Arctic energy balance

[Aluminium]

Estimations of energy input/output. Many factors affect the ice. It's important to determine if the factor deserves attention. Often simple calculations can help understand the order of magnitude.

1 km³ of ice is equivalent to 3*10¹⁷ J or approximately 1 Tsar Bomba.

Warm air advection.

50 km^3 per day by WAA would be extremely serious, yes. If it'd happen... You sure you got the number right?

My assumptions: 1 km x 1000 km x 5 m/s of air, 1 kJ/ (kg * °C), 1.2 kg/m³, 10°C, 10 g/m³ of water vapor, 2.3 MJ/kg. I saw warmer and wetter events. Infrared radiation provides effective interaction between snow/ice surface and wet air mass.

River discharge.

200 000 m³/s of water with 10°C may melt 2.3 km³ of ice per day or 70 km³ per month. It may be significant with high efficiency. But it's not enough to determine strong melting season.

[F.Tnioli]

Alright, now that we're not limited by main thread's requirements to omit excessive detail - i ask you to bring me the whole equation, sir. Mantissas of the numbers you gave don't quite fit the end result of "5.0" mantissa. Something's not clicking in. I'd like to figure out where i am not seeing what i need to see - or perhaps where you've slipped even if a little.

Because you know, like i said, 50 km^3 / day is quite crazy, yes? Thanks in advance!

[Phoenix]

My assumptions: 1 km x 1000 km x 5 m/s of air, 1 kJ/ (kg * °C), 1.2 kg/m³, 10°C, 10 g/m³ of water vapor, 2.3 MJ/kg. I saw warmer and wetter events. Infrared radiation provides effective interaction between snow/ice surface and wet air mass.

1 km x 1000 km - this is both horizontal dimensions of a front, with 5 m/s wind speed? What about thickness of it then - 1 m? If you'd elaborate a bit more, i'd be grateful for sure.

P.S. For clarity and lurkers, let's note here that 1 kJ/kg*K is air specific heat; 1.2 km/m^3 is air mass, rounded, per m3; 2.3 MJ/kg is specific latent heat of vaporisation of water, which is released whenever vapor turns back into liquid. There is also 334 kJ/kg, which is latent heat of fusion of ice - the amount of energy it takes to melt 1 kg of it. And, of course, 50 km^3 is no less than roughly 50,000,000,000,000 kg of it (50 trillions kilograms).

Copy and paste from melting season thread.

Small difference in the calculation of weight of ice which is ~8% less dense than liquid water at 0.919 gram / cc. So 45.95 trillion kilograms.

[Aluminium]

More detailed about WAA.

Let's imagine a 1000 km width 1 km thick front. Wind speed is 5 m/s, temperature is 10°C, 100% humidity (10 g/m³ of water vapor). Density is 1.2 kg/m³, specific heat capacity is 1 kJ/(kg*°C), specific heat of vaporization is 2.3 MJ/kg. Total energy density is 1.2 kg/m³ * 1 kJ / (kg*°C) * 10°C + 2.3 MJ/kg * 10 g/m³ = 35 kJ/m³. Flux of air is 1 km * 1000 km * 5 m/s = 5*10⁹ m³/s. Total power is 35 kJ/m³ * 5*10⁹ m³/s = 1.75*10¹⁴ J/s = 1.5*10¹⁹ J/day.

This energy may be received by ice or dissipate into space. Both ways are probably significant.

[Hefaistos]

Estimations of energy input/output. Many factors affect the ice. It's important to determine if the factor deserves attention. Often simple calculations can help understand the order of magnitude.

Albedo would be such a factor.

Figure byline: "Figure S6. All-sky albedo as a function of latitude for 5-year averages during the observational period and for the estimated change under a transition to ice-free conditions from a 1979 baseline state. The average value over the Arctic Ocean during 1979-1983 is shown as a grey circle on the right side of the gure, with Arctic Ocean spatial averages associated with the 3 curves also included for reference."

From "Radiative Heating of an Ice‐Free Arctic Ocean", Kristina Pistone,Ian Eisenman, Veerabhadran Ramanathan
First published: 20 June 2019
https://doi.org/10.1029/2019GL082914

[pleun]

More detailed about WAA.

This energy may be received by ice or dissipate into space.

Or simply stay in the airmass. Think you should calculate the energy difference between two points in time.

[Hefaistos]

Some valid input might be gathered from this paper:
"Trends in the CERES Dataset, 2000–13: The Effects of Sea Ice and Jet Shifts and Comparison to Climate Models"
Dennis L. Hartmann and Paulo Ceppi

From the Abstract

The Clouds and the Earth’s Radiant Energy System (CERES) observations of global top-of-atmosphere radiative energy fluxes for the period March 2000–February 2013 are examined for robust trends and variability. The trend in Arctic ice is clearly evident in the time series of reflected shortwave radiation, which closely follows the record of ice extent. The data indicate that, for every 106 km2 decrease in September sea ice extent, annual-mean absorbed solar radiation averaged over 75°–90°N increases by 2.5 W m−2, or about 6 W m−2 between 2000 and 2012.  ...."


Full paper:
https://journals.ametsoc.org/doi/10.1175/JCLI-D-13-00411.1

[F.Tnioli]

More detailed about WAA.

Let's imagine a 1000 km width 1 km thick front. Wind speed is 5 m/s, temperature is 10°C, 100% humidity (10 g/m³ of water vapor). Density is 1.2 kg/m³, specific heat capacity is 1 kJ/(kg*°C), specific heat of vaporization is 2.3 MJ/kg. Total energy density is 1.2 kg/m³ * 1 kJ / (kg*°C) * 10°C + 2.3 MJ/kg * 10 g/m³ = 35 kJ/m³. Flux of air is 1 km * 1000 km * 5 m/s = 5*10⁹ m³/s. Total power is 35 kJ/m³ * 5*10⁹ m³/s = 1.75*10¹⁴ J/s = 1.5*10¹⁹ J/day.

This energy may be received by ice or dissipate into space. Both ways are probably significant.

And then you mean 1.5*10^19 J/day / 3.34^5 J/kg = 4.491^13 kg/day, which with density of sea ice ~910 kg/m3, we get 4.491*10^13 kg/day / 9.1*10^2 kg/m3 * 10^9 m3/km3 = 4.93*10^1 km3/day. I.e. 49.3 km3/day. Ok, got the math, close indeed to just call it 50.

But assumtptions? Man...

That amount of energy is not just being spent "either to space or to ice". It's not just half of IR "ends up" going up to space, which would already make it 25 km3/day; there are other huge "cuts" to the number, i think:

- only rather small fraction of that vapour will change to liquid through the _whole_ track of the front over the ice, because lots of it will remain as a gas. Never anything close to 0% humidity in summer Arctic, from what i see. So, good portion of that water vapour will remain largely unaffecting the ice if we talk 1km-high air column, so gotta cut resulting ice loss accordingly. Hell to estimate it - we can agree to "halve it again" for starters, so 12.5 km3/day?

- even "downwards" half of IR won't all be absorbed by the ice: good fraction will be reflected (even though in IR ice is less reflective than in optical - it still somewhat is), and good portion of that reflected IR will end up going up into space in addition to the half which was heading there to begin with. So, perhaps dropping it further from 12.5 km3/day to say 10 km3/day?

- for 1-km thick layer, "angled" downwards IR will suffer athmospheric opacity, especially longer wavelengths (this page has some good general info about). Thing is, the way i understand it, when an IR photon is being absorbed by air - sometimes it will result in another IR photon generated, but not always, because often times that IR photon's energy ends up being spent to increase kinetic energy of the atom which absorbed it (thus temperature increase). Thing is, with 1km-tall layer, density difference start to play a role: going "up" is noticeably easier for IR radiation than going "down", simply because there is "less per meter of altitude" air as you go up (lower density). When "re-iterated" great many times by IR exchanges, this difference becomes quite significant, and the whole process is really mind-boggling in complexity overall. For 1km-thick air layer - it'd take proper physicist, not me, to even napkin this; so again, can only offer to imagine the figure halved once again, so from 10 to 5 km3/day.

So, you see, those things combined, quite possibly would drop the resulting figure by an order of magnitude or so. 5 km3/day is still significant, but not seriously crazy anymore, right?

[F.Tnioli]

Copy and paste from melting season thread.

Small difference in the calculation of weight of ice which is ~8% less dense than liquid water at 0.919 gram / cc. So 45.95 trillion kilograms.

This particular detail may need further correction for whenever one would be willing to calculate with precision: i read it's not 0.919, but 0.910 as reported on this page. In-situ density of 0.90...0.94 for below the waterline averages to 0.920, but seriously lower above-waterline density still drops overall ice density measurably below 0.919.

Thus i am generally using 0.91 - when not going "roughly" for round numbers of 1000 kg / m3, that is.

[Aluminium]

Assumptions was not about the most powerful event. I remember +16°C at 850 hPa with 40 kg/m² of water vapor. Air masses from the Arctic are much colder, about -5...-15°C.

There is greenhouse effect and 7 km of air above this 1 km thick layer.

[F.Tnioli]

Under "assumptions" i meant this line: "This energy may be received by ice or dissipate into space. Both ways are probably significant". Not the numbers. Sorry for being misleading about it initially. See, the assumptions in this line - i think are far insufficient, as per my larger post above; not detailed enough even for napkin calc of the kind.

As for numbers you just noted, sure, you already mentioned in melting season topic that numbers you used initially were far from warmest / wettiest event of the sort. The strongest ones would do well over 100 km3/day ice loss by your initial napkin math, i recon; do your observational efforts confirm this magnitude of melt from those athmospheric fronts, though? I doubt. But if you say they do, i'll pay attention for sure.

added: oh and about 7 kilometers of greenhouse effect on top of near-surface 1-km-thick layer of the athmosphere: it's more of a sink than insulation, i think. Gets seriously colder with altutude in the troposphere. Soaks lots of heat into itself.

[Aluminium]

I think, 7...8 km upward is harder way for photons than 0...1 km downward.

There was strong one in mid-June 2019.

[igs]

This thread has the kind of discussions i like, just see to it that nobody makes it a main thread 

[F.Tnioli]

I think, 7...8 km upward is harder way for photons than 0...1 km downward.
...

I think you think wrong. See, it depends, if to be precise, on what kind of IR photon we talk about. Some IR photons will get absorbed and not re-emitted after travelling merely 30 cm through near-surface air. Many others will suffer same fate after travelling merely few meters. Like i said above, "re-iterated great many times by IR exchanges". I said above this gets complicated, right? Can't be helped. Physics.

Yet some other IR photons will freely travel both down to surface _and_ through all the 7...8 km above (actually, all the way to space). And of course, there are yet other kinds of IR photons which on average gets absorved every few dozens meters, every few hundreds meters, every few kilometers - all kinds of 'em. Depends on wavelength - IR photons are actually very big and diverse "zoo".

You can see table 1 on page 1526 of this fragment for confirmation of the above and some further detail. Oh and that same table 1 also perfectly illustrates my above words about air density playing a role within 1-km column in troposphere: as you can see in the 2nd half of that table, at 150 mb mean free path of an IR photon with wavelength most easily captured by specificaly CO2 moleculae - is massively higher than at surface (1st half of the table).

So you see, this whole deal is exactly _why_ i was assuming, initially, that we're talking 1-meter-thick air layer. If it's just 1 meter, then some napkin energy transfer via IR can be calculated; but for 1-km, with those "short mean free path" kinds of IR photons requiring on average thousands (because omni-directional each time) re-emissions? Nope, this gets more like liquid dynamics than anything else, meaning it's not doable on a napkin in general. I think.

[Aluminium]

Re-emission will not necessary be at the same wavelength. Energy may stay in the atmosphere for a time but it means more back radiation to surface. Suitable wavelength exists for any height. Convection tends to warm up upward. Radiation is isotropic.

[Phoenix]

I think, 7...8 km upward is harder way for photons than 0...1 km downward.

There was strong one in mid-June 2019.

This chart shows a increase in daily melt rate of 200 km3 in response to a well documented Siberian heat wave. Is there any reasonable alternative explanation for the rapid melt increase other than WAA from Siberia ?

Eyeballing, I would say approximately 600 km3 of incemental melt vs average over the two week duration of the depicted event. The next logical steps would be to try and determine how much WAA is included in the average and how much insolation based melt is dependent on a WAA trigger event which reduces albedo.

[binntho]

I think, 7...8 km upward is harder way for photons than 0...1 km downward.

There was strong one in mid-June 2019.

This chart shows a increase in daily melt rate of 200 km3 in response to a well documented Siberian heat wave. Is there any reasonable alternative explanation for the rapid melt increase other than WAA from Siberia ?

How do you know it was in response to a heatwave in Siberia? And have you looked at other explanations? Coincidence is not causation.

Looking at Nullschool and Worldview for the same dates shows winds from Siberia and high air temps over ESS and Chukchi. But also very clear skies over ESS and Chuckhi. The Arctic itself was under a minor low pressure area, but the air seems to have been very dry at the time which may have led to cloud levels that lets a fair bit of solar energy through, and isolates against outgoing infrared radiation as well. I don't know if that was the case, the clouds, as usual, are a wildcard.
 
1. Hypthetical explanation: clear skies and/or high cloud during maximum insolation coincides with a Siberian heatwave and may well share some of the causes, i.e. slow-moving winds and clear skies.

2. Hypothetical explanation: The very real  warm air advection from Siberia clears away the ice  in the ESS and Chukchi (which wouldn't have been there in the first place if it wasn't for being sheltered all winter by Siberia). The appearance of open water under clear skies and massive insolation excelerates melt.

But the biggest problem with your claims Phoenix is that the amount of heat that can concievably be carried by air from Siberia in over the Arctic is far too small to be able to cause the observed melt.

a. Latent heat of ice 333 kJ/kg, specific weight of ice 919 kg/m3, 333 * 919 = 306.027 kJ/m3

Roundly speaking, 3E8 kJ/m3 * 200 km3* 1.000.000.000  m3/km3=6E19 kJ needed to melt 200km3 of ice, 60 billion billion.

b. Windspeed at the time was around 15 km/hour and the Siberian front (EES and Chukchi) is some 2000 km. Ef we generously count the lowest 5 meters of air (assuming some turbulance), we are looking at perhaps 3600 km3 of air in 24 hours.

c Specific heat of air is 1kJ/kg, specific weight 1,3 kg/m3, so specific heat per volume = 0,8 kJ/m3. Temperatures in the EES at the time were around 3 degrees Centigrade, so if we assume that the air loses 5 degrees of heat solely into the melting of ice, the daily volume of air from Siberia would supply 3600 * 1.000.000.000 * 5 kJ = sligtly less than 2E13 J.

d) 200km3 of ice needs 6E19 J to melt, warm air from Siberia supplies 2E13 or 0,00003% of the energy needed.

Of course my calculations could be wildly off, I've been through them a couple of times and I am honestly very surprised to see how little effect the WAA from Siberia has directly according to these calculations. So perhpas somebody could check them.

Phoenix, you should as a minimum be able to do these calculations yourself before making claims such as those above.

[gerontocrat]

Nice to see one of my graphs still in circulation.

[Phoenix]

Nice to see one of my graphs still in circulation.

Thank you for creating it as well as the abundance of other content you provide.

[Aluminium]

a. Latent heat of ice 333 kJ/kg, specific weight of ice 919 kg/m3, 333 * 919 = 306.027 kJ/m3

Roundly speaking, 3E8 kJ/m3 * 200 km3* 1.000.000.000  m3/km3=6E19 kJ needed to melt 200km3 of ice, 60 billion billion.

b. Windspeed at the time was around 15 km/hour and the Siberian front (EES and Chukchi) is some 2000 km. Ef we generously count the lowest 5 meters of air (assuming some turbulance), we are looking at perhaps 3600 km3 of air in 24 hours.

c Specific heat of air is 1kJ/kg, specific weight 1,3 kg/m3, so specific heat per volume = 0,8 kJ/m3. Temperatures in the EES at the time were around 3 degrees Centigrade, so if we assume that the air loses 5 degrees of heat solely into the melting of ice, the daily volume of air from Siberia would supply 3600 * 1.000.000.000 * 5 kJ = sligtly less than 2E13 J.

d) 200km3 of ice needs 6E19 J to melt, warm air from Siberia supplies 2E13 or 0,00003% of the energy needed.

a. 6E19 J or 6E16 kJ is in result.

b. I think, thicker layer should be considered. Back radiation is comparable with solar globally and more effective to warm up snow or ice.



c. 2E13 kJ or 2E16 J is in result.

d. Very thin layer without water vapor was considered. Yes, It's negligible.

[F.Tnioli]

Re-emission will not necessary be at the same wavelength. Energy may stay in the atmosphere for a time but it means more back radiation to surface. Suitable wavelength exists for any height. Convection tends to warm up upward. Radiation is isotropic.

Radiating - is isotropic. Radiation, as in "the process of energy transfer over distance by means of infrared emissivity", though  - is not entirely isotropic in this case, which was whole point above. Mean free path "upwards" is a bit longer than mean free path "downwards" for IR in general. For wavelengths with said path being short relative to air layer in consideration this leads to significant effect, which affects IR overall.

It's like liquid - say, a river - "preferring" to flow towards lower grounds. Except in this case, IR "prefers to flow up". Sort of. Can't put it any simpler.

And about same wavelength, - the spectrum is quite static, actually, and is defined by air temperature, only. You may find some quite surprising details on this page if you're interested about "how it really works". Great read, IMO.

[Aluminium]

Interesting. I have got some knowledge about radiation from astrophysics but it was not too detailed about properties of atmospheric gases.

[Simon]

Of course, we need to remember than the Arctic energy budget may not be typical for the planet as a whole as radiative emissivity varies with temperature, T^4 in fact. Thus a lot less radiation escapes into space than elsewhere on the planet.

[F.Tnioli]

This varies great deal seasonally. Polar night, it's surely low. But as soon as there is no ice, it gets big soon enough. Anyone in Alaska / Siberia will tell you: when it's polar day, thing can get pretty hot in the Arctic. Over 30C is nothing exceptional summer-time. Ice and to less extent open ocean prevent that, but Arctic as a whole still generates plenty hot air from its lands.

[binntho]

a. Latent heat of ice 333 kJ/kg, specific weight of ice 919 kg/m3, 333 * 919 = 306.027 kJ/m3

Roundly speaking, 3E8 kJ/m3 * 200 km3* 1.000.000.000  m3/km3=6E19 kJ needed to melt 200km3 of ice, 60 billion billion.

b. Windspeed at the time was around 15 km/hour and the Siberian front (EES and Chukchi) is some 2000 km. Ef we generously count the lowest 5 meters of air (assuming some turbulance), we are looking at perhaps 3600 km3 of air in 24 hours.

c Specific heat of air is 1kJ/kg, specific weight 1,3 kg/m3, so specific heat per volume = 0,8 kJ/m3. Temperatures in the EES at the time were around 3 degrees Centigrade, so if we assume that the air loses 5 degrees of heat solely into the melting of ice, the daily volume of air from Siberia would supply 3600 * 1.000.000.000 * 5 kJ = sligtly less than 2E13 J.

d) 200km3 of ice needs 6E19 J to melt, warm air from Siberia supplies 2E13 or 0,00003% of the energy needed.

a. 6E19 J or 6E16 kJ is in result.

b. I think, thicker layer should be considered. Back radiation is comparable with solar globally and more effective to warm up snow or ice.

c. 2E13 kJ or 2E16 J is in result.

d. Very thin layer without water vapor was considered. Yes, It's negligible.

Thanks for pointing out the discrepancies in units. It should have read kJ all the way down. Furthermore I can see that where I say "Chukchi" it should read "Laptev".

The result is so vastly less than anything that could be accounted as likely - i.e. there is no way that the heat wave in Siberia on 10th June 2019 could have melted 200km3 of ice that day, as Phoenix claimed (not suggested, but a straight faced claim).

I am aware that no water vapor was included in my calculation. Nullschool shows that the air was very dry on that day, and anyway the difference in heat capacity of air at varous levels of humidity seems to make hardly any real difference. The specific heat of water wapor is 1.82 kJ/kg compared to ~1 kJ/kg for air. At 5 degrees C and 50% humidity, the proportion of water wapour to air is still very low, less than 1% so making hardly any difference, well within the error margin of ~1 kJ/kg.

The thickness of the air is of course debatable, I used 5 m as a guess, but even if we were include all 10km of the troposhere, the difference is still only 2000 fold, going from 0,00003% to 0,06% of the energy needed to melt 200 km3, totally unrealistic presumption but not getting us anywhere near a real effect.

Also if we raise the temperature, even to the max of 32 degrees, the highest that has ever been recorded within the Arctic circle, the difference is still only 6 fold.

The heatwave in Siberia and the very impressive looking WAA from Siberia in over the EES and  Laptev on 10th of June 2019 had the potential to melt 0,006 km3 of ice. THe EES and Laptev together are around 1.6 million km2, implying a melt of 4mm on that day if all the melt happened in those to seas. Which is not negleglible but nowhere near enough to have any real impact in the Arctic as a whole.

So my conclusion seems to stand: WAA from the continents does not have anywyhere near the capacity to cause any significant melt in the Arctic.

[Aluminium]

The thickness of the air is of course debatable, I used 5 m as a guess, but even if we were include all 10km of the troposhere, the difference is still only 2000 fold, going from 0,00003% to 0,06% of the energy needed to melt 200 km3, totally unrealistic presumption but not getting us anywhere near a real effect.

2E16 J is 0.03%. 1 km thick layer has 200 times more heat capacity and was at least 3 times warmer. This event was not dry. Water vapor contains approximately 2 times more energy. Result rises to 2E19 J or 30%. Thicker layer contains even more energy and wind speed was higher.

I made some calculations before with the same order of magnitude in result.

[Phoenix]

https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/2015GL064373

Geophysical Research Letters
Warm‐air advection, air mass transformation and fog causes rapid ice melt
June 2015

Abstract
Direct observations during intense warm‐air advection over the East Siberian Sea reveal a period of rapid sea‐ice melt. A semistationary, high‐pressure system north of the Bering Strait forced northward advection of warm, moist air from the continent. Air‐mass transformation over melting sea ice formed a strong, surface‐based temperature inversion in which dense fog formed. This induced a positive net longwave radiation at the surface while reducing net solar radiation only marginally; the inversion also resulted in downward turbulent heat flux. The sum of these processes enhanced the surface energy flux by an average of ~15 W m−2 for a week. Satellite images before and after the episode show sea‐ice concentrations decreasing from > 90% to ~50% over a large area affected by the air‐mass transformation. We argue that this rapid melt was triggered by the increased heat flux from the atmosphere due to the warm‐air advection.

(from conclusion - An extra 20 W m−2 surface heating would theoretically melt an additional ~4–5 cm of ice over 7 days)

[binntho]

The thickness of the air is of course debatable, I used 5 m as a guess, but even if we were include all 10km of the troposhere, the difference is still only 2000 fold, going from 0,00003% to 0,06% of the energy needed to melt 200 km3, totally unrealistic presumption but not getting us anywhere near a real effect.

2E16 J is 0.03%. 1 km thick layer has 200 times more heat capacity and was at least 3 times warmer. This event was not dry. Water vapor contains approximately 2 times more energy. Result rises to 2E19 J or 30%. Thicker layer contains even more energy and wind speed was higher.

I made some calculations before with the same order of magnitude in result.

To begin with, there was an error in the units, it should have been 2E16 kJ which adds 3 zeroes to the calculation. So the calculation is correct, the units were in error. I'll hurry and correct the original post!

Secondly, Nullschool shows very little precipitable water in the air coming from Siberia at the time.

Thirdly, even if water vapor has two times the heat capacity of air, the total amount of water wapor that air can hold is very low, and at arctic temperatures, it is well below 1%. Even at tropical temperature levels, 100% humidity translates into about 4% water by weight of the air column.

It took me a few seconds to find on the omniscient Internet that at 5 degrees, the maximum carrying capacity of air is less than 0,07 kg/m3, so 0,7% per weight.

Water vapour makes practically no difference inn our case - 99% * 1 + 1% * 2 = 1,01.

I don't know why you think that we should count a full 1km in this case. The air that touches the surface is the air that transfers heat to the ice. Radiative transfer is most likely negligible in this scenario, so again - given som turbulence at the surface I think that counting the bottom 5 meters is of the correct order of things.

Finally, Nullschool for the day shows that air temps leaving the Siberian shore at around 15 degress C. I used the temperatures over the EES as a basis for my calculation, and estimated that 5 degrees C were available to be used for melt. I'll admit that the 15 degrees is better, raising the final number to:

Only 0,001% of the energy needed to melt 200km3 of ice was supplied by the WAA.

[binntho]

https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/2015GL064373

Geophysical Research Letters
Warm‐air advection, air mass transformation and fog causes rapid ice melt
June 2015

Abstract
Direct observations during intense warm‐air advection over the East Siberian Sea reveal a period of rapid sea‐ice melt. A semistationary, high‐pressure system north of the Bering Strait forced northward advection of warm, moist air from the continent. Air‐mass transformation over melting sea ice formed a strong, surface‐based temperature inversion in which dense fog formed. This induced a positive net longwave radiation at the surface while reducing net solar radiation only marginally; the inversion also resulted in downward turbulent heat flux. The sum of these processes enhanced the surface energy flux by an average of ~15 W m−2 for a week. Satellite images before and after the episode show sea‐ice concentrations decreasing from > 90% to ~50% over a large area affected by the air‐mass transformation. We argue that this rapid melt was triggered by the increased heat flux from the atmosphere due to the warm‐air advection.

(from conclusion - An extra 20 W m−2 surface heating would theoretically melt an additional ~4–5 cm of ice over 7 days)

Interesting article, but the melt is caused by changes in radiation, not the heat of the WAA.

In my original post, the cloudiness of the entire Arctic except for the EES and Chukchi caused me to wonder what the net radiative effect was,  i.e what type of clouds they were. Also given that the Arctic was governed by a high-pressure system at the time, I would have expected clear skies and not the clouds that were actually visible.

So this paper is saying that the clouds were in fact lowlying fog that enhanced melt significantly. Both assumptions are interesting, the fog being the result of the "humid" air from Siberia, and that fog over ice increases rates of melting.

Going back to Nullschool, the "humidity" of this event was very low, similar to that found over the Sahara at the same time. So where do they get their "moist air" from? Or is Nullschool so totally wrong?

Finally, if 5cm of ice melts over the entire arctic over the 7 day period, that translates into 500 km3, while the actual melt at the same time was around 1500 km3. So this "massive" event could concievably count for 1/3 of the melt at the time.

[Phoenix]

https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/2015GL064373

Geophysical Research Letters
Warm‐air advection, air mass transformation and fog causes rapid ice melt
June 2015

Abstract
Direct observations during intense warm‐air advection over the East Siberian Sea reveal a period of rapid sea‐ice melt. A semistationary, high‐pressure system north of the Bering Strait forced northward advection of warm, moist air from the continent. Air‐mass transformation over melting sea ice formed a strong, surface‐based temperature inversion in which dense fog formed. This induced a positive net longwave radiation at the surface while reducing net solar radiation only marginally; the inversion also resulted in downward turbulent heat flux. The sum of these processes enhanced the surface energy flux by an average of ~15 W m−2 for a week. Satellite images before and after the episode show sea‐ice concentrations decreasing from > 90% to ~50% over a large area affected by the air‐mass transformation. We argue that this rapid melt was triggered by the increased heat flux from the atmosphere due to the warm‐air advection.

(from conclusion - An extra 20 W m−2 surface heating would theoretically melt an additional ~4–5 cm of ice over 7 days)

Interesting article, but the melt is caused by changes in radiation, not the heat of the WAA.

Do you not understand that heat is thermal radiation?

https://en.wikipedia.org/wiki/Thermal_radiation

When solar radiation is absorbed by the earth's crust, it is changed in wavelength to long wave infrared radiation and reflected back into the atmosphere. AGW is happening because that long wave radiation frequency is absorbed and redirected by ghg molecules. It's getting hotter here on earth because we are retaining more long wave radiation.

[Aluminium]

To begin with, there was an error in the units, it should have been 2E16 kJ which adds 3 zeroes to the calculation.

3600 km³ * 5kJ/m³ = 2E13 kJ = 2E16 J = 0.03% * 6E19 J. I am still sure this is correct result in your calculations. Though I doubt in assumptions.

Secondly, Nullschool shows very little precipitable water in the air coming from Siberia at the time.

Thirdly, even if water vapor has two times the heat capacity of air, the total amount of water wapor that air can hold is very low, and at arctic temperatures, it is well below 1%. Even at tropical temperature levels, 100% humidity translates into about 4% water by weight of the air column.

Specific heat of vaporization is 2.3 MJ/kg. Given this, water vapor can transfer more energy than dry air.

Radiative transfer is most likely negligible in this scenario

It sounds like solar radiation is negligible for the ice. Seriously, both have similar power but visible light is mostly reflected and longwave infrared is mostly absorbed. Back radiation strongly (T⁴) depends on air temperature.

[binntho]

Do you not understand that heat is thermal radiation?

No, heat is not thermal radiation. Your question shows a lack of undersanding of thermodynamics. But I must admit that I have been using the word "heat" with lack of precision.

So let us split it into more formal categories / definitions:

1) Heat transfer is the transfer of energy from one system to another. There are basically two methods of transferring thermal energy, radiative transfer and conduction (some Americans add friction as a method of transferring heat, but that's of no consequence).

Radiative transfer of energy is how the sun warms the earth (and our skin when we are out in the sun). Conduction is how air (and water) primarily transfer heat internally and to other bodies, although some radiative transfer is always ongoing (and grows with added energy, but it needs well above "weather" temperatures before radiative heat transfer overcomes conductive heat transfer).

Heat transfer is sometines referred to with the single word "heat" only. So in that sense, thermal radiation is "heat" but not the other way around, since conduction is another form of heat transfer, and the use of "heat" can mean more than heat transfer.

2) The energy contained in a system, and which we measure by e.g. sticking a thermometer into it (which by the way works by conduction) is also called "heat". As in if air is at this and this temperature then it contains so and so amount of heat.

Strictly speaking we should use the word "energy" here, and in all my calculations when I use kJ
 (kilo Joules) for latent heat or specific heat and the melting potentiality of air at this or that temperature, it is in fact "energy" we are talking about.

3) In my calculations I have shown that the Warm Air Advection contains nowhere near enough "heat" (meaning energy) to have any significant effect on melting. The paper that you have linked does not claim that the energy carried by the WAA has done any melting. So the paper does not disagree with my calculations.

4) The paper claims that the WAA created circumstances whereby the efficiency of the thermal radiation from the sun was kicked into overdrive. That may or may not be true, I have no idea, but I do question the use of "moist" air when Nullschool shows dry air, comparable to the Sahara at the same time. Perhaps they are right, and the air was "moist" (although where did the moisture come from?), or perhaps Nullschool has it wrong, or perhaps I am misunderstanding Nullschool. Makes no real difference to this discussion.

[binntho]

To begin with, there was an error in the units, it should have been 2E16 kJ which adds 3 zeroes to the calculation.

3600 km³ * 5kJ/m³ = 2E13 kJ = 2E16 J = 0.03% * 6E19 J. I am still sure this is correct result in your calculations. Though I doubt in assumptions.

I am unable to change my original posting to correct the "Chukchi" for "Laptev" and "J" for "kJ". But I have been through my calculations again and found several errors. The final conclusion is however still valid.

Let's start with the ice:


Latent heat is the amount of energy needed to melt ice at melting point. Of course, more energy is needed in the real world, since any large scale melt event will also have to heat up a significant amount of ice.

The calculations were made in response to Phoenix' claim that one day's WAA had melted 200km3 of ice, hence the last line. Which by the way is off from my previous calculations by a factor of 1000! 

So for the air:


Specific heat is the amount of energy released when temperature goes down by 1 degree C (actually defined the other way around, the amount of energy needed to heat by 1 degree C).

As has been pointed out, humidity does not alter the result in any significant way. Air at 15 C and 100% humidity is only 1% water (or 10 g/kg), so if the air had been at 100% humidity (which is a wild overestimate), the final number would still go up by only 1%.

So finally for the air mass on that fateful day, the 10th of June 2019. I am assuming a temperature of 15 degrees, and the thicness of the layer of air able to conduct energy to the surface at a generous 5 meters. Wind speed is around 15km/hour as the air leaves the coast according to Nullschool.



So here I have the second major mistake found in my earlier calculations when I assumed 3600 km3 of air, while the actual number is 36.

I end with exactly the same percentage as in this post which was calculated in a different way, which increases my confidence in the result being correct.

Secondly, Nullschool shows very little precipitable water in the air coming from Siberia at the time.

Thirdly, even if water vapor has two times the heat capacity of air, the total amount of water wapor that air can hold is very low, and at arctic temperatures, it is well below 1%. Even at tropical temperature levels, 100% humidity translates into about 4% water by weight of the air column.

Specific heat of vaporization is 2.3 MJ/kg. Given this, water vapor can transfer more energy than dry air.

I have never disputed that water vapour can carry more than double the amount of energy as dry air. But the point that I seem to be making all the time is that even at the maximum possible humidity of 100%, only 1% of the airmass is water vapour. So it effectively makes no difference.

So was this a dry or a humid event? We have both referenced Nullschool showing "Total Precipitable Water". How that relates to humidity I have no idea. Given that the Worldview images of EES and Laptev was clear as can be on that day (the 10th of June 2019), I doubt that there was much if any condensed water in the air, so the appellation "moist" seems most misguided.

Nullschool puts most of the wind coming in over EES at 10-14 kg/m2 total precipitable water, but over the Laptev up to 20 kg/m2. As I have no idea what means in the real world, I compared with some dry places on the planet, and found that the Sahara desert fitted the bill pretty well, ranging from 5 to 18 kg/m2. So I would tend to assume that this was an extremely dry event, dry as a desert.

Radiative transfer is most likely negligible in this scenario

It sounds like solar radiation is negligible for the ice. Seriously, both have similar power but visible light is mostly reflected and longwave infrared is mostly absorbed. Back radiation strongly (T⁴) depends on air temperature.

Well, I certainly did not mean to say that solar radiation was negligible.

But I did intend to say that a warm airmass will release its heat by conduction primarily, with radiative heat loss being very small and in effect negligible. This is however outside my knowledge of physics, so I am not able to give any calculations.

Having said that, I am pretty sure that radiative thermal transfer of air only starts to become significant in comparison to conduction at much higher temperatures than are found in the atmosphere.

[Aluminium]

Tables are convenient. One last thing is wind front. 5 m = 0.005 km. And we have 0.1% in result.

10-14 kg/m² of water vapor is a high number for the Arctic. May begins with below 5 kg/m³ usually.

Considering thicker layer, strong WAA contains enough energy to melt 200 km³/day. It does not mean that only the WAA did 200 km3 per day. But there is a way to transfer significant part of this energy to make the ice darker and more vulnerable for visible light. Combined effect of the WAA and clear sky seems to explain extreme melting that day.

Having said that, I am pretty sure that radiative thermal transfer of air only starts to become significant in comparison to conduction at much higher temperatures than are found in the atmosphere.

looking at the main image here, conduction is actually negligible compared to radiation.

[Freegrass]

We have both referenced Nullschool showing "Total Precipitable Water". How that relates to humidity I have no idea. Given that the Worldview images of EES and Laptev was clear as can be on that day (the 10th of June 2019), I doubt that there was much if any condensed water in the air, so the appellation "moist" seems most misguided.

Nullschool puts most of the wind coming in over EES at 10-14 kg/m2 total precipitable water, but over the Laptev up to 20 kg/m2. As I have no idea what means in the real world, I compared with some dry places on the planet, and found that the Sahara desert fitted the bill pretty well, ranging from 5 to 18 kg/m2. So I would tend to assume that this was an extremely dry event, dry as a desert.

Good discussion! I'm learning stuff in relation to our other discussion on the Newbie thread. But way to many numbers here... 

I found this. Hope that helps you understand TPW. I don't get it either. I just find it a useful Nullschool tool that shows both heat and moisture, the stuff hungry storms love...

The vertical distribution of atmospheric water vapour is quite uneven since most of it is concentrated in regions below 700 hPa (~ 3 km). The concept behind TPW is to get an absolute measure of the water content of the air. It strongly differs from the more familiar relative humidity as the latter depends on the capacity of the air to hold water and hence on its temperature.

Assuming a vertical column of air reaching from the ground to the top of the atmosphere with a base of 1 m2, the TPW content of this column equals the amount of water if all water vapor was condensed. The commonly used units are [kg/m2] reflecting the weight of the condensed water or [mm] if water is accumulated on the bottom of the column.

This page has some calculations on it, so best to open this link to the article for more.

[Freegrass]

One year time lapse of total precipitable water: 2016

[igs]

Even though I do not agree with your conclusion as to WAA I want to thank you a lot for the post with the quoted number.


I learned a lot from that post, very much appreciated.

[pleun]

@bintho : in the above calculations, I believe I miss the energy of condensation of the water vapour in the air mass. The air is likely to reach dew point on the ice I think?

[Glen Koehler]

Thanks Freegrass.  That Total Precipitable Water video explains a lot of ecosystem/habitat variation around the planet.  In particular the grasslands of central and northern Asia, and why the polar regions are considered deserts despite being dominated by water ice.  It also demonstrates better known moisture habitat relationships like the Amazon and central African rain forests.  And finally it demonstrates the intermittent and somewhat random, but over time, reliable variation between wet and dry that supports agricultural regions in the mid-latitudes.

[Freegrass]

You're welcome Glen. Amazing that you were able to see all that... 

Did you see the 2019 video that shows up at the end? That one is actually better to view as the globe isn't spinning. But I only noticed it after I posted the 2016 video already. It's also located in the Nullschool Youtube channel.

[binntho]

Having spent a fitful night tossing and turning (blackout, thunderstorms, mosquitos etc.) I couldn't help but feel that I had mad several major mistakes in my calculations. Being too busy to go through them again this morning, I'd like to state here and now that I wouldn't be surprised if I was off by a factor of 100 in the final numbers, although even that would not change the basic conclusion.

[binntho]

@bintho : in the above calculations, I believe I miss the energy of condensation of the water vapour in the air mass. The air is likely to reach dew point on the ice I think?

You are absolutely right, I am aware of this missing factor and I'm not at all sure how to estimate it. Phoenix had a link to a paper on another big warming event in early August 2014 where they claimed that the air was very "moist" (I think they meant "humid") and that the water vapour condensed into a dense fog which then increased the effects of insolation significantly. Which is interesting, since one of the perennial questions on this forum is ast to what effects the often observered fog can have on the ice.

Not knowing the humidity of the air is a problem, but if we assume 50% humidity (probably way too high, but never mind) at 15 degrees C then that translates into 5 g vapour per kg (or m3) of air. Total amount of water vapour in 36 km3 can then be calculated as 180.000 tons of water. The latent heat of water vapour is 2,260 kJ/kg, so the total latent heat of 180.000 tons is 4E12 kJ. Which is truly a massive amount of energy!

This is not surprising, everybody who has ever seen the different melting rates from a relatively humid wind as opposed to a dry wind would not be surprised.

But even if we were to add this number to my previous calculations, it would only change the final conclusion by a factor of 10, i.e. increasing the percentage contribution of the WAA on 10th June 2019 to the melt of 200km3 to a number that is still less than 1%.

[binntho]

Tables are convenient. One last thing is wind front. 5 m = 0.005 km. And we have 0.1% in result.

10-14 kg/m² of water vapor is a high number for the Arctic. May begins with below 5 kg/m³ usually.

Considering thicker layer, strong WAA contains enough energy to melt 200 km³/day. It does not mean that only the WAA did 200 km3 per day. But there is a way to transfer significant part of this energy to make the ice darker and more vulnerable for visible light. Combined effect of the WAA and clear sky seems to explain extreme melting that day.

Having said that, I am pretty sure that radiative thermal transfer of air only starts to become significant in comparison to conduction at much higher temperatures than are found in the atmosphere.

looking at the main image here, conduction is actually negligible compared to radiation.

I can see that you found the error that I feared in my dreams!

The only real contention in all of this is the thickness of the air used in the calculation, and that in the end rests on the relative effects of conduction vs. radiation.

At "weather" temperatures, radiation is truly a wimp when it comes to transferring heat. You can test this on your self - you do not feel heat radiating from a window even if it's freezing inside and 20 degrees outside. But you feel the temperature of the air as soon as you touch it.

You do not feel the heat of thermal radiation from a hot tub of water - but you do feel the heat of the steam settling on your skin. You do start to feel the heat of radiation from the sides of a freshly boiled kettle, but still not enough for you to stop touching it by mistake and then fealing the real heat transfer of conduction.

So if you think that thermal radiation from a 15 degree C body of air in one day is enough to melt any ice at all, let alone any significant amount, then please show some evidence!

[Aluminium]

Thermal radiation from a 15°C black body is 391 W/m². Thermal radiation from a 0°C black body is 315 W/m². Difference is 76 W/m² or 2 cm/day of ice.

[binntho]

Thermal radiation from a 15°C black body is 391 W/m². Thermal radiation from a 0°C black body is 315 W/m². Difference is 76 W/m² or 2 cm/day of ice.

I am unable to make sense of this. Are you saying that 76 W/m2 melts 2cm/day of ice? And how can you say that a black "body" has a thermal radiation of X W/m2. What size? What mass? How long does it last?

Thermal radiation at 400 W/m2 is the average insolation in the Tropics! Does this sound likely? That standing next to a 15 degree C blackbody would warm you as much as lying out in the sun on the Equator? Does this sound likely.

Solar insolation at peak in the Artctic reaches 500 W/m2 and would melt 13.5 cm/day according to those numbers, or around 1300 km3 per day. Does that sound reasonable?

A nonsensical hypothethical: If a warm body of air at 15 degrees is able to lose all of it's energy through thermal radiation in one day, then the atmosphere would be quickly plummeting towards 0 degrees K.

Even if we substitute "all" with "a significant amount" then we would still be hurtling towards a mega ice-age.

Air at 15 degrees may be radiating heat in all directions, and "black body" radiation may be significant, but I suspect that most of that radiation is caught internally by the air itself. The effect of this thermal radiation on the surface is going to be exceedingly small.

[Aluminium]

Do you feel atmospheric pressure? I don't. But this pressure is 1 kg/cm²!

Effect of insolation is strongly reduced by high albedo and atmosphere. But thermal radiation from the ice is not and should be exceeded somehow for any melting.

Those numbers are about radiation from surface of body or border of atmospheric layer.

[binntho]

Do you feel atmospheric pressure? I don't. But this pressure is 1 kg/cm²!

Effect of insolation is strongly reduced by high albedo and atmosphere. But thermal radiation from the ice is not and should be exceeded somehow for any melting.

Those numbers are about radiation from surface of body or border of atmospheric layer.

I am still absolutely convinced that thermal radiation is negligible when it comes to transfer of heat from a body of air to the surface (e.g. ice). Conduction is how things happen in the real world, with the huge exception of the sun of course.

As for your pressure comparison, it falls at the first hurdle: We would very certainly feel if the pressure went away. We do not feel any difference from being in the vicinity of a 15 degree or 0 degree or 30 degree black body - or not. Do you feel radiative heat from other people? Try walking up to a person and away again - could you feel the difference in thermal radiation from a 37 degree C "black or otherwise coloured or not" body?

The only thermal radiation we are likely to feel (both when it appears and when it disappears) is from the sun, from fire, and from objects significantly warmer than atmospheric temperatures.

Thermal radiative heat transfer from air is negligible no matter what numbers you band around, neither of us has the abillty to work this out mathematically, specifically: How much thermal radiation is received by the surface from a 15 degree C airmass passing overhead. My contention is that it is exceedingly small (otherwise the entire airmass would be cooling rapidly).

All experience and common sense says that it is negligible, and claiming that the thermal radiative transfer from a 15 degree airmass passing overhead is in the same as the heat of the sun in the tropics is ridiculous.

I've had quite a lot of experience in standing under a 15 degree C airmass in deeply overcast weather and it does not feel warm. Where I am now, if I stick my head outside at this very moment (12:38) I recieve a full 400 W/m2 tropical insolation. And I can very well tell you that it feels extremely much hotter!

[Aluminium]

Gravitation and radiation are two rulers in our world.

To cool by 1 degree, 1 m² of the atmosphere should radiate 10 MJ. It requires 7 hours with 400 W/m².

Usually, thermal radiation around is almost isotropic just like atmospheric pressure. We are used to it.

[oren]

Binntho heat radiation depends on the difference in temperature. Of course you can't feel other people when you are the same temperature. But the ice is at 0^o.

[binntho]

Binntho heat radiation depends on the difference in temperature. Of course you can't feel other people when you are the same temperature. But the ice is at 0^o.

I'm not at all sure that heat radiation depends on the difference in temperature. Heat transfer does, yes. But radition is an intrinsic quality of the radiating body, and is not dependent on the surroundings. It is however dependent on the energy content of the body, the warmer it gets the more thermal radiation it will emit.

A human body at 37 degrees should according to Aluminium broadcast a thermal radiation in the region of 500 W/m2. Are you saying that we would not feel that if we stood next to them? Of course we would! And how much energy would they be losing through all this radiation?

Standing next to a lump of ice, a person would radiate enough energy to melt several centimeters of that ice. Well, le'ts make some calculations: If a standard westerner were to stand naked next to a 2 m2 lump of ice, and melt 5 cm off the surface, or 100000 cm3, i.e. 100 kg. The latent heat of ice at 333 kJ/kg equates this to 33.300 kJ or almost 8000 Calories.

If Aluminium's numbers were correct, we would each of us be losing energy at a rate of some 15.000 Calories per day. So bring on the beer and the strawberries in cream!

Consider also: if thermal radiation was anywhere near as important a method of heat transfer as conduction or convection at "weather" temperatures, then insulation would not work. Wearing clothes would not keep you warm. Insulating your house would not keep you warm. Using double or triple layered glass would not be worth the effort. The whole idea of insulation is to stop or retard convection and conduction, but most of the time it does not take thermal radiation into account at all. And yet it works surprisingly well!