Either a large, constrained cylinder of water or a large volume of open water (perhaps in the ocean) could be persuaded to form a vortex by pumping energy in. The body of water would be acting as a fluid version of a flywheel, storing energy.

If one then removed the input energy, would the vortex continue for a significant time or would it quickly lose all its energy by heating up the water? Would storing energy in the rotational kinetic energy of a fluid be horribly inefficient compared to pumped-hydro storage?

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    $\begingroup$ To me this doesn't sound very different from pumping energy in a ferris wheel for storage. $\endgroup$ – Bzazz Sep 6 '18 at 10:41
  • $\begingroup$ @Bzazz Are you thinking of a flywheel? Ferris wheels are the big amusement park rides. I don't think they are usually used for energy storage. $\endgroup$ – JMac Sep 6 '18 at 10:44
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    $\begingroup$ I did mean ferris wheel. $\endgroup$ – Bzazz Sep 6 '18 at 11:00
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    $\begingroup$ They don't, that's the point. $\endgroup$ – Bzazz Sep 6 '18 at 11:14
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    $\begingroup$ The usual design to store energy in the ocean is to build a dyke and pump the water out; energy is recovered by letting the water flow back in through a turbine. (It's probably not a surprise that this is a Dutch idea. Hard to do pumped hydro storage when you have no mountains) $\endgroup$ – MSalters Sep 6 '18 at 13:52

Yes, this would be horribly inefficient compared to pumped hydro or even a regular flywheel.

With a rotating fluid, there's a lot of viscosity. This viscosity generates heat and slows the fluid down. You would be able to offset this somewhat if you kept the container for the fluid moving with the fluid itself; but even then I believe there would still be significant losses from internal viscous effects. You would then also just have what is essentially a fluid filled flywheel, at which point you have to ask yourself why you didn't go with a solid material of higher density and likely better strength/stability.

The advantage of pumped hydro over this vortex system would be because a pumped hydro system doesn't require constant motion. It stores the energy as potential, so it only loses energy to viscosity when it is moving the fluid to the storage tanks. The less you have to move the fluid, the less you will convert energy to heat through viscosity.

It's an interesting idea, but I don't think it would be practical.

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    $\begingroup$ "You would be able to offset this somewhat if you kept the container for the fluid moving with the fluid itself; but even then I believe there would still be significant losses from internal viscous effects." But once all of the fluid is at rest relative to the container, then there will no longer be any losses to viscous effects, right? (Assuming that the container continues to rotate at a constant velocity.) $\endgroup$ – Tanner Swett Sep 6 '18 at 15:16
  • $\begingroup$ @TannerSwett That part is what I was having trouble with one way or the other in my head. I do understand what you're saying; but I'm not sure if that would ever really be the case. The middle of the flow should lag behind in that case, and there would be losses keeping it up to speed with the container walls. I'm pretty sure the curvature would also cause viscous losses. $\endgroup$ – JMac Sep 6 '18 at 16:01
  • $\begingroup$ My thinking is, in the container's frame of reference, the kinetic energy of the water has to decay towards zero. As long as the water is still moving, it must be losing kinetic energy, and there's no way it can gain kinetic energy, since the container isn't moving in this reference frame. Is my reasoning sound? $\endgroup$ – Tanner Swett Sep 6 '18 at 16:10
  • $\begingroup$ Or if you have to use water, why not just pump the water up to somewhere high, and then when you need to get the energy back using the falling water to turn some turbines. $\endgroup$ – Shufflepants Sep 6 '18 at 20:03
  • $\begingroup$ @Shufflepants That’s pumped hydro. $\endgroup$ – JMac Sep 6 '18 at 23:27

As mentioned by other writers, water viscosity will rapidly convert any kinetic energy into heat. However, efficient energy storage could be attainable with superfluids. Superfluids are related to Bose-Einstein condensates. (but they are not always interchangeable) Their special property is having absolutely no viscosity, so a whirlpool of superfluid will rotate indefinitely. The primary issue with this is the cryogenic temperatures required. Helium-4, for example, requires temperatures below 2.17K to reach superfluidity. The energy required to create and maintain such temperatures could exceed the amount of energy stored, making the system useless. Harvesting the energy could prove problematic. Since there isn't viscosity, a turbine would experience no drag. (to be precise, very little drag from residual viscosity)

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    $\begingroup$ When you want to extract the energy, you can let it go above the critical temperature. Very odd idea, but fun! $\endgroup$ – Bzazz Sep 6 '18 at 13:00
  • $\begingroup$ Interesting thought, though that might exacerbate the problem since now the fluid has to be heated and cooled $\endgroup$ – Dayrius Tay Sep 6 '18 at 14:21
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    $\begingroup$ @DayriusTay I assume the heating portion could be accomplished by stopping the cooling. Critical temperature is so low that you would only have to worry about the best way to expose it to more atmospheric conditions. $\endgroup$ – JMac Sep 6 '18 at 14:29
  • $\begingroup$ No viscosity doesn't exactly mean no drag. A non viscous fluid still can move a turbine because it has mass and inertia and changing its direction or stopping it causes a reaction that can be used to extract energy. $\endgroup$ – Pere Sep 6 '18 at 14:41
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    $\begingroup$ @Pere I don't think that's right: a non-viscous fluid won't move a turbine, because, although it has mass and inertia, the turbine doesn't change its direction or stop it. My understanding is that if a jet of superfluid hits an airfoil (or any object) which is submerged in the superfluid, then the jet will flow around it and continue at the same velocity. See: physics.stackexchange.com/questions/311069/… $\endgroup$ – Tanner Swett Sep 6 '18 at 16:24

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