After the essays in May on mirrors and light bulbs, I’ve been regularly poked and prodded via email for not wanting to engage “the slayers” anymore, or to do that “third experiment” I mentioned in May. I long ago concluded by my experiences afterwards with “the slayers” that it is a waste of time and effort to try to explain anything to them. Curt Wilson, who did the second experiment and was planning to do the third, has come to the same conclusion, as have many others.
I have to give them credit though, they are entertaining. When I saw this profoundly ridiculous rebuttal (reflectional denial) at their headquarters while arguing over Willis’ Steel Greenhouse post, I just had to share it.
LOL! That’s the “slayers” in nutshell right there. No better example of the absurdity of their position exists in my opinion. Epic.
WUWT regular, Duke physicist Dr. Robert G. Brown has been trying to talk some sense into them over at Principia Scientific. I keep telling him he’s being sucked into a time and energy sink like gravity around a neutron star. Just as it is a good policy to steer clear of neutron stars, so it is with these folks who are incapable of assimilating the real world of physics, but live in an alternate reality of absurd second law constructs.
So, that’s why I’m not bothering anymore, when you have reflection denial statements like the one above, why engage in a pointless dialog with the hopelessly lost who don’t want to learn anything? Thank goodness for my spam filter.
For those that might care, keeping the filament of a lightbulb within its optimum temperature range increases its life, by limiting hotspots and thus tungsten evaporation. Putting an incandescent bulb into a reflector housing not designed for it will in fact increase the filament temperature, increasing tungsten evaporation and deposition on the inside bulb glass surface.
See: http://www.lightingassociates.org/i/u/2127806/f/tech_sheets/FAQs_Reflector_Design__Why_is_it_important_.pdf
Tungsten evaporation from hotspots is why standard incandescent bulbs eventually fail.
Greg House 2:55pm: My clip is intended to be brief but capture context. In this case, a thermos flask “under usual conditions” does contain a vacuum thus the context was carefully preserved in my clip.
The reflective coating however is not useless. And remember radiation travels thru a vacuum. Under intense competition worldwide now, manufacturers would not bother with reflective coatings but in case they are wrongfully expending shareholder money, consider your story above where you write: “Reflected radiation like any radiation can warm…”
Here, if you follow thru with my recommendation and drop the term “warm” and reduce confusion toward science preciseness short of formulas try these for a thermos flask:
“Reflected radiation like any radiation can slow the cooling of the hot drink in the thermos…”
“A vacuum in a thermos flask like any vacuum can slow the cooling of the hot drink in the thermos…”
Both of which cause thermos buyers to part with some cash or credit.
You are right despite your confusing wording in that neither the reflected radiation nor the vacuum insulation can increase the energy in the thermos flask. And eventually the hot drink reaches equilibrium with the surroundings (aka room temp.).
Notice also, thermos temperature does not run away from reflections onto the energy source, refuting your 7/19 5:02pm claim above about GHE falseness. Nature protects a thermos from runaways with non-perfect reflectors and non-perfect vacuum insulation in much the same way as it does the atm. A difference is the location of the energy source (inside vs. outside), another difference is forced vs. unforced.
Gary Hladik says:
July 22, 2013 at 3:38 pm
“Just as the temp of a constantly heated house goes up when insulation is added (thus “slowing” conductive energy transfer to the environment) a radiation shield “slows” radiative transfer to the environment.”
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“A radiation shield “slows” radiative transfer to the environment” in some cases, but it can never change the temperature of the source, this is physically impossible, see my demonstration above.
The first point about house insulation is misleading and blatantly wrong, unfortunately warmists use it quite often. Insulation neither warms nor cools, it creates only a barrier and the effect can be both cooling and warming, depending on air temperatures inside and outside.
Gary Hladik 3:38pm: The thermos discussion in context is unforced (once filled) so I think it works (grad T of the hot drink always neg.), what do you think?
The light bulb, furnace, etc. are forced. You are right in that forced context. In the forced context, even earthshine falling on the sun increases the sun’s temperature a bit more than without the earthshine (like the bulb & furnace experience) but the system doesn’t run away in temp. reality refuting the 7/19 5:02pm post claim about GHE falseness.
Trick says:
July 22, 2013 at 4:17 pm
“Nature protects a thermos from runaways with non-perfect reflectors…”
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I have already suggested that you made a calculation similar to mine above assuming only 50% reflection. Was it so difficult? Then you would have come to the same absurd result: runaway warming and export of energy created out of nothing.
Thermos does not work by reflecting radiation, this effect is physically impossible.
Greg House says (July 22, 2013 at 4:32 pm): ‘A radiation shield “slows” radiative transfer to the environment” in some cases, but it can never change the temperature of the source, this is physically impossible, see my
demonstrationmiscalculated thought-less experiment above.’This is physically possible, see the real live examples that I referenced above.
They’re guaranteed not to melt your face off, too! 🙂
Trick says (July 22, 2013 at 4:38 pm): “The thermos discussion in context is unforced (once filled) so I think it works (grad T of the hot drink always neg.), what do you think?”
Yup.
Greg House 4:50pm:
Here I believe you jump back from unforced thermos system to forced (internal battery) system in your 7/19 5:02pm post refuted by tjfolkerts calc.s 7/19 5:57pm and I on 7/21 7:19pm.
You use an unnatural perfect reflector to come to the wrong natural conclusion about the GHE. A perfect reflector does not exist in nature so there are no runaways, nature always protects us.
Your reality forced system w/imperfect natural reflector will equilibrate energy in and out at a higher temperature than without the reflector; tjf does the correct calc.s. Your system, made real in an experiment, will not runaway.
Nature shows us a thermos does function in part by reflecting radiation non-perfectly, this thermos effect is physically possible with no runaway.
What I would do if I were you is take my suggestion seriously and rewrite your 7/19 5:02pm post without using these terms: warm, warms, warmer, warming, warmed. Just use energy and temperature terms. You will reach a different conclusion once you eliminate your confusion caused by these inadequate, imprecise “warm” terms. By doing it yourself, you really will learn your conclusion about the false GHE can be easily refuted.
Trick says:
July 22, 2013 at 6:12 pm
“You use an unnatural perfect reflector to come to the wrong natural conclusion about the GHE. A perfect reflector does not exist in nature so there are no runaways, nature always protects us.”
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I have told you twice, if you assume only 50% reflection, you will get the same absurd outcome. Get it? Just not in 2, but in 4 seconds. 50% reflector is not perfect, is it? Satisfied now?
Greg House says:
July 21, 2013 at 8:01 pm
Tsk Tsk says:
July 21, 2013 at 5:00 pm
“In order for your “example” to work you have to have effectively instantaneous heating and by extension instantaneous cooling.”
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Last time I checked the notion of “greenhouse effect” suggested warming, not cooling.
I understand you desire, you know, to somehow insert cooling into the absurd “greenhouse effect” to avoid it’s absurd consequence, but you can not have it. Blame the IPCC&Co for that, not me.
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At this point I should probably just say, “I know you are, but what am I?” since that has been the extent of your replies, but again I’m bored so I’ll try once more.
The “cooling” that you continue to dwell upon is your heating. More specifically I am simply stating that the heat transfer is symmetrical, i.e. the mechanisms that heat the plate(just radiation in this case) are the same mechanisms that cool the plate. It’s really only through your integration error that you arrive at your conclusion, so let’s try once more to enable your understanding by building your problem up step by step:
1) Begin with the blackbody plate you specified and infinitesimally thin so only the two large sides need be considered.
2) Place the body in empty space surrounded by no background radiation, i.e. 0K
3) Apply 800W of power production so that it comes up to temperature in equilibrium with its surroundings. Per your example before this is instantaneous (or at least is complete within your time step) but it works for real systems as well.
4) Remove the power source. What happens? Since you want the plate to absorb all incoming radiation immediately that also means that it emits that radiation just as easily. And in your tortured problem you’ve stipulated the boundary condition that the plate MUST radiate away all of that additional heat in the next time step. Let’s come back to the real (but still idealized) plate again. The reality is that the plate is emitting 800W. Next we remove the power supply. In the immediate instant (Planck time?) after we do that the plate emits 800W, but that lowers the internal energy and hence the temperature of the plate to:
Tnew=Told – E/m/cp
Where Tnew and Told are the current and preceding step temps; m is the plate mass; cp is the specific heat of the plate and E is the energy (heat/radiation) emitted by the plate which is just power * time or:
E=area*sigma*T^4*t
Where sigma is the S-B constant; T is the present temperature; and t is the time step. Really the E, P, and t should all be dE, dP, and dt but let’s indulge ourselves.
Now let’s return to your example where you say that the plate which absorbs the radiation has to emit that same radiation on the next step. If it doesn’t you don’t get your amplification. This is the “cooling” that you keep tripping on. But if it emits that energy then its temperature reverts to what it was before it received that energy because if it did not then you would be making energy. In order for that to not be the case you have to argue that an object cannot cool by radiation. Good luck with that.
So let’s do your calculation properly accounting for the fact that an object which emits loses the energy that it emits.
Step Energy Produced Total Emission Reflected Energy Energy Lost
1 800 800 0 400
2 800 1200 400 600
3 800 1400 600 700
4 800 1500 700 750
5 800 1550 750 775
6 800 1575 775 787.5
7 800 1587.5 787.5 793.75
8 800 1593.75 793.75 796.875
9 800 1596.875 796.875 798.4375
10 800 1598.4375 798.4375 799.21875
11 800 1599.21875 799.21875 799.609375
Sum 8800 16000.78125 7200.78125 8000.390625
So by the 11th time step we have produced a grand total of 8800J of energy. We have emitted (lost to space) 8000J and we have retained 800J which is exactly the amount that we produce each cycle. However, because we have saved the output of previous cycles in the reflection loop we have a total of 800J resident in the system at any given time step. Without the reflector we would have no energy resident in the system. Again, the reality is that the system is continuous and these time steps are very crude but we’ve at least made everything self-consistent.
But here’s the more interesting part of the problem: the temperature. The emission to empty space is still 800J(W) but the area radiating to space is half the original area since the other half is interacting with the reflector. So P=area*sigma(T^4):
Pold=800=Pnew => Pold=area_old*sigma(Told^4)=area_new*sigma(Tnew^4)=Pnew
Tnew/Told=(area_old/area_new)^1/4
and area_old=2*area_new, so:
Tnew/Told=(2)^1/4 which just so happens is a (global) warming.
Since you won’t like the derivation that way, then do it from a total plate emission perspective:
Pnew=2*Pold (from the tabulation) => Tnew^4=2*Told^4 => Tnew/Told=(2)^1/4.
So again, your entire supposed falsification is based upon the fact that you didn’t add properly. You gloss over that by claiming that hot objects emit therefore they keep emitting the same energy. But that itself is a horrible error.
Mods, if the text table doesn’t come through don’t worry about it.
Gary Hladik says:
July 22, 2013 at 5:31 pm
“…see the real live examples that I referenced above.”
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You referenced claims. They contradict basic physics and simple math, see my demonstration above. We know all that, thermos, blanket, radiation shield. Whatever some people might imagine, back radiation can not warm the source. Physically impossible.
You other line “why would manufacturers use reflective layer inside a thermos flask” has nothing to do with physics and math. Manufacturers are known to be, let’s say, not particularly smart in some cases. Not so long ago some of them tried to buy a patent on a perpetual motion machine, a real life example. Should be a proof of existence of a perpetual motion machine, right?
Trick says:
July 21, 2013 at 7:19 pm
…
In your example, nature protects against your proposed runaway with reflectors that are not perfect. Made real, your 7/19 5:02pm system will not runaway; it will equilibrate with surroundings at a higher temperature than before your reflector was placed. The real equilibrium temperature will not be infinite (as tjf tries to show you) even with a darn good internal battery. Nature protects.
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No. The system is perfectly functional and capable even with perfect reflectors and blackbodies. Nature doesn’t have to protect against anything other than bad addition and the circular logic that comes from the insertion of a violation of the 1st law of thermo into a “proof” that a very real effect violates the 1st law of thermo…
Tsk Tsk says:
July 22, 2013 at 7:29 pm
“… bad addition and the circular logic that comes from the insertion of a violation of the 1st law of thermo into a “proof” that a very real effect violates the 1st law of thermo…”
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The only thing I “inserted” was the assumption of the “greenhouse effect” as presented by the IPCC, “warming by back radiation”. Blame the IPCC.
GH 7:06pm:
“I have told you twice, if you assume only 50% reflection, you will get the same absurd outcome. Get it?”
No, b/c you have confused yourself with the “warm” terms, there is no absurd outcome in nature with 50% reflection. You got it wrong once. Now twice. The system temperature with reflector will equilibrate with surroundings at a higher T, it will not runaway to an absurd outcome. If it would, power up the USA. Want to go 3for3?
If not, rewrite your 7/19 5:02pm post using temperature and energy terms only. You will quickly find where you slip up.
Tsk Tsk 7:29pm: “The system is perfectly functional and capable even with perfect reflectors and blackbodies.”
Not a real system. I will have to disagree in real nature on entropy grounds, no entropy increase is only theoretical and used for learning. With real nature allowing a perfect reflector or perfect insulation, a forced system will have no equilibrium, there will be no energy out, it will runaway, Greg’s face WILL melt. I would turn my furnace on once a winter and when it cycles off, it is off for good. Might have to turn on the A/C when the winter sun comes thru the south windows.
In an unforced system, the thermos will just sit there keeping your coffee hot ‘til the next millennium. Not so bad.
But you won’t be able to go around a curve in your car without friction, and brakes? None. Engine RPM, no limit, unbalance ends its life quickly. Friction can be your friend, nature protects the innocent.
Not sure what you mean by a very real effect violates 1st law.
Gary & Trick (and others) discuss the idea that “Reflected radiation like any radiation can slow energy transfer but cannot raise the temperature above that of the original source…”
People often seem to miss that there is a SECOND, INDEPENDENT energy source in the cases of interest. The furnace has heating elements (even hotter than the chamber of the furnace). The earth has the sun (even hotter than the surface of the earth).
It is quite true that neither back radiation nor the insulation of your house nor the vacuum flask can BY ITSELF raise the temperature of the furnace chamber or the contents of the vacuum flack or the surface of the earth. For an unheated object, these forms of insulation (conductive insulation and/or radiative insulation) merely slow the cooling toward the temperature of the surroundings. Without the sun, the earth would slowly cool toward 3 K — the GHGs can do nothing more in this case than slightly prolong the time it takes to cool.
But for the earth’s surface (or the analogous chamber of the furnace) has an active heater included (ie the sun or the heating elements of the furnace). The sun is actively trying to warm the earth’s surface; deeps space is continuously trying to cool it. The steady-state temperature (or more precisely, the quasi-steady-state temperature since there are variations during the day and the year) is the result of the competition between warming due to the sun and cooling due to space. The GHGs limit the effectiveness of the cooling part of the interactions. With the same energy input (eg the sun or the heating elements), but less energy output (eg radiation shields or GHGs ), the only possible result is a warmer surface or furnace chamber.
Energy is conserved. Entropy increases. The net flow of energy is always from warmer to cooler. All is good in the universe.
.
Greg House says:
July 22, 2013 at 8:07 pm
Tsk Tsk says:
July 22, 2013 at 7:29 pm
“… bad addition and the circular logic that comes from the insertion of a violation of the 1st law of thermo into a “proof” that a very real effect violates the 1st law of thermo…”
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The only thing I “inserted” was the assumption of the “greenhouse effect” as presented by the IPCC, “warming by back radiation”. Blame the IPCC.
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No, you inserted a violation of conservation of energy by having the plate remain hot after it emits the received reflected energy. You have to explain how the same energy can be in two places at once. Good luck.
Tsk Tsk 7:15pm: Nice work, glad you are bored.
As I tried to show GH 7/21 at 7:19pm, I agree with you here, in that GH doesn’t yet get that the atm. being T>0 has to radiate at the surface as you write:
“In order for that to not be the case you have to argue that an object cannot cool by radiation. Good luck with that. “
Trick says:
July 22, 2013 at 8:19 pm
Tsk Tsk 7:29pm: “The system is perfectly functional and capable even with perfect reflectors and blackbodies.”
Not a real system. I will have to disagree in real nature on entropy grounds, no entropy increase is only theoretical and used for learning. With real nature allowing a perfect reflector or perfect insulation, a forced system will have no equilibrium, there will be no energy out, it will runaway, Greg’s face WILL melt. I would turn my furnace on once a winter and when it cycles off, it is off for good. Might have to turn on the A/C when the winter sun comes thru the south windows.
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No it will not. There is energy out. He placed a reflector only on one side. The other side is still free to emit. Once the transient is overcome the system will again emit 800W to space. In the interval from the introduction of the reflector to the steady state the system will actually emit less energy to space than 800W.
Tsk Tsk 8:19pm; “He placed a reflector only on one side”
GH placed a perfect reflector so it doesn’t emit from the other side at all. The perfect reflector T=0K which is just what GH misses in nature of GHE. The other side is NOT free to emit in GH construct. Of course, in nature it will have a T>0 and emit as you show.
Using a plate with infinitesimal mass in Greg’s example is problematic. It is better to use some finite mass — perhaps an object with a heat capacity of 100 J/K. Then either take small time steps (1 second should work OK here) or actually use calculus to get an exact answer. Assuming that things change 100’s of degrees in each time steps will give poor results.
In this case, the plate with a 800 W heater would radiate 400 W/m2 from the two sides = 289.8 K (for a blackbody) with no reflector in place (and ~ 0 K surroundings to avoid any other “back-radiation”)
Add the reflector .. and suddenly 400 W/m^2 of reflected light comes back from one side.
1ST SECOND: the object at 298 K will emit 400 + 400 J and absorb 800 +400. The net gain is 400, which warms it up by (400 J/K / 100 J) = 4 K to 294.
2ND SECOND: the object at 293.8 K will emit 422 + 422 J and absorb 800 +422. The net gain is (800+422) – (422+422) = 378J , which warms it up by 3.78 K.
3RD SECOND: the object at 297.6 K will emit 445 + 445 J and absorb 800 +445 . The net gain is (800+445 ) – (445 +445 ) = 355J , which warms it up by 3.55 K.
You can write your own spreadsheet to finish it up. after 10 seconds, it has warmed to 320 K, with a net loss of 200 J/s. After 30 seconds, it is 341 K losing a net (800+ 739) – (739-739) = 61 J/s
After 75 seconds, it is 344.6 K, emitting 800 + 800 from the two sides. It is also absorbing 800 (form the heater) + 800 (from reflection). Balance has been achieved.
Trick says:
July 22, 2013 at 8:50 pm
Tsk Tsk 8:19pm; “He placed a reflector only on one side”
GH placed a perfect reflector so it doesn’t emit from the other side at all. The perfect reflector T=0K which is just what GH misses in nature of GHE. The other side is NOT free to emit in GH construct. Of course, in nature it will have a T>0 and emit as you show.
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Go re-read the problem statement. The is only reflection on one side of the power plate. The other side is free to radiate into empty 0K space.
Tsk Tsk says:
July 22, 2013 at 7:15 pm
“The “cooling” that you continue to dwell upon is your heating. More specifically I am simply stating that the heat transfer is symmetrical, i.e. the mechanisms that heat the plate(just radiation in this case) are the same mechanisms that cool the plate.”
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OK, let me try it again. The blackbody in my example is initially at a stable temperature, no cooling, no warming. Then comes the reflector and I assume that there is “greenhouse effect”, that is “warming by back radiation”. So, it is not a reality, it is an assumption. My intention is to find out, if thisassumption would lead to an absurd outcome. A typical “reductio ad absurdum” approach.
Now, there can not be any cooling, only warming, because it is what the “greenhouse effect” is supposed to do. Crucially important here is to take into consideration that our body a)gets continuously warmer because of back radiation and b)therefore radiates according to it’s increasing temperature more and more and gets accordingly more and more back radiation and so on. Again, this is not real, this goes only under the assumption of the “greenhouse effect”.
As a result, the system blackbody-reflector exports more energy than it has at its disposal. This is absurd and proves the assumption (“greenhouse effect”) false (see my calculation above).
You calculation is wrong, because you ignore that the blackbody’s temperature in my example increases constantly (“greenhose warming”). Again, this is not real, this is the assumption. I assume the “greenhouse effect” and stick to it doing the calculation, so simple is that.
Your “coolings” in the process of constant warming are not a part of the “greenhouse effect”, nor are they a part of reality. You invented an absurd thing, congratulations, it is even “better” than the “greenhouse effect”.
Using a plate with infinitesimal mass in Greg’s example is problematic. It is better to use some finite mass — perhaps an object with a heat capacity of 100 J/K.
Which is why you have to contort things and make the plate give up all of the heat it received on the previous cycle.
Then either take small time steps (1 second should work OK here) or actually use calculus to get an exact answer. Assuming that things change 100′s of degrees in each time steps will give poor results.
…
1ST SECOND: the object at 298 K will emit 400 + 400 J and absorb 800 +400. The net gain is 400, which warms it up by (400 J/K / 100 J) = 4 K to 294.
2ND SECOND: the object at 293.8 K will emit 422 + 422 J and absorb 800 +422. The net gain is (800+422) – (422+422) = 378J , which warms it up by 3.78 K.
I tried writing the progression down with a cp of 1 and you do indeed get ridiculous temp swings with a time step of 1sec. The second step is where he makes his mistake. He wants that step to be 400+422 for total reflection and then he just keeps compounding the error.
Greg House says (July 22, 2013 at 7:23 pm): “You referenced claims.”
No, no, I referenced a real live peer-reviewed paper and actual products that people have made, at least two of which (thermos and dichroic bulb) you can actually buy. (I suppose you could also buy a compact vacuum furnace, for the right price). Wait! Three real products (cue the Spanish Inquisition): the infrared microbolometer! The unsubstantiated “claims” around here are yours.
Or do you have a peer-reviewed paper (paywalled or otherwise) saying the so-called “greenhouse effect” will melt your face? (The Journal of Pink Unicorns doesn’t count) 🙂
Tsk Tsk says:
July 22, 2013 at 8:28 pm
“No, you inserted a violation of conservation of energy by having the plate remain hot after it emits the received reflected energy. You have to explain how the same energy can be in two places at once.”
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I understand that the step by step illustration might be difficult for some people. Following you logic, even before we introduce the reflector, when our blackbody emits 800 per second, it’s temperature drops to the absolute zero every second because, you know, it has emitted those 800 and has nothing more, then comes the next 800, the body emits and its it and it’s temperature drops to the absolute zero again and so on. Wait, we can take a millisecond instead of a second and it will be the same. So, according to your approach, a body at a stable temperature, let us say 333K is not actually constantly at 333K, but it’s temperature is jumping back and forth like that: 333-0-333-0-333-0-333-0-333-0-333-0- and so on.
Congratulations, this absurdity beats even the “greenhouse effect”.
So, again, the absurd result I got from the assumption of the “greenhouse effect” is to blame on the notion of the “greenhouse effect”. The nature does not work this way, of course.