# How rapid would air loss be in the event that a supermassive torus is punctured?

Artist's conception of a Stanford Torus. [Source]

Consider a Stanford Torus of radius $4km$, with the inner radius of the torus ring being $1km$. Disregard structural stability issues. To simulate $1g$ of centripetal acceleration, it would rotate at around $0.5 rpm$ generating a rotational speed of ~$200m/s$ at the edge of the torus. This structure is then filled with air to maintain $1 atm$ of pressure, with a mixture of gases matching Earth's own atmosphere.

What would happen if the hull is punctured? How rapidly would air escape from the structure? How would this variable change depending on the size and location of the puncture? How long would it take for the entire structure to fully decompress?

• Homework problem? – Organic Marble Nov 4 '15 at 18:14
• @OrganicMarble This is a problem far more complicated than what my school would give as homework ;) No, it's a project I'm working on. I personally don't know the mathematics that go into calculating air venting through a hole of any certain size, so I came here for help. – Vedant Chandra Nov 4 '15 at 18:23
• Try googling for compressible flow equations and/or choked flow. The wikipedia article on choked flow has the equation you seek, then you can calculate the rest for yourself. – Organic Marble Nov 4 '15 at 18:33
• "How rapid would air loss be in the event that a supermassive torus is punctured?" For a small puncture, in a structure of that size? Minimal. At least, that is what the engineering students who worked on the initial designs determined. – Andrew Thompson Nov 4 '15 at 22:14
• First off, the loss would be quite proportional to the puncture size. You can't expect the same results from a millimeter micrometeorite and from a space shuttle crashing into it and creating a 300-meter long breach as the rotation drags it through the coating. – SF. Nov 6 '15 at 10:06

Plugging my values into the Wolfram Alpha Venturi flow equation calculator, I got my result:

320,000 litres per second


Formula:

$$Q = (\pi (D_1^2 \sqrt{\frac{2 (P_1-P_2)}{\rho}}\over(4 \sqrt{\frac{D_1^4}{D_2^4}-1))}$$

Q | flow rate
rho | fluid density
P_1 | upstream pressure
P_2 | downstream pressure
D_1 | upstream pipe diameter
D_2 | downstream pipe diameter
(Q is the flow rate as measured by a Venturi flow meter)

I cannot verify this answer as correct.

• Got a link? Can you show the formula? – kim holder Nov 5 '15 at 19:49
• What value did you use for the size of the puncture? – called2voyage Nov 5 '15 at 20:02
• @called2voyage I used the larger diameter as 1000 metres, and the smaller diameter as 1 meter. – Vedant Chandra Nov 5 '15 at 20:45
• 1 meter does not sound like a small puncture, and 1000 meters is huge. – called2voyage Nov 5 '15 at 20:48
• Are you using this: wolframalpha.com/input/?i=Venturi+flowmeter&lk=3 ? – called2voyage Nov 5 '15 at 21:23