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re: multiple fans or a/c more frugal?
18 jun 2003
joelncaryn wrote:
>>more mass can help store coolth.
>
>if you can get it to swing fast enough.
...1/2" drywall has about 0.5 btu/f-ft^2. with an r0.67 slowly-moving
airfilm resistance, rc = 0.67x0.5 = 0.335 h, about 20 minutes, so it can
quickly swing to "room temp," reducing the drywall-to-room-air temp diff
by a factor of 1-e^3 = 0.95, ie 95% in an hour. making the room air temp
close to the outdoor temp is a different problem. a c cfm fan adds about
c btu/h-f of conductance in series with the drywall capacitance, raising
the overall time constant.
>the low here last night was 76f at 6:15 am. my house with gypboard walls
>and about 500 sq. ft. of saltillo flooring tends not to get any cooler
>than the 4 am temperature, which is currently about 80f. which is where
>my a/c is set anyway.
maybe you need a smart whole house fan controller (tm) with an optional
fountain control output, as well as the fan and ac control outputs. then
again, it sounds like your house doesn't have much thermal mass, maybe
2k btu/f for 4,000 ft^2 of 1/2" drywall plus 500 btu/f for the flooring.
with 500 btu/h-f of thermal conductance to outdoors, rc = 2500/500 = 5h.
nrel says the long-term average daily min is 72.9 f in june in phoenix,
with a 24 hour 88.2 average and a 103.5 average daily max (ie an average
96 f "daytime" temp) and an average 0.0056 humidity ratio. you might cool
your mass to 75 f and see it rise to 96+(75-96)e^-12/5 = 94 over a 12 hour
average day, with warmer indoor air. how many 2-liter soda bottles would
you have to add to the house to keep it below 80 f?
better to use wet surfaces, as steve baer says, except in monsoons. you
might make an attractive shaded 4' diam x 8' tall cooling gabion in the
back yard with some welded-wire fencing and rocks. a 3" spherical rock has
113 in^2 of surface and 113 in^3 of volume. the 174k in^3 cylinder might
hold 0.6x174k/113 = 922 rocks with 724 ft^2 of surface with 40% voids.
you might use 2 $35 1984 dodge omni automobile radiators upstairs for
coolth distribution. thus, viewed in a fixed font:
tc ti
| 1600 btu/h-f | 500 btu/h-f
*------www------*------www------ 103.5 f
| <--
--- i 1/1600 + 1/500 = 1/381
-
|
-
the vapor pressure of outdoor air pa = 29.921/(1+0.62198/0.0056) = 0.267 "hg,
so i = 100x724(pw-pa) = (103.5-tc)381 btu/h, using an ashrae approximation
for the heat loss from a swimming pool. pw = e^(17.863-9621/(460+tc)), using
another approximation for the vapor pressure near the gabion with water temp
tc. in rankine degrees, the equation for i becomes 190pw = 614.2-tc, or tc
= 9621/(23.1-ln(614.2-tc)) =... 519 r or 59 f, after a few iterations, so
i = 17026 btu/h and the indoor air temp ti = 59+17026/1600 = 70 f. at 100% rh,
70 f air has a vapor pressure of 0.748 "hg, so, with no additional indoor
humidity, the indoor rh = 0.267/0.748 = 36%. this might be on the cool side.
maybe you only need one automobile radiator, or some sort of indoor fountain.
nick
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