re: passive solar cooling in the desert
10 sep 2004
auntie em wrote:
>in the not so distant future my husband and i will be locating to
>south texas, where it is desert.
that could be good for evaporative cooling...
>...we are expecting to live off grid with only a generator for emergency use.
you might enjoy a few pv panels and batteries.
>i am told that the summers there are brutal. a month of highs in the
>100+ degrees is not at all unusual.
desert lows might be comfy. open the house at night. close it up
during the day...
>can any of you guys give me ideas about other ways to cool a house
how about "less"?
indoor misting with a vent fan for outdoor air works like a swamp cooler,
when 4500(wi-wo)>to-ti. the left side is the amount of latent cooling per
cfm of vent air. the right is the amount of sensible heating per cfm of
vent air. as they become equal, swamp cooling and misting no longer work.
the ashrae comfort zone has an efficient corner at ti = 80.2 f and
wi = 0.012 (at 56% rh), which makes to < 134.2-4500wo. nrel says phoenix
has an 88.2 f average temp in june, with an average daily min and max of
72.9 and 103.5 and average humidity ratio wo = 0.0056 pounds of water per
pound of dry air in june, so misting should help until the outdoor temp
rises to to = 134.2-4500(0.0056) = 109 f.
evaporating a pound of water takes about 1000 btu, and air weighs about
0.075 lb/ft^3, so a house with no internal heat gain or unshaded windows
and a 200 btu/h-f conductance that's 80.2 f indoors with wi = 0.012 when
it's 88.2 f and wo = 0.0056 outdoors needs 1000x60c0.075(0.012-0.0056)
= 28.8c = (88.2-80.2)(200+c) btu/h of cooling, including cooling c cfm of
air from 88.2 to 80.2, so c = 77 cfm (not much), with 4.5c(0.012-0.0056)
= 2.2 pounds or 0.27 gph of water, which might come from a $5 0.5 gph
mister mister nozzle and a solenoid valve scrounged from an old washing
machine in series with a 80.2 f room temp thermostat and a $5 humidistat
that turns on a $12 window exhaust fan when the rh rises to 56%.
we can make 77 cfm move through the house with a ft^2 vents with an 8' height
difference if 77 = 16.6asqrt(8(88.2-80.2)), ie a = 0.58 ft^2, eg a 1 ft^2 vent
to the attic, to keep it cooler. the 2.2 pounds of water per hour might come
from plants in an indoor greywater wetland. with enough plants, the thermostat
may never turn on the mister. in an air-leaky house, the humidistat may never
turn on the fan. this could work with no electricity.
herbach and rademan (800) 848-8001 http://www.herbach.com sell a nice
brass $4.95 navy surplus humidistat, item number tm89hvc5203, with a
20-80% range, a 3-6% differential, and a 7.5a 125v switch that can be
wired to open or close on humidity rise.
pw = 1.033 "hg at 80.2 f and 100% rh, and pi = 0.566 "hg with wi = 0.012.
ashrae says a square foot of pool evaporates 100(pw-pi) = 47 btu/h, so we
can evaporate 2.2 pounds of water per hour with 2200/47 = 47 ft^2 of wet
surface, counting droplets in air, or less, if we evaporate close to the
house air inlet. masonry floors and walls or an indoor cylindrical rock
gabion with a fountain pump (in a toilet tank?) with high thermal conductance
and mass might allow more efficient cooling with cooler night air and keep
the house mass cool with the fan off during the heat of the day. the gabion
might build up an interesting mineral encrustation that helps evaporate water.
when a man is wrong and he will not admit it, he almost always becomes angry.