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.

--
Perry E. Metzger perry@...