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Old 02-25-2011, 06:01 PM   #1
400d
 
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ok I know this is not about beer, but still couldn't find better place to post it. I have a mead that is fermenting for more than a month. It's 5 liters (1.32 gallons)

Original gravity was 1.135 and after a month I added 0.5 liters (0.132 gallons) of water to top up because the head space could cause oxidation after a long time.

now, how do I know what is the exact OG and how do I calculate % alcohol after this dilution?

thanks!
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Old 02-25-2011, 06:10 PM   #2
BendBrewer
 
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1.32*135 = 178.2

178.2 / 1.452 = 122.7

og = 1.123 (Rounded up)

 
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Old 02-25-2011, 06:22 PM   #3
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Quote:
Originally Posted by BendBrewer View Post
1.32*135 = 178.2

178.2 / 1.452 = 122.7

og = 1.123 (Rounded up)
what is 1.452 in this calculation?
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Old 02-25-2011, 06:25 PM   #4
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Your total volume. (1.32 gallons) + (0.132 gallons)

 
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Old 02-25-2011, 06:26 PM   #5
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Quote:
Originally Posted by BendBrewer View Post
Your total volume. (1.32 gallons) + (0.132 gallons)
ok thanks a lot!
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Old 02-25-2011, 06:27 PM   #6
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Quote:
Originally Posted by 400d View Post
what is 1.452 in this calculation?
it's a simple relationship

original gravity / original volume = new gravity / new volume.

OG = 1.135 (use 135)
OV = 1.32 gal
NV = 1.32 + .132 as you stated, so 1.452
solve for NG

 
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