Subject: A thermal design case study for a warmstore shed

From: [email-masked] (Nick Pine)
Date: 7 Jun 1996 11:02:48 -0400

Here's a guided meander of a thermal design for a 100% solar-heated 8 x 12'
shed in the Philadelphia area, one that might sit on a lawn and be used for
8 hours a day in the winter, for a workshop, studio, playhouse or retreat.
Since it would only be used during the day, perhaps not at all on some days,
we might put the thermal mass overhead and use a low-thermal-mass solar air
heater, an "igloo heat trap" on the south wall with no dampers or power 
required to warm the thermal mass in the attic, and a low-power attic fan
to bring down heat to the low-thermal-mass room when needed...

These days, many "100% solar houses" are very airtight and well-insulated,
and costly, with little glazing or thermal mass, and they are largely heated
by the internal electrical use and bodies of their occupants. Suppose we go
in some other directions, away from superinsulated houses, and try to see
how much thermal mass and glazing and how little insulation we can use in
this solar structure with no other form of heat...

Let's start by assuming the shed is 12' tall, with a ceiling height of 8'
and a 4' tall attic above that, with twenty 55 gallon drums full of water
in the attic, each 2' diameter x 3' tall:

                                                   12'
                                       -------------------------
                     view from above  |  D   D   D   D   D   D  |
                                      |                         |
So, this structure might have about   |  D   D  ceiling  D   D  |
10,000 pounds of water in the attic   |                         | 8'
with R10 attic insulation, eg 2" of   |  D   D    fan    D   D  |
foam, and R1 walls below, eg 3/8"     |                         |
T-111 wooden walls. The lower 8' of   |  D   D   D   D   D   D  |
the south wall would be R2, since      ------------------------- 
it would have glazing over the wood,
with an air gap.
                       8'                        96 ft^2
                ---------------        -------------------------
               |               |      |           R10?          |
               |      R10?     |      |                         |
               |    32 ft^2    |      |         48 ft^2         | 4'
               |               |      |                         |
               |---------------|      |-------------------------|
               ||              |      |                         |
               ||              |      |                         |
               ||              |      |                         |
               ||      R1?     |      |            R2?          | 8'
               ||    64 ft^2   |      |          96 ft^2        |
               ||              |      |                         |
   reflecting  ||              |      |                         |
      pool?    ||              |      |                         |
           -----------------------   --------------12'-------------
                   east view                   south view           

The first thing to check might be whether the solar glazing can keep the room
warm on an average December day, with some sun. The room has a total R1 wall
area of 96 + 64 x 2 = 224 ft^2, and another 96 ft^2/R2 for the south wall, for
a total thermal conductance of 224ft^2/R1 + 96ft^2/R2 = 272 Btu/hr-F, not
counting the room ceiling or attic walls or attic ceiling. On a typical 6 hour
December day in Philadelphia, when it's 36 F outside and the south wall gets
1,000 Btu/ft^2/day of sun, the room might lose 8hr(76-36)272 = 87K Btu of
heat, and the lower 8' x 12' of glazed wall might receive about 96 x 1,000
= 96K Btu. Looks good so far.

IF the water is 130 F after a long string of average December days, it can
supply about (130-80)20x500 = 500K of heat for the room below, enough for
about 5 days without sun, keeping the room warm for 8 hours a day with the
ceiling fan. That seems OK...

But it's marginal, thermally-speaking, and 5 tons is a lot of overhead water,
requiring lots of space, and strong ceiling joists and walls. Suppose we add
a little more insulation to the walls below.

Let's add 2" of Styrofoam to the outside of the north wall, with some latex
paint to protect it from the sun. (Styrofoam on the inside would lower the
thermal mass of the room, but it could be a fire hazard, and it would reduce
the floorspace.) Now the room has 96ft^2/R10 + 96ft^2/R2 + 2x64ft^2/R1 = 186
Btu/hr-F, so on an average winter day it needs 8 (76-36)186 = 60K Btu to stay
at 76 F for 8 hours, so we need 300K/25K = 12 drums in the attic to supply
heat for 5 days without sun, ie 3 tons vs 5 tons of water upstairs, at an
added cost of about $50 for the foam. Seems like a good tradeoff. Note that
no heat is lost from the room to the attic during the day or night, because
the attic is always warmer than the room, as designed. And the room will
gain, not lose heat through the south wall on an average winter day.

Let's check that. In full sun, if the drum temp is 130 F, and the south wall
is about 130 F, and the room temp is 70 F, about (130-70)96 ft^2/R1 = 5760
Btu/hr of heat will flow thru the south wall. Where will it go? Out the other
3 walls of the room, which have a thermal conductance of 96ft^2/R10 (for the
north wall) + 128ft^2/R1 (for the side walls) = 137.6 Btu/F, so we won't have
to open the windows and waste any solar heat if the outdoor temp is less than
Ta, where (70-Ta)137.6 = 5760, ie Ta = 70 - 5760/137.6 = 28 F. But the average
winter temp is 36 F, so we may have to waste a lot of the sun that comes 
through the glazing by venting, especially on warm sunny winter days. So
let's put R10 insulation inside the south wall, eg foil-faced foamboard,
tucked between the 2x4 wall studs on 2' centers. Then the heat that flows in
when the wall is 130 F and the room is 70 F will be 576 Btu/hr (more, if the
studs are exposed), and we can collect solar heat without wasting any by
venting the room, until the outdoor temp rises to 70-576/137.6 = 66 F, at
which point we hardly need heat anyway. Insulating the south wall will also
help with natural cooling in summertime.

With that R10 in the south wall, we have a thermal conductance for the room
of 2x96ft^2/R10 + 2x64ft^2/R1 = 147 Btu/hr-F, so on an average December day
the room would need 8(76-36)147 = 47K Btu to stay warm, so we'd need 236K/25K
= 10 drums in the attic to supply heat for 5 days with no sun, ie 5,000 vs
6,000 pounds of water upstairs, adding another $50 in foam. Still seems like
a good tradeoff... 

How big do the joists for the 8' x 12' ceiling have to be to support 5,000
pounds, ie 52 pounds per square foot? If the wood has a max fiber stress
rating in bending of f=1,000 psi and the joists are 16" on center, with an
L=8' span, the total load on a beam is W = 52 psf x 4/3' x 8' = 555 pounds.
The ceiling might be open, with one or two 12' boards running lengthwise to
keep the joists from twisting under the weight of the drums. Bending moment
M=WxLx12/8=6,660 in-lb, section modulus S=M/f=6.6 in^3, and a joist width
b=1.5" would require a joist depth d=sqrt(6S/b)=5.13". So we might use 2x6s. 

Let's add 2 more 4' x 8' sheets of foam to the outside of the sidewalls, so
the total thermal conductance of the room becomes 2x96ft^2/R10 + 2x32ft^2/R1
+ 2x32ft^2/R10 = 89.6 Btu/hr-F. Now on an average day the room needs about
8(76-36)90 = 29K Btu to stay warm, with no sun, so we need about 150K/25K = 6
drums in the attic to supply heat for 5 days without sun, ie 3K vs 5K pounds
of water upstairs, spending another $32 on foamboard, which still seems like
a good tradeoff... The room's air infiltration is becoming a larger part of
the thermal conductance now: assuming the attic is absolutely airtight (we
don't want any air leaks there), the 768 ft^2 of room below with 1 Air Change
per Hour would have an effective thermal conductance of about 768/60 cfm =
13 Btu/F. This wasn't very important before, but it has now grown to 15% of
the total thermal conductance. 

At this point, we might remove some of the air heater glazing, and add more
insulation and air-infiltration-proofing, and begin to turn this into a
superinsulated building, but let's put in 4' of south clerestory windows 
instead, for heat and daylight (and drama :-), with some interior insulating
shutters, hinged at the top edge, that fold up and back to the north when the
structure is occupied, and hang down vertically to cover the windows when it
is not. So the 4'x12' of 2" Styrofoam at the top of the south wall becomes
4'x12' of R10 foil-faced foamboard, hinged at the top edge. Single layer
polycarbonate glazing costs about $1/ft^2, lasts many years, and comes in
sheets 49" wide and 0.020" thick, from Replex Plastics at (800) 726-5151.

We now have 144 ft^2 of solar glazing, but only 6 55 gallon drums, each one
having only 25 ft^2 of surface, so the glazing/thermal mass area ratio is
6x25/144=1.04. This is not a good passive system, because with 300x144
=43,200 Btu/hr or about 13 KW of full sun, with an R-value of 2/3 for the
still air film at a drum surface, the air-water temperature difference will
be about 43,200x2/3/150 = 192 F, so the attic will be very warm, and a lot
of the heat in the attic will flow thru the R10 attic insulation, wasting
a lot of the sun that comes through the glazing.

If the thermal mass had 10 times the glazing area, the temp difference across
the thermal mass surface would only be 19 F, which is better. We could switch
to smaller containers of water, eg 50 8 gallon plastic paint pails with lids,
which are about 12" in diameter and 16" tall, for a glazing/thermal mass area
ratio of 2:1, or 720 2 liter soda bottles for an area ratio of 5:1. Another
way to reduce the thermal mass surface resistance is to run the fan to help
collect sun when nobody's in the room. If this could be done without moving
heat down to the room, it might be a good idea. With a 5 mph airflow, the
thermal conductance of the plastic buckets would increase from about 3/2 to
3/2 + 5/5 = 2.5 Btu/hr, according to Fig 1 on page 22.1 of the 1993 ASHRAE
Handbook of Fundamentals. If the buckets had a rougher surface, like stucco,
their conductance would rise to 2 + 5/2 = 4.5. Using roughened paint pails
(how to do that?) might allow saving materials, if they were suspended from
Dacron ropes going from stud to stud overhead, with no ceiling joists. But
suppose we sit them on an attic floor instead, say along the 12' north wall,
stacked 2 high, sitting in a 4' wide x 6" deep EPDM rubber trench full of
rainwater from the roof, perhaps with some 2 liter soda bottles in between.
The pail bottoms would take up 19.6 ft^2 of the 48 ft^2 trench bottom, leaving
28.4 ft^2 x 1/2 = 14.2 ft^3 or 113 gallons or 900 pounds of water in the
trench. It rains about 4" a month in Philadelphia. We might collect rainwater
>from  the flat roof, and run it down into this trench...

Now how about natural cooling in summertime? NREL's Solar Radiation Manual
for Buildings says July is the hottest month in Philadelphia, with an average
daytime high of 86 F, 24 hour average of 77 and average night low of 67, and
an average windspeed of 8 mph. The air contains an average of 0.0133 pounds
of water per pound of dry air, and NREL's metric Solar Radiation Data Manual
for Flat Plate and Concentrating Collectors says this corresponds to 70%
humidity. At what temperature? The ASHRAE HOF says 65 F air with 0.01327
pounds of water per pound of dry air has 100% humidity, which may be why we 
often see dew on the grass on a summer morning. Perhaps this condensation of
water from air limits the nightime temperature drop during the summer in
Philadelphia. When a pound of water condenses from air, it releases about
1,000 Btu of heat, and this heat helps keep the air warm. (Does the same thing
happen in the winter? Is there a sticky point, a temporary plateau, as the
air temperature tries to drop below 32 F, where it has to freeze a lot of
water on the way down? Weather scientist Richard James says no, but I wonder.
It only takes 144 Btu to freeze a pound of water, so freezing may limit the
lower air temp less than condensation, especially in places away from large
bodies of freezing water. 

Suppose we have 50 8 gallon plastic paint pails in the attic, each having
a surface area of about 6 ft^2, and about 4,100 pounds of water with a
thermal mass of about 4,100 Btu/F and a surface area of about 330 ft^2.
If this water is cool, it will cool the room below, since cold air sinks,
and the water will become warmer during a summer day. Suppose the water were
77 F at the end of a summer day, and we opened some vents to the outside
to allow 67 F night air to flow through the attic and cool the water, and
closed up the attic again the next day. If the night air moves slowly, the
rough-surfaced containers of water will have an air film conductance of
about 2 Btu/hr-F, so during the first hour exposed to night air they might
lose about (77-67)330ft^2x2=6,600 Btu of heat, about the same as a small
window air conditioner, but using no power :-) If we add a fan, and somehow 
raise the average air velocity past the pails to say, 4 mph, their thermal
conductance will increase to about 2 + 4/2, or 4, so in the first hour, they
might lose 12,000 Btu of heat while we use 100 Watts of electricity. The
water might cool about 3 F (12,000/4,100) in the first hour. 

The water trench under the pails might also lose heat to the night air by
evaporation (not much, on an average night in Philadelphia.) Page 664 of
Duffie and Beckman's _Solar Engineering of Thermal Processes_ (Wiley, 1991,
2nd edition) has this equation for relative heat loss from a pond by
evaporation:

     Qc,pa   0.46(Tp -Ta)  p
     ----- = ------------ ---
      Qe       Pwp - Pa   760

where Qc,pa is the loss of heat from the pond by convection, in Watts/m^2-K,
      Qe is the loss of heat from the pond by evaporation, in Watts/m^2-K,
      Tp and Ta are the Celsius temperatures of the water and air, and 
      Pwp and Pa are the vapor pressures of the pond surface and air, and 
      p is the barometric pressure, with all pressures in mm Hg.

The book says this ratio of convective to evaporative loss is essentially
independent of windspeed. We already have an estimate for the convective 
loss using a fan, 2 Btu/hr-F, but that's in US units. Let's convert this
nice metric equation above to US units, since most US HVAC engineers are
more familiar with US units. We can start by ignoring the absolute pressure,
since Philadelphia is at sea level, and divide by a factor of 1.8 to make
the ratio come out the same, since there are 1.8 F degrees in each C degree.
Since this is a heat loss ratio, we don't need to worry about the difference
between Btu/hr and Watts. So we have:

     Qc,pa   0.46(Tp -Ta)/1.8 
     ----- = ----------------
       Qe        Pwp - Pa  

where Tp and Ta are the Farenheit temperatures of the water and air, and 
      Pwp and Pa are the vapor pressures of the pond surface and air in mm Hg.

There are 25.4 mm in an inch, so 1 mm Hg is equivalent to 1/25.4" Hg in US
units. Multiplying the denominator of the equation by 25.4, we get a nice
decimal constant:

     Qc,pa   0.46(Tp -Ta)/1.8   0.01(Tp-Ta)
     ----- = ---------------- = -----------
       Qe     25.4(Pwp - Pa)      Pwp-Pa

where Qc,pa is the loss of heat from the pond by convection, in Btu/hr-F,
      Qe is the loss of heat from the pond by evaporation, in Btu/hr-F,
      Tp and Ta are Farenheit temperatures of water and air, and 
      Pwp and Pa are vapor pressures at pond surface and air in inches Hg.

Let's cover the flat roof with a single piece of EPDM rubber, with a 2" board
around the edge, under the rubber, to make a shallow roof pond, so we can pump
some water up from the attic pond through this roof pond on a summer night. 

Since this won't work very well on dewy Philadelphia nights, where we don't
need much cooling anyway, let's temporarily move this shed to Abilene, Texas,
where seldom is heard a discouraging word, and happy Christians spend long
hours listening to air conditioners purr (around indoor campfires? :-), while
avoiding Dark Gulf War Thoughts. July is the hottest month in Abilene, on 
average, with a nightime average low of 73 F and a 24 hour average temp of 84
and an average daytime high of 95, and the humidity ratio of the air is
0.0130 #w/#da. This is almost the same in August, ie hot. The average wind
is about 10 mph (at night, too?) Table 2 on page 6.4 of our trusty HOF says
that air with that humidity ratio is saturated at about 64.5 F, so we now
have a chance to use swamp cooling; 73 F air can hold 0.017575 #w/#da, at a
water vapor pressure of 0.81882" Hg, so that nightime air would have a
relative humidity of about 74%, and a water vapor pressure of 0.74x0.81882
= 0.606" Hg, if I'm doing this right. Water vapor in 77 F moist air has a
pressure of 0.93589" Hg, so on an average 73 F night, with a 10 mph wind,
the 77 F pond might lose (77-73)8'x12'(2+10/2) = 2,688 Btu/hr by convection,
and the ratio above would be 0.01(77-73)/(0.93589-0.606) = 0.0132, so the
pond would lose 2,688/0.0132 = 204K Btu/hr by evaporation, like 41 window
air conditioners...

Wow. Is that right? Can we evaporate 204 pounds of water in an hour, ie 0.45
gallons per minute, on an average night in August in Abilene? There isn't much
water in Texas, so we'd probably want to drain a roof pond during the day, but
Abilene has lots of electricity made from natural gas. A rational Texan with
an AC with a COP of 2, paying 10 cents/kWh of electricity might want to import
evaporative cooling water from a neighboring state at 10 cents/6800 Btu, ie
1.5 cents per pound or 12 cents per gallon, or send the other state gas or
electricity in exchange for water. A rational neighboring state might want to
trade water for gas, especially if they get most of the water back as rain.
Should the tall, dramatic and proud triangular tower at Abilene Christian
University have a pond on its flat roof, if Faith without Works is Dead? :-)

Baruch Givoni's _Passive and Low Energy Cooling of Buildings_ book says that
even a dry roof can be 10 C cooler than the surrounding air, on a calm clear
night, because of night sky radiation. A wind would warm up the roof...

                      *           *            *

Let's move the shed back to Philadelphia now, where pacifist Mennonites
listen to oil burners purr in their basements most of the year, and
people want new houses to look like they were built 100 years ago. 

Suppose we left the roof pond filled up during the day in the summer here,
and stretched a piece of greenhouse shadecloth over it, so the roof is
close to the 67.5 F wet bulb temperature 24 hours a day...

If we left the water in the roof pond in winter, with an R1 layer of bubble
pack on top, it would receive 530 Btu/ft^2/day of sun in December, which would
raise the average temperature of the roof water from 36 F to T, which would
be higher because of the sun, as well as the heat the roof pond receives from
the 130 F pails thru R10 insulation: 24(T-36) = 24(130-T)/R10+530 ==>
T = 1706/26.4 = 64.6 F.

The roof pond would also limit the lower roof temperature for a time, in very
cold weather. Once the 1024 pounds of water on the roof begins to freeze,
it will take 147K Btu to freeze it solid, eg 36 hours at -10 F.

Melting 3' of snow equivalent to 4" of ice off the roof by pumping up some
water from the attic would take about 295K Btu, vs. the available heat from
the attic: 4100 Btu/F x (130-32) = 402K. Once the attic reached 32 F, it
would require another 600K Btu to freeze it solid. If the outdoor temp were
- 10 F, this would happen in t hours, where

  t(32-(-10))/256ft^2/R10 = 600K, ie t = 558 hours, or 23 days.

My house has an old hand-dug well, with water 9' below the ground. Using this
55 F water to melt snow off the roof or raise the roof temp in colder times
is better than wasting our solar store for that purpose. Melting 3' of snow
requires moving up 295K/(55-32) = 12.8K pounds of water, ie 1600 gallons, eg
3.3 gallons per minute for 8 hours. Keeping the roof at 32 F when it's -10 F
outside without freezing the water requires about 4K Btu/hour, ie 0.36 gpm.
Lifting that water 12' requires less than 1 watt of power, while saving over
1 kW of heat, which seems like an energy bargain.

This roof seems consistent with the spirit of section 1610 of the 1993 BOCA
code, which specs a snow load rating of only 12 psf for continuously-heated
flat greenhouse roofs with R-values of less than 2, where I live. So the roof
might have to support 12 psf of snow, 11 psf of water, and 2 pounds of OSB,
etc, ie 25 psf. If the roof joist wood has a max fiber stress in bending of
f=1,000 psi and the joists are 16" on center, with an L=8' span, the beam
load is W = 25 psf x 4/3' x 8' = 267 pounds, M=WxLx12/8=3200 in-lb, section
modulus S=M/f=3.2 in^3, and joist width b=1.5" requires depth d=sqrt(6S/b)
=3.57". So we might use 2x4s for the roof. 

Another way to make the roof warmer is to add a peaked roof, with a more or
less vertical transparent south roof and a straight or parabolic north roof,
reflective on the underside, eg some foil-faced foamboard with kerfs on the
concave side, screwed to some kerfed and curved north roof rafters, to
concentrate another 1000x4x12'= 48K Btu/day of sun down into a 2' wide
shallow water trench along the north edge. Perhaps half of this heat would
end up in the water, with the remainder heating the space under the roof.

A 10' long x 4" diameter PVC pipe might make a nice water heater, tucked up
under the roof peak, with insulation above, but none below.

An underground warmstore might be better here, so the shed is only about 12'
tall at the peak. The attic would only serve as an active collector, with 9
55 gallon drums in a 8' x 8' x 4' deep hole under the floor, lined with a
single 16 x 16' piece of EPDM rubber, folded up like a Chinese restaurant
take-out box, then 1' of earth, then another 12' x 12' layer of rubber, to
make a 6' x 6' x 3' deep tub, which might have the 9 drums standing up in a
3x3 array, with 6" of water in the tub, and a 6" layer (4 ft^3) of pieces of
closed-cell foam above that. One might heat the drums by pumping up water
through the attic trench during the day, and heat the room by opening a
small damper in the 8x8' insulated frame floor to let some warm air out.
This might be a good way to make a 24-hour vs. 8-hour heated shed, since
the storage water temp would be higher, because of the concentrator.

So where are we, in terms of the steady state energy budget for this shed? 
Suppose it has a flat roof, with a roof pond. When last we looked, the 
room's thermal conductance was 2x96ft^2/R10 + 2x32ft^2/R1 + 2x32ft^2/R10
= 89.6 Btu/hr-F. Hmmm. Over half of that is the bare wood 4x8' areas of the
sidewalls. Suppose we insulate them, so we have an overall room conductance
of 320 ft^2/R10 = 32. On an average December day, the room will need about
8(76-36)32 = 10K Btu to stay warm, and our 4,100 pounds of water in the attic
will keep it warm for about 4100(130-80)/10K = 20 days without sun, if it
starts out at 130 F.

Let's check the steady-state water temperature. On an average December day,
the south wall admits about 144K Btu of heat when the shed is occupied, with
the clerestory thermal shutters open, and the 3 non-south walls need about
7K Btu to keep warm. If the attic water temp is T, 8(T-36)48ft^2/R1 Btu flow
through the windows, the south wall glazing loses about 6(T-36)96ft^2/R1 of
heat to the outside world, 24(T-36)160ft^2/R10 flows out through the attic
walls, and 24(T-64.6)96ft^2/R10 flows out through the roof. So, if the energy
that flows into the structure during an average day equals the energy that
flows out of the structure,

  144K = 7K + 8(T-36)48 + 6(T-36)96 + 24(T-36)16 + 24(T-64.6)9.6
       = 7K + 384T      + 576T      + 384T       + 230.4T
            - 13824     - 20736     - 13824      - 14884
       = 1574.4T - 56268, so

     T = (144K + 56K)/1574 = 127 F.

Let's try double-glazing the windows:

  144K = 7K + 8(T-36)24 + 6(T-36)96 + 24(T-36)16 + 24(T-64.6)9.6
       = 7K + 192T      + 576T      + 384T       + 230.4T
            - 6912      - 20736     - 13824      - 14884
       = 1382.4T - 49356, so

     T = (144K + 49K)/1382 = 140 F. That's better.

This looks like an R10 100% solar house :-) Ignoring air infiltration. A
bigger house should have better thermal performance, with the same R10 walls.
We probably can't make the water much hotter without a selective surface.

This 1000 Btu/ft^2/day was measured with a 20% bare ground reflectivity,
I think. A shallow reflecting pool on the south side of the shed would keep
vegetation from blocking the sun and bounce about 60% of the sun onto the 
shed when frozen, vs 20%, so we'd have 1.6/1.2 = a third more sun. 

In that case, T = (4/3x144K + 49K)/1382 = 174 F. Seems like a good idea...

For how many average sunless days can the shed stay warm? The attic has
a thermal conductance of 256ft^2/R10 = 25.6 with the shutters closed,
and the room need 10K Btu/day to keep it warm for 8 hours, and
when 1 pound of water cools 1 F, it releases 1 Btu, so...

Day  Twater  Qattic   Qhouse   Qtotal    DT = Utotal/4100

0    130 F   58K Btu  10K Btu  68K Btu   16.5 F
1    113.5   48K      10K      58K       14
2    99.5    39K      10K      49K       ...
3    87.5    32K      10K      42K
4    77.3    25K      10K      35K
5    68.7    ...

Not too bad, but let's try changing this around a bit, by hinging the
shutters along the upper edge of the 4' x 4' x 12' attic warmstore, instead
of along the upper edge of the windows. Then the room conductance becomes
about 32 + 48ft^2/R2 = 56, and it needs 8(66-36)56 = 13,440 Btu to stay at
66 F every day, and the attic conductance becomes 17.6 Btu/hr-F, so...

Day  Twater  Qattic   Qhouse     Qtotal     DT = Utotal/4100

0    130 F   40K Btu  13.4K Btu  53.4K Btu  13 F
1    117     34K      13.4K      48K        11.6
2    105.4   29K      13.4K      43K        ...
3    95      25K      13.4K      38K
4    85.6    21K      13.4K      34K
5    77.2    17K      13.4K      31K
6    69.7    ...

That's better. We could improve it more by pumping 0.2 gpm of 55 F water
through the roof pond to keep the roof at 46 F when the sun isn't shining. 
Or we could use R20 or R30 walls. 

In the summer, with some shadecloth over the south wall, the room might 
start out in the morning at 66 F, and the 4 12' walls would have a thermal
conductance of 480 ft^2/R10 = 48 Btu/hr-F. If the daytime high of 86 F
lasted 8 hours, the 4100 lb of water might absorb 8(86-66)48 = 7680 Btu, 
which would raise the water temp about 2 F. If it were 86 F for 24 hours,
the water temp would rise to 72. After a few days of this, we might want
to pump up some of that 55 F groundwater into the attic pond. Keeping the
room at 71 F would require (86-71)48 = 720 Btu/hour, ie 720/(71-55) = 45
pounds of water per hour, or about 0.1 gpm. 

The lower part of the shed south wall might look like this:

             view from inside               east view

                                               R10
      |      foil-faced foam      |        |   wall   |
      |                           |        |          |
      |---------------------------|  ---   |----------|
      |  air  |            | air  |        |/    .    <== cooler attic air  
      |  out  |            | out  |        |     .    .       
      |---------------------------|   |    |     /------------ceiling------
      |       |     air    |      |        |     s    warmer air to attic ==>
      |       |  to attic  |      |   |    |     s
      |        ------------       |        |     s     |
      |        vent area Av       |   |    |     s     |
      |                           |        |     s     |
      |                           |   |    |     s     |
      |                           |        |     s     |
      |                           |   |    |     s     |
      |                           |        |     s     | R10 wall
      |                           |   |    |     s     |
      |                           |        |     s     |
      |       glazed area A       |   H    |<-G->s<-G->|
      |                           |        |     s     |      room
      |                           |   |    |     s     |
      |                           |        |     s     |
      |                           |   |    |     s     |
      |                           |        |     s     |
      |                           |   |    |     s     |
      |                           |        |     s     |
      |                           |   |    |     s     |
      |                           |        |           |
      |                           |   |    |\        / |
       ---------------------------   ---    ---------------

This is a simple passive air heating system a la Norman Saunders, PE, or
Steve Baer, in which the sun shines through an outer layer of glazing, onto
a porous absorber surface, with an air gap of dimension G between the glazing
and the absorber, and another between the absorber and the wall. Cooler air
>from  the attic warmstore would be persuaded to flow  down  between the cold
glazing and the absorber, then  up  through the air gap between the absorber
and the wall. This is thermally more efficient than the typical system in
which the solar heated air is in contact with the cold outer glazing.

This R10 wall would have 2" of foil-faced foamboard behind a piece of dark-
painted plywood. No backdraft dampers are needed, because at night the cold
air forms a stagnant pool that just stays in the air heater pocket, since
cold air sinks. 

The width of gap G determines the airflow resistance, in part. Steve Baer
says G might be about H/15, eg 6" for an 8' high wall. Vent area Av might be
about 5% of the glazed area, eg 12' long x 6" tall for a 12' long x 8' tall
glazed wall. The absorber might be one or more layers of black aluminum window
screen, with an absorbing area up to 5 times the window area, large enough 
that most of the heat from the absorber is transferred to the flowing air, vs
having a small-area hot absorber that loses a lot of heat to the glazing by
radiation. One measure of success is glazing and absorber temperatures, the
lower the better. One way to measure these temps is to use an Exergen D501
scanning thermometer (about $1000.)

One way to look at the airflow is to blow some smoke into the collector. The 
useful heat output of the collector is proportional to the product of the
air velocity and the temperature difference between the input and output air.
Steve suggests that 50% solar collection efficiency is a good target.

If the air temp next to the glazing were 136 F on a winter day when it was
36 F outside, and 300 Btu/ft^2/hr arrived inside this shallow sunspace, the  
heat loss through one square foot of R1 glazing would be (136-36)1ft^2/R1
= 100 Btu/hour, and the solar collection efficiency would be 67%.

So... one might build solar houses like this, ie houses that heat themselves
with the sun, and make domestic hot water, with no other form of heat.

Nick

Some useful rules of thumb:

Q (cfm) = 16.6 Av sqrt(H delta T), for a chimney, in which
               Av is the smaller vent area in square feet,
	       H is the chimney height in feet, and
	       delta T is the input/output air temp difference (F).

U = 0.174E-8 Ac (T+460)^4  Btu per hour is re-radiated by an Ac ft^2 surface
               at temperature T (F).  

Full sun is about 300 Btu/ft^2/hour.

1 Btu will heat about 55 ft^3 of air 1 degree F.