bakins
Well-Known Member
After seeing various discussions on BIAB, I decided to attempt to calculate the maximum efficiency possible for BIAB ( brew in a bag).
I used the info I gained from http://braukaiser.com/wiki/index.php?title=Troubleshooting_Brewhouse_Efficiency
I had a conversation with Kai and he said my numbers looked reasonable. Any errors, however, are my fault, not his.
All water measurements are at room temperature. (If not you have to account for expansion).
Lets assume a basic 10 lb grain bill and that the grain has an 80% extract potential (pretty normal). Assume we mash in with 8 gallons of water. Now the math is much more straight forward with metric numbers, so I'll convert.
10lb = 4.5 kg
8 gallons = 30.24 L
Our maximum pre-boil gravity, since we are not sparging, assuming 100% conversion efficiency (which is doable):
4.5 * 0.8 * 100 / (4.5 * 0.8 + 30.24) = 10.64 Plato or about 1.043 SG
Assuming, the "normal" grain absorption of .125 gallons/pound (1.05 L/kg), we should collect 6.75 gallons (25.5 L).
So, our efficiency "into the kettle" is:
100 * 25.5 * 1.043 * (10.64/100) / (4.5 * 0.8) = 78%
(Note: depending on how you round, the numbers fluctuates. I did this in a spread sheet and didn't copy all the decimals to this post).
One of the claims of BIAB is lower grain absorption rate (because the bag can "squeeze" a little more wort from the grain). So, just lowering it to .1 gallon/pound, we should collect 7 gallons (26.5L). So our efficiency would be:
100 * 26.5 * 1.043 * (10.64/100) / (4.5 * 0.8) = 81%
I'm sure this is full of errors, but wanted to get a discussion going. Please correct me. I just know that I don't get the horrible efficiencies that people say I should when I BIAB.
I used the info I gained from http://braukaiser.com/wiki/index.php?title=Troubleshooting_Brewhouse_Efficiency
I had a conversation with Kai and he said my numbers looked reasonable. Any errors, however, are my fault, not his.
All water measurements are at room temperature. (If not you have to account for expansion).
Lets assume a basic 10 lb grain bill and that the grain has an 80% extract potential (pretty normal). Assume we mash in with 8 gallons of water. Now the math is much more straight forward with metric numbers, so I'll convert.
10lb = 4.5 kg
8 gallons = 30.24 L
Our maximum pre-boil gravity, since we are not sparging, assuming 100% conversion efficiency (which is doable):
4.5 * 0.8 * 100 / (4.5 * 0.8 + 30.24) = 10.64 Plato or about 1.043 SG
Assuming, the "normal" grain absorption of .125 gallons/pound (1.05 L/kg), we should collect 6.75 gallons (25.5 L).
So, our efficiency "into the kettle" is:
100 * 25.5 * 1.043 * (10.64/100) / (4.5 * 0.8) = 78%
(Note: depending on how you round, the numbers fluctuates. I did this in a spread sheet and didn't copy all the decimals to this post).
One of the claims of BIAB is lower grain absorption rate (because the bag can "squeeze" a little more wort from the grain). So, just lowering it to .1 gallon/pound, we should collect 7 gallons (26.5L). So our efficiency would be:
100 * 26.5 * 1.043 * (10.64/100) / (4.5 * 0.8) = 81%
I'm sure this is full of errors, but wanted to get a discussion going. Please correct me. I just know that I don't get the horrible efficiencies that people say I should when I BIAB.