Nader Chalfoun and Christopher Trumble had an interesting Tucson cool tower story in the Spring 2005 U Oregon "Connector" architecture newsletter.
It's nice to avoid the energy of a swamp cooler blower. Can we also avoid the large structure (sacrificing architectural drama) and use less water, based on weather conditions, with constant comfort and better controls?
How about testing an alternative? Evaporate water inside a house and also run a small exhaust fan as needed to remove water vapor from the house. The most efficient corner for evaporative cooling in the ASHRAE 55-2004 comfort zone is at 80 F and w = 0.012, approximately.
We (not me, with recent flooding in PA :-) might turn on a small indoor swamp cooler with a thermostat when the indoor temp rises to 80 F and turn on an exhaust fan with a humidistat when the indoor RH rises to
54% (w = 0.012 at 80 F.)
With enough green plants in a house, the cooler might seldom turn on. With enough air leaks, the exhaust fan might seldom turn on.
Or run a soaker hose with pressurized water from a solenoid valve (which might come from a dead washing machine) over a floorslab in an existing house or under a floorslab with a vapor barrier under the hose in a new house. The slab's thermal mass might store coolth for more efficient cooling with cooler night air below the comfort temp near the floor.
During the day, a slow ceiling fan with a room temp thermostat and an occupancy sensor could provide efficient cooling as needed. The fan could provide comfort cooling and raise the acceptable RH all the way to 100% at
81 F with v = 0.5 m/s, according to ASHRAE-55's BASIC program, altho that might cause mold, on a continuous basis. The slab could also lower the mean radiant temp. A low-e ceiling and walls could radiate less to the slab when nobody's home, conserving stored coolth.
NREL says Tucson has an average humidity ratio wo = 0.0054 in June, with a 67.9/99.6 F daily min/max. An 80 F house with a 400 Btu/h-F thermal conductance and 4K Btu/h of internal gain might need (99.6-80)400+4K = 8240 Btu/h of cooling at 3 PM.
Evaporating P lb/h of water makes 1000P Btu/h, and cooling C cfm from 99.6 to 80 F to make up for required exhaust air takes about (99.6-80)C Btu/h, and 1000P = 8240+(99.6-80)C with 0.075 lb/ft^3 air and P = 60C0.075(wi-0.0054) and wi = 0.0120 makes P = 0.0297C and 29.7C = 8240 + 19.6C, so C = 816 cfm and P = 24.2 lb/h of water, with a net cooling of 8240/24.2 = 340 Btu/lb.
How many pounds of water per hour would a cool tower need to achieve the same 80 F at 54% RH inside this house?
Ps = e^(17.863-9621/(460+80)) = 1.047 "Hg at 80 F and 100% RH, so A ft^2 of 80 F damp floorslab in 80 F air at 54% RH might evaporate 0.1APs(1-0.54) = 0.048A lb/h of water, (mis)using an ASHRAE swimming pool formula, ie
502 ft^2 of slab might evaporate 24.2 lb/h.
At 81 F and 100% RH indoors, 1000P = (99.6-81)400+(99.6-81)C and wi = 0.0233 and P = 0.0808C, so 80.8C = 7440 + 18.6C, so C = 120 cfm and P = 9.7 lb/h with 7440/9.7 = 770 Btu/lb of net cooling. This could work even in August, when conventional swamp cooling wouldn't, with wo = 0.0117 and Tdp = 61 F. It might precool a slab faster and more efficiently than simple AC.
An 80 F slab under 67.9 F air with wo = 0.0054 and Pa = 29.921/(0.62198/wo+1) = 0.257 "Hg might evaporate 0.1A(Ps-Pa) = 0.0789A lb/h and lose (80-67.9)1.5A = 18.2A Btu/h of sensible heat, for a total of 97.1A Btu/h. With enough air, a 1000 ft^2 slab might lose 24hx8240Btu/h = 198K Btu in 198K/97100 = 2 hours on a June night, with 198K/158 = 1255 Btu/lb of net cooling.
Nick