If you put a gas in a tube, and spin the tube [edit: along the axis of the cylinder], you will get a centrifuge.

What's the maximum pressure differential between edge and centre?

Asked another way: how close can we get to perfect vacuum, in a human-scale cylinder, with gas in the cylinder initially at atmospheric pressure or higher, and without liquidizing the gas or moving the cylinder walls at relativistic speeds? What're the variables we need to know?

Would Bernoulli’s Equation for a Rotating Fluid be the right way to tackle a compressible gas?

Cylinder diameter, and gas molecules per inch of the cylinder seem two obvious variables. With infinite diameter and a zero gas, we end up with a fairly good vacuum, which is why I ask about the biggest pressure differential, even though getting close to perfect vacuum in the centre of the cylinder is the goal.

You can increase the friction against the walls with fins or whatever, and start with the gas injected at an angle at high speed, or whatever explicitly-declared other handwaves necessary to maximize the differential in your answer.

  • $\begingroup$ Have you tried to model this yourself? It just involves a differential force balance radially in conjunction with the ideal gas law, subject to the constraint that the total mass of gas is fixed. $\endgroup$ – Chet Miller May 18 at 2:08
  • $\begingroup$ @ChetMiller I fear my knowledge of fluid dynamics is below highschool level, and math not much higher, but I'm willing to learn, given such pointers. It feels like you've already made some logical leaps from your deeper understanding, leaving me several steps behind. Can we ignore the lack of drag between the cylinder and the gas, and between the gas near the outside and those molecules in the center of the near-vacuum? Can we assume equilibrium in a rotating reference frame is the same as being static with a force gradient? Does ideal gas law hold for near-vacuum? $\endgroup$ – Dewi Morgan May 18 at 4:39
  • $\begingroup$ You mention: 'you can increase the friction'. While increasing the friction would help to arrive at the final state faster, I think it's useful to be aware that increasing the friction is not necessary. Any fluid that is being rotated around some axis at a constant angular velocity eventually settles in a state of solid body rotation. A state of solid body rotation is arrived at when convection has decayed to zero. $\endgroup$ – Cleonis May 18 at 7:45
  • $\begingroup$ @Cleonis response has answered most of your questions. $\endgroup$ – Chet Miller May 18 at 11:44

To a first approximation, imagine you have a gas in the tube at a particular temperature. That gives you the average $KE$ and therefore $v$. Then predict what acceleration would be necessary for items at that speed to not be able to reach a particular distance "up".

Assume $\text{N}_2$, a tube $1\text{m}$ from center to edge, and kept at $300\text{K}$.

$$E_k = \frac{3}{2} \frac{RT}{N_A} = 6.21 \times 10^{-21}\text{J}$$ $$v^2 = \frac{2E_k}{m} = 267000\text{m}^2/\text{s}^2$$ $$a = \frac{v^2}{2x} = 1.34 \times 10^5 \text{m}/\text{s}^2 = 13600g$$

Since many molecules are going faster than the average speed, you'd need a fair bit more than 13 000 g to keep the gas confined to a 1m region. For a rotating tube, the end would be subject to that acceleration with a rotation speed around 60 revolutions per second.

You're going to need a higher speed than that since not all the gas will be at the bottom of the tube, and you're going to have a long tail for the speed distribution of the molecules. 10 times that acceleration will probably get you a measurable difference in pressure.

  • $\begingroup$ Ah, I shoulda specified the rotational axis! I meant rotating along the axis of the cylinder, rather than perpendicular to it! Still, I think this gives essentially the same result, assuming a tube of 1m radius... right? $\endgroup$ – Dewi Morgan May 18 at 1:08
  • 2
    $\begingroup$ Mostly true. But spinning around the axis makes it harder for the gas to interact with the cylinder. You now only have drag from the wall which is much less efficient. $\endgroup$ – BowlOfRed May 18 at 1:32
  • $\begingroup$ That's what I was thinking. And as you near vacuum, the drag linkage between the wall and the gas in near-vacuum of the centre drops to approximately nil. And at some point, the gas at the edge becomes liquid. Which is why I don't think anything simple like Bernoulli will work here. Hence asking here. $\endgroup$ – Dewi Morgan May 18 at 3:51

The eventual rotation of the gas as a virtually rigid body (as per @Cleonis response) is equivalent to creating artificial gravity in the radial direction, with "gravitational acceleration" equal to $\omega^2 r$, where $\omega$ is the angular rate of rotation. So the differential force balance in the radial direction is given by the "barotropic equation," $$\frac{dp}{dr}=\rho \omega^2r$$where $\rho$ is the gas density at radial location r. This is determined from the ideal gas law: $$\rho=\frac{pM}{RT}$$If we combine these two equations, we obtain: $$\frac{1}{p}\frac{dp}{dr}=\frac{M\omega^2}{RT}r$$This equation can be integrated with respect to r to yield: $$p=p(0)\exp{\left(\frac{M\omega^2r^2}{2RT}\right)}$$where p(0) is the pressure at r = 0. The average pressure $\bar{p}$ in the gas is obtained by integrating between r = 0 and r = a, the radius of the cylinder: $$\pi a^2\bar{p}=\int_0^a{2\pi r p(r)dr}=\pi a^2p(0)\left[\frac{e^{\xi}-1}{\xi}\right]$$where $$\xi=\frac{M\omega^2a^2}{2RT}$$Therefore, the pressure at the center of the cylinder is related to the average pressure by: $$p(0)=\bar{p}\left[\frac{\xi}{e^{\xi}-1}\right]$$By conservation of mass, the average pressure is the same as the original pressure which existed in the cylinder prior to rotation. So the above equation can readily be used to calculate the pressure (i.e., the amount of vacuum created) at the center of the cylinder during rotation.

  • $\begingroup$ Thank you so much for fully defining all your terms :) I think my only question now is, what does the [square bracket] operator mean? $\endgroup$ – Dewi Morgan May 19 at 16:58
  • $\begingroup$ It's just a parenthesis for multiplication. Let's see you make the calculation now for a sample case that you dream up. $\endgroup$ – Chet Miller May 19 at 17:00
  • $\begingroup$ Lessee... I'm gonna have to make assumptions about units. angularVelocity('ω') = 2pi rad/s ; molarMassOfGas('M') = 30 g/mol (~air) ; cylinderRadius(a) = 1 m ; idealGasConstant(R) = 8 J/K/mol ; absoluteTemperature(T) = 300 K (~25C, room temp) ; averagePressure(pbar) = 100,000 Pa (~1 atm). Throwing those in, I get tempVar(ξ)= pi*pi/40, and centerPressure(p(0)) ~= 90,000. $\endgroup$ – Dewi Morgan May 19 at 18:47
  • $\begingroup$ Now, I'm not sure I used the right units, and I failed to get Wolfram Alpha to confirm units, but I'm hoping that's ~90k Pa, which gives a 10% drop in the centre of a 1m radius cylinder of air rotated at 1Hz. That looks within the right order of magnitude, but rather more than I'd intuitively expect. $\endgroup$ – Dewi Morgan May 19 at 18:47
  • $\begingroup$ Also, if I'm interpreting it right, p(0) is always a factor of pbar so as suspected can never be truly zero, but since it scales with e^ω^2, it suggests that it'd fall FAST. Plugging in 20pi rad/s I get an output of 0.00004747368 ... that's definitely lower than I'd intuitively suspect, and if true, makes me wonder why we don't use centrifuges to make vacuum chambers. Where'd I go wrong? :D $\endgroup$ – Dewi Morgan May 19 at 18:56

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