This morning I was struck by an odd thought:

  1. Hydroelectric power stations get energy by dropping water from a height. The greater the height difference between top and bottom, the more energy can be gained.
  2. To evaporate 1 gram of a boiling liquid, the amount of energy required is constant.
  3. Once evaporated, the gaseous substance will rise as high as possible in the air due to buoyancy (well, assuming it's less dense than air). In other words, no extra energy is necessary to raise it high.

Put these three together, fiddle with the substances you use for the "air" and "liquid" and you should be able to raise the evaporated liquid high enough that after condensing at the top and falling back down it produces more energy than was required to evaporate it in the first place.

Obviously this won't work because perpetual motion machines can't work, but I don't know which of my above assumptions is wrong or what other factors would come into play to make this impossible.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Mar 11 '20 at 12:33

The scenario you describe is more or less how hydroelectric plants operate in the first place, but they use a power source- the sun- to do the evaporation work and to generate the winds that move the humid air around.

If there were no energy input from the sun, evaporation and global circulation of the atmosphere would stop, rains would stop, and hydro power plants would stop once their reservoirs ran dry.

This demonstrates that an evaporation-based perpetual motion machine cannot work.

  • $\begingroup$ My idea was to use the energy produced by the hydroplant to power evaporation. To get enough energy, just raise the ceiling (where the condensation happens) high enough. $\endgroup$ – Vilx- Mar 10 '20 at 20:38

Once evaporated, the gaseous substance will raise as high as possible in the air due to buoyancy (well, assuming it's less dense than air). In other words, no extra energy is necessary to raise it high.

This seems to be the key mistake. It is not correct that the column can be arbitrarily tall. Since it is only the vapor pressure of the working fluid that matters we can get rid of the air and focus on the working fluid only. I will call it water, but the principles hold for other fluids.

In the hydrostatic case the vapor pressure is given by the weight of the vapor above. Vapor, like anything with mass, falls in gravity unless supported by pressure below. Therefore, the height that it can rise is limited by the height of the water vapor column. The weight of the water vapor column both limits the height that it can rise as well as determining the pressure at the bottom.

The pressure at the bottom is important because part of the enthalpy of vaporization is the $P\Delta V$ work done when the liquid expands into the vapor phase at the bottom. This work is the mechanical work which raises the column of vapor back to its original pre-condensation height.

The mechanical work done by letting the liquid water fall to the bottom is equal to the mechanical $P\Delta V$ work required at the bottom (because the center of mass is the same before and after). Since that is just a portion of the enthalpy of vaporization the overall process will require more energy than generated, even at maximum height.


Here is a conceptual reason why without math using extremes.

First some statements, then the actual reasoning: There are no completely isolated systems in the universe, and every system loses a positive amount of energy as heat into the environment. Heat flows from hotter to colder, and there are always colder regions out there until the universe reaches thermal equilibrium (ignoring fluctuations), which won't happen for trillions of years. The entropy of a thermally insulated system cannot decrease and is constant if and only if all processes are reversible. Because there are no perfectly thermally insulated systems, you see the flow of heat and thus increase of entropy in the universe. But reversible systems allow for the backwards flow of heat and entropy, so we need to establish at least one irreversible process. Once we have it, the above all comes together to give the entropic arrow of time, and thus we can say every process or system loses energy to the universe as whole. So you need to input energy to a system to keep it equilibrium (with itself, not the environment).

As far as irreversible processes. This is a very interesting topic, and I have asked questions on SE before for clarification. Until I get a response, I will appeal to Susskind's belief that there are no irreversible processes without the multiverse. Even in an accelerating, expanding universe like ours, we still have reccurances due to the finiteness of our "box"/horizon. This guarantees a backwards arrow of time at some far future point. So we wouldn't have true, complete irreversibility. Enter Susskind's argument that Koleman-De Luccia processes of bubble universes forming and collapsing within eternal inflation produce a truly complete irrervisbile process. My understanding of his reasoning is that the rate of universes being spawned is so great in such a multiverse, it out paces the number of universes currently in recurrances, which is why we can use self-sampling to say why we don't see any recurrances in our universe. Susskind calls this process a fractal flow. The total makeup of the multiverse is thus trending toward more and more fresh (and thus not in reccurance yet) compared to the aging ones. The multiverse trending toward this is the ultimately the arrow of time, and ultimately the irreversible process needed to completely explain why we witness only a singular direction of time - and a roundabout way of answering your question.

  • $\begingroup$ This is... awesome... and completely misses my question. ^^D $\endgroup$ – Vilx- Mar 10 '20 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.