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Temperature gradient of air in a very large volume

  1. Sep 10, 2013 #1
    Hello, I am new here and took a look through the forms, if there is a better place for this question, feel free to move it there.

    I have been busting my noggin trying to google the answer to this problem I have. (Which isn't school/homework related)

    I am looking to determine an approximate temperature gradient of a large contained body of air. Think something in the rage of 800 to 2000 meters in vertical height. Thinking on the problem myself I understand it will probably be complicated, depending on the energy being added to the system, turbulence of the air, density, etc.

    Assumptions I'm making:
    Temperature known at the top
    Temperature known at the bottom
    Little to no added turbulence
    Normal earth-type air

    Things I would like to know:
    How humidity and pressure effect the system.
    Energy flow from the hotter (assuming) top of the system to the cooler, bottom of the system.
    Pressure gradient of the system.
    And obviously from the title, temperature gradient from the top to bottom of the system.

    Any help in solving this problem would be awesome, I imagine the answer won't necessarily be clear cut. Though it may be simpler than I thought.
     
  2. jcsd
  3. Sep 10, 2013 #2
    Temperature gradient is defined as the difference in temperature between two points divided by their distance. In your case the distance is the height of the container
     
  4. Sep 10, 2013 #3
    Maybe I used the wrong word than, what I'm looking for is the temperature at every point between the top and bottom of the container.
     
  5. Sep 10, 2013 #4

    jfizzix

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    If we can assume this volume of air is an ideal gas under the influence of (local) gravity ([itex]F=mg[/itex] and not the universal law) then the only temperature gradient of the volume of air is in the vertical direction and is given by the dry adiabatic lapse rate.

    [itex]\frac{dT}{dz} = -\frac{mg}{k_{B}}\frac{\gamma-1}{\gamma}=\text{const}[/itex]

    where:
    [itex]m[/itex] is the average mass of the oxygen and nitrogen etc.. molecules in the volume of air
    [itex]k_{B}[/itex] is Boltzmann's constant
    [itex]\gamma = \frac{7}{5}[/itex], the ratio of heat capacities at constant volume and at constant pressure
    [itex]g[/itex] is the acceleration due to gravity

    This equation comes from the barometric equation relating the change of pressure with height due to the weight of the air above
    [itex] \frac{d P}{dz}=-mg\frac{P}{k_{B}T}[/itex]
    along with the assumption that the entropy per unit volume is constant in equilibrium.

    In this situation, we have the identity
    [itex]\frac{d P}{P} = \frac{\gamma}{\gamma -1}\frac{d T}{T}[/itex]
    which gives us the relation
    [itex]\frac{d P}{dz} = \frac{P}{T}\frac{\gamma}{\gamma -1}\frac{d T}{d z}[/itex],
    and the rest is algebra.

    If you put in the relevant values for these constants, you get a dry adiabatic lapse rate of about [itex]\frac{-9.4 K}{km}[/itex]

    There's also something called the wet adiabatic lapse rate, which is the lapse rate at 100% humidity, and then the true lapse rate is somewhere in between the two.

    Getting more accurate results can be done with models of non ideal gases, but this is as far into meteorology as an undergrad thermodynamics course (as opposed to fluid mechanics) is likely to go.

    Hope this helps:)
     
  6. Sep 10, 2013 #5
    Amazing!

    Give me some time to digest the equations and play with the numbers and I'll probably be back with a couple follow up questions.

    Thank you! :)
     
  7. Sep 10, 2013 #6
    I guess I disagree a with jfizzix. The adiabatic lapse rate applies only to situations in which the top is colder than the bottom, so that natural convection can be a major factor.

    If the temperature at the top and bottom of the container are fixed, the side walls are insulated, and, as you said, the temperature at the bottom is colder than at the top, then the temperature will vary linearly with vertical position:

    T = Tbottom+(Ttop-Tbottom)z/h

    where h is the total height of the container. With this temperature profile, the gradient provided by duato is correct. The energy flow rate (per unit cross sectional area) from the hotter top of the system to the colder bottom of the system is equal to the temperature gradient times the thermal conductivity of air:

    q=ka(Ttop-Tbottom)/h

    The humidity and pressure have very little effect on the thermal conductivity and, thus, on the temperature gradient and heat flux.

    You also what to know the pressure variation in the system. This is determined by the hydrostatic equation:

    [tex]\frac{dp}{dz}=-ρg[/tex]

    where ρ is the gas density. For an ideal gas,
    [tex]ρ=\frac{pM}{RT}[/tex]
    where T is the absolute temperature and M is the molecular weight.

    Substituting, we get:
    [tex]\frac{dp}{dz}=-\frac{pM}{RT}g[/tex]
    or, equivalently,
    [tex]\frac{d(\ln{p})}{dz}=-\frac{Mg}{RT}[/tex]
    You need to substitute the linear temperature profile into the right hand side of this equation and integrate with respect to z. The constant of integration is determined by whether you are specifying the pressure on the bottom, the pressure on the top, the average pressure, or the mass of gas in the container.

    Chet
     
  8. Sep 10, 2013 #7
    I think there are at least two regimes here. If the bottom is very hot relative to the top you will get convection and turbulence, and there won't be a stable equilibrium distribution of temperature. If the bottom is cooler than the top, or the bottom is slightly warmer than the top but not enough so that air in contact with the bottom becomes less dense than the air above, then there won't be convection and there will be a nice constant temperature gradient, as Chestermiller says.

    The adiabatic lapse rate is relevant for determining the temperature difference where the transition from stability to instability occurs. I do not believe that the temperature gradient will be given by the adiabatic lapse rate in general.

    Since the original question was about the top being warmer, the temperature gradient would be linear. Note that this is not a sensible model of the atmosphere.
     
  9. Sep 11, 2013 #8

    jfizzix

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    A few points:

    It is true that if this volume of air were far away from other gravitating bodies, and the walls were held at fixed temperatures, that the temperature gradient would be governed by Fourier's law of conduction and the overall equilibrium distribution given by Poisson's equation (provided the self-gravitation of the air is insignificant, which is is at a cubic kilometer).

    The lapse rate comes from underlying assumptions that gravitational potential energy increases linearly with height (an approximation), and in virtue of this, the temperature decreases linearly with height. By default, the "top" of the cube would be colder than the "bottom" of the cube if the air in the cube were in equilibrium.

    It could certainly happen to be the case that the top is hotter than the bottom (also known as a thermal inversion), but this is an unstable state, and heat will flow, and the air will convect until equilibrium is reached usually by way of severe weather.

    The assumptions that go into its derivation are that:
    -the force of gravity is constant over the volume
    -the atmosphere is an ideal gas in thermodynamic and hydrostatic equilibrium (no convection)
    -there is no net heat flow into or out of the volume (also part of thermal equilibrium)

    These assumptions require that the temperature increase linearly in the direction of the gravitational force (down).

    All this being said, this is an approximation, and to talk about anything other than equilibrium, fluid mechanics and thermal transport theory are required.

    As a side point, the lapse rate for humid air being different than the lapse rate for dry air is a useful tool in understanding when clouds will and will not form, and how powerful they will be. You can get conditions for a stable or unstable atmosphere and make judgements accordingly.
     
  10. Sep 11, 2013 #9
    Actually, I disagree with almost all of this, and I've had a lot of experience with atmospheric science to back it up. If you would like a list of publications, i would be glad to furnish it.

    First of all, at steady state, even in a gravitational field, if the top temperature is fixed to be hotter than the bottom, and the system is at steady state, the temperature will be described by Laplace's equation, not Poisson's equation. Poisson's equation would only apply if heat were being generated inside the enclosure, which, from the problem statement, it is not. The temperature profile will decrease linearly from the top to the bottom. Also, at steady state, there will be heat flow into the volume at the top, and an equal amount of heat flow out of the volume at the bottom.

    Thermal inversion is a stable state, not an unstable state. The hotter air is on the top, and the colder air is on the bottom, so there is no tendency for natural (buoyancy driven) convection to occur. Thermal inversion becomes a problem in urban areas where local atmospheric pollution (e.g., photochemical smog/ozone) becomes trapped near the ground as a result of the absence of natural convection (and associated vertical mixing) to remove the pollution. This is a big problem for the Los Angeles basin. Eventually, weather conditions and winds on the meso scale disrupt the inverted temperature profile, and re-establish the negative vertical gradient in temperature so that the associated natural convection can remove the pollutants.

    Under ordinary atmospheric conditions, the lapse rate is associated with hotter parcels of air near the surface rising through the atmosphere, from the higher pressure region near the surface to the lower pressures higher up. This is accompanied by adiabatic expansion of the parcels, and associated cooling. That's why they call it the adiabatic lapse rate.

    In the problem as described by the OP, the top surface is hotter than the bottom surface, and this establishes a stably stratified temperature profile, which, at steady state, is accompanied by no natural convective air flow and no adiabatic lapse rate.
     
  11. Sep 12, 2013 #10

    jfizzix

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    Oops.. Laplace's equation is what I meant (i.e. Poisson's equation without sources).

    I think we interpreted the question differently. I looked at the assumption that the temperature at the top and bottom being known as something that is not necessarily held constant. It appeared to me that the OP assumed or expected that the top would be hotter, but did not set up the problem that this had to be the case.

    For my part, I was assuming that you had a closed box of air in thermodynamic equilibrium. Away from gravitational sources, there would be no variation in the temperature, or pressure. With a local source of gravity pointing down, the new equilibrium state would be where the temperature decreases linearly with height.

    Thermal inversions are indeed stable on small-ish time scales. Hot air at constant pressure is less dense than cold air, so the hot smog would certainly persist for awhile above the comparatively cool ground. However, air is also a poor thermal conductor, and if you waited long enough (all other things being equal), the thermal inversion would dissipate, and eventually thermodynamic equilibrium would be reached (ideally).

    Because there is an equilibrium pressure gradient, there must also be an equilibrium temperature gradient (no exceptions?).
    In every parcel of air in the box, no net energy is flowing in or out of it because the (infinitesimal) work done by the pressure gradient on the parcel is equal and opposite to the (infinitesimal) heat flux due to the temperature gradient.

    From the fourth law of thermodynamics (the Onsager Reciprocal relations), we know that sufficiently close to equilibrium, the heat flow per unit of pressure difference is equal to (a constant times) the matter flow per unit of temperature difference. Since there is no convection in equilibrium, there can be no heat flow either.

    I'm not an atmospheric scientist. If I'm incorrect in my application on thermodynamics, I would be grateful to be corrected.
     
  12. Sep 12, 2013 #11
    This is wrong. Without an external temperature difference, the contents can only be in equilibrium if the temperature of the box is the same everywhere.
    Even if there is no convection, you still have conduction and radiation to equalize the temperature. If a temperature gradient would spontaneously form, you could run
    a heat engine and break the second law of thermodynamics.
     
  13. Sep 12, 2013 #12
    We definitely interpreted the question differently. The problem statement said:

    Temperature known at the top
    Temperature known at the bottom
    Energy flow from the hotter (assuming) top of the system to the cooler, bottom of the system.

    I assumed that the temperature at the top of the system was fixed at a value hotter than the temperature at the bottom of the system, and that heat was flowing at steady state from the top to the bottom. Under these circumstances, the system is not at thermal equilibrium, even though it can still be at hydrostatic mechanical equilibrium.

    I hope the clarifies things a little.

    Chet
     
  14. Sep 13, 2013 #13

    jfizzix

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    A stable temperature gradient doesn't break the second law of thermodynamics because it can only exist in the presence of an external potential (gravity). For the same reason, a stable pressure gradient also doesn't break the second law.
     
  15. Sep 13, 2013 #14
    Yes it does. you can use a heat conductor or radiation to get the temperature difference to the same altitude, and then you can run a heat engine off it, extracting energy and breakting the second law of thermodynamics.
     
  16. Sep 13, 2013 #15

    jfizzix

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    It's an interesting idea, but the second law would not be broken in this case. If you had an ideal heat conductor connecting the cold air at the top of the box to the warm air at the bottom of the box, each point on the conductor would equilibrate to the temperature directly surrounding that area, and the heat engine wouldn't work because the temperature difference wouldn't be large enough compared to the difference in gravitational potential energy.

    If you insulated the sides of the heat conductor, so that only the top and bottom were in thermal contact with hot and cold reservoirs, the effect of gravity still cannot be neglected. However, I think it might be possible to run a heat engine under these ideal circumstances with the right (possibly nonexistent) materials, but the second law would still not be broken since at no point is the entropy of this closed system decreasing.
     
  17. Sep 13, 2013 #16
    This discussion has been great. I didn't have time to really check out the first post and I knew right away it was not quite what I was looking for. However I thought it contained enough details I could work from, and probably still does.

    The following might be a big jumbled as I'm writing as I'm reading the replies.

    I did mess up a little bit on how I worded the question. I am assuming a higher temperature towards the top.

    Chestermiller had it mostly right with his interpretation. Since hot air rises I would expect there to be a region at the top where it tended to stick at the same time there being a gradual equalization of the energy in the system. That is one of the things I'm trying to get a better hold of, what kind of energy transfer rate will there be going downwards, and will it be aided by any sort of inherent turbulence (no external force mixing the air).

    I guess I'm looking for somewhat of an equilibrium condition where there is a section of hot air holding itself up top under its own buoyancy, say around 80 degrees. If the air at the top was 1,000 degrees at the start of the conditions (only much less volume, to keep the total energy of the system the same across the two examples) I imagine there would be a rapid mixing phase until it reached a more stable state, with a larger volume of cooler air residing at top). Or would that not be true, in that it would still simply slowly distribute the energy under known conditions?

    What I meant to imply by the top and bottom temperatures being known is that there would be a heat source and heat sink at each end of the system to maintain the two temperatures constant, to induce some sort of steady state, and understand what the energy flow rate would be like. Along with, just to restate, what kind of temperature distribution there would be.

    Thinking about this problem more I wonder if it is simpler than I expected. Could it be that you could look at the system as if they were immobile particles throughout with a known thermal conductivity. The thermal conductivity would not be constant throughout the column of air as the it would go down with the rise in temperature as a result of the loss in density. I suppose there would be no reason for rapid turbulent mixing as there is no temperature that doesn't fit the gradual decrease in pressure. If air suddenly got more dense at 90 degrees for example, it would want to fall rapidly when it reached that point and cause turbulence. One thing that would complicate it is maybe upward pressure caused by the warm air onto the warmer air above it.

    I'm going to leave it there, currently the more I think on it the more confused I get. I look forward to future insights into this.
     
  18. Sep 13, 2013 #17
    Reading Chestermiller's response for the fourth time (heh, thick skull), I believe he's spot on to what I want, pending any misunderstandings I have.
     
  19. Sep 13, 2013 #18
    If you can run a heat engine, you would extract energy from the volume of air, you could let int heat in the coldest part from the surroundings, resulting in a device that would produce work forever by cooling the surroundings.

    The box doesn't have the lowests possible entropy, if heat is transported from the hot to the cold part,the entropy of the whole box goes up because dQ/T_low - dQ/T_high is positive, so the box is clearly not in an equilibrium.
     
  20. Sep 13, 2013 #19
    Thanks Wilber Force. Now that you mentioned that you were looking for the steady state temperature- and pressure profile, that kind of locks it in.

    A couple of things: The thermal conductivity of an ideal gas is independent of pressure (density), and air behaves like an ideal gas at about 1 atm. Secondly, you never mentioned that the temperature could be as high as 1000C. If the top is a 1000 C and the bottom is several hundred degrees colder, you need to consider radiative heat transfer in addition to conductive heat transfer. For distances and pressure that you are assuming, the air probably could not be considered transparent to infrared radiation, and this would have to be taken into account.

    Chet
     
  21. Sep 13, 2013 #20

    jfizzix

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    What has been said about heat could easily be said also about pressure. Air naturally wants to flow from high pressure to low pressure, and that process would raise the entropy as well. Could we conclude then that the system is not in thermodynamic equilibrium because the pressure is not everywhere the same? The answer is no, because again the conditions in the box are not everywhere the same.

    The entropy of the air in the box is as high as it can be given that it has this external gravitational potential acting on it.

    Consider the kinematic example:
    -you throw an atom up inside the box against gravity; the higher it goes, the slower its speed until it starts falling back down again (or hits the top of the box if you threw it hard enough) just like any object would.
    -If you threw up a whole bunch of different atoms at different speeds and directions, they would all slow down the higher they went and either fall back down on their own, or hit the top and fall down.

    It certainly makes sense then that the average speed of atoms under the influence of gravity decreases the higher you go. This means that the average kinetic energy must also decrease the higher you go.

    The average kinetic energy per atom in an ideal gas is proportional only to its temperature.
    [itex]\frac{U}{N}=\frac{3}{2}k_{B}T[/itex]

    Therefore, (in this situation) the temperature must go down as you increase in height; this is the equilibrium state (such as it is) of the ideal gas under gravity.
     
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