I keep finding myself in a position where I feel I should explain, but I am at a loss as to why I should have to, because I am discussing this with someone who is supposed to have been a physics major.
You pointed out to MIT's derivation of energy transfer between infinite gray bodies. It does not apply here because (a) we have specifically defined areas, they are not infinite, and (b) that derivation makes use of Kircchoff's law which does not apply in Spencer's challenge.
This is a very simple but essential concept: I am not going to agree to calculations that were derived based on a physical principle that does not apply to the problem. This is a very basic concept. I do not agree with trying to solve a problem starting with invalid assumptions.
Once again, no. I never said that all surfaces were at the same temperature. I've already explained that the final outer temperature of the enclosing shell doesn't happen at the same time as the initial temperature of the heated plate. Initially, the heated plate is at 150F and the enclosing shell is cooler than 100F. But because power in > power out, the plates slowly warm to a new steady-state.
You did say it, quite clearly. I quoted you twice and linked to your web page. LATER you changed your tune. I can accept that you changed it later, but you did say it.
Further, in your link there, you say:
Jane's insistence that "a non-zero difference is all we need" between the heated plate's initial temperature of 150F and the enclosing plate's final temperature of ~150F was interesting. In this thought experiment [archive.today], the enclosing plate was initially cooler than 100F.
It is interesting, because it is the heart of the matter. Since, according to the S-B relation, if we are using gray bodies as you have several times insisted we use, the direction of net energy transfer via radiation at any given time is determined solely by the temperature difference and nothing else. Therefore (this is elementary logic), a non-zero difference in radiant temperature *IS* all we need, if we are using gray bodies, to determine which body is transferring energy to the other. At no time in this experiment are the temperatures equal, so net heat transfer is always in one direction and only one direction.
You then go on to say above:
By the time the outer temperature of the enclosing shell is ~149.6F (accounting for area differences), the heated plate is ~233.8F.
But you do not give any justification for this answer, you just throw it out there.
Earlier, you explained in some detail that you had calculated this number for thermal equilibrium using Kircchoff's radiation law. But as I've explained many times now, there is no thermal equilibrium so Kircchoff's law does not apply. Try again.
It feels as though I'm explaining to a high-school student who has never seen a physics problem before. Since the enclosing, passive plate is at all times cooler than the heat source, and therefore NET heat transfer is only outward from the heat source to that plate, then the only net energy input to the source remains the original electrical input.
So let's put 2 + 2 together here, which is really quite simple. We don't even need any math (but I welcome VALID math, if you can think of any to offer):
[1] Initial electrical input to heat source does not change.
[2] Since all other components of the system at all times remain cooler than the heat source, net power transfer to the rest of the system is invariably FROM the source TO the rest of the system. Nothing has been introduced to change that.
[3] Therefore, net input of power (energy per unit time) to the source remains a CONSTANT.
[4] Therefore, since any temperature of the source that is higher than the initial radiative equilibrium (150 deg. F) represents higher power output from that same source, any such higher temperature would violate conservation of energy.
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
NO, we do not agree with that, because as you state yourself that equation is derived from Kircchoff's radiation law, which does not apply here. Sheesh. How many different ways must I explain this?
Since HP basically got out of the calculator business, the HP 50G, which in my opinion is a better calculator anyway, has been available to the public in software form for free. It uses the actual ROM code from the 50G, which HP donated to the public domain.
You have to look around a bit, but versions are available for Mac, PC, and Linux.
Unlike the TI models, straight math can be entered in algebraic or RPN mode, and formulas can also use the "formula writer".
I've always like TI, and I have nothing against them, but over the years, having used various TI and comparable HP models, I've invariably thought the HP was superior.
If you replaced "businesses" with "large corporations", and "99%" with "85%", I might agree.
Our economy is still mostly driven by small business. Most small businesses and government do not have their hands in each other's pockets like big corporations do. Which of course is part of the problem. Nobody should.
My university took the attitude that computer science was an engineering discipline. You need to understand the theory, because the theory helps you classify and interpret that problem you're dealing with. But as an engineer, you're also on the hook for the process of design, and the actual design itself.
Certainly they do not want to just teach you a programming language, because that's like teaching a mechanical engineer the tools and settings for a single CAD program. Or an architect just how to draw blueprints.
But they still called it a computer science degree because I guess the world assumes people with a "software engineering" degree don't understand theory? I don't know why they made that choice.
I basically agree with you but the complaint seems to be that many people are coming out of university without the basic programming skills needed for real-world jobs. I'm not claiming it's true, I'm just saying that's the complaint.
Meh. It seems there is no end to clarifying. But it is also simplifying, in a sense, because it is eliminating irrelevant sidetracks:
I can conceive of a situation in which an enclosing, passive plate, of specific dimensions, might manage to be the same temperature on the inside and the outside in these circumstances. But I'm not going to bother getting sidetracked trying to do the calculations to either prove it or disprove it, because if it ever arose at all it would be a very rare special case, and whether it does or not is irrelevant to the central point.
To be even more clear, because I want to eliminate all misunderstands, this statement that I made above is incorrect:
Great. Except that it doesn't pertain to Spencer's challenge for several reasons. First, the chamber walls in Spencer's experiment are not "empty" space, but a material body that is being actively refrigerated, while the "enclosing passive plate" is being heated on the other side. So that plate is not in radiative equilibrium with the chamber wall or with anything else for that matter. In fact that would be impossible. There are other reasons why that description does not match Spencer's challenge, but that is irrelevant for now. One is enough.
Mea culpa. The outside of the enclosing passive plate would eventually reach radiative equilibrium with the chamber walls. But not thermal equilibrium. Further, the inside of the passive heated plate would reach radiative equilibrium with the heat source. But not thermal equilibrium in that case either. Nor, for that matter, is that same plate in thermal equilibrium even with itself, since realistically its inside and outside surfaces must be at different temperatures, in order to be at radiative equilibrium with those opposing surfaces.
Because I was incorrect to state that there is no radiative equilibrium, I was incorrect to state that a roughly analogous situation does not apply to Spencer's experiment. The opposing surfaces do reach radiative equilibrium. But it is still not very relevant here, because thermal (and therefore thermodynamic) equilibrium still does not exist.
Thermodynamic equilibrium is when every object is in thermal and radiative equilibrium as its surroundings.
Not a single object in Spencer's described challenge -- at any time -- meets these criteria. When everything is in thermal equilibrium -- as you have noted -- they are all at the same temperature. That never happens here.
The passive plate MAY be said to reach radiative equilibrium at some point... I stated that incorrectly before, and I apologize for doing so. I'm correcting it here so we don't have any misunderstandings.
There is no thermal equilibrium. Period. None. There MAY (and eventually would) arise a condition of radiative equilibrium for the (enclosing, passive, however you want to describe it) plate. But the other objects (heat source and chamber walls) do not meet this criteria because they are heated/cooled by means that may be other than radiative. "The system" is not in radiative equilibrium.
Without thermal equilibrium (which unequivocally does not occur here), Kircchoff's law does not apply, except perhaps in coincidental specific cases. It may not be assumed.
Those ESA absorptivities are for absorption of sunlight. Consider the first diagram here which shows that 6000K sunlight has much shorter wavelengths than the radiation from objects at the temperatures we're considering. In fact they hardly overlap. But the emissivities are for radiation emitted by much cooler objects. That's one reason why those ESA emissivities aren't equal to their absorptivities.
I repeat that these are under conditions of Earth-absorptive-surface insolation, which was what Spencer's experiment was supposed to emulate. But whatever.
But the emissivities are for radiation emitted by much cooler objects.
No, they are not. They are reported for the previously stated conditions: 1367 W/m^2 in incident radiation. But again: whatever. I already stated that this is not important enough to argue the point. I'm not conceding your point, but I'm willing to move on with gray bodies.
Once again, I never said that. In reality, I said that both sides of a thermal superconductor are at the same temperature. This was the source of much of the misunderstanding here, and you strongly objected to the notion of a thermal superconductor. Again, that's why I calculated the small temperature difference across an aluminum shell with finite conductivity.
Electric input of 509 W/m2 is constant and the walls are held at 0ÃF (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150ÃF (339K).
You were referring to "the second plate", as opposed to the "heated plate". That corresponds to what I have been calling the "passive" or "enclosing" plate. And you further referred to a supposed thermal equilibrium that doesn't exist.
Which fantasy would you prefer we believe? A thermal superconductor that makes no sense in this context, or an equilibrium which does not exist in this context? And you don't have the excuse that you meant "steady state", because the figure you gave would only be appropriate for actual equilibrium.
But enough of old arguments. Let's move on.
I was using this definition: "When incoming solar energy is balanced by an equal flow of heat to space, Earth is in radiative equilibrium and global temperatures become relatively stable."
Great. Except that it doesn't pertain to Spencer's challenge for several reasons. First, the chamber walls in Spencer's experiment are not "empty" space, but a material body that is being actively refrigerated, while the "enclosing passive plate" is being heated on the other side. So that plate is not in radiative equilibrium with the chamber wall or with anything else for that matter. In fact that would be impossible. There are other reasons why that description does not match Spencer's challenge, but that is irrelevant for now. One is enough.
Dr. Spencer disagrees: "Eventually the second plate will also reach a state of equilibrium, where its average temperature (letâ(TM)s say 100 deg. F) stays constant with time."
It is unfortunate that Spencer plays almost as fast-and-loose with terms as you do. That is a steady state. It is NOT "equilibrium". They are different things.
If you don't particularly mind, could we finally take the very first step in this calculation? Please?
Yes, I mind very much. There is no point in doing any calculations at all until we rid you of the false assumptions you have been making about this experiment (as I have been trying to do). They h
A better question, how much did they make selling this data?
Whether laws are heeded by corporations is dependent on a simple formula: what's to be gained by ignoring the law / (chance to get caught * fine). Unless that's below 1, the law becomes simply a cost factor to do business.
And even that equation is grossly unethical and doesn't backfire as often as it should. But it can, as Ford found out in the Pinto fiasco.
This is a ridiculously small settlement. How much is that per person? $3.70? AND -- this is just as big of a problem -- will ANY of those people who were actually harmed see any of that money?
This is what corporate cronyism (or what some people call "market capture") is all about. Government revolving door. It's a travesty and a tragedy. And it's also why I won't do business with Verizon.
I wouldn't say that learning to code necessarily outweighs a degree. But I do think university courses are too heavily focused on theory, and not enough practical application.
They complement each other. The big problem here (having gone though both, most but not all of the college being quite a while ago) is that a computer programmer back in the day HAD to know theory well, because programming was hard work! Input/output was so slow that you had to get it right the first time. Often you would present your code to somebody at a window to run on the mainframe, and if you were lucky you got a printout (!!!) the next day. If you got it wrong, a whole day was down the tubes.
Memory and storage were always in short supply, and CPU time was expensive. So everything had to be optimized. Sometimes for speed, sometimes for size, somethings a compromise of both. Theory was everywhere and you had to use it.
Heavy on theory, short on practice model that university CS was built upon, out of necessity. And they've kind of stuck with it, because universities are slow to change such things.
But I would also say that it is not a waste of time. As a practical programmer, theory will get you far. Look! De Morgan's Theorem just let me reduce those 5 lines of code to 2. You may not need to know linear algebra to work on sets of numbers, but if you do, hey, check it out. Now our program is half the size and our memory usage is down by 2 orders of magnitude.
So I don't think either one replaces the other. They complement each other. But I do think universities could concentrate, at least for their BS programs, a bit more on practical programming and just a bit less on theory.
In contrast, you're citing ESA figures from page 32 which are at 0K (-273C). But nothing in this experiment is anywhere near that cold.
Nonsense. They are figures at at incident radiation of 1367 W/m^2, which is sunlight at 1 AU, for the very reason that it is an approximation of Earth insolation. So in fact it would make a good representative example of what Spencer's model is supposed to be all about. Or do you (like Spencer) claim that space is "cold"?
But since you want to try to mischaracterize everything I say, and pick it all apart for reasons of your own, have it your own way. This is simply not very important.
You were right when you said [slashdot.org] the best we can realistically do is graybodies where emissivity = absorptivity. Otherwise we'd need to derive a new equation where heat transfer is an integral over wavelengths. In other words, we'd have to recreate MODTRAN [wikipedia.org]. I simply don't have time for that.
As I mentioned before, ESA gives observed values for integrated emissivity and absorptivity for aluminum. This is a good approximation and it is used in the real world for aluminum in a vacuum. If you really insist on gray bodies that's up to you; but I do not acknowledge that there is any legitimate reason to NOT use reasonable approximations of integrated absorptivity and emissivity.
We might be talking past each other. What you're calling steady-state is what I'm calling equilibrium. Radiative thermodynamic equilibrium doesn't require all surfaces to be at the same temperature, it simply means that temperatures don't change with time. At radiative equilibrium, power in = power out, which also means irradiance in = irradiance out.
You USED this before to ASSUME all surfaces were at the same temperature! I quoted you saying it in a post above, and you referenced that passage just the other day. In fact this was the source of much of the misunderstanding here. I did not understand why you were assuming some of the things you were assuming, and so my conclusion was that you were just messing with me. (And I am still not convinced that you were not.) THERE IS NO RADIATIVE EQUILIBRIUM HERE. THERE IS NO THERMAL EQUILIBRIUM HERE. None. You may not assume them.
I've already shown that MIT used Kirchhoff's law to derive heat transfer between gray bodies. I've already shown that Goodman 1957 tested the gray body approximation (Kirchhoff's law) and found that it's valid for aluminum at the temperatures in this experiment.
Note that my definition of equilibrium is consistent with this one: "In physics, radiative equilibrium is the condition where a steady state system is in dynamic equilibrium, with equal incoming and outgoing radiative heat flux and negligible heat transfer by conduction and convection."
In other words, irradiance in = irradiance out at radiative thermodynamic equilibrium. We're just using different words to describe the same concept.
Kircchoff's law (and MIT's example) both assume no bodies involved are storing thermal energy, and there is thermal equilibrium. In fact that is how Kircchoff's law is derived: technically Kircchoff's law only applies at thermal equilibrium. MIT was free to apply it in their example because th
Earlier, when I saw your mentions of equilibrium, I thought you were referring to the steady-state that would eventually be achieved. But even though you mentioned Kircchoff's law, it didn't sink in to my brain that you were referring to actual, literal equilibrium.
Uh-uh. As they say in my neck of the woods: it ain't happenin'.
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium. I promise to provide public-readable versions of my Sage worksheet from now on.
This is one of the whole problems with your analysis. THERE IS NO THERMODYNAMIC EQUILIBRIUM IN THIS EXPERIMENT. There is a steady-state, but no actual equilibrium. That is not possible, because we are actively pumping heat in at one "end", and pumping it out of the other.
Since one of the requirements of thermodynamic equilibrium is that all surfaces be at the same temperature, it will never be achieved because the experiment requires that the outside wall be maintained at 0 deg. F, yet we are still pumping significant heat in to the center.
Therefore Kircchoff's law does not apply to this experiment, and no situation arises in which the temperatures are the same everywhere, or the emissivities vs absorptivities. There is a steady-state arising from active (but constant) exchange. But there is no equilibrium.
As you said, the best we can realistically do is graybodies where emissivity = absorptivity. If you'd like to use a different emissivity just let me know, and we can both independently calculate the required electricity to check each other's answers.
After considering the situation I changed my mind. Since we are discussion what is supposed to be a real model of a real situation, we can use real emissivity and absorptivity. And the emissivity of aluminum (as you pointed out yourself some time ago) is different from the absorptivity by a factor of about 3. The ESA figures are observed figures for aluminum plates in near-vacuum, so those figures would appear to be perfect.
And as I stated before, I am busy and I don't have time to figure out your nomenclature right now. That's why I wanted to agree on one.
I do have one more comment I want to make tonight, though. I will reply to the relevant post of yours. It is pointless to continue 3 separate threads at once.
Nothing further can be inferred from their interest other than their serious and factual opinions that climate change remains a true and actual risk that our planet faces.
That's a real laugh. Despite your protests, there is a hell of a lot more that can be inferred from their interest. Your assertion that genuine risk assessment is the only factor in their decisions is the more unlikely scenario.
I already did. Global warming is a risk that actuaries can't actually calculate, but because the government says it's true they can claim it as a risk anyway. They can do their best to do some math on it, but ultimately they're basing it on something that isn't proven.
Historical risk assessment is based on statistics. There are no statistics about the hazards of global warming. Media and government have been proclaiming more "catastrophic weather events", but there haven't actually been any. Florida just set a record for number of years without a hurricane. The U.S. East coast is close to setting a record for number of years without a major hurricane. (Sandy was not "major". It only did as much damage as it did because it took an unusual path through expensive terroritory.)
Global cyclonic energy has been in a 40-year lull, and actually at a 30-year record or near-record low. Despite the drought in California this year there have been much worse droughts in California in living memory.
There isn't enough carbon dioxide in the entire atmosphere to even come close to "acidifying" the ocean if all of it were dissolved. Just ask a chemist.
So where is all this devastation the actuaries are supposed to calculate? Not only are there no statistics, there isn't even any devastation.
I had reasons for choosing the variable names that I chose. I am well aware that they are not according to convention. But for just one factor out of several, neither is Slashdot's character handling. Again, just for one example, I used capital E for emissivity rather than epsilon because it shows up well here. And rather than using upper- and lower-case characters for one body vs another, for example, an upper-case letter with subscript() works just fine. I have a couple of other reasons as well, I didn't just say this arbitrarily. At least this way when you refer to what you call the "heated plate" I know which one you mean without ambiguity.
Regardless, you are already skipping ahead. What do you want to use for material? We might as well use the same material throughout. So if you want to use aluminum for source, passive plate, and walls that is fine with me.
We know then, from ESA that the emissivity of aluminum in vacuum is approximately 0.15, and absorptivity 0.05.
I have been too busy to work through this this evening. I'll return tomorrow, if I'm not still too busy. I haven't even looked at your other posts.
Slashdot Mirror
I keep finding myself in a position where I feel I should explain, but I am at a loss as to why I should have to, because I am discussing this with someone who is supposed to have been a physics major.
You pointed out to MIT's derivation of energy transfer between infinite gray bodies. It does not apply here because (a) we have specifically defined areas, they are not infinite, and (b) that derivation makes use of Kircchoff's law which does not apply in Spencer's challenge.
This is a very simple but essential concept: I am not going to agree to calculations that were derived based on a physical principle that does not apply to the problem. This is a very basic concept. I do not agree with trying to solve a problem starting with invalid assumptions.
Once again, no. I never said that all surfaces were at the same temperature. I've already explained that the final outer temperature of the enclosing shell doesn't happen at the same time as the initial temperature of the heated plate. Initially, the heated plate is at 150F and the enclosing shell is cooler than 100F. But because power in > power out, the plates slowly warm to a new steady-state.
You did say it, quite clearly. I quoted you twice and linked to your web page. LATER you changed your tune. I can accept that you changed it later, but you did say it.
Further, in your link there, you say:
Jane's insistence that "a non-zero difference is all we need" between the heated plate's initial temperature of 150F and the enclosing plate's final temperature of ~150F was interesting. In this thought experiment [archive.today], the enclosing plate was initially cooler than 100F.
It is interesting, because it is the heart of the matter. Since, according to the S-B relation, if we are using gray bodies as you have several times insisted we use, the direction of net energy transfer via radiation at any given time is determined solely by the temperature difference and nothing else. Therefore (this is elementary logic), a non-zero difference in radiant temperature *IS* all we need, if we are using gray bodies, to determine which body is transferring energy to the other. At no time in this experiment are the temperatures equal, so net heat transfer is always in one direction and only one direction.
You then go on to say above:
By the time the outer temperature of the enclosing shell is ~149.6F (accounting for area differences), the heated plate is ~233.8F.
But you do not give any justification for this answer, you just throw it out there.
Earlier, you explained in some detail that you had calculated this number for thermal equilibrium using Kircchoff's radiation law. But as I've explained many times now, there is no thermal equilibrium so Kircchoff's law does not apply. Try again.
It feels as though I'm explaining to a high-school student who has never seen a physics problem before. Since the enclosing, passive plate is at all times cooler than the heat source, and therefore NET heat transfer is only outward from the heat source to that plate, then the only net energy input to the source remains the original electrical input.
So let's put 2 + 2 together here, which is really quite simple. We don't even need any math (but I welcome VALID math, if you can think of any to offer):
[1] Initial electrical input to heat source does not change.
[2] Since all other components of the system at all times remain cooler than the heat source, net power transfer to the rest of the system is invariably FROM the source TO the rest of the system. Nothing has been introduced to change that.
[3] Therefore, net input of power (energy per unit time) to the source remains a CONSTANT.
[4] Therefore, since any temperature of the source that is higher than the initial radiative equilibrium (150 deg. F) represents higher power output from that same source, any such higher temperature would violate conservation of energy.
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
NO, we do not agree with that, because as you state yourself that equation is derived from Kircchoff's radiation law, which does not apply here. Sheesh. How many different ways must I explain this?
Since HP basically got out of the calculator business, the HP 50G, which in my opinion is a better calculator anyway, has been available to the public in software form for free. It uses the actual ROM code from the 50G, which HP donated to the public domain.
You have to look around a bit, but versions are available for Mac, PC, and Linux.
Unlike the TI models, straight math can be entered in algebraic or RPN mode, and formulas can also use the "formula writer".
I've always like TI, and I have nothing against them, but over the years, having used various TI and comparable HP models, I've invariably thought the HP was superior.
If you replaced "businesses" with "large corporations", and "99%" with "85%", I might agree.
Our economy is still mostly driven by small business. Most small businesses and government do not have their hands in each other's pockets like big corporations do. Which of course is part of the problem. Nobody should.
My university took the attitude that computer science was an engineering discipline. You need to understand the theory, because the theory helps you classify and interpret that problem you're dealing with. But as an engineer, you're also on the hook for the process of design, and the actual design itself.
Certainly they do not want to just teach you a programming language, because that's like teaching a mechanical engineer the tools and settings for a single CAD program. Or an architect just how to draw blueprints.
But they still called it a computer science degree because I guess the world assumes people with a "software engineering" degree don't understand theory? I don't know why they made that choice.
I basically agree with you but the complaint seems to be that many people are coming out of university without the basic programming skills needed for real-world jobs. I'm not claiming it's true, I'm just saying that's the complaint.
Meh. It seems there is no end to clarifying. But it is also simplifying, in a sense, because it is eliminating irrelevant sidetracks:
I can conceive of a situation in which an enclosing, passive plate, of specific dimensions, might manage to be the same temperature on the inside and the outside in these circumstances. But I'm not going to bother getting sidetracked trying to do the calculations to either prove it or disprove it, because if it ever arose at all it would be a very rare special case, and whether it does or not is irrelevant to the central point.
Great. Except that it doesn't pertain to Spencer's challenge for several reasons. First, the chamber walls in Spencer's experiment are not "empty" space, but a material body that is being actively refrigerated, while the "enclosing passive plate" is being heated on the other side. So that plate is not in radiative equilibrium with the chamber wall or with anything else for that matter. In fact that would be impossible. There are other reasons why that description does not match Spencer's challenge, but that is irrelevant for now. One is enough.
Mea culpa. The outside of the enclosing passive plate would eventually reach radiative equilibrium with the chamber walls. But not thermal equilibrium. Further, the inside of the passive heated plate would reach radiative equilibrium with the heat source. But not thermal equilibrium in that case either. Nor, for that matter, is that same plate in thermal equilibrium even with itself, since realistically its inside and outside surfaces must be at different temperatures, in order to be at radiative equilibrium with those opposing surfaces.
Because I was incorrect to state that there is no radiative equilibrium, I was incorrect to state that a roughly analogous situation does not apply to Spencer's experiment. The opposing surfaces do reach radiative equilibrium. But it is still not very relevant here, because thermal (and therefore thermodynamic) equilibrium still does not exist.
Thermodynamic equilibrium is when every object is in thermal and radiative equilibrium as its surroundings.
Not a single object in Spencer's described challenge -- at any time -- meets these criteria. When everything is in thermal equilibrium -- as you have noted -- they are all at the same temperature. That never happens here.
The passive plate MAY be said to reach radiative equilibrium at some point... I stated that incorrectly before, and I apologize for doing so. I'm correcting it here so we don't have any misunderstandings.
There is no thermal equilibrium. Period. None. There MAY (and eventually would) arise a condition of radiative equilibrium for the (enclosing, passive, however you want to describe it) plate. But the other objects (heat source and chamber walls) do not meet this criteria because they are heated/cooled by means that may be other than radiative. "The system" is not in radiative equilibrium.
Without thermal equilibrium (which unequivocally does not occur here), Kircchoff's law does not apply, except perhaps in coincidental specific cases. It may not be assumed.
Those ESA absorptivities are for absorption of sunlight. Consider the first diagram here which shows that 6000K sunlight has much shorter wavelengths than the radiation from objects at the temperatures we're considering. In fact they hardly overlap. But the emissivities are for radiation emitted by much cooler objects. That's one reason why those ESA emissivities aren't equal to their absorptivities.
I repeat that these are under conditions of Earth-absorptive-surface insolation, which was what Spencer's experiment was supposed to emulate. But whatever.
But the emissivities are for radiation emitted by much cooler objects.
No, they are not. They are reported for the previously stated conditions: 1367 W/m^2 in incident radiation. But again: whatever. I already stated that this is not important enough to argue the point. I'm not conceding your point, but I'm willing to move on with gray bodies.
Once again, I never said that. In reality, I said that both sides of a thermal superconductor are at the same temperature. This was the source of much of the misunderstanding here, and you strongly objected to the notion of a thermal superconductor. Again, that's why I calculated the small temperature difference across an aluminum shell with finite conductivity.
Yes, you did say that, and anybody who wants to can read it on your website. And you wrote it BEFORE any discussion with me of "thermal superconductors". I will quote it again here:
Electric input of 509 W/m2 is constant and the walls are held at 0ÃF (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150ÃF (339K).
You were referring to "the second plate", as opposed to the "heated plate". That corresponds to what I have been calling the "passive" or "enclosing" plate. And you further referred to a supposed thermal equilibrium that doesn't exist.
Which fantasy would you prefer we believe? A thermal superconductor that makes no sense in this context, or an equilibrium which does not exist in this context? And you don't have the excuse that you meant "steady state", because the figure you gave would only be appropriate for actual equilibrium.
But enough of old arguments. Let's move on.
I was using this definition: "When incoming solar energy is balanced by an equal flow of heat to space, Earth is in radiative equilibrium and global temperatures become relatively stable."
Great. Except that it doesn't pertain to Spencer's challenge for several reasons. First, the chamber walls in Spencer's experiment are not "empty" space, but a material body that is being actively refrigerated, while the "enclosing passive plate" is being heated on the other side. So that plate is not in radiative equilibrium with the chamber wall or with anything else for that matter. In fact that would be impossible. There are other reasons why that description does not match Spencer's challenge, but that is irrelevant for now. One is enough.
Dr. Spencer disagrees: "Eventually the second plate will also reach a state of equilibrium, where its average temperature (letâ(TM)s say 100 deg. F) stays constant with time."
It is unfortunate that Spencer plays almost as fast-and-loose with terms as you do. That is a steady state. It is NOT "equilibrium". They are different things.
If you don't particularly mind, could we finally take the very first step in this calculation? Please?
Yes, I mind very much. There is no point in doing any calculations at all until we rid you of the false assumptions you have been making about this experiment (as I have been trying to do). They h
A better question, how much did they make selling this data?
Whether laws are heeded by corporations is dependent on a simple formula: what's to be gained by ignoring the law / (chance to get caught * fine). Unless that's below 1, the law becomes simply a cost factor to do business.
And even that equation is grossly unethical and doesn't backfire as often as it should. But it can, as Ford found out in the Pinto fiasco.
This is a ridiculously small settlement. How much is that per person? $3.70? AND -- this is just as big of a problem -- will ANY of those people who were actually harmed see any of that money?
This is what corporate cronyism (or what some people call "market capture") is all about. Government revolving door. It's a travesty and a tragedy. And it's also why I won't do business with Verizon.
I wouldn't say that learning to code necessarily outweighs a degree. But I do think university courses are too heavily focused on theory, and not enough practical application.
They complement each other. The big problem here (having gone though both, most but not all of the college being quite a while ago) is that a computer programmer back in the day HAD to know theory well, because programming was hard work! Input/output was so slow that you had to get it right the first time. Often you would present your code to somebody at a window to run on the mainframe, and if you were lucky you got a printout (!!!) the next day. If you got it wrong, a whole day was down the tubes.
Memory and storage were always in short supply, and CPU time was expensive. So everything had to be optimized. Sometimes for speed, sometimes for size, somethings a compromise of both. Theory was everywhere and you had to use it.
Heavy on theory, short on practice model that university CS was built upon, out of necessity. And they've kind of stuck with it, because universities are slow to change such things.
But I would also say that it is not a waste of time. As a practical programmer, theory will get you far. Look! De Morgan's Theorem just let me reduce those 5 lines of code to 2. You may not need to know linear algebra to work on sets of numbers, but if you do, hey, check it out. Now our program is half the size and our memory usage is down by 2 orders of magnitude.
So I don't think either one replaces the other. They complement each other. But I do think universities could concentrate, at least for their BS programs, a bit more on practical programming and just a bit less on theory.
I also suspect that this "standard" would be unworkable with some established de facto industry standards that aren't likely to change.
The long underline was an editing mistake.
In contrast, you're citing ESA figures from page 32 which are at 0K (-273C). But nothing in this experiment is anywhere near that cold.
Nonsense. They are figures at at incident radiation of 1367 W/m^2, which is sunlight at 1 AU, for the very reason that it is an approximation of Earth insolation. So in fact it would make a good representative example of what Spencer's model is supposed to be all about. Or do you (like Spencer) claim that space is "cold"?
But since you want to try to mischaracterize everything I say, and pick it all apart for reasons of your own, have it your own way. This is simply not very important.
You were right when you said [slashdot.org] the best we can realistically do is graybodies where emissivity = absorptivity. Otherwise we'd need to derive a new equation where heat transfer is an integral over wavelengths. In other words, we'd have to recreate MODTRAN [wikipedia.org]. I simply don't have time for that.
As I mentioned before, ESA gives observed values for integrated emissivity and absorptivity for aluminum. This is a good approximation and it is used in the real world for aluminum in a vacuum. If you really insist on gray bodies that's up to you; but I do not acknowledge that there is any legitimate reason to NOT use reasonable approximations of integrated absorptivity and emissivity.
We might be talking past each other. What you're calling steady-state is what I'm calling equilibrium. Radiative thermodynamic equilibrium doesn't require all surfaces to be at the same temperature, it simply means that temperatures don't change with time. At radiative equilibrium, power in = power out, which also means irradiance in = irradiance out.
You USED this before to ASSUME all surfaces were at the same temperature! I quoted you saying it in a post above, and you referenced that passage just the other day. In fact this was the source of much of the misunderstanding here. I did not understand why you were assuming some of the things you were assuming, and so my conclusion was that you were just messing with me. (And I am still not convinced that you were not.) THERE IS NO RADIATIVE EQUILIBRIUM HERE. THERE IS NO THERMAL EQUILIBRIUM HERE. None. You may not assume them.
I've already shown that MIT used Kirchhoff's law to derive heat transfer between gray bodies. I've already shown that Goodman 1957 tested the gray body approximation (Kirchhoff's law) and found that it's valid for aluminum at the temperatures in this experiment.
Note that my definition of equilibrium is consistent with this one: "In physics, radiative equilibrium is the condition where a steady state system is in dynamic equilibrium, with equal incoming and outgoing radiative heat flux and negligible heat transfer by conduction and convection."
In other words, irradiance in = irradiance out at radiative thermodynamic equilibrium. We're just using different words to describe the same concept.
No, we aren't, and you are incorrect. Actual thermodynamic equilibrium DOES require that there is no radiative transfer, and you aren't going to get it both ways. A steady-state is NOT the same thing as equilibrium. In Spencer's challenge, thermodynamic equilibrium does not exist. Radiative equilibrium does not exist. At no time are ANY of the surfaces in this experiment at the same temperature, and there is constant radiative transfer between bodies. This is another example of how you have played fast-and-loose with terminology. You do not get to re-define equilibrium any way you choose. Just no.
Kircchoff's law (and MIT's example) both assume no bodies involved are storing thermal energy, and there is thermal equilibrium. In fact that is how Kircchoff's law is derived: technically Kircchoff's law only applies at thermal equilibrium. MIT was free to apply it in their example because th
Earlier, when I saw your mentions of equilibrium, I thought you were referring to the steady-state that would eventually be achieved. But even though you mentioned Kircchoff's law, it didn't sink in to my brain that you were referring to actual, literal equilibrium.
Uh-uh. As they say in my neck of the woods: it ain't happenin'.
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium. I promise to provide public-readable versions of my Sage worksheet from now on.
This is one of the whole problems with your analysis. THERE IS NO THERMODYNAMIC EQUILIBRIUM IN THIS EXPERIMENT. There is a steady-state, but no actual equilibrium. That is not possible, because we are actively pumping heat in at one "end", and pumping it out of the other.
Since one of the requirements of thermodynamic equilibrium is that all surfaces be at the same temperature, it will never be achieved because the experiment requires that the outside wall be maintained at 0 deg. F, yet we are still pumping significant heat in to the center.
Therefore Kircchoff's law does not apply to this experiment, and no situation arises in which the temperatures are the same everywhere, or the emissivities vs absorptivities. There is a steady-state arising from active (but constant) exchange. But there is no equilibrium.
As you said, the best we can realistically do is graybodies where emissivity = absorptivity. If you'd like to use a different emissivity just let me know, and we can both independently calculate the required electricity to check each other's answers.
After considering the situation I changed my mind. Since we are discussion what is supposed to be a real model of a real situation, we can use real emissivity and absorptivity. And the emissivity of aluminum (as you pointed out yourself some time ago) is different from the absorptivity by a factor of about 3. The ESA figures are observed figures for aluminum plates in near-vacuum, so those figures would appear to be perfect.
And as I stated before, I am busy and I don't have time to figure out your nomenclature right now. That's why I wanted to agree on one.
I do have one more comment I want to make tonight, though. I will reply to the relevant post of yours. It is pointless to continue 3 separate threads at once.
Nothing further can be inferred from their interest other than their serious and factual opinions that climate change remains a true and actual risk that our planet faces.
That's a real laugh. Despite your protests, there is a hell of a lot more that can be inferred from their interest. Your assertion that genuine risk assessment is the only factor in their decisions is the more unlikely scenario.
Yes, I don't disagree. Part of my point was that too many patents are being awarded for things that should never even have been really considered.
EFF has a "stupid patent of the month" section on their website.
I already did. Global warming is a risk that actuaries can't actually calculate, but because the government says it's true they can claim it as a risk anyway. They can do their best to do some math on it, but ultimately they're basing it on something that isn't proven.
Historical risk assessment is based on statistics. There are no statistics about the hazards of global warming. Media and government have been proclaiming more "catastrophic weather events", but there haven't actually been any. Florida just set a record for number of years without a hurricane. The U.S. East coast is close to setting a record for number of years without a major hurricane. (Sandy was not "major". It only did as much damage as it did because it took an unusual path through expensive terroritory.)
Global cyclonic energy has been in a 40-year lull, and actually at a 30-year record or near-record low. Despite the drought in California this year there have been much worse droughts in California in living memory.
There isn't enough carbon dioxide in the entire atmosphere to even come close to "acidifying" the ocean if all of it were dissolved. Just ask a chemist.
So where is all this devastation the actuaries are supposed to calculate? Not only are there no statistics, there isn't even any devastation.
You could at least get my name right. You're not basing that on my comments, you're basing it on someone else's.
Nope. That one did not work.
I had reasons for choosing the variable names that I chose. I am well aware that they are not according to convention. But for just one factor out of several, neither is Slashdot's character handling. Again, just for one example, I used capital E for emissivity rather than epsilon because it shows up well here. And rather than using upper- and lower-case characters for one body vs another, for example, an upper-case letter with subscript() works just fine. I have a couple of other reasons as well, I didn't just say this arbitrarily. At least this way when you refer to what you call the "heated plate" I know which one you mean without ambiguity.
Regardless, you are already skipping ahead. What do you want to use for material? We might as well use the same material throughout. So if you want to use aluminum for source, passive plate, and walls that is fine with me.
We know then, from ESA that the emissivity of aluminum in vacuum is approximately 0.15, and absorptivity 0.05.
I have been too busy to work through this this evening. I'll return tomorrow, if I'm not still too busy. I haven't even looked at your other posts.
Well, I'll be darned. It works now. I'll see if others are working now too. Like
I tried those before here a long time ago and it didn't work. But maybe they changed things. So here's another try.
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