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Direct gain in Oregon?

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Posted by nicksanspam on June 12, 2007, 12:50 pm
Kat from Portland OR wants to:

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.

1. An 8' direct gain cube with R80(!) walls and ceiling and an 8'x8' R4
south window with 50% transmission and conductance G = 4x8x8/80+8x8/R4
= 19.2 Btu/h-F and 0.5x8x8x470 = 15,040 Btu = 24h(T-40)G would have
T = 72.6 on an average day. If the indoor temp droops to 60 F over 5
cloudy days, it needs a 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, eg an 8'x8'x3' thick concrete
north wall with some holes for airflow.

2. If we figure the sun arrives over (say) 2 hours on an average day and
the cube has R20 walls, including an R20 wall between the living space
and an R2 window with 80% transmission to make a low-mass isolated
sunspace that's cold at night and on cloudy days (when it loses little
heat to the outdoors) and 0.8x8x8x470 = 24,064 Btu = 2h(T-40)8x8/2
[for the window during the day] + 22h(T-40)8x8/22 [the window at "night"]
+ 24h(T-40)3x8x8/20 [for the other 3 walls], T = 107 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 = 4x8x8/20+8x8/22 = 15.7 Btu/h-F,
C = 246x15.7 = 3864 Btu/F, with lots of heat transfer surface. Better.

3. 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!) with 24,064 Btu = 2h(T-40)8x8/2
+ 22h(70-40)8x8/22 + 24h(T-40)8x8/20 [the ceiling] + 24h(70-40)3x8x8/20
[for the other 3 walls] makes T = 148 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)/(148-75)
= 1004 Btu/F. Better.

4. If the cube is 70 F for 12 hours and 50 with a night setback thermostat
and a 60 F average temp, 24,064 Btu = 2h(T-40)8x8/2 + 22h(60-40)8x8/22
+ 24h(T-40)8x8/20 + 24h(60-40)3x8x8/20 makes T = 172 F. If it cools to 75
over 5 days at 123 F average and loses 5dx24h(123-40)8x8/20 = 31872 Btu
while the rest of the cube loses 5dx24h(60-40)4x8x8/20 = 30720 Btu, then
C =(31872+30720)/(172-75) = 645 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 radiator) 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 77 gallon
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).


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