YES!!! This is a different equation! It's not an equation for heat transfer! It's the Stefan-Botlzmann RELATION between radiative power out and temperature for gray bodies. It is used for calculating RADIATIVE POWER OUT versus TEMPERATURE and vice versa. It is not for heat transfer and I'm not using it for heat transfer. YOU are the one who is getting them confused, not me. This other equation shows that radiative power is dependent ONLY on emissivity and temperature. It does not depend on other bodies. For the third time (today): it's a temperature vs. power equation, not a heat transfer equation. Further, "electrical" is your own addition. The equation is for power. It doesn't specify "electrical". [Jane Q. Public, 2014-09-22]
My equation for electrical power is different than the equation for radiative power out, which is why it's bizarre that Jane keeps using the equation for radiative power out to determine electrical power. That's what I've been trying to tell Jane: we don't disagree about the equation for radiative power out. The equation for radiative power out is simply a part of the equation for conservation of energy: power in = power out through a boundary where nothing inside is changing. That's why we need to use a heat transfer equation to determine electrical heating power, not just an equation for radiative power out.
... it is not necessary to account for cooler bodies in the temperature versus power out equation.... The second equation you cited above is the STANDARD equation for calculating radiative power out of a gray body. I showed you where it was in Wikipedia. It also just happens to be in my heat transfer textbooks. The answer is 82.12 W/m^2. It is the textbook answer. It isn't going to change. Why don't you look it up in a textbook and discover that for yourself?... Radiative power out of the warmer body is dependent ONLY on emissivity and thermodynamic temperature. Anything else violates the second law of thermodynamics. It isn't controlled or mitigated by nearby cooler bodies.... [Jane Q. Public, 2014-09-22]
I've already agreed that it's not necessary to account for cooler bodies in the temperature versus power out equation. Again, we're not disputing the equation for radiative power out. We're disputing the equation describing conservation of energy around a boundary drawn around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
This doesn't violate the equation for radiative power out. It simply uses that equation to account for the power flowing out of the boundary, and uses that same equation for radiative power to describe radiative power flowing into the boundary.
... I will repeat: I did not and do not claim that no radiation is absorbed. Just no net radiative power. Any that does get absorbed is just re-transmitted... [Jane Q. Public, 2014-09-22]
Jane's been calculating the required electrical heating po
Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't repeatedlyobject to including a term for radiation from the chamber walls in his calculation of required electrical power. Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.
NO!!! This is just plain bullshit. I do NOT object to a term for electrical power.... I don't object to a term for "electrical power" and never have.... [Jane Q. Public, 2014-09-22]
I never said Jane objected to a term for "electrical power". I said Jane repeatedlyobjects to including a term for radiation from the chamber walls in his calculation of required electrical power. And Jane continues to do this:
... I simply asserted a physical truth: in our isolated system, the electrical power to the heat source, called for by Spencer, has zero dependency on the chamber walls. It is this nonsense dependency on the chamber walls that I have disputed, nothing else. That is a violation of the Stefan-Boltzmann law.... My only objection is your insistence that the power input to the heat source is somehow related to radiation from the chamber walls. If these are treated as gray bodies: just no. That's a violation of Stefan-Boltzmann. [Jane Q. Public, 2014-09-22]
Ranting about imaginary violations of the Stefan-Boltzmann law won't help Jane understand physics. It might help Jane to draw a boundary around the heat source and think carefully about exactly why Jane keeps ignoring the heat radiated in from the chamber wells. Accounting for that radiation doesn't "violate the Stefan-Boltzmann law" but ignoring it violates conservation of energy.
... The power output is not dependent on the chamber walls, therefore the power input is not dependent on the chamber walls.... [Jane Q. Public, 2014-09-21]
Why does Jane think the second part follows from the first? It doesn't. For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls. Maybe Jane could explain why he wrote "therefore" when his reasoning fails to describe even a simple black body problem? (Keep in mind that the gray body equation has to reduce to the black body equation when emissivities = 1.)
Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.
NO!!! Repeat, for about the 100th time now: no NET radiative power input from cooler objects. That is ALL I have claimed, and it's a direct result of the Stefan-Botlzmann radiation law. Why do you keep disputing textbook physics laws? Stop lying. Because that's all you're doing now. [Jane Q. Public, 2014-09-22]
Jane/Lonny Eachus can capitalize "NET" all he wants, but it doesn't change the fact that
... What's ridiculous is your constant repetition of this bullshit idea. Yes, the cooler walls radiate inward but they have no effect whatsoever on the heat source. ALL of that radiation is reflected or scattered by the heat source. (It is not transmitted because we're dealing with diffuse gray bodies of significant mass.)... [Jane Q. Public, 2014-09-21]
It's truly surreal to watch Jane repeatedly double-down on nonsense which Jane claims is too ridiculous even for Sky Dragon Slayers (as if that were possible!).
... You took a badly-worded sentence or two and jumped on them as though Latour made a mistake. But his only mistake was wording a couple of sentences badly. He does in fact NOT suggest that warmer objects absorb no radiation, and he has written as much many times.... You have refuted NOTHING but a couple of unfortunately-worded sentences, which Latour himself publicly corrected shortly after that post appeared.... [Jane Q. Public, 2014-07-27]
Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't repeatedlyobject to including a term for radiation from the chamber walls in his calculation of required electrical power. Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.
... shortly after Latour published that blog post, it became clear that the language he used implied that no radiation at all was absorbed by the warmer body. So a reader could not reasonably be blamed for inferring that. But Latour quickly apologized for the unfortunate wording and corrected himself to make it very clear he was referring to net, not absolute, heat transfer.... [Jane Q. Public, 2014-07-27]
Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body. Otherwise Jane's calculation of the required electrical power would include a term for radiation from the chamber walls. Since Jane adamantly insists that this term can't be included, Jane's calculation assumes that no radiation at all is absorbed by the source. None. Zero.
... Repeat: this ASSUMPTION of yours that the chamber walls must be accounted for in the power requirement of the heat source is a direct violation of the Stefan-Boltzmann law. There are no 2 ways around it. Established physics (the Stefan-Boltzmann law) says that the radiative power out (and therefore power in) of a gray body is dependent ONLY on emissivity and thermodynamic temperature. It is completely unrelated to any nearby cooler bodies.... [Jane Q. Public, 2014-09-21]
Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary. But "power in" through a boundary around the heat source looks like this:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
Jane refuses to account for the chamber wall radiative "power in" which would only be true if the source didn't absorb any of that radiation. Zero.
If you are sincere (you certainly haven't been acting like you are), then you must be postulating some kind of "tractor beam" effect that allows the chamber wall to "suck" power out of the heat source from a distance. I assure you that at least at out current level of technology, we have not managed to build such a sucking device. The heat source radiates out what it radiates out, and nothing around it is "sucking" any power from it. Although you seem to be doing your very best at "sucking" my time away over stupid bullshit. [Jane Q. Public, 2014-09-21]
That's ridiculous, Jane. I'm just noting that the chamber walls are hotter than 0K, so they emit radiation into a boundary around the heat source. Therefore Jane's wrong to ignore that radiation when applying the principle of conservation of energy:
... Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source.... [Jane Q. Public, 2014-09-15]
It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source.
So the only "heat source" where we could validly ignore the radiation from the chamber walls would be a perfectly reflective "bobble" from Vernor Vinge's Marooned in Realtime. I assure you that at our current level of technology, we haven't managed to build such a device. And even if we could, it wouldn't be a heat source.
... No NET radiative energy. I did not claim "none at all", and I have repeatedly pointed this out to you. Just no NET transfer from cooler to warmer.... [Jane Q. Public, 2014-09-20]
Jane's equation claims "none at all":
electrical power per square meter = (s)*(e)*Ta^4
Since Jane's equation for required electrical power doesn't even include a term for radiation from the chamber walls, Jane's equation wrongly says that no radiation at all is absorbed by the source. None. Zero.
It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source. Slayer "physics" are incoherent nonsense.
... Before that it can't, because Ta^4 - Tb^4 is a positive number so no net radiant energy is absorbed by (a) from (b). That means all the way up to the exact point thermal equilibrium is achieved, all radiant power is a result of electrical power, therefore the power input and power output are constant. It is not a "gradual" process.... [Jane Q. Public, 2014-09-20]
So Jane claims:
electrical power per square meter = (s)*(e)*Ta^4
The actual answer is:
electrical power per square meter = (s)*(e)*(Ta^4 - Tb^4)
Since Jane refuses to include a term accounting for radiation from the chamber walls, Jane's equation is saying that no radiation at all is absorbed by the warmer source. Why?
... Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source.... [Jane Q. Public, 2014-09-15]
Of course it is! Again, this is just Sky Dragon Slayer nonsense. Absorption doesn't work like Slayers imagine. It's controlled by the surface's absorptivity, which doesn't change if the source is slightly warmer or cooler than its surroundings. All that's required for the source to absorb radiation (from warmer or colder objects) is having absorptivity > 0. Since the source has absorptivity = 0.11, some radiative power from the chamber walls is absorbed by the heat source.
Jane's been regurgitating Slayer nonsense for years:
... Warmer objects cannot, and do not absorb lower-energy radiation from cooler objects.... [Jane Q. Public, 2012-11-20]
Then how do uncooled IR detectors see cooler objects? How did we detect the 2.7K cosmic microwave background radiation with warmer detectors?
... explain how radiation that is of a LOWER "black-body temperature" will be absorbed by a body of a HIGHER black-body temperature.... [Jane Q. Public, 2013-05-30]
... An object that is radiating at a certain black-body temperature WILL NOT absorb a less-energetic photon from an outside source. This is am extremely well-known corollary of the Second Law.... [Jane Q. Public, 2013-05-30]
No, that's a Slayer fantasy. On the atomic scale, absorption of radiation doesn't depend on temperature because individual atoms don't have temperatures. Only very large groups of atoms have temperatures. Individual photons also don't have temperatures. Very large groups of photons from a 10C warm object have slightly different average wavelength curves than a -10C cold object, but they're very similar. This means that even if temperature somehow applied at the atomic scale of absorbing individual photons, an atom couldn't tell if a photon came from the 10C warm object or the -10C cold object.
... You took a badly-worded sentence or two and jumped on them as though Latour made a mistake. But his only mistake was
... If body (b) is brought up to the same temperature as (a), Ta^4 - Tb^4 = 0, and no net heat transfer takes place. Although radiant power output of (a) at that temperature doesn't change, as a corollary of that same law.... [Jane Q. Public, 2014-09-19]
If Ta = Tb, no electrical heating power is required. But radiant power output of (a) doesn't change. So radiant power output can't be equal to electrical heating power. Using conservation of energy, can you write down an equation which yields the required electrical heating power given Ta and Tb?
Since the emissivity for every object in our system is the same, power output is proportional to the T^4. Period. End of story. Draw your boundary around the heat source. Power in = power out (your own principle). Therefore the power in is 41886.54 Watts, which is the power initially being radiated out. SPENCER stipulated that this power is held constant. It wasn't my idea. It's a condition of the experiment. [Jane Q. Public, 2014-09-19]
No. Once again, in this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate...."
Jane's even stumbled across this point:
... Of course it wouldn't need a separate heat source if its environment were maintained at 150 degrees.... [Jane Q. Public, 2014-09-15]
Of course! That's why the variable Jane's holding constant isn't the electrical power supplied to the separate heat source. If Jane can realize that there's no need for a separate heat source if its environment were maintained at 150 degrees, why can't Jane see that his equation for required electrical power doesn't reflect this obvious fact?
... you have confirmed that you have not abandoned your incorrect (and actually quite ludicrous) version of heat transfer, which violates the Stefan-Boltzmann radiation law on its very face.... [Jane Q. Public, 2014-09-15]
... or maybe we disagree about which variable to hold constant.
Instead of holding electrical heating power constant, Jane held the source's radiative power output constant. That held source temperature constant and forced electrical heating power to change. Solving this problem using both sets of boundary conditions shows that Jane's solution forces electrical heating power to drop by a factor of two after the shell is added.
These two sets of boundary conditions are very different, just like Neumann boundary conditions are different from Dirichlet boundary conditions. Upon hearing that a disagreement might be caused by holding different variables constant, a real skeptic might consider working the problem again while holding that other variable constant. But Jane can't even admit there's a difference between holding electrical heating power constant and holding the source's radiative power output constant. Jane even insists he held electrical heating power constant, despite the evidence.
So Jane won't solve this problem with the electrical heating power constant. That's unfortunate, because it's critical:
"... critical to the whole experiment is that, like the sun heating the surface of the Earth, there is energy being continuously pumped into the system from outside...."
1. Holding electrical heating power constant while adding an enclosing shell is like doubling CO2 while holding solar heating power constant, then calculating how much Earth's surface warms.
2. Holding source temperature constant while adding an enclosing shell is like doubling CO2 while holding Earth's surface temperature constant, then calculating how much solar heating power would have to drop to keep Earth's surface temperature constant.
Even if Jane doesn't want to solve that first problem, he should recognize that it's different from the second problem Jane actually solved.
"In thermodynamics, where a surface has a prescribed heat flux, such as a perfect insulator (where flux is zero) or an electrical component dissipating a known power."
"In thermodynamics, where a surface is held at a fixed temperature.
Dr. Spencer's thought experiment placed Neumann boundary conditions on the source and Dirichlet boundary conditions on the chamber walls. Instead, Jane placed Dirichlet boundary conditions on the chamber walls and the source.
In other words, the electrical heating power is determined by drawing a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
... Since emissivity doesn't change the input required to heat source to achieve 150F is constant, regardless of where it comes from. But as long as the walls of the chamber are cooler than the source, NONE of the power comes from the chamber walls... [Jane Q. Public, 2014-09-15]
But if the chamber walls are also at 150F, they're not cooler than the source and the input required to heat the source to 150F is zero.
... do you still maintain that after the enclosing passive sphere is inserted, the central heat source raises in temperature to approximately 241 degrees F? You haven't said anything about that in a while, so I'm just checking. [Jane Q. Public, 2014-09-15]
Once again, if the electrical heating power is held constant, the heat source has to warm. Once agin, Jane's heat source keeps the source temperature constant by halving its electrical heating power. Jane/Lonny Eachus might ask himself why his required electrical heating power goes down by a factor of two after the enclosing shell is added.
... Of course it wouldn't need a separate heat source if its environment were maintained at 150 degrees.... [Jane Q. Public, 2014-09-15]
In other words, the electrical heating power is determined by drawing a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F. The answer is YES, and here is why: You are proposing to bring the whole system up to a level of higher thermodynamic energy, rather than just the heat source. And you are somehow proposing that it doesn't take more energy to do that. But of course it does. The power required to bring the heat source up to 150F remains the same, because the Stefan-Boltzmann law says it has to be. But NOW, you are ALSO bringing the walls up to that higher temperature, and THAT would require even more power (because of the slightly larger surface area). [Jane Q. Public, 2014-09-15]
Again, that's completely ridiculous. I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant.
For example, you could simply place the vacuum chamber somewhere with an ambient temperature of 150F. That would require zero power, but once again it doesn't matter even if the vacuum chamber were on Pluto. Because that power never crosses the boundary.
Either way, as long as the chamber walls are held at 150F, the heat source would need absolutely no electrical heating power to remain at 150F. Zero. Period.
You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F. The answer is YES... [Jane Q. Public, 2014-09-15]
Here's our disagreement. Conservation of energy demands that a heat source at 150F requires no electrical heating power inside 150F vacuum chamber walls.
You're either disputing conservation of energy, or you're not calculating the actual electrical heating power. If you're calculating the actual electrical heating power, your calculation has to account for radiation from the chamber walls because it passes in through that boundary. That's why the electrical heating power would be zero if the chamber walls were also at 150F!
Nonsense. This is textbook heat transfer physics. We have a fixed emissivity. Therefore, according to the Stefan-Botlzmann radiation law, the ONLY remaining variable which determines radiative power out is temperature. NOTHING else. That's what the law says: (emissivity) * (S-B constant) * T^4. That's all. Nothing more. This makes it stupidly easy to calculate the radiative power out, and therefore the necessary power in. [Jane Q. Public, 2014-09-15]
It's "stupidly easy" to calculate radiative power out and power in through what boundary? The boundary you're describing has to include the source's radiative power passing out through it, without including radiative power from the chamber walls passing in. I think that's impossible, but feel free to explain exactly where such a boundary would be drawn.
One question only: do you agree with the Stefan-Boltzmann relation: power out P = (emissivity) * (S-B constant) * T^4 ?? No more bullshit. "Yes" if you agree that equation is valid, or "No" if you deny that it is valid. Just that and no more. I'm not asking your permission. I'm just trying to find out whether you're actually crazy or just bullshitting. [Jane Q. Public, 2014-09-15]
Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law. But I've obviously failed to communicate that the power from the chamber walls has to pass in through that boundary, so you're only using half the equation to calculate the electrical heating power.
The REASON there would not be as great a power DIFFERENCE if the chamber walls were also at 150F, is that the walls would themselves be radiating more power out, so there would be less heat transfer (in that case 0). It is NOT, as you assert, because the heat source would be using less power. That's false, by the S-B equation. Its power output remains the same because (Spencer's stipulation) the power input remains the same. The reason my solution does not violate conservation of energy, is that the power consumption of the chamber wall is allowed to vary. THAT is where the change takes place, not at the heat source. Again, this is a stipulation of Spencer's challenge. Once again: power out of heat source remains constant, because P = (emissivity) * (S-B constant) * T^4. There is nothing in these conditions that changes this at all. Therefore, BECAUSE the power out and power in at the heat source remain constant, so does the temperature. It's all in that one little equation. [Jane Q. Public, 2014-09-15]
Once again, no. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
"Power in" has to include the radiative power passing in through the boundary. Otherwise energy isn't conserved, because power in = power out through any boundary where nothing inside that boundary is
I am disputing nothing of the sort. As I have explained many times now, you are not drawing your lines properly. You keep making the same bullshit assertions, after I have proved them false. Why do you do this? You're just going to look that much more foolish later. [Jane Q. Public, 2014-09-13]
You're either disputing conservation of energy, or you're not calculating the actual electrical heating power. If you're calculating the actual electrical heating power, your calculation has to account for radiation from the chamber walls because it passes in through that boundary. That's why the electrical heating power would be zero if the chamber walls were also at 150F!
Can we agree that the required electrical heating power would be zero if the chamber walls were also at 150F?
It's so adorable that Jane keeps insisting that Jane kept the power constant, even after I showed that Jane's calculation was only able to hold the source temperature constant after the enclosing shell was added by halving the actual electrical heating power.
It's also adorable that Jane keeps ignoring the fact that his "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.
NO!!! I have told you 5 or 6 or maybe more times now, this is a VIOLATION of the very straightforward Stefan-Boltzmann law. How it applies in this situation is quite straightforward, and not at all as complex as you are making it out to be. Radiant power output of a gray body is calculated using ONLY the variables: emissivity and temperature. THAT IS ALL. There is no other variable dealing with incident radiation, or anything else. When the system is at radiant steady-state, power out (and therefore power in) are easily calculated, and I have calculated them. Further, Spencer's "electrical" input power was to the heat source, not to the whole system. YOUR OWN PRINCIPLE: power in = power out. Now you're trying to contradict yourself and say it meant something else. It's just bullshit. You're squirming like a fish on a hook. You just don't seem to realize you have already been flayed, filleted, and fried in batter. You're owned, man. [Jane Q. Public, 2014-09-13]
No. Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:
So are you disputing that power in = power out through a boundary where nothing inside that boundary is changing with time? Or are you disputing that the radiation from the chamber walls passes through a boundary drawn just inside them?
And again, if you keep ignoring that "power in" half of the equation that all Sky Dragon Slayers miss, you'll have to keep wondering why your "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.
... It's not that I don't agree. You might come up with the right answer for some sub-calculation. I don't know, I don't care, and I'm not even going to bother to check, much less agree. The issue is that I have already solved the problem, and arrived at the correct answer (within reasonable limits). So I don't HAVE to agree or disagree with you. I've already done it, according to the correct textbook-approved physics. AND (unlike you) I checked my work and it checks out. And unlike your answer it doesn't violate conservation of energy.... [Jane Q. Public, 2014-09-13]
I just showed that Jane/Lonny Eachus solved the "correct answer" to a different question. Instead of holding the electrical heating power constant like Dr. Spencer did, Jane/Lonny held the source temperature constant. In that case, the electrical heating power required to keep the source at 150F drops by a factor of two after the enclosing shell is added. This shows that holding the electrical heating power constant like Dr. Spencer did is different than holding the source temperature constant like Jane/Lonny did.
... SIMPLE CALCULATION, which I have already shown several times: power "sufficient" to heat the heat source under initial conditions to 150F: 41886.54 Watts. Power input at the source remains constant.... [Jane Q. Public, 2014-09-13]
No, in your example the electrical heating power drops by a factor of two after the enclosing shell is added. And once again, your calculation of the power sufficient to heat the heat source would be exactly the same if the chamber walls were also at 150F. But the right answer there is zero, because an electric heater wouldn't be necessary. Is this really so hard to understand, or are you deliberately spreading misinformation?
... For a given gray body, its thermodynamic temperature is related ONLY to emissivity, radiant power output, and the S-B relation (emissivity)* (S-B constant) * T^4. PERIOD. That's physics.... [Jane Q. Public, 2014-09-13]
And that's why what you're calculating isn't Dr. Spencer's electrical heating power, because it should be "zero" if the chamber walls are also at 150F.
... I repeat: given your OWN "draw a border around it" thermodynamic reasoning, the power input (whether it is electrical, chemical, or something else) must equal that output. That's physics. You're trying to bring in energy from elsewhere, but it isn't relevant to this calculation AT ALL; it is erroneous thinking. Power input is specified to be constant. Calculating the total power in initial conditions is, as I stated before, "dirt simple". Specified emissivity is known: 0.11. Temperature is known: 338.71K. Solving for the above we get 82.12 W/m^2. We already have ALL the information needed to calculate this, given the Stefan-Boltzmann relation (above), relating these numbers. Nothing else is required, and in fact trying to introduce other factors is ERROR. That is what the accepted science says.... [Jane Q. Public, 2014-09-13]
If you draw a boundary around the heated source, you have to account for the 0F chamber walls because they're radiating power in through the boundary. Otherwise you're not actually calculating Dr. Spencer's electrical heating power, or you misunderstand conservation of energy.
So it seems like in your interpretation, Dr. Spencer's challenge is basically: "Assuming the source temperature is held fixed, does the source temperature change after a passive plate is added?"
If the power input to the heated sphere is fixed, then the power output in the form of radiant temperature is fixed: (epsilon)(sigma)T^4. It's physics! It doesn't matter how you try to squirm and twist this. You have been owned. End of story. [Jane Q. Public, 2014-09-13]
Jane, didn't it seem odd that you interpreted Dr. Spencer's challenge to mean "Assuming the source temperature is held fixed, does the source temperature change after a passive plate is added?"
How is that different than asking "Assume x = 150 forever. Will x change?"
Isn't that a silly question? Shouldn't you at least consider the possibility that you've misinterpreted "power input to the heat source"?
No. Holding constant the electrical power heating the source is very different than holding constant the source temperature. Like Jane, let's assume the source temperature is constant (rather than the electrical heating power) and use Jane's equation and notation:
... we have 4 surfaces, which I will call 1, 2, 3, 4 moving outward, so 1 is the surface of the heat source, 2 the inside of the hollow sphere, 3 the outside of the hollow sphere, and 4 the chamber wall. T3 for example would be radiative Temperature of surface 3.... [Jane Q. Public, 2014-09-10]
Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:
Now add the hollow sphere and draw a boundary between the source (T1=150F) and the inside of the hollow sphere (T2). A different "electricity_final" flows in, and heat transfer p(12) flows out.
Now draw a boundary between the outside of the hollow sphere (T3=T2) and the chamber walls (T4=0F): "electricity_final" flows in, and heat transfer p(34) flows out. Since power in = power out:
So if the source temperature is held constant at 150F, adding the hollow sphere reduces the necessary electrical heating power to keep the source at 150F by a factor of two, from 55.6 to 27.8 W/m^2.
... input power at steady-state is fixed, and a value that we already know: 41886.54 W.... [Jane Q. Public, 2014-09-12]
Again, we disagree about what's held fixed. That value you keep calculating isn't the constant electrical power heating the source.
In this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate...."
In my interpretation, Dr. Spencer's challenge is basically: "Assuming an electric heater pumps energy at a constant rate to the source, does the source temperature change after a passive plate is added?"
You've repeatedlynoted that there are no other factors involved in calculating your 82 W/m^2 (41886.54 W) value. So if it's held fixed, the source temperature is also held fixed.
So it seems like in your interpretation, Dr. Spencer's challenge is basically: "Assuming the source temperature is held fixed, does the source temperature change after a passive plate is added?"
... Power input to the heat source is constant. It is sufficient to heat the source to 150 deg. F (338.71K). Given the known temperature, and the emissivity, we compute the power out with (epsilon)(sigma)(338.71^4) = 82.12 W/m^2. Using that radiant emittance and the fixed, agreed upon area we get 41886.54 Watts total radiated power output.... [Jane Q. Public, 2014-09-11]
Once again, the constant electric power is sufficient to heat the source to 150F when it's surrounded by chamber walls at 0F. That's the initial condition in the experiment that we agreed on. Your "82 W/m^2" value isn't the constant electrical power sufficient to heat a 150F source inside 0F chamber walls.
Again, if you want to see why your calculation doesn't yield the power input to the heat source, just ask what power input would be necessary if the chamber walls were also at 150F. In that case Dr. Spencer's electric heater wouldn't be necessary, so that power input would be zero.
Since your "82 W/m^2" calculation can't do that, it's not the electric heater power that's held constant. On the other hand, your 55.6 W/m^2 calculation would be zero if the chamber walls were at 150F. So it represents the constant electrical power in your analysis. Hold it constant as Dr. Spencer said, and you'll obtain the correct solution if you correctly apply the principle of conservation of energy.
Any heat transfer which doesn't cross the boundary can't be included because it can't change the total amount of energy inside the boundary.
PRECISELY! Here you are confirming, once again, my explanation of how you got it wrong.... [Jane Q. Public, 2014-09-11]
No, I explained why you can't add heat transfer from heat source to the inside of the enclosing plate to the heat transfer from the outside of the enclosing plate to the wall to get 55.6 W/m^2 from the shell to the chamber walls. Again, that's because any heat transfer which doesn't cross the boundary can't be included because it can't change the total amount of energy inside the boundary.
Any heat transfer which doesn't cross the boundary can't be included because it can't change the total amount of energy inside the boundary.
PRECISELY! Here you are confirming, once again, my explanation of how you got it wrong. You assumed the total radiant power output of the heat source was also being put out by the outside of the hollow sphere, through the "boundary" you drew around it. BUT... as I very clearly explained, that is not so. The hollow sphere has TWO surfaces, of nearly equal area. So the power output at the outside surface is actually only approximately HALF of what you thought it was. Because your calculations (I still have them) assume 511.346 m^2 when the actual radiating surface area is 511.346 m^2 + 511.186 m^2 = 1022.53 m^2. [Jane Q. Public, 2014-09-11]
No. I've assumed that the electrical power heating the source to 150F inside 0F chamber walls is constant. (Note that this constant rate would be zero if the walls were at 150F.) That's the assumption we disagree on. I never assumed the total radiant power output of the heat source was also being put out by the outside of the hollow sphere. Maybe the fact that we disagree about what's held constant (the electrical heating power to keep the source at 150F inside 0F chamber walls) is leading to yet another miscommunication?
... The formula for radiant power is (e * s) * area * T^4. Period. This is according to the Stefan-Boltzmann law, and no other variables are required at steady-state. The initial temperature of the heat source was 150F, or 338.71K. So we agreed that the input power to the heat source is sufficient for the equation (e * s) * (heat source area) * 338.71^4. The power input doesn't change.... the total power output (and therefore power input) at the heat source, in initial conditions, was (we agreed on this) 82.12 W/m^2 * 510.065 m^2 = 41886.54 Watts. Power in = power out.... [Jane Q. Public, 2014-09-11]
No. We've never agreed that the unchanging power input (my "constant electrical heating power") is "82 W/m^2". I've repeatedlyfailed to explain that the constant electrical heating power would only be "82 W/m^2" if the chamber walls were 0K blackbodies.
In this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate...."
Note that the constant rate of Dr. Spencer's electric heater would equal zero if the chamber walls were also at 150F. So any calculation of this crucial constant rate would also need to be zero in the case of chamber walls at 150F.
Since Jane's "82 W/m^2" value isn't the constant electrical heating power needed to keep the source at 150F inside 0F chamber walls, it isn't held constant. Here's where Jane actually calculated the constant electrical power heating the source inside 0F chamber walls:
... Calculate initial (denoted by "i") heat transfer from heat source to chamber wall. We are doing this only to check our work later.... = 55.5913 [W/m^2]... [Jane Q. Public, 2014-09-10]
So Jane's source needs 55.6 W/m^2 of constant electrical heating power to stay at 150F inside 0F chamber walls. This value is held constant. After the enclosing shell is added and temperatures stabilize, conservation of energy demands that net heat transfer out equals Jane's 55.6 W/m^2. Does it?
... you should at least have tried drawing your boundary around your own goddamned heat source, both for initial conditions and your final result, to check your work. But you didn't. What you got was a universe-busting violation of conservation of energy.... [Jane Q. Public, 2014-09-11]
No, I drew that boundary for both initial and final conditions to guarantee conservation of energy. In fact, I repeatedlysuggested that you check your work by drawing a boundary between the source and the enclosing shell at your proposed steady-state temperatures, then calculating power in = power out using the original constant electrical power you calculated before the source was enclosed.
Let's do that:
Jane's constant electrical power of 55.6 W/m^2 flows into that boundary. At steady-state, power in = power out. But power out through that boundary is the
... if you need to "draw a boundary", it needs to be drawn around the passive plate itself. We have already firmly established that your "boundary" around the heat source and the "enclosing shell" is even thermodynamically incorrect.... [Jane Q. Public, 2014-09-11]
Good grief. How predictably ridiculous. All boundaries where nothing inside changes have power in = power out. Seriously. All of them. That's why I tried to convince you that this general principle is true, but obviously we'll have to agree to disagree.
We have already shown that your particular application of "drawing boundaries" here was a MISAPPLICATION of the principle you are trying to use.... [Jane Q. Public, 2014-09-11]
Jane agreed that the general principle is true that power in = power out through a boundary where nothing inside the boundary is changing. But now that this general principle contradicts Slayer dogma, Jane considers it a misapplication.
Jane might wonder why cooler power wasn't included in "power in = power out": because it just moves energy from one point outside the boundary to another point that's also outside the boundary. In the same way, energy moved from one point inside the boundary to another point that's also inside the boundary isn't included in the equation describing conservation of energy.
... You should have drawn your shell around THAT, and that alone. And you should at least have tried drawing your boundary around your own goddamned heat source, both for initial conditions and your final result, to check your work. But you didn't. What you got was a universe-busting violation of conservation of energy.... [Jane Q. Public, 2014-09-11]
Ironically, power in = power out through all the boundaries I've constructed. That includes the boundary around my own "goddamned heat source". But that's not true for Jane's solution, because it violates conservation of energy.
... The rate of energy transfer from surface 1 to 2 is (p12) = (e*s) * ( T1^4 - T2^4 ). And T1 is known!... (e*s) * ( 338.71^4 - T2^4 )... [Jane Q. Public, 2014-09-10]
The enclosed source temperature at steady state is known to be 338.71 K (150F)? No, absolutely not. The chamber wall temperature is constant at 0F, and the electrical power heating the source is constant. But the enclosed source temperature is only constant in Jane's PSI Sky Dragon Slayer bizarro world.
Jane assumed the source's final enclosed steady state temperature was exactly the same as before it was enclosed. Surprise, Jane found that the source didn't warm! As a result, he got nonsensical answers and had to invent a new energy conservation law where power adds to the energy inside a boundary even if it never crosses that boundary.
We have already shown that your particular application of "drawing boundaries" here was a MISAPPLICATION of the principle you are trying to use.... [Jane Q. Public, 2014-09-10]
Jane agreed that the general principle is true that power in = power out through a boundary where nothing inside the boundary is changing. But now that this general principle contradicts Slayer dogma, Jane considers it a misapplication.
Jane might wonder why cooler power wasn't included in "power in = power out": because it just moves energy from one point outside the boundary to another point that's also outside the boundary. In the same way, energy moved from one point inside the boundary to another point that's also inside the boundary isn't included in the equation describing conservation of energy.
... You should have drawn your shell around THAT, and that alone. And you should at least have tried drawing your boundary around your own goddamned heat source, both for initial conditions and your final result, to check your work. But you didn't. What you got was a universe-busting violation of conservation of energy.... [Jane Q. Public, 2014-09-10]
Ironically, power in = power out through all the boundaries I've constructed. That includes the boundary around my own "goddamned heat source". But that's not true for Jane's solution, because it violates conservation of energy.
... The rate of energy transfer from surface 1 to 2 is (p12) = (e*s) * ( T1^4 - T2^4 ). And T1 is known!... (e*s) * ( 338.71^4 - T2^4 )... [Jane Q. Public, 2014-09-10]
The enclosed source temperature at steady state is known to be 338.71 K (150F)? No, absolutely not. The chamber wall temperature is constant at 0F, and the electrical power heating the source is constant. But the enclosed source temperature is only constant in Jane's PSI Sky Dragon Slayer bizarro world.
Jane assumed the source's final enclosed steady state temperature was exactly the same as before it was enclosed. Surprise, Jane found that the source didn't warm! As a result, he got nonsensical answers and had to invent a new energy conservation law where power adds to the energy inside a boundary even if it never crosses that boundary.
But again, did you ever consider drawing a boundary between the source and the enclosing shell at your proposed steady-state temperatures, then calculating power in = power out using the original constant electrical power you calculated before the source was enclosed?
Slashdot Mirror
My equation for electrical power is different than the equation for radiative power out, which is why it's bizarre that Jane keeps using the equation for radiative power out to determine electrical power. That's what I've been trying to tell Jane: we don't disagree about the equation for radiative power out. The equation for radiative power out is simply a part of the equation for conservation of energy: power in = power out through a boundary where nothing inside is changing. That's why we need to use a heat transfer equation to determine electrical heating power, not just an equation for radiative power out.
I've already agreed that it's not necessary to account for cooler bodies in the temperature versus power out equation. Again, we're not disputing the equation for radiative power out. We're disputing the equation describing conservation of energy around a boundary drawn around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
This doesn't violate the equation for radiative power out. It simply uses that equation to account for the power flowing out of the boundary, and uses that same equation for radiative power to describe radiative power flowing into the boundary.
Jane's been calculating the required electrical heating po
I never said Jane objected to a term for "electrical power". I said Jane repeatedly objects to including a term for radiation from the chamber walls in his calculation of required electrical power. And Jane continues to do this:
Ranting about imaginary violations of the Stefan-Boltzmann law won't help Jane understand physics. It might help Jane to draw a boundary around the heat source and think carefully about exactly why Jane keeps ignoring the heat radiated in from the chamber wells. Accounting for that radiation doesn't "violate the Stefan-Boltzmann law" but ignoring it violates conservation of energy.
Why does Jane think the second part follows from the first? It doesn't. For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls. Maybe Jane could explain why he wrote "therefore" when his reasoning fails to describe even a simple black body problem? (Keep in mind that the gray body equation has to reduce to the black body equation when emissivities = 1.)
Jane/Lonny Eachus can capitalize "NET" all he wants, but it doesn't change the fact that
It's truly surreal to watch Jane repeatedly double-down on nonsense which Jane claims is too ridiculous even for Sky Dragon Slayers (as if that were possible!).
Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't repeatedly object to including a term for radiation from the chamber walls in his calculation of required electrical power. Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.
Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body. Otherwise Jane's calculation of the required electrical power would include a term for radiation from the chamber walls. Since Jane adamantly insists that this term can't be included, Jane's calculation assumes that no radiation at all is absorbed by the source. None. Zero.
Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary. But "power in" through a boundary around the heat source looks like this:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
Jane refuses to account for the chamber wall radiative "power in" which would only be true if the source didn't absorb any of that radiation. Zero.
That's ridiculous, Jane. I'm just noting that the chamber walls are hotter than 0K, so they emit radiation into a boundary around the heat source. Therefore Jane's wrong to ignore that radiation when applying the principle of conservation of energy:
It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source.
So the only "heat source" where we could validly ignore the radiation from the chamber walls would be a perfectly reflective "bobble" from Vernor Vinge's Marooned in Realtime. I assure you that at our current level of technology, we haven't managed to build such a device. And even if we could, it wouldn't be a heat source.
Jane's equation claims "none at all":
electrical power per square meter = (s)*(e)*Ta^4
Since Jane's equation for required electrical power doesn't even include a term for radiation from the chamber walls, Jane's equation wrongly says that no radiation at all is absorbed by the source. None. Zero.
It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source. Slayer "physics" are incoherent nonsense.
So Jane claims:
electrical power per square meter = (s)*(e)*Ta^4
The actual answer is:
electrical power per square meter = (s)*(e)*(Ta^4 - Tb^4)
Since Jane refuses to include a term accounting for radiation from the chamber walls, Jane's equation is saying that no radiation at all is absorbed by the warmer source. Why?
Of course it is! Again, this is just Sky Dragon Slayer nonsense. Absorption doesn't work like Slayers imagine. It's controlled by the surface's absorptivity, which doesn't change if the source is slightly warmer or cooler than its surroundings. All that's required for the source to absorb radiation (from warmer or colder objects) is having absorptivity > 0. Since the source has absorptivity = 0.11, some radiative power from the chamber walls is absorbed by the heat source.
Jane's been regurgitating Slayer nonsense for years:
Then how do uncooled IR detectors see cooler objects? How did we detect the 2.7K cosmic microwave background radiation with warmer detectors?
No, that's a Slayer fantasy. On the atomic scale, absorption of radiation doesn't depend on temperature because individual atoms don't have temperatures. Only very large groups of atoms have temperatures. Individual photons also don't have temperatures. Very large groups of photons from a 10C warm object have slightly different average wavelength curves than a -10C cold object, but they're very similar. This means that even if temperature somehow applied at the atomic scale of absorbing individual photons, an atom couldn't tell if a photon came from the 10C warm object or the -10C cold object.
If Ta = Tb, no electrical heating power is required. But radiant power output of (a) doesn't change. So radiant power output can't be equal to electrical heating power. Using conservation of energy, can you write down an equation which yields the required electrical heating power given Ta and Tb?
No. Once again, in this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."
Jane's even stumbled across this point:
Of course! That's why the variable Jane's holding constant isn't the electrical power supplied to the separate heat source. If Jane can realize that there's no need for a separate heat source if its environment were maintained at 150 degrees, why can't Jane see that his equation for required electrical power doesn't reflect this obvious fact?
... or maybe we disagree about which variable to hold constant.
Instead of holding electrical heating power constant, Jane held the source's radiative power output constant. That held source temperature constant and forced electrical heating power to change. Solving this problem using both sets of boundary conditions shows that Jane's solution forces electrical heating power to drop by a factor of two after the shell is added.
These two sets of boundary conditions are very different, just like Neumann boundary conditions are different from Dirichlet boundary conditions. Upon hearing that a disagreement might be caused by holding different variables constant, a real skeptic might consider working the problem again while holding that other variable constant. But Jane can't even admit there's a difference between holding electrical heating power constant and holding the source's radiative power output constant. Jane even insists he held electrical heating power constant, despite the evidence.
So Jane won't solve this problem with the electrical heating power constant. That's unfortunate, because it's critical:
"... critical to the whole experiment is that, like the sun heating the surface of the Earth, there is energy being continuously pumped into the system from outside. ..."
1. Holding electrical heating power constant while adding an enclosing shell is like doubling CO2 while holding solar heating power constant, then calculating how much Earth's surface warms.
2. Holding source temperature constant while adding an enclosing shell is like doubling CO2 while holding Earth's surface temperature constant, then calculating how much solar heating power would have to drop to keep Earth's surface temperature constant.
Even if Jane doesn't want to solve that first problem, he should recognize that it's different from the second problem Jane actually solved.
To see this difference, solve a problem with Neumann boundary conditions:
"In thermodynamics, where a surface has a prescribed heat flux, such as a perfect insulator (where flux is zero) or an electrical component dissipating a known power."
... then solve the same problem with Dirichlet boundary conditions:
"In thermodynamics, where a surface is held at a fixed temperature.
Dr. Spencer's thought experiment placed Neumann boundary conditions on the source and Dirichlet boundary conditions on the chamber walls. Instead, Jane placed Dirichlet boundary conditions on the chamber walls and the source.
But if the chamber walls are also at 150F, they're not cooler than the source and the input required to heat the source to 150F is zero.
Once again, if the electrical heating power is held constant, the heat source has to warm. Once agin, Jane's heat source keeps the source temperature constant by halving its electrical heating power. Jane/Lonny Eachus might ask himself why his required electrical heating power goes down by a factor of two after the enclosing shell is added.
In other words, the electrical heating power is determined by drawing a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
Right?
Again, that's completely ridiculous. I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant.
For example, you could simply place the vacuum chamber somewhere with an ambient temperature of 150F. That would require zero power, but once again it doesn't matter even if the vacuum chamber were on Pluto. Because that power never crosses the boundary.
Either way, as long as the chamber walls are held at 150F, the heat source would need absolutely no electrical heating power to remain at 150F. Zero. Period.
Here's our disagreement. Conservation of energy demands that a heat source at 150F requires no electrical heating power inside 150F vacuum chamber walls.
It's "stupidly easy" to calculate radiative power out and power in through what boundary? The boundary you're describing has to include the source's radiative power passing out through it, without including radiative power from the chamber walls passing in. I think that's impossible, but feel free to explain exactly where such a boundary would be drawn.
Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law. But I've obviously failed to communicate that the power from the chamber walls has to pass in through that boundary, so you're only using half the equation to calculate the electrical heating power.
Once again, no. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
"Power in" has to include the radiative power passing in through the boundary. Otherwise energy isn't conserved, because power in = power out through any boundary where nothing inside that boundary is
You're either disputing conservation of energy, or you're not calculating the actual electrical heating power. If you're calculating the actual electrical heating power, your calculation has to account for radiation from the chamber walls because it passes in through that boundary. That's why the electrical heating power would be zero if the chamber walls were also at 150F!
Can we agree that the required electrical heating power would be zero if the chamber walls were also at 150F?
It's so adorable that Jane keeps insisting that Jane kept the power constant, even after I showed that Jane's calculation was only able to hold the source temperature constant after the enclosing shell was added by halving the actual electrical heating power.
It's also adorable that Jane keeps ignoring the fact that his "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.
No. Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:
electricity_initial = p(14) = (e)(s) * ( T1^4 - T4^4 )
So are you disputing that power in = power out through a boundary where nothing inside that boundary is changing with time? Or are you disputing that the radiation from the chamber walls passes through a boundary drawn just inside them?
And again, if you keep ignoring that "power in" half of the equation that all Sky Dragon Slayers miss, you'll have to keep wondering why your "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.
I just showed that Jane/Lonny Eachus solved the "correct answer" to a different question. Instead of holding the electrical heating power constant like Dr. Spencer did, Jane/Lonny held the source temperature constant. In that case, the electrical heating power required to keep the source at 150F drops by a factor of two after the enclosing shell is added. This shows that holding the electrical heating power constant like Dr. Spencer did is different than holding the source temperature constant like Jane/Lonny did.
No, in your example the electrical heating power drops by a factor of two after the enclosing shell is added. And once again, your calculation of the power sufficient to heat the heat source would be exactly the same if the chamber walls were also at 150F. But the right answer there is zero, because an electric heater wouldn't be necessary. Is this really so hard to understand, or are you deliberately spreading misinformation?
And that's why what you're calculating isn't Dr. Spencer's electrical heating power, because it should be "zero" if the chamber walls are also at 150F.
If you draw a boundary around the heated source, you have to account for the 0F chamber walls because they're radiating power in through the boundary. Otherwise you're not actually calculating Dr. Spencer's electrical heating power, or you misunderstand conservation of energy.
Jane, didn't it seem odd that you interpreted Dr. Spencer's challenge to mean "Assuming the source temperature is held fixed, does the source temperature change after a passive plate is added?"
How is that different than asking "Assume x = 150 forever. Will x change?"
Isn't that a silly question? Shouldn't you at least consider the possibility that you've misinterpreted "power input to the heat source"?
No. Holding constant the electrical power heating the source is very different than holding constant the source temperature. Like Jane, let's assume the source temperature is constant (rather than the electrical heating power) and use Jane's equation and notation:
Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:
electricity_initial = p(14) = (e)(s) * ( T1^4 - T4^4 ) = (e)(s) * (8908858139.78) = 55.5913 W/m^2
Now add the hollow sphere and draw a boundary between the source (T1=150F) and the inside of the hollow sphere (T2). A different "electricity_final" flows in, and heat transfer p(12) flows out.
electricity_final = p(12) = (e)(s) * ( T1^4 - T2^4 )
Now draw a boundary between the outside of the hollow sphere (T3=T2) and the chamber walls (T4=0F): "electricity_final" flows in, and heat transfer p(34) flows out. Since power in = power out:
electricity_final = p(34) = (e)(s) * ( T2^4 - T4^4 )
Combine these two equations:
T1^4 - T2^4 = T2^4 - T4^4
Solve for:
T2 = T3 = 305.47K = 90.176 deg. F.
electricity_final = 27.8 W/m^2.
So if the source temperature is held constant at 150F, adding the hollow sphere reduces the necessary electrical heating power to keep the source at 150F by a factor of two, from 55.6 to 27.8 W/m^2.
Can we agree on that?
Again, we disagree about what's held fixed. That value you keep calculating isn't the constant electrical power heating the source.
In this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."
In my interpretation, Dr. Spencer's challenge is basically: "Assuming an electric heater pumps energy at a constant rate to the source, does the source temperature change after a passive plate is added?"
You've repeatedly noted that there are no other factors involved in calculating your 82 W/m^2 (41886.54 W) value. So if it's held fixed, the source temperature is also held fixed.
So it seems like in your interpretation, Dr. Spencer's challenge is basically: "Assuming the source temperature is held fixed, does the source temperature change after a passive plate is added?"
Is that right?
Once again, the constant electric power is sufficient to heat the source to 150F when it's surrounded by chamber walls at 0F. That's the initial condition in the experiment that we agreed on. Your "82 W/m^2" value isn't the constant electrical power sufficient to heat a 150F source inside 0F chamber walls.
Again, if you want to see why your calculation doesn't yield the power input to the heat source, just ask what power input would be necessary if the chamber walls were also at 150F. In that case Dr. Spencer's electric heater wouldn't be necessary, so that power input would be zero.
Since your "82 W/m^2" calculation can't do that, it's not the electric heater power that's held constant. On the other hand, your 55.6 W/m^2 calculation would be zero if the chamber walls were at 150F. So it represents the constant electrical power in your analysis. Hold it constant as Dr. Spencer said, and you'll obtain the correct solution if you correctly apply the principle of conservation of energy.
No, I explained why you can't add heat transfer from heat source to the inside of the enclosing plate to the heat transfer from the outside of the enclosing plate to the wall to get 55.6 W/m^2 from the shell to the chamber walls. Again, that's because any heat transfer which doesn't cross the boundary can't be included because it can't change the total amount of energy inside the boundary.
No. I've assumed that the electrical power heating the source to 150F inside 0F chamber walls is constant. (Note that this constant rate would be zero if the walls were at 150F.) That's the assumption we disagree on. I never assumed the total radiant power output of the heat source was also being put out by the outside of the hollow sphere. Maybe the fact that we disagree about what's held constant (the electrical heating power to keep the source at 150F inside 0F chamber walls) is leading to yet another miscommunication?
No. We've never agreed that the unchanging power input (my "constant electrical heating power") is "82 W/m^2". I've repeatedly failed to explain that the constant electrical heating power would only be "82 W/m^2" if the chamber walls were 0K blackbodies.
In this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."
Note that the constant rate of Dr. Spencer's electric heater would equal zero if the chamber walls were also at 150F. So any calculation of this crucial constant rate would also need to be zero in the case of chamber walls at 150F.
Since Jane's "82 W/m^2" value isn't the constant electrical heating power needed to keep the source at 150F inside 0F chamber walls, it isn't held constant. Here's where Jane actually calculated the constant electrical power heating the source inside 0F chamber walls:
So Jane's source needs 55.6 W/m^2 of constant electrical heating power to stay at 150F inside 0F chamber walls. This value is held constant. After the enclosing shell is added and temperatures stabilize, conservation of energy demands that net heat transfer out equals Jane's 55.6 W/m^2. Does it?
No, I drew that boundary for both initial and final conditions to guarantee conservation of energy. In fact, I repeatedly suggested that you check your work by drawing a boundary between the source and the enclosing shell at your proposed steady-state temperatures, then calculating power in = power out using the original constant electrical power you calculated before the source was enclosed.
Let's do that:
Jane's constant electrical power of 55.6 W/m^2 flows into that boundary. At steady-state, power in = power out. But power out through that boundary is the
Good grief. How predictably ridiculous. All boundaries where nothing inside changes have power in = power out. Seriously. All of them. That's why I tried to convince you that this general principle is true, but obviously we'll have to agree to disagree.
Jane agreed that the general principle is true that power in = power out through a boundary where nothing inside the boundary is changing. But now that this general principle contradicts Slayer dogma, Jane considers it a misapplication.
Jane might wonder why cooler power wasn't included in "power in = power out": because it just moves energy from one point outside the boundary to another point that's also outside the boundary. In the same way, energy moved from one point inside the boundary to another point that's also inside the boundary isn't included in the equation describing conservation of energy.
Ironically, power in = power out through all the boundaries I've constructed. That includes the boundary around my own "goddamned heat source". But that's not true for Jane's solution, because it violates conservation of energy.
The enclosed source temperature at steady state is known to be 338.71 K (150F)? No, absolutely not. The chamber wall temperature is constant at 0F, and the electrical power heating the source is constant. But the enclosed source temperature is only constant in Jane's PSI Sky Dragon Slayer bizarro world.
Jane assumed the source's final enclosed steady state temperature was exactly the same as before it was enclosed. Surprise, Jane found that the source didn't warm! As a result, he got nonsensical answers and had to invent a new energy conservation law where power adds to the energy inside a boundary even if it never crosses that boundary.
Jane agreed that the general principle is true that power in = power out through a boundary where nothing inside the boundary is changing. But now that this general principle contradicts Slayer dogma, Jane considers it a misapplication.
Jane might wonder why cooler power wasn't included in "power in = power out": because it just moves energy from one point outside the boundary to another point that's also outside the boundary. In the same way, energy moved from one point inside the boundary to another point that's also inside the boundary isn't included in the equation describing conservation of energy.
Ironically, power in = power out through all the boundaries I've constructed. That includes the boundary around my own "goddamned heat source". But that's not true for Jane's solution, because it violates conservation of energy.
The enclosed source temperature at steady state is known to be 338.71 K (150F)? No, absolutely not. The chamber wall temperature is constant at 0F, and the electrical power heating the source is constant. But the enclosed source temperature is only constant in Jane's PSI Sky Dragon Slayer bizarro world.
Jane assumed the source's final enclosed steady state temperature was exactly the same as before it was enclosed. Surprise, Jane found that the source didn't warm! As a result, he got nonsensical answers and had to invent a new energy conservation law where power adds to the energy inside a boundary even if it never crosses that boundary.
But again, did you ever consider drawing a boundary between the source and the enclosing shell at your proposed steady-state temperatures, then calculating power in = power out using the original constant electrical power you calculated before the source was enclosed?