Solutions for Ballistics: The Theory and Design of Ammunition and Guns, 3rd Edition Carlucci (All Chapters included)
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Module
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 and...
Ballistics: The Theory and Design of
Ammunition and Guns 3rd Edition
Solutions Manual Part 0
S
R
Donald E. Carlucci
PE
Sidney S. Jacobson
PA
** Immediate Download
ED
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** All Chapters included ✅
M
1
,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?
lbf
Answer p = 292
in 2
Solution:
S
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)
R
pV = mg RT (IG-4)
Rearranging, we have
PE
mg RT
p=
PA
V
Here we go
1 kg 1 kgmol
(10)g (8.314)
kJ
(737.6)ft − lbf (12)in (1000)K
kgmol K 252 kg C H N O kJ ft
ED
1000 g
p= 8 2 9
(10) in 3
6
lbf
p = 292
M
in 2
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.
C6 H8 N 2 O9 → 4H 2 O + 5CO + N 2 + C(s)
Then for each constituent (we ignore solid carbon) we have
2
, N i T
pi =
V
So we can write
kJ 1 kgmolC6 H8 N2O9
(4) kgmolH O (8.314) (1000 ) K
(10) g C6 H8 N2O9 1 kg C6 H8 N2O9
kgmol C H N O
2
kgmol - K 252 kg 1,000 g C H N O
C H N O
= 6 8 2 9 6 8 2 9
p
( 10) in3 1 kJ 1 ft
6 8 2 9
H 2O
737.6 ft − lbf 12 in
lbf
pH2O = 1,168
in 2
S
kJ kgmolC6 H8 N2O9
(5) kgmolCO (8.314) (1000)K 1
(10) g C6 H8 N2O9 1 kg C6 H8 N2O9
kgmolC H N O kgmol - K 252 kg C H N O 1,000 g C H N O
R
p = 6 8 2 9 6 8 2 9 6 8 2 9
3
CO
1 kJ 1 ft
(10) in
−
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 in 3/lbm
3
, lbf
Answer: p = 314.2
in 2
Solution:
This problem is again straight-forward except for those pesky units – but we’ve done this
before. We start with equation (VW-2)
p(V − cb) = mg RT (VW-2)
Rearranging, we have
S
mg RT
p=
V − cb
R
Here we go
1 kg 1 kgmol
(10)g (8.314)
kJ
PE
(737.6) ft − lbf (12)in (1000)K
1000 g kgmol K 252 kg C H N O kJ ft
6 8 2 9
p=
( ) 3 − ( ) 1 kg ( )lbm( ) in 3
10 g
1000 g
2.2
kg
32.0
lbm
10 in
PA
lbf
p = 314.2
in 2
ED
So you can see that the real gas behavior is somewhat different than ideal gas behavior at
this low pressure – it makes more of a difference at the greater pressures.
M
Again 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 again have to use
Dalton’s law for the partial pressures of the gases. Again, assuming no air in the vessel
we write the decomposition reaction.
C6 H8 N 2 O9 → 4H 2 O + 5CO + N 2 + C(s)
Then for each constituent (again ignoring solid carbon) we have
Ni T
pi =
(V - cb)
So we can write
4
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