Real Steel Greenhouse Effect

13 years ago, amateur scientist Willis Eschenbach developed a thought experiment that he hoped would very simply illustrate how the Greenhouse Effect works.

The main claim is that the addition of a steel shell surrounding a planetary surface will cause the inner surface to emit TWICE (235 -> 470 W/m2) the radiation as compared to not having steel shell. This should significantly raise the surface temperature from ~254K to ~302K.

Is this true? No.

Willis gives us the freedom to construct any power source with any chosen radius. I will choose a mini nuclear reactor wrapped in a steel housing, with the total surface area being 1 m2. The inner radius of steel housing is 75% of total radius.

A = No Steel Shell, B = With Steel Shell

Let us go through the equations to make set up Willis’ initial scenario (A):

The nuclear power reactor is ONLY capable of making its wall 254.041K – to meet Willis’ initial criteria. It is not capable of anything greater, because nuclear reactions are fixed. No varying levels of downstream radiation will enable nuclear fission reactions to generate more joules.

Now let us see what happens when we add a steel shell (B):

I will give Willis credit for doing a good job of demonstrating the real greenhouse effect:

Outgoing radiation is halfed and T2 (our “surface”) has increased from 253.726K to 253.884K, a very feeble gain.

The problem with Willis’ approach is that he doesn’t reduce outgoing radiation and relies on his heat source to crank up … when there is no physical way it can do so.

Summary:

T#Willis AWillis BReality AReality B
T2253.726K (235W)301.732K (470W)253.726K (235W)253.884K (235.588 W)
T3253.726K (235W)213.490K (117.393 W)
Summary Table (W=W/m2)

So there you have it. The real steel greenhouse effect managed to raise the surface temperature by 0.062%.

Subsequent additions of steel shells will keep raising the surface temperature (T2), in an inverse asymptotic fashion approaching nuclear reactor wall temperature (T1).

Enjoy 🙂 -Zoe

Published by Zoe Phin

https://phzoe.com

84 thoughts on “Real Steel Greenhouse Effect

  1. Zoe, I can’t check your maths. Let’s hope Willis shows up to defend his argument. Intuitively, steel is a good conductor. But a steel shell would stop the atmosphere convecting. Convection elevates a molecule to a height where there is less material that is capable of absorbing radiation directly to space. Ozone is a large molecule that efficiently absorbs wave lengths that are most commonly emitted by the Earth system and its presence gives rise to an increase in gas temperature above an altitude of 10,000 metres in the mid latitudes. There is no evidence that the enhanced emission by the molecules warmed by ozone (nitrogen and oxygen) actually warms the layers beneath the point where that temperature increase occurs.

    The supposed warming effect of CO2 fails to take into account the efficiency of convective processes.

    Willis should spend a little time living in a sea container in an environment where air temperature falls to near zero overnight.

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    1. Well, um, there is no atmosphere in this example. No convection at all.

      Merely adding conductive matter on top of our “surface” would also raise temperature, but not as much as a detached shell via radiative means.

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  2. I don’t think it’s fair to call Willis an amateur scientist,
    nor would it be fair to use that title for you.
    You both present scientific articles, so are scientists
    by my definition of scientist, even if not paid for your work.
    There are plenty of people with science degrees
    paid to do reckless climate scaremongering. They
    are called “scientists”, but always wrong wild guesses
    of the future climate are not science.

    No need to print the following
    portion of my comment:

    Your three posts in three days is like a miracle.
    Every reader wishes you would post more often.
    Once in a week, or two, would be fine
    if you had the time. It’s nice to be “in demand”.

    I publish a climate science and energy blog
    where I present the best articles I’ve read that
    day. So far over 319,000 page views.
    I have only recommended two other blogs
    in the past few years, that have consistently
    good articles. Your blog is also consistently
    good, and I’d recommend it … but …
    you don’t have a regular publishing
    schedule, so if people visit on my
    recommendation, and see nothing new
    for a month or two, they’ll get mad at ME.
    This already happened with two friends
    who I told about your late 2021
    global sea ice area articles
    — both of them thought you had
    “gone out of business”
    after those two articles.

    If you had a regular schedule, such as a
    a post every week, or every two weeks,
    or even once a month, that would allow me
    to recommend your blog.
    I get 300 to 400 page views a day.

    PS: If you are anything like my wife,
    you are now mad at me for giving
    you advice. If so, please tale it out
    on Willis E. ha ha

    Richard Greene
    Bingham Farms, Michigan
    http://www.elOnionBloggle.Blogspot.com

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    1. I’m not mad. I’m a just very busy gal. I’d love to have more posts, but, as you can see, I do original research. Original research is very time consuming. When I don’t have the time, I make errors.

      I don’t want to have a blog that just links to other articles. Neither do I want to have boiler plate posts about updates to climate data.

      Right now I have just 2 ideas for next posts. After that, I will again have to scour for something original.

      Ideally I’d like to share everything I know, but I think that will be boring for most people. My computer skills aren’t interesting to people. Those posts get like no views.

      Liked by 1 person

      1. So, let me annoy you even more with
        an example. Let’s say you have the time
        to do 12 posts in a year. Your readers
        would be happy if they knew Zoe would
        be publishing something in the first week
        of every month. That would mean NOT
        publishing three articles in three days,
        as your just did. Those three articles could
        have been published over three months.
        When readers don’t see a post for a long
        time, or there is no predictable schedule,
        they stop looking. … I know genius can’t
        be rushed, but if you publish three articles
        in three days, and perhaps nothing else
        for the next month, readers disappear.
        That’s just my two cents, which is
        probably worth one cent after inflation.

        Liked by 1 person

        1. This is not a job. I’m not on a schedule. When I have free time and desire, I write articles. Making this a scheduled task is making it like work, of which I have plenty.

          Thanks for the suggestion tho. Let’s see what happens.

          Liked by 1 person

        2. Zoe:
          Sorry, I jumped to the conclusion that
          having more readers was your goal.
          That’s apparently not a priority.
          I also failed to explain my ideas well.
          There is no need for you to do more work,
          or to work at different times.
          Your publishing schedule does not
          have to match your writing schedule.
          Your work is just as valuable if presented
          the day it is finished, or many months
          later.

          I had recommended your blog to two
          friends in late 2021 after reading two articles
          on Global Sea Ice Area. With no new articles
          in the next 2+ months, both friends deleted
          your blog from their Bookmarks list, and never
          visited again. Which I thought was a shame.

          My own blogs serve a different purpose.
          I merely share the best articles I’ve read
          that day, and rarely write my own articles.
          I used to write articles for my for-profit
          newsletter ECONOMIC LOGIC, from
          1977 to 2020. That was a lot of work.

          Now I use my blogs as an antidote
          for social and mass media censorship,
          and parroting government press releases.
          So I want more readers, and assumed that
          desire applied to everyone with a blog
          or website.
          I was just trying to be helpful.
          Richard Greene
          Bingham Farms, Michigan

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        3. Nah. Publishing is the instant reward for writing. I have 7 posts I never published. I’m still writing them. lol. Actually they are crap, and I should delete them. And that happens every time I take that approach. If I start writing, it must be finished today or early tomorrow, or never.

          Don’t apologize. Thank you for the advice.

          I didn’t even think anyone would read my blog.

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  3. So the outgoing radiation is halved, while the source output stays the same. Please explain how T2 remains largely unchanged. I follow the math, but to me it is counter intuitive.

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    1. What were you expecting? Willis result or no warming at all?

      The net flow became ~118W, down from 235W. Less cooling … warmed up.

      A net flow of 0W between surface and shell would mean the surface and shell is same temperature as nuclear reactor wall. This is impossible but presented for guidance.

      Nothing in this system will exceed nuclear wall temperature.

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      1. I was expecting no warming. To me it is intuitive that no part of the system can achieve a higher temperature than the nuclear wall temp. It is also intuitive that the source output remains constant. So I guess I was expecting the cooling to remain the same, but after some condideration I agree that it must reduce. Radiation is inefficient

        Liked by 1 person

        1. Just to clear things up … I never doubted the REAL “greenhouse effect” (really: thermal resistence effect) of preventing cooling leading to a warming. I just never believed warming can exceed the SOURCE. I still don’t.

          Liked by 1 person

        2. The greenhouse effect depends upon stability in the medium. The atmosphere is anything but stable. Its highly mobile. Cooling ultimately depends on radiation to space but in the interim, it is very efficiently maintained via conduction and displacement, ie convection. The part of the globe that receives the most intense radiation, inside the tropics of Capricorn and Cancer, is temperature limited due to the efficiency of plant transpiration, evaporation from open bodies of water, the movement of ocean currents and most of all, convection. Accordingly, the tropics radiate much less energy to space than is received directly from the sun due to the temperature decline as gas expands as it is elevated. It is here that the tropopause, where the air is coldest, due to simple decompression and the impact of convection in displacing ozone upwards, that the air is as cold as over the poles in winter.

          As the Hadley cell is energized by convection, the descent and compression of air in the mid latitudes is enhanced. Here, the air is dry, its greenhouse potential reduced and radiation to space proceeds with relative freedom. Dry air is not conducive to the production of cloud. The temperature of the air in the mid latitudes, is dependent, as it is everywhere else, on the partial pressure of ozone that is elevated by convection in extratropical (polar) cyclones to the upper limits of the atmosphere, from where it is transported to the mid latitudes, affecting cloud cover.

          This is a complex system, incapable of being mathematically modelled. The most sophisticated minds think in terms of an ‘open system’ whereby the atmosphere is susceptible to change according to change in the wider solar and intergalactic influences. Sophistication is gained via a study of the geography and history of the atmosphere.

          Greenhouse effect. Propaganda. A simple idea, easily assimilated, that is maintained in order to secure an economic, social and political objective. A scam.

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  4. The greenhouse effect is what it is. The claim that CO2 is a control button for the climate i something different. Most scientists do not dare to challenge this view, in fear of destroying their own reputation. Roy Spencer is known for having said that since the climate system is so complicated, we do not know whether humans have caused 10% or 90% of the observed warming. Of late though, he feels obliged for some reason, to state that: “I still provisionally side with the view that warming has been mostly human-caused”. He can’t back it up, but safe to preserve ones reputation since there is doubt. And this is how it goes, decade after decade. I hope this is the last decade. The truth must come, be it in favor of the alarmist or the sceptic side.

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    1. CO2 causes between 0% and 100%
      of the global warming, with both
      0% and 100% unlikely, in my opinion.

      I gave Roy Spencer a hard time on his blog
      about guessing CO2 had a major effect
      when the correct answer is no one knows.
      I liked his 10% to 90% range.
      But then I felt bad when later found out UAH
      had been defunded and he and Christy
      were providing UAH data on a voluntary basis
      for the past few years. Hard to criticize a
      scientist who does that.

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  5. Firstly, Willis Eschenbach’s thought experiment does NOT reflect how a real greenhouse works and neither does it reflect how the man-made (fictional) climate change greenhouse effect works – as these both require an atmosphere, whereas the thought experiment requires a vacuum within the infinitely small gap between the inner sphere and the outer shell. However, the thought experiment DOES correctly demonstrate how a colder object (the outer steel shell) can make a hotter object (the inner sphere) even hotter. I admit to once being a Joe Postma supporter in believing that a colder object could never make a hotter object even warmer – but I was wrong.

    The thing that everyone who refutes Willis Eschenbach’s thought experiment overlooks is the TIME required to establish the new steady-state temperature of the outer shell, once the outer steel shell is introduced. From the moment that the outer shell is introduced until the out shell reaches it’s steady-state temperature, not all of the energy that is constantly being generated by the inner sphere is actually being radiated away from the outer surface of the outer shell. That energy, which being generated by the inner sphere but which is not being radiated away from the outer surface of the outer shell does NOT disappear. Rather, it manifests itself as kinetic energy (and thus temperature increases) in BOTH the inner sphere and the outer shell. No ‘additional’ energy is required to be created by the power-source within the inner sphere to raise the temperatures of the inner sphere and outer shell. Rather, it is only the regular energy that the power source constantly generates (but which has not been immediately radiated away from the outer surface of the outer shell), which is responsible for the temperature increases in BOTH the inner sphere and the outer shell.

    To help further, let me give you an analogy, using the transfer MONEY instead of ENERGY. I know that many dislike analogies, but this one is good, so please bear with me and let the (necessarily long) story unfold….
    A person has been raised to believe that they should give away 10% of their wealth (which is represented by the funds held in their checking account) to an external charity, each month.

    On the first day of the first month, when this person starts their first paid job (say at a monthly rate of $1000), they have zero dollars in their checking account, so they give away nothing to charity. At the end of “month one”, they receive their $1000 salary paid into their checking account.

    At the commencement of “month two”, they now have $1000 in their checking account and so give away a $100 check to charity. At the end of “month two”, they again receive their $1000 salary paid into their checking account.

    At the commencement of “month three” they now have $1900 in their checking account and so give away a $190 check to charity. At the end of “month three”, they again receive their $1000 salary paid into their checking account.

    At the commencement of “month four” they now have $2710 in their checking account and so give away a $271 check to charity. At the end of “month four”, they again receive their $1000 salary paid into their checking account.

    You will see that eventually their checking account will reach $10,000 and they will give away $1,000 each month to charity, and they will earn $1,000 for the next month – the state of financial equilibrium has been reached. This is like the equilibrium state of ‘inner sphere without the outer shell in situ’ in the thought experiment.

    Many years pass and this hardworking and generous person rears a child and this child is also raised with a very charitable nature. However, unlike their hardworking parent, this child (now entering adulthood) does not become an independent worker in their own right but instead, remains wholly dependent upon their generous parent for their entire income. The parent, believing that “charity begins at home”, continues to give 10% of their wealth away but now, all of their charitable giving goes to their dependent child, rather than the external charity.

    On the first day of the first month (donation day), when the dependent child leaves home, the dependent child has no wealth in their own checking account and so gives no charitable donation to any external recipient and similarly gives no charitable donation to their hard-working parent either. However, the parent has $10,000 in their checking account, so the parent now gives away 10% of their wealth (as a $1,000 check) posted to their dependent child – and, on arrival, this check is paid into the dependant child’s own checking account, so the dependant child’s checking account jumps to $1,000 dollars, whilst the parent’s checking account drops to $9,000. However, by the end of the first month, the parent is paid their regular $1,000 salary so the parent’s own wealth is again restored to $10,000.

    At the commencement of “month two” (donation day), the parent posts another $1,000 check to the dependent child. For the first time, the dependent child similarly posts a check for 10% of it’s own wealth (the $1,000 gratefully received last month) to an external charity and also posts a check for 10% of it’s own wealth back to their generous parent (the child, like the shell in the thought experiment, has two directions for giving). So, the child posts a check for $100 to an external charity and posts a check for $100 check back to their parent (leaving the child $800). The two checks sent between the parent and the dependent child always cross in the post. On their arrival, the dependent child will now have $1,800 whilst the Parent will now have $9,100 but, by the end of “month two” the parent also receives their regular $1,000 paycheck, so the parent’s own wealth is restored, but this time to $10,100.

    At the commencement of “month three” (donation day), the parent posts a $1,010 check to the dependent child. Similarly, the dependent child posts a check for 10% of it’s own wealth (now $1,800) to an external charity and also posts a check for 10% of it’s own wealth back to their generous parent. So, the child posts a check for $180 to an external charity and posts a check for $180 check back to their parent (leaving the child $1,440). The two checks sent between the parent and the dependent child again cross in the post. On their arrival, the dependent child will now have $2,450 whilst the Parent will now have $9,270 but, by the end of “month three” the parent also receives their regular $1,000 paycheck, so the parent’s own wealth is restored, but this time to $10,270.

    At the commencement of “month four” (donation day), the parent posts a $1,027 check to the dependent child. Similarly, the dependent child posts a check for 10% of it’s own wealth (now $2,450) to an external charity and also posts a check for 10% of it’s own wealth back to their generous parent. So, the child posts a check for $245 to an external charity and posts a check for $245 check back to their parent (leaving the child $1,960). The two checks sent between the parent and the dependent child again cross in the post. On their arrival, the dependent child will now have $2,987 whilst the Parent will now have $9,488 but, by the end of “month four” the parent also receives their regular $1,000 paycheck, so the parent’s own wealth is restored, but this time to $10,488.

    You will find that, eventually, the parent’s checking account will grow to reach $20,000 and that the parent (whilst still earning a salary of only $1,000 per month) will be required give away $2,000 each month to the dependent child. The dependant child’s checking account will grow to reach $10,000 and the dependent child will give a $1,000 check to their ‘external’ charity and will give a $1,000 check to their parent – the new state of financial equilibrium has been reached. This is like the steady state of ‘inner sphere and the outer shell in situ’ in the thought experiment.

    All the ‘additional’ money in the system is accounted for (it only ever came from the gainful employment of the working parent and yet, by the introduction of a wholly dependent child, the parent has, over time, become wealthier – twice as wealthy in fact – whilst still only earning the same $1,000 salary each month. Neither the parent nor their dependent child has fraudulently created any fake money. The wealth held in the two checking accounts is entirely genuine – but the (wholly dependent) yet generous child’s “back-giving” has allowed the parent’s own wealth to increase.

    A dollar, when given by a poorer person to a richer person, must inevitably make the richer person one dollar richer. However, the richer person will always ‘outgive’ to the poorer person – the net flow of dollars is always from the wealthier parent to the poorer child.

    The same is true with the energy conveyed by the photons that are emitted from a colder object toward a warmer object – the radiant energy from these photons will be thermalized by the warmer object (like the dollar from a poor person, each Joule has to be accounted for). However, the warmer object will always ‘outgive’ the amount of energy it gives to colder object. The net flow of energy is always from the hotter surface to the colder surface.

    The colder object does not prevent the hotter object from emitting all of it’s radiant exitance at the level prescribed by the S-B law. Similarly, the hotter object does not prevent the colder object from emitting all of it’s radiant exitance (on both it’s surfaces) at the level prescribed by the S-B law.

    Finally, the steel greenhouse thought-experiment is based upon a shell having a radius that is only just larger than the radius of the sphere – that requirement in the thought experiment is necessary to ensure that all of the back-radiation from the inside surface of the shell falls upon the surface of the sphere – none is ever allowed to reach another area upon the inside surface of the shell. By this stipulation, the energy held in the sphere is now double that which would have been held if the nearby shell was absent and hence the temperature is now greater by a factor of the fourth root of two i.e. 1.189 times greater than if the nearby shell was absent. If the radius of the shell is not just slightly larger than that of the sphere but significantly larger, then the effect of back-radiation is significantly diminished. A larger radius for the shell can again be explained in the “charitable parent and dependent child” analogy: if the Charitable Parent had three dependent children (instead of one) then the parent would continue to give 10% of their wealth away each month which would be shared equally between the three dependent children. Each of the three dependent children would again give 10% of their wealth to external charities and would also give 10% of their wealth as generous giving back to their internal family but this time, the parent would only receive one third of that which it would have got back had with one child because each of the three children distribute that internal giving as one third to the parent and then one third to the other sibling#1 and also one third to other sibling#2 (and as the two other siblings do exactly the same thing then the inter-sibling transfers count as zero net effect). In summary, the parent does get more wealth because of the existence of three dependent children but not as much as would be the case with just one dependent child i.e. as the radius of the shell increases in proportion to the radius of the sphere, the temperature-increasing (wealth) effect of the back-radiation from the inside surface of the shell upon the external surface of the sphere becomes less.

    As I said earlier, ‘back radiation’ between the sphere and the shell is only significant in a vacuum. As soon as a gas is present in the gap between the sphere and the shell, the standing temperature difference between the sphere and the shell is reduced (the gas molecules act like a ‘resistive’ short-circuit). As the density of gas molecules is increased, the ‘resistive’ short circuit will tend to become a true ‘short circuit’ where the sphere temperature and the shell temperature are exactly the same.

    Liked by 1 person

  6. Zoe,

    Regarding the caveat: Yes, if we are splitting hairs, then you are correct: the radius of the inner sphere (and hence surface area) must always be smaller than that of the outer shell and so the Watts/M^2 radiant emittance (and hence S-B temperature) must always be greater on the surface of the inner sphere. However, in my defence, in the opening paragraph I did state that “…the thought experiment requires a vacuum within the infinitely small gap between the inner sphere and the outer shell”. That ‘infinitely small gap’ allows for the Radius of the inner shell to approximate to the Radius of the outer shell. That is important for the mathematics used in the thought experiment.

    To assist you to imagine the thought experiment more faithfully, try imagining the inner shell to comprise a steel ball bearing with a 2mm diameter which is tightly surrounded by 999 steel shells each 1mm thick acting as laminations. You have now created a steel sphere of radius 1000mm. Now, imagine that you have miraculously ‘vanished’ the penultimate lamination from within the sphere and you have created an inner sphere with a radius of 998mm and a vacuum gap of 1mm and an outer steel shell which has an inner radius of 999mm and an outer radius or 1000mm. Once you’ve then miraculously inserted the radio-active power source into the centre of the of the inner sphere then you have constructed (in your head of course) a reasonable working model for the thought experiment to operate, with both the inner sphere and the outer shell each having a radius of approximately 1 metre.

    If the vacuum gap was instead allowed to be filled with gaseous molecules then the inner shell would no longer represent it’s S-B temperature (being 1.189 times greater than the Kelvin temperature of the outer shell), as Radiation would no longer be the only energy transport mechanism. Instead, convection would allow more energy to move across the air gap to the outer shell and thus causing the surface temperature of the inner shell to be reduced closer to (although not exactly the same as) that of the outer shell. And the more gas molecules place in that air-gap then the more effective the convection would be. That was the point that I was trying to make. To help further: if that penultimate lamination were to be re-inserted i.e. miraculously ‘unvanished’, then the energy transport mechanism would then comprise only conduction, such that the temperature of the laminations which previously comprised the inner shell would be identical to the surface temperature of the outer surface (assuming steel has perfect conductance).

    As I have said, several times, the mathematics used by Willis does require the radius of the inner sphere to be the ‘same’ as the radius of the outer shell. If the radius of the inner sphere was allowed to be 1 and the radius of the outer shell was allowed to be 2 then simple trigonometry can be used to show that only 1/3 of the ‘back-radiation’ from the inner-surface of the outer shell will strike the surface of the inner sphere such that the S-B temperature of the surface of the inner sphere will only be the fourth root of 1.3333 times greater than that of the Kelvin temperature of the outer shell i.e. the ability of the outer shell to affect the surface temperature of the inner sphere is substantially reduced.

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  7. Zoe, I have found a mistake in your calculations. In the third line of the section with the steel shell, you plug in 254 K for T1. However, this value (254 K) was calculated from the case without the steel shell, so it no longer applies when you add the steel shell. You have to treat T1 as a variable.

    As a result of using the old value for T1, your solution with the steel shell no longer obeys energy conservation. The outer shell emits less energy to space than the nuclear reactor is producing. So energy is piling up in the system, and it will get hotter.

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    1. Wrong. T1 is not and can not be a variable, for the reason I explained. Nuclear reactions are FIXED! You will not get more energy from the fission of fuel based on anything else down the line.

      I’m not an expert on nuclear physics, but that is absolutely true.

      As a thought experiment, take a regular 9V battery, and create a heat source by directly connecting the terminals. The max power you get is P=IV= 0.5 A × 9V = 4.5 W.

      Let’s assume the battery has infinite “fuel”, voltage never drops, and you have perfect conversion to heat. You can couple the “short” to a metal shell with an inner surface area of 1m^2.

      So, you will have the same experiment, but the inner metal shell gets 4.5W/m^2. If we imagine the shell to have zero thickness, it will also emit 4.5W/m^2 to space.

      But let’s add some thickness. Now we got 3 W/m^2 emerging to space.

      This is where the problem starts. Will adding another shell cause more than 4.5 W/m^2 to be available at the inner wall? Will the battery crank out more power? No, of course not.

      The max temperature of the inner wall is limited to ~94K. It is max, it is fixed. Nothing else you put down the line will cause it to rise.

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      1. While the energy density of the fuel is fixed, the rate of reaction, and therefore the rate of energy released, and therefore the equilibrium temperature, are definitely not fixed, and depend on factors such as density of the fuel, neutron flux, and even the speed of the neutrons. If the rate of reaction were fixed, then we wouldn’t be able to control the output of reactors, or make nuclear weapons, nor would be have to worry about meltdowns.

        At equilibrium, the total rate of energy emitted by the outer shell must equal the total rate of energy being absorbed by the inner shell, and this rate of energy must match the power output of the reactor. while the rate of energy transport across a medium is proportional to the difference in temperature on either side and inversely proportional to the thickness.

        Putting it all together, the power output is given by the reactor. The temperature of the outside part of the shell is then determined by the power output and the radius using the Stefan-Boltzmann Law. And the temperature of the inside part of the shell is given by the the power output, the outer shell temperature, and Fourier’s Law.

        If the temperature of the reactor were fixed, then you have a violation of either the first or second laws of thermodynamics, since either you have energy disappearing or the reactor heating the inner part of the shell higher than its own temperature, depending on if you set the inner part no higher than the reactor, or if you have the same total emission on the outer shell as the power output of the reactor.

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        1. I gave a link. The link shows diagrams with a maintained room temperature. The temperature is not adjusted to that of what’s outside. The fuel rate is adjusted to maintain the enclosed (inside building) temperature.

          So whether it’s hot or cold outside, the reactor’s surface is set to room temperature. I just use that same assumption. What’s inside the steel housing is the room, and its temperature is maintained.

          What would happen if the flux, rather than temperature, was maintained constant and the sink (people’s usage of generated electricity) was abruptly halted? Oopsie

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        2. Which means that the power output of the reactor is adjusted to keep the temperature the same, and so doesn’t address the point that with a given power output, T1 will be different with the shell versus without.

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        3. A reactor is set to keep T1 stable. It is DANGEROUS to keep the flux stable.

          I think you misunderstand stability. The logic circuit is not set to make stable whatever new temperature may arise by keeping the flux the same. It is set to a specific temperature. If the room temperature in a real reactor goes to 40C for some unknown reason, that is also dangerous, and the reactor will reduce fuel further so that the wall sensor is back at 20C, or whatever is the standard.

          I don’t know why that’s hard to understand. My home central heating/cooling system is set to a fixed temperature: 71F. When it gets too hot, the AC goes on. Too cold, the heat goes on. My system is not set to produce a fixed flux, and neither is a normal nuclear reactor.

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        4. It’s not that it’s hard to understand, it’s that you’re not comparing apples to apples anymore when you vary the flux.

          The entire point of the thought experiment is that with a given flux, T1 will change. If the flux is no longer given, then it’s a completely different thought experiment.

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        5. No, the entire point of the thought experiment is that T2 will change. T1 doesn’t exist in Willis’ set up.

          No existing nuclear reactor ever built creates a constant flux.

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        6. Willis sets the flux at a constant amount, in his case 235 W/m^2, and lets the temperature change as determined by the Stefan-Boltzmann law, Fourier’s law, and the conservation of energy.

          You vary the flux to force the temperature the same, so it’s no longer addressing the same points.

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        7. Right. His nuclear reactor is a unicorn. Mine is a horse with an attached horn.

          The central claim of the greenhouse gas theory is that outgoing flux is reduced. Is it not? Reducing emissivity should lead to a warming. But Willis keeps the emission the same. So who is misleading?

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        8. Close; the core principle of the greenhouse effect is that the ratio of surface flux to escaping flux is reduced via absorption and emission of IR by greenhouse gasses, e.g. only a certain percentage actually passes through. It’s the Beer-Lambert law applied to the atmosphere.

          At equilibrium, the first law of thermodynamics dictates that the escaping flux is equal to the incoming flux. So the outgoing flux is determined entirely by the incoming flux.

          If you compare two systems with the same flux, one without the reduction and one with the reduction (either via Fourier’s law or the greenhouse effect), the latter will have a higher surface temperature than the former, just from those two above principles.

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        9. lol. but you kept the albedo fixed!

          You have 235 initially (current GHG levels) because albedo is set to ~0.3 from the ~340 we get from the sun.

          Let’s say we introduce so much GHGs that Earth now emits 117.5.

          Since the albedo is determined to be just (shortwave in – longwave out) / shortwave in

          All that will happen is that albedo will increase to ~0.655.

          The solar received at the surface will be ~117.5.

          There’s only a warmup because you adjust some parameters and not others.

          I account correctly that new fluxes will be 117.5 and 117.5, but you want to keep 235 and 235 as if it’s some sacred fixed number.

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        10. Albedo is the ratio of reflected flux / incoming flux of visual (shortwave) radiation. Emissivity is the ratio of emitted thermal radiation to that of a perfect blackbody. What you described is the imbalance ratio, which is zero for equilibrium.

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        11. Thought you’d go there.

          “Albedo is the ratio of reflected flux / incoming flux of visual (shortwave) radiation”

          Yeah, and in order to satisfy the equilbrium at TOA, reflected shortwave = incoming shortwave minus outgoing longwave.

          So if you reduce outgoing LW, you increase reflected shortwave … albedo goes up.

          The amount of shortwave reaching the surface is reduced.

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        12. That’s not true at all. Reflected shortwave = incoming – absorbed shortwave flux. Absorbed shortwave flux is only equal to outgoing longwave flux at equilibrium. If outgoing longwave flux does not match absorbed shortwave flux, then it’s no longer in equilibrium and the temperature changes until it’s equal due to the first law of thermodynamics.

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        13. Again, you keep albedo fixed as if it’s a conserved quantity. It’s not.

          Now you’re suggesting that GHGs do not in fact reduce outgoing radiation. And yet the textbook math requires it.

          Look at my response to Jarle.

          “is only equal to outgoing longwave flux at equilibrium.”

          Yeah, and a new equilbrium will be established, and there is no fixation to the old.

          We’re not even talking about the real albedo – what determines surface insolation. Only 48% of insolation reaches the surface, not 70%.

          On Venus, albedo is 80%! and only 1-2% of TOA insolation reaches the surface.

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        14. Zoe wrote “So if you reduce outgoing LW, you increase reflected shortwave … albedo goes up.”

          There is no physical mechanism for that to happen.

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        15. Really? Filling the atmosphere with more molecules that can absorb in shortwave such as H2O and CO2 will not affect albedo? No more clouds will be created if we add water vapor galore?

          Funny how Venus has 80% albedo, and only 1-2% insolation reaches the surface …

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        16. CO2 has an inconsequential absorption of shortwave. Clouds are liquid water, not water vapor. Water vapor increases as a feedback to increased temperature. Increased water vapor’s major and immediate effect is absorption of longwave. Increased water vapor does allow for increased clouds, but different kinds of clouds have differing balances of effects on longwave retention and shortwave reflection. Any effects on shortwave albedo are distinctly secondary in time, causality, and influence.

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        17. And yet … while only 48% of the sun reaches Earth’s surface, only 1-2% reaches Venus’ surface.

          CO2’s absorption bands and transmissivity are also set by temperature and pressure. Venus has an incredible new set of absorption bands that are not seen on Earth, including in the shortwave portion.

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        18. It is completely possible to reduce the longwave emission without touching albedo, e.g. with CO2 alone. Water vapor concentration is entirely dependent on the Clausius–Clapeyron relation.

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        19. Right now at TOA:

          340 SW IN = 105 SW OUT + 235 LW OUT

          Now we reduce LW out to 230

          340 > 105 + 230

          Imbalance. Now what happens?

          Don’t contradict yourself a 2nd time.

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        20. I already answered this a few comments ago:

          “If outgoing longwave flux does not match absorbed shortwave flux, then it’s no longer in equilibrium and the temperature changes until it’s equal due to the first law of thermodynamics.”

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        21. Be more clear. We’re not talking about temperature changes at the bottom. We are talking about TOA fluxes. Are you saying that 230 LW returns to 235?

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        22. Zoe asked “Now we reduce LW out to 230. 340 > 105 + 230. Imbalance. Now what happens?”

          Energy accumulates in the atmosphere–lower in the atmosphere first, because there is more LW absorption happening by the air above there. LW heading toward space is intercepted by greenhouse gases between it and space, then transferred by collision to surrounding molecules of all types, and some of it reradiated in all directions. The now-warmer greenhouse gases increase their radiation because they have more energy.

          That’s more outgoing radiation trying to squeeze through the obstacle course of greenhouse gases between those radiating molecules and space. Failing to squeeze through, that extra radiation is absorbed by those intermediate molecules, thereby increasing the energy in those intermediate molecules.

          The proportion of LW radiated toward space, that actually makes it to space without being intercepted, switches location from being radiated by greenhouse gases lower in the atmosphere, toward those higher in the atmosphere. But those gases higher in the atmosphere radiate less than the gases lower in the atmosphere, because the former are colder. That lesser radiation makes those higher gases retain more absorbed energy, thereby warming.

          Eventually the greenhouse gases in the lower layers radiate enough, so even with their difficulty of making it all the way to space, enough LW does make it to space to for their outflow to balance the inflow from the Sun.

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        23. Skip the drama. Why can’t you just give a simple answer?

          Let’s try again:

          340 > 105 + 230

          Imbalance. Now what happens?

          Are you saying things at TOA will return to:

          340 = 105 + 235

          YES or NO

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        24. The first law of thermodynamics dictates that the temperature will increase, thereby increasing the emitted surface flux and so the TOA flux, until the TOA flux matches the incoming absorbed flux.

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        25. Yes. The surface temperature increases as required by the first law of thermodynamics, du = dq – dw. Since the surface emitted flux is proportional to T^4, it will also increase, and since the TOA flux is proportional to the surface flux, it will also increase, until flux-in = flux-out.

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        26. OK, so 230 will return to 235, you claim

          Physics works in real time. We don’t need to wait for a change in 5 W/m^2. Same principle should work for a change in 0.0000000000000000001 W/m^2.

          So, how can you say it’s possible to lower OLR when it will instantaneously return to its previous state?

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        27. Nobody said anything about instantaneous. That first law of thermodynamics is a differential equation, du = dq – dw where du is the differential of internal energy, dq the differential of heat, and dw the differential of work.

          Since no work is being done on the atmosphere, dw = 0, so du = dq.

          For most materials, like the air, u = CT, so du = CdT, where C is the specific heat capacity. And since dq is the total heat flow, dq = (P_absorbed – P_emitted) dt = (P_absorbed – (1/λ) εσT^4) dt, where A εσT^4 is the Stefan-Boltzmann law for a greybody, and λ is the ratio of surface emission to TOA flux (e.g. the strength of the greenhouse effect).

          This means that the rate of change of the temperature, dT/dt = (P_absorbed – (1/λ) εσT^4) / C.

          Equilibrium occurs when dT = 0, which happens when P_absorbed = (1/λ) εσT^4, rearranging gives T = (λ*P_absorbed/εσ)^(1/4) as the equilibrium temperature.

          With a gradual change in λ, there is a gradual change in T.

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        28. Why do you have to be so obtuse? Can you stick to just one thing? We can discuss temperature later.

          You said that it’s possible to reduce OLR, and at the same time claim it will be restored. So any minute change is quickly offset.

          Bringing in heat capacity doesn’t change this either because it’s active in both OLR reduction and subsequent return. So you can just drop it from your argument.

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        29. I’m giving you the physics and explaining the mechanism behind the outgoing flux eventually matching the incoming flux, and why it’s not instantaneous.

          It takes time for the flux to match, as described in the math above, so a faster change in λ leads to a greater imbalance between incoming and outgoing flux. If you change λ fast enough, you increase the imbalance faster than it can equalize.

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        30. “If you change λ fast enough, you increase the imbalance faster than it can equalize.”

          Very much false. It takes same time to unbalance as rebalance.

          Why? You explained yourself: The IR-active gases are surrounded by non-IR active gases. While IR active gases can drop temperature (and emission) faster, they are still surrounded by gases that can’t. Those non-IR gases will cool slower. They can only effectively cool by bumping into IR gases. And guess what? They will transfer their energy to the IR gas to emit, thereby slowing down the IR gas cooling.

          It’s quite amusing that you only consider heat capacity one way but not the other way.

          Heat capacity works both ways exactly the same. Which is why I urged you to drop this.

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        31. That doesn’t follow, since λ is determined entirely by Schwarzschild’s equation.

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        32. Uhm, what does Schwarzchild have to do with this?

          Emission equations can only deal with IR active substances.

          You want to warm nitrogen and oxygen by conduction on the rebalancing part, but can’t comprehend nitrogen and oxygen warming H2O/CO2 by conduction when they cool too fast on the imbalancing side.

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        33. Schwarzschild’s equation gives you the transmittance of a reactive medium. dI(v) = nσ(v) (B(T,v) – I(v)) ds, where I(v) is the intensity of incident radiation at frequency v, n is the density of the reactive medium, σ(v) is the cross-section at frequency v, B(T,v) is the Planck emission at temperature T and frequency v, and s is the path.

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        34. Schwarzschild’s equation is given in terms of spectral radiance. You can simply integrate over the relevant range if need be to get the transmittance in a band, including the entire spectrum.

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  8. It seems you are making an energy balance argument, but instead of calculating the difference between energy in and out, you are just stating that T1 can’t change because of energy balance.

    But let’s do the math. In your first calculation, you correctly set that the energy produced by the nuclear reactor must equal the energy leaving from r1. by conduction, q=4 pi k r1 r2 (T1-T2)/(r2-r1) = 235 W.

    In the second calculation, you no longer set this condition. Instead, you arrive at T2=253.884 K, and you set T1=254.041 K . Now if we calculate conduction, we get q=4 pi k r1 r2 (T1-T2)/(r2-r1) = 117.5 W. This is exactly half as much as before. But the nuclear reactor is still producing 235 W.

    So the reactor is still producing 235 W, but only 117.5 W is escaping from radius r1 by conduction. Energy balance is not satisfied.

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  9. Zoe, just to make sure we’re on the same page: which of these points do you agree with?

    1) the reactor produces a constant output power, regardless of the environment, as you have just said. This is 235 W

    2) in your second calculation, only 117.5 W leave the reactor through conduction

    3) the reactor is producing more energy than it is getting rid of, so it must be getting hotter

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    1. As I said, the temperature goes up by 0.158K.

      The reactor is not producing more energy than it is getting rid of, it is producing the same energy that it is not getting rid of a little bit more.

      You have a different solution? What is it?

      Like

    2. Oh I see what you’re saying. It’s simple: conduction is a loss. Your nuclear reactor is split between powering the local matter and emitting to space.

      Like

  10. You didn’t treat conduction as a loss in the first calculation (no shell), so why do so in the second?

    In my solution to the second calculation, I have almost the same equations as you, but I do exactly what you did in the first calculation and set the energy produced by the reactor equal to the conduction away. Then your first line looks like

    235 W=4 pi k r1 r2 (T1-T2)/(r2-r1) = 4 pi r2^2 sigma (T2^4-T3^4) = 4 pi r2^2 sigma T3^4

    This leads to (using your numbers so 4 pi r2^2=1 m^2)

    T3 = (235 W/m^2/sigma)^.25
    T2 = 2^.25 T3
    T1 = T2 + (235 W) (r2-r1) / (4 pi k r1 r2)

    From here just plug numbers into the first equation, to get T3 = 253.73 K

    Plug this into the second equation to get T2 = 301.73 K

    Plug this into the first equation to get T1 = 302.05 K

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    1. If you’re that clever why did you fall for my trick?

      I was describing a real world nuclear reactor, but you believed my false description of a fake reactor.

      https://www.nuclear-power.com/nuclear-power-plant/nuclear-reactor/

      “Almost all of the current reactors built to date use thermal neutrons to sustain the chain reaction. These reactors contain neutron moderator that slows neutrons from fission until their kinetic energy is more or less in thermal equilibrium with the atoms (E < 1 eV) in the system."

      In real world reactors, the flow is actually cut off to obtain equibrium temperature of environment.

      The first equations tell what the operators would set as that equilbrium temperature to.

      O

      The other type of reactor is a fast reactor. In this scenario the flow increases.

      So you see, the trick is that Willis used a power source that doesn't actually produce a constant flux 🙂

      Then I distracted you with the battery example. But that would actually work as you said. But you didn't catch on.

      I like to that you discovered that I'm a tricky gal. There are a few other little physics/description tricks on this site.

      I never however manipulate data. That is 100% as is. If I made a calculation mistake, then it's a mistake. I provide the code to catch it. I don't play around in this area.

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  11. Nice trick. My calculation used a constant power source, because that was the original problem statement from Willis, and more importantly, that is a more analogous situation to the earth. The amount of sunlight incident on the earth does not change based on the earth’s temperature.

    But if you are assuming the power source adjusts its power to maintain a constant temperature, regardless of the environment, of course the result is that you will get a constant temperature. In that case your calculation may be more appropriate.

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    1. Yes, correct. Willis stated a constant power source, but used a physically unconstant power source.

      T1 is constant, but T2 changes, with a normal nuclear reactor.

      I’m not assuming discarding the environment. The environment IS the inner wall of steel housing. The designers made the device work as initial equations show.

      I’m glad you took the bait. I like discussions. Thank you very much!

      Like

    2. Always good to have a thoughtful discussion, thanks.

      I think the next natural question is: tricks aside, let’s say we actually have a constant power source, for example a heating resistor with controlled P=I*V. Would you then agree with my solution?

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      1. Yes, it’s correct. Although I don’t fully understand the electrical ramifications of having two metals separated by an infinitely tiny gap. Probably a complication there. But otherwise, yes.

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  12. I might be confused, but isn’t Q measured in Watts, and the flux of 235 is W/m^2? So for calculating T1, shouldn’t you be setting the left hand side of the equation equal to 235 x the surface area of the outer sphere? And would this not significantly increase T1?

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  13. Actually, if you look closely at this supposedly scientifically accurate animation, the temperature at the point of emission stays the same, which implies unchanging emissivity??

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    1. The basic one layer model warm-up formula is:

      S(1-a)/(1-o/2)

      Where S is ~340, a = 0.3, and o is opacity (1 minus emissivity) = 0.78

      So the sun’s 340 W/m^2 becomes ~390 at the surface.

      If opacity is increased from 0.78 to 0.81 (emissivity decreased by 0.03)

      The result becomes 400 W/m^2 at the surface.

      Well, that’s the basic textbook theory anyway.

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  14. All you’ve done is succeeded in showing that you did not understand the problem set up by Willis. The internal source is intended to supply a constant energy flux. Willis’ solution to the problem is correct.

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      1. That’s not *his* power source. That’s YOUR straw man. Did you even read what he wrote?

        “For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square metre of surface area. How is it heated from the interior? Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.”

        He tells you to heat the interior in any manner that you like so long as it supplies a constant 235 watts per square metre of surface area.

        You didn’t do that.

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        1. Exactly. His power supply is a unicorn. I use a real power supply.

          He needs the output of a brown dwarf. But he doesn’t use a brown dwarf:

          “For our thought experiment, imagine a planet the size of the Earth”

          He needs something at least 10x the size of Jupiter.

          Like

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