I want to dilute a wort with an OG: 1.047 to 1.040 in order to have a final ABV 5%. I dont know how to determine in advance the volume of water needed to hit the desired gravity points. So

Got 100 mL of a wort with an OG: 1.081 (from LME, approx. 40 grams) and added 30 ml of water. The gravity was 1.060. Kept adding water as follows:

Volume (mL) Gravity

---------------- ----------

30 1.060

40 1.055

50 1.050

70 1.044

90 1.038

110 1.035

130 1.031

150 1.028

170 1.026

200 1.023

230 1.020

260 1.018

300 1.016

Found that the correlation is logarithmic.

Log (Vol) = -22.203xOG + 25.014

Going back to my problem, with an OG: 1.047,

Vol = 10^(-22.203x1.047 + 25.014) = 58.5 mL

For 1.040,

Vol = 10^(-22.203x1.040 + 25.014) = 83.7 mL

The difference between the two is 83.7 mL 58.5 mL = 25.2 mL.

This suggests that I need to add 25.2 mL of water/100 mL wort (at OG: 1.047) to get the desired gravity points (40).

Since my knockout volume is 5.5 gallons, there are 5.5 x 3785 = 20,817.5 mL, which, divided by 100 mL, the result is 208.175.

Therefore,

208.175 x 25.2 mL = 5,246.01 mL/3785 = 1.386 gallon = 1.4 gallons of water.

If my calculations are correct, this means that I can setup a mash with a target FG: 1.060 and a knockout volume of 7 gallons. To reduce 1.060 → 1.040, I would need to add 3.75 gallons of water (53.6 mL/100 mL wort) to have a total final volume of 10.75 gallons enough for 4 boxes of beer.

This may be a very nice way to control FG and thus have a more consistent ABV between batches.

Please let me know your thoughts.

Thanks, Nil