Wing Commander Physics and Bernoulli's Principle?

[Bigwillyd]

"Wing Commander Physics" and Bernoulli's Principle?

Hey guys, I was wondering if you could lend me a hand on how to deal with a particular situation I have. I have an assignment regarding bad physics in movies, and the movie I chose is really an easy one to bash: "Wing Commander" (1999). If any of you have seen it, it's a real joke. So far I've it's been more than easy to mathematically disprove the movie's horrible physics, but the "cargo bay depressurization" scene is rather tricky.

If you havn't seen the movie, I'll bring you up to speed on this particular scene: A small tear opens up in one of the ship's cargo bay's bulkheads and naturally as Bernoulli's Principle states the air inside the cargo bay is blown out into space through the hole at very high speeds as the two pressure systems try to balance out. What I'm trying to find out here, is a very rough appriximation of the velocity of air being blown out of the oriface.

I've gathered the givens which I believe are needed:
Pressure in space = 0 (approximately)
Pressure inside the cargo bay= 101325 Pascals (atm)
Density of air inside the cargo bay = 1.2 kg/m^3 (average density at 20 deg. C in Earth atmosphere)
Velocity of air flowing out of the oriface = ?

Ultimately what I need here is a very rough equation which will allow me to get velocity. If any of you guys can help me tackle this bad boy, a link to another thread where this situation has already been discussed, or maybe even just a URL on how to solve this myself, I would be very appreciative. Again, all I need is a simple equation to give me a very rough approximation; no integrals please!

Thanks again, -Willy

 

[russ_watters]

The is a basic Fluids 101 question. The answer is that you can apply a simplified version of Bernoulli's equation equating the static pressure in the ship (real spaceships aren't pressurized to 1 atm, but that's ok for your scenario) to the velocity pressure of the air in the orifice. So then Bernoulli's equation reduces to:

p=1/2 Rho * V^2

Rho is the density of the air. Btw, this is the real equation. Simple enough for ya!?

[edit] If you care about possible complications, you can say that in an orifice, there is a certain efficiency and loss due to the sides of the orifice not being optimized for good flow. But if the orifice is large enough, the air in the middle, at the very least, will follow Bernoulli's equation almost exactly.

 

[FredGarvin]

You can complicate it further by the fact that the delta P across the orifice is more than enough so the flow would most likely be choked at the orifice.

 

[russ_watters]

Is it? Huh, I thought of that but didn't do the calculation so I didn't realize it was that high. (click-click-click) Yeah, how 'bout that! Not too much over, though.

Also, I just checked and the space shuttle and ISS run at atmospheric pressure, but reduce to 10.2psi for a day before spacewalks to avoid the bends. Reading further, the Apollo craft operated at 2 psi and 100% oxygen (after Apollo 1, it switched during ascent from an oxygen/nitrogen mixture at sea level pressure). Gemini used 5.2 psi.

 

[Ben Niehoff]

Also, I just checked and the space shuttle and ISS run at atmospheric pressure, but reduce to 10.2psi for a day before spacewalks to avoid the bends.

So, when they reduce to 10 psi, where does the extra air go? Because presumably, they would want to re-pressurize again later...

 

[Bigwillyd]

Thanks for the equation Russ. I plugged everything in and got a big number for speed like 450 m/s! In reality the air would most likely become choked at the oriface and reduce this value, but at the immediate start before the cabin air pressure decreases it should roughly still apply. Where P=101325 Pa and p=1.2 kg/m^3,

101325=(1/2)(1.2)v^2
v=449.76 m/s I believe.