Johnathan: Units of radiant energy transfer were quite confusing to me until I (hopefully) mastered them. The sun’s peak output is about 500 nm (0.5 um). However, a negligible fraction of photons will be emitted with a wavelength of exactly 500.0000 nm. We can talk about how much power is emitted between 499.9995 and 500.0005 nm: 26.882 W/m2/sr. Or we can talk about the power emitted between 499.5 and 500.5 nm: 26,882 W/m2/sr. However, since Planck’s function doesn’t change appreciably over either wavelength range, both measurement are the same when reported in units of W/m2/sr PER NM: 26,882 W/m2/sr/nm. This is often called “spectral intensity”, the intensity at a particular wavelength. However, Planck’s function does vary modestly over the range 400-600 nm, so dividing the total power emitted over this wavelength range by 200 nm gives a lower, misleading value

In other words, to get the power delivered over a range of wavelengths where the Planck function varies, you need to integrate the “spectral intensity” over that range to get the “intensity” of the radiation in W/m2/sr. And you need to numerically integrate in wavelength increments small enough that Planck’s function is effectively constant. When you integrate over all wavelengths, there is a definite integral available – the Stefan-Boltzmann equation for a blackbody (W = oT^4). For non-ideal emitters or “gray bodies, we add an emissivity fudge factor (W = eoT^4).

Spectral intensity can also be reported in units of W/m2/sr/cm-1 or W/m2/sr/Hz.

Planck’s Law is for the intensity of radiation emitted in all directions – over a sphere of 4Pi steradians, which is why “sr” present in W/m2/sr/um. This is appropriate for emission by GHG molecules in the atmosphere. For radiation emitted by the surface of the earth, radiative cooling to space (OLR), or DLR or SWR arriving at the surface of our planet, we usually think in terms of the “flux” of radiation perpendicular to a surface in units of W/m2 or possibly the “spectral flux” (W/m2/um). We talk about fluxes perpendicular to a surface because “viewing angle” is critical to radiative transfer of power (Lambert’s cosine law). When you convert the intensity (W/m2/sr) emitted from a point on a surface over a hemisphere (2Pi steradians) into the flux emitted perpendicular to the surface, you pick up factor of Pi. 1 W/m2/sr becomes 3.14159 W/m2 radiated perpendicular to the surface. When considering GHGs emitting in all directions in the atmosphere, 2 W/m2/sr becomes 3.14159 W/m2 traveling upward (OLR), 3.14159 W/m2 traveling downward (OLR) and horizontal fluxes that cancel. The is called the “two-stream” approximation. Climate science usually deals with incoming or outgoing flux (W/m2) perpendicular to the surface, not intensity (W/m2/sr).

The maximum spectral intensity of thermal IR at 288 degK is at 10 um and 8.1 W/m2/sr/um. The maximum spectral flux is about 25 W/m2/um. This is the spectral flux reported by Modtran.

http://climatemodels.uchicago.edu/modtran/

Contrast it with the spectral intensity (8.1 W/m2/sr/um) reported by this online Planck calculator:

]]>Nope. It’s 8.3 W/m²/μ on the graph or from the Planck equation for wavelength adjusted for distance. If the peak of a Planck curve is 8, than the area under the curve is going to be a lot higher than 8. Your problem seems to be using the S-B equation. The total area is on the order of 1,000 W/m² at the Earth’s surface for ~6,000K minus the atmospheric absorption and the 1/r² loss from the distance between the surface of sun and the Earth’s orbit. At the top of the atmosphere it’s 1380W/m² plus or minus a couple of watts depending on where the Earth is in it’s orbit around the sun.

]]>The units are different. There is no conflict. Both numbers, assuming they were calculated correctly are right.

]]>I think you’re mixing units. The units of the Planck distribution for wavelength is W/m²/μm. The units for the Stefan-Boltzmann equation are W/m². There is no conflict between the two.

]]>That is not an answer.

My answers – feel free to dismiss them as ‘opinions’ if yours differ – are

1. About 14%

2. About 0%

What are your answers?

Getting a straight answer here is about as difficult as getting a Creationist to explain away the existence of ancient fossils.

]]>Mark wrote: “That is working backwards from the wrong end. The amount is IR an object emits bears no relation to radiation it has absorbed to reach its temperature.”

Applying conservation of energy to a problem often allows one to deduce the equilibrium or steady state for a problem without worrying about transient details. If the stratosphere has a stable temperature, the power being absorbed by the stratosphere must be equal to the power being emitted by the stratosphere. Assuming we know the temperature and composition of the stratosphere and troposphere, radiative transfer calculation allow us to calculate the upward and downward LWR fluxes into and out of the stratosphere. Conservation of energy allows us to deduce how much power is delivered to the stratosphere by SWR; the net flux must be zero.

The amount of IR an object emits is a function of its temperature and emissivity, BUT emissivity and absorptivity are equal (Kirckhoff’s Law). The temperature of the stratosphere depends on absorption and emission of LWR as well as absorption of UV. For optically thin layers like the stratosphere, emissivity and absorptivity are proportional to the concentration of GHGs and their absorption cross-sections. Radiative transfer calculations are great for summing up absorption by and emission from many optically thin layers of stratosphere. (If they handled SWR as well as LWR, they would directly calculate absorbed UV for you, but the only software accessible to me only deals with SWR.)

Mark wrote: “There is no significant exchange of thermal energy between stratosphere and troposphere, or else there would not be this temperature minimum at the tropopause.”

Yes, but there is significant exchange of LWR energy between the troposphere and the stratosphere. Stratospheric ozone is involved in both absorption of UV and absorption and emission of LWR. The temperature of the stratosphere is very sensitive the the amount of CO2 and is predicted to drop about 10 K when CO2 doubles (from increased radiative cooling by CO2).

Mark asked: “Also, I thought that only GHGs can absorb and emit IR. If the stratosphere is just N2, O2 and O3, and can’t absorb IR, how can it emit any?”

Today’s stratosphere is 400+ ppm CO2 and 1.8 ppm methane, just like the troposphere. Water vapor is about 3 ppm, the mixing ratio at the tropopause. To be more precise, some of the methane is oxidized to CO2, increasing the water vapor modestly. However, we don’t need to worry about all of these details, since programs like Modtran have been loaded with the best mixing ratio and temperature data available at the time. However, we do need to worry that the appropriate temperature and mixing ratio data are used and these values vary different regions of the planet.

I think that your calculation is based on the hypothesis that the internal energy of the stratosphere (heat capacity * temperature) is related to the internal energy of troposphere and 1 m of ocean lying below it. Equipartition of energy is important in some situations, but it isn’t clear why this should be the case in this situation. And the assumption of 1 m of ocean appears somewhat arbitrary. The mixed layer of the ocean about (about 50 m) is stirred by winds.

]]>Well, facts are facts. Energy comes in as radiation and is converted to mainly thermal energy, plus chemical energy (in plants) or electric energy (by solar panels).

Frank, the stratosphere is on average only 4% or so cooler than the troposphere, so multiply up mass x specific heat capacity x temperature and you find that the thermal energy in the stratosphere is about 14% of the thermal energy in the troposphere plus top bit of ocean (my top one metre is just an estimate for the amount that is the same temperature as the air above it).

Therefore, we must assume that the stratosphere and the ozone in it convert about 14% of incoming radiation energy to thermal energy. Solar radiation is the only source of energy that it has. There is no significant exchange of thermal energy between stratosphere and troposphere, or else there would not be this temperature minimum at the tropopause.

There’s no need to average incoming solar between day and night in this context, 14% is 14%.

“Therefore the tropical stratosphere emits 10.5 W/m2 more LWR than it absorbs”

That is working backwards from the wrong end. The amount is IR an object emits bears no relation to radiation it has absorbed to reach its temperature.

I could warm up two identical objects to 50C. I have warmed, one by putting it in warm water and the other by putting it in front of a three-bar electric fire. The first hasn’t absorbed any radiation whatsoever, the second one clearly has. But the two objects will be emitting the same amount of IR.

Also, I thought that only GHGs can absorb and emit IR. If the stratosphere is just N2, O2 and O3, and can’t absorb IR, how can it emit any?

]]>DWP, re my question 2, I refer you to the first diagram in your post <a href=https://scienceofdoom.com/2010/02/11/co2-%E2%80%93-an-insignificant-trace-gas-part-six-visualization/here.

It shows 390 W/m2 up from ground and 265 W/m2 going into space. means that 1125 W/m2 is reflected-trapped-absorbed by the atmosphere. If we assume it’s only CO2 doing the trapping, that seems pretty conclusive evidence of GHG theory, until you remember clouds.

I agree that clouds appear to block-reflect-absorb all IR coming up from the land/ocean. And when I said “terrestrial” I meant upwards IR from clouds as well. Perhaps I should have mentioned that expressly. So we can ignore IR emitted by land/oceans under cloud cover and start again with IR emitted from upper surface of clouds.

I used your approach and did the more complicated calculation with a weighted average IR emitted by land/oceans and clouds, you get upward radiation 240 W/m2, which is what goes to space. So there’s no obvious “missing” or “trapped” IR.

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