Does the atmosphere cool with altitude due to gravity?

[moatilliatta]

The work of the total process is zero. The calculation in the link results in a net energy production because W₂ and W₄ do not include the volumetric work done by pushing the air out of the vessel or into the vessel. They just cover the expansion or compression and are therefore be valid in an evacuated environment only only.

Hi DrStupid,

I'm very glad someone is having a go at this. I'm not sure I'm reading you correctly, but I am pretty sure the vessel does not expand. I think its quite clear that the energy is harnessed from air moving through expanders that are stuck to side of the elevator shaft. There is no volume to be expelled.

I could easily be misunderstanding you so please elaborate.

 

[votingmachine]

I think that is not quite the set up they describe. The vessel is a fixed volume. The vessel is attached at the bottom for "filling", then detached, and attached again at the top for "emptying". I follow the math, but the intuitive sentence for step 2b and 4b elude me.

I still half-think there is a heat engine going on. As a mental image, picture a box with a gossamer divider. Say the top pressure is 0.5 atmospheres and the bottom is 1.0. The divider is at the end of the box and the air inside is at 0.5. At the bottom, the expanding 1.0 atmosphere air moves the divider to 1/2 the volume, and fills the other side. That makes the box full of 1 atmosphere air. Then at the top the divider slides to the end again, as the air expands.

At the bottom, the isothermal expansion absorbs heat, to keep the temperature constant as 1.0 atmosphere air expands, which would ordinarily cool. At the top the expanding gas again absorbs heat. To have an isothermal expansion, heat has to be added.

I'm not sure that the answer cannot be non-zero. It seems like it ought to be zero. But if it is not I think the heat flow explains it.

 

[mic*]

It IS a heat engine in a guise where the external heat added that drives a delta-P (to do work) is subsituted for gravity driving the delta-P.

So the true question is can we harness gravity like this? I am questioning all the idealised elements of the elevator.

I suspect the 2nd law of thermodynamics isn't really in jeopardy.

 

[willem2]

I think that is not quite the set up they describe. The vessel is a fixed volume. The vessel is attached at the bottom for "filling", then detached, and attached again at the top for "emptying". I follow the math, but the intuitive sentence for step 2b and 4b elude me.

You need to push the outside air away. The usual formula for the work gained from isothermal expansion, assumes that the gas is in a cylinder pushing against a piston with vacuum on the other side. If there's air on the other side, you need to do work to push it away.
Note however that the work on the outside air at the top and at the bottom cancel, so you could just ignore this.

At the bottom, the isothermal expansion absorbs heat, to keep the temperature constant as 1.0 atmosphere air expands, which would ordinarily cool. At the top the expanding gas again absorbs heat. To have an isothermal expansion, heat has to be added.

So you could have a process that takes or produces no work and cools the whole elevator?
The expansion at the top produces work and removes heat, and the compression at the bottom takes work and removes heat. This is only OK as long as there is no temperature difference, since otherwise the 2nd law would be violated. In the real atmosphere, the elevator would take work, cool the top and heat the bottom more.

I'm not sure that the answer cannot be non-zero. It seems like it ought to be zero. But if it is not I think the heat flow explains it.

The problem with the calculations on the Isothermal Elevator page is in

W_{exp} = \Delta nRT ln (\frac {P_{final}}{P_{start}})

used in the calculation for the expansion or compression of gas.

I don't know where he gets the Δn from.

 

[moatilliatta]

The problem with the calculations on the Isothermal Elevator page is in

Wexp=ΔnRTln(PfinalPstart)​

W_{exp} = \Delta nRT ln (\frac {P_{final}}{P_{start}})

used in the calculation for the expansion or compression of gas.

I don't know where he gets the Δn from.

Yes,you must be right. So if that is the wrong formula, what is the right formula for the process?

 

[willem2]

Yes,you must be right. So if that is the wrong formula, what is the right formula for the process?

Just

W_{exp} = n RT ln (\frac {P_{final}}{P_{start}})

You let gas in a cylinder expand and do work against a piston until the pressure inside the cylinder is equal to the pressure outside. Then you and open a hole in the piston, and move it back to reduce the volume to the original, but this doesn't take or produce any work.

 

[moatilliatta]

Just

Wexp=nRTln(PfinalPstart)​

W_{exp} = n RT ln (\frac {P_{final}}{P_{start}})

You let gas in a cylinder expand and do work against a piston until the pressure inside the cylinder is equal to the pressure outside. Then you and open a hole in the piston, and move it back to reduce the volume to the original, but this doesn't take or produce any work.

What I mean is for the expansion into a non-vacuum. If that formula is used in place, it would be an even better perpetual motion machine!

 

[votingmachine]

The thing I can't accept in the non-zero answer is that it is a non-zero answer for any arbitrarily small height difference. There is a theoretical free bit of energy between the air a meter apart. And the air would just move.

Now there is a great deal of atmospheric movement, convection up and winds around. I live in Salt Lake City, which is in a bowl surrounded by mountains. We get a temperature "inversion" in the winter when ground air temperatures get low, and there is no air exchange with clean high altitude air ... no thermal movement and the pollution builds up. But with ordinary atmospheric conditions, air is moving up and down. Since clean air matters, I pay attention to the fronts and the weather, especially when we get a month wit a stable inversion. There are stable systems where energy can be extracted (stable temperature gradients in the ocean, for example). The atmosphere allows for wind extraction.

But the atmosphere seems like it is too dynamic for a stable energy gradient to exist. JMO.

EDIT: It has the intuitive feel of being an efficient elevator that lifts mass, and then the mass is dropped harvesting the energy efficiently ... and a net gain on the cycle. And I think the gain comes because you expand the same air twice. At the bottom you expand some air into the vessel, absorbing heat, and then again at the top, you expand the same air again, absorbing heat. And that seems wrong intuitively, as though the piston exhaust cycle powers a second piston. But if there is heat absorbed from the environment ... I guess even that works.

It does seem that the atmosphere would not really have a stable energy gradient available though ... the self correction seems too simple.

And again, I have to wonder what the math says about the temperature at different heights. If you calculate a temperature at which the process does not have a net energy surplus, what temperature profile do you get. So at a height, you bring down cold air, which has to be warmed at the bottom (free), and then you raise a bottle of warm air, but you can't expand isothermally, but only can warm at the ambient temperature. So at a height with lower pressure and density, what temperature would be net energy neutral?

 

[moatilliatta]

The thing I can't accept in the non-zero answer is that it is a non-zero answer for any arbitrarily small height difference. There is a theoretical free bit of energy between the air a meter apart. And the air would just move.

I can't accept an above zero answer because it is perpetual motion!

The logic of DrStupid didn't make sense until willem2 explained it differently. Afterwards I got a pen and paper out and was able to determine that W2 + W4 is equal to half what Vanquish Opprobrium had calculated, same conclusion as DrStupid but using a different route. If everyone is happy with DrStupid calculations, I won't bother showing mine (LaTex is confusing ).

 

[votingmachine]

Just

W_{exp} = n RT ln (\frac {P_{final}}{P_{start}})

You let gas in a cylinder expand and do work against a piston until the pressure inside the cylinder is equal to the pressure outside. Then you and open a hole in the piston, and move it back to reduce the volume to the original, but this doesn't take or produce any work.

How is the "n" in this formula different than the "delta-n" in the original formula?

They use the change in moles in the vessel. What is the "n" you use?

I thought the delta-n followed from the integral of PdV from V1 to V2, and then a substitution using the ideal gas law.

I can't accept an above zero answer because it is perpetual motion!

The logic of DrStupid didn't make sense until willem2 explained it differently. Afterwards I got a pen and paper out and was able to determine that W2 + W4 is equal to half what Vanquish Opprobrium had calculated, same conclusion as DrStupid but using a different route. If everyone is happy with DrStupid calculations, I won't bother showing mine (LaTex is confusing ).

I don't accept any conclusion that implies perpetual motion with a surplus energy. But I still think that the entire process is not perpetual motion. It has that feel, but the heat flows from the atmosphere are hidden. I would love to see your calculations. I still am not clear on all the steps of the cycle. If you want to just summarize, that is fine also.

 

[mic*]

I thought the delta-n followed from the integral of PdV from V1 to V2, and then a substitution using the ideal gas law.

Think about this integral. If n changes wrt V then you need to integrate n.dV... This is not done.

 

[votingmachine]

Think about this integral. If n changes wrt V then you need to integrate n.dV... This is not done.

Ahhh. Of course. It is the total gas (moles) in the vessel holds at high pressure that are being expanded at the top, and in the reverse at the bottom.

The top and bottom gas work still seems the same, but what would "n" be at the bottom? The atmosphere is expanding, pushing the "delta-n" back into the vessel. It seems like it should e the same "n" at top and bottom.

Changing to "n" from "delta-n" does not move the work total to zero though. If the top and bottom are the same "n":
W=2nRTln(Pt/Pb) (Pt=pressure at top, Pb= pressure at bottom)
n=PbV/RT

W=2PbVln(Pt/Pb)
put in the exponential relating Pt and P
W=(2Pb)VMgh/RT

Meanwhile the work for raising and lowering was:
W=(Pt-Pb)Mgh/RT

 

[mic*]

You're right, it doesn't change the total work to zero.

The true vessel is the whole atmosphere... So... mmm. I'll leave that for now but I think you can see where all the number crunching may lead there...

The (calculated) work done by moving air up and down is real. It is equal to the work you would need to ADD/DO to lift a heavy packet of air minus what is recovered by sinking a light packet of air. (Plus real world losses, friction etc.)

 

[votingmachine]

This still seems like perpetual motion. I've googled around to look at heat pumps to try to find some insight and I do think there is a bit of heat pump energy being hidden, but it still seems like perpetual motion, and therefore there cannot be surplus work.

I come back to a piston engine analogy. The hot gas expands in the piston, generating some work. But due to this trick of isothermal expansion, the gas in the piston is just as full of energy at the end as at the beginning. There is work available from expanding gases ... it is everywhere. But in this weird scenario, the expanding gas ends up in the vessel with as much energy as the beginning.

I agree that real world inefficiencies would dwarf the energy gain anyway, even if you took these calculations as correct.

 

[DrStupid]

I think we have a consensus that the gas remaining after decompression at the top doesn't contribute to the total work of a complete cycle. It is lowered, compressed, lifted up and decompressed again within a closed process without net energy change. The isobar volumetric work against the outer pressure also cancels out.
Thus there are only two relevant processes left: lifting up compressed air from the bottom to the top and decompression of this air at the top. The work done in the first process is

W_{lift} = n \cdot M \cdot g \cdot h

During the decompression the volume changes from an initial volume Va to the final volume

V_b = \frac{{p_a \cdot V_a }}{{p_b }} = V_a \cdot \exp \left( {\frac{{M \cdot g \cdot h}}{{R \cdot T}}} \right)

and the pressure changes according to

p = \frac{{p_a \cdot V_a }}{V} = \frac{{n \cdot R \cdot T}}{V}

That results in the work

W_{\exp } = \int\limits_{V_a }^{V_b } {p \cdot dV} = - n \cdot M \cdot g \cdot h

I still do not see any net energy production.

 

[votingmachine]

I think the difference between what you are saying and the elevator page said is at the bottom cycle. And I have trouble with that one. There is agreement on the work of the raising and lowering. And I think there is agreement on the top expansion. But what about the work at the bottom? Here is a quick powerpoint assembled GIF. The "pistons" are just boxes and I did not think ahead to make the top process as the same height, but put the shapes in a cycle ... just imagine the drawing tilted 45-degrees. The temperature is the same everywhere, with heat flowing from almost an infinite sink in each case.

If there is too much confusion in this GIF, I can re-make a better on, but it seems good enough. I keep thinking that there is no heat flow at the bottom, and an inward flow of heat at the top. And heat flow is the only way I see to justify a cycle with net energy production over the gravity work.

 

[DrStupid]

But what about the work at the bottom?

At the bottom there are two processes where work is done:

1. The compression of the air inside the vessel - in your picture from the original volume V and top pressure p/2 to the Volume V/2 and bottom pressure p.
This work is compensated by the expansion of the same air back to the original Volume V and top pressure p/2 at the top.

2. The intake of environmental air with volume V/2 and pressure p.
This work is compensated by pushing out the same (but already expanded) air with volume V and pressure p/2 at the top.

There is no other work at the bottom.

I keep thinking that there is no heat flow at the bottom, and an inward flow of heat at the top. And heat flow is the only way I see to justify a cycle with net energy production over the gravity work.

That would be a perpetual motion of the second kind. Do you really think violating the second law of thermodynamics is that easy?

 

[votingmachine]

At the bottom there are two processes where work is done:

1. The compression of the air inside the vessel - in your picture from the original volume V and top pressure p/2 to the Volume V/2 and bottom pressure p.
This work is compensated by the expansion of the same air back to the original Volume V and top pressure p/2 at the top.

I agree with this accounting. By this accounting, there is an invisible divider, and there is a fixed bit of air that compresses and expands, and it is just a reversible process with equal and opposite energy amounts.

At the bottom there are two processes where work is done:
2. The intake of environmental air with volume V/2 and pressure p.
This work is compensated by pushing out the same (but already expanded) air with volume V and pressure p/2 at the top.

There is no other work at the bottom.

This part eludes me. The intake is not energy absorbing, it is spontaneous. The expansion at the top is also spontaneous. The question is if those two are greater in energy than the energy in raising the gas via the elevator.

That would be a perpetual motion of the second kind. Do you really think violating the second law of thermodynamics is that easy?

I've said several times that I know perpetual motion with energy production is impossible, and that the real world has inefficiencies. So no, I would say I've already answered that in the negative may times.

My first thought at seeing the process was that it was a perpetual motion machine and impossible. One math error has been found in the expansion formula, but that error leaves a net work. I see the step-wise division of the air between vessel contents and the in-and-out air as a useful visualization, but it adds some confusion as well.

I won't be surprised to finally see an insight that says the work adds to zero. I just am struggling to find that insight. If you have that insight, it may seem like I am arguing for a perpetual motion device ... I am not. I am merely unable to see what the right breakdown of the problem is. It is very possible that I am breaking it down incorrectly ... I've been wrong before and I'll be wrong again. I don't have a horse in this race other than the general curiosity raised by reading the initial proposal. I regard the proposal as non-productive for energy, even as written. It has too many parts with 100% efficiency assumed, and what looks like a small yield even with those assumed efficiencies. But as a puzzle, it still is interesting as to why they chose the math they did, which leads to a non-zero answer, and what the right math is, and what the right answer is. You are mistaken if you think this is my proposal, or that I am advocating it as proof-of-concept.

 

[votingmachine]

I should clarify the perpetual motion remarks. When I say that it looks like perpetual motion, I am speaking intuitively. There cannot be perpetual motion with an energy output. If the math ends up saying there is an energy output, it is because we are not looking at an energy source properly. I know that the impeller on a hydroelectric dam may be in motion, perpetually, but that is not perpetual motion, but motion from an energy source, the water height.

I still think that there really can't even be an energy source. If there is a viable energy difference between two altitudes with the same temperature, there is no barrier to the energy flowing already, and equalizing those energies. It takes the dam to build up the energy source for hydroelectric, and there is no dam between all the different altitudes.

Since the air is cooler at any realistic altitude, that seems likely to be the factor that equalizes the energy (if the elevator conclusively has a difference to harvest).

 

[DrStupid]

The intake is not energy absorbing, it is spontaneous.

The intake is not energy absorbing but energy providing. You can harvest this energy with a piston that is pushed inside by the incoming air.

The expansion at the top is also spontaneous.

I do not talk about the expansing but about pushing the air outside away. This isobar process is not spontaneous and consumes energy. This energy is provided by the expansion and it is equal to the work of the isobar intake of air at the bottom.

 

[votingmachine]

The intake is not energy absorbing but energy providing. You can harvest this energy with a piston that is pushed inside by the incoming air.
I do not talk about the expansing but about pushing the air outside away. This isobar process is not spontaneous and consumes energy. This energy is provided by the expansion and it is equal to the work of the isobar intake of air at the bottom.

Got it. I understand what you are saying now. I will have to go back and look at your math later.

EDIT: The thing that I have found confusing (and is once again confusing to me) is that the work you are describing already seems accounted for in an isothermal expansion formula. The expanding high pressure air in a cylinder expanding can generate energy. The energy it generates is during the process of moving aside a volume of low pressure air.

As I said, I am confused again, but I will look at it later.

 

[gleem]

Unless I missed something in the previous responses wouldn't the absorption of the sun's radiant energy by the surface of the Earth and the insulative effect of the air surrounding it account for a substantial ( if not most of ) gradient of temperature with altitude? Also with the re-radiation of the heat trapped in air with the subsequent returning of only some (half?) of the heat back to the Earth and the loss of heat through the atmosphere above (density decreasing with altitude) also contribute to heat loss and thus temperature decrease with altitude?

 

[votingmachine]

Unless I missed something in the previous responses wouldn't the absorption of the sun's radiant energy by the surface of the Earth and the insulative effect of the air surrounding it account for a substantial ( if not most of ) gradient of temperature with altitude? Also with the re-radiation of the heat trapped in air with the subsequent returning of only some (half?) of the heat back to the Earth and the loss of heat through the atmosphere above (density decreasing with altitude) also contribute to heat loss and thus temperature decrease with altitude?

That is the explanation I have always seen. Certainly that has to be a factor.

In googling around I found a comment that the molecules of a gas are moving randomly, but that they still have mass. Moving up, they get slowed by gravity, moving down, they get accelerated by gravity. So it might be possible to conclude that air molecules that have moved to the bottom are more accelerated than ones that are moved to the top.

That seems intuitively sound, but I like to see actual proof ... so I had not quoted that idea to this thread ... I've no idea of the source now. It might be nonsense ...

And besides the thread turned into analysis of a rather weird thought experiment.

 

[jbriggs444]

That is the explanation I have always seen. Certainly that has to be a factor.

If the temperature gradient resulting from surface heating from below combined with purely conductive heat transfer is greater than the relevant adiabatic lapse rate then you will have convection. Warm air from the surface will rise through the layers above, even though it cools as it rises. The result of convection is then to reduce the temperature gradient. *voila*. The resulting temperature profile matches the adiabatic lapse rate.

 

[moatilliatta]

Votingmachine,

Excuse the late response; I was had a conference last week and have been flat out catching up with things.

At the top, the n_bottom is expanded and preforms work, but not without clearing room for the gas in the atmosphere, which requires work.

At the bottom, n_top has to be compressed and that requires work. The energy harnessed comes from expanding the atmosphere into the void created in the vessel when n_top has been compressed.

The compressing and expanding of the atmosphere cancel out, leaving the sum of the two expansion to be f(n_bottom) - f(n_top) which equals f(n_delta).

This is exactly half of what was calculated in the medium post and result in the device producing zero energy.

 

[moatilliatta]

I've just cleaned up my inbox and saw an email from medium linking me to this:

https://medium.com/criminal-clouds/the-drop-down-cycle-part-3-of-3-af7d54b3142b#.ljgxauq3z

Looks like VO came up with a new version of part 3 a couple of weeks back. I've only had a brief look and am yet to work out how its wrong.

Maybe someone here can work it out?

 

[moatilliatta]

I've had some time to look at this new one. Here is a spreadsheet I've made:

https://docs.google.com/spreadsheets/d/19kqXiOmVe0kTX7JdSvBFoSfhlE2uITBScq2tn82udZg/edit?usp=sharing

I was thinking this thing would produce net zero like the last, if the height of the tall column formula was bungled. It doesn't appear to be the case, unless Wikipedia doesn't have the right formulas .

 

[kyle Bacon]

Ugh, so much maths... it eats my brain.

After several months, It still doesn't make any sense why gravity wouldn't just cool gas particles as they move against it.

Look at the Atmospheric Escape wiki page which says "The more massive the molecule of a gas is, the lower the average velocity of molecules of that gas at a given temperature, and the less likely it is that any of them reach escape velocity". https://en.wikipedia.org/wiki/Atmospheric_escape

If the gas particles just maintained their velocity/temperature as moved upwards, they would just fly off into space!

 

[Drakkith]

After several months, It still doesn't make any sense why gravity wouldn't just cool gas particles as they move against it.

If the situation was solely about how the kinetic energy of a gas particle behaves as it moves away from a source of gravity, then you would probably be right. But there are many other effects to take into account, such as convection, radiation, etc.

 

[moatilliatta]

Let me try to distil this new device down to something simple for Kyle.

The barometric formula for an isothermal body of gas is P_top=P_basee^-Mgh/RT

Because temperature is found in the denominator of the negative exponent, the higher the temperature, the less pressure falls for a given increase in altitude. Hence the hotter column must be taller for the columns to have a "proportional pressure" between the tops and bottoms (e.g. the tops are half the pressure of the bottoms). This height difference is then exploited by an elevator mechanism which lowers gas from the top of the hot column, to the top of the cool column.

Once you get your head around its principles, it seems quite elegant. Yet elegance can be deceptive...