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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 :-)


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

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