Ballistics: The Theory and Design of
Ammunition and Guns 3rd Edition
Solutions Manual Part 0
Donald E. Carlucci
Sidney S. Jacobson
** Immediate Download
** 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 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 p = 292 lbf2
in
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 = m g RT (IG-4)
Rearranging, we have
m g RT
p=
V
Here we go
1 kg kJ 1 kgmol ft − lbf
(10)[g ]
(8.314 ) (737.6 ) (12) in (1000)[K ]
1000 g kgmol ⋅ K 252 kg C6 H8 N 2O9 kJ ft
p=
(10)[in 3 ]
lbf
p = 292 2
in
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.
C 6 H 8 N 2 O 9 → 4H 2 O + 5CO + N 2 + C(s )
Then for each constituent (we ignore solid carbon) we have
, N i ℜT
pi =
V
So we can write
kgmol H 2O kgmol C 6 H 8 N 2O 9 1 kg
(4)
(8.314 )
kJ
(1000)[K ] 1 [
(10) g C 6 H 8 N 2 O 9 ] 1,000
C6 H8 N 2O9
kgmol C 6 H 8 N 2O 9 kgmol - K 252 kg C 6 H 8 N 2O 9 g C6 H8 N 2O9
p H 2O =
[ ]
(10) in 3 1 kJ 1 ft
737.6 ft − lbf 12 in
lbf
p H 2O = 1,168 2
in
kgmol CO kgmol C6 H8 N 2O9 1 kg C 6 H8 N 2O9
(5)
(8.314)
kJ
(1000)[K ] 1 [
(10) g C6 H8 N 2O9 ]
kgmol C6 H8 N 2O9 kgmol - K 252 kg C6 H8 N 2O9 1,000 g C6 H8 N 2O9
p CO =
[ ]
(10) in 3 1 kJ 1 ft
737.6 ft − lbf 12 in
lbf
p CO = 1,460 2
in
kgmol N 2 kgmol C6 H8 N 2O9 1 kg
(1)
(8.314)
kJ
(1000)[K ] 1 [
(10) g C6 H8 N 2O9 ] 1,000
C6 H8 N 2O9
p N2 = kgmol C6 H8 N 2O9 kgmol - K 252 kg C6 H8 N 2O9 g
C6 H8 N 2O9
[ ]
(10) in 3 1 kJ 1 ft
737.6 ft − lbf 12 in
lbf
p N 2 = 292 2
in
Then the total pressure is
p = p H 2O + p CO + p N 2
lbf lbf lbf lbf
p = 1,168 2 + 1,460 2 + 292 2 = 2,920 2
in in in in
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
,Answer: p = 314.2 lbf2
in
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 ) = m g RT (VW-2)
Rearranging, we have
m g RT
p=
V − cb
Here we go
1 kg kJ 1 kgmol ft − lbf
(10)[g ]
(8.314 ) (737.6 ) (12) in (1000)[K ]
1000 g kgmol ⋅ K 252 kg C6 H8 N 2O9 kJ ft
p=
3
[ ]
(10) in 3 − (10)[g ] 1 kg (2.2) lbm (32.0) in
1000 g kg lbm
lbf
p = 314.2 2
in
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.
C 6 H 8 N 2 O 9 → 4H 2 O + 5CO + N 2 + C(s )
Then for each constituent (again ignoring solid carbon) we have
N i ℜT
pi =
(V - cb )
So we can write
, kgmol H 2 O kgmol C6 H8 N 2O9 1 kg C 6 H8 N 2O9
(4)
(8.314)
kJ
(1000)[K ] 1 [
(10) g C6 H8 N 2O9 ]
kgmol C6 H8 N 2O9 kgmol - K 252 kg C6 H8 N 2O9 1,000 g C 6 H8 N 2O9
p H 2O =
(10) in 3 − (10)[g ] 1 kg (2.2) lbm (32.0) in 1 kJ 1 ft
3
[ ]
1000 g kg lbm 737.6 ft − lbf 12 in
lbf
p H 2O = 1,257 2
in
kgmol CO kgmol C 6 H 8 N 2O 9 1 kg C 6 H 8 N 2 O9
(5)
(8.314 )
kJ
(1000)[K ] 1 [
(10) g C 6 H 8 N 2 O 9 ]
kgmol C 6 H 8 N 2O 9
kgmol - K 252 kg C 6 H 8 N 2O 9 1,000 g C 6 H 8 N 2 O9
p CO =
(10) in 3 − (10)[g ] 1 kg (2.2) lbm (32.0) in 1 kJ 1 ft
3
[ ]
1000 g kg lbm 737.6 ft − lbf 12 in
lbf
p CO = 1,571 2
in
kgmol N 2 kgmol C 6 H 8 N 2O 9 1 kg C 6 H 8 N 2 O 9
(1)
(8.314 )
kJ
(1000)[K ] 1 [
(10) g C 6 H 8 N 2O 9 ]
kgmol C 6 H 8 N 2O 9 kgmol - K 252 kg C 6 H 8 N 2O 9 1,000 g C 6 H 8 N 2 O 9
p N2 =
(10) in 3 − (10 )[g ] 1 kg (2.2 ) lbm (32.0 ) in 1 kJ 1 ft
3
[ ]
1000 g kg lbm 737.6 ft − lbf 12 in
lbf
p N 2 = 314 2
in
Then the total pressure is
p = p H 2O + p CO + p N 2
lbf lbf lbf lbf
p = 1,257 2 + 1,571 2 + 314 2 = 3,142 2
in in in in
Problem 3 – A hypothetical “air mortar” is to be made out of a tennis ball can using a
tennis ball as the projectile. The can has a 2-1/2” inside diameter and is 8” long. If a
tennis ball of the same diameter weighs 2 oz. and initially rests against the rear of the can,
to what air pressure must one pressurize the can to in order to achieve a 30 ft/s launch of
the tennis ball? Assume that the tennis ball can be held against this pressure until
released, that it perfectly obturates and also assume an isentropic process and ideal gas
behavior with γ = 1.4 for air.
, Solution: First we need to get some parameters in order. We start with a chamber
volume. Since the tennis ball rests against the bottom of the can, the chamber volume is
the volume of a cylinder of 2-1/2” diameter by 1-1/4” length minus the volume of half a
sphere of 2-1/2” diameter.
The empty chamber has a volume of
π
U= (2.5)2 [in 2 ](1.25)[in ] = 6.136[in 3 ]
4
π π
Vsphere = d3 = (2.5)3 [in 3 ] = 8.181[in 3 ]
6 6
Vsphere
V0 = U −
2
[ ]
V0 = (6.136 ) in 3 −
(8.181) [in 3 ] = 2.045[in 3 ]
2
There are several ways to solve this problem – the simplest one is to follow the method I
described in the notes. We can determine our fake chamber length through
V0 = Al
V0
l=
A
l=
(2.045)[in 3 ] = 0.417[in ]
π
(2.5)2 [in 2 ]
4
We shall assume muzzle exit of the projectile occurs as the equator of the tennis ball
passes through the plane of the end of the can. Then the travel of the projectile is the
length of the can less half the diameter of the tennis ball
L = (8)[in ] −
(2.5) [in ] = 6.75[in ]
2
The formula for muzzle velocity given an isentropic expansion of air was provided in the
notes as equation (IG-28)
2mg RTi l (γ −1)
Vm =
m p (1 − γ )
[(l + L)(1−γ )
− l (1−γ ) ] (IG-28)
Ammunition and Guns 3rd Edition
Solutions Manual Part 0
Donald E. Carlucci
Sidney S. Jacobson
** Immediate Download
** 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 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 p = 292 lbf2
in
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 = m g RT (IG-4)
Rearranging, we have
m g RT
p=
V
Here we go
1 kg kJ 1 kgmol ft − lbf
(10)[g ]
(8.314 ) (737.6 ) (12) in (1000)[K ]
1000 g kgmol ⋅ K 252 kg C6 H8 N 2O9 kJ ft
p=
(10)[in 3 ]
lbf
p = 292 2
in
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.
C 6 H 8 N 2 O 9 → 4H 2 O + 5CO + N 2 + C(s )
Then for each constituent (we ignore solid carbon) we have
, N i ℜT
pi =
V
So we can write
kgmol H 2O kgmol C 6 H 8 N 2O 9 1 kg
(4)
(8.314 )
kJ
(1000)[K ] 1 [
(10) g C 6 H 8 N 2 O 9 ] 1,000
C6 H8 N 2O9
kgmol C 6 H 8 N 2O 9 kgmol - K 252 kg C 6 H 8 N 2O 9 g C6 H8 N 2O9
p H 2O =
[ ]
(10) in 3 1 kJ 1 ft
737.6 ft − lbf 12 in
lbf
p H 2O = 1,168 2
in
kgmol CO kgmol C6 H8 N 2O9 1 kg C 6 H8 N 2O9
(5)
(8.314)
kJ
(1000)[K ] 1 [
(10) g C6 H8 N 2O9 ]
kgmol C6 H8 N 2O9 kgmol - K 252 kg C6 H8 N 2O9 1,000 g C6 H8 N 2O9
p CO =
[ ]
(10) in 3 1 kJ 1 ft
737.6 ft − lbf 12 in
lbf
p CO = 1,460 2
in
kgmol N 2 kgmol C6 H8 N 2O9 1 kg
(1)
(8.314)
kJ
(1000)[K ] 1 [
(10) g C6 H8 N 2O9 ] 1,000
C6 H8 N 2O9
p N2 = kgmol C6 H8 N 2O9 kgmol - K 252 kg C6 H8 N 2O9 g
C6 H8 N 2O9
[ ]
(10) in 3 1 kJ 1 ft
737.6 ft − lbf 12 in
lbf
p N 2 = 292 2
in
Then the total pressure is
p = p H 2O + p CO + p N 2
lbf lbf lbf lbf
p = 1,168 2 + 1,460 2 + 292 2 = 2,920 2
in in in in
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
,Answer: p = 314.2 lbf2
in
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 ) = m g RT (VW-2)
Rearranging, we have
m g RT
p=
V − cb
Here we go
1 kg kJ 1 kgmol ft − lbf
(10)[g ]
(8.314 ) (737.6 ) (12) in (1000)[K ]
1000 g kgmol ⋅ K 252 kg C6 H8 N 2O9 kJ ft
p=
3
[ ]
(10) in 3 − (10)[g ] 1 kg (2.2) lbm (32.0) in
1000 g kg lbm
lbf
p = 314.2 2
in
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.
C 6 H 8 N 2 O 9 → 4H 2 O + 5CO + N 2 + C(s )
Then for each constituent (again ignoring solid carbon) we have
N i ℜT
pi =
(V - cb )
So we can write
, kgmol H 2 O kgmol C6 H8 N 2O9 1 kg C 6 H8 N 2O9
(4)
(8.314)
kJ
(1000)[K ] 1 [
(10) g C6 H8 N 2O9 ]
kgmol C6 H8 N 2O9 kgmol - K 252 kg C6 H8 N 2O9 1,000 g C 6 H8 N 2O9
p H 2O =
(10) in 3 − (10)[g ] 1 kg (2.2) lbm (32.0) in 1 kJ 1 ft
3
[ ]
1000 g kg lbm 737.6 ft − lbf 12 in
lbf
p H 2O = 1,257 2
in
kgmol CO kgmol C 6 H 8 N 2O 9 1 kg C 6 H 8 N 2 O9
(5)
(8.314 )
kJ
(1000)[K ] 1 [
(10) g C 6 H 8 N 2 O 9 ]
kgmol C 6 H 8 N 2O 9
kgmol - K 252 kg C 6 H 8 N 2O 9 1,000 g C 6 H 8 N 2 O9
p CO =
(10) in 3 − (10)[g ] 1 kg (2.2) lbm (32.0) in 1 kJ 1 ft
3
[ ]
1000 g kg lbm 737.6 ft − lbf 12 in
lbf
p CO = 1,571 2
in
kgmol N 2 kgmol C 6 H 8 N 2O 9 1 kg C 6 H 8 N 2 O 9
(1)
(8.314 )
kJ
(1000)[K ] 1 [
(10) g C 6 H 8 N 2O 9 ]
kgmol C 6 H 8 N 2O 9 kgmol - K 252 kg C 6 H 8 N 2O 9 1,000 g C 6 H 8 N 2 O 9
p N2 =
(10) in 3 − (10 )[g ] 1 kg (2.2 ) lbm (32.0 ) in 1 kJ 1 ft
3
[ ]
1000 g kg lbm 737.6 ft − lbf 12 in
lbf
p N 2 = 314 2
in
Then the total pressure is
p = p H 2O + p CO + p N 2
lbf lbf lbf lbf
p = 1,257 2 + 1,571 2 + 314 2 = 3,142 2
in in in in
Problem 3 – A hypothetical “air mortar” is to be made out of a tennis ball can using a
tennis ball as the projectile. The can has a 2-1/2” inside diameter and is 8” long. If a
tennis ball of the same diameter weighs 2 oz. and initially rests against the rear of the can,
to what air pressure must one pressurize the can to in order to achieve a 30 ft/s launch of
the tennis ball? Assume that the tennis ball can be held against this pressure until
released, that it perfectly obturates and also assume an isentropic process and ideal gas
behavior with γ = 1.4 for air.
, Solution: First we need to get some parameters in order. We start with a chamber
volume. Since the tennis ball rests against the bottom of the can, the chamber volume is
the volume of a cylinder of 2-1/2” diameter by 1-1/4” length minus the volume of half a
sphere of 2-1/2” diameter.
The empty chamber has a volume of
π
U= (2.5)2 [in 2 ](1.25)[in ] = 6.136[in 3 ]
4
π π
Vsphere = d3 = (2.5)3 [in 3 ] = 8.181[in 3 ]
6 6
Vsphere
V0 = U −
2
[ ]
V0 = (6.136 ) in 3 −
(8.181) [in 3 ] = 2.045[in 3 ]
2
There are several ways to solve this problem – the simplest one is to follow the method I
described in the notes. We can determine our fake chamber length through
V0 = Al
V0
l=
A
l=
(2.045)[in 3 ] = 0.417[in ]
π
(2.5)2 [in 2 ]
4
We shall assume muzzle exit of the projectile occurs as the equator of the tennis ball
passes through the plane of the end of the can. Then the travel of the projectile is the
length of the can less half the diameter of the tennis ball
L = (8)[in ] −
(2.5) [in ] = 6.75[in ]
2
The formula for muzzle velocity given an isentropic expansion of air was provided in the
notes as equation (IG-28)
2mg RTi l (γ −1)
Vm =
m p (1 − γ )
[(l + L)(1−γ )
− l (1−γ ) ] (IG-28)
