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.