How quickly would a space ship vent air?

Question

In a lot of scifi media, holes are often blown into the sides of ships, sending objects or people tumbling into space. I am curious about how to generalise this as an equation, but cursory research has found me no way to calculate these forces.

Imagine a room of arbitrary size, temperature and air pressure. A hole of Xm² surface area opens to an infinitely large vaccuum.

How can i calculate:
A. The pressure in the room after y seconds?
B. The velocity and displacemnt of an object of defined mass and surface area that was within the room after y seconds, ignoring aerodynamics?

Answer 1

There are online calculators that will provide a reasonable result if your leak is relatively small. Fpr example https://toolbox.tlv.com/global/TI/calculator/air-flow-rate-through-orifice.html suggests that a 1 atmosphere room venting into space through a 10 mm diameter hole will lose approximately 1.5 m^3 of air per minute. So a 5x5x5 m cube would lose a bit more than 1% of its pressure in the first minute. Air loss will be approximately proportional to remaining pressure. So pressure loss rate would would slow down over time.

That rate actually matches my real-world experience where I am witnessing the reverse situation. I.e. filling a vacuum space of ~ 1 m^3 via a 10 mm diameter port with air. That takes approximately 1 minute, to fill the chamber with air.

The approximation probably breaks down if the hole is too large. For example, the same calculator suggests that the hypothetical 5x5x5 m room would take 20 ms to lose half its pressure if one wall was completely removed. That seems a little optimistic to me.

Wrt forces and velocities etc. If it is a big hole, then the air (and anything in it that isn't tied down) will potentially accelerate to around the speed of sound in the vicinity of it. Forces on a body will be similar, but worse than those experienced in a large tornado. Probably not much chance of holding on to anything to avoid being sucked out if you were anywhere near it.

On the other hand, at risk of losing the digit, you could plug a 10 mm hole with your finger. If you, more sensibly, used the flat of your palm to seal it while someone arrived with a suitable patch, you might get away with no worse than a nasty blister.

Answer 2

To answer A, lets start from basic ideal gas law equation,

$$ \tag 1 p = m\frac {RT}{VM} ,$$

Now differentiate with respect to time both sides (assuming $T=\text{const}$),

$$ \tag 2 \frac{dp}{dt} = \frac {RT}{VM}\cdot \frac{dm}{dt},$$

$\dot m$ is gas mass flow rate (negative in this case) through the hole, so lets sub it into (2),

$$ \tag 3 \frac{dp}{dt} = -\frac {RT}{VM}\cdot \rho v A,$$

As per same ideal gas law,- density of gas is simply $\rho=pM/RT$, using it in (3) and canceling terms we get,

$$ \tag 4 \frac {dp}{dt} = -\frac {pvA}{V} ,$$

Re-arranging (4) and then integrating both sides, while resolving integration constant in LHS

$$ \tag 5 \int \frac {dp}{p} = - \int_0^t \frac {vA}{V} dt,$$

gives (assuming constant gas flow speed $v$, which not necessarily true, but as a first approximation),

$$ \tag 6 \ln\left(\frac{p}{p_0}\right) = - \frac {vA}{V} t,$$

This gets us to the final equation of pressure in the room with hole vs time : $$ \tag 7 p = p_0 \exp\left(- \frac {vA}{V} t\right) ,$$

where $A$ is hole area, and $V$ is room volume.

Exercises for the post author:

1. Do not assume temperature a constant, since when room will be open to the vacuum, room temperature should equalize to the vacuum temperature, which is $2.7K$,- if we take as a reference CMB. In this case one needs to use product rule of differentials in (2) and then somehow to resolve $dT/dt$, i.e. finding function of $T(t)$.

2. In integration (5),- do not assume that gas flow speed from the room hole into vacuum is constant, since when pressure will decrease, gas outflow speed $v(t)$ should also decrease too and so after finding flow speed profile vs time we need to use that expression in (5) integration, instead of just assuming it constant and moving it out of integral.

3. Discard ideal gas law and use real gas equation, like Van der Waals equation to see what happens when gas molecules are considered not a point particles.

In the end,- real $p(t)$ dependence can be a way more complex and probably follows exponential law only for the first moments of flow, then it should fallback to other mechanisms.