[Greenbuilding] Air Conditioning for Phoenix

[Nick Pine]

"Rob Tom" <ArchiLogic at yahoo.ca>

> Air conditioners, swamp coolers, stone slingers, solenoid-operated
> watering hoses...

Simple, low-energy stuff...

The ASHRAE comfort zone has an efficient evaporative corner at Ti = 80.2 F
and wi = 0.012 (at 56% RH.) 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 humidity ratio
wo = 0.0056 pounds of water per pound of dry air in June. Evaporation can
help until the outdoor temp rises to To = 134.2-4500(0.0056) = 109 F.

At 1000 Btu/lb of water and 0.075 lb/ft^3 of air, 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 and 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 the cooling of C cfm of outdoor 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 solenoid valve scrounged from an old washing machine in
series with a 80 F room temp thermostat and a humidifier, and a humidistat
that opens a 2-watt motorized damper when the indoor RH rises to 56%.

If 2 A ft^2 vents with an 8' height diff move 77 = 16.6Asqrt(8(88.2-80.2))
cfm, ie A = 0.58 ft^2. The 2.2 pounds of water per hour might come from
a soaker hose on a slab or plants in an indoor wetland...

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, or less, if close to the house air inlet. Masonry floors and walls
or an indoor cylindrical rock gabion with a fountain pump with high thermal
conductance and mass could allow more efficient cooling with cooler night
air and keep the house mass cool with the fan off during the day. A mineral
encrustation might add more surface that helps evaporate water.

If the solenoid valve operates for 0.27/5gpm = 0.054 minutes per hour
and uses 10 watts, ie 0.009 Wh (0.031 Btu) of electrical energy and
produces 1584 Btu of net cooling, the COP is 1584/0.031 = 52,000,
and the EER is 176,000 :-)

> Rather than an AC unit or a mechanical swamp cooler, perhaps BIT thinks
> about installing a rainwater harvesting system and a Green roof.

Or maybe a white roof, eg white pebbles over EPDM, with water sprinkled
over the pebbles at night for cooling. Plants provide shade and cool the air
around them, but they don't directly cool a building below them. Why waste
water cooling outdoor air? Pebbles can provide shade and a low roof thermal
conductance, with mere point contacts. With enough roof surface and house
insulation, the water below the pebbles can be close to the dew point vs
the wet bulb, for excellent indirect evaporative cooling.

Nick