Does a Greenhouse Operate through the Greenhouse Effect?

August 11th, 2013 by Roy W. Spencer, Ph. D.

real-greenhouse
One of the oft-cited objections to the term “greenhouse effect” is that it is a misnomer, that a real greenhouse (you know, the kind you grow plants in) doesn’t work by inhibiting infrared energy loss. It is usually claimed that a real greenhouse works by inhibiting convective heat loss by trapping the sun-heated air inside.

While working on a new website devoted to answering greenhouse questions, I decided to examine this issue. What piqued my interest was a couple quick back-of-the-envelope calculations that (1) for a glass covered greenhouse, the downward infrared (IR) emission from the roof should be about 100 W/m2 more than from a clear sky (a pane of glass is high emissivity, and opaque to broadband infrared), and (2) the realization that a greenhouse generates its own convection from the roof because the glass heats up, so convective air currents inside have their heat conducted (albeit inefficiently) through the glass, then the warm glass of the roof causes its own convection.

Add to this the fact that greenhouses are usually vented, which means they lose heat convectively anyway that by-passes the greenhouse structure.

So, the question is: Does a greenhouse work more from infrared heating (the “greenhouse effect”), or more from the inhibition of convective heat loss?

First, let’s examine some approximate energy fluxes for vegetation in the summertime. These are only rough estimates, and there are rather large variations in these depending on cloud cover, etc., and to be meaningful they need to represent a day+night average (infrared fluxes are orange; solar are yellow; convective are blue):
Greenhouse-effect-without-greenhouse
For simplicity, the calculated IR emission from solid surfaces assumes an emissivity of 1. The downward sky infrared is consistent with the BSRN network measurements during the warm season. Note that the energy fluxes have to sum to zero for temperature equilibrium, and we will ignore the photosynthetic storage of energy in plants which is very inefficient.

Now, with a greenhouse in place, we assume the average temperature of the interior rises, and that the glass roof reaches a temperature intermediate between the inside and outside air temperatures:
Greenhouse-effect-in-greenhouse

What really changes a lot is the downwelling IR, increasing from the sky value of 350 W/m2 to 450 W/m2, an increase of 100 W/m2. Convective heat generated (but temporarily “trapped”) within the greenhouse increases substantially, from 208 without the roof to 275 with the roof, for an increase of 67, which further heats the air, which in turn is helping to heat up the roof.

But notice that the convective heat loss by the greenhouse roof (200 W/m2, inferred as a residual) is only 8 W/m2 less than if the greenhouse was not there (208 W/m2). In contrast, the extra IR energy “input” (actually, reduced IR “loss”) is twelve times as large (100 W/m2) as the reduction in the convective loss (8 W/m2).

Of course, changing any of the assumed numbers will change the result. But, assuming I haven’t made a fundamental mistake, I think you would find that the “greenhouse effect” will consistently be larger than the convective inhibition effect.

So, maybe the greenhouse effect really does work like a real greenhouse. Again, the basic issue is this: replacing the downwelling sky radiation with a roof that is opaque to infrared (but still transparent to sunlight) represents a huge decrease in the IR energy loss by the vegetation, whereas the greenhouse roof still generates convective heat loss nearly as large as if the greenhouse wasn’t there.

I’m open to ideas, and better estimates of energy fluxes on this subject. The problem is actually surprisingly difficult one to think through. There are many energy fluxes involved (I haven’t even addressed energy losses out the side of the greenhouse) and the trick is to know which are the important ones and which ones can be ignored for the purposes of a rough estimate.

For example, the emissivity of glass is less than 1, but what that means is that it “traps” even more IR energy inside because it partly reflects the higher levels of IR the warmer vegetation is emitting upward.

What do the experts say about all this? I’m sure this problem has been analyzed before, probably in great detail, by multiple aggie graduates in their theses. Unfortunately, a Google search on “greenhouses” and “energy budget” is hopelessly cluttered with pages related to the Earth’s greenhouse effect (wow! how did that happen?)

If anyone is aware of studies done on the energy budget of greenhouses (of the agricultural kind), I would appreciate a reference or two. But until someone finds a serious error in the above analysis, I’d say we might need to admit that the “greenhouse effect” is pretty accurately named.

UPDATE: It appears the debate was brought up in the literature by R. Lee (“The Greenhouse Effect”, J. Appl. Meteorology, 1973) who has been referenced by many as showing a greenhouse does not work through the greenhouse effect, but he curiously admits the analogy “is correct only with respect to the glass, not with respect to the space enclosed”. Well, duh. That’s the point…the glass produces a greenhouse effect. In any event, his paper was refuted by Edwin Berry (Comments on “The Greenhouse Effect”, J. Appl. Meteorology, 1974) who showed several problems with Lee’s analysis.

So, I guess I’m left wondering…where did the oft-cited claim that a greenhouse does not operate through a greenhouse effect come from?


437 Responses to “Does a Greenhouse Operate through the Greenhouse Effect?”

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  1. Massimo PORZIO says:

    I’ve a question:

    You wrote “the glass roof reaches a temperature intermediate between the inside and outside air temperatures”

    Yes but, shouldn’t the external side be quite colder than the interior due to the glass thermal resistance?
    In that case the upward radiation could be far less than than the downward radiation, not the same as you supposed.

    Here in North Italy we have double glasses windows to increase the efficiency of the house thermal budget, but AFIK the thickness of the glasses plays an important role in thermal dispersion.

    I missed something?

    Have a nice day.

    Massimo

    • Yes, the inside of the glass will be slightly warmer than the outside. I don’t think this will change the fundamental conclusion, though.

    • James Goodone says:

      As for the glass roof allegedly additionally warming the inside on the picture above (450 W/m²), this is a fundamental mistake, sorry.

      The source of energy on the picture above is the Sun with the power of 300 W/m², that means each m² receives 300 Joules every second. This is the only supply to the system “greenhouse”. That means that all the recipients TOGETHER, surface, glass, plants, workers and whatever inside receive those 300 Joules times the area of the section every second. Together. Nothing more has been delivered to the system.

      For example, if a worker takes something, then the surface under his feet does not get the sunlight, because he covers this part of the surface with his feet.

      Regardless of how those 300 Joules per 1 m² section area are distributed, the sum is always 300, it can not exceed 300.

      • Please, James, that is the same old tired slayer argument.

        Yes, the ultimate *source* of energy is the sun. But if the total *downward* energy flux was only 300 W/m2, then the temperature would never rise above about 35 deg. F (assuming an infrared emissivity of 0.9)…even colder if you assume convective losses.

        Downward energy fluxes from the sky of over 300 W/m2 are routinely measured by the BSRN network…for instance, here are near-real time measurements in the Netherlands:
        http://www.knmi.nl/bsrn/

        How do you explain such measurements? A HAARP-chemtrail conspiracy?

        Energy-in does NOT determine temperature. Temperature represents a balance between energy input and energy loss, and the glass inhibits the loss of IR by emitting back to the surface close to what was emitted from the surface to the glass.

        • James Goodone says:

          Well, about conspiracy or not, let’s put it aside. Anyway, such a measurement must be wrong, exactly like if someone claimed he measured 2+2 and got 5.

          As for the supposed 300 W/m2 flux from the Sun, if it can only heat the ground up to 35 deg. F but in fact the ground is warmer, then the given number 300 is false.

          I mean, you can not get around the math by referring to allegedly correct measurements. This is a contradiction and math beats the claimed measurements. Math wins.

          Back to our problem, let us simplify that and say the section is exactly 1 m² and consider the period of time of only 1 second. So, the Sun has sent 300 Joules to the absolutely cold (0K) system. Now tell me, how it is mathematically possible, that after this 1 second there is more than 300 Joules in the system.

          Whatever each of the objects in the system has got, the sum can not exceed 300. There is simply nothing more there.

          • Steve Reddish says:

            James Goodone says “I mean, you can not get around the math by referring to allegedly correct measurements. This is a contradiction and math beats the claimed measurements. Math wins.”
            Then you proceed to use 300 W/m2 as the energy flux input to the greenhouse. Do you not realize the solar input was determined by actual measurement of the Sun’s effect on the temperature of a receiver of a certain size and mass? This measurement was then used to mathematically calculate the solar flux. Measurements provide the data for the calculations. Measurements are also used to confirm results of calculations. Math doesn’t win when it contradicts actual measurement. In that case the numbers input into the math calculation must be wrong. So, if the ground is warmer than 300 W/m2 could make it, then there must be some additional source of heat energy.

            2ndly, to prove your point you make the stipulation that a temperature measurement be taken 1 second after the Sun begins to shine on your test section. That you insist that the measurement be taken immediately means you must be aware that the temperature would rise as time passes, until equilibrium is reached. How does this rise in temperature with passing time happen, if not by the accumulation of some of the sun’s energy? Surely the level of the accumulated energy is affected by energy lost from the system. The 450 W/m2 is the heat not lost – trapped by reflection from the greenhouse grass, which would otherwise have immediately radiated away from the system. As far as the ground is concerned, the IR reflected downward by the glass is in addition to the energy coming directly from the Sun.

            300 W/m2 and 450 W/m2 are rates of energy flux. Temperature is a measurement of the level of the accumulated energy over time. Allow enough time for equilibrium to be reached. Rate of energy loss must also be figured into your math if you expect to get the correct answer for resulting temperature of your test section. Don’t forget to actually measure its temperature to check your math. Successful prediction is the proof of a hypothesis.

          • James Goodone says:

            Steve Reddish says August 17, 2013 at 10:06 PM
            “So, if the ground is warmer than 300 W/m2 could make it, then there must be some additional source of heat energy. …
            to prove your point you make the stipulation that a temperature measurement be taken 1 second after the Sun begins to shine on your test section. That you insist that the measurement be taken immediately means you must be aware that the temperature would rise as time passes, until equilibrium is reached.”

            Formally, there are 2 possibilities: either 300 is a false number or indeed there must be some additional energy supply to the system.

            Since we know that there is no second invisible sun or anything like that, the only possibility would be 300 is a false number.

            If you mean back radiation can be treated as an additional energy supply, then no, it can not, because this assumption is equivalent to creating energy out of nothing.

            As for equilibrium, let’s say it is reached after 1 second. Or, if you wish, after N seconds. This does not change the calculation: after N seconds the surface in the picture above would get (300+450)xN = 750N Joules altogether, although the Sun has delivered only 300N. 250N Joules have been created out of nothing in the picture.

          • Steve Reddish says:

            James Goodone says
            “If you mean back radiation can be treated as an additional energy supply, then no, it can not, because this assumption is equivalent to creating energy out of nothing.”

            The 300 W/m2 is a continuous input to the ground, at least during most of the day. The temperature of the ground rises as it receives input from the sun. As the ground’s temperature rises, the amount of IR energy it emits rises. If all IR emitted by the ground escapes, the ground’s temperature stabilizes at the temperature where energy emitted (and conducted) away balance the energy being received from the Sun only.
            If some of the IR emitted by the ground is subsequently reflected back to the same ground by a pane of IR reflecting glass placed nearby, the ground’s temperature will then rise until it is emitting IR at a rate to balance the energy input from the Sun plus the reflected IR. This additional IR is not created “out of Nothing”, it is emitted by ground that has been heated by energy received from the sun earlier that day.
            You see, James, your math is failing to match the real world because you did not account for the solar energy that is stored in the ground as heat. You were assuming the ground’s temperature is a product of only the current direct solar input.

          • James Goodone says:

            Steve Reddish says August 19, 2013 at 12:31 AM
            “If some of the IR emitted by the ground is subsequently reflected back to the same ground by a pane of IR reflecting glass placed nearby, the ground’s temperature will then rise until it is emitting IR at a rate to balance the energy input from the Sun plus the reflected IR. This additional IR is not created “out of Nothing”, it is emitted by ground that has been heated by energy received from the sun earlier that day.”

            ============================

            Steve, I perfectly understand the idea of back radiation warming also known as “greenhouse effect”. It is exactly what we see in the Dr.Spencer’s picture above.

            The problem of the concept is that if you start counting Joules which is energy, you will see immediately that such an effect is mathematically impossible.

            To make it easier, forget any storage and start with the equilibrium as in the picture above. What we see is that for any given time N a part of the System receives more Joules than the whole system has received. 750N Joules is more than 300N Joules. That means that 450N joules has been created out of nothing. Energy has been created out of nothing, Steve. No problem with that?

            In other words, you can not distribute 300 whatever so that someone gets 750. So, the process we see in the picture is not a part the real world.

          • JG, if you as a body at 300 K are sitting next to a block of dry ice at say 200 K and you interpose a block of copper at say 250 K between you and the dry ice, the copper will warm you even though it’s colder than you. This is because it decreases the amount of heat you’re losing to the dry ice.

            Your reasoning ignores this possibility.

          • James Goodone says:

            Vaughan Pratt says August 21, 2013 at 3:51 AM
            “JG, if you as a body at 300 K are sitting next to a block of dry ice at say 200 K and you interpose a block of copper at say 250 K between you and the dry ice, the copper will warm you even though it’s colder than you. …”

            The notion “cold warms hot” is false. It becomes obvious, if you start counting energy. Maybe you did not understand the calculation above, I do not know, must be simple. Anyway, “cold warms hot” always leads to creation of energy out of nothing. I guess you understand that this is impossible.

        • James Goodone says:

          Roy W. Spencer, Ph. D. says:
          August 12, 2013 at 11:01 AM
          “Energy-in does NOT determine temperature. Temperature represents a balance between energy input and energy loss, and the glass inhibits the loss of IR by emitting back to the surface close to what was emitted from the surface to the glass.”

          Roy, let us do some math then and check if it is possible.

          Let us consider 10 seconds period of time and again assume 1 m² section. The system (your greenhouse) got 10×300=3000 Joules of energy from the Sun, no other energy supply to the system being available. According to your picture, the surface got (300+450)x10=7500 Joules. This is clearly more, then 3000 that came into the system. The same goes for any period of time.

          Your setup simply can not work this way.

          It would be logical if you asked yourself where exactly you made a mistake, but first you need to accept the fact that simple calculation proves your version impossible.

          • @JG: you need to accept the fact that simple calculation proves your version impossible.

            Simple correct calculation would do that, but your calculation is not correct, as I explained above.

          • James Goodone says:

            Vaughan Pratt says August 21, 2013 at 3:53 AM
            “…your calculation is not correct, as I explained above.”

            No, you did not deal with my calculation at all, you just expressed the idea “cold warms hot”.

        • Ned Nikolov says:

          Roy,

          You are right – the down-welling atmospheric infrared radiation is 346 W m-2 as a global average, while the solar radiation absorbed by the entire Earth-atmosphere system is only 239 W m-2. Where does the additional flux of 107 W m-2 come from (remember that the air has a very small heat storage capacity)? … In your greenhouse example above, you have 350 W m-2 coming down from the free atmosphere towards the greenhouse roof. However, you do not have such an extra long-wave flux heating the Earth atmosphere from Space! So, the question is really how the absorbed 239 W m-2 by the Earth system becomes 346 W m-2 in the lower atmosphere. Radiative transfer alone cannot explain this!

          Do an experiment with a multi-layered RT model using a representative vertical distribution of the IR absorbers in a standard atmosphere as input (NCAR has some sample data sets). Start with a uniform vertical temperature profile (zero laps rate) and spin the model until downward and upward ratiative fluxes equilibrate and create what is known as the radiative equilibrium temperature profile. Examine the calculated downward IR flux just above the surface. You will find out that it is always less or equal to the total absorbed shortwave flux, but never larger than the latter! In other words, the observed extra 107 W m-2 cannot be traced down to the Sun and are not the results of transformation of solar energy through IR absorption and re-radiation by the atmosphere … How can you explain this?

      • Leonard Weinstein says:

        Roy,
        That is not the slayer argument. Their argument (at least the ones I argued with) seem to be that there is no absorption of the back radiation. I have been arguing with them a lot, and their claims are wrong. However, it is true the only NET energy to the ground (average) is 300 W/m^2. You are correct that back radiation slows net radiation loss and thus results in the increase in temperature, and thus increase in convected loss to make the difference, but that is not the same as back radiation as a heat transfer source, it is a resistance technically (otherwise you violate the 2nd law). I think the difference in our position is just how you define terms.

        I ran across a blog a while back (I don’t have it available) that claimed a study was made on an enclosed a volume with an opaque top and a second comparable volume with a clear glass top, and they both came to nearly the same internal and ground temperature. I do not know if that is correct, but it implies that trapped convection is the dominate mechanism. If there is no ventilation to the exterior, that is approximately true, but I am sure there is some slight difference to the two cases.

        • James Goodone says:

          Leonard Weinstein says:
          August 12, 2013 at 5:57 PM
          “However, it is true the only NET energy to the ground (average) is 300 W/m^2. You are correct that back radiation slows net radiation loss and thus results in the increase in temperature, … that is not the same as back radiation as a heat transfer source, … (otherwise you violate the 2nd law)”

          Leonard, I can smell a contradiction in your argumentation.

          If you say NET energy to the ground (average) is 300 and the temperature is higher than those 300 can cause, then sorry, but this is nonsense.

          To get a higher temperature you need additional energy. So, back radiation must deliver this additional energy. But then you are saying this would be according to your own words a violation of the 2nd law…

          Leonard, you can not have back radiation warm something and not transfer energy to it at the same time. This is a second contradiction.

          • Interposing a 250 K object between a 300 K object and a 200 K object warms the 300 K object while cooling the 200 K object. How does this contradict the 2nd law of thermodynamics?

          • James Goodone says:

            Vaughan Pratt says August 21, 2013 at 3:57 AM
            “Interposing a 250 K object between a 300 K object and a 200 K object warms the 300 K object while cooling the 200 K object. How does this contradict the 2nd law of thermodynamics?”

            I did not refer to the 2nd law of thermodynamics, it was Leonard Weinstein.

            My argumentation is that your “cold warms hot” is equivalent to creation of energy out of nothing and therefore false.

  2. Svend Ferdinandsen says:

    The problem is far more complicated like in the real outside “greenhouse”. You might have clouds and the balances are quite different between night and day.
    If the greenhouse is difficult to calculate, then imagine the real world.

  3. Ball4 says:

    There are some ref.s in the literature in this link which covers your subject very well in prose. If long link doesn’t work, google “delicate shade of purple while sputtering”.

    Basically of course, the answer to your question about how a farmer’s greenhouse works is “It depends” on geometry, location et. al. whether radiation or convection is the driver.

    http://books.google.com/books?id=CZuNCZqtZZUC&pg=PA83&lpg=PA83&dq=%22The+atmospheric+science+community+seems+to+be+divided%22&source=bl&ots=97bJUeuBNi&sig=0eRgpu4IaYOuEH0ebNlTIDVfzaY&hl=en&ei=cfFLTo-QIM_SiALujf2GAQ&sa=X&oi=book_result&ct=result&resnum=2&sqi=2&ved=0CB0Q6AEwAQ#v=onepage&q=%22The%20atmospheric%20science%20community%20seems%20to%20be%20divided%22&f=false

  4. seeker says:

    I think the energy budget/temperature analysis would be quite simple.

    At the most basic, a greenhouse works by trapping hot air, as you said. The air inside gets heated from solar heating of the surfaces inside the greenhouse.

    So now, if there is additional heating coming from the internal IR, the IR from the roof or whatever, then your premise is that this should cause more heating. More heating meaning higher temperature than the solar input.

    So, if the solar input/absorption internally is known, you can calculate the expected maximum temperature. If you add in IR from the roof or wherever, then a temperature higher than that should be expected. The math is quite simple really. Of course, emissivity is a different issue than IR amplification.

    • Roy Spencer says:

      “So, if the solar input/absorption internally is known, you can calculate the expected maximum temperature.”

      No, energy input alone does not determine temperature…temperatures change in response to imbalances between energy input and energy output.

      I’ve covered this repeatedly. For example:
      http://www.drroyspencer.com/2009/04/a-global-warming-cookbook-what-causes-temperature-to-change/

      • seeker says:

        Roy said: “No, energy input alone does not determine temperature”

        Roy said: “temperatures change in response to imbalances between energy input and energy output”

        So then, calculate the energy inputs and outputs and see if the temperature inside a greenhouse can exceed the maximum temperature of the solar input, properly accounting for emissivity. The reason to calculate the input/absorption and related temperature is so that you know it, so that you can test it.

        • Tim Folkerts says:

          seeker says: “…see if the temperature inside a greenhouse can exceed the maximum temperature of the solar input”.

          You are now adding a second misconception to the one Roy just addressed. The “temperature” of thermal radiation is the same as the temperature of the source that created the radiation.
          * a 300 K blackbody (eg a typical object on earth) creates a spectrum of photons this is characteristic of a 300 K object = “300 K photons”.
          * a 5780 K blackbody (eg the sun) creates a spectrum of photons this is characteristic of a 5780 K object = “5780 K photons”.

          We are NOT concerned about exceeding the “maximum temperature of the solar input” = 5780 K. The 2nd law of thermodynamics simply tells us that there is no way (eg no combination of mirrors or lenses or IR opaque glass) to warm anything above 5780 K.

          [The “temperature of sunlight” is NOT 255 K or 394 K or some number in between. This number is the equilibrium temperature of some specific OBJECT under some specific energy balance scenario involving 5780 K sunlight and ~ 300 K “earthlight” and 3 K “spacelight”.]

          • Gary Hladik says:

            “The 2nd law of thermodynamics simply tells us that there is no way (eg no combination of mirrors or lenses or IR opaque glass) to warm anything above 5780 K.”

            You’re kidding, right? At 5780 K an object radiates about 63 million watts per square meter, or about 63,000 times the sunlight falling on 1 square meter (sun directly overhead) or 63,000 “suns”. Now I understand practical considerations limit sunlight concentration to a bit over 20,000 suns

            http://www.nrel.gov/docs/legosti/fy97/23377.pdf

            but that’s an optical limitation, not the 2nd Law. I’m guessing that a solar furnace at Mercury’s orbit, where sunlight is about nine times as intense, could exceed 5780 K (assuming we had a target that could take the heat).

          • Leonard Weinstein says:

            This reply is to Gary Hladik.
            NO.
            The Sun is an extended size object. The closer you are to it, the larger it appears, and the larger it’s image will be. You can collect more power with a given sized collector if you are closer, but the power per area in the image does not increase, since the image size also increases, and you just can’t squeeze it to a higher power density.

          • Tim Folkerts says:

            Gary says: “You’re kidding, right?”
            No I am not kidding.

            Gary says: “but that’s an optical limitation, not the 2nd Law. ”
            No, that is BOTH. if you completely surround an object with a 5780 radiator, the object will equilibrate at 5780 K. A furnace on Mercury would still be limited to 5780 K. A furnace at the surface of the photosphere would STILL be limited to 5780 K.

            You are proposing that heat could go spontaneously from from a “cool” 5780 K object (the sun’s surface) to a warmer object (the solar furnace that is above 5780 K). This is a CLEAR violation of the 2nd Law.

        • Gary Hladik says:

          Tim,

          Focus the light from 1 square meter at 5780 K onto one side of a thin disk with area 1/2 square meter. Focus 1 square meter of another object at 5780 K onto the other side. All emissivities are 1. Total power to disk is about 126 million watts, total radiating area = 1 square meter, disk temperature is about 6877 K.

          BTW, using non-imaging optics, apparently we can get sunlight brighter than the surface of the sun here on Earth:

          http://www.readcube.com/articles/10.1038/346802a0?locale=en

          • Tim Folkerts says:

            Gary says: “Focus the light from 1 square meter at 5780 K onto one side of a thin disk with area 1/2 square meter. “
            I challenge you to describe a way to actually do this.

            For instance, if the two surfaces are squares, you could try to build a “square funnel” (or “square frustum”) using perfect mirrors. This would prevent any light from escaping out to space. But a very quick look shows that this would reflect some of the light back to the “emitting” surface, rather than to the “receiving” surface. Without doing the integral, I am 100% convinced that only 1/2 of the emitted light will reach the second surface.

            It is not as simple as you assumed.

            ******************************

            The article you link to is interesting! I will have to think more about that. Somehow they are violating my understanding of the 2nd Law (or they are not doing what they claim).

          • Gary Hladik says:

            “I challenge you to describe a way to actually do this.”

            Well, the article I linked to apparently found a way to do far more. For our case, if a solar furnace can focus the sun >20,000 times, I figure we can focus a light source 2X. šŸ™‚

            Note that we could make the case more realistic by assuming the sources are a more down-to-earth temperature, say 1000 K.

            BTW, isn’t this much like the case of the dichroic light bulb, in which the filament heats itself by its own reflected infrared radiation? This bulb would be a case of “spontaneous heat flow” (via radiation) between objects of the same temperature.

          • Tim Folkerts says:

            Gary, you are comparing apples and oranges here.

            You can double the incoming solar radiation with a single plane mirror. You can triple it with another plane mirror.

            The sun only covers about 1/100,000 of the hemisphere of the sky, so you can easily increase the input 1,000x with a lens or parabolic mirror. But there should be no way to do any better than to focus in sunlight from every direction. Thus there should be no way to warm anything above 5780 K with sunlight no matter how clever you are with lenses & mirrors.

            ******************************************

            NOTE 1: This is intimately connected to why “focusing the 350 W/m^2 of back radiation” cannot work. In this case, the thermal IR is ALREADY coming from every direction. If you add a mirror, you are adding some IR by blocking some OTHER IR.

            NOTE 2: This is why their paper was so surprising to me. They are seemingly violating the 2nd Law.

          • Gary Hladik says:

            Tim,

            You’re perfectly correct about the difficulties with imaging optics. As I mentioned before, the best solar furnace I’ve found online produces only a bit over 20,000 “suns”, and as Leonard Weinstein pointed out, imaging optics can’t do much better. The paper I referenced, however, was one of several by that group claiming much higher amplification using non-imaging optics.

            I think we’ve been looking at this the wrong way. Sunlight has roughly the spectrum of a black body at 5780 K (about 63 million watts per square meter), but its “temperature” at Earth’s surface is only about 1,000 watts per square meter, tops. If sunlight’s “temperature” were actually 5780 K, a piece of paper exposed to sunlight would burst into flame. Note that I then routinely “violate the 2nd Law” whenever I use my pocket magnifier to start a fire or scorch an ant. šŸ™‚

            Likewise, the all-infrared light from a carbon dioxide laser would seem to come from a “cool” source, yet the “temperature” of the light in watts per square meter can be high enough to cut steel. Focusing the light of multiple lasers on a tiny target can produce temperatures high enough to initiate nuclear fusion, even though none of the lasers themselves reach that temperature. Maybe it would help to think of Cooke et al’s “non-imaging optics” as a kind of non-laser laser. šŸ™‚

          • Gary Hladik says:

            Tim,

            Long story short, I’ve been reading up on the subject and thinking about what you’ve written, and I’m now pretty sure you’re right: you can’t focus the light from a hot object onto a smaller target such that the target becomes hotter than the source. If you’re interested the final straw was a comment here

            http://www.halfbakery.com/idea/Portable_20solar_20concentrator_20with_20high_20concentration

            by “redmun” June 25 2006.

            Non-imaging optics complicate the situation, but I suspect one can only come closer to the same temp as the source without exceeding it. Thanks for taking the time to discuss.

            BTW, I think the “oranges” case (multiple sources) runs into he same problems.

  5. Scott Scarborough says:

    The story I heard is that about 100 years ago experiments were run with glass and crystallized salt sheets (salt transmits all wavelengths involved in the problem). A box with the salt sheet over it heated up more than a box with a sheet of glass over it when exposed to the sun. That would suggest that the green house effect would be primarily due to the blocking of convection and the salt sheet box would get hotter because slightly more energy was allowed in from the sun.

  6. Mike Flynn says:

    I believe that Professor Woods in 1909 demonstrated by experiment that the “greenhouse effect” was the result of reducing convective “heat loss”.

    Recently, (2011), Professor Nasif Nahle confirmed Woods’ results.

    My own experiments (quite some years ago), seemed to show that replacing a near vacuum with CO2 results in a cooling of a body receiving its energy from a radiant heat source, rather than heating.

    I assume this is due to energy loss as the CO2 absorbs some radiant energy, and then reemits it in all directions. This, of course, results in the target receiving less energy than is the case where there is no impediment to the photons emitted by the source on their way to the target.

    Dr Spencer, I am a little curious. You say that the glass roof is opaque to infrared, so how does the incoming IR from the Sun penetrate the glass? If you say that the the IR is absorbed by the opaque glass, and then is reradiated in all directions, I point out that this occurs with IR from any source, internal or external. It results in less IR reaching the target than without the glass.

    My point is that the glass heats up in the Sun, as it absorbs energy. It then then has no choice but to radiate this additional energy away, until it becomes the same temperature as its surroundings. If you place your hand (or a thermometer, even better), against a car window exposed to the Sun, you will notice that it is much hotter than the ambient temperature in your air conditioned car.

    I guess it’s a question of whether you want to verify “thought experiments” with real ones.

    You will find you can drop a real greenhouse (hot house) temperature really fast, by opening the peak vents (at the apex) a little bit, and lifting the sides at the base a little bit. That’s what they do commercially when the temperature gets too high.

    Live well and prosper,

    Mike Flynn.

    • James Goodone says:

      The article by Wood can be easily found on the web: http://is.gd/exp1909

    • James Goodone says:

      According to the experiment by Wood the trapped radiation produces no effect.

    • Roy Spencer says:

      The glass probably would absorb the small fraction of the solar radiation that is in the infrared, which will raise its temperature (during the day, not the night). This means the glass will be at a higher temperature and will emit even MORE infrared toward the inside of the greenhouse.

      The result? Even a stronger greenhouse effect than I was assuming.

      Yes, you can cool the inside of the greenhouse rapidly by evacuating the warm air. But that does not help explain why the air warmed in the first place.

      • seeker says:

        “But that does not help explain why the air warmed in the first place.”

        The air warms in the first place because solar input heats the internal surfaces…

      • Phillip Bratby says:

        Surely it’s obvious. You only have to feel the surfaces to know the sun is heating them. You then only have to look and you can see the convective currents risng from the surfaces. I can’t comprehend why you cannot understand this heating process.

    • Tim Folkerts says:

      Mike says: “You say that the glass roof is opaque to infrared, so how does the incoming IR from the Sun penetrate the glass?”

      You have to remember that “infrared” covers a huge range — from 0.7 um to 1,000 um. The solar IR is almost entirely from 0.7 – 4 um, while the thermal IR from the earth, sky, etc is mostly from 4-50 um. Thus it is quite possible that a material can transmit solar IR but absorb thermal IR. (Typical window glass has such characteristics. http://www.astro.virginia.edu/~mfs4n/ir/glass.jpg)

    • Mike Flynn says:

      Tim Folkerts,

      This is what Dr Spencer said ” . . . (a pane of glass is high emissivity, and opaque to broadband infrared) . . .”

      I foolishly assumed he meant what he said, in the absence of any specific frequency range.

      You will note I asked Dr Spencer a question based on his statement. I prefer discussions of his nature to be more specific in terms of wavelength (rather than vague terms such as “solar IR”, “thermal IR”, “broadband IR”, and so forth.

      If you choose to believe that placing a sheet of glass over something on the ground (magnifying glasses excepted, obviously!), will cause it to increase its temperature, then good for you.

      Dr Spencer’s peculiar question asking how the interior of the greenhouse warms, was, I assume, a form of sarcasm which I failed to appreciate. I can only refer the good Doctor to Feynman’s discussions of the interaction between light (as used by Feynman – includes the entire radiation spectrum).

      In any case, I am always surprised that supposedly reputable scientists demonstrate such an aversion to experiment. A couple of commonly available and cheap glass aquaria, a couple of temperature sensors with remote readouts, a sheet or two of the appropriate IR transparent/opaque materials, and there you go!

      As Feyman observed, you can’t fool Nature. Maybe you can prove him wrong.

      Live well and prosper,

      Mike Flynn

      • steveta_uk says:

        If you choose to believe that placing a sheet of glass over something on the ground … will cause it to increase its temperature, then good for you.

        What an odd thing to day. Why would you not beleive it? I know from direct experience when working on windows frames that when a glass window was left for some time in my lawn, the grass below got much hotter, and condensation formed on the lower side of the glass from water leaving the heated soil below.

        If you doubt this, why not simply try it? This is basic science – if you doubt it, go outside today and try it!

        • Massimo PORZIO says:

          You missed Mike point.

          He was highlighting that it isn’t the glass itself which warm the below, in fact he added: “magnifying glasses excepted, obviously!”
          It’s the glass own temperature, which build up because of all the complex thermodynamic interactions inside the greenhouse.

          Have a nice day.

          Massimo

  7. Scott Basinger says:

    Now you’re thinking like an engineer.

  8. Mike Flynn says:

    Apologies to all. My mistake. The Professor’s name was Wood, not Woods.

    Thanks.

    Live well and prosper.

  9. D'Avila Tarcisio says:

    I’m open to ideas, and better estimates of energy fluxes on this subject. The problem is actually surprisingly difficult one to think through. There are many energy fluxes involved (I haven’t even addressed energy losses out the side of the greenhouse) and the trick is to know which are the important ones and which ones can be ignored for the purposes of a rough estimate.
    Roy S.

    The largest energy reserve of these greenhouses is located in water content of the atmosphere retained by the shape of the roof. In a normal atmosphere we find 1.6 g to 5.9 g of liquid water per kg of air, to form a cloud invisible to evaporate retains 11.28 kJ of latent heat while the air requires 1.0 kJ per degree centigrade.

    • D'Avila Tarcisio says:

      Ops.
      Must say per square meter.
      scuseme

    • D'Avila Tarcisio says:

      Dr. Roy
      An experiment to understand the greenhouse effect of the atmosphere and then to analyze the functioning of the “greenhouse”.
      Point the IR thermometer to space (Zenith) in times of radio probing 12Z and 00Z.
      Compare the indicated temperatures from the temperature at the intersection of isopleths state dew point at ground level with the line state temperature and find that the IR flux is equal to the flow temperature of the blackbody at the cloud base temperature multiplied by emissivity factor ranging between 0.72 and 0.99, depending on the cloud cover.
      0.72 for blue sky, 0.85 for gray clouds and 0.99 for dark clouds.
      This shows that the greenhouse effect is governed only by the amount of liquid water or ice in the atmosphere.
      Eg here in Manaus got variation of 301 w/m2 to 424 w/m2 between blue sky and dark cloud.

  10. dallas says:

    “So, maybe the greenhouse effect really does work like a real greenhouse. ”

    Pretty much. The biggest difference is that the atmosphere actually absorbs solar energy. So with a greenhouse, you also have surrounding “greenhouse” effect. If you consider the planetary boundary layer as the roof, you would have 330Wm-2 SW absorbed in the house and 150 Wm-2 absorbed by the roof resulting in a Day surface energy of ~ 405 Wm-2. At night, the surface energy would decay to 330Wm-2 if the house were fairly well insulated. The greater the thermal mass inside the house or the higher the insulation value, the greater the retained energy.

    You can also get condensation on both roofs that impact performance šŸ™‚

  11. Arfur Bryant says:

    Easy way to find out:

    If you think the radiative effect is greater than the anti-convection effect, then simply remove all the side walls/windows/doors from the greenhouse, just leave the roof in place – and then see how well the greenhouse works!

    Simples!

    • Tim Folkerts says:

      A variation of this experiment is done all the time using a structure known as a “carport”. It is common for cars under an unheated roof to not have frost in he morning, while nearby cars exposed to the sky will have frost. The difference is the downward radiation from the roof (which is greater than the downward radiation from the sky).

      So the effect DOES exist and is easily observed, even with the walls removed.

      This doesn’t, however, answer the question of how large the “radiation effect” is vs the “reduced convection effect”. That would require a much more detailed analysis.

      (Google “frost carport” for several explanations)

      • Kristian says:

        As always you get this wrong, Tim.

        You say: “The difference is the downward radiation from the roof (which is greater than the downward radiation from the sky).”

        No. The difference is the temperature of the roof compared to the temperature of the night sky. T_roof is higher than T_sky and so is closer to T_windshield. That means the temperature gradient is a fair bit less steep from the car up, so when it cools at night it does so a lot more slowly, because the heat flux away from it is reduced.

        But yes, radiation is indeed involved. The result remains the same.

        • James Goodone says:

          Kristian says:
          August 12, 2013 at 2:34 PM
          “But yes, radiation is indeed involved.”

          Kristian, could you please elaborate how exactly radiation is indeed involved? Many thanks in advance.

          • Kristian says:

            Sure. Like this:

            P/A = ε*σ* (T_car^4 – T_roof^4) as opposed to
            P/A = ε*σ* (T_car^4 – T_sky^4)

            As you know, the radiative cooling rate (P/A) of a warm
            object slows down (gets smaller) if the temperature difference to its surroundings gets smaller.

            This is not the only process at work here, though.

          • James Goodone says:

            Kristian says:
            August 14, 2013 at 3:40 AM
            “the radiative cooling rate (P/A) of a warm
            object slows down (gets smaller) if the temperature difference to its surroundings gets smaller.”

            Kristian, “gets smaller” and “slows down” are 2 different things. It is like driving with the same speed but a shorter distance or driving the same distance but with decreased speed. I am surprised that you mixed it up.

            Known formulas do not imply any decreased speed. I guess you mix up the Newton’s law of cooling about cooling an object BY the surrounding air (conduction and convection) and radiative cooling, (let’s say in vacuum) where surrounding objects have no influence on the SPEED of cooling. They have, of course, influence on the AMOUNT of cooling, that is on at what temperature the cooling stops.

            You “gets smaller” is correct, but your “slows down” is incorrect.

          • Kristian says:

            I feel we’ve had this conversation before … James.

            The rate of radiative energy loss (J/s) from the car’s windshield is smaller with the roof above than with the sky above. Because the roof is warmer than the sky. That’s directly derived from the S-B equation.

            This means that the windshield loses internal energy (cools) at a slower pace with roof than without.

            That’s all I’m saying …

          • Tim Folkerts says:

            I gotta smile ….

            ****************************

            TIM: The difference is the downward radiation from the roof (which is greater than the downward radiation from the sky).

            KRISTIAN: As always you get this wrong, Tim.

            JAMES: Could you please elaborate?

            KRISTIAN:
            P/A = ε*σ* (T_car^4 – T_roof^4) as opposed to
            P/A = ε*σ* (T_car^4 – T_sky^4)

            *****************************

            IE, Kristian explains that the difference is due to radiation of the roof vs the radiation of the sky, but that I was wrong when came to the same conclusion earlier.

          • Kristian says:

            Yes, I know that you don’t see the difference, Tim. I’ve known that for a long time.

            Makes me smile … Chuckle, in fact.

      • Tim Folkerts says:

        Kristian, as usual I find at least bits of your contributions that are good. We agree that “radiation is indeed involved”.

        It is not the temperature per se that matters (and definitely not the temperature gradient). The air temperature around the car is the same whether or not there is a roof overhead (since the air is free to circulate). So conduction will be the same; the only difference is the radiation.

        1) Without a roof, little radiation is coming down (from the cold air and from even colder outer space). So the net radiation is upward and relatively large.

        2) With a roof, more radiations is coming down. As you point out, the roof is the temperature of the ‘sort of cold’ air near the ground (warmer than higher air, and much warmer than space), so it will provide much more downward IR than in (1). Thus there will be a smaller net loss of IR, and slower cooling.

        Conduction and/or convection are not important, since they will be the same in both cases.

      • Arfur Bryant says:

        Tim,

        The radiative effect MIGHT exist sufficiently to be observed but, as you rightly pointed out later in your post, the question is whether or not the radiative effect is greater than the anti-convection effect. THAT was the point of my post. The lack of frost on the car COULD be due to the insulating effect of the roof. If you raised the roof by twenty feet, would there be frost because the radiative effect is no longer effective or because there is no insulation?

        As usual with these debates, the answer very often depends on more than one or two factors. Personally, I doubt that the radiative effect of greenhouse glass is greater than the anti-convective effect.

    • Bryan Woodsmall says:

      I don’t think this experiment would tell us anything. The walls/windows/doors do far more than produce an anti-convection effect. They stop the wind, which if allowed to blow through will continually replace the air before it has a chance to heat up (even at low wind speeds).

  12. Tim Folkerts says:

    Roy,

    I know you have an IR thermometer handy (and even a FLIR imager). I would suggest you collect some actual data from an actual greenhouse. Measure the temperature of the floor, the inside of the roof, the outside of the roof, the inside of the walls, the outside of the walls, the sky overhead. Measure air temp inside and out with a regular old thermometer. Ideally you could do this in a variety of conditions (sunny vs cloud; hot day vs cool day; vents open vs vents closed).

    Then you could draw a much better “energy budget of a greenhouse”, rather than have to “assume the average temperature of the interior rises, and that the glass roof reaches a temperature intermediate between the inside and outside air temperatures”.

  13. Fulco says:

    Noor van Andel did a nice experiment with different kind of greenhouses in his daugthers backyard. He used both IR transparent en IR opaque coverings in both dry and wet greenhouses.

    The greenhouse effect is in the covering not the type of covering.
    See yourself.

    you can find his results halfway at:
    http://klimaatgek.nl/wordpress/broeikastheorie/
    the diagram is called “simple greenhouse experiment”
    This is a Dutch site but you can have it translated to English, see the top section.

    • Roy Spencer says:

      Yes, it looks like his experiment does come up with close to 10 deg. C warming just due to the IR effect.

      Another good experimental setup, the best I have seen, comes up with an even bigger temperature increase…(and it refutes Wood’s original 1909 experiment, which was not very well set-up as far as I can tell):

      http://boole.stanford.edu/WoodExpt/

      • Fulco says:

        Roy,

        The best way to do this experiment is to dig holes into the ground and place the thermometer in it and cover the holes with different materials. If you place the boxes above the ground you need to correct for ambient heating. What strikes me most is that it does not matter much if you use an IR transparent or an IR opaque roof. The most important way to loose heat is bij convection and evaporation not by IR radiation. And as you remarked the IR warming only accounts for about 10 deg. C.

        Fulco

      • Massimo PORZIO says:

        Dr.Spencer,
        I would like to know your opinion about this research which instead complies with Prof.Wood experiment:

        http://www.biocab.org/Experiment_on_Greenhouses__Effect.pdf

        It looks well done to me and much detailed respect to the one you reported about Prof.Pratt.

        Have a nice day.

        Massimo

        • Good question, Massimo. I’ve only read through Stage 1 of Nahle’s experiment, and it is a reasonably good setup. I would have insulated all of the boxes so that the conductive heat losses out the sides would have been minimized, since in the real world there are no average heat losses in the horizontal direction on a global basis. But still, I’m surprised that Nahle didn’t see an effect. It’s interesting that Pratt’s results directly contradict Nahle’s results. I’d like to do my own experiment at some point. I would use nested Styrofoam coolers, but the rest of the setup would be similar to these other investigators. Luckily, the experimental setup is simple enough that many people could do it.

          • Ball4 says:

            I observe one has to carefully watch the peas in the Nahle tests but not in Pratt’s test.

            In first segment of the Nahle test, only the PE film box has the added white glass wool insulation. Apparently that is just the insulation needed to get the near box temperature up to roughly the same as in the no white glass wool boxes. No recording of wind speed is given.

            In segment 2, Nahle reduces the PE in near box to one thickness from two in segment 1 and removes the white wool insulation. In addition there is now a terminal wind speed 7.2kmph given, apparently these peas are needed to get roughly the same temperatures in the 3 boxes with the holed box apparently getting a little more breeze inside so now 3C cooler than unholed box for this segment.

            Still, there may be even more Nahle peas to watch for. Dr. Craig Bohren book clip did tell us watch for breezes and wall thicknesses (i.e. your POV influenced by thick walled GH in calm spots vs. thin walled GH in windy spots).

          • Mike Flynn says:

            Dr Spencer,

            I find it even more interesting that Pratt was unable to reproduce his initial results. The reasons are maybe a little embarrassing (he should be aware how to set up an experiment properly), but you can find them with little effort. Pratt has become silent on his promised follow up.

            Lord Kelvin went to his death with absolute faith in his calculation of the age of the Earth. He was wrong. He was still a brilliant man. Intellect and education are not necessarily protection against being wrong. Kelvin was at least convinced that his belief in the caloric theory was wrong, after seeing the results of Joule’s experimental work. It took him a few years, however. Such is belief.

            I know you are not going to take much notice. It doesn’t worry me – I’d rather be happy than right!

            Live well and prosper,

            Mike Flynn.

          • Massimo PORZIO says:

            I wondered too about Prof.Nahle’s choice of using the white wool insulation only on the PE box.
            It looks like a provocation to me, because he dealt with the white wool question in the second segment, where he considered the wind speed too.

            A critic I would make to Prof.Nahle is that he mixed too many changes a time in the various phase, but it could be a constraint due to the long time needed to run the experiments.

  14. Massimo PORZIO says:

    Hi Dr.Spencer,
    yesterday night I wrote about the glass thermal resistance and its possible efficiency in keep insulated the greenhouse from the external temperature.
    Indeed it seems that increasing the glass thickness is not so cheap since the thermal transmittance of a 4 mm thick glass is about 5.8W/(m^2*K) and increasing its thickness the thermal transmittance stabilizes to an asymptotic value of about 5.4W/(m^2*K) (the values are for vertical glass sheets, but the horizontal positioning shouldn’t change that value so much).
    Note that (if I’ve not misunderstood how it works, that i could be because I’m not skilled in this matter), this make a delta T of 51K for your outgoing flux of 300W/m^2 at the glass interfaces.
    Anyways, for buildings thermal efficiency computation, I found this online calculator:

    http://www.thermalcalconline.com/u-value-calculator/surface-resistance/surface-resistance.html

    For the “select surface type” use the last selection, which is “External surface (Heat flow any direction)”.
    Instead of thermal transmittance they use its inverse, that is the surface thermal resistance. For matching a standard 4mm glass sheet, set the surface emissivity to 34%.
    In that conditions without any winds @25°C you get the exact thermal resistance of a glass having the thermal conductance of 5.8W/(m^2*K) (0.17(K*m^2)/W.
    What I would like you note in that calculator is that, even if you add a little wind as 4m/s (14.4km/h) the thermal resistance falls below one third of the starting value.
    So with your 300W/m^2 flux, the effective delta T reduced from 51K to 15K.
    IMHO, this seems to make the thermodynamic path the great driver of the insulation in buildings, not the radiative one.

    As said, I could be wrong, I’m just an electronic engineer and may have misunderstood something.

    Have a nice day.

    Massimo

    • Massimo PORZIO says:

      Hoops!.
      I missed a piece in my previous post.
      If you change the emissivity from 100% to 0% the maximum change in the thermal resistance for a worst case temperature of 70°C is 0.08 to 0.25(K*m^2)/W that is one third, which is comparable to the very low wind discussed above.

      Again, have a nice day.

      Massimo

  15. Bert Walker says:

    Controlled Environment Systems ABE 483/583
    Dr. Gene A. Giacomelli
    Professor & Director Controlled Environment Agriculture Center Department of Agricultural and Biosystems Engineering Shantz Building,

    Lecture #7 Greenhouse Energy Balance

    At:

    http://ag.arizona.edu/ceac/sites/ag.arizona.edu.ceac/files/Lecture%20_7%20GH%20Energy%20Balance.doc.pdf

  16. I’ve added the following update to the above post:

    It appears the debate was brought up in the literature by R. Lee (“The Greenhouse Effect”, J. Appl. Meteorology, 1973) who has been referenced by many as showing a greenhouse does not work through the greenhouse effect, but he curiously admits the analogy “is correct only with respect to the glass, not with respect to the space enclosed”. Well, duh. That’s the point…the glass produces a greenhouse effect. In any event, his paper was refuted by Edwin Berry (Comments on “The Greenhouse Effect”, J. Appl. Meteorology, 1974) who showed several problems with Lee’s analysis.

    So, I guess I’m left wondering…where did the oft-cited claim that a greenhouse does not operate through a greenhouse effect come from?

    • Tim Folkerts says:

      I might have to try these experiments for myself. I have data collection hardward (but I should get some better/smaller thermocouples).

      Besides some of the experiments described above, some other options include
      *intentionally ventilating the walls of the containers.
      *intentionally ventilating the top (perhaps like this … http://images.reachsite.com/e9bd0308-9812-4bdc-b01c-f28bebdebbf1/media/312014/medium/312014.PNG?gen=1)
      * adjusting the emissivity & albedo of the containers.

      • Massimo PORZIO says:

        Hi Tim,
        I don’t believe that it is so much important to change the inner albedo, because the whole issue is about the LWIR radiative balance compared to the thermodynamic heat flow.

        If I did it, I would place a well known constant power supplied heather inside each box instead and I would keep them in darkness to avoid any influence of other energies coming from the environ.

        So you don’t have to wait for a particular sunny day with a specific humidity to allow your experiment replication.

        Have a nice day.

        Massimo

      • Tim Folkerts says:

        Massimo,

        Your set-up does have some distinct advantages — better control of the input in particular is quite useful. The biggest disadvantage that I can see is mostly “psychological”. Some people will dismiss the experiment because it doesn’t “seem” like it corresponds to sunlight heating the earth.

        • Massimo PORZIO says:

          Yeah. I know what you mean! :):):)

          Anyways let us know if you take the time to do something about it.

          Have a nice day.

          Massimo

  17. Bryan says:

    This is an interesting paper especially as it comes from a source with no “spin” on the AGW debate.

    The way I read the paper is it gives massive support for the conclusions of the famous Woods experiment.

    Basically the project was to find if it made any sense to add Infra Red absorbers to polyethylene plastic for use in agricultural plastic greenhouses.

    Polyethylene is IR transparent like the Rocksalt used in Woods Experiment.

    The addition of IR absorbers to the plastic made it equivalent to “glass”
    The experiment was carried out over three years.

    The results of the study show that( Page2 )

    …”IR blocking films may occasionally raise night temperatures” (by less than 1.5C) “the trend does not seem to be consistent over time”
    If the graphs are studied there is at times a small anti – greenhouse effect.
    At times the greenhouse temperature drops below the ambient outside temperature (an effect noticed elsewhere).

    http://www.hort.cornell.edu/hightunnel/about/research/general/penn_state_plastic_study.pdf

    • Interesting paper. But there appears to be no information on the infrared opacity of the films they are using. 6 mil polyethylene is pretty thick, and the manufacturer has no info online about how much IR “block” there is in their treated sheeting than their regular sheeting. In other words, there might not be much difference in the 2 types of sheeting in the infrared.

      Also, it appears that the solar transmission between the two sheets is also different, which adds another variable.

      This experiment seems to settle it for glass versus very thin polyethylene:

      http://boole.stanford.edu/WoodExpt/

      • Massimo PORZIO says:

        I’m not convinced that Prof.Pratt settled the glass vs the thin polyethylene issue with those experiments.
        I say that because of your very same explanation of how the glass should work in a greenhouse.
        You said that it warm at an average temperature between the outer and the inner temperature. This two temperatures are strictly dependant by the PE thermal resistance multiplied by the net power flux passing through it.
        Being PE not LWIR active, it doesn’t means that it is not a solid. So such an heated mirror which doesn’t emit any IR because of it’s very low emissivity, but which is anyways hot and heat its surrounding air, the PE sheet heats the air above and below according to its two surfaces temperature.

        IMHO, placing two PE sheet with a 6mm gap, it probably lead to the very same temperature as if Prof.Pratt moved the temperature sensor 6mm away from the first PE sheet (plus the the tiny insulation added by the second PE sheet of course).

        Have a nice day.

        Massimo

        • Yes, maybe you are right considering Nahle’s experimental results. More people need to do this experiment.

          • Massimo PORZIO says:

            I fully agree, that should be how science works.

            Have a nice day, thanks for this interesting thread.

            Massimo

        • Kristian says:

          I do find it disturbing (and odd) that Pratt covers the polyethylene box with a layer of 0.015mm thickness and with a thermal conductivity (k) of 0.42-0.51 W/(m*K) (PEHD), the glass box with a layer 159 times thicker and with k = 1 W/(m*K), and the acrylic box with a layer 635 times thicker and with k = 0.25 W/(m*K).

          How are those initial terms in any way equal?

      • Bill Hunter says:

        the flaw in Pratt’s experiment is the conductivity of the IR clear saran wrap is many times that of the glass covered “greenhouse effect” box.

        With the air convecting inside, the saran wrap box will lose heat several times faster than the glass box. Wood used rock salt which is similar to the conductivity of glass and got temperature results very much the same.

        My claim is conductivity controls the rate of cooling, not backradiation. So Pratt’s variation doesn’t surprise me in the least.

        I think the important thing to note is that objects in a radiation field with no atmosphere will either warm uniformly to the temperature of the radiation source as adjusted by the inverse square law, or the warm side will and a temperature gradient will exist in the object.

        If you put an excruciatingly hot ball of tar in a box the box will get hot but the tar is not going to get hotter.

        What confuses many is you can warm the surface of the tar by putting it in a box but only if you first allow a temperature gradient to form in the surface of the tar.

        Again its conductivity creating this temperature gradient and setting the rule for the transfer of heat from the core of the ball to the surface.

        It is clear backradiation warms nothing. the whole issue boils down to whether a 5800K sun 93 million miles away can warm something more than 278.5K

        As I see it, the earth at that temperature will be radiating 341 watts from the surface, so 341 watts going into the surface will not warm it. We get that result by the very same rule that slows radiation loss between objects.

        Now I could be wrong about that but I think somebody has to come up with a bit more than a thought experiment to support it. I would think Jozef Stefan would say no.

        • Tim Folkerts says:

          1) Yes, conduction difference between the glass & rock salt & think plastic wrap are important to consider. I understand that the boundary layer of air on either side of the divider is an important part of the overall thermal insulation. Since this air is the same in all cases, this would reduce the relative difference in the various cases.

          2) “My claim is conductivity controls the rate of cooling, not backradiation.” Can you back up this claim? Clearly any sort of energy in or out will affect the temperature. Can you show that conduction is an order of magnitude larger than radiation?

          3) “It is clear backradiation warms nothing.” That is not clear to me — especially since imprecise language can make statements like this open to misunderstanding. All photons carry energy, and all energy that gets absorbed by an object “warms” the object.

          * If 341 W/m^2 of sunlight is the only input to a blackbody, then that blackbody will tend toward 278.5 K. (Put the object in a vacuum to remove issues with conduction and convection for now).
          * If 341 W/m^2 of “back radiation” (thermal IR from the air/clouds/buildings/plants) is the only input to a blackbody, then that blackbody will tend toward 278.5 K
          * If 341 W/m^2 of sunlight AND 341 W/m^2 of thermal IR are both input to a blackbody, then that blackbody will tend toward 331 K.

          If the sun by itself would warm something to 278K, but the sun assisted by backradiaton would warm it to 331K, it would seem perverse to say the backradiation didn’t warm the object.

          • Bill Hunter says:

            “If the sun by itself would warm something to 278K, but the sun assisted by backradiaton would warm it to 331K, it would seem perverse to say the backradiation didn’t warm the object.”

            Its sloppy logic. Greenhouse gases can be a condition necessary for surface warming and they probably are. But being a condition necessary falls short of a condition sufficient and if thats the case then greenhouse gases may determine how much warming will be realized.

            Greenhouse gases could have their potential fully realized or they may not. We might call that negative feedback but thats really a misnomer because the so called feedback may indeed be a condition apriori. Like in first a solar heat pump operates via gravity and the diurnal cycle and greenhouse gases modify the results near surface all or in part. If only in part then more greenhouse gases might be cause of more warming. If realized in full then maybe we need a thicker atmosphere to get more.

            You say the sun might provide the energy. But the sun cannot provide the energy by virtue of the law that was applied to identify the backradiation “potential” warming in the first place.

            The concept of potential is critical to get right. In electricity resistance and potential defines energy flow. Stefan Boltzmann defines potential for objects of a certain temperature and the inverse square law defines that potential with regards to diffusion of energy with distance as a variable.

            The 2nd law of thermodynamics prevents a cooler object from warming a warmer object and the combination of Stefan Boltzmann and the inverse square law limit the warming the sun can provide.

            Since the sun does not warm the surface and greenhouse gases cannot, I think the air is warmed by another mechanism.

            Perhaps a solar heat pump modified by latent heat. In this system the amount of greenhouse gas could critical but we just don’t know. The simple formula appears to have been destroyed. The simple formula used by models to match blocking of CO2 to recent historical temperature change has proven to be a failure. There is no reason to hang on to even part of the calculation since the system obviously does not work in such a simple manner and it is basic to that simple manner even the minimum warming is achieved (i.e. either the gases do it directly or the sun does it directly). It seems neither do it at least directly.

            In fact it could be dangerous to hold on to the simple concept especially if skeptics are going to be persecuted and labeled with derogatory names. Its like a edict to never look outside of the box for the correct answer. Fortunately as Galileo and many others have proven such edicts have always been unsuccessful, but how many did we burn at the stake who were essentially right?

            Dragon Slayer is hardly a derogatory name, but its probably not healthy to not even have a conversation about it. So kudos to Roy for putting up with it.

            Yes there is a chance the simple answer is correct but there seems to be no certainty of it. I say that as the power source does not seem to be easily identifiable. In electricity or heating and air conditioning you need to match a tested and identifiable power source to the energy needs as a very basic and essential step in designing a workable solution. In this case one has to wonder if such a system is likely to emerge as the process of designing it seems to more closely resemble the Tower of Babel than anything else.

      • Bill Hunter says:

        One follow up. Clearly greenhouse gases play a role or the surface could not warm to radiate 390watts. However we do confound “climate temperature” with “surface temperature”. Greenhouse gases may make surface temperature the equivalent of atmosphere temperature and serve to prevent the loss of heat from the surface. But anyway you cut it greenhouse gases are the determinate for the warming of the surface but may play an important role in maintaining what does warm it.

    • David L. Hagen says:

      Roy
      Great exploration
      In their 2009 paper exploring greenhouse physics, Gerhard Gerlich and Ralf D. Tscheuschner (2009)note:

      An experimentum crucis therefore is to build a glass house with panes consisting of NaCl or KCl, which are transparent to visible light as well as infrared light. For rock salt (NaCl)such an experiment was realized as early as 1909 by Wood [112-115]:
      There appears to be a widespread belief that the comparatively high temperature produced within a closed space covered with glass, and exposed to solar radiation, results from a transformation of wave-length, that is, that the heat waves from the Sun, which are able to penetrate the glass, fall upon the walls of the enclosure and raise its temperature: the heat energy is re-emitted by the walls in the form of much longer waves, which are unable to penetrate the glass, the greenhouse acting as a radiation trap.

      [112] R. W. Wood, Note on the Theory of the Greenhouse”, Philosophical magazine 17, 319-320 (1909)
      [113] M. D. H. Jones and A. Henderson-Sellers, History of the greenhouse effect”, Progress in physical geography 14 (1), 1-18 (1990)
      [114] J. Schloerer, “Climate change: some basics”, http://www.faqs.org/faqs/sci/climate-change/basics/

  18. Fred-M says:

    Can the glass roof reflect infrared from the interior back into the greenhouse?

    That should be a pretty easy experiment as long as one has access to IR detectors.

    If so, the energy balance would be shifted pretty radically to the interior.

    • Fred, yes window glass is partly reflective in the IR, and any detailed calculations would need to take that into account.

      • Fred-M says:

        I think one would also have to consider the wavelength of the IR as it appears the longer lengths are reflected more and the wavelength would change with the temperature of an emitter.

  19. crosspatch says:

    It seems to me that greenhouse works differently than the greenhouse in an atmosphere. If our atmosphere were non-convective, it would operate much as has been modeled. For example, if your real greenhouse had glass all the way from the top to the surface of the dirt, you would find a “hot spot” in the middle of it. But our atmosphere is convective and any “hot spot” that tries to form in the mid-troposphere would convect upwards and cool.

    Expecting to find a “hot spot” in the middle of a convective column seems just nuts to me. It would only work in a magical universe where convection gets turned off. How could one maintain a “hot spot” in the mid-troposphere even if they wanted to create one?

    • crosspatch, the hotspot is a very slight bulge in the average temperature profile, which normally falls off rapidly with height at an average rate of 6.5 deg. C/km or so. The hot spot leads to a slight difference in the lapse rate of the upper troposphere and lower troposphere. We see it clearly during El Nino, but not in the long term warming.

      • crosspatch says:

        Ok, I will buy a lapse rate anomaly. For some reason I was thinking literally a spot with higher temperature, most likely because of the way it has been oversimplified for presentation in the mass media.

        But in any case, I sincerely don’t believe the earth’s atmosphere works like a conventional greenhouse. One other reason is that a greenhouse has a fixed surface area, the tropopause doesn’t. The tropopause would be the “top” of our greenhouse. If you warm the troposphere, the tropopause rises and when it rises, it’s diameter becomes larger which increases its effective surface area which would, to my mind, increase the rate of heat radiation out of it. So maybe it is stratospheric CO2 that has more impact on surface temperature than tropospheric CO2 does because convection will take heat above most of the tropospheric CO2 in a storm but it can’t rise above stratospheric CO2. Also, I would think cooling of the stratosphere would have the same net impact as warming of the troposphere. If the stratosphere cooled, the troposphere would rise. The higher the troposphere goes, the more atmospheric CO2 can be bypassed by convection.

        The problem seems to be the way data are presented. For example, we see data that show stratospheric cooling but it is hard to determine from what is widely available where that cooling is taking place. Is it taking place in the tropics or at the poles? Is it polar summer cooling or polar winter cooling? It probably makes a large difference.

  20. Geoff wood says:

    This is a quote from,

    Controlled Environment Systems ABE 483/583
    Dr. Gene A. Giacomelli
    Professor & Director Controlled Environment Agriculture Center Department of Agricultural and Biosystems Engineering Shantz Building,

    Lecture #7 Greenhouse Energy Balance

    At:

    http://ag.arizona.edu/ceac/sites/ag.arizona.edu.ceac/files/Lecture%20_7%20GH%20Energy%20Balance.doc.pdf

    Page 15 of 18 PDF,

    “Radiation heat losses are directly related to the physical properties of the cover material. These include the emissivity and the transmissivity (in the infrared and thermal wavebands) of the covering material. The emissivity is a material property which defines its ability to emit radiation energy that it has absorbed. Energy absorbed by the cover from inside the greenhouse and emitted to the outdoor environment as long wave radiation provides a heat loss from the greenhouse environment. The larger the emissivity, then the greater the rate of radiation heat loss. The transmissivity is a material property which determines its ability to transmit radiation energy. In this case, it is not the visible radiation which is of concern, but infrared and thermal radiation. This energy is directly transmitted by the cover from inside the greenhouse and it creates a heat loss from the greenhouse environment. The larger the transmissivity for infrared or thermal radiation, the greater the rate of radiation heat loss.”

    Now, I would hope that that statement would have been proof read. It contains two statements of interest, the first,

    “The larger the emissivity, then the greater the rate of radiation heat loss.”

    The second,

    “The larger the transmissivity for infrared or thermal radiation, the greater the rate of radiation heat loss”

    Now, as we know, optical transmission is the inverse of opacity. Opacity can be of the form of absorption, reflection or scattering. As we are not considering the latter two then opacity in this case is absorption. As we know the absorptivity for any substance = emissivity.

    Therefore, the above author, through lengthy wording concedes that whether the greenhouse cover transmits or absorbs it does not ‘trap radiation’ as both lead to cooling, in agreement with Wood’s experiment.

    Food for thought.

  21. Fred-M says:

    From my little googling it appears that there should be some reflection of infrared from the interior glass surface (and then re-absorption in interior objects) in your energy balance.

    That seems like it might be a pretty significant thing to leave out although I do not have any numbers.

  22. Mike Smith says:

    But wait…

    I thought the [greenhouse effect] science was settled?

  23. Geoff wood says:

    The flux from a 5800K source through entropic rules will spontaneously downgrade within a 288K environment. Radiative heat transfer being the result of ‘differences’. The Solar flux containing many photons that are ‘not of this world’. Radiative heat transfer is a net flux expressed in W/m^2 from a warmer to cooler object. The only time you can use Watts is in the netted flux. All other calculations are relative to 0K which is not relevant in ‘this’ coupled thermal system. Within this system, the same molecules transfer heat from a ‘limited’ energy pool and therefore ‘compete’ through modification and evolution of the thermal gradients through all available heat transfer modes. The calculated back radiative fluxes are ‘not available for work or power’. They cannot be realised. They do not have a ‘power spectrum’ that ‘can’ spontaneously ‘upgrade’ to effect a warmer emitter. Borrow a 5m parabolic dish and point it at clear blue sky. NO HEAT WILL BE PRESENT AT THE FOCUS on Earth’s surface. The thermal distribution of accumulated photons will not disobey entropy. They will not spontaneously go up a thermal gradient. They do not have any high energy photons to produce the thermal distribution of a source higher than the emitting source. They do not obey simplistic addition. Electromagnetic ‘waves’ CAN cancel.

    The same applies within your greenhouse model Roy. The 450’W’ calculated cannot be realised. Use it to boil water like you can will the ‘lesser’ solar input.

    SB, actually says, if the temperatures are similar, then very little radiation in Watts is transferred, as it is the ‘difference’ that is the netted flux in Watts, and only that.

    “We know, by immutable measurement, how heat travels”.

    • wayne says:

      Brilliant! Not one statement you made that I can disagree with. Geoff wood, if your not a physicist you class right up there with them.

  24. Willis Eschenbach says:

    Dr. Roy, I salute you for making another attempt to clarify some of the misconceptions about the greenhouse effect, and I wish you luck.

    I did the same thing in The R. W. Wood Experiment, as usual with mixed results. Some folks are determined not to get it. There’s a whole post somewhere on the web devoted to “debunking” my post, it’s great.

    People seem to think that somehow the greenhouse effect violates the Second Law. Nothing could be further from the truth.

    My best wishes to you and thanks for all the good works.

    w.

    • Alex says:

      I like you Willis. But why do you need to be such a fawning a-hole?

    • Bill Hunter says:

      You gotta love the logic there. Since the greenhouse effect exists obviously it doesn’t violate any laws of physics. If it doesn’t exist obviously it also does not violate any laws of physics.

      An interesting extension of the “I think, therefore I am” logic.

      It works nicely because one does not then need to explain anything else. . . .unless of course you want to predict anything from it.

  25. Tim Folkerts says:

    Let me quote Wood himself on the fact that his experiments are applicable to real greenhouses, but may not be applicable to other situations like the earth:

    “I do not pretend to have gone very deeply into the matter, and publish this note merely to draw attention to the fact that trapped radiation appears to play but a very small part in the actual cases with which we are familiar.”

  26. Alex says:

    Does it get hot in a tin shed in the sun?

    • Mike Flynn says:

      Alex,

      Never. It is well known that zinc anodised steel is totally opaque to all known wavelengths of infrared. Therefore, no infrared can penetrate the interior of the shed. The aforementioned tin is also totally opaque to visible light. Therefore, no visible light can penetrate, and be reemitted as infrared.

      Obviously, the best way to store ice is in a tin shed. The cold radiation from the ice will be reflected internally, keeping he ice cold.

      If you have an envelope handy, you can quickly verify the calculations.

      Just remember to keep your shed door closed. Otherwise the ice might melt.

      Live well and prosper,

      Mike Flynn.

      • Alex says:

        I get it. So people working in these sheds are the cause of their own heat stroke. If they weren’t emitting heat that was reflected back on them then they wouldn’t suffer. Perhaps they should have cool thoughts instead.

        • Mike Flynn says:

          Alex,

          Hence the expression “chill out, man!”

          Live well and prosper,

          Mike Flynn.

        • Arfur Bryant says:

          So it’s true then, back-radiation really will cause us all to burn up!

          OMG, it’s worse than we thought… šŸ™‚

      • John K says:

        Hi Mike,

        You wrote:

        “Therefore, no visible light can penetrate, and be reemitted as infrared.”

        Why can’t visible light be absorbed by the tin roof instead of penetrating through it? Once the tin roof absorbs the visible light energy, the tin roof can re-emit the energy as infrared. Perhaps I’m missing something please help us to understand.

  27. Tom in Florida says:

    I no expertise in thermodynamics, engineering or anything technical about this issue. I do know that if I park my car in the sun with the windows closed the interior surfaces that are in the direct light get very hot and the inside air is also very hot. However, if I use a sunshade on the inside of the windshield and even with the windows closed, the surfaces do not get hot and the air inside does not get hot. I bring this up because the amount of sunlight energy that comes through the windshield is the same but the outcome is different. So it has to be that the air warms only from the heating of the surfaces inside the car. Now where does all the sunlight energy go when it is reflected off the sunscreen? It stays inside the car but does not heat the air as much. Is it because the material of the sunscreen is highly reflective and therefore does not absorb the sunlight energy thus does not then emit IR? If so, isn’t that showing that it is the trapped IR that heats the air inside the car and nothing else?

    • Alex says:

      IR is not trapped. Heat is trapped (for use of a better word). IR is the radiation that comes off a heated body. IR moves around till it finds an appropriate host and ‘mingles’ if it wants to. Excuse the anthropomorphism.

    • Mike Flynn says:

      Tom in Florida,

      I’ll sit this one out. I’m interested in the responses you get.

      I have a question for you – if energy comes in through the windshield, and is reflected (no change in wavelength), do you think it’s possible that it can go out the same way it came in.

      Just for fun, walk in front of your windshield, and see if the Sun reflected from your internal sunshield makes you feel warmer.

      I agree with your observations. What do others think?

      Live well and prosper,

      Mike Flynn.

  28. Dr A Burns says:

    How is the figure 442 W/m2 calculated ?
    70% of the Earth’s surface is covered with clouds. Where clouds are present I calculate a radiation loss of 0.0054 W/m2 (based on a 2000m cloud base, 20 deg C temp diff, E = 0.6 ).
    The 442 W/m2 figure appears to be radiation from Earth to space at 2 deg K, ignoring any back radiation ? This is only applicable to 30% of the Earth’s surface.

    • Alex says:

      It’s all algorithms and assumptions. Based on measurements made a long time ago with inferior equipment. Even the modern era ‘scientists’ don’t have a clue. They don’t understand how the sensors work on the equipment they use. Just check equipment manufacturers for IR readers and then check the sensor they use and the manufacturers specs. Its all about assumptions and constant calibration and updating slope algorithms. I don’t trust any of the so-called scientists of today. I worked in a resin lab when I was a kid. there were several tests for determining the melting point of a resin. If you mentioned a melting/softening point you had to mention which test you used, because they were different depending on the method you used.

    • Tim Folkerts says:

      The calculations are simple:
      75 F = 297 K
      5.67e-8 * 297^2 = 441 W/m^2 (slight round-off errors)
      (And this assumes the leaves are blackbodies for thermal IR.)

      The emitted IR is ~442 W/m^2 in this case. The absorbed IR was assumed to be 350 W/m^2 (a typical sort of value for the actual atmosphere). So there was a net loss of 92 W/m^2.

      (NOTE: the emitted photons depend on the temperature of the object, not the temperature of the surroundings. So the 4421 W/m^2 is independent of the surroundings. The *net* thermal radiation would depend on the surroundings.)

      ***************************************

      The net radiation loss will be much larger than 0.0054 W/m^2. First of all, clouds have an emissivity of very nearly 1 for thermal IR. The 277 K clouds would emit ~ 334 W/m^2 (assuming blackbody). The warm air in between would boost this number a bit, but there would still be a net transfer of several W/m^2 .

      With no clouds, the downward radiation would be much less. And the net upward radiation would be larger.

      • Alex says:

        You used ‘assume’ 3 times

        • Mike Flynn says:

          Alex,

          You’re too clever by half! How do you expect to prove a non existent effect without a few minor and unimportant assumptions?

          Do you really expect people involved in such a complicated field as the study of average temperatures, to accurately measure anything?

          If the measurement doesn’t accord with their “theory”, the measurement is obviously wrong. Therefore, measurements are irrelevant. It’s a pity that Nature refuses to fall into line, from time to time.

          Live well and prosper,

          Mike Flynn.

        • Tim Folkerts says:

          Alex, I’m not sure what your specific objection is. All calculations (and all experiments for that matter) make assumptions. I am simply stating (a small number of) my assumptions so they are clear.

          Is there any specific assumption that you object to?

          **********************************

          Mike Flynn says “If the measurement doesn’t accord with their “theory”, the measurement is obviously wrong. Therefore, measurements are irrelevant. It’s a pity that Nature refuses to fall into line, from time to time.”

          Fascinating! Much of this discussion has been about measurements that conflict with each others; some support one conclusion and some support the opposite conclusion. Apparently you are quite happy to ignore the experiments where nature doesn’t fall into line with *your* expectaions.

          • Alex says:

            My problem is with the word “assumption”. I see assumption as being equal to speculation. I frankly, don’t have ‘my’ expectations. I am always looking for ‘truth’. I am not calculating things (though I am capable). I am just dubious about things presented to me as fact when I see them as conclusions based on doubtful information. It may be your presentation that gives me doubts. I have no doubt that you are a sincere person who believes in what you say

          • Mike Flynn says:

            Tim Folkerts,

            I was referring to people who insist the globe is warming, when it appears it isn’t, even though CO2 levels are rising. These people then have to hypothesise “missing heat” rather than accept what Nature is telling them. I may have been imprecise in my choice of words.

            Also, people who refuse to accept experimental measurements that appear to contradict the CO2 “greenhouse effect”. If you can’t come up with reproducible experimental evidence to support your case, maybe there isn’t any. Possible?

            I know it’s all irrelevant anyway, so I’ll depart.

            Live well and prosper,

            Mike Flynn.

          • Arfur Bryant says:

            Guys,

            The biggest assumption is that Arrhenius was right…

            Further, it is assumed that one can take the known radiative properties of a single molecule of CO2 and transpose that property into having a significant effect in the globe’s atmosphere.

            Against these assumptions we have observed data which basically tells us that there is no correlation with increased CO2 and warming – particularly the sort of warming ‘assumed/speculated’ by the pro-cAGW theorists.

            And the magic roundabout continues…

          • Kristian says:

            Well, on Mars the atmospheric radiative warming effect of CO2 on the surface simply doesn’t exist. It is at least NIL, likely negative. And Mars’s atmosphere has a stated scale height of 11 km – should be plenty enough of gas column to make a difference.

            Mars’s atmosphere also contains about 8 times the absolute number of CO2 molecules of Earth’s atmosphere, and these are even packed more tightly together, into a total volume about 2.5 times smaller – so the thermal radiation (IR) emitted from the Martian surface has to run the gauntlet on its way to space of an average density of CO2 molecules per volume og ‘air’ above the ground that’s 20 times as high as on Earth.

            And how high up does this lift the Martian ERL? Zero kilometers. The S-B-calculated effective emission temperature of Mars might very well be higher than the actual mean surface temperature. Which would place the ERL … underground!

            At the same time, note that the Martian atmosphere weighs down about 170 times lighter on the ground than Earth’s atmosphere does on its surface. Just like Venus’s atmosphere weighs down more than 90 times heavier on its surface than what our atmosphere does on ours.

  29. Thanks, Dr. Spencer. Very interesting article.
    I am posting 3 quotations and a link to this article in my climate and weather pages (English and Spanish).

  30. Chewage says:

    It would seem that the density and composition of the integrated air (moisture content) would be a crucial flux density determining factor, for both incoming & outgoing…

  31. Dr A Burns says:

    Oops … I redid the calc and it should be around 62 W/m2 for a 20 deg C temperature difference.

  32. Bill Hunter says:

    My opinion is we have the whole concept wrong. We confound radiation with conduction. We expect 400w/m2 radiant sources to warm 17c surfaces (which are radiating 400w/m2 at the radiant source-or elsewhere- for zero net energy exchange).

    If one has difficulty seeing that imagine a small vacuum spaced between the surface and the atmosphere. Bottom line is while the atmosphere might trap radiation the surface is then radiating 400w/m2 away from the surface.

    There is no getting around the claim that backradiation has to be the source of the additional energy if such a surface is going to warm, and that violates the 2nd law. Any argument denying this is just a game of watching a pea get flicked around shell to shell.

    Imagine 2 slabs of steel all in equilibrium with their environment everything is 17c and radiating 400w/m2. Between the slabs and between the slabs and the walls of the environment.

    Now we open a window to 0K space on one side. What happens? well the slab next to the windows starts losing heat. Nothing is happening anywhere else because they are still in equilibrium. A temperature gradient begins to form in the cooling slab and soon the opposite surface that was in equilibrium is now starting to cool drawing heat from the other slab at the rate of conduction through the slab but not faster than what the cool side of the cool slab can radiate to space.

    All substances conduct heat faster than they can radiate energy for a given temperature difference and only when they get thick enough they will conduct less.

    Conduction controls the flow of heat through objects and a 400watt energy source cannot warm a 17C surface and neither can conduction as heat is flowing the other way.

    The key is temperature gradients and they can only exist inside of objects. Thus a 6,000C sun can only warm a surface to more than 17C if it is closer or in contact with the surface. The inverse square law determines what is closer.

    The idea that the sun can heat objects warmer than what Stefan Boltzmann and the inverse square law says solar rays can heat an object is a violation of known physics.

    My opinion is the answer lies in potential energy, latent heat, the diurnal cycle, greenhouse gases, the mass and heat capacity of the atmosphere, the emissivity of the atmosphere as compared to the emissivity of the surface, and gravity. I don’t think its simple but requires all that stuff. . . .and its about as tricky as several thousand year old passive solar heating technology can be.

    • Alex says:

      It doesn’t matter whether you are wrong or right. No-one is listening. You are not Obama or a world leader. There are vested interests that want us to continue this discourse. While we are talking about this shit they continue making money. When the bubble bursts they will tell us they knew all along. And you will believe them. The human race is hopeless. Everyone wants to win. When they tell you it’s for humankind. Don’t believe them.
      Apart from that, humanity is great. I might get laid by a sophomore this year. Therein lies your lesson.

      • Kristian says:

        Alex,

        Well said.

      • Arfur Bryant says:

        Alex,

        I agree with Kristian – well said.

        But it is worse than that, the current hype about global warming/ climate change is nurtured by the media. Fear sells newspapers/magazines/TV programmes. If it should pass (and I think it will) that the warmist ‘scientists’ are shown to be wrong, then the media will simply change their tack and right more reports/TV time on ‘…how the scientists got it so wrong.’

        They will continue to sell newspapers and it will all have been a complete waste of time and resources.

    • Kristian says:

      Bill Hunter,

      I’ve been banging my head against the wall of trying to explain this simple fact for a long time now to the likes of Tim Folkerts. But they seemingly don’t want to get it. They’re not even making the effort.

      Leonard Weinstein, on this very thread, keeps up the good old tradition of incessantly moving the pea around. He says:

      “You are correct that back radiation slows net radiation loss and thus results in the increase in temperature, and thus increase in convected loss to make the difference, but that is not the same as back radiation as a heat transfer source, it is a resistance technically (otherwise you violate the 2nd law). I think the difference in our position is just how you define terms.”

      Can you believe such an incongruous (I would almost say ‘schizoid’) statement?!

      How can they not see it themselves what they’re doing here?!

      “It is resistance technically (otherwise you violate the 2nd law).”

      So in order for them not to ‘violate the 2nd law’ on paper, they just call it ‘resistance’ rather than what it really is.

      Nothing is resisted! This is not about resisting the outgoing. That doesn’t happen. It is all about adding to the incoming!

      Just take a look at Spencer’s first diagram. What do you see?

      300 W/m^2 coming IN from the Sun
      +
      350 W/m^2 coming IN from the atmosphere

      442 W/m^2 of BB radiation going OUT from the surface
      =
      208 W/m^2 of convective loss going OUT from the surface

      He is not restricting any energy from leaving the surface at all. What he is doing is adding extra energy (apparently separate/independent from the solar) to the surface to increase the outgoing. Like you correctly point out:

      “There is no getting around the claim that backradiation has to be the source of the additional energy if such a surface is going to warm, and that violates the 2nd law. Any argument denying this is just a game of watching a pea get flicked around shell to shell.”

      That is exactly the game they’re playing.

      Yes, all the energy coming in from the Sun is free to leave the surface again by way of radiation, completely unobstructed or delayed by any physical means.

      The outgoing radiation is NOT restricted. It is NOT slowed down.

      And still the surface warms! It is obvious from Spencer’s diagram that the energy doing the warming is the seemingly ‘additional’ energy from the atmosphere. Extra energy is ADDED.

      But everyone knows that adding energy to an object to raise its temperature can only mean a positive transfer of HEAT (or work).

      Everyone knows this. The GHErad proponents too. So they try to cover up what they’re actually doing. And they do it by way of semantics. By claiming ‘resistance to outgoing’ rather than ‘adding to incoming’. They cleverly hide behind the ‘net’ term.

      But resisting the outgoing and increasing the incoming are NOT the same process. They are NOT interchangeable.

      They claim it’s a matter of ‘how you define terms’. NO! It is a matter of physics. You simply cannot add energy from a colder object to a warmer object with the end result of making the warmer object even warmer. End of story.

      It can’t be done. It is the most blatant violation of the 2nd Law around.

    • Tim Folkerts says:

      “The idea that the sun can heat objects warmer than what Stefan Boltzmann and the inverse square law says solar rays can heat an object is a violation of known physics.”

      Let’s focus on that one statement — which I (and apparently Alex & Kristian as well) agree with. This is a statement of conservation of energy. The “object” (the earth as a whole, as seen from space), cannot be warmed above 255 K (on average) by the sun. At this temperature, the earth (as a whole) is absorbing 341 W/m^2 (inverse square law) and emitting 341 W/m^2 (Stefan Boltzmann). Energy is conserved.

      From this starting point, there are a few different ways to explain the fact that the surface is warmer on average than 255 K. You give several of the key ideas. To me, the most direct and intuitive explanation is that GHGs high in the atmosphere cool the top of the atmosphere below 255 K, emitting few photons to space. As a consequence, other places (the land & water at the bottom) must be above 255 K to conserve energy.

      Some people might prefer to say that this warming of the surface causes more IR photons to flow up and down. Some might say that more IR photons flowing up and down cause the surface to be warmer. This is semantics.

      (Downward IR photons from the atmosphere to the surface)
      is inseparable from
      (the surface is above 255 K)
      is inseparable from
      (GHGs exist in the atmosphere)

      • Bill Hunter says:

        I am in agreement except that the Stefan Boltzmann conversion of 341 watts/m2 equals 278.5K, not 255K.

        I think climate science has tinkered with the size of the greenhouse effect as well. They seem to podge it when calculating the actual absorption rate of entire system and then suggest the average temperature should be lower because the surface is closer to a blackbody.

        This is probably correct as far as warming the surface via radiation goes but it ignores other methods of bulk transfer of heat through the atmosphere like via potential energy.

        I am convinced the simple backradiation model is just a bunch of hooey. When I learned heating and air (and electricity also) I had to shun all these pretty pictures of how things work and just think in terms of heat and energy flow. If you think that way it works. Everything else is mental gymnastics that don’t amount to a hill of beans. I think it goes directly to the difference between conduction and radiation.

        I think there is backradiation but it does not control flow.

        Flow is controlled by the conductivity of the article placed in the radiation path and if there are vacuum breaks the flow can be less but not exceed Stefan Boltzmann for a given temperature.

        The average temperature of objects is completely meaningless in this view. The only meaningful information is the temperature gradient that forms between the first absorption and the eventual reemission on the other side of the object.

        If there is no temperature gradient in that object then there is no obstruction of flow.

        If there is a temperature gradient then the average temperature of the object is meaningless.

        • Tim Folkerts says:

          “I am in agreement except that the Stefan Boltzmann conversion of 341 watts/m2 equals 278.5K, not 255K.

          I was clearly over-tired when I wrote that! Thanks for catching that. 341 W/m^2 is the energy that arrives but 239 W/m^2 is the energy that is actually absorbed and the amount that is again emitted as thermal IR.

          **********************************

          I agree with most of what you say in this last post. A simple model based purely on back-radiation would indeed be “hooey”. The temperature gradient is controlled in the troposphere by the lapse rate — convection increases quickly and easily to ensure the lapse never gets more than 10 K/km (or ~6 K/km when the air is saturated with H2O). So any useful model must include the lapse rate and the physics that leads to the lapse rate.

          I disagree that “If there is a temperature gradient then the average temperature of the object is meaningless”.
          1) since we live at “ground level”, the average temperature at ground level is quite meaningful for people.
          2) the “average temperature” (suitably averaged) determines the radiation that escapes to space, and will adjust until the average outgoing IR equals the average incoming solar power.
          Certainly, the average temperature does not tell us everything; nor does the gradient. Both are important.

          • Jos says:

            In essence that is the point of why the “greenhouse effect” (GHE) does not work like a greenhouse: the GHE works by virtue of the lapse rate. A true greenhouse does not work by virtue of a lapse rate, regardless of whether IR back radiation plays a role in explaining the physics of a true greenhouse.

            In your radiative calculations you implicitly assume the presence of (A) atmospheric layering and (B) that those layers are only connected by radiation. That is not true, in the real atmosphere as well as greenhouse convection plays a role (wet convection and dry convection, respectively). Including convection makes the calculations inherently much more complex.

            Or, in other wording, without consideration of convection you will get a lapse rate (= decrease of temperature with altitude), but it will be with of wrong magnitude and shape (lapse rate then is not constant with height in the tropophere). Only including convection will result in the near constant lapse rate. It is not for nothing that the climate is described as a radiative-convective system, not a pure radiative system …

  33. Alex says:

    Thank you Roy
    I appreciate your sense of humour in getting this chat going

  34. GaelanClark says:

    Okay, I must be missing the obvious….
    Dr.Spencer, you are with the UnivAL…yes? You know how to write research grants…yes?
    Then get this set up and analize this from the glass down with measurement devices every inch in order to get every layer detected.
    Please, enough of the back of the envelope.

    • I have already posted results from an excellent experimental setup which gave results supporting my view:

      http://boole.stanford.edu/WoodExpt/

      Unfortunately, those results conflict with the results of Nahle which showed no “greenhouse effect”:
      http://www.biocab.org/Experiment_on_Greenhouses__Effect.pdf

      Looks like more people will need to perform the experiment. I need to order a couple temperature data loggers before I can perform it.

      • Ball4 says:

        Dr. Spencer – Observe only the PE covered box in segment 1 of your second link has extra insulation unlike the same insulation used in all the boxes of your 1st link. There is no conflict; just an experimental demonstration that environmental radiative and convective energy balance can easily alter one’s POV on farmer’s greenhouse “working” theory as Dr. Bohren pointed out in my book clip above.

        • Massimo PORZIO says:

          Hi Ball4.
          I’m not sure I understand what you mean (excuse me, I’m Italian and English is not my language).

          Do you find a reason for that extra insulation on the PE box only?

          What do you mean with “There is no conflict; just an experimental demonstration that environmental radiative and convective energy balance can easily alter one’s POV on farmer’s greenhouse “working” theory as Dr. Bohren pointed out in my book clip above”?

          Have a nice day.

          Massimo

          • Ball4 says:

            Massimo 1:51pm: ?

            Dr. Spencer points out this apparent conflict: 1st experiment finds glass & acrylic covered boxes warmer by 15 to 20 degrees than the PE covered box. The 2nd experiment segment 1, purporting to be a replication of the 1st (“… to repeat the experiment of Professor Pratt…”), finds all 3 boxes with no significant temperature differences after ~1hr.

            The conflict is resolved (for me) by photo 06 inspection showing the 2nd link exp. has obvious added white glass wool insulation on the PE covered box walls thereby enabling an increase in its measured temperature to roughly equal the other two boxes vs. no added insulation in the 1st link photo.

            I do not find a reason for the added white glass wool insulation on the 2nd link experiment’s PE box as you ask.

            I am satisfied that these modern experiments are different (by the added glass wool insulation at least) therefore their results are different (i.e not in conflict). I have not researched which experiment best replicates Prof. Wood’s experiment “packed in cotton” long ago (linked above 8/11 7:00pm).

          • Massimo PORZIO says:

            Ok,

            I get your point.

            Thank you.

            Have a nice day.

            Massimo

  35. Lindsay says:

    Interesting discussion
    Some time ago I raised a question of measuring the impact of CO2 in a greenhouse, Two identical greenhouses one with standard 400 ppm and another with 1600 ppm (often done in the real world because plants grow better).
    Is there a difference in temperature between the two greenhouses, with the same incoming radiation?
    Theoretically there should be a slight difference, but none of the greenhouse operators seem to have noticed any difference.
    Perhaps an experiment to test for sensitivity !

    • The effect would be tiny because the path length is so small. The broadband IR absorption/emission effects of CO2 are very small on temperature until you get path lengths of 1 km or more.

      • Will Pratt says:

        Since when did path length do anything but increase entropy?

        It is mass which determines how much energy is converted to heat.

        As the path length increases so too does entropy, yet the total available energy stays the same.

        Hence this is why the “GHE” violates the the laws of thermodynamics.

        • @WP: It is mass which determines how much energy is converted to heat.

          Indeed, but the mass of CO2 in a greenhouse is tiny compared to the mass of the glass. How do you reconcile that with the rest of what you say?

          • Will Pratt says:

            @VP: Sorry, excuse my ignorance but I simply cannot follow what you are actually asking me to reconcile. Would you be so kind as to elaborate for me?

    • Tim Folkerts says:

      The effect of CO2 in the atmosphere depends on 1) the temperature difference that develops from the top to the bottom of the atmosphere and 2) large enough distances for the CO2 to absorb significant amounts of IR (10’s of meters). In a relatively small enclosed space that is nearly the same temperature everywhere, GHGs would have minimal effect.

      • Massimo PORZIO says:

        I would responded that the whole CO2 effect is completely overcame by the glass greenhouse effect…

        Or do I mģss something?

        Massimo

  36. Brian D says:

    If one were to visit a greenhouse, one would see the glass is usually painted somewhat with white paint to reduce the amount of SW entering in. Maybe this was mentioned above, but this would change some of the numbers being used in real world greenhouse operations. Experiments need to be done in them to get a good idea of what is really happening. That’s true science, right?

    • yes, exactly. I don’t think this has been addressed in the experiments that have been done. If you use glass, most glass has a visible transmissivity of only 0.7 to 0.9 at most, from what I have read. For direct solar illumination, you are then automatically losing up to 100 to 300 W/m2, which will lower temperatures, opposite to the IR effect you are trying to measure. For a two-box experiment, one might want to choose a tinted polyethylene film that “shadows” about the same as glass.

      • Bill Hunter says:

        the problem with polyethylene film is it has a different rate of conductivity of glass. Robert W. Wood, at John Hopkins University in his famous 1909 experiment used rocksalt (transparent to IR) and glass (opaque to IR) he found:

        1. the rocksalt greenhouse warmed faster than IR opaque greenhouse demonstrating very significant absorption of incoming sunlight in the IR spectrum. If its significant it should be accounted for.

        2. he found that the two greenhouses warmed to the same temperature, suggesting the IR opaque greenhouse also absorbs light going out equalizing the effects on incoming and outgoing.

        So here is a way to have your greenhouse effect but it may not amount to anything and may only if you ignore incoming absorption.

        I would suggest this non-intuitive outcome can be potentially be explained geometrically. A surface of a sphere absorbs light from a 1368w/m2 source at the rate of 341w/m2 because of the geometric orientation of the surface. But we know the case is that the atmosphere is exposed to twice the radiation the surface its as the light is either on or off owing to its 3 dimensional transparency. Its getting 1368watts/m2 half the day long.

        I realize there is discussion of Plancke and Wien calculations involved in all this but I have never seen a study on the matter and I find it hard to take the word of anybody in this debate as it has risen to political and religious dimensions.

        Trenberth’s 2009 energy budget has the surface absorbing only 161 watts of 184 watts shining on it giving an emissivity of .875.

        The atmosphere absorbs 78 watts of the 341 watts shining on it giving an emissivity of .229.

        According to Stefan Boltzmann the atmosphere should be 278.5 K(341watts as the root) and the surface should be 238.7K(184 watts as the root).

        But via conduction, convection, internal radiation, and laws of thermodynamics the atmosphere will warm the surface to 278.5 K via all the means at its disposal.

        that puts us about 9.5 degrees short. Error’s from a lack of representative surface stations might over estimate the surface temperature by 2 degrees or more (thats what I understand the reference network is suggesting for the US). If thats the case maybe its a smaller warming.

        What I think we are looking for is an area in the atmosphere that is 278.5 degrees with the lapse rate revolving around that temperature, higher below and lower above that gives us the 278.5 average we should expect from the light that shines on the atmosphere. Everything else seems manufactured by some blurry process and the processes by which all this revolves may indeed be established by gas laws and not backradiation. At least thats how it works for sophisticated passive solar heating design within structures. Greenhouse gases may or may not have a role in raising the average temperature but very clearly they do have a role.

        In my opinion the role is to limit both warming and cooling rates but not to effect the average rate which will be achieved whether slowly or quickly based upon the total amount of gases in the atmosphere. But what do I know? Practically nothing!

        • Tim Folkerts says:

          Just a quick suggestion as you move forward with your understanding of the physics here:
          emissivity is a function of wavelength.

          White paint has a high reflectivity (and low emissivity) for visible light. But that same paint can (and often does) have low reflectivity (and high emissivity) for IR. Clouds are the same.

          Since sunlight and “earthlight” are almost entirely in different wavelength ranges, an object can have completely different emissivities for the two ranges.

          • Bill Hunter says:

            Emissivity may be a function of wavelength but conductivity and convection is not. So if you rob some heat from a highly emissive surface via evaporation/conduction and convection and put it in a less efficient emitter, you have effectively sequestered some heat.

        • Paul in Boston says:

          If you read Wood’s experimental description carefully you will find that in the final design he placed a glass sheet above both of the models to aborb any incoming IR. This was to restrict the measurement to IR generated within the models by the incident visible light.

  37. Svend Ferdinandsen says:

    An interresting discussion.
    In my opinion the clouds are the real greenhouse working horse.
    When ever i point my IR thermometer to the sky i get below -20C (reading LO) unless the sky is cloudy.
    From that i infer that clouds radiates much more energy back than the CO2 or water vapor could ever do. Clouds at the same time shadows the radiation from the sun (in the day time) so it is a delicate balance and it seems not to be resolved by the climate models.
    But basically it is OK to compare the atmosphere with a grenhouse, you just have to add some workers painting the glass sheets/cleaning them white and opening the ventilations from time to time.

    • Tim Folkerts says:

      Thanks for the smile — “painting” and “cleaning” clouds up in the atmosphere. It’s an apt analogy, and an important part of the “greenhouse effect”.

    • Massimo PORZIO says:

      Hi Svend.

      I’m not sure, but maybe that your IR thermometer is blind to the 15um CO2 band, so it doesn’t measure the CO2 radiation at all.
      Usually those devices are engineered to receive IR emissions in the 8-14um range only. I suppose they do it just to reject the CO2 absorption band, but mine it’s just a supposition.

      Have a nice day.

      Massimo

      • Will Pratt says:

        15µm has a corresponding temperature of -80ŗ C.

        The same temperature that CO2 freezes in our atmosphere.

        Coincidence?

      • @Massimo: I’m not sure, but maybe that your IR thermometer is blind to the 15um CO2 band, so it doesn’t measure the CO2 radiation at all.

        Quite right. By rejecting wavelengths that CO2 radiates strongly in, IR thermometers can give a more accurate reading of the temperature of objects that are sufficiently far away that the intervening CO2 would otherwise throw off the reading.

        The flip side is that they’re not that great as fire detectors that work by detecting the CO2 byproduct of combustion.

    • D'Avila Tarcisio says:

      Hi Svend

      I post in this topic,
      August 11, 2013 at 6:44 PM
      Complain this issue bat It is not minus 20C degris.
      Problable you take a incorrect read.
      Hou countri is you? I`m from Manaus 3 degris sounther
      Thanks
      Tarcisio DĄvila

  38. Don K says:

    Interesting problem. Unfortunately, I’ve always been rotten at physics, so I’m probably wrong. But if you were to go into a greenhouse at night with your IR thermometer and measure the temperature of the glass, might that not tell you a bit about greenhousing? If the temperature of the inside of the glass is close to outside ambient then warmth is probably being retained because of a lack of convection in the greenhouse? If the glass is nearer inside ambient, then greenhouse IR emission could be a factor?

  39. Bri says:

    This is all such fun I think you should get the 73 climate models and you had in your previous post and apply them to a green house. They after all can predict the temperature of the entire planet so a green house should be simple.

    Just put in the data you have and you will know everything about future green house temperatures, no need do all these experiments just trust the computer models.

  40. Dr. Strangelove says:

    Roy,
    I think there’s something wrong with your calculation of convective heat loss. Without greenhouse, the air is moving while the vegetation is exposed to sun. There must be a greater temperature difference between air and vegetation. Hence faster convective heat transfer. With greenhouse, the air inside is stationary and cannot cool quickly. The temperature differential between air and vegetation will be smaller. Hence, slower convective heat transfer.

    Temperature inside greenhouse will rise to attain equilibrium. But this is due to slower convective loss. The radiative heat transfer is catching up to balance it. So it is convection that’s driving the rise in temperature. That’s my interpretation of the phenomenon. You have to conduct experiments to verify.

  41. A small amount of solar radiation in the “mid-IR” range, mostly medium to long IR-B and nearby IR-C wavelengths, has somewhat significant presence in solar radiation reaching Earth’s surface, and has significant absorption by typical window glass.

    This appears to me to require a small adjustment of the “greenhouse energy budget”, with small addition of input into the glass from solar radiation, and small decrease of input into the glass from below. I would favor that decrease being in the area of convection from below, due to greenhouses being said to decrease convectional and related evaporative cooling of what’s under them.

  42. Dr A Burns says:

    Roy,

    Do you have any comment on the calculation of 62 W/m2 radiation loss for a 20 deg C temperature difference between the Earth’s surface and the bottom of a cloud ?

    Clouds cover 70% of the Earth’s surface. At night the radiation loss would be even less. This is far less than your stated figure of 442 W/m2.

  43. Dr. Strangelove says:

    Roy,
    Your diagram is wrong. The roof of the greenhouse is not in equilibrium. There’s 150 W/m^2 net heat inflow. It will be continuously heating. I believe the correct diagram should be as follows:

    The equilibrium temperature of 350 W/m^2 infrared inflow is 44 F. Therefore, the glass should be cooler than air outside, which is at 65 F. The air is radiating at 406 W/m^2. The convective arrow should be reversed. Air is heating glass until they are both at 65 F. The glass is absorbing 406 – 350 = 56 W/m^2 from air.

    Now glass is also radiating at 406 W/m^2. This will split into two: up and down, 203 W/m^2 each. The down radiation will hit the vegetation. Note this is lower than 350 W/m^2 without the greenhouse. Therefore, the heating of the greenhouse is not due to increased infrared but suppressed air cooling inside.

    • Dr. Strangelove says:

      The radiation hitting the vegetation is 300 + 203 = 503 W/m^2. The temperature will rise to 93 F. Since air inside is now hotter than outside. Heat flow will reverse from vegetation to glass to air outside. The glass temperature is the equilibrium temperature between 93 F and 65 F.

    • Massimo PORZIO says:

      Hi Dr.Strangelove.

      Even if I agree with your previous post, where you argue that the convective heat loss inside the greenhouse should reduce, not increase as per Dr.Spencer diagram (in fact the T delta reduced from 10°F(75-65) to 7°F(85-75), I don’t see where your get the 150W/m^2 of residual net inflow at the outer glass surface.

      Instead, my point is always the same that I exposed before. That is, the temperature of the two surfaces of the glass could be not so insignificant as supposed, and they are a function of the glass thermal resistance multiplied by the outgoing heat flux which it’s the only constant flux of the system and which it will be always 300W/m^2 (the Sun driving flux).

      In my opinion at equilibrium inside the greenhouse the IRs are fully absorbed and partially reflected back without almost any convective heat loss, they keep hot the inner.
      But that doesn’t mean that the greenhouse works because of the IR “back-radiation” as you correctly say.
      This because the “thermal flywheel” is the outside glass temperature, which is exposed to the great mass of the outer air. The inner glass temperature (and consequently its IR “back-radiation”) is fixed by the difference from the outer glass temperature minus the glass thermal resistance multiplied by the 300W/m^2 outgoing flux.

      Even in my opinion, that external 65°F of the glass surface should not be a fixed value as you suppose (if I understood you right), it could be higher depending on the wind speed which is not driven by the glass temperature which is supposed to be a minor mass compared to the whole surrounding atmosphere.
      So, the real greenhouse works mainly because of suppression of the inner convective loss, but also because of the radiation loss depending on the ability of the external wind speed to keep the glass temperature the closest to the air temperature.

      If you read my post of August 12, 2013 at 8:32 AM, you can find a link to a thermal resistance calculator.
      As I already said there, the thermal resistance is a function of emissivity (that is absorptance), but the wind speed is more influential in its computation.
      So, its seems to me that Prof.Wood wasn’t so wrong in thinking that the IR play a minor role in a real greenhouse.

      Have a nice day.

      Massimo

      • Massimo PORZIO says:

        Hoops!!
        Again, I should reread better before press the “submit comment”.

        My:
        “The inner glass temperature (and consequently its IR “back-radiation”) is fixed by the difference from the outer glass temperature minus the glass thermal resistance multiplied by the 300W/m^2 outgoing flux.”

        Want to be:
        “The inner glass temperature (and consequently its IR “back-radiation”) is fixed by the outer glass temperature PLUS the glass thermal resistance multiplied by the 300W/m^2 outgoing flux.”

        If it was true, I would invented a very cheap air conditioner!
        šŸ™‚ šŸ™‚ šŸ™‚

    • Tim Folkerts says:

      Dr. Strangelovesays: “The equilibrium temperature of 350 W/m^2 infrared inflow is 44 F. “

      There is no ‘equilibrium with to respect to only one type of energy’. You have to use the NET inflow; the glass panels don’t care how they got the energy.

      Each square meter of the top (in the example in the top post) is receiving 350 W/m^2 from IR and 300 W/m^2 conducted up through the glass. It is losing 200 via convection, so in equilibrium (or more specifically in steady-state) it must be losing 450 W/m^2 of thermal IR. It is THIS number that determines the temperature of the glass, not the incoming 350 W/m^2. (You are of course welcome to argue that Dr Roy’s assumptions are poor, but this is a different issue).

      • Massimo PORZIO says:

        Tim,
        I agree with you, we should talk about “thermal steady state” not “thermal equilibrium”.

        I often use the bad words.

  44. Alex says:

    I was clearly too subtle with my comment about a tin shed heating up in the sun. Has anyone considered the dirt on the outside of the glass? or the condensation on the inside of the glass/PE? How does that effect things? All SW does not convert to LW because it went through glass. For god’s sake, I can see colours through glass, from the outside. The solar energy heats up the greenhouse as well as anything inside it. Eventually the energy leaves by convection, radiation, conduction or through absorbed latent heat in the moisture in the air that is vented. Is the greenhouse full or empty? Are they seedlings or full grown plants? What is the total biomass enclosed? Has anyone heard of transpiration of water from plants? Do plants have a temperature? Every living thing is a heat producer due to biological processes.
    Regardless of all of the above the net energy entering the system cannot be multiplied. You can play all kinds of silly games with figures but you can’t take a total then break it down to various components and then decide to add some of these components to the total. That’s just too silly for words.

    • Simon says:

      Alex. So true!. If such a multiplication were possible, there would be a mass of power stations built using the principle by now. As there is not a single one, I think that is ample demonstration that any notion of ‘extra’ radiation from the atmosphere is too stupid for words.

      Think of it this way too. If the energy from the sun has been expended heating the atmosphere, it cannot also heat the greenhouse, and what energy gets through to heat the greenhouse cannot also have heated the atmosphere. To put it another way: “1 unit of energy cannot do 2 units of work”. All energy derives from the sun, and the atmosphere cannot magically conjure up some more out of thin air!

      Think of it also like this with regard to the much vaunted ‘back-radiation’, if 1 unit of energy is emitted from the surface, and IF a fraction, 0.x is radiated back, the sum at the surface is now -1 + 0.x, i.e. less than 1 or cooler, and NOT +1 + 0.x, i.e. warmer. How much simpler does the maths need to be?

      There’s no other way to view it, but that Spencer’s proposition is totally laughable.

      • Tim Folkerts says:

        “Think of it also like this with regard to the much vaunted ‘back-radiation’, if 1 unit of energy is emitted from the surface, and IF a fraction, 0.x is radiated back, the sum at the surface is now -1 + 0.x, i.e. less than 1 or cooler, and NOT +1 + 0.x, i.e. warmer. How much simpler does the maths need to be?”

        No, think of it like this instead …

        If 1 unit of sunlight is absorbed by the surface and 1 unit of energy is emitted, then energy is balanced and the temperature will hold steady.

        If 1 unit of sunlight is absorbed by the surface and 1 unit of energy is emitted and IF a fraction, 0.x is radiated back, then 1+x is getting absorbed and only 1 is being emitted, and the surface will warm up until the extra 0.x units are being radiated by the now-warmer surface.

        • Arfur Bryant says:

          Tim,

          I believe this to be the crucial factor in the cAGW debate.

          Your back-radiation has to be absorbed for net energy gain in order for your 1+0.x to work.

          Why would the surface absorb ‘for net gain’ the radiation it has just emitted?

          According to you, if there is a closed room with one working electric bar fire inside it, and the room reaches ‘thermal equilibrium’, then the simple addition of another electric fire – without it being switched on! – will make the room warm up due to the re-emitted radiation from the ‘off’ fire adding to the temperature of the ‘on’ fire.

          Can this really be what you mean?

          • Tim Folkerts says:

            Why *wouldn’t” it absorb it??? Photons get absorbed. they deposit their energy.

            I think you critical misunderstanding here is ” the radiation it has just emitted”. The radiation was not emitted by the object itself. It was emitted by the surroundings using the energy of the atoms in the surroundings.

            As an analogy, suppose you pay me $100 for fixing your plumbing. I now have $100. Is there any reason I can’t take some of that money to pay you $20 for mowing my lawn? Is that $20 somehow still “your money” since you originally paid it to me and hence can’t be given to you?

            ********************************

            As for your electric heater … let’s suppose it is a flat panel 1 meter square with a constant electrical power input, which equilibrates at 200 C on the surface.

            Now put a thin later of insulation over the two sides. The sides will STILL have to equilibrate to 200 C in order to transfer heat at the same rate to the room. But because there is a thermal gradient across the insulation, the original surface will be considerably warmer than 200 C. Does this surprise you?

            Same thing if we surround the two sides by a thin covering with a vacuum layer between the original surface and the new surface. The outer surface will be 200 C. To drive heat from the original surface to the new surface there once again has to be a thermal gradient. The interior will warm up above 200C — which can only be due to “back-radiation” since there is no conduction or convection.

            Can this really be what you mean? That objects covered with insulation (either physical insulation or a vacuum space) will NOT warm up when power keeps being supplied at a constant rate?

          • Arfur Bryant says:

            Tim,

            I disagree.

            “Why wouldn’t it absorb the backradiation?” I didn’t say that. I asked whether or not the backradiated energy would be absorbed by the surface for net energy gain. That is the crucial factor. The energy emitted from the hot surface is greater than the energy re-emitted back from the atmosphere. So why would the surface gain energy from its own previously emitted radiation? If I add two glasses of water each at 5 deg C to each other, I don’t get one big glass of 10 deg C. There is no net gain in terms of thermal energy.

            You are trying to confuse the issue with your analogy of money. Applying that analogy to your theory, I would pay you $100 so you would keep the $100 AND pay me $20 (because you said [“If 1 unit of sunlight is absorbed by the surface and 1 unit of energy is emitted, then energy is balanced and the temperature will hold steady.”] Where did you get the extra $20? The analogy is wrong.

            When you say [“IF a fraction, 0.x is radiated back, then 1+x is getting absorbed and only 1 is being emitted, and the surface will warm up…”] then you are invoking a theoretical addition of energy which is either unlikely or impossible.

            AS for my heater, no just tick to the question and don’t try to change the goalposts. According to you, any addition in the room which can absorb and re-emit the radiation from the ‘on’ fire will increase the temperature of the ‘on’ fire. I maintain that the addition of an ‘off’f ire will not cause the ‘on’ fire to get warmer. It just doesn’t work that way.

            If you are now asserting that CO2 acts like a layer of insulation, then you really need to apply that to the real world. Increasing a layer of insulation by 0.028% would not have a significant effect. You cannot compare a doubling of CO2 to wrapping the planet in a space blanket. To understand what effect adding 0.028% would have, you would have to know – categorically – what portion of the GHE is currently contributed by CO2.

            Do you know that? If so, please tell me and provide supporting evidence.

          • Tim Folkerts says:

            Arfur,

            If you can’t see the huge conceptual difference between adding two glasses of water and adding two photons to a single object, then I don’t think I can help you there.

            ******************************

            For the money analogy to be more complete, you have to consider you getting paid $100 per day and paying me $100 per day; your bank account will stay constant. If I now give you an extra $20 per day (and it makes absolutely no difference whether I got that money from you originally or from somewhere else), your bank acct will start to increase. I don’t have to “create” any money to do that.

            Simuilary, with no atmosphere in the way, the surface would gain ~ 240 joules every second (in an average square meter of surface) from sunlight and would also lose ~ 240 joules every second to space.

            With an atmosphere those 240 Joules can be absorbed before escaping — the atmosphere is NOT “invoking a theoretical addition of energy”. They are real joules that simply go into the atmosphere instead of to space.

            The atmosphere now HAS ITS OWN ENERGY. It will deal with its own energy according to the laws of physics. One of those laws says that objects emit photons based on their temperature. Another law says that the net energy flowing by such photons must always be from a warmer object to a colder object. So the atmosphere will be getting more from the ground than it gives back, but it can and does and must give some energy to the ground.

            *****************************

            The geometry is such that your “off fire” will have minimal effect. I gave a counter example where the geometry was maximized. Your example would give a small (but real) effect Mine would give a much larger effect. It is simply a matter of the size, not the existence. If you admit that my examples would have an effect, then you have to admit that your example would have smaller but real effect.

            ******************************

            You are now “changing the goalposts” dramatically with your discussion of CO2, but I will add few comments to this NEW topic. First, I never asserted that CO acts “like insulation” per se. Both can help keep things warm, but they act via very different mechanisms.

            ” Increasing a layer of insulation by 0.028% would not have a significant effect.”
            OK … you want to talk specifically about insulation. Consider two objects separated by a few sheets of “copper insulation”. I’ll add 0.01% of some other material like sheets of paper. It turns out that paper conducts about 0.01% as well as copper, so adding 0.01% of extra insulation has about doubled the insulating value.

            N2 is like the copper — it has basically no resistance to the flow of heat (photons in this case). CO2 is like the paper — a good resistance to the flow of heat.

            Or a different analogy, the air around my house is insulation for my house. Adding even a thin layer of plastic sheeting to the air (say 10 cm from the house) will cut down on convection and greatly increase the insulting value. Once again, a small addition makes a huge difference.

            “To understand what effect adding 0.028% would have, you would have to know – categorically – what portion of the GHE is currently contributed by CO2. “
            I haven’t claimed to know just how much effect CO2 has — I only claim that it must have SOME effect. Determining the actual sensitivity is a challenging proposition.

            Are you willing to state categorically that CO2 has NO effect on surface temperature?

          • Arfur Bryant says:

            Oh, come on Tim.

            You accuse me of failing conceptual vision and then you dig your failed analogy deeper by saying you add $20 to my account but not worrying where the money came from! Of course it matters where the money came from! You have added $20 to my account but, in the real world, the extra $20 has, according to you, come from the initial $100 I gave you!

            Stop using analogies. Just explain how the lower energy radiation is absorbed FOR NET GAIN by the warmer surface!

            There is no mechanism whereby the surface can get warmer than it was originally . Seriously, please provide some sort of real-world (not modelled) evidence that this is possible. Whether or not the atmosphere gives some energy to the ground is actually irrelevant if that energy is not absorbed for net energy gain.

            As for CO2, I am not changing the goalposts. You started talking about back-radiation. It is the back-radiation from CO2 that fires the CAGW debate. Therefore the effect of CO2 is absolutely fundamental to this debate.

            You claim that back-radiation from CO2 has an effect but you don’t know how much. Well, that is fair.

            In answer to your question [“Are you willing to state categorically that CO2 has NO effect on surface temperature?”], my answer is no, I am not. I am, however, willing to state categorically that there is NO EVIDENCE that CO2 has any effect on real-world surface temperature!

            Is that fair enough?

  45. lemiere jacques says:

    something is very entertaining..listening people explaining that climate system is simple..
    a simple framehouse is soooo complicated..
    well add water vapour..dust on the glass…drops and droplets and so on…and night and day..add a tank of water… and why not breathing plants..
    i need a supecomputer to make calculations!

  46. Mike Flynn says:

    Tim,

    Put some water at 100C into a device which is specifically engineered to reflect as much radiation as possible back to the object, a Dewar flask.

    Do all your calculations, but prepare to be amazed when the temperature of the water falls.

    Yes, we know that the Earth”s surface warms when exposed to the Sun’s rays. We also know that it cools in the absence of sunlight.

    There is a temperature gradient from the Earth’s core to the surface of the crust, and thence to outer space. If you can’t accept that this shows that the Earth hasn’t finished cooling, then obviously you will be a Warmist to the day you die.

    You are in good company. Lord Kelvin refused to believe anyone who claimed the Earth was more the 40 million years old. He refined his calculations to such an effect that in later years he claimed that the Earth could not possibly be more than 20 million years old.

    But anyway, before I leave again,

    Live well and prosper.

    Mike Flynn.

    • Tim Folkerts says:

      I am mightily confused by what you wrote, Mike!

      1) “Put some water at 100C into a device … “
      You are missing “the sun” in your analogy. If the sun stopped shining, then the earth would continuously cool like the pots of water (with or without GHGs). The analogy should be more like an un-insulated coffee pot and an otherwise identical insulated coffee pot. Plug them both in and the insulated coffee pot gets hotter with the same energy input. I will not be at all amazed when all my calculations agree with this intuitive result.

      2) “If you can’t accept that this shows that the Earth hasn’t finished cooling, then obviously you will be a Warmist to the day you die.”
      How does one relate to the other???

      Of course the (interior of the) earth has not “finished cooling”. Around 0.1 W/m^2 of geothermal energy is “leaking” out of the surface (from the estimates I have seen). The interior is indeed slowly cooling and has been for billions of years. This is not enough to make a big difference in the earth’s surface temperature — certainly not enough to warm the surface from 255 K to 288 K.

      3) “You are in good company.”
      First, you did’t know what company I was in. Second, Kelvin made a perfectly sound calculation — based on the understanding of physics at the time. I do not at all mind being in the company of a great scientist who made a valiant (but ultimately incorrect) effort to apply know physics to a new problem.

  47. James Goodone says:

    Tim Folkerts says August 15, 2013 at 2:55 PM:
    “Simuilary, with no atmosphere in the way, the surface would gain ~ 240 joules every second (in an average square meter of surface) from sunlight and would also lose ~ 240 joules every second to space.

    With an atmosphere those 240 Joules can be absorbed before escaping — the atmosphere is NOT “invoking a theoretical addition of energy”. They are real joules that simply go into the atmosphere instead of to space. […) So the atmosphere will be getting more from the ground than it gives back, but it can and does and must give some energy to the ground.”

    This is exactly what can not happen: the atmosphere can not raise the energy supply to the surface above what the sun would give the surface in absence of the atmosphere, the sun being the ONLY source of energy for the system surface+atmosphere.

    Otherwise it would mean creating energy out of nothing, which is clearly nonsensical.

    The math is simple, I have done it already for the Dr.Spencer greenhouse picture: http://www.drroyspencer.com/2013/08/does-a-greenhouse-operate-through-the-greenhouse-effect/#comment-86062, http://www.drroyspencer.com/2013/08/does-a-greenhouse-operate-through-the-greenhouse-effect/#comment-86062 and http://www.drroyspencer.com/2013/08/does-a-greenhouse-operate-through-the-greenhouse-effect/#comment-86094

    • Tim Folkerts says:

      There a couple things I would say here … mostly related to the need to choose “systems” and then apply the rules to the systems appropriately.

      For example, earlier you said “This [300 W/m^2 of sunlight] is the only supply to the system “greenhouse”.”
      No. The system “greenhouse” is in thermal contact with the system “sun”. It is also in thermal contact with the system “atmosphere”. (It is also in thermal contact with the system “ground”, but we choose to assume that interaction is minimal.)

      The “greenhouse” can get photons (ie energy) from either of the systems “sun” or “atmosphere”. The 2nd Law will assure that the net energy exchange is always from the sun to the greenhouse (300 sun->greenhouse and 0 greenhouse->sun), and from the greenhouse to the atmosphere (450 greenhouse-> atmosphere and 350 atmposphere->greenhouse). No law of physic prevents photons from traveling from the atmosphere to the greenhouse.

      “the atmosphere can not raise the energy supply to the surface … the sun being the ONLY source of energy for the system surface+atmosphere.”
      Here you switched systems in the middle of your thought — from “surface” to “surface+atmosphere”. I would agree that the sun cannot raise the “effective temperature of the surface+atmosphere as seen from space” above 255 K, and that is what is seen.

      **************************************

      “As for the supposed 300 W/m2 flux from the Sun, if it can only heat the ground up to 35 deg. F but in fact the ground is warmer, then the given number 300 is false.”

      You seem to be on the wrong side of experimental data. The average sunlight absorbed the earth is ~ 240 W/m^2. The average at the surface is only ~ 160 W/m^2. These could be off a few W/m^2, but not too much. We have a clear paradox — you suggest that average sunlight must be significantly more than 300 W/m^2, but the actual amount is significantly less than 300 W/m^2.

      The obvious answer is that the sun is not the only source of energy to the “surface” — the other system called “atmosphere” also contributes.

      Remember — be careful with systems. The sun IS the only source of energy for the “surface+atmosphere” system, but it is NOT the only source for the “surface” system.

  48. Jim Butts says:

    We all have experienced the greenhouse effect when returning to a car which has been sitting out in the sun with the windows up. The high temperature inside is easily explained with simple physics concepts. Sunlight is coming from the surface of the sun which is a hot gas at about 6000 degrees Kelvin and this electromagnetic energy is mostly at wavelengths in the visible light region around 0.5 x 10-4 cm (0.5 microns). Glass is transparent to this radiation (by design so we can see through it) so the sunlight goes right into the car and is absorbed by the interior materials which heat up somewhat above where they would be if the windows were open and air were freely circulating keeping things more or less at the outside air temperature. The materials inside the car will heat up and radiate energy but since they are more or less at “room” temperature (~ 300K ) they will radiate energy at wavelengths 20 times longer than the visible light wavelengths in the neighborhood of 10 microns (wavelength of radiated energy is inversely proportional to absolute temperature) and these wavelengths are strongly absorbed by the glass. With the windows up, energy is trapped inside and the temperature inside goes up. The inside temperature will be more or less uniform due to the thermal radiation exchange between all inside surfaces and due to convective heat flow of the air inside the car. But the energy and temperature inside cannot go up indefinitely; a point will be reached where the temperature differential from inside to outside will cause heat to flow from inside to the outside of the car. The path of least resistance is through the glass windows. Heat conduction through the glass will increase until the energy flow out of the car is equal to sunlight energy entering the car. If we assume that the total glass area which is transporting heat out is twice as large as the glass area that is admitting sunlight then we have:

    Energy flux out = μΔΤ/Δh = V/2

    where μ is the thermal conductivity of glass (~ 1 Watt/m2/(degree C/m)), ΔΤ is the temperature drop from the inside to the outside of the glass, Δh is the thickness of the glass (~ 0.5 cm = 0.005 m for typical car windows), and V is the solar visible flux through the atmosphere (~ 1000 Watts/m2 on a clear day). Thus the temperature drop through the glass is:

    ΔT = VΔh/2μ = 1000×0.005/(2×1) = 2.5 deg C = 4.5 deg F.

    At the outside surface of the glass the energy balance requires:

    V/2 = (σT4 + HU )

    where T is the temperature of the outside surface of the glass, σT4 is the blackbody radiation flux from the outside surface of the glass (σ = 5.67 x 10-8 Watts/m2/(deg K)4 , and HU is the energy flux outward due to conductive/convective cooling by the air. If the air is very still so that cooling by the air is very small we get the maximum glass temperature,

    T = (V/2σ)1/4 = (1000/2×5.67 x 10-8 )1/4 = 306 K = 33 C = 91 F

    and the temperature inside would be 91 + 4.5 = 95.5 degrees F. The other extreme would obtain when it was windy and convective cooling keeps the external glass surface near the external air temperature in which case the inside temperature would be just 9 degrees F greater than the external air temperature. The real situation is somewhere in between these two limits. I found by experiment that on a clear still day when the external air temperature was 72 F the internal temperature in my car was 89 F a difference of 17 degrees.

    So to recap, the greenhouse effect is the trapping of heat energy inside the car (or actual greenhouse) which occurs because the visible light can get in through the glass, is absorbed by the upholstery and other materials inside the car (or the plants in an actual greenhouse) and is reradiated as long wavelength infrared radiation which bounces around (actually absorbed and reradiated by) all the interior surfaces, but cannot get back out through the glass. The temperature inside increases until the heat energy conduction out through the glass equals the visible light energy coming in through the glass.

  49. James Goodone says:

    Jim Butts says August 15, 2013 at 6:47 PM
    “So to recap, the greenhouse effect is the trapping of heat energy inside the car (or actual greenhouse) which occurs because the visible light can get in through the glass, is absorbed by the upholstery and other materials inside the car (or the plants in an actual greenhouse) and is reradiated as long wavelength infrared radiation which bounces around (actually absorbed and reradiated by) all the interior surfaces, but cannot get back out through the glass.”

    And now let’s see what happens really.

    1. The sun is capable of warming the surface up to a much higher temperature, than we usually observe, the lower temperature we observe is the result of air cooling, air cools the surface by conduction and convection.

    2. If we reduce the convective cooling by closing windows etc. (but still allow the sunlight get inside, like through windows), we reduce decrease in temperature and get as a result a higher temperature.

    This is what happens in cars, homes and greenhouses.

    “Bouncing radiation” providing the inside with additional energy is equal to creating energy out of nothing and therefore a complete nonsense, sorry.

    • Massimo PORZIO says:

      James,
      I’m skeptical too about the “back-radiation” heating, but that’s not the case. The glasses are emitting the IR of their own inside surface temperature, they are not “back-radiating” anything.
      In fact, if they was “back-radiating” the process should be immediate, since the radiation runs at the speed of light (or little less in matter). The reality tell us that, in greenhouses (like a car with windows up), the temperature rises in very long times if compared to the speed of light.
      The car glass just reduce the net outgoing flow of IR from the internal surfaces, so that they heat up themselves and the air inside and the internal surface of the car glasses too. This is a positive feedback which is compensated by the thermal energy loss of the car glasses which depends on their thermal resistance (or the thermal conductivity).

      ““Bouncing radiation” providing the inside with additional energy is equal to creating energy out of nothing and therefore a complete nonsense, sorry”
      It’s not the bouncing radiation that heat the inside, it’s the fact that the inside objects reduce their net outgoing IR flux.

      • James Goodone says:

        Back radiation or not, on Dr.Spencer’s picture things inside the system get more energy for any given period of timethan the whole system gets from the Sun for the same period of time. This is impossible.

        • Ball4 says:

          Define arrows down as + in 1 sec. period of time over 1m^2 greenhouse (GH).

          +300 solar rad. + 450 terrestrial rad. – 475 terrestrial rad. – 275 terrestrial conv. = 0

          GH energy budget IS balanced so IS possible steady state. 100% (solar + terrestrial) energy originally from sun fusion process using up mass. Unless you want to add the 0.1 joule from underground, earth orig. energy + fission process.

          • James Goodone says:

            Ball4 says August 16, 2013 at 5:12 PM
            “Define arrows down as + in 1 sec. period of time over 1m^2 greenhouse (GH).
            +300 solar rad. + 450 terrestrial rad. – 475 terrestrial rad. – 275 terrestrial conv. = 0
            GH energy budget IS balanced so IS possible steady state.”

            Your “0” and “balanced” are completely irrelevant to the question how high or low the temperature inside the greenhouse is. Just replace your “+450” with “+950” and “-475” with “-975”. You will get the same 0 and “balanced” as a result, but a very different temperature inside. In fact, you can get ANY temperature you wish on the picture by choosing the numbers you like. Despite the same energy supply from the Sun (300) you can “make” the inside even hotter than the Sun and still have “0” and “balanced” as a result. I hope you understand now that your “0” and “balanced” are irrelevant.

            The surface temperature depends on how much it gets, and this can not be more then what the Sun delivers. I have shown this already on this thread, this is a very simple math. By the way, the calculation for Trenberth picture the IPCC referred to in their reports would be similar with the same result: impossible.

          • Ball4 says:

            James G. 5:48pm says: “Just replace your “+450″ with “+950″ and “-475″ with “-975″”

            You mean like on Venus? No, James the 450 and 475 are roughly fixed for earth’s steady state energy balance (note I wrote terrestrial) but maybe not impossible to simulate with a searchlight pointed down into the GH. These are not spatially and temporally avg.d just roughly a local city on EARTH at a certain time of day with clear sky probably. Would be a good learning experiment for James to build the boxes (it is pretty easy on EARTH) and report what happens on overcast days, moving them around to say Kapuskasing and Phoenix, and do other tests James can just think up, maybe point them DOWN! Advancement in knowledge is good.

            Note to Dr. Spencer: Add a note to your top post that your GH is built on Earth, James G. seems easily confused wandering around the solar system. Someone else built their GH on the moon IIRC. Replace James Venusian style +950 with +0 or starlight there? ~Yep or a searchlight.

  50. Stevek says:

    What about greenhouse in a vacuum. No air, no wind. Hotter than non greenhouse ?

    • James Goodone says:

      Stevek says August 15, 2013 at 8:10 PM
      “What about greenhouse in a vacuum. No air, no wind. Hotter than non greenhouse ?”

      No, colder, if you mean a non-greenhouse in vacuum, too. Because glass roof blocks some solar IR, so less sunlight would get into the greenhouse.

    • Tim Folkerts says:

      Yes!

      A greenhouse on the moon would be warmer than the rocks around it. Since the moon spins slowly we can assume the surface is pretty close to the BB temperature. Lets imagine a place where the rocks are absorbing 1000 W/m^2 and emitting 1000 W.m^2 out to space. If this region is covered by a sheet of glass, the *glass* will emit 1000 W/m^2 upward to space — and also 1000 W/m^2 downward*

      To conserve energy, the surface is absorbing 1000 W/m^2 of sunlight and 1000 W/m^2 of thermal IR, so it will warm up until it emits 2000 W/m^2 upward.

      This has all been rehashed in numerous threads in numerous blogs.

      ————————————

      * The glass will actually emit slightly more than 1000 W/m^2 due to the temperature gradient across the glass.
      * The calculations assume clear glass that blocks all thermal IR.

      • James Goodone says:

        Tim Folkerts says August 15, 2013 at 8:51 PM
        “Lets imagine a place where the rocks are absorbing 1000 W/m^2 and emitting 1000 W.m^2 out to space. If this region is covered by a sheet of glass, the *glass* will emit 1000 W/m^2 upward to space — and also 1000 W/m^2 downward*
        To conserve energy, the surface is absorbing 1000 W/m^2 of sunlight and 1000 W/m^2 of thermal IR, so it will warm up until it emits 2000 W/m^2 upward.”

        Your additional warming is impossible, unless 1000=1000+1000 is correct. OK, let me simplify it for you: it is equivalent to 1=2.

        Again, the system “1m² surface + glass” receives every second 1000 Joules energy from the sun. This makes 10,000 in 10 seconds. In your version a part of the system (the surface) receives 20,000 Joules in the same 10 seconds, that is twice as much for the same period of time.

        So, in your version a part of the system receives MORE energy than the whole system for the same period of time. One does not need to have a degree to understand that this is impossible.

        • Tim Folkerts says:

          To make an analogy ….

          Suppose you get paid $1000 per day and you pay out $1000 per day (and have $10,000 in the bank). At the end of every day you have $10,000 Money is “conserved”.

          Now suppose I intervene. I intercept that $1000, pay out only $200, give you $400 back, and pocket $400. I didn’t create any money. I simply found a way to spend less, so that you and I both end up with a little more.

          If we keep this up for a while, pretty soon you have $20,000 and I have $10,000. We could even change the rules. You could now give me $2,000 per day, I could pay out $1000, and give back $1000. You would not be getting any richer now; I would not get any richer; but we would both have accumulated more money.

          All of this comes from just the $1,000 per day you are getting paid “from the outside”. No new dollars were created by me. The “excess” came from the fact that for a while, I let less than $1000 leave the system, so I could store up the extra.

          [The analogy could be made more accurate by making rules like “everybody pays out 10% of their current balance each day” or even “everybody pays out 5.67e-8 * (current balance)”. ]

          The “excess” always comes from a temporary decrease in the amount that is allowed to leave.

          • James Goodone says:

            Tim Folkerts says August 15, 2013 at 10:17 PM
            “To make an analogy ….”

            You can claim whatever analogy you like, but the fact remains: in your version of the “greenhouse effect” a part of the system gets more than the whole system for the same period of time, FOR ANY PERIOD OF TIME.

            I guess, absolute everyone understands that this is impossible. Including you.

          • Tim Folkerts says:

            There are LOTS of topics in science that many people are dead sure about — and are dead wrong about.

            If you think my explanation (and Roy’s) is wrong, then what specific law of physics does it violate?

            There are two key laws here, and both are satisfied.
            * No energy (or money in the analogy) was ever created or destroyed, so Conservation of Energy is obeyed.
            * The net flow of energy is always from hotter to cooler, so the 2nd Law of Thermodynamics is obeyed.

            If you *still* think that “everybody” thinks your are right, I would challenge you to go to a physics professor at any local university and see if he agrees with you or with me.

            [NOTE: there are still lots of *other* details about just how hot each part would be, and just how much energy would be moving around by various means, etc, but those are questions of degree, not questions of fundamental impossibility.]

      • Massimo PORZIO says:

        I don’t agree Tim.

        Without any glass thickness and thermal resistance you fall in the paradox highlighted by Dr.Strangelove about Dr.Spencer greenhouse fluxes diagram.

        That is, when you consider the glass “back-reflecting” 1000W/m^2, you must imply that it already emits the same value to the outer space too because it has absorbed the radiation and heated up.
        So, your setup is never balanced because the outgoing radiation is 2000W/m^2, while the incoming is only 1000W/m^2.

        Note that the system has no steady state, because the net flux between the glass and the ground must be 1000W/m^2, and glass must stabilize at a temperature above 0K, so the outgoing flux to the outer space wil be always greater than the incoming flux.

        IMHO it works as follows:
        1) the glass will reach the very same temperature of the ground around the greenhouse on its upper surface.
        2) the lower surface stabilize at a temperature equal to the upper surface plus the temperature delta due to the thermal resistance of the glass multiplied by the solar flux (1000W/m^2).
        3) the ground inside the greenhouse reach a temperature equivalent to a flux which is the sum of the fluxes of the lower glass surface plus the incoming solar flux.

        Have a nice day.

        Massimo

  51. Dr. Strangelove says:

    @Massimo

    350 – 200 = 150 W/m^2
    That’s how I got it. 350 is incoming infrared. 200 is outgoing convective loss from roof (which is suppose is the glass). 450 cancels out as heat flow in and out of the glass are equal. 65 F glass temp. is not constant if wind speed is varying. If you assume constant wind speed, glass temp. will eventually attain equilibrium.

    @tim
    I know there’s no equilibrium with respect to one type of energy. You have to think why I did it that way. The phenomenon is dynamic. It starts in thermal non-equilibrium and moves to equilibrium as time goes on. So there is no one correct temperature because temperature is changing over time. You have to specify the temp. T1 and time t1. You can graph it. Each point in time corresponds to a particular point in temperature until you reach equilibrium. At equilibrium, temp. is constant irrespective of time.

    The reason why I computed 44 F is to determine the direction of convective heat flow from time t = 0 to t1 when temp. T1 = 65 F. In other words, now we know that radiative heat transfer alone cannot attain equilibrium. You need convective heat INFLOW (the direction is important). But this is from time t to t1. Thereafter, the heat flows could change again if the equilibrium is disturbed.

    • Massimo PORZIO says:

      I get your point.
      But it is because Dr. Spencer assumed that the glass as an infinitesimal thickness and zero thermal conductivity.

      • Dr. Strangelove says:

        No need to assume zero thickness or zero thermal conductivity. Constant temperature of glass can be attained if input and output heat flows are equal.

        • Massimo PORZIO says:

          yes, but that’s not a greenhouse at all, or it’s a greenhouse at steady state by night.

  52. Dr. Strangelove says:

    Roy,
    The 93 F equilibrium temperature of vegetation assumes the temperature of air inside and vegetation are the same. Hence, no convective heat transfer. If we assume there’s temperature differential and vegetable is at 85 F, then there is convective heat transfer from vegetable to air inside at 28 W/m^2

    Your proposition that greenhouse heating is due to increased infrared inside the greenhouse is true when outside air temperature is above 118 F, which corresponds to infrared over 300 W/m^2.

    However, such high air temperature is found only in deserts and we don’t put greenhouses in desert because it’s too hot anyway. Therefore, practically all greenhouses do not heat by increased infrared.

    • Dr. Strangelove says:

      Correction:
      Your proposition that greenhouse heating is due to increased infrared inside the greenhouse is true when outside air temperature is above 141 F, which corresponds to infrared over 350 W/m^2.

  53. Jim Butts says:

    Maybe I should make it even simpler. Energy gets into the greenhouse via visible light. Energy gets out of the greenhouse only via conduction through the glass. These must be equal at equilibrium and this determines the temperature differential through the glass.

    • Dr. Strangelove says:

      The glass is also radiating infrared inside the greenhouse. When air is moving outside, heat flows from glass to air outside by convective heat transfer. Conduction is very slow. The greenhouse will overheat if heat flows by conduction only. Radiation and convection are more important.

      • Massimo PORZIO says:

        I guess that Jim was talking of conduction about the glass path only.
        In a previous his post, he stated
        “The temperature inside increases until the heat energy conduction out through the glass equals the visible light energy coming in through the glass.”
        And he also highlighted how the final temperature of his computations was not fixed, but dependent by the outer wind conditions.

        Have a nice day.

        Massimo

    • James Goodone says:

      So, “these must be equal at equilibrium” regardless of how high or low the temperature inside is, right?

      Maybe you have lost focus a little bit. We know that the temperature inside a greenhouse is higher than outside, the whole debate is about what causes it.

      • Massimo PORZIO says:

        It’s just wording:
        I often use “equilibrium” (wrongly of course), instead of “steady state”.

        Have a niced day.

        Massimo

  54. KevinK says:

    Dr. Spencer wrote;

    “So, I guess I’m left wondering…where did the oft-cited claim that a greenhouse does not operate through a greenhouse effect come from?”

    With all due respect, the “greenhouse effect” is only caused by the restraint of convection, period. Textbooks that claim otherwise are incorrect (yes, textbooks are often wrong, what a shock).

    This restraint of convection “robs” the volume outside of the greenhouse of heat energy. The restraint of convection simply concentrates (temporarily) thermal energy inside the greenhouse. Once the energy source goes away (i.e. sunset) the concentrated energy eventually spreads to the volume outside the greenhouse. Perhaps slowly in the summer months, but quite quickly in the winter. This is why greenhouses are not very useful further North.

    The energy returning to the ground from the glass IS NOT (NEVER EVER, NO WAY, NO HOW) “extra” energy that adds to (sums with, is in addition to, etc) the sole energy source which is the Sun.

    You cannot simply sum up ”alleged” energy fluxes (incorrectly expressed in units of POWER i.e. Watts/m^2) and get a correct answer. Sure you can add them and get an answer but it is incorrect.

    I know some folks think I’m a lunatic on this topic, but you cannot ignore the speed at which these energy flows travel. To correctly analyze this subject you need to remember that the radiation is flowing through the system at the speed of light (well, very nearly the speed of light in a vacuum) and the heat traveling via convection and conduction is traveling at the “speed of a slug” (in comparison), quite a bit slower.

    The energy from the glass surface is simply emitted back towards the ground for another pass through the system at the speed of light. Back and forth it bounces, ground to glass to ground to glass, etc, etc, each time some is lost from the outer surface of the glass to space. These bounces simply delay the flow of radiative energy fluxes by a very short period of time. This time period is determined by the travel time (speed of light times the height of the greenhouse (a few microseconds), and the delay as the energy flows through the glass (a few milliseconds)).

    This is why it is perfectly possible to make a totally useful greenhouse (the real plant growing kind) out of plastic films that are substantially transparent to LWIR. It is done all the time, the plastic films are inexpensive and quite a bit tougher than glass.

    Again, I respectfully suggest that you study how a multi-layer optical anti-reflection filter works. It is a very similar geometry (thin layers that reflect some energy back towards a surface) but it only reduces reflected energy (i.e. “makes more energy stay here”) with the presence of optical interference (constructive and destructive).

    To correctly analyze this you need to follow a small packet of energy (a milli-Joule for example) as it flows through the system alternating as visible EMR, thermal energy, LWIR, thermal energy, LWIR, etc, etc.

    Cheers, Kevin.

    • Ball4 says:

      KevinK 9:44pm: “With all due respect, the “greenhouse effect” is only caused by the restraint of convection, period.”

      I’m curious given this statement then how you explain the experiment of Prof. Platt linked by Dr. Spencer 8/13 9:23am above. The PE covered box has exactly the same “restraint of convection” yet it runs much cooler steady state than the acrylic and glass covered boxes.

      This Pratt experiment is repeated in the 2nd link except for the PE covered box having added insulation over the other two and now running essentially the same temperature steady state as the acrylic and glass covered boxes.

      I would conclude from these modern experiments that conductive, convective and radiative energy transfer all play a role in the steady state temperature of a greenhouse on the earth surface. Their relative contributions depend on environment at the location of the greenhouse & greenhouse construction, essentially the same conclusion as Dr. Bohren reported given the literature & modern experiment by the mid-1980s in the link I posted above 8/11.

      • Mike Flynn says:

        Ball4,

        Prof Pratt was unable to replicate his initial results. Other researchers have pointed out various things that Pratt neglected, that initially led him to false conclusion.

        It’s instructive to learn what he overlooked. It’s a bit long to put here.

        So far, no one has been able to demonstrate CO2 warming anything at all, by the supposed “greenhouse effect”.

        All attempts to prove it seem to indicate its non existence.

        Live well and prosper,

        Mike Flynn

      • Ball4 says:

        Mike Flynn 7:02pm: “Prof Pratt was unable to replicate his initial results.”

        Where did you find the link or report of this? The 2nd link Dr. Spencer listed 8/13 9:23am independently replicated Prof. Pratt’s experiments with added insight due the exp. mod.s reported. Watch the pea carefully though. Insulation is added where Dr. Pratt had no added insulation.

        What did Prof. Pratt neglect? What false conclusion did Prof. Pratt make?

        “…no one has been able to demonstrate CO2 warming anything at all…”

        http://en.wikipedia.org/wiki/File:TyndallsSetupForMeasuringRadiantHeatAbsorptionByGases_annotated.jpg

        “Air from which water vapor and carbon dioxide had been removed deflected the galvanometer dial by less than 1 degree, in other words a detectable but very small amount (book, page 80-81).”

        Tyndall’s book is here, will take some time to read. Or find a good modern text on Atm. Radiation. See the modern absorption/emission charts for CO2. Compare them at the surface and at satellite orbit. The difference in CO2 lines is existent.

        http://archive.org/details/contributionsto02tyndgoog

        • Will Pratt says:

          Nowhere in any of Tyndall’s experiments does he measure, or even attempt to measure, an increase in temperature of the gas under observation.

          NEVER, NOT ONCE!

        • Ball4 says:

          Will 3:34pm – John Tyndall did indeed attempt and succeed to measure and quantify the difference in temperature due to radiant energy absorption and emission in different gases including water vapor and CO2 over 150 years ago. His book calmly describes the process, there is no need to shout. Tyndall was able to demonstrate CO2 warming by the supposed “greenhouse effect” (GHE). Read the experiments documented in the book, it’s free. Many others since have improved on his techniques, they are proven repeatable. Try it.

          Today specialized radiometers and spectrometers on earth’s surface looking up and on board satellites looking down are used to measure and quantify earth’s supposed GHE. Mr. Wizard used an improvement: Crookes radiometers, you must have missed that show. What do you know: work! just from invisible IR radiation. They demonstrate radiative energy transfer just like in the theoretical GH in the top post and in the atm. at large. And even other wavelengths; IR being the hottest part of the spectrum per Herschel in 1800 – found using just a thermometer & prism in the atm.! Fascinating; you may learn a lot if you try.

          • Will Pratt says:

            Ball4 5:55pm

            Perhaps you haven’t read John Tyndall’s “Contributions to Molecular Physics in the Domain of Radiant Heat”.

            I have.

            Nowhere does Tyndall measure, attempt to measure or document, any increase in the temperature of any of the gases in the tube.

            He merely assumes it, as do you.

            The reason he does not and couldn’t have done so, is quite simple. It’s called “local thermal equilibrium”. The effect of which would have run counter to any point he was trying to falsely to imply.

            Water vapour and indeed CO2 or any other so called “GHG” for that matter, can only ever induce cooling, such is the nature of “radiative energy transfer”, passive feedback mechanisms and thus reality.

            Or perhaps you could point to the thermometer on the tube which is measuring the temperature of the gas inside in this diagram: Tyndall’s Setup

            No…I didn’t think so!

            As is clear from Tyndall’s equipment, his brass tube is thermo-coupled to the ambient temperature of the local environment because it is not insulated in any way. Thus the gas in the tube is also directly thermo-coupled to the ambient temperature of the local environment and therefore directly subjected to effects of local thermal equilibrium.

            This is why there is no thermometer measuring the temperature of the gas inside the tube.

            The thermopile, the only heat sensor present in the setup, is detecting the heat source, not the gas. According to Tyndall the thermopile detects a drop in energy reaching it when the “GHG” is introduced into the tube.

            As I said, nowhere does Tyndall attempt to measure a corresponding increase in the temperature of the gas because, for the reasons I have given, there would not be one.

            Or, perhaps you are unaware, unlike Tyndall, of the nature of passive feedback mechanisms. Perhaps you don’t understand that passive feedback mechanisms are always negative and lead to stability.

            So called “GHG’s” in the atmosphere, are all passive feedback mechanisms and therefore, negative feedback mechanisms. Water vapour, being Tyndall’s and the IPCC’s most powerful “GHG” is the most easily observed negative feedback mechanism we have in nature. Or perhaps you don’t understand what a “Maritime Climate” means.

            Let me help you out there. A “Maritime Climate” is one where higher than average atmospheric water vapour content, due to close proximity to the ocean, leads to cooler summers and milder winters than would generally be the case at those latitudes further inland. A negative water vapour feedback mechanism. A stabilising effect. A negative feedback mechanism works against the prevailing conditions. It works against the warming of the summer and it works against the cooling of the winter. It works against the warming of the day and the cooling of the night.

            Simple truth’s yes! Fascinating, indeed! You may learn some small part of the reality of nature, if you try!

          • Ball4 says:

            Will – You really do need to read & quote directly in Tyndall’s own words from Tyndall’s book you name to discuss Tyndall’s experiments, your own words will not do. Don’t just arm wave. That won’t get you a seat on the bus, you have to show the driver the money.

            Will claims: “Nowhere does Tyndall measure, attempt to measure or document, any increase in the temperature of any of the gases in the tube.”

            Page 90 Tyndall writes to document: “My first care was to determine the difference of temperature within the experimental tube….I then examined, by an extremely sensitive thermometer, the increase of temperature produced by the admission of dry air into the tube…”.

            This is one “where” Tyndall does “measure, attempt to measure or document, any increase in the temperature of any of the gases in the tube.”

            Will then asks me “…perhaps you could point to the thermometer on the tube.” I don’t have to, Tyndall just did. Use Tyndall’s own words from the source; internet wiki or even random posters may not be up to his standards. Always, I mean always, go to the source material. We all learn that way, even you.

          • Will Pratt says:

            Pardon me, I made the assumption that all would know I was referring to so called “GHG’s”

            Allow me to rephrase that;

            Nowhere does Tyndall measure, attempt to measure or document, any increase in the temperature of any of the gases now falsely referred to as “greenhouse gases”, in the tube.

          • Will Pratt says:

            So appart from deliberately quoting me out of context in order to split hairs, what exactly is your point?

        • Will Pratt says:

          Ball4 5:55 PM

          “John Tyndall did indeed attempt and succeed to measure and quantify the difference in temperature due to radiant energy absorption and emission in different gases including water vapor and CO2 over 150 years ago.”

          You don’t seem to be able to find a quote to substantiate that claim in “Contributions to Molecular Physics in the Domain of Radiant Heat”.

          And you won’t either. No one has ever been able to show it because it does not happen. 150 years on and we are all still waiting.

          Still keep looking. The entire treatise is a terrible bore, however you may actually learn something. Like what a complete fraud John Tyndall actually was.

  55. Mike Flynn says:

    Tim,

    I wrote : –

    2) “If you can’t accept that this shows that the Earth hasn’t finished cooling, then obviously you will be a Warmist to the day you die.”

    You responded : –

    How does one relate to the other???

    Of course the (interior of the) earth has not “finished cooling”. Around 0.1 W/m^2 of geothermal energy is “leaking” out of the surface (from the estimates I have seen). The interior is indeed slowly cooling and has been for billions of years. This is not enough to make a big difference in the earth’s surface temperature — certainly not enough to warm the surface from 255 K to 288 K.

    Can I politely ask two questions?

    1. If the Earth is, as you put it, “leaking energy”, does this mean that its surface temperature is :

    A) rising
    B) falling
    C) not changing.

    2. When was the Earth’s surface ever the 255K you claim that it has warmed from. It’s no use now claiming that the Earth’s surface “would have” cooled to 255K in the absence of your GHG’s. slowed cooling is not he same as warming.

    I say again – the Earth has not finished cooling. Once the Earth has become isothermal throughout, apart the transitory and ephemeral diurnal surface variations, then we can say cooling has ceased. Until then now.

    I wish you well.

    Live well and prosper,

    Mike Flynn.

    • Tim Folkerts says:

      “1. If the Earth is, as you put it, “leaking energy”, does this mean that its surface temperature is :

      A) rising
      B) falling
      C) not changing.”

      D) impossible to determine from the information given.

      During the winter, my house is always “leaking energy” to the world around. To know if the temperature on an exterior wall is rising or falling or staying the same requires a knowledge of ALL the energy transfers. We need to know if more energyt is leaving the surface to the air, or arriving at the surface from the interior.

      “2. When was the Earth’s surface ever the 255K you claim that it has warmed from.”

      You are being too literal. Going back to the house analogy, I could calculate that my house might be 5C on a cold the winter with the furnace running but insulation and, and then measure it to be 20 C with insulation. I don’t have to actually pull the insulation off to say the insulation “warms the house by 15 C”.

      Further more, the earth has indeed had a much lower surface temperature in the past — during the glacial periods or the “snowball earth”. So the earth has warmed from lower temperatures to the current higher temperature. The continuously decreasing heat flow from the interior could not be responsible for such rebounds.

      • Tim Folkerts says:

        Let me rewrite that a little to correct and clarify.

        I could calculate that my house should be 5 C on a cold winter day with the furnace running but with NO insulation. If I measured the house on that day WITH insulation, it might be 20 C.

      • John K says:

        Hi Tim,

        Thank you for the post you made good points. However, you wrote:

        “Further more, the earth has indeed had a much lower surface temperature in the past — during the glacial periods or the “snowball earth”.”

        True, but as I’ve indicated in previous posts the period immediately prior to the rapid glacial onslaught proved enormously warmer. Extensive geological evidence including the very existence of permafrost point to a period in which the polar regions supported lush, green vegetation and very large animal life forms including elephant sub-species (Mastadons, mammoths), rhinoceros, wolves, humans etc. All these were found in regions far too cold to support anything like these creatures today. The polar ice-caps likely didn’t even exist. Different and conflicting attempts have been made to date the remains (DNA analysis, carbon-dating) and while we don’t have any absolute chronometers all the dates obtained by such guesses place them within a few thousand to tens of thousands of years. In any case, the glacial period commenced rapidly freezing and preserving many forms of vegetation and animal life. Given historical precedents the planet remains much cooler than periods prior to the recent glaciation.

        Thanks again for the post.

      • Mike Flynn says:

        Tim,

        I asked about the Earth. Your analogy is not useful. To know whether the temperature of a wall is rising, falling or staying the same, does not require anything more complicated than a temperature measuring device. If the calculated and the measured temperatures differ, which would you be inclined to use?

        I am not sure how one can be “too literal”.

        Once again, your analogy lacks utility. Insulating a cold, unoccupied house, or a block of ice, will not warm it one bit. In fact, the greater the insulation, he slower the rate of warming from external radiation.

        The glacial periods were geographically restricted. There is no evidence at all that the Earths average surface temperature has ever been less than it is now.

        Higher, yes, obviously, if the Earth was created molten. Lower, physically impossible, given that the crust is generally overlaid by the hydrosphere and the atmosphere etc.

        I think we may as well agree to disagree.

        Live well and prosper,

        Mike Flynn.

        • Tim Folkerts says:

          Mike says: “Insulating a cold, unoccupied house, or a block of ice, will not warm it one bit. In fact, the greater the insulation, he slower the rate of warming from external radiation.”

          And comparing an un-heated block of ice to a heated earth is an analogy that lacks even more utility. People come back to this analogy again and again and again, thinking that it somehow argues against CO2 and/or insulation assisting the active heater to further raise the temperatures. Only an analogy with an active heater is of any use, since the earth is actively being heated by the sun.

          • Mike Flynn says:

            Tim,

            I am not sure where you get the idea that I have neglected to include he Sun. However, this whole exchange indicates that avoiding facing up to reality, and concocting an artifiicial scenario that favours one’s ideas, achieves nothing.

            I will continue to adhere to facts. If you have any verifiable experimental evidence to support your position, I would be glad to consider it.

            Of course, you haven’t, and neither has anybody else. Hence all the ridiculous analogies, and spurious calculations. Even the self proclaimed expert climatologists have achieved precisely nothing of use, despite all their supposed “research”.

            Time will tell, and I don’t expect to be grovelling in mortification any time soon.

            As I have said, let us agree to disagree, until you can provide a fact (or maybe two).

            Live well and prosper,

            Mike Flynn

  56. Mike Flynn says:

    Tim,
    Sorry. Couple of typos. Bit of a hurry.

    “Until then now.” should read “Until then, not.”

    “. . . not he same as warming.”, should read “. . . not the same as warming.”

    I’m off.

    Live well and prosper,

    Mike Flynn.

  57. papijo says:

    I agree with the above comment from Kevin, but would like to add a remark: the convection exchange is not correctly calculated in the R. W. Spencer calculation. This exchange coefficient should vary dramatically depending on the wind speed, the ground / leaves humidity, the air humidity, the actual exchange surface (total leaves surface vs ground surface), etc. The “green house effect” is in my opinion just the consequence of the variation of the convection coefficient !

  58. papijo says:

    An addition to my above remark is that the calculation done by R. Spencer should be valid in the case not of a green house, but just in the case of a “small” shad with glass cover, but no walls

  59. Max™ says:

    Wait, Roy… are you saying cracking a window a little won’t lead to a car being cooler than it would be if they were all the way up, because the convective losses through the window gap aren’t important compared to the radiation inwards by the windows/roof/other warmed surfaces?

  60. Massimo PORZIO says:

    Sorry, I prefer to repeat this post about the “greenhouse on the moon”, because it’s a little buried in the many post above, and I would like to know Tim’s opinion (and the opinion of all the others of course).

    I don’t agree with you Tim.

    Without any glass thickness and thermal resistance you fall in the paradox highlighted by Dr.Strangelove about Dr.Spencer greenhouse fluxes diagram.

    That is, when you consider the glass “back-reflecting” 1000W/m^2, you must imply that it already emits the same value to the outer space too, because it has absorbed the radiation and heated up.
    So, your setup is never balanced because the outgoing radiation is 2000W/m^2, while the incoming is only 1000W/m^2.

    Note that the system has no steady state, because the net flux between the glass and the ground must be 1000W/m^2, and glass must stabilize at a temperature above 0K, so the outgoing flux to the outer space will be always greater than the incoming flux.

    IMHO it works as follows:
    1) the glass on its upper surface will reach the very same temperature of the ground around the greenhouse.
    2) the lower surface stabilize at a temperature equal to the upper surface plus the temperature delta due to the thermal resistance of the glass multiplied by the solar flux (1000W/m^2).
    3) the ground inside the greenhouse reach a temperature equivalent to a flux which is the sum of the fluxes of the lower glass surface plus the incoming solar flux.

    Have a nice day.

    Massimo

    • Tim Folkerts says:

      Massimo, I agree with your assessment. I was taking the limit where the glass was infinitesimally thin so your Point 2 didn’t matter. A quick calculation suggests that a 1 mm thick glass sheet would have a temperature difference of about 1 K for the given circumstances. If we assume a 10 K difference across the glass, then the downward radiation would be ~ 1100 W/m^2 instead of 1000 W/m^2. So the ground would be receiving and emitting 1000 + 1100 = 2100 W/m^2 instead of 2000 W/m^2. This would raise the surface temperature from ~433K to ~ 438K.

      So the thickness of the glass and its thermal conductivity makes a measurable difference, but not a hugely critical impact.

      (Please check the numbers if you want. I did this rather rapidly.)

    • Mike Flynn says:

      Massimo PORZIO,

      I agree with you that I also don’t agree with Tim. May I suggest that the glass on its upper surface will only be the same temperature as the surrounding ground if it possesses precisely the same physical characteristics relating to photon interaction as the surrounding ground.

      I am not looking to be controversial. I hope you understand.

      Live well and prosper,

      Mike Flynn.

      • Massimo PORZIO says:

        Hi Mike,
        nice to read you again here after a long time.

        Yes you are right, the temperature of the upper surface of the glass layer should be corrected by its emissivity in the LWIR range respect to the surrounding ground emissivity.
        I missed it indeed.

        Have a nice day.

        Massimo

      • Tim Folkerts says:

        Mike, besides assuming that glass was thin, I was ALSO assuming that the glass and rocks were the same emissivity. For simplicity in the calculations, I was using ε=1 for both. Tables like this (http://www.thermoworks.com/emissivity_table.html) suggest that glass and most minerals are in the range 0.9 < ε < 1, so I was not far off in any case.

        For "back-of the envelope" calculations that are appropriate for blogs like this, it is easiest to start with ideal conditions. Such idealizations suggest the bare surface should be ~ 364 K; the glass will warm the surface by ~ 70 K. More detailed calculations would indeed have to take into account the emissivities, reflectivities, insolation, thermal conductivities, etc. This could change the 70 K rough estimate considerably, but it will NOT make the effect 0 K.

        Other than these details (that effect the *magnitude* but not the *existence* of the warming due to the glass cover), I am in agreement with Massimo, so I curious where specifically you disagree with me and agree with Massimo.

    • John K says:

      Hi Massimo Porzio,

      Thank you for a good post. However, I take issue with your claim:

      “1) the glass on its upper surface will reach the very same temperature of the ground around the greenhouse.”

      Hmmh! Glass being largely transparent to visible light will allow most of the spectrum to pass through unobstructed. Someone pointed out in a previous post that green-house glass is not completely transparent and an absorptive material will often be added to it. Nevertheless, the vast majority of the visible spectrum will pass through the glass to be absorbed by the largely opaque ground or plants. The ground will absorb a great deal more energy than the glass unless the ground is unusually reflective, painted white or with special thermal paint that doesn’t absorb as much energy or simply covered. Please explain why you believe the temperatures should be similar.

      Thanks, I’m interested to know your thoughts.

      • Massimo PORZIO says:

        Hi John,
        thank you for considering my issue.

        In that setup, I believe that it will reach that temperature because the glass is warmed by the LWIR coming from below and converted from the UV/VIS/SWIR by the ground underneath the glass itself (I ignore in this case the glass warming due to the incoming UV and some SWIR absorbed by the glass itself).
        The glass upper surface outgoing value is a constrain, that because the system to be in steady state must return to the outer space the same amount of the energy, it previously received from the Sun, that is those 1000W/m^2. This is a fixed value, and it’s the very same value of energy flow in the glass which fixes the lower glass surface temperature.

        My further consideration is:
        What does it happen there, if the lunar greenhouse was built of clear NaCl sheets?
        My answer is:
        the greenhouse doesn’t work at all (except for the tiny absorptance of NaCl in the UV/Vis bands).

        And I note how on this Earth we are still arguing if there is any difference between build a greenhouse of glass or NaCl, while on the Moon the results are so clearly different.
        IMHO Prof.Wood was right, in an atmosphere the thermodynamic path works much better than the radiative path.

        Have a nice day.

        Massimo

        • For a greenhouse of negligible height, I would agree with Massimo that, at least theoretically, the upper surface of the glass roof should be at the same temperature as the ground outside the greenhouse. The lower surface of the roof would however be hotter due to the temperature gradient across the glass resulting from upward heat flow. As my experiment with a salt window shows, the temperature gradient would be less with a roof transparent to FIR though not zero by any means.

          The walls of a typical greenhouse make a difference in that they permit more heat to be removed by convection, so one could expect the outside of the greenhouse to be a bit colder than the ground outside the greenhouse. Whether the lower surface of the roof would be hotter or colder than the ground outside would then depend on the circumstances.

          • Massimo PORZIO says:

            I agree with you Prof. Pratt,
            but maybe you missed that we were talking about an hypothetical greenhouse on the Moon.
            Since there is no atmosphere there, there is no heat exchange with the exterior too, so the radiation of the roof must always match the Sun incoming radiation at thermal steady state. And for the same reason, ignoring it’s UV/Vis/SWIR absorption, a greenhouse made of NaCl (or halite) should have the structure at 0K (theoretically of course).

            Refining the lunar setup, one could argue that the incoming radiation arrives normal to one radiation plane only, because of the very long distance from the source (the Sun). While the LWIR converted by the rocks below will spread over the whole structure of the greenhouse (roof and walls) so the temperature should be far less than the surrounding rocks.

            Have a nice day.

            Massimo

  61. Peter Norman says:

    I have an evacuated tube solar collector sitting on my roof. It heats my hot water when the sun shines. If I move the solar panel into the greenhouse and fill the greenhouse with Co2, will this same solar panel then collect more energy through the “greenhouse effect”?

    • Very nice question, Peter. I would say the simplest way to answer it is to ask about the mass of CO2 involved.

      On the one hand you’re proposing 100% CO2 as against the 0.04% of atmospheric CO2, a factor of 2500.

      On the other hand you’re looking at about 3 m of gas vs. about 8 km of atmosphere (taking into account that density decreases with altitude), a factor of 2666.

      Sounds like about a wash to me. If increasing CO2 from 300 ppmv to 400 ppmv adds 1 °C to the global temperature, then one might imagine that increasing the CO2 level of your greenhouse from 75% to 100% might also add 1 °C.

      The fallacy in this line of reasoning is that whereas 8 km of atmosphere has a lapse rate of 10 °C/km that therefore drops the temperature by 80 °C, the 3 m of your greenhouse has no such temperature drop. Lapse rate is the key to CO2-induced global warming. Hence the reasoning underlying the greenhouse effect for Earth’s atmosphere does not transfer to your greenhouse.

  62. Fuzzy says:

    This document has lots of greenhouse design references:

    http://www.mnproject.org/resourcecenter/Solar%20Greenhouse%20Resource%20List.pdf

    You can find additional information by Binging “principles of greenhouse design”, but you will get some garbage.

    By the way David Koch, a bogie man of the AGW crowd, has a bachelors and masters degree from MIT in chemical engineering, so he can be expected to be knowledgeable in thermodynamics and other aspects of physical chemistry. His father was a pioneer in refinery technology.

  63. Mike Flynn says:

    Tim,

    You said : –

    “Other than these details (that effect the *magnitude* but not the *existence* of the warming due to the glass cover), I am in agreement with Massimo, so I curious where specifically you disagree with me and agree with Massimo.”

    You disagree with Massimo about the *existence* of the warming due to the glass cover, as I read it.

    I agree with him.

    Live well and prosper,

    Mike Flynn.

    • Tim Folkerts says:

      Maybe we need to get him to clarify. When Massimo said “the ground inside the greenhouse reach a temperature equivalent to a flux which is the sum of the fluxes of the lower glass surface plus the incoming solar flux.” that is a pretty clear indication of a temperature rise for the ground inside. This means roughly doubling the flux relative to the uncovered rock, so the interior will be much warmer.

      • Mike Flynn says:

        Tim,

        I assume Massimo meant what he said. I cleared up one minor point with him. I am not sure what you don’t understand.

        If you ask him a specific question, I assume he will give you a specific answer. Sorry I can’t be of more help.

        Live well and prosper,

        Mike Flynn.

        • Tim Folkerts says:

          But “a temperature equivalent to a flux which is the sum of the fluxes” would be a temperature much higher than a temperature equivalent to just the flux from the sun. Therefore he IS concluding that the ground in the greenhouse is much warmer than the ground outside the greenhouse.

      • Massimo PORZIO says:

        Hi to both.
        I guess it’s just a philosophical quest, and it was indeed.
        I wrote:
        “Note that the system has no steady state, because the net flux between the glass and the ground must be 1000W/m^2, and glass must stabilize at a temperature above 0K, so the outgoing flux to the outer space will be always greater than the incoming flux.”
        Tim, I understand your point that the increase of temperature on the lunar greenhouse is little, but without the glass thickness the system falls in a paradox: it emits more energy than what it receives.

        I hope this doesn’t start a controversial discussion about who is right or wrong on this little detail.

        And I finally could be wrong, because I still haven’t found any real proof that longer wavelength radiations enter a body who is emitting shorter ones.

        Have a nice day.

        Massimo

        • Tim Folkerts says:

          It is always important to be careful with approximations, especially when taking things to limits. In this case, the idealization is that the glass must be thick enough to block the IR, but thin enough to conduct readily. In this limit, the temperature increase in the greenhouse is 10’s of Kelvins, which I would not call “little”

          As long as the limits are taken appropriately, there is no reason to get an paradox here.

          ********************************************

          “I still haven’t found any real proof that longer wavelength radiations enter a body who is emitting shorter ones.”
          I am really not sure how to answer that one. You could look at definition of a blackbody — a surface that both emits and absorbs all wavelengths perfectly. Even for real objects, they must absorb any given wavelength exactly as well as they emit it. This absorption is independent of the photons it is actually emitting.

          Or think about a 300 K object and a 290 K object. Both emit 6 um photons and 12 um photons. You would have to conclude both
          1) the 290 K object (which emits 6 um photons) cannot absorb 12 um photons from the 300 K object)
          2) the 300 K object (which emits 6 um photons) cannot absorb 12 um photons from the 290 K object)

          Even more problematic is that there is no shortest wavelength emitted by an object (the probability just gets smaller). That 300 K object occasionally emits 4 um photons; a bit less often it emits 2 um photons; even less often it emits 1 um photons. Now (using your assumption) the 300 K object cannot absorb much of the energy from the sun!

          • Massimo PORZIO says:

            Hi Tim, maybe I’ve been not clear.
            You concluded that I assumed that a “300 K object cannot absorb much of the energy from the sun”, but I didn’t.
            Or better, I don’t believe I wrote something that could lead you to that conclusion. If I did it, it’s because I probably used wrong words that I believe have a meaning in English that probably is not that.
            What I wrote is that I don’t find until today any explanation of how could a longer wavelength photon be absorbed by a solid or liquid molecular structure which has its molecules already vibrating so that they emit photons at shorter wavelength.
            I know that the VIS photon can enter a solid exciting its outer electrons and successively convert that radiation into LWIR by perturbing the molecular bindings and making the structure “vibrating” and emitting at longer wavelengths.
            But arguing that at a certain temperature photons of longer WL can be absorbed because they fall in the Planck’s BB probabilistic curve, it doesn’t explain how those photons should move the curve to shorter WL.

            Being an electronic engineer, I learnt that light emitting diodes work by electrically exciting the semiconductor junction, the electrons which are in the valence band are moved to the conduction band which is highly unstable, so they

          • Massimo PORZIO says:

            Sorry… I inadvertently pushed on the [Submit Comment] button šŸ™
            Here I continue my previous post.

            I was saying:

            Being an electronic engineer, I learnt that light emitting diodes work by electrically exciting the semiconductor junction, the electrons which are in the valence energy band are moved to the conduction band which is highly unstable, so they fall back to the valence band and doing that they emit photons at the wavelength determined by the specific mobility of the semiconductor electrons.
            These device are bidirectional, that is they work also as narrow band detectors, that is they can be used to produce a current when they absorb photons which have WL included in their typical BB curve.
            You can try to produce current with a green LED exciting it with a yellow one, but the current that you get is generated by the photons which have WL included in both the BB of the two LEDs. In no way the receiving green LED changes is BB curve, even if you submit it to a great yellow photon flux.
            Of course the phenomenon of thermal heating by radiation is completely different, and I would like to have a physical explanation which highlight me how a longer WL photon could move the BB curve to shorter WL in a solid.
            Only that.

            Have a nice day.

            Massimo

          • Massimo’s request, “I would like to have a physical explanation which highlight me how a longer WL photon could move the BB curve to shorter WL in a solid” is a fair one. Here’s how I’d answer it.

            Consider a black body BB at temperature 300 K, in equilibrium with its environment. Total energy leaving BB equals that entering it.

            Suppose further that the environment is not uniform in temperature but that in a certain direction the environment is at 100 K (whence for equilibrium there must be other directions where the environment exceeds 300 K).

            Now introduce a new object G at temperature 200 K so as to block BB’s view of the 100 K source.

            This increases the net energy flux into BB, thereby raising its temperature. There is no violation of the 2nd law of thermodynamics because the environment as a whole, which before G arrived was effectively at the same temperature as BB, is now slightly hotter than BB as a result of introducing G.

            The rise occurs despite the face that G is radiating at a lower temperature than BB.

            In the case of the greenhouse effect, BB is the Earth and its environment is very nonuniform because the heat is coming mainly from the Sun. Space is much colder even than 100 K, namely 2.7 K. G is any greenhouse gas, e.g. water vapor, CO2, etc. When G is introduced it increases the net energy reaching Earth’s surface. This additional downwards longwave radiation or DLR, commonly referred to as back radiation, therefore raises the temperature of the surface of the Earth. There is no violation of the second law of thermodynamics because the environment as a whole, including the Sun, becomes hotter than the Earth when G is introduced. More precisely, more energy now enters the Earth’s surfce than leaves it, and the surface warms accordingly until equilibrium is restored.

          • Massimo PORZIO says:

            Hi Prof. Pratt.
            Yes I already read about that explanation, but this explanation assumes that the energy coming from the body colder than the average BB is at least warmer than the coldest localized place of the BB object itself.
            AFIK the CO2 15um band temperature should be around -80°C so I should find a place on Earth below that temp to absorb it.
            Or am I missing something?

            Have a nice day.

            Massimo

          • @Massimo: AFIK the CO2 15um band temperature should be around -80°C so I should find a place on Earth below that temp to absorb it.
            Or am I missing something?

            Wow, this thread of Roy’s is producing some great questions, thanks for this one, Massimo.

            The connection between wavelength and temperature is not as simple as you suggest. Take a look at the three boldface entries in the “288 K planet” table at

            http://en.wikipedia.org/wiki/Planck%27s_law#Percentiles

            These give respectively the wavelength, neutral, and frequency peaks of Planck’s law of thermal radiation at 288 K. These are the peaks obtained by plotting Planck’s law as a function of wavelength, log(wavelength) (same peak as log(frequency) since log(wavelength) = -log(frequency)), and frequency respectively. The article discusses these.

            One might imagine that as temperature T increases, Planck’s law decreases on the low frequency side of these peaks as a result of the peak moving towards higher frequencies/shorter wavelengths.

            This is false: every point of Planck’s law increases with increasing T.

            The idea that a low frequency photon cannot add any energy to a body whose temperature is “at” a higher frequency (depending on which peak you prefer for that judgment) is based on unsound reasoning. Low frequency photons are not “reflected” by high-temperature objects but are absorbed by them. Any one photon might not add much energy thereby, but you have to look at the total energy flux from the low-temperature sources in order to determine the thermal energy added to the high-temperature object absorbing that flux.

          • Massimo PORZIO says:

            Hi Prof. Pratt.
            Thank you for considering my questions.
            you wrote:
            “One might imagine that as temperature T increases, Planck’s law decreases on the low frequency side of these peaks as a result of the peak moving towards higher frequencies/shorter wavelengths.”
            I never thought that, the contrary, I’m asking from any proof that adding any energy in the longer WL of the BB curve it changes temperature of the same BB curve.
            I know that as its temperature rise the Planck’s BB curve expands to higher frequency instead of shift to higher frequency.

            Effectively, I used the wrong words when I wrote:
            “I still haven’t found any real proof that longer wavelength radiations enter a body who is emitting shorter ones.”

            So I can easily understand that photons at -80°C can be absorbed by a 25°C BB, but is there any experimental proof that the BB increases its temperature?

            The only case I heard of photons frequ

          • Massimo PORZIO says:

            Damned… I did it again… I involuntarily pushed the [Submit Comment] button again!

            I was saying that the only case I heard of photons frequency increase (or in that case multiplication), it was about green/violet diode-pumped solid-state lasers. But in that case the efficiency is very low.

            Have a nice day.

            Massimo

  64. Bill Hunter says:

    Maybe this has already been asked but I see the surface inside the greenhouse receiving 750watt/m2 (300 + 450 for all us math challenged folks)

    Stefan Boltzmann suggests that the surface inside the greenhouse should be 339K or 66C or 150F.

    But it is notated as 85F. Why?

    • Tim Folkerts says:

      275 W/m^2 is carried away by convection in his scenario, so only 475 is carried away by radiation, not the entire 750. 475 W/m^2 gives about 85 F

  65. Alex says:

    Old school science, you know;specific heat, thermal conductivity, latent heat, gas laws, were determined long before any consideration was given to ‘back radiation’. These laws and numbers were determined by gentlemen scientists who meticulously measured and worked out formulae that we use every day in our daily lives and has been the basis of the world we see outside our window now. To reiterate: they factored in all outside influences to achieve their numbers, so any back radiation or still unknown effect was included in their calculations. Now with the ‘new wave’ scien(tologists) where everyone is a ‘quantum mechanic’, we start getting all sorts of calculations based on fallacies.
    I suggest that some of these people should go back to high school and learn something instead of trying to pass an exam. I speak with some authority in this regard because I have been teaching dumb-shit students in a university for 10 years.
    If u want to set new standards that seperate radiation when determining the specific heat of a substance , then be my guest. You can also do the same for heat conductivity in a substance. Whatever you do ,don’t try to convince me to include radiation on top of formulae that already include that effect.
    Its time to leave the Unicorns and My Little Pony where they belong. In the hands of 12 year old virgin girls.
    I’m not interested in imaginary glass that has no properties, neither am I interested in shells around an imaginary planet or blackbodies that have no properties like mass and composition.

    • Arfur Bryant says:

      Alex,

      Could you direct me to a text book or paper which will once and for all explain whether or not a photon from a lower energy molecule will add to the energy level of the absorbing molecule?

      Everything have read so far indicates that there would be no net gain of energy but I am willing to learn more about it.

      Thanks,

      Arfur

      • Arfur Bryant says:

        Sorry, I probably meant ‘receiving’ molecule, not absorbing molecule – as there may be no absorption. (?)

        • Alex says:

          I can’t direct you to a specific book for information but if you want information on electrons moving to different levels of energy after absorption of photons and also emitting photons on dropping to lower levels then I suggest reading up on lasers. These books focus specifically on absorption and emission. Some of it can be heavy going but you will obtain a lot of focused knowledge in the area. You will discover that its the outer shell electrons that emit photons(light) and inner shell electrons that require higher energies to shift and emit X-rays. It’s a complex area with lots of maths but if you can absorb 30% of the information you will be well on your way to ask intelligent questions. Knowledge is about asking intelligent questions. I hope that was helpful.

    • Tim Folkerts says:

      But ‘back radiation’ = “thermal radiation” which is very ‘old school science’.
      * Herschel
      * Kirchhoff
      * Boltzmann
      * Stephan
      * Rayleigh
      * Planck

      I realize that the science of the 1800’s is more ‘comfortable’, but that is no reason to think that the advances of the last 150 years are taking us backwards.

      • Alex says:

        I’m not sure if you are agreeing with me or disagreeing with me. There have certainly been great advances in science in the last 100+ years. I am referring to the ‘old’ science that gave us the values of specific heat, thermal conductivity, etc. These values incorporated the visible world around us. For example: when the specific heat of Iron, water etc. were determined they incorporated everything. There was external heat applied and certain outcomes were measured. After calculation there was a number applied that was referred to as specific heat of the ‘thing’. I don’t know the exact method used but I could probably find it online. But I could bet my left testicle that there was no mention of back radiation. These people actually measured things. They may have got it wrong as far as some detail like back radiation but they just looked at it as energy in, energy retained or whatever and applied a number that we can now use in our daily lives. Sorry, engineers who make the world go around ,use. The foolish old school scientists incorporated everything including back radiation into their calculations without knowing it. So now I am uncertain about the specific heat of water etc. because as a new age man being aware of the radiation emitted from everything I can’t be sure if the old guys got it right. Did they include everything? Perhaps the heat retention of 1 gram of water is incorrect. So the specific heat of water is wrong. The formulae need to be redone. Too complicated for me. I will simply assume that they got it right and included everything known and unknown at that time (more comfortable) and do calculations clearly separating the magic numbers of specific heat etc. from the new numbers of back radiation and being extremely careful not to add the two, thereby getting erroneous outcomes. Nothing worse than mixing metaphors.

        • Tim Folkerts says:

          I think you are assuming too much here.

          For example, heat capacity is often performed in a metal container that is insulated from the surroundings. The container and material to be studied are at the same temperature so that conduction, convection and thermal radiation are minimized. An electric heater is turned on to deliver a known amount of energy and the temperature change is recorded. It is important to do this quickly so that conduction, convection and radiation don’t have time to play a big role.

          Back-radiation was not “incorporated” in the formulas; back-radiation (and conduction and convection) were actively minimized so that the true results could be determined using ONLY the heat from the electral heater.

          It is exceedingly odd that you claim that they must have incorporated all this, yet also claim that this is all too complicated for you to understand.

      • Arfur Bryant says:

        Ok Tim, another chance to explain your earlier statement…

        [“If 1 unit of sunlight is absorbed by the surface and 1 unit of energy is emitted and IF a fraction, 0.x is radiated back, then 1+x is getting absorbed and only 1 is being emitted, and the surface will warm up until the extra 0.x units are being radiated by the now-warmer surface.”]

        So please point me to any publication which effectively states that a photon emitted from a lower energy source will be absorbed by a higher energy molecule.

        Please.

        • Tim Folkerts says:

          This is getting beyond my ability to fathom what sort of “proof” would satisfy you. (For one thing, many skeptics will semi-automatically dismiss any articles that disagree with them because they must have been written by evil CAGW warmistas).

          * The emissivity of an object (at a given temperature)depends on the wavelength. Period. It does not depend on the temperature of the source of the photons. A 10 um photon will be absorbed by a 300 K chuck of iron with a certain probability. The iron has NO WAY TO KNOW if that photon came from the sun or from a block of dry ice. The two photons are identical and therefore MUST be absorbed in the same way.

          Only complex mental gymnastics akin to epicycles will make your interpretation (whatever it might precisely be) work out to give results that agree with reality.

          • Arfur Bryant says:

            Tim,

            {“This is getting beyond my ability to fathom what sort of “proof” would satisfy you.”]

            It is quite easy to satisfy me. Just answer the question. Stop hand-waving. Stop trying to deflect away from the point by talking about what sceptics think. Don’t try to interpret my question any other way.

            YOU are the one who asserted that it would be possible for a surface to absorb back-radiated photons FOR NET ENERGY GAIN. I’m just asking you to provide some sort of evidence that your statement could be true.

            It is not simply about absorption. It is about whether that absorption can lead to the surface warming up.

            You made the assertion. You can hardly blame me for asking you to support it.

            If you have a problem with what you perceive as ‘mental gymnastics’, then maybe you shouldn’t make statements which you later have trouble explaining.

            Don’t make out it is I who has the problem here. I am happy for you to be as technical as you wish. I’ll try to keep up. However, what you say will have to make sound, reasonable, logical sense. I am happy for you to provide ‘results that agree with reality’. Please do so.

            By the way, I don’t react well to sanctimony. Keep it balanced and stick to the point.

          • Tim Folkerts says:

            “It is not simply about absorption. It is about whether that absorption can lead to the surface warming up.”

            That’s simply conservation of energy!

            With “x” amount of energy from the sun, the surface will warm up until it is emitting “x” amount of thermal IR.

            With “x” amount of energy from the sun and “y” amount of energy come from some other object, the surface will warm up until it is emitting “x+y” amount of thermal IR. See, this is why I can’t understand your objections. Do you doubt that “x+y” amount of energy will make something warmer than “x” amount of energy” ???

            **********************************

            [And to head off another common argument, energy has no ‘memory’. A water molecule in a cloud doesn’t say “the energy I have came here from sunlight that was turned into thermal energy by the ground and then got absorbed by me, so I am not allowed to now send a photon back to the ground”. The water molecule simply as energy, some of which will occasionally be emitted as thermal IR that hits the ground and gets absorbed.]

          • Kristian says:

            Arfur,

            He’s got you exactly where he wants you.

            As long as you agree to his axiomatic, conceptual truth that energy being radiated from a cooler object (‘back-radiated’ or otherwise) can and will be absorbed by a warmer object which is constantly being supplied with energy from its own separate heat source, then you cannot win. Because then you will in the end have to admit that his concocted “x OUT, but x+y IN” situation is real. And you will have to go along with him in believing that the laws of thermodynamics are not universal; they can be circumvented … Like he says: “Do you doubt that “x+y” amount of energy will make something warmer than “x” amount of energy” ???” No. And as we all know, energy making an object warmer, increasing its internal energy content, is called HEAT (if not work). So he’s feeding HEAT back from the cool object to the warm object. He’s not calling it heat, of course. But that’s what it is. Because the end result is the same. ‘If it looks like a duck, swims like a duck …’

            And I’m afraid that you will have to come to the same conclusion as he does …

          • Arfur Bryant says:

            Tim,

            [“With “x” amount of energy from the sun and “y” amount of energy come from some other object, the surface will warm up until it is emitting “x+y” amount of thermal IR. See, this is why I can’t understand your objections. Do you doubt that “x+y” amount of energy will make something warmer than “x” amount of energy” ???”]

            You STILL are either unable or unwilling to answer the simple question; Where is it published that the energy from a cooler object can be absorbed FOR NET ENERGY GAIN by the receiving surface/molecule?

            The ‘y’ amount you speak of was originally part of the ‘x’ amount from the surface. Show me any publication which states that the ‘y’ energy can be absorbed AND increase the temperature of the surface from whence it originally came.

            You are being evasive and it is obvious that you do not wish to support your earlier assertion.

          • Arfur Bryant says:

            Kristian,

            No, he hasn’t got me where he wants me because I do not accept his ‘conceptual truth’. Tim has done this before. He makes an assertive statement then he fails to deliver any validation of that statement in the form of evidence.

            This then becomes a game of who blinks first. Unfortunately for Tim, he was the one who made the original assertion with his [“…then 1+x is getting absorbed and only 1 is being emitted, and the surface will warm up…”] statement.

            That he now tries to wriggle out of providing some sort of evidence merely indicates a lack of integrity. He wants to have his cake (making the assertion) and he wants to eat it (not backing that assertion up). That might work on some people but not with me.

            You think (know) that a cooler object can’t heat a warmer object. Tim thinks (asserts) it can.

            I just want proof either way.

            A simple retraction on his behalf would suffice as an alternative to providing the evidence.

          • Kristian says:

            Arfur,

            I’m glad;)

          • Tim Folkerts says:

            Where is it published that the energy from a cooler object can be absorbed FOR NET ENERGY GAIN by the receiving surface/molecule?
            Every thermodynamics textbook. Can you find a published article or book that says a 10 um photon can be absorbed by a surface with no gain in energy?

            To me, your demand is similar to demanding that I find a reference that states that gravity pulls on red cars. No one is going to publish that, because everyone who knows about gravity knows that it must and does pull on red cars. Anyone who knows thermodynamics knows photons go both ways. Take an upper level physics thermodynamics class. See what the book and the prof say. See if there is some way to make your hypotheses fit with 200 years of thermodynamics.

            “He’s not calling it heat, of course. But that’s what it is.”
            No it is NOT heat! (Not in the strict thermodynamic sense of heat). “Forward-radiation” is energy. “Back-radiation” is energy. Energy moves from hot to cold and from cold to hot. “Heat” is the net result of the various energy flows. The 2nd Law stipulates than that the net flow (= the heat) must be from the warmer to the cooler object.

            Time to bow out here. You two are not convincing me and I am not convincing you two.

          • Kristian says:

            Tim,

            What’s NOT heat, but works exactly like heat? That’s a real good one …!

            It’s your ‘back-radiated energy flux’ that does the extra warming, here. And that alone. None of the other two fluxes contribute. X comes in. X goes out. No restriction there. You add Y from the atmosphere to warm the surface. You openly admit this. Again and again. You’re practically flaunting it around.

            But as we all (?) know, energy added to an object with the result of making it warmer, i.e. increasing its internal energy (U), is by definition HEAT (or work). There is no way around it. It doesn’t matter how much you twist and turn, Tim. We understand you’re trying to hide behind the ‘net’ term. But that doesn’t change the fact about what you’re actually doing. You still end up feeding the warmer object with HEAT from the cooler object.

            In your world the atmosphere is a second, separate source of energy for the surface. This is quite an astounding position to hold. Do you seriously not see it yourself?

          • Arfur Bryant says:

            Your inability to provide evidence is duly noted.

            “Go read a textbook” is the sort of answer given by someone who really doesn’t know.

            All the best…

          • Will Pratt says:

            Tim,

            Take a one inch frozen water ice cube radiating at 273 K in an environment which is also radiating at 273 K (say somewhere in the Arctic circle) and now place next to it a 1 inch cube of frozen CO2 radiating at 193.15 K.

            What happens to the water ice cube? Nothing.

            Now place a 1 ton block of frozen CO2 (dry ice) radiating at 193.15 K next to the water ice cube. What happens to the water ice cube? Again nothing.

            Now try one thousand, one hundred thousand or even a million 1 ton blocks of frozen CO2 all radiating at 193.15 K around the water ice cube radiating at 273 K and what happens to that little 1 inch water ice cube?

            Answer, exactly NOTHING.

            The fact is you can melt a million tons of dry ice radiating at 193.15 K right on top of a 1 inch water ice cube and all that energy released by that 1 million tons of dry ice (frozen CO2) will not and cannot make it warmer.

            It will however make it much cooler for a time.

          • Will Pratt says:

            So called “Back-Radiation” is a completely bogus concept.

          • Ball4 says:

            Will 4:24pm: “What happens to the water ice cube? Nothing.”

            Unforced, before the CO2 shows up, in the bath at 273K, the ice cube & bath glow with IR radiation cooling off say to deep space though your lying eyes cannot see the glow.

            In the bath with the CO2 cube now placed nearby glowing somewhat cooler in the IR, the ice cube absorbs the IR glow of the CO2 cube in its view and starts cool off a little slower than before, meaning to glow a little more in the IR than it would have before the CO2 cube showed up. Since your lying eyes cannot see this happening, you erroneously conclude nothing happens.

            Do not trust your lying eyes, trust a Crookes radiometer that will be turning a little faster than before but also slowing with the ice cube AND the CO2 cube glow cooling off to deep space in line of sight than w/o the CO2 cube placed. See John Tyndall’s work, and esp. William Herschel’s as a cite. Note in the bath in the arctic on the surface, convective and conductive energy transfer are also in operation.

            I wrote that fast, read Tyndall & Herschel, come back & check my work, pls. Get a Crookes radiometer and perform this experiment. Inside the Artic circle, I advise wear a good goose down coat, it will warm you more than you would be in shorts & tank top. Cite Eddie Bauer.

          • Will Pratt says:

            Oh dear, that’s embarrassing.

            Think it through.

            Deep space cannot be the sink in this example because the sun is always shinning somewhere on Earth maintaining an average temperature of 273 K at that location in this example. This means that the only heat sink we need be concerned with is the environment at 273 K.

            However, even if we could shut the sun off, both the CO2 and the water ice cube drop to 3 K. The ice would quickly become as cold as the CO2 and never could be warmed by it under any circumstances in this example.

            Please try to think, it is so important to think, don’t you think?

          • Ball4 says:

            Will 12:04pm – “Deep space cannot be the sink in this example…”

            Part of the IR glow of Will’s example ice cube, CO2 cube and their environment will indeed escape to the sink of deep space at the speed of light in Arctic’s cold, dry air. Will can find out why that is for homework.

            Will says “…the sun is always shinning somewhere on Earth maintaining an average temperature of 273 K at that location in this example.”

            I can assure Will that the earth is not flat; earth is a spheroid that is rotating. There will be at least a diurnal temperature cycle, weather and seasonal (temporal) effects on the 273K avg. at the location on earth in Will’s example, 273K avg. will not be “maintained” it will vary monthly by GHCN measurement; I will let Will look up the definition of “diurnal”, “temporal” and GHCN also for homework. Moving from unforced to the more complex forced steady state will not change the IR glow physics of Will’s cubes.

            It is obvious Will hasn’t studied this niche science to any great extent; I was misinformed that Will has actually read Tyndall’s named book.

          • Will Pratt says:

            Ball 9:43 AM

            Clearly no point responding to you or trying to discuss anything with you as your Dunning Kruger syndrome will inevitably lead to your need to resort to ad-hom and taking all my points out of context.

            You automatically forfeit all your arguments by doing so. A clear sign that you have lost before you started. Please refrain from replying to my comments. They are not for the likes of you.

  66. Alex says:

    My earlier tirade was because I wanted to calculate the temperature of a piece of iron on the moon that was receiving a certain amount of energy. It would have been simple if I used basic high school science. If I did that now I would be eviscerated, torn apart and otherwise insulted by various groups ranging from CAGW, lukewarm, to sceptics.

  67. Kasuha says:

    The main difference is not in the way how the temperature is retained, both atmospheric greenhouse effect and real greenhouse work with insulating the ground to suppress energy flow.

    The main difference is how the temperature of the ground changes. In real greenhouse we don’t add more layers of glass or raise the roof to strenghten the insulation when we want to increase the temperature. We shut vents to suppress circulation with outside. Or open them when we need to lower the temperature.

    In atmosphere, venting the air to some cool outside is not an option. Changes in ground temperature are achieved by changing the insulation effect itself.

    And that’s what I believe is meant by the statement that atmospheric greenhouse effect does not work like in a standard greenhouse. Because it’s not about the state but about the change.

  68. I’ve just (belatedly) been reading these comments including those about my attempts in 2009 to duplicate Wood’s 1909 experiment. As it looks like Roy’s readers would be interested in what I’ve done since 2009, I’ve added an additional section to the end of my page at

    http://boole.stanford.edu/WoodExpt/

    The new section is titled “August 2010: Preliminary Results with a Salt Window.” Hopefully it’s more or less self-explanatory.

    • Massimo PORZIO says:

      Thank you Prof.Pratt for participating to this thread.

      You wrote:
      “If one pictures each window as two resistors in parallel, one for (thermal) conduction and the other for radiation, the observed drops are consistent with both windows having similar thermal conductances but with the salt window’s radiation resistance being much lower.”

      I’ve one question for you.

      Before the experiment do you know that the thermal conduction of the two windows is the same or do you inferred it from your measurement?

      In the second case, if I have to do my conclusion, I see that the box bottom (which I suppose was the black body emitter of the LWIR) stayed almost at the same temperature, so the supposed back-radiation did almost nothing.

      I’m just an electronic engineer, so maybe I misunderstood something, and I could be wrong here of course.

      Have a nice day,

      Massimo

      • Massimo PORZIO says:

        Dear Prof. Pratt.

        I had lunch and I thought a little more to your August 2010 experiment.

        Maybe I had been too hurry in my previous post, because indeed I don’t know what is the size of the window surface compared to the whole internal box size.
        So it’s possible that that 1°C could be the result of the reduced “back-radiation” of the inner surface of window.
        To evaluate your experiment you should give us more details on the box size and its building.

        One thing left me still dubious instead, it is that the outer surface temperature of the glass window is higher than the salt one.
        So comparing the two boxes behaviour, we have more energy inside the “glass box”, but at the same time we have more energy exiting from it.

        It seems that the Sun incoming energy in the “glass box” was greater than the one incoming into the “salt box”, or that those boxes were not reaching the thermal steady state yet.

        Have a nice day.

        Massimo

      • Hi Massimo. I said “similar” rather than “same” since they’re not the same. Thermal conductivity is measured in watts per meter per degree (W/m/K). As can be seen at

        http://www.gly.uga.edu/railsback/PGSG/ThermalCond&Geothermal01.pdf

        halite (NaCl) has a thermal conductivity of around 5 W/m/K. The table at

        http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

        gives the corresponding number for window glass (what I’m using) as around 1.

        Radiation conductivity is not at all similar since 1/4″ thickness of glass completely blocks far-infrared (FIR) while the same thickness of optically transparent salt passes essentially all FIR at wavelengths below around 16 um. With the latter, the only loss is by reflection at the front and back surfaces, of around 4% at each surface, so 8% loss total.

        Window glass at visible wavelengths exhibits the same behavior: 8% is reflected back by the two surfaces but the interior of the glass absorbs a negligible amount of visible light. You need a foot-thick slab of glass in order to see any significant light absorbed.

        Regarding “almost the same temperature,” note that increasing CO2 from 300 to 400 ppmv has only raised the global temperature by about a degree, corresponding to a climate sensitivity of 1/log_2(400/300) = 2.4 degrees per doubling of CO2. Many people feel that a mere 2 or 3 degrees rise in temperature is no big deal, so by that standard you are correct in calling a mere one degree difference “almost the same.”

        However the imbalance in my boxes, which are tricky to manufacture repeatedly, may be creating an experimental error that makes this one-degree difference meaningless. I therefore need to redo the experiment with more attention to balance between the boxes. This will take some time to get right.

        • Massimo PORZIO says:

          Dear Prof. Pratt.

          If I was you, I’ll start calculating the thermal conduction of glass and halite so that you have at least the certainty that temperature delta across them is known.
          For doing that one could make a sandwich like measurement tool having this structure bottom to top:
          1) a self-adhesive heater mat like this:
          http://uk.rs-online.com/web/p/silicone-heater-mats/0731366/
          2) an aluminium plate
          3) a heat-sink silicone pad like this:
          http://dx.com/p/mc-2-0t-diy-thermal-conductive-silicone-pad-for-heat-sink-blue-400-x-200-x-2-0mm-118686
          4) The window under test
          5) a heat-sink silicone pad like the one above
          6) an aluminium plate

          I would use 4 plastic clips to produce a pressure amongst the sandwich layers so that the silicone pads work right, avoiding any air bubble on both their surfaces.

          Keeping the environment temperature constant and laying the “sandwich” on an adequate thermal insulating base (made of white glass wool for example) you could measure the temperature of the upper aluminium plate for both the windows under test.
          This setup should give you a good certainty that the temperature difference is due to the different thermal conductivity of the two windows under test, avoiding any radiative path.

          Have a nice day.

          Massimo

      • Hi Massimo. The window is the same size as the horizontal cross-section of the box’s interior, which is exactly cubical in shape. This is already more information than Wood provided for his experiment, so I can’t tell how faithfully I’ve replicated his experiment.

        Regarding the energy leaving the top of each window, you can’t infer this from the temperature of the top of the salt window because the FIR passing through the salt window does not heat it at all, since the salt window absorbs a negligible amount of FIR. The salt window is being heated entirely by the hot air inside the box, independently of the quantity of radiation passing through it. The glass window however is being heated by the FIR it absorbs as well as by the air inside.

        In equilibrium, and assuming the cotton packing is doing an adequate job of insulating, the energy leaving the top of the boxes should equal that entering them. This will be mainly via radiation and convection. It would be a good exercise to work out the relative proportions of each based on the top temperatures I measured, assuming convective losses are linear in temperature and radiative losses are in proportion to the fourth power of temperature. (At FIR wavelengths glass looks black while salt looks transparent.)

        • Massimo PORZIO says:

          Ok, I get your point.
          Effectively the temperature on top say nothing about the outgoing energy flux for the salt window.

    • Mike Flynn says:

      Vaughan Pratt,

      Have you any updated results since this?

      I agree with Massimo PORZIO that the recorded temperatures show a distinct lack of “greenhouse effect” due to “back radiation”.

      Am I right in assuming you are unable to do any better than this? It’s not very convincing, is it.

      I’m not trying to be “narky”, but it surprises me that your experiments are the absolute best that the climatological community can come up with, after the expenditure of millions, if not billions of dollars.

      At least you actually did an experiment. For that, I salute you.

      Live well and prosper,

      Mike Flynn.

      • Hi Mike. I’m not trying to do better or worse than anything, I’m just reporting on what I observed.

        Furthermore the question Roy raised is not climate but the extent to which IR trapping by greenhouses raises their temperature. These are not necessarily the same thing.

        If you want to make a connection with climate then you have to ask what a difference of one degree at the bottom and six degrees at the top in this particular experiment translates to in climate terms. You can’t just say that replacing salt by glass makes no difference at all, you need to quantify the observed differences in climate terms. For starters, what is the relationship between a 1/4″ window and an atmosphere many kilometers thick? And which temperatures inside the box are most relevant to that analogy: the bottom (which you and Massimo immediately focused on!), the top, the average over that range, or something else? Note that no FIR is intercepted until the radiation reaches the bottom of the window.

        Regarding “millions if not billions of dollars,” if billions are being spent on research into the greenhouse effect you have my full attention. šŸ™‚

        • Mike Flynn says:

          Vaughan Pratt,

          I hope you don’t take offence.

          One thing that may be overlooked from time to time is what happens to IR “blocked” by window glass. My understanding is that if the glass absorbs IR (and it does, reflecting but a portion, depending on the angle of incidence, the structure of the glass “surface”, coating treatments, etc.), then like any other matter, the glass will “heat”, and a temperature rise will occur.

          As quick practical test, the morning sun is shining through my window, and has been so for an hour or so.
          The glass feels considerably ” warmer” to the touch than the sills, or the wall containing the window. However, this perception is of course affected by the different conductivities of the materials in question.

          Luckily, I have to hand a remote IR thermometer. This gives me a readout of 36C for the interior surface of the window glass.The adjacent matte paint sill surfaces, and adjacent wall surfaces are 27C.

          I am aware that the thermometer assumes a standard emissivity, and I have rounded the readings. I should point out that the window is louvred, and partially open, so I have assumed that the air temperature on both sides of the glass is roughly equivalent. The room that contains the window is well ventilated, and the house is not artificially heated or cooled.

          I am inferring that the window glass temperature has risen due to the absorption of radiant energy from Sun. This inference seems to be supported by well established measurements used by industry when calculating heating and cooling requirements etc.

          I still see no “greenhouse effect” (in the “back radiation” sense) demonstrated by your experiments. Any measurements you have are easily explained using “normal” physics. Obviously, it is quite difficult to measure convective and advective movements in a relatively transparent gas.

          Once again, as Feynman said “You can’t fool Nature.”

          And at the moment, Nature seems to be telling us that the supposed correlation between CO2 levels and “global temperature” is absent.

          Live well and prosper,

          Mike Flynn

          • @MF: Luckily, I have to hand a remote IR thermometer. This gives me a readout of 36C for the interior surface of the window glass.The adjacent matte paint sill surfaces, and adjacent wall surfaces are 27C.

            Sounds like your non-window measurements are only of the surfaces in the shade. The interior surface of your window glass will pick up the radiation from the sunlit surfaces in the room as well as the shaded surfaces. So for completeness you should also report the temperatures of the various sunlit surfaces. These might conceivably account for some of the 36 °C temperature of the interior surface of the window glass.

            Glass starts to become opaque to IR at around 2 or 2.5 um. About 10% of insolation is above that wavelength, which may account for some of the heating of the glass. Glass is so transparent at shorter wavelengths that these should have no thermal impact at all.

        • Mike Flynn says:

          Vaghan Pratt,

          I just noted that you said : –

          “Regarding “millions if not billions of dollars,” if billions are being spent on research into the greenhouse effect you have my full attention. ”

          Please note I didn’t say anything about the amount of money spent on research. I referred to expenditure, without implying that anything at all was spent on research involving actual physical experiments. As far as I can see, yours is the sole attempt to “prove” the basis for the assumption that additional CO2 in the atmosphere can somehow “warm” anything at all.

          This appears, prima facie, a pretty poor return for the time, effort, and money expended by the climatological fraternity. And the results? You may perceive some benefit, I see none at all.

          Live well and prosper,

          Mike Flynn.

          • @MF: As far as I can see, yours is the sole attempt to “prove” the basis for the assumption that additional CO2 in the atmosphere can somehow “warm” anything at all.

            Who said anything about CO2? This thread is in response to Roy’s question at the top, “Does a greenhouse work more from infrared heating (the “greenhouse effect”), or more from the inhibition of convective heat loss?” This question concerns only glass, not CO2. Likewise with my experiment, which is solely to try to duplicate Wood’s experiment and not to say anything about CO2.

            Although many since 1909 have questioned both the soundness of Wood’s experiment and the validity of his finding, if it were to turn out nonetheless that Wood was right about greenhouses, that still wouldn’t show anything about atmospheric CO2. The Wikipedia article on the greenhouse effect claims, “the mechanism is named after the effect of solar radiation passing through glass and warming a greenhouse, but the way it retains heat is fundamentally different,” citing Wood as its primary source for this claim (primary in the sense that Wikipedia’s other sources for this claim also cite Wood).

  69. Karl J says:

    After reading through all comments I get the following picture.

    High energy photons that in principle can heat the earth to some 6000 degrees, if there were enough of them, enters the atmosphere. On the way down some are reflected back again, and some are absorbed by various gasses like water vapor, reemitted as less energetic photons radiated in all directions. Some of the ones that reach the ground are absorbed by plants and stuff, heating the stuff, which radiates away in all directions as low energy photons.

    Then you put up a green house where the air is heated by the stuff at the floor. The green house hinders convection. After a while the energy flow to/from the green house reaches equilibrium. The low energy photons produced inside the green house is absorbed by things with lower temperature such as the glass and then emitted again as even less energetic photons. One thing I don’t get is what happens if a photon has to low energy, coming from an object with low temperature and hitting an object with higher temperature, does it just bounce off or what?

    Without a green house the air near the ground is heated by contact with the ground and sea. The heated air is diluted mainly by convection. Convection takes time so the air near the ground will be much warmer than air at higher altidudes. The air itself also radiates low energy photons in all directions. But since it is warmer near the ground the net flow of low energy photons will be outwards, eventually to space, where theese photons will have a characteristic signature of a certain temperature.

    And in principle back radiation can happen if there are objects with low enough temp that can absorb the back radiated photons.

    • Karl: Given a hot object at temperature T and a colder one at temperature S, both absorbing to the same extent, the net heat transferred from the former to the latter is proportional to T^4 – S^4. This formula is based on the assumption that the quantity of radiation passing from one object to another depends solely on the temperature of the radiator and is independent of the temperature of the absorber. In your terminology, “cold” photons are absorbed just as effectively by hot objects as by cold ones.

      That air near the ground is warmer is due in the first instance to lapse rate, which is at most 10 degrees per km of altitude for dry air and about half that for moist air saturated to dew point.

      Convection does take time however, as you say. At night the lapse rate goes negative over the bottom km or so, due to the ground cooling the bottom layer. The bottom 2 km of atmosphere filters out this diurnal fluctuation with the result that the atmosphere above 2 km shows relatively little temperature variation between night and day. There is however a pronounced seasonal fluctuation between summer and winter, an effect evident at typical ski slopes which are free of snow in summer.

      When these fluctuations are averaged out over more than a year, the impact of global warming is an increase in temperatures at all altitudes within the bulk of the troposphere, i.e. up to 10-15 km. In particular a rise of 1 C at the surface in the course of a century will in principle be accompanied by an equal rise at 10 km elevation. This assumes equilibrium, which may not be a good assumption when the temperature is rising more rapidly.

      Venus has a lapse rate close to that of Earth, gaining around 8 C per km decrease in altitude over the bottom 60 km, almost perfectly linearly, see e.g. page 4 of the pdf at

      http://www.gaknowledge.org/handle/META/13137

      At 60 km Venus’s atmosphere is at 0 C, increasing by some 480 C when descending to the surface, even though insolation at Venus’s surface is only around 20 W/m2 (due to thick haze and clouds from ~50 km on up) as against Earth’s ~1000 W/m2 at high noon. I’ve found these similarities and differences helpful in understanding at least some aspects of global warming on Earth.

      • Tim Folkerts says:

        Vaughan,

        Its seem that the experiments could be made MUCH simpler by using an electric heater, rather than the sun. The method of heating the “bottom of the box” really doesn’t matter. This would
        * eliminate issues of reflection of sunlight from multiple surfaces
        * eliminate variations in from one box to the next
        * eliminate variations over time.
        * allow controlled experiments with different power inputs.
        * allow covers that are not transparent to light (eg metal)

        In fact, I think I will try a few experiments along those lines. It would make an excellent student project, since nothing is terribly complicated or expensive.

        • Indeed, Tim, and I’d considered doing it that way myself back in 2009. However my main goal was to duplicate Wood’s experiment as faithfully as possible in order to see if I got the same results, and to understand the limitations of his approach.

          I’d say the main limitation is that lapse rate plays no role in Wood’s experiment but plays an important role in the atmospheric greenhouse effect. One should therefore expect some difference between glass and salt windows, but not as big an effect as with the atmosphere.

          • Mike Flynn says:

            Vaughan Pratt,

            I’m probably about to have my nose bloodied, but . . .

            The lapse rate is merely that which results from having a surface temperature greater than the 3-4K of outer space. Obviously, the atmosphere in contact with the surface is warmer than that closer to a lower temperature radiative sink, ie the near vacuum of outer space.

            There is a temperature gradient. It cannot be otherwise. Your experiment shows a temperature gradient within your boxes. The rate of this gradient of course, like the free atmosphere, is not simply calculated.

            To try to blame your lack of proof positive on a mysterious lack of lapse rate, without being able to explain it, might indicate to a dispassionate observer, that it is easier to apply Occam’s razor.

            In this case, if the effect can’t be demonstrated by experiment, and doesn’t accord with observation (in that CO2 levels are rising, and even the most furiously adjusted data set temperatures haven’t, then there is obviously a case for no GHE as the simplest explanation.

            Mind you, I would rather be happy than right. You?

            Live well and prosper,

            Mike Flynn.

          • Massimo PORZIO says:

            Hi prof. Pratt.
            are you sure that you got a lapse rate in the box?
            I mean, you stated that the thermometer under the window was attached to its lower surface.
            What I’m arguing is that the air in the box could be almost at the same temperature and only a little space below the windows could have a temperature gradient.
            Otherwise, I don’t imagine why the hot air below doesn’t mix with the above by convection.

            As an old professor of physics used to say to his students: “be careful that any thermometer reads the temperature of itself.”

            Have nice day.

            Massimo

          • Massimo PORZIO says:

            Hi Prof. Pratt.

            With reference to my previous post, I did some very rough and simplified computations about the measured temperature in your experiment.

            In particular I would like to infer where, below the lower window surface, the temperature could rise to a level similar to the bottom of the box.

            To infer that I used the given flux of outgoing heat and the air thermal resistance.

            You stated that the glass window was 1/4″, so it was 6.35mm.

            This means that the gradient for the given thermal resistance of glass is (54.4°C-38.7°C)/6.35mm = 2.47°C/mm.

            But since I want to compute the gradient in air for the very same flux of heat which passes through the window and air behaves more similar to the salt window in this context, so I use salt window gradient instead (48.2°C-36.5°C)/6.35mm = 1.84°C/mm.

            Considering that, as you also said, air has about 40 times the thermal resistance of glass, multiplying that gradient by 40 I have 73.6°C/mm.

            The maximum measured difference from the box bottom and the window lower surface is for the salt window box and in detail it is 74.3°C-48.2°C = 26.1C, so suffice just few mm (or even fraction of mm, maybe about 26.1/73.6 = 0.35mm) below the window surface the inner box temperature could be homogeneous and equal to the box bottom temperature.

            Of course I could have done some macroscopic errors, so the above could be just garbage, but I still believe that until today Wood’s experiment conclusion can’t be considered wrong (or right of course).

            Have a nice day.

            Massimo

          • @MF: The lapse rate is merely that which results from having a surface temperature greater than the 3-4K of outer space.

            It’s not that simple. Adiabatic expansion plays a central role in the concept of lapse rate as it applies to the atmosphere. If you’re picturing it as working like Ohm’s law, it’s nothing like that, it’s quite a delicate concept. For now I suggest reading the Wikipedia article of that name. If that doesn’t make the concept completely obvious I’ll have a go at writing my own account.

            To try to blame your lack of proof positive on a mysterious lack of lapse rate

            First, there is no mystery in the fact that lapse rate as it applies to the atmosphere and global warming plays no role in Wood’s experiment. Since lapse rate is on the order of 10 °C/km, the box would have to be 30 feet high before one could observe even 0.1 °C degree change in temperature due to lapse rate. What I suspect both you and Massimo have in mind by “lapse rate” in Wood’s experiment is more like Ohm’s law, more precisely Fourier’s law as the thermal counterpart of Ohm’s law.

            Second, you’re the one claiming “lack of proof positive,” which you do by arguing that a mere degree of difference is insignificant. Yet presumably you would claim that the last 15 years of global temperature have been flat instead of climbing 0.2 °C as forecast 15 years ago.

            How can you consistently claim that a 0.2 °C difference is significant when judging flatness but a one degree difference is not? You would have to argue that the centimeter-scale Wood’s experiment should somehow magnify the kilometer-scale greenhouse effect by a factor of more than five. How do you propose to argue that?

          • Massimo PORZIO says:

            Dear Prof. Pratt,
            maybe I misread your post when I wrote about the lapse rate.
            I believed that you were convinced that a there was a lapse rate in the box… Sorry, but I’m Italian and sometimes I do embarrassing misunderstandings reading a post in English.
            I never thought that a lapse rate could be measured in the very low height of the walls of a greenhouse, the same for your experiment boxes.
            What I wanted to highlight in my previous post is that an experiment based on a cubic box where only one side is the input/output window maybe, probably can’t never be useful to establish if and how much the IR contribute to warm the inner surfaces of the box. Because in a cubic box the window represent on 1/6th of the radiating inner surface flux, and since temperatures change at the 4th power radix of the flux the measurements could be useless.

            Have a nice day.

            Massimo

          • @Massimo: Because in a cubic box the window represent on 1/6th of the radiating inner surface flux, and since temperatures change at the 4th power radix of the flux the measurements could be useless.

            Did you mean the 1/4-th power?

            The fluxes cancel except at the window, which is therefore the only part of the experiment that matters. If you hold the window area constant while multiplying the volume of the box by say one thousand, what difference would that huge multiplication make? The energy entering and leaving the window should remain the same.

        • Massimo PORZIO says:

          Hi Tim,
          in my post on August 18, 2013 at 9:57 AM.
          I explained my opinion about a way to extrapolate the heat conductive part of the thermal resistance of the material used for the greenhouse.
          I would like to have your opinion about it.
          I think that knowing the heat conductivity should be a good point to evaluate the overall effect of the radiation in a successive experiment.

          Have a nice day.

          Massimo

        • Bryan says:

          Tim Folkerts

          An even simpler test of Roy Spencer’s and Vaughan Pratt’s belief that a glass greenhouse explains a 33K so called greenhouse effect.

          Take a glass greenhouse to the Moons equator.
          It must work the same magic there
          The 390K maximum would then increase to 423K.

          Since there is a vacuum inside and outside the greenhouse there is no convection to confuse the issue.

          • Tim Folkerts says:

            In concept, doing the experiment on the moon would be simpler. In practice it would be nearly impossible.

            One option would be to use a large vacuum chamber on earth (eg http://www.nasa.gov/multimedia/imagegallery/image_feature_1281.html), or even a light bulb with a vacuum inside. In these cases, it is important to remember that “3 K outer space” is replaced by “290 K earth” – no conceptual difference but a huge practical difference.

          • Bryan makes an excellent point about the vacuum. However that difference is easily eliminated on Earth by resting the greenhouse roof flat on the ground so as to trap no air in between. This is because the thickness of the vacuum (i.e. the height of the roof) in the Moon-based greenhouse seems to play no role in influencing the temperature, and therefore can be set to zero without changing the moonhouse temperature.

            In the case of Wood’s setup this amounts to an experiment with a flat black surface on which is placed a smaller pane of glass. At equilibrium three temperatures are measured: above the glass, below the glass, and beside the glass (i.e. on the uncovered portion of the black surface). Massimo has argued, rather convincingly in my opinion if we assume emissivity of 1 for all surfaces, that the first and third of these should be about equal, and it seems pretty obvious that the second should be considerably greater, though I’m not aware of anyone having done that particular experiment.

            There is a very important second benefit to Bryan’s Moon-based experiment: no atmosphere above the greenhouse to heat it by back radiation! Wood’s experiment does not take this back radiation into account, which cannot pass into his glass-topped box (nor into a greenhouse) but can pass into the salt-topped one, thereby heating the interior of the latter but not the former. Given that the daily average shortwave insolation at the surface is 168 W/m2 while the daily average longwave back radiation from the atmosphere is nearly twice that, namely 324 W/m2, as per Figure 7 of Kiehl and Trenberth 1997 at

            http://www.geo.utexas.edu/courses/387h/PAPERS/kiehl.pdf

            this rather undermines Wood’s assumption that the only difference between the two boxes is the additional heat lost by the salt-topped box.

            This particular defect of Wood’s experiment was first noticed a mere five months after Wood’s paper, namely by Charles Greeley Abbot, then the director of the Smithsonian Astronomical Observatory and later Secretary of the Smithsonian, in the July 1909 issue of the same journal Wood’s paper appeared in. Assuming back radiation corresponding to 0 °C (namely 314 W/m2, a little less than K&T’s 324 W/m2) Abbot calculated that radiation considerations alone show that the salt cover will be at least 65% as efficient as the glass cover in keeping the interior of the box warm. He also pointed out that since the glass window will be hotter than the salt one this difference is further narrowed due to the greater convection of heat from the top of the glass window. Abbot concludes “In view of these figures we may agree with Professor Wood that a salt cover is nearly as efficient as a glass one for a ‘hot-box,’ although it would seem strange that he observed no difference at all. Perhaps in the spite of the glass filter the cover-glass obstructed the entering sun rays more than salt.”

            @Bryan: An even simpler test of Roy Spencer’s and Vaughan Pratt’s belief that a glass greenhouse explains a 33K so called greenhouse effect.

            While I can’t speak for Roy, I believe no such thing, and have never said anything remotely like it. Furthermore I.M. Pei’s glass pyramid over the entrance to the Louvre proves beyond question that a glass roof cannot possibly add 33K to the inside temperature. For if it did, then on a summer day with the outside at 30 °C the inside would be at an impossibly hot 63 °C. Given the long lines to get into the Louvre its visitors would drop dead from heat exhaustion before they could ever reach the safety of the air-conditioned interior!

            That said, when I visited the Louvre in April 2010 I was struck by the blast of air greeting those entering the pyramid. It was an enormous relief to eventually get off the escalator and away from the heat of the pyramid. While it was nowhere near 63 °C it was very uncomfortable nonetheless.

            On my next visit I will try to measure the temperature of the inside of the pyramid’s glass with an IR thermometer. While I don’t expect it to be 33 K warmer than a typical non-glass ceiling, I would be surprised if it were no warmer.

            Wood’s experiment and I.M. Pei’s much larger glass pyramid both fall far short of duplicating the lapse rate of the Earth’s atmosphere, which over just the bottom 10 km of the troposphere integrates to over 80 K difference! They therefore cannot be expected to exhibit anything like the same 33 K warming caused by Earth’s greenhouse gases.

            But just because we cannot expect a greenhouse to add 33 K to the temperature is no reason to infer that IR-trapping cannot warm a greenhouse significantly, which is the question raised by Roy at the start of this thread. Obviously inhibiting convection doesn’t add 33 K to the greenhouse either, so comparison with Earth’s atmosphere doesn’t answer the question of whether IR trapping plays a significant role in warming greenhouses (which is how I’d prefer to word Roy’s question given that convection plays no role in keeping Earth cool, only radiation to space).

          • Massimo PORZIO says:

            Hi Prof.Pratt.
            you wrote:
            “There is a very important second benefit to Bryan’s Moon-based experiment: no atmosphere above the greenhouse to heat it by back radiation! Wood’s experiment does not take this back radiation into account, which cannot pass into his glass-topped box (nor into a greenhouse) but can pass into the salt-topped one, thereby heating the interior of the latter but not the former.”

            I believe that Wood did it instead, he use probably the wrong word, but he wrote:

            “When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass.”

            If I’m not wrong in translating that statement to my first language, he seen the salt window box warming faster, so he concluded that it was because of the LWIR coming from the Sun. He just misinterpreted the source of that downward LWIR. Now we know that it comes from the heated atmosphere.

            Then Prof. Wood added:
            “In order to eliminate this action the sunlight was first passed through a glass plate.”

            That is, he added what Tim some posts ago called the “carport” to avoid the LWIR from the above.

            And he concluded:
            “There was now scarcely a difference of one degree between the temperatures of the two enclosures. The maximum temperature reached was about 55 oC. From what we know about the distribution of energy in the spectrum of the radiation emitted by a body at 55 o, it is clear that the rock-salt plate is capable of transmitting practically all of it, while the glass plate stops it entirely. This shows us that the loss of temperature of the ground by radiation is very small in comparison to the loss by convection, in other words that we gain very little from the circumstance that the radiation is trapped.”

            IMHO here lays Prof. Wood error.
            The “carport” not only blocked the downward LWIR of the atmosphere, but by paradox it enhanced the LWIR impinging the two windows because that “carport” made of glass was a good emitter of LWIR due to its own temperature.
            Which I assume was at steady state, that is at the same value of the air at ground, or little more because of the underneath boxes (that depends upon the distance between the carport and the boxes).

            Even in my opinion the reduction from 65°C to 55°C that he experienced, it could be due not to the blocking of the LWIR, but because of the UV/Vis/SWIR absorption of the glass as Dr. Spencer argued many post above.

            Have a nice day.

            Massimo

          • Looks like we’re getting close to agreement here.

          • …except for the following:

            @Massimo: “When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass.”

            If I’m not wrong in translating that statement to my first language…

            The glass only stops waves longer than about 2.5 um. Most of the back radiation from the atmosphere is more like 10 um, and dwarfs any longwave radiation from the Sun. Wood is quite explicit that he’s talking about longwave radiation from the Sun, which is negligible compared to longwave radiation from the atmosphere.

          • Massimo PORZIO says:

            @Prof. Pratt.

            I deduced that Prof.Wood wrongly believed that the 10um waves came from the Sun instead that they came from the atmosphere GHGs as we know they do, that because he explicitly said that the glass blocked them, so he shouldn’t refer to the Sun’s LWIR shorter than 2.5um.

            Anyways I could have misread him because of my translation limits.

  70. Erl Happ says:

    Roy,
    As a believer in the ‘greenhouse effect of CO2 in the atmosphere’are you surprised that global temperatures have plateaued since 1998 and while CO2 concentration in the atmosphere has steadily increased?

    • Erl,
      CO2 is not the only influence on temperature, which varies considerably without any help from CO2. Are you surprised that the temperature decreased by 0.3 C in the decade 1940-1950?

      http://www.woodfortrees.org/plot/hadcrut3vgl/from:1940/to:1950/trend

      Obviously that wasn’t caused by falling CO2, it must have had some other cause. How do you know we’re not seeing a repeat of whatever phenomenon caused that cooling? Or some other phenomenon?

      • Arfur Bryant says:

        Hi Vaughan!

        Long time, no chat…:)

        A few thoughts on your post above:

        1. [“CO2 is not the only influence on temperature, which varies considerably without any help from CO2.”]
        Correct, and the logical extension of that argument is that ALL of the warming seen since 1850 (IPCC start of records) could have been caused by something other than CO2.

        2. [“Are you surprised that the temperature decreased by 0.3 C in the decade 1940-1950?”]
        No, why would anyone be?

        3. [“Obviously that wasn’t caused by falling CO2, it must have had some other cause.”]
        Yes, obviously, since CO2 concentration did not fall. However, the rise of CO2 between 1940 and 1950 would have been less than 10 ppm. Since 1998, the rise of CO2 has been about 30 ppm. So, if the CO2 = cAGW theory/hypothesis/postulation/assertion has even the slightest chance of being correct, one would reasonably expect the warming to continue and even accelerate.

        4. [“How do you know we’re not seeing a repeat of whatever phenomenon caused that cooling? Or some other phenomenon?”]
        We don’t. Neither do you. However, equally we don’t know that CO2 caused any warming either. The ‘other phenomenon’ (or phenomena) you speak of could equally be a combined set of factors which, arranged in different ratios, either cause warming, or cooling, or flattening of temperature.

        CO2 doesn’t have to have anything to do with any of it!

        Fond regards,

        Arfur

        • Tim Folkerts says:

          I find it fascinating that the “logical extension” of “CO2 is not the only influence” becomes “CO2 doesn’t have to have anything to do with”.

          This is the logical eq

        • Tim Folkerts says:

          I find it fascinating that the “logical extension” of “CO2 is not the only influence” becomes “CO2 doesn’t have to have anything to do with”.

          This is the logical equivalent of “tuning on the 11th light bulb in a room doesn’t help me see better since there is already plenty of light. Since the last light doesn’t matter, therefore none of the other 10 light bulbs matter, either.”
          šŸ™‚

          • Tim Folkerts says:

            Dang — somehow I double posted (or 1.5x posted).

          • Arfur Bryant says:

            Yep, nice try Tim.

            [“…which varies considerably without any help from CO2.”]

            THAT was the part that the logical extension followed.

            [“…the logical extension of that argument is that ALL of the warming seen since 1850 (IPCC start of records) could have been caused by something other than CO2.”]

            See?

            If, as Vaughan says, temperature can vary considerably “without any help from CO2” then, logically, any observed warming in any observed period COULD have been due to those factors (other than CO2) which affect temperature. I didn’t say it WAS; I said it COULD.

            That is called objectivity.

            šŸ™‚

          • Well, sure, the whole thing could have been merely the result of an experiment by Arcturans on Earth’s geophysics. So what?

            The fact that the temperature declined substantially during 1880-1890 and again 60 years later during 1940-1950 raises the distinct possibility of this happening yet again a further 60 years later, namely during 2000-2010.

            And if some other periodicity is also contributing to this, e.g. a 20 year cycle observed over 160 years, then even more so.

            While there’s certainly no guarantee of this, it’s a heck of a lot more plausible than that all these fluctuations are merely the result of Arcturan interference, or whatever other mechanism one wants to propose to explain ostensibly random fluctuations.

          • Arfur Bryant says:

            Vaughan…

            [“The fact that the temperature declined substantially during 1880-1890 and again 60 years later during 1940-1950 raises the distinct possibility of this happening yet again a further 60 years later, namely during 2000-2010.”]

            Thank you for agreeing my point. The warmings and coolings of the 1850 – 2013 period need not have anything to do with CO2.

            Unless you are implying that all the warmings are due to CO2 and all the coolings/flattenings are due to other factors?

        • Hops says:

          Regarding the greater rise of CO2 since 1998, bear in mind that the increase of temperature with CO2 level is a log function, not a linear one. There is also a lot of inertia in the system as land and water absorb heat. We are not even close to reaching the equilibrium of the current CO2 level.

          It is as if we just built the greenhouse and are waiting for it to fully warm up as predicted. A passing cloud caused a pause in the warming trend, and the skeptics say the greenhouse isn’t working…

          • @Hops: bear in mind that the increase of temperature with CO2 level is a log function, not a linear one.

            For small x, ln(1+x) ~ x. As long as the changing part of CO2, namely x, remains small, around 100/280 today, compared to the natural part, namely 280/280 = 1, temperature will continue to rise as a linear function of x.

            And since the changing part of CO2 is growing exponentially with time, namely as exp(t), this means that for the time being temperature relative to its preindustrial value will grow exponentially with time.

            When the changing part of CO2 greatly exceeds the natural part, if x continues to be modeled as exp(t), ln(1 + x) will start to approximate t instead of exp(t), a linear function. But that’s still a few centuries away, if ever.

            In the meantime relative temperature is going to grow at more or less the exponential rate of growing CO2 emission.

          • Arfur Bryant says:

            Vaughan…

            [“In the meantime relative temperature is going to grow at more or less the exponential rate of growing CO2 emission.”]

            Eh? How can you say that when you have just stated:

            [“CO2 is not the only influence on temperature, which varies considerably without any help from CO2.”]

            You cannot ASSUME a relationship which has no real-world evidential support.

            And see these…

            http://www.climate4you.com/images/CO2%20MaunaLoa%20MonthlySince1958.gif

            http://www.climate4you.com/images/HadCRUT4%20MAATand3yrAverage%20Global%20NormalisedFor1979-1988.gif

            So where do you get exponential growth in both CO2 and temperature?

            Just askin’…

          • Arfur Bryant says:

            Hops,

            [“There is also a lot of inertia in the system as land and water absorb heat. We are not even close to reaching the equilibrium of the current CO2 level.”]

            1. Please provide evidence for this inertia lag. Real, observed evidence. Not models.

            2. How long is the lag?

            3. Does the lag apply equally to the start of warmings as it does to the end of warmings?

            An answer to point 2 would be most appreciated. As we are now cooler than 1998, would you suspect the length of lag is at least 15 years? In which case, what caused the significant warming 15 years before 1975 AND 1910 which were the start of equal amounts of warming?

            Objectivity. It’s a killer…

          • beng says:

            ***
            Hops says:
            August 21, 2013 at 5:06 AM

            There is also a lot of inertia in the system as land and water absorb heat. We are not even close to reaching the equilibrium of the current CO2 level.
            ***

            There is inertia in the system, but added energy has to be stored somewhere to contribute to that inertia. Shortwave solar penetrates the ocean & is partially stored there, but added longwave from CO2 backradiation doesn’t penetrate water/land significantly (a few MMs) and can’t be stored significantly.

            Not all added watts are equal — from increased solar SW (albedo changes, etc), some is stored & produces a time lag in achieving equilibrium, but LW GHG backradiation doesn’t.

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  72. Fulco says:

    The question was:

    Does a greenhouse work more from infrared heating (the “greenhouse effect”), or more from the inhibition of convective heat loss?

    I think we had the answer a long time ago:

    It is almost 100% from inhibition of convective heat loss and a few degrees extra if we block the IR outgoing radiation.

    The whole exercise here is about those extra degrees, not the original question.

    • dallas says:

      No, it is not 100% convection. If you add a radiant barrier to an existing home, which already has insulation and a vapor barrier, the radiant barrier will improve the quality of the overall insulation. How much the radiant barrier improves insulation efficiency depends on a number of factors though.

      http://blog.radiantbarrier.com/wp-content/uploads/2012/02/radiant-barrier-research-ashrae.pdf

      There are people that actually test this stuff. The problem with CO2 is that you have a fair radiant barrier to start with, initial water vapor and CO2. Adding 3.7Wm-2 of radiant barrier to an existing ~150Wm-2 of existing radiant barrier has less return on investment than just adding a radiant barrier. Since the initial barrier was less than 50% efficient, you can’t expect the 3.7Wm-2 add on to be 100% efficient. You would get about the same or a little less efficiency out of the add on layer.

      Lindzen and Kimoto had a simple way to ballpark the impact, use the estimated total impact, that irritating DWLR value of about 334 Wm-2. Increasing that to 337.7 Wm-2 would produce ~ 0.8C of insulation improvement. It doesn’t matter how that 3.7Wm-2 was produced, 3.7Wm-2 would improve insulation by ~0.8C.

      If you don’t like DWLR, consider the average energy of the thermal mass, 334.5Wm-2 for the ~4C degree average ocean temperature. See how that works? Source 334.5 Wm-2 return 334.5 Wm-2. If you don’t like using the oceans since they only cover 71% of the surface, then shift your frame of reference to the total are, .71*334.5=237 Wm-2. Now you have an imaginary surface somewhere above the atmospheric boundary layer where you can have fun trying to keep track of “Effective” surfaces and “Effective” radiant layers. Adding 3.7 Wm-2 without adjusting for the actual “surface” let’s you get big scary numbers to impress your friends.

      But why make it complicated, it is only adding a thin sheet of additional insulation. Call ASHREA

      • Fulco says:

        So your answer is Yes not No.

        • Vaughan Pratt says:

          That was my take too. Furthermore Earth has no convective loss at all, though does that count as “100%”?

          • dallas says:

            Vaughan, Earth doesn’t have convective loss but the greenhouse, from the true surface to 100 to 3000 meters, below the atmospheric boundary layer has convective loss. There are several layers that have to be considered because there is much more to life than CO2. If CO2 didn’t regulate “surface” temperature, instead of convection you would have atmospheric erosion is a pretty large way. The maximum SST is limited to about 300K which is probably about the maximum past temperature of the Earth where there was likely some erosion of the atmosphere to get us where we are today. Pure speculation on my part, but it is interesting that R specific at 300 C and 1000 millibars kinda matches “sensitivity”.

            Now if you want to use that “effective” surface with that “effective” 237Wm-2, you can get bigger numbers, but I don’t know many that live in the mid troposphere.

          • Capn (if that’s you since you talk the HVAC talk), what you say is just right enough to point up why this whole subject is so complex.

            Just to make sure we’re on the same page, let me reword your argument. Convection can bring warm air at say 3 km up to say 6 km where it has a better shot at radiating its thermal load to space due to fewer GHGs intercepting it.

            If that’s anything like what you had in mind, here’s my counterargument to your argument, in two parts.

            (i) When the environmental lapse rate (ELR) equals the adiabatic lapse rate (ALR), convection adjusts the temperatures of the moving parcels so as to maintain the status quo. A warm parcel convected to a higher elevation becomes a cold parcel and vice versa, whence convection does nothing to help cool planet Earth.

            (ii) Global warming is currently proceeding at a mere 0.16 °C per decade, a number that has itself been slowly increasing over the decades. A decade is long enough that mere conduction (very slow) starts to compete with convection (slow in still/stable air but dominating at every hurricane). This can be quantified with Ficke’s Law, but the upshot is that the whole atmosphere warms whether or not convection changes anything, which (i) implies that it doesn’t.

            If you had something else in mind I’d love to understand it, especially if it has an explanation in words of few syllables.

            @dallas: I don’t know many that live in the mid troposphere.

            I hear you. Neither do I.

          • dallas says:

            Vaughan, Starting with your second point,

            ” This can be quantified with Ficke’s Law, but the upshot is that the whole atmosphere warms whether or not convection changes anything, which (i) implies that it doesn’t.”

            Ficke’s law would have boundary / thermocline issues. Diffusion/conduction are important but weak processes so density gradient and thermal gradients would require “special” treatment. You can get 90C in a solar pond by limiting heat transfer across a density gradient. In the lower atmosphere, the temperature of condensation doesn’t change appreciably as long as there is fairly constant cloud condensation nuclei. That is one of the reasons I push thermal envelops instead of adiabatic cylinders or generic diffusion slabs. The atmospheric boundary layer is effectively the glass of the greenhouse with a water condensation temperature thermostat. So the “Whole” atmosphere doesn’t warm uniformly, neither does the”Whole” ocean on relevant time scales since latent approaches a point where latent/convective loss equals heat gain.

            If you like you can consider the ABL a water vapor ground plane for thermal radiation with temperature/energy fixed at 0C (316 Wm-2). With 4C (334.5Wm-2) as the source or black body energy for the oceans and 0C (316Wm-2) the source or black body energy of the moist air envelop, Ficke is kinda Phucked.

          • dallas says:

            Vaughan, on the secular trend. Since ln(2) curves are common in nature you have to be a bit careful using an ln(2) curve to “explain” things. There are a few newer studies out on Solar and Tropical ocean dynamics. So I borrowed Steinhilber TSI and Oppo Indo-Pacific Warm Pool SST reconstructions to show how those pesky longer term lags can make life difficult.

            https://lh5.googleusercontent.com/-zpvEmGcB7mg/UhTIuFIoFfI/AAAAAAAAJQE/Lsl8-W9sr9I/s720/Solar%2520and%2520IPWP.png

            Century scale pseudo-oscillations and multi-century recovery periods should not be unexpected on planetary scales. Since the rate of ocean heat uptake is producing about +0.8C per 316 years, once you chunk in a volcano here and there you have yourself a pretty significant secular trend. Woods Hole has even modeled what they call the Pacific Centennial Oscillation (PCO), Brierley has modeled and estimated the impact of meridional and zonal SST gradients to be 3.2C and 0.6C respectively, so one the whole I would say that Callandar is the Guy to beat.

        • dallas says:

          Yes, there is a greenhouse effect. There is also an atmospheric effect that deals with the latent and sensible plus 75Wm-2 about of solar absorption in the atmosphere. The atmospheric effect is about half and the greenhouse effect about half. Of the greenhouse effect, water vapor is about 2/3 and CO2 equivalent about 1/3, so CO2 is about 1/6 of the total effect. Callandar pretty much nailed it.

          What drives the Sky Dragons nuts is the atmospheric effect. 75Wm-2 solar, 88Wm-2 latent and 24 Wm-2 sensible are absorbed by the atmosphere at and above the atmospheric boundary layer which is like the glass of the main greenhouse. The atmosphere can’t absorb much more because the specific heat capacity of the atmosphere can’t contain much more. Specific heat capacity is related to gravity so they jump all over some “new” discovery. The Greenhouse Effect, at night, just tries to maintain the atmospheric energy. It comes close, the average global diurnal temperature range is about 4C. That would be about the absolute maximum for a perfect greenhouse Earth, there is just not enough atmospheric mass to go much higher.

          Of course this is coming from a retire HVAC guy, not like a real scientist, more like Guy Callandar, a steam engineer, which if you are into steam, you would know you can’t increase steam quality unless you increase containment pressure šŸ™‚

          • @dallas: Of the greenhouse effect, water vapor is about 2/3 and CO2 equivalent about 1/3, so CO2 is about 1/6 of the total effect.

            Static or dynamic? If water vapor is increasing only a little and CO2 a lot, wouldn’t that make CO2 the major contributor to global warming as distinct from global warmth?

          • dallas says:

            Static. Dynamically, CO2 would have a larger impact at lower average temperature in order to regulate, but water vapor still does the majority of the work. CO2 has the most impact where is can maintain or increase liquid water with less impact as water vapor increases and no significant impact where there is no water vapor, below -20C or so.

  73. KevinK says:

    Dr. Spencer,

    With respect, you sure have “poked” at a hornet’s nest here. What an interesting conversation. Only a few years ago the “Greenhouse effect” was accepted as a “given” theory. Now it has clearly been “demoted” back down to a garden variety hypothesis. Boy, there are sure lots of those around aren’t there?

    Again, I will suggest that folks interested in this somewhat esoteric topic study two optic devices; the integrating sphere and the optical delay line. Both are well understood by folks using them to produce real results that match predictions. The integrating sphere has been around for decades and clearly produces what a climate scientist would term as 100 percent “radiative forcing” (BTW a term that after many decades of productive optical engineering experience I have never heard of outside of the “climate science” community) and still does not produce “net energy gains” and is well known to simply delay the flow of radiative energy through the system. Given the very high speed of light this delay is miniscule.

    Ok, I also need to suggest that folks study the multilayer optical interference anti-reflection filter technology as well. Lots of interesting parallels there; reflections (absorptions/re-emissions from a surface), reabsorptions from a nearby layer (CO2 for climate science, MgF for Optics) and ultimately the only mechanism that “makes more energy stay here” is the presence of Constructive/Destructive Optical interference.

    The optical delay line is a man (or woman) made invention that intentionally delays the flow of radiative energy through a designed system. As the speed of electronic circuits has increased exponentially over the last decades its use has been largely unnecessary. But it still accomplished its intended result; delay the flow of radiation by causing the radiation to make multiple passes through a system at nearly the speed of light (just a slight bit slower outside of a vacuum).

    Here is an alternative hypothesis that I think the climate science community might benefit from; the “greenhouse effect” simply acts to delay the flow of energy through the complex Sun/Earth/Atmosphere/Universe System by causing energy to make multiple passes through the system at the speed of light (it only varies slightly between a vacuum and a normal atmosphere).

    This is simply a hybrid “thermal/optical” delay line. NOTHING NEW HERE, it has been observed and designed on purpose before. The delay can simply be calculated from the the distances involved (a few miles to TOA) and the velocities involved (about 1 nanosecond per foot) and amounts to a few tens of milliseconds. This could possibly (under the very best of circumstances) be observed by monitoring the response time of the gases in the atmosphere. For example if we had data that demonstrated that the gases in the atmosphere warmed up faster when the “GHG” are higher we might possibly be able to confirm this effect. However the historical temperature databases do not contain this data. And tree rings are useless in this regard.

    Cheers, Kevin.

    • Kristian says:

      An interesting post, KevinK. Thanks.

      I have no problem with the concept of ‘a delay in energy flow through the system’. However, this phenomenon doesn’t delay or restrict the output of radiative energy from the actual surface emitting it. Only this energy’s journey from there and out to space.

      The radiation going out (some of it, at least), like (some of) the radiation coming in, does however contribute to the general warming of the atmosphere, besides convection and latent heat transfer, and hence does help affect the temperature of the surface through influencing the temperature gradient going up. The ‘problem’ here, though, is that the GHGs through this mechanism in fact on average aid the convectional flow from surface to TOA. They work to facilitate faster upward flow –> cooling of the surface and each individual atmospheric layer.

      What is really slowing the throughput of energy from surface to TOA is the weight of the atmosphere on the surface, restricting convective (and evaporative) heat loss.

      This is the real ‘atmospheric greenhouse effect’.

      I would say that the most significant contribution of our GHGs (and primarily, then, H2O) to the Earth system temperature regime is the slowing down of cooling at night and of warming at daytime. They do this, not by way of ‘back radiation’, but by their absorptive properties (holding the thermal energy in the atmosphere) and the many other properties of water – cloud reflection, massive heat capacity (slow response), storage and release og latent heat upon change in temperature (and/or pressure).

      Interestingly, the atmosphere above Earth’s tropical/equatorial rainforests contains much, much more GHGs than the atmosphere above the world’s tropical/subtropical deserts. They both lie within the zone of highest incoming solar at TOA. But allowing for differences in altitude the deserts are consistently 3-5 degrees Celsius warmer on an annual basis than the rainforests.

      In far extratropical regions, continental (arid) regions are usually significantly colder annually than maritime (humid) regions (although they are also normally situated significantly higher above sea level). The solar input is of less significance.

      • Kristian’s comment points up the need to simplify, simplify, simplify when arguing about climate.

        There is however a lower bound on simplification, namely the role of adiabatic expansion as it applies to lapse rate.

        The lapse rate of the troposphere is around 10 °C/km. Where adiabatic expansion and compression enters is that no matter how violent or gentle the convection, the lapse rate cannot be changed by convection!

        Anyone trying to reason about the role of convection in cooling the Earth who does not take adiabatic expansion and compression into account is going to arrive at a wildly incorrect result.

        • Arfur Bryant says:

          Vaughan,

          I think this is a fair comment of yours:

          [“Anyone trying to reason about the role of convection in cooling the Earth who does not take adiabatic expansion and compression into account is going to arrive at a wildly incorrect result.”]

          However, to be accurate, the Environmental Lapse Rate of the Troposphere is actually 6.5C/Km.

          It is the Dry Adiabatic Lapse Rate that is nearly 10C/Km.
          And the Saturated Adiabatic Lapse Rate is about 5C/Km.

          I’m sure you are aware of this distinction. The ELR can be different for each individual location on the surface of the planet but it is globally averaged at 6.5C/Km. Likewise, the DALR and SALR are (as far as I know) constant across the globe. So you can’t just say “the lapse rate is 10C/Km”.

          Obviously the effect of these Adiabatic LRs on any convection will depend on whether the air is saturated or unsaturated. So, you’re right, it’s not a simple subject. I suspect the impact of the ALRs is more to do with weather than climate.

      • KevinK says:

        Kristian wrote;

        “However, this phenomenon doesn’t delay or restrict the output of radiative energy from the actual surface emitting it. Only this energy’s journey from there and out to space.”

        Very correct observation. Now the next important consideration comes into play. The Stephan-Boltzman equation only predicts the amount of energy emitted by a SURFACE (spectral shape and absolute values). It says NOTHING about how quickly the surface can be recharged with additional energy (thermal in this case) so the surface can continue to emit that amount of energy.

        Here is the important distinction; the emitted energy flows away from the surface at the “speed of light” while the surface is only “recharged” with thermal energy at the “speed of heat” (i.e. the thermal diffusivity, a crude system level proxy for the speed of heat).

        So if you examine the thermal diffusivity of several typical materials you will find that water (~70% of the Earth’s surface) has a much slower speed of heat than the speed of light.

        Thus, the radiated energy travels away from the surface at a much higher speed and is “long gone” (almost “gone with the wind”, but in fact at a MUCH higher speed) before more thermal energy can make its way up from the water just below the surface of the Oceans.

        The “missing heat” is long gone, with a slight delay (a few tens of milliseconds, perhaps a few seconds at most). Since the “frequency” of the energy input (sunrise/sunset) is about 86 million milliseconds this delay has NO EFFECT on the average temperature of the Earth.

        This delay cannot be observed when the input to the system is “steady state” (i.e. sunlight) with the tools now available (FLIR cameras and satellites) but it exists never the less.

        Cheers, Kevin.

  74. torontoann says:

    Hey, ho…2nd law…knickers-in-a-twist time.

    The emission and absorption of black-body radiation
    are reversible processes. (Use a mirror!) Therefore
    there is never any change of entropy in the Universe
    as a whole from these processes alone, and net energy flow
    depends on details of the setup. For one thing you have to be
    very specific about the areas of any emitting and absorbing
    surfaces. It is the surfaces that count immediately, not
    the bodies themselves

    It is easy enough to make a cool (in the sense of faint) body heat a warm (in the sense of bright) body by radiation: Employ a converging lens, so that effectively a large area of the faint body sends all its radiation to a small area of the bright body.

    The 2nd law always applies – because it applies to everything.
    However it does not always provide a ‘clinching rule’ such that further thought is unnecessary.

    • Ball4 says:

      torontoann 8:11am – “The emission and absorption of black-body radiation are reversible processes.”

      This is exactly why there are no perfect black bodies yet found in the universe. BBs are good for learning tools though.

    • Tim Folkerts says:

      torontoann says: “The emission and absorption of black-body radiation are reversible processes.”

      No! They are not at ALL reversible (other than one trivial situation). If a blackbody ~ 300 K absorbs a single photon from the sun, it will in turn emit a bunch of IR photons (about 20 of them). The process of absorbing a single high energy photon and emitting a bunch of low energy photons definitely increases the entropy of the universe! If the process were reversible, we could take 20 IR photons, let them be absorbed by the blackbody @ 300 K, and have it emit a single visible light photon.

      The one trivial case would be when both the object emitting the photons and the object absorbing the photons are the same temperature. Then the process is reversible and the entropy will not increase.

      “It is easy enough to make a cool (in the sense of faint) body heat a warm (in the sense of bright) body by radiation … “
      A valiant try, but alas this is incorrect (or at least misleading). “Faint” and “cool” are not synonyms — they are separate ideas. The farther you get from the sun, the fainter it will be, but the sun is STILL 5780 K no matter how far away you go. You will never truly make a cool body heat a warmer body (in the thermodynamic meaning of heat as “net thermal energy transfer”).

    • Ball4 says:

      Tim 1:16pm: Well, this thread is fun wandering all over the place.

      “The one trivial case would be when both the object emitting the photons and the object absorbing the photons are the same temperature. Then the process is reversible and the entropy will not increase.”

      I don’t think so. There can be no truly reversible process in nature or what you describe would be perfect insulation. In your example, I propose to you that finding your example to actually occur in nature would take infinite amount of time.

      What do you think? You describe a perfect BB calibration. There are none.

  75. torontoann says:

    Of course. But the interesting point is that an
    approximation to a reversible process exists which
    can be discussed without needing to invoke the very
    abstract fictions, of “infinitesimal reversible steps”
    and final thermal equilibrium.

    Also, of course, the main components of the earth’s skin do
    act to make the earth’s surface an almost perfect blackbody – with respect to infrared radiation, at least.

  76. torontoann says:

    “…make the earth’s surface an almost perfect blackbody…”

    But not all areas at the same temperature, I add!

  77. pochas says:

    An important concept is that of Local Thermal Equilibrium (LTE, code for “at the same temperature”). If you consider a greenhouse a unit in LTE, then its temperature and the temperature of everything inside is determined by the ability of the glass to radiate/conduct/convect. The IR from inside the greenhouse transmitted by the glass simply “fills in” the radiation spectrum from the glass so that the radiation from the whole greenhouse looks approximately like blackbody radiation. So for all practical purposes you can consider a greenhouse to be a blackbody radiator with additional major terms to account for convection from the outside of the greenhouse to the atmosphere. Thats why the Woods experiment did not find a dependence of the inside temperature of the box on the window material (salt, glass, or glass and salt). The whole Woods box with its its various window assemblies comes to Local Thermal Equilibrium. So a greenhouse in fact affects its inside temperature entirely by limiting convection.

    CO2 has its effect if any in the radiating zone of the atmosphere above 14000 ft altitude, and any resulting temperature elevation is transmitted to the surface via the lapse rate. I say “any temperature elevation” because I suspect that radiation from individual water bands does not follow the Stefan-Boltzmann law. These bands all develop population inversions (lase) so that as radiators they “turn on” at different temperatures. That being the case at lower temperatures with few bands active the sensitivity to a temperature increase would be high, resulting in a pronounced negative feedback factor which would decrease as temperature increases. This would align with Miskolczi’s empirical finding of a constant atmospheric tau over many years of observation, regardless of CO2 concentration. Any effect from CO2 simply activates a few more water bands in the radiating zone.

    http://www.opticsinfobase.org/ao/abstract.cfm?uri=ao-22-23-3701

    http://www.friendsofscience.org/assets/documents/E&E_21_4_2010_08-miskolczi.pdf

    • Ball4 says:

      pochas 1:42pm: “…Local Thermal Equilibrium (LTE, code for “at the same temperature”).”

      LTE does not require same temperature, Maxwell-Boltzmann distribution still applies at LTE.

      “So a greenhouse in fact affects its inside temperature entirely by limiting convection.”

      Not entirely, as you write conduction/radiate work on T also in the greenhouse (GH). See Prof. Pratt’s 3 box experiment in LTE linked by Dr. Spencer 8/13 9:23am above. The PE covered box has pretty much the same limited convection but a temperature within is measured substantially different. The construction variable and the environment of the GH both matter for temperature as the other link shows.

  78. Massimo PORZIO says:

    Dear Prof. Pratt.
    I reply here because your post is many post above and you could miss it.

    You wrote:

    “Did you mean the 1/4-th power?”

    Yes, I don’t know the way you call y^(1/x), anyways it seems you understand. Sorry again for my English.

    “The fluxes cancel except at the window, which is therefore the only part of the experiment that matters. If you hold the window area constant while multiplying the volume of the box by say one thousand, what difference would that huge multiplication make? The energy entering and leaving the window should remain the same.”

    I don’t believe you are completely right here, at least considering the radiative energy path and considering what a greenhouse is useful to.
    A greenhouse is useful to warm its floor and the air inside it.
    The incoming solar flux is almost one direction oriented because the radiator is very far (the Sun), but as that energy is absorbed by the bottom of your boxes its radiative part is spreaded by the bottom cardboard also to the walls cardboards, so the LWIR flux is distributed on a larger area.
    Thus all the inner surface of the box become the radiator, which finally define the radiative part of the temperature of the air inside.
    There is no doubts that only the upper windows is the exit gate of the energy and that at the steady-state through that gate is the only place where the outgoing energy balance the incoming one, but the air inside is warmed by all the box surfaces, at least for the radiative path.

    In a cubic box like yours, the air is warmed by a radiator which receive the flux on one side (the bottom) and almost immediately spread that radiative energy to the walls using a lambertian surface radiation pattern (the convective heat should almost all transferred to the air at this bottom radiator instead, at least at the beginning of the warming
    where convection works better).

    For this reason I guess that Wood was right measuring the temperature with the bulbs of the thermometers inside the boxes (I presume) at their bottom.

    Of course maybe I’m completely wrong, because that’s not my field. In this case I apologize to having take your time reading me.

    I formerly argued that except for few mm below the windows the temperature inside the boxes should be almost homogeneous.
    This measurement (say at 1/2 the box height), could be the next step in your experiments, to prove Wood right or wrong.

    Have a nice day.

    Massimo

    • Tim Folkerts says:

      Just a brief comment — measuring the temperatures accurately will itself be a challenge. Whatever probes that are used will have their own emissivity for IR and reflectivity for visible light.
      * A black thermometer will get warmer than a white thermometer in sunlight, even if the surrounding air is the same temperature.
      * A high emissivity will help warm the thermometer if there are warm wall around, and help cool the thermometer if there are cool walls around.

      You would almost want some sort of miniature “Stevenson screen” around each thermometer. But that would make it impossible to attach the thermometer directly to the glass or the walls.

      • Massimo PORZIO says:

        Hi Tim,
        I fully agree with you.

        My next consideration about Prof. Pratt experiment is that those measured temperature below the salt window probably was more than the 6°C below the one he measured respect the glass window, because the glass window has almost the same emissivity of the TMP05 thermometric device package (epoxy resin), while the salt window has not. He probably get a more realistic temperature if, under the salt window, he covered the Tmp05 with a small piece of adhesive aluminium foil, to avoid any IR absorption.

        Have a nice day.

        Massimo

  79. Nullius in Verba says:

    Roy, I don’t know if you’ll read this, but just in case…

    The website you linked to says in several places that the greenhouse effect can’t be studied in the lab because the absorption of IR by GHGs is so weak. That doesn’t quite follow, because a greenhouse material doesn’t have to be a gas.

    Liquid water absorbs and re-emits thermal IR within a few millimetres, but is nearly transparent to visible light. So a shallow pool of water is about 20,000 times more powerful a greenhouse material than CO2. Energy from sunlight shining into a pool will be absorbed at the bottom, will radiate as IR, and will be almost immediately absorbed and re-radiated in all directions by the water, each additional few millimetres of water providing another ‘greenhouse’ blocking the radiation from the layer below.

    It’s a perfect lab model for the radiative greenhouse effect. And you can test the relative roles of convection and radiation by suppressing convection in the pool of water and watching the temperature climb. It’s called a ‘solar pond’, and is used as a practical energy generating technique.

    • Kristian says:

      How do you ‘suppress’ convection in a pool of water that undergoes heating? The only way I can think of is by reducing the cooling from the surface to an absolute minimum. But even this would hardly stop convection/advection from taking place until the bulk ended up completely isothermal, which in practice would never happen.

      And the thing is, suppressing convection is not the same as removing the effects of convection in order to single out any radiative effects. Quite the opposite. This is why so many people seem to confuse effects on temperature specifically arising from ‘suppressed’ or ‘perturbed’ convective flow with radiative effects.

      Whenever you suppress convection (be it in air or water) you enhance the temperature effect of the absorbed heat as a result.

      So for instance, the so-called GHGs in the atmosphere, by continuously letting the upper troposphere cool radiatively to space, help promoting convective flow from BoA to ToA which in turn helps cooling the surface.

      Also, you don’t need a bottom of a pool for the Sun to heat the water. Almost 65% of the energy from the solar radiation (that which comes in the mid and near IR range of the spectrum) is already absorbed within the first 5 metres of the water column. And as soon as the water starts warming, it starts moving, preferably upwards. And you have convection. Just like in the atmosphere.

      • Nullius in Verba says:

        “How do you ‘suppress’ convection in a pool of water that undergoes heating?”

        You very carefully fill the pond with a strong salt solution at the bottom and fresh water on top. The lower layers are denser, because of the dissolved salt, and so even when warmer than the water above they are not buoyant.

        http://en.wikipedia.org/wiki/Solar_pond

        “So for instance, the so-called GHGs in the atmosphere, by continuously letting the upper troposphere cool radiatively to space, help promoting convective flow from BoA to ToA which in turn helps cooling the surface.”

        Yes, but there is an impediment to that convective cooling because convection stops short when the adiabatic lapse rate limit is met.

        If air was incompressible, then the adiabatic lapse rate would be zero and the atmosphere would be close to isothermal. Convection would instantly carry any heat build-up away from the surface. It would make no difference whether it radiated from the top or the bottom.

        But the adiabatic lapse rate *is* non-zero, so it does make a difference where it is radiated from.

        “Also, you don’t need a bottom of a pool for the Sun to heat the water.”

        Yes, I know. But I wanted to make the analogy clear with the atmospheric greenhouse effect. The atmosphere *does* have a bottom.

        • Kristian says:

          Yes, but there is an impediment to that convective cooling because convection stops short when the adiabatic lapse rate limit is met.”

          Sure. But that is not by virtue of the GHGs. Only by convection itself. The GHGs do their best to steepen the gradient. Just like solar heating of the surface during the day does. So the presence of the GHGs in our atmosphere, if anything, makes convection (and hence surface cooling) ever so slightly more efficient. And that is after having first deprived the global surface of 45% of its potential heat input from the Sun.

          “But the adiabatic lapse rate *is* non-zero, so it does make a difference where it is radiated from.”

          Why does it matter where it is radiated from? The only ambition of the Earth system as a whole is to release as much energy back to space as it receives from the Sun. It does this from the BoA all the way up to the ToA. The location of the mean level of this is irrelevant. That is simply situated wherever the mean temperature of the Earth system ensemble at any one time happens to be. And this is set from the surface up (heated by the Sun), not from some mean radiating level in the middle of the convective troposphere down. That is turning the situation (the direction of causation) completely on its head. It is letting the tail wag the dog. And it finds no support in the real-world data. Neither from Earth nor from Mars. It is a purely hypothetical mechanism with no empirical back-up evidence for any claimed effect on surface temperatures whatsoever.

          You all seem to be forever ignoring the elephant in the room here, consumed rather by this endless nitpicking on minor details and side issues (that is, anything to do with thermal radiation within the troposphere). What really determines the mean global surface temperature of our planet, besides the insolation from the Sun, is the atmospheric weight on the surface, restricting the upward (buoyant) acceleration of near-surface heated air and the free evaporation of water vapour from the surface. Read KevinK’s posts above. The delay on thermal radiation’s escape from the ground through the troposphere and out to space is truly miniscule and its effect on surface temperatures are practically nonexistent next to this convective/evaporative suppression effect.

          • Nullius in Verba says:

            “Why does it matter where it is radiated from?”

            The temperature at the surface (Ts) is the effective radiative temperature (Tr) plus the adiabatic lapse rate (ALR) times the average height from which energy is radiated to space (h).

            Ts = Tr + ALR * h

            The air at altitude h has to be at such a temperature as to emit the same total energy that the Earth absorbs (Tr). If it was warmer it would emit more than the Earth received and the Earth would cool. If it was cooler it would emit less and the Earth would warm.

            Why does it matter where this is? Because the further away it is from the Earth’s surface, the bigger the temperature difference the adiabatic lapse rate produces between them. If h increases, so does the surface temperature.

            Insolation and albedo affect Tr, humidity affects ALR, and GHGs affect h.

          • Tim Folkerts says:

            Well said, Nullius. šŸ™‚

          • Kristian says:

            Yes, I realise this is what the hypothesis claims.

            I will just have to reiterate:

            “The only ambition of the Earth system as a whole is to release as much energy back to space as it receives from the Sun. It does this from the BoA all the way up to the ToA. The location of the mean level of this is irrelevant. That is simply situated wherever the mean temperature of the Earth system ensemble at any one time happens to be. And this is set from the surface up (heated by the Sun), not from some mean radiating level in the middle of the convective troposphere down. That is turning the situation (the direction of causation) completely on its head. It is letting the tail wag the dog. And it finds no support in the real-world data. Neither from Earth nor from Mars. It is a purely hypothetical mechanism with no empirical back-up evidence for any claimed effect on surface temperatures whatsoever.”

            The tropospheric temperature profile is naturally set by the lapse rate with the surface as its point of departure, because the surface is where the solar heat is absorbed and released back up to warm the troposphere. That’s the way the heat flows through the system. The surface warms first, then the troposphere. Progressively upwards.

            It is not like the power density flux radiated from the Earth system as a whole to space (239 W/m^2) has to be emitted from or has to have anything to do with the specific layer within the ensemble whose physical temperature happens to match the Stefan-Boltzmann-calculated emission temperature of our planet when seen from space. That temperature (the 255K) is simply what the satellites would compute when reading/detecting such a power density flux coming out from our planet. But that flux contains all the radiation released from the Earth system from BoA to ToA, most of it way above the ‘effective radiating level’ at 5 km. The mean is just the mean, somewhere in the middle.

            Remember, the tropopause holds an average global temperature of around 210K. And still the intensity of the thermal radiation flux flowing out from it is 239 W/m^2. According to its physical temperature it should have been a mere 110 W/m^2.

            What I’m trying to say is: The physical temperature of atmospheric layers doesn’t have anything to do with the radiative flux going through them and out to space. The two are non-related.

          • Nullius in Verba says:

            Re-iterating it doesn’t help.

            The temperature of the radiating surface (not the solid surface) tends towards the temperature at which it radiates the input energy. It has to be at the output and not the input because it is the output that is adjusting itself to maintain equilibrium, not the input.

          • Kristian says:

            “Re-iterating it doesn’t help.”

            No, apparently not, because you seem to be ignoring what I’m writing anyway, the arguments I put forward.

            Again, yes I know this is what the hypothesis is saying. And
            again I’m telling you that it’s not how the world is
            observed to work. It is purely a theoretical construct.

            The total energy output from Earth to space at any one
            time is not determined by the physical temperature of some
            specified layer of air a few kilometres above the surface. The two are not related.

            Why do you believe the clearly warped idea that it is somehow the 255K temperature of a gaseous ‘surface’ somewhere between the BoA at 288K and the ToA at 210K that happens to be the one responsible for sending out the 239 W/m^2 from Earth to space? That there’s a connection at all between the actual radiative flux streaming through and from this surface and its physical temperature? (You know the Stefan-Boltzmann law does not even apply to gases.) And that somehow this fixed-temperature ‘surface’ pulls the lapse rate ladder along with it, up and down, controlling the surface temperature instead of being controlled by it?

            We all know that what the Earth needs to do is shed the
            239 W/m^2 that it absorbs back out to space. And this flux is not bound to some specific layer with a temperature of 255K simply because this happens to be the emission temperature of Earth when seen from space based on the output flux of … 239 W/m^2. No, the flux adds up along the way from BoA to ToA. It is released throughout the atmospheric column, and directly from the surface. It is a total flux. The final flux to flow out through the ToA. The heat moved up there past the so-called ‘effective radiating level’ to a large extent by way of convection rather than by radiation. The final radiative flux thus goes out NOT through the radiating level at 5 km above the surface, but through the ToA, on average 45K colder and at least 7 kilometres higher up. Above the convection top.

            Sorry, but the total energy output from Earth to space is determined by the solar input and whatever the delayed convective response manage to put back out from the surface. The thermal radiation is simply released whenever and from whatever level it needs to be released, being the instrument ultimately carrying the heat out of the system. It doesn’t ‘trap’ the heat even when the GHGs absorb it. On the contrary. What ‘traps’ or ‘holds on to’ the heat on its way back out is the relative sluggishness of convective flow and the thermal mass (heat capacity) of the Earth system (land, ocean, air). The GHGs do not impede any heat transport. They facilitate it. They let the swift radiative heat transfer process take over the job from its slow convective counterpart. Getting that energy shipped upward and back out to space. At nearly the speed of light …

            It is a bit like the observation of the raised tropopause.
            A raised tropopause is a result of a warmer surface
            >> warmer troposphere. It is not the cause of
            a warmer troposphere >> warmer surface.

            Read this. It’s an enlightening and refreshing, empirically
            based analysis. Unlike the armchair hypothesizing (fiction)
            of surface warming by raised radiating levels.

            http://chiefio.wordpress.com/2012/12/12/tropopause-rules/

          • Nullius in Verba says:

            “Why do you believe the clearly warped idea that it is somehow the 255K temperature of a gaseous ‘surface’ somewhere between the BoA at 288K and the ToA at 210K that happens to be the one responsible for sending out the 239 W/m^2 from Earth to space? That there’s a connection at all between the actual radiative flux streaming through and from this surface and its physical temperature?”

            Because that’s how it works. The details are slightly more complicated, but the effect is the same.

            The temperature at all levels is convectively coupled, so emission at any level affects the temperature of the whole. The overall rate of emission is temperature-dependent, intermediate between the warmest and coldest, as the warmest parts emit a bit more than the average and the coolest parts emit a bit less than the average. The effect is the same as if the whole lot had emitted as if at an intermediate temperature, at an intermediate level.

            When boundaries are a bit fuzzy, it’s generally the case that the effect is similar to that of a sharper boundary somewhere in the middle of the fuzzy range. It’s just a more extreme example of the normal state of affairs – all materials are a little bit translucent.

            “(You know the Stefan-Boltzmann law does not even apply to gases.)”

            Yes it does. Why do you think it doesn’t?

          • Kristian says:

            “Because that’s how it works.”

            Yes, of course. How silly of me.

            Funny. This is what always happens with you people. You end up retreating into your little corner of simply reciting your articles of faith. Like incantations on repeat.

            And of course employing the famous ‘eyes shut, fingers in ears, going la-la-la-la-la’ tactic alongside.

            You know you cannot support your position with observational evidence of any kind from the real world out there, the actual Earth system. So you don’t even bother trying. All you have are fancy diagrams on pieces of paper. Clever ideas about how the world simply MUST work, because … it must.

            And yet you keep invoking them as if they were empirically established as the eternal, absolute Truth on the matter.

            Ah, so you just KNOW that’s how it works, do you? How exactly do you ‘know’ this, Nullius? Because some authority figure once told you so? Because you read it in a book somewhere and what’s written in books HAS to be true, because otherwise, hehe, they wouldn’t publish it, now would they? Is that it?

            Have you had a look at temperature data from the surface of Mars, Nullius? What is the radiative warming effect on the surface of the atmospheric CO2 in the Martian atmosphere? How high up does the atmospheric content of CO2 (8 times that of Earth’s atmosphere packed into a total volume about 2.5 times smaller) lift the effective radiating level off the ground up there?

            And why are tropical/equatorial rainforests consistently cooler on an annual basis (by 3-5 degrees Celsius) than tropical/subtropical deserts when the former has an atmosphere simply loaded with GHGs retaining ‘heat’ and the deserts hardly any?

          • Kristian says:

            ““(You know the Stefan-Boltzmann law does not even apply to gases.)”

            Yes it does. Why do you think it doesn’t?”

            Because specific layers or parcels of gas inside a gas volume do not constitute BB surfaces. They do not radiate according to their temperature to the outside of that volume. Just like inside a volume of liquid water. The Stefan-Boltzmann law applies only to the surfaces of ideal emitters (black or gray bodies), like stars or, in situations on Earth, to other downright ‘hot’ objects and/or objects that are very much warmer than its surroundings (the T^4 relation). In other situations the errors from S-B equation calculations would become large.

            http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

          • Kristian says:

            If you have a volume of pure O2 or N2 and you warm it up to say 255K, according to you, then, it would by the S-B law radiate 239 W/m^2 to its surroundings. No matter what. It is a gas with a temperature, after all, so the S-B law applies to it, doesn’t it?

            But if so, what do we need the GHGs for? Then the O2 and N2 of our atmosphere can radiate all its absorbed heat perfectly fine to space by themselves it seems …

          • Nullius in Verba says:

            “Funny. This is what always happens with you people. You end up retreating into your little corner of simply reciting your articles of faith. Like incantations on repeat.”

            The feeling’s mutual, I assure you. šŸ™‚

            “Ah, so you just KNOW that’s how it works, do you? How exactly do you ‘know’ this, Nullius? Because some authority figure once told you so?”

            Nope. By fitting together individual experiments and observations to figure out a mental model of how all the individual pieces work, and then fitting the parts together.

            That’s how engineering works – you don’t have to build a bridge to see if it’ll stand up, you can calculate what will happen. You know the theory and properties of all the components, and you know how they interact, so it’s generally a straightforward matter to figure out how they work in combination.

            “What is the radiative warming effect on the surface of the atmospheric CO2 in the Martian atmosphere? How high up does the atmospheric content of CO2 (8 times that of Earth’s atmosphere packed into a total volume about 2.5 times smaller) lift the effective radiating level off the ground up there?”

            About 5 K.

            About 1 km.

            (ALR = 5 K/km on Mars. Partial pressure of CO2 on Mars is about 15 times that on Earth, or 4 doublings. On Earth each doubling gives 1 K warming from a 150 m rise in emission altitude. 4 doublings would give 4 K warming from a 600 m rise on Earth. And there’s no water vapour on Mars, which is the dominant GHG on Earth. So take Earth’s GHE, knock off the water vapour component, add 4 K, and then reduce the lapse rate. It’s a lot more complicated than that, of course – there’s direct surface emission, higher density of cold gases, pressure broadening, dust/aerosols, different spectral windows, etc. to consider. But we’re in the same ballpark.)

            “And why are tropical/equatorial rainforests consistently cooler on an annual basis (by 3-5 degrees Celsius) than tropical/subtropical deserts when the former has an atmosphere simply loaded with GHGs retaining ‘heat’ and the deserts hardly any?”

            Because the lower humidity over deserts results in a steeper adiabatic lapse rate.

            Ts = Tr + ALR * h

            If ALR is bigger, the surface temperature will be higher.

            Also, heat is transported horizontally by convection, so the equator is cooled and the deserts warmed by horizontal transfer not accounted for in the simple 1D model. The 1D model can only tell you about the global average, it’s somewhat more limited in what it says about local conditions.

            “Because specific layers or parcels of gas inside a gas volume do not constitute BB surfaces.”

            Ah, I see. You meant that non-black bodies don’t obey SB. So why did you say “gases”?

            Or do you think there’s something special about the behaviour of atoms at surfaces that distinguishes them from those in the bulk of the material?

          • Kristian says:

            “About 5 K.

            About 1 km.

            (ALR = 5 K/km on Mars. Partial pressure of CO2 on Mars is about 15 times that on Earth, or 4 doublings. On Earth each doubling gives 1 K warming from a 150 m rise in emission altitude. 4 doublings would give 4 K warming from a 600 m rise on Earth. And there’s no water vapour on Mars, which is the dominant GHG on Earth. So take Earth’s GHE, knock off the water vapour component, add 4 K, and then reduce the lapse rate. It’s a lot more complicated than that, of course – there’s direct surface emission, higher density of cold gases, pressure broadening, dust/aerosols, different spectral windows, etc. to consider. But we’re in the same ballpark.)”

            Nope. ALR = 2.5 K/km on Mars. S-B-calculated emission temperature as seen from space is 210K. Estimated mean global surface temperature is 210K, but more likely lower (that is, based on actual measurements from various points on its surface). Your figures are based on summer and/or ‘tropical’ measurements.

            The effective radiating level on Mars is either on the ground or below it …

          • Kristian says:

            “Because the lower humidity over deserts results in a steeper adiabatic lapse rate.

            Ts = Tr + ALR * h

            If ALR is bigger, the surface temperature will be higher.”

            And h should be bigger over the rainforests. Much bigger. That’s the whole point with your GHG effect, isn’t it?

            No, the deserts are warmer than the rainforests for a very simple reason: They receive and absorb much more solar heat during a full year. Because of less moisture and clouds (H2O).

            “Also, heat is transported horizontally by convection, so the equator is cooled and the deserts warmed by horizontal transfer not accounted for in the simple 1D model.”

            This only works for ocean currents. Not for atmospheric convection. Atmospherically, the equator cools vertically, not horizontally, by convection. The equatorial zone would receive air coming IN from the subtropics (where the warmer deserts are).

          • Kristian says:

            “Nope. By fitting together individual experiments and observations to figure out a mental model of how all the individual pieces work, and then fitting the parts together.”

            And may I ask you specifically what ‘individual experiments and observations’ from the real Earth system you base your ‘mental model’ on?

            There is no point having a ‘mental model’ or a hypothesis, as beautiful as it might be, if you cannot observe it to work in nature.

            “That’s how engineering works – you don’t have to build a bridge to see if it’ll stand up, you can calculate what will happen.”

            Yes, but only after building on other people’s empirical observations and experiences. There was probably quite a lot of trial and error in building the first bridges, don’t you think? The problem with your model is that there has been no period of empirical trial and error to see if it works at all, to see if it actually has something to it, measured against reality.

            It is still all theory.

          • Kristian says:

            “Ah, I see. You meant that non-black bodies don’t obey SB. So why did you say “gases”?”

            No, I meant gases. Stop evading the issue and answer me this:

            If you have a volume of pure O2 or N2 and you warm it up to say 255K, according to you, it would by the S-B law radiate 239 W/m^2 to its surroundings (and same with every internal parcel). No matter what. It is a gas with a temperature, after all, so the S-B law applies to it, doesn’t it?

            But if so, what do we need the GHGs for? Then the O2 and N2 of our atmosphere can radiate all its absorbed heat perfectly fine to space by themselves it seems …

      • Tim Folkerts says:

        Google is your friend. Google ‘solar pond’ and you will find things like “However the salinity gradient forms a density gradient that increases with depth, and this counteracts the temperature gradient, thus preventing heat in the lower layers from moving upwards by convection and leaving the pond.” It all has to do with the salt in the water.

        Kristian says: “So for instance, the so-called GHGs in the atmosphere, by continuously letting the upper troposphere cool radiatively to space, help promoting convective flow from BoA to ToA which in turn helps cooling the surface.
        You are missing half of the physics here. Since I know you have heard the correct explanation many times before, it is not worth going into detail here as to why Conservation of Energy ensures us that GHGs must indeed WARM the surface, not COOL the surface.

  80. Bryan says:

    Nullius in Verba says in post above

    “Liquid water absorbs and re-emits thermal IR within a few millimetres, but is nearly transparent to visible light.”

    I would agree, so that about 50% of solar energy is absorbed at surface and is largely responsible for the water evaporation.
    This energy is stored as latent heat of vaporisation.

    However you then go on to give a wiki link which contradicts your own (correct) statement.

    “If the water is relatively translucent, and the pond’s bottom has high optical absorption, then nearly all of the incident solar radiation (sunlight) will go into heating the bottom layer.”

    That the pond bottom absorbs the remaining 50% of shorter wavelengths and heats the surface to 80C is to be expected in equitorial regions.
    (Theoretical maximum about 123C)

    Also the IR released from pond bottom is absorbed near the pond bottom by radiative insulation effect causing warm water to form in that area.
    That the atmosphere and water cool the sun facing Earth surface is quite clear.
    The solar pond reduces this cooling but is not some kind of independent energy source.
    This relates in a general way to the thread theme of the glass greenhouse being similar to the atmosphere effect.

    Who would buy a glass greenhouse to cool their plants by day?

    Its clear that the greenhouse theory of the atmosphere is a failed conjecture and should be abandoned.

    • Nullius in Verba says:

      It depends what you mean by “cooling”. It’s “cooling” in the sense that heat is being transported away from the surface. It’s not “cooling” in the sense of being cooler than it would be without the ‘greenhouse’.

    • Tim Folkerts says:

      1) Thermal IR is absorbed and emitted almost entirely within the first few microns. Since there is much more thermal IR leaving than arriving, IR would (and does) create a cool skin on water. Solar IR is shorter wavelength and penetrates deeper. Only about 20% of sunlight would be absorbed in the first cm. How much of the remaining 80% would get to the bottom would depend a great deal on how deep the pond is.
      http://www.lsbu.ac.uk/water/images/watopt.gif
      http://en.wikipedia.org/wiki/Planck%27s_law#Percentiles
      http://scienceofdoom.files.wordpress.com/2010/12/ocean-temp-profiles-2-kawai-wada-2007.png?w=500

      ************************************************
      2) The “theoretical maximum” that sunlight can warm something is 5780 K — the same temperature as the sun.

      The maximum that unconcentrated, full-strength sunlight alone could warm a blackbody is indeed about 121 K. But we know that only ~ 1000 W/m^2 reaches the surface, and we just established that only about 1/2 of that (500 W/m^2) reaches the bottom of the pond. But 500 W/m^2 would only heat the bottom of the pond to 33 K.

      It seems that *something* else must be aiding the sunlight — ie “back-radiation” and “the greenhouse effect”

  81. Bryan says:

    Tim Folkerts says:

    “Thermal IR is absorbed and emitted almost entirely within the first few microns. Since there is much more thermal IR leaving than arriving, IR would (and does) create a cool skin on water. Solar IR is shorter wavelength and penetrates deeper. Only about 20% of sunlight would be absorbed in the first cm. How much of the remaining 80% would get to the bottom would depend a great deal on how deep the pond is.”

    Agreed

    “ The “theoretical maximum” that sunlight can warm something is 5780 K — the same temperature as the sun.”

    Who was assuming that lenses and mirrors were involved?
    So this would appear to be an off topic remark.

    “The maximum that unconcentrated, full-strength sunlight alone could warm a blackbody is indeed about 121 K.”

    Agreed

    “ But we know that only ~ 1000 W/m^2 reaches the surface, and we just established that only about 1/2 of that (500 W/m^2) reaches the bottom of the pond.”

    You contradict yourself here as above you say 80% could reach the bottom of a shallow pool.

    So my original post still stands in that there is nothing magical about warm water at around 80C near an equatorial Sun facing surface.

    Solar IR absorbed on way down plus surface contact conduction and surface up radiation will heat the water.
    So there is no need to speculate as you seem to be doing in your last paragraph that the water is a separate energy source.

    “It seems that *something* else must be aiding the sunlight — ie “back-radiation” and “the greenhouse effect”

    This is just wishful thinking on your part with no convincing evidence provided.

  82. Tim Folkerts says:

    You contradict yourself here as above you say 80% could reach the bottom of a shallow pool.
    The 80% number is for a pool only few cm deep, with clear water. A ‘realistic’ pond a few meters deep would be much less than 80%. At least according to the graphs I found. Besides, 800 W/m^2 woulds STILL be less than the 90C observed temperatures.

    So there *is* something magical about 90 C water.

    *********************************

    PS. When you say “Solar IR absorbed on way down plus surface contact conduction and surface up radiation will heat the water”, do you really mean that sunlight absorbed by the upper cooler layers will flow down to the deeper warmer layers? Isn’t that exactly the sort of thing that slayers say is impossible? šŸ™‚

  83. Bryan says:

    Tim Folkerts

    The figure of 90C has no link to when where it was obtained.
    Can you provide some hard details of this claim.

    Here is what this poor link (even by pathetically low wiki standards) actually says;

    “The energy obtained is in the form of low-grade heat of 70 to 80 °C compared to an assumed 20 °C ambient temperature.”

    Notice the ‘assumed’ 20C temperature.
    Where can one assume this?

    Also the link says (incorrectly) practically all the solar radiation reaches the bottom of the pond.

    Why do folk like yourself clutch at such poor evidence?

    Is it because no robust evidence of a greenhouse effect exists?
    I look forward to Roy Spencer’s experiment

    You say

    “PS. When you say “Solar IR absorbed on way down plus surface contact conduction and surface up radiation will heat the water”, do you really mean that sunlight absorbed by the upper cooler layers will flow down to the deeper warmer layers? Isn’t that exactly the sort of thing that slayers say is impossible?”

    Firstly I am not a slayer or part of any ‘group think’ faction.
    I think for myself after weighing up the evidence in a rational way.
    They will have to speak for themselves however absorption of the Suns IR in water causing a temperature rise is pretty uncontroversial.

    .

    • Tim Folkerts says:

      Bryan, I am not “clutching” at any particular evidence here — solar salt ponds are a tangent to a tangent here. Even so, it is easy to find other links that list temperatures of 70-90 C for the ponds. Also, the “assumed” 20 C is in the section for calculating the maximum theoretical efficiency. They could have given the efficiencies for all sorts of temperatures (in fact, they do with the equation). “20 C” is simply an example that might be typical around the world. In most places, you can find a cold reservoir somewhere around 20 C.

      You seem to be doing a bit of “clutching yourself. On the one hand, you agree that the wiki page over-estimates the energy at the bottom of the pond, but then you need the over-estimated power levels to get the bottom to even 60-80 C. Or perhaps you do agree that thermal radiation from cooler objects (the higher layers of water) could help raise the temperature of hte lowe, warmer layers.

      As to “the greenhouse effect”, every thermodynamics textbook supports the idea that ALL energy inputs — including thermal IR from colder objects — will contribute to the energy (and hence the temperature) of an object. Only by “clutching” at convoluted (and incorrect) interpretations of fundamental thermodynamics can people ‘refute’ impact of “back-radiation”.

      I too look forward to Dr Spencer’s results, and plan to add some of my own as well.

  84. Nullius in Verba says:

    The 90 C comes from the limitation that allowing the pool to boil would disturb the density gradient. The temperature itself wasn’t intended as a demonstration that the effect could only occur through a greenhouse effect – I’ve given up trying to argue with the doubters, I was addressing my remarks to people like Roy who know that there is one.

    But I’m not convinced by the argument that you can expect the bottom of a pool in the tropics to be heated to ~120 C by sunlight alone. I’m guessing that’s calculated by assuming a flat plate perpendicular to the sunlight at the top of the atmosphere that is a perfect black body on the sunlit side and a perfect reflector on the other side. The solar constant is around 1366 W/m^2 which is about what a black body at 120 C emits. If it gets no energy input from anywhere else (i.e. the rest of its surroundings are at absolute zero) then it would equilibrate at this temperature. However, I don’t consider this a realistic scenario for the bottom of a pond.

    The solar constant at the surface is around 1000 W/m^2, the ground is generally at an angle to the sun, and isolated objects radiate from all sides. (Non-isolated objects conduct heat away.) And at the bottom of a pond, you’re not getting much of the solar IR, which is a significant percentage, and there is reflection, absorption, etc. as noted. Judging from the way I can see the white tiles on the bottom of a 3m deep swimming pool, I’m highly dubious that half the visible component will be absorbed in such a short distance, but it’s certain that it’ll be a lot less than 1000 W/m^2. I’d guess around 600 W/m^2 perpendicular, and a quarter of that for a spherical isolated object. If the rest of the object’s surroundings were at absolute zero, it would be chilly indeed.

    Given that the surroundings are not at absolute zero, it’s going to be a lot warmer, but that will depend on the details, and is besides the sort of thing the doubters wouldn’t accept anyway, given that it involves energy input from cooler surroundings.

    I don’t think there’s any point in trying to convince people of the radiative greenhouse effect with thought experiments, if they’re determined not to change their minds. I’m dubious that Roy’s attempts at experiment will work, either. If you can’t see why it works, you’re probably not going to. And it’s a waste of time anyway, since the convective greenhouse effect works by a completely different principle.

  85. Bryan says:

    Tim Folkerts you say

    “Even so, it is easy to find other links that list temperatures of 70-90 C”

    Could you supply even one robust link to the 90C claim?

    “20 C is simply an example that might be typical around the world. In most places, you can find a cold reservoir somewhere around 20 C.”

    But not coupled to a natural solar heat pond of 90C.

    The maximum solar irradiation can reach as high as 1412W/m2 in early January because of Earths elliptical trajectory.

    It seems to me clutching at straws if you pick an atypical set of conditions and claim a general effect.

    Nullius in Verba says

    “besides the sort of thing the doubters wouldn’t accept anyway, given that it involves energy input from cooler surroundings.”

    Stamp and label all doubters as identical makes it easy for a lazy reply.
    For instance I doubt that the CO2 driven greenhouse effect is a real problem and yet I agree that two way energy flows occur between cold and hotter objects.

    Prior to 1990 you will not find any reference to the so called atmospheric ‘greenhouse effect’ in any thermodynamic or physics text book.

    Is that because it does not exist?

  86. Kristian says:

    Bryan, you say:

    “The solar pond reduces this cooling but is not some kind of independent energy source. This relates in a general way to the thread theme of the glass greenhouse being similar to the atmosphere effect.”

    If one just reads the wiki article on the solar pond, it says plainly how this warming by restricting free cooling is accomplished. Thermal radiation isn’t even mentioned. It is simply assumed (as always) to be the magical component by Nullius and Tim here …

    It is not. It is just what I said, suppression of convective flow that does the trick. Just like with regular greenhouses. And just like with the real ‘greenhouse effect’ of the atmosphere.

    The article states:

    “There are 3 distinct layers of water in the pond:

    # The top layer, which has a low salt content.

    # An intermediate insulating layer with a salt gradient, which establishes a density gradient that prevents heat exchange by natural convection.

    # The bottom layer, which has a high salt content.

    If the water is relatively translucent, and the pond’s bottom has high optical absorption, then nearly all of the incident solar radiation (sunlight) will go into heating the bottom layer.”

    And here it comes. The warming mechanism:

    “When solar energy is absorbed in the water, its temperature increases, causing thermal expansion and reduced density. If the water were fresh, the low-density warm water would float to the surface, causing a convection current. The temperature gradient alone causes a density gradient that decreases with depth. However the salinity gradient forms a density gradient that increases with depth, and this counteracts the temperature gradient, thus preventing heat in the lower layers from moving upwards by convection and leaving the pond. This means that the temperature at the bottom of the pond will rise to over 90 °C while the temperature at the top of the pond is usually around 30 °C.

    The solar heat slowly builds in the bottom layer. Because it cannot escape up through the intermediate layer … by way of convection.

    Q.E.D.

    Isn’t it a bit cute how they always try to ‘remove the effects of’ convection when what they really do is suppressing or perturbing free convective flow, apparently thinking that the resulting accumulation of ‘heat’ must thus somehow arise from radiative effects …?

    This is exactly the misconception that has led to the misinterpretation of all of these similar type experiments since Fourier jumped to his conclusion in 1824 on De Saussure’s results. And they’re still all doing it to this day …

  87. Nullius in Verba says:

    “Isn’t it a bit cute how they always try to ‘remove the effects of’ convection when what they really do is suppressing or perturbing free convective flow, apparently thinking that the resulting accumulation of ‘heat’ must thus somehow arise from radiative effects …?”

    Thermal conductivity or resistance depends on all three mechanisms (convection, conduction, radiation) working in parallel. The temperature build-up is simply due to the total combined thermal resistance between the heated body and the cold sink.

    The particular question being discussed here is whether an IR-opaque barrier increases the radiative component of the thermal resistance. Which if you suppose radiative resistance is a significant component of overall resistance, and other factors held equal, will increase overall resistance – but that’s besides the point for the moment. The simple question is: does an IR-opaque barrier increase the radiative component of thermal resistance? That’s what the slayers have been arguing cannot be.

    And to test and measure this particular heat transport mechanism, you first have to control for the other heat transport mechanisms.

    If we can get that point established, then there is some hope of the debate moving on to whether radiative resistance is actually significant in various circumstances, like the oceans or the atmosphere.

    But for the time being, all such arguments are lost in the roar of background noise of people saying there can be no such thing as radiative resistance. Nobody will listen to the right alternative, because of all the wrong alternatives milling around. And in the meantime, the campaigners for climate catastrophe can more easily portray all sceptics as cranks who deny simple radiative physics.

    We have to make progress one step at a time. As I said, the point of me proposing it wasn’t to *prove* to a doubter that the radiative greenhouse effect exists, it was to offer a lab model in which the physics could be explored for someone who already knows that it does. Trying to argue that this doesn’t prove anything is a straw man. It wasn’t meant to.

    • Kristian says:

      Nullius,

      You cannot ‘control for the other heat transport mechanisms’ by suppressing them. That’s all I’m saying. You end up misinterpreting the resulting buildup of heat as a radiative effect. It is not.

      Didn’t you read the link you yourself provided? I even quoted from it, where it specifically stated the actual warming mechanism of the solar pond. And surprise, surprise!Suppression of convective flow.

      You and Roy don’t know that the radiative GHE is real at all. No one does. If you did, you would have been able to show it unequivocally a long time ago. But the effect remains purely theoretically deduced and has never been shown to be net positive, significant, not even to exist at all. It is no more than a conjecture. Still, what you both do is assume its existence.

      ‘To know something’ and ‘to assume something’ are not the same.

      Yes, we all agree that GHGs absorb IR within certain segments of the EM spectrum and that this in and by itself will help warm the atmosphere. But the atmosphere would warm with or without the presence of GHGs. It’s not like they’re a prerequisite for atmospheric warming. There are other mechanisms out there, mechanisms that by natural necessity are put to work as soon as the atmosphere is in place on top of the sun-heated surface. The GHGs are a prerequisite for (efficient) atmospheric cooling to space, though. Also, the atmosphere with GHGs warms a lot more from the incoming solar radiation that it does from the outgoing terrestrial radiation. But mostly by convective processes (latent and sensible heat transfer from the surface).

      The GHG IR absorption, however, is NOT the ‘effect’ we’re discussing. And you know that. If anything, it is simply a physical property making up the first part of a hypothesized mechanism for warming.

      The proclaimed GH effect is supposed to be the ‘extra’ warming of the surface below the atmosphere. And there is absolutely no empirical evidence anywhere to suggest and hence no justification whatsoever to assert that such an effect does in fact exist. I’ve already explained to you what the atmospheric GHGs primarily do. First, they block 45% of the potential incoming heat from the Sun from ever reaching the surface. Then they help stimulate the convective heat loss going back out from that same surface, by letting the atmosphere cool radiatively from aloft to space.

      The atmospheric warming effect on the global surface has got nothing to do with its content of GHGs. Quite the contrary. They would on average clearly help cool the surface.

      The atmospheric warming effect on the global surface is rather the result of the weight of the atmosphere on the surface putting a constant restriction on the free convective/evaporative heat loss from the surface.

      And thus it could quite aptly be named ‘THE GREENHOUSE
      EFFECT’.

      You are the ones insisting on not listening to the right alternative, what actually produces the effect, because of your absolute obsession with a wrong/irrelevant alternative.

  88. Nullius in Verba says:

    The build-up of heat is the result of the blocking of *all* the heat transport mechanisms, acting in combination. Any one of them can nullify the effect.

    The Wikipedia page says what it does because of the specific comparison being made – the pool with a density gradient versus the one without. The reason for the difference *between those cases* is the suppression of convection. But if you compared a pool with water versus one without, the difference is due in large part to an IR-opaque material blocking radiation.

    The radiative greenhouse effect (the subject of Roy’s post) is a simple matter of radiative physics – opaque bodies emit and absorb radiation, the amount emitted increases with temperature. Those are effects that have been verified thousands of times. There are simple everyday observations that confirm them. The problem is that if someone is determined not to believe it, they can construct ever more bizarre alternative physics to explain the conflicts away, and you wind up having to re-demonstrate the whole of physics from the ground up. It’s entertaining and potentially useful as an exercise for a while, but there’s only so much time anyone is willing to expend on the game. It’s OK if someone comes up with something new, but I haven’t seen anything interestingly new out of the slayers for a couple of years now. I don’t mind them, but it’s no longer interesting to me.

    The convective greenhouse effect is a different mechanism. The adiabatic lapse rate (due, if you like, to the weight of the atmosphere) sets the *gradient* but it *doesn’t* set the intercept. Without GHGs, you would get the same gradient, but the surface would be at -20 C. The atmosphere above it would still cool with altitude but starting from a lower point would be even colder. If the surface was any warmer than that, it would emit more energy than the Earth was absorbing, and would cool down.

    Without GHGs, the surface radiating to space would cool the Earth very effectively.

  89. Kristian says:

    “Without GHGs, the surface radiating to space would cool the Earth very effectively.”

    See, this is what you proponents of the radiative GHE don’t get.

    Without an atmosphere altogether the surface would cool the Earth to space by radiation very effectively. And the surface would on average be much colder than with an atmosphere.

    But introducing an atmosphere into the mix changes this simple situation quite dramatically. Like I said, you can’t have an atmosphere, a medium of any kind, around a sun-heated planet and not have convection. It will quite naturally carry a major portion of the energy absorbed by the surface from the sun away from the surface and it starts doing this as soon as the surface begins to warm from the incident solar radiation.

    But the weight of the atmosphere on top of the surface resists the upward flow of surface-heated air, so the energy will pile up at and below the surface (which of course has a thermal mass in which to store the absorbed energy it cannot release fast enough) until the kinetic level (the temperature) is sufficiently high to create enough buoyancy to balance the incoming with the outgoing.

    Energy will of course always be able to escape the surface without restriction through radiation (EM waves). It might be slowed somewhat on its hurried journey to space. But it will never pile up at or below the surface, because it will still always be able to leave the surface whenever needed without hindrance. Not so with conduction/convection/evaporation through molecular heat transfer.

    Yes, in the end radiation is the meachanism by which all the energy received by the planet will have to be ejected back into space. But that has no bearing on the surface temperature of the planet. The surface temperature of the planet is what it needs to be to adequately drive the global convective circulation considering mean solar input and atmospheric weight.

    Real Earth equivalent: The global surface has a mean radiative HEAT loss of 52-53 W/m^2 (according to Stephens et al. 2012) and at the same time a convective (sensible+latent) heat loss of 112 W/m^2. In total 164-165 W/m^2. Why? Because that is the mean heat input it absorbs from the Sun. So that is also all it needs to rid itself of. Nothing more, nothing less. But its physical average temperature is ~288K.

    So this surface of 288K does not need to shed more heat than what it receives – 165 W/m^2 – to maintain balance. And so you see, there is no S-B connection between the physical temperature of that surface and its heat loss, either radiative, convective or both. And why is this? Because the surface of the Earth is not an ideal emitter radiating into a vacuum at 0 K, which is the situation the Stefan-Boltzmann equation you use for your ‘effective radiating level’ was meant for: P/A = ε*σ*T^4. The surface of the Earth is rather immersed in a massive ocean of air in a gravitational field. Gas laws apply.

    • Nullius in Verba says:

      “It will quite naturally carry a major portion of the energy absorbed by the surface from the sun away from the surface and it starts doing this as soon as the surface begins to warm from the incident solar radiation.”

      It will carry it away from the hotter parts of the surface and towards the colder portions. The net transfer to/from the surface as a whole will be zero.

      “But the weight of the atmosphere on top of the surface resists the upward flow of surface-heated air,”

      I think this may be at the root of the disagreement.

      Convection is cyclic. What goes up in one place must come down somewhere else, and the air moving down provides the force that lifts the air elsewhere up. It’s like a wheel – one side of the wheel goes up and the other side goes down, and the wheel rolls without effort. If we didn’t have any viscosity or frictional forces, it would take an arbitrarily small force to start the atmosphere rolling in the same sort of way.

      There is no significant resistance to upward-rising air from the weight of the atmosphere – it’s driven by the air pressure from the surrounding areas where the air is falling. The reason for the limitation on convection is that air is compressible, and gases get hotter when compressed and colder when allowed to expand. This temperature change means that convection stops not when it becomes isothermal, but when the cooling of rising air due to decompression matches the vertical temperature gradient. Convection stops cooling the surface when it reaches this gradient, not when all the excess heat is gone.

      It’s caused by pressure differences, and hence in a sense by “the weight of the atmosphere”, but it’s not the weight “holding it down”. The air moves freely, but cools or warms as it does so.

      I don’t know if that helps. But I can at least see why you might have thought it worked that way.

      • Kristian says:

        “It will carry it away from the hotter parts of the surface and towards the colder portions. The net transfer to/from the surface as a whole will be zero.”

        I assume we are now talking about the planet with an atmosphere without GHGs in it. The one that is warmed by convective heat transfer during the day (ideally), and which thus has a temperature way higher than space, but which still does not radiate according to this temperature, which in fact hardly radiates anything at all. Isn’t that right, Nullius? Or does it radiate according to its temperature as per the S-B law?

        Well, if the first is it, the atmosphere will have to somehow convect the transferred energy back down to the surface during the night (ideally). Otherwise, energy would start piling up in the atmosphere.

        So in such a scenario, yes, the net energy transfer between surface and atmosphere by convection would be zero. (Not at all so for Earth, though, but I guess you know that – it’s after all thanks to the GHG’s, isn’t it?)

        But this is not to say the convective engine doesn’t need surface heating and fairly high surface temperatures in order to run properly, much higher than an input of 165 or 239 W/m^2 would give.

        A heavier atmosphere requires more lifting power to move the same amount of energy away from the surface per unit of time, hence more kinetic energy and higher temperatures.

        A planetary surface which receives and absorbs the same amount of J/s/m^2 from the sun as Earth’s surface does, but with a lighter atmosphere on top, would not need to maintain a mean temperature as high to rid itself of energy as fast as Earth. It would be cooler. A planetary surface with the same solar input but with a heavier atmosphere would need to keep a higher mean surface temperature in order for energy to be carried away as fast as on Earth. It would be warmer.

        You say:

        “There is no significant resistance to upward-rising air from the weight of the atmosphere (…)

        It’s caused by pressure differences, and hence in a sense by “the weight of the atmosphere”, but it’s not the weight “holding it down”. The air moves freely, but cools or warms as it does so.”

        We’re already at balance, Nullius. The surface is at an elevated temperature. Because energy wouldn’t manage to leave the surface by convection/evaporation as fast as it came in at freezing temperatures, so energy would pile up and raise the temperatures until we had our balance. Upward acceleration of surface-heated air is adequate because the surface temperature is high enough. It wouldn’t have been had the mean surface temperature been lower or the weight of the atmosphere been any different. That’s the effect. The more surface heating (or the greater the gradient in temp/pressure, but these are normally related), the stronger the convection/evaporation. And the more energy is carried away from the surface.

    • Tim Folkerts says:

      Kristian applies some faulty logic:
      “Without an atmosphere altogether the surface would cool the Earth to space by radiation very effectively. …

      But introducing an atmosphere into the mix changes this simple situation quite dramatically.”

      ONLY IF THAT ATMOSPHERE HAS AN IMPACT ON IR RADIATION!

      If the atmosphere were transparent to IR, then the cooling by IR would remain the same as before — there would be no “quite dramatic” change. The IR photons would leave the surface and head out to space as if the atmosphere were not there at all.

      If the atmosphere can absorb and radiate IR photons, then the atmosphere would indeed have a quite dramatic effect. Convection is indeed a big part of that effect. The concentration of GHGs –along with the overall ‘weight’ — matter. But of course, we are now back to requiring GHGs, so we have
      * convection without GHGs = little or no effect
      * convection with GHGs = dramatic effect.

      [The one significant effect that the atmosphere would have would be to “even out” the temperature swings, cooling the day side and warming the night side. This would have a slight warming effect that would bring the average temperature up toward ~255 K, but no higher. Conservation of energy ensures that the average would nto get any higher than ~ 255 K.]

  90. Ball4 says:

    “But the weight of the atmosphere on top of the surface resists the upward flow of surface-heated air, so the energy will pile up at and below the surface (which of course has a thermal mass in which to store the absorbed energy it cannot release fast enough) until the kinetic level (the temperature) is sufficiently high to create enough buoyancy to balance the incoming with the outgoing.”

    Kristian – Can you write out or do you have a ref. for an equation(s) showing physics of this description, those words sound like they can be put in eqn. form. I haven’t run across this in basic atm. thermo and radiation text books that I have read thru.

    “..this is what you proponents of the radiative GHE don’t get.”

    For me, because I cannot find the equations supporting what you write in words. Nullius 4:49pm wrote out the text book equations I can find easily but in words so I do get that version.

    • Kristian says:

      So, Ball4, that means you’ve never heard of perfectly natural and basic phenomena like ’atmospheric pressure’, ’buoyancy’, ’natural convection’, ’vapour pressure’, ’dew point’ and ’the Clausius-Clapeyron relation’ (pressure/temp relationship) …?

      The basics of these things you can find described and explained in most elementary physics or chemistry textbooks. What you can not find anywhere except within climate science is the model presented by Nullius in Verba here. It is nothing but a conjured-up theoretical mechanism without observational backing passing itself off as an explanation for an effect that is really caused by something else entirely, namely the basic gas phenomena mentioned above.

      This should all be fairly common knowledge. But I could give you a run-through …

      Like I said, the atmospheric warming effect on Earth’s global surface is obviously caused simply by the atmosphere weighing down on it.

      About atmospheric pressure:

      Atmospheric pressure is the force per unit area exerted on a surface by the weight of air above that surface in the atmosphere of Earth (or that of another planet). (…) as elevation increases, there is less overlying atmospheric mass, so that atmospheric pressure decreases with increasing elevation.”

      The weight of the atmosphere on the surface is given by its mass (its total content of matter) (m) times Earth’s specific gravitational acceleration (g). This is equal to the classical Newtonian F = ma. In other words, the atmosphere’s weight is a force constantly pressing down on the surface from above, expressed by its pressure. The heavier the atmospheric weight, the greater the constant downward force and the greater the atmospheric pressure. The density of the air is given by its mass per volume: ρ = m/V.

      About buoyancy:

      In science, buoyancy is an upward force exerted by a fluid, that opposes the weight of an immersed object.

      http://upload.wikimedia.org/wikipedia/commons/7/74/Buoyancy.svg

      (…) This can occur only in a reference frame which either has a gravitational field or is accelerating due to a force other than gravity defining a “downward” direction (that is, a non-inertial reference frame).”

      “Objects in water are buoyed up because the pressure acting up against the bottom of the object exceeds the pressure acting down against the top. Likewise, air pressure acting up against an object in air is greater than the pressure above pushing down. The buoyancy, in both cases, is equal to the weight of fluid displaced – Archimedes’ principle holds for air just as it does for water.”

      ”Any object that has a mass that is less than the mass of an equal volume of air will rise in air – in other words, any object less dense than air will rise.

      “Unlike water, the atmosphere has no discernible surface (there is no “top”). Furthermore, unlike water, the atmosphere becomes less dense with altitude. Whereas a cork will float to the surface of water, a released helium-filled balloon does not rise to any atmosphere surface. With regards to how high a balloon will rise, a balloon will rise only so long as it displaces a weight of air greater than its own weight. Air becomes less dense with altitude, so, when the weight of displaced air equals the total weight of the balloon, upward acceleration ends.

      A given volume of air at the surface of the Earth will rise as soon as its upward force (its buoyancy) exceeds the downward force exerted by the atmospheric weight on top of it. This occurs when our volume becomes less dense than the air above it. How is this accomplished? You heat it. When it warms, the gas molecules within the volume start moving faster (more kinetic energy), which means they spread out. The air expands -> its density and hence its pressure falls.

      Well, here comes the crucial point: How fast will this volume of air be able to lift away from the surface?

      This depends primarily on two things: 1) the atmospheric weight, and 2) the temperature (kinetic level) of the volume of air. To maintain a constant rate of upward acceleration from the surface, these two factors need to follow each other: If one changes, the other one needs to change too, in the same direction. This relates to the ideal gas law: T = PV/nR. If you increase the temperature of a gas, keeping the net substance flow constant, you will be able to maintain its volume (which is ‘under pressure’) by increasing its outward pressure.

      In other words, for a given volume of air to be able to rise as fast from the surface with a heavy atmosphere on top as a similar volume of air with a lighter atmosphere on top, the former needs to be … warmer than the latter. It needs to contain a higher level of kinetic energy.

      As you can understand, under equal pressure conditions, a volume of air heated more will rise faster (its upward acceleration will be greater) and will thus be able to reach higher before it finally comes to a halt to eventually start its descent. This is the main reason why the tropopause (~ convection top) is so high in the tropics and so much lower toward the poles -> surface heating.

    • Kristian says:

      About natural convection:

      “Natural convection is a mechanism, or type of heat transport, in which the fluid motion is not generated by any external source (like a pump, fan, suction device, etc.) but only by density differences in the fluid occurring due to temperature gradients. In natural convection, fluid surrounding a heat source receives heat, becomes less dense and rises. The surrounding, cooler fluid then moves [in] to replace it. This cooler fluid is then heated and the process continues, forming a convection current; this process transfers heat energy from the bottom of the convection cell to [the] top. The driving force for natural convection is buoyancy, a result of differences in fluid density. Because of this, the presence of a proper acceleration such as arises from resistance to gravity, or an equivalent force (arising from acceleration, centrifugal force or Coriolis effect), is essential for natural convection.”

      === === ===

      About vapour pressure:

      ”Vapor pressure or equilibrium vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The equilibrium vapor pressure is an indication of a liquid’s evaporation rate. It relates to the tendency of particles to escape from the liquid (or a solid).”

      The vapour exerts its (partial) pressure to all sides, upwards too, against the atmospheric pressure. In this sense, it works much like regular buoyancy. The warmer the water, the higher its vapour pressure (kinetic level) and the more water molecules (and hence energy/heat) will manage to escape the surface.

      About evaporation:

      ”The ability for a molecule of a liquid to evaporate is based largely on the amount of kinetic energy an individual particle may possess. Even at lower temperatures, individual molecules of a liquid can evaporate if they have more than the minimum amount of kinetic energy required for vaporization.”

    • Kristian says:

      About the boiling point of water and atmospheric pressure:

      “Clean fresh water boils at about 100°C (212°F) at standard atmospheric pressure. The boiling point is the temperature at which the vapor pressure is equal to the atmospheric pressure around the water. Because of this, the boiling point of water is lower at lower pressure and higher at higher pressure. This is why cooking at elevations more than 1,100 m (3,600 ft) above sea level requires adjustments to recipes.”

      So, if the atmospheric pressure (weight) is lower, a lower temperature is needed for water to boil. Conversely, if the atmospheric pressure is higher, the water needs to get hotter before it can boil. This is because the higher the pressure from above, the higher the vapour pressure pushing against it from below needs to be in order to overcome it. So the temperature (the kinetic level) needs to rise. This applies also to evaporation rates at lower temperatures. The higher the atmospheric pressure, the warmer the water needs to be to maintain the rate of evaporation. This relates to the saturation point of water vapour (dew point).

    • Kristian says:

      About water vapour pressure (pressure/temp relationship), saturation, evaporation and condensation:

      The vapour pressure of water is the pressure at which water vapour is saturated. At higher pressures water would condense. The water vapour pressure is the partial pressure of water vapour in any gas mixture saturated with water. As for other substances, water vapour pressure is a function of temperature and can be determined with Clausius–Clapeyron relation.”

      We have two water surfaces at equal temperatures and under equal atmospheric pressure, evaporating into the air above, meaning they’re both below the saturation pressure point and above the dew point temperature, where evaporation would stop. Then we increase the atmospheric pressure above one of the evaporating surfaces but maintain its surface temperature. What happens? We lower the saturation point and raise the dew point. The two get closer to each other. So the air above the high pressure surface now has a higher level of saturation than the low pressure surface, and evaporation rates slow down accordingly (if the pressure increase is of such magnitude that the saturation point in fact meets the dew point, it stops altogether).

      Given equal temperatures, the air with the highest pressure will saturate first. So the surface feeding it with water vapour will have to warm to keep up its evaporation rate.

    • Kristian says:

      You can also state it more plainly:

      ”Factors influencing the rate of evaporation (…)

      # Concentration of the substance evaporating in the air: If the air already has a high concentration of the substance evaporating, then the given substance will evaporate more slowly. (…)

      # Pressure: Evaporation happens faster if there is less exertion on the surface keeping the molecules from launching themselves. (…)

      # Temperature of the substance: If the substance is hotter, then its molecules have a higher average kinetic energy, and evaporation will be faster. (…)”

      With a higher atmospheric pressure on the surface, less heat would escape the surface through evaporation because the saturation point would be lower. This means that with a heavy atmosphere on top, the water surface (like the ocean) would have to be warmer to maintain a rate of heat loss by evaporation equal to that of an ocean with a lighter atmosphere on top. It needs to produce a higher vapour pressure. At equal atmospheric pressures (and equal wind stress, which after all is also an important factor), the warmer ocean surface would rid itself of more energy/heat per unit of time through evaporation than the cooler one.

      There is nothing mysterious or self-invented about any of this … These are standard priciples.

      So you see, this is how the atmosphere – by its sheer weight on the surface – forces the mean surface temperature to be higher than what our specific instantaneous heat input flux from the sun could ever manage to maintain by itself. Of course, it helps having an Earth system with a massive heat capacity (mostly the oceans) to store and build the absorbed energy to obtain the required temperature to stay dynamically balanced.

    • Ball4 says:

      Kristian 5:05am: You are on a roll. Couple points for more study where Kristian asserts:

      “1) the atmospheric weight, and 2) the temperature (kinetic level) of the volume of air. To maintain a constant rate of upward acceleration from the surface, these two factors need to follow each other: If one changes, the other one needs to change too, in the same direction. This relates to the ideal gas law: T = PV/nR”

      So…. “two factors..in the same direction…This relates to the ideal gas law: T = PV/nR.”

      It may come as a complete surprise to Kristian but this is not the case we observe in surface weather station data, the sum of which is climate. I have been told many times that cold air is denser than warm air. This statement is only true if accompanied by the assertion that pressure is constant.

      Kristian – Go to weatherspark dot com. Click on a city you like, I chose Madison, Wisc., USA. Click on temperature and pressure. Daily or quarterly.

      According to Kristian’s formula clipped above, “two factors need to follow each other”. Notice very plainly there is no such obvious IGL relation between pressure and temperature. Sometimes temperature goes up and pressure goes DOWN. Sometimes temperature goes up and pressure goes UP. This is real data Kristian.

      Moral: For atm. thermo study, IGL is not a good predictor of anything unless it can be supplemented by a constraint: air density goes up as temperature goes down provided that pressure is constant.

      These density & IGL are the only equations Kristian provides and can be so easily used to show his words are imprecise to observed data.

      For me to “get” Kristian’s version of the “radiative GHE”, Kristian needs to follow up with more eqn.s to support Kristian’s original imprecise assertion 5:47am, 3:22pm that:

      “..the upward flow of surface-heated air, so the energy will pile up at and below the surface..”

      Until then, the text book version “radiative GHE” of Dr. Spencer, Nullius 4:49pm, 9:19am, Tim Folkerts and Ball4 and various diverse atm. thermo text authors will stand.

      • Kristian says:

        Riiight …

        Well, have a good life, Ball4.

        • Will Pratt says:

          The “radiative GHE” hypothesis requires that the atmosphere is heated by the ground directly below it.

          The radiosonde data provides clear empirical evidence that over most of the Earths solid surface for most of the year, the ground is actually cooler than the air above it. This empirical evidence proves that the ground cannot be responsible for the temperature of the air directly above and therefore proves the so called “GHE” hypothesis is false.

          • Tim Folkerts says:

            Will Pratt says: “The radiosonde data provides clear empirical evidence that over most of the Earths solid surface for most of the year, the ground is actually cooler than the air above it.”

            Could you provide some support or clarification for this claim? I am quite sure that there is a lapse rate of ~ 6 K/km over most of the earth, with the top of the troposphere being much cooler than the bottom of the troposphere.

            Are you talking about the temperature 2 m above the ground? or 2 km? or 200 km?

          • Will Pratt says:

            I think you will find that the normal lapse rate is 5.5 K/km.

            I am referring to the inversion in the first 4′. You know, the reason why Stevenson Screens are always a minimum 4′ off the ground.

            It is possible from the radiosonde data to determine exactly where the ground is warmer and therefore heating the air above and also where the ground is cooler and therefore not heating the air above. From this data it is clear that not very much of the Earths solid surface ever gets warm enough to heat the air directly above it.

            Very unfortunate for the “GHE” hypothesis which is entirely based on the premise that ground heats the air directly above.

            There are of course many other empirical observations which show this premise to be ridiculous.

          • Tim Folkerts says:

            “em>”I think you will find that the normal lapse rate is 5.5 K/km.”
            No. The “standard” lapse rate is 6.49 K/km, which was chosen because it is close to the observed average lapse rate.

            “I am referring to the inversion in the first 4′.”
            I’m not an expert here, but my understanding is that such inversions often occur at night, but that during the day the surface is usually warmer than the air higher up.

            ” This empirical evidence proves that the ground cannot be responsible for the temperature of the air directly above … “
            Well, of course the ground is just ONE of MANY factors that affect the air temperature. No one factor could be considered solely responsible for air temperature.

            ” … and therefore proves the so called “GHE” hypothesis is false.”
            This seems like a huge non-sequitur.

            “The “radiative GHE” hypothesis requires that the atmosphere is heated by the ground directly below it. “
            The GHE hypothesis is typically about the effect of the atmosphere on the ground, not vice versa (although the two are clearly related). The “radiative GHE” requires that the GROUND absorbs energy from the ATMOSPHERE. This energy (added to the energy directly from the sun), raise the temperature of he surface higher than it would be with only the energy input from the sun.

            In any case, standard estimates (eg Trenberth * Kiehl) conclude there is a net energy transfer (ie “heat”) of 17 W/m^2 to the atmosphere from “thermals” (heat conducted from the surface to the atmosphere and then convected upwards) and a net transfer of 63 W/m^2 from thermal IR. So the surface (as a whole) does heat the atmosphere, exactly as you say must happen. šŸ™‚

          • Will Pratt says:

            Correct, 6.5 K/km is what I thought I’d written, sorry.

            “I’m not an expert here, but my understanding is that such inversions often occur at night, but that during the day the surface is usually warmer than the air higher up.”

            No, wrong. You are very confused.

            I am talking about the influence of the earths solid surface on the atmosphere directly above it. If you remove the only source of heat, ie the sun, how could the surface produce a temperature inversion?

            What happens is when the sun goes down, the atmosphere and the air above move towards equilibrium.

            The temperature inversion in the first 4′ of air above the surface is at its maximum at the diurnal peak temperature. The hottest part of the day. It is caused by the sun and is the result of the difference in mass between the solid surface and the air directly above it.

            This is empirical evidence of two things. a) That the ground does not heat the air above it and b) that the atmosphere is heated directly by incoming EMR.

            “The GHE hypothesis is typically about the effect of the atmosphere on the ground, not vice versa (although the two are clearly related). The “radiative GHE” requires that the GROUND absorbs energy from the ATMOSPHERE. This energy (added to the energy directly from the sun), raise the temperature of he surface higher than it would be with only the energy input from the sun.”

            No it isn’t. You have arbitrarily selected a small section of the “GHE” hypothesis out of context and mixed that in with the fact that the atmosphere is in-fact heated directly by incoming EMR to make your point. Yet you then quote from the discredited “Trenberth * Kiehl” energy budget diagram to claim there is net heating from the surface to the atmosphere. Yet in this same diagram there is NO direct heating of the atmosphere by incoming EMR.

            You are a very confused individual.

            If you knew “GHE” hypothesis you would know that according to that hypothesis the atmosphere is not heated directly by the incoming EMR at all. As I have pointed out, there is no direct heating from the incoming EMR in the “Trenberth * Kiehl” energy budget diagram you have referenced.

            Yet the fact that there is a temperature inversion in the first 4′ above the ground which increases towards the diurnal peak, the hottest part of the day, is empirical evidence that the atmosphere is indeed heated directly by EMR.

            When you combine this empirical evidence with the empirical evidence from the radiosonde data that shows most of the Earths solid surface is generally cooler than the air directly above it, you have clear indisputable proof that the so called “Greenhouse Effect” hypothesis is ridiculous, dangerous pseudoscientific nonsense.

          • Will Pratt says:

            Typo:

            What happens is when the sun goes down, the atmosphere and the air above move towards equilibrium.

            Should read:

            What happens is when the sun goes down, the surface and the air above move towards equilibrium.

          • Tim Folkerts says:

            The more I read of your latest reply, the more it confused me!

            “If you knew “GHE” hypothesis you would know that according to that hypothesis the atmosphere is not heated directly by the incoming EMR at all.”
            That is not a part of the GHE hypothesis. Many simple models assume that atmosphere is perfectly transparent to make the math more straightforward, but this is not essential. These are the “high school” models to introduce the concepts, but other models can and do include absorption by the atmosphere. The only thing that is needed is that sunlight gets in more easliy than thermal IR gets out.

            “Yet in this same diagram [Trenberth & Kiehl]there is NO direct heating of the atmosphere by incoming EMR.”
            Yes, there is! There is “78 W/m^2 of incoming solar radiation “Absorbed by atmosphere. So *part* of the heating of the atmosphere is directly by solar radiation, and the rest is from the warm surface of the earth.

            “If you remove the only source of heat, ie the sun, how could the surface produce a temperature inversion?”
            The surface is an effective radiator of thermal IR; the atmosphere is less effective. When the sun goes down, the ground cools more effectively than the atmosphere. This can easily leave the ground cooler than the air above it.

            “I am talking about the influence of the earths solid surface on the atmosphere directly above it.”
            Actually, you are specifically DENYING the influence of the earth’s solid surface. I am saying there IS an influence of the surface — that the surface is warmed effectively by sunlight, and that the warm surface heats the cooler air above. You are saying the atmosphere is heated directly by sunlight, AND that this heating is so effective that the atmosphere transfers net energy to the surface.

            ************************************

            The ultimate issue here is the temperature inversion you claim in the first 4′. Can you provide evidence or references to support your claim?

            Sources I found (like facstaff.unca.edu/dmiller/Chapter%2003-Air%20Temperature.ppt) suggest the air is typically hottest NEAREST the surface and cools in the first meter.

          • Will Pratt says:

            [email protected] 3:38 pm

            To your responses

            1. Incorrect. The “GHE” hypothesis absolutely does require an atmosphere which is entirely transparent to incoming EMR and that is exactly how it has been presented to the public. Moving goalposts at some convenient point after the fact and then returning to the transparent claims when ever you choose to is a reflection on your state of mind not on the reality of atmospheric physics.

            2. 78 W/m2 has a corresponding maximum temperature of -80.56ŗ C. If the atmosphere already has an average temperature of 15ŗ C then how much extra heating can be provided by -80.59ŗ C ???

            As I said, the diagram you insist on referencing shows NO direct “heating” of the atmosphere from incoming EMR.

            3. Quote: “The surface is an effective radiator of thermal IR; the atmosphere is less effective. When the sun goes down, the ground cools more effectively than the atmosphere. This can easily leave the ground cooler than the air above it.”

            Again with the selective logic to fit your own fictional narrative Tim.

            Or perhaps you forget that the advantage that the atmosphere has over the solid surface is convection.

            Or perhaps you forgot that non-convecting air is an insulator.

            No the ground does not cool more effectively than the atmosphere above it because it has an atmosphere above to insulate it and because of convection. This is why frosts tends to form mostly at dawn when the atmosphere begins to warm above the surface temps from the early morning incoming EMR.

            4th and final point.

            Quote: “Actually, you are specifically DENYING the influence of the earth’s solid surface. I am saying there IS an influence of the surface — that the surface is warmed effectively by sunlight, and that the warm surface heats the cooler air above. You are saying the atmosphere is heated directly by sunlight, AND that this heating is so effective that the atmosphere transfers net energy to the surface.

            Now you are putting words in my mouth Tim.

            I am saying that the sun heats both the surface and the atmosphere but that because the atmosphere has much less mass than the surface, a detectable temperature inversion occurs close to the ground, in the first 4′ and increases towards the diurnal peak, the hottest part of the day.

            I discovered this effect based on a prediction I made using such logic. I have since confirmed this prediction with empirical observation. This is how science is done Tim. Not through logical fallacies arbitrarily selected to suit your own fictional narrative. And when you fear you are no longer sounding credible you resort to putting words in peoples mouths, the last resort of a dishonest man.

            I can see that you are a very busy bee here on Roy’s blog Tim. Yet you display a complete lack of honesty and integrity.

            I therefore have no desire to respond any further to your comments, I suggest to try your best to refrain from responding to mine.

            Thanks.

            W.

          • Tim Folkerts says:

            ”1. Incorrect. The “GHE” hypothesis absolutely does require an atmosphere which is entirely transparent to incoming EMR and that is exactly how it has been presented to the public.”
            You are describing the introductory explanation – the “high school” version. An entirely transparent is no more required than entirely frictionless surfaces are required in a first year physics course. Both are just idealizations that make things easier for beginners.

            ”2. 78 W/m2 has a corresponding maximum temperature of -80.56ŗ C. If the atmosphere already has an average temperature of 15ŗ C then how much extra heating can be provided by -80.59ŗ C ???
            This is a persistent misunderstanding. Sunlight comes from an object @ 5780 K; the “maximum temperature” of sunlight is 5780 K. The “-81C” temperature is the temperature of a blackbody 1 AU from the sun which is heated only by the sun. The atmosphere is 1) not a blackbody and 2) is also heated by thermal IR from the ground. The “-81 C” temperature is pretty meaningless in this context.

            ”As I said, the diagram you insist on referencing shows NO direct “heating” of the atmosphere from incoming EMR.
            Since heating is defined in thermodynamics as an energy flow from a hotter object (the sun) to a cooler object (the atmosphere) due to that temperature difference, then the diagram shows EXACTLY 78 W/m^2 of “direct heating of the atmosphere”.

            3. … No the ground does not cool more effectively than the atmosphere above it because it has an atmosphere above to insulate it and because of convection. This is why frosts tends to form …”
            The atmosphere also has outer space above it, and the surface can radiate some IR directly though the atmosphere to space. This allows a cooling mechanism not available to air near the surface. Frost can form on the ground or on cars even when air temperatures remain above freezing because the surfaces cool below the temperature of the air.

            Now you are putting words in my mouth Tim.
            Let me clarify. You claim that warmer air exists in contact with cooler ground during the day (“a detectable temperature inversion occurs close to the ground, in the first 4′ ”). The logical extension is that there must be heat flowing from the hotter air to the cooler ground. If “the atmosphere transfers net energy to the surface “ are NOT your words, they should be!

            I discovered this effect based on a prediction I made using such logic. I have since confirmed this prediction with empirical observation. This is how science is done Tim.
            Yes, that is indeed how science is done. Then the results are compared to other results (and to theory). If there is disagreement, then the differences must be worked out. In this case, I find lots of references that say the ground tends to be hotter than the air just above the ground during the day.
            http://scienceofdoom.com/2010/04/09/sensible-heat-latent-heat-and-radiation/
            http://www.bom.gov.au/climate/how/faq-data.shtml
            http://www.erh.noaa.gov/rah/education/eduit.html

            Using completely different approach, the “shimmering water” seen over roads on hot days is caused by the air directly above the pavement being hotter (and hence having a higher speed of light) than the air slightly higher up. Such mirages would not exist if air got warmer as you got higher above the surface of the road.

            I don’t doubt that temperature inversions do occasionally occur during the day, but evidence suggests they are unusual, and not due the reasons you describe.

            I can see that you are a very busy bee here on Roy’s blog Tim.
            Yeah, I had a little extra free time lately, and I enjoy adding my scientific knowledge from time to time. I guess we will have to leave it to others to decide the relative integrity and/or scientific correctness of our various contributions.

          • Will Pratt says:

            [email protected] 10:49 AM

            I point out that 78 W/m2 has a corresponding temperature of -80.56ŗ C and could not produce any heating in our atmosphere.

            you counter with:

            Quote:”The “-81 C” temperature is pretty meaningless in this context.”

            Followed by

            Quote: “then the diagram shows EXACTLY 78 W/m^2 of “direct heating of the atmosphere”.

            You are not just dishonest Tim, you are a calculated sophist tying yourself up in knots. That is clear for all to see.

            I’ve asked you once and I will ask you again, please stay off my posts. I do not wish to converse about science with a sophist. What is the point in that?

            Thank you.

            W.

          • Tim Folkerts says:

            Let me repeat my original question:

            “Will Pratt says: “The radiosonde data provides clear empirical evidence that over most of the Earths solid surface for most of the year, the ground is actually cooler than the air above it.”

            Could you provide some support or clarification for this claim? “

            You state this as if it is common knowledge. This is the starting point for everything else you claim.

            Can you provide a link or reference that presents ‘clear empirical evidence’ of your claim?

  91. Dr A Burns says:

    Roy,

    Do you have any comment on the calculation of 62 W/m2 radiation loss for a 20 deg C temperature difference between the Earth’s surface and the bottom of a cloud ?

    Clouds cover 70% of the Earth’s surface. At night the radiation loss would be even less. This is far less than your stated figure of 442 W/m2.

    • Ball4 says:

      Dr A Burns 12:05am: Comment on the 442 W/m^2 downwelling.

      Please go to:

      http://www.esrl.noaa.gov/gmd/grad/surfrad/dataplot.html

      Click on Goodwin Creek, MS data ~closest site to Dr. Spencer “box” tests, click on downwelling infrared.

      Then “plot data”. For 8/25, 2013 I observe a Noon (LST) oscillation between ~440-450 W/m^2.

    • Tim Folkerts says:

      442 W/m^2 is the ‘gross’ IR from the surface upward in the diagram. 350 W/m^2 is the ‘gross’ from the sky (clouds & GHGs). The ‘net’ IR would be 442-350 = 92 W/m^2. *This* is the number that would correspond to your 62 W/m^2. So the two numbers are in rough agreement.

      As to your specific 62 W/m2 — it seems a little low. This would work for a ground temperature of 249 K (assuming the ground and the clouds are blackbodies and that the atmosphere in between is not a significant emitter (not an especially good assumption)).
      T _______ Power
      249 _____ 218
      229 _____ 156

      Difference = 62 W/m^2

      For warmer ground the difference is higher. For example
      T _______ Power
      290 _____ 401
      270 _____ 301

      Difference = 100 W/m^2

      How did you get 62 W/m^2?

      ***********************************
      Since all of the number in his example are ‘typical’ values, then they seem to all be in the right ballpark.

  92. Kelvin Vaughan says:

    How does a lower frequency electro magnetic wave from a cool object increase the motion of a molecule of a hotter object which is vibrating at a greater frequency? Logic suggests to me that it would actually slow the vibration down. The cooler object will cool the hotter object.

  93. torontoann says:

    “The frequencies” refer to completely different phenomena.
    The e-m frequency refers to the rate at which
    a field intensity is changing. The frequency
    of the molecule refers to the rate at which
    the molecule is going through a cycle of
    movements caused by a local field.

    The cooler object cools ITSELF by sending away some energy.
    There are three places energy can be:

    in the cooler object,
    in the hotter object,
    in the e-m waves.

    The cooler object does not “know anything” about the energy it has lost.

    The recipient of that energy is not cooled. That energy will heat ANYTHING which absorbs it.