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I think about a rotating torus (simplified tire) filled with ideal gas. Mass of gas is $m$ and molar mass is $M$. Pressure in non rotating torus is $p_0$. Temperature is constant $T$. Inner radius of torus is $r$ and outer radius is $R$.

Then the cylinder begins to rotate with angular velocity $\omega$

How can I derive difference of pressures near the wall between inner wall and outer wall? Do I need more data (for a simple model)?

If someone knows how to make it numerically, It would be also interesting for me.

I would also appreciate if someone could tell which book should I read to know more about solving such problems.

Bernhard
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user46147
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Actually, in your rotating torus your are mimicking gravity, which is point outward. I.e. the outside of the torus acts as a floor. The centrifugal acceleration would be $g\approx\omega^2 R$. This $g$ plays the same role as the gravitational acceleration on liquid or gas pressures under normal gravity conditions, so you can say $\Delta p = \rho g h$, or in your system $\Delta p =\rho \omega^2 R r$.

Actually, I have made quite a lot of assumption on the way, which can make it more difficult if you want to include these effects

  1. $r\ll R$, such you can assume the pseudo-gravity to be constant in the inner tube.
  2. The system is in equilibrium, i.e. the gas is moving with the tube, and does not need additional acceleration.
Bernhard
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  • Are you absolutely sure about your answer? It seems that units don't fit. – user46147 Jul 21 '14 at 20:39
  • @user46147 Which unit did do you think are incorrect? $\omega$ in $1/s$, $R$, $r$ in $m$, $g$ in $m/s^2$ and $p$ in $Pa$. – Bernhard Jul 21 '14 at 20:42
  • I think in your first assumption you meant to say that R-r << R (or, practically equivalently, r), correct? Because the pseudo-gravitational acceleration varies with the radius, you would want the walls of the torus to be very close together relative to its radius. Other than that, I think this is a very good approximation if the system is in equilibrium. – Hausdorf Jul 21 '14 at 20:43
  • @Hausdorf I understand the confusion. I had this in mind: http://www.onlineconversion.com/images/object_surfacearea_torus.png – Bernhard Jul 21 '14 at 20:47
  • Oh that makes more sense. Thanks for the clarification. – Hausdorf Jul 22 '14 at 16:19
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For a proper answer I think you will need to look at research papers on gas centrifuges which look at radioisotope separation of gases for enrichment of nuclear fuels.

I see two big problems with Bernhards answer (for some reason my comment was deleted before, that's quite disgraceful that you can't question the validity of a science answer...)

Apart from the assumption that the effective gravitational force is constant (which it isn't as it goes from the max to zero at the center), what about your assumption that the gas is in-compressible? As the density of air is dependent on pressure/altitude, it's mass is variable, hence you cannot use a formula with a single value for 'm'. As a result your approximations would be way off.

If my comment gets deleted again, then this place is a joke. How can you delete a response from someone pointing out critical flaws in an answer. I presume the point of a public forum is to prevent the spread of misinformation, not to blindly accept an answer because someone has reputation points...

Paul
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I did this calculation last night for a friend who is interested in centripetal compressors. The ideas behind Bernard's solution will help you understand conceptually, but he misses the point that you have an ideal gas. Density isn't constant and will be greater as you move outward. You will need to do an integration from inside radius r to outside radius R. For an incompressible fluid like water at radius r take the pressure to be P and at r+dr it is P+dP. The net force is a radially inward force causing the centripetal acceleration and is due to the difference in pressure. F=AdP=mw^2r. The mass can be taken as ρ times volume which is ρAdr. Hence dP=ρw^2rdr. To get the pressure difference you would integrate this with ρ constantand find that for an incompressible fluid in increases as r^2. Now for an ideal gas density can be obtained from the ideal gas law. PV=NkT. ρ=(MW)N/V=(MW)P/kT. Using this we can create a differential equation. dP=ρw^2rdr=((MW)P/kT)w^2rdr. this means dP/P=(MW w^2/kT )rdr. Integrating both sides ln Pout-ln Pin =Constant r^2dr. I think you can finish from there. Hope my lack of formatting is okay. Best, Nathan Heston

Nathan Heston
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