Thermodynamics Air Cannon

  • Thread starter RazerM
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  • #1
RazerM
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Homework Statement


1.75 J is required to fire a spherical pellet of radius 8.75 mm at a maximum velocity of 100 m/s

Find appropriate values for the inital pressure and volume and final pressure and volume


Homework Equations


Adiabatic Expansion
[tex]P_{1}V_{1}^{\gamma}=P_{2}V_{2}^{\gamma}[/tex]

[tex]\gamma=1.40\qquad P_{2}=10^5\text{ Pa}[/tex]

[tex]W_{\text{total}}=\frac{P_{1}V_{1}-P_{2}V_{2}}{\gamma-1}[/tex]

[tex]
\begin{align*}
W_{\text{total}} &= W_{\text{atm}}-W_{\text{useful}} \\ \frac{P_{1}V_{1}-P_{2}V_{2}}{\gamma-1}&=10^{5}(V_{2}-V_{1})-1.75
\end{align*}
[/tex]

The Attempt at a Solution


[tex]
\begin{align*}
V_{2}&=V_{1}\left( \frac{P_{1}}{P_{2}} \right)^{\frac{1}{\gamma}}\\
V_{2}&=V_{1}\left( \frac{P_{1}}{10^{5}} \right)^{\frac{1}{1.4}}\\
\end{align*}
[/tex]
It is then possible to rearrange work equation into
[tex]V_{1}\left( [P_{1}+4\cdot 10^{4}]-\left[ 1.4\cdot 10^{5}\left( \frac{P_{1}}{10^{5}} \right)^{\frac{1}{1.4}} \right] \right)=0.7[/tex]

[tex]\text{Then using an arbitary guess for }V_{1}\text{ and an iterative process to find }P_{1}\text{ I found }[/tex]
[tex]
V_{1} = 10^{-5}\text{ m}^{3}[/tex]
[tex]
P_{1} = 420465\text{ Pa}[/tex]
[tex]
V_{2} = 2.79\cdot 10^{-5}\text{ m}^{3}
[/tex]
Giving a ridiculous barrel length of
[tex]
\begin{align*}
L &= \frac{V_{2}-V{1}}{\pi r^{2}} \\
L &= 0.0744\text{ m}
\end{align*}
[/tex]


I'm not asking for a full solution; just a hint as to how to find out values that satify the criteria of work and have a reasonably small V_1 so that there can be many shots for say a 0.001m^3 tank
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
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The pellet travels down a cylindrical barrel of length L whose cross-sectional area is [itex]A = \pi r^2[/itex] where r = 8.75 mm. So the final volume is LA. The minimum final pressure is 1 atm. Choose a reasonable length for the barrel, say .75 m. Find an initial volume and pressure such that the work done in adiabatically expanding to volume LA is equal to 1.75 J.

AM
 
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