Light Bulb Back Radiation Experiment
Guest essay by Curt Wilson
In the climate blogosphere, there have been several posts recently on the basic principles of radiative physics and how they relate to heat transfer. (see yesterday’s experiment by Anthony here) These have spawned incredibly lengthy streams of arguments in the comments between those who subscribe to the mainstream, or textbook view of radiative heat transfer, and those, notably the “Skydragon Slayers” who reject this view.
A typical statement from a Slayer is that if “you have initially a body kept at a certain temperature by its internal source of energy”, that if another body at a lower temperature is placed near to it, that the radiation from this colder body could not increase the temperature of the warmer body, this being a violation of the 2nd Law of Thermodynamics. They continue that if this were possible, both objects would continually increase the other’s temperature indefinitely, which would be an obvious violation of the 1st Law of Thermodynamics (energy conservation).
This is part of a more general claim by Slayers that radiation from a colder body cannot transfer any energy to a warm body and lead to a higher temperature of the warm body than would be the case without the presence of the colder body.
It occurred to me that these claims were amenable to simple laboratory experiments that I had the resources to perform. A light bulb is a classic example of a body with an internal source of energy. Several Slayers specifically used the example of reflection back to a light bulb as such an example.
In our laboratory, we often have to do thermal testing of our electronic products so we can ensure their reliability. Particularly when it comes to power electronics, we must consider the conductive, convective, and radiative heat transfer mechanisms by which heat can be removed from these bodies with an “internal source of energy”. We have invested in good thermocouple measurement devices, regularly calibrated by a professional service, to make the temperature measurements we need.
We often use banks of light bulbs as resistive loads in the testing of our power electronics, because it is a simple and inexpensive means to load the system and dissipate the power, and it is immediately obvious in at least a qualitative sense from looking at the bulbs whether they are dissipating power. So our lab bench already had these ready.
If you want to isolate the radiative effects, the ideal setup would be to perform experiments in a vacuum to eliminate the conductive/convective losses. However, the next best thing is to reduce and control these to keep them as much alike as possible in the different phases of the experiment.
So, on to the experiment. This first picture shows a standard 40-watt incandescent light bulb without power applied. The lead of the thermocouple measuring device is taped to the glass surface of the bulb with heat-resistant tape made for this purpose. The meter registers 23.2C. In addition, a professional-grade infrared thermometer is aimed at the bulb, showing a temperature of 72F. (I could not get it to change the units of the display to Celsius.) Note that throughout the experiment, the thermocouple measurements are the key ones.
Next, the standard North American voltage of 120 volts AC (measured as 120.2V) was applied to the bulb, which was standing in free air on a table top. The system was allowed to come to a new thermal equilibrium. At this new equilibrium, the thermocouple registered 93.5C. (The IR thermometer showed a somewhat lower 177F, but remember that its reported temperature makes assumptions about the emissivity of the object.)
Next, a clear cubic glass container about 150mm (6”) on a side, initially at the room temperature of 23 C, was placed over the bulb, and once again the system was allowed to reach a new thermal equilibrium. In this state, the thermocouple on the temperature of the bulb registers 105.5C, and the outer surface of the glass container registers 37.0C (equivalent to body temperature).
The glass container permits the large majority of the radiative energy to escape, both in the visible portion of the spectrum (obviously) and in the near infrared, as standard glass is highly transparent to wavelengths as long as 2500 nanometers (2.5 microns). However, it does inhibit the direct free convection losses, as air heated by the bulb can only rise as far as the top of the glass container. From there, it must conductively transfer to the glass, where it is conducted through the thickness of the glass, and the outside surface of the glass can transfer heat to the outside ambient atmosphere, where it can be convected away.
The next step in the experiment was to wrap an aluminum foil shell around the glass container. This shell would not permit any of the radiative energy from the bulb to pass through, and would reflect the large majority of that energy back to the inside. Once again the system was allowed to reach thermal equilibrium. In this new state, the thermocouple on the surface of the bulb registered 137.7C, and the thermocouple on the outer surface of the glass registered 69.6C. The infrared thermometer is not of much use here due to the very low emissivity (aka high reflectivity) of the foil. Interestingly, it did show higher temperatures when focused on the tape on the outside of the foil than on the foil itself.
Since adding the foil shell outside the glass container could be reducing the conductive/convective losses as well as the radiative losses, the shell was removed and the system with the glass container only was allowed to re-equilibrate at the conditions of the previous step. Then the glass container was quickly removed and the foil shell put in its place. After waiting for thermal equilibrium, the thermocouple on the surface of the bulb registered 148.2C and the thermocouple on the outside of the foil registered 46.5C. The transient response (not shown) was very interesting: the temperature increase of the bulb was much faster in this case than in the case of adding the foil shell to the outside of the glass container. Note also how low the infrared thermometer reads (84F = 29C) on the low-emissivity foil.
Further variations were then tried. A foil shell was placed inside the same glass container and the system allowed to reach equilibrium. The thermocouple on the surface of the bulb registered 177.3C, the thermocouple on the outer surface of the foil registered 67.6C, and the infrared thermometer reading the outside of the glass (which has high emissivity to the wavelengths of ambient thermal radiation) reads 105F (40.6C).
Then the glass container was removed from over the foil shell and the system permitted to reach equilibrium again. The thermocouple on the surface of the bulb registered 176.3C and the thermocouple on the outside of the foil registered 50.3C.
All of the above examples used the reflected shortwave radiation from the aluminum foil. What about absorbed and re-emitted longwave radiation? To test this, a shell of black-anodized aluminum plate, 1.5mm thick, was made, of the same size as the smaller foil shell. A black-anodized surface has almost unity absorption and emissivity, both in the shortwave (visible and near infrared) and longwave (far infrared). Placing this over the bulb (without the glass container), at equilibrium, the thermocouple on the bulb registered 129.1C and the thermocouple on the outside of the black shell registered 47.0C. The infrared thermometer read 122F (50C) on the tape on the outside of the shell.
The power source for this experiment was the electrical input. The wall voltage from the electrical grid was steady at 120.2 volts. The electrical current was measured under several conditions with a professional-grade clip-on current sensor. With the bulb in open air and a surface temperature of 96.0C, the bulb used 289.4 milli-amperes of current.
With the bulb covered by a foil shell alone and a surface temperature of 158.6C, the bulb drew slightly less, 288.7 milliamperes.
Summary of Results
The following table shows the temperatures at equilibrium for each of the test conditions:
| Condition | Bulb Surface Temperature | Shell Temperature |
| Bulb open to room ambient | 95C | — |
| Bulb covered by glass container alone | 105C | 37C |
| Bulb covered by glass container and outer reflective foil shell | 138C | 70C (glass) |
| Bulb covered by outer reflective foil shell alone | 148C | 46C (foil) |
| Bulb covered by inner reflective foil shell inside glass container | 177C | 68C (foil) |
| Bulb covered by inner reflective foil shell alone | 176C | 50C |
| Bulb covered by black-anodized aluminum shell alone | 129C | 47C |
Analysis
Having multiple configurations permits us to make interesting and informative comparisons. In all cases, there is about a 35-watt (120V x 0.289A) electrical input to the system, and thermal equilibrium is reached when the system is dissipating 35 watts to the room as well.
I used a low-wattage (40W nominal) bulb because I had high confidence that it could take significant temperature increases without failure, as it has the same package design as much higher-wattage bulbs. Also, I would not be working with contraband high-wattage devices 😉
The case with the glass container alone is the important reference case. The glass lets virtually all of the radiant energy through, while inhibiting direct convection to the room ambient temperature of 23C. Conductive/convective losses must pass from the surface of the bulb, through the air under the container, to and through the glass, and then to the room atmosphere, where it is conducted/convected away. Under these conditions, the bulb surface temperature is 105C, which is 10C greater than when the bulb can conductively dissipate heat directly to the room atmosphere.
Compare this case to the case of the larger foil shell alone. The foil shell also inhibits direct conductive/convective losses to the room atmosphere, but it will not inhibit them to any greater extent. In fact, there are three reasons why it will inhibit these losses less than the glass container will. First, the material thermal conductivity of aluminum metal is far higher than that of glass, over 200 times greater (>200 W/(m*K) versus <1.0 W/(m*K)). Second, the foil, which is a small fraction of a millimeter thick, is far thinner than the glass container, which is about 4 mm thick on average. And third, the surface area of the foil is somewhat larger than the glass container, so it has more ability to conductively transfer heat to the outside air.
And yet, the surface of the bulb equilibrated at 146C under these conditions, over 40C hotter than with the glass container. With conductive/convective losses no less than with the glass container, and very probably greater, the only explanation for the higher temperature can be a difference in the radiative transfer. The glass container lets the large majority of the radiation from the bulb through, and the foil lets virtually none of it through, reflecting it back toward the bulb. The presence of the foil, which started at the room ambient of 23C and equilibrated at 46C, increased the temperature of the bulb, which started at 105C on the outside (and obviously warmer inside). The reflected radiation increased the temperature of the bulb, but did not produce “endless warming”, instead simply until the other losses that increase with temperature matched the input power of 35 watts.
Interestingly, the foil shell without the glass container inside led to a higher bulb temperature (148C) than the foil shell with the glass container inside (138C). Two layers of material around the bulb must reduce conductive/convective losses more than only one of them would, so the higher temperature must result from significantly more reflected radiation back to the bulb. With the glass inside, the reflected radiation must pass through two surfaces of the glass on the way back to the bulb, neither of which passes 100% through.
Another interesting comparison is the large foil shell that could fit outside of the glass container, about 160mm on a side, with the small foil shell that could fit inside the glass container, about 140mm on a side. With the large shell alone, the bulb temperature steadied at 148C; with the smaller shell, it steadied at 176C. With all direct radiative losses suppressed in both cases, the difference must come from the reduced surface area of the smaller shell, which lessens its conductive/convective transfer to the outside air at a given temperature difference. This is why halogen incandescent light bulbs, which are designed to run hotter than standard incandescent bulbs, are so much smaller for the same power level – they need to reduce conductive/convective losses to get the higher temperatures.
All of the above-discussed setups used directly reflected radiation from the aluminum foil. What happens when there is a barrier that absorbs this “shortwave” radiation and re-emits it as “longwave” radiation in the far infrared? Can this lead to higher temperatures of the warmer body? I could test this using black-anodized aluminum plate. Black anodizing a metal surface makes it very close to the perfect “blackbody” in the visible, near-infrared, and far-infrared ranges, with absorptivity/emissivity (which are the same at any given wavelength) around 97-98% in all of these ranges.
With a black plate shell of the same size as the smaller foil shell, the bulb surface temperature equilibrated at 129C, 24C hotter than with the glass container alone. Once again, the thin metal shell would inhibit conductive/convective losses no better, and likely worse than the glass container (because of higher material conductivity and lower thickness), so the difference must be from the radiative exchange. The presence of the shell, which started at the room ambient of 23C and increased to 47C, caused the bulb surface temperature to increase from 105C to 129C.
Another interesting comparison is that of the smaller foil shell, which led to a bulb surface temperature of 176C and a shell temperature of 50C, to the black plate shell of the same size, which led to a bulb surface temperature of 129C and a shell temperature of 46C. While both of these create significantly higher bulb temperatures than the glass container, the reflective foil leads to a bulb surface temperature almost 50C higher than the black plate does. Why is this?
Consider the outside surface of the shell. The foil, which is an almost perfect reflector, has virtually zero radiative absorptivity, and therefore virtually zero radiative emissivity. So it can only transfer heat to the external room by conduction to the air, and subsequent convection away. The black plate, on the other hand, is virtually the perfect absorber and therefore radiator, so it can dissipate a lot of power to the room radiatively as well as conductively/convectively. Remember that, since it is radiating as a function of its own temperature, it will be radiating essentially equally from both sides, there being almost no temperature difference across the thickness of the plate. (Many faulty analyses miss this.) The foil simply reflects the bulb’s radiation back to the inside and radiates almost nothing to the outside. This is why the infrared thermometer does not read the temperature of the foil well.
The electrical voltage and current measurements were made to confirm that the increased temperature did not come from a higher electrical power input. The current measurements shown above demonstrate that the current draw of the bulb was no higher when the bulb temperature was higher, and was in fact slightly lower. This is to be expected, since the resistivity of the tungsten in the filament, as with any metal, increases with temperature. If you measure the resistance of an incandescent bulb at room temperature, this resistance is less than 10% of the resistance at its operating temperature. In this case, the “cold” resistance of the bulb is about 30 ohms, and the operating resistance is about 415 ohms.
Let’s look at the dynamic case, starting with the thermal equilibrium under the glass container alone. 35 watts are coming into the bulb from the electrical system, and 35 watts are leaving the bulb through conductive losses to the air and radiative losses to the room through the glass. Now we replace the glass with one of the metal shells. Conductive losses are not decreased (and may well be increased). But now the bulb is receiving radiant power from the metal shell, whether reflected in one case, or absorbed and re-radiated back at longer wavelengths in the other. Now the power into the bulb exceeds the power out, so the temperature starts to increase. (If you want to think in terms of net radiative exchange between the bulb and the shell, this net radiative output from the bulb decreases, and you get the same power imbalance.)
As the temperature of the bulb increases, both the conductive losses to the air at the surface of the bulb increase (approximately proportional to the temperature increase) and the radiative losses increase as well (approximately proportional to the 4th power of the temperature increase). Eventually, these losses increase to where the losses once again match the input power, and a new, higher-temperature thermal equilibrium is reached.
I originally did these tests employing a cylindrical glass container 150mm in diameter and 150mm high with and without foil shells, and got comparable results. In the second round shown here, I changed to a cubic container, so I could also create a black-plate shell of the same shape.
It is certainly possible that improvements to these experiments could result in differences of 1 or 2C in the results, but I don’t see any way that they could wipe out the gross effect of the warming from the “back radiation”, which are several tens of degrees C.
All of these results are completely in line with the principles taught in undergraduate engineering thermodynamics and heat transfer courses. The idea that you could inhibit net thermal losses from an object with an internal power source, whether by conductive, convective, or radiative means, without increasing the temperature of that object, would be considered ludicrous in any of these courses. As the engineers and physicists in my group came by the lab bench to see what I was up to, not a single one thought for a moment that this back radiation would not increase the temperature of the bulb.
Generations of engineers have been taught in these principles of thermal analysis, and have gone on to design crucial devices and infrastructure using these principles. If you think all of this is fundamentally wrong, you should not be spending your time arguing on blogs; you should be out doing whatever it takes to shut down all of the erroneously designed, and therefore dangerous, industrial systems that use high temperatures.
Conclusions
This experiment permitted the examination of various radiative transfer setups while controlling for conductive/convective losses from the bulb. While conductive/convective losses were not eliminated, they were at least as great, and probably greater, in the cases where a metal shell replaced the glass shell over the bulb.
Yet the bulb surface temperature was significantly higher with each of the metal shells than with the glass shell. The only explanation can therefore be the radiative transfer from the shells back to the bulb. In both cases, the shells were significantly cooler than the bulb throughout the entire experiment, both in the transient and equilibrium conditions.
We therefore have solid experimental evidence that radiation from a cooler object (the shell) can increase the temperature of a warmer object (the bulb) with other possible effects well controlled for. This is true both for reflected radiation of the same wavelengths the warmer body emitted, and for absorbed and re-radiated emissions of longer wavelengths. The temperature effects are so large that they cannot be explained by minor setup effects.
Electrical measurements were made to confirm that there was not increased electrical power into the bulb when it was at higher temperatures. In fact, the electrical power input was slightly reduced at higher temperatures.
This experiment is therefore compatible with the standard radiative physics paradigm that warmer and cooler bodies can exchange radiative power (but the warmer body will always transfer more power to the cooler body). It is not compatible with the idea that cooler bodies cannot transfer any power by radiative means to warmer bodies and cause an increase in temperature of the warmer body.
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UPDATE: The Principia/Slayers group has post a hilarious rebuttal here:
http://principia-scientific.org/supportnews/latest-news/210-why-did-anthony-watts-pull-a-bait-and-switch.html
Per my suggestion, they have also enabled comments. You can go discuss it all there. – Anthony
Curt, If you’re willing, you might go back and redo your measurements (if you don’t have them already) with the cylinder, if you allow the hot air to escape, adding and removing the foil should change the filiment resistance, showing the effect of radiance only.
edim, out of interest on the ice cubes what does happen. way beyond my thinking.
I come here to learn something, so ask in my own way, sorry if that offends.
Bryan says:
Bryan: I find it boring and useless to argue about terminology. The fact is that increasing the level of greenhouse gases will increase the Earth’s surface temperature. Most reasonable people would say that “warming” is not a bad way to describe the fact that the temperature is increased. Yes, the heat flow is from the warmer Earth to the colder atmosphere but the fact is that the Earth is at a higher temperature as a result of changes in the composition of the atmosphere.
“And why do you think foil is an insulator? Would it be because it’s used in vacuum flasks, and insulation sheets, and space blankets, and all the other applications where it is there to reflect IR? It would have no effect on convection, and minimal effect on conduction, so what else is there?”
My God! Where do I start?
1. Vacuum flasks- reduces heat loss, but does not make the contents warmer.
2. Insulation sheets-useful because it creates a barrier to convection.
3. Space blankets- for extreme heat from the sun because it is reflective, for warmth it is irrelevant as long as it impedes loss of heat due to convection. In which case a plastic bag would be ok.
I wonder what Myrrh makes of this technology:
http://www.wi-charge.com/technology.php?ID=25
Alex, your comprehension failure is a bit sad.
1. Vacuum flask. Poor in hot tea. Wait 4 hours. Measure the temp. Is this “warmer” than the tea left in the uninsulated teapot? If “YES”, then that’s the “warmer” we are talking about – not hotter than before, just warmer than without the insulation!”
2. Why have a foil surface on a foam insulation sheet? The foam prevents 100% of convection anyway? The foil reflects IR, making it ever better!
3. Look up space blankets – you clearly don’t know what they are.
Jeez.
I like the mixture of thermos-flask and emissivity analogies, Hypothetically; If we filled the flask with boiling hot water, and at one end of the flask we made a transparent hole, we could convert the thermal radiation into a portable inferred laser. And if we continued to add “energy” we could make the laser more powerful. /jk
Bryan:
An object “warms” if the temperature of the object increases. If something I did “caused” the object to increase in temperature, by that action I “warmed” the object. The mechanism by which I “warmed” the object here is by impeding the flow of heat away from the object, rather than by adding heat energy to the object.
‘Alex, your comprehension failure is a bit sad.’
You are right
1. The tea gets warmer
2. I will go back to the insulation suppliers (who I bought the materials for my house from) and tell them that they misinformed me.
3. I have used space blankets and plastic bags , but I guess I’m an idiot and don’t know warm from cold.
Thanks for letting me know about my lack of comprehension, I appreciate that.
I will now give up this failure of an experiment and try to reproduce Pictet’s experiment. ‘Cold waves ‘ have always appealed to my logic. My following project will be the investigation of phlogiston and the philosopher’s stone.
richard says (May 29, 2013 at 6:47 am): “two ice cubes, one a lot colder than the other, if you place them next to each other what happens.”
Depends on the experimental conditions, doesn’t it? Put the ice cubes in a working freezer, and they both cool to freezer temp. Put them in a vented sea level room at 20 degrees C and they both warm (from the outside in) to zero degrees C and melt. Presumably the cooler one would reach melting temp faster when placed next to a cube warmer than itself, and more slowly when placed next to a cooler cube, but good luck measuring the effect with all the complications of air convection, room radiation, and phase change.
Instead of considering such chaotic thought experiments, why not focus on more controlled situations, e.g. Curt’s experiment, Anthony’s experiment, the Pictet Experiment referenced above, or (best in my opinion) Dr. Spencer’s “Yes, Virginia” thought experiment here:
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/
The Pink Unicorn Brigade has expounded at length on all the reasons this relatively simple experiment wouldn’t work as advertised, so it would seem to be the definitive proof of their “theories”. No doubt their version of the experiment is already in progress. 🙂
To Roger and others who say that only the filament temperature matters: The last time I checked, the bulb’s glass surface was a physical object, and therefore subject to the laws of thermodynamics. The earth’s surface does not have its own source of power that maintains its temperature; that power comes from the sun (and a miniscule amount from the earth’s core). But we care about that temperature, and analyze it endlessly.
The claim of the Slayers that I was seeking to test was very universal. The claim was that radiation from a body of lower temperature could under no circumstances increase the temperature of a body at higher temperature. A single counter-example would refute this universal claim. The surface of the bulb was at 95C at the end of the control setup. Bodies at 23C (one reflective, one “black”) were placed around it, sending radiation back to it (with other transfers controlled for). The temperature of the surface increased markedly, way past any possible measurement or setup error. The main thing I was interested in with the electrical measurements was to rule out the possibility that the increased temperature was due to increased electrical power in.
Steve C says (May 29, 2013 at 2:22 am): “As usual, this topic has generated far more heat than light (pun fully intended). As far as I can see, it will continue to do so forever until Slayer and Antislayer sit down together and design an experiment satisfactory to both sides”
I hate to sound like a broken record, but once again I have to bring up Dr. Spencer’s “Yes, Virginia” thought (so far) experiment. Dr. Spencer says it will work one way, Pierre LaTour (for one) says it will work another. There’s your “satisfactory” experiment.
Problem is, the Pink Unicorns have to do the experiment themselves, because as we’ve seen here they don’t accept “warmist” tests. For what I think are obvious reasons, they won’t.
‘A single counter-example would refute this universal claim. ‘
Sorry, proof is needed . Your experiment proved nothing. Try again. This time focus on anything connected with the filament and the slightest changes involved with that. That is the heart of it.
The point was to see if the source will increase in temperature if its own heat was reflected on it.
Many measurements of different situations were taken. Unfortunately we did not know what happened to the source. The closest measurement to the source was the surface of the globe filled with an inert gas. That’s a lot of ‘stuff’ between the source and the measurement.
Curt says (May 28, 2013 at 11:27 pm): “I agree with you on this one. In my analysis, I stated, “With all direct radiative losses suppressed in both cases, the difference must come from the reduced surface area of the smaller shell, which lessens its conductive/convective transfer to the outside air at a given temperature difference.”
Have you considered view factor? The bigger the shell, the more its radiation “misses” the bulb and hits itself instead. The smaller shell emits more of its radiation to the bulb and less to itself.
Alex:
Now were that heart felt it would mean something. As it is, sarcasm just serves to highlight the magnitude of your ignorance.
Maybe not to somebody who doesn’t know how a reflective-lined dewier works, or foil-lined insulation, or a mylar blanket.
There is no external cure to willful ignorance.
I agree with Curt that the argument that only the filament matters is a false argument.
Let the bulb represent the surface of Earth, which is being heated by some source, be it a filament or the Sun. The point of the test was to show that the surface of the bulb warmed in response to a change in configuration that the PSI groups claims would have no effect.
Asking whether the filament warms is akin to asking if the GHG warms the center of the Earth too.
Whether the filament is warmed is not germane to the original experiment. Of course it is, and that it is can be inferred from the change in resistance of the filament.
But it really is a non sequitur.
Alex says: “Unfortunately we did not know what happened to the source. ”
Unfortunately, you missed the NUMEROUS comments (and the original post as well ) that mentioned an increasing resistance of the filament, which means an increasing temperature.
richard says (May 29, 2013 at 4:39 am): “is this used in industry to extract more heat out of a cooler body, sounds wonderfully efficient.”
That’s refrigeration, a different subject.
Two radiating bodies in the cold of space? Is the light I see from the Moon IR? It seems to me that the light reflected from the full moon to the earth should prove the Greenhouse Effect. The data that I would look for would be the ambient temperature of the surface of the moon, as measured by the astronauts, as well as the measurement at some height above the surface. The earth and the moon are at the same relative distance from their heat source. The Moon does not have an atmosphere and the earth does. We “simple” people would understand this data and any argument. Against it would prove the ignorance/avarice of any trolls.
Curt’s test set-up (and I want to make sure this is clear, I really appreciate his experiment and the spirit of his exploration of thermodynamics) is a perfect testbed for demonstrating GHG heating. Turn on the lamp for an hour, then turn it off for an hour and repeat for as many cycles as you like. Now, add anything reflective or insulating that is analogous to “back radiation” heating to increase the average temperature of the bulb by 33C–which should be easy considering how intense the lightbulb energy source is compared to the Earth’s outgoing IR. To be fair, like the atmosphere, there should be nothing done to constrict convection.
I think this may have started with Derek Alker’s “Do moths make a light bulb burn brighter?”
http://principia-scientific.org/publications/New_Concise_Experiment_on_Backradiation.pdf
though Siddons really puts his foot in it as Anthony highlights; “a light bulb facing a mirror does not heat up”: http://wattsupwiththat.files.wordpress.com/2013/05/psi_siddonscapture.png
My number one objection to popular PSI thinking is the idea that the atmosphere increases night time cooling:
“Postma’s model equations also showed that, overnight, the majority of atmospheric cooling occurs directly at the surface and that the amount of cooling overnight was at least ten-times the value expected without a theoretical backradiation delay in cooling. Cooling at the surface is actually enhanced overnight rather than impeded, and there is no sign of delayed overnight cooling occurring at all.”
http://climateofsophistry.com/2013/05/13/slayers-putting-up-not-shutting-up/
I had a long email exchange with Joe Postma and Carl Brehmer about this a while back, and supplied sufficient evidence to refute it, but I guess it went in one ear and out the other.
Roger Clague says:
May 29, 2013 at 6:37 am
A change of 0.0007. That is 0.0007/ 0.2894 x 100%
= 7/2894 x 100%
= 0.24%
This is within the accuracy of the meters used to measure V and A.
I suppose that the voltmeter has a better accuracy than 0.1%, but even if the ampèremeter is less accurate, if the drop is repeatable (the ampères go back to about the same values with and without cover), then the accuracy is of no interest: it simply proves that the filament is heating up with a covered bulb. Not much, but observed.
@Michael Moon
“The Temperature of the Filament is the only temperature that matters here. Did the Filament warm from back-radiation? Did heat actually “flow uphill?””
Michael you are confusing heat conduction (which does not flow from colder to hotter objects) with electromagnetic radiation (IR) just as Postma has done. Radiation does not ‘flow’ in any particular direction based on the temperature somewhere else, only based on the temperature of the emitting object and its emissivity.
Shining a flashlight (torch) onto a green wall shows us a green wall. The light leaving the flashlight does not know that the wall is green and only transmit green wavelengths. In the same way, a radiative emitter just radiates. A filament absorbing that radiation has a higher temperature because it absorbs the incoming radiation whatever its temperature. It is not conducting, it is radiating.
The slight drop in power consumption evidenced in the experiment was cause by the absorbed heat: the hotter filament has a higher electrical resistance and the current flow drops, even as the light output increases. Cause? absorbed back radiation.
Just for the interest of readers, the emissivity of a surface changes very little with temperature. Just because the filament is glowing white hot does not mean it has a lower emissivity. A black filament radiates more effectively than a grey one, at the same elevated temperature.
How about using a frame covered in cling film/saran wrap/low density polyethylene Placed around the bulb and suspended in mid air – should stop convection but not impede IR.and not change conduction.
Then aluminium foil over same frame surrounding bulb and suspended in mid air should stop convection reflect IR.and not change conduction.
Use DC for bulb power (don’t need high temperature – not good for the plastic!) Keep the power into bulb constant (manually) to keep losses the same..
Record the resistance of the filament (V/I) when stable and at fixed power under both conditions
Record the air ambient in the room
In plastic case IR will escape to background. IR from background will add to heat of filament
In aluminium case some IR will be reflected back to bulb and some will heat aluminium and hence environment and IR from background will be reflected back to background.
A tip for using thermal cameras.
As someone pointed out the bulb showed the reflection of the operatives. The bulb is therefore reflecting background.
All objects need to have reflections removed and if possible the emittance of the object normalised.
Dr Spencer suggested a white paint. I have used a greay undercoat spray. – only a very fine layer is needed. Until you nomalise the emittance you cannot compare temperatures of different items in the same image.