Re: Ceiling mass air heater options

On 29 Nov 2003 09:06:28 -0500,
[EMAIL PROTECTED] wrote:

>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.
>
Cube 2: You might do another iteration with the
average temperature of the air heater, Tah, set to
99F, the average of 70F and 128F, and keep going
until Tc is nearly equal to 70+2(Tah-70). 

David