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Alcohol Dilution


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Guest mdepew

Hey Will,

I was fooling around with a spread sheet as I was reading this. Is this a fair representation of what you were talking about?

Notwithstanding the fact that method (2) is the prescribed method, I was referring to the somewhat more intuitive approach (for me) which is to compute how much alcohol and water will be present in the final proof, and how much you already have in the given proof, and compute the differences.

Later, I'll show a shortcut.

Your example:

500mL @ 160 pf -> 1000mL @ 80 pf

160 pf is 80% alcohol by volume, so we can break it down to its component parts.

100% alcohol weighs 6.6097 LBS / Gallon

100% water weighs 8.32823 LBS / Gallon

for the 160 proof (using Table 6)

.80 x 6.6097 = 5.29 LBS / Gallon alcohol

.2287 x 8.32823 = 1.91 LBS / Gallon water

for the 80 proof

.40 x 6.6097 = 2.64 LBS / Gallon alcohol

.6342 x 8.32823 = 5.28 LBS / Gallon water

Let's look at the alcohol:

5.29 / 2.64 = 2.00 - so every unit of volume at 160 proof will make 2 units volume of 80 proof.

For simplicity, let's think through making 2 gallons - we'll adjust the quantity later.

2.00 * 5.29 = 10.58 LBS water total in the final batch.

Every time I use one gallon of 160 proof, I'm also contributing 1.91 lbs of water toward the amount I need in the final product. So:

10.58 - 1.91 = 8.67 LBS water extra needed.

So let's fire-up the scale:

the 160 proof weighs 5.29 + 1.91 LBS per gallon, so measure-out 7.2 LBS of 160 proof stuff.

Now add water until we have 2 * (2.64 + 5.28) = 15.84 LBS total.

You can scale this to any size batch, or convert to SI units using the coefficients I posted earlier (but don't use coefficients you find in the scientific literature - use the TTB values).

That walks us through the details or mechanics of what's happening, but did you notice the shortcut we could have taken?

Once we know the ratio of alcohol to alcohol (given proof / ending proof) we can just multiply that ratio by the starting quantity, which equals the ending quantity we will be making, then look-up the weight of each gallon of the ending proof (using Table 5), multiply that by the ending quantity, then add water until we get that weight. Duck soup.

Like this:

8 gallons of 140 proof -> 80 proof to bottle:

140 / 80 = 1.75

so 8 gallons times 1.75 will make 14 gallons of final stuff.

80 proof weighs 7.93 LBS / gallon (Table 5)

14 * 7.93 = 111.02 LBS total

Now tare your scale and dump the 8 gallons of 140 proof. According to Table 5, that will weigh 7.41 LBS/Gal or 59.28 LBS (though we don't really need to know that). We simply add water until the whole thing weighs 111 LBS.

See, no screwin' around...

Will

Ethanol.xls

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Will posted the following quiz on 8 Dec 2009 but nobody has posted an answer yet "When you feel like it, conquer this one: You're diluting 350 LBS of 135 proof to 86 proof for bottling, so you figure you'll be making 73.6 gallons of finished product, which will weigh 580.6 LBS. You tare the scale, dump the stuff into the proofing tank and turn-on the water...but then you get distracted. By the time you regain consciousness, the scale reads 603.2 LBS.

What's the present proof?

How do you fix it? "

My answer is present proof = 82.99

so we now add 34.36 lbs of 135 to bring it back up to 86 proof.

Final reading on scales will be 637.56 lbs dry.gif

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There has been a number of articles published on alcohol dilution (some in early and now reprinted treatises) which we have used to expand an Alcohol dilution Table. The table used the Gauging Manual Table 6 and calculations to set a number of key dilutions for the early distillery worker. We expanded this table and have found other algorithms that can be of use to the distiller and brewer.See here for the table: http://alcbevtesting.com/wp-content/uploads/2009/05/alcoholdilutiontable1.pdf

Furthermore, we encourage distillers to work more carefully with density measurements and the use of OIML (Legal Metrology) tables now freely available on-line to convert density to alcohol by weight and volume values. Nathan and I (Gary) at Brewing and Distilling Analytical Services will be happy to assist further on these issues.

Gary (Spedding).

gspedding@alcbevtesting.com

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  • 11 months later...

Gary,

You seem to have a grip on this stuff. Would you care to take a stab at the problem posted in 2009, and restated in post #27 directly above yours?

We're still hoping to find someone who knows how to compute this stuff and is willing to share the approach.

Will

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When you feel like it, conquer this one:

You're diluting 350 LBS of 135 proof to 86 proof for bottling, so you figure you'll be making 73.6 gallons of finished product, which will weigh 580.6 LBS. You tare the scale, dump the stuff into the proofing tank and turn-on the water...but then you get distracted. By the time you regain consciousness, the scale reads 603.2 LBS.

What's the present proof?

How do you fix it?

My solution to Wills challenge

From Table 4 we know 135 proof contains 0.18093 ProofGallons/lb,

so in 350 lbs we have 350 X 0.18093 = 63.3255 PGs

After adding water the mass is now 603.2 lbs,

but no more PGs were added so we have 63.3255 / 603.2 = 0.10498 PG/lb

From Table 4 we see this PG/lb corresponds to 83 proof.....................= First part of answer

In the 603.2 lbs 83 proof we have 31.6615 gal alc + 47.2939 gal water –from Tables

To fix it add X GALLONS of 135 proof

Each gallon 135 contains 0.675 gal alcohol + 0.3602 gal water {(Table 6) / 100}

Final gallons when fixed will be

31.6615 + 0.675X galls alcohol and 47.2939 + 0.3602X galls water

1 gal 86 proof has 0.43 gals alc and 0.6056 gals water, 0.43/0.6056 = 0.71 (alcohol to water)

31.6615 + 0.675X / 47.2939 + 0.3602X =0 .71

(edited) 0.71(47.2939 + 0.3602X) = 31.6615 + 0.675X

33.58103 + 0.25576X = 31.6615 + 0.675X

1.191 = .419X

X = 4.5773 gallons ……………………. 135 proof = 0.13402 gall / lb

4.5773 / 0.13402 = 34.16 lbs of 135 to be added

Scales should now read 34.16 + 603.2 = 637.36

I quite enjoyed the mental challenge, I proved to myself I understand how it is done, but never again

It is so much easier with a program I use called Alcodens.

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Is there a formular that can be used to determine the information on Table 6 (pounds per gallon) or Table 7 (temp correction) that could be used in a spread sheet to eliminate the need to reference the tables when making calculations?

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@HedgeBird

With a moderate amount of data entry to create VLOOKUP tables, it could be incorporated into the spreadsheet posted earlier. I've had that on the to-do list for some time, but haven't done it.

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PeteB:

Part 1 is correct, nicely done.

Part 2 is probably in the zone, but did you use .7(47....) or .71(47...)? that's an error of around 1.5%. Also, since you're "cheating" by using Alcodens, are you correcting back to 15.56C (60F)? Note that if you do this in wt%, you won't have to worry about temperature.

Pressure9pa & HedgeBird:

Contact me via PM, and I'll send you the scanned tables in excel format. I would post them here, but they're not pretty enough yet.

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It is 12 months since I did the calcs so I can't remember, but I did not take the easy route and use Alcodens with any of the the posted calcs, only tables.

I also solved the same problem using weights and got the same answer (slight rounding errors)but found there were less calculations in this case when I used volumes.

Normally weights are easier but not in this example.

Like I said, it was 12 months ago and I have basically erased from my memory the use of TTB tables.

Proof gal, wine gal, imperial gal, farenheight, proof temperature they play tricks with my brain when I get tired. WHEN ARE YOU GUYS GOING METRIC?

I assume the .7 or .71 refers to "in a vacuum" or "in air", I fairly sure I would have used the vacuum columns.

EDIT OK Will, I found what you were refering to, 0.71 is the ratio of alcohol to water and is correct

I copied-typed the equation from an Excel spreadsheet and made a typo, sorry.

The maths is correct, just the copy has the 1 missing.

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Here is a slightly different way of solving the second part

edit (sorry if I have confused some but I missed off the first 5 lines when I cut and pasted, all here now)

From Table 4 we know that 83 proof has 0.12648 WineGallons/lb

We have 603.2 lbs, so there must be 76.2927 WGs

From Table 6 we see that 83 proof contains 41.5 volumes of alcohol and 61.99 vols of water

76.2927 *41.5/100 = 31.6615 and 76.2927 * 61.99/100 = 47.2939

  • We have 31.6615 gallons alcohol and 47.2939 gallons water in the 603.2 lbs 83 proof

– Let the mass of 135 proof spirit added be X lbs

From Table 4: 135 proof is 0.13402 WG/lb

ð We have to add 0.13402 * X wine gallons of 135 proof

From Table 6: 100 gals of 135 proof is 67.5 gals of alcohol and 36.02 gals of water (divide these gals by 100 for 1 gal)

This means we are adding

0.675 * 0.13402 * X = 0.09046 * X gallons of alcohol

And

0.3602 * 0.13402 * X = 0.04827 * X gallons of water

So the total alcohol we will have, in volumetric terms, will be 31.6615 plus 0.09046X gallons

The total water will be 47.2939 plus 0.04827X gallons

Also from Table 6 we know that 86 proof contains 43 vols alcohol and 60.56 vols water. This tells us the volumetric ratio of alcohol to water is 43/60.56

So (31.6615 + 0.09046X) / (47.2939 + 0.04827X) = 43/60.56 = 0.7100

31.6615 + 0.09046X = 0.71 * (47.2939 + 0.04827X)

0.05619X = 1.9172

X = 34.12 lbs

Final weight is now 34.12 + 603.2 = 637.32 lbs

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PeteB:

Part 1 is correct, nicely done.

Part 2 is probably in the zone, but did you use .7(47....) or .71(47...)? that's an error of around 1.5%. Also, since you're "cheating" by using Alcodens, are you correcting back to 15.56C (60F)? Note that if you do this in wt%, you won't have to worry about temperature.

Pressure9pa & HedgeBird:

Contact me via PM, and I'll send you the scanned tables in excel format. I would post them here, but they're not pretty enough yet.

Thanks for the offer. Sent a pm this evening.

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. Also, since you're "cheating" by using Alcodens, are you correcting back to 15.56C (60F)? .

Just answering this part of your query, When using Alcodens with "Proof" units the program automatically sets at 60F.

There is a temperature correction window also which corrects hydrometer readings taken at other than 60F, but for this problem that was not needed.

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PeteB:

Note that if you do this in wt%, you won't have to worry about temperature.

With this version I have converted everything to weight (mass) before I start the calcs.

It requires more calculations but it may make it clearer to some people

From table 6, 100 gallons of 83 proof contains

41.5 gal of pure alcohol and 61.99 gallons of water ( at 60 F)

1 gal pure alcohol weighs 6.6097 lbs

1 gal pure water weighs 8.32823 lbs

100 gal 83 pr will contain 41.5 X 6.6097 = 274.3026 lbs alcohol

and 61.99 X 8.32823 = 516.267 lbs water

in a total weight of 790.5695 lbs

Proportion of alcohol by weight is 274.3026/790.5695 = 0.346968

Proportion of water by weight is 516.267/790.5695 = 0.653032

We have 603.2 lbs of 83 proof which contains 603.2 X 0.3469 = 209.273 Lbs alcohol

And 603.2 X 0.65303 = 393.8745 lbs water

Now what are the proportions of alcohol and water in the 135 pr we are to add?

Table 6 again, 100 gals of 135pr contains 67.5 gal alcohol and 36.02 gal of water

Convert to lbs as above and we get 0.597952 of alcohol and 0.402048 of water

We also need to know the proportions of alcohol and water in the final proof of 86 proof

Table 6, 86 pr = 43.0 alc and 60.56 of water, multiply these gallons by lbs per gal as above

And we get weight proportions of 0.360419 for alcohol and 0.639581 for water

Now we add X lbs of 135 pr to the 603.2 lbs of 83 proof to get 86 proof

X lbs of 135 pr is made up of 0.597952*X lbs alcohol and 0.402048*X lbs water

We now have 209.273 lbs of alcohol from the 83 proof plus 0.597952*X lbs alcohol from the 135 pr

Ie. 209.273 + 0.597952*X of alcohol

And 393.8745 + 0.402048*X lbs water

The ratio of alcohol to water, in the 2 lines just above, is the ratio of alc to water in the 86 proof

86 proof is 0.360419 of alcohol

and 0.639581 of water

therefore (209.273 + 0.597952*X) / (393.8745 + 0.402048*X) = 0.360419 / 0.639581 = 0.563523

209.273 + 0.597952*X = 0.563523*(393.8745 + 0.402048*X)

209.273 + 0.597952*X=221.9573 + 0.226563*X

0.597952*X - 0.226563*X = 221.9573 - 209.273

0.371389*X = 12.6843

X = 12.6843 / 0.371389 = 34.15365 lbs of 135 proof

Including the original 603.2 and the final weight will be 637.35 lbs

same answer again

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PeteB:

, but did you use .7(47....) or .71(47...)? that's an error of around 1.5%.

Will, I am not sure what you meant by the above.

I have now re-familiarised myself with using TTB tables so I might be able to answer now.

EDIT a few hours later---------- OK Will, I found what you were refering to, 0.71 is the ratio of alcohol to water and is correct

I copied-typed the equation from an Excel spreadsheet and made a typo, sorry.

The maths is correct, just the copy has the 1 missing.

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Right, I presumed that it was a typo when you got the same answer using weight. I prefer weight because measuring volume is so problematic, but if everything is corrected back to 60F, then the results should be equal...as you've shown.

My approach is similar to yours, but I OCRed Table 6, then used columns 2 & 3 to compute Alcohol and Water weights per gallon, and from that the weight of the mixture, and finally the weight% of alcohol for each proof. Using wt%, it's rather trivial to compute the present proof using the LOOKUP function in Excel. Simply search down the wt% column and find the value that's just below what you're searching for, and interpolate.

Once that was done, I admit I got lazy, and rather than code the algebra, I just do a binary search (test & iterate) for the amount of product to add to get the proof I'm looking for. Takes less than a minute. Your results and mine differ by a few percent, and that's significant.

Didn't I send you the spreadsheet a year ago or more?

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Hi Will, I did not receive your Excel spreadsheet last year, if you sent it, it has got lost in the never-never.

I would still like to see it, just for the brain exercise. info@belgrovedistillery.com.au

I know you are a fan of blending by weight (mass). I totally agree that the final physical blending is much easier and more accurate when done by mass, (for small distilleries) but the calculations are not necessarily easier. In the last example above you can see it took me about twice as many lines to do the calcs if I converted to mass at the start. The fewest calculations were in the first example where I used volumes right to the second last line, then did just one conversion to mass.

The TTB table 6 should have extra columns in mass (I assume that is what your spreadsheet has)

For others reading this post, if blending is done by weight (mass) on a good set of scales, then the temperatures of the components does not matter.

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Ok, so has anyone else tried to solve Will's challenge. No one has even said if they think I am correct--or not.

I must say it was a good excercise for my old brain, I have heard mental exercises help stave off alzheimer's.

Open the following attached PDF if you would like to see the easy way to solve it with AlcoDens

Wills challenge solved with AlcoDens - edited.pdf

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  • 2 weeks later...

My contact in the Australian Excise department told me they accept AlcoDens as a calculation tool.

The quote below is from the help menu of the current version of AlcoDens

"We believe the accuracy of the conversions is more than sufficient for doing process design calculations. However, because of the differing regulations in different countries, you will have to evaluate for yourself if it is good enough for trade and excise requirements in your area.

AlcoDens has been tested against tabulated data and the following maximum and typical errors have been found when calculating densities from known concentrations. Obviously the maximums are the worst results we found - there may be worse cases but they are unlikely to be significantly different from those shown below.

We believe the accuracy of the conversions is more than sufficient for doing process design calculations. However, because of the differing regulations in different countries, you will have to evaluate for yourself if it is good enough for trade and excise requirements in your area.

AlcoDens has been tested against tabulated data and the following maximum and typical errors have been found when calculating densities from known concentrations. Obviously the maximums are the worst results we found - there may be worse cases but they are unlikely to be significantly different from those shown below.

Temperature Strength Max.Error Average Error

Celsius Mass% kg/m3 kg/m3

10 - 40 0 - 25 0.046 0.018

10 - 40 25 – 100 0.030 0.012

40 - 100 0 - 100 0.380 0.090

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