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Re: State of the Art in Heating at High Altitudes

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Posted by nicksanspam on January 7, 2006, 9:14 pm

Less could work, esp since insulation works 5% better per 1000' of elevation.

Maybe not, since warm air rises.

That's 20-year-old "mass and glass," vs the state of the art! :-)

Expensive and hard to seal at the edges. Better to circulate warm air
between the living space and a low-thermal-mass sunspace with lots of
south glazing during the day and let the sunspace get cold at night.

Cooling and dehumidification are unlikely needs at that altitude.

NREL says 660 Btu/ft^2 falls on the ground and 1240 falls on a south wall
on an average 20.6 F December day with a 34.9 high in Eagle, CO at 6513'.
The average temp in July is 66.6, with an 86.0 daily high and a 47.2 low
and humidity ratio w = 0.0079 (very dry.)

A 40'x60'x8' tall house with 192 ft^2 of R4 windows and R48 (12" SIP)
walls and ceiling and 0.2 ACH would have 192/4 = 48 Btu/h-F of window
conductance + 1408/4829 = 29 for walls + 2400/48 = 50 for the ceiling
plus about 0.2x40x60x8/60 = 64 for 64 cfm of air leakage, totaling 191.
With 600 kWh/mo (2843 Btu/h) of indoor electrical use, it would only
need (65-20.6)191-2843 = 5637 Btu/h of heat, ie 135K Btu/day or 676K
for 5 cloudy days in a row. Solar gain through the windows is gravy.

A $ square foot of R1 polycarbonate solar siding or sunspace glazing
with 90% solar transmission might gain 0.9x1240 = 1116 Btu/day and lose
6h(120-27)1ft^2/R1 = 558, for a net gain of 558 Btu/day, so we could heat
the house with 135K/558 = 242 ft^2 of glazing, eg an 8'x30' wall.

We could store heat for 5 cloudy days in a row in P pounds of water
cooling from 120 to 80 F, where 676K = (120-80)P, so P = 16900 lb,
eg 263 ft^3 in a 2' deep x 13' diameter welded-wire fence tank with
a plastic film liner, which might also contain a $0 1"x300' PE pipe
coil to make hot water for showers.


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