Sneak Peak Video of the 
New Solar Hydrogen Home DVD
Coming SOON!

Download Over 100Meg of
FREE Hydrogen Video
Ride in the Famous H2 Geo
Click Here

re: water trombe-michael wall
12 nov 2004
yatseck  wrote:

>...i'm new to the subject of passive solar systems in architecture.

you might enjoy the appended notes. we are working on a more metric
version for the 2005 ises conference in orlando, fl.

>i've decided to place the passive indirect or isolated gain (not sure
>yet) solar systems on the second floor of my building. what types of
>air circulation system would be the best to use in order to provide
>warm air to the lower story?

you might store heat from thermosyphoning air panels in a massy 2nd floor
ceiling and move warm air down to the first floor with a 2 foot x 2 foot
vertical duct and blower and return it to the 2nd floor via a stairway.
during heat storage, hot air from the panels might move from south to north
through ceiling spancrete channels and flow back to the panel entrances,
as in the barra system, or it might pool under a solid concrete ceiling
with a low-e coating beneath, with 10x more ceiling than panel surface, or
it might heat a thick layer of water in flat polyethyelene film ducts over
a plywood ceiling or a thinner layer of water that merely collects vs stores
heat, with a low-power circulating pump and a copper coil heat exchanger
in a large unpressurized tank on the ground floor, and a fan coil unit to
keep the ground floor warm.
 
>question2: i've been thinking about making a modification to the,
>already modified, trombe-miachel wall, which consist of a row
>(oriented w-e, south exposure) of thermal mass short walls/rectangular
>pillars rotated at an angle of -45deg. (longer side to se - providing
>shade in morning hours and warming up, and letting the evening sun
>inside). what if the wall was a huge waterglass, instead of opaque
>brick/concrete? methinks: better thermal storage, transparency, and
>the opportunity to empty the wall before a long period of cloudy days
>in order to avoid the reversed thermall process=heat loss.

trombe walls collect heat well, but they lose a lot at night and on
cloudy days, because of the glazing's low thermal resistance. making
water stand up isn't easy. where would you empty it to?

nick

notes for ases workshop on solar house heating and natural cooling techniques

portland, or  july 9, 2004

written by nick pine, with steve baer, drew gillett, and rich komp. 

debatable conclusions

1. heat flows like electricity.
2. solar heat can be 100 times cheaper than solar electricity.
3. superinsulated houses have to be very small or very large.
4. direct gain houses can be improved.
5. indirect gain can be more efficient.
6. we might store heat in the ceiling.
7. we might have a separate cloudy-day heat store.
8. low temp heat storage and distribution are difficult.
9. shurcliff's lung might be a good air-air heat exchanger.
10. greywater heat exchangers can help.
11. we might also gather heat from pvs.
12. smart ventilation can be helpful.
13. swamp cooler controls can be improved.

0.0 introduction

the us has 5% of the world's population and consumes 26% of the world's
energy. house heating and cooling accounts for about one third of that.
in 1980, "envelope house" inventor tom smith said: 

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

this workshop aims at improving that grasp, which we can control better
than our us cheap energy policy... if we paid related costs of healthcare
and air pollution and gulf wars at the pump, gasoline would be a lot more
expensive. drew says this writeup needs exercises for the reader. ok:

exercise 0.1: the us consumed 20 million 42 gallon barrels
of oil per day in 2003. what was the real cost per gallon? 
(debatable answers appear at the end of these notes. :-)

most people think "electricity" when they hear "energy," even though most
houses need more heating energy than electrical energy (the ratio is 1:1
in hawaii and 5:1 in vermont.) it's easy to shrink the small electrical
slice of the home energy pie with compact fluorescent (cf) lights and more
efficient appliances. solar heat can be very inexpensive compared to solar
electricity. pv panels at $3 per peak watt cost 150x more than polycarbonate
glazing at $1/50w = 0.02/pw. and sunspaces add floorspace to a house.

a square foot of "solar collector" only collects about $1/year at today's
oil prices, so anything (except pvs??? :-) that costs more than $10/ft^2 
(half of that being labor) and only collects energy with no other useful
purpose seems economically-doomed...

exercise 0.2: should we a) replace a 60 w bulb with a 14 watt cf or
b) buy 60-14 = 46 additional watts of pv power or c) caulk the house?

most of us "know" how to design passive solar houses with well-established
rules of thumb, but let's relax and take a fresh look from a standpoint
of basic physics...

   berlin is a nice town and there were many opportunities for a student to
   spend his time in an agreeable manner, for instance with the nice girls. 
   but instead of that we had to perform big and awful calculations.

       konrad zuse, inventor of the 1936 z1 computer

overview

this is a workshop on "ohm's law for heatflow" with applications to solar
water and house heating and natural cooling. we'll also discuss a simple
greywater heat exchanger, a $60 300' piece of 1" plastic pipe coiled inside
a 55-gallon drum. with hot water bursts of 13 gallons or less, it could be
97% efficient. if it is, why bother with solar hot water?

we'll provide arithmetic tools and data and strategies needed to site-build
effective house heating and cooling systems using inexpensive materials and
skills. we expect workshop participants to have some familiarity with high
school algebra.

we'll discuss power, energy, heatflow, and overnight and cloudy day heat
storage at the high-school math and physics level, with insulation values
and heat capacities of materials, simple equations involving time constants,
evaporative and night ventilation cooling, passive and low-energy solar
heating, climate data, and schemes for houses that are 100% solar-heated
and naturally cooled, by design. we'll provide a calculator (steve baer
says "throw away your calculator." :-) and a cd-rom. promising techniques
include solar closets, trickle collectors, "pancake houses," soap bubble
foam insulation, and solar attics, including systems to collect heat and
electricity from water-cooled standard pv panels. 

i'm an ee by training, steve baer is a well-known solar inventor,
rich komp is president of the maine solar energy association and
a pv author with a pchem phd, and drew gillett is a professional
engineer with civil engineering and architectural degrees.

disclaimer

some of the techniques we describe are experimental. some have never been
tried. we do not accept responsibility for their safety or functionality. 

1. power and energy

energy is the stuff we pay for, measured in joules or watt-hours or
kilowatt-hours (kwh) or calories or "british thermal units" (btu), no
longer used in britain :-) the british now use joules or kwh. a btu is
a quantity of heat, about the same as the energy in a kitchen match or
a mouse-hour. one btu can heat one pound (16 ounces) of water one degree f.

exercise 1.1: how many btu are needed to heat 8 ounces of water
from 50 to 200 f to make a cup of tea?

power is the rate of energy flow over time. a mere number, vs the stuff we
pay for. energy is power times time. one watt-hour of energy is equivalent
to 3.41 btu. if energy were miles traveled, power would be miles per hour.
if energy were a paycheck, power would be an hourly rate of pay.

exercise 1.2: how long would it take to heat the tea water with
a 300 w immersion heater?

we might check this with an immersion heater and a watch and a $100 raytek
ir thermometer. or a hobo from onset computer corp (1-800-loggers.) their
$119 battery-powered u12-013 hobo is about the size of a matchbox. it can
record 43,000 12-bit samples at 1 second to 18 hour intervals of its own
temperature and relative humidity (rh), with jacks for 2 more temperature
probes or other devices on cables, and upload them to a pc spreadsheet via
a usb port.

people often confuse power and energy, as in "my house uses lots of power"
(vs energy) or "my furnace capacity is 50,000 btu," vs btu/h. power is
measured in watts or kw. unlike energy, it can't be used or consumed.

people confuse heat and temperature, too. a bathtub full of hot water
contains a lot of useful house heat, compared to a candle, but the candle
is much hotter. a lower minimum usable temperature increases useful heat.
temperature is a measure of heat intensity. a 12-volt 100 amp-hour 50 pound
automobile battery stores 267 times more energy (12vx100ah = 1200 wh) than
a 9-volt 500 milliamp-hour (9vx0.5ah = 4.5 wh) 2 ounce transistor radio
battery, at a lower voltage (ie "electrical temperature.") the $40 battery
can store about 200 kwh over its lifetime, at 20 cents/kwh. a $1 cubic foot
of water cooling from 130 to 80 f stores (130f-80f)64btu/f = 3200 btu, ie
about 1 kwh, with a much longer lifetime and simpler i/o.

2. rich komp, ohm, and newton 

rich komp (who is still alive) says heat moves by conduction (a hot frying
pan handle), convection (including air movement), radiation (the sun brings 
about 1000 watts per square meter or 300 btu per hour per square foot on
a clear day at noon in the sahara), and phase change (144 btu melts a pound
of ice and 1000 btu evaporates a pound of water.)

about 300 years ago, isaac newton said the amount of heat that flows through
a wall is proportional to its area and the temperature difference from one
side to the other and its thermal conductance. about 100 years later, georg
ohm said the same about electricity: v = ir, ie a current i in amps times a
resistance r in ohms produces a voltage difference v. electrical conductance
is measured on "mhos," which is ohms spelled backwards (or "siemens" which
is snemeis spelled backwards :-) if 6 amps flows through a 2 ohm resistor,
we'll see v = ir = 6ax2ohms = 12 volts across it. another example: 120 volts
across 48 ohms makes i = e/r = 120v/48ohms = 2.5 amps flow, with electrical
power p = ie = 120vx2.5a = 300 watts.

exercise 2.1: if 24 a flows through a 12 v headlamp, what's the resistance?

3. thermal ohms?

ohm's law for heatflow (aka newton's law of cooling) uses temperature instead
of voltage difference. heatflow is measured in units of power, in watts or
btu per hour. there's no such thing as a thermal "ohm." the closest thing is
the us "r-value" stamped on foamboards and batts in hardware stores.

beadboard (expanded white polystyrene coffee cup material) has an r-value of
4 (ft^2-f-h/btu) per inch. blue or pink or green styrofoam board has r5 per
inch. so does air, for downward heatflow. the slow-moving indoor and windier
outdoor air films near a single layer of glass have a combined r-value of
about 1. a smooth square foot in slow-moving air loses about 1.5 btu/h-f,
with a u1.5 or an r2/3 airfilm resistance. a rough square foot in v mph air
loses about 2+v/2 btu/h-f. larger objects have lower film conductances. 

tiny cold soap bubbles (1/16" at 50 f) have an r-value of about r3 per inch.
bill sturm's calgary greenhouse filled the space between two polyethylene
film covers with air during the day and soap bubble foam insulation at night
and measured an 82% propane energy savings with and without the foam on -20 f
nights, without much solar energy storage. he thinks about heating poly film
refugee shelters this way in a cold climate.

fiberglass is r3.5/inch, or half that, if it contains 2% moisture (which it
might, in a cold climate, with some holes in a vapor barrier), or even less,
if air flows around or through it, in very cold places. but r-values are not
quite thermal ohms. to find the thermal resistance of a wall, we need to
divide the r-value by the wall area. an 8'x10' r20 wall has a resistance
of r20/(8'x10') = 0.25 f-h/btu. we might call this "0.25 fhubs" or "4 buhfs."
we can add buhfs for different kinds of house walls and windows, ie resistors
in parallel, and add fhubs for series layers of wall insulation.

exercise 3.1: if 6 btu/h of heat power flows through a wall with
a 12 f temp diff, what's the wall's thermal resistance in fhubs?

exercise 3.2: what's the thermal conductance of an 8'x16' r10 wall?

if it's 70 f indoors and 30 f outdoors, (70f-30f)/0.2f-h/btu = 200 btu/h
of heat power will flow through 1 ft^2 of r5 wall. but stud walls have
lower stud resistances (r1/inch "thermal bridging") in parallel with the
insulation resistance and heat collecting stud fins on each side of the wall
in the form of sheathing and drywall, which lowers the overall r-value.

ignoring the fins, we can look at a 10.66 ft^2 16"x8' wall section with r13
insulation framed with r1 per inch 2x4s on all edges as a "stud conductor"
with 2(16/12+8)1.5/12 = 2.33 ft^2 of surface in parallel with a 10.66-2.33
= 8.33 ft^2 "insulation conductor." the studs have a 1/(r1x3.5") = u0.2857
u-value and the insulation has 1/r13 = u0.0769. the combined conductance
is 2.33x0.2857+8.33x0.0769 = 1.306 btu/h-f, for an effective r-value of
10.66ft^2/1.306buhf = r8.2, vs r13, without the studs.

structural insulated panels (sips, glued plywood-foam-plywood sandwiches)
have less thermal bridging and fewer air leaks. so do walls with plywood
i-joists used as studs and loose-fill insulation (as used by me al eggen, 
ktl14@mindspring.org) and walls with a layer of foamboard insulation over
the studs. an 8'x24' r24 6" sip wall with a 40 f temperature difference
would allow (70f-30f)8'x24'/r24 = 320 btu/h of heatflow.

metric (european, canadian, australian...) u-values are 5.68 times bigger
than us u-values. a metric u1 window has 1 w/m^2c of thermal conductance.
it's like a low-loss us r5.68 window. a 1 m^2 metric u1 window with 20 c
air on one side and 0 c air on the other would pass (20c-0c)1m^2x1w/m^2c
= 10 watts of heat power. 

exercise 3.3: how much heat flows through a 2mx6m metric u0.5 wall with
a 20 c temperature difference?

exercise 3.4: how much heat would flow through a 3'x4' us u0.25 window
with 70 f air inside and 30 f outside?

exercise 3.5: how much heat flows through a 24'x32' r64 ceiling with
a 40 f temperature difference?

4. a three-dog house?

now we can talk about superinsulated houses. they need to be small, like
sleeping bags. can a person heat a house with the help of a few dogs? 
an ashrae- (american society of heating, refrigeration and ac engineers)
standard 154 pound 19.6 ft^2 human at rest makes about 350 btu/h, like
a 100 w light bulb...

exercise 4.1 how big can an l' r10 cube be, if it's 70 f inside and
30 f outside, with six outdoor faces, and it's heated by one person?

with r20 walls, l = sqrt(350/12) = 5.4'. better, but cramped...

exercise 4.2: how big can it be with the help of a 100 btu/h dog? with
two dogs? with three dogs? with three dogs and r50 strawbale walls?

if internal heat gain grows with volume, superinsulated apartment houses
have minimum sizes. if a family of 8 1.65 btu/h ashrae-standard 0.046 lb
mice in a 6" cubical room make 8x1.65 = 13.2 btu/h-ft^3, what's the smallest
cubical "mouse motel" with r20 walls that can stay 70 f on a 30 f day? if
13.2l^3 = (70-30)6l^2/r20, l = 12/13.2 = 0.91', with 48 mice in 6 rooms.

exercise 4.3: how big is l, with r10 walls?

germans build "passive houses" like this (for people), with a "passive
house institute" that certifies them. jackson lab in bar harbor, maine
is heated by 972,000 mice, even on -11 f cloudy days :-) a 6.61 pound
ashrae-standard cat makes 68.02 btu/h. a 50.0 pound normally-active
ashrae dog makes 354.9 btu/h. a 1,000 pound jersey calf makes 4,000 btu/h.
larger creatures with lower surface-to-volume ratios produce less heat
per pound. in "the return of the solar cat" (patty paw press, 2003) pe
jim augustyn mentions doe's secret $1.6 trillion 100 megaton solar cat
demonstration project "designed to mimic a solar cat's basic method for
capturing energy on a grand scale." 

a frugal 100 kwh per month of indoor electrical use (vs the us average of
833 kwh/mo) would add 100kwhx3412/(30dx24h) = 474 btu/h to a person's 300
btu/h, making an internal heat gain of 474+300 btu/h = (70f-30f)6l^2/r20,
which makes l = sqrt(774/12) = 8.03 feet, for a cube with r20 walls. with
r50 strawbales, 474+300 btu/h = (70-30)6l^2/r50 makes l = 12.7 feet max.

exercise 4.4: if n families each use 774 btu/h in 1024 ft^2 of floorspace
inside an l' r20 cube, what's l, min?

but some people define a "solar house" as "one with no other form of heat,"
not even electrical usage or creatures-in-residence... 

5. the sun, and weather

suppose the r20 8' cube has an a ft^2 r2 window that admits 200 btu/h-ft^2
of sun. if 200a = (70f-30f)(a/2+(6x8'x8'-a)/r20) = 18a+768, we can make it
70 f indoors on a 30 f day with an a = 4 ft^2 window, but the temperature
would instantly drop to 30 f when night falls, unless the cube contained
some thermal mass. of course the sun doesn't put a constant 200 btu/h-ft^2
into the window all day. this is a crude but useful approximation. 

exercise 5.1: what's the steady-state 24-hour temperature of an 8' r20 cube 
with a *huge* thermal mass and a 4 ft^2 window with 80% solar transmission,
if 1000 btu/ft^2 falls on the window over a long string of average 30 f days?

keeping it warm at night requires a larger south window. the nrel (national
renewable energy lab) solar radiation data manual for buildings (the "blue
book" at http://rredc.nrel.gov) implies that january is the worst-case month
for solar house heating in boulder, co, when 1370 btu/ft^2 of solar heat
falls on a south wall on an average 29.7 f day, and 1370/(65-29.7) = 38.8.
december is the worst-case month for solar house heating in portland, or,
where solar heating is harder, with 470/(65-40.2) = 19.0. january is the
worst-case month in albuquerque, where solar heating is much easier, with
1640/(65-34.2) = 53.2.

pe norman saunders says the "worst-case month" is the one with the least
solar heat per degree day, on average. this is usually december or january.
aerospace engineers often use "worst-case" specs instead of saying "i know
you are going to be very happy with this airplane" or "it will go very far
and very fast and carry a lot of weight." :-)

people often wishfully overestimate passive solar performance, saying things
like "we only burn a few cords of wood per year." how vague. what's "a few"?
what kind of wood? a 4'x4'x8' cord of oak with 20% moisture content burned
at 60% efficiency delivers 158 therms of heat (158x10^5 btu), about the same
as 158 gallons of oil. white cedar can only deliver 72 therms. and wood is
ongoing work, compared to solar heat. was it a cold year or a warm year or
an average year?

what's the solar heating fraction of this house?

   every day is sunny for george and charlotte britton of lafayette hill.
   their 2,900-square-foot house is blessed with energy bills 20 percent
   lower than one of comparable size... the design incorporates "passive"
   solar principles. there are large double pane windows and sliding glass
   doors on the south side. inside, tile floors and a trombe wall absorb
   the sun's heat during the day and radiate it at night... a fireplace
   on the south wall of the living area provides additional heat during
   colder months... britton says "we have a fire every day of the winter."

solar houses with no other form of heat are 100% solar-heated, with no doubt.
then it's just a matter of comfort, or temperature swing. you might buy an
inexpensive 100% solar-heated tent with a -20 to 120 f temperature swing :-)
a house that can keep itself warm in the worst-case month should do fine in
other months. should america's cup boats have outboard motors? should solar
architects pay clients backup heating fuel bills, if any?

nrel's "blue book" contains long-term monthly average solar weather data
for 239 us locations. nrel's typical meteorological year (tmy2) hourly data
files can be useful for simple solar heating spreadsheet simulations with
equations from high-school physics. pe howard reichmuth says a house that
can pass a 1 year tmy2 simulation with no backup fuel requirement can also
survive a 30-year hourly simulation for the same location. but a calculator
seems like enough, with simple worst-case monthly average weather data. can
we connect these basic insights and understandings with leed criteria which
seem to have more to do with feng-shui than fossil fuels? people cared more
about heating bills in 1975.

i wish we could circulate hot air between a house and sunspace and stop the
airflow at night with an energy-10 simulation, but it looks like that's not
part of the present package (available from www.sbicouncil.org) or future
energy-10 improvements. i still think of sbic (aka psic) as crooked masonry
salesmen out to cripple performance of solar houses by loading up sunspaces
with thermal mass, but people can change.

an 8' r20 cube might be 70 f (eg 75 at dusk and 65 at dawn) over an average
boulder december day with an a ft^2 r2 window with 80% transmission that
admits 0.8x1370 = 1096 btu/ft^2. if 1096a/24h = (70-30)(a/2+(6x64-a)/r20)
= 18a+768, a = 28 ft^2. some thermal mass and a 4'x8' window would do, with
a cube conductance of 32ft^2/r2+(6x64-32)/r20 = 34 btu/h-f, about half window
and half walls, like an early los alamos labs passive solar test cell.

exercise 5.2: what's a in portland, or (where 470 btu/ft^2
falls on a south wall on an average 40.2 f december day)?

6. thermal mass and direct gain, aka "direct loss"

water is cheap and heavy, but easily moved. when moving, it has a low thermal
resistance. a square foot of 1" drywall (a "c1" board foot) stores 1 btu/f,
like a pound of water. a 95 lb cubic foot of dry sand with a 0.191 btu/f-lb
specific heat can store 0.191x95 = 18 btu/f. adding 40% water would increase
this to about 24 btu/f-ft^3 and let it support something above. moving water
would greatly lower its r0.4 per inch resistance. a cubic foot of concrete
stores about 25 btu/f, like wet sand, vs 64 for a cubic foot of water or
steel, but concrete's r0.2 per inch resistance makes it hard to move heat
in and out. a 32 pound 8"x8"x16" hollow concrete block with lots of surface
has a specific heat of 0.16 btu/f-lb and stores about 0.16x32 = 5 btu/f. 

an r fhub cube outside c btu/f of thermal capacitance has a "time constant"
rc = c/g in hours (dimensionally, (f-h/btu)/(btu/f) = h.) in rc hours, the
indoor-outdoor temp difference decreases to about 1/3 (e^-1) of its initial
value. if rc = 24 hours and t(0) - 75 f, t(h) = 30+(75-30)e^(-h/24) after h
30 f hours and t(d) = 30+(75-30)e^(-d) after d days, where e^x is the inverse
of the natural log ln(x) function on a casio fx-260 calculator. t starts at
75 f when e^(-0) = 1 and ends up at 30 f (the outdoor temp) much later,
when e^(-oo) = 0. between those times, the exponential factor gradually
squashes the initial temperature difference of (75-30) = 45 degrees.

exercise 6.1: what's t(d) after 1 day?

a liter of water weighs 2.2 pounds, so it stores 2.2 btu/f. n 2-liter bottles
inside the cube would make rc = 4.4n/34/24 = 0.00539n days. starting at 75 f,
t(d) = 30+(75-30)e^(-d/0.00539n) = 30+(75-30)e^(-185d/n) d days later. when
65 = 30+(75-30)e^(-185d/n), ln((65-30)/(75-30)) = -185d/n, and n = 738d.

storing heat for 1 day requires 738 2-liter bottles, 2 days takes 1496, and
so on. if clear and cloudy days were like coin flips, storing heat for 1 day
would make the cube's solar heating fraction 50%, with 75% for 2 days, 88%
for 3, 94% for 4, and 97% for 5. the chance of 5 cloudy days in a row would
be like the chance of 5 tails in a row, ie 2^-5 = 0.03. real weather has more
persistence, like a 3-state (clear, average, cloudy) markov chain.

but 738x5 = 3,690 is 4 pickup trucks full of bottles, and 16,236 pounds of
water is a good ballast foundation :-) and pet bottle walls leak water vapor.
they might need topping up once a year. we can fit about 9 4" diameter by 12"
long bottles into a cubic foot, so they would occupy 3690/9 = 410 ft^3 of the
cube's 8^3 = 512 ft^3, ie 80% of the living space, rather intrusively. this
fraction shrinks with larger cubes with larger potential heat-storage volume
to heat-losing surface ratios.

exercise 6.2: what's the time constant of an "empty" 8' cube with a 1/2"
layer of drywall inside r20 foamboard insulation? if it's 70 f indoors
and 30 f outdoors when we turn off the furnace at 10 pm, what will the
temperature be at 7 am? what would it be if we add 1000 concrete blocks? 

exercise 6.3: what would it be if we filled it with water?

7. indirect gain and night insulation

why look out a black window at night? with an r20 wall between the cube and
a "low-thermal-mass sunspace" containing a south window, we can circulate
warm air between the sunspace and living space during the day and stop the
air circulation at night, so we can have the best of both worlds, the daytime
gain and views and of the window without nighttime and cloudy-day heat loss,
and the window and thermal mass can be smaller, compared to direct gain, but
it's hard to store solar heat from warm air, compared to mass in direct sun,
because of the high airfilm resistance at the mass surface.

exercise 7.1: can we use less than the 3690 bottles in section 6, with
indirect gain? how many are needed to store heat for 5 cloudy days?

we can stop air circulation at night with a one-way passive plastic film
damper hung over a vent hole in the r20 wall, with a mesh that only allows
the film to swing open in one direction. i think doug kelbaugh invented this
"7-cent solution" in princeton in 1973. drew thinks he described it in the
proceedings of the first passive solar conference in albuquerque. pe norman
saunders says they need inspecting every two weeks for folded or torn or
hung-up films, but automatic systems are nice, in general. over the years,
people usually stop moving movable insulation that requires twice-a-day
people power, even the outdoor shutter over the wall of steve baer's house.

gary reysa  (who should be
at this workshop) built and tested a fine barn heater with plastic film
dampers, described at http://users.montanadsl.net/~reysa/.

david delaney  invented (on paper) a "flow-organizer"
that might replace a plastic film damper. it would allow a downgoing cool
indoor airstream near a window to cross an upgoing warm airstream behind
an absorber mesh.

for room temperature control, the damper might be in series with an automatic
foundation vent like leslie-locke's $12 8"x16" afv-1b. its louvers open as
air temperature rises, but the bimetallic coil spring that opens them can be
reversed to close the louvers as air temp rises, and we can adjust its soft
threshold temperarture by turning the spring mounting screw. nasa satellites
use "deep-space coolers" that open to radiate heat as needed.

a very low power motorized damper and thermostat might control the room air
temp more accurately. honeywell's $50 6161b1000 damper motor uses 2 watts
when moving and 0 watts when stopped. if it runs 1 minute per day, that's
0.03 wh. we might power it with a 10 milliwatt pv cell and a rechargeable
battery and some low-power electronics. drew says tamarack tech and thermal
technology (john schnebley) made a pv-powered shutter like this. rich komp
would like to help make them in nicaragua.

fans use more power, but they can raise the solar collection efficiency of
a sunspace, compared to natural thermosyphoning air, and a fan can work with
smaller vent holes with less conduction loss to the sunspace at night.

8. airflow and heatflow

one btu can raise the temperature of 55 cubic feet of air 1 f, so 1 cubic
foot per minute (cfm) of airflow with a temperature difference of 1 degree
moves about 1 (60/55) btu per hour of heat. ashrae says a person needs 15
cfm of outdoor air to stay healthy. if this air just leaks through a house,
that adds about 15 btu/h-f to the house thermal conductance, unless the house
has some sort of air-air heat exchanger that preheats incoming cold air with
outgoing warm air.

bill shurcliff proposed attaching a "lung" (picture a giant bellows) to
the outside of a house, with a fan that periodically inflates and deflates
it with house air, thus turning all the cracks and crevices in the house
envelope into bidirectional heat exchangers with latent heat recovery, as
in a camel's nose. an "infinite virtual lung" might divide a house into two
partitions with a fan between them that periodically reverses. this could
be efficient, done slowly, with lots of heat exchange area. (an exhaust fan
or stack-effect chimney with scandinavian-style "breathing walls" seems less
efficient, with no heat recovered from exhaust air.) 

if every 10'x10' wall section has a 4x10'x6"deepx1/32"wide perimeter crack
with 40/32/12 = 0.104 ft^2 of cross-sectional area, a 4096 ft^2 envelope
would have 4096/100x0.104 = 4.27 ft^2. if 30 cfm flows through it, the air
speed would be 2x30/4.27 = 14 fpm. if each section has 40x0.5x2 = 40 ft^2
of heat exchange surface, cmin = 30 and ntu = au/cmin = 4096/100x40x1.5/30
= 82 and e = 1-exp(-82). very close to 100% efficiency :-) 

lbnl tested a finite lung in the 80s, but the lung volume was small compared
to the stud cavity volume, so there was little fresh air exchange. they aren't
able to test it now, since their funding was cut by a factor of four. someone
else might, with a humidistat and grainger's $73.25 4tm66 16" reversible fan
(61 watts at 3240 cfm on low speed) and their 2a179 $88.15 programmable cycle
timer and its $4.37 5x852 octal socket.

what is the mean size and standard deviation of house envelope cracks?
a few big holes or cracks could make this work poorly. are envelope cracks
small in volume all the way through, or do they admit air to a large stud
cavity? what about freezing and condensation?

most us houses leak a lot more than 15 cfm/occupant. an old house might leak
2 air changes per hour (ach), eg 2x2400x8/60 = 640 cfm for a 2400 ft^2 one-
story house. a new us house might leak 1 house volume per hour. an "energy
star" house might leak 0.5 ach. nisson and dutt's superinsulated house book
suggests a 0.2 ach target. jeff christian at oak ridge has built some full-
size 0.04 ach test houses. people measure air leaks by pressurizing and
depressurizing a house to 50 pascals (0.00725 psi) with a blower door. they
measure the airflow in cfm and divide that by 20 to estimate the natural
air leakage in wintertime.

exercise 8.1: what's the thermal conductance of a 30'x40'x8' tall house
with r30 walls and an r60 ceiling and 8% of the floorspace as r4 windows
and 0.5 ach?

http://www.odpm.gov.uk/stellent/groups/odpm_buildreg/documents/page/odpm_
breg_600356.pdf (with a line-wrap) is a 1999 uk report. page 2 says a 1981
canadian housing development holds the world record for low air leakage.
page 5 mentions the world's tightest voluntary standard, canada's ideas
(post r2000) 0.15 m^3/h per m^2 of envelope, tested at 50 pa, which makes
for a natural air leakage of about 2.5 cfm, or 0.008 ach for a 2400 ft^2
1-story house :-)

   the blower door as we know and love it today springs from technology
   first used in sweden in 1977, where it was actually a blower window.
   the idea migrated to the united states with ake blomsterberg, who came
   to princeton university to do research in 1979...

   the princeton researchers decided to mount the fan in a door because door
   sizes are more uniform than windows... with the help of the blower door,
   the researchers found that hidden leaks accounted for a greater proportion
   of air leakage in a home than the more obvious culprits, such as windows,
   doors, and electrical outlets--a giant leap forward in our understanding
   of how a house operates...

   a monster blower door is being used to test large residential and
   commercial buildings in canada. the super sucker is a whopping 55,000 cfm
   fan that is 40 ft long and 5 ft in diameter. it is transported to the site
   on a flatbed trailer, and it takes a team of five people [with safety
   belts?] to hook it up to a pair of double doors and perform the test.

                      from: http://hem.dis.anl.gov/eehem/95/951109.html

one empirical formula says an h foot chimney with a ft^2 vents at the top
and bottom and an average temp ti (f) inside the chimney with an outside temp
to has q = 16.6asqrt(hdt) cfm of airflow, where dt = ti-to. the heatflow in
the airstream is qdt = 16.6asqrt(h)dt^1.5 btu/h, approximately.

a square foot of r1 sunspace glazing with 90% transmission might gain 0.9x1370
= 1233 btu over 6 hours on an average december day. it could be one layer of
clear flat replex polycarbonate plastic, which comes in 0.020"x49"x50' rolls
and costs about $1.50 per square foot, or one layer of corrugated dynaglas
"solar siding," which costs about $1/ft^2. both last at least 10 years. we
might make an 8' long x 8' radius quarter-cylindrical sunspace with 3 $2 1x3
beams on 4' centers. i've made these beams by bending 2 12' 1x3s into an 8'
radius, held together with 1x3 spacer blocks and deck screws every 2'.

this could be a food and heat producing greenhouse like bill yanda's or
tom lawand's or pe howard reichmuth's ecotope greenhouse near seattle
(which has a steep transparent south wall and a reflective parabolic north
wall that concentrates sun into a water trench), but greenhouses tend to
be more humid than houses, and they need to stay warmer at night to avoid
freezing plants, and it's hard to insulate them at night and still provide
light for the plants. filling the space between two plastic film covers
with air during the day and soap bubbles at night is one way to do this. 
engineer bob quist in toronto has developed a standard "replacement foam
insulation" system for venlo glass greenhouses (the most popular brand
in the netherlands.)

with a ft^2 of sunspace glazing, we can keep an r20 cube 70 f on an average
day if 1233a = 6h(70-30)a/r2 (the daytime sunspace loss) + 18h(70-30)a/r20
(the nighttime sunspace loss) + 24h(70-30)(6x64-a)/r20 (the loss from the
rest of the cube), which makes a = 16.4 ft^2. a 4'x5' window would do, with
a cloudy-day cube conductance of 6x64/20 = 19.2 btu/h-f.

to size the sunspace vents, we might figure that 0.9x250 = 225 btu/h-ft^2
of peak sun enters a square foot of r1 sunspace glazing and 70 f air near
the glazing (on the south side of a dark screen north of the glazing, with
warmer air north of the screen) loses (70-30)1ft^2/r1 = 40 btu/h, for a net
gain of 185 btu/h-ft^2, or 3.7k btu/h (1.1 kw) for 20 ft^2. if 70 f room
air enters the sunspace through the lower vent and exits into the house at
say 120 f, through the upper vent, the average temp inside the sunspace is
97.5 f, and 3.7k btu/h = 16.6asqrt(8')(97.5-70)^1.5 makes a = 0.55 ft^2.
we might use 1 ft^2 vents with an 8' height difference. 

9. less mass with more a larger temp swing?

we might warm ceiling mass (like "pugging" with soil in scotland) with
hot air from a sunspace. ignoring sunspace vent and ceiling resistances,
on an average january day in boulder, our 8' r20 cube with a ceiling at
temperature t (f) and 8'x8' of r1 sunspace glazing might collect 1233x64
= 78912 and lose about 6h(t-30)64ft^2/r1 from the glazing during the day
+ 18h(65-30)64ft^2/r20 from the glazing at night + 24h(t-30)64ft^2/r20
from the ceiling + 24h(65-30)3x64ft^2/r20 from the other walls at night,
which makes t = 137 f, if gain equals loss on an average-day.

if the ceiling conductance to slow-moving air below is 1.5x64 = 96 btu/h-f,
and the cube needs (70-30)19.2 = 768 btu/h of heat, the ceiling must be at
least 70+768/96 = 78 f to provide this. if the average temp is (137+78)/2
= 107.5 f over 5 cloudy days, the cube loses about 5x24h((65-30)4x64ft^2/r20
+ (107.5-30)64ft^2/r20) = 83.5k btu of heat, so we need 83.5/(137-78) = 1416
btu/f of ceiling mass, eg 1416/64 = 22.1 pounds (4.15") of water per square
foot of ceiling.

or less, using a more accurate differential equation. at ceiling temp t,
i = (t-30)64ft^2/r20 + (65-30)4x64ft^2/r20 = 3.2t + 352 btu/h flows from the
ceiling mass c, so dt/dt = -i/c = -3.2t/c - 352/c, and -352/3.2 = -110, so
t(t) = -110 + (t(0)+110)e^(-3.2t/c). after 5 days, with t(0) = 137 f, t(120)
= 78 = -110 + 247e^(-384/c) makes ln((78+110)/247) = - 384/c, so c = 1407
btu/f, or 22.0 pounds (4.12") of water per square foot of ceiling. 

exercise 9.1: how many inches of water are needed in december in portland,
when 470 btu/ft^2 falls on a south wall on an average 40.2 f day, for an
8' r32 cube with an 8'x8' r2 sunspace window with 80% solar transmission? 

keeping the heat in the ceiling allows the cube to be cooler when vacant, 
so stored solar heat can last longer, provided we reduce the ceiling's
downward heatflow by radiation...

exercise 9.2: how might this change if the cube only needs 8h/day of heat?

solar pioneer harold hay built houses with roofponds and movable insulation. 

10. radiation

gustav robert kirchoff (1824-1887) said all radiation hitting a surface
must be transmitted, absorbed, or reflected. radiation obeyed. t+a+r=1.
kirchoff's identity says emissivity equals absorptivity for gray surfaces
with constant emissivity that doesn't depend on wavelength.  

a surface emits se(t^4) of heat flux by radiation, where t is an absolute
temperature in rankine (f+460) or kelvin (c+273) degrees, and s is the stefan-
boltzman constant, 0.1714x10^-8 btu/ft^2-r^4 or 5.660x10^-8 w/m^2-k^4, and 
e is the surface's "emissivity," which varies from 0 (shiny) to 1 according
to shininess. most natural surfaces are close to 1, but mirrorlike surfaces
have emissivities close to 0.

the net heatflow from a surface at t1 degrees facing a t2 degree surface
is se(t1^4-t2^4). for example, a square foot of 80 f trombe wall might
lose 0.1714x10^-8((80+460)^4-(30+460)^4) = 47 btu/h by radiation on a 30 f
night. in still air, it might lose 1.5(80-30) = 75 btu/h by radiation and
convection combined. a "selective surface" can absorb well at short solar
wavelengths (<3 microns) and radiate poorly at longer heat wavelengths
(10 microns at 80 f.) a 100% solar-heated house might have thick sips and
radiant floors and a big enough well-insulated hot water tank to store heat
for 5 cloudy days in a row and lots of $12/ft^2 vertical unglazed stainless
steel es solar roof collectors with durable selective surfaces from
www.energie-solaire.com. they might be 0.959-8.91(0.375)-0.047(0.375)^2
= 62% efficient at an average collector temp of 30 c and a 0 c outdoor temp
in full sun (800 w/m^2.)

exercise 10.1: what's the temperature of a surface
with e = 0.2 in 300 btu/h-ft^2 sun on a 20 f day?

   the most serious mistake was making the outer container of the receiver
   of plywood. we thought that the plywood would be sufficiently insulated
   from the copper panel which was the receiver proper, that it would not
   get too hot. the copper panel was separated from the plywood by 4" of
   fiberglass insulation. nevertheless, the plywood caught fire and the unit
   was completely destroyed. we suppose this is a success, of sorts.

   the copper panel which was plated with chrome black to provide a
   selective surface originally had a copper tube fastened to the back by
   a high melting point soft solder. when we first attempted to operate
   the unit, the soft solder melted, and the tubing became detatched from
   the panel. we attempted to repair this failure by silver soldering
   copper bars to the copper tube and screwing the bars to the plated
   copper sheet. this worked, after a fashion...

   the glass window was originally tempered glass. this shattered due to
   thermal shock. we replaced it with ordinary window glass. this cracked
   due to thermal shock, but we were able to hold it in place well enough
   to make some measurements... the measurements we were able to make
   before the fire generally confirmed our thinking concerning the design...

   from "a solar collector with no convection losses," (a downward-facing 
   receiver over a 4:1 concentrating parabolic mirror), by h. hinterberger
   and j. o'meara of fermilab, in "sharing the sun," a joint ases/sesc
   conference, august 15th-20th, winnepeg, volume 2, pp 138-145. 

if tb is their average temperature, the "linearized radiation conductance"
between two surfaces is 4setb^3. a 96 f ceiling exposed to a 70 f room with
tb = 543 r has 4setb^3 = 1.097 btu/h-f-ft^2, roughly r0.9. if our cube needs
(70f-30f)19.2btu/h-f = 768 btu/h, e(96-70)64x1.097 = 768 makes e = 0.42. we
might make about 60% of the ceiling a low-e (0.05) foil surface and the rest
an ordinary surface and let radiation warm the room on an average day and use
a slow ceiling fan and a thermostat to bring warm air down on cloudy days.
a 72 f ceiling would supply 0.42(72-65)64ft^2x1.097 = 206 btu/h of radiant
heat. the other 672-206 = 466 btu/h might come from q cfm of airflow.
viewed in a fixed font: 

        1/96  1/q                              (96+q)/(96q)
72 f ---www---www---65 f   equivalent to   72------www------65
       ----------->                            ----------->
        466 btu/h                               466 btu/h

where (72-65)96q/(96+q) = 466, so q = 217 cfm. grainger's $120 48" 315 rpm
86 w 21k cfm 4c853 ceiling fan might move 217 cfm at 217/21kx315 = 3.3 rpm
with 86(217/21k)^3 = 100 microwatts, according to the fan laws :-) large slow
fans can be very efficient and quiet. the gossamer ceiling fan developed by
danny parker at the florida solar energy center can move 1907 cfm with 9.1
watts, so it could move 217 cfm with 1 watt, running 11% of the time.

11. even less mass?

on an average day, we might only store 13.8k of overnight heat in c btu/f
in the ceiling, with t(6) = 120+(72-120)e^(-6x96/c) = 120-48e^(-576/c) and
(t(6)-72)c = 18,432, so c = 384/(1-e^(-576/c). c = 384 on the right makes
c = 494 on the left, and plugging that in on the right again makes c = 558,
596, 620, 635, 644, 649, 653, 655, 657, and 657 (whew!), so it looks like we
can store overnight heat with 657/64 = 10.3 psf (about 2") of water above
the ceiling, with t(6) = 120-48e^(-576/657) = 100 f. 

the 8' cube needs 4x24(70-30)19.2 = 73,728 btu for 4 more cloudy days. this
might come from a "solar closet" (see paper on cd) inside a sunspace (the
heat lost from the closet air heater glazing efficiently ends up in warm
sunspace air that heats the cube :-) with about 73,728/(120-70) = 1475 lb
or 184 gal. or 23 ft^3 of water cooling from 120 f to 70 f over 4 days.

kallwall's "solar battery" was an early solar closet. tom hopper built an
insulated box containing fiberglass water cylinders with double-glazed
south wall and an insulating curtain (5 layers of aluminized mylar) that
automatically rolled down over the glazing at night. bill shurcliff (in his
1980 brick house thermal shutters and shades book) said this remarkable 
mylar curtain (with spacers that unfurled when it deployed) was invented
by the insulating shade co. of branford cn, and demonstrated at a 9/9/77
exhibition in hartford, cn, and featured in the january 1979 issue of
popular science monthly. 

with 1792 pounds of water in 56 10"x10"x13" 4-gallon ropak plastic tubs,
we can supply 73,728 btu as it cools from 120 to 120-73728/1792 = 78.9 f,
stacking the tubs 7-high and 4-wide and 2-deep in 2'x4'x8' tall closet that's
completely surrounded by insulation, with an air heater with its own closet
vs sunspace glazing over the closet's insulated south wall and one-way dampers
in that wall. with 56x4x10x13/144 = 202 ft^2 of tub surface and 300 btu/h-f
(buhfs) of water-air thermal conductance, we have diagram 1 again with a
78.9-70 = 8.9 f temperature difference and a (300+q)/(300q) resistor, so
q = 121 cfm.

with an 8' height difference and 121 cfm = 16.6asqrt(8'(78.9-70)),
we need 2 vents with a = 0.86 ft^2 for natural airflow into the cube.

as an alternative, the cloudy-day heat might come from 3072 pounds of water
inside a 2'x4'x8' tall "shelfbox" with a 2'x4'x2' tall water tank below 18
2'x4' wire shelves on 4" centers, with 2" of water inside a $20 continuous
piece of poly film duct folded to lay flat on each shelf and a small pump to
circulate tank water up through the duct as needed. the tank might have a
pressurized tank inside to make hot water for showers, with the help of an
efficient external greywater heat exchanger, eg 300' of 1.17" od plastic pipe
pushed into two coils inside a 35"x23.5"" id 55 gallon drum. the outer coil
can be longer, with 27 turns and a 23.5-1.17 = 22.3" diameter and a 5.85'
circumference and 27x5.85 = 157' of pipe. if the inner coil perfectly nested
inside the turns of the outer one, its diameter would be 20.3", with a 5.31'
circumference and 143' of pipe. pushing the pipe around inside the drum is
awkward but doable with two keystone kops trying to avoid kinks.

12. a greywater heat exchanger 

about 15 years ago, eric olsen at earthstar manufactured a heat exchanger
with a copper coil in a 55 gallon drum to recover heat in kidney dialysis
centers. drew says professor jane davidson at u minnesota is an expert on
plastic heat exchangers. she says "the efficiency you calculate is actually
effectiveness which is very different from efficiency. a high effectiveness
does not mean a good heat exchanger!" i wonder what she means by that.

i'm using a new $35 55 gallon lined steel drum with a strong removable lid
(because the drum might end up under 2' of greywater head, with the inlet
and outlet above the lid) and bolt ring and a 3/4" bung and a 2" bung with
a 3/4" threaded knockout, with 100 psi/73.4 f pipe from pt industries at 
(800) 44 endot. their pbj10041010001 1"x300'100psi nsf-certified pipe is
tested to 500 psi. the price is $59.99 from any true value hardware store.
lowes sells the rest of the hardware. it's all installed through the lid,
so the drum itself has no holes:

sales        total
#       qty  price   description

25775    1   $5.73   24' of 1.25" sump pump hose (for greywater i/o)
105473   1    1.28   2 ss 1.75" hose clamps (for greywater hose) 
54129    2    3.24   1.25" female adapter (greywater barb inlet and outlet)
23859    2    2.36   1.25x1.5" reducing male adapter (bulkhead fittings)
75912    1    0.51   2 1.25" conduit locknuts (bulkhead fittings)
28299    1    1.53   2 1.25" reducing conduit washers (")
22716    1    1.36   1.5" pvc street elbow (horizontal greywater inlet)
23830    1    2.98   10' 1.5" pvc pipe (for 3' greywater outlet dip tube)

the parts above are greywater plumbing ($18.99.)

23766    4    1.28   3/4" cpvc male adapter (for 1" pipe barbs)
23766    2    0.64   3/4" cpvc male adapter (fresh water i/o)
42000    2    3.84   3/4" fip to 3/4" male hose adapter
23813    1    1.39   10' 3/4" cpvc pipe (for 1"x3/4" fresh water outlet)
23760    2    0.96   3/4" cpvc t (fresh water i/o)
22643    2    0.86   3/4" cpvc street elbow (fresh water i/o)
         4      -    1" 3/4" cpvc pipes (fresh water i/0)
         1      -    3' 3/4" cpvc pipe (fresh water inlet)
22667    2    2.56   2 ss 1.125" hose clamps (fresh water i/o)
219980   1    4.87   10.1 oz dap silicone ultra caulk (bulkhead fittings) 
150887   1    3.94   4 oz primer and 4 oz pvc cement

parts above are fresh water plumbing. subtotal $39.33.

26371    1    6.83   1500 w electric water heater element
22230    1    2.31   1" galvanized t ("nut" for heating element)
61294    1   11.76   single element thermostat with safety
136343   1    0.56   5 10-24x3/4" machine screws (mount thermostat with 3)
33368    1    0.37   5 #10 ss flat washers (mount thermostat with 3)
198806   1    1.38   10 #0 rubber faucet washers (mount thermostat with 3)
8763     1    0.67   5 10-24 ss nuts (mount thermostat with 3)

the above would make a standalone water heater, if needed. grand total: $63.21.

for 4 10 min showers per day and 20 minutes of dishwashing at 1.25 gpm we
might heat 75 gallons of 55 f water to 110 with 8x75(110-55) = 33k btu with
about 10 kwh worth about $1/day at 10 cents/kwh. if the "heat capacity flow
rate" cmin = cmax = 75gx8/24h = 25 btu/h-f and the pipe coil has a = 300pi/12
= 78.5 ft^2 of surface with u = 10 btu/h-f-ft^2 (for an hdpe pipe wall with
slow-moving warm dirty water outside and 8x300pi(1/2/12)^2 = 13 gallons of
fresh water inside), the "number of heat transfer units" for this counterflow
heat exchanger ntu = au/cmin = 78.5ft^2x10btu/h-f-ft^2/25btu/h-f = 31.4, so 
the "efficiency" e = ntu/(ntu+1) = 97% for hot water usage in bursts of less
than 13 gallons. this works best with equal greywater and cold water flows, 
with either a 110 f water heater setting (preferable), or the heat exchanger
output feeding the cold water shower inlet as well as the water heater.

the hazen-williams equation says l' of d" smooth pipe with g gpm flow has a
0.0004227lg^1.852d^-4.871 psi loss. at 1.25 gpm, the pressure drop for 2 150'
coils of 1" pipe might be 0.0004227x150x(1.25/2)^1.852x1^-4.871 = 0.03 psi.

if greywater leaves a shower drain and enters the heat exchanger at 100 f and
fresh water enters at 50 f, the fresh water should leave at 50+0.97(100-50)
= 98.5 f. warming it further to 110 f would take 8x75(110-98.5) = 6.9k btu/day
with 2 kwh worth about 20 cents/at 10 cents/kwh, for a yearly savings of about
($1-0.20)365 = $292, or more, with a tighter shower enclosure and higher drain
temperature. the 1500 w heater might operate 2kwh/1.5kw = 1.3 hours per day.
we might wrap the drum with 3.5" of fiberglass and a 4'x8' piece of foil-
foamboard with 7 4' kerfs (knife cuts partially through the board) to make
an octagon and aluminum foil tape to cover the kerfs and hold it closed. 

13. other promising solar house heating schemes

harry thomason's trickle collectors were used in hundreds of houses. pump
water up to the ridgeline of a metal roof and trickle it down under a glass
cover to a gutter, then down to a large tank on the ground buried in rocks
to help with water-air heat distribution. he also trickled water over bare
asphalt shingle roofs at night for evaporative cooling. dick bourne at davis
energy systems also does this sort of thing, and lives in a roofpond house. 

these days, a hydronic floor with a low water-air thermal resistance in
an airtight house with lots of insulation might be simpler and use less
power for heat distribution. this system needs lots of collector pump power,
and it looks like the roof cover has to be polyethylene film or glass,
since polycarbonate quickly degrades in warm water vapor. we might inflate
a polyethylene film cover over a dark asphalt roof when water is flowing
between them or it's very windy outdoors (>15 mph) or the roof is very hot
(>150 f), with straps over the poly film to prevent outward ballooning.

trickle collectors can only heat water to 80 or 90 f on cold winter days,
so heat storage and distribution tend to be inefficient. an efficient solar
heating system might keep a house warm on an average day with a hydronic
floor and low temp water from a trickle collector and lots of insulation,
and keep the house warm for 5 cloudy days with some higher temp water from
a well-insulated and seldom used cloudy-day store. 

exercise 13.1: what's the min floor temp required to keep a house with a
200 btu/h-f conductance and a 2400 ft^2 radiant floor 70 f on a 30 f day?

zomeworks may soon have a more efficient "double-play" system with plastic
tubes under closely-spaced metal roof standing seams, with a polycarbonate
cover or a selective surface. dawn solar in new hampshire is also working
on something like this.

zomeworks "cool cells" connect water pipes under the ceiling with roof ponds
or radiators that lose heat by natural thermosypnoning at night, which stops
during the day. a system with water in weaker flat poly ducts under the
ceiling might use a 10 watt fountain pump to keep a roofpond full at night
vs thermosyphoning. a roofpond that's dry during the day would consume less
water and kill biological growths with solar heat.

donald wright's habitat for humanity houses near safford arizona use large
hydronic solar collectors with fiberglass window screen between two hypalon
rubber layers for more uniform water flow. but they pump the warm water under
a slab beneath lots of sand with a high thermal resistance, which seems like 
a big mistake, thermally-speaking. adirondack houses also work this way, but
they seem to be mostly wood-heated. kachadorian's solar slab is a good idea,
but i doubt it would work with thermosyphoning air, as he claims. and with 
no insulation above, it can't be much warmer than 70 f without overheating
the occupants. rich komp calls his earlier slab a hypocaust, as in hollow
roman wood-fired floors.

a thomason "pancake house" might have a draindown polyethylene film roof pond
and an underfloor poly film pillow for heat distribution. a "concentrating
solar attic" might have a transparent steep south roof and a north roof that
approximates a parabola with 4 or 5 line segments (we want to avoid line foci)
aimed slightly above the southern horizon to reflect 2-3 suns down into a
4'-wide water trough along the attic floor near the north wall. the trough
might be 30" round polyethylene film greenhouse air duct (about 40 cents per
linear foot) that lays flat to 48", with 1-2" of water inside during the day.
this might lay on top of standard photovoltaic panels on the attic floor. we
need careful layout or some sort of optical mixing for uniform illumination
to avoid inefficient shadows and bright spots on the pvs. cooled to 120 f,
they should have a long lifetime under 2-3 suns. your milage may vary.

don booth built a beadwall attic with waterbed heat storage in concord, nh.

14. natural cooling

less electrical usage helps, eg reflected sunlight at 200 lumens per watt
instead of fluorescents at 50 lumens per watt. these days, the illuminance
of a well-lit building is 50 footcandles (fc)...

  to increase the "demand" for more power plants, the power and lighting
  industries were very effective in promoting ever-increasing illumination
  levels and incorporating them into codes. in some cases "required" light
  levels were so high that artificial lights had to be used all the time,
  regardless of the external climate and availability of natural light.
  such lights often heated buildings to the extent that air conditioning
  was needed year-round. 
                           from sunlighting as formgiver for architecture,
                           by william m. c. lam, van nostrand, 1986

...we might provide 50 fc = 50lumens/ft^2x64ft^2 = 3200 lumens with 3200/50
= 64 watts of fluorescents or 16 watts of sunlight. bright sun is about 8000
fc, so we only need about 50/8000x64ft^2 = 0.4ft^2 of window for this, ie 0.6%
of the floorspace as windows, with no transmission or reflection loss, if the
sun could be distributed uniformly, or 2%, counting losses and nonuniformity.
or 4%, to have enough light on overcast days.

our 8' cube might have a 1.28 ft^2 clerestory window 8' long and 2" tall on
the south side of the roof near the peak, with a reflective roof underneath
and an overhang that's reflective underneath and a reflective north ceiling. 
we might want taller windows with overhangs to admit all beam sun at noon on
the winter solstice and exclude it all on the summer solstice.

at noon on 12/21 in the northern hemisphere, sun elevation emin = 90-lat-23.5
degrees. at noon on 6/21, emax = 90-lat+23.5 degrees. a horizontal overhang
that projects p feet from a south wall d feet above the top of the glass of an
h foot tall window can completely shade the window on 6/21 and admit all sun
on 12/21 if tan(emin)=d/p and tan(emax)=(h+d)/p, so p=h/(tan(emax)-tan(emin))
and d = ptan(emin). for example, at 40 n. lat, emin = 26.5 degrees and emax
= 73.5, so an h = 8' tall window needs a p = 2.7' overhang d = 1.2' above
the top of the glass. 

shading and light-colored walls help. plants on trellises come to mind.
walls and roofs have more insulation than windows, but they also have much
more surface. some mobile homes have old tires on flat roofs. it pays (@$1)
to collect old tires where i live in pa. a simple green roof might have weeds
growing in gravel over epdm rubber. it might cool better and use less water
without the weeds, which only cool air. attics and sunspaces need venting.

night ventilation can help, if a house has lots of internal thermal mass
and lots of insulation. ventilate with cool night air and button the house
up during the day and let the thermal mass keep it cool. nrel says july is
the warmest month in boulder, with average 73.5 f days and daily lows and
highs of 58.6 and 88.2 and an average humidity ratio w = 0.009. this is the
number of pounds of water vapor per pound of dry air. it is more constant
over a day than the relative humidity (rh), which depends how much water
the air can hold, which depends on the air temperature. 

"comfort" depends on temperature, humidity, air velocity (faster is cooler),
activity (sleeping vs wrestling) and clothing (three-piece suits vs shorts and
a t-shirt.) the ashrae comfort zone relates temperature and humidity ratio.
experiments have found that people in developed countries are "comfortable"
from about 67 to 81 f, with w = 0.0045 to 0.012. t = 89.4-1867w is a "constant
comfort" line diagonally down through the zone. with w = 0.009, t = 72.6 is 
most comfortable. can we keep our cube in the zone with night ventilation?

we might put n 2-liter water bottles inside the cube and remove the roof
to make a "cold trap" at night with no lower temperature limit, or put
a motorized damper near the top that lets warm air flow out of the cube
every night until the inside air temp drops to 70.6 f.

if we model an average boulder july day as a square wave with 12 hours at
(58.6+73.5)/2 = 66 f followed by 12 hours at (73.5+88.2) = 81 f (not very
close to the real sine wave, but maybe more useful than solar architects'
rules of thumb), we have c = 4.4n btu/f in series with its 1.83n thermal
conductance to slow moving air and a 19.2 bhuf cube-to-outdoor conductance
that's shunted at night with a q cfm airflow conductance.

                 1/19.2         1/(1.83n)    | |
81/66 ------------www--------------www-------| |---|
         |                 |                 | |
         |        1/q      |               
          ---/ ---www------              4.4n btu/f

as the bottles warm, 74.6 = 81+(70.6-81)e^(-12/rcc) makes the charge time
constant rcc = 24.7 h = 4.4n/19.2, approximately, so n = 108 (recall kim
basinger's weight in "batman"?) and c = 475 btu/h. as the bottles cool,
70.6 = 66+(74.6-66)e^(-12/rcd) makes rcd = 19.2 h. each bottle has about
1.2 ft^2 of surface with a 1.5x1.2 = 1.8 bhuf conductance, so 108 have a
198 bhuf conductance or a 5.06 millifhub thermal resistance. if rcd = 19.2,
the total series resistance r to outdoors is 19.2/475 = 0.0404 f-h/btu,
and 1/q = 0.0404-0.00506 = 0.035 fhub. q = 28.6 cfm = 16.6asqrt(8'(70.6-66))
makes a = 0.28 ft^2 for natural airflow. we could cool 2" of water above
the ceiling with a gable vent or a thermal chimney above a flat roof. 

we could also use a fan (an exhaust fan that moves the motor heat outdoors.)
in pablo laroche and murray milne's tiny ucla test houses with thermal mass,
a "smart whole house fan controller" turned on a fan when outdoor air was
cooler than indoor air. "enthalpy economizers" do this for large buildings,
but seldom for houses. with a humidity sensor, we might heat as well as cool
a house by ventilation, and avoid condensation on mass surfaces, and bias
the house temperature into the upper part of the comfort zone in wintertime
and the lower part in summertime in order to store more heat or coolth in
the house mass. brand electronics may soon sell such a fan controller.

we might also cool 2" of water above the ceiling with a roof pond, as in a
zomeworks thermosyphoning "architectural cool cell." phil niles says a t (f)
pond in to (f) air loses 1.63x10^-9((t+460)^4-a(to+460)^4) btu/h-ft^2 by
radiation, where a = 0.002056tdp+0.7378, with a dew point temp tdp (f). with
v mph of wind, it also loses qc = (0.74+0.3v)(t-to) btu/h-ft^2 by convection,
and it evaporates qe = b(t-twb)-qc, where b = 3.01(0.74+0.3v)((t+twb)/65-1),
and twb (f) is the wet bulb temperature.

the pond radiates more heat to the sky if the air contains less moisture,
with a higher dew point, since water vapor is a "greenhouse gas" that blocks
radiation. air at the dew point temperature is saturated with water vapor.
the relative humidity is 100%. we can find the approximate dew point by first
finding the vapor pressure of water in air on an average july day in boulder
with humidity ratio w = 0.009. pa = 29.921/(1+0.62198/w) = 0.427 inches of
mercury ("hg--29.921 "hg is 1 atmosphere.) then we use a clausius-clapeyron
approximation (don't ask) to find the temperature corresponding to that
pressure, at 100% rh. if pa = e^(17.863-9621/tdp), ln(pa) = 17.863-9621/tdp,
so tdp = 9621/(17.863-ln(pa)) = 514 r or 54 f in boulder in july. 

this makes a = 0.7399 in the formula above, so a 70.6 f pond in 58.6 f night
air would lose 1.63x10^-9((70.6+460)^4-0.7399(58.6+460)^4) = 42 btu/h-ft^2.
nrel says the average july windspeed in boulder is 8.1 mph, so a pond would
lose (0.74+0.3x8.1)(70.6-58.6) = 38 btu/h-ft^2 by convection. an 8'x8' roof
pond would lose 8x8(42+38) = 5120 btu/h by radiation and convection, more than
a 5,000 btu/h window air conditioner. keeping the cube 72.6 f on an 81 f day
in boulder only requires 12h(81-72.6)19.2 = 1935 btu, so we don't need to
evaporate water from this roof pond. it might have a polyethylene film cover,
since poly film is essentially transparent to radiation.

the dew point temperature only depends on the amount of water in an air
sample, vs its temperature. a little water inside a perfectly-insulated cup
might find itself at the dew point temp, as water evaporates from the surface
and water vapor molecules diffuse to the top of the cup, while the air above
the water acts as a good insulator (about r5 per inch for downward heatflow.) 
which water depth will keep it coolest in an r5 3" diameter x 6" tall cup?  
as we fill the cup, the air layer insulates less, but more water evaporates,
according to fick's law, because the concentration gradient increases as the
air layer thins...

when the water gets to the top of the cup, we'd expect to see it at the wet
bulb temperature, when its heat loss by evaporation equals its heat gain by
convection. a perfect swamp cooler would make air at this temperature. in
1926, i.s. bowen said a pond's ratio of heat loss by evaporation to heat gain
by convection equals 100(pp-pa)/(tp-ta), regardless of windspeed. this ratio
is -1 at the wet bulb temp. with pa = 0.009, and ta = 58.6 f, and tw (r),
100(e^(17.863-9621/tw)-0.427) = 58.6+460-tw, so tw = 9621/(22.47-ln(561.3-tw)).

the wet bulb temp is easy to find on a calculator. plugging in tw = 518.6 r 
(58.6 f) on the right of the equation above makes tw = 514 on the left. doing
this again makes tw = 517, 515, 516.2, 515.6, 515.9, 515.7, and 515.8 (55.8 f),
between the dew point and dry bulb temperatures. so an uncovered roof pond
would lose qe = b(t-twb)-qc, where b=3.01(0.74+0.3x8.1)((70.6+55.8)/65-1)
= 9.01 and qe = 9.01(70.6-55.8)-38 = 95 more btu/h-ft^2 by evaporation, like
a 10k btu/h window ac. this could be useful in phoenix. one simple ashrae
swimming pool formula says q = 100(pw-pa) btu/h-ft^2, regardless of air temp.
pw = e^(17.863-9621/(460+70.6)) = 0.764 "hg and pa = 0.427 "hg makes q = 34
btu/h-ft^2.

indoor misting with a vent fan for outdoor air works like a swamp cooler,
when 4500(wi-wo)>to-ti. the left side is the amount of latent cooling per
cfm of vent air. the right is the amount of sensible heating per cfm of
vent air. as they become equal, swamp cooling and misting no longer work.

the ashrae comfort zone has an efficient corner at ti = 80.2 f and
wi = 0.012 (at 56% rh), which makes to < 134.2-4500wo. nrel says phoenix
has an 88.2 f average temp in june, with an average daily min and max of
72.9 and 103.5 and average humidity ratio wo = 0.0056 pounds of water per
pound of dry air in june, so misting should help until the outdoor temp
rises to to = 134.2-4500(0.0056) = 109 f.

evaporating a pound of water takes about 1000 btu, and air weighs about
0.075 lb/ft^3, so a house with no internal heat gain or unshaded windows
and a 200 btu/h-f conductance that's 80.2 f indoors with wi = 0.012 when
it's 88.2 f and wo = 0.0056 outdoors needs 1000x60c0.075(0.012-0.0056)
= 28.8c = (88.2-80.2)(200+c) btu/h of cooling, including cooling c cfm of
air from 88.2 to 80.2, so c = 77 cfm (not much), with 4.5c(0.012-0.0056)
= 2.2 pounds or 0.27 gph of water, which might come from a $5 0.5 gph
mister mister nozzle and a solenoid valve scrounged from an old washing
machine in series with a 80.2 f room temp thermostat and a $5 humidistat
that turns on a $12 window exhaust fan when the rh rises to 56%. 

we can make 77 cfm move through the house with a ft^2 vents with an 8' height
difference if 77 = 16.6asqrt(8(88.2-80.2)), ie a = 0.58 ft^2, eg a 1 ft^2 vent
to the attic, to keep it cooler. the 2.2 pounds of water per hour might come
from plants in an indoor greywater wetland. with enough plants, the thermostat
may never turn on the mister. in an air-leaky house, the humidistat may never
turn on the fan.

herbach and rademan (800) 848-8001 http://www.herbach.com sell a nice
brass $4.95 navy surplus humidistat, item number tm89hvc5203, with a
20-80% range, a 3-6% differential, and a 7.5a 125v switch that can be
wired to open or close on humidity rise.

pw = 1.033 "hg at 80.2 f and 100% rh, and pi = 0.566 "hg with wi = 0.012.
ashrae says a square foot of pool evaporates 100(pw-pi) = 47 btu/h, so we
can evaporate 2.2 pounds of water per hour with 2200/47 = 47 ft^2 of wet 
surface, counting droplets in air, or less, if we evaporate close to the
house air inlet. masonry floors and walls or an indoor cylindrical rock
gabion with a fountain pump (in a toilet tank?) with high thermal conductance
and mass might allow more efficient cooling with cooler night air and keep
the house mass cool with the fan off during the heat of the day. the gabion
might build up an interesting mineral encrustation that helps evaporate water.

exercise 14.1: how many gph and cfm and square feet are needed if the house
has 1000 btu/h of internal heat gain and 400 btu/h-f of conductance and
outdoor air enters the top of a gabion and leaves via the sides? 

in arid places like phoenix, why not distill the water we drink
in a greenhouse and filter the rest through a high-temp wetland?

nrel says 1020 btu/ft^2 of sun falls on the ground and 1550 falls on a south
wall on an average 53.6 f january (the worst-case month for solar heating)
day in phoenix. a 16'x16'x8' tall greenhouse would get 0.8(256x1020+128x1550)
= 368k btu of sun. distilling 10 gallons of potable water at 1,000 btu/lb
(or a lot less, with steve baer's multi-effect still) would leave 368k-83k
= 284k btu. with 600 ft^2 of r1.2 surface, the greenhouse conductance would
be 500 btu/h-f, so 284k btu could keep it 53.6+284k/(24hx500) = 80 f (27 c)
on an average day.

we might put a outer layer of plastic film over 4 16' long x 8' tall outer
bows and another layer over 4 inner bows, with a slot at the top to allow
warm moist air to enter the spaces between the glazings and condense water
into east and west troughs on the ground made from a fold in the outer layer,
with a slot in the inner layer above the trough to allow air to recirculate.

if 80 f air enters the top at 100% rh with wt = 0.0226 and 53.6 f air leaves
the bottom at 100% with wb = 0.0088, distilling 10 gallons in 24 hours means
24x60c0.075(wt-wb) = 1.48c = 10x8.33, so c = 56 cfm = 16.6asqrt(8x(80-53.6)),
for an upper vent area a = 0.23 ft^2, eg 3 4" holes. the inner films might
have 3 2" holes near the ground. the troughs might also collect rainwater,
if any. the greenhouse might have a gravel bed over an epdm rubber layer
to make an artificial wetland.

metcalf and eddy say an average family produces 400 gallons per day of
wastewater with a 200 mg/l bod concentration, ie about 0.7 lb/day, and it
needs about 175 ft^2 of surface for treatment. sherwood reed says blackwater
contains more than 75% of the nitrogen in only 40% of the flow. eliminate
that with a composting toilet or solar poop cooker, and we might only treat
240 gallons per day (0.908 m^3/day) with 0.175 lb of bod, ie 86 mg/l.

can we purify this to 10 mg/l (a tertiary sewage treatment standard)
and recycle it for non-potable uses? pe sherwood reed's natural systems
for waste management and treatment book (mcgraw hill, 2nd edition, 1995)
has a detailed design procedure for constructed wetlands, aka "attached-
growth biological reactors." they work well at higher temperatures.

equation 6.38 on page 226 of reed's book says the rate constant for
bod removal in subsurface-flow wetlands at 20 c is k20 = 1.104d^-1.
equation 6.34 makes k27 = 1.104x1.06^(27-20) = 1.66d^-1 at 27 c. we
need to reduce the bod from co = 86 to ce = 10 mg/l with 256 ft^2
(23.8 m^2) of surface. if the bed is "medium gravel" (d10 = 32 mm)
with a 0.38 porosity, it needs to be 0.908ln(co/ce)/(k27x23.8x0.38)
= 0.13 meters deep, ie 5.1 inches. we might add gravel or divide
the bed into 3 series cells for better plug flow and performance.

what will grow in this rainforest? orchids? mangos? bananas? :-)

if we must use an air conditioner, why not mist the air around the hot coils 
or trickle rainwater over them? d. y. goswami at u. florida measured a 20%
gain in ac efficiency as a result of misting the air around the hot coil.
i've measured 20% by trickling rainwater over a window unit.

a more sophisticated cooling system with no water consumption or outdoor air
exchange might have a multi-effect licl solar still on the roof that absorbs
water vapor at night from the pond below and distills water out of the licl
solution during the day. this might also store cloudy day heat, with a wet
basement floor to evaporate water in wintertime, and a licl pond on the first
floor that acts as a "chemical heat pump." 

nick

nicholson l. pine                      system design and consulting
pine associates, ltd.                                (610) 489-1475
821 collegeville road                           fax: (610) 831-9533
collegeville, pa 19426                     email: nickatece.vill.edu

computer simulation and modeling. high performance solar heating and
cogeneration system design. bsee, msee, sr. member, ieee. registered
us patent agent. web site: http://www.ece.vill.edu/~nick 

answers to exercises

0.1: if the current gulf war costs $600 billion per year, and we consume
20millionx42gx365d = 307 billion gallons per year, a gallon of oil costs
600billion/307billion = $1.95 plus the pump price. i've just heard 38,000
people per year die prematurely because of pollution from oil burning. at
$1 million per death, that would add $38 billion, ie another 12 cents...
what's the cost of acid rain? drew says low pump prices that only reflect
part of the real costs send the wrong market message to consumers, who
would conserve more energy if they paid the whole cost at the pump.

0.2: c) $20 worth of caulk might halve the air leaks and the heating bill.

1.1: c = 1/2 pound x 1 btu/lb-f = 0.5btu/f and dt = 200f-50f, so we need
cdt = 75 btu, if it's done quickly, so most of the heat stays in the
water and little heat flows into the room.

1.2: 75x60m/h/(300x3.41) = 4.4 minutes.

2.1: 12v/24a = 0.5 ohms.

3.1: r = t/i = 12f/6btu/h = 2 f-h/btu (fhubs.)

3.2: 8'x16'/r10 = 12.8 btu/h-f (buhfs.)

3.3: 20cx2mx4mxu0.5 = 80 watts.

3.4: it would pass (70f-30f)3'x4'x0.25btu/h-f-ft^2 = 120 btu/h of heat.
an r8 hurd heat mirror window would only pass 60 btu/h-ft^2. c-sash makes
low-e storm windows...

3.5: 40fx24'x32'/r64 = 480 btu/h.

4.1: 350 btu/h = (70f-30f)6l^2/r10 makes l = sqrt(350/24) = 3.82 feet, max.

4.2: with 1 dog, l = sqrt(450/12) = 6.1'. with 2, l = sqrt(550/12) = 6.8'
...3 make l = 7.4'. with r50 strawbales, 650 btu/h = (70-30)6l^2/r50 makes
l = 11.6 feet. an lxlx8' tall "cube" has 650 = (70-30)(2l^2+4x8l)/r50, so
l^2+16l-406.25 = 0, and l = (-16+sqrt(16^2-4(-406.25))/2 = 13.7'.

4.3: l = 24/13.2 = 1.8'

4.4: 774l^3/8/1024 = (70-30)6l^2/r20 makes l = 127' min, eg a 128'
16-story cube, something like the borg, or a large anthill. 

5.1: if 0.8x1000x4ft^2/24h = 133 = (t-30f)(4/2+(6x8'x8'-4)/r20) = 12(t-30),
t = 30 + 133/12 = 41 f. 

5.2: if 0.8x470 = 376 btu/ft^2-day of solar energy comes in the window and
376a/24h = 15.7a = (70-40.2)(a/2+(6x64-a)/r20) = 13.4a + 572 btu/h leaves,
a = 249 ft^2, which is larger than the 64 ft^2 south wall of the cube, so
we can't design a direct gain 100%-solar-heated house in portland this way,
without more insulation or mice or people or indoor electrical energy use.

6.1: t(1) = 30+(75-30)e^(-1) = 46.6 f.

6.2: c = 6x8x8/2 = 192 btu/f. g = 6x8x8/20 = 19.2 btu/h-f. rc = c/g
= 10 hours. t(9) = 30+(70-30)e^(-9/10) = 46.3 f. adding 1000x5 btu/f makes
c = 5019 and c/g = 261 h and t(9) = 30+(70-30)e^(-9/261) = 68.6 f, which
is good for direct gain, but bad for night setback savings. 

6.3: rc = 8x8x8x64/19.2 = 1707 hours makes t(9) = 69.8 f. t(5 days) = 67.3,
and after 2 weeks, t = 67.0. this is a good direct gain house, for a fish. 

7.1: on a cloudy day, with no sunspace airflow, the r20 cube in section 5
with an r20 wall between the cube and the sunspace would have a conductance
of 6x64/r20 = 19.2 vs 34 btu/h-f, with rc = 4.4n/19.2/24 = 0.00955n days.
starting at 75 f, t(d) = 65 = 30+(75-30)e^(-5/0.00955n) 5 days later, so
-523.6/n = ln((65-30)/(75-30)), and n = 2083.

8.1: the 0.08x2400 = 192 ft^2 of windows have 192ft^2/r4 = 48 btu/h-f.
the 2(30+40)8-192 = 928 ft^2 of walls add 928/30 = 31, the ceiling adds 
2400/60 = 40, and 0.5x2400x8/60 = 160 cfm, or about 160 btu/h-f, over
10x the 15 cfm/occupant ashrae ventilation standard. the total is 279,
of which 57% comes from air leaks and 17% comes from windows. why do
houses need windows?

9.1: ignoring sunspace vent and ceiling resistances, an 8' r32 cube
with a ceiling temp t (f) and 8'x8' of r2 80% sunspace glazing might
collect 0.8x470x64 = 24064 btu and lose about 6h(t-40.2)64ft^2/r2 from
the glazing during the day + 18h(65-40.2)64ft^2/r32 from glazing at night
+ 24h(t-40.2)64ft^2/r32 from the ceiling + 24h(65-40.2)3x64ft^2/r32 from
the other walls at night, which makes t = 121.9 f, if gain equals loss
on an average-day.

if the ceiling conductance to slow-moving air below is 1.5x64 = 96 btu/h-f,
and the cube needs (70-40.2)4x64/32 = 238 btu/h of heat (less than 1 human),
the min ceiling temp is 70+238/96 = 72.5 f. the average is (121.9+72.5)/2
= 97.2. if the cube loses 5x24h((65-40.2)4x64ft^2/r32+(97.2-40.2)64ft^2/r32)
= 37.49k btu over 5 days, we need 37.49k/(121.9-72.5) = 758.9 btu/f, eg
758.9/64 = 11.9 pounds (2.22") of water per square foot of ceiling.

or less, using the differential equation. at ceiling temperature t,
i = (t-40.2)64ft^2/r32 + (65-40.2)4x64ft^2/r32 = 2t + 118 btu/h flows
from c, so dt/dt = -i/c = -2t/c - 118/c, and -118/2 = -59. after 5 days,
t(t) = 72.5 = -59 + (121.9+59)e^(-2x120/c), so c = 240/ln(131.5/180.9)
= 752 btu/f, ie 752/64/64x12 = 2.20" of water.

9.2: i = (t-40.2)64ft^2/r32 + 8/24(65-40.2)4x64ft^2/r32 = 2t-14.3
btu/h flows from c, so dt/dt = -i/c = -2t/c + 14.3/c, and 14.3/2 = 7.1.
t(t) = 72.5 = 7.1 + (121.9-7.1)e^(-2x120/c) makes c = 240/ln(65.4/114.8)
= 427 btu/f, ie 427/64/64x12 = 1.25" of water.

10.1: 300 btu/h = 0.1714x10^-8x0.2(t^4-(460+20)^4) makes t = 982 r or 522 f.

13.1: i = (70-30)200 = 8k btu/h. t = 70 + i/(2400x1.5) = 72.2 f.

14.1: if 28.8c = 1000+(88.2-80.2)(400+c), c = 202, with 4.5c(0.012-0.0056)
= 5.8 pounds or 0.70 gph of water. pw = 1.033 "hg at 80.2 f and 100% rh, and
pi = 29.921/(0.62198/0.0056+1) = 0.267 "hg with wi = 0.0056 and 100(pw-pi)
= 77 btu/h-ft^2 make 5800/77 = 76 ft^2 of wet surface. a 3" sphere has about
4pir^2 = 0.196 ft^2 of surface, or 0.196x4^3 = 12.6 ft^2/ft^3, so we need 7.1
ft^3 of 3" rocks, eg a 2' diameter x 3' tall hollow gabion with a $10 10 watt
fountain pump and a float valve in the bottom.




I got ALL of these 85 Solar Panels for FREE and so can you.  Its in our Ebook
Ready for DOWNLOAD NOW.

Site Meter