Arguments
Trenberth talks about energy flows and global warming
What the science says...
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Trenberth's views are clarified in the paper "An imperative for climate change planning: tracking Earth's global energy". We know the planet is continually heating due to increasing carbon dioxide but that surface temperature sometimes have short term cooling periods. This is due to internal variability and Trenberth was lamenting that our observation systems can't comprehensively track all the energy flow through the climate system. |
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Climate Myth...
Trenberth can't account for the lack of warming
in one e-mail, a top "warmist" researcher admits it’s a "travesty" that "we can’t account for the lack of warming at the moment." As it happens, the writer of that October 2009 e-mail—Kevin Trenberth, a lead author of the warmist bible, the 2007 Intergovernmental Panel on Climate Change (IPCC) report—told Congress two years ago that evidence for manmade warming is "unequivocal." He claimed "the planet is running a ’fever’ and the prognosis is that it is apt to get much worse." But Trenberth’s "lack of warming at the moment" has been going on at least a decade. (Michael Fumento)
This has been most commonly interpreted (among skeptics) as climate scientists secretly admitting amongst themselves that global warming really has stopped. Is this what Trenberth is saying? If one takes a little time to understand the science that Trenberth is discussing, his meaning becomes clear.
If you read the full email, you learn that Trenberth is actually informing fellow climate scientists about a paper he'd recently published, An imperative for climate change planning: tracking Earth's global energy (Trenberth 2009). The paper discusses the planet's energy budget - how much net energy is flowing into our climate and where it's going. It also discusses the systems we have in place to track energy flow in and out of our climate system.
Trenberth states unequivocally that our planet is continually heating due to increasing carbon dioxide. This energy imbalance was very small 40 years ago but has steadily increased to around 0.9 W/m2 over the 2000 to 2005 period, as observed by satellites. Preliminary satellite data indicates the energy imbalance has continued to increase from 2006 to 2008. The net result is that the planet is continuously accumulating heat. Global warming is still happening.
Next, Trenberth wonders with this ever increasing heat, why doesn't surface temperature continuously rise? The standard answer is "natural variability". But such a general answer doesn't explain the actual physical processes involved. If the planet is accumulating heat, the energy must go somewhere. Is it going into melting ice? Is it being sequestered deep in the ocean? Did the 2008 La Niña rearrange the configuration of ocean heat? Is it all of the above? Trenberth wants answers!
So like an obsessive accountant, Trenberth pores over the energy budget, tallying up the joules accumulating in various parts of the climate. A global energy imbalance of 0.9 W/m2 means the planet is accumulating 145 x 1020 joules per year. The following list gives the amount of energy going into various parts of the climate over the 2004 to 2008 period:
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Land: 2 x 1020 joules per year
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Arctic sea Ice: 1 x 1020 joules per year
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Ice sheets: 1.4 x 1020 joules per year
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Total land ice: between 2 to 3 x 1020 joules per year
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Ocean: between 20 to 95 x 1020 joules per year
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Sun: 16 x 1020 joules per year (eg - the sun has been cooling from 2004 to 2008)
These various contributions total between 45 to 115 x 1020 joules per year. This falls well short of the total 145 x 1020 joules per year (although the error bars do overlap). Trenberth expresses frustration that observation systems are inadequate to track the flow of energy. It's not that global warming has stopped. We know global warming has continued because satellites find an energy imbalance. It's that our observation systems need to be more accurate in tracking the energy flows through our climate and closing the energy budget.
So what may be causing the discrepancy? As the ocean heat data only goes to 900 metre depth, Trenberth suggests that perhaps heat is being sequestered below 900 metres. There is support for this idea in a later paper von Schuckmann 2009. This paper uses Argo buoy data to calculate ocean heat down to 2000 metres depth. From 2003 to 2008, the world's oceans have been accumulating heat at a rate of 0.77 W/m2. This higher trend for ocean heat would bring the total energy build-up more in line with satellite measurements of net energy imbalance.
A subsequent study by Balmaseda, Trenberth, and Källén (2013) determined that over the past decade, approximately 30% of ocean warming has occurred in the deeper layers, below 700 meters. This conclusion goes a long way to resolving the 'missing heat' discrepancy. There is still some discrepancy remaining, which could be due to errors in the satellite measurements, the ocean heat content measurements, or both. But the discrepancy is now significantly smaller, and will be addressed in further detail in a follow-up paper by these scientists.
Summary
So to summarise, Trenberth's email says this:
"The fact is that we can't account for the lack of warming at the moment and it is a travesty that we can't."
After reviewing the discussion in Trenberth 2009, it's apparent that what he meant was this:
"Global warming is still happening - our planet is still accumulating heat. But our observation systems aren't able to comprehensively keep track of where all the energy is going. Consequently, we can't definitively explain why surface temperatures have gone down in the last few years. That's a travesty!"
Skeptics use Trenberth's email to characterise climate scientists as secretive and deceptive. However, when one takes the trouble to acquaint oneself with the science, the opposite becomes apparent. Trenberth outlines his views in a clear, open manner, frankly articulating his frustrations at the limitations of observation systems. Trenberth's opinions didn't need to be illegally stolen and leaked onto the internet. They were already publicly available in the peer reviewed literature - and much less open to misinterpretation than a quote-mined email.
Intermediate rebuttal written by dana1981
Update July 2015:
Here is a related lecture-video from Denial101x - Making Sense of Climate Science Denial
Last updated on 14 July 2015 by pattimer. View Archives
Mods,
Please note that the link to the Trenberth paper is broken and should point to
http://www.cgd.ucar.edu/staff/trenbert/trenberth.papers/EnergyDiagnostics09final2.pdf
FOUND - Missing Global Warming from Science Magazine
Major climate data sets have underestimated the rate of global warming in the last 15 years owing largely to poor data in the Arctic, the planet's fastest warming region. A dearth of temperature stations there is one culprit; another is a data-smoothing algorithm that has been improperly tuning down temperatures there. The findings come from an unlikely source: a crystallographer and graduate student working on the temperature analyses in their spare time.
http://www.sciencemag.org/content/344/6182/348.summary
[JH] Hot-linked the url.
How does Trentbergh explain that it is hotter on the International Space Station (ISS) (121°C facing the sun) than on the Earth's surface (14°C average), while his energy budget claims that the Earth's surface emits more energy as infrared (396 W.m-2) than the total incoming solar irradiance (340 W.m-2) ?
ab @33,
Trenberth would have little difficulty answering. The answer is simply physics.
The 121°C temperature derives from the Stefan-Boltzmann Law which defines how much energy a hot surface will radiate. To be in balance with continuous sunlight (1,366Wm^-2), a surface normal to the sunlight incidence with emissivity=1 and zero-reflection would be in equilibrium with a temperature equal to 121°C.
The Earth, of course is only illuminated by the sun during the day and that is normal (directly overhead) only at noon. The result is an average level of sunlight reaching the Earth being 25% the constant normal solar level. That would result in an equilibrium temperature of about +5°C, except about a third of sunlight is reflected back into space so the global solar warming averages one sixth the constant normal level requiring an equilibrium temperature of -18°C. But the surface is warmed not just by the sun but also by the atmosphere. The average surface temperature therefore will be hotter than that value. With the Earth's GHGs the surface temperature is +14°C. And being hotter, it will on average radiate more than a surface warmed solely by the sun, even without any reflection.
I think that answers both your questions. The answer is 'physics'.
ab, assuming experienced physicists have got it all wrong because you dont understand it takes some hubris. Physics is not a smorgasbord. You cant pick and choose which bits you want. Before you can declare that observations show current physics has it wrong, you need to settle down with serious textbook and learn what the actual theory this. Sounds to me like you are spending too much time with ilk of PSI and not enough time with a real textbook. Nothing in climate science violates physics. If you think it does, then the problem is with your understanding not the science.
Thanks MA Rodgers, your answer is very helpful.
So the answer is that the ISS, approximated as a rectangular plane, receives 1366 W m-² from the sun when facing the sun, while on average, Earth's system, approximated as a sphere, receives only 341 W m-² from the sun.
But then, how do you explain that 1366 W m-² received from the sun by the ISS rise the temperature to 121°C, while an average of 987 W m-² on the surface gives a mean temperature of 10-14°C on Earth, which would be the temperature of the ISS on the ground, right ?
So, it suggests that more than 100°C degrees of difference between the ISS and Earth's surface is caused by a difference of only 399 W m-² of radiation (1366 - 987), that is, the same amount of radiation than emitted IR radiations from the ground ( 396 W m-² )...
If only 400 W m-² would lead to an increase of 100°C on Earth, what would be the powerful cooling mechanism which doesn't appear into the Earth's Energy Budget ?
And also, by what processes or how does radiation build up and accumulate into the atmosphere in such a way that there is as much backradiated radiation (333 W m-²) than emitted ground radiation ( 396 W m-² ), than incoming solar radiation (341 W m-²) ?
@ scaddenp (35), it is not what Trenberth is saying: for him climate science violates physics, and that's why he said that "we can't account for the lack of warming"...
Indeed, if you calculate the emitted IR from the ground by the Stefan-Boltzmann Law as pointed out by MA Rodger in @34, for 1m², 14°C and a black body of emissivity 1, you'll find a radiative power of 385,5 W m-², approximatively equal to the one of the Earth's Energy Budget for IR emitted by the ground only (396 W m-²).
But according to the same Energy Budget, any body on the earth surface is subjected to 161 W m-² from the sun, 396 W m-² from the ground and 333 W m-² from the atmosphere backradiation, that is, at least 890 W m-² without taking into account heat from thermals, evapotranspiration and latent heat.
So according to the energy budget, the mean temperature should be, for a total radiation of 890 W m-², approximatively 81°C according to the Stefan-Boltzmann Law ...
[DB] Sloganeering, falsehoods and evidenceless assertions snipped.
FYI, I have corresponded personally with Dr. Trenberth and the words you put into his mouth are indeed falsehoods you made up.
Last warning to adhere to this venues Comments Policy. No more shall be given.
ab @36,
Your final three paragaphs are predicated on the assumption that you present a valid assessment in your second paragraph, but you do not.
From the various numbers you are using, it is plain that you are using the data presented this diagram to calculate your 987Wm^-2. The error you make in calculating that 987Wm^-2 is that half the quantity is downward radiation and (most of) the rest is upward radiation. Such energy fluxes cannot heat the same surface. So, just as is required in converting the solar TOA energy flux (from a 'plain' value of 1,366Wm^-2 warming the disc of the Earth into an 'areal' value of 341Wm^-2 warming the globe), we have to convert the energy flux warming the earth-parked International Space Station into upward and downward. You need to divide by 2.
In more detail, the surface latent heat transfer is unlikely to be warming the ISS (or its own latent heat transfer cooling the ISS) while we should perhaps expect the sensible heat loss to be, like the radiative fluxes, from both upper and lower surfaces. The resultant equilibrium temperature would this be a tad lower than the surface.
MA Rodger @ 38,
Yes, I'm taking the data from Trenberth's energy diagram which is quoted by the IPCC in their reports.
Let's forget a moment about the ISS, and just consider a physical black body on Earth's surface, whatever it is.
As I pointed out in @37, according to the IPCC diagram, any physical body on the surface of the Earth is, on average, subject to 161 W m-² coming from the sun, 396 W m-² coming from the ground and 333 W m-² coming from the atmosphere backradiation, that is, at least 890 W m-² without taking into account secondary heat sources from the diagram.
So, according to Stefan-Boltzmann Law, such radiation as absorbed by a black body would rise its temperature to 81°C... However, the mean temperature on Earth's surface is 15°C, not 81°C...
There is a 66°C difference between the model and the measured temperature at the surface of Earth, and as pointed out by Trenberth, we can't account for the lack of observed warming compared to the modeled radiative flow on the surface.
ab @39,
And what is the surface area of this body?
Let is keep it simple and say it it has no side area being a flat plate 1m x 1m. What is the surface area of such a body? Remember it will be radiating from the entire area. Being flat we can ignore the sides. Its top being warmed by the 494W sun+atmosphere is 1m x 1m = 1 sq m. Its bottom being warmed by the 396W surface flux is 1m x 1m = 1 sq m. Total surface area = 1 + 1 = 2 sq m. And this is the area being radiated from due to the body's temperature. So the body has to shed (494+396)/2=890W/2=445W/m^-2. So without any sensible heat loss, its blackbody temperature would be +24°C not +81°C.
MA Rodger @ 40,
Of course, I was implicitely supposing a blackbody of one square meter: a flat rectangular blackbody plane of 1m² as you pointed out.
Let's say a plane that is floating 1 meter above the earth's surface, one face pointing to the sky and the other pointing to the ground, and let's give it a name: "thermometer".
So this blackbody "thermometer" is going to receive from the sky as much as 1m² of earth's surface, as 1meter above the ground is negligible here to make any difference.
And because heat is transmitted within the ground by conduction, this blackbody "thermometer" will also receive as much from the ground than what is indicated into the IPCC diagram, as any difference would be negligible for 1m².
So, the thermometer will receive on its top 161 W m-² from the sun and 333 W m-² from the atmosphere backradiation, and on its bottom 396 W m-² from the ground.
That is, it receives 496W m-² from the top and 396 W m-² from the bottom, that is, in total, it receives 890 W m-².
As it is a blackbody, in thermal equilibrium, it is also going to emit as much as it receives, that is, also 890 W m-², in equal quantity at the top and the bottom: 445 W m-².
Now, the thermal equilibrium of that blackbody thermometer is based on its total energy, right ? It is not based on just the top or the bottom, it is based on the total energy it receives which is also equal to the total energy it emits.
And that total energy is 890 W m-², from which we can infer the temperature of the blackbody thermometer in thermal equilibrium: 80,8°C, according to the according to the Stefan-Boltzmann Law.
That is: a 66°C difference from the model compared to the measured temperature on Earth's surface (15°C), and not just a 10°C difference like you concluded in @40, which is already a considerable difference !
ab @41,
Your sixth paragraph needs correcting. This hypotheitcal body does not recieve 890Wm^-2. It recieves 860W. It has a top surface (area 1 sq m) recieving 494Wm^-2=496W. It has a bottom surface (area 1 sq m) recieving 396W^m-2=393W. The body is thus recieving 494W+396W=860W. It has a total surface area (top+bottom) of 2 sq m so will need to radiate 445Wm^-2 to be in equilibrium. You do manage to present this 445Wm^-2 value in your seventh paragraph but the erroneous 860Wm^-2 rears its head in the ninth paragraph even though there was never a flux of 860Wm^-2, only a total (up + down) flux of 860W acting on a body with a surface area of 2 sqm.
MA Rodger @ 42,
My apologies, I have messed up with numbers because I was looking at two different Trenberth's Energy Budget diagrams from different dates and with updated values.
Let's take the most recent one from Earth's Global Energy Budget:
Absorbed by Surface: 161 W m-²,
Back Radiation: 333 W m-²,
Surface Radiation: 396 W m-²,
Top of the blackbody thermometer: 161+333 = 494 W m-²,
Bottom of the blackbody thermometer: 396 W m-²,
Total received energy: 494 +396 = 890 W m-²,
Total emitted energy at thermal equilibrium: 890 W m-²,
Temperature of the thermometer at thermal equilibrium : 80,8°C (Stefan-Boltzmann Law)
Difference with measured temperature: 80,8 - 15 = 65,8°C
We are reasonning here in terms of fluxes of energy, not in terms of energy.
The thermometer received two fluxes from the top, one flux from the bottom. You combine (add) all those fluxes together in order to get the total flux for the thermometer. That total flux is what is received, and also what is emitted, because the thermometer is a blackbody. Think of the "radiative balance" of the Earth.
After that, how much energy is actually emitted depends on the surface that is considered, that is, in this particular case of a rectangular plane: 445 W m-² at the top and 445 W m-² at the bottom, in equal quantities.
ab @43,
I don't think I can put it any simpler. This hypotheitcal body does not recieve 890Wm^-2. It recieves 890W. It has a top surface (area 1 sq m) recieving 494Wm^-2=496W. It has a bottom surface (area 1 sq m) recieving 396W^m-2=393W. The body is thus recieving 494W+396W=860W. It has a total surface area (top+bottom) of 2 sq m so will need to radiate 445Wm^-2 to be in equilibrium.
Or perhaps put it this way. If it were a 1m cube it would radiate 445Wm^-2 over its surface area of 6 sq m = 2,670W. If it were just half a metre thick, it would radiate 1,780W over 4 sq m. And a quarter of a metre - 1,335W. And that level of radiation would continue to reduce as the thickness reduces until at 1mm thick it would radiate 892W over 2.004 sq m. So when it is effectively without depth, why would it not be then radiating 890W over 2 sq m = 445Wm^-2? And if that is the radiative intensity, (identical to the cube) what is its temperature?
MA Rodger @43,
I can not put it any simpler either and it is very simple physics: of course the blackbody thermometer receives 890 W in the lapse of time that is considered, which is the period between Mars 2000 and May 2004 for the diagram, but here we are talking of fluxes of energy, we are not concerned with the energy itself.
A flux, or flow, represents an emission or a reception of a certain amount of energy per unit of surface in time.
So what is the total flux of energy coming in ? 890 W m-²
And what is the total flux of energy coming out ? also 890 W m-² because it is a blackbody: it emits as much energy as it receives, or in other words, it has the same emission and reception flux.
The only difference between the reception flux and the emission flux being that the blackbody radiates its energy according to its own surface: both 445 W m-² at the top and at the bottom, whereas the reception flux comes in three different fluxes: two at the top and only one at the bottom.
But the energy is conserved, because both incoming and outgoing fluxes are equal: 890 W m-².
Otherwise, if you say that the total flux of energy going out is only 445 W m-² you are violating the law of conservation of energy: there is more energy going into the system than energy going out of the system.
Let's look at it from the energy point of view if you wish: there is 890 W entering into the system, you output 890 W, that is 445 W on the top and 445 W on the bottom.
Now, advance in time, and do the same thing: you receive again 890 W, and you ouput again 445 W on the top and 445 W on the bottom.
That is, you have created a flux of 890 W coming in and two fluxes of 445 W going out, one at the bottom and one at the top.
Those two output fluxes are generated from one square meter of surface, thus the fluxes that you have created are of 445 W m-².
That is simple physics of fluxes.
Ab,
This is a perfect example of why it is a waste of time for posters to do their own calculations.
As MA Rodger has pointed out, the surface you are discussing has two faces. One is 1 m2 pointing up and the other is 1 m2 pointing down. That means it has a total surface area of 2 m2. This absorbs 890 Watts total and radiates 495 W/m-2 up and 495 W/m-2 down. The area of the object is 2 m2.
Your calculation has a gross math error and you do not understand MA Rodgers explaination. Since you cannot do basic physics calculations without gross math errors, why should I think you can explain the greenhouse effect to me? You think the IPCC has made a mistake but the issue is you cannot do math correctly.
If you are interested in learning how to do the calculation correctly, if you ask nicely there are several people at SkS who are willing to show you how it is done. If you want to lecture us on how smart you are you need to find a venue where they cannot do the math properly.
michael sweet @ 46,
My calculations have no error and the area of the object is of no importance at all:
Once again, one is not dealing with energy but with fluxes of energy.
Kindly reread my commentary above (@45).
The results are that IPCC's radiative model has a 66°C divergence with observed temperatures.
Unfortunately, here, most posters do not seem to be familiar with physics of fluxes and confound fluxes with energy exchange.
You can not comprehend radiative fluxes with simple energy exchange considerations.
It is like wanting to do quantum mechanics with classical physics.
[DB] Inflammatory snipped.
ab @45,
Reading your comment @45 suggests you have a problem, not with physics, but with the simplest geometry. You write "That is, you have created a flux of 890 W coming in and two fluxes of 445 W going out, one at the bottom and one at the top. Those two output fluxes are generated from one square meter of surface, thus the fluxes that you have created are of 445 W m-²."
But it is not "one square metre of surface."
If I have a flat object measuring 1m x 1m. It has two surfaces, a top surface and a bottom surface. Both top and bottom surfaces have a surface area of 1 sq m. Thus the flat object with its two surfaces has a total surface area of 2 sq m.
ab is trying to pull a G&T type of argument. Not interesting. Just trying to play games.
[PS] Looks more like skydragon to me.
"and the area of the object is of no importance at all".
Um, sorry ab, wrong. Area is absolutely important.
You aren't clearly enough distinguising between flux per unit area, and flux.
In your simple plate example, the plate absorbs 890 watts, from where-ever. It then needs to radiate 890. Since it has a total surface area of 2 m^2 it thus has to radiate 445 per square meter. And the SB equation calculates flux per square meter. So feed 445 into the SB calculator and you get around 14.
The mistake you seem to be making, at least implied, is assuming the SB calculator gives you flux (in watts) when it is actually flux per square meter.
If the plate had a surface area of 10 m^2 and absorbed 890 watts from where ever, it would need to radiate at 89 watts/m^2 to be in balance and the SB calculator would give a much lower temperature than 14 C.
Area matters totally unless you keep all your calculations on a per square meter basis.