-1
$\begingroup$

enter image description here

Let's say we would fill a long tube with water. Let's say the tube is 1km in height.

At the bottom of this tube, the pressure should be much higher than the top. Also, the tube would lead into a container at the bottom as shown in the picture.

Using some kind of mechanism, we would be able to seal the water within the tube from the container at the bottom. Could be valves or some other shutter mechanism depicted as the red line in the picture.

Then we would open the yellow valve. The water being at a much higher pressure than the pressure outside, should shoot up and be able to rotate the wheel as drawn in the picture.

The water would be collected and be sent back into the container. Then we open up the red valves/shutter and let it pressurize again to repeat the process.

Now i am not looking for a perpetuum mobile, but i do not see why this wouldn't work either, so i am trying to guess where the energy would come from.

Since energy is extracted from the water, my best guess would be the water is getting colder, but since this isn't a completely closed system, the water should heat up back again(repressurized) from the atmosphere surrounding it.

Of course i could be wrong (probably am) but i cannot see where else the energy would come from. Any ideas?

Ignore the green shutter. It's just for some fine tuning not really adding to the experiment much.

$\endgroup$
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Jan 13 '16 at 18:00
4
$\begingroup$

Some energy would come from the part where "the water would be collected and sent back into the container", which you somewhat glossed over.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I cannot see why. The water would simply drop back from a greater height back into the container. In fact, you could actually have it do work in the process. The container would have less pressure afterwards, so you wouldn't need to "force" the water back in, if that is what you are thinking. $\endgroup$ – user2367657 Jan 13 '16 at 15:13
  • $\begingroup$ @user2367657 Water compresses (slightly) under pressure. If you want all the water back in the container, you have to compress it again. If you don't want to compress it again, then not all the water will fit in the container. $\endgroup$ – user253751 Mar 6 '19 at 4:10
3
$\begingroup$

There are two possibilities: water (or whatever fluid you're using) is compressible or it's incompressible.

If it is incompressible, when the red valve is closed the tank pressure drops to zero, and opening the yellow valve produces no effect on the generator.

If it is compressible, its pressure remains constant when the red valve is closed, and it expands out of the tank when the yellow valve is opened. We're good so far. However,

The water would be collected and be sent back into the container.

requires that the water be recompressed in order to fit in the tank. This compression process will take (and it's not a coincidence) exactly as much energy as the expansion out of the tank produced. Assuming no temperature change between the beginning and end of the experiment.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "requires that the water be recompressed in order to fit in the tank". I don't think it requires to be recompressed. The container/tank at the bottom was under higher pressure when the red valve/shutter is closed. Meaning that the water actually would need MORE space/a bigger tank under normal pressure. So choosing the right sized tank, the non pressured water should fit just exactly in the size of the tank afterwards at normal pressure. With the green shutter i told you to ignore, containing some pressured water still, you could fine tune it, but not really needed with a proper container size $\endgroup$ – user2367657 Jan 13 '16 at 15:26
  • 1
    $\begingroup$ @user2367657 - Goddess give me strength. No, it won't fit back in the tank without compression. It expanded, remember? And "fine tuning" by adding water via the green valve will work fine as long as you realize that that will take energy as well, and the total will still result in zero energy production. $\endgroup$ – WhatRoughBeast Jan 13 '16 at 15:32
  • $\begingroup$ Ok, even though i don't like your attitude, i will explain again. Let's take a closed metal container with water inside we were able to fill in without any compression. Now heat this container/water inside up. The water heats up/gets pressurized and would if we open some shutter shoot out of this container. Hope you agree up until here. Now when the water gets cold again, we can fill it back in without having to compress it. Right? How is my example different from this? $\endgroup$ – user2367657 Jan 13 '16 at 15:39
  • $\begingroup$ It's advisable not to talk down to the people who are trying to help you. You're now unlikely to get any more helpful comments or answers, and that's your fault. $\endgroup$ – Matt Jan 13 '16 at 15:51
  • 1
    $\begingroup$ @user2367657 - "Now when the water gets cold again, we can fill it back in without having to compress it. Right? " Right. And until you pump energy back into it by heating it, it won't expand again. Right? And guess how much energy that will take. If you're thinking, "Exactly as much as I got out when it expanded and cooled off", you've begun to see the light. $\endgroup$ – WhatRoughBeast Jan 13 '16 at 16:03
1
$\begingroup$

In order to send it back to the container you have to do work against gravity. The energy obtained by rotating wheels would be from the kinetic energy of the fluid which came into existence because of the work we did. The energy from wheel would equal the work we did to pour the fluid back in. So energy gain would be 0. Moreover the efficiency of the wheel will never be 100% and you will end up using energy than producing it.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That was already answered by me. You would actually gain work from letting water drop from a greater height back into the now unpressurized container, so unless you can be more detailed this does not count as an answer. $\endgroup$ – user2367657 Jan 13 '16 at 15:31
  • $\begingroup$ It's difficult to be more detailed when you have described in exactly zero detail how you intend to get your water back to the top of your container. Or, even, to a greater height than the top of your container. $\endgroup$ – Matt Jan 13 '16 at 15:53
  • $\begingroup$ The water shoots up, spins the wheels, gets collected by some bucket or whatever else up there. When no more water shoots out from the bottom, we just let it drop back in from the bucket back into the container by opening another shutter or whatever else. I did not explain this in detail because it can be done in a million different ways and is not really adding anything to the experiment. $\endgroup$ – user2367657 Jan 13 '16 at 15:57
  • $\begingroup$ What is it that makes you think the water will shoot up to a height higher than that of the top of the container? $\endgroup$ – Matt Jan 13 '16 at 16:02
  • $\begingroup$ Higher pressure. What makes you think it would not shoot out? $\endgroup$ – user2367657 Jan 13 '16 at 16:14
0
$\begingroup$

You will not be able to produce any useful energy with this system. The previous answers have picked up on a number of the issues...

  • Water is nearly incompressible, so the gain in pressure that you are counting on is too small to do any significant work. Don't forget that the work required to fill a 1km tall pipe with water is part of the system.

  • The mechanical part of the system will always require more energy than it provides.

  • The water that leaves the vessel as you release the valve will not fit back into the vessel, though as I stated the amount that will not fit back into the vessel will be very little due to the lack of compressibility. Your response to WhatRoughBeast is different, to answer your question. The model you present compresses the water with the valve open, and closes the valve with the water in a compressed state, in your response you are pressurizing the water after the valve is closed by adding heat, these are not the same. In order to return the water to the vessel, you would have to apply the force that the column of water had provided prior to closing the valve, thus doing the work that you had released.

Your responses demonstrate that you are not open to considering the information being provided as answers, my suggestion is that you build a scale model and go into the lab, you call it your experiment so why not try it.

As another suggestion, your drawing looks an awful lot like a hydroelectric power plant, lots of engineers and scientists have worked to make these systems as efficient as possible, looking at what that industry has done may give you some clarity.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Your first part is irrelevant. The experiment starts with a tube full of water. The question is if this process can continue indefinitely, getting energy out of the system which is NOT just potential energy from the water dropping down. Hence, if there is also energy involved drawn from the atmosphere as the water heats up again through the air/sun. What i got from the answers so far is that i have to refine the experiment to make it more clear. $\endgroup$ – user2367657 Jan 13 '16 at 17:29
  • $\begingroup$ It boils down to if when using a container at the bottom with water that is not pressurized yet through the tube, and THEN opening the red valve. IF after opening the red valve or shutter the water in the tube would drop exactly by the amount required to pressurize the container water. Hence, transforming potential energy of the water in the tube into kinetic energy of the water in the container. If it does not drop by exactly that amount, then the energy has to come from other than sources as well, other than just potential energy. $\endgroup$ – user2367657 Jan 13 '16 at 17:32
  • $\begingroup$ If the water in the column drops to pressurize your vessel, and then you release it, when the water runs back into the vessel it would either not fit back into the now unpressurized vessel because the volume expanded, or it would have to push the water in the column back to the original height. Judging from your new explanation the paddle wheel does nothing to get at what you want to determine and you really want to measure the temperature change of the water do to the pressure change. Again, give it a try and see what happens. $\endgroup$ – blaydes Jan 13 '16 at 18:07
  • $\begingroup$ See above. What you describe would be true, only if the increase in pressure(kinetic energy) would come ONLY from water in the tube dropping(permanently), hence potential energy from the water in the tube. IF however, the water in the tube does not drop by that amount, then the kinetic energy to pressurize the water in the container would also come from atmospheric energy(sun, air heating up the tube). Therefore, water molecules in the tubes passing kinetic energy to the container molecules just to be heated up again by the sun/air. Which would also mean that the water in the tube would cool. $\endgroup$ – user2367657 Jan 13 '16 at 18:24
0
$\begingroup$

Conservation of energy works pretty well.

Assume that instead of a squirting fountain and a paddle wheel, you had a cylinder and a piston to allow the compressed water to do work.

The work done would be $\int PdV$.

At the end of the expansion the water in the lower container is at atmospheric pressure. Now we can either repressurize by opening the red valve (in which case the water column drops and we use up potential energy) or we push the water back in - in which case we have to do exactly the same amount of work as was done when the water expanded, plus a bit more to overcome friction and other losses.

If we decide to let the water cool down before compressing the cylinder, then a certain amount of heat would escape from the liquid in the lower container; although this would allow the water to be "sucked back in", we would have a vessel of unpressurized water now. Pressurizing it from the column would require a loss of potential energy; pressurizing by heating would require heat energy.

The question then becomes - did it take more energy to heat the liquid than we extracted when we cooled it down? The answer is "yes" - because when a compressible liquid expands, it cools down a little bit. That is more obvious in a gas but it occurs in a liquid as well. So during the expansion you loss pressure and temperature. And even if you could extract the rest of the heat in the liquid perfectly after the expansion, when you put it back you would have less pressure than before.

Nature is ruthless. No free lunch.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Sounds simple... if you ignore the environment that is. Sun light/air both transferring heat/energy from the pipe into the water. Think of the same experiment but instead of water, use some ideal gas on a planet with much higher gravity than earth. Would the gas-level(as in water level inside the pipe) after you extracted pressure/energy from the container and then re-pressurized it by opening the shutter, remain the same (lower level) at all times, or would it balance itself with the environment it is within to a different level (higher as i assume), given you gave it enough time? $\endgroup$ – user2367657 Jan 13 '16 at 23:59
  • $\begingroup$ Or to give a more extreme example. Do the same with the gas, but no shutters at all. Just the pipe and the container. But this time we use two systems. One system pipe+container we fill up with the same amount of gas, but cool it down which should give us a lower level. The 2nd system pipe+container we heat the gas which should give us a higher level. Now give both systems enough time within the same environment (sunlight+air). Wouldn't they both strafe towards the same balance within the environment and consequently both end up having the same gas-level within the pipes in both Systems? $\endgroup$ – user2367657 Jan 14 '16 at 0:28

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