Leakage of Air in a Spacecraft

[flyusx]

Homework Statement

A spacecraft measuring 8 m by 3 m by 3 m is struck by space junk leaving a circular hole of radius 4 mm. About how much time is left to patch the leak? Assume an ideal gas and STP (1 mol of atoms occupies 22.4 L).

Relevant Equations

Leakage: $$\frac{NA\bar{v}}{4V}$$
Boltzmann Distribution: $$\Pr(v)=4\pi\left(\frac{m}{2\pi k_{B}T}\right)^{\frac{3}{2}}v^{2}\exp\left(-\frac{mv^{2}}{2k_{B}T}\right)$$

The leakage equation is pretty straightforward to use once I found the average velocity: N is the number of atoms, A is the area of the hole, v-bar is the average velocity and V is the volume of the container.

To find the average velocity, I used the Boltzmann distribution. I set the temperature be 273.15 K. For mass, I used $$\frac{28.96 g}{1 mol}\frac{1 mol}{N_A atoms}\frac{1 kg}{1000 g}=\frac{4.808921063\cdot10^{-26} kg}{atom}$$I then used the Boltzmann distribution to find the average velocity by taking $$\frac{\int_{0}^{10000}v\Pr(v)\text{d}v}{\int_{0}^{10000}\Pr(v)\text{d}v}=446.8783872 m/s$$From the leakage formula, I just took the area of the hole to be $$A=\pi r^{2}=\pi\left(4\cdot10^{-3}\right)^{2}$$, N to be the number of atoms in a mol and V to be the volume of the container (multiplied by $10^{-3}$ to convert to cubic metres). This returned a leakage rate of $1.509739807\cdot10^{23}$ atoms per second.

To find the number of atoms in the spaceship, I used 1.293 kg/m^3 as the average density of air. $$\frac{1.293 kg}{m^{3}}72m^{3}\frac{28.96 g}{1 mol}\frac{1 mol}{N_A atoms}\frac{1 kg}{1000 g}=1.935902021\cdot10^{27}\text{ atoms}$$Dividing the total number of atoms by the leakage rate returns 213.7 minutes (over 3 hours). The textbook says the correct answer is about 2 hours.

What am I doing wrong?

 

[berkeman]

I don't know if it helps, but the leakage rate should decrease as the pressure inside the spacecraft drops due to leakage. I'm not able to tell if you've accounted for that in your work.

Also, if I'm in that spacecraft, I'd want to patch the hole before the air pressure inside dropped to zero. Does the problem statement say what the pressure can drop to in that "time to patch the hole"?

 

[flyusx]

I don't know if it helps, but the leakage rate should decrease as the pressure inside the spacecraft drops due to leakage. I'm not able to tell if you've accounted for that in your work.

Also, if I'm in that spacecraft, I'd want to patch the hole before the air pressure inside dropped to zero. Does the problem statement say what the pressure can drop to in that "time to patch the hole"?

I'm sorry, I should've been clearer. We ignore the drop in leakage rate and take the time to patch as until the ship is empty. Physics textbooks and their simplifying assumptions, I suppose.

 

[Steve4Physics]

Hi @flyusx. A few thoughts....

Homework Statement: A spacecraft measuring 8 m by 3 m by 3 m is struck by space junk leaving a circular hole of radius 4 mm. About how much time is left to patch the leak? Assume an ideal gas and STP (1 mol of atoms occupies 22.4 L).

A key word here is 'About'. It indicates that you don't need to work to great precision (such as $446.8783872 m/s$).

Relevant Equations: Leakage: $$\frac{NA\bar{v}}{4V}$$

Not familiar with that equation – can you give explanation or reference (in particular for the '4')?

The leakage equation is pretty straightforward to use once I found the average velocity: N is the number of atoms, A is the area of the hole, v-bar is the average velocity and V is the volume of the container.

The average velocity of the particles is zero because velocity is a vector and directions are random in 3D! I guess you mean average speed.

For an approximate calculation of this sort, you would typically use the RMS (root mean square) velocity or the average speed. There are simple, standard formulae for RMS velocity and for average speed which you may be expected to use (rather than using the Boltzmann distribution and integrating!).

We ignore the drop in leakage rate and take the time to patch as until the ship is empty. Physics textbooks and their simplifying assumptions, I suppose.

Ignoring the drop in leakage rate is (IMO) a terrible (unphysical) thing to do! Also, it wasn't mentioned in the original (Post #1) question, indicating that you didn't post the full/accurate question.

 

[flyusx]

Hi @flyusx. A few thoughts....


A key word here is 'About'. It indicates that you don't need to work to great precision (such as $446.8783872 m/s$).


Not familiar with that equation – can you give explanation or reference (in particular for the '4')?


The average velocity of the particles is zero because velocity is a vector and directions are random in 3D! I guess you mean average speed.

For an approximate calculation of this sort, you would typically use the RMS (root mean square) velocity or the average speed. There are simple, standard formulae for RMS velocity and for average speed which you may be expected to use (rather than using the Boltzmann distribution and integrating!).


Ignoring the drop in leakage rate is (IMO) a terrible (unphysical) thing to do! Also, it wasn't mentioned in the original (Post #1) question, indicating that you didn't post the full/accurate question.

Thank you very much. I haven't a clue as to where 1/4 comes from. My textbook effectively states that it's 1/4 for 'complicated math reasons we won't discuss here' and that a full derivation can be found in a statistical mechanics textbook. You are correct that I meant average soeed, not velocity (and not rms speed for the textbook explicitly states average speed).

I included all the decimal points because I was working inside Maple and would always just round off the answers at the very end instead of at intermediate steps.

I agree that ignoring the drop in leakage rate doesn't make sense. The problem, which I should have stated, states 'state what simplifying assumptions you make' which in my opinion is very vague.

 

[Steve4Physics]

I agree that ignoring the drop in leakage rate doesn't make sense. The problem, which I should have stated, states 'state what simplifying assumptions you make' which in my opinion is very vague.

First note that different ‘reasonable assumptions’ will lead to different answers. In fact 2 hours and 3 hours are pretty close! But your approach is based on incorrect physics so wouldn't gain marks.

The solution should start with a clear statement of any necessary reasonable assumptions. The assumption that the rate of particle loss is constant is (IMO) unreasonable.

My solution would be something like this (hope I’m not giving too much away):

Assumptions:

1) The occupant has no breathing apparatus but can remain conscious and work until the pressure is about 0.5atm.

2) The internal temperature remains constant so the internal pressure is proportional to the remaining number of air particles.

Strategy:

Use the given formula for rate of particle loss: $\frac {dN}{dt} = - \frac {NA \bar v}{4V}$. This describes a simple exponential decay process. $N(t)$ is easily found by integration.

Average particle speed is given by standard equation $\bar v = \sqrt {\frac {8RT}{\pi M}}$.

Using the equation for $N(t)$ find the time for the $N$ to halve (which corresponds to pressure falling from 1atm to 0.5atm). If you know how to find half-life from a decay constant, this will save a line or two of working.

If you can, work symbolically and evaluate at the end.

Edited as text and LaTeX got badly corrupted somehow.

 

[haruspex]

The assumption that the rate of particle loss is constant is (IMO) unreasonable.

Agreed; what the question is effectively asking for is the number of molecules divided by the initial rate of loss of molecules.

Here’s how I reasoned:
Let the width of the box in the x direction be $s_x$, the velocity of molecule $i$ in the x direction have magnitude $v_{ix}$, and the box volume be $B$; similarly y, z.
Let the hole be in a wall normal to the x axis. The hole's fraction of that wall is $\frac{As_x}V$.
(Typo fixed.)
Frequency at which a given molecule hits where the hole will appear is $\frac{v_{ix}}{2s_x}\frac{As_x}V=\frac{v_{ix}}{2}\frac{A}V$.
Rate of loss of molecules = $\Sigma_i\frac{v_{ix}}{2}\frac{A}V$.
If the overall speed of a molecule is $u_i$ then, on average, its x direction speed will be $\frac 12u_i$.
Average (over the molecules) rate of loss = $\frac{\bar{u_{i}}}{4}\frac{A}V$.
But note that $\bar{u_i}$ is not the RMS speed. Rather, it is calculated by the Boltzmann formula in post #1, which is there evaluated at 447m/s.
Based on this as the constant rate of loss, the time to lose all the molecules is $4\frac V{A\bar{u_i}}$, or about 3 hours.

 

[Steve4Physics]

what the question is effectively asking for is the number of molecules divided by the initial rate of loss of molecules.

Maybe! The (very badly posed) question simply asks "About how much time is left to patch the leak?" which, in the absence of any further information, is open to widely different interpretations.

Let the hole be in a wall normal to the x axis. The hole's fraction of that wall is $\frac{As_i}V$.

Typo'? $\frac{As_i}V$ should be $\frac{As_x}V$.

If the overall speed of a molecule is $u_i$ then, on average, its x direction speed will be $\frac 12u_i$.

Wouldn't the average x-direction speed be $\frac 1{\sqrt 3}u_i$ because $\vec u_i$ has three orthogonal components which are equal on average?

 

[gmax137]

I don't understand why this is being worked as a statistical mech problem. Wouldn't the air be driven out the hole by the pressure difference between the spacecraft (Po=1 atm) and the vacuum of space?

I tried using choked (sonic) flow (343 m/s at STP) through the 4 mm hole; I get 4.6 hours to empty the spacecraft. Assuming the velocity doesn't drop off. I was surprised that my result is longer than the ~3 hours being calculated above.

 

[Steve4Physics]

I don't understand why this is being worked as a statistical mech problem.

I'd guess it's a teaching-exercise set in the context of the material being taught.

Wouldn't the air be driven out the hole by the pressure difference between the spacecraft (Po=1 atm) and the vacuum of space?

Yes - initially. The initial pressure-difference is 1atm but immediately starts to fall.

I tried using choked (sonic) flow (343 m/s at STP) through the 4 mm hole; I get 4.6 hours to empty the spacecraft. Assuming the velocity doesn't drop off. I was surprised that my result is longer than the ~3 hours being calculated above.

I'd guess the exit velocity is more or less constant. But since the density of the air inside falls, the mass flow-rate will also fall.

 

[gmax137]

But since the density of the air inside falls, the mass flow-rate will also fall.

Yes, but that means that my 4.6 hours is a minimum time required. I am surprised that the time based on molecules bumping around and happening to hit the hole is less. Or have I misunderstood the posts above mine.

 

[haruspex]

Maybe! The (very badly posed) question simply asks "About how much time is left to patch the leak?" which, in the absence of any further information, is open to widely different interpretations.

Except that post #3 adds that the leak rate is to be treated as constant.

Typo'? $\frac{As_i}V$ should be $\frac{As_x}V$.

Yes, thanks.

Wouldn't the average x-direction speed be $\frac 1{\sqrt 3}u_i$ because $\vec u_i$ has three orthogonal components which are equal on average?

In two dimensions, the average value of $\cos(\theta)$ between $0$ and $\pi/2$ is $2/\pi$, not $1/\sqrt 2$. In 3D, we can use Archimedes' observation that a band on the surface of a sphere between $x$ and $x+\delta x$ has the same area as its projection onto the enclosing cylinder sharing the x axis.
https://mathspace.co/textbooks/syllabuses/Syllabus-451/topics/Topic-8309/subtopics/Subtopic-109173

Wouldn't the air be driven out the hole by the pressure difference between the spacecraft (Po=1 atm) and the vacuum of space?

Sure, but that’s just another approach. The molecular view is valid, and arguably more obviously so. Using concepts like pressure and energy is a macroscopic view, and can have limitations. If the answers don’t match, I would trust the reductionist view.