Major Errors Apparent in Climate Model Evaporation Estimates

Yeah, whatever.
Squealing little lukewarmers were given their chance. They blew it.
Evaporation rates wrong? This is news?
Evaporation is the primary way the atmosphere cools the oceans. How does the atmosphere cool? Radiative gases.
Just get over yourselves.
Sure, I am a nasty piece of work. That doesn’t stop me being right 😉
“There are no good and bad people in the world. There are only ever and always the bad people, it’s just that some of them are on different sides.” – T Pratchett
An unending sea of evil, shallow in most places, but deeper, oh so much deeper in others……
I may be a monster, by I am MY monster.
(Anthony, I can take out Monckton, Willis and Dr. Brown, yah got anything or anyone else?*)
*I am always the smartest guy in the room.**
** Dependant on room size. Serving suggestion only. Results may vary.

I understand this blog to support negative feedback from phase change in H2O at the surface and radiation of the energy from the top of the atmosphere.
The author observes that models that support global warming incorrectly posit an energy constraint on evaporation.
I conclude that this is what forces them to introduce positive feedback to make their model work.
I think my conclusion follows from the data and discussion presented here. Any comment?

“6% / C” seems a great over simplification. The following is for fresh water:
E = 5([Tc+ 18]**2.5– r[Ta+ 18]**2.5)(V + 4) x 10**-6
where:
E = evaporation rate, kg/m2/h
r = relative humidity/100T
Ta= air temperature, °C
Tc= water surface temperature, °C
V = wind velocity, km/h.

If the author is correct, then this is a pretty big deal. Infact, this could be one of the most significant events in the debate against AGW.

This might be a little tangential, but it is also connected to precipitation, and specifically the resolution being too coarse to pick up rainstorms.
While discussing the Lovejoy 99% paper it struck me that a lot of the measurement accuracy is obtained by removing inhomogeneities by statistical techniques. However, if you consider that things that lower the surface temperature (rainstorms) are smaller scale than the rest of of the climate (i.e. there are more places, covering a larger area, where it is raining rather than not raining at any given time). What’s more, this evaporation problem will also occur over land, because you will have increased evaporation where it has recently rained.
In this context, what you get by doing that is just a high pass filter, and the more “accurate” result, the higher the skew to the positive side will be. If that is correct (and do correct me if I am wrong) it is not just in the models that suffer from this problem, but the measurements too.

This doesn’t pass a smell test, if evaporation increases 6% then precipitation must increase a similar amount. Taking into account that average rainfall across the surface is a meter per annum and the specific heat of evaporation and change in potential energy between the surface and 3 km a forcing of at least 5.5W per meter squared would be required to break even on the energy budget to do this. Since the total forcing is only 3.7 Watts per square meter, and the imbalance only 0.6W per square meter, there is clearly insufficient energy in CO2 related reflected IR to sustain the 6% increase in the hydrological cycle, even at 5.5W per square meter, if there was a 6 % increase in evaporation then the cooling effect would completely cancel the warming, so it seems to me that a driving energy of considerably more than 5.5W per square meter would be required to sustain such an increased evaporation AND warm the atmosphere at the same time. (Of course then we should add in the other losses, such a heating gigatonnes of liquid water, melting a few gigatonnes of ice, sustaining increased photosysthesis, greater storm energies and all the other magical effects that are supposed to happen – apparently without expending any of that magical 0.6W of heating that’s supposed to be causing these effects
I need to understand where you think the energy is coming from to sustain this warming in the presence of such a huge increase in evaporation.

“It seems to me almost emblematic of those “scientists” inside the AGW bubble world that they would not bother to get the science and the maths right on these things as that really would reveal an inconvenient truth.”
I have oten wondered if that is the sad and sorry case. They are certainly quite capable of correcting their errors to make sure those laughable models be much more accurate, but they resist, why?
Why do they continually demonstrate their complete ineptitude and bias by not fixing those models to reflect true current results, instead of their own biased, fantasy driven ever higher temperatures. It is just beyond comprehension. Does it refer to that old saying “better to let them think you’re stupid than to open your mouth and remove all doubt?’. I am perplexed by this entire scenario. Even when faced with the obvious facts like the examples above they will still continue on their ignorant and deliberate of scientific principles, theories and conduct.

Either I’m confused, or several commenters are confused (or both).
The article assumes a constant relative humidity, so the evaporation would need to be greater and would cause more water to STAY in the atmosphere. This water cannot then cause surface reheating by additional rainfall because by definition it is staying up there. (This doesn’t mean the water cycle can’t go up or down separately, just not with this water. BTW, alarmists need to demonstrate water cycle changes using data and not models.)
So the net effect of 6% more evaporation per degree C, is a negative feedback which isn’t being caught by the GCMs.
If I’m wrong, please tell me. Thanks.

Slow to follow:
Here’s a [pointer] to a calculator for the evaporation from a body of water. Inputs are rel humidity, abs humidity, air temp, water [surface] temp, and wind velocity.
See http://www.engineeringtoolbox.com/evaporation-water-surface-d_690.html
Any commenters who haven’t worked the examples, nor done the math, are simply muddying the waters of human knowledge. Arguments to authority, ad hominum (IEEE is an engineer…), are bogus as always.

Mods – Correction to previous post – water surface temp

Trenberth himself acknowledges that the climate models ignored evaporative cooling from Tropical Cyclones of 1.13 Wm^2 see the last Fig in my post at
http://climatesense-norpag.blogspot.com/2013/02/its-sun-stupid-minor-significance-of-co2.html
I said in this post from about a year ago
“The modelling community and the IPCC have both recognized that they have a problem. For example both Hansen and Trenberth have been looking for the missing heat and generating epicycle type theories to preserve their models. Hansen thinks it might have something to do with aerosols and Trenberth first wanted to hide it down the deep ocean black hole. Death Train Hansen is a lost cause as far as objective science is concerned but Trenberth has always been a more objective and judicious scientist and has recently made excellent progress in discovering a real negative feedback in the system. see
http://www.cpc.ncep.noaa.gov/products/outreach/proceedings/cdw31_proceedings/S6_05_Kevin_Trenberth_NCAR.ppt
”
Note PPT slides 38,39 and 40
He says this TC cooling of 1.13Wm^2is important relative to the CO2 radiative forcing of 1.5 Wm^2 and has not been included in the climate models . I’m sure this negative feed back was not included in the latest AR5 report sensitivity calculations..

I looked for a mention of the Clausius-Clapeyron relationship (equation) and did not find any. Is this not an application?
Ian

Richard: I’ve been interested in analyzing radiative forcing from a surface energy balance perspective rather than the balance across the tropopause. The instantaneous increase in DLR (from RTE) associated with 2XCO2 is only 0.8W/m2 at the surface, while the change at the tropopause is 3.7 W/m2. The 0.8 W/m2 at the surface will be amplified by water vapor feedback; call this 0.8*f W/m2.)
The surface can get rid of this excess energy by two mechanisms: warming until the surface radiates an additional 0.8*f W/m2 or by increased evaporation. From a surface perspective, climate sensitivity is controlled by the ratio of these two routes to escape the surface. However, the rate of evaporation is mostly controlled by wind speed, not the surface temperature of the water. This is because the layer of air immediately above the surface of the ocean is saturated with water vapor and the slow step in is mixing that saturated air into the boundary layer. (We can experience this difference leaving a pool on a warm windy day vs a warm still day.) I’m not sure how GCM handle this problem, but their failure to predict the full increase in rainfall with GW is proof that they haven’t got moisture transport into and out of the boundary layer right.

Electromagnetic forces are strong compared to gravity. To move a water molecule from liquid to gas phase 1 micron above the surface needs about the same amount of energy as moving it to an elevation of 15 miles against gravitational gradient. Evaporation matters.

Global average precipitation is 990 mm/annum, evaporation should be the same. Precipitation over land is 715 mm/year. Of this 210 mm (31,500 km³) is lost as river runoff to sea, the rest is evaporated. Therefore evaporative cooling is significant even over land, over oceans it is only about twice as much.
Also, river runoff indicates a 16 W/m² latent heat import by the global land area.
Moreover, about 90% of water droplets in clouds are evaporated in mid air (at the cloud base), before falling to the surface as precipitation. Water vapor rises again, then recondenses at a higher elevation releasing its latent heat content. Therefore water plays an even more significant role in moving heat to higher elevations than usually assumed. Also, this is how clouds stay afloat in spite of liquid or frozen water being heavier than air.
It is not easy to calculate rate of evaporation, because it does not only depend on temperature and relative humidity, but also on area of water-air interface. In calm weather it is the surface area of oceans, but as soon as wind is strong enough to generate spray, interface area is increased by many orders of magnitude.
Over land effective interface area is mostly set by vegetation. In forests total leaf area alone is much higher than the area covered by trees (it is the same in grasslands). On top of that internal surface of stomata (microscopic cavities on backside of leaves) is even higher and they are opening and closing on demand, therefore rate of evaporation is regulated. At higher atmospheric CO₂ concentration there is less need to open up those breathing holes, so evaporation per unit leaf area is decreasing. However, it is compensated by an increasing total leaf area, which tends to maintain higher primary production of biomass with the same water consumption as before.

Greg Goodman: Several people have pointed out that you can’t heat a ventilated body of water with IR. It simple increases evaporation. Unless I’ve missed by scanning the article too quickly this does not seem to be accounted for.
Do you have references for that? I may have asked before and missed the answer; if so, I apologize.

Richard J. Petschauer,
That was a most illuminating post. Thank you.
I look forward to your response to RACookPE1978; addressing the distribution of temperatures, radiant fluxes, and evaporative changes, rather than calculations based spatio-temporal distributions. However, I think the main point that the GCMs are off is likely to withstand scrutiny.
Is your work available in pdf format? I can easily copy/paste, but I always fear (this may seem absurd) transcription errors.

Told you so.

@William Howard Astley
No, I think my comment holds. Sure, if the temperature increases by 1 degree and the evaporation increases by 6%. What does that do to the energy budget given that you must evaporate 30,600,000,000,000 tonnes of water and raise it to 3km high What does that cost in energy? Given the additional energy for a doubling of CO2 is IIRC 3.7W per square meter, there is insufficient energy to sustain even a 1 degree rise. The situation is non-physical, as there is insufficient additional energy to sustain the temperature rise given the evaporation rate. To evaporate that much water you need more energy than the change in energy that was supposed to be driving it – You are energy saturated – ergo, doubling CO2 cannot raise temperature 1 degree.
Add to that the parts of that extra retained energy that goes into the atmospheric window, melting ice, heating oceans, plant growth , wind, waves and numerous other unclaimed losses, the liklihood of a non zero sensitivity diminishes quickly, and may even be negative, for example, in plant growth, the CO2 acts both as a reactant and a catalyst – the small increase in CO2 partial pressure leads to larger increases in the consumption of shortwave energy, and increases in transpiration extracting more energy than the CO2 puts in This is likely to have a nett near surface cooling effect

JohnWho says:
April 15, 2014 at 6:06 am
“He says, as he posts from a phone booth.”
——————————————————
Well I did say “dependant on room size” 😉
The judges would also have accepted “broom closet”.
But broom closet or ball room makes little difference when up against lukewarmers. They are playing at a disadvantage. Desperately running around looking for a “sciencey” sounding solution that allows “warming but far less than we thought”. Lukewarmers cannot even contemplate the idea that they are as wrong as the believers and that radiative gases don’t warm at all.
The problem is indeed evaporation. The oceans are primarily cooled by evaporation. Without this they would become a giant evaporation constrained solar storage pond with temperatures topping 80C. An atmosphere without radiative gases cannot provide this cooling as it has no effective way of cooling itself.
Climastrologists have calculated the temperature of the oceans to be -18C in the absence of DWLWIR and atmospheric cooling and claimed that our atmosphere must be warming our oceans. Empirical experiment shows that figure to be out by 98C. The atmosphere is therefore cooling our oceans. And the only effective cooling mechanism for the atmosphere is radiative gases.
AGW is a physical impossibility, but that’s not the answer lukewarmers want.

bobl: Since the total forcing is only 3.7 Watts per square meter, and the imbalance only 0.6W per square meter, there is clearly insufficient energy in CO2 related reflected IR to sustain the 6% increase in the hydrological cycle, even at 5.5W per square meter, if there was a 6 % increase in evaporation then the cooling effect would completely cancel the warming, so it seems to me that a driving energy of considerably more than 5.5W per square meter would be required to sustain such an increased evaporation AND warm the atmosphere at the same time.
Not disputing you, but the author, Richard J. Petschauer, did not actually claim that doubling the CO2 concentration would increase the global mean temp by 1 C and raise the rate of evaporation by 6%. What he did was mimic some of the equilibrium type assumptions commonly used (mean temp as equilibrium, uniform surface and TOA, maintaining the same relative humidity, etc) and show how much difference is made in the standard computations by changing the relationship of vaporization change to temp change from that used by IPCC to a more realistic value.
I don’t believe that any of the equilibrium models are accurate enough to use for planning, and the GCMs are to date clearly too inaccurate to support planning. But if you take equilibrium calculation seriously, then this post shows that the inaccurate function for vaporization used in the GCMs is probably an important contributor to their inaccuracy. That’s how I read it.
I have posed the question: If doubling CO2 increases downwelling LWIR by 3.7 W/m^2, how much of the energy transfer goes to warming the water and near-surface air, and how much goes to vaporizing the water (in the non-dry parts of the Earth surface)? His calculations and your calculations suggest that most has to go to vaporizing the water, and little to temperature increase.

Steven Mosher says:
April 15, 2014 at 4:52 pm
————————————
“Radiative gases don’t warm.”
I’m well aware the the claim is “slow the cooling rate”. Semantics won’t save you.
“They lower the rate of energy loss to space.”
Without radiative gases our atmosphere would have no radiative cooling ability at all. So for the atmosphere they increase the energy loss to space. An atmosphere without radiative gases would have its temperature driven by surface Tmax, but would have no effective cooling mechanism. Empirical experiment proves conduction back to the surface ineffective.
“When we add ghgs we raise the erl. The higher erl means the earth radiates from a higher colder altitude.”
There is no such thing as ERL or EEH. These are a mathematical fiction used only by climastrologists. The simplest empirical experiment disproves the very concept of an average ERL. Go outside on different days in different weather conditions and scan the sky with an IR thermometer. The atmosphere is radiating in ever changing patterns from ever changing altitudes.
“Over 100 years ago we predicted an increase in co2 would result in a warmer planet. So far that prediction is correct”
Correlation is not causation. The warming since the little ice age started well before any significant human CO2 emissions.
The bottom line is this –
The sun heats our oceans.
The atmosphere cools our oceans.
Radiative gases cool our atmosphere.
AGW is a physical impossibility.
Steven, the climastrologists have made a fist-biting error in the very foundation of their claims. They used “blackbody” calculations to claim the oceans in the absence of DWLWIR or atmospheric cooling would have a Tmean of -18C. This is totally and utterly wrong. Our deep transparent oceans are not a “blackbody”, they are a “selective coating” over 71% of the lithosphere. Empirical experiment proves that the oceans in the absence of DWLWIR and atmospheric cooling would be at 80C or beyond. That’s a 98C error for 71% of the Earth’s surface! You can’t laugh that one off.
You may object to my use of crude language, but there is no other way to put it. 97% of climate “scientists” are assclowns!

Richard Petschauer
You get 6%/C if relative humidity remains constant. But it will decrease because evaporation cools the water. So water temperature drops as well as its vapor pressure. But air remains warm and its absolute humidity at saturation point increases with temperature. You get initial increase in evaporation then decrease. You average the two. 6%/C if only increase and no decrease.

I think the point of the 6-7% is actually the measured evap increase/C. The point being the models being set at 2%/C guarantees they don’t follow reality. We know the models don’t follow reality and this is merely a reason why. Trying to say that measured reality is impossible is pretty much what the CAGW folks do daily. Obviously in local conditions water temp does in fact increase over time and then decrease, this drives the water cycle. The static annual average of global temps used to measure GW mean nothing when it comes to actual local conditions. The truly astonishing thing is how little the very meaningless annual average varies.
v/r,
David Riser

Janice Moore says:
April 15, 2014 at 3:30 pm
———————————-
Janice,
yes, it is the water that’s going to get them in the end 😉
It seems they forgot that there is a big puddle of it 4-5 km deep covering 71% of the planet’s surface.
When you look at what the maths would actually mean, the Church of Radiative Climatology is actually claiming that our atmosphere is acting to warm the oceans. I think those big fluffy white things in the sky rather indicate that our atmosphere is cooling the oceans. I would have thought some of the climastrologists might have noticed, but apparently not…

Richard Petschauer,
Let me paraphrase, in order to avoid energy limitation arguments, you have used an assumption of available driving energy of 10W / square meter, not caring where it comes from.
The point I’m trying to make though is that in reality we energy constrained. We are told that a doubling of CO2 produces a driving energy of 3.7W per meter squared not 10W. What I was wanting you to do was to constrain your solution to 3.7 watts and calculate the maximum temperature change, and evaporation when that constraint is applied, or alternately calculate the energy required for an increase of 1C given that evaporation increases 6 % either way this makes for a quick comparison with IPCC models. I am not suggesting that the energy constraint would reduce the evaporation.
Eg the IPCC says that a doubling will cause 3.3 degrees of warming and by your calcs an attendant 20% increase in evaporation. Given that the energy is constrained to 3.7 Watts per square meter what is the revised temperature rise and evaporation rate. I’d assume that if we calculated the maximum extra evaporation that could be sustained by the excess energy, one could use the temperature/evaporation relation to tell us the temperature rise, all other losses ignored. By my estimate rather than 3.3 degrees and 20% we would get about 0.5 C temperature rise, and 3% evaporation for a doubling. That’s a big difference, a headline.

I will try to cover a number of comments and questions here
Regarding evaporation and precipitation:
It is usually agreed that globally over about a months average, precipitation will approximately equal evaporation since the atmosphere can’t hold that much water. But the sources of the evaporation and the destinations of the precipitation can be separated by very great distances. Evaporation is easier to calculate (if local temperatures, humidities and wind speeds are known), but precipitation is easier to measure, except some data over oceans may be lacking. What causes which? I would think evaporation is the main driver; however precipitation causes downdrafts of drier air that can enhance evaporation locally. Cloud formation is difficult to simulate, as is the onset of precipitation, except in the very short term such as in weather forecasting.

From Mike Jonas:
Confirmation by Dr Susan Wijffels on ABC (Australian Broadcasting Corporation) “Catalyst” program ..
http://www.abc.net.au/catalyst/stories/3796205.htm
“We’re already starting to detect and see big changes in the extreme events. And we’ve only really warmed the Earth by 0.8 of a degree. If we were to warm the Earth by 3 or 4 degrees, the changes in the hydrological cycle could be near 30 percent. I mean, that’s just a huge change, and it’s very hard for us to imagine.“
It seems extraordinary unreasonable to assume that precipitation could increase like that without there being a similar increase in evaporation.

My Response:
We won’t get increases as high as 3 to 4 degrees. Now they are admitting evaporation & precip rises 7.5 to 10%, but the models only use 2.5% to justify the warming. The increased temperature rise of 3 to 4 degrees can’t happen, because the increased evaporation will cut in almost in half and the lack of positive cloud feedback another large reduction.

Old England says:
April 15, 2014 at 1:24 am
Interesting, and the effect of convection on the calculations is?

My Response: Convection is the H in the equations. Here we use the estimate at 24 W/m squared. If it changes with surface temperature, it would be treated as an additional feedback factor the same way as evaporation. One would need to its change per C of surface temperature change. Since the atmosphere temperature tracks the surface, I would not think this is significant.

Joe Born says:
April 15, 2014 at 1:27 am
Is conductive loss so low that Equation (1) can ignore it?

My Response: Yes. Air that does not move has very little actual conduction; it is a fairly good insulator. Air moves heat by parcels of air moving either through convection (rising air thermals) or cooling air dropping. This net is the term H in the equation. Winds also move much heat from one area to another, but that does not enter into the global heat balance, but it helps distribute the heat and help make a global heat model work .

Frederick Colbourne says:
April 15, 2014 at 1:27 am
I understand this blog to support negative feedback from phase change in H2O at the surface and radiation of the energy from the top of the atmosphere.
The author observes that models that support global warming incorrectly posit an energy constraint on evaporation.
I conclude that this is what forces them to introduce positive feedback to make their model work.
I think my conclusion follows from the data and discussion presented here. Any comment?

My Response: I think they had the water vapor positive feedback there already. But I think you have a point in the case of the cloud feedback, which as I pointed out was simply manufactured so the simple energy balance models will give the same temperature rise from 2x CO2 as the models do (for the nominal value of about 3C; the maximum value of 4.5C is just because there is so much variation among the models). So just eliminating the cloud feedback (that’s probably a little negative with more water vapor) and correcting the evaporation rate feedback, the feedback multiplier drops from 2.48 to 0.91. And elimination of the uncertainty of the climate models could cut the maximum by about 30%.
My present project is to check on the large positive water vapor feedback claimed using both the IPCC estimate of its increase and that based on a large amount of data how it has varied across the globe.

bobl says: (See last Para)
April 15, 2014 at 6:04 am
This doesn’t pass a smell test, if evaporation increases 6% then precipitation must increase a similar amount. Taking into account that average rainfall across the surface is a meter per annum and the specific heat of evaporation and change in potential energy between the surface and 3 km a forcing of at least 5.5W per meter squared would be required to break even on the energy budget to do this. Since the total forcing is only 3.7 Watts per square meter, and the imbalance only 0.6W per square meter, there is clearly insufficient energy in CO2 related reflected IR to sustain the 6% increase in the hydrological cycle, even at 5.5W per square meter, if there was a 6 % increase in evaporation then the cooling effect would completely cancel the warming, so it seems to me that a driving energy of considerably more than 5.5W per square meter would be required to sustain such an increased evaporation AND warm the atmosphere at the same time. (Of course then we should add in the other losses, such a heating gigatonnes of liquid water, melting a few gigatonnes of ice, sustaining increased photosysthesis, greater storm energies and all the other magical effects that are supposed to happen – apparently without expending any of that magical 0.6W of heating that’s supposed to be causing these effects
I need to understand where you think the energy is coming from to sustain this warming in the presence of such a huge increase in evaporation.

My Response: First, I use 6% of 78 or about 4.68 W per meter squared, not 5.5. Secondly, you perhaps missed what I said twice in my essay, that the 6% applies to the temperature rise after the negative feedback and its temperature reduction process is complete. Assuming there are no other feedbacks (that would change the 3.7 value), my equation (3) shows 3.7 gives a surface temperature rise of 0.98 C before feedback. To get this, change k from 195 / (195 + 324) to (195-3.7) / (195-3.7 +324) from the change in G calculate a change in temperature (about 0.184 times the change). IPCC gets 3.7 / 3.2 or 1.16C. In my paper I gave a negative feedback factor for a 6% change in E of -0.741, which gives a feedback multiplier of 1 / (1 – F) or 1/ 1.741 or 0.5744. So the final temperature rise is drops to 0.98 x 0.5744 or 0.56C and the evaporation is 6% x 0.5744 times 78 (the original E) or about 2.68 W per meter squared larger. For a 0.56 temperature rise from the surface I get an increase in radiation of 3.04 W per meter squared for a total increase of 3.04 + 2.68 or 5.72 W per meter squared. But how can 3.7 W per meter squared at the top of the atmosphere cause a 5.72 increase at the surface?
Well in this case D increased by the same amount and supplied the energy. It’s the greenhouse multiplier. The same thing that causes 390 – 40 through the atmospheric window per meter squared leaving the surface when the net incoming solar heat is only about 235 W per meter squared. Note 350 / 235 = 1.50 and 5.72 / 3.7 = 1.55. The difference is a small additional increase in the amount through atmospheric window.
More comments to follow later (RJP)
[block quotes added to increase clarity. Mod]

I will try to cover a number of additional comments and questions here
Bob Shapiro says:
April 15, 2014 at 7:12 am
“Either I’m confused, or several commenters are confused (or both).
The article assumes a constant relative humidity, so the evaporation would need to be greater and would cause more water to STAY in the atmosphere. This water cannot then cause surface reheating by additional rainfall because by definition it is staying up there. (This doesn’t mean the water cycle can’t go up or down separately, just not with this water. BTW, alarmists need to demonstrate water cycle changes using data and not models.)
So the net effect of 6% more evaporation per degree C, is a negative feedback which isn’t being caught by the GCMs.
If I’m wrong, please tell me. Thanks.
Minor point re the acceleration of the water cycle – there can be a great deal of condensation on the water at night without ‘precipitation’ and it goes unnoticed. In other words the argument that the 6% increase is ‘too much’ should be examined in the light of all routes for returning water vapour to the sea.”
My Response:
It is well known that warmer air tends to hold more water vapor, how much is in question. The climate models assume or estimate that relative humidity (the fraction of the saturated value) will remain “approximately” constant. Data shows it actually drops a little, but I want to use the climate model’s values. Less humidity will cause a greater increase in evaporation with temperature, so I am being conservative. One the water vapor increase to what ever level it seeks, than additional evaporation will force increased cloud formation and then precipitation. In the long run of a few months, it is generally agreed that the average will equal evaporation. It has been estimated that a particular water vapor molecule only stays in the atmosphere from about 7 to 30 days. And precipitation reduce the humidity locally with dry downdrafts to help increase evaporation. Regarding the condensation at nighttime that does occur as you point out, that is already estimated in the present net value of about 78 to 80 Watts per square meter average globally. We just assume it will increase in the same proportion that the evaporation will increase, which seems a reasonable.
Greg Goodman says:
April 15, 2014 at 8:12 am
“Several people have pointed out that you can’t heat a ventilated body of water with IR. It simple increases evaporation. Unless I’ve missed by scanning the article too quickly this does not seem to be accounted for.”
My Response:
I do not believe this. While it will increase the evaporation, but only after first warming it. What’s special about IR heating. My microwave warms my coffee and some may evaporate.
Matthew R Marler says:
April 15, 2014 at 1:52 pm
“reg Goodman: Several people have pointed out that you can’t heat a ventilated body of water with IR. It simple increases evaporation. Unless I’ve missed by scanning the article too quickly this does not seem to be accounted for.
Do you have references for that? I may have asked before and missed the answer; if so, I apologize.”
My Response: See comments above on same subject.
george e. smith says:
April 15, 2014 at 1:03 pm
“Well I am not a believer; but I applaud Richard’s effort.
It’s the latent heat; his G that bothers me.
Please stop saying that latent heat warms the atmosphere; it doesn’t.”
My response:
The latent heat warms the atmosphere, but only after it is turned into sensible heat when clouds form. I took it for granted that this was well understood. Thanks for pointing that out. Technically, it is not latent heat any more, but the process of converting it into a liquid during cloud formation releases the heat. This is how an air conditioner works. Except the heat is just moved outside your house. A person can feel it leaving the condenser. With the planet, this is nature’s air conditioning system. The big difference is that the energy to run it is free. It comes from the sum. And the heat is “pumped” upwards by heat rising in the atmosphere because of the temperature drop with altitude until its cold enough to reach the dew point. And part of this warming migrates further upward and increases the radiation to outer space, a cooling action and negative feedback. And the warmer atmosphere radiates the rest of the added heat back to the surface (the atmosphere can’t store heat) which offsets part of the cooling and restores the energy balance needed for evaporation. The value, k, in my equation (3) provides the fraction of the added heat that leaves in the upward direction.
Matthew R Marler says:
April 15, 2014 at 1:52 pm
“reg Goodman: Several people have pointed out that you can’t heat a ventilated body of water with IR. It simple increases evaporation. Unless I’ve missed by scanning the article too quickly this does not seem to be accounted for.
Do you have references for that? I may have asked before and missed the answer; if so, I apologize.”
See comments above on same subject.
Matthew R Marler says:
April 15, 2014 at 2:05 pm
“Richard J. Petschauer,
That was a most illuminating post. Thank you.
I look forward to your response to RACookPE1978; addressing the distribution of temperatures, radiant fluxes, and evaporative changes, rather than calculations based spatio-temporal distributions. However, I think the main point that the GCMs are off is likely to withstand scrutiny.
Is your work available in pdf format? I can easily copy/paste, but I always fear (this may seem absurd) transcription errors.”
My Response: I plan to address RACookPE1978 later. Short answer: The globe and weather system have a good way of averaging these things out as we exprect they will continue to do so.
****
I JUST MADE A PDF of the paper with a few added clarifications one page 1.
FOR A COPY EMAIL TO: rjpetsch@aol.com
****
Janice Moore says:
April 15, 2014 at 3:21 pm
. . . .
“{Note: THE FOLLOWING IS A BONA FIDE QUESTION about the above article — please, if you have the time to explain to a non-tech like I, I would really like to read your explanation. If it is a poorly worded question, please re-word and answer the more useful one. Thanks!}
Question:
Given the above quotes and also this fact: there is no evidence that CO2 causes temperature increases in an open system like the earth,
can Petschauer accurately and with a meaningful level of confidence assert this: “… latent energy for evaporation comes from the temperature of the water itself. … reduced net energy from increased CO2 still warms the surface, … .”
{bolding is to emphasize what I am asking about, here, affect on earth’s climate zones of CO2}
In a highly controlled laboratory experiment, okay. But, on WHAT PROOF, i.e., what MECHANISM, does he base this grand assertion about CO2 and the oceans of the earth?”
My Response: Yes, there is no proof by actual data that mode CO2 will cause the planet to warm. But I have seen what is reported to be satellite data that shows a drop in infrared heat leaving the planet in the same 12 to 18 micron range where CO2 has been shown to absorb and radiate heat. And more CO2 will reduce the heat more in this region according to well established spectral physics. I have run these myself, but they show only about a drop of 2.53 W per square meter for a doubling of CO2, not the 3.7 generally accepted value.
(See:http://climateclash.com/does-the-tropopause-limit-carbon-dioxide-heat-trapping/).
(A stratospheric correction could increase my estimate to about 2.9). The band of CO2 is partial saturated so the heat loss turns out to follow a slow logarithmic function which means each doubling (or any percent increase) will cause the same amount of heat loss. The question is how much the surface temperature will rise per Watt per meter squared of addition heat loss to space. More heat loss will cause the atmosphere to warm enough to increase the heat enough to offset the loss and this warmer atmosphere will reduce the heat from the surface, causing it to warm. The IPCCs and many others divide the 3.7 by 3.2 (or multiply it by about 0.315) to get 1.16 C rise before certain feedbacks alert this. My model’s equation (3) gives me about 0.264 C / per Watt or 2.9 x 0.264 or about only 0.77 C before any feedbacks, which I estimate are much less than the IPCC’s.
No real proof, only estimates but with some good basis I think. By the way, at current annual growth rates of CO2 in the atmosphere of about 0.54% (based on data) it will take about 130 years to double. I think the climate models still use 1%, which gives 70 years. The 1% number ignores the absorption of the CO2 by the planet. This absorption also means the added CO2 will not stay in the atmosphere for over 1000 years as often quouted, if we ever reduce the CO2 emissions.
More comments to follow later (RJP)

“””””….george e. smith says:
April 15, 2014 at 1:03 pm
“Well I am not a believer; but I applaud Richard’s effort.
It’s the latent heat; his G that bothers me.
Please stop saying that latent heat warms the atmosphere; it doesn’t.”
My response:
The latent heat warms the atmosphere, but only after it is turned into sensible heat when clouds form. I took it for granted that this was well understood. Thanks for pointing that out. Technically, it is not latent heat any more, but the process of converting it into a liquid during cloud formation releases the heat. …..
Well my response is that NO latent heat energy becomes available from water vapor, until the surrounding atmosphere drops down to the dew point, at which time, given any kind of substrate to nucleate on, the COLDER atmosphere can start removing the energy necessary to allow the phase change to the liquid phase to occur. At no time during liquefaction, does the local Temperature EVER increase, due to the condensation process going on, and if it did then the condensation would stop due to the second law of thermodynamics, not permitting “heat” to flow unaided from a cold source (the water vapor) to a hot sink; aka your warmed up atmosphere.
The transport of the latent heat energy out of the atmosphere is by way of conduction or convection to EVEN COLDER layers of air at higher altitudes. No air ever gets a higher Temperature from latent heat of condensation.
A refrigerator gets rid of “heat” energy by compressing the working fluid which makes it hotter, and then the fluid is cooled by running it through a radiator which is cooled by outside air,
There is NO latent heat involvement, in Cooling the refrigerant by the outside radiator. Automobile engines get rid of “heat” energy by the exact same process, as does a refrigerator; y air cooling of a hot radiator.

RACookPE1978 says:
“April 15, 2014 at 7:07 am
Thank you for your efforts.
But – and you knew a “but” was coming didn’t you? 8<)
If we start with balance at the surface and again solve for G, we get the same result that also forces balance at the atmosphere.
Does that not "require" the usual assumption of a single flat-earth, flat-plate, one-sided, "average' earth at a single constant distance from the sun in a "perfect" equilibrium" with space?
Rather, how does your work change if you require the use of a "spotlight" solar heating on a rotating 24 hour sphere? Solar heating is NOT an uniform average of 324 watts over 24 hours, but is a constantly-changing value between 200 watts (at 07:00) to over 1150 watts (noon) to 0 (after dark, before sunrise.)
Today, for example, day-of-year 105 (15 April) the amount of solar energy on each square meter of the earth's surface at the equator each hour of the day on a clear day is NOT 324 watts, but the following:
01:00 0
03:00 0
05:00 0
07:00 184
09:00 747
11:00 1087
12:00 1131
13:00 1087
15:00 747
17:00 184
19:00 0
21:00 0
23:00 0
So, your “equilibrium model” must evaporate, radiate (long wave), convect, and conduct (from the upper 2 millimeters of water to the depths), not 342, but 1131 watts (at the equator, at noon.) But, over night, an adequate model must change to evaporate, long wave radiate, convect, and conduct 0.0 SW solar radiation, and cool the ocean using the available stored energy from the previous day.
But the earth at any latitude is not radiated evenly either, and the north-south distribution invalidates a simplified flat-plate radiated evenly at equilibrium mode as well. Same day-of-year as above, same one square meter flat surface at noon as above, same clear skies in direct sunlight as above, but at each latitude.
+85 178
+80 278
+70.6 467 (Southern Edge of 2012 Arctic Sea Ice this doy)
+70 479
+67.5 527 (Arctic Circle)
+60 666
+50 830
+40 965
+30 1065
+23.5 1110 (Tropic)
+20 1127
+10 1150
000 1131 (Equator)
-10 1073
-20 977
-23.5 935 (Tropic)
-30 846
-40 684
-50 499
-60 299
-64.4 211 (Northern Edge of 2011-2013 Antarctic sea ice extents this doy)
-67.5 150 (Antarctic Circle)
-70 104
-80 0
-85 0 "
My response:
You provide a lot of data thatr I assume is correct. But the earth and atmosphere stores a lot of heat and the atmosphere and oceans to some extent move it from heat to cold places
Here are some examples:
I live in the Minneapolis area, very close to the 45% latitude.
On June 21 with the most sunlight, the ratio to day and night solar radiation is infinite. But the difference between the average high (81) and low (61) is only 20F. These values convert to 300.4 and 289.6 degrees K. Using the Stefan-Botzmann radiation equation, W = 5.670e-8 T^4, one get radiations of about 461 and 396 Watts per square meter, a difference of only 16%.
Looking at winter vs, summer and using June and December, the hours per day of sunlight drops from about 16 to 8 hours per day. Beside this, the sun’s angle at noon goes from to 65 to 22 degrees. Using the ratios of the hours and the sine of these two angles gives a combined ratio of (16 x 0.907) / (8 x 0.374) or 4.85 to 1. This ignores the cooling effect of the reflection from winter snow that would increase the ratio. The average daily temperature drops from 71 to 13 F. This gives radiations of about 427 and 269 Watts per square meter, a ratio of 1.59 compared to a ratio of heat from the sun of 4.85.
Now lets look at different latidues. Comparing the tropical atmosphere with a latitude of 15 degrees with the mid latitude of 45 degrees, the data I have show an average annual surface temperatures of 299.7 K and 283.2. Using the cosine of the angles, the ratio 0.966 and 0.707 is 1.37. The radiations are 442 and 365, a ratio of 1.21. So here we get less difference, but some heat is indicated in moving from the warmer climates to the colder ones.
I never said my model requires that the earth is flat. Only that there is an average global values of heat transfers between the three levels and outer space and the only way heat can move in or out of the planet is by radiation to/from space. There is some non-linearity between radiation and temperature, but not much. For example at 15C, the radiation is 390.1 Watts per square meter while the average of 5C and 25C is 392.9, a difference of only 0.7%.

Dr. Strangelove says:
April 15, 2014 at 6:47 pm
“You get 6%/C if relative humidity remains constant. But it will decrease because evaporation cools the water. So water temperature drops as well as its vapor pressure. But air remains warm and its absolute humidity at saturation point increases with temperature. You get initial increase in evaporation then decrease. You average the two. 6%/C if only increase and no decrease.”
My Response:
But moist air rises since H2O has a molecular weight of 18 compared to air at about 29. And the cooler water is denser so it sinks. Both of these refresh the surface at the interface even when there is no wind and/or wave action, which usually exist, and accelerate this. Perhaps there is some small drop as you describe. I use 6% since there is some date to support this and the calculated value for constant RH is about 6.5%. Also, I just discovered data by Dai (2006) from over 15,000 stations from 1975 to 2005 that shows on a global basis humidity increases 5% / C, indicating a drop in RH from about 70% to about 69%. This slight change increases evaporation rate to 10% / C. Details: vapor pressure at 17 C = 19.36 mbar; at 18C, 20.63. So (20.63 – 20.63 x 0.69) / (19.36 – 0.7 x 19.36) = 1.10. At a typical 70% humidity, small changes in RH make a big difference. I think the IPCC considers a drop from 70% to 69% as still being “consistent with” constant RH. Repeating the calculation at 50% humidity, a drop to 49% for a 1 C rise gives an increase in evaporation of 8.7% / C. So my 6% seems conservative.
Curious George says:
April 15, 2014 at 8:15 pm
“There is an excellent reason why at least one IPCC model (CAM5) has a problem with evaporation: it treats a latent heat of water evaporation as a constant, independent of temperature (in reality, it is 3% lower at 30 C than at 0 C) – http://judithcurry.com/2013/06/28/open-thread-weekend-23/#comment-338257. Modelers take it philosophically: why should a latent heat, of all things, be treated correctly?:
My Response:
I looked at that. It seems they are referring to the latent heat per kg of weight. This varies a little with temperature. My physics book shows for water it drops from 595.4 to 538.7 calories per gram when the temperature increases from 0 to 100 C. That is only a change of about 0.1% per C, which is not a large concern it seems.
Terry says:
April 15, 2014 at 9:46 pm
Richard Petschauer
In your analysis you hold H as a constant. How realistic is that. Also your calcs indicate that for an increase in E there is no increase in U. Surely U must increase with an increase in E and H that both result from the increase in evaporation, and it is U that is actually important since it is the only pathway for the evaporative losses to escape the system.
My Response:
If H varies, that should be treated as a separate feedback. One would need to know how much it changes per C of surface temperature change. Then use the same method as for E. In fact you could scale it from the feedback factor of –0.741 for a 0.06 x 78 or 4.68 Watts per square meter. This would be additional negative feedback that combines with others algebraically. It is probably not much since the atmosphere temp will increase with the surface.
Regarding the increase in U, it is small because for both the cases with and without changes in evaporation, E, the forcing at the to of the atmosphere is constant at 10 W per meter squared. Note in all cases after the climate warms to offset the 10 w loss to outer space (which our model provides), the sum of U and W, (the amount through the atmospheric window that goes directly from the surface to outer space) always total 235, the net incoming solar after the reflections from clouds and the surface. W varies with G as shown in equation (6) in the paper.
bobl says:
April 15, 2014 at 10:54 pm
“Richard Petschauer,
Let me paraphrase, in order to avoid energy limitation arguments, you have used an assumption of available driving energy of 10W / square meter, not caring where it comes from.
The point I’m trying to make though is that in reality we energy constrained. We are told that a doubling of CO2 produces a driving energy of 3.7W per meter squared not 10W. What I was wanting you to do was to constrain your solution to 3.7 watts and calculate the maximum temperature change, and evaporation when that constraint is applied, or alternately calculate the energy required for an increase of 1C given that evaporation increases 6 % either way this makes for a quick comparison with IPCC models. I am not suggesting that the energy constraint would reduce the evaporation.
Eg the IPCC says that a doubling will cause 3.3 degrees of warming and by your calcs an attendant 20% increase in evaporation. Given that the energy is constrained to 3.7 Watts per square meter what is the revised temperature rise and evaporation rate. I’d assume that if we calculated the maximum extra evaporation that could be sustained by the excess energy, one could use the temperature/evaporation relation to tell us the temperature rise, all other losses ignored. By my estimate rather than 3.3 degrees and 20% we would get about 0.5 C temperature rise, and 3% evaporation for a doubling. That’s a big difference, a headline.”
I think you understand it. The 6% / C applies after the final temperature change. The 3 C rise will not stand, but I don’t get as low as 0.5 C that you do. My case # 3 is close to what I get, ignoring other feedbacks which I as a rough estimate include in the 10 W forcing which was to represent the 3.7 multiplied by the IPCC or climate models feedbacks. For 10 W they 10/32 or 3.125 C. My model ignoring evaporation changes gets 2.7 C. That is case 2 in my Table 1. With evaporation increased to 6% per C (case 3), the rise drops to 1.57 C. A breakdown of the energy transfers follows.
In case 2 with no change in evaporation, G (surface radiation) goes up from 390 to 404.85, an increase of 14.85 (all values in Watts per square meter). (14.85 is larger than 10 because of the “greenhouse multiplier”). For case 3 with 6% rise in evaporation / C, the temperature rise drops from 2.7 C to 1.57 C, and G goes down to 398.57 a decrease of 6.28 from the 404.85 value. The evaporation (E) goes up from 78 to 85.4 a gain of 7.40. So where does this small energy gain of 7.40 – 6.28 or 1.12 come from? Look at the change in D, the down radiation from the atmosphere. It increased from the latent heat condensing and increased from 338.85 to 339.97 or 1.12, exactly matching increase in heat leaving the planet. These numbers were not fudged, but came directly from equations (4) through (8). And the “In minus Out” numbers shown under the table show that all energy balances at all the three levels are maintained. For a 3.7 W per meter squared change, just scale these numbers by 3.7 / 10 or 0.37.
By the way the evaporation latent heat did not increase 6% / C from the original 2.7 C of warming. This seems to be a major confusion point although the paper mentions this in three places, but you seem to understand this. The 6% increase is from the reduced rise of 1.57 after the cooling action. For case 3, 1.57 x 0.06 x 78 = 7.34. The table shows an increase of 78 to 85.4 or 7.4, slight difference from nonlinearities.
By the way, while 10 W forcing is intended to include the approximate added forcing from the positive feedback multipliers, the correct way is to include sum the feedbacks
and create a single feedback multiplier. The negative feedbacks have a large effect then in offsetting the positive. For example a positive feedback of 0.5 and a negative feedback of 0.5 cancel each other out. However, the positive feedback multiplier is 1 /( 1 – 0.5 ) or 2 while the negative one is 1 / ( 1 – (-0.5)) or 1 / 1.5 or 0.67. But 2 x 0.67 give a feedback multiplier of 1.33, not the correct value of 1. That is why in my paper, I show the drop in the feedback factor from 2.48 down to 0.91. For a 3.7 W my model fives a temp rise before feedback of 0.98 C (others and the IPC get about 1.15 C). So for a final with no changes except in those in the paper, I get 0.98 x 0.0.91 or 0.89 C.
Now I am in the process of using the Hitran spectral absorption database and my energy balance to estimate water vapor feedback now estimated at 1.8 W per square meter which is a feedback factor of +0.5625 C / C.
I also have published a paper with much detail on estimating if the 3.7 w forcing for 2 x CO2 is correct. (See:http://climateclash.com/does-the-tropopause-limit-carbon-dioxide-heat-trapping/). It shows only 3.38 that drops to an average 2.53 when typical clouds are included. However, the paper does not include a stratosphere cooling correction that is supposed to be included. Adding 15% for this gives a value of 2.9 W, still less than the 3.7 now widely accepted. Replacing the 3.7 with 2.9, I get after feedback a temperature rise of 0.70 C from doubling CO2 .