Do metal springs really store sig work as potential?

[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.