tisdag 21 november 2017

Talk at Climate Sense 2018

I have put up a preliminary version of my talk at Climate Sense 2018 for inspection.
Upplagd av kl.
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26 kommentarer:

  1. Dirk Gentley21 november 2017 kl. 18:12

    There seems to be some strange ideas about radiative transport in your slides.

    If two bodies exchange energy through radiative transport, you are describing the interaction of two open systems.

    When dealing with open systems, to have a reasonable expression of the first and second law of thermodynamics, the laws must necessarily be local.

    That is not the case of your expression

    \( Q = \sigma (T_2^4 - T_1^4) \),

    that is a factorisation in the limit that we are dealing with idealised materials. This factorisation can not be physical in a fundamental sense since it would violate the necessary condition of local thermodynamical laws.

    If we go back to the 19th and first half of the 20th century, the physics of nonequilibrium thermodynamics was not fully developed. They lacked the machinery back then to fully treat this situation. So, exactly what Planck thought on the subject is not that important, since he passed in 1947.

    SvaraRadera
  2. Claes Johnson22 november 2017 kl. 07:35

    Yes, Planck is important, in particular because his radiation law is misinterpreted as two-way heat transfer warm-to-cold and cold-to-warm leading to the misconception of massive "back radiation" from colder atmosphere to warmer Earth surface as the basic element of the "greenhouse effect". The version I present is idealized to two perfect black bodies but contains the basic element of one-way transfer warm-to-cold and as such serves to exhibit the fundamental misconception behind the hoax of then "greenhouse effect" as the mission of my talk.

    SvaraRadera
  3. Dirk Gentley22 november 2017 kl. 09:20

    But the idealised law must be consistent with a non-idealised situation where you have a \(\sigma_1(t)\ \ne \sigma_2(t)\) (time dependence from e.g. a first order phase transition).

    How do you factorise

    \(Q = \sigma_2(t)T^4_2 - \sigma_1(t)T^4_1\).

    SvaraRadera
  4. Claes Johnson22 november 2017 kl. 20:20

    An answer is given by the model analyzed at https://computationalblackbody.wordpress.com

    SvaraRadera
  5. Dirk Gentley22 november 2017 kl. 23:10

    I don't see how that model manage to get around the necessary requirement that the energy conservation must be local.

    You are treating a special case, where the bodies have homogeneous temperatures.

    The more general situation must be able to treat a spatial temperature variation in the bodies, and then, of course, there must be local energy conservation. This has the implication that the situation needs to be treated in the usual way when dealing with non-equilibrium thermodynamics.

    The systems needs to be separated into small enough subsystems, in which relaxation processes are fast enough so that the subsystem can be treated as being in thermal equilibrium.

    But here is the key, the interaction of a subsystem in one of the bodies, are the averaged interaction with all the other subsystem in which the subsystem is in radial contact with. In the general case all these subsystems have varying temperatures, in the transient towards thermal equilibrium, given that the material has a thermal conductivity.

    For that factorisation to be general

    \(Q = \sigma(T_2^4 - T_1^4)\)

    you have to assume that the energy interaction, fundamentally, is of a non-local nature.

    So, no that would not give an answer.

    SvaraRadera
  6. Dirk Gentley23 november 2017 kl. 11:12

    To clarify the point.

    You are proposing an artificial and idealised setting, to argue for constraints on the underlying physics. But the constraints proposed are not compliant with a more general setting, that is also the physical one (i.e. not an idealisation). Of course the physical situation has precedence.

    In general it would be ludicrous to throw away local conservation laws. And that, of course, includes local conservation of heat flux.

    SvaraRadera
  7. Claes Johnson23 november 2017 kl. 18:58

    Of course the model I study has local energy conservation. The focus is on one-way heat transfer by radiation, which is physical, as compared to two-way which is non-physical yet serves as the basic vehicle of the so called greenhouse gas effect.

    SvaraRadera
  8. Dirk Gentley24 november 2017 kl. 07:12

    There is a one-way heat transfer when writing

    \(Q = \sigma T^4_2 - \sigma T^4_1 \).

    I don't really know what a two-way heat transfer would be since that is impossible, by definition. I hope you understand why, it is right at the basics of thermodynamics, since thermodynamics concerns about averaged quantities, not the microscopic interactions.

    SvaraRadera
  9. Claes Johnson24 november 2017 kl. 17:54

    Look att the Kiehl-Trenberths energy budget, and you will see two-way radiative heat transfer.

    SvaraRadera
  10. Dirk Gentley25 november 2017 kl. 01:08

    I suppose that you are referring to this one?

    https://chriscolose.files.wordpress.com/2008/12/kiehl4.jpg

    Where do you see two-way transfer?

    If the whole earth is considered as a system, then there is a heat flux of \(0.9 W/m^2\) to the earth.

    If the ground level is considered, there is also a heat flux of \(0.9 W/m^2\) into the ground.

    It's a very simple consequence of the first law of thermodynamics. The first law of thermodynamics can be stated as

    \(\Delta U = Q - W \)

    where \(Delta U\ = E_{in} - E_{out}) is the change in internal energy. If this expression is differentiated and treated in the usual way one does with a control volume, the first law can be written (in the absence of work)

    \(\frac{\partial E_{in}}{\partial t} - \frac{\partial E_{out}}{\partial t} = \oint_S \vec{\phi_q} \cdot \vec{dS}\)

    where \(\vec{\phi_q}\) is the heat flux and \(S\) the surface to the control volume.

    SvaraRadera
  11. Claes Johnson25 november 2017 kl. 14:32

    Yes, there is 333 W/m2 "back radiation"' from cold atmophere to warm Earth surface.

    SvaraRadera
  12. Dirk Gentley25 november 2017 kl. 17:06

    Shoot, I messed up the closing of one of the LaTeX statements.
    Let's rewrite from the first law as I remember it.

    ...
    The first law of thermodynamics can be stated as

    \( \Delta U = Q - W \)


    Where \( \Delta U = E_{in} - E_{out} \) is the change in the internal energy. If this expression is differentiated and treated in the usual way one does with an open control volume, the first law can be written (in the absence of any work \( W \)) as

    \( \frac{\partial E_{in}}{\partial t} - \frac{\partial E_{out}}{\partial t} = \oint \vec{\phi}_q \cdot d\vec{S} \)

    where \( \vec{phi}_q \) is the heat flux and \( S \) the surface to the control volume.
    ...

    Notice that the heat flux \( \vec{\phi}_q \) is only defined on the boundary and is directly related to the change in internal energy. That is also why the heat flux always is one-directional by definition, the rate of change through a surface element can of course not be positive and negative at the same time.

    The 333 W/m\( ^2 \) that you mention is an energy flow, but not an heat flux since it isn't the change in internal energy of a system. The person who made the energy balance figure seems to be mindful of this discrepancy, as it says energy flows and not heat fluxes/flows.

    SvaraRadera
  13. Claes Johnson25 november 2017 kl. 19:01

    You seem to swallow the misleading confusion behind "back radiation". Heat is energy and thus heat flux is energy flow. If you don't see this, you have to rethink your argument, whatever it is.

    SvaraRadera
  14. Anonym26 november 2017 kl. 23:24

    What do you mean with unstable in the case of a two way transfer? What is it that becomes unstable?

    SvaraRadera
    Svar
    1. Claes Johnson29 november 2017 kl. 07:41

      This is like quickly running back and forth, which will make you fall, as compared to stably standing still.

      Radera
    2. Anonym29 november 2017 kl. 10:20

      I don't really follow.

      How do you estimate the time scales on which the system gets dizzy?

      And to be clear, over all, do you mean that the colder body doesn't radiate towards the hotter body?

      Radera
    3. Claes Johnson1 december 2017 kl. 13:30

      There is no transfer of heat energy from the cold to the warm body. No transfer.

      Radera
  15. DL27 november 2017 kl. 10:10

    But looking at the energy budget diagram, the 333 Wm-2 is accompanied with an upward 356 Wm-2. So the 333 Wm-2 can't possibly heat the ground. It must give a net positive in the ground direction for heating the ground.

    The surplus 0.9 Wm-2 seems to come from the 161 Wm-2 that is from a much hotter source, the sun :)

    SvaraRadera
    Svar
    1. P. Tuvnes30 november 2017 kl. 11:39

      Uncertainties, Errors In Radiative Forcing Estimates 10 – 100 Times Larger Than Entire Radiative Effect Of Increasing CO2 - See more at: http://notrickszone.com/2017/03/13/uncertainties-errors-in-radiative-forcing-estimates-10-100-times-larger-than-entire-radiative-effect-of-increasing-co2/#sthash.g7IZbDQT.dpuf
      Then the "surplus/imbalance" is completely incredible.

      Radera
  16. Günter Heß29 november 2017 kl. 20:11

    @Claes Johnson
    You seem not to understand that energy flux is not heat flux.
    This explains your confusion.
    Dirk gently is correct

    SvaraRadera
    Svar
    1. Claes Johnson1 december 2017 kl. 13:33

      What kind of energy flux are you then talking about measured to 333 W/m2 ?

      Radera
  17. P. Tuvnes3 december 2017 kl. 09:43

    Check the Planck curves for EM radiation. Low temperature objects has lower frequency than high temperature objects. Temperature and frequency are related. Low frequency/temperature radiation does simply not have a frequency range that can increase the frequency of a hotter body, and can thus not make it hotter. However, the cold object radiation will reduce the cooling rate of the hot object by interfering with the lower frequency radiation from the hot object. In a way reducing the effective radiation (cooling) from the hot object.

    SvaraRadera
    Svar
    1. Anonym5 december 2017 kl. 12:24

      You are missing that a black-body thermalize the radiation it absorbs. Internal conversion processes during this thermalization (e.g. phonon collision) can convert two lower energy states to a higher energy state, that is later relaxed at a higher frequency.

      Radera
    2. P. Tuvnes6 december 2017 kl. 07:47

      Anonym's assertion will imply that low frequency radiation from a low temperature object will make a hotter "blackbody" hotter still. Not credible.

      Radera
    3. Mike7 december 2017 kl. 00:21

      Petter Tuvnes, you wrote

      "Temperature and frequency are related."

      How is temperature and frequency related?

      Are they not both independent variables in Planck's radiation law?

      Radera
  18. P. Tuvnes9 december 2017 kl. 21:54

    A certain temperature has a maximum frequency, thus temperature and frequency are not completely independent. check the Planck curves.

    SvaraRadera

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