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re: geothermal heat storrage?
2 sep 2001
nonkie  wrote:

>you want to use solar energy for heating a home, you have always one big
>problem :

would that there were only one!

(this is the "hortatory subjunctive" tense in english. does flemish have
an equivalent? :-)

> in summer time you have a lot of son heat, but you don't need any home
>in winter the opposite is true.
> solution: heat storage.

that's one solution, but i think seasonal heat storage makes less sense
than making the collectors large enough to gather enough sun to heat
the home on an average day in the worst-case month (cold and cloudy, with 
the least kwh/m^2/(tin-tout) on a vertical wall), and store enough heat
to keep the house warm for a cloudy week.

the national renewable energy laboratory (nrel... does belgium have an
equivalent?) says january is the worst-case month for solar house heating
where i live near philadelphia, with a 30-year average of 3.15 kwh/m^2-day
of sun on a south wall and -0.9 c outdoors and 3.15/(20-(-0.9)) = 0.151,
vs dec with 2.84/(20-2.1) = 0.159 and feb with 3.41/(20-0.6) = 0.175.

suppose we live in a 3 meter cube with metric r3 (us r17) insulation,
with no air leaks or internal heat gain and 54m^2/r3 = 18 w/c of thermal
conductance. in phila, it needs (20-(-0.9)18 = 376 watts or 9029 wh/day
of heat on an average january day. let's cover part of the south wall with
2 layers of r0.2 polycarbonate glazing with 90% solar transmission and
collect 2.55 kwh/m^2-day of sun over 6 hours and lose 6h(20-(-0.9))/r0.4
= 314 wh/m^2 to the outdoors (a 2.24 kwh/m^2 net gain) on an average
january day, and fill the glazing cavity with r3 foam at night.

we need to store 376x18 = 6.8 kwh on an average day and 5dx9029 = 45 kwh
of heat during a cloudy week with the wall foamed. say sun shines through
the south wall onto a water wall with 2 layers of glazing on the south
side and insulation on the north. on an average day, at temp t, the water
collects 0.9^4(3.15) = 2067 kwh/m^2 of solar heat and loses 24h(t-20)/r0.4
to the room, so t = (2067+1800)/60 = 54 c (or maybe more, if the floor
in front of the wall is white.) we need about 6.8/3.15 = 2.2 m^2 of south
water wall surface, and the water wall loses 85 w to the room, so we need
another (376-85)x6h/2240 = 0.8 m^2 of south window with no water behind
it to warm the room directly over 6 hours.

say we use a small fan and thermostat to circulate room air around the
water wall under the glazing and insulation. at 10 mph, its airfilm
conductance is about 2(2+10/2) = 14 btu/h-f-ft^2 or 80 w/m^2c, counting
both sides, so the minimum usable usable water temp is 20+376/80/2.2
= 22 c, and 1 m^3 stores 1.16 kwh with a 1 c temp diff., so we need
45kwh/(54-22)/1.16/2.2 = 0.55 m (22") of depth. 

seasonal storage in a desiccant like lithium chloride might make sense,
with a much smaller volume than hot water and and no loss of heat over time.
to recover the heat from the concentrated solution, we might sprinkle a
basement floor with water to make it colder and warm the upstairs, or use
a small evacuated groundwater heat exchanger. this might also provide year-
round hot water and summertime air conditioning and dehumidification. 


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