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How would pressure of an ideal gas be distributed over the inside of a capsule (a cylinder with semi-spheres on the ends)? What about the strain on the material? Is there a general formula for how much?

PixelArtDragon
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1 Answers1

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If the cylinder is stationary, it is according to the hydrostatic pressure equation:

$\vec{\nabla} p = \rho \vec{g}$

You can derive this equation by eliminating all velocities in the Navier-Stokes equations. If gravity is oriented in the negative $y$ direction, the equation becomes:

$\frac{dp}{dy} = -\rho g$

EDIT:

If you integrate this equation you get an expression for the pressure at any point in the cylinder:

$p = -\rho g y + C$

where $C$ is a constant that you pick to satisfy boundary conditions. Basically the points on the surface of the capsule that are closer to earth have higher pressure than the points that are further away. (Think about what your ears feel if you dive to the bottom of a swimming pool.)

As for the distribution over the semi-sphere ends, the pressure acts normal to the surface. Using this fact you can integrate the equation for the pressure over these surfaces to get the forces.

OSE
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  • This doesn't say how it would be distributed over the different parts. I'm looking for how much force would be on the two different types of sections. – PixelArtDragon May 13 '13 at 15:05
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    Even after the edit it doesn't help. Your formula is for liquids; I specified an ideal gas. – PixelArtDragon May 13 '13 at 15:19
  • @Garan It is the same for liquids and gases as long as the continuum assumption applies. Gases are typically much less dense than liquids so the pressure at the bottom of the capsule won't be much different than at the top. – OSE May 13 '13 at 15:28
  • The Navier-Stokes equations don't apply to what I'm asking about. They refer to some form of motion; I'm referring to a static system. – PixelArtDragon May 13 '13 at 15:33
  • Here is another example of this phenomenon with air instead of water: If you start on the bottom floor of a tall building and ride the elevator to the top, your ears will pop due to the change in pressure as you go up. – OSE May 13 '13 at 15:36
  • The Navier-Stokes equations are simply conservation of momentum applied to a fluid, i.e. $\vec{F} = m\vec{a}$. Newton's second law is used all the time for static problems. Just set $\vec{a}= 0$. Eliminating the terms with velocities in the Navier-Stokes equations does the exact same thing. – OSE May 13 '13 at 15:38
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    This answer may be more complicated than the questioner intends---for small objects you generally ignore the vertical variation due to gravity in questions that specify an ideal gas---but it is correct. The pressure is constant plus a small correction. The thing that may be confusing the OP is that thought the pressure is everywhere (almost) the same the strain on the enclosing material is not. But that is not the question. – dmckee --- ex-moderator kitten May 13 '13 at 16:20