Subject: Re: Trombe wall efficiency

From: [email-masked] (Nick Pine)
Date: 20 Jun 1995 06:19:23 -0400

Nick Pine <[email-masked]> wrote:
 
>>Is the air between the glazing and the insulation only at 70 degrees?
 
>To a first approximation... Not much more than that. I calculated the
>temperature rise for some "solar siding" recently, and it came out to
>be about 7 degrees F. To calculate this, you can look at the sunspace
>as a thermal chimney, and use the empirical formula CFM = 16.6 Av x
>sqrt((Tu-Tl)h), where Av is the vent area at the top and the bottom in ft^2,
>Tu and Tl are upper and lower sunspace temperatures in degrees F, and
>h is the sunspace height in feet. And you might figure that in full sun,
>the chimney is receiving about 300 Btu/ft^2/hour, and that 1 Btu heats
>up about 5 ft^3 of air 1 degree F (which tells you how much the sun
>warms a certain CFM, which in turn determines the CFM that flows.)

Oops. Let me do this again: this solar siding is about 16' high, and it has
a continuous 1 1/2" wide airspace at the top and the bottom, so a 1' wide
section would have cfm = 16.6 Av sqrt((Tu-Tl)h), where Av = 1.5/12 x 1'
and Tu-Tl is about U/cfm, and h = 16', and

U = 300 Btu/hr x 16 ft^2 - (68F-32F) x 16 ft^2/R1  =  4200 Btu/hr,
    full sun received      loss thru single glazing

so cfm = 539/sqrt(cfm) or cfm = 539^(2/3) = 66 cfm per horizontal linear foot
of siding, per 16' tall x 1' wide section of solar chimney. Not bad... But,
the temperature rise is about 4200/66 = 64F, not 7F, as I miscalculated above.
So if the air in the room behind it is 68F, it looks like the air will emerge
at the top of the siding at about 132F.

And the solar collection efficiency of this siding appears to be about
4200/(311 Btu/hr x 16 ft^2) = (66 cfm x 64F)/(311 x 16) = 84%. Not bad...

>In the design I just finished, it seems that the air leaving the house at
>68F (Tl) will return to the house at 75F (Tu.) The average air temp in the
>sunspace will be about 72F, but in this case, most of the warmed air that
>returns to the house will be North of some greenhouse shadecloth, not in
>contact with the cold glazing, and most of the unwarmed house air flowing
>out of the house will be South of the shadecloth, between the shadecloth
>and the glazing. So the air that is exposed to the cold glazing will be
>the 68F unwarmed house air, before it flows from South to North, sideways,
>through the shadecloth. This is good, but it doesn't matter much.

The above paragraph is still true, mostly :-)

Here's a small picture, a view from the East, say:

---------------       ^
132F <=O= <|   |      up        32F                    311 Btu/hr/ft^2
       |  <| ^ |                                 < --- Sun
   |   | ^<|   |<-Polycarbonate plastic
   |   |  <| ^ |
 house | ^<|<--|--80% shadecloth
  16'  z   z   z i                          South
       |  <| ^ |
   |   |<--|---|--Original house wall                  (clear enough? :-)
       | _<| ^ |
 68F ==I====>  |
-------------------------------------------
       |  3"   |

The plastic flap dampers are located at I and O, in the above picture,
and they look like this:

      .                      .
    p .                      .p
   p  .                      . p 
  p   .  <===  O             .  p  ===> I     
 p    .                      .   p
p     .                      .    p
      ^ hardware cloth ------^     
^                                 ^
|__ 1 mil polyethylene film ______|

The top one (O) will have a plywood door, hinged at the top, over the
plastic film, to close it off in summertime. When you close the upper
inner door, a link will open an upper outer door from the glazing cavity
to the outside, to make this work as a solar chimney in summertime,
cooling the room inside with some natural ventilation, and keeping
the sun off the original wall... If the house were air-conditioned,
the summer cooling air input could come from the outside, at the
bottom of the South wall. 

This house won't have a solar closet, but in general, a house with a
sunspace (or solar siding) and a solar closet might look like this:

   East view                 ^                      South view
---------------              up              ----------------------- 
               |d                           |     |                 |
               | d <--clear polycarbonate   |     |                 |
               |  d                         |     |    sunspace     |
    house      |   d       South            |     |                 |
 -  -  -   ----|    d                       |     |         --------|
          | SC |     d                      |     |        |  solar |
          |    |  SS  d                     |     |        | closet |
          |    |       d    srp?            |     |        |        |
--------------------------------        --------------------------------

               ^--dark surfaced, insulated wall   

--------------- ------- ..............
	  |    |       |              .
          | SC |       |              .
          |    |       |   shallow    .
          |    |  SS   |  reflecting  .       South             
   house   ----|       |    pool?     .
               |       |              .
               |       |              .
               |------- ..............
               |    ^  
               |    |__ dark, low-thermal-mass floor
---------------

Nick

---

From: [email-masked] (Nick Pine)
Date: 19 Jun 1995 17:27:22 -0400

Will Stewart, <to whom I would reply by email, were his address either
mentioned in his posting, or not obscured by our dumb listserver>, wrote:

>>  A "Trombe wall" with insulation on the outside, and 1 square foot of
>>South-facing single-glazed area and an R-value of 20, will receive about
>>1000 Btu/day of heat on an average 32F December day, where I live.
>
>How will this heat be absorbed?

The sun will heat up some surfaces in the low-thermal-mass sunspace of this
"Trombe wall," (which is not really a Trombe wall, but an improvement on a
Trombe wall) by either striking a piece of greenhouse shadecloth or striking
the dark insulated back wall of the sunspace. The shadecloth or back wall of
the sunspace will heat up the air in the sunspace, and during the day, the
warmed sunspace air will go into the house through some passive plastic film
dampers, as invented by Doug Kelbaugh in Princeton in about 1978, described
on page 47 of the 1982 Time-Life book, _Energy Alternatives_. During the
night, the sunspace will get icy cold (UNlike a Trombe wall, which uselessly
heats the outdoors all night through the glazing) and the warm air in the
house will not flow backwards through the sunspace, because the plastic film
dampers will prevent that.

>What about using glazing that has R values over 4 and using less area for
>the trombe wall?

That would probably help a conventional Trombe wall, but in this kind of
passive air heater, not much energy is lost through the glazing at night,
because the space gets cold at night. Glazing with R-values greater than 4
may be more expensive and transmit less sun than simpler single or double
glazing, in this case. If you look at the numbers below, for this air heater,
only a small fraction of the energy collected is lost over a day, only while
the sun is actually shining. So higher R-value glass can't improve it a lot,
and may hurt it, by making it more expensive and cutting down on the amount
of sun that gets through the glass. A conventional Trombe wall (with no
quotes) would have to be about 25 times bigger in area to supply the same
amount of heat as this sort of passive air heater system, I think.

>> ....If the
>>room behind it has a constant temp of 70F, and the sun shines 6 hours a
>>day, on the average, the energy that leaks out of the glass will be about
>>6 hours x (70F-32F) x 1 ft^2/R1 = 228 Btu during the day,

[ie only about 30% of the net energy collected per day is lost through
the glass, so improving the glass can increase the efficiency by about 30%,
at most, not counting losses due to decreased solar transmission.]

This is similar to the new Conserval air heater, which uses a thin aluminum
absorber plate with holes and transverse airflow, to achieve a measured 80%
(quite high for an air heater) solar collection efficiency, with no glazing
at all. The Conserval people say that adding some glazing would cut their
efficiency to 60%. And high-R glazing would cost more and further lower
efficiency, with its lower solar transmission. Of course if you have no
glazing, you have to start with ambient air temperature, eg warming up 32F 
outside air to 39F air, which isn't much good for passive solar heating...
So it's better to have some glazing, I think, even if that cuts down on
the solar collection efficiency, so you can start with 68F air and heat
it up to 75F, or more.

>Again, how does the heat energy get into the house.

Through holes in the insulated wall between the low-thermal-mass sunspace
and the house. These holes have simple dampers which automatically open up
during the day and close at night.

>>and 18 hours x (70-32) x 1 ft^2/R20 = 34 Btu at night, a net gain of 1000
>>-228 -34 = 738 Btu/day.
>
>What happens at night, with no sun?  Does the house get cold?

That depends on the design. It may be desirable to have the house cool off
at night, when most people are asleep in bed, to save energy, and reduce
the heat storage requirements for long strings of cloudy days. 

One way to store heat in the house is to insulate the "Trombe wall" on
the outside and leave it uninsulated on the inside, so that heat that
comes into the house during the day, via the warm air that comes into the
house, is stored in the wall during the day, from the inside of the house.
In this case, a small mass wall area may overheat the room behind it.
Trombe walls are good at not overheating during the day, since the
mass is exposed to the sun directly. Unfortunately, this means the
mass is exposed to cold glazing at night.

Another way to store heat in the house is by filling it up with
concrete furniture :-) Lots of it... And masonry walls and floors
and big piles of rocks here and there. ("Here, come sit on this boulder.")
Remember the Dick Van Dyke show with the boulder in the living room?

If the house is superinsulated, with lot of thermal mass inside, which
has lots of surface area, the daily temperature swing will be almost zero.
The house will be like my friend's underground house. He gets ready for
winter in November, by closing the windows, to make the inside temperature
rise from 70F to 72F, over two weeks... 

A better way is to somehow isolate the warmstore from the living area,
so that the living temperature can be controlled separately, and so
that the warmstore temperature can vary widely, and it can be charged up
to MORE THAN ROOM TEMPERATURE on sunny winter days, using lots of South
glass area. You can't live in a 130F room, but it is desirable to be able
to charge up a heat battery to higher than room temperature, in order to
maximize its discharge life, when the sun goes in for a few days. This can
be done by making another insulated wall between the mass wall and the
living space, and circulating the sunspace air between the mass wall and 
that inside wall during the day.

Even better, I think, is to make the back wall of the sunspace an
ordinary insulated frame wall, except for a part of it, which is glazed,
with a passive air heater and high-thermal-mass "solar closet" behind that.

The solar closet would ideally be full of sealed containers of water,
that are heated by airflow from the air heater on the south side. It
would be insulated all-round, including on the air heater side. The
warm air would get into the solar closet from its air heater in the same
way that the warm air from the sunspace gets into the house, via holes
at the top and bottom of the insulated South wall, equipped with simple
plastic flap backdraft dampers.

Containers of water are better than masonry for thermal mass, because
you can store about 3 times more heat in a cubic foot of water than in 
a cubic foot of masonry, and water has a lower thermal resistance than
masonry, which makes it easier to get heat into and out of water quickly.
Water may also be cheaper than masonry, eg in 55 gallon drums. Let me
say that again: the thermal store can be about a third as big, if it's
water instead of masonry. And containers of water can have smoother airflow
around them, with less air resistance than rockpiles, so the fans or blowers 
can use less electrical power, to transfer heat to or from air. ("To" usually 
requires a higher heat transfer rate, since the sun shines less than half
the day in the winter.) A solar closet can have a more compact shape than
a mass wall, so a given amount of insulation can be more effective, with
the lower surface to volume ratio of a solar closet, compared to a mass wall.

>Many passive solar houses that do not store the solar energy gained overheat
>in the daytime and become cold at night.

True. They are badly designed, and one answer to this problem has been
to invent rules of thumb like, "the South-facing glass area should not
exceed 25% of the floor area." This prevents overheating (although opening
a window is one solution to that), but it may also limit the house to make it,
say, 30% solar heated, at most, especially in cloudy climates. I think it's
better to use lots of South-facing glass and put the extra heat into a heat
battery on sunny days. Of course a large enough house with small enough
South-facing glass needs no heat storage at all.

>The trombe wall is one answer to this problem.

A fairly inefficient one. Think of a Trombe wall as a car that gets 2 miles
per gallon, or goes 10 mph. Not bad in the days when there were no cars...
(Trombe walls were actually invented by Morse, long before Trombe, in 1881.)

>Another answer is to have a masonry wall, table/bench, or floor of some sort
>inside the residence

Ah, concrete furniture! :-)

>(possibly containing phase change materials)

Phase change materials have their problems. For instance, Glauber's salts
get tired, and have to be stirred up mechanically, and only work over a 
limited temperature range, as I recall.

>to even out the daily swings in energy gain.

Yes, to even out daily temperature swings. (Did I mention that I can sell 
you a 100% solar heated house, in any climate, with no windows, even, if
you give me the freedom to pick the indoor temperature swing? :-)

BUT, you don't want to live inside the heat battery, unless you are a Venusian.
If you live inside the heat battery, you can't charge it up to more than about
80F, which severely limits the usable heat you can store in the heat battery,
vs a heat battery that can be charged up to, say 130F, and discharged down
to a usable temperature of 80F, which is about where people-occupied heat
batteries run out of sweat.

When the sun goes in for a few days, if you live inside the heat battery,
you have little control over the house temperature. It just gets colder and
colder. Whereas with a solar closet, you can open or close a damper with 
a thermostat, even a setback thermostat, to let some warm air into the
house as needed... The higher temperature of a solar closet or warmstore
(Norman Saunders' attic version) can also provide a large fraction of the
hot water needs of a house, if it has a water heater within, with a preheat
pipe near the ceiling.

>>  A standard unvented Trombe wall (Table IV-14b of Mazria's book says
>>vented ones don't work much better) with a very large uninsulated thermal
>>mass right behind the glass and an R-value of, say 2 (roughly 1' of
>>masonry), would have an average temperature at the outside wall surface
>>of about 32F + R1 x (70F-32F)/(R2+R1) = 45F, if there were no sun. If you
>>add a heatflow of 1000 Btu/day of sun to that model, falling on the
>>outside of the wall, the outside wall surface will have an average
>>temperature of about 45F + 1000/24 x (R=2/3) = 72.4F, which contributes
>>24 hours x (72.4F- 70F) x 1 ft^2/R2 = 29 Btu/day to the room behind the
>>wall.
 
>Of course, these are average values, and the actual values vary according
>to time of day, and interior heat loss.

Sure, but I think the average values are reasonably accurate, and
represent what we care about.

>These calculations truly require calculus, which Nick as an engineer
>will readily admit,

Nonono. I won't admit that :-) The average passive solar house in this
country is so ill-designed that even applying simple arithmetic, with
assumptions that are within 2:1 of real values, can make a dramatic
performance improvement. Given the present situation, I think we should
forget about calculus completely for a few years, and then perhaps start
nit-picking over the 12th decimal place, once our Pentiums get repaired:-)

A lot of engineers don't know how to do things on the back of an envelope.
They believe that if an answer is not good to 1%, it is useless. Or that
something that is 75% efficient, and costs $10 per square foot, is a lot
better than something that is 60% efficient, and costs $1 per square foot.
It is, in a sense...

A lot of architects do simplified beam calculations now. Why not simplified
passive solar calculations? We need to get better at knowing what mathematical
details can be ignored, and which ones are important... This may change
as the art evolves.

Voltaire said "The best is the enemy of the good," which I interpret
to mean that if you keep looking for the best solution all your life,
you may never get around to implementing a good solution.

>...working out the formulas can be tedious!  :-)

There's that, too :-) And in a lot of cases, it's better to build things
and measure them than try to calculate them. High-powered math is only
useful if you have high-powered physical models to go with it, and in
many cases, those don't exist.

>>  So the "improved Trombe wall" above, (actually an air heater with the
>>  thermal storage inside the house) is more than 25 times as efficient
>>  (738/29) at collecting and keeping heat in the room behind it, than
>>  the usual Trombe wall. This is somewhat oversimplified, of course...
>
>I'm curious to hear more about this once you address the above points.
>Don't stop innovating, continuously refine!

I'm trying. I'd like to see more of these built, so we can measure how
they work and improve them, and learn some newer tricks :-)

Nick

Here are a couple of earlier postings describing how this works...

Article: 2080 of alt.architecture.alternative
From: [email-masked] (Nick Pine)
Subject: One way to build a high-performance passive solar house
Date: 9 Feb 1995 08:56:54 -0500 [updated 4/25/95]
Organization: Villanova University

Step 1.

Look up the average outdoor temperature in December, where you live.

The nice new, free book, _Solar Radiation Data Manual for Flat-Plate
and Concentrating Collectors_, from the National Renewable Energy
Laboratory at (303) 275-4099, has this information, as well as some
solar information, for many places in the United States. (They also
have a CD, with *hourly* solar weather data for 247 US cities over the
last 30 years, that might be used for passive solar house simulations.)

For example, where I live, in the northeast, the average December
temperature is about 32 degrees F, and the average amount of sun falling
on a south wall is about 1,100 Btu per square foot per day.

Step 2. 

Estimate how many Btu/day you need to heat your house at that average
outdoor temperature.

For example, if you have a 30' x 30' 2-story house, you have about
3,000 square feet of walls and roof. If the R-value of that surface is,
say, 30, it takes about 3,000/30 = 100 Btu per hour to heat the house to
68 degrees, if it is 67 degrees outside. Subtract the average outdoor
temperature from the indoor temperature and multiply by 24 hours, then
multiply this by the sum of each surface area divided by its R-value.

Our example house takes (68-32) x 24 x 100 = 86,000 Btu per day to heat. 
Call it a hundred thousand, a nice round number that is about the same
amount of heat as a gallon of oil burned in an old oil burner.

Step 3. 

Calculate how much south-facing glazing you need on your low-thermal-mass
sunspace, to supply that amount of solar heat, on an average day.

Where I live, each square foot of south facing wall receives about 1,000
Btu/day, another nice round number. If the low-thermal-mass sunspace has an
insulated low-thermal-mass wall between it and the house, with a big window
fan in one window, to suck most of the warm air into the house during the day,
and you let the sunspace get icy cold at night, the sunspace will be about
68 F during the day, as a first approximation. So the heat lost from each
square foot of glazing in the sunspace to the outside, during the day, in our
example house, will be about (68-32) x 5 hours, or 180 Btu, if the sun shines
for 5 hours on a winter day. So each square foot of sunspace glass provides
about 800 Btu/day to the house, net, on an average winter day. Another way
to figure this is that each square foot of south-facing glazing collects the
heat-equivalent of 1-2 gallons of oil per year. If your present oil bill
indicates that you need a thousand square feet of glazing or more, perhaps
more house insulation and caulking are needed before solar heating, or you
could just use the sunspace to reduce the oil heat, with no thermal store.

Our example house would need about 100,000/800 = 125 square feet of glazing
in the low-thermal-mass sunspace glass to keep it warm on an average day.
Say, an 8' high x 16' wide window, with an insulated wall between that window
and the rest of the house. The sunspace might be quite shallow, eg some
cost-effective "solar siding," instead of, say, vinyl siding, as in the solar
closet wall below. Or it might be a bit deeper, extending out from the house
8', and used for growing ripe, red tomatoes and basil in December, perhaps
with a couple of extra hours of 90% high-pressure sodium and 10% mercury
vapor light, and a little heat leaked from the house on freezing nights.

Step 4.

Take a guess at how many cloudy days there are in a row in December, where you
live, and what the outdoor temperature is during those days. In many places,
cloudy days are warmer than sunny days in December.

(If you wanted to be more precise, the weather bureau, or perhaps NREL or NOAA
could help. Or you could calculate Gaussian statistics using the CD data.)

Let's say that our example house is in a climate with 5 cloudy days in a row,
in December, and that the average temperature during those days is 32 F.

Step 5.

Calculate how many 55 gallon drums full of water you need to keep the house
warm for that cloudy day period.

In our example house, this would be 5 x 100,000 = 500,000 Btu, about the same
as 5 gallons of oil. If the water in the drums is hot, say 130 F, and the
drums can keep the house warm until the water cools to, say, 80 F, then
each drum stores about 25,000 Btu, about the same as a quart of oil. So
5 gallons divided by one quart is 20 drums (talk about apples and oranges :-)

How do you keep the drums that hot? You build an insulated solar closet behind
the sunspace, in the house, with an air heater as part of the insulated wall
between the sunspace and the house, with a transparent vertical cover of glass
or transparent "solar siding," eg Dynaglas or Replex ((800) 726-5151)
polycarbonate plastic, which costs about a dollar a square foot, and comes
in long sheets, about 4' wide.

Behind that siding, you staple some 80% greenhouse shadecloth, which costs
about 14 cents per square foot, leaving a 1" air gap between the siding and
the shadecloth, and you leave another 1" air gap between the shadecloth and
the 3 1/2" of fiberglass insulation in the 6" wall, and you put small vents
(about 1% of the overall area, eg 1 square foot in our example house) at the
top and bottom of this air heater, to allow warm air from the solar closet to
flow into the outside air gap through the vent hole at the bottom of the air
heater, _through_ the shadecloth, and back through the upper vent hole into
the insulated solar closet. The vent holes should have plastic-film backdraft
dampers to keep the air from flowing when the sun is not shining. These might
be made from chicken wire and a thin (1 mil) plastic film like that used for
dry cleaner bags. 

The inside wall of the closet could be the foil face of the fiberglass
insulation. The floor might be dirt, covered with a layer of plastic.
There should be an air space between the drums and the walls of the
solar closet, to allow air to circulate around the drums. There should
also be a vent to the house, to be slightly opened on cloudy days. This
could be an electric air damper controlled by a thermostat. (Another
approach might be to put the 55 gallon drums on a strong attic floor as
an overhead "warmstore," a la Norman Saunders, for new construction.)

In our example house, if the 2' diameter x 3' long drums were stacked up
horizontally, 4 high, the solar closet would be 8' high x 10' long x 4' deep.
I would make it 6' longer, and use the non-drum space for a sauna. The
solar closet should also have 3 1/2" of fiberglass insulation in its ceiling,
ie the second floor of the house, and in its back (north) wall, inside the
house. Note that most of the "waste heat" from this solar closet ends up
in the house via the sunspace from the front glazing, when it is collecting
heat, and via most of the insulated surface the rest of the time. Note
that the solar closet does not usually provide heat for the house, except
during cloudy day periods, so it stays hot like a stagnant solar collector.

The sauna might have a very small woodstove, for burning newspapers, junk
mail, old paper towels, college committee recommendations, letters from
congressmen, and press releases announcing amazing new price breakthroughs
in photovoltaic technology.

Article: 2127 of alt.architecture.alternative
From: [email-masked] (Nick Pine)
Subject: Passive solar house evolution
Date: 13 Feb 1995 09:49:42 -0500
Organization: Villanova University

Question 1: If you put a 55 gallon drum full of water in a 2' square x 4' tall
uninsulated box in the shade in Philadelphia, in January, what would the
average temperature of the drum be? 

Answer 1: According to the NREL _Solar Radiation Data Manual_, the average
January temperature in Philadelphia is -0.9 degrees C, or about 30 degrees F.

Q2: What would the average box temperature be if it were in the sun,
in Philadelphia, in January, and if it were painted white?

According to the 1993 ASHRAE Handbook of Fundamentals, the "sol-air
temperature," or equivalent air temperature, Te, of a vertical surface
in the sun, is the outdoor temperature + 0.15 x U, IF the surface is painted
a light color, and the amount of sun falling on the surface is U Btu/hour.

According to the NREL book, a south-facing wall in Phila in January receives
about 3.3 kWh/m^2/day, ie 3300 x 3.41 Btu/10.76 ft^2/m^2 = 1000 Btu/ft^2/day
of sun. This is an average daily insolation of about 40 Btu/ft^2/hour. So
assuming the sun only shines on the south side of the box, if it were painted
white, the average temperature of the south side would be 30 + 0.15 x 40 = 36
degrees. The interior temperature should be the average temp. of all of the
outside surfaces, I think. Each face of the box has a surface area of 8 ft^2,
and the top and bottom have areas of 4 ft^2, so the average drum temperature
should be about (8x36+3x8x30+2x4x30)/(8+3x8+4+4) = 1248 / 40 = 31.2 F, 1.2
degrees warmer than the drum in the shade.

Q3. What would the drum temperature be if the box were painted black?

The ASHRAE HOF says that the sol-air temperature of a DARK vertical surface
is the outdoor temperature + 0.3 * U, so if the black box were receiving 
the same amount of sun as the white box, the average temperature of the
south wall would be 30 + 0.3 x 40 = 42 F. So you can raise the effective
average outdoor temperature of a white south-facing wall in Philadelphia by
an average of 6 degrees, in January, just by painting it darker. Not bad... 

In the above case, the average temperature of the drum/box would be 

(8x42+3x8x30+2x4x30)/40 = 32.4 F, just above freezing.

Q4. Suppose the box had a single-pane glass south wall, with no insulation?

In this case, the solar energy, Ein, that goes into the box would be about

Ein = 2 x 4 ft^2 x 1,000 Btu/ft^2/day = 8,000 Btu/day,

assuming the glass transmits 100% of the solar energy.

If the drumwater has an average temperature of Tw, and the walls and glass
front of the box have an R-value of 1, the energy that goes out of the box
in one day is Eout = (Tw-30) x 24 hours x 40 ft^2/R1. If energy is conserved,
ie Ein = Eout, then Tw = 30 + 8,000/(24x40) = 38.3 degrees F. An improvement.
Six degrees warmer than the unglazed box...

Q5. How about if we add R-14 insulation to the other three sides and the top
and the bottom?

In this case, the solar energy that goes into the box is the same, but
the solar energy that goes out of the box, into the outside air, is

Eout = (Tw-30) x 24 hours x (8 ft^2/R1 + 32ft^2/R14) = (Tw-30) x 247, so if
Eout = Ein, then Tw = 30 + 8,000/247 = 62 degrees F, a livable temperature,
when the sun is shining, which begins to decrease when the sun stops shining.
This is often as far as passive solar house designs go, with the house getting
colder and colder on cloudy days, as a lot of heat leaks out of the south-
facing windows. One of the problems with this design is that you have to
*live* inside the "heat battery," so you can't make it too warm. 

Q6. But then suppose we make the glazed side an ideal air heater, so it
collects the sun's heat during the day, but the drum is insulated at night?

Now Eout = Es + Eother, where Es is the heat lost through the ideal south wall
air heater. Say the sun shines for 6 hours a day in January... Then
Es = (Tw-30) x 6 hours x 8 ft^2/R1 + (Tw-30) x 18 hours x 8 ft^2/R14.

Eother is the heat lost through the east, north and west walls of the box,
as well as the top and bottom. Eother = (Tw-30) x 24 hours x 32 ft^2/R14. 

So, if the energy into the box equals the energy out of the box, then

(Tw-30)(48+10+55) = 8,000, so Tw = 30 + 8,000/113 = 103 degrees F. But wait!
This is no good... This passive solar house has too much south-facing glass!
It overheats! At this point, the thing to do is open the windows in January,
or (better) move out of the little box with the drum in it, and build a house
behind it, using the hot water in the drum as a heat battery for cloudy days.

If the drum and its air heater are inside a sunspace, and the air from the
sunspace heats the house during sunny-day periods, the heat lost from the south
side of the air heater will help heat the house during sunny day periods. If
the house is built around an insulating solar closet containing the warm drum,
the heat lost from the drum will help heat the house too... So the heat that
leaks out of the heat battery during sunny-day periods is not wasted.

Q7. Now suppose we make the box 8' tall instead of 4' tall, so that the air
heater collecting area is 16 ft^2, and the drum on top still gets all the
collected heat, but none of the collected heat goes out through the lower half
of the box or the uninsulated glass at night?
                                                 g: glass       giii
Ein = 16 ft^2 x 1,000 Btu/day = 16,000 Btu/day.  D: drum        giDi
                                                 i: insulation  giii
Eout = Es + Eother                                              giii

       Es = (Tw-30)x6x16ft^2/R1 + (Tw-30)x18x8ft^2/R14 = (Tw-30)(96+10).

       Eother = (Tw-30)x24x32/R14 = (Tw-30)x55, as before.

So Ein = Eout ==> Tw = 30 + 16,000/(96+10+55) = 131 degrees. This is getting
interesting... We could not possibly live inside this solar closet, except
for a few minutes at a time, as a sauna, but it is good to have a heat battery
like this sitting around, charged up to a high temperature, because that will
make the useful heat that we can get out of it last for a long time, during
periods of cloudy days. Along with a bit more glazing, one might also put an
electric water heater inside this closet, and preheat its cold water input
with about 20' of 1 1/4" copper pipe running along the ceiling of the closet,
to heat water for taking showers, etc... If the input water is preheated, and
the water heater is in a 130F room, the electric heating element should rarely
turn on.

Q8. Suppose we used two layers of glazing instead of one, above?

Ein would be the same, in this simple model.

Es would be about (Tw-30)(48+10), and Eother would be the same.

So Tw = 30 + 16,000/(48+10+55) = 174 degrees F.

Q9. Then suppose we add a reflecting pool or shutter in front, which
increases the solar input by 50%?

Tw = 30 + 16,000x1.5/(48+10+55) = 242 F. (Which of course, would make
the water steam. Maybe these drums should be full of sand. But then you
would need about 3 times more of them, since masonry has about a third
the heat capacity of water, and a higher thermal resistance.)

Q10. But wait, we were going to put the solar closet inside the sunspace,
right? So during the day, when the solar air heater is working, the south
side of the air heater will be exposed to, say, 68 F house air, not 30 F air,
and the other walls of the solar closet will also be exposed to 68 F air,
not 30 F outside air. So what would the "water temperature" be in this case?

It's roughly the same little calculation, using 68 F instead of 30 F:

Tw = 68 + 16,000x1.5(48+10+55) = 280 F.

Q11. Then suppose we lay the drum/box down horizontally, and put the air
heater side at the focus, under an ideal 4:1 reflective linear parabolic
concentrator? Or use an R-14 movable reflective shutter to cover the glass
over the drum when the sun is not shining?

Ein = 16,000 x 4 = 64,000 Btu/day.

Eout does not change.

So Tw = 68 + 64,000/(48+10+55) = 634 degrees F.

...

So, it seems to me that it isn't too hard to arrange for a passive solar house
to have a "solar closet" with a few high temperature 55 gallon drums full of 
water, a "heat battery" that can be discharged in a controlled way, to provide
heat for a house during cloudy days...

Nick

---

From: [email-masked] (Steve Paschke)
Date: 21 Jun 1995 14:28:07 GMT

> Here's a small picture, a view from the East, say:

Wouldn't black painted metal be better than shadecloth at absorbing heat
and radiating the heat absorbed?

I've built a unit several years ago with painted beverage cans with mild
success.  I think that a large heat sink (made from scrap sheet metal)
would be most effective.

-- 
  *   /\  *  ^  /\  * /\         
\/\  /  \   / \/  \/\/  \   *
 \ \/    \ /   \  /  \   \       [email-masked]
 *\/      /     \  *   *  \     
    *  * /       \               standard disclaimer goes here
 *        (  _  )    *
   *      ( | | )        *
**********(_|_|_)**************

---

From: [email-masked] (Nick Pine)
Date: 21 Jun 1995 16:04:11 -0400

---------------       ^
 
>and radiating the heat absorbed?

The idea here is that the cool air rises up through the passage between
the glazing and the shadecloth, moves ***through*** the shadecloth,
from right to left, becoming warmed in the process, as in the Conserval
wall, and goes back into the house at the top. If the black-painted metal
had no holes in it, it would probably not work as well as the shadecloth.
If it did, it might work better, like the Conserval wall.

>success.  I think that a large heat sink (made from scrap sheet metal)

Lots of absorbing surface area is nice, altho it's curious that
in a linear model, it looks like that doesn't matter at all.

Nick

---

From: [email-masked] (Nick Pine)
Date: 23 Jun 1995 17:27:17 -0400

jason almeter <[email-masked]> wrote:
 
>Is there a group of retailers that usually carry shadecloth?  I would
>like to try an experiment, but would like to find a local supplier.
>Would greenhouses or nurseries have this stuff?

You could try some local nursery or greenhouse suppliers, but the best
price I found was from Stuppy in Kansas City, MO, at (800) 877-5025 voice/
(800) 423-1512 fax. I just got a 16' x 32' piece of their item # 23691
80% shadecloth, with large grommets and hems on the edges. The cost was
$71 as I recall, incuding UPS shipping from Atlanta. Delivery time was
about 10 days.

Nick

---

From: [email-masked] (Nick Pine)
Date: 23 Jun 1995 17:08:47 -0400

Woodrow W. Baker <[email-masked]> wrote:
>Nick Pine ([email-masked]) wrote:
>
>: The idea here is that the cool air rises up through the passage between
>: the glazing and the shadecloth, moves ***through*** the shadecloth,
>
>expanded lath (expanded metal used on stucco and some rockwork usually 
>embedded in the cement) painted black is what I have been told works best.

Steve Baer mentions that 4 or 5 layers of metal lath work nicely for
natural convection, in _Sunspots_. I think the Jersey Devil used something
like that in an early solar house. "Air hair" (fiberglass insulation,
painted black, with transverse flow?) or black fabric air filter material
should work too. But shadecloth is very cheap and durable, and in a
linear model, the heat that comes into the glazing has nowhere else to
go but out the R1 glazing or into the airflow...

Nick

---

From: [email-masked] (Nick Pine)
Date: 7 Jul 1995 10:12:30 -0400

Johnathan Sawyer <[email-masked]> writes:

>I am building a house outside Boulder Co with an unvented  16 x 28 ft Trombe
>wall. The masonry wall is 12" thick and the glazing is single pane...

Pretty far into it, huh? 32 tons of masonry... Perhaps it isn't too late to
soup it up by adding some vents, and some insulation between this Trombe wall
and its glazing? :-)

>I am using an expensive ($3.20/sqft) selective coating made by an
>outfit in New Jersey.

I'd be interested to know their name, address and phone number. I wonder
how this $tuff can be attached to a masonry wall. What will the R-value be
at the interface where it attaches, if the selective surface is a metal
film with an adhesive backing? If the interface R-value is not 0 (and it
may be difficult to attach it well to roughish masonry), the film may
heat up a lot more than the wall, when the sun is shining. 

>They claim this coating gives the wall an
>equivalent of a R12 insulation value at night.

Sounds great, altho pricey, but have you asked about the temperature range?
Trombe walls usually have low temperatures, since they lose heat to the
outside world and the inside of the house, 24 hours a day, through fairly
low thermal resistances, and selective coatings usually help with high
temperature radiation losses. The formula for radiation loss is

Ur = 0.174 x 10^-8 A e ((Tw+460)^4 - (Ta+460)^4), where U is in Btu/hr,
     A is in ft^2, e is a number called emissivity, which is close to 1
     for a non-selective surface and perhaps 0.1 for a selective surface,
     and Tw and Ta are wall and outside surrounding temps in degrees F.
     (I may be doing this wrong, and I would be pleased if someone were to
     post better calculations, vs. posting talk about better calculations.) 

This is on top of convection loss, which is roughly R2, with a still
air surface and a single pane of glass and an airspace. R-values are
supposed to include some radiation loss, but roughly speaking, the total
loss per square foot is Ut = Ur + Uc, where Uc is (Tw-Ta)/R2, so if it's
0 degrees F outside, and the wall has temperature Tw, and e = 0.1,

Tw,   Ut,  Uc, Urs, Urs/Ut and Reff (the effective R-value) would be:

70    41   35  6    15%        1.7 = (70F-0F)/41
80    47   40  7    15%        1.7      ...
90    53   45  8    15%        1.7
100   59   50  9    15%        1.7

>I wonder how the thermal calculations will come out using R12 instead
>of R2 heatloss.

Much better, but it seems to me that convection will make up such a large part
of the total heat loss that the real picture will be about R = 1.7, 24 hours
a day, so the wall temp in the above case, with, say 1000 Btu/ft^2/day falling
on the wall, if it were perfectly insulated on the inside, would be such that 

1000 = 24 Ut, so Ut = 42, and the wall would have a temp of about 70F,
according to the above table.

But of course the wall is not perfectly insulated on the inside--the masonry
has a thermal resistance of about R2. Then again it may be warmer than 0 F
outside, most of the time... In this case, it looks to me like the thermal
resistance of the wall doesn't matter much during sunny days, since it will
have about the same temp as the inside of the house. It won't add much heat
to the house, but it won't be a thermal disaster either, unless the sun goes
in for a few days in a row. Another 32 ton car that gets 2 mpg, while going
10 mph, max :-) Perhaps it can still be helped, tho, into the 20th century.

Nick

---

From: [email-masked] (Nick Pine)
Date: 8 Jul 1995 08:25:23 -0400

Leo Hendry  <[email-masked]> wrote:
>>The formula for radiation loss is
>>Ur = 0.174 x 10^-8 A e ((Tw+460)^4 - (Ta+460)^4), where U is in Btu/hr...

>Are American R values something other than the m2K/W we use, or are you
>mixing units in a near-masochistic manner?

There's actually no mixing above, just pure infernal English units, as
Norman Saunders, PE, says. He uses MKS in this country, and perhaps this
is one reason so few people know of his work. Our colonial R-values are in
hr-ft^2-F/Btu, about 6 times bigger than the R-values in the rest of the
world. We are thinking of exporting them, which will reduce world energy
consumption by 83%, if we don't export the rest of the units. 

>What's the matter with metres, degrees Kelvin and watts?

They are nice for electrical engineering, but it seems to me that people who
build things in this country are already confused or ignorant enough about
passive solar heating mathematics and performance, that it isn't a good idea
to go switching the system of units around as well, yet. (Or perhaps it would
be a good idea (?), to help people get a fresh perspective, and unlearn what
they think they know, but I think it would just further isolate the solar
ranters who use numbers from the builders and spenders who use numbers.)

One basic grounding is that our R-values are what you see stenciled on big
rolls of insulation for sale in stores in this country. Most people here
think R-19 is pretty good, vs R-3, and superinsulated houses with R-40 walls
are pretty good, and 68 is a good house temperature. People here still measure
things in feet, too.

And most people don't give much of a hoot about energy conservation anymore.
It would be nice if passive solar house heating were to aquire the same hobby
status as sailing, something a body does that a body is not obliged to do, a
capital-intensive sport where performance matters. I've seen lots of passive
solar palaces that use 70% or more of the yearly backup heat of a conventional
home the same size. Would that for every million dollar earthship of such ilk,
there were 10 slick $200K houses with smiling owners enjoying close to 100%
solar space and hot water heating. 

I'm used to this system right now. I know a lot of the numbers by heart, and
it's easy to look at new numbers and recognize patterns right away. This would
take a while to relearn in a new system, just for me, ignoring the additional
problem of communicating with others via a new language barrier. To me, a
kitchen match has 1 Btu, and a gallon of oil and a ccf of natural gas contain
about 100K Btu, a therm, and a cord of wood is a hundred gallons of oil. And
windows and still air films have an R-value of 1, and a pound of water needs
1 Btu to warm up 1 degree, as do 50 ft^3 of air. 

>(Posting talk about better calculations)

We are not amused...

Nick