Lots of questions, which is good, but so many I’m not sure I can answer them all.

In your earlier questions (August 12, 2010 at 5:56 pm) about the maths.. the full solution (using lots of substitutions) is tedious to write out but hard to figure out without me writing it out.

The integration is from χ* = 0 at the top of atmosphere to a value we want to find out at z=0.

F↑ – F↓ = F0, where F0 is the solar energy absorbed, averaged over the surface area.

From August 12, 2010 at 8:15 pm, the missing link here is the radiation to space takes place at higher altitudes as the concentration of “greenhouse” gases increases.

The grey model is just a very simple conceptual model to help visualize what happens in the troposphere. Convection actually takes over as the heat transfer mechanism to reduce the lapse rate from its radiative equilibrium.

More “greenhouse” gases = the same convective lapse rate but a higher average radiation location to space. A higher location is therefore colder, and therefore less radiation is emitted from the earth’s climate system. Let’s call the new location, L*.

Therefore (with less radiation emitted from the climate system to space), in this simplified model the world heats up, and with the same convective lapse rate the temperature at L* will become higher. Eventually, L* will be hot enough so that the radiation into space matches the solar radiation absorbed.

All things being equal of course.

The grey model plus convection plus the concept of height of emission into space helps explain some basics of our climate.

And why increasing concentration of “greenhouse” gases is worthy of further study.

]]>F↑ = (χ*+2). F0/2

F↓ = χ*. F0/2

The IPCC thinks we know the radiative forcing of GHG’s to within +/-10%. I assume that we know the absorption spectrum of CO2 to within +/-10%, but converting mixing ratios to radiative forcings appears to require that we pick an the altitude/optical density. If I double X* from 0.04 to 0.08, I get predicted temperature rise of 2.1 degK and if I double from 0.01 to 0.02, I get 0.5 degK. If I double from 1 to 2, the temperature goes up from 27 degK. This implies that forcing varies with altitude.

Questions: 1) When converting the optical properties of GHG’s to radiative forcing, does one need to pick an optical density/altitude? 2) If one does pick an altitude, how do we know that the correct altitude has been chosen? If a fixed lapse rate is going to translate a temperature rise at the tropopause back to the surface of the earth, we better start at the correct altitude.

]]>My earlier analysis (8/9 18:45) about whether the F vs X* graph should be linear bogus. How do you get from the integrated/flattened equations:

dF↑/dχ* = F↑ – πB missing a minus sign?

-dF↓/dχ* = F↓ – πB

to B = (χ*+1). F0/2π?

I’m not sure if F↑ and F↓ are magnitudes or have different signs for different directions. Does F↑ + F↓ = F0 or F↑ – F↓ = F0? Or -F0? Looks like we may need to integrate over X* at some point, but the limits aren’t obvious.

I may understand the following:

nB(T) = eoT^4 = (X*+1)F0/2

T = [(X*+1)F0/2eo]^1/4 gives cause and effect

X* = 2eoT^4/F0 – 1 gives your OD vs T graph

[F0/eo]^1/4 may be the equilibrium temp of the surface without GHGs, Teq. Teq = 255 degK for e = 0.7

T = Teq * [(X*+1)/2]^1/4

T(X*=0) = Teq/(2^1/4) = 84% of Teq = 214 degK

T(X*=1) = Teq

T(X*=3) = Teq*(2^1/4) = 119% of Teq = 303 degK

T(X*=2.26) Teq*(1.13) = 288 degK

T(X*=5) = Teq*(1.32) = 336 degK

The last temperature is about right for earth with no convection, but decreasing water vapor with altitude isn’t compatible with a uniform X*. It is becoming apparent why gray models lack some feature I expect to see.

]]>The sea is not an airplane wing.

Moving air is not an insulator.

Moving the gas removes heat effectively.

(My CPU is grateful.)

Cooling my CPU with the same volume of CO2 instead of ordinary air would cool it a little better because of higher heat capacity alone.

]]>Oh for an edit function: …so low that a temperature… not …at emperature…

]]>In a world where convection and conduction is significant, the temperature of the ground will almost always be different than the air immediately above it, depending on what you mean by immediately. The thermal conductivity of air is so low that at emperature gradient of tens of degrees per meter must exist in the boundary layer consisting of the first few millimeters above the surface (or more if the local wind velocity is very low) to obtain heat flows on the order of tens of watts/m2. Convection only occurs above this boundary layer so all sensible heat transfer between the air and the ground is, in fact, by conduction only. Convection acts to thin the boundary layer which increases the temperature gradient and the rate of sensible heat transfer.

]]>Question for SOD: Did you accidentally treat B as a constant rather than a variable that depends on temperature, B(T)?

No, B(T) is a function of χ and therefore of z:

B = (χ*+1). F0/2π

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