Yet another solar heated swimming pool
Subject: Yet another solar heated swimming pool
From: email@example.com (Nick Pine)
Date: 24 Mar 1996 13:53:02 -0500
.geodesic: 4449 alt.architecture.alternative:6498 sci.environment:88954 alt.home
.repair: 20777 sci.engr.heat-vent-ac:5811 alt.solar.thermal:643
Article: 643 of alt.solar.thermal
Newsgroups: sci.energy, alt.energy.renewable, bit.listserv.geodesic, alt.architect
Organization: Villanova University
Xref: newz.oit.unc.edu sci.energy:47660 alt.energy.renewable:14645 bit.listserv
So, I'm looking at the $450 15' diameter x 42" deep swimming pool in the 1993
Spring/Summer J C Penny catalog, which would contain about 619 cubic feet of
water, ie about 5,000 gallons or 40,000 pounds of water...
Solar pool heating is easy, compared to other kinds of solar heating, since
temperatures are low, and there's a built-in thermal mass. What would it take
to keep this pool warm all winter, or to make it into a huge outdoor hot tub,
in the 5,500 F degree day climate where I live, near Philadelphia?
The biggest problem with solar pool heating is the cover. "Solar pool covers"
have low R-values, usually not more than R1, and they are tinted blue, which
keeps the sun out :-) Suppose we somehow cover the pool with a fairly rigid
R-20 cover, and keep it closed all the time, except when the pool is in use.
The cover might be a 20' x 20' x 6" thick panel made with 2x6s on 4' centers,
with 6" of fiberglass insulation inside, and 1/4" flakeboard or galvanized
sheet metal on top, with some plastic-coated shiny aluminum foil on the bottom.
Innovative Insulation in Arlington, Texas, sells some shiny tough foil for
mobile home roofs in 4' wide rolls for about 50 cents/ft^2. The pool cover
might be hinged with a pipe along the north edge, with a counterweight, so
it could easily be tilted up to the south at about a 45 degree angle. This
would reflect some winter sun into the pool and reduce the wind when the cover
is open, but the main pool heating would come from a 4' high x 20' long piece
of vertical plastic glazing along the south edge, with an air gap behind that
and some dark-colored straw bales behind that, to act as a solar air heater.
Suppose the pool were surrounded by straw bales, each 16" x 16" x 36", costing
$2 each, eg 40 straw bales under the pool, and another 70 bales around all
the sides of the pool, to make a 4' high wall around the pool, 6" taller than
the pool, with an average R-value of 20 (?) The bales around the perimeter
could be stacked up and mortared together like giant concrete blocks to make
a deck for the pool with a wide flat edge, and the soft underside of the cover
might deform and sit on the deck to make a good seal when it is closed.
The straw bales under the pool itself could be spread apart to make 8 air
ducts, each 6" wide x 16" deep, running from north to south along the ground,
to allow solar-warmed air from the south glazing to travel across the top
of the pool, under the cover, down the north back wall of the pool, between
the pool and the bales, and under the pool from north to south. This might
work with passive plastic film gravity backdraft air dampers in holes at the
top and bottom of the straw bales behind the glazing, but a PV-powered (-:)
fan would make it work better. The whole ground area should be covered with
some sort of vapor barrier, eg poly film, to keep moisture out of the straw,
and the tops of the ducts below the pool should have some bridging, eg 1/8"
of newspaper with 3 layers of chicken wire on top of that, and an inch of sand
mix cement over that. Let's ignore the heat loss from the ducts to the ground.
A shallow reflecting pool made from a single piece of 10' wide EPDM rubber
roofing material over a 6" earth berm around the edge would augment the sun
when frozen and keep weeds from growing up in front of the glazing.
It might look like this, not to scale: ^ (open)
gggggggggggggggggg g..................P _
gggggggggggggggggg <-- S g ppppppppppppppp P
gggggggggggggggggg g pppp water pppp P 4'
gDDDDDDDDDDDDDDDDg r ~~~~~~~~~~ g ppppppppppppppp P
|<--- 8' --->| |<--- 15' --->| P
|<------ 18' ----->|
p cgsssssssssssssssssss .
p is the pool r cgs ppp s .
s is straw p EPDM cgs p-c- (D) -c-p HcccccCW
g is glazing . rubber cgs (D) s .
c is the cover . cgsp -c- (D) -c- pHcccccCW
D are hot air ducts . 20' cgsp (D) ps .
rp is a reflecting pool cgsp -c- (D) -c- pHcccccCW
H is a hinge . cgs (D) s .
P is a post . cgs p-c- (D) -c-p HcccccCW
CW is a counterweight. cgs ppp s .
. cgsssssssssssssssssss .
|<-- 8' -->|<--- 20' --->|
Suppose that in winter, 1,000 Btu ft^2/day makes it into the single layer
flat polycarbonate glazing, augmented 50% by the frozen reflecting pool
on an average day in December. Then the solar energy input would be
Ein = 1,000 x 1.5 x 4' x 20' = 120K Btu/day,
Suppose that during 6 hours of sun on an average day in December,
the outdoor temp is 36 F, and the pool has some sort of R0 cover that
just prevents evaporation, and the air heater air and the pool water
have the same temperature T.
Then the energy that flows out of the pool during an average December day is
Eout = 6 hours (T-36) 80 ft^2/R1 from the south glazing during the day
+ 18 hours (T-36) 80 ft^2/R20 from the south glazing at night
+ 24 hours (T-36) 400 ft^2/R20 up through the cover all day
+ 24 hours (T-36) 240 ft^2/R20 out through the ENW sides all day
= (6x80/1 + 18x80/20 + 24x400/20 + 24x240/20) (T-36)
= (480 + 72 + 480 + 288 ) (T-36)
= 1320 (T-36), so if Ein = Eout for an average day, we have
1320 (T-36) = 120K or
T = 36 + 120K/1320 = 126.9 degrees F.
Looks good... :-) Suppose we keep the pool at 106 F, eg by allowing some of
the average 30 gallons of rain per day that fall on the pool cover in PA to
flow through the pool and into the reflecting pool, which would keep the pool
cleaner with fewer chemicals.
How much will the pool cool on a average December day with no sun?
Eout = 24 hours (106-36) 80 ft^2/R20 from the south glazing all day
+ 24 hours (106-36) 400 ft^2/R20 up through the cover all day
+ 24 hours (106-36) 240 ft^2/R20 out through the ENW sides all day
= (24x70x80/20 + 24x70x400/20 + 24x70x240/20)
= 6,720 + 33,600 + 20,160 = 60,480 Btu.
Since 1 pound of water loses 1 Btu when it cools 1 degree F, the pool temp
after a day without sun would be roughly 106 - 60,480/40,000 = 104.5 F.
How many 36 F days without sun would it take for the pool to reach 70 F?
The pool has a thermal resistance of R = R20/720 ft^2, so its RC time constant
is RC = R20/720x40,000 Btu/F = 1,111 hours or 46 days.
With no sun at all, and the cover closed, the pool temperature should be about
T = 36 + (106 - 36) exp(-t/46), over many days, so if T = 70,
34 = 60 exp(-t/46), or 0.57 = exp(-t/46), or t = - 46 ln(0.57) = 26 days.
If it were 10 below zero outside for t days without any sun, the pool
might just begin to form a thin layer of ice on top when
32 = -10 + (106-(-10)) exp(-t/46) ==> t = -46 ln(42/116) = 16.7 days.
Freezing it solid from top to bottom would take another 144 Btu/lb x 40K lb
= 5.76 million Btu, as the pool loses 36K Btu/day, not counting the thermal
resistance of the ice layer, ie at least another 159 days with no sun, at
minus 10 degrees F.
It's a snap to save energy in this country. As soon as more people become
involved in the basic math of heat transfer and get a gut-level, as well as
intellectual, grasp on how a house works, solution after solution will appear.