Mechanical Engineering

# Rotating drum

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Guys & Girls,

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 end 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!

Edited by Carlo Pescatori
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Dear Carlo,

Thank you for an interesting problem. I have attached my solution as a PDF file. It is certainly not guaranteed; there could certainly be algebraic errors in it. I encourage you to check it over very carefully before applying it.

Note that as I have used the Greek letter rho, it has units kg/m^3, not per liter as you did.

I'd like to know more about the origins of this problem. It would be nice if you would send me a message telling me more about where it came up.

DrD

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Dear DrD,

Some background: this is a real life application of a fast spinning centrifuge. G-forces at the outer diameter of the drum are 45000m/s^2, so the liquid will nearly a cylinder. The centrifugal forces 4500x higher than the gravitation, so it is safe to leave it out of the equation for this application.

Thanks again, much appreciated!

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Dear All,

For any of you who may have read my previous post on this question, I have to tell you that it has an error. I left out the tangential forces on the fluid element, and thus got a wrong result.

I hope to have a better result for you in a few days, but for now, do not make any use of the result posted earlier. Sorry about that!

DrD

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It should be a cylinder of air at any speed with a meniscus at the flat edges if there were zero gravity no?

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