There are other things in life than these list jousts , illuminating
as they may sometimes be ; some take precedence .

In your example , which I applaud as an excellent exercise , you
analyze a very different situation than that which we were
discussing : balls with uniform surface properties . Equivalently ,
rotating spheres like the earth . Your example does point out one effect
that should affect earth temperature : change in albedo over the
poles as opposed to equatorial regions and would be , in fact , a
negative feedback on the expansion of ice caps . That's my unanalyzed
intuition in any case . Confirm or disconfirm if you like .

Perry E. Metzger wrote:
> Bob Armstrong <bob@...> writes:
>>> Imagine a white metal sphere and a black metal sphere out in space at
>>> about our distance from the earth. One half of each sphere is
>>> illuminated by the sun and absorbs energy. One half is facing the rest
>>> of the universe and is radiating it out by black body
>>> radiation. Assume good heat transfer within the sphere. If both
>>> radiate out freely on the same curves, but one absorbs more slowly,
>>> then you end up with the one that absorbs more slowly at a lower
>>> asymptotic temperature. Again, you haven't considered the entire
>>> system.
>> That is exactly the ultimate experiment I propose and will approximate .
>> Kirchoff says they will be the same asymptotic temp and so do I .
>
> You do, he does not.
>
> Here is Kirchoff's law:
>
> At thermal equilibrium, the emissivity of a body (or surface)
> equals its absorptivity.
>
> So lets analyze the situation, shall we? Since you keep invoking poor
> Mr. Kirchoff, we'll use his law over and over.
>
> Assume that we have two thin plates in space, one meter square, aimed
> with one side squarely facing the sun, and the other side facing
> space, at the same distance from the sun. For purposes of simplicity,
> assume that space is at 0K instead of 3K -- it makes little difference
> other than making the calculation simpler. Lets also assume the
> distance from the sun is 1AU, so we can just use the earth's solar
> constant.
>
> We arrange for the front of one plate to be a perfect black body, and
> for the the front of the other to reflects 90% of the incident
> radiation. Let us further assume (for simplicity) that the rear of
> each plate is a perfect black body (i.e. is painted exactly like the
> front of the first plate). This will simplify our calculation.
>
> The first plate is a perfect black body and therefore absorbs all
> incident radiation. At equilibrium, therefore, it will be absorbing
> 1367W of energy. Based on Kirchoff's law, once we achieve thermal
> equilibrium, the object must also be radiating 1367W of energy. The
> surface area of the object is about 2m^2 (consider both the front and
> the back), so it has to be emitting 683.5W/m^2.
>
> The thermodynamic temperature of a black body can be measured by its
> radiation by the Stefan-Boltzmann law, which states that the radiant
> flux is equal to the Stefan-Boltzmann constant times the fourth power
> of the temperature. Its value is 5.6704E-8.
>
> So we divide 683.5 by 5.670400E-8 and take the fourth root. I'll save
> you the trip to a calculator and note the value is 331.3K.
>
> Now, lets consider the plate with one side 90% reflective and one side
> a perfect black body. Total absorption is, by definition, 1367/10 or
> 136.7W. Total emission, by Kirchoff's law, must again be 136.7W. We
> have to now emit from both sides -- but the corollary of Kirchoff's
> law, we emit only 1/10th as much through the side that is 90%
> reflective*, so we have the black body side emitting 124.3W/m^2 or
> so. (We can't measure temperature directly from the other side because
> it is not a black body -- it is emitting about 12.4W/m^2 or so).
>
> We now note that by the Stefan-Boltzmann law, a black body emitting
> 124.3W/m^2 is at a temperature of 216.4K.
>
> Please note that 331K and 216K are NOT the same temperature. (By the
> way, for the Kelvin and Celsius challenged, 331K is about 136F, and
> and 216K is about -71F -- different temperatures indeed.)
>
> (*that's a slight lie -- no substance reflects in all bands equally,
> so the emission in the much colder temperature band will be different
> (and likely higher, resulting in an even lower temperature), but we'll
> ignore that for this exercise, just as we ignored the fact that there
> are no perfect black bodies.)
>
> Anyway, if you have better math to present, please tell me about
> it. Feel free to correct anything I did wrong.
>
>> Otherwise you could transfer energy laterally ( not front to back )
>> between them making a heat engine . That doesn't compute .
>
> Er, these objects CAN act as a heat engine, if you like. There is
> nothing wrong with that -- it doesn't violate the laws of
> thermodynamics. Heat engines are just fine in an environment where the
> heat is moving from a high heat spot (the sun) to a low heat region
> (the entire universe).
>
>>>>> First, we need to know the number we're starting with. The solar
>>>>> constant is 1367W/m^2.
>>>> Of course , the solar constant is not 4 or even 3 digits constant .
>>>> In fact at least half and estimate range to .8 and .9 and greater
>>>> of total variance in mean earth temperature has been found explained
>>>> by variation in solar radiance . I claim it will eventually be
>>>> understood to be 1.00 .
>>>
>>> I have my doubts on that. That would assume that there is no
>>> greenhouse effect at all, and there very clearly is. That would also
>>> assume there is no variance in albedo, and there very clearly is.
>> Do a little web surfing or go to my http://cosy.com/views/warm.htm for
>> links . The variability in solar output and it's correlation with
>> earth's temperature is an established fact .
>
> So what? The question is whether there is a 100% correlation between
> variance in the earth's temperature and solar radiance, not whether
> there is a majority correlation.
>
>> I used to think quite a bit about albedo because there are such extreme
>> changes in it . How can the earth not get trapped in an ice age when
>> that positively feedsback increases in the reflectivity of the planet
>> so much ?
>
> Large scale environmental models are good ways of examining that problem.
>
>> I have come to believe that the energy density at this distance
>> from the sun is the only determinant of mean temperature ,
>
> We all have our personality quirks.
>
> If you would like to bet money on this, even after I showed you the
> math, please, by all means, bet me on it. We will need to come up with
> an adequate statement of what we are betting on, of course.
>
>>>> Of course in the same 24 hours an average nuke will produce 24,000
>>>> MWh versus your 3MW .
>>> 3MWh is one acre. The average nuclear installation covers many many
>>> acres. If it covers a mere 8 acres (not unreasonable including
>>> security buffers etc) you have the same output.
>
> I note that you didn't catch my math error here, and it was a huge
> one-- 24GWh vs 24MWh. Still, 8,000 acres isn't really that much
> space. The world is filled with wastelands.
>
> Consider also that there is plenty of room in orbit, and you get both
> an extra 25% (the amount of sunlight blocked by the atmosphere and
> clouds), and the sun shines 24x7, and you can aim perpendicularly at
> the sun at all times in orbit. Moving the power to the ground via
> microwaves loses maybe 20% but you've more than made up for that here
> already...
>
>>> Also, we *will* ultimately produce multiple absorber panels, which
>>> will have much higher efficiency still.
>> I see your .3 efficiency being reached in labs . 0.2 seems to be
>> about SoA commercially .
>
> At the moment, yes, because of cost. However, manufacturing techniques
> only improve with time, and with them, costs fall.
>
>> But the talk is of doubling the lab efficiencies to .6 or more .
>
> Not yet really. No one knows how to make a multiple absorber -- yet.
>
>> That would change the area required to equal a moderate nuke from
>> 19 to 6 square miles .
>
> My number is 12.5 square miles for a "worst case" 24GWh/day setup --
> the latitude of NY in wintertime. Nearer the equator is much
> better. If we do get to .6 (multiple absorbers), we'd need about 6.25
> square miles at that "worst case" latitude.
>
>>> Consider also that
>>>
>>> a) This is just a way of collecting energy. We can store it and
>>> transport it in a variety of ways.
>>> b) The price will only go down with time, and the efficiency will only
>>> go up.
>> Pumping water is hard to beat .
>
> At the moment, yes. There are better things on the horizon. We can
> also move power around very efficiently over extreme distances if
> we're willing to use superconducting cables.
>

--
Bob Armstrong -- http://CoSy.com -- 719-337-2733
NoteComputing Environment : http://CoSy.com/CoSy/
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