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Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

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: Re:FYI (Score 1) 595

by tbannist (#47922683) Attached to: Extent of Antarctic Sea Ice Reaches Record Levels

But 3 consecutive years of expansion would be....

Good news?

It's not happening, though. This year is really, really close to last year so it's more like a 2 year rebound from a new record low. If we're really lucky, 2012's minimum extent record will stand for a decade or longer. That would be good news for us, but I don't expect it to.

Comment: Re:The protruding lens was a mistake (Score 1) 398

by Blakey Rat (#47922545) Attached to: Apple Edits iPhone 6's Protruding Camera Out of Official Photos

My Lumia 1020 has one, I was wary at first, but honestly it's not a big deal at all. (Having one, that is. Lying about it on the website-- not good.)

It also takes better pictures than my dedicated camera, which is only a few years old. I think it's worth the tradeoff.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

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) 155

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) 155

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) 155

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) 155

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: Have they Denied? (Score 2, Interesting) 196

NSA officials were unable to find any evidence Snowden ever had.

This is essentially the "I do not recall" equivalent of paperwork investigations.

The essential question here is whether the NSA can conclusively deny that Snowden never raised concerns at the agency. Since if he did raise concerns, he probably would have raised them to people personally, a document search is not nessesarily going to uncover whether he did.

What will uncover this conclusively is a simple interview of NSA and affiliate company employees and especially supervisors who worked with Snowden. But since such a set of interviews would either a) reveal that he did raise concerns, b) involve people having to sign their names to untruths, or most unlikely c) reveal he really raised nothing, then I think it's easier for the NSA to just pretend that a half-assed email server word search constitutes an appropriate investigation.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

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) 155

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) 155

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?

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