At 9:46 PM 3/31/5, Robin van Spaandonk wrote: >In reply to Horace Heffner's message of Sun, 13 Mar 2005 17:53:41 >-0900: >Hi Horace, > >I'm having some trouble understanding this formula. If it's meant >to give the relationship between the absolute height of the water >surface at any radius, then it seems to say that at w=0, h= h0, >i.e. h0 is the height of stationary water in the tank. > >>Correction follows. Sorry! >> >>The shape of the final equilibrium surface is: >> >> h = (w^2/2g) x R^2 + h0 > >However when the water rotates, a dip forms at the middle, which >can drop right down to the floor of the tank at sufficiently high >w. However, according to the formula, for any w > 0, h > h0 for >all R, since the first term is always positive. Therefore, the >formula either doesn't represent what I thought it was meant to >represent, or it doesn't describe reality.
It apparently doesn't mean what you think it means, and that's my fault. The surface given by: h = (w^2/2g) x R^2 + h0 is only meant to represent a *final* state (or initial state) surface where no water is falling at all and the rotational velocity at every radius is a fixed constant w. This really only applies to the case where the water is in a tank which rotates so viscosity is not important except maybe for considerations of how the vortex *reaches* this final state wherein the angular velocity is constant at every radius. Viscosity is not an issue when this defines the initial state of a rotating tank assuming the angular velocity is desired to be constant at every radius in the initial state. I provided this because it shows precisely why water may not go down the drain at all, and how a final state can arise in which some or all the water might not be able to go down the drain The case for an "active" vortex tank that is well in progress of emptying, where the tangential speed is proportional to 1/r, and the vertical speed to 1/r as well, is shown in Feynman's *Lectures on Physics*, Vol II, 40-10 ff. Fig. 40-12 is a great drawing of such a tank, showing the surface contour of a sample vortex. In this case the surface is a rotation of: h = k/R^2 + h0 which I noted earlier. I've undoubtedly confused the situation by mixing assumptions regarding viscosities and angular velocity distributions, and looking at initial, intermediate, and final states separately. I thought I was making the concept simpler to follow by these assumptions. I see no way to make analysis of the intermediate states simple. The initial and final states are fairly easy to analyze though, assuming in those states w = K*R for some constant K. Regardless the function used to represent angular velocity, say w = F(R), a *final* surface results which has a perimeter height and a radius at h=0 that is equal to the hole radius. Given a function w = G(R) that represents the initial angular velocity as a function of R, there is a corresponding surface. If the radius of that surface at h=0 is larger than the drain radius then no water can go down the hole at all, and F(R) = G(R). You expressed concern about how it is water can not go down the drain. This describes why. The surface contour does not extend within the drain radius. If there is not sufficient energy from gravity (mgh) to make the water go down the drain, it can not go down the drain (assuming a rotating tank or zero viscosity.) If the water is viscous, however, the viscosity gradually reduces the angular velocity of water remaining in the tank and thus it all can go down the drain. However, the assumption that rotational energy plus heat increases by more than the potential energy being expended is then invalid. There is therefore no reason to think free energy is available. Regards, Horace Heffner
