Solutions Manual For Ballistics Theory and Design of Guns and Ammunition 3rd Edition By Donald E. Carlucci; Sidney S. Jacobson 9781138055315 ALL Chapters .

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Solutions Manual
S o l u t i o n s M a n u a l F o r B a l l i s t i c s T h e o r y a n d D e s i g n
o f G u n s a n d A m m u n i t i o n 3 r d E d i t i o n B y D o n a l d E .
C a r l u c c i ; S i d n e y S . J a c o b s o n 9 7 8 1 1 3 8 0 5 5 3 1 5 A L L
C h a p t e r s .
Complete Solutions Manual , All Chapters are
included. Ballistics: The Theory and Design of Ammunition and Guns 3rd Edition Solutions Manual Part 0 Donald E. Carlucci Sidney S. Jacobson 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 
2inlbf292p
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) RTm pgV (IG-4) Rearranging, we have VRTmpg
Here we go   
3ONHC
in10K 1000ftin12kJlbfft6.737kgkgmol
2521
K kgmolkJ314.8gkg
10001g10
9286
















p

2inlbf292p
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. sC N CO5OH4 ONHC2 2 9 2 8 6  
Then for each constituent (we ignore solid carbon) we have VTNpi
i
So we can write    





























inft
121
lbfftkJ
6.7371in10gkg
000,11g10kgkgmol
2521K 1000K- kgmolkJ314.8kgmolkgmol
4
3ONHCONHC
ONHC
ONHCONHC
ONHCOH
OH92869286
9286
92869286
92862
2p

2 OHinlbf168,1
2p
   





























inft
121
lbfftkJ
6.7371in10gkg
000,11g10kgkgmol
2521K 1000K- kgmolkJ314.8kgmolkgmol5
3ONHCONHC
ONHC
ONHCONHC
ONHCCO
CO92869286
9286
92869286
9286p

2 COinlbf460,1 p
   





























inft
121
lbfftkJ
6.7371in10gkg
000,11g10kgkgmol
2521K 1000K- kgmolkJ314.8kgmolkgmol
1
3ONHCONHC
ONHC
ONHCONHC
ONHCN
N92869286
9286
92869286
92862
2p

2 Ninlbf292
2p
Then the total pressure is 2 2 N CO OH p p pp 




2 2 2 2inlbf920,2inlbf292inlbf460,1inlbf168,1p
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