With no energy transfer, the total energy (per unit mass)of the particle would be constant.īefore exit: pressure energy = H z and kinetic energy = tang vel squ / 2g (radial velocity = zero)Īfter exit: pressure energy = zero and kinetic energy = total vel squ /2g = (tang vel squ radial vel squ) / 2g This is meant to indicate that the rate of mass exit from the cylinder is very small as to allow neglecting change of liquid level and also change of momentum and energy content when mass moved to a different radius.Īlso the cylinder would be perceived to have zero thickness such that a particle can be considered at the same radius before and after it exits the cylinder. The problem statement indicated the hole size is very small. The radial velocity energy will be equal to the total liquid height above the hole, thus the radial velocity $v=\sqrt $. How can I calculate the exit velocity using Bernoulli and the maximum altitude of the fluid in the parabola? I know that Bernoulli is applicable only to streamlines, but I can't even imagine the course of these. More precisely it is a revolution paraboloid defined by this equation: My attempt to solve the exercise was: as a first step, I calculated the shape of the surface using a non-intertial frame of reference with a centripetal force.ĭoing this I obtained that the surface is paraboloid shaped. Now we have to calculate the radial velocity of the fluid when it leaves the cylinder. We have as data that the lower part of the surface is at $z=H$. The hole is small, thus we can neglect the velocity of the surface of the fluid. The cylinder has a small hole in the wall almost in the bottom. We have a cylinder of radius $R$ rotating with respect to the vertical axis with angular velocity $\omega $. I´m a little confused with a fluid-dynamics exercise.
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