Environmental Engineering Reference
In-Depth Information
Table 3.13 Formation enthalpy of 100 g of explosive
Component
Mass
(100 g/g)
The amount of material
(mol)
Formation enthalpy
(J mol 1 )
HNO 3
70.56
1.12
173.0
C 6 H 5 NO 2
28.0
0.2276
11.2
H 2 O
1.44
0.08
285.9
Solution:
Let M = 100, then, these basic data as listed in Table 3.13 can be obtained
according to the known conditions.
Composition of each element in the 100 g liquid explosives is
C 1.366 H 2.418 N 1.348 O 3.895 . And this explosive has a slightly negative oxygen balance
and belongs to that of c b/2
2a < 0. Thus, the following calculation can be done
according to c b/2
2a < 0 and Eqs. 3.9 , 3.10 and 3.12 ,
5Jg 1
Q ¼
5993
:
x ¼
14
:
04
C 0 ¼
2
:
4526
Thus,
0 : 5
v D ¼
33
:
0
ð
5993
:
5
Þ
þ
243
:
2
14
:
04
1
:
40
335 m s 1
¼
7
;
v (measured) = 7,300 m s 1
e 0 : 5461 1 : 4
r ¼
1
:
25
þ
2
:
4526
1
¼
2
:
56
2
10 6
1
:
40
ð
7335
Þ
P C J ¼
¼
21
:
16 GPa
ð
Þ
þ
:
1
2
56
The calculated values and the measured values of several liquid explosives are
listed in Table 3.14 . The experimental data of explosion pressure are rarely reported
and the measured data are related to experimental methods with signi
cantly large
error, therefore, only the explosion velocity data are listed in the table for
comparison.
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