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Thread: Rotating drum

  1. #1
    Associate Engineer
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    Apr 2020
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    Rotating drum

    Guys & Girls,
    I need your help.
    Problem is this: Suppose I have a round drum, rotating about its main vertical axis. Top and bottom are closed off. The drum is partially filled with a liquid. How do I calculate the force that the liquid exerts on the top and bottom lid of the drum. Let's ignore gravity.
    I have attached 2 calculations I did. The picture I found somewhere on the web.
    Which one of the two is wrong and why? Or are they both wrong???
    Looking forward to hear from you!
    Attached Images Attached Images

  2. #2
    Technical Fellow jboggs's Avatar
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    If you're ignoring gravity their both wrong. Without gravity there would be no liquid at the bottom. The hole in the middle would go all the way from the top to the bottom. A perfect cylinder.

  3. #3
    Associate Engineer
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    Thanks for your reply.
    Sure I will get a perfect cylinder shape. This is a real life application, where g-forces at the drums outer diameter are 4500xg. So the 1 g vertical is much smaller and can be ignored.
    Question is: what is the correct formula formula for the force that this liquid cylinder puts on the two lids of the drum?

  4. #4
    Associate Engineer
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    Nov 2016
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    Determine the centrifugal force assuming there is really not gravity.

    https://www.engineersedge.com/physic...ugal_force.htm

  5. #5
    Technical Fellow jboggs's Avatar
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    Without gravity the force on the top and bottom will be a result of the pressure of the water in contact with them. That pressure will be created by the centrifugal force the water experiences. It probably will also vary from the inner surface of the water to the outer surface just as, under gravity, water pressure increases with depth.

  6. #6
    Banned
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    May 2020
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    To be honest, I have the same problem and I have no idea how to solve it, but I think they will tell me here)

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