Home/BlogIn Part I of this series, I mentioned how Wood’s (1909) “greenhouse box” experiment, which he claimed suggested that a real greenhouse did not operate through “trapping” of infrared radiation, was probably not described well enough to conclude anything of substance. I provided Wood’s original published “Note”, which was only a few paragraphs, and in which he admitted that he covered the issue in only cursory detail.
Wood’s experiment was not described well enough to replicate. We have no idea how much sunlight was passed through his plate of rock salt-covered box versus the glass-covered box. We also don’t know exactly how he placed another glass window over the rock salt window, which if it was very close at all, invalidated the whole experiment.
I also mentioned two more recent experiments which came to totally opposite conclusions: one showed a substantial temperature rise in a glass covered box versus one covered with an IR-transparent material, the other did not.
Here I’ll present first results from my own backyard experiment. Ideally, one wants to have identical boxes in terms of their absorption of sunlight and resistance to conductive heat loss. We want to measure just the effects of IR-transparent and IR-opaque materials placed over the boxes on their energy budgets, as measured by a temperature difference of the air trapped within the boxes.
I used inexpensive Styrofoam coolers purchased from WalMart, doubled-up and sealed with transparent packing tape to trap the air space between them. The insides of the coolers were completely lined with adhesive metal tape from Lowes, then sprayed with 3 coats of Krylon #1502 flat white paint, which has a measured IR emissivity of at least 0.99:
Temperatures within the boxes are monitored with K-type probes using a 3-probe Extech SD200 thermometer datalogger, set to record temperatures every 2 minutes, as well as the ambient air temperature in the shade of one of the boxes. The probe tips in the boxes were in the shade near one end, half way between the top and bottom of the coolers:
Both boxes were covered with kitchen plastic wrap (0.5 mil polyethylene), which is approximately 90% transparent to IR. Both coverings are sealed around their periphery with clear packing tape to make the boxes approximately air-tight. A sheet of 0.22 in thick plexiglass was placed about 1 inch above one of the boxes:
The first thing I discovered is that, without the plexiglass, it was difficult to get the two boxes to run at the same temperature, one being a little warmer (by 5 deg. F or more) as see in the thermal imager photo:
I believe this was the result of the spray paint…the warmer one had a slightly more textured painted surface, slightly darker in appearance (and thus absorbing more sunlight) while the cooler box had somewhat smoother paint, with a slightly brighter appearance.
The effect of the plexiglass is to block the IR coming out of the box, as is clearly seen in the following FLIR images of the two boxes, where the one on the left has the plexiglass sitting above the plastic wrap (indicated temperatures are for the crosshair locations):
After monitoring temperatures and deciding that one box was going to run a little warmer even without the plexiglass present, I decided the best way to see if the plexiglass caused IR warming was to place it over one box, then over the other box, switching boxes every 10 minutes. I figured I would then see how the temperature difference between the two boxes changed as a result. I did this for 2 hours, from 3 p.m. until 5 p.m., with the following temperature readings taken every 2 minutes (gridlines, not the arrows, indicate when the plexiglass was swapped):
The effect of the plexiglass can then be most clearly seen when we then plot the temperature difference between the two boxes over that 2 hour period:
Now we clearly see the warming effect of the plexiglass. Even though the plexiglass only passes 92% of the visible sunlight, which by itself should cause cooler temperatures, its presence over one box causes that box to warm relative to the other box (or, you can say its absence causes the other box to run cooler).
This is how the “greenhouse effect” works. Even though the plexiglass is at a cooler temperature than the inside of either of the 2 boxes (just as the atmosphere is at a cooler temperature than the solar-heated surface of the Earth), its presence causes warmer temperatures in the box it is placed over.
I don’t believe this is being caused by suppression of convective heat loss from the plastic wrap because I had considerable air space under the plexiglass and there was a light breeze ventilating that air (see UPDATE, below).
Of course, there are many different ways the experiment could be structured. I could have used black paint instead of white, which would have caused higher temperatures, but I wanted the experiment to produce temperatures closer to those seen naturally. I hope that there is enough detail above for others to replicate what I have done, if they wanted to.
Finally, it should be mentioned that using an experiment like this to demonstrate the fundamental mechanism of the greenhouse effect is somewhat difficult because one is trying to produce a marginally increased greenhouse effect over that which is already present. The sky is already largely opaque to the transfer of IR radiation, and so such an experiment tries to measure the incremental warming effect of a solid surface (the plexiglass) over and above that already being produced by downwelling IR from the sky.
UPDATE: Since there is concern expressed that the plexiglass might be inhibiting convective heat loss from the top of whichever box it is placed over (even through there is a 1+” inch air space for ventilation), here are the temperatures of the two boxes from last evening (during which I swapped the plexiglass a couple of times) and during the night:
Importantly, note that even when the interior of the box is cooler than the ambient temperature, the plexiglass has a warming influence. This is better revealed in a plot of the temperature difference between Box 1 and the ambient air temperature:
So, since convection can only transport heat from warmer to colder temperatures, convective inhibition cannot explain the warming effect of the plexiglass. It must be an infrared effect.
Good job, but … I think that you don’t give enough consideration to the convective heat transfer.
1 – Did you checked the wind orientation effect when you found different temperatures between the 2 boxes. My guess: the wind was coming from the top right direction in the photo !(rather than a difference in the paint texture)
2 – A plexyglass cover is placed only one inch above the box. I suppose that the air trapped between the cover and the box is much warmer than the ambiant air above the second box, thus reducing the box convective heat loss. I would rather suggest to have an IR tranparent cover (your plastic wrap ?)above the 2nd box, at the same distance (ideally, the plexiglass and plastic wrap should have the same heat conduction properties but the test would be too difficult to realize in a backyard experiment)
Your second idea is a good one…next time I run the experiment I will put a second layer of plastic wrap in place of the plexiglass so the convective loss from the first layer is similar between the two boxes.
I ran the boxes in different orientations…one ran warmer the other no matter what I did. It looked slightly darker by eye, which means it was absorbing more sunlight.
In the graphs, I would suggest putting the arrows right at the time where the plexiglass is switched. Or perhaps add some other bar that shows the entire periods when the plexiglass is in each position. My first thought when looking at your graphs is that the temperature swing started BEFORE the plexiglass was moved!
I also find it odd that Box 2 seems to be much less effected by the plexiglass. Do you have any explanation for that?
Yes, the graph labeling could have been clearer.
I think Box 2 needs more spray paint, to the point where the surface gets slightly rough, to increase it’s IR emissivity. If the foil tape isn’t coated really thick, the aluminum reflects some of the IR, which I think is why its temperature wasn’t as responsive to the presence of the plexiglass.
Nice job. It seems like it would be easier and more uniform to leave the coolers as is and use an identical target like a floor tile or metal sheet (bread pan)for the IR source. It would be kind of fun to adjust the distance of the plexiglass from the source to find the peak impact.
This was unfortunately another experiment where suppression of convection led to higher temperature. No conclusion about “greenhouse effect” by back radiation is therefore possible.
But both coolers had the same “suppression of convection” ie the plastic wrap over the cooler. The plexiglass above the cooler did not suppress convection, since the sides were open. The difference must be due to the “back-radiation” from the plexiglass.
Papijo suggested above that a “blank cover” of plastic wrap could be put above the “other” container that did not have the plexiglass — just to be even more certain that the physics conditions relating to convection adn conduction would be the same. Do you agree then that any difference must be due to radiation?
having the side open does not compensate for an obstruction. If you have enough crosswind perhaps the error will be small. But from the center of the box the path is how much longer to get around the obstruction vs through it?
the test at night to see if convection was obstructed ignores the fact that convection does not occur at night. Convection only runs from late morning to late afternoon.
And of course we expect a “greenhouse effect” at night.
The question is whether there is a greenhouse effect during the day.
To determine that you need to ensure both the control box and the experimental box have identical conductivity and convection values.
You also need to ensure that IR transmittance through a 3 dimensional IR blocker is at the same angle as average sunlight on the surface of a sphere. I think the plexiglass over the box must be have sunlight shine through it at a 30 degree incline from the plane of the plexiglass.
I will be fascinated to hear what underlying physics supports the contention that “convection does not occur at night”. That assertion will revolutionize the home insulation industry and turn the modern appliance industry on their ear. I had no idea that my wife’s fancy new oven could tell when it was nighttime. There was certainly nothing in the sales brochure disclosing that the convection function would only work in daytime. Should I sue?
The truth is that convection is a physical process that is utterly unrelated to time of day. Convection happens in darkened rooms (and tightly closed ovens). Convection happens in bright summer daylight and in the middle of a winter night.
Convection happens anytime a fluid has a heat imbalance and a sufficient cross-sectional area for the fluid to establish a convection cell. Warm air rises and cold air sinks but the currents can’t (easily) pass through each other so the circulation self-organizes into cells. As long as there is enough horizontal space for the circulation patterns to form, you will get convection. For atmosphere, the necessary gap is about an inch. (That’s why double-paned glass works as an insulator – the gap between the panes is too small to allow the formation of the necessary circulation patterns.)
Note: The one-inch gap described above IS sufficient for convection cells because that is a vertical gap. Convection suppression depends on horizontal restriction. Convection cells can be very short; there is just a limit on how skinny they can be.
OK does convection is not an important part of nighttime cooling work better for you?
Thank you for a polite reply to what on second reading was an unnecessarily snarky comment by me.
As an unqualified statement, I still don’t like that wording though. Convection can be every bit as important for nighttime heat transfer as for daytime heat transfer. But if you were trying to say that “convection stops being important when the temperature in the box drops below ambient”, then I have to agree. Convection does require that the hot air be lower than the cool. If that’s what you were originally saying, my apologies.
But if that’s what you were trying to say then I am confused why you don’t think it was an adequate control to demonstrate that the obstruction factor was negligible. The plexiglas in day had a positive effect (which at that point in the experiment could have been radiative or convective). The plexiglas at night had the same positive effect (when it could not have been convective) so the effect in day must have also been radiative.
James, apparently you did not read to the end of the post.
Roy Spencer says August 26, 2013 at 9:55 AM
“James, apparently you did not read to the end of the post.”
You are right, Roy, I did not. Now I did and this did not change my opinion.
The temperature of the thing that suppress convection is secondary. You can use a cold thing and still get higher temperature by suppressing convection. In short, that thing would be an obstacle for air circulation thus reducing convective cooling of the object below that thing.
If it was reducing air circulation, it would then be reducing the HEATING of the cold box. Again, convection can only transport heat from warmer temperatures to colder temperatures.
Roy Spencer says August 26, 2013 at 10:16 AM
“If it was reducing air circulation, it would then be reducing the HEATING of the cold box.”
The air circulation reduces heating of both boxes as long as the air is colder than the boxes, and this was the case.
An additional obstacle further reduces this convective cooling, so you can expect relative higher temperature therefore.
James, I think the key point is that both boxes warm because the convection is eliminated from the interior of the box to the rest of the atmosphere. But one box warms MORE than the other. This DIFFERENCE cannot be due to suppressed convection.
Sigh. That’s the point James…the ambient air temperature was WARMER than the air in the box. Look at the last plot in the above post.
Roy Spencer says August 26, 2013 at 10:44 AM
“Sigh. That’s the point James…the ambient air temperature was WARMER than the air in the box. Look at the last plot in the above post.”
Roy, let me quote you then: “even when the interior of the box is cooler than the ambient temperature, the plexiglass has a warming influence.”
Of course, anything warmer than the interior of the box would have a warming influence on the interior of the box by radiation. What you’ve got is this: a warmer ambient air warms the plexiglass and the warmer plexiglass in turn warms the colder interior of the box (by radiation).
This is certainly not a greenhouse effect, is it?
So, summing up, you have either a “suppressed convection effect” or “warm warms cold effect”. Still no reason to conclude on “greenhouse effect”. In this last respect the experiment failed, I would say.
Just to avoid misunderstanding, I’d like to tell you that it is impossible to prove “greenhouse effect” experimentally, because it has been proven non-existent analytically.
I recommend re-reading my comments here: http://www.drroyspencer.com/2013/08/does-a-greenhouse-operate-through-the-greenhouse-effect/. Simple math. Assumption of heating by back radiation will always be equivalent to a part of the system receiving more energy than the whole system for any period of time, which is obviously impossible.
An alternative to show James is wrong (or correct) would be to have 2 plexiglass sheets with holes drilled in them at regular intervals. Place both sheets above one of the containers at 1 and 2 inches. Cover the other container with the solid sheet. James would be proven wrong if there was no difference in temperature and correct if there was a difference in temperature.
Thanks for running the experiment and sharing the results Dr. Spencer.
Tim Folkerts says August 26, 2013 at 10:42 AM
“James, I think the key point is that both boxes warm because the convection is eliminated from the interior of the box to the rest of the atmosphere. But one box warms MORE than the other. This DIFFERENCE cannot be due to suppressed convection.
Certainly not due to suppressed convection inside the boxes, right.
I meant cooling the boxes by the outside air. Introducing an extra thing more or less close to the box would “spoil” the convective cooling.
At night, aren’t these the factors at play with the styrofoam boxes?
* IR Radiation from the box.
* IR Radiation the atmosphere to the box.
* Adiabatic heating at the surface from the collapsing air column that is no longer supported by the sun.
Covering with plexiglass just demonstrates that the box is generating more IR than the less dense atmosphere which shouldn’t be surprising I’d think.
James, I did re-read most of your comments .. and they are interesting but ultimately misguided. You miss the idea that the GHGs can collect outgoing energy and — for a while — prevent it from escaping.
You said “In other words, you can not distribute 300 whatever so that someone gets 750. So, the process we see in the picture is not a part the real world.”
Consider a bare blackbody planet that gets 300 J each second each square meter uniformly over the surface. It will warm until the surface is radiating 300 W/m^2 = 270 K.
Now add a cold (0 K) shell around the planet (transparent to sunlight; blackbody for thermal IR) with a heat capacity of 300 J/kg*K. The surface initially absorbs 300 W/m^2 of power and emits 300 W/m^2 of IR. All of the IR gets absorbed by the shell. It warms up 1 K in one second. Since it is so cold, it basically radiates nothing — either to space or to the ground (which means the whole system is gaining energy) . The seconds go by .. eventually it has reached the point where it is radiating 10 J/m^2/s up to space and down to the ground.
Now the ground is receiving 310 J/m^2/s .. so it will start to warm up, and start to emit more than 300 J/m^2/s.
…
The process continues until
* the shell is radiating 300 J/s/m^2 to space and 300 J/s/m^2 to the ground, and receiving 600 J/s/m^2 from the hot ground.
* the ground is receiving 300 J/s/m^2 from the shell and 300 J/s/m^2 from the sun, and emitting 600 J/s/m^2 to the shell.
I never had to create any energy; energy is conserved each and every second. So I can — with a very simply model — have only 300 J each second from the sun but have 600 J to the ground. Multi-layer models will push the numbers up even higher.
PS Look at this spreadsheet I made a while ago. Play with the numbers. You can make the temperatures and heat capacities anything you want.
https://docs.google.com/spreadsheet/ccc?key=0AgM8XE4GABYQdHVZTDZLblNOaTBpSkxEckxUQXdSMEE&usp=sharing
Tim Folkerts says August 26, 2013 at 10:21 PM
“James, … You said “In other words, you can not distribute 300 whatever so that someone gets 750. So, the process we see in the picture is not a part the real world.” Consider a bare blackbody planet … The process continues until … So I can — with a very simply model — have only 300 J each second from the sun but have 600 J to the ground.”
Tim, I skipped your process, because I am not interested in demonstrating where exactly you made mistakes. It is not our topic.
I can only repeat that the assumption of the back radiation warming a.k.a. “greenhouse effect” is equivalent to a part of the system receiving more energy than the whole system for any period of time, which is obviously impossible.
So, if you mean it is possible that a part of the system receives more energy than the whole system for any period of time, that is for the same period of time of course, then I can not help you.
James Goodone 10:46pm: “I can only repeat that the assumption of the back radiation warming a.k.a. “greenhouse effect” is equivalent to a part of the system receiving more energy than the whole system for any period of time, which is obviously impossible.”
Only obviously impossible if & due to James Goodone not adding the energy in and energy out arrows correctly. I showed you how energy in and energy out correctly balances including your “back radiation” term in the thread you link 8/16 5:12pm.
The energy balance eqn. (sometimes called heat eqn.) is an eqn. you need to learn how Dr. Spencer’s boxes work as well as the sun/earth surface/atm. work in real life LTE:
1) Put an energy control volume (cv) around the system of interest
2) Count accurately enough: energy into cv – energy out of cv = m*Cp*DT/dt W/m^2
At LTE, DT/dt=0. If you don’t get this you might believe looking into a mirror will melt your face. It won’t, the dT/dt term equilibrates with surroundings slightly higher T, no runaway just like earth and Dr. Spencer’s boxes. (Note: OT but Carrie Fisher claims if SHE looks in the mirror, she owes George Lucas $2 for use of copyright.)
Think about conduction and convection effects on Dr. Spencer experiment. Change the coolers into thin cardboard box on a windy day (7.8km/hr) like Prof.s Pratt & Nahle that came to the same result as Dr. Spencer. Temperatures inside change. It all adds up, in those cases, insulation, conduction and convection were important drivers too. The energy balance eqn. used with basic and testable physics will let you predict the results of the new experiment close enough to learn things.
Ball4 says August 27, 2013 at 12:06 PM
“Only obviously impossible if & due to James Goodone not adding the energy in and energy out arrows correctly. I showed you how energy in and energy out correctly balances including your “back radiation” term in the thread you link 8/16 5:12pm.”
I remember your argumentation and can only repeat my answer I gave you that it was irrelevant to my point. Your energy in and energy out balance (=0) can not be a proof for any heating by back radiation, because it would be the same 0 with any assumed amount of back radiation as well as completely without any back radiation. You simply missed the point, which was about back radiation warming being equivalent to a part of the system absorbing more energy for any period of time than enters the whole system for the same period of time, which is obviously impossible.
Secondly, “back radiation” term is not mine. The IPCC used this term in the 2nd report and referred to it as the source of additional warming. Dr.Spencer only reproduced the idea.
James 2:14pm – My point is Earth’s near surface atm. control volume (cv) balance as I wrote it doesn’t work with 0 for back radiation, which is a measured amount for earth. Dr. Spencer uses roughly the correct measured amount of back radiation for his local site.
“Your energy in and energy out balance (=0) can not be a proof for any heating by back radiation..”
True of course, back radiation affects the 2nd “out” cooling term not the 1st “in” heating term
for DT/dt; the atm. lapse rate generally keeps the atm. cooler than surface, only the sun (always on) heats up the system thru the 1st “in” term.
The correct amount of earth’s atm. radiation toward the surface cv is not 0 or any assumed amount since it is measured routinely & can be avg.d to a global mean, that is why you get an incorrect impossible situation with your “..back radiation warming being equivalent to a part of the system absorbing more energy for any period of time.” Balance does not mean this & this doesn’t happen. The local site cv balance is close to as shown as I calculated for you using Dr. Spencer’s arrows as in the top post of the link you provided (for one sec. if you want to use joules).
Back radiation is a different amount in the 2nd “out” cooling term for Venus and Mars due atm. opacity differences but still not 0 (well, on Mars back radiation pretty close to zero as that atm. opacity is so nearly transparent).
Ball4 says August 27, 2013 at 3:43 PM
“James 2:14pm – My point is Earth’s near surface atm. control volume (cv) balance as I wrote it doesn’t work with 0 for back radiation, which is a measured amount for earth. Dr. Spencer uses roughly the correct measured amount of back radiation for his local site.”
Ball4, I have already addressed the issue of “measurements” in that other thread. You can not correctly measure 2×2=5. The back radiation heating is equally false for the reasons I presented above. Whoever “measured” it, no back radiation heating is possible.
James 4:06pm – Yes, that’s true to text book stuff, as you write “back radiation heating is equally false”. Earth’s back radiation doesn’t “heat” i.e. affect “energy in” balance term as I wrote, that “energy in” term heating all comes from sun. Earth’s back radiation that is measured (by ESRL NOAA, downwelling IR) is in earth’s cooling term “energy out” for balance to = 0. Back radiation can generally do no heating of the surface (1st term) as it comes from a cooler source as you say (there are local temperature inversions that complicate matters).
Dr. Spencer experiments include earth’s back radiation day & night in the top post. Since Dr. Spencer’s experiments are local over time, the “energy in” from the sun varies (locally to 0 even) as does the “back radiation” (never to 0 though) in earth’s “energy out” term. DT/dt and T varies from earths local balance changing. Draw a cv around the coolers and count energy arrows to get a 0 balance in LTE or find +/- DT/dt as his data loggers show graphically.
Note to Dr. Spencer – how about showing a foam cooler cartoon w/the approximate W/m^2 arrows like the greenhouse post?
Ball4 says August 27, 2013 at 8:58 PM
“James Goodone 5:57pm: “I see. No heating by back radiation”
Yes. Back radiation is not part of the “energy into cv” term which is the heating term from hot sun only. Back radiation doesn’t “heat” as you agree, repeatedly.
…you added up Dr. Spencer arrows incorrectly. Once added correctly, back radiation is necessary for balance at 288K.”
Well, now it is getting nasty, in a known warmists tradition: distortion, obfuscation, distortion, obfuscation and so on.
I did not agree with your point, I sarcastically criticized it. No heating by back radiation is possible for the reasons I presented above, but the IPCC claims exactly that: heating by absorption of back radiation, see the link above.
Tim Folkerts says:
August 26, 2013 at 10:21 PM
The process continues until
* the shell is radiating 300 J/s/m^2 to space and 300 J/s/m^2 to the ground, and receiving 600 J/s/m^2 from the hot ground.
* the ground is receiving 300 J/s/m^2 from the shell and 300 J/s/m^2 from the sun, and emitting 600 J/s/m^2 to the shell.
Why did you arbitrarily stop at that point? The shell is now receiving 600 J/s/m^2 from the ground, and 300 J/s/m^2 from the sun. It will warm up further and re-radiate back the the ground again, which will warm further, and so on.
You need to patent this device. You could run a steam turbine with a candle and a number of strategically placed surrounding shells.
Your concept works in an alternate universe and fictional thought experiments, but not in the real world where the laws of physics apply.
Ball4 says August 27, 2013 at 5:02 PM
“James 4:06pm – Yes, that’s true to text book stuff, as you write “back radiation heating is equally false”. Earth’s back radiation doesn’t “heat” i.e. affect “energy in” balance term as I wrote, that “energy in” term heating all comes from sun. Earth’s back radiation that is measured (by ESRL NOAA, downwelling IR) is in earth’s cooling term “energy out” for balance to = 0. Back radiation can generally do no heating of the surface (1st term) as it comes from a cooler source as you say (there are local temperature inversions that complicate matters).”
I see. No heating by back radiation, but due to this “non-heating” back radiation the temperature just stays 33C above what the Sun can possibly do, right? Come on, how stupid do you think the readers are?
Anyway, the IPCC refer to back radiation from the so called “greenhouse gases” as “absorbed by the Earth surface” claiming this responsible for higher temperature than solar heating could possibly do. Exactly this absorption is impossible for the reasons I presented above.
This is from the 2nd IPCC report: http://imgur.com/gDRQL15 where they refer to back radiation as absorbed by tge surface. Let’s do some simple calculation. The reflected part of the solar radiation subtracted and considering only 1 m2 of the surface to simplify the calculation, for any given period of time of N seconds 235N Joules (from the Sun) enters the system “atmosphere+surface”, but the surface absorbs 168N+324N=492N Joules. So, again, the part of the system gets more than the whole system (492N>235N), which is clearly impossible.
Another funny thing is that according to this IPCC explanation the so called “greenhouse gases” radiate only to the surface but not to space. I wonder, why would the “greenhouse gases” radiate only in one direction? Like in order to not “spoil” the nice “0”-balance?
The atmospheric greenhouse effect has nothing to do with the workings of a garden greenhouse, and this is well understood.
The atmospheric greenhouse effect has also not much to do with “IR backradiation to the earth” although it is indirectly part of the balance.
The atmospheric greenhouse effect, as it is understood physically, comes from two effects:
– the convection-determined lapse rate. This lapse rate has to do with the composition of the air, latent heat and so on, and is also dependent on local temperature, but doesn’t depend on any backradiation or not: it is a convective feedback mechanism which instores the lapse rate. If the actual temperature gradient is lower than the adiabatic lapse rate, convection stops, and if the actual temperature gradient is higher than the lapse rate, convection sets in, to restore it.
– the position of the “highest totally black layer” in the atmosphere. Grossly one radiation length deep in the atmosphere when coming from outside. This is function of wavelength, and of course, of the concentration of greenhouse gases. The higher the concentration of greenhouse gases, the higher in the troposphere this “last black layer” is situated.
This “last black layer” is the radiator of IR radiation into space, the ONLY cooling mechanism of the earth.
As such, this “last black layer” temperature is determined by the total amount of heat the earth has to lose to space and the black body law. So for a given solar irradiation and albedo, this temperature is fixed.
Now, for a given lapse rate, the higher this “last black layer” is situated at fixed temperature, the hotter it has to be “down under” because down under is deeper.
That is the atmospheric greenhouse effect. It has nothing to do with a backyard greenhouse.
Patrick says August 28, 2013 at 12:51 AM
“The atmospheric greenhouse effect has nothing to do with the workings of a garden greenhouse, and this is well understood.
The atmospheric greenhouse effect has also not much to do with “IR backradiation to the earth” although it is indirectly part of the balance.
The atmospheric greenhouse effect, as it is understood physically, comes from two effects: …”
Apart from what you describe as “atmospheric greenhouse effect” being nonsense, it is clearly not the “greenhouse effect” as it is described by the IPCC in their reports and FAQs.
The IPCC “greenhouse effect” is warming of the Earth surface by back radiation from “greenhouse gases”.
I suggest you find a different term for your effect to avoid confusion.
Anyway, the whole topic here is about alleged warming by back radiation.
@ James:
The IPCC “explanation” of the greenhouse effect using backradiation is somewhat childish. Of course the backradiation does play a role, but it is indirect. What I have described it the actual physical mechanism of the “greenhouse effect” in the atmosphere: it is due to the rising of the “last black layer” in the troposphere, and due to the lapse rate.
The division in energy transport mechanism (convection, radiation, conduction) within the atmosphere is a complicated affair, and is of course influenced by the presence of greenhouse gasses, but the point is that it doesn’t matter, because the lapse rate is the result of a feedback process that ADAPTS the convection heat transport to whatever other things happen such as “backradiation”, in order to restore the (wet) adiabatic lapse rate.
In other words, you can do it the hard way, by trying to establish different heat fluxes and different heat budgets and so on, and get confused a lot, or you can do it the simple way: using the lapse rate. As there is a feedback mechanism adapting to instore the lapse rate, this is much easier.
No matter what complicated heat transport there is in the atmospheric column, with more convection, and less radiative upward and downward fluxes, at the end of the day, the lapse rate is established. And then, what counts, is what the atmosphere can radiate outward on its top level. That radiation is determined by the temperature of the “upper black layer”, and the height of the position of that upper black layer is determined by the concentration of greenhouse gasses: the more there is of it, the shorter the absorption length and the higher is the last opaque layer. The temperature of THAT opaque layer determines the outgoing IR flux in space, and that flux must (in equilibrium) be equal to the total amount of received heat (that is, everything the earth receives from the sun and doesn’t scatter in the visible – albedo). That “top layer temperature” is hence fixed. And if it lies higher, applying the lapse rate, you arrive at higher temperatures down under.
Patrick says: August 28, 2013 at 3:32 AM
“@ James: The IPCC “explanation” of the greenhouse effect using backradiation is somewhat childish. Of course the backradiation does play a role, but it is indirect. What I have described it the actual physical mechanism of the “greenhouse effect” in the atmosphere: …”
I know the narrative and can only repeat that it is completely unscientific nonsense. The IPCC must have well understood that, which is why they chose a different narrative that at least has the appearance of being valid. Their back radiation warming can indeed fool some people, especially when it is backed up by an international body.
Yours however, apart from being nonsense is absolutely irrelevant politically and has nothing to do with the topic of this thread.
The “back radiation” is not totally wrong, but it is irrelevant in the net greenhouse effect in the atmosphere. Of course the VERY FACT that the atmosphere is opaque for some IR bands comes from the fact that there is absorption, and hence, re-emission, of which half is downward and half is upward. And to get the last absorbing (and emitting) layer “up there” you need to have absorption all along, and hence also “backradiation”. So yes, it is true that greenhouse gases cause IR radiation to go “back down”. But it is not because of this that the surface gets warmer.
The NET upward energy flux in the atmosphere is constant throughout the entire atmosphere, and is equal to what is radiated to outer space. In equilibrium, this is also equal to the received solar radiation.
Part of this upward flux is in the visible, and compensates directly the incoming visible flux from the sun: the albedo. So there’s no point taking this with you.
Another part of this upward flux is IR radiation from the surface, where the atmosphere is transparent. If there were no greenhouse gasses, that would be it. The atmosphere wouldn’t contribute to any heat transport, as the IR radiation from the surface would go out directly.
So we can set apart the direct outgoing visible, and directly outgoing IR from the surface to space in the transparent bands. It doesn’t have anything to do with the atmosphere.
The interesting part is where the atmosphere is opaque. We can make the simplification that the atmosphere is totally opaque to some wavelengths, and completely transparent to others, to simplify things. We only consider that part where the atmosphere is opaque. When I talk about IR, I mean this part (not the part that goes directly to space).
Now, if the atmosphere were, for some or other reason, isothermal, then the fact that the atmosphere were opaque wouldn’t affect heat transport. Instead of having DIRECT IR going through a transparent atmosphere, the atmosphere would absorb it and re-emit it at any layer, and the upper layer would emit it to space. You would have the atmosphere and the earth surface at the same temperature.
However, you cannot (even if there were no gravity) have an isothermal atmosphere in which there is only radiative transport. A temperature gradient would set in.
You can clearly see this because each “black layer” would emit as much downward as upward, and receive as much from above or from below, apart from the upper layer, that emits downward and upward, and only receives from below. So another heat transport mechanism must be active, such as quick mixing, which means, heat transport by convection.
In a rapidly convective mixed atmosphere without gravity, there is no net heat transport by radiation between the layers (although they emit a lot upward and downward, but they receive as much from above and from below). The only heat transport is by the mixing convection, which supports the entire heat flux upward, until it is radiated from the last layer.
No matter how thick such an atmosphere is, there’s no “greenhouse” effect: the top layer is just as hot as the surface, and the heat flux is exactly the same as if the atmosphere were transparent, only: convection transports the heat to the upper layer, where it is radiated into space.
Now comes in gravity. Due to gravity, pressure is higher below than above, and due to adiabatics, a gas bubble going up will cool and a gas bubble going down will heat up. On top of that, if there are condensible gases, they might condense and there’s latent heat and so on: we have the lapse rate.
As long as there is convection, the lapse rate will be installed.
As now in an opaque atmosphere the layers get cooler when you go up, upper layers radiate less than lower layers, and there is hence a NET UPWARD IR flux between the layers. So the total upward flux of heat is now partially taken by the convection, and partially by the radiative balance between the layers.
Note that the lapse rate depends on gravity and the adiabatic properties of the atmosphere, not so much the fact that they are IR opaque ; on the other hand, if the atmosphere were not IR opaque, there was no heat transport to be done, and there wouldn’t be any convection either. The atmosphere would then become isothermal because of conduction if there was absolutely no radiative absorption.
But at the end of the day, the only way to get the heat into space is by radiating it out, and that’s done by the highest “black” layer.
Basic energy balance text book construct from 1st law which works for top post coolers:
1) Put an energy control volume (cv) around the system of interest, the near surface atm.
2) Count accurately enough: energy into cv – energy out of cv = m*Cp*DT/dt W/m^2
James Goodone 5:57pm: “I see. No heating by back radiation”
Yes. Back radiation is not part of the “energy into cv” term which is the heating term from hot sun only. Back radiation doesn’t “heat” as you agree, repeatedly.
“..but due to this “non-heating” back radiation the temperature just stays 33C above what the Sun can possibly do, right?”
Yes, the back radiation is part of the 2nd term “energy out of cv”. Balanced at 288K with back radiation; if back radiation turned off to 0 in radiative transfer theory, texts show us balanced at 255K.
“Come on, how stupid do you think the readers are?”
Not very, they can mostly see the 1st law energy balance and read text books like me.
“Exactly this absorption is impossible for the reasons I presented above.”
No, I have shown you added up Dr. Spencer arrows incorrectly. Once added correctly, back radiation is necessary for balance at 288K.
“..235N Joules (from the Sun) enters the system “atmosphere+surface”, but the surface absorbs 168N+324N=492N Joules. So, again, the part of the system gets more than the whole system (492N>235N), which is clearly impossible.”
No, clearly possible because you missed some cv cooling terms (some arrows) in the picture you linked. Use 1 sec. 235 Joules (from the Sun) enters the cv system “atmosphere+surface” this is the heating term but the surface has the cooling terms also, the 2nd term, the “energy out of cv” term. This second cooling term is across bottom (-24-78+390-324) so get your arrows added up correctly, you are not stupid. Again, + is defined away from sun heating, or down toward surface.
heating – cooling = 0 = DT/dt LTE
energy into cv – energy out of cv = m*Cp*DT/dt W/m^2 = 0
+168 joules into cv from sun – (-24-78-390+324) joules out of cv = 0 ?
+168 – 168 = 0 checks. Balanced. Mean T=288K
If you want to include the 67 absorbed by atm. in the cv then:
(168+67) heating – (-24-78-390+257) cooling = 0. Balanced. Possible. Back radiation is necessary for balance but it doesn’t “heat” the near surface atm. in the 1st term as you agree, back radiation ONLY affects the 2nd cooling term.
“…the so called “greenhouse gases” radiate only to the surface but not to space. I wonder, why would the “greenhouse gases” radiate only in one direction? Like in order to not “spoil” the nice “0″-balance?”
James Goodone here misses the 165 “emitted by atmosphere” the 30 “emitted by clouds” and the 40 “window” = 235 in the other direction to space.
Ball4 says August 27, 2013 at 8:58 PM
“James Goodone 5:57pm: “I see. No heating by back radiation”
Yes. Back radiation is not part of the “energy into cv” term which is the heating term from hot sun only. Back radiation doesn’t “heat” as you agree, repeatedly.
…you added up Dr. Spencer arrows incorrectly. Once added correctly, back radiation is necessary for balance at 288K.”
Well, now it is getting nasty, in a known warmists tradition: distortion, obfuscation, distortion, obfuscation and so on.
I did not agree with your point, I sarcastically criticized it. No heating by back radiation is possible for the reasons I presented above, but the IPCC claims exactly that: heating by absorption of back radiation, see the link above.
P.S. Sorry for posting that twice, I posted it in a wrong place.
Ball4 says August 27, 2013 at 8:58 PM
““…the so called “greenhouse gases” radiate only to the surface but not to space. I wonder, why would the “greenhouse gases” radiate only in one direction? Like in order to not “spoil” the nice “0″-balance?”
James Goodone here misses the 165 “emitted by atmosphere” the 30 “emitted by clouds” and the 40 “window” = 235 in the other direction to space.”
No, it is in the IPCC picture, not in reality, that “greenhouse gases” radiate only in one direction: to the surface. Look at this picture again (http://imgur.com/gDRQL15): from “greenhouse gases” goes 324 to the surface (back radiation) and is absorbed by the surface. Since gases radiate in all directions, the same 324 should go to space, but it does not in the IPCC picture.
You can not obfuscate this.
Now the funny thing: if you put this missing “324 to space” in place and do simple math, you will get more energy coming out of the system “atmosphere+surface” than comes into the system from the Sun. Another ridiculous outcome.
You understand the implications, don’t you?
James 9:36pm – The boxes in the top post demonstrate lots of energy flow arrows too. Nature gets them right; James can too. I urge James to think thru the boxes above.
“..the same 324 should go to space, but it does not in the IPCC picture.”
James – it doesn’t rain in space. So no. This can’t happen. And doesn’t.
Think about the components of the 324. Only radiation gets out to space, the 235. Not the latent heat (LH) or the thermals, they can’t get out, their energy dumped in atm. Here are the components making up the 324 back radiation necessary to balance at the surface to 288K:
324 = 24 thermals + 78 LH + 67 solar absorbed + 155 atm. radiation all in joules over 1 sec. Balanced.
The 155 is the radiation that James seeks going both ways: up to space (in the 235) and down to surface from just the atm. from its mean temp. You have to do a little work to understand the cartoon to see thru obfuscation. Energy flow can be superposed since it is linear, as in the boxes above. Can’t superpose temperature (goes as 4th power).
“more energy coming out of the system “atmosphere+surface” than comes into the system from the Sun.”
No.
+ 342 in heating – 107 reflected – 235 out cooling = 0 LTE
Just examine the picture in detail James; some answers you can find on your own, just with a little work & text book study.
“You understand the implications, don’t you?”
Yes. 288K with back radiation, 255K with back radiation of atm. gases set to 0 theoretically. Consistent with box experiment in top post.
Ball4 says August 28, 2013 at 6:39 AM
“James 9:36pm – “..the same 324 should go to space, but it does not in the IPCC picture.”
James – it doesn’t rain in space. So no. This can’t happen. And doesn’t. Think about the components of the 324.”
There are no components of the 324. In the IPCC explanation the 324 is back radiation from the “greenhouse gases” to the surface. Since gases radiate in all directions, there must be the same 324 to space, but it is not there.
So, for any given period of time N (and 1 m² surface to simplify the calculation) we have 2 impossible outcomes in the IPCC “greenhouse effect” model:
a) the surface absorbs (324N+168N=492N Joules) more energy than enters the system (235N Joules or 235N+24N=259N Joules if we include “thermals”), which is obviously impossible and
b) More energy (324N+40N=364N Joules) leaves the system than enters the system (235N Joules or 235N+24N=259N Joules if we include “thermals”), which is obviously impossible as well.
You can not escape that.
James Goodone 2:51pm: “You can not escape that.”
Remember Ball4 12:06pm, 8:58pm above wrote 1st law:
1) “Put an energy control volume (cv) around the system of interest
2) Count accurately enough: energy into cv – energy out of cv = m*Cp*DT/dt W/m^2
At LTE, DT/dt=0.”
What I cannot escape is James skipping my Step 1, then mixing up thermo control volumes in accounting for earth energy balance at LTE. Each of the foam boxes in top post has similar energy accounting needs. Once you establish control volumes, then do CPA level accounting and make some educated estimates of energy flows, the mean Teq. of the boxes will be found fairly close to Dr. Spencer thermometer T fields. This is well known basic text book thermo.
Here’s what James confusingly says mis-accounting for each energy control volume (cv) drawn around surface system (A) and the energy control volume (cv) drawn around (surface+atm.) (B).
I will substitute more specific cv (A) and cv (B) for James confused undefined word “system”. Which is exactly where James incorrectly adds things up then finds obviously impossible results. Per James & modified for the specific cv’s crossed by the named energy number:
a) the cv (A) absorbs (324N+168N=492N Joules) more energy than enters the cv (B) or + part of energy loss (cooling) from cv (A) (235N Joules or 235N+24N=259N Joules if we include “thermals”), which is obviously impossible and
b) More energy (324N+40N=364N Joules) leaves the cv (A) than enters the cv (B) or + part of energy loss from cv (A) ( (235N Joules or 235N+24N=259N Joules if we include “thermals”), which is obviously impossible as well.
Yes, this IS obviously impossible because James doesn’t establish & use thermo control volumes correctly since cv (A) and (B) are obviously and incorrectly mixed up by James accounting.
In the CPA profession, this would be James confusing accounts receivable with payables; debits confused with credits. Getting accounts to balance is what CPAs do James.
The IPCC does balance each cv correctly, its accounting does NOT confuse cv (A) and cv (B) as James does using just the undefined term “system”. The IPCC balances both control volumes as I already showed. N=1sec. No mixing up cv’s allowed in thermo James.
Surface, Step 1: define cv (A) surrounding surf., Step 2:
+168 in heating cv (A) from sun – (24+78+390-324) out cooling cv (A) = 0. Balanced joules accounting for cv (A).
Surface+atm., Step 1, define cv (B) surrounding (surf. and atm.), Step 2:
+ 342 in heating cv (B) from sun – 107 reflected out cv (B) – 235 out cooling cv (B) = 0 LTE. Balanced joules accounting for cv (B).
James – This is CPA level accounting, your undefined “system” leads you to incorrectly balance the energy books finding obviously impossible results. You can not escape that.
Urge James to learn & try the CV concept for the boxes in top post. CV are free, draw them any way you like then do the CPA level accounting.
Dallas, the IR emissivity of styrofoam is only 0.6, thus it needs a high emissivity coating. You can’t use bare metal for a target for the same reason, too reflective in the IR.
Dr. Roy, The low emissivity of styrofoam shouldn’t matter, you are looking for an up/down impact. Increasing the emissivity of the sytrafoam is like creating a black body cavity which would amplify the effect, but that could be misconstrued. A dark bread pan has an emissivity near .80 – .90 and an unglazed floor tile is close to the same. Then you have more of a greenhouse comparison with just floor and roof interaction. You could also use hot water bottles, to simulate heat capacity. A regular reddish hot water bottle should have an emissivity near 0.9 which is not bad or you could use water bottles.
I don’t think there is any way you can not show some radiant impact, but it would be neat to show how important specific heat capacity is in maintaining diurnal temperature range.
Oh, I wasn’t trying to mimic a greenhouse per se. Just trying to observe the temperature-elevating effect of an infrared absorber/emitter (the plexiglass) placed between a heated object (the inside of the box) and the “cold” sky. I’m also curious to see how cold I can get the air in the box at night.
How cold? That is more interesting. I will have to ponder that one.
the air in the box can’t get any colder than the effective radiating temperature of the sky, and even that assumes zero heat conduction into the box, which is not possible. So, this morning the coldest box reached 58 deg. F, which would correspond to a maximum downward IR flux of 385 W/m2. In a previous experiment I found that the air in the box seldom gets lower than 6-8 deg F below ambient, and that’s what I’m seeing again.
Yeah, I would guess about dew point as a minimum. It would be nice to have fan powered psychrometer.
Roy,
Building on what Papijo suggested earlier, you might try one OTHER variation where the plastic wrap is removed from both boxes. Then use only the loose-fitting covers ~ 1″ above the boxes. This will allow free convection, but the plexiglass-covered box should still be warmer than the plastic wrap-covered box. (you would have to be careful with wind in this case, since slight changes in airflow could make significant differences.
yes, that is a possibility, too.
I am puzzled by the results showing the inside temperature of an enclosed box being lower than the ambient outside temperature.
This is despite having the benefit of the supposed ‘heat trapping’ Plexiglas.
Any possible explanation?
Bryan, yes, in the evening the Earth’s surface cools radiatively to temperatures less than the ambient air temperature, but there is still some mixing occurring with warmer air from higher up which keeps surface temperatures from falling as much. Also, the heat capacity of the ground keeps releasing heat built up during the day.
Both of these warming effects are almost totally absent inside the insulated box. So the air in the box falls to a lower temperature than ambient, as the air is radiatively cooled by the high-emissivity surfaces inside the box.
Its cool how a basic understanding of the physics involved gives clear, consistent predictions that are in line with observations 🙂
Roy Spencer says
“but there is still some mixing occurring with warmer air from higher up which keeps surface temperatures from falling as much. Also, the heat capacity of the ground keeps releasing heat built up during the day.”
The effect of wind mixing (largely horizontal effect) and ground conduction on the ambient temperature could be tested for by using a third identical box with no film lid.
However with trying to link a glass greenhouse to an atmospheric greenhouse is not helped if an actual greenhouse gets colder at night while claiming an atmospheric greenhouse will increase night temperatures.
Dear Dr Roy
I see that these tests are not taking into account the regulatory function of water on the temperature. Probably the presence of plexigles is changing condensation / evaporation of water. I suggest a drop of water dripping plexigles and the other in plastic and measure the diameter of the drop in the material. With this we will know if their behavior is the same. Otherwise there will be change in convection even without temperature change.
?? But there is no water involved with *either* box.
Dr. Roy
In the atmosphere has water vapor, about 20 g per kg of air and liquid water approximately 4 g per kg of air. The phase shift of these components alter the behavior of the system between 540 cal / g and 0.24 cal / g and that needs to be taken into consideration as this phase shift is that the greenhouse effect keeps running overnight. If you notice your notes. The temperature variation with and without plexigles is equivalent a exchange between the ambient temperature and dew point. Probably the greater adherence of the water to plexigles.
I don’t see how water can be a factor here. Yes, the atmosphere in the box had an ambient moisture load. But both boxes had the same moisture load and both boxes were tightly sealed with plastic wrap, trapping the moisture load inside the boxes for the entire experiment. Since the temperature measurements were taken inside the boxes, any effects of evaporation/condensation should have been identical.
No condensation of dew on the plexiglas was reported but as it was suspended an inch above the entire mechanism, I don’t see how the added thermal mass on the plexiglass could plausibly have affected the temperature inside the sealed boxes. And even if that did matter, it would be a strictly nighttime effect – there is no significant condensation in daylight.
Roy I used a thermal imaging camera with bubble wrap (1cm air bubles in 2 layers of plastic for packaging)
http://climateandstuff.blogspot.co.uk/2013/06/more-fun-with-thermal-imagin-camera.html
It seems better at IR transmission than cling film
2 layers of cling film = 4 layers of bubblewrap = 8 layers of plastic
using 2 sheets of bubble wrap (bubbles interlocked) will give less IR attenuation and significantly more conductive attenuation than cling film
This was of course measured only over the FLIR IR camera bandwidth of 7.5 to 13um band (missing IR from 13 to 50um)
Interesting! I was wondering about bubble wrap. I’ll give it a try. I also read that the plastic used for shrink wrap is more transparent than cling wrap. Thanks for the tip.
Did you measure the temperature of the plexiglass? If that’s elevated it would skew the results too.
The temperature of the plexiglass stays at an intermediate temperature between ambient air and the interior of the box…it has to do that, as it absorbs and emits IR. That’s part of it’s radiative effect, so it really can’t be considered a “skewing” of the results. The plexiglass has a higher physical temperature, and thus radiating temperature, than the clear sky…that’s why it keeps the inside of the box warmer.
Skew was the wrong word to use. The question is what’s heating it? The air, i.e. conduction, IR from above, i.e. the sun, , or IR from below, i.e., the box. The AGW models rest on it being heated from below. Wood used the plate glass to filter out the IR from above to test if the heating came from below.
I’d suggest taking a run swapping the plexiglass back-and-forth at night. Let the sun warm up the boxes but keep the plexiglass in the shade until after sundown so it stays at air temperature. Now see how big the effect is.
Finally, by measuring the temperature of the plate you can estmate the heat fluxes and whether the rise you see is consistant with a purely radiative effect.
Roy
What do you think will constitute as sound evidence for the radiative atmospheric greenhouse effect?
3 competing schools of thought.
A. 33K effect for the whole atmosphere.
B. No effect whatever possible.
C. A very small (almost negligible) effect, say around 1K.
Your box height is not given but lets say 20cm.
Then if A is correct and taking the lapse rate as 9.8K/km then vertical change in box temperature = 0.002K without adding the Plexiglas.
In climatology classes a slab is used to represent the atmospheres radiative effect.
So the Plexiglas plate is a good analogue.
If the Plexiglas has emissivity of one then this is a greater analogue fraction than the atmosphere.
Because of the window effect’ the total atmosphere figure would be 0.9.
Is it unreasonable then to expect the measured difference with the plate present to no plate to be not less than 33K?
My own guess would be C.
I don’t think there is any way to convert the experiment results into what it means for the atmospheric greenhouse effect, other than this qualitative statement: infrared absorbers/emitters placed between a heated object and its cooler surrounds will cause the heated object to reach a higher temperature.
Hi Dr. Spencer,
the graph say more or less what I expected for that setup.
Just one question, why do you completely lined the insides of the coolers with adhesive metal tape?
Its good thermal conductivity could be the culprit of the difference in temperature between the two boxes.
IMHO spraying over that good thermal conductor a paint, which thermal conductivity is probably lower, make the thickness of that paint critical for the effective total thermal conductivity of the inner surfaces.
The box with more paint distributed less heat via conduction to the metal tape beneath it, while the one with less paint transferred more heat to the walls which having a shorter path to the top window probably lead to the difference that you measured.
Have a nice day.
Massimo
I don’t think that thermal conductivity is a significant component of what temperatures are reached. The important thing is for IR radiation being virtually the only cause of temperature changes within the box.
That’s funny, I would say it is the conductivity of the Glad wrap causing more difference. Wrap one layer around your fist, ** now now don’t burn yourself **, light a match or lighter underneath, the heat is instant, high conductivity because it is so thin. Try that with plexiglass, very slow, you really don’t feel the lighter at all.
Seems this does provides a layer of insulation in the plexi-over-box case than without.
Roy, you need to measure the exactly same thing as you were with varying gap between the plexiglass and the box top. You used a 1″ gap, try and measure at 3″, then 5″, then 7″ above and plot THOSE differences to see if you can get a convergence to say by that relation what “no plexiglass but the radiative effects” would be. Make any sense?
Many are saying this is mixing effects of the conductivity and convection and I agree… but you should be able to measurably compensate for this as described above.
@Wayne
I was talking about the measured temperature difference between the to boxes covered with the same PE wrap.
And my hypothesis is exactly that since so thin PE wrap have a little thermal resistance, the heat transfer between the inside air and the box walls could be more significant than what Dr.Spence thinks instead.
I already had been critical about the use of so thin wrap “window”, and I fully agree with you that I would placed a K sensor on the two faces of the plexiglass window to evaluate the temperature drop along its thickness, which I guess it’s not irrelevant.
I always said in my previous posts about the Dr.Spencer’s first backyard experiment, that I don’t believe that the PE wrap really stops the effects of the convective cell inside the box.
Have a nice day.
Massimo
For a more consistent “coating” on the interior, you could use some adhesive backed material (black cloth gaffer’s tape or white cloth tape) with the desired emmisivity.
(yes for those keeping score I did suggest fixing it with duct tape)
This crude backyard experiment effectively shows the unsurprising result that the inhibition of radiative cooling of a body by a proximate cooler body raises the first body’s temperature above what would be otherwise experienced. But since the cooler object can be heated by various mechanisms, a putative GHE would be experienced with an atmosphere of ANY composition heated by conduction and convection.
The key to temperature control in situ is MOIST convection, which often results in potential temperature increasing with altitude in the tropics. Since evaporation increases exponentially with T, it would be interesting to see what effect placing pans of water covering 71% ot the boxes’ bottoms would have upon the results. I believe it would show that the attribution of the observed elevation of global surface T to GHGs alone is not substantiated.
This modified experiment, of course, would not be definitive, because insolation-reducing cloud formation as a result of moist convection could not be induced within the boxes.
Irrelevant to your experiments, I suppose, but modern greenhouses use polycarbonate panels.
Dr Spencer, with respect;
You have created two nominally identical radiative collectors. You have done a nice complete job of eliminating conductive and convective heat transfer mechanisms.
However you do admit that one “collector” is more efficient than the other; “The first thing I discovered is that, without the plexiglass, it was difficult to get the two boxes to run at the same temperature, one being a little warmer (by 5 deg. F or more) as see((n) sic) in the thermal imager photo:”. As an aside I suspect the dimensional tolerances on foam boxes from Walmart™ are not too tight, in optical radiation work ¼ inch is HUGE, we routinely use laser trackers to place radiation detectors at the correct and very accurate place (within tenths of millimeters) to eliminate 1/r^2 and cosine errors.
So, it is fair to say that the average efficiency of your collectors is; Eavg = (box 1 temp + box 2 temp)/2 (units are Degrees F/1 unit of sunshine). This assumes that both collectors received the same amount of sunshine (a fair assumption I agree).
So the efficiency of one box is; E1 = (b1 + b2)/2 + 2.5 degrees (2.5 degrees being 1/2 of your measured bias), the average efficiency of the other box is; E2 = (b1 + b2)/2 – 2.5 degrees. It is important to remember these bias terms; you cannot simply dismiss them later in your analysis. And of course the “average efficiency” is not the “absolute efficiency” of your collectors, determining that would require quite a bit more instrumentation (spectroradiometers, standard radiation soruces, etc.).
You did not make clear in your post which box is more efficient (#1 or #2) but this is important.
For example, if you subtract the results from the less efficient box (-2.5 degrees) from the more efficient box (+2.5 degrees) you effectively double the bias error by +5 (+2.5 minus -2.5 = +5). However if you perform the subtraction in the other order you get an error of –5 (-2.5 minus +2.5 = -5). Bias errors do not “average out”.
It would be interesting to see your second plot (vertical axis labeled: “Box 1 minus Box 2 Temperature Diff…”) plotted again as “Box 2 minus Box 1……”, just saying…
So you have an interesting experiment that appears to produce results that are within the error bounds you observed (a bias of 5 degrees) and is inconclusive. Until you have repeatable radiation collectors that are within a few degrees (or less) of each other I do not think we can conclusively decide that radiation has been “trapped”.
So we have; Woods and Nahle 2, Spencer and Pratt 2, a draw.
But it does show the difficulties involved in measuring optical radiation, if you check the NIST website you will see that even if you do everything as they recommend you should not expect absolute radiation measurements better than a few percent. And of course the resulting temperature is a direct result of the absolute optical radiation received at a surface.
Cheers, Kevin.
Kevin
You can’t add upp reports more than you can add upp scientist when deciding what is science and what is wrong.
Dr. Spencers experiments are supposed to provide the definite empirical answer to the greenhouse mechanism: Mainly convection restriction or mainly IR “reflection”.
Theory and simulation are on the IR “reflection” side, the experiments will show if this can be verified convincingly in a controlled experiment similar to Wood, Nahle and Pratts. As I see the issue all 3 can’t be right, the results diverge to much.
By showing the Acrylic IR-Absorber/emitter/convection-blocker reduces the heat loss in both boxes and that a similar, IR transparent convectionblocker has much less effect on the heatloss in both boxes the difference in IR absorbtion (your Efficiency) and potential dimension effects between boxes cancel.
I’m prepared to accept i’m wrong on this if the measurements shows convection dominates over IR trapping. But I’m curious, how many “convectionists” will remain unconvinced, suspecting cheating in these experiments or demand a new setup using full size greenhouses with glass and cesium fluoride Windows?
Sigmundb you say
“But I’m curious, how many “convectionists” will remain unconvinced, suspecting cheating in these experiments or demand a new setup using full size greenhouses with glass and cesium fluoride Windows?”
This experiment has been done.
http://www.hort.cornell.edu/hightunnel/about/research/general/penn_state_plastic_study.pdf
For real large commercial greenhouses.
Some with IR blocking plastic
Some with IR transparent plastic.
Findings
Mostly near identical temperatures
Occasionally at night IR blocking gave a small increase (less than 1.5C)
Occasionally at night IR blocking gave a small decrease.
Sometimes the temperatures inside the greenhouses were less than the outside temperature.
Conclusion
It makes no sense to pay extra for IR blocking material to be added to greenhouse plastic covers.
Now how many Greenhouse Theory Enthusiasts will be dissapointed by these results?
Thanks for the link Bryan,
it had interesting data and report their findings as
“On some, but not all nights, the IR blocking materials provided extra protection by way of retaining heat inside the high tunnel at night. Graph 7 is an example of such a cooling event, occurring right after sundown, which makes timing of row cover application very important. The IR plastics do not prevent this cooling from occurring, however they decrease the effect by 2 or 3 degrees”.
Unfortunately the Experiment was designed to determine the films effect on plant growth and not the underlying physics of the greenhouse heat retention. With so little data available on the VIS-IR transmission spectra or diffusivity (tried googling Tufflite, Dupont IR , Smartlite Red,Tufflite IR and other greenhouse films, no luck)its impossible to draw from it what to expect from a Dr. Spencers setup.
But I noticed IR absorbing greenhouse films are sold at a premium on the promise they keep the greeenhouse cooler during the day and hotter during the Nights. 🙂
Sigmundb
You appear to be a greenhouse theory dogmatist and you are not prepared to look objectively at any experiment that contradicts your dogma despite your claim to be open minded.
Here is what the report says on page 2 which anyone can check
“Graphs 5 and 6 illustrate that although IR blocking films may occasionally raise night temperatures by 0.5 – 3 degrees (F) in tunnels covered, the trend does not seem to be consistent over time.”
So your greenhouse effect does not occur most of the time and happens on a few random occasions.
The graphs also show on some nights there is an anti – greenhouse effect
Sigmundb you say
“But I noticed IR absorbing greenhouse films are sold at a premium on the promise they keep the greenhouse cooler during the day and hotter during the Nights.”
This is because some people ignore science like the report above.
Some people buy filter cigarettes thinking that the filter will stop them getting lung cancer.
I suppose the gullible will always fall for advertisers hype
Dear Sigmundb, with all due respect; I HAVE NEVER ACCUSED ANYONE OF CHEATING……
From my extensive experience with measuring (in absolute traceable units to a national/international standard) optical radiation I have learned it is indeed very very difficult and error source abound (spectral content, reflectance (albedo), distances, angles, material properties, etc., etc.).
In my line of work we produce extensive (man years of work) thermal models that attempt to predict the temperatures a surface will reach under different conduction/convection/radiation conditions. A model that matches experimental results within 5 degrees F is a RARITY.
I am indeed sorry if I am not convinced as I apparently should be (since I am a “convectionist”, whatever the heck that means) but these simple experiments are so filled with potential error sources that I honestly dismiss them as “eye candy”. The simple fact that four experiments (Woods, Nahle, Pratt and now Spencer) show temperature divergences of 5 degrees or so tells me that nobody has yet perfected the experimental techniques necessary to reduce the “Radiative Greenhouse Effect” from a hypothesis to a solid theory that I would use to design anything that anybody’s life depended on.
The null hypothesis is still what should be used for any policies, i.e. the climate of the Earth is still very poorly understood (as clearly demonstrated by the climate science communities complete failure to predict ANYTHING) and we simply do not know enough at this point in history to do anything.
Cheers, Kevin.
Hi Kevin
First let me apologize, it was not my intention to express an expectation you in particular would find all kind of excuses for not accepting certain outcomes of Dr. Spencers experiment. What I meant to say is that I expect some posters to remain unconvinced no matter what the results are.
Secondly, let me admit that the main mechanism that keeps real greenhouses warmer then the ambient appears not to be the Greenhouse (with capital G) effect, this is evident from the data in the link Bryan provided: The difference in temperature between ambient and any greenhouse was much larger then the difference between greenhouses with different films.
But – Dr. Spencers setup is designed to maximise the temperature effects of the Greenhouse effect: Low thermal mass, well insulated, high IR emissivity and “focused ” on the sky. I expect a big difference in energy loss between an IR blocking and IR transmitting window and that is the Greenhouse effect to me.
The interactive nature of the experiments allow Dr. Spencer to adjust the setup to adress concerns like yours and I look forward to the next installments.
Wbr, Sigmund
“But – Dr. Spencer’s setup is designed to maximise the temperature effects of the Greenhouse effect: Low thermal mass, well insulated, high IR emissivity and “focused ” on the sky.”
Yes indeed I agree, but by maximizing the “alleged” radiative greenhouse effect Dr. Spencer has also unwittingly maximized the error sources. I do not believe he did this on purpose, nor did Woods, Nahle or Pratt. I do believe all these folks are acting honestly (i.e. NOT CHEATING), but may have overlooked error sources in their particular version of the experiment.
This is the true nature of empirical science, someone proposes a hypothesis and a way to prove it (and also a way to disprove it) then others try it as well. Almost anybody can replicate Ohm’s law in the lab, if 1/2 of the folks trying to prove Ohm’s law came up with different results it would not be a law, would it ?
These experiments are in fact quite difficult to perform with traceability to absolute standards and folks that believe they are just “kitchen table physics” have never actually tried to measure absolute optical radiation, it is in fact quite difficult.
So we still have four “similar” (after a first look) experiments that vary by about 5 degrees. I suggest that the current “noise floor”(the minimum real knowledge we can obtain about an effect) is currently about 5 degrees (F). Any experiment that shows that amount or less of warming is inconclusive.
Cheers, Kevin.
A real greenhouse indeed does not trap IR radiation. THe glass absorbs the IR, realizes it as heat and loses this rapidly by conduction to the outside colder air. The warming of the greenhouse is simply the solar light being absorbed by the floor and walls, conversion to heat and then heating of the air by conduction and convection. It’s really not all that complicated, how come people keep thrashing with it? IR radiation is NOT heat.
Charles, you said it yourself: “THe glass absorbs the IR, realizes it as heat”… IR is one method of heat transference. Since all objects emit IR at a rate relative to surface temperature, as the temperature of the greenhouse interior rises, the IR it emits rises. If the glazing absorbs this IR, its temperature becomes higher than it would have if it was being heated solely by conduction from contact with the warm air inside the greenhouse. The IR emitted from the underside of the glazing then consequently transfers heat down to the interior of the greenhouse. Thus the interior of the greenhouse becomes warmer than it would have if the IR it initially emitted had freely radiated away.
The point at issue here is whether glazing that is IR absorbent (thereby retaining the energy of the otherwise escaping IR emitted by the greenhouse interior)would result in a warmer greenhouse than glazing that allowed that same IR to radiate away.
Doubters have raised 1 main objection:
Perhaps the IR absorbent glazing used in this experiment skewed the results by inhibiting the loss of heat by conduction/convection. One way the test can be fixed to address this question is by covering (at the same gap) one cooler with a non-IR absorbing glaze of similar heat conduction characteristics as the plexiglass.
Of course, proving greenhouses are warmer with IR absorbing glaze doesn’t prove reducing human caused CO2 emissions will stop global warming. It has stopped already.
2 James goodone:
When you play tennis each side surely receives the ball all the time, yet the ball conservation law is never broken in the process.
Roy,
I don’t understand your experiment. It appears the plastic-covered box is warmer (153 F) than glass-covered box (97.5 F) So how can you conclude that IR is the cause? The plastic-covered box represents no greenhouse since plastic is transparent to IR. The conclusion should be no greenhouse is warmer. This is because it has more IR.
Without the greenhouse, the IR from sun is 350-500 W/m^2. Add the IR from air at 85 F and 0.83 emissivity, that’s 471 W/m^2. Half of that (235 W/m^2) will go to the ground. Total IR is > 585 W/m^2.
The temperature of glass is < 135 F. At 0.93 glass emissivity, we get < 625 W/m^2. Half of that (312 W/m^2) will go to the ground. You see the higher IR without greenhouse. That explains the higher temperature.
In the real greenhouse, it is warmer because of suppression of convective air cooling. This is not captured by your experiment because the plastic-covered box also has suppressed convective cooling similar to the glass-covered box. No greenhouse should have free flowing air. No side walls, no roof.
Dr. Strangelove
The 153F is the temperature inside the Box (assuming Dr.Spencer has the correction for paint emissivity and saran wrap absorbtion right)as measured by the thermal camera (FLIR)looking into the box. The 97.5F reading for the glass- (acrylic)covered box is from the IR image of the acrylic plate (showing the Acrylic plate = air/ambient temperature), it is not from the thermocouples inside the box.
With regard to the energy balance you must consider all forms of energy, not just IR. Then you have sunny, haze, light clouds, heavy clouds, day, night, dusk, dawn.
The temperature difference graph is inconclusive. Sometimes it has negative values which means glass-covered box is cooler. 10 minutes is probably too short to attain equilibrium. Temperature is changing over time because it is not in equilibrium. Try longer time period until the temperature stabilizes. Then see which box is warmer.
We can consider only IR because we are testing the hypothesis that IR is the cause of warming. If IR is lower in a greenhouse, then whatever is causing the warming is not IR. Suppressed convective cooling is the conventional answer. I believe it is true, which we can validate with experiment and calculation.
Do American scientists still use the Fahrenheit scale, or only for the convenience of their readers?
No, scientists use deg. C and K. I used deg F for convenience of the readers.
Since the supposition is the 1 inch gap does not inhibit convection, why inhibit convection at all?
Its not clear what role the saran wrap plays in this experiment except to ensure the results come out above ambient due to restricting convection.
Its not clear what you mean by “entering shade” the average inclination of the sun shining through our atmosphere is at 4 pm. Absorption of incoming IR is going to be change depending upon inclination.
The question is if you take the saran wrap off does the plexiglass, taking care to hold it to ensure it does not restrict convection, cool or warm the surface? Keeping in mind of course that a doubling of CO2 isn’t 100% transparent to sun light either.
The experiment as run, roughly counting squares covered by plexiglass I come up with 7 squares warming and 5 squares cooling. Not very convincing! Especially considering how warm this plexiglass panel must be. Did you get any figures on how much IR it blocks?
Just a question:
how come your boxes became cooler than ambient air at night ? Where does the fridge effect come from ?
The boxes are radiating away energy to the night sky. It is very common for the earth’s surface temperature, particularly on a clear night, to be noticeably below the air temperature just above the surface.
I very often see frost on the ground at sunrise when the air temperature is +3 or 4 C. The pumps on my swimming pool system automatically turn on when the air temperature hits 4C or below to prevent the possibility of water freezing in the pipes.
Curt
The effect of air in enclosures like greenhouses often being actually colder than ambient air temperature at night is well documented.
This fact directly contradicts the greenhouse theory which claims that greenhouses (atmospheric and glass) make their contents warmer at night
Bryan 4:48pm – “The effect of air in enclosures like greenhouses often being actually colder than ambient air temperature at night is well documented.”
Agree, but doesn’t necess. contradict theory, greenhouse experiments CAN be found warmer than the outside air if research the GH literature. Journal of Applied Meteorology 1963, Vol 2, p. 793 Kirby Hanson writes: “during January 1961 the minimum temperature averaged 2.4F lower inside than outside” a small PE covered greenhouse. Here is a documented example of a greenhouse effect in reverse: the greenhouse suppresses mixing of warmer surrounding air than inside it.
Correct GH theory has to truly account for convective, radiative, and conductive energy transfer across the control volume. Thin wall, windy GH locations could be different inside T field than thick wall, calm ones.
Bryan:
Glass greenhouses work primarily by suppression of convection. In the daytime, with absorption of solar radiation, this serves to keep the temperature inside the greenhouse higher than the outside, where greater convective losses can occur.
In the nighttime, there is net radiative loss, and convection serves to warm the surface. This is why you will hear on weather reports that protected (from wind) valleys will have the lowest overnight temperatures. It is why there are fans in fields such as vineyards and citrus groves that are turned on when threatened by frost. The glass greenhouse’s suppression of convection would be expected to have a cooling effect in nighttime conditions such as these.
In my engineering heat transfer classes (in the 1970s), the rule of thumb I learned was to treat the clear night sky as having an effective (near blackbody) radiating temperature of -20C (253K) for purposes of calculating radiative heat transfer of outdoor equipment. If there were no radiatively active gases (no atmospheric “greenhouse effect”), things would be radiating directly to space, which has an effective radiating temperature of 3K.
Indeed. So the greenhouse content at night is radiatively coupled with the effective IR radiator above. As there no convection, nor much conduction, the only non-adiabatic process is the IR balance. So at night, it is essentially coupled to the effective temperature of the IR radiator in the atmosphere (opaque part) and the black space (transparent part). This is effectively colder than the ambient air in most cases, so the thermal coupling through radiation with this colder IR body will decrease the temperature.
On the other hand, if you put an IR radiator at ambient temperature (plexiglass) at the window, the box is IR coupled to this radiator, and will tend to the temperature of this radiator, which is kept at ambient air temperature.
So essentially the box gets hotter at night because you put a hot radiator above it…
I have another remark, and that is the relevance of the effects. If we are ready to accept that the IR contribution is a 10 degree effect, while “blocking the convection” is a 45 degree effect when the sun is shining (the temperature inside the box with the wrap on it versus ambient air), this experiment would indicate that the main contribution to the working of a greenhouse is, as is usually accepted, due to the limitation of convection.
To really test this, you should have taken another box in identical conditions, but without the foil on it, to quantify the effect of convection.
Personally, I have even difficulties accepting the IR contribution, because the difference in radiated IR in the box with and without plexi has to do with the IR radiated from the plexi, which is determined by the plexi temperature, versus the IR radiated without the plexi, which is the atmospheric IR radiation (where the atmosphere is opaque) and direct solar IR radiation. It is true that the atmospheric IR radiation will be somewhat colder than ambient air, because it comes down from the “lowest fully black layer” which is at some height in the troposphere, and hence at lower temperature than at ground level. But I wonder whether this is not compensated by the IR directly from the sun, which you block with the plexi.
My personal impression is rather that the conductive and convective thermal resistance of just a foil is lower than the construction of foil, air gap, and plexiglass plate. IR might play a role, but it cannot be very important, as long as your plexiglass is not hot.
Dr. Spencer, I’ve been wondering about water feedback. A type of experiment I’d like to see is one that compares heating a water and air environment using SW vs LW radiation.
Here’s one possible idea. Two aquariums with foil place temperature and humidity sensors. Heat each using the same amount of power, one using LW radiation in the water, the other using SW radiation from above.
Nice job showing the actual ‘greenhouse’ effect with the plexiglass. I suppose the real question is does a thicker (more massive and equivalent to more CO2) plexiglass warm the containers more than a thinner sheet? Is it a measurable difference?
Wouldn’t that difference be the actual ‘man made greenhouse” effect?
Dr. Roy wrote, “infrared absorbers/emitters placed between a heated object and its cooler surrounds will cause the heated object to reach a higher temperature.”
Can you then tell me why, when pressure effects are discounted, Venus is no hotter than it “should” be just from being closer to the sun?
I mean, it seems like the ultimate experiment has already been done — a 96% CO2 atmosphere — and Venus at altitude 49 km, where the atmospheric pressure equals Earth’s at sea level, is only 337degK, 17% (1.17 being the fourth root of (93/67)^2) more than Earth’s 288degK.
Thanks.
mellyrn 2:04pm – “Can you then tell me why…?”
Yes this Venus stuff is very interesting and if you go back and check the papers where they worked out Venus’ T vs. h profile from satellite radio occultation, the form of the IGL used was p=density*R*T, the appropriate form of the IGL for atm. app.s.
So quite plainly, if you select for Venus “where the atmospheric pressure equals Earth’s at sea level” and measure the density at that location on Venus, then the mean T can simply be found as you write to be equal the T factored from orbital and albedo differences. To within the weather differences at that location anyways. Very neat. R has to be of the right form also. These occultation T vs. h experiments were confirmed in the ballpark from in situ thermometers.
So…can we calculate theoretically from what is demonstrated by Dr. Spencer et.al. experiments, what would happen if the cooler set up in top post were placed on Venus surface? Other than some if it just melts? I would answer yes. If someone cared enough to do the work, managed the materials not to theoretically melt for the right time frames & since no one wants to be there replicating.
“This is how the “greenhouse effect” works. Even though the plexiglass is at a cooler temperature than the inside of either of the 2 boxes (just as the atmosphere is at a cooler temperature than the solar-heated surface of the Earth), its presence causes warmer temperatures in the box it is placed over. ”
A solar pond has cooler water above hotter water under the surface. Due to using saltwater and density gradient of fresher and salter water. And no panes of glass or sheets of plastics are required. Nor are gases required- though having some gases involved are unavoidable in terms of attempting to isolated them.
So, is a solar pond a greenhouse effect?
And isn’t the fact of having a planet covered mostly by water an important aspect related to warming caused by the greenhouse effect?
Is it not possible that the largest greenhouse effect is not caused by gases, but rather caused by liquid water of the Earth’s vast oceans?
The way a solar pond works is by preventing convection of heat. And it’s trapping heat- the energy of sunlight passing thru water and warms beneath the surface of the water. Water is poor conductor of heat, and density gradient stops warm water rising. One could also look at in terms of radiant energy- liquid water blocks a lot wavelengths- but this is true of any rock.
In terms of radiant energy the difference with water is the majority of energy of sunlight is transparent. So energy goes thru it, and has harder time leaving it.
It’s a trap!
It seems the main problem of the greenhouse effect is it’s all about greenhouse gases, whereas I would say most of greenhouse effect is actually caused by a greenhouse liquid- water.
“It seems the main problem of the greenhouse effect is…”
It seems the main problem of the greenhouse effect theory is …
Wiki:
“The greenhouse effect is a process by which thermal radiation from a planetary surface is absorbed by atmospheric greenhouse gases, and is re-radiated in all directions.”
And:
“By their percentage contribution to the greenhouse effect on Earth the four major gases are:[21][22]
water vapor, 36–70%
carbon dioxide, 9–26%
methane, 4–9%
ozone, 3–7% ”
http://en.wikipedia.org/wiki/Greenhouse_effect
gbaikie asks
“So, is a solar pond a greenhouse effect?”
But in fact answers himself
“The way a solar pond works is by preventing convection of heat.”
So it does not work by trapping radiation but by stopping convection just as R W Wood found back in 1909
Bryan, your answer is too simple.
The GHE requires blocking of IR by a layer (or multiple layers) that is thermally isolated from the ‘ground’ layer.
A perfectly isolated blocking layer gives the maximum ‘greenhouse effect. This would occur with absolutely no atmosphere between the ground and the IR absorbing layer.
A fluid between the ground and the blocking layer will reduce the GHE. if that fluid only conducts but does not convect, the GHE will be fairly large (eg the salt pond). If the fluid convects moderately well, the GHE will be moderate size (eg the atmosphere). If the fluid convect very well (eg fresh water) then the GHE will be very small.
So yes, convection is important, but it is not the whole story. It is the *interplay* or convection and IR blocking that matters. (Another example: a “solar pond” with an IR transparent fluid would not warm up, whether or not there was convection.)
Tim Folkerts
You seem to have a direct equivalence between IR active molecules and the greenhouse effect.
I think most greenhouse advocates would say there is a bit more to the ‘effect’ than that.
But you do seem to agree that R W Wood would not be surprised by the solar pool.
“A perfectly isolated blocking layer gives the maximum ‘greenhouse effect. This would occur with absolutely no atmosphere between the ground and the IR absorbing layer. ”
I seems a perfect “greenhouse” would a reflecting layer. And if not perfect, multiple reflecting layers.
But then again, I was saying and still saying I am not sure what this so called “greenhouse effect” is, if “greenhouse effect” can only mean it involves just gases, then I believe gases under normal condition don’t reflect- only something with a surface can reflect- so liquids and/or solids.
Now, a maximum greenhouse effect is also an interesting concept. What exactly is a maximum greenhouse effect? Should get visions of Earth becoming like Venus?
Hot and very uniform temperature? Is it how hot it gets, or is it how constant the temperature get despite a very long night? Or must it be both.
Perhaps it’s not a matter of how hot it gets, perhaps it’s about how much heat is stored??
So, for example, Earth ocean hold more heat energy than Venus’ atmosphere. Does that make Earth ocean [a liquid] have more of greenhouse effect than 92 atm mostly CO2 atmosphere in a earth-like gravity planet?
Or is an unfair comparison because Earth ocean is so much more massive than Venus atmosphere. Does the most maximum greenhouse effect only count when one compare it to equal mass?
If btw, what is the comparison joule to joule of 92 atm venus’s atmosphere and 92 atm Earth’s ocean? Or should just figure it, kg to kg?
So anyways 92 atm of Earth ocean is 3036 foot depth. The ocean darkness. “[Aphotic zone] formally defined as the depths beyond which less than 1% of sunlight penetrates”
And:
“The zone between 200 meters (656 feet) and 1,000 meters (3,280 feet) is usually referred to as the “twilight” zone, but is officially the dysphotic zone. In this zone, the intensity of light rapidly dissipates as depth increases. Such a miniscule amount of light penetrates beyond a depth of 200 meters that photosynthesis is no longer possible.
The aphotic, or “midnight,” zone exists in depths below 1,000 meters (3,280 feet). Sunlight does not penetrate to these depths and the zone is bathed in darkness. ”
http://oceanservice.noaa.gov/facts/light_travel.html
So down 92 atm of ocean there still some sunlight- kind of similar to Venus’ surface.
Ocean temperature in tropics:
“Thermoclines are strongest in the tropics, where the temperature of the epipelagic zone is usually above 20°C. From the base of the epipelagic, the temperature drops over several hundred meters to 5 or 6°C at 1,000 meters.”
http://en.wikipedia.org/wiki/Deep_sea
So in 800 meter goes from 20 to 5 C. So say 12 C average. and 200 meters to surface 20 to 28 C- so 24 average.
So say 14- 15 C average for down to 3036′. So 288 K.
Joule per kg to heat kg of 0 K water to 288 K:
-100 C ice has specific heat of 1.389 kJ/kgK
And -10 C is 2.000 kJ/kgK
To make easy 0 K to 270 K around average 1.600 kJ/kgK-
432 kJ per kg
To melt ice: 334 kJ/kg
and 270 to 288: 18 times 4.1 = 73 KJ per kg. Total:
839 KJ per kg.
Venus. Just consider CO2:
175 K 0.709 KJ per kg
375 K 0.918 KJ per kg. So say .82 KJ times 200 is
164 KJ per kg
Latent heat of vaporization: 571.08 kJ/kg
So 735.08 KJ/kg to raise from 0 K to 375 K
How warm is the 92 atm of Venus atmosphere?
Surface temperature is:
“Average temperature: 737 K (464 C)
Diurnal temperature range: ~0 ”
http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html
Let’s say it’s 650 K.
375 K 0.918 KJ per kg
650 1.102 KJ per kg
Say .95 KJ per kg times 275 is
261.25 + 735.08 = total of 996.33 KJ per kg
Earth ocean average to 3036′: total joules per kg: 839 KJ
Venus total atmosphere average per kg: 996.33 KJ
So in terms of equal masses, Venus atmosphere holds more joules of heat than Earth ocean of same mass.
But simply due to Earth’s more massive ocean, Earth entire oceans has about 3 times of much joules of energy as compare to the atmosphere of Venus.
I think I have a nearly ideal ‘inexpensive, amateur backradiation experiment’.
See the image linked below. An electric heater (red) is placed on a sheet of styrofoam. The heater is surrounded by a plastic wrap enclosure 1-2 cm above & beside the heater to limit conduction and convection. Above this is another sheet of plastic wrap, about 1-2 cm above the lower plastic wrap. This will create a 1-2 cm space through which a fan will gently blow air — which will keep this intermediate layer at room temperature.
Turn on the power to the heater and let it come to steady-state– somewhere around 100 C for example.
Then lower a room temperature sheet of high-emissivity material (eg plexiglass or painted metal) and suspend it close above the top layer of plastic wrap.
If the ‘greenhouse effect’ is real, then
* the suspended sheet will warm above room temperature due to IR radiation (NOT due to conduction or convection).
* the sheet will start radiating IR toward the heater at a rate higher than the ‘background radiation’.
* the heater will warm ABOVE its previous steady-state temperature.
* remove the suspended sheet, and the temperature of the heater should cool to the previous temperature.
Just to be crystal clear …
1) the room = ‘space’, which provides a ‘cosmic background radiation’.
2) the heater = ‘earth’
3) the electrical power to the heater = ‘solar power input’
4) the suspended sheet = GHGs
5) there is no way that conduction or convection could be responsible for the elevated temperature of the heater due to the layer of room temperature air between. Basically, the layers of uniform temperature air are a ‘poor mans’s vacuum’.
https://docs.google.com/file/d/0BwM8XE4GABYQWUJGdkl6WXc2LU0/edit?usp=sharing
Tim Folkerts
Google on my windows 7 notebook reports a problem when trying to download link.
Roy,
Remove the air gap in the boxes to eliminate convection inside the box. The objective is to test if IR is causing the warming inside the box. This would simply the problem as temperature will depend only on radiative heat transfer and convection in the glass and plastic. Styrofoam is a good insulator so we can neglect heat transfer through the sides of the boxes.
Try using only one box, with and without glass cover. See which is warmer. Wait until the temperature stabilizes. We want to get the equilibrium temperature. Then use two boxes, one with glass cover, another without. See which is warmer. I predict box without glass cover will be warmer in both cases.
Using this simple box “greenhouse” with Styrofoam walls and glass cover, no air gap. I calculated what air temperature is needed so IR inside is > 585 W/m^2. This is the total IR from sun (350) plus air at 29 C (235). When this condition is met, it means the box is heating by increased IR vs. no box.
Energy balance of air inside box
300 + Rg – Rf = 0
300 W/m^2 is sunlight, Rg is down radiation of glass, Rf is up radiation of box floor
Energy balance of glass
585 – C + Rf = 0
585 W/m^2 is down IR to glass, C is convective heat transfer in glass
Convective heat transfer equation
C = k (Tg – Ta)
k is convective heat transfer coefficient of glass = 25 W/m^2/K, Tg is temp. of glass, Ta is temp. of air outside
Stefan-Boltzmann law of radiation
Rg = (e o Tg^4)/2
Rg is down IR of glass, e is emissivity of glass = 0.93, o is Stefan-Boltzmann constant
Solving these four equations simultaneously we get:
Ta = (2 Rg/eo)^0.25 – (885+Rg)/k
When Rg > 585 it means IR inside box is higher than no box. This is the condition needed to heat box by IR
Solving the above at Rg = 585 we get Ta = 327 K or 54 C
Therefore, air temp. outside must be > 54 C to heat the box by increased IR.
Assumptions in this calculation:
Heat conduction is negligible
Temperature of box floor equal to temp. of air inside box
System is in thermal equilibrium that’s why energy balances equal zero
If I remove IR from air, assuming it’s already included in the 350 W/m^2 IR from sun, we get Ta = 299 K (26 C)
So if air temp. is > 26 C we have greenhouse warming due to increased IR.
At noon and clear sky IR from sun is 500 W/m^2 while light is 440 W/m^2, we get Ta = 313 K (40 C)
It looks like Dr. Spencer has a point. At warm air temp. above 26-40 C it’s possible to get enhanced IR greenhouse warming.
Thanks, Dr. Spencer.
Good experiment that shows how difficult it is to get clear-cut, yes/no results in the lab.
More clarity comes from your good effort.
I may add that inexpensive Styrofoam coolers probably vary in their insulating performance due to a loosely controlled fabrication process; foam density and included air pockets come to mind.
Suggestion to find out whether a greenhouse’s temperature rise comes mainly from blocking convection, or mainly from the IR radiation balance (which is, I think, the question):
compare a box where you obstruct the IR balance but allow for convection with a box where you obstruct the convection, but allow for the IR balance.
The box where you obstruct convection but allow for IR balance, you already have: it is with the plastic film on it.
To make a box where you obstruct the IR balance but allow for convection, you could place a large plexi plate almost vertically over the open box. If the plate is high enough, it will cover most of the solid angle as seen from the box to the sky, hence obstructing the IR exchange with the sky, but if the inclination from vertical is small enough, it will almost not obstruct convection.
I bet on higher temperatures with the film than with the open box under the optical sky obstructor, which would indicate that the main heat trapping mechanism in a greenhouse is due to convection obstruction and not by IR backradiation, but if you really want to know, you can do the experiment.
I would also tell you NOT to paint the interior of the box white. That is minimising the visual solar heating and hence putting the balance artificially maximally to the IR balance, because the paint is black in IR, and white in visible. To do an honest test, your box should have a less-than-100% albedo in the visible, and have colored painting.
I think there is a problem with comparisons between panes and gases [greenhouse gases].
A glass or plastic window reflects light. Look a windows at night time- they act mirror like if there is darkness outside.
Yes if you radiating into 2 K cosmic background universe, you radiate more energy towards the blackness of the universe as compared to radiating towards a nearby planet you are orbiting.
But if some part of galaxy where there vast oceans of gases- Molecular clouds which are say 50 K. If one close to or inside them, does it have much effect on amount energy radiated and does matter if they are greenhouse gases or not?
Btw, apparently there enormous very hot gas cloud which surrounds our galaxy:
“Astronomers have discovered a cloud of gas engulfing our Milky Way galaxy that weighs as much as all the stars inside our galactic home. If the size and mass of this cloud is confirmed, it may solve a longstanding astronomical mystery, experts say.”
And:
“Initial signs of our galaxy’s halo came from the Chandra observatory, which observed eight objects shining brightly in X-ray light, and found that some of this light was being absorbed by charged oxygen atoms around the Milky Way. Scientists determined that this absorbing gas is between 1 million and 2.5 million Kelvin (1.8 million and 4.5 million degrees Fahrenheit) — a few hundred times hotter than the surface of the sun.
“We know the gas is around the galaxy, and we know how hot it is,” Anjali Gupta, lead author of a paper reporting the findings in The Astrophysical Journal, said in a statement. “The big question is, how large is the halo, and how massive is it?” ”
So as saying seems to be problem using a pane of transparent
material to represent what greenhouse gases do. Because reflection insulate radiating heat loss. And because solid 2 dimensional surface radiates directionally- and radiates dissimilar a gas.
You are absolutely right. There is even not any relationship between a pane and an absorbing layer in the atmosphere, for different reasons.
The first reason is that the atmospheric greenhouse effect comes about the difference in temperature (set by the lapse rate and the height) of the place we are interested in (the surface of the earth) and the thing that lets the heat escape (the black top of the atmosphere). The temperature gradient between the surface and the “emitter” does not depend upon the amount of heat that needs to be transported, but only on the height and the lapse rate. The more heat that needs to be transported, the more convection will set in, to transport the heat.
The second reason is that the outgoing heat has only one final way to get out: radiation (into space). There’s no conduction, and there is no convection in space. Radiation is the only way to cool.
The atmospheric greenhouse effect comes about because of two things:
– the “radiator” into space is at a certain height (determined by the amount of greenhouse gases) and radiation is the only ultimate heat loss, so it follows Wien’s law.
– there is a given delta-T between the radiator temperature and the surface, given by the height and the lapse rate.
The heat transport between the surface and the radiator above is complicated, split into contributions of convection and radiation, with different amounts of re-radiation and everything you want, but it doesn’t matter: in the end the heat gets up at the radiator and is radiated into space.
The important thing to note is that the delta-T doesn’t depend on the actual net upward heat flux (if the flux has to increase, this will be done by more convection without altering the lapse rate). So you don’t have to care about the intermediate different pathways of the heat in its way up: the delta-T doesn’t depend on it.
This is entirely different in a garden greenhouse. The garden greenhouse can loose heat through radiation, but also by conduction and convection, because the heat aborber is the atmosphere on top of outer space. So yes, radiation will play a certain role, but it will turn out to be minor in comparison to other heat transport mechanisms, mainly convection.
By obstructing convection, you cut off the most important heat transport vector. Of course you get convection inside the garden greenhouse, and now you will be heating the wall that stops convection. That wall will then conduct heat, radiate heat, and also loose heat by convection on the outside. Playing on the thermal resistance of the window will limit the thermal loss by conduction, but this was in any case very small (even though a very thin window, the amount of heat that is conducted is much smaller than a convective current can transport).
Making the window opaque to IR will also make you win somewhat, because now instead of seeing the upper atmosphere and outer space, which is cooler, you will look at a somewhat warmer window. But radiation is not the main heat vector at earth’s surface anyway.
I think that the fluctuations of temperature difference you show in your “Box 1 minus Box 2” figure is a result of the radiation from the plexiglass in deed. The IR from the plexiglass reduces the cooling rate of the box which it covers and thus increases temperature slightly since it is still warmed by the surroundings. This is an effect that radiation from a cooler object can have on a warmer. However, this effect is so small in the real athmosphere that it is impossible to measure any effect on the global temperature, ref. your book “The great global warming blunder” where you report your effort to try to measure the effect of CO2, but cannot find any. CO2 has no real measurable effect on global temperature or climate. Forget CO2, – it is a plant nutrition and beneficial for life.
Petter, we cannot just forget CO2. While it may be true that plants incorporate the carbon in CO2 into carbohydrates, those carbohydrates, such as cellulose and lignin, eventually decay and releast CO2. Beyond a certain point, we cannot increase the biomass, especially given that humans continue to destroy forests.
If CO2 were the limiting factor in plant growth, we’d have more plants instead of higher CO2 levels.
The impact of faster plant growth could have interesting implications in and of itself. Imagine a time when oil is $200 a barrel, every plant grows faster, and it thus becomes too expensive to mow grass and prune the trees that line rural roads. A sort of Gaia’s revenge…
More CO2 and the natural global warming we have had since the little ice age have the beneficial effect that there is now more vegetation in the higher mountains, the sahel is greener and with the help of fertilizer more food is produced. Warm periods are better for the life on earth than cold, but the sun (the main driver of climate) is now probably entering a “cold” phase. Enjoy the interglacial as long as we can.
Dr. Spencer, I admit that on reading of your planned experiment, I was expecting that your results wouldn’t be very convincing to the doubters. They always seem to raise objections that are hard to easily dismiss, at least to my non-expert mind.
I was pleasantly surprised to see that you added two features to the experimental procedure I wasn’t expecting: swapping the Plexiglas plate every ten minutes, and running the experiment both during the day and at night. Combined with your graphic presentation of the measurements, I think you really nailed it.
Impressive work! Thank you!!
Don’t you find it odd that both boxes are cooler at night than the ambient air.
The assumption must be that air warmed by IR radiation from the earth isn’t allowed into the boxes by the cellophane wrap.
I’ll also wager that the plexiglass weighs more than the Styrofoam boxes, but even if it doesn’t, it still represents a heat sink that you are holding over your heat probe.
Am I to believe there is significant IR radiance from a styofoam box at night? What is a styofoam box’s IR transmittance?
As for the daytime, if you are able to measure a difference in temperatures merely based on the spray painted texture of the boxes, than surely an ever so slightly opaque chunk of plexiglass acting as a heat sink, and converting some sunlight into IR just above your heat probe, will raise the box temperature too.
Right! Defenders of the greenhouse theory in defense of charges of violating the 2nd law of thermodynamics claim the warming is caused by slowing of cooling. You put plexiglass over the box at the outside ambient temperature, the slowing in cooling should be massive by virtue of the plexiglass at +30C replacing a cold upper atmosphere that is -30C or less.
We all know that should be the case. We also know this panel decreases in incoming radiation. But that should not be treated as a defect with the plexiglass, incoming sunlight is 50% IR so if you block IR you are going to block sunlight.
My suggestion of having the sunlight go through the plexiglass at 30 degree inclination from the plane of the plexiglass is to emulate average sunlight going through the atmosphere. Everybody instead tries to minimize incoming absorption but that should give you a greenhouse effect!
You can see it in Roy’s results when the sun was high, with the results disappearing around 4pm when the sun is roughly at the 30degree incline. (assuming he did not tilt the glass)
Additionally, regardless of their absolute temperature, the fact that the two styofoam boxes cool at EXACTLY the same rate, at night, argues against any “greenhouse effect” as a result of the plexiglass anyway.
Dear Dr. Roy
The temperature difference observed with or without a plexiglass cover due to reduction in the area available to the external convective motion caused by the presence of plexiglass. The box no plexiglass, this area is approximately 0.0875 m^2 while the box with plexiglass this area is 0.0304 m^2.
To avoid this error my suggestion is to use two sandwiches of two glass plates, one laced with a IR-transparent polyethylene film and the other with IR-opaque film. With this we will have the effect of convection control with identical performance in both boxes.
Dr. Spencer, with respect, I have a few observations that you may (or may not) find helpful;
First, you need to match the collection efficiency of your “radiation to heat” collectors very closely (goal is within a degree F, or so). Since the dimensions of the boxes are not “precise” you may need to compress the bottom of one of the boxes (smack it down a little bit with a hammer) to increase the 1/r^2 losses in one box to match the other box. Until the efficiency of each box (degrees F/unit of sunshine) are equal you are “chasing your tail”. This radiation to temperature conversion is a fundamental effect necessary to demonstrating the “GHE” accurately. It is not something that can be “massaged” by post processing the temperature data.
Second, you really need to change ONLY the IR energy flow through your system. It may not be apparent but the plexiglass ™ “window” you inserted into your system has a significant thermal capacity when compared to the thermal capacity of the air you have trapped in “the box”. It is essential to change (as far as practicable) ONLY the IR transmission characteristics of the “window” you insert. You need to find two materials (i.e. plastic films) that only differ in their IR transmission characteristics and they need to have (as closely as possible) identical thermal “mass/capacity”, reflectance, thickness, etc. You need to attempt to change ONLY the IR transmission of the “windows” and NOTHING else.
Finally, it is not clear from your post but I suggest you discard any temperatures derived from “FLIR” cameras, these devices are highly dependent on the spectral content of the observed radiation and also on assumptions about the emissivity of the measured surface. They are rarely good to an absolute temperature better than 5 degrees (F) or so.
And you should be aware that in the thermal testing world differences of a few degrees are considered “the noise”. Unlike measuring voltages where accuracies on the order of millivolts are considered routine.
And I do wonder why if the “GHE” can “force” the oceans of the world into thermal equilibrium with the “GHG’s” by some alleged 30 degrees C, why your nearly perfect greenhouse can only raise the temperature of air (with a thermal capacity at least a hundred times less) by 5 degrees or so ???
I.E. it takes a whole lot less energy to raise the temperature of air than the temperature of water so if the “GHE” is as effective as posited your air in the box should be at 200 degrees or so ???
Perhaps your observations are merely an as yet misunderstood experimental error ?
Thanks for your efforts.
Cheers, Kevin.
“why if the “GHE” can “force” the oceans of the world into thermal equilibrium with the “GHG’s” by some alleged 30 degrees C why your nearly perfect greenhouse can only raise the temperature of air (with a thermal capacity at least a hundred times less) by 5 degrees or so ???”
Because earth’s radiative balance is around 15 C. In Roy’s experiment, ambient temperature is already at 30 C. So the enhanced IR in the experiment is over and above the atmospheric GHE. And the atmosphere is 100 kilometers thick while the glass is a few millimeters thick.
“it takes a whole lot less energy to raise the temperature of air than the temperature of water so if the “GHE” is as effective as posited your air in the box should be at 200 degrees or so ???”
The ave. temp. of the ocean is lower at 3 C than 15 C radiative balance temp. Air in the box is cooling by conduction to glass and convection to air gap. Earth cannot cool by conduction and convection to space.
Oh another thing, the atmosphere is transparent to SWIR from sun, glass is not. Glass blocks the SWIR. Atmosphere is more effective in trapping heat.
Roy,
Though my calculations show that at air temp. above 26-40 C there’s increased IR inside the greenhouse (box). It doesn’t follow IR is the cause of greenhouse warming. Air is largely transparent to IR and light. Air inside the greenhouse is heated by conduction from the floor and cooled by conduction to the walls and roof.
Even with increased IR inside greenhouse, if you remove the walls, it will not warm. Because air under the glass roof is constantly being replaced by cool air from outside. On the other hand, the greenhouse will warm even with reduced IR if you put back the walls.
Therefore it is not the increased IR but the confinement of the air inside greenhouse that’s causing the warming. The conventional explanation of suppressed convective cooling is partly correct. But a more accurate explanation is prevention of mass transfer of air.
@Dr. Strangelove
For what it is worth, I fully agree.
Have a nice day.
Massimo
Massimo PORZIO,
And I agree with you.
So then there were three.
Live well and prosper,
Mike Flynn.
Dr S, I think you should consider a few expansions on what you wrote.
“Air inside the greenhouse is heated by conduction from the floor … “
But the floor is heated (in part) by the IR absorbs. Since you agree there is increased IR in the greenhouse (ie box in this case), then the floor of the greenhouse will be warmer and will be better at warming the air as well.
“air under the glass roof is constantly being replaced by cool air from outside.”
The box only models one small part of the whole GHE for the earth. The “real” GHE encompasses the whole earth and whole atmosphere (so the box really out to encircle the whole earth). For the earth as a whole, there is no “air from outside”. The air can only move around “within the box” because the ‘box’ is everywhere.
“But a more accurate explanation is prevention of mass transfer of air.”
There is no mass transfer of air to/from the earth as a whole. The warmed air doesn’t leave — it simply moves to cooler areas and warms them.
************************
In this model, the plastic wrap needs to be thought of as the “Top of Atmosphere”. Everything outside the box is “space” providing a “cosmic background radiation” at ~ 270 K. The plexiglass is then like a thin shell of GHG’s at or above the “top of the atmosphere”.
Hi Tim.
“But the floor is heated (in part) by the IR absorbs. Since you agree there is increased IR in the greenhouse (ie box in this case), then the floor of the greenhouse will be warmer and will be better at warming the air as well.”
Exactly the contrary, the bottom warms more and emits more LWIR because of the reduced convection. If you were right there was a violation of conservation of energy at that bottom surface. That because, you supposed to get more LWIR emission and at the same time you supposed to heat the air more, that’s an energetic paradox.
The point is: if the air inside the box wasn’t constrained in that place by the box walls, then that LWIR flux emitted by the bottom surely reduces because of the huge heat “extraction” due to the convective air path. It’s just a question of conservation of energy on that surface.
In the world of matter (made of solids, liquids or gases) the LWIR emitted is the residual of energy that matter doesn’t exchange by convection or conduction.
The ground is constantly doing the work of throw away from itself the atmospheric molecules. It’s true that that energy is constantly returned to the ground by the fall back of the same air molecules, but it’s true only at the thermal steady state. As the ground increases its temperature, it firstly “exports” that excess of absorbed energy to the air trying to throw away its molecules, and after the thermal steady state is reached, the LWIR flux for that state is emitted. During the heating of the atmosphere the LWIR must be lower than the steady state, otherwise the conservation of energy fails during that thermal transition.
In few words, first came the temperature then the radiation.
“In this model, the plastic wrap needs to be thought of as the “Top of Atmosphere”. Everything outside the box is “space” providing a “cosmic background radiation” at ~ 270 K. The plexiglass is then like a thin shell of GHG’s at or above the “top of the atmosphere”.”
AFIK many climate model (if not all), compute a warming effect of GHGs to the ground and a cooling effect to the TOA. I don’t see any cooling under the box window, so your parallel doesn’t work.
I believe that a greenhouse has nothing to do with our atmosphere.
“There is no mass transfer of air to/from the earth as a whole. The warmed air doesn’t leave — it simply moves to cooler areas and warms them.”
Yes, but there is between lower and higher troposphere so convection could move great masses of air from the place where the GHGs warm the ground to the place where the GHGs cool the top.
I repeat, IMHO a greenhouse is a complete different system compared to out atmosphere. The first works because of a strong suppression of the convective thermal transfer, the second works at the altitudes near the ground almost only by that.
Of course, as always told, this is my modest opinion, so I could be wrong.
Have a nice day.
Massimo
“Exactly the contrary, the bottom warms more and emits more LWIR because of the reduced convection.”
I was comparing the box WITH the Plexiglas to the box to the box WITHOUT the Plexiglas. The INITIAL warming was due to reduced convection. The EXTRA warming with the Plexiglas is due to the EXTRA IR absorbed by the walls. Similarly, a real greenhouse, both effects (reduced convection and backradiation) should play a role.
I do agree that there are huge differences between 1) Dr Spencer’s boxes, 2) a real greenhouse, and 3) the earth’s atmosphere. As such, this model here can do no more than highlight a few key issues. In this case, the key issue is “adding backradiation from a cooler plexiglas sheet warmed the warmer box.”
Hi Tim,
I apologize, I misunderstood your former post.
I believed that you were saying that the bottom increase of LWIR was the direct cause of the air warming.
Have a nice day.
Massimo
Dr. Roy
You deserve two new initials beside your name, CH.
Cat Herder
There is none so blind as he who will not see.
Thanks much.
Dr Spencer
I have an other question: How do you measure the “ambient temperature” ? Is there any possibility that the probe be influenced by the radiation from the sky or from the ground ? It should be possible to measure the “true air temperature” by using a few opaque isolating screens.
I like your: “true air temperature”.
If you are arguing that it doesn’t exist indeed, I agree.
The “air temperature” should be the result of a standardized method of measurements.
AFIK NOAA gave the rules, but it seems that often rules are made to be violated:
http://www.surfacestations.org/
Have a nice day.
Massimo
Now, if you want to show the result of Beer’s law, “double” your “CO2” by doubling the plexiglass, and then tripling it. That would be an excellent exposition….
Regards,
Brendan
@tim
“then the floor of the greenhouse will be warmer and will be better at warming the air as well.”
True. But this is only heat inflow. What about heat outflow? It is greatly reduced because heat conduction in closed box is much slower process than mass transfer of air in open box.
Analogy is car’s engine. It is heated by fuel combustion and cooled by the radiator. If you increased the engine speed, the radiator malfunctioned and the engine overheated. What is the cause of overheating? Increased engine speed or radiator malfunction? We know we can increase engine speed without overheating if the radiator is working properly. So we conclude the cause is radiator malfunction. Though increased engine speed actually increased the heat inflow.
Your other comments show the difference between a greenhouse and earth’s atmosphere. So the GHE cannot be simulated in a real greenhouse.
If Dr Spencer takes his apparatus inside, all will agree that all of its parts will eventually come to ‘room temperature’, regardless of glass windows, opaque windows, no windows. This is known as Kirchoff’s Law and follows from the second law of thermodynamics. The apparatus is now at Local Thermal Equilibrium, i.e., all temperature differences removed and all temperature driving forces disappear. How is this possible in a case where there is a window that transmits some photons and blocks others? Apparently the fact that some photons are blocked makes no difference as far a equilibrium temperature is concerned. Could it be that photons that are blocked are simply reradiated when equilibrium is achieved, at a rate that exactly cancels the blocking effect?
Out-of-doors equilibrium is never achieved because of constantly changing radiative conditions, so dynamic effects arising from device construction are possible. You could theoretically generate power by connecting a thermopile between two dissimilar devices and rectifying the energy flows between them. But I wouldn’t expect to get rich doing it. Although I did sell the Brooklyn Bridge once.
Dear Mr pochas
This is not the biggest problem.
The problem is the ability to store thermal energy.
The atmosphere stores 12.6 MJ per m^2 per degree centigrade, only sensible heat and more 90.88 Mj of latent heat and discharging 240 w/m^2 takes 139 hours to deplete.
The system of Dr. Roy is discharged at 0.65 seconds. What is observed is the system flow that penetrates the walls and roof. It have not the soil warm for to sustent the flux.
Pochas,
Well said. You are absolutely correct, in my (worthless) opinion.
This whole “blocking”, “storing”, “trapping” nonsense seems to be based on the caloric theory of heat.
May I respectfully point out to all that the caloric theory of heat was demolished by Joule (amongst others) by the simple expedient of performing simple qualitative experiments with extremely primitive equipment, and limited resources.
Wood’s experiment is a good example of using what was available a century ago, to establish a result which still stands. His use of a generally IR opaque sheet of glass interposed between the Sun and his boxes is an excellent example of “nullius in verba”. In other words, just because someone says the atmosphere is opaque to long wave IR, assume they are wrong, but set up your experiment to ensure that it will work as if the atmosphere was, indeed, opaque to long wave IR.
A rather clever experiment, in my (worthless) opinion. Rather like thinking about heat generated when boring a cannon, and then asking the proponents of caloric heat theory to “please explain”.
Live well and prosper,
Mike Flynn.
pochas says: “If Dr Spencer takes his apparatus inside, all will agree that all of its parts will eventually come to ‘room temperature’, regardless of glass windows, opaque windows, no windows.”
I would agree — if there are no local sources of heat. For example, if there is a light bulb that is turned on, then clearly the filament will not be in equilibrium with the room — in fact there will be a temperature gradient away from the bulb. In this case, it is also possible for the ‘hot’ photons from the light bulb to warm a box above the “room temperature” of the walls & floors and tabletops. But of course, then we are back to the ‘radiative greenhouse effect’.
“How is this possible in a case where there is a window that transmits some photons and blocks others? “
Well, if the rest of the room really is at equilibrium (eg no sunlight through windows or lightbulbs) then there will only be ‘one type’ of photon = thermal photons with a spectrum described by blackbody radiation due to the temperature of the room. So we don’t need to worry about any other ‘type’ of photon.
@pochas
“Out-of-doors equilibrium is never achieved because of constantly changing radiative conditions, so dynamic effects arising from device construction are possible. You could theoretically generate power by connecting a thermopile between two dissimilar devices and rectifying the energy flows between them”
This is not only theoretical. It is actually being done but not using thermopile. OTEC and geothermal heat pumps are based on this principle of extracting energy from temperature differential in the environment.
I have posted a response to Dr. Spencer, which is still in moderation. That post has a number of links to similar empirical experiments I have designed and run. The first of these is a cleaner version of the experiment Dr. Spenser has shown, involving vacuum pump and peltier chips. Those that replicate it will find that Dr. Spencer is correct and the two shell radiative model works.
The following five experiments show why the two shell radiative model does not work for a moving gas atmosphere in a gravity field, and why the net effect of radiative gases in our atmosphere is cooling at all concentrations above 0.0ppm. This is the core reason all the climate models fail against observation.
If that comment clears moderation, I would be happy to provide further diagrams and instruction to those wishing to replicate the experiments.
Would be good to see those experiments. BTW the GHE is directly observable. We can measure the LWIR coming down to the ground. We know this came from the atmosphere not the sun. Yes the atmosphere is cooling the surface but if the cooling slows down, the surface will warm.
Hi Konrad,
if you still see your comment in moderation, maybe that Dr.Spencer stopped to read this thread.
Try to re-post it changing the slash in the URLs with other unconventional symbols such as backslashes or a pipes. this should bypass the automatic moderation on the post.
I’m very interested in your experiments, thank you for did them.
Have a nice day.
Massimo
Dr. Stangelove and Massimo,
below is a repost of my comment in moderation. Links have “http//:” replaced with “#” and “.” replaced with “*”. My apologies for the inconvenience. I will follow this post with a simple summary of what the experiments show.
————————————————————————————————
Dr. Spencer,
firstly my congratulations on actually running an empirical experiment and making it simple enough for others to replicate. Your results are essentially correct, however the experiment can be designed better to eliminate gas conduction issues. Here is the version I use –
If you have access to both high torr vacuum pump and peltier cooling chips the experiment build is easy. Build two evacuated test chambers similar to this one –
#i44.tinypic*com/2n0q72w*jpg
– with internal matt black target plates and external SW illumination. An exploded view of the internals here –
#i43.tinypic*com/33dwg2g*jpg
– the only difference between the chambers is the matt black foil layer in chamber 1 between the target plate and the -25C base plate. Shown in cut away here –
#i43.tinypic*com/2wrlris*jpg
It does not effect the experiment that the target plate is illuminated from the “back”. Controlled sources of energy external to the chambers is important. The target plate in chamber 1 does reach a higher temperature.
Secondly, I believe you are still wrong. There is no net warming effect from radiative gases in our atmosphere. Radiative gases act to cool our atmosphere at all concentrations above 0.0ppm.
Here are five further simple experiments that I have built and run which together disprove the global warming hypothesis –
Experiment 1. Effect of incident LWIR on liquid water that is free to evaporatively cool.
Incident LWIR can slow the cooling rate of materials. Climate scientists claim that DWLWIR has the same effect over oceans as it does over land, and this is shown in many Trenberthian energy budget cartoons. Does the ocean respond to DWLWIR the same way as land?
– Build two water proof EPS foam cubes 150mm on a side and open at the top.
– Position a 100mm square aluminium water block as LWIR source 25mm above each cube.
– Position two small computer fans to blow a very light breeze between the foam cube and the water blocks.
– Insert a probe thermometer with 0.1C resolution through the side of each cube 25mm below the top.
– Continuously run 80C water through one water block and 1C water through the other.
– Fill both EPS foam cubes to the top with 40C water an allow to cool for 30 min while recording temperatures.
– Repeat the experiment with a thin LDPE film on the surface of the water in each cube to prevent evaporative cooling.
Here is an early variant of this experiment in which IR from cooling water samples was reflected back to the water surface – #i47*tinypic*com/694203*jpg
Experiment 2. Radiative cooling properties of CO2
CO2 can both absorb and radiate IR. Some of the energy CO2 is radiating to space is from intercepted outgoing IR from the Earths surface. Most of the net energy CO2 radiates to space is acquired from latent heat from condensing water vapour and conductive contact with the Earths surface. Could the radiation of energy from the atmosphere to space acquired by surface conduction or release of latent heat balance the energy intercepted from surface IR?
– Build two EPS foam boxes 250 x 250mm and 100mm deep, open at the top.
– Make a small 5mm hole in the bottom corner of each box to ensure constant pressure
– Place an identically sized matt black 200 x 200 x 2mm aluminium target plate in the base of each box.
– At one side of the interior of each box position a IR and SW shielded tube 200mm long containing a small circulation fan to cycle all the gas in the box through the tube.
– Position a thermometer probe with 0.1C resolution in each tube.
– Seal the top of each box with a frame double glazed with thin LDPE film.
– At equal distances above each box position a 50w halogen light source with sealed glass face.
– Use small computer fans to cool the glass face of each halogen globe to minimise LWIR emission.
– Fill one box with air and the other with CO2
– Wait for box temperatures to equalise then illuminate each target plate with the SW source.
– Record gas temperatures during 30min of heating for each box.
– Switch off the halogens and record gas temperatures during cooling.
Here is image of equipment for experiment 2. Bike tyre inflater cartridges are an easy source of dry CO2 – #i49*tinypic*com/34hcoqd*jpg
Experiment 3. The role of energy loss in convective circulation.
In describing convective circulation in the atmosphere the role of heating low in the atmosphere is often emphasised. Does cooling at altitude have an equally important role in convective circulation?
– Get a large glass container of hot water and mix a ¼ teaspoon of finely ground cinnamon into it.
– Wait until Brownian motion slows till the suspended particles are barely moving.
– Now suspend a beer can full of ice water in the top 50mm of the hot water to one side of the clear container.
– Observe any circulation patterns developing in the hot water.
Experiment 4. Convective circulation and average temperature in a gas column.
Most AGW calculations are for linear fluxes into and out of a static atmosphere. However the gases in our atmosphere move. Should these linear flux equations have been run iteratively on models with discrete moving air masses? The height of energy gain and loss in a gas column effects convective circulation. Does this effect the average temperature of a gas column?
– Build two sealed EPS foam boxes, 1000mm wide, 200mm deep and 1000mm high.
– Penetrate each box with a number of thin aluminium water heating and cooling tubes
– In box 1 position heating tubes on the lower right hand side and cooling tubes on the upper left hand side. Keep the heating tubes as close to the lower interior surface as possible.
– In box 2 position heating tubes on the lower right hand side and cooling tubes on the lower left hand side. Keep the heating and cooling tubes as close to the lower interior surface as possible.
– Make small thermometer probe holes in the face of each box in a number of different horizontal and vertical positions.
– Position 0.1C resolution thermometer probes in identical positions in each box.
– Start 1C water running through the cooling tubes in each box and 80C water running through the heating tubes in each box at around 1 litre a min. Record temperatures over 30 min.
– Cut water flow and equalise the temperature in each box. Reposition the thermometer probes and re run the experiment until a circulation pattern and average temperature can be obtained for each box.
Here is a diagram of the initial experiment – #i48*tinypic*com/124fry*jpg and an image of a later small variant in which the strength of cooling can be altered at the top and bottom of the gas column – #tinypic*com/r/15n0xuf/6
Experiment 5. Surface to gas conductive flux in a gravity field.
Climate scientists have claimed that under an atmosphere without radiative gases the radiative cooling of the surface will be greater (see also experiment 1). Does this mean the conductive cooling of the atmosphere in contact will be significantly higher? Is it correct to model the conductive flux between the atmosphere and the surface with the atmosphere modelled as a single body without moving gases?
– build two small EPS foam tubes with internal volume 75 x 75mm by 200mm high open at one end.
– For tube 1 cover the open top with LDPE film
– For tube 2 cover the open base with LDPE film
– on each tube attach a battery pack and a small 5V computer fan blowing across the outside of the cling film.
– On tube 1 add small legs on one side to tilt it to around 5 degrees off vertical.
– On tube 2 attach 50mm legs to allow its fan to move air freely across the cling wrap base
– Make multiple thermometer probe entry points along each tube for K-type probes from a dual probe thermometer.
– Place the thermometer probe position equal distance from the cling film for each tube.
– Equalise the internal temperature of each tube to room temperature by turning each tube cling film down and running the fans for 15 minutes.
– Now orientate the tubes so tube 1 has cling film at the top and tube 2 has cling film at the base.
– Place them on a shelf in a refrigerator with the fans running and close the door with the thermometer units outside.
– Use the probe differential button on the thermometer to observe the temperature differential between the tubes develop as they cool from room temperature over about 2 min.
– Remove the tubes from the refrigerator and allow them to equalise to room temperature again, move the thermometers to new positions and repeat the cooling run. Do this a number of times to build up a picture of the temperature differential at various distances from the cling wrap in each tube at the 2 minute mark.
Build the tubes small enough to fit within your refrigerator. If you have wire shelves, place a plate under each tube – #oi49*tinypic*com/akcv0g*jpg
Dr. Spencer, you should take particular note of experiment 5. It demonstrates why the bulk of a theoretical non radiative atmosphere that has trended isothermal would have it’s temperature driven by surface Tmax not surface Tav.
Dr. Strangelove and Massimo,
The reason why adding radiative gases to the atmosphere will cause cooling rather than warming is not due to problems with radiative physics except the unusual effect demonstrated in Experiment 1. The problem with the AGW hypothesis lies in calculating for radiative exchange only while largely ignoring gas conduction and the role of radiative gases in the fluid dynamics of a gas atmosphere in a gravity field.
The first experiment shown (un-numbered) is a cleaner version of the experiment Dr. Spencer built. This is a two shell radiative model, simulating radiative exchange between surface, atmosphere and space for both a radiative and non radiative atmosphere. The results are essentially the same as Dr. Spencer claims. The target plate in chamber 1 gets hotter due to the additional radiative foil layer between the target plate and cooling plate. The radiative physics behind AGW concerns are basically correct.
Experiment 1 demonstrates a condition for which the radiative two shell model fails. The gas / liquid interface at 1 bar gas pressure for liquid water that can evaporatively cool is a special condition. Incident LWIR does not effect the cooling rate of liquid water in the same manner as other materials. The effect of absorption of incident LWIR on the cooling rate of liquid water cannot be calculated from the emissivity of the liquid. That is to say downwelling LWIR is a reality, however it has little effect over 71% of the earth’s surface.
Experiment 2 serves to remind climate blog readers that CO2 both absorbs and emits LWIR. Many concentrate on the ability of radiative gases to absorb and thermalise surface IR, while ignoring the role these gases play in releasing energy from the atmosphere to space.
Experiment 3 demonstrates that energy loss from the top fluid column can drive vertical convective circulation. Many blog readers associate convective circulation with heating only.
Experiment 4 is important. It shows that for tall gas columns, varying the height of energy gain and energy loss can effect the average temperature of the gas column, even though the energy input and output from the gas columns may be identical. Smaller scale experiments give differing results, as entrainment occurs allowing vertical circulation to continue even though energy input and output may be at the base of the column. As the column gets larger and taller, entrainment breaks down and layering occurs. In Experiment 4 with gas columns at 1m tall, box 2 runs hotter. The larger the experiment gets, the more marked the difference between the two columns. This is due to the fact that gas conduction does not scale. A tropospheric gas column could be considered to be 10~15 km tall.
Experiment 5 shows one of the common errors of climate scientists, failing to understand that surface to gas conductive energy flux is biased by gravity. The surface is far better at conductively heating the atmosphere than it is at conductively cooling it.
At the time of conducting these experiments, I was unaware that Dr. Spencer had written a post in 2009, covering a number of these points. He correctly claimed that a theoretical non-radiative atmosphere would not have strong vertical convective circulation below the tropopause, as without radiative cooling at altitude hot gases could not lose buoyancy and subside. He also correctly claimed that the observed tropospheric lapse rate would disappeare in this stagnant atmosphere as gas conduction would drive it isothermal.
Where I disagree with Dr. Spencer is his claim the the temperature of this theoretical isothermal atmosphere would be set by surface Tav, and this surface Tav would be significantly lower than present. Experiment 1 demonstrates that the surface Tav under a non radiative atmosphere would
not be as low as Dr. Spencer assumed. Experiment 5 demonstrates that the temperature of the theoretical isothermal atmosphere would be driven by surface Tmax not surface Tav. This means that the bulk of an atmosphere without radiative properties would actually run far hotter than our current atmosphere. Therefore radiative gases act to cool our atmosphere, they do not have a net warming effect. The claim that our atmosphere is 33C warmer than it would be without radiative gases is incorrect.
Radiative gases do warm our lower atmosphere, and do slow the cooling of un-vegetated land surface. However the role these gases play in atmospheric cooling-
1. energy loss to space from the upper atmosphere
2.driving convective circulation and mechanical energy transport from the surface
3. the resulting vertical circulation reducing average temperatures through pneumatically induced lapse rate
– far outweigh their warming effects.
Hi Konrad,
first of all thank you for take your time to do those experiments which surely are more convincing than thousand simulations.
I’m not able to open any of the pictures you linked, because for an unknown reason I can’t access the tinypic.com web-site at all.
I’ll try again tomorrow.
Anyways, maybe you have an answer to a question I have from a long time unsolved.
You stated that a GHGs free atmosphere must be isothermal. Well, what I can’t figure out is how could that atmosphere have molecules which run at the very outer layer absolutely tangent to the atmosphere, getting their energy from the layers below.
I mean, in gases only the molecular KE is the temperature, since the molecules have to stop their vertical run at a certain point (otherwise they escape from the planet gravity field), the only way they have to keep their very same KE is to convert the vertical motion in horizontal one, but how do they can do it?
I’m unable to realize how an object (even small as a molecule) could receive a vertical push and transform it to an horizontal run.
Have a nice day.
Massimo
Massimo,
As to a non radiative atmosphere being isothermal, this is simply a generalisation for what would happen to the bulk of the atmosphere below the level currently called the tropopause.
Dr. Spencer has previously indicated (2009) that he also believes the bulk of the troposphere would trend isothermal in the absence of radiative gases, with only a thin near surface layer experiencing extreme diurnal temperature variation.
What many do not understand is that the observed lapse rate below the tropopause is a result of strong vertical circulation of gas exhibiting a vertical pressure gradient in a gravity field. With vertical circulation comes pneumatic heating and cooling. Horizontal gas conduction equalises the resulting vertical temperature changes in rising and descending gas across the troposphere.
Strong vertical convective circulation in the troposphere depends on radiative gases. Radiative energy loss at altitude allows heated gases that have risen to lose buoyancy and subside. If this radiative ability is removed, conductively heated gases could rise, but not descend. Further heated gases rising simply layer under previous buoyant air masses without over turning. (overturning and entrainment can be observed in small experiments, but as experiment 4 shows, increasing the height of a gas column causes divergence from these results and layering occurs.)
In such a stagnant atmosphere, gas conduction would slowly equalise the temperature of the bulk of the atmosphere, causing it to trend isothermal.
Experiment 1 shows that the surface Tav under such as theoretical non radiative atmosphere would be higher than Dr. Spencer supposed. Experiment 5 shows that a figure closer to surface Tmax should be used to determine the temperature of such a non radiative atmosphere. Together this means that the bulk of such a non radiative atmosphere would be far hotter than our current atmosphere. This would mean that the net effect of radiative gases is cooling at all concentrations above 0.0ppm
D’Avila Tarcisio says:
August 30, 2013 at 1:40 PM
“The problem is the ability to store thermal energy.”
Correct. We have a situation in which the equilibrium temperature fluctuates up and down daily and the apparatus is simply trying to follow the equilibrium temperature.
Mike Flynn says:
August 30, 2013 at 5:42 PM
“Well said. You are absolutely correct….
This whole “blocking”, “storing”, “trapping” nonsense seems to be based on the caloric theory of heat.”
Thanks for the kind words, but my archaic thought habits continue to include the concepts of heat capacity, thermal conductivity, enthalpy, entropy, system time constants, thermal spectra, etc.
Tim Folkerts says:
August 30, 2013 at 5:45 PM
“Well, if the rest of the room really is at equilibrium (eg no sunlight through windows or lightbulbs) then there will only be ‘one type’ of photon = thermal photons with a spectrum described by blackbody radiation due to the temperature of the room.”
Exactly, at equilibrium the thermal distribution is blackbody (to a close approximation) regardless of the material and in the Woods-type experiment we are never far from thermal equilibrium, even though the inside apparatus temperature may deviate from ambient due to restricted convection.
To Dr Spencer:
I’m a little puzzled about your statement that the test enclosure temperature going below ambient at night somehow shows the “Greenhouse Effect” Perhaps I misunderstand. I have two temp transmitters outside, one mounted at 1 ft height, the other at 6 ft height. At night the low rtb reads below the high rtb. In daytime the situation reverses. I attribute this to reduced convection around the lower rtb. Are you not looking at the same phenomenon in your apparatus? At night the apparatus radiates to space but there is no convection inside the enclosure to mix the cold inside air with the ambient. Am I missing something?
“We have a situation in which the equilibrium temperature fluctuates up and down daily and the apparatus is simply trying to follow the equilibrium temperature.”
Which apparatus? The various “greenhouses”? They do NOT simply try to follow ‘the equilibrium temperature’.
For one thing, there is no ‘equilibrium temperature’. Even ignoring the variations with time, the solar photons are ‘at 5780 K’, while the thermal IR from the sky is ‘rather cold’. A better way to say it would be that each object tries to achieve its own thermal balance. During the day, a black surface will balance at a higher temperature than a white object; a high emissivity object will balance at a lower temperature than a low emissivity object; a box covered with plexiglas will balance at a higher temperature than a similar box covered with saran wrap.
“… in the Woods-type experiment we are never far from thermal equilibrium, even though the inside apparatus temperature may deviate from ambient due to restricted convection.”
If the whole system is at equilibrium (eg a dark, closed room in a climate-controlled building) then restricted convection cannot make the interior of the box warmer. Only with the inclusion of “hot photons” (eg from a light bulb or the sun) will make the interior warmer.
Dr. Spencer,
I have a question concerning your last plot of Box1 minus ambient in your UPDATE section. When I compare it to the data in the preceding plot, shouldn’t Box1 (red) minus ambient (black) be a larger difference at the BEGINNING of the experiment and taper down to near ZERO at the end of Box1 plexiglas? The last plot looks like Box1 minus Box2 instead. If not, where am I confused?
Nice one, Roy!
I think you’ve invented double glazing!
This is a very good teaching tool, not just because of the application of physical principles, but also experimental design. Regarding the former, has anyone commented on the rise in the ambient temperature before sunrise? Notice how the box temperatures both rise concurrently. Does this represent conduction through the styrofoam?
Also, there was a suggestion by Genghis at comment 28 and Brendan at comment 44 to increase the thickness of the plexiglass analogous to adding CO2 to the atmosphere. Will box 1 minus box 2 remain the same or show a Beer’s Law effect?
Would be interesting to see how the results wouldchange if two sheets of plexiglass were used, both stacked closely and with a gap from each other.
How is this experiment supposed to show us anything if you can’t even get two boxes, equal in design/construction, to act the same way? Is one box shading the other? Is one box protecting the other from wind, etc.? Is the density of the Styrofoam the same? Did you weigh them? How can the slight difference in the texture of the paint on one be causing the drastic difference? Is the paint the same thickness on each? Something is missing from the equation.
This reminds me of the problem I have with the “global” temperature. I have several millimeters that are accurate to the 3rd decimal point. However, they display to 5 and in some cases 6 decimal points (Being a mathematician I ask why as that is garbage) Then, they only have a specified (rated) repeatability of 4 decimal points. That means that the last displayed digit(s) is absolutely worthless, and should be ignored – always. Yet all of the “global” temperatures take all of these digits to the last readable digit and factor them into this meaningless number (it is not an average of the world – it can’t be.) It has too many unknown and inaccurate numbers averaged together.
My spell checker magically corrected multi-meter to millimeter. please use the term multi-meter.
@konrad
“Therefore radiative gases act to cool our atmosphere, they do not have a net warming effect. The claim that our atmosphere is 33C warmer than it would be without radiative gases is incorrect.”
The first sentence is correct. The second is wrong. It seems you could not believe that slowing down the cooling can increase the temperature (warm) of the object that is being cooled.
A simple example to illustrate the point. We all know the car’s radiator has a cooling effect on the engine. Reduce the radiator water to one-fourth the normal level. The engine will heat up. Is the radiator warming the engine? No. It’s still cooling the engine but the cooling rate has slowed down.
The 33 C warmer atmosphere is based on earth’s radiative energy balance as viewed from space given the observed average surface temperature. Convective and conductive heat transfers are neglected because earth can only transmit heat to space by radiation. The calculation is very basic radiation physics and uncontroversial.
Dr. Strangelove,
You clearly have not conducted similar empirical experiments yourself. Even worse, you clearly have not compared my two shell radiative experiment with Dr. Spencer’s sub-standard offering. Sorry to disappoint, but despite Dr. Spencer being far better at radiative physics than I, when it comes to designing and conducting empirical experiments, Dr. Spencer is not in my league.
The error in the the AGW hypothesis does not lay in radiative physics*, but rather gas conduction and fluid dynamics.
Your argument-
“Convective and conductive heat transfers are neglected because earth can only transmit heat to space by radiation.”
– Gives a clear indication of where every climate pseudo scientist got it wrong. We are concerned about atmospheric temperatures. The atmosphere is a mobile fluid mass in a gravity field. When it comes to near surface temperatures convective and conductive energy transfers cannot be ignored. No if, no but, no maybe. After all, radiative gases govern the speed of convective circulation below the tropopause. Convective circulation drives mechanical energy transfer, the primary mechanism for transporting energy away from the surface.
Lets review the “do nots” of atmospheric modelling –
A. Do not model the “earth” as a combined land/ocean/gas “thingy”
B. Do not model the atmosphere as a single body or layer
C. Do not model the sun as a ¼ power constant source without diurnal cycle
D. Do not model conductive flux to and from the surface and atmosphere based on surface Tav
E. Do not model a static atmosphere without moving gases
F. Do not model a moving atmosphere without Gravity
G. Do not model the surface as a combined land/ocean “thingy”
H. Do not apply SB equations to a moving gaseous atmosphere (the davidmhoffer rule 😉
Dr. Strangelove, you appear to be arguing that adding radiative gases to the atmosphere will reduce the atmospheres radiative cooling ability. Have you thought this through?
*there is a small error in the radiative physics used by AGW believers. Incident LWIR does not effect the cooling rate of liquid water in the same manner it effects other materials (Experiment 1. No, don’t argue, just do it.). But as I often say, given that this only effects 71% of the earth’s surface, as problems with the AGW hoax go, it’s probably not a “biggy” 😉
[“Therefore radiative gases act to cool our atmosphere, they do not have a net warming effect. The claim that our atmosphere is 33C warmer than it would be without radiative gases is incorrect.”]
“The first sentence is correct. The second is wrong.”
Q. How can the first sentence be correct and the second sentence wrong?
The first sentence states that radiative gasses do not (net) warm the atmosphere.
The second sentence states that radiative gasses do not warm the atmosphere.
What is the problem?
Arfur, the main problem is that the original statement and the critique are both too vague and open to misinterpreation.
**************************************
Radiative gases act to cool our atmosphere when they radiate to space.
Radiative gases act to warm our atmosphere when they absorb IR that came from the ground..
They [radiative gases] do not have a net warming effect on the earth as seen from space.
They [radiative gases] do have a net warming effect on the earth’s surface and the bottom of the atmosphere..
They [radiative gases] have a net cooling effect on the top of the atmosphere..
*************************************
Since the radiative gases cool the top of the atmosphere and warm the bottom, there is no simple answer for the net effect on the atmosphere as a whole. I’m sure it could be calcualted based on various models, but it is not a very useful thing to know.
The claim that the overall atmosphere is 33 C warmer with GHGs is clearly wrong. The claim that the bottom of the atmosphere is 33 C warmer with GHGs is basically correct (give or take a few details and approximations).
“Since the radiative gases cool the top of the atmosphere and warm the bottom, there is no simple answer for the net effect on the atmosphere as a whole.”
Tim,
what happens when you heat the bottom of a fluid column in a gravity field while cooling the top? What is the primary mechanism for transporting energy from the surface to the upper atmosphere? Convective circulation.
What is the primary mechanism for energy loss from the upper atmosphere? Radiative gases.
Radiative gases therefore have a critical role in reducing near surface temperatures during the day. They have a more minor role in increasing surface Tmin at night, but only over land.
Adding radiative gases to the atmosphere will simply increase the speed of convective circulation below the tropopause. Due to the diurnal cycle this Rayleigh Bernard circulation pulses at the surface level, with the Rayleigh number for air masses having to be exceeded after dawn each day for air masses to break away. All that doubling CO2 in the atmosphere would do is reduce the average time to the Rayleigh number being exceeded by a second or two. Average near surface temperatures would fall, but this would be by an immeasurably small amount.
Go back and look at the “basic physics” of the “settled science”. Did they calculate the change in radiative flux for increasing radiative gases only without calculating the change that would also occur in the speed of convective circulation? That one mistake invalidates the whole of the AGW hypothesis.
I should have been a little more clear and said:
“Since the radiative gases cool the top of the atmosphere and warm the bottom, there is no simple answer for the net effect on the AVERAGE TEMPERATURE OF THE atmosphere as a whole.”
I agree with much of what you then say about convection, but I would suggest you are simplifying a little too much when you claim: “Adding radiative gases to the atmosphere will simply increase the speed of convective circulation below the tropopause.” Adding radiative gases will ALSO increase the height from which the IR is emitted to space. This means the gas rising by convection must rise higher before it can dump energy to space. And subsequently, the falling air must warm more on the way down because it is falling farther.
We have two effects that in part counteract each other. Deciding which “wins” by looking at the fluid dynamics would be non-trivial. But from conservation of energy the answer becomes clearer. With more GHGs, the higher top of atmosphere will emit less IR to space because it is radiating from a colder place. To balance out the constant input from the sun, *somewhere* else emit more IR to space. That ‘somewhere’ would be the surface, since it is the other main source of IR to space besides the ToA.
Tim,
this is sounding like the old effective radiating level or ERL argument. It won’t wash. Hot gases rising through the atmosphere are always radiating more strongly than the gases at the altitude they are rising through. They are always radiating above the level of maximum IR opacity in the atmosphere.
Tim,
I think you are erecting a strawman argument when you distinguish between the ‘overall’ atmosphere and the ‘bottom’ of it.
The AGW climate debate is about the GHE as caused by radiative forcing of radiative gasses. The GHE temperature measured is the near-surface temperature.
When you say ‘radiative gasses’ do you mean non-condensing gasses or do you include water vapour? Whether you do or not makes a difference.
Water vapour has not increased in the last thirty years (I can’t find data before 1983) but non-condensing GHGs have. Therefore, in this regard, water vapour can be ignored when it comes to the loose term ‘radiative gasses’ for the purpose of this small section of the debate.
So your comment…
[“They [radiative gases] do have a net warming effect on the earth’s surface and the bottom of the atmosphere…”]
…warrants investigation.
What is the GHE today? Let us assume the IPCC (and Konrad is wrong) is correct and it is 33C
Fact: the near-surface temperature rise (Enhanced GHE?) since 1850 (start of IPCC data) is 0.8C (+/-0.1). A small increase (2.5% of the 1850 GHE).
Fact: the measured increase of non-condensing radiative gasses is at least 40% (from appx 40% CO2, over 60% CH4 etc). A big increase.
Fact: no-one (repeat no-one) can say with any evidential support how much of the 0.8C rise is due to the increase in nGHGs
Therefore all ANYONE can say with any credibility is that the large increase in radiative gases (non-condensing) MAY have added an UNKNOWN portion of 0.8C to the planet’s GHE if such a thing exists. (I prefer Atmosphere Effect).
Folk MUST realise that the damning of the ‘radiative forcing = cAGW’ argument is right here in the observed data. There is just NO correlation in the rise of a supposed (assumed) ‘key radiative gas player since 1850 and the meagre increase in surface temperature. If the radiative gasses you refer to were truly effective, the temperature would be higher. It is no more complicated than that. Therefore, it is the theory that is wrong. Konrad is right on this.
I know some folk will talk about ‘feedbacks’ and ‘inertia lag’ as if these factors exist (although they will not provide evidence) but what they have to take into account is that the current ‘enhance GHE’ takes into account all feedbacks and any lag. The observed data do not support the hypothesis of either a warming radiative effect or inertia lag and feedback.
Therefore my question regarding Dr Strangelove’s comments is still valid. Both parts of Konrad’s comment say the same thing, so how can the first part be correct and the second part wrong?
Regards,
Arfur says: “Therefore my question regarding Dr Strangelove’s comments is still valid. Both parts of Konrad’s comment say the same thing, so how can the first part be correct and the second part wrong?”
My point still stands as well. There are legitimate interpretations of the ambiguous comment and the reply that where the first part is correct and the second part wrong.
“Folk MUST realise that the damning of the ‘radiative forcing = cAGW’ argument is right here in the observed data.
I would suggest you are introducing your own strawman here. There are two different issues.
1) Do GHGs have any affect on surface temperature?
2) If the answer to (1) is ‘yes’, how much effect has the last 50 ppm CO2 had on the climate.
To me, there is no possible way to answer ‘no’ to question 1. The fundamental thermodynamics are simply too well established. Question 2 is much more interesting and subtle. For one thing, the first 50 ppm of CO2 would have a much larger impact than the last 50 ppm. Additionally, feedbacks or changes in the sun or changes in cosmic rays or changes in land use could all serve to hide the effect of recent changes in CO2.
Or put another way, I think it is open to debate whether climate sensitivity is 0.1 C/doubling vs 5 C/doubling, but I see no possible way to conclude that the sensitivity is identically zero.
Ok, Tim, your comment above is a fair comment overall.
I see no point in arguing over the ambiguity of Konrad’s comment. You think it is ambiguous and I think it was clear. Say no more. I was arguing more with Dr Strangelove’s comment anyway.
As to your more interesting comments:
[“There are two different issues.
1) Do GHGs have any affect on surface temperature?
2) If the answer to (1) is ‘yes’, how much effect has the last 50 ppm CO2 had on the climate.”]
1) You say there is no way to answer ‘no’ to Q1. I agree in principle with the following major caveat: GHGs can radiate to molecules (in this case surface molecules) but the only way the radiation will be absorbed for net energy gain is if the surface molecules are cooler than the GHGs. So, yes, there may be a few places on the surface of the globe where warm air radiates – and that radiation is absorbed – to a cooler surface. However, taken globally, GHGs do not affect the surface. temperature.
2) As the answer to 1) was effectively ‘no’, this question is mute. However. I will answer it anyway. I don’t know. I agree with your running logic but you ASSUME the initial proposition – that CO2 has a significant effect. If so, then of course the logarithmic nature of its effect will incur a progressively decreasing effect. The data still supports that this effect has by now reached practically zero.
[“Additionally, feedbacks or changes in the sun or changes in cosmic rays or changes in land use could all serve to hide the effect of recent changes in CO2.”]
Only if you assume your initial proposition as above. You completely ignore the logical option that the lack of observed correlated warming COULD be due to the complete INEFFECTIVENESS of CO2 as a warming agent.
In other words, is there no warming because it’s hidden or because it was never there?
I ask you again to look back to 1850. Was there a radiative GHE then? With 280ppm CO2? And yet a large increase to 400ppm CO2 has hardly made any difference – if at all.
Even if the CO2 effect is not zero, the data suggests it is perilously close to zero.
Regards,
Arfur
Why are we still having this debate? The atmosphere obviously slows the path of IR to space. However it doesn’t warm the surface to any higher temperature in doing so (although the atmosphere itself may be warmer). The greenhouse analogy is a poor one and a dam on a river would serve better. The river is dammed and so the water behind it rises whilst the water being let through the dam is less than that flowing to it. When the dam gates are opened to allow an equal flow of water through the dam as flows to it, the water level behind the dam stays constant. At no time does the source of the river have to increase in flow to maintain the level of water in the dam. So is like the atmosphere; whilst the IR may need to flow in a round about path (progressing in a two steps forward and one step back fashion) it is still essentially all flowing one way and one way only. The notion that it is all finding its way back to the surface and increasing its temperature directly is pure and utter nonsense. Anything else would be akin to perpetual motion.
Good question, Iansview.
My answer is that we are still having this debate because there are a lot of people – some with lots of letters after their name – who prefer to believe that the ‘theory’ of ‘CO2 radiative forcing = cAGW’ is correct without any real evidence to support it.
Hence they prefer to try to make the pieces fit, a bit like using a hammer to make the jigsaw puzzle complete, instead of re-visiting the ‘theory/hypothesis/postulation/whatever’.
The debate just keeps going round in a never ending whirlpool of assumptions and dogma – while the planet just keeps disagreeing with the ‘experts’.
🙂
Just one dam analogy after another. 🙂
Seriously, though, this is a pretty good analogy, but the conclusions are a bit muddled. (I would go with an analogy of a large storage take instead).
The water is analogous to energy; the depth is analogous to temperature. The tank (earth) receives a fairly constant input of high-pressure water (energy). The water will start to fill the tank, but there are “leaks” that allow water to leave the tank (eg IR to space).
IR gases ‘plug up’ some of the leaks because certainly wavelengths of IR energy can no longer escape to space. This will necessarily cause the water to start backing up and getting deeper until the remaining leaks can emit the same energy.
“At no time does the source of the river have to increase in flow to maintain the level of water in the dam.”
And furthermore, at no time does the source have to increase in flow to raise the level of the water if some of the leaks are plugged. Ie the same amount of sun light that would warm a “bare earth’ to 255 K can also raise the temperature to 288 if the outgoing IR is blocked by the atmosphere.
“So is like the atmosphere; whilst the IR may need to flow in a round about path (progressing in a two steps forward and one step back fashion) it is still essentially all flowing one way and one way only.”
In the original analogy, we should more precisely consider the GHGs not as ‘blocking’ the IR, but rather shunting it off to another tank — a tank that is ALSO leaking to space.
So the analogy is that water is poured into one tank. Instead of that tank draining to the ground, it drains to ANOTHER tank which drains to the ground. If the tanks are identical, the water in the second of two tanks will be as deep as the water in the lone tank. The first tank of the pair will be *deeper* than the lone tank.
Tim,
Simple question…..whilst Oxygen and Nitrogen may not block IR from leaving the atmosphere do they not radiate energy to space just like the ‘radiative’ gases?
Roy,
Sorry to say that the Plexiglass sheet does inhibit a lot of convection, because it is way to close to the plastic wrap compared to the dimensions of the sheet. Try to stay under the same surface/gap ratio glass “umbrella” under the sun, you will shorty notice it.
The only way you can come out of something reliable with our experiment is to stick to Wood experiment, by using a KCl salty sheet and a glass sheet – instead of the plastic wrap – with the same thickness. Further more, these two materials have very similar thermal conductivity so that the convection/conduction thermal exchanges will nearly identical.
Also, you should measure the temperature of the bottom (the earth surface temperature).
Dr. Spencer,
“For example, 3mm clear glass: 83% of solar radiation is transmitted, 8% reflected and 9% is absorbed of which 3% is radiated inside and 6% outside.”
http://www.silentwindows.com.au/measuring_the_benefits/
You should check the temps of both sides of the plexi to find out how much difference there is in temp as a real Greenhouse is warmer inside and the plastic or glass will radiate more inside due to this temp differential!! You mention there was a breeze so there probably was not as large a differential in you experiment.
Since GHG’s have no BIAS to where they radiate this is a small but absolute difference between a physical Greenhouse and the atmosphere.
In case it hasn’t been posted by someone else here is a Green House Experiment that found varying effects for IR materials:
http://www.hort.cornell.edu/hightunnel/about/research/general/plastic_comparisons_reid.pdf
h/t Carrick
KuhnKat
KuhnKat
This slightly later report from the same source came to a different conclusion.
They found that occasionally at night a slight 1 to 3 degree F warmer effect from IR blocking polyethylene but it was not consistent over time.(page 2)
Sometimes it was the other way round with temperatures less than ambient.
They concluded that it made no sense to pay extra for IR blocking plastic.
This report makes uncomfortable reading for the greenhouse effect enthusiasts.
No 33K greenhouse effect
At best a occasional negligible effect (maximum 1.5K) but on most nights no effect whatsoever
http://www.hort.cornell.edu/hightunnel/about/research/general/penn_state_plastic_study.pdf
Bryan,
I guess you don’t understand the term variable!! Sorry if I was too technical for you.
These reports also did not quantify the effect of moisture on the covers whether glass, plexi, or film. A report I am not finding again showed that film treated with a chemical to decrease the surface tension of water allowing it to slide off or be a thinner layer was substantially more effective than IR treated film!!
KuhnKat
I fully understand the term variable.
There seems to be as many versions of the ‘greenhouse effect’ as advocates of this hoax.
How’s that for variable?
If the greenhouse effect is simply the fact that some gases are more IR active than others then everyone (almost) could agree.
However the ‘greenhouse effect’ that is sold to the public, is of a majic radiative effect causing a 33K increase in the Earth’s surface temperature is bogus.
R W Woods 100 year old experiment still stands.
Typo Majic should read magic
konrad
“The error in the the AGW hypothesis does not lay in radiative physics*, but rather gas conduction and fluid dynamics.”
What exactly is the error? You don’t believe the 33 C greenhouse effect? Do you know how the 33 C was computed? By computing earth’s radiative equilibrium temperature without an atmosphere and comparing it with earth with atmosphere. How can gas conduction and fluid dynamics play any part when there’s no atmosphere?
In the earth with atmosphere scenario, the surface temperature is based on thermometer measurements. Where’s the error in gas conduction and fluid dynamics? The thermometer can only measure radiation, not conduction and convection? Ridiculous.
“When it comes to near surface temperatures convective and conductive energy transfers cannot be ignored.”
Only a pseudo scientist thinks convective and conductive energy transfers are ignored. Surface temperatures are measured by thermometers. What is computed using radiative transfer equations is TOA fluxes because no more conduction and convection at TOA.
Conduction and convection can only move heat within the atmosphere. What determines earth’s energy balance is radiative heat transfer. You cannot remove a 5 W/m^2 energy imbalance at TOA by conduction and convection.
“Incident LWIR does not effect the cooling rate of liquid water in the same manner it effects other materials”
Clearly pseudo science. Countless experiments have proven liquid water absorbs IR at 6-25 microns. When radiation is absorbed by water, it is converted to thermal energy. Have you never observed a swimming pool heated by the sun? I will waste time anymore. This is very basic stuff.
BTW water is nearly perfectly transparent to light so all that heating in your swimming pool is almost entirely due to IR. Before we try to overturn climate science, let’s try to learn first the basics.
Uh, wouldn’t that mean the sunlight incident upon the bottom/sides of the pool does almost nothing?
I mean, about half of sunlight is IR anyways, but I think you were saying that the pool is heated by what… IR from the air above it?
Dr. Strangelove,
nowhere did I claim that incident IR was not absorbed by liquid water. What I claimed was that incident IR does not slow the cooling rate of liquid water, a very different thing. Long wave IR is absorbed but cannot penetrate more than 10 microns of the surface. This is a small percentage of what is known as the skin evaporation layer. All that incident LWIR does to liquid water is “trip” some molecules in the skin evaporation layer in phase change to water vapour sooner than they otherwise would. It has no effect on the cooling rate of the water below this layer.
Type is cheap, if you doubt my words, build and run Experiment 1 listed above for yourself. If experiment 1 looks too difficult, you can try a simpler experiment. Try heating a plastic container of water with a hair dryer. Try by pointing the hair dryer at the surface of the water, then try pointing it at the side of the plastic container. This should at least teach you about the skin evaporation layer.
Downwelling LWIR in does not effect the cooling rate of liquid water in the same manner that it effects other materials. Calculating the effect of LWIR over the oceans based on the emissivity of liquid water was just one of the mistakes of the AGW believers.
You ask –
“Do you know how the 33 C was computed? By computing earth’s radiative equilibrium temperature without an atmosphere and comparing it with earth with atmosphere.”
Please go back and read the list of atmospheric modelling “do nots” I listed above. I believe first on the list is “do not model the earth as a combined surface, ocean and atmosphere thingy” Why not? Because you get the wrong result for atmospheric temperatures. You must model the atmosphere as a moving fluid body with a pressure gradient to get the right answer. Radiative physics alone can not do this.
Nitrogen and Oxygen are the “greenhouse” gases on this planet. CO2 and H2O are the broken panes in the green house.
Konrad 6:44am – I agree with you that “Type is cheap, if you doubt my words, build and run…” an earth sized experiment.
You are obviously & easily confused by your small non-ISO experiments and lack of ISO attention to their details and their theoretical basics. Improve with ISO data from experts. Do not look at GHCN surface temperature data just yet.
Here is an experimental plan for your further testing that will beat Dr. Spencer backyard stuff above (that proves your assertions incorrect in their own way), the plan will be tough, takes sacrifice and commitment but will be rewarding and you will find the truth:
1) Measure the emissivity of actual in situ earth’s land and ocean surface, sampling enough variety to make statistics work for you. This has been done by experts, once done, you can check their work. I recommend using a portable radiometer looking down maybe about 3-5′ above earth’s surface.
2) Measure the atm. emissivity at surface. Travel to dry, arctic regions and humid, tropic regions looking up from surface with your precise, calibrated instrumentation. Sample enough locations to make statistics work for you. This has been done by experts, once done, you can check their work.
3) Save on the cost of a satellite, use NASA. Use accurate data reduction techniques of satellite (ERBE) public data from March 2000 to May 2004 to find earth’s albedo from the upwelling SWIR and temporally & spatially avg. This has been done by experts, once done, you can check their work.
4) Use the same satellite, same period, but check the incoming solar SWIR data, and temporal and spatial avg. This has been done by experts, once done, you can check their work.
NB: Enroute in doing this, looking also at upwelling LWIR measurment, you will find this ERBE time period reports an imbalance of around 6 W/m^2 out of some 340. Deal with it, look up several science papers that do so & find ways to understand the issues.
5) The easy part. Sit in an armchair with several modern texts on atm. thermo, including radiative, convective, conductive energy transfer. Touch up on the atm. thermo physics. Read up also on statistics and esp. “confidence intervals” (CI).
6) Use a simplified, basic 1st thermo law formula you find in all these texts to compute atm. near surface mean temperature from all the measurements you have & is computed out at ~289K about 1.5m above the mean surface. Compute the CI.
7) Then, and only then, compare to the GHCN mean temperature March 2000 to May 2004 and find they measured mean T ~288K about 1.5m above the mean surface. Compute the CI.
Then, in the face of all this ISO data, compare your results to your 6:44am assertion: “Because you get the wrong result for atmospheric temperatures.”
Improve this assertion by more precisely referring to atm. near surface temperature where humans live and breathe.
Find being wrong by 1K out of 289K is not too shabby even given the CI. As an engineer I can accept that result while examining CI, make it useful and be done with it; as a scientist I will apply for research funding to find why the 1K difference exists and how to reduce the CI.
“You are obviously & easily confused by your small non-ISO experiments and lack of ISO attention to their details and their theoretical basics.”
Go back and look at the experiments I listed and in particular the three construction diagrams for the two shell radiative experiment. Compare it to Dr. Spencer’s effort. Get the picture?
The experiments labelled 1 to 5 are indeed simple. These are collection of experiments that were posted over time typically in response to AGW believers at the Talkshop. Some of these are variants of more sophisticated designs that have been designed so they can be replicated using simple equipment by other blog readers. I do not expect everyone to have access to water pumps, high torr vacuum pumps or Q-Cell insulation. The physics they demonstrate is so basic that greater sophistication is not required.
Before I deal with “experts”(call to authority), “GHCN”(corrupt data) or planet scale experiments (unavailable technology), perhaps you could give your yes or no responses to the following six very simple questions –
1. Do radiative gases such as H2O and CO2 both absorb and emit IR radiation? Yes or No?
2. Are Radiative gases critical to strong vertical tropospheric convective circulation? Yes or No?
3. Does altering the quantity of radiative gases alter the speed of tropospheric convective circulation? Yes or No?
4. Is convective circulation the primary mechanism for transporting energy from the lower atmosphere to the upper atmosphere? Yes or No?
5. Are radiative gases the primary mechanism for energy loss to space from the upper atmosphere? Yes or No?
6. Does LWIR emitted from the atmosphere in the 15 micron band significantly effect the cooling rate of liquid water that is free to evaporatively cool? Yes or No.
Six honest and direct Yes or No answers. That is all that is required.
Konrad 8:48am: “Six honest and direct Yes or No answers. That is all that is required.”
Fair enough; I posted only in response to Konrad’s assertion ““Because you get the wrong result for atmospheric temperatures.” to show the text book “wrong result” is off by only 1K out of 289K all with measured data and using just the 1st thermo law: conservation of energy for the earth near surface thermo system.
Here, Konrad asks questions way beyond this scope but they are interesting and arguably somewhat relevant to top post.
Off the top of my head I can answer a couple, will find a cite for the harder ones, other informed, critical posters might help if it takes me some time.
1. Do radiative gases such as H2O and CO2 both absorb and emit IR radiation? Yes or No?
Yes, observations indicate all matter absorbs and >0K emit IR radiation. H2O and CO2 are matter.
2. Are Radiative gases critical to strong vertical tropospheric convective circulation? Yes or No?
No. Critical for convection to occur is situation where a fluid is heated from the bottom in a gravity field. Minor infrared active gases will show strong vertical convection too. Of course Konrad’s meaning of “strong” and “critical” may be different than mine. For an interesting discussion of atm. free and forced convection principles vs. conduction, esp. near no slip boundaries see Bohren 1998 p.354. Not a fallacious appeal to authority: author is authority & cites orig. work, there is consensus on convective energy transfer, not in context of deductive judgment.
Beg patience on others huge scope while I search for a decent science ref.
Ball4 says:“Critical for convection to occur is situation where a fluid is heated from the bottom in a gravity field.”
I would add the idea (perhaps so obvious that it doesn’t need to be said) that there must be cooling (ie heat flow back out of the atmosphere) *somewhere*. With no cooling, the heating would eventually warm the atmosphere to the point where heat would stop flowing into the atmosphere.
There are two places where the atmosphere can cool effectly
1) at the top via GHGs
2) via contact with cool ground (either lighter color ground or on the night side).
If GHGs were removed so that (1) didn’t occur, I am pretty sure there would still be convection, but it would be very different in nature (I hate to make any specific predictions in fluid dynamics, becasue it it usually WAY more difficult than it seems).
So (IMHO) GHGs are critical for convection how/where we currenly observe it on earth, but are not critical for convection in general.
Tim 1:39pm – There is no perfect insulation in nature, N2 and O2 would still radiate away IR energy to sink of deep space at TOA creating the sink you seek without any stronger infrared active gas in the atm. I agree earth’s vert. convection speeds would certainly be different given 0 ppm IR active gas.
Konrad 8:48pm – “3. Does altering the quantity of radiative gases alter the speed of tropospheric convective circulation? Yes or No?”
Yes. Only to the extent added earth infrared active gas ppm affects temperatures at certain heights (+ Tmean near surface, -Tmean at stratosphere). The max. work done by buoyancy on a parcel’s vertical velocity named convective potential energy (cvp) is a function of that temperature difference and ln p, where for initial parcel v=0, the max. vert. velocity the parcel may attain is (2*cvp)^0.5. For instance, between 700mb and 200mb, find cvp ~1800 J/kg for 60 m/sec max. updraft, not the condition most pilots want to fly thru. Max. recorded cvp ~7000 J/kg reported. This highest amount might result in pieces of airplane falling out of cloud as sometimes reported. Cite Bohren 1998 p. 316 for the interesting discussion.
This change in convection due IR gas ppm would not alter the overall LTE balance or surface mean temperature (as Tim posted 2:32pm) because no convective energy transfer gets to space; all the convection energy dumps into the atm. as the sink. TOA radiative balance is unaffected so mean near surface T=288K unaffected by convection speed increase (or decrease).
Ball4 says: “Tim 1:39pm – There is no perfect insulation in nature, N2 and O2 would still radiate away IR energy to sink of deep space …
Certainly true. But N2 & O2 are several orders of magnitude less effective at radiating IR, so they would radiate orders of magnitude less power, resulting in several orders of magnitude weaker convection. Whether the IR to space was reduced to ~ 1 W/m^2 or exactly to zero W/m^2, the effect on convection would be dramatic.
PS. This website has a great too for seeing the effectiveness of the various gases to radiate IR.
http://www.spectralcalc.com/spectral_browser/db_intensity.php
Tim,
You state –
“There are two places where the atmosphere can cool effectly
1) at the top via GHGs
2) via contact with cool ground (either lighter color ground or on the night side).”
These are indeed atmospheric cooling mechanisms, but as experiment 5 shows number 2 could not be considered very effective. Gravity moves colder air to the surface, and due to the slow speed of gas conduction, this biases the conductive flux between the surface and atmosphere. The surface is far better at conductively heating the atmosphere than it is at cooling it. This is demonstrated by experiment 5.
Many climate scientists, sadly Dr. Spencer included, calculate conductive flux between the surface and atmosphere based on surface Tav. Experiment 5 demonstrates why you should never do this. Eliminating the diurnal cycle and using averages may work for radiative exchange calculations, but it will give the wrong answer for conductive flux between surface and atmosphere for a moving gaseous atmosphere in a gravity field.
As to the fluid dynamics question, experiment 4 demonstrates why this is important. Two gas column can have similar amounts of energy entering and exiting each column, but the differing heights of entry and exit within the columns results in very different average temperatures between the two columns. The gas column with cooling at altitude runs cooler. Radiative gases are the only mechanism for cooling at altitude in our atmosphere.
konrad
If you admit IR is absorbed by water and you observed the swimming pool exposed to sun is heated, the logical conclusion is IR do slow down the cooling rate of water. Otherwise the pool will not warm at all because it is cooling by convection to the air.
Your hairdryer experiment cannot prove your point because it heats mainly by convection between hot air and water surface. Not by radiative heat transfer. Rather than imagining useless experiments, just read scientific literature describing countless experiments on IR spectroscopy. The warming effect of IR on water was not invented by AGW believers. You can actually feel the warm water in the swimming pool exposed to the sun for a long time.
People who say radiation physics give the wrong results are ignorant of the subject because it gives the right results.
“The formal radiative transfer equations for the atmosphere are not innovative or in question – they are in all the textbooks and well-known to scientists in the field.”
“Experimental results closely match theory – both in total flux values and in spectral analysis. This demonstrates that radiative transfer is correctly explained by the standard theory.”
Read this material from atmospheric radiation textbook.
http://scienceofdoom.com/2010/11/01/theory-and-experiment-atmospheric-radiation/
Dr. Strangelove,
Water is heated by SW radiation fro the sun. You can observe SW being adsorbed when you go scuba diving. Take a waterproof colour chart with squares of colour from the red to blue spectrum. As you descend the red squares will quickly go dark before 10m. At 50m only blue light is reaching that depth. What is happening to the lower frequency photons that don’t make it to 50m? They are being absorbed and thermalised. This is how the sun heats our oceans.
Your claim that incident IR heats water is incorrect. Experiment 1 listed above proves this. I have built several variants of this experiment. It works. If you doubt that incident LWIR has no effect on the cooling rate of water that is free to evaporatively cool, I would suggest you build and run the experiment yourself.
You have cut and pasted this –
“The formal radiative transfer equations for the atmosphere are not innovative or in question – they are in all the textbooks and well-known to scientists in the field.”
You appear to be making a common mistake, believing that I am disputing radiative physics. As the first radiative two shell experiment I posted shows, I am in no way disputing radiative physics, but rather it’s misapplication to atmospheric physics. The miscalculation of the effect of incident LWIR on liquid water is but one example.
The critical error in the global warming hypothesis lies in fluid dynamics and gas conduction. Climate scientists have calculated changes in radiative flux at differing levels of the atmosphere without simultaneously calculating the changes in the speed of mechanical energy transport that will result. Quite simply they did not calculate the increased speed of strong vertical tropospheric convective circulation that would result from increased concentrations of radiative gases in the atmosphere. This is the reason the climate models have all failed against empirical observation, the net effect of radiative gases in our atmosphere is cooling at all concentrations above 0.0ppm.
Konrad, I share you concern that fluid dynamics is hugely complicated and difficult to deal with. OTOH, I question some of your conclusions.
“Climate scientists have calculated changes in radiative flux at differing levels of the atmosphere without simultaneously calculating the changes in the speed of mechanical energy transport that will result.”
That is a VERY broad claim! Are you *sure* than no climate models include some sort of fluid dynamics? I suspect that many of them have some sort of fluid dynamics built in, including convective energy flow. If nothing else, many models include the lapse rate, which means they at a minimum indirectly include convection.
“the net effect of radiative gases in our atmosphere is cooling at all concentrations above 0.0ppm.”
I just can’t see how you can claim this. Without radiative gases, the surface temperature would average ~ 255 K — a number that is easy to calculate. We have radiative gases and we have an average surface temperature above 255 K. And we have a simple theory that says radiative gases should warm the surface.
Now, I could agree if you wanted to claim that the net effect of radiative gases in our atmosphere is cooling OF THE TOP OF THE ATMOSPHERE (and warming at the surface level).
Additionally, I could agree that the the net effect of CONVECTION in our atmosphere is cooling (at the surface level, and warming at the top).
But if you are claiming that the net effect of radiative gases in our atmosphere is cooling AT THE SURFACE LEVEL, then the grand experiment that is Planet Earth says you are most definitely wrong.
Tim,
general circulation models do indeed include computational fluid dynamics. I am talking about the early calculations that concluded that the net effect of radiative gases was atmospheric warming. In these calculations mechanical energy transport is included as a flux, but its speed is not altered simultaneously with altered concentrations of radiative gases. This is a critical error. This error has been transferred into the parameters driving GCMs.
“And we have a simple theory that says radiative gases should warm the surface. “
Radiative gases do slow the cooling of the land surface (not the oceans) when viewed in isolation. But in driving tropospheric convective circulation, and radiating the energy mechanical transferred to altitude out to space, they also play a critical role in cooling the surface and the atmosphere as a whole.
The question is the net effect of radiative gases in the atmosphere. Radiative gases do play a role in raising surface Tmin at night. But they significantly reduce surface Tmax during the day due to their critical role in tropospheric convection.
Earth has a higher surface Tav than the moon (even when the 2 week lunar day is accounted for) not because of the radiative gases, but because we have an insulating gas atmosphere. Without radiative gases energy can still enter the atmosphere via conduction, but as shown in experiment 5 it cannot be effectively cooled by surface conduction. Without radiative gases we cook, and that’s before we even get to radiative super heating of N2 and O2 in a stagnant atmosphere.
“Earth has a higher surface Tav than the moon (even when the 2 week lunar day is accounted for) not because of the radiative gases, but because we have an insulating gas atmosphere. ”
This is wrong on two fronts.
1) Everything else being equal, a rapidly rotating world will have a HIGHER average temperature than a slowly rotating world. The faster rotation will keep the day side a little cooler and the night side a little warmer. But since radiation is proportional to T^4, the net result is that a planet with more uniform temperatures will be a higher average temperature. If the moon rotated faster, its average temperature would go UP!
2) The radiating surface can be no warmer than ~ 255 K (given current albedo & emissivity.) Whatever temperature the atmosphere might be, the actual land/water surface could average no more than 255 K if the atmosphere did no radiating. That is simple conservation of energy. So *maybe* (but I doubt it) we would have warm heads 2 m above the ground, but the ground itself would be frigid (ie ~ 255 K average around the world).
Tim Folkerts,
I see you denying the reality of our atmosphere!!
HAHAHAHAHAHAHAHAHAHAHAHAHAHA
Tim,
your are quite correct, if the moons rotation were increased to 24 hours it should have a higher average temperature. Currently average temperatures on the moon, -77c, are far lower than blackbody calcs would indicate, because of it’s slow rotation. Adding a non radiative atmosphere would increase this. I am aware that the radiative greenhouse hypothesis claims that such a non radiative atmosphere could not raise the average temperature above the theoretical blackbody temperature. And the message from that is do not apply SB equations to a moving gaseous atmosphere.
As to the second part of your response I can offer you a simple thought experiment. (LWIR transparent potassium chloride lenses are not strong enough to build it for real.)
Take two matt black iron balls 250mm in diameter. Heat them to 100C. Place both balls inside LWIR transparent spherical shells (unobtainium required). Fill shell A with 1bar pressure of nitrogen gas. Empty shell B to high vacuum. Place both spheres in deep space away from stars and separated from themselves. Allow the iron balls to cool. Which cools faster?
The ball in shell A will cool slower. The Nitrogen gas will be heated conductively by the iron ball and hold the heat, as nitrogen is a poor radiator. As the iron ball cools, the Nitrogen gas will conductively return the energy it has stored, slightly slowing the cooling rate of the ball.
This is one effect our atmosphere has on the surface of the planet. It stores energy and moderates surface temperature fluctuations.
Now repeat the experiment with shell B filled with CO2. Which sphere cools faster?
KuhnKat, since I have been extensively discussing and affirming the effects of our very real atmosphere, your comment is patently absurdly.
If you have a point, then make it. Is there some PARTICULAR aspect of our atmosphere that you think we should be paying attention to?
Konrad,
I think your latest analysis is corerct.
I also think that the cooling rate is a bit of a red herring here.
*****************************************
Consider adding a ~ 72 W heater to the iron balls. That works out to ~ 360 W/m^2 (if I got the math right). That in turn works out to a blackbody temperature of 282 K = 9 C.
So for a bare iron ball, no matter what the initial temperature, the surface will eventually settle in to a steadtystate value of 9 C.
Adding an atmosphere that is transparent to IR would not change this number. It will increase the time it takes to reach 9 C a little, but the final temperature of the surface of the iron will be 9 C in either case. The whole atmosphere of N2 would eventually settle in to 9C due to conduction.
Adding some CO2 would cause the steadystate temperature of the surface to be above 9 C. You would need a fairly big, fairly dense atmosphere to see a significant effect, but any amount of CO2 would cause some warming of the surface of the iron. The atmosphere would settle in to a temperature gradient at steadystate, starting above 9 C at the surface of the sphere, and dropping to below 9C at the outermost layers.
Konrad 1:04am- “Which sphere cools faster?”
You can use the 1st law energy balance to solve your problem exactly as Tim (et. al.) shows for the earth system theoretically reducing the atm. IR gas emission/absorption towards surface Tmean 255K as earth atm. opacity reduces from less IR active gas. N2,O2 won’t absorb as much (SW+LW) IR nor need to emit as much LWIR.
The convective vertical speeds wouldn’t change much based on the formula I posted above 3:46pm and in any event would have no effect on surface Tmean balance any way b/c convective thermal IR energy doesn’t reach space, it is all dumped off in the atm. radiation bath as is rain.
@Ball4
I’m not sure I understand you:
“The convective vertical speeds wouldn’t change much based on the formula I posted above 3:46pm and in any event would have no effect on surface Tmean balance any way b/c convective thermal IR energy doesn’t reach space, it is all dumped off in the atm. radiation bath as is rain.”
Are you arguing that convection doesn’t cool the ground at all?
As almost anybody here agree, atmospheric convection happens only in presence of GHGs, because only they “extracts” heat from the above via the LWIR radiation path. That high altitude “extracted” energy is “shielded” by the underneath higher concentration of GHGs, so it has much more probability to exit to the outer space than when it was at the ground (in that case the “shield” is above it).
So IMHO, Konrad should be right, vertical convective cells should work very well as a ground cooler.
Instead, I still don’t agree with Konrad (and many others here) that an atmosphere without any GHGs would be isothermal.
I never said that without GHGs the atmosphere still shows convective processes, I agree with them that the convective effect must be suppressed. What I’m saying is that any molecule in a planetary gravitational field is attracted by the planet. So when a molecule starts its vertical trip at ground because of an initial upward KE impulse, it converts its KE to its PE as it “gains” altitude, at a certain altitude it stops to raise and then start to fall down converting back its PE to its KE, and finally it release the very same KE initial impulse to the ground.
This is elementary physics which should apply to gaseous molecules too. The start impulse is the vibrational impulse given by the temperature of the ground to the molecule of gas which touched the ground. And since in gases only the KE is the temperature, my opinion is still that even in a GHGs free atmosphere there must be a lapse rate because at the TOA KE reach the value of zero.
Otherwise I would like to know out how to compute the TOA altitude in absence of GHGs.
In my opinion, Konrad’s well done experiments say nothing about that. Because they demonstrate only that in a gravitational field the convective path is locked when the source of heat is placed above the sink. The little distance between the two points in his experiment, can’t be useful to appreciate if any KE as been transformed in PE and the temperature eventually reduced with altitude.
Have a nice day.
Massimo
Massimo 3:26am: “Are you arguing that convection doesn’t cool the ground at all?”
This is imprecise question. Convection doesn’t affect global Tmean. No convection term shows up in the usual 1st law basic energy balance for global Tmean=288K (what energy goes up by convection, that energy comes down – remains in balanced system of interest). You could write the surface balance convective term in there on both sides of the surface balance (in&out) before cancelling out though. I seldom see that, sometimes I have done so for clarity. Trenberth09 has the convective energy returning to surface as a 17 component of the total 333 W/m^2.
The root cause being convection doesn’t dump thermal energy into deep space, convection just dumps energy into the atm. If vertical convection speeds could reach escape velocity then convection would affect surface atm. Tmean but an earth updraft going to deep space sink isn’t observed nor is it calculated this can happen.
“atmospheric convection happens only in presence of GHGs”
No, not if you mean GHG as infrared active gas. A container of N2 in a gravity field would show convection if heated from the bottom; would always cool at top (even TOA) & downdraft since there is no perfect insulation.
“..the convective effect must be suppressed..”
Only by any difference in T(z) field. The vertical speed of convection (updraft velocity) can be calculated from buoyancy, hydrostatic & KE consideration, it depends on temperature difference over a pressure interval and R, not the amount of infrared active gas. Probably a better discussion for a severe weather topic but Dr. Spencer box experiments do include local convection.
This MS paper discusses some of the CAPE basics of updrafts/downdrafts with values, basic topic in text books too.
http://www.nssl.noaa.gov/users/brooks/public_html/papers/cravenbrooksnwa.pdf
“there must be a lapse rate because at the TOA KE reach the value of zero.”
Here you enter the round robin debate of non-isothermal atm. when GHG free which will explode this thread OT, so I resist comment.
Hi Ball4
first of all thank you for considering my question, I’m not a climate scientist, just an engineer and maybe I write incorrect statements which reflect my (probably wrong) opinion.
You wrote:
“No, not if you mean GHG as infrared active gas. A container of N2 in a gravity field would show convection if heated from the bottom; would always cool at top (even TOA) & downdraft since there is no perfect insulation.”
I’m not sure, are you saying that N2 cool because it radiate someway to the outer space?
Have a nice day.
Massimo
Massimo 2:25pm – “I’m not sure, are you saying that N2 cool because it radiate someway to the outer space?”
Since I had earlier mentioned fluid heated at the bottom in a gravity field (e.g. N2) is critical for convection, Tim then earlier pointed out there has to be a way to cool; either convection, conduction and/or radiation to deep space will afford a sink – which is always available – because there is no perfect insulation.
Tim 4:31pm – Has to sneak in a comment about “how a gas would behave in a “perfectly insulated container” even though Dr. Spencer’s WalMart containers are not perfect. So I will sneak in that text books teach only in the absence of gravity field will Tim’s isothermal “correct answer” actually be correct.
Tim,
“Everything else being equal”
You can’t use the current albedo and other STATE criteria on a planet that has a faster spin. You have to back up and try and figure out what would have developed with that different constant.
I think the discussion has about run its course. Just a few quick comments about my understanding of convection.
We can discuss thought experiments about atmospheres with no IR active gases (and no clouds or dust to radiate IR). We could consider thought-experiments about this using perfectly insulated containers.
But this is fairly pointless. Conduction through atmospheres will be less than 1 mW/m^2. So heating of even 1 mW/m^2 (with 1 mW of cooling somewhere else) will overwhelm conduction and drive convection.
The atmospheres of all known planets and moons (that have significant atmospheres) have GHGs (CO2, CO, H2O, CH4 and/or NH3). All atmospheres will have a little dust. Even N2 & O2 will radiate a LITTLE IR. In addition, all planets & moons rotate, which will also drive convection.
Similarly, trying to decide how a gas would behave in a “perfectly insulated container” is fairly pointless (other than to understand the underlying physics). (The correct answer is that for a gas with a measurable pressure, the temperature will indeed be constant throughout the column.) But again, even 1 mW/m^2 would be enough to render a real life experiment moot.
Tim,
sorry for the slow response, I have been somewhat preoccupied with the Australian federal election. The news is that the party that introduced the carbon tax has been fired. Out of a cannon. Into the sun.
While I agree that thought experiments have limited utility, the one I suggested does serve one purpose. It illustrates why SB equations should not be applied to planets with liquid oceans or gas atmospheres.
As to the though experiment, I have worked out a way that the experiment could be conducted for real on earth at surface pressures. The IR transparent shell of unobtainium can be replaced by a shell of mono-crystaline inconel or duplex 2205 anodised matt black on the interior. On the inner face a geodesic structure of syntactic foam (Q-Cells and epoxy) could support individual potassium chloride salt lenses, with vacuum between the lenses and metal shell. The metal shell would then need external cryogenic cooling. Sadly I would not be able to build this myself as it would require “climate science” dollars 😉
Tim, read back over this thread. How do “Dr. Strangelove’s” assertions about the effect of SW on liquid water stack up against mine? How does ball4’s understanding of Rayleigh- Bernard circulation systems compare to mine? I can show through empirical experiment that without radiative gases our atmosphere would super heat and most of it would boil off into space. You should acknowledge that the AGW hypothesis depended on the misapplication of SB equations to a planet with liquid oceans covering most of the surface and a gaseous atmosphere in a gravity field.
Konrad 2:55am: “How does ball4′s understanding of Rayleigh- Bernard circulation systems compare to mine?”
It is Rayleigh-Bénard circulation. At 9/4 3:46pm I’ve provided Konrad some hints of the atm. physics foundation how to compute R-B vertical velocity.
Konrad 4:04pm: “…(climate models) did not calculate the increased speed of strong vertical tropospheric convective circulation that would result from increased concentrations of radiative gases in the atmosphere.”
I just showed Konrad how to compute the convective vertical speeds in troposphere. Now Konrad would be better off in discussions build on that, understand and apply the basics.
“…the net effect of radiative gases in our atmosphere is cooling at all concentrations above 0.0ppm.”
1st thermo law with measured data determines this hypothesis is not correct for atm. near surface as Dr. Spencer’s experiments in top post demonstrate.
Ball4,
I would highly recommend that you read this 2009 post by Dr. Spencer –
http://www.drroyspencer.com/2009/12/what-if-there-was-no-greenhouse-effect/
He also agrees that radiative gases are critical to tropospheric convective circulation, and without this gas conduction would send the bulk of the atmosphere isothermal.
My experiments show that while this is correct, Dr. Spencer has the wrong temperature for the resulting isothermal atmosphere. It would in fact be far hotter than present. In fact due to molecular super heating of stagnated N2 and O2 at altitude by IR,SW & UV, much of an atmosphere without radiative gases would boil off into space. This effectively invalidates the radiative green house hypothesis.
Konrad 4:41pm: Thanks, that link is constructive.
Note Dr. Spencer is discussing :…this thought experiment is an atmosphere that is heated from below by the ground absorbing sunlight, but the atmosphere has no way of cooling.…eventually, the entire atmosphere would reach a constant temperature with height.”
This is a bit imprecise, ambiguous: heated from below, no way of cooling, constant temperature with height?
For an infinite opaque to LWIR atm. yes, you can get any earth surf. mean temperature you want up to sun’s radiating T at 1a.u. (assuming the infinite opaqueness lets in SWIR). The gravity loss of KE to PE would still be in operation though and an atm. T gradient would have to exist given by poisson eqn. So convection and natural buoyancy would exist & it is not clear how Dr. Spencer gets around that physics. There is no perfect insulation so this sort of boiloff won’t/can’t exist in nature, a balance would be attained around 303K for an atm. emissivity of 1.0. Maybe up to 345K with more opacity at today’s mean surface p; increase p by adding more atm. mass, get more opacity and find higher mean surface temp.s.
What we are usually discussing though is not so extreme; regards reducing infrared active gas ppm as reducing the atm. opacity (both SWIR and LWIR) to near 0 (surf. Tmean=255K) from a natural mean atm. emissivity of near 0.8 (surf. Tmean=288K) while holding albedo constant.
Conversely, the greenhouse effect added IR gas ppm being responsible for raising opacity from near 0 (surf. Tmean=255K) to near natural 0.8 emissivity (surf. Tmean=288K) always in the presence of convection. Like the Plexiglas in the top post experiment.
The net effect of radiative gases then in our atmosphere raises the atm. opacity at all concentrations above 0.0ppm which the 1st law shows contradicts your hypothesis for the surface “…the net effect of radiative gases in our atmosphere is cooling at all concentrations above 0.0ppm.” Proof: the Plexiglas didn’t cool its WalMart container.
Dr. Spencer ref.s the Lindzen paper discussing absence of convection which then ref.s the Manabe 1961 paper which doesn’t discuss the conditions causing Dr. Lindzen’s 345K (72C) that I could find.
“People who say radiation physics give the wrong results are ignorant of the subject because it gives the right results.”
The skeptic world is still waiting for the evidence that a doubling of CO2 has any effect at all on global temperature, let alone the 3 to 5 degrees C predicted by radiation physics.
Hmmm, haven’t seen this asked yet, does changing the height of the plexiglass affect the outcome?
The data shows that the box with the Plexiglass above it is clearly warmer than the one not. Let us accept this data. It is a real result.
The explanation, we are told, is that the plexiglass somehow returns the IR back to the box increasing its energy. OK, how does it do that?? Does it absorb the IR and emit or does it reflect?? This is important.
If it reflects the IR this has little to do with our atmosphere and we can ignore the experiment.
If it absorbs and reemits does this really confirm the efficacy of the IR as explained by the good Doctor?? I say it does not.
The “trapping” of IR is a minor pip in this experiment. The plexiglass has a black body radiation based on its temperature. That radiation has wider spikes than the BB of our atmosphere due to its material density and more varied frequencies. The atmospheric BB radiation has many HOLES in it compared to a solid. The plexiglass is radiating far more energy at the box at more frequencies than just the IR bands that water vapor and CO2 cover!!
No one bit on my arm waving so let me add some numbers.
The emissivity of Plexiglass is about .86.
The emissivity of the atmosphere is alleged to be about .7.
This is simple blackbody physics.
Aug anomaly?
Crikey! I consider myself an AGW skeptic, but I never realized there was any dispute about the concept of the greenhouse effect.
I always thought that the dispute was about the significance of the greenhouse effect, from rising CO2 levels.
Dr. Spencer explains very clearly that there are negative feedbacks to any increased warming from rising CO2 levels, such as the albedo effect of increased cloud cover (resulting from increased evaporation of water due to any warming).
Such concepts make sense to me. However, the fact that some of you seem to be questioning whether or not the concept of the greenhouse effect itself is real, I find a bit disturbing.
Surely that is the very first thing that should be established in this very complex issue of climate change.
The basis for the greenhouse effect is important because it is used to justify the predictions of global warming due to increasing CO2 and, by extension, catastrophic consequences. If a greenhouse effect occurs with any atmosphere even one without IR absorbing gases, then CO2 must cool, not warm, as Konrad is arguing.
I probably shouldn’t do this, but why do people keep saying isothermal when speaking of a gas column in a gravity well at equilibrium?
Isentropic is not isothermal when gravity is involved, but it is the only state where work could not be extracted, and thus maximizes entropy for said gas column.
For what it’s worth I fully agree.
Have a nice day.
Massimo
A system does not tend toward EQUAL entropy for all subsystems. It tends toward MAXIMUM entropy for the whole system. This could well mean than some moles of the system have higher entropy than others. In an isentropic system, it is possible that entropy in one part could decrease while entropy in another part increases. As long as the NET change is positive, the new, non-isentropic system would be prefered.
But systems DO tend toward isothermal. That is pretty much the definition of “thermal equilibrium”. The derivations of isentropic columns of gas (ie cooler at the top) all start with “adiabatic processes”. In other words they disallow the movement of heat within the system, so they specifically disallow thermal equilibrium.
So …
“Isentropic is not isothermal when gravity is involved …”
YES
“but it is the only state where work could not be extracted”
NO
“and thus maximizes entropy for said gas column”
NO
Simple counter example. Take two columns of DIFFERENT gas that are the same temperature at the bottom. They will have DIFFERENT temperatures at the top for the isentropic (non-isothermal) case. Use this temperature difference to run a heat engine.
See, every time the isothermal/isentropic debate comes up, I’ve got a little mental aide I thought up.
I picture a box full of ping pong balls in a spaceship, the balls are evenly distributed within the box, their motion is even and randomized, corresponding to an isothermal state well enough… oh and for simplicity the box is secured to the wall that would correspond to a floor while the ship was accelerating.
Everything ok so far?
Ok, well, the isothermal position is like taking that box of ping pong balls in a spaceship, and then being told that if you fire up the rockets in the ship the balls will remain in exactly the same arrangement within the box, despite the ship accelerating at 1 G.
Or, alternatively I suppose it would be like opening a box of ping pong balls that has been sitting undisturbed for an arbitrarily long period of time and finding the balls are bouncing about and evenly distributed inside, and that they spontaneously arranged themselves into this state no less!
“Ok, well, the isothermal position is like taking that box of ping pong balls in a spaceship, and then being told that if you fire up the rockets in the ship the balls will remain in exactly the same arrangement within the box, despite the ship accelerating at 1 G.”
No, when you fire the rockets, there will still be a “density gradient” even for the isothermal case. It will jsut be a different density gradient than the isentropic case. The ‘bottom’ will still compress in either scenario.
The question is “how does the average KE of a pingpong ball at the top compare to the average KE of a pingpong ball near the bottom’. It is certainly counterintuitive to many people, but the answer is ‘they are the same’ (under reasonable assumptions about perfect collisons, perfect insulation in the walls, etc).
What is the potential energy of the ping pong balls near the top after the rockets are fired?
Is the kinetic energy and potential energy the same for the balls at the bottom as for those at the top?
KE is the same everywhere; PE greater at the top.
So total energy is higher at the top, couldn’t you make use of that then?
My question Tim is:
What happen if you remove the top of the box?
What could stop the escaping upward trajectory of those ping-pong balls, despite the rockets is accelerating in that very same direction?
Have a nice day.
Massimo
“What happen if you remove the top of the box?
What could stop the escaping upward trajectory of those ping-pong balls …?”
The short answer is “nothing”. 🙂
Just like with a planet, some of the gas molecules could and would escape to space. The rate that they escape will dpend on the temperature, the escape speed, and the mass of the molecules.
For our moon, pretty much any common gas can escape easily — hence no atmosphere. For the earth, only the lighest molecules can escape — hence no H2 or He in our atmosphere.
Even though the KE is still large at the top of an equilibrium (ie isothermal) gas in a gravitational field, the density is small, so there are not many molecules there that could escape.
IMHO the problem is that all the ping-pong balls would escape because in that case, the first time the speed vector of any single ball points up, they have no way to return back to the floor.
Really, I’m not convinced by the isothermal atmosphere.
Even in my opinion (if I understand what you meant) your question “how does the average KE of a pingpong ball at the top compare to the average KE of a pingpong ball near the bottom” is:
The KE will be the same for any ball which has the very same distance from the floor, the upward balls are decelerating, the downward ones are accelerating of the very same 1G.
The equilibrium is given by the same number of upward and downward balls which pass through any single layer parallel to the floor.
The density will decrease as the layers are far from the floor because of the reducing transit speed in the two vertical direction as the horizontal planes are higher.
Of course, maybe I’m missing something and I could completely wrong about that.
Have a nice day.
Massimo
It’s funny you say “at the top of an equilibrium column […] in a gravitational field” and then slip in “ie isothermal” as if we aren’t supposed to notice you’re trying to assert that a column of gas in equilibrium in a gravity well can only be isothermal.
Then you say that escape will be minimal, if a column of gas suspended in a gravity field is unable to trend towards an isentropic state then there is no reason to think it will relax and form a density gradient.
Indeed one would expect it to behave as though it were not in a gravity well at all, it sounds almost like you want to call isentropic equilibrium “isothermal” at times, but not enough to suggest you actually mean to do so.
“the first time the speed vector of any single ball points up, they have no way to return back to the floor.”
No, that is not quite right. A specific molecule does not always have the same KE. A specific molecule will indeed loose KE as it gains PE.
Maybe this will help. Consider a cross-secton of the column that is above where the pingpong balls usually go. But occasional balls do make it that high. These ball are self-selected to be exceptionally high kinetic energy balls (otherwise they would not have gotten that high to begin with). Now, they have lost some of that exceptional kinetic energy as they rise up, but that will actually leave them with exactly the average energy.
If you look higher, fewer balls will get there, but these are the once that had even more exceptional energy to start with. And again the average KE of these very few balls will be the same as the average KE at the lower layers.
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|̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦̊◦̥̊◦| |◦˳̥◦ ◦ ˚|
|̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦| |˳̥̻̻˳ ˚ ◦˚|
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|̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦| |˳̥̻̻ ̻ ˚˳̥̻̻◦˚|
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|̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦| |̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦̥̊◦|
——— ———
0G 1G
Ok, I’m going to screenshot this because I’m certain the character mapping will break, but hey, worth a shot, if it comes out all garbled, well, that’s what the pic is for: http://i341.photobucket.com/albums/o396/maxarutaru/zerotoonegee_zpsbfd9e0b7.png
The 0G image has the balls or whatever loosely floating in the box bumping around a bit, with a smooth and even distribution of velocities to represent an isothermal state.
The 1G image is what happens after the box is accelerated and the velocity distribution is perturbed by the bottom imparting kinetic energy to the lower grouping of balls, which collide with those above them, and so on.
The energy imparted onto any individual object by collisions with the floor or other bodies will not be sufficient to overcome the steady acceleration of the box itself, so the possibility of escape is very low to nil.
The distribution of velocities while the box is being accelerated will be highest in the more densely packed region near the floor due to the constrained phase space, while the upper regions will have more widely spaced bodies and a range of velocities tapering off towards the limits where the energy imparted is sufficient to outpace the vast majority of the other bodies for a time, but remains insufficient to avoid being overtaken by the still accelerating box.
Where did I limit the evolution of this system to be adiabatic, as you claimed I would?
The “temperature” of the lower regions would increase, the density would rise, so the ever more densely packed and excited bodies would have to “cool” by ejecting other bodies upwards due to the lack of a role for IR radiation in this system.
@Tim,
I really trying to understand what you mean with “Now, they have lost some of that exceptional kinetic energy as they rise up, but that will actually leave them with exactly the average energy.”
I don’t thing there is nothing of exceptional in the KE of those molecules, they just received their energy from the ones beneath, and since the supposed system is inelastic, the upper molecules should exchange their KE with the KE of lower ones; and in my opinion the net exchange should be fixed by their altitude from the floor, since any molecule behave the same way converting its KE to PE and vice versa.
My point is that for gases KE is the molecular mass multiplied by the square of the molecules speed divided by two, so I can’t imagine any other of the two possibilities:
1) the very last molecules at the top (the one without any other molecules over their upward trajectory) changes voluntarily their directions from vertical to horizontal and then return to vertical downward keeping their speed constant (highly improbable, when not impossible to me).
2) the very same molecules of above progressively reduce their KE in favor of their PE and a certain point they stop at all their run, before starting again their trip downward. In this case the temperature fall to zero at the upper boundary, because only KE in gases is the temperature.
I’m unable to imagine other behaviors of those very last upper molecules, if you have one other plausible I’m glad to read about it.
Have a nice day.
Massimo
It appears we are now discussing an analogy of a side issue of a tangent. 🙂
I think it is time to admit that we simply don’t have a reasonable chance of resolving the issue in a forum like this. We have all presented some plausible arguments, but none of this is a substitute for really working through an undergraduate thermo text — followed by a graduate thermo text.
For fun, I would suggest you all look at http://www.falstad.com/gas/
Try turning gravity way up and turning the heat on fairly low. You should see that the molecules at the top of the box still have a distribution of speeds (colors) similar to the distribution at the botom.
Hi Tim,
nice the applet that you linked.
Thy this:
reduce the simulation speed to allow you to see the molecules moving not too fast. Set the molecule count to the minimum (1 molecule), at this point set the maximum gravity.
Then activate the heater and adjust it until the vibrating ground became red (so that the molecule doesn’t reach the top of the box).
Look what it happen to the parameter kT.
It seems to me the k is constant, and T well follows the altitude.
Have a nice day.
Massimo
“Set the molecule count to the minimum (1 molecule) … activate the heater …”
Unfortunately, there is (at least) one bit of bad programming in the simulation.
The “heater” is not like a real heater. With a real heater, there would be a distribution of energies of the atoms in the heater, and a distribution of energies for the particles that hit the heater.
This “heater” gives every particle that hits the heater exactly the same energy after the collision. Every particle gets exactly kT worth of energy, but some should get more and some should get less. It is only the collisions with other particles that then randomizes the energy distribution.
In this case, you are right, that the KE drops as you go up. But this is not how a real system would work. Here, you pretty much HAVE to use a large number of particles to acheive the Maxwell-Boltzmann distribution of kinetic energies for the particles flying around.
Hi Tim,
I agree that that simulator has something wrong.
This night I did some play with it and I seen some molecules popup from the vacuum!!!
Not really a “real simulation”.
Have a nice day.
Massimo
To clarify my original example, take the applet so helpfully linked by Tim and set it as follows:
Stopped, Single Gas, Equal Speeds, Gravity all the way up, fiddle with the speed of simulation/count however you want, uncheck Stopped.
System evolves from a single speed:
http://i341.photobucket.com/albums/o396/maxarutaru/startingstate_zpsb6af5c3f.png
To one with a distribution of speeds and a distinct density gradient:
http://i341.photobucket.com/albums/o396/maxarutaru/evolvedstate_zpsf8084d8d.png
Whether I turn on low or low-med heat, let it run for a loooong time: http://i341.photobucket.com/albums/o396/maxarutaru/haditrunningawhilenow_zpsfa9c7b17.png
Or whatnot, I just don’t see how this is isothermal: http://i341.photobucket.com/albums/o396/maxarutaru/Ijustdontseeit_zpsd83654a4.png
What am I missing here?
Max, the images you link to actually appear to be WARMER at the top than the bottom!
If you look at the bottom ~ 100 particles in any of the pictures, you only find a few (2-4) white particles (= fastest speeds). If you look at the top ~100 particles, you will see MORE of the whites (6-12). The top part has a higher fraction of the fastest moving particles.
So you are right, it doesn’t look isothermal … it looks hotter at the top!
*************************************
Actually, that is an artifact of the programming. The particles only update the color when they collide. This means that the particles that rise a long way (eg many of the white particles near the top) are often going slower than the color suggests.
The two effects — it looks too warm, but the colors artificially make it look too warm — presumably cancel out to give a truly isothermal distribution with height.
Ok, from this picture: http://i341.photobucket.com/albums/o396/maxarutaru/haditrunningawhilenow_zpsfa9c7b17.png
Cropped 1229 pixels from the particle box to leave it 40 pixels high after counting the lowest ~100 there:
http://i341.photobucket.com/albums/o396/maxarutaru/bottom100ish_zps2f301395.png
………………………………………
45/100
Cropped 648 pixels from the particle box to leave it 621 pixels high after counting the highest ~100 there:
http://i341.photobucket.com/albums/o396/maxarutaru/toponehundred_zps84287245.png
…………………………..
32/100
Note that I had that applet running for several minutes on the highest simulation speed, hence the name “haditrunningawhilenow” on the image.
“So you are right, it doesn’t look isothermal … it looks hotter at the top!
*************************************
Actually, that is an artifact of the programming.”
I would say this is an excellent replay of the Gorebull Warming conflict.
Warmers say it is getting warmer, look at our model. Sceptic looks at the model, plays with it, and starts asking inconvenient questions. Except in real life the warmer does not admit any errors in the models!! 8>)
In this vein, Tim still claims it shows isothermal when there is no basis for the claim other than that is how he thinks it is programmed and that is how he believes the actual physics works!!
Of course the basis for the greenhouse effect is important. I’m just flabbergasted that this process has not been established without doubt.
As I’ve mentioned, I can appreciate the difficulty of being certain regarding the significance of any additional greenhouse warming due to CO2, because we’re talking about very tiny percentages of CO2 in the atmosphere, and water vapour in its entirety is a far more significant greenhouse gas than the ‘two fifths of one tenth of one percent’ of CO2 that we currently have.
However, if the basic concept of the greenhouse effect is in doubt, then that makes a mockery of the entire AGW edifice.
VincentRJ.
I liked his statement.
But the discussion here is off course it was not defined yet what is the greenhouse effect.
Firstly we have to say that the greenhouse is intended to maintain the temperature at 24 degrees centigrade average temperature that is optimal for most traditional cultures.
To fulfill this role the greenhouse should store thermal energy during the hours of sunlight and releases it at night keeping the temperature above room temperature. This leads directly to the idea of latent heat, water vapor and this has not yet entered into the discussion.
cont ……
the experiment of dr. Roy, of Wood and of the others show here do not reproduce the thermal capacity of a real greenhouse. A real greenhouse has a column two meters in height for each square meter of window through which energy enters or exits.
So these experimentor are inconclusive.
Konrad
You can do a simple experiment at home to test whether LWIR can heat water or the water will just evaporate. Your flat iron emits LWIR 8-20 microns at 220 C. The flux at 700 W and 0.31 emissivity of iron is 217 W. Get a bowl with the same area as the flat iron. Pour a glass of water (250 ml) to the bowl and put a thermometer in the water.
Switch on the iron and place it one inch above the water surface. The heat of vaporization of water is 2260 J/g. The iron emitting 217 W of LWIR can theoretically evaporate 250 ml of water in 43 minutes. Wait for 45 minutes and see what happens.
Hypothesis 1: No change in water temperature and it completely evaporates in 45 minutes
Hypothesis 2: Water temperature rises to 100 C and it boils after 45 minutes
Take your pick. I pick Hypothesis 2
Dr. Strangelove, you haven’t tried this have you? Not scoring too many points at the empirical experiment game….
Is any part of the sky radiating at 220C? Only above active volcanoes 😉
Re-read the instructions for experiment 1 I listed above.
Now insulate the top edge of your bowl with Kaowool with an outer coating of aluminium foil, shiny side out. Make sure no part of the foil or ceramic wool contacts the water, but also that the edge of the bowl is not exposed to IR from your totally inappropriate source. Use a probe thermometer through the side of the bowl 10mm below the water line. Water proof the thermometer penetration. Ensure no part of thermometer and cable is exposed to IR.
Now what other part of experiment 1 did you miss? That’s right, the small computer fans! Without these, gas conduction will interfere with the experiment. You must maintain a very light breeze between the water surface and IR source to allow evaporative cooling and prevent gas conduction from the IR source to the water.
Your hypothesis 2 is not correct. If you had actually bothered to conduct a properly designed experiment you would know this. Try again.
If you’re familiar with radiation physics, you would know that a hot body at 220 C is radiating the same IR wavelengths 8-20 um as those observed in the atmosphere. The IR photons have the same absorption, emission and thermalization effects on water regardless of the temperature of the source. IR energy depends on wavelength not the radiative flux.
Computer fan provides convective cooling. You should eliminate that in the experiment because you are trying to measure radiative heat transfer. Once you have proven that IR can heat water, it doesn’t matter if convective cooling is greater than IR heating because that will only slow down cooling rate and still affect temperature of water surface.
A good experimenter, which you claim to be, knows that he must isolate each cause in order to measure the effect of each cause separately. Do the experiment and see for yourself.
Dr. Strangelove,
The computer fans were used in experiment 1 listed above to keep conductive cooling constant in each phase of the experiment and to eliminate gas conduction between the IR source and the surface of the water.
If you wish to use a domestic iron for your IR source you can. (the water block in experiment 1 ran near 90C). You can eliminate the computer fans simply by keeping the iron 50mm from the surface of the water and ensuring the base of the iron is at a 10 to 15degree angle from horizontal. This will allow hot air to convect away from the underside of the iron without excessive gas conduction between the iron and the water surface.
If your thermometer probe is 10 mm below the water surface and no part of the water container itself is exposed to IR from the iron, you will find that IR does not heat water.
Try the experiment again with a sheet of microwave safe LDPE cling wrap floated onto the surface of the water. (do not get water on the upper side of the cling wrap). Now IR will heat the water. As Dr. Spencer has pointed out, cling film is largely transparent to LWIR. The water can still cool by radiation and conduction, but evaporative cooling has been eliminated. If water is free to evaporatively cool, incident IR will not heat it nor slow its cooling rate.
I have run similar experiments many times, the results are consistent. Climate scientists quite simply got it wrong. They reached for the text book when they should have reached for the iron 😉
Well, let’s stop this discussion and just do the iron experiment without computer fan. Report the result if you wish. Don’t worry about heat conduction. The thermal conductivity of air is 0.025 W/m/K. It’s negligible compared to the radiation flux.
Keep the iron parallel to and 1 inch above water surface. Get rid of convection or you will invalidate the result of experiment. Thinking IR had no effect when in fact convective cooling cancelled it. As I said, once you detected the IR effect, it doesn’t matter if in reality convective cooling is greater than IR heating. It will reduce the cooling rate.
This is not a climate model. This is basic radiation physics. Physicists got it right. Or else you can write your own textbook.
I can tell you haven’t actually run this experiment 😉
In experiment 1 (listed above so others can replicate) the light breeze across the water surface also simulates real ocean conditions. That is why there are two test containers, one with an strong IR source, one without. The air flow over each container is identical. Only the one thing we want to test, IR, varies.
Other than “climate science” textbooks, can you point to a single text that claims incident IR heats liquid water? Physicists didn’t get it wrong, climate pseudo scientists got it wrong.
To Dr. Strangelov and Mr. Konrad.
My experiments agree with Mr. Konrad.
An aluminum cup with a little water, exposed in the infrared radiation.
After 10 minutes reaches the temperature (Dp+IR)/2.
A bottle of water (100%),capped and exposed gto IR radiation will heat up (Tamb+IRR)/2 them, for physical IR heats the water but for a climatologist no.
In order to understand the „isothermal equilibrium condition“ of a gas column in a gravitational field one needs to consider the following „Gedankenexperiment“.
Let’s start with a gas column of isotopically pure nitrogen gas in a gravitational field. And we start with a state corresponding to the isentropic situation, temperature gradient and density gradient. Now, we replace the top layer that should consist at least of 1 mol nitrogen to have a good statistic with isotopically pure carbon monoxide of 1 mol in order to be able to neglect the influence of different mass. Now, we start and wait what develops.
Occasionally, the carbon monoxide will be evenly distributed, which in turn means that carbon monoxide molecules from the top have been exchanged with nitrogen molecules from the bottom.
What does this in turn mean for the velocity distribution.
In the beginning the molecules in each layer are characterized by the Maxwell-Boltzmann (MB)-distribution according to the different temperature in the different layers.
We start in our thought experiment with an amount of substance carbon monoxide at the top and exchange it by molecular diffusion with an amount of substance of nitrogen on the bottom. We in turn ask the question how does this exchange change the local velocity distribution.
Carbon monoxide molecules in the top layer with the colder MB-distribution have an equal probability to move to the bottom layer by molecular diffusion across their MB-distribution. They move to the bottom and experience individually an increase of their translational energy or velocity, but are now evenly distributed across the local MB-distribution on the bottom with the higher temperature. This first step therefore does not change the local temperature at the bottom or the top.
In contrast, nitrogen molecules from the bottom layer have a different probability across their local MB-distribution to reach the top layer by molecular diffusion. Molecules in the bottom layer with a higher velocity have a higher probability to reach the top layer.
This means that on average molecules from the fast tail of the MB-distribution in the bottom layer will experience a decrease in translational energy or velocity and get exchanged to the top layer. However, this still increases the number density in the fast tail of the MB-distribution in the colder top layer. This process therefore results in an increase of the temperature in the top layer and a decrease of the temperature in the bottom layer due to molecular diffusion.
This goes on until the MB-distribution in the bottom layer and in the top are the same, carbon monoxide is evenly distributed and we have in equilibrium in a gravitational field an isothermal gas column.
This is valid for molecular diffusion as the mixing process.
This situation changes as soon as turbulent mixing comes into play as the mixing process. In the case of turbulent mixing, the “equilibrium state” is the isentropic state with a temperature gradient.
Best regards
Guenter
That is a good point, I admit I default to assuming turbulent mixing is taking place, which would indeed be why I find the isothermal outcome rather improbable to say the least.
The nitrogen molecules that rise will cool, just as the CO molecules that sunk, warmed. No isothermal state will result, and no turbulence is needed.
Individual molecules do not cool as they rise, rather their translational energy decreases.
This is different.
Cooling and warming is a concept that can only be applied to an ensemble of molecules.
So won’t an ensemble of molecules, having lower average translational energy than the molecules of another ensemble, also have a lower temperature? Why wouldn’t the diffused CO N2 mixture also become isentropic, and have the same density gradient, and a slightly different temperature gradient due to different specific heats for CO and N2?
” … a slightly different temperature gradient due to different specific heats for CO and N2″
But then we could run a heat energy off the temperature difference! (This is more obvious if we keep the two gases in different columns that are the same temperature at the bottom.). The perpetual temperature difference at the top would allow perpetual heat flow from one gas to the other at the top — with NO input of energy = perpetual motion!
Tim 9:22am: “But then we could run a heat energy off the temperature difference!”
No since the temperature difference results from isentropic state – as stated, the gravity field column has achieved max. entropy point. The isolated column of gas would have suffered “heat death” mixing itself and no useful work can then be extracted even with the top-bottom temperature differential in LTE.
“The isolated column of gas would have suffered “heat death” mixing itself and no useful work can then be extracted even with the top-bottom temperature differential in LTE.”
You misunderstood my proposed scenario.
The adiabatic lapse rate is g/cp. For H2 this is (9.8 m/s^2) / (14.3 kJ/kg*K) = 0.7 K/km. For Argon, it is (9.8 m/s^2) / (0.52 kJ/kg*K) = 18.8 K/km
Make two insulated columns 1 km tall. The bottoms of both columns are held at 300 K. The top of the H2 column will be (according to you) 299.3 K. The top of the Ar column will be 281.2 K.
Once this supposed equilibrium state has been achieved, connect a small heat engine to the tops of the two columns (you will have to cut back a little bit of the insulation). You can extract energy from the temperature difference at the top. If the temperature difference drops too far, then either 1) put the insulation back for awhile or 2) use a smaller engine. In either case, you can KEEP extracting work from the heat engine indefinitely with no energy input. The tops will (according to you) continue to sort themselves out to different temperatures as the “equilibrium condition”.
Tim 3:43pm – Still no PM.
Guenter started this subthread with a state corresponding to insentropic, implying gas column isolation from all forcings.
You scenario an Ar column and a H2 column but it doesn’t matter, you could have two universes that suffered “heat death” max. entropy, unchanging entropy or isentropic. Once both (perfectly) insulated gas columns (or universes) are left alone to mix constituents to isentropic max. entropy conditions even with temperature differences, then you cannot “extract energy” from either of them. To do so you will have to decrease entropy, the only way that can happen is by a forcing. There is no PM.
When you cut back a little insulation and connect the heat engine, the connecting device is at a different T or thermally connected to different T then you have introduced a forcing to at least 1 column. The heat engine will work, then stop when the connection and both columns again achieve isentropic conditions if left alone to each thoroughly mix.
The problem you are having also is evidently the ambiguous thermostat keeping “bottoms of both columns are held at 300 K”. Then, yes, energy can enter if the thermostat turns on, so this is not PM, but you supposed insulated columns so confusing.
“The adiabatic lapse rate is g/cp.”
Then the temperatures you computed in the Ar and H2 columns are only approximations off maybe 1-2% at 1km; for the exact ideal temperature T(p), use the poisson eqn. p(z).
I clearly disagree with much of your conclusions, but just as clearly, we will not resolve them here since we are both still convinced after giving the key points of our positions and hearing the key points of the other.
I’m curious how you overlooked the point about the thermostat keeping the bottoms at the same temperature… if you have two columns with the same density profile and temperature but different gases then the columns will be different heights.
If they are the same height and the same temperature at the top and bottom, the density profiles will be different.
If the density profiles and height are the same then the temperature will be different at the top and bottom.
You’re trying to artificially fix a variable and then claim it invalidates an argument which doesn’t fix that variable.
Not in general. If your ensemble is large enough and MB distributed. Yes.
But this is not the point.
In equilibrium or in a stationary state the molecule current from bottom to top must be equal the molecule current from top to bottom.
If you have an isolated, isochoric gas column in a gravitational field molecular diffusion will finally lead to an isothermal equilibrium, because for the process of molecular diffusion fast molecules in the bottom layer have a higher probability to reach the top layer than slower molecules in the bottom layer.
Therefore, the isothermal state is the most probable state, where molecular diffison is the dominating process.
This is different in an open system with „turbulent mixing“ as the dominating process. Turbulent mixing in a system requires import and export of energy. Per example heating on the bottom and cooling on the top.
Both systems are very different with respect to boundary conditions and the processes that dominate the respective stationary state.
Hi Guenter,
first of all thank you for your well written post about MB mixing above.
You write:
“In equilibrium or in a stationary state the molecule current from bottom to top must be equal the molecule current from top to bottom.”
This is compatible with a lapse rate atmosphere too.
At any layer the average speed of the upward molecules is the same as the one of the downward, what it changes is the sign of their acceleration. So the two molecular currents are perfectly balanced.
You continue with:
“If you have an isolated, isochoric gas column in a gravitational field molecular diffusion will finally lead to an isothermal equilibrium, because for the process of molecular diffusion fast molecules in the bottom layer have a higher probability to reach the top layer than slower molecules in the bottom layer.
Therefore, the isothermal state is the most probable state, where molecular diffison is the dominating process.”
Of course only the fastest molecules have chances to reach the TOA (or the ones which receiving bumps from below gets more average energy from the underneath molecules), but the simple fact that all those molecules at a certain point have to stop their raising trip, implies a reduction of their KE in favor of their PE. That is a temperature reduction.
I don’t really understand how do you reach your conclusion. Could you please explain me why do you link the starting speed of the bottom molecules to the whole atmospheric column speed?
I could agree with you if the ground wasn’t a thermal flywheel at the higher temperature as it is. That is if the only energy in the system was the one of the gases, and that energy was the only one who heats/cools the ground. But in the reality it’s vice versa. It’s the ground which heat the air above. The gases would cool the ground because of the MB distribution, but can’t do that because of their very lower density.
Be patient, I’m just an electronic engineer and the last time I seriously dealt with thermodynamics was a long time ago at school.
Have a nice day.
Massimo
Fast molecules from the bottom will be trading KE for PE as they move upwards in a gravity well, won’t they?
Would they not slow down as they move upwards?
Would faster and slower molecules moving down not accelerate?
Hi Max.
Yes, I agree with you and Ball4.
But maybe we are wrong, I would like to understand that point of the MB distribution which should lead to an isothemal steady state, that I don’t really understand.
Have a nice day.
Massimo
Moving the bulk of this response down here so it is easier to find:
From this picture, with a 1269 pixe high particle box: http://i341.photobucket.com/albums/o396/maxarutaru/haditrunningawhilenow_zpsfa9c7b17.png
I cropped the top 621 pixels after counting about 100 there: http://i341.photobucket.com/albums/o396/maxarutaru/toponehundred_zps84287245.png and count 32/100 white.
I also cropped the bottom 40 pixels after counting about 100 there: http://i341.photobucket.com/albums/o396/maxarutaru/bottom100ish_zps2f301395.png and count 45/100 white.
Max, I fear we are pushing that “gas in a box” simulation (http://www.falstad.com/gas/) beyond what it was designed for.
In particular, the ‘heater’ gives all the particles that hit it the exact same speed — which is NOT how a real heater would work. All the particles that have just left the ‘heater’ will have the same energy that must get thermalized by collisions with other particles.
Looking at the first image you gave, there is a distinct ‘spike’ in the speed distribution (the histogram at the very bottom) in the “white” region. This spike is “white particles” that have just left the heater and haven’t yet collided with other particles. This will produce an artificially large # of “white particles” right near the bottom.
**********************************
In any case, the simulation can easily have a change in density of 4-10 between the bottom and the top. If a lapse rate was present, it should be easily visible. The top molecules are NOT especially different in color, and hence are not especially different in temperature.
Hi Tim,
“In any case, the simulation can easily have a change in density of 4-10 between the bottom and the top. If a lapse rate was present, it should be easily visible. The top molecules are NOT especially different in color, and hence are not especially different in temperature.”
If you do the simulation with one molecule as I suggested with the heather just a little hot to avoid that the molecule reaches the top, you can see how the molecule doesn’t change its color, but the parameter kT changes. With the molecule jumping to about half the maximum height I seen kT starting from 7.8 reduced to almost zero and returned to 7.8, but the molecule maintained its color. It seems that it changes its color only when bumps something.
I don’t believe that simulator is good for say anything, at least by its molecules color.
Have a nice day
Massimo
From the top, select for “Setup; 1 Gas,” One Moving Molecule, Check the box by “Heater”, set Simulation Speed at about 10%, slide the molecule count all the way to the left to see just one molecule, put the color scale bar in the middle, set the heater temperature to about 10% and the gravity to about 40%. Now, watch kT as the molecule moves up and down.
Did this before I went back and noticed you specific the top and bottom 100, rather than top and bottom half, so I had been counting just the dots that looked closest to white, typed a period for each to track/check my mental count against.
Bottom half: ………………………………………………………………………………………………………………………………………………………………… 183 of 916
Top half:
…………………………. 31 of 83
So 214 of 999 were white, about 21%, 18% in the bottom, 3% in the top.
Bright yellow in the bottom half:
…………………………………………… ……………… 69 of 916 (51 in the bottom quarter)
Dark blue in the bottom half:
………………………………………………………………………………… …………………. 115 of 916 (93 in the bottom quarter)
Bright yellow in the top half:
……. .. 9 of 83 (2 in the top quarter)
Dark blue in the top half:
…………. .. 15 of 83 (2 in the top quarter)
252 of 916 in the bottom half were white/bright yellow.
40 of 83 in the top half were white/bright yellow, 10 of which were in the very top quarter.
I agree that the sim has bounds which can be broken, a case in point is turning heat on, all the way down, and gravity all the way down, the particles will settle to the bottom and then spontaneously begin spawning in the upper sections and falling down.
Whether you use the heater or just start with the equal speed 1 particle, sim speed all the way down, experiment with different grav/particle counts -> raise the speed to max, you don’t get an isothermal state, just a smooth distribution of velocities and a density gradient.
Hi Massimo,
you need to be patient with me. I try to explain a result
of statistical thermodynamics that has been derived by Boltzmann, Arnold Sommerfeld and others. In an isochoric, isolated gas column in a gravitational field the equilibrium distribution is a unique Maxwell-Boltzmann distribution, multiplied at every height by a factor of exp(mgh/kT). This result implies homogeneous temperature throughout the column.
It is way above my knowledge in statistical thermodynamics to derive a different result by myself. I only try to understand the result even though I am equally puzzled by the questions you and Max ask.
I think the discussion in the internet is crowded by homebrew models that neglect and select important points and boundary conditions.
Therefore I appreciate that you and Max try to understand.
The boundary condition isochoric, isolated and gravitation and molecular diffusion only is very different from the boundary condition, open system and heating from below and cooling on the top with turbulent mixing. Mixing these two different situations will certainly lead to error, because they cannot be compared.
My model with the higher probability of a fast molecule on the bottom to reach the top is my way of deriving why the unique equilibrium MB distribution above comes about starting out with a temperature gradient. In a situation with a temperature gradient we have a different MB distribution in each layer.
I can phrase it macroscopically.
The molecular diffusion coefficient however is a function of temperature. This does not go away by applying a gravitational field. So in our starting condition with the temperature gradient and the density gradient of hydrostatic balance. the molecular diffusion current up is higher than the molecular diffusion current down, because of the different temperatures.
The result is molecular diffusion homogenizes the temperature gradient with and without a gravitational field in an isolated and isochoric gas column.
This is very different in an open system that is heated from below and cooled on top with turbulent mixing in a gravitational field. Turbulent mixing in this situation does not homogenize the temperature. Turbulent mixing homogenizes dry static energy.
I think personally it is more rewarding trying to understand these differences first, before claiming one has disproven the existence of the greenhouse effect.
Again, therefore I appreciate it that you and Max try to understand.
But also, I may be wrong. However, my knowledge is not good enough to claim that Boltzmann or Arnold Sommerfeld were wrong.
Best regards
Guenter
Hi Guenter,
first of all thanks for your post.
I agree with you about almost all you wrote.
In this context I’m not trying to disprove the greenhouse effect, I’m just looking for how much the vertical temperature gradient could be due to other effects. Since it seems to me a little overestimated the effect of GHGs to justify all the dry lapse rate.
I’ve also one doubt about the use of MB distribution in a gravitational field, even not warmed by the below. Because I read in a physics book that one of the conditions “sine qua non” for apply the MB distribution is that the system must be isotropic.
My question is:
could the space subject to a gravitational field be considered an isotropic system?
And more, I read that when Boltzmann deduced the column density, he assumed that KE = mgh which is the average KE of the column not the KE of all the single molecules.
Thus, why should T considered the constant temperature of all the molecules trough the height of the column? Should’n it be considered the average temperature of the whole column too?
The more I read, the more I fell confused.
Have a nice day.
Massimo
Hi Massimo,
a single molecule does not have a temperature.
The equilibrium is characterized by the following:
The temperature of the column is T.
The average kinetic energy is homogeneous throughout the column.
But the column is not isotropic, because we observe a density gradient due to gravitation.
Best regards
Günter
Hi Guenter,
I’m not sure where you get that a single molecule doesn’t have any temperature.
Maybe you are arguing that it’s very difficult to be measured, but this physicist forum seems assert that also a single molecule have temperature indeed.
http://www.physicsforums.com/showthread.php?t=80111&page=2
So, I don’t understand why many molecules should behave differently than a single one, and in this case the worst is that it seems that many molecules “stores” in some way KE at the TOA???
In my opinion, what Boltzmann meant with e^(mgh/kT) was that was the energy acquired at the ground by any single molecule which rebound there. And the MB distribution should be recomputed at any gravitational significant level to get the real distribution there. Note that if you do that with a temperature lapse rate included in the T term, the pressure at ground doesn’t change so much, it changes only at very high altitudes.
Thank you anyways for your kindly reply.
Have a nice day.
Massimo
Hi Massimo,
What is the temperature of a single molecule that has a velocity of 300 m/s?
It is not possible to assign a unique temperature to a single molecule.
Temperature is a macroscopic concept.
Best regards
Günter
Maybe I’m wrong here, but let me explain my point.
I believe that a molecule by itself could be your case, but in a gravitational field is an another story. In the second case, the molecule is not alone, it’s interacting with the mass of the body which generated the gravitational field (the planet in our case).
In this case its KE should be its temperature too.
My point is: when that molecule it is thrown away from the vibrating ground it received a KE which is a function of the ground temperature, and when it returns to the ground because of the gravitational field it release the KE in the form of heat back to the ground.
If the ground has cooled a little during the molecule “trip”, the single molecule falling down should warm up a little the ground anyways (even if for an infinitesimal value) . And since it’s a thermodynamic exchange in this case the molecule must have a temperature higher than the ground.
Of course, as already said maybe I’m wrong, but I don’t find my hypothesis so impossible. If I missed something I would like to know.
Thank you for your patience.
Have a nice day.
Massimo
Günter 9:54pm: “The equilibrium is characterized by the following: The temperature of the column is T. The average kinetic energy is homogeneous throughout the column.”
The Poisson derivation I mentioned earlier in the thread for the exact ideal temperature(p) profile in a tall perfectly isolated unsaturated (dry) column of air in gravity field at equilibrium results from looking into many basic meteorology texts:
T(p) = To * (P(z)/Po)^R/Cp
Those terms should be familiar, if not ask. The M-B distribution derivation requires the assumption of no external forces on the molecules so M-B is not strictly valid in a g field.
For a column of air in 0g field then yes, p=constant so this reduces to T(p) = To constant and M-B holds.
Add a gravity field g and P(z) is no longer constant as this standard eqn. for T(p) results where the M-B distribution is no longer strictly valid. Left alone, the convective column mixes itself to equilibrium at max. entropy with the ideal exact Poisson T(p) profile. For a citation of the formal math I like to use Bohren 1998 sec. 4.4 where T(p) and entropy maximization is worked in detail. As Tim wrote above, can’t resolve it here.
Since the ideal exact T(p) in g field varies as thermometer is raised (as Massimo deduces) throughout z height, then the parcel avg. KE cannot be homogenous with gravity however the column enthalpy IS constant (the conserved quantity over z).
Run the numbers for earth, see what T(p) profile you find for the standard atm. P(z). A good exercise. Shows why many just use adiabatic g/Cp for the approx. lapse rate. And the standard lapse is not majorly different in T either.
Hi Ball4,
that is what I explained above.
If you have a mechanism for turbulent mixing you get the temperature gradient.
But you need energy input and output for this stationary state.
This is Bohren’s statement. He emphasizes mixing by convection
In an isolated isochoric column there is in equilibrium no mechanism for turbulent mixing by convection because we have no heating or cooling.
Gravitation doesn’t change that.
Mixing by molecular diffusion which equals conduction in solids or fluids does lead to homogeneous temperature.
Regards
Guenter
Guenter 2:09pm – “Mixing by molecular diffusion which equals conduction in solids…”
Bohren p. 167: “(Mixing) in the atmosphere and other fluids differs fundamentally from energy transfer in solids, in which mass motion is absent.”
Tim is right, can’t settle this here. In Bohren’s eqn.s and detail discussion “turbulent” is not mentioned. The eqn. I cited (Bohren eqn. 4.149) stands for both no gravity and gravity conditions. Pressure constant in the column (no gravity) leads to homogenous temperature; pressure not constant (with gravity) leads to temperature gradient T(p) shown for ideal exact solution at equilibrium.
Place two IR transparent spheres with an iron inner core in a vacuum chamber one filled with dry CO2 the other with dry N2. Heat the cores with an IR laser.
Which iron core cools faster?
Which gas cools faster?
Repeat the experiment this time add 1% H2O to the gas mixture.
Hi Ball 4,
we might not settle this
Bohren writes:
But in the atmosphere energy transfer in an isolated layer is dominated by convection rather than conduction. So his derivation is valid for a layer with convective motion or as I called it “turbulent mixing”.
But he does have a problem. Isolation and convection is mutually exclusive and Bohren does not mention a mechanism that maintains his convective equilibrium. So one needs to assume that he assumes heating from below, cooling on top. But this in turn would not qualify for an isolated layer. So again he does have open ends and we would need to ask him.
In the absent of convection with only molecular diffusion respective conduction Bohren has made no derivation.
Others have, like Boltzmann or Coombes and Laue
They have derived the result for a column of gas
https://www.vttoth.com/CMS/physics-notes/72-on-the-barometric-formula
Regards
Guenter
Guenter 9:58pm: An interesting discussion. I read Bohren as this being settled. Comments still open so continuing…
Guenter: “…one needs to assume that (Bohren) assumes heating from below, cooling on top. But this in turn would not qualify for an isolated layer. So again he does have open ends and we would need to ask him.”
Bohren 1998 p. 164: “…consider an isolated layer of atmosphere…neither heated nor cooled by radiation nor by interactions with adjacent air (or ground). What is the equilibrium temperature profile in this layer?”
In more detail Bohren 1998 poses: “What temperature and pressure profile maximizes the total entropy subject to the constraint of constant enthalpy?”
Bohren 1998 answer: “Entropy maximization requires the equilibrium temperature…to decrease with height…” as shown in the Poisson T(p) formula I wrote out; T(p) becoming constant only with constant pressure P(z)=Po (at 0 g).
The Toth link based on the Coombes&Laue 1984 paper assumes a distribution function f(z,v,t) and then shows this assumed distribution to be isothermal. The problem is they do NOT show this assumed f(z,v,t) is max. entropy state, it is just one possible solution that is NOT proven to be at max. entropy equilibrium. In fact, entropy cannot be max. in Toth (or CL1984) assumed f(z,v,t) with gravity field. Bohren 1998 does maximize entropy in the column with gravity, a more basic and general solution. Very cool.
Guenter 2:01pm: “What is the temperature of a single molecule that has a velocity of 300 m/s?”
For H2, by strict definition of temperature, 7.28 Kelvin. The temperature we (and thermometers) feel is macroscopic but the strict definition of temperature shows can calculate T for any number of molecules in a parcel even just 1.
Do not be impeded by assuming only molecular diffusion (conduction) when convection (mass motion) is possible. Bohren 1998 p. 167: “…conduction in solids can impede our understanding of atmospheric convection.”
Hi Massimo,
you write:
“And since it’s a thermodynamic exchange in this case the molecule must have a temperature higher than the ground. Of course, as already said maybe I’m wrong, but I don’t find my hypothesis so impossible. If I missed something I would like to know.”
You apply thermodynamic concepts to single molecules and you miss the following:
You repeatedly miss that thermodynamic is a macroscopic axiomatic theory.
It can only be applied for macroscopic ensembles of molecules. Thermodynamic concepts like temperature can not be applied to single molecules. If you do that your hypothesis will be based on wrong assumptions
and you will reach the wrong conclusion.
Best regards
Guenter
Hi Guenter,
maybe that you are right.
I know that thermodynamic as been built over the direct experience of what happen in nature, so it’s the result of approximation and axioms. Anyways, what intrigue me is this change of behavior of matter in the microscopic world.
About your link about Coombes and Laue, I don’t know much of statistic, but I read:
“As per the equality established earlier, this also means that
f(z2,v2,t2)=Ce^(mg(z2−z1)/RTe−mv(2)^2/2RT).
The first part of the right-hand side is a function of positions only. The velocity-dependent part of the distribution (hence, the relative probabilities of different velocities) is the same as before. In other words, the temperature is the same, regardless of the altitude; only the density of the medium varies with height.”
I see that the velocity part has changed indeed: now is v(2) not v(1).
Again, I miss something?
Thank you for your interesting posts.
Have a nice day.
Massimo
Roy,
One of these days I’m going to put together the pieces I already have to do the definitive experiment. Instead of sunlight, where you continually have to reorient the boxes or mount them on a solar tracker, which I don’t have, I will use an electrically heated plate and measure the power dissipated in the heater coils.
I’ll mount the box upside down several feet above a floor so convection will be minimized. It will be inside so ambient won’t vary much and no wind. The box is insulated with at least six inches of fiberglass contained within a box constructed of 1/2 inch polystyrene foam board. The interior side walls of the 12 inch deep interior box are covered with highly reflective metallized film (space blanket).
Then I can run the box with no cover, a thin polyethylene cover and a glass cover and the input power will remain constant. I’ll have multiple thermocouples constantly measuring the heated plate, the temperature profile in the box and the cover temperature.
Just wondering about the cooling during night-time.
Why is the box cooler than ambient air ??
Radiation losses to a clear night sky ? If so it is very much like covering tha car glasses with a blanket to avoid frost on the windows. I do not see that as a new discovery – rather something people has known for a long time. One can reduce radioation losses by covering an object with a poor IR emitter.
My second concern is the temperature of the plexiglass during day-time. Is it a good IR absorber (of suns IR) ? Does the plexiglass become hotter than the box ?