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re: looking for info on pool solar heating in nj
9 sep 1996
john w. moffett wrote:
>hello all,
hello john,
>i live in new jersey with a well shaded pool. have nice sunny roof and
>would like to use solar to warm up the pool by 10 degrees.
how about floating two dozen $16 2' x 8' x 2" thick pieces of styrofoam on the
pool, and putting a layer of 5 cent/ft^2 black polyethylene film on your sunny
roof, and putting another layer of clear film over that, attached with some
75 cent/linear foot aluminum extrusion clamps, and drilling a 1/8" hole every
6" in a 1 1/2" pvc pipe along the ridgeline, to spray some pump filter water
between the two plastic films, which runs down into the gutter and back into
the pool, whenever the sun is shining and the pool is not warm enough?
the weak point in solar pool heating is usually the cover, which is a poor
insulator, compared to a house wall. a 24' x 32' 80 f pool with an outdoor
temp t might lose 24(80-t)768ft^2/r10 btu/day through a 2" styrofoam cover,
and a rooftop with a south-facing area of 20' x 32' might gain 640k btu/day
of winter sun, and lose (6 hr)(80-t)640ft^2/r1, so if ein = eout,
24(80-t)76.8 = 6(80-t)640, or equivalently,
4(80-t)76.8 = (80-t)640, or equivalently,
(80-t)76.8 = (80-t)160, or equivalently,
76.8 = 160.
hmmm. what is wrong with this solar arithmetic? is it true that solar heating
does not work, as many people say?
aha, i forgot the sun...
24(80-t)76.8 = 640k - 6(80-t)640, or equivalently,
(80-t)76.8 = 26.7k -6(80-t)26.7, or equivalently,
(80-t)(76.8+160) = 26.7k, or equivalently,
(80-t)236.8 = 26.7k, or equivalently,
80-t = 26.7k/236.8, or equivalently,
80-t = 112.8, or equivalently,
t = 80 - 112.8 = - 32.8.
so, could this pool stay 80 f when the outdoor temp is minus 33 f? probably
not, since there would be some evaporation of water and heat losses between
the floating panels, and losses to the cool ground around the sides, and 55 f
r10 ground under the bottom, but still... not bad.
how hot could we make the pool, with this arithmetic? in philadelphia,
in december, with an average outdoor temperature of 36 f? tp-36 = 112.8,
so tp = 112.8 + 36 = 148.8 f. could we make it a 104 f hot tub?
how much would it cool over a cloudy week? if it were 4' deep, and most of
the heat loss were through its r10 cover, it would hold c = (24x32)x4x64,
about 200,000 pounds of water, with a thermal resistance r = r10/(24x32), so
rc = r10/(24x32)x(24x32)x4x64 = 10x4x64 = 2560 hours or 106 days or 15 weeks.
over one week, it might cool to 36 + (104-36)exp(-1/15) = 99.7 f.
nick
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