```re: 100% solar heated house for cold climate 2 dec 2004 >the pipes under the ceiling might only have enough surface and thermal >capacitance to collect and store heat for an average winter day, with >an average amount of sun, and no pumping. this would help make the air >heaters more efficient, if the pipe water cooled to 80 f or so by dawn >on an average winter day. or cooler. >we might only pump water through the pipes on clearer and milder or >cloudy days. the pump needn't use much power if the up and down pipes >are always full, with the end of the down pipe underwater. eric hawkins tried this and found the collector stayed primed for days on end, without an expansion tank above it. >how many pipes of what diameter and how big a crawlspace tank would we >need for that 24x24x8' building in albany, with r32 walls? a daily heat store that starts every day cold and a larger and hotter cloudy day store that's just trickle charged can be efficient... the air heater size and house insulation determine the constant air temp over 6 hours on an average december day and the amount of heat needed for overnight heat storage, in one crude model. if that 24x24x8' building with r32 walls and ceiling and floor were 70 f for 8 hours per day and 40 at night, it would need about (70-47.4)576/r42 + (70-30.7)1728/r32 = 2432 btu/h during the day and (40-47.4)576/r42 + (40-26.5)1728/r32 = 628 at night, a total of 29.5k btu/day. if 0.9x700x8x24 = 121k btu enters an r1 south wall cover over 6 hours on an average day and 121k = 6h(t-30.7)192/r1 + 29.5k, we might get t = 30.7+(121k-29.5k)/6/192 = 110 f air out of the heater for 6 hours... a 4'x16' plywood or wire mesh shelf under the ridge with a poly film duct on top might keep the building 40 f with duct water at 70+628/(64x2x1.5) = 43 f. how much water? we need to store about 2x2432+16x628 = 14.9k btu/h of overnight heat. if the shelf reached 110 f by dusk and 43 by dawn, we would only need 14.9k/(110-43) = 222 pounds of water, less than an inch deep. but rc = c/g = c/(64x2x1.5) = 0.00521c makes tmax = 110+(43-tmax)e^-6/(rc) = 110+(43-tmax)e^-1152/c and c(tmax-43) = 14.9k, ie tmax = 43+14.9k/c, so 43+14.9k/c = 110+(43-(43+14.9k/c)e^-1152/c, ie c = 222(1+e^-1152/c), plugging in c = 225 on the right makes c = 223.327 on the left. repeating makes c = 223.277, then 223.275, so it looks like an inch of water will do. the building needs 8hx2432+16hx628 = 29.5k btu on a 26.5 f cloudy day. if 5x29.5k = c(109.6-43), c = 2215 pounds of water, eg a 4'x8'x1' deep tank under the floor. nick ```  I got ALL of these 85 Solar Panels for FREE and so can you.  Its in our Ebook Ready for DOWNLOAD NOW. 