The percentage of alcohol and of extract in beer can be obtained, and the original gravity deduced, if the specific gravity and the refractive index are known. H. Tornoe elaborated the method at the instance of the Norwegian Government, and a description of it has been given by Ling and Pope.1

Addition of alcohol to water causes a diminution of the specific gravity of the latter, and an increase in its refractive index. The extractive matter of beer gives an increase of both. Let c1 be the amount by which the refractive index is raised by each 1 per cent. of alcohol; c2 the corresponding increment for each 1 per cent. of extract; c3 the amount by which the specific gravity is lowered by 1 per cent. of alcohol, and c4 the increment for 1 per cent. of extract. Let, further, 8 be the specific gravity of the beer, and r its refractive index, both referred to water as unity; whilst A and E respectively denote the percentages of alcohol and extract in the beer. Then we have: and

r

=

1

+

c2E

+

c1A.

S

=

1

+

C4E

-

c3A.

From these two equations the values of A and E are readily obtained: -

A

=

c4(r-l)-c2(S-l)

.......(1).

C1C4+C2C3

E

=

c3(r-1) + c1(S-1)

.......(2).

C1C4+C2C3

If the values of the constants c1 c2, c3, c4, are determined once for all, these equations give A and E, the percentage of alcohol and extract in the beer, when 8 and r are obtained on the sample.

C1 c2, etc., are, however, not strictly constant, as the variations produced in the refractive index and specific gravity are not exactly proportionate to the quantity of alcohol or extract present. Hence tables are constructed, by means of which, after determining the specific gravity and the refraction of a sample, the percentage of alcohol and of extract can be read off.

Determinations of the values of c1 c2, c3, and c4 have been made by Earth.2 The figures obtained, substituted in equations (1) and (2), give the formulae: -

A

=

759.8

(r- 1 )

-

292.3

(S-l).

E

=

336.6

(r-1)

+

130.3

(S-l).

1 J. Inst. Brewing, 1901, 7, 170. 2 Zeitsch. ges, Brauw., 1905, 28, 303.

Earth's results have been checked by J. Race1 who finds that with samples containing up to about 45 per cent. of alcohol the formula; give satisfactory results, but with higher percentages the tendency is to give rather low values for A and E. For such samples Race prefers the expressions: -

Fig. 46. immersion refractometer, with rotating water.bath. A, prism, which dips into the liquid in the glass beneath; B, bath with thermometer.

Fig. 46. - immersion refractometer, with rotating water.bath. A, prism, which dips into the liquid in the glass beneath; B, bath with thermometer.

A =

778

(r-1)

-

290

(S - l).

E =

350

(r - 1)

+

130

(5 - 1).

Tornoe used a modification of Hallwach's prism for determining the refraction values, but the Zeiss immersion refractometer or Pulfrich's instrument is more convenient.

1 J. Soc. Chem. Ind., 1908, 21, 544,

When the percentage of alcohol is obtained, the original gravity is readily calculated. From the former, by reference to ordinary alcohol tables, we find the specific gravity of aqueous alcohol containing the same proportion of alcohol as is present in the beer. Deducting this from 1000, the spirit indication is found, and thence the "degrees of gravity lost."

The spirit indication plus the specific gravity of the beer gives the extract gravity. Adding to this the degrees of gravity lost, we obtain the original gravity.

Example: Suppose we have a beer with sp. gr. 10095 and a refraction reading 39.0 by the Zeiss immersion refractometer at 60° F. Remembering that in the expression r - 1 the unit is the refractive index for water, 1.33335 at the above temperature, the calculation is as follows: -

Reading 39.0 = ref. index (v. Table, p. 288).................

1.34237

Less ref. index of water................................................

1.33335

r . 1 =

0.00902

S . 1 = 1.0095 . 1 = 0.0095.

Hence, using Barth's coefficients, the percentage of alcohol is given by:

A = 7598 X 000902 - 2923 X 00095, = 4.12 per cent. (by weight).

From the ordinary alcohol tables, this corresponds with a specific gravity 9926.

.'. spirit indication

=

7.4

= 32.86 degrees of " gravity lost."

sp. gr. of sample

=

1009.5

.'. Extract gravity

=

1016.9

Add gravity lost

32.86

.. Original gravity

=

1049.76

In practice, lengthy calculations are obviated by the use of special tables which correlate the immersion-refractometer readings directly with the specific gravity and the original gravity.

One such table is given below. It is used as follows: Let R denote the refraction reading of the beer by the immersion instrument, and D the specific gravity minus 1000. Deduct D from R.

Then corresponding with the value of R - D thus obtained, the table gives a number, N, which added to the specific gravity of the beer gives the original gravity of the sample.

R.D.

N.

R . D.

N.

17

4.6

30

40.1

18

7.4

31

42.7

19

10.3

32

45.2

20

13.1

33

47.7

21

15.9

34

50.1

22

18.7

35

52.6

23

21.4

36

55.1

24

24.2

37

57.6

25

26.9

38

60.1

26

29.6

39

62.6

27

32.2

40

65.0

28

34.9

41

67.5

29

37.5

42

70.0

Example: - Let the specific gravity be 1008.6, and the refraction reading 33.5.

Then D = 8.6, and R - D = 24.9. From the table, by interpolation, the value of N is found to be 26.6. Adding this to the specific gravity, we get 1008.6 + 26.6 = 1035.2, the required original gravity. For regular use the interpolation values are, of course, calculated out and tabulated once for all.