Example Problem with Complete Solution

4C-4 : Muzzle Velocity of a Bullet Fired From an Air Pistol
6 pts
An air pistol contains compressed air in a small cylinder, as shown below. Assume that the volume of the air in the cylinder behind the bullet is 1 cm3 and that it is at a pressure of 1 MPa and a temperature of 27°C immediately before firing. The bullet, mB = 15 g, acts as a piston. It isintially held in place by a pin. When the trigger is pulled, the pin retracts, the air expands isothermally and the bullet moves along the barrel of the gun. If the air pressure in the cylinder is 0.1 MPa just as the bullet leaves the gun barrel, estimate the following:
           
a.) The mass and final volume of the air in the cylinder.        
b.) The work done by the air within the cylinder on the bullet and the work done ON the ambient air (outside the gun) by the bullet.
 
c.) The  velocity of the bullet when it leaves the gun barrel (muzzle velocity).  
                   
Boldly assume that the process is a quasi-equilibrium process.
   
  Figure 1.

Read :
  • We must assume that the process is a quasi-equilibrium process.  This is not a great assumption, but it does yield a reasonable 1st approximation of the muzzle velocity of the bullet.
  • We can use the compressibility EOS to show that, despite the molar volume, the gas behind the bullet actually behaves as an ideal gas.  This fact allows us to solve the problem using the Ideal Gas EOS.
  • We can determine the work done on the bullet by the air behind it using the relationship for boundary work done by an ideal gas as it expands isothermally.
  • The bullet does work on the surrounding air in a constant pressure process.  So, we can evaluate this work term using the formula for isobaric compression of an ideal gas.
  • Finally, we can apply the 1st Law to the bullet.  There is no heat exchanged and no change in the internal energy or potential energy of the bullet.  The only remaining terms are the two work terms we already know how to determine and the change in kinetic energy.  The initial velocity is zero, so the only unknown left in the 1st Law equation is the final velocity of the bullet as it leave the barrel of the gun !
Given: V1 =
1.0E-06
m3 P2
100
kPa
P1
1000.0
kPa
T1 =
27
°C TC =
132.5
K
mB
0.015
kg PC
3770
kPa
Find: V2 =
???
m3 Won surr
???
J
Won bullet
???
J v2 =
???
m/s

Assumptions:
  • For purposes of computing the work done on the bullet, you may treat the air inside the cylinder as an ideal. This is not entirely accurate because the initial pressure is so high.
  • Assume that the air in the barrel is initially in an equilibrium state.
  • Assume that the air in the barrel as the bullet leaves the gun is also in an equilibrium state.
  • Assume the process is isothermal.
  • For estimation purposes, assume that the process is a quasi-equilibrium process.  This assumption will yield the maximum muzzle velocity that the bullet could attain.
  • Assume that the gas within the system is a closed system until the bullet leaves the gun.
Part a.) Begin by using the initial state to determine the number of mole of air inside the barrel.  This remains constant until the bullet leaves the gun and that is the time interal in which we are interested.
Use the compressibility EOS : Equation.
Eqn 1
or :
Equation. Eqn 2
Reduced temperature and pressure are required in order to use the compressibility charts to determine the compressibility, Z :
Equation. Eqn 3 Equation. Eqn 4
R =
8.314
J/mol-K Z1 =
1
(Because PC
is so high)
TR1 =
2.27
PR1 =
0.265
n =
4.01E-04
moles
TR2 =
TR1
Z2 =
1
(Because PC
is so high)
PR2 =
0.027
Since Z = 1 throughout the process, it is safe to treat air as an ideal gas throughout this process.
This is a surprise since the molar volume is 2.5 L/mole and that is less than 5 L/mole.
The process is assumed to be isothermal and we discovered that the air could be treated as an ideal gas.  Therefore:
 
Equation. Eqn 5 or : Equation. Eqn 6
Answer: V2 =
1.00E-05
m3
m = n MW Eqn 7 MWair =
29
g/mole
Answer: m =
0.0116
g
Part b.) Next, we can calculate the work done by the air, on the bullet using the work equation derived for isothermal processes like this one:
 
Answer: Equation. Eqn 8 Won bull 2.303 J
The bullet does work on the surrounding air against a constant restraining pressure , Patm.
Therefore:
Answer: Equation. Eqn 9 Won surr
0.900
J
Part c.) In order to determine the muzzle velocity of the bullet, we must determine the change in the kinetic energy of the bullet as a result of the net amount of work done on it.
We can do this by applying the 1st Law, using the bullet as our system.
Equation. Eqn 10
Assumptions: - If the temperature of the bullet remains constant, then its internal energy does not change.
- Changes in the gravitational potential energy of the bullet are negligible, especially if the gun is fired horizontally !
- Heat transfer to or from the bullet is negligible if the the process is isothermal.
Solution : The net work is the difference between the work done by the bullet and the work done by the bullet.
 
Equation. Eqn 11
Wnet =
-1.403
J
 
Answer: Solving for v2 : Equation.
Eqn 11
v2 =
13.7
m/s
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