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Post by icefisher on Feb 7, 2016 5:57:15 GMT
Andrews take, and I hate to say it Harvard University too, on how the greenhouse effect operates is by using the Engineering toolbox net radiation loss chart as a law rather than a tool.
I know how the tool is used. Andrew apparently does not. So lets explore this a bit.
Andrew maintains that a non-insulated surface will warm to 1/2 the temperature of a heated surface (assuming a field of view of 1.0). At this point in time the non-insulated surface will radiate half the radiation received away from the experiment into deep space where we are assuming its at absolute zero and it will radiate 1/2 its energy back to the heated surface. This is the only mathematical proportion that supports Andrews point of view.
OK so if the surface was instead 100% insulated only then will it warm to the full blackbody temperature and it will radiate all the heat back.
OK so Woods went looking for even one degree of this heating. His experiment wasn't perfect but by blocking all IR on a nice warm day of say 80 degrees (300k) that heated to say 110 degrees (316k) by restricting convection he should have been able to generate. Now 316K has a radiant emittance of 565w/m2.
Now keep in mind if insulation value of the cover of the greenhouse is above zero the amount of heating is increased. So Woods should have been able to generate 350k or more out of the greenhouse blocking the IR. That would be an increase of 34K or a temperature of 170F or a 60f increase.
But Woods IR blocking greenhouse actually got not even one degree warmer. Yeah the experiment may have been far from perfect but it seems implausible that Andrews. and God forbid Harvard University's model is correct.
Now if you double glaze the greenhouses to stem heat loss by conduction to say r-2 the Woods experiment should pull another 15k to 196f for an 80f gain.
Anybody still buying Andrew's model? Shiit!!! greenhouses would be deadly! Imagine the death rate of gardeners! And talk about the home heating you could get out of a small greenhouse window on the south side of your house. Yeah I am skeptical.
Actually I have designed and installed some greenhouse warming systems on homes. the restriction on convection provides a lot of daytime heat for a home, but not anywhere near what Andrew thinks it is. And of course the big caveat with these greenhouse systems is they don't work at night and actually cause more heat loss at night if you don't have insulated covers for the greenhouse. But you can get some modest nighttime success by using what is called a Trombe Wall. A Trombe Wall can be constructed of stone, masonry, or columns of water. The excess daytime heat is absorbed into the walls and will provide a warm wall for the house at night. They work modestly well especially with moveable insulation for the greenhouse and excellent insulation in the remaining walls of the home.
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Post by Andrew on Feb 7, 2016 7:39:28 GMT
Andrew maintains that a non-insulated surface will warm to 1/2 the temperature of a heated surface (assuming a field of view of 1.0). At this point in time the non-insulated surface will radiate half the radiation received away from the experiment into deep space where we are assuming its at absolute zero and it will radiate 1/2 its energy back to the heated surface. This is the only mathematical proportion that supports Andrews point of view. OK so if the surface was instead 100% insulated only then will it warm to the full blackbody temperature and it will radiate all the heat back. Icefisher I have totally no idea what that text means. 1. The only way you can 100% insulate a heated surface is when the surface is inside the heater. I have now told you that many times, and yet here you are repeating the same stupidity. 2. No doubt there are other errors if the text can ever be understood by another person to enable the audit process to be completed ----------- Either way you failed this audit. |
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Post by icefisher on Feb 7, 2016 7:53:23 GMT
Andrew maintains that a non-insulated surface will warm to 1/2 the temperature of a heated surface (assuming a field of view of 1.0). At this point in time the non-insulated surface will radiate half the radiation received away from the experiment into deep space where we are assuming its at absolute zero and it will radiate 1/2 its energy back to the heated surface. This is the only mathematical proportion that supports Andrews point of view. OK so if the surface was instead 100% insulated only then will it warm to the full blackbody temperature and it will radiate all the heat back. Icefisher I have totally no idea what that text means. 1. The only way you can 100% insulate a heated surface is when the surface is inside the heater. I have now told you that many times, and yet here you are repeating the same stupidity. 2. No doubt there are other errors if the text can ever be understood by another person to enable the audit process to be completed Its your theory that when no insulation exists it will only heat half way. The implication of that is it would take perfect insulation to heat all the way because of your math and the inability to deal with heat losses through the surface. My theory is its a crock! I think it will heat to the full temperature no matter what insulation is behind it. |
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Post by Andrew on Feb 7, 2016 7:55:04 GMT
Its your theory that when no insulation exists it will only heat half way. What do you mean it will heat half way?  ? What is my theory??? |
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Post by icefisher on Feb 7, 2016 9:05:37 GMT
Its your theory that when no insulation exists it will only heat half way. What do you mean it will heat half way? ? What is my theory??? your theory is a non-insulated blackbody surface receiving 341 watts from a 278.5K (+5.5C) heated blackbody surface with a field of view equal to 1.0 will heat up so that each of its surfaces will be radiating 170.5watts/m2. A blackbody surface emitting 170.5 watts per second has a temperature of 234K(-39c). Your model says this would slow cooling by the heated surface by 50%. All the other calculations above just figure out what the warming would be with but one such layer. . . .like in the Woods Experiment, though the Woods experiment was not quite entirely uninsulated suggesting if anything more heat would be returned to the heated surface. I am just saying its implausible to me that Woods got nothing when he should have gotten 34k or 60F according yours and Harvard University's model. Sure maybe he botched it. But I haven't seen anybody else achieve anything. I saw a saran wrap as the IR transparent cover get a small amount of additional warming but saran wrap is about 40 times less insulating that a sheet of glass and despite the results were anemic and possibly entirely attributable to the lack of insulation of the saran wrap. The saran wrap if anything should have come up more than 60f short of the ir blocking glass. |
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Post by Andrew on Feb 7, 2016 10:00:46 GMT
What do you mean it will heat half way? ? What is my theory??? your theory is a non-insulated blackbody surface What do you mean by a non-insulated surface? |
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Post by icefisher on Feb 7, 2016 10:18:25 GMT
your theory is a non-insulated blackbody surface What do you mean by a non-insulated surface? One molecule thick surface Andrew |
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Post by Andrew on Feb 7, 2016 10:36:07 GMT
What do you mean it will heat half way? ? What is my theory??? your theory is a non-insulated blackbody surface receiving 341 watts from a 278.5K (+5.5C) heated blackbody surface with a field of view equal to 1.0 will heat up so that each of its surfaces will be radiating 170.5watts/m2. A blackbody surface emitting 170.5 watts per second has a temperature of 234K(-39c). Your model says this would slow cooling by the heated surface by 50%. All the other calculations above just figure out what the warming would be with but one such layer. . . .like in the Woods Experiment, though the Woods experiment was not quite entirely uninsulated suggesting if anything more heat would be returned to the heated surface. I am just saying its implausible to me that Woods got nothing when he should have gotten 34k or 60F according yours and Harvard University's model. Sure maybe he botched it. But I haven't seen anybody else achieve anything. I saw a saran wrap as the IR transparent cover get a small amount of additional warming but saran wrap is about 40 times less insulating that a sheet of glass and despite the results were anemic and possibly entirely attributable to the lack of insulation of the saran wrap. The saran wrap if anything should have come up more than 60f short of the ir blocking glass. objects cool at different rates depending on the temperature of their distant surroundings. Something that humans have known about for around 200 years. ie. The warmer temperature of a distant object causes the heat losses of an object to be reduced. If you want to create thought experiments with particular conditions then you need to explain why you are objecting to the results you get when all we have to work with are the known laws of physics. You seem to be getting your knickers in a twist by talking about 'Andrews theory' Your objections to the known laws of physics have nothing to do with me whatsoever.And to remind you, 4 years ago you were objecting to climate scientists laughable calculations where it has been clear since day one of this conversation that you do not like the known laws of physics. |
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Post by icefisher on Feb 7, 2016 10:38:51 GMT
its ok if disavow it all andrew. but the appeal to authority impresses me. . . .not.
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Post by Andrew on Feb 7, 2016 11:20:40 GMT
its ok if disavow it all andrew. but the appeal to authority impresses me. . . .not. If you want to think you have some better ability to imagine than all of the best minds of the last two hundred years that is fine by me, but why do you have to bring me into it?Why are you constantly taking about my theory?If you do not like the laws of physics you should be able to say what you are objecting to. Instead you call me a liar and a cheat. You tell me I change goal posts. You tell me there is magnification where there is none at all. You even bizarrely object to experiments that use electricity to heat objects and demand natural methods are used and then you object to them too. Endlessly you object to the laws of physics and yet you are utterly unable to state the basis of your objection.And when i point that out to you, you tell me i am appealing to authority.  |
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Post by icefisher on Feb 7, 2016 14:11:26 GMT
its ok if disavow it all andrew. but the appeal to authority impresses me. . . .not. If you want to think you have some better ability to imagine than all of the best minds of the last two hundred years that is fine by me, but why do you have to bring me into it?Why are you constantly taking about my theory?If you do not like the laws of physics you should be able to say what you are objecting to. Instead you call me a liar and a cheat. You tell me I change goal posts. You tell me there is magnification where there is none at all. You even bizarrely object to experiments that use electricity to heat objects and demand natural methods are used and then you object to them too. Endlessly you object to the laws of physics and yet you are utterly unable to state the basis of your objection.And when i point that out to you, you tell me i am appealing to authority. What law of physics does my theory break Andrew? |
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Post by icefisher on Feb 7, 2016 18:44:02 GMT
It really does no good to whine about the breaking of the laws of physics without specifying what law has been broken. The figures for what one might expect from a greenhouse effect that is estimated by science to be 33K fits quite nicely with the expected effect from the scaled down Woods experiment that I gave the spectral numbers for. The model endorsed by Andrew, Harvard University and many others states that for each layer of a full greenhouse effect you get an identical increase of 50% of the original wattage as the model for the atmosphere blocks 50% of the light with each absorption.
The multilayered models Andrew and I worked over the years in this forum displayed that 50% increment in each layer. So a greenhouse with complete IR blocking should achieve what I calculated it should achieve namely 34K.
Now the world's greenhouse effect of 33K is darned close to one layer. As know the atmmospheric window is only about 20 watts out of 390 watts so the 33K is about one degree smaller than a complete blockage.
Everything fits the popular theory like a freaking glove except for the complete failure of Woods Experiment to produce even one degree. Doesn't that get anybody's attention???
34K expected, zero achieved!
See guys I have my big boy pants on and can do font enlargments too!! LOL!
OK I can see Woods getting losses in his system, but insulation is not zero and the model even with zero insulation is supposed to work via radiation alone!!!
If anybody has done an experiment with the detail and care that say the folks working on the high energy ionization cloud formation project and gotten success out of it I fully expect it would be plastered on just about every lamppole in the nation.
Yet all we get is appeals to authority. Einstein obviously not only understood the complexity of light he understood the lack of thinking by most scientists on the topic.
All these fifty years of conscious brooding have brought me no nearer to the answer to the question, “What are light quanta?”. Nowadays every Tom, Dikk and Harry thinks he knows it, but he is mistaken. (Einstein shortly before his death 1954)
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