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Solutions for Ballistics: The Theory and Design of Ammunition and Guns, 3rd Edition Carlucci (All Chapters included)
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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...

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  • October 1, 2024
  • 560
  • 2024/2025
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MEDCONNOISSEUR©2024 ✅💯👌😒




Ballistics: The Theory and Design of
Ammunition and Guns 3rd Edition
M
ED

Solutions Manual Part 0
C
O

Donald E. Carlucci
N

Sidney S. Jacobson
N
O
IS
SE

** Immediate Download
** Swift Response 👌
U

** All Chapters included ✅
R




1

,MEDCONNOISSEUR©2024 ✅💯👌😒




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
in2 
M

Solution:
ED

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)
C

pV = mg RT (IG-4)
O

Rearranging, we have
N

mg RT
p=
N

V
O

Here we go  
    ft − lbf 
(10)g 1  kg (8.314) kJ  1   kgmol (737.6 
) ( 12)
in 
(1000)K
1000 g kgmol  K    
IS

     252 kg C H N O   kJ  ft 
p=
( 8 2 9 



 
10) in 3
SE

6




 lbf 
p = 292
in2 
U

You will notice that the units are all screwy – but that’s half the battle when working
R

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 → 4H2 O + 5CO + N 2 + C(s)

Then for each constituent (we ignore solid carbon) we have




2

,MEDCONNOISSEUR©2024 ✅💯👌😒


NiT
pi =
V

So we can write    kgmolC H N O 
(4) kgmol (8.314) kJ (1000)K 1  (10)g   1  kg C H N O
6 8 2 9

kgmol H O  kgmol - K     C H N O 1,000  g 6 8 2 9 6 8 2 9 

   252  kg C H N O   
2

p = C H N O     CHNO 
    6 8 2 9 

(10)in3  1  kJ  1 
6 8 2 9 6 8 2 9
ft 
HO 2
    
 737.6 ft − lbf 12 in 
 lbf 
M

= 1,168
pH2 O in2 
    1 kgmolC6 H8 N2O9 
(5) kgmol (8.314) kJ
(1000)K (10) g    1 kg 
ED


 kgmol
CO  kgmol - K 

 252  kg
 C H N O
6 8 2 9

1,000  g

C6 H8 N2 O9 
      CHNO 
p CO = 
C H N O C H N O
9   1    6 8 2 9 
10 in3  kJ  1  ft 
6 8 2 6 8 2 9




C

( )   −   
 737.6 ft lbf  12 in 
O

 lbf 
= 1,460
pCO
N

in2 
    1 kgmolC6 H8 N2O9 
(1) kgmol (8.314) kJ
(1000)K (10) g    1 kg 

N

   252  kg 
kgmol - K  1,000  g
C6 H8 N2O9 
kgmol
N 2 C H N O
6 8 2 9

      CHNO 
pN = 
C H N O C H N O
9   1 kJ  6 18 2 9 ft   6 8 2 9 
O


2 6 8 2
10 in3  
( )    −   
IS

 737.6 ft lbf  12 in 

 lbf 
= 292
SE

p
N2 in2 

Then the total pressure is
U

p = pH 2O + pCO + pN 2
R

 lbf   lbf   lbf   lbf 
p = 1,168 +1,460 + 292 = 2,920
in2  in2  in2  in2 


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 in 3/lbm




3

, MEDCONNOISSEUR©2024 ✅💯👌😒



lbf 
Answer: p = 314.2
in2 

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)
M

Rearranging, we have
ED

mg RT
p=
V − cb

Here we go
C

    1  kgmol 
(10)g 1  kg (8.314) kJ (737.6)ft − lbf (12)in (1000)K
     
O

 1000  g  kgmol  K  252 kg C H N O   kJ3  ft 
p=   
 6 8 2 9 


( ) 3 − (10)g 
 1  
 
N

 kg (2.2)lbm(32.0)  in 
10 in  1000 g  kg  lbm
N

       
O

lbf 
p = 314.2
IS

in2 

So you can see that the real gas behavior is somewhat different than ideal gas behavior at
SE

this low pressure – it makes more of a difference at the greater pressures.

Again please note that this result is unlikely to happen. If the chemical composition were
U

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
R

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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