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Re: swimming pool solar covers vs. "pills"

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Re: swimming pool solar covers vs. "pills" nicksanspam 04-16-2007
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Posted by on April 16, 2007, 5:01 am



>... some day it's going to warm up and I'll want to open my in-ground
>swimming pool. In order to extend the season (I live in Maryland),
>I'm thinking about either a solar blanket or a "pill".

NREL says 1550 Btu/ft^2 of sun falls on the ground on an average 53.4 F
April day in Baltimore. The average humidity ratio w = 0.0052, with
V = 10.4 mph average windspeed, which makes the airfilm conductance
about 2+V/2 = 7.2 Btu/h-F-ft^2.

>The latter is apparently a substance that slows down evaporation,
>which presumably is the main form of head loss.

ASHRAE says an uncovered A ft^2 pool loses about 100A(Pw-Pa), where Pw
and Pa are pool and air vapor pressures. Pw = e^(17.863-9621/(460+Tp))
"Hg (100% RH at Tp (F)), approximately, and Pa = 29.921/(1+0.62198/w).

>The pills are not supposed to be as efficient as the blankets (perhaps
>the blankets have a slight insulating value...

Maybe R1 with 80% solar transmission, for a clear vs blue blanket.

>... I'm leaning towards the pill, but I'm skeptical about how well
>they work. And it's hard for me to do a scientific comparison,
>since I only have one pool :-).

You might weigh a pill and chop it up or dilute it several times,
homeopathic-like, to treat water in one dark plastic bucket (does
the coverage formula include depth?), and put a pool cover sample
over another bucket of the same kind...

If the pill completely eliminates evaporation and you wrap both buckets
with a few layers of bubble wrap with black plastic on the outside, the
covered one might gain 1240 Btu/ft^2 = 24h(Tc-53.4)1ft^2/R1, which makes
Tc = 105 F :-) The pill version might have 1550 = 24h(Tp-53.4)1ft^2x7.2,
which makes Tp = 62 F.

These will both work better if you paint your pool black
or let it get muddy, like a natural lake in the sun.

Nick


Real Goods Solar, Inc.
Posted by on April 16, 2007, 6:18 am


>NREL says 1550 Btu/ft^2 of sun falls on the ground on an average 53.4 F
>April day in Baltimore. The average humidity ratio w = 0.0052, with
>V = 10.4 mph average windspeed, which makes the airfilm conductance
>about 2+V/2 = 7.2 Btu/h-F-ft^2...

>The pill version might have 1550 = 24h(Tp-53.4)1ft^2x7.2,
>which makes Tp = 62 F.

It might be about 59 without the pill...

20 W=.0052'humidity ratio
30 PA=29.921/(1+.62198/W)'ambient vapor pressure ("Hg)
40 FOR TP=58.78 TO 58.8 STEP .01'pool temp (F)
50 PW=EXP(17.863-9621/(TP+460))'pool vapor pressure
60 GAIN=1550-24*(100*(PW-PA)+(TP-53.4)*7.2)'net gain (Btu/day)
70 PRINT TP,GAIN
80 NEXT

58.78 2.795166
58.79 .6347656
58.8 -1.525879

Nick


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