Hi Doug,

Long time no see.

SoD,

I suspect you need to add ‘pseudo scatter’ to your screening criteria.

]]>No it doesn’t. The back radiation is pseudo scattered without any of its energy being converted to thermal energy in the warmer surface.

]]>lot and don’t seem to get anything done. ]]>

Thanks SoD. I see what you mean.

]]>Susan,

You are correct, but the humidity differential is a function of temperature. So it’s implied in the description.

It might be clearer in the formula..

The “bulk aerodynamic” formula:

LH = LρC_{DE}U_{r}(q_{s}-q_{a})

which says Latent Heat flux = latent heat of vaporization x density x “aerodynamic transfer coefficient” x wind speed at the reference level x ( humidity at the surface – humidity in the air at the reference level)

And the “real formula”:

LH = Lρ<w’q’>

Latent heat flux = latent heat of vaporization x density x covariance of upwards velocity variations with specific humidity variations

Obviously, the specific humidity changes at the surface due to the temperature increase.

]]>Samedi,

You seem to confirm that nobody bothered to experimentaly determine those total emissivities under 300 K? Do you think it’s worthless? Why?

Do you understand what total emissivity is?

Emission of thermal radiation, E:

E = ∫ ε(p,T,λ).B(λ,T)

where the integration is across all wavelengths

where the variables are as previously described. From this calculation you can work out a “total emissivity”.

E = ε_{total}T^{4}

many decades of work have confirmed that ε is primarily a very strong function of λ with a dependence on p,T that I previously noted.

When you have these you can calculate total emissivity for the specific conditions under consideration.

If you are asking has anyone validated the measurements in the real atmosphere the answer is yes, lots of measurements.

Because the atmosphere absorbs and emits radiation knowing “total emissivity” for a given path length with changing pressure and temperature is not actually very useful.

If you had a table of “total emissivity” from the surface to TOA as a function of surface temperature, atmospheric temperature profile, concentration of GHGs how would you use it?

The reason people don’t build these massive tables is because they don’t serve any use. Instead the useful building blocks for atmospheric emission are the extensive tables of absorption coefficients for each gas at each wavelength. These can be used.

Besides, you’re talking about line width alone. We still miss two dimensions to get to absorbtion.

You can see more about that question at:

Visualizing Atmospheric Radiation – Part Six – Technical on Line Shapes – I’ve extracted a few pages from a textbook and explained in some detail

Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Twelve – Curve of Growth

As the line spreads out at higher pressures the absorption in the middle of the line is reduced. As the line “sharpens” at lower pressures the absorption in the middle of the line is increased. There is a formula for it and the calculations of radiative transfer – numerical integration across all wavelengths – takes this into account.

In the HITRAN database:

S = intensity is the total absorption strength under the curve

γ_{air} = Air-broadened half-width – this parameter allows you to plot the curve

n_{air} = temperature-dependence exponent for γ_{air} (this is usually the power 0.5 for T0/T as explained in the earlier comment)

SoD,

sorry my sentence about decreasing emissivity and fourth power emission was indeed very badly turned… Too much in a hurry this morning. Just wanted to say, yes, it’s a bit like having the exponent less than 4.

You say, they well know that. Maybe, but what in practice?

You seem to confirm that nobody bothered to experimentaly determine those total emissivities under 300 K? Do you think it’s worthless? Why?

Take another example of things which are even more well known: how many people say that, as the surface has an albedo of 0,07, you put away the 7% reflected part, but when they compute their BB emission, do they consider a grey body emissivity of 0,93? Basic question, but I’m sorry this is not clear at all in most papers, litterally they seem to have their objects absorbing like grey bodies but emitting like black bodies. So the reader is suppose to guess what they actually have done?

You replied with a formula, OK but it’s not measurement. Besides, you’re talking about line width alone. We still miss two dimensions to get to absorbtion.

“*The total line strength stays constant. So you are “spreading out” the absorption line, but the area under the curve stays constant.*”

I’m are still missing something? Conservation of the area is not conservation of the total emissivity, it all depends on where you put the mean level. Integrating something between 0,2 and 0,4 or between 0,7 and 0,9 won’t give you the same. In terms of emissivity, one mm² on the bottom has not the same weight as one mm² on the top. Hmm I’m having difficulties, and English after 11 pm…

And of course, as you well noticed yourself, the change in emissivity – a fortiori your one quarter mention (bandwidth ratio) – still gives a fain idea of the change in terms of emission.

Take a look, e.g. at pages 32-33 (190-191) of this set of papers, I’m sure you’ll see what i mean this time…

PP,

I don’t want you to waste your time on that, my remark about HITRAN line numbers was certainly on a marginal issue.

]]>The equation for total emissivity hasn’t changed. Modest, second edition* is* recent as those things go. Its publication date is March 21, 2003. As far as I know, there hasn’t been a revolution in radiative heat transfer theory since then. The major advance has been the replacement of Hottel’s empirical graphs with calculations (Leckner, e.g.) using either full line-by-line or moderate resolution band emission calculations all traceable back to HITRAN.

Samedi

Maybe your readers also wonder sometimes if one is not prisonier of the conventions when emissivity is defined relatively to a black body whereas gases are completely different from grey bodies.

Gases are completely different from grey bodies.

What’s your point?

Gases are not black bodies – which means emissivity = 1. Gases are not grey bodies – which means emissivity(λ) = constant.

Emissivity is a physics definition. It is a **wavelength dependent** property which defines how a body emits radiation compared with a black body.

As you may have noticed yourself, my question was about total emissivities VS temperatures, and Hottel and Leckners curves are mainly decreasing, which is not exactly comparable to a fourth power law. Could you try and be serious sometime?

You were unhappy with me explaining the difference between emissivity and emission, because “*Don’t let your readers think you mock me, I know what emissivity is and they also know.*”

If you have a point, what is it?

Emissivity is not emission. Emission will change exactly to the 4th power of temperature if total emissivity is a constant:

Rout = ε_{total}T^{4}.

Usually total emissivity, ε_{total}, is not a constant as a function of temperature. Therefore emission will not change exactly as the 4th power of temperature.

What is constant is ε(λ) and Planck’s law. The rest follows naturally.

I am being serious. I cited some equations. Your questions that follow indicate you don’t understand the implications of these basic equations. That’s not mocking you. It just seems like you don’t understand them.

– The fact that emissivity varies with temperature is something that everyone in atmospheric physics already knows.

– The fact that atmospheric radiation does not vary exactly according to the 4th power of temperature is something that everyone in atmospheric physics already knows.

– No one in atmospheric physics uses Hottel’s empirical curves because they were calculated for a different application and aren’t usable in atmospheric conditions.