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
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.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
Mosher writes “Radiative gases are nothing much more than leaky reflective surfaces. They retard the return of energy to space.”
Ouch. Worst analogy ever.
Radiative gasses intercept the IR and capture it by vibrating more (the modes of capture are well understood) and then collide with a non-GHG (eg O2 or N2) and transfer that energy towards that molecule and hence warm the atmosphere.
Mosher likened it to “The silver lining on a thermos does not warm the coffee”
No, but the silver lining reflects all of the radiation back towards the coffee to minimise energy loss. Yes, to keep it warm. GHGs by comparison dont reflect anything back towards the ground. They capture IR and warm the atmosphere. They capture energy from the atmosphere and radiate both up and down for no net effect.
AGW “warming theory” comes from the TOA (as you later mentioned)
@ur momisugly Curious George (8:15pm) “a 3% error…” — Thank you for that jaw-dropping, helpful, statistic. Good for perspective!
********************************
Dear Mr. Petschauer,
Thank you for a fine article (even if I still have a Q about the CO2 remark in it). It is well-written and full of highly relevant information. Your careful, informed, analysis shows. That article took a lot of your time, no doubt. We are blessed that you shared it with us.
And GOOD FOR YOU to come back and answer questions or to clarify.
Gratefully,
Janice
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.
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.
Mods:
I was about to suggest to Mr. Petschauer that he use LaTeX to get over his typographical problems above. If you speak LaTeX, all you do is put the math between dollar signs, as you do in LaTex, with the exception that after the first dollar sign you insert “latex” and a space.
The problem is that in the past it’s been best to try it first on the test page (saving your work before you submit), whereas today that page kept telling me that it could not accept my submissions (even without the LaTeX). So I’ll submit LaTeX here without testing it first:
[Your values, your expression with latex worked in the above, but for “Everybody else” – those of us less skilled than you for example – the site policy is to “test” such functions and processes FIRST on the site’s Test page. Mod]
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.
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.
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.
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 .
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.
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]
cementafriend says:
April 15, 2014 at 6:31 am
“I believe IEEE stands for Institute of Electrical and Electronics Engineers. Electrical engineers normally have a good grasp of mathematics but I am not so sure of their grasp of chemical engineering science. I have come across electrical engineers who have no idea of process control because they do not understand engineering science such as reaction kinetics, fluid dynamics or heat & mass transfer. Mention of physics instead of engineering science makes me suspicious. Instead of humidity you should be using partial pressures-just think of the lower . . . .
Kiehl & Trenberth that paper would never be published in an engineering journal.”
My Response
First, I am using partial pressure in the change of evaporation as a function of water temperature and air humidity. By the way, this is physics not chemistry.
Second, I am not sure what the comments about Kiehl & Trenberth mean. I think the paper I refer to and use as a reference “Earth’s Annual Global Mean Energy Budget” from 1997 is one of the best resources I have seen on the subject. It is better than the 2009 update because it explains how and where the data came from and because the numbers are balanced at all three levels: planet, atmosphere and the surface. The 2009 version has an unbalance at the surface of 0.9 W per meter squared (which I think was intentional to show future warming) and is hard to understand if you have not read the first version. But if a person uses the new numbers in my model, the results change very little. Evaporation latent heat goes up a little from 78 to 80 W per meter squared and surface radiation from 390 to 396.
Third, regarding, my electrical engineering degree, I took a year of separate courses of college chemistry and physics and happen to get 3 A’s in each not to mention the same in a year’s of advanced calculus in grad school. I am not using any chemistry in my global warming work, but I am with regard to physics. I still have the excellent textbook that has several chapters on heat including partial pressure of water vapor, latent heat and humidity calculations. And in my retirement, I have spent the last 6 years doing literature research and my own computer calculations on CO2 and the feedbacks and how they relate to global warming. As far as engineers go as opposed to “scientists”, we like to solve problems in the simplest, practical, yet sound way, and not spend a lifetime on the same subject. I think the climate people made a major mistake many years ago in dropping improving the simple energy balance models in favor of the complex, bottoms up across the planet for the next 100 years approach, while thinking they could accurately simulate the complex non linear, chaotic climate system just to tell us how much doubling CO2 would warm the average temperature across the globe. Industry would never attack a problem this way. On the other hand maybe they wanted to make things extra complicated so outsiders could not question their work. And it’s not a bad gig. Only your buddies understand what you are really doing and it will take 15, 20 or more years before anyone knows if you are close to the correct answer. Then one can make up excuses. Plus all that government funding while you are helping save the planet. Real “science” in action. And now these same people who can’t even get close to the warming from CO2 are trying to tell how many ways this little warming, they did not predict, will “disrupt the planet”. It’s time for a little common sense. Not even an engineering degree needed to figure this out, and I think the public has. Hopefully, finally the media will too.
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)
Richard Petschauer says:
April 16, 2014 at 9:13 pm
———————————–
Richard,
in looking at issues of evaporation, you are looking in the right area. Oceans cover 71% of the planet’s surface and evaporation is their primary cooling mechanism.
The approach you are using seems overly complex , in that you are trying to model the water cycle and then test what altered levels of radiative gases may do.
There are far simpler questions that can be asked as a “sanity check”.
The first of these is –
“given 1 bar atmospheric pressure, is the net effect of our radiative atmosphere over the oceans warming or cooling?”
Or you could ask –
“How hot would our oceans get in the absence of DWLWIR and atmospheric cooling?”
Or you could ask –
“could a non-radiative atmosphere cool our oceans, as it would have no effective means of cooling itself?”
I would note here that the “basic physics” of the “settled science” claims that DWLWIR is warming our oceans from -18C to 15C. Some climastrologists go so far as to claim that -18C is for the “surface” without an atmosphere. Given that the primary cooling mechanism for 71% of the planet’s surface is evaporation, do their claims pass the most basic “sanity check”?
“””””….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.
I am sorry George, we disagree on the role transfer of latent heat throught change of state. Do you believe in cooling by evaporation? If so, it makes sense to think that the reverse process applies. Otherwise we have no conservation of energy.
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%.
Matthew R Marler says:
April 15, 2014 at 5:47 pm
“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.”
My Response:
You can get similar information from Table 1 by comparing cases 2 and 3 where all the details are shown. Case 2 has a temperature rise of 2.7C from the 10 Watts per square meter at the top of the atmosphere (IPCC claims about this much starting with 3.7 if you include their feedback multipliers). This case has 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 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. 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, a slight difference from nonlinearities.
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 .
@richard Petschauer April 17, 2014 at 1:40 pm: “That is only a change of about 0.1% per C, which is not a large concern it seems.” I agree with your numbers, not with your easy dismissal. At 30 degrees C (the highest surface temperature of tropical seas) the latent heat is 3% lower. That means that with the same supply of energy the model will evaporate 97 kg of water instead of correct 100 kg. Clouds can only form when a relative humidity reaches 100%, so the model will underestimate clouds.
.. and of course it will also underestimate the effect of water on IR radiation.
Reply to Curious George from
April 17, 2014 at 7:02 pm
The 0.1% should apply to the difference in the average global temperatire of 15C and 30C, so it is about half of what you estimate or 1,5%. Also this was based of the differences of radiation that go as the 4th power of the absolute temperature. But all this is already happening. I was concerned with changes from one surface temperature to another. Also if my estimate of evaporation increase is only 1.5% less than 6%, or 5.91% that is not too bad.