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Real perpetual motion?? Options
 
SWIMfriend
#1 Posted : 6/5/2011 5:17:24 AM

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I only have a year of college physics, so I'm not FULLY prepared to debunk this--but I do have enough knowledge to LOVE to debunk the infinity of perpetual motion machine videos on youtube. This seems a bit different.

The principle is this: A pipe of water (up to ten meters high, at standard AP), with the bottom open and submerged in a pool of water, and the top closed. The water will remain in the pipe, due to air pressure on the surface of the pool. If you place a buoyant object into the tube (it would take a tiny bit of energy to push it the TINY BIT under water to get it in the tube), the object will then float to the top (and you could steal a tiny bit of energy from that phase--which would slow the ascent. Or maybe you wouldn't do that, so that you could use the HEAVIEST object possible--the object would be, like, styrofoam, with metal attached). THEN, when it gets to the top, you close a valve, open the top, take the object out, and you could gain all the energy produced by DROPPING the object 10 meters. Then close the top, open the valve, and do it all again (he discusses using two valves--I don't see a problem with his "valve thinking" ).

To me, it seems like, to DISPROVE that this could work, you would have to show that the energy used in pushing the buoyant object a tiny bit underwater, plus the energy to WORK THE VALVES, would HAVE TO BE more than you could acquire from dropping the object. To me it doesn't seem clear that those costs would HAVE TO be more (please note, OF COURSE it might be IMPRACTICAL; but such things aren't disproved because it can be shown they're impractical. They're disproved because they CANNOT produce more energy than they would use).

Just curious if any physicists or engineers in the audience can easily demonstrate why this COULD NOT provide more energy than it uses to work (in theory).

Here's the video:

BTW, the guy is actually quite creative. Here's a video of his "ten cent" solar collector.

Also, here's a "buoyancy and perpectual motion" debunking page--but it doesn't seem to address exactly this type of idea.
 

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The Traveler
#2 Posted : 6/5/2011 1:54:51 PM

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I think I see why this cannot work, it has to do with the displacement of water by the object floating to the top.

As I understand it the tube will have two valves, one at the bottom and one at the top. Once the object has risen to the top the bottom valve is closed, after which the top valve is opened and the object is taken out.

At the moment you take the object out, air will be in the spot where the object was. So the water that was displaced by the object has to be put back at the top to get the next object rise to the complete top again. If you don't add that water then the next object will not reach the top since there will be a little bit of air there. More objects would mean more air until the tube is completely empty.

Logical thinking makes me state that: The energy that the object gains by rising to the top is the same energy that you need to put the water back at the top to replace the air. This to make sure that the next object floats to the complete top again.

Of course you could say "take an object that has more mass per volume" but the moment it has an theoretically energy advantage that object would sink and not float. Pleased

So in practice this will bleed energy since the valves need to be opened/closed and the objects put in/taken out.


Kind regards,

The Traveler


 
gibran2
#3 Posted : 6/5/2011 2:59:58 PM

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Traveler got it right: As soon as you open the top, you’re allowing air into a reduced-pressure system – the floating object displaces an equal mass of water (part of the object will still be submerged, so it’s not equal volume). As a result, the water level will be lower. The next time you open the top, the level will drop a bit more. Finally, the level in the pipe will match the level in the pool.

Even if we ignore the energy required to work the valves, there is still energy lost as the pipe loses water: it takes energy to get water into the pipe in the first place, and conservation of energy tells us that the potential energy produced by the rising object is lost by loss of water in the pipe.

Another way to look at it – each time the top valve is opened and the object is removed, an amount of water equal in mass to the mass of the object is displaced. So, if we don’t want the water level to gradually fall, we could add that water back. But as soon as we add it back, we balance the gain. We must use energy to lift the mass of water to the top of the pipe, and that energy is equal to the potential energy gained by the rising object. Can’t get around that pesky conservation of energy.
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SWIMfriend
#4 Posted : 6/5/2011 6:10:29 PM

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The Traveler wrote:
I think I see why this cannot work, it has to do with the displacement of water by the object floating to the top.

As I understand it the tube will have two valves, one at the bottom and one at the top. Once the object has risen to the top the bottom valve is closed, after which the top valve is opened and the object is taken out.


I'm missing something, in the need to have two "valves." I don't see why one isn't enough. And instead of a "valve" let's call it a "plate."

First, imagine a square or rectangular tube, instead of a circular one. And the only "valve" is a flat piece of metal that rests in a square slot that extends out from the square tube filled with water.

When the object gets to the top (and I don't see why there can't be an airspace at the top, with air at STP*, so that STP air is just exchanged with other STP air when you take the top cap off to remove the object), you slide the piece of metal out from it's extended slot, into the tube area, where it would be supported around the edges by a ledge built it at that height. If it's done right, you then have a full tube of water below the plate, and above the plate is a fraction of water, the object (only partly submerged in it), and some STP air above it. When you take off the cap and lift the object out, the SAME QUANTITY of water and STP air should be in the top section, both before and after you remove the object. You then put the cap back on, and retract the plate back into its slot area, out of the way of the main tube (the slot area would allow the plate to rest in it with some slight extra room, and would in any case be "underwater" at all times--and would not allow transfer of air (or an air pressure effect) to affect the water in the tube below that point)

It that scenario, it seems like conditions would always remain the same inside the tube--before and after taking out an object

* I'm concluding that a fraction of air at STP could remain at the top of the column based on the fact that, when he showed the 10 foot glass pipe in use, it had NO VACUUM SPACE at the top (ala a barometer)--because ambient air pressure was enough to support the column--and indeed it would support a column of slightly under 10 meters at sea level, IIRC.

I don't see why there can't be a permanent air space at the top of the column, that would not change because of displacement or from air pressure changes by taking off and replacing a top cap (as long as the plate-valve was in place).


Of course, the contraption I describe would be IMPRACTICAL. But the main energy requirements are lifting the object a small distance OUT of the water, and forcing it a small distance, against it's buoyancy, to put it BACK into the pipe. It's difficult to see how that would be more energy than the energy gained by dropping the object 10 meters. But I'll admit I'm not certain about STP air remaining at the top of the column through a cap-on/cap-off procedure--with plate-valve in place to prevent air pressure from entering at the top. In that situation, I don't think that displacement is an issue (there's always the small problem of compressibility of the water--surely the water at the top of the column should be SLIGHTLY less dense than the water at the bottom, no?)

I'm almost thinking that this is more of a "Maxwell's Demon" sort of problem that the standard "Perpetual motion" sort of problem. Perpetual motion machines are EASILY MADE if natural forces present could be turned on and off at no cost (typically if gravity or an magnetic force could be turned on and off). A Maxwell's Demon situation requires only INTELLIGENT MANIPULATION to get energy.


EDIT: Hmmm. I'm rethinking this. I'm puzzled. Even as I described it, you would somehow HAVE TO allow extra air to come into the top area after opening the cap, because the volume of the space does HAVE TO change when you removed the object, and it HAS TO come from air--meaning you would have to allow more air in the top section after a trial.

This is puzzling, because I can't picture where the "change" takes place. Well...AFTER you slid the plate-valve across, the object would NO LONGER BE BUOYANT, because there would no longer be water pressure UNDERNEATH IT. But that shouldn't change anything...YES. I guess it's true: there's just NO POSSIBLE WAY of preventing MORE AIR from coming into the system with each trial.

Yet...that seems to be a PRACTICAL PROBLEM (although an insurmountable one) rather than an ENERGY problem. That's very interesting. The interesting event happens when you slide the plate--at that point the object is no longer "buoyant." This only makes the scenario MORE interesting to me, actually.

Right though: the reality is that any air at the top of the column would have to COMPRESS as soon as the object entered the pipe--there's just more volume in the pipe when the object is in there. Of course the water in the entire POOL also rises when you push the buoyant object under the water. To me this is EXTREMELY INTERESTING because it doesn't seem that "energy conservation" foils you so much as the ABSOLUTE REALITY that you can't retrieve the object without "spoiling" the unique condition of the system.

In a way, this is like all PP problems, where, for them to work, forces have to magically turn on and off--in this case, air pressure would have to go from zero to STP and back again, when taking the top cap off, in order to avoid more air coming into the top with each trial. To me that's a VERY interesting "undoing" of this scenario.
 
SWIMfriend
#5 Posted : 6/5/2011 7:04:54 PM

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OK. There is a way around it: Have an airspace, and then, BEFORE taking the cap off, lift the object up INTO the airspace, and then slide the plate-valve exactly on top of the water.

But to do that you would end up COMPRESSING the air in the airspace (when the object displaced AIR). And that compression would use energy which (as these things always do) would equal the energy you would get out. But it's not easy to see exactly WHERE or HOW that it would work out that they're equal: the energy you get out relates to harvesting from GRAVITY (dropping the object), the energy you need to use is to compress a gas. It's dificult to see the energy "equilibrium" where one ends up equaling the other.

EDIT: Of course, the air pressure that holds up the column only EXISTS because gravity has compressed the air to that extent...
 
SWIMfriend
#6 Posted : 6/5/2011 7:53:43 PM

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A variation. Suppose you used a (magical) buoyant object that slowly dissolved, so that a thin monofilament line could be attached to it, and it could do work as it buoyed up the column, by lifting a weight, through pulleys, outside the column.

Then, other than the cost of reattaching new, dissolvable buoyant objects to the monofilament each time, there would be no valve issue or air issue, etc.

Of course, I haven't factored in the energy cost of "recrystallizing" the buoyant object--but let's say for sake of argument, that it's not important.

The point is, I don't think you could RELATE the chemical/energy costs involved with producing the object, to the machine output harvest from raising the weight each time. It would only be "coincidence" if the synthesis cost of the buoyant object was more than the energy harvested.

Of course this is an INSANELY impractical idea. But I would have to wonder if it contains some ABSOLUTE restriction to say that net energy could positively not be gained by such a scenario. If the cost of synthesis is not included, this seems to be the only setup I've seen that doesn't require mechanical energy to be SUPPLIED that equals the mechanical energy that is HARVESTED.
 
Garfield
#7 Posted : 6/6/2011 3:16:38 AM
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The good old problem or task of perpetual motion.

It can not be achieved on this way.Stop

First of all you cannot produce energy out of nothing, it can only be converted. This means you can only "steal" it from another system.

On the other hand the second law of thermodynamics which is often (mis)used to debunk "free" energy also fails! Stop
Why?
A very simple reason: The law is based on closed systems and this is BS, because there are no closed systems in our universe. Everything interacts with everything else, so there is always a transfer of energy and/or inner and outer order between systems.

The conclusion here is that overunity machines are possible in principle.
One example which is alredy in use all over the planet, is a heat pump, they normaly work with an efficiency of labor of more than 100%. This means the heat output is higher than the electrical input. But the additional heat is not produced, it´s just extracted from the surrounding. So it´s a bit warmer in your house and a bit colder outside.

Gravity would be the last source for free energy I would investigate into, because your whole aparatus is sitting in the same gravitational environment and as long as your not able to shield off the gravitational influence on parts of the system all attempts must fail.Rolling eyes

Hydrogen electrolysis and magnetism are much more likely to give a working result.

But again you cannot produce energy! So the main question will be where does the excess energy comes from?
We might plug into a life-support system of mother earth or we are reducing the orbital energy of our planet just to give some examples. :evil: Very bad ideas.

I think we will have such technologies somewhen in the future, hopefully they will be used wisely. If our population now would use them on a large scale: OMG

In Lak´ech

Garfield
“Coincidences are what are left over after you've applied a bad theory.” P.W. Bridgman
 
SWIMfriend
#8 Posted : 6/6/2011 3:38:15 AM

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The purpose of the thread wasn't to invoke the law of conservation of energy; it was to try to find EXPLANATIONS of exactly where the law applies in specific examples.

Can you explain, for example, why it MUST take more energy to convert a dissolved object into a solid object (that would be buoyant) and then use it in the device I described? I doubt it would work, but I'm aware of NO way to look at thermodynamics, and energy systems, that says that "re-crystallizing" such an object from a solution MUST use more energy than, say, the buoyancy force that object could generate over 10 meters. There's simply no thermodynamic connection between the two processes--yet they COULD be connected to do work (in theory, I'm NOT saying this could work in practice).

In fact FINDING a thermodynamic connection between those two processes (if it were even possible, which I doubt) would probably involve a completely new set of ideas about thermodynamics.
 
Garfield
#9 Posted : 6/7/2011 3:56:29 AM
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SWIMfriend wrote:
The purpose of the thread wasn't to invoke the law of conservation of energy; it was to try to find EXPLANATIONS of exactly where the law applies in specific examples.

Can you explain, for example, why it MUST take more energy to convert a dissolved object into a solid object (that would be buoyant) and then use it in the device I described? I doubt it would work, but I'm aware of NO way to look at thermodynamics, and energy systems, that says that "re-crystallizing" such an object from a solution MUST use more energy than, say, the buoyancy force that object could generate over 10 meters. There's simply no thermodynamic connection between the two processes--yet they COULD be connected to do work (in theory, I'm NOT saying this could work in practice).

In fact FINDING a thermodynamic connection between those two processes (if it were even possible, which I doubt) would probably involve a completely new set of ideas about thermodynamics.


Sorry that I didn´t answered exactly on this example.
Here are my thoughts on it.

First of all the gravitational force and the buoyancy force are coupled, thats why I wrote the part on gravity.
In addition the buoyant object looses power from the friction in the hydrodynamic system, this reduces the theoretical output of this system.
When it comes to the theoretical recrystallisation process keep in mind that many if not most of these processes are endothermic, this means they need an external energy supply. This can be heat and/or chemicals which you have to add in order to do the work.
The object which should be buoyant must be less dense than water, right?! So the potential force which converts to the kinetic force during the fall ist not very high and I don´t know a single "crystallisation" process which need less power than that. The buoyancy force is simply too week to get an excess of energy here.
And if such a process would exist it would form the object at the top of your column and not at the bottom where you would need it to happen.Crying or very sad

In Lak´ech

Garfield
“Coincidences are what are left over after you've applied a bad theory.” P.W. Bridgman
 
SWIMfriend
#10 Posted : 6/7/2011 5:49:45 AM

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Garfield wrote:
First of all the gravitational force and the buoyancy force are coupled, thats why I wrote the part on gravity. In addition the buoyant object looses power from the friction in the hydrodynamic system, this reduces the theoretical output of this system.


If one disregards the problems of "what to do" with the buoyant object at the top of its path, then the apparatus EASILY produces net energy: the only energy cost is the slight work needed to push the buoyant object slightly beneath the water, to put it in the pipe. If the object simply "disappeared" at the end of it's run, another object could be sent in the same way, compounding the net energy gain. It could work simply by a pulley system that cranked up a weight against gravity as the buoyant object floated up the column.


Garfield wrote:
When it comes to the theoretical recrystallisation process keep in mind that many if not most of these processes are endothermic, this means they need an external energy supply. This can be heat and/or chemicals which you have to add in order to do the work.


I'm not trying to make a claim that recrystallization would work in reality. What I'm saying is that:

1) IF recrystallization cost was low, the system would EASILY produce net energy. The buoyant object (no matter HOW SMALL it's buoyancy) would easily create a net gain of energy against the very small bit of energy needed to submerge it only, say, a centimeter. AND...

2) The point of bringing up the recrystallization is that there is NO ENERGY CONNECTION between the net gain that could be produced by the gravity-based/air pressure-based system that raises the object vs. the energy cost of the recrystallization. There is no RULE or SYSTEM CALCULATION from which it could be claimed that the recrystallization MUST use at least the net energy that the process would generate. There's (obviously) no connection. That's my point.

Usually, perpetual motion machines fail because the inventor cannot see the forces that work AGAINST his device, but only the net positive forces. Of course those forces MUST balance--and so such machines can be dismissed without explanation.

But this case is different. If the object "disappeared" at the end of the run, there is no need for a CIRCLE of forces (such as in gravity machines, where things must go up AND down). In this case, there is a permanent force available (the air pressure, CAUSED by gravity) already set up to generate as much work as you please (as long as you can supply new buoyant objects). AND, since it is not circular, NO FORCE IS NEEDED TO RESET THE APPARATUS. A simple pulley system could utilize the buoyant force to raise a weight with each trip up the pipe (and it could be sent as high as you please, or as many weights could be lifted to (whatever height) as MANY TIMES as you please. And the only COST (not counting the cost of the object "disappearing" and "reappearing" ) is the tiny fraction needed to submerge the buoyant object each time.

NOR is it necessary that the components that dissolve be THEMSELVES less dense than water--it's only necessary that the object they form INTO is less dense (and can't quickly become water-logged).

I'm not saying that it would WORK. What I'm saying is that there is NO FUNDAMENTAL THERMODYNAMIC REASON that the energy required to dissolve and re-constitute the buoyant object MUST utilize more energy than the apparatus could generate. Nothing in CHEMISTRY "knows" or has any physical connection to this gravity apparatus. It's only a COINCIDENCE that (almost certainly) more energy is required to re-constitute the object than could be gained from the apparatus.

Two examples:

1) If you have a "gravity machine" from which you harvest energy as a weight falls in a gravitational field, there is NO ESCAPE from the fact that you must then LIFT another weight to the same height to continue the function. You simply can't get a net positive result.

2) If you use a magnet, a metal object that is PULLED by the magnet must then be PUSHED with equal force to get it away from the magnet. And so there can be no net positive result (and if the metal object "disappeared" in this case, you would STILL have to force each metal atom away from the magnet with, I think, the same force. There's no way to use the "disappearing trick" here).

This is the only example I've seen where it could actually work if only the object simply DISAPPEARED (dissolved). "WORK" gets done by this apparatus simply by "starting it" (unlike, e.g., dropping a weight, which requires FIRST raising the weight).
 
Tsehakla
#11 Posted : 6/7/2011 1:13:24 PM

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SWIMfriend wrote:

I'm not saying that it would WORK. What I'm saying is that there is NO FUNDAMENTAL THERMODYNAMIC REASON that the energy required to dissolve and re-constitute the buoyant object MUST utilize more energy than the apparatus could generate. Nothing in CHEMISTRY "knows" or has any physical connection to this gravity apparatus. It's only a COINCIDENCE that (almost certainly) more energy is required to re-constitute the object than could be gained from the apparatus.

Are you forgetting about entropy and the Second Law of Thermodynamics?

see: wikipedia:Entropy
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SWIMfriend
#12 Posted : 6/7/2011 5:45:06 PM

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Tsehakla wrote:
SWIMfriend wrote:

I'm not saying that it would WORK. What I'm saying is that there is NO FUNDAMENTAL THERMODYNAMIC REASON that the energy required to dissolve and re-constitute the buoyant object MUST utilize more energy than the apparatus could generate. Nothing in CHEMISTRY "knows" or has any physical connection to this gravity apparatus. It's only a COINCIDENCE that (almost certainly) more energy is required to re-constitute the object than could be gained from the apparatus.

Are you forgetting about entropy and the Second Law of Thermodynamics?

see: wikipedia:Entropy


I'm not forgetting anything. I'm not making any STATEMENTS about the energy costs of recrystallization--other than there's no PHYSICALLY CAUSAL reason that it must be more than...some arbitrary apparatus where it could be useful.

The interesting part for me is that it would only be necessary to get a buoyant object "out of the way" for this apparatus to have a net energy gain.
 
Tsehakla
#13 Posted : 6/7/2011 10:27:43 PM

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So, basically, you are saying that if thermodynamics didn't get in the way [of getting the object out of the way]... it would work.

The "fundamental thermodynamic reason..." is the Second Law. Recall that entropy has units of energy/temperature, so energy is involved in the equation!

You can argue with thermodynamics, but you will lose every time.
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SWIMfriend
#14 Posted : 6/7/2011 11:10:12 PM

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Tsehakla wrote:
So, basically, you are saying that if thermodynamics didn't get in the way [of getting the object out of the way]... it would work.

The "fundamental thermodynamic reason..." is the Second Law. Recall that entropy has units of energy/temperature, so energy is involved in the equation!

You can argue with thermodynamics, but you will lose every time.


The point is, for the criticism to be "relevant" you need to calculate the energy required for "recrystallization" IN TERMS OF the energy gained by floating up to the top. If you can't DO THAT (which you can't--not "directly" ) then the refutation isn't of the APPARATUS. Instead it's only "bad luck" that recrystallizations of buoyant materials use more energy than can be gained by this type of gravity-buoyancy driven contraption.

See what I mean? There's no thermodynamic REASON that the recrystallization has to cost more energy. It just HAPPENS TO (we're supposing).


It might be interesting to consider this example: Suppose you did the experiment on the bottom of the ocean--where you had several miles of buoyancy to utilize. I wonder if you could pre-calculate that it would cost most to assemble a buoyant material ON THE BOTTOM than you could gain from it floating to the top. What if the ocean were a few THOUSAND miles deep?

I don't think a buoyant material would HAVE TO actually contain air (which you could show would require energy to get down there). I think it could be possible that you could fabricate a material (from elements/molecules MORE dense than water, let's say) that are less dense than water in your new, buoyant molecule/polymer. Are you saying that you could CALCULATE IN ADVANCE that it MUST TAKE more energy to fabricate that material there on the bottom of the ocean, than could be gained by it buoyantly floating to the surface--no matter HOW DEEP the ocean was? Remember that pressure is generally an INDUCEMENT to forming higher energy chemical bonds (i.e., it might help to make ANYTHING, generally speaking, to be at higher pressure--i.e., it might help to supply activation energies or configurations...just saying).
 
Tsehakla
#15 Posted : 6/8/2011 1:06:47 AM

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SWIMfriend wrote:
The point is, for the criticism to be "relevant" you need to calculate the energy required for "recrystallization" IN TERMS OF the energy gained by floating up to the top. If you can't DO THAT (which you can't--not "directly" ) then the refutation isn't of the APPARATUS. Instead it's only "bad luck" that recrystallizations of buoyant materials use more energy than can be gained by this type of gravity-buoyancy driven contraption.

See what I mean? There's no thermodynamic REASON that the recrystallization has to cost more energy. It just HAPPENS TO (we're supposing).

No, I don't see it.

You can't just look at the "energy" balance--which is where I think you are going wrong--both enthalpy and entropy enter into the equations... and, yes, it could be calculated.


Quote:
It might be interesting to consider this example: Suppose you did the experiment on the bottom of the ocean--where you had several miles of buoyancy to utilize. I wonder if you could pre-calculate that it would cost most to assemble a buoyant material ON THE BOTTOM than you could gain from it floating to the top. What if the ocean were a few THOUSAND miles deep?

Feet, miles, or even thousands of miles... the calculation would be the same, and the result would be thermodynamically unfavourable because of the entropy loss.

Quote:
I don't think a buoyant material would HAVE TO actually contain air (which you could show would require energy to get down there). I think it could be possible that you could fabricate a material (from elements/molecules MORE dense than water, let's say) that are less dense than water in your new, buoyant molecule/polymer. Are you saying that you could CALCULATE IN ADVANCE that it MUST TAKE more energy to fabricate that material there on the bottom of the ocean, than could be gained by it buoyantly floating to the surface--no matter HOW DEEP the ocean was? Remember that pressure is generally an INDUCEMENT to forming higher energy chemical bonds (i.e., it might help to make ANYTHING, generally speaking, to be at higher pressure--i.e., it might help to supply activation energies or configurations...just saying).

"I" couldn't calculate it because I'm about 20yrs out of practice ( Smile ), but if someone did you would find that either the pressure is effectively cancelled out of the equation, or doesn't enter into it.
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SWIMfriend
#16 Posted : 6/8/2011 2:17:21 AM

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Maybe I AM misunderstanding your inclusion of entropy--which I originally thought you were referencing only in regard to the matter of crystals vs. solutions.

The point I really feel you're not hearing clearly is that, in discussing situations like this, the real heart of the matter is considering that it's a "system," and realizing that the work CREATED never gets to equal the work USED to SET UP the process of creation--in the garden variety perpetual motion machine.

Sure, you can have a system where you drop a weight from a ladder, and say "look at this, I just got some free energy!" But the person saying that wouldn't realize that it takes PRECISELY the same amount of energy to carry the weight UP the ladder, as you would harvest by it falling (excluding all friction and similar effects).

In this case you're presented with an apparatus that will produce some work for you merely by "feeding in" an element. And the element, that produces the work, actually travels AGAINST gravity.

The problem is (in the way I decided to talk about it) you're left with the "logical problem"--NOT energy problem--of the object staying up there at the top of the column, getting in the way of sending more objects up. NOTE, that if you COULD take the object out, you could get even MORE energy from it by DROPPING IT from the height it had reached!

[Just for the sake of being complete, the "imaginary apparatus" I have in mind would have a piece of monofilament attached to the buoyant object, which passes down through the tube, around a pulley, then up a height, around another pulley, and then is attached to a weight. As the buoyant object floats to the top of the tube, the attached weight (which together with the buoyant object produce a "combined object" that is still less dense than water) is then lifted against gravity, and held in place when the buoyant object hits the top of the tube. If the buoyant object now DISAPPEARED (and I gave the example of dissolving), the monofilament could drift down and another buoyant object could be attached and the same operation repeated. The only energy (ignoring all the connection of parts, friction, etc.) is the very proportionately SMALL amount needed to push the buoyant object slightly underwater to put into the tube]

Now, I'll repeat this again, to try to EMPHASIZE IT; and this is what you should address: NOWHERE in that apparatus is there ANY CONNECTION between the amount of energy that could be harvested using a disappearing buoyant object AND the chemical/thermodynamic concerns regarding composing/decomposing the buoyant object. There's simply no CAUSAL CONNECTION between those two processes, and the only reason it probably WOULD NOT WORK is that, BY MERE COINCIDENCE, it takes more energy to produce buoyant objects than happens to be generated by this apparatus.

Again, I think the example of undersea use is apt. If one imagines a frictionless and weightless monofilament line (there's nothing wrong with such imagining in such problems--this is a theoretical discussion, and if it can be shown that one could do better than breakeven IDEALLY, then something substantial has been accomplished!), then a buoyant object going up several miles (or let's say a THOUSAND MILES) could lift an object on the pulley system a THOUSAND MILES. Sure, we can pretty easily surmise, in the real world, that it would take more energy to manufacture the buoyant object for my surface experiment of a column of water ten meters tall; but a column of water a THOUSAND MILES TALL?? I'm not so sure.

And let's consider this: I'm sure that creatures who live on the sea floor synthesize some hydrocarbon dense molecules--like fat. Those molecules--or a polymer of those molecules, are no doubt less dense than water. I doubt that such creatures have to use a GREAT DEAL more metabolic energy to produce them at depth than creatures on land use (it's probably the same).

AND HERE'S THE MAIN POINT: Since (let's say) it would take the SAME ENERGY to produce such a polymer at the ocean floor as it would take to produce that molecule in my pool of water supporting the ten meter water tubes--BUT FAR MORE ENERGY COULD BE HARVESTED from the apparatus I describe, if it utilized a thousand-mile length column than a ten meter column--I think it's clear that the PRODUCTION OF THE BUOYANT OBJECT is not a RELATED PART of the apparatus system. That's why I say that it's just BAD LUCK if it takes more energy to produce the buoyant object--the amount of energy it DOES TAKE is NOT RELATED TO THE APPARATUS--so the calculation of that energy cannot be the UNDOING OF THE APPARATUS.

I don't see how I can spell it out any more explicitly than that. If it doesn't reach you I don't think I can do any better. Perhaps you can reach ME, and explain where I'm THEORETICALLY WRONG. But to do that, you would have to show that the thermodynamics of the creation of the object would have to be CONNECTED in some way to the energy produced by the apparatus. And I don't see how you could do that, because there IS no connection.
 
Tsehakla
#17 Posted : 6/8/2011 3:04:53 AM

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Sounds like magic to me--things don't just disappear (at the surface) or appear (at the bottom). If you want to ignore reality like that then you can theorize anything...
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SWIMfriend
#18 Posted : 6/8/2011 3:21:59 AM

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Well, systems where there is NO CHANGE in objects simply can't work to produce work from static forces--any force that acts to ADVANTAGE on the object, also has to work equally to DISADVANTAGE. It took this example to give me the idea that if you considered MODIFYING the object (on which forces act), then the story might change.

In this case, if you're assuming that the object can turn into ions (or small molecules) that can diffuse through water, then you can (in theory) retrieve your object with having to subject it to the forces that sap the system.

The thermodynamics of dissolving and recrystallizing the object don't "know anything about" the apparatus the object is used in, that derives work from the object in a purely MECHANICAL manner.

The energy involved in the object chemistry is used only to solve a "logical" problem in the mechanical apparatus--not an energetic problem. The two processes are connected only "incidentally."
 
Entropymancer
#19 Posted : 6/8/2011 7:02:10 AM

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SWIMfriend, it does indeed sound like you're arguing against the second law of thermodynamics... there are no real perpetual motion machines. Still, this did make for some interesting mental exercise. Here's my take on it, for what it's worth. I don't know if it answers the question you're asking, but I think it does.

Basically what it boils down to is this: You can use this device to harvest a finite amount of work from the energy in the system (the water reservoir), basically using the reservoir as a non-rechargeable battery. Or you could open the system up to energy input from the outside world to make it sustainable; it's still not perpetual motion, but more along the lines of a solar panel or wind turbine (i.e. harnessing energy that in one form or another comes from the sun and using it to do work).

First, let's consider the flow of energy in the case of a single ball. First, we push the buoyant object (of arbitrary volume V0) down some arbitrary small distance h0; this requires energy equal to the change in potential energy of raising the same volume (V0) of water through a distance equal to V0/A. Let's call this energy U1. The object then rises to the height of the water level in the tube, gaining potential energy as its height increases; let's call this U2. When this happens, the object is essentially displacing the same volume of water, losing potential energy; let's call this change in potential energy U3 (it will have a negative value). Finally there will be some energy lost to friction and turbulence and whatnot; let's call this change in energy U4 (again, this value will be negative). Because the object is buoyant, i.e. has a smaller mass per volume than water, |U3| > |U2|.

When we sum these terms, we will be left with some remaining energy, U5=U1+U2+U3+U4; it is this U5 that sets the upper boundary on work that can be done by the system (by running a pulley with it or whatever you please)

The question then becomes, how do we run multiple cycles? Where are we going to run into limits to how much work can be obtained?

In the first case presented, with a valve at the top and bottom of the tube, it's pretty simple. With each cycle, water is displaced from the tube, so each cycle raises the object a smaller and smaller distance, obtaining less and less energy. Sooner or later, to set the system back to it's starting point, we'll have to add more water to the tube, and in so doing we are paying energy for the work we've obtained. We aren't getting free energy from the system, we're borrowing and paying it back later.

The dissolving buoy case is more interesting, but still presents limits. In the simplest case, we will be limited by the height of water in the tube. As more and more buoys dissolve, we are adding volume to the system, which will ultimately raise the water level with respect to the bottom of the tube. As the initial depth to which we must push the buoy increases the amount of useful work we can do decreases, until the energetic cost of pushing the buoy down far enough to get into the tube becomes prohibitive, not worth it for the amount of work obtained.

But then, we could imagine that the tube is itself anchored by a buoyant platform, so that as the water level rises, the depth to reach the opening of the tube remains constant. In this case there is still a theoretical limit; as the tube rises, the atmospheric pressure decreases, which in turn decreases the height of water in the tube, until eventually the distance which the buoy rises through the tube is too short to do a useful amount of work for the amount of energy it takes to get the buoy into the tube. Alternatively, the reservoir might reach full saturation before that happens, at which point the buoyant object can no longer dissolve, so the tube fills up with buoys and our machine can do no more work.

It is at this point that we would need to open the machine up to the elements to make it sustainable. Imagine that the reservoir has a porous surface, and that the rate of rainfall is balanced with the rate at which the reservoir drains and the rate at which buoys are dissolved, so we could indefinitely maintain the same height of water and the same concentration of dissolved buoy. Now the machine will run indefinitely (if we don't run out of soluble buoys), but it's not a perpetual motion machine; the energy required to bring the machine back to its precise starting state is simply provided by rainfall and drainage, just as solar-powered machines get their energy from the sun.

Another interesting thought to consider is a buoy material with a pressure sensitive phase change. Imagine a substance which will spontaneously solidify into a material less dense than water given sufficient depth (high pressure) and concentration, and (when it floats up) will dissolve slowly enough that it can operate a cycle in the tube before it dissolves/melts. Sounds perfect, right? Still, in a closed system this will not run forever. Due to entropy, energy will be lost in each cycle of liquid->solid-?liquid phase changes. Sooner or later, the temperature of the water will become too hot and the substance will be stuck in liquid form, unable to solidify and operate the machine. And again, the only way around this is to open the system up to the elements, using the surrounding environment as a heat sink to maintain the reservoir at constant temperature. Again, this would provide sustainable constant motion... but not a perpetual motion machine.
 
SWIMfriend
#20 Posted : 6/8/2011 9:32:13 AM

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Entropymancer wrote:
As more and more buoys dissolve, we are adding volume to the system


I'm assuming that we're reassembling the object from it's solute material. So the same material is used over again.

Entropymancer wrote:
Due to entropy, energy will be lost in each cycle of liquid->solid-?liquid phase changes. Sooner or later, the temperature of the water will become too hot and the substance will be stuck in liquid form, unable to solidify and operate the machine.


I'm saying suppose the ENTIRE ENERGY REQUIREMENT for the object phase changes, entropy, etc., is LESS than can be produced by the apparatus. Of course I know (just from practical knowledge) that it WOULDN'T be; but my point is that doesn't HAVE TO BE THE CASE--it only HAPPENS to be the case by "coincidence."

[If you're saying the water is going to heat up, then that's GREAT--we can include that heat as part of the production. I'm assuming that the NET energy use (and thus the net temperature change) would be negative]

This is my MAIN POINT, but no one seems to really be absorbing it.

If I said "Hey, I have a great idea for perpetual motion! We'll take CO2 and H2O and chemically combine it into sugar, and then BURN the sugar to make heat to run an engine." It's EASY to explain directly from thermodynamic principles why that's ridiculous.

But there's no CAUSAL CHAIN between the chemistry of a buoyant object, and the AMOUNT of energy it can produce from MECHANICAL means, from the static force of the pressure in the column. In this case, you can't say "Well, it wouldn't work, because the energy from the mechanical system IS WHAT MAKES the chemical change. The chemical change is an ENTIRELY DIFFERENT SYSTEM--which will simply have a measurable energy cost. And that cost won't in ANY WAY be related to the mechanical goings on of the apparatus. That means it's just COINCIDENCE whether the chemical cost happens to be higher or lower than the mechanical output.

The chemical cost is related to the strength of CHEMICAL BONDS (and to the entropy changes involved)--and that process simply "knows nothing" about the mechanical apparatus we've built.

But I do thank you guys for reading through all this and coming up with your analysis. IMO though, none of the answers really overturns the idea from first principles (except The Traveler pointed out how the first scenario fails "practically"--and I like your description of it being an energy "resevoir" that is used up, in terms of it gaining air and losing efficiency).
 
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