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Air Collector - Black Heat Paint vs Selective Paint? - Page 6

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Posted by Iain McClatchie on January 18, 2006, 12:19 am
 
Morris> This has led me to consider the entire absorber area as an
Morris> "input aperture"

Yep.  You take the minimum size window through which all that light
can squeeze, and that's your input aperture.  Note that it's a bit
smaller than the glass cover face because the panel is not usually
face-on to the sun.

Your output (emission) aperture is the size of the glass face of the
panel.  It's bigger than the input aperture and that size difference
cuts directly into your selectivity ratio.

Morris> I think that spacing the vanes more closely may "close the
Morris> window" at least somewhat on the return path

Imagine you are a speck-sized insect at some point on your cover
glass.  The sight lines through half the solid angle around you
terminate on hot metal surfaces.  If the emission from those surfaces
is the same in all directions, then the total heat you receive does
change depending on which way those surfaces point.  Nor does the
number of surfaces, or their distance from you matter.  The only
things that matter are the emissivity coefficients of those surfaces,
and the temperatures of those surfaces.

More vanes is not going to close any window.  But go ahead and
experiment, don't let me stop you!

Iain> If you want to get high utilization of the solar energy crossing
Iain> your aperture, you'll need absorptivity > 0.2.

Let me expand.  You want total absorptivity better than 0.8 or so,
otherwise you're just wasting your aperture.  But the actual surface
can have lower absorptivity if it reflects any incoming light a lot.
With absorptivity of 0.2, 5 reflections gets you 67% total
absorptivity,
and 10 reflections gets you 89%.  0.2 is a little low, 0.3 is better.
But note that once you have lots of reflections, there's hardly any
difference between 0.3 and, say, 0.9:
 0.3, 10 reflections = 97% absorptive
 0.9, 10 reflections = 100% absorptive

Stagnation temperature: if you don't cool the collector with air or
water,
to what temperature does it rise?

Selectivity: ratio of absorptivity and emissivity.

Actually, these words absorptivity and emissivity are sort of bogus,
and may be leading to your difficulty with the idea of "closing the
window".

Unless we're talking about clever fiber-optic devices costing six
figures, light is bidirectional.  At any given wavelength, for a smooth
surface, the emissivity coefficient (fraction of black-body radiation
emitted) is the same as the absorptivity coefficient.

But at different wavelengths, some materials reflect/absorb
differently.
Green paint, for instance, does not absorb or emit green light well,
but does absorb and emit red light well.  If you got green paint hot
enough, in a dark box, and it didn't chemically decompose or change
state, it would not look green, it would look purple.

The Sun's surface temp is around 6000 K, and the light from there
is mostly around 400nm wavelength.  Both numbers are very
approximate.  But the point is that they are also very different than
temperature down here.  Here a solar collector might have a 375 K
surface, and emit primarily at some wavelength in the many microns.
http://scienceworld.wolfram.com/physics/WiensDisplacementLaw.html

Selective surfaces are shiny (non absorptive, not emissive) at 375 K
colors, and black (absorptive, emissive) at 6000 K colors (yellow).

Morris>  What I decided was that I could live with the reflectivity so
Morris> long as there was a really poor quality exit path (to the
Morris> outside world).

You're almost right, and barking up the right tree, but here you got it
backwards.

There is nothing you can do about the emissivity aperture or
coefficient.  Maybe if you can get the temperature of the outer part
of the collector down, that will help a little, but I wouldn't expect
much.

On the other hand, your reflection idea does a *lot* about the
absorptivity coefficient. Since you get ten tries at catching each
photon, you don't have to be particularly good at it.

So, since you can't help emissivity with geometry, let your material
selection solve that problem: get some shiny stuff.  Let your
geometry force it to absorb.

Morris> Indeed, but we're talking about serious cost increase here.
Morris> Ultimately the product must be inexpensive enough for
Morris> ordinary people to purchase as a practical means of buying
Morris> down current heating costs without having to mortgage the
Morris> farm.

Once again I think you're barking up the right tree.

Morris> FWIW, it took a year to get the maximum operating
Morris> temperature down from 180F to 165F.

What do you mean?

In any case, good luck, keep going.


Posted by Morris Dovey on January 18, 2006, 9:22 am
 
Iain McClatchie (in
1137543585.723439.18470@g44g2000cwa.googlegroups.com) said:

| Morris> This has led me to consider the entire absorber area as an
| Morris> "input aperture"
|
| Yep.  You take the minimum size window through which all that light
| can squeeze, and that's your input aperture.  Note that it's a bit
| smaller than the glass cover face because the panel is not usually
| face-on to the sun.

Essentially - though the entire internal area is painted with the
flattest, blackest paint I could find - in an effort to discard as
little as possible.

| Your output (emission) aperture is the size of the glass face of the
| panel.  It's bigger than the input aperture and that size difference
| cuts directly into your selectivity ratio.

Agreed (your vocabulary lesson is taking hold). I had actually been
thinking of this in different terms. I recognized the glazing
aperture; and had also been considering a (hypothetical) surface that
touched the front edges of all the vanes - and was attempting to wrap
my head around behavior at and behind that plane in an effort to make
the energy flow across that plane as unidirectional as reasonably
possible.

| Morris> I think that spacing the vanes more closely may "close the
| Morris> window" at least somewhat on the return path
|
| Imagine you are a speck-sized insect at some point on your cover
| glass.  The sight lines through half the solid angle around you
| terminate on hot metal surfaces.  If the emission from those
| surfaces is the same in all directions, then the total heat you
| receive does change depending on which way those surfaces point.
| Nor does the number of surfaces, or their distance from you matter.
| The only things that matter are the emissivity coefficients of
| those surfaces, and the temperatures of those surfaces.

Sounds reasonable. I think I'll need to take a bit of time to digest
this completely. More surfaces mean more metal/air exchange surface,
which may assist the process of removing energy and lower the vane's
temperatures. I'd really like to develop a better gut feel for these
behaviors - and I'd like to accumulate enough data to build a good
mathematical/software model.

| More vanes is not going to close any window.  But go ahead and
| experiment, don't let me stop you!

I won't. :-) I think the experimentation is necessary to my
understanding of what's going on.

| Iain> If you want to get high utilization of the solar energy
| crossing Iain> your aperture, you'll need absorptivity > 0.2.
|
| Let me expand.  You want total absorptivity better than 0.8 or so,
| otherwise you're just wasting your aperture.  But the actual surface
| can have lower absorptivity if it reflects any incoming light a lot.
| With absorptivity of 0.2, 5 reflections gets you 67% total
| absorptivity,
| and 10 reflections gets you 89%.  0.2 is a little low, 0.3 is
| better. But note that once you have lots of reflections, there's
| hardly any difference between 0.3 and, say, 0.9:
|  0.3, 10 reflections = 97% absorptive
|  0.9, 10 reflections = 100% absorptive
|
| Stagnation temperature: if you don't cool the collector with air or
| water, to what temperature does it rise?

Strange as it may seem, I've never tried to find out - almost all of
my recent efforts have been directed at improving the airflow and
bring the operating temperature _down_. A year ago (before the most
recent changes) the discharge temperature seemed to peak just above
180F for a 6' tall test panel.

| Selectivity: ratio of absorptivity and emissivity.

Thank you. This would seem to be a key vocabulary item that I lacked.

| Actually, these words absorptivity and emissivity are sort of bogus,
| and may be leading to your difficulty with the idea of "closing the
| window".

Perhaps so - but the difficulty may stem from simple ignorance (and
the resultant lack of understanding) as much as anything else.
Fortunately, I know how to solve this problem. :-)

| Unless we're talking about clever fiber-optic devices costing six
| figures, light is bidirectional.  At any given wavelength, for a
| smooth surface, the emissivity coefficient (fraction of black-body
| radiation emitted) is the same as the absorptivity coefficient.

Ok. I was with you up to "black-body radiation". My desk dictionary
says that a black-body is "an ideal surface or body that can absorb
completely all the radiation striking it." Presumably this ideal body
will absorb and emit equal amounts of energy. I can see that there
would/could be serious problems if there were not a balance; but other
than "output = input", does it have other qualities? Importantly, does
it emit radiation in exactly the same form that it absorbed?

| But at different wavelengths, some materials reflect/absorb
| differently.
| Green paint, for instance, does not absorb or emit green light well,
| but does absorb and emit red light well.  If you got green paint hot
| enough, in a dark box, and it didn't chemically decompose or change
| state, it would not look green, it would look purple.
|
| The Sun's surface temp is around 6000 K, and the light from there
| is mostly around 400nm wavelength.  Both numbers are very
| approximate.  But the point is that they are also very different
| than temperature down here.  Here a solar collector might have a
| 375 K surface, and emit primarily at some wavelength in the many
| microns.
| http://scienceworld.wolfram.com/physics/WiensDisplacementLaw.html

[Bookmarked. I'm not going to absorb that in a single session.]

| Selective surfaces are shiny (non absorptive, not emissive) at 375 K
| colors, and black (absorptive, emissive) at 6000 K colors (yellow).

Hmm. I think I want highly absorbtive (and consequently highly
emissive) vanes at wavelengths of highest energy content here at the
planet's surface. Is the 400nm wavelength still a good centerpoint
after atmospheric filtering; and if so, do I not want black vanes?

| Morris>  What I decided was that I could live with the reflectivity
| so Morris> long as there was a really poor quality exit path (to the
| Morris> outside world).
|
| You're almost right, and barking up the right tree, but here you
| got it backwards.
|
| There is nothing you can do about the emissivity aperture or
| coefficient.  Maybe if you can get the temperature of the outer part
| of the collector down, that will help a little, but I wouldn't
| expect much.

You just put your finger on one of the things I'm hoping that I can
accomplish by maximizing the airflow. The coolest air is that at the
front of the vane. I'm not sure what the front-to-back temperature
differential is; but I'm certain it exists.

| On the other hand, your reflection idea does a *lot* about the
| absorptivity coefficient. Since you get ten tries at catching each
| photon, you don't have to be particularly good at it.
|
| So, since you can't help emissivity with geometry, let your material
| selection solve that problem: get some shiny stuff.  Let your
| geometry force it to absorb.

Good stuff. I'm about to embark on an interesting (slightly bizarre,
but interesting) adventure with analytic geometry, aluminum ribbons, a
laser pointer, and lots of cigarette smoke.

| Morris> Indeed, but we're talking about serious cost increase here.
| Morris> Ultimately the product must be inexpensive enough for
| Morris> ordinary people to purchase as a practical means of buying
| Morris> down current heating costs without having to mortgage the
| Morris> farm.
|
| Once again I think you're barking up the right tree.
|
| Morris> FWIW, it took a year to get the maximum operating
| Morris> temperature down from 180F to 165F.
|
| What do you mean?

I felt that my 180F maximum discharge temperature for a six-foot panel
was too much warmer than the outside temperature and might be
resulting in unnecessary losses. I set the goal of lowering that
maximum by 15F by improving the airflow to more efficiently remove the
energy from the panel without noticably increasing panel costs. It was
a challenge.

| In any case, good luck, keep going.

Thanks for your considerable help as well as your good wishes.

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto



Posted by Iain McClatchie on January 18, 2006, 7:45 pm
 Morris,

Once you've got the vanes shaped right, they'll appear black whether
you paint them flat black or leave them shiny.  Until you see that I
don't think you're going to buy my idea of shiny chrome vanes.

(What's cheaper, btw: shiny chrome or shiny aluminum?  I'm
concerned the aluminum is going to get dull over time, and the major
car manufacturers proved that chrome stays shiny forever back in
the 50s.)


A perfect black body would absorb 100% of all the radiation directed at
it.  It would also emit radiation, depending only on its temperature,
following Planc's black-body radiation curve.  In other words, what
comes out has nothing to do with what goes in.  The curve shows
radiation intensity versus wavelength.  The black-body curve has two
salient features:

1) Total area goes up as the fourth power of temperature.  That means
that a little rise in temperature makes a big rise in output radiation.

2) The curve basically has a single hump in the middle.  The peak of
that hump shifts to smaller wavelengths as the black body gets hotter.
If you could see all wavelengths (instead of just the narrow
visible spectrum), you would see houses glowing, just like you can
see red-hot metal glowing.  Heat metal very hot and it glows white-
hot.  The color shift due to temperature is how you experience this
curve.  Every temperature has a corresponding color, and
temperatures like 160 F have colors you can't see, in the deep
infrared.

Anyway, a black body absorbs and emits different amounts of energy.
This is not a serious problem -- it's how you build absorbers!

The whole point of all this text is that you can build a much better
collector than a black body.  If your collector is shiny in the
infrared, it will have a low coefficient of absorption/emission for
160 F colors, and so will lose less heat.  That's why I want you to
try shiny strips of material.


Great.  You'll need to do this to work out the geometry.  I'm very
excited to hear about side-by-side comparisons of collectors, one
shiny and one flat black.


Posted by Morris Dovey on January 19, 2006, 7:37 am
 Iain McClatchie (in
1137613552.316057.240080@o13g2000cwo.googlegroups.com) said:

| Morris,
|
| Once you've got the vanes shaped right, they'll appear black whether
| you paint them flat black or leave them shiny.  Until you see that I
| don't think you're going to buy my idea of shiny chrome vanes.
|
| (What's cheaper, btw: shiny chrome or shiny aluminum?  I'm
| concerned the aluminum is going to get dull over time, and the major
| car manufacturers proved that chrome stays shiny forever back in
| the 50s.)

It would appear that, for the purposes of conducting a trial, aluminum
is _much_ less expensive. A quick web search turned up a number of
possibilities. My guess is that most of what I saw is either
clear-coated (high probability) or clear-anodize (lower probability).
Neither one is likely to have anywhere near an ideal surface.

Beyond the testing stages, there may be more durable solutions using
other metals and/or other surface treatments. Chrome plating isn't a
particularly attractive option in terms of energy consumption,
environmental issues, or continuous processes. (Know anyone capable of
extruding continuous 1.5" x 0.0025" stainless steel ribbon with a
mirror-bright surface? Me too.)

You're right - I will remain skeptical until I see the effect with my
own eyes. On the other hand, I don't have any problem with the notion
that I have much to learn - and some of my most enjoyable experiences
have been tagged with: "Well, would you just look at /that/!"

|| Ok. I was with you up to "black-body radiation". My desk dictionary
|| says that a black-body is "an ideal surface or body that can absorb
|| completely all the radiation striking it." Presumably this ideal
|| body will absorb and emit equal amounts of energy. I can see that
|| there would/could be serious problems if there were not a balance;
|| but other than "output = input", does it have other qualities?
|| Importantly, does it emit radiation in exactly the same form that
|| it absorbed?
|
| A perfect black body would absorb 100% of all the radiation
| directed at it.  It would also emit radiation, depending only on
| its temperature, following Planc's black-body radiation curve.  In
| other words, what comes out has nothing to do with what goes in.
| The curve shows radiation intensity versus wavelength.  The
| black-body curve has two salient features:
|
| 1) Total area goes up as the fourth power of temperature.  That
| means that a little rise in temperature makes a big rise in output
| radiation.
|
| 2) The curve basically has a single hump in the middle.  The peak of
| that hump shifts to smaller wavelengths as the black body gets
| hotter. If you could see all wavelengths (instead of just the narrow
| visible spectrum), you would see houses glowing, just like you can
| see red-hot metal glowing.  Heat metal very hot and it glows white-
| hot.  The color shift due to temperature is how you experience this
| curve.  Every temperature has a corresponding color, and
| temperatures like 160 F have colors you can't see, in the deep
| infrared.
|
| Anyway, a black body absorbs and emits different amounts of energy.
| This is not a serious problem -- it's how you build absorbers!

I'm beginning to suspect that Max was a fairly bright cookie with a
particularly twisted sense of humor. It'd appear that a black body can
only be a truly black body in an otherwise empty universe. :-)

My grasp of the theory is tenuous (at best) and will take some amount
of time and effort for me to absorb. Perhaps the experimentation will
help that along.

| The whole point of all this text is that you can build a much better
| collector than a black body.  If your collector is shiny in the
| infrared, it will have a low coefficient of absorption/emission for
| 160 F colors, and so will lose less heat.  That's why I want you to
| try shiny strips of material.

Why /not/ try it? There isn't a large downside - and there's an
absolutly huge "Wow!" factor involved if/when a flat collector with
shiny vanes works even just a bit better than an identical collector
with black vanes.

|| Good stuff. I'm about to embark on an interesting (slightly
|| bizarre, but interesting) adventure with analytic geometry,
|| aluminum ribbons, a laser pointer, and lots of cigarette smoke.
|
| Great.  You'll need to do this to work out the geometry.  I'm very
| excited to hear about side-by-side comparisons of collectors, one
| shiny and one flat black.

Be patient. It'll take a while - more than a day and less than a year.
If you get really impatient, bring work clothes. Bookmark my web
site - it has a map and phone number.

Gotta go finish up a set of vacuum clamps for a an eager customer in
Oregon - he doesn't know it; but he's helping to finance what might be
the next generation of home heating appliances.

Talk to you later.

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto



Posted by Jeff Thies on November 3, 2005, 6:33 pm
 Gary wrote:

Look up Stefan-Boltzman law of radiation. The radiation is proportional
to the 4th power of the absorber minus the 4th power of ambient. So,
larger temperature differentials yield much higher losses. E=K(T 4th -
To 4th). (Those are absolute temps, ie Kelvin)

   I'm sure that's at least a large part of why solar air collectors are
much more efficient at lower temperature differentials.

   Cheers,
Jeff

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