Water Rimming inside drum

[Tagger]

How do you calculate the point where water inside a copper drum begins to "rim" (attaches to the ID of the drum) due to centrifugal force. The drum starts rotating at 180 RPM to a top speed of 1300 RPM. The ID of the drum is 15.5 inches and 28 inches long. Water enters/exits using a dual (inlet/outlet) rotating joint along the axis of rotation. It is a brake where the water cools the drum. Does the mass of the water relate to the speed of the drum (i.e. the more mass of water the faster the drum needs to rotate for the water to rim). I guess it is dependent on the friction coefficient inside, etc. The drum ID wall does become dirty over time, and there is no exact answer.
Thanks

 

[Ralph2]

Good question....
This is something I have wondered about as well. This is the principal of a centrifuge and they frequently have accelerator bars to help lift the water (or any liquid) and prevent it from pooling in the bottom. From my experience you can "flood" the system and then not be able to spin fast enough to establish a "wall" of liquid. So that in that case the mass (volume) is working against you.

The reason I would like to know the solution is that we frequently spin babbitt bearings to re furbish them. This process requires spinning at a high enough speed to not centrifuge out the components of the babbitt yet fast enough to build a even wall of liquid babbitt. As this process happens up to 600 degrees F there is some degree of danger. The worst scenario is for the wall to collapse, causing massive unbalance as well as allowing the pool to overflow the inlet dam.

Ralph