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From: peien@bud.peinet.pe.ca (PEI Environmental Network)
Newsgroups: misc.rural
Subject: Re: Passive Solar Heating/alternative energy
Date: 5 Mar 1994 00:30:12 -0400
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I'm in the midst of building a passive solar house. It was designed by Don
Roscoe, a solar architect from Nova Scotia. His designs have been featured
in Harrowsmith magazine (Canadian version) 3 times. I don't have the issue
numbers handy, but can get them if you're interested. I will get 60 - 70 %
of my heat from the sun. His houses are no more expensive to build than any
other house and obviously, I highly recommend his designs. The magazine
articles give details on how the houses work.
Sharon.
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Newsgroups: alt.solar.thermal,alt.architecture
Subject: Re: solar house information
Date: 28 Feb 1995 04:15:13 -0500
>...here are the 4 basic solar systems:
>DG, Direct gain, which means sun-facing glass into the living space with
>thermal storage inside. Simplest and cheapest, but...
A poor performer, often reducing the backup heat for a house by less
than 30%, or even increasing the backup heat requirement in very
cloudy areas. Movable insulation helps, but you still have to live
inside the heat battery...
>AG, Attached greenhouse...
A low-thermal-mass sunspace, NOT a greenhouse full of plants, containing
thermal mass, is a better performer, by as much as a factor of three.
>TW, Trombe wall, a system developed by a Frenchman in which a thick
>concrete wall is placed on the sun-side with glass on the outside.
Another poor performer, compared to a low-thermal-mass air heater on
an insulated south wall. I put Trombe walls on my house 20 years ago,
when I didn't know any better.
>WW, Water wall, in which water-filled prisms (oil drums, plexiglass
>cylinders, etc) are just inside the sun-glass and act as thermal mass.
Again bad, in cloudy areas. If the water is just inside the glass, the heat
will leak out through the glass all night and during cloudy day periods.
>H2O is _the_ best storage mass (in efficiency, at any rate.
Agreed. Water is also cheap, and it can be easily moved around.
>Another option is less direct, in which you run pipes through solar
>panels and use that water to heat the house (radiant or baseboard or
>whatever);
Very expensive, and those panels up on the roof have lots of thermal
losses, which do nothing to heat the house, compared to a water heater
in the sunspace. And they may freeze, or heat up to 400 degrees if the
pump fails, ie during collector stagnation, which may break the glass covers.
It is better to have a solar closet as a deliberately stagnant collector,
IMHO.
>another includes a rockbed, in which heat is gathered and stored in a
>rockbed near the house, and the heated air is circulated up to the house
>(thermosiphoning).
How about putting the rockbed INSIDE the house, in an insulated
closet, on the first floor, and using thermosyphoning to heat it from
an air heater in the sunspace in front of it, and using 55 gallon
drums full of water instead of rocks. Drums are free, and last at least
20 years, and they are easy to stack up. Water has three times more
thermal mass than rocks, and the air spaces are bigger, so natural
convection or a small high-temp fan (eg Grainger's 4C861) works better
to heat these "rocks."
>That's a brief summary; just about any solar book will cover these.
I guess we need a new book. Perhaps John Allred will write one...
Nick
Date: 3 Mar 1995 12:39:35 -0500
From: Nick Pine <nick@vu-vlsi.ee.vill.edu>
Subject: Re: Nick's lecture
>I would be interested in finding out more about your lecture and designs,
>I don't suppose that you'll be in the Melbourne FL area at all?
I might get down that way some time. Would you like me to lecture there?
I've only been to Florida a couple of times, once for a business meeting
in a posh Naples hotel (nice sailing :-) and once to visit my aunt in
Homestead, who has since moved closer to Miami. If you paid me to come,
I'd be a lot more likely to show up :-)
>Would it be possible to get copies of your lecture and slides?
Maybe. Most of my lecture is comments on slides of local buildings,
with green film pen markings on the back of each slide to show how they
might be improved with solar and insulation retrofits, and comments
about the costs of materials and numerical estimates of the amount of
energy each retrofit would save. I hand out a few copies of postings,
and tape a few large architectural drawings of my designs on the wall,
and I bring along 3 empty 55 gallon drums, a K-mart radiant kerosene
heater used as an infra-red solar simulator and 3 linear parabolic
reflectors made of Thermo-Ply, foil-faced foam and 3M SA-85 solar
reflective film, to aim at the simulator to make a sort of infra-red
whispering gallery, boiling water in a coke bottle at the focus, a
high temp fan and speed control, a bimetallic-coil-spring-operated
automatic foundation vent and a hair dryer to make it work, after I
adjust it for a higher temp range, 200 ft^2 of shadecloth, 140 slides,
12 viewgraphs, 50 lbs of books, some 1 x 3s, large pieces of Thermo-Ply,
various kinds of thermometers and light meters, a laser level, some 12'
pieces of Dynaglas, etc., etc. So this is not a very portable show,
as I have done it so far. I'll be doing a small and different version
again in Dallas, TX next week, at the Monolithic Dome convention,
and then at 3 or 4 local schools and universities soon after that...
Nick
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Newsgroups: alt.architecture.alternative,alt.solar.thermal
Subject: Re: RADIANT SLAB HEAT IN PASSIVE SOLAR HOUSE
Date: 3 Mar 1995 08:04:53 -0500
>>I need some advice on hot water heat. The insulated slab on my passive
>>solar house is too cold after several days of no sun. The woodstove warms
>>the room, but the floor remains cool. We want the new addition to be more
>>comfortable. Could hot water coils in the slab solve the problem? No
>>contractor in my area knows anything about the subject. Can someone
>>please direct me to some information sources.
>...Now, since you're passive solar already, it
>probably makes a lot of sense to use a drum (solar closet) as your
>boiler (as per Nick Pine's notes); with proper nighttime insulation,
>you'll be able to hold the heat in your drum for days after the sun is gone.
I went a little further than that in the design for a new addition to
my house. It will have a 44' wide x 20' tall sunspace, with a 20' wide x 48'
long, shallow reflecting pool made of EPDM rubber roofing material in
front of that. About 20% of the sunspace glazing is 4' x 6' single-pane
sliding glass door replacement panels mounted on butyl tape with aluminum
cap strips, and 80% of the glazing, eg the sunspace roof, is Dynaglas.
All of the glazing is on a 4x6 frame on 4' centers, so some future owner
can replace the Dynaglas with glass if he/she prefers that. Behind one end
of the sunspace is a new 16' x 16' 2-story room with a balcony, with a
plywood floor with 2 x 4's on 2' centers and 3 1/2" of fiberglass insulation
and Thermo-Ply underneath, and under the balcony is the solar closet/sauna,
with 20 drums behind double glass and a 16' x 8' nightime-insulating/daytime-
reflecting shutter operated by a garage door opener. The center point of
the balcony will be held up with a wood truss, with a kingpost going up to
the center of the roof.
The flat roof of the new room will also be 2x4s on 2' centers, with
1/2" OSB and EPDM rubber and a deck on top, and 3 1/2" of fiberglass
insulation under that, and Thermo-Ply under that, as the finished ceiling.
Under the 16 x 16' plywood ground floor is 3 1/2" of insulation and 32 more
drums, resting on dry clay soil on top of a piece of Tyvek. A foot below that
is a layer of polyethylene and then the undisturbed soil. The drums actually
rest on 4x4 sleepers on staggered 2' centers, running north-south, so air
can flow under the drums perpendicular to their long axis, as well as east-
west thru the circular triangular air spaces. 2 x 4 posts support the plywood
floor 2' above the sleepers, every 4'. A Grainger 4C861 high-temp, low-power
fan with a variable speed control will make all this work, I hope :-)
The east wall of the new room will have some solar siding, in a 2 x 6 wall
with 3 1/2" of insulation and some shadecloth and air gaps, with natural
convective airflow... I am interested to see how this will work. If it
doesn't work very well, I may have to change the design, or get Lee to
come over and do some incantations.
In full sun, this new addition and attic will be receiving solar energy
at a rate of about 566 horsepower, or 422 kiloWatts, not counting the
reflecting pool. I think it will cost about $20K to build. That's about
5 cents per peak watt, for all you photovoltaic people.
>That said, I should point out that Mac Wells is against this type of
>radiant heating, and instead uses sniffer ducts and air tubes through
>the floor -- sucks warm air from the ceiling down into the floor,
>conserving the heat in the air and prolonging the thermal "charge" in
>your slab. No leaks, and your floor stays at a more median temperature.
That's pretty much what I'll be doing, as well, only in more of a closed
system. The attic may be a more open sniffer duct system like that. I
may collect attic heat at the peak and push it down to the house through
the unused chimneys at both ends, with high temp fans. I'll be replacing
my 32' long x 16' slant height, south-facing 45-degree-pitch roof with
Dynaglas. This will be a one good place to put some water heating drums.
I sent all my plans to Mac for review, BTW, with a $95 check, per his
_Designing Your Natural House_ book. His rates have gone up. Last time
he helped me, 20 years ago, I only sent him $20.
>-- Malcolm Wells, who is the nicest guy I've ever met.
He has a nice sense of humor, too.
Nick
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Subject: Solar closets in Alberta
Date: 2 Mar 1995 10:45:36 -0500
Sherwood Botsford was asking how warm the drums in his solar closet might be
in frozen Alberta... Let's find the drumwater temp, Tw, for an average day.
I don't know much about Alberta weather, but when I was there for three days
last year in January, the temp was about -12 F and cloudy, since it snowed
about an inch a day, and nobody talked about this weather at all, and they
kept driving their cars on the highways at 80 mph on the hard-packed snow :-)
On page 851 of my second edition, 1991 copy of Duffie and Beckman's _Solar
Engineering of Thermal Processes_, I see that the average temperature in
January in Edmonton, Alberta is -15 C, or 5 F. (McMurray looks a bit colder.)
The average amount of sun is about 620 Btu/ft^2/day, I think, which is about
60% of the average sun in Phila. Suppose we have a 10' wide x 8' tall x 4'
deep solar closet inside a sunspace, with a 20' long x 8' wide solar closet
collection area on the north side of the sunspace, and that the closet is
airtight, and insulated on the sunspace side with 12" of fiberglass insulation
behind the air heater, and that the closet has 6" of insulation everywhere
else, inside the house. Suppose that the house temp is a constant 68 degrees,
and the sunspace is 68 F when the sun is shining and the passive air heater
is operating, for 6 hours a day, and that the sunspace is 5 F for the other
18 hours each day.
Then:
The amount of daily sun that gets into the solar closet, assuming single-
pane glazing for the outside of the sunspace and double-pane for the
air heater, with 10% transmission loss for each layer of glazing, is
Ein = 8' x 20' x .9 x .9 x .9 x 620 Btu/day = 72,317 Btu/day.
The frontal area of the solar closet itself is 8' x 10', and it has an
R-value on the front side of about 40, and the rest of the closet has
a surface area of 224 ft^2 with R20 insulation, so the front of the
closet has an area/R-value of SARf=80/40 = 2, and the rest of the closet
surface has a SARb of 224/20 = 11.2.
The amount of heat that leaves the solar closet each day is
Eout = Eout(day) + Eout(night), where
Eout(day) = (6 hours)((Tw-5))(8'x20'/R2) + (Tw-68)(SARb))
= 6((Tw-5)(80)+(Tw-68)(11.2)) = 6(80Tw-400+11.2Tw-762)
= 6(91.2Tw-1162) = 547Tw - 6,972,
and Eout(night) = (18 hours)((Tw-5)xSARf + (Tw-68)x11.2)
= (18)(2Tw - 10 + 11.2Tw - 762)
= (18)(13.2Tw - 772) = 238Tw - 13,896,
So if Ein = Eout,
547Tw + 238 Tw = 72,317 + 13,896 + 6,972,
or equivalently, 785Tw = 93,185, from which one can easily deduce that
Tw = 118.7 degrees F.
Hmmm. Not very warm. How about we add an 8' x 20' white or shiny aluminum
or frozen water or snow, horizontal reflector to bounce 80% more sun onto
the south glass of the solar closet?
Then, 785Tw= 72,317x1.8 + 13,896 +6,972, so Tw = 192.4 F.
That's better. Sherwood will doubtless point out at this point that that
8' wide horizontal reflector won't work very well, since 1/tan(14 degrees)
is 4, but then we are just playing here, not actually building something...
Nick
From: "Jason M. Roth" <jr4q+@andrew.cmu.edu>
Subject: Re: solar house information
Date: Mon, 27 Feb 1995 13:44:46 -0500
>: For my son's school project, he must design and build a model of
>: a solar heated house. Can anyone suggest a source of information
>: or plans?
>Check the local library for anything by Malcolm Wells, if you want the
>underground option. Basically you want most of the glass facing the
>equator, perhaps a greenhouse on the south wall. You need mass to store
>the energy, such as a thick stone or concrete wall, or water tanks, or
>Glauber's salt. The heat will be radiated back into the house at night.
>You want earth berms and coniferous trees on the north side to shelter
>from the winter wind. You could also put solar water heating and
>photovoltaic panels on the south side of the roof. The floor plan could
>be similar to a high-rise apartment with all the windows on one side.
>Brian Calvert (mcalvert@spartan.ac.brocku.ca)
Of course it's nice to see someone else recommending Mac, but at least
one point: there's no need for all the windows to be on one side;
indeed, the non-sun side will give the clearest, glare-free views, not
to mention cross-ventilation. To go one more step, here are the 4 basic
solar systems:
DG, Direct gain, which means sun-facing glass into the living space with
thermal storage inside. Simplest and cheapest, but relatively difficult
to control and requiring detailed design.
AG, Attached greenhouse, which means a sun-facing, glassed-in space
adjacent to the living space; this does a number of things, inc.
improving (reducing) infiltration, allowing control (usually vents into
living spaces), providing a greenhouse, and giving privacy.
TW, Trombe wall, a system developed by a Frenchman in which a thick
concret wall is placed on the sun-side with glass on the outside. The
wall gets hot, the air between the wall and the glass gets super-hot
(150+), and vents exist top and bottom to allow air to circulate.
Super-efficient, traditional-looking from the inside, a bit goofy on the
outside.
WW, Water wall, in which water-filled prisms (oil drums, plexiglass
cylinders, etc) are just inside the sun-glass and act as thermal mass.
Kind of weird, but H2O is _the_ best storage mass (in efficiency, at any
rate).
Another option is less direct, in which you run pipes through solar
panels and use that water to heat the house (radiant or baseboard or
whatever); another includes a rockbed, in which heat is gathered and
stored in a rockbed near the house, and the heated air is circulated up
to the house (thermosiphoning).
That's a brief summary; just about any solar book will cover these.
ASHRAE puts out publications that can tell you how much of each system
you need to make up a given # of BTUs in a given climate. Good luck
Date: 8 Mar 1995 08:58:52 -0500
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Organization: Villanova University
> The power required to operate a fan varies with the cube of the amount of
>air moved (usually in CFM, cubic feet per minute).
I think we might add, "for a given duct size." If we make the air ducts large,
and the airspeed low, we can make the dynamic air pressure and duct friction
and fan power as low as we want, practically-speaking. For instance, 100' of
a 2' x 2' duct with a cross sectional area of 4 ft^2, carrying 2,000 CFM and
500 linear feet per minute (lfm) of air, has a back pressure of only about
0.01" of water, according to my 1993 ASHRAE HOF.
This goes well with the performance of the 16" Grainger 4C861 1600 rpm (max),
high-temp (311 F), high-priced ($359), very reliable (3 year guarantee),
low-power (240 watts, max, with no speed control) fan, which is rated to
deliver 2765 CFM in free air, and 2190 CFM at 0.20 inches of water.
And this pressure and fan power is quite low compared to many hot air heating
systems using more powerful blowers with tiny cheap ducts. One way to make big
ducts is to blow air between multiple studs or joists or rafters... So if
we want to minimize fan power, and use fans instead of more energy-consuming
blowers, we should make the ducts big and the air velocities and rpm's low.
The last two things also minimize acoustical noise.
Norman Saunders recommends that new houses have a 2' x 2' airshaft from
top to bottom, so solar heat can be stored upstairs in the winter, and
easily transported downstairs, and cool air from the basement floor can
easily be transported upstairs in the summer. He says this allows easy
air movement of 1 m^3/sec, which transfers 1 kW/degree Kelvin, or was that
10 degrees Kelvin? Let's see. One Btu can heat 55 ft^3 of air 1 degree F...
Nick
Date: 8 Mar 1995 05:05:31 -0500
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Sender: london@sunsite.unc.edu
Subject: Re: An insulated ceiling pond house -Reply
>>So here's a more conventional house with an insulated pond in the ceiling...
> There is a major problem with this in earthquake areas (most of the
>U.S.). A ceiling and/or roof needs to be light weight and act as a
>membrane to hold the walls together. Adding a pond overhead would
>make a dangerous situation as it would add to the forces trying to tear
>the walls apart and, if it fell, would crush those under it.
Thanks for the thought, Robert. I didn't think about that at all. There
is probably some solution, though, like putting the water under the floor,
thus sacrificing natural convection and heat trapping with overhead water,
or designing the building to take earthquakes, eg as a concrete/foam dome
with an integral ferro-cement lenticular diaphram tank above, or putting in
some sort of giant rip cord, like the ones in hot air balloons, that opened up
to let the water out quickly when an earthquake happened...
At the moment, I like the idea of hot water, sewage and rainwater under the
floor, in EPDM rubber/Thermo-Ply tank sandwiches, with a high-temp, low-power
fan to heat it by flowing 500 lfm air underneath...
Nick
Date: 8 Mar 1995 05:34:06 -0500
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Subject: Re: How much glass for solar gain?
> Guidelines for South facing glass:
> Radical: as much as possible
I agree. Well, enough to heat the house 100%, anyway, as a low-thermal-mass
sunspace, with a smaller glazed area inside in front of a high-thermal-mass
solar closet/sauna for cloudy days. And not just glass. Alternative sunspace
glazing include polycarbonate plastic (Dynaglas or Replex) or even good old
5 cent per square foot, 3-year greenhouse poly film...
> Now, forget the guidelines.
Good idea...
> Consider 5-20 gallons water in direct gain per
>square foot of glazing. Too hot, too much temp flux -
>add more mass; too cold - reduce mass, add movable
>insulation. Get Mazria: The Passive Solar Energy Book.
This works pretty well in the southwest, but I'm fairly convinced that
in places with a few cloudy days in a row, direct gain is bad, unless
there is movable insulation. Or at least it is much better to use passive
or low-power active air heaters on the south walls, with a good insulated
wall behind them, and a higher-than-living-space temperature heat battery,
inside the house, that can be well-insulated, vs a masonry floor behind
a big uninsulated picture window in the living room.
Nick
Date: 3 Mar 1995 12:39:35 -0500
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Subject: Re: Nick's lecture
>I would be interested in finding out more about your lecture and designs,
>I don't suppose that you'll be in the Melbourne FL area at all?
I might get down that way some time. Would you like me to lecture there?
I've only been to Florida a couple of times, once for a business meeting
in a posh Naples hotel (nice sailing :-) and once to visit my aunt in
Homestead, who has since moved closer to Miami. If you paid me to come,
I'd be a lot more likely to show up :-)
>Would it be possible to get copies of your lecture and slides?
Maybe. Most of my lecture is comments on slides of local buildings,
with green film pen markings on the back of each slide to show how they
might be improved with solar and insulation retrofits, and comments
about the costs of materials and numerical estimates of the amount of
energy each retrofit would save. I hand out a few copies of postings,
and tape a few large architectural drawings of my designs on the wall,
and I bring along 3 empty 55 gallon drums, a K-mart radiant kerosene
heater used as an infra-red solar simulator and 3 linear parabolic
reflectors made of Thermo-Ply, foil-faced foam and 3M SA-85 solar
reflective film, to aim at the simulator to make a sort of infra-red
whispering gallery, boiling water in a coke bottle at the focus, a
high temp fan and speed control, a bimetallic-coil-spring-operated
automatic foundation vent and a hair dryer to make it work, after I
adjust it for a higher temp range, 200 ft^2 of shadecloth, 140 slides,
12 viewgraphs, 50 lbs of books, some 1 x 3s, large pieces of Thermo-Ply,
various kinds of thermometers and light meters, a laser level, some 12'
pieces of Dynaglas, etc., etc. So this is not a very portable show,
as I have done it so far. I'll be doing a small and different version
again in Dallas, TX next week, at the Monolithic Dome convention,
and then at 3 or 4 local schools and universities soon after that...
Nick
Date: Thu, 23 Feb 1995 13:56:53 -0500
Subject: Solar Hotwater
>I do not think that our situation is unique, there must be others in the
north who have marginal solar in the winter. Do you have any suggestions or
better ways to deal with this problem?
We don't have solar hot water in the summer but in the winter we get a lot of
help from a heat exchanger attached to our woodstove. This circulates
(thermosyphon) to a hot water holding tank, then to an Aquastar that accepts
preheated water.
The man who makes the add-on-boiler is:
Joseph Lyman.
21 Park Street
Rutland, VT
05701
802.773.8133.
Date: 21 Feb 1995 02:18:57 -0500
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
>Many years ago I read about Steve in The Mother Earth News - Steve was the
>driving force behind a company called Zomeworks. Is Steve still out there
>- and is Zomeworks still in business?
Yes :-) Steve BAER still seems to be alive and well and very busy, and even
answers mail sometimes. Altho Zomeworks mostly makes PV-related stuff
these days, which Steve attributes to the fact that "energy is just too
cheap." Or perhaps it's that PV's are trendy or glamorous. (This is
fiddling while the middle-east burns, I think.) Steve's drumwall steel
drums wore out after 20 years, so he has now replaced them with plastic
ones, which he expects will last longer.
Speaking of which, yesterday I saw an ad in a free ad paper around
here that read "Plastic 55 gallon drums, $5 each. Used once to contain
aloe product." And a local meat packing plant called me, in response
to my "drums wanted" ad in the paper. They are now saving me some (free :-)
plastic-coated steel drums, which they use (~40 a month) for soap,
vegetable oil and glycol...
A couple of interesting phone numbers are: (215) 945-0444 for the $144
Cuppson "Lazer level," a low-power semiconductor laser attached to a
carpenter's level, that shoots a 1/8" red dot at least 60' (500' or so
indoors or at night), that can be used to adjust parabolic reflectors
for solar ovens, etc, made from 3M SA-85 film, and 3M's Solar Optical
Products number, (612) 733-1898, for the SA-85 reflective film itself
(when is Real Goods going to carry this stuff?) 3M's address is
3M Center/St. Paul, Minnesota 54144-1000. Their general phone number
seems to be (612) 733-1110.
To adjust your oven, etc. reflector, you might put it in a darkish room and
move the laser level along to make parallel rays, where you want the sun to
be, and look to see where the reflection spot ends up, and how big and how
bright the spot is (the brightness/reflectivity might be measured with a
digital lightmeter, I suppose, eg the $100 Grainger version.) I've noticed
that the 3M film makes a much smaller spot, than say, foil-covered foamboard,
alumimum foil or Thermo-Ply. Yesterday I managed to boil some water in
a pepsi bottle at the focus of a 3M reflector a foot away from my 10K
Btu/hour radiant kerosene solar simulator...
Nicholson Pine
Subject: Re: El Cheapo tracker
Kerry Miller <ASTINGSH%KSUVM.BITNET@cmsa.Berkeley.EDU> wrote:
->> I drilled a hole through each in the center, and welded them
->> into the ends of the 2X2.
->Since its a vapor you're trying to get from one tube to the
->other, wouldn't it be better to have your holes above center?
First, I have better luck when I am not trying to weld shut, an unvented
container. I mean, the material was stout enough to accommodate the
stress, and still have a good weld, maybe.... I never tried. But, there
has to be holes in the tank somewhere...
Actually, the idea is to drive the water/antifreeze mix from one side to
the other. It seemed using solely propane, did not have the mass to
overcome the friction. I went for the *duel* system. The 50/50 mix
approximates the mass of freon. (If it works with freon, it'll work with
water.... The anitfreeze is for rust inhibition and freeze protection.
The setup is basically, a *U* (or a liquid manometer). As the sun
advances past the shade trough it hits the east tank. The temperature
increases. Pick your favorite Law. (Perfect Gas Law PV=nRT, or the
popular combo Datlton/Charles P(i)T(i)=P(ii)T(ii) + P(i)V(i)=P(ii)V(ii).)
The expansion of the propane shifts the mass to the west tank.
It doesn't matter where the holes are, just as long as the fluid can shift
for the whole E/W swing of the gizmo, without gurgling and going out of
balance.
(Now, I don't know if this would work with propane only in the gas state,
or if there is the actual need for some liquid propane in the system.
I haven't thought that out. I'm not ready to labor this out on paper, and
I don't want to burden you with me pondering the matter in such a raw
state...it would be hell to follow, and I'd look stupid.)
-> Another thought: if said holes were located near the angle-
->iron cross braces instead of in the end plates, the copper
->tubing would be better protected. I know the risk is small,
->but I know Mr Murphy well...
-> (OTOH, if somebody reaching
->for the frame "to see if it really moves" cracks the tube, maybe
->it would be appropriate to flood them in propane..)
You are absolutely right. This is a problem. The tube is just sticking out the
re, aching for attention.
I was moving my beast one day, and it slipped (it's heavy!);
I grabbed the tube!, just as the other end hit the ground, absorbing
most of energy. <Whew!>
However, being dowsed with the tank contents is probable only
in a rupture. Most likely, unless someone wails on it, is breaking
the seal at the flare and the contents leak out over a period of a few
minutes to a year (just a wild guess).
I think by moving the lower angle down to protect the tube is a
good idea. This move might even help with weight distribution on
the N/S axis.
->Too, what was your conclusion about the depth of threads for
->the gas fittings?
It seemed the walls of the 2X2 were too thin to accommodate the threads
of the fittings to a comfortable degree.
The rule of thumb is, 5 threads is all that's needed, or some such thing.
The problem is I don't remember if if that refers to straight thread, or
pipe thead. I also don't remember if that's 5 threads *minimum*, or after
5 threads you're just screwing around.
So, I'm just one step ahead of knowing nothing.
->> I took three pieces of plate steel. Welded two of them to the
->> pivoting support, with the third piece in between, bored a hole
->> and ran a bolt. This allows for seasonal adjustment.
->
->I didnt get this at all, or see it in the .jpg... are you saying
-> there is a lateral (N-S) tilt adjustment?
Yes there is a tilt adjustment. However, my last art class was in 7th
grade...let me try to describe it another way.
Take three pieces of cardboard; stack them together; pull the middle
piece out half way; at a point in the center of where these three pieces
overlap, stab a pencil through all three pieces. Grab the middle piece
of cardboard with your left hand, and the other two with your right.
(Or the other way around, it doesn't make any difference.)
What is in your left hand is affixed to the earth,...your right, to the gizmo.
(This Procedure, however, cannot be changed!;-) The arc is your
latitude adjustment, (N/S).
One item that has been on my mind. I would like to find a cheap way to
increase the bearing surface on the pivoting axis. But I run out of room
using 1 1/4" angle irons.
->BTW, what's the significance of the "Flat flies straight..." note?
The *Flat flip flies straight, tilted flip curves. Experiment*...well aside fro
m
being the instructions for a frisbee... means, the thing flies. If you can't
work it one way, try another. (...or, does it mean, it will work one way,
but if your off, it will smash into your neighbor's window?)
Either way, I thought it was pretty good advice, considering it's on the
bottom of a toy...plus, I get silly and obtuse when I work on things late
into the night.
This is, after all, backwoods engineering.
bigdog@netins.net *** With a homepage coming to a URL near you ****
...just a lying dog, chewing on a stick in the mud...
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Subject: On temperature control
Here is a quote from Steve Baer's book, _Sunspots_:
What is it like to live in houses where the temperature changes
during the day? What is it like to do without a thermostat to
control the temperature within a degree or two? It is only very
recently that there have been thermostats for controlling the
temperature in houses, and still today almost everyone alive in the
USA has spent some time in buildings without automatic thermostats.
I believe that it is perfectly satisfactory to have the temperature
change during the course of each day, from a high in the afternoon
to a low in the morning, and to have the temperature change from
week to week according to how cloudy or sunny it is. The variations
in temperature keep your blood circulating.
What extremes of temperature within a house are comfortable? In a
dry climate like Albuquerque, I believe yearly lows and highs of 55 F
and 85 F are perfectly easy to live with inside a house--especially
if you have warm spots such as fireplaces or stoves to stand next to
when it is chilly. But what is the advantage of having temperature
variation within the house? The advantage of _not_ going to great
lengths--as most present day heating and colling systems do--to
achieve something that you don't really need or enjoy that much. Now
that all of us are plagued with the pollution resulting from the
overabundance of devices we have purchased, perhaps government or
church groups should sponsor a series of "you don't need it" commercials.
Instead of the bright uniformed "service personnel" of the Ace Air-
Conditioning Company briskly delivering and installing the latest gadgets,
the commercials would show the expensive equipment misused: a bored
housewife growing geraniums in her new dishwashing machine; a small child
casually dismantling a TV-stereo combo with a claw hammer...
Reptiles need mammal houses. The reptile is at a disadvantage because
he cannot regulate his body temperature, but, instead, equilibrates
near to the temperature of his surroundings. If it is cold he cannot
move fast. The regulatory function of the mammal is a great advantage,
since he can keep his body temperature constant.
Does this apply to houses and temperature regulation? Is it the same
kind of improvement when a thermostat and gas heating system are
installed? If the temperature outside one's body--the temperature of
the house--is regulated to within 1/2 degree F, of what use is the
sophisticated temperature regulating metabolism of the mammal?
Obsession with temperature control seems more like Reptile Technology
than Mammal Technology. The reptile badly needs it--the mammal does not.
This leads to the general question of what view one should take of
equipment manufactured to do for you what your body is equipped and
prepared to do for itself. Certainly we are all grateful for the
discovery of fire, but the thermostat--I don't know. A person's body
has already incorporated the muscles, organs, etc., to steer him
through dangers and difficulties. Yet we cleverly make them
unnecessary by an entirely new level of design and invention. What
is the result of this? The now unnecessary organs are not removed
from the body; instead they are simply unemployed--hanging around,
so to speak, in one's body, talking to the brain, being fed by the
heart and bloodstream.
For the utmost in design I can imagine the equipment manufacturers'
surgical teams removing now unnecessary organs with the installation
of their automatic control systems. Perhaps the now outdated glands
and organs could be sold to reptiles on another planet.
To: renew-energy@ces.ncsu.edu
Date: 18 Feb 1995 04:39:27 -0500
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Subject: Affordable Comfort Conference
Would anyone know when the Affordable Comfort Conference is in PA this year?
April? Howard Reichmuth will be speaking there on instrumentation for energy
conservation in houses. Howard is an engineer, and one of the designers of
the Ecotope concentrating greenhouse...
To: renew-energy@ces.ncsu.edu
Date: 18 Feb 1995 07:52:53 -0500
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Subject: Indirect-fired, electric-backup water heater?
Would anyone know who manufactures a nice, inexpensive, standard, reliable,
indirect-fired water heater (ie a 60 gallon insulated tank with a some
copper tubing in the bottom, to put boiler water through to heat the tank)
that also has some built-in 220V electric heaters for backup heating?
To: renew-energy@ces.ncsu.edu
Date: Fri, 17 Feb 1995 15:29:11 GMT
From: hatunen@netcom.com (DaveHatunen)
Subject: Re: Thermocouples
>> Again: thermocouples require two junctions, and the emf is a function
>> of the temperature differential between them. So if you already have a
>> source of higher temperature, then, with the other junction at ambient,
>> you can produce a small amount of energy. But if you do not have a
>> temperature differential, then thermocouples are useless.
>In many cases you have already temperature differentials.
>This is the nice thing with thermocouples
>Sources of surplus high temperatures you have for example in the case of
>power stations or even by some solar applications let alone some
>industries as the steel industry. The question is the efficiency and especiall
y
>the costs per kWh.
Quite true. But I believe the original poster was under the impression
that somehow they generated an emf from only the one temperature.
Once you have a temperature differential there are all sorts of ways to
derive a little power; thermocouples are probably the worst choice
except under very singular circumstances.
********** DAVE HATUNEN (hatunen@netcom.com) **********
Date: Thu, 24 Nov 1994 06:51:57 -0800 (PST)
From: David Godwin <dgodwin@freenet.vancouver.bc.ca>
Subject: Re: radiator heat
To: Malcolm Stebbins <Malcolm.Stebbins@125-53.psybbs.durham.nc.us>
Cc: renew-energy@twosocks.ces.ncsu.edu
On 16 Nov 1994, Malcolm Stebbins wrote:
> RA> I highly advise against the use of Polybutylene tubing in under floor
> RA> applications. It has been only recently that homes piped with this
> RA> material have won lawsuits against the manufacturer because the stuff
> RA> doesn't hold up. It starts deteriorating from the moment water hits
> Boy! did you just make my day - I've got it in my concrete floor; guess I'll
> just wait for the springs to start popping up. Any easy fixes to the
> problem??
>
>From articles that I have read, the deterioration comes from contact with
the chlorine or chlorides in "city" water. If you changed the water in
your heating system to non-chlorinated water, you should be ok. Also add
an anti-oxinator (sp?) to the water.
David Godwin dgodwin@Freenet.vancouver.bc.ca
604 351 4812
FAX 604 733 3340
To: renew-energy@twosocks.ces.ncsu.edu
Date: 22 Nov 1994 10:59:08 -0500
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Subject: Another drumwall water heater?
Here is another possible design for a drumwall that might be used
as a solar water heater.
It would use 30 drums, stacked up horizontally in a square array, 20'long x 6'
tall, perhaps behind 120 square feet of single pane glass, with a reflective
insulating shutter that lifts up from the top edge of the drums at a 45 degree
angle, and a reflective floor, inside a shed with a south wall made of,
eg Dynaglas.
It might look like this, not to scale:
......................................... ... .
. . . .
. . south east . . .
. . d. . .
. . y. day. .
......................................... .. 10' n. H...... . ..
. D . D . . . . . . . . . a. n. . .
......................................... <--S g. i...... .
. . . . . . . . . . . 6' l. g. . . 6'
......................................... a. h...... .
. . . . . . . . . . . s. t. . .
............................................................................
| 20' |
The D's are drums, H is a horizontal hinge... Day and night shutter positions
are shown...
The numbers below again indicate drum water temperatures in degrees F, from
a small simulation using average Philadelphia weather conditions in January
(ambient temp, 32 degrees, 1200 Btu/day of sun on a south-facing wall,
T^4 heat loss) while extracting 100K Btu/day of domestic hot water from
the wall, using some food-grade, *lined* drums plumbed in series, on
the top row, pressurized with cold water on the way to the input of an
electric hot water in a house. It looks like 4 heat exhanger drums are enough.
Nick
10 'Todd Hall drumwall solar water heater?
20 'est. water temps for rectangular drumwall with 45 degree upward shutter
30 NH=10'number of horizontal drums
40 NV=3'number of vertical drums
50 ND=NH*NV'total number of drums
60 PI=4*ATN(1)'pi...
70 AD=(2*(23/2)^2*PI+23*PI*35)/144'drum area in ft^2
80 DIM T(NV, NH)'array of drum water temperatures
90 OPEN "drumout" FOR OUTPUT AS #1
100 'find steady-state temps
110 FOR R=1 TO NV'1 is top row, nv is bottom row...
120 FOR D=1 TO NH'1 is west drum in row...
130 T(R,D)=150'initialize drum temps
140 NEXT:NEXT
150 AREA=4*NH*NV'effective area of drumwall surface
160 SUNH=1.85*AREA*1000/ND/24'hourly sun energy/drum, inc. by shutter refl.
170 ES=15*5*3*8*55'energy for 15 5 min, 3 gpm, 110 degree showers/day
180 UH=ES/24'useful hourly heat output of drumwall
190 GLOSSF=1.74E-09*6*AREA/ND/24'factor for heat lost thru glazing per drum
200 AMRADF=(32+459)^4'factor for radiation from surroundings to drumwall
210 CP=55*8'thermal mass of a drum full of water
220 CLS
230 HMAX=192
240 PRINT #1,"Charging mode, with sun..."
250 PRINT "Charging mode, with sun..."
260 DT=.1'simulation time step
270 FOR H=0 TO HMAX+.01 STEP DT'hours of use
280 TECON=0'initialize total convection heat from below
290 LB=UH*DT/(T(1,NH)-55)'water moved thru system per time step
300 GOSUB 740
310 'adjust top drum row temperatures
320 FOR D=1 TO NH'drum number in top row
330 TD=T(NV,1)-T(1,D)'temp diff between top drum and lower drums
340 IF TD>0 THEN ECON=TD*AD ELSE ECON=0'convection heat from below
350 TECON=TECON+ECON'accumulate total hourly convection heat
360 HEATFLOW=(SUNH+ECON-((T(1,D)+459)^4-AMRADF)*GLOSSF)*DT
370 T(1,D)=T(1,D)+HEATFLOW/CP'adjust top row drum temp
380 IF T(1,D)>T(NV,1) THEN T(1,D)=T(NV,1)'limit upper drum temp
390 NEXT D
400 ECONA=TECON/(NH*(NV-1))'heatflow to top row from average drum
410 FOR R=2 TO NV'adjust bottom row temps
420 FOR D=1 TO NH
430 HEATFLOW=(SUNH-ECONA-((T(R,D)+459)^4-AMRADF)*GLOSSF)*DT
440 T(R,D)=T(R,D)+HEATFLOW/CP
450 NEXT D
460 NEXT R
470 IF INT(10*H+.05) MOD 480= 0 THEN GOSUB 800'print drum temp array
480 NEXT H
490 PRINT #1,
500 PRINT
510 PRINT #1,"Storage mode, with no sun..."
520 PRINT "Storage mode, with no sun..."
530 FOR H=DT TO 168+.01 STEP DT'hours of use
540 TECON=0'initialize total convection heat from below
550 'adjust top drum row temperatures
560 LB=UH*DT/(T(1,NH)-55)'water moved thru system per time step
570 GOSUB 740
580 FOR D=1 TO NH'drum number in top row
590 TD=T(NV,1)-T(1,D)'temp diff between top drum and lower drums
600 IF TD>0 THEN ECON=TD*AD ELSE ECON=0'convection heat from below
610 TECON=TECON+ECON'accumulate total convection heat
620 T(1,D)=T(1,D)+ECON*DT/CP
630 IF T(1,D)>T(NV,1) THEN T(1,D)=T(NV,1)'limit upper drum temp
640 NEXT D
650 IF INT(10*H+.05) MOD 240= 0 THEN GOSUB 800'print drum temp array
660 ECONA=TECON/(NH*(NV-1))'heatflow to top row from average drum
670 FOR R=2 TO NV'adjust bottom row temps
680 FOR D=1 TO NH
690 T(R,D)=T(R,D)-ECONA*DT/CP
700 NEXT D
710 NEXT R
720 NEXT H
730 END
740 T(1,1)=(LB*55+(CP-LB)*T(1,1))/CP'move 55 degree water into top left drum
750 IF T(1,1)<55 THEN T(1,1)=55'limit lower drum temp to 55
760 FOR D=2 TO NH'move water across top row
770 T(1,D)=(LB*T(1,D-1)+(CP-LB)*T(1,D))/CP'move water from top drum d-1-->d
780 NEXT D
790 RETURN
800 PRINT #1,
810 PRINT
820 PRINT #1,"Day:";INT(H/24+.01)
830 PRINT"Day:";INT(H/24+.01)
840 FOR RS=1 TO NV'display drum temperatures
850 FOR P=1 TO NH-1
860 PRINT #1,INT(T(RS,P)+.5);TAB(6*P);
870 PRINT INT(T(RS,P)+.5);TAB(6*P);
880 NEXT P
890 PRINT #1,INT(T(RS,NH)+.5)
900 PRINT INT(T(RS,NH)+.5)
910 NEXT RS
920 RETURN
Charging mode, with sun...
Day: 0
149 150 150 150 150 150 150 150 150 150
150 150 150 150 150 150 150 150 150 150
150 150 150 150 150 150 150 150 150 150
Day: 2
95 120 136 146 152 156 156 156 156 156
156 156 156 156 156 156 156 156 156 156
156 156 156 156 156 156 156 156 156 156
Day: 4
96 121 137 146 153 156 156 156 156 156
156 156 156 156 156 156 156 156 156 156
156 156 156 156 156 156 156 156 156 156
Day: 6
96 122 137 147 153 157 157 157 157 157
157 157 157 157 157 157 157 157 157 157
157 157 157 157 157 157 157 157 157 157
Day: 8
97 122 138 148 154 158 158 158 158 158
158 158 158 158 158 158 158 158 158 158
158 158 158 158 158 158 158 158 158 158
Storage mode, with no sun...
Day: 1
89 112 128 138 145 149 149 149 149 149
149 149 149 149 149 149 149 149 149 149
149 149 149 149 149 149 149 149 149 149
Day: 2
84 104 118 128 135 139 141 141 141 141
141 141 141 141 141 141 141 141 141 141
141 141 141 141 141 141 141 141 141 141
Day: 3
80 97 110 119 125 129 132 132 132 132
132 132 132 132 132 132 132 132 132 132
132 132 132 132 132 132 132 132 132 132
Day: 4
75 90 101 109 115 120 123 124 124 124
124 124 124 124 124 124 124 124 124 124
124 124 124 124 124 124 124 124 124 124
Day: 5
71 83 93 100 106 110 113 115 115 115
115 115 115 115 115 115 115 115 115 115
115 115 115 115 115 115 115 115 115 115
Day: 6
67 77 85 92 97 101 104 106 107 107
107 107 107 107 107 107 107 107 107 107
107 107 107 107 107 107 107 107 107 107
Day: 7
64 72 78 84 88 91 94 97 99 99 <--hot water temp
99 99 99 99 99 99 99 99 99 99 supplied to house
99 99 99 99 99 99 99 99 99 99 after a week w/o
sun, in January
To: renew-energy@twosocks.ces.ncsu.edu
Date: 22 Nov 1994 16:24:01 GMT
From: cgood@magnus.acs.ohio-state.edu (Charles W Good)
Subject: Wood Heat
I have seen wood heat flamed (excuse the pun) a lot on this newsgroup, and I
would be interested in a discussion of the pros and cons. I have been heating
with wood for 17 years. The fuel accounts for about 90% of my home heating.
Consider the following:
1- Although wood smoke is polluting, it probably isn't as polluting in some
ways as fossil fuel smoke. Wood contains almost no sulphur and very little
nitrogen. Thus wood smoke has no sulphur dioxide and very little in the way of
nitrous oxides. This means no acid rain.
2- Trees and other forms of biomass naturally decay slowly, releasing all their
carbon hydrogen and oxygen to the environment as carbon dioxide and water.
Burning biomass fuels just speeds up this process, but the end result is the
same. There is no difference in the amount of carbon dioxide released to the
atmosphere whether biomass burns or decays. Burning fossil fuels adds EXTRA
carbon dioxide to the atmosphere which can affect global warming. Burning
biomass fuels does not add extra carbon dioxide over and above what is
naturally cycled between the atmosphere and biomass via photosynthesis (removes
carbon dioxide) and respiration (adds carbon dioxide).
3- I know that deforestation is a big problem in many parts of the world and
has been for hundreds of years. I have read, however, that currently forest
land east of the Mississippi in the USA has been gradually increasing. This is
due to the abandonment of marginal farmland that is difficult to farm with big
machines. An example is in southeast Ohio in the area that includes Wayne
State Forest near Athens Ohio. This is in spite of the increased use of fuel
wood. East of the Mississippi in general (there are local exceptions) I don't
think deforestation is a problem. WOOD IS A RENEWABLE FUEL!
4- Money spent purchasing wood fuel goes into the local economy, not over to
Saudi Arabia to make some rich person there even richer.
5- Wood fuel costs less (if you have to buy it) than almost any other form of
domestic heating fuel. The only fuel that has a comparably low cost is *yuck*
coal. This cost factor is one of the main reasons I heat with wood. If I had
a new house I would design it from the beginning to take advantage of solar for
much of its heat. However, I own a nice old victorian house built in 1895, so
my solar options are limited.
Regards,
Charles W. Good
cgood@lima.ohio-state.edu
P.S. I have a PV system that powers everything in my bedroom, even though I
live right next to grid power lines. I have the PV system not because it is
cheap but because "its the right thing to do" for the environment.
Newsgroups: alt.architecture.alternative
From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Subject: solar heating explorations (an update)
Organization: Villanova University
Date: Fri, 17 Jun 1994 13:33:25 GMT
This is about the same as the last version, until about problem 25...
Nick
Philadelphia weather assumptions...
W1. Amount of sun falling on a south facing wall on an average day in
January: 1100 Btu/ft^2/day
W2. On a clear day, the amount of sun falling on a surface aimed at the
sun is about 300 Btu/ft^2/hr or 1 Kw/meter^2
W3. Number of hours of sun in January, on a clear day: 6 hours
W4. Average temperature in January: 32 degrees F
W5. 99% of the time, in Philadelphia,
winter temperatures are > 10 degrees F, and
summer temperatures are < 93 degrees F.
W6. A Philadelphia heating season has about 5500 degree days, ie if you
add up the temperature differences between the outside temperature
and an assumed inside temperature of 68 degrees, for each day of an
average heating season, the total is 5500.
W7. Average windspeed in January: 10 mph.
Other assumptions...
F1. The R-value of one pane of glass is about 1 hr-degree F-ft^2/Btu
F2. The R-value of an air film at a surface is about 1/(1+v/2),
where v is the air velocity in mph.
F3. The R value of a still water film at a surface is approximately 1/60.
F4. Each pane of glass transmits about 90% of the energy falling on it.
F5. Heat capacity of water: Cp = 1 Btu/degree F/pound
(This is the definition of a Btu, which happens to be about the
amount of energy in a kitchen match.)
F6. It takes about 1000 Btu to evaporate a pound of water
F7. 1 Ft^3 of water weighs 62 pounds
F8. Heat capacity of air: Cp = 1/55 Btu/degree F/ft^3
(ie, with 1 Btu you can heat up a pound of water 1 degree F, or 55
cubic feet of air 1 degree F. Air is easier to heat up than water.)
F9. A gallon of oil contains about 130,000 Btus, which when burned in a
70% efficient oil burner, makes about 100,000 Btus of heat.
F10. A cord of wood has a volume of 4' x 4' x 8', and it contains the
equivalent of about 100 gallons of oil.
F11. A south-facing double glazed window with 1 ft^2 of glass is
equivalent to about one gallon of oil/year, in house heating.
F12. 1 watt is equivalent to 3.41 Btu/hr.
F13. 1 person is equivalent to 100 watts. 1 cat is equivalent to 20 watts.
Equations...
E1. Heat flow--"Ohm's law" for heat...
U = delta t * delta T * sum [Area(i)/(R-value(i)]
{Btu = degrees F * hours * sum of areas (ft^2)/R-values}
(Heat flow = temperature difference/thermal resistance)
E2. Heat storage--how much heat does a material absorb or release as it
warms up or cools down delta T degrees?
U = delta T * Cp * V
{Btu = degrees F * heat capacity * volume}
E3. Cooling rate of a heat storage battery, which is initially delta T
degrees above the surrounding temperature...
delta T(t) = delta T(0) * exp (-t/RC)
-hours/time constant
{degrees F = initial degrees F * e }
E4. Counterflow heat exchangers
For a heat exchanger with heat capacity rates of the cold liquid or
gas of Cc (in Btu/hr/degree F), and Ch for the hot liquid or gas,
with no phase changes occuring, eg no condensation or evaporation,
let Cmin be the minimum heat capacity rate, and Cmax the other rate.
For example, if 2 gpm of cold water is cooling 200 cfm of air,
Cc = 2 gpm x 60 m/hr x 8 lb/g x 1 Btu/lb/F = 960 Btu/hr/F, and
Ch = 200 cfm x 60 m/hr x 1/55 Btu/ft^3/F = 218 Btu/hr/F, then
Cmin = Ch.
Define a parameter Z = Cmin/Cmax, eg Z = 218/960 = .227 above.
If the heat exchanger has effective area A and an average thermal
conductivity U (=1/R), define the number of heat transfer units as
NTU = AU/Cmin, eg NTU = (47)(1/(1/60 + 1/5))/218 = .995 above,
if the heat exchanger above has an area of 47 ft^2, with still water
on one side and air at 8 mph on the other.
Now define the effectiveness E of the heat exchanger as
1- exp(-NTU/(1-Z)) NTU
E = ---------------------, or E = ------- for Z = 1.
1- Z exp(-NTU/(1-Z)) 1+ NTU
Thi - Tho
Finally, E = ------------ when Ch = Cmin, or
Thi - Tci
Tco - Tci
E = ------------ when Cc = Cmin,
Thi - Tci
where Th is the temp of the hot fluid,
Tc is the temp of the cold fluid, and
the subscript i indicates the entering condition and
the subscript o indicates the leaving condition.
Problems...
1. Find the steady state temperature of the inside of the box below
----------------
| R5 |
-------------- | <--Non-glazed part of box
black surface--->| | | surface area: 5 ft^2
.a| infinite | |
.i| thermal | |
.r| mass |R | Ambient temp: Ta = 32 degrees
1 ft^2 glass-->. | |5 |
.g| Ti = ? | |
1100 Btu/day --> .a| | |
.p| | |
. | | |
-------------- |
| R5 |
----------------
Hint: During a 24 hour day, Energy in = energy out.
the sun puts about 1000 Btu into the box, and
the box loses heat thru the glass and the walls.
And temperature of the infinite thermal mass never changes.
2. Assuming the water in the box below starts out at the same temperature
as the average temperature of the infinite thermal mass in the box
above, find the temperature of the water after 3 days without sun.
----------------
| R5 |
-------------- | <--Non-glazed part of box
black surface--->| | | surface area: 5 ft^2
.a| 50 pounds | |
.i| of water | |
.r| |R | Ambient temp: Ta = 32 degrees
1 ft^2 glass-->. | |5 |
.g| Tw = ? | |
0 Btu/day --> .a| | |
.p| | |
. | | |
-------------- |
| R5 |
----------------
Hint: During a 24 hour day,
the sun puts 0 Btu into the box, and
the box loses heat mainly thru the walls...
Use fact F2 and E1 and E2, or F2 and E3.
3. Find the steady state temperature of the inside of the box below. In
this case, instead of losing heat all night from the water to the
outside air thru the glass, the water stays insulated all night, and the
sunspace containing the solar collector gets cold. Then during the day,
the solar collector pump runs for 6 hours, and the thermal battery is
heated. During this time (only) the sunspace is heated up to the thermal
battery temperature, and so more heat is lost through the glass.
------------------
water-type solar | R5 |
collector with ---------------- | <--Non-glazed part of box
black surface---->0 | | | surface area: 5 ft^2
.a0 | infinite | |
.i0 | water | |
.r0R | mass |R | Ambient temp: Ta = 32 degrees
1 ft^2 glass-->. 05 | |5 |
.g0 | Tw = ? | |
1100 Btu/day --> .a0 | | |
.p0 | | |
. 0 | | |
---------------- |
| R5 |
------------------
Hint: During a 6 hour day,
the sun puts about 1000 Btu into the box, and
the box loses heat thru the glass
and the walls during the day, and
thru the walls, mainly, during the night.
The temperature of the infinite thermal mass never changes.
And Energy in = Energy out.
4. Assuming the water in the box below starts out at the same temperature
as the average temperature of the infinite water mass in the box above,
find the temperature of the water after 3 days without sun.
------------------
water-type solar | R5 |
collector with ---------------- | <--Non-glazed part of box
black surface---->0 | | | surface area: 5 ft^2
.a0 | 50 pounds | |
.i0 | of water | |
.r0R | |R | Ambient temp: Ta = 32 degrees
1 ft^2 glass-->. 05 | |5 |
.g0 | Tw = ? | |
0 Btu/day --> .a0 | | |
.p0 | | |
. 0 | | |
---------------- |
| R5 |
------------------
5. Find the steady state temperature of the inside of the box below, which
is the box above, with all linear dimensions scaled up 10X.
------------------
water-type solar | R5 |
collector with ---------------- | <--Non-glazed part of box
black surface---->0 | | | surface area: 500 ft^2
.a0 | 50,000 | |
.i0 | pounds of | |
.r0R | water |R | Ambient temp: Ta = 32 degrees
100 ft^2 glass-->. 05 | mass |5 |
.g0 | | |
110,000 Btu/day-->.a0 | Tw = ? | |
.p0 | | |
. 0 | | |
---------------- |
| R5 |
------------------
Hint: During a 6 hour day,
the sun puts about 100,000 Btu into the box, and
the box loses heat thru the glass
and the walls during the day, and
thru the walls, mainly, during the night.
And the temperature of the large thermal mass hardly changes.
6. Assuming the water in the box below starts out at the same temperature
as the average temperature of the infinite water mass in the box above,
find the temperature of the water after 3 days without sun.
------------------
water-type solar | R5 |
collector with ---------------- | <--Non-glazed part of box
black surface---->0 | | | surface area: 500 ft^2
.a0 | 50,000 | |
.i0 | pounds of | |
.r0R | water |R | Ambient temp: Ta = 32 degrees
100 ft^2 glass-->. 05 | mass |5 |
.g0 | | |
0 Btu/day --> .a0 | Tw = ? | |
.p0 | | |
. 0 | | |
---------------- |
| R5 |
------------------
7. Find the steady state temperature of the inside of the box below, which
is the box above, with an extra layer of glass and better insulation.
--------------------
water-type solar | R20 |
collector with ------------------ | <--Non-glazed part of box
black surface-----> 0 | | | surface area: 500 ft^2
.a.a0 | 50,000 | |
.i.i0 | pounds of | |
.r.r0R | water |R | Ambient temp: Ta = 32 degrees
100 ft^2 glass-->. . 02 | mass |2 |
.g.g00 | |0 |
110,000 Btu/day-->.a.a0 | Tw = ? | |
.p.p0 | | |
. . 0 | | |
------------------ |
| R20 |
--------------------
Hint: During a 6 hour day,
the sun puts about 90,000 Btu into the box, and
the box loses heat thru the glass
and the walls during the day, and
thru the walls, mainly, during the night.
The temperature of the large thermal mass hardly changes.
8. Assuming the water in the box below starts out at the same temperature
as the average temperature of the large water mass in the box above,
find the temperature of the water after 3 days without sun.
--------------------
water-type solar | R20 |
collector with ------------------ | <--Non-glazed part of box
black surface-----> 0 | | | surface area: 500 ft^2
.a.a0 | 50,000 | |
.i.i0 | pounds of | |
.r.r0R | water |R | Ambient temp: Ta = 32 degrees
100 ft^2 glass-->. . 02 | mass |2 |
.g.g00 | |0 |
0 Btu/day --> .a.a0 | Tw = ? | |
.p.p0 | | |
. . 0 | | |
------------------ |
| R20 |
--------------------
9. Now the passive solar house above is getting quite warm inside. So why
don't we make the heat battery separate from the house, but inside the
house, so that it stays warm better, and the heat losses from the heat
battery keep the house warm...
How many Btus per day are necessary to keep the inside of the house
below at 68 degrees F, for 24 hours a day in January?
Floor plan:
32'
----www---wwww---www----
| | Assume the house is 32' tall,
w w ie it has 3 floors,
w w
w w and it has R20 insulation on all
| | exterior surfaces, except for the
| |32' windows, which each have a surface
| | area of 12 ft^2, and an R-value of 8,
w w
w w and the air infiltration rate is 0
w w air changes per hour.
| |
----www---wwww---www----
10. How much oil will it take to keep this house at 68 degrees F, 24 hours
a day, throughout an average Philadelphia heating season, under the
assumptions above?
11. Assume the above house has an air-infiltration rate of 0.2 air changes
per hour (a very tight house, compared to the air infiltration rate for
average new houses of about 1 air exchange per hour, or older houses
with 2 air exhanges per hour.)
How much additional oil is necessary to keep this house at 68 degrees
during a Philadelphia heating season, owing to non-zero air infiltration?
12. Assume the above house uses 500 KWh of electricity per month (a modest
amount), and that it contains two people and two cats, and that all of
the heat from the electricity used, and the people and cats goes into
heating the house.
How many fewer Btus are necessary to keep this house at 68 degrees during
an average January day in Philadelphia, owing to internal heat generation?
13. How many total Btus will it take to keep this house at 68 degrees for
an average January day in Philadephia, including air infiltration, but
excluding electricity consumption, and heat from people and cats?
14. If the occupants of the house above use hot water only for showers, and
they take 4 10-minute 3 gpm showers a day, and the incoming cold water
temperature is 55 degrees F, and the shower temperature is 110 degrees
F, how many extra Btus are required for domestic water heating per day.
15. Assume that the house above, with air infiltration, has a sunspace, and
during sunny days in January, the house is heated entirely by the
sunspace, with 78 degree air flowing freely into the house from the
sunspace during the day, and assume that at night, the sunspace is
thermally insulated from the house. The sunspace gets cold at night,
while the house stays warm overnight, because it has enough inherent
thermal storage in the building materials to stay warm overnight, at
approximately the daytime temperature.
Under those assumptions, how much single pane sunspace glass is needed
to heat the house to 68 degrees, during periods of sunny days in January?
16. If the sunspace gets very hot, compared to the house, collecting solar
energy from it will be less efficient. How big a fan is needed to
circulate air into the house, keeping the sunspace temperature less
than 78 degrees, on an average January day, in CFM?
17. What will the approximate day/night temperature swing be in the above
house, if the daily thermal storage mainly resides in the walls, which
are made of gypsum wall board, weighing about 1 lb/ft^2, with a heat
capacity of about 0.25 Btu/lb/F, ie, if the house temperature is, say
75 degrees at 3 PM one sunny January afternoon, what will it be the
next morning at 9 AM, when the sun comes up again?
18. Suppose we make the south wall of the house out of heavy cement block,
insulated on the outside of the house, to increase its thermal
capacity. Heavy cement block weighs about 65 lb/ft^2 and has a heat
capacity of about 0.2 Btu/lb/F. What will the morning temperature be?
19. Suppose that, while the house is warm, say 68 degrees, and the sun
is still shining, we somehow divert the airflow from the sunspace so
that it circulates through the holes inside the cement block wall,
instead of circulating through the house, to raise the south wall
temperature. How much warmer than 68 degrees will we have to make the
cement block wall, in order to keep the house itself at 68 degrees
until morning, as the heat leaks out of the south wall into the house?
20. Draw a diagram of how that diversion might be accomplished...
21. The cement blocks above are 8" x 8" x 16", with two holes in each
block, each hole being about 8" x 5" x 2". The R value of the cement
is about 2/inch. What is wrong with this system?
22. Calculate the temperature rise required for overnight heat and the
temperature rise during charging, if the heat battery is made from
a double wall, with an airspace inbetween, of lighter 8" x 8" x 16"
cement blocks, each having three 8" x 6" x 3" holes. The double wall
would have a weight of about 75 lb/ft^2.
23. Now assume that the sun stops shining for 10 days, so we have to use a
big water-based heat battery to heat the house. Ignoring losses from
the heat battery itself, how many cubic feet of water are required in
the heat battery to keep the house warm for 10 days, if the initial
temperature is 130 degrees F, and the minimum usable temperature of the
heat-battery water is 80 degrees F?
24. Assuming the water is stored inside the house, in a 4' deep x 12'
wide x L' long box, lined with a single piece of EPDM rubber roofing
material, with R20 insulation all around it, what would the self-
discharge rate be, in degrees F per day, starting from 130 F, when the
heat battery is not being used to heat the house?
25. How many ft^2 of unglazed water-type solar collector, located in the
78 degree sunspace, will be necessary to keep the heat battery fully
charged, when not in use?
26. How much will that solar collector area increase if the heat battery
also has to supply domestic hot water, as above, both on sunny days and
over 10 days without sun?
27. If the hot water comes from a 20' long x 8" diameter heat exchanger
made of PVC pipe, located in the 130 degree heat battery, how long will
it take to heat a pipeful of incoming 55 degree water to 110 degrees?
28. Adjust the model above, to find a new required sunny day sunspace glass
area, solar collector area, and storage volume, based on the facts that
the heat battery losses also heat the house and the losses from the
solar collector in the sunspace adds heat to the sunspace.
29. How much water, in gallons per minute, will have to be pumped through
the water-type solar collector, to keep the temperature rise below 10
degrees F, on an average January day?
30. How would the above sunspace glass area and solar collector area change
if the sunspace were double glazed and the collector had an additional
layer of glass over it?
31. How much would the collector area and heat battery volume have to
change above, if the water system provided all the heat for the house
during the night, while the air system only heated it when the sun was
shining, with no air-type heat battery?
32. What would the be the most economical operating temperature for the
collector (perhaps not 130 degrees F), in the last problem, if heat
battery storage costs $1.50/ft^3 and collectors cost $4/ft^2 and
sunspace glazing costs $2/ft^2? And how much heat battery volume,
collector area and sunspace glazing would be used at that temperature?
33. If the water-type collector above consisted of a 32' x 16' 2:1
concentrating parabolic reflector, made of 90% reflective 3M SA-85
aluminized self-adhesive mylar film, costing $1/ft^2, adhered to
galvanized metal costing $ 0.50/ft^2, over a shallow 32' long x 8'
wide water trough made of EPDM rubber costing $ 0.40/ft^2, covered with
glass costing $2/ft^2, and the sunspace glazing were Dupont Tedlar
PVF, costing $0.10/ft^2, how would the system cost change?
34. If the incoming water temperature in the counterflow heat exchanger
example in equation E4, above, is 55 degrees F, and the incoming air
temperature is 90 degrees F, what will the outgoing air temperature be?
35. If an automobile radiator and its electric fan can keep an engine cool
when it is burning 2 gallons of oil per hour, when the radiator water is
190 F and the ambient temp is 90 F, how much heat can it remove from
80 degree air, when using 55 degree well water?
36. Assume that one adds a circular freestanding fireplace, with a double
wall chimney that acts as an air-air heat exchanger, with an inside
pipe diameter of 6" and an outside pipe diameter of 8", and that all of
the house air infiltration takes place through this chimney. How do the
above glass areas and storage volume change?
37. How many Btus of heat will the above fireplace add to the house, if it
is burning one cubic foot of wood per hour?
Answers... (?)
1. sum (A(i)/R(i)) = R=1 ft^2/R=1 + 5 ft^2/R=5 = 2
glass front insulated walls
U = delta t * delta T * sum (A(i)/R(i))
= 24 hours * (Ti-32) * 2
During a day the energy lost is U, and the energy gained is 1000 Btu,
and they are equal, so
Ti = 32 + 1000/(2*24) = 52.83 degrees F.
2. delta T(t) = delta T(0) * exp(-t/RC)
and R = 1/(sum (A(i)/R(i)) = 1/2 and C = 50 and t = 72 hours,
and delta T(0) = 52.83 - 32 degrees F, so
delta T(t) = (52.83-32) * exp(-72/(1/2*50)) = 1.17 degrees F, so
after three days, Tw = 32 + 1.17 = 33.17 degrees F.
3. Energy out = Edayout + Enightout
Edayout = (Tw-32)*(6 hours)*(1/1+5/5)
Enightout = (tw-32)*(18 hours)*(6/5)
Energy IN = Energy OUT ==>
1000 = (Tw-32)*(6*2+18*6/5) ==> Tw = 32 + 29.76 = 61.76 degrees F.
4. R = 1/(sum (A(i)/R(i)) = 6/5 and C = 50 and t = 72 hours,
and delta T(0) = 61.76 - 32 degrees F, so
delta T(t) = (61.76-32) * exp(-72/(5/6*50)) = 5.29 degrees F, so
after three days, Tw = 32 + 5.29 = 37.29 degrees F.
5. Energy out = Edayout + Enightout
= (Tw-32)*(6 hours)*(100/1+500/5)
+ (Tw-32)*(18 hours)*(600/5), so
100,000 = (Tw-32)*(6*200+18*600/5) ==>
Tw = 32 + 29.76 = 61.76 degrees F.
6. R = 1/(sum (A(i)/R(i)) = 5/600 and C = 50,000 and t = 72 hours,
and delta T(0) = 61.76 - 32 degrees F, so
delta T(t) = (61.76-32) * exp(-72/(5/600*50,000)) = 25.04 degrees F,
so after three days, Tw = 32 + 25.04 = 57.04 degrees F.
7. Energy out = Edayout + Enightout
= (Tw-32)* (6 hours)*(100/2+500/20)
+ (Tw-32)*(18 hours)*(600/20)
Energy IN = Energy OUT ==>
90,000 = (Tw-32)*(6*75+18*30) ==> Tw = 32 + 90.91 = 122.91 degrees F.
8. R = 1/(sum (A(i)/R(i)) = 20/600 and C = 50,000 and t = 72 hours,
and delta T(0) = 122.91 - 32 degrees F, so
delta T(t) = (90.91) * exp(-72/(1/30*50,000)) = 87.07 degrees F,
so after three days, Tw = 32 + 87.07 = 119.07 degrees F.
9. Window area = 12*30 = 360 ft^2
sum (A(i)/R(i)) = 360/R=8 + (6*32*32-360)/R20
windows insulated walls
= 45 + 289.2 = 334.2
U = delta t * delta T * sum (A(i)/R(i))
= 24 hours * (68-32) * 334.2 = 289,000 Btu.
10. Gallons of oil = degree days * 24 * sum (A(i)/R(i))/100,000
= 5500 * 24 * 334.2 /100,000
= 441 gallons
11. U = delta T * Cp * volume [E2]
Cp * volume = 1/55 *.2 * 32*32*32 = 119 Btu/F/hr
Gallons of oil = 5500 degree days * 24 * 119/100,000
= 157 gallons
12. Animal energy = (200 watts + 40 watts) * 24 hr/day *30 days/mo
people cats
= 173KWh/mo
Internal energy = electrical energy + animal energy
= (500KWh/mo + 173KWh/mo) *3.41 Btu/wh/30 days/mo
= 57,000 Btu/day.
13. From the answers to problems 10, 11 and 12,
Btu/day = 24 * (68-32) * (334.2+119) - 57,000
= 335K Btu/day (3.35 gallons of oil/day, in January)
14. Volume of hot water = 4 showers * 10 min/shower * 3 gpm * 8 lb/gal
= 960 lbs, so
Btu/day = (110-55) * 960 = 52,800.
15. The house requires about 335,000 Btus/day to keep it at 68 degrees in
January, so Ein-Eout must be at least 335K Btu for the sunspace
(ignoring the effect that the sunspace insulates the house wall a bit
more.) If the glazed area of the sunspace is At,
Ein = 1000 * At Btu/ft^2/day, and Eout = (78-32) * 6 hours * At/R=1,
so we need 1000At-(78-32)(6)At = 335,000, which gives a sunspace area
of At = 463 ft^2, or 15' x 32', a lean-to greenhouse, 32' tall,
covering a bit bit less than half of the south wall area.
16. Moving 335,000 Btu over 6 hours is equivalent to a heat flow of
U = 335,000/6 = 55.8K Btu/hr, or 930 Btu/min, so from formula E2,
930 Btu = (78-68 degrees F) * 1/55 Btu/F/ft^3 * CFM, so
CFM = 5115, which can be provided by one (or two, for reliability)
large window fans, such as Grainger catalog number 2C614 fans, which
have three speeds, and which each deliver 7005 CFM at 835 rpm at the
highest speed, while consuming 364 watts each.
17. The house as described has an approximate heat capacity of
C = 6*32*32' * 1 lb/ft^2 * 0.25 Btu/lb/F = 1536 Btu/F, and
R = 20/(6*32*32), so the time constant of the house is RC = 5 hours, so
if the temperature of the house is 75 degrees when the sun goes down,
18 hours later,
Ti = Ta + (Tinitial-Ta) exp( -t/RC)
= 32 + (75-32) exp(-18/5) = 33.17 degrees F.
18. The house now has an approximate heat capacity of
C = 5*32*32' * 1 lb/ft^2 * 0.25 Btu/lb/F
+32*32' * 60 lb/ft^2 * 0.2 Btu/lb/F = 13568 Btu/F, and (still)
R = 20/(6*32*32),
so RC = 44 hours, so if the temperature of the house is 75 degrees when
the sun goes down, the morning temperature, 18 hours later, is
Ti = Ta + (Tinitial-Ta) exp( -t/RC)
= 32 + (75-32) exp(-18/44) = 60.6 degrees F.
19. The cement block wall would have a heat capacity of about
C = 32*32' * 65 lb/ft^2 * 0.2 = 13,312 Btu/lb/F,
and the house loses about 335K*18hr/24hr = 251K Btu when the sun is not
shining, so by equation E2, the south wall would have to be charged up
to delta t = 251K/13,312 = 18.9 degrees more than 68 degrees, ie about
87 degrees, to provide constant temperature heat for the rest of the night.
20. The diversion to charge up the block wall might be accomplished like this:
----------------------------
x
<==== x <--motorized damper, eg Grainger cat # 3C727
airflow to house x
with damper open | |x sunspace air flows down through
| |x wall when damper is closed
house | wall |x sunspace and fan is on...
| |x
| |x <--insulation on outside of wall
airflow from house | |x
with damper open | f. gravity
====> a . damper, open when ===>
n . fan is on...
-----------------------------
21. The problems are 1) we need to transfer 251K Btus in 6 hours to the
wall, which is a rate of 42K Btus/hr, and the average distance the heat
has to travel thru the cement to heat it up is about 2", and the area
exposed to hot air inside the blocks is about 1800 ft^2, so the
temperature rise of the air is about 42K*2*2/1800 = 93 degrees while
charging the wall, which makes the sunspace almost 180 degrees when
finished, which makes the heat losses thru the glass very large, and 2)
the 5000 cfm has to flow thru the cement block holes, which have an
area of 3.33 ft^2, so the air velocity would be 1500 fpm, which
probably needs a blower instead of a fan, because of the pressure
required to drive the air through the holes.
22. The double block wall has to be charged up to 16.3 degrees warmer than
the house, to hold enough overnight heat. Each of the 2304 blocks has
an exposed surface area of about 3.41 ft^2, and the average distance
heat has to travel is about 1/2" thru the cement, so the temperature
rise of the air during charging would be about 42K*2*.5/(2304*3.41) =
5.3 degrees. With a 3" air gap between the walls, the 5000 cfm would
flow thru an area of about 35 ft^2, so the air velocity would be about
150 fpm, more reasonable for a fan.
23. 10 days * 335,000 Btu/day = (130-80) * V * 62, so V = 1080 ft^3.
24. The box would be 1080/(4*12) = 22.5' long, so the heat loss per
day, with the water at 130 degrees F, would be about
U = (130-68)*24 hours*2*(4*12+4*22.5+12*22.5)/R=20 = 60,700 Btu/day,
which would cool the 1080 ft^3 of water by
delta T = 60,700/(1080*62) = 0.9 degrees F/day.
25. To keep the heat battery fully-charged requires a net 60,700
Btu/day, and with a water-collecting surface of Ac ft^2, under one
layer of glass, with still air next to the water-type collector,
Ein = 1000 * Ac, and Eout = (130-78) * 6 hours * Ac/R=1, so
we need 1000 Ac - (130-78) * 6 Ac = 60.7K, or
Ac = 88 ft^2, eg a 32' wide x 3' tall collector plate.
26. Supplying hot water requires another Ahw ft^2 of collector (ignoring
the fact that a somewhat larger heat battery is required, which has
slightly larger losses), where
1000 Ahw - (130-78) * 6 * Ahw = 52,800 Btu/day (from problem 14), so
Ahw = 77 ft^2, eg another 32' wide x 3' tall collector plate.
27. The pipe contains 54 gallons of water, and it has a surface area of 42
ft^2, with an R-value of 2 * 1/60, so the RC time constant of this
heater is 54*8*2/60/42 = .343 hours, so a delta T of 130-110 = 20
degrees F would require a time t such that
20 = (130-55) * exp(-t/.342), so t = .342 ln (20/75) = .453 hours.
28. The new combined sunspace needs to collect a total of
335K - 60.7K Btu/day to heat the house on sunny days, and
52.8K + 60.7K Btu/day to maintain the heat battery temperature, or
------
387.8K Btu/day total.
If the temperature in the sunspace is 78 degrees when the sun is
shining,
Ein - Eout = 1000 At - (78-32) * 6 hours * At/R=1 = 387.8K, from which
At = 536 ft^2, eg a total sunspace glazed area on the south side of the
house, 32'tall by 17' wide.
The water-type solar collector needs to collect a total of
52.8K + 60.7K = 113.5K Btu/day, and assuming the air in the sunspace is
fairly still, it has an R-value of 1 at its bare surface, so it loses
(130-78)*6*Aw/R=1 = 312 Aw Btu/day to the sunspace, but 1000 Aw of sun
energy falls onto its surface, per day, therefore, we need a total
water collecting area such that
1000 Ac - 312 Ac = 113.5K ==> Ac = 165 ft^2, eg 32' x 5' of water
collecting area.
The storage volume should not change much, since the house storage
requirement decreased by 60.7K/day, but the water heating storage
requirement increased by 52.8K/day, since the storage volume and losses
were calculated above.
29. Moving 113.5K Btu over 6 hours is equivalent to a heat flow of
U = 113.5K/6 = 18.9K Btu/hr, or 315 Btu/min, so from formula E2,
315 Btu = (10 degrees F) * 1 Btu/F/lb * 8 lb/gal * gpm, so
gpm = 3.94, which can be provided by a Grundfos pump, Grainger catalog
number 2P079, which can deliver about 5 gpm at 12' of head, while
consuming about 100 watts of electrical power.
30. The combined sunspace still needs to collect a total of 387.8K Btu/day, and
Ein - Eout = 1000 *.9 * At - (78-32) * 6 hours * At/R=2 = 387.8K, from
which At = 508 ft^2, eg a total sunspace glazed area on the south side
of the house, 32'tall by 16' wide.
The water-type solar collector needs to collect a total of
52.8K + 60.7K = 113.5K Btu/day, and assuming the air in the sunspace is
fairly still, it has an R-value of 2 from collector surface to the
sunspace, so it loses about
(130-78)*6*Aw/R=2 = 156 Aw Btu/day to the sunspace, but 900 *.9* Aw of
sun energy falls onto its surface, per day, therefore, we need a total
water collecting area such that
810 Ac - 156 Ac = 113.5K ==> Ac = 174 ft^2, eg 32' x 5' of collecting area.
31. Assuming the size of the heat battery does not change much, the
sunspace area still needs to be 536 ft^2, eg 32' tall by 17' wide.
The water-type solar collector needs to collect a total of 53.2K
Btu/hr, when the sun is shining, and the sunspace needs to provide
about 11.4K Btu/hr of 78 degree hot air for the house. Still assuming
that the air in the sunspace is fairly still, the water collector loses
(130-78)*Ac/R=1 = 52 Ac Btu/hour to the sunspace, but 1000/6 Ac Btu/hr
of sun energy falls onto its surface, so we need a total water-type
collecting area such that
167 Ac - 52 Ac = 53.3K Btu/hr ==> Ac = 463 ft^2, eg 32' x 15' of
collecting area.
32. The optimum temperature is about 150 degrees F, using a small computer
program in this linear model. In this case,
storage volume = 926 ft^3, ie 4' x 12' x 19', storage cost = $1400,
sunspace area = 463 ft^2, ie 32' x 14', glass cost = $ 926,
collector area = 463 ft^2, ie 32' x 14', collector cost = $1854,
------
and total cost = $4180.
The total cost increases about a hundred dollars, as the water
temperature changes from 150 to 130 or 160 degrees.
33. The cost of the water collector above is about $1000, and the cost of
the collector in the last problem was $1854, so we save $854 in
collector costs, but since the sunlight falling into the collector has
an intensity of 167 *2*.9*.9 = 271 Btu/ft^2/hr, the collector can
work at a water temperature Tw, such that
53.3K = 271 Ac - (Tw - 78) * Ac/R=2, which implies that Tw can be 204
degrees F, in this linear model, so the storage volume only needs to be
about (150-80)/(204/80)*926 ft^3 = 540 ft^3, costing $800 instead of
$1400, and the Tedlar vs. glass sunspace glazing would save another
$850, saving a total system cost of abour $2300.
The parabola might have an equation of x=y^2/32, where y is the height
of the reflector and x is the distance from the base, measured
forward to the sunspace glazing. So at a height of 8', the reflector
would be 2' in from the back wall. It would need to be a fairly
empty space, to collect sunlight, and it would be nice if one
could walk on the glass floor, perhaps made of 3/16" tempered
glass sliding door replacement panels, supported on 2' centers,
resting on 2 x 6's with the EPDM rubber draped over them. If UV
transparent Tedlar were used, this could be a tanning space...
Both plastic films should last a long time. Both Dupont and 3M
require large minimum orders for these films, about $4000's worth.
They are not yet in the Real Goods catalog.
34. In the above example,
1- exp(-NTU/(1-Z)) 1 - exp(-.995/(1-.227))
E = --------------------- = ------------------------------
1- Z exp(-NTU/(1-Z)) 1 - .227 exp(-.995/(1-.227))
= .603, and
Thi - Tho
E = ------------, since Ch = Cmin, so
Thi - Tci
Tho = Thi - E * (Thi - Tci) = 90 - .603 * (90 - 55)
= 68.9 degrees F.
35. The auto radiator and its fan remove about 260K Btu/hr from the
engine, with an air-water temperature differential of 100 degrees,
so they should remove about 65,000 Btu/hr from 80 degree air, with
a 25 degree temperature differential. A small window air
conditioner removes about 6,000 Btu/hour from the air, although it can
make the air cooler and dryer.
36. The house air infiltration is .2 * 32*32*32 = 6554 cfh, (109 cfm), so
Cc = 6554 * 1/55 = 119 Btu/hr/F, and
The area between the chimney walls is
pi*((8/2)^2-(6/2)^2)/144 = .1527 ft^2, so the air velocity is about
109 cfm/.1527 ft^2/60 sec/min/88ft/sec*60 mph = 8.11 mph on either side
of the inside chimney wall, so the U value in the NTU equation is
U = 1/Rinside + 1/Routside = 2/(1+8.11/2) = 10.11,
and the area of the heat exchange surface, the inside chimney wall, is
A = pi * 6"/12"/ft * 30' = 47 ft^2, so
NTU = AU/Cmin = 47 * 10.11/119 = 4, and E = NTU/(1+NTU) = 4/(1+4) =.8,
since the same amount of air is flowing up the inside and down the
outside of the chimney, so Cmin = Cmax, and Z = 1, so the temperature
of the incoming air to the house would be
Tco = Tci + E (Thi - Tci) = 32 + .8 (68-32) = 60.8 degrees F,
instead of 32 degrees F, so the air infiltration loss would be
(68-60.8) * 1/55 Btu/ft^3/F * 109 cfm * 60 m/hr * 24 hr/day
= 20,547 Btu/day, vs. the 102,816 Btu/day from problem 13,
so the daily Btu requirement would be reduced from 335K Btu/day to
335K - (102,816-20,547) = 252,731 Btu/day, ie about 25%, so the glass
area and storage volume can be reduced by about 25%.
37. By fact F10, a cubic foot of wood contains about
1/128*100*130,000 = 101,562 Btu of energy. If this is all used to heat
up 109 cfm of air, the flue temperature will be
Thi = 68 + 101,562*55/109*60 = 922 degrees F, and if E = .8, as above,
the temperature of the air entering the house will be
Tco = 32 + .8 (922 - 32) = 744 degrees F,
so, ignoring the heat given off directly from the fireplace, the
chimney air-air heat exchanger will add at least
U = (744 - 68) * 1/55 * 109 cfm * 60 m/hr = 80,393 Btu/hr to the house.
From: peien@bud.peinet.pe.ca (PEI Environmental Network)
Newsgroups: misc.rural
Subject: Re: Passive Solar Heating/alternative energy
Date: 5 Mar 1994 00:30:12 -0400
Organization: PEINet, Charlottetown, Prince Edward Island, Canada
I'm in the midst of building a passive solar house. It was designed by Don
Roscoe, a solar architect from Nova Scotia. His designs have been featured
in Harrowsmith magazine (Canadian version) 3 times. I don't have the issue
numbers handy, but can get them if you're interested. I will get 60 - 70 %
of my heat from the sun. His houses are no more expensive to build than any
other house and obviously, I highly recommend his designs. The magazine
articles give details on how the houses work.
Sharon.