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storing heat in house walls
22 oct 2001
r. h. turner, geng liu, y. a. cengel, and c. p. harris of the u nevada/reno
wrote "thermal storage in the walls of a solar house," in the asme journal
of solar energy engineering, november 1994, vol. 116, pp 183-193. 

they suggest putting 4 3" pvc water pipes in each 2x4 16" interior wall
cavity of a house ("the 7.5' long pipes were glued and capped at one end,
completely filled with water, and then capped at the other end... it was
necessary to drill a small hole in the second 3" diameter end cap to allow
air to escape during cap installation, and then seal the hole...")

they used drywall on both sides, and blew 40 cfm of solar-warmed air in at
the top, with "active charge" and "passive discharge" and ducts connecting 
to top and bottom slots in the walls. 

they say this takes up less space than a rock bin and the walls can supply
useful radiant heat. they go on to suggest the warm air might come from an
"integrated-roof air heater" with a selective surface under plastic glazing. 

this could be more passive and efficient and inexpensive with exposed pipes
(at least on one side) and a sunspace on the ground. we could put 5 4" pipes
between 2x6 studs on 24" centers. we might paint them red, a la pompidou.

a 4"x7.5" pipe holds c = 42 btu/f (pounds) of water, with 7.9 ft^2 of area
and g = 11.8 btu/h-f of thermal conductance in slowly-moving air, so
rc = c/g = 3.57 hours.

a 32'x32'x8' tall house with r20 walls and r40 ceiling and 80 ft^2 of r4
windows and 0.5 air changes per hour has 80ft^2/r4 = 20 btu/h-f of window
thermal conductance plus 944/20 = 23.6 for walls plus 1024/40 = 25.6 for 
ceiling, with 0.5x32x32x8/60 = 68 cfm of air leakage, a total conductance
of approximately 137 btu/h-f, so it needs (60-30)137 = 4110 btu/h to stay
60 f indoors on an average 30 f philadelphia january morning. 

suppose we keep the house air 70 f for 6 hours on an average january day.
if the pipes are 60 f at dawn, 18 hours (5 time constants) after dusk,
they will charge to 70-(70-60)exp(-6/3.56) = 68 f by dusk. with n pipes,
rc = 42n/137 for discharge, and 60 = 30+(68-30)exp(-18x137/(42n)) makes 
n = 248 pipes in 99' of interior walls. that seems a bit large. let's
use 64' of walls with 160 pipes with 6720 btu/f, and add 4000 btu/f for
2" of concrete slab. then rc = 10720/137 = 78 hours, and the mass will
cool to 30+(68-30)exp(-18/78) = 60 f by morning. 

we need to get 24h(65-30)137 = 115k btu/day of heat from the sunspace,
or 19.2k btu/h over 6 hours. with 2'x4' vents at the top and bottom and
6' of height difference, 19.2k = 16.6x2'x4'sqrt(6)dt^1.5, so dt = 15 f,
ie the sunspace needs to be 85 f to keep the house 70 f. if it's 8' deep
with an insulated roof and a single layer of 8' r1 vertical south glazing
with 90% solar transmission, a square foot of glazing will gain about
900 btu/day and lose about 6h(85-30)/r1 = 330 btu, for a net gain of 570,
so the sunspace needs to be 115k/8'/570 = 25' long.

we might make hot water and keep the house warm on cloudy days with heat
from a reflective parabolic trough in the attic.

nick




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