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re: alternative energy desalinization 30 mar 2001 can we distill a pound of water with less than 1000 btu of sun, using a heat exchanger inside an inexpensive plastic film greenhouse, with vegetable plants or reeds for the evaporating surfaces? plants evaporate mostly during the day, in sun. we might want a rockpile and a pump to make this work at night as well. suppose the greenhouse has 4 concentric poly film ventilation ducts near the top with indoor and outdoor air flowing in opposite directions... say outdoor air flows in through a 12" duct and indoor air flows out through the space between the 12" duct and an 18" duct, and outdoor air flows in between the 18" duct and a 24" duct, and indoor air flows out between the 24" duct and a 30" duct. we might move the air with a single fan, in an airtight greenhouse. these heat transfer surfaces total 14.1 ft^2 (and $2) per foot, eg 14,140 ft^2 for a 100' greenhouse. slow-moving air on one side and condensation on the other makes e = 21.2k/(21.2k+c), approximately, with a c cfm airflow in this condensing counterflow exchanger. e = 0.955 for c = 1000 cfm, so 70 f indoor air would leave the greenhouse at 70-0.955(70-30) = 31.8 f on a 30 f day. suppose we keep the greenhouse at 70 f day and night, and fill the space between its two layers of poly film cover with soap bubbles at night, so the main heat loss is conduction for 6 hours per day in january, as well as sensible and latent heat exchanger losses over 24 hours. nrel says a south wall in phila gets 1000 btu/ft^2 of sun on an average 30.4 f january day, with 620 on a horizontal surface. the average outdoor humidity ratio wa = 0.0025 pounds of water per pound of dry air. a shallow frozen reflecting pool might add 30% to the south sun. a 14'x100'x9' tall hoophouse ($568.25 from schwarz and sons in wilmington :-) with two layers of poly with 0.81 solar transmission and r1.2 during the day would collect about 0.81x100'(620x14'+1.3x1000x9') = 1.65 million btu and lose about 6h(70-30)21'x100'/r1.2 = 420k btu by conduction, leaving 1.23 million btu per day or 51k btu/h for the heat exchanger. with no heat exchanger, evaporating p lb/h of water takes about 1000p btu/h. air at 70 f and 100% rh has partial pressure pg = 0.748 "hg and humidity ratio w = 0.016. a c cfm airflow removes 60c(0.075)(0.016-0.0025) = 0.061c pounds of water per hour, so we need c = 16.5p to remove it, and heating that air from 30 to 70 f requires about 40c btu/h. so, without the heat exchanger, we have 51k btu/h = 1000p + 40x16.5p = 1660p, and p = 31 pounds per hour, max. the heat exchanger sensible loss is c(70-e(70-30)-30) = 40c^2/(21.2k+c), with about 1000btu/lbx60cx0.075lb/ft^3(w-wa) = 4500c(w-wa) of latent loss, where w is the humidity ratio of the saturated indoor air leaving the heat exchanger to the outdoors. at temp tout, pw = exp(17.863-9621/(460+tout)) "hg and w = 0.62198/(29.921/pw-1). tout = 70-21.2k/(21.2k+c)(70-30). putting this together, c = 3515 makes tout = 35.7 f which makes pw = 0.213 "hg and w = 0.00446, with a sensible heat loss of 20k btu/h and a latent loss of 31k btu/h (93 gallons per day.) we might slow down the fan to save electrical energy and conserve water when there is less to evaporate. suppose we have 50k ft^2 of evaporating surface, about 36 ft^2/ft^2 of greenhouse, eg both sides of 5 ft^2 of leaves and a 4" layer of 1/2" rocks on the impervious epdm floor. ashrae says p = 0.1(50k)(0.748-pg) lb/h of water will evaporate into greenhouse air at partial pressure pg, and w = wa+p/(60x3515x0.075) makes pg = 29.921/(1+0.62198/w), which makes p = 199 lb/h or 597 gallons per day in january :-) nick 10 c=3515'airflow (cfm) 20 e=21200/(21200+c)'heat exchanger efficiency 30 tout=70-e*(70-30)'outgoing air temp (f) 40 pw=exp(17.863-9621/(460+tout))'outgoing vapor pressure ("hg) 50 w=.62198/(29.921/pw-1)'outgoing humidity ratio 60 sens=40*c^2/(21200+c)'sensible heat loss (btu/h) 70 lat=1000*60*c*.075*(w-.0025)'latent heat loss (btu/h) 80 print tout,sens,lat 90 a=1/(60*c*.075)'airflow (lb/h) 100 s=50000!'evap surface (ft^2) 110 b=.0025+.62198+2.9173*a*s'quadratic term 120 c=.0025*2.9173*s-.046524*s' " 130 p=(-b+sqr(b^2-4*a*c))/2/a'distillation rate (lb h20/h) 140 wg=.0025+a*p'greenhouse air humidity ratio 150 rh = 100*wg/.016'rh of greenhouse air (%) 160 print s,p,rh 35.68886 19996.32 31001.69 50000 198.9603 94.24054 |