About a year ago, researcher Willis Eschenbach proposed a simple problem at WUWT that relates conduction and radiation in order to show that, at steady state, Cold Side Radiation (CSR) equals Conductive Heat Flux (CHF), and that therefore my articles (here and here) “must” be in error. Here’s a brief outline:
|Absorptivity = Emissivity = 0.95||Thermal conductivity: k = 0.8 W/(m*K)|
|Volume = 1 m³ … A = L = 1||No radiation to / from stars|
|Adiabatic Wall on 4 of 6 sides||Input = 1360 W/m² (~Sun at Earth Distance)|
Willis’ solution setup was:
We have two conditions that must be met at steady-state. First, the amount of energy entering the block must be equal to the amount of energy leaving the block. The amount entering is equal to 1360 * epsilon, which I’ve said is the emissivity (and thus the absorptivity) at all frequencies. The amount leaving the block is equal to sigma epsilon (T_hot^4 +T_cold^4). So the first equation is:
sigma epsilon (T_hot^4 + T_cold^4) == 1360 epsilon [eqn1]
The second condition at steady-state is that the flow through the block has to be equal to the flow out of the cold side. The flow through the block is k (T_hot – T_cold), and the cold side radiation is sigma epsilon T_cold ^4, so the second equation is:
sigma epsilon T_cold^4 == k (T_hot – T_cold) [eqn2]— Willis’ Comment
The majority of commenters had pretty much the same idea. I will call Willis et al.‘s solution Wal for short, so as not to single out poor Willis.
Let’s clean up the equations:
This is actually a complicated simultaneous equation to solve by hand. I liked Greg’s gnuplot solution the best:
|Sunny (Hot) Side||383.338 K||1163.2 W/m²|
|Far (Cold) Side||221.410 K||129.5 W/m²|
So is this correct?
I’m afraid not. One of Wal’s two criteria is an insistence that the block must radiate out to empty space. The implicit justification for this is Stefan-Boltzmann’s Law, as Wikipedia explains:
That’s all good and well, but I don’t recall Boltzmann doing or analyzing experiments with radiation into the void of space, but only inside cavities. Are cavities matter or space?
And what is up with j*? What does the star mean? Wikipedia doesn’t go into that very important detail. But it does accidentally hint at it:
The fact that the energy density of the box containing radiation is proportional to T^4 can be derived using thermodynamics.
A box you say? Interesting. The star probably indicates a potential, not guaranteed radiation.
It is common for textbooks to completely leave out the detail and context of the development of Stefan-Boltzmann’s Law (SB Law). For example:
This gives the impression that a body will radiate into the void the space with nothing in sight. And so people who don’t know any better might assume that’s the case!
But how did that chapter begin?
Oh, it begins with a hot object inside a vaccum chamber! Later on …
Kirchoff’s Law is also derived from a “small body contained in a large isothermal enclosure”.
As you’re probably aware, SB Law is intimately connected to Planck’s Law. In fact, it’s just the integral over all wavelengths using Planck’s spectral density formula. So it’s very important to know how Planck derived his formula.
You caught it? “in a cavity”, “no net flow of energy between the body and its environment”.
But, if a body was forced to emit radiation to empty space, (without getting equal radiation in return) that would violate the “no net flow of energy between the body and its environment” rule. This is obviously not a problem “in a cavity”.
Planck’s law describes the unique and characteristic spectral distribution for electromagnetic radiation in thermodynamic equilibrium, when there is no net flow of matter or energy. Its physics is most easily understood by considering the radiation in a cavity with rigid opaque walls. Motion of the walls can affect the radiation. If the walls are not opaque, then the thermodynamic equilibrium is not isolated. It is of interest to explain how the thermodynamic equilibrium is attained. There are two main cases: (a) when the approach to thermodynamic equilibrium is in the presence of matter, when the walls of the cavity are imperfectly reflective for every wavelength or when the walls are perfectly reflective while the cavity contains a small black body (this was the main case considered by Planck); or (b) when the approach to equilibrium is in the absence of matter, when the walls are perfectly reflective for all wavelengths and the cavity contains no matter.— Wikipedia
All I see is cavity, cavity, cavity. Where’s the radiation to nowhere? Nowhere. Let’s see the starting steps of Planck’s Law derivation:
The derivation of Planck’s Law starts off by considering how much photon energy can fit in a box. As you can see, the dimension of the box matters! The intensity of photons in each wavelength is constrained by length (L)!
Only later is the volume of the box taken out to find the spectral density. The key thing to note is that you can only derive Planck’s Law by using Wave Theory, never Particle Theory. And you need these waves to go from matter to matter, never matter to nothing. If you try to think about waves going from matter to nothing, why would there be a constrained by length (L)? There would be no such constraint in this case. But if there’s no L constraint, then there should be no difference between intensity at different wavelengths. Then Planck’s Law would be wrong, the blackbody curve would be wrong, and SB Law would be wrong. They’re not wrong. There’s simply no emission from matter to NO thing.
For matter not enclosed in such a cavity, thermal radiation can be approximately explained by appropriate use of Planck’s law.— Wikipedia
The appropriate use of Planck’s Law and SB Law would incorporate the View Factor(s) between 2 or more surfaces.
The source gives a lot of examples of different view factors for different problems.
The main point is that actual heat transfer occurs between matter, never matter and nothing.
I like the resister analogy. Notice the space resistance. According to Wal’s theory, empty space has a view factor of 1, thus their space resistance is 1/A, or just 1 by chosen parameters. In actuality, empty space has a view factor of ZERO, and thus the resistance to emit to empty space is infinite. Thus it doesn’t happen.
What Wal would like to do is treat space as a surface with a view factor of 1, and drop the T₂ ( and ε₂ ) terms.
Let’s think about why this is physically inappropriate. If space really had a view factor of 1, every textbook heat transfer problem would have arrows drawn out in every direction from the hotter object, not just toward another surface. All those arrows would indicate a radiative drain on the hotter object. Every textbook would thus be incorrectly draining the hotter object and deceiving its students. Do you think textbooks are deceiving their students by offering a simplified view that neglects the “obvious” radiation to space?
Or maybe the more likely explanation is this radiation to empty space doesn’t happen?
Let’s see what NASA scientists think. Here’s a link to their paper: A METHOD FOR THE THERMAL ANALYSIS OF SPACECRAFT, INCLUDING ALL MULTIPLE REFLECTIONS AND SHADING AMONG DIFFUSE, GRAY SURFACES.
Do you see the emission to empty space? Me neither. It doesn’t exist.
Now I know this paper is from 1970, but it’s the best I could find on the topic. Being right doesn’t change with time, so there’s no need to update the fundamentals. If you can find emission to space inside NASA’s formulas elsewhere, please let me know.
If the lack of emission to space in this document is wrong, then NASA’s scientists are underestimating the cooling, and thus risking the lives of astronauts. All of these simultaneous equations would give the wrong results if the emission to empty space actually occurred. Luckily it doesn’t, and astronauts landed on the moon and returned safely because of it.
I hope I have convinced you that there’s no emission (cooling) to empty space. Matter only cools to matter matters!
Let’s move on. The other problem with Wal’s theory is their understanding of what Steady State implies in their problem.
They think that steady state means you set the conductive flux equal to the radiative output of the cold side. Wrong!
You need to set the heat transfer of the input to the conductive flux.
They do not satisfy the third criteria. The “heat in” is not 1360 W/m², that’s only the input in the 1st second to a block at 0 Kelvin.
There is no heat out at all on the space side, as we just discussed.
The proper solution to the problem was the minority opinion:
Both sides will come to the same temperature of 393.534 Kelvin. We satisfy all steady state criteria. Disagree? Look again, the 2nd criteria says “can” not “must”.
Why was this example important? Because mainstream climate scientists also feel that greenhouse gases block radiation to nothing just because they sense a lot on a remote satellite, which is not nothing. 🤣. They too are in error, but this discussion will come later.
Enjoy 🙂 -Zoe
I’m not suggesting an object can only cool if another object is present, only radiatively so. If an object’s internal energy is lowered, then it cools. Internal processes are not perpetual, they will degrade over time.
A paper with no mention of heat transfer to space.