[Greenbuilding] Pond cooling and clay dehumidification

[nick pine]

NREL says 1890 Btu/ft^2 of sun falls on the ground on a 76.7 F average July 
day in Phila with an 86.1 and 67.2 high and low and average 81.4 and 72 day 
and night temps and w = 0.0133 and an 8 mph windspeed.

An 80 F house with a 200 Btu/h-F thermal conductance that uses 300 kWh/month 
of electricity indoors would need 24h(76.7-80)200+34.1K = 18.3K Btu/day of 
cooling.With 100 cfm of natural air leakage and 2 gallons per day of 
evaporation from occupants and w = 0.0120 indoors (inside ASHRAE's 55-2004 
standard comfort zone), we need to remove 
24hx60x100x0.075(0.0133-0.0120)+16.6 = 30.6 pounds of water vapor per day.

Fig. 5 of 
http://www.agmcontainer.com/desiccantcity/pdfs/Desiccant%20performance.pdf 
says the maximum moisture content of bentonite clay (eg clumping kitty 
litter) is 15% at 113 F and 5% at 77 F, ie MC = 0.364-0.00278T(F), eg 0.142 
at 80 F. If the clay is solar heated and dried at 113 F during the day and 
cooled to 80 F at night, the house needs 30.6/(0.142-0.05) = 334 pounds (9.1 
cubic feet) of clay (heh heh.) If air from a 1000 Btu/h-F car radiator with 
a 20 watt fan cools the solar-heated clay bed at night, the house needs 
(18.3K+30.6K)/24h = 2038 Btu/h of cooling, with an 80-2038/1000 = 78 F temp 
water supply.

The average vapor pressure of the July air Pa = 29.921/(1+0.62198/w) = 0.626 
"Hg and the dew point Tdp = 9621/(17.863-ln(Pa))-460 = 64.9 F and the wet 
bulb Twb(R)= 9621/(22.47-ln(76.7+460+100Pa-Twb)) = 528.3 R (68.4 F.) Phil 
Niles says a 78 F shaded pond in 76.7 F air with a 64.9 dew point and a 68.4 
F wet bulb temp loses Qr = 1.63x10^-9((78+460)^4-a(76.7+460)^4) = 18.7 
Btu/h-ft^2 by radiation, where a = 0.002056x64.9+0.7378. Qc = 
(0.74+0.3x8mph)(78-76.7) = 4.1 Btu/h-ft^2 by convection, and Qe = 
b(78-68.4)-Qc = 109.5 by evaporation, where b = 
3.01(0.74+0.3x8)((78+68.4)/65-1), totaling 132.3, so it looks like a 
2038/132.3 = 15.4 ft^2 pond could provide the 78 F water.

If A ft^2 of R1 glazing with 90% solar transmission evaporates 30.6 pounds 
of water at 113 F and 0.9x1890A = 30.6K + 8h(113-81.4)A, 21 ft^2 of pond 
glazing would work, eg a 4'x8' clay bed with 3.4" of cat litter. We might 
use more cat litter to store more dryness for non-average days and use more 
glazing in an equilateral A-frame with an insulated north wall for winter 
air heating.

Nick