urbaf2

140 CHEMISTRY AND TECHNOLOGY OF EXPLOSIVES

with a mixture:

HNO 3 32%

H 2 SO 4 60%

H 2 O 8%

then for each 100 kg of mixed acids, i.e. for 32 kg of HNO 3 , x kg of water is formed.

18 x 32

x = = 9.14 kg = [H 2 O]f

63

hence

D.V.S. =

60

8 + 9.14

= 3.50

This is the “theoretical” value of D.V.S. In practical work allowance should be made for the pre-

sence of the excess of HNO 3 used. The above calculations may be made with theoretical refe-

rence to either hydrocarbon or nitric acid. When 95% of the theoretical hydrocarbon is used, the

hydrocarbon factor is calculated. :

C 6 H 6 78

HNO 3

=

63

=1.24

Hydrocarbon factor = 1.24 x 95% = 1.18

18x100

Water of nitration per 100 kg of hydrocarbon is =

78

= 23.08

HNO 3 used in the mixed acid is:

1.18*32 = 37.76 kg HNO 3 per 100 kg of mixed acid.

Hence the water of nitration per 100 kg of mixed acid =

37.76 x 23.08

=

100

8.71

60

D.V.S. = =

8 + 8.71

3.59

Instead of 95% of hydrocarbon, 105% of nitric acid can be used:

HNO 3 63

C 6 H 6

= = 0.808

78

Nitric acid factor = 0.808 x 105% = 0.848.

HNO 3 used in mixed acid is:

32

= 37.74 kg HNO 3 per 100 kg of mixed acid.

0.848

This gives approximately the same figure for the water of nitration per 100 kg of mixed

acid :

37.74 x 23.08

100

8.71 and D.V.S. = 3.59

(according to Groggins [1]).

The D.V.S. should be as high as possible. For example if benzene is nitrated

with two acid mixtures A and B, both containing the same quantity of water but

having different D.V.S. values, nitration with the A mixture, for which D.V.S.