So
here is my theoretical situation:
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
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