Pressure of stagnant fluid volume in rotating cylinders

In summary, the fluid pressure over the radial distance from the OD of the rotating shaft to the ID of the non-rotating housing decreases rapidly as the Reynolds number increases.
  • #1
RedCoat999
4
0
I have a high speed rotating shaft (~30,000 rpm). Outside the shaft I have a relatively stagnant volume of fluid surrounded by a non-rotating cylinder (housing).

I am looking to find a 'simple' laminar flow model to describe the pressure of the fluid fat any point radially from the OD of the rotating shaft to the ID of the non-rotating housing wall. I have researched online and found many references to Taylor vortices and the like, but this is too high-level.

Thank you in advance.
 
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  • #2
RedCoat999 said:
I have a high speed rotating shaft (~30,000 rpm). Outside the shaft I have a relatively stagnant volume of fluid surrounded by a non-rotating cylinder (housing).

I am looking to find a 'simple' laminar flow model to describe the pressure of the fluid fat any point radially from the OD of the rotating shaft to the ID of the non-rotating housing wall. I have researched online and found many references to Taylor vortices and the like, but this is too high-level.

Thank you in advance.

The simplicity is proportional to the assumptions you make.

The assumption of laminarity depends on how large is your Reynolds number, which should be defined upon the inner linear velocity and the gap between cylinders. Turns out to be that the laminar solution becomes unstable depending how large is the so-called Taylor Number. For some Taylor Number a bifurcation appears and Taylor cells or Taylor Vortices can be seen in laboratory. The flow is no longer laminar and the assumption of no radial and no vertical velocity is no longer valid.

The easiest model is the viscous model. Assuming low Reynolds numbers and low Taylors numbers, you only have azimuthal component of the velocity. The equations of motion are reduced to the radial laplacian of the azimuthal component to be zero, and the radial gradient of pressure to be compensated by the centrifugal force. Such a solution is as I said unstable for large Taylor numbers.

PS: the surrounding fluid is not stagnant at all. It is moving because of the viscous transport of momentum, and for long times after the start, the viscous effect has reached the whole fluid in the gap.
 
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  • #3
Clausius, I got to tell you something. Even when it's a subject that I know nothing about (as with most of your answers), it is an absolute delight to read your posts. You, my friend, are elegant.
 
  • #4
Taylor Instability

Danger said:
Clausius, I got to tell you something. Even when it's a subject that I know nothing about (as with most of your answers), it is an absolute delight to read your posts. You, my friend, are elegant.

:blushing: That's the nicest thing that a friend has told me in a long, long time (let's not talk about girls).
 
  • #5
Get a room you two.

Nice post Clausius.
 
  • #6
FredGarvin said:
Get a room you two.
I don't think that's what he meant by not talking about girls. :tongue:
 
  • #7
I have a high speed rotating shaft (~30,000 rpm). Outside the shaft I have a relatively stagnant volume of fluid surrounded by a non-rotating cylinder (housing).
By that, I would take one to mean, no axial flow.

As Clausius indicated, the fluid is from from stagnant, but there will be boundary conditions, and a radially dependent azimuthal velocity component between the inner rotating cylinder and the stagnant one.

In velocity of the fluid on the inner surface must be equal to the linear velocity of the rotating shaft, i.e. [itex]v_{\theta}[/itex](r) = [itex]\omega[/itex]R1, r = R1, where [itex]\omega[/itex] is the angular velocity of the rotating cyliner and R1 is the radius. At the outer surface, [itex]v_{\theta}(r)[/itex] = 0, for r = R2. Then one solves the momentum equation for the fluid.
 
  • #8
FredGarvin said:
Get a room you two.

Nice post Clausius.


Yeah, is Holidays Inn ok with you Danger? :rofl:

I forgot to say that here the Taylor Number is just the square of the Reynolds.
 
  • #9
Clausius2 said:
Yeah, is Holidays Inn ok with you Danger? :rofl:
I prefer Super-8, but I'm too old to be picky. :biggrin:
 
  • #10
Thank you Astronuc and Clausius2 for your responses. Since I am more of a nuts and bolts engineer can either of you point me to some equations for the 'simple' model (laminar flow) to solve for pressure over the radius?
Thanks again.
 
  • #11
I have found some equations on page 111, §3-2.3 of "Viscous Fluid Flow", 2nd Ed. Following their methodology and using MathCad to cheat at the inegration! I am getting the pressure dropping off very quickly (and going negative) from the non-rotating OD (housing). The Reynolds numbers are very high (100000 near the OD, and over 1 million at the OD). I am thinking I need to bug my busy CFD buddies to get a solution.
 
  • #12
RedCoat999 said:
I have found some equations on page 111, §3-2.3 of "Viscous Fluid Flow", 2nd Ed. Following their methodology and using MathCad to cheat at the inegration! I am getting the pressure dropping off very quickly (and going negative) from the non-rotating OD (housing). The Reynolds numbers are very high (100000 near the OD, and over 1 million at the OD). I am thinking I need to bug my busy CFD buddies to get a solution.

The pressure must increase radially, and it will be negative depending on the reference pressure you chose.

I don't have any information to work out the Reynolds number, but if it is 100000 according to your calculations, simply your viscous flow solution does not exist on reality, it would become unstable before reaching any steady state.

Try to do the same with fully turbulent flow...may be it hasn't been done ever (congratulations!)
 
  • #13
What kind of fluid? Gas (air) or liquid?

At 30,000 rpm, I don't imagine liquid would be practical.

Also what is the gap between the OD of the shaft and ID of housing?
 
  • #14
The liquid has a density of ~50 lb/ft^3 and viscosity of ~1.5 cs. The gap between the shaft and the housing is ~0.5 inch
 

1. What is the pressure of a stagnant fluid volume in a rotating cylinder?

The pressure of a stagnant fluid volume in a rotating cylinder is dependent on the radius of the cylinder, the angular velocity of rotation, and the height of the fluid column. It can be calculated using the equation P = ρω²r², where P is pressure, ρ is the density of the fluid, ω is the angular velocity, and r is the radius of the cylinder.

2. How does the pressure change with increasing rotation speed?

As the rotation speed increases, the pressure in the stagnant fluid volume also increases. This is because the centrifugal force acting on the fluid also increases, causing it to push outward and increase the pressure on the walls of the cylinder.

3. Does the shape of the cylinder affect the pressure of the stagnant fluid volume?

Yes, the shape of the cylinder does affect the pressure of the stagnant fluid volume. For a given rotation speed, a narrower cylinder will have a higher pressure than a wider one, as the centrifugal force acting on the fluid is greater in the narrower cylinder.

4. What happens to the pressure if the fluid volume is not perfectly stagnant?

If the fluid volume is not perfectly stagnant, there will be additional forces acting on the fluid, such as friction and turbulence. This can cause the pressure to vary slightly from the calculated value, but the overall trend of increasing pressure with increasing rotation speed will still apply.

5. Can this equation be applied to any type of fluid?

The equation P = ρω²r² is derived from the principles of fluid mechanics and can be applied to any type of fluid, as long as the assumptions of stagnant flow and uniform rotation are met. However, for non-ideal fluids, such as highly viscous or compressible fluids, additional factors may need to be taken into account.

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