How to calculate how much top up water to add to reach desired gravity?

[Elysium]

Hey guys,

I actually have 2 questions that I'd like to receive answers to:

1, How can I calculate the amount of water (top up water to the wort after boil) required to get a desired final gravity? For instance if I want final gravity 1.050 and 1.5 gallons in the fermentor, then how much wort will I need to have after the boil and at what gravity to reach that 1.050 gravity and 1.5 gallons of fermentation size?

2, How can I calculate the amount of grains to get to a specific gravity (high gravity such as 1.095 that can be diluted after the boil) after the boil?

 

[billl]

1) You convert to points of sugar. Gravity points x gallons = points of sugar and that remains constant. 1.5 gallons of 1.050 wort is 50 x 1.5 = 75 points. You need 75 points from whatever you are starting with, so 1 gallon of 1.075 or 0.75 gallons of 1.100.

2) That depends on your systyem and your grains. Most grains are about 36 points per lb and most people get around 75% into the fermenter - about 27 points. If you need 75 points for your total batch, that is about 75/27 = 2.78 lbs.

Really though, if you are mashing, you should be doing a full boil and not a concentrated/partial boil. It's easier and more predictable. For a 1.5 gallon batch, kettle space and stove power should not be an issue.

 

[andrewf1985]

(Volume x OG)/required OG = X

X-volume = volume to add

 

[Elysium]

(Volume x OG)/required OG = X

X-volume = volume to add

Thanks for the reply.

I actually know what my final gravity should be (1.050) and what my capacity is (I can brew 9 liters), and I wanna top it up with 6 liters of water to reach a final size of 15 liters. So, all I need to do is what the final gravity of the wort in the kettle should be.

If my calculation is correct.....
(2.37 gallons*X)/50 (from 1.050) =3.96 gallons then in my kettle I willl need a 1.083 beer to reach 1.050 after the dilution. I think that is too high and I screwed it up.

 

[ajdelange]

I actually know what my final gravity should be (1.050) and what my capacity is (I can brew 9 liters), and I wanna top it up with 6 liters of water to reach a final size of 15 liters. So, all I need to do is what the final gravity of the wort in the kettle should be.

You want 15 L of beer of specific gravity 1.050. Assuming this is 20 °C SG referenced to 20 °C water then the density of the beer is 1.050*.998203 = 1.04811 g/cc. 15 L of it will weigh 15000*1.04811 = 15721.7 grams.

Wort of 20/20 SG = 1.050 corresponds to 12.3877 °P (see below for how to get this) i.e. it contains 12.3877% extract by weight so your beer will need 12.3877*15721.7/100 = 1947.56 grams of extract in both the diluted (15L) and concentrated (9L) volumes.

Note that as the finished 15721.7 grams of beer contains 1947.56 grams of extract it must contain 15721.7 - 1947.56 = 13774.14 grams of water.

Now turning our attention to the 9L and following what we did for the 15 L the amount of extract in the 9L is

E = 9*.998203*SG*P(SG)/100 = 1947.56 grams

SG is the specific gravity of the 9 L wort and P is the degrees Plato corresponding to the SG. We must solve this equation for SG. It is usually rearranged to

9000*.998203*SG*P(SG)/100 - 1947.56 = 0

SG is converted to Plato using the official ASBC (EBC) formula

P(SG) = (((135.997*SG -630.272)*SG + 1111.14)*SG - 616.868)

The easiest way to do this is designate a cell on an Excel spreadsheet to hold trial SG numbers, another to contain P(SG) and a third 9000*.998203*SG*P(SG)/100 - 1947.56

Thus
A1: 1.07
A2: = (((135.997*A1 -630.272)*A1 + 1111.14)*A1 - 616.868)
A3: = 9000*.998203*A1*A2/100 - 1947.56

If you put 1.07 into A1 then A2 will show 17.06 °P and A3 -308.06
If you put 1.09 into A1 then A2 will show 21.57 °P and A3 +164.51

These as we want A3 to be equal to 0 it is clear that SG must be between 1.07 and 1.09. If you know how to use the Solver ask it to set cell A3 to 0 by varying A1. It quickly sets cell A1 to 1.0830 with A2 showing 20.02 and A3 displaying 0. The required SG for the 9 L is 1.083. Surprise! the same result you got by using the approximate (points) method. But lets look a little further.
We have now 9L of beer or density 0.998203*1.0830 and so its mass must be
9000*0.998203*1.0830 = 9729.48 grams. As the 15L volume weighs 15721.7 grams to do the dilution we need to add 15721.7 - 9729.48 = 5992.22 grams of water which is, at 20 °C 5992.22/998.203 = 6003.01 mL, a wee bit more than 6 L but hardly a significant bit.

So given that the computed final gravity is, to 4 decimal places, identical to what you get by the points method and the extra water required is only 3 mL why did I go through all this?

1. To be honest, to see if I can still explain it in terms of robust conservation of mass i.e to run some of the rust out of the hinges here
2. In the hopes that some readers will be interested
3. To illustrate that the points method is not precise but close enough for government work in most cases.
4. To show how brewing scientists approach this problem

If my calculation is correct.....
(2.37 gallons*X)/50 (from 1.050) =3.96 gallons then in my kettle I willl need a 1.083 beer to reach 1.050 after the dilution. I think that is too high and I screwed it up.

Doesn't look that way.


PS: Didn't realize this was the beginners forum. This doesn't really belong here.