Posted by nicksanspam on October 17, 2006, 3:14 pm
I wonder where you live in Texas... 940 Btu/ft^2 of sun falls on the ground
on an average 48.8 F January day in Austin with an outdoor humidity ratio wo
= 0.0052 pounds of water per pound of dry air, so 1 ft^2 of 80 F unshaded pool
with an R1 cover would gain 0.9x940 = 846 Btu and lose 21.5h(80-48.8)1ft^2/R1
= 671 Btu/day of sensible heat and 1.5hx100(Pw-Pa) Btu/day of latent heat,
where Pw = 1.05 "Hg at 80 F and 100% RH and Pa = 29.921/(1+0.62198/wo) = 0.248
"Hg. The total loss would be 791, less than 846, so you might make it more
than 80 F with solarcover.com's $5 14'x28' 411428 clear cover, and make it
warmer still with 2 of them, or use their $9 16'x32' 411632 cover with water
under 2' of dark coping during the day.
Darkening the pool sides and bottom would also help.
Posted by Solar Guppy on October 17, 2006, 3:46 pm
You are not accounting for the loss on the other 5 sides ( the ground )
My pool in Florida, it takes 300 sf of solar thermal panels ( thermoplastic
type of pool heating ) to keep the water ~80
Texas is a bit cooler in Jan than my location so it would require even more
Posted by nicksanspam on October 17, 2006, 4:44 pm
This is a ball park calc... 90% of the heat loss is from the top. Here's
a more detailed calc: say it's above-ground... 696 ft^2 of R10 sides and
bottom would lose 24h(80-48.8)69.6 = 52.1K Btu/day. The cover would gain
12x28x846 = 284.3K Btu/day and lose 12x28x791 = 265.8K Btu/day. At 80 F,
the 12% shortfall is 284.3K-265.8K-52.1K = 33.6K Btu/day, not much, but...
If 1 ft^2 of R1 cover with 90% solar transmission collects 846 Btu and loses
6h(80-48.8)1ft^2/R1 = 187 Btu/day, the net gain is 659, so another 33.6K/659
= 51 ft^2 of cover over a dark coping could provide the extra 33.6K Btu...
solarcover.com's $9 16'x32' 411632 cover could provide an extra 116K Btu.
You seem to have left out a few details :-) How big is it? In ground?
Shaded? Covered? With a clear solar cover or a blue one?
Posted by Solar Guppy on October 17, 2006, 5:44 pm
Why do you say 90% is from the top? , the other 5 sides will have ~25 degree
delta to the ground, area wise wouldn't this double the losses
My pool is 7500 gallon in ground, about 28x14x5 , but its not a rectangle ,
more hour glass shaped. no cover .. the panels can keep up with the losses
except for those rare weeks of sub 70 degree weather
Point was your not accounting for over 65% of the surface area from a BTU
loss stand point
Posted by nicksanspam on October 17, 2006, 7:58 pm
That's how it usually works out, esp with no cover.
But the OP's new pool might have R10 sides, and some people soil has
a built-in R10 thermal resistance to downward heatflow, and the deep
ground temp in Austin is 68.6 F, vs the average 48.8 January temp,
and the pool warms the ground under it.
"No cover" makes a big difference, esp in a dry climate.