Calcualting Motor Torque

You're spinning a liquid. If the drum has no internal vanes, The liquid will come up to its speed with much delay. The starting torque -- friction aside -- will be little more than for an empty drum.

Jerry

You could obtain a worst-case answer of the torque during the pullout period by assuming that the drum and contents are solid. The result will be somewhat greater than the case where the fluid slips, but is that an issue? If you design with too small small a margin you're likely to find that the setup lacks grunt, at least at times.

Without knowing much about your application, it looks like the pullout torque requirement would be a lot higher than the running torque if your mechanics are good, so sizing for pullout is probably going to be OK.

This also depends on your drive type. Fixed frequency AC induction motors aren't happy for long at speeds significantly below their synchronous speed, so if you're considering such a drive you'll need to ensure that it can pull out fast enough. There's plenty of information on the web on motor protection.

Eh? You're describing an 8 gallon tank.

How quickly does the thing need to spin up? What sort of bearing drag do you expect? What do you define as "a decent speed" -- all of these things make a significant difference in your answer.

When I started working at my Dad's shop one of my jobs involved rolling

55 gallon drums around. I couldn't tell you just how big they are in inches, but I can close my eyes and see one, and feel just how much heft it takes to tip a full one up on it's rim to roll it around.

...

I feel silly for not picking up on that. 36*16*pi/231 = 7 5/6 gallons. That comes out to about 65 pounds; just over a cubic foot.

...

Jerry

Have you ever dealt with one filled with CCl_4. That stuff is *dense*. About 1.6 * water.

Jerry