[Greenbuilding] passive solar homes - floors

[Nick Pine]

Kat writes:

> I've got a client of my very own (my first!) and I'm trying to convince 
> them to go with the simplest passive solar system - that of direct gain...

Eeewww. It's very hard to make a direct gain house with a high solar heating 
fraction in Portland. December is the worst-case month, when 470 Btu/ft^2 of 
sun falls on a south wall on an average 40 F day.

If an 8' direct gain cube has R80(!) walls and ceiling and an 8'x8' R4 south 
window with 50% solar transmission and thermal conductance G = 
4x8x8/80+8x8/4 = 19.2 Btu/h-F and 0.5x8x8x470 = 15,040 Btu = 24h(T-40)19.2, 
then T = 72.6 on an average day. If the indoor temp droops to 60 F over 5 
cloudy days, it needs a natural time constant RC 
= -120h/ln((60-40)/(72.6-40)) = 246 hours, ie capacitance C = RCxG = 
246x19.2 = 4716 Btu/F, eg 4716 pounds of water or 189 ft^3 of concrete.

If we figure the sun arrives over (say) 2 hours on an average December day 
and the cube has R32 walls, including an R32 wall between the living space 
and the window to make a low-mass isolated sunspace that gets cold at night 
and on cloudy days (when it loses little heat to the outdoors) and 15040 = 
2h(T-40)8x8/4 [for the window during the day] + 22h(T-40)8x8/38 [for the 
window at "night"] + 24h(T-40)3x8x8/32 [for the other 3 walls] makes T = 111 
F on an average day. Too hot! If we keep it 72.6 F by ventilation and it 
droops to 60 over 5 cloudy days with a conductance G = 4x8x8/32+8x8/38 = 9.7 
Btu/h-F, then C = 246x9.7 = 2386 Btu/F, with lots of heat transfer surface. 
Better.

With higher temp mass under a shiny ceiling, we can keep the room air 
precisely 70 F with a slow ceiling fan and a thermostat for 24 hours a day 
(wow, room temperature control!) and change to R20 walls and an R2 window 
with 80% solar transmission, so 0.8x64x470 = 24,064 Btu = 2h(T-40)8x8/2 [for 
the window during the day] +22h(70-40)8x8/20 [for the window at "night"] + 
24h(T-40)8x8/20 [for the ceiling] + 24h(70-40)3x8x8/20 [for the other 3 
walls] makes T = 147 F :-) If the ceiling mass cools to 75 F over 5 cloudy 
days with an average temp of 111 F and it loses 5dx24h(111-40)8x8/20 = 27229 
Btu while the rest of the cube loses 5dx24h(70-40)4x8x8/20 = 46080 Btu, then 
C = (27229+46080)/(147-75) = 1018 Btu/F. Better.

If we make the cube 70 F during the day and 50 at night with a setback 
thermostat and a 60 F average temp,  24,064 Btu = 2h(T-40)8x8/2 
+22h(60-40)8x8/20 +24h(T-40)8x8/20 + 24h(60-40)3x8x8/20 makes T = 168 F. If 
it cools to 75 over 5 cloudy days  with an average temp of 122 and it loses 
5dx24h(122-40)8x8/20 = 31488 Btu while the rest of the cube loses 
5dx24h(60-40)4x8x8/20 = 30720 Btu, then C = (31488+30720)/(168-75) = 669 
Btu/F. Better.

If we don't like hot water under shiny ceilings, we can put a duct heat 
exchanger (eg MagicAire's SHW2347 2'x2' horizontal unit) at the top of a 
vertical duct and let hot sunspace air thermosyphon down the duct without 
mixing with room air while it heats an unpressurized tank on the ground, 
with a low-power pump.

We could refine this with a simple simulation using NREL's TMY2 hourly 
weather data file for a typical year in Portland (WBAN 24229).

Nick