alt.energy.homepower

>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------------- -
>
>1. 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
> --- --- ---Btu/F
> - --- ---
> | | |
> - - -
>
>2. 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 -
> | |
> - -
>
>3. 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...
>
>We could do this more efficiently and simultaneously with Big Fins
>in the sunspace, but that would cost more money.

With constant weak sun at 48KBtu/h/480ft^2 inside the sunspace, 64 ft^2
of 140 F bare solar collectors or Big Fins would absorb 6400 Btu/h and
lose about (140-T) Btu/h, so we'd have something like this:

-6.4K = 3 Btu/h
--- 1/278
|---|-->|---------------- T -----------------------www--- T
--- | | 14+3 /278 = 143F
| - ---
1/ = 1/278 | - -
| - |
14 F ---www------------- -

and this:

1/278 T 1/800
-----------www--------------www-----
| 143 F | | 140
--- w ---
- w 1/96 - radiator
| 6400 Btu/h w |
- --- | -
|--|-->|-------|
--- | 140
---
- collector
|
-

which is equivalent to this:

1/278 T = 141 1/896
-----------www----------------www-----
| 143 F I--> | 140
--- (143-140)/(1/278+1/800) ---
- = 589 Btu/h -
| |
- -

so the radiator and collector would gain 6400+589 = 6989 Btu/h, 30% more
than the radiator alone. But sun isn't constant. It might be 50% diffuse
in December, so we might heat water whenever the house needs are met and
the collector output is at least 140 F in the first two phases above,
as well as the third. This can also work well with poor collectors, eg
copper pipes banged into grooves in thin brown aluminum coil stock,
since their heat losses to sunspace air become radiator gains.

Engineer Nathan Hurst will talk about his Mazda radiator experiments in
Australia at the Pennsylvania Renewable Energy Festival on 9/22-23/07,
.

Nick