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This question already has an answer here:

I recently came up with the following concept. It is very simple, and may have been thought of before. A picture says more than a thousand words, so here is it explained in a picture:

Conveyor belt using densities of gases to keep it in motion Note that water was used to make the example easier to understand. Another gas (denser than Gas A and B) could be used instead, resulting in less friction than when using water.


At first glance, it seems that this machine could run for an indefinite amount of time.

But I do not deem it possible to break the law of energy conservation. However, I have a hard time finding out what kind of force would cause this machine to slow down and stop.

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marked as duplicate by Bernhard, Jim, ACuriousMind, Kyle Kanos, Qmechanic Sep 7 '14 at 16:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Is "perpetuum mobile" a common term for perpetual motion machine? I'd never heard of it before, and the wikipedia article is about music. $\endgroup$ – BlueRaja - Danny Pflughoeft Oct 31 '13 at 15:31
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    $\begingroup$ @BlueRaja-DannyPflughoeft Here in the Netherlands I have heard that term quite often, pronounced per-pe-tu-um-mo-bi-lay. $\endgroup$ – orlp Oct 31 '13 at 17:50
  • $\begingroup$ Perpetuum Mobile is Latin, and translated it indeed means 'perpetual motion'. People like Da Vinci already experimented with perpetual motion in the time that all science and literature in general still written down in Latin, and since then the name stuck. $\endgroup$ – Qqwy Oct 31 '13 at 17:53
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    $\begingroup$ possible duplicate of What is the fallacy in this infinite motion machine? $\endgroup$ – Bernhard Sep 7 '14 at 12:09
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If gas A and gas B are of different density, then the situation sketched is not in equilibrium: the water level on the side of the light gas will be higher. There, the containers are moving down, and you have to push your containers through this net difference in level. You do need to put in energy here, which is probably the piece that you are trying to gain per cycle...

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In addition to Bernhard's answer, just because three gases (Gas A,B and air - which is itself a mixture of nitrogen, oxygen, and other gases) have different densities, it does not mean they will remain seperated when in a container. In fact, as entropy of the system increases over time, Gas A, B and air will make an even (if heterogeneous) mixture.

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It does not work for the following reason. Let's look at the right side, where the containers float to the top.

When a new container enters at the bottom (from water to gas B), it pushes some gas B away, so it can occupy the space. The pushed away gas B has nowhere to go, but up. Gaining height means potential energy. That is energy used, to make the machine go perpetual (read: this piece of energy is "consumed"/used/lost, just to make the container enter the gas B area). But since there is no energy input, existing energy is used up for that. So what you "gain" by floating up ("free" energy pushing the containers upwards), you "lose" when new containers enter the gas B compartment.

(the same is true for the left side, just "in opposite")

So the machine has no "extra" energy source. Its existing energy (if you start observing when the machine is spinning) will be used to fight friction and by that warming up the gases and eventually stop. Like every other perpetuum mobile (@BlueRaja - Danny Pflughoeft, yes, that is a common name for this, maybe less so in english, see http://de.wikipedia.org/wiki/Perpetuum_Mobile)

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  • $\begingroup$ Also part of the reason is what Bernhard explained, of course. $\endgroup$ – David Balažic Oct 31 '13 at 17:51
  • $\begingroup$ @Boluc Papuccuoglu - instead of gas A and gas B some liquid could be used, that would stay separate (think water and oil)... (commenting here, as I can not comment other peoples answers) $\endgroup$ – David Balažic Oct 31 '13 at 17:55
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This is just a restatement of David Balažic's answer.


We agree, right at the outset, that this perpetual-energy-source machine won't work, and are only trying to narrow down where exactly it fails. But, neglecting all dissipative forces like friction and viscosity, the machine may very well work as a perpetual motion machine - when you attach a dynamo the whole thing will have to come to a halt.


First thing to do is to simplify the system so as to avoid misleading explanations, though they may contribute partially to the overall failure. So, we'll remove the air and water portions (or make the thickness of those 0) - just two layers left and right. Also, we'll not worry about the gases mixing at the common surfaces. We'll leave the (impossible) construction of this oversimplified setup to engineering.

Also let's restrict our discussion to a single container. The perpetual motion argument presented in the question will go through for this case also, without the added complication of containers driving one another.


Right side

What happens here is similar to leaving a hollow plastic ball at the bottom of a tube of water. There is a net flow of mass downward (Gas B downward, container upward). The loss of PE is converted into KE of the container. No problems here.

Left side

Ordinarily to restore the system to the initial condition, we'll have to push the container through Gas B (against the increasing pressure gradient). But instead, we've decided to cheat by dropping the container through Gas A (against a much more slowly increasing pressure gradient - one overcome by the weight of the container itself) and then relying on the speed developed while travelling upward through Gas B and downward through Gas A to take the ball back into the Gas B (across the pressure barrier).

Claim is that the work required to overcome this barrier is the energy gain achieved elsewhere. An clean way of seeing this is the following:

Work done to break the potential barrier $W$ is given by $$W = (\rho_B-\rho_A)ghV$$ where $V$ is the volume of the container. KE gained when moving through Gas B $KE_2$ is given by $$KE_B = (\rho_B-\rho)ghV$$ where $\rho$ is the density of the container. Similarly, $$KE_A = (\rho-\rho_A)ghV$$ Now, clearly $$KE_A + KE_B = W$$

In other words, in the absence of all dissipative forces, the container will barely make it back into Gas 2 (unless the container had a non-zero initial velocity in the right side to begin with). Add a dynamo, and the machine will stop immediately (or after the initial KE has all been lost in the dynamo).


Final words:

  1. I say clean, because of some of the oversimplifications involved. The work done in breaking the pressure barrier is actually work done to lift volume $V$ of Gas B by height $h$ minus energy gained by the drop of volume $V$ of Gas A by height $h$. But our oversimplified system has no room for change in heights of the gas columns. I've ignored this and conveniently used the formula for pressure at the bottom of a fluid column open at the top (which is not applicable here) so as to not muck up the discussion. But this where some of the other answers come into the picture.

  2. This is kind of interesting. It seems that our little trick of dropping the container through Gas A has accomplished perpetual motion at least, if not perpetual energy generation. But I claim that nothing has been accomplished and perpetual motion has always been there to begin with. Can you see why? What happens to a hollow plastic ball placed at the bottom of a tube of water in the absence of dissipative forces?

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