Posted by nicksanspam on April 6, 2005, 7:53 pm
<William P. N. Smith> wrote:
You might build something like Gary Reysa's "solar pond," eg an 8'x16'x1'
deep excavation lined with 2" of styrofoam under plastic film with a 2x4
frame at the top and another 2" of Styrofoam floating on top under a layer
of plastic film under a layer of EPDM, with an airspace and a single layer
of polycarbonate over that, and a 4' reflective wall to the north. Water
would flow between the plastic film and the EPDM during the day and drain
back into the pond. The pond would heat pressurized cold water with 800'
of 1/2" plastic pipe in flat floating spirals.
In Phila, something like this might collect 0.9x8x16x620 = 71.4K Btu
from above plus 0.9x0.9x4x16x1000 = 51.8K from the reflective wall on
an average 34 F January day. At 130 F, it might lose 6h(130-34)8x16/R1
= 73.7K, for a net gain of 49.5K Btu/day.
Posted by nick pine on April 9, 2005, 2:28 pm
Deeper would be nice, for more cloudy-day heat storage, and
we don't really need that much "excavation," just a perimeter
trench inside an earth berm formed from the soil in the trench.
It might have a 4' deep x 2' wide trench.
And we might get away with (only) perimeter insulation lining
the bottom and outside wall of the trench.
This is similar to the "Freshwater Floating-Collector-Type Pond"
described by M. Sokolov and A. Arbel in Solar Energy Vol 44 No. 1.,
pp 13-21, 1990.
It would work better in January as an 8' iscoceles A-frame
with polycarbonate on the 60-degree-sloped wall and foil-faced
foamboard for the north wall, with reflective Mylar greased
to the board. The polycarbonate would shed snow and keep the
How much? The pond might collect about 0.9x8x16x250 = 28.8K Btu/h in
full sun. A 10 F dT means moving 28.8K/10 = 2880 pounds of water per
hour, ie about 6 gpm with a low-head pump, or less, if it pumps cool
stratified water up from the trench.
Gary paid $0 for a 400' roll of this pipe. It's 0.722" OD, so a
a 4' OD x 1' ID spiral with 2.75Pi ft^2 would contain 196' of pipe.
In Phila, the A-frame version might collect 0.9x4x16x620 = 35.7K Btu
from above plus 0.9x0.9x6.93x16x1000 = 89.8K from the reflective wall
on an average 34 F January day and lose 6h(130-34)8x16/R1 = 73.7K,
for a net gain of 51.8K Btu/day.
Over 5 cloudy days, the pond might provide 250K Btu of hot water and
lose 120h(120-30)8x16/R20 Btu through the cover. With a 110 F minimum
usable water temp, it needs an average 2.4' depth for 5 cloudy days.
How much hot water would it produce in June, when the sun rises and
sets behind the back wall?
Posted by nick pine on April 10, 2005, 5:33 pm
I wonder what you mean by that. Tankless performance seems unimportant,
if it is rarely used. If we store solar heat for 5 cloudy days in a row
and cloudy days are like coin flips, the tankless only gets used 2^-6
= 1/64 of the time.
Perimeter trench math: 2' deep x 1' wide can make an 8'x16' pond with
an undisturbed middle that holds 2'x8'x16' = 256 ft^3 in the upper 2'
plus 1x2x2(8+14) = 88 ft^3 in the trench, a total of 344 ft^3, vs the
311 ft^3 needed for 5 cloudy days.
Grainger's $01 5P428 pump might be nice. It can move 13 gpm with 4'
of head, using 120 V at 0.53 amps.
Posted by sparky on April 10, 2005, 7:03 pm
That cloudy day business only works if you live where there are lot's
of cloudy days!
Posted by nick pine on April 10, 2005, 11:04 pm
We might run vinegar through it once in a while, but my point was
that the tankless heater efficiency becomes unimportant if it is
rarely used. If it only provides 1/64 (1.6%) of the heating energy,
the difference between 50 and 100% tankless efficiency is very small,
compared to the overall efficiency.