It's not a "black body" source, it's a "gray body" source, as per our agreement when this discussion first started. And I showed you my equations not just once but many times. You're just lying again. [Jane Q. Public, 2014-10-03]
Again, Jane's gray body equation has to reduce to the black body equation when emissivity = 1, so this is a way to check Jane's work. But since Jane seems convinced that checking his work is "lying" let's write down both equations simultaneously.
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Now use the Stefan-Boltzmann law to describe the radiative terms, one at a time. First for Jane's gray body:
Because "radiative power in from chamber walls" is emitted by graybody walls at temperature T4, the Stefan-Boltzmann law says:
gray electrical heating power + (e*s)*T4^4 = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Now for Jane's black body check:
Because "radiative power in from chamber walls" is emitted by blackbody walls at temperature T4, the Stefan-Boltzmann law says:
black electrical heating power + (s)*T4^4 = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Jane probably won't write down an equation describing electrical heating power for a blackbody source, so I'll try to guess at Jane's reasoning.
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".
Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]
Draw a boundary around the blackbody heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
That reference shows the object (i.e. chamber wall) temperature has an effect on the temperature controlled cavity (i.e. source). Which Jane denies:
Via a QUANTUM EFFECT, you fucking moron.
... [Jane Q. Public, 2014-10-03]
Charming. As I just explained, IR detectors don't have to depend on quantum effects. Classical mainstream physics allows a temperature-controlled source to detect IR from the cooler chamber walls as follows:
electricity = (e*s)*(T1^4 - T4^4)
If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).
If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.
If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.
If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.
If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.
... Further, I repeat for about the 100th time that I do not deny that some radiation is absorbed; but then it's just re-emitted. Sometimes, in a non-gray body, in a slightly different form. And ALL of that is straw-man irrelevancy, since no NET radiation absorption occurs from colder bodies to warm, which was the subject under discussion.
... [Jane Q. Public, 2014-10-03]
If you don't deny that some radiation is absorbed, then it should be very easy to write down a simple equation describing the required electrical heating power (not the radiative power out) of a blackbody source.
I don't need to "agree" with you about anything. I've already demonstrated how TEXTBOOK PHYSICS proved you wrong. That doesn't require any kind of "agreement". I'm just wondering when you're going to stop the dishonesty and admit you were wrong. The whole world is going to see it soon anyway, so you might as well "come clean", as they say. [Jane Q. Public, 2014-10-03]
Jane, if we can't agree on the meaning of the term "NET", why are you still capitalizing the word "NET"? Screaming the word louder and louder is unlikely to be productive.
1. Can we agree that net heat transfer through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net heat transfer always contains terms in both directions?
3. Can we agree that just because "radiative power out" > "radiative power in", that doesn't mean "radiative power in" = 0?
If Jane answers "no" to any of those three yes/no questions... why?
I don't need to "agree" with you about anything. I've already demonstrated how TEXTBOOK PHYSICS proved you wrong. That doesn't require any kind of "agreement". I'm just wondering when you're going to stop the dishonesty and admit you were wrong. The whole world is going to see it soon anyway, so you might as well "come clean", as they say. [Jane Q. Public, 2014-10-03]
If you're so confident that you're right, why not prove it by taking a few seconds to write down a simple equation describing the electrical heating power required to keep a blackbody source at 150F inside 0F chamber walls. Remember that "electrical heating power" is different than "radiative power out". Also remember that blackbodies can only absorb radiation, not reflect or scatter it. Finally, remember that the graybody equation has to reduce to the blackbody equation when emissivity = 1.
... Almost Latour's entire thesis is that S-B law says net heat transfer is either 0 or in one direction, from the hotter area to the colder. If the roles are reversed, and the colder item becomes the hotter, then the sign changes and the net heat transfer is still only in one direction... from hotter to colder.
... [Jane Q. Public, 2014-07-29]
... At no time in this experiment are the temperatures equal, so net heat transfer is always in one direction and only one direction.
... [Jane Q. Public, 2014-09-04]
... HEAT TRANSFER is always in one direction.
... [Jane Q. Public, 2014-09-07]
... There is heat transfer which is energy, which represents NET flow in one direction.
... there IS a net, non-zero flow of energy (heat transfer) THROUGH that boundary in one direction from the hollow enclosing plate to the chamber wall. This is a net, non-zero quantity. [Jane Q. Public, 2014-09-08]
... According to the S-B equation itself, net heat transfer is either 0, or only in one direction. Yes, we are talking NET here.
... [Jane Q. Public, 2014-09-10]
... If the chamber walls were hotter than the central source, then heat transfer would be in the other direction (because the sign of the solution to the equation above changes), and only THEN are you getting net heat transfer TO the central sphere.
... [Jane Q. Public, 2014-09-15]
... Another requirement of the S-B law, and also of thermodynamics: since EVERY other object in the system is at a lower temperature than the heat source, NET heat transfer is in ONLY one direction: from hotter to colder. Therefore, no energy is flowing "backward" to boost the output of the heat source.
... [Jane Q. Public, 2014-09-19]
... When A is warmer than B, (Ta^4 - Tb^4) yields a positive number. Which means all NET radiative energy transfer goes from A to B. That is clearly indicated by the minus sign, and is further dictated by the Second Law of Thermodynamics. There is no NET energy going from B to A. Only when B is hotter than A does any NET energy transfer in the other direction.
... [Jane Q. Public, 2014-10-01]
... You could not NOT understand it, unless you are 100% clueless about what the term NET means.
... [Jane Q. Public, 2014-10-01]
It's beginning to seem like we disagree about the meaning of the term "NET".
1. Can we agree that net heat transfer through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net heat transfer always contains terms in both directions?
3. Can we agree that just because "radiative power out" > "radiative power in", that doesn't mean "radiative power in" = 0?
If we can agree on all those points, that's great. Maybe this will help Jane write down a simple equation describing the electrical heating power required to keep a blackbody source at 150F inside 0F chamber walls. Remember that "electrical heating power" is different than "radiative power out". Also remember that blackbodies can only absorb radiation, not reflect or scatter it. Finally, remember that the graybody equation has to reduce to the blackbody equation when emissivity = 1.
On the other hand, if Jane answers "no" to any of those three yes/no questions... why?
You didn't bother to read my reference on pyrometers, did you?
... [Jane Q. Public, 2014-10-01]
That reference shows the object (i.e. chamber wall) temperature has an effect on the temperature controlled cavity (i.e. source). Which Jane denies:
... Radiation from the cooler walls has no effect on the heat source whatsoever. This is a basic requirement of thermodynamics!
... [Jane Q. Public, 2014-09-21]
No, that's Sky Dragon Slayer nonsense. If radiation from the cooler walls really had no effect on the heat source whatsoever, the IR thermometer wouldn't work because the cooler object temperature would have no effect on the temperature controlled cavity whatsoever.
When the source temperature is held constant, its required electrical heating power is an IR thermometer.
Here's one way to see that: draw a boundary around a heated aluminum source. It's heated by constant electrical power flowing in. Aluminum cold walls at some unknown temperature T4 also radiate power in. The source at 150F (T1 = 338.7K) radiates power out. At steady-state, power in = power out. Using the equation which neglects reflections:
electricity = (e*s)*(T1^4 - T4^4)
If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).
If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.
If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.
If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.
If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.
That's why when the source temperature is held constant, its required electrical heating power is an IR thermometer. At least, it's a thermometer when using mainstream physics. But Jane's equation is:
electricity = (e*s)*T1^4 (Jane's equation)
Since Jane's equation doesn't depend on the chamber wall temperature, uncooled IR detectors can't see cooler objects in Janeland. And we couldn't possibly have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How? This must be inexplicable to Slayers who are brainwashed into believing that:
... 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]
No. Again, mainstream physics shows that electrical heating power gradually decreases to zero as the chamber wall temperature increases. That's how uncooled IR detectors can see cooler objects.
Once again, Jane insists electrical heating power = (e * s) * (Ta^4). Once again, Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.
NO. That is NOT what I claimed, and that is not what I am claiming. That isn't even misunderstanding, it's just a lie. You HAVE TO understand this by now. You could not NOT understand it, unless you are 100% clueless about what the term NET means. I do not claim "no" radiation is absorbed. To repeat once again: no NET power from radiation is absorbed. Those are 2 completely different claims. You keep saying I claim the former, when I've actually only claimed the latter. And by now, there can be no remaining misunderstanding about that. You are simply lying. Again. [Jane Q. Public, 2014-10-01]
That's not what you claimed? Jane, you've been insisting for months that electrical heating power per square meter = 82 W/m^2, regardless of the chamber wall temperature. That means Jane insists that electrical heating power per square meter = (e * s) * (Ta^4).
If Jane were only writing down the Stefan-Boltzmann equation to calculate radiative power out, then he'd be able to honestly say that he's only claiming that net radiative power flows from warm to cold. But calculating "electrical heating power" requires writing down a heat transfer equation with power in = power out. Since Jane refuses to include a term for "radiative power in" in his heat transfer equation, he's wrongly saying the source absorbs no radiative power at all.
... Power output of A at a given temperature Ta is independent of B. Changing the temperature of B (as long as it remains cooler) does not affect the power output of A. This is exactly where you have been getting it wrong, by trying to use a heat transfer equation rather than a power output equation. This is textbook stuff, and you're getting it wrong. Period.
... [Jane Q. Public, 2014-10-01]
Jane, I've repeatedly explained that there's a big difference between electrical heating power, and radiative power out. Calculating "radiative power out" just requires writing down the Stefan-Boltzmann law. Calculating "electrical heating power" requires drawing a boundary around the heat source at steady-state, and setting power in = power out.
Once again, Jane's completely backwards. He needs to use a heat transfer equation, not an equation for "power out" only. This is probably because Jane doesn't understand the difference between electrical heating power and radiative power out. And since Jane seems convinced that he's right and physicists are wrong, he'll probably never be able to recognize his error, let alone correct it.
When A is warmer than B, (Ta^4 - Tb^4) yields a positive number. Which means all NET radiative energy transfer goes from A to B. That is clearly indicated by the minus sign, and is further dictated by the Second Law of Thermodynamics. There is no NET energy going from B to A. Only when B is hotter than A does any NET energy transfer in the other direction. A high-schooler can easily understand this. It's simple subtraction. [Jane Q. Public, 2014-10-01]
When A is warmer than B, more heat flows from A to B than vice-versa. Once again, this doesn't mean we can ignore the heat flowing from B to A. And that's exactly what Jane's doing, by insisting that:
... 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]
... because T(p) < T(s), no matter now much of the radiation from P strikes S, no net amount is absorbed; it is all reflected, transmitted, or scattered according to S-B.
... [Jane Q. Public, 2014-09-04]
Once again, no. This Slayer talking point can't possibly apply to blackbodies. How does Jane rationalize ignoring "radiative power in" when the blackbody source can only absorb that radiation, not reflect or scatter it? Once again, remember that the gray body equation has to reduce to the blackbody equation when emissivity = 1.
Can Jane write down a simple equation describing the electrical heating power of a blackbody source, like I did? If Jane would at least try to do that, he might learn about the difference between "radiative power out" and "electrical heating power" and he might learn why it's impossible to ignore "radiative power in" when the blackbody source can only absorb that radiation, not reflect or scatter it.
... Venus proves nothing about CO2-based warming on Earth. If you ASSUME it's causing warming here, then you can ASSUME it causes warming there, in proportion. Such assumptions prove nothing.
... [Jane Q. Public, 2014-09-26]
Again, if CO2 isn't the reason, then why is Venus hotter than Mercury? This isn't an assumption, it's a real-world example which any true skeptic should ponder before dismissing mainstream physics in favor of Sky Dragon Slayer brainwashing. Is Venus hotter than Mercury because of CO2, gray Oreos, or basketball player gloves?
Maybe the Slayers could explain how uncooled IR detectors see cooler objects?
Straw-man. Our argument involved gray bodies, not detectors of specific wavelengths or electronics that take advantage of specific quantum effects. But I have an answer anyway: they measure DIFFERENCES, not absolute radiation.
... [Jane Q. Public, 2014-09-28]
This isn't a quantum effect. The reason IR detectors measure DIFFERENCES, not absolute radiation, is because electrical heating power = (e * s) * (Ta^4 - Tb^4). If that weren't true, there would be no way to detect this difference, so uncooled IR detectors wouldn't be able to see cooler objects. And we couldn't have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How?
... 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]
... I have NOT been claiming that no radiation from a cooler body is absorbed by a warmer body. What I claimed, I repeat, is that no NET radiative energy transfer occurs from cooler bodies to warmer. That concept does not conflict with the ability of infrared cameras or pyrometers to detect "cooler" radiation. Energy can be absorbed and re-emitted... and often (for non-gray-bodies) it is re-emitted in different wavelengths. But the fact remains that there is still no NET energy transfer from cooler to warmer. If there were, it would violate the second law of thermodynamics. My argument has always been about NET heat transfer. I have explained to you many times that I do NOT claim no radiation from cooler bodies is ever absorbed. My argument is, and has been, about NET.
... [Jane Q. Public, 2014-09-28]
Once again, Jane insists electrical heating power = (e * s) * (Ta^4). Once again, Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.
If Jane would reconsider conservation of energy and include a term for "radiative power in", then Jane could honestly say he was only claiming that no net radiative power is absorbed by the source. Until then, Jane's equation claims that no radiation is absorbed by the source at all. And since Jane seems to think he's only saying no "NET" radiative power is absorbed, Jane will probably never be able to recognize his error, let alone correct it.
... And further, contrary to your own assertions, since the NET energy transfer from cooler bodies is ZERO, it is not included in the "radiative power out" term of heat transfer equations. Which is a concept that (apparently, if we assume you're being honest, which I doubt) you have had supreme difficulty getting through your head.
... [Jane Q. Public, 2014-09-28]
Once again, it's not included in the "radiative power out" term of heat transfer equations because it's included in the "radiative power IN" term.
I'm having supreme difficulty getting your concept through my head because it's Sky Dragon Slayer nonsense. The fact that more heat flows from warm to cold than vice-versa doesn't mean we can ignore the smaller amount of heat flowing from cold to warm. In fact, as I've repeatedly stressed, ignoring that heat violates conservation of energy.
Here's one way to see that: draw a boundary around a heated blackbody source. It's heated by constant electrical power flowing in. Blackbody cold walls at 0F (T4 = 255.4K) also radiate power in. The source at 150F (T1 = 338.7K) radiates power out. At steady-state, power in = power out:
electricity + (s)*T4^4 = (s)*T1^4 (Eq. 1J.2)
But Jane's equation is:
electricity = (s)*T1^4 (Jane's equation)
Because Jane's equation completely ignores radiative power flowing in, Jane's equation violates conservation of energy. How does Jane justify this violation?
... 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]
... because T(p) < T(s), no matter now much of the radiation from P strikes S, no net amount is absorbed; it is all reflected, transmitted, or scattered according to S-B.
... [Jane Q. Public, 2014-09-04]
Once again, no. This Slayer talking point can't possibly apply to blackbodies. How does Jane rationalize ignoring "radiative power in" when the blackbody source can only absorb that radiation, not reflect or scatter it? Once again, remember that the gray body equation has to reduce to the blackbody equation when emissivity = 1.
... In incorrectly "solving" Spencer's challenge, YOU ignored basic textbook methods and math to get your answer. You used an imaginary "khayman80" method of arriving at your answer, which not only contradicts everything engineering textbooks say about heat transfer, your methodology directly contradicts the Stefan-Boltzmann radiation law, even though you used it yourself in calculations. Talk about hypocrisy. I repeat: I checked your final "answer" for temperature of the heat source and it violates both the Stefan-Boltzmann law and the second law of thermodynamics. [Jane Q. Public, 2014-09-28]
Good grief, Jane. Once again, my solution doesn't violate the Stefan-Boltzmann law or the second law of thermodynamics. But it's fascinating that Jane/Lonny Eachus keeps regurgitating this baseless Slayer talking point. It seems like Prof. Brown was right to say that arguing with Slayers is a pointless waste of time.
... Venus proves nothing about CO2-based warming on Earth. If you ASSUME it's causing warming here, then you can ASSUME it causes warming there, in proportion. Such assumptions prove nothing.
... [Jane Q. Public, 2014-09-26]
Again, if CO2 isn't the reason, then why is Venus hotter than Mercury? This isn't an assumption, it's a real-world example which any true skeptic should ponder before dismissing mainstream physics in favor of Sky Dragon Slayer brainwashing. Is Venus hotter than Mercury because of CO2, gray Oreos, or basketball player gloves?
... Funny, because he's contradicting just about every argument behind the whole idea of AGW. I like how he makes these claims but isn't able to show how it actually works. He claims you can show warming via back-radiation WITHOUT the S-B equation? When it is absolutely fundamental to the very "energy transfer" he is asserting? What garbage.
... [Jane Q. Public, 2014-09-28]
That's odd. Just yesterday Jane had no argument with Prof. Brown. Now Jane claims that Prof. Brown is spreading "garbage" that contradicts just about every argument behind the whole idea of AGW. But Jane certainly isn't arguing with Prof. Brown or Dr. Shore or even me. Perish the thought.
Jane, don't you see how ironic it is to accuse three physicists (and the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society, etc.) of "FUCKING UP" their physics, while also claiming that:
.. I'll take the physicists' word for it, thank you very much.
.. [Jane Q. Public, 2012-07-05]
.. I consult "the experts". When it's a question of physics, for example, I look to references from physicists, not climatologists. After all, physicists are "the experts" when it comes to physics. [Jane Q. Public, 2013-11-15]
If Jane wanted to be consistent, he'd have to retract one of these claims. Jane is either a Sky Dragon Slayer accusing physicists of "FUCKING UP" their physics, or he believes physicists are "the experts" when it comes to physics and Jane takes the physicists' word for it. But not both!
Also, STOP sock-puppet modding down my comments. THAT'S AGAINST SLASHDOT'S RULES and it's just plain an asshole thing to do. [Jane Q. Public, 2014-09-28]
I haven't used moderator points in over a year. But the fact that Jane is so convinced I am that he's cussing and screaming in ALL CAPS is emblematic of Jane's reasoning problems, just like when Jane was absolutely convinced that I'm a six-headed hydra.
... As usual, you distort reality. Prof. Brown had nothing in the way of refutation or rebuttal or even retort to my second comment? Don't you find that interesting? I do.
... [Jane Q. Public, 2014-09-26]
It's not that interesting that Prof. Brown decided to ignore Jane/Lonny Eachus, given that he later said:
"Wow, Joel, I gotta say (after reading some of the replies on this thread) that this really is pointless. These folks have no conception of the FIRST law of thermodynamics, let alone the second. The argument for warming doesn't even require mentioning the SBE, it only requires the first law, the second law, and a monotonic relation between temperature difference in ANY channel and the rate of energy transfer in that channel, subject to very broad constraints.
But seriously, just a waste of time. When people just make stuff up and reject the contents of ELEMENTARY textbooks on the subject because they just don't like the conclusion those contents lead to, how can you argue with them? If somebody tries to solve the light bulb problem while pretending that it doesn't primarily cool via radiation and completely ignoring radiation, what can you do?
Get them to say "oops"?
Never happen. It's a religious issue, not a scientific one."
In other words, Prof. Brown gave up trying to educate Slayers like Jane/Lonny Eachus because it's a "waste of time."
... As for Joel Shore, again he was mis-applying an equation for heat transfer when he should have been using the equation for radiant power out. Both you and Shore insist on mis-applying this equation in a way that violates the Second Law of Thermodynamics. It's rather amusing that you brought him up, because you both FUCKED UP YOUR PHYSICS in a similar way.
... [Jane Q. Public, 2014-09-26]
That's odd. Just yesterday Jane had no argument with Dr. Shore. Now Jane claims that Dr. Shore "FUCKED UP" his physics.
... As for Joel Shore, again he was mis-applying an equation for heat transfer when he should have been using the equation for radiant power out. Both you and Shore insist on mis-applying this equation in a way that violates the Second Law of Thermodynamics. It's rather amusing that you brought him up, because you both FUCKED UP YOUR PHYSICS in a similar way.
... Engineers the world over do the math the way I did. So far that hasn't resulted in you either freezing or burning to death in your home. If they're all crazy, you might want to ask yourself why. [Jane Q. Public, 2014-09-26]
Physicists have "FUCKED UP" their physics, and only the Slayers can save the day! Or maybe the Slayers are crackpots. How could anyone tell, unless maybe Dr. Shore explained that:
"Actually, the idea that radiation goes only from the warmer to colder objects is an invention of the Slayers. It appears nowhere in the physics literature. I don't know about the exact history of our understanding, but my physics textbook from 1983 (Serway, "Physics for Scientists and Engineers", after introducing the law P = sigma*A*e*T^4 says
"A body radiates and also absorbs electromagnetic radiation at rates given by Eq. 17.11. If this were not the case, a body would eventually radiate all of its internal energy and its temperature would reach absolute zero. The energy that the body absorbs comes from the surroundings, which also emit radiant energy. If the body is at a temperature T and its surroundings are at a temperature T_0, the net power gained (or lost) as a result of radiation is given by
P_net = sigma*A_*e*(T^4 - T_0^4) (17.12)
When a body is in equilibrium with its surroundings, it radiates and absorbs energy at the same rate and so its temperature remains constant. When a body is hotter than its surroundings, it radiates more energy than it absorbs, and so it cools..."
Maybe the Slayers could explain how uncooled IR detectors see cooler objects? Using the equations Dr. Shore and I are using, the source's required electrical heating power depends on the chamber wall temperature. That's how uncooled IR detectors can see cooler objects. But Jane's "Slayer physics" insists that the chamber wall temperature doesn't affect the source's electrical heating power. So how do uncooled IR detectors see cooler objects? In particular, how did we detect the 2.7K cosmic microwave background radiation with warmer detectors?
... Venus proves nothing about CO2-based warming on Earth. If you ASSUME it's causing warming here, then you can ASSUME it causes warming there, in proportion. Such assumptions prove nothing.
... [Jane Q. Public, 2014-09-26]
If CO2 isn't the reason, then why is Venus hotter than Mercury? This isn't an assumption, it's a real-world example which any true skeptic should ponder before dismissing mainstream physics in favor of Sky Dragon Slayer brainwashing. Is Venus hotter than Mercury because of CO2, gray Oreos or basketball player gloves?
The Slayers have their own incoherent answer, while mainstream physics is presented by the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.
.. Be a man for a change and admit it.
.. [Jane Q. Public, 2014-09-15]
.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it.
.. [Jane Q. Public, 2014-09-19]
... Time to act like a man and admit that you were wrong.
... [Jane Q. Public, 2014-09-24]
Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.
... These are just straw-man arguments, as usual. I have no argument with these other physicists. It was about Spencer's challenge and how YOU got it wrong, nothing more. Have you asked them, personally, about Spencer's experiment? (No, you haven't, or you would know you were wrong.)
... [Jane Q. Public, 2014-09-25]
Does Jane have the memory of a goldfish? Of course Jane has argued with these other physicists. Jane personally asked Prof. Brown about Sky Dragon Slayerism, but wasn't able to "educate" him. Lonny Eachus personally asked Dr. Joel Shore about Sky Dragon Slayerism, but wasn't able to "educate" him. And now Jane/Lonny Eachus fantasizes that these physicists agree with his Sky Dragon Slayerism? Maybe Jane/Lonny Eachus should read those exchanges again, and notice that Prof. Brown and Dr. Shore told Jane/Lonny Eachus the same things I am. That's because Prof. Brown, Dr. Shore and I are simply reiterating elementary mainstream physics.
... Bringing up OTHER arguments like greenhouse gases won't win THAT argument for you. You have already lost it.
... [Jane Q. Public, 2014-09-25]
How bizarre. The whole reason Slayers deny that an enclosed source warms is because that implies greenhouse gases can't warm the surface:
.. the CO2-warming model rely on the concept of "back radiation", which physicists (not climate scientists) have proved to be impossible. I'm happy to leave actual climate science to climate scientists. But when THEIR models rely on a fundamental misunderstanding of physics, I'll take the physicists' word for it, thank you very much.
.. [Jane Q. Public, 2012-07-05]
That's why Jane, Dr. Latour and the rest of the Slayers disagree with the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.
Again, how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?
If Sky Dragon Slayers could answer these questions without resorting to gray Oreos or basketball player gloves, physicists might take the Slayers more seriously.
.. Be a man for a change and admit it.
.. [Jane Q. Public, 2014-09-15]
.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it.
.. [Jane Q. Public, 2014-09-19]
... Time to act like a man and admit that you were wrong.
... [Jane Q. Public, 2014-09-24]
Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.
... I mean, didn't it send up a red flag when you took your answer and fed it back into standard heat transfer equations and it didn't balance? Oh, that's right... you didn't. But I did.
... [Jane Q. Public, 2014-09-24]
Completely backwards, as usual. I've already shown that my solution keeps electrical heating power constant. Once again, Jane's solution halved the electrical heating power. Jane didn't notice this because he calculated net transfer incorrectly, which led him to the absurd conclusion that Jane was only off by about 0.1% when Jane was actually off by ~100%.
... because ALL of the incoming cooler radiation is reflected or scattered, and no NET amount is absorbed... [Jane Q. Public, 2014-09-24]
Good grief, Jane. How did the Sky Dragon Slayers brainwash you into endlessly regurgitating this nonsense? Once again, radiation is absorbed by any surface with absorptivity > 0. Jane's either hopelessly confused about the very term "NET" which he keeps capitalizing, or Jane/Lonny Eachus has betrayed humanity by deliberately spreading civilization-paralyzing misinformation.
Again, how do Slayers think we detected the 2.7K cosmic microwave background radiation with warmer detectors? How do Slayers think uncooled IR detectors see cooler objects? Again, why do Slayers think Venus is hotter than Mercury?
... I'm not arguing with you now and I'm not going to again. You're either a fool or a liar, and I do not care which. I have already proved it and I intend to publish that for the world to see. Along with textbook explanations and diagrams showing exactly where and how you went wrong. [Jane Q. Public, 2014-09-24]
Again, Jane/Lonny Eachus actually means that he intends to show where mainstream physics "went wrong" according to the Sky Dragon Slayers. There are many ignorant, stupid physicists that Jane/Lonny Eachus needs to educate: Prof. Brown, Dr. Joel Shore, the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society, etc.
.. Be a man for a change and admit it.
.. [Jane Q. Public, 2014-09-15]
.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it.
.. [Jane Q. Public, 2014-09-19]
... Time to act like a man and admit that you were wrong.
... [Jane Q. Public, 2014-09-24]
Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.
... I used the proper equation for radiative power, which at steady-state doesn't depend on other bodies. So there is no "difference" term. Just temperature. That's simple physics. You are trying to use a heat transfer equation to calculate power out of a single body at known temperature. That's just plain WRONG.
... [Jane Q. Public, 2014-09-24]
No, Jane tried to use an equation that only calculates radiative "power out" when Jane needs to use an equation for heat transfer that calculates radiative "power out minus power in".
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".
Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]
Jane's accounting for "power out" without including a term for "power in". That's not A = A, it's A = 0 because one of the terms has been ignored. It's led Jane to the absurd conclusion that electrical heating power doesn't depend on the cooler chamber wall temperature. If that's the case, then how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?
... All the radiation going IN from the cooler body just goes right back OUT again, making the NET radiation crossing your boundary from the cooler body zero. If that were not so, then you'd have net energy being transferred from a cooler body to a hotter one, which is a violation of the second law of thermodynamics. As I've explained to you many times now. You're just plain wrong.
... [Jane Q. Public, 2014-09-24]
This is complete gibberish, Jane. Power radiated in from the chamber walls needs to be accounted for using one term. Power radiated out from the source needs to be accounted using another. Once again, accounting for power flowing in doesn't violate the second law of thermodynamics or somehow imply net energy transfer from cool to hot, no matter how many times Jane wants to assert that nonsense. However, failing to account for power flowing in does violate conservation of energy, because power in = power out through any boundary where nothing inside is changing.
So Jane refuses to retract his absurd claim that view factors vary as the radius ratio, which violates conservation of energy. A cynic might have expected as much, given how Jane flagrantly violates conservation of energy by incoherently ignoring radiative power passing in through a boundary around the heat source.
I made no such claim, you liar. As you well know, the view factor from the surface of the inner sphere to the inner surface of the outer sphere is 1. The calculated view factor from the outer sphere to the inner was 0.9998... [Jane Q. Public, 2014-09-24]
Jane made no such claim? Jane keeps making that absurd claim! Again, the link I've repeatedly given Jane shows that for smaller radius R1, F21 = (R1/R2)^2 = 0.9978.
If the view factor varied as the radius ratio like Jane claims, energy really wouldn't be conserved. The view factor has to vary as the area ratio, which is the square of the radius ratio.
Jane's campaign of educating ignorant, stupid physicists about physics has only just begun. Jane still needs to educate Prof. Brown and Lonny Eachus still needs to educate Dr. Joel Shore.
No, I don't need to educate either one. They can both pick up a textbook on heat transfer and see that I am correct. I'm not arguing with them. [Jane Q. Public, 2014-09-24]
Of course Jane argued with Prof. Brown and wasn't able to "educate" him. Of course Lonny Eachus argued with Dr. Joel Shore and wasn't able to "educate" him. Why not, Jane? Do those physicists not have heat transfer textbooks, or are they just ignorant and stupid?
... I did NOT make broad claims in this recent exchange about "greenhouse gas" or any such thing. So I'm not arguing with those other people. I simply showed YOU to be wrong.
... [Jane Q. Public, 2014-09-24]
But Jane does make broad claims:
.. the CO2-warming model rely on the concept of "back radiation", which physicists (not climate scientists) have proved to be impossible. I'm happy to leave actual climate science to climate scientists. But when THEIR models rely on a fundamental misunderstanding of physics, I'll take the physicists' word for it, thank you very much.
.. [Jane Q. Public, 2012-07-05]
Jane/Lonny Eachus insists that an enclosed source doesn't warm, which means CO2 emissions couldn't cause warming. That's why Jane/Lonny Eachus needs to educate the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.
.. Be a man for a change and admit it.
.. [Jane Q. Public, 2014-09-15]
.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it.
.. [Jane Q. Public, 2014-09-19]
... Time to act like a man and admit that you were wrong.
... [Jane Q. Public, 2014-09-24]
Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.
.. By the Stefan-Boltzmann radiation law, the chamber walls add no net power in. It just goes right back out through your boundary again. How many times must I explain this to you?
.. [Jane Q. Public, 2014-09-23]
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out". For instance:
There is no net "radiative power in" from cooler to hotter. It's against the second law of thermodynamics, and it violates the S-B radiation law: (e * s) * (Ta^4 - Tb^4). [Jane Q. Public, 2014-09-23]
That's exactly the equation Jane should be using to calculate electrical heating power! It has separate terms for "power in" and "power out" so it can describe power entering and exiting a boundary. If Jane would use that equation, he'd honestly be only saying there is no net "radiative power in" from cooler to hotter.
Instead, Jane insists that electrical heating power = (e * s) * (Ta^4). Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.
There is nothing more to say. You have been proved wrong. You can write books about your nonsense "physics", and it won't make your bullshit theory any more correct.
.. The textbooks all say you're wrong. Goodbye. [Jane Q. Public, 2014-09-23]
So Jane refuses to retract his absurd claim that view factors vary as the radius ratio, which violates conservation of energy. A cynic might have expected as much, given how Jane flagrantly violates conservation of energy by incoherently ignoring radiative power passing in through a boundary around the heat source.
.. I honestly -- and I mean that: honestly -- don't believe you could be this stupid and possess a degree in physics.
.. [Jane Q. Public, 2014-09-15]
.. I only replied on the off-chance that you really were ignorant and could be educated.
.. [Jane Q. Public, 2014-09-20]
Jane's campaign of educating ignorant, stupid physicists about physics has only just begun. Jane still needs to educate Prof. Brown and Lonny Eachus still needs to educate Dr. Joel Shore.
Then, Jane/Lonny Eachus needs to educate the "ignorant" and "stupid" American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.
.. the CO2-warming model rely on the concept of "back radiation", which physicists (not climate scientists) have proved to be impossible. I'm happy to leave actual climate science to climate scientists. But when THEIR models rely on a fundamental misunderstanding of physics, I'll take the physicists' word for it, thank you very much.
.. [Jane Q. Public, 2012-07-05]
.. I consult "the experts". When it's a question of physics, for example, I look to references from physicists, not climatologists. After all, physicists are "the experts" when it comes to physics. [Jane Q. Public, 2013-11-15]
All those professional physics societies agree that our CO2 emissions are causing warming, which Slayers like Jane/Lonny Eachus deny. Jane's claimed that physicists are "the experts" when it comes to physics, and that Jane "takes the physicists' word for it." I'm skeptical.
.. Be a man for a change and admit it.
.. [Jane Q. Public, 2014-09-15]
.. Be a man and admit the truth.. You've been owned, man. BE enough of a man to admit it.
.. [Jane Q. Public, 2014-09-19]
Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.
.. I set out to have a scientific discussion, not to argue about your religion. [Jane Q. Public, 2014-09-21]
As Jane expands his endless campaign to educate ignorant, stupid physicists about physics, Jane might set out to have a scientific discussion without writing things like this:
".. non-person.. disingenuous and intended to mislead
"Jesus, you're a dumbshit.
".. what a despicable human being you are
".. you were "hanging yourself", as the saying goes. Hoist by your own petard.
But Jane's equation is different:
electrical power per square meter = (s)*(e)*Ta^4YES!!! 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 power, which requires using a net heat transfer equation to describe power in = power out through a boundary around the source. Because Jane's equation doesn't even include a term for "radiative power in", Jane's equation does claim that no radiation is absorbed at all.
If Jane would reconsider conservation of energy and include a term for "radiative power in", then Jane could honestly say he was only claiming that no net radiative power is absorbed by the source. Until then, Jane's equation claims that no radiation is absorbed by the source at all.
But Jane's equation is nonsense, because absorption is controlled by absorptivity. So we could only ignore the power radiated from the chamber walls 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.
Look at your S-B equation above. What does it say? No net radiative power is absorbed by warmer bodies from cooler bodies. You said so yourself. But NOW, you're claiming that it is. You contradict yourself.
... [Jane Q. Public, 2014-09-22]
No, I said the source has to absorb some radiation as long as it has absorptivity > 0. I never said the source would absorb more radiation than it emitted. In fact I said the opposite "happens automatically".
... The chamber walls neither transfer any of their net radiative power to the heat source, nor do they cause the net radiative power of the heat source to be any less. They have NO EFFECT. Net energy flows only FROM the heat source to the walls, and the temperature of the walls effects heat transfer only, not radiative power of the heat source.
... [Jane Q. Public, 2014-09-22]
If the temperature of the walls affects heat transfer, they also affect how much electrical heating power is required to keep the source at 150F. Note that I said "electrical heating power" and not "radiative power out" because these are two very different things. Calculating "radiative power out" just requires writing down the Stefan-Boltzmann law. Calculating "electrical heating power" requires drawing a boundary around the heat source at steady-state, and setting power in = power out.
... Do you think we're all idiots? Power in = power out. Your Newmann and Dirichlet boundary conditions are just more straw men. We don't need them to find the answer to this. Plain old algebra works just fine, because everything is at steady-state. So knock off the bullshit, because I see right through it, and so will the others I show this to.
... [Jane Q. Public, 2014-09-19]
Of course boundary conditions are needed to find the answer, because they determine what "plain old algebra" is used, even at steady-state. We're all applying Dirichlet boundary conditions to the chamber walls, but Jane mistakenly applied them to the source as well, instead of the correct Neumann boundary conditions. Jane also continues to wrongly insist that Jane held electrical heating power constant as well as holding source temperature constant. So apparently in Janeland there's no difference between Neumann and Dirichlet boundary conditions. If that's true, why do physicists and engineers use different names for Neumann and Dirichlet boundary conditions?
... The areas in his equation were unnecessary
... Therefore the areas were irrelevant and about all he accomplished with his large equation was to further confuse the issue. ... [Jane Q. Public, 2014-09-10]
After I originally solved a simple equation without areas, Jane objected that neglecting areas was a "fucking logical error!". That's why I had to solve the more accurate large equation with areas, even though I warned Jane that it wouldn't substantially change the answer.
The large equation with areas was also necessary because:
... 788.01 W != 721.44 W (!!!) Power is not conserved.
... the inner surface of the cavity has twice as much area, so the total power radiated is twice as much. Power is not conserved. ... [Jane Q. Public, 2014-09-07]
Jane confused himself about areas so badly that he claimed "power is not conserved". So I explained that Wikipedia’s equation correctly takes into account areas and view factor.
Wikipedia's equation conserves energy because the view factor from chamber walls to enclosed source equals the area ratio. If the view factor didn't vary exactly like that, energy really wouldn't be conserved.
But the chamber wall completely encloses the source, so its view factor is 1.
No. If the surfaces are numbered 1, 2, 3, 4 as I did in my solution, F12 = F34 = 1. In the other direction (as you already know, and so do I) it is R1/R2, where R1 is the smaller diameter. F21 = F43 = 0.9989. [Jane Q. Public, 2014-09-19]
As I said, the view factor from enclosed source to chamber walls is 1. If Jane wants to calculate the view factor in the other direction, the link I've repeatedly given Jane shows that for smaller radius R1, F21 = (R1/R2)^2 = 0.9978.
If the view factor varied as the radius ratio like Jane claims, energy really wouldn't be conserved. The view factor has to vary as the area ratio, which is the square of the radius ratio.
... I'd say that Wikipedia's equation is more correct because it includes area and view factor, which MIT's equation does not.
... [Jane Q. Public, 2014-09-19]
If only I'd mentioned that repeatedly.
... The equation you are trying to use there is a partial equation for heat transfer, not radiant power output. They're not the same things. The proper equation for power out given radiant temperature is right there in the above paragraph. It can be found in any heat transfer textbook and many physics books. Didn't you notice that MIT's equation is essentially the SAME equation as Wikipedia's heat transfer equation, except for areas? I sure did. Why didn't you notice that?
... [Jane Q. Public, 2014-09-19]
Of course I noticed that they're both net heat transfer equations, which is why I used them both in the same way to get nearly identical answers. I'm using MIT's and Wikipedia's equations because they yield radiative "power out minus power in". These net heat transfer equations are the proper equations for applying conservation of energy to a boundary around the source.
In contrast, Jane's clinging to an equation for "power out" and incoherently trying to justify ignoring "power in" through that boundary.
... I will make use of only ONE of your assumptions: that the enclosing plate (hollow sphere) is, due to thermal conductivity, approximately the same temperature on both sides. It's only 1mm thick after all, and the thermal conductivity of aluminum was a stipulation of yours so it will be the same to a couple of decimal places, give or take. So the answer won't be exact, but it will be reasonably accurate. Certainly close enough to demonstrate the concept.
... [Jane Q. Public, 2014-09-10]
When I approximated the enclosing shell as a thermal superconductor, Jane insisted that there's no way to demonstrate anything with it without leading to a contradiction, and that it was nothing but misdirection and a fantasy ultimate straw-man argument.
When Jane approximates the enclosing shell as a thermal superconductor, it's reasonably accurate and certainly close enough to demonstrate the concept.
A cynic might suspect double standards.
... I have already explained how your "boundary" assumed that all the power was output from the outside of the enclosing sphere.
... you neglected to account for the fact that the hollow sphere has TWO surfaces it is radiating from. You left out half the m^2 in A, so your figure for W/m^2 was off by very nearly 100%. Q.E.D. [Jane Q. Public, 2014-09-11]
Once again, Jane's completely wrong. When I held the source temperature constant, I reproduced Jane's result. So we're actually disagreeing about what to hold constant. If Jane's hilarious "Q.E.D." were correct, I wouldn't have been able to reproduce Jane's result simply by changing what variable I held constant.
... Add them together for the total heat transfer: 27.7832 + 27.7813 = 55.5645 total heat transfer. This checks against our initial calculation which was 55.5913. The difference is only 0.0268, or about 0.1%. Close enough for what we're doing.
... [Jane Q. Public, 2014-09-10]
Ironically, Jane's off by ~100%. Again, Jane's total heat transfer dropped to 27.8 W/m^2 after the shell was added, so Jane's meaningless 55.6 W/m^2 value is ~100% higher than the actual value.
... Factor out (e*s) from both sides. (Despite khayman80's assertion that we cannot do this, yes we can. It is the same scalar and the same constant on both sides.)
... [Jane Q. Public, 2014-09-10]
Once again, I never asserted that. In fact, I repeatedly showed Jane an equation derived by factoring out the sigmas and epsilons from both sides. Only Jane/Lonny Eachus could repeatedly quote that equation and even agree with it, then accuse me of asserting the opposite.
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.
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 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:
... 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 Jane's equation assumes warmer objects absorb no radiation from colder objects. Here's an equation which only says there's no NET radiative power input from cooler objects:
electrical power per square meter = (s)*(e)*(Ta^4 - Tb^4)
The above equation satisfies conservation of energy and says there's no NET radiative power input from cooler objects.
But Jane's equation is different:
electrical power per square meter = (s)*(e)*Ta^4
Jane's equation doesn't just say there's no NET radiative power input from cooler objects. That happens automatically. Jane's equation violates conservation of energy by completely ignoring the term describing radiative "power in" from the chamber walls. So Jane's equation says warmer objects absorb no radiation from colder objects.
But Jane's equation is nonsense, because absorption is controlled by absorptivity. So we could only ignore the power radiated from the chamber walls 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.
Jane/Lonny Eachus can capitalize "NET" all he wants, but it doesn't change this fact. Unless Jane/Lonny Eachus would like to correct his equation for required electrical heating power and derive an answer other than 82 W/m^2?
If the aborigine drafted an IQ test, all of Western civilization would presumably flunk it. -- Stanley Garn
