Posted by Duane C. Johnson on February 9, 2004, 2:50 am
I guess the simple answer is I don't know precisely.
( or even imprecisely.)
1 transparent layer.
No air insulator spaces.
Ambient temperature is 26F
Barn temperature is 40F
Temperature rise is 14F
Not a sophisticated collector but doesn't need to be
for such a low temperature rise.
I have done some experiments with a test collector that
operated at 0% efficiency. It was made with 2 glass layers
and an aluminum flat black painted absorber all spaced
with 3/4" gaps. I used 6" of iso board insulation.
This high performance collector has:
2 layers of transparent glass insulation.
2 layers of stagnant air insulation.
Stagnant air is the primary transparent insulation.
This collector can reach a temperature rise of about
340F above ambient.
This is what I call Stasis temperature. At stasis
the collection efficiency is 0% as no heat is being
If all the heat removed until the absorber temperature
is the same as ambient the efficiency is 100% although
Most collectors operate between these extremes.
Gary's collector has no stagnant air insulation.
This would seriously limit the Stasis temperature rise.
My guess is Gary's stasis temperature rise would be
1/4 to 1/3 rd that of my example.
Let's assume the value is 1/3 of 340F or 110F.
In the case of a 14F rise the collector efficiency
(110F - 14F) / 110F
96F / 110F = 87%
The part missing in your calculation is losses due to
re-radiation. In this case radiation loss is not much
as the collector temperature rise above ambient is low.
I usually make a small test model of a collector
and cover to see how they perform. This method is
not rigorous but quite sufficient for me.
Then add the reflective augmentation and see what happens.
Home of the $5 Solar Tracker Receiver
Powered by \ \ \ //|
Thermonuclear Solar Energy from the Sun / |
Energy (the SUN) \ \ \ / / |
Red Rock Energy \ \ / / |
Duane C. Johnson Designer \ \ / \ / |
1825 Florence St Heliostat,Control,& Mounts |
White Bear Lake, Minnesota === \ / \ |
USA 55110-3364 === \ |
(651)426-4766 use Courier New Font \ |
firstname.lastname@example.org (my email: address) \ |
http://www.redrok.com (Web site) ===
Posted by nicksanspam on February 9, 2004, 2:04 pm
The left side of the equation is the aamount of sun that passes
through the collector glazing on an average December day, and
the right is the heat loss through the collector glazing over
6 hours, when I assumed it was warm and working, plus the heat
loss of the barn. If 82.2K/0.9 = 91.3K Btu of sun were to fall on
the glazing and it were to provide 24h(40.4-21.2)160 = 73.7K Btu
of useful heat, the collector efficiency would be 73.7K/91.3K = 81%.
I assumed the barn and outdoor temps were constant and the
air temp next to the glazing was the same as the barn temp,
while the collector was working, ie barn air rises up between
the screen and mesh, then gets heated as it passes from south
to north through the warm mesh. I also assumed the air temp
inside the glazing was the same as the outdoor temp during
the 18 hour non-collection time.
R1, with 90% solar transmission...
I think of that R1 as coming from the slow-moving air layer
near the inside of the glazing. It's really closer to R2/3, with
a moving air layer outside the glazing with less resistance,
depending on wind, and there's radiation loss too, and the
glazing absorbs some (2%?) of the sun, which lowers the heat
loss from the inside, and each of the two air-glazing interfaces
reflects some (4%) of the sun, and some of that is rereflected
from inside to outside, but I try to make these calculations
Really Simple. R1, period :-) We can worry about the 2nd decimal
place after our houses are 90% solar heated.
Increasing the space between the glazing and screen decreases
the air velocity and inside film conductivity (about 1.5+V/5
Btu/h-F-ft^2 for a smooth surface like glass in V mph air,
or 2+V/2 for a rough surface like stucco.)
During the day...
Bigger might be better.
I'd say it has slow-moving air, with some insulation,
We could estimate how much. His 4% vent area keeps
the temp rise low, which makes it efficient. More vent
area would raise efficiency, but it would also raise
the heat loss at night, with R1 vent vs R19 wall area.
With the vents blocked... Crudely speaking, I'd say
a square foot of R1 glazing transmitting 90% of 250
Btu/h ("full sun") with no airflow or heat loss or
heat storage behind it would lose dTx1ft^2/R1 to the
outdoors, so dT = 0.9x250 = 225 F. Two layers would
gain 0.9x0.9x250 = dTx1ft^2/R2, so dT = 405 F. The
real temp rise would be less, because of reradiation.
With the vents blocked? :-)
The screen also lowers re-radiation.
It seems more "rigorous" than these calculations,
but also more work :-)
Posted by Gary on February 9, 2004, 3:36 am
Based on your comments, I will increase the spacing between the glazing and
the siding to at least 6 inches. I take it that H/15 inches is not the
ideal -- i.e. that more is better.
I will also try to grow the collector height and width as much as my
aesthetics boss allows :-)
I also plan to build the top vents such that a single horizontal duct could
be used to connect them all at a later date -- this would allow a single
duct fan to be used to provide forced circulation. But, I want to try the
free convection first.
Thanks for the details on the flapper valve design.
I hope to get it all done in the next couple weeks.
I'll post a message to let the group know how it works out.
Posted by nicksanspam on February 9, 2004, 2:17 pm
Yes, if you can make the air move more slowly, but you start to lose more heat
through the "roof" and sidewalls. You might also use translucent siding, eg
barn skylight panels with the same corrugation pattern as the metal siding,
and leave the wall cavity uninsulated behind the collector, and insulate
inside the inside wall.
Good... maybe 24 or 32 vs 16'.
You are welcome. You might have one long piece of plastic film that covers
all the holes, with a vertical gap between the top of the holes and the flap
hinge so it requires less bouyant air pressure to open, at a smaller angle.