Suppose we have two tanks connected by a tube with a tap in the middle. One tank is filled with water to a height of $h$ and the other is empty. The tap is then opened and the water 'levels itself'. I had problems with understanding how such a system brings about a loss of $\frac{mgh}{4}$ in the water's potential energy, but I do understand very well now. However, isn't the water here a body under the sole effect of gravity? Shouldn't, provided no energy is lost to the surroundings or converted to internal energy, mechanical energy be conserved in such a way that the loss in K.E. is gain in P.E. and vice versa? That is, the kinetic energy used to transport the water will all eventually be turned into P.E, provided no energy 'losses' occur. I know that every body strives to reach a state of low potential energy, which is why water flattens out in contact with certain surfaces, but I'm still confused as to what becomes of the lost P.E. and would appreciate help. Thanks.
2 Answers
If you had an ideal fluid (with zero viscosity), then the difference in potential energy would appear as kinetic energy of the fluid (mostly the one in the originally empty tank). In other words, you would have some sort of fluid motion in that tank (probably some large-scale vortices, and possibly others, depending on the parameters of the experiment; you could also have turbulent flow, generating a cascade of smaller and smaller-scale vortices) which, in the absence of viscosity, would persist forever. With viscosity, kinetic energy is going to be dissipated and turned into heat and any motion would eventually cease in the asymptotic limit.
And, yes, again depending on exactly how this experiment is conducted, you could have an oscillating solution where potential energy is converted into kinetic energy and back into potential energy, analogous to what happens with a mechanical pendulum. However, in most cases your fluid system has many, many more degrees of freedom than a simple mechanical pendulum, and energy will consequently spread within a far more complex, potentially even infinite-dimensional configuration space.
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$\begingroup$ With the final system state being static with the same water level in both tanks there is no final kinetic energy. This is a question about a simple system, not a complex one. $\endgroup$ Commented Dec 26, 2016 at 23:04
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$\begingroup$ You might want to re-read both the original question and my answer to it. $\endgroup$– PirxCommented Dec 26, 2016 at 23:47
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$\begingroup$ I think your answer is right; it was in fact my view that developed when I considered the question and final arrived at my own answer. $\endgroup$ Commented Dec 27, 2016 at 0:46
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$\begingroup$ just a bit rusty. unfortunatly cannot undownvote. $\endgroup$ Commented Dec 27, 2016 at 0:55
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$\begingroup$ Don't worry about it. But, I'm surprised you can't reverse the downvote. What happens if you click on the down arrow by my post, which should appear blue for you? $\endgroup$– PirxCommented Dec 27, 2016 at 1:03
The potential energy will go down because the water level in the final state is lower than in the initial state (aka some water has fallen). This provides the energy released. This energy will go into friction, thermal or turbulant losses, or could be used to drive a turbine.
If there isn't enough energy to ovecome these losses, (a more unusual situation), there will be not be enough flow to achieve the final state.
Whether it flows or not instantaioneously can be determined by resolution of forces, pressure of the water vs resistive forces.
