Calculate thickness of a pressurized vessel

gioscarab

Hi guys.
I am trying to figure out how thick should be an aluminium pressurized vessel with a internal pressure of 0.125 pa or 1/8 bar. The shape of the box is a cylinder 130cm base x 170cm height. The maximum pressure difference is 0.125 bar inside -> 0 bar outside.
I know that i need to know the joint efficiency of the vessel, but i would estimate that generally.

gioscarab

Sorry i made a mistake writing data:
Internal pressure: 0.125 bar
Minimum external pressure: 0 bar
I forgot: The vessel will be made by laser cutted aluminium sheet, welded with TIG technology.
Please help.

Bandit127

I am assuming gauge pressure. So the outside is at normal air pressure and not a vacuum?

0.125 bar might be the operating pressure but what is the maximum permissible pressure? And what do you intend to design to make sure the maximum pressure is not exceeded?

If the pressure is absolute (i.e. your 0 bar is actually a vacuum) then you might want to plan for a leak to atmosphere from outside the system and, therefore, a pressure of 1 bar in the vessel.

If it will not exceed 0.5 bar then it does not fall in to the EU definition of a Simple Pressure Vessel.

gioscarab

Hi I am very pleased by your answer.
I am intellectually designing only as a proof of concept a vessel that has a minumum internal pressure of 0.125 bar (like 10.000m in atmosphere) with a minimum outside pressure of 0.012 bar so near vacuum. The maximum permissible pressure difference from inside to outside, with always less pressure outside could be 1.5x 0.125 bar -> 0.012 bar. I also propose aluminium material, and also TIG tenchnology for welding.

My goal is to estimate weight of the bare vessel structure.
I am not considering g forces and stress to put on the object, i am only need a estimation for thickness, after that i could estimate also weight and so on, i am stuck here :P

Q_Goest

The wall thickness for a cylindrical tube can be estimated from S = Pr/t
where
S = stress
P = Pressure (internal)
r = radius
t = thickness

Assuming a stress of 10,000 psi, the wall thickness is only around 0.005", so about the thickness of a human hair. The ends of the cylindrical vessel however, will be a bit more difficult if you have flat heads. To minimize thickness of the head material, they need to be spherical.

Basically, you end up with a balloon made out of aluminum. You could use aluminum foil, but there would be no way to weld it together. Glue would work. But the resulting balloon would have no structural strength and would collapse and tear without pressure inside it. Note that the Atlas rocket and its upper stage Centaur was actually built this way. They were a lot thicker but were rated for much higher pressure as well. The similarity though is that both your balloon and this rocket would collapse under their own weight without pressure inside.

tasp77

I recall a similar problem posed many years ago (late 70s) that was quite similar except for scale.

The pressure was 14.7PSI, and the diameter of the cylinder was a kilometer as I recall.

(yeah, it was in regards to a Gerard O'Neill space habitat).


IIRC, the thickness turned out to be around a meter, but I don't remember if that had a safety factor.


(and after 30 some years, I might be misremembering some of the details)

Fun recalling an old problem, thanx for the memories!