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Need to cool bedroom (and heat pool)

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Posted by Bill Rowland on July 4, 2003, 7:21 pm
Here's my situation: First of all, we live in Sacramento, CA, so there is
no shortage of solar energy coming in our direction. The master bedroom of
our house sticks out to the southeast, exposed on three sides. There is no
cross-ventilation, just a patio door on one side. The outside of the
southeast back wall is stucco, gable end, 20 feet long, with very little
shading from trees. In short, this room would make a very good greenhouse,
but as a bedroom it sucks!

Running the length of this wall, about three feet away, is a swimming pool,
containing about 20,000 gallons of cool water. After living here for 10+
years, I suddenly had a brilliant (??) idea: why not set up a system to
pump water from the pool through pipes, or some other solar collection
device, placed against the wall; with the primary objective of keeping the
room cooler, and a side-effect of heating the pool?

Since I know nothing about this sort of thing, I'm looking for any advice
on the best way to do this, or if it even makes sense at all. Here is my
initial idea, with advice from some slightly more knowledgeable friends:

Use four sheets of thin plywood, providing an 8' x 16' support. Using black
pvc pipe, mount a grid of piping onto the plywood. The input pipe would
take water to the top at one end, pipes would zigzag back and forth
horizontally, with heated water coming out into the other end of the pool.

I would use a very low-flow pump to pull water out of the pool (run the
input pipe down into the deep end to get the coldest water). The pump could
be on a timer, or solar-activated switch. The 3-foot wide sidewalk connects
directly from the pool to the house, so there is no way to run pipes
underneath. I would have to run input and output pipes across the sidewalk,
and probably use some kind of threshold to cover them. I would like to do
this with very minimal structural impact (nailing, screwing) to the house,
sidewalk, etc., in case it turns out to be a failure!

Mount the apparatus a few inches away from the wall, to provide an airspace
for additional insulation. I believe I could secure the top to 2x4's
fastened to the eave beams and to the attic vent opening in the middle of
the wall, without attaching to the stucco. The bottom could be anchored
into the gap where the pool drain strip runs.

Other thoughts and questions:

- I assume that copper would absorb heat better than pvc, but would
probably be more expensive, and more hassle to work with: cutting,
sortering rather than gluing, etc. ???

- I know that such devices are usually placed on roofs rather than against
walls, but my theory is that the attic airspace and insulation, and attic
exhaust fan, are helping with heat from above, and the heat-absorbing wall
a few inches from our bed is the best opportunity for heat reduction. Sound
like a good theory?

- I'm not sure if this idea would really work that well for cooling the
room. It might work well for heating the pool, but don't know how
completely the wall would have to be covered to divert enough heat from the
room to be helpful. How close together would pipes need to be to be

- Perhaps paint the front of the plywood silver to help reflect more heat
back onto the pipes? Maybe even mount something like gutter pipe behind
each pipe to provide a dish-reflection effect?

- Are there solar devices that I could buy that would be more efficient
and/or less hassle than building my own system with pipes? And at what

- With pipes running horizontally, would the water just flow down through
the system too fast to absorb much heat? It seems that running pipes
vertically would require much more pumping power.

Thanks in advance to anyone who can provide guidance!!

(Remove "NOSPAM" from email name to reply back directly.)

Posted by Nick Pine on July 5, 2003, 12:07 pm

And it's dry, so evaporative cooling works well.

Like this (viewed in a fixed font)?  Or this?

 ---------------------------          ----------------
|               |           |        |                |
|               |    BR     |        |                |
|               |    20'    |        |                |
|     house     |-----------         |    house       |
|               |    SE              |                |----------
|               |   pool             |                |          |
|               |                    |                |    BR    |
|               |                    |                |    20'   |
 ---------------                      ---------------------------
In this case, the house                                   pool
might shade the pool...

Sounds like you need to shade the wall, eg grow something over the wall,
eg grapes or trumpet vines in planters, or hang some greenhouse shadecloth
over the wall or shade it with a wood screen. You might also shade the roof
or ventilate the attic or use night ventilation--move cool outdoor air
through the house at night and button it up during the day.

Those objectives seem incompatible. You might pump some pool water through
a heat exchanger in the bedroom, eg a 1984 Dodge Omni automobile radiator,
but that would work best with no pool cover or pool heating. Maybe it's
better to separate these objectives: heat the pool with a solar pool cover
and/or sinkable foamboard rafts, and cool the bedroom (if required after
shading) with an evaporative system.

NREL says July is the warmest month in Sacramento, with average daily min,
24h, and max temps of 58.1, 75.7, and 93.2 F and w = 0.0087 humidity ratio.
Pa = 29.921/(1+0.62198/w) = 0.413 "Hg and 460+58.1+100Pa = 559.4 R, so the
Rankine wet bulb temp Tw = 9621/(22.47-ln(559.4-Tw)) = 500, 523, 509, 518,
513, 516, 514, 516, maybe 515 R, ie 55 F, after a few iterations.

You might get 55 F water from a 4'x20'x2' tall EPDM-rubber-lined tank along
the NW wall of the bedroom with rocks above the water level. (Would it be
a bit colder, approaching the 9621/(17.863-ln(Pa))-460 = 53 F dew point?)
You might pump some of the water over the NW roof at night...

If the bedroom has, say, 1000 ft^2 of exterior R10 surface with
100 Btu/h-F of conductance, it might be 70 F on a 190 F day :-)

       <-- I   70 F                  I = (70-55)800 = 12K Btu/h,
       1/800   |     1/100
55 F ---www----*------www--- T    so T = 55+12K(1/800+1/100) = 190 F.
       Dodge        bedroom


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