A Note: Global Warming Threads

Discussion in 'Earth Science' started by Tristan, Aug 27, 2004.

  1. Trippy ALEA IACTA EST Staff Member

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    10,890
    Nope, no misunderstanding.

    Here you're talking about the way heat is distributed accross the surface of the body.
     
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  3. Andre Registered Senior Member

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    889
    Yes that's the next step. We should be looking at a few different cases, to see the difference, a perfect black body, the current moon, and maybe a moon with a hypothetical nitrogen atmosphere, to simulate our null hypothesis.

    Case 1. A theoretically perfect (grey) black body redistributes all absorpted radiated energy evenly over its surface; instanteneously, as it is hypothetically a perfect conductor. None of the energy is stored at heat below the surface because it has hypothetically a zero heat capacity. This gives an uniform radiation equilibrium temperature everywhere, the for Earth so often mentioned -18C with the disc-sphere surface factor 1/4.

    Case 2. If we look at the moon, there is practically no heat redistributed over its surface and sunlit places will approach the isolated local equilibrium radiation temperature according to Stefan Boltzman (times cosine for lattitude and longitude but obviously without the factor 1/4); the hottest place being above the boiling point of water. The rate of heating is limited to the heat capacity of the surface and some energy penetrates to deeper layers.

    At nightside outwelling radiation cools the surface, and it's only the heat energy stored below the surface that limits the rate of cooling.

    Case 3. If we give the moon a hypothetical nitrogen atmosphere, we're very close to the hypothetical null hypothesis. Now the atmosphere takes up heat energy from the surface at the sunny side, which is redistributed within the nitrogen atmosphere by convection and advection. The amount of energy that enters the atmosphere via radiation is negliglible because nitrogen is virtually transparent for infrared radiation. This also means that the energy leaving the atmosphere via infrared radiation is equally negliglible, which brings us to that earlier mentioned "toe curling counter intuitive" situation.

    But does this all make sense so far?
     
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  5. Trippy ALEA IACTA EST Staff Member

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    I don't see the point in considering the moon. The moon is irrelevant to the question.

    This is similar to what I had in mind, but I'm not quite there yet in terms of having time to write anything up. Although I will contend one thing, I think some of your assumptions in your N-only atmosphere are wrong, for example, your scenario focuses on microscale effects and completely ignores macro-scale effects. It has no consideration, of, for example, hadley cells, which transfer heat from the equator to the poles, and convection cells which would transfer heat from the day side to the nightside.

    I would also contend that there is a relevant case that you're not considering, but I'll get to that as well.
     
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  7. Andre Registered Senior Member

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    889
    But it greatly helps visualizing as an true, real, empiric example how the solar heat is distributed on a real (non-black-) body. But we can take an imaginary concrete ball or anything in solar orbit and rotating

    well I did say:


    Obviously what goes up must come down and the hadley cell is a convection cycle but we did not get to the details here yet.

    So shall I wait or continue?
     
  8. Trippy ALEA IACTA EST Staff Member

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    10,890
    I said that once I had a straight answer from you on the question I was asking I would continue my elaboration further when I had the time. I've also told you that this week has not been a kind week to me one way or another.
     
  9. iceaura Valued Senior Member

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    24,295
    There is no null hypothesis here. You have to have an active hypothesis, a postulated explanation for something, first. That requires an argument be presented, the null hypothesis being the basis of a counterargument. You haven't done that.

    Since you have no arguments of any kind, and present no hypothesis to deal with, there's nowhere for a null hypothesis to have a role if you did present one. The entire exercise is empty. But it takes up a lot of bandwidth, doesn't it? The entire thread subject has vanished. So kudos, of a kind.
     
  10. Andre Registered Senior Member

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    889

    Maybe this YES was too concealed?

    But by all means take all the time you need. I'll be here.
     
  11. Trippy ALEA IACTA EST Staff Member

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    No, it wasn't.
     
  12. Andre Registered Senior Member

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    889
    Anyway, I wasn't sure if it was my turn to react on:

    But since we were spending time on less relevant things in another thread I guess it is.

    So we take it from heat exchange of an inert atmosphere without radiation.

    Consider this, convection alone in a (nearly) non-radiating atmosphere follows (nearly) the law of preservation of energy. So if an nitrogen parcel rises, it expands and cools dry adiabatically with these relationships. Consequently when the same parcel is forced downwards again, to be replaced by more convecting nitrogen, it will heat up dry adiabatically at the same rate due to compression. Now, in the period before equilibrium, the problem is that this descending -heating- nitrogen is soon to be warmer than the surroundings and hence stops descending by the logic of buoyancy. We see similar effects in the real atmosphere, which is called subsidence inversion. So initially it would not reach ground levels. Consequently, the convection is maintained by replacing the updrafting air by cooler air below the subsidence inversion. This effect would make the energy one way transfer -at least in the transient state- more effective, as the warmer nitrogen is not even returning close to the surface, where it could exchange energy by direct conduction. So again, the atmosphere would warm despite the lack of greenhouse effect.
     
    Last edited: Nov 17, 2013
  13. Trippy ALEA IACTA EST Staff Member

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    No, not really. We're waiting on me to find the time in amongst my carbon world commitments to tackle what I consider will be a particularly in depth and complex post. I had hoped to do so over the course of the weekend but it seems the universe had other plans.
     
  14. Andre Registered Senior Member

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    889
    Fair enough. Obviously this mechanism has been discussed elsewhere and so far it has been conceded -by a routine modeler- that such would be very hard to quantify.

    For that we still have another element to consider. The only way that the nitrogen atmosphere can cool due to conduction to the surface of our (non-black)body. So if cool nitrogen is constantly generated at the surface at the night side, we should look at the behavior of cold and warm air masses meeting at frontal systems in relation to this convection cycle, I think that for a complete picture, it's better to have a brief look. Most importantly:

    This should illustrate -in the inert nitrogen atmosphere- what obstructs the warm gas aloft to reach the surface again, further obstructing the cooling of the atmosphere.
     
  15. Trippy ALEA IACTA EST Staff Member

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    10,890
    No, see, this is one of the mistakes you have made. Conduction to the surface is not the only way the atmosphere can cool.

    When a parcel of air is heated, it rises.

    As a parcel of air rises it expands.

    As a parcel of air rises and expands, it expends internal energy (heat) to do work to push the surrounding atmosphere aside and cools (according to a dry adiabat - 9.8k/km).

    It continues to rise, expand, and cool following a dry adiabat until its temperature and pressure are in equilibrium with the surrounding air.

    Moist adiabats cool slower than dry adiabats because the parcels of air derive energy from the latent heat of water condensing to do work.
     
  16. Andre Registered Senior Member

    Messages:
    889
    I think that's exactly what I said. Note that's not about the temperature but about the energy.

    Oh and incidentely, part of that is also about potential energy. As a high energy parcel of gas rises, the temperature goes down due to expansion but also due to gaining potential energy, but that returns in the decend exactly like the restoration of the temperature due to dry adiabatical compression. But in all these changes the total energy within the parcel of gas remains constant

    Don't think in terms of temperature but in terms of energy.

    So I should have said there:

    "The only way that the nitrogen atmosphere can lose energy due to conduction to the surface of our (non-black)body."
     
  17. Trippy ALEA IACTA EST Staff Member

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    I don't, and I still don't.

    If only I had thought of that...

     
  18. Andre Registered Senior Member

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    Now where did this energy go? I guess we agree that you can't 'destroy' energy. And where is it, when you force this parcel of air down again? What would that do to the temperature of the parcel of gas?

    Hint: check the meaning of adiabatic
     
    Last edited: Nov 18, 2013
  19. Trippy ALEA IACTA EST Staff Member

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    10,890
    I addressed that as well:
    In order for the parcel of air to sink, it must be compressed. Work must be done upon it. Work is done by the parcel of air as it rises, and done upon the parcel of air as it sinks. I trust you've used a bicycle pump at least once in your life? But this leads us to one of your next mistakes, which I shall get to when I again find time.

    I'm familiar with the meaning of the term, thankyou, I've studied thermodynamics on three seperate occasions in three seperate topics (Physics, Chemistry, and Geology). I may utterly despise the topic, but it has been well drummed into me.
     
  20. Andre Registered Senior Member

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    889
    So would that change anything on where the energy went? And what would change that to the notition that energy enters the non radiation atmosphere by convection but only leaves by conduction with the surface?
     
    Last edited: Nov 18, 2013
  21. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    QUESTION:
    Are a disk and sphere at the same temperature (when it is steady) and if not, which is hotter and what is the temperature ratio?

    To define the question well, I hope, both are opaque solids; the disk is extremely thin (neglect any real thickness for answer); and have the same diameter, approximately that of the Earth, and are made of the same “infinite” thermal conductivity material (no surfaces variation of temperature) and both are in a perfect vacuum. Note that both sides of the disk radiate so the radiating area of the sphere is twice that of the disk.

    The light source is a point, with approximate the solar spectrum, on the axis of the disk that has approximately the same intensity at 1 AU as the sun does. I. e. assume, for convenience, 1000W / square meter strikes the closest point of both to the source; but for convenience when answering, neglect fact that for both, some parts of each are in very slightly less intense light as they are slightly more than 1 AU from the source.

    Use a system of units such that power radiated from a unit area is P = T^4.

    Assume for the surface with angle of incidence A = 0 that 100% of the flux striking there is absorbed, but as A goes to 90 degrees the absorption decreases linearly to zero and the scattering (or reflection, which ever term you prefer) become 100% of the flux “striking” there.

    I. e. the absorption coefficient, a, is given by a = 1 - (A / 90) for A between 0 and 90.

    The answer, that I have given before (to half the question) is that the disk is hotter due to a much larger energy flux scattered away and would still be hotter even if there were no scattering (a =1) due to having twice the radiating area.

    In that a = 1 case, the absolute temperature of the sphere is the fourth root of 2 lower than that of the disk. For the linearly decreasing a case, one must integrate differential rings about the normal point to get the average absorption, which is only slightly less than 1 for the disk but significantly less for the sphere. There is some a = f (A) for which the average absorption of the sphere is half that of the disk. In that case, the sphere's temperature is square root of 2 lower than that of the disk.

    Trippy seems to assume there is no reflection or scattering even when the angle incidence is almost 90 degrees on the sphere. That is not true for any real material, not carbon or any black paint:
    So I hope he will answer my question or at least tell what makes his just quoted comments correct. I am sorry I was too busy to reply sooner.
     
  22. Trippy ALEA IACTA EST Staff Member

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    10,890
    This is a leg pull right Billy?

    I'll start at the end of your post.
    It's because your focus is on the wrong words, allow me to illustrate.
    Pay attention, I'm going to be repeating that more than once.

    That's not the question being asked in any context of the discussion, not only that, but it is completely irrelevant to the discussion. The question was regarding the use of the stefan boltzman law, and the treatment of a nonideal blackbody (strictly, greybody) as an ideal one. I asked the question because the calculation rests on the assumption that total power in equals total power out, as required by the conservation of energy. There are exceptions to this, however, they involve internal energy sou

    None of this is even remotely relevant to the discussion.

    This is actually almost relevant.

    Ummmm....

    The rest of us use the albedo of the object rather than making assumptions such as this.

    Say what? The rest of us could take the cosine of the angle multuplied by the absorption coefficient, because that would be the accurate way of calculating.

    But Billy, I said:
    Do your disc and sphere have identical properties? No. One has an isotropic albdeo that varies as a gradient, the other does not.

    In your very contrived case, your disc and sphere no longer have identical properties, where I specifically and explicitly stated:
    The difference being that one displays a variation in its albedo as a function of radial distance across its surface where the other does not.

    Now, if you disagree with this:
    I suggest you write your thesis, submit it for peer review, and collect your Nobel Prize.
     
  23. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I said they had ALL properties identical, as made of same material including thermal conductivities, and a variation of absorption coefficient with angle of incidence, quite like ALL real materials have. I.e. I. e. the absorption coefficient, a, is given by a = 1 - (A / 90) for A between 0 and 90 degrees is very reasonable (except for your assumed "Non-existium"). I.e.

    You assert that areas, being struck by the same cross sectional areas of a beam, absorb the same energy (and implying they scatter the sane fraction) is not true of any real material.

    Perhaps consideration of a sharply pointed (apex angle, of say 5 degrees) cone, made of black carbon, illuminated by two equal beams parallel to cone axis, one striking the flat bottom diameter end at 0 degree incidence and the other the cone surface at 85 degree angle of incidence everywhere it strikes, will help you understand that the variation of absorption is a strong function of angle of incidence, much like I postulated above, (and in my "question post") and CANNOT be ignored when speaking of the energy absorbed from say 1 square cm of the beam.
     

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