2
$\begingroup$

Based on our current knowledge of CO$_2$ and current instrumental techniques, how big would a laboratory sample of air with 400ppm of CO$_2$ need to be in order to quantify the back radiation from CO$_2$?

I am envisioning a large container with many cubic metres of gas, probably thermally isolated longitudinally and with a radiative IR input at one end (A) plus measuring apparatus at the other end (B) and some distributed thermistors throughout. We know the thermal input at A, measure the thermal output at the B and measure the trapped thermal energy via the thermistors.

I would like to get a result with something like +/- 1% accuracy. Would this be possible? Assume I have a budget of at least $10,000,000.

Improve this question
$\endgroup$
2
  • 7
    $\begingroup$ I suspect this can be done with less gas volume and much less money $\endgroup$
    – Gimelist
    Apr 17, 2017 at 5:32
  • 1
    $\begingroup$ this question might be among the top ten of hardest to answer,there is so many things that needs to be solved to get a correct measurment. $\endgroup$
    – trond hansen
    Apr 21, 2017 at 11:05

4 Answers 4

Reset to default
5
$\begingroup$

The thing is, we don´t have any instruments(to my knowledge) to measure the intensity of that radiation. But, we measure the intensity of atmospheric emission from space with satellites, which is seen as the "effective temperature". Logically this is the same as the net radiation downwards. And heat transfer then confirms that no net radiation arrives at the surface. Which is equal to "no radiation at all". Because "net" radiation is the only radiation that can heat anything.

There are lots of people claiming that measurements done with IR sensors and pyrgeometers confirm backradiation. This is not true and it is based on a misunderstanding of the function of the device, as well as misapplication of the stefan-boltzmann equation. Both IR-thermometers and pyrgeometers use the s-b equation in combination with a thermocouple or thermopile(pyrgeometer). They are both a kind of more advanced and more sensitive thermometers. The thermopile consist of a sensor measuring a gradient across itself indirectly by changes in resistance acting on a current through the thermopile. So, by determining the gradient in the thermopile, it can sense the amount of heat transferred to the surroundings using the s-b equation. When someone claims measurements show "back-radiation", they refer to measurements done with these devices. As most interested people understand, a measurement of heat transfer away from the device, cannot show incoming radiation from a colder atmosphere. If you measure something hotter than the temperature of the thermopile, sure, then you have high accuracy. But a measurement of "back-radiation" from the atmosphere is per definition a measurement of something colder than the device. Because the atmosphere is almost always colder(exception is inversion, but that is not relevant). So any measurement of backradiation made from the surface, is a measurement of nothing. It is simply what the device doesn´t measure coming from the atmosphere. This is what I mean with "misapplication" of the s-b equation. It can not, and should not, be used that way. That is, to add flux densities that are not calculated as "net" transfer.

In regards to your suggested experiment. Unless the gas is hotter than the instruments(if using IR-instruments), you will measure a heat sink, not any backradiation. And, yes, that is what greenhouse gases really are. Heat sinks. The earth emit its heat to the ultimate bottomless heat sink in space, the only observed infinity, and greenhouse gases are added heat sinks along the highway for heat, that leads to 3 Kelvin space.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ I am aware of the limitations on using pyrgeometers as indeed anyone will be if they bothered to actually read the manufacturers manuals and understood how they work and what they can and cannot do. It amazes me that even well educated professors think they can just grab some instrument out of the suppliers catalogue and use it in a lab experiment without even thinking about how said instrument works.(or RTFM). Some clever people sure are dumb ! Nice to see you are not one of them. $\endgroup$
    – user7733
    May 14, 2017 at 22:59
3
$\begingroup$

Check out www.arm.gov an ongoing set of Atmospheric Radiation Measurement experiments at many locations around the globe that measures many aspects of the atmosphere including atmospheric back radiation using infra-red spectroscopy. This work started in the mid 90's. Some of the cleanest data comes from the https://www.arm.gov/tour/north-slope.html North Slope of Alaska ARM research station.

Here are some articles on the research.

https://www.nature.com/articles/nature14240

https://newscenter.lbl.gov/2015/02/25/co2-greenhouse-effect-increase/

Improve this answer
$\endgroup$
0
$\begingroup$

It seems like you use a tube of sufficient volume that the effects of the interior walls are de minimus on the air in the tube, and long enough for there to be some temperature gradient top to bottom. Then you need a "piston" made of wire mess of a material with high emissivity. The wire mesh is so that the piston has de minimus effect on convection. Finally you need a hot plate at the bottom that produces a known amount of radiation and thermometer close to the plate to monitor temperature. Then you measure the temperate with "no piston" in the tube, and compare with the piston in the tube at various distances from the plate.

It seems this would enable one to prove/disprove back-radition.

Improve this answer
$\endgroup$
-1
$\begingroup$

"Logically this is the same as the net radiation downwards".

No, this is not the case. The atmosphere is almost black and opaque at certain wavelengths such as 15um (CO2). Then only the external layers radiate, mostly. The rest of the atmosphere transport heat by convection.

The atmosphere is black because an absorbed IR photon energy will excite a CO2 molecule, for example, for a time which is 1000 x bigger than mean time between collisions in the case of low altitude. The molecule will deexcite through collision first.

Due to the gravity gradient, the top of the atmosohere is low pressure and mean time between collisions is much longer. We can see CO2 radiation from a thicker layer, this is the little glitch in the middle of the 15um band dip.

But at low altitude, only 10m of atmosphere will backradiate and the energy density is low because the temperature difference with earth is small.

The backradiation diagram from Nasa in ipcc AR4 is simply wrong. This happens (remember Mann hockeystick scandal).

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Hi Backradek, welcome to Earth Science. I think you have mispreresented radiative-convective equilibrium, at least your point goes against established scientific consensus. All atmospheric layers radiate (Stefan-Boltzmann Law), which transports heat when layer temperatures differ, greenhouse effect is (more or less) because TOA radiation is from colder layers than the surface. Can you cite sources supporting your claim that IPCC AR4 quoted energy diagram is wrong? $\endgroup$
    – gerrit ♦
    Feb 10, 2020 at 14:08
  • $\begingroup$ @gerrit "Established scientific consensus" means nothing in sciences. Science is not based on consensus and not on authority sophisms. Climate science is new, polluted by politics, and very far from any kind of convergence even if this had some meaning. The SB and Kirchhoff laws apply to a black body facing space, not to convecting gasses. Basic physics tell us that the pressure gradient in the atmosphere leads to a dissymetry in the CO2 reemission (see my other post), leading to an IR flux towards high altitudes. This flux of course has a cooling effect. $\endgroup$
    – Backradek
    Feb 12, 2020 at 14:27
  • $\begingroup$ This flux has a cooling effect on the atmosphere and a heating effect on the surface. You are right that scientific consensus can be wrong, if you claim it to be wrong you need to provide strong (peer reviewed) evidence. Stefan-Beltzmann and Kirchhoff laws apply always, everything radiates, including convecting gases (but radiation is not the only means of heat transport, hence the radiative-convective equilibrium). Your other answer didn't answer the question and was converted to a comment, unfortunately the second half of the post was lost in the conversion (flag that post for access). $\endgroup$
    – gerrit ♦
    Feb 12, 2020 at 16:07
  • $\begingroup$ How could this flux from the surface cool the atmosphere and heat the surface ? Violation of the laws of thermodynamics. Energy goes from hot to cold. What heats the atmosphere is for example the dip around 15um in the emission spectrum of Earth, given the fact that the total average emitted power must be equal to the sun average input. Yes everything radiates. If you use SB law for a thing which is not a blackbody and not in equilibrium and not facing space, then you have a little problem, no ? This is empirical at best. In the case of Earth radiating to space, then I have less of a problem. $\endgroup$
    – Backradek
    Feb 13, 2020 at 18:26
  • $\begingroup$ You misunderstood what I meant by "cool the atmosphere and heat the surface". Of course, directly speaking, radiation cools the surface and heats the atmosphere; I mean relative to the absence of those greenhouse gases. The presence of greenhouse gases heat the surface, the presence of greenhouse gases cools the upper trophosphere. Does not violate laws of thermodynamics. $\endgroup$
    – gerrit ♦
    Feb 13, 2020 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.