Ballistics Solutions Manual: The Theory and Design of Ammunition and Guns, 3rd Edition

Ballistics:TheTheoryandDesign of A
. . . . . .




mmunition and Guns 3rd Edition
. . . .




SolutionsManual Part 0 . . .




Donald E. Carlucci
. . .




Sidney S. Jacobson
. .




** Immediate Download
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** Swift Response
. .




** All Chapters included
. . .

,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.co
mposition.. If.the.process.takes.place.in.an.expulsion.cup.with.a.volume.of.10.in3,.assuming.ide
al.gas.behavior,.what.will.the.final.pressure.be.in.psi?
lbf.

Answer. . p.=.292
.



in2. . 

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
.   . 1. . kgmol. 
(10)g. 1... k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. = 
 
6.

(10) in 3.
8




.lbf. 
p.=.292
in2. . 

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.t

o.happen..If.the.chemical.composition.were.reacted.we.would.have.to.balance.the.reaction.eq
uation.and.would.have.to.use.Dalton’s.law.for.the.partial.pressures.of.the.gases.as.follows..Firs
t,.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

, NiT
p.i =.
V

So.we.can.write   kJ  kgmolC6.H8.N.2O9. 
(4) .
kgmolH. O (8.314) 
(1000)K. 1. . (10) g C H N. O .. . 1. kg
.
.

C H .N. O . 
  6 8 2 9


kgmol.C. H. N. O. .  kgmol.-.K.   
6. 8. 2 9.
1,000. .g C. H. N. O. . 
   C . H . N . O
p =
2
6. 8. 2. 9. 
252  . kg 6. 8. 2. 9   6. 8. 2. 9. . 
H.2O 
(. ) . 3.  1  kJ .1. .ft.
10. in.   . .. . 


.737.6.ft. −.lbf..12.in.
lbf.
=.1,168
pH2O in2. .
    1. .kgmolC6.H8.N2O9. 
(1000)K (10)g
kJ
(5) (8.314)  1. . kg.C6.H8.N2O9. 
..
kgmolCO . .


       





  C. H. N. O
.kgmol kgmol.-.K 252 kg 1,000 g
6. 8. 2. 9


 C. H. N. O. .      C. H. N. O    C. H. N. O. . 
CO 
p = ( ) 

 6. 18. 2.9ft.  
.
6. 8. 2. 9. 3. 6. 8. 2. 9. .


10. in . .. . 



1  kJ
pCO. lbf.
=.1,460 .737.6.ft.−.lbf. .12.in.
.



 in2. ..  
  kgmol 1. . kg 
(1)   
kgmol (8.314) kJ
(1000)K 1  . .. C6.H8.N2O9
(10)g  . . C6.H8.N2O9.

N2
         
C6.H8.N2O9
.kgmol kgmol.-.K 252 .. kg 1,000 g


( )
N2

 C. H. N. O. .   .    C. H. N. O    C. H. N. O. . 
p = 




6. 8. 2. 9.    kJ  6. 18.. 2..9ft.  
1  6. 8. 2. 9. . 
pN. . =. 292 lbf 
10. in3. .  .   
.737.6.f. t.−.lbf.1. 2.in.
in2. .
2




2 2



Then.the.total.pressure.is
p.=. pH.O.+l. bpfCO+. pN .lbf. lbf.  lbf. 
p.=.1,168.in2. . ..+.1,460 +.292 =.2,920
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.in3/lbm

, l. bf. 
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)

Rearranging,.we.have

mg.RT
p.=.
V.−.cb

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.
p. =   6. 8. . 2. 9. 

( 10 
) . .in3. −. ( .10).g..1000
1 ..kg.. 2.2. . lbm
. g. (
32.0.
)kg. . .( lbm).in. . .  
3. .


      

l. bf. 
p.=.314.2
in2. . 

So.you.can.see.that.the.real.gas.behavior.is.somewhat.different.than.ideal.gas.behavior.at.this.lo
w.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.react
ed.we.would.have.to.balance.the.reaction.equation.and.would.again.have.to.use.Dalton’s.law.f
or.the.partial.pressures.of.the.gases..Again,.assuming.no.air.in.the.vessel.we.write.the.decomp
osition.reaction.

C6.H8.N.2.O9. →.4H.2.O.+.5CO.+.N.2. +.C(s)

Then.for.each.constituent.(again.ignoring.solid.carbon).we.have
i
Ni.T.
p. =.
(V.-.cb)
So.we.can.write