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combining a solar water heater and attic warmstore
18 oct 1996
                      ........ r20                    .     r20       ---
                       6 drums                  . drum drum .
polyethylene       .g b  b                 .    .collar beam.    .---
film sunspace --> . g r20                  .    . solar air .    .    11'
                 .  g     polycarbonate    .    .   heater  .    . 8'
   pond         .   g <-- single glazing   .    .           .    . r
            .....................      ...................................
                                           |          12'        |

during the day, the air from the drum space would slide down the outside of
the south wall, south through a hole at the bottom of an air heater, and up
through a 6" air gap between the dark north wall of the sunspace and a single
layer of polycarbonate glazing, and back into the drumspace. at night, the
air heater would get cold, and the heat would stay trapped in the attic.
steve baer and norman saunders, pe, have been building systems like this 
for the last 30 years.
suppose there are 6 plastic 55 gallon drums in the attic. 

a few numerical questions: 
1. what would the average stagnation drumwater temperature be in december?  

with a frozen pond to the south, about 1300 btu/ft^2/day would fall on the
sunspace glazing, of which about 1050x8'x12' = 100k btu would enter the air
heater, on an average december day in philadelphia, with a 24 hour average
outdoor temperature of 36 f. so if the drumwater has temp t, and the sunspace
air is 70 f, and the solar energy that flows into the drumwater equals the
water heat loss, on an average day, we have  

100k btu =   6 hr(t-70)8x12/r1   from the air heater during the day
           +24 hr(t-36)10x12/r20 from the roof, 24 hours a day
         = 576t - 40320 + 144t - 5184 = 720t - 45504, or
t = (100k+45.5k)/720 = 202 f.

ok, altho radiation losses would make this temperature closer to 130 f,
without a selective surface, and this doesn't consider house heating or
hot water consumption, yet. 

2. if the average drumwater temp were 110 f, what is the minimum average
r-value needed for the cabin walls, so the drums can keep the cabin at 68 f 
for 8 hours per day for 5 days without sun? 

the drums store about 6(110-80)55x8 = 80k btu, and the cabin walls have an
area of about 350 ft^2, so ignoring the heat loss through the roof, we have
80k = 5x8x(68-36)350/r = 450k/r, so r = 450k/80k = 5.6. suppose we use an
r10 average. r20 for the walls themselves, and window losses that make the
total no less than r10. then the cabin needs 8hr(68-36)350ft^2/r10 = 9k btu
to stay warm on an average 8 hour day.

3. would the sunspace really be 70 f, during the day?

the heat loss from the air heater would be about (110-70)8x12/r1 = 3840
btu/hour, and the 12'x8'x8' curved sunspace has about 400 ft^2 of r1 glazing,
and the avg daytime high temp in phila in december is 43 f, for an approximate
sunspace temp tss = 43 + 3840xr1/400 = 52.6 f. we could put less energy into
the water to make the average sunspace temp 70 f, so we can heat the cabin
with sunspace air, by impeding the air heater airflow so the air temp goes up 
and we lose 12k vs 4k btu/hr. then tss = 43 + 12k/400 = 73 f. better. but
the sunspace loss would increase to 6hrx12k = 72k btu/day. we could double
glaze the sunspace to cut this in half... or just remove 9k btu/day of 110 f
air from the air heater, ie 40 cfm, and leave the sunspace cooler. it seems
to me a human or plant might be fairly warm in a 53 f room full of still air,
if exposed to direct sun.

or we could use a larger sunspace. we might add on another curved poly film
12'x 8' sunspace to the east, with a poly film covered strawbale wall to the
north. this would collect an additional 100k btu of sun per day, and increase
the sunspace glazing area to about 700 ft^2. how much leftover energy will
there be if we keep the whole sunspace at 70 f on an average day? 

200k - 6(70-43)700ft^2/r1 = 87k btu/day.

4. how many 4 minute 2 gpm showers can we take on an average day, while
heating 32 f water to 110 f? 

each shower takes 4x2x8(110-32) = 5k btu, so with n showers/day, we have

87k btu =  24 hr(110-36)10x12/r20  from the roof, 24 hours a day
         + 9k btu                  for space heating
	 + 5k n btu                for water heating
         = 10.6k + 9k + 5k n, so n = (87k - 19.6k)/5k = 13.

did i do this arithmetic correctly?

this would work better with 55 f well water. 

off to do more hammering, on this cool and blustery day.


it's a snap to save energy in this country. as soon as more people become
involved in the basic math of heat transfer and get a gut-level, as well as
intellectual, grasp on how a house works, solution after solution will appear.

                                          tom smith, 1980

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