No. I've repeatedly told you that power in = power out demands that an unheated inner sphere will be at exactly the same temperature as the chamber walls.
That isn't quite what you said. This is what you said:
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.
I already understand this, and I mentioned it myself in the post above. My point was that it does not translate directly into power in = power out at a boundary just inside the cavity surface. It most certainly does not if the bodies are not in thermal equilibrium, which again I must point out this system is not in. See my reference again. By the way, the author is Incropera, not "Incopora". Slip of the keyboard, there.
As for the rest, I am out of time right now and will reply tomorrow if I have more time.
Just one last closing comment tonight, though: I am aware that energy in a system must be conserved. But "system" is not anywhere you choose to draw a line. In the case of heat transfer, energy does not have to be conserved between two bodies at different temperatures. That was what Incorpora was saying in his book. And that is why I balk at your "conservation of energy just inside the surface".
I will do you a favor here, and say: don't bother to go calculating the energy, either.
The problem is that an analysis of this kind, based on the assumption that power-in = power-out, is doomed to fail except in coincidental cases. Even conservation of energy can give very misleading results.
The black body example I gave shows why your "energy conservation just inside the surface" won't work. Aside from just "view factor" and a few other things, a certain amount of the power in (often a very significant amount) just ends up going right back out, but you often don't see that in the formulas.
Quote from one of my references, "Fundamentals of Heat and Mass Transfer", by Inropera, et al., 6th edition, 2006, p13. I have to type this in by hand from the book so any typographical errors are mine. Emphasized words have been capitalized.
Relationship to Thermodynamics
At this point it is appropriate to note the fundamental differences between heat transfer and thermodynamics. Although thermodynamics is concerned with the heat interaction and the vital role it plays in the first and second laws, it considers neither the mechanisms that profide for heat exchange nor the methods that exist for computing the RATE of heat exchange. Thermodynamics is concerned with EQUILIBRIUM states of matter, where an equilibrium state necessarily precludes the existence of a temperature gradient. Although thermodynamics may be used to determine the amount of energy required in the form of heat to pass from one equilibrium state to another, it does not acknowledge that HEAT TRANSFER IS INHERENTLY A NONEQUILIBRIUM PROCESS. For heat transfer to occur, there must be a temperature gradient and, hence, thermodynamic nonequilibrium. The discipline of heat transfer therefore seeks to do what thermodynamics is inherently unable to do, namely, to quantify the RATE at which heat transfer occurs in terms of the degree of thermal nonequilibrium. This is done via the rate equations for the three modes...
Heat transfer requires a temperature gradient, and therefore thermodynamic non-equilibrium (as we established early on). I was hoping you would catch on that this also implies that power-in = power-out is not necessarily true, and in fact that is probably a very rare exception.
Therefore, you aren't going to prove anything with this approach. I wanted to stop you before you wasted more of your time.
Further, the above example of the black body suspended in the black-body cavity at thermal equilibrium shows why your "conservation of energy just inside the heated plate surface" is more complex than you make it out to be.
No, before you jump all over my black body example, I am aware that view factor has to be taken into account.
But that is actually part of my point: a simple power-in = power-out view is not always the right answer.
It is true that the interior of the cavity is radiating twice as much power out. There is a view factor involved, which may account for the difference. But the view factor does not involve power output of the radiating body. We know what that is. Much of it is being re-absorbed by the interior of the cavity, true. But it shows how power-in = power-out calculations can easily mislead.
Perhaps it would be more informative if you calculate ENERGY in and ENERGY out, since that is what is actually conserved.
You seem to keep forgetting that (A) power is a RATE, not a unit of energy, and (B) we are not at thermal equilibrium.
Classical example from Wikipedia: running up the stairs requires more power than walking up the stairs, because more energy is expended per unit time. (Granted, the time period is also shortened, but it still illustrates that they are not the same.)
Let me give you a physics example: We have a gray-body hemisphere, emissivity 0.5, with radius of the flat surface 1.00 m and temperature T of 200K. We do NOT assume thermal equilibrium.
The area of the curved part is 6.28 m^2, of the flat part is pi, so the total area is 9.42 m^2.
We have incident radiation hitting the flat surface of 229.64 W/m^2. (We draw a "boundary" around our hemisphere, so that is our "system", and the incident radiation is the only "power" input.)
Our total input -- our ONLY power input -- is 229.64 W/m^2 * pi = 721.44 W.
Total power output in this case is emittance + unabsorbed incident radiation, which would normally be "radiosity", except radiation is only being absorbed on one surface. (I.e., our "view factor" F is only 1/3.)
Since emissivity = 0.5, total "reflected" (i.e. unabsorbed) radiation is 360.7 W
Total power out then is 45.36 W/m^2 * 9.42 m^2 = 427.29 W + 360.72 W = 788.01 W.
788.01 W != 721.44 W (!!!) Power is not conserved.
Obviously this does not represent radiative steady-state.
Now let's take an even simpler example: a black body sphere of surface area 1 m^2 inside a spherical "black body cavity" with area of 2 m^2, at thermal equilibrium.
Obviously, since radiative power = (epsilon)(sigma)T^4, both surfaces are radiating the same power in W/m^2.
However, the inner surface of the cavity has twice as much area, so the total power radiated is twice as much. Power is not conserved.
If you tried to argue that the increased power would warm up the interior sphere, then you're no longer in thermal equilibrium.
So... are you suggesting that if I hollowed out enough of a mountain to make a hollow rock sphere (assume the rock is diffuse gray body) 1000 m diameter, suspended a 1m dia. sphere of the same rock in the center, and evacuated the cavity: the inner sphere is going to get much hotter than the surrounding rock?
Power in = power out would seem to demand that very thing.
Let me rephrase what I was saying: at least theoretically, the power at the chamber wall is allowed to vary, in order to keep the temperature at 0 degrees F.
But, if we draw a boundary around the system, and assume that the ONLY power in is what we put in, and the ONLY power out is what is removed, then of course it must be conserved.
I was simply expressing my concern that your electricity figure may not be properly observing those boundaries. If your electricity figure is simply power in - power out, I fail to see why you need to calculate it in such a fashion. I think it is an unnecessary complication and potential source of error.
Never mind. I will back up on that, at least for now.
If we assume that power output of the exterior surface of the enclosing shell is the same as it was from the heat source under initial conditions (something I am not yet ready to stipulate, since we are not at thermal equilibrium), I calculate a temperature, using my own shown above, at 338.49K.
However, I want to make this clear: I am not convinced that your power in = power out assumption is correct in this case, because we have a refrigerated outer shell, which also consumes power (we do not yet know how much), which keeps things OUT of thermal equilibrium. We are adding power in the center, and we are removing power at the outside. But because of Spencer's conditions, I am not convinced at this point that we can assume power is conserved.
If everything were at thermal equilibrium, I would be convinced. But at the very least, we would have to calculate the difference between power consumed by the refrigerator on the outside, between initial and final conditions. Do we have enough information to do that?
What is the ambient temperature? What is the volume of the chamber wall?
We already know what the radiated power output of the heat source is, given the initial conditions, which I calculated via a far simpler and unambiguous equation which we know to be relevant: (epsilon)(sigma)T^4.
No "electricity" needed. Your "electricity" figure is NOT the "power out" of the heat source. It is a figure for total power consumed that I do not agree applies in this instance, since we have a refrigerator on the outside which also consumes power.
To put it another way, your "electricity" figure is not power output of source it is a figure for a DIFFERENCE, which I do not agree applies in this instance.
Again, using (epsilon)(sigma)T^4:
Radiative emittance of heat source under initial conditions: 82.12 W/m^2
You already agreed with this figure.
Total radiative power out = (82.12 W/m^2) * (510.065 m^2) = 41886.54 W
You are contradicting yourself. Either this is the correct figure, or it is not.
Just this year, a Federal judge ruled that the FAA has no authority to regulate drones outside of navigable airways. (Which are clearly specified on aviation charts.)
The FAA has appealed the ruling, but since the judge appears to have ruled on solid Constitutional grounds, I doubt very much they'll win the appeal.
It's just a fact: FAA doesn't have jurisdiction over everything in the air. All of their authority is based on the Federal ability to regulate manned interstate airplane flight.
At the original steady-state without the shell, the net radiative power leaving the source equals the constant electrical power heating the source. This constant power doesn't change even after the shell is inserted.
Yes, this was one of the reasons I took the time to calculate the irradiance = radiative power output / m^2.
So you object to calculating heat transfer via radiation by using radiative transfer equations?
By the way: upon looking at the situation more closely, I found that applying the S-B law directly does not apply in this exact situation. (I looked because it gave a much different answer than the one I had already calculated.) It applies when a body is radiating to its ambient surroundings, not between two bodies. (We don't have "ambient" surroundings per se... just vacuum between 2 bodies.)
My first method of calculating the heat transferred was the correct one:
( (epsilon) * (sigma) * T(s)^4 * Area(s) ) - ( (epsilon) * (sigma) * T(w)^4 * Area(w) )
Factoring out (es) you get ( T(s)^4 * Area(s) ) - ( T(w)^4 * Area(w) ), which only holds of course when e = a.
We understand that I'm still working on the initial conditions, correct? There is no "enclosing plate" at this point.
You can also ignore the latter part involving absorbed radiation for now. It isn't really relevant to anything I am doing at this time.
It would seem that the next logical step would be to calculate the net power loss of the heat source to its surroundings under these conditions, applying the Stefan-Boltzmann law directly. Using the same variable names I used earlier:
power = (se) * radiating area * (T(s)^4 - T(w)^4) = (se) * 510.065 * (338.71^4 - 255.37^4) =
It is not radiant flux, or radiant energy. Nor is the irradiance (I'll use that term from now on for clarity) of the heat source I calculated dependent on the chamber wall in any way. I repeat: the only variables for calculating this value for a gray body are temperature and emissivity. It is independent of any other object, and it is independent of absorbed incident radiation.
It's not a matter of the equation. I repeat: you jumped way ahead, and used a shortcut. That's not what I am doing, and that's the source of the problem here.
I've only calculated the radiant power value of one surface so far! I haven't even calculated the second yet. So how you could you possibly think I had calculated net heat transfer?
I'm not trying to calculate your "electricity" value. I don't want to calculate your electricity value. I'm starting from the basics, and working all the way through. I'm not using your shortcut. Is this clear?
Do you have a problem with my formula for calculating radiant power of a gray body surface at a given temperature? If not, I will continue. I repeat: it isn't dependent on any heat transfer, the only variables are emissivity and temperature. With that, and the Stefan-Boltzman constant, you can calculate the radiant power value.
I was not trying to calculate electricity. I was merely calculating radiative power out of the heat source at a known temperature.
The next step (because, as I have repeatedly stated, I am going slowly and carefully, in a step-wise manner) would have been to calculate the radiative power out of the chamber walls.
I was not calculating any heat transfer. On the contrary: heat transfer is dependent on those two values. I hadn't even calculated the second value yet.
I have been very straightforward and clear about what *I* am doing.
Furthermore, your value for electricity is completely irrelevant to the problem at hand, which was to calculate internal temperatures under the given conditions.
In these initial conditions, I'm stating that the radiant power of the heat source is 82.12 W/m^2.
I admit that I made an error somewhere, and plugged in the wrong number. 29.4 / (6.24 * 10 ^-9) is indeed 4711538461.55, and the 4th root of that is 261.99K, or 11.91 degrees F.
However: as you have said, let's be clear: you throw around terminology fast and loose, and you have refused to show your calculations, which leads to misunderstandings. So: what, then, do you claim that 29.4 W/m^2 figure represents and why?
I don't give a rat's ass about "electricity" at this point. The radiative power, (we are using units of W/m^2) of a gray body surface depends only on its temperature, its area, its emissivity, and the Stefan-Boltzmann constant. I do not need to know what the net heat transfer is to calculate this value (which IS 82.12 W/m^2 at the surface of the heat source under the given conditions). The total radiated power in Watts is: W/m^2 times the area. This is all pure textbook stuff. It matters not a whit at this point what that radiation strikes AFTER it is emitted. Even considering that right now is premature.
If I misunderstood, and your 29.4 W/m^2 represents something other than radiative power at the surface of the heat source at 338.7K, then please state clearly in plain terms what it IS supposed to represent, so we can move on.
The next step in the problem I am analyzing, because as I stated I am doing this in a careful stepwise manner precisely to avoid these misunderstandings is to calculate the radiative power of the chamber walls at 0 deg. F, or 255.37K.
But let me be clear: I don't give a damn about electricity at this point. In fact, I don't give the slightest damn whether the sphere is heated by steam, or an internal campfire, or burning unicorn farts. We know the power required for a gray body of the given area and emissivity to have a radiative temperature of 338.7K.
That is all I was saying. Nothing else. Trying to assume what I'm doing with that number before I do it is magical thinking.
I wasn't TRYING to calculate heat transfer. I was only calculating radiative power out. That was a necessary step to THEN going on to calculate the radiative power of the chamber walls, and THEN calculating the net heat transfer.
The TEMPERATURE of the heat source does not depend on heat transfer. It is an independent variable. On the contrary: the net heat transfer depends on the temperature, not the other way around.
We have already seen that you have mis-applied your equation and arrived at a power out figure that gives an incorrect value of temperature, which is an already known value.
Your "refutation" is disproved, almost before we've properly begun.
QED
I don't care if you don't understand that. Other people will.
You have not only made assumptions that don't actually apply, you have thrown around equations without carefully considering how they apply to the clearly stated problem. I said this in the beginning, and I have proven it now.
We already know, at the given steady-state, that the heat source is at 338.7K. We already know it is a sphere with a surface area of approximately 510 m^2. You have insisted we assume that it is a gray body. Given those two numbers, the emissivity, and the Stefan-Boltzmann constant, we calculate the power out: 82.12 W/m^2.
We do not need to do anything else at this point. We already know that this is the temperature GIVEN any heat transfer to the walls... which we have not calculated yet. I was going to go on to do that, but it isn't necessary now. We've seen that you're already wrong.
This was precisely why I insisted we do this slowly and carefully, and explain our steps. Because I knew you were doing it incorrectly (I said so) and that it would show up in the calculations. I did not expect it quite this soon, but there it is.
YOU may not understand that I have already proved your "refutation" wrong, but I assure you that other people will have no such difficulty. And they will have ample opportunity to see this, because I'm going to post it all online.
They are NOT typically used to "prevent piracy" or "spread of malware" or "website access", I don't even see the use case here. I think the OP is confusing full packet capture with layer 7 application state firewalls, which ARE used for the above.
Um, wrong.
Deep-packet inspection was used routinely by the large ISPs to throttle certain kinds of traffic, until the FCC made them stop. This was just a couple of years ago.
Maybe not "full" packet inspection, but it was deep packet inspection, so they could distinguish, for example, packets of BitTorrent traffic from packets containing streamed video from YouTube.
Slashdot Mirror
Damn. Finger slipped again.
No. I've repeatedly told you that power in = power out demands that an unheated inner sphere will be at exactly the same temperature as the chamber walls.
That isn't quite what you said. This is what you said:
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.
I already understand this, and I mentioned it myself in the post above. My point was that it does not translate directly into power in = power out at a boundary just inside the cavity surface. It most certainly does not if the bodies are not in thermal equilibrium, which again I must point out this system is not in. See my reference again. By the way, the author is Incropera, not "Incopora". Slip of the keyboard, there.
As for the rest, I am out of time right now and will reply tomorrow if I have more time.
Just one last closing comment tonight, though: I am aware that energy in a system must be conserved. But "system" is not anywhere you choose to draw a line. In the case of heat transfer, energy does not have to be conserved between two bodies at different temperatures. That was what Incorpora was saying in his book. And that is why I balk at your "conservation of energy just inside the surface".
The problem is that an analysis of this kind, based on the assumption that power-in = power-out, is doomed to fail except in coincidental cases. Even conservation of energy can give very misleading results.
The black body example I gave shows why your "energy conservation just inside the surface" won't work. Aside from just "view factor" and a few other things, a certain amount of the power in (often a very significant amount) just ends up going right back out, but you often don't see that in the formulas.
Quote from one of my references, "Fundamentals of Heat and Mass Transfer", by Inropera, et al., 6th edition, 2006, p13. I have to type this in by hand from the book so any typographical errors are mine. Emphasized words have been capitalized.
Relationship to Thermodynamics
...
At this point it is appropriate to note the fundamental differences between heat transfer and thermodynamics. Although thermodynamics is concerned with the heat interaction and the vital role it plays in the first and second laws, it considers neither the mechanisms that profide for heat exchange nor the methods that exist for computing the RATE of heat exchange. Thermodynamics is concerned with EQUILIBRIUM states of matter, where an equilibrium state necessarily precludes the existence of a temperature gradient. Although thermodynamics may be used to determine the amount of energy required in the form of heat to pass from one equilibrium state to another, it does not acknowledge that HEAT TRANSFER IS INHERENTLY A NONEQUILIBRIUM PROCESS. For heat transfer to occur, there must be a temperature gradient and, hence, thermodynamic nonequilibrium. The discipline of heat transfer therefore seeks to do what thermodynamics is inherently unable to do, namely, to quantify the RATE at which heat transfer occurs in terms of the degree of thermal nonequilibrium. This is done via the rate equations for the three modes
Heat transfer requires a temperature gradient, and therefore thermodynamic non-equilibrium (as we established early on). I was hoping you would catch on that this also implies that power-in = power-out is not necessarily true, and in fact that is probably a very rare exception.
Therefore, you aren't going to prove anything with this approach. I wanted to stop you before you wasted more of your time.
Further, the above example of the black body suspended in the black-body cavity at thermal equilibrium shows why your "conservation of energy just inside the heated plate surface" is more complex than you make it out to be.
No, before you jump all over my black body example, I am aware that view factor has to be taken into account.
But that is actually part of my point: a simple power-in = power-out view is not always the right answer.
It is true that the interior of the cavity is radiating twice as much power out. There is a view factor involved, which may account for the difference. But the view factor does not involve power output of the radiating body. We know what that is. Much of it is being re-absorbed by the interior of the cavity, true. But it shows how power-in = power-out calculations can easily mislead.
Correction to one of the equations above. Total power out then is (45.36 W/m^2 * 9.42 m^2 = 427.29 W) + 360.72 W = 788.01 W.
Perhaps it would be more informative if you calculate ENERGY in and ENERGY out, since that is what is actually conserved.
You seem to keep forgetting that (A) power is a RATE, not a unit of energy, and (B) we are not at thermal equilibrium.
Classical example from Wikipedia: running up the stairs requires more power than walking up the stairs, because more energy is expended per unit time. (Granted, the time period is also shortened, but it still illustrates that they are not the same.)
Let me give you a physics example: We have a gray-body hemisphere, emissivity 0.5, with radius of the flat surface 1.00 m and temperature T of 200K. We do NOT assume thermal equilibrium.
The area of the curved part is 6.28 m^2, of the flat part is pi, so the total area is 9.42 m^2.
We have incident radiation hitting the flat surface of 229.64 W/m^2. (We draw a "boundary" around our hemisphere, so that is our "system", and the incident radiation is the only "power" input.)
Our total input -- our ONLY power input -- is 229.64 W/m^2 * pi = 721.44 W.
Total amount absorbed = 721.44 * 0.5 = 360.72 W.
Emittance is 0.5 * sigma * T^4 = 0.5 * 5.67 * 10^-8 * 1600000000 = 45.36 W/m^2
Total power output in this case is emittance + unabsorbed incident radiation, which would normally be "radiosity", except radiation is only being absorbed on one surface. (I.e., our "view factor" F is only 1/3.)
Since emissivity = 0.5, total "reflected" (i.e. unabsorbed) radiation is 360.7 W
Total power out then is 45.36 W/m^2 * 9.42 m^2 = 427.29 W + 360.72 W = 788.01 W.
788.01 W != 721.44 W (!!!) Power is not conserved.
Obviously this does not represent radiative steady-state.
Now let's take an even simpler example: a black body sphere of surface area 1 m^2 inside a spherical "black body cavity" with area of 2 m^2, at thermal equilibrium.
Obviously, since radiative power = (epsilon)(sigma)T^4, both surfaces are radiating the same power in W/m^2.
However, the inner surface of the cavity has twice as much area, so the total power radiated is twice as much. Power is not conserved.
If you tried to argue that the increased power would warm up the interior sphere, then you're no longer in thermal equilibrium.
So... are you suggesting that if I hollowed out enough of a mountain to make a hollow rock sphere (assume the rock is diffuse gray body) 1000 m diameter, suspended a 1m dia. sphere of the same rock in the center, and evacuated the cavity: the inner sphere is going to get much hotter than the surrounding rock?
Power in = power out would seem to demand that very thing.
Energy is always conserved.
Of course it is.
Let me rephrase what I was saying: at least theoretically, the power at the chamber wall is allowed to vary, in order to keep the temperature at 0 degrees F.
But, if we draw a boundary around the system, and assume that the ONLY power in is what we put in, and the ONLY power out is what is removed, then of course it must be conserved.
I was simply expressing my concern that your electricity figure may not be properly observing those boundaries. If your electricity figure is simply power in - power out, I fail to see why you need to calculate it in such a fashion. I think it is an unnecessary complication and potential source of error.
Never mind. I will back up on that, at least for now.
If we assume that power output of the exterior surface of the enclosing shell is the same as it was from the heat source under initial conditions (something I am not yet ready to stipulate, since we are not at thermal equilibrium), I calculate a temperature, using my own shown above, at 338.49K.
However, I want to make this clear: I am not convinced that your power in = power out assumption is correct in this case, because we have a refrigerated outer shell, which also consumes power (we do not yet know how much), which keeps things OUT of thermal equilibrium. We are adding power in the center, and we are removing power at the outside. But because of Spencer's conditions, I am not convinced at this point that we can assume power is conserved.
If everything were at thermal equilibrium, I would be convinced. But at the very least, we would have to calculate the difference between power consumed by the refrigerator on the outside, between initial and final conditions. Do we have enough information to do that?
What is the ambient temperature? What is the volume of the chamber wall?
We already know what the radiated power output of the heat source is, given the initial conditions, which I calculated via a far simpler and unambiguous equation which we know to be relevant: (epsilon)(sigma)T^4.
No "electricity" needed. Your "electricity" figure is NOT the "power out" of the heat source. It is a figure for total power consumed that I do not agree applies in this instance, since we have a refrigerator on the outside which also consumes power.
To put it another way, your "electricity" figure is not power output of source it is a figure for a DIFFERENCE, which I do not agree applies in this instance.
Again, using (epsilon)(sigma)T^4:
Radiative emittance of heat source under initial conditions: 82.12 W/m^2
You already agreed with this figure.
Total radiative power out = (82.12 W/m^2) * (510.065 m^2) = 41886.54 W
You are contradicting yourself. Either this is the correct figure, or it is not.
Just this year, a Federal judge ruled that the FAA has no authority to regulate drones outside of navigable airways. (Which are clearly specified on aviation charts.)
It doesn't matter whether the use is commercial.
The FAA has appealed the ruling, but since the judge appears to have ruled on solid Constitutional grounds, I doubt very much they'll win the appeal.
It's just a fact: FAA doesn't have jurisdiction over everything in the air. All of their authority is based on the Federal ability to regulate manned interstate airplane flight.
At the original steady-state without the shell, the net radiative power leaving the source equals the constant electrical power heating the source. This constant power doesn't change even after the shell is inserted.
Yes, this was one of the reasons I took the time to calculate the irradiance = radiative power output / m^2.
So you object to calculating heat transfer via radiation by using radiative transfer equations?
By the way: upon looking at the situation more closely, I found that applying the S-B law directly does not apply in this exact situation. (I looked because it gave a much different answer than the one I had already calculated.) It applies when a body is radiating to its ambient surroundings, not between two bodies. (We don't have "ambient" surroundings per se... just vacuum between 2 bodies.)
My first method of calculating the heat transferred was the correct one: ( (epsilon) * (sigma) * T(s)^4 * Area(s) ) - ( (epsilon) * (sigma) * T(w)^4 * Area(w) )
Factoring out (es) you get ( T(s)^4 * Area(s) ) - ( T(w)^4 * Area(w) ), which only holds of course when e = a.
I objected to thermal superconductors because they led to contradiction.
I am curious: how do you propose to calculate the outer temperature first?
No, I have no intention of finishing today. I am busy and I have been putting in what little time I have had to "spare".
Please explain why conductivity is relevant. We are examining the system in steady-state.
The plate is inserted into the system colder than the heat source (Spencer's stipulation).
I didn't assume a black body. But whatever.
We understand that I'm still working on the initial conditions, correct? There is no "enclosing plate" at this point.
You can also ignore the latter part involving absorbed radiation for now. It isn't really relevant to anything I am doing at this time.
It would seem that the next logical step would be to calculate the net power loss of the heat source to its surroundings under these conditions, applying the Stefan-Boltzmann law directly. Using the same variable names I used earlier:
power = (se) * radiating area * (T(s)^4 - T(w)^4) = (se) * 510.065 * (338.71^4 - 255.37^4) =
(6.24 * 10^09 W) / (m^2 *K^4) * 510.065 * (13,161,702,663.0 - 4,252,844,523.22) = 28387.68 W
Very well. As I say I'm doing intermediate calculations for later.
So initial irradiance of the heat source at 338.7K is 82.12 W/m^2
Initial radiative output of heat source at 338.7K is 82.12 W/m^2 * 510.065 m^2 = 41886.54 W
Irradiance of the outer wall at 255.37K = (se)T^4 = (6.24 * 10^-9 W/m^2) / K^4 * 255.4^4 = 26.55 W/m^2
We agreed on gray bodies, so absorptivity = emissivity = 0.11.
The "view factor" from the spherical heat source to the chamber wall is 1. All radiated output intercepts the wall.
Incident radiation on chamber wall: 41887W / 512.469 m^2 is 81.73 W/m^2
81.73 W/m^2 incident radiation * 0.11 absorptivity = 8.99 W/m^2 absorbed.
So transfer from source to walls is = 8.99 W/m^2 absorbed * 512.469 m^2 = 4607.09 W
This makes perfect sense, since the areas are not that much different and the absorptivity is only 0.11.
Are we in agreement so far? I know I'm taking the long way around. I said I would.
If you want to be technical, what I calculated is variously called irradiance, radiant emittance, radiosity, or radiant exitance. And sometimes "emissive power", and probably by other names too.
It is not radiant flux, or radiant energy. Nor is the irradiance (I'll use that term from now on for clarity) of the heat source I calculated dependent on the chamber wall in any way. I repeat: the only variables for calculating this value for a gray body are temperature and emissivity. It is independent of any other object, and it is independent of absorbed incident radiation.
It's not a matter of the equation. I repeat: you jumped way ahead, and used a shortcut. That's not what I am doing, and that's the source of the problem here.
I've only calculated the radiant power value of one surface so far! I haven't even calculated the second yet. So how you could you possibly think I had calculated net heat transfer?
I'm not trying to calculate your "electricity" value. I don't want to calculate your electricity value. I'm starting from the basics, and working all the way through. I'm not using your shortcut. Is this clear?
Do you have a problem with my formula for calculating radiant power of a gray body surface at a given temperature? If not, I will continue. I repeat: it isn't dependent on any heat transfer, the only variables are emissivity and temperature. With that, and the Stefan-Boltzman constant, you can calculate the radiant power value.
If you want to be clear, then let's be clear:
I was not trying to calculate electricity. I was merely calculating radiative power out of the heat source at a known temperature.
The next step (because, as I have repeatedly stated, I am going slowly and carefully, in a step-wise manner) would have been to calculate the radiative power out of the chamber walls.
I was not calculating any heat transfer. On the contrary: heat transfer is dependent on those two values. I hadn't even calculated the second value yet.
I have been very straightforward and clear about what *I* am doing.
Furthermore, your value for electricity is completely irrelevant to the problem at hand, which was to calculate internal temperatures under the given conditions.
In these initial conditions, I'm stating that the radiant power of the heat source is 82.12 W/m^2.
I admit that I made an error somewhere, and plugged in the wrong number. 29.4 / (6.24 * 10 ^-9) is indeed 4711538461.55, and the 4th root of that is 261.99K, or 11.91 degrees F.
However: as you have said, let's be clear: you throw around terminology fast and loose, and you have refused to show your calculations, which leads to misunderstandings. So: what, then, do you claim that 29.4 W/m^2 figure represents and why?
I don't give a rat's ass about "electricity" at this point. The radiative power, (we are using units of W/m^2) of a gray body surface depends only on its temperature, its area, its emissivity, and the Stefan-Boltzmann constant. I do not need to know what the net heat transfer is to calculate this value (which IS 82.12 W/m^2 at the surface of the heat source under the given conditions). The total radiated power in Watts is: W/m^2 times the area. This is all pure textbook stuff. It matters not a whit at this point what that radiation strikes AFTER it is emitted. Even considering that right now is premature.
If I misunderstood, and your 29.4 W/m^2 represents something other than radiative power at the surface of the heat source at 338.7K, then please state clearly in plain terms what it IS supposed to represent, so we can move on.
The next step in the problem I am analyzing, because as I stated I am doing this in a careful stepwise manner precisely to avoid these misunderstandings is to calculate the radiative power of the chamber walls at 0 deg. F, or 255.37K.
But let me be clear: I don't give a damn about electricity at this point. In fact, I don't give the slightest damn whether the sphere is heated by steam, or an internal campfire, or burning unicorn farts. We know the power required for a gray body of the given area and emissivity to have a radiative temperature of 338.7K.
That is all I was saying. Nothing else. Trying to assume what I'm doing with that number before I do it is magical thinking.
I wasn't TRYING to calculate heat transfer. I was only calculating radiative power out. That was a necessary step to THEN going on to calculate the radiative power of the chamber walls, and THEN calculating the net heat transfer.
The TEMPERATURE of the heat source does not depend on heat transfer. It is an independent variable. On the contrary: the net heat transfer depends on the temperature, not the other way around.
We have already seen that you have mis-applied your equation and arrived at a power out figure that gives an incorrect value of temperature, which is an already known value.
Your "refutation" is disproved, almost before we've properly begun.
QED
I don't care if you don't understand that. Other people will.
You have not only made assumptions that don't actually apply, you have thrown around equations without carefully considering how they apply to the clearly stated problem. I said this in the beginning, and I have proven it now.
We already know, at the given steady-state, that the heat source is at 338.7K. We already know it is a sphere with a surface area of approximately 510 m^2. You have insisted we assume that it is a gray body. Given those two numbers, the emissivity, and the Stefan-Boltzmann constant, we calculate the power out: 82.12 W/m^2.
We do not need to do anything else at this point. We already know that this is the temperature GIVEN any heat transfer to the walls... which we have not calculated yet. I was going to go on to do that, but it isn't necessary now. We've seen that you're already wrong.
This was precisely why I insisted we do this slowly and carefully, and explain our steps. Because I knew you were doing it incorrectly (I said so) and that it would show up in the calculations. I did not expect it quite this soon, but there it is.
YOU may not understand that I have already proved your "refutation" wrong, but I assure you that other people will have no such difficulty. And they will have ample opportunity to see this, because I'm going to post it all online.
I calculated the radiant power out from a surface of ~510 m^2 at 150 degrees F, using the canonical textbook formula for doing so.
I did nothing more. I did not need to do anything more. The system is in a steady-state and the temperature is known.
Wait... I did do something more. I also showed that your own calculation was incorrect.
What is the difference between a person with computerized artificial legs, and a person with a memory chip in her skull?
Answer: none. Both are people.
They are NOT typically used to "prevent piracy" or "spread of malware" or "website access", I don't even see the use case here. I think the OP is confusing full packet capture with layer 7 application state firewalls, which ARE used for the above.
Um, wrong.
Deep-packet inspection was used routinely by the large ISPs to throttle certain kinds of traffic, until the FCC made them stop. This was just a couple of years ago.
Maybe not "full" packet inspection, but it was deep packet inspection, so they could distinguish, for example, packets of BitTorrent traffic from packets containing streamed video from YouTube.