Subject: armoire solaire du jour

From: [email-masked] (Nick Pine)
Date: 30 May 1996 08:26:13 -0400

A very-well-insulated house, not to scale:
                                                 No air infiltration
                                  30'            No internal heat generation
     30 F                ---------------------- 
     1,000 Btu/ft^2/day |         R30          | R30 average walls and ceiling
       in December      |                      |
                        |R30                   |     ==> heat loss on average
       top view         |                   R30|         December day =
                     R1 |        2-story       |         24h(70-30)3000/R30
   sun              ----|         house        |         = 96K Btu.
                   |    <==                    |  30'
                 R1|    | R10    surface area  |     Each square foot of
                   |    |---     ~2,000 ft^2   |         sunspace collects
              12'  | SS |   |       walls +    |         1000 Btu/day and 
                   |   R| S |10'  1,000 ft^2   |         loses 6(80-30)
                   |   2| C |R10    ceiling    |         = 300, for a net
Hmmm. If the       | 8'0| 4'|                  |         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     |    D ==>                  |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                 |         55 ft^2 should make
sidewalls...       |   2|SC |1 8'              |         up for the sidewall
                   |   0|   |0                 |         loss...
         --------------------------------------------------
                             west view
Larger detail:

                   |            |      daytime airflow within closet
                   |            |     /
     sun           |            |--------------      
                   |   80 F     | D ==>        |  D is a one-way damper
                   |   sunspace | |  20        |
                   |            | |  drums     |
                   |            | |            | <--insulated wall
                   |            | |  130 F     |
                   |            | |    SC      |
                   |            | |            |
      reflective?  |reflective? |   <==        |
          ---------------------------------------------------
                                  ^--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
20*25 ft^2 of drum area, ie a glazing/thermal mass area of 6.25:1, so
one might want to use a small fan to circulate air thru the solar closet
when the sun is shining. Grainger's 4C688 560 cfm 36 W 149 F fan might
be nice, with a thermostat set to 130 F in a glass box to turn it on.
If the closet were taller or had a larger ratio of container/glazing
area, 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  - 2880
	    +  72T  - 2160
	    + 538T  - 38K

            = 1090T - 42,672

or    1090T = 122692 or T = 122692/1090 = 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:

      1090T = 80K x 1.33 + 42692 = 149K Btu/day, so T = 137 F :-)

Nick