seasonal house heating SOLAR.KnowledgePublications.com

seasonal house heating
4 aug 2000
yes, it seems the heat store would work better under the house, and
we should count whatever winter sun can be collected, but why would
we be interested in annual storage if there's any winter sun at all?
maybe collectors cost more than storage, and we can use more storage
volume and less collection area if we store heat in warmer months
with more sun and less house heating load...

consider a simplified case with these assumptions: 130 f max and 80 f
min water temps, no heat loss during storage (under the house) or during
(concentrated) solar collection, and a house with a 200 btu/h-f thermal
conductance and no windows or internal heat gain.

where i live, it would lose 30dx24h(70-36)200 = 4986k btu in december,
the worst-case month for solar heating, when the average outdoor temp
is about 36 f. nrel says a 1-axis ew concentrator can gather at most
i = 2.0 kwh/m^2-day of sun with 100% efficiency, or 30dx0.9^2x317ia
= 6933ia btu/ft^2 per month using attic glazing with 90% transmission
and a 90% reflector and 90% target glazing. if the solar energy that
flows into the house equals the heat energy that flows out in december,
4986k = 6933(2.0)a, so we need a = 353 ft^2 of collectors, and thermal
storage for about 5 cloudy days, ie (130-80)8g = 5dx24h(70-36)200,
eg g = 2,040 gallons of water.

over november (average temp 46 f with 2.3 kwh/m^2-day of beam sun)
and december, the house needs 30dx24h(70-46)200 + 4896k = 8352k btu,
so we need 6933(4.3)a = 8352k, so a = 280 ft^2. if the store begins
at 130 f in december, we need 6933(2.0)280+(130-80)8g = 4896k, so
g = 2,528 gallons of water. 

over october (56 f and 2.9 kwh/m^2-day), november and december, we need
30dx24h(70-56)200 + 8352k = 10368k btu = 6933(7.2)a, so a = 208 ft^2
and 6933(4.3)208+(130-80)8g = 8352k, so g = 5,400 gallons.

including september (68 f and 3.0 kwh), the house needs no heat,
but we can store 10368k btu = 6933(10.2)a, so a = 147 ft^2, and
6933(7.2)147+(130-80)8g = 10368k, so g = 7,575 gallons.

including august (3.3 kwh) makes 10368k btu = 6933(13.5)a,
so a = 111 ft^2 and 6933(10.2)111 + (130-80)8g = 10368k, 
and g = 6,296 gallons, theoretically speaking.

including july (3.4 kwh) makes 10368k btu = 6933(16.9)a,
so a = 88 ft^2 and 6933(13.5)88 + (130-80)8g = 10368k, 
and g = 5,329 gallons...

and june (3.5 kwh) makes 10368k btu = 6933(20.4)a, so
a = 73 ft^2 and 6933(16.9)73 + (130-80)8g = 10368k, 
and g = 4,537 gallons...

nick