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A solar cave
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Subject: A solar cave
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From: clarion@dorsai.org (Clarissa Coffey)
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Date: Sun, 30 Oct 1994 14:33:14 GMT
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Article: 1279 of alt.architecture.alternative
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Newsgroups: alt.architecture.alternative, alt.energy.renewable
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Organization: The Dorsai Diplomatic Mission NYs Computer Consulate
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Xref: bigblue.oit.unc.edu alt.architecture.alternative:1279 alt.energy.renewable:2692
Rex Roberts talked about caves dug into south-facing hills...
Suppose cavepersons had had glass:
. ^
. | up
.
R10 earth at . south ==>
55 degrees F .
.
/ g If the enterprising caveperson had insulated
Rc / g the inside of the cavern with leaves and
------------------g W g mud, until, say, Rc = 10, and the daytime
| .----- . temp of the cavern had been 65 degrees F
| 20 x 20' cavern | . and the nighttime temp had been 55, and
| 10' tall | . he or she had desired to warm the cave
| | . for, say, 5 days without sun, with an
--------Rf-------- . average sunnytime steady-state floor
|vaulted -- stone | . temp of, say 90 F, and mammoth skins
| ----- ----- | T . piled on top of the floor to make
|--- floor ---| . a sunny day R-value of Rf (which
. . . skins could be moved aside to
. Rc . . . decrease the R-value of the floor
. g during sunless times), how much
Note that since .Rc g glass area, Ag, would have been
the upper edge of the . g Ag needed, and what would Rf and
glass is lower than the . g the average floor thickness T
bottom of the floor, the floor .Aa g have had to have been,
stays warm during sunless times, . g ignoring the clerestory
because of this igloo-like heat trap. . g cave window, W?
(How did the caveperson insulate around . g
the edge of the floor?) g south ==>
.
The daily heatloss of the cavern would have been .
.
Lday = (8 hours)(1200 ft^2/R20)(65-55) ~ 5K Btu. 32 F outside temp
.
During sunny times, the sun would shine onto the .
absorbing surface Aa, assumed equal to Ag, and some .
heat would be lost thru Rc to the earth underneath... .
.
The daily net heat resulting from each square foot of .
glass might have been on the order of .
.
Eg = 1000 - 8(90-32)/R1 - 8(90-55)/R20 .
sun glass loss absorber back loss .
= 500 Btu/ft^2/day .
.
So keeping the cavern warm might have required only about .
10 ft^2 of glass... A 3.16 x 3.16' single-glazed window, or .
an early sliding-glass door... .
.
The mammoth-skin floor resistance would want to have been about .
.
Rf = (90-65)(400 ft^2)/(5000/8) = R-16. .
.
During sunless days, the skins would be moved around or a trapdoor .
in the floor would have been opened to reduce this, to get more heat .
out of the floor... .
.
Assuming the floor would have had a minimum R-value of 1, per square foot .
(both sides), the required heat transfer rate would have been about 5000/8 .
Btu/hour, from 800 ft^2 of floor, which leads to a minimum temperature .
differential, after 5 days, of about 1 degree F.
So if the masonry had held about 22 Btu/ft^3/degree F, the floor thickness
might have wanted to be about
T = (5 days)(5000 Btu/day)/((90-66)(22)(400 ft^2)) = .12'
Hmmm. Ferro-cement...
Other modern improvements might include making the whole thing above ground,
including some throwaway diet-pepsi bottles filled with water in the floor,
making the absorbing surface a parabolic reflector, and making the floor a
wall.
Nick Pine (nick@ece.vill.edu)