Solutions for Ballistics: The Theory and Design of Ammunition and Guns, 3rd Edition Carlucci (All Chapters included)
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Course
Ballistics
Institution
Ballistics
Complete Solutions Manual for Ballistics, The Theory and Design of Ammunition and Guns, 3rd Edition By Donald E. Carlucci, Sidney S. Jacobson ; ISBN13: 9781138055315. (Full Chapters included Chapter 1 to 21)....1. Introductory Concepts.
2. Physical Foundation of Interior Ballistics.
3. Analytic a...
Ballistics: The Theory and Design of
Ammunition and Guns 3rd Edition
Solutions Manual Part 0
Donald E. Carlucci
Sidney S. Jacobson
** Immediate Download
** Swift Response
** All Chapters included
,2.1 The Ideal Gas Law
Problem 1 - Assume we have a quantity of 10 grams of 11.1% nitrated nitrocellulose
(C6H8N2O9) and it is heated to a temperature of 1000K and changes to gas somehow
without changing chemical composition. If the process takes place in an expulsion cup
with a volume of 10 in3, assuming ideal gas behavior, what will the final pressure be in
psi?
Answer p = 292 lbf2
in
Solution:
This problem is fairly straight-forward except for the units. We shall write our ideal gas
law and let the units fall out directly. The easiest form to start with is equation (IG-4)
pV = m g RT (IG-4)
Rearranging, we have
m g RT
p=
V
Here we go
1 kg kJ 1 kgmol ft − lbf
(10)[g ]
(8.314 ) (737.6 ) (12) in (1000)[K ]
1000 g kgmol ⋅ K 252 kg C6 H8 N 2O9 kJ ft
p=
(10)[in 3 ]
lbf
p = 292 2
in
You will notice that the units are all screwy – but that’s half the battle when working
these problems! Please note that this result is unlikely to happen. If the chemical
composition were reacted we would have to balance the reaction equation and would
have to use Dalton’s law for the partial pressures of the gases as follows. First, assuming
no air in the vessel we write the decomposition reaction.
C 6 H 8 N 2 O 9 → 4H 2 O + 5CO + N 2 + C(s )
Then for each constituent (we ignore solid carbon) we have
, N i ℜT
pi =
V
So we can write
kgmol H 2O kgmol C 6 H 8 N 2O 9 1 kg
(4)
(8.314 )
kJ
(1000)[K ] 1 [
(10) g C 6 H 8 N 2 O 9 ] 1,000
C6 H8 N 2O9
kgmol C 6 H 8 N 2O 9 kgmol - K 252 kg C 6 H 8 N 2O 9 g C6 H8 N 2O9
p H 2O =
[ ]
(10) in 3 1 kJ 1 ft
737.6 ft − lbf 12 in
lbf
p H 2O = 1,168 2
in
kgmol CO kgmol C6 H8 N 2O9 1 kg C 6 H8 N 2O9
(5)
(8.314)
kJ
(1000)[K ] 1 [
(10) g C6 H8 N 2O9 ]
kgmol C6 H8 N 2O9 kgmol - K 252 kg C6 H8 N 2O9 1,000 g C6 H8 N 2O9
p CO =
[ ]
(10) in 3 1 kJ 1 ft
737.6 ft − lbf 12 in
lbf
p CO = 1,460 2
in
kgmol N 2 kgmol C6 H8 N 2O9 1 kg
(1)
(8.314)
kJ
(1000)[K ] 1 [
(10) g C6 H8 N 2O9 ] 1,000
C6 H8 N 2O9
p N2 = kgmol C6 H8 N 2O9 kgmol - K 252 kg C6 H8 N 2O9 g
C6 H8 N 2O9
[ ]
(10) in 3 1 kJ 1 ft
737.6 ft − lbf 12 in
lbf
p N 2 = 292 2
in
Then the total pressure is
p = p H 2O + p CO + p N 2
lbf lbf lbf lbf
p = 1,168 2 + 1,460 2 + 292 2 = 2,920 2
in in in in
2.2 Other Gas Laws
Problem 2 - Perform the same calculation as in problem 1 but use the Noble-Abel
equation of state and assume the covolume to be 32.0 in3/lbm
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