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Re: Trombe wall efficiency



Will Stewart  <AE%SJSUVM1.BITNET@cmsa.Berkeley.EDU, not!> wrote:
 
>>Oops. Let me do this again: this solar siding is about 16' high, and it has
>>a continuous 1 1/2" wide airspace at the top and the bottom, so a 1' wide
>>section would have cfm = 16.6 Av sqrt((Tu-Tl)h), where Av = 1.5/12 x 1'
>>and Tu-Tl is about U/cfm, and h = 16', and
>>
>>U = 300 Btu/hr x 16 ft^2 - (68F-32F) x 16 ft^2/R1  =  4200 Btu/hr,
>>    full sun received      loss thru single glazing
>
>why are you using 68F instead of an average?

Because, Will, like I said before, the air that is near the cold glazing is
mostly unwarmed house air, with a temp of 68F. This siding has two vertical
airflow passages, each 1 1/2" wide (Norman Saunders improves on this by having
one passage, divided by a diagonal porous absorber.) The unwarmed house air 
passes vertically up through the Southern passage, from South to North through
the shadecloth, and further up and North through the Northern passage, closer
to the house, and farther from the glazing.

But go ahead and use an average if you like. It doesn't make an enormous
difference. The average temp is 100F, which would make the 4200 above about
3200, and lower the solar collection efficiency from about 85% to 66%. Who
cares? Houses need siding anyhow, and the solar collection efficiency of
vinyl siding is about 0, only slightly worse than a conventional Trombe wall,
and it doesn't do much to keep the sun from heating up the house in the summer,
or provide any natural summertime ventilation, either.

>Is the height 16' or the square footage 16 ft**2(i.e., a 1' wide glazing)?

Yes. As I said, I was talking about a 16' high x 1' wide piece of siding.

>It's been a few years since I had fluid mechanics, so I'm having trouble
>following your CFM estimates. 

I have the benefit of never having taken a course in fluid mechanics :-)
You probably learned how to do this more painstakingly. I did, however,
read Bill Yanda's book, _The Food and Heat Producing Solar Greenhouse_,
which has that nice little solar chimney formula in the back, on page 189,
cfm = 16.6 Av sqrt((Tu-Tl)h), which roughly agrees with the formula in
the 1993 ASHRAE HOF book, cfm = 407 Av sqrt((Tu-Tl)/(459+T)h), where T
is the average temp, as I recall. The two formulas agree when T is 142F.

The only other thing you need to "know" :-) is that 1 Btu heats about 55 ft^3
of air 1 degree F, which for large values of 55, means that Tu-Tl = U/cfm.
Substitute all this in the above equation, do a little algebraical cranking,
and voila, cfm=k^(2/3), or as you might say, cfm=xk**(0.6666666666666667) :-)

>Don't forget they have to overcome the plastic damper, no matter how light.

They, the teeny air molecules? If this bugs you a lot, perhaps you could
use a small fan, or a more diaphanous damper, or a manual damper or a Grainger
motorized damper controlled by a thermostat, or some Zomeworks Skylids (tm). 

>What is the heat loss from the damper area?

Not much, comparatively speaking.

>What are the kinetic energy losses around the damper?

I don't know. I suppose you could write a thesis about that. Or look it up in
the ASHRAE HOF, or go build one... Or better, a hundred. Do you want to help,
or stand around displaying your erudition and throwing darts at this idea?
As I said before, some solar engineers recognize that it's often better to 
build something and measure it, than to calculate forever and ever.

>Are you only considering radiative losses?

Not even. Just R-value losses. Simple arithmetic.

>If not, are you assuming laminar flow on your convection losses at the
>glazing and on your heat gain at the shadecloth?

I assume that a still air film has an R-value of 1. And that the air in
the sunspace is still. (Is it? 66 cfm passing through 3" x 1' is 264 fpm,
which would make an air film R-value of about 1/(1+264/176) = 0.4. But
the plastic glazing is horizontally corrugated, so the air near the plastic 
is probably still (0 velocity, if laminar flow), but the plastic has more
than 1 ft^2 of area/ft^2 of glazing, because of the corrugations, etc.
Where do we stop? With some reasonable assumptions and simple models.)

And that the glazing has an R-value of 1 on the inside surface, 0 in the
middle, and 0 on the outside surface. To a first approximation, the heat that
comes in through the glazing has nowhere to go but into the air, and into the
house. If you want to refine this, see Duffie and Beckman's _Solar Engineering
of Thermal Processes_, 2nd edition, 1991, figure 6.4.4(b), page 261, which
says that the R-value of a single piece of glass varies from about 0.75 to
1.5 for an absorber plate temperature of 0 to 200C. I call that R1. Recall
Voltaire: "The best is the enemy of the good." Recall Keynes, "In the long
run, we are all dead." As I said, passive solar houses are so disastrously
ill-designed now that I don't think it pays to get too picky about the math,
in improving their design. A little arithmetic goes a long way, for now. 

>Won't the shadecloth at the top transfer less heat to the air and more to the
>glazing due to the smaller tempature differential?

Sure. Or maybe it will just run hotter than the shadecloth below. As I said,
to a first approximation, the solar heat that enters the glazing has nowhere
else to go, except into the house. BTW, you mispelled "temperature" :-)

Regards,

Nick