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Probably many, most or all of my readers wonder why I continue with this theme when it’s so completely obvious..

Well, most people haven’t studied thermodynamics and so an erroneous idea can easily be accepted as true.

All I want to present here is the simple proof that thermodynamics textbooks don’t teach the false ideas circulating the internet about the second law of thermodynamics.

So for those prepared to think and question – it should be reasonably easy, even if discomforting, to realize that an idea they have accepted is just not true. For those committed to their cause, well, even if Clausius were to rise from the dead and explain it..

On another blog someone said:

Provide your reference that he said heat can spontaneously flow from cold to hot. And not from a climate ‘science’ text.

I had cited the diagram from Fundamentals of Heat and Mass Transfer by Incropera and DeWitt (2007). It’s not a climate science book as the title indicates.

However, despite my pressing (you can read the long painful exchange that follows) I didn’t find out what the blog owner actually thought that the writers of this book were saying. Perhaps the blog owner never grasped the key element of the difference between the real law and the imaginary one.

So I should explain again the difference between the real and imaginary second law of thermodynamics once again. I’m relying on the various proponents of the imaginary law because I can’t find it in any textbooks. Feel free to correct me if you understand this law in detail.

The Real Second Law of Thermodynamics

1a. Net heat flows from the hotter to the colder

1b. Entropy of a closed system can never reduce

1c. In a radiative exchange, both hotter and colder bodies emit radiation

1d. In a radiative exchange, the colder body absorbs the energy from the hotter body

1e. In a radiative exchange, the hotter body absorbs the energy from the colder body

1f. This energy from the colder body increases the temperature compared with the case where the energy was not absorbed

1g. Due to the higher energy radiated from a hotter body, the consequence is that net heat flows from the hotter to the colder (see note 1)

The Imaginary Second Law of Thermodynamics

2a. – as 1a

2b.  – as 1b

2c.  – as 1c

2d.  – as 1d

2e. In a radiative exchange, the hotter body does not absorb the energy from the colder body as this would be a violation of the second law of thermodynamics

Hopefully everyone can clearly see the difference between the two “points of view”. Everyone agrees that net heat flows from hotter to colder. There is no dispute about that.

What the Equations Look Like for Both Cases

Now, let’s take a look at the radiative exchange that would take place under the two cases and compare them with a textbook. Even if you find maths a little difficult to follow, the concept will be as simple as “two oranges minus one orange” vs “two oranges” so stay with me..

Here is the example we will consider:

 

Radiant heat transfer

Radiant heat transfer

 

We will keep it very simple for those not so familiar with maths. In typical examples, we have to consider the view factor – this is a result of geometry – the ratio of energy radiated from body 1 that reaches body 2, and the reverse. In our example, we can ignore that by considering two very long plates close together.

E1 is the energy radiated from body 1 (per unit area) and we consider the case when all of it reaches body 2, E2 is the energy radiated from body 2 (per unit area) and we consider that all of it reaches body 1.

We define Enet1 as the change in energy experienced by body 1 (per unit area). And Enet2 as the change in energy experienced by body 2 (per unit area).

Radiation Exchange under The Real Second Law

E1 = εσT14; E2= εσT24 (Stefan-Boltzmann law)

Enet1 = E2 – E1 = εσT24 – εσT14

Enet2 = E1 – E2 = εσT14 – εσT24

Therefore, Enet1 = -Enet2

Under The Imaginary Second Law

Enet1 = – E1 = -εσT14

Enet2 = E1 – E2 = εσT144 – εσT24

Therefore, Enet1 ≠-Enet2 ; note that ≠ means “not equal to”

This should be uncontroversial. All I have done is written down mathematically what the two sides are saying. If we took into account view factors and areas then the formulae would like slightly more cluttered with terms like A1F12.

In the case of the real second law, the net energy absorbed by body 2 is the net energy lost by body 1.

In the case of the imaginary second law, there is some energy floating around. No advocates have so far explained what happens to it. Probably it floats off into space where it can eventually be absorbed by a colder body.

Alert readers will be able to see the tiny problem with this scenario..

What the Textbooks Say

First of all, what they don’t say is:

When energy is transferred by radiation from a colder body to a hotter body, it is important to understand that this incident radiation cannot be absorbed – otherwise it would be a clear violation of the second law of thermodynamics

I could leave it there really. Why don’t the books say this?

Engineering Calculations in Radiative Heat Transfer, by Gray and Müller (1974)

Note that if the imaginary second law advocates were correct, then the text would have to restrict the conditions under which equation 2.1 and 2.2 were correct – i.e., that they were only correct for the energy gain for the colder body and NOT correct for the energy loss of the hotter body.

Heat and Mass Transfer, by Eckert and Drake (1959)

Note the highlighted area.

Basic Heat Transfer, M. Necati Özisik (1977)

Note the circled equations – matching the equations for the “real second law” and not matching the equations for the “imaginary second law”. Note the highlighted area.

Heat Transfer, by Max Jakob (1957)

Note the highlighted section, same comment as for the first book.

Principles of Heat Transfer, Kreith (1965)

Note the highlighted sections. The second highlight once again confirms the equation shown at the start, that under “the real second law” conditions, Enet1 = – Enet2. Under the “imaginary second law” conditions this equation doesn’t hold.

Fundamentals of Heat and Mass Transfer, Incropera and DeWitt (2007)

Note the circled section. This is false, according to the advocates of the imaginary second law of thermodynamics.

And the very familiar diagram shown many times before:

 

From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)

From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)

 

Conclusion

There are some obvious explanations:

1. Professors in the field of heat transfer write rubbish that is easily refuted by checking the second law – heat cannot flow from a colder to a hotter body.

2. Climate science advocates have crept into libraries around the world, and undiscovered until now, have doctored all of the heat transfer text books.

3. (My personal favorite) Science of Doom is refuted because these writers all agree that net heat flows from the hotter to the colder.

4. Look, a raven.

Relevant articles – The Real Second Law of Thermodynamics

Notes

Note 1 – Strictly speaking a hotter body might radiate less than a colder body – in the case where the emissivity of the hotter body was much lower than the emissivity of the colder body. But under those conditions, the hotter body would also absorb much less of the irradiation from the colder body (because absorptivity = emissivity). And so net heat flow would still be from the hotter to the colder.

To keep explanations to a minimum in the body of the article in 1e and 1f I also didn’t state that the proportion of energy absorbed by each would depend on the absorptivity of each body.

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In the three part series on DLR (also known as “back radiation”, also known as atmospheric radiation), Part One looked at the network of stations that measured DLR and some of the measurements, Part Two reviewed the spectra of this radiation, and Part Three asked whether this radiation changed the temperature of the surface.

Very recently, on another blog, someone asked whether I thought “back radiation” heated the ocean. I know from a prominent blog that a very popular idea in blog-land is that the atmospheric radiation doesn’t heat the ocean. I have never seen any evidence for the idea. That doesn’t mean there isn’t any..

See note 1 on “heat”.

The Basic Idea

From what I’ve seen people write about this idea, including the link above, the rough argument goes like this:

  • solar radiation penetrates tens of meters into the ocean
  • atmospheric radiation – much longer wavelengths – penetrates only 1μm into the ocean

Therefore, solar radiation heats the ocean, but atmospheric radiation only heats the top few molecules. So DLR is unable to transfer any heat into the bulk of the ocean, instead the energy goes into evaporating the top layer into water vapor. This water vapor then goes to make clouds which act as a negative feedback. And so, more back-radiation from more CO2 can only have a cooling effect.

There are a few assumptions in there. Perhaps someone has some evidence of the assumptions, but at least, I can see why it is popular.

Solar Radiation

As regular readers of this blog know, plus anyone else with a passing knowledge of atmospheric physics, solar radiation is centered around a wavelength of 0.5μm. The energy in wavelengths greater than 4μm is less than 1% of the total solar energy and conventionally, we call solar radiation shortwave.

99% of the energy in atmospheric radiation has longer wavelengths than 4μm and along with terrestrial radiation we call this longwave.

Most surfaces, liquids and gases have a strong wavelength dependence for the absorption or reflection of radiation.

Here is the best one I could find for the ocean. It’s from Wikipedia, not necessarily a reliable source, but I checked the graph against a few papers and it matched up. The papers didn’t provide such a nice graph..

Absorption coefficient for the ocean - Wikipedia

Absorption coefficient for the ocean - Wikipedia

Figure 1

Note the logarithmic axes.

The first obvious point is that absorption varies hugely with the wavelength of incident radiation.

I’ll explain a few basics here, but if the maths is confusing, don’t worry, the graphs and explanation will attempt to put it all together. The basic equation of transmission relies on the Beer-Lambert law:

I = I0.exp(-kd)

where I is the radiation transmitted, I0 is the incident radiation at that wavelength, d is the depth, and k is the property of the ocean at this wavelength

It’s not easy to visualize if you haven’t seen this kind of equation before. So imagine 100 units of radiation incident at the surface at one wavelength where the absorption coefficient, k = 1:

Figure 2

So at 1m, 37% of the original radiation is transmitted (and therefore 63% is absorbed).

At 2m, 14% of the radiation is transmitted.

At 3m, 5% is transmitted

At 10m, 0.005% is transmitted, so 99.995% has been absorbed.

(Note for the detail-oriented people, I have used the case where k=1/m).

Hopefully, this concept is reasonably easy to grasp. Now let’s look at the results of the whole picture using the absorption coefficient vs wavelength from earlier.

Figure 3

The top graph shows the amount of radiation making it to various depths, vs wavelength. As you can see, the longer (and UV) wavelengths drop off very quickly. Wavelengths around 500nm make it the furthest into the ocean depths.

The bottom graph shows the total energy making it through to each depth. You can see that even at 1mm (10-3m) around 13% has been absorbed and by 1m more than 50% has been absorbed. By 10m, 80% of solar radiation has been absorbed.

The graph was constructed using an idealized scenario – solar radiation less reflection at the top of atmosphere (average around 30% reflected), no absorption in the atmosphere and the sun directly overhead. The reason for using “no atmospheric absorption” is just to make it possible to construct a simple model, it doesn’t have much effect on any of the main results.

If we considered the sun at say 45° from the zenith, it would make some difference because the sun’s rays would now be coming into the ocean at an angle. So a depth of 1m would correspond to the solar radiation travelling through 1.4m of water (1 / cos(45°)).

For comparison here is more accurate data:

From "Light Absorption in Sea Water", Wozniak (2007)

From "Light Absorption in Sea Water", Wozniak (2007)

Figure 4

On the left the “surface” line represents the real solar spectrum at the surface – after absorption of the solar radiation by various trace gases (water vapor, CO2, methane, etc). On the right, the amount of energy measured at various depths in one location. Note the log scale on the vertical axis for the right hand graph. (Note as well that the irradiance in these graphs is in W/m².nm, whereas the calculated graphs earlier are in W/m².μm).

From "Light Absorption in Sea Water", Wozniak (2007)

From "Light Absorption in Sea Water", Wozniak (2007)

Figure 5

And two more locations measured. Note that the Black Sea is much more absorbing – solar absorption varies with sediment as well as other ocean properties.

DLR or “Back radiation”

The radiation from the atmosphere doesn’t look too much like a “Planck curve”. Different heights in the atmosphere are responsible for radiating at different wavelengths – dependent on the concentration of water vapor, CO2, methane, and other trace gases.

Here is a typical DLR spectrum (note that the horizontal axis needs to be mentally reversed to match other graphs):

Pacific, Lubin (1995)

Pacific, Lubin (1995)

Figure 6

You can see more of these in The Amazing Case of Back Radiation – Part Two.

But for interest I took the case of an ideal blackbody at 0°C radiating to the surface and used the absorption coefficients from figure 1 to see how much radiation was transmitted through to different depths:

Figure 7

As you can see, most of the “back radiation” is absorbed in the first 10μm, and 20% is absorbed even in the first 1μm.

I could produce a more accurate calculation by using a spectrum like the Pacific spectrum in fig 6 and running that through the same calculations, but it wouldn’t change the results in any significant way.

So we can see that while around half the solar radiation is absorbed in the first meter and 80% in the first 10 meters, 90% of the DLR is absorbed in the first 10μm.

So now we need to ask what kind of result this implies.

Heating Surfaces and Conduction

When you heat the surface of a body that has a colder bulk temperature (or a colder temperature on the “other side” of the body) then heat flows through the body.

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.

The equation for heat transfer by conduction:

q = kA . ΔT/Δx

where k is the material property called conductivity, ΔT is the temperature difference, Δx is the distance between the two temperatures, and q is the heat transferred.

However, conduction is a very inefficient heat transfer mechanism through still water.

For still water, k ≈ 0.6 W/m.K (the ≈ symbol means “is approximately equal to”).

So, as a rough guide, if you had a temperature difference of 20°C across 50m, you would get heat conduction of 0.24 W/m². And with 20°C across 10m of water, you would only get heat conduction of 1.2 W/m².

However, the ocean surface is also turbulent for a variety of reasons, and in Part Two we will look at how that affects heat transfer via some simulations and a few papers. We will also look at the very important first law of thermodynamics and see what that implies for absorption of back radiation.

Update – Does Back-Radiation “Heat” the Ocean? – Part Two

Reference

Light Absorption in Sea Water, Wozniak & Dera, Atmospheric and Oceanographic Sciences Library (2007)

Notes

Note 1 – To avoid upsetting the purists, when we say “does back-radiation heat the ocean?” what we mean is, “does back-radiation affect the temperature of the ocean?”

Some people get upset if we use the term heat, and object that heat is the net of the two way process of energy exchange. It’s not too important for most of us. I only mention it to make it clear that if the colder atmosphere transfers energy to the ocean then more energy goes in the reverse direction.

It is a dull point.

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Following discussions about absorption of radiation I thought some examples might help illustrate one simple, but often misunderstood, aspect of the subject.

Many people believe that radiation from a colder atmosphere cannot be absorbed by a warmer surface. Usually they are at a loss to explain exactly why – for good reason.

However, some have the vague idea that radiation from a colder atmosphere has different wavelengths compared with radiation from a warmer atmosphere. And, therefore, that’s probably it. End of story. Unfortunately for people with this idea, it’s not actually solved the problem at all..

The specific question I posed to one commenter some time ago was very specific:

If 10μm photons from a 10°C atmosphere are 80% absorbed by a 0°C surface, what is the ratio of 10μm photons from a -10°C atmosphere absorbed by that same surface?

It was eventually conceded that there would be no difference – 10μm photons from a -10°C will also be 80% absorbed. This material property of a surface is called absorptivity and is the proportion of radiation absorbed vs reflected at each wavelength.

Basic physics tells us that the energy of a 10μm photon is always that same, no matter what temperature source it has come from – see note 1.

Here’s an example of the reflectivity/absorptivity of many different materials just for interest:

 

Reflectivity vs wavelength for various surfaces, Incropera (2007)

Reflectivity vs wavelength for various surfaces, Incropera & DeWitt (2007)

 

Clearly materials have very different abilities to absorb /reflect different wavelength photons. Is this the explanation?

No.

The important point to understand is that even though radiation emitted from different temperature sources have different peak wavelengths, there is a large spread of wavelengths:

The peak wavelength of +10°C radiation is 10.2μm, while that of the -10°C radiation is 11.0μm – but, as you can see, both sources emit photons over a very similar range of wavelengths.

Scenarios

Let’s now take a look at the proportion of radiation absorbed from both of these sources.

First, with the case where the surface absorptivity is higher at shorter wavelengths – this should favor absorbing more energy from a hotter source and less from a colder source:

The top graph shows the absorptivity as a function of wavelength, and the bottom graph shows the consequent absorption of energy for the two cases.

Because absorptivity is higher at shorter wavelengths, there is a slight bias towards absorbing energy from the hotter +10°C source – but the effect is almost unnoticeable.

The actual numbers:

  • 43% of the -10°C radiation is absorbed
  • 46% of the +10°C radiation is absorbed

So let’s try something more ‘brutal’, with all of the energy from wavelengths shorter than 10.5μm absorbed and none from wavelengths longer than 10.5um absorbed (all reflected).

As you can see, the proportion absorbed of the energy from the hotter source vs colder source appears very similar. It is simply a result of the fact that +10°C and -10°C radiation have almost identical proportions of energy between any given wavelengths – the main difference is that radiation from +10°C has a higher total energy.

The actual numbers:

  • 22% of the -10°C is absorbed
  • 27% of the +10°C is absorbed

So – as is very obvious to most people already – there is no possible surface which can absorb a significant proportion of 10°C radiation and yet reflect all of the -10°C radiation.

And If There Was Such a Surface

Suppose that we could somehow construct a surface which absorbed a significant proportion of radiation from a +10°C source, and yet reflect almost all radiation from a -10°C source.

Well, that would just create a new problem. Because now, when our surface heats up to 11°C the radiation from the 10°C source would still be absorbed. And yet, the radiation is now from a colder source than the surface. Red alert for all the people who say this can’t happen.

Conclusion

The claim that radiation from a colder source is not absorbed by a warmer surface has no physical basis. People who claim it don’t understand one or all of these facts of basic physics:

a) Radiation incident on a surface has to be absorbed, reflected or transmitted through the surface. This last (transmitted) is not possible with a surface like the earth (it is relevant for something like a thin piece of glass or a body of gas), therefore radiation is either absorbed or reflected.

b) The material property of a surface which determines the proportion of radiation absorbed or reflected is called the absorptivity, and it is a function of wavelength of the incident photons. (See note 2)

c) The energy of any given photon is only dependent on its wavelength, not on the temperature of the source that emitted it.

d) Radiation emitted by the atmosphere has a spectrum of wavelengths and the difference between a -10°C emitter and a +10°C emitter (for example) is not very significant (total energy varies significantly, but not the proportion of energy between any two wavelengths). See note 3.

The only way that radiation from a colder source could not be absorbed by a warmer surface is for one of these basic principles to be wrong.

These have all been established for at least 100 years. But no one has really checked them out that thoroughly. Remember, it’s highly unlikely that you have just misunderstood the Second Law of Thermodynamics.

See also: The Real Second Law of Thermodynamics

Intelligent Materials and the Imaginary Second Law of Thermodynamics

The First Law of Thermodynamics Meets the Imaginary Second Law

The Amazing Case of “Back Radiation” – Part Three

and Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics

Note 1 – Already explained in a little more detail in The Amazing Case of “Back Radiation” – Part Three – the energy of a photon is only dependent on the wavelength of that photon:

Energy = hc/λ

where h = Planck’s constant = 6.6×10-34 J.s, c = the speed of light = 3×108 m/s and λ = wavelength.

Note 2 – Absorptivity/reflectivity is also a function of the direction of the incident radiation with some surfaces.

Note 3 – For those fascinated by actual numbers – the energy from a blackbody source at -10°C = 272 W/m² compared with that from a +10°C source = 364 W/m² – the colder source providing only 75% of the total energy of the warmer source. But take a look at the proportion of total energy in various wavelength ranges:

  • Between 8-10 μm  10.7% (-10°C)   12.2% (10°C)
  • Between 10-12 μm  11.9% (-10°C)   12.7% (10°C)
  • Between 12-14 μm  11.2% (-10°C)   12.4% (10°C)
  • Between 14-16 μm   9.8% (-10°C)     9.5% (10°C)

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Many people often claim that the atmosphere cannot “heat” the earth’s surface and therefore the idea of the inappropriately-named “greenhouse” effect is scientifically impossible. The famous Second Law of Thermodynamics is invoked in support.

Let’s avoid a semantic argument about the correct or incorrect use of the word “heat”.

I claim that energy from the atmosphere is absorbed by the surface.

This absorbed energy has no magic properties. If the surface loses 100J of energy by other means and gains 100J of energy from the atmosphere then its temperature will stay constant. If the surface hasn’t lost or gained any energy by any other means, this 100J of energy from the atmosphere will increase the surface temperature.

I also claim that because the atmosphere is on average colder than the surface, more energy is transferred from the surface to the atmosphere compared with the reverse situation.

Let’s consider whether this violates the real second law of thermodynamics..

The Conceptual Problem

In Heat Transfer Basics – Part Zero a slightly off-topic discussion about the “greenhouse” effect began. One of our most valiant defenders of the imaginary second law of thermodynamics said:

An irradiated object can never reach a higher temperature than the source causing the radiation

I have demonstrated previously in The First Law of Thermodynamics Meets the Imaginary Second Law that a colder body can increase the temperature of a hotter body (compared with the scenario when the colder body was not there).

In that example, there was more than one source of energy. So, with this recent exchange in Heat Transfer Basics it dawned on me what the conceptual problem was. So this article is written for the many people who find themselves agreeing with the comment above. As a paraphrased restatement by the same commenter:

If the atmosphere is at -30°C then it can’t have any effect on the surface if the surface is above -30°C

Entropy Basics and The Special Case

Entropy is a difficult subject to understand. Heat and temperature are concepts we can understand quite easily. We all know what temperature is (in a non-precise way) and heat, although a little more abstract, is something most people can relate to.

Entropy appears to be an abstract concept with no real meaning – nothing you can get your hands around.

The second law of thermodynamics says:

Entropy of a “closed system” can never reduce

Before defining entropy, here is an important consequence of this second law:

Increasing entropy means that heat flows spontaneously from hotter to colder bodies and never in reverse

This fits everyone’s common experience.

  • Ice melts in a glass of water
  • A hot pan of water on the stove cools down to room temperature when the heat source is removed
  • Put a hold and cold body together and they tend to come to the same temperature, not move apart in temperature

And all of these are easier to visualize than a mathematical formula.

What is entropy?  I will keep the maths to an absolute minimum, but we have to introduce a tiny amount of maths just to define entropy.

For a body absorbing a tiny amount of heat, δQ (note that δ is a symbol which means “tiny change”), the change in entropy, δS, is given by:

δS = δQ / T, where T is absolute temperature (see note 1)

It’s not easy to visualize – but take a look at a simple example. Suppose that a tiny amount of heat, 1000 J, moves from a body at 1000K to a body at 500K:

Example 1

The net change in energy in the system is zero because 1000J leaves the first body and is absorbed by the second body. This is the first law of thermodynamics – energy cannot be created or destroyed.

However, there is a change in entropy.

The change in total entropy of the system = δS1 + δS2 = -1 + 2 = 1 J/K.

This strange value called “entropy” has increased.

Notice that if the energy flow of 1000J was from the 500K body to the 1000K body the change in entropy would be -1 J/K. This would be a reduction in entropy – forbidden by the real second law of thermodynamics. This would be a spontaneous flow of heat from the colder body to the hotter body.

Updated note Sep 30th – this example is intended to clarify the absolute basics.

Think of the example above like this – If, for some reason, in a closed system, this was the only movement of energy taking place, we could calculate the entropy change and it has increased.

The example is not meant to be an example of only one half of a radiative energy exchange. Just a very very simply example to show how entropy is calculated. It could be conductive heat transfer through a liquid that is totally opaque to radiation.

The Special Case

The simplest example demonstrating the second law of thermodynamics is with two bodies which are in a closed system.

Let’s say that we have a gas at 273K (Body 1) and a solid (Body 2) surrounded by the gas. The solid starts off much colder.

What is the maximum temperature that can be reached by the solid?

273K

Easy. In fact, depending on the starting temperature of the solid and the respective heat capacities of the gas and solid, the actual temperature that both end up (the same temperature eventually) might be a little lower or a lot lower.

But the temperature reached by the solid can never get to more than 273K. For the solid to get to a temperature higher than 273K the gas would have to cool down below 273K (otherwise energy would have been created). Heat does not spontaneously flow from a colder to a hotter body so this never happens.

This defining example is illuminating but no surprise to anyone.

It is important to note that this special case is not the second law of thermodynamics, it is an example that conforms to the second law of thermodynamics.

The second law of thermodynamics says that the entropy of a system cannot reduce. If we want to find out whether the second law of thermodynamics forbids some situation then we need to calculate the change in entropy – not use “insight” from this super-simple scenario.

So let’s consider some simple examples and see what happens to the entropy.

Simple Examples

What I want to demonstrate is that the standard picture in heat transfer textbooks doesn’t violate the second law of thermodynamics.

What is the standard picture?

 

From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)

From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)

 

This says that two bodies separated in space both emit radiation. And both absorb radiation from the other body (see note 2).

The challenging concept for some is the idea that radiation from the colder body is absorbed by the hotter body.

We start with Example 1 above, but this time we consider an exchange of radiation and see what happens to the entropy of that system.

Example 2

What I have introduced here is thermal radiation from Body 2 incident on Body 1. We will assume all of it is absorbed, and vice-versa.

According to the Stefan-Boltzmann equation, energy radiated is proportional to the 4th power of temperature. Given that Body 2 is half the temperature of Body 1 it will radiate at a factor of 24 = 2 x 2 x 2 x 2 = 16 times less. Therefore, if 1000J from Body 1 reaches Body 2, then 62J (1000/16) will be transmitted in the reverse direction. However, the exact value doesn’t matter for the purposes of this example.

So with our example above, what is the change in entropy?

Body 1 loses energy, which is negative entropy. Body 2 gains energy, which is positive entropy.

δS1 = -(1000-62)/1000 = -938 / 1000 = -0.94 J/k
δS2 =(1000-62)/500 = 938 / 500 = 1.88 J/K

Total entropy change = -0.94 + 1.88 = 0.94 J/K.

So even though energy from the colder body has been absorbed by the hotter body, the entropy of the system has increased. This is because more energy has moved in the opposite direction.

There is no violation of the second law of thermodynamics with this example.

Now let’s consider an example with values closer to what we encounter near the earth’s surface:

Example 3

This isn’t intended to be the complete surface – atmosphere system, just values that are more familiar.

Surface:        δS1 = -(390-301)/288 = -89 / 288 = -0.31 J/k
Atmosphere: δS2 = (390-301)/270 = 89 / 270 = 0.33 J/K

Total entropy change = -0.31 + 0.33 = 0.02 J/K.

So even though the temperatures of the two bodies are much closer together, when they exchange energy, total entropy still increases.

Energy from the colder atmosphere has been absorbed by the hotter surface and yet entropy of the system has still increased.

Now, the example above (example 3) is an exchange of a fixed amount of energy (in Joules, J). Suppose this is the amount of energy per second (Watts, W) or the amount of energy per second per square meter (W/m²).

If the atmosphere keeps absorbing more energy than it is emitting it will heat up. If the earth keeps emitting more energy than it absorbs, it will cool down.

If example 3 was the complete system, then the atmosphere would heat up and the earth would cool down until they were in thermal equilibrium. This doesn’t happen because the sun continually provides energy.

The Complete Climate System

The earth-atmosphere system is very complex. If we analyze a long term average scenario, like that painted by Kiehl and Trenberth there is an immediate problem in calculating the change in entropy:

 

From Kiehl & Trenberth (1997)

From Kiehl & Trenberth (1997)

 

[Note from Sep 28th – This section is wrong, thanks Nick Stokes for highlighting it and so delicately! Preserved in italics for entertainment value only..] If we consider the surface, for example, it absorbs 492 W/m² (δQ = 492 per second per square meter) and it loses 492 W/m² (δQ = -492 per second per square meter).

Net energy change = 0. Net entropy change = 0.

Why isn’t entropy increasing? We haven’t considered the whole system – the sun is generating all the energy to power the climate system. If we do consider the sun, it is emitting a huge amount of energy and, therefore, losing entropy. But the energy generation inside the sun creates more entropy – that is, unless the second law of thermodynamics is flawed.

[Now the rewritten bit]

Previous sections explained that calculations of entropy “removed” (negative entropy) are based on energy emitted divided by the temperature of the source. And calculations of entropy “produced” are based on energy absorbed divided by the temperature of the absorber. In a closed system we can add these up and we find that entropy always increases.

So the calculation in italics above is incorrect. Change in entropy at the surface is not zero.

Change in entropy at the surface is a large negative value, because we have to consider the source temperature of the energy.

So as Nick Stokes points out (in a comment below), we can draw a line around the whole climate system, including the emission of radiation by the sun (see example 4 just below). This calculation produces a large negative entropy, because it isn’t a closed system. This is explained by the fact that the production of solar energy creates an even larger amount of positive entropy.

Example 4

The Classic Energy Exchange by Radiation

I was in the university library recently and opened up a number of heat transfer textbooks. All of them had a similar picture to that from Incropera and DeWitt (above). And not a single one said, This doesn’t happen.

In any case, for someone to claim that an energy exchange violates the second law of thermodynamics they need to show there is a reduction in entropy of a closed system.

But one important point did occur to me when thinking about this subject. Let’s reconsider our commenter’s claim:

An irradiated object can never reach a higher temperature than the source causing the radiation

As I pointed out in the The Special Case section – this is true if this is the only source of energy. Yet the surface of the earth receives energy from both the sun and the atmosphere.

If the colder atmosphere cannot transfer energy to a warmer surface, and the second law of thermodynamics is the reason, the actual event that is forbidden is the emission of radiation by the colder atmosphere. When the colder atmosphere radiates energy it loses entropy.

After all, the entropy loss takes place when the atmosphere has given up its energy. Not when another body has absorbed the energy.

Our commenter has frequently agreed that the colder atmosphere does radiate. But he doesn’t believe that the surface can absorb it. He has never been able to explain what happens to the energy when it “reaches” the surface. Or why the surface doesn’t absorb it. Instead we have followed many enjoyable detours into attempts to undermine any of a number of fundamental physics laws in an attempt to defend “the imaginary second law of thermodynamics”.

Conclusion

Entropy is a conceptually difficult subject, but all of us can see the example in “the special case” and agree that the picture is correct.

However, the atmosphere – surface interaction is more complex than that simple case. The surface of the earth receives energy from the sun and the atmosphere.

As we have seen, in simple examples of radiant heat exchange between two bodies, entropy is still positive even when the hotter body absorbs energy from the colder body. This is because more energy flows from the hotter to the colder than the reverse.

To prove that the second law of thermodynamics has been violated someone needs to demonstrate that a system is reducing entropy. So we would expect to see an entropy calculation.

Turgid undergraduate books about heat transfer in university libraries all write that radiation emitted by a colder body is absorbed by a hotter body.

That is because the first law of thermodynamics is still true – energy cannot be created, destroyed, or magically lost.

Other Relevant Articles

The Amazing Case of “Back Radiation” – Part Three

The First Law of Thermodynamics Meets the Imaginary Second Law

Intelligent Materials and the Imaginary Second Law of Thermodynamics

Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics

Notes

Note 1 – There are more fundamental ways to define entropy, but it won’t help to see this kind of detail. And for the purists, the equation as shown relies on the temperature not changing as a result of the small transfer of energy.

If the temperature did change then the correct formula is to integrate:

ΔS = ∫Cp/T. dT (with the integral from T1 to T2) and the result,

ΔS = Cp log (T2/T1),   this is log to the base e.

Note 2 – This assumes there is some “view factor” between the two bodies – that is, some portion of the radiation emitted by one body can “hit” the other. Just pointing out the obvious, just in case..

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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?

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During a discussion about Venus (Venusian Mysteries), Leonard Weinstein suggested a thought experiment that prompted a 2nd article of mine. Unfortunately, it really failed to address his point.

In fact, it took me a long time to get to grips with Leonard’s point and 500 comments in (!) I suggested that we write a joint article.

I also invited Arthur Smith who probably agrees mostly with me, but at times he was much clearer than I was. And I’m not sure we are totally in agreement either. I did offer Leonard the opportunity to have another contributor on his side, but he is happy to write alone – or draw on one of the other contributors in forming his article. The format here is quite open.

The plan is for me to write the first section, and then Arthur to write his, followed by Leonard. The idea behind it is to crystallize our respective thoughts so that others can review them, rather than wading through 500+ comments. What was the original discussion about?

It’s worth repeating Leonard Weinstein’s original thought experiment:

Consider Venus with its existing atmosphere, and put a totally opaque enclosure (to incoming Solar radiation) around the entire planet at the average location of present outgoing long wave radiation. Use a surface with the same albedo as present Venus for the enclosure. What would happen to the planetary surface temperature over a reasonably long time? For this case, NO Solar incoming radiation reaches the surface. I contend that the surface temperature will be about the same as present..

Those who are interested in that debate can read the complete idea and the many comments that followed. During the course of our debate we each posed different thought experiments as a means to finding the flaws in the various ideas.

At times we lost track of which experiment was being considered. Many times we didn’t quite understand the ideas that were posed by others.

And therefore, before I start, it’s worth saying that I might still misrepresent one of the other points of view. But not intentionally.

Introductory Ideas – Pumping up a Tyre

This first section is uncontroversial. It is simply aimed at helping those unfamiliar with terms like adiabatic expansion. Unfortunately, it will be brief with more Wikipedia links than usual.. (if there are many questions on these basics then I might write another article).

So let’s consider pumping up a bicycle tire. When you do pump up a tire you find that around the valve everything can get pretty hot. Why is that? Does high pressure cause high temperature? Let’s review two idealized ways of compressing an ideal gas:

  • isothermal compression – which is so slow that the temperature of the gas doesn’t rise
  • adiabatic compression – which is so fast that no heat escapes from the gas during the process

Pressure and volume of a gas are inversely related if temperature is kept constant – this is Boyle’s law

.

 

Isothermal compression, Thermal Physics - Schroeder

Isothermal compression, Thermal Physics - Schroeder

 

Imagine pumping up a tire very very slowly. Usually this isn’t possible because the valve leaks.

If it was possible you would find that the work done in compressing the gas didn’t increase the gas temperature because the heat increase in the gas would equalize out to the wheel rims and to the surrounding atmosphere.

Now imagine pumping up a tire very quickly. The usual way. In this case, you are adding energy to the system and there is no time for the temperature to equalize with the surroundings, so the temperature increases (because work is done on the gas):

 

Adiabatic Compression, Thermal Physics, Schroeder

Adiabatic Compression, Thermal Physics - Schroeder

 

The ideal gas laws can be confusing because three important terms exist in the one equation, pressure, volume and temperature:

PV = nRT or PV = NkT

where P = pressure, V = volume, T = absolute temperature (in K), N = number of molecules and k = Boltzmann’s constant

So the two examples above give the two extremes of compression. One, the isothermal case, has the temperature held constant because the process is very slow, and one, the adiabatic case, has the energy leaving the system being zero because the process is so fast.

In a nutshell, high pressures do not, of themselves, cause high temperatures. But changing the pressure – i.e., compressing a gas – does increase the temperature if it is done quickly.

Introductory Ideas – The “Environmental Lapse Rate” and Convection

Equally importantly, adiabatic expansion reduces the temperature in a gas.

If you lift air up in the atmosphere quickly then it will expand and cool. In dry air, some simple maths calculates this expansion as a temperature drop of just under 10K per km. In very moist air, this temperature drop can be as low as 4K per km. (The actual value depends on the amount of moisture).

Imagine the (common) situation where due to pressure effects a “parcel of air” is pushed upwards a small way, say 100m. Under adiabatic expansion, the temperature will drop somewhere between 1K (1°C) for dry air and 0.4K for very moist air.

Suppose that the actual atmospheric temperature profile is such that the temperature 100m higher up is 1.5K cooler. (We would say that the environmental lapse rate was 15K/km).

In this case, the parcel of air pushed up is now warmer than the surrounding air and, therefore, less dense – so it keeps rising. This is the major idea behind convection – if the environmental lapse rate is “more than” the adiabatic lapse rate then convection will redistribute heat. And if the environmental lapse rate is “less than” the lapse rate then the atmosphere tends to be stable against convection.

Note – the terminology can be confusing for newcomers. Even though temperature decreases as you go up in the atmosphere the adiabatic lapse rate is written as a positive number. Just imagine that the temperature in the atmosphere actually decreases by 1K per km and think what happens if the adiabatic lapse rate is 10K per km – air that is lifted up will be much colder than the surrounding atmosphere and sink back down.

Now imagine that the temperature decreases by 15K per km and think what happens if the adiabatic lapse rate is 10K per km – air that is lifted up will be much warmer than the surrounding atmosphere (so will expand and be less dense) and will keep rising.

All of this so far described is uncontentious..

The Main Contention

Armed with these ideas, the main contentious point from my side was this:

If you heat a gas sufficiently from the bottom, convection will naturally take place to redistribute heat. The environmental “lapse rate” can’t be sustained at more than the adiabatic lapse rate because convection will take over. This is the case with the earth, where most of the solar radiation is absorbed by the earth’s surface.

But if you heat a gas from the top (as in the original proposed thought experiment) then there is no mechanism to create the adiabatic lapse rate. This is because there is no mechanism to create convection. So we can’t have an atmosphere where the environmental lapse rate is greater than the adiabatic lapse rate – but we can have one where it is less.

Convection redistributes heat because of natural buoyancy, but convection can’t be induced to work the other way.

Well, maybe it’s not quite as simple..

The Very Tall Room Full of Gas

Leonard suggested – Take an empty room 1km square and 100km high and pour in gas at 250K from the top. The gas doesn’t absorb or emit any radiation. What happens?

The gas is adiabatically compressed (due to higher pressure below) and the gas at the bottom ends up at a much higher temperature.

Another way to think about adiabatic compression is that height (potential energy) is converted to speed (kinetic energy) because of gravity – like dropping a cannon ball.

We all agree on that – but what happens afterwards? (And I think we were all assuming that a lid is placed over the top of the tall room and the lid effectively stays at a temperature of 250K due to external radiation – however, no precise definition of the temperature of the room’s walls and lid was made).

My view – over a massively long time the temperature at the top and bottom will eventually reach the same value. This seemed to be the most contentious point.

However, in saying that, there was a lot of discussion about exactly the state of the gas so at times I wondered whether it was fundamental thermodynamics up for discussion or not understanding each other’s thought experiments.

In making this claim that the gas will become isothermal (all at the same temperature), I am assuming that the gas will eventually be stationary on a large scale (obviously the gas molecules move as their temperature is defined by their velocity). So all of the bulk movements of air have stopped.

Conduction of heat is left as the only mechanism for movement of heat and as gas molecules collide with each other they will all eventually reach the same temperature – the average temperature of the gas. (Because external radiation to and from the lid and walls wasn’t defined this will affect what final average value is reached). Note that temperature of a gas is a bulk property, so a gas at one temperature has a distribution of velocities (the Maxwell-Boltzmann distribution).

The Tall Room when Gases Absorb and Emit Radiation

We all appeared to agree that in this case (radiatively-absorbing gases) that as the atmosphere becomes optically thin then radiation will move heat very effectively and the top part of the atmosphere in this very tall room will become isothermal.

Heating from the Top

The viewpoint expressed by Leonard is that differential heating (night vs day, equatorial vs polar latitudes) will eventually cause large scale circulation, thus causing bulk movement of air down to the surface with the consequent adiabatic heating. This by itself will cause the environmental lapse rate to become very close to the adiabatic lapse rate.

I see it as a possibility that I can’t (today) disprove, but Leonard’s hypothesis itself seems unproven. Is there enough energy to drive this circulation when an atmosphere is heated from the top?

I found two considerations of this idea.

One was the Sandstrom theorem which considered heating a fluid from the bottom vs heating it from the top. More comment in the earlier article. I guess you could say Sandstrom said no, although others have picked some holes in it.

The other was in Atmospheres (1972) by the great Richard M. Goody and James C. Walker. In a time when only a little was known about the Venusian atmosphere, Goody & Walker suggested first that probably enough solar radiation made it to the surface to initiate heating from below (to cut a long story short). And later made this comment:

Descending air is compressed as it moves to lower levels in the atmosphere. The compression causes the temperature to increase.. If the circulation is sufficiently rapid, and if the air does not cool too fast by emission of radiation, the temperature will increase at the adiabatic rate. This is precisely what is observed on Venus.

Venera and Mariner Venus spacecraft have all found that the temperature increases adiabatically as altitude decreases in the lower atmosphere. As we explained this observation could also be the result of thermal convection driven by solar radiation deposited at the ground, but we cannot be sure that the radiation actually reaches the ground.

What we are now suggesting as an alternative explanation is that the adiabatic temperature gradient is related to a planetary circulation driven by heat supplied unevenly to the upper levels of the atmosphere. According to this theory, the high ground temperature is caused, at least in part, by compressional heating of the descending air.

In the specific case of the real Venus (rather than our thought experiments), much more has been uncovered since Goody and Walker wrote. Perhaps the question of what happens in the real Venus is clearer – one way or the other.

What do I conclude?

I’m glad I’ve taken the time to think about the subject because I feel like I understand it much better as a result of this discussion. I appreciate Leonard especially for taking the time, but also Arthur Smith and others.

Before we started discussing I knew the answers for certain. Now I’m not so sure.

_____________________________________________________________________

By Arthur Smith

First on the question of convective heat flow from heating above, which scienceofdoom just ended with: I agree some such heat flow is possible, but it is difficult. Goody and Walker were wrong if they felt this could explain high Venusian surface temperatures.

The foundation for my certainty on this lies in the fundamental laws of thermodynamics, which I’ll start by reviewing in the context of the general problem of heat flow in planetary atmospheres (and the “Very Tall Room Full of Gas”). Note that these laws are very general and based in the properties of energy and the statistics of large numbers of particles, and have been found applicable in systems ranging from the interior of stars to chemical solutions and semiconductor devices and the like. External forces like gravitational fields are a routine factor in thermodynamic problems, as are complex intermolecular forces that pose a much thornier challenge. The laws of thermodynamics are among the most fundamental laws in physics – perhaps even more fundamental than gravitation itself.

I’m going to discuss the laws out of order, since they are of various degrees of relevance to the discussion we’ve had. The third law (defining behavior at zero temperature) is not relevant at all and won’t be discussed further.

The First Law

The first law of thermodynamics demands conservation of energy:

Energy can be neither created nor destroyed.

This means that in any isolated system the total energy embodied in the particles, their motion, their interactions, etc. must remain constant. Over time such an isolated system approaches a state of thermodynamic equilibrium where the measurable, statistically averaged properties cease changing.

In our previous discussion I interpreted Leonard’s “Very Tall Room Full of Gas” example as such a completely isolated system, with no energy entering or leaving. Therefore it should, eventually at least, approach such a state of thermodynamic equilibrium. Scienceofdoom above interpreted it as being in a condition where the top of the room was held at a given specific temperature. That condition would allow energy to enter and leave over time, but eventually the statistical properties would also stop changing, and then energy flow through that top surface would also cease, total energy would be constant, and you would again arrive at an equilibrium system (but with a different total energy from the starting point).

That would also be the case in Leonard’s original thought experiment concerning Venus if the temperature of the “totally opaque enclosure” was a uniform constant value. The underlying system would reach some point where its properties ceased changing, and then with no energy flow in or out, it would be effectively isolated from the rest of the universe, and in its own thermodynamic equilibrium. However, Leonard allows the temperature of his opaque enclosure to vary with latitude and time of day which means that strictly such a statistical constancy would not apply and the underlying atmosphere would not be completely in thermodynamic equilibrium. I’ll look at that later in discussing the restrictions imposed by the second law.

In a system like a planetary atmosphere with energy flowing through it from a nearby star (or from internal heat) and escaping into the rest of the universe, you are obviously not isolated and would not reach thermodynamic equilibrium. Rather, if a condition where averaged properties cease changing is reached, this is referred to as a steady state. Under steady state conditions the first law must still be obeyed. Since internal statistical properties are unchanging, that means the system must not be gaining or losing any internal energy. So in steady state you have a balance between incoming and outgoing energy from the system, enforced by the first law of thermodynamics.

If such an atmospheric system is NOT in steady state, if there is, say, more energy coming in than leaving, then the total energy embodied in the particles of the system will increase. That higher average energy per particle can be measured as an increase in temperature – but that gets us to the definition of temperature.

The Zeroth Law

The zeroth law essentially defines temperature:

If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other.

Here thermal equilibrium means that when the systems are brought into physical proximity so that they may exchange heat, no heat is actually exchanged. A typical example of the zeroth law is to make the “third system” a thermometer, something that you can use to read out a measurement of its internal energy level. Any number of systems can act as a thermometer: the volume of mercury liquid in an evacuated bulb, the resistance of a strip of platinum, or the pressure of a fixed volume of helium gas, for example.

If you divide a system “A” in thermodynamic equilibrium into two pieces, “A1” and “A2”, and then bring those two into physical proximity again, no heat should flow between them, because no heat was flowing between them before separating them since neither one’s statistical properties were changing. I.e. Any two subsystems of a system in thermodynamic equilibrium must be in thermal equilibrium with each other. That means that if you place a thermometer to measure the temperature of subsystem “A1”, and find a temperature “T” for thermal equilibrium of the thermometer with “A1”, then subsytem “A2” will also be in thermal equilibrium with “T”, i.e. its temperature will also read out as the same value.

That is, the temperature of a system in thermodynamic equilibrium is the same as the temperature of every (macroscopic) subsystem – temperature is constant throughout. The zeroth law implies temperature must be a uniform property of such equilibrium systems.

This means that in both “Very Large Room” examples and for my version of Leonard’s original thought experiment for Venus (with a uniform enclosing temperature), the thermodynamic equilibrium that the atmosphere must eventually reach must have a constant and uniform temperature throughout the system. Temperature in the room or in the pseudo-Venus’ atmosphere would be independent of altitude – an isothermal, not adiabatic, temperature profile.

The zeroth law can actually be derived from the first and second laws – this is done for example in Statistical Physics, 3rd Edition Part 1, Landau and Lifshiftz (Vol. 5) (Pergamon Press, 1980) – Chapter II, Thermodynamics Quantities, Section 9 – “Temperature” – and again the conclusion is the same:

Thus, if a system is in a state of thermodynamic equilibrium, the [absolute temperature] is the same for every part of it, i.e. is constant throughout the system.

The Second Law

The first and zeroth laws tell us what happens in the cases where the atmosphere can be characterized as in thermodynamic equilibrium, i.e. actually or effectively isolated from the rest of the universe after sufficient period of time that quantities cease changing. Under those conditions it must have a uniform temperature. But what about Leonard’s actual Venus thought experiment, where there are constant fluxes of energy in and out due to latitudinal and time-of-day variations in the temperature of the opaque enclosure? What can we say about the temperatures in the atmosphere below given heating from above under those conditions? Here the second law provides the primary constraint, and in particular the Clausius formulation:

Heat cannot of itself pass from a colder to a hotter body.

A planetary atmosphere is not driven by machines that move the air around, there are no giant fans pushing the air from one place to another. There is no incoming chemical or electrical form of non-thermal energy that can force things to happen. The driving force is the flux of incoming energy from the local star that brings heat when it is absorbed. All atmospheric processes are driven by the resulting temperature differences. Thanks to the first law of thermodynamics each incoming chunk of energy can be accounted for as it is successively absorbed, reflected, re-emitted and so forth until it finally leaves again as thermal radiation to the rest of the universe. In each of these steps the energy is spontaneously exchanged from a portion of the atmosphere at one temperature to another portion at another temperature.

What the second law tells us, particularly in the above Clausius form, is that the net spontaneous energy exchange describing the flow of each chunk of incoming energy to the atmosphere MUST ALWAYS BE IN THE DIRECTION OF DECREASING TEMPERATURE. Heat flows “downhill” from high to low temperature regions. The incoming energy starts from the star – very high temperature. If it’s absorbed it’s somewhere in the atmosphere or the planetary surface, and from that point it must go through successfully colder and colder portions of the system before it can escape to space (where the temperature is 2.7 K).

There can be no net flow of energy from colder to hotter regions. And that means, if the atmosphere below Leonard’s “opaque enclosure” is at a higher temperature than any point on the enclosure, heat must be flowing out of the atmosphere, not inward. The enclosure, no matter the distribution of temperatures on its surface, cannot drive a temperature below it that is any higher than the highest temperature on the enclosure itself.

So even in the non-equilibrium case represented by Leonard’s original thought experiment, while the atmosphere’s temperature will not be everywhere the same, it will nowhere be any hotter than the highest temperature of the enclosure, after sufficient time has passed for such statistical properties to stop changing.

The thermodynamic laws are the fundamental governing laws regarding temperature, heat, and energy in the universe. It would be extraordinary if they were violated in such simple systems as these gases under gravitation that we have been discussing. Note in particular that any violation of the second law of thermodynamics allows for the creation of a “perpetual motion machine”, a device legions of amateurs have attempted to create with nothing but failure to show for it. Both the first and second laws seem to be very strictly enforced in our universe.

Approach to Equilibrium

The above results on temperatures apply under equilibrium or steady state conditions, i.e. after the “measurable, statistically averaged properties cease changing.” That may perhaps take a long time – how long should we expect?

The heat content of a gas is given by the product of the heat capacity and temperature. For the Venus case we’re starting at 740 K near the surface and, under either of the “thought experiment” cases, dropping to something like 240 K in the end, about 500 degrees. Surface pressure on Venus is 93 Earth atmospheres, so in every square meter we have a mass of close to 1 million kg of atmosphere above it. [Quick calculation: 1 earth atmosphere = 101 kPa, or 10,300 kg of atmosphere per square meter, or 15 pounds per square inch. On Venus it’s 1400 pounds/sq inch.] The atmosphere of Venus is almost entirely carbon dioxide, which has a heat capacity of close to 1 kJ/kgK (see this reference). That means the heat capacity of the column of Venus’ atmosphere over 1 square meter is about one billion (109) J/K.

So a temperature change of 500 K amounts to 500 billion joules = 500 GJ for each square meter of the planetary surface. This is the energy we need to flow out of the system in order for it to move from present conditions to the isothermal conditions that would eventually apply under Leonard’s thought experiment.

Now from scienceofdoom’s previous post we expect at least an initial heat flow rate out of the atmosphere of 158 W/m² (that’s the outgoing flow that balances incoming absorption on Venus – since we’ve lost incoming absorption to the opaque shell, this ought to be roughly the initial net flow rate). Dividing this into 500 GJ/m² gives a first-cut time estimate for the cooling: 3.2 billion seconds, or about 100 years. So the cool-down to isothermal would be hardly immediate, but still pretty short on the scale of planetary change.

Now we shouldn’t expect that 158 W/m² to hold forever. There are four primary mechanisms for heat flow in a planetary atmosphere: conduction (the diffusion of heat through molecular movements), convection (movement of larger parcels of air), latent heat flow (movement of materials within air parcels that change phases – from liquid to gas and back, for example, for water) and thermal radiation. The heat flow rate for conduction is simply proportional to the gradient in temperature. The heat flow rate for radiation is similar except for the region of the atmospheric “window” (some heat leaves directly to space according to the Planck function for that spectral region at that temperature). Latent heat flow is not a factor in Venus’ present atmosphere, though it would come into play if the lower atmosphere cooled below the point where CO2 liquefies at those pressures.

For convection, however, average heat flow rates are a much more complex function of the temperature gradient. Getting parcels of gas to displace one another requires some sort of cycle where some areas go up and some down, a breaking of the planet’s symmetry. On Earth the large-scale convective flows are described by the Hadley cells in the tropics and other large-scale cells at higher latitudes, which circulate air from sea level to many kilometers in altitude. On a smaller scale, where the ground becomes particularly warm then temperature gradients exceeding the adiabatic lapse rate may occur, resulting in “thermals”, local convective cells up to possibly several hundred meters. If the temperature difference between high and low altitudes is too low, the convective instability vanishes and heat flow through convection becomes much weaker.

So as temperatures come closer to isothermal in an atmosphere like Venus’, except for the atmospheric “window” for radiative heat flow, we would expect all the heat flow mechanisms to decrease, and convection in particular to almost cease after the temperature difference gets too small. So we might expect the cool-down to isothermal conditions to slow down and end up much longer than this 100-year estimate. How long?

Another of the thought experiment versions discussed in the previous thread involved removing radiation altogether; with both radiation and convection gone, that leaves only conduction as a mechanism for heat flow through the atmosphere. For an ideal gas the thermal conductivity increases as the 2/3 power of the density (it’s proportional to density times mean free path) and the square root of temperature (mean particle speed). While CO2 is not really ideal at 93 atmospheres at 740 K, using this rule gives us a rough idea of what to expect – at 1 atmosphere and 273 K we have a value of 14.65 mW/(m.K) so at 93 atmospheres and 740 K it should be about 500 mW/(m . K). For a temperature gradient of 10 K/km that gives a heat flux of 0.005 W/m². 500 GJ would then take about 1014 seconds, or 3 million years.

So the approach to an isothermal equilibrium state for these atmospheres would take between a few hundred and a few million years, depending on the conditions you impose on the system. Still, the planets are billions of years old, so if heating from above was really the mechanism at work on Venus we should see the evidence of it in the form of cooler surface temperatures there by now, even if radiative heat flow were not a factor at all.

The View From a Molecule

Leonard in our previous discussion raised the point that an individual molecule sees the gravitational field, causing it to accelerate downwards. So molecular velocities lower down should be higher than velocities higher up, and that means higher temperatures.

Leonard’s picture is true of the behavior of a molecule in between collisions with the other molecules. But if the gas is reasonably dense, the “mean free path” (the average distance between collisions) becomes quite short. At 1 atmosphere and room temperature the mean free path of a typical gas is about 100 nanometers. So there’s very little distance to accelerate before a molecule would collide with another; to consider the full effect you need to look at the effect of collisions due to gas pressure along with the acceleration by gravity.

An individual molecule in a system in thermodynamic equilibrium at temperature T has available a collection of states in phase space (position, momentum and any internal variables) each with some energy E. In the case of our molecule in a gravitational field, that energy consists of the kinetic energy ½mv² (m = mass, v = velocity) plus the gravitational potential energy = gmz (where z = height above ground). The Boltzmann distribution applies in equilibrium, so that the probability of the molecule being in a state with energy E is proportional to:

e(-E/kT) = e(-(½mv² + gmz)/kT).

So the Boltzmann distribution in this case specifies both the distribution of velocities (the standard Maxwell-Boltzmann distribution) and also an exponential decrease in gas density (and pressure) with height. It is very unlikely for a molecule to be at a high altitude, just as it is very unlikely for a molecule to have a high velocity. The high energy associated with rare high velocities comes from occasional random collisions building up that high speed. Similarly the high energy associated with high altitude comes from random collisions occassionally pushing a molecule to great heights. These statistically rare occurences are both equally captured by the Boltzmann distribution. Note also that since the temperature is uniform in equilibrium, the distribution of velocities at any given altitude is that same Maxwell-Boltzmann distribution at that temperature.

Force Balance

The decrease in pressure with height produces a pressure-gradient force that acts on “parcels of gas” in the same way that the gravitational force does, but in the opposite direction. At equilibrium or steady-state, when statistical properties of the gas cease changing, the two forces must balance.

That leads to the equation of hydrostatic balance equating the pressure gradient force to the gravitational force:

dp/dz = – mng

(here p is pressure and n is the number density of molecules – N/V for N molecules in volume V). In equilibrium n(z) is given by the Boltzmann distribution:

n(z) = c.e(-mgz/kT);

for the ideal gas pressure is given by p = nkT, so the hydrostatic balance equation becomes:

dp/dz = k T dn/dz = k T c (-mg/kT) e(-g m z/kT) = – mg c e(-mgz/kT) = – m n(z) g

I.e. the Boltzmann distribution for this ideal gas system automatically ensures the system is in hydrostatic equilibrium.

Another approach to this sort of analysis is to look at the detailed flow of molecules near an imaginary boundary. This is done in textbook calculations of the thermal conductivity of an ideal gas, for example, where a gradient in temperature results in net flow of energy (necessarily from hotter to colder). In our system with gravitational force and pressure gradients both must be taken into account in such a calculation. Such calculations are somewhat complex and depend on assumptions about molecular size and neglecting other interactions that would make the gas non-ideal, but the net effect must always satisfy the same thermodynamic laws as every other such system: in thermodynamic equilibrium temperature is uniform and there is no net energy flow through any imagined boundary.

In conclusion, after sufficient time that statistical properties cease changing, all these examples of a system with a Venus-like atmosphere must reach essentially the same isothermal or near-isothermal state. The gravitational field and adiabatic lapse rate cannot explain the high surface temperature on Venus if incoming solar radiation does not reach (at least very close to) the surface.

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By Leonard Weinstein

Solar Heating and Outgoing Radiation Balances for Earth and Venus

The basic heating mechanism for any planetary atmosphere depends on the balance and distribution of absorbed solar energy and outgoing radiated thermal energy. For a planet like Earth, the presence of a large amount of surface water and a relatively optically transparent atmosphere to sunlight dominates where and how the input solar energy and outgoing thermal energy create the surface and atmospheric temperatures. The unequal energy flux for day and night, and for different latitudes, combined with the planet rotation result in wind and ocean currents that move the energy around.

Earth is a much more complex system than Venus for these reasons, and also due to the biological processes and to changing albedo due to changes in clouds and surface ice. The average energy balance for the Earth was previously shown by Science of Doom, but is shown in Figure 1 below for quick reference:

 

From Kiehl & Trenberth (1997)

From Kiehl & Trenberth (1997)

 

The majority of solar energy absorbed by the Earth directly heats the land and water. Some water evaporation carries this energy to higher altitudes, and is released by phase change (condensation). This energy is carried up by atmospheric convection.

In addition, convective heat transfer from the ground and oceans transfers energy to the atmosphere. It is the basic atmospheric temperature differences from day and night and at different latitude that creates pressure differences that drive the wind patterns that eventually mix and transport the atmosphere, but the buoyancy of heated air from the higher temperature surface areas also aids in the vertical mixing.

This energy is carried by convection up into the higher levels of the atmosphere and eventually radiates to space. The combination of convection of water vapor and surface heated air upwards dominates the total transported energy from the ground level. In addition, some of the ground level thermal energy is radiated up, with a portion of the thermal radiation passing directly from the ground to space. Water vapor, CO2, clouds, aerosols, and other greenhouse gases also absorb some of this radiated energy, and slightly increase the lower atmosphere and ground temperatures with back radiation effectively adding to the initial radiation, resulting in a reduced net radiation flux. This results in a higher temperature atmosphere and ground than without these absorbing materials.

Venus, however, is dominated by direct absorption of solar energy into the atmosphere (including clouds) rather than by the surface, so has a significantly different path to heat the atmosphere and ground. Venus has a very dense atmosphere (about 93 times the mass as Earth’s atmosphere), which extends to about 90 km altitude before the tropopause is reached. This is much higher than the Earth’s atmosphere.

Very dense clouds, composed mostly of sulfuric acid, reach to about 75 km, and cover the planet. The clouds have virga beneath them due to the very high temperatures at lower elevations. The clouds and thick haze occupy over half of the main troposphere height, with a fairly clear layer starting below about 30 km altitude. Due to the very high density of the atmosphere, dust and other aerosol particles (from the surface winds and possibly from volcanoes) also persist in significant quantity.

The atmosphere is 96.5% CO2, and contains significant quantities of SO2 (150 ppm), and even some H2O (20 ppm), and CO (17 ppm). These, along with the sulfuric acid clouds, dust, and other aerosols, absorb most of the incoming sunlight that is not reflected away, and also absorb essentially all of the outgoing long wave radiation and relay it to the upper atmosphere and clouds to eventually dump it into space.

A sketch is shown in Figure 2, similar to the one used for Earth, which shows the approximate energy transfer in and out of the Venus atmosphere system. It is likely that almost all of the radiation out leaves from the top of the clouds and a short distance above, which therefore locks in the level of the atmospheric temperature at that location.

The surface radiation balance shown is my guess of a reasonable level. The top portion of the clouds reflects about 75% of incident sunlight, so that Venus absorbs an average of about 163 W/m², which is significantly less than the amount absorbed by Earth. About 50% of the available sunlight is absorbed in the upper cloud layer, and about 40% of the available sunlight is captured on the way down in the lower clouds, gases, and aerosols.

Thus an average of solar energy that reaches the surface is only about 17 W/m², and the amount absorbed is somewhat less, since some is reflected. The question naturally arises as to what is the source of wind, weather, and temperature distribution on Venus, and why is Venus so hot at lower altitudes.

Venus takes 243 days to rotate. However, continual high winds in the upper atmosphere takes only about 4 days to go completely around at the equator, so the day/night temperature variation is even less than it would have otherwise been. Other circulation cells for different latitudes (Hadley cells) and some unique polar collar circulation patterns complete the main convective wind patterns.

The solar energy absorbed by the surface is a far smaller factor than for Earth, and I am convinced it is not necessary for the basic atmospheric and ground temperature conditions on Venus. Since the effect on the atmospheres of planets from absorbed solar radiation is to locally change the atmospheric pressure that drive the winds (and ocean currents if applicable), these flow currents transport energy from one location and altitude to another. There is no specific reason the absorption and release of energy has to be from the ground to the atmosphere unless the vertical mixing from buoyancy is critical. I contend that direct absorption of solar energy into the atmosphere can accomplish the mixing, and this along with the fact that the top of the clouds and a short distance above is where the radiation leaves from, is in fact the cause of heating for Venus.

We observed that unlike Earth, which had about 72% of the absorbed solar energy heat the surface, Venus has 10% or less absorbed by the ground. Also the surface temperature of Venus (about 735 K) would result in a radiation level from the ground of about 16,600 W/m².

Since back radiation can’t exceed radiation up if the ground is as warm or warmer than the atmosphere above it, the only thing that can make the ground any warmer than the atmosphere above it is the ~17 W/m² (average) from solar radiation. The ground absorbed solar radiation plus absorbed back radiation has to equal the radiation out for constant temperature. If the absorbed solar radiation were all used to heat the ground, and the net radiation heat transfer was near zero (the most extreme case of greenhouse gas blocking possible), the average temperature of the ground would only be about 0.19 K warmer than the atmosphere above it, and the excess heat would need to be removed by surface convective driven heat transfer. The buoyancy would be extremely small, and contribute little to atmospheric mixing.

However, the net radiation heat transfer out of the ground is almost surely equal to or larger than the average solar heating at the ground, from some limited transmission windows in the gas, and through a small net radiation flux. The most likely effect is that the ground is equal to or a small amount cooler than the lower atmosphere, and there is probably no buoyancy driven mixing. This condition would actually require some convective driven heat transfer from the atmosphere to the ground to maintain the ground temperature. Since the measured lower atmosphere and ground temperature are on the dry adiabatic lapse rate curve projected down from the temperature at the top of the cloud layer, the net ground radiation flux is probably close to the value of 17 W/m². This indicates that direct solar heating of the ground is almost certainly not a source for producing the winds and temperature found on Venus. The question still remains: what does cause the winds and high temperatures found on Venus?

The main point I am trying to make in this discussion is that the introduction of solar energy into the relatively cool upper atmosphere of Venus, along with the high altitude location of outgoing radiation, are sufficient to heat the lower atmosphere and surface to a much higher temperature even if no solar energy directly reaches the surface. Two simplified models are discussed in the following sections to support the plausibility of that claim. This issue is important because it relates to the mechanism causing greenhouse gas atmospheric warming, and the effect of changing amounts of the greenhouse gases.

The Tall Room

The first model is an enclosed room on Venus that is 1 km x 1km x 100 km tall. This was selected to point out how adiabatic compression can cause a high temperature at the bottom of the room, with a far lower input temperature at the top. This is the type of effect that dominates the heating on Venus. While the first part of the discussion is centered on that room model, the analysis is also applicable for part the second model, which examines a special simplified approximation of the full dynamics on Venus.

The conditions for the tall room model are:
1) A gas is introduced at the top of a perfectly thermally insulated fully enclosed room 1 km x 1 km x 100 km tall, located on the surface of Venus. The walls (and bottom and top) are assumed to have negligible heat capacity. The walls and bottom and top are also assumed to be perfect mirrors, so they do not absorb or emit radiation.

2) The supply gas temperature is selected to be 250 K. The gas pours in to fill all of the volume of the room. Sufficient quantity of gas is introduced so that the final pressure at the top of the room is at 0.1 bar at the end of inflow. The entry hole is sealed immediately after introduction of the gas is complete.

3) The gas is a single atom molecule gas such as Argon, so that it does not absorb or emit thermal radiation. This made the problem radiation independent. I also put in a qualifier, to more nearly approximate the actual atmosphere, that the gas had a Cp like that of CO2 at the surface temperature of Venus [i.e., Cp=1.14 (kJ/kg K) for CO2 at 735 K]. Cp is also temperature independent.

The room height selected would actually result in a hotter ground level than the actual case of Venus. This was due to the choice of a room 100 km tall. The height to 0.1 bar for Venus is only about 65 km, which would give a better temperature match, but the difference is not important to the discussion. A dry adiabatic lapse rate forms as the gas is introduced due to the adiabatic compression of the gas at the lower level. The value of the lapse rate for the present example comes from a NASA derivation at the Planetary Atmospheres Data Node:
http://atmos.nmsu.edu/education_and_outreach/encyclopedia/adiabatic_lapse_rate.htm
The final result for the dry adiabatic lapse rate is:

Γp = -dT/dz|a = g/Cp (1)

In the room at Venus, this results in:

TH =Ttop +H * 8.9/1.14 (2)

Where H is distance down from the top.

Ttop remains at 250 K since it is not compressed (not because it was forced to be at that temperature by heat transfer), and Tbottom=1,031 K due to the adiabatic compression.

Two questions arise:
1) Is this dry adiabatic lapse rate what would actually develop initially after all the gas is introduced?
2) What would happen when the system comes to final equilibrium (however long it takes)?

The gas coming in would initially spread down to the bottom due to a combination of thermal motion and gravity, but the converted potential energy due to gravity over the room height would add considerable downward velocity, and this added downward velocity would convert to thermal velocity by collisions. Once enough gas filled the room to limit the MFP, added gas would tend to stay near the top until additional gas piled on top pushed it downwards, and this would increasingly compress the gas below from its added mass. The adiabatic compression of gas below the incoming gas at the top would heat the gas at the bottom to 1,031 K for the selected model. The top temperature would remain at 250 K, and the temperature profile would vary as shown by equation (2). Thus the answer to 1) is yes.

Strong convection currents may or may not be introduced in the room, depending on how fast the gas is introduced. To simplify the model, I assume the gas flows in slowly enough so that the currents are not important. It is quite clear that the temperature profile at the end of inflow would be the adiabatic lapse rate, with the top at 250 K and bottom at 1,031 K. If the final lapse rate went toward zero from thermal conduction, as Arthur postulated, even he admits it would take a very long time. The question now arises- what would cause heat conduction to occur in the presence of an adiabatic lapse rate? i.e., why would an initial adiabatic lapse rate tend to go toward an isothermal lapse rate if there is no radiation forcing (note: this lack of radiation forcing assumption is for the room model only). The cause proposed by Arthur and Science of Doom is based on their understanding of the Zeroth Law of Thermodynamics. They say that if there is a finite lapse rate (i.e., temperature gradient), there has to be conduction heat transfer. This arises from not considering the difference between temperature and heat. This is discussed in:
http://zonalandeducation.com/mstm/physics/mechanics/energy/heatAndTemperature/heatAndTemperature.html (difference between temperature and heat)

When we consider if there will be heat conduction in the atmosphere, we need to look at potential temperature rather than temperature. This is discussed at: http://en.wikipedia.org/wiki/Potential_temperature
The potential temperature is shown to be:

(3)

This term in (3) is general, and thus valid for Venus, if the appropriate pressures are used.

A good discussion why the potential temperature is appropriate to use rather than the local temperature can be found at:
https://courseware.e-education.psu.edu/simsphere/workbook/ch05.html

This includes the following statements:

  • “if we return to the classic conduction laws and our discussion of resistances, we note that heat should be conducted down a temperature gradient. Since we are talking about sensible heat, the appropriate gradient for the conduction of sensible heat is not the temperature but the potential temperature. The potential temperature is simply a temperature normalized for adiabatic compression or expansion”
  • “When the environmental lapse rate is exactly dry adiabatic, there is zero variation of potential temperature with height and we say that the atmosphere is in neutral equilibrium. Under these conditions, a parcel of air forced upwards (downwards) will stay where it is once moved, and not tend to sink or rise after released because it will have cooled (warmed) at exactly the same rate as the environment”.

The above material supports the claim that there would be no movement from a dry adiabatic lapse rate toward an isothermal gas in the room model.

If the initial condition was imposed– that the lapse rate was below the dry adiabatic lapse rate, it is true that the gas would be very stable from convective mixing due to buoyancy, and the very slow thermal conduction, which would drive the temperature back toward the dry adiabatic lapse rate, could take a very long time (in the actual case, it would be much faster due to even small natural convection currents generally present). However, there is no reason for any lapse rate other than the dry adiabatic lapse rate to initially form, as the problem was posed, so that issue is not even relevant.

The final result of the room model is the fact that a very high ground temperature was produced from a relative cool supply gas due to adiabatic compression of the supply that was introduced at a high altitude. This is actually a consequence of gravitational potential energy being converted to kinetic energy. Once the dry adiabatic lapse rate formed, any small flow up or down stays in temperature equilibrium at all heights, so this is a totally stable situation, and would not tend toward an isothermal situation.

If there were present a sufficient quantity of gas in the present defined room that radiated and absorbed in the temperature range of the model, the temperature would tend toward isothermal, but that was not how the tall room example was defined.

Effect of an Optical Barrier to Sunlight reaching the Ground

The second model I discussed with Science of Doom, Arthur, and others, relates to my suggestion that if an optical barrier prevented any solar energy from transmitting to the ground, but that the energy was absorbed in and heated a thin layer just below the average location of effective outgoing radiation from Venus, in such a way that the heat was transmitted downward through the atmosphere, this could also result in the hot lower atmosphere and surface that is actually observed. The albedo, and the solar heating input due to day and night, and to different latitudes, was selected to match the actual values for Venus, and the radiation out was also selected to match the values for the actual planet.

This problem is much more complicated than the enclosed room case for two reasons.

The first is that it is a dynamic and non-uniform case. The second is due to the fact that the actual atmosphere is used, with radiation absorption and emission, and the presence of clouds. The lower atmosphere and surface of the planet Venus are much hotter than for the Earth, even though the planet does not absorb as much solar energy as the Earth. It is closer to the Sun than Earth, but has a much higher albedo due to a high dense cloud layer composed mostly of sulfuric acid drops. The discussion will not attempt to examine the historical circumstances that led up to the present conditions on Venus, but only look at the effect of the actual present conditions.

In order to examine this model, I had postulated that if all of the solar heating was absorbed near the top of Venus’s atmosphere, but with the day and night and latitude variation, the heat transfer to the atmosphere downward would eventually be mixed through the atmosphere and maintain the adiabatic lapse rate, with the upper atmosphere held to the same temperature as at the present. Since the atmosphere has gases, clouds, and aerosols that absorb and radiate most if not all of the thermal radiation, this is a different problem from the tall room.

However, it appears that the radiation flux levels are relatively small, especially at lower levels, so the issue hinges on the relative amount of forcing by convective flow compared to net radiation up (which does tend to reduce the lapse rate). I use the actual temperature profile as a starting point to see if the model is able to maintain the values. Different starting profiles would complicate the discussion, but if the selected initial profile can be maintained, it is likely all reasonable initial profiles would tend to the same long-term final levels.

The assumption that the solar heating and radiation out all occur in a layer at the top of the atmosphere eliminates positive buoyancy as a mechanism to initially couple the solar energy to the atmosphere. However, the direct thermal heat-transfer with different amounts of heating and cooling at different locations causes some local expansion and contraction of different locations of the top of the atmosphere, and this causes some pressure driven convection to form.

The pressure differences from expansion and contraction set a flow in motion that greatly increases surface heat transfer, and this flow would become turbulent at reasonable induced flow speeds. This increases heat transfer and mixing to larger depths. The portions cooler than the local average atmosphere will be denser than local adiabatic lapse rate values from the average atmosphere, and thus negative buoyancy would cause downward flow at these locations.

As the flow moved downward, it compressed but initially remains slightly cooler than the surrounding, with some mixing and diffusion spreading the cooling to ever-larger volumes. At some level, the flow down actually passes into a surrounding volume that is slightly cooler, rather than warmer, than the flow, due to the small but finite radiation flux removing energy from the surrounding. At this point the downward flow stream is warming the surrounding. The question arises: how much energy is carried by the convection, and could it easily replace radiated energy, so as to maintain a level near the dry adiabatic lapse rate?

A few numbers need to be shown here to best get an order of magnitude of what is going on. Arthur has already made some of these calculations. The average input energy of 158 W/m² applied to Venus’s atmosphere would take about 100 years to change the average atmospheric temperature by 500 K. This means that to change it even 0.1 K (on average) would take about 7 days. Since the upper atmosphere at low latitudes only takes about 4 days to completely circulate around Venus, the temperature variations from average values would be fairly small if the entire atmosphere were mixed.

However, for the model proposed, only a very thin layer would be heated and cooled under the absorbing layer. Differences in net radiation flux would also help transfer energy up and down some. This relatively thin layer would thus have much higher temperature variation than the average atmosphere mass would (but still only a few degrees). The pressure variations due to the upper level temperature variations would cause some flow circulation and vertical mixing to occur throughout the atmosphere. The circulating flows may or may not carry enough energy to overcome radiation flux levels to maintain the dry adiabatic lapse rate. Let us look at the near surface flow, and a level of radiation flux of 17 W/m².

What excess temperature and vertical convection speed is needed to carry energy able to balance that flux level. Assume a temperature excess of only 0.026 K is carried and mixed by convection due to atmospheric circulation. Also assume the local horizontal flow rate near the ground is 1 m/s. If a vertical mixing of only 0.01 m/s were available, the heat added would be 17 W/m², and this would balance the lost energy due to radiation, thus allowing the dry adiabatic lapse rate to be maintained.

This shows how little convective circulation and mixing are needed to carry solar heated atmosphere from high altitudes to lower levels to replace energy lost by radiation flux levels in the atmosphere, and maintain a near dry adiabatic lapse rate. It is the solar radiation that is supplying the input energy, and adiabatic compression that is heating the atmosphere. As long as the lapse rate is held close to the adiabatic level with sufficient convective mixing, it is the temperature at the location in the atmosphere where the effective outgoing radiation is located that sets a temperature on the adiabatic lapse rate curve and adiabatic compression determines the lower atmosphere temperature.

Since the exact details of the heat exchanges are critical to the process, this optically sealed region near the top of the atmosphere is a poorly defined model as it stands, and the question of whether it actually would work as Arthur and Science of Doom question is not resolved, although an argument can be made that there are processes to do the mixing. However, the real atmosphere of Venus absorbs almost all of the solar energy in its upper half, and this much larger initial absorption volume does clearly do the job. I have shown the ground likely has little if any effect on the actual temperature on Venus.

Some Concluding Remarks

The initial cause for this discussion was the question of why the surface of Venus is as hot as it is. A write-up by Steve Goddard implied that the high pressure on Venus was a major factor, and even though some greenhouse gas was needed to trap thermal energy, it was the pressure that was the major contributor. The point was that an adiabatic lapse rate would be present with or without a greenhouse gas, and the major effect of the greenhouse gas was to move the location of outgoing radiation to a high altitude. The outgoing level set a temperature at that altitude, and the ground temperature was just the temperature at the outgoing radiation effective level plus the increase due to adiabatic compression to the ground. The altitude where the effective outgoing radiation occurs is a function of amount and type of greenhouse gases. Steve’s statement is almost valid. If he qualified it to state that enough greenhouse gas is still needed to limit the radiation flux, and keep the outgoing radiation altitude near the top of the atmosphere, he would have been correct. Thus both the amount of atmosphere (and thus pressure and thickness), and amount of greenhouse gases are factors. Any statement that greenhouse gases are not needed if the pressure is high enough is wrong (but this was not what Steve Goddard said).

Another issue that came up was the need for the solar energy to heat the ground in order for the hot surface of Venus to occur. I think I made reasonable arguments that this is not at all true. While there is a small amount of solar heating of the ground, the ground is probably actually slightly cooler that the atmosphere directly above it due to radiation, and so there is no buoyant mixing and no heating of the atmosphere from the ground other than the small radiated contribution. The main part of solar energy is absorbed directly into the atmosphere and clouds, and is almost certainly the driver for winds and mixing, and the high ground temperature.

The final issue is what would happen if most (say 90%) of the CO2 were replaced by say Argon in Venus’s atmosphere. Three things would happen:

1) The adiabatic lapse rate would greatly increase due to the much lower Cp of Argon.

2) The height of the outgoing radiation would probably decrease, but likely not much due to both the presence of the clouds, and the fact that the density is not linear with altitude, and matching a height with remaining high CO2 level would only drop 10 to 15 km out of the 75 to 80, where outgoing radiation is presently from. If in fact it is the clouds that cause most of the outgoing radiation, there may not be any drop in outgoing level.

3) The radiation flux through the atmosphere would increase, but probably not nearly enough to prevent the atmospheric mixing from maintaining an adiabatic lapse rate. Keep in mind that Venus has 230,000 times the CO2 as Earth. Even 10% of this is 23,000 times the Earth value.

The combination of these factors, especially 1), would probably result in an increase in the ground temperature on Venus.

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In Part One we looked at a thought experiment designed to make it easier to understand a basic principle – that inner surfaces of systems can have higher temperatures and therefore higher values of radiation than the outside of a system.

We looked at a PVC hollow sphere out in the vastness of space:

The only reason for creating that experiment in its unreal environment was to make the basic physics and the corresponding maths easier to follow.

In that example, with the heat source of 30,000 W, the final equilibrium values (with emissivity = 0.8) were:

T1 = 423 K and T2 = 133 K

In this case the outer wall radiates 30,000 W out into space and so the system is in equilibrium.

Yet the inner wall is radiating 1,824,900 W.

You can see the maths in Part One. From a few comments on this blog, and some comments I saw elsewhere, clearly many people still have problems with it.

That’s a good thing. It’s good because it means that the simple problem is doing its job. It is simple enough that almost everyone realizes the calculated temperatures are correct. It is simple enough that the maths can be followed. But it exposes an idea that many can’t accept – total radiation from a surface inside the system is much higher than the energy source.

The solution is quite simple.

One of the commenters, John N-G, pointed out:

Your point might be driven home more emphatically by noting that the inner surface absorbs only 23.8 W/m2 from the super-light-bulb, yet in the 3m-thick example it emits the afore-mentioned 1452 W/m2 (and also absorbs 1452 W/m2 of wall-emitted radiation).

The inner surface is also in energy balance.

We can consider the values as total energy per second (W), or as energy per second per unit area, (W/m2). We will stay with the total energy because these are the values we have already been working with.

  • Energy in = 30,000 W (from the energy source) + 1,824,900 W (from the inner surface surface)
  • Energy out = 30,000 W (conducted through the PVC wall to the outer surface) + 1,824,900 W (radiated from the inner surface)

No energy is created or destroyed in the system encompassed by the inner surface.

Dynamic Results

From some comments related to this and other articles, explaining how equilibrium is reached might help. Once again, the 1st law of thermodynamics is used to calculate the dynamic situation. If there is net energy added to an element of the system then its temperature increases.

The approach to the calculation is a simple numerical model, which divides the sphere wall radially (into 50 equal “slices”):

The inner wall receives 30,000 W. It radiates an amount dependent on temperature but also receives that same amount from the inner wall surface (therefore the net heat received is always = 30,000 W).

The outer wall radiates according to its temperature.

And for all of the “slices” of the wall in between the heat flow is according to the temperature differential (equation in Part One), and the heat gained is according to the simple equation:

dQ = mc.dT    – where dQ = change in energy, m = mass, c= specific heat capacity and dT = change in temperature

Spherical symmetry makes the calculation much easier than if it was a box.

Conclusion

If the first law of thermodynamics meant that no inner surface could radiate at a higher value than the outer surface of the system then everything would be at the same temperature.

In the PVC hollow sphere example the inner wall would have to be at 133K – the same as the outer wall. This would mean that no heat could flow from the inner wall to the outer wall – or that Fourier’s law was wrong. And lagging hot water pipes was also “a big con”.

We all know that is not the case – well, maybe not everyone has heard of Fourier, but everyone knows about insulation.

The reason why so many people think that Trenberth and Kiehl’s diagram is flawed is because of an incorrect understanding of the first law of thermodynamics.

If Trenberth and Kiehl’s diagram violates the first law of thermodynamics then so does my PVC sphere. There’s just the small matter of trying to explain how the inner surface would stay at 133K.

At least there is one Get Out of Jail Free card. As a commenter on another blog put it:

The atmosphere cannot both behave like a PVC blackbody and an ideal gas

Analogies prove nothing. But for those brave enough to consider that they might be wrong, I hope the PVC hollow sphere provides some illumination.

Update: Do Trenberth and Kiehl understand the First Law of Thermodynamics? Part Three – The Creation of Energy?

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In Part One we took a look at what data was available for “back radiation”, better known as Downward Longwave Radiation, or DLR. And we saw that around many locations the typical DLR was in the order of 300 W/m2, and it didn’t decrease very much at night.

In Part Two we looked at several measured spectra of DLR which clearly demonstrated that this radiation is emitted by the atmosphere.

In this article we will consider what happens when this radiation reaches the ground. The reason we want to consider it is because so many people are confused about “back radiation” and have become convinced that either it doesn’t exist – covered in the previous two parts – or it can’t actually have any effect on the temperature of the earth’s surface.

The major reason that people give for thinking that DLR can’t affect the temperature is (a mistaken understanding of) the second law of thermodynamics, and they might say something like:

A colder atmosphere can’t heat a warmer surface

There are semantics which can confuse those less familiar with thermal radiation.

If we consider the specific terminology of heat we can all agree and say that heat flows from the warmer to the colder. In the case of radiation, this means that more is emitted by the hotter surface (and absorbed by the colder surface) than the reverse.

However, what many people have come to believe is that the colder surface can have no effect at all on the hotter surface. This is clearly wrong. And just to try and avoid upsetting the purists but without making the terminology too obscure I will say that the radiation from the colder surface can have an effect on the warmer surface and can change the temperature of the warmer surface.

Here is an example from a standard thermodynamics textbook:

From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)

From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)

Probably this diagram should be enough, but as the false ideas have become so entrenched let’s press on..

Note that this topic has been covered before in: Intelligent Materials and the Imaginary Second Law of Thermodynamics and The First Law of Thermodynamics Meets the Imaginary Second Law

The First Law of Thermodynamics

This law says that energy is conserved – it can’t be created or destroyed. What this means is that if a surface absorbs radiation it must have an effect on the temperature – compared with the situation where radiation was not absorbed.

There’s no alternative – energy can’t be absorbed and just disappear. However, as a technical note, energy can be absorbed into chemical bonds or phase changes of materials. So you can put heat into ice without changing the temperature, while the ice turns into water. Of course, energy is still not lost..

Therefore, if your current belief is that radiation from a colder atmosphere cannot “change the temperature” of the hotter surface then you have to believe that all of the radiation from the atmosphere is reflected.

Or, alternatively, you can believe that the first law of thermodynamics is flawed. Prove this and your flight to Sweden beckons..

Bouncers at the Door – or Quantum Mechanics and the Second Law of Thermodynamics

One commenter on an earlier post asked this question:

But if at the surface the temperature is higher than in the atmospheric source then might the molecules which might have absorbed such a photon be in fact unavailable because they have already moved to a higher energy configuration due to thermal collisions in the material which contains them?

Many people have some vague idea that this kind of approach is how the second law of thermodynamics works down at the molecular level.

It (the flawed theory) goes like this:

  1. the atmosphere emits “a photon”
  2. the photon reaches the surface of the earth
  3. because the temperature of the surface of the earth is higher the photon cannot be absorbed – therefore it gets “bounced”.

Except it’s not physics in any shape or form – it just sounds like it might be.

Let’s review a few basics. It’s important to grasp these basics because they will ensure that you can easily find the flaw in the many explanations of the imaginary second law of thermodynamics.

“They all look the same to me” – The Energy of a Photon

This part is very simple. The energy of a photon, E:

E = hν = hc/λ

where ν = frequency, λ = wavelength, c = speed of light, h = 6.6 ×10−34 J.s (Planck’s constant).

You can find this in any basic textbook and even in Wikipedia. So, for example, the energy of a 10μm photon = 2 x 10−20 J.

Notice that there is no dependence on the temperature of the source. Think of individual photons as anonymous – a 10μm photon from a 2,000K source has exactly the same energy as a 10μm photon from a 200K source.

No one can tell them apart.

Wavelength Dependence on the Temperature of the Source

Of course, radiation from different temperature sources do have significant differences – in aggregate. What most, or all, believers in the imaginary second law of thermodynamics haven’t appreciated is how similar different temperature Planck curves can be:

Blackbody radiation curves for -10'C (263K) and +10'C (283K)

Blackbody radiation curves for -10'C (263K) and +10'C (283K)

Notice the similarity between the 10°C and the -10°C radiation curves.

Alert readers who have pieced together these basics will already be able to see why the imaginary second law is not the real second law.

If a 0°C surface can absorb radiation from 10°C radiation, it must be able to absorb radiation from -10°C radiation. And yet this would violate the imaginary second law of thermodynamics.

What determines the ability of a surface to absorb or reflect radiation?

Absorptivity and Reflectivity of Surfaces

The reflectivity of a surface is a measurement of the fraction of incident radiation reflected. It’s very simple.

This material property has a wavelength dependence and (sometimes) a directional dependence. Here is a typical graph of a few materials:

Reflectivity vs wavelength for various surfaces, Incropera (2007)

Reflectivity vs wavelength for various surfaces, Incropera (2007)

As you can see, the variation of absorptivity/reflectivity with wavelength is very pronounced. Notice as well in this diagram from a standard textbook that there is no “source temperature” function. Of course, there can’t be, as we have already seen that the energy of a photon is only dependent on its wavelength.

Just to be clear – radiation incident on a surface (irradiation) can only be absorbed or reflected. (In the case of gases, or very thin surfaces, radiation can also be “transmitted” through to the other side of the material or gas).

Conclusion from Basic Physics

So from basic physics and basic material properties it should be clear that radiation from a colder surface cannot be all reflected while at the same time radiation from a warmer surface is absorbed.

And if any radiation is absorbed it must change the surface temperature and therefore violate the (imaginary) second law of thermodynamics.

You have to ditch something. I would recommend ditching the imaginary second law of thermodynamics. But you can choose – instead you could ditch the first law of thermodynamics, or the basic equation for the energy of a photon (make up your own), or invent some new surface properties.

While considering these choices, here’s another way to think about it..

If All the “Back Radiation” Was Reflected..

So let’s suppose you still think that all of the radiation from the atmosphere, all 300W/m2 of it, gets reflected.

That presents a problem even bigger than the tedious physics principles articulated above. Why is that?

Well, let’s take the earth’s average surface temperature of around 15°C (288K) and the typical emissivity of various surface types:

Emissivity vs wavelength of various substances, Wilber (1999)

Emissivity vs wavelength of various substances, Wilber (1999)

As you can see, the emissivity is pretty close to 1. So for a temperature of 15°C the Planck curve will be pretty close to a blackbody, and the total surface emitted radiation (“flux”) is given by the Stefan-Boltzmann equation of σT4 – so in the typical surface case:

j = 390 W/m2

Now we have to add the reflected surface radiation of 300W/m2. So the upward radiation from the surface will be around 690 W/m2.

Here is one result from a very thorough experiment, The Energy Balance Experiment, EBEX 2000 (reference below):

Upward and downward radiation measurements, EBEX 2000, Kohsiek (2007)

Upward and downward radiation measurements, EBEX 2000, Kohsiek (2007)

The location was a cotton field of 800m × 1600m at coordinates 36°06’ N, 119°56’ W, approximately 20 km south-south-west of the town of Lemoore CA, USA. In this experiment, radiation measurements were taken at nine sites across the field along with wind and humidity measurements in an attempt to “close the energy budget” at the surface. Downward measurements were taken at a few of the sites (because the values wouldn’t change over a small distance) while upward measurements of both shortwave and longwave were taken at every site. Some sites measured the same value with two instruments from different manufacturers.

As you can see the upward longwave measurement is around 400-500 W/m2. The paper itself doesn’t record the temperature on that day, but typical August temperatures in that region peak above 35°C, leading to surface radiation values above 500 W/m2 – which is consistent with the measurements. [Update – the peak temperature measured at this location was 35°C on this day – thanks to Wim Kohsiek for providing this data along with the temperature graph for the day]

Temperature for 14 August 2000, from Wim Kohsiek, private communication

Temperature for 14 August 2000, from Wim Kohsiek, private communication

And so here it is the theoretical upward longwave radiation from the temperature graph (=σT4):

Calculated (theoretical) upward radiation, 14 August 2000

Calculated (theoretical) upward radiation, 14 August 2000

As you can see, this upward radiation calculation matches what was measured.

If the surface reflects all of the downward longwave radiation then the upward longwave measurement for these temperatures should be in the region of 700-800 W/m2.

There is a great opportunity for some enterprising people who still think that DLR is all reflected and not absorbed – buy a decent pyrgeometer and take some upward surface measurements to demonstrate that the whole science community is wrong and upward surface measurements really are 80% higher than everyone thinks. If you can afford an FT-IR to do a spectral analysis you will be able to prove your theory beyond a shadow of doubt – as the spectrum will have those characteristic CO2, O3 and water vapor peaks that were shown in DLR spectra in Part Two.

Conclusion

DLR is emitted by the atmosphere, reaches the surface and is absorbed by the surface. This absorption of energy changes the surface temperature.

The physics behind this are very basic and have been known for around 100 years.

Proving that the surface doesn’t absorb DLR should be a walk in the park for anyone with a small amount of cash. But only if it’s true.

The world we live in does absorb DLR and adding 300W/m2 to the surface energy budget is the reason why the surface temperatures are like they are.

Further reading – Do Trenberth and Kiehl understand the First Law of Thermodynamics?

Darwinian Selection – “Back Radiation”

References

The Energy Balance Experiment EBEX-2000. Part III: Behaviour and quality of the radiation measurements, Kohsiek et al, Boundary Layer Meteorology (2007)

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Do Trenberth and Kiehl understand the First Law of Thermodynamics?

Yes

But many people claim that they don’t after reviewing their well-known diagram from their 1997 paper, Earth’s Annual Global Mean Energy Budget:

From Trenberth and Kiehl (1997)

From Trenberth and Kiehl (1997)

The “problem” is – how can the absorbed solar energy be 235W/m2 when the radiation from the surface is 390W/m2? Where is this energy coming from?

Clearly they have created energy and don’t understand basic physics!

There have been many comments to that effect on this blog and, of course, on many other blogs.

Instead of pointing out that many of these values can be easily measured and checked, we will turn to a simple experiment which might help the many who believe the answer to the title is “No“.

The Thought Experiment

Picture, for the practical among you, building some kind of simple heat chamber in your garage. A wooden or a plastic box, with a light bulb in the center. You want to test whether some new gizmo really works at the high temperatures claimed. Or you want to find the melting point of gold.

The principle is simple – the thicker the material and the higher the energy from the bulb – the hotter it will get inside the heat chamber.

As a method of simplifying the calculations, my chamber will be spherical (because it makes the maths easier than when it is a box) and we will place it in the vastness of space and assume that the ambient temperature is 0.0K. Again, this is just to make the maths simpler to understand.

The inner radius of the sphere is 10m, and the thickness of the wall is 3m. (In a followup comment or post I will show how the values change with x).

The material used for this experiment is PVC which has a thermal conductivity of 0.19 W/m.K – I’ll explain a little more about this parameter in a moment. Probably down at such low temperatures the thermal conductivity won’t be this value but it doesn’t matter too much. We will also assume that the emissivity = 0.8.

You can see on the diagram that the outer surface temperature is T2, the temperature inside the sphere is T1 and the “ambient” is 0K. We don’t yet know what T1 or T2 is, we want to find that out.

In the center, we have our super-light-bulb, which radiates 30,000W. It is mysteriously powered, perhaps it is a nuclear device, or just electric with such a thin power cord we can’t detect it – we don’t really care.

Now – the first law of thermodynamics – energy cannot be created or destroyed. So for our thought experiment, the system receives 30,000W. The “system” is the entire PVC sphere, and everything it encompasses, right to the outer surface. No other source of energy can be detected.

Start Your Engines

Now that we have turned on the energy source the inside of the sphere will heat up. It has to keep heating up until the energy flow out of the sphere is balanced by the energy being added inside the sphere.

How does heat flow out from the center of the sphere?

  • First, by conduction to the outer surface of the sphere
  • Second, by radiation from the outer surface of the sphere to the vastness of space

Both of these processes are governed by very simple equations which are shown in the maths section at the end. Here, I will just attempt to explain conceptually how the processes work.

We start with consideration of the complete system and after equilibrium is reached the energy gained will be equal to the energy lost.

Energy gained, q = 30,000W = Energy lost

(Note that we are considering energy per second). For a rigid stationary body in the vastness of space, the only mechanism for losing energy is radiating it. All bodies radiate according to their temperature and a property called emissivity. Using the Stefan-Boltzmann law, we can calculate that:

Outer surface temperature, T2 = 133K

With this temperature, at an emissivity of 0.8, the whole sphere is radiating away 30,000W.

Time Out

So at this point, surely everyone is in agreement. We have calculated the steady-state temperature of the outer surface of the sphere as 133K. We can see that the mysterious energy source of 30,000W is balanced by the outgoing 30,000W radiated away from the outer surface.

The first law of thermodynamics is still intact and no one has to fight about anything..

Systems check?

But What’s the Story Inside?

We also want to calculate T1, the temperature of the inner surface. This is also very easy to calculate. The only mechanism for transferring heat from the inner surface (where the energy source is located) to the outer surface is by conduction.

The maths is below but effectively heat is transferred through a wall when there is a temperature differential between two surfaces. The higher the differential, the more heat. And the property of the material that affects this process is called the thermal conductivity. When this value is high – like for a metal – heat is transferred very effectively. When this value is low – like for a plastic – heat is transferred much less efficiently.

Once the system is generating 30,000W internally the inner wall temperature will keep rising until 30,000W can flow through the wall and be radiated away from the outer surface.

If we use the simple maths to calculate the temperature differential we find that it is 290K.

That is, to get 30,000W to flow through a hollow sphere with inner radius 10m and outer radius 13m and conductivity of 0.19 W/m.K you need a temperature differential of 290K.

Which means that the inner surface is 423K.

Everyone still ok?

What is the Radiation Emitted from the Inner Surface?

With an emissivity of 0.8 and a surface temperature of 423K, the inner surface will be radiating at 1,452 W/m2.

So the total radiation from the inner surface will be 1,824,900W.

What??? You have created energy!!!

Before bringing out the slogans, find out which step is wrong. If you can’t find an incorrect step then perhaps you should consider the possibility that this system is not violating the first law of thermodynamics.

Well, perhaps everyone is comfortable with the idea that with sufficient insulation you can raise the inner temperature of a box or sphere to a very high value – without having to build a power station.

In any case, the system is not creating energy. Inner surfaces are receiving high amounts of radiation while also emitting high amounts of radiation – they are in balance.

And of course, this has nothing whatever to do with the earth’s climate system so everyone can rest easy..

Update – Do Trenberth and Kiehl understand the First Law of Thermodynamics? – Part Two

Do Trenberth and Kiehl understand the First Law of Thermodynamics? Part Three – The Creation of Energy?

And new article on the real basics – Heat Transfer Basics – Part Zero

Maths

The System

In equilibrium, the outer surface of the sphere has to radiate away all of the heat generated internally. This is the first law of thermodynamics.

The internal energy source,  q  = energy radiated from the outer surface

q = εσT24.4πr22 – the Stefan-Boltzmann equation for emitted energy per m2 x the surface area

where r2 = radius of the outer surface = 10 + 3 = 13

If q = 30,000W, r2 = 13m and ε = 0.8

T2 = q / (εσ.4πr22)1/4 and so T2 = 133K

If you recalculate back using the Stefan-Boltzmann law you will find that 133K with an emissivity of 0.8 radiates at 14.2 W/m2 (corrected-thanks to John N-G) and if you multiply that by the surface area of 4×3.14×132 = 2,124 m2 you find the emitted energy = 30,000W.

Conduction through the Sphere

The equation of heat conduction is very simple:

q = -kA . ΔT/Δx

This is for a planar wall. For a hollow sphere the equation is quite similar:

qr = -kA . dT/dr = – k (4πr2) . dT/dr and the important point is that qr is a constant, independent of r

After a small amount of maths we find that:

qr = 4πk . (T1 – T2) / (1/r1 – 1/r2)

So for the values of k = 0.19, r1 = 10, r2 = 13:

T1 – T2 = 290, therefore, T1 = 423K

Internal Radiation

Therefore, the radiated energy from the inner surface will be 1,452 W/m2 or a total of 1,824,900W (= εσT14.4πr12).

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I didn’t think that a Part Two would be needed after the initial installment – The Sun and Max Planck Agree

Anyway, I always appreciate commenters explaining why the article hasn’t done its job, and so following various comments, hopefully, this article can address the deficiencies of the first.

I recommend reading the First Article before diving into this one. The main thrust of the article was to explain that solar radiation and terrestrial radiation have quite different “signatures”, or properties, which enable us to easily tell them apart.

For reader unfamiliar with how radiation varies with wavelength, here are two “blackbody” curves:

 

Blackbody radiation curves for -10'C (263K) and +10'C (283K)

Blackbody radiation curves for -10'C (263K) and +10'C (283K)

 

What you can notice is that the curve for 10°C radiation is – at all wavelengths – greater than the curve at -10°C radiation. The peak radiation for both of these curves occurs around 10μm. This is Wien’s law where the peak occurs at:

λpeak = 2898/T

where λpeak is the wavelength of peak radiation. So for 283K (10°C) the peak wavelength = 10.6μm. But radiation – as you can see – occurs at all wavelengths.

The total radiation from a body varies in proportion to T4, and here is the comparison of solar (at 5780K) and terrestrial radiation (at 260-300K):

Where is the terrestrial radiation? It can’t be seen on a linear plot, it is so small.

Let’s see it on a log plot:

For those not used to seeing log plots, check out the left hand side axis – the peak of the 300K radiation is around 1000x lower than the solar radiation at that wavelength (each major division corresponds to a factor of 10). And the spectral intensity of the higher temperature radiation exceeds the value of the lower temperature radiation at every wavelength.

This is the radiation measurement you would get from a spaceship parked just off the surface of the sun (for the solar radiation at 5780K) and just off the earth (for the terrestrial radiation of 260 – 300K).

The earth is some distance away from the sun and so the radiation at the top of the earth’s atmosphere will be reduced quite a lot.

If we consider the radiation from the sun expressed as per m2 then the solar radiation incident on the earth’s atmosphere will be reduced by a factor of (rsun / distancesun-earth)2 – this is known as the inverse square law – and there is a nice explanation at Wikipedia.

The amount the solar radiation is reduced for top of atmosphere = (696×106/150×109)2 = 1 / 2152 = 1 / 46,225.

(If we calculated it the other way, we would work out the total solar radiation from the whole surface area and conclude that this value must be divided by 2×109).

So if we look at the incident solar radiation at top of atmosphere (TOA) compared with terrestrial radiation:

We can see that the cross-over is quite small.

And this is not the whole story. Some solar radiation is reflected from the earth (on average around 30%). And depending on the angle of the sun from the zenith, the amount of solar radiation per m2 will be reduced accordingly.

So the above graph is the best case.

If we take the average of 30% of solar reflected, and at an angle from the zenith of 45°, we will have these curves:

You can see the crossover is very low.

Just for completeness, here it is on a linear plot, with the crossover section clear:

Hopefully, it’s clear, hopefully, there is no possibility of confusion – solar radiation can be easily distinguished from terrestrial radiation within the earth’s climate system.

If we measure radiation > 4μm we can conclude it is terrestrial and if we measure radiation < 4μm we can conclude it is solar.

Of course, at night, it is even clearer.

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