Rate of Air Loss Through a Hole in a Spaceship

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SUMMARY

The discussion focuses on calculating the rate of air loss from a cylindrical space vehicle when a meteoroid punctures a hole in it. The derived formula for the rate of loss of air molecules is given as $$\left| \frac{dN}{dt} \right| \approx \frac{a \left| v_x \right|_{avg}}{2V} N$$, where 'a' is the hole's area, 'v_x' is the average x-velocity of the air molecules, and 'V' is the volume of the cylinder. The solution involves considering the fraction of molecules passing through a representative slice of the cylinder and adjusting for the hole's position. The discussion also highlights the importance of using the average velocity of the molecules for accurate calculations.

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This discussion is beneficial for aerospace engineers, physicists, and students studying fluid dynamics or thermodynamics, particularly those interested in the effects of structural breaches in spacecraft.

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Homework Statement



We want to calculate the rate of air loss from a space vehicle (module) if a meteoroid punches a hole in i
t. Assume the module is sealed off from other modules. It is shaped like a cylinder roughly 4 m in diameter and 10 m long. The hole's area is a cm2. The hole is punched in one of the end faces of the cylinder.

Show that the rate of loss of air molecules through the hole is

$$\left| \frac{dN}{dt} \right| \approx \frac{a \left| v_x \right|_{avg}}{2V} N$$

2. Homework Equations [/B]

The Attempt at a Solution



Here's my thinking on this. Please point out any reasoning flaws.

I take a representative slice of the cylindrical module, of width dx, and ask how many molecules dN' pass through this slice in a time dt. Since the molecules are assumed to be very many in number and their motion is assumed to be randomly distributed, the fraction of all the molecules in this slice can be taken as the same as the fraction of the volume of the whole module taken up by this slice. So:

$$\frac{dN'}{N} = \frac{Av_x dt}{V}$$

where vx is the average x-velocity of a molecule. But I know that I will only want half of this, since the escape hole is only on the one side. So I divide by 2:

$$\frac{dN'}{N} = \frac{Av_x dt}{2V}$$

Now I don't want all of these molecules, either. I only want the fraction that will pass through the hole. The expression I have so far is how many would be lost if the entire cylinder end were open, i.e. just one big hole. So I fix this by multiplying by the fraction a/A, where A is the area of the entire end of the cylinder.

So my final result is
$$ \frac{dN'}{dt} = \frac{a v_x N}{2V}$$

And since N' are the only molecules that can be lost (since molecules passing through other slices aren't lost) I can say dN' = dN. Which gives me the result I want.

My only reservation here is that I haven't dealt carefully with the averaged velocity. [/B]
 
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So if ##v_x## if the velocity of a random particle, it wouldn't have the average value. It might be too high or too low. I think you need the average velocity to be correct most of the time, as it were. I think that's why they say it's approximately the average. I suppose if there is any sound in the capsule, there will be areas with molecules traveling at different speeds.
 

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