Adding water post boil

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MikEH86

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Good day everyone.

I am working on my first 5gallon SMaSH beer and realised my kettle is not big enough so I added some water after the boil the problem is that I forgot to take OG after adding the water. I was wondering if there is any math one could do to figure what the OG was after adding water. I added 3 liters or .792 us.gallons and the OG was 1.050 after boil. Or pre water add.

While we are already on this how does one figure your efficiency from the grain. Mine will be a little lower due to it being home malted.

Pic is just for fun, going into secondary. And a pic of FG
 

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You need to know the original volume as well, but you can definitely work it out. You take the old volume divided by the new volume, and multiply your gravity points (whatever is after the decimal in your OG reading, so 50 in this case).

Based on your picture, it looks like the volume is now 21.5L, so let's assume your volume when you took the reading was 18.5L and the 3L addition brought it to 21.5L. 21.6/18.5 = 0.86. Multiply that by 50 and you get 43 gravity points in the diluted wort, so 1.043 OG.
 
start with the equation: V1 * C1 = V2 * C2 (link)
  • V1 = 5 gallons - .792 gallons = approximately 4.2 gallons (before adding water)
  • V2 = 5 gallons (after adding water)
  • C1 = 50 OG (before water additions)
  • C2 = ?? OG (after water additions)
make the equation useful for this situation:

C2 = v1 * c1 / v2

C2 = 4.2 * 50 / 5
 
You need to know the original volume as well, but you can definitely work it out. You take the old volume divided by the new volume, and multiply your gravity points (whatever is after the decimal in your OG reading, so 50 in this case).

Based on your picture, it looks like the volume is now 21.5L, so let's assume your volume when you took the reading was 18.5L and the 3L addition brought it to 21.5L. 21.6/18.5 = 0.86. Multiply that by 50 and you get 43 gravity points in the diluted wort, so 1.043 OG.

Ah you know what I did was I thought you had to have 23 liters for 5 US gallons but 23 liters is 6 US gallons while 23 liters is also 5 gallons confusing hate living in Canada some times because we have both systeme.
They teach metric in school but then you go out and get a job and it's all Imperial anyways back on subject. Before adding water I had 20 liters or 5.28 US gallons lol.


So before adding water was 20 liters so I would do 23/20 = 1.15 so would I mutiply 1.15 by 50 or .15*50
 
Close, it's 23/20. Or go with the equation from @BrewnWKopperKat. The units don't matter (gallons vs liter) because they cancel each other out:

C2 = v1 * c1 / v2

C2 = 20 * 50 / 23

C2 = 43.48


So, still basically 1.043 OG.
 
Ah you know what I did was I thought you had to have 23 liters for 5 US gallons but 23 liters is 6 US gallons while 23 liters is also 5 gallons confusing hate living in Canada some times because we have both systeme.
They teach metric in school but then you go out and get a job and it's all Imperial anyways back on subject. Before adding water I had 20 liters or 5.28 US gallons lol.


So before adding water was 20 liters so I would do 23/20 = 1.15 so would I mutiply 1.15 by 50 or .15*50
Glad I live where 5 gallons is really 5 gallons 😎
 
Close, it's 23/20. Or go with the equation from @BrewnWKopperKat. The units don't matter (gallons vs liter) because they cancel each other out:

C2 = v1 * c1 / v2

C2 = 20 * 50 / 23

C2 = 43.48


So, still basically 1.043 OG.

Thanks for the help! Math is not my strong point!
 
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