So I have a friend who ages his own small batches of whiskey. Since he wants to make them in the bourbon tradition, he uses them once and then can't use them again. Good deal for me.
So per advice from this forum, I made a stout for my first attempt with these barrels. Because these barrels are so small, I intend to oak two liters of the batch and then blend it back into the main batch. The big question is, how long do I need to age the two liters in order to pick up enough oak/whiskey flavor for the whole five gallon batch. I did some calculations, made some assumptions and came up with a number, but I'd like you all to check my assumptions (and my math if you don't mind). So, here we go:
If I want my beer to develop a particular flavor profile, I will have to over-age the small amount of beer I put in the barrels to get that flavor profile.
In order to get my 5 gallon batch to taste like a commercial beer might after being aged 6 months in a 53 gallon barrel, I did these calculations:
53 gallons = approx 200 liters
6 months = 26 weeks = 182 days
200 liters /182 days = 1liter /x days
To solve for x, you cross multiply.
200x = 182
Divide both sides by 200 and you get x = .91 days
So, to oak 1 liter it would be about 1 day to get the flavor profile commercial brewers get in 6 months of aging in a 53 gallon barrel. It would be 2 days to get a 1 year age effect.
This assumes that the rate of oak/whiskey flavor is a direct ratio with size of the barrel, surface area to beer, etc.
Now, to get the 5 gallon batch to taste like 6 months of aging I have to over oak the two liters. in order to get the desired flavor profile for the whole batch, for which i will use the number 100 (as in 100%), I have did these calculations:
2 liters at .91 days = 100%, but I have 17 more liters at 0
Once I mix them, the mixture becomes diluted.
2 liter /19 (whole batch)= .105
.105 times 100 and you get 10.5%
10.5% if aged .91 days. How many days?
10.5% / .91 days = 100% / x days
10.5x = 91 Divide both sides by 10.5 which gives me 8.6 days
Now, I know that every barrel will age differently and each beer will hide or show oak/whiskey flavors differently. But this could give me a guide for how long I would need to age the beer in the barrels as long as my big assumption about the direct ratio of size to time in gaining oak flavor.
What do you think?
So per advice from this forum, I made a stout for my first attempt with these barrels. Because these barrels are so small, I intend to oak two liters of the batch and then blend it back into the main batch. The big question is, how long do I need to age the two liters in order to pick up enough oak/whiskey flavor for the whole five gallon batch. I did some calculations, made some assumptions and came up with a number, but I'd like you all to check my assumptions (and my math if you don't mind). So, here we go:
If I want my beer to develop a particular flavor profile, I will have to over-age the small amount of beer I put in the barrels to get that flavor profile.
In order to get my 5 gallon batch to taste like a commercial beer might after being aged 6 months in a 53 gallon barrel, I did these calculations:
53 gallons = approx 200 liters
6 months = 26 weeks = 182 days
200 liters /182 days = 1liter /x days
To solve for x, you cross multiply.
200x = 182
Divide both sides by 200 and you get x = .91 days
So, to oak 1 liter it would be about 1 day to get the flavor profile commercial brewers get in 6 months of aging in a 53 gallon barrel. It would be 2 days to get a 1 year age effect.
This assumes that the rate of oak/whiskey flavor is a direct ratio with size of the barrel, surface area to beer, etc.
Now, to get the 5 gallon batch to taste like 6 months of aging I have to over oak the two liters. in order to get the desired flavor profile for the whole batch, for which i will use the number 100 (as in 100%), I have did these calculations:
2 liters at .91 days = 100%, but I have 17 more liters at 0
Once I mix them, the mixture becomes diluted.
2 liter /19 (whole batch)= .105
.105 times 100 and you get 10.5%
10.5% if aged .91 days. How many days?
10.5% / .91 days = 100% / x days
10.5x = 91 Divide both sides by 10.5 which gives me 8.6 days
Now, I know that every barrel will age differently and each beer will hide or show oak/whiskey flavors differently. But this could give me a guide for how long I would need to age the beer in the barrels as long as my big assumption about the direct ratio of size to time in gaining oak flavor.
What do you think?