Subject: A solar cave

From: [email-masked] (Clarissa Coffey)
Date: Sun, 30 Oct 1994 14:33:14 GMT

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 ([email-masked])