# Wort Dilution ### Help Support Homebrew Talk: #### Nil

##### Well-Known Member
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 dont 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 &#8594; 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.

Thanks, Nil OP
N

#### Nil

##### Well-Known Member
Data reposted below:

Volume (mL) Gravity
---------------- ----------
30&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;..1.060
40&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;..1.055
50&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;..1.050
70&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;..1.044
90&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;..1.038
110&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;&#8230;&#8230;1.035
130&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;1.031
150&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;&#8230;1.028
170&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;&#8230;1.026
200&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;&#8230;1.023
230&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.....&#8230;1.020
260&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;1.018
300&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;...&#8230;&#8230;1.016

#### McKnuckle

##### Well-Known Member
This would have also helped:
http://www.brewersfriend.com/dilution-and-boiloff-gravity-calculator/

The basic formula is:
TotalVolume = Volume1 + Volume2
TotalGravity = Volume1/TotalVolume*Gravity1 + Volume2/TotalVolume*Gravity2

Your funky equation is probably due to measurement error with both the volumes you added and your hydrometer. I don't think it's non-linear as you are proposing. But I'd like to be educated if I'm wrong.

• Nil

#### Sniderzj

##### Member
Hi Nil,
I admire your empirical efforts! Alas, there is an easier way, and it's the same way you calculate mash efficiency. Simply multiply your gravity "points" by the gallons you have , then divide by your new target gravity points and you'll get your volume....subtract that from the volume you are starting with and you've got how much water to add. So if you have 7 gallons at 1.060' that's 60 points, so 7*60=420, now you want a 1.040 beer or 40 points, so divide 420 by 40 and you get 10.5. Subtract that from your original 7 gallons and you need to add 3.5 gallons. It's all a ratio.

• McKnuckle and Nil
OP
N

#### Nil

##### Well-Known Member
The equation appears to work based on the more basic calculations. Long way to get the same value.

Thanks, Nil #### ajdelange

##### Well-Known Member
The key to solving problems like this is the ability to convert between the Plato and Specific Gravity scales. In order to do this the Kaiserliche Normal-Eichungskommission was launched at the turn 18th century tasked with producing a table of precisely (6 decimal places) measured specific gravities of solutions of sucrose of known composition by weight. The resulting tables are found in texts on brewing, in the ASBC's Methods of Analysis, the EBC's Analytica etc. Tables are kind of a PITA so ASBC fit the data in the table (afters massaging it around to apparent SG 20°/20°) with a polynomial

°P = -616.868 + 1111.14*S -630.272*S*S + 135.9975*S*S*S

in which S is the 20/20 apparent specific gravity and °P is the grams of sucrose in 100 grams of solution i.e. the percentage strength by weight.

With this available to you your problem is solved in the following steps:
Find the extract content of the wort at hand by use of the polynomial: Plato(1.047) = 11.6737
Find the weight of 100 mL of this wort: w = (100*0.998203*1.047) = 104.5 grams

At this point you know that 100 mL of your wort contains (11.6737/100)*104.512 = 12.2 grams of extract and 104.512 - 12.2 = 92.312 grams of water (though you don't really care about that).

Now Determine the strength by weight of the desired diluted wort from the polynomial: Plato(1.040) = 9.99359 %

At this point you know that you want your 12.2 grams of extract to be 9.99359% of the total mass of the diluted wort which is clearly 12.2/0.099359 = 122.787. As the mass of 100 mL of the original wort is 104.5 grams you must add 122.787 - 104.5 = 18.281 grams of water to each 100 mL. Convert that to mL at whatever temperature you will measure it out.

Plato(1.060) = 14.7414
Weight 100 mL: 100*1.060*0.998203 = 105.81
Weight extract: 14.7414*105.81/100 = 15.5979 grams
Plato(1.040) = 9.99359
Weight diluted to target: =15.5979/0.099359 = 156.985
Water needed/100 mL = 156.985 - 105.81 = 51.175 grams

Were you working at 20 °C that would be 51.2671 mL of water to be added to 100 mL wort or 51.2671% Thus to 7 gallons you would add 7*0.512671 = 3.5877 gal. Note that this is pretty close to the answer obtained by the approximate method suggested in #4.

One final comment and that's on the final volume. You would have, after the dilution, 156.985 grams of wort (per original 100 mL) of density 0.998203*1.040 = 1.03813 and hence volume 156.985/1.03813 = 151.219 cc. The volume increase is thus 51.219% and your original 7 gallons would become 7*1.51219 = 10.5853

OP
N

#### Nil

##### Well-Known Member
Thanks.

I did a mash that produced a wort with an OG: 1.058, yet my target was 1.053. I did the calculation and needed 2 gallons of water. I added first 2L and got the desired OG.

Maybe I am missing something.

#### McKnuckle

##### Well-Known Member
I don't know what you did wrong, but it's about a 10:1 ratio that you needed. For every 1 gallon at 1.058, you'd add 0.1 gallons of water to achieve 1.053. You would have needed 20 gallons of 1.058 wort for 2 gallons of water to have brought it down as desired.

Did you do the simple math provided earlier, or attempt to exercise the rocket science posted immediately above? #### ajdelange

##### Well-Known Member
Plato(1.058) = 14.2739 grams/100 grams
Wt 100 mL = 1.058*0.998203*100 = 105.61 gram @ 20 °C
Extract /100 mL = 14.2739*105.61/100 = 15.0747 gram
Plato(1.053) = 13.098 grams/100 grams
Total Wt Diluted wort = 15.0747/.13098 = 115.092
Additional water required = 115.092 - 105.61 = 9.482 gram/100 mL.
Additional water per liter original =10*9.482/0.998203 = 94.9907 mL/L @ 20*C

Rocket science?

#### McKnuckle

##### Well-Known Member
Not really, just overkill for what is really just a simple equation of proportions.

I said it's "about" a 10:1 ratio, and that is true. 10:1 would bring 1.058 down to 1.05273 which is 1.053 to any hydrometer on the planet. The precision of your comprehensive method is wonderful to behold but it is unnecessary for practical use.

(This argument will go nowhere so I will bow out...)

#### ajdelange

##### Well-Known Member
You shouldn't think of it as an argument but rather as a discussion. It isn't actually a simple matter of proportions as the relationship between extract content and volume is not linear. It is almost linear and that's why a simple proportion based calculation gets you pretty close to the correct answer. What it is, in fact, is a simple matter of conservation of mass. As this is the Brew Science forum I think there are people here that would want to know about that. Some of those folks have training in STEM disciplines and I'm supposing that they would appreciate having the full explanation which is consistent with the calculation methods used by professional brewers. I'm guessing that the OP, who went to the trouble to make a set of measurements on wort of various strengths, would be interested in this.

Some people (and I'm obviously one) are accuracy freaks. My thinking is that there are plenty of other error sources in brewing calculation and measurement and this doesn't have to be one. I don't ever do a dilution calculation as I posted above except when trying to explain how it it done. It's programmed into a spreadsheet and the additional effort required of me to use it is no greater than had I programmed the 'simple proportion' method.

You do raise a good point in saying that it isn't necessary for most people using cheap hydrometers but some use narrow range hydrometers and some use density meters. They are found in pretty small breweries these days. Thus I really think the bottom line is that the readers should be exposed to the more precise method and decide for themselves whether they want to use it or not.

• doug293cz

#### McKnuckle

##### Well-Known Member
Fair enough. I think it's a good idea to present both approaches to people. It is interesting to see more exhaustive, precise methods that leverage all of the relevant science. But many people post to the science forum to learn practical methods as well, and may glaze over with so much math when it's not strictly necessary.

I use a few cells in a spreadsheet for dilution calculations - it applies the simple formula I posted on page one of this thread. Given the accuracy of measuring devices, and the lack of requirement that gravity (or other dilution) be calculated to an extreme degree of precision in homebrewing, it truly works fine.

I retract my snarky "rocket science" comment but hopefully you see the pragmatism of my approach also. Cheers

• doug293cz

#### doug293cz

##### BIABer, Beer Math Nerd, ePanel Designer, Pilot
Staff member
Mod
... Tables are kind of a PITA so ASBC fit the data in the table (afters massaging it around to apparent SG 20°/20°) with a polynomial

°P = -616.868 + 1111.14*S -630.272*S*S + 135.9975*S*S*S

in which S is the 20/20 apparent specific gravity and °P is the grams of sucrose in 100 grams of solution i.e. the percentage strength by weight.

...
@ajdelange : Is there an analytical inverse for the above equation? The commonly seen: SG=1+(&#730;P/(258.6-((&#730;P/258.2)*227.1))) is not an exact inverse (at least not if you do a round trip &#730;P --> SG --> &#730;P calculation.)

Also, what is the range (&#730;P) of applicability the &#730;P --> SG curve fit equation above?

Brew on #### ajdelange

##### Well-Known Member
Yes, -616.868 + 1111.14*S -630.272*S*S + 135.9975*S*S*S - °P = 0 can be solved in closed form for S but as a practical matter the easiest thing to do is let Excel solver find the solution for you. If writing code root bisection finds the right root very quickly which brings us to the answer to your second question: it is valid up to 20 °P. Beyond that one has to get data from elsewhere and the best place is the ICUMSA polynomial which has lots more coefficients but takes you as high as you would ever want to go and also lets you specify any temperature you want but consequently has lots of coefficients and is harder to program or enter into a spreadsheet. For simpler calculation above 20 °P I have done a fit to data in the CRC tables under the constraint that the fit have the same value and slope at 20 °P as the ASBC curve.

You can, of course, always take the Plato table and make °P the independent variable coming up with a polynomial fit for SG in powers of °P but as is the case with the Lincoln equation it is not an exact inverse.

If you want more details post again.

#### ajdelange

##### Well-Known Member
Another easy way to invert is to estimate the SG as 1 + 4*°P/1000 and then use Newtons method to refine that estimate. Because the relationship is so close to being linear one or two iterations is all you need as the following Excel example shows.

Desired °P 12
1st guess 1.048
Calc °P 11.91208076
1st err 0.087919244
1st Deriv 238.1880353
1st cor 0.000369117
2nd guess 1.048369117
Calc °P 11.99997239
2nd err 2.76101E-05
2nd Deriv 238.0384527
2nd corr 1.1599E-07
3rd guess 1.048369233
Calc °P 12
3nd err 2.72848E-12
3nd Deriv 238.0384057
3nd corr 1.14624E-14
4th guess 1.048369233

This doesn't require the Solver and is thus easier to use but you'd want to put these cells off the main page somewhere or hide them.

I usually write the polynomial

°P = (((135.9975*S -630.272)*S + 1111.14)*S-616.868)

as habit from the old days when operations count mattered and thus the derivative is, in that same form

d°P/dS = ((3*135.9975*S -2*630.272)*S + 1111.14)

In case you haven't used Newton's method in so long that you've forgotten it you simply add the error divided by the derivative to the current estimate to get the new estimate.

#### doug293cz

##### BIABer, Beer Math Nerd, ePanel Designer, Pilot
Staff member
Mod
Yes, -616.868 + 1111.14*S -630.272*S*S + 135.9975*S*S*S - °P = 0 can be solved in closed form for S but as a practical matter the easiest thing to do is let Excel solver find the solution for you. If writing code root bisection finds the right root very quickly which brings us to the answer to your second question: it is valid up to 20 °P. Beyond that one has to get data from elsewhere and the best place is the ICUMSA polynomial which has lots more coefficients but takes you as high as you would ever want to go and also lets you specify any temperature you want but consequently has lots of coefficients and is harder to program or enter into a spreadsheet. For simpler calculation above 20 °P I have done a fit to data in the CRC tables under the constraint that the fit have the same value and slope at 20 °P as the ASBC curve.

You can, of course, always take the Plato table and make °P the independent variable coming up with a polynomial fit for SG in powers of °P but as is the case with the Lincoln equation it is not an exact inverse.

If you want more details post again.
AJ, Thanks for the responses. I am very much interested in learning about the ICUMSA polynomial. I have a mash/lauter simulator spreadsheet in which I have to convert volume @ °P to weight. And since for really thick mashes, first runnings °P can be in the high 20's, I'd like to have a method with reasonable accuracy beyond 20°P. Right now I do all calcs for 20°C volume, but I would still like to know how to calculate wort density vs. temp & concentration, rather than assume the density vs. temp is the same as water.

Do you have a link for the °P <--> SG tables in machine readable form?

Another easy way to invert is to estimate the SG as 1 + 4*°P/1000 and then use Newtons method to refine that estimate. Because the relationship is so close to being linear one or two iterations is all you need as the following Excel example shows.

Desired °P 12
1st guess 1.048
Calc °P 11.91208076
1st err 0.087919244
1st Deriv 238.1880353
1st cor 0.000369117
2nd guess 1.048369117
Calc °P 11.99997239
2nd err 2.76101E-05
2nd Deriv 238.0384527
2nd corr 1.1599E-07
3rd guess 1.048369233
Calc °P 12
3nd err 2.72848E-12
3nd Deriv 238.0384057
3nd corr 1.14624E-14
4th guess 1.048369233

This doesn't require the Solver and is thus easier to use but you'd want to put these cells off the main page somewhere or hide them.

I usually write the polynomial

°P = (((135.9975*S -630.272)*S + 1111.14)*S-616.868)

as habit from the old days when operations count mattered and thus the derivative is, in that same form

d°P/dS = ((3*135.9975*S -2*630.272)*S + 1111.14)

In case you haven't used Newton's method in so long that you've forgotten it you simply add the error divided by the derivative to the current estimate to get the new estimate.
Thanks for the Newton method review, but I think I'll stick to "Goal Seek" (I use LibreOffice Calc rather than Excel) since it's right there waiting for me.

And, operation count and algorithm efficiency still matter for problems that take days/weeks to run on petaflop super computers. And even I've run apps that max out a 4 core, 8 thread, 3GHz x64 for hours at a time, although none of my spreadsheets are anywhere near that intense.

Brew on #### ajdelange

##### Well-Known Member
I am very much interested in learning about the ICUMSA polynomial.
Here are the coefficients for sucrose:
{385.85074,-45.9244,60.198,-51.1,19.86,-13.03435,7.5699,-13.008,15.8,0,-3.6663,6.2667,-4.907,0,0,0,0,0,0,0}

Arrange those into a matrix of 5 rows by 4 columns filling the columns first. Now post multiply that matrix by a column vector containing powers of c (c, c^2,c^3,c^4). c is the number of grams of sucrose in 1 cc of solution measured at 20 °C. Dot the resulting vector with a vector of powers of (T-20)/100 where T is the temperature you are interested in °C. It's a little weird but that's the way ICUMSA does it. Now add the density of water at the temperature of interest and the result, p(c,T), is the density of the solution of concentration c at temperature T.

Multiplying the density by the mass fraction (°P/100) gives back the grams of sucrose per cc. Thus to determine density from °P we solve
p(c,20)*(°P/100) - c = 0 for c and then insert c into p(c,T) to calculate density (or specific gravity). This is the method used to calculate ICUMSA Table A.

The coefficients for glucose are:
{382.9716,-55.1502,75.5366,-44.059,0,-23.0838,19.0140,-30.9314,0,0,0.1218,4.2741,0,0,0,-0.7704,0,0,0,0}

For fructose:

{390.5399,-74.4141,81.7245,-77.138,32.396,-22.1864,15.5436,-44.6412,42.824,0,-0.0829,5.3744,4.9248,0,0,-1.7937,0,0,0,0}

And for invert sugar:

{386.7294,-64.7839,78.6570,-60.676,16.241,-22.9485,17.6392,-37.8841,21.588,0,1.7684,4.2073,2.5197,0,0,-2.3475,0,0,0,0}

Do you have a link for the °P <--> SG tables in machine readable form?
I can send them to you as a text file. PM your e-mail. And, it turns out, they go up to 1.083 SF (21.7 °P).

Edit: You will also need the ICUMSA water density:

(((((-281.03006e-12*t +105.84601e-9)*t-46.241757e-6)*t-7.9905127e-3)*t+16.952577)*t +999.83952)/(1+16.887236e-3*t)

where t is in °C

[Edit 2 Feb:] There were two minor errors in the sucrose and glucose polynomials.

#### ajdelange

##### Well-Known Member
Another note: If you don't care about temperatures other than 20°C you can obviously just use the coefficients in the first row multiplied by c and its powers.

density = 0.998203 +385.85074*c -13.03435*c^2 -13.03435*c^3

for sucrose, for example.

#### doug293cz

##### BIABer, Beer Math Nerd, ePanel Designer, Pilot
Staff member
Mod
AJ,

Thanks for the info above and the Plato <--> SG table data.

I assume there are no coefficients for maltose because its hygroscopic nature makes accurate weight measurements next to impossible. So, which of the other sugars is most like maltose w.r.t. density vs. concentration & temp?

Brew on #### ajdelange

##### Well-Known Member
I assume there are no coefficients for maltose because its hygroscopic nature makes accurate weight measurements next to impossible.
I think it's because ICUMSA (International Commission for Uniform Methods in Sugar Analysis) serves the beet and cane sugar industries and their consumers and not maltsters or theirs. No doubt that maltose is tricky to handle, though.

So, which of the other sugars is most like maltose w.r.t. density vs. concentration & temp?
When you get coded up and start playing with numbers you will see that the differences between the sugars are very small. At 10 °P

&#8226;print DensA(10,20,1,sucrose)
1.04003
&#8226;print DensA(10,20,1,glucose)
1.03961
&#8226;print DensA(10,20,1,fructose)
1.04044
&#8226;print DensA(10,20,1,invert)
1.04002
(These numbers are 20/20 apparent specific gravities)

The differences are in the 4th decimal place in addition to which brewers use the sucrose based Plato scale even though their worts typically contain little sucrose. It's probably safe to assume that the magnitude of the difference between maltose and glucose is about the same as the difference between sucrose and glucose.

EDIT: In looking at that again it seems to suggest correcting a maltose determination by the maltose/glucose difference. That's not what I am suggesting. What I am suggesting is just using the sucrose tables for everything.

#### doug293cz

##### BIABer, Beer Math Nerd, ePanel Designer, Pilot
Staff member
Mod
Here are the coefficients for sucrose:
{385.85074,-45.9244,60.198,-51.1,19.86,-13.03435,7.5699,-13.008,15.8,0,-3.6663,6.2667,-4.907,0.0,0,0,0,0}

Arrange those into a matrix of 5 rows by 4 columns filling the columns first. Now post multiply that matrix by a column vector containing powers of c (c, c^2,c^3,c^4). c is the number of grams of sucrose in 1 cc of solution measured at 20 °C. Dot the resulting vector with a vector of powers of (T-20)/100 where T is the temperature you are interested in °C. It's a little weird but that's the way ICUMSA does it. Now add the density of water at the temperature of interest and the result, p(c,T), is the density of the solution of concentration c at temperature T.

Multiplying the density by the mass fraction (°P/100) gives back the grams of sucrose per cc. Thus to determine density from °P we solve
p(c,20)*(°P/100) - c = 0 for c and then insert c into p(c,T) to calculate density (or specific gravity). This is the method used to calculate ICUMSA Table A.

...
Went to put the sucrose coefficients into a matrix, and discovered there are only 18 instead of 20. Is this an error, or am I supposed to just 0 fill?

Brew on 