Albedo & Reflected Solar Energy (in the Arctic) 5

[racookpe1978]

I am trying to program a set of spreadsheets to predict the solar heat energy available to the ocean and ice surfaces in the far north: that is, above 70 degrees latitude.

Solar incident levels (the angle of the sun above the horizon) are easy to predict for any day of the year. This equation also provides the hours of sunlight per day, which ranges from 0 through 24 depending on day of year.

Absorption in the atmosphere can be approximated from the thickness of atmosphere that the light energy passes. Cloud reflection and amount of haze in the air are significant have to be addressed, but that will be later.

I have not found a reliable reference for albedo - but have read hundreds of times with little more than "ice reflects 97% of the sun's energy" or "the (open) ocean absorbs 90% of the inbound energy." These might be adequate for some uses, but are only good for very high (if not vertical) incident angles: which will only happen if I were on an ideal tropical island below a perfectly clear sky on the equinox in a dead calm.

(1) So, at very low angles (below 20 degrees), what is the best reference to specify what proportion of the sun's energy is absorbed (by ice and by open water) and what proportion is reflected?

(2) In the real world case of moderately to very calm seas (waves between 6 inches and 1 foot), does the open-water albedo change significantly from laboratory conditions?

 

[IRstuff]

This page seems to have some good stuff:

http://www.philiplaven.com/p20.html

Note that water's Brewster's angle is about 53°.

http://en.wikipedia.org/wiki/Fresnel_reflection

shows that above that sizable portions of the visible spectrum is reflected. This graph shows ideal water:

However, that's only relevant in the visible. Water is very dark in the infrared, but that's mostly low net emission. We can see solar glint beyond 3.5 um, so there's still a bit of reflectivity at that wavelength.

Snow ostensibly has similar behavior at the crystal level. However, unpacked snow is appears somewhat Lambertian, since the low physical density allows light to rattle around.

TTFN

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[racookpe1978]

Seems like I'd want to
(1) assume (or declare) that I'm using the unpolarized (circular or yellow line) average condition for sunlight (that is, the light won't be polarized until after it has reflected up),
(2) assuming "laboratory" water (flat, no waves)
(3) that I'd want to break the analysis into at least 2.5 degree sections: there is a significant change in absorption with angle after 70 degrees should not be ignored or "approximated" out by using just one number, even 10 degree incident angle sections.

NSIDC gives no values for sea ice reflection (nor land ice) in their web pages, though they caution "many" areas of the sea ice extents have snow covering the ice up to 1-2 meters. Should it be similar to water?

 

[IRstuff]

I'd assume that unpacked snow is mostly Lambertian. As it melts and refreezes, it'll behave more like flat, liquid, water, I think.

You may consider RF'ing this posting and have the moderator move it to forum386 or post there, asking for responses here. You still might not get that many responses, since it's bit of an academical problem, i.e., there's probably a professor, somewhere, who's REALLY knowledgeable about this subject, but isn't a member of ET.

TTFN

FAQ731-376
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[MintJulep]

Poke around here:

http://www.crrel.usace.army.mil/

 

[GannetAM]

If you have not already, you should checkout

http://nsidc.org/seaice/processes/albedo.html

Also, checkout

http://www.gi.alaska.edu/~eicken/he_teach/GEOS615icenom/albedo/albedo and climate.htm

 

[GannetAM]

Thank you for this thread

At home now where tons of my research on the Arctic is available.

The following is from Bitz, Cecilia M. & Shawn J. Marshall; 2011; Modeling of the Cryosphere, in the ENCYCLOPEDIA OF SUSTAINABILITY SCIENCE AND TECHNOLOGY, section on Climate Change Modeling and Methodology; Available at

http://www.atmos.uw.edu/~bitz/514_2011/Bitz_Marshall_2011.pdf

"Surface energy balance, radiation, surface albedo, and melt ponds
The net flux entering the top surface of land and floating ice, soil, or snow is a sum of radiative and turbulent heat fluxes:
F(T)|top = Fr(1 ? ?) ? I0 + FL ? ??T4 + Fs + Fe, (6)
where Fr(1??) is the net downward solar irradiance above the top surface, ? is the surface albedo, Io is the solar irradiance that penetrates the top surface, FL is the downward longwave irradiance, ??T4 is the upward longwave irradiance (for T in Kelvin), ? is the emissivity, ? is the Stefan-Boltzmann constant, Fs and Fe are the downward sensible and latent heat fluxes,
respectively.
In many ice and snow models, the surface albedo is a function of various quantities such as temperature, snow grain size, snow age, impurities, snow depth, ice thickness, and melt pond coverage. Often shortwaver radiation is absorbed in the ice interior based on Beer’s law, although Beer’s law is inappropriate in materials with depth dependent surface albedo parameterizations. Usually the temperature dependence of the surface albedo is a proxy for modeling melt pond, grain size, and/or surface scattering characteristics. These relatively crude methods are being revamped considerably in models at this time.
A better way is to design a highly interdependent set of physics for radiative transfer, ponding, and liquid infiltration. Ideally one would have radiative transfer account for multiple-scattering and be based on intrinsic optical properties that vary with impurity concentrations, snow grain size, ice bubbles, and brine pockets. Ponds would accumulate water above sea level when there is insufficient hydraulic connectivity to drain meltwater, and they would accumulate below sea level when there is hydraulic connectivity is high enough to allow liquid water to rise up from below and flood the surface.

Also, see attached

 

[racookpe1978]

Good detail, and much more reading to do. 8<)

Many thanks for your time, I value your aid. Please check the "melt" file, the link is failing at this end.


That equation is more eleborate than the very simplified approach used at NSIRD in the papers I've read there (and in several of college "physics/earth science descriptions I have read on-line).

? ??T4 ... T in Kelvin of course, but at the surface of the ice exposed to the air, right?

Can Fr be approximated by the solar constant for dry air at the equator times a linear approximation as an attenuation factor for the many extra kilometers of atmosphere those rays must cross to reach the ice at sea level at 80 north?

None of those papers at NSIRD that I've found used the actual solar incidence angle found in the high Arctic at actual time of year. Most papers, actually, used an even more simplified model that had no correction for the sun's angle at all, and only assigned two albedo values: one for water with the light directly overhead, and one for "ice" with the sun in the same position directly overhead.

In the Arctic in mid September (when the sea ice extents are historically at their minimum, which will be when the maximum amount of ocean water is exposed to the sun) the sun is (at most!) only 33.5 degrees above the horizon at noon.

 

[IRstuff]

a nitnoid, "??T4 is the upward longwave irradiance (for T in Kelvin), " should be "radiance" not irradiance, since that term represents the Planck blackbody emission.

Fr can be approximated by the solar constant with the angle correction to account for the fact that the solar constant is a wattage/m^2, and that power is spread out across a larger area. Note, however, Fr should also include the irradiance from the atmosphere itself, which is the scattered blue light. The atmospheric transmission, however, needs to cranked through something like Modtran or Lowtran to get the correct transmission, depending on the angle and visibility. The linear scaling might be OK, but some of it is not linear, like the scattering of blue light by air molecules.

Note that the last two terms include convection and condensation/evaporation, and might be quite small if the air is cold enough.

TTFN

FAQ731-376
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[IRstuff]

Oops, my bad, linear scaling is incorrect anyway. It's an exponential factor, i.e., if you take the solar constant at the surface and divide by the solar constant at the edge of the atmosphere, you get the net transmission through the air mass. That should equal exp(-alpha*D), where D is the thickness of the air mass, and alpha is the extinction coefficient. You would then apply exp(-alpha*D_artic) to get the transmission at the lower sun angle. That's basically Beer's Law:

http://en.wikipedia.org/wiki/Beer's_law

TTFN

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[GannetAM]

I found the source for the melt attachment checkout pages 32 through 35

http://www.iac.ethz.ch/education/bachelor/climate_systems/notizen/The-Cryosphere.pdf

 

[GannetAM]

Here are some more, I believe you find helpful. They are from an online textbook I've been studying.
Citation as Goosse Hugues., P.Y. Barriat, W. Lefebvre, M.F. Loutre and V. Zunz, (date of view). Introduction to climate dynamics and climate modeling. Online textbook available at

http://www.climate.be/textbook.

http://stratus.astr.ucl.ac.be/textbook/chapter1_node16.html

http://stratus.astr.ucl.ac.be/textbook/chapter2_node6.html

http://stratus.astr.ucl.ac.be/textbook/chapter4_node8.html

http://stratus.astr.ucl.ac.be/textbook/chapter2_node3.xml

 

[racookpe1978]

Thank you!

I will check the second set tonight; your cryosphere data matches (and confirms) the other lists that I have found.

 

[racookpe1978]

My boss is rejecting this:

Wants to use the simplification that
albedo ocean = 0.07, emissivity ocean = 0.98
albedo sea ice = 0.80, emissivity ice = 0.99

For a solar constant = 1370 w/m^2

Thus, he claims
summer (open ocean) heat absorbed = (1-.98)(1-0.07)1370 = 27.2
and
winter (sea ice present) heat absorbed = (1-0.99)(1-0.80)1370 = 2.74

I have got him to concur with a month-month correction for solar constant, and for a air mass correction.

Values for ice and water 1.33 and 1.31 are very close, and, from the Fresnel chart for average light, I see that 0.35 of the energy is reflected for both sea ice and doe open ocean at 80 degrees incidence angle. Thus, if I ignore air mass losses) maximum energy absorbed open ocean = 1370*(1-0.35)(1-0.07)=828 w/m^2
That is, energy absorbed = energy available at the surface *(percent not reflected due to the Fresnel reflection) * (amount reflected by the surface albedo (its color and roughness).
His function for transmittal of the absorbed light is irrelevent. It's as if he were using a formula for panes of glass, not a flat ocean surface.

 

[IRstuff]

Well, his results fly in the face of known properties of the oceans, which is that they absorb a truckload of heat, the same heat that fuels hurricanes. His equations treat emissivity as if it were reflectivity, which is not correct, in addition to mixing spectral regions. Typically, albedo refers to the visible light and adjacent bands, while emissivity generally refers to the thermal bands, i.e., wavelengths above 2.5 um, which us sticklers for accuracy refer to as real "infrared."

So in the infrared bands emissivity is numerically equal to absorption, since blackbodies and graybodies don't transmit. So, without that term, most of the energy is absorbed, which fits how hurricanes get their energy from warm ocean water. So, based on ideal blackbody behavior, 90% of the solar constant comes below 1.6um wavelength. The emissivity comes into play in the radiated emission, which, assuming a 17°C ocean temperature and 230K sky temperature results in about 238W/m^2 of radiated loss of the absorbed heat.

The overall heat balance in the atmosphere is a pretty involved subject.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[racookpe1978]

So... The fight will continue. Thanks for the confirmation that I've to keep plugging at this.

I'm going to schedule some time down south at the university in their physics dept: clearly, he isn't going to accept my equations with a fight, and it appears that he's "pre-judged" against what I've found. Odd, you'd figure that if I get assigned (take on) a project, then he wouldn't fight the results of what I find.

 

[IRstuff]

Yep, we could spend hours discussing the vagaries of bosses. Just imagine if your general manager walked into your office and argued with you about those equations. We had one like that; he lost his entire division to another GM because he was too busy playing in the sandbox instead of minding the store.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize