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One popular trope in science fiction is the astronaut who (deliberately or inadvertently) punctures an air tank or something and goes rocketing off.

How would we calculate the amount of force that would be exerted by eg a 1cm round puncture in a vessel containing breathable air at 500mbar?

This is basically a cold gas thruster, but the equations I found for those assume you know the exhaust velocity, whereas in this case we know only the pressure inside the vessel and the fact that there's vacuum outside. Bernoulli's principle falls down in this case since the gas is decompressing (running the numbers assuming constant density gave me an exit velocity exceeding the speed of sound).

So what is the right way to approximate the newtons you'd get from blowing a hole in an airlock? (For the simplicity of the thought experiment let's say the hole is perfectly round and the gas is regular air and we can ignore turbulence.)

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  • $\begingroup$ I don't know much about fluid flow, but "puncture" sounds like a traumatic event (e.g., punctured by a meteoroid), and that suggests that the shape of the nozzle is going to be unpredictable. That's unfortunate because the shape of the nozzle is going to play a significant role in determining both the magnitude and the direction of the impulse. $\endgroup$ – Solomon Slow Oct 5 '15 at 21:29
  • $\begingroup$ @jameslarge I meant it to connote "a perfectly circular aperture suddenly and magically appears" to avoid getting caught up in complications around nozzles and their shapes and gradually opening valves and so on. $\endgroup$ – Crashworks Oct 5 '15 at 21:31
  • $\begingroup$ @jameslarge It's a problem in that the Bernoulli equations are incorrect at fluid speeds > 0.3mach. $\endgroup$ – Crashworks Oct 5 '15 at 21:36
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    $\begingroup$ As a round approximation, it would just be the pressure times the area of the hole. Of course, if it's off-center, which it will be, the astronaut will also spin wildly. $\endgroup$ – Mike Dunlavey Oct 6 '15 at 0:19
  • $\begingroup$ @MikeDunlavey Indeed he would. ;) $\endgroup$ – Crashworks Oct 6 '15 at 1:15
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A good approximation would be to assume that the air leaving the hole is exiting at approximately sonic velocity, unless the hole miraculously opens up in the shape of a converging/diverging nozzle. Under that assumption, the force involved is equal to the mass flow rate per second multiplied by the velocity of the exiting fluid stream. Since the mass flow rate is density multiplied by cross-sectional area of the hole, multiplied by exit velocity, the force will decrease as the tank pressure falls, and it should decrease as an exponential decay. Note that the assumption of sonic velocity is supposedly good until the ratio of pressures (high/low) goes below approximately 2:1, which means that the assumption should hold until the tank "runs dry".

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If we assume that the air in that conditions is a perfect gas, we can compute the exhaust velocity.

Let's assign the state ① to the condition before the hole and ② to the condition outside the perforation.

In the hypothesis of energy conservation from state ① to state ②,the total enthalpy per unit mass is hence conserved:

$$ h_0 = c_pT + \frac{u^2}{2} $$

being $c_p$ the specific heat at constant pressure.

If the total enthalpy is conserved we can write: $$ h_{01} = h_{02} \rightarrow c_pT_{01} = c_pT_2 + \frac{u_2^2}{2} $$

$T_{01}$ is the total (or stagnation) temperature in the state ①.

We can now isolate the velocity of the gas that is flowing out from the hole $u_2$: $$ u_2 = \sqrt{2c_p(T_{01}-T_2)} = \sqrt{2c_pT_{01}\left( 1-\frac{T_2}{T_{01}} \right)} $$

with minor rearrangements using the equation of adiabatic for the temperature ratio: $$ \frac{T_2}{T_{01}} = \left(\frac{p_2}{p_{01}}\right)^\frac{k-1}{k} $$

we get the final equation for the exhaust velocity: $$ u_2 = \sqrt{2c_pT_{01}\left(1-\left(\frac{p_2}{p_{01}}\right)^\frac{k-1}{k}\right)} $$

if we are in space we can assume that: $$ p_2 \simeq 0 $$

Now, we need to know what are the values of $p_{01}$ and $T_{01}$. They are so-called total quantities and are related to the speed of the gas we're considering at that state:

$$ T_{01} = T_1\left( 1 + \frac{k-1}{2}M_{01}^2 \right) $$ $$ p_{01} = p_1\left( 1 + \frac{k-1}{2}M_{01}^2 \right)^\frac{k}{k-1} $$

We can moreover assume that the velocity of the air inside the vessel is negligible, so that the Mach number $M_{01}$ is $\simeq$ 0, therefore the total pressure and total temperature are the same as the static ones.

The only unknown we have now is the static temperature inside the vessel $T_1$. To compute it we need to know at least the density of the gas inside the vessel or, equivalently, volume and mass of the air.

$$ T_1 = \frac{p_1}{\rho_1 \mathcal{R}_\text{air}}$$

Once got the exhaust velocity the thrust is proportional to it.

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  • $\begingroup$ Sorry, what are the units of the (u^2)/2 term in the first enthalpy equation? $\endgroup$ – Crashworks Oct 7 '15 at 0:28
  • $\begingroup$ @Crashworks the units are $\frac{J} {kg}$. The enthalpy is intended per unit of mass. $\endgroup$ – DiTTiD Oct 7 '15 at 1:09

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