(Simple?) Spinning Fluidmass Question . . . 3

[AlbertG]

Hello Folks.

I believe that this will be a REALLY simple affair to one of the anointed ;o)

OK.

We have a square tube 1" in diameter X 2' long spinning at 3,000 RPM ; filled with H₂0.

Metric: The spinning radius is .3 meters (r); the watermass weight (for the .3 meter radius) is .1944 kg (m).

On the basis of the following common equation used to figure centripetal force vs RPM

F_c (N) = 4 m pi² n² r / 60

we get a figure of about 345,021 N as the result.

However, (if I read my tealeaves right) this basic arithmetic assumes that all of the mass in question is concentrated at the distal end of r; giving nothing to the fact that the mass is uniformally distributed along r from the center of rotation outward.

So, how can this be done to accurately determine the restraining force required for the given water column as would actually be experienced at the distal end of r?

Thanks!

 

[GregLocock]

Hmm, does a force of 35 tons even feel in the ballpark?

well, you've got two choices. Integration by calculus, or integration by approximation, say by taking 1mm steps outwards, in a spreadsheet. Might sound a bit tricky, wait until you try it with a compressible fluid!



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[zekeman]

You integrate (M/L)w^2xdx from 0 to L/2 since it is spinning at the center of a 2' length.
This yields for each side the restraining force at the end
(M/L)w^2L^2/8=Mw^2L/8

 

[IRstuff]

The inertia for a hollow cylinder is mass*len^2/4, which would be equivalent to your cylinder with mass/3. The mass you cite is mass/2.

The centripetal force equation is mass/3*v^2/r, where v = angular rate*r. so the equation should be mass/3*rate^2*r

rate = 3000 rpm = 50 rps = 314 rad/s
r = len/2 = 0.3048 m
mass/3 = 0.131 kg

force = 3943 N

I see at least one error; if your rate is squared, the factor of 60 should be squared as well, as that's the conversion from min to sec.

TTFN

FAQ731-376

 

[zekeman]

"The inertia for a hollow cylinder is mass*len^2/4, which would be equivalent to your cylinder with mass/3. The mass you cite is mass/2.

The centripetal force equation is mass/3*v^2/r, where v = angular rate*r. so the equation should be mass/3*rate^2*r"

The problem has nothing to do with inertia, The centripetal force, as I pointed out, has a factor of 1/2, not 1/3. If you integrate rdr you get r^2/2. Inertia integrates r^2dr yielding r^3/3

 

[SomptingGuy]

I got Mw^2L/8 also. Does coriolis come into it too somewhere? My 1st year dynamics is a bit rusty.

- Steve

 

[AlbertG]

"Hmm, does a force of 35 tons even feel in the ballpark?"

Actually, it looks closer to 39 tons (if I counted my fingers and toes correctly); and I'm glad I'm not the only one who smelled fish ;o)

"...integration by approximation..."

Thought about that; but I wasn't sure if there was a well-known formula for grabbing the numbers here without brute force or calculus (didn't want to reinvent the wheel)...

Thank you all for your helpful input!

However, I may be WAAAAY off here, but something else occurred to me:

It would seem as though we'll not be needing to factor in about half of the radius' fluidmass in consideration of the fact that anything which falls outside of a centered triangular section of the water column (extending outwardly from the pivot) should be working radially against the walls of the restraining tube.

In other words, the section of fluid which acts directly on the distal end of the tube is better characterized as a "pie slice;" such as one would anticipate in the case of a filled rotating cylindrical housing.

Yes?

So, if this is indeed the case, are there any ideas as to how we might form an (integration by approximation) equation to eventually solve for F_c?

In any event, you all might've guessed that I'm a bit rusty here; so please bear with me for a while if you can.

Thanks again!

 

[IRstuff]

In other words, the section of fluid which acts directly on the distal end of the tube is better characterized as a "pie slice;" such as one would anticipate in the case of a filled rotating cylindrical housing.

Yes?

No, that's not how water pressure works. For a non-flowing fluid, pressure is uniform throughout the volume.

As for the force calculation, yes, you can integrate amd get Mw^2L/8, but the equivalent inertia approach resulted in Mw^2L/6, which is close enough for estimation.

If the OP's calculation is corrected for the factor of 60 error, the resultant force is 5750 N. Corrected for the distributed mass results in 2875 N

TTFN

FAQ731-376

 

[AlbertG]

@IRstuff:

Thanks for the encapsulation.

So, to "sum" up, what we're looking at is correcting the original equation to eliminate the minutes to seconds error and dividing the result by 2, to allow for the distributed mass along the column. By doing so, a reasonably accurate answer will be obtained indicating the distal restraining force required by the spun water column at n RPM. Correct?

Finally, if this is so, I'd like to ask just one more question related to this line of thinking: Just how would we use the corrected original equation to solve for a homogeneous mass of water centrifugally slung against the outer wall of a rotating cylindrical housing (think spinning barrel; not tube) at n RPM? Could this setting involve a triangular section as I supposed?

In other words, could one take a fixed 1" square (outer wall footprint) "pie slice" type segment of the fluidmass in question and solve for that one element using the corrected original equation? Would this accurately represent the local restraining force required of the outer wall? If so, how would we correct for the distributed mass in this situation?

Just interested . . .

In any event, thanks again for all the help.

 

[IRstuff]

You would simply do the integral that others have suggested. The same basic approach holds.


TTFN

FAQ731-376

 

[GregLocock]

The pressure is not constant in the entire volume as it is an accelerating system.

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[AlbertG]

"No, that's not how water pressure works. For a non-flowing fluid, pressure is uniform throughout the volume."

"The pressure is not constant in the entire volume as it is an accelerating system."

It would seem to me that GregLocock is right here; as the fluidmass will have a gradient of pressures from pivot to outer margin.

As to my latest question, I still need just a wee bit of help in making this all tick over. Specifically, will evaluating for a radial slice of the fluidmass in the spinning cylinder context which I mentioned accurately represent the local restraining force required of the outer wall?

As a working example which could be quickly evaluated, let's take the case of a flat, closed cylinder 1" high by 2' in diameter, filled with water; and spinning at 3,000 RPM.

On the basis of what I have gleaned so far, the "raw" restraining force might be termed as 2,875 N (not adjusted for mass distribution) for a one square inch (outer wall footprint) "pie slice" triangular segment of the homogeneous fluidmass in question. Is this so?

Again, if it is, would multiplying this figure (2,875 N) by 70.7% accurately represent the local restraining force (2,033 N) required of the outer wall?

Thanks again...

 

[GregLocock]

I get

Force on end = 1/2*rho*A*R^2*omega^2


omega is in rad/sec, everyhting else needs to be in consistent units. It does at least look dimensionally correct.

If that is right then the calculus is a one liner, based on dF=(rho*A*dr)*r*w^2 and the answer is just under 3 kN on a 1 inch square

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.