In Part One we saw how the ocean absorbed different wavelengths of radiation:
- 50% of solar radiation is absorbed in the first meter, and 80% within 10 meters
- 50% of “back radiation” (atmospheric radiation) is absorbed in the first few microns (μm).
This is because absorption is a strong function of wavelength and atmospheric radiation is centered around 10μm, while solar radiation is centered around 0.5μm.
In Part Two we considered what would happen if back radiation only caused evaporation and removal of energy from the ocean surface via the latent heat. The ocean surface would become much colder than it obviously is. That is a very simple “first law of thermodynamics” problem. Then we looked at another model with only conductive heat transfer between different “layers” in the ocean. This caused various levels below the surface to heat to unphysical values. It is clear that turbulent heat transport takes place from lower in the ocean. Solar energy reaches down many meters heating the ocean from within – hotter water expands and so rises – moving heat by convection.
In Part Three we reviewed various experimental results showing how the temperature profile (vs depth) changes during the diurnal cycle (day-night-day) and with wind speed. This demonstrates very clearly how much mixing goes on in the ocean.
The Different Theories
This series of articles was inspired by the many people who think that increases in back radiation from the atmosphere will have no effect (or an unnoticeable effect) on the temperature of the ocean depths.
So far, no evidence has so far been brought forward for the idea that back radiation can’t “heat” the ocean (see note 1 at the end), other than the “it’s obvious” evidence. At least, I am unaware of any stronger arguments. Hopefully as a result of this article advocates can put forward their ideas in more detail in response.
I’ll summarize the different theories as I’ve understood them. Apologies to anyone who feels misrepresented – it’s quite possible I just haven’t heard your particular theory or your excellent way of explaining it.
Hypothesis A – Because the atmospheric radiation is completely absorbed in the first few microns it will cause evaporation of the surface layer, which takes away the energy from the back radiation as latent heat into the atmosphere. Therefore, more back-radiation will have zero effect on the ocean temperature.
Hypothesis B – Because the atmospheric radiation is completely absorbed in the first few microns it will be immediately radiated or convected back out to the atmosphere. Heat can’t flow downwards due to the buoyancy of hotter water. Therefore, if an increase in back radiation occurs (perhaps due to increases in inappropriately-named “greenhouse” gases) it will not “heat” the ocean = increase the temperature of the ocean below the surface.
For other, more basic objections about back radiation, see Note 2 (at the end).
I believe that Part Two showed that Hypothesis A was flawed.
I would like to propose a different hypothesis:
Hypothesis C – Heat transfer is driven by temperature differences. For example, conduction of heat is proportional to the temperature difference across the body that the heat is conducted through.
Solar radiation is absorbed from the surface through many meters of the ocean. This heats the ocean below the surface which causes “natural convection” – heated bodies expand and therefore rise. So solar energy has a tendency to be moved back to the surface (this was demonstrated in Part Two).
The more the surface temperature increases, the less temperature difference there will be to drive this natural convection. And, therefore, increases in surface temperature can affect the amount of heat stored in the ocean.
Clarification from St.Google: Hypothesis = A supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation
An Excellent Question
In Part Three, one commenter asked an excellent question:
Some questions from an interested amateur.
Back radiation causes more immediate evaporation and quicker reemission of LWR than does a similar amount of solar radiation.Does that mean that the earth’s temperature should be more sensitive to a given solar forcing than it would be to an equal CO2 forcing?
What percentage CO2 forcing transfers energy to the oceans compared to space and the atmosphere?
How does this compare with solar forcing?
Is there a difference between the effect of the sun and the back radiation when they are of equal magnitude? This, of course, pre-supposes that Hypothesis C is correct and that back radiation has any effect at all on the temperature of the ocean below the surface.
So the point is this – even if Hypothesis C is correct, there may still be a difference between the response of the ocean temperatures below the surface – for back radiation compared with solar radiation.
So I set out to try and evaluate these two questions:
- Can increases in back radiation affect the temperature of the ocean below the surface? I.e., is Hypothesis C supported against B?
- For a given amount of energy, is there a difference between solar forcing and back radiation forcing?
And my approach was to use a model:
Oh no, a model! Clearly wrong then, and a result that can’t fool anyone..
For a bit of background generally on models, take a look at the Introduction in Models On – and Off – the Catwalk.
Here is one way to think about a model
The idea of a model is to carry out some calculations when doing them in your head is too difficult
A model helps us see the world a bit more clearly. At these point I’m not claiming anything other than they help us see the effect of the well-known and undisputed laws of heat transfer on the ocean a little bit more clearly.
Ocean Model
The ocean model under consideration is about a billion times less complex than a GCM. It is a 1-d model with heat flows by radiation, conduction and, in a very limited form, convection.
Here is a schematic of the model. I thought it would be good to show the layers to scale but that means the thicker layers can’t be shown (not without taking up a ridiculous amount of blank screen space) – so the full model, to scale, is 100x deeper than this:
Figure 1
To clarify – the top layer is at temperature, T1, the second layer at T2, even though these values aren’t shown.
The red arrows show conducted or convected heat. They could be in either direction, but the upwards is positive (just as a convention). Obviously, only a few of these are shown in the schematic – there is a heat flux between each layer.
1. Solar and back radiation are modeled as sine waves with the peak at midday. See the graph “Solar and Back Radiation” in Part Two for an example.
2. Convected heat is modeled with a simple formula:
H=h(T1-Tair), where Tair = air temperature, T1 = “surface” temperature, h = convection coefficient = 25 W/m².K.
Convected heat can be in either direction, depending on the surface and air temperature. The air temperature is assumed constant at 300K, something we will return to.
3. Radiation from the surface:
E = εσT4 – the well-known Stefan-Boltzmann equation, and ε = emissivity
For the purposes of this simple model ε = 1. So is absorptivity for back radiation, and for solar radiation. More on these assumptions later.
4. Heat flux between layers (e.g. H54 in the schematic) is calculated using the temperature values for the previous time step for the two adjacent layers then using the conducted heat formula: q” = k.(T5-T4)/d54, where k= conductivity, and d54 = distance between center of each layer 5 to the center of layer 4.
For still water, k = 0.6 W/m.K – a very low value as water is a poor conductor of heat.
In this model at the end of each time step, the program checks the temperature of each layer. If T5 > T4 for example, then the conductivity between these layers for the next time step is set to a much higher value to simulate convection. I used a value for stirred water that I found in a textbook: kt = 2 x 105 W/m.K. What actually happens in practice is the hotter water rises taking the heat with it (convection). Using a high value of conductivity produces a similar result without any actual water motion.
For interest I did try lower values like 2 x 10³ W/m.K and the 1m layer, for example, ended up at a higher temperature than the layers above it. See the more detailed explanation in Part Two.
5. In Part Three I showed results from a number of field experiments which demonstrated that the ocean experiences mixing due to surface cooling at night, and due to high winds. The mixing due to surface cooling is automatically taken account of in this model (and we can see it in the results), but the mixing due to the winds “stirring” the ocean is not included. So we can consider the model as being “under light winds”. If we had a model which evaluated stronger winds it would only make any specific effects of back radiation less noticeable. So this is the “worst case” – or the “highlighting back radiation’s special nature” model.
Problems of Modeling
Some people will already know about the many issues with numerical models. A very common one is resolving small distances and short timescales.
If we want to know the result over many years we don’t really want to have the iterate the model through time steps of fractions of a second. In this model I do have to use very small time steps because the distance scales being considered range from extremely small to quite large – the ocean is divided into thin slabs of 5mm, 15mm.. through to a 70m slab.
If I use a time step which is too long then too much heat gets transferred from the layers below the surface to the 5mm surface layer in the one time step, the model starts oscillating – and finally “loses the plot”. This is easy to see, but painful to deal with.
But I thought it might be interesting for people to see the results of the model over five days with different time steps. Instead of having the model totally “lose the plot” (=surface temperature goes to infinity), I put a cap on the amount of heat that could move in each time step for the purposes of this demonstration.
You can see four results with these time steps (tstep = time step, is marked on the top left of each graph):
- 3 secs
- 1 sec
- 0.2 sec
- 0.05 sec
Figure 2 – Click for a larger image
I played around with many other variables in the model to see what problems they caused..
The Tools
The model is written in Matlab and runs on a normal PC (Dell Vostro 1320 laptop).
To begin with there were 5 layers in the model (values are depth from the surface to the bottom edge of each layer):
- 5 mm
- 50 mm
- 1 m
- 10 m
- 100 m
I ran this with a time step of 0.2 secs and ended up doing up to 15-year runs.
In the model runs I wanted to ensure that I had found a steady-state value, and also that the model conserved energy (first law of thermodynamics) once steady state was reached. So the model included a number of “house-keeping” tests so I could satisfy myself that the model didn’t have any obvious errors and that equilibrium temperatures were reached for each layer.
For 15 year runs, 5 layers and 0.2s time step the run would take about two and a half hours on the laptop.
I find that quite amazing – showing how good Matlab is. There are 31 million seconds in a year, so 15 years at 0.2 secs per step = 2.4 billion iterations. And each iteration involves looking up the solar and DLR value, calculating 7 heat flow calculations and 5 new temperatures. All in a couple of hours on a laptop.
Well, as we will see, because of the results I got I thought I would check for any changes if there were more layers in my model. So that’s why the 9-layer model (see the first diagram) was created. For this model I need an even shorter time step – 0.1 secs and so long model runs start to get painfully long..
Results
Case 1: The standard case was a peak solar radiation, S, of 600 W/m² and back radiation, DLR of 340 with a 50 W/m² variation day to night (i.e., max of 390 W/m², min of 290 W/m²).
Case 2a: Add 10 W/m² to the peak solar radiation, keep DLR the same. Case 2b – Add 31.41 W/m² to solar.
Case 3a: Keep solar radiation the same, add 3.14 W/m² to DLR. This is an equivalent amount of energy per day to case 2, see note 3. Case 3b – Add 10 W/m² to DLR.
Many people are probably asking, “Why isn’t case 3a – Add 10 W/m² to DLR?”
Solar radiation only occurs for 12 out of the 24 hours, while DLR occurs 24 hours of the day. And the solar value is the peak, while the DLR value is the average. It is a mathematical reason explained further in Note 3.
The important point is that for total energy absorbed in a day, case 2a and 3a are the same, and case 2b and 3b are the same.
Let’s compare the average daily temperature in the top layer, 1m, 10m and 100m layer for the three cases (note: depths are from the surface to the bottom of each layer; and only 4 layers of the 5 were recorded):
Figure 3
The time step (tstep) = 0.2s.
The starting temperatures for each layer were the same in all cases.
Now because the 4 year runs recorded almost identical values for solar vs DLR forcing, and because the results had not quite stabilized, I then did the 15 year run and also recorded the temperature to the 4 decimal places shown. This isn’t because the results are this accurate – this is to see what differences, if any, exist between the two different scenarios.
The important results are:
- DLR increases cause temperature increases at all levels in the ocean
- Equivalent amounts of daily energy into the ocean from solar and DLR cause almost exactly the same temperature increase at each level of the ocean – even though the DLR is absorbed in the first few microns and the solar energy in the first few meters
- The slight difference in temperature may be a result of “real physics” or may be an artifact of the model
And perhaps 5 layers is not enough?
Therefore, I generated the 9-layer model, as shown in the first diagram in this article. The 15-year model runs on the 9-layer model produced these results:
Figure 4
The general results are similar to the 5-layer model.
The temperature changes have clearly stabilized, as the heat unaccounted for (inputs – outputs) on the last day = 41 J/m². Note that this is Joules, not Watts, and is over a 24 hour period. This small “unaccounted” heat is going into temperature increases of the top 100m of the ocean. (“Inputs – outputs” includes the heat being transferred from the model layers down into the ocean depths below 100m).
If we examine the difference in temperature for the bottom 30-100m deep level for case 2b vs 3b we see that the temperature difference after 15 years = 0.011°C. For a 70m thick layer, this equates to an energy difference = 3.2 x 106 J, which, over 15 years, = 591 J/m².day = 0.0068 W/m². This is spectacularly tiny. It might be a model issue, or it might be a real “physics difference”.
In any case, the model has demonstrated that DLR increases vs solar increases cause almost exactly the same temperature changes in each layer being considered.
For interest here are the last 5 days of the model (average hourly temperatures for each level) for case 3b:
Figure 5
and for case 2b:
Figure 6
Pretty similar..
Results – Convection and Air Temperature
In the model results up until now the air temperature has been at 300K (27°C) and the surface temperature of the ocean has been only a few degrees higher.
The model doesn’t attempt to change the air temperature. And in the real world the atmosphere at the ocean surface and the surface temperature are usually within a few degrees.
But what happens in our model if real world situations cool the ocean surface more? For example, higher temperatures locally create large convective currents of rising hot air which “sucks in” cooler air from another area.
What would be the result? A higher “instantaneous” surface temperature from higher back radiation might be “swept away” into the atmosphere and “lost” from the model.. This might create a different final answer for back radiation compared with solar radiation.
It seemed to be worth checking out, so I reduced the air temperature to 285K (from 300K) and ran the model for one year from the original starting temperatures (just over 300K). The result was that the ocean temperature dropped significantly, demonstrating how closely the ocean surface and the atmosphere (at the ocean surface) are coupled.
Using the end of the first year as a starting temperature, I ran the model for 5 years for case 1, 2a and 3a (each with the same starting temperature):
Figure 7
Once again we see that back radiation increases do change the temperatures of the ocean depths – and at almost identical values to the solar radiation changes.
Here is a set of graphs for one of the 5-year model runs for this lower air temperature, also demonstrating how the lower air temperature pulls down the ocean surface temperature:
Figure 8 – Click for a larger image
The first graph shows how the average daily temperature changes over the full time period – making it easy to see equilibrium being reached. The second graph shows the hourly average temperature change for the last 5 days. The last graph shows the heat which is either absorbed or released within the ocean in temperature changes. As zero is reached it means the ocean is not heating up or cooling down.
Inaccuracies in the Model
We can write a lot on the all the inaccuracies in the model. It’s a very rudimentary model. In the real world the hotter tropical / sub-tropical oceans transfer heat to higher latitudes and to the poles. So does the atmosphere. A 1-d model is very unrealistic.
The emissivity and absorptivity of the ocean are set to 1, there are no ocean currents, the atmosphere doesn’t heat up and cool down with the ocean surface, the solar radiation value doesn’t change through the year, the top layer was 5mm not 1μm, the cooler skin layer was not modeled, a number of isothermal layers is unphysical compared with the real ocean of continuously varying temperatures..
However, what a nice simple model tells us is how energy only absorbed in the top few microns of the ocean can affect the temperature of the ocean much lower down.
“It’s obvious“, I could say.
Conclusion
My model could be wrong – for example, just a mistake which means it doesn’t operate how I have described it. The many simplifications of the model might hide some real world physics effect which means that Hypothesis C is actually less likely than Hypothesis B.
However, if the model doesn’t contain mistakes, at least I have provided more support for Hypothesis C – that the back radiation absorbed in the very surface of the ocean can change the temperature of the ocean below, and demonstrated that Hypothesis B is less likely.
I look forward to advocates of Hypothesis B putting forward their best arguments.
Update – Code files saved here
Notes
Note 1 – To avoid upsetting the purists, when we say “does back-radiation heat the ocean?” what we mean is, “does back-radiation affect the temperature of the ocean?”
Some people get upset if we use the term heat, and object that heat is the net of the two way process of energy exchange. It’s not too important for most of us. I only mention it to make it clear that if the colder atmosphere transfers energy to the ocean then more energy goes in the reverse direction.
It is a dull point.
Note 2 – Some people think that back radiation can’t occur at all, and others think that it can’t affect the temperature of the surface for reasons that are a confused mangle of the second law of thermodynamics. See Science Roads Less Travelled and especially Amazing Things we Find in Textbooks – The Real Second Law , The Real Second Law of Thermodynamics and The Three Body Problem. And for real measurements of back radiation, see The Amazing Case of “Back Radiation” -Part One.
Note 3 – If we change the peak solar radiation from 600 to 610, this is the peak value and only provides an increase for 12 out of 24 hours. By contrast, back radiation is a 24 hour a day value. How much do we have to change the average DLR value to provide an equivalent amount of energy over 24 hours?
If we integrate the solar radiation for the before and after cases we find the relationship between the value for the peak of the solar radiation and the average of the back radiation = π (3.14159). So if the DLR increase = 10, the peak solar increase to match = 10 x π = 31.4159; and if the solar peak increase = 10, the DLR increase to match = 10/π = 3.1831.
If anyone would like this demonstrated further please ask and I will update in the comments. I’m sure I could have made this easier to understand than I actually have (haven’t).
The Science of Doom 
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Out of curiosity, what proportion of outputs goes to the atmosphere, and what proportion goes to the deep ocean? And does the source of increased radiation make a difference to these proportions?
The amount of heat moving into the ocean depths is much lower than convection and radiation out to the atmosphere (in this model).
For the 2nd 4-year model run in Figure 3, the value of heat flow into the ocean depths = 1.92 x 10^4 Joules/m^2 per day – for both the solar increase and the equivalent DLR increase.
Convected = 5.64 x 10^6 Joules/m^2 per day.
Radiated = 4.11 x 10^7 Joules/m^2 per day.
You can see an example of this in the 3rd graph (bottom left) in Figure 8 – where the legend “Depth” is the daily heat value moving into the ocean depths.
Confusingly, given my notation in the original model description, R is the radiated out value.
Then S = solar, DLR = back radiation and Conv = convected heat.
By the way, I just added Figures Numbers to the article.
I just read this. Probably not what TC is asking about (model?,) but I’ve been looking for information on greater than 2000 meters, and found it interesting:
“The deep ocean accounts for a significant fraction of the global energy imbalance (Levitus et al. 2005; Johnson et al. 2007). Johnson et al. (2007) estimated that the deep ocean could add an additional 2-10% to the upper ocean heat content trend, which is likely to grow in importance as the anthropogenic warming signal propagates to increasing depth with time. …”
This article deserves a much closer look that I can give it right now but I’ll just mention Hypothesis D which I believe has been missed.
Hypothesis D is that according to Stefan-Boltzmann, the ocean must radiate energy according to its SST according to the familiar
E = εσT4
The energy in this radiation is made up from some of the energy leaving the bulk of the ocean plus some from DLR that was absorbed in the first few microns and then immediately re-radiated upwards.
Something like this…
E = εσT4 = DLR Energy Re-radiated + Energy radiated from the bulk.
Hence more DLR means more immediate re-radiation and therefore less energy used from the ocean bulk to satisfy the S-B law and the ocean heats up.
Of course the DLR also results in increased evaporation and so the ratio of energy used for increased evaporation vs that immediately re-radiated effectively determines the rate at which the ocean bulk energy is radiated away and consequently the rate at which the ocean warms.
I guess I’ll make further comments on this article once I’ve had a chance to consider it properly.
Isn’t this a situation that can be trivially tested without any sums or models at all?
Why not have two deep cooking pots full of room-temperature water and a digital thermometer probe at the bottom.
Heat one with an IR lamp (try to ensure environment doesn’t get any incidental heating)
What happens?
That’s not really the same. Take two pots of water and put immersion heaters in the bottom of each pot set to the same power level. Measure the temperature. Now put a heat lamp over one of the pots and measure again. Or take two pots of water at higher than ambient temperature and put a heat lamp over one of them and measure the cooling curves. Want to bet on the results?
Even simpler, pure cold tap water into two well-insulated cups. Cover one with a lid so to restrict the interaction of water with the ambient air. I’d be willing to bet almost any amount of money that the one without the lid warms faster.
Anyone care to explain where the additional energy in the warmer cup comes from if not the air?
This is a tough situation to model convincingly. Some example: 1) Your atmosphere has a fixed temperature (300 degK in one case), but you assume that DLR from that atmosphere is the same as DLR from the earth’s atmosphere. Most DLR photons arriving at the surface of the ocean are from altitudes where it is colder than 300 degK. On the other hand, upward radiation from the surface of your ocean is determined by the temperature of that ocean. 2) How much colder and denser does water higher in the ocean really need to be before convection begins to supplement and then overwhelm conduction? Is this process is accurately modeled by a linear equation with a constant slope and zero intercept? 3) Unlike the real ocean, your model doesn’t have a skin layer that is usually cooler than the water below.
Observations showing that the skin (top tens of um) of the ocean is usually colder than the water below convincing demonstrate that increasing DLR will warm the upper ocean and that Hypothesis A and B are untenable. Despite the enormous amount of energy deposited by DLR into the skin of the ocean (compared with its heat capacity), this skin layer is usually slightly colder than the water below! Why? The skin layer is the source of: 1) all upward LWR from the ocean AND 2) latent heat of evaporation AND 3) any conductive heat loss from the water to the air. Since the part of the atmosphere exchanging photons with the surface of the ocean is generally cooler than the surface (you model handles this poorly), LWR is a net loser for the skin layer even before evaporation and conduction are considered. Only a small fraction of solar SW radiation is deposited in the skin layer (and only during sunny days), and this usually isn’t enough to make up the deficit. Since the skin layer usually needs to tap into solar energy deposited deeper in the ocean to make up its deficit, the skin layer will obviously make use of any extra DLR from an enhanced greenhouse effect. Only during the short periods around noon on sunny days – when the skin layer is warmer than the water below – is there any point in worrying about how long energy from DLR remains in the skin layer and what fraction is lost upward rather than downwards (warming the ocean).
SOD: Rather than question the limitations your models, I’d learn more from seeing how well the IPCC’s models actually reproduce the daily and seasonal changes in ocean temperature with depth. The ARGO network seems to be generating some surprising results. How good (or bad) are the models we are using to drive policy decisions (rather than debates on the web)?
I’d like to see your code if possible?
I think your following assumption might cause your model problems…
“In this model at the end of each time step, the program checks the temperature of each layer. If T5 > T4 for example, then the conductivity between these layers for the next time step is set to a much higher value to simulate convection. ”
Because real convection involves momentum and so any rising water particularly in the topmost layers of your model wont be artificially limited by temperature.
TimTheToolMan:
Here’s v3.5 of my Matlab code used to generate the results.
—————— code ———————-
% DLR and solar forcing on the ocean
% to investigate the impact if DLR does and doesn’t heat the ocean
% this version to look at solar heating below the surface and its effect
%
% v2 considers convection if temperature inversion occurs
% v3 looks at heat flows to compare changing heat flow for changing DLR
% and changing solar
% also tries to improve the algorithm (v3.2) by reducing the vector to
% a manageable size and just retaining hourly figures instead of every
% second. See p19 of workbook2 (plus p98 of book1)
% v3.3 improves algorithm by calculating heat movement between layers only
% once, v3.2 had the up & down calc for each layer in turn – duplication
% v3.4 minor plotting improvements
% v3.5 find FE issues when add more layers close to surface, requiring
% small time steps – so speed up with precalc of thicknesses and distances
% and investigate further
% Time properties
nperiod=24*5; % number of *periods*
periodsec=3600; % number of seconds in a *period* (= 1 hour currently)
tstep=.2; nstep=periodsec/tstep; % time step in seconds; number of time steps in a period
t=tstep:tstep:(nstep+1)*tstep; % create time vector with nstep values for one *period*
realt=periodsec:periodsec:nperiod*periodsec; % create time vector of real time in secs
% Radiation properties & radiation daily cycle
smax=600; % solar max radiation at surface
dlrav=340; dlramp=50; % average DLR; max & min variation of DLR
s=smax.*sin((1:1:1441).*2*pi/1440); % create solar vector for 24 hr period, every min
sneg=find(s1e4
% heat(j)=1e4;
%end
heats(j,p)=heats(j,p)+heat(j)*tstep; % accumulate summary in J/m^2
end
heat(nlayers+1)=kinf.*(Tinf-T(j,i-1))./dist(nlayers+1); % special case for
% ocean depths into bottom layer (note value will be negative as
% heat flows the other way)
heat(1)=h.*(T(1,i-1)-Tair); % convective heat from surface
% accumulate summary stats in J/m^2 per period
heats(nlayers+1,p)=heats(nlayers+1,p)+heat(nlayers+1)*tstep;
heats(1,p)=heats(1,p)+heat(1)*tstep; % accumulate summary in J/m^2
rads(1,p)=rads(1,p)+s(daymins)*tstep; % “”
rads(2,p)=rads(2,p)+dlr(daymins)*tstep; % “”
for j=1:nlayers; % j is the step through the depths
% radiation in and out
if j==1 % top layer
Ea=ab*dlr(daymins)+s(daymins)*sol_prop(j); % absorb DLR & solar
Ee=em*SB(T(j,i-1)); % Ee is radiation out via SB law
rads(3,p)=rads(3,p)+Ee*tstep; % accumulate energy out
else
Ea=s(daymins)*sol_prop(j); % absorb solar radiation proportion for that depth
Ee=0; % no longwave from below top layer
end
dT=(heat(j+1)-heat(j)+Ea-Ee).*tstep./c(j); % heat cap
T(j,i)=T(j,i-1)+dT; % new temperature
end
if T(1,i)>1000 % FE problem
disp([‘Terminated at period ‘ num2str(p) ‘ and i = ‘ num2str(i)]);
return
end
% now check for convective overturning and use the results in the next
% iteration
for j=2:nlayers
if T(j,i)>T(j-1,i)
kk(j)=kt; % turbulent or stirred conductivity
else
kk(j)=ks; % still conductivity
end
end
end
% now collect the summary data
Ts(:,p)=T(:,nstep); % collect snapshot data
kks(:,p)=kk; % capture conductivity for each layer for checking
for avi=1:nlayers
Tav(avi,p)=mean(T(avi,:)); % probably more elegant way to do this
end
end
% create daily min,max and average values
ndays=nperiod*periodsec/86400; % number of days in the analysis
if ndays>1 % if number of days is less than 1 not much point
nvaluesday=nperiod/ndays; % number of samples per day to check
Tdmax=zeros(nlayers,ndays); % Tdmax is daily max value
Tdmin=zeros(nlayers,ndays); % Tdmin =daily min
Tdav=zeros(nlayers,ndays); % Tdav = daily mean
heatstats=zeros(5,ndays); % main heat flows, 1=solar,dlr,Eout,conv,5=base
for i=1:ndays
for j=1:nlayers % calculate daily max, min and average temp for each layer
Tdmax(j,i)=max(Tav(j,(i-1)*nvaluesday+1:i*nvaluesday));
Tdmin(j,i)=min(Tav(j,(i-1)*nvaluesday+1:i*nvaluesday));
Tdav(j,i)=mean(Tav(j,(i-1)*nvaluesday+1:i*nvaluesday));
end
% calculate daily heat flows: solar, DLR, rad out, conv, heat into
% depth
heatstats(1,i)=sum(rads(1,(i-1)*nvaluesday+1:i*nvaluesday));
heatstats(2,i)=sum(rads(2,(i-1)*nvaluesday+1:i*nvaluesday));
heatstats(3,i)=sum(rads(3,(i-1)*nvaluesday+1:i*nvaluesday));
heatstats(4,i)=sum(heats(1,(i-1)*nvaluesday+1:i*nvaluesday));
heatstats(5,i)=sum(heats(nlayers+1,(i-1)*nvaluesday+1:i*nvaluesday));
end
end
%avperiods=24*7; % number of periods to create an average over – must be an
% even multiple of nperiods at this stage
%avnum=round(nperiod/avperiods); % number of values to create
% ————– PLOTTING RESULTS ————————-
textlegall=cell(nlayers,1);
textleg=cell(nlayers,1);
for i=1:nlayers
textlegall(i)={[num2str(d(i+1)) ‘m’]}; % creates legend for graph of all depths
end
dplotix=1:nlayers; % but can be a different definition
for i=1:length(dplotix)
textleg(i)={[num2str(d(dplotix(i)+1)) ‘m’]}; % selected legend defined by dplotix
end
plottime=24*5; % max number of periods to plot
if plottime>nperiod
plottime=nperiod;
end
% ——plot average daily temp (Tdav) for all layers
subplot(2,1,1), plot(1:ndays,Tdav)
xlabel(‘Days’)
title(‘Average daily temperatures’)
grid on
legend(textlegall);
%legend(‘Av top’,[‘Av @’ num2str(d(6)) ‘m’],[‘Av @’ num2str(d(10)) ‘m’])
%subplot(2,1,1), plot(realt(1:plottime)/3600/24,Tav(dplotix,1:plottime))
%xlabel(‘Days’)
%grid on
%legend(textleg);
% ———plot average hourly temp (Tav) for all layers listed in dplotix
subplot(2,1,2), plot(realt(end-plottime+1:end)/3600/24,Tav(dplotix,end-plottime+1:end))
xlabel(‘Days’)
title(‘Average hourly temperatures’)
grid on
legend(textleg);
% Data to display
disp([‘Time step = ‘ num2str(tstep) ‘ S= ‘ num2str(smax) ‘ DLR= ‘ num2str(dlrav)]);
disp(‘Daily average temperatures’);
disp(num2str(Tdav(:,end)));
esum=heatstats(1,end)+heatstats(2,end)-heatstats(3,end)-heatstats(4,end)+heatstats(5,end);
disp(‘The missing heat on last day’);
disp(num2str(esum));
% ———plot daily heat flows into and out of ocean slab
subplot(2,2,3), plot(1:ndays,heatstats’/1e3)
xlabel(‘Days’)
ylabel(‘kJ/m^2’)
legend(‘S’,’DLR’,’R’,’Conv’,’Depth’)
% ———plot daily total heat imbalance for slab
esum=heatstats(1,:)+heatstats(2,:)-heatstats(3,:)-heatstats(4,:)+heatstats(5,:);
subplot(2,2,4), plot(1:ndays,esum/1e3)
title(‘Total J/m^2.day imbalance’)
xlabel(‘Days’)
ylabel(‘kJ/m^2’)
There is either a typo in this code, or it got mangled.
Line “29” of the code has
sneg=find(s1e4
Anyway you could archive all of the code in a zip file so we could download and run it ourselves?
Thanks for spotting it, something went wrong in the copy and paste.
I can’t upload a zip file unfortunately, not supported in wordpress.
I can load the Matlab file into word and save as a .doc file and upload that.
I will do an experiment (elsewhere) to see what happens with this kind of file, then add a comment at the end of the comments plus link the files in the article itself.
Files here.
Hi SoD,
Thanks for the code.
I haven’t used MatLab (but am a programmer so can read code)…so what is sol_prop()? I assume its adding heat to each layer according to some sort of curve but cant see it actually defined in what you’ve provided.
The other fairly significant shortcoming to this model is that when DLR is at a maximum, DSR is at a minimum and vice-versa (and I dont mean day-night variations either) because its cloud that enhances DLR to the levels you’ve quoted.
So as far as I can tell your model assumes lots of cloud varying to none on a 24 hour cycle that coincides with “night” which will tend to mean ocean heat losses are minimised because “night” is also when the overall cooling to a minimum (whatever that may be) will occur.
And I think the whole cloud (DLR) variation is fairly poorly modelled. Still… we do what we can do, eh.
Am I right in saying your model will never perform “downwards” convection where the surface is colder than the bulk?
for j=2:nlayers
if T(j,i)>T(j-1,i) % Tim – Only looks at deeper layers warmer
kk(j)=kt; % turbulent or stirred conductivity
else
kk(j)=ks; % still conductivity
end
The ocean_solar2() function:
Its purpose it to lookup the amount of solar radiation absorbed in each layer of the ocean.
Here is the code. (I haven’t removed the plotting function so not all of it is required for this call).
The graph it produces is found in Part One.
I added a “round to 100%” (% ensure total absorption = 100%) because only about 99% is absorbed in the first 100m and I wanted to conserve energy in the model.
—– ocean_solar2() —————-
function [ strans ] = ocean_solar2(d)
% Solar radiation transmitted vs depth
% this version to create only the radiation transmitted vs depth
% input vector is set of depths to consider
% output (strans) is ratio (0 < ratio < 1) transmitted between each depth
d=d';
ploton=0; % change to 1 if want to plot
allin=1; % =1 if we want 100% absorbed – last depth gets the extra
dw=0.05; % change in wavelength for each step – need to ensure "integrals" work
ww=(0.1:0.05:4)'; % wavelengths to evaluate
alb=0.3; costheta=1; % take into account reflection of solar, and later, angle from zenith
rs=planckm(ww,5780)'.*((1-alb)*costheta/215^2); % solar radiation at TOA
rstotal=sum(rs)*dw; % rstotal = total solar radiation
k=ocean_absorb(ww); % get the lookup table of absorption coefficients vs wavelength
[kk dd]=meshgrid(k,d);
rss=repmat(rs',length(d),1);
rd=exp(-kk.*dd).*rss; % length(d)x79 of energy transmitted to each depth
% by depth & wavelength
st=sum(rd').*dw./rstotal; % sums across wavelengths for total energy at each depth
for i=1:length(st)-1
strans(i)=st(i)-st(i+1); % strans gets difference between each layer
% caution as strans(1) is the radiation between
% d(2) and d(1) -amended 2Jan2010 to match T(1)
% is temp between d(2) and d(1)
end
if allin==1 % ensure total absorption = 100%
strans(length(d)-1)=strans(length(d)-1)+(1-sum(strans));
end
if ploton==1
plot(d,sum(rd').*dw./rstotal.*100)
grid on
set(gca,'xscale','log')
title('Ocean Transmission of Solar Radiation by Depth');
xlabel('Depth, m');
ylabel('% transmitted');
end
end
——— code end ————–
You will see that it calls up a planckm function and ocean_absorb():
——- planckm() —————-
function ri = planckm( w,t )
% planck function for matrix of wavelength, w and temp, t
% result in spectral intensity
[ww tt]= meshgrid(w,t);
ri = 3.742E8./(ww.^5.*(exp(1.439E4./(ww.*tt))-1));
end
——–code end —————-
——– ocean_absorb() ———–
function [ k ] = ocean_absorb( w )
% Creates ocean absorptivity vector, k, in, m^-1 for a wavelength vector, w,
% in um. Uses a short lookup table and “interp1”.
% Used graph from Wikipedia
% http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water
% values match Soloviev 2007 and Stamnes
% this data allows interpolation to find the absorption coefficient in m-1
% against wavelength in um
% old one la=[.2 .4 .6 1 1.25 1.5 2 2.5];
la=[.1 .2 .3 .4 .5 .6 .7 .8 1 2 3 4 6 10 15 20 30];
aa=[8e3 10 .2 .06 .02 .1 .7 3 60 1000 1e6 1e4 2e5 8e4 4e5 3e5 1e5];
% old one aa=[1 .03 1 6 100 2000 10000 8000];
k=interp1(la,aa,w);
end
——–code end —————-
Hi SoD, cracking series and a subject I’ve explored before without finding any really solid conclusions, but there are some nuggets around.
TimTheToolMan:
“E = εσT4 = DLR Energy Re-radiated + Energy radiated from the bulk”
One I thing I did discover is that the bulk cannot radiate energy, as noted by Frank when he said:
“The skin layer is the source of… …all upward LWR from the ocean”. It has to be, because any IR emitted from lower down is re-absorbed within a few microns. Only the top 10 microns can emit anything to the air.
So Tim, the formula should actually read:
“E = εσT4 = Energy radiated from skin layer (with no contribution from the bulk). Obviously though energy does get transferred out of the bulk, but it has to be conducted or convected to the skin layer to escape the ocean.
This brings me on to a link I want to share to a post on this topic at Realclimate: http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/
The basic premise is that additional DLR warms the skin layer a bit, which means that heat transfer from below decreases because it decreases both conduction (obviously) AND the overnight natural convection that SoD described.
I believe that this is the situation we should be studying. A correspondent of mine adheres to hypothesis A (I think) and disregarded out of hand the evidence I presented that increased DLR results in a measureable, observed increase in skin temperature; this in itself disproves the idea that all DLR is immediately used up in evaporation, which would result in no observed warming at all.
I think he disregarded the results because they’re from Realclimate, which is a shame, they’re just the messanger. We did spend a good deal of time discussing it; his theory of extremely efficient climate homeostasis depends on it.
One thing I think this series has shown is that regardless of how the energy moves around in the ocean skin/mixed layer, you can’t maintain ocean temps without a contribution from DLR. From the Realclimate article, it seems the most likely mechanism is that additional DLR slows the flow of energy from the mixed layer, so the mixed layer gets to ‘keep’ more of the energy it absorbs from the sun.
Hope you find the link useful SoD.
I’m sorry, I realise I have made a syntax boo-boo above!
The fact that my correspondent disregards the evidence does not disprove the idea that all DLR is immediately used up in evaporation!
The evidence, posted at Realclimate, is what does that. Ahem.
TimTheToolMan:
Both the solar and DLR peak at the same time. Both sine functions.
It would be interesting to make the DLR peak later in the day. That would be more realistic.
Here is a random DLR extract – found in The Amazing Case of “Back Radiation” – Part Two:
You can see some more samples there and in Part One.
And lots of data in Darwinian Selection – Back Radiation, one sample here:
Sorry, now I realize what you are actually saying..
DLR will be lower when solar is higher, due to clouds..
Well, some nice graphs anyway in my last comment.
TimTheToolMan (from January 8, 2011 at 2:33 am):
Following on from your comments there, I have used some fairly arbitrary numbers for solar and DLR.
I have now enhanced my Matlab program a bit so I can run 3 simulations in one hit and summarize the results. So fairly easy to find what happens if say solar is S W/m^2 and DLR is X W/m^2 +/- Y.
Well – a bit painful because the model has to be run for long enough to reach equilibrium..
Also I have just been playing around with changing the “turbulent conductivity” (Kt) – I tried 2×10^4 instead of 2×10^5.
9 levels, tstep=0.1s, 365 days for: std; Solar+31.4159; DLR+10
I haven’t dug into the details very deeply but the results are similar for the DLR and Solar increase.
Correct.
If the layer above (j) is warmer than the layer below (j+1) then kk = ks = 0.6. This means heat flows very slowly – by conduction.
If the layer above is colder than the layer below then kk = kt (2×10^5). This means heat flows very fast – by convection.
I’ve examined the detailed behavior and as soon as the temperature is higher below the heat rapidly moves up.
The only question is whether there is a more realistic model which doesn’t require modeling turbulent ocean flows.
For example, perhaps I could slow down the movement of heat by adding a time lag.
Well, there are 100s of improvements that could be made.
Here are some values produced by ocean_absorb2() for values of d I am using right now (9 layers):
d=
0
0.0050
0.0200
0.0500
0.2000
1.0000
3.0000
10.0000
30.0000
100.0000
sol_prop=
0.2368
0.0927
0.0532
0.0704
0.1094
0.0981
0.1343
0.1251
0.0801
To Stu N
Regarding “E = εσT4 = DLR Energy Re-radiated + Energy radiated from the bulk
…
One I thing I did discover is that the bulk cannot radiate energy, as noted by Frank when he said: ”
Just to clear up any confusion here, I dont mean radiated directly from the bulk, as you say all radiation occurs at the skin, rather the equation is a measure of where the energy is supplied from to satisfy the S-B law.
So…Some energy comes from the bulk (via convection and conduction to the skin) and some comes from the DLR (which naturally happens at the skin).
To SoD
What I’d like to see is your model being able to “naturally” create the cold skin 😉
Tim
Noted, though slap on the wrist for sloppy use of ‘radiated’ 😉
Reading a comment from Gavin Schmidt at Realclimate, you can have two cases. First is where the ocean is losing heat to the atmosphere – you have a cool skin and additional DLR slows energy transfer upwards from the mixed later. Or you can have the ocean gaining heat from the atmosphere, in which case there is a warm ‘skin’ and heat is transferred downwards.
In both cases additional DLR results in higher ocean temps, which is of course the right result and a win for physics.
My correspondent at one point said that additional IR speeds up evaporation so much it would actually cool the ocean! I retorted that you should then be able to freeze a tray of water by shining an IR lamp on it…
“Or you can have the ocean gaining heat from the atmosphere, in which case there is a warm ‘skin’ and heat is transferred downwards.”
I’ve never seen any references to a warm skin in the literature? Do you have a reference to his quote?
And in fact I think I’ve found the quote from RC
…” The skin temperature in areas where the ocean is being heated, is generally warmer than the bult temperature, but in this example, the ocean is losing heat to the atmosphere (the skin SST is less than bulk T). “… -gavin
As I said the literature doesn’t mention an ocean warm skin ever and there is much said about the ocean’s cool skin. I suspect Gavin was talking out of his …ignorance.
Either that or he knows something that isn’t well documented. If a warm skin is common, I’d certainly like to see a reference.
Part Two doesn’t seem to deal with evaporation at all, merely radiation, convection and conduction.
How then can it be said to deal adequately with Hypothesis A ?
Stephen Wilde:
Re-reading Part Two, I don’t think I explained very well what hypothesis I was reviewing.
The point is, if DLR only goes into evaporation then what will be the equilibrium temperature of the ocean.
So it’s exactly focused on Hypothesis A.
I gave hypothesis A a thorough workout some time ago because I see it as critical to the plausibility of AGW theory as much as regards timing of any effects as well as regards its existence.
See here:
http://climaterealists.com/index.php?id=4245
“Greenhouse Gases Can Cause Cooling !”
Let me know whether you can find a flaw.
“TimTheToolMan
And in fact I think I’ve found the quote from RC
…” The skin temperature in areas where the ocean is being heated, is generally warmer than the bulk temperature, but in this example, the ocean is losing heat to the atmosphere (the skin SST is less than bulk T). “… -gavin”
Gavin seems unaware of the Knudsen layer which is what responds to IR.
The Knudsen layer warms, the skin layer (1mm deep) cools further and energy is drawn up faster not slower.
See the article of mine that I refer to above.
“The Knudsen layer warms, the skin layer (1mm deep) cools further and energy is drawn up faster not slower.
See the article of mine that I refer to above.”
There is no evidence of this Stephen, it’s just something you’ve hypothesised. And as I mentioned before, if what you say is true then increasing the DLR will lead to the following:
Increased DLR -> warming of the knudsen (or evaporation) layer -> evaporation that takes more energy than the DLR provides -> flux of energy upwards from skin layer and below -> cooling of the bulk ocean.
Which is phycially impossible because then, as I said, it should then be possible to freeze water by shining an IR lamp onto it…
Moving on, I feel SoD has now adequately explained how hypothesis A is covered, by saying that IF only solar heats the ocean and all energy from DLR is used for evaporation, there is an energy imbalance and the oceans will lose so much heat they freeze in a few years.
And my final point, Tim, the ‘warm skin’ effect, I presume Gavin does know what he’s talking about. Here’s a useful page which I must have missed first time I looked at this topic: http://ghrsst-pp.metoffice.com/pages/sst_definitions/
“SSTskin measurements are subject to a large potential diurnal cycle including cool skin layer effects (especially at night under clear skies and low wind speed conditions) and warm layer effects in the daytime (although the latter are not shown in Figure 1).”
Anyway, I look forward to part five and hope SoD investigates this ‘skin’ effect which to me seems to be the crux of the matter.
1) “Which is phycially impossible because then, as I said, it should then be possible to freeze water by shining an IR lamp onto it…”
No.
The process of evaporation takes the energy it needs from wherever it is most readily availale.
Thus once the water reaches the temperature of the air above the necessary energy will be taken from both air and water so the ocean will never freeze. Indeed it will never cool below the temperature of the air above it.
2) ““The Knudsen layer warms, the skin layer (1mm deep) cools further and energy is drawn up faster not slower.
See the article of mine that I refer to above.”
There is no evidence of this Stephen, it’s just something you’ve hypothesised.”
No.
The activity in the Knudsen layer is what causes the existing 1mm deep cooler layer. That is a fact and not mere hypothesising by me. Thus if there is more activity in the Knudsen layer the cooler layer must become cooler and/or deeper. I simply extrapolate from observed and recognised fact.
Note that the Knudsen layer is neither air nor water. It is simply a haze of air molecules and water molecules that are in the process of changing state. Yet it is the temperature of that layer that the sensors interpret as a surface warming notwithstanding the further cooling of the ocean skin below it.
The error being made is in conflating the Knudsen layer with the ocean sakin.
“Yet it is the temperature of that (the Knudsen) layer that the sensors interpret as a surface warming notwithstanding the further cooling of the ocean skin below it.”
This is not correct.
The same link I provided in my previous post states that the temperature of the Knudsen layer is unmeasurable and the ocean skin temperature, the top 20 microns, is the top-most measurement we can get. For comparison I’ve seen the Knudsen layer described as ‘several mean free paths thick’ which in the atmosphere is about 70 nanometres and would be much less in water. So we’re looking at the Knudsen layer orders of magnitude thinner than the skin layer.
For reference here the skin layer is the layer capable of emitting IR. The subskin layer is the layer that, under cool skin conditions, has a lower temperature than the ocean below because of energy loss from the skin layer.
Stu N said:
“and the ocean skin temperature, the top 20 microns, is the top-most measurement we can get”
But the ocean skin layer is that 1mm layer which is 0.3C colder than the ocean bulk. Imm is at least an order of magnitude greater than 20 microns.
The top 20 microns will be a measurement of activity in the Knudsen layer rather than of the temperature in the 1mm deep layer.
We need to get a better handle on what is meant by the ocean skin. As I said before, the error is in not separating it adequately from the Knudsen layer.
Lets take an extreme example.
Suppose the sun gets very hot with lots more IR than we see at present. Assume the solar IR can take the Knudsen layer to boiling point. What will happen to the 1mm deep cooler layer between the Knudsen layer and the ocean bulk?
It will get colder and deeper as it tries to balance the energy required by the evaporative/boiling process in the Knudsen layer with the energy being supplied from below via convection and conduction.
The hotter the Knudsen layer gets the more the energy requirement will outstrip what can come up from below and the ocean skin will get deeper and colder with the shortfall supplied by the air as a last recourse. The availability of energy from the air limits the coldness and depth of the 1mm ocean skin.
The heat of the Knudsen layer may well propogate down a few microns but that just deepens the Knudsen layer. 20 microns down and the sensors register increased warmth. Meanwhile, unnoticed, that 1mm layer below is getting colder.
The ocrean bulk probably does not change at all because the deepening and/or cooling of the 1mm ocean skin layer restores the equilibrium between ocean bulk and ocean skin.
One must treat the Knudsen layer and the ocean skin as separate entities. It appears that established climatology fails to do so.
Um, Second Law violation. You cannot have a net thermal energy flow from colder to hotter without doing work as in a heat pump. All the energy for boiling the Knudsen layer is being supplied from the input solar energy. There is no need for energy transfer from below. In fact, in accord with the Second Law, the subsurface will warm too.
Probe thermometers and heat lamps are cheap. Put a heat lamp on a dimmer to minimize visible radiation over a pot of water and see if the layer just below the surface of the pot gets cooler. IR thermometers aren’t expensive either and can be used to measure the surface temperature. The water in the rest of the pot will warm too, if it’s well insulated. But in a small pot, turbulent eddy diffusion will be small and the process will be slow.
DeWitt,
You have forgottenthe evaporative process which involves a change of state. An evaporating molecule operates as a pump extracting energy from water or air depending on availability. Thus no Second Law violation.
The water inthe body of the pot will warm from conduction from the container and not from IR heating onto the water surface. Evaporation from the top will leave a cooler layer below if the temperature in the room is cool enough but if the room air temperature is higher than the water temperature all the energy will be taken from the air and the cooler layer will not develop. In the open air it is rare for the air to be warmer than the water except around land masses.
Turbulent eddy diffusion takes energy up from the ocean rather than down from the surface because warm water is lighter than colder water. Hence increasing cold with depth. Thus your analogy fails.
http://ghrsst-pp.metoffice.com/pages/sst_definitions/
Stephen did you read this link?
Sorry Stu, I missed that through concentrating on the rest of your post.
I’ll adapt my wording to those definitions so that we are back on the same page. Will respond shortly.
OK.
The SST(int) warms up from IR impact. Presumably that is the Knudsen layer.
The SST(skin) also warms up and the temperature is recorded by the sensors.
The SST(subskin) cools as energy is drawn upwards. It has been measured as globally averaging 0.3C cooler than the ocean bulk below it.
All that has happened there is that the region above the cooler 1mm deep layer has been subdivided into the Knudson layer which heats up a lot and a few microns below it which heats up a little less.
The process I described remains intact but additionally we see the reason why the sensors record warming despite that cooling lower 1mm deep layer.
Doesn’t that actually answer your earlier objection?
There is still a cooler layer that cools more if evaporation increases but now we see that the temperature measured by sensors is above that layer, is not part of it and therefore records a misleading warmth while the sub skin cools unnoticed as I said:
“The heat of the Knudsen layer may well propogate down a few microns. 20 microns down and the sensors register increased warmth. Meanwhile, unnoticed, that 1mm layer below is getting colder.”
Just so.
Can you find any evidence that the subskin is NOT cooled by increased evaporation ?
After all, evaporation causes it to exist in the first place.
I’ll come back to this later, but first:
“Can you find any evidence that the subskin is NOT cooled by increased evaporation ?”
If the reason for increased evaporation is anything other than an increase in DLR, the answer should be ‘no evidence’. For example, drier air (at the same temperature) over the ocean would increase evaporation and cool the subskin layer.
However, increase DLR and keep everything else the same, then I don’t think the subskin will cool. Sure you’re adding energy, but you simply cannot make the increase in energy carried away by evaporation greater than the energy you added with the DLR. I agree with DeWitt, that’s a violation of the second law.
Consider it this way. Say you have a device that can capture and use all the latent energy of evaporation from a 1 metre-square patch of ocean surface. Never mind that such a device cannot exist, this is a thought experiment!
Now shine 50 watts of (additional) infrared onto your patch of ocean. You will note an increase in evaporation and hence an increase in the latent heat flux. You are able to capture all of this latent heat, and if you capture more than 50 watts, you have created free energy, violated the 2nd law and made the universe mad.
By the way, way back when didn’t I point you towards observations showing that DLR is usually considerably higher (~400W/m2 in the tropics) than the latent heat flux (averaging ~200W/m2 in the tropics)?
See http://ams.confex.com/ams/pdfpapers/52793.pdf, for example. Does this not show that not all DLR goes into evaporation?
Stephen Wilde,
So let’s see. We’ve added sufficient IR to heat the surface layer to 100 C, but now all the heat of vaporization of that surface layer comes from the layer below which is colder? Bzzzt. Wrong answer. The heat of vaporization comes from the same source as the heat used to elevate the surface temperature n the first place, the incoming IR radiation. I didn’t forget about the heat of vaporization. You have separated processes that are, in fact, inseparable. Second Law violation. I used up all my patience on Jeff Glassman at Judith Curry’s. You’ve now joined him, in record time, in my burn before reading file.
Bryan, where are you. You should love this.
Stephen Wilde on January 8, 2011 at 10:02 am:
Can you help me understand what you are saying?
Here is a diagram of the various movements of energy at the ocean surface:
Note: I’ve ignored the energy moving up by conduction/convection from beneath the surface because it arrives from S (solar energy).
Are you saying:
a) L = DLR, or
b) L >= DLR, or
c) something else?
It seems like you are saying a) or b), but that might be my incorrect interpretation.
If you are saying either a) for example then:
S = R + E
where S= solar, R= emission of thermal radiation from the surface, E = sensible heat
Therefore, R = S – E
So, best case, with E=0, R = εσTs^4 = S
Therefore, Ts = (S/εσ)^(1/4)
And given that S, averaged over the surface area of the planet, equals something between 170 W/m^2 and 240 W/m^2, Ts will be a lot colder than we currently seem to measure.
Hopefully, you can clarify which bits you agree with.
“C, but now all the heat of vaporization of that surface layer comes from the layer below which is colder? Bzzzt. Wrong answer.”
OK, if you believe that essentially all the energy of evaporation comes from the DLR, how do you explain the existence of the colder skin?
And in what form is the energy that leaves the ocean through the skin?
And to Stu N (appologies to Gavin), Fair enough… technically the skin can be warmer than the bulk but the skin is (always) still colder than the subskin and that doesn’t invert as far as I’m aware.
Furthermore, the increase in subskin temperature over the bulk temperature is as a result of DSR and not DLR because it only happens when DSR is happening.
SoD, Regarding your question to Stephen
“Can you help me understand what you are saying?”
I would like to answer it in terms of Hypothesis D, and that is that some of the energy used in evaporation is used from the DLR and some is used from your “L” as energy from the ocean.
Sufficient energy is used from the ocean “L” to drive the skin temperature downwards to the point we could measure.
Additional to that, sufficient energy is used from the DLR to cause the rate of evaporation that we could measure.
The rest of the energy from the DLR that is re-radiated back up towards the atmosphere is part of the ocean’s requirement to radiate according to S-B law.
The rest of this requirement to radiate energy comes from the ocean skin as your “R”.
Tim I think you’ve answered your own question. You asked:
“OK, if you believe that essentially all the energy of evaporation comes from the DLR, how do you explain the existence of the colder skin?”
And answered it with:
“The rest of this requirement to radiate energy comes from the ocean skin as your “R””
Because the ocean skin cools by emitting radiation R, you could conceivably have a scenario where all DLR –> L and the cooler skin temperature is solely due to R.
In practice this would not happen and Tim is right to say that L takes contributions from energy already in the ocean (essentially from S) and from DLR.
Re : Tim I think you’ve answered your own question.
I know I answered my own question and I’m quite happy with my own hypothesis and explanation but I’m not sure yet what DeWitt Payne thinks. He could think that evaporation is pretty much entirely driven by DLR.
Afterall he says this… “The heat of vaporization comes from the same source as the heat used to elevate the surface temperature n the first place”
So…I want to know what DeWitt Payne thinks.
I haven’t been through everyone’s comments in detail yet.
I notice that some people think that the cool skin is easy to understand..
I’ve just had a reread of the fascinating 1996 paper: The behavior of the bulk-skin sea surface temperature difference under varying wind speed and heat flux, GA Wick, WJ Emery, LH Kantha & P Schlussel Journal of Physical Oceanography – unfortunately behind a paywall. If I sum up their work I could say that it’s very hard to explain the skin effect properly.
Here’s a 2005 paper which is freely available – Cool-skin simulation by a one-column ocean model, by Chia-Ying Tu and Ben-Jei Tsuang. I’m just about to start reading it but I thought the intro might tempt a few other people to read it (emphasis added):
TimTheToolMan on January 9, 2011 at 2:32 am:
I don’t understand your Hypothesis D. If what I have written below shows that I have misunderstood it, maybe you can have another attempt at explaining it, or show a few equations so there’s no misunderstanding..
Energy in comes from S and DLR.
So energy sources for L can only be from S or DLR.
Obviously, in the short term, the heat stored in the ocean can and will be used. But for energy balance considerations, R (radiation), L (latent heat) and E (sensible heat) can only be “funded” by S & DLR.
(Unless the hypothesis includes the idea that the average direction of L or E is from the atmosphere to the ocean).
“But for energy balance considerations, R (radiation), L (latent heat) and E (sensible heat) can only be “funded” by S & DLR.”
I would say mostly right except the ocean energy isn’t funded by DLR because it cant be absorbed in anything other than the topmost molecules.
So S funds the ocean energy.
DLR “funds” the energy at the skin which in turn partially “funds” the energy that S-B says the ocean must radiate.
Does that make sense?
TimTheToolMan on January 9, 2011 at 4:04 am:
On Hypothesis D..
I still don’t know how it is different from Hypothesis B. Or whether you are endorsing Hypothesis C.
When you say:
I agree that DLR is absorbed in the top few μm.
I propose a theory that because an increase in DLR increases the surface temperature it reduces the heat (from solar radiation) flowing up to the ocean surface.
This theory is consistent with DLR being absorbed in the top few μm.
Can an increase in DLR change the ocean heat?
I propose “yes” = Hypothesis C.
Advocates of Hypothesis B say “no”.
So I’m still unclear on Hypothesis D.
Stephen Wilde:
I see your comment on another blog – under “#339 Posted : 08 January 2011 08:48:28(UTC)”:
Leaving aside the cool skin for a later comment – and referencing my earlier question of January 9, 2011 at 1:25 am – it appears you are saying that the atmosphere supplies the energy to the ocean surface on average?
So on a globally annually averaged basis:
i) S + DLR + L = R + E
where L is positive from the atmosphere to the ocean
OR
ii) S + DLR + E = R + L
where E (sensible heat) is positive from the atmosphere to the ocean.
Version i) seems very confused as the argument of Hypothesis A is that DLR all goes towards latent heat. Whereas version i) implies that latent heat generally flows to the ocean.
So you can’t be meaning that.
Version ii) would mean that the atmosphere was generally warmer than the ocean so you can’t mean that.
I can’t think of a version iii)
Maybe your comment didn’t come out as you intended?
I look forward to your clarification on my first question of January 9, 2011 at 1:25, and this one.
More on the cool skin for those who want to learn more.
I just found that one of the “most-cited articles” on the subject is available free:
Saunders, Peter M., 1967: The Temperature at the Ocean-Air Interface. J. Atmos. Sci.
Or is this the link?
And if that doesn’t work go to the source page and click on the pdf.
SoD “So I’m still unclear on Hypothesis D.”
Firsly, hypothesis B as you’ve stated it doesn’t (obviously) allow for ocean heating. But its the closest of the three to hypothesis D.
I disagree with Hypothesis C on the grounds that your increase in SST meaning drecreased energy flow across that boundary sounds utterly contrived.
The boundary is (potentially) microns wide and the heat doesn’t need to cross it, rather it is radiated from there. In a sense, the skin is the “destination” for the energy rather than something to be negotiated on the way to somewhere else.
According to Stefan-Boltzmann when the temperature increases, the energy radiated increases to the power of 4. Hypothesis C is fighting that one too.
Furthermore there is no evidence of what the increase in SST actually means and I would say it means the hook that is the temperature gradient of the first few cm of the ocean simply shifts a little left or right but doesn’t change shape appreciably.
Either way, its unknown what happens specifically. The only measurements I’m aware of are those of Minnett who has SST and 5cm below the surface which misses the interesting bit entirely.
Hypthesis D isn’t difficult. Its a lot like Hypothesis B except the radiated energy lost by the ocean (which cools it) is partly made up of that re-radiated DLR. So the DLR slows down the rate of energy loss from the ocean by accounting for some of the radiation that the ocean MUST constantly lose according to the S-B law.
Hypothesis D doesn’t need to assume anything about how the temperature gradient changes at the top of the ocean, nor does it care.
Would a diagram help?
TimTheToolMan:
A diagram would help.
When I’ve understood the hypothesis I can comment on it.
But re a couple of your comments:
Perhaps you’ve misunderstood Hypothesis C. Or in fact, misunderstood my explanation of what happens in the model.
The movement of heat from 1m (for example) up to the surface is dependent on the temperature difference between 1m below and the surface.
Is this contrived?
If not, then what is strange or contrived about a higher surface temperature reducing heat flows from 1m to the surface?
The model didn’t set out to contrive a result. I put the equations for heat flow, for solar absorption, for Stefan-Boltzmann and ran a simulation.
You have seen the code.
What equation would you have instead for heat flow between each layer.
No, Hypothesis C has that one exactly as the formula.
Again, you can see it in the code:
Ee=em*SB(T(j,i-1)); % Ee is radiation out via SB law
— and here is the code for SB() —-
function r=SB(t)
% Stefan-Boltzmann function for absolute temperature
r=5.67E-8.*t.^4;
—– code end —-
“The movement of heat from 1m (for example) up to the surface is dependent on the temperature difference between 1m below and the surface.”
Sure if the movement of heat was a limiting factor then you might have a point. But is it? Convection is very efficient at moving heat upwards and as I pointed out, has momentum.
Compare this to what hypothesis C is fighting and that is that the increase in SST means an increase in energy radiated by the ocean according to Stefan-Boltzmann.
I know you tried to account for it in the model, but I feel that your model just isn’t modelling convection well enough to be relevant. Plus of course you’re missing effects of wind, currents, evaporation and so on.
So can I see why you think that hypothesis C works? Yes. But for the reasons I’ve given above, I’m far from convinced.
And…IMO Hypothesis D is much simpler and accounts for everything.
scienceofdoom asked:
“Are you saying:
a) L = DLR, or
b) L >= DLR, or
c) something else?”
Consider one IR photon at a time because each photon interacts with a water molecule at SST(int).
So every time a photon hits a water molecule it adds to the energy carried by that molecule and brings forward the timing of the evaporative event that would have occurred anyway.
When evaporation occurs more energy is drawn from the surroundings than is added to the molecule by the photon. That extra energy now in latent form is drawn from both water and air in proportions dependent on their relative temperatures. Generally in the open air above ocean surfaces most is drawn from the water because the water has been heated by the solar shortwave that has previously penetrated it and is therefore mostly warmer than the air above.
The energy is taken from the water faster than it comes up from below so that cooler layer develops and the faster
evaporation is the cooler and deeper that subskin layer becomes.
Thus there is a barrier to energy going deeper and the faster is evaporation the stronger is the barrier.
Note that we are not provoking evaporation ab initio. What happens is that every IR photon brings forward the timing of evaporation and the evaporation then requires more energy than the photon provided for a net cooling effect on the local environment.
If it were as you say then that cooler layer could not develop. The surplus energy that you propose would indeed heat that cooler layer and its temperature would approach that of the ocean bulk so it would dissipate.
In fact however if evaporation causes the cooler layer (is there any doubt about that?) then more evaoration must intensify it which is the opposite response to what you require.
How you want to deal with that by way of complex equations is up to you but the logic and the facts are clear.
StuN said:
“but you simply cannot make the increase in energy carried away by evaporation greater than the energy you added with the DLR. I”
A single photon of DLR will often be enough to switch a single water molecule from water to vapour. Remember that water molecules in Skin(int) are already moving towards the evaporative event even without any DLR. We are not considering a situation where the DLR has to initiate the entire evaporative event ab initio. If that were the case then you could be right.
So what we see here is a situation of ongoing evaporation where the DLR just brings forward the timing. The more one brings forward the timing the more intense is the evaporative process and the stronger and deeper the cooler sub skin becomes.
If there were any energy left over from the extra DLR then the subskin would warm up and the cool layer dissipate. That does not happen.
Consider the diurnal range of solar IR input. When the sun is shining the subskin coolness is more intense than when the sun is not shining or at night. That is because there is more evaporation in daylight.
It’s odd that this seems to be inadequately dealt with by established science when the principle of evaporation being a net cooling process is so well known.
Stephen Wilde from January 9, 2011 at 11:46 am:
I don’t think you can call this question of mine:
“a) L = DLR, or
b) L >= DLR, or
c) something else?”
– Complex.
The first law of thermodynamics is a handy law. Writing out the energy balance makes it easy to see whether the first law is complied with – or not.
Let’s find out if your theory is supported by the first law – or violates the first law.
Your ideas do appear to violate the first law – and the fact that you want to “Consider one IR photon at a time..” demonstrates that this appearance may be correct.
So according to you the process of evaporation is a violation of the First Law ?
And according to Stu and DeWitt evaporation is a violation of the Second Law ?
Weird.
Evaporation is clearly impossible then. You win.
Stephen, this comment is specifically in response to
“Note that we are not provoking evaporation ab initio. What happens is that every IR photon brings forward the timing of evaporation and the evaporation then requires more energy than the photon provided for a net cooling effect on the local environment.
If it were as you say then that cooler layer could not develop. The surplus energy that you propose would indeed heat that cooler layer and its temperature would approach that of the ocean bulk so it would dissipate.”
and
“A single photon of DLR will often be enough to switch a single water molecule from water to vapour. Remember that water molecules in Skin(int) are already moving towards the evaporative event even without any DLR. We are not considering a situation where the DLR has to initiate the entire evaporative event ab initio. If that were the case then you could be right.”
Now, not much DLR is absorbed at SSTint, because this is the Knudsen/evaporation layer and as I described earlier is orders of magnitude thinner than the skin layer in which all DLR is absorbed, (something like 200nm compared to 20microns for most of it and 0.1mm for all of it).
So clearly the majority of photons do not result in evaporation by striking a molecule and so causing it to reach or get closer to its ‘tipping point’. Most photons are absorbed lower down and cause molecules to evaporate indirectly by providing energy to the skin layer, which is then available to the evaporating molecules via convection and conduction.
I have no concept of how often a photon will ‘tip’ a molecule to evaporate, but I don’t think it’s a high proportion because one thing I do have a concept of is that the density of photons in a 400W/m2 beam of DLR is far greater than the number of water molecules capable of absorbing that DLR. Hang tight while I work it out!
Very simply (and possibly wrongly as well, we’ll see!), the energy of a photon is E=hv, where h is Planck’s constant and v is the frequency. We know a typical wavelength of the DLR might be 10 microns (and since the dependence is linear I wont bother being more accurate). The frequency then is v=c/wavelength where c = speed of light.
v = 3×10^8 x 1×10^-5 = 3×10^3 hertz.
So the energy of each photon, with h = 6.6×10^-34:
E = 6.6×10^-34 x 3×10^3 = 1.98×10^-32 joules.
Each second, 400 joules of energy arrives at each m2 of the surface in the form of
400/1.98×10^-32 = 200×10^34 photons.
Which is a lot of photons. Now all that remains for me to make my point is calculate how many molecules could absorb this. 1m2 x 0.1mm depth gives a mass of water of 0.001 kg.
The number of water molecules per mole of water is Avogadro’s number: 6×10^23. The molecular mass of water is 18, so a mole weighs 18 grams or 0.018kg. So there are 0.001/0.018 = about 0.06 moles in our mass of water. This means there are 6×10^23*0.06 = 3.6×10^22 molecules in the mass of water.
The difference is 12 orders of magnitude, massive! The evaporative flux of molecules must be much, much, MUCH lower, or we’d be losing the top 0.1mm of the ocean and then some every second of every day, and that ain’t right.
So what does that leave us with? Your statement that “A single photon of DLR will often be enough to switch a single water molecule from water to vapour” does not look to be true, and evaporation is far more likely to happen when collisions randomly transfer enough energy to a molecule to make it evaporate. A good chuck of that energy arrived as DLR but it hasn’t caused spontaneous evaporation. I believe I’ve shown this to be the case by demonstrating that the incoming radiation is a DENSE stream of LOW ENERGY photons.
So while each individual evaporation event is endothermic, very few are spontaneous upon absorption of a photon and so most energy goes into warming the skin layer; it just so happens that the skin layer is still cooler because this is where all the energy for R comes from and it is also cooled by L; but increasing L beyond any increase in DLR is definitely still a violation of the second law!
I hope you can see how your comments are in error, since the vast, vast majority of incoming photons do NOT cause spontaneous evaporation.
Correction to the above:
400/1.98×10^-32 = 200×10^34 photons is wrong. It should be:
400/1.98×10^-32 = 2×10^34
Whoops, and the mass of water is 0.0001kg, not 0.001kg. So the order of magnitude difference between photons per second and molecules is 13, not 12.
And one more, to my final conclusion:
I hope you can see how your comments are in error, since the vast, vast majority of incoming photons do NOT cause spontaneous evaporation OR even bring the molecules they strike close to the point of evaporation.
Bugger, I have one more major correction to make.
v = 3×10^8 x 1×10^-5 = 3×10^3 hertz should be
v = 3×10^8 / 1×10^-5 = 3×10^13 hertz!
So E = 6.6×10^-34 x 3×10^13 = 1.98×10^-21 joules
And 400/1.98×10^-21 = 2×10^23 photons.
And there are 3.6×10^21 molecules.
SO only 2 orders of magnitude difference in the end. Apologies for my errors. However, giving the absorbing layer a depth of 0.1mm is quite generous since most is absorbed in the top 0.01mm – so it could be 3 orders of magnitude.
It’s now quite close so if I have made further errors they affect the results of my analysis, so although it was interesting I think other lines of evidence are stronger. SoD, sorry for ‘flooding’ but I’d rather this be correct 🙂
“the vast, vast majority of incoming photons do NOT cause spontaneous evaporation OR even bring the molecules they strike close to the point of evaporation.”
So not only is the process of evaporation in violation of the Laws of Thermodynamics but also that observed 1mm cooler layer cannot really be there either because of all that surplus energy from unused DLR.
I give up.
No Stephen, the big problem with your reasoning is that you don’t consider ALL the energy going in and out! Net energy flow is vital.
“So not only is the process of evaporation in violation of the Laws of Thermodynamics but also that observed 1mm cooler layer cannot really be there either because of all that surplus energy from unused DLR.”
The top layer emits more upward longwave than it emits! That contributes to the cooler layer – there is a net emission of longwave from the skin layer.
Whoops, really having an off day – the top layer emits more longwave than it ABSORBS. Sorry.
“The top layer emits more upward longwave than it emits! That contributes to the cooler layer – there is a net emission of longwave from the skin layer.”
Well that refers to radiation and not evaporation. I know that radiation is a contributor to the cooler layer in addition to evaporation.
So if one increases EITHER radiation OR evaporation the cooler layer gets even cooler and goes even deeper.
The remaining bit missing from your above calculations (assuming you’ve corrected all the other errors now:)
is the extra energy required by the change of state from water to vapour.
When you include that you should then see the net cooling effect.
Well as per SoD’s diagram, the equilibrium energy equation is as follows:
S + DLR = E + R + L
Now, if one increases evaporation and/or upward radiation while keeping the left hand side (LHS) the same, then yes there is clearly an imbalance and the skin temperature will cool.
However, we’re considering an increase in DLR. To move towards equilibrium, either R and E increase (R and E being functions of temperature), L increases or all three increase.
You say:
“The remaining bit missing from your above calculations
is the extra energy required by the change of state from water to vapour.”
Well first my rambling calculation above isn’t to do with the equilibrium equation, but anyway this change of state energy is fully accounted for by L.
And if you increase L beyond the increase in DLR, as you seem to be suggesting happens, the ocean is now losing more energy than it gains.
Are you saying that when you increase DLR by 10W/m2, L increases by >10W/m2? You seem to be saying this when you declare that there is a ‘net cooling effect’. This is a crucial point that needs clarifying.
I’m still undecided on the specific point but I think I see why you think that the extra L may be limited by the extra DLR but that still gives zero heating of the ocean does it not ?
I don’t think you can do better than show that extra L is much the same as extra DLR. I don’t see how you can say there is any surplus energy left over.
If there were then that cooler layer would dissipate.
BUT returning to your equation:
Why should L be limited by the energy content of S or DLR ? L is a matter of the physical characteristics of water rather than the size of the energy input that provokes evaporation.
At lower pressure L will occur with less S or DLR than at a higher pressure so how can L be limited in the way you suggest ?
There is always a predisposition to evaporate even in the complete absence of S or DLR because of the pressure and density differential between air and water. That should always skew the balance towards the negative it seems to me but you require a positive outcome from more S or DLR.
If all the word’s oceans were a few molecules deep they would evaporate in a flash with no iinput from S or DLR.
If zero S and zero DLR still results in L then ANY extra energy added from S or DLR results in a continuing negative balance hence the recognised net cooling effect of evaporation.
Whether the negative balance gets bigger or not with increased S or DLR doesn’t really matter. It certainly does not get any less negative
However you play it and whatever imperfections there may be in my narrative there is no way that you can turn S and DLR positive in the face of L hence no ocean heating is possible
S in your equation should be limited to solar infra red longwave because that is the portion of the solar spectrum involved at the ocean surface.
Solar shortwave does of course heat the ocean but that is not generated by GHGs and so should be excluded from any discussion of GHG effects.
“S in your equation should be limited to solar infra red longwave because that is the portion of the solar spectrum involved at the ocean surface.
Solar shortwave does of course heat the ocean but that is not generated by GHGs and so should be excluded from any discussion of GHG effects.”
Not true – solar ‘longwave’ still counts as ‘shortwave’ radiation because almost all of it is less than 4 microns wavelength. By contrast almost all terrestrial (aka longwave) radiation is more than 4 microns. Wherever SoD talks about ‘solar radiation’ this is inclusive of the solar IR. Put simply, solar IR, being a shorter wavelength than terrestrial IR, penetrates further into the ocean.
Besides, why the heck should we exclude solar from our considerations? As discussed by SoD, convection and conduction move energy of solar origin around (including upwards) to where it contributes to the surface energy balance. If you leave it out you get an energy imbalance, and the wrong answers.
TTTM,
I was referring to the thought experiment of an increase in DLR sufficient to boil the surface layer. Once the temperature reaches the boiling point, additional energy results in evaporation rather than a temperature increase. It’s no different than turning up the burner on a pot of boiling water. The temperature doesn’t go up, the rate of evaporation does. (That’s not strictly true if you measure the temperature at very high precision.)
In the real world, as you point out, the ocean during the day is almost always losing heat from incoming solar radiation to the atmosphere by convection and radiation. For heat to transfer from below the surface to the surface, the skin must be colder than the water below it. If that were reversed, heat would go the other way. Increasing DLR does not change that fact.
Turbulent mixing can move heat against a temperature gradient in the atmosphere when there’s a horizontal temperature gradient to provide the energy needed to do the work. That’s the mechanism that would create a near adiabatic temperature profile in the atmosphere in a perfectly transparent atmosphere on a spherical planet. But that’s not going to happen to any significant extent in the ocean because the compressibility of water is far less and water is far more viscous than air.
One more comment regarding the following.
“The movement of heat from 1m (for example) up to the surface is dependent on the temperature difference between 1m below and the surface.
Is this contrived?”
If you accept that DLR cannot heat anything below the topmost few molecules then there is no temperature difference between 1m below and the surface.
At least there is no difference as far as convection goes. Consider a bit of water at 1m depth. What does it see above it? Precisely the same as the “no extra DLR” case and convection procedes “normally”.
This is exactly the same view all the water sees above it until the last 10 microns where it does see a temperature difference and so convection starting at 11 microns down is going to be effected.
And yes, this is very much a contrived result because convection cant stop suddenly at the last 10 microns in reality but it sure can in a model…
That’s only true if there were no solar radiation. This was argued on the Venus threads too. If the atmosphere of Venus were perfectly opaque then it would be isothermal. But the Venusian atmosphere isn’t perfectly opaque, neither is the ocean.
“Once the temperature reaches the boiling point, additional energy results in evaporation rather than a temperature increase.”
No.
In the case of DLR to the top of a water surface the topmost molecules can hit boiling point for little or no temperature effect on lower molecules.
The situation is quite unlike heating an entire pot of water. To do that one has to place the source of heat at the base and inside the water as in a kettle.
Thus ANY extra heat applied to a water surface increases evaporation WITHOUT heating the bulk of the water. That outcome is only possible because the DLR only affects the topmost molecules and the subsequent increase in evaporation has a net cooling effect.
As regards the Laws of Thermodynamics there is no violation because all the energy remains in the system but becomes locked in as latent heat rather than sensible heat.
Thus sensible heat APPEARS to have vanished but it hasn’t really it is just in latent form.
But latent heat cannot register on sensors (it is not sensible) so the effect from our point of view is a reduction in temperature despite the system overall containing the same amount of energy.
This stuff was tought in detail in the 1960s but not so much now it seems.
Stephen, if you haven’t already pls see my latest post in reply to your previous. I’m covering energy rather than temperature to attempt a more complete explanation.
By the way, incoming radiation does not bring any molecules to ‘boiling point’ because temperature is a statistical measure and on the scale of individual molecules has little meaning. What you mean is that some molecules gain enough energy to escape the liquid and become water vapour. In doing so they take energy from their immediate surroundings and leave the liquid cooler (the energy is, as you rightly state, now in the form of latent heat; plus the liquid is losing its most energetic molecules which also contributes to cooling).
However you will only get NET cooling if the increase in energy lost via evaporation is not compensated for by an increase in energy gained via other sources. DLR is one such source and a great deal of photons are absorbed and don’t cause evaporation, thereby ‘topping up’ the energy of the liquid and preventing it from cooling. That’s why earlier on I said that all DLR could theoretically go to L (immediately or eventually), but this would NOT result in a net cooling. The temperature would stay the same. It’s an unphysical scenario but demonstrates what I’m talking about.
I think I’ve covered that in my reply above.
Of course I accept the point abot ‘boiling’ since in effect evaporation is the same as boiling but at a lower temperature. I was just seeking a graphic image.
I’ve taken on board the point about NET cooling and dealt with it above but your problem is that somehow you need to get to NET warming DESPITE increased evaporation.
I don’t think you can overcome that probem.
I think I just nailed it.
The balance between the process of evaporation and the amount of energy required to provoke it is pressure dependent.
Thus a higher atmospheric pressure will cause water to boil at a higher temperature than 100C and a lower atmospheric pressure will allow water to boil at less than 100C.
So if evaporation is a net cooling process it must be the case that at current Earthly atmospheric pressure the latent heat energy required by the evaporative process is greater than the energy required to provoke it.
If Earth’s atmosphere produced a greater pressure then one could envisage a scenario whereby the reverse would be the case.
At current atmospheric pressure ALL energy that fails to get past the evaporative barrier is used up in evaporation for a net cooling effect.
As I said many times before it is indeed the case that the energy required by the evaporative process is greater than the energy required to provoke it.
Hypothesis A would appear to be valid.
Stephen Wilde on January 9, 2011 at 2:53 pm:
A handy literary device.
It matches up nicely with your essay (January 9, 2011 at 11:46 am) about the life of the photon.
Just in conclusion for old-fashioned folk like myself, do you believe your model complies with the first law of thermodynamics?
If your answer is “yes” then you do need to answer the earlier very simple question so we can all see the implications. That’s the only way we can find out .
If you don’t answer this specific point, well, it’s clear what your answer is.
Of course the First Law is complied with. The energy is still there in the system but in latent form and so it does not register on sensors.
This is equilibrium energy balance we are talking about.
Energy in – energy out = heating of system
If the system is not heating up or cooling down, energy in = energy out.
Even better we can find the numerical value of energy in – energy out and work out how fast heating or cooling is taking place.
So back to my original question of January 9, 2011 at 1:25 am.
Answer the multiple choice question and complete the equation.
It’s not enough to say “of course” – you have to demonstrate it.
TimTheToolMan:
We will compare Case 1 – “Now”, with Case 2 – “Increased DLR”.
The comments below only apply at a certain time, not all of the time. They will apply just a little bit earlier in the solar heating cycle in Case 2 compared with Case 1.
So let’s consider the water 1um below the “DLR layer”.
Before case 2 (the increase in DLR) it would have become more buoyant than the “DLR layer” and convection would have taken place.
Now it doesn’t, so 1um below “DLR layer” the water is warmer just below the “DLR layer”.
Now let’s consider the water 2um below the DLR layer.. In case 1 it would have become more buoyant than the 1um layer and convection would have taken place.
In fact, if I had a model which had 1000 layers which performed the same you could make the same criticism. That’s the thing about finite element analysis models, they have unphysical isothermal layers.
We could calculate the net buoyancy – gravitational forces on a heated layer and calculate the distance/time required to bring a moving parcel of water to a stop.
In any case, it’s important to understand what actually happens in the model.
—————
Layer 1
————–
Layer 2
————–
etc
Layer 2 heat will move to layer 1 “quickly” if layer 1 is colder than layer 2. It moves and stops each time step.
Layer 3 heat will move to layer 2 “quickly” if layer 2 is colder than layer 3.
This is how the “signal” from layer 1 is “transmitted” to further down layers.
In the model the heat from 1m down (actually 3m-10m) only “sees” the heat at 3m (actually 1m-3m).
“Now let’s consider the water 2um below the DLR layer.. In case 1 it would have become more buoyant than the 1um layer and convection would have taken place.”
No. Because water 2um below the DLR layer sees above it (for the next 1um) exactly the same as the same as the no-DLR situation. The same goes for all the water below that too.
By saying water below the DLR layer is effected by the DLR layer, you’re actually projecting the heat gradient of the warmer DLR layer right through the gradient through which convection is taking place.
Here is how the temperature profile looks where
convection will be effected…
Temp ————>
10um————-|
11um————|
12um———–|
13um———-|
14um———|
15um——–|
16um——-|
But here is how we believe it looks on account of DLR being absorbed in the first 10um and the fact the skin layer is cold.
10um————-|
11um–|
12um—|
13um—-|
14um—–|
15um——|
And you’re still ignoring the fact convection has momentum.
scienceofdoom
My answer is c)something else, as explained above extensively.
The energy in the system stays the same but some of it changes form and location. Parts of the system do change their sensible heat content for a change in temperature but that happens all the time.
Therefore no violation of any of the Laws of Thermodynamics.
Perhaps you should clarify why you think the simple processes involved in evaporation somehow violate those laws ?
scienceofdoom
Try:
L > energy required to provoke it at current atmospheric pressure. Hence it’s net cooling effect.
The energy required to provoke it comes from a variety of sources, Solar longwave IR, DLR, upward conduction and convection from below (formerly solar shortwave) conduction from the air above.
L is opportunistic and always takes energy from where it is most readily available so as to always seek to restore equilibrium between sea surface and surface air temperatures.
But surely you know all this ? It was common knowledge 50 years ago.
Ooh we’re so close to being in agreement Stephen!
Of course L cools the ocean surface; it cannot add any energy to it. And yes, the energy for L comes from a variety of sources. So far so good. (NB: calling evaporation a ‘net cooling’ process; the ‘net’ is superfluous because evaporation by itself can cause cooling only! It is necessary to say, for example, that the surface longwave exchange is a net cooling process because it could potentially be a net warming process if DLR > R).
What I don’t see is how, if you add some extra energy, L somehow might increase by more than that measure of energy. Such an exchange would result in a NET cooling of the surface despite the addition of more energy. You need to make up your mind on this and reject the possibility that adding 10W/m2 of DLR results in a greater than 10W/m2 increase in L. That’s just not possible.
But then again, there is a set of observations showing that if you add some extra energy in the form of DLR, the part of the ocean that we are able to measure by its IR emission (the skin) warms up a little. As you yourself have said, L takes energy from where it is most available. Because L happens in a shallower layer than IR absorption, the additional energy provided by the IR is the most available, and not the energy from BELOW the ocean skin.
You could have two scenarios – all the additional DLR gets used by the additional L and the ocean skin doesn’t warm up, and the conduction/convection of energy from below is unaffected. OR, not all the additional DLR gets used by the additional L and the ocean skin does warm up a bit (as per the observations), which then does affect the rate of conduction/convection of energy from below.
The scenario that is simply impossible is the one you espoused earlier – that the skin layer warms and the subskin cools – because that’s refuted by your own claim that “L is opportunistic and always takes energy from where it is most readily available”.
Do you see what I mean?
“What I don’t see is how, if you add some extra energy, L somehow might increase by more than that measure of energy. Such an exchange would result in a NET cooling of the surface despite the addition of more energy.”
Well I see this as a side issue because it is not critical from my point of view where the energy comes from to cause a cooling effect. It’s enough for me for the DLR effect to be zero overall. You have a bigger problem in trying to show a surplus.
As long as L > than the energy required to provoke it as it is at current atmospheric pressure then adding more energy from ANY source results in cooling rather than the warming required by AGW theory.
So it doesn’t really matter where L draws the energy from there is more cooling with more evaporation.
However energy such as DLR directed straight into the evaporative region and absorbed there is likely to go first before the process resorts to other sources. Hence the complete eradication of that DLR in the first instance.
“The scenario that is simply impossible is the one you espoused earlier – that the skin layer warms and the subskin cools – because that’s refuted by your own claim that “L is opportunistic and always takes energy from where it is most readily available”.
Yet the skin layer does show warming and the sub skin is cooler than the ocean bulk Observations rule and as we have seen already the science on these issues is vague and inconclusive so I wouldn’t use the term ‘impossible’.
My guess would be that active evaporation processes do generate sensible heat before it disappears into latent heat.
“As long as L > than the energy required to provoke it as it is at current atmospheric pressure then adding more energy from ANY source results in cooling rather than the warming required by AGW theory.”
What about my earlier point? Each individual molecule that evaporates takes energy with it and results in a cooling of water. But what of all the DLR that is absorbed and does not immediately result in evaporation? It exists, and I know it exists, because DLR is typically 1-200W/m2 greater than L in the tropics.
As you correctly state, evaporation would happen even if there was no DLR, assuming the air was unsaturated. Let’s call that value Lo and assign it the arbitrary value 100W/m2. Now, ramp up DLR from zero to a typical tropical value of 400W/m2.
You would not see L increase to 500W/m2! Regardless of your starting value of L, and I can’t stress this enough, you will NOT get a like for like increase as you ramp up DLR. This would result in L > DLR and this is NOT what is observed. Therefore not all additional DLR –> L.
Stephen Wilde:
Hypothesis A:
From the choices I provided:
a) L = DLR, or
b) L >= DLR, or
c) something else?
You selected (c).
c) must be L < DLR
Which is contrary to hypothesis A.
So I think you have convincingly demonstrated how difficult it is to support hypothesis A. You don't believe it yourself.
Stephen Wilde on January 9, 2011 at 9:12 pm:
Yes, I know this. Although to be picky, latent heat transfer is dependent on the humidity of the air above the ocean. For example, with a temperature difference but saturated air, L will be zero.
You can see some of this in Clouds and Water Vapor – Part Two.
More importantly, what is the relationship between your statement and hypothesis A?
Latent heat transfer is dependent on a lot of things locally but the overall scenario is as I say.
As regards hypothesis A it is broadly correct but if being picky I would refine it somewhat.
For example, it gives the impression that the DLR goes straight to latent heat whereas in fact the DLR is only part of a package of energy sources from which L draws what it needs for an overall negative effect.
What exactly do you mean when you say ‘overall negative effect’?
Stephen Wilde on January 9, 2011 at 9:42 pm:
I think you have demonstrated something important.
You believe hypothesis A “broadly” but add the fact that “DLR is only part of a package of energy sources from which L draws what it needs for an overall negative effect.”
Earlier you chose c) from this set of options:
a) L = DLR, or
b) L >= DLR, or
c) something else?
So you believe that L>=DLR but chose option c).
The prosecution rests.
Defense m’lud: having been in discussion with Stephen about this before, his statement does not imply he agrees with L>=DLR.
Where does the latent heat of vapourisation come from? It’s not exclusively DLR as Stephen says – particularly as he correctly states that evaporation will happen even if there’s no DLR or S with unsaturated air. So his statement is independent of the magnitude of DLR relative to L. It remains perfectly true regardless of whether L = DLR or L >= DLR or L <= DLR.
“Stu N
What exactly do you mean when you say ‘overall negative effect’?”
More energy into the evaporative layer from ANY source
will provoke increased evaporation and overall cooling of the local environment wherever L is drawn from.
“More energy into the evaporative layer from ANY source
will provoke increased evaporation and overall cooling of the local environment wherever L is drawn from.”
Unless the flux of energy that provokes additional L is greater than that increase in L, in which case there will be warming instead. I believe I have shown this to be the case in my post at 10:15pm: https://scienceofdoom.com/2011/01/06/does-back-radiation-%e2%80%9cheat%e2%80%9d-the-ocean-%e2%80%93-part-four/#comment-8683
Don’t start with ‘evaporation is always a net cooling process’ because all that means is that the ocean surface is cooler than it would otherwise be if there was no evaporation. It is perfectly possible for the ocean surface to warm up when the increase in incoming energy outstrips the increase in L.
The ocean surface will only warm up if the incoming energy gets past the evaporative layer (DLR from GHGs doesn’t) or as in the tropics the sun overwhelms the evaporative response locally and temporarily.
What about night time in the tropics. Why do you think it is so humid ?
That does not apply to DLR from GHGs though, only to the greater power of the sun.
The ocean surface will only warm up if the incoming energy gets past the evaporative layer (DLR from GHGs doesn’t)
YES IT DOES!! I made this point earlier, more than once. This image from part one of this series is the most telling:
https://scienceofdoom.files.wordpress.com/2010/10/dlr-absorption-ocean-matlab.png?w=499&h=526
The evaporative layer is several mean free paths thick. Mean free path at sea level ~ 68nm. So let’s say depth of evaporative layer ~200nm, 2×10^-7m. So with only 20% of the DLR absorbed by the top 1×10^-6m (1 micron), probably 95% of the DLR makes it past the evaporative layer.
scienceofdoom said:
“So you believe that L>=DLR but chose option c).”
No.
I chose option c) something else and then said:
“L > energy required to provoke it at current atmospheric pressure. Hence it’s net cooling effect.”
Not Guilty.
StuN said:
“What about my earlier point? Each individual molecule that evaporates takes energy with it and results in a cooling of water. But what of all the DLR that is absorbed and does not immediately result in evaporation? It exists, and I know it exists, because DLR is typically 1-200W/m2 greater than L in the tropics.”
We are considering global averages. At current atmospheric pressure then globally L> is greater then the energy required to provoke it.
Any excess at the tropics would get shifted poleward and increase L relative to DLR away from the tropics.
But remember that the issue here is not the sun. We are considering the effect of DLR from increased GHGs which is a different kettle of fish as you pointed out to me before.
“Any excess at the tropics would get shifted poleward and increase L relative to DLR away from the tropics.”
The point is Stephen that there can be an excess. That’s all I need to demostrate; in fact I would expect an increase in L relative to DLR away from the tropics. Considering the full cycle of things;
Additional DLR from rising GHG levels is a global effect. In the tropics oceans absorb energy on average before it is transported poleward by currents and generally given up to the atmosphere.
The additional DLR causes even more excess energy in the tropics because L is not capable of keeping up (as the numbers demonstrate) so the shortfall is made up by R. In short, the ocean warms up. This does not mean the dissolution of the cool skin layer; the skin effect may reduce slightly when DLR is higher, but in effect the whole mixed layer is warmer due to knock-on effects on heat transfer.
Anyway, because the ocean’s warmed up, the currents carrying heat away from the tropics are warmer too. Once in the cooler, drier high latitudes L is sure to increase because the ocean is warmer and therefore more energy is available to fund L. At this point I am happy to let you describe a speeding up of the hydrological cycle, but perhaps not here because it’s quite off topic. My response would be: yes, it’s sped up. But it’s done little to nothing to prevent the ocean warming up (that is, a key part of AGW has happened). The ocean warmed up first, that’s what speeds the hydrological cycle up. The effect cannot prevent the cause.
So is there an excess value for DLR over L at the tropics when one considers the diurnal cycle and as one moves away from the tropics, the seasonal cycle ?
The sun only shines for half the day and is low in the sky for part of it. Are you just referring to a short period of peak insolation which is offset later in the day ?
Note that the ITCZ is at the tropics and it responds very fast to reduce solar DLR later in the day.
If any excess at the tropics is matched by an increased deficit of DLR over L elsewhere then there is no net accretion to ocean heat content.
Indeed during any period of excess the only bit that warms is the evaporative layer. The energy does not transmit downwards. It stays at the top until the sun sinks and is then released either in the same locality or nearby. That is why the tropics are so humid. Lots of evaporation at night. Sometimes the sea steams.
So one would get a faster water cycle from increased equator/pole differentials at the surface but a zero net effect on global ocean bulk temperature.
A zero global effect on the ocean bulk from increased DLR due to the evaporative process cancelling it out does not preclude local or temporary variations involving the evaporative layer alone.
To support AGW you have to demonstrate net warming of the ocean bulk globally from increased DLR from GHGs alone. An increase in latitudinal temperature differentials as a result of surface phenomena is not enough because it is that increase in surface differentials that prevents an ocean bulk effect from the extra DLR cause.
What evidence do you have that the bulk ocean gets any warmer as a result of surface heating at the equator from more GHGs alone ?
All the warmth in tropical oceans (indeed all oceans) is acquired from solar energy getting past the evaporative barrier. DLR from GHGs does not get through.
“Note that the ITCZ is at the tropics and it responds very fast to reduce solar DLR later in the day.”
Erm, just skimming here, but there’s no such thing as solar DLR.
DLR is almost entirely, like >99%, emitted by the atmosphere. The L stands for longwave, and almost everything emitted by the sun is counted as shortwave. The dividing line is 4 microns. There is very little solar radiation of longer wavelength and very little terrestrial radiation of shorter wavelength.
Also, have you spotted my post showing that yes, more than 90% of DLR gets past the evaporative layer?
Let me pull it together – my recent posts are scattered up and down this page in various threads.
Stephen says DLR from GHGs doesn’t get past the evaporative layer. IT DOES. See https://scienceofdoom.com/2011/01/06/does-back-radiation-%e2%80%9cheat%e2%80%9d-the-ocean-%e2%80%93-part-four/#comment-8690.
Stephen says the energy for L comes from a variety of sources. He’s right, and you can also have a latent heat flux even if there was no incoming solar or longwave radiation, assuming an unsaturated atmosphere. Using this knowledge I have demonstrated that L is unable to keep up with DLR (see https://scienceofdoom.com/2011/01/06/does-back-radiation-%e2%80%9cheat%e2%80%9d-the-ocean-%e2%80%93-part-four/#comment-8683).
This means that some DLR sticks about in the topmost layer of the ocean, making it warmer than it would otherwise be (not necessarily eradicating the cool skin of course!). Adding DLR makes E, R and L all increase, but E increases with temp and R can ONLY increase when temp increases. This affects heat transfer from below.
I am happy that I have demonstrated that DLR can warm up the skin layer without cooling of the subskin layer as Stephen claimed. Seems his biggest misconception is that all the DLR is absorbed in the evaporative layer – it isn’t.
I am going to put a mathematical clarification into Hypothesis A (and referencing this diagram:
Hypothesis A:
L ≥ DLR
Because the atmospheric radiation is completely absorbed in the first few microns it will cause evaporation of the surface layer, which takes away the energy from the back radiation as latent heat into the atmosphere.
Therefore, more back-radiation will have zero effect on the ocean temperature.
End of Hypothesis A
The mathematical clarification says in an equation what the words are meant to say.
People are free to present hypothesis A2 etc with mathematical clarifications.
StuN said:
“I am happy that I have demonstrated that DLR can warm up the skin layer without cooling of the subskin layer as Stephen claimed. Seems his biggest misconception is that all the DLR is absorbed in the evaporative layer – it isn’t”
I regard both SST(int) and what is now called SST(skin) as the evaporative layer. There may have been some confusion on that.
In the past the skin was regarded as the 1mm deep 0.3C cooler layer which is now called the sub skin.
Measurements show that the SST (skin) layer warms up but the sub skin does not.
The sub skin is 0.3C cooler than the ocean bulk below it so there can be no transmission of energy downward through it as a result of DLR.
The sub skin warms up when DLR induced evaporation and upward radiation is lowest and cools down when DLR induced evaporation and upward radiation is strongest.
The coolness of the sub skin is a result of upward radiation AND evaporation. Increase the rate of either and the sub skin cools more.
I do not see that it has been shown that increased DLR into the SST(int) and SST(skin) causes that lower cooler skin layer to warm up.
AGW theory does require it to actually warm up. I take the position that it either stays the same or cools further when DLR increases. I don’t mind either way.
AGW theory requires a warming of the local environment when DLR increases yet when DLR increases evaporation increases and evaporation always cools the local environment.
So which is it ? Does more DLR cause the sub skin to warm up, cool down or stay the same ?
“I regard both SST(int) and what is now called SST(skin) as the evaporative layer. There may have been some confusion on that.”
Why? Evaporation only happens in the very top most layer, which is much thinner than the skin or subskin layers. As per a link I posted way back*, the skin is the bit that is able to absorb/emit longwave, and the subskin, I suppose, can be described as the transition between the skin and the mixed layer.
*http://ghrsst-pp.metoffice.com/pages/sst_definitions/
Look at their schematics: it is entirely possible for the skin to warm up while the subskin does not, and the skin will still be cooler than the subskin. This would reduce the flux of energy upwards from the mixed layer and as per the original argument this means the mixed layer ‘keeps’ more of its energy. It is only necessary for GHGs to warm the skin layer for this to happen; the energy from GHGs does not need to penetrate the mixed layer in order for the mixed layer to warm up.
“AGW theory requires a warming of the local environment when DLR increases yet when DLR increases evaporation increases and evaporation always cools the local environment.”
Sorry I might have to facepalm for a sec here.
DLR warms stuff up. Evaporation cools stuff down. As per my posts above, it is possible for DLR to increase more than evaporation, and so the warming from the DLR beats the cooling from evaporation, leading to a warming whereby the system is moving towards equilibrium by increasing temperature and hence increasing sensible heat flux and emitted longwave radiation. What the heck could possibly be wrong with that explanation?
StuN said:
“Evaporation only happens in the very topmost layer, which is much thinner than the skin or subskin layers”
True but the skin just below it appears to act as a reservoir for energy that is unable to penetrate lower.
“subskin, I suppose, can be described as the transition between the skin and the mixed layer”
I agree with that.
“it is entirely possible for the skin to warm up while the subskin does not, and the skin will still be cooler than the subskin”
In day to day situations that must be true but in extremis I do not think it would hold because the energy in the skin layer seems unable to pass down to the subskin.
“This would reduce the flux of energy upwards from the mixed layer and as per the original argument this means the mixed layer ‘keeps’ more of its energy.”
This is what really matters. The flux from the mixed layer to the subskin will only reduce if the subskin warms.
There are three possibilities:
i) Energy from the skin can pass down to the subskin and warm it.
ii) The evaporative process having increased draws more energy up from the subskin to compensate for a shortfall in available latent energy from the skin upwards so the subskin cools further.
iii) There is enough energy in the skin to supply the latent energy needed and the subskin temperature remains unchanged.
To substantiate AG W theory you need to demonstrate i).
“As per my posts above, it is possible for DLR to increase more than evaporation,”
That is of no account unless you can warm the subskin and slow upward energy transfer from the ocean bulk. Simply creating a reservoir of energy in the skin is not good enough.
The equilibrium temperature of the system only changes if you change the temperature of the ocean bulk. If the ocean bulk does not change temperature then there will be surface effects only and they will be negated by a faster water cycle for no change in equilibrium temperature.
The water cycle changes from extra DLR will be miniscule compared to ocean and solar induced natural variability.
So, with more DLR does the subskin warm, cool or stay the same ?
“True but the skin just below it appears to act as a reservoir for energy that is unable to penetrate lower.”
The skin is not really a reservoir, more a river. Because there is a temperature gradient there, and the skin is cool, energy is always flowing into the skin where it can escape to the atmosphere.
This means that in general there is a flux of heat, via convection and conduction, into the skin layer from below, because the skin is cooler than the subskin, which is cooler than the mixed layer (let’s consider only night time for simplicity here).
Now consider a situation where DLR increases, and all of the following variables increase too: L, R, E and skin temp. The last 3 increase because L does not go up as much DLR does; I have shown this to generally be the case, at least in the tropics.
This means that the skin is not as cool. The subskin doesn’t see any of the energy from the DLR directly – that’s all absorbed in the skin. But now that the skin has warmed up, the temperature gradient across the subskin begins to decrease, and hence so does upward transport of heat energy via convection/conduction.
I’ll plump for the first answer: increase DLR and the subskin warms, because the subskin temperature depends on the skin temperature.
Note that this doesn’t require i) to be true. Energy is not flowing down from the skin to the subskin, because the former is still cooler than the latter. In fact I’ll go for secret answer number iv)
iv) because there is enough energy arriving at the skin to supply the latent heat AND warm up the skin at the same time, less energy is drawn up from below. The temperature gradient decreases and the subskin warms.
Then why does the subskin cooler than the ocean bulk exist in the first place ?
On the basis of your explanation either it would never have existed or the general background DLR would have got rid of it long ago by offsetting the radiative portion of its causation.
And there are many sources that specifically say that evaporation contributes to its existence so more evaporation must enhance it.
It exists simply because the skin is cooler than the mixed layer. There cannot be a temperature discontinuity, there must be a gradient.
Therefore make the skin less cool and the gradient decreases. This means that the subskin warms up, though the effect is greatest at the top of the subskin.
This can’t be right can it?
Hypothesis A2
Disregard S because the solar energy goes into the ocean beyond the evaporative layer and the skin so in relation to the ocean surface flux solar energy can be better represented by upward conduction and convection otherwise there is double counting.
Thus:
Conduction + Convection (from the ocean)+ DLR
= Radiation + Evaporation + Convection + Conduction (to the air).
More DLR increases Evaporation and that increases the amount of latent heat required and also increases upward Convection because water vapour is lighter than air.
It also appears that more DLR warms the ocean skin (but not the subskin) so Conduction and Radiation increase too.
The combined increases in Latent heat requirement, increased upward Radiation, Convection and Conduction outweigh the increase in DLR so to make up the shortfall extra energy is drawn from the subskin which cools to a temperature lower than that of the ocean bulk.
The subskin descends into the ocean bulk until equilibrium is restored between the energy being pulled out and the solar energy entering the ocean.
The net effect is a reduction in the equilibrium temperature of the bulk ocean because solar energy is being held by the oceans less long than would have been the case with less DLR.
There must be something wrong but what ?
Note that any solar effect on the evaporative layers SST(int) and SST(skin) is accounted for as part of DLR and the rest of solar input is accounted for by the returning energy from upward Convection and Conduction so S has no additional place in hypothesis A2.
Since at current atmospheric pressure the energy required to provoke evaporation at SST(int) is always less than the energy extracted from SST(int) by evaporation in the form of latent heat the net outcome of all the reactions in SST(int) is always cooling.
Hence the creation of the cooler layer of the subskin and the temperature discontinuity at the subskin and mixed layer interface.
ANY extra energy arriving in SS(int) from any source will only ever enhance the net cooling effect within SST(int).
The accumulation of extra warmth in SST(skin) can only be due to the formation of an energy reservoir in SST(skin) that is awaiting processing in SST(int).
Due to the net cooling effect of events within SST(int) and the existence of the lower temperature discontinuity between SST subskin and the mixed layer it is not possible for the warmth in SST(skin) to be indicative of a net warming process within ST(int).
One has to use the word ‘net’ (contrary to what Stu suggests) becuase DLR warms up and Evaporation cools down but what we are after is the final outcome of the total reactions involving DLR and Evaporation.
That is always net cooling within SST(int) under present atmospheric pressures.
I think I can take one further step.
The ‘backing up’ of energy in SST(skin) pending processing in SST(int) seems to be a natural phenomenon yet still we see a temperature discontinuity lower down between ocean bulk (the mixed layer) and SST subskin.
Now if SST (skin) temperature increases for whatever reason that should have an effect on upward energy transfer from below all other things being equal. Therefore AGW could be correct in that a warmer SST(skin) reduces upward energy floiw from below to increase ocean bulk temperatures.
However when SST(skin) gets warmer all other things are not equal. At the same time the reactions in SST(int) increase for a greater net cooling effect within SST(int).
So from extra human GHGs we see both an increase in net cooling in SST(int) AND a temperature increase in SST(skin).
The net outcome is an increase in the speed of energy throughput for the entire system which offsets what would have otherwise been a slowing down of throughput from the ocean by virtue of the greater energy content in SST(skin).
Thus a zero effect on energy transfer from the oceans as a result of more DLR and the consequent maintenance of that temperature discontinuity between subskin and ocean bulk.
So, rather than warming or cooling the oceans the effect of DLR is zero as per opition iii) above
“iii) There is enough energy in the skin to supply the latent energy needed and the subskin temperature remains unchanged.”
Any flaws ?
“Stu N
It exists simply because the skin is cooler than the mixed layer. There cannot be a temperature discontinuity, there must be a gradient.”
Everything I have read suggests a discontinuity.
What makes your scenario implausible to me is that under current atmospheric pressure the energy required to provoke evaporation is less than the latent heat required to effect it so there really cannot be the surplus you suggest.
Anyway, we both have logical and clear positions and to resolve the issue we now need data but I’ve not yet found anything suitable or accurate enough.
Off to bed now. Thank you for a civilised discussion.
I wonder what scienceofdoom will do with hypothesis A2 🙂
Hi Stephen
can you point me to your resources that suggest a temperature discontinuity please?
And yes, when a molecule evaporates it takes lots of energy with it. But (ignoring solar for a sec) there is still DLR absorbed that doesn’t get used up in evaporation. The question is, what happens to that energy?
BTW, did you catch my post correcting your use of the term ‘solar longwave’? For all intents and purposes there is no such thing. https://scienceofdoom.com/2011/01/06/does-back-radiation-%e2%80%9cheat%e2%80%9d-the-ocean-%e2%80%93-part-four/#comment-8693
Hi Stu,
I think you must have meant at the subskin/skin interface whereas I was referring to the mixed layer/subskin interface so I think we were at cross purposes.
Yes I did take note of the solar point, thank you. If you look at my hypothesis A2 you can see that that led me to replace S in scienceofdoom’s equation with upward conduction and convection from the ocean.
As for what happens to the surplus DLR from above well all it does is replace the energy taken from other sources for a zero effect overall.
If you look through my comments following that of 2.21 am and follow the logic I think you will find that with yoiur help I have squared the circle.
To summarise:
“The cool subskin exists because there is a net cooling from Evaporation in the absence of any DLR at current atmospheric pressure.
When DLR is applied the increase in energy loss from SST(int) speeds up the flow and offsets the increase in energy gain in SST(skin) which slows down the flow for a zero net effect on the rate of flow from the mixed layer to the subskin and thence out to space.
So however much DLR is applied from whatever source the rate of energy flow out of the ocean does not change. That rate of flow remains completely dependent on the atmospheric pressure which under current conditions causes evaporation to have a net cooling effect.”
As far as I can see I have resolved all outstanding issues and dealt with your various objections from along the way.
Whether you agree or not is another issue 🙂
TimTheToolMan on January 10, 2011 at 12:28 am:
I don’t understand your comment.
But can we separate out two effects:
1. Momentum – the effect here is that a “parcel of water” that has begun moving upwards because of its buoyancy will continue moving upwards even when it encounters exactly equal density water – due to its momentum.
This is what you are saying?
2. The bit I don’t understand in your last comment (or others).
Because you are taking issue with my model and explanation you appear to be saying that (for some reason) the increase in temperature of the top few microns due to a (hypothetical) increase in DLR will have zero affect on the convection from below.
Sorry for putting words in your mouth, it is my way of getting clarification.
Is that what you are saying?
If so can you explain why?
If not, can you explain exactly what changes when the temperature of the top few microns increases.
Maybe the diagram you promised? If the drawing in the comment was the diagram I don’t understand the point you are making.
And back to point 1, when I understand your point 2 we can see how to investigate the effects of momentum.
Stu N on January 8, 2011 at 3:35 am:
Thanks for the link, I found this comment the most illuminating – more so than the article:
As regards the existence of the cooler subskin I think I can now pin that down too.
It represents the net cooling effect of evaporation at current atmospheric pressures in the ABSENCE of any DLR at the surface.
It is a constant and unavoidable minimum baseline for ocean to air energy flow set up by the physical characteristics of the system.
If one then starts to introduve DLR whether from natural or human sources the subskin never changes for the reason I set out above so the energy flow from the oceans doesn’t change either.
Essentially DLR both increases net cooling within SST(int) and increases the energy reserve in SST(skin) but because the two effects cancel out by each varying the speed of throughput equally but in opposite directions there is never any effect on the subskin or the speed of energy flow from the oceans.
Voila.
Stephen Wilde on January 10, 2011 at 2:21 am:
So you have reproduced the equation that everyone has to agree with, discarded clarification “L ≥ DLR” (January 9, 2011 at 11:28 pm) of Hypothesis A, and claimed that changes in L can go into DLR , Convection, Conduction and Radiation from the surface – but not change the conduction and convection from the ocean depths (effectively S=solar, from my diagram) – thereby embracing Hypothesis B.
You haven’t proved it. You’ve just claimed it. And there is a lot wrong with it.
Increases in DLR can definitely cause an increase in latent heat, sensible heat, and radiation – and will.
But usually adding energy doesn’t cause more to be removed (local positive feedback).
Usually when you add energy to a surface it moves to a new higher equilibrium temperature, not a lower equilibrium temperature.
And what causes latent heat flux?
There is an equation which mathematically describes the value of latent heat flux:
where L=latent heat of vaporization, ρ= density of air, qs = specific humidity at the surface, qr = specific humidity at a reference height, usually 10m, CDE = empirically measured aerodynamic humidity transfer coefficient (typically around 10^-3 over the ocean), Ur = wind speed at the reference height.
This is found in Global Physical Climatology, Hartmann, Academic Press (1994), but is also found in almost identical form in all standard textbooks.
There is a similar equation for sensible heat, but with a different coefficient and temperature difference replacing humidity difference.
The equations aren’t:
ΔR + ΔL + ΔS = ΔDLR
Usually the surface just above the ocean is already saturated. The reason why other terms appear in the equation above is that without some movement of air, latent heat flux, L = 0. Because the atmosphere is saturated.
So the equation has a term for the humidity 10m (or some other reference height) above the ocean surface, and the wind speed (because that is the main mechanism for moving the less saturated air to the surface).
That’s why the claim that increases in DLR automatically cause it all to disappear (and possibly more) is flawed. L & S depend on the movement of air above, the temperature and the humidity. R depends only on the surface temperature change (Stefan-Boltzmann law).
I think that my subsequent posts resolve the outstanding issues.
The cool subskin exists because there is a net cooling from Evaporation in the absence of any DLR at current atmospheric pressure.
When DLR is applied the increase in energy loss from SST(int) speeds up the flow and offsets the increase in energy gain in SST(skin) which slows down the flow for a zero net effect on the rate of flow from the mixed layer to the subskin and thence out to space.
So however much DLR is applied from whatever source the rate of energy flow out of the ocean does not change. That rate of flow remains completely dependent on the atmospheric pressure which under current conditions causes evaporation to have a net cooling effect.
“1. Momentum – the effect here is that a “parcel of water” that has begun moving upwards because of its buoyancy will continue moving upwards even when it encounters exactly equal density water – due to its momentum.
This is what you are saying?”
Correct. Or even warmer water such as may be found in the top 10 microns.
So the only part of the ocean where DLR will effect convection is the last 10 microns and then the momentum of the convection only needs to get the water into that region where it will then radiate or evaporate.
The whole rest of the ocean depths is unaffected by the DLR increase for convection. So its the top 10 microns vs tens or even hundreds of meters of unaffected convection and the top ten microns is the destination anyway.
You suggest I’m saying “you appear to be saying that (for some reason) the increase in temperature of the top few microns due to a (hypothetical) increase in DLR will have zero effect on the convection from below.”
And you’d be pretty much right. As I said before, I can see why you believe it but my gut feeling is that the ability of DLR to affect ocean temperatures will be very small indeed. Essentially none compared with the warming mechanism in hypothesis D. And I dont think your model models it.
So back to hypothesis D…the effect of increased DLR in that hypothesis is immediate and direct. An increase in DLR DIRECTLY effects the amount of radiation the ocean radiates and hence how quickly it cools.
Hopefully I’ll get a diagram to you soon but I’m still debating with myself the best way to show it since my words dont seem to have worked…
Just to head people off at the pass because I know they love to take things out of context, when I say
“my gut feeling is that the ability of DLR to affect ocean temperatures will be very small indeed.”
I mean it in the context of DLR effecting the ocean skin temperature which in turn effects the ocean’s ability to convect heat to to skin as per my ongoing discussion with SoD.
I’ll try one last time before resorting to Visio and finding a place to host a picture… This story ignores evaporation because when you see the mechanism its easy to see where evaporation fits in….
Imagine Stefan and Boltzmann looking down upon the ocean and saying “You must radiate according to your surface temperature …and in this case I see that it must be 100W/m2 !”
DLR is being absorbed at the skin and some is being re-radiated. The re-radiated bit looks up at Stefan and Boltzmann and says, “Hey I’m doing my bit, see I’m radiating 40W/m2”
The ocean looks up at Stefan and Boltzmann and says “Sigh, I guess than means I have to provide 60W/m2 to keep you happy then, dont I”
Does that help you to see the mechanism?
You’re very close. It’s the “some is being re-radiated” that’s the problem. The DLR is absorbed and the surface emits. There is no re-radiation. Strictly speaking there is a small probability that a molecule at the surface that has just absorbed a photon will emit again before it can transfer the energy by collision with other molecules, but that probability is very small, < 1E-04. The surface would emit in the absence of any DLR. The DLR would be absorbed in the absence of emission. The difference between the DLR absorbed and the surface emission is the amount that has to be supplied.
“You’re very close. It’s the “some is being re-radiated” that’s the problem. The DLR is absorbed and the surface emits. There is no re-radiation.”
There are two options for DLR absorbed at the surface. It can either be radiated again or it can be used to perform evaporation. Thats it. Its not part of some “consolidated ocean heat figure”
Now technically there are other paths off the surface such as conduction to the atmosphere (but not the bulk) or mixing down through wave action…but the vast majority is going to be radiation or evaporation.
TimTheToolMan on January 10, 2011 at 1:46 pm:
No, I’m not sure what you are saying. You can email the picture to me if you like: domain gmail.com, the first bit is scienceofdoom and I will host it here.
With your latest comment I think – well, if that’s more or less your point of view then what you are saying is more DLR causes less energy to leave from the ocean depths. But I didn’t think that was your point of view.
Anyway, increasing energy input in one form causes a new equilibrium temperature. This new temperature will be dependent on how energy out varies with temperature.
Rereading your comment, scratch my 2nd paragraph, as I’m just not sure of your point.
A before and after diagram would help.
Stephen Wilde on January 10, 2011 at 1:19 pm:
No they don’t.
Can you write out a the equations that you believe govern energy movement through the ocean surface.
That way your idea can be first, understood and second, tested.
scienceofdoom said:
“Anyway, increasing energy input in one form causes a new equilibrium temperature. This new temperature will be dependent on how energy out varies with temperature”
There need be no new equilibrium temperature for the system as a whole (as opposed to just the ocean skin) if energy output in another form cancels it out and if the process involved is not wholly temperature dependent as is the case with evaporation.
More DLR does two things:
i) It increases the temperature of SST(skin) which, all other things being equal, would reduce the rate of flow from the subskin as you suggest
but all other things are not equal
ii) At the same time it increases evaporation and therefore net cooling in SST(int) which increases the flow through the lower layers again thus cancelling i) for a zero net effect on the rate of energy flow from the subskin.
The evaporative process in changing energy from one form to another and in doing so accelerating throughput again overrides the ‘normal’ application of Fourier’s Law and thereby cancels out the anticipated slowdown from a warmer skin alone.
More DLR therefore fails to achieve a net slowdown in energy throughput and the equilibrium temperature of the subskin and bulk ocean fails to rise despite the rise in temperature of the ocean skin.
The reason is that the energy exchange involved in evaporation is not solely temperature dependent
It is pressure that determines whether the energy required to provoke evaporation exceeds or not the latent energy required by the process and if so by how much.
We have already determined that at Earth’s current atmospheric pressure there is always a net deficit to the local environment when evaporation occurs.
So the existing pressure regime permits changes in the rate of energy flow independently of temperature such that the warmer skin does not change the equilibrium temperature of the ocean bulk.
That is also how evaporation contributes to the creation of the temperature inversion between ocean bulk and subskin in the first place so nothing novel is being proposed.
In fact one can regard the evaporative effects between SST(int) and SST(skin) as a second stage repeat of the evaporative effects causing the inversion between the ocean bulk and the subskin.
In both cases energy flow is being accelerated upwards independently of temperature.
The combined effect is to prevent DLR from changing the equilibrium temperature of the ocean bulk.
“Can you write out a the equations that you believe govern energy movement through the ocean surface.”
No I cannot unfortunately because I am not a trained scientist. My field is words and concepts rather than equations but I’m quite sure it could be done by someone suitably experienced.
My purpose here is to clarify things in my own mind in the face of criticism and I am satisfied that I have made progress here.
As for testing, the observations are what they are and as long as the verbal narrative fits the observations then that is enough for me at the moment.
I’ll try another tack, hoping not to bore everyone.
1) Evaporation is primarily pressure and density dependent so on Earth the energy required to provoke it is always less than the energy taken out of the local environment when it occurs.
2) That establishes an energy flow gradient towards the point at which evaporation is occurring, in this case the ocean surface.
3) That gradient overrides all the rules that govern conduction and convection. They are primarily temperature driven and are of no account in the face of evaporation.
4) The dominance of evaporation in the ocean to air energy flow scenario results in two apparently perverse phenomena that are contrary to the rules of conduction and convection.
5) Firstly the upward energy gradient caused by evaporation at the ocean surface pulls energy up from the subskin faster than it can be delivered to the subskin by convection and conduction from the ocean bulk. That causes the temperature inversion between ocean bulk and subskin.
6) Secondly that same upward energy gradient caused by evaporation pulls energy out of the heated ocean skin fast enough to prevent the warmer temperature of that skin from slowing down the energy flow from the subskin below.
7) The net outcome is a balance between DLR plus energy from below reaching the ocean surface and a variety of types of energy leaving the ocean surface with neither a surplus nor a deficit.
8) Note that because evaporation on Earth is always a net cooling effect there needs to be an adjustment to achieve that balance which includes solar energy input to the ocean. That is the function of the cooler subskin through which energy has been accelerated up to the skin to ensure that there is sufficient energy in the skin (neither too much nor too little) to exactly supply the amount of energy required for the evaporative process despite the net cooling effect of evaporation.
So more or less DLR from any source can never affect the rate of energy flow from ocean bulk through the subskin to the skin. That will always be set at the rate of flow that would obtain if there were no DLR at all.
None of this is novel thought. It is simply a common sense application of well known physical principles to observations.
It puzzles me that it is not obvious to those far more experienced in scientific matters than me.
“With your latest comment I think – well, if that’s more or less your point of view then what you are saying is more DLR causes less energy to leave from the ocean depths. But I didn’t think that was your point of view.”
My point of view explicitely excludes DLR “general absorbtion” and consequent heating of the ocean. DLR isn’t an input of energy “into” the ocean. DLR doesn’t heat the ocean.
DLR strikes the surface and then re-radiates or DLR strikes the surface and provides energy for evaporation. Hence DLR is responsible for decreased cooling only. Never heating.
This, I think, is where my view differs from others.
“DLR strikes the surface and then re-radiates or DLR strikes the surface and provides energy for evaporation. Hence DLR is responsible for decreased cooling only. Never heating.”
I think I agree. But DeWitt has a good point that immediate re-radiation is very unlikely, but likewise collisional energy transfer is unlikely to carry the energy from absorbed DLR out of the skin layer, so it is still available to make up the ‘Stefan-Boltzmann’ requirement.
Hence if you increase DLR, as you say, you decrease cooling. You reduce the amount of energy being drawn up from below to make up R, and if you have a source heating everything from subskin and below, like the sun, more of that energy hangs about to warm the mixed layer. And, if I’ve understood it correctly, this is what the Realclimate article claims is happening.
“But DeWitt has a good point that immediate re-radiation is very unlikely”
There may be an average time for the energy from individual photons to kick around on the surface before radiating again, maybe of the order of milliseconds to seconds but its a bit of a mute point.
“collisional energy transfer is unlikely to carry the energy from absorbed DLR out of the skin layer”
This is conduction and it cant because that would violate the laws of thermodynamics by taking energy from a colder place to a warmer one. Refer back to the hook shaped temperature profile if needed.
Convection is in the wrong direction too.
Basically there is no physical way for the energy to get to the bulk (short of wave action for example)
” if I’ve understood it correctly, this is what the Realclimate article claims is happening.”
I dont think so. If its the RC article of the Minnett experiment you’re talking about, then thats more along the lines of the SoD hypothesis where increased DLR causes an increased temperature gradient at the surface which in turn makes it more difficult for the heat to pass through.
Is that the article you mean?
The RC article claims a decreased temperature gradient as a result of absorbing more DLR, not increased. From the article:
“Reducing the size of the temperature gradient through the skin layer reduces the flux [of energy from below the skin layer to the atmosphere]”
So long as the skin remains cooler than the ocean below it (the hook shape you refer to), the absorbed DLR wont directly heat the mixed layer/bulk ocean (same diff). It does however reduce the amount of energy flowing out of the mixed layer.
Do we agree on this?
“The RC article claims a decreased temperature gradient as a result of absorbing more DLR, not increased. From the article:
Do we agree on this?”
Yes sorry, reduced, not increased – more uniform like the picture I drew some posts above. My bad.
But essentially its an argument of reducing the amount of energy that can leave the ocean by making it harder to get out. This ignores the fact that firstly getting the heat to the surface might not be a limiting factor and secondly increased surface temps mean increased emission anyway.
Their whole hypothesis is dubious in my opinion.
I probably wasn’t clear on whether I agreed whether the RC (and SoD) temperature gradient hypothesis must mean increased ocean temperatures and the answer is no I dont agree.
Here is an analogy.
Suppose you receive shipment of a truckload of apples a day. You eat as many as you can but really hardly make a dent in them and most of the truckload is sent back at the end of each day.
So now suppose the truck only comes and goes every two days. You still eat like crazy but still at the end of the second day the truck still goes back pretty full.
There is no evidence to suggest heat reaching the surface is a limiting factor to keep it “optimally” radiating. Indeed the shape of the hook suggests to me that its always “full”.
So will the RC (and SoD) hypothesis result in a significant reduction in cooling? The answer is unknown and I think not.
SoD has tried to show it with his model but so far the model is severely lacking and I dont think answers the question.
Your analogy has left me more confused, sorry. What do the apples represent? Who is the eater? Hmm, maybe I should not be commenting here after a couple of fine scotches 😉
Look, the way I’d put it is as follows: the energy that gets out of the ocean in the form of L, E and R all has its final port of call in the top 10-20 microns of the ocean. The source of this energy is a combo of DLR and energy conducted/convected upwards from the bulk ocean. If DLR is able to provide more of it, energy conducted/convected upwards provides less of it.
But energy input from the sun hasn’t changed. If less energy is being conducted/convected upwards then it accumulates in the bulk ocean. And, as per my previous posts on temp gradients, if the temp at the bottom of the skin layer (i.e. top of mixed layer) warms, then energy flow out of the mixed layer again increases as the ocean heads towards a new, warmer equilibrium.
” maybe I should not be commenting here after a couple of fine scotches ”
I think thats the best time. For anything really.
“What do the apples represent?” The apples are the energy, you as ther eater represents radiating them away.
The truck is the convection to get them to you…
Maybe I could rephrase it in terms of single malt scotches?
I think I agree with your summary except this suggestion “If less energy is being conducted/convected upwards ” which isn’t necessary or in my opinion significant.
Energy (ie warmer water) can convect upwards all it likes but it can only leave* through the skin via evaporation or radiation which (I believe) is the limiting factor.
*usual caveats on other minor mechanisms.
“I think I agree with your summary except this suggestion “If less energy is being conducted/convected upwards ” which isn’t necessary or in my opinion significant.
Energy (ie warmer water) can convect upwards all it likes but it can only leave* through the skin via evaporation or radiation which (I believe) is the limiting factor.”
Hmm. As you said earlier, upwards is the only way it goes because of the temp gradient. Once it’s up there, enough of it has to be leaving the ocean in order to maintain the temperature gradient that caused it to get there in the first place.
So it can’t convect upwards ‘all it likes’, because if it does so without the energy then leaving the ocean, it will eventually reverse the temp gradient and convect back down again.
“So it can’t convect upwards ‘all it likes’, because if it does so without the energy then leaving the ocean, it will eventually reverse the temp gradient and convect back down again.”
This is why I reckon its limited to the radiation from the surface and not the convection. The convection will happily keep energy energy available to the surface as its emitted.
And again, remember its only the top 10nm of the ocean where there is any question of “decreased” convection.
Stephen, re: discontinuities, it’s possible you’ve misunderstood what it means.
A discontinuity in the temperature gradient means, in the simplest terms, that if you were to plot a graph of temperature with depth, it would not result in a smooth curve but would have a ‘break’ in it, see the examples here: http://en.wikipedia.org/wiki/Discontinuity_%28mathematics%29
I have a feeling you think it means the temperature gradient reverses, as in the ‘daytime’ graph of this link: http://ghrsst-pp.metoffice.com/pages/sst_definitions/ and perhaps if you had not encountered the term before, that’s understandable.
I first mentioned it when you asked why the cool subskin exists; I said there cannot be a discontinuity because you see that if the surface is 0.5C cooler than 1mm below the surface, there will be a gradient such that, eg, 0.5mm below the surface it will be 0.25C cooler than at 1mm depth. In this case the gradient is a consequence of heat transfer by convection/conduction. Let me see if I can draw an example.
Continuous temperature profile:
Surface \
\
\
\
\ 1mm
Discontinuous temperature profile:
Surface |
|
|
|_____
|
|
| 1mm
In meteorology (and many other fields), there are often very sharp gradients that might appear to be discontinuities when considering a large scale, eg. the thermocline in the ocean, but on the macroscale at least (i.e ignoring the molecular scale so that terms like ‘temperature’ have some meaning) they do have a defined gradient.
Examples of where there is a discontinuity would include things like refraction, whereupon entering a new medium a beam of light immediately has its direction of travel altered. There is a discontinuity in its path. I think that’s a legit example; any physicist disagree? I wont mind being corrected 😉
Damn, formatting fail. Ignore my examples there, though they looked alright when I was writing my comment 😉
Just look at the examples in the link, the give the same general idea (first image in particular).
I had the impression that the boundary between the subskin and the ocean bulk was a sharp discontinuity because it is a temperature inversion.
No matter. My latest post is more important.
I think it would be fruitful to look very closely at the interface between SST(int) and SST(skin).
It is necessary to get a clear idea as to exactly why the higher temperature of SST(skin) fails to slow down the rate of energy flow from the subskin below.
A. The Default situation
i) Evaporation occurs primarily because of pressure and density differentials between water and air. Thus it will occur even if both water and air are at the same temperature. The process of evaporation is not dependent on any temperature differential. There are other influences that will increase or decrease the rate of evaporation but they need not concern us here.
ii) At Earth’s atmospheric pressure the energy required to provoke evaporation is always less than the energy taken from the local environment when evaporation occurs so we need to analyse exactly where the deficit can be provided from.
iii) In the absence of DLR it is taken from the water below because the water is generally warmer than the air hence the development of a layer of cool water 1mm deep and 0.3C cooler than the ocean bulk below.
B. When DLR is added to the mix.
i) DLR in itself does nothing. Before it can warm anything it must be absorbed by a water molecule.
ii) When DLR impacts the water surface some molecules will evaporate immediately and others will need to wait a moment to acquire enough additional energy.
iii) Those which are in the process of evaporating form SST(int). Those which are busily acquiring energy form SST(skin). The molecules in SST(skin) steadily gain more energy and move upward towards SST(int). In the process they gain more energy and become warmer with sensible energy that registers on our sensors.
C. The response to DLR once evaporation from DLR begins.
i) The molecules in SST(int) evaporate producing a local energy deficit. The energy most readily available is in the nearest molecules of SST(skin) so a flow of energy is set up from SST(skin) to SST(int)
ii) That energy flow is upward so the additional energy being supplied to the molecules in SST(skin) cannot flow downward to increase the temperature of the subskin.
iii) We then have both energy AND individual molecules moving upwards towards SST(int)
iv) The DLR cannot penetrate beyond SST(skin) so ALL the DLR gets absorbed by molecules in that region and ALL those molecules in due course find their way to SS(int). Thus there is no surplus energy from DLR left over to warm the subskin and even if there were it is flowing in the wrong direction.
v) Meanwhile remember that there is a net deficit to deal with when evaporation occurs. If ALL the DLR is now in molecules that are going to move upward and evaporate it can only be provided by a cascade of energy from molecule to molecule up through SST(skin).
vi) But at the bottom of that cascade where SST(skin) has it’s interface with the subskin there is still going to be that deficit. That remaining deficit must be accounted for and it already has been provided by the pre-existing upward flow of energy from the ocean bulk below which is always present even in the absence of DLR
vii) Additionally that energy is already of the correct quantity to make up the deficit because the DLR is ALL accounted for in the process of accelerated evaporation leaving the background equilibrium undisturbed.
Thus DLR in any quantity or from any source cannot alter the equilibrium temperature of the oceans.
Actually query Part C para vii)
Any ideas on that ?
Wouldn’t the extra evaporation from the DLR mean that the background flow of energy isinsufficient ?
Scrub that query. The extra evaporation from the DLR would be self limiing in that if it is used to cover part of the deficit then the amount of evaporation would decrease so only the background deficit needs to be covered.
Stephen and TTTM,
Water is a black (grey) body. Backradiation is energy. What you are saying is that a body receiving some amount of energy B, that is greater than some amount of energy A, will have exactly the same temperature when it is receiving A as when it is receiving B. That’s just silly.
Please, feel free to conduct Steve’s experiment after taking DeWitt’s bet from 3 days ago.
Also, do you have an alternate source for the energy represented by the ocean’s increasing heat content?
See my comments at 12.23 a m.
You ave not interpreted my words correctly. The sensors tell us there is a rise in temperature. The question is whether or not that slows down energy transfer from the bulk ocean.
I explain why it does not because the energy transfer mechanism in command is evaporation which does not obey the rules of convection and conduction.
Stephen Wilde
January 10, 2011 at 8:46 pm:
That’s why your ideas – as for example from January 10, 2011 at 8:39 pm – are flawed.
The mechanisms that govern the flow of heat via radiation, conduction, convection and latent heat movement are all described mathematically and describing what you believe is happening without reference to the actual values of heat transfer means that you will be – with 99.999% certainty – wrong. That’s because the chance of hitting on the right values for all the different effects by chance is extremely low.
And chance is what it is when you have a vague idea about a number of inter-related mechanisms but no concrete value for any of them.
For anyone else reading these comments and wondering whether Stephen is on to something I recommend reading that particular comment from January 10, 2011 at 8:39 pm and asking yourself:
– “how is Stephen Wilde so sure about the relative values of each of these elements?”
It’s great to have ideas, but ideas that can’t be tested are not science.
Stephen Wilde, you are welcome to write your ideas here, but I think there is little point me responding to them.
All that is necessary in connectionwith most scientific questions is to ascertain direction of trend for a phenomenon and any absolute limits on the behaviour of a phenomenon.
Words and concepts are enough for that. Equations are fine if you have data and wish to go into all the aspects of the behaviour of a phenomenon but for most of this climate stuff the data is not available, the direction of trends is not known and the absolute limits on behaviour are not clear.
My comment at 12.23 a. m. deserves proper consideration.
TimTheToolMan on January 11, 2011 at 12:28 am (and a few preceeding comments):
1. Energy absorbed is energy absorbed – once DLR is absorbed it is indistinguishable from any other source of energy in water molecules.
2. There is an equation which describes how energy is radiated from the surface, as well as equations which describe other mechanisms of heat transfer.
3. More energy input of course leads to a higher equilibrium surface temperature but calculating the new equilibrium temperature as a result of more energy input involves solving some simultaneous equations.
Which is what is interesting about the whole discussion.
On the one hand it seems obvious to people that the subject is so simple that my analysis is wrong, yet on the other hand it “much more complex” than I have described.
Solving non-linear simultaneous equations isn’t an easy process. I needed a maths program to do it.
And I agree that my model is quite simplistic.
But if the “correct” answer is so obvious it should be easy for others to demonstrate different results.
That doesn’t mean I don’t think the momentum question doesn’t need to be resolved.
But it’s important to break apart the different elements of the problem.
“1. Energy absorbed is energy absorbed – once DLR is absorbed it is indistinguishable from any other source of energy in water molecules.”
Obviously. I make the distinction because I’m making the point that DLR doesn’t “enter the ocean” and the ocean provides the required energy to satisfy S-B.
This is quite distinct from a proportion of the DLR from the moment its incident on the surface (or within moments anyway) plus energy from the ocean.
I think thats a rather important distinction to be clearly making in a hypothesis because if DLR radiation is logically kept distinct from ocean supplied energy then it cant be added to ocean heat in any way.
re: Equations and solving them.
Yes, of course. The solution to ocean heating is very complex because it incorporates most aspects of the climate including clouds, evaporation/rainfall, wind and aerosols none of which are well understood.
“That doesn’t mean I don’t think the momentum question doesn’t need to be resolved.
But it’s important to break apart the different elements of the problem.”
Unless you disagree with Hypothesis D, and I haven’t heard you disagree nor give reason to disagree, then the obvious answer to ocean heating is staring you in the face.
Why not investigate that first and then look at subtle temperature gradient/convection adjustments once you have got the obvious one out of the way first?
“For anyone else reading these comments and wondering whether Stephen is on to something I recommend reading that particular comment from January 10, 2011 at 8:39 pm and asking yourself:
– “how is Stephen Wilde so sure about the relative values of each of these elements?”
Simple.
Observations demonstrate the relative values.
The precise quantities need not be known, only the relative values. I am not saying anything that would not have been readily understood and accepted 50 years ago.
Everyone knows or should know that the energy transfer values in the phase changes of water trump all other day to day physical processes.
Read my post at January 11, 2011 at 12:23 am and tell me where it is wrong.
Stephen Wilde,
I’m going to break my rule because this one’s too good to pass up:
Several of us here are suitably experienced yet you refuse to believe us when we say that it can’t be done.
If you’re so good at words and concepts, define entropy and give us your interpretation of how it relates to the Second Law of Thermodynamics. No cribbing from Wikipedia.
It never ceases to amaze me how someone who admits he has no training can claim to have found something that tens of thousands of people who have trained for years and earn their daily living from their training have somehow missed for decades.
DeWitt,
Intellectual snobbery is so unbecoming.
Read my post at January 11, 2011 at 12:23 am and tell me where it is wrong.
Stephen Wilde:
It is hard to find anything falsifiable.
That means there are no specific claims to test.
Let me give you an example of a falsifiable claim:
[included in the claims would be a diagram preferably, or an explanation of what the “top layer” meant, e.g. the 10um surface layer, then other parameters like the period of time under consideration and so on]
That would at least be a start because then your claim could be evaluated against well-known theory.
Statements like:
Are not statements that I can even understand. What does “provoke evaporation” mean?
In fact evaporation occurs spontaneously when an air surface that is not saturated is above water. The rate of evaporation is determined by a number of parameters including the air temperature and the specific humidity.
Is “we need to analyse exactly where the deficit can be provided from..” a claim?
If it is a claim it is false.
In fact we don’t need to analyze exactly where the deficit can be provided from. Internal energy in the water is the source. So heat moves from the ocean to the air when the air is unsaturated and that removes heat from the ocean.
If the ocean surface receives no energy it will cool down. If the ocean surface receives more energy than is removed it will heat up. If the air stays saturated immediately above the ocean surface the latent heat flux will be zero.
Where have you accounted for it?
I see no numbers. I see no energy balance calculations.
Somehow you have determined that whatever increase in DLR takes places actually causes more heat to leave the ocean via evaporation.
But there are no rate equations. There are no
Eout = sigma x T^4 + Cl.L.(qs-qa)…
calculations
For example, if the ocean temperature increases from 25’C to 30’C and the specific humidity of the air above the ocean = 25g/kg:
– what is the change in emission of thermal radiation?
– what is the change in latent heat flux?
– what is the change in sensible heat flux?
Can you do tell me the answer? Without a calculator? I need you to give me a result.
Which one changes the most? By how much?
How will you do it?
Researchers in the 1800’s found it necessary to write down equations and use numbers because saying:
– didn’t work out too well.
You need to draw a diagram. The diagram needs temperatures, length scales, heat flows and a few equations.
If you can’t write that down, why on earth do you think you have “solved” the problem?
Try flying a spaceship to the moon. It’s not so hard. It can be done without equations. Just work out roughly how long you think it might take in days, have a stab at how much the moon moves over that time, swing the launch pad over roughly that far in that direction and press the “Go” button.
They don’t need those big computers. It’s just simple gravity.
Ok, but which bit of my narrative is inaccurate ? Your failure to understand the terms of expression does not necessarily indicate that I am wrong for example:
You asked:
“What does “provoke evaporation” mean?”
Well I started off by pointing out that under current atmospheric pressure and with no temperature differential evaporation occurs routinely with no ‘provocation’. However a suitable change in the basic conditions or an additional energy input will obviously ‘provoke’ more evaporation than would otherwise occur.
So if the basic conditions remain the same and more energy is added then more evaporation will be ‘provoked’ and the basic conditions on Earth dictate that for a unit of extra energy provoking that additional evaporation the amount of energy required in latent form will be greater for a net cooling effect in the local evironment.
That is not ‘rocket science’. It is basic, well known and accepted science. I am with the historical mainstream on all this but you are not.
Do you have an equation that suggests that any of the assumptions I have made or outcomes I have described are incorrect ?
Since my narrative is dependent on several assumptions it would be readily falsifiable on a number of grounds.
All my narrative consists of is a series of descriptions of real world events, easily observable, and linked together. in a logical sequence.
Which portion(s) are a false description of the observed reality ?
Everything I have set out is established science, or was some 50 years ago.
Have we actually gone backwards ?
Are your equations just a collection of irrelevancies and obfuscation, missing the most basic points about real world events ?
The spaceship to the moon analogy is not remotely comparable.
50 years ago any number of scientific professionals, even properly educated schoolchildren would have set out the narrative that I have set out and the scientific establishment would have accepted it as obvious from known scientific principles at the time.
Kindly read it again and tell me which portions are wrong and why.
Even if only one aspect is wrong the rest crumbles so it is worth your while trying.
And if you are right I will be grateful because I will be better educated.
“If the ocean surface receives no energy it will cool down. If the ocean surface receives more energy than is removed it will heat up.”
Although I expect you’re just making a point rather than describing an actual climatic state, I’d just like to point out that DLR doesn’t excede the radiative output from the ocean unless externally provided (which doesn’t happen in nature)
DLR “comes from” the radiative output of the ocean. The ocean radiates up, the clouds and GHGs catch it and radiate it back down minus losses to space primarily.
The sun cant provide any significant IR because ocean surface IR is maximised when there are clouds and the sun’s contribution to IR at that time is minimised.
The only practical exception I can think of might be precisely correct conditions right on the coastline where land based IR results in DLR that is greater than the ocean emission right at the shoreline.
Just something to keep in mind when contemplating ocean heating…
Yes, I’m just making a point in passing as part of explaining the real basics.
Your “narrative” fails because, among other confusion, it appears to claim (note 1) that latent heat flux increases the same, or more than, (note 2) the increase in DLR.
Based on what principle?
When you can provide the answers to this question from earlier:
Then we will have made progress.
Why?
Because you will have realized that to give the answer to how much latent heat flux and how much radiated heat flux and how much sensible heat flux changes you can’t do it with words.
You need to calculate.
When you provide the answer to my question we can move forward.
By the way, if you can’t answer it, what makes you think you have the right answer to the more complex question behind the article?
Note 1 – I think this is what it says but the claims are vague so I can’t be sure.
Note 2 – as note 1
SoD said:
“Is “we need to analyse exactly where the deficit can be provided from..” a claim?
If it is a claim it is false.
In fact we don’t need to analyze exactly where the deficit can be provided from. Internal energy in the water is the source. So heat moves from the ocean to the air when the air is unsaturated and that removes heat from the ocean”
You are utterly wrong.
Evporation takes the eneregy it needs from the most readily available source.
If the air is warmer than the water more will be take from the air and vice versa.
On balance in the real world energy is taken mostly from the water because in most situations and at most times on Earth the water is warmer than the air due to solar input.
At least you accept that when there is evaporation the net flow is from ocean to air but you do not follow through. On that basis more evaporation means faster flow from the ocean does it not ?
Yet more DLR means more evaporation and you then try to say that the flow from the ocean slows down.
Don’t you see the illogicaliy of that ?
In fact as I say the effect of DLR on the flow from the ocean is zero. Initially I though it could be increased but now that I have analysed the facts I see that there can be no change either way.
DLR does not and cannot affect the rate of energy flow from the oceans. Period.
It is for you to prove that more DLR causes a slowdown. You have not done that, nor has Realclimate or anyone else either by way of equations or observed data because the observed warming of the skin layer does not necessarily slow down the rate of flow due to the countervailing effect of increased evaporation higher up
So do not preach to me about the appropriate means of expression in scientific matters.
I’m not sure whether this was directed to me or Tim but SoD said:
“Your “narrative” fails because, among other confusion, it appears to claim (note 1) that latent heat flux increases the same, or more than, (note 2) the increase in DLR.
Based on what principle?”
My reply:
Evaporation is a net cooling process. Capice ?
I can see why evaporation was not adequately dealt with in your initial lengthy treatise.
I’ll expand on this a bit:
In practice the latent heat flux turns out to be the same as the extra DLR because some of the energy provided by the DLR to the molecules in the skin goes to address the deficit caused by the extra evaporation induced by the DLR.
That better describes why the net effect of the extra DLR on the energy flow from the ocean is zero.
The DLR provides the energy needed both to provoke more evaporation and supply the deficit from the extra evaporation which it induces.
That helps to clarify points vi and vii in my narrative. Because the DLR effect is a zero sum process the only remaining deficit to be accounted for is the natural background one from ‘unprovoked’ evaporation.
Hence the upward flux from the ocean stays exactly the same despite the higher temperature in the skin.
I was unsure as to whether I had explained that fully enough so I thank you for the opportunity.
SOD, Steve, TTTM: Some readers seem to be having difficulty because of the traditional practice of separately considering DLR and upward long-wavelength radiation (ULR). DLR-ULR is the net loss from the skin of the ocean due to long wavelength radiation. DLR and ULR are often both 300+ W/m^2 in magnitude. ULR is trivial to calculate using S-B, but DLR is problematic. (Therefore, SOD uses measured values for DLR appropriate for some circumstance.) I’d like to suggest that we can often estimate net long wavelength radiation (NLR) more accurately than we can calculate the difference between (observed) DLR and (calculated) ULR.
Why is the idea of NLR used so infrequently? For convection, we could separate the amount of energy transfered vertically into updrafts and downdrafts. We don’t; convection is always NET convection. When the air is saturated with water vapor, loss of water molecules from the ocean to the air is exactly balanced by the gain from the air to the water. When the air is partly saturated (80% for the boundary layer), more water molecules leave the ocean than the air, but almost as many still go from the air to water. We always think about evaporation as NET evaporation, not the difference between separate upward and downward fluxes. When energy is transfered by conduction, some collisions transfer energy from cooler molecules (slower moving molecules) to hotter (faster-moving) ones, but many more collisions transfer energy the other direction. We would never dream of calculating the energy flux from each type of collision and then calculating the difference. Conduction always refers to NET energy transfer from hot to cold. However, thanks to S-B’s law, we rarely think of LW radiation in terms of NET energy transfer from hot to cold. NLR = o(T2^4 – T1^4). The sign of NLR is more obvious from this equation than the difference between DLR (measured in a few locations) and ULR (calculated from S-B for a specific location or temperature).
How do we estimate NLR for the skin layer of the ocean? First, we must remember that the surface of the ocean emits some photons at wavelengths (the “window”) where GHG’s can’t absorb them. After accounting for the escaping photons, I like to picture GHG molecules at particular locations in the atmosphere exchanging photons with water molecules in the skin of the ocean. If it is colder in the atmosphere than in the skin of the ocean (the usual situation), there will be a net loss of energy from the ocean to the atmosphere. The higher the altitude of the GHG molecules exchanging photons with the ocean, the colder the GHG’s and the greater the NLR. Every kilometer of altitude increases the temperature difference by about 7 degK or about 10% in W/m^2. (A 1% decrease in temperature is a 4% decrease in radiation according to the Stefan-Boltzmann law.) The concept of escaping and “exchanging” photons can be merged if we remember that space is filled with radiation from the big bang with an effective temperature of a few degK.
Since 70% of the Earth is ocean, Trenberth’s global energy balance diagram gives us a starting estimate for NLR for the ocean. ULR is 396 W/m2. DLR is 333 W/m^2. NLR is 63 W/m^2; 40 W/m2 of which escapes directly to space from the skin of the ocean. The other 23 W/m^2 is due to the temperature difference between the skin of the ocean and the atmosphere with which it exchanges photons. This suggests that, if ocean and air temperature were the same, the average DLR photon is emitted about 1 km above the ocean’s surface. Trenberth’s numbers are global averages. Most importantly, when we consider which way photons must be flowing (net from hot to cold), the local atmosphere must be significantly warmer than the local ocean skin for NLR to be positive.
NLR is a big loser (ca 70 W/m2) for the skin of the ocean. Evaporation is a big loser for the skin of the ocean (80 W/m2 globally according to Trenberth, presumably more for the ocean than land). The skin of the ocean might even freeze from these loses, if it weren’t for two factors: a) Conduction and convection from water immediately below the skin layer warms the skin layer. b) A small fraction of SWR is absorbed directly by the skin layer during daytime. In Part One, we learned this fraction was about 20%. In the tropics at noon on a sunny day without wind, SWR (1365 W/m^2 hits the upper atmosphere) can deposit 200+ W/m^2 into the skin layer and make the skin layer warmer than the water immediately below. For these few hours, the only way energy from AGG-enhanced DLR can penetrate the bulk of the ocean is by conduction and conduction probably isn’t fast enough. Therefore, the skin layer could warm until it radiates away the excess energy from AGG-enhanced DLR without warming the bulk of the ocean. When these special circumstances don’t apply – ie most of the time – the skin layer will be the cooler than the water below (and will absorb the extra DLR without immediately returning its energy to the atmosphere by evaporation or ULR) OR wind will mix the skin layer with the water below (allowing extra DLR to warm the bulk of the ocean).
Frank:
All good stuff but misdirected.
Everything you say may well be true but it all relates solely to things that affect the rate of evaporation or the rates of other energy fluxes independently of the DLR/evaporation issue.
All that would go on just the same without any DLR at all so it has no bearing on the issue under discussion.
Mind you I think you’ve got the effect of more wind wrong. By increasing surface area it increases evaporation for a faster upward flow. It doesn’t ‘mix’ any surface energy to a lower level.
Anyway. lets make this really simple.
i) DLR causes more evaporation.
ii) In the process it warms the skin layer.
iii) At the same time it cools the interactive layer.
iv) The energy flow effects of ii) and iii) cancel out for a zero effect on the natural upward energy flow.
Anyone who disagrees please provide proof rather than conjecture or kindly give up.
Everything I have ever seen on this subject bangs on about the dire effect of ii) for humanity and the planet but never mentions the opposing force of iii).
Time to put up or shut up.
Steve: I agree that wind increases the rate of evaporation. (The air next to the ocean surface is saturated with water vapor and the wind speeds up the rate at which diffusion and convection can move this water vapor elsewhere in the atmosphere.) HOWEVER, in SOD’s Part III of this series, you can find graphs showing that wind ALSO mixes the top few meters of the ocean and that daily temperature gradients appear when the wind is weak or absent.
i) Does DLR cause more evaporation? No: When sunlight is weak or absent (ie most of the time), the rate at which water can evaporate from the skin of the ocean is limited by the rate at which energy reaches the skin layer from the deeper ocean. At RC (sorry for the biased source – I disagree with some of the reasoning in their post), you can see how the little the TEMPERATURE DIFFERENCE between the skin layer and the water below varies with 100 W/m^2 changes in DLR, even when sunlight is present. From that data, it is obvious that the temperature of the top few meters controls the temperature of the skin layer and therefore its evaporation and its radiation, not DLR. When 100 W/m^2 changes in DLR has little effect on this temperature difference, will a few W/m^2 of AGG-enhanced DLR make a difference? Of course not. http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/
ii) Does DLR warm the skin layer? Not enough to make up for energy lost to ULR and evaporation. Extra DLR simply reduces the energy the skin layer must obtain from the water below to make up for its energy deficit.
Using global averages, the skin layer is losing 80 W/m^2 through evaporation and 63 W/m^2 through NET LWR! If AGG-enhanced DLR reduces the loss from NET LWR to “only” 60 W/m^2 with AGG’s, the skin layer will not get warmer? It will simply require less energy from elsewhere (the next 10 m of the ocean) to MAINTAIN its current temperature. To properly understand the consequences of AGG-enhanced DLR on the skin layer, you need to take this huge energy deficit into account.
Hi Frank,
I didn’t intend to get drawn back here because SoD and others really have no grip on the nature of evaporation and so their comments are pretty much valueless in my opinion.
Nevertheless you seem sincere so I’ll address a few of your points as follows:
i) You said:
“Does DLR cause more evaporation? No: When sunlight is weak or absent (ie most of the time), the rate at which water can evaporate from the skin of the ocean is limited by the rate at which energy reaches the skin layer from the deeper ocean.”
Well DLR is nothing to do with sunlight. It is actually the downward energy generated by the temperature of the sky and is created by GHGs in the air. It applies day and night but obviously is blocked by cloud. Anyway if energy from that DLR reaches the ocean surface then it must accelerate evaporation because it adds energy to water molecules that are already moving towards evaporation naturally and so brings forward the timing of their evaporation.
If energy is being added to those molecules by DLR then why would the rate of evaporation be limited by the energy coming up from below? The water molecules do not care where the energy comes from and they are getting an extra dollop from the DLR. So your assertion cannot be right.
Realclimate completely ignores the thermal effect of the evaporative process occurring in the interactive layer above the skin so their account is misleading and valueless.
ii) You said:
” Extra DLR simply reduces the energy the skin layer must obtain from the water below to make up for its energy deficit.”
The extra DLR adds energy to water molecules and brings forward the timing of evaporation. It is correct that once accelerated evaporation occurs a lot of the DLR is used to cover the extra energy deficit. What happens therefore is that the deficit arising from DLR induced evaporation is also covered by DLR so that all the DLR is used up either in more evaporation or reducing it’s own deficit.
But that is a zero sum game. The final effect of the extra DLR after the evaporative effects it induces is zero.
Thus there is no change to the rate of upward flow and the ocean bulk is not affected either way.
Give it a bit more thought. I found it difficult too and those with an agenda will never see it.
Best Wishes.
Frank,
Regarding your statement ” When these special circumstances don’t apply – ie most of the time – the skin layer will be the cooler than the water below (and will absorb the extra DLR without immediately returning its energy to the atmosphere by evaporation or ULR)”
Where do you propose this absorbed energy goes?
Hello Tim,
We are obviously in agreement on this, thank God.
These guys really have been by passed by several hundred years of scientific endeavour.
They do not seem to know that under Earth’s current atmospheric pressure the bonds of energy that bind water molecules together contain more energy than that which is required to break them apart.
The higher the atmospheric pressure more energy is required, the lower the atmospheric pressure less is required. Just as the boiling pont of water changes as pressure changes. They don’t seem to deny that do they ? The amount of energy binding the molecules is a constant but the effort needed to break them is not.
They accuse me of suggesting something that violates the Laws of Thermodynamics yet the reverse is the case.
If the heating of the skin layer were NOT exactly offset by the cooling of the interactive layer when extra energy is added then that would be the real violation of those Laws.
As you ask Frank, where does he think that ‘extra’ energy goes.
It just goes either into more evaporation or to reduce the deficit from the extra evaporation the DLR has already induced. No surplus left overall.
And some pretty ‘reputable’ scientists these days (mostly so called climatologists) do not have a handle on these elementary well established principles about the phase changes of any material, not just water.
TTTM: Extra DLR from AGGs absorbed by the skin layer will reduce the energy deficit of the skin layer. The energy deficit of the skin layer is supplied by the water immediately below (the next 10 m or so) which is warmed by absorbing SWR. If the skin layer requires less energy from the next 10 m, the next 10 m will warm.
Every summer, extra seasonal SWR from the sun (and some extra DLR from the warmer air) is absorbed by the top 10 m is partially distributed through roughly the top 100 m of the ocean. Any SWR no longer needed by the skin layer (because of enhanced DLR) should also increase the temperature of the top 100 m within a few months. Extra energy arriving at the surface takes years to millennia to reach depths substantially lower than 100 m. One extra W/m^2 of DLR is 31.5*10^6 J/m^2/yr. Dividing by 100 m gives 0.315*10^6 J/m^3/yr and water has a heat capacity of 4.2*10^6 J/m^3. If my math is correct, this produces an initial warming rate of a little less than 0.1 degK per year (a rate which will slow as warmer water evaporates more quickly and radiates more).
WHEN the skin layer is warmer than the water below (and not mixed by the wind), there is no obvious mechanism (other than very slow conduction) for AGG-enhanced DLR to penetrate the ocean. The skin layer will warm until its energy flux is balanced by increased radiation and evaporation (and a slight increase in conduction). However, limited observations show and common sense calculations suggest that the skin layer is warmer than the water below only for a small fraction of the time (and never at night).
Frank,
Regarding “TTTM: Extra DLR from AGGs absorbed by the skin layer will reduce the energy deficit of the skin layer. The energy deficit of the skin layer is supplied by the water immediately below (the next 10 m or so) which is warmed by absorbing SWR. If the skin layer requires less energy from the next 10 m, the next 10 m will warm.”
I kind…of agree. Its a requirement for the skin layer to radiate according to Stefan-Boltzmann law and the energy must come from somewhere.
In my opinion it comes in part from the DLR (ie. all of the DLR that isn’t used for evaporation) and in part from the ocean bulk which makes up the “deficit”.
So I guess I reckon you have you reasoning backwards and its the ocean energy that makes up the deficit not the DLR.
…although having said that, that could also be what you’re saying so I think we probably have agreement. Except I still dont understand what you mean by absorbed with out necessarily being radiated again to satisfy S-B.
If you believe the DLR that you say is absorbed then doesn’t radiate and somehow gets “into” the ocean then I’m going to disagree with that.
One more post on this actually directed to SoD because he specifically made the point about this…
…if you could put little nametags on the individual bits of energy that come from the various photons (Both DLR and DSR) then you wont* find any nametags of DLR energy within the ocean.
*caveat, wave action mixing down would put a few there but essentially none.
I do have to respond to this bit though…
“However, limited observations show and common sense calculations suggest that the skin layer is warmer than the water below only for a small fraction of the time (and never at night).”
Actually the belief is that the skin layer is warmer than the bulk most of the time during the day…as per here
Its colder than the subskin and the subskin is warmer than the bulk too.
Basically I see this as the result of convection (from the DSR warmed water below) which increases the temperature up to the last mm or so and then the cooling above that is as a result of energy loss at the surface and additional energy loss from evaporation which pushes the temperature down.
I’m sure there is already an equation describing the relationship between atmospheric pressure, the strength of the bonds between molecules and the energy required to break them.
I’ve never looked for it because the principle has been established science for a very long time before AGW decided to ignore it.
Let SoD et al do some work themseves instead of trying to make us ‘prove’ long established science from the beginning.
Basically, AGW theory involving alleged warming of the oceans from increased DLR is itself in breach of the Laws of Thermodynamics.
Wikipaedia isn’t the best source but this fits the bill:
http://en.wikipedia.org/wiki/Heat_of_vaporization
“the molecules in liquid water are held together by relatively strong hydrogen bonds, and its enthalpy of vaporization, 40.65 kJ/mol, is more than five times the energy required to heat the same quantity of water from 0 °C to 100 °C (cp = 75.3 J K−1 mol−1).”
So the energy required to heat the water to 100C (provoking evaporation) is only one fifth of the energy required when the evaporation actually occurs.
The other four fifths is extracted from the local environment for a net cooling effect.
Now, how exactly is DLR going to warm the oceans with some sort of left over surplus ?
Or to achieve reduction of the upward energy flow from the ocean when there is a deficit like that to address first ?
Stephen Wilde,
So is deflecting a legitimate question. What’s entropy?
As for your post, I’ll give you Wolfgang Pauli’s response: “It isn’t even wrong.”
To be wrong, your statement has to be falsifiable. It isn’t because it’s all hand waving.
H2O molecules constantly leave and return to the surface at all temperatures, including when the water is frozen. An increase in temperature caused by adding heat to the water increases the rate of molecules leaving the surface. The temperature does not decrease anywhere in the system. The heat of vaporization of the water that evaporates is always less than the heat added to the system if that is the only change in the system. The temperature never drops. That’s required by the Second Law of thermodynamics. What’s the Second Law?
http://en.wikipedia.org/wiki/Heat_of_vaporization
“An alternative description is to view the enthalpy of condensation as the heat which must be released to the surroundings to compensate for the drop in entropy when a gas condenses to a liquid.”
Thus the enthalpy of vapourisation can be described as the heat which must be absorbed from the surroundings to compensate for the rise in entropy wnen a liquid vapourises to form a gas.
You are wrong. I think you are confusing evaporation with condensation.
Wrong about what? Was my description of the details at the molecular level of the equilibrium between water and water vapor incorrect? Your quote is irrelevant to that. Meanwhile try reading this Wikipedia article on chemical equilibrium. Then answer these questions:
How much energy must be added to convert 1 kg of water at 100 C to 1 kg of water vapor at 100 C?
How much energy must be removed to condense 1 kg of water vapor at 100 C to 1 kg of water at 100 C?
What’s entropy?
What’s the Second Law of Thermodynamics?
You might also want to read this article on Evaporation and this article on Microscopic Reversibility.
The existence of back radiation is confirmed experimentally, so there is not much to argue about. This also means that back radiation must be taken into account when discussing the temperature development of land and ocean.
I imagine the total process as follows:
At night time we are dealing with the “radiator cooling” problem, which is of a common knowledge from the everyday life. This is most simple to handle by assuming some temperature step between the surface (“radiator”) and the air layer of some thickness close to the surface. At the opposite side, this imaginary air layer is in contact with the next one being at somewhat lower temperature and so on all along through the mesosphere.
In the case of dry air and without CO2, the cooling of the radiator is given by h*(T-Ta) where T and Ta are temperatures of the surface and the air layer, respectively, at the given time t. h describes the heat transport from the surface to the layer by radiation and convection. Convection will rise temperature of the layer while radiation will pass through. The heat exchange at the other side of the layer will be given by h*(Ta-Tb) where Tb is the temperature of the next layer and so on.
If we now add water vapor and/or CO2 to the air then temperature of the air layer will become Ta’>Ta due to the absorption of a part of the upward radiation by vapor and/or CO2, which will reduce the heat losses from the surface to the layer since we now have h*(T-Ta’). The cooling process will slow down and the resulting temperature of the “radiator” will be somewhat higher and so on for the consecutive layers.
At daytime, energy delivered by the back radiation adds to the irradiation from the Sun, which increases somewhat the surface temperature as compared to that without the back radiation.
It should be observed that the back radiation exists both with and without the water vapor and/or CO2, the difference is only that the back radiation is slightly higher in the latter case being now related to Ta’.
The increase of temperature of the surface (and below) due to the increase of the back radiation is dependent on the heat capacity of the material of the surface (and below). The increase will be therefore rather small especially in the case of water.
And the effect of evaporation ?
Evaporation will lead to the decrease the surface temperature and to increase of the temperature of the air layer. But it should be kept in mind that there is still a continuous heat flow to the next air layer and so on along the mesosphere.
I would like also mention that when calculating the influence of water vapor or CO2 one ought to calculate not with the total back radiation but only with that part that is due to water vapor and/or CO2. The back radiation always exists since a (gas) sample radiates energy in all directions. The water vapor and/or CO2 will only contribute somewhat the total amount of this back radiation. This effect is more pronounced at land when comparing the passage of areas of low and high pressures or when comparing desert climate with the tropical one.
Ernest,
Yes it does lead to a decrease in the temperature of the interactive layer where the process of evaporation is actually going on.
However the extra DLR that causes that evaporation also heats up the ocean skin layer below the interactive layer.
Nonetheless the upward energy flux that you mention does indeed continue uninterrupted because both effects offset one another in energy flow terms otherise the Laws of Thermodynamics would be violated.
Mind you the evaporation does not warm up the air layer because the extra energy is in latent form and therefore does not register on temperature sensors. The energy is still present, it just doesn’t affect the temperature.
Stephen Wilde
I don’t think that it is correctly to use the words “heats up” in connection to the back radiation. There is heat exchange between the ocean skin layer and the air layer. And as long as the temperature of the ocean skin layer is higher than that of the air layer, the net heat flow will go from the ocean skin layer in the direction to the air layer.
The addition of water vapor and CO2 to the air layer will reduce somewhat the total net flow, which means that the cooling of the ocean skin layer will be slightly slower. Thus it is not a question of heating the ocean skin layer.
Evaporation of water molecules takes energy from the skin layer and transports it to the air layer – this can be compared to the convection process but without the movement of the mass back into the ocean skin layer as it would be in the case of the real convection. And, of cause, the water molecules that escape from the ocean skin layer have usually higher kinetic energy than the mean kinetic energy corresponding to the given temperature. These water molecules are from the higher energy wing of the Maxwell energy distribution function and they will increase the population of the molecules with the higher kinetic energy wing in the Maxwell energy distribution function of the molecules in the air layer.
Let’s reduce the complexity of the problem.
Take a cube of dimension 1 m. Make the bottom and sides perfectly reflective so no energy can enter or leave. Make the top a black body maintained at constant temperature. Put 500 kg water in the cube at 290 K and set the temperature of the top to 290 K. There is now nothing in the cube but water and water vapor.
1. What’s the water vapor pressure and mass?
2. What’s the rate of change of water vapor pressure?
3. What’s the rate of change of water temperature?
Increase the temperature of the top to 300 K.
4. Describe what happens to the temperature of the water and water vapor pressure. Extra credit for being quantitative.
5. After a very long time, what’s the water vapor pressure, mass and water temperature?
DeWitt and Ernest,
No offence meant but I don’t have the will or the time to correct your confusions.
The points made here have helped me to refine the way in which I should express my descriptions of the established science and I will make good use of it elswhere.
Adios.
No offense taken. In fact, you’ve made my day. That’s the funniest thing I’ve read in some time.
Or the knowledge.
TTTM: We seem to be in agreement that the skin layer doesn’t received enough energy from DLR and needs to obtain energy from the water below to replace the energy lost by evaporation and emission of LW radiation (ULR). However, I’ve been trying to make the point that the simple, convincing way to reach this conclusion is via NET long wavelength radiation. Like everyone else, SOD wants to subtract ULR (calculated for an arbitrary ocean temperature of 300 degK or 290 degK) from DLR (who knows where these numbers came from: most likely land, possibly sunny ocean at 290 degK, possibly cloudy ocean at 300 degK, possibly some average that may even be reasonable for one – but not both – of these two temperatures). Calculating the difference between two such large numbers is a risky proposition.
For all other energy transfer processes (convection, evaporation, conduction) we don’t separate the upward and downward fluxes, we deal with the NET flux. The 2nd Law of Thermodynamics places a constraint on NET flux mediated by LWR; net flux must flow from warmer to colder. The photons absorbed/emitted by the skin of the ocean originated/ended up in locations (including space) that are usually colder. We don’t need S-B! We don’t need to measure DLR! We don’t need to worry about whether we are above a tropical (300 degK) ocean or mid-latitude ocean (290 degK) or a non-frozen section of the Arctic Ocean; in all three cases the ocean is usually warmer than the atmosphere roughly a kilometer above (an average origin for DLR) and much warmer than space.
NLR is negative for the skin layer; evaporation is negative. When SWR is weak or absent, the skin layer must be draining lots of energy from the water below – energy that entered the water as SWR.
The question to ”scienceofdoom”.
What about the direction of the arrows labeled H43, H54 and H65 in Figure 1? Shouldn’t they have been given in the opposite direction, from the layer with higher temperature to the layer with the lower one?
Frank:
You are right about the problem.
There is no right answer for a 1-d model, due to the huge movement of heat from equator to poles. So anywhere near the equator loses heat in a “2-d direction”. And anywhere near the poles gains heat in the same way.
I picked the global average value of DLR and the diurnal variation was picked by looking at a lot of the DLR values from the BSRN network.
The important point is that I let the temperatures at each level in the ocean trend to equilibrium before comparing results from DLR vs solar increases.
I could try the same result with a much higher solar and lower DLR to see what happens.
There is of course a relationship between DLR and upwards radiation (R). But R-DLR isn’t a “constant”. Although in a 1-d model with a fixed amount of water vapor it would be.
If you want to pick a relationship between DLR and any other variable – or a value of DLR and solar – I will be happy to see what model results we get.
I expect they will be the same. But then I didn’t expect DLR and solar to give (almost) exactly the same results. So who knows. But I might as well produce results for a model that someone would like to see.
On my comment “But R-DLR isn’t a “constant”. Although in a 1-d model with a fixed amount of water vapor it would be.”.
– not instantaneously, and different locations would have different “constants”. Not a very well thought out comment really.
Ernest:
It’s just an arbitrary choice – in this case more for coding reasons at the time.
In any case, more heat moves upward by convection & conduction than moves downward by conduction. This is because solar energy is deposited by radiation and causes buoyancy-driven convection. Whereas heat conduction downward is very low due to the low conductivity of water.
But I don’t think that was really the reason. I just drew out the finite element map and identified all of the parameters, worked out the equations and wrote that into the code:
I would like to specify my question above. Figure 1 shows the situation at daytime with the irradiation from the Sun heating the ocean. This will give higher temperature at the surface of ocean than below and as the consequence the net heat flow into the ocean. It would be more clarifying to present two diagrams one for the daytime and one for the nighttime when there are effective heat losses from the surface into the air layer above.
Steve: Thanks for recognizing my sincerity. I’m personally appalled at how politics has perverted climate science. Too many climate scientists writing for the IPCC have abandoned Steven Schneider’s “tell the truth, the whole truth, and nothing but — which means that we must include all the doubts, the caveats, the ifs, ands, and buts” for his recommendation to “offer up scary scenarios, make simplified, dramatic statements, and make little mention of any doubts we might have”. HOWEVER, the politicization of science doesn’t invalidate science – it just means we need to sort through and test what scientists are telling us far more carefully than usual. Unfortunately, personal prejudices about climate change legislation and the motivations of some scientists interfere with this search for “the scientific truth”. I find SOD a good place to explore climate science, except that too much of SOD’s time is spent repeatedly addressing flawed science (such as “DLR can’t warm the ocean” or G&T) popularized by skeptical websites. Correcting this science is certainly easier and safer for SOD than challenging misleading science from RealClimate or the IPCC’s SPMs. One of the best things about SOD is that the proprietor and some well-informed commenters are kind enough to help me when I get off-track and they force me to clarify fuzzy thinking. So I’ll make a final try to explain how increased DLR warms the ocean without penetrating more than the skin layer.
If the US government raises my taxes, will the government have more money to spend? The obvious answer is yes. However, the US government is running a huge deficit, so my increased taxes could go to reducing the amount of money the government borrows, instead of increasing its spending.
If AGG-enhanced DLR deposits a few more W/m2 of energy in the skin layer of the ocean, will more evaporation and radiation be emitted by the skin layer? The obvious answer is yes. However, the skin layer is also running a huge deficit most of the time. It’s getting energy from the water below the skin layer (that is warmed by SWR during the day). Extra DLR from GHGs could: a) increase the temperature of the skin layer -causing more evaporation and radiation from the skin layer OR b) reduce the amount of energy flowing into the skin layer from below. You favor hypothesis a). I favor hypothesis b). What does data show? Does the skin layer get warmer when more DLR is present? If it doesn’t get warmer, there is no reason to postulate that DLR will increase evaporation or emission. (Collisions re-distribute the energy from DLR among the molecules of the skin layer faster than emission or evaporation can remove it. aka local temperature equilibrium.)
http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/ If you look at the data in Figure 2 of the RealClimate post (and ignore their discussion), what happens to the DIFFERENCE between the temperature of the ocean skin and water immediately below? Is the naturally-occurring 100 W/m^2 variation in observed DLR warming the skin layer (compared the the water below) so that it will cause more evaporation and emission? Yes, but every extra W/m2 of DLR reduces the difference by only 0.002 degK. A few W/m2 of AGG-enhanced DLR will warm the skin layer relative to the water below by <0.01 degK, certainly not enough to increase evaporation and emission significantly. In Figure 2, most of the 100 W/m2 variation in DLR is being used to reduce the amount of energy flowing into the skin layer from below (Hypothesis B). The temperature of the skin layer is mostly controlled by the temperature of the water below, not by DLR.
(In other subject, clouds block DLR by absorbing it, but they also emit DLR. Clouds are composed of tiny droplets of water or particles of ice, so they emit DLR based on the emissivity and temperature of these substances. Their emissivities are higher than the emissivity of the surrounding atmosphere, so DLR is enhanced under cloudy skies, not "blocked".)
Frank. In my opinion the nighttime back radiation partly affects the rate of the evaporation from the surface and partly reduces the amount of heat flowing into the skin from below. This is due to the slowing down of the cooling of the ocean skin layer due to the reduction of the net flow of heat by radiation. Evaporation is a function of temperature. If we reduce the decrease of temperature of the ocean skin layer we will maintain evaporation on the higher level. Simultaneously, the slowing down of the decrease of temperature in the ocean skin layer reduces the heat transport from below.
Ernest: Please look at Figure 2 of the RealClimate post. DLR increases with changes in weather by as much as 100 W/m^2 – energy which is only absorbed by the skin layer. Where did the extra DLR go? It can’t have gone into evaporation or emission without warming the skin layer first. Almost all of that 100 W/m^2 is consumed by the skin layer’s energy deficit. Globally, the skin layer’s average deficit is bigger than 100 W/m2.
Remember, the surface is in balance only when we include the average SWR reaching the surface (ca 160 W/m2). Unfortunately, about 80% of that SWR gets deposited beneath the skin layer. Some is deposited only 1-10 mm below the surface (100-1000 times further than DLR penetrates) and can get to the surface (in hours?) by conduction. Some is deposited meters below where convection may be needed to keep up with daily cycle.
Fran, you said:
“Extra DLR from GHGs could: a) increase the temperature of the skin layer -causing more evaporation and radiation from the skin layer OR b) reduce the amount of energy flowing into the skin layer from below. You favor hypothesis a). I favor hypothesis b). What does data show? Does the skin layer get warmer when more DLR is present? If it doesn’t get warmer, there is no reason to postulate that DLR will increase evaporation or emission.”
The skin layer does get warmer with more DLR but at the same time the interacting layer above it gets cooler because of the deficit created by increased evaporation taking place in that interacting layer.
One photon of energy input causes 5 photons to be absorbed as latent heat. See the link about the enthalpy of vapourisation.
In terms of upward energy flow the cooler interacting layer pulls energy upward exactly as much as the warmer skin slows it down for a zero net effect on the upward rate of flow from the ocean. Otherwise the Laws of Thermodynamics would be violated.
There is no DLR left over because it all disappears in the evaporative events that it induces. With a negative ratio of 5:1 that is inevitable.
The thing is that if the DLR all disappears with a zero effect on upward flow then everything else remains as it would have been in the absence of more DLR.
Thus consideration of all the other heat transfer mechanisms and phenomena is irrelevant.
DLR has no effect on ocean heat content. Period.
It really is that simple. All the other thrashing about is pointless obfuscation.
Realclimate and all the other AGW proponents simply miss out the cooling evaporative process in the interacting layer and argue their point only on the basis of the physics of radiation, conduction and convection. That is what all the above complexity is about but it will never be right until they get the evaporative process right. In this situation the only thing that matters is the thing they have left out.
If they do get the evaporative process right then they are truly stuffed.
Stephen Wilde.
If you are saying that back radiation cannot heat an ocean (or land) by itself then you are absolutely correct. Without the heat flow from the surface there would be no back radiation since it is the upward radiation, convection and heat conduction that are responsible for the heating of the atmosphere which, in turn, results in the emission of radiation by atmosphere. Without atmosphere the surface of the ocean or land would lose
o(T^4 – Ts^4) (1)
where Ts is the temperature of the space (about 4K) while in the presence of the atmosphere the heat losses are
hc*(T – Tl) (2)
and
o(T^4 – Tl^4) (3)
where (2) represents the heat transfer by convection (inclusive conduction) through the air layer and (3) corresponds to the net flow due to the heat exchange by radiation, Tl being the mean temperature of the air layer. Both T and Tl refer to the temperatures at the given time t. (2) and (3) are often given by
h*(T – Tl) (4)
where h refers now to the heat transfer by both convection and radiation.
Now, Tl can never be higher than T (except for the possible local temperature inversion which can happen due to the other effects). And since Tl is lower than T then there is no heating of the ocean or land by the back radiation itself. However, there is slowing down of the cooling of the surface of the oceans or lands as compared to the case (1).
The presence of the additional effect, evaporation, influences certainly the time relation between T and Tl but will not change the conclusion that the back radiation does not heat ocean or land by itself. In fact, evaporation can be worked into h in (4).
The total power density responsible for the heating of the surface at the latitude a is
F*0.7*cos(a)*cos(wt) + h*(T(t) – Tl(t)) + cm(dT/dt) + nC(dT/dt) (5)
where the first term represents the irradiation from the Sun (which is time dependent due to the rotation of the Earth represented by w), the third term is the heat stored/recovered from the material in and below the surface layer absorbing the radiation and the fourth term is the heat stored/recovered from the atmosphere where n is the number of moles and C is the molar heat capacity of air at constant pressure. If one desires so, the second term can be split into separate terms involving convection and radiation in accordance to (2) and (3) as well as the additional term involving evaporation in the case of oceans. The integration over the time of (5) will give the total gain or loss of energy through the given area of the ocean or land at the given period of time (as a day, season, yea, decade, and so on). The change of the composition of air will influence Tl(t) and possibly c and C in (5), the whole discussion being mainly concerned to the question of “how great the influence is due to the given agent?”. In order to answer this question one has to establish how much the given agent influences Tl (and thus the back radiation), c and C.
By the way, the integration of cos(a)*cos(wt), see (5), over the period of rotation and all latitudes gives the factor 0.203 and not 0.25 as it is normally used in the climate calculations. By using 0.203 and the Moon albedo of 0.88 one gets the mean temperature of the Moon to be 259K as compared to the measured value 250K. The inclusion of the heat capacity of the material constituting the soil of the Moon brings these numbers even closer. On the other hand, the use of the factor 0.25 gives about 271K, which is obviously too high. When applying the factor 0.203 to the case of the Earth one gets 240K and if taking the heat capacity of H2O into account, one comes to the temperature of about 200 – 220 K without the atmosphere. Astonishing, these latter numbers are very close to the temperature at the end of the mesosphere, which includes the most of the mass of the atmosphere.
“Their emissivities are higher than the emissivity of the surrounding atmosphere, so DLR is enhanced under cloudy skies, not “blocked”.)”
Yes I know. The net effects of clouds has been puzzling me and I think it should be distinguished in some way from DLR effects.
After all a cloudier world is a cooler world is it not ?
If one takes the cloud analogy to its logical conclusion then DLR should have a net cooling effect but even I don’t aver that at the moment though I did regard it as a possibility earlier in this discussion.
All I say at present is that DLR doesn’t affect the ocean bulk and that the atmospheric effect is insignificant.
“The temperature of the skin layer is mostly controlled by the temperature of the water below, not by DLR. ”
Depends how much DLR there is. On those figures not much and the human part a lot less so nothing much to be concerned about.
Thanks to Carrick (January 11, 2011 at 4:06 pm) for pointing out that something got mangled in the pasting of code into the comments.
Here are the Matlab files, saved as word docs:
DLR_oceanheating_3_5 – the main file
ocean_solar2 – Solar radiation transmitted vs depth
ocean_absorb – Creates ocean absorptivity vector, k, in, m^-1 for a wavelength vector
planckm – calculates the spectral intensity vs wavelength and temperature
TimTheToolMan from January 10, 2011 at 10:58 pm (and earlier at January 10, 2011 at 1:46 pm):
You proposed hypothesis D, which I didn’t really understand. But offered a diagram.
I think the diagram will help. I would still like to understand your perspective.
Frank,
“For all other energy transfer processes (convection, evaporation, conduction) we don’t separate the upward and downward fluxes, we deal with the NET flux.”
My hypothesis D requires them to be considered separately because although the net result at one instant may be the same, the contributions come from different parts of the ocean.
And hence when factors effect the DLR , they can only effect the DLR portion of the ULR and when factors effect the DSR, they only effect the DSR portion of the ULR.
I think this is an important distinction to be making because as one varies, the other also varies in response and that will have ramifications on how the “ocean system” behaves but that fact isn’t represented looking at the net energy flow alone.
TTM said:
“And hence when factors effect the DLR , they can only effect the DLR portion of the ULR and when factors effect the DSR, they only effect the DSR portion of the ULR”
So if they must be kept seperate, as they must, is it not obvious that the DLR takes care of itself, disappears completely in more evaporation and radiation and providing for the evaporative energy deficit that it causes itself with no knock on effect on anything ?
And what factors affect the DLR anyway ? It simply disappears in latent heat with nothing left over. What other factors could affect that ?
Other factors do affect DSR and the ULR response to DSR but only because DSR by passes the evaporative response by penetrating the ocean more deeply. However those factors would apply anyway in the complete absence of DLR and so should not be considered at all.
Yet here you are, all tying yourselves in knots over stuff that would happen anyway without any DLR at all and which has no effect on DLR and which DLR cannot affect.
And the bottom line is that every equation submited so far omits the evaporative cooling effect in the interactive layer above the ocean skin.
Everyone is setting out calculations that represent Latent Heat (L) as the same as the input required to provoke evaporation.
Evaporation of water takes out of the local environment (but it stays in the system in latent form) 4 times more energy than the sensible energy required to heat the same amount of water to 100C. See that link on the enthalpy (energy required to break a molecular bond) of vapourisation of water molecules
That doesn’t require more input from below either as some have suggested because the process is self limiting. If DLR photons get soaked up to cover the deficit caused by DLR induced evaporation then the process runs out of DLR photons and the evaporation cannot increase further so no more is needed from below.
Thus DLR photons come in, for every one that contributes to evaporation 4 more photons vanish into latent heat until the DLR is all gone. Even the extra radiant energy caused by the warmer skin disappears into the interacting layer.
The heating of the skin is exactly matched by the cooling from extra evaporation and the net effect is zero.
Everything else remains as it would have been without any DLR at all.
DLR minus extra evaporation and radiation = zero.
Everything else unchanged.
Stephen,
“So if they must be kept seperate, as they must, is it not obvious that the DLR takes care of itself, disappears completely in more evaporation and radiation and providing for the evaporative energy deficit that it causes itself with no knock on effect on anything ?”
All I know is that whatever radiation comes from the DLR doesn’t have to come from the ocean and so in that respect the ocean warms.
It is possible that whatever proportion of the DLR that funds evaporation, requires precisely the same amount of energy to be drawn from the ocean to further fund that evaporation to nullify that heating effect but its not an obvious result to me.
So basically at this stage, I would propose the series of effects that happen at the surface and include convection, conduction and anything else that might be relevant and then form equations based on those.
Solve them and you’ll have an answer. If the solution of ocean heating turns out to be independant of DLR then you’ll have an answer that supports your hypothesis.
Well Tim, it isn’t my hypothesis.
It was established science 50 years ago.
I’m sure someone must have demonstrated it at some time so I’ll just have to look for it but to many even today it is self evident from the basic Laws of Thermodynamics and no one tries to ‘reinvent the wheel’.
It is only AGW proponents who have come up with the alternative conjecture but have never proved or demonstrated it.
That conjecture is being substantiated by comparing the passing over of a cloud with the effect of DLR but the position regarding passing cloud cover is far more complex for a nmber of reasons and so is not good evidence in support of their conjecture.
Tim, just look at these definitions.
http://ghrsst-pp.metoffice.com/pages/sst_definitions/
The ocean skin a few nanometers deep is all that warms, not ‘the ocean’.
It is only a few nanometers deep because that is the full extent of DLR penetration. The energy gets no lower and the interactive layer above pulls the energy straight up again to convert it ALL into latent heat.
The whole Realclimate ocean skin theory is disinformation without any evidence in support and just designed to protect AGW theory from the implications.
The oceans control air temperatures so if the ocean bulk cannot be heated by DLR then AGW dies as regards any change in global equilibrium temperature.
I accept that DLR still affects the equilibrium temperature of the AIR but that is another matter.
The only climate effect of that is a miniscule shift in air circulation which brings the surface air temperature swiftly back to the sea surface temperature for a tiny unmeasurable climate change but no temperature change.
Natural shifts in air circulation are so large that the CO2 effect will never be discernible.
Slight clarification required.
I am saying that the temperature of the ocean bulk does not change and nor does the upward energy flow from bulk ocean to skin BUT
The temperature of the skin does change, the rate of evaporation and radiation changes and the speed of energy transfer to space increases to compensate.
So there is a change in equilibrium temperature from ocean skin upwards.
That does result in a miniscule shift in climate to compensate which I have conceded elsewhere.
However, not having changed the temperature of the ocean bulk the whole process is rendered miniscule because the natural climate changes induced by the sun and ocean bulk varying independently are comparatively huge.
The thing is that to have a significant climate effect in the face of natural variability one really does have to change the temperature of the ocean bulk but even if extra DLR were achieving that it would be many thousands of years before it became noticeable in any way.
As it happens the ocean bulk does not change temperature and the ‘missing heat’ problem of Trenberth et al is some evidence of that.
And if the ocean bulk were taking some of the energy from extra DLR it would supply such an enormous buffering effect that the effects of human emissions on the climate would be deferred for millennia and would still be swamped by natural variability even then.
By that time we would have solved our energy and population problems or be in difficulty for other reasons anyway.
Strangely it is in the interests of AGW proponents to concede the ocean heating issue because they can then argue that the entire effect of extra DLR goes straight to atmospheric effects.
They still have to show it to be significant though.
Let’s make a rough estimation.
According to the satellite measurements, the mean temperature of the lower part of mesosphere is increasing by approximately 0.167K per 10 years which is related to the increase of CO2 in the atmosphere.
This would give us the increase of mean temperature of the air layer above the ocean surface from 300K to 300.167K during the ten years period. This corresponds to the increase of the back radiation by approximately 1 W/m^2 during the 10 years period or by 0.1 W/m^2 a year, for example from 390 W/m^2 to 390.1 W/m^2 for the peak value.
Is this anything to worry about?
That’s odd, Ernest.
http://www.voanews.com/english/news/science-technology/Australian-Scientists-Probe-Distant-Clouds-With-Giant-Antarctic-Laser-103849314.html
“Our atmospheric dynamics are such that as we’ve got a warming troposphere – which is where we live – as that warms that in fact is interlinked with a phenomenon called global cooling up in the mesosphere above 50 kilometers,”
Do you think it could have gone into reverse 10 years ago when the sun became less active ?
Stephen Wilde.
The satellite measurements of the temperature of the atmosphere up to day are presented in
http://www.ssmi.com/msu/msu_data_description.html.
Temperature of the Lower Troposphere is increasing by approximately +0.163K/decade, temperature of the Middle Troposphere changes very little while the stratosphere shows a cooling trend by approximately -0.306K/decade.
Why this is so is the question for the debate but I could roughly explain such a behavior as follows:
A (small) part of the outgoing radiation that previously was passing through the mesosphere goes now to increase the temperature of the lower part of the atmosphere. 50% of this absorbed energy increases the back radiation and 50 % is emitted up. The up going radiation (which is contributing to the thermal equilibrium of the stratosphere) is now slightly lower than previously which results in the lowering of the temperature of the Stratosphere. The difference in magnitude might be prescribed to the difference in the initial temperature of the mesosphere as compared to that of the stratosphere. But please remember that we are talking about very small changes of the total upward radiation, of the order of 0.1 W/m^2 per year as compared to the upward radiation of about 459 W/m^2 corresponding to 300K at the surface of the Earth.
Well, I don’t know if it is so and if it holds quantitatively but this, may be naïve, explanation seems to be logical to me, at least as long as somebody comes with a better explanation.
Frank,
It’s not the difference between the skin and the water immediately below that causes an increase in evaporation and emission. It’s the absolute temperature. The difference (temperature gradient) is what controls the rate of heat flow upward from the solar radiation deposited below the surface. The amount of solar energy hasn’t changed, so the temperature gradient won’t change either.
But at steady state, energy in must equal energy out. An increase in average DLR, all other things being equal, must cause an equal increase in the sum of radiative and convective heat loss so that the net heat flow from the ocean to the atmosphere and space remains constant. That requires the absolute temperature of the surface to go up by ~ 1 K in the absence of feedbacks for radiation only and somewhat less than that for a combination of increased radiation and convection.
How about a combination of increased radiation, convection and evaporation ?
But anyway, you seem to be agreeing with me that the upward energy flow to the skin stays the same and that changes are from the skin upwards.
Or have I missed something ?
Yes you’ve missed something. If the temperature gradient of the water doesn’t change and the skin temperature goes up, it means the temperature of the water below the skin must go up too.
For the sake of argument, let’s say the gradient is linear and can be described by the equation T(z) = a*z + Tsurf where z is the depth in meters as measured from the surface and T is Kelvin. The value of a, which has the units K/m has been shown to be a constant. Increase Tsurf and you increase T(z) for all z. Which is what we’ve been trying to tell you all along.
dQ/dt = a*k where k is the thermal conductivity of water and dQ/dt is the heat flow in W/m2. If a changes, than so does dQ/dt. But dQ/dt is determined by the rate of solar energy absorption, which for the problem at hand is constant.
DeWitt: Since DLR is only absorbed by the skin layer, the temperature difference between the skin layer is sensitive to which hypothesis is correct: Hypothesis A, energy from higher DLR warms the skin layer enough to be lost as enhanced evaporation and emission; or Hypothesis B, energy from higher DLR reduces the skin layer’s demand for energy from below. Environmental changes (air temp, clouds, humidity?) during data collection cause natural variation in DLR of almost 100 W/m2 without significantly changing the temperature difference between the skin and below. When DLR is high, most of the extra energy is NOT being used to warm the skin layer enough to leave as evaporation or emission (without penetrating the bulk of the water). When DLR is high, the extra energy stays in the skin layer, reducing the need for energy import from below. When DLR is high, less of the SWR absorbed below is used supplying the deficit of the skin layer and the “surplus” energy is used to warm the ocean.
This analysis only applies when the skin layer is losing more energy to emission and evaporation than it gets from DLR plus the roughly 20% of SWR that doesn’t make it past the skin layer. Around noon on a sunny day in the tropics, 20% of SWR could be enough to convert the skin layer’s usual deficit into a surplus. Under those circumstances, extra DLR should warm the skin layer ABOVE the temperature of the water below enhance evaporation and emission.
This diagram is from TimTheToolMan:
Click for a larger view
And this is different in principle from saying DSR + DLR – R – L = ΔH how? Splitting R and L into components is uninformative considering it’s impossible to measure or calculate the value of the components without making arbitrary assumptions.
“Splitting R and L into components is uninformative considering it’s impossible to measure or calculate the value of the components without making arbitrary assumptions.”
There are a lot of things that are impossible to measure in our climate but that doesn’t mean we can ignore the detail because as soon as you start to simplify your understanding, you risk misinterpreting results.
TimTheToolMan:
Comments and questions on the diagram.
1. What are the constraints on R1 & R2? And L1 & L2? If I rewrote the equations without them what problem occurs with your Hypothesis D?
E.g. (from your diagram):
L = L1 + L2 = (DLR-R1) + (DSR-R2-ΔH)
L = DLR + DSR – R – ΔH
2. In the diagram as it stands the surface temperature cannot change.
So what is the relationship for surface temperature change? Or, putting it another way, which energy balance equations change in your diagram as heat is gained or lost in the surface layer?
3. And once surface temperature is introduced what determines ΔH? Or, putting it another way, what is the relationship between ΔH, and changes in surface temperature?
SoD : “What are the constraints on R1 & R2?”
Well the sum of them equates to the energy radiated by the ocean according to S-B.
Individually R1 must be less than or equal to DLR and R2 is less than the energy radiated by the ocean according to S-B.
SoD : “What are the constraints on L1 & L2?”
Again the sum of them equates to the energy lost by the ocean as a result of evaporation.
Individually L1 must be less than or equal to DLR and L2 is less than the energy lost by the ocean as a result of evaporation.
SoD : “If I rewrote the equations without them what problem occurs with your Hypothesis D?”
If you just use R and L then the hypothesis no longer forbids DLR directly effecting dH and can now be used to heat the ocean directly in equations of heat transfer.
Why does this matter if mathematically it amounts to the same thing? Well imagine there is a feedback associated with DLR. This only effects a portion of R and not the whole thing. If its non-linear as most of them are, this will probably lead to an incorrect result if applied to all of R.
SoD : “2. In the diagram as it stands the surface temperature cannot change.”
Why do you say that? dH changes and therefore T changes. I’m not trying to specify every equation here and so I haven’t related dH and T specifically… I think you expect too much from me 😉
SoD : “putting it another way, which energy balance equations change in your diagram as heat is gained or lost in the surface layer?”
Heat isn’t just lost FROM the surface layer, it is lost from R2 and L2 which come from the bulk. Heat is transmitted AT the surface layer…
SoD : “And once surface temperature is introduced what determines ΔH? Or, putting it another way, what is the relationship between ΔH, and changes in surface temperature?”
To calculate the surface temperature changes given the energy lost from the bulk is possible but non-trivial. I’m not trying to solve that here but instead simply stating that energy changes in the bulk will result in surface temperature changes.
Once DLR gets into the skin layer how could it NOT contribute to increased evaporation?
Since evaporation is an ongoing process occurring steadily even without DLR then ANY extra energy however small is going to add to the energy content of molecules that are already progressing towards evaporation and thereby make them evaporate earlier than they otherwise would have done.
Then once they do evaporate they suck energy out of the local environment in accordance with the enthalpy of vapourisation thus removing more of that DLR than the molecule absorbed in the first place.
And so they suck the DLR energy upwards out of the skin layer as fast as it is injected by DLR.
How then can the pre existing general upward energy flow be slowed down?
It is like a tributary joining a river. The incoming water from the tributary does not slow down the rate of the upstream flow.
Thus incoming DLR adds to the volume going down the ‘river’ but doesn’t slow the river down.
And as with a river the only differences are observed downstream from the junction point.
In the case of the ocean surface the junction point is the interface beteen the subskin and the skin because DLR can get no further.
TimTheToolMan on January 13, 2011 at 11:01 pm:
You have defined R1 as the radiation of energy “previously supplied” by DLR.
Does this have any physical meaning?
The surface radiates according to its temperature. The surface temperature changes according to energy imbalances.
So what does “R=R1+R2” add to our understanding of how temperature changes and energy moves?
If we take a 10um layer and work out energy changes and therefore temperature changes we can do it without knowing what R1 is.
True? or False?
TimTheToolMan on January 13, 2011 at 11:01 pm:
What you have done with your equations is prevented the surface from heating or cooling by ignoring the relevant terms for temperature change. If I add them back in we see a clearer picture, and also why your putative 2 separate terms for radiation just obscures rather than clarifies the picture.
With reference to your diagram let me define:
Ts = temperature of the “skin layer”
ds = depth of skin layer
Tb = temperature of “bulk ocean”
db = depth of bulk
cp = specific heat capacity of water
ρ = density of water
C = convected/conducted heat from bulk to skin layer
In your diagram:
ΔH = change in heat content of the bulk ocean
We’ll add a similar term for the surface:
ΔHs = heat content of skin layer
Neither ΔH nor ΔHs are needed in the final result, just makes it easier to follow the equations
Note for other readers less mathematically inclined: change in heat content = energy in – energy out..
ΔH = DSR – C (note the inclusion of the convected term)
Note for other readers: change in temperature =change in heat content / heat capacity..
ΔTb = ΔH/ρcpdb
So: ΔTb = (DSR – C) / ρcpdb
Similarly – Note for other readers: change in heat content = energy in – energy out..
ΔHs = DLR + C – (R1+R2+L1+L2)
Note for other readers: change in temperature = change in heat content / heat capacity..
ΔTs = ΔHs / ρcp.db
What is convection, C ?
C= f(Ts, Tb) : means “a function of Ts, Tb”
We know R1 + R2 = R = εσT^4
We could use a “bulk equation” for L if we knew the air temperature & specific humidity + wind speed at reference heights.
This means we can solve the equation for the two temperatures so long as we know the function C.
If we know this function we have 3 equations and 3 unknowns.
We can see in this formulation how dividing R into R1 and R2 doesn’t help solve the equations. Unless there is some actual formula for one of the terms, of a dependency on other variables.
What is much more important is to include the terms for change in temperature of the skin layer and bulk ocean.
Without these terms we cannot calculate the magnitude of convection, and in fact the only way the original diagram allows a solution is if the temperature of the skin layer never changes.
Notes:
1. We are ignoring the solar component absorbed in the skin layer, but for now let’s assume it is negligible.
2. We are ignoring the conducted heat down into the ocean depths, but for now assume it is neglible.
Both of these terms are included in the model.
SoD : “What you have done with your equations is prevented the surface from heating or cooling by ignoring the relevant terms for temperature change.”
I have ignored the relation between dH and T in my diagram, thats true. Beyond acknowledging it exists, of course.
My diagram ought to represent T as being T + dt
where dt = f(dH)
But you’re making up the result when you say it depends on convection so that you can then go on to suggest a rate of convection has an impact.
C= f(Ts, Tb) : means “a function of Ts, Tb”
As per out continuing discussion my view is that your f(Ts, Tb) is, for all intents and purposes, independent of Ts.
SoD : “What is much more important is to include the terms for change in temperature of the skin layer and bulk ocean.”
I agree that there needs to be included a relationship between dh and T and its going to be complex but I dont agree that its going to depend on your convection equation to any great extent.
“As per out continuing discussion my view is that your f(Ts, Tb) is, for all intents and purposes, independent of Ts.”
I should clarify what I mean here. Your f(Ts, Tb) is actually better represented by f(g(depth))
where g() represents a function returning temperature at a depth and f() represents related convection.
…and now Ts as the surface temperature, as distinct from subsurface temperature, becomes less important as it relates to 10nm out of the entire depth.
Its difficult to apportion the importance of that 10nm convection “reduction” mathematically but I think this may help.
SoD : “You have defined R1 as the radiation of energy “previously supplied” by DLR.
Does this have any physical meaning?”
Part of hypothesis D is that radiation supplied by DLR is immediately* used by evaporation or re-radiated back up.
*immediately really means afer a very short time as there is no storage in the skin and energy cant move to anywhere else from there.
SoD : “So what does “R=R1+R2″ add to our understanding of how temperature changes and energy moves?”
R2 is one of the components of energy loss from the ocean. Its value depends on R1 and T and any effect that modifies R1 or T. Similarly for L2.
SoD : “If we take a 10um layer and work out energy changes and therefore temperature changes we can do it without knowing what R1 is. True? or False?”
You’re going to need to define what you mean by “work out”. What are you measuring?
TimTheToolMan on January 14, 2011 at 2:31 am:
Ah ha.
Well Hypothesis C has exactly the same implicit physics.
And it is described in a way which is physics:
a) The surface radiates according to its temperature.
b) If the surface receives more energy than it radiates it will heat up – thereby radiating even more energy – until it is back in energy balance.
This is the real mechanism and this is included in the model which provides Hypothesis C as the result.
What isn’t physics (at least so far unknown physics) is that the surface can “re-radiate everything it receives immediately”. It is through temperature increase that the surface radiates more.
Let’s take a look at a 10μm layer which absorbs most of the DLR – about 85%.
The heat capacity of this layer (per m^2) = 4.2 x 10^6 x 10 x 10^-6 = 42 J/K.
So picture a surface at 398K (25’C) for example.
And add a DLR imbalance of 100 W/m². What happens?
Does the skin layer respond by immediately increasing its radiated output by 100 W/m² ?
No.
In the absence of any other change in heat flow (an unphysical assumption but let’s use this as a thought experiment), instead this increases the temperature by (initially) 100/42 = 2.4 K per second.
The temperature of the skin layer will reach equilibrium at 313K, where it will be radiating out 100 W/m² more than before “the change”.
Understanding this and why it is different to Hypothesis D is the key to understanding the problem.
And here is the graph of how the temperature would respond:
Temperature is the mechanism by which energy imbalances are sorted out.
SoD : “What isn’t physics (at least so far unknown physics) is that the surface can “re-radiate everything it receives immediately”. ”
This is where we’re going to have a disagreement then. Because …it can. DLR is always less than the ocean emission, never more. And so therefore the DLR is always only a fraction of what the ocean MUST radiate by S-B.
SoD : “If the surface receives more energy than it radiates it will heat up – thereby radiating even more energy – until it is back in energy balance.”
I mentioned this before, suggesting you were simply making a point. Now perhaps you could describe the situation where this happens as its no longer a point made in passing.
No points for describing what happens at the shoreline 😉
TimTheToolMan:
You said:
“..there is no storage in the skin..”
Incorrect, there is. The skin has a heat capacity just like any other body. 10μm of ocean skin has a heat capacity of 42 J/K per m²
“.. DLR is always less than the ocean emission, never more. And so therefore the DLR is always only a fraction of what the ocean MUST radiate by S-B..”
This is not relevant to the calculation of radiation of energy from the ocean surface. Or the calculation of surface temperature.
Does the ocean radiate according to the Stefan Boltzmann equation?
– Yes
Does the temperature of the ocean skin change according to energy in – energy out?
– Yes
Is the change in temperature defined by (Ein-Eout)/Heat capacity?
– Yes
If you agree with the “Yes” answers then the equations I have described are correct.
You have some conceptual issue which makes you want to write the equations differently because of how the ocean might respond to changes in one component of incoming energy.
Unless you can identify some problem with the equations I wrote, the only issue you have is that you don’t think that convection from below the surface is related to the surface temperature.
That is, your Hypothesis D is ONLY related to the question of convection. Unless you can find a flaw in the equations written.
More on the convection question later.
SoD and TTTM
All well and good but what about the issue of fluid mechanics as in the river analogy?
River water is warmer lower down but water doesn’t flow uphill because of that. The direction and rate of flow is determined by gravity and the volume by the quantity of water.
In the ocean and atmosphere interface the direction and rate of flow is determined by pressure and the volume by the phase changes of water.
Pressure determines the background upward energy flow in the absence of DLR.
When one adds DLR it just behaves like a tributary joining a river because pressure having set the background upward rate of flow (rather than gravity) the phase changes of water ensure that only the volume (quantity) of the energy flow changes from ocean skin upward.
So no change in the upward flow from the ocean bulk below the ocean skin.
“Incorrect, there is. The skin has a heat capacity just like any other body. 10μm of ocean skin has a heat capacity of 42 J/K per m²”
Well duh. Its 4.2J I think you’ll find from your figures.
SoD : “The heat capacity of this layer (per m^2) = 4.2 x 10^6 x 10 x 10^-6 = 42 J/K.”
A millionth of a million is one, not 10.
Thats sufficiently small to change instantaneously.
SoD : “This is not relevant to the calculation of radiation of energy from the ocean surface. Or the calculation of surface temperature.”
Of course it is. You’re talking about hypothetical heating scenarios that dont happen in nature. Its one of the limitations of what DLR can do and its important (I think) to understand and accept that.
Let me put it this way, if you calculate energy flows at the ocean in detail with cloud and aerosol feedbacks and wind and so on and you use R and L rather than R1, R2 and L1,L2 are you confident that your answer will be correct?
“SoD : “The heat capacity of this layer (per m^2) = 4.2 x 10^6 x 10 x 10^-6 = 42 J/K.”
A millionth of a million is one, not 10.”
Sorry, my bad, you have an extra 10 multiplier in there I missed. Still with hundreds of W/m2 for DLR, even 42J is not much!
I calculated it myself from scratch and I get the heat capacity of the skin (10um) as being 0.42J/K
Number of cm3 (ie grams) = 100*100 * 10 x 10^-6 = 0.1
Heat capacity = 0.1 * 4.2J/K = 0.42J/K
And clearly I shouldn’t be doing this now…because I’ve missed a 100 term on my 10^-6 myself so you’re right SoD, it is 42J/K
The flow of river water doesn’t slow down either but that is what you are proposing.
The limitations of what DLR can and can’t do are clearly stated in the equations that I have already described.
I can do no more.
If you can point out the mistake in the equations then I can take a look.
If you can point out a limitation in the equations then I can take a look.
Yes.
Take a textbook on energy transfer – like this one by Profs. Lienhard, you will find that even though there might be multiple sources of energy, the equation for the transfer of heat only depends on the temperature of that surface – along with the relevant material properties. You will not find a partitioning of where the energy came from – because it is not relevant.
If you mean – do I know the value of aerosol feedbacks and wind at any given location – then the answer is “no”.
But if you prescribe or calculate or measure these values then the equations can be solved in exactly the way shown.
I will return to your conceptual issue in a separate comment and propose a way to think about it..
TTTM,
This is only true if there’s a flow of energy in and out. Your arbitrary separation of DLR and DSR into components split between radiation and convection is unnecessary for the energy balance calculation. You still haven’t explained why you think you need to do this and what new information it provides.
DSR + DLR – R – L = DSR + DLR – R1 – R2 – L1 – L2 = ΔH
How do the relative magnitudes of R1 and L1 vs. R2 and L2 make any difference?
If DLR or DSR increases and R and L are unchanged, ΔH increases. That increase in ΔH leads to an increase in temperature and which causes an increase in R and L until ΔH is zero again. As long as DSR is non-zero, R + L will be greater than DLR. This isn’t exactly news. In a closed, insulated system, though, DLR and R would be equal and L would be zero. There would, however continue to be a flux of molecules into and out of the surface, maintaining the partial pressure of water vapor above the surface at a constant value. Evaporation isn’t ‘provoked’. It’s part of a continuous process linked with condensation.
If you’re trying to say that DLR has no effect at all because “it’s immediately re-radiated” that is a misconception. For one thing, there is no significant “re-radiation”. All the DLR is absorbed and thermalized. Radiation is due to collisional excitation. Evaporation is due to the fraction of molecules with sufficient kinetic energy to escape the surface (See the Boltzmann distribution). Other molecules of water vapor collide with the surface and are captured. L is the difference between evaporation and condensation. So DLR is intimately linked to the temperature of the skin layer. If DLR had no effect at all then L + R = DSR and the surface temperature would be a lot colder because it wouldn’t have to emit as much energy.
L and R are functions of temperature only. If DSR and DLR were suddenly removed, the surface would continue to transfer energy to the atmosphere and space by radiation and convection. The temperature would, of course, go down and R and L would drop too. The sum of DSR and DLR is why R , L and temperature are approximately constant when averaged over a year.
SoD : “Unless you can identify some problem with the equations I wrote, the only issue you have is that you don’t think that convection from below the surface is related to the surface temperature.”
Lets try a different tack. If DLR changes, how do you express the changes to both retained ocean heat AND evaporation?
Surely at that point you need to split R into R1 and R2 for the purposes of calculation/consideration anyway?
TimTheToolMan
On conceptual problems..
..of wanting to partition upwards radiation into the amount from DLR and the amount from solar (via ocean convection).
I can’t solve your conceptual problem – i.e., how you are thinking about it – but I can suggest a solution. Only based on how you appear to be thinking about it, of course..
In this example we will ignore L1 and L2. Not because they are unimportant but for the purposes of simplifying the explanation. If someone is still bothered by this, imagine that the relative humidity over this patch of the ocean is 100% and the wind is zero.
You have:
. R = R1 + R2 = εσT4
Now R1 “comes from” DLR, and R2 “comes from” the ocean.
Let’s suppose that the surface is at 298K and it has an emissivity of 1 so it is radiating:
. R= R1 + R2 = 447 W/m² [note 1]
Now suppose that R1 = 300 W/m², so R2 = 147 W/m².
This means that DLR = 300 W/m². (Actually, from the definition, the fact that DLR = 300 W/m² means that R1 = 300 W/m², not the other way round.)
Now suddenly the DLR increases by 100 W/m², R1 (apparently) jumps to 400 W/m².
This is a key question. This is a question which will demonstrate whether or not you understand heat transfer.
Note 1: If having an emissivity of 1 because this means a “blackbody” is concerning to anyone, just pick an emissivity of your own choice and multiply the results by the emissivity – the final point will be exactly the same.
Sod : “Now suppose that R1 = 300 W/m², so R2 = 147 W/m².
This means that DLR = 300 W/m². (Actually, from the definition, the fact that DLR = 300 W/m² means that R1 = 300 W/m², not the other way round.)”
Actually if DLR is 300 W/m² then R1 has an upper bound of 300 W/m² and doesn’t equal it because L1 also takes a portion of DLR. Or put simply some DLR goes to evaporation and some goes to radiation. So unless I know what happens to evaporation as a result of the extra 100 W/m², I cant answer the question numerically.
SoD : “What happens to R2 ?
This is a key question. This is a question which will demonstrate whether or not you understand heat transfer.”
Pardon me if I find that statement a little ironic. So what do you mean R1 increases? R1 isn’t a quantity that changes on its own, it only changes in proportion to all 4 variables R1,R2,L1 and L2.
Nevertheless I guess its possible to logically change R1 and look at how the other variables must change too…
The amount of radiation the ocean radiates remains constant and so R2 must decrease. If you ignore L then R2 decreases to 47 W/m²
This then eventually heats the ocean (Depending again on the result of L2) and R2 will eventually increase to compensate.
TimTheToolMan:
I already stated – for reasons of simplicity – L = 0.
The atmosphere is saturated and no less humid air is moving in. If the idea that L = constant or zero seems impossible then you can say.
And I don’t want to press you into a corner before we review the answer, so let’s put it like this:
– if DLR increases by 100 W/m² in a fraction of a second, somehow, what happens to R1 and R2?
And if you believe it is impossible that:
a) L=0, or
b) dL/dt = 0
then what happens to L1 and L2?
If it is possible, even though unlikely, that L=0, then please leave L out of the picture for now.
While people are pondering:
– from the previous comment.
Here is a way to think about the convection issue (from January 14, 2011 at 2:56 am and 3:12 am):
and
Let’s first take the simple example of conduction across a body. It could be a block of metal, a block of PVC, or a body of water. To make it simple to visualize (and the maths easy), you can think of a very long, very wide shape with parallel faces.
This is possibly a Zeno’s paradox type of problem.
The simplified equation for heat flux across this body:
. q” = k.ΔT/Δx [eq. 1]
. where q” = heat flux in W/m²; k = conductivity, a material property; ΔT = temperature difference across the two faces; Δx = distance between the two faces
Let’s take the case of a long wide slab of PVC 1m thick, with the bottom face held at 0’C and the top face held at 100’C.
Thermal conductivity of PVC = 0.19 W/m.K.
ΔT = 50’C, Δx = 1m
. So q” = 0.19 x 100 = 19 W/m²
Now we increase the top face to 110’C.
. So q” = 0.19 x 110 = 21 W/m²
The temperature of the top face is just contained in a tiny sliver of PVC.
Surely it can’t affect the conductivity of heat travelling from 1m away through this insulating material – or, if it does, it will have only the very slightest effect..
What really happens is the temperature gradient across the merest sliver of PVC causes heat flow. That heat flow causes the “other side of the sliver” to heat up. The resulting temperature different to the “next sliver” causes a little more heat flow..
And finally, there is a constant heat flux through the material and a temperature gradient that looks like this:
The real equation is (for 1d heat flow):
. q” = k.dT/dx [eq. 2]
. where dT/dx = the rate of change of temperature with respect to distance
In the case where we “inject heat” at different thicknesses in the material, eq.1 can’t be used anymore and eq 2 has to be used.
And instead of a straight line relationship between distance and temperature, there will be a more complex relationship.
This is similar to the case of ocean heating by solar radiation.
Convection is different from conduction, but both present the same conceptual problems.
How does a change in surface temperature at one location cause a change in heat flow at another location?
In the case of buoyancy-driven convection, the equation for motion of a parcel of heated water is:
. a = g(ρ2 – ρ’)/ρ’
. where ρ2 = ρ + (dρ/dz)water.δz
. where ρ’ = ρ + (dρ/dz)parcel.δz
where a = acceleration, ρ = density, ρ2 = density of surrounding water, ρ’ = density of parcel, z = vertical distance
The movement of a parcel of water is only related to how much it heats up relative to the body of water it is resting in.
But that moves heat upwards – which then has the same effect on the sliver of water above. And so on.
If there is any doubt about how the temperature of the top layer of water can affect convection from below – is this any different from questioning how conducted heat can be affected by the change in temperature of the top layer ?
“If there is any doubt about how the temperature of the top layer of water can affect convection from below – is this any different from questioning how conducted heat can be affected by the change in temperature of the top layer ?”
Look, I’m not going to argue there is an effect of some sort if you agree that…
a. Its likely to be very small for the 10um ocean skin
b. You haven’t actually modelled it correctly yet because convection involves momentum.
Additional to this I’d like to point out that at night (According to the Wiki diagram anyway – make of that what you will) the temperature gradient is essentially flat or put another way, all the “additional” heat has left due to convection. So if convection reduction was heating the ocean during the day, where is the energy in the morning?
Let us summarize the discussion and check where we are. The question was what happens to the temperature of the ocean due to the changes of the back radiation. And since the changes of the back radiation are attached to the changes of the composition of air (more exactly to the increase of CO2 in the atmosphere to make the problem interesting not only for scientists but also for politicians) the question might be reformulated as “what happens to the temperature of the ocean concurrently with the increasing deposition of CO2 into the atmosphere?”.
Well, if we take into consideration the total (bulk) amount of water in the oceans then the answer is that “we will not see any changes at all even if we are waiting in thousands of years”. This is due to the large heat capacity of water and the enormous mass of water in the oceans. So fishes should not worry in the future (except to be caught in the fisher net).
But what about the surface area both at lands and oceans? If we accept the results presented by “science of doom” in this poster, then some effect for the surface temperature might be expected – at least theoretically – if we simplify the problem to minimum (i.e. by omitting the contribution of evaporation, winds, etc). The inclusion of the omitted contributions will most truly reduce the theoretically calculated small rise of the surface temperature rather than increase it.
Now, what is the reason for the refusal of the oceans and lands to respond more pronounced to the changes of the CO2 concentration in the air? The answer might be seen from the equation h*(T – Tl) representing the heat exchange between the ocean skin layer and the adjacent air layer. Here, h involves all types of heat exchange as radiation, convection, conduction and evaporation (evaporation must be taken into account also in the case of lands as for example wet soils, vegetation, lakes, etc). T is temperature of the surface layer and Tl is temperature of the adjacent air layer. The increase of CO2 gives both the higher value of Tl due to the increased absorption of the outgoing surface radiation and the higher value of T after the radiation from the Sun is “switched off” at the evening. But since the changes of these temperatures due to the increase of CO2 are very small the resulting effect will be negligible. In principle, you do not need to specify how much the different heat transport processes contribute to the temperature changes as long as you can find the total value of h.
This can be easily confirmed by the experiment with the radiator facing an open room (i.e. with the opposite wall removed) and by simulating the heating of the radiator so that the input of heat corresponds to the irradiation from the Sun in accordance to F*0.7*cos(a)*cos(wt), where F = 1367 W/m^2, (a) is the latitude and w is the angular velocity equal to the velocity of rotation of the Earth. Since the outside temperature is changing from day to day one can establish how the variation of the temperature of the air influences the heat development on the surface of the radiator. However, this experiment will not show the dependence on evaporation and water convection under the ocean skin layer, sorry for that.
It appears from our discussion and the posters presented by “science of doom” show that CO2 acts on the climate system only through
a) the additional absorption of the outgoing surface radiation and, hence, the additional rice of the average temperature of the air, as far in the rate of +0.163K/decade if we attach all this trend to the increase of CO2 (which might be true only partly);
b) increased back radiation due to the increased average temperature of the air;
c) the (relatively small) rise of the surface temperature of lands and oceans at the daytime; and
d) the somewhat higher ambient temperature after the sunset which might cause that the surface temperature will end up at somewhat higher temperature.
If so, then we can put numbers behind the qualitative discussion on the influence of CO2 on the climate of the Earth since we can now relate the additional heat available due to the absorption of the outgoing radiation by CO2 to the total heat available due to the irradiation from the Sun and the back radiation previously to the enriching of the atmosphere by CO2.
The recalculation of 0.0163K/year gives the increase of the power density due to the increase of CO2 in the order of magnitude of 0.1W/m^2 per year as compared to the total available average power density of 550 W/m^2 (which includes the mean value of the irradiation from the Sun and the back radiation).
Is anybody impressed by the numbers related to the enhancing of the atmosphere by CO2?
Now, there arises also the question of the data for the temperature increase of the surface of the Earth which can be found in the different reports. As “science of doom” has pointed out, there are no (or rather few) real measurements of the surface temperature. The most of the thermometers measuring temperature at the given position are placed in the air above the ground (please, disregard from what I am writing below if I have misunderstood how the direct stationary temperature measurements are performed). So the direct measurements give the temperature of the air above the surface and not that of the surface itself. The averaging of the results of such measurements over the total space of the Earth might be given in two ways, either as
T(t) = SUM(Ti(t)*dSi)/SUM(dSi) (1)
where Ti(t) is the temperature of the area element dSi and the summation is performed over all the area elements where the temperature has been measured, or by
T(t) = INTEGRAL(Ti(t)*dSi)/INTEGRAL(dSi) (2)
if we can cover all the area of the Earth by the measurements from the satellites.
Of course, the measurements in accordance to (2) are more reliable, not least because the method one is not giving the complete picture of the average surface temperature at the given time t and because the method (1) opens the possibility to manipulate the data.
The averaging in time can be performed in the similar way as (1) and (2), namely
T = SUM(T(t)*dt)/SUM(dt) (3)
where dt represents the time period at which the measured temperature could be treated as being constant, T(t) is given above in (1) and (2) and the summation is performed over a given period of time (day, month, season ,year, decade and so on) or as
T = INTEGRAL(T(t)*dt)/INTEGRAL(dt) (4)
the integration being performed over the desired period as in (3).
The measurements performed by the apparatus placed on satellites give the results in accordance to (2) and (4) and are thus of the high degree of reliability. According to these results, the temperature of the lower part of the mesosphere increases at the rate of +0.163K/decade.
Thanks Ernest,
That seems to suggest three things of significance:
i) The temperature effect of more DLR is from the ocean skin upwards and does not involve an increase in the temperature of the ocean bulk.
ii) If the ocean bulk did take some of the strain as here proposed by others the buffering effect would be so huge that we could forget about AGW for millennia.
iii) Even limiting any warming effect to the ocean skin and air above, it is utterly insignificant compared to natural variability from events such as El Nino/La Nina or from multidecadal variations in the levels of solar activity such as those from MWP to LIA to date.
A lot of effort to give birth to a gnat.
That would be troposphere not mesosphere, I think.
Thank you, DeWitt Payne.
The data are for troposphere, of course, not mesosphere. I am flying to high. I hope I will not crash.
TTTM
Please explain what you mean by “convection involves momentum”. What are the physical consequences? How exactly is SoD’s model incorrect.
“Please explain what you mean by “convection involves momentum”. What are the physical consequences? How exactly is SoD’s model incorrect.”
I thought this was obvious. Convection involves water physically moving, not just heat moving as in conduction. Anything with mass and movement has momentum. In this case the water need to only move “into” the last few microns to be radiated.
Specifically SoD’s model is incorrect in that the top layer (which I believe was 5mm and is the “radiating layer”) assumes that heat only convects to it if its cooler than the lower layer.
I’m suggesting this is a gross simplification that has significant ramifications on the result.
Reality has a permanent* cool skin compared to the water directly beneath it and so be permanently convecting into the radiating region.
*usual caveats for exceptions, in this case when there is high wind the skin breaks down…but then evaporation is generally high.
An interesting thought.
If the extra DLR were warming the skin layer and that warming were then reducing the upward flow of energy then the temperature of the skin layer would NOT be changing.
Instead, the energy added from the DLR would be offset by the energy in the ocean bulk that would NOT be rising up to reach the skin layer. The ocean bulk would heat up INSTEAD of that energy progressing to the skin layer.
If the skin layer does indeed warm up (as observed) then the DLR is ADDING to the energy that is coming up from below.
Thus the energy coming up from below cannot be slowing down.
The analogy of the river and tributary is therefore accurate.
DLR cannot be reducing the upward energy flow from the oceans.
The reason being that the upward flow of energy is always driven by pressure differentials just as gravity drives the flow of a river.
“Ernest
Thank you, DeWitt Payne.
The data are for troposphere, of course, not mesosphere. I am flying too high. I hope I will not crash.”
That explains my earlier post querying the mesospheric temperature trends.
It happens to us all. There are a couple of errors in my earlier posts which I would have liked to go back and amend.
DeWitt,
It is often better to know a little about a lot than a lot about a little.
Back in the 60s the most important thing we were taught was how the bits of the world’s physical systems fitted together. Paramount were the water and carbon cycles and especially the perverse logic of the phase changes of water on which every organism depends.
Nowadays there are scientists who are deeply knowledgeable in very narrow specialities but who have no idea how it all fits together in the wider world
I think that is why various sectors of science, especially climate science, are a confused mess.
Any science that requires interdisciplinary expertise is similarly affected.
TimTheToolMan on January 14, 2011 at 7:19 am:
I didn’t write my question very well.
R= 447 W/m² because the surface is at 298K.
DLR = 300 W/m².
According to the definition of R1 and R2 this means that R1 = 300 W/m² and R2 = 147 W/m².
So when – in a fraction of a second) – DLR increases by 100 W/m² to 400 W/m², by definition R1 increases by 100 W/m² to 400 W/m², and by definition the value R2 must go to 47 W/m².
You are correct and I am a bad question writer. Because here is the important point which I thought I was asking..
Yet, in the short terms, the first few seconds, the amount of energy which is moved by convection from the ocean bulk to the surface actually hasn’t changed. Regardless of “R2 changing”.
The term C = convection (introduced in my equations of January 14, 2011 at 2:16 am) cannot change immediately.
What drives convection? Buoyancy differences driven by temperature differences.
What has happened is R is still the same = 447 W/m²
The “definition” R2 has dropped from 147 W/m² to 47 W/m² and what does this mean?
Nothing at all. The term R2 is a term with no meaning. The amount of energy moving into the skin layer by convection stays the same and has no relationship to R2.
In fact what happens is that the amount of energy into the skin layer has increased by 100 W/m² and the amount of energy leaving the skin layer hasn’t changed.
The skin layer heats up. As its temperature changes the value of R increases, and the value of C – well, something happens to it due to the temperature of the skin layer increasing.
[edited my question immediately after posting]
Do you think C does change immediately, in the first few seconds (my real question I meant to write) ?
And do you think that C doesn’t change at all as the temperature of the skin layer increases?
Eventually the skin layer reaches a new equilibrium where energy in = energy out. It is the changing temperature which:
a) increases energy radiated
b) reduces heat flowing in from the bulk by convection and conduction (or with a colder bulk, increases the conducted heat into the bulk
c) increases heat loss to the air
And changes in “R2” has no relationship to any of these.
Sod : “Nothing at all. The term R2 is a term with no meaning. The amount of energy moving into the skin layer by convection stays the same and has no relationship to R2.”
Are you sure of your logic here?
SoD : “In fact what happens is that the amount of energy into the skin layer has increased by 100 W/m² and the amount of energy leaving the skin layer hasn’t changed.
The skin layer heats up. ”
…because…no I dont think so. You’re probably referring to the Minnett experiment here (but please enlighten me if you’re not) and the skin layer heats up relative to the temperature of the ocean 5cm below it, not as an absolute independent to the rest of the ocean.
In fact I’m going to suggest that when the ocean changes from sunny to cloudy, then the amount of DSR decreases but so does the amount the ocean can radiate from its bulk (as per ongoing discussion) and therefore convection will increase the warmth at the top of the ocean where it cant get out as effectively.
So for a time at least, increased DLR will result in an increased temperature gradient near the surface. Once its been cloudy for long enough (or indeed gets dark) then I’m going to suggest the ocean heads towards its night time vertical but with a cool skin temperature profile.
So what does this mean?
No I dont think convection (C) changes immediately. I think it effectively banks up the heat higher in the ocean and thus makes the entire top few cm warmer until the new equilibrium can be established.
This has the obvious consequences of increased emission from the surface (due to S-B) which may or may not offset the increase in DLR.
SoD : “And changes in “R2″ has no relationship to any of these.”
I think your statement
“In fact what happens is that the amount of energy into the skin layer has increased by 100 W/m² and the amount of energy leaving the skin layer hasn’t changed.”
…is wrong because I think that R2 actually decreases by 100W in line with keeping the radiation from the surface constant and I think your assertion this is merely a “definition” is non-sensical.
What EXACTLY do you mean when you say this? Where does the extra 100W DLR go?
TimTheToolMan on January 14, 2011 at 1:34 pm:
Well that is a very good point.
Every night the top few meters of the ocean mix. The surface continues to lose energy via radiation after the sun sets, and so the surface cools. So it sinks. Replaced with water from just below – ie the layer just below rises.
Then that cools. And so on.
Therefore, all of the energy gets mixed throughout the ocean layers. The model gets this correct – see figure 5 and 6 for example.
This was the point of Part Three in the series and is also noted under Ocean Model point 5 in this article.
SoD said:
“The surface continues to lose energy via radiation after the sun sets, and so the surface cools. So it sinks.”
I don’t think that bit can be right.
Even at night evaporation continues and the process of evaporation doesn’t just supplement convection and convection and radiation, it dominates completely so even if no convection or conduction or radiation occurs the upward flow of energy continues from warm lower down (due to past solar input) to cool at the top. In fact they all do continue and in turn supplement the evaporative effect.
That is the result of the pressure differential which only requires one fifth energy input to cause an evaporative event that then absorbs an additional 4 times that initial requirement.
So no water sinks. Energy keeps flowing upward to maintain surface warmth as fast as energy is lost to the air.
Otherwise that 1mm layer below the skin which is 0.3C cooler than the ocean bulk cannot exist, yet it does.
The reason it exists is that upward evaporation and radiation from the surface pulls energy out of that 1mm layer faster than conduction and convection can bring energy up from below.
Thus there can be no downward convection or conduction within that 1mm layer. Tat layer has to dissipate first but it does not. It is a global and permanent feature.
It effectively seals off the bulk ocean from a reversed (downward) energy flow.
Understand evaporation correctly and it will all slot into place (I hope).
SoD
When you refer to convection in the oceans you also need to factor in density variations thus:
“Oceanic convection is also frequently driven by density differences due to varying salinity. It is possible for relatively warm, saline water to sink, and colder, fresher water to rise, reversing the normal transport of heat.”
Surface water tends to be less saline and colder water of low salinity is less dense and so rises potentially reversing the ‘normal’ situation.
That is another reason why evaporation and radiation can successfully maintain that 1mm cooler layer above the ocean bulk without downward mixing.
I mention it to show that your attempts at equations will fail because you are not getting all the signs right as a result of the physics of evaporation and ocean salinity which you are not taking into account.
Anyway, suffice it to say that your work is flawed and downward mixing doesn’t happen.
For those just joining the discussion down here I make some comment on Stephen Wilde’s (recent) entertainingly unconventional ideas:
From January 14, 2011 at 10:36 pm:
Even if no radiation occurs?
Radiation takes place continually from the surface of the ocean. And radiation from the atmosphere takes place continually to the surface of the ocean.
Evaporation, as taught in thermodynamics, takes place with a dependency on the humidity of the air. If the air over the ocean is saturated no evaporation takes place. If the air is dry a lot of evaporation takes place.
“So no water sinks” ??
During the day the solar radiation heats below the surface which causes buoyancy – moving the heat upwards with it. This is convection.
At night this solar heat source stops. But the surface keeps losing heat. And so the surface cools. And so the surface sinks.
Here are some experimental results from Diurnal Cycling-Observations and Models of the Upper Ocean Response to Diurnal Heating, Price and Weller, Journal of Geophysical Research (1986):
Click for a slightly larger image
You see the temperature curves in the 2nd graph. This clearly shows that during the night the top 30m is very well-mixed – because this part of the ocean is all at the same temperature.
From Observation of large diurnal warming events in the near-surface layer of the western equatorial Pacific warm pool, by Soloviev and Lukas, Deep Sea Research (1997):
“Thirty-five vertical temperature profiles during the warming and cooling phases of the diurnal cycle (Fig. 6) were obtained by free-rising profiler at low wind speed.”
How did this happen?
I believe SW meant, “even if no radiation from the ocean were to occur,” a counterfactual.
“… during the night the top 30m is very well-mixed …”
The question, again, is not the top 30m but the top few microns vs the top few millimeters vs the top few meters.
Note that in your next article on this subject, you quote:
“Away from the interface the temperature gradient is quickly destroyed by turbulent mixing. Thus the cool-skin temperature change is confined to a region of thickness, which is referred to as the molecular sublayer.”
In other words, the crucial layers in question apparently for the most part maintain their temperature gradient regardless of turbulent mixing. Note your graph on the day vs. night gradient, which strongly suggests that, as SW asserts, IR impingement on the molecular (micron) layer continues to cause evaporation, which continues to cool the millimeter layer.
I don’t understand your objection. Are you seriously trying to maintain that evaporation does NOT massively dominate IR emission as a cooling mechanism of the tropical ocean?
SoD,
This question is OT to the current post, but i don’t see any vehicle on your website to pose general questions.
Some climate scientists are claiming that more extreme weather events are occurring than in the past, and that the primary reason is because the atmosphere contains more water vapor due to the increase in the average global atmospheric temperature.
Could you at some time in the future prepare a post that addresses any measured increases in atmospheric water vapor, if any, and also what increase we should have based on an approximate 1C reported increase since 1880? I think the 1C increase is as measured at the earth’s surface, so not sure how useful that can be used for calculating increases in the total atmosphere.
In the interest of full disclosure, my “sense” is that such a small temperature increase would not increase water vapor significantly enough to cause a statistically significant increase in numbers and/or severity of extreme events, especially since most of the warming has reportedly been in the high northern latitudes where temperatures are well below 0C where the water vapor saturation value vs temperature curve is pretty flat.
I’m sure you can fill some gaping holes in my understanding of the subject. I would think there would be others interested in the science surrounding this subject.
Not suggesting you drop everything you have planned, but consider it for a future posting.
John: I happened to see today where Trenberth 2010 says that the carrying capacity of the atmosphere increases by 7% for every 1 degK increase in temperature. http://www.int-res.com/prepress/c00953.html
I also calculated an answer of 6.7% from one form of the Clausius-Clapeyron eqn:
ln(P1/P2) = (dH/R)*[(1/T2)-(1/T1)]
using temps of 288 and 289 degK; dH (heat of vaporization of water) of 45,000 J/mole (10% higher than at 100 degC) and R = 8.31 J/mol/degK; P1 and P2 are the partial pressures of water vapor.
Carrying capacity is the maximum amount of water vapor the atmosphere can hold in equilibrium with liquid water at a given temperature. However, just because a warmer atmosphere CAN hold more water vapor – where it is in equilibrium with liquid water – doesn’t mean that it DOES hold more water where there is no reason to assume that equilibrium exists. Relative humidity varies widely and decreases with altitude. You can see what AR4 WGI says about observed changes in water vapor in Section 3.4.2, but do note that they don’t include a quantitative change with uncertainty. (Nor do they discuss what the absence of amplified warming in the upper tropic troposphere implies about the change in water vapor at this location.) http://www.ipcc.ch/publications_and_data/ar4/wg1/en/ch3s3-4-2-2.html
Thank you Frank. Looking at the water saturation vs temp curve, 7% looks about right for 288K to 289K. I had read AR4 some time ago but didn’t remember reading about the change in water vapor. I’ll go back and read section 3.4.2. The biggest problem, I believe, with the IPCC reports is the lack of discussion of statistical uncertainties.
My recollection is that the abstract of the most convincing experiment mentioned by the IPCC says their data is “consistent” with the hypothesis of constant relative humidity and “not consistent” with hypothesis of constant absolute humidity. This is a far cry from asserting that water vapor feedback is X+/-Y. (Or asserting that the observed water vapor feedback is consistent with certain AOGCMs and not with others.)
Since some parts of the atmosphere should be in equilibrium with liquid water, it won’t be surprising to learn that there is some water vapor feedback.
SOD: On 1/15 at 12:03 you wrote: “During the day the solar radiation heats below the surface which causes buoyancy – moving the heat upwards with it. This is convection.” SWR from the sun deposits more energy in the top 10 um skin layer than in any other 10 um layer lower in the ocean. Heating below the surface (from the sun) alone therefore can’t “cause” buoyancy because deeper water always gets less energy from the sun than shallower. Loss of energy from the skin layer to the atmosphere is the only mechanism that makes the surface cooler than the water below. This cooling allows conduction and then convection to transfer energy from below. (This minor correction doesn’t invalidate anything else you’ve said.)
I get the feeling that some of your readers may be ignoring local temperature equilibrium and assuming that the energy from a photon of DLR can be immediately returned to the atmosphere as the kinetic energy of a newly-evaporated water molecule. This would allow evaporation to be increased by enhanced DLR WITHOUT a higher skin temperature. By coincidence, a 10 um DLR photon delivers 2*10^-20 J and an average water molecule needs 7.5*10^-20 J to evaporate.
SoD,
I am aware of the diurnal variations, seasonal variations and regional upwelling and downwelling.
However they are all solar driven because solar shortwave gets past the evaporative layer and so those phenomena are irrelevant for prresent purposes.
The issue here is whether additional DLR alone onto the surface skin causes mixing locally and independently of those background variations to achieve an additional downward transfer of energy to the ocean bulk.
It cannot do so because the 1mm deep layer above the ocean bulk and 0.3C cooler than the ocean bulk below (the subskin) effectively insulates the skin layer from the ocean bulk.
If the subskin layer were first to dissipate and the upper boundary of the ocean bulk were to move up to meet the warmed skin layer then of course you would be right but that never happens.
That subskin cooler than the ocean bulk is omnipresent and blocks downward energy transfer from DLR alone.
Frank said:
“Loss of energy from the skin layer to the atmosphere is the only mechanism that makes the surface cooler than the water below. This cooling allows conduction and then convection to transfer energy from below.”
Yes. As I said, the energy is drawn up by cooling at the top. Hence the presence of that subskin where energy has been pulled out faster than it has been replaced from below.
The net effect of more DLR being zero for reasons I have set out the subskin remains intact despite the warmed skin and the upward energy flow continues as before.
If the upward energy flow were to be reduced by a warmer skin then the energy NOT reaching the skin from below would offset the extra energy being added to the skin and the skin would then NOT warm.
Since the skin DOES warm the upward flux is getting ADDED to the preexisting upward flux so there is no accretion of extra energy to the ocean bulk.
And no extra mixing occurs from that effect because the subskin remains intact having failed to dissipate.
“SWR from the sun deposits more energy in the top 10 um skin layer than in any other 10 um layer lower in the ocean”
So how then does it turn out that the ocean has a 1mm cooler layer at the top above the ocean bulk and below the skin ? There should be increasing warmth right up to the very top. There is not.
The reason is that the energy is lost from the top faster than it arrives from the sun.
That is only achieved because of evaporation requiring more energy than is needed to induce it.
The budget is only balanced because some solar energy goes in deeper and returns to make up the deficit.
Extra DLR has the same effect but does not also provide additional energy to penetrate more deeply and make up its own deficit so that deficit also has to come from the DLR thus there can be no DLR left over to warm the system.
If I can’t make progress with these attempts then I’ll just have to give up.
SW quotes F: “SWR from the sun deposits more energy in the top 10 um skin layer than in any other 10 um layer lower in the ocean”
SW: So how then does it turn out that the ocean has a 1mm cooler layer at the top above the ocean bulk and below the skin ? There should be increasing warmth right up to the very top. There is not. The reason is that the energy is lost from the top faster than it arrives from the sun.
F: Partially right. The main source of energy to the skin layer is actually DLR. On the average, DLR at the surface is twice SWR from the sun and about 80% of SWR is absorbed below the skin layer and subskin layers. At night, DLR is the only radiant energy source. The 80% of SWR that is absorbed below the skin and subskin requires a temperature gradient (cooler skin) before it flow to the surface by any mechanism.
Energy is lost from the skin by
a) radiation – which depends only the skin temperature according to S-B.
b) evaporation – which also depends on skin temperature AND the humidity of the air above the skin layer. Increased DLR does NOT directly eject water molecules into the air, it must FIRST raise the temperature (average kinetic energy) of the skin layer. Without a wind or convection to mix the air, the only way water vapor escapes from near the ocean-air interface is by diffusion, which is very slow. Even when the air is saturated and evaporation is zero, the skin layer will still be colder than below because upward radiation (to the cooler atmosphere and space) must (by the 2LoT) be larger than DLR (from the cooler atmosphere).
c) Presumably a small amount of conduction (which can be ignored).
SW: That [a cooler skin layer?] is only achieved because of evaporation requiring more energy than is needed to induce it.
F: A statement that doesn’t appear to make sense. The rate at which water molecules in the skin layer escape to the atmosphere is determined by their average kinetic energy, ie their temperature. (It doesn’t make any difference where their energy came from). The rate at which water molecules leave the atmosphere for the ocean is determined by how many water molecules are in the air near the ocean. When the air is saturated with water vapor at any temperature, the exit and entry rates are equal and NO NET evaporation occurs. A cooler skin layer doesn’t require evaporation, especially at night.
SW: The budget is only balanced because some solar energy goes in deeper and returns to make up the deficit.
F: Perfect.
SW: Extra DLR has the same effect but does not also provide additional energy to penetrate more deeply and make up its own deficit so that deficit also has to come from the DLR thus there can be no DLR left over to warm the system.
F: Confusing. A few W/m2 of AGG-enhanced DLR is much too small to make up for the skin layer’s energy deficit. The Figure on the RC Post shows that even natural weather-induced variation in DLR of 100 W/m2 doesn’t eliminate the skin layer’s energy deficit, it still remains cooler than below. This does make sense: Globally, the skin layer’s deficit is roughly equal to SWR (160 W/m2) minus the 20% of SWR absorbed by the skin layer. The skin layer is usually running such a big deficit that its temperature moves in parallel to the temperature below and usually is cooler SO THAT energy can flow upward.
Since natural and anthropogenic changes in DLR don’t change the skin temperature relative to the water below, those changes simply increase or decrease the amount of energy that must flow upward from below. This is most clearly seen at night. When higher than average DLR (a foggy night or AGGs) reduces the demand the skin’s energy deficit places on the water below, the water below loses – at a slower rate – the SWR energy it stored (in the form of a higher temperature) during the day. In the morning the water below will be warmer than it would have been with lower DLR (a clear night or no AGGs). By this mechanism, higher than average DLR (absorbed by the skin layer) results in warmer water below.
When the skin layer is warmer that the water below, the skin layer does not have a deficit and any extra energy in DLR will warm the skin layer further and be lost by radiation or evaporation.
If I can’t make progress with these attempts then I’ll just have to give up.
Thank you Frank. That is helpful.
It shows that we are at cross purposes because you are not using the Definitions previously linked to.
The relevant layering is as follows:
i) An interacting layer at the very top where all the evaporative action is going on and which is not measurable for temperature.
ii) A layer 20 nanometers deep (now known as the ocean skin) which does warm up and is usually the part measured by our sensors. Basically a holding area for water molecules that are gaining energy from DLR plus upward convection and conduction prior to evaporation
iii )Subskin (formerly known as the skin as per your usage but not mine) which is about 1mm deep, which is about 0.3C cooler than the ocean bulk.
iv) ocean bulk.
There are other deeper layers but they need not concern us here.
To resolve the issue we need to know what happens to the temperature of the subskin when DLR changes.
Do you know of any data on that ?
If it warms then upward energy flow is being reduced by DLR. If it cools then upward energy flow is being increased by DLR. If it remains the same then the net effect of DLR is zero as I contend.
For the purpose of ascertaining the net effect of DLR we need to know the temperature of the subskin and NOT the temperature of the skin.
The revised nomenclature is causing confusion. Until recently just like you I took the skin to be what is now called the subskin.
As for the evaporation point it seems clear to me why evaporation is a net cooling process:
Wikipaedia isn’t the best source but this fits the bill:
http://en.wikipedia.org/wiki/Heat_of_vaporization
“the molecules in liquid water are held together by relatively strong hydrogen bonds, and its enthalpy of vaporization, 40.65 kJ/mol, is more than five times the energy required to heat the same quantity of water from 0 °C to 100 °C (cp = 75.3 J K−1 mol−1).”
So the energy required to heat the water to 100C (provoking evaporation) is only one fifth of the energy required when the evaporation actually occurs.
The other four fifths is extracted from the local environment for a net cooling effect.
Now, how exactly is DLR going to warm the oceans with some sort of left over surplus ?
Or to achieve reduction of the upward energy flow from the ocean when there is a deficit like that to address first
But despite everything I might say the issue under discussion here is now so contentious that the sources are contradictory.
I think the best thing to do is to find or await data that clearly and unambiguously demonstrates the temperature trends in the subskin when the amount of DLR changes.
Once we know that this issue will be resolved.
Stephen Wilde draw our attention to the fact that evaporation energy per water molecule corresponds in average to 4 photons of the wavelength 10 um, see also Frank on January 15, 2011 at 10:42 am presenting numbers 2*10^-20 J for energy of the photon and 7.5*10^-20 J for average evaporation energy.
I would like to direct your attention to the relation between the energy of a “green” photon of about 4*10^-19 J to 2*10^-20 J of the 10 um photon. It is obvious that the corresponding energy relation is 1 to 20. Thus, if the green photon is absorbed by water or soil at daytime then this will lead to the emission of 20 low energy photons at nighttime. But if the green photon is not converted into heat then there will be no need for emission of 20 photons at nighttime. In other words, water or soil will be colder (now, it is not much in the case of only one green photon but what if we have much more of such ones?).
You have surely guessed already at what direction I am going. Photosynthesis captures a green photon and converts its energy into the chemical one. This converted energy is stored in the cells of plants or algae. This stored energy will be released with time by the combustion or the moldering process but this will take its time.
Assume now that we cut forests or let domestic animals convert green fields into desert. This will certainly result in the increase of temperature of the region. In fact, the satellite observations of situation in Africa have shown that when people move in towns the previously eroded lands covers very quickly by vegetation and the temperature of the land drastically decreases. The “global warming” disappears at least in this particular region.
“Stephen Wilde draw our attention to the fact that evaporation energy per water molecule corresponds in average to 4 photons of the wavelength 10 um,”
Not quite.I pointed out that it took only one fifth of the energy taken up by the evaporative process to induce it (the enthalpy of vapourisation which is pressure dependent.)
I haven’t heard that about a ‘green’ photon before.
SoD and all,
On googling the issue I find that the measurement of the subskin (layer iii) temperature in relation to the temperature of the ocean bulk (layer iv) is a subject of immediate ongoing interest:
http://portal.acm.org/citation.cfm?id=1541411
and
http://www2.hawaii.edu/~jmaurer/sst/
Thus we can take this no further until the results are known.
Either the subskin warms along with the skin above or it doesn’t.
Steve: To understand the effects of DLR on the ocean, you only need two compartments: the skin of the ocean that absorbs “all” of the DLR and the water below that absorbs “none” of the DLR. (Draw the line wherever you want: 95%, 99%, 99.9%.) The skin layer receives energy from: DLR (all), SWR (about 20% of the total), below by convection and conduction. The skin layer loses energy by radiation (according to S-B, which we use to measure its temperature), evaporation, and (insignificant) conduction. The ocean below receives energy from SWR (about 80%) and loses energy to the skin layer by conduction and convection. This simple model is all you need to determine that the skin layer is usually running large deficit and that increases in DLR are used to reduce the energy that must be imported into a cooler skin layer from the water below – leaving water below warmer than it would have been otherwise.
Why make the problem more difficult by breaking up the “water below” into multiple compartments or separate an “interacting” layer from the skin layer. The energy flux between these additional compartments is complicated – changing with hour of the day and with the mechanism of energy flow to the skin layer (conduction or convection). None of these details change the fate of DLR absorbed by a cooler skin layer that is receiving energy from below. There is no mechanism that will allow additional energy to escape to the atmosphere from the skin layer WITHOUT FIRST WARMING THE SKIN LAYER. The data in the RC post, other sources and estimated fluxes show that the skin layer is almost always slightly cooler than the water immediately below. Therefore both compartments must warm in parallel or the skin layer has not warmed.
Your references don’t change anything: The abstract of the first reference doesn’t even mention the word DLR. The second reference is about different methods to MEASURE the temperature of the top compartments of the ocean: If infrared wavelengths similar to DLR are used, the temperature sensed originates in the layer that absorbs DLR. Microwaves penetrate deeper and therefore are emitted from deeper than the skin layer, so they report on the average temperature of a thicker layer at the top of the ocean than infrared. As long as we are confident that the skin layer of the ocean is cooler than the water below – from observation or estimated energy flux – we know that excess DLR will reduce the energy flux from below.
Why can’t the skin layer be much colder than the water below? This would allow the skin layer to warm without the water below warming. Sorry, these layers are in contact with each other. Convection and conduction automatically reduce any temperature gradient that is too steep.
Why can’t the skin layer be warmer than the water below and get even warmer when extra DLR arrives? Under these circumstances, the warmer skin layer can lose the extra DLR upward without warming any water below. This can happen when DLER + the 20% of SWR that is absorbed by the skin layer is enough to fund all of the skin layer’s emission and evaporation. It probably does happen around noon on calm sunny days in the tropics. In this case, the 80% of SWR that gets through the skin layer and absorbed by the water below has no where to go and ends up as heat – ie the ocean is warming rapidly. As the skin layer warms, its losses increase. As the water below warms, it losses remain the same (except for an increase in the trivial amount of energy lost to slow downward conduction). Soon the temperature skin will be slightly cooler than the water below and energy will flow upward.
SOD: Maybe you will find this new approach useful. No one else is likely to be reading this far. It reminds me of Catch 22.
1) When DLR increases, the skin layer can’t increase its emission or evaporation until it warms.
2) A huge amount of SWR is absorbed every day in a few meters below the skin layer.
3) That energy can only get from the ocean to the atmosphere (and then to space) when the skin layer is cooler than the water immediately below. (Upward convection and conduction aren’t allowed when the skin layer is warmer. Radiation is impossible.)
4) For this reason, the skin layer must usually be slightly cooler than the water immediately below.*
5) If the skin layer is usually slightly cooler than the water below, extra DLR reduces the need to import energy from the water below to the skin layer.
6) Extra DLR therefore warms both the skin layer and the water below in parallel.
QED: Enhanced DLR warms the ocean, without penetrating more than the skin layer.
More on Point 4) a) When the sun is shining and the skin is warmer than the water below, the water below is warming rapidly from SWR and the usual situation will soon be restored. Warming of the skin layer, but not the water below, will slow as evaporation and radiation increase with rising temperature. b) When it is dark and the skin layer happens to be warmer than the water below, radiative cooling and or evaporation will soon restore the usual situation. c) If the skin layer is much colder than the water below, convection will soon restore the usual situation.
Frank.
You describe the processes on the surface of the sea precisely as I imagine them by myself. The difference is that in my opinion the word heating means that something that is colder becomes warmer as when you are putting water on a warm plate of a cooker. But if you let warm water to cool down in the warmer room instead of cold outside than you achieve the slowing down of the cooling of the water.
As to evaporation, “Heat must be supplied to a solid or liquid to effect vaporization. If the surroundings do not supply enough heat, it may come from the system itself as a reduction in temperature.” – according to Britannica.
In the actual case, the surrounding supplies heat by both the shortwave and the longwave radiation by daytime and the longwave radiation by nighttime. But at nighttime the most of energy of evaporation is coming from the heat stored in water below the surface and the surface skin layer itself, asd I understand it.
What about the air layer above the surface? If it is colder than the surface skin layer it is not expected to contribute to the energy of evaporation by more than the longwave radiation. However, I have found also the information that “in the case of vegetation, the evaporation from the leaves after raining might be so intensive during a short period of time that it will exceed the available net radiation energy. Energy is then taken from the heat stored in the air”.
However, I don’t know if this might be applicable to the surface of the oceans or to soils without vegetation.
Ernest: My scientific training usually prompts me to think at the molecular level; molecular dynamics rather than thermodynamics. A molecular view of evaporation may clarify some things you describe or question than Britannica’s “heat must be supplied” formulation.
Water molecules in a liquid have a temperature (average kinetic energy) and a Boltzmann distribution of energies. Some of the molecules in the surface layer of molecules with greatest energy escape the network of non-covalent interactions between liquid water molecules and enter the gas phase, leaving the rest with less kinetic energy on the average, ie cooler. If the temperature is higher, this happens faster.
The reverse process is also happening all the time: Water vapor collides with the surface of liquid water and rate of entry depends on the concentration of water vapor in the air. Saturated air returns water molecules to the liquid phase as fast as they leave. There is no net transfer – evaporation – to saturated air, but exchange continues on the molecular level.)
On the average, the evaporation transfers about 45 kJ/mol of energy from the liquid phase to the gas phase at 20 degC (and 40.65 kJ/mol at 100 degC). Due to this loss of energy, the temperature of the remain water has dropped ( 1/heat capacity = 0.24 degK/(kJ/kg) ) OR the lost energy must be imported by: conduction (collisions transferring kinetic energy), convection (a new group of molecules with a new average energy), or absorbing radiation. Usually we are concerned with energy (and molecular) transfer per unit time, technically power, making all of the Joules into Watts.
Sod
Interesting and well run discussion.
It is generally conceded that a doubling of CO2 has a forcing of 3.7 w/m2 which the IPCC says will likely increase temp. by 3.0C at equilibrium.
All the anthropogenic GHG’s (CO2, Methane, NO2, etc.) had a forcing of 2.7 w/m2 in 2005 according to the IPCC. By ratio this should produce a rise in air temp. of 2.2C at equilibrium.
In 2005 all the AGHGs had produced a temp. increase of between 0.2C and 0.5C depending on who you believe. This is a long way short of the 2.2C above and to square the circle the IPCC are now proposing that 80% to 90% of the energy trapped by AGHGs is going into heating the bulk ocean.
Is this consistent with your position or are they being a little ambitious?
Incidentally their initial position was that transient temps. would increase by 0.2C per decade with a further 0.1C per decade at equilibrium. Unless I’m completely mixed up, that’s 33% into the oceans which only went to 80% to 90% after temps. stopped rising post 2000.
One more question.
The IPCC uses the fact that a 7w/m2 produced 5C increase in temp. during the last ice age as evidence that 3.7 w/m2 for CO2 x 2 will produce 3C of warming. This is based on DSR which penetrates along way into the ocean.
Why would this temp. sensitivity be assumed for DLR which hardly penetrates the oceans at all and has a greater proportion of it’s energy taken up in evaporation (I believe but could be wrong), is radiated back to space and elseware more quickly and therefore likely has a much smaller turnaround time in the ocean ?
Keep up the good work.
Bob: Aerosols and the amount of time it takes the earth to reach a new equilibrium temperature are part of the “standard” explanation for the inconsistencies you note. See Ramanathan 2008 (www.pnas.org/content/105/38/14245.full) for an explanation that disturbs this skeptic more effectively than most. It would be nice if someone were willing to show a plot of aerosol emissions and estimated aerosol forcing vs. time to support their assertions, but no one wants to show the large uncertainties associated with aerosols.
There may be some confusion in the data you cite for the last ice age. For one version, you might try Annan (2006) http://www.jamstec.go.jp/frsgc/research/d3/jules/GRL_sensitivity.pdf
None of this stuff says climate sensitivity IS 3.0 degC for 2XCO2; it gives a RANGE for climate sensitivity of 1.5-4.5 degC, plus the possibility of bias as data is selected for inclusion in these calculations. One answer inside this range is “correct”. Future climate won’t be determined by Mother Nature picking a value from within this range every year.
John Phillips on January 15, 2011 at 4:40 am:
It’s a very interesting idea to pursue.
Also in line with Frank’s comment from earlier, it’s much easier to deal with simpler subjects than challenging ones.
That doesn’t mean it isn’t worthwhile and interesting to deal with simple subjects. Many people have legitimate questions about these.
I hope to create a balance between more challenging subjects and the “easier stuff”.
Thank you SoD. I always look forward to your posts and the discussions that follow. I’m not usually knowledgeable enough to engage constructively in any of the debates, but I learn a lot.
Sorry for the poor grammar and spelling in my comments, I used to press on the “Post comment” button before rereading what I have written or after making changes in the text.
I don’t think that anybody questions the huge impact of water evaporation on the climate. The condensation of water vapor is the largest single contributor to the heating of the atmosphere. In its cycle, evaporation of water is cooling the surface (and hence even the bulk below), delivering heat to the air through the absorption of longwave radiation and condensation and cooling then the surface again when reaching it in the form of precipitation. And we were considering here only the thermal effects of this water cycle, as far.
CO2 comes into the picture mainly due to the photosynthesis as ell as due to the contribution to the longwave back radiation. However, the role of CO2 in the photosynthesis process is the most important issue of CO2. It is so important that the planters increase artificially the amount of CO2 in the air inside their greenhouses in order to intensify the growing of plants.
The life on Earth needs four elements, which are: “the green photons (i.e. photons from the visible part of the electromagnetic radiation)”, H2O, CO2 and the heat amount keeping temperature of the system within the temperature range where the photosynthesis can take place.
The circulation of CO2 in and out of the atmosphere is due mainly to the photosynthesis which takes CO2 molecules out from the atmosphere and the returning of these molecules to the atmosphere by means of combustion and moldering. The returning process is what actually happens when we breathe during all our entire life, for example.
So the question is: “Does the increase of CO2 in the atmosphere benefits the life on the Earth or if it is the danger to it, instead?”. And if the ocean “breathe” out to the atmosphere some more of CO2 being stored in the ocean, then is this a good sign or bad one or mostly insignificant?
1. A greenhouse is an artificial and controlled environment.
2. If you do a search for impacts on vegetation and plant species regarding CO2. The results are mixed. Some species do better, others don’t.
Changes to climate may increase invasive species of trees and increase diseases that are detrimental to native species.
More comments on the “green” photons, vegetation and temperature.
Everybody has the personal experience from the childhood on how it is pleasant to spring barefooted at a hot day on the grassland and play with a boll. But when the boll rolls outside the lawn and we must step on a stone then we will quickly step back. And if it is asphalt that the boll has rolled over then we will not repeat the same mistake once again of stepping on the asphalt barefooted. We will keep this knowledge in memory precisely as we keep in memory our experience of touching a hot plate when being a child.
Why the grassland or forest is colder than stones, sand or asphalt is quite evident. Plants absorbs high energy photons, which results in the combining of CO2 and H20 into CH2O and O2. This reaction goes through the intermediate state but it is not important in this connection except that this reaction will not taking place outside the plants even if we have both CO2 and H2O in the air.
Beside the conversion of energy of photons into chemical energy (which effectively prevents the heating of the surface), the plants evaporate a great deal of water into the atmosphere. This contributes to the cooling of the plants and lowers additionally the temperature of the lawn or forest.
Conclusions? Stop the erosion of lands and cutting of forests if you want to restrict the mankind’s negative influence on the climate. The increase of CO2 in the atmosphere will only help us in this endavour.
Good post Ernest.
I would add that here in Britain it has become fashionable to pave the front garden.
Lots more concrete and asphalt placed elsewhere further reduce photosynthesis.
Another problem with this practice is the reduction of the environment to cope with heavy rain.
The main urban drainage systems were not designed with this in mind.
Consequently they are being overwhelmed by the rapid runoff of rain.
“Conclusions? Stop the erosion of lands and cutting of forests if you want to restrict the mankind’s negative influence on the climate. The increase of CO2 in the atmosphere will only help us in this endavour.”
As pointed out, the issues are far more complicated.
Some plant life may benefit from CO2 others won’t.
Stopping the cutting of forests will be beneficial, however that issue is in addition to cutting CO2 emissions.
http://onlinelibrary.wiley.com/doi/10.1111/j.1365-3040.1991.tb01450.x/abstract
“…photosynthetic capacity is often reduced after long-term exposure to elevated CO2”
http://www.springerlink.com/content/32370807846477k5/
“High CO2 grown plants had lower photosynthetic capacity than 350 mgrl l-1 grown plants when measured at each CO2 concentration. Reduced photosynthetic rates were correlated with high internal (non-stomatal) resistances and higher starch levels. It is suggested that carbohydrate accumulation causes a decline in photosynthesis by feedback inhibition and/or physical damage at the chloroplast level.”
Frank,
As I pointed out previously:
http://en.wikipedia.org/wiki/Heat_of_vaporization
“the molecules in liquid water are held together by relatively strong hydrogen bonds, and its enthalpy of vaporization, 40.65 kJ/mol, is more than five times the energy required to heat the same quantity of water from 0 °C to 100 °C (cp = 75.3 J K−1 mol−1).”
So the energy required to heat the water to 100C (provoking evaporation) is only one fifth of the energy required when the evaporation actually occurs.
The other four fifths is extracted from the local environment for a net cooling effect.
and you then said:
“On the average, the evaporation transfers about 45 kJ/mol of energy from the liquid phase to the gas phase at 20 degC (and 40.65 kJ/mol at 100 degC). ”
So we are agreed about how much energy has to come from somewhere to provide the necessary energy for the change of state from liquid to water.
However I don’t see why the energy deficit caused by DLR has to come from the water or radiation. Part can come from the air and/or incoming DLR (via a top down warming of the water molecules).
Anyway, my question is :
Why exactly would there be any slowdown of upward energy flux when the incoming DLR would provide both the energy for increased evaporation AND cover the energy deficit for the extra evaporation that the DLR induces ?
And why should any of the incoming DLR remain unused so as to replace upward flowing energy and thereby slow down the upward energy flow?
As far as I can see the upward energy flux need not be affected at all.
The incoming DLR provokes more evaporation and exactly covers its own deficit and no more for a zero effect, surely ?
Following on from the above I’d like a simple answer to a simple question.
If an incoming DLR photon warms the bulk ocean by substituting itself within the ocean skin for a package of energy that would otherwise have moved up to the ocean skin then how could that DLR photon also warm the ocean skin ?
Either the incoming DLR warms the bulk ocean OR it warms the skin.
It cannot do both.
Yet observations show that the skin does warm.
Because the skin warms then upward evaporation and radiation increases INSTEAD of any reduction in energy flow from the bulk ocean.
If anyone can overcome that conundrum to increase the temperature of both ocean bulk and ocean skin simultaneously from incoming DLR photons then I’d like to hear the explanation.
We are not here dealing with a closed system whereby energy flows between two materials at different temperatures until they each arrive at an equilibrium with each other.
Instead we are dealing with an open system with a flow of energy through the system.
That flow of energy prevents any equilibrium ever being reached between the components of the system because the equilibrium is set by the rates of flow of energy in and energy out and not by absolute temperature.
DLR energy therefore cannot be in two places at one and the same time because all it does is add to the volume of the energy flow ‘downstream’ from the point at which it is absorbed.
Previously, on two occasions I suggested the analogy of a tributary joining a river. No one has yet challenged the validity of that analogy.
Now in theory one could suggest that the DLR warms up the ocean bulk first and then the warmer ocean bulk warms up the ocean skin but for that to be the case the DLR would need to somehow get past the ocean skin. It does not.
Any answers ?
The only possibility I can think of is to suggest that DLR energy adds more to the system than is lost by increased upward evaporation and radiation.
That issue is made impossible by the enthalpy of vapourisation that I have referred to several times previously to little effect.
The increased evaporation must always use up ALL the DLR with nothing left over to add to the system.
If someone wants to challenge this point then please explain exactly how additional upward evaporation and radiation could leave anything left over from the incoming DLR given the enthalpy of vapourisation of water.
Let me provide a clue.
The enthalpy of vapurisation is positive at 5:1.
That means that the amount of energy required to break the bonds between water molecules is 5 times greater than the energy required to warm the molecule to a point where it will evaporate (assuming that the air is unsaturated).
That value is pressure dependent. As we should all know a rise or fall in atmospheric pressure changes the boiling point of water. So it is too for the temperature at which evaporation will occur.
The lower the atmospheric pressure the less energy is required to break the bonds. However the energy required (and taken up from the surroundings) when the bonds break is a constant.
The pressure of the Earth’s atmosphere is such that given the physical properties of water molecules the enthalpy of vapourisation is always positive.
Now if we are to see any DLR energy left over after evaporation has taken place the enthalpy has to be negative, say, 1:5.
It isn’t and if it were then solar energy could never escape the oceans.Instead of being absorbed energy would be released and the local surroundings would get increasingly hot.
Indeed in our universe water could never exist in liquid form because solar energy would always heat the molecules and their surroundings to above the point at which they could exist in liquid form.
Thus the whole proposition that DLR can heat the oceans is without merit unless someone can show otherwise.
Ernest,
Grass is cooler than pavement because grass contains water. A moist surface can exchange far more energy with the atmosphere with a small change in temperature because the latent heat of vaporization of water is very high. A dry surface like pavement can only lose energy by sensible heat transfer and radiation. The thermal conductivity of moist soil is also higher than dry soil or pavement.
DeWitt Payne
I agree. I have mentioned previously two of the mechanisms that make grass cooler than pavement. One of them is the evaporation from the blade of grass. Evaporation is cooling the blade (or the leaf in the case of a plant or tree). This cooling mechanism can be partly controlled by the vegetation. The other one is photosynthesis that requires the presence of “visible” photons, water and carbon dioxide plus the other stuff that makes it possible to complete the Calvin Cycle. As it is described in
http://en.wikipedia.org/wiki/Photosynthesis
“Photosynthesis changes the energy from the sun into chemical energy and splits water to liberate O2 and fixes CO2 into sugar”. If being absorbed by the pavement, this energy of the photon would be converted into heat, instead, rising thus the temperature of the pavement.
You have added two additional important elements. One is the water content in the plants in connection to the thermal properties of water. Due to the high specific heat of water one needs to supply the water sample with more energy in order to rise the temperature by one degree than in the case of some other material with the same mass but the lower heat capacity. This is given by
dQ = c*m*dT (1)
i.e. if c is higher, then also dQ must be higher in order to get the same dT with the same m. Thus the same number of the longwave photons will give a smaller rise of temperature of water as compared to the material in the pavement.
And there is the additional effect related to the difference in the heat conductivity between the medium containing water and the dry medium, as you have correctly mentioned in your post.
All these point on the prevalent importance of water and vegetation as compared to the other agent contributing to the climate of the Earth.
By the way, equation (1) is useful to explain how CO2 is influencing the temperature of the air. The addition of CO2 does not change c or m of the sample of air. However, by absorbing a part of the outgoing longwave radiation from the surface, CO2 delivers some additional quantity dQ to the air sample and thus gives rise to the additional increase of temperature dT. But even here the influence of water (which in the case of water involves both the latent heat and the absorption properties of longwave radiation) is predominant for the regulation of temperature of the atmosphere as compared to that of CO2. Since the temperature of condensation of CO2 is much lower than that of H2O, the latent heat of CO2 does not enter the equations for heating the atmosphere as it is in the case of H2O.
(This is of common knowledge, of course, but I wrote this for the case if someone is not clear with the mathematics behind the impact of CO2 on the temperature of air).
Frank (in responce to https://scienceofdoom.com/2011/01/06/does-back-radiation-%e2%80%9cheat%e2%80%9d-the-ocean-%e2%80%93-part-four/#comment-8957 )
Your description of evaporation mechanism on the molecular level is in consistence with the theory of molecular physics, as I can see that. And there should not be any difference in the final results independently which of the approaches is applied. The discussion has however moved from the general problem of the exchange of heat between the air and ocean (which involves all the available mechanisms) to the specifics of one of them, evaporation. This is of cause unavoidable if one wants to cover the subject as completely as possible. In other words, instead of using the empiric relation h(T – Tl) where h, being experimentally determined, involves all the channels for the exchange of heat, one splits the problem into the detailed discussion on how the different channels contribute to the final results. And by that you open the Pandora’s Box, since the different effects might (but not must) lead to the opposite effects that partly cancel each other.
Anyway, evaporation is very important in this connection. Evaporation is leading normally to the lowering of the temperature of the liquid phase since the evaporation drains the liquid from the molecules of higher energy. This energy will be delivered, after some time, to air when the water molecules in the gas phase condense into droplets.
The decrease of temperature of the surface layer will initiate, normally, the heat flow from the lower region of the water phase to the surface. However, if radiation from outside (independently on its source) delivers the energy to the surface layer then temperature of the surface layer will not decrease as in the absence of this extra energy. And if the deliverance of the radiation energy is intensive, then this might even cause both the increase of the evaporation rate (here on the behalf of the incoming radiation without the need of energy supply from below, which is the point of Stephen Wilde if I understand it correctly) and the increase of the temperature of the surface layer which is expected to initiate the flow of heat into the bulk of liquid phase. This is governed by
(F – P) + H = LE
where F is the absorbed power density, P is power density loosed to the surrounding by radiation, convection and heat conduction, H is the power density stored into the liquid phase (sensible heat), E is the evaporation rate and L is the latent energy of evaporation. Equation is not mine, it can be found in literature in one or another form.
Evaporation is thus competing with radiation, convection and conduction on the “robbing” the liquid phase from energy delivered from outside (this “stolen” energy would otherwise be taken by H).
When approaching the saturation condition, the intensity of evaporation planes out. At daytime the increase of the inflow of radiation energy into the liquid phase from outside intensifies the evaporation while at night time the reduction of the inflow of energy to the liquid phase reduces the energy amount that compensate the energy losses due to evaporation.
One could continue the description of the evaporation effect by the description of formation of clouds, fog, mist, frost, rain, etc but this would lead us beyond of subject.
Stephen Wilde claims (if I am correct) that the evaporating layer is warmer than the layers both beneath and above it and that this is the fact confirmed experimentally. This was a new information for me and I suppose that the scientists that are specialized in this area has a plausible explanation such as the one presented by Stephen. I don’t dare to put my fingers in-between.
Ernest wrote: “And if the deliverance of the radiation energy is intensive, then this might even cause both the increase of the evaporation rate … and the increase of the temperature of the surface layer” and “Evaporation is thus competing with radiation, convection and conduction on the “robbing” the liquid phase from energy delivered from outside” and “the increase of the inflow of radiation energy into the liquid phase from outside intensifies the evaporation”. These phrases reflect careless thinking about cause. The amount of DLR does not determine the rate of evaporation or upward radiation from the skin layer. Neither does competition. The TEMPERATURE of the water (its average kinetic energy) determines the rate of both evaporation and radiation. (The evaporation rate is also effected by the humidity of the atmosphere, only “upward” component of evaporation is controlled solely by temperature.) Evaporation and emission don’t drop if the skin layer is running a deficit, both continue at the rates determined by temperature UNTIL temperature has changed.
So, we must know about temperature before we can say anything sensible. The amount of incoming DLR can change the temperature – but first you need to consider all of the energy fluxes into and out of the skin layer, including the amount of energy consumed in raising of lowering the temperature (via heat capacity). We can’t calculate some of these fluxes accurately (especially convection), so we run the risk of spouting meaningless gibberish don’t base reasoning on the MEASURED temperature of the skin layer (from its infrared emission). So, I refer you again to the RC post with MEASUREMENTS showing that that skin layer is usually a few tenths of a degC cooler than the water below. This temperature difference doesn’t change significantly with 100 W/m2 changes in DLR – unless the water below also changes temperature. Since DLR doesn’t change SKIN TEMPERATURE relative to bulk temperature, extra DLR is not lost by radiation and evaporation – it indirectly warms the bulk of the ocean. http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/
Sorry, the sigh before H in the equation above is incorrect. It should be F – P – H = LE. It can be hazardous to write out equations from memory if you are not dealing with the subject in your everyday life (and you are not thinking when writing them).
It’s similar as when writting mesosphere instead of troposphere as I did in the previous posts leaving thus my personal contribution to the general confusion in the question of the climate.
“Stephen Wilde claims (if I am correct) that the evaporating layer is warmer than the layers both beneath and above it and that this is the fact confirmed experimentally.”
The evaporating layer is the haze of interacting molecules above the skin.
Apparently it has not been possible to measure the temperature of it but since it is where the evaporative cooling takes place I would expect it to be cooler than the skin.
The skin is the place of maximum warming whilst molecules acquire energy and move upwards towards that interactive layer.
The skin could be warmer or cooler than the subskin bellow depending on the intensity of the irradiation above.
The subskin is always cooler than the ocean bulk below it.
The issue as I see it is whether the evaporative cooling in the interacting layer is strong enough to negate the effect of the warming in the skin so as to cancel out any effect on the upward background energy flow from the subskin.
For the reasons I have tried to explain that must be the case.
The test can only come when we can measure the temperature of that 1mm subskin.
Either it warms or it cools or it does neither. I contend that the temperature of the subskin remains unaffected by DLR.
SoD & DeWitt Payne, Apologies for not posting several days…FYI I’ve now replied to your posts from 14th. I dont know whether you saw them so I thought I’d specifically flag it.
TTTM,
After a long time, replying in line is inconvenient. Better to append to the end.
If the top layer is warmer than the layer below, then heat is convected/conducted away from the layer, not to it. Momentum has nothing to do with this. It is required by the Second Law. A heat pump can move energy against a temperature gradient, but only by doing work and increasing entropy somewhere else more than the entropy is reduced locally.
El Nino and La Nina are caused by movement of large warm and cool masses of water. In that case, momentum is indeed important. On the micro scale, what’s happening is more like turbulent mixing, also called eddy diffusion.
“If the top layer is warmer than the layer below, then heat is convected/conducted away from the layer, not to it. Momentum has nothing to do with this.”
In this case it does because its the target layer. It cannot be convected or conducted away, the only option is radiated and thats precisely what happens.
However in order for the energy to be radiated, it needs to get there and this is by convection (look at the temperature profile of the skin down to the subskin…its perfect for convection)
De Witt,
“The second law of thermodynamics states that in general the total entropy of any system will not decrease other than by increasing the entropy of some other system. Hence, in a system isolated from its environment, the entropy of that system will tend not to decrease. It follows that heat will not flow from a colder body to a hotter body without the application of work (the imposition of order) to the colder body.”
In the case under discussion:
i) The entropy of the system up to the ocean skin decreases by increasing the entropy of the system from ocean skin upwards.
ii)The necessary ‘work’ is performed by the process of evaporation.
Thus heat can flow from the colder subskin to a warmer skin ( or from the colder subskin to an even colder skin) by virtue of the ‘work’ performed by the evaporative process thus complying with the second law.
I am near bald. I can asure you all that at night my head does not warm from DLR if I take off my hat.
Harold: DLR should cause your head should cool more slowly when you take your hat off and stabilize (reach equilibrium) at a higher temperature. Performing a real experiment of this type isn’t very practical since you can’t turn DLR on and off. The large heat source in your head and the fact that hats insulate your head (as well as screen it from DLR) introduce further complications. The general observation that the temperature is warmer on cloudy winter nights than on clear nights is attributed to the fact that low clouds emit more DLR than clear sky.
The prominent skeptic, Roy Spencer, has several posts on detecting DLR in your backyard. http://www.drroyspencer.com/2010/07/first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature/
http://www.drroyspencer.com/2010/08/help-back-radiation-has-invaded-my-backyard/
SOD has written several nice posts about measurement of DLR.
Frank
……”the fact that hats insulate your head (as well as screen it from DLR)”………
Surely the hat provides an even higher level of DLR than the atmosphere!
I don’t know. First you need to tell me the temperature and emissivity of the hat compared to the atmosphere.
The temperature of the hat is very likely to be almost as high as Harold’s head.
The emissivity of the hat is probably higher than the 1% of the atmospheres that has radiating gases with restricted spectrum.
The increased density of hats molecules will produce a more continuous spectra than the band spectra of CO2 and H2O.
Stephen Wilde
….”The necessary ‘work’ is performed by the process of evaporation.”……
This seems to imply that if wear damp clothes you will be warmer than with dry clothes.
Or to heat up a bottle of water you could cover it with a damp cloth.
Have you any evidence of this?
[…] Comments « Does Back Radiation “Heat” the Ocean? – Part Four […]
Hurrah: Steve wrote (1/17, 5:54): If an incoming DLR photon warms the bulk ocean by substituting itself within the ocean skin for a package of energy that would otherwise have moved up to the ocean skin then how could that DLR photon also warm the ocean skin?
Steve is completely right, that DLR photon can’t do both. It does NOT warm the skin of the ocean. It does “substitute itself within the ocean skin for a package of energy that would otherwise have moved up to the ocean skin”. The latter process is what warms the bulk of the ocean even though the DLR is only absorbed by the skin of the ocean.
Why doesn’t the DLR photon doesn’t warm the skin of the ocean if it is absorbed by the skin? Steve has confused warmth and energy. “Warm” means to change the TEMPERATURE and a warmer skin layer certainly will return energy from extra DLR to the atmosphere by increasing radiation or evaporation. However, it is the ENERGY from a DLR photon – NOT WARMTH – that is added to the skin layer. The total energy – and therefore the temperature – of the skin layer will only increase IF the net energy flux is positive! When DLR photons “substitute within the ocean skin for a package of energy that would otherwise have moved up to the ocean skin”, the flux is unchanged. The TEMPERATURE of the skin layer remains unchanged. Evaporation and radiation are unchanged. The energy from extra DLR photons is not immediately returned to the atmosphere. DLR warms the ocean.
How can we be certain that DLR photons “substitute within the ocean skin for a package of energy that would otherwise have moved up to the ocean skin”. We look at Figure 2 from the RC post which shows that the temperature difference between the skin layer and the water below is almost unchanged by a 100 W/m^2 change in DLR. Massive amounts of DLR do not warm the skin layer (where they are absorbed) relative to below, so these photons must be “substituting within the ocean skin for a package of energy that would otherwise have moved up to the ocean skin”. Is there any other explanation? (Remember the vertical axis shows the temperature difference. If both the skin layer and the water below warm, there may be no change in the difference. Both layers are probably warmer when DLR is higher.) http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/
TimTheToolMan on January 17, 2011 at 1:00 pm:
(Responding to my comments of January 14, 2011 at 9:09 pm)
Yes. It’s basic heat transfer. If one external forcing changes, then the other external forcings don’t somehow “instantly re-balance” to ensure a constant forcing.
What happens is more energy is stored, causing a temperature increase.
And a temperature increase (or decrease) is the “ultimate re-balancing mechanism”.
Or the “mediation mechanism for heat flows”.
It is temperature changes which then affect the flows of energy in or out of a “body” or layer.
And the final question:
It goes into heating the layer.
Heating the layer causes an increase in radiated energy.
And also reduces the flow of energy into the layer from below (or increases the flow of energy out of the layer to below – depending on the sign of the temperature difference).
It’s possible I have misunderstood your latest comments. Half of your comment implied you might understand (and agree with) what I have explained, but then the other half stated that you didn’t agree, leaving me unsure.
If – and only if – I have correctly understood your comments, then hypothesis D is a hypothesis which relies on unknown mechanisms of heat transfer. Which would explain why I couldn’t understand hypothesis D from the start.
Technical note: When I write about “layers” – in reality, as partly described in my comment of January 14, 2011 at 10:00 am, there is only a continuous temperature change. Thinking about layers is just a mental device to help us visualize the problem.
SoD : “Yes. It’s basic heat transfer. If one external forcing changes, then the other external forcings don’t somehow “instantly re-balance” to ensure a constant forcing.
What happens is more energy is stored, causing a temperature increase.”
No. You’ve already calculated the heat capacity of the skin and its 42J/K. So in other words it takes less than one second to raise the temperature of the skin by 2K for your example of 100W DLR increase.
Minnett measured around 0.5K temperature difference between max and min DLR (about 100W change) and so this equilibrium would be reached in about 0.25 Seconds
So what happens to the extra 100W DLR after 0.25 seconds?
And I need to respond to this bit too…
SoD : “It goes into heating the layer.
Heating the layer causes an increase in radiated energy.”
What mechanism do you propose the warming skin (because that what you’re saying) passes the energy to the bulk?
I’ll just head you off at the pass to save time.
Radiation? No because water is opaque to IR and it is stuck at the surface wrt radiation.
Convection? No because the skin layer is cold atthe top and warms with depth. This is the reverse of the temperature gradient needed to convect.
Conduction? No because for the same reasons of cooler skin layer, conduction would violate the laws of thermodynamics by moving heat from a cooler place to a warmer one.
…but feel free to speculate a mechanism! No doubt you have something in mind.
One of the important points for the actual dispute is if the increase of evaporation from the surface of the oceans prevents the rise of temperature of oceans, which would be expected due to the increase of the back radiation (which, in turn, is due to the change of the composition of the atmosphere).
In other words,
*we can observe the increase of CO2 in atmosphere above the ocean,
*CO2 absorbs some part of the outgoing radiation from the surface of the ocean which increases somewhat the temperature of the air
*The increasing of temperature causes the (slight) increase of the (already existing) back radiation
*This (now increased) back radiation is absorbed by the surface skin layer of the ocean which means that the energy delivered by the back radiation to the surface skin layer is now slightly higher
*This additional energy will now be distributed over the channels that are participating in the heat transfer from the absorbing surface skin layer to both the air above the skin layer and the bulk of the ocean.
The question on how much the increase of CO2 can contribute to the increase of back radiation is simple to answer if we can measure the increase of the temperature of atmosphere. We can treat the atmosphere as an object facing by one side the surface of the ocean. Atmosphere emits in the direction to the ocean o*Ta^4 W/m^2 where Ta is the mean temperature of the “air object”. If the temperature of the air increases from Ta to Ta + DTa then the surface skin layer will receive the increased power density equal to o*[(Ta + dTa)^4 – Ta^4] W/m^2. dTa is measured to be equal to 0.0163K/year if we continue to enrich the atmosphere with CO2 at the same rate as during the past decades. The mean temperature of the troposphere is 256.5K if we take into consideration the linear gradient across the troposphere from about +17 deg C at the surface to about -50 deg C at the end of the troposphere. Thus, we can expect the increase of the diurnal back radiation from the average 245.433 W/m^2 this year to the average 245.495 W/m^2 next year (i.e. by 0.062 W/m^2) and so on.
Thus, how the addition of 0.062 W/m^2 to the diurnal irradiation into the surface skin layer will be distributed over the thermal channels transporting heat out from the absorbing skin layer. These channels are the heat conduction, convection, radiation and evaporation in the direction to the air and heat conduction and convection on the water side of the surface skin layer.
And here we are at impasse as long as we do not come to a consensus on the question on “how the evaporation process is responding to the changes in both the diurnal irradiation and the diurnal variation of temperature and humidity of the air above”. All the other channels can be nicely described if the contribution from evaporation is determined.
Frank said:
“Steve is completely right, that DLR photon can’t do both. It does NOT warm the skin of the ocean. It does “substitute itself within the ocean skin for a package of energy that would otherwise have moved up to the ocean skin”. The latter process is what warms the bulk of the ocean even though the DLR is only absorbed by the skin of the ocean. ”
But Frank the skin DOES warm up. It is the subskin that doesn’t warm up (as far as I can tell from logic alone given that the necessary measurements do not yet exist).
I mentioned before that you need to be clearer about the different layers because they all behave differently.
Steve wrote: “But Frank the skin DOES warm up” Steve obviously did bother to look at Figure 2, which shows that the skin layer did not warm – relative to the water below – as DLR changed by 100 W/m2. Just because the skin layer absorbs DLR doesn’t mean it will get warmer than it currently is. The average kinetic energy of the skin layer – not one single flux contributing to that energy – determines its temperature (and emission and evaporation). When DLR substitutes for energy from below (your expression), the skin layer doesn’t warm.
What do the terms skin and subskin really refer to? For the record, here are the definitions you mentioned above. These are functional terms derived from the process of temperature measurement, not actual physical layers. A thermometer or other temperature sensor has a substantial physical size and heat capacity that interferes with measuring the temperature of small quantities or locations of water. By analyzing the emission from the ocean, we can use S-B to calculate the temperature of the water that emitted photons.
“The skin SST (SSTskin) SSTskin is defined as the radiometric temperature measured by an infrared radiometer operating in the 10-12 micrometer spectral waveband. As such, it represents the actual temperature of the water across a very small depth of approximately 20 micrometers….”
“The sub-skin SST (SSTsub-skin) represents the temperature at the base of the thermal skin layer. The difference between SSTint and SSTsubskin is related to the net flux of heat through the thermal skin layer. For practical purposes, SSTsubskin can be well approximated to the measurement of surface temperature by a microwave radiometer operating in the 6-11 GHz frequency range, but the relationship is neither direct nor invariant to changing physical conditions or to the specific geometry of the microwave measurements. SSTsubskin is the temperature of a layer ~1mm thick at the ocean surface.”
Notice that the skin temperature is a component of the subskin temperature. We don’t have a technique that allows us to determine the temperature of the interactive layer (if there are enough collisions to define a temperature), but the interactive layer is included in the “skin temperature.
To understand the fate of DLR, we ONLY need to know if changing DLR is changing the temperature of the part of the ocean that absorbs DLR. That is the skin layer. When the subskin layer shows a different temperature profile from the skin layer layer, it isn’t directly because of DLR. A lot of visible sunlight is absorbed during the day by water that is part of the subskin layer. This causes a maximum near the depth labeled subskin seen in daytime vertical temperature profiles. But this maximum doesn’t change the fate of DLR – the energy in those photons – as you describe it – “substituents for energy from below.” This will always be the case when the skin layer is cooler than the water immediately below.
The measurements performed from the vessel Tangaroa (thanks Frank for the link http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/ )
show that the influence of the back radiation on the temperature of the surface skin layer does exist, as expected, and that this influence is very tiny, also as expected.
The measurements have been performed for the increase of the back radiation by approximately 100 W/m^2 due to the appearance of clouds above the ocean as compared to the cloudless sky. The results indicate the reduction of the temperature difference between the bulk temperature of water (5 cm below the surface) and the temperature of the surface skin layer (which is approximately 1 mm thick), by approximately 0.002K/(W/m^2) as compared to the temperature difference without the additional increase of back radiation. Thus at the cloudy diurnal period the temperature difference between the bulk and the skin layer will be 0.2K less (the skin layer will be 0.2K warmer) than that at the cloudless diurnal period.
This decrease of the temperature difference will result in the decrease of the heat flow from the region of 5 cm deep to the surface of the ocean. The ocean is thus continuously loosing energy into the air (and hence also to space) but these heat losses are recompensed by the irradiation from the sun (plus back radiation) during the daytime. At the cloudy day the losses are lesser but the irradiation is lesser, too.
Now, these results tell us nothing about the temperature of the water layer just below the surface skin layer which give us no hint on the problem of evaporation that is a part of our discussion in this Forum.
However, we can apply the Taranoga results to the problem of CO2 in atmosphere. Namely, if the temperature of the air is increasing by 0.0163K each year due to the increase of the CO2 content in the air (which corresponds to the addition of 0.062 W/m^2 per year to the back radiation) then the difference between the water temperature 5 cm down and the temperature of the surface skin layer will reduce by 0.002K/(W/m^2)*0.062W/m^2 = 0.000124K if the increase of CO2 content in air is continued at the present rate.
We can now use the relation F = P/A = -k*[(DT – dT)/dx – DT/dx) = k*dT/dx in order to estimate how much power density F will be retained within the water mass of the volume V = 1m^2*0.05m = 0.05 m^3 due to the reduction of the initial temperature difference, DT, by dT = 0.000124K. We don’t need to know DT itself since it cancels in the square parentheses. k = 0.62 W/mK, dx = 0.5 m. This gives the average value of F = 0.00154W/m^2 which means that the temperature of the volume V of water will stabilize at somewhat higher temperature due to the fact that it loses now less energy to air.
The increase of temperature of the water volume V is given by Q = c*m*$T where we are looking for $T. c = 4180 J/kgK, m = 1000 kg/m^3 * 0.05m^3 = 50 kg. Q = F*1m^2*3600s*24 = 10.7 J, where 3600s*24 gives the number of seconds per 24 time period determining the average reductiion of the temperature gradient across 0.05m hight.
This gives $T = Q/(c*m) = 0.00005K for the increase of the average bulk temperature of water (from surface and up to 5 cm deep) due to the increase of the back radiation by 0.062 W/m^2.
That would be 0.00005K/day or 0.018 K/year. That’s only ~1/3 of the the total temperature change required to balance the increase in DLR. And it goes up again the next year so the temperature change would be 0.018 + 0.009 the next year, etc. After a while, the rate would be on the order of 0.036 K/year. That starts to be significant after a few decades.
DeWitt Payne.
Why you are involving day after day increase during the year? 0.062W/m^2 is the average increase of the back radiation per year. The next year the back radiation will increase by the next 0.062W/m^2 and so on as long as the emission of CO2 into the atmosphere is continued at the same year rate as today.
In other words, the stepwise increase of the back radiation by the additional 0.062W/m^2 per year results in the stepwise increase of the temperature of water (within the layer of 5 cm) by 0.00005K per year.
This is the same as if the diurnal irradiation from the Sun is higher by 0.062W/m^2 this year as compared to the year before or if you are increasing the power delivered to your radiator by some additional amount, say 1W, this year and the same next year and so on. This will give you some additional increase of temperature in the room by some additional deg C, but this increased temperature will be the same every day during all the year and it will increase again only the next year when you have increased the input by the next additional 1W.
Of course, you can go over to a smooth weakly (or daily or in seconds) increase of CO2 and as a consequence to the weakly (or daily or in seconds) increase of back radiation. But then you must use the number of W/m^2 per weak, day, second and not per year.
Frank on January 18, 2011 at 10:11:
Winding my way through the work that has been done on water vapor, I did post some results from Sun & Oort – Clouds and Water Vapor – Part Three. Their results indicate a water vapor response between constant relative humidity and constant absolute humidity. Plenty of caveats attached, see the article.
Ernest,
You calculated the energy transfer for one day (24*3600 seconds), not one year. Unless your ocean is only 0.5 m deep, it won’t equilibrate in one day. And radiative balance at the surface won’t be achieved by a change in temperature at the surface of 0.0005 K. So the next day the temperature will go up some more, etc. Meanwhile the DLR is also increasing. You’re off by several orders of magnitude.
Dewitt Payne.
I am using the results of the measurements presented by the scientists on Tangaroa. They have measured changes of the temperature difference at the surface and 2 cm below due to the changes of the back radiation by approximately 100 W/m^2. The changes in the back radiation are caused by clouds.
There is no information presented about the doubling of the effect during two successive cloudy days or about the thrice increase after three cloudy days which would point on the accumulation of the effect. I understand therefore their results so that with the 100 W/m^2 greater back radiation the upper ocean layer “works” in one regime around some mean value of the surface skin layer while under the sunny day with less back radiation the same layer oscillates around another mean value of the surface layer. When temperature of the skin layer becomes higher than the temperature 5 cm below then we have the heat flow down (the daytime regime) while at night the temperature of the skin layer becomes less than that 5 cm below and the ocean loses energy to the air.
Of course, there might be some delay in the response of the system to the changes of the intensity of radiation. This will give the stabilization of the mean temperature at some higher respectively lower value when the “steady state” regime is achieved. This is why “science of doom” works with up to 15 years program. However, my intentions were to show how small effect can be expected due to the increase of CO2 in atmosphere. This can be also deduced from the comparing the results between the case 1 and case 3b in “science of doom” calculations where the increase of the back radiation by 10 W/m^2 resulted in the increase of the temperature by about 0.32K, see Figure 4. If adopting these results, the increase of the back radiation by 0.062 W/m^2 would give the increase of temperature by about 0.002K, assuming the linear relations. Even this value is very low. However, please remember that the values presented by “science of doom” are calculated ones and are given for the simplest conditions. The decrease of the temperature gradient due to the increase of the back radiation by 100 W/m^2 as presented by Tangaroa is, on the other hand, the experimental one being obtained under the real, natural conditions.
DeWitt Payne
You wrote: “You’re off by several orders of magnitude”. I have get the value 0.00005K while it should be 0.002K if applying the “science of doom” results. The reason for this discrepancy is that I have taken into account heat conduction, 0.62 W/mK, exclusively, while “science of doom” has included also the heat transport by convection, 25 W/K m^2. Convection is 40 times intensive than heat conduction and the reduction of convection by diminishing the temperature gradient between the bulk and surface will give 40 times higher reduction of losses. The reduction of the temperature of the bulk will be correspondingly 40 times less than in the case of the pure heat conduction. The increase of the temperature (when adding convection) is, thus, 0.00005K*40 = 0.002K. After 15 years this increase will reach also the deeper water, compare the results of “science of doom”.
But I would point out that the most significant for this discussion is how small the change of the absolute value of the bulk temperature of the ocean is due to the contribution of CO2 to the back radiation. This is assuming that CO2 contributes with 0.062 W/m^2 per year. The value of 0.062 W/m^2 per year is based on the results of the satellite measurements showing that the temperature of the atmosphere increases by 0.0162K/year, which is assigned by the climatologists to the enriching of the atmosphere by CO2.
Where is there a calculation of the thermal effect of increased evaporation at the ocean surface ?
Stephen Wilde
If we assume that the back radiation leads indeed to the decrease of the gradient through the upper part of the ocean then we will have the decrease of the flow of heat from the bulk to the surface.
However, if the water layer just below the ocean skin layer has the lower temperature than the surface skin layer (which might be due to the mechanism that you are offering in connection to evaporation) then the decrease of the temperature difference between the bulk and surface skin layer will not affect the heat flow from the ocean. The increase of the temperature of the surface skin layer would indicate in such a case only that the evaporation “sucks” energy from the adjacent layer below (or, in other words, pumps it up to the region of higher temperature characterizing the surface skin layer) and transports this energy to air in the form of both the latent heat and the other transport channels. This would mean that
a) the temperature of the layer below the skin layer remains the same as primarily to the increase of the back radiation;
b) the heat flow from the bulk to this adjacent layer is not affected;
c) the increase of the temperature of the skin is thus of no importance as to the magnitude of the heat flow from the bulk to air;
d) the bulk temperature of the ocean is not affected since evaporation continues to pump the heat up from the adjacent layer below to the skin layer as before;
e) the heat that is pumped up to the skin layer is transferred to the air through the radiation, convection, conduction and evaporation as before;
f) all the additional energy coming from the increase of the back radiation is thermalized within the skin layer and then emitted back to the air by increasing radiation, convection, conduction and evaporation into the air so that none part of it is delivered to the ocean – otherwise it would give us the decrease of the cooling of the bulk temperature of the ocean;
g) this would indicate also on that the increased back radiation will have some effect on the bulk temperature only if the saturation condition for the evaporation is achieved;
h) the temperature relation between the skin and the adjacent water layer would exist also without back radiation (i.e. if the air above would be removed) but the temperature of the surface skin layer would be much less since the heat losses to the space above would be much intensive resulting in the much rapid cooling of the bulk.
This is how I see the logic behind your description of the effect of evaporation on the cooling of the ocean and the interaction between evaporation and back radiation, but maybe I have misunderstood all the thing.
Nevertheless, even if some part of the increased back radiation does reduce the cooling of the bulk, this effect is very small.
Thank you Ernest, that appears to be a fair summary and is in accordance with that which I have always understood as regards the power of the evaporative process.
Can anyone demonstrate it to be false ?
I’m still waiting for a convincing demonstration that the ocean bulk does in fact absorb heat (from increased Downward Longwave Radiation) from more GHGs after accounting fully for the thermal effects of increased evaporation caused by the very same GHGs.
Some warming of the ocean skin does seem to occur but the evidence is lacking as to whether the subskin warms cools or stays the same as a result.
Since evaporation and radiation both combine to cause the subskin to be cooler than the ocean bulk below then one would have thought that an increase in either would intensify the cooling of the subskin rather than warm it.
However on balance I think that the extra DLR just results in a zero effect on the ‘normal’ upward energy flow because all the DLR would be used up in enhancing the rate of evaporation and accounting for the energy deficit caused by that enhancement of evaporation by virtue of the enthalpy of vapourisation (vapourisation has a net cooling effect).
I haven’t yet seen any evidence that a DLR warmed ocean skin does actually slow down the rate of energy flow from the subskin below it.
It is often said that because the temperature gradient (from subskin to skin) changes then the rate of upward energy flow must slow down but that would not be the case if the enhanced rate of evaporation speeds up the rate of flow again to negate the expected slowdown from a decreased gradient.
I think this is a critical issue for the entire AGW construct because if the extra DLR from more GHGs cannot add to ocean heat content then only the effects on the air need be considered and that would be insignificant in the face of oceanic control of surface air temperatures.
If one takes a look on the temperature variation in a very long time perspective (millions of years) then it seems to be obvious that the temperature of the Earth and, hence, that of the oceans is decreasing with the start from Cretaceous and up to nowadays, of course with some up en down going oscillation. This can be seen in http://math.ucr.edu/home/baez/temperature/temperature.html#65Myr presented by John Baez.
The general tendency is thus the cooling of water in the oceans. The precession of the axes of the Earth causes some swing of the temperature up and down but the general tendency is the decrease of temperature and this tendency seems to be retained.
I wander if there is an explanation to this general down going trend?
I think that all has to do with the location of the continents and global circulation. Put a large land mass at a pole, you get a permanent ice cap. That’s an increase in albedo and a loss of energy in. Close the Isthmus of Panama and block circulation from the Pacific to the Atlantic and you increase heat transport to the high latitudes where it is radiated away more efficiently, an increase in energy out. The result is a decrease in heat content leading to the current conditions.
The uplift of the Tibetan Plateau has likely played a part as well. The resulting annual monsoon might have an effect on energy balance too. It could also have increased the rate of weathering and helped to draw down atmospheric CO2.
DeWitt Payne. Thank you for the information.
We can thus assume that the movement of continents in relation to each other is responsible for the long term climate trends. This movement opens or closes the channels through which heat from the inner parts of the Earth reaches the surface contributing thus to energy received from the Sun.
All the other agents (collisions with asteroids, volcanoes, precession, etc) are mostly modifying the temperature profile during some longer or shorter periods of time giving thus deviations from the main temperature trend. After every such a disturbance, temperature returns back to the “main line” and continuous as if nothing has happened. Isn’t it so?
An additional comment concerning the heat developed within the surface skin layer as the result of increase of the back radiation. This extra heat might be expected to go both to the air layer above and to the water layer below. However, the transport of heat into the air might go through four channels, which are radiation, convection of the air in the air layer, heat conduction and evaporation. To the contrary, the heat transport to the adjacent water layer can be realized only by the heat conduction. The conclusion is that the most of heat is delivered very effectively back to air preventing thus any major increase of the temperature of the skin layer. This might explain why the experimentally determined decrease of the temperature different between the bulk and surface layer is so small even if the back radiation increases be 100 W/m^2 (compare the Tangaroa results).
Is it all over then ?
Ernest seems to have got the picture but if there is some fatal flaw then I’d like to know.
As was pointed out to you earlier in the thread, your conception of heat transfer violates the second law of thermodynamics. Since you can’t seem to grasp this very basic concept, further discussion is pointless.
Nonsense.
I explained why in an earlier post to which you did not respond.
The Second Law is fully complied with.
Unless, that is,you can somehow demonstrate that the evaporative process is impossible because it breaches the Second Law.
Over to you.
De Witt, Your focus on the Second Law seems utterly misconceived. I found this:
“Consider simply a black bucket of water initially at the same temperature as the air around it. If the bucket is placed in bright sunlight, it will absorb heat from the sun, as black things do. Now the water becomes warmer than the air around it, and the available energy has increased. Has entropy decreased? Has energy that was previously unavailable become available, in a closed system? No, this example is only an apparent violation of the second law. Because sunlight was admitted, the local system was not closed; the energy of sunlight was supplied from outside the local system. If we consider the larger system, including the sun, available energy has decreased and entropy has increased as required. ”
In the case under discussion additional DLR external to the system being considered has been introduced so as to provoke more evaporation with the consequences that I described. There is no breach of the Second Law.
Irrelevant to your argument. What you said above:
Wrong. It will get warmer as heat transfers from the hot surface to the colder layers below. All the energy for evaporation will come from the incoming radiation. And not all the radiant energy will be used for evaporation. Some of it will transfer downward and warm the water below the surface because there will be a huge temperature gradient between the surface and the layer below.
To do as you describe is a violation of the second law with heat transferring from colder to warmer. It would be like putting a bucket of water in the sun and having the top warm and the bottom cool. Not going to happen.
You still fail to grasp that evaporation and condensation aren’t two separate and independent processes. They are the same process running in different directions. At equilibrium the rates of evaporation and condensation are equal but not zero.
If you remove the air saturated with water vapor from above the surface of the water and continuously replace it with air at the same temperature but lower humidity, the water will cool. But only because you’re doing work by moving the air. And the surface will cool faster than the bulk.
Conversely, if you add heat to the system, two things will happen, the water will warm and the rate of evaporation will increase until the water vapor partial pressure gets high enough so that the rate of condensation again equals the rate of evaporation. But the water will always be warmer as long as heat is being added.
You also begged off answering my questions to you in this comment:
https://scienceofdoom.com/2011/01/06/does-back-radiation-%e2%80%9cheat%e2%80%9d-the-ocean-%e2%80%93-part-four/#comment-8801
The problem is simple and so is the solution. I’m still waiting for an answer, although I’m not holding my breath.
“And not all the radiant energy will be used for evaporation.”
That seems to be where you go wrong. How could any be left over in light of the enthalpy of vapourisation ?
“It would be like putting a bucket of water in the sun and having the top warm and the bottom cool”
The bottom is indeed cooler than the top. The ocean deeps are much colder than the surface. That does not prevent upward movement of solar energy out of the ocean rather than downward to the depths.
“It will get colder and deeper as it tries to balance the energy required by the evaporative/boiling process in the Knudsen layer with the energy being supplied from below via convection and conduction.”
I subsequently changed my position on that to propose no change because the DLR provides the energy for both extra evaporation and the energy shortfall with nothing left over.
You need to demonstrate how ANY DLR can be left over after the evaporation rate has increased.
No, that’s where you go wrong. You completely neglect the fact that the temperature gradient between a warm surface and a cooler bulk requires that energy be transferred so not all the incoming radiation can cause evaporation. Note that’s cause evaporation. Evaporation isn’t some spontaneous process that can somehow suck energy out of a system in violation of the second law.
But the bottom is not cooler than it was before the bucket was put in the sun. That cannot happen. Every part of the water in the bucket will warm. The different parts will warm at different rates, but no part will have a temperature lower than the initial temperature.
Solar energy leaving from the ocean surface does not come from the cooler deep ocean. It can’t because that would violate the second law. Look at the daytime temperature profile in The Cool Skin of the Ocean. The temperature increases with depth and then decreases. Heat is therefore being transferred in both directions, from the peak just below the surface to the surface and also from the peak to deeper in the ocean. When the sun goes down, the transfer is from near the surface to the surface, not from the cooler deep ocean. The temperature just below the skin is nearly constant because that layer is well mixed.
You need to demonstrate how ANY DLR can be left over after the evaporation rate has increased.
I have. It’s required by the second law of thermodynamics. The evaporation rate cannot increase unless the temperature increases. That temperature increase must come from the DLR so 100% of the the increase in DLR cannot be used for evaporation. You have failed to demonstrate that all of any increase in the DLR is used up by evaporation. That is an assertion on your part based on no evidence at all.
Heat does transfer from the warmer upper part of the ocean to the deeper cooler part, not the other way around as you claim, but it’s balanced by flows of cold water descending into the deep ocean near the poles.
” Evaporation isn’t some spontaneous process that can somehow suck energy out of a system in violation of the second law.”
It can and does suck energy out of one part of a system (the ocean skin layer) and place it in another part of the system (water vapour in the air).
If extra DLR brings forward the timing of evaporation as it does for every water molecule affected then every time a molecule does evaporate it takes another 4 times as much energy with it as the energy required to provoke the evaporative event (the enthalpy of vapourisation).
Thus there can be nothing left over.Since there is nothing left over the natural background flux from the ocean remains unaffected despite the warmth in the skin layer.
There is no breach of the Laws of Thermodynamics.
Since there appears to be no data as regards subskin temperatures (only of the skin and 5cm down) this issue cannot be resolved until someone does the necessary measuring.
Try explaining to me why the subskin is 0.3C cooler than the ocean bulk below. If you were correct there would be a continuous and smooth gradient to the surface skin and no 1mm deep cooler layer between the skin and the ocean bulk.
Something sucks out the energy from that 1mm deep layer faster than it can rise up from below or be injected by incoming solar shortwave. That ‘something’ is upward radiation and evaporation. So more upward radiation or evaporation must cool it more.
However DLR does not cool it more because the DLR provides its own energy both for the extra evaporative events and to cover the inevitable deficit thus leaving the background energy flow unaffected.
Once again you demonstrate your lack of understanding of thermodynamics. What you say is impossible. It is a violation of the second law of thermodynamics. It is exactly equivalent to the construction of a perpetual motion machine.
There is no spontaneous sucking in a system at steady state. There is no bringing forward of timing. There is only energy flow from hotter to colder and changes in the rates of flow. We have the hot sun depositing energy deep into the cooler water. That energy then flows out to even colder deep space. Net energy can only flow if there is a temperature difference and the net flow must be from hotter to colder. That’s required by the second law. Energy in must equal energy out plus energy retained by the first law. If energy retained increases, the temperature goes up and conversely. The surface temperature has to be high enough that the difference between the DLR and upward transfer of energy by radiation and convection is equal to the net deposition of solar energy minus the retained energy. Any increase in DLR must then result in an increase in energy retained and an increase in temperature, not just at the surface but in the bulk as well. No amount of hand waving about heat of vaporization can change these facts.
“Any increase in DLR must then result in an increase in energy retained and an increase in temperature, not just at the surface but in the bulk as well. No amount of hand waving about heat of vaporization can change these facts.”
Why should the energy retained need to heat the ocean bulk in order to comply with the second law ?
It is sufficient for the retention of extra energy to be from ocean skin upwards and the process of evaporation ensures just that.
In climate terms that extra energy retention manifests itself in a faster water cycle and a miniscule unmeasurable shift in the air circulation systems.
Your narrative ignores the special features of evaporation completely.
What special features? That evaporation somehow violates the second law? Nope, doesn’t happen.
http://www.webpages.ttu.edu/brecaldw/Refrigerator.htm
“Professor: So a heat pump is the same way. A heat pump is just anything that moves heat the wrong way against a temperature gradient – moving heat uphill in a sense. The hard part is that we can’t hold heat itself in a bucket and carry it around like we can with water. To get heat to move uphill we have to use some clever tricks of thermodynamics. And one of the laws of thermodynamics is that any pump that moves heat “uphill” will require at least as much energy as the heat you move.
Student: Oh yeah, I learned in my chemistry class that what you just said is one of the ways of stating the second law of thermodynamics.”
Thus extra DLR onto surface water molecules acts as an evaporative heat pump where energy is made to flow ‘uphill’ and which requires exactly as much energy as the heat moved. Therefore no DLR left over and a zero effect on the natural upward flow of energy from ocean to air despite the warmed skin layer.
For there to be any energy left over to allow warming of the ocean bulk would be the true breach of the second law because that would entail requiring less energy than the heat moved (energy left over) which the second law prohibits.
To pump heat, you have to do work. Absorption of radiation does no work. Your example fails.
The phase change from liquid to vapour does the ‘work’.
Stephen Wilde,
Since you continue to ignore my questions about the simple system containing only water, here’s the qualitative answer:
In the 1 m cube there is initially 5,000 kg water at a temperature of 293.15 K and 0.5 m3 empty space. The top wall is maintained at 293.15 K. The vapor pressure of water at 293.15 K is 2348.6 Pa which means there will be 8.6795 g of water vapor. Not correcting the heat of vaporization for the change in temperature from 0 C to 20 C, that required 8.6795g*2.502E3 J/g = 2.172E4 J. The heat capacity of water is 4183 J/kg K so the temperature change from the evaporation of the water is 2.172E4/(5000 kg * 4183 J/kgK) = 0.001 K which I will neglect.
Since the water vapor pressure equals the saturation vapor pressure, there is no net evaporation or condensation so there is no flow of energy from the top wall to the water and water vapor or vice versa by radiation or conduction nor to or from the water vapor to the water and the temperature of all components will not change with time.
To make things more exciting I’ll raise the temperature of the top wall from 293.15 K to 373.15 K. Two things happen immediately, the temperature of the water vapor goes up by conduction and absorption of radiation from the top wall and so does the temperature of the surface of the water.
Now the system is no longer at equilibrium. The water vapor pressure is below the saturation value of 101325 Pa at 373.15 K (that isn’t quite right either, but close enough) so there will be net evaporation of water from the surface as it warms from absorption of radiation and conduction from the water vapor. Note that the surface will never cool below the initial 293.15 because if it did, it would be immediately re-heated by condensation of water vapor which would then be above the saturation vapor pressure. Similarly, the water below the surface cannot cool below the initial temperature because then heat would be flowing from cool to hot in violation of the second law.
If we allow the vapor space to equilibrate, it will contain 294.1965 g ( all those significant figures aren’t justified as there are second order changes like the expansion of the liquid water with temperature) of water as the vapor for a net increase of 285.5 g and a flow of energy from the top wall (not the water) of 285.5 *2502 = 7.143 E5 J. This is actually an overestimate because the heat of vaporization decreases with temperature. Now we have 0.5 m3 of water vapor at 373.15 K in contact with water that must then have a surface temperature equal to 373.15 K. But the bulk temperature of the liquid water is still 293.15 K. There must be heat flow by conduction from the surface to the bulk because heat must flow from hotter to colder. The initial flow of heat will be very high because there is a large temperature gradient at the boundary. In fact, the surface initially won’t reach 373.15 K because of this heat flow. In fact, it won’t be exactly equal to 373.15 until all energy flow ceases and the system is at equilibrium.
The rate of heat flow will then decrease over time as the bulk warms and the temperature difference (gradient) becomes smaller. Finally, the entire system will be at 373.15 K and heat flow will stop. The total flow of heat will be 4183 * 80 * 5000 = 1.67 E9 J, again approximate because the heat capacity isn’t constant with temperature. Thus the principle of heating by thermal radiation from above is demonstrated.
In the real world, though, the Teff of the DLR is lower than the surface temperature so there will be a net flow by radiation from the surface to the atmosphere and space, which is required to remove the net energy deposited by solar radiation. DLR still increases the surface temperature above the value for DLR = 0 for all the usual reasons.
I ignored your question because it postulates a scenario entirely unlike the real world.
A cube containing water is an indequate analogy for the real world where the seas are open to the skies, the atmospheric pressure dictates that there will always be evaporation in the absence of saturation and saturation is very rare and very short lived.
In the real world one can influence the rate of evaporation either by reducing pressure or by increasing the energy content of the water (amongst other ways such as increased air movement and humidity changes).
If the energy source that adds energy to the water cannot get past the evaporative region (such as DLR) then ALL of it is used up creating extra evaporation leaving no surplus and failing to reduce the upward energy flow from the bulk ocean.
The skin temperature does indeed increase but the added heat is whisked away upward as fast as it is formed for a zero net effect on the upward flow.
Nothing will persude me otherwise apart from real world data proving that the subskin layer does indeed warm up as more DLR is added.
I am content to await it and so should you be.
Stephen Wilde says of DeWitt Payne’s excellent question:
Exams for undergraduate science courses always have questions with scenarios entirely unlike the real world.
So do good textbooks.
And for very good reason – it ensures that the student really understands the subject.
Do they know how to apply the basic elements of theory? Or not?
Once the student fully understands each element then they can combine them to solve problems which are more “real world”.
Without this grounding then any attempt to understand real world scenarios are likely to be fatally flawed.
Such is science.
Very different from writing stories.
Then you will be well able to say exactly what relevance DeWitt’s question has to the current discussion ?
“If we allow the vapor space to equilibrate”
In the real world open to space there is no ‘equilibration’.
The pressure and density differentials between water and air are such that in the absence of saturation the upward energy loss from evaporation never stops and the net global position is never one of saturation.
Thus one never reaches a point at which downward energy transfer can occur.
Hence the following never happens and the analogy becomes pointless:
“Finally, the entire system will be at 373.15 K and heat flow will stop.”
Then DeWitt says:
“In the real world, though, the Teff of the DLR is lower than the surface temperature so there will be a net flow by radiation from the surface to the atmosphere and space, which is required to remove the net energy deposited by solar radiation. DLR still increases the surface temperature above the value for DLR = 0 for all the usual reasons.”
But again he has left out the cooling effect of the ever present evaporation. DLR does increase the skin temperature but the increased evaporation takes away the energy that came from the DLR before it can either progress downward or slow down the upward rate of flow from the natural background processes.
The net flow (which is required to remove the net energy deposited by solar radiation) continiues unaffected.
Unless someone can demonstrate with data that the subskin gets warmer and not just the skin.
Is there a subskin in that cube which is cooler than the bulk below ?
No there is not because it is a closed system in which after a while the cool subskin dissipates and the energy can then conduct downward. That never happens (except locally and temporarily or via circulatory subduction) in open air.
If in the open air DLR caused the subskin to dissipate then I would agree with the closed cube analogy but that is not the real world.
Stephen Wilde on January 26, 2011 at 7:30 am:
The point is clear. As explained in many places, for example, on January 11, 2011 at 2:39 am and January 11, 2011 at 10:53 pm.
If you don’t understand the subject, why are you telling us that you have the answer?
If you do understand the subject you will be able to provide an equation that describes the factors affecting the process of evaporation.
And you will be able to answer DeWitt’s question.
You can’t answer his question because you don’t understand the subject. Not because it’s not relevant.
I can’t speak Japanese. Nothing wrong with that. Until I start explaining to a native Japanese speaker that their pronunciation and grammar are wrong.
I believe that would go beyond simple arrogance and into the category of chutzpah, and not in the good sense.
Stephen,
When we say we know more about this than you and list credentials, it’s not snobbery, it’s the simple truth.
Stephen
* 50% of solar radiation is absorbed in the first meter, and 80% within 10 meters
* 50% of ”back radiation” (atmospheric radiation) is absorbed in the first few microns (μm).
A further translation of the strange language of SoDspeak might help
When he says 50% of ”back radiation” (atmospheric radiation) is absorbed in the first few microns (μm). he really means that the surface if it is warmer than the atmosphere will radiate less than it would if it was a bare rock planet.
Yes Bryan,
Bare rock planets and enclosed cubes not open to space.
Just the ticket for dealing with evaporation.
And of course evaporation is impossible because it violates the Second Law of Thermodynamics.
You still don’t understand. Evaporation by itself cannot violate any of the laws of thermodynamics. It is a thermodynamic process. It’s your conception of how evaporation works and the resulting heat transfer that violates the Second Law.
Stephen Wilde,
Now that I’ve analyzed the static case for water in a box, let’s make things more complicated and more like the real world. Take the same 1 m cube with five sides perfectly reflective and the top a black body maintained at constant temperature containing 5,000 kg of water. I’ll put a resistive heating coil just above the bottom. Let the coil have a total resistance of 10 ohms and pass a current of 4.0988 amperes through the coil resulting in a dissipation of 168 W. The question is now, what temperature does the top of the box have to be to maintain a water surface temperature of 18 C? For the purposes of this analysis, we’ll pretend that water can’t freeze.
This one is a lot harder. I’ll give you some time to think about it.
The oceans do not have a resistive heating coil just above the bottom.
I think you are just trying to divert attention from inconvenient truths.
I’m simplifying the problem to make it easier to understand and solve. The sun on average deposits 168 W/m2 into the ocean. This is no different in principle from having no sun and a heat source below the surface. Crawl before you walk, walk before you run.
Let’s make it 15 C rather than 18 C. Here’s a hint, the temperature of the top of the box will be less than 15 C and greater than -22.75 C.
A heat source below the surface sets up a temperature profile of warm at the source to cool at the top in an undisturbed gradient.
The sun heats most at the surface so the gradient is cold at the bottom to warm at the top.
An utterly different scenario.
Now the sun would be expected to set up an undisturbed gradient from cold at the bottom to warm at the top but it does not because upward radiation from the surface plus energy drawn upwards by evaporation at the surface creates a layer 1mm deep near the surface (the subskin) which is 0.3C cooler than the water below it.
Thus there is a temperature inversion caused in part by evaporation whereby energy is drawn out of the oceans faster than energy coming up from below plus solar shortwave entering into the cooler subskin layer. The point of equlibrium is reached 1mm below the surface where the subskin joins the ocean bulk.
Nothing in your equations deals with that phenomenon and it is absent from your thought models.
So far you have directed me to a model that is not open to the sky. Then when challenged on that point you come up with a model that treats the oceans like a domestic kettle and throughout this thread you ignore or deny the process of evaporation and in particular the well known effects of the enthalpy of vapourisation.
I think you are wasting my time.
You keep harping on the lower temperature of the deep ocean and conveniently forgetting that the top 1 m of the ocean where much of the solar radiation is absorbed is measurably warmer than the skin layer both day and night. See this graph (again). If it weren’t then absorbed heat wouldn’t transfer back to the surface, which it does. The temperature of the deep ocean is irrelevant and caused by something else entirely.
Here’s another hint to the solution of the heated closed cube problem:
Having a black body for the top of the box is a convenient substitute for the effective temperature of the atmosphere as observed from the surface.
Cite please. If I’ve ignored anything, and I haven’t, it’s your interpretation of the effects of the enthalpy of vaporization that are well known only to you. In my solution to the static closed cube problem, I use the enthalpy of vaporization of water twice.
Oh, and cooling with depth (warming as the pressure decreases) is a temperature inversion, not warming.
There is now some serious effort being put into this issue as here:
http://climateclash.com/2011/01/19/climate-science%E2%80%99s-blind-spot-%E2%80%93-evaporation-cooling/
I found this interesting comment:
“This blind spot seems to be widespread in climate science: Ray Pierrehumbert in the previous post, the recent paper by Lacis and Schmidt, Judith Curry’s blog, the Scienceofdoom blog, all concentrate to excess on radiative heat transport.”
They don’t seem to pin down the basic issue investigated in this thread as yet.
Perhaps we should just await further data and better science. Clearly the relevant science is incomplete at present despite it’s critical significance for AGW theory.
Next hint: Total heat transfer, radiative plus latent cannot exceed 168 W. The rate of heat transfer by radiation and latent heat increases as the temperature at the top decreases given the assumption of constant water surface temperature.
“The sun on average deposits 168 W/m2 into the ocean. ”
and
” radiative plus latent cannot exceed 168 W.”
You should say that radiative plus latent plus convective plus conductive from ocean to air cannot exceed the sun’s 168 W/m2 plus DLR assuming the system is in equilibrium.
Thus if one increases DLR then one also increases radiative plus latent plus convective plus conductive as Ernest pointed out previously.
All that can be accommodated by energy transfers from ocean skin upwards without affecting the flow from ocean bulk to ocean skin.
You may as well stop playing games and get to your point.
You need to demonstrate that the energy flow from ocean bulk to ocean skin slows down or reverses. Equations nor theory (and I regard equations and theory as just alternative and equally valid means of expression) don’t ‘prove’ that either way. Only data will do that and no one has measured the subskin temperature changes (if any) in response to DLR changes.
They have only measured the temperature of the skin layer and a point 5cm down in the ocean bulk. Not good enough.
The problem with presenting an example as a work in progress is that all your errors are made in public. I need to change one of the conditions in the example. There needs to be air in the head space of the box. That does several things, it allows me to assume a linear gradient of water vapor pressure from top to bottom at steady state (I think), prevents the water from boiling and allows me to adjust the magnitude of the diffusion coefficient of water vapor in air by changing the air pressure so I can get the result I want. It adds the complication of sensible heat transfer by the air. I’m going to avoid some of the complication by assuming that all transfer in the head space, mass and heat, is by diffusion with no convection, even though the temperature gradient is going to be too high to be stable to convection. I also don’t care about the temperature gradient in the water. I’m fixing the skin temperature of the water and don’t care about the details of how that’s maintained.
Yes I find that quite painful too but it’s a helpful stimulus to concentration.
You are right to want to introduce the matter of air pressure because in the real world it is the atmospheric pressure that dictates the amount of energy required to provoke evaporation. That amount of energy is pressure dependent but the enthalpy of vapourisation is a constant so the lower the pressure the more energy evaporation pulls out of the local surroundings as compared to the amount of energy required to provoke it.
That explains too why evaporation always occurs under all situations on Earth as long as the air is not saturated. Under current Earthly atmospheric prsssure the enthalpy of vapourisation for water is ALWAYS larger than the energy required to provoke evaporation.
Now on Earth convection/air movement constantly refreshes the air/sea interface so saturation is a rare and short lived situation and so there is never equilibriation as I told you before. That is another inadequacy of your box analogy.
As regards the skin layer you are right to not care how the temperature is maintained. However once the skin temperature is set you then need to consider what forces act on it.
Solar energy on the skin can be largely ignored because solar energy mostly if not all goes past the skin into the ocean bulk.
The primary energy input to the skin layer is DLR whether natural or anthropogenic.
However DLR never gets past the skin layer so although it adds to the skin temperature it also adds to the rate of evaporation and as evaporation has a net cooling effect (the enthalpy of vapourisation is GREATER than the energy required to provoke evaporation) more (formerly) DLR energy is pulled out of the local environment than is required to provoke that EXTRA evaporation.
Thus ALL the DLR disappears for no net effect on the natural background upward energy flow DESPITE the warming skin.
You referred me to this:
https://scienceofdoom.files.wordpress.com/2010/12/ocean-temp-profiles-2-kawai-wada-2007.png?w=490&h=441
Now note that the cooling towards the top exists both day and night despite the huge change in energy input from above. That suggests that the subskin coolness is a constant feature independent of what happens from above. For your scenario to work the cooler subskin layer would have to dissipate to produce a warming gradient right up to the surface. The subskin layer has to disappear before one can slow down the rate of upward energy transfer or reverse it but it seems that it never does disappear or even reduce even under the stress of daytime solar input.
The existence of that persistent subskin layer 1mm deep and 0.3C cooler than the ocean bulk below is observational evidence that whatever goes on in the atmosphere has no effect on the natural background upward energy flow. The reason is that in response to increased downward DLR the upward energy transfer rate increases to negate the effect one would otherwise expect from the warmed skin layer
The logic is sound but I would be prepared to accept actual observational evidence to the contrary but we need actual evidence of SUBSKIN temperature changes. At present we only have evidence of SKIN and ocean BULK temperatures.
Nevertheless the creators of the graph seem content to show that persistent cooler subskin day and night despite huge downward energy changes so who am I to argue with them ?
Slight correction, I said this:
“The subskin layer has to disappear before one can slow down the rate of upward energy transfer or reverse it but it seems that it never does disappear or even reduce even under the stress of daytime solar input.”
Actually a reduction in the depth or intensity of the cooler subskin layer would be enough to suggest a reducing rate of upward energy flow. However the diagram shows no such reduction day or night.
It’s gone suddenly quiet so I’ll ask a simple question.
Why is the temperature gradient from the very surface to 1mm deep (where the subskin joins the ocean bulk) identical both by day and by night despite the huge change in energy input from above ?
All that changes between day and night is the gradient below 1mm which is clearly solely a result of changed solar shortwave input.
Neither variations in solar input nor in DLR from above 1mm deep appears to make a scrap of difference to the gradient from 1mm depth to the top.
Sure it changes the absolute temperature but not the gradient.
Yet it is the gradient that matters for the background flux from ocean bulk to the subskin and not the absolute temperature.
So, on the basis of that diagram, changes in energy input from solar or DLR make not a scrap of difference to the gradient from 1mm upwards and therefore cannot be affecting the energy content of the bulk ocean at all.
Now look. You have already said in earlier parts of this post that IR penetrates seawater to a depth of only a few microns.
You have also said that the top few millimeters in a still ocean are actually around a quarter of a degree cooler than the lower mixing layers, due to evaporation.
Very well; there is a difference in behavior between the molecules-thick epidermis, the mm-thick skin, and the rest of the upper mixing layer.
We also know that the heat capacity of seawater is so much greater than that of air that the top three meters of global ocean have the same capacity as the entire planetary atmosphere, and that the “mixing layer” being discussed is at least thirty times that depth.
Yet your model says nothing at all about the only layer of the ocean that could possibly be effected by incident IR: the micron layer.
Further, when mixing occurs, the first mixing will be between the micron layer and the mm layer — but we already know a) that in fact the mm layer is actually cooler than the water below it; and b) that the heat capacity of a layer measured in microns is three orders of magnitude less than that of one measured in millimeters.
This account remains incoherent and unphysical, model or no.
Craig Goodrich on February 7, 2011 at 4:31 am:
You are just telling us that you don’t understand the mechanism I have already described. Because the model contains basic and simple physics.
The main conceptual problem that people have with the heat transfer mechanisms is not understanding that a hotter skin/subskin reduces heat flow from below.
It’s not so difficult to understand but this appears to be the common conceptual problem. And repeating “how can it happen?” means that you haven’t understood it.
The physics in the model is a very simplified version of real life.
If the solution you propose is simple then it should be very easy to identify the flaw in the mechanisms.
If instead you are proposing that the model is too simple and that heat transfer mechanisms in the ocean are more complex and this is why you take issue with the results – then you need to describe the physics of your model.
If you can’t identify the flaw but don’t like the results – well, it’s clear where the difficulty lies.
“a hotter skin/subskin reduces heat flow from below.”
Not if another more effective process takes the extra energy upward and away from the hotter skin it doesn’t.
That is about as simple as it can be.
And that must be happening to produce the subskin cooler than the bulk ocean below.
You have to reduce the cooling gradient in the subskin to demonstrate any reduction in upward flow.
The graph shows that the gradient stays the same day and night so it can’t even be altered by direct solar input to the skin layer. The whole temperature profile moves to the warm side but the gradient in the subskin stays exactly the same as at night.
Thus no effect on the upward rate of flow.
Steve, I have really enjoyed reading this entire thread. The ‘kids’ can certainly learn a thing or two from traditional physics and from those taught traditional physics such as you. They don’t teach it anymore, unfortunately. Hence why traditional classical scientific thought and the scientific method is foriegn to them. As I note that you said above, specialization has created a wall between cross disciplinary knowledge – I totally agree. Makes them easier pickings for the IPCC. Will be reading more threads here and hope to find more of your input. It is very enlightening and refreshing. Long live traditional science!
The MatLab model is quite incomplete. It’s missing the Knudsen layer which whose depth is the free path lenght of 3 water molecules. It also fails to model viscosity which is the dominant force at the surface. A water bug would sink in your model.
Models direct experiments. Models do not replace experiments.
scienceofdoom
That’s not how it works. YOU need to verify the model by comparing its predictions to experiment results. Lacking that the model output remains nothing more than unconfirmed hand waving.
Moreover I quickly identified at least two flaws in the matlab model. It does not model the Knudsen layer and it ignores viscosity.
The model needs to be validated by experiment. This and only this will transform it from hypothesis to theory. This post-modern reliance on computer models is an affront to the scientific method.
It it variously claimed that ocean temperatures are rising.
But by what mechanism/s? How much is due to
– back-radiation ?
– slower cooling of the oceans into the atmosphere, due to a lower temperature gradient ?
Montalbano,
That kind of questions have no definitive answers. Oceans are warming because the net energy flow is towards oceans, but telling which component is the reason or the most important reason for that has no unique answer.
Depending on the point of view various answers may be justified but my view is that none of them is really worth presenting. There are other related questions that can be answered at least in principle, but this one cannot.
Your statement “Solar energy reaches down many meters heating the ocean from within” is incorrect. Energy diagrams show a mean of only 168W/m^2 of solar radiation reaching the surface. Even a black asphalt surface receiving such would only be “warmed” to something colder than 40 degrees below freezing point – as you can confirm with Stefan Boltzmann calculations. Furthermore, because the flux varies a lot, the actual mean temperature would be lower again because of the T^4 relationship in S-B. But the ocean is not a flat asphalt sheet anyway. The radiation is spread over many layers in the thermocline and what is absorbed in each layer (only a fraction of that 168W/m^2) is far to low to raise the temperature of any such layer to the observed temperatures, even the colder temperatures at the base of the thermocline. Even if some deep layer were warmed, the warming down there would not cause the surface to warm: the energy would eventually only get to the surface by following isotherms to the polar regions. The whole concept of direct radiation supposedly heating the surface to observed temperatures is wrong and a totally different paradigm involving entropy maximization and thermodynamics at the molecular level is now known to be what explains planetary surface temperatures being higher than what can be explained with radiation. You cannot count back radiation in S-B calculations and it is obviously different from solar radiation as seen by the penetration depths in water. Even if you did “count” 390W/m^2 mean flux into the surface, that comes nowhere near explaining a mean surface temperature of 288K because of the variation in the flux and the T^4 relationship in S-B.
Hi Doug,
Still trying to peddle your nonsense I see.
There is no question that hypothesis B and C are one and the same and correct, the slight warming of the skin causes the ocean to retain more heat it got from DSR, but the fact remains that DLR from CO2 can not enter the ocean. It can not be stored in the ocean. The ocean is not capable of slowing or delaying the warming of CO2 on the troposphere.
Thus the supposed theory of ‘ocean heat uptake’ proposed by the IPCC to explain the lack of tropospheric response to CO2 is utterly false.
CC,
Sunlight heats the ocean during the day, penetrating to a depth on the order of 100m in clear water. That heat is lost to the atmosphere and to space at night. If the DLR from the atmosphere is higher, the rate of heat loss is lower unless the surface temperature goes up to increase surface OLR. That’s how increasing DLR ‘heats’ the ocean. It doesn’t do it directly. It can’t because DLR is less than surface OLR.
ConcernedCitizen,
But if what you say is true, the emission of IR from the surface can not cool the ocean. As DeWitt points out, sunlight does penetrate rather deep, so it surely heats the ocean. And cooling by evaporation is not sufficient to balance the sunlight absorbed. So why doesn’t the ocean boil?
The answer, of course, is that the IR absorption and emission in the top mm or so actually matters, as DeWitt says.
Concerned Citizen wrote: “but the fact remains that DLR from CO2 can not enter the ocean”.
DLR is absorbed by the skin layer of the ocean – roughly the top 10 um – but that doesn’t make the skin layer immediately “boil” all of the heat from DLR into the atmosphere. The skin layer is cooler than the water immediately below, not hotter!
If LWR penetrates only the top 10 um; then LWR must all be emitted from the top 10 um too. On the average, 333 W/m2 is absorbed, but 390 W/m2 is emitted. The skin of the ocean is also the source of evaporation (averaging about 80 W/m2). For this reason, the skin layer of the ocean is almost always cooler, not hotter, than ocean a few cm below. In a sense, the ocean is heated from below the skin layer by SWR and cooled from the skin layer by net LWR and evaporation. Convection and conduction transfer heat upward into the skin layer to produce a steady state. If more DLR from more CO2 enters the skin layer, less heat from absorbed SWR will need to flow upwards, and the ocean will get warmer.
If you don’t believe these numbers, try some common sense. A large amount of SWR is absorbed by the ocean. That energy can’t escape as LWR or evaporation unless it reaches the skin layer. According to the 2LoT, heat only flows from hot to cold. For heat to flow into the skin layer, the skin layer must be colder than the bulk of the ocean immediately below.
The temperature of the top roughly 50 meters of the ocean (the mixed layer) rises and falls with the seasons. Your suggestion that radiation (the sum of SWR and LWR) can’t heat the ocean is ludicrous given the annual rise in ocean temperature every summer. And SWR (annual average of about 160 W/m2) certainly can’t warm the ocean alone – with upward LWR and evaporation totaling an average of 490 W/m2.
The interesting question is how much heat is reaching below the mixed layer and what mechanisms are moving that heat. Conduction is far to slow. The bottom of the ocean is rough, so horizontal currents can produce vertical mixing. In any case, the ARGO buoys are now monitoring heat flux below the mixed layer and their data says the ocean below the mixed layer IS warming – but not as fast as climate models predict.