shaded ponds for cooling houses SOLAR.KnowledgePublications.com

shaded ponds for cooling houses
20 jul 1998
i've been hearing how it's so hot and dry in texas, and wondering if
anyone is using a pond for cooling a house, say an 8' wide x 16' long
x 1' deep pond along the north side of a house, made with a $50 18'
folded piece of 10' wide epdm rubber draped over an earth berm, with
a fan-coil unit inside the house (like an auto radiator with a fan)
with a small submersible pump circulating pond water through it.

duffie and beckman's _solar engineering of thermal processes_
(wiley, 1991, 2nd edition) has an old (bowen, 1926) ratio for
the heat loss from a pond by evaporation, independent of windspeed: 

     qc,pa   0.46(tp -ta)  p
     ----- = ------------ ---
      qe       pwp - pa   760

where qc,pa is the heat loss from the pond by convection, in watts/m^2-k,
      qe is the heat loss from the pond by evaporation, in watts/m^2-k,
      tp and ta are the celsius temperatures of the water and air, and 
      pwp and pa are the vapor pressures of the pond surface and air, and 
      p is the barometric pressure, with all pressures in mm hg.

the upside-down sea-level english version has a nice decimal constant:

       qe      25.4(pwp-pa)    100(pwp-pa)
     ----- = --------------- = -----------
     qc,pa   0.46(tp-ta)/1.8      tp-ta

where qc,pa is the heat loss from the pond by convection, in btu/hr-f,
      qe is the heat loss from the pond by evaporation, in btu/hr-f,
      tp and ta are farenheit temperatures of water and air, and 
      pwp and pa are vapor pressures at pond surface and air in inches hg.

july is the hottest average month in abilene, texas, with an average low
of 73 f and a 24-hour average of 84 and an average high of 95. the average 
humidity ratio is w = 0.013 pounds of water per pound of dry air. the
average wind is about 10 mph (at night, too?), and an air film over a
square foot of rough dry surface has a thermal conductance of about
2+v/2 btu/h-f, where v is in mph...

table 2 on page 6.4 of the trusty 1993 ashrae hof says air with w = 0.013
is saturated at about 64.5 f, and 73 f saturated air has w = 0.017575 at
a vapor pressure of 0.81882" hg, so abilene's 73 f air would have a
relative humidity of about 74% and a vapor pressure of 0.74x0.81882
= 0.606" hg, if i'm doing this right. water vapor in 77 f saturated air
has a pressure of 0.93589" hg, so on an average july night, that 77 f
pond might lose (77-73)8'x16'(2+10/2) = 3,584 btu/h by convection and
100(0.93589-0.606)/(77-73) = 8.25 times more by evaporation, a total of
3,584x9.25 = 33k btu/h, like 6 window acs. abilene's elevation makes
the air pressure 13.8 psi, adding another window ac. 

wow. is that right? can that pond evaporate 35 pounds of water per hour,
ie 0.074 gallons per minute on an average july night in abilene?

nick