re: solar collector design for a solar closet? SOLAR.KnowledgePublications.com

re: solar collector design for a solar closet?
9 nov 2002
>if it's 20 f outdoors and 70 indoors, a superinsulated house with a
>conductance of 200 btu/h-f needs 18h(70-20)200 = 180k btu of overnight
>heat on an average day...

or less, if it's 70 f at dusk and 60 at dawn. and some of this heat
might come from electrical energy use, eg 68.2k btu/day at 600kwh/mo.

>how many 4"x10' pipes would the airshaft need for the house above, if
>the air inside were 100 f for 6 hours per day? ignoring the mass of the
>house itself (drywall, etc.), 56 btu/f and 10 ft^2 at 1.5 btu/h-f-ft^2
>makes rc = 3.7 h for each pipe, so tmax = 100+(tmin-100)exp(-8/3.7)
>= 88.5+0.115tmin. the house needs (70-20)200 = 10k btu/h.

oops... (60-20)200 = 8k btu/h at dawn. with n pipes, 8k = (tmin-60)10x1.5n,
so tmin = 60+533/n. if 180k btu = 56n(tmax-tmin), so 3214 = 35.4n-472, and
n = 104. big oops. tmin = 65.1, and tmax = 96.0. the pipes would fit in a
4'x4'x10' box with 5.75 ft^2 of cross-sectional airspace. a vertical box
might thermosyphon sunspace air, or use a large, slow, low-power fan at the
top moving 2300 cfm of sunspace air down among the pipes at 400 feet per
minute. if the box is horizontal, under the floor, running north-south, with
a vertical airshaft at the north end, it might have a fan at the south end
blowing air back into the sunspace. a 2-story house might have 52 20' pipes
around 3 sides of a 4'x4' 2-story airshaft with a 35"x39" empty space
in the middle.

that's a lot of pipes and fans and floorspace. maybe it's better to build
an anna-edey-style water-wall in the sunspace with a bubblewall over it for
overnight heat storage. or use zomeworks skymats and ceiling pipes, or a
hydronic floor with thomason-style trickle collectors on the roof or a solar
attic with a clear corrugated polycarbonate south roof and a linear parabolic
down-reflector for the north roof and a 4'-wide x 32'-long water heating
trough on the attic floor near the north wall, with 2" of water in a 4' flat
greenhouse uv poly duct over some pv panels in 2 or 3 suns. 

>>flat-plate collector facing south at fixed tilt=90, solar radiation,
>>           dec  
> sun        810   btu/ft^2-day |
  beam sun   457   btu/ft^2-day |
> avg. min   11.2  f            |
> avg. 24h   21.2  f            | nrel data for helena, mt
> avg. max   31.3  f            |
> avg. day   26.3  f during a 6h solar collection time?
  
a square foot of r1 vertical sunspace glazing with 90% solar transmission
could collect 0.9x810 = 729 btu/day (or maybe 30% more, with snow or a
frozen pond in to the south.) it might lose about 6h(100-26.3)/r1 = 442
of that, leaving 287. if the house needs 24h(65-21.2)200 = 210k btu/day, 
it might have 210k/287 = 732 ft^2 of sunspace glazing. seems like a lot. 

reducing the sunspace temp from 100 to 80 f increases the net sunspace gain
from 287 to 407 btu/ft^2-day and lowers the glazing requirement to 516 ft^2.
still large. adding a pond or white surface reflector increases the gain to 
626, for 335 ft^2 of vertical glazing. nrel says 390 btu/ft^2-day falls on
a horizontal surface, so we might tilt the glazing back atan(390/810) = 26
degrees. if the glazing is 8' tall and touches the south wall of the house
at the top and it's 4' away at the bottom, a linear foot might collect 
0.9(4x390+1.3x8x810) = 8986 btu/day and lose 6h(80-26.3)9ft^2/r1 = 2882 btu,
for a net gain of 6104, and 210k/6104 = 34 linear feet of sunspace with
310 ft^2 of glazing. not much less than the vertical glazing...

we might heat the house for 6 hours with 80 f sunspace air while circulating
100 f air through the airshaft from a space behind an inner sunspace glazing
near the north wall of the house, or steal some hot air from a solar closet
air heater inside the sunspace.

the closet needs to store about 5x210k = 1050k btu for 5 21.2 f cloudy days
in a row. it might have 1050k/(130-80) = 21k pounds or 2,625 gallons or 48
55 gallon drums of water stacked 2-high and 4-deep inside an 8'x12'x8' tall
box with a 48x25/(8'x12') = 12.5 mass/glass surface ratio, with most of the
box inside the house. on an average day, it might gain 0.9x0.9x1.3x810x8'x12'
= 82k btu of sun and lose about 6h(130-80)8'x12'/r1 = 29k from the glazed
south wall to the sunspace air and about 82k-29k = 53k through 512 ft^2 of
box surface, so we need 24h(130-70)512/r = 53k, ie the closet walls need
r14 insulation.

the closet might be cheaper and more compact with 328 ft^3 of waterwalls vs
drums. it might contain 6 8'x8'x1' waterwalls running north-south with 4"
spaces between them, in an 8'x8'x8' box with a 6x2x64/64 = 12 mass/glass
surface ratio. these shelves for 55 gallon plastic film drum liners could
be made with 2x4s and bolts and 2"x4" welded wire fencing. 

nick

http://www.ece.vill.edu/~nick