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armoire solaire du jour
30 may 1996
a very-well-insulated house, not to scale:

                                                 no air infiltration
                                  30'            no internal heat generation
     30 f and            --------r30----------- 
     1,000 btu/ft^2 of  |                      | r30 average walls and ceiling
     sun on the south   r       top view       |
     wall on an average 3                      r     ==> heat loss on average
     december day...    0                      3         december day =
                        |        2-story       0         24h(70-30)3000/r30
   sun --->         -r1-|         house        |         = 96k btu.
                   |    <==                    |  30'
                   r    |        surface area  |     each square foot of
                   1    |r10     ~2,000 ft^2   |         sunspace collects
              12'  | ss |   |       walls +    |         1000 btu/day and 
                   |    r s r 10' 1,000 ft^2   |         loses 6(80-30)
                   |    2 c 1       ceiling    |         = 300, for a net
hmmm. if the       | 8' 0 4'0                  |         gain of 700, so
sunspace has        ---- ----------------------          we need 96k/700
128 ft^2 of r1                                           = 137 ft^2 of 
sidewalls and       r10-----------r30----------          sunspace here, 
a r10 roof, it     |    m ==>                  |r        eg 12' x 16' x 8'
will lose another  |    |   sunspace heats     |3        deep... (?)
6(80-30)128/r1     | ss |       house on an    |0
= r38k of heat     |    |r10    average day.   | 16'     <==yes, another
thru the           r    r s r                  |         55 ft^2 should make
sidewalls...       1    2 c 1  8'              |         up for the sidewall
                   |    0   0                  |         loss...
         --------------------------------------------------
                             west view

larger detail:
                   |            r
                   |            3      daytime airflow within closet
            30 f   |            0     /
     sun           |            |--------r10---      
                   |   80 f     | d ==>        m  m is a motorized damper
                   |   sunspace | |  20        |  d is a one-way damper
                   r            | r  drums     |
                   1            | 2            | <--insulated wall
                   |            | 0  130 f     |
                   |            | |    sc      r
                   |            | |            1
      reflective?  |reflective? |   <==        0
          ---------------------------------------------------
                                  ^--insulated wall

we need to store about 500k btu for 5 days w/o sun, ==> 500k/25k = 20
55 gallon drums full of water, cooling from 130 f to 80 f, which will
fit in a space about 4' wide x 8' tall x 10' long.

in this case, the closet glazing would be 8' x 10' and there would be 20x25
ft^2 of drum area, ie a glazing/thermal mass area of 6.25:1, so we might want
to use a small fan to circulate air thru the solar closet when the sun is
shining. grainger's $61 4c688 560 cfm 36 watt fan with a maximum temperature
spec of 149 f might be nice, with a $6 2e247 snap-disc thermostat in a glazed
box to turn on the fan at 130 f. if the closet were taller or had a 10:1
container area/glazing area ratio, it might not need a fan.

if the solar closet has r20 insulation on the south side and r10 on the
other sides, and it isn't needed to keep the house warm on an average night,
and it doesn't make hot water, here's how to find the water temp t inside...

80x1000 btu = 6 hr(t-80)80ft^2/r1     for the south wall, day
            +18 hr(t-30)80/r20                  "         night
            +24 hr(t-70)224ft^2/r10   for the other 5 walls, night

ie      80k = 480t  - 2,880
            +  72t  - 2,160
            + 538t  - 38k

            = 1,090t - 42,672

or   1,090t = 122,692, or t = 122,692/1,090 = 112.5 f.

so, this closet could use a reflector in front or a little more glazing
or a little more insulation... let's try a reflector to get 33% more sun:

     1,090t = 80k x 1.33 + 42,692 = 149k btu/day, so t = 137 f :-)

nick



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