Forgot your password?
typodupeerror

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47960791) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... What's ridiculous is your constant repetition of this bullshit idea. Yes, the cooler walls radiate inward but they have no effect whatsoever on the heat source. ALL of that radiation is reflected or scattered by the heat source. (It is not transmitted because we're dealing with diffuse gray bodies of significant mass.) ... [Jane Q. Public, 2014-09-21]

It's truly surreal to watch Jane repeatedly double-down on nonsense which Jane claims is too ridiculous even for Sky Dragon Slayers (as if that were possible!).

... You took a badly-worded sentence or two and jumped on them as though Latour made a mistake. But his only mistake was wording a couple of sentences badly. He does in fact NOT suggest that warmer objects absorb no radiation, and he has written as much many times. ... You have refuted NOTHING but a couple of unfortunately-worded sentences, which Latour himself publicly corrected shortly after that post appeared. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't 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.

... shortly after Latour published that blog post, it became clear that the language he used implied that no radiation at all was absorbed by the warmer body. So a reader could not reasonably be blamed for inferring that. But Latour quickly apologized for the unfortunate wording and corrected himself to make it very clear he was referring to net, not absolute, heat transfer. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body. Otherwise Jane's calculation of the required electrical power would include a term for radiation from the chamber walls. Since Jane adamantly insists that this term can't be included, Jane's calculation assumes that no radiation at all is absorbed by the source. None. Zero.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47957813) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... Repeat: this ASSUMPTION of yours that the chamber walls must be accounted for in the power requirement of the heat source is a direct violation of the Stefan-Boltzmann law. There are no 2 ways around it. Established physics (the Stefan-Boltzmann law) says that the radiative power out (and therefore power in) of a gray body is dependent ONLY on emissivity and thermodynamic temperature. It is completely unrelated to any nearby cooler bodies. ... [Jane Q. Public, 2014-09-21]

Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary. But "power in" through a boundary around the heat source looks like this:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Jane refuses to account for the chamber wall radiative "power in" which would only be true if the source didn't absorb any of that radiation. Zero.

If you are sincere (you certainly haven't been acting like you are), then you must be postulating some kind of "tractor beam" effect that allows the chamber wall to "suck" power out of the heat source from a distance. I assure you that at least at out current level of technology, we have not managed to build such a sucking device. The heat source radiates out what it radiates out, and nothing around it is "sucking" any power from it. Although you seem to be doing your very best at "sucking" my time away over stupid bullshit. [Jane Q. Public, 2014-09-21]

That's ridiculous, Jane. I'm just noting that the chamber walls are hotter than 0K, so they emit radiation into a boundary around the heat source. Therefore Jane's wrong to ignore that radiation when applying the principle of conservation of energy:

... Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. ... [Jane Q. Public, 2014-09-15]

It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source.

So the only "heat source" where we could validly ignore the radiation from the chamber walls would be a perfectly reflective "bobble" from Vernor Vinge's Marooned in Realtime. I assure you that at our current level of technology, we haven't managed to build such a device. And even if we could, it wouldn't be a heat source.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47957481) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... No NET radiative energy. I did not claim "none at all", and I have repeatedly pointed this out to you. Just no NET transfer from cooler to warmer. ... [Jane Q. Public, 2014-09-20]

Jane's equation claims "none at all":

electrical power per square meter = (s)*(e)*Ta^4

Since Jane's equation for required electrical power doesn't even include a term for radiation from the chamber walls, Jane's equation wrongly says that no radiation at all is absorbed by the source. None. Zero.

It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source. Slayer "physics" are incoherent nonsense.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47955263) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... Before that it can't, because Ta^4 - Tb^4 is a positive number so no net radiant energy is absorbed by (a) from (b). That means all the way up to the exact point thermal equilibrium is achieved, all radiant power is a result of electrical power, therefore the power input and power output are constant. It is not a "gradual" process. ... [Jane Q. Public, 2014-09-20]

So Jane claims:

electrical power per square meter = (s)*(e)*Ta^4

The actual answer is:

electrical power per square meter = (s)*(e)*(Ta^4 - Tb^4)

Since Jane refuses to include a term accounting for radiation from the chamber walls, Jane's equation is saying that no radiation at all is absorbed by the warmer source. Why?

... Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. ... [Jane Q. Public, 2014-09-15]

Of course it is! Again, this is just Sky Dragon Slayer nonsense. Absorption doesn't work like Slayers imagine. It's controlled by the surface's absorptivity, which doesn't change if the source is slightly warmer or cooler than its surroundings. All that's required for the source to absorb radiation (from warmer or colder objects) is having absorptivity > 0. Since the source has absorptivity = 0.11, some radiative power from the chamber walls is absorbed by the heat source.

Jane's been regurgitating Slayer nonsense for years:

... Warmer objects cannot, and do not absorb lower-energy radiation from cooler objects. ... [Jane Q. Public, 2012-11-20]

Then how do uncooled IR detectors see cooler objects? How did we detect the 2.7K cosmic microwave background radiation with warmer detectors?

... explain how radiation that is of a LOWER "black-body temperature" will be absorbed by a body of a HIGHER black-body temperature. ... [Jane Q. Public, 2013-05-30]

... An object that is radiating at a certain black-body temperature WILL NOT absorb a less-energetic photon from an outside source. This is am extremely well-known corollary of the Second Law. ... [Jane Q. Public, 2013-05-30]

No, that's a Slayer fantasy. On the atomic scale, absorption of radiation doesn't depend on temperature because individual atoms don't have temperatures. Only very large groups of atoms have temperatures. Individual photons also don't have temperatures. Very large groups of photons from a 10C warm object have slightly different average wavelength curves than a -10C cold object, but they're very similar. This means that even if temperature somehow applied at the atomic scale of absorbing individual photons, an atom couldn't tell if a photon came from the 10C warm object or the -10C cold object.

... You took a badly-worded sentence or two and jumped on them as though Latour made a mistake. But his only mistake was wording a couple of sentences badly. He does in fact NOT suggest that warmer objects absorb no radiation, and he has written as much many times. ... You have refuted NOTHING but a couple of unfortunately-worded sentences, which Latour himself publicly corrected shortly after that post appeared. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't 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.

... shortly after Latour published that blog post, it became clear that the language he used implied that no radiation at all was absorbed by the warmer body. So a reader could not reasonably be blamed for inferring that. But Latour quickly apologized for the unfortunate wording and corrected himself to make it very clear he was referring to net, not absolute, heat transfer. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body. Otherwise Jane's calculation of the required electrical power would include a term for radiation from the chamber walls. Since Jane adamantly insists that this term can't be included, Jane's calculation assumes that no radiation at all is absorbed by the source. None. Zero.

It's truly surreal to watch Jane repeatedly double-down on nonsense which Jane claims is too ridiculous even for Sky Dragon Slayers (as if that were possible!).

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47947973) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... If body (b) is brought up to the same temperature as (a), Ta^4 - Tb^4 = 0, and no net heat transfer takes place. Although radiant power output of (a) at that temperature doesn't change, as a corollary of that same law. ... [Jane Q. Public, 2014-09-19]

If Ta = Tb, no electrical heating power is required. But radiant power output of (a) doesn't change. So radiant power output can't be equal to electrical heating power. Using conservation of energy, can you write down an equation which yields the required electrical heating power given Ta and Tb?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47943463) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Since the emissivity for every object in our system is the same, power output is proportional to the T^4. Period. End of story. Draw your boundary around the heat source. Power in = power out (your own principle). Therefore the power in is 41886.54 Watts, which is the power initially being radiated out. SPENCER stipulated that this power is held constant. It wasn't my idea. It's a condition of the experiment. [Jane Q. Public, 2014-09-19]

No. Once again, in this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."

Jane's even stumbled across this point:

... Of course it wouldn't need a separate heat source if its environment were maintained at 150 degrees. ... [Jane Q. Public, 2014-09-15]

Of course! That's why the variable Jane's holding constant isn't the electrical power supplied to the separate heat source. If Jane can realize that there's no need for a separate heat source if its environment were maintained at 150 degrees, why can't Jane see that his equation for required electrical power doesn't reflect this obvious fact?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47922765) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... you have confirmed that you have not abandoned your incorrect (and actually quite ludicrous) version of heat transfer, which violates the Stefan-Boltzmann radiation law on its very face. ... [Jane Q. Public, 2014-09-15]

... or maybe we disagree about which variable to hold constant.

Instead of holding electrical heating power constant, Jane held the source's radiative power output constant. That held source temperature constant and forced electrical heating power to change. Solving this problem using both sets of boundary conditions shows that Jane's solution forces electrical heating power to drop by a factor of two after the shell is added.

These two sets of boundary conditions are very different, just like Neumann boundary conditions are different from Dirichlet boundary conditions. Upon hearing that a disagreement might be caused by holding different variables constant, a real skeptic might consider working the problem again while holding that other variable constant. But Jane can't even admit there's a difference between holding electrical heating power constant and holding the source's radiative power output constant. Jane even insists he held electrical heating power constant, despite the evidence.

So Jane won't solve this problem with the electrical heating power constant. That's unfortunate, because it's critical:

"... critical to the whole experiment is that, like the sun heating the surface of the Earth, there is energy being continuously pumped into the system from outside. ..."

1. Holding electrical heating power constant while adding an enclosing shell is like doubling CO2 while holding solar heating power constant, then calculating how much Earth's surface warms.

2. Holding source temperature constant while adding an enclosing shell is like doubling CO2 while holding Earth's surface temperature constant, then calculating how much solar heating power would have to drop to keep Earth's surface temperature constant.

Even if Jane doesn't want to solve that first problem, he should recognize that it's different from the second problem Jane actually solved.

To see this difference, solve a problem with Neumann boundary conditions:

"In thermodynamics, where a surface has a prescribed heat flux, such as a perfect insulator (where flux is zero) or an electrical component dissipating a known power."

... then solve the same problem with Dirichlet boundary conditions:

"In thermodynamics, where a surface is held at a fixed temperature.

Dr. Spencer's thought experiment placed Neumann boundary conditions on the source and Dirichlet boundary conditions on the chamber walls. Instead, Jane placed Dirichlet boundary conditions on the chamber walls and the source.

In other words, the electrical heating power is determined by drawing a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Right?

No. Not right. Since emissivity doesn't change the input required to heat source to achieve 150F is constant, regardless of where it comes from. But as long as the walls of the chamber are cooler than the source, NONE of the power comes from the chamber walls, because of that minus sign in the equation above. Nothing has changed in that respect, and that's what the Stefan-Boltzmann law requires. The only time that changes is if the walls are at an equal temperature, in which case heat transfer is 0 and you can begin to use "ambient" temperature as input. You are still supplying the same input power, you are just supplying it a different way. 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]

Note that conservation of energy through a boundary around the source leads directly to an equation describing the electrical power required to keep the source at temperature T1 inside chamber walls at temperature T4. This equation is valid for T1 > T4, T1 = T4, and T1 < T4. Jane might wonder why he can't derive a single equation which works for all these cases.

Again, warming the chamber walls is like partially closing the drain on a bathtub where water is flowing in at a constant rate. This raises the bathtub water level simply by reducing the water flow out. In exactly the same way, a source heated with constant electrical power warms when the chamber walls are warmed because that reduces the net power out.

... 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]

Are you REALLY the moron you make yourself out to be? NET radiation from a cooler surface that passes the boundary is reflected, transmitted, or scattered and passes right back out through the boundary. This is a corollary of the Stefan-Boltzmann radiation law, which states that NET heat transfer is always from hotter to cooler. ... by that same law, it just passes right back out again because the same NET amount of radiative power that crosses the boundary and intercepts the smaller sphere is either reflected, transmitted, or scattered. ... [Jane Q. Public, 2014-09-15]

... Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. ... [Jane Q. Public, 2014-09-15]

Hopefully these are just more badly-worded sentences because they all require absorptivity = 0. But these gray bodies have emissivity = absorptivity = 0.11. Furthermore, the gray body equation has to reduce to the black body equation for emissivity = absorptivity = 1. In that case there are no reflections, just absorption.

Once again, a heated blackbody source is 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)

Since Jane's proposed equation is missing the "(s)*T4^4" term, it doesn't reduce to this simpler Eq. 1J.2 for blackbodies where (e) = 1. So it's wrong.

It's also ironic that Jane claims to account for reflections, because:

... Calculate initial (denoted by "i") heat transfer from heat source to chamber wall. We are doing this only to check our work later. Using the canonical heat transfer equation for gray bodies...
p(i) = (e)(s) * ( T1^4 - T4^4 ) ... [Jane Q. Public, 2014-09-10]

... You are ignoring (e*s) * (Ta^4 - Tb^4). Anything other than what I described does not add up. ... (e*s) * (Ta^4 - Tb^4) ... [Jane Q. Public, 2014-09-15]

That equation is true for blackbodies with emissivity = 1, which is why it's consistent with my equation 1.

But for gray bodies it's just an approximation because it ignores reflections. After obviously failing to explain that we need to account for reflections, I decided to agree to disagree. For two gray bodies interacting with small view factors (e.g. Earth's tiny view factor of the Sun) reflections can be safely neglected. But the chamber wall completely encloses the source, so its view factor is 1. That's why MIT's equation is more accurate here: it accounts for reflections.

Again, here's MIT's equation using Jane's new variable names:

p(i) = (s)*(T1^4 - T4^4)/(1/(e) + 1/(e) - 1) (Eq. 2J.2)

Luckily this disagreement isn't important because it just shifts the emissivity values. We can translate because plugging emissivity = 0.058 into Jane's equation yields the same net heat transfer as MIT's equation with emissivity = 0.11. Furthermore, my black and gray body calculations yielded identical enclosed steady-state temperatures, so those don't depend on emissivity.

But after using Jane's equation in pointless attempts to illustrate more fundamental problems in Jane's analysis, I wanted to stress once again that MIT's equation is more appropriate for enclosing chamber walls because it accounts for reflections.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47914831) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... Since emissivity doesn't change the input required to heat source to achieve 150F is constant, regardless of where it comes from. But as long as the walls of the chamber are cooler than the source, NONE of the power comes from the chamber walls... [Jane Q. Public, 2014-09-15]

But if the chamber walls are also at 150F, they're not cooler than the source and the input required to heat the source to 150F is zero.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47914733) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... do you still maintain that after the enclosing passive sphere is inserted, the central heat source raises in temperature to approximately 241 degrees F? You haven't said anything about that in a while, so I'm just checking. [Jane Q. Public, 2014-09-15]

Once again, if the electrical heating power is held constant, the heat source has to warm. Once agin, Jane's heat source keeps the source temperature constant by halving its electrical heating power. Jane/Lonny Eachus might ask himself why his required electrical heating power goes down by a factor of two after the enclosing shell is added.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47914709) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... Of course it wouldn't need a separate heat source if its environment were maintained at 150 degrees. ... [Jane Q. Public, 2014-09-15]

In other words, the electrical heating power is determined by drawing a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Right?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47914127) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F. The answer is YES, and here is why: You are proposing to bring the whole system up to a level of higher thermodynamic energy, rather than just the heat source. And you are somehow proposing that it doesn't take more energy to do that. But of course it does. The power required to bring the heat source up to 150F remains the same, because the Stefan-Boltzmann law says it has to be. But NOW, you are ALSO bringing the walls up to that higher temperature, and THAT would require even more power (because of the slightly larger surface area). [Jane Q. Public, 2014-09-15]

Again, that's completely ridiculous. I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant.

For example, you could simply place the vacuum chamber somewhere with an ambient temperature of 150F. That would require zero power, but once again it doesn't matter even if the vacuum chamber were on Pluto. Because that power never crosses the boundary.

Either way, as long as the chamber walls are held at 150F, the heat source would need absolutely no electrical heating power to remain at 150F. Zero. Period.

You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F. The answer is YES... [Jane Q. Public, 2014-09-15]

Here's our disagreement. Conservation of energy demands that a heat source at 150F requires no electrical heating power inside 150F vacuum chamber walls.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47913475) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

You're either disputing conservation of energy, or you're not calculating the actual electrical heating power. If you're calculating the actual electrical heating power, your calculation has to account for radiation from the chamber walls because it passes in through that boundary. That's why the electrical heating power would be zero if the chamber walls were also at 150F!

Nonsense. This is textbook heat transfer physics. We have a fixed emissivity. Therefore, according to the Stefan-Botlzmann radiation law, the ONLY remaining variable which determines radiative power out is temperature. NOTHING else. That's what the law says: (emissivity) * (S-B constant) * T^4. That's all. Nothing more. This makes it stupidly easy to calculate the radiative power out, and therefore the necessary power in. [Jane Q. Public, 2014-09-15]

It's "stupidly easy" to calculate radiative power out and power in through what boundary? The boundary you're describing has to include the source's radiative power passing out through it, without including radiative power from the chamber walls passing in. I think that's impossible, but feel free to explain exactly where such a boundary would be drawn.

One question only: do you agree with the Stefan-Boltzmann relation: power out P = (emissivity) * (S-B constant) * T^4 ?? No more bullshit. "Yes" if you agree that equation is valid, or "No" if you deny that it is valid. Just that and no more. I'm not asking your permission. I'm just trying to find out whether you're actually crazy or just bullshitting. [Jane Q. Public, 2014-09-15]

Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law. But I've obviously failed to communicate that the power from the chamber walls has to pass in through that boundary, so you're only using half the equation to calculate the electrical heating power.

The REASON there would not be as great a power DIFFERENCE if the chamber walls were also at 150F, is that the walls would themselves be radiating more power out, so there would be less heat transfer (in that case 0). It is NOT, as you assert, because the heat source would be using less power. That's false, by the S-B equation. Its power output remains the same because (Spencer's stipulation) the power input remains the same. The reason my solution does not violate conservation of energy, is that the power consumption of the chamber wall is allowed to vary. THAT is where the change takes place, not at the heat source. Again, this is a stipulation of Spencer's challenge. Once again: power out of heat source remains constant, because P = (emissivity) * (S-B constant) * T^4. There is nothing in these conditions that changes this at all. Therefore, BECAUSE the power out and power in at the heat source remain constant, so does the temperature. It's all in that one little equation. [Jane Q. Public, 2014-09-15]

Once again, no. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

"Power in" has to include the radiative power passing in through the boundary. Otherwise energy isn't conserved, because power in = power out through any boundary where nothing inside that boundary is changing with time.

... EVEN IF we accepted your idea that the "electrical" power required to be input to the heat source is dependent on the temperature difference between the heat source and chamber wall (a violation of the S-B law), you still contradict yourself because your answer of a hotter heat source would still then require MORE power, because the difference is greater. But that is not allowed by the stated conditions of the experiment, and you keep glossing over that simple check of your own work which proves it wrong. So no matter how you cut it, your answer is wrong, by your own rules. ... [Jane Q. Public, 2014-09-15]

Once again, no. I've already shown that the electrical power in my solution remains constant.

Once again, that's because I'm correctly applying the principle of conservation of energy to determine the electrical heating power.

It seems like we can't agree that "power in" includes the radiative power passing in through a boundary around the heat source. Is that because you disagree that power in = power out through any boundary where nothing inside that boundary is changing with time? Or is it because you disagree that the radiative power from the chamber walls passes in through a boundary around the heat source?

The REASON there would not be as great a power DIFFERENCE if the chamber walls were also at 150F, is that the walls would themselves be radiating more power out, so there would be less heat transfer (in that case 0). It is NOT, as you assert, because the heat source would be using less power. ... [Jane Q. Public, 2014-09-15]

That's absurd. A 150F plate surrounded by 150F chamber walls wouldn't need an electrical heater at all. Period. The electrical heating power would be exactly zero. Maybe you're mistaking "electrical heating power" with "radiative power out"? Or maybe you're missing half the equation necessary to calculate the required electrical heating power, and it's leading you to bizarre conclusions?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47898793) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

I am disputing nothing of the sort. As I have explained many times now, you are not drawing your lines properly. You keep making the same bullshit assertions, after I have proved them false. Why do you do this? You're just going to look that much more foolish later. [Jane Q. Public, 2014-09-13]

You're either disputing conservation of energy, or you're not calculating the actual electrical heating power. If you're calculating the actual electrical heating power, your calculation has to account for radiation from the chamber walls because it passes in through that boundary. That's why the electrical heating power would be zero if the chamber walls were also at 150F!

Can we agree that the required electrical heating power would be zero if the chamber walls were also at 150F?

... I held the power constant, just as Spencer stipulated. ... [Jane Q. Public, 2014-09-13]

It's so adorable that Jane keeps insisting that Jane kept the power constant, even after I showed that Jane's calculation was only able to hold the source temperature constant after the enclosing shell was added by halving the actual electrical heating power.

It's also adorable that Jane keeps ignoring the fact that his "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47898707) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

NO!!! I have told you 5 or 6 or maybe more times now, this is a VIOLATION of the very straightforward Stefan-Boltzmann law. How it applies in this situation is quite straightforward, and not at all as complex as you are making it out to be. Radiant power output of a gray body is calculated using ONLY the variables: emissivity and temperature. THAT IS ALL. There is no other variable dealing with incident radiation, or anything else. When the system is at radiant steady-state, power out (and therefore power in) are easily calculated, and I have calculated them. Further, Spencer's "electrical" input power was to the heat source, not to the whole system. YOUR OWN PRINCIPLE: power in = power out. Now you're trying to contradict yourself and say it meant something else. It's just bullshit. You're squirming like a fish on a hook. You just don't seem to realize you have already been flayed, filleted, and fried in batter. You're owned, man. [Jane Q. Public, 2014-09-13]

No. Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:

electricity_initial = p(14) = (e)(s) * ( T1^4 - T4^4 )

So are you disputing that power in = power out through a boundary where nothing inside that boundary is changing with time? Or are you disputing that the radiation from the chamber walls passes through a boundary drawn just inside them?

And again, if you keep ignoring that "power in" half of the equation that all Sky Dragon Slayers miss, you'll have to keep wondering why your "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 168

by khayman80 (#47898663) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... It's not that I don't agree. You might come up with the right answer for some sub-calculation. I don't know, I don't care, and I'm not even going to bother to check, much less agree. The issue is that I have already solved the problem, and arrived at the correct answer (within reasonable limits). So I don't HAVE to agree or disagree with you. I've already done it, according to the correct textbook-approved physics. AND (unlike you) I checked my work and it checks out. And unlike your answer it doesn't violate conservation of energy. ... [Jane Q. Public, 2014-09-13]

I just showed that Jane/Lonny Eachus solved the "correct answer" to a different question. Instead of holding the electrical heating power constant like Dr. Spencer did, Jane/Lonny held the source temperature constant. In that case, the electrical heating power required to keep the source at 150F drops by a factor of two after the enclosing shell is added. This shows that holding the electrical heating power constant like Dr. Spencer did is different than holding the source temperature constant like Jane/Lonny did.

... SIMPLE CALCULATION, which I have already shown several times: power "sufficient" to heat the heat source under initial conditions to 150F: 41886.54 Watts. Power input at the source remains constant. ... [Jane Q. Public, 2014-09-13]

No, in your example the electrical heating power drops by a factor of two after the enclosing shell is added. And once again, your calculation of the power sufficient to heat the heat source would be exactly the same if the chamber walls were also at 150F. But the right answer there is zero, because an electric heater wouldn't be necessary. Is this really so hard to understand, or are you deliberately spreading misinformation?

All life evolves by the differential survival of replicating entities. -- Dawkins

Working...