re: need to cool bedroom (and heat pool)
5 jul 2003
bill rowland wrote:
>here's my situation: first of all, we live in sacramento, ca, so
>there is no shortage of solar energy coming in our direction.
and it's dry, so evaporative cooling works well.
>the master bedroom of our house sticks out to the southeast,
>exposed on three sides.
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...
>the outside of the southeast back wall is stucco, gable end, 20 feet long,
>with very little shading from trees...
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.
>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?
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.