Friction concentric rotating cylinders 1

[Adrian77]

Hello All,

I am trying to calculate the friction (torque) between two concentric cylinders.
Inside one is roating at 750 rpm, diameter is 826 mm and diametrical clearance is 4 mm.
Outside cylinder is static. Length 98 mm.
I guess it should be quite large but couldn't find any reference.
Maybe you can give some advice or reference.

Thanks in advance.

 

[nickelkid]

Adrain77,

This may not be the best forum for this post. You may want to repost this in "Mechanical engineering other topics"

 

[BigInch]

Is there some fluid between the two cylinders?

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"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[Adrian77]

Yes, I am sorry .

I forgot to mention that there is sea water in between.

Adrian

 

[hydtools]

If there is clearance, no contact, there is no friction.

Ted

 

[BigInch]

Use the viscosity of the fluid to calculate the shear force per unit area of rotating surface. Take that force/unit area and multiply by the area of the rotating cylinder, then multiply by the radius to get the torque at the shaft.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[Adrian77]

Hi biginch,
Thanks for your comment but I have already done that. I expect higher torque due to the boundary layer. I guess this cannot be calculated like a normal surface in free flow.

 

[BigInch]

normal surface in free flow?

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[btrueblood]

It is a variant of Couette flow, known as Taylor-Couette flow. Tried wiki, but they don't give equations.

My fluids book gives the following (only valid for laminar region, so you will have to look up Taylor instabilities and determine if your particular rot. speeds/Reynolds numbers put the flow in a turbulent region or not):

T = Torque per unit length
w = rotary speed of inner cylinder, in rad/sec
u = kinematic viscosity (units N-m/s^2)
Rs = radius of inner shaft
Ro = radius of outer wall

T = 4*pi*u*w*(Rs)^2 /[1 - (Rs)^2/(Ro)^2]

If there is flow within the annulus, you have an entirely different flow regime. Also, the above ignores end wall effects, which tend to increase the drag...

...basically, for a real-world case, there is no good way to calculate the flow a priori, and you should find references to test data (or correlations to test data) to get better numbers. But, the flow you describe has been studied a great deal over the past centuries, and you should be able to find a lot of info. in a good engineering library.

 

[Bribyk]

Is the water acting as lubricant to form a hydrodynamic bearing or is the inner cylinder supported by other means? You could try Petroff's Equation for Bearing Friction and see how well this lines up with your gut and the Taylor-Couette results.

 

[unclesyd]

Your question is a variant of the flow region in a Tesla turbine or pump. Here some info on how this flow at the rotating element is handled.

Here is a paper on a rotor-stator disc flow that analyzes the flow region at the rotating disc.

http://journals.cambridge.org/downl...46a.pdf&code=7a65521e76e66d7ead870197ef071dd8

Here is a practical application of this principle in the form a Discflow pump. I'm looking around for my old literature that gave horsepower requirements on these pumps.

http://www.discflo.com/technology.htm

http://www.advancedfluid.com/discflo/concepts.htm

 

[sailoday28]

I believe btrueblood is correct with Coette flow.
Another approximation is to consider the gap so small compared to the radius in order to use the following simplification:

v=r*omega where r is the radius of the inner cylinder and omega the rotational speed, radians per unit time.

Shearing force is
tau=mu*dv/dy where mu is the dynamic viscosity and dv/dy is the velocity gradient between the cylinders.

dv/dy =v/clearance where clearance is the gap

Follow the suggestion of biginch to calculate the friction.

To calculate torque: The intensity of the shear, tau is equal to the torque divided by the raidus and the area of either cylinderical surface.

My reference for the above is "Elementary Mechanics of Fluids, by Rouse.

 

[btrueblood]

Sailoday,

Your analysis is good and our equations mostly agree, but in the equation I gave (which is derived using polar coordinates) there is a term that goes as 1/gap^2, not 1/gap as in your equation. Dunno which is right...

In the derivation, the book shows a thin annular layer of fluid, and notes that if the net torque on the thin layer is zero (i.e. steady state operation), then d(r^2*tau)/dr = 0, and goes on to derive from that tau = u*r*d(v/r)/dr ...

But I got stuck when they say the net torque on a thin ring of fluid is zero...I think they are right (i.e. the torque acting on the shaft is equal and opposite to that acting on the case, so the torque on the id of a thin layer equal the torque on the od), but haven't ground out the full derivation.

BTW, my "book" is Roberson & Crowe, Engineering Fluid Mechanics, 2nd ed., Houghton Mifflin Co, 1980.

 

[sailoday28]

btrueblood quotes are from your previous input
"then d(r^2*tau)/dr = 0, "
From Navier-Stokes with constant properties and neglecting elevation effects.
0 = d/dr[1/r(d/dr(rV)]
Where V is velocity in rotational direction, theta
r is radial direction
Integrating twice, noting that at Rs, V=wRs
at Ro, V=0

rv=w[r^2-Ro^2]/[1-(Ro/Rs)^2]
v/r=w[1-Ro^2/r^2]/[1-(Ro/Rs)^2]
" and goes on to derive from that tau = u*r*d(v/r)/dr ...
above should have a negative sign
d(v/r)= 2wRo^2/r^3/[1-(Ro/Rs)^2]
so that tau=2uwRo^2/r^2/[1-(Ro/Rs)^2]
tau|at Rs= tau=2uwRo^2/Rs^2/[1-(Ro/Rs)^2]

Torque =-Area*(tau at Rs)*moment arm
Torque = 2Pi*Rs*L* 2uwRo^2/Rs^2/[1-(Ro/Rs)^2]* Rs
Torque= 4Pi*uw*Rs^2*L * Ro^2/Rs^2/[1-(Ro/Rs)^2] or
Torque = 4Pi*uw*Rs^2*L/[(Rs/Ro)^2-1]

Note, above is your equation, however, Length, L of annulus must be considered.

Regards

 

[btrueblood]

Nice, thanks sailoday.