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inexpensive 100% solar-office heating with warm air collection
15 jun 2003
a 10'x14' office with r20 walls and ceiling and 8 ft^2 of r2 windows and
15 cfm of natural air leaks would have a thermal conductance of about 15
btu/h-f for air plus 10'x14'/r20 = 7 btu-h-f for the ceiling plus 8/2 = 4
for the windows plus (2(10+14)8-8)/20 = 19 for the walls, about 45 total.
 
january is the worst-case month for solar house heating in providence, ri,
with 27.9 f average days with an average daily max of 36.6. the average
temp during a 6 hour "solar collection day" might be (27.9+36.6)/2 = 32.2.
at an average 65 f indoors, the office would need 24h(65-27.9)45 = 40k btu
per day of heat. with no internal electrical usage or solar heat gain,
it would need 200k btu for 5 27.9 f cloudy days in a row. 

say we collect solar heat in warm air from a low-thermal-mass sunspace on
the south wall and store it in a 2'x8'x8' tall box with a 2'x8'x4' tall
water tank below and 12 2'x8' shelves above which act as an air-water heat
exchanger with a slow-moving air to water thermal conductance of about
12x2'x8'x1.5x2 = 576 btu/h-f...

say we drape a 100' length of 12" round poly film greenhouse duct ($38) in
a zigzag over 100' of 3' wide 2"x4" mesh welded-wire fencing ($20), with a
2x4 at each end to make the water depth 2", and a low-power pump to circulate
water between the shelves and the tank, which has plywood walls and a 10'x16'
piece of epdm rubber folded up like a chinese takeout box for a liner.

the office needs (70-27.9)45 = 1895 btu/h on an average day. the shelf water
needs to be 1895/576 = 3.3 f warmer than the air in the box to provide this.
with natural convection and a 6' height difference between a 4 ft^2 supply 
vent with a 2-watt motorized damper and an 4 ft^2 return vent and a dt box-
to-room air temp diff, 1895 = 16.6x4ft^2'xsqrt(6)dt^1.5 makes dt = 5.1 f,
so the shelf water temp needs to be at least 70+3.3+5.1 = 78.4 f to keep
the office 70 f. 

the 2'x4'x8'+2'x2'x8' = 96 ft^3 of water in the box has about 96x64 = 6,144
btu/f of thermal capacitance. if it stores heat for 5 cloudy days in a row,
200k = 6144(twa-78.4), and the water temp twa needs to be at least 110.1 f
on an average day. 

nrel says 1,020 btu/ft^2 of sun falls on a south wall on an average jan day
in providence, 410 falls on east and west walls, and 590 falls on a horizontal
surface. a 4' deep x 8' tall x 12' wide sunspace with r1 polycarbonate walls
and roof with 90% transmission would gain 0.9(2x4x8x410+8x12x1020+4x12x590)
= 152.5k btu on an average day. the 208ft^2 of glazing could supply the daily
average office heating need of 40k btu at an average sunspace air temp ts,
where 152.5k = 6h(ts-32)208ft^2/r1+40k, so ts = 122 f. 

say the office has a 30" diameter x 14' long poly duct containing t inches
of water resting on foil on a 4'x14' piece of welded wire fence above the
collar beams, and we keep the office 70 f for 6 hours per day using 122 f
air from the sunspace which warms the overhead water enough to store 100%
of the overnight heat the office needs on an average day, with the help of
a slow ceiling fan and a thermostat.

on an average day, the sunspace needs to supply 40k btu over 6 hours at
6667 btu/h. with natural convection and a 6' height difference between an
a ft^2 supply vent with a 2-watt motorized damper and an a ft^2 return vent,
6667 = 16.6asqrt(6')(122-70)^1.5 makes a = 0.4 ft^2. we might use a 1 ft^2
sunspace damper. 

on an average day, we want to store about 40kx18h/24h = 30k btu in the
overhead duct water with heat capacitance c and water-air conductance
g = 30pi/12x14x1.5 = 165 btu/h-f. at dawn, the 60 f office needs about
(60-27.9)g = 1445 btu/h, with a min water temp 60+1445/165 = 69 f. with
t (f) water at dusk and 69 f water at dawn, c(t-69) = 30k. t = 69+30k/c
and t = 122+(69-122)e^(-6/(rc)) with rc = c/165 makes c = 566/(1-e^-990/c)
btu/f. plugging in 566 on the right makes 685 on the left, then 740, 768,
...then 794, with t = 106.8 f. with t" of water, duct width w = 30pi/2-t
inches, and c = 64x14'tw/144 = 6.22(47.1t-t^2) = 794, so t = 2.9 inches.

on an average day, with water temp tc and r30 insulation, the 2'x8'x8'
box would lose about 18h(tc-30)192ft^2/r30 btu to the sunspace and supply
little heat to the office. if cloudy days are like coin flips, the box
needs to supply 40k btu about half the time, ie about 20k btu/day at
a charging rate of about 20k/6h = 3.3k btu/h, so the air near the shelves
needs to be 3.3k/576 = 5.8 f warmer than the shelf water during charging.
with the shelves behind 64 ft^2 of their own glazing behind the sunspace
glazing and a 4 ft^2 airflow path in the shelf-heating loop, the air in
the shelf heater needs to be dt f warmer than the air near the shelves...
3.3k = 16.6x4sqrt(6')dt^1.5, so dt = 7.4. if 1020 btu/day of sun passes
through two 90% r1 glazings, 826x64ft^2 = 6h(tc+5.8+7.4-122)64ft^2/r1 +
18h(tc-27.9)192ft^2/r30 + 20k, so tc = 156 f :-), assuming (conservatively)
that the shelves gain no more heat on a clear day than an average day. 
we can probably use less sunspace or box glazing, esp with reflective
ground to the south, eg whitewashed rocks.

the shelves would cool by about 40kbtu/2048btu/f = 19.5 f over a cloudy
day and might rewarm the following day. leaving them cool is desirable for
efficient solar heat collection. the circulating pump might run just enough
to keep the lower tank warm on an average day, and more on a cloudy day... 
if we turn on the pump when the shelf water is at least 5 f warmer than the
tank water, it needs to move p pounds of water per day, where 20k = 5p,
and p = 4k pounds per day or 0.35 gpm. taco's $101 60 watt 006-bt4-1 pump
(grainger item 5p429) can move 5 gpm with a 6' head, so it might only run
1.7 hours per day, consuming 36 kwh per year at a cost of $3.60. 

the ashrae handbook of fundamentals says the 99% "winter design temp" in
providence is 5 f, which determines the required heating system capacity.
the office needs (70-5)45 = 2925 btu/h to stay 70 f on that very cold day,
when the shelf water needs to be 2925/576 = 5.1 f warmer than the air in
the box, and 2925 = 16.6x4ft^2'xsqrt(6)dt^1.5 makes dt = 6.9 f. at 5 gpm,
2925 = 5x8x60dtw, so dtw = 1.2 f, and the tank water temp needs to be at
least 70+5.1+6.9+1.2 = 83.2 f on that day. we might turn on the pump when
the shelf water temp is 5 f more than the tank water temp or the shelf
water temp is less than 83.2... 

nick




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