Consider an 8' cube with walls and ceiling with R-value R and an Ac ft^2
air heater with a screen absorber that keeps 70 F air near R1 glazing
with 90% solar transmission, as in the Barra system. If 1000 Btu/ft^2 of
sun falls on the south wall over 6 hours on an average 30 F January day
in Philadelphia, we might store overnight heat with water in pipes just
under the ceiling at about 60 F min and 140 F max, averaging about 100 F
on an average day. If 0.9x1000Ac = 6h(70-30)Ac/R1 + 2h(70-30)64x64/R
+ 16h(50-30)4x64/R + 24h(100-30)64/R, Ac = 318/R. An R10 cube might have
a 4'x8' collector.

With L' of 3" pipe with pipe surface A = Pi3"/12"L = 0.785L ft^2 and pipe-
air conductance Gp = 1.5A = 1.18L Btu/h-F and capacitance Cp = 3.1L Btu/F
and cube conductance G = 5x64/R10 = 32 Btu/h-F at night and dawn heat loss
Qdawn = (50-30)32 = 640 Btu/h and heat storage requirement Ed = 2h(70-30)32
+16h(50-30)32 = 12.8K Btu/day, the minimum usable pipe water temp Twmin
= 50+Qdawn/Gp = 50+542/L, eg 59 F with L = 60' of pipe.

T (F) air near the ceiling over 6 hours and a 4'x8' R1 air heater with 90%
solar transmission and 70 F air near the glazing makes an hourly gain of
0.9x1000x32/6h = 4800 Btu/h = (70-34)32ft^2/R1[glazing]+(70-34)3.5x64ft^2/R10
[3.5 walls]+(T-Tw)1.5A [into the pipes]+(T-34)64ft^2/R10 [ceiling], which
makes T = (2597+TwL)/(L+5.44).

If dTw/dt = (T-Tw)Gp/Cp = 989/(L+5.44)+0.381Tw(L/(L+5.44)-1), dTw/dt+CTw=d
and Twmax = d/c+(Tmin-d/c)e^-(6c) = 477+(Tmin-477)e^-(6x0.381(1-L/(L+5.44))).
(Twmax-Twmin)Cp = Ed makes L = (L-1.27)e^(-12.46/(L+5.44))+11. L = 50 on
the right makes L = 49.92 on the left, then 49.85, and 49.77, so 50' of pipe
will do, with Tmin = 60.8 and Tmax = 144.5 and Cp = 155 pounds of water.

Notes:

1. It may not be easy to keep 70 F room air between the screen absorber
and the cold glazing by natural convection. A fan might help.

2. A system like this with room air entering the heater may need to control
collector airflow to avoid room overheating. If (T-70)cfm = (70-30)3.5x64/10,
cfm = 896/(T-70), which will decrease as T increases over an average day.

3. Starting the day with 60.8 F vs warmer water keeps the air heater cool
and allows solar collection to be more efficient.

4. To avoid overheating the room by radiation at 144 F, the ceiling needs
a low-e surface facing the floor.

5. A shed or cathedral ceiling could be more efficient than a horizontal
one, since it would trap warm air near a smaller surface under the ridge,
with less heat loss to the outdoors.

5. A slow ceiling fan with an occupancy sensor and a room temp thermostat
could bring down warm air from the ceiling as required. Keeping warm air
trapped under the ceiling requires controlling air leaks and convection
currents arising from glazing and insulation differences on walls. Room
air would tend to rise near a warmer wall and sweep across the ceiling
(undesirably mixing with hot air) and fall near a cooler wall.

6. A lower-mass structure could cool more quickly to 50 F at the end of
the day, making it more efficient than a higher-mass structure.

7. A less passive system with higher performance might have a pump
and an underfloor tank to store heat for 5 cloudy days in a row.
More insulation would decrease the required tank size.

8. Two pumps and two tanks could be more efficient, with cool water in one
and warm in the other, and a hose between them. We could pump cool water up
into the pipes at the end of the day to cool the cube to 50 F sooner.

9. We might have fin-tube vs PVC pipes near the ceiling.

10. And more fin-tube or a flat poly film water duct under the floor,
with a central thermal chimney, instead of the slow ceiling fan.

Nick