4 of the 5 Great Lakes about to freeze over

marque2

If it freezes over the whole top, it will be 0 degC, at least at the surface. And this is unusual. In 2012, only 10% of all the lakes had ice cover this time of year compared to 85% today.

No. no. It is far worse than that.
Look at an “average” open water lake, from bottom to surface to air temperature a few feet above (2 meter air temperature in the parlance.)
Rock below the bottom (-5 meters: like a cave, the dirt and rocks stay at 8 deg C 5 meters below the surface in Ontario)
Bottom dirt and mud = 4 deg C
Bottom water = 4 deg C (notice it went up?)
Middle water = 3-4 deg C (natural circulation, assumes no water flow)
Upper water = 2-3 deg C
Water surface = 1-2 deg C
Air film (average of water surface and 2 meter air temp) = -10 deg C
Air temp = -20 deg C
Heat lost (from the lake water) is by convection, evaporation, LW radiation (to the sky and space) proportional to 273 + 1 = 274 K ^ 4th
So, the air heats up (because it crosses a comparatively “hot” 1 degree surface, roughened by wind and waves, and it gains water vapor’s energy and enthalpy.
Now, if the lake freezes over … Assume the air temperature is still at -10 deg C.
Rock below the bottom (-5 meters: like a cave, the dirt and rocks stay at 8 deg C 5 meters below the surface in Ontario)
Bottom dirt and mud = 4 deg C
Bottom water = 4 deg C (no real difference down here)
Middle water = 3-4 deg C (natural circulation, assumes no water flow into the lake)
Upper water = 2-3 deg C
Water, right at ice bottom = 0-1 deg C
Bottom of ice = 0-1 deg C (water is freezing to bottom of ice, adding thickness below)
Top of ice = -18
Air film (average of ice surface and 2 meter air temp) = -19 deg C
Air temp = -20 deg C
But the air is “dry”! No evaporation! Heat is being radiated into the sky/space from a surface now -18 + 273 = 255 ^ 4th power.
In addition to radiation losses being reduced, much less energy is lost from the water below because the new ice cover insulates the water from the moving air, (so the film coef for convection goes down), the energy that used to “evaporate” water (heating air and changing its energy with that extra enthalpy by the water mass that used to go “up” into the air) is now inside the freezing water below the ice instead.
So, after freeze over, the downstream air is much colder, much drier, with much less humidity to deposit as snow and ice elsewhere.
That now colder air downstream cools the ground further “south and east” more than the previous humid air does with some additional effects. (Snow doesn’t melt as quickly south and east, the air down there is drier and clearer so more energy hits the still-frozen ice and snow to be instantly reflected back into space, there is less water vapor for new clouds to form so more energy is reflected from the still-lingering ground ice than from the missing upper atmosphere clouds (and, since the clouds are missing, the direct sunlight albedo and penetration are different, etc etc etc.)

Kevin Kilty

Last year, Superior had ice in places until the last week in June, maybe later. The water remains exceptionally cold. This provides a nice running start to ice build-up the following year unless some unusually warm summer weather, or a warm winter, breaks the cycle. We shall see.

True. We do not know what this coming summer will bring, not what next year portends.
Now, I am perhaps the stongest critic of Sereze’s “Arctic Death Spiral” . I will make no bones about how wrong that concept is. With ONE caveat. Sereze – and any one who quotes him or uses the same words! – is dead wrong about the “Arctic” death spiral:
Sereze’s Death Spiral is commonly explained as follows: Global warming causes regional air to heat up – to be hotter than it would otherwise be. That “warm air” causes (local) ice heating and (local) ice melt; (local) ice melt causes more open water which absorbs more sunlight (locally); more sunlight absorbed into the open water (because of water’s darker albedo compared to ice and snow) causes (locally) warmer water which causes more (local) ice to melt. Thus, there is more open water and more sunlight absorbed.”
Is he correct?
Well, his theory is correct – but ONLY where the available sunlight hitting the newly melted waters,a nd being absorbed into the newly opened ocean waters (or lake waters) is greater than the amount absorbed and reflected by the formerly frozen snow and ice!
At what latitude does a Death Spiral take place, and when does it take place?
Seems like a simple requirement of the problem, but Sereze refuses to report where that difference occurs.
See, if there is no sea ice (no lake ice) there can be no Death Spiral. If the sea ice (or land ice) copletely melts every summer now, then the theory of “thinner ice (one summer) means thinner ice (the next spring and summer) and thus “the thinner ice will melt faster the next summer.” Well, if all of the ice completely melts each summer – even if only for one day – then ALL of the next winter’s ice needs to re-freeze the next winter.
It is ONLY the combined effect of “What is the difference in energy absorbed by open water and that absorbed by ice covering that water?” AND “What is the available energy reflected from open water and from ice covering that water?” THAT is the quesion Sereze refuses to address.
Well, there is no sea ice and no lake ice in the tropics, so his theory is irrelevent to CAGW. And to global warming in general.
There is almost no waters covered by ice in the southern hemisphere – certainly none down to Cape Horn at -56 south. So, ONLY the ANtarctic Sea Ice matters down there.
And, when you run the numbers, the Antarctic sea ice IS EXPANDING, and the greater area of Antarctic Sea Ice IS reflecting more energy every year by TerrorWatts from the earth.
Sereze is correct!
More Antarctic sea ice DOES cool the planet!
More sea ice one year DOES increase the residual sea ice left over to re-freeze the next year! (The Antarctic sea ice anomaly DOES keep getting larger each year!
Much more energy IS reflected by sea ice between 67 south latitude and 58 south latitude every day than is absorbed by the previously open water!
Now, what was true between 68-69-70 south latitude and 58 south latitude is even MORE TRUE up north.
At 42 north, Lake Eire. More lake ice = more energy reflected, less energy absorbed.
At 42-45 north, Lake Michigan. Same thing. More lake ice = more energy reflected from the region, colder regional weather
At 46-47-48 north, Lake Superior? Same thing. More lake ice = more energy reflected from the region, colder weatehr, more lake ice area and thicker lake ice left over each summer (almost! – Not quite yet!) to refelct energy later each spring and reflect energy earlier the next winter.
At 60 north? Hudson Bay, Bering Sea. Yes, Sereze is correct. These still melt every year in the summer, so there has never (lately) been any multi-year thick ice remaining year-to-year. But, less sea ice still means more solar energy could be absorbed, more sea ice means more energy reflected.
(Hudson Bay has the same physics and albedoes as down south at the Antarctic’s sea ice edge: The Antarctic sea ice is “cleaner, and has a different albedo characteristic summer melt period, so you have to run the hour-by-hour math for every summer day. But three things differ between Hudson Bay and the Antarctic Sea ice.
1. There is a 1.2 Mkm^2 limit on HudsonBay ice. It cannot get any larger. It cannot get any smaller than today either! When “all” Hudson Bay sea ice melts every year, there is no more to melt. (Obvious, isn’t it?) The only thing that “more” global warming can do is melt that final bit if HudsonBay sea ice a bit earlier in the year. But it will STILL freeze each winter. Then be present to melt again the next year. Globally, there is no feedback in planetwide heat gian frmo entirely losing all of Hudson Bay’s tiny 1.2 Mkm^2 area.
2. As mentioned, Hudson Bay is tiny. The “excess” sea ice around Antarctic at latitudes 60 south and 61, and 62, and 62, and 64 etc. has routinely exceeded Hudson Bay in size recently. And that excess sea ice anomaly (the excess sea ice down south) continues ALL year to cool the planet.
3. Hudson Bay is melting during June-July-August each year, when the sun is at its lowest heat energy at Top of Atmosphere. Antarctic sea ice is instead reflecting energy all year, but the Antarctic summer is at a time when the earth is closer to the sun, and the earth is receiving more energy from the sun every second of every 24 hour day. By September, at the equinox when the Antarctic sea ice is reflecting 5 times the solar energy as what the edge of the Arctic sea is receiving, Hudson Bay (and the Bering Sea) are both completely melted. There is no sea ice ever present then.
4. (I miscounted above.) There is NO LIMIT to the area possible for Antarctic Sea Ice to expand into. It can keep going every year without boundary, ever closer to the Equator and ever-more reflecting more energy back into space. Not so for Hudson Bay. Once frozen, it can get no larger. Once the Arctic Ocean freezes over down to the Canadian coast and the Siberian coastlines, it can get no higher in sea ice area. So, there is an absolute maximum amount of solar energy that can EVER be reflected (or absorbed) in the Arctic Ocean, Hudson Bay, or Bering Seas.
In the Arctic Ocean, howeer, the sea ice is much further north. Its southern border is a rough circle well beyond the Arctic Circle at 71 – 76 – 82 north latitudes. At THOSE latitudes, Sereze’s theory is reversed.
There is never enough solar energy present to be dominant in heat absorbed/heat lost/heat reflected in any month other than May-June-July!
The other nine months of the year?
More Arctic sea ice = less heat lost from the Arctic Ocean = a warmer planet.
Less Arctic sea ice = more heat loss => A cooler planet.

AndyG55

When the incident angle of the sun’s energy is less than about 8 to 10 degrees, water is almost totally reflective.

No, not really.
There have been a number of different equations for the albedo of the open ocean that have been proposed over time. The only one that has stood up to actual measurements over open water at different wind speeds is Pegau and Paulson, 2001 using data (as usual) from the SHEBA over-winter Arctic sea ice study. W. Scott Pegau, Clayton A Paulson; Annals of Glaciology 33, 2001.
The rest are below:
Briegleb, 1986
Ebert and Curry, 1993
Grenfell and Perovich, 1984
Katsaros , McMurdie, Lind and DeVault, 1985
Maykut and Perovich, 1987
Perovich, 2002
Payne, 1972 (ancestor of many of these)
Preisendorfer and Mobley, 1986
Saunders, 1967
Simpsonn and Paulson, 1979
Coakley, 2003 (repeats from the above)
Kloe and Stoffelen, 2008
Jin, Charlock, Smith and Rutledge, 2004 (also uses Payne’s Chesapeake Lighthouse Tower, but updates it with dramatic differences between direct and diffuse albedo’s from a new mounting platform)
Rutledge, Schuster 2004
Jin, Charlock, and Rutledge, 2002
Long and Ackerman, 2000
Notice any pattern here? There are several others as well we don’t have to go into. Notice that Fresnel equations are useless (misleading even!) on the open ocean with live waves and real measurements.
Regardless, Pegau and Paulsen duplicate the measured albedo’s of direct radiation to the lowest solar elevation angle, and provide measurements for wave and wind speed corrections that do work. In my opinion, their values are a little bit too low, but they at least provide measured data to comment upon. Above 30 degrees SEA, maintain the default 0.066 values.
1. For diffuse radiation, the albedo is the “standard” “everybody knows it” value of 0.066 for all solar elevation angles. If the solar radiation is diffused through clouds, use this albedo. ONLY.
But first, reduce the remaining solar radiation as for direct radiation on a clear day because the atmosphere still attenuates the energy received at top-of-atmosphere.
Then, reduce the amount of radiation by 70% because the clouds reflect sunlight off of the top of the clouds back into space.
Then, if storms (high winds for example) reduce the radiation by the energy lost in the clouds themselves. Usually, the final radiation is 10% of the top-of-atmosphere values or less.
For direct radiation, clear skies.
Calculate the earth’s declination angle, the day angle mu_day, and the hour angle to get the solar elevation angle.
Calculate the air mass for that solar elevation angle, and assign (or select) an appropriate atmospheric clarity value.
You might be given a solar elevation angle directly. Regardless, you will have a solar elevation angle.
(If you have a solar zenith angle, SZA = 90-SEA )
Define mu = sin (SEA) = cos (SZA)
Define w = wind speed (meters/sec)
Albedo = [ 0.026/ (mu^1.7 + (-0.0002*w^2 + 0.0076*w + 0.0266) ]+ [ 0.15 *( mu – 0.1) * (mu – 0.5) * (mu – 1) ]
Thus, you get the following. Note that over 30 degrees SEA (60 degrees SZA) you end up back at Wikipedia’s default 0.066 “standard” albedo.
This is why the default Wikipedia albedo is “adequate” for equatorial latitudes going as far as 45 – 50 degrees, and for midday albedo at certain times of the year in the Arctic summer.

		Wind=0	Wind=3	Wind=9	Wind=12	(All winds)
		0	3	9	12
		Direct	Direct	Direct	Direct	Diffuse
SEA    Sin(SEA)	Albedo	Albedo	Albedo	Albedo	Albedo
3	0.052	0.779	0.476	0.301	0.269	0.066
5	0.087	0.613	0.409	0.274	0.247	0.066
7	0.122	0.478	0.345	0.245	0.223	0.066
9	0.156	0.378	0.290	0.216	0.200	0.066
11	0.191	0.304	0.245	0.191	0.178	0.066
13	0.225	0.250	0.209	0.169	0.159	0.066
15	0.259	0.209	0.180	0.149	0.141	0.066
20	0.342	0.142	0.128	0.112	0.108	0.066
30	0.500	0.078	0.073	0.067	0.066	0.066
40	0.643	0.048	0.046	0.043	0.042	0.066