At 4:28 PM 4/5/5, Robin van Spaandonk wrote: >In reply to Horace Heffner's message of Thu, 31 Mar 2005 23:33:55 >-0900: >Hi Horace, > >Thanks. I have now derived the formula for myself, so I understand >where it comes from, and what the various constants mean. I have >also applied the same derivation principle to an active vortex >that it constantly being "topped up" to maintain a constant level. >The result for a vortex with no initial angular velocity
Uh, if there is no initial angular velocity the water merely runs down the drain. Not sure what you mean here. >can be >found at >http://users.bigpond.net.au/rvanspaa/vortex-shape.mcd >and >http://users.bigpond.net.au/rvanspaa/vortex-shape.gif Glad to see you are making progress. > >I'm still thinking about how to correctly introduce an initial >angular momentum. I may try it with a fixed angular velocity at >the rim, and see what happens. (This is what one would get with >tangential addition of water as in one of your previous drawings). The thing that strikes me as most important is the equation of the surface countour. If you accept that all the water above the drain initially (instantly) goes down the drain, then what is left is a body of water with a surface countour. Once that surface contour reaches equilibium due to more water running down the drain (if that is possible), then, provided there is no viscosity, what remains under the surface contour (i.e. toward the tank wall and below the surface) really doesn't matter, because things are in equilibium. Such a surface is monotonically increasing in value with increase in radius. This is because inward force which permits radius reduction is due to pressure, and pressure is due to the height of the highest particle less the height of the given particle. There can be no relative maxima or minima on that surface. After equilibrium is reached, any molecule added at the rim, regardless its tangential velocity, can be visualized as merely rolling down that surface. When the molecule reaches a point where more energy is required to reduce radius than is available from falling, the molecule stops and the surface countour is modified to a new equilibrium point. This concept could be used to develop a finite element model of the continuous process. Conservation of energy (COE) and conservation of angular momentum (COAM) are intrinsic to this method however, so the results will only show what COE and COAM predict, to the granularity chosen. It might be easiest to start with a tank with the drain open and in equilibrium, and where angular velocity is the same everywhere, i.e. h = (w^2/2g) x R^2 - 2g/( w^2 x (R1)^2) where g is the acceleration due to gravity, w is the initial angular velocity everywhere in the water and tank, R1 is the drain radius, and R is a given radius. This initial condition is comparatively easy to achieve fairly accurately experimentally by rotating the tank. Measurements can easily show if the equation is right, ignoring surface tension effects at the edges. From that point it is possible to add water to the rim and see what happens, but viscosity will play a role from that point on. It might be interesting to make a hard tank surface that corresponds to the above formula. > >BTW the initial restriction imposed on the angular velocity by the >radius needing to be less than the drain radius doesn't appear to >be serious. IOW even an initial angular velocity that produces >just a slight dip in the surface would already be sufficient to >yield OU according to the first document I posted. I'm not sure what this paragraph means. >However, I'm now having second thoughts about the validity of that >first document (http://users.bigpond.net.au/rvanspaa/vortex.gif). Good! Regards, Horace Heffner
