Ballistics: The Theory and Design of

Ballistics: The Theory and Design of

Ammunition and Guns 3 rd Edition

Solutions Manual Part 0

Donald E. Carlucci Sidney S. Jacobson

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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 in 3 , assuming ideal gas behavior, what will the final pressure be in psi?

Answer       = 2 in lbf 292p

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) RTmp g =V

(IG-4)

Rearranging, we have V RTm p g =

Here we go

() ( )( ) ()()

()

3 ONHC in10 K1000 ft in 12 kJ lbfft 6.737 kg kgmol 252 1 Kkgmol kJ 314.8 g kg 1000 1 g10 9286           

−

                                 =p       = 2 in lbf 292p

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.()sCNCO5OH4ONHC

229286

+++→

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

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V TN p i i  =

So we can write

() ( ) ()() 

()

                  −                                                 = in ft 12 1 lbfft kJ 6.737 1 in10 g kg 000,1 1 g10 kg kgmol 252 1 K1000 K-kgmol kJ 314.8 kgmol kgmol 4 3 ONHC ONHC ONHC ONHC ONHC ONHC OH OH 9286 9286 9286 9286 9286 9286 2 2 p       = 2OH in lbf 168,1 2 p

() ( ) ()() 

()

                  −                                                 = in ft 12 1 lbfft kJ 6.737 1 in10 g kg 000,1 1 g10 kg kgmol 252 1 K1000 K-kgmol kJ 314.8 kgmol kgmol 5 3 ONHC ONHC ONHC ONHC ONHC ONHC CO CO 9286 9286 9286 9286 9286 9286 p       = 2CO in lbf 460,1p

() ( ) ()() 

()

                  −                                                 = in ft 12 1 lbfft kJ 6.737 1 in10 g kg 000,1 1 g10 kg kgmol 252 1 K1000 K-kgmol kJ 314.8 kgmol kgmol 1 3 ONHC ONHC ONHC ONHC ONHC ONHC N N 9286 9286 9286 9286 9286 9286 2 2 p       = 2N in lbf 292 2 p

Then the total pressure is 22 NCOOH pppp ++=

      =       +       +       = 2222 in lbf 920,2 in lbf 292 in lbf 460,1 in lbf 168,1p

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 in 3 /lbm 3 / 4

Answer: 

     = 2 in lbf 2.314p

Solution:

This problem is again straight-forward except for those pesky units – but we’ve done this before. We start with equation (VW-2)

• ) RTmcbp

g =−V

(VW-2)

Rearranging, we have cb RTm p g − = V

Here we go

() ( )( ) ()()

()() () ()

                              −           

−

                                 = lbm in 0.32 kg lbm 2.2 g kg 1000 1 g10in10 K1000 ft in 12 kJ lbfft 6.737 kg kgmol 252 1 Kkgmol kJ 314.8 g kg 1000 1 g10 3 3 ONHC 9286 p

      = 2 in lbf 2.314p

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.

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.()sCNCO5OH4ONHC

229286

+++→

Then for each constituent (again ignoring solid carbon) we have

• )cb

TN p i i -V  =

So we can write

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