# Phase change floatation cycle

1. Jul 16, 2012

### Ambforc

Consider the following theoretical cycle. There should be a flaw somewhere, but I could not find it yet and will appreciate it if someone can point it out to me. A lot of ideal assumptions are made, such as reversible heat transfer, but it is not intended to be practical, as long as it adheres to allowable theoretical ideal assumptions.

Two identical expandable vessels regulated to be at independent constant pressures using some means, i.e. connected to an external constant pressure source, contain equal amounts of a saturated fluid. These vessels are insulated in the sense that no heat is lost to the surroundings, however, heat can be transferred between them and provided by an external source, therefore they are not strictly adiabatic. Both vessels are placed in a liquid body of significant depth. The liquid is assumed to have no appreciable change of density with pressure (height).

Step 1:

One vessel starts of with all fluid in a saturated liquid form (vapour fraction 0). The combined density of the vessel and fluid in liquid form allows it to sink to the bottom of the liquid body. The second vessel starts with all fluid in the saturated vapour form (vapour fraction 1). Since both vessels are at equal pressure, the second vessel is expanded significantly without an increase in its mass, causing its overall density to be very much less than the density of the body of liquid it is immersed in, and it floats on top.

A thermally conductive link is made between the two vessels, allowing thermal energy to be transfered between the two vessels. In order for heat transfer to occur, a thermal gradient will be needed. This can be accomplished by dropping the regulated pressure of the vessel at the bottom of the body of liquid by an infinitesimal small amount, causing some vapour to flash and the temperature of the fluid to drop infinitesimally. (To the same degree that the gradient it infinitesimal, the heat transfer will take an infinite amount of time to complete, but I do not believe that is a crippling condition, I will return to this later.) Thermal energy is now transferred from the fluid at the top of the liquid body to the fluid at the bottom of the liquid body until the fluid in the first vessel is saturated liquid and the fluid in the second vessel is satrurated vapour.

Step 2:

The vessel originally at the top of the body of liquid now has an overall density that is greater than the density of the liquid, and it sinks to the bottom. Connecting that vessel to a pulley system allows it to do work on the pulley consisting of the net downward force times the distance. If the vessel is dropped infinitesimally slowly, the drag force can be eliminated for simplicity, but this is not a necessary condition.

Likewise, the vessel now filled with vapour at the bottom of the liquid column rises to the surface and can do work on a pulley system.

Step 3:

The vessel now at the top of the liquid body needs to be restored to the initial condition for the vessel at the top of the body of water mentioned in Step 1. This is done by increasing the pressure regulated pressure to its initial amount before it was dropped in Step 1 to allow for heat exchange to take place. Some of the fluid will condense and latent heat needs to be supplied until all of the fluid is in the vapour state again. The amount of heat required will be infinitesimal for an infinitesimal pressure drop, and will be larger for a larger pressure drop in Step 1.

Now, as long as the heat required to restore the system to its original state is less than the work done on the pulleys in Step 2, this cycle is outputting net work, which can't be right. Since these two steps are not directly related in any way, there appears to be no obvious reason why they should cancel out, and they do not appear to do so if all processes are taken to occur infinitely slow, maintaining reversibility.

I will appreciate comments. Please let me know if something is unclear. If a more quantitative argument is preferred, I will supply it as soon as possible, but I expect that most people acquianted with thermodynamics will get the basic idea.

2. Jul 16, 2012

### CWatters

It takes energy to phase change the medium and to some extent that's dependant on pressure.

3. Jul 17, 2012

### Ambforc

Yes, of course. However, I fail to see how that answers my question.

4. Jul 17, 2012

### CWatters

A thermally conductive link is made between the two vessels, allowing thermal energy to be transfered between the two vessels. In order for heat transfer to occur, a thermal gradient will be needed. This can be accomplished by dropping the regulated pressure of the vessel at the bottom of the body of liquid by an infinitesimal small amount,

Can it? Wouldn't the thermal gradient created quickly dissapear as soon as a little heat is transferred?

I very much doubt it would keep going "until the fluid in the first vessel is saturated liquid and the fluid in the second vessel is satrurated vapour".

Are the vessels rigid?

If not then you would have to overcome the pressure on the outside of the one at the bottom AND then reduce it further to cause the content to evaporate.

If it is rigid why would one sink? The mass and volume is constant regardless of internal state.

5. Jul 17, 2012

### Ambforc

Since the pressures are slightly different in the two vessels, the saturation temperature is also different, and there will be a temperature difference in the two vessels as long as they remain saturated (have both phases) and do not become subcooled or superheated.

The constraints on the vessel is that it should be able to change the volume of the compartment containing the saturated fluid so that it is always at a constant pressure, regardless of its vapour fraction. Furthermore, it should be possible to control the pressure on the compartment containing the saturated fluid at a constant level, this pressure being at least equal to or higher than the pressure experienced at the bottom of the water body.

One way this can be accomplished is to visualize a rigid U-shaped shell with a diaphragm connected to expandable piston-like devices connected to a constant pressure source (see attached drawing).

At the bottom, when the liquid in the vessel is busy evaporating, it will have to do work against the constant applied pressure. However, at the same time the vessel busy condensing is contracting in size. Except for the slight pressure difference to get the thermal gradient, the change in enthalpy (as apposed to just internal energy) between the two vessels is equal, therefore the fluid in the bottom vessel will receive enough energy to do the required work against the externally applied pressure.

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6. Jul 18, 2012

### CWatters

I'm not sure what to say. You still have to do work to keep the pressure lower in the submerged vessel and that's against the external liquid pressure otherwise there is no increased bouancy.

7. Jul 18, 2012

### Ambforc

I don't have to keep the pressure lower in order to get a vapour than the pressure at the bottom of the liquid body as it is at a temperature such that the saturation pressure is higher than or equal to the maximum pressure exerted by the liquid body above the vessel.

Thanks for trying anyway.

8. Jul 18, 2012

### jbriggs444

Well, let me have a go at it.

It appears that you are describing a perpetual motion machine and asking where you've gone wrong. The obvious answer is that you are using constant pressure and buoyancy in the same apparatus. But those two notions are mutually exclusive.

Let me go over the description more carefully...

Your initial post describes two expandable vessels at independent constant pressures.

"Expandable" would seem to mean that they are like balloons.
"Constant pressure" would seem to mean that they are rigid.

You elaborate with a drawing showing a somewhat cryptic piston arrangement. You have what is purportedly a constant pressure source pushing on part of one side of the piston and you have the ambient fluid pushing on another part of that side. Clearly, this arrangement will not guarantee that the working fluid on the opposite side of the piston is under a fixed pressure.

In Step 1: you use the assumption of equal pressure in the working fluid to conclude that the vessel which contains vapor occupies a greater volume than the vessel which contains liquid.

But the setup of the problem as I have understood it contradicts the assumption of equal pressure in the working fluid. So you'll have to describe the setup more carefully.

9. Jul 19, 2012

### Ambforc

Thank you very much, I believe that is the answer. Too bad I missed it, it was rather obvious as you say.