In technical operations of reducing spirits from a higher to a lower strength it is common to disregard the effect of contraction when no great accuracy is required. The calculations are then much simplified. From the principle that the volume is inversely as the strength, the volume which the reduced spirit will have is calculated. The difference between this and the original volume is taken as the volume of water to be added. A few examples will make the method clear.

Example (1). Given 54.5 gallons of spirit at strength 8.5 over proof; how much water is required to reduce it to 3.5 over proof ?

Here the given strength = 108.5 per cent. of proof spirit, and the required strength = 103.5. Since the volumes are inversely as the strengths, the volume of the reduced spirit will be

54.5 X 108.5/103.5 = 57.1 gallons.

The volume of water to be added is therefore taken as 57.1 - 54.5 = 2.6 2 6 gallons. The exact quantity should be 2.76 gallons, if contraction were allowed for, but the result is sufficiently accurate for technical purposes.

Example (2). A much greater discrepancy, however, may occur when the two strengths differ more widely. Take, for instance, the example already worked out (p. 279), viz., how much water is required to reduce 100 gallons of spirit at 60 o.p. to a strength of 20 o.p.?

Here the volume of the spirit after reduction to the lower strength will be: -

100 x 160/120 = 133.3 gallons, and the quantity of water to be added would be taken as 133.3 - 1000 = 333 gallons.

The accurate quantity would be 36.2 gallons, as already shown.

Example (3). Given volume 80 gallons, at strength 10 over proof; how much water must be added to reduce the spirit to 25 under proof ?

Here the percentage strengths are respectively 100 + 10 and 100 - 25, or 110 and 75; and the resulting volume of the reduced spirit is

80 X10/75 = 117.3 gallons.

The quantity of water to be added is therefore taken as 117.3 - 80.0 = 37.3 gallons. The correct quantity would be 386.

Example (4). It is sometimes convenient to calculate the equivalent proof gallons before proceeding to find the final volume of the reduced spirit, as in the following instance: -

Given 50 gallons of spirit at 12 over proof, and 20 gallons at 10 under proof; how much water must be added to the mixture of these in order to reduce the strength of the whole to 25 under proof ?

First calculate each quantity to proof gallons: -

50 gallons at 12 o.p. =50 X 112/100 = 56.0 gallons at proof strength.

20/70 „ 10 u.p. = 20 x 90/100 = 18.0

Total . . == 740

Then 74.0 at proof = 74.0 x 100 / 75 = 98.7 „ 25 under proof.

Hence the water to be added is taken as 98.7 - 70.0 = 28.7 gallons.

Dilution of alcohol from given strength to required lower strength.

 Required lower strength. Given higher strengths, per cent. by volume 96. 95. 94. 93. 92. 91. 90. 85. 80. Volume of water in c.c. to be added to 1 litre of alcohol. 95 12 94 25 12 - - - - - - - 93 38 25 12 92 50 38 25 12 91 64 51 38 25 12 90 77 64 50 38 25 13 - - - 85 147 133 119 106 92 79 66 - - 80 224 209 195 180 166 152 138 68 __ 75 310 295 279 263 249 234 219 145 72 70 408 391 374 359 342 326 310 231 153 65 520 502 484 467 449 432 415 331 246 60 650 630 613 592 573 555 536 440 354 55 800 730 759 739 71!) 698 678 579 480 50 981 960 936 914 892 869 847 739 631 45 1200 1175 1151 1126 1101 1077 1053 932 813 40 1472 1444 1417 1389 1362 1335 1308 1173 1040 35 1818 1787 1756 1725 1694 1663 1632 1480 1329 30 2279 2241 2205 2169 2133 2098 2062 1886 1711 25 2916 2873 2830 2788 2745 2703 2661 2451 2243 20 3872 3820 3767 3715 3662 3610 3558 3298 3040 15 5466 5397 5328 5259 5190 5121 5053 4710 4369 10 8658 8556 8453 8351 8249 8147 8045 7536 7029