,
,
,
, 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.
,
,
, 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.
