Solution Manual - for Ballistics: Theory and Design of Guns and Ammunition, Third Edition 3rd Edition by Sidney S. Jacobson, Donald E. Carlucci, All Chapters |Complete Guide A+

uytrewuytrew



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

Solutions Manual Part 0
ST

Donald E. Carlucci
SO
Sidney S. Jacobson
LU
** Immediate Download
** Swift Response
** All Chapters included
TI
O
N

, uytrewuytrew



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
TE
psi?
 lbf 
Answer p = 292
 in 2 

Solution:
ST
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 = mg RT (IG-4)
SO
Rearranging, we have

mg RT
p=
V

Here we go
LU
1  kg    1  kgmol 
(10)g   (8.314)
kJ
  (737.6)ft − lbf (12)in (1000)K
 1000  g   kgmol  K  252  kg C H N O   kJ   ft 
 
2 9 
p=  3 8
 
6

(10) in
TI
 lbf 
p = 292
 in 2 

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

, uytrewuytrew


Ni T
pi =
V

So we can write
    1   kgmol C 6 H 8 N 2 O 9 
(4) kgmolH O (8.314 )
kJ
 (1000 )K    (10 ) g  C6 H8 N 2O9   1  kg C6 H8 N 2O9
 

 kgmol C H N O   kgmol - K   252  kg C H N O  1,000  g C H N O 
2


=  6 8 2 9     6 8 2 9 
 
p
( ) 3  1  kJ  1  ft 
6 8 2 9
TE
H 2O

10 in    

 737.6 ft − lbf  12 in 

 lbf 
p H 2O = 1,168
 in 2 
ST
    1   kgmol C 6 H 8 N 2 O 9 
(5) kgmolCO (8.314) kJ
(1000)K  (10)g  1  kg C6 H8 N2O9 

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

( ) 
CO
=
SO
p
    
 
 1  ft 
6 8 2 9 6 8 2 9 6 8 2 9
3
10 in

 kJ
1
pCO 
= 1,460
 lbf   737.6  ft − lbf  12 in 
 in 2    kgmol C 6 H 8 N 2 O 9 
   kg C6 H8 N2O9 
 kgmol (8.314) kJ
(1000)K 1  (10) g   1
LU
(1) 2
N
   kgmol - K     C6 H8 N2O9   
kgmol 252 kg 1,000 g


( ) 
N2

 C H N O     C H N O    C H N O 

TI
p =



    
O
6 8 2 9 6 8 2 9 6 8 2 9


 1 
10 in 3  737.6 
kJ 
  
1 ft
p N 2 = 292 lbf  ft − lbf  12 in 
 in 2 
N
2 2



Then the total pressure is
p = pH O + lbf
pCO + pN  lbf   lbf   lbf 
p = 1,168 in 2  + 1,460 + 292 = 2,920
  in 2   in 2   in 2 
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

, uytrewuytrew



 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)
TE
p(V − cb) = mg RT (VW-2)

Rearranging, we have
ST
mg RT
p=
V − cb

Here we go
1  kg    1  kgmol 
(10)g   (8.314)
kJ
   (737.6) ft − lbf (12)in (1000)K
SO
 1000  g   kgmol  K  252  kg C H N O   kJ   ft 
 
 6 8 2 9 
p=
10 in
 
( ) 3 − ( )  1  kg ( )lbm ( )  in 3 
 10 g    2.2   32.0 lbm 
  1000  g   kg   
LU
 lbf 
p = 314.2
 in 2 

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.
TI
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.
O
C6 H8 N 2 O9 → 4H 2 O + 5CO + N 2 + C(s)
N
Then for each constituent (again ignoring solid carbon) we have

N i T
pi =
(V - cb)
So we can write