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Suppose this example:

enter image description here

If I only apply a rule of $F = (P_1-P_2)A$ then every layer of the second box must resist less force right.

But I feel must be something wrong with this reasoning, because if that's true we could build a recipient with vacuum inside made only with a thin paper. We just need to create multiple layers of papers with a small pressure gradient between them, until we reach the $0$ atm.

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  • $\begingroup$ Well, yes, you can do that, making a horribly complex mess that needs exquisite control to pump down. And all that surface area is a real problem for real vacuum systems. And you never get to actual 0 pressure. Holding off 1 atm just isn’t that hard from a mechanical design perspective. It isn’t even considered a pressure vessel by ASME standards. $\endgroup$
    – Jon Custer
    Dec 30, 2019 at 16:03
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    $\begingroup$ With all those layers of paper... how much would it weigh compared to just making it from a solid of the same material that withstands the same pressure? I'm not sure if this would really be that helpful. $\endgroup$
    – JMac
    Dec 30, 2019 at 17:22
  • $\begingroup$ paper was an extreme, but what if we have only two or three layers, or metal, for applications like the vacuum airship: en.wikipedia.org/wiki/Vacuum_airship they say is impossible because even diamond is not strong enough $\endgroup$
    – Enrique
    Dec 30, 2019 at 17:26
  • $\begingroup$ @Enrique You would have to see if those designs actually used less weight than an un-layered design of the same mass. That's more engineering than physics though. Although each layer does reduce the stress on the material, it's not clear if that reduced stress can offset the increase in mass required for additional layers. Also, thin layers are prone to failure from buckling. $\endgroup$
    – JMac
    Dec 30, 2019 at 17:30
  • $\begingroup$ Actually that was the question (that's why I also painted with different widths the boxes in my example). I guess it all depends if it's a linear relation or not. I mean, if we increment the pressure difference by 2 then we need to increment the size of the wall by 2 too? $\endgroup$
    – Enrique
    Dec 30, 2019 at 18:10

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