Subject: A high-performance, cost-effective, solar-heated pool [update]

From: [email-masked] (Nick Pine)
Date: 14 May 1995 08:14:39 -0400

[When I calculated how warm the air in the greenhouse over the pool would be
during the day, after the previous posting, it turned out to be less than 80F,
so it seems better not to elevate the pool and circulate greenhouse air under
the pool. The text below reflects this simpler version, vs. the previous one.]

Solar pool heating should be easy, since temperatures are low, it is usually
done only in mild weather, and there is a large inherent water heat battery.
The May/June issue of Solar Today, published by the American Solar Energy
Society, has a primer on conventional solar pool heating that says: "In many
areas of the country, solar pool heating is the most economically attractive
solar technology available today."

But there is still room for improvement, especially for year-round heating.
Yesterday, a physics teacher told me about a remarkable house that was next
to a swimming pool, which was under a conventional greenhouse. In winter, the
engineer/owner pumped some of the swimming pool water into the house, to heat
the house. This was done years ago, in PA, and it's probably all gone now,
along with the engineer. I wonder how it worked. A water-source heat pump?

Here is one way to heat a pool:

1. Buy a 15 x 30' rectangular steel pool from J C Penny, for about $2000.
Or build one out of 2 x 6s, with ferrocement over 4' wide chicken wire. 

2. Make a 15 x 30' rigid pool cover out of 2" of beadboard and 2 x 4s, with a
layer of Thermo-ply or perhaps foil-faced foamboard underneath, sprayed with
clear urethane, shiny side down, and another layer of Thermo-ply on top, shiny
side up. This would be hinged along the 30' north edge of the pool, with a few
counterweights hanging over that edge. Put an R1 solar pool cover underneath.

3. Buy a 21' x 36' commercial plastic film greenhouse with extra-long ground
stakes, for about $1000, from Stuppy at (800) 877-5025 or E C Geiger at
(800) 432-9434, and put it up over the pool. Or build an A-frame over the
pool, with Dynaglas for the south side and ferrocement over kerfed 16' 2 x 4s,
on 4' centers, for the curved north side. Attach a $139 1/3 HP Sears garage
door opener with all of its safety features and limit switches to the top
of the greenhouse, to lift up the pool cover to an angle of about 45 degrees,
when the sun is shining and the pool needs heat. 

Why this should work:

In the Philadelphia area, in January, the average amount of sun that comes
into the greenhouse, Ein, is about 12' (height) x 36' x 1000 Btu/ft^2/day, ie
430K Btu/day, over about 6 hours, ie 72K Btu/hour. The average outdoor temp
in this area in January is about 30F. Say the pool is kept at 80F, in January,
and the dry ground under the pool has an R-value of 10, for downward heatflow,
and an average temperature of 50F, and the sides of the pool are covered with
an inch of R4 beadboard.

Let's find out what the greenhouse air temperature, Tg, is at night. Assume 
the greenhouse is covered with plastic film with an R-value of 1, and the pool
cover has an R-value of 12, when it is closed. Here is an electrical analog,
ignoring the bottom of the pool:

                      Rtop (night)   Tg    Rg
            80F-------wwww-----------|-----wwww-----30F
                 |                   |   
                 |    Rsides         |
                  ----wwww----------- 
		 
           pool------->|<----greenhouse---->|<---outdoors

Rtop (night) = R12/(15 x 30) = 0.02666... Rsides = R4/360 ft^2 = 0.01111...,

and Rg = R1/(area of endwalls + area of half cylinder)
       = R1/(pi x (21'/2)^2 + pi x 21' x 36' / 2) = 0.00065.

This simplifies to a schematic like this:

                      Rp (night)     Tg    Rg
            80F-------wwww-----------|-----wwww-----30F

where Rp (night) is the resistance of Rtop (night) and Rsides in parallel,
ie Rp (night) = 1/(1/Rtop (night) + 1/Rsides) = 0.00784.

To find Tg at night, we can use Ohm's law for heatflow to first find the
heatflow from the pool to the outdoors, (80-30)/(Rp (night) + Rg) = 5887
Btu/hour. Then multiply that by Rg, to find the temperature drop across the
plastic film walls of the greenhouse, which is 5887 x 0.00065 = 3.83F. Tg is
this number plus 30F, ie, Tg = 30 + 3.83 = 33.83F, just above freezing.

The pool's solar heat gain during the day depends on the air temperature in
the greenhouse when the sun is shining, and an analogous electrical circuit
looks like this:

                      Rtop (day)     Tg    Rg
            80F-------wwww-----------|-----wwww-----30F
                 |                   |
                 |    Rsides         |
    |    ---     |----wwww-----------|
  | |---|-->|----|                   |
    |    ---     |                   |
         60K     w                  ---
        Btu/hr   w Rbot            | ^ |  12K
                 w                 | | | Btu/hr
                 |                  ---
                 |                   |
                50F                 --- 
                                     -

The sun is a current source, putting most of its energy into the pool, ie, 
I'm assuming that all of the energy that falls on the underside of the raised
cover, 12' high x 30' long x 1000 Btu/day over 6 hours, or 60K Btu/hr, is
reflected down into the pool (this assumption is dubious, because the aluminum
foil face of Thermo-Ply is not like a mirror, and also because some of the
morning and afternoon sun gets reflected sideways, not down into the pool.)
The sun that doesn't fall on the bottom of the cover puts another 12K Btu/hour
into the greenhouse itself. The schematic above can be simplified, in order to
find the greenhouse air temperature Tg, by ignoring the sun going into the
pool directly, and ignoring the heat loss from the bottom of the pool, for
the moment, since we assumed that the pool temperature is 80 degrees.

With the pool cover open, Rtop (day) = R1/(15' x 30') = 0.00222...

This leads to the simplified schematic below:

                      Rp (day)       Tg    Rg
            80F-------wwww-----------|-----wwww-----30F
                                     |
                                     |
                                    ---
                                   | ^ |  12K
                                   | | | Btu/hr
                                    ---
                                     |
                                    --- 
                                     -

where Rp (day) = 1/(1/Rtop (day) + 1/Rsides)
	       = 1/(450 + 90) = 0.00185.

This can be further simplified by removing the sun for a moment, to find
an equivalent circuit (Thevenin) for what is left of the circuit above:

                       Rp (day)      Tg    Rg
             80F-------wwww----------|-----wwww----30F
      
In the equivalent circuit below, the resistance Rt is the parallel combination
of Rp (day) and Rg, ie Rt = 1/(1/Rp (day) + 1/Rg) = 1/(540 + 1538) = 0.000502.

                       Rt            Tg
              Tt-------wwww----------|

Tt is the Thevinin equivalent temperature. To find Tt, we can find the
heatflow out of the pool again, (80-30)/(Rp (day) + Rg) = 20K Btu/hr, and
multiply that by Rg to find the temperature difference across the plastic
film wall of the greenhouse again, 20K Amps x .00065 Ohms = 13 volts, oops,
degrees, and add that to the outside temperature of 30 degrees, so Tt = 43F.

Now let's turn the sun back on:

                       Rt            Tg 
             43F-------wwww----------|
                                     |
                                     |
                                    ---
                                   | ^ |  12K
                                   | | | Btu/hr
                                    ---
                                     |
                                    --- 
                                     -

The greenhouse air temperature Tg is 43F plus the temperature drop across Rt,
ie Tg = Tt + Rt x 12K Btu/hr = 43 + 0.000502 x 12K = 49.2 degrees F. So the
air in the greenhouse will be rather cool when the sun is shining, since most
of the sun's heat will be going into the pool. This is efficient.

Will the pool be able to stay at 80 degrees F, with the assumptions thus far?

Will Dudley Doright save Nell from the clutches of Snidely Whiplash?

Let's plug in the greenhouse air temperature that we just calculated, and see
if the pool will be gaining or losing heat, if the greenhouse air is 49.2F:

                      Rtop (day)     49.2  Rg
            80F-------wwww-----------|-----wwww-----30F
                 |                   |
                 |    Rsides         |
    |    ---     |----wwww-----------|
  | |---|-->|----|                   |
    |    ---     |                   |
         60K     w                  ---
        Btu/hr   w Rbot            | ^ |  12K
                 w                 | | | Btu/hr
                 |                  ---
                 |                   |
                50F                 --- 
                                     -
     
Or more simply: 

                      Rtop (day)     49.2
            80F-------wwww-----------|
                 |                   |
                 |    Rsides         |
    |    ---     |----wwww-----------|
  | |---|-->|----|                   
    |    ---     |                  
         60K     w                 
        Btu/hr   w Rbot          
                 w             
                 |           
                 |        
                50F       

Or still more simply:

                      Rp (day)       49.2F
            80F-------wwww-----------|
                 |  
                 |    where Rp (day) = .00185
    |    ---     |
  | |---|-->|----|                   
    |    ---     |                  
         60K     w                 
        Btu/hr   w Rbot = R10/(15' x 30') = 0.0222...         
                 w             
                 |           
                 |        
                50F       

In the above situation, 60K Btu/hour of heat from the sun is flowing into
the pool, and (80-50)/Rbot = 1350 Btu/hour is flowing from the pool bottom
into the ground, and (80-49.2)/Rp (day) = 16.6K Btu/hour is flowing from the
open pool top and sides into the greenhouse, so the pool will gain heat at
a rate of 60K - 16.6K - 1.35K = 42K Btu/hour on an average January day, with
some sun, ie it will gain about 6 x 42K = 252K Btu, on an average January day. 

How much heat will the pool lose on an average January day?

    6 hours x 16.6K = 100K,  from the top and sides during daylight,
 + 18 hours x 5887 Btu/hour = 106K, from the top and sides at night,
 + 24 hours x 1350 Btu/hour = 32.4K, from the bottom into the ground.

Since the sum of the pool's daily losses is 238K Btu, which is less than the
pool's daily heat gain of 252K Btu, we should be able to keep the pool at 80F.

If the 12,000 gallon pool were closed up, while the sun didn't shine for a few
days, it would initially lose heat at a rate of about 24 x 5887 = 141K Btu/day
through the top and sides, and 32K Btu/day through the bottom, so the water
temperature in the pool would initially decrease by about

(141K + 32K Btu)/(12,000 gal x 8lb/gal) = 1.8 degrees F per day.

Possible enhancements include a white surface or a reflecting pond on the
ground along the south edge of the greenhouse, a layer of vertical glazing
along the 30' south side of the pool, tomato plants, an indoor herb garden,
orange trees, lights, fountains, music, insulation for the north greenhouse
roof, and an extra layer of glazing or UV-transparent Tedlar film for the
south side, to allow tanning in January. 

This might also make a good sewage treatment system, with some venting. 

Who will be the first to try this...?

Nick