At 7:39 AM 3/14/5, Robin van Spaandonk wrote: >In reply to Horace Heffner's message of Fri, 11 Mar 2005 17:22:03 >-0900: >Hi, >[snip] >Given the complexity of the equations for ASCII representation, I >have placed a Mathcad document (24 kB) and a "gif" version thereof >(36 kB), for readers without Mathcad, on my web page at >http://users.bigpond.net.au/rvanspaa/vortex.mcd and >http://users.bigpond.net.au/rvanspaa/vortex.gif respectively. > >Most of the variables at the top of the page are irrelevant >(inherited from my standard template).
This is great stuff! I hope, for the sake of the archives, one of us takes the time to post this in ascii when all is done, and tabularly descirbe each variable. I suppose I can insure that happens. This could be a useful analysis to refer to for various things. I also hope escribe continues in existance. I see you are opting to analyze the zero viscosity fluid version, which is very handy. So, to obtain the energy or momentum of what went down the drain you need only integrate inside the boundary consisting of the edge of the tank or the surface of the remaining water, whichever is smaller at a given height. To obtain the energy and angular momentum of the water that remains you need only intergate those values outside that surface and inside the tank. You might find the following analysis useful for your next phase. The shape of the final equilibrium surface is: h = (w^2/2g) x R + h0 where h is height, w is angular velocity, g = 9.80665 m/s^2, and R is radius. Using R1 as the radius of the hole, R2 as radius of of tank, we have h = 0 at the radius R1 when equlilbrium is established and no more water can go down the hole. So: 0 = (w^2/2g) x R1 + h0 h0 = 2g/( w^2 x R1) The final height Hf of the water at the edge of the tank is thus: Hf = (w^2/2g) x R2 + 2g/( w^2 x R1) Bravo and keep cooking! Regards, Horace Heffner
