I am designing a rotating heat exchanger for the purpose of exposing sufficient ambient heat from the atmosphere to liquid air to return it to a gas, while spinning in a range where centrifugal force keeps the exterior of the device clear of ice.
This device (as currently planned) consists of two copper discs, each 8 inches in diameter and 3/8 inch thick rotating about a hollow axle. These are a mirror image of each other each with a groove on the "inside" forming a cylindrical tunnel in the finished part once the halves are joined together. Now, clearly, great hydraulic pressures would form under high rotational speeds which could exceed the strength of the he bond joining the disks allowing the disks to separate. My question is roughly how much hydraulic pressure would be created under the following circumstances:
For ease of understanding, let's assume the "tunnel" to be a tube 3/16 inch in inside diameter and the liquid air to weigh .9 of a comparable quantity of water. The tube would extend from the hollow axle (zero radius) to a point 3 1/2 inches from this point at which point it would make a 180 degree 3/4 inch radius bend and continue parallel and on the same plane to the original path for one inch. At this point it would be assumed to have absorbed sufficient heat to have returned to gaseous form. ( I am quite willing to ignore the weight of this gaseous air for this calculation)
I would very much like an answer in PSI for 15,000 rpm, with an explanation of how it was arrived at. If it is inconvenient to work in "inches", it is ok to round it out a little and use metrics.
thanks much,
bill_michaels