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Ceiling mass air heater options



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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