Are you aware the book is endorsed by RealClimate.org?

]]>No.

]]>Have you read Ray Pierrehumbert’s book?

]]>“As best as I can tell, you have invented a new law of thermodynamics. I summarize it in 3 points.

1. For every body space is the ultimate boundary of the encompassing system.

2. Radiation can only leave and enter the ultimate boundary via radiation.

3. Therefore, convective heat flux from any body within the system must be zero.

Is this your hypothesis?

I agree with steps 1 & 2.

You have invented step 3 and it appears you think it is obvious. But it’s not true and this is why heat transfer textbooks are full of conduction and convection calculations.

The surface of the earth has a net convective loss to the atmosphere. This is returned via radiation.

We can measure the radiation imbalance. We can demonstrate the convective heat transfer from the surface.

Yes, the surface would be hotter if there was no convection. That is because convection is a net heat transfer from the surface to the atmosphere. If convection = 0, the surface would be hotter.

The fact that, at the ultimate boundary of the climate system, heat is only transferred via radiation has no bearing on what heat transfer mechanisms operate at the surface.”

I agree the surface has net convective loss to the atmosphere, which significantly cools the surface, but if the surface is cooler because of convection, how is it also returned via radiation as you claim above? Unless there is net energy loss, I don’t see how the surface can be cooler as a result of convection. Can you clarify what you mean by this?

The reason why the surface is cooler due to convection is kinetic energy from the surface is being traded off for radiative energy emitted at the surface, requiring the surface to emit less radiation to maintain equilibrium output power (239 W/m^2) at the TOA, as a lot of the convected energy from the surface to the atmosphere is ultimately being radiated out to space. In other words, without convection, the surface would need to emit more and subsequently be warmer as a result to maintain equilibrium. Because we are talking about the system in the steady-state, all of the kinetic energy trade offs from the surface to the atmosphere are automatically embodied in the surface emission of 385 W/m^2, which is just the net surface energy flux in and of itself.

So in regards to no. 3 above, NO that is not ‘my hypothesis’.

]]>“[“He says “A further constraint of Conservation Of Energy is that the global net non radiative flux between the atmosphere and the surface must be zero in the steady state if the radiative flux is also zero.”]

And yet if net radiative flux is not zero then non-radiative flux is also not zero. As I have already explained, numerous times.

Conservation of energy = “conservation of energy” ≠ “conservation of radiation”.

He says “if” and offers no proof.

Here are the correct equations, using his notation and adding the missing terms:

1. For TOA:

F0 = solar radiation absorbed by the climate system, therefore at TOA, radiation leaving must balance

F0 = 239 = 385.T + (1-F).A ….[eq 1]

so far so good.

2. Surface:

Energy out = energy in

⇒ 385 + K = 239(1-Fs) + F.A ….[eq 2]

where K is net convected flux from the surface

and Fs = fraction of solar radiation absorbed by the atmosphere.

3. Atmosphere:

Energy out = energy in

⇒ A = 385(1-T) + K + 239.Fs ….[eq 3]

To prove K = 0 requires proving it.

To prove Fs=0 requires proving it.”

I’ve spent some time re-examing this issue. I realize that maybe it’s not clear that G White’s model is not a depiction of the actual behavior, but the net result equivalent of the actual behavior at the boundaries of the surface and the TOA. Do you know this? Perhaps you do.

At any rate, you seem to have agreed that the 385 W/m^2 emitted from the surface and 239 W/m^2 at the TOA are all radiative fluxes, yet you object to K and Fs equaling zero in the steady-state. How could they not be zero? Where is the energy from K or FS going? It’s either going back to the surface in some form as part of the net incoming flux of 385 W/m^2, or it’s being radiated out to space as part of the 239 W/m^2 flux leaving at the TOA.

Post albedo energy, Fs, “absorbed by the atmosphere” (mostly by clouds) that ends up being radiated out to space without ever reaching the surface is simply trading off energy emitted from the surface absorbed by clouds that would otherwise be leaving the planet, and subsequently falling back to the surface kinetically mostly in the form of precipitation. So indirectly, the full post albedo gets to the surface by requiring less of what’s absorbed by the atmosphere from the surface to be leaving the planet.

How else do you explain the graph below, which clearly shows that even when absorption is 100%, half is still escaping to space. If the full post albedo did not ultimately get to the surface one way or another, this would not be possible, would it?

]]>Fair enough. You can’t say I didn’t try.

]]>RW,

I have no background or education to be able to answer your question. I can only offer a guess that makes sense to me – the heat you’re speaking of moves from surface to atmosphere, then travels by wind within the atmosphere from tropic to higher latitudes. At that point, the return of heat to the surface is still primarily radiative. Again, this is just a complete guess on my part. I don’t pretend to know anything about this aspect.

What I do feel comfortable with is saying that Trenberth’s energy balance diagram makes complete sense to me. My studies of heat transfer in chemical engineering forced me to be very conversant with energy balance calculations. This is a very simple system where all the mechanisms make sense, all the numbers balance, despite your attempts to make them unbalance. I don’t see any chance of us reaching a common understanding.

]]>Whoops, I forgot to provide the link to paper:

]]>Take a look at this paper, specifically on pages 19 and 20. G. White is not the only researcher deriving essentially the same energy flux relationships, where 239 W/m^2 gets to the surface and half of what’s absorbed from the surface emitted of 390 W/m^2 goes to space and half goes back to the surface.

I think we can at least agree that all of the energy at and below the surface originated from the Sun, correct (excluding an infinitesimal amount from geothermal)???

Also, if the surface is emitting 390 W/m^2 and 70 W/m^2 of this goes to space through the ‘window’, there is only 320 W/m^2 left to be absorbed by the atmosphere (390 – 70 = 320). If only 98 W/m^2 (165 – 67 = 98) of the 320 W/m^2 is emitted to space, that leaves only 222 W/m^2 available for downward emitted LW. If 102 W/m^2 from latent heat and thermals is subtracted from 222 W/m^2, there is only 120 W/m^2 left, which is short of the net flux of 390 W/m^2 required at the surface; 168 directly from the Sun + 122 from the atmosphere = 290 W/m^2 (390 W/m^2 required). There is not enough energy to account for all the fluxes.

On the other hand, if the 67 W/m^2 gets to the surface, it all balances out perfectly. 168 W/m^2 directly from the Sun arrives at the surface. 324 W/m^2 of ‘back radiation’ minus 102 from latent heat and thermals = 222 W/m^2 net arriving at the surface (168 + 222 = 390); 222 – 67 from the Sun = 155, and 155 + 168 + 67 = 390 W/m^2, the exact amount emitted at the surface. 390 – 70 – 165 = 155 W/m^2, the exact amount coming back from the atmosphere, and of course 165 + 70 = 235 W/m^2, the exact amount leaving. And finally, 168 + 67 = 235, the exact amount arriving from the Sun.

]]>I don’t see how the DLR value could be lower when the atmosphere was transparent to shortwave. In order to reduce DLR you’d need to lower the temperature of the atmosphere, but I can’t see how this could happen if you absorb all the shortwave on ground level instead of partly in the atmosphere.

]]>