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