The required electrical power to keep the heated plate at 150F is completely independent of the chamber wall temperature?
No. That is not what I wrote. You are drawing a conclusion that does not follow from my actual words. Making assumptions again.
It is dirt simple to show you are wrong.
The initial conditions, with the surface of the heat source at 338.7K, ARE A STEADY-STATE. The radiative transfer between the surface of the heat source and the chamber wall is already accounted for. You are trying to account for it twice. It is easy to show this.
The temperature of this surface is a GIVEN, initial steady-state condition. It is known, and a constant at this time. YOU are trying change it, and give one surface 2 different temperatures at the same time.
Proof: all we have to do is plug your value for radiative power output back into the known, canonical equation for radiative temperature.
Temperature is the 4th root of ( (power in W/m^2) / (se) ). So using your calculated value: 4th root of ( (29.399) / ((6.24 * 10^-9 W/m^2) / K^4) ) = 4th root of 3749839743.59 = 247.46K = -14.24 degrees F.
However, we already know what this temperature is, because it's a given:: 150 deg F (338.7K).
Your value gives a wrong answer. Your methodology contradicts itself, which is what I have been saying all along.
Plug my 82.12 W/m^2 back into the same canonical equation for radiant temperature for a given radiative power output, and the answer comes out just as it should: 150 degrees F.
If you can't even get the initial conditions right, we might as well stop here.
One way to see this is to consider how much power the electrical heater would need if the chamber walls were also at 150F. The correct answer is zero watts, because the heated plate wouldn't lose net heat to walls at the same temperature. But since your expression doesn't depend on the chamber wall temperature, you wouldn't be able to obtain the correct answer of zero in that case.
No, I am not wrong, you are. You are describing a radiative power difference, or net transfer.
That is not what I was doing. I was simply calculating the net power output of the heat source at 150 deg. F using the textbook example of how to do that.
It is not a difference. It is a constant radiative power output that depends on NOTHING else but temperature and emissivity, and the S-B constant.
It doesn't matter what temperature an opposing surface is at. I'm calculating the power output of THIS surface, at THIS temperature. As long as the temperature OF THIS SURFACE remains the same, the radiative power output remains the same. The way to calculate it is well-known and I have clearly stated it in my calculations.
YOU are contradicting yourself: "Power out = power in", you said. Right?
I have calculated the radiative power output using nothing more than area (~ 510 m^2), radiative temperature (338.7K), the emissivity you gave (0.11), and the well-known and proven relation:
Radiative power out (in W/m^2) = emissivity * sigma * T^4, where sigma is the Stefan-Boltzmann constant.
This is the textbook solution. Please show where it is incorrect. Simply asserting that it is incorrect is not sufficient.
It's certainly subjectively good. But I think it's also an important case for monitoring data wise to see the objective value of these things. My hope is that it's a net positive for every important metric, because an even slightly mixed bag of results could be enough to talk a lot of departments out of the idea.
I agree. That's why I think it's important that ground rules be firmly established.
For one thing, camera use must not be discretionary. It must be used every time there is an interaction with citizens. Because otherwise, there is too much potential for them to be used only when it is in their favor, and at other times, "Oops, I forgot to turn it on."
So if a camera is not turned on, or data is missing or shown to be deleted, a serious inquiry should be made to determine the actual reason why.
Why do I insist on this? Because I was once a victim of "missing" camera footage. I was told everything was being recorded, and the light on the camera was on. But when it came time to go to court and testify, they claimed there was no recording and it had "never existed".
Which was complete bullshit, of course.
Never mind what it was all about. It was a non-criminal charge and despite their bullshit stories I was not convicted. But "I forgot to turn it on" is too easy of a bullshit abuse of authority.
You are a bit too eager to "pounce" with your solution. There was no need to repeat it 3 times. "I say it 3 times" does not make your analysis correct. Only correct analysis will. I will proceed from this post.
Once again, you zoomed ahead and did not proceed carefully. You are overlooking things. If you want this to be an actual solution of the problem, then let's do the problem completely. I have stated several times that I am only willing to do this if you agree to do it thoroughly.
Once again, energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At steady-state, that rate is zero because the system doesn't change. So at steady-state, power in = power out.
I agree, as long as you can legitimately draw a boundary around your system. So let's start out by drawing appropriate boundaries around out initial system.
I agree with your calculations for area. However I do not agree with your calculation for initial power output (which corresponds to our constant input).
We know the initial temperature of the heat source: 338.7K.
The acknowledged formula for finding radiative power from temperature is just (sigma epsilon)T^4. There are no other factors involved, so our power input is equivalent to the power output of the heat source. There is no reason to not assume perfect efficiency here. You don't show your work here, just an unreadable Sage file, so I don't know where the discrepancy lies.
This formula for radiative power output from radiative temperature and the corresponding formula for temperature from power both make use of (sigma epsilon), and epsilon is a scalar, so I will abbreviate it to (se) and pre-calculate it to make later calculations easier (I am using an HP, not Sage):
So here now is the reason for this preliminary setup: in our initial steady-state, heat source is 338.7K so total power output from the heat source (and therefore constant power input to the system) is just (se) * 338.7^4 * area
I asked for a simple yes or no. You did not give me a simple yes or no, you had to bury it in the middle of about 500 words about other things. If you had simply written "yes", one word, you would have saved both of us about 20 minutes.
I am entirely convinced that you are deliberately trying to make things difficult.
SO... you have agreed. Now we may proceed with calculations. But I will not, more, tonight.
But -- and this is no joke -- I expect you to show and describe your work. We start from extant conditions and proceed from there, in a step-wise manner. My intent, as it was before, is to minimize confusion and misunderstandings, which it is pretty obvious have been rampant. So... as is often the case in programming, if we proceed slowly in a step-wise manner, we achieve the end result all the sooner, because errors don't crop up.
I was serious about this: you have had a tendency to make assumptions and use terms loosely. We have had LOTS of misunderstandings here, and I am willing to chalk some of them up to communications problems, but not all of them.
So let's go slowly, and start simply. The first step is to define the initial conditions. We have a fixed power input, which heats the "source" (and I am calling it that for good REASONS), which we are assuming is a spherical body. Specific dimensions are up to you, I really don't care very much, although I have to wonder why you picked radii that seem so inconvenient... but on the other hand I admit that I did not do any calculations at this point using your dimensions
So, since we are starting fresh, describe your equation now, what the variables represent, and how you arrived at the formula. I suspect that it is a bit premature.
NO. See my comment above. One more bullshit comment like this, and as I said, I will just call you a clown and few reasonable people will disagree.
Will you, yes or now, agree to the definition of Spencer's experiment, as I have already described and asked you now about 3 times.
That is the FIRST step in solving any problem. I will not agree to any math regarding the problem until we have an agreement about what the problem is. Anything else is nonsense.
ONCE THAT IS DONE, I agree to move on with calculations about the problem before us.
BEFORE you do the math, the problem must be defined. No reasonable scientist would disagree with that. So far, you have refused at every turn to define the problem, even though this is the well-known first step to proving anything.
If you continue to just bullshit your way around, as I have stated I will declare you in default and damned few reasonable people would disagree.
You're sidestepping again, and you're ignoring the important point, again.
This is a sidetrack, about an OLD comment you made that was stupid and inaccurate. You can argue about THAT comment until the cows come home and die of old age, and you still won't be correct. BUT IT'S IRRELEVANT TO THE POINT NOW AT HAND.
If you want to move on, see my other comments which describe the problem. If you continue to refuse to agree to the definition of the problem, I will (with every justification in the logical world) declare you in default.
STOP THE BULLSHIT. It is pretty obvious what it would actually take to refute Latour. I have described in my other comment a statement of the actual challenge. If you refuse to agree with my description of the problem, without having a reasonably valid objection, I shall (with perfect justification) declare you in default.
If your time is really short as you say, then just abandon those recent sidetracks (which is in your own best interest anyway) and let's continue.
If you agree to the problem description I laid out above, in this comment, just say yes. Then we can continue.
But I will repeat, and keep repeating as long as necessary, that unless we agree on what it would take to refute Latour, then there is no point in going further. You can violate thermodynamics all you want, and it doesn't prove a damned thing.
Cute. I've repeatedly explained ad nauseum that neglecting area ratios is an approximation. I've already shown how tiny the effects are for Earth's area ratio. Does this mean you don't intend to perform even the simplest calculation to confirm this? Why don't we check to see wrong these approximations are, by actually doing some calculations? Finally? Please?
THERE'S NOTHING "CUTE" ABOUT IT!
IT'S AN ACCURATE ASSESSMENT OF YOUR ERROR!
This is not "approximation", it's fucking logical error! JESUS CHRIST, man, you can't talk your way around this.
Once again, no. I've repeatedly explained [slashdot.org] that the outer surface of the enclosing passive plate is never at the same temperature as the heat source.
And once again, NO, it doesn't matter. I don't care at what time you assume this to be. Because we have a constant power input which has radiative power output at X W/m^2, which is required to reach the radiative temperature of 150F.
You then (at ANY time, I don't care when) claim that a larger surface is at the same temperature, which requires the same amount of W/m^2. But you have more m! So the total power output is greater than your input.
There is no way to weasel out of this, man. You're trying to output more power than you're putting in. This isn't even 11th-grade physics.
Let's try it at something more like your level:
You have 200 beans equally distributed among 10 squares. If you now take those beans, and divide them equally among 25 squares of the same size, how many beans do you now have per square?
I've been explaining for over a month that the heated plate (aka Jane's "source") warms after it's enclosed.
Yes, you've been explaining it but you haven't been backing it up. See my other comment. Once we agree on the statement of the problem, we can move forward. I haven't been trying to block you, I've only been trying to get you to unambiguously agree to terms.
If Jane agrees that the heated plate (aka Jane's "source") warms after it's enclosed, then that's great news! In that case, we can all agree that the mainstream physics describing the greenhouse effect is accurate, obeys the laws of thermodynamics, and proves that the Sky Dragon Slayers are wrong.
No. Jane does not agree that "the heated plate" (if by that you mean THE HEAT SOURCE), gets warmer when it's enclosed. THIS IS WHAT I'VE BEEN TRYING TO TELL YOU.
You need to use more precise language. In plain English, you have one heat SOURCE in this experiment, and THE OTHER plate is being "heated" by it. So the "heated plate" is the passive plate.
You insist on using confusing terminology, and wonder why other people have a hard time with it. Jesus, I'm glad you weren't one of my physics profs.
I suggested a standard terminology and variable names to avoid exactly this problem. You have refused to use them. That's your goddamned problem, and you don't get to complain about it.
I'm really looking forward to showing this latest exchange to my friends.
It's just so hard to work my way though your maze of comments, some of which are correct, and others which are provably bullshit (see my other comments above).
THE FIRST STEP in mutually working your way through a problem is agreeing on the statement of the problem, as you well know. THAT is why I have not willing to get into calculations yet. That and nothing else. So let's make sure we agree on the statement of the actual experiment. I want you to acknowledge these in so many words, before I am willing to move forward. Because I'm just plain tired of your bull.
1. Conditions apply as per Spencer's experiment:
2. Constant input power sufficient to heat the source to a radiative temperature at radiative equilibrium of 150F.
3. Outer chamber walls actively cooled to a maintained radiative temperature of 0F.
4. Passive plate (which you insist in your variant of Spencer's challenge fully encloses heat source, with vacuum between).
5. Passive plate is introduced at a temperature that is cooler than the heat source.
6. Any other conditions that were actually contained in Spencer's challenge but not mentioned here.
7. Your own condition: you insist on using gray bodies, because you claim you're dying (or something of the sort, you really didn't specify so I won't assume), you don't have the spare 15 minutes or so it might take to do approximate, more real-world calculations, and want to use Kirchhoff's law (although it really isn't necessary) to make your life easy.
8. I reserve the right to re-visit this same scenario using more real-world materials, emissivities and absorptivities.
Are we agreed on these conditions? I want a simple yes or no. Anything else, and I am not willing to continue.
Once again, I never said that all surfaces were at the same temperature.
Look, let's get this straight: what you actually meant was completely ambiguous for several reasons. You used the term "equilibrium" and you said:
Electric input of 509 W/m^2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).
Will you acknowledge that no matter what you meant, this is still wrong? If the enclosing passive plate must radiate out the same power as the enclosed heat source, it cannot be at the same temperature, because radiated power is measured in W/m^2, and there are more m^2 in the enclosing passive plate. Therefore (SIMPLE MULTIPLICATION), because there is greater area they could not be at the same temperature and radiate outward the same power.
No matter how you try to bullshit your way around this, it is still WRONG.
As I said: "Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out."
I explicitly said a system in "equilibrium" doesn't change, which Jane calls "steady-state".
I call it "steady state" because that is how radiative equilibrium is defined. It is a condition in which radiative transfer between elements of the system remains constant. I prefer to make sure this is kept distinct from thermal equilibrium.
I just want to make sure we're all talking about the same things. Because in the past, we sure as heck were not.
Why would Jane think mentioning Kirchhoff's law means that I'm somehow trying to claim that all surfaces are at the same temperature? I first mentioned Kirchhoff's law by linking to MIT's explanation:
Stop being obtuse. You were throwing around the term "equilibrium" rather loosely, and at one point you mentioned that "at equilibrium" the outer surface of the enclosing passive plate must be at the same temperature as the surface at the heat source:
Electric input of 509 W/m^2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).
Now, if it isn't a reasonable conclusion that by this you meant thermal equilibrium, I don't what is, since you are claiming the heat source and the passive plate are the same temperature. Of course upon closer examination it could not have been actual thermal equilibrium, because you also mention the walls are at 0 degrees F, but then what actual kind of "equilibrium" you were referring to is worse than ambiguous, because at radiative equilibrium the stated condition is impossible. So why the hell are you trying to blame me for being confused? The condition you described is impossible, so how do you expect me to know what "equilibrium" you mean?
"... the relation "absorptance = emittance" is known as Kirchhoff's Law. It implies that good radiators are good absorbers. It was derived for the case when "body temperature = cavity temperature" and is not strictly true for all circumstances when the temperature of the body and the cavity are different, but it is true if the absorptance and emittance are not functions of wavelength. This situation describes a 'gray body.'... "
Very well. You refuse to use real materials with measured absorptivities and emissivities, and insist on using gray bodies and Kirchhoff's law, neither of which actually represent Spencer's experiment in anything like the real world, even though it was intended to be a model of the real world.
I get that. But I want to make sure everyone else gets it too.
I offered to use real materials with measured emissivities in the approximate temperature range we are talking about (though you refuse to acknowledge that), but you refuse to use them.
Just so we know where we stand. I have already explained to you that there is no need to resort to gray bodies, and that we have plenty of information to calculate more realistic, real-world results. But whatever. You refuse to do anything but what you want to do, so let's just go with it for now. But I reserve the right to re-visit this issue.
I've been explaining for over a month that the heated plate warms after it's enclosed. I realize you don't agree, which is why I'm trying in vain to get you to finally perform a single, solitary calculation of your own.
I never said I disagree with this. Please find where I said that. On the contrary; I definitely agree that it warms. In fact it must: Spencer stipulated that it was to be inserted when it was colder than the heat source.
I have been doing calculations. I just haven't been showing them here, because there isn't any point yet. Before there is any point to showing calculations, we must agree that certain conditions either do or do not exist, given the parameters of this experiment. That is what I am trying to sort out with you now.
It is the "50g mode" from the Emu48 project. The authors claim that it is authentic ROM, they thank HP for donating it to the project, and I have yet to find that it does anything differently from what is described in the 50G user's manual. And I've definitely been putting it through its paces. I don't claim to have tried everything in its repertoire... I haven't had need for literally everything it's got, and that would take a long time anyway.
Sigh. No, again I must correct myself. I would apologize for the confusion, but I'm not the one who has been causing it. I've just been getting mixed up in the maze of your assumptions.
Your second sentence above does not assume a thermal equilibrium. What it does, though, is violate conservation of energy.
If the power input to the object we call the heat source is constant, and it is the only net power input to the system (the outer walls are refrigerated), then we have a contradiction.
Enough power (for illustration we can assume your figure of 509W/m^2 but I haven't checked it for accuracy) is being input to warm the heat source to an initial temperature of 150 deg. F.
I am aware that the enclosing passive plate will absorb power and convert it to thermal energy. But even you admit that it remains cooler than the heat source.
Your figure of 233 deg. F radiant temperature at what you called "equilibrium" represents a constant radiative power output from the heat source greater than its initial power output at 150 deg. F. Where is this additional power coming from?
S-B law says all the heat transfer in the system under discussion is outward from the heat source. So from whence comes this magical additional power you have calculated?
Huh? Of course we need the emissivities to model gray body heat transfer. If you'd like to solve the simpler problem of black body plates, then we can set the emissivities to 1, but I thought you wanted to skip that simpler problem.
Pardon me. That was a mistake on my part. I was thinking of specific wavelength emissivities and absorptivities, and was conflating that with your epsilons in my head. It was late and my thinking was muddled.
By all means, let's use emissivities and absorptivities. But you'll still have to modify your equation if that is based on the one you borrowed from MIT. I repeat that Kircchoff's law does not apply here.
What we have left is rather simple, except for "view factor". The view factor of the enclosing plate for radiation outward from the heat source is 1 or very close to it. The view factor of the heat source for radiation emitted by the enclosing plate is more complicated.
Not that it matters in the latter case, since 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.
And just to hammer it home, here it is again, as a direct quote from your website, except that I have replaced the "degree" symbol with "deg." to compensate for Slashdot's character handling:
For the moment, letâ(TM)s pretend the enclosing shell is a thermal superconductor, so its inner temperature Tc1 is also 149.6 deg. F (338.5K). Energy conservation at equilibrium just inside the enclosing shell shows that the heated sphere will warm to an equilibrium temperature of 233.8 deg. F (385.3K)
Here are two invalid assumptions in two consecutive sentences. The first postulates a thermal superconductor (which is neither necessary or relevant), and the second assumes a THERMAL equilibrium that does not exist.
Slashdot Mirror
The required electrical power to keep the heated plate at 150F is completely independent of the chamber wall temperature?
No. That is not what I wrote. You are drawing a conclusion that does not follow from my actual words. Making assumptions again.
It is dirt simple to show you are wrong.
The initial conditions, with the surface of the heat source at 338.7K, ARE A STEADY-STATE. The radiative transfer between the surface of the heat source and the chamber wall is already accounted for. You are trying to account for it twice. It is easy to show this.
The temperature of this surface is a GIVEN, initial steady-state condition. It is known, and a constant at this time. YOU are trying change it, and give one surface 2 different temperatures at the same time.
Proof: all we have to do is plug your value for radiative power output back into the known, canonical equation for radiative temperature.
Temperature is the 4th root of ( (power in W/m^2) / (se) ). So using your calculated value: 4th root of ( (29.399) / ((6.24 * 10^-9 W/m^2) / K^4) ) = 4th root of 3749839743.59 = 247.46K = -14.24 degrees F.
However, we already know what this temperature is, because it's a given:: 150 deg F (338.7K).
Your value gives a wrong answer. Your methodology contradicts itself, which is what I have been saying all along.
Plug my 82.12 W/m^2 back into the same canonical equation for radiant temperature for a given radiative power output, and the answer comes out just as it should: 150 degrees F.
If you can't even get the initial conditions right, we might as well stop here.
One way to see this is to consider how much power the electrical heater would need if the chamber walls were also at 150F. The correct answer is zero watts, because the heated plate wouldn't lose net heat to walls at the same temperature. But since your expression doesn't depend on the chamber wall temperature, you wouldn't be able to obtain the correct answer of zero in that case.
No, I am not wrong, you are. You are describing a radiative power difference, or net transfer.
That is not what I was doing. I was simply calculating the net power output of the heat source at 150 deg. F using the textbook example of how to do that.
It is not a difference. It is a constant radiative power output that depends on NOTHING else but temperature and emissivity, and the S-B constant.
It doesn't matter what temperature an opposing surface is at. I'm calculating the power output of THIS surface, at THIS temperature. As long as the temperature OF THIS SURFACE remains the same, the radiative power output remains the same. The way to calculate it is well-known and I have clearly stated it in my calculations.
YOU are contradicting yourself: "Power out = power in", you said. Right?
I have calculated the radiative power output using nothing more than area (~ 510 m^2), radiative temperature (338.7K), the emissivity you gave (0.11), and the well-known and proven relation:
Radiative power out (in W/m^2) = emissivity * sigma * T^4, where sigma is the Stefan-Boltzmann constant.
This is the textbook solution. Please show where it is incorrect. Simply asserting that it is incorrect is not sufficient.
Best you point that thing away from the station until we can figure out what's going on.
Those damned satellite thieves are getting bolder every day.
Note to future astronauts: be sure to roll the windows up and take your keys when you leave the module.
I'm not sure how that would play out in Postgres, but if you're using MySQL or a variant like MariaDB, you can
GRANT ALL ON *.* TO 'username'@'*' IDENTIFIED BY 'password';
Works for me.
Besides: despite what "the reviewers" say, I don't think "the age of the smartwatch" is quite here yet.
And it won't be, until it does something really useful on its own, rather than being a smartphone accessory.
It's certainly subjectively good. But I think it's also an important case for monitoring data wise to see the objective value of these things. My hope is that it's a net positive for every important metric, because an even slightly mixed bag of results could be enough to talk a lot of departments out of the idea.
I agree. That's why I think it's important that ground rules be firmly established.
For one thing, camera use must not be discretionary. It must be used every time there is an interaction with citizens. Because otherwise, there is too much potential for them to be used only when it is in their favor, and at other times, "Oops, I forgot to turn it on."
So if a camera is not turned on, or data is missing or shown to be deleted, a serious inquiry should be made to determine the actual reason why.
Why do I insist on this? Because I was once a victim of "missing" camera footage. I was told everything was being recorded, and the light on the camera was on. But when it came time to go to court and testify, they claimed there was no recording and it had "never existed".
Which was complete bullshit, of course.
Never mind what it was all about. It was a non-criminal charge and despite their bullshit stories I was not convicted. But "I forgot to turn it on" is too easy of a bullshit abuse of authority.
Once again, you zoomed ahead and did not proceed carefully. You are overlooking things. If you want this to be an actual solution of the problem, then let's do the problem completely. I have stated several times that I am only willing to do this if you agree to do it thoroughly.
Once again, energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At steady-state, that rate is zero because the system doesn't change. So at steady-state, power in = power out.
I agree, as long as you can legitimately draw a boundary around your system. So let's start out by drawing appropriate boundaries around out initial system.
I agree with your calculations for area. However I do not agree with your calculation for initial power output (which corresponds to our constant input).
We know the initial temperature of the heat source: 338.7K.
The acknowledged formula for finding radiative power from temperature is just (sigma epsilon)T^4. There are no other factors involved, so our power input is equivalent to the power output of the heat source. There is no reason to not assume perfect efficiency here. You don't show your work here, just an unreadable Sage file, so I don't know where the discrepancy lies.
This formula for radiative power output from radiative temperature and the corresponding formula for temperature from power both make use of (sigma epsilon), and epsilon is a scalar, so I will abbreviate it to (se) and pre-calculate it to make later calculations easier (I am using an HP, not Sage):
(se) = ((5.67 * 10^-8 W/m^2) / K^4) * 0.11 = (6.24 * 10^-9 W/m^2)/K^4
So here now is the reason for this preliminary setup: in our initial steady-state, heat source is 338.7K so total power output from the heat source (and therefore constant power input to the system) is just (se) * 338.7^4 * area
Therefore radiative power (W/m^2) = (6.24 * 10 ^-9) * 338.7^4 = 82 W/m^2
So then the total power of the heat source is 82 W/m^2 * 510.064 m^2 = 41.886 * 10^3 W
This does not seem like an unreasonable figure for heating a 12+m dia. sphere with 510 m^2 surface area to 150F.
So who is wrong and why?
No point in going further until we straighten this out.
I asked for a simple yes or no. You did not give me a simple yes or no, you had to bury it in the middle of about 500 words about other things. If you had simply written "yes", one word, you would have saved both of us about 20 minutes.
I am entirely convinced that you are deliberately trying to make things difficult.
SO... you have agreed. Now we may proceed with calculations. But I will not, more, tonight.
But -- and this is no joke -- I expect you to show and describe your work. We start from extant conditions and proceed from there, in a step-wise manner. My intent, as it was before, is to minimize confusion and misunderstandings, which it is pretty obvious have been rampant. So... as is often the case in programming, if we proceed slowly in a step-wise manner, we achieve the end result all the sooner, because errors don't crop up.
I was serious about this: you have had a tendency to make assumptions and use terms loosely. We have had LOTS of misunderstandings here, and I am willing to chalk some of them up to communications problems, but not all of them.
So let's go slowly, and start simply. The first step is to define the initial conditions. We have a fixed power input, which heats the "source" (and I am calling it that for good REASONS), which we are assuming is a spherical body. Specific dimensions are up to you, I really don't care very much, although I have to wonder why you picked radii that seem so inconvenient... but on the other hand I admit that I did not do any calculations at this point using your dimensions
So, since we are starting fresh, describe your equation now, what the variables represent, and how you arrived at the formula. I suspect that it is a bit premature.
I will check in again tomorrow.
NO. See my comment above. One more bullshit comment like this, and as I said, I will just call you a clown and few reasonable people will disagree.
Will you, yes or now, agree to the definition of Spencer's experiment, as I have already described and asked you now about 3 times.
That is the FIRST step in solving any problem. I will not agree to any math regarding the problem until we have an agreement about what the problem is. Anything else is nonsense.
ONCE THAT IS DONE, I agree to move on with calculations about the problem before us.
BEFORE you do the math, the problem must be defined. No reasonable scientist would disagree with that. So far, you have refused at every turn to define the problem, even though this is the well-known first step to proving anything.
If you continue to just bullshit your way around, as I have stated I will declare you in default and damned few reasonable people would disagree.
You're sidestepping again, and you're ignoring the important point, again.
This is a sidetrack, about an OLD comment you made that was stupid and inaccurate. You can argue about THAT comment until the cows come home and die of old age, and you still won't be correct. BUT IT'S IRRELEVANT TO THE POINT NOW AT HAND.
If you want to move on, see my other comments which describe the problem. If you continue to refuse to agree to the definition of the problem, I will (with every justification in the logical world) declare you in default.
STOP THE BULLSHIT. It is pretty obvious what it would actually take to refute Latour. I have described in my other comment a statement of the actual challenge. If you refuse to agree with my description of the problem, without having a reasonably valid objection, I shall (with perfect justification) declare you in default.
If your time is really short as you say, then just abandon those recent sidetracks (which is in your own best interest anyway) and let's continue.
If you agree to the problem description I laid out above, in this comment, just say yes. Then we can continue.
But I will repeat, and keep repeating as long as necessary, that unless we agree on what it would take to refute Latour, then there is no point in going further. You can violate thermodynamics all you want, and it doesn't prove a damned thing.
Cute. I've repeatedly explained ad nauseum that neglecting area ratios is an approximation. I've already shown how tiny the effects are for Earth's area ratio. Does this mean you don't intend to perform even the simplest calculation to confirm this? Why don't we check to see wrong these approximations are, by actually doing some calculations? Finally? Please?
THERE'S NOTHING "CUTE" ABOUT IT!
IT'S AN ACCURATE ASSESSMENT OF YOUR ERROR!
This is not "approximation", it's fucking logical error! JESUS CHRIST, man, you can't talk your way around this.
X / 3 = X / 3.
X / 6 < X / 3
End.
Once again, no. I've repeatedly explained [slashdot.org] that the outer surface of the enclosing passive plate is never at the same temperature as the heat source.
And once again, NO, it doesn't matter. I don't care at what time you assume this to be. Because we have a constant power input which has radiative power output at X W/m^2, which is required to reach the radiative temperature of 150F.
You then (at ANY time, I don't care when) claim that a larger surface is at the same temperature, which requires the same amount of W/m^2. But you have more m! So the total power output is greater than your input.
There is no way to weasel out of this, man. You're trying to output more power than you're putting in. This isn't even 11th-grade physics.
Let's try it at something more like your level:
You have 200 beans equally distributed among 10 squares. If you now take those beans, and divide them equally among 25 squares of the same size, how many beans do you now have per square?
Show your work.
I've been explaining for over a month that the heated plate (aka Jane's "source") warms after it's enclosed.
Yes, you've been explaining it but you haven't been backing it up. See my other comment. Once we agree on the statement of the problem, we can move forward. I haven't been trying to block you, I've only been trying to get you to unambiguously agree to terms.
If Jane agrees that the heated plate (aka Jane's "source") warms after it's enclosed, then that's great news! In that case, we can all agree that the mainstream physics describing the greenhouse effect is accurate, obeys the laws of thermodynamics, and proves that the Sky Dragon Slayers are wrong.
No. Jane does not agree that "the heated plate" (if by that you mean THE HEAT SOURCE), gets warmer when it's enclosed. THIS IS WHAT I'VE BEEN TRYING TO TELL YOU.
You need to use more precise language. In plain English, you have one heat SOURCE in this experiment, and THE OTHER plate is being "heated" by it. So the "heated plate" is the passive plate.
You insist on using confusing terminology, and wonder why other people have a hard time with it. Jesus, I'm glad you weren't one of my physics profs.
I suggested a standard terminology and variable names to avoid exactly this problem. You have refused to use them. That's your goddamned problem, and you don't get to complain about it.
I'm really looking forward to showing this latest exchange to my friends.
It's just so hard to work my way though your maze of comments, some of which are correct, and others which are provably bullshit (see my other comments above).
THE FIRST STEP in mutually working your way through a problem is agreeing on the statement of the problem, as you well know. THAT is why I have not willing to get into calculations yet. That and nothing else. So let's make sure we agree on the statement of the actual experiment. I want you to acknowledge these in so many words, before I am willing to move forward. Because I'm just plain tired of your bull.
1. Conditions apply as per Spencer's experiment:
2. Constant input power sufficient to heat the source to a radiative temperature at radiative equilibrium of 150F.
3. Outer chamber walls actively cooled to a maintained radiative temperature of 0F.
4. Passive plate (which you insist in your variant of Spencer's challenge fully encloses heat source, with vacuum between).
5. Passive plate is introduced at a temperature that is cooler than the heat source.
6. Any other conditions that were actually contained in Spencer's challenge but not mentioned here.
7. Your own condition: you insist on using gray bodies, because you claim you're dying (or something of the sort, you really didn't specify so I won't assume), you don't have the spare 15 minutes or so it might take to do approximate, more real-world calculations, and want to use Kirchhoff's law (although it really isn't necessary) to make your life easy.
8. I reserve the right to re-visit this same scenario using more real-world materials, emissivities and absorptivities.
Are we agreed on these conditions? I want a simple yes or no. Anything else, and I am not willing to continue.
Once again, I never said that all surfaces were at the same temperature.
Look, let's get this straight: what you actually meant was completely ambiguous for several reasons. You used the term "equilibrium" and you said:
Electric input of 509 W/m^2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).
Will you acknowledge that no matter what you meant, this is still wrong? If the enclosing passive plate must radiate out the same power as the enclosed heat source, it cannot be at the same temperature, because radiated power is measured in W/m^2, and there are more m^2 in the enclosing passive plate. Therefore (SIMPLE MULTIPLICATION), because there is greater area they could not be at the same temperature and radiate outward the same power.
No matter how you try to bullshit your way around this, it is still WRONG.
As I said: "Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out."
I explicitly said a system in "equilibrium" doesn't change, which Jane calls "steady-state".
I call it "steady state" because that is how radiative equilibrium is defined. It is a condition in which radiative transfer between elements of the system remains constant. I prefer to make sure this is kept distinct from thermal equilibrium.
I just want to make sure we're all talking about the same things. Because in the past, we sure as heck were not.
Why would Jane think mentioning Kirchhoff's law means that I'm somehow trying to claim that all surfaces are at the same temperature? I first mentioned Kirchhoff's law by linking to MIT's explanation:
Stop being obtuse. You were throwing around the term "equilibrium" rather loosely, and at one point you mentioned that "at equilibrium" the outer surface of the enclosing passive plate must be at the same temperature as the surface at the heat source:
Electric input of 509 W/m^2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).
Now, if it isn't a reasonable conclusion that by this you meant thermal equilibrium, I don't what is, since you are claiming the heat source and the passive plate are the same temperature. Of course upon closer examination it could not have been actual thermal equilibrium, because you also mention the walls are at 0 degrees F, but then what actual kind of "equilibrium" you were referring to is worse than ambiguous, because at radiative equilibrium the stated condition is impossible. So why the hell are you trying to blame me for being confused? The condition you described is impossible, so how do you expect me to know what "equilibrium" you mean?
"... the relation "absorptance = emittance" is known as Kirchhoff's Law. It implies that good radiators are good absorbers. It was derived for the case when "body temperature = cavity temperature" and is not strictly true for all circumstances when the temperature of the body and the cavity are different, but it is true if the absorptance and emittance are not functions of wavelength. This situation describes a 'gray body.' ... "
Very well. You refuse to use real materials with measured absorptivities and emissivities, and insist on using gray bodies and Kirchhoff's law, neither of which actually represent Spencer's experiment in anything like the real world, even though it was intended to be a model of the real world.
I get that. But I want to make sure everyone else gets it too.
I offered to use real materials with measured emissivities in the approximate temperature range we are talking about (though you refuse to acknowledge that), but you refuse to use them.
Just so we know where we stand. I have already explained to you that there is no need to resort to gray bodies, and that we have plenty of information to calculate more realistic, real-world results. But whatever. You refuse to do anything but what you want to do, so let's just go with it for now. But I reserve the right to re-visit this issue.
I've been explaining for over a month that the heated plate warms after it's enclosed. I realize you don't agree, which is why I'm trying in vain to get you to finally perform a single, solitary calculation of your own.
I never said I disagree with this. Please find where I said that. On the contrary; I definitely agree that it warms. In fact it must: Spencer stipulated that it was to be inserted when it was colder than the heat source.
I have been doing calculations. I just haven't been showing them here, because there isn't any point yet. Before there is any point to showing calculations, we must agree that certain conditions either do or do not exist, given the parameters of this experiment. That is what I am trying to sort out with you now.
I should clarify:
It is the "50g mode" from the Emu48 project. The authors claim that it is authentic ROM, they thank HP for donating it to the project, and I have yet to find that it does anything differently from what is described in the 50G user's manual. And I've definitely been putting it through its paces. I don't claim to have tried everything in its repertoire... I haven't had need for literally everything it's got, and that would take a long time anyway.
I'll do better than that.
Here is a screenshot from my Mac.
sigma, K, F, and T are variables I stored for some thermodynamic calculations I was doing earlier.
Sigh. No, again I must correct myself. I would apologize for the confusion, but I'm not the one who has been causing it. I've just been getting mixed up in the maze of your assumptions.
Your second sentence above does not assume a thermal equilibrium. What it does, though, is violate conservation of energy.
If the power input to the object we call the heat source is constant, and it is the only net power input to the system (the outer walls are refrigerated), then we have a contradiction.
Enough power (for illustration we can assume your figure of 509W/m^2 but I haven't checked it for accuracy) is being input to warm the heat source to an initial temperature of 150 deg. F.
I am aware that the enclosing passive plate will absorb power and convert it to thermal energy. But even you admit that it remains cooler than the heat source.
Your figure of 233 deg. F radiant temperature at what you called "equilibrium" represents a constant radiative power output from the heat source greater than its initial power output at 150 deg. F. Where is this additional power coming from?
S-B law says all the heat transfer in the system under discussion is outward from the heat source. So from whence comes this magical additional power you have calculated?
Huh? Of course we need the emissivities to model gray body heat transfer. If you'd like to solve the simpler problem of black body plates, then we can set the emissivities to 1, but I thought you wanted to skip that simpler problem.
Pardon me. That was a mistake on my part. I was thinking of specific wavelength emissivities and absorptivities, and was conflating that with your epsilons in my head. It was late and my thinking was muddled.
By all means, let's use emissivities and absorptivities. But you'll still have to modify your equation if that is based on the one you borrowed from MIT. I repeat that Kircchoff's law does not apply here.
What we have left is rather simple, except for "view factor". The view factor of the enclosing plate for radiation outward from the heat source is 1 or very close to it. The view factor of the heat source for radiation emitted by the enclosing plate is more complicated.
Not that it matters in the latter case, since 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.
For the moment, letâ(TM)s pretend the enclosing shell is a thermal superconductor, so its inner temperature Tc1 is also 149.6 deg. F (338.5K). Energy conservation at equilibrium just inside the enclosing shell shows that the heated sphere will warm to an equilibrium temperature of 233.8 deg. F (385.3K)
Here are two invalid assumptions in two consecutive sentences. The first postulates a thermal superconductor (which is neither necessary or relevant), and the second assumes a THERMAL equilibrium that does not exist.