Major Errors Apparent in Climate Model Evaporation Estimates

Guest essay by Richard J. Petschauer, Senior Member IEEE

The physics of evaporation has complications related to what happens at the water / air interface such as wind speed and wave action. However if these factors remain constant, how evaporation changes with temperature and humidity can be estimated with well-known equations based on how water vapor pressure varies with temperature. For example, at a typical ocean temperature of 17 C, it should increase about 6.5% / C if the water vapor increases to maintain relative humidity, that the climate models indicate. If the surface air tracks the water within ± 2 C, the rate varies from 6.2% to 6.9% / C. Data over oceans by Wentz et, al (2007) report values of about 6% / C.

But the complex computer climate models show averages of only about 2.5% / C. There are no claims of reduced wind speeds or wave action or increased relative humidity to explain this. However many papers on the subject claim that the available energy is limiting evaporation in these models. But physics theory tells us that the latent energy for evaporation comes from the temperature of the water itself. The latent heat leaving the surface cools it and deposits heat in the atmosphere, part of which escapes to outer space. This combination causes negative feedback. The reduced net energy from increased CO2 still warms the surface, but this energy can’t be separated from what aids the final increased evaporation. A 6% / C increase applies to the water after the negative feedback is complete. Do the climate models ignore this cooling and feedback process?

A typical paper on this subject is one by O’Gorman and Schneider (2008) that defines this energy balance constraint that is supported by many other climate model references. Their equations (8) and (9) correctly show that an increase in latent heat transfer from evaporation must equal a reduction in the net surface radiation heat loss, assuming the loss from convection plus sensible heat and the net solar surface absorption all remain constant. However, this cannot provide a solution.

Let E = the latent heat loss from the surface due to evaporation, G the outward surface radiation and D the downwelling radiation from the atmosphere to the surface. Net surface radiation loss = G – D.

For a reduced radiation heat loss,

D E = D D – D G (1)

However, the developers of the climate models seem to be confusing independent and dependent variables. Evaporation is the driver or forcing agent controlled by the physics at the surface, and G and D must respond to a change in it. If the surface temperature rises, the additional latent heat lost at the surface will cause an offsetting decrease in the temperature and thus G. And the latent heat deposited in the atmosphere warms it and increases the downwelling radiation, D (and the outgoing radiation). We now have a feedback process at work. Equation (1) can only be used as a check after a correct solution is found to new values of E, D and G after the feedback process is complete. It appears there is a serious error in how climate models estimate evaporation as indicated in the rest of this paper.

We have developed a dynamic three level energy balance model (reference 1) with updates as described later that can be used for a number of forcings and feedbacks including the response to changes in evaporation and the cooling of the surface and the warming of the atmosphere.

The results are shown on the next page. No energy constraints of evaporations are seen.

As shown in Figure A1 in the appendix, we define S as the net incoming solar flux after albedo, A the absorption of the net solar flux by the atmosphere, G the surface radiation, W the surface radiation through the atmospheric window, H the convection from the surface, E the latent heat from surface evaporation (both H and E transfer heat to the atmosphere), U the atmosphere upward outgoing longwave radiation, and D the atmosphere downwelling longwave radiation to the surface. For this estimate the following values are fixed: S = 235; A = 67; H = 24. These and the baseline values shown in Table 1 are from Kiehl and Trenberth (1997) with average cloudy conditions of 60% coverage net considering overlaps.

From eqs (4 to 8) on the next page, Table 1 compares the baseline case with three having large forcings of 10 Wm-2 at the top of the atmosphere. One case has no evaporation changes, while two have rate changes of 6% and 10% / C. D T is calculated from the changes in G from the baseline.

The In minus Out fluxes are equal at all three levels for all the cases with each parameter used at least twice. No problem in finding the energy to support evaporation; the surface gave up some by cooling and the down radiation, D increased. Note that the increase in E is based on the final reduced temperature rise. In all cases D E = D D – D G, measured from the baseline.

Table 1 – With large TOA forcings no energy constraints on evaporation changes.

Increase in E follows that estimated from temperature change and the specified change %.

For example in case 3, E » Eo + r D T Eo = 78 + 0.06 x 1.57 x 78 = 85.34 » 85.40 shown.

Ignoring the drop in D T, from the value of 2.70 the increase in E to 85.40 is only 3.5% / C.

TOA forcing & evap change

D T – C

G

W

E

U

D

1) 0 & 0 (Baseline)

0

390

40

78

195

324

2) 10 Wm-2 & 0 % / C

2.70

404.85

41.52

78

193.48

338.85

3) 10 Wm-2 & 6% / C

1.57

398.57

40.88

85.40

194.12

339.97

4) 10 Wm-2 & 10% / C

1.23

396.69

40.69

87.63

194.31

340.32

In – Out: Case 1

TOA = S – W + U = 235 – 40 – 195 = 0

Atmosphere = A + G – W + H + E – U – D = 67 + 390 – 40 + 24 + 78 – 195 – 324 = 0

Surface = S – A + D – G – H – E = 235 – 67 + 324 – 390 – 24 – 78 = 0

In – Out: Case 2

TOA = S – W + U = 235 – 41.52 – 193.48 = 0

Atm = A + G – W + H + E – U – = 67 + 404.85 – 41.52 + 24 + 78 – 193.48 – 338.85 = 0

Surface = S – A + D – G – H – E = 235 – 67 + 338.85 – 404.85 – 24 – 78 = 0

In – Out: Case 3

TOA = S – W + U = 235 – 40.88 – 194.12 = 0

Atm = A + G – W + H + E – U – D = 67 + 398.57 – 40.88 + 24 + 85.4 – 194.12 – 339.97 = 0

Surface = S – A + D – G – H – E = 235 – 67 + 339.97 – 398.57 – 24 – 85.4 = 0

In – Out: Case 4

TOA = S – W + U = 235 – 40.69 – 194.31 = 0

Atm = A + G – W + H + E – U – D = 67 + 396.69 – 40.69 + 24 + 87.63 – 194.31 – 340.32 = 0

Surface = S – A + D – G – H – E = 235 – 67 + 340.32 – 396.69 – 24 – 87.63 = 0

Note the increase in E follows that estimated from the temperature change and the specified change %. For example in case 4, E » Eo+ r D T Eo = 78 + 0.10 x 1.23 x 78 = 87.59 » 87.63.

No energy constraint is seen and all energy balances at the three levels are maintained.

The details of the calculations for the above table follow. The basic equations for energy balance at all three levels are from our paper, reference (1). For balance at the top of the atmosphere,

S = k (A + H + E) + k Ga + G (1 – a) (2)

Refer to Figure A1 in the appendix. S is the net incoming solar after albedo, k is the fraction of the total heat absorbed by the atmosphere that is radiated upward (here 0.3757), and a is the fraction of the surface longwave radiation absorbed by the atmosphere including clouds (here 0.8974).

Solving for G,

G = [S – k (A + H + E)] / (1 – a + ak) (3)

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.

To determine the feedback factor for E, add to it the increase caused by a 1 C surface temperature change and convert the change in G to a temperature change. For a 6% increase of 78, E becomes 82.68, the new value of G is 386.0014 Wm-2, down from 390, and provides a temperature change of –0.741 C which equals the feedback factor, the temperature change before additional feedback. With no other feedbacks, the feedback multiplier is M = 1 / (1 – F); here we get M = 0.5744. The temperature change of –0.741 would produce another change of -0.741 x –0.741 or +0.549, followed by (–0.741)3 or –0.406 then (–0.741)4 or +0.301, etc which sum converges to a final temperature drop –0.4256 C which also equals M x F or -0.741 x 0.5744.

As an alternate to using a feedback factor and a way to check it, the above equation for G can be modified to allow E, the evaporation rate, to vary with the change of surface temperature implied from the change in G, the surface radiation. Then the solution for the new surface radiation is,

G » [S – k (A + H) – k (E0 – r E G0 Tr)] / (1 – a + ak + k r E0Tr) (4)

Where r = the fractional rate of change / C of surface evaporation, E0 the initial evaporation, G0 the initial surface radiation, and Tr the temperature change rate factor at G0 which is T0 / (4 G0) with T0 the initial surface temperature. At 288 K or 15 C, Tr = 0.1846 C / W m-2. Here we get M = 0.581.

(Equation 3 is more accurate. The two values of M are very close for smaller forcings)

The final value of evaporation latent heat,

Ef = E0 + r TrE0 (G – G0) (5)

The temperature change uses the inverse of the Stefan-Boltzmann equation for G and G0.

The final value of W, Wf = G (1 – a) (6)

The final value of U, Uf = S – Wf (7)

The final value of D, Df = G + A + H + Ef – S (8)

The parameter a is the fraction of the surface longwave radiation absorbed by the atmosphere. Here it is 0.8974 or 1 – W0 / G0, whereW0 = 40, the amount through the atmospheric window and G0 = 390. The value k is the fraction of the total heat radiated from the atmosphere that is upward outgoing radiation. So k = U / (U + D). For our baseline k = 195 / (195 + 324) = 0.3757. To impose a forcing R at the TOA, k = (195 – R) / (195 – R + 324). Unless a or k is the value being perturbed, the equations above require the baseline values for a and k. For other values of a and k, partial derivatives are needed as described in the appendix.

It appears the climate models are grossly underestimating the negative feedback from latent heat transfer. For case 3 in the table above, the feedback multiplier of 1.57 / 2.70 = 0.581 implies a feedback factor for a change in evaporation of 6% / C of –0.720 C / C. This corresponds to the IPCC value for water vapor of 1.8 Wm-2 / C divided by their value of l of 3.2 to give a feedback factor of +0.562 C / C.

If we use the IPCC value of only 2.5% / C for evaporation changes, our feedback factor of –0.720 drops to –0.308. This compares fairly closely to the IPCC lapse rate feedback factor of –0.262 C / C, based on their value of –0.84 Wm-2 / C.

If one just wanted the feedback factor, equation (2) is more accurate. As described above, for a 6 % evaporation change rate, it gives a feedback factor of –0.741

The IPCC has a positive cloud feedback of 0.69 Wm-2 / C with a very large range. But it is not based on reduced clouds with warming, but as a residual of the amount of warming the models can not explain by the other feedbacks (Soden and Held (2006), p 3357, paragraph 2). So this is not a true estimate of cloud feedback. Eliminating it and replacing the lapse rate feedback with our evaporation feedback cuts the IPCC feedback multiplier from 2.48 down to 0.910.

The three level energy balance model used here is dynamic since it handles balance simultaneously at all three levels: the planet, the atmosphere and the surface. With atmospheric CO2 content increasing very slowly, only about 0.54% per year, there is more than enough time for the normal weather systems to move and distribute the small additional heat across the globe as it always has done in the past. So a simple improved global energy balance should be adequate. Another benefit of the three level model is that it can also handle changes in downwelling radiations. For both increased CO2 and water vapor, besides decreasing outgoing radiation, they will also increase downwelling radiation since these emission levels will move down to warmer temperatures. Present models that must refer everything to the outgoing radiation at the top of the atmosphere have a problem with this.

The use of spectral radiance tools for the atmosphere in both outward and downwelling directions under clear and cloudy conditions can handle the effects of CO2 and the significant water vapor feedback, including its negative feedback component of absorbing incoming solar radiation. These tools, available to all, can greatly improve accuracy and replace the present complicated unreliable computer models which, besides overestimating climate sensitivity, have large ranges of uncertainty of about ± 50%.

Richard J. Petschauer

Email: rjpetsch@aol.com

References

1) http://climateclash.com/improved-simple-climate-sensitivity-model/

2) Kiehl, J. T., and K. E. Trenberth (1997): Earth’s Annual Global Mean Energy Budget. Bull. Amer. Meteorol. Soc., 78: 197-208

3) Wentz, F. J., L. Ricciardulli, K. Hilburn and C. Mears (2007): How much more rain will global warming bring? Science, Vol 317, 13 July 2007, 233-235

4) Soden, B.J., and Held, I.M. (2006): An assessment of climate feedbacks in coupled ocean-atmosphere models. J. Clim.19: 3354–3360.

5) O’Gorman, P. A., and Schneider, T (2008): The Hydrological Cycle over a Wide Range of Climates Simulated with an Idealized GCM. Amer. Meteorol. Soc., 1 August 2008, 3815-2831

=============================================================

Appendix

Figure_A1

From Figure A1, the present balanced conditions before any perturbation changes are (all in W m-2):

S = 342 – 77 – 30 = 235; A = 67; H = 24; E = 78; G = 390; W = 40; a = (390 – W) / 390 » 0.8974

where W is the amount through the atmospheric window, and k = 195 / (195 + 324) » 0.3757.

From Figure A1 it can be seen that for balance of heat flux in and out at the TOA,

S = k (A + H + E) + kGa + G (1 – a) (A1)

Solving for G,

G = [S – k (A + H + E)] / (1 – a + ak) (A2)

With the above base value in equation (A1), G = G0 = 390 Wm-2 corresponding to a surface at 14.9853 C. To perturb any value, change it and calculate a new G and from that a temperature change.

To impose a forcing R at the TOA, k = (195 – R) / (195 – R + 324). Unless a or k is the value being perturbed, the equations above require the baseline values for a and k. For other values of these, partial derivatives as shown below

At the lower part of the atmosphere,

G/ E = G/ H = ∂G/A = –∂G/D = – k / (1 a + ak) (A3)

At the top of the atmosphere for longwave radiation only,

G/ U= (k – 1) / (1 a + ak) (A4)

For change in net solar, S, shortwave incoming radiation, the forcing is substantially larger than for longwave radiation:

From changes in the solar strength,

G/ S= (1 – kA / S) / (1 a + ak ) (A5)

From changes in albedo,

  G/ S » 1 / (1 a + ak) (A6)

For changes in evaporation with the present value of k or a different one, equation (A3) is used to get a feedback factor. The present value of k assumes the division of changes in radiation leaving the atmosphere are in the same ratio as the present total values. This ends up with a smaller temperature change at the upward outgoing emission level than that of the downwelling level. Changing the ratio of U / D to (U / D) 0.75 results in equal temperature changes and increases k from 0.3757 to 0.4059. From forcing at the TOA from CO2 at a typical emission level of about 10 km, one would think that the upward emission level temperature would increase more than the lower level. This suggests a value of k greater than 0.4059 of about 0.42 to 0.43.

The value a is simply a function of the fraction of emission through the atmospheric window and the estimated net fractional cloud cover, Cc. For changes to be compatible with this baseline with 60% cloud cover for different cloud coverage,

a = 1 – 100 / 390 (1 – Cc) (A7)

This implies a clear sky atmospheric window of 100/390 or 25.6%. Based on spectral radiance runs with Hitran 2008, a closer value of 22.8% results. Then,

a = 1 – 0.228 (1 – Cc) (A8)

For 60% cloud coverage, a = 0.9086, up from 0.8974.

Changing k to 0.4059 and a to 0.9086, increases the evaporation feedback factor from = –0.741 to –0.765.

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William Astley
April 15, 2014 7:12 am

In reply to:
Old England says:
April 15, 2014 at 1:24 am
Interesting, and the effect of convection on the calculations is ?
William:
The following is the calculated affect on global warming of a change in evaporation from the same author warming for a doubling of atmospheric CO (nominal) = 0.84C with IPCC assumed positive feedback from clouds, 0.73C with neutral feedback from clouds, and 0.64C with negative feedback from clouds.
http://climateclash.com/improved-simple-climate-sensitivity-model/
http://climateclash.com/files/2011/02/PetT1b.jpg
Table 1 – Surface Temperature Increases for 2xCO2 with feedbacks (Min/Nominal/Max)
1) IPCC(2007) with constant evaporation and Ts =3010_______ 2.00/3.20/4.5 C
2) IPCC(2007) with constant evaporation and Ts =2645_______ 1.60/2.29/2.89 C
3) Same as above with 2.5%/C evaporation change rate______ 1.25/1.33/1.58 C
4) Same as above with 6%/C evaporation change rate________ 0.69/0.84/0.97 C
5) Same as above with no (William: Neutral) cloud feedback __0.61/0.73/0.82 C
6) Same as above with negative cloud feedback____________0.55/0.64/0.71 C

Bob Shapiro
April 15, 2014 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.

William Astley
April 15, 2014 7:28 am

In reply to:
bobl says:
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.
William:
The temperature of the planet will warm less than 1C for a doubling of CO2. There will not be 6% increase in evaporation.
Your comment suggests using forcing in watts to force the evaporation rate of the oceans. That confuses the issue. The evaporation of ocean is controlled by temperature at the surface of the ocean (and the variables noted in the next sentence.). The amount of water evaporation is not a free variable that can be changed to tune the general circulation model.
Physics controls the evaporation rate of water based on the controlling variables temperature, relative humidity above the water, wind speed, and wave action. The evaporation rate is not a free variable that can be change by the IPCC in their general circulation models to give the answer they want.
William:
The following is the calculated affect on global warming of a change in evaporation from the same author warming for a doubling of atmospheric CO (nominal) = 0.84C with IPCC assumed positive feedback from clouds, 0.73C with neutral feedback from clouds, and 0.64C with negative feedback from clouds.
http://climateclash.com/improved-simple-climate-sensitivity-model/
http://climateclash.com/files/2011/02/PetT1b.jpg
Table 1 – Surface Temperature Increases for 2xCO2 with feedbacks (Min/Nominal/Max)
1) IPCC(2007) with constant evaporation and Ts =3010_______ 2.00/3.20/4.5 C
2) IPCC(2007) with constant evaporation and Ts =2645_______ 1.60/2.29/2.89 C
3) Same as above with 2.5%/C evaporation change rate______ 1.25/1.33/1.58 C
4) Same as above with 6%/C evaporation change rate________ 0.69/0.84/0.97 C
5) Same as above with no (William: Neutral) cloud feedback __0.61/0.73/0.82 C
6) Same as above with negative cloud feedback____________0.55/0.64/0.71 C

wsbriggs
April 15, 2014 7:48 am

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.

Crispin in Waterloo
April 15, 2014 7:48 am

The ultimate conclusion is that thunderstorms are the main mechanism for cooling of the oceans. I think the assumption that the oceans are 17 is wa-ay wrong. Evaporation is far higher where the oceans are warmer because that is where more insolation lands.
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.
Burns
++++++
E = 5([Tc+ 18]^2.5– r[Ta+ 18]^2.5)(V + 4) x 10^-6
where: E = evaporation rate [kg/m^2/h]; r = relative humidity/100T [units?]; a= air temperature [°C]
Tc= water surface temperature [°C] ; V = wind velocity [km/h]
+++++
I was unable to turn that into a spreadsheet.
r = % or …?
What is Ta ? Air Temp in C?
What is 100T ?
Thanks

wsbriggs
April 15, 2014 7:49 am

Mods – Correction to previous post – water surface temp

Greg Goodman
April 15, 2014 8:03 am

“The IPCC has a positive cloud feedback of 0.69 Wm-2 / C with a very large range. But it is not based on reduced clouds with warming, but as a residual of the amount of warming the models can not explain by the other feedbacks (Soden and Held (2006), p 3357, paragraph 2). So this is not a true estimate of cloud feedback. Eliminating it and replacing the lapse rate feedback with our evaporation feedback cuts the IPCC feedback multiplier from 2.48 down to 0.910.”
OMG, they really are just making it up. This is totally a fudge factor.
I think tropical sensitivity to a change in radiation is near zero once the climate has settled which takes 2-3 years after an impact like a major volcano. Extra-tropics are more sensitive but buffered to a fair extent by exchanges with the tropics due to major ocean gyres.
I think these values are a lot more credible.
Sorry don’t have time to plough through the maths but it looks well thoughtout.

April 15, 2014 8:03 am

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..

Greg Goodman
April 15, 2014 8:12 am

One thing that is important in all this is that downwards IR ( radiation does not “well” in any direction ) is absorbed in the top 100 microns of the water surface.
It thus must heat this thin film and it is just where evaporation happens and it is this temperature which is relevant to the effect of the radiation.
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.
It would seem that the above account would apply to visible and UV which can penetrate deeper into the bulk of the water.
However, I think this is an excellent article, I shall come back and pour over it when I have more time.

William Astley
April 15, 2014 8:19 am

William:
A ‘smell’ test to confirm the conclusion that warming due to a doubling of atmospheric CO2 will be significantly less than 1C is Idso’s calculations. Idso uses 10 different natural phenomena where a known forcing change caused by different natural phenomena is used to calculate the planet’s sensitivity to a change in forcing. Idso’s best estimate for the warming for a doubling of atmospheric CO2 from that calculation is 0.4C.
http://www.mitosyfraudes.org/idso98.pdf
CO2-induced global warming: a skeptic’s view of potential climate change by Sherwood B. Idso
Over the course of the past 2 decades, I have analyzed a number of natural phenomena that reveal how Earth’s near-surface air temperature responds to surface radiative perturbations. These studies all suggest that a 300 to 600 ppm doubling of the atmosphere’s CO2 concentration could raise the planet’s mean surface air temperature by only about 0.4°C. Even this modicum of warming may never be realized, however, for it could be negated by a number of planetary cooling forces that are intensified by warmer temperatures and by the strengthening of biological processes that are enhanced by the same rise in atmospheric CO2 concentration that drives the warming.
This author reaches a similar conclusion that the planet will warm significantly less than 1C for a doubling of atmospheric CO2 due to enhanced evaporation.
http://typhoon.atmos.colostate.edu/Includes/Documents/Publications/gray2012.pdf
The Physical Flaws of the Global Warming Theory and Deep Ocean Circulation Changes as the Primary Climate Driver
….These two misrepresentations result in a large artificial warming that is not realistic. A realistic treatment of the hydrologic cycle would show that the influence of a doubling of CO2 should lead to a global surface warming of only about 0.3°C – not the 3°C warming as indicated by the climate simulations….
But this pure IR energy blocking by CO2 versus compensating temperature rise for radiation equilibrium is unrealistic for the long-period and slow CO2 rises that are occurring. Only half of the blockage of 3.7 Wm-2 at the surface should be expected to go into temperature rise. The other half (~1.85 Wm-2) of the blocked IR energy to space will be compensated by surface energy loss to support enhanced evaporation. This occurs in a similar way as the earth’s surface energy budget compensates for half its solar gain of 171 Wm-2 by surface to air upward water vapor flux due to evaporation.
Note in Figures 1 and 2 that the globe’s annual surface solar absorption of 171 Wm-2 is balanced by about half going to evaporation (85 Wm-2) with the other half (86 Wm-2) going to surface to atmosphere upward IR (59 Wm-2) flux and surface to air upward flux by sensible heat transfer (27 Wm-2). Assuming that the imposed extra CO2 doubling IR blockage of 3.7 Wm-2 is taken-up and balanced by the earth’s surface as the solar absorption is taken-up and balanced, we should expect a direct warming of only ~ 0.5°C for a doubling of the CO2. The 1°C expected warming that is commonly accepted assumes that all the absorbed IR goes to balancing outward radiation (through E = σT4) with no energy going to evaporation. This is not realistic. These two figures show how equally the surface solar energy absorption (171 Wm-2) is balanced by a near equal division between temperature rise (enhancing IR and sensible heat loss) and energy loss from surface evaporation. We should assume that the imposed downward IR energy gain for a doubling of CO2 at the surface will likely be similarly divided. Such a division will cause an enhancement of the strength of the hydrologic cycle by about 2 percent (or 1.85 Wm-2 of extra global average evaporation over the ~ 85 Wm-2 energy equivalent of current evaporation).
This analysis shows that the influence of doubling atmospheric CO2 by itself (without any assumed positive feedback) leads to only very small amounts of global warming.

RACookPE1978
Editor
April 15, 2014 8:20 am

wsbriggs says:
April 15, 2014 at 7:48 am
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.

Other users are reminded that – unlike many equations – there are as many different models and computerized linear approximations for evaporation losses from open water sources from as many different labs, universities, docs, post-docs, pre-docs, and graduate thesis as could be funded or approved by as many different departments and universities that ever existed. Papers have been written on these losses by as many different people as can apply for grants and thesis topics.
I recommend the ONLY evaporation/heat loss/heat transfer equations and models that anybody uses are those who simultaneously list the experimental conditions: latitude, relative humidity, area, weather conditions (air temperature, wind measurements, cloudy/clarity, radiation levels, relative humidity and wet bulb temperatures, measurement dates, level records, water temperature AT A MINIMUM) .
Then, once the reader is shown the actual conditions that “validate” each different approximation (model) he or she can evaluate whether that particular model is going to be valid for any particular situation. In particular, I distrust greatly “generic” computer models that do not list what very few conditions exist that validate the model.
For example: I’ve checked several different models that claim they predict the heat equilibrium for various swimming pools under various conditions; indoor closed cover, outdoor, indoor open room, outdoor exposed to sunlight, etc. Nice models, but they differ between equal conditions by as much as 100%! (50 watts/meter^2 losses compared to over a 100 watts/meter^2) ! Further, a “swimming pool” model may be valid for average conditions in swimming pool conditions: 80-100 deg F air temperatures, relative humidity of 30 – 70%, sunshine skies at 34-45 degrees latitude in the summer, etc. Does that mean they are accurate for any square meter of Arctic ocean water at 2 deg C, winds = 10 m/sec, air temperature of -15 deg C at a relative humidity of 30% under cloudy skies at latitude 78 north?
You mentioned IEEE – an good organization usually valid. But, even the ASHRAE – the specific heating and Air Conditioning Engineering Society (ie, heat exchange and cooling specialists working in a field SPECIFICALLY using relative humidifies and air temperatures!) have been accused in various papers of mis-calculating evaporation heat losses in their long-used “standard model” by over 30%.

Ian L. McQueen
April 15, 2014 8:23 am

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

Greg Goodman
April 15, 2014 8:32 am

Dr Norman Page links Trenberth’s: http://www.cpc.ncep.noaa.gov/products/outreach/proceedings/cdw31_proceedings/S6_05_Kevin_Trenberth_NCAR.ppt
TC flux climatology

globally this is 0.36 and 1.13 W/m2
vs CO2 1.5 W/m2
It matters !
and it’s not included in the models.
========
It’s worse than we thought.
Non detrended AMO and hurricane energy (ACE) :
http://climategrog.wordpress.com/?attachment_id=215

Frank
April 15, 2014 9:27 am

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.

jorgekafkazar
April 15, 2014 9:38 am

“No energy constraints of evaporations are seen.”
Something is wrong, then.

Berényi Péter
April 15, 2014 10:40 am

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.

William Astley
April 15, 2014 12:42 pm

Another smell test that we have found a significant modeling ‘climategate’ fudge is other papers discussing the issue and a response from one or more of the climategate cabal.
This paper explains how the general circulate models suppress evaporation of the oceans.
http://adsabs.harvard.edu/abs/2008AGUFMGC43A0709R
Muted precipitation increase in global warming simulations: A surface evaporation perspective by Ingo Richter and Shang-Ping Xie
One of the important consequences of a rise in global atmospheric temperatures is the increase of the atmosphere’s capacity to hold water vapor in accordance with the Clausius-Clapeyron (CC) equation. Under the assumption of constant relative humidity, this implies an increase in specific humidity at the rate of ∼7% per Kelvin of atmospheric warming, a prediction roughly borne out by observations and model simulations.
If, in addition, the atmospheric circulation remained approximately unchanged, we would expect precipitation to increase at a similar rate as water vapor. This, however, is not what is predicted by a wide array of climate models, which put the rate of precipitation increase at just 2%/K [Held and Soden, 2006].
One way the models can achieve this muted precipitation response is through a slowdown of the tropical circulation so that the decrease in upward velocity partially offsets the increase in atmospheric moisture [Emori and Brown, 2005]. Such a slow down is confirmed for the simulated Walker Circulation [Vecchi and Soden, 2007] and, to a lesser extent, for the Hadley Circulation [Lu et al., 2007]. [3]
The muted precipitation increase and slowing of the tropical circulation constitute a consistent response to greenhouse gas (GHG) forcing among models but it is not obvious why the climate system should behave in this particular manner. In fact, recent observational studies by Wentz et al. [2007] and Allan and Soden [2007] suggest that the actual rate of precipitation increase might be significantly higher than what is simulated by climate models. Similarly, a recent analysis of surface heat flux using merged satellite and reanalysis data [Yu and Weller, 2007] indicates a rate of latent heat flux increase that is much higher than in the models.

Berényi Péter
April 15, 2014 12:52 pm

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.

george e. smith
April 15, 2014 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.
Now Ice remains ice even at zero deg..C unless, and until, you add 80 calories per gram of ice to cause the phase change from ice to water still at zero deg. C That’s what latent heat is. And water does not become ice until something colder sucks out that 80 calories per gram at zero deg. C to allow the phase change.
The same thing happens at the other end. Water remains liquid even at 100 deg. C until something hotter supplies it with about 590 calories per gram to cause the phase change to water vapor.
And conversely, water vapor remains vapor until something COLDER sucks out that 590 calories per gram to allow it to condense. NOTHING GETS WARMER !! in the process.
The water when it precipitates out as rain or whatever, does not bring any “latent heat” back down to the ground. That was already sucked out by the colder upper atmosphere to get the water.
At Temperatures below 100 deg. C the evaporation of water requires a somewhat larger latent heat absorption. Common sense tells me the amount is equal to the 590 value at 100 deg. C plus another calorie per gram, for each degree below 100 deg C you are at. Maybe it’s not that simple, but that would be my bet.
People have to stop thinking that latent heat is a source of Temperature rise.
When you put your hand in steam and get scalded, it IS NOT that the latent heat of condensation (590 cal per gram) raises the Temperature of your skin above 100 deg C, it is simply that you get a dose of 590 plus the difference between the 100 deg. C stam Temperature, and your 98.6 deg. F body Temperature.
The amount of “heat” energy dumped in your skin is much higher than hot water, but the Temperature is no higher than 100 deg. C
Please stop saying that latent heat warms the atmosphere; it doesn’t.

Matthew R Marler
April 15, 2014 1:52 pm

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.

Matthew R Marler
April 15, 2014 1:57 pm

Berényi Péter
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.
Do you have references for those numbers? I’d like something to down load to my file on this topic.

Matthew R Marler
April 15, 2014 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.

April 15, 2014 2:41 pm

I got back from taking the taxes the post office and was looking forward to enjoying the author’s sound defense of this post.
What a disappointment.

aaron
April 15, 2014 2:50 pm

Told you so.

April 15, 2014 2:51 pm

I will try to answer some of the questions in segmentsi
Evaporation.
My early work was on evaporation related to indoor swimming pools based on ASHRAE (American Society of Heating, Refrigerating and Air Conditioning Engineers) empirical equations. I used them in conjunction of a device I designed, installed and patented to automatically adjust indoor room humidity without outside temperature and reduce condensation damage in the winter. These equations have the same property regarding how water temperature and air humidity affect changes in evaporation assuming things such as wind speed and wave action if they do not change (if they do that should be treated as and additional separate feedback). In only depends on the vapor pressure of water as a function of temperature and the temperature and relative humidity of the surface air. Water vapor pressure can be very accurately measured and I use Bolton’s equations [p = 6.112*exp((17.67*t) / (t+243.5)) mbars] which is a fit to real data and is more accurate at sea level that the general purpose Clausius – Clapeyron equation. Using Bolton’s I get going from 17C to 18C an increase ratio of 1.0652 and with the C-C equation 1.0640. But I am using only 6% as a conservative estimate. The question was raised if 17 C is too low. Using 30C drops it to 1.590, but that is too warm for an average. Near the equator, the wind speeds are less so it cuts its importance. Regarding wind speed, it is important, but at 100% relative humidity, net evaporation is zero for any speed. And as shown in indoor swimming pools with very little wind, the evaporation is certainly not zero. Incidentally, the water drop in inches per day is independent of the pool shape or area and can be used to measure evaporation after a few days. With no wind the water-air interface is still refreshed. Water vapor has a molecular weight of 18 compared to air at about 29, so the lighter, moist air will rise and be replaced with less humid air. And as the evaporation cools the water at the interface, its density increases so it will drop and force the warmer water below to rise. Wind speed will accelerate this.
The equation for indoor swimming pools uses an empirical “activity factor” multiplier which can greatly effect the final value.
Regarding Dr Burns comments “about 6% / C is too simple”. I “stated with constant with and constant relative humidity”. So what does he get with his empirical equation? If I understood it correctly, I got values ranging from 7.3% / C at 17 to 18C and at 5.29 %/ C from 30C to 31C which seem to be changing too fast with temperature.

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