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Max. Applied force on outside wall of spinning drum

  1. Feb 8, 2013 #1

    mpf

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    I have a 36" OD aluminum drum within another 48" OD aluminum drum creating 6” channel between the two diameters. The total height of the drum assembly is 16” This design project that I am working on requires me to fill this channel up with 12” of water which equates to about 342.5 lbs of water. I am going to spin this drum at 17.91 RPM which equates the outside radius (24”) to have a tangential velocity of 3.7483 ft/s. Now to my question…..I am looking for the maximum pressure caused by the centripetal force on the outer ring of the drum while the drum is spinning. Logically the maximum force will be at the 12 depth mark because that’s where the most amount of pressure is located.

    My way of doing it: I found the ambient pressure at a 12in depth which was 0.43363psi. From the f=pa eq. I then picked an area at the bottom of the drum which was 6” wide (channel width) x 1” giving me an area of 6 in^2. Took the f=pa eq. and multiplied it by the ambient pressure at 12” and got 2.60178 lbs. Plugged that in to the f=ma eq. to get my weight given that area and got .0808lbs. Put that bad boy into the centripetal force eq. given my tan. Velocity which is 3.7483 ft/s and the radius which is 2ft and got .5674lbs of force at any location around the outside of the drum at a 12” depth. I know the load is a distributed load so you can solve with an integral for the whole drum wall or use a finite elemental analysis program to solve it but I tried to cut corners and get a ball part estimate. Can someone either validate this or tell me I’m way wrong and help me out. Thanks much
     
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  3. Feb 8, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    Where does centripetal force appear?
    Every element of mass dm adds ##\omega r^2 dm## to the total force acting on the outside, which is just (pressure)*(surface area).
     
  4. Feb 11, 2013 #3

    mpf

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    So are you saying my way strategy was incorrect? I'm just a little confused how to apply your response.

    Thanks
     
  5. Feb 11, 2013 #4

    mfb

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    I don't understand what you did. In particular, I don't understand how you accounted for the rotation of the setup.

    Another issue: Did you consider the variable water level height?
     
  6. Feb 11, 2013 #5

    mpf

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    I basically found an area of water at the 12" depth and, found its mass and used that mass in the centripital force eq. Used my tan. velocity of 3.7 ft/s at the 24" radius and solved for centripital force. And I did not consider varaible water level height because I am looking for a ball park estimate. I calculated earlier that the max water level will be around 13.1" so I guess I could use that instead of my 12" h.

    Again, I don't think this is the correct way to go about this problem, it's just the only way I thought might work so if you have any ideas how to solve please help.

    Thanks
     
  7. Feb 11, 2013 #6

    mfb

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    Let's see:
    Outer radius 61cm, outer velocity 1.14m/s, outer acceleration 2.14m/s^2 or about 1/4 g. Ok, let's neglect the different water locations due to a different water level.

    If we consider some small segment, all relevant forces are inwards/outwards only. In the same way, it is possible to consider the whole water, just with radial forces.

    Without rotation, average pressure on the wall outside is same as pressure in a depth of 6'' or 15.25cm, this gives about 1500N/m^2, increasing from 0 to 3000N/m^2 from the surface to the bottom.

    With rotation, water at a radius of r feels a force of ##r \omega^2## outwards. Integrating over the whole volume, this gives
    $$\int_r^R \rho r'^2 \omega^2 dr' = \frac{1}{3} \rho \omega^2 (R^3-r^3)$$
    Using r=45.72cm as inner radius, this gives 154N/m, to distribute over 2*pi*R, therefore I get an additional pressure of 40N/m^2. Hmm, looks negligible.
     
  8. Feb 11, 2013 #7

    mpf

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    Terrific, thank you
     
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