84 gallons per hour, it really doesn't matter if you have 7 gallons in the pot or 3, you still boil off .84 gallons per hour. You should use volume per hour boiloff as opposed to % boiloff.
 
Ya I guess ur right. I got threw off b/c most software uses %. But looking at it from a volume only standpoint makes sense now.
 
It may help to think if it in terms of the fact that a given amount of heat is required to vaporize a given mass of water. For water it takes a little less than 44 kJ to evaporate a mole (18 grams). So if you have a heat input to the kettle of 44 kJ/min you can potentially evaporate 18 grams of water per minute. Doesn't matter if the kettle contains a liter or a hectoliter. This, of course, neglects heat losses which may differ some with volume.
 
I love how AJ can take a complicated topic and "boil" it down to its simplest form. :mug:
 
I don't know why they use %. It is a constant rate as long as the boil intensity stays about the same.
 
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