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How can I calculate the velocity of air that would come out of a pierced can of compressed air if I assume that the pressure inside the can is 7 atm and the size of the hole is that of a regular nail?

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The flow approaching the exit hole inside the can will converge toward the exit hole, primarily with spherical symmetry. In spherical coordinates, the radial Navier Stokes equation (neglecting viscosity) will be: $$\rho v \frac{dv}{dr}=-\frac{dp}{dr}$$where $\rho$ is the density. Therefore, $$vdv=-\frac{dp}{\rho}$$Assuming adiabatic reversible expansion in the converging flow, we have $$\frac{p}{\rho^{\gamma}}=const$$or, equivalently, $$\frac{\rho}{\rho_0}=\left(\frac{p}{p_0}\right)^{1/\gamma}$$So, $$\frac{dp}{\rho}=\frac{p_0^{1/\gamma}}{\rho_0}\frac{dp}{p^{1/\gamma}}$$So, integrating, we have: $$\frac{1}{2}v^2=\frac{\gamma}{(\gamma-1)}\frac{p_0}{\rho_0}\left(1-\left(\frac{p_f}{p_0}\right)^{(\gamma-1)/\gamma}\right)$$Substituting $\rho_0=\frac{p_0M}{RT_0}$yields $$\frac{1}{2}v^2=\frac{\gamma}{(\gamma-1)}\frac{RT_0}{M}\left(1-\left(\frac{p_f}{p_0}\right)^{(\gamma-1)/\gamma}\right)$$If this exceeds the speed of sound at the exit conditions, then the exit velocity will be the speed of sound.

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  • $\begingroup$ How valid is your assumption of negligible viscosity? $\endgroup$
    – nluigi
    Aug 21, 2018 at 7:08
  • $\begingroup$ Pretty good as lone as the predicted velocity does not exceed the speed of sound. Otherwise, there will be a shock wave. $\endgroup$ Aug 21, 2018 at 12:50
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The method proposed below is rather primitive and I am not sure how accurate the result would be, but that's how I would attempt to do it, if I did not want to deal with equations.

You can ballpark the initial velocity (flow rate) using a calculator (like this) for the air flow through an orifice.

As the air is coming out of the can, the internal pressure and the flow rate would decay.

If you want to chart the flow rate decay curve over time, you can calculate the flow rate for a number of time steps by figuring the percentage of the air loss and, correspondingly, the pressure drop in each step, based on the flow rate determined in the previous step and the volume of the can.

For instance, if, based on the initial flow rate, you determine that, in $1$ second, $10$% of the air mass has escaped the can, you can assume that the pressure inside the can has dropped from $7$atm to $6.3$atm and use this new pressure level to calculate the flow rate for the next step.

Then you can actually perform the test, measure the time it takes for all excessive air to escape and compare it with the time indicated by the curve. If the difference is significant, you can adjust the diameter of the orifice entered into the calculator or use a calculator based on a better model until the curve fits the test.

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