Do metal springs really store sig work as potential?

[Steve Harris]

This is a good one.

We all know that in an ideal gas, there is no isothermal potential energy stored when you do work on the gas. Compress the gas: all the work appears as heat, and when you cool to ambient, it's all gone. Compressed ideal gas stores no work EXCEPT as heat.

Now, less well known is that elastomers like rubber, within limits and to a good first approximation, act just like ideal gasses. They don't store elastic energy as bond potential. Which means if you stretch a rubber band, ALL the work you do on it appears as heat, and NONE of it goes into the classic kind of elastic potential energy that you're thinking of.

In both the band and gas gases, this has an interesting corollary when they are forced to DO work-- they get cold and absorb heat from the environment. In fact, they absorb enough heat to equal the energy of the work they do.

WHAT?? You say? What about entropy? You can't just turn ambient heat willy nilly into useful F x D work! Well, you can in a system like this where it's not reversible. When you compressed the gas or stretched the rubber, you made changes in entropy which allow you now to extract the work again using ambient heat, and let those entropy changes pay the price for you, as dG = TdS. Neat, eh?

Now my question. What about metal springs? My faith suffered horribly enough with rubber bands! Now in reading, I find that cold-working of metals, well within their Hooke limits, produces a lot of heat. In fact, so much, that most of the work is not left in the metal. Wups. Is that true of springs, too? Do they heat up when you do work on them? What fraction of the work appears as heat? 10%? 98% I want to KNOW!

So I went to the net and looked, and it's such a morass of student texts and theory discussions and lack of experiments, that I suspect that not many people actually DO know the answer. Most of what's written out there is from the viewpoint of profs writing what they think SHOULD happen. Stretch a spring and it *should* store the energy and not get hot. Let it do work, and the work SHOULD be extracted from pure elastic potential between the metal atom fields. And so on. Baloney. I think that CAN happen, but mostly, in real life do not. But can't prove it.

Now, keep in mind that in answering this question, ignore how the heat is stored. We all know that in solids like metals (including springs) any heat is partitioned 50% into kinetic and 50% potential, at the microscopic level. But we seek the answers after all this has been allowed to dissipate into the environment. Does a spring which has been compressed and allowed to cool to ambient, then made to do work, cool down? Or does it just sit there at the same temp it already had, having drawn entirely on special potential energy "reserves"? You think you know, but do you REALLY?

Steve Harris

 

[Mentz114]

This is an interesting question. In a good spring most of the work put into compressing it can be recovered, so it must have been stored as potential. I rely on personal experience and observation for this. In a crystal the potential could be due to a change in charge distribution.

Another related quesion is - if the work done in compressing a spring is stored as potental, does the potential gravitate ?

 

[Steve Harris]

This is an interesting question. In a good spring most of the work put into compressing it can be recovered, so it must have been stored as potential. I rely on personal experience and observation for this. In a crystal the potential could be due to a change in charge distribution.

Another related quesion is - if the work done in compressing a spring is stored as potental, does the potential gravitate ?

Ah, but the fact that the energy can be "recovered" does NOT mean it's stored as potential (I mean as field potential). I can recover the energy I store in compressed gas (after I let it cool) and NONE of that energy is stored as potential. It's all "stored" as decreased entropy, which later let's me "recover" it from the ambient heat in the room. In which case it's certainly NOT stored as anything that gravitates. The compressed gas has the same temp as the gas before I compressed it, so its mass is the same. Steve

 

[Cyrus]

Your post confused me.

A spring stores its energy elastically within the bonds. When I place a weight on a spring and hold it down so the spring can't 'spring' back up, the bond length between atoms inside the spring decreases. This causes an increase in the repulsive force between the atoms that pushes back up on the mass. As long as the force of the mass is not so large as to overcome the repulsive forces and break the bonds, i.e. go to strain hardening or softening of the spring, it will alway spring back no matter how long you wait.

Springs don't dissipate heat into the environment just sitting there held down with a weight because they don't ever heat up to begin with.

Now in reading, I find that cold-working of metals, well within their Hooke limits, produces a lot of heat.

Cold working a material occurs when you go into the plastic region of the material. That is by no means within the elastic limit.

http://www.engineersedge.com/material_science/cold_hot_working.htm

Also:

Which means if you stretch a rubber band, ALL the work you do on it appears as heat, and NONE of it goes into the classic kind of elastic potential energy that you're thinking of.

Last time I checked, rubber bands spring back when you let go. All the energy clearly did not go into heat.

I think that CAN happen, but mostly, in real life do not.

What do you mean by something "can happen, but mostly, in real life dont. It if does not happen, it does not happen. Theory invalid.

 

[Mentz114]

Steve, so you said. But if you can recover the energy without a heat engine - it probably was not stored as heat. I think Cyrus's picture is right. Redistribution of the the charges can increase the potential energy when the material is strained.

 

[Steve Harris]

Your post confused me.

A spring stores its energy elastically within the bonds. When I place a weight on a spring and hold it down so the spring can't 'spring' back up, the bond length between atoms inside the spring decreases. This causes an increase in the repulsive force between the atoms that pushes back up on the mass. As long as the force of the mass is not so large as to overcome the repulsive forces and break the bonds, i.e. go to strain hardening or softening of the spring, it will alway spring back no matter how long you wait.

Springs don't dissipate heat into the environment just sitting there held down with a weight because they don't ever heat up to begin with.

>COMMENT: How do you know? Have you ever checked? I would have said the same about an elastic band until I held a stretched one against my face, and found it warmer. I spring may be the same.




Last time I checked, rubber bands spring back when you let go. All the energy clearly did not go into heat.

>COMMENT: Irrelevent. An (ideal) gas compressed in a cylinder springs back when you let it. (Ideal-- forget Joule-Thompson effect and intramolecular forces; do this in the limit of a thin gas, or with helium). NONE of the energy you put in, goes into heat in the long run. In the short run (adiabatic) it all does, but if you let it leak out of the system, the gas THEN sits there as a sort of stored-energy reservoir, ready to do work. But the stored energy is in ambient heat, and work done draws the energy out of ambient thermal reserve, and the gas gets colder while it does the work. The "storage" of work-capacity here is in entropy, not in fields. So also in rubber bands. I'm merely asking is the same applies to metals, and if so, what fraction. Obviously ALL springs heat up a LITTLE when you stretch them, but what fraction of the work appears as heat? And if ANY does, then THAT is really NOT work stored as potential, but rather as entropy-fiddling, as with the gas and the rubber-band.

Steve

 

[Integral]

We all know that in an ideal gas, there is no isothermal potential energy stored when you do work on the gas. Compress the gas: all the work appears as heat, and when you cool to ambient, it's all gone. Compressed ideal gas stores no work EXCEPT as heat.

"stored work" ? some how that term just does not seem right. I assume you mean stored energy.

What am I missing? Let me fill a scuba tank with compressed air. As the compressor does work during the filling process there is a significant amount of heat which must be removed from the system. But how can you say there is no stored energy? Knock the valve off the end of the tank an tell me there is no stored energy?

So fill me in, what am I missing?

 

[Steve Harris]

"stored work" ? some how that term just does not seem right. I assume you mean stored energy.

What am I missing? Let me fill a scuba tank with compressed air. As the compressor does work during the filling process there is a significant amount of heat which must be removed from the system. But how can you say there is no stored energy? Knock the valve off the end of the tank an tell me there is no stored energy?

So fill me in, what am I missing?

===

There is no stored energy in a scuba tank (in the limit of ideal behavior), but there is (now) only stored ability to do work, by turning some of the tank gas heat, into energy. You've made that possible that by moving entropy into a particular place (at a cost), outside the tank. But the work you do to get the gas into the tank all goes into heat that goes away long before you want to use the gas for work, so it's not located anymore in the tank, so it's not stored in the tank as energy, in ANY sense. NOT as potential. The tank and associated gas doesn't weigh any more by the amount E= deltaM*c^2, etc. It does right after you fill it, and it's still hot. But not after that heat has leaked away. Yet the capacity to do work remains, and it's not stored as potential energy. Just as potential to do work as free energy dG = TdS.

 

[Integral]

So how do YOU define potential energy?

 

[Steve Harris]

Potential energy

Well, it's the Lagrangian plus the kinetic energy :). Energy stored when work is done against a force field (this does NOT happen when you compress a classical gas or stretch a rubber band). Seriously, it's all those kinds of energy which show up associated with mass and gravitational fields and stress, but which are NOT due to motion. In the scuba tank that you've put compressed gas into and cooled back to ambient energy, there isn't any extra kinetic OR potential energy. The thing's ability to do work is not due to any stored potential energy, but rather due to its extra order (which happens in compression) which allows it to transform thermal energy (which in the case of an idea gas is entirely kinetic energy) into work. No potentials are actually involved at all (again, in the ideal gas limit, where by definition there are no potential interactions between molecules, and they just bounce off each other).

Steve

 

[Mentz114]

Steve, this is very interesting. Is the stored energy in the increased kinetic energy of the gas molecules ? No, I guess that is thermal energy...

 

[cesiumfrog]

Integral: If you consider the ideal gas as consisting of rigid spheres, or even non-interacting point particles, then the only natural definition of the total energy is to sum each particle's kinetic energy. Since there aren't any interaction terms, the total energy of the ideal gas can not depend on the density of particles (and there is nothing that can appropriately be described as potential energy).

So if you take two different volumes, each containing an equal quantity of an ideal gas and each being at the same temperature (so all particles have the same average kinetic energy individually), then each must also contain the same total amount of energy (over its respective volume). Then, you can notice that in the more compressed volume the particles reflect off the walls more often, so if you relax the confinement the two volumes will do different amounts of work on their respective containers (and afterwards, the more initially compressed one which will have less total energy than the other). Stored work is the appropriate term since, while both stored the same amount of energy, their energy is unrepresentative of their potential to do work. The ocean (being at hundreds of degrees Kelvin) has plenty of energy, yet you can't power your boat with that. I understand this is just basic thermodynamics.

Before you knock the valve off your scuba tank, tell us where you think the energy is stored. See why "energy" is defined as "the property in a system that does not change over time", rather than "the potential to do work"?

 

[Mentz114]

There has to be increased potential energy in the walls of the container to with stand the extra collisional forces. I don't think one can consider the gas separately from the container.
If you do work on something and can recover the work, then energy must have been stored somewhere. I don't believe thermodynamics has the answer.

 

[berkeman]

===

There is no stored energy in a scuba tank (in the limit of ideal behavior), but there is (now) only stored ability to do work, by turning some of the tank gas heat, into energy. You've made that possible that by moving entropy into a particular place (at a cost), outside the tank. But the work you do to get the gas into the tank all goes into heat that goes away long before you want to use the gas for work, so it's not located anymore in the tank, so it's not stored in the tank as energy, in ANY sense. NOT as potential. The tank and associated gas doesn't weigh any more by the amount E= deltaM*c^2, etc. It does right after you fill it, and it's still hot. But not after that heat has leaked away. Yet the capacity to do work remains, and it's not stored as potential energy. Just as potential to do work as free energy dG = TdS.

Um, not true. A full scuba tank does weigh more than an empty one (that's the reason your bouancy changes as you drain the tank at a given depth while you dive). The delta is the extra mass of the extra air that is stored at pressure. It has nothing to do with the temperature of the tank. I'm not sure where you got the idea that a hot scuba tank weighs more than a room temperature one.

And if you accidentally drop a full scuba tank on a concrete floor, you will definitely see the stored potential energy of that pressure released and converted to kinetic energy.

 

[Steve Harris]

There has to be increased potential energy in the walls of the container to with stand the extra collisional forces. I don't think one can consider the gas separately from the container.
If you do work on something and can recover the work, then energy must have been stored somewhere. I don't believe thermodynamics has the answer.

Well, if you use a tank of old (re-cooled to ambient temp) gas to do work, then the initial work you did when you compressed the gas was "stored" as heat. Whether that's heat still in the gas, or heat which escaped since you compressed it, is a matter of taste. Certainly it's conserved. But only thermodynamics (and the idea that you decreased the gas entropy when you compressed it) allows you to USE that heat, to do work. If you allow the gas to expand and do work and don't supply the heat back from the surroundings, the gas itself will cool to below ambient temp. That energy has to come from somewhere. But it's not a reversible situation. Once the gas has expanded, you have to get it back into the bottle.

Steve

 

[Doc Al]

This is a good one.

We all know that in an ideal gas, there is no isothermal potential energy stored when you do work on the gas. Compress the gas: all the work appears as heat, and when you cool to ambient, it's all gone. Compressed ideal gas stores no work EXCEPT as heat.

Now, less well known is that elastomers like rubber, within limits and to a good first approximation, act just like ideal gasses. They don't store elastic energy as bond potential. Which means if you stretch a rubber band, ALL the work you do on it appears as heat, and NONE of it goes into the classic kind of elastic potential energy that you're thinking of.

In both the band and gas gases, this has an interesting corollary when they are forced to DO work-- they get cold and absorb heat from the environment. In fact, they absorb enough heat to equal the energy of the work they do.

WHAT?? You say? What about entropy? You can't just turn ambient heat willy nilly into useful F x D work! Well, you can in a system like this where it's not reversible. When you compressed the gas or stretched the rubber, you made changes in entropy which allow you now to extract the work again using ambient heat, and let those entropy changes pay the price for you, as dG = TdS. Neat, eh?

Got references for these claims?

 

[Mentz114]

Steve,
what you say in post #15 is undoubtedly true.
However, my gut feeling is that although the gas seems to have acquired the ability to do work, this is bestowed by the container and may not be a property of the gas. I'm still thinking about this ...

 

[cesiumfrog]

There has to be increased potential energy in the walls of the container to with stand the extra collisional forces. [..] I don't believe thermodynamics has the answer.

Wow, Mentz! Historically, thermodynamics is one of the most important fields of physics (it's where we get the whole conservation of energy thing to begin with!), and I think it played a big part in the inception of all the modern fields of physics, yet you're still willing to contradict it (and on it's home ground no less, the like of steam engines)?

Assume the container is "rigid". Work (done on the container) is proportional to force (ie. pressure) and distance. Since the container doesn't flex, no work is done on it, and nor does it transfer energy to the gas afterwards (except thermally).

Um, not true. A full scuba tank does weigh more than an empty one

The idea is to compare the compressed air with the same quantity of uncompressed air, not with a vessel containing a different quantity.

 

[Mentz114]

...yet you're still willing to contradict it

I wouldn't dare !

You misunderstand me. I mean to say it is incomplete - it does not completely explain this situation, or any other.

I should also point out that there's no such thing as a "rigid" container ( your quotes).

My whole interest in this scenario is 'where is the energy' - not just the heat.

 

[berkeman]

Wow, Mentz! Historically, thermodynamics is one of the most important fields of physics (it's where we get the whole conservation of energy thing to begin with!), and I think it played a big part in the inception of all the modern fields of physics, yet you're still willing to contradict it (and on it's home ground no less, the like of steam engines)?

Assume the container is "rigid". Work (done on the container) is proportional to force (ie. pressure) and distance. Since the container doesn't flex, no work is done on it, and nor does it transfer energy to the gas afterwards (except thermally).

The air being pushed into the rigid container isn't moving the walls as work is done by the compressor, it's pushing against the air inside.

The idea is to compare the compressed air with the same quantity of uncompressed air, not with a vessel containing a different quantity.

PV=nRT. I stand by my previous comments.

 

[cesiumfrog]

Got references for these claims?

I mean to say [thermodynamics] is incomplete - it does not completely explain this situation, or any other.

I think the OP raised an interesting question, but this thread first seems to be debating thermodynamics. Can I encourage some revision of that topic? It should at least be in any first year general university physics textbook (and if it isn't, discard that textbook).

The air being pushed into the rigid container isn't moving the walls as work is done by the compressor, it's pushing against the air inside.

An ideal gas can be modeled as point particles that do not interact in any manner whatsoever (unlike a real gas where there may actually be some potential energy involved in coulomb repulsion). Yes, the compressor does work on the particles it pushes against (increasing their kinetic energy, which may be interpreted as saying the compressed gas is at a higher temperature initially), however this energy can be returned to the environment (say, the gas will radiate more than the environment does, until they reach thermal equilibrium). Incidentally, attributable to relativistic effects (the mass-energy of the radiated photons) the tank loses mass as it cools. The compressed ideal gas then has the exact same total amount of energy (and mass) as if it were uncompressed, despite also having greater capability to do work (though, conserving energy, doing that work will take kinetic energy from the particles).

 

[Steve Harris]

Got references for these claims?

This much is Chem 111. You have a text for that course? Look under adiabatic expansion of ideal gas.

Steve

 

[Doc Al]

This much is Chem 111. You have a text for that course? Look under adiabatic expansion of ideal gas.

Very amusing, but I'm talking about your claims about elastomers not storing PE when stretched.

 

[Mentz114]

I think the OP raised an interesting question, but this thread first seems to be debating thermodynamics. Can I encourage some revision of that topic? It should at least be in any first year general university physics textbook (and if it isn't, discard that textbook).

True. ( I'm re-reading a text on Thermodynamics and Kinetic theory now).
Can we get back to springs and solids ( like the walls of our container) ?

 

[cesiumfrog]

Very amusing, but I'm talking about your claims about elastomers not storing PE when stretched.

This is interesting. It may be evident that elastic absorbs some heat as it relaxes, but it seems likely that another component of the energy is stored as electrostatic potential in the shapes of the chemical bonds. Is this an equipartition thing?

 

[Steve Harris]

Very amusing, but I'm talking about your claims about elastomers not storing PE when stretched.

For an ideal elastomer dE/dl = 0, where l is the length change on work load. I kid you not. See http://individual.utoronto.ca/alisonmcguigan/teaching_and_supervision/elasticity_lectures_student_lecture_1.doc

That means all the work done on the elastomer ends up as heat (or work done BY the elastomer is subtracted from its internal heat). Which means NO work done on the rubber ends up as potential. And no work done BY the rubber COMES from stored potential. The mechanism is often used as a student demo of TdS free energy, where the rubber is behaving like an ideal gas.

Steve

 

[cesiumfrog]

How should we visualise an ideal elastomer, on the microscopic scale?

 

[Q_Goest]

We all know that in an ideal gas, there is no isothermal potential energy stored when you do work on the gas.

There's no such thing as "isothermal potential energy". There is potential energy. And isothermal is a term used to describe a process.

Compress the gas: all the work appears as heat, and when you cool to ambient, it's all gone. Compressed ideal gas stores no work EXCEPT as heat.

I think what you're pointing to is the fact that for an ideal gas, the process of isentropic compression will result in the gas getting hotter, and this heat energy is equal to the work input. Is that right? Once you cool it back to 'ambient' as you say, the entropy of the gas has decreased significantly. So although the amount of energy removed in the form of heat after this process of isentropic compression and isobaric cooling is complete, is equal to the amount of work input, the entropy is significantly reduced. The air now has the capacity to do work by expanding that it did not have at the lower pressure - even though the amount of heat removed equals the amount of work put in!

Now, less well known is that elastomers like rubber, within limits and to a good first approximation, act just like ideal gasses. They don't store elastic energy as bond potential. Which means if you stretch a rubber band, ALL the work you do on it appears as heat, and NONE of it goes into the classic kind of elastic potential energy that you're thinking of.

I disagree. Elastomers store and release energy similar to metals with the exception being that some energy may be lost as bonds between the polymers that must slide past each other when enough stress is placed on the molecules or perhaps when cross-links are broken. It's this 'sliding' between the long polymer molecules that creates the heat in a rubber band, not the storing or discharging of energy.

In both the band and gas gases, this has an interesting corollary when they are forced to DO work-- they get cold and absorb heat from the environment. In fact, they absorb enough heat to equal the energy of the work they do.

For a gas that does work through an isentropic process, it does indeed get cold. You are correct there. But this isentropic process is adiabatic, so obviously this process does not absorb any heat from the environment to do work.

WHAT?? You say? What about entropy? You can't just turn ambient heat willy nilly into useful F x D work! Well, you can in a system like this where it's not reversible.

An isentropic process is reversible by definition.

When you compressed the gas or stretched the rubber, you made changes in entropy which allow you now to extract the work again using ambient heat,

Isentropic compression or expansion is adiabatic. No heat from ambient is exchanged.

Now my question. What about metal springs? My faith suffered horribly enough with rubber bands! Now in reading, I find that cold-working of metals, well within their Hooke limits, produces a lot of heat. In fact, so much, that most of the work is not left in the metal. Wups. Is that true of springs, too? Do they heat up when you do work on them? What fraction of the work appears as heat? 10%? 98% I want to KNOW!

Metal springs which extend and contract at stress levels below their elastic limit do not get hot - they do not store thermal energy. For a spring, the force they produce is linearly related to the spring constant, so regardless of whether the spring is being compressed or extended, the force is the same in that position. In other words, they loose no force (no energy) during the change in shape.
- Conclusion: The spring would need to change force at some point during the compression or extension in order for it to have changed potential energy to heat, but this does not occur!

Does a spring which has been compressed and allowed to cool to ambient, then made to do work, cool down?

See above.

Note: Air is fairly close to an ideal gas at low pressure (less than about 1000 psi) and roughly ambient temp. Let's do an analysis of air being compressed and then expanded.

Air isentropically compressed from 0 psig at 70 F to 100 psig will be heated to 489 F which requires 101.5 Btu/lbm of work.

Air isentropically expanded from 100 psig at 70 F will cool to -167 F and put out 56.1 Btu/lbm of work.

The amount of energy released is only 55% of that stored assuming the above processes, which is why using compressed air as a media to transmit energy is so inefficient. On the other hand, a spring which is compressed and forced to store energy can essentially release all of that energy on being extended.

 

[Integral]

How do you define potential energy:

Well, it's the Lagrangian plus the kinetic energy :)

Ok, so now you need to define the Lagrangian...

How in the world does a 1st year physics student work with something that you need 4th year concepts to define? I find that completely unsatisfactory. It seems to me that you are simply making up definitions to fit your misbegotten argument. Drop it.

Potential energy is a very fundamental concept, it does not have connection to any SPECIFIC mode of "storage".

Halliday and Resnick define it, in conservative systems, as "energy of configuration". As the state (configuration) of a system changes so does the form of the system energy.

So a spring mass system has a state (configuration) where there is no kinetic energy, but the total system energy is still available. This is potential energy.

A scuba tank full of compressed air is capable of doing work therefore it has energy of configuration, or potential energy.

Once again potential energy is NOT defined in terms of mechanisms. So discussion of mechanisms is NOT equivalent to a discussion of potential energy.

To argue that a system does not have potential energy because you do not understand how the energy is stored is simply incorrect.

 

[Cyrus]

Well, it's the Lagrangian plus the kinetic energy :)

Errr? The lagrangian is defined as L=T-V. So, why do you want to add the kinetic energy to itself? As you stated it, I *Think* you mean potential is Kinetic energy *MINUS* the lagrangian. But even this is a pointless definition. Why are you bringing this up?

You have a bad habbit of using poor definitions in your wording, and its confusing the hell out of all of us . Please use standard terms when you describe your conepts, because its making it hard to understand what you really mean.

Im not saying your right or wrong, just that its super difficult to understand you.

 

[Cyrus]

There is no stored energy in a scuba tank (in the limit of ideal behavior), but there is (now) only stored ability to do work, by turning some of the tank gas heat, into energy...Well, if you use a tank of old (re-cooled to ambient temp) gas to do work, then the initial work you did when you compressed the gas was "stored" as heat.But the work you do to get the gas into the tank all goes into heat that goes away long before you want to use the gas for work, so it's not located anymore in the tank, so it's not stored in the tank as energy, in ANY sense. NOT as potential. The tank and associated gas doesn't weigh any more by the amount E= deltaM*c^2, etc. It does right after you fill it, and it's still hot. But not after that heat has leaked away. Yet the capacity to do work remains, and it's not stored as potential energy. Just as potential to do work as free energy dG = TdS

Ummm, no. You stored that energy in the form of pressure. You increased the pressure, therefore you increased the energy density inside the cointainer. Its not 'heat'. You can have a pressurized tank at room temperature and it clearly maintains its energy within the tank. It does not leak energy out as it reaches ambient temperature.

Pressure is a force per unit area, it has nothing to do with temperature. I can fill up an empty bottle. Right after its been filled,the temperature might be slightly above ambient. I can come back a few days later and the contents inside the tank will now be at ambient temperature. But the pressure inside the tank is still way above ambient pressure. In fact, the change in pressure will tell you how much energy was lost as the tank reached equilibrium.

Also,

The tank and associated gas doesn't weigh any more by the amount E= deltaM*c^2, etc

Are you saying that a filled tanks weight goes back down to its original value after its cooled off? If so, this is clearly wrong.


see: http://hyperphysics.phy-astr.gsu.edu/hbase/press.html#ed

It seems like you read most of your physics off of wikipedia.

 

[Cyrus]

The air being pushed into the rigid container isn't moving the walls as work is done by the compressor, it's pushing against the air inside.



PV=nRT. I stand by my previous comments.

Yep, its called the flow work of the fluid element.

 

[cesiumfrog]

So how do YOU define potential energy?

Energy stored when work is done against a force field

even this [expressed in terms of the lagrangian] is a pointless definition. Why are you bringing this up?

I'm guessing Steve used the Lagrangian because the meaning (of V) is undisputed in that context.

A scuba tank full of compressed air is capable of doing work therefore it has energy of configuration, or potential energy.

By way of contradiction: if the "potential work" in a scuba tank (full of compressed air) is actually potential energy (stored in the pressure itself of the air) then, when this pressure is used to perform work (converting the pressure potential energy to, say, kinetic energy of a turbine) then (by conservation of energy) the air still should have the same kinetic energy (temperature) as before. However, the air is in fact colder (its total kinetic energy has decreased as much as the turbine's energy has increased) which is absurd.

Are you saying that a filled tanks weight goes back down to its original value after its cooled off? If so, this is clearly wrong.

Please do enlighten us.

----

I've pondered that ideal elastic for a bit. Consider collection of (nonpointlike) asymmetric/prolate particles, with an interaction potential that demands (except at extreme temperatures, obviously) that the particles remain in contact with one another without any preferance to the specific orientation:

Normally, these particles would be reorientating themselves randomly. If the bulk material is stretched, all the particles would have to tend to align with each other. (The material would also release heat, because otherwise particles that had been just spinning idly would now have enough kinetic energy to strain those interaction-potential bonds more than before.) If the material is relaxed, its particles will gain degrees of freedom, and absorb energy back from the environment (again invoking equipartition theorem).

This suggests that in a real material, the elasticity results more from the arrangement of polymers (which have less statistical freedom when held partially straightened) and, more specifically, free rotation of symmetric bonds - rather than coulomb-opposed (ie. potential energy related) bending/stretching of bonds.

On the other hand, if you consider stretching apart the plates of a charged capacitor, this case clearly does alter the potential energy (rather than having a thermal effect). At first glance the metallic spring looks more similar to this latter case than to the ideal elastic, but (“once bitten, twice shy”) I have to suspect it could be shown otherwise.

 

[bgwowk]

For whatever it's worth, I have a PhD in physics, so maybe I can help explain this. Steve Harris and cesiumfrog are correct that a compressed tank of ideal gas has no potential energy. It has only the kinetic energy determined by the temperature of the gas, which for a given temperature is the same whether the gas is compressed or uncompressed.

Nobody denies that a tank of compressed gas has potential to do work, but this potential is not potential energy, it is Free Energy. In fact the potential of a tank of compressed gas to do work is a good example of why the idea of Free Energy was created in thermodynamics. The Helmholtz Free Energy is defined as

U - TS

where U is the internal energy (pure kinetic energy for an ideal gas), T is the temperature, and S is the entropy. Both compressed and uncompressed gas at the same temperature have the same internal energy, U. But the compressed gas has a much smaller TS, and therefore much larger Free Energy. In essence, the compressed gas is in a better thermodynamic configuration to do work. We say it has a higher "thermodynamic potential", but this is not potential energy. At a given temperature, the actual energy content of a compressed gas and container are the same as before compression. Really. The low TS (high free energy) just makes it easier for the gas to give up its kinetic energy quickly, such as in an adiabatic explosion of a scuba tank.

A one dimensional analogy is instructive. Consider a ball bouncing between two walls.

| <------ B ------> |

Now consider the same ball moving at the same speed between two closer walls.

| <-- B --> |

Which system has the greater energy? Answer: They both have the same energy, which is purely the kinetic energy of the ball. There is no potential energy, just as there is no potential energy in a stored compressed gas. But the lower entropy of the ball bouncing between the narrow walls means it can do lots of work fast on the narrow walls if the walls become non-rigid and move apart.

 

[Mentz114]

Thanks, Doc, I reckon that's sorted out then.

Isn't the potential for the compressed gas to do work dependent on the difference in pressure between it and, say, the atmosphere ?

 

[vanesch]

This is an interesting discussion, and the OP raised an interesting point that BTW never occurred to me, but I think he's right!

Indeed, the internal energy of an ideal gas is function only of temperature:
u = u(T), and not of the other thermodynamical variable (pressure, entropy, density, whatever).

As pointed out before, u is the macroscopic version of the hamiltonian, and given that an ideal gas has no potential interaction energy, it is purely kinetic energy, which is purely determined by the temperature.

So, what if we use a gas bottle as a spring ? We can do that in two ways: we can do it relatively quickly (like, in looking at the oscillations of a weight on a piston), or we can do it slowly (fill a tank, and come back a few days later).
The first one is adiabatic, the second one is isothermal.

In the adiabatic case, the potential energy of the gas-spring is stored in the gas, and this is simply done by the increase in temperature. So this is the simple "storage of energy in the medium as 'potential' energy". Only, it is not really microscopically potential energy, but rather internal energy in this case ; nevertheless, macroscopically, we can call this the "potential energy of the gas spring" as long as we don't "look inside".

However, the isothermal case is more interesting. As pointed out by the OP (and I never realized this until reading his post ), the fact that there is still "pressure to do work" in the tank after an isothermal compression (with heat loss to the environment exactly equal to the amount of work done on the "spring") is a very peculiar property of gasses, and is a thermodynamic effect. The gas works indeed as a heat engine but we don't realize it!

If the expansion (the "work done by the gas spring"), after being at room temperature again, is adiabatic, then the gas TAKES energy from its own energy content, lowers its temperature to expand. Clearly, this can only be done if the gas is not at 0K! (that's why there are no ideal gasses at 0K). So we have the gas acting both as a heat engine, and as a heat reservoir.

If the expansion is slow, and isothermal, then the gas acts as a heat engine, but the environment acts as the heat reservoir.

But in both cases, the tank with compressed gas is a heat engine, which transforms heat into work. It is not the restoration of stored "potential" energy from the compression, as in a conservative force field.

Now, as to gravitational effects: the gravitational source is the internal energy, and will hence be given by u(T). That means that, during adiabatic compression, the gas heats, has more internal energy u, and will have hence a (minuscule) increase in gravitational mass, relativistically speaking. This is because the molecules are moving faster in the COG of the tank, and hence the relativistic mass will increase slightly.
When the gas cools, its weight will decrease (very very tiny effect in reality!).
When the gas will expand adiabatically, it will cool down below room temperature, and have even less weight (relativistically speaking).

So, indeed, in agreement with the OP, there is NO storage of energy in a compressed tank by the pressure. There is only a decrease in entropy, which allows a "one-shot" thermal engine to extract heat from the environment and to do work with it.

EDIT: I see that bgwowk said about the same (left the editor open on my computer and forgot to submit the text above... and went to a meeting).

 

[vanesch]

On the other hand, if you consider stretching apart the plates of a charged capacitor, this case clearly does alter the potential energy (rather than having a thermal effect). At first glance the metallic spring looks more similar to this latter case than to the ideal elastic, but (“once bitten, twice shy”) I have to suspect it could be shown otherwise.

I guess that if the "constant of elasticity" (in Hooke's law) is the same for slow (isothermal) and for fast (adiabatic) motion, then one can say that there is some form of local storage of the work done (which could be called, in broad terms, "potential energy" although as we saw with the adiabatic ideal gas spring, was in fact microscopic kinetic energy).

 

[Cyrus]

Please do enlighten us.

You put mass into the bottle when you filled it. It has to weigh more than the empty tank, and it has to have more pressure than the outside. Thats the whole point of storing it in a pressure vessel. You did work to fill this bottle in the form of flow work. It took work to push the air against the already pressurized air inside the botttle.

The pressure is an energy density per unit volume. If the tank is 1L and you increased the pressure ten fold, then that energy is sitting there inside that bottle waiting to do work. It never went away. I get the feeling steve is saying that the energy inside the bottle goes away but the order remains and that it is this order that 'unorders' itself when you open the bottle, thus outputting work. As I said, its the energy in the form of pressure, which never left the bottle, which is doing that.

 

[vanesch]

You put mass into the bottle when you filled it. It has to weigh more than the empty tank, and it has to have more pressure than the outside. Thats the whole point of storing it in a pressure vessel. You did work to fill this bottle in the form of flow work. It took work to push the air against the already pressurized air inside the botttle.

I think you misunderstood what the weight increase was about: it was about a relativistic effect (which is, for practical pressure vessels, unmeasurable).

Consider a vessel with a piston. The amount of gas doesn't change when pushing on the piston. Put it on a superduper balance. Now, when the gas is relaxed, it has (bottle + gas + piston) a certain mass (found by the weight) M0.

Now, compress the gas using the piston. This will first heat the gas, as the work you've done on the piston goes into internal energy of the gas, which is u(T). If you weight it now, it will have total mass M = M0 + dM, where dM is the work you've done divided by c^2 (this is not measurable in practice). The energy is now in the increased thermal energy of the gas molecules, which go slightly faster, and hence have a slightly greater relativistic mass (not rest mass).

Now, let the vessel (piston down) cool down to ambient temperature. Guess what ? If you weight it now, the mass will be M0 again. dM is gone, and is in fact dissipated in the environment.

Now, let the piston expand and do some work (lift a weight or something). If you weight the vessel now, the mass will be M0 - dM'. This is because the gas cooled itself down in order to expand. It worked as a thermal engine.

Let the vessel now heat up again to ambient temperature. The mass will again be M0.

The pressure is an energy density per unit volume. If the tank is 1L and you increased the pressure ten fold, then that energy is sitting there inside that bottle waiting to do work. It never went away.

Then how come that the gas COOLS (looses thermal energy) when it expands adiabatically ? If the "pressure energy" was there all along, it shouldn't need to pick heat energy to do its work, right ?

No, the pressure gives you the thermodynamic possibility to convert thermal energy (u(T)) into work.

 

[Cyrus]

Consider a vessel with a piston. The amount of gas doesn't change when pushing on the piston. Put it on a superduper balance. Now, when the gas is relaxed, it has (bottle + gas + piston) a certain mass (found by the weight) M0.

Ok, I see what your saying. But I was referring to a scuba tank. In a scuba tank, it is not physically represented by a closed amount of mass being compressed. There is a definite, finite, measurable amount of increase in mass to the system.

I am not arguing with what your saying,...Im saying I am talking about a totally different system. I do agree with what you are saying about YOUR system though.

I was responding to Steves comment:

But the work you do to get the gas into the tank all goes into heat that goes away long before you want to use the gas for work, so it's not located anymore in the tank, so it's not stored in the tank as energy, in ANY sense.

I am saying that in THAT case the energy is inside the tank in the form of pressure. Yes, no, maybe so?

 

[bgwowk]

Isn't the potential for the compressed gas to do work dependent on the difference in pressure between it and, say, the atmosphere ?

Once a gas expands so that its pressure is equal to ambient pressure, it can do no more useful work. But for air in a scuba tank that starts off at 100 times atmospheric pressure, whether expansion is stopped at one atmosphere or zero atmospheres (vacuum) makes little difference in total work done.

 

[bgwowk]

I was responding to Steves comment:

"But the work you do to get the gas into the tank all goes into heat that goes away long before you want to use the gas for work, so it's not located anymore in the tank, so it's not stored in the tank as energy, in ANY sense."

I am saying that in THAT case the energy is inside the tank in the form of pressure. Yes, no, maybe so?

Negative. The energy content of an ideal gas depends only on temperature, not pressure. Here's what happens: As you compress gas into a scuba tank, the work done on the gas appears as heat in the gas. As long as the gas remains hot, it still contains the energy you put into it by compressing it. But as the gas cools back down to room temperature, ALL the energy you put into the gas by compressing it leaves as heat. All of it.

By being compressed, the gas contains no extra energy. But its low entropy makes it ideally positioned to do useful work. The energy of any work it does during expansion comes at expense of cooling the gas or the environment around it. Although the tremendous ability of compressed gas to do work SEEMS like stored energy, it really isn't. Put compressed gas to work doing work, and it will cool the room around it. It is physically impossible to do any process with compressed gas that will heat a room because there's no actual excess energy in the gas.

 

[Cyrus]

So what about the notion of pressure being the energy density per unit volume?

 

[vanesch]

So what about the notion of pressure being the energy density per unit volume?

Consider a tank of water. Now, push on a tiny piston, do a tiny bit of work, and raise the pressure in the big tank to, say, 100 bars. The volume of the water almost didn't change, and the amount of work you did was ridiculously low. Nevertheless, you have the same volume and pressure as with the gas tank.

 

[vanesch]

I suppose that a good way to view the "potential work a tank under pressure" can do, is similar to the potential work 100 liters of hot water and 100 liters of icy-cold water can do.
The total energy of the hot and cold water is the same as the energy of 200 liters of lukewarm water, but the first system has the thermodynamic potential to do work, while the second doesn't.
So we're slowly drifting to Helmholtz free energy as our notion for "potential energy" :-)

 

[bgwowk]

So what about the notion of pressure being the energy density per unit volume?

For an ideal gas, pressure is indeed proportional to energy density per unit volume. That's because at constant volume, pressure is proportional to temperature which is proportional to energy. And at constant temperature, pressure is proportional to number density which is proportional to energy density.

Note though that increasing the pressure of a gas at constant temperature doesn't increase the total energy content of the gas. That's because although the energy density increases with pressure, the volume decreases by the same amount, so volume * energy density remains constant.

vanesch is right that with a tank of compressed gas at ambient temperature, we are dealing with a system with a thermodynamic potential to do work, not stored energy of compression. You could say that the compressed gas is storing its thermal energy in a smaller space, and that by being more dense the energy is more "available" to do mechanical work. The mathematical manifestation of that is decreased entropy, and increased Free Energy.