Subject: Re: Converting An old House to Renewables

From: [email-masked] (Nick Pine)
Date: 1 Jun 1995 05:29:51 -0400

Steve Paschke  <[email-masked]> wrote:

>If your thinking of re-siding soon, add 2" of rigid foam to the exterior
>of your house. 

How about some solar siding? Put Dynaglas or Replex single wall corrugated
clear polycarbonate sheets on the house, with an inch air gap behind
them, a layer of 80% greenhouse shadecloth, another inch air gap, and
holes at the top and bottom of the wall, with plastic flap dampers, so
that in the wintertime, house air flows up between the glazing and the
shadecloth, sideways thru the shadecloth, and back into the house. In
summer, you'd want to vent air from the top of the wall to the outside,
and vent house air into the wall at the bottom, if the house is not
air-conditioned, otherwise vent outside air into the wall at the bottom.

>  The most cost effective solar power at present is solar hot water heating.

Really? I used to spend a lot more on space heating than water heating, 
and passive solar space heating is usually more efficient, since the
temperatures are lower, and cheaper, since there are no moving liquids,
pumps, heat exchangers, plumbing, etc.

>If your going to change your water heater, buy a solar heater.

Or better yet, perhaps, try making a water heater as described in Q7
of the second posting below...

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.

Nick

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 or fin tube 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