| 4C-4 : | Muzzle Velocity of a Bullet Fired From an Air Pistol | 6 pts |
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| 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. | |||||||||
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Read : |
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| 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: |
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| 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 : |  |
Eqn 1 |
or : |
 |
Eqn 2 | ||||
| Reduced temperature and pressure are required in order to use the compressibility charts to determine the compressibility, Z : | |||||||||
 |
Eqn 3 |  |
Eqn 4 | ||||||
| R = | 8.314 |
J/mol-K | Z1 = | 1 |
(Because PC is so high) |
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| TR1 = | 2.27 |
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| PR1 = | 0.265 |
n = | 4.01E-04 |
moles | |||||
| TR2 = | TR1 |
Z2 = | 1 |
(Because PC is so high) |
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| PR2 = | 0.027 |
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| 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: | |||||||||
 |
Eqn 5 | or : |  |
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: |  |
Eqn 8 | Won bull | 2.303 | J | ||||
| The bullet does work on the surrounding air against a constant restraining pressure , Patm. | |||||||||
| Therefore: | |||||||||
| Answer: |  |
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. | |||||||||
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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. | ||||||||
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Eqn 11 | ||||||||
| Wnet = | -1.403 |
J | |||||||
| Answer: | Solving for v2 : |  |
Eqn 11 |
v2 = | 13.7 |
m/s | |||
Example Problem with Complete Solution
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