Lubos Motl over at The Reference Frame has an interesting post about CO2 sensitivity. He shows that there is an upper limit to how much we can expect the temperature of a body to increase when it is required to radiate an extra 1 W/m2. So far as we can tell, he implies that this radiation-based limit restricts the amount by which the Earth's surface will warm up if it is required to rid itself of an extra 1 W/m2 of heat.
But the greenhouse effect, so far as we understand it, occurs because the Earth's surface cannot radiate its heat directly into space, but instead must pass its heat to the atmosphere by convection, conduction, and radiation. The heat will be radiated when it reaches an altitude where the atmosphere becomes transparent to long-wave radiation. Black-body radiation arguments apply to the atmosphere at this transparent altitude, but not to the Earth's surface. We have not finished our analysis of the greenhouse effect, but we suspect that this transparent altitude corresponds to the tropopause, at about 12 km, where the temperature of the atmosphere stops decreasing.
In order to force heat up through 12 km of air to the tropopause, the Earth must be warmer than the tropopause. In our previous post, we showed how doubling of the concentration of a black impurity in the atmosphere might raise the altitude of transparency from 12 km to 18 km, thus increasing by 50% the distance that the Earth's heat must pass in order to reach the altitude at which it is radiated. The amount by which the Earth's surface temperature would increase under such circumstances is dominated by convection, and is only weakly dependent upon black-body radiation.
By "CO2 sensitivity", we understand climatologists to mean the amount by which the Earth's surface will warm up if we double the concentration of CO2 in the atmosphere. We don't know of any calculation of CO2 sensitivity that is consistent with our understanding of the greenhouse effect. We have looked at a half-dozen over the years, and all are similar and casual. Implementing such calculations in a computer programs and adjusting them until they fit our climate history is, to us, unconvincing. Our long series of posts on the greenhouse effect is motivated by our desire to produce a calculation for CO2 sensitivity that makes sense to us.
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Hi,
ReplyDeleteSince I'm making no headway with Lubos, I'll leave him alone now.
Were there any other questions you wanted to ask about the adiabatic lapse rate explanation for the greenhouse effect?
Welcome, Paint Me Skeptical,
ReplyDeleteI am very interested in what you have to say about expansion and cooling in the atmosphere. Before we proceed, let us establish a baseline from which to discuss the greenhouse effect, so that we avoid the kind of confusion that I think exists in our discussion with Lubos.
Consider a planet with a transparent atmosphere. The planet itself is black. There is no greenhouse effect. Is this atmosphere the same temperature all the way through, or does it get warmer as you go higher, or does it get cooler as you go higher?
If you want to see my answer to this question before you produce your own answer, you will find my answer here:
http://homeclimateanalysis.blogspot.com/2010/01/radiative-symmetry.html
Once we agree about the temperature profile of the planet without a greenhouse effect, I'll be confident of understanding your explanation of adiabatic lapse rate.
It depends whether there is enough heat input to maintain convection, or whether radiation from the surface is sufficient.
ReplyDeleteIf there is no convection, this situation can only be stable with any hot air resting above the cold, and then with the only heat transfer mechanism being conduction, the atmosphere will approach a constant temperature throughout over a period of a few centuries, as you describe in your other post.
For an example of this sort of situation, consider Earth at night, with no wind to stir things up. After a while, the surface cools enough for convection to stop, for hot air to rise above cold, and you get a temperature inversion in which the temperature rises with altitude. It's a well-known phenomenon. The stratosphere works a bit the same way.
On the other hand, if convection is still active, then it will push the temperature up against the boundary of stability - the adiabatic lapse rate.
So with sufficient heat input, the temperature will still drop with altitude. Convection is an active process, like a refrigerator that is turned on. Turn it off, and the temperature equalises. Turn it on, and a constant temperature difference is maintained. Like a refrigerator, it requires energy input to drive it, and in the atmosphere's case, this is supplied by the temperature differences that drive convection.
But it is important to note that the adiabatic lapse rate only sets the gradient of the straight line, it doesn't set the intercept. Thus on a planet with no greenhouse effect, the surface is at the radiative equilibrium temperature, and the atmosphere is colder. With a greenhouse effect, the equilibrium temperature resides high in the atmosphere, the surface is warmed relative to it, and the atmosphere above cooled.
A point on terminology - heat is not the only way to change the temperature in thermodynamics. You can also do work on a gas. This conflicts with the everyday use of language in which a rise in temperature is synonymous with 'heating'. But in thermodynamics this isn't generally true.
The word 'adiabatic' means without transfer of heat. No heat flows in or out of the gas as it rises and falls. The internal energy (temperature) is changed by the work done on it by compressive forces. (Think of it as like storing energy in a spring.)
Are you with me so far?
I think we agree about the transparent atmosphere, but one last question about it, to make sure. In the case of the transparent atmosphere, let us ignore day and night for the moment, and consider uniform and constant radiation on the surface. In this case, the atmosphere will reach one uniform temperature with or without convection. If the planet is hotter than the atmosphere, convection will begin the transfer of heat to the atmosphere. If the planet is colder, conduction will have to do all the work. Do you agree with that?
ReplyDeleteThe night-time temperature inversion is interesting. Thank you for that.
Now, you're saying that temperature of the atmosphere is, to the first approximation, dictated by the relationship between pressure and temperature for adiabatic expansion. I like that idea ver much.
In adiabatic expansion, we have P^(γ−1)T^(−γ) = constant, where γ is the ratio of the specific heat capacity at constant pressure to the specific heat capacity at constant volume, which is 1.4 for nitrogen. So, let us consider a large body of warm gas rising from the Earth. It expands adiabatically from 100 kPa to roughly 30 kPa at an altitude of 10 km. In the process, ignoring all other effects, it cools from (roughly) 300 K to 213 K. What happened to the air that it displaced? The displaced air descended at the same time, and was compressed, during which it warmed up from 213 K to 300 K.
But your point is that we can plot T versus P with altitude because we can assume that large bodies of gas expand adiabatically.
So, given that P is equal to the weight of the atmosphere above per square meter, and T is related to P, and we have PV/T, I can solve for T and P with altitude. I will do that.
You go on to say that the above solution will work up to a point, but at some altitude, there is a "boundary of stability". What is this boundary?
You say, "Thus on a planet with no greenhouse effect, the surface is at the radiative equilibrium temperature, and the atmosphere is colder." This I don't accept. If the atmosphere is cold, heat will pass to it from the planet until it is no longer cold. All convection will do is accelerate the distribution of heat. There is no means for the upper atmosphere to lose heat, because the transparent atmosphere radiates no heat. But I think you already agreed to that point, so maybe I'm misunderstanding you.
So, when you say, "Are you with me?" I think my answer is "Pretty much."
We're nearly there. I need to re-emphasise the point about work versus heat in changes of temperature. In thermodynamics, this is expressed by the first law of thermodynamics dE = dQ-dW, where E is the energy, Q is the heat, and dW is the work done. For a gas, the work is PdV, the pressure integrated over the change in volume.
ReplyDeleteTo change the temperature of a gas, you can either transfer heat into or out of it from its surroundings, or you can apply forces to compress it. No heat flows in doing this.
So the air next to the surface is at the same temperature as the surface, and no heat flows into it. As the wind blows and the air rises, it does work on its surroundings and loses energy. No heat can flow, because all the air in direct contact with it is doing exactly the same thing, and so is at exactly the same temperature.
The distribution of heat from the surface into the atmosphere by convection is in addition to this underlying rise and fall of temperature.
You may also be interested in the meteorological concept of "potential temperature", which is a sort of temperature adjusted to account for the pressure change. The potential temperature in a turbulent well-mixed atmosphere does equalise.
This situation can only be maintained if the atmosphere is in constant motion keeping it mixed. Otherwise, conduction (or more strictly speaking, diffusion) will as you say transfer heat from the surface to the colder air.
If you're willing to take the time, there's a good video of a university lecture discussing the matter here: http://geoflop.uchicago.edu/forecast/docs/lectures.html (Warning! BIG files!) See the second part of chapter 5 in particular - "Why it's colder aloft". You can jump straight in without the earlier lectures - it's all kept at a pretty simple level of explanation.
I think you'll find that your adiabatic lapse rate applies to an atmosphere in which the speed with which heat is leaving out the top by radiation is far greater than the speed with which heat is being distributed by conduction and by mixing.
ReplyDeleteIn the case of the transparent atmosphere, there is no heat loss by radiation, so I invite you to think through your case again.
You say, "No heat can flow, because all the air in direct contact with it is doing exactly the same thing, and so is at exactly the same temperature." Now, you must be aware that there is no such thing as perfect adiabatic expansion. Sometimes we can get 99% adiabatic expansion because the enthalpy change in the gas is far greater than the heat crossing the boundary.
In the case of the transparent atmosphere, if some air goes up, some other air goes down, and at some point there is a boundary, and at that boundary there is mixing and friction. Furthermore, there is conduction, which is the source of all the convection that would occur in a transparent atmosphere, so to say that conduction is negligible, when it is conduction that starts the convection, cannot be true.
So, when there is no heat leaving the top of the atmosphere, the situation changes. Convection serves to mix up the air, and after a few centuries, the air is at the same temperature throughout. It's at the same temperature as the planet.
"I think you'll find that your adiabatic lapse rate applies to an atmosphere in which the speed with which heat is leaving out the top by radiation is far greater than the speed with which heat is being distributed by conduction and by mixing."
ReplyDeleteI don't think that's correct, but it may help us to understand if you explain why you think so.
Adiabatic changes of temperature only require air to be moving vertically and for pressure to vary with altitude. Are you saying that the air could not move vertically if no heat was being lost from the top? If air was rising at the equator, descending at the poles, and thus doing no more than transferring heat from one part of the surface to another with no losses along the way, what would stop it?
No net heat is being distributed into the atmosphere - its potential temperature is constant throughout. Heat only enters and exits the system at the surface. The changes in air temperature with altitude do not involve any movement of heat.
"so to say that conduction is negligible, when it is conduction that starts the convection, cannot be true."
Air has a thermal conductivity of 0.024 Watts per metre-Kelvin. So to transfer 100W/m^2 across a metre of air by pure conduction requires a temperature difference of 4167K. To transfer the same power across ten kilometres of air would require a temperature difference 10,000 times bigger.
Seriously, air isn't a good conductor of heat.
What normally happens is that the heat is conducted into the surface boundary layer, which is roughly a couple of millimetres thick, and from there convection and bulk motion take over.
We agree on several important points.
ReplyDeleteFirst, we agree that there is no heat lost out the top of the transparent atmosphere. Second we agree that convection can take place in the transparent atmosphere, whenever the air at the surface of the planet is cooler than the planet. Third, we agree that air is a poor conductor of heat.
Suppose you have air in a cylinder at ambient temperature and you expand it quickly. It expands adiabatically because there's no time for heat to pass through the walls of the cylinder. But if you expand it infinitely slowly, heat does have time to pass through the walls, and the expansion is isothermal. If you expand and compress air in this cylinder repeatedly and quickly, after a long time the fact that its average temperature is below ambient (we started with expansion from ambient temperature and pressure) means that heat passes into the air. Now, when we expand it to ambient pressure again, it's volume is greater. Alternately, if you expand it to its original volume, it is warmer.
Air may be a poor conductor of heat: its conductivity is ten thousand times less than aluminum. But that just means it takes ten thousand times longer for it to transport heat in the same geometry. In the case of the planet with a transparent atmosphere, we have millions of years for equilibrium to establish itself.
So, in the case of the transparent atmosphere, the air going up and down and up and down will eventually reach thermal equilibrium with all the other air. It will be at the same temperature. Give it a million years if you like.
Here's another way of looking at it. Suppose you began with a transparent atmosphere that was uniformly the same temperature as the surface of the planet. There would be no way to cool down the top of the atmosphere because this would amount to heat passing from a cooler body (the top atmosphere) to a hotter body (the planet surface), which violates the second law of thermodynamics. In the case of the transparent atmosphere, the top of the atmosphere can lose heat only by conduction and convection to the surface.
By this argument, we see that the stable state of a transparent atmosphere is uniformly warm.
Regardless of the initial state of the atmosphere, mixing by convection, combined with the slow action of conduction, and the significantly greater action of mixing at boundaries, will cause the atmosphere to approach its stable state of uniform warmth.
I look forward to hearing what you have to say about that, and thank you for coming here and debating with me.
"There would be no way to cool down the top of the atmosphere because this would amount to heat passing from a cooler body (the top atmosphere) to a hotter body (the planet surface), which violates the second law of thermodynamics."
ReplyDeleteFirstly, as I said, adiabatic changes of temperature do not involve a transfer of heat.
Second - can you explain how a refrigerator manages to violate your version of the second law, transferring heat from it's cold interior to the warm exterior?
You're very welcome. I like a good debate, as it allows me to try out different ways of explaining the idea. I'm still learning in that regard.
Your hypothetical perfect "adiabatic change" does not exist any more than the frictionless plane exists, or the incompressible fluid. The adiabatic change is a theoretical construction that can be applied in certain cases. It is your job to show us that heat flow through the boundaries of the expanding gas is negligible over the course of a million years, so that your application of the adiabatic simplification is justified. This you have not done.
ReplyDeleteThe refrigerator does not move heat "of itself" from the cooler to the hotter. It does so with the help of mechanical work. In the case of the transparent atmosphere, there is source of work with which to pump heat from a cooler to a hotter body.
"Your hypothetical perfect "adiabatic change" does not exist any more than the frictionless plane exists, or the incompressible fluid."
ReplyDeleteAgreed. But like frictionless planes and incompressible fluids, it is a sufficiently close approximation to be useful.
"The refrigerator does not move heat "of itself" from the cooler to the hotter. It does so with the help of mechanical work. In the case of the transparent atmosphere, there is source of work with which to pump heat from a cooler to a hotter body."
Exactly. Convection relies upon mechanical work, externally supplied by surface temperature differences between one place and another, to pump the working fluid around the atmosphere.
So would you like to re-word your version of the 2nd law to take that into account?
Let's keep the geometry simple to start with.
ReplyDeleteConsider an infinite flat plane heated by a distant sun. The plane has reached thermal equilibrium at 300 K. We cover it with a transparent atmosphere that is uniformly at 300 K. Your job is to show us how it is that the upper part of this atmosphere is going to cool down, through some process of adiabatic expansion and compression that I don't understand.
Meanwhile, in anticipation of this mysterious adiabatic process, I will build a column of aluminum from the infinite plane to the upper atmosphere, and insulate it perfectly. At the top of the column I place a huge heat sink. According to you, the upper atmosphere is going to cool down. If it does, I will use the heat of the earth to boil a refrigerant and turn a turbine to create work. I will condense the refrigerant with my cold aluminum column, and set up a sustained work-generating cycle. This cycle is driven by the cooling of the upper atmosphere you claim will take place through your mysterious adiabatic process.
Now, we replace the sun with a mirror, and insulate the infinite plane on the underside. No heat goes in or out, but I am still generating work with my turbine. I transport the work out of the bottom of the flat plane with a shaft, and hey presto, I have a perpetual motion machine of the first kind. I have work coming out for nothing.
So, please proceed with your calculations. If you're right, we won't have to burn any fossil fuels any more, and this whole AGW problem will go away.
Certainly. First, as convection proceeds and air from high altitude descends to the ground, it increases in temperature by compression. You could think of this as a sort of conversion of gravitational potential energy into kinetic, although maybe that's misleading.
ReplyDeleteThe hot gas is now much hotter than it was, hotter than the initially uniform temperature of atmosphere and surface. So as it comes into contact with the surface it transfers heat to it, which is radiated from the surface into space. The gas then moves along the surface in the form of wind until it reaches a warmer spot where it rises again, but colder than it was before since energy has been radiated away.
Your aluminium column is a simple heat engine, akin to extracting work from the difference between the inside and outside of your refrigerator. It is of course a fraction of the input that is driving the fluid into circulation. The energy input driving convection is your source, and the motion will only last as long as this external source does.
The greenhouse effect relies on supplied work, to drive convection. Part of this work can be extracted. It is no more 'perpetual motion' than wind power is.
If the input stopped, as happens in real life at night, then convection stops and you return to the situation you were describing earlier where the air cools from the bottom up.
It may be worth going back to the earlier questions and trying to answer them. How does a refrigerator work? How does it conform to the second law of thermodynamics? Because the situations are closely analogous.
I think we'd best agree to disagree on this one. Nevertheless, I am grateful to you for pointing out the role of adiabatic expansion in dictating the temperature profile when heat is passing out through the top of the atmosphere, and I will be making use of that in future posts.
ReplyDeleteAs you wish. Although I do hope you'll continue to think about it.
ReplyDeleteBut for future reference, I'd be interested to know why you think adiabatic expansion only occurs when heat is passing out of the top of the atmosphere. It might help me avoid future misunderstandings in other conversations.
I won't pursue the matter any further, if you do not ask me to.
I took your belief in cooling of the upper transparent atmosphere and I used it to construct a perpetual motion machine of the first kind. I proved that you cannot possibly be right. I spent considerable time doing so. Your only answer is to repeat your original statement and suggest that I need to study how a refrigerator work. Well, fine. You are welcome to your beliefs. I have done all I can.
ReplyDeleteCorrection: The closed planetary system I described, with the mirror above and the insulation below, is a perpetual motion machine of the Second kind, not the First. The first kind produces work forever and violates the First Law of Thermodynamics by creating energy from nothing. The second kind transforms heat into work with no cold reservoir, thus causing entropy to disappear and violating the Second Law of Thermodynamics. In the case of the closed planetary system, we start with the entire system at 300 K, and your magical adiabatic process causes this heat to be transformed into work.
ReplyDeleteMy response was to simply take your excellent construction and apply it to a refrigerator.
ReplyDeleteThe inside is colder than the outside and this temperature difference is maintained permanently, so if you open the door a crack and insert your heat engine, you can generate usable energy forever in exactly the same way. Your perpetual motion argument would therefore purport to prove that refrigerators are impossible.
The intention was to demonstrate that it wasn't a perpetual motion machine, by providing a more intuitively familiar example of the same physics in which the flaw in the argument could be more clearly seen. Or alternatively, by getting you to expand your argument, to locate the misunderstanding. But you chose not to expand on the argument, but instead to drop the debate.
However, I think I begin to understand the problem. I suspected it earlier, from your truncated version of the second law of thermodynamics - which says that heat does not spontaneously flow from colder to hotter without an application of work.
I think you are probably treating convection as a spontaneous equalisation of temperatures (which in small volumes of air with no pressure differential it usually is) and you are not counting the convection-driven motion of the air as supplied mechanical work. As such, you applied a specialised form of the 2nd law that is only valid in this case.
In much the same way that wind power can extract usable work from this source indefinitely, the temperature differences that arise through the same convective effect could be used to supply power in much the same way. You're not extracting energy from nowhere, you're extracting it from convection flows, and without convection the source dries up.
It takes a continual power input from the sun to keep the atmosphere convecting and to maintain the vertical temperature difference.
It would have been a lot easier to understand that if you had answered the questions on the 2nd law and how you thought adiabatic temperature changes would be prevented if you had convection in a transparent atmosphere. But nevertheless, I thank you for your time. I found it very useful.
As you say, you are welcome to your beliefs. I have done all I can. :-)
Thank you.
Your example with the refrigerator does not apply.
ReplyDeleteIf you open the refrigerator door and use the cold inside as a heat sink for a turbine, the refrigerator's compressor must pump the additional heat out of the refrigerator. The work you get from the turbine cannot be greater than the work required to drive the compressor. If 100% efficient, the system will be closed, and it will pump heat around in a circle, it will not produce any work or heat for the outside world.
Your adiabatic process allows us to build a machine that converts heat into work with 100% efficiency. That's impossible.
Our posts crossed.
ReplyDeleteIf the entire planet is at 300K, with no cold sink, how would convection occur?
Is it your argument that that the situation is impossible because you are talking about the non-convective case here?
In that case, I agree. If there is no convection, this situation can only be stable with any hot air resting above the cold, and then with the only heat transfer mechanism being conduction, the atmosphere will approach a constant temperature throughout over a period of a few centuries, as you describe in your other post.
Does that help?
"If you open the refrigerator door and use the cold inside as a heat sink for a turbine, the refrigerator's compressor must pump the additional heat out of the refrigerator."
ReplyDeleteAgreed.
"The work you get from the turbine cannot be greater than the work required to drive the compressor."
Agreed.
"If 100% efficient, the system will be closed, and it will pump heat around in a circle, it will not produce any work or heat for the outside world."
It will produce work derived from the electricity supply that runs the refrigerator.
"Your adiabatic process allows us to build a machine that converts heat into work with 100% efficiency. That's impossible."
No it won't. It's not converting heat into work, it's converting the energy of convection into work, via a somewhat indirect mechanism.
It's exactly the same physics. Replace 'convection' with 'the refrigerant pump', 'air' with 'refrigerant', the 'compressor' with 'gravity', the top of the atmosphere with the inside of the refrigerator, and the surface of the planet with the radiator fins at the back. The sun is replaced by the supply of electricity.
A gas is driven round a cycle of compression and expansion in which it cools where it has expanded and warms where it is compressed. It takes externally supplied work to do so.
Does that help any?
"It will produce work derived from the electricity supply that runs the refrigerator."
ReplyDeleteThe work produced by the turbine cannot be greater than the work required by the compressor. Do you agree with that?
My construction still stands: a planetary perpetual motion machine built out of your spontaneous adiabatic heat pump theory. You cannot be correct. Once you accept this, I will show you, with calculations, why your proposed adiabatic movements cannot take place in the one-dimensional, uniform planetary system I described above.
Okay: we have our starting point.
ReplyDelete"the atmosphere will approach a constant temperature throughout over a period of a few centuries, as you describe in your other post."
Good. I'm going on vacation now, back Tuesday. After that, I'd like to discuss the case where there is an infinitely small gradient in the surface temperature of the planet. How does the atmospheric temperature vary with position in the presence of an infinitely small gradient, over an infinitely large distance?
Yes, I agree with that.
ReplyDeleteAnd likewise, the work produced by your aluminium tower cannot be greater than the work required to drive convection. Do you agree with that?
You talk of, "The work required to drive convection." Work = Force × Distance. Which force are you referring to? Over what distance does it act?
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeleteAnd I apologize for deleting all three copies of your comment, after encountering a similar problem. Here is the text of your comment, restored with the help of my browser back button.
ReplyDeletePaind me sceptical said...
"Depends where in the chain of events you're talking about.
"The force is air pressure multiplied by the surface area of a 'parcel' of air. The distance over which it acts is the increase or decrease in radius of the parcel. Multiplying these gives the integral PdV, pressure integrated over the change in volume.
"Or if you mean the forces driving convection generally, on the large scale, these are the result of large and small scale weather/pressure systems, the Hadley cells, that result from the differential heating of day and night, equator and pole, sunshine and cloud, dark and light coloured land, land and sea. The pressure force is integrated over the distances taken to accelerate the air when the wind blows.
"Having revisited what you said before, I think that some confusion has arisen from your idealised example, where you both made the atmosphere transparent and removed all differential heating. I had assumed you intended to study the consequences of the transparent atmosphere, but I missed the implications of the uniform surface temperature. This would result in no convection, and I had already said that in the absence of convection that your static solution would apply.
"The transparency of the atmosphere has no qualitative influence on the question, the uniformity of the atmosphere's temperature does not matter, but the idealisation to a uniform surface temperature/heating does. I apologise for missing that.
"Have we made any progress?"
Let us introduce a temperature difference in the body. Let us assume the gradient of received heat from the sun is a W/m^2/m in one direction, say x. Let h be a distance that represents the height of the atmosphere. We could choose for h the height the atmosphere would be if it were an incompressible fluid (see Upper Gas post). The surface of the body is black at all wavelengths. Assume it is a perfect insulator. The atmosphere is transparent.
ReplyDeleteWe agree that for a = 0 W/m^3, no convection will take place in the atmosphere and the atmosphere will be uniformly at the same temperature as the body surface.
Let us consider a > 0. In terms of h, the specific conductivity of air, α, and the viscosity of air, γ, at what value of a will convection commence?
Doesn't the line pass through the origin?
ReplyDeleteWhich line are you talking about?
ReplyDeleteConvection as a function of a.
ReplyDelete"Convection" is not a scalar quantity that I can plot against a. There is no convection when a = 0 W/m^3, if that's what you mean. But there will be no convection for a less than some threshold that is a function of h, γ, and α.
ReplyDeleteWhy?
ReplyDeleteWhen a = 0 we agree there is no convection and the atmosphere is uniformly warm. You claim that for some value a > 0 there is so much convection that temperature varies with altitude as if it were expanding adiabatically as it goes up. In the one case, temperature at 10 km is 300 K and in the other it's 220 K. What happens for in-between values of a?
ReplyDeleteTake an empty water tank with no outlet, and drip water into it until it overflows. At zero drips per second, the tank remains empty. Turn the tap on full, and the tank soon ends up full and overflowing. What happens for the cases in between?
ReplyDeleteI have to admit, I haven't thought very deeply about this isothermal transparent atmosphere, so I could well be wrong. But I can't see any immediate reason why it wouldn't convect.
Without convection, the surface can only lose heat by radiation and conduction. If the heat input is increased by any non-zero amount, the temperature will rise, the temperature of the air in contact with it will rise, and it's density will drop. There will be a net buoyant force. And while viscosity may damp the resulting acceleration, I don't see how it can stop it.
So there will be convection. Where it will go from there I haven't thought about deeply.
That's assuming a static, isothermal atmosphere, of course. If you have hot air on top of cold, then a small enough warming of the cold air doesn't necessarily disturb things. Is that what you mean?
In your water tank example, evaporation takes place, so that there is a minimum flow rate at which the tank will begin to fill. At lower flow rates, it does not fill at all.
ReplyDeleteMy preliminary calculations suggest that for non-zero viscosity, there is NO convection for all values of a. Furthermore, the atmosphere warms up slightly as you ascend. But my calculation does not take into account the onset of turbulence, which is strongly dependent upon viscosity and the thickness of the atmosphere.
Is that it? "My calculations suggest..."?
ReplyDeleteIn my water tank example, no evaporation takes place because the air is at 100% relative humidity. You understood the point I was making, why try to get round it?
But please do carry on. I'm interested to see where you're going with this.
Okay, your tank example assumes that the air is saturated with water. What assumptions do you make for a, h, γ, and α in order for your claim about convection to be true? I'm hoping you will thrill us with some calculations of your own.
ReplyDeleteViscosity is the result of diffusion. When a fluid undergoes shear flow, molecules from the fast moving layer diffuse into the slower layers, speeding them up, and molecules from the slower moving layers diffuse into the faster layers, slowing them down. The momentum is spread around - it doesn't disappear.
ReplyDeleteViscosity isn't like friction. There isn't some lower limit of force below which it can hold things still. A gas has no elasticity.
If the surface at different locations is at a different temperature, the air in contact with them will be at a different temperature, will therefore have a different density, will therefore be subject to a net buoyant force, and force equals mass times acceleration.
At which step do you think the above chain goes wrong?
There are pieces missing from your description. When I try to fill in the missing pieces, I arrive at contradictions. Furthermore, I have no idea what you mean by "a gas has no elasticity" or "viscosity isn't like friction" or "net buoyant force".
ReplyDeleteOn this website, I build all my arguments around equations, diagrams and graphs. I do not expect my reader to fill in missing pieces. I do not expect them to translate qualitative statements into equations. If I can't describe something with equations, graphs, and diagrams, I don't expect my reader to take it seriously.
In the spirit of my own posts, I'd like you to describe, in detail and with equations, what happens to an element of air at every stage of your hypothetical convection cycle. Once you have done that, you will have given me something to consider and comment upon.
No, I don't think so. The description is clear enough for someone with an average grasp of physics to follow. Buoyancy and elasticity are well-known concepts.
ReplyDeleteYou say you don't expect readers to fill in missing pieces, but you have alluded twice now to "calculations" and "contradictions" without any indications of what you mean, other than that it has something to do with viscosity. And you have asserted with an appearance some confidence that one can have differential heating of the surface without convection but not said how, or why you think so.
I'm curious as to what you mean, but not excessively so. Do you want to continue the discussion?
I invite you to explain to me in more detail, with the help of mathematical equations, what happens to an element of air in your hypothetical transverse convection cycle. Once you have done that, I can comment upon your hypothesis. But I am not interested in further qualitative or philosophical discussion of the subject because I simply have no idea what you are talking about.
ReplyDeleteThat's an unanswerable question, because convection is turbulent and chaotic - the result of the Navier-Stokes equation. There is no way of predicting what happens to an element of air!
ReplyDeleteAnd I think you know perfectly well what I'm talking about. Your discussion above indicates that you are reasonably competent at physics, and the basics of convection aren't complicated. I'm not sure why it matters so much to you that there should be some static solution in this hypothetical situation; it seems an odd thing to make a stand on.
OK, suppose for the sake of argument that the air on this almost uniformly lit sphere is not only transparent and initially isothermal, but elastic, so that it can resist a force without flowing. What now? What significance does it have?
Take a look at my most recent post, Atmospheric Convection, for the sort of calculations I'm talking about. I'll follow that with a calculation of the work available from the convection cycle, and what happens to that work in the atmosphere.
ReplyDelete