alt.energy.homepower

At an average 65 F indoors, we need 24h(65-14)104 = 127K Btu of heat.
If a frugal 300 kWh/mo of indoor electrical use provides 34K Btu and
a 4'x8'x4'-tall R30 140 F plywood heat storage tank in the house loses
24h(140-65)128ft^2/R30 = 8K, we need another 85K Btu of solar heat.

If the house is 70 F for 12 hours per day and 60 at "night" and sunspace
air keeps it 70 F for 6 hours with 6h(70-14)104 = 35K Btu, we might store
85K-35K = 50K Btu of "overnight heat" in 4000 Btu/F of inherent mass with
4K(70-60) = 40K Btu, and 50K-40K = 10K Btu in 384 ft^2 of shiny ceiling
mass C that cools from 80 to 60 F, with C = 10K/(80-60) = 500 Btu/F.

If sun arrives at a constant 48K Btu/h and we subtract 35K/6h =
for sunspace air that warms the house, we can model the sunspace
itself like this, with air temp T, viewed in a fixed font:

Btu/h
--- 1/278
|---|-->|---------------- T -----------------------www--- T
--- | | 14+ /278 = 166F
| - ---
1/ = 1/278 | - -
| - |
14 F ---www------------- -

If RC = 4500/278 = hours with lots of airflow, we can charge
the house and ceiling from 60 to 70 F in - ln((70-166)/(60-166))
= hours:

1/278
---------www----------------------- T 60->70 F
| 166 F | 4000 | 500
--- --- ---
- --- ---
| | |
- - -

With RC = 500/278 = hours, we can charge the ceiling alone from
70 to 80 in another - ln((80-166)/(70-166)) = hours, leaving
hours like this

1/278 T = 147 1/800
-----------www--------------www-----
| 166 F I --> | 140
--- (166-140)/(1/278+1/800) ---
- = 5364 Btu/h -
| |
- -

with an 800 Btu/h-F auto radiator and fan at the top of a duct that returns
air to the sunspace without mixing with room air and stores I =
Btu/day of 140 F water in the tank. Subtracting the 8K tank loss leaves
. With 60% greywater heat recovery, a pressurized $60 1"x300' PE pipe
coil in the tank could make /( ) = 36K Btu/day of hot water.

We could do this more efficiently and simultaneously with Big Fins
in the sunspace, but that would cost more money.

Nick