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a 16' cube 7 mar 2001 nrel says 1000 btu/ft^2 of sun falls on a south wall and 425 falls on east and west walls in phila, pa on an average 30.4 f january day, when a 1-axis ew concentrator can collect 729 btu/ft^2 over 3 hours. with a 90% reflector and 2 layers of r1 clear polycarbonate glazing with 90% solar transmission, we might collect 0.9x0.9x0.9x729x16'x16' = 136k btu/day with a 16' cube with a parabolic reflector inside. with a focal line at x=f=4' south of the base of the north wall and equation y^2=4fx=16x, the reflector would bounce dawn sun down onto a 4'x16' horizontal target at the base of the north wall. the focus moves closer to the north wall during the day until the reflector begins to shade the base of the wall at a 45 degree sun elevation (at noon on april 4 and september 28, at 40 n latitude.) we would still have lots of hot water in summertime. with a 4' strip near the north wall dedicated to solar collection and 12'x16' = 192 ft^2 of clear floorspace, the cube could serve another purpose as well, eg a dramatic greenhouse or workplace. the target might have a 0.020" clear polycarbonate cover and a plywood bottom with an epdm rubber liner and 2x3 sides. it might thermosyphon hot water through a 4'x8'x16' 4,096 gallon 16 ton tank above, supported by 13 2x4 studs on 2' centers along the north wall and the north 4' of the east and west walls and the reflector, with 9 more 2x4 studs on 4' centers around the east and west and south walls, and a 2x4 "foundation" (80' of 2x4s laid flat on the ground over poly film.) we need 64 ft^2 of polycarbonate for the target plus 256 ft^2 for the south wall plus 4/3x16'x16' = 341 ft^2 for the east and west wall area under the reflector. that's a total of 661 ft^2, ie $992 at $1.50/ft^2. rimol greenhouse systems (877-rimol-gh) sells it for $290 plus $10 ups shipping per 49"x50' roll. we might use 512 ft^2 of osb for the roof and north wall and 448 ft^2 for the reflector and target bottom and 384 ft^2 for the tank sides and bottom and 172 ft^2 for the east and west walls above the reflector. that's 1,516 ft^2, or $284 at $6 per 4x8 sheet. gluing 384 ft^2 of nielsen's reflective mylar ($0.09/ft^2 in 4'-wide rolls from http://www.snomo.com/mylar.html) to the underside of the reflector adds $35 to the materials cost. we could line the tank with a single folded $192 20'x32' piece of epdm rubber, and cover the roof with another $77 16'x16' piece, with a $20 4'x16' piece for the target liner. the tank might have 256 ft^2 of 3" polyisocyanurate board (eg $10 atlas energy-shield "r7.2" double-foil 4'x8'x1" boards) for another $240, with lots of loose fill (eg bags of leaves) under the bottom and on the south side, which would make rc = 32768btu/fxr21.6/256ft^2 = 2765 hours or 115 days. we might approximate the parabolic shape with 9x24' = 216 ft of 2x4s on 2' centers with a dozen sawkerfs in each bow. the roof might need 144 ft more 2x4s, with another 352 ft for the sides and 80 ft for the foundation. that's 792 ft, $198 at 25 cents per foot. the total of the materials costs above is $2038. if the tank water is 150 f and we want to keep the target temp 160 f max and 46.7k btu/h of heat power falls on the target and we ignore the heat lost to the warm air in the cube, we need a thermosyphoning water flow q ft^3/s, where 3600s/hx64btu/f-ft^3q(10f) = 46.7k, so q = 0.0203 ft^3/s, about 10 gpm. water weighs about 63.74 - 0.0158t lb/ft^3, with t in degrees f. the difference in density caused by the average temperature difference between up and down pipes causes a pressure difference proportional to the height of the water column, which makes water flow through the resistance of the pipe loop. with 16' of height and a 10 f temperature difference, dp = 2.53dt lb/ft^2. here's a formula for laminar flow in a pipe with radius r and length l in feet and pressure difference dp: q = pir^4dp/8mul ft^3/s. mu is the viscosity, about 8.17x10^-6 lb-s/ft^2 for 150 f water. we might figure the water flows through 24' of pipe, 16' up and 8' down, which makes q = 0.0203 = pir^4(2.53)/(8x8.17x10^-6x24'), so r = 0.0447' or 0.54", eg a 1.5" pipe (or larger, with some fitting head losses.) if 0.9x1000x16'x16' = 230k btu of sun enters the cube from the south wall on an average day and 0.9x425x4/3x16'x16' = 131k enters the east and west wall area beneath the reflector, a total of 361k btu enters the cube, including about 16'x16'x531 = 136k btu of beam sun. removing this from the daily total, say 225k btu enters over 6 hours at 37.5k btu/h. the cube has about 600 ft^2 of surface below the reflector. with loose insulation above, the conductance to outdoors is about 600 ft^2/r1. if 136k/3 = 45.3k btu/h hits the target for 3 hours when it has an average temp t and 1.5x4'x16' = 96 btu/h-f of conductance to the air in the cube which has temp tc, we have something like this (in courier font): 30.4 f ----www-----------------tc -------www-----tc 1/600 | | 1/600 | --- 93 f ----- | equivalent to - |------| --> |-------| | ----- | _ 37.5k/h < < 1/96 . . . . . . . . . < ----- | |------| --> |-------*-------t = 160 f ----- | 45.3k/h ----- 32,768 btu/f ----- | _ removing the 1/96 resistor makes the upper thevenin equivalent temp tc = 93 f. using that equivalent and replacing the 1/96 resistor, (160-93)/(1/600+1/96) = 5,545 btu/h flows from target to cube air, making tc = 93+5545/600 = 102 f. too hot, so we need to vent in full sun. if tc = 70 f, (160-70)x96 = 8,640 btu/h flows from target to cube air and the sun provides (93-70)x600 = 13,800 btu/h. the total is 22,400. with a 40 f indoor-outdoor temp difference, we need about 22.4k/40 = 560 cfm of outdoor airflow to dump the heat. using an empirical chimney formula, 560 = 16.6asqrt(16'x40f), so we need a = 1.33 ft^2 of vent area at the top and bottom. we could probably use more than that in summertime, even with noontime shading. on an average day, with 70 f air inside, we might collect beam sun in 150 f water with an efficiency of 100(45.3k-8640)/45.3k = 81%, ie the cube might collect 0.81x136k = 110k btu/day and lose about 24h(150-30)256ft^2/r21.6 = 34k btu (cooling 34kbtu/32768btu/f = 1 f), so it could provide 110k-34k = 76k btu per day of hot water in january at $2038/(110k/3.41) = 6.3 cents per peak watt. at 150 f, with a minimum usable temperature of 120 f, the tank would store about (150-120)32768 = 983k btu. with a 135 f average temp, it would lose about 24h(135-30)256ft^2/r21.6 = 30k btu to the outdoors on a cloudy day, so it could provide 76k btu of hot water for about 983k/(30k+76k) = 9 cloudy days in a row. adding bubblewalls to the cube would allow the tank to provide cloudy-day heat by circulating hot water through the target using a small pump and a thermostat. we might increase the thermal efficiency and make electricity at the same time by laying 4'x16' of pvs on the target floor and collecting 130 f water at about 3 suns. nick |