Sunday, October 13, 2013

A direct gain rant

John Canivan writes:
>Dug What happens if you have a week without sun in December?
If it's a direct gain house with sun shining through windows onto a massy floor, you freeze or wave your hands and say it's really warm inside when it isn't, or wear sweaters and arctic army pants and sit in one small room with an electric space heater. Direct gain house owners and builders and designers often fool themselves and others.
What's the solar heating fraction of the Britton's house?
>Every day is sunny for George and Charlotte Britton of Lafayette Hill. The Britton's 2,900-square-foot house is blessed with energy bills 20 percent lower than one of comparable size... The design of the house incorporates "passive" solar principles. There are large double pane windows and sliding glass doors on the south side. Inside, tile floors and a Trombe wall absorb the sun's heat during the day and radiate it at night... A stone fireplace on the south wall of the living area provides additional heat during colder months. Britton said "We have a fire every day of the winter."
Direct gain (aka "direct loss") houses in cold cloudy places rarely have solar heat fractions greater than 50%, according to Passive Solar Institute (aka Sustainable Building Institute) Guidelines... 30% is more common, ie the sun only provides 30% of the house heating over an entire year, but indirect gain houses can have solar heating fractions of 90% or more in the month of January. 
An 8' 70 F direct gain cube with R20 walls and ceiling and floor and A ft^2 of R4 south windows with 50% solar transmission needs 24h(70-30) (A/4+(384-A)/20) Btu of heat on an average 30 F January day in Phila with 1000 Btu/ft^2 of sun on a south wall... 0.5x1000A = 24h(70-30)(A/4+(384-A)/20) makes A = 41 ft^2, with a total cube conductance G = 41/4 + (384-41)/20 = 27.4 Btu/h-F, right? (Or would you like to argue about that? :-)
With a 4" 533 Btu/F concrete floorslab, time constant RC = C/G = 19.5 hours. With no sun, 300-year-old high-school physics says the direct gain cube temp drops to 30+(70-30)e^(-24/19.5) = 41.7 F after 24 hours and 30+(70-30)e^(-48/19.5) = 33.4 after 48 hours, and so on. That's MISERABLE solar house heating  performance! (Would you like to argue about that? :-)
OTOH, an 8' 70 F indirect gain cube with R20 walls and ceiling and floor with a 384/30 = 19.2 Btu/h-F conductance needs (70-30)19.2 = 768 Btu/h or 18.4K Btu/day. A T (F) 64 ft^2 radiant ceiling heated by thermosyphoning sun-warmed air (as in the Barra system) with a 1.5x64 = 96 Btu/h-F slow-moving airfilm conductance can keep the cube 70 F if (T-70)96 = (70-30)320/20, ie T = 77 F. If the cube has 8'x8' of R2 air heater glazing with 80% solar transmission (eg 2 $42 4'x8' sheets of twinwall polycarbonate) over the R20 south wall,
0.8x8'x8'x1000 = 51.2K Btu/day

= 6h(T-30)64ft^2/R2 for the air heater during the day
+ 18h(70-30)64ft^2/R22 for the south wall at night
+ 24h(T-30)64ft^2/R20 for the ceiling all day
+ 24h(70-30)4x64ft^2/R20 for the other walls
makes T = 167 F.
We can keep the cube exactly 70 F for 5 cloudy days in a row with a slow ceiling fan and a room temp thermostat and 5x18.4K/(167-77)/62.33 = 16.4 ft^3 of water cooling from 167 to 77 F, with an approximate 1-2^-5 = 0.97 solar heating fraction. But that's an overestimate, since the ceiling can be 77 vs 167 F most of the time.

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