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Let's consider the following system shaped as in the picture below, in which the only fluid contained is water at room temperature.

gravitational condensation system

As far as I understand, the water should be in an equilibrium between its liquid and gaseous phases. While some of the liquid water at the bottom continuously evaporates due to vapor pressure, some of the water vapour molecules will cluster into droplets, causing condensation. Solid surfaces --such as the ceiling and walls of this system-- are likely sites for this condensation because they reduce the energy barrier that needs to be overcome for this nucleation to take place.

However, when I try to bring gravity into the equation, I'm struck by what seems to me like a remarkable asymmetry. Any water droplets condensed against the ceiling of the container have greater potential gravitational energy than the liquid molecules at the bottom. The stalactite-esque spire protruding from the ceiling takes advantage of water's surface tension to direct a trickle of water onto a tiny waterwheel below, powering a tiny turbine.

Going in the other direction, any evaporated water molecules that end up condensed against the ceiling seem to do so without any input of external energy. Gas molecules will travel in any direction throughout a container, spontaneously reaching the upper regions merely through their own energetic brownian motions, trading heat for gravitational energy if you will; while apparently decreasing entropy of the entire system over time, violating the 2nd law of thermodynamics while summoning Maxwell's Demon.

That can't be right, right?

NB: It deserves mention that condensation produces heat, whereas evaporation consumes heat. The resulting temperature differences should remain constant though, given that convection and conduction would keep the system in thermodynamic equilibrium between the sites of evaporation and condensation. Using thermally conductive materials in-between top and bottom (e.g. copper container walls) is just one measure that can be taken to minimize the temperature difference of this equilibrium.

Will
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    The water doesn't evaporate "due to vapor pressure", vapor pressure is created due to the evaporated water. Browmian motion refers to smaller molecules suspending larger particles in a fluid. The water molecules actually are less massive than either nitrogen or oxygen molecules. – bpedit Feb 08 '17 at 04:03
  • @bpedit I think you are wrong on both accounts. At the very least, I'm pretty sure that Brownian motion takes place in any fluid, liquid or gas. – Will Feb 08 '17 at 07:51
  • You can be "pretty sure" or you can actually do some research. – bpedit Feb 08 '17 at 15:31
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    @bpedit Oh, you mean like reading at least the first sentence of the wikipedia article on Brownian motion: "Brownian motion or pedesis (from Ancient Greek: πήδησις /pέːdεːsis/ 'leaping') is the random motion of particles suspended in a fluid (a liquid or a gas) resulting from their collision with the fast-moving atoms or molecules in the gas or liquid."? – Will Feb 08 '17 at 22:15
  • The "particles" referred to are far larger than the "fast moving atoms or molecules". Finish reading that article. We're talking about particles of visible, at least with a microscope, size. If you have access to a microscope, put a tiny bit of India ink in a drop of water on a slide to observe Browian motion. – bpedit Feb 10 '17 at 15:53
  • @bpedit Not true. A "particle" refers to anything in size down from an electron up to the size you refer to. Just because a fluid doesn't have any large particles suspended in it, doesn't mean the motion of its small particles (e.g., water molecules) is any less Brownian. – Will Feb 10 '17 at 22:42

3 Answers3

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At a given temperature, in your liquid water-air system, equal numbers of water molecules will enter the air from the liquid as return to the liquid from the air. The system will be in equilibrium and the air will be "saturated" with water vapor.

There are two ways that condensation will form on your ceiling. If the air is supersaturated with water then your "nucleation" sites will facilitate condensation. But conditions in this system are not those that would result in supersaturation. The second way for condensation to occur is for the ceiling to be colder than the air.

If condensation occurs on the ceiling, much of the latent heat of condensation will be transferred to the ceiling therefore warming it. To continue the condensation process, you will have to keep the ceiling cool requiring expenditures of energy from outside the system. Your system is closed but it is not isolated in thermodynamic speak.

Furthermore, as the condensing water loses heat to the ceiling, the system cools. This will result in a lowered equilibrium vapor pressure, that is, less water in the vapor state. To make matters worse, the lower temperature of your system will require an even greater lowering of the ceiling's temperature to maintain condensation.

As far a entropy, you must, in addition to events within your system, consider those happening outside to power the refrigeration process.

Hopefully you understand the turbine you may be running inside the system won't even come close to powering the refrigerator outside!

bpedit
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  • Thanks for your answer. Actually, there is no air in the system I had in mind (First sentence: "Consider the following system shaped as in the picture below, in which the only fluid* contained is water at room temperature."*), but it probably doesn't matter much either way. – Will Feb 08 '17 at 07:45
  • More importantly though, I am not convinced -as of yet- that condensation necessarily requires cooling. See this answer, for example: "In conclusion... with a mesh-like microstructure made of material with appropriate surface tension with water, you can condense even without cooling.". – Will Feb 08 '17 at 07:46
  • But even more importantly, if cooling must take place, why does that cooling need to be done from outside the system? Doesn't evaporation require an amount of heat equal to what the condensation produces? Given that the vapor itself can circulate that heat, they should cancel each other out. – Will Feb 08 '17 at 07:46
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    @WilliamBudd. Even if condensation were o take place initially, do you imagine it would continue as the surface becomes hotter than the vapor? – bpedit Feb 08 '17 at 15:38
  • @WilliamBudd. It's true that the liquid would cool if evaporation exceeds recapture. This makes your scenario even more unlikely. – bpedit Feb 08 '17 at 15:46
  • The container walls and the water are all in contact with each other. Any temperature difference would be limited due to conduction and convection creating a thermal equilibrium. This temperature difference definitely would cause the cycle to slow down somewhat, but I don't see how that would stop it altogether. – Will Feb 08 '17 at 22:22
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    If the ceiling is wet and warmer than the dew point of your vapor, no condensation will occur. The answer Willioam bud links claims you need to cool the surface. If the surface tension is so high that you significantly lower the necc. vapor pressure for condensation, it is for that sam reason unlikely that drops will fall. The ceiling will stay wet. This added in hope it adds to answer, feel free to incorporate and delete this comment. – mart Feb 10 '17 at 08:33
  • @mart I think you meant to write "claims you don't* need to cool the surface". Anyway, I am puzzled by your claim that "it is for that same reason unlikely that drops will fall"*. Fog catching nylon nets have been used for many years in Chili. How do you think the droplets on those nets end up collected in the water container below? Yes, gravity, and nothing else. – Will Feb 10 '17 at 23:10
  • @WilliamBudd. Conditions conducive to fog are supersaturated air. This you do not have in your system. – bpedit Feb 10 '17 at 23:34
  • Whether or not drops will form and whether or not drops will fall are two separate issues. It's true that I don't have supersaturated air in this system, because as I pointed out in the first comment to your answer I don't have air in this system. – Will Feb 11 '17 at 00:01
  • Fog is already condensed. supersaturation has nothing to do with wether air is present or not. – mart Feb 13 '17 at 09:07
  • @mart Nothing to do with it? (super)saturation is a state of solution, i.e., a homogeneous mixture composed of two or more substances. There is only one substance in our location though: water vapor. Maybe you have a point somewhere, but it's not getting across so far... – Will Feb 15 '17 at 02:55
  • My point is not getting through because your understanding of vapors is lacking. Steam is supersaturated if the partial pressure of the vapor is higher than the actual vapor pressure at that temperature. – mart Feb 15 '17 at 08:14
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Why doesn't the gravitational energy in this system of evaporating and condensing water violate the second law of thermodynamics?

this is the second law :

The second law of thermodynamics states that the total entropy of an isolated system always increases over time, or remains constant in ideal cases where the system is in a steady state or undergoing a reversible process.

italics mine.

In reality creating an isolated system is an approximate process, one has to assume that external to the system conditions do not affect the system. In the statement of your question you already have opened the system to gravity, so it is not a closed system and the force of the second law does not apply.

This can be understood in the statistical formulation of entropy

[This definition] describes the entropy as being proportional to the natural logarithm of the number of possible microscopic configurations of the individual atoms and molecules of the system (microstates) which could give rise to the observed macroscopic state (macrostate) of the system. The constant of proportionality is the Boltzmann constant.Specifically, entropy is a logarithmic measure of the number of states with significant probability of being occupied

entropystat

Intoducing gravity into the problem introduces gravitons, the carriers of gravitational waves, and each gravitational interaction of a graviton with a putative drop generates extra microstates. As these come from the mass of the earth the system by construction is not isolated so that the second law does not apply.

Now for the content of the question: at best , if it is true that condensation can happen at a fixed temperature with special materials, as you state in a comment to bpedit, you are transforming thermal energy to gravitational energy to kinetic energy, and it might go on for a long time like those birds drinking water perpetually, until dissipation stops them. Dissipation would be the cooling from removing the tails of the distribution, and also the black body radiation cooling the system .

The distributions of kinetic energy of the water and of the vapor over it have long tails. It is the molecules from the tails that evaporate from the water and allow the droplets to reach the ceiling,

maxwedistr

i.e. acquire gravitatiional potential, and form the droplets on the ceiling surface (hypothesis that this can happen at constant temperature for special materials).

When a molecule from the tail condenses into a drop, the average temperature of the gas drops by that tiny amount because it is no longer contributing in the average that defines the temperature. The same had happened when the molecule left the liquid. When the drop falls, all the molecules acquire back the kinetic energy and if they drop in the water the steady temperature is maintained. If they hit the propeller of the turbine they give up the kinetic energy , and when they fall back into the liquid they do not restore the temperature to the previous value, because their kinetic energy, leaving with the evaporation has not been returned. So slowly the temperature falls, because it is connected with the root mean square of the velocities in the liquid.

So thermal energy is turned into gravitational energy which is turned into the kinetic energy of the turbine, so the temperature will fall to the point that no longer droplets can form on the ceiling. (depending on the material). If such a material does not exist,the other answers are adequate.

anna v
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  • +1, because I like how this answer addresses "the big picture". I don't have enough time at the moment to check if I agree with all its content though. Will do within a few days. If you have time, could you expand on your last sentence? For one, I'm not sure what you mean by "removing the tails of the distribution". – Will Feb 10 '17 at 22:52
  • Thank you for expanding your answer (and sorry about getting back here so late). You've certainly made a good case for why this system does not violate the 2nd law. There's just one last thing that still puzzles me -- with regard to "So thermal energy is turned into gravitational energy which is turned into the kinetic energy of the turbine, so the temperature will fall to the point that no longer droplets can form on the ceiling.". This sentence must indeed clearly be true for a closed system, but I'm not convinced the same is necessarily true for the open case... – Will Feb 15 '17 at 02:29
  • If the temperature of the system as a whole drops, it will elicit heat transfer from whatever medium surrounds the container to its inside, effectively cooling the surroundings while establishing a constant temperature inside (and at the roof of) the container. What the temperature gradient of this equilibrium would be is an engineering question, but I don't see that the temperature anywhere along this gradient would necessarily rule out the possibility of condensation occurring there. Now we have a machine that may continue to cool its surroundings until the heat death of the universe? – Will Feb 15 '17 at 02:44
  • I do not see why not. It will be a not very efficient heat engine. – anna v Feb 15 '17 at 05:04
  • I think I've narrowed my problem down now. A heat engine requires both a "hot sink" and a "cold sink", right? All along it seemed to me that the cold sink was missing. The surrounding environment acts as a hot sink, and externally there is indeed no cold sink. But your take is that (part of) this heat engine itself acts as a cold sink, made possible by the external influence of gravity, right? I supposed it just never occurred to me that a cold sink could be hidden inside a heat engine itself -- in plain sight if you will. – Will Feb 15 '17 at 07:29
  • more or less yes . Not a classical heat engine as you note, but it does use heat to turn into kinetic energy with gravity an intermediary – anna v Feb 15 '17 at 08:39
  • Turning thermal energy from a system at a uniform temperature into kinetic energy (directly or not) violates the second law of thermodynamics. – user253751 Feb 05 '20 at 11:27
  • @user253751 the law depends on "isolated systems" as I have emphasized in my answer. I do not know what you mean by "directrly or not" , it is not part of the definition. – anna v Feb 05 '20 at 11:46
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Think about how you could adjust the operation of your machine if you temporarily assumed it were not a closed system:

  • You could speed up the operation of your machine by heating the water and/or cooling the ceiling. In that case your machine is a typical heat engine, with energy being transferred from the hot water to the cold ceiling by convection and the spinning of the turbine a side-effect.

  • You could slow down, or stop, your machine by cooling the water and/or heating the ceiling. That's how the rear-window defroster on your car works, after all.

That means that somewhere between those two temperature gradients is a configuration where your machine doesn't run at all. If you set it up and leave it closed, it will eventually reach this equilibrium configuration and stop.

Now it is possible that, since gravity is involved, the equilibrium configuration isn't actually at uniform temperature. For instance, if the chamber were ten miles high, the water vapor molecules near the ceiling would have less average kinetic energy than those near the bottom, and a lower effective temperature. But, like all perpetual motion proposals, it will only run for a while at best.

Community
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rob
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  • Thanks for your answer. A temperature difference slows down the cycle, but will not stop it; provided that they remain in-between freezing and boiling point. Vapor pressure describes an approximately logarithmic relationship between temperature and pressure. In/decrementing one parameter does not drop the other to zero. Here's the relevant diagram. – Will Feb 09 '17 at 02:31
  • Also, your last sentence lacks argumentation. As it stands, it reads as: "This is not perpetual motion, because perpetual motion doesn't exist.", which is circular reasoning (regardless whether the assumption holds or not). – Will Feb 09 '17 at 02:33
  • Two replies, reverse order. (2) I don't think that appealing to the second law of thermodynamics is circular reasoning. If the system is closed, its total possible entropy is finite, and its entropy tends to increase over time. Eventually it'll fluctuate up to some maximum and stop there. I don't have to calculate those entropies to know that the limits exist. – rob Feb 09 '17 at 02:47
  • (1) Are you really proposing there is no temperature gradient with a hot ceiling that would prevent condensation? I think you're mistaken. Limiting to freezing/boiling isn't reasonable --- if the maximum-entropy endpoint of the system has a layer of ice on the water surface, that's what'll eventually happen. – rob Feb 09 '17 at 02:47
  • (1) Sure, it seems highly likely to me that a ceiling at or near 100 C would probably prevent all condensation. However, I see no reason why it needs to become that hot, provided proper heat transferring means are installed between the ceiling and the colder liquid surface near the bottom (as I mentioned in my question above). Surely a copper heatsink would prevent such a huge temperature gap between its ends? – Will Feb 09 '17 at 03:04
  • No heatsink can overcome the second law of thermodynamics. The convection and condensation is a way to move heat and entropy around within the system; it's temporary. – rob Feb 09 '17 at 03:18
  • There are two separate aspects to do this conundrum. (A) The heat produced by condensation can be transferred to the liquid to be used for evaporation, all within the system. This is not temporary. It is continuous. (B) So where would the energy leaving the system through the waterwheel generator come from? It would enter the system from the surrounding environment as heat. The system would effectively cool the environment over time. Note that I never said this was a closed system. – Will Feb 09 '17 at 03:24
  • If it's not a closed system, then you have an engineering question about efficiency, not a conceptual question about the second law of thermodynamics. – rob Feb 09 '17 at 03:29
  • From an outside perspective this is a black box that takes heat from its surroundings and converts it into electricity. How is that not a question about (the validity of) the second law of thermodynamics? – Will Feb 09 '17 at 03:33
  • It is not a closed box i.e. isolated system and the second law is testable only in isolated system. All living matter violates the second law after all, but it is not a closed system. – anna v Feb 10 '17 at 07:12
  • @annav This system would generate more electrical energy than the Gibbs free energy contained in the surroundings for its given temperature -- in violation of the 2nd law. Also, living matter does not violate the 2nd law. – Will Feb 10 '17 at 08:03
  • @WilliamBudd If you do not see that biological processes make order within the skin of a live organism take crystalization. The crystal has high order i.e. small entropy but it was produced in an open system, so crystals are not a violation of the second law. – anna v Feb 10 '17 at 11:42
  • @annav Interesting examples ...but I'm going to abandon arguing that point here because I'm afraid I'd just end up hopelessly off-topic and lost in semantics... – Will Feb 10 '17 at 23:18