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re: question about heating a home
14 dec 1995
nick pine wrote:
:if we make the cube an epdm-rubber-lined plywood box, 8' on a side, with a
:foot of insulation, ie r-40 sides, with a 6' cube of water weighing 13,500
:pounds inside, then rc = 40/(6x8x8)x13500 = 1872 hours or 78 days. if the
:water began at 100 f, after 1 hour in a 50 f room it would have a temp of
:50 + 50 exp(-1/1872) = 50 + 50 x 0.99947 = 99.97 f. it would cool to 68 f
:in t = -rc ln((68-50)/(100-50)) = 1913 hours or 80 days...
:we might make the south face of this cube dark and glaze it with thin plastic
:over an air gap, and add a $100 10" x 10" honeywell 64ls damper and 6161b1000
:motor that opens to the inside of the cube when the sun shines, and replace
:the epdm rubber tank with 18 sealed plastic 55 gallon drums full of water...
:the average outdoor temp in december in philadelphia is about 32 f. on an
:average december day, 1000 btu/ft^2 of sun falls on a south wall, in about
:6 hours. how hot will the cube get, sitting outdoors, or inside the house,
:behind a window?
crudely assuming that the water and the air in the passive air heater and
the "solar closet" all have the same temperature t when the sun is shining,
the r40 wall insulation value includes thermal bridges, the glazing transmits
all of the sun and none of the heat by radiation, and it has an r-value of 1,
then on an average december day, the energy that goes into the box is
ein = 1000 btu/day x 64 ft^2 = 64k btu/day,
and the energy that leaves the box is
eout = 6 hours x (t-32f) x 64 ft^2/r1 for the south face, daytime
+ 18 hours x (t-32f) x 64 ft^2/r40 for the south face, nightime
+ 24 hours x (t-32f) x 64 ft^2 x 5/r40, for the other 5 cube faces.
so if ein = eout, then
64k = 64 x (6 + 18/40 + 24x5/40) x (t-32), or equivalently
1k = (6 + 9/20 + 3) x (t-32), or equivalently
t = 32 + 1000/9.45 = 32 + 105.8 = 137.8 f.
if this ideal cube were sitting inside a house, the water temperature would be
t = 68f + 105.8f = 173.8 f. after a week, the water temperature would be
t(168 hrs) = 68f + (173.8f-68f) x exp(-168hr/1872hr) = 164.7 f.
if there is snow in front of the window, or some other 50% reflective surface,
the water temperature would be
t = 68 + 1.5 x 105.8 = 226.7 f.
if the ideal cube were a long epdm-rubber-lined plywood box full of water with
a transparent horizontal top and an insulating/reflecting movable cover, lying
along the north wall in the line focus of a 5:1 linear parabolic concentrating
80% reflector, the "water temperature" would be something like
t = 68 + 5 x 0.8 x 105.8 = 491.2 f :-)
and the water would be quite warm even on this gray day, with 3" of fresh
wet snow on the ground outside.
nick
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