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Inside the atmosphere, I guess the atmospheric pressure negates the pressure exerted by the gas that's inside a gas cylinder. However, when taken to a vacuum (space), Will the gas cylinder burst due to the pressure exerted on its internal walls by the gas that is present inside it?

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    $\begingroup$ If one extra atmosphere of pressure difference causes a gas canister to burst, it is either a very, very unsafe gas canister, or someone is filling it far in excess of its rated pressure. $\endgroup$
    – notovny
    Nov 11, 2021 at 12:40
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    $\begingroup$ Re, "...atmospheric pressure negates the pressure...inside a gas cylinder." Depends what you mean by "negates." The pressure that the cylinder must contain is the difference between the absolute pressure inside and the absolute pressure outside. At Earth's surface, the pressure acting on the outside is in the neighborhood of 15 psi or 100 kPa. The pressure inside a high-pressure gas bottle can be thousands of psi (tens of MPa). $\endgroup$ Nov 11, 2021 at 12:41

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The stresses in the wall of a gas cylinder can be estimated by modeling it as a thin-walked pressure vessel. The so-called hoop stress is $\Delta Pr/t$, where $\Delta P$ is the pressure difference, $r$ is the radius of the cylinder, and $t$ is the wall thickness. For a cylinder filled to a pressure of 100 bar, for example, the move from Earth’s surface (1 bar) to space (0 bar) would increase the stress by only 1%. As noted in the comments, the cylinder would have to already be at the threshold of failure for this increase to be important. Related scenarios can be considered for different values of pressure and factor of safety.

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  • $\begingroup$ Why it is merely geometric and does not depend on the canister material? Or it is just intended to be, to be used in conjunction with the material characteristics? $\endgroup$
    – Alchimista
    Nov 11, 2021 at 14:43
  • $\begingroup$ Yes, you compare the output stress with a material characteristic (in this case, the strength). It's very typical for mechanics-of-materials problems such as this one to depend on the loading, the geometry, and the material. This offers various ways to achieve success (e.g., change the load type, optimize the geometry, select an appropriate material). $\endgroup$ Nov 11, 2021 at 18:06

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