Rate of air pressure drop in a reservoir with a hole to a vacuum

AI Thread Summary
The discussion centers on estimating the time it takes for air pressure in a space capsule to drop by 30% after a micrometeorite creates a small hole. Key points include the capsule's volume and surface area calculations, the significance of the hole's size relative to the capsule, and the assumption that external pressure is zero. Participants suggest that the problem likely involves exponential decay and diffusion principles, while debating the appropriateness of Fick's law and gas dynamics equations. The conversation highlights the complexity of modeling the pressure drop, considering factors like the speed of sound and adiabatic conditions, with a call for further insights on solving the differential equations involved.
superiority
Messages
9
Reaction score
0

Homework Statement


A space capsule, which may be treated as a sphere of radius 10m, is hit by a micrometeorite which makes a hole of diameter 2mm in its skin. Estimate how long it will take for the air pressure to drop by 30%. Avogadro's number is about 6 \times 10^{23}

Homework Equations


PV^\gamma = k
J = -\D \frac{\partial \phi}{\partial x}
J = -\D\nabla\phi
\frac{\partial \phi}{\partial t} = D \frac{\partial^2 \phi}{\partial x^2}
Flux = -P \cdot A \cdot (c_1 - c_2)

The Attempt at a Solution


No gas would enter. Eventually, all the gas would leave. The conditions of the gas inside the capsule are unspecified; call pressure P, temperature T.

Volume of gas is \frac{4000\pi}{3} \approx 4190 \textrm{m}^3. Surface area of capsule is 400\pi \approx 1260 \textrm{m}^2. Area of hole is \approx \pi \times 10^{-6} \approx 3.14 \times 10^{-6} \textrm{m}^2. Area of hole is \approx 2.5 \times 10^{-7} \% of surface area of capsule.

Pressure outside the capsule is 0, volume of space may be treated as infinite (i.e. gas leaving the capsule will not be obstructed).

I'm pretty sure that the rate of transfer would be an exponential decay.

I'm not sure whether Fick's law for one dimension or many dimensions is appropriate; it's really only a movement in the direction perpendicular to the whole that will result in gas leaving the capsule. I think this is a diffusion problem, but I stuck adiabatic gas expansion law there just in case it's relevant.

I believe what I'm trying to find is \frac{\partial \phi}{\partial t}, the rate of change of concentration with respect to time. \frac{\partial \phi}{\partial t} = -\frac{\partial J}{\partial x} (I think). J is the amount of gas passing through the hole every second, which is 2.5×10-7% of all gas hitting the sides of the capsule every second.

I'm afraid that from here, I have no clue where to go. I have a hunch that I would integrate something to find the total proportion of gas hitting the sides every second, work out the proportion that would leave through the hole based on that, assume that loss of gas to space results in instantaneous uniform pressure drop throughout the capsule, get some equation out of that and solve for 30% drop in pressure. But I'm stuck. Any pointers?
 
Last edited:
Physics news on Phys.org
Perhaps you can determine the mass flow rate through the hole by considering with which very well-known characteristic speed (that is a function of pressure and density) the air would leave the hole with. Having the instantaneous mass rate should allow you to put up a differential relationship for pressure and (if you model the expansion as non-isothermal) temperature.

While I do agree with you that you will most likely end up with an exponential decay, I'm not sure you can get it from just looking at how likely air molecules are to hit the hole instead of the wall, since the mean free path of air at standard pressure is way lower than 10m, so air molecules are far more likely to hit each other than hitting the wall or the hole.
 
P = \frac{1}{3}\rho {v_{rms}}^2
\rho = \frac{m}{V}
P = \frac{m}{3V} {v_{rms}}^2
v_{rms} = \sqrt{\frac{3PV}{m}}
In some amount of time dt, if A_h is the area of the hole, the amount of mass dm that passes through the hole is
dm = A_h\cdot \sqrt{\frac{3PV}{m}}dt
Rearranging:
\sqrt{\frac{m}{P}}dm = A_h\cdot \sqrt{3V}dt
But I can't figure out how to write P in terms of m.
 
Wouldn't \frac{m}{P} actually be a constant? But that doesn't seem right at all.

Edit: Volume, not mass. Dammit.
 
Last edited:
I am by no means expert here, but assuming the volume of the air expands through the hole with the instantaneous speed of sound (like an expanding piston) and it does so adiabatically (pV^\gamma = p_0V_0^\gamma), I get a differential equation of the form

\dot{V} = A_h \sqrt{\frac{\gamma p_0 V_0^\gamma}{m}} V^{(1-\gamma)/2}

which seems solvable allowing the pressure to be calculated as a function of time. However, this also assumes \gamma is constant, which may be false when temperature varies.

Perhaps others here can provide some better advice on how to model and solve this problem?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top