re: solar options SOLAR.KnowledgePublications.com

re: solar options
30 jan 2003
>over 5 cloudy 30 f days, the house needs 5x24(65-30)86 = 361k btu, not far
>from the 8k(100-70) = 240k we could store if the rest of the ceiling cooled
>from 100 to 70 f over 5 days. with a lower rate of heatflow, it can operate
>closer to the upper and lower temp limits than the overnight store. we can
>also put more water in those ducts, with a lower surface-to-volume ratio
>and a longer time constant. and having stratified storage, we might stratify
>collection as well by adding another mesh to the sunspace to collect warmer
>cloudy-day air from the back and cooler overnight air from the front.
>how would all this work, arithmetically? 

an unstratified system looks simpler. the house needs 3010 btu/h or 72.2k btu
on an average day. with r10 insulation above, 1024 ft^2 of basement ceiling
mass can supply this with no airflow at 65+10/1024x3010 = 94.4 f. with airflow
on both sides, it can supply 3010 btu/h at 65+3010/(1.5x2x1024) = 66 f. over
5 cloudy days, (94.4-66)c = 361k makes c = 12,711 btu/f, ie 12.4 psf, or 2.3"
of water under the ceiling. 

warming the water underneath for 6 hours per day with t (f) sunspace air,
6h(t-94.4)1024x1.5 = 24x3010 makes t = 102.2 f. on an average jan day in
phila, a square foot of south glazing with 90% solar transmission would
gain 900 btu. at 102 f, it might lose about 6h(102-30)1ft^2/r1 = 432 for
a net 468 btu gain and a 72.2k/468 = 154 ft^2 wall warmer. with some fin-
tube pipe and an extra water preheater and a greywater heat exchanger, an
8'x24' wall-warmer might heat the house and most of the water needed for
showers. counting indoor electrical use and the outdoor air and the south
wall of the house being warmer during the day would help.

nick