Ballistics Theory and Design of Guns and Ammunition, 3e Donald Carlucci, Sidney Jacobson (Solution Manual) Latest Edition Complete Guide A+

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




Solutions Manual Part 0 #hg #hg #hg




Donald E. #hg




#hg Carlucci Sidney #hg




#hg S. Jacobson
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** Immediate Download
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** Swift Response
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** All Chapters included
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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 


SCHOLARVAULT

, 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  mg RT (IG-4)

Rearranging, we have

mg RT
p
V

Here we go  
   
10g 1  kg 8.314 kJ  1  kgmol 737.6ft  lbf 12in 1000K
1000 g kgmol  K     
     252 kg C H N O   kJ  ft 
2 9 
p  
 
6

10 in3
8




 lbf 
p  292
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




SCHOLARVAULT

, NiT
pi # h g  # h g
V

So #hgwe #hgcan #hgwrite    kgmolC6 #hgH8 #hgN#hg2O9 # h g 
41 kgmol
# h g 
kg #hg
8.314  kJ 1000K#hg # h g 1 # h g #hg 10 g #hg # h g 
#hg     C
# h g C H #hgN#hg O # h g
 
6 8 2 9
H #hgO


kgmol # h g kgmol #hg- #hgK#hg
#hgH8 #hgN #hg2O 9 # h g
#hg 252 # h g kg
1,000 # h g #hgg
C #hgH #hgN #hgO #hg #hg     C # h g H # h g N # h g O   C # h g H # h g
#h g O # h g # h g 

  6 # h g 8 # h g 2 # h g 9 # h g 
2
  6 # h g 8
2 # h g 9 # h g # h g 
p
# h g 6 # h g 8 # h g 2 # h g 9



H#hg2O  #hg  # h g 3 #hg
 1  kJ  1 # h g #hgft #hg
#hg 
in # h g   # h g #hg#hg # h g # h g 
10 # h g 
 #hg737.6#hgft # h g  #hglbf #h g #hg12 #hgin#hg

lbf #hg

#hg1,168
pH2O in#hg2#hg
    1 # h g #hgkgmolC6 #hgH8 #hgN2O9 # h g 
5 kgmolCO 8.314 kJ
1000K #hg # h g
10g  # h g # h g 1 #hg

kg#hgC6 #hgH8 #hgN2O9 # h g 
# h g

       
 




  C #hgH
#hgkgmol kgmol #hg-#hgK 252 kg 1,000 g
6 # h g 8 # h g 2 # h g 9


 C #hgH #hgN #hgO #hg #hg    C # h g H # h g N # h g O   C # h g H # h g
# h g N # h gO


# h g O # h g # h g 
CO


p 
6 # h g 8 # h g 2 # h g 9 # h g 
 # h g   3 #h g      6 #hg8 # h
#h g 9 1# h g # hg  ft#hg
6#hg 8 # h g 2#hg 9




 10 # h g in # h g # h g  # h g # h g 




 1  kJ
pCO 
#hg1,460 #hg737.6#hgft #h g #hglbf # h g #hg12 #hgin #hg
# h g


lbf #hg
#hg



 in#hg2 #hg #hg  1000K#hg # h g 1 # h g 
10g 
   kJ
1  kgmol 8.314 C6 #hgH8
#hg N2O9
# h g # h g 1 C6 #hgH8 #hgN2O9 # h


kgmol kg
2
#hg
# h g # h g

 
N
       
C6 #hgH8
kgmol
#hg kgmol #hg- 252 #hg #hg kg #hgN2O9
1,000 g
#hgK





N2

 C #hgH #hgN #hgO #hg #hg  # h g
    C #hgH #hgN #hgO    C #hg

  # h g N #hgO # h g # h g 
p 







6 # h g 8 # h g 2 # h g 9 # h g   6 # h g 8 # h g 2 # h g 9   6 # h g 8

# h g 2 # h g 9 # h g # h g  1
lbf 10 in#hg3 #hg kJ # h g 1# h g #hgft#hg

# h g


pN # h g # h g  #hg737.6#hg#hgft #hg #hglbf #hg #hg12 #hgin#hg
 
# h g 292






SCHOLARVAULT

, in#hg2 #hg
2






2 2



Then #hgthe #hgtotal #hgpressure #hgis
p #hg # h g pH#hgO#hg #hg#hglbf lbf lbf # h g 
p  # h g pN
#hglb fCO#hg
 # h g 

p #hg#hg1,168 #hgin#hg2 # h g  #hg#hg1,460  #hg292  #hg2,920
in#hg2 #hg in#hg2 #hg in#hg2 #hg
2.2 Other Gas Laws #hg #hg

Problem #hg2 #hg- # h g Perform #hgthe #hgsame #hgcalculation #hgas #hgin #hgproblem #hg1 #hgbut #hguse #hgthe
3
#hgNoble-Abel #hg equation #hg of #hg state #hg and #hg assume #hg the #hg covolume #hg to #hg be #hg 32.0 #hg in /lbm




SCHOLARVAULT