torsdag 27 oktober 2011

Two-Way Transfer of Heat as OLR/DLR Violates the 2nd Law


In a sequence of posts on radiative heat transfer between two bodies of different temperature, I have compared two views with deep historical roots:
  1. One-way transfer from hot body to cold body (Pictet)
  2. Two-way transfer with net transfer from hot to cold (Prevost),
where 2. is used to support CO2 alarmism in the form of DLR/backradiation.

1. satisfies the 2nd law of thermodynamics, but what about 2.?

Well, two-way transfer is commonly viewed as two opposite streams of photons, which are not considered to interfere with each other, and thus must viewed to be independent. But this means that one of these independent streams of photons concerns transfer from cold to hot and thus violates the 2nd law.

This is a simple argument showing that the mantra of DLR/backradiation lacks rationale, by violating the 2nd law.

The argument can be dressed up in more precise mathematical form, as shown in Computational Thermodynamics and Mathematical Physics of Blackbody Radiation.

10 kommentarer:

  1. I suggest to complete the table of options as follows:
    1) Sf > Si --> irreversible process
    2) Sf = Si --> reversible process
    3) Sf < Si --> when external work is done
    Michele

    SvaraRadera
  2. Let me start to thank you Claes. I really enjoy these kind of discussions where you have to think hard about the situation and try to understand what is happening and you are a big help in pushing this struggle forward for me.

    After some thought I come up with a suggestion of an experiment to see if there is one or two way radiation.

    Imagine two identical spheres with radius $r$. They are both considered to be blackbodies. They will have an outgoing radiation $q_i = A_i \sigma T_i^4$, $i\in\{1,2\}$ and they are regulated so that they will never lower their temperature below the start value. We place them in a large cavity with low temperature compared to the spheres.

    We imagine that they are separated by a distance $x$, that is far enough so that each one will look like a disk for the other one. Then we have a ratio $R = \frac{\pi r^2}{4\pi x^2}=\frac{r^2}{4x^2}$ that corresponds to the correction factor we need to use to see how much radiation falls from each one upon the other one.

    $1)$ If there is only one way radiation, there can be no radiation between the spheres and temperature remains.

    $2)$ If there is two way radiation then the absorbed energy will be re-emitted or the temperature will rise. Lets say that the spheres re-emit and then wait for a new equilibrium. We get a new outgoing radiation $q'_1 = q_1 + R q'_2$ for body one, and similar for body two $q'_2 = q_2 + R q'_1$. Due to symmetry, $q'_1 = q'_2$ and we can solve for $q'_1$.

    This gives $q'_1 = \frac{q_1}{1-R}$. Using Stefan-Boltzmann we see that $\frac{T'_1}{T_1}=(1-R)^{-1/4}$. The temperature will rise an amount governed by the geometrical factor, shorten $x$ and the temperature will rise even more. And as an extension one could try to find more suitable geometries, maybe parallel plates or something like that.

    Claes, is there something you disagree with?

    Sincerely,
    Dol

    SvaraRadera
  3. I don't think an absence of any temperature rise will convince any believer about the non-existence of two-way flow. Do you?

    SvaraRadera
  4. Lets try with a different syntax for poor Latex

    Let me start to thank you Claes. I really enjoy these kind of discussions where you have to think hard about the situation and try to understand what is happening and you are a big help in pushing this struggle forward for me.

    After some thought I come up with a suggestion of an experiment to see if there is one or two way radiation.

    Imagine two identical spheres with radius $r$. They are both considered to be blackbodies. They will have an outgoing radiation $q_i = A_i \sigma T_i^4$, $i\in\{1,2\}$ and they are regulated so that they will never lower their temperature below the start value. We place them in a large cavity with low temperature compared to the spheres.

    We imagine that they are separated by a distance $x$, that is far enough so that each one will look like a disk for the other one. Then we have a ratio $R = \frac{\pi r^2}{4\pi x^2}=\frac{r^2}{4x^2}$ that corresponds to the correction factor we need to use to see how much radiation falls from each one upon the other one.

    $1)$ If there is only one way radiation, there can be no radiation between the spheres and temperature remains.

    $2)$ If there is two way radiation then the absorbed energy will be re-emitted or the temperature will rise. Lets say that the spheres re-emit and then wait for a new equilibrium. We get a new outgoing radiation $q_{n,1} = q_1 + R q_{n,2}$ for body one, and similar for body two $q_{n,2} = q_2 + R q_{n,1}$. Due to symmetry, $q_{n,1} = q_{n,2}$ and we can solve for $q_{n,1}$.

    This gives $q_{n,1} = \frac{q_1}{1-R}$. Using Stefan-Boltzmann we see that $\frac{T_{n,1}}{T_1}=(1-R)^{-1/4}$. The temperature will rise an amount governed by the geometrical factor, shorten $x$ and the temperature will rise even more. And as an extension one could try to find more suitable geometries, maybe parallel plates or something like that.

    Claes, is there something you disagree with?

    Sincerely,
    Dol

    SvaraRadera
  5. Claes wrote:
    I don't think an absence of any temperature rise will convince any believer about the non-existence of two-way flow. Do you?

    I don't know if you can see the correct statements since Latex seem to have trouble with my primes. So I tried a re-post with different syntax.

    The important thing is that if we see a rise in temperature, it falsifies one way flow.

    Sincerely,
    Dol

    SvaraRadera
  6. There could not be any absence of temperature rise. Evidently, the rising temperature would not convince you that your theories are disastrously mistaken.

    SvaraRadera
  7. The radiation does travel both ways, Back radiation is real, HOWEVER, it is impossible for back radiation to heat the body it originally came from.

    That is why the GHE from back radiated heat violates the 2nd law.

    SvaraRadera
  8. DLR of course exists but it can´t heat the earth directly because it is reradiated. The greenhouse gases will on the other hand be warmed by the earth because they are not fully transparent and they can´t reradiate all heat they absorb from the earth. And the more greenhouse gases the warmer they get. And this will make the earth temperature rise acc to the SB law, if the input from sun is constant,

    SvaraRadera
  9. How does a body know where the radiation originally came from?

    SvaraRadera
  10. May be our thinking about the thermal power emitted by the earth surface has to be made clearer.
    The atmosphere behaves as an impedance for the alternating current because of the energy storage elements that it contains (the well-known GHG) which periodically reverse the direction of the energy flow.
    We have to take into account that: The portion of power flow that, averaged over a complete cycle of the AC waveform, results in net transfer of energy in one direction is known as real power (also referred to as active power). That portion of power flow due to stored energy, that returns to the source in each cycle, is known as reactive power. (http://en.wikipedia.org/wiki/Electric_power)
    See for example " the attached figure".
    We can say that the DLR exists only if we refer to the instantaneous values of the reactive power which have not effect upon the system surface-atmosphere because its high thermal inertia. So, the system above simply feels the mean value of the active power (the net transfer of energy in one direction) whereas the mean value of the reactive power is nil.
    Michele

    SvaraRadera