re: my grid power usage is now down to 10 kwh per month
5 jun 2001
mike fritz wrote:
>i live in the desert, phoenix arizona. unfortunately, i only live in an
>apartment, so there is so much i can do. for example, i cannot install
>a swamp cooler... putting a panel on the balcony, wouldn't work, our
>apartment faces south...
you might lay foamboard on a north balcony and cover it with plastic film
to make an 8'x12' shade pool for evaporation. maybe flood it only at night,
and use a tank or fountain or fan-coil unit (eg an auto radiator with fan)
inside to cool the house.
swamp coolers always cool if the rh is less than 100%, with the wet bulb
temp as a lower limit, close to the shade pool temp. but moving outdoor
air into a house is less efficient than just cooling the air that is in
the house, and it introduces extra humidity which makes you feel warmer,
compared to dry air at same temperature.
>here in phoenix, the humidity is usually about 6 - 10%.
nrel's 30-year average for june in phoenix shows an average 88.2 f,
with an average daily low and high of 72.9 and 103.5 and an average
humidity ratio w = 0.0056 pounds of water per pound of dry air, so the
partial water vapor pressure pa = 29.921/(1+0.62198/w) = 0.267 "hg.
pw = exp(17.863-9621/(460+72.9)) = 0.826 "hg at 72.9 f and 100% rh,
and 1.367 and 2.202 "hg at 88.2 and 103.5, which makes rh = 100pa/pw
= 32, 20, and 12% at those temps, assuming the amount of water in
outdoor air stays constant over 24 hours, which it tends to do.
>during the monsoon season... switch over to the air conditioner.
>my uncle has both, and i am impressed with the swamp cooler,
>it can drop the temperature by 40 - 50f.
you might use both, preferring evaporation. at night, the wet bulb
temp tw and partial pressure pw near the pond surface would follow
bowen's 1936 equation, 100(pw-pa)/(tw-ta) = -1, so 100pw = 99.6-tw,
and tw = 55.4 f, after a few iterations, ie you could keep your
apartment 55.4 f all day and night on an average june day, with
a perfect indoor heat exchanger and enough insulation and water.
to supply 5k btu/h of cooling for 12 hours (60k btu/day) of cooling,
like a window ac, the pool needs to lose, say, 10k btu/h over 6h on
an average night, ie 10k/(8'x12') = 104 btu/h-ft^2. ashrae says pools
lose about 100(pw-pa) btu/h-ft^2, which makes pw = pa+1.04 = 1.309, so
tw = 9621/(17.863-ln(1.309))-460 = 86.8 f, with 20% humidity indoors.
you could still turn on an indoor fountain to make the air cooler.
the pond would lose more heat by night sky radation. at the dew point,
tdp = 9621/(17.863-ln(0.267))-460 = 41.5 f or 5.3 c, and the ambient
temp ta = 72.9 f or 295.8 k. duffie and beckman have a equation for sky
temp ts = ta(0.711+0.0056tdp+0.000073tdp^2+0.013cos(15t))^0.25 = 275.4 k
or 35.8 f, with t = 3 hours since midnight, so a square foot of 86.8 f
pool would lose an additional 0.1714x10^-8((460+86.8)^4-(460+35.8)^4)
= 49.7 btu/h-ft^2 by radiation, making it equivalent to a 7.5k btu/h ac.
for lower indoor temps or more cooling capacity, you might flow water
down over the south balcony wall at night, circulating it back with a
sump pump in a gutter, and even blow air over the wall with a fan...
why don't you try something like this and let us know how it works?
i don't worry much about ac or water in pa, having pumped 2,400 gallons
per day over 5,000 ft^2 of greenhouses and 2 acres of vegetables over
the last drought summer. i just finished a bamboo octet truss to support
another 120 tomato plants...
it's a snap to save energy in this country. as soon as more people
become involved in the basic math of heat transfer and get a gut-level,
as well as intellectual, grasp on how a house works, solution after
solution will appear.
tom smith, 1980