Do Global Energy budgets make sense. ???

[Derek]

From this thread, and, THANK YOU Richard111.
http://www.globalwarmingskeptics.info/fo...age-1.html

Post 15

(11-29-2010 11:48 AM)Richard111 Wrote:  Q_C, thanks for that maths link. Will be very helpfull.

I've got the e-book and it was well worth the effort and the money just to read the following quote and get my mind all screwed up:

Quote:As for the famous minus 18C surface temperature that earth is supposed to have without the greenhouse effect, that figure assumes a blackbody surface absorbing about 239W/m^2 "on average." But check the Kiehl-Trenberth chart. Due to clouds and other obscuring factors, the actual surface average is given as only 168W/m^2. That figure corresponds to minus 40C on the surface, meaning that it has to rise by 55 degrees, not 33, in order to reach the accepted average of plus 15. Anyone who tells you, then, that the 'greenhouse effect' makes the earth's surface 33 degrees warmer is merely confessing his (or her) own ignorance.

Check; minus 40C equals 233.15K. Put that into the formulae
E = 1.0 * sigma * T^4 (for a 'blackbody', hah!)
where sigma equals 0.000000057 (5.7e-8)
and T equals 233.15
and you get 168W/m^2.

Here is a chart showing that 168W/m^2:

http://www.globalwarmingskeptics.info/fo...ml#pid6199

If you use an average global absorbance/emissivity figure of 0.7 (i.e. a very poor 'greybody') you can get that -18C for the surface at 168W/m^2 so did Kiehl-Trenberth do something like that? Thats a very low figure, 0.7, more ususally around 0.9. This of course means that any 'greenhouse' effect must be that much stronger to push the temperature up by 33 degrees. Ah, well, more puzzles to figure out.

Post 25

(12-01-2010 04:19 AM)Derek Wrote:  

(11-29-2010 11:48 AM)Richard111 Wrote:  Check; minus 40C equals 233.15K. Put that into the formulae
E = 1.0 * sigma * T^4 (for a 'blackbody', hah!)
where sigma equals 0.000000057 (5.7e-8)
and T equals 233.15
and you get 168W/m^2.

Just wondering aloud Richard111.
Is this difference the difference between the sphere's average temp and the discs average temp.
(the disc has the same peak and average temp, the sphere has different temps for max, min, and average)
Or, put another way, the spheres average temp (which is what I think you have calculated - more correctly than K&T incidentally) is lower than a discs average temp
(because a half lit sphere has 4 times the area which is lit than that of a same diameter lit disc -Alan described this as diluted in the book).

So, you need to multiply the sphere's average temp you have calculated by the square root of two, to get the sphere's peak temp.
I think this means you use, T equals 329.724 for your calculation.
Which (should be) is equal to the discs average temp (remember this is K&T "disc world" we are looking at here, not the real world).

Does that make it 33 rather than 55. ?

Post 26

(12-01-2010 06:36 AM)Richard111 Wrote:  Derek, that is simply a check. Nothing hidden, nothing implied. T is just the temperature in degrees Kelvin (you must not use celcius in these calcs). The point is the K-T budget figures don't agree with the explanation.

If a blackbody has a temperature of -40C (233.15K) then it will radiate 168W/m^2. But the global surface temperature of the planet should be -18C BEFORE the greenhouse effect pushes up the temperature by 33 degrees to +15C. But to have -18C you need 240W/m^2. And for +15C you need 390W/m^2.

None of those figures make sense. So K-T must be applying a greybody fudge factor. Why don't they say what they are doing?

None of the above will actually apply to a rotating planet where half the planet is always sheltered from incoming energy. I must do much more study on that aspect before I make any comment. I do love this book!

Anyway, just for fun now. I've often posted about my time in Singapore and noticed how small are the daily temperature changes compared to say the Karroo in South Africa. It seems the answer is water vapour. That wonderful greenhouse gas H2O. At or near the equator there is so much water vapour in the air that it blocks a lot of incoming solar radiation preventing the temperature rising as much as it would in a dry desert region. But at night it absorbs the upwelling longwave surface radiation and slows down the rate of cooling at night, again as compared to a desert region. Now the deserts have the same amount of CO2 as the equator but desert temperature monitors do not show any effect of the increasing CO2 because CO2 does not have the traction of H2O. If you don't already know, look up why deserts occur where they do both north and south of the equator. Fascinating planet we live on.

quick edit: found my signature in the book at location 1205

Post 27

(12-01-2010 07:34 AM)Derek Wrote:  Richard111, I think you may want to read post 25 again.

Disc and sphere average temps for the same input, they are not the same.
- that is the "fudge factor" you were looking for, as I understand your post 15.
ie K&T = "disc world" average temp, you calculated a real sphere average temp (which is lower).
To check if I'm right multiply 233.15K by the square root of two, and run your calculation again.
(That converts sphere average temp to, sphere max temp, which is the same as disc average temp.)

I did not use celcius anywhere. ?
Is there a difference between a 33 and 55 degrees Celcius or a 33 and 55 degrees Kelvin change in temp. ?

Later edit, after dog walk... - The "disc world" calculation is the "fudge factor", because it
a) Gives a single, and seemingly constant figure to explain, simply easier.
and
b) It gives a false sense of "continuity" or "compatability" with GMTs.
(It is probably easier to model as well, not that the grid structure of models could cope with sphere "greenhouse effect" variations as they should be calculated.)

I recently looked again at MLO with respect to GISS, I could not make a comparison, because
one set was too smooth (MLO) and the other too noisy (GISS).
I suggest something similar applies here if you used sphere calculations.

Disc calculation = 33 degrees, "constant" and "everywhere all the time" difference to be explained.

Sphere calculation = 55 degrees AVERAGE (on the lit side), with a minimum effect of 33 degrees at one spot only, where the sun is directly over head.
Everywhere else the effect is different, averaging 55 degrees, which means at the lit edges there would be even more of a difference.
I do not know if there is an easy way to calculate the maximum difference, if anyone can help in this respect that would be greatly appreciated.
Then there is night time on a rotating planet, that you do not get on a disc... That is an even bigger difference no doubt (water jacket anyone).

For example, if you were stood on the equator at mid summer over the length of the day the effect / difference to explain would vary considerably as the planet rotated.
From a minimum of 33 degrees (sun over head, mid day), and a maximum of (working guess) 75 degrees (dawn and dusk),
whilst averaging in between these times 55 degrees...
There is also night time to take into account...

So, if you had to marry this effect up to some global temperature data sets, and then model them,
there really is no contest, "disc world".

In the end Alan Siddons makes the valid point, if calculated correctly (for a sphere rather than a disc) the "greenhouse effect" should be
an average of 55 degrees (on the lit side, even more on the dark side) and vary considerably,
not be a constant 33 degrees "everywhere" and "all the time".

It is easy to see why "they" went for, and defend so strongly "disc world",
without "disc world" they are simply lost.