several suitable and robust options, the validity of any one method can be tested against another, thus providing confidence in the appropriate distillery operations and reportable values. Inputting the example values from above into equation 1 and setting equal volumes of the 95% alcohol and water to be used:
% w/w diluted = 92.41% ×
0.8114 × 50 0.8114 × 50 + 0.998203 × 50
% w/w diluted = 92.41% ×
40.57 40.57 + 49.91
= 41.43% w/w
Where % w/w is also simply noted as ABW – alcohol by weight. Taking the mass value of the diluted alcohol from such a mixture to the OIML tables provides a density value of 0.9322 g/mL and then taking this to the density vs. ABV table shows the ABV to be 48.94% (not a simple additive 50%) for the resulting mixture when stabilized to 20 °C. It is important to note that the mixing of alcohol and water is an exothermic reaction; the solution warms up as heat is liberated and therefore the sample must be brought back to a temperature of 20 °C to ensure accurate measurements. Here 48.94% ABV is obtained, compared to a value of only 47.5% as would be obtained by a simple non-contracted volume type dilution—a 1.44% ABV difference! The conversion from the mass to volume value can also be done mathematically, instead of relying on the OIML tables, but requires the specific gravity of the sample (not the density) and the relative density of pure alcohol. Consult the authors for more on this topic. Again, Travagli showed that the equations above can be used when non equal volumes of water and alcohol are used. His example shows 45.5L of 95% alcohol, when mixed with 9.5L of water, results in an 80% ABV solution. Using the approaches discussed here and consulting the OIML tables, the reader might want to check this out in practice going through the intermediate ABW stages to obtain the final answer. Using mass relationships in the alcohol world can be very useful.
CALCULATING THE AMOUNT OF WATER NEEDED TO DILUTE A GIVEN VOLUME OF ALCOHOL TO A LOWER PERCENTAGE Of more use to those wishing to dilute matured spirits to bottling strength (proofing) the following guidelines and equations will be useful. Again, for modern formulations the density of the sample might be reflective of not only the original alcohol water mix but also sugars and flavorings used, so the actual specific gravity of the sample may be at play here. Thus a distillation in the lab of a well-mixed and fully equilibrated sample may still be required after making dilutions to obtain the true final alcohol value and the actual extract content. The usual proofing rules on obscuration WWW.ARTISANSPIRITMAG.COM
solids if between 400-600 mg/100 mL will not interfere as much with these calculations as there is still essentially a hydroalcoholic mixture at play here (the multitude of congeners only adding a fractional amount to the density). The reader should consult the TTB website for the proof obscuration method and regulations. However, they should be aware of the composition of their products. TTB Gauging procedures and Proofing/Proof Obscuration details may be found here:
www.ttb.gov/foia/gauging_manual_toc.shtml#27:22.214.171.124.25.4.504.1 As noted above, tables exist showing the amount of water to add to a given volume of alcohol of specific concentration to obtain a desired final lower concentration. In addition to that approach, an equation can be applied which acts independently of the final volume of the hydroalcoholic solution desired. It involves a simple algebraic resolution, introduced here: EQUATIO N3
%v/v conc. × Vconc. = %v/v dilute × Vdilute
From equation 3 can be derived the formula that allows the determination of the final volume of a hydroalcoholic solution which undergoes the volumetric contraction as discussed above:
xV diluted =
% v/v conc. % v/v dilute
Example (volumes in milliliters):
xV diluted =
100mL = 237.5mL
This value is very important in cross-checking the final alcohol concentration following mixing of the defined volumes of alcohol and water (and will be used in the example below). Now, multiplying the volume by its density leads to the final weight of the diluted solution and then the water of dilution to be added is determined: EQUATIO N5
water to be added = ywater = xVdilute × ρ dilute – Vconc. × ρ conc.
Where y = volume of water to add and ρ = the respective densities of concentrated alcohol solution and the final diluted sample. Again, tables can be referenced for these values. For this example, plugging in the numbers* gives 144.0 mL** water to add to 100 mL 95% alcohol by volume solution to yield an apparent 244 mL of final solution at 40% ABV. Note: the volume contraction is 244 – 237.5 = 6.5 mL or 2.66%. The final alcohol concentration is now corrected for in making this mix of alcohol and water, and voila, an accurate sample of known alcohol strength has been made! Of course, it should be tested to be certain. *For this example, the number inputs for equation 5 need to be for 95% and 40% ABV solutions and the respective densities and 237.5 mL as xVdilute from equation 4 and 100 mL for Vconc. **There
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