The case of two different fluxes!

Today I will be investigating the difference between two different types of fluxes that elude almost every climate scientist. Let’s take a typical simple conduction problem you can find in every high school or college textbook:

Figure 1: Conduction through a concrete block

We need to figure out the temperature and radiative emission on the colder side (Tc and Ec).

The formula for conduction is:

         Q         k * A * (Th-Tc)
Power = --- = q = -----------------
         t               L

We can rearrange this equation to suit our needs:

           q * L
Tc = Th - -------    (Equation I)
           A * k

And now we solve for Tc:

            2  *  8            16
Tc = 75 - ----------- = 75 - ------ = 75 - 25 = 50
           0.8 * 0.8          0.64

Tc is 50°C.

What is the conductive heat flux inside the concrete block? That is given simply by:

                        q      2 W
Conductive Heat Flux = --- = -------- = 2.5 W/m²
                        A     0.8 m²

And how much radiation is emitted from the right hand side (Ec)? To figure that out we have to apply Stefan-Boltzmann’s Law:

Cold-side Radiation = Ec = εσ(Tc)⁴   (Equation II)

First we need to know the emissivity of concrete. According to these experts, it is between 0.85 and 0.95, so we will use 0.9. Now we solve:

Ec = 0.9 * 0.00000005670367 * (273.16+50)^4 = 556.58 W/m²

As you can see we have two different flux (W/m²) figures: 2.5 and 557 W/m². We need appropriate labels for them so we don’t confuse them:

Conductive Heat Flux through the medium (CHF)2.5 W/m²
Cold-Side Radiation from the medium (CSR)557 W/m²

We can easily see that despite the small CHF through the medium, the emergent CSR is 557/2.5= 223 times larger.

Let’s now combine Equation I and II to summarize what is going on:

         /       q * L  \ 4
Ec = ε*σ*| Th - ------- |
         \       A * k  /

Remembering that CHF is just q/A, we reduce further:

CSR = εσ(Th-CHF*L/k)⁴

It becomes obvious now that CHF and CSR have an inverse relationship. The higher the CHF, the lower the CSR, and the lower the CHF, the higher the CSR.

Why is the distinction between CHF and CSR important?

Professors Davies and Davies have done a good job in measuring Earth’s heat flux:

We present a revised estimate of Earth’s surface heat flux that is based upon a heat flow data-set with 38 347 measurements, which is 55% more than used in previous estimates.

We conclude by discussing our preferred estimate of 47 TW, (rounded from 46.7 TW given that our error estimate is ± 2 TW)

It’s unfortunate that they call their measurement “surface heat flux”, when in reality they measure heat fluxes at various depth ranges which they don’t disclose, and then average that. They measure CHF. The value they get for averaged CHF is 46.7 TW divided by Earth’s surface area: 47×10¹² / 510.1×10¹² = 91.6 mW/m².

91.6 mW/m² is a very small number when compared to the average insolation we receive at the surface, which according to NASA’s official energy budget is 163.3 W/m². 0.0916/163.3 = 1783 times smaller. Perhaps you’ve heard of the idea that the sun supplies 99.95% of our energy? Well guess what? All of this is nonsense, because CHF is irrelevant!!! What is relevant is CSR!

What Davies & Davies should have done is measure all the averaged parameters (Th, k, L) and not just q or q/A (CHF). We need to know the actual CSR before we can start comparing it to insolation. This is key, because without CSR, climate scientists have a completely erroneous view of the way things really are.

Assuming k = 1 and A=1, we examine all the possible temperatures that produce Davies’ CHF (q/A) of 91.6 mW/m².

Depth = 110 metersDepth = 10 meters°C/m
10°C0.84°C-0.0916
100°C90.84°C-0.0916
1000°C990.84°C-0.0916
10000°C9990.84°C-0.0916
All result in Conductive Heat Flux of 91.6 mW/m²

Let’s think about this: Does it matter whether it’s 0.84°C or 9990.84°C 10 meters below your feet? Of course it does! But you can’t tell the difference using CHF. Only using CSR can we tell the true radiation emerging out of the earth!

We will now transform the above table into CSR, using emissivity = 0.93643 (why this number?).

Depth = 110 metersDepth = 0 metersCSR
10°C-0.076°C295.31 W/m²
100°C89.924°C922.81 W/m²
1000°C989.924°C135,150 W/m²
10000°C9989.924°C589,112,205 W/m²

Makes a big difference, right? It would be interesting to know the actual Earth-wide averaged CSR, but we will leave that for another day. (Update: That day came: Measuring Geothermal, CSR = 294 W/m²)

In short summary, most climate scientists are clueless about the difference between CHF and CSR and therefore erroneously greatly underestimate the power of geothermal. It may even be the biggest scientific scandal of our time!

Yours Truly, -Zoe

Update

Go on to read Measuring Geothermal (1) and Measuring Geothermal (2).

98 thoughts on “The case of two different fluxes!

    1. The atmosphere is not a ball or meteor that falls to the ground. Can’t you see that the atmosphere is aloft and not falling? If it did fall, you would have a one time heat up and its dissipation via radiation. One time!

      Liked by 1 person

  1. You cannot create energy in this experiment. You completely forget about conduction into the air at each end. Assuming you are adding new energy by radiation at the left end and assuming it raises the temperature of the left end above the ambient temperature then there will be some conduction back to the air and so Stefan-Boltzmann calculations do not give the equilibrium temperature achieved at the left end by the radiation because that surface does not have black-body or grey body characteristics as it loses some energy by other than radiation. The same applies for the right hand end which cools partly by conduction to the air and partly by radiation. So all you calculations are wrong.

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    1. I presented a simple textbook problem. If the conductive material is inside a vacuum tube, then there is no conduction or convection – only radiation. But even without a vacuum tube, conduction is tiny and even on all sides, and therefore doesn’t discredit the gradient that will form.

      Liked by 1 person

  2. Let me start by saying I don’t have this level of education in physics or math but I do have a skepticism of any science that is being used by people and organizations that aren’t scientists to push agendas about social outcomes.
    A lot of the predictions (all of them?) that have come from the IPCC have turned out to be wrong so the modelling they use has to be called into question.
    For this reason, Zoe, I applaud your efforts to bring to the public sphere this type of granularity concerning the way climate scientists actually use the physics to produce their findings.
    Personally I find it an affront when I’m continuously told “the science is settled” and to shut up because I’m not just parroting what the spokespeople are saying from organizations like the IPCC.
    Keep up the good work, Zoe. I like seeing the equations and the methodology explained. It’s something that’s lacking in most of the other sites I come across. They tend to get bogged down in arguing about historical factors/readings (of various core samples, tree rings, etc) and their significance in the interplay/influence with the advent of ice ages, warming periods, and so forth.
    Cheers.

    Liked by 2 people

  3. Hey Zoe, I came to your site from this thread at Principia Scientific https://principia-scientific.org/the-circular-reasoning-greenhouse-gas-theory-argument/#comments. I got a sense you had something important to offer but you weren’t getting much respect in this thread. Most of the discussion went way over my head – I’m not competent at math/science/physics (high school dropout) but I’m very curious to find out what is true or not about climate change… I agree that there would be geothermal energy emitting from the earth. It’s a living being – stuff grows out of it. I was reminded of this when in a packed train yesterday on a hot day – a man with a beard was shoved up next to me and I could feel his body emitting heat. So this intuitively makes sense. But how can a layperson understand what you’re saying from a conceptual level? Is it even possible?… If my understanding is correct, Postma and many of the people you were in dialogue with in the said thread were claiming there is NO heat emissions from the earth? Btw, I’ve consumed many of Postma’s videos so I’m broadly aware of his work. But I haven’t quite grasped what you say he’s missing… thanx, Nicky

    Liked by 1 person

    1. Thank you.

      “Postma and many of the people you were in dialogue with in the said thread were claiming there is NO heat emissions from the earth.”

      If only they read and understood this article…

      Postma has blocked me from any possible way to contact him. He used to like me, then he got jealous or something.

      Postma thinks he found a great paradigm, but he can’t even explain why it’s ~15C at the surface.He really can’t. Well, you saw my arguments.

      My argument in this post boils down to this:

      Planck’s and Stefan-Boltzmann’s Law are derived from radiation oven experiments where the oven was at near perfect equilbrium, i.e. the conductive heat flux was nearly 0 W/m^2, yet nearly 0 W/m^2 is NOT what emerges.

      The best analogy I can think of is spending. You spend $80/day every day. How much money can you take out of the bank every day?

      Postma and most climate scientists will say $80/day. I say more information is needed. That’s the difference.

      Liked by 2 people

      1. Thanks for the reply. I’ve re-read this post a few times in an attempt to absorb more. Though I’m not sure if I’m over hurdle one yet… Is the example experiment meant as an analogy to illustrate heat coming FROM the earths core? And emitting to the surface?… Or am I stuck already?… Feel free to link me to relevant posts to catch up.

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      2. Zoe,

        I have only just discovered your blog. It’s great. I am a retired electronics engineer. Your calculation of the temperature of the cold end is absolutely right. Every electrical engineer uses this calculation many times during their career. I am looking forward to reading the rest of your posts.

        Liked by 1 person

        1. Great. Thats what we need in climate science. People with electrical engineering experience. There is much more electrical stuff going on than the mainstream wants to acknowledge.

          There is a fellow calling himself WXCycles at Joannes place. To me he’s the gold standard in observational reportage of what is going on between the lower stratosphere and the upper troposphere and the consequences. Well the thing is he proves day after day my prejudices of the way electricity moves through the atmosphere, on its way to the deep earth. But he doesn’t see it. So I started calling him Epicycles. Despite that blind spot I would encourage everyone to follow his work.

          I won’t be on Joannes any more, but earlier I was leaving comments showing the implications of his observations. Thats if you wanted to confirm to yourself that his observations prove the movement of electrical energy.

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        2. I understand electrostatic discharge, induction, etc. It may even be possible that what I call geothermal, is just capacitor disharge. I just don’t have a source of data I can mine that will reaffirm this. Gravity may even be electrostatic cling. I don’t know. And because I don’t know I, I’m limited in what I can say with confidence.

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  4. Thank you for your reply, Zoe. I think I’m comprehending what you’re saying; that the CHF and CSR inverse relationship suggests more heat is being emitted from the surface (via the core) into the atmosphere than is estimated by main stream science (or not even considered?), and the sun is emitting less heat than is estimated because of the suns low CHF towards the surface?… or something like that?

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    1. You’re very welcome 🙂

      Solar and Geothermal need to be compared apples to apples. Either CHF to CHF, or CSR to CSR. When scientists compare Earth CHF to Solar CSR, they are committing a fallacy.

      One wonders whether they’re conscious of what they are doing. By focusing on Earth CHF, they ignore Earth CSR, which gives room for the fallacious Greenhouse Effect.

      It is my contention that the so called Greenhouse Effect is nothing but Geothermal CSR flipped upside down.

      They flipped what geothermal does into what they claim greenhouse gases do.

      CHF is based on a vertical differential.
      CSR is an absolute.

      The current state of climate science is completely disconnected from reality. Their assumptions are fallacious assertions.

      Liked by 1 person

  5. Okay I think I get what you’re saying. With regard to this following statement…

    “Solar and Geothermal need to be compared apples to apples. Either CHF to CHF, or CSR to CSR. When scientists compare Earth CHF to Solar CSR, they are committing a fallacy.”

    Are you saying if scientists compared the suns and earths CHF they should get the same surface temperature as they would if they compared the suns and earths CSR?

    Liked by 1 person

    1. CHF is based on a vertical differential. It is the steepness of a thermal gradient.

      CSR is an absolute. Best to use that.

      If you took two ice cubes (at 0 celcius) and pressed them together: They will emit 315 W/m^2 from all sides (assuming they were in a vacuum). Because they are the same temperature, CHF between them is 0 W/m^2. CSR is 315 W/m^2.

      Now imagine you expose 165 W/m^2 of solar energy on to them.

      Consensus science would add 165 to 0 and get 165 W/m^2

      Rational science would add 165 to 315 and get 480 W/m^2.

      Notice that in the article, there are many temperatures that produce the same CHF.

      You could be melting in hot lava, and consensus scientists would be saying that can’t happen because the temperature difference between the top of the lava and a meter below is very little. At least they would say that if they thought CHF was relevant to the situation. They don’t do that for someone drowning in lava, but for the climate that’s exactly what they do.

      Think of it this way: if the CHF between earth’s core and your feet was 0, it would mean you’re standing on >5000 degree ground.

      CHF is a differential, and CSR is an absolute.

      I hope that helps.

      Liked by 1 person

      1. So perhaps think of CHF as the heat lost through a length of resistor, and Cold Side Radiation CSR as the voltage and current coming out of the other side, ready and available to do heat work in the atmosphere?

        Btw – your insight is fantastic.

        Did you study in Russia at all? Last time that I came across such logical clarity was in Russia.

        Liked by 1 person

        1. Yes, it’s so simple a climate scientist could understand it, IF they wanted to. But they prefer to think that’s it’s GHGs that are “raining down” all this energy from their tiny mass in the atmosphere. We live in a crazy clown world.

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  6. This is an interesting way to think on it, intuitive and rational. When I make any attempt to describe the idea to, for example – a farmer – the response is often something like “hmm, well the earth is supposed to be hot inside, I wonder …” but the more educated, (i.e schooled), the person is, the more resistant or argumentative they are, whichever pill they’ve chosen to swallow on the warming issue. Maybe it’s as simple as: people who heat their homes with wood-stoves are just more intelligent about how these things really function.

    Anyways, thanks for the arithmetics and coding tips, and please just keep on shining that light into the fog.

    Liked by 1 person

    1. Thank you very much.

      It is indeed simple to understand.

      If you ask a climate scientist how much radiation a frying pan emits, he will not use conduction physics to give a small answer. He will use the correct physics. But when it comes to climate, by analogy, he’ll be measuring the thickness of the pan, the temperature difference between top and bottom, and the thermal conductivity (k) factor. He’ll get the wrong answer, then claim you can’t cook eggs without emission from GHGs.

      It’s just stupid and sad. They are indoctrinated fools. It’s really difficult for some people to accept that the experts are this grossly misinformed about the basics. Even I couldn’t believe it at first. I also couldn’t believe that I personally uncovered this huge scandal.

      We are being led by the blind.

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  7. Hi Zoe
    What you are doing here is great and you seem to be attracting some people who can properly check your work. Please email me privately. Let’s talk and work at this together.
    Rosie

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  8. I don’t understand what your figures are in the following equation:

    Ec = 0.9 * 0.00000005670367 * (273.16+50)^4 = 556.58 W/m²

    if the equation is supposed to be filled-in values for the following equation:

    Cold-side Radiation = Ec = εσ(Tc)⁴ (Equation II)

    How does Tc become (273.16+50) ? — if Tc is 50?

    And why, if I’m seeing this correctly, would you use Celsius in the Stefan Boltzmann equation, when that equation requires the use of Kelvin?

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      1. It would have been much more clear, if you had indicated this, because, as is, it looks like that number comes out of nowhere, or that you are trying to add kelvin and celsius of two different temperatures. But, okay, I see what you did there, but now for the bigger issue:

        I don’t know what the heck you are doing in those two tables. What is all that? Sorry, if I seem dense, but it really is not clear what you are doing there.

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        1. I’m showing the proper relationship between CHF and CSR. CSR is relevant to energy budget, and CHF is not.

          Boltzmann and Planck had a CHF of ZERO, and what’s emitted on the far side is not ZERO.

          Using CHF to claim geothermal is tiny, is as stupid as claiming that the sun is irrelevant because its CHF is tiny. One must compare apples to apples, and not oranges to apple seeds.

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  9. Using CHF to claim geothermal is tiny, is as stupid as claiming that the sun is irrelevant because its CHF is tiny. One must compare apples to apples, and not oranges to apple seeds.

    Why would anybody even claim that the CHF from the sun is tiny, when the concept is not even applicable to any significant level? For all practical purposes, the CHF from sun to Earth does not exist, whereas with Earth-core/Earth-surface, it DOES. CHF is irrelevant for the sun, whereas for Earth, it is not.

    Why are you comparing the nonexistence of a concept for one system to an existent concept for another system, to claim that the comparison of the two systems is stupid? You set up an illogical premise in order to set up a stupid comparison. That’s comparing apples to elves. (^_^)

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    1. What do you mean, Robert?
      Are you saying the hot sun does not produce a conductive heat flux in the atmosphere? That hot sun?

      Are you saying the sun does not have a CHF within itself? Have you bothered to check the CHF within the sun itself to see how small it is?

      Once you see how small the CHF is within the sun will you then say the sun can’t heat anything? HYPOCRITE

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      1. No, what I am saying is that, in the space between the sun and Earth, there is NO MEDIUM to even speak of conduction. The space between the sun and Earth cannot conduct heat.

        Flux is a radiation calculation determined over an astronomical distance where no conducting medium exists for heat transfer. This was the context within which I was addressing your analogy.

        Let’s step back a second from your mistaken assessment of my hypocrisy (^_^) to look your Figure 1 again.

        In your Figure 1, you focus on Cold-Side-Radiation (CSR or Ec) and Conductive-Heat-Flux (CHF), but you do not give any attention to Hot-Side-Radiation that results from the 75C temperature on the other side.

        You use the Stefan Boltzmann equation to figure Cold-Side Radiation (Ec), as follows: Ec = εσ(Tc)⁴

        Can we not also figure Hot-Side-Radiation (Eh) similarly? — Eh = εσ(Tc)⁴

        We convert 75C to Kelvin (using your conversion figure of 273.16), which gives us 75 + 273.16 = 348.16K

        To be clear, 75C = 348.16K

        And, using the Stefan Boltzmann equation as you did to find Cold-Side-Power, we can do the same to find Hot-Side-Power, which I will denote (consistent with your notation) as Eh, which gives us:

        Eh = 0.9 * 0.00000005670367 * 348.16^4 = 748 W/m^2

        So, now we could create a table showing:

        Hot-side radiation to the medium ………………… 748 W/m^2
        Conductive heat flux through the medium …….. 2.5 W/m^2
        Cold-side radiation from the medium …………… 557 W/m^2

        For your table showing only Conductive Heat Flux and Cold-Side-Radiation, you noted, “We can easily see that despite the small CHF (Conductive Heat Flux) through the medium, the emergent CSR (Cold Side Radiation) is 557/2.5= 223 times larger.”

        … to which I could add:

        “We can also easily see that despite the small CHF (Conductive Heat Flux) through the medium, the entering HSR (Hot Side Radiation) is 748/2.5 = 299 times larger.”

        By not focusing on the hot side of the conducting medium in terms of flux, you consider only the Conductive Heat Flux and Cold Side Radiation in terms of W/m^2, which gives the first impression that you are pointing out something remarkable. This is what was throwing me, I think, because there is nothing really that remarkable about comparing a HEAT-TRANSFER flux to a standing flux — because they are two different things. One is a TRANSFER of power, while the other is a sustained power caused by the greater power on the other side feeding it via this transfer.

        We now see an entering flux of 748 W/m^2 TRANSFERRING power at a rate of 2.5 W/m^2 to sustain an emerging flux of 557 W/m^2 on the other side of the conducting medium. That 557 W/m^2 is NOT a heat TRANSFER flux. Rather, it is a sustained flux that the heat TRANSFER flux ENABLES from the GREATER sustained flux of 748 W/m^2 on the other side.

        In different words, a temperature of 75C on one side of the conducting medium sustains an ENTERING power of 748 W/m^2, which power transfers THROUGH the medium at a rate of 2.5 W/m^2 (2.5 joules per second per meter squared) to produce a temperature of 50 C on the other side that sustains an EMERGENT power of 557 W/m^2.

        How this 557 W/m^2 flux might transfer (or not) its own power to another medium depends on what medium it goes to next and what that medium’s temperature might be (colder? … hotter?), which is unspecified in the example.

        I do not see how all this makes geothermal heat transfer any more of a factor in what keeps Earth’s temperature in a habitable range. Again, sun power dwarfs it many times over.

        DISCLAIMER:Any hypocrisy is unintentional, at best, and just a misunderstanding at worst.

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        1. Robert,

          “No, what I am saying is that, in the space between the sun and Earth, there is NO MEDIUM to even speak of conduction.”

          Earth’s atmosphere is not between sun and earth?

          Does not the sun ray’s hit Earth’s surface and thereby create a CHF in Earth’s atmosphere?

          Does not the Sun have a CHF within it just before surface EM emission?

          “We now see an entering flux of 748 W/m^2 TRANSFERRING power at a rate of 2.5 W/m^2 to sustain an emerging flux of 557 W/m^2 on the other side of the conducting medium. That 557 W/m^2 is NOT a heat TRANSFER flux. Rather, it is a sustained flux that the heat TRANSFER flux ENABLES from the GREATER sustained flux of 748 W/m^2 on the other side.”

          Yes, kind of! I think you’re finally starting to get it!

          The CHF is a small tiny value that represents flux lost per distance times material conductivity (k).

          You can take that metal bar and stick it into the Earth vertically, and I think you will finally understand it (after some value changes). You will understand that Earth is more than capable of sustained 0C at the surface.

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  10. Zoe,
    Does this mean that the Bond Albedo for the Earth is incorrect? The calculation for Bond Albedo assumes that the outgoing radiation is reflected solar irradiance however your theory suggests that the calculation might be ignoring a significant extra contribution from the Earth itself. This would presumably lower the Bond Albedo (less solar radiation reflected but the total is still 0.304 of incoming as the Earth’s output is added in.)

    Liked by 1 person

    1. My impression is that Bond Albedo is limited by optical depth from space.

      If you look at the “official” energy budget, the surface only receives from the sun 163.3 of 340.3 W/m2. So the “albedo”, or whatever you would call this, must be 0.52, not 0.304.

      Your thoughts on this?

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  11. Zoe, you start from a false declaration of your exercise supposedly found in any text book. ( You do not support that claim either ). Maybe you need to check the details in your text book.

    You assert the rate of heat flow in the block as a given ( 2 watts ) and derive Tc. So far so good. You then make the illegitimate jump to assuming that this is consistent with the tiny surface area emitting the heat flow necessary to assure this level of heat flow. Nothing of the sort follows. That bit was not in your text book, you made that up yourself.

    If you chose a different arbitrary heat flow of say 1W , you get a totally different answer. This should tell you you have a problem.

    If you want to impose an arbitrary heat flow, you can do so as long as you don’t make any assumptions about what boundary conditions are on the “cold end”.

    Alternatively you can specify how the heat is extracted from the cold end ( for example by Plank radiation into a void at absolute zero ) and work backwards to find out what the heat flow would be in the block of concrete in this situation. That would give physically consistent results.

    One of the most fundamental axioms of physics is the conservation of energy. If you have one heat flux on one side of a boundary and a different heat flux on the other side you have a violation of conservation. That is the second indication you have a made a mistake.

    You are little too sure of yourself and you as yet limited knowledge of physics. You should be a little more humble before trying to “shame” people who know a bit more than you do.

    Have fun learning.

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    1. “If you chose a different arbitrary heat flow of say 1W , you get a totally different answer. This should tell you you have a problem.”

      From my research, I found concrete has a k value of anywhere between 0.5 and 1.1, so I chose 0.8. This can’t change.

      Yes, I suppose I could have changed the dimensions to suit you, but then you’d complain to change it back to what I have now.

      Is your argument that CSR = CHF?

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      1. Thanks for the reply. I am not questioning the value of k you used, nor your dimensions. It’s a little strange that you make such a claim, maybe you should start by quoting what I wrote that claimed that.

        What I was referring to is where do you get the 2 watts? That is an arbitrary choice, not research.

        You do not address the key issue that you assume, without justification, that this value is consistent with Planck radiation being the means of evacuation of the heat flow from the cold end. There is nothing in such text book example calculations which specifies what the heat sink at that end is. Again you need to check hypothetical “text book” or maybe link to such an exercise to justify the claim. We can then look at what is specified at that end.

        It seems that because it is not specified you are incorrectly assuming it is empty space and thus Planck radiation is what is happening. That is fundamentally where you are going wrong. It is simply not defined. In order to freely chose the heat flow as such an exercise may do, you have to leave that undefined, as I explained.

        My argument is that conservation of energy must be maintained across the boundary of the surface at the right hand side : your “cold end”.

        You seem to think that you can have “two different fluxes”. I am pointing out that if the power flux on either side of that boundary is different you have a violation of the the conservation of energy at that boundary. If you think that does not apply, you need to say why not. Choosing to ignore conservation is not an option.

        I hope you will pay a little more attention this time and come up with a more coherent reply.

        Liked by 1 person

        1. “You seem to think that you can have “two different fluxes”. I am pointing out that if the power flux on either side of that boundary is different you have a violation of the the conservation of energy at that boundary.”

          Again, is your argument that CSR = CHF? YES or NO

          “If you think that does not apply, you need to say why not.”

          Total energy = Translational Energy (Conduction) + Vibrational Energy (EM Radiation potential)

          Equipartition Theorem

          You don’t conserve subtypes of energy, only total energy.

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        2. “You don’t conserve subtypes of energy, only total energy” . Agreed.

          Nothing is moving here apart from molecular vibration, so there is no kinetic energy other than thermal. Equipartition between the different degrees of freedom seems to be a total red herring, it’s all thermal and accounted for.

          So where is the other energy you think is being ignored to balance the difference between the heat flux in and the radiated flux out of that boundary, if you consider they can be different?

          Your example says heat flow is steady ( in equilibrium ) and one dimensional, so the temperature profile of the bar is constant : no parts are getting warmer or cooler, no net change in the heat content of bar of concrete and no losses in the other two dimensions.

          We both agree that total energy has to be conserved, so where is the difference between 557 and 2.5 W/m^2 going?

          Yes or No ?

          I’m open to the idea they may be different if you can account for the discrepancy and not violate conservation of energy.

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        3. Greg,
          If Area (A) is in dimension (x,y), then
          L and K are in the z-dimension. You see it?

          All you see is two things using W/m^2 and you think they must be equal because two subtypes of energt must have their heat flows conserved, for some reason.

          “Nothing is moving here apart from molecular vibration, so there is no kinetic energy other than thermal. Equipartition between the different degrees of freedom seems to be a total red herring, it’s all thermal and accounted for.”

          I read this as gibberish, no offense.

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        4. Greg,
          Set your stove for 400F. Place a frying pan on top. After 30 minutes, what flux are you getting out of the top of the frying pan? (don’t over complicated this with pedantics)

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        5. “L and K are in the z-dimension. You see it?”

          If K is what you previously called k , it is an omni-directional property of the material and does not have a direction, but L yes. Again you are trying to pretend that I’m questioning things that I have not even referred to. There is no disagreement about direction, you are just trying cloud the issue.

          All you see is two things using W/m^2 and you think they must be equal because two subtypes of energt must have their heat flows conserved, for some reason.

          What are these “subtypes” you keep bringing up? I’ve asked you to explain what other energy you think is being ignored and you ignore the question. It seems you do not know, maybe there isn’t one !

          I read this as gibberish, no offense.

          I’m just trying to make sense of your pseudo scientific gobbledygook , like the subtypes and the irrelevant throwing in of “Equipartition Theorem” without making any clear point. Looks like you are just trying hide behind a few scarey sciencey terms in the hope you will look more learned, instead of making any coherent point. No offense.


          We both agree that total energy has to be conserved, so where is the difference between 557 and 2.5 W/m^2 going?

          With your frying pan ( ignoring conducted and convective losses via the air), at equilibrium the heat input from the flame will equate to the radiated EM flux from the pan. Fluxes are equal energy is conserved. Or maybe you think it is giving off 200 times more EM energy than the flame is producing ?

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        6. Now Greg, please tell me the conductive heat flux through your frying pan. And please don’t tell me it’s the same as the emergent radiation. Do use the conduction formula. No excuses for not having the k and L values.

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        7. “If K is what you previously called k , it is an omni-directional property of the material and does not have a direction.”

          Really? Because k is in units W / (m * K). What direction do you think m is in?

          Omni directional would imply a cubic m, No, genius?

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        8. Look , I’m not interested in diverting this discussion into a juvenile and imprecise frying pan analogy.

          Please address the fundamental question it seems you are so intent on ignoring:
          We both agree that total energy has to be conserved, so where is the difference between 557 and 2.5 W/m^2 going to / coming from?

          If you cannot align your ideas with the fundamentals of science you need to just give up , stop insulting others and go back to school.

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        9. I don’t have a “theory” to test. I’m not the one trying to rewrite the fundamental laws of physics.

          Now you have refused to answer that question for the fourth time, I hear your embarrassed silence LOUD AND CLEAR.

          Your “fans” , whoever they are, may fall for this but no one over the age of 20 is likely to buy it.

          Now I don’t want to defend climate scientists, they are by 97% a bunch of lying, disingenuous grant seeking activists, so full of hubris they think their meagre science skills mean they are qualified to run the planet. They may even under estimate geothermal , though not for the reasons you suggest.

          However that is not ” the biggest scientific scandal of our time” , that accolade without doubt belongs to the CO2 scam.

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        10. I’ll help you out. You previously said:

          “at equilibrium the heat input from the flame will equate to the radiated EM flux from the pan. Fluxes are equal energy is conserved.”

          OK, so here you seem to suggest that input and output fluxes are equal, so the top and bottom temperatures must also be equal.

          Now what is the conduction formula?

          Cond = -k (Ttop – Tbottom) / (pan height)

          Since Ttop = Tbottom, Cond = 0

          Now what is CHF?
          CHF = Q/(A*dt) = k (Tbottom – Ttop) / (pan height)

          CHF is also 0.

          Thank you for playing 🙂

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        11. You are playing all by yourself. If you are not able to apply science to your initial example nor able to address your obvious logical blunders even when they are repeatedly pointed out to you , I’m not surprised that you make an equal mess of your frying pan analogy.

          You are displaying the same mentality as the average flat-earther. You refuse to engage in logical discussion or apply basic science, yet you are convinced you have made some major scientific “discovery” that the rest of the world is unable to see. This is also the problem with discussing climate with your generation.

          You know nothing but arrogantly think you know it all and everyone else is stupid. This is akin to thinking that the rest of the world is mad. At some stage you need to realise that it is not the rest of the world which has a problem.

          Have fun growing up. The arrogance of youth will ( eventually ) be tempered by experience and you will ( hopefully ) begin to acquire a little wisdom and humility.

          BTW if you can get 557 W/m^2 radiated from 2W/m^2 thermal input you have solved the worlds energy needs. I suggest you apply for a patent immediately. You will be rich beyond your dreams.

          Good luck.

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  12. PS the expression εσT⁴ is the total radiation of a body in all directions into space , ie integrated over 4 pi steradians ( an enclosing sphere ) not a small planar surface emitting in a limited direction.

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  13. @Greg: “Please address the fundamental question it seems you are so intent on ignoring: We both agree that total energy has to be conserved, so where is the difference between 557 and 2.5 W/m^2 going to / coming from?”

    The amount of energy from the CSR (= 557 w / m2) ultimately comes from the temperature of 75C (> 557). The energy flow in the material – the CHF – actually has no independent energy value.

    The only thing the heat flow does is determine the decrease in temperature on the other side (CSR temperature). As the source, the HSR is the only independent energy carrier.

    Every comparison, between CHF and HSR / CSR – under the law of conservation of energy – is like comparing apples with pears.

    Liked by 1 person

    1. Jack, you are falling for the trap which Zoe set for herself in pretending that the mythical “text book” example, which she made up herself, is physically meaningful.

      Her first diagram could be from a text book except that she added the ” E=? “. There is no text book on such a problem which makes any linkage to the false assumption that the heat sink on the “cold end” S-B radiation. She refuses to even deal with that once it is pointed out.

      If she won’t deal with that, it stops being a scientific discussion and she joins the flat-earther mentality.

      The “hot end” has a given heat source which is assumed to supply ( by whatever unspecified means ) enough heat energy to maintain the given temperature; and no more ! It’s a given.

      Similarly at the “cold end”, there is a heat sink which is removing the arbitrary given heat energy flux flowing through the test block. There is no assumption about how that is happening, it is just one of the given conditions of the exercise. Recall that the heat flow is purely arbitrary and is pulled from the air, it is not derived from anywhere, it could equally be given as 2GW or as 2mW.

      The whole of this ridiculous waste of time is because she spuriously and incorrectly assumes that because nothing is stated about the physical conditions on the cold end she can assume it is radiation to space which is evacuating all the heat. That is the fundamental error which leads to the astounding “revelation” that you can create 557 W/m^2 from 2W/m^2. It’s baloney. Sadly. ( I’d love to have some of that if it worked !) .

      Ironically, in the other thread she states another world changing reinventions of physics that “nothing radiates to space”, radiation can only land on another body. In other words an IR photon has to know in advance of being emitted , what path it will take and where it will end up and ensure is has the necessary tourist visas when it arrives. I won’t list how many scientific principals that stupidity defies.

      As Zoe and I agreed, total energy has to be conserved. That’s like having a law of conservation of the number of fruit when comparing apples and oranges. Like her, you want to have “different” sub-energies which are not conserved.

      Her calculation of Tc is fine and the calculation of Ec for a temperature at that body is fine. The logical error is in thinking that two calculations describe the same physical set up. They don’t. The fact that Ec is vastly greater means that it is NOT the means by which heat is leaving that end when you have a 2W heat flux. It would require some material with a finite thermal resistance conducting heat away more slowly or a highly polished end which radiates much less energy.

      The whole mess is a false problem created by a false assumption which she is unable and unwilling to address.

      Regrettably this will not be the killer argument which kicks the stool from beneath fat butts of deceitful climatologists and solves the whole climate question with another single cause. Climate is immensely complex and does not depend , even “mostly” on any one cause or variable.

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        1. I said I was not interested in your ridiculous frying pan analogy. Since you can not even follow the basics on an idealised rectangular block you need to follow your own advice before “working your way up” to frying pans. You made more of a mess of that than your initial example.

          I see no point in explaining that one to you as well since as soon as your are stuck you will just ignore questions and divert to another issue again.

          You really are rather ridiculous, but that is your birth right to be so. Far be it from me to spoil your fun.

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        2. What frying pan “analogy”? There is no analogy.
          If you can’t tell me the CHF through the frying pan, it looks like you’re a coward that likes to insult people while pretending to teach them. Are you a low life, or what?

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      1. With all due respect for your opinion. I don’t agree with you, unfortunately only a handful of people are able to understand this. The majority is by your side.

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        1. Everyone is entitled to their opinion. You are not entitled to your own version of physics. There are millions of people across the world who understand this kind of basic high school physics. There is only a handful of people who are NOT able to understand this and think you can make 557 W from 2.5W.

          I’m sorry I was not able to explain it to you in a way you could understand.

          There will always be a tiny clique who think they have invented perpetual motion, free energy or proof that the earth is flat. This makes them very special and they will fight tooth and nail to remain special. This is a psychological need which renders them immune to logic and reason.

          I can do nothing for you. I bid you farewell and leave you to your own magic corner of the universe where the laws of nature do not apply.

          You all seem to be what H2G2’s Ford Prefect would describe as “mostly harmless”.

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        2. “Everyone is entitled to their opinion. You are not entitled to your own version of physics.”

          Greg I’m supporting you on one point only. But what you are saying here is utter bullshit. Because modern physics is a very obvious psychological operation. Its virtually a religion unto itself with its own demi-gods, who in sober reality weren’t even all that bright. Most of the alleged laws of physics don’t pan out. Modern cosmology has a creation myth, more irrational than all others, even those involving tortoises …. they adopted an idiotic view of stars without seeming to realise that the lightest substances naturally float to the top.

          So take a few potshots at Zoe where you think she has something not quite right. But don’t be sticking up for twentieth century physics. It was a gigantic fraud from tip to stern. Only the stuff that we have lab evidence for is settled science. And sometimes not even then. If your gurus can jump to the conclusion that stars are made of whatever floats to the top, then consider what other idiocy they are locking in as gospel?

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  14. @greg: I cannot change who I am and I cannot change what is very clear to me. Again, even if I am completely wrong, grant me my own vision, after all, I grant you yours too. We call this respect.

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    1. @Jack: I’m not suggesting you change “who you are” but you can do something about what seems very clear to you but may not necessarily be correct.

      There is little of this basic level of physics which is a matter of personal opinion. It’s the sort of thing which can be established by logical reasoned discussion.

      Sadly some refuse to do this out of pride or belligerence, some others are truly not capable of following such a discussion. The latter should defer to those who can. Though on the subject in hand, that is really not that hard and should be within the grasp on most people who are prepared to spend some time reading and informing themselves.

      There is no disrespect in pointing out that you are not allowed your own version of physics. The only way you get that right is to come up with something equally coherent which works as well or better than the current paradigm.

      We are in dire need of such a saviour, since modern physics has got it self into a blind alley. However, that does not include problems like basic thermal conduction.

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  15. @Zoe: Could you give us a short impression about your thoughts regarding the radiation direction of photons. I am particularly interested in the existence of only one-way radiation instead of the generally accepted two-way traffic.

    This is because the current main team vision is based on two-way radiation BUT where heat always flows in one direction. In the context of the energy balance, the one or two-way approach therefore makes no difference. Nevertheless, I would like to gain a deeper understanding on this subject.

    If emissions are always only one-way traffic – hot to cold – then that also invalidates the entire concept of back-irradiation. What is your opinion about this? And .. what if there is no cold object?

    Could you also post a copy of your response on this climategate.nl? See the link below.

    https://www.climategate.nl/2020/01/wetenschap-2/comment-page-9/#comments

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    1. “@Zoe: Could you give us a short impression about your thoughts regarding the radiation direction of photons.”

      Jack you ought to not be thinking of photons. A volley of particles is not compatible with a series of waves.

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      1. ” instead of the generally accepted two-way traffic.”
        That is NOT the generally accepted view.

        ” A volley of particles is not compatible with a series of waves.”

        Newton’s billiard ball corpuscular light has been superceded by the idea of a wave packet. A pulse of wavelets which propagate as a localise concentration but not with a hard boundary like a ball.

        https://en.wikipedia.org/wiki/Wave_packet

        You can compare this to the profile of the concentric rings which spread out when you throw a stone into a pond.

        I’m sure Jack will be able to get his head around that idea. Hope that helps.

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        1. Looks like wordpress is trying to spam filter that gif link. You may need to copy it by hand.


          You can compare this to the profile of the concentric rings which spread out when you throw a stone into a pond.

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        2. Google “wave packet”.

          You may also be interested in the soliton wave. Look it up since links get held.

          A soliton is a non-dispersive wave in a linear channel which maintains its profile in time. It contains and thus transports energy. There is no constant amplitude continuous sine wave , nor a standing wave oscillating between two end points. It may be seen as analogous to a “particle” hence its name ending in “-on”.

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        3. Greg,
          Your gif shows a “tether” that was always there. You were supposed to prove how the “tether” forms from one side and attaches to the other side. That’s what your Corpuscular Theory claims: that the wave is created without attachment on the other end (and this depletes kinetic energy from the source). What happens when there is no other side?

          Your gif validates my theory, not yours.

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        4. Whats the difference between waves and wave packets? The difference is that the wave packets concept is still going to be idiocy if it has been formed under conditions of enforced aether denial. Anything coming out of aether denial will be idiotic. Why wave packet? Why not just waves? Waves are what we measure when we find the wave-length for light, so waves it is. No need to get fancy about it. A wave packet would only occur if you set it up that way. To be a quick burst of light.

          And in any case a wave packet is not a photon. So we want to be very clear that the photon concept has been sent to the fires, as an artifact of a process, which came out of enforced aether denial, as a control mechanism.

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  16. Your scenario is interesting but incomplete. You are correct that “As you can see we have two different flux (W/m²) figures: 2.5 and 557 W/m².” But what OTHER fluxes might there be; what other conduction, convection, and/or radiation to/from the cold end? This is where you need specific more details about the situation.

    The surroundings will affect how well the cold side actually sheds heat. For a simpl,e example, if the surroundings are a vacuum at …
    A) -273 C, then the cold end will indeed shed Ec = 557 W/m^2 of thermal IR to the surroundings, and Es = radiation from the surroundings back to the cold surface.
    B) 50 C, then the cold end will shed ZERO net thermal IR to the surrounding (Ec = 557 out, but also Es = 557 back in)
    C) 75 C, then the cold end will be ABSORBING net thermal radiation (Ec = 557 out, Es = 750 in for a net gain of 193 W/m^2
    D) 49.7 C, then cold end will shed 2.5 W/m^2 net thermal IR. (Ec = 557 out, Es = 554.5 in)
    The cold side will always emit Ec = 557, but it can ALSO absorb various amounts or radiation, Es, depending on the temperature of the surroundings. Any absorbed thermal IR from the surroundings should also be included in your analysis!

    Scenario A is the simplest, and fits your narrative most closely. “As you can see we have two different flux (W/m²) figures: 2.5 and 557 W/m². ”
    The proper conclusion here is that since much more energy is leaving than arriving, the cold end will be cooling! As you note, “It becomes obvious now that CHF and CSR have an inverse relationship” so as the end cools, CHF will increase while CSR decreases. Eventually the cold end will settle at a temperature much cooler than 50 C, and CHF will equal CSR.

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  17. Hey, Zoe.

    When I look at graphics of the temp-gradient, then I see clearly the power must come from inside. It is very hot inside and 0C on the outside. It is almost a straight line on the graph! That tells it all, to me, it is. It looks to me as if the earth is for over 95 % its own stove. https://www.google.nl/search?q=temperatuur+aarde+diepte&client=opera&tbm=isch&source=iu&ictx=1&fir=OwgvEUID_8yo1M%253A%252CrudKepTzdeYEiM%252C_&vet=1&usg=AI4_-kQ-G8YBZwoipgFnE5ZBTGSKxRUmwg&sa=X&ved=2ahUKEwiizKeT2J3oAhUB6qQKHRcNBOwQ9QEwAHoECAkQAw#imgrc=k1CDtyF6VIRkGM&imgdii=x-N-0qtKHc0OtM
    So, there is something very strange happening in ‘politics’, but so it is. The Powers that Be dictates politics.
    Thats it, I think. We better leave it for what it is.

    Liked by 1 person

        1. Geothermal heat loss from the planet to the atmosphere is less than 1 watt per square metre. it is about 166,000,000 megaWatts. This is not much power considering everything else. It is low enough so that the core temperature of the Earth only drops one degree per billion years. When the Sun burns out and dies, the earth’s core will still be hotter than the Sun’s surface.

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        2. Jarle,
          The temperature in both your ears are the same. Therefore the conductive heat flux between your ears is 0 W/m^2. Does that mean your ears can’t emit a potential 527 W/m^2 (37C)? Even if it’s 20C outside, your ears are not going to emit 0 W/m^2.

          The conductive heat flux is a gradient (differential) measure, and tells you nothing about what can emerge.

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        3. Zoe,
          I have no problem with the absolute potential. I was just curious as to how comparable the two fluxes are (163 from sun vs. 294 from geothermal). Could you please elaborate a bit on this?

          Liked by 1 person

        4. They are perfectly comparable. Remember, this 294 is not necessary for emission. The point is that geothermal provides the surface molecules with such an equivalent of Kinetic Energy. Geothermal makes the surface molecules randomly move AS IF there was a 294 radiant source shining upon it. The sun merely adds to that.

          The conductive heat flux in the sun’s photosphere is also tiny, but again, that’s not what emerges.

          Geothermal deniers want to confuse things in order to make room for their junk theory.

          The greenhouse effect is just geothermal flipped upside down. Cool scam, right?

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      1. 1. The conductivity of the slab can’t transport that much heat.

        2. The hot side radiates even more heat.Where does that come from?

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        1. 1. What do you mean conductivity can’t transport that much heat?

          2. There’s obviously an energy source that is not shown.

          I have videos in the other link I provided.

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        2. In your picture you state “q=2W” in the concrete slab. I take it that you used a heat transport equation to compute it, I trust your number. So your slab transports 2W of heat from the hot end to the cold end. At the cold end, it radiates 0.8[m2]*557[W/m2] = 446W. Only 2W are coming from the hot end through the slab. Where are 444W coming from?

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        3. The 2W is a differential of kinetic energy levels divided by time. It is not an absolute energy level as can be emitted to a zero potential on the cold side.

          “transports 2W of heat from the hot end to the cold end.”

          Yes, and that 2W is an ADDITION to what is already there.

          The 2W is a FLOW that gets added to the top of an existing stack, so to speak.

          The temperature of both your ears are the same, therefore the conductive heat flux through your head is 0 W/m^2. Does that mean your ears can’t warm anything, like an ice cube at 315 W/m^2? Or would you ignore ZERO conductive heat flux and focus on the 527 W/m^2 (37C) your ears can emit?

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        4. I your picture, 2W is a thermal flux. Don’t confuse it with kinetic energy – you don’t consider a moving slab. I thought you meant a steady state, with a temperature not changing, and with all heat flows balanced. What heat flow is already there, and why is it missing from your equations?

          Please attempt to make sense. I have better things to do.

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        5. Kinetic Energy means the molecules are moving. They’re moving faster in one place than another. The difference has nothing to do with the absolute on the cold side.

          The conductive heat flux in the sun is also small, therefore the sun can’t emit solar-thermal energy?

          I make perfect sense, but some don’t want to accept reality.

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        6. Hi there Curious G ol’ buddy

          I have been following this conversation from afar for some weeks. I think you are wasting your time. I was susrprisd to find you still here instead of the regular haunts. It is not possible to discuss a balanced heat flow scenario if one is missing fundamentals of heat transfer theory. You can tell by the questions the others have no background, even high school level, in physics. I don’t deride people for not taking physics, but there is a basic level of understanding required to understand a system involving radiation, thermal mass and conduction in thermal equilibrium. There is a reason why thermal engineering training is 4 years long.

          Zoe:

          Rgarding
          https://phzoe.com/2019/11/13/how-much-does-co2-absorb/comment-page-1/#comment-814

          The question of how much CO2 absorbs is framed incorrectly. When trying to figure out “how much the atmosphere warms” when CO2 is present (which is the underlying motivation) the question has to include both how much is absorbed and how much is emitted. There is no net heating if there is no net gain by something within the system. The net heating of the CO2 is zero.

          Your chart showing that 5% is absorbed is incorrect. That atmosphere doesn’t exist. It shows none if the overlap with water vapour which is BY FAR the dominant GHG. What really matters for policy discussion is what happens if the CO2 concentration increases – how much additional warming is produced. That is a “nett” question, not and “absorption” question. At present concentration the answer is “approximately zero”.

          The rest is noise and wind. The lower atmospheric temperature is dominated by the surface temperature of the oceans. Oceans are not heated by back-radiation from CO2 or water vapour or anything else. They are heated by sunlight at wavelengths above IR, factored for cloud cover and ozone. Anything that affects cloud cover such as GCR’s, water vapour change and the availability of aerosol particles serve as cloud condensation nuclei changes cloud cover. The difference between a glaciation and an inter-glacial is 2.5% cloud cover. Nothing else matters much at all. The great minds of climate science like Fairbridge, Singer and Soon have been pointing this out for decades, literally. The physics hasn’t changes just because newbies found new funding sources.

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  18. You don’t even understand basics. I give you a simple arithmetic problem you can find in every elementary school textbook: Alice got 5 dollars yesterday. How much does she have today?

    Your “typical simple conduction problem” is ill-defined. You don’t specify conditions on the right side of the slab. Just like my Alice example. Then you start distinguishing heat and kinetic energy. Phooey. Bye, Smarty Pants.

    Like

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