This post “follows” on from Heat Transfer Basics and Non-Radiative Atmospheres and Do Trenberth and Kiehl understand the First Law of Thermodynamics? and many other posts that cover some basics.
It’s clear from comments on this blog and many other blogs that a lot of people have difficulty understanding simple scenarios because of a lack of understanding of the basics. Many confident (but erroneous) comments state that particular scenarios can never occur because they violate this or that law..
I know from my own experience that until a concept is conceptually grasped, a mathematical treatment is often not really helpful. It might be right, but it doesn’t help..
So this post has a number of examples that paint a picture. It has some maths too.
Enough examples might help some readers unfamiliar with thermodynamic concepts grasp the essence of some heat transfer basics.
Ignore the details if you like and just check the results from each example. Some maths is included to make it possible to check the results and understand the subject a little better.
Conduction
We will use the example of a “planar wall”. What exactly is that?
It’s a wall that extends off to infinity in both directions.
For those thinking this is some kind of climate madness, it’s simply physics basics – draw up a problem with simple boundary conditions and find the answer. If we start with some massively complex problem that approximates the real world then unfortunately there will be no conceptual understanding. And this article is all about conceptual understanding. Start with simple problems and gradually extend to more complex problems.. (The wall can just be a long wall if that makes you happier).
Example One is a wall, made out of PVC, with both sides held at a constant temperature (probably by a fluid at a constant temperature pumped over each side).
Example One – constant temperature conduction
We want to calculate the heat flux (heat flow per unit area) travelling through this wall of PVC.
The basic equation of heat conduction is:
q = kA . ΔT/Δx (see note 1)
where ΔT is the temperature difference, Δx is the thickness of the wall, A is the area, k is the conductivity (the property of the material) and q is the heat flow.
To make things slightly easier we consider heat flux – heat flow per unit area, q”:
q” = k . ΔT/Δx
For PVC, k=0.19W/m.K
And for the case of this wall, T1 = 50°C, T2 = 10°C and therefore ΔT = 40°C
So,
q” = 0.19 x 40 / 2 = 3.8 W/m²
In this example, because the system is holding both surfaces at a constant temperature we have a constant (and continuous) flow of heat between surfaces.
You can see that not much heat is flowing because PVC is a very good insulator.
Example Two – constant temperature conduction, thinner wall
q” = 0.19 x 40 / 0.2 = 38 W/m²
So with the wall 10x thinner, the heat flux is 10 times greater. Hopefully, for most, this is intuitively obvious – put thinner insulation on a hot water pipe and it loses more heat; wear a thinner coat out in the cold and you get colder..
If we changed the PVC for metal then the heat flow would be very much higher, as metal conducts heat very effectively.
Now what’s supplying the heat? The liquid or gas being pumped over the higher temperature surface to keep it at that temperature.
Note that these surfaces will be radiating heat. However, this doesn’t affect the calculation of conducted heat between the two surfaces.
In simple terms, heat flow due to conduction depends on the temperature difference, the material and the dimensions of the body.
Example Three – no temperature differential
Now both sides of the wall are held at the same temperature,
q” = 0.19 x 0 / 2 = 0 W/m²
This is very simple, but obviously confuses some people, including some visitors to this blog. It is temperature difference that drives conduction of heat. If there is no temperature difference, there will be no conduction.
In these three examples we have constrained the temperature on each side to see what happens to heat flow. Now we will change these boundary conditions.
Conduction and Radiation
Example Four – constant heat supply one side, fixed temperature the other
This is example two but with a constant heat supply instead of a constant temperature on one side.
This example is now more complex. The right side of the wall is held at a constant temperature of 10°C, as with the first few examples, but the other surface of the wall now has a constant input of heat and we want to find out the temperature of that surface.
The heat source for the left side is incident radiation. We will assume that the proportion of radiation absorbed (“absorptivity”) is 80% or 0.8. And we will assume that the emissivity of the surface is also 0.8. See note 2.
How do we now calculate the surface temperature T1?
It’s quite simple in principle. We use the first law of thermodynamics – energy cannot be created or destroyed. And we will calculate the equilibrium condition – which is when steady-state is reached. This means no heat is being retained to increase the temperature.
So all we have to do is balance the heat flow terms at the surface (the left surface). Let’s take it step by step.
Energy absorbed from radiation:
Ein(absorbed) = Ein x 0.8
This is because 80% is absorbed and 20% is reflected, due to the material properties of PVC.
For energy balance, once the surface has reached a steady temperature:
Ein(absorbed) = q” + Eout (see the diagram)
q” is the heat flux through the wall, and Eout is the radiated energy. At this point we are assuming no convection (perhaps there is no atmosphere for example) to keep things simple.
Hopefully this is quite a simple concept – the heat absorbed from radiation is balanced by the heat radiated from the surface plus the heat conducted through the wall.
We can calculate the energy radiated using the well-known Stefan-Boltzmann equation,
Eout = εσT4
where ε = emissivity (0.8 in this example), σ is the Stefan-Boltzmann constant (5.67 x 10-8) and T is absolute temperature in K (add 273 to temperature in °C).
Except we don’t yet know the temperature.. Still, let’s put it all together and see what happens:
Ein(absorbed) = q” + Eout
Now put the numbers in that we know:
500 x 0.8 = 0.19 x ΔT / 0.2 + 0.8 x 5.67×10-8 x T14
Now ΔT is the temperature difference between T1 and T2. T2 is held constant at 10°C so ΔT=T1-10. However, the first term on the right is expressed in °C and the second term in absolute temperature (K). We will express both as absolute temperature, so ΔT = T1-283.
So now the equation is:
400 = 0.19 x (T1-283) / 0.2 + 0.8 x 5.67×10-8 x T14
The important point to note for those a little bewildered by all the numbers – we have used the first law of thermodynamics, the equation for emission of radiation and the equation for conducted heat and as a result we have an equation with only one unknown – the temperature.
This means we can find the value of T1 that satisfies this equation. (See note 3 for how it is found).
T1 = 302.80 K = 29.65°C
And using this value, conducted heat, q” = 18.7 W/m² and radiated heat, Eout = 381.3 W/m².
In this case, there is a lot more heat radiated compared with conducted.
Example Five – as example four with a thinner wall
With a much thinner wall or a much higher conductivity the balance changes.
Changing the thickness of the wall from 0.2m to 2mm, keeping everything else the same and so using exactly the same equations as above, we get:
T1 = 284.24 K = 11.09°C
Note that this means the temperature differential across the wall has reduced to only (just over) 1°C.
And using this value, conducted heat, q” = 103.6 W/m² and radiated heat, Eout = 297.4 W/m².
Example Six – as example four with increased “colder” temperature
Now with the 0.2m wall (example four) we increase the temperature of the colder side, from 10°C to 25°C.
Using the same maths we find that the temperature, T1, has increased:
T1 = 305.16K = 32.01°C
An increase of 2.36°C.
Most people are probably asking “why this example? it’s obvious that increasing the temperature of one side will lift the other..”
However, many people believe that a colder atmosphere cannot affect the temperature of a warmer surface. See, for example, The First Law of Thermodynamics Meets the Imaginary Second Law. This reasoning is due to a misunderstanding of the second law of thermodynamics.
However, as conduction is quite familiar and more intuitive I expect that this example will be more easily accepted. And perhaps this last example will help a few people to see that a colder body can affect a warmer body without violating any laws of thermodynamics.
Conclusion
In Do Trenberth and Kiehl understand the First Law of Thermodynamics? I presented a hollow sphere in space with a heat source at its center. Some people were (and still are) convinced that there is something wrong with the results from that example. One person (at least) is convinced that the inner surface must be at the same temperature as the outer surface.
The only correct approach to calculating heat transfer and temperatures is to apply the relevant equations of conduction, convection and radiation to the particular problem in question.
- Conduction of heat is proportional to the temperature difference across a material
- Radiation of heat is proportional to the 4th power of (absolute) temperature of a surface
- The first law of thermodynamics is used to solve these problems: energy in – energy out = energy retained, for any particular part of a system that you analyze
Many people rely on intuition for determining whether a solution is correct. However, intuition is not as reliable as applying the basic equations of heat transfer.
Note that the example in Do Trenberth and Kiehl understand the First Law of Thermodynamics? uses exactly the same equations and approach as the examples here. If these six examples are correct you will have trouble finding the flaw in the hollow sphere example.
Notes
Note 1 – Conventionally the equation of heat conducted has a minus sign because heat travels in the opposite direction to the temperature gradient. And for the purists, the more general equation of conduction is:
q” = -k∇T
where ∇T is the three dimensional version of the “change of T with respect to distance”
Note 2 – Emissivity = Absorptivity at a particular wavelength (and direction for “non-diffuse” surfaces). In the case of a surface receiving radiation and emitting radiation there is no reason why these two values should be the same. This is because the incident radiation will be at one wavelength (or range of wavelengths), but the wavelength of emission depends on the temperature of the surface.
Note 3 – One simple way to find the value that satisfies the equation is to plot the equation for a wide range of temperatures and look up the temperature value where the result is correct. This is what I did here. It is the work of a minute with Matlab.
Update – added Sep 15th
A graph of temperature vs wall thickness for Examples Four & Five (with T2 = 10°C):
Update – Added Sep 15th
3D graph for examples 4 to 6 – of how T1 varies with wall thickness and T2:
Click for a larger image
Update – added Sep 16th
3D graph of how T1 varies with emissivity. First, when absorptivity = emissivity:
Click for a larger image
Now with absorptivity (the proportion of incident radiation absorbed) set at 0.8, while the emissivity varies.
Click for a larger image
Notice that when the absorptivity and emissivity are equal the temperature T1 is pretty much independent of the actual value of emissivity/absorptivity – why is that?
And when emissivity varies while absorptivity is fixed (and therefore absorbed energy is fixed) the temperature T1 is pretty much independent of emissivity for very thin walls – why is that?
The Science of Doom 
[…] Heat Transfer Basics – Part Zero […]
And now for the obvious objection, as any lava lamp owner will be able to attest, the equilibrium assumption is hard to apply, to the real world. Isn’t it?
Thanks very much for this posting. It’s so simple even I can understand it, and can now re-read the previous postings with far more understanding.
Maurizio Morabito:
Every physics and engineering course studies equilibrium results.
The reason is that it provides a basic understanding of the outcome. It helps to identify the variables that will change the steady state outcome and by how much.
You could equally argue that any calculation of physics/engineering problems is futile because the material properties are hard to gauge accurately. Or the boundary conditions are impossible to know accurately.
Any yet, this is why physics and engineering uses equilibrium conditions as a staple. It informs us of the steady state outcome. And therefore, we have a good approximation of the actual real world result.
In the real world, science and engineering uses it because it is informative.
Next dynamic/transient response is covered. This requires yet more knowledge – the initial conditions, the heat capacities (if we are considering heat transfer) and so on. These are also flawed, as real-world knowledge is imperfect.
Still, I believe it is better to know the solution approximately than to know nothing in a quest for perfection. Others might disagree.
Hold it… who mentioned “futility”???
The objection is about the relevance of the “equilibrium” approximation. One could say there is no malnutrition as the daily amount of produced food is on average more than enough. But in the real world, whilst knowing of how to compute that average is not futile, it’s less relevant than grasping the dynamics of the whole system.
It’s Maurizio here again, after logging in to WordPress
Here’s an even more… relevant example 🙂
Antioxydants are well-known for their action in protecting cells in the lab (against free oxygen and its effects). However, even if studying them in controlled conditions is far from futile, their importance as anti-aging agents in the real world appears so far to have been overestimated.
Well SoD in example 4
..”At this point we are assuming no convection (perhaps there is no atmosphere for example) to keep things simple.”…
It seems you have worked out the temperature of a vacuum.
in this case being T1 = 302.80 K
Well done.
I’m sure a Nobel prize cannot be far away!
It is not the temperature of vacuum numbnuts, it is the temp of the left side of the plate (wall/surface/whatever)
Mohamed Iqbal Pallipurath says
“it is the temp of the left side of the plate”
As SoD made clear in his reply (see below).
Such a long post to indicate that insulation works.
Hope you were convinced.
The radiation source is unknown.
The less creative folk who visit the site (like myself) will take a lot of convincing that heat can spontaneously move from a colder to a hotter surface.
Others will gladly accept that heat does just that and offer up the greenhouse theory as proof.
So its always a good idea to get clarification
As to whether a vacuum has a temperature there is also some debate.
Some say that background interstellar radiation gives around 2.7K
Others say it has no temperature
Others give an operational definition of temperature as that which is measured by an accurate thermometer.
In this case an accurate IR thermometer would read the same as the outer wall or slightly higher if more inclined towards the radiation source.
But welcome to the discussion even if it has taken you over three years to form an opinion .
Its seldom that a solid gold wingnut like yourself makes a contribution.
Bryan,
So many words to say nothing at all.
Precisely where in this article, or anywhere on the website, is it ever stated that net energy, or in your terminology, heat, flows spontaneously from a colder to a hotter surface? That’s entirely your straw man.
In example 4 we have a flux of 400 W/m² into the surface of the PVC on the left side from an input of 500W/m². That surface then radiates back to the left, but at a rate lower than it absorbs and conducts heat to the right to the constant temperature surface. The energy flows balance at a left surface temperature of 302.8 K, which is warmer than the 283.2 K of the right surface, and lower than the effective temperature of the incoming radiation of 306.44K. I see no flow of heat from colder to hotter, only from hotter to colder. You get exactly the same result if you use only one way flow. If you don’t, you’re not doing the calculation correctly.
The source of the incoming radiation is irrelevant. It could be a black body spectrum at 306.44K, monochromatic at any wavelength or a complex spectrum as long as the absorptivity of the PVC for the incoming radiation was 0.8.
Bryan:
I worked out the temperature of the surface. Surfaces radiate into a vacuum or a gas or a liquid.. they aren’t picky.
Anyway, welcome to part zero. Tell me, are you happy with example six, where the colder surface has increased the temperature of the warmer surface?
10 X the thickness of material X does not reduce the flux by 10 times. Are you going to clarify this further?
Another point is this: you spent a great deal of time showing how heat from cooler sources is absorbed by the warmer source next to it. How does this reconcile with the examples you are working with in this post?
hunter:
Example 2 compared with example 1 has a conducted heat flux change of 10x for a thickness change of 10x.
Example 5 compared with example 4 has a conducted heat flux change of 5.5x for a thickness change of 10x.
I guess this is what you are talking about?
Conducted heat flux is always equal to k . ΔT/Δx (see note 1 in the article).
In the case of example 4 vs example 5, the temperature of one side has been modified because of the radiation of heat from the surface. Conducted heat is still calculated by that same formula but surface temperature has been changed by radiation. The same formula was still used to calculate conducted heat.
Does this make sense?
I’m not sure I understand your question. Guessing at the idea..
The examples in previous article used the first law of thermodynamics to calculate temperature changes as a result of emission of radiation from cooler bodies in the vicinity of hotter bodies (e.g. The First Law of Thermodynamics Meets the Imaginary Second Law ).
Example 6 in this article shows a similar situation but this time using conduction – where a colder surface affects the temperature of a hotter surface..
Well, maybe you can be more specific about your question.
SoD
Check your figure for T1 of example 4.
I make it 397.8K
In general there seems to be a great deal of confusion of terms.
This is not helped when technical definitions are used loosely.
The difference between an electric blanket and an ordinary blanket can give a useful illustration.
Both will provide insulation.
An ordinary blanket stops heat loss from a person by reducing heat loss by conduction convection and radiation.
However it will never raise the temperature of the person.
An electric blanket on the other hand can raise the temperature of the person as heat can flow from the blanket to the person.
What the second law states is that heat can never flow from a lower to a higher temperature without work being done.
As you know a great number of misguided people assumed that when Gerhard Gerlich and Ralf Tscheuschner made this statement they thought that they were implying that the atmosphere did not radiate.
If they had taken time to read their paper carefully they would not have come to this foolish conclusion.
Heat has the thermodynamic capacity to do work.
If in any example you are in any doubt, try obtain work.
If you can its heat, if not, its not.
Bryan:
Perhaps I made a mistake. Why don’t you explain your working and we can find out. Wrong formula? Mistake in calculation?
So your answer about example six is?
SoD
Accepting your working for example 4 to;
“So now the equation is:
400 = 0.19 x (T1-283) / 0.2 + 0.8 x 5.67×10-8 x( T1)^4”
400 = 0.95(T1 – 283) + 0.8 x 5.67×64.14
421.05 = T1 – 283 + 290.94/0.95
421.05 = T1 – 283 + 306.25
=> T1 = 397.8K
In example 6 you have set conditions so that the radiation will produce the higher temperature on surface of T1
There is no doubt in anyone’s mind that the Earth with an atmosphere will be warmer than one without.
What I have tried to communicate is that this does not mean that the atmosphere will heat the Earth Surface. Rather it means that the atmosphere will reduce the heat loss from the Earth surface.
Since you are solving for T1, how do you get 64.14 as T1^4/1E8?
In fact, if you use your value of T1, then 397.8^4/1E8 = 250.4, not 64.14. Plugging the equation into Excel and using Solver, the answer I get is T1 = 302.7773K.
ATTN : SOD
RE: R Factor
The R factor is parameter that relates to the retardation of heat loss by various materials used for insulation.
For example, pink fiberglass batts for walls with 2 x 4 studs has an R value of 12. For walls with 2 x 6 studs the batts are thicker and have an R value of 20.
During the lunar night which is 2 weeks, the surface temperature drops to -150 deg C. The coldest temperature ever recorded during winter in Antarctica at the south pole is about -90 deg C.
Since the period of darkness in winter there is much longer than the lunar night, we would anticipate that the temperature at the south pole would be lower than moon to the first approximation. Since the atmosphere insulates the surface from heat loss, the temperature only drops to -90 deg C.
Using a google search, I found an article which calculated for the earth’s atmosphere a R factor of 0.85.
The polar cell (air rising about 60 degS, moving south and descending) brings energy into the Antarctic. Ocean currents import energy to the edge of the continent.
DeWitt Payne, SoD
Yes my mistake I used value of T2 for the T^4 part.
Harold Pierce Jr:
According to Wikipedia “The R-value is a measure of thermal resistance [1] used in the building and construction industry”
This is the material property that defines the effectiveness of heat transfer by conductivity. You can see it in this article, it is called “thermal conductivity” – this is the general term in use in thermodynamics.
The atmosphere has a very low conductivity (0.025W/m.K) as do most gases and so conductivity can be neglected in the atmosphere, except at the boundary layer with the earth’s surface.
The two other methods of heat transfer are convection – bulk movement of fluid transporting heat, and radiation. These two are the important ones.
The atmosphere has a large interaction with the radiation emitted from the earth’s surface. Convection has a huge impact on the heat transfer from the earth’s surface, without convection the surface temperature would be much higher.
But ten times the insulation does not insulate ten times as well.
Hunter,
True, but then it has long been established that effect of an increase in a GHG is logarithmic with respect to the level of increase.
Bryan wrote (9/13 11 am): “An ordinary blanket stops heat loss from a person by reducing heat loss by conduction convection and radiation. However it will never raise the temperature of the person.”
1) Unless a blanket is impossibly skin tight everywhere, a blanket presumably won’t stop most emission of radiation (eoT^4) by a person’s skin. (After the inner surface of the blanket absorbs that radiation, some of the energy is conducted through the blanket and some returns to the person as radiation (and some may go to raising the blanket’s temperature.) 2) Air is a fairly poor conductor of heat, so it isn’t clear that a blanket conducts less heat from skin than air conducts. In many cases, insulation is mostly due to small pockets of air in the insulating material and not the poor thermal conductivity of the insulator itself. 3) A blanket does stop convection.
If you cover a dead person with a blanket, you won’t make him warmer, you’ll just slow down the rate at which his temperature approaches ambient temperature. If you cover a live person with a fixed rate of internal energy production (from metabolism of food), that person’s temperature will rise from the previous equilibrium temperature until the rate of heat loss matches the rate of internal heat generation. If you need experimental proof, first wear only your swimsuit and then winter clothes outside in the winter and see if hypothermia (core body temperature <35 degC) sets in. (Before that point, your body will shiver to increase internal heat generation and make other changes capable of compensating for modest increases in heat loss.) The energy needed to raise the person's temperature certainly came from the person himself – not the blanket – but that doesn't change the fact that the person IS warmer. (The earth appears to be similar, except that its source of energy is external, the sun, rather than internal.)
@ Frank 1611 today.
At secondary school. I was taught that the earth had a molten rock centre with a core of molten iron. evidence was in the form of earthquakes and active volcanoes. I was at school a long time ago. Have things changed?
It’s my vague understanding that the energy released by collisions as the earth was formed was enough to melt most of the solid (accounting for the spherical shape slight flattened by spin) and allow the heaviest elements (mostly iron) to sink to the center. The heat generated by decomposition of radioisotopes keeps earth very hot inside and the outer core molten. The rate at which that heat reaches the surface of the earth by conduction and convection is so slow that the energy input from inside the earth can be neglected compared to the sun.
Frank
The trouble with a “homely” example like electric and ordinary blankets and a person is that the illustration can be stretched until it loses the point.
If however I had said perhaps a bronze statue at say 80c placed in a room at say 20c with and without a blanket, the respective temperature time graphs would both show falling temperature.
The rate of temperature fall of the blanket covered statue would be less than that of the bare statue.
All three methods of heat transfer would be from the higher temperature to the lower temperature at all times.
Dear SoD,
The examples 4-6 are not fully specified. We also need to know the temperature of the blackbody radiation source T(BB). This temperature will influence the “color” of the emitted radiation, that is, its radiation wave-length, or frequency, spectrum. If T(BB) is higher than T1, heat tranfer will occur, but not otherwise.
The wall surface T1 will also emit radiation and will re-radiate the entire input (500W/m2) if its temperature is eual to T(BB) and the wall will therefore not be heated. If T1 is higher than T(BB) (this corresponds to the atmosphere/earth case), T1 (the earth) will radiate all incident radiation from T(BB) (atmosphere) plus additional energy depending on the temperature difference, that is, >500W/m2, and the wall (the earth) will thus again not be heated. Only in the case when T(BB) is higher than T1, heat will be radiativelly transferred to the wall. You can find basic and easily digested information on the physics of blackbody radiation and radiative heat transfer at: http://claesjohnson.blogspot.com/
Your examples highlight the classical mistakes being made all the time (including by the IPCC), based on lack of understanding of the physics of blackbody radiation and radiative heat transfer.
Kind regards, JanS
First, who said the source was a black body? It could be monochromatic microwave radiation and the calculations would still be correct if the absorptivity of the surface for the incident radiation was 0.8. According to your logic, you shouldn’t be able to boil water in a microwave oven.
Second, if it were a black body, the exact temperature of the source of radiation would depend on the geometry. For the case of an infinite parallel plane as the radiator, then a flux of 500 W/m2 specifies a minimum temperature of 306.4 K by the S-B equation. For any other geometry, such as a distant luminous sphere, the source will have a higher temperature. If you think you can provide a counter example where the black body source temperature is less than 306.4 K and still have an incident flux of 500 W/m2, have at it.
I wouldn’t be so quick to refer to Claes Johnson as a source of expert information either.
Third, if T2 is higher than 306.4 K, then heat flow in the PVC layer is in the opposite direction and T1 is also higher than 306.4 K but less than T2.
In a microwave oven molecules having a dipolar moment, such as water molecules, are made to oscillate and thereby absorb the microwave energy. This has nothing to do with blackbody radiation. Other substances (without dipolar moment) will not absorb microwave radiation.
I assumed that the incident power density (or energy flux density) in the examples 4-6 had a blackbody wavelength or frequency distribution since this is relevant to the climate discussion and since the examples 4-6 in the blog post use the Stefan-Boltzmann blackbody radiation law to estimate the T1 temperature. The SB law is only valid for blackbody radiation and not for monochromatic radiation.
Clearly, the minimum (corresponding to parallel surfaces) blackbody temperature, with the incident radiative power density (500W/m2) and emissivity (0.8) given in the examples, comes down to a blackbody temperature of the level you indicate (306.4K). This corresponds to my case, where T(BB)>T1, entailing that heat will be radiatively transferred from T(BB) to T1 and T1 will thus increase.
So far so good. I, however, saw the examples as ‘educative’ such that one may consider the effects of the incident power density taking higher or lower values than the one indicated. I therefore considered the case where the incident radiative power density has a lower value, such that the minimum blackbody temperature becomes lower that T1, that is, T(BB)<T1. Here T(BB) may be seen as the colder atmosphere that interacts with the warmer surface of the earth (T1)). In this case, radiative (but also convective and latent heat (evaporation/condensation)) heat will be transferred from T1 to T(BB). entailing that T1 will decrease. The incident radiative energy from T(BB) will, in accordance to the second law of thermodynamics, in this case be re-radiated (lossless) from T1 to T(BB) and will thus not contribute to heating T1.
JanS
You can see my questions to Claes Johnson in the comments of Blackbody: Transformer of Radiation.
He does agree that the Stefan-Boltzmann equation is the integral of the Planck equation across all wavelengths and directions.
This equation calculates the emission of thermal radiation from a surface without any reference to the surrounding sources of radiation. It only requires knowledge of the temperature of the surface and the emissivity.
However, Claes Johnson doesn’t believe this equation is correct. As it is just the integral of the Planck equation he needs to explain why Planck’s equation is wrong.
In claiming “Your examples highlight the classical mistakes being made all the time (including by the IPCC), based on lack of understanding of the physics of blackbody radiation and radiative heat transfer.“, perhaps you can pick up the baton where Claes Johnson didn’t and explain what is wrong with Planck’s equation for the emission of thermal radiation.
It’s a big claim – 100 years of physics has failed to spot the problem. I’ll be fascinated to see your evidence..
I referred to Claes Johnson since he is an experienced applied mathematician with a good insight into the ‘mysteries’ of physics. He does use non-conventional techniques in solving physical problems, such as blackbody problems and has therefore met resistance from ‘pure’ physicists, which does not necessarily mean that he is wrong. I will not interfere in the discussion between you and Claes, but suggest that you again contact him to sort out you questions. An open mindset related to the problem is helpful in such a discussion.
Sorry SoD, I know you don’t like closed answers, but I’m going to paraphrase something Dr. Schneider said in his last TV appearance, “No JanS, that’s just wrong.”
I followed the discussion between SoD and C. Johnson, and the most standing out thing Claes said was, “A is incomplete because the receiver is not specified. …. One cannot speak about only emission, at least I can’t. ”
I’m no expert on thermodynamics, but what Claes is suggesting is akin to saying that an excited molecule will not emit a photon unless whatever the receiver will be is known, whether that receiver is a micron away or 100 light years away. That’s just plain wrong. And, if he gets that wrong, I’m not terribly interested in trying to sort out what he got right.
It takes a little longer in Excel, but here’s the plot of T1 and q vs T2:
Bryan wrote (9/13 11 am): “As you know a great number of misguided people assumed that when Gerhard Gerlich and Ralf Tscheuschner made this statement they thought that they were implying that the atmosphere did not radiate.” G&T are offended at the lack of rigor in climate science (a frustration I share after sorting out some fallacies with regard to slab atmospheres and Venus). However, in their 2010 reply to their critics (www.skyfall.fr/wp-content/gerlich-reply-to-halpern.pdf), G&T write in Section 3.2:
Once again, we never claimed — allegedly with reference to Clausius — that a colder body does not send radiation to a warmer one. Rather, we cite a paper, in which Clausius treats the radiative exchange. The correct question is, whether the colder body [that radiates less intensively than the warmer body] warms up the warmer one. The answer is: It does not.”
1) Unfortunately, G&T do not explain what happens to the energy in the photons that are emitted from the colder body to the warmer one. 2) They do not appear to explicitly address the objections of Halpern, Rahmstorf, or Gregory (Gregory cited below):
“2nd law is always a statement on net heat flows. To consider only one part of the exchange is incorrect.”.
The discussion in this section focus on flaws in the approach of Ozawa, but does not conclude with a statement that with a statement contradicting Gregory’s net heat flow explanation. Although I certainly can’t follow all of the discussion, the equations G&T cite from their own work (∆S1 + ∆S2 > 0) appear to support the net flow approach. (I assume I am wrong about this.)
Frank
First of all thank you for the link to G&Ts reply to Halpern et al.
I have not had a chance to read it – much appreciated.
I pointed out to Joel Shore (during an exchange at WUWT) long before the comment was published that he (or they) had drawn an unreasonable interpretation of G&Ts paper in that they thought G&T were implying that the atmosphere did not radiate to a warmer Earth surface.
My own view is that the word “Heat” should only be used for the net flow.
My only quibble with G&T is that on one page they talk of “net heat flow”
My understanding of such a term would be if two heat sources were flowing into one surface.
Do G&T think that: a) the second law is violated whenever one postulates that energy is transferred from the colder atmosphere to the warmer ground or b) such a transfer is allowed as long as more energy flows from the ground to the atmosphere than vice versa? G&T are vague on this subject. For example, the abstract for their response says:
“In particular, it is not true that we are “trying to apply the Clausius statement of the Second Law of Thermodynamics to only one side of a heat transfer process rather than the entire process”
G&T certainly enjoy complaining about how the system, subsystems, and surroundings are defined in various published descriptions of greenhouse theory and whether the second law applies to energy flow or radiation as well as heat flow. If transfer of energy by radiation lies outside the scope of the second law, why did they raise the issue of the second law? If transfer of energy or radiation falls within the scope of the second law, why complain about which word is used?
You claim that the best way to tell if heat is involved is to see if work can be done. If I put a photovoltaic cell in the path of radiation, work can be done. Am I correct in saying radiation is heat and subject to the second law?
Frank
..”You claim that the best way to tell if heat is involved is to see if work can be done. If I put a photovoltaic cell in the path of radiation, work can be done.” …..
Yes, if the cell is used to point at say the Sun the radiation produces electrical energy and all manner of work can be done.
Perhaps you mean however, what happens if it is pointed to the cold sky at night?
In this case more radiation will leave its surface than lands on it.
Depending on the type of instrument, the cell is attached to and whether there was a battery included an indication of flux change might be detected.
JanS on September 15, 2010 at 10:39 am:
If you believe in and promote his theories then we look to you to support your claims. What makes you think he has a good insight?
He has probably met resistance from everyone who has ever studied the subject and everyone who has done research in the subject. For good reason. His ideas aren’t supported in any textbook you can find.
That doesn’t necessarily mean he is wrong. But it does mean that you would be expected to provide evidence for the claim.
It certainly means that if you write comments like:
-that people are going to ask you to support them.
I did ask Claes to explain why, but his best seemed to be a dinner party analogy and an “I don’t know”.
Claes hasn’t grasped physics basics, OR has uncovered an amazing new fundamental flaw in Planck’s theories that will revolutionize the field.
I’m just going with the odds here, but will be happy to review new evidence when (if) presented.
Bryan from September 13, 2010 at 12:28 pm:
And Lo! There was much rejoicing.
So more radiatively-active gases in the atmosphere will – all other things being equal – increase the temperature of the earth’s surface?
SoD
……”So more radiatively-active gases in the atmosphere will – all other things being equal – increase the temperature of the earth’s surface?”……..
Lets remember here that less than 1% of the atmosphere radiate significantly in the Infra Red.
Of the molecules that do there are on average about 26 H2O molecules to each CO2.
Water has game changing property from Ocean extent to Latent Heat contributions.
Roy Spenser currently is featured on WUWT with an article saying that there is good evidence that the overall feedback effect is cooling rather than heating.
Negative feedback does not mean cooling. It means that the warming would be less than in the case of zero feedback. Positive feedback means the warming would be greater than if the feedback were zero.
Bryan (9/15 1:43pm): To avoid sounding like a CAGW’er, let’s not dodge SoD’s awkward question with vague comments about the percentage of GHGs in the atmosphere. What mechanism permits all warming to be avoided after increasing a GHG reduces the rate of energy loss to space? The only mechanism I can see would be for the GHG to also reduce incoming radiation. However, the purpose of the “two peaks” graphs that G&T abhor (for their undisclosed scaling factors) is to show that this doesn’t happen.
And let’s not misrepresent Spenser, who actually said:
“While it seems rather obvious intuitively that a warmer world will have more atmospheric water vapor, and thus positive water vapor feedback, I’ve just listed the first 5 reasons that come to my mind why this might not be the case. I am not saying that’s what I necessarily believe. I will admit to having waffled on this issue over the years, but that’s because there is evidence on both sides of the debate.”
Feedback is the response to the warming forced by an increasing GHG. Positive feedback will amplify any warming you chose to accept. Negative feedback will diminish any such warming, but not change it to cooling. If negative feedback could could change a warming to a cooling, the earth would be unstable towards runaway global cooling (an interesting idea given the snowball earth hypothesis).
Frank
….”What mechanism permits all warming to be avoided after increasing a GHG reduces the rate of energy loss to space? The only mechanism I can see would be for the GHG to also reduce incoming radiation. “……..
I have seen arguments both for and against the net effect of CO2 in the atmosphere.
A small overall heating effect in the troposphere.
A cooling effect above the troposphere.
Seems most likely.
The preponderance of water vapour which also has phase change effects makes separating the variables affecting temperature very difficult.
The fact that there is no experimental proof that the increase in CO2 in the last 150years is in any way linked to the 0.7C rise in temperature certainly leaves me unconvinced about a CO2 greenhouse effect.
I added a graph of how T1 varies with wall thickness and T2.
(The version in the body of the article has a clickable link to a larger graphic).
I added two more graphs in the article, reproduced here (but clickable for larger versions in the article)
3D graph of how T1 varies with emissivity.
First, when absorptivity = emissivity:
Second, with absorptivity (the proportion of incident radiation absorbed) set at 0.8, while the emissivity varies:
Notice that when the absorptivity and emissivity are equal the temperature, T1, is pretty much independent of the actual value of emissivity/absorptivity – why is that?
And when emissivity varies while absorptivity is fixed (and therefore absorbed energy is fixed) the temperature T1 is pretty much independent of emissivity for very thin walls – why is that?
Frank
…..”Unfortunately, G&T do not explain what happens to the energy in the photons that are emitted from the colder body to the warmer one.”…….
Yes I agree this is an area we would all like to get a handle on.
The hot surface is pouring out photons at high intensity yet is also subject to a lower intensity stream of photons.
I get the picture of a few spectators arriving late at a football stadium trying to get in as the crowds stream out.
The lower frequency lower intensity photons have each been issued a ticket from Kirchhoff which they insist is valid.
Some people claim they know exactly what happens.
For them it produces a heating effect accounting for a 33 celsius rise in the temperature of the atmosphere.
I have followed with interest the discussion around the articles started by Tom Vonk on WUWT about photon absorption.
Several contributors who appeared to know exactly what goes on have come to exactly opposite conclusions.
G&T will start from the classical exposition of equilibrium thermodynamics.
Two reservoirs, T1 at a higher temperature and T2 at a lower temperature.
Between these two reservoirs several examples of heat transfer are analysed as can be found in any Physics Textbook.
All textbooks agree that heat cannot be transferred in a continuous cycle from the colder reservoir to the hotter one.
Are the Earth Surface and the Atmosphere two such reservoirs?
Can heat be taken from the Earth Surface and transferred to the atmosphere without an appreciable drop in the Earth Surface temperature?
If your answer to the two questions above is yes then you will agree with G&T.
Some will argue that the Sun /atmosphere/Earth interactions are so turbulent that non equilibrium thermodynamics must be used to make any sense out of the situation.
SoD likes to look at the blackbody spectrum of the hotter source and the colder sink.
He does this for the solar(5800K) and terrestial (290K)spectra and the radiative gases plus the spectra shift gives 33C warming.
Curiously JanS agrees that the “colour”(temperature of source) is important and says therefor the temperature of the source in Example 4 above should be specified .
However this is countered by DeWitt Payne who gives the example of a microwave oven to reinforce a view that it is the magnitude rather than the frequency of the radiation that is important.
So Frank the matter is far from simple but if I was a betting man I would put my money on G&T but keep an open mind.
Re Bryan (9/16 9:00 am):
According to the scientific method, the greenhouse hypothesis becomes greenhouse theory as it withstands repeated experimental challenges (and theoretical challenge from the second law) and as competing hypotheses are found to be inconsistent with observation. I’ll deal with your points paragraph-by-paragraph and save space by not quoting each point.
If my memory is correct, SoD has presented data showing that about the right number of photons (or ticket holders) arrive at the earth’s surface at night (not from the sun), with wavelength and intensity consistent with DLR emitted by GHGs in the colder atmosphere. Since the downward flux is more than 50% of the upward flux, it can’t be due to scattering of the upward flux.
All of Tom Vonk’s blather contains the caveat that no radiation/energy enters or leaves the gas he discusses. Of course GHG’s can’t warm up (or cool down) the bulk of the atmosphere containing them under these circumstances – energy is conserved. In the relevant world, energy enters and exists. It’s hard to take most science that appears at WUWT seriously when they post this kind of stuff and sensible comments are drowned out by the cheering section.
As noted previously, G&T never unambiguously state that the second law invalidates all schemes with two-way flux of photons and net flux of energy from warmer to colder. If they explicitly make such a statement in a respected, peer-reviewed journal and their argument survives the inevitable replies, greenhouse theory (or the second law) will be dead. However, G&T seem to object to aspects greenhouse theories that are less fundamental than two-way flux.
Heat can be taken from the surface of the earth without the surface temperature dropping. This happens every day when the sun is shining. Once the sun goes down, surface temperature drops.
As best I can tell, greenhouse theory survives and no competing hypotheses exist.
Frank
> it withstands repeated experimental challenges
Such as? (question)
Omnologos: “Greenhouse” theory (a misleading name) was devised to explain why the average surface of the earth was much warmer than necessary to radiate away the power delivered to the surface and the lower atmosphere by the sun. Greenhouse theory POSTULATES that GHG’s in the atmosphere absorb some outgoing infrared radiation and re-emit half of that energy towards the surface of the earth (“Downward Longwave Radiation or DLR). Challenge #1: Experiments show the existence radiation with appropriate spectrum and intensity to be DLR from the atmosphere. See SOD’s posts on DLR. Challenge #2: DLR is present at night, so the sun is not the source. Challenge #3: DLR doesn’t have the right intensity or wavelength to come from other stars. Challenge #4: There is too much DLR for the radiation to be scattered OLR from the surface of the earth. Challenge #5: Is transfer of energy from the colder atmosphere to the warmer surface by DLR inconsistent with the the second law of thermodynamics when accompanied by a larger transfer of energy in the opposite direction by OLR? G&T don’t make this claim – they seem to object to other things THEY choose to include within the scope of greenhouse theory. This challenge probably hasn’t been completely resolved. In the absence of other explanations (do you have any?), there is strong experimental evidence for the existence of DLR and no reason to believe the atmosphere is NOT the source.
Accepting greenhouse theory does NOT require one to accept all of the IPCC “consensus”. If you read SoD’s posts on DLR, you’ll learn that even when we send a radiosonde up to determine the exact temperature and humidity of the atmosphere, we can’t calculate the intensity of DLR with enough accuracy (ca 1%) so that the discrepancy between theory and observation isn’t comparable with the forcing expected for 2X CO2. That tells us there are good reasons to doubt the reliability of GCMs, which must predict temperature, humidity, and then DLR.
thank you Frank
It’s good to read about experiments, really 😉
As for Tom Vonk, if we’re not free to be wrong, then we’re not free to be right either.
Actually, if we are suppressing views because they are “wrong”, we end up as literal slaves of whatever happens to appear “right” at the moment. Hardly the way to progress science…
omnologos: This is scienceofdoom, not politicsofdoom, religionofdoom, philosophyofdoom, lawofdoom. The goal of science is to learn what is right, not to be free to be wrong! In pursuit of knowledge, we DO have the freedom to follow our own ideas, but that freedom comes (as usual) with responsibility – the responsibility to listen with An open mind, to make constructive criticism (a challenge for me), to defend ideas on their scientific, not political merits, and to publicly admit when our ideas are incompatible with reliable observations or a well-established scientific theory. (Well-established theories are occasionally toppled, but that is unlikely to happen here at scienceofdoom.) It may seem to happen at Climateaudit, but the toppling actually begins after McIntyre publishes in a peer-reviewed journal AND successfully responds to peer-reviewed criticism of his work.
I seriously doubt that Andy Watts lets Tom Vonk post at WUWT so that Tom can be “free to be wrong” in front of Andy’s audience. Tom’s mistakes were subtle, but he didn’t engage with his critics. In his first post, Tom defines LTE (local temperature equilibrium):
“If the particle stays long enough in a small volume to interact with other particles in this small volume , for example by collisions , then the particle will equilibrate with others.”
An emitted photon usually doesn’t remain inside the small volume where its source GHG molecule is colliding frequently with its neighbors. Photons usually escape LTE, partly because they (being massless) can’t exchange energy by collision. A photon can be lost to space, reach the surface of the earth or be absorbed by a region of the atmosphere with a different temperature; all of which can change temperature. But Tom insists:
“Because the number of excited molecules in a small volume in LTE must stay constant , [it] follows that both processes emission/absorption must balance .”
When temperature changes because of net loss or gain of photons, the number of excited molecules does not remain constant. Tom doesn’t allow gas to gain or lose energy by radiation – and conservation of energy ensures that CO2 can’t heat up N2. The bulk of Tom’s posts heroically describes the mechanical details of that energy conservation, unfortunately with mistaken assumptions about photons.
The behavior of radiation interacting with a volume of air is described by the Schwarzschild equation, a subject SoD was kindly suggested when I rejected slab models of the atmosphere. When B(T) = I (I is now intensity, not a personal pronoun), incoming and outgoing radiation are in equilibrium, but this is not the same thing as LTE – I comes from OUTSIDE, B(T) depends on INSIDE temperature. The probability of emitting a photon is determined by the Planck function, B(T), not the Boltzmann distribution. http://www.sundogpublishing.com/AtmosRad197.pdf
Frank
….”Heat can be taken from the surface of the earth without the surface temperature dropping. This happens every day when the sun is shining. Once the sun goes down, surface temperature drops.”……
This is at the centre of the second law debate.
Is the surface of the Earth a big enough reservoir?
If it is not, then no other reservoir is available to test the second law constraints for almost any given example.
The average temperature of the Ocean surface is about 16C
If we take 1Km radius section of calm water through several cycles exchanging energy with the atmosphere say at night for one hour IMHO that the change in Ocean surface temperature would be unmeasurable.
If the loss of heat is unmeasurable then since we all agree that the energy from the atmosphere to Earth (being even less) would certainly be unmeasurable.
Two aspects show the direction of heat transfer.
1. Can work be done
2. Has the transfer resulted in a temperature rise.
On rare occasions the air temperature can be higher than the ground temperature.
When this happens the energy transfer capacity from Atmosphere to ground is not sufficient to stop energy transfer from Earth Surface resulting a drop in temperature and sometimes ground frost.
In pages 27 to 30 of their paper G&T made a calculation of the Terrestial Radiation.
Its in conflict with the IPCC K&T figure.
Strangely enough Halpern et al did not seek to contest this.
Further G&T say;
1. All the radiation figures given by IPCC are far too high.
2. That the energy transfer values cannot be calculated from the spectrographs or radiative transfer equations as given.
3. The use of the Stephan Boltzmann equation is invalid when applies to gases.
It seems that points 1,2and 3 above would be open to test by experiment.
However we would need to ensure that the measuring instruments were calibrated independently of the Stephan Boltzmann equation or we would end up in a circular argument.
Byran: Thanks for the reply, but the debate seems to be getting far afield. My points were that solid experimental evidence for DLR exists (SoD has posted some) and neither G&T nor Vonk convince me (for reasons I have explained) that the atmosphere is not the source.
I hated some aspects of greenhouse theory when they were presented as slab atmospheres radiating as grey bodies. Look at my link to Schwarzschild equation, it contains the real physics. I think SOD posted on this subject when he calculated the temperature profile for a gray atmosphere.
Frank
Thanks for the link to Schwarzschild equation.
I will check it out.
I hope I did not give the impression that I found Tom Ronk convincing.
My point was there seemed to be a number of people who went into considerable detail in backing up their point of view during this item.
TT,VT,VV interactions between CO2 and N2 were analysed by seemingly knowledgeable people.
However they came to exactly opposite conclusions.
I was interested in the debate to find out what happens at a micro level with CO2s interactions in the atmosphere.
I’m still looking.
Bryan,
Look closely at the graphs on pages 27-30 of solar vs Earth surface radiation. G&T are comparing brightness at the surface of the sun to brightness at the surface of the Earth. But the sun is 150 million km from the Earth and brightness falls as r-2. That’s why the solar constant for the Earth is 1378 W/m2 not 63.3E6 W/m2. This is a flaw so basic as to be mind boggling. G&T’s graphs are also in cgs units rather than SI, which makes no sense either. I believe that answers point 1.
See also: http://bccp.lbl.gov/Academy/pdfs/InvSquareLaw.pdf
For point 2, their assertion is simply wrong. One calculates the energy values by integrating under the spectrum whether measured or calculated. The spectrometers are calibrated against black body radiation at two different temperatures so the readings are in power units. Total IR up and down near the surface can also be measured with an instrument called a pyrgeometer. Measurements are in reasonable agreement with calculations.
See: https://scienceofdoom.com/2010/07/17/the-amazing-case-of-back-radiation/ plus parts two and three.
For point 3, they are indeed correct. But see the reply to point 2. One doesn’t need to use the Stefan-Boltzmann equation when you have a measured or calculated spectrum to integrate.
I made a mistake in point 1. G&T do in fact correct for the distance from the sun, but they don’t correct for hemispherical illumination, atmospheric absorption or albedo, which is still a mind boggling error. So they are comparing solar irradiance at the top of the atmosphere with the sun directly overhead to emission from the earth’s surface. Hemispherical illumination reduces the average solar irradiance at the surface by a factor of 4 and albedo and absorption by a factor of 2.04. So rather than the radiation at the ground being 3.46 times weaker than the incoming radiation, the surface radiation is 2.36 times greater than the incoming solar radiation.
DeWitt Payne
The ratio of the Solar Radiation arriving at the Earths Atmosphere to that of the upward Terrestial Radiation is probably one of the least falsifiable figures.
I think I will spend some time on this.
I have just printed off the paper byKramm, Dlugi and Zelger .
They go into the matter in great detail and seem to back G&T.
I have found in the past that G&T have been considerably underestimated.
Although I have not had a chance as yet to come to a hard conclusion on this point; it seems to me unlikely that two theoretical physicists are likely to put on print several times the same conclusion, if its an obvious error.
Bryan,
Lets look at totals rather than per unit area. The amount of solar radiation intercepted by the Earth is the area of an ellipse with the equatorial and polar radii of the Earth:
Etot = Fo * pi* R1*R2
Fo = 1365.2 W/m^2
R1 = 6378137 m
R2=6356752.3m
Etot = 1.7389E17 W
But that’s at the TOA. The amount that gets to the surface has to be reduced by the amount reflected and the amount absorbed by the atmosphere. The Bond albedo which is the fraction of the total EM energy from the sun that is reflected is 0.2986 (TFK09) and the fraction absorbed by the atmosphere before it gets to the surface in the UV and near IR is 0.2285 (TFK09). 1-0.2986-0.2285 = 0.4729 is then the amount transmitted to the surface. So:
Esurf = 1.7389E17*0.4729 = 8.2226E16 W
The amount radiated by the surface of the Earth, Eu, is the surface area of the Earth times the sigmaT^4 (ignoring emissivity). If we take T = 289.1 K and area = 5.101E14 m2:
Eu = 2.020E17 W
To put it in the same form as G&T:
Es/Eu = 0.407
That’s almost an order of magnitude different from the 3.46 ratio calculated by G&T. The errors, particularly the fundamental geometric error, are inexcusable for physicists with their reputation. It also knocks their whole argument into a cocked hat, not counting all their other errors. Note that even if we ignore the amount of incident radiation absorbed by the atmosphere, the Es/Eu ratio is still less than one. A ratio less than one is a requirement for a greenhouse effect to be present.
DeWitt Payne
After reading the G&T paper it is now clearer in my own mind that they are talking about maximum values.
They do this as background for a standard calculation For the “glasshouse effect”.
They are not using the calculation to contest IPCC values.
This is made clear in conclusion 2.3.5.
If we take your figure of 1365W/m2 TOA and divide by BB Earth Surface figure 396W/m2 we get
3.45 which is almost exactly the same as the G&T figure.
Comparing TOA solar zenith insolation with surface emission is comparing two very different things, apples and oranges in the idiom and is a logical fallacy. The ratio of 3.46 is, therefore, meaningless.
This is because the full 1365 W/m2 does not reach the surface because of reflection and absorption, only illuminates half the globe at a time and illuminates a hemisphere rather than a flat disk. So a glass greenhouse (or a car) doesn’t ever see 1365 W/m2 even if it were at the equator at noon in the desert on a clear day on the equinox. For half the day, it doesn’t see the sun at all. The surface, OTOH, radiates 24/7 and is the surface of a sphere. The surface temperature of 290 K is the planetary average, not the surface temperature at the equator on a clear day at noon on the equinox.
Claiming that graphs which represent solar energy scaled to reflect these facts as being scientific misconduct is, in fact, scientific misconduct or sleight of hand at best. I could, with far more justification, say that G&T’s Figure 11 is an obscene picture since it is physically misleading.
The (large) part of the paper that concentrates on how a glass covered greenhouse works is a very stinky red herring. In spite of all their quotes, no one actually says that the atmospheric greenhouse effect works exactly the same way as a glass greenhouse.
The inference of a greenhouse effect results from the observation that the surface receives less energy from sunlight than it radiates, as demonstrated by the calculations above and measured with pyrheliometers. That energy has to come from somewhere. It is also an observed fact that the atmosphere radiates toward the ground at about the rate necessary to make up the difference as measured by pyrgeometers. It is possible to calculate the IR emission spectrum of the atmosphere as well as measure much of it with high resolution FT-IR spectrophotometers. The calculated and observed spectra are in good agreement. This isn’t, of course, proof. One can’t prove a scientific hypothesis or theory. But it’s good evidence that the theory works. There is absolutely nothing in G&T that falsifies these observations and hypotheses.
Bryan,
Another thing about real greenhouses: In a tunnel style greenhouse with an IR transparent cover like polyethylene the surface can become colder on a clear night than the outside air. By blocking convection, you force a temperature inversion.
From: http://www.hort.cornell.edu/hightunnel/about/research/general/penn_state_plastic_study.pdf
DeWitt Payne
G&T are simply using readily calculated values to obtain a quick result for what they sarcastically call a “standard cookbook recipe” for undergraduates.
This is made clear in conclusion 2.3.5.
They then went on to examine the experiment by Woods.
Thanks for the link to the polyethylene tunnel.
These links away from the sometimes heated climate change debate are all the more believable since they have no “axe to grind”.
Please tell me how all the discussion by G&T on greenhouses falsifies the atmospheric ‘greenhouse’ effect. It’s window dressing as far as I can tell and has no relevance. If you can’t do that, then show me where G&T actually falsify the atmospheric greenhouse effect. I haven’t been able to find it.
Do you see the relevance of the plastic covered tunnel greenhouses to Wood’s experiment? A clue: What do you think the temperature of a salt plate covered box would be at night compared to a glass covered box. How efficient would you expect a kitchen oven to be if the transparent part of the door were IR transmitting instead of IR opaque?
The temperature during the day is the least important part of a greenhouse. What really counts is how cold it gets during the night. A box with highly absorbing walls and low thermal mass should get to about 160 F during the day if the peak insolation at the surface is 800 W/m2, which is not unusual at mid latitudes on a clear summer day. But that same low thermal mass means that if allowed to freely radiate in the IR, it will cool quickly as well.
DeWitt Payne
Ive now had a chance to read your link to the poly tunnels.
The link to the construction of the tunnels seems to be in a process of changing .
This means that I have no idea what the sides of the tunnels were made of.
As I read it 3 of the films had IR blockers(like glass) one Tufflite Control was transparent to IR(like Rocksalt).
During the day it made almost no difference to the internal temperature of the tunnel.
One would have thought that in terms of classical glasshouse theory that the IR blockers would have produced a higher temperature.
At night the IR blockers showed inconsistent results sometimes having a small effect at other times no effect.
The surprising thing to most people is that at night all the tunnels followed a pattern of dropping below the outside ambient temperature.
Perhaps the still air inside cut off from the outside convective changes gave rise to this result.
In general these results are consistent with the Woods experiment.
Reasons to be cautious
Two of the films were kept constant throughout the study and two were changed.
The two left for two years had different composition and time dependent degradation may be different.
It’s a tunnel. The shape is a half cylinder like a Quonset hut so it’s all made from the same material.
See: http://www.northerngreenhouse.com/ideas/how_to/rebar.htm
Of course it doesn’t make a difference during the day. That’s not the point. The point is that a long wavelength transparent greenhouse cover does make a difference to the average temperature inside the greenhouse. So Woods was wrong because he only looked at the daytime peak temperature and not a full 24 hour cycle.
DeWitt Payne
From your chosen source on polytunnels;
In the conclusions paragraph
….Although IR blocking films MAY OCCATIONALLY raise night temperatures by 0.5 to 3 degrees F…..the trend does NOT SEEM to be consistant over time.
This set of results if true, are a vindication of Woods experiment.
Nobody argued that there might not be a small residual radiative effect.
But it appears to be either non existant or very small.
Bryan is a wonderful resource for this blog – helping to demonstrate how weak the arguments are against the inappropriately-named “greenhouse” effect.
Bryan is still to explain what happens to the DLR (aka “back-radiation”) when it reaches the surface.
As explained in The Amazing Case of “Back Radiation” – Part Two and Part Three.
And previously in The First Law of Thermodynamics Meets the Imaginary Second Law
It’s clear that the DLR is absorbed and this affects the temperature.
If Bryan – or his icons, G&T – knew an alternative answer they would have told us by now.
scienceofdoom, DeWitt Payne
G&T. Some incontrovertible points.
1.Navier Stokes equation throws up a number of insoluble partial differential equations which when applied to the atmosphere means any computer model of the atmosphere is a joke.
2.The “greenhouse theory” and its account of how it produces a 33c rise in temperature do not seem to have any logical link.
3. Woods experiment stands.
4. The atmospheric slab model is now being discarded by the more discerning advocates of the greenhouse theory.
5. The Halpern et al comments totally missed their target which is unfortunate as no proper debate has taken place.
6. As it stands the G&T paper has shaken off any serious criticism.
7. As you scan the Internet on this topic any contributor who declares a professional background in heat transfer physics has come out in support of G&T.
8. The impact of G&Ts paper can be observed, since its publication there have been far fewer new diagrams showing HEAT moving from the atmosphere to the Earth Surface.
Much more attention is being paid to the correct definitions of; Heat,Work,Internal Energy, Temperature,Infra Red Radiation and so on
Is that enough to be going on with?…………
Bryan,
1. Irrelevant to the topic at hand. You don’t need a GCM to calculate the local radiative/convective balance. You can measure it.
2. That’s an assertion by you. Just because you don’t understand or aren’t willing to accept the argument doesn’t mean it isn’t logical.
3. Irrelevant. Wood’s experiment has nothing whatsoever to do with the atmospheric greenhouse effect.
4. If you mean the single slab model, that was always a toy pedagogic model meant to demonstrate the fundamentals, not to make detailed calculations. Textbooks on atmospheric radiation still have sections devoted to a slab model for just this reason. See for example Petty, Grant W. A First Course in Atmospheric Radiation, chapter 6.4.3, Simple Radiative Models of the Atmosphere: Single Layer, Nonreflecting Atmosphere. You can read it Amazon by searching on 6.4.3 in Look inside the book and selecting the link to page 139.
5. Assertion. Missed what target?
6. Only in your mind. There has been little serious criticism of G&T because working scientists don’t consider it to be worth their time. A criticism would be normally rejected by most serious journals as a waste of space. Valueless papers are almost always ignored, not rebutted. That’s why you only see this being discussed in the blogosphere.
7. Anecdotal so not in the least probative and also an appeal to authority, which is delightfully ironic. If you’re going to make that sort of a statement, you need to cite a few references so a neutral observer can evaluate the expertise of the cited examples.
8. Show me a diagram from a serious scientist that ever showed net heat flow from the atmosphere to the surface except when a temperature inversion was present and the air above the surface was warmer than the surface.
None of that addresses the question I asked you. I’ll rephrase: Where do G&T show that the surface of the Earth receives sufficient energy from direct sunlight to balance the energy lost by radiation and convection? I haven’t been able to find it. It is precisely this point that is successfully addressed by atmospheric radiation, i.e. the greenhouse effect.
DeWitt Payne
G&T. Some incontrovertible points.
1.Navier Stokes equation throws up a number of insoluble partial differential equations which when applied to the atmosphere means any computer model of the atmosphere is a joke.
…………
These computer models are used to make projections for the IPC C.
Although the models have been a dismal failure, the IPCC are still using them to force a major disruption of the world economy
……….
2.The “greenhouse theory” and its account of how it produces a 33c rise in temperature do not seem to have any logical link.
………..
Try to go from your IR poly tunnels (probably non existent) radiative effect to a 33C claimed rise in the Earths Atmosphere
…….
3. Woods experiment stands.
…….
4. The atmospheric slab model is now being discarded by the more discerning advocates of the greenhouse theory.
…………..
So what number of slabs would you suggest?
……………
5. The Halpern et al comments totally missed their target which is unfortunate as no proper debate has taken place.
…………..
The main point of the comment paper was that they said G&T believed the atmosphere did not radiate.
G&Ts original paper, the Halpern et al Comment and G&Ts reply are all now available to download freely
…………..
6. As it stands the G&T paper has shaken off any serious criticism.
……………
7. As you scan the Internet on this topic any contributor who declares a professional background in heat transfer physics has come out in support of G&T.
…………..
Two straight off Fred Staples Terry Oldberg and another in the G&T blog on Climate Research News.
Three others I have come across.
However I have never come across any of the opposite persuasion
…………..
8. The impact of G&Ts paper can be observed, since its publication there have been far fewer new diagrams showing HEAT moving from the atmosphere to the Earth Surface.
Much more attention is being paid to the correct definitions of; Heat,Work,Internal Energy,Temperature,Infra Red Radiation and so on.
………….
A recent lapse however was when Trenberth asked “whats happened to the missing heat”
Bryan,
1. GCM’s are used to estimate the sensitivity of the climate to a change in CO2 concentration. The magnitude of the current atmospheric greenhouse effect is based on measurements, not GCM calculations. Just because there is no analytic solution to the Navier-Stokes equations doesn’t mean that any solution is impossible. A system with more than two gravitationally bound bodies doesn’t have an analytic solution either. In fact, it’s chaotic. So does that mean that you can’t calculate the future positions of the planets in the solar system? No. It just means that the predictions get worse the farther ahead in time you try to make them. This point is irrelevant to whether a greenhouse effect exists in the Earth’s atmosphere, i.e. whether the surface is warmer than it would be if the atmosphere were transparent to LW IR radiation.
More later.
You continue to ignore my question.
My Ph.D s in Electroanalytical Chemistry. That means I learned to solve the diffusion equation. The diffusion equation has exactly the same form as the heat equation. So you can now start a list of those who know something about heat transfer that think the G&T paper is, shall we say, incontrovertibly flawed and that the fundamentals of the atmospheric greenhouse effect theory are correct.
Bryan,
2. I guess if wouldn’t seem logical if you didn’t think that DLR is absorbed by the surface. But the incontrovertible fact is that the surface is warmer on average than if it were a black body with no atmosphere receiving the same amount of sunlight. The 33 degree calculation is based on a simplified toy model of a sphere with what amounts to a superconducting surface so that every point on the surface is at the same temperature all the time. That’s, of course, not correct so 33 degrees is too low. But given the high heat capacity of the surface and the atmosphere of the Earth, it’s a fair approximation and the error amounts to no more than 3 degrees.
3. You have yet to explain the relevance of Woods’ experiment to the reality of the atmospheric greenhouse effect. At best, it suggests that greenhouse is a misnomer for the atmospheric effect, but it is completely irrelevant to whether the surface of the planet is warmer than if there were no atmosphere present. If you restrict the analogy to the interior of a greenhouse is warmer than it would be if the greenhouse weren’t present and the surface of the planet is warmer than it would be if the atmosphere were transparent in the LW IR, then the name is still valid. Just because the actual mechanisms of the two are different doesn’t mean the effect doesn’t exist.
Shame that we can’t cook up 8 red-herrings to make one explanation pie.
Perhaps we should add some more of Bryan’s own red herrings and see what kind of dish we get:
– the DLR is Rayleigh scattering
– the DLR is “mostly reflected”
– the DLR values are too high
– Kirchhoff’s law doesn’t work
– Stefan-Boltzmann’s law is wrong
– it doesn’t matter because the ocean is so massive
4. It depends on the resolution you want. Miskolczi uses 150 layers in his line-by-line program HARTCODE, IIRC, SpectralCalc uses 15. MODTRAN uses 32, or at least it gives you the temperature and flux at 33 altitudes including the surface in the text output. This is SOP for using numeric analysis to solve problems that don’t have an analytic solution. A lot of interesting problems don’t have analytic solutions. You solve them by breaking position and time into small steps. Then you calculate the change at each discrete position with each time step based on what happened in the adjacent positions in the previous time step. Miskolczi, btw, does not dispute the existence of the greenhouse effect, his claim is that the climate sensitivity is very small.
5. I’m not surprised that Halpern, et.al. make what looks to you like a mistake. There is so much hand-waving, straw men and irrelevant arguments in G&T that it’s difficult to know what they are saying. Whether they say the atmosphere is radiating or not, as near as I can tell, they say that it doesn’t make a difference to the surface temperature. That’s called a distinction without a difference. That makes the Halpern et.al. argument correct in principle whether it is correct in detail or not.
6. That’s an assertion on your part. Very few knowledgeable people agree with you. And of course I define, like you, knowledgeable people to be people who agree with me.
7. I can’t find Fred Staples. From Terry Oldberg’s comment here: https://scienceofdoom.com/2010/03/15/the-imaginary-second-law-of-thermodynamics/#comment-1183 , he clearly doesn’t understand radiative heat transfer for all his hand waving about thermodynamics.
8. Trenberth’s comment refers to the heat from the model calculated mean radiative imbalance between incoming and outgoing radiation. The models predict that there should be ~0.8 W/m2 more incoming than outgoing radiation at the TOA. The fact that this isn’t observed in the ARGO ocean heat content measurements is a problem for the models, not with whether there is a greenhouse effect or not. I have no sympathy for modelers who think that their complex models, which contain lots of assumptions on the size of the poorly understood forcings other than from greenhouse gases (which are well understood), like aerosols in particular, are more accurate than measurements in the real world. The historic aerosol forcings used in the models are used to tune the models to approximate the temperature history. The IPCC AR4 puts very large error bars on these forcings which means the net forcing from all sources also has a large range of uncertainty. But again, this has nothing to do with the existence of an atmospheric greenhouse effect that G&T claim to falsify, only on the sensitivity to changes in forcings.
DeWitt Payne,scienceofdoom.
Several people including Halpern, SoD and now DeWitt Payne feel they can comment on the G&T paper without having to read it properly.
In my opinion this has resulted in misguided comments that produce random noise but little else.
On the DLR
Keep in mind say a Maxwell Boltzmann distribution of the number and speed of molecules in a sample of gas at a particular temperature.
Another graph at a lower temperature would show the amplitude (number rising) but the average speed decreasing.
If we add 10litres of water at 30c to a bath of water at 50c although we have added more energy the average speed hence temperature goes down.
Now I know that photons do not have any mass however certain features may be helpful.
If we now look at the BB spectrum of radiation leaving the Earth surface centred around 15C and compare it to atmospheric radiation with a smaller amplitude arriving at the surface centred around say -30C.
What happens if we combine the two spectra into a composite.
The amplitude would increase but the average energy per photon would decrease.
Would the addition of the smaller amplitude compensate from the decrease in average energy per photon?
What would happen to the characteristic temperature of the surface?
Amplitude is not the only factor important when we consider radiation.
For instance if we shine a power of 1w of radiation of 1.09um on silicon we will improve its conductivity (photoconductivity) .
However if we shine 1000w at a wavelength of 1.2um or more, no such effect occurs.
Illustrating that quantum effects also have to be considered.
Which brings us neatly back to SoDs present article.
I think that the post from JanS September 15, 2010 at 10:37 am was accurate and deserves a reply.
Although probably JanS has given up and is no longer viewing.
@Bryan
“The amplitude would increase but the average energy per photon would decrease.”
This is incorrect. Look at two BB spectra. At a lower temp the intensity at ALL wavelengths goes down. This is because there is no conservation of photons. Will the “average energy per photon” decrease? Yes, but just because you are averaging together two different distributions when you calculate such a thing. The actual measured spectrum would actually have two peaks – the photons do not equilibrate amongst themselves. If they did then we could never see emission or absorption spectra.
Bryan:
What happens to it when it reaches the earth’s surface?
This is a question I have asked you many times and you never answer it.
We measure over 300W/m^2 of DLR. What happens to it when it reaches the earth’s surface?
Simple question. What’s the answer?
SoD
In case you missed it!
If we now look at the BB spectrum of radiation leaving the Earth surface centred around 15C and compare it to atmospheric radiation with a smaller amplitude arriving at the surface centred around say -30C.
What happens if we combine the two spectra into a composite.
The amplitude would increase but the average energy per photon would decrease.
Would the addition of the smaller amplitude compensate from the decrease in average energy per photon?
What would happen to the characteristic temperature of the surface?
You seem to have statistics packages with nice graphing features why don’t you put some numbers on the above scenario and find out!
Bryan:
That’s not an answer.
a) Is the DLR absorbed?
b) Is it reflected?
c) Does it vanish?
I know you won’t answer the question because your answer would either support mainstream science (the inappropriately-named “greenhouse” effect) or support a position easily falsified.
So please provide yet more red herrings.
scienceofdoom
I think it is an answer.
Lets put in some numbers
Earth BB emission 396W/m2 at temperature 15C
Incident radiation from atmosphere at night say 200W/m2 at temperature of -30C
This is maintained for one hour
Combine the two spectra with no losses due to scattering.
At the new steady state.
The surface now emits the combination.
What is the power?
More importantly what is the temperature?
Bryan,
The surface will not emit the combination. It will emit a new BB spectrum for the new temperature
Tom W.
Thanks for your post.
I’m trying to answer SoDs oft repeated question of what happens to say 200W/m2 of back radiation when it lands on the Earth Surface.
I think SoD thinks that the 200W/m2 simply increases the temperature of the already warmer planet surface.
I think that the frequency spread is more important than the amplitude.
See example 4 where SoD works out the temperature of T1
with radiation of unspecified origin through a vacuum.
I think that the post from JanS September 15, 2010 at 10:37 am was accurate.
What do you think?
The maximum temperature produced by the -30c atmosphere ion the Earth surface is -30c at best.
Because of the Oceans it is never likely that the Earths temperature will drop as low as that.
So the answer to the question does the atmospheric radiation will raise the temperature of the planet surface is no it does not.
Bryan
Concerning the radiation of unspecified origin in the original thought experiment:
I don’t feel that the temp. of the radiating body is important and here is why. Whether or not the radiation is absorbed only depends on the absorbtivity (spell?) of the receiving surface. Absorptivity (got it right!) is general is dependent on wavelength. For most substances the absorptivity is pretty much constant over IR wavelengths. This is because of the high density of vibrational and (to some extent) rotational levels in this range. In heterogeneous substances you may even have a pretty good approximation of BB absorption in visible wavelenghts due to the large number of electronic levels. In short, we require a special set of circumstances for the specific spectrum of incoming radiation not to be absorbed by the receiving body.
It also occurs to me to mention that the temp. of the receiving body will not impact its ability to absorb radiation of whatever spectrum except in some very contrived materials and radiation sources. In other words the energy levels of the receiving body do not “fill up” in any earthly setting.
Bryan,
What have I said about the specific details of G&T that is contradicted in the text. Cite please.
While you’re at it, please quote chapter and verse from G&T where they actually falsify the atmospheric greenhouse effect.
You can only combine light sources that point in the same direction. A spectrometer pointed up only sees the atmosphere. One pointed down only sees the surface. Talking of somehow combining their spectra is meaningless.
No it doesn’t. A non-reflective opaque surface emits a black body spectrum with characteristics defined by the surface temperature and emissivity only. Any incident radiation is absorbed The power and temperature question cannot be answered without knowing the heat capacity of the surface. But I can give you an example calculation:
Take a well mixed layer of water 0.1 m deep at an initial temperature of 15 C. The heat capacity per m2 is then 418.7 kJ and the heat content is 15*418.7 = 6280.5 kJ/m2. If the surface of the water sees DLR at 200 W/m2, then the net radiative heat loss rate is 190 W/m2. The temperature and heat loss rate will drop over the course of an hour. There’s an analytic solution, IIRC, but I’ll do it numerically in 10 sec intervals because it’s quicker than looking up the solution. For simplicity I’ll use an emissivity/absorptivity of 1.00
The answer is that the temperature of the water at the end of one hour will be 13.40 C, the water surface will be radiating 381.5 W/m2 or a net loss rate of 181.5 W/m2. If the DLR is 300 W/m2, the temperature at the end of one hour would be 14.24 C, 386 W/m2, 86 W/m2 respectively. If the layer thickness is reduced to 0.01 m, for 200 W/m2 DLR, the temperature, gross and net radiation flux will be: 1.79C, 323.3 W/m2, 123.3 W/m2. Note that the shape of the spectrum of the DLR is not relevant as long as the absorptivity of the surface is constant for the full range of wavelengths of the DLR.
Bryan,
In your reply to Tom W. you miss the point again. Of course the surface cools when the sun goes down. Nobody disputes that. If all the surface ever saw was 200 W/m2 DLR, then it would indeed cool to a temperature of approximately -30 C. But it doesn’t because the sun is up for half the day. Over the course of a day the surface sees incoming radiation from the sun and the atmosphere and warms during the day and cools at night. But because it sees radiation from the atmosphere 24/7, it doesn’t cool as fast at night as it would if the atmosphere didn’t emit.
Your calculation would assumes that conduction and convection within the water has stopped.
If the water was deep enough I think after one hour the temperature would be little changed.
Bryan,
The same principle applies if the surface saw only sunlight. That’s 235 W/m2 average after correcting for albedo, or a surface temperature of -19.5 C. But it doesn’t. It sees ~500 W/m2 for atmosphere + sunlight, loses ~100 W/m2 by convection and ~400
W/m2 by radiation.
Tom W.
….It also occurs to me to mention that the temp. of the receiving body will not impact its ability to absorb radiation of whatever spectrum except in some very contrived materials and radiation sources. In other words the energy levels of the receiving body do not “fill up” in any earthly setting…..
I’m afraid your simply wrong here Tom.
An irradiated object can never reach a higher temperature than the source causing the radiation.
Reason; Since radiation travels in straight lines, the irradiated object would become the net radiator once its temperature reached that of the source.
When you read the literature about focusing an extended source we find even getting to the source temperature quite difficult.
Therefor the post from JanS September 15, 2010 at 10:37 am was accurate.
DeWitt Payne
” You have yet to explain the relevance of Woods’ experiment to the reality of the atmospheric greenhouse effect.”
Well in a glasshouse where there is an undisputed “greenhouse effect” the radiative effect is so small as to be almost unmeasurable.
There is very little indication that this effect could be responsible for a 33C rise in average temperature.
Irrelevant. No one ever said that the mechanism of the atmospheric greenhouse effect is the same as that for a greenhouse used for growing plants.
Bryan,
I went back to G&T and read as much as I could stomach without becoming nauseous or throwing something at the computer screen. Their entire case rests on semantics. They don’t like the terminology so they misinterpret what they see and then claim it doesn’t exist. Section 3.7.2 A note on “radiation balance” diagrams is indicative:
That is a straw man argument. Those sections do not, in fact, explain why the arrows cannot be interpreted as integrated intensities. Just because arrows are used doesn’t mean that the quantity indicated is a vector. They only talk about the definition of a specific intensity and how you integrate it. There’s also irrelevant hand waving about the inapplicability of the Stefan-Boltzmann equation to the atmosphere, but nobody ever said that the real atmosphere was gray except in a toy model. There’s nothing there that says you can’t integrate over the observed spectrum and get an integrated intensity or use a some form of pyrometer to actually measure it. There’s also no comprehension that the numbers in the energy balance diagrams represent measurements of physical quantities made in the field.
If you take their argument at face value an IR thermometer couldn’t work. What’s actually being measured by an IR thermometer, btw, is not the temperature, it’s the integrated IR intensity. That’s converted to temperature using something similar to the Stefan-Boltzmann equation corrected for the wavelength range covered by the instrument. I’m sure they would say I’m misinterpreting them. But since they go out of their way to misinterpret everyone else, I don’t see the problem.
DeWitt Payne
What have I said about the specific details of G&T that is contradicted in the text. Cite please.
Your comments have been of a generalised nature such as
“There has been little serious criticism of G&T because working scientists don’t consider it to be worth their time. A criticism would be normally rejected by most serious journals as a waste of space. Valueless papers are almost always ignored, not rebutted. That’s why you only see this being discussed in the blogosphere.”
“There is so much hand-waving, straw men and irrelevant arguments in G&T that it’s difficult to know what they are saying. Whether they say the atmosphere is radiating or not, as near as I can tell, they say that it doesn’t make a difference to the surface temperature.”
This was the only specific question I could find
“None of that addresses the question I asked you. I’ll rephrase: Where do G&T show that the surface of the Earth receives sufficient energy from direct sunlight to balance the energy lost by radiation and convection? I haven’t been able to find it. It is precisely this point that is successfully addressed by atmospheric radiation, i.e. the greenhouse effect.”
The various radiative calculations used for the Earth Surface and Atmosphere are disputed throughout the text.
Best to read the paper yourself and then come up with specific criticism.
Bryan,
Look at what I said:
“…the temp. of the receiving body will not impact its ability to absorb radiation…”
I didn’t say anything about emitting radiation and it’s the balance of the two that determines final temp. You come so close in your comment when you say that it will become a net radiator.
It actually wouldn’t be net anything at equilibrium temperature. An object that is a net radiator is cooling down (unless it has an internal power source). An object that is a net absorber is heating up. Yes the temp. of the sending body is the max. temp that the receiving body could achieve but it has nothing to do with the spectra. It only has to do with the fact that a body will not be a net radiator (without an internal power source).
Bryan,
As I said, Wood’s experiment looks only daytime heating. The Romans in Palestine made ice in the desert by making an inverse greenhouse. A pan of water was put in a hole in the ground with straw insulation on the sides and bottom. Sunlight was blocked during the day by polished shields and straw insulation. Then the shields were removed and the water exposed directly to the night sky. A thin layer of ice formed even though the outside air temperature was well above freezing. Do you think that would still happen if the hole were covered at night by an LW IR opaque cover like ordinary glass? The atmospheric greenhouse effect is all about reducing heat loss at night, not raising the temperature during the day. In fact, one of the signatures of ghg warming is that the difference between max and min daily temperatures decreases over time because the min temperature increases faster than the max.
That should be: …one of the signatures of warming caused by an increase in ghg’s is…
Bryan,
As I’ve said before, many of the disputed figures, like the energy balance diagram cited above, are based on measurements, not calculations.
DeWitt Payne
If you follow the link below you can see that this Pyrgeometer from the most prestigious supplier is still giving huge errors
Fairly dodgy procedures are suggested which they say will allow them to back correct previous false readings.
The instrument is calibrated assuming that SB equation is applicable for the atmosphere.
The instruction manual has the SB equation included in case the operator gets it wrong.
Call me cynical if you like but I would need more evidence than this to suggest we dislocate the world economy.
http://huey.colorado.edu/LTER/datasets/meteorology/pyrgeometer.html
Bryan from September 20, 2010 at 10:25 pm:
If that was the only source of course.
However, as demonstrated in The First Law of Thermodynamics Meets the Imaginary Second Law a colder source can increase the temperature of a warmer body.
You haven’t demonstrated what is wrong with the model yet, and never will.
It’s very simple. If you add energy to a body it will increase in temperature until a new equilibrium is reached where energy in = energy out.
Anything different would violate the first law of thermodynamics.
Basic Thermodynamics
If you have Body 1 at 300K and it is in proximity to Body 2 then if and only if Body 2 has no other source of energy then obviously Body 2 must reach an equilibrium below 300K.
Bryan’s law:
If Body 1 is at 300K and it is in proximity to Body 2 then it can affect Body 2’s temperature while Body 2 is below 300K. But no matter how much energy Body 2 is receiving from other sources Body 1 will instantly stop affecting Body 1’s temperature after Body 2 gets above 300K. We haven’t worked out the mechanism how this happens yet, but watch this space.
Well the only choices I can think of for ensuring Bryan’s law is true are:
1. The first law of thermodynamics is flawed.
2. The instant the temperature of Body 2 goes above the temperature of Body 1 radiation from Body 1 is magically diverted somewhere else. Unfortunately Body 1 doesn’t stop radiating – because we can measure it. Unfortunately Body 2 doesn’t reflect it – because we can measure that this doesn’t happen. Magic.
It’s a basic misunderstanding – you are applying the result from a simple problem to a more complex one without understanding why the simple problem has the result that it does.
SoD
What you are suggesting seems to me like;
Lets say 200W/m2 from the atmosphere at -30c lands on the Earth Surface.
Lets use energy units 200Joules per second on one square metre.
This energy is totally absorbed and results in an infinitesimal INCREASE in temperature of the surface.
Although in practice this would be impossible to measure you think that if you could it would show this exact value
Lets call this infinitesimal increase in temperature dT
dT consists of 200J of photons centred around 11.91um peak.
After absorption this 200J is transformed into 200J of energy at 10.91um peak
I would contend that although this complies with Ist law it contradicts the 2nd law of thermodynamics.
SoD
The planet surface temperature used above was 15C
SoD correction writing this in haste!
Wavelength peak at 15C is 10.05um
Bryan:
Explain how it contradicts the 2nd law of thermodynamics.
scienceofdoom
The “quality” of the energy has been improved.
200J of photons centred around 11.91um have been transformed into 200J centred about 10.05um.
The entropy has decreased which is forbidden by the 2nd Law.
Bryan,
Think about this – heat something up in the microwave – a monochromatic source of radiation (12cm). Say it heats to 100C. If the object heated is approx. a BB then the resulting spectrum emitted by the object will have a peak at about 8um.
Tom W.
See Jans comment on microwaves
…..”In a microwave oven molecules having a dipolar moment, such as water molecules, are made to oscillate and thereby absorb the microwave energy. This has nothing to do with blackbody radiation. Other substances (without dipolar moment) will not absorb microwave radiation.”…….
I don’t think going down that road will be helpful.
Obviously a machine
Obviously not a blackbody radiation emitter either.
Also a refrigerator which does extract heat from a colder source and dumps it at a higher temperature requires a heat pump to function.
My point is not that a microwave is a BB emitter, it’s monochromatic. Only that the heated body (if IT is a BB) will emit a BB spectrum that does not depend on the spectrum of the incoming radiation. Once the energy goes in it gets spread out among all the different degrees of freedom of the system and redistributed according to the Planck distribution.
I just saw your comment about the “quality” of a spectrum. Do you mean entropy?
@Bryan,
“The “quality” of the energy has been improved.
200J of photons centred around 11.91um have been transformed into 200J centred about 10.05um.
The entropy has decreased which is forbidden by the 2nd Law.”
By “quality” and the rest of your comment it seems that you mean to say that the lower T radiation field (11.91 um) has higher entropy than the higher T radiation field (10.05um).
This took me quite a bit of thought.
Check out this link:
Click to access PlanckStory.pdf
See that dS/dU = 1/T > 0
where U is the mean energy of a radiation field mode. The entropy increases with increasing energy of the field. This makes sense in the following way. Low frequency modes are always available but higher energy ones will not be at a given temperature. But as the total E increases these higher energy modes become available and therefore there are more ways for the energy to be distributed which is the definition of entropy.
So I think the higher T field will have a higher entropy and there is no Second Law problem here. Not that there necessarily would be anyway since we have neglected the entropy changes of the emitting and receiving bodies.
Tom W
You have it back to front here.
High temperature source has lower entropy state than low temperature sink.
That why a hot cup of coffee will gradually cool of its own accord in a cooler room.
However it will never of its own accord extract heat from the cooler room to increase its temperature.
Entropy of the universe is always increasing never decreasing as energy gets degraded into lower quality forms.
No. Hot things – in general have more entropy than cool things. Entropy is not a substance that flows from sources to sinks. A hot cup of coffee in a cold room will not get warmer at the room’s expense because the decrease in entropy of the room would not be offset by the increase of entropy of the coffee.
Remember dS/dU = 1/T. This is a general thermodynamic principle (one that I didn’t immediately remember until I looked at the article on Planck). Since 1/T is always positive it means that entropy always increases with increasing internal energy. If this sometimes seems counterintuitive, remember that you always need to take the whole system into account.
A hot body that is cooling is decreasing in entropy, but this is more than offset by the increase in entropy of the rest of the universe.
Bryan,
Your pyrgeometer correction link is dated in 1996. EBEX was in 2000. The errors listed are small compared to the measured quantity, on the order of 10%, and point 1 in the link doesn’t apply to EBEX at all as the temperature was well above -25 C. Point 2 only applies during the day and the experimenters and manufacturer were aware of the problem by then. The same goes for point 3 about the battery voltage. A 10% error is important for modeling, but it is too small to invalidate the existence of the effect. Atmospheric radiation would still be a major part of the surface energy balance as required by the greenhouse effect.
Cite please. The instrument measures the total incident radiation between 3.5 and 50 micrometers. It may be calibrated using a black body source, but the measurement in no way depends on the observed source having a black body spectrum. The output is in W/m2, not temperature.
That’s policy, which is another subject entirely. What we’re discussing here is why you don’t believe an atmospheric greenhouse effect exists at all.
DeWitt Payne
I simply picked this one up first at random from Google.
A few months ago I spent some time on Google looking into pyrgeometers and their problems.
Try it yourself if you have the time.
What I found were a number of citations of people claiming solutions for correcting the large errors which seem to be related to this instrument.
They are generally fitted with an offset knob to get them reading “properly”.
I have a picture of a young researcher (mortgage and young family) charged to take readings to confirm some global warming hypothesis.
Would it not be human nature to use the offset and guided by SB equation get some “acceptable” answers.
I had a dialog with a contributor to Physics Forums who assured me that she regularly used the SB equation for calculations of the sky and clouds.
To my mind there is a danger of a circular argument developing here.
@Bryan
I defer to your greater knowledge of pyrgeometers but I think just about every precision lab instrument I’ve ever seen has an offset knob. Especially useful if you don’t trust other people to do your calibration.
Bryan,
Here is a link to the pyrgeometer calibration method used by ARM:
Click to access reda-i.pdf
As I expected, they use a black body source at different temperatures to calibrate the instrument. They also use different instrument temperatures in the process. The temperatures used to calculate radiant energy from the S-B equation are temperatures measured in the instrument, Td, the dome temperature and Tr, the receiver temperature, which is calculated from Tc, the case temperature.
You can always calculate a brightness temperature from radiant power using the S-B equation. The two are interchangeable. That doesn’t make the brightness temperature correspond to any real temperature. It’s just a convention.
I went outside last night and pointed my $50 IR thermometer at the sky. The temperature reading was 12.5 F = -16.25 C = 256.95K That corresponds to a radiant power of 247 W/m2 or about what you’d expect. The indicated temperature does behave as the S-B equation because it’s calculated from radiant power using the S-B equation, not because the radiant power of the atmosphere can be calculated from a single measured temperature using the S-B equation. As usual, G&T stand reality on its head.
Your paranoia is showing. Besides, you’d have to stand next to the instrument 24/7 and twiddle the knob. It would be easier to just completely fake the data. This sort of thing does happen, the Korean stem cell guy is an example, but it doesn’t happen often because you will eventually get caught when someone somewhere tries to replicate your results.
Bryan said, September 21, 2010 at 7:23 am:
I asked:
And Bryan responded:
I don’t think you understand entropy. Perhaps it will be easier to explain if I write an article..
Tom W
In the example I gave to SoD the process suggested that 200J of longer wavelength photons could be transformed into 200J of shorter wavelength photons.
This increase in the “quality of the energy” is forbidden by the 2nd Law.
This example once is a bit artificial since we cannot look at the heat transfer in two separate parts.
However this is what SoD wants us to do.
This is OK since this discussion is itself based on a not fully physically defined Ex4,5,6.
“What if” scenarios can be educationally useful.
My combining of the two spectrums was intended similarly.
DeWitt Payne used a very good laser analogy in a Post on the Air Vent site on CO2 thermalisation.
SoD is interested about what happens to the energy of the photons absorbed by the Warmer Earth.
However the answer is not what he anticipated.
ENERGY DEGRADATION.
The energy that enters the Earth system—not only solar radiation but also geothermal and tidal energy—ultimately leaves the system. As shown by the second law of thermodynamics, however, the energy that departs the Earth system will be in a degraded form compared with the energy that entered it.
In a steam engine, water in the form of steam goes to work to power gears or levers. In the process, it cools, and the resulting cool water constitutes a degraded form of energy. Likewise, the ashes that remain after a fire or the fumes that are a by-product of an internal combustion engine’s operation contain degraded forms of energy compared with that in the original wood or gasoline, respectively. In the same way, Earth receives short-wavelength energy from the Sun, but the energy it radiates to space is in a long-wave length form.
All physical bodies with a temperature greater than absolute zero emit electromagnetic energy in accordance with their surface temperatures, and the hotter the body, the shorter the wavelength of the radiation. The sunlight that enters Earth’s atmosphere is divided between the visible portion of the spectrum and the high-frequency side of the infrared portion. (Note that the Sun emits energy across the entire electromagnetic spectrum, but only a small part gets through Earth’s atmospheric covering.) Earth, with an average surface temperature of 59°F (15°C), is much cooler than the Sun, with its average surface temperature of about 10,000°F (5,538°C). The radiation Earth sends back into space, then, is on the low-frequency, long-wave-length side of the infrared spectrum.
Read more: http://www.scienceclarified.com/everyday/Real-Life-Earth-Science-Vol-2/Energy-and-Earth-How-it-works.html#ixzz10Fh1BvD0
Bryan,
As an alternative to the microwave oven example, suppose that 200J is delivered by a 10.6 micrometer CO2 laser beam focused to a very small area. Are you saying that any material at the focal point could not emit photons at a wavelength shorter than 10.6 micrometers? That would seem to fly in the face of the fact that CO2 lasers can cut steel.
But there is no violation of the second law even if the 200 J comes from a lower temperature black body because the entropy change has to include the whole system, which means whatever is providing heat to the the other black body. For the Earth, that heat source is the sun at a temperature of greater than 5,000 K, so entropy still increases for the system as a whole if the surface of the Earth warms.
DeWitt Payne
….”that heat source is the sun at a temperature of greater than 5,000 K, so entropy still increases for the system as a whole if the surface of the Earth warms.”…..
I am always reluctant to throw the Sun into a discussion unless it is absolutely necessary.
Of course the entropy loss of the Sun is so enormous that it will drown out any proposition good or bad.
Just think of the nonsense that could be argued where entropy decreases and the retort is “well not, if we include the Sun”
…..”suppose that 200J is delivered by a 10.6 micrometer CO2 laser beam focused to a very small area.
In the example I followed with SoD the two objects are in normal thermal contact.
With the usual thermodynamic rules applying.
I think the introduction of a laser here is unhelpful here.
It can only muddy the water and will not help tease out the underlying physics of the problem.
Bryan,
And if doubling CO2 causes the average surface temperature of the Earth to increase from 288 to 290K, it will still be on the long wavelength side of the IR spectrum. As observed from far out in space, the Earth will still radiate ~235 W/m2 corresponding to a brightness temperature of ~254 K. The spectrum will change a little, but the total power won’t. I don’t see the problem.
The example I posted to SoD covered the fate of the radiation from the colder atmosphere to the warmer Earth.
It appears that it cannot increase the temperature of the surface even infantesimally.
Bryan,
Heat transfer basics. Net heat flow between two parallel planes at different temperatures T1 and T2 is equal to sigma*(T1^4-T2^4). If net heat flow is fixed then T1 depends on T2. Say the heat flow is 235 W/m2 and T2 = 0K. T1 = 253.7K Now let’s increase T2 so that plane is radiating 200 W/m2. T2 = 243.7 K. If T1 doesn’t increase then the net heat flow is 35 W/m2 which violates the boundary condition of net heat flow = 235 W/m2. At a net heat flow of 235 W/m2 and T2 = 243.7, T1 = 296 K. That’s a bit more than infinitesimal. The energy radiated by the second plane does not warm the first plane. That is accomplished by the energy flowing into the first plane that has to be radiated away.
Bryan,
To get even more basic: Why does a blanket keep your skin warmer than no blanket? The blanket itself is cooler than your body so it can’t possibly warm you according to your logic.
DeWitt Payne
….”To get even more basic: Why does a blanket keep your skin warmer than no blanket? The blanket itself is cooler than your body so it can’t possibly warm you according to your logic.”……
Lets say a bronze statue at say 80c placed in a room at say 20c with and without a blanket, the respective temperature time graphs of statue would both show falling temperature.
The rate of temperature fall of the blanket covered statue would be less than that of the bare statue.
All three methods of heat transfer would be from the higher temperature to the lower temperature at all times.
That’s a straw man argument. There is no heat source in the bronze statue as there is in your body from food metabolism or the sunlight absorbed by the surface of the earth. In your 20 C room, we have a bronze statue containing an electrical heater running at constant power. With no blanket, the average temperature of the surface of the statue is 30 C. Now put a blanket over the statue but don’t change the power of the heater. What happens to the average temperature of the statue? It goes up. How much depends on the insulation characteristics of the blanket. But at steady state, the power flow from the surface of the blanket into the room will equal the power flow from the heater.
DeWitt Payne
Thanks for the link to the pyrgeometer calibration method used by ARM.
It will be interesting to see if their accuracy is confirmed over time.
However much of the AGW argument unfortuanately does not rest on experimental verification.
Speculative calculation seems to play an important part.
Take what would appear to be a very simple problem the average radiation from the Earth.
K&T(IPCC3) gives it as 390W/M2
This is exactly the value calculated from SB for a perfect black body at 15C temperature.
By IPCC4 it had been pointed out from a statistical analysis of average temperatures that it would need to be given as 396W/m2
Note, not by experment, but again by calculation and assumption of perfect black body properties!
The concept of radiative balance is also unphysical.
There is no reason to believe that in any ten years period of the planets history the Earth was in perfect balance with the Sun.
Rather the norm I think is for the Earths surface temperature to slowly rise and fall.
There is indeed no requirement for radiative balance. But there is a requirement for conservation of energy. If there is a radiative imbalance, the heat content will go up or down depending on the direction of the imbalance. Temperature is proportional to heat content so it will vary as well. Probably the best direct measure of heat content is ocean heat content, as the oceans are the largest reservoir by far. There is no question whatsoever that the ocean heat content has increased over the last 100 years. There are multiple lines of evidence this is true. The exact value of the rate and total amount of increase is in question, but not the sign. There is no source of energy other than the sun that is of sufficient magnitude to account for the increase. There is also no evidence that the solar radiance has changed sufficiently to account for the increase. That leaves the characteristics of the emission and absorption of the Earth as the primary cause. It’s possible that some of the increase could have been caused by a reduction in planetary albedo. It’s also possible that some of the increase has been caused by an increase in atmospheric absorption of LW IR from increased CO2 and other greenhouse gases.
DeWitt Payne
…”There is indeed no requirement for radiative balance. But there is a requirement for conservation of energy. If there is a radiative imbalance, the heat content will go up or down depending on the direction of the imbalance. Temperature is proportional to heat content so it will vary as well.”…
I agree with the above but my point is the rise of 0.7C over the last 150 years is perhaps typical variation rather than something unusual.
[…] Heat Transfer Basics – Part Zero a slightly off-topic discussion about the “greenhouse” effect began. One of our most […]
Bryan:
Check out The Real Second Law of Thermodynamics
[…] Conduction is driven by temperature differences. Once you establish a temperature difference you inevitably get heat transfer by conduction – for example, see Heat Transfer Basics – Part Zero. […]
[…] A is the area, k is the conductivity (the property of the material) and q is the heat flow. (See Heat Transfer Basics – Part Zero for more on this […]
[…] January 2, 2011 by scienceofdoom A while ago we looked at some basics in Heat Transfer Basics – Part Zero. […]
Which do you use visualize program? Could you give software’s name above ? i need it
Matlab
Thank you BEP
If you don’t have Matlab, Octave ( http://www.gnu.org/software/octave/ ) is an open source program that is supposed to run most Matlab programs.
[…] open system). A warming atmosphere will heat the oceans. A more technical explanation can be found here. Technical explanation of a cooler body heating a warmer body based purely on radiative […]
Scienceofdoom: In Example One, you used delta T=40 C in calculation, and k=0.19W/m.K. Shouldn’t delta T=40+273=313K?
Nope. ΔT is 50°C-10°C = 40°C or 323K – 283K = 40K.
sir, we have one vacuum oven we are maintaining 150 degree inner chamber but it is outer body transfer heat almost 40 degree please give me suggestion.
and then it and
The thermal conductivity of such liquids is given in Table A.7.
● Note that the values of k are much larger than those of the nonmetallic liquids.
Insulation Systems
● Thermal insulations consist of low thermal conductivity materials combined to achieve an even lower system thermal conductivity.
● In conventional fiber-, powder-, and flake-type insulations, the solid material is finely dispersed throughout an air space.
● Such systems are characterized by an effective thermal conductivity, which depends on the thermal conductivity and surface radiative properties of the solid material, as well as the nature and volumetric fraction of the air or void space.
● Heat transfer through any of these insulation systems may include several modes: conduction through the solid materials; conduction or convection through the air in the void spaces; and radiation exchange between the surfaces of the solid matrix.
● The effective thermal conductivity accounts for all of these processes, and values for selected insulation systems are summarized in Table A.3.
2.2.2 Other Relevant Properties
● Thermophysical properties include two distinct categories:
○ transport properties and
○ thermodynamic properties.
● The transport properties include the diffusion rate coefficients such as k, the thermal conductivity (for heat transfer), and ν, the kinematic viscosity (for momentum transfer).
● The thermodynamic properties pertain to the equilibrium state of a system. Density (ρ) and specific heat (cp) are two such properties used extensively in thermodynamic analysis.
The Volumetric Heat Capacity (ρcp)
● The product ρcp (J/m3.K), commonly termed the volumetric heat capacity, measures the ability of a material to store thermal energy.
● Many solids and liquids, which are very good energy storage media, have comparable heat capacities (ρcp > 1 MJ/m3.K).
● Because of their very small densities, gases are poorly suited for thermal energy storage (ρcp 1 kJ/m3.K).
● Values of densities and specific heats are provided in tables of Appendix A for a wide range of solids, liquids, and gases.
The Thermal Diffusivity (α)
● In heat transfer analysis, the ratio of the thermal conductivity to the heat capacity is an important property termed the thermal diffusivity α, which has units of m2/s:
● It measures the ability of a material to conduct thermal energy relative to its ability to store thermal energy.
● Materials of large α will respond quickly to changes in their thermal environment, while materials of small α will respond more slowly, taking longer to reach a new equilibrium condition.
A Final Note
● The accuracy of engineering calculations depends on the accuracy with which the thermophysical properties are known.
● Selection of reliable property data is an integral part of any careful engineering analysis.
● Study Example 2.1, pp. 68-69.
2.3 THE HEAT DIFFUSION EQUATION
Major Objective in a Conduction Analysis
● A major objective in a conduction analysis is to determine the temperature field (distribution) in a medium resulting from conditions imposed on its boundaries.and prof. dr. mircea orasanu with prof horia orasanu
Importance of Know
If there are temperature gradients, conduction heat transfer will occur across each of the control surfaces.
● The conduction heat rates perpendicular to each of the control surfaces at the x, y, and z coordinate locations are indicated by the terms qx, qy, and qz, respectively.
● The conduction heat rates at the opposite surfaces can then be expressed as a Taylor series expansion where, neglecting higher order terms,
(2.11a)
(2.11b)
(2.11c)
● In words, Eq. (2.11a) simply states that the x component of the heat transfer rate at x + dx is equal to the value of this component at x plus the amount by which it changes with respect to x times dx.
Thermal Energy Ge
Partial differential equations arise whenever the function to be found varies with more than one independent variable. In fluid mechanics, solid mechanics, heat transfer and mass transfer, the variables of interest (velocity, stress, temperature, and concentration, respectively) will vary with three spatial dimensions and with time. Take, for example, a one-dimensional time-dependent mass transfer problem. The concentration, depends on both , the spatial dimension, and time. The differential equation is written in terms of partial derivatives ( ) as opposed to total derivatives ( ). The operator means that the derivative with respect to is taken while all other variables are held constant.
Partial Differentiation:
If a variable, such as concentration, depends on both space and time, one must evaluate the total derivative to find a rate of change. If and are the variables of interest, then the total derivative is:prof dr mircea orasanu and prof horia orasanu As a specific example, let the temperature field increase linearly with and sinusoidally in time so that . Let the particle move with constant velocity in the direction, so that . The correct expression for the rate of change of temperature is:in the temperature field with time, i.e. . The partial derivative of this field with respect to is zero, so the first term on the right hand side of Eq. 2 is zero. However, there must be a change in temperature because the particle is moving into a region of higher as increases. The rate of that change is the second term on the right hand side of Eq. 2, which becomes:
The first term on the right hand side is the rate of change of temperature at a given point caused by the changing temperature field. The second term is the rate of change in temperature that occurs because the particle moves along the direction and temperature increases with . To demonstrate clearly that the second term is necessary, consider what happens when there is no change
The rounded operator ( ) is a necessary element of this equation and cannot be substituted for the straight operator ( ), which would otherwise mean ”the rate of change of along a given pathway.” The second term on the right hand side of Equation 1 cannot be evaluated until the pathway is described. Thus, the curved operator ( ) is used to denote that the derivative is to be taken with respect to the variable while the variable is held constant.
To gain physical insight into the concept of partial differentiation, consider the temperature to which a particle is exposed as it moves in the x-direction. Assume that the ambient temperature changes with both the x position and time, as in Figure 1. The electric field changes along the x-direction, but also changes with time, so that . One wishes to find the rate at which the temperature surrounding the particle changes as the particle moves through a path .
As expected, the temperature changes at a rate proportional to the velocity of the particle.
A sample partial differential equation for mass transfer is:
To solve an ordinary differential equation, one seeks specific functions, such as or that satisfy the equation. In contrast, solutions to partial differential equations are determined by the argument of the functional form. For example, in Equation 3, any function whose argument is
This result and Equation 7mbine to yield:
It follows that is the inverse Fouirer transform of the equation above.
.
Once this inverse Fourier transform is evaluated, the boundary conditions for and can be applied.
The two-dimensional steady state heat equation for a thin rectangular plate with time independent heat source shown in Figure 3.9-1 is the Poisson’s equation
= f(x,y) (3.9-1)
The heat equation for this case has the following boundary conditions
u(0,y) = g1(y), u(a,y) = g2(y), 0 < y < b
u(x,0) = f1(x), u(x,b) = f2(x), 0 < x < a
The original problem with function u is decomposed into two sub-problems with new functions u1 and u2. The boundary conditions for the sub-problems are shown in Figure 3.9-2. The function u1 is the solution of Poisson’s equation with all homogeneous boundary conditions and the function u2 is the solution to Laplace’s equation with all non-homogeneous boundary conditions. The original function u is related to the new functions by
u = u1 + u2
The function u2 is already evaluated in section 3.8 where
u2(x,y) = Ansin sinh + Bnsin sinh
Cnsin sinh + Dnsin sinh
To complete the solution of the Poisson’s equation for the problem in Figure 3.9-1, we only need to treat Poisson’s equation with zero boundary condition shown in Figure 3.9-2.as prof dr mircea orasanu and prof horia orasanu then
We have solved the linear homogeneous PDE by the method of separation of variables. However this method cannot be used directly to solve nonhomogeneous PDEA macroscopic object can often be approximated by a "particle", which has a mass and position in space. A particle has one physical parameter, its mass, and three translational degress of freedom because it can move in 3-dimensional space.
The equations of motion of a particle can be generalized to a system of particles. Such a system is defined by mass parameters, and has translational degrees of freedom. Its configuration at any time can be represented by points in 3-dimensional space, or by a single point in dimensional configuration space.
When the size and shape of a macroscopic object matters, it can often be approximated by a "rigid body". A rigid body is a system of particles in which every pair of particles has fixed relative displacement. This is an approximation because the smallest parts of objects are atoms which do not have definite positions according to quantum theory. It is also an approximation because a change in position of one particle cannot affect the position of another particle instantaneously according to the theory of relativity.
Suppose that the rigid body is made of particles or "atoms", how many degrees of freedom does it have, and how many physical parameters are needed to describe it? Its location and orientation are completely fixed by specifying the positions in space of any three non-collinear particles. A rigid triatomic molecule, which can translate and rotate but not vibrate, has 6 degrees of freedom, 3 translational and 3 rotational. Therefore a rigid body also has 6 degrees of freedom. The configuration . The mechanics of rigid bodies was developed in great detail by Leonhard Euler in his book "Theoria motus corporum solidorum seu rigidorum". He introduced many important concepts including moment of inertia and Euler angles. His student Lagrange later applied the Lagrangian method to rigid body motion.
Space-Fixed and Body-Fixed Coordinate Systems
A rigid body can move in space like a free particle, and can rotate, tumble, spin and twist in three indpendent directions at the same time! To simplify this complex motion it is very useful to introduce two coordinate systems.
The space-fixed coordinate system is chosen in an inertial reference frame in which Newton's equations of motion are valid. These equations are exactly the -body equations of Topic 1.
The body-fixed coordinate system is an origin and three orthogonal axes (unit vectors) that are fixed to the body and rotate, tumble, spin and twist along with it. This is definitely not an inertial coordinate system, so Newton's equations must be modified by non-inertial accelerations, centripetal, Coriolis, etc.
Dr. Butikov's Gyroscope Precession Applet
as
The KT energy balance diagram says that about 20 W/m2 of sensible heat moves from the surface to the atmosphere. Does the air above the ocean need to be colder for this sensible heat transfer to take place?
I have heard informally that surface air temperature (SAT) is currently warmer than SST in AOGCMs. That seems to be supported by this Figure at Climateaudit:
https://climateaudit.org/2009/06/15/tas-vs-tos/
AOGCMs predict that this difference will grow. See Figure 8 in Ritcher and Xie:
https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2008JD010561
If so, are AOGCMs violating the 2LoT?
Frank,
The 20 W/m² of sensible heat transfer is a global average, not a local average. I haven’t looked at the actual numbers, but my guess would be that much, if not most of the heat transfer over land is sensible and effectively all the heat transfer over the ocean is latent. You’re not going to get significant latent heat transfer over the Sahara or other deserts, for example.
Thanks for the replies, DeWitt. If there is no sensible heat transfer over the oceans, then there will need to be more than 50 W/m2 over land to compensate. Here is what KT97 said about sensible heat. Clearly they and the sources they cite believe sensible hear transfer occurs over the oceans.
“The remaining heat flux into the atmosphere from sensible heat is deduced as a residual from the condi- tion of the global energy balance at the surface,
SW – LW – LH – SH = 0.
Employing the surface budget values described above of a net shortwave flux of 168 W/m2, a net longwave flux of 66 W/m2, and a latent heat flux of 78 W/m2 implies a sensible heat flux of 24 W/m2. Sensible heat flux can also be determined from a bulk formula (e.g., Sellers 1965; Budyko 1982) and these values can be checked with more recent estimates over the oceans using the Comprehensive Ocean-Atmosphere Data Set (COADS) as analyzed by da Silva and Levitus (1994). The global ocean and annual mean sensible heat flux from the latter is 10 W/m2 versus 11 W/m2 from both Budyko and Sellers.
For the global and annual mean (including land) Budyko gives 17 W/m2 and Sellers gives 18 W/m2. However, the uncertainties in these values are consid- erable. Gleckler and Weare (1995) estimate errors in the bulk flux sensible heat of ±5-10 W/m2 with systematic errors of order 5 W/m2.”
https://journals.ametsoc.org/doi/pdf/10.1175/1520-0477%281997%29078%3C0197%3AEAGMEB%3E2.0.CO%3B2
Since land surfaces in sunlight in the absence of wind can get really hot (especially when they are dark), I agree with you and this article that sensible heat transfer over land may be greater than over the ocean. However, as best I can tell, the net flux is outward over both land and ocean.
The 2009 version of the global energy balance contains the amount of sensible heat transfer determined by reanalysis. Do re-analyses and AOGCMs disagree about the relationship between SST and SAT? Do the former attempt to describe the physics of heat transfer across the interface more fully?
https://journals.ametsoc.org/doi/pdf/10.1175/2008BAMS2634.1
I left out a key word in my reply, net, as in the net heat transfer. Sensible heat transfer could well be from air to water, but because the heat of vaporization of water is so large, the net transfer is from water to air. Think about blowing warm dry air over cool water. What happens? Both the air and the water cool. The lower temperature of the water shows that the net transfer of energy is from water to more humid air even though there may well be some sensible heat transfer from air to water. There is no violation of the 2lot because the entropy still increases.
Net heat transfer is the reason I posted my question. Does each mechanism of heat transfer need to flow from hot to cold or just the total?
AFAIK, in the case of radiation, the physics of absorption and emission guarantees that the net flux is from hot to cold no matter what else is going on. (As long as LTE exists.)
AFAIK, in the case of sensible heat – heat transfer by collision? – on a molecular level, kinetic energy is on the average going to be transferred from faster-moving to slower-moving molecules. On macroscopic scale, that will be from hot to cold.
Convection merely moves matter and the internal energy it contains, including latent heat. I’m not sure how to analyze convection in terms of the second law.
All three major processes of heat transfer seem to be independently following their own rules. This suggests to me that each mechanism must transfer heat from hot to cold, but I’m be happy to understand why this view is incorrect.
DeWitt wrote: “Think about blowing warm dry air over cool water. What happens? Both the air and the water cool.”
I instinctively think about this at a molecular level. Do those molecules of water vapor escaping the surface of the ocean have some way of knowing that they are carrying latent heat? Yes, they expend energy escaping from surface tension and most are probably moving slower than the average of the neighboring gas molecules. That would mean they cool the air immediately above the surface. (Of course, loss of the most energetic water molecules at the surface cools the “skin layer” of the ocean, too) So, thinking at a molecular level, I reach the same conclusion as you do, except that the air blowing over the ocean doesn’t need to be “warm” and any relative humidity less than 100% will permit evaporation (or “net evaporation”, given that there is really a two-way flux).
Let’s compare a wind that is much colder than the ocean (say from Antarctica blowing north) with a wind that is much warmer than the ocean (say Santa Ana winds blowing from the California desert over the cold water off Southern California). Sensible heat transfer (assuming it is mediated by collisions) must be from warmer to cooler, from the ocean to the air off Antarctica and the air to the ocean off Southern California. In both cases, there will be evaporation that transfers latent heat to the atmosphere.
So far, all of these rationals seem to say the air above the ocean must be cooler than the ocean itself for sensible heat flow to occur.
So far, however, I’m only thinking in terms of heat flowing from hot to cold and not expressing my arguments in terms of entropy: dS = dq/T. And there is a deltaH and deltaS of evaporation. Formulating concepts in these terms is challenging for me.
I appreciates your posing questions Frank.
Change in latent heat is dependent on winds. Winds are dependent on temperature differences. And it looks like scientists have got it all wrong. Latent heat is increasing more than they thought. Why do model miss this?
According to Ritcher and Xie (thank you for the reference): “The muted precipitation increase and slowing of the tropical circulation constitute a consistent response to greenhouse gas (GHG) forcing among models but it is not obvious why the climate system should behave in this particular manner. In fact, recent observational studies by Wentz et al. [2007] and Allan and Soden [2007] suggest that the actual rate of precipitation increase might be significantly higher than what is simulated by climate models. Similarly, a recent analysis of surface heat flux using merged satellite and reanalysis data [Yu and Weller, 2007] indicates a rate of latent heat flux increase that is much higher than in the models. While these observational studies are still subject to measurement errors and natural variability in short data records (see Lambert et al. [2008] and Previdi and Liepert [2008] for a discussion of the results of Wentz et al.), it is important to achieve a physical understanding of the reasons behind the muted precipitation response in model simulations.”
NK: I don’t have time to read your references, but I do know that the precipitation data used by Wentz (2007) has been refined and the increase is now consistent with 2%/K. With 20/20 hindsight, accurately measuring the change in precipitation or temperature over several decades is challenging enough. Can we convincingly tell if the increase in precipitation is 2%/K or 7%/K?
After years of fooling myself, I finally realize that convection can’t address the fundamental issue in AGW – getting heat past rising GHGs in the upper troposphere and out to space. Convection can’t operate when the upper atmosphere is too warm compared with the surface. Relative humidity rises, suppressing evaporation, and precipitation doesn’t increase at 7%/K. If it did, 7%/K of -80 W/m2 of latent heat would be -5.6 W/m2/K. Planck feedback is only -3.2 W/m2/K. In the absence of massively negative cloud feedback, there is no way precipitation can rise 7%/K, because there is no way to get that much released latent heat out of the upper troposphere by radiative cooling to space. However, the difference between the 2%/K and 3%/K is 0.8 W/m2/K. That is enough to convert an ECS of 3.6 (-1.0 W/m2/K) into an ECS of 2.0 (-1.8 W/m2/K).
I’m still hoping for answers about whether thermodynamics requires average SAT to be lower than SST if sensible heat is being transported out the ocean.
Another reference (from 2010): Evaporation Change and Global Warming: The Role of Net Radiation and Relative Humidity
David J. Lorenz Eric T. DeWeaver Daniel J. Vimont
Not in line with the “refined” data.
But who knows? Perhaps all measurements are one huge mess.” Low accuracy in the estimation of moisture-exchange components and the need to improve old estimates and develop new evaporation and precipitation databases based on satellite data are noted.” From : Variability of Evaporation and Precipitation over the Ocean from Satellite Data
V. N. MalininEmail authorS. M. Gordeeva, 2018
Some numbers:
Vapor trend 1988 – 2012 (RSS global oceans 60S to 60N): 1,14 % pr decade.
Temperature increase 1988 – 2012 (woodfortrees) 0,4 deg C
Vapor trend 25 years is 2,85 %, and pr deg C it will be 7,1 % increase.
I know that this is quite unrealistic, but will 2% be more accurate?
Ref:
http://www.remss.com/research/climate/#Atmospheric-Temperature
“Over the oceans, we can monitor decadal-scale changes in the total amount of water vapor in the atmosphere using our merged water vapor product, derived from measurements made by SSM/I, SSMIS, AMSRE, and WindSat. ”
“The bottom panel shows the ratio of the vapor trend to the TLT trend. Climate models suggest that this ratio should be about 6.2%/K. All combinations of satellite dataset show larger ratio, suggesting that either the measurements show too much moistneing, or too little warming.”
So Clomate Models rule at RSS.
Frank,
Yes. Otherwise it violates the Second Law.
Why would it not be? Consider the North Atlantic and North Pacific in the winter. Air coming off the sea ice at -30°C or so is going to be a whole lot colder than the surface temperature of open water, which can’t be colder than the freezing point. Sure there are going to be places where the net sensible heat transfer is from air to water, as where the wind blows from heated land to the ocean. But on average, the absorption of solar radiation by the ocean and release at night is going to result in the ocean surface being warmer than the air above it.
as
are possible