[Greenbuilding] A combined solar closet in Pittsburgh

[nick pine]

Young Ben would like to live inexpensively on a farm near Pittsburgh...

http://pittsburgh.craigslist.org/fod/4062701580.html offers a 12' by 56' 2 
bedroom mobile home with a new steel roof, a front kitchen, and a large 
bath, in very good condition for $3,900.00.

http://pittsburgh.craigslist.org/cto/4039312791.html lists "Antique CLASS 
C - 22' MOTOR HOME for sale - pictures sent to serious inquires - inspected 
and ready-to-go anywhere you want. Mechanically VERY RELIABLE, needs TLC 
(tune-up) - everything works. Great for camping, hunting, or just fix it up 
for the ultimate tail-gator mobile. Comes with kitchen, refrigerator, stove, 
oven, microwave, furnace, holding tanks (for gray and black water), full 
bathroom, overhead bed, couch, chair and dinette table. It has a 350 small 
block engine (that is very reliable), and transmission also very reliable, 
no slips or bangs, very smooth. Very easy to drive and is still being driven 
daily. Selling for $1500, or best offer."

 Here's one for $100... http://pittsburgh.craigslist.org/for/4056583831.html 
"1974 Holly Park mobile home for sale. Must be moved. Serious inquiries 
only; sold as-is. You move it. Needs repairs and attention, but would make a 
great hunting cabin."

 Ben has also been offered the free use of a pop-up camper. None of these 
structures have much thermal insulation, but they could be solar-heated if 
enclosed in a strawbale structure inside an inexpensive plastic film 
greenhouse. Ben might be happy living inside a strawbale structure without a 
pop-up camper, with a strawbale bed and a strawbale table and a sawdust 
toilet.

 An 8'x12'x8'-tall room with about 536 ft^2 of R55 dry 2' strawbale walls 
would have thermal conductance G = 536/55 = 9.7 Btu/h-F... 10 cfm of fresh 
air would raise the conductance to about 20. Keeping it 70 F on an average 
31.5 F December day in Pittsburgh would only require (70-31.5)20 = 770 
Btu/h.

http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/US/code/pvwattsv1.cgi 
says 1.7 kW/m^2 of sun falls on a vertical south wall and 1.91 falls on a 
south wall with a 45 degree slope on an average December day with a 35.1 
average daytime temp. Suppose a 6 cent/ft^2 4-year 14'x24' R1 sloped 
polyethylene film south wall with 90% solar transmission and 100% longwave 
IR transmission connects the top edge of the south room wall to the ground 
10' south of the room wall and transmits 0.9x14'x24'x1.91x317 = 185K Btu and 
keeps the room 70 F for 24 hours and radiates heat from the ground inside 
the wall and the south room wall at absolute Rankine temp Ta to an outdoor 
temp Ts (R) and 185K = 24h(70-31.5)20 + 6hx0.1714E-8V(Ta^4-Ts^4)480ft^2, 
where V = 1/4 is the approximate view factor of the radiating surfaces. (A 
more expensive and durable polycarbonate south wall would block longwave 
IR.)

 Duffie and Beckman's 2006 Solar Engineering of Thermal Processes book has 
equation (3.9.2) for sky temperature Ts = 
Ta[0.711+0.0056Tdp+0.000073Tdp^2+0.013cos(15t)]^(1/4), where Ts and Ta are 
in degrees Kelvin and Tdp is the dew point temp in degrees Celsius and t is 
the number of hours from midnight.

 Ta = 35.1+460 = 495.1 degrees R, ie 495.1/1.8 = 275.1 degrees K. NREL says 
the humidity ratio w = 0.0030 pounds of water per pound of dry air on an 
average December day in Pittsburgh, which makes the partial pressure of 
water in air Pa = 29.921/(0.62198/w-1) = 0.145 "Hg, which makes the dew 
point Tdp = 9621/(17.863-ln(Pa)) = 486 R, ie 486-460 = 26 F, ie -3.3 C. With 
t = 12 hours (noon), cos(15t) = -1, so Ts = 
275.1[0.711-0.0056x3.3+0.000073x3.3^2-0.013)]^(1/4) = 253.2 K, ie 1.8x253.2 
= 455.7 R, ie -4.3 F, 39 F degrees less than the air temperature.

 ... 185K = 24h(70-31.5)20 + 6hx0.1714E-8V(Ta^4-455.7^4)480 makes 166.5K = 
1.234E-6(Ta^4-455.7^4), so Ta = 649.5 R, ie 190 F. A quarter-cylindrical R1 
film wall with a 10' radius and a 24' length and 377 ft^2 of surface with 
190 F air inside and 35.1 F (495.1 R) air outside would lose 
6h(190-35.1)377ft^2 = 350.4K Btu/day to the outdoor air, which is more than 
the solar input, so 190 is too hot.

 If 185K = 6h(T-495.1)377 + 1.234E-6(Ta^4-455.7^4), ie T = 576.9 - 
5.46E-10(Ta^4-455.7^4). Plugging in T = 540 R on the right makes T = 554.0 
on the left. Repeating makes T = 549.0, 550.8, and 550.2 R, ie 90.2 F, warm 
enough to solar heat the room on an average day.

 If the room is 70 F at dusk and 60 at dawn, 18 hours later, and 60 = 
31.5+(70-31.5)e^(-18/RC), time constant RC = -18/ln((60-31.5)/(70-31.5)) = 
60 hours = C/G, and C = 60hx20Btu/h-F = 1200 Btu/F of room temp thermal mass 
with lots of surface can store average-day heat, eg 1200/5 = 240 8"x8"x16" 
hollow concrete blocks, but they would take up a lot of space in the room. 
Putting 2 750 ml glass bottles of Pellegrino water into the holes in each 
block would raise its capacitance to about 9 Btu/F and lower the number of 
blocks to 1200/9 = 133.

 But it seems more interesting to combine daily and cloudy-day heat storage 
in a single higher-temp glazed Solar Closet 
http://www.ece.villanova.edu/~nick/solar/solar.html with 8'x8' of R1 
vertical polycarbonate south air heater glazing over an insulated south 
wall. If the room is 70 F for 16 hours per day and 60 F for 8 hours and we 
count Ben's 300 Btu/h body heat for 8 hours and ignore the south room wall 
heat gain from the 90.2 sunspace air during the day, we need to store 
8h((60-31.5)20-300)+16h(70-31.5)20 = 14.5K Btu of average-day heat. With 
lots of surface, we can store this heat in C Btu/F of daily mass cooling 
from T (F) to 60, with (T-60)C = 14.5K, ie T = 60+14.5K/C.

 If the air heater glazing transmits a constant 0.9^2x1.7x317/6h = 73 
Btu/h-ft^2 of sun with a 90.2+73xR1 = 163 F Thevenin equivalent temperature 
and the daily mass warms to T = 163-(60-163)e^(-6/RC) F in 6 hours of sun, 
and we replace T with 60+14.5K/C, 60+14.5K/C = 163-(60-163)e^(-6/RC), ie 
14.5K/C = 103(1-e^(-6/RC), ie 141/C = 1-e(-6/RC), and an R = 1/64 glazing 
resistance makes 141/C = 1-e(-384/C), ie C = 141/(1-e^(-384 /C)). Plugging 
in C = 160 on the right makes C = 155.1. Repeating makes C = 153.9, then 
153.7 Btu/F, eg 31 hollow sparsely-stacked concrete blocks cooling from 
60+14.5K/155 = 154 to 60 F on an average night.

 But wait! An 8"x8"x16" hollow concrete block with 2 4"x4"x8" holes has 384 
in^2 for the solid faces + 192 in^2 for the two faces with holes + 256 in^2 
for the holes, totaling 832 in^2, ie 5.78 ft^2. With a 1.5 Btu/h-F-ft^2 
slow-moving airfilm conductance, one block's airfilm conductance is 8.67 
Btu/h-F, and 31 blocks have a 269 Btu/h-F film conductance.

 If the blocks absorb 14.5K/6h = 2417 Btu/h from hot air, the air will be 
2417/269 = 9 F warmer than the blocks. And the air in the air heater has to 
be warmer than the air near the blocks for thermosyphoning to occur. One 
empirical chimney formula says cfm = 16.6Asqrt(HdT), where A is the area of 
the chimney opening in square feet and H is the chimney height in feet and 
dT (F) is the temperature difference between the chimney and outdoor air. 
With 6"x8' slots at the top and bottom of an 8' tall air heater, A = 4 ft^2 
and H = 8'. Heatflow in Btu/h is cfmxdT, approximately, so Btu/h = 2417 = 
16.6x4xsqrt(8)dT^(3/2) makes dT = 12.9^(2/3) = 5.5 F. With 2 1 ft^2 vents to 
the room and an 8' height difference, 770 = 16.6x1xsqrt(8)dT^(3/2) makes dT 
= 16.4^(2/3) = 6.4 F.

 Given these effects, the daily heat store has a smaller temp swing, so it 
needs more capacitance... If  T = 66.4+14.5K/C = 
148.5-(66.4-148.5)e^(-384/C), ie 14.5K/C = 82.1(1-e^(-384/C), ie C = 
177/((1-e^(-384/C)). Plugging in C = 200 on the right makes C = 207 on the 
left. Repeating makes C = 210.0, then 210.9 Btu/F, eg 42 hollow concrete 
blocks cooling from 135.4 to 66.4 F. Or fewer blocks, since 42 would have 
more surface than 31, with smaller charging and discharging temp drops.

 Heating the room directly for 6 hours with 90.2 F air from the sunspace 
instead of air from the daily store would also reduce the number of blocks 
required in the daily store. And a thermosyphoning air-air heat exchanger 
could help. If 31.5 F air falls down through the corrugations of 64 ft^2 of 
8 mm Coroplast and 10 cfm of 65 F air rises up between the Coroplast faces 
with possible condensation and freezing, NTU = AU/Cmin = 
128ft^2x0.75Btu/h-F-ft^2/10Btu/h-F =  9.6, and E = 9.6/10.6 = 0.906, and Tco 
=  31.5+E(65-31.5) = 61.9 F, and Tho = 65-E(65-31.5) = 34.6 F, with 46.7 and 
49.8 F average air temps in the hot and cold chimneys and a 3.1 F difference 
between the averages. Not much, for thermosyphoning air. So maybe this heat 
exchanger needs 2 small DC fans that only run when the indoor RH or CO2 
concentration exceed 60% or 1000 ppm, with an Arduino controller that can 
also shut off the cold air fan if condensation in the outgoing air passage 
begins to freeze, eg http://www.youtube.com/watch?v=wexdNx_StRc

 Without these refinements, the room needs 5x14.5K = 72.5K Btu for 5 cloudy 
days in a row, with an approximate solar heating fraction of 1-2^-5 = 0.97. 
This could come from 72.5K/(135-70) = 1108 Btu/F of cloudy-day mass with 
lots of surface cooling from 135 to 70F, eg 2 vertically-stacked 450 Btu/F 
steel 55 gallon drums with plastic film liners surrounded by a layer of 
rocks inside a 3'-diameter x 6' tall cylindrical welded-wire ag fence 
gabion.

If the drums are well-insulated with strawbales, trickle-charging them hot 
won't require much daily overflow hot air from the daily air heater, which 
could first heat the daily store to 135 with thermosyphoning air and a 
passive one-way plastic film damper, then heat the cloudy store behind the 
daily store with more thermosyphoning air and another film damper.

For more fun stuff, see http://pineassociatesltd.blogspot.com/

Nick