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ceiling mass air heater options
29 nov 2003
would an unstratified mass (eg a poly film duct containing 1" of water over
a shiny ceiling, with insulation above that) be warmer with david delaney's
flow separator or 70 f house return air filling a sunspace/air heater?

consider 2 steady-state 8' r10 cubes with 64 ft^2 of r2 glazing with 80%
solar transmission, after a long string of 30 f average january days in
phila, with 1000 btu/ft^2 of sun on a south wall...

1.  --------------
   |fs|    tc     |          cube 1 has a flow separator and a transpired
   | .|           | r10      mesh collector. we might assume the air hear the 
   | .|           |          glazing is about the same as the ceiling temp tc. 
s  | .|   70 f    | 30 f     this sunspace might be uncomfortably hot, but the
   | .|           |          amount of airflow is not limited by the cube temp.
   | .|           |          how does the ceiling keep the cube 70 f? 
   | .|           |        
    --------------        

2.  --------------
   | .     tc     |          cube 2 has a ventilation slot at the top and 
   | .|           | r10      a transpired mesh collector and a potentially-
   | .|           |          unreliable motorized damper at the bottom which
s  | .|   70 f    | 30 f     is controlled by a thermostat that keeps the 
   | .|           |          cube air 70 f during the day. we might assume
   | .|           |          the air near the glazing is about 70 f. we might
   | .d           |          need another layer of glazing south of the mesh
    --------------           to ensure this. the ceiling might keep the cube
                             70 f with a thermostat and a slow ceiling fan.

in each case, the glazing transmits 0.8x1000x64tft^2 = 51.2k btu/day of sun. 

in case 1,

   51.2k = 6h(tc-30)64ft^2/r2+18h(70-30)64ft^2/r12   [south wall]
         +24h(tc-30)64ft^2/r10                       [ceiling]
         +24h(70-30)4x64ft^2/r10                     [remainder of cube.]

which makes tc = 96 f, if i did that right. 

in case 2,

   51.2k = 6h(70-30)64ft^2/r2+18h(70-30)64ft^2/r12   [south wall]
         +24h(tc-30)64ft^2/r10                       [ceiling]
         +24h(70-30)4x64ft^2/r10                     [remainder of cube.]

which makes tc = 128 f.

nick




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