A 16' cube
Nick Pine
on east and west walls in Phila, PA on an average 30.4 F January day,
when a 1-axis EW concentrator can collect 729 Btu/ft^2 over 3 hours.
With a 90% reflector and 2 layers of R1 clear polycarbonate glazing
with 90% solar transmission, we might collect 0.9x0.9x0.9x729x16'x16'
= 136K Btu/day with a 16' cube with a parabolic reflector inside.
With a focal line at x=f=4' south of the base of the north wall and
equation y^2=4fx=16x, the reflector would bounce dawn sun down onto a
4'x16' horizontal target at the base of the north wall. The focus moves
closer to the north wall during the day until the reflector begins to
shade the base of the wall at a 45 degree sun elevation (at noon on
April 4 and September 28, at 40 N latitude.) We would still have lots
of hot water in summertime.
With a 4' strip near the north wall dedicated to solar collection and
12'x16' = 192 ft^2 of clear floorspace, the cube could serve another
purpose as well, eg a dramatic greenhouse or workplace.
The target might have a 0.020" clear polycarbonate cover and a plywood
bottom with an EPDM rubber liner and 2x3 sides. It might thermosyphon
hot water through a 4'x8'x16' 4,096 gallon 16 ton tank above, supported
by 13 2x4 studs on 2' centers along the north wall and the north 4' of
the east and west walls and the reflector, with 9 more 2x4 studs on 4'
centers around the east and west and south walls, and a 2x4 "foundation"
(80' of 2x4s laid flat on the ground over poly film.)
We need 64 ft^2 of polycarbonate for the target plus 256 ft^2 for the
south wall plus 4/3x16'x16' = 341 ft^2 for the east and west wall area
under the reflector. That's a total of 661 ft^2, ie $992 at $1.50/ft^2.
Rimol Greenhouse Systems (877-RIMOL-GH) sells it for $290 plus $10 UPS
shipping per 49"x50' roll.
We might use 512 ft^2 of OSB for the roof and north wall and 448 ft^2
for the reflector and target bottom and 384 ft^2 for the tank sides and
bottom and 172 ft^2 for the east and west walls above the reflector.
That's 1,516 ft^2, or $284 at $6 per 4x8 sheet.
Gluing 384 ft^2 of Nielsen's reflective Mylar ($0.09/ft^2 in 4'-wide
rolls from http://www.snomo.com/mylar.html) to the underside of the
reflector adds $35 to the materials cost.
We could line the tank with a single folded $192 20'x32' piece of EPDM
rubber, and cover the roof with another $77 16'x16' piece, with a $20
4'x16' piece for the target liner. The tank might have 256 ft^2 of 3"
polyisocyanurate board (eg $10 Atlas Energy-Shield "R7.2" double-foil
4'x8'x1" boards) for another $240, with lots of loose fill (eg bags of
leaves) under the bottom and on the south side, which would make
RC = 32768Btu/FxR21.6/256ft^2 = 2765 hours or 115 days.
We might approximate the parabolic shape with 9x24' = 216 ft of 2x4s
on 2' centers with a dozen sawkerfs in each bow. The roof might need
144 ft more 2x4s, with another 352 ft for the sides and 80 ft for the
foundation. That's 792 ft, $198 at 25 cents per foot. The total of
the materials costs above is $2038.
If the tank water is 150 F and we want to keep the target temp 160 F max
and 46.7K Btu/h of heat power falls on the target and we ignore the heat
lost to the warm air in the cube, we need a thermosyphoning water flow
Q ft^3/s, where 3600s/hx64Btu/F-ft^3Q(10F) = 46.7K, so Q = 0.0203 ft^3/s,
about 10 gpm.
Water weighs about 63.74 - 0.0158T lb/ft^3, with T in degrees F. The
difference in density caused by the average temperature difference
between up and down pipes causes a pressure difference proportional
to the height of the water column, which makes water flow through the
resistance of the pipe loop. With 16' of height and a 10 F temperature
difference, dP = 2.53dT lb/ft^2.
Here's a formula for laminar flow in a pipe with radius r and length
L in feet and pressure difference dP: Q = Pir^4dP/8MuL ft^3/s. Mu is
the viscosity, about 8.17x10^-6 lb-s/ft^2 for 150 F water. We might
figure the water flows through 24' of pipe, 16' up and 8' down, which
makes Q = 0.0203 = Pir^4(2.53)/(8x8.17x10^-6x24'), so r = 0.0447' or
0.54", eg a 1.5" pipe (or larger, with some fitting head losses.)
If 0.9x1000x16'x16' = 230K Btu of sun enters the cube from the south
wall on an average day and 0.9x425x4/3x16'x16' = 131K enters the east
and west wall area beneath the reflector, a total of 361K Btu enters
the cube, including about 16'x16'x531 = 136K Btu of beam sun.
Removing this from the daily total, say 225K Btu enters over 6 hours
at 37.5K Btu/h. The cube has about 600 ft^2 of surface below the
reflector. With loose insulation above, the conductance to outdoors
is about 600 ft^2/R1. If 136K/3 = 45.3K Btu/h hits the target for
3 hours when it has an average temp T and 1.5x4'x16' = 96 Btu/h-F
of conductance to the air in the cube which has temp Tc, we have
something like this (in Courier font):
30.4 F ----www-----------------Tc -------www-----Tc
1/600 | | 1/600
| --- 93 F
----- | equivalent to -
|------| --> |-------| |
----- | _
37.5K/h <
< 1/96 . . . . . . . . .
<
----- |
|------| --> |-------*-------T = 160 F
----- |
45.3K/h ----- 32,768 Btu/F
-----
|
_
Removing the 1/96 resistor makes the upper Thevenin equivalent temp
Tc = 93 F. Using that equivalent and replacing the 1/96 resistor,
(160-93)/(1/600+1/96) = 5,545 Btu/h flows from target to cube air,
making Tc = 93+5545/600 = 102 F. Too hot, so we need to vent in full
sun. If Tc = 70 F, (160-70)x96 = 8,640 Btu/h flows from target to
cube air and the sun provides (93-70)x600 = 13,800 Btu/h. The total
is 22,400. With a 40 F indoor-outdoor temp difference, we need about
22.4K/40 = 560 cfm of outdoor airflow to dump the heat. Using an
empirical chimney formula, 560 = 16.6Asqrt(16'x40F), so we need
A = 1.33 ft^2 of vent area at the top and bottom. We could probably
use more than that in summertime, even with noontime shading.
On an average day, with 70 F air inside, we might collect beam sun
in 150 F water with an efficiency of 100(45.3K-8640)/45.3K = 81%,
ie the cube might collect 0.81x136K = 110K Btu/day and lose about
24h(150-30)256ft^2/R21.6 = 34K Btu (cooling 34KBtu/32768Btu/F = 1 F),
so it could provide 110K-34K = 76K Btu per day of hot water in January
at $2038/(110K/3.41) = 6.3 cents per peak watt.
At 150 F, with a minimum usable temperature of 120 F, the tank would
store about (150-120)32768 = 983K Btu. With a 135 F average temp, it
would lose about 24h(135-30)256ft^2/R21.6 = 30K Btu to the outdoors
on a cloudy day, so it could provide 76K Btu of hot water for about
983K/(30K+76K) = 9 cloudy days in a row. Adding bubblewalls to the
cube would allow the tank to provide cloudy-day heat by circulating
hot water through the target using a small pump and a thermostat.
We might increase the thermal efficiency and make electricity at
the same time by laying 4'x16' of PVs on the target floor and
collecting 130 F water at about 3 suns.
Nick
Zentuck
Mtnguy (CO)
"Nick Pine" <ni...@ufo.ee.vill.edu> wrote in message
news:985r4h$q...@ufo.ee.vill.edu...
Jeepers!
>> 983K/(30K+76K) = 9 cloudy days in a row. Adding bubblewalls to the
>> cube would allow the tank to provide cloudy-day heat by circulating
>> hot water through the target using a small pump and a thermostat.
>>
>> We might increase the thermal efficiency and make electricity at
>> the same time by laying 4'x16' of PVs on the target floor and
>> collecting 130 F water at about 3 suns.
>>
>> Nick
>
> Man o man, I do love your posts.
What the hell did he say?
--
RANDOM THOUGHT FEED©
I assume full responsibility for my actions, except the ones that are
someone else's fault.
Chuck Simmons
>
> NREL says 1000 Btu/ft^2 of sun falls on a south wall and 425 falls
> on east and west walls in Phila, PA on an average 30.4 F January day,
> when a 1-axis EW concentrator can collect 729 Btu/ft^2 over 3 hours.
> With a 90% reflector and 2 layers of R1 clear polycarbonate glazing
> with 90% solar transmission, we might collect 0.9x0.9x0.9x729x16'x16'
> = 136K Btu/day with a 16' cube with a parabolic reflector inside.
The parabolic reflector bothers me. The problem is that if the parabola
is accurate at all, there will be quite high temperatures if an
efficient absorber is placed at the focus. A spherical reflector would
produce lower temperatures but experience with less than 3 foot diameter
segmented spherical reflectors indicates that the temperature might be
high enough to cause significant radiation loss from any object placed
at the focus. There is no point to the reflector unless an efficient
absorber is located at the focus. This is the one problem you glossed
over completely but it needs attention because you seemed to make an
assumption of 100% absorbtion. The absorber efficiency depends on its
temperature since it will radiate as well as absorb.
Chuck
--
... The times have been,
That, when the brains were out,
the man would die. ... Macbeth
Chuck Simmons chr...@webaccess.net
Zentuck
>
> What the hell did he say?
He basically explained how much energy is available to use. Then
explained the best ways to use it. Provided the numbers involved so
that the person wishing to do the project can do a cost benefit
analysis. By providing such a detailed post the builder can assess the
finished product and see if it is working at a reasonable efficiency.
Such a detailed post is a wonderful resource given freely by an engineer
that understands what is involved and how to quantify the project.
For me it is refreshing to see detailed answers that can be evaluated
and checked. While I do find much value in the posts that have a less
technical nature they can not compare to this stuff.
I thought it was clear and concise. But hey I like numbers that can be
crunched!
Jeepers!
That dictum is usually applied to hypotheses that run counter to accepted
paradigms.
--
RANDOM THOUGHT FEED©
I had a linguistics professor who said that it's man's ability to use
language that makes him the dominant species on the planet. That may be, but
I think there's one other thing that separates us from animals. We aren't
afraid of vacuum cleaners.
Neon John
Yes, but with all the assumptions, it was little more than an
academic exercise in GIGO. For someone who's going to actually try
to heat with solar, a more practical-oriented answer would be much
more useful. Instructions for how to make a few simple measurements
along with advice on what steps to take to constrain the unknowns
would be really handy.
Following my own advice, here's what I did back in '74 to heat my
1200 sq ft house cuz I was a student too poor to afford utility
heat.
Right in the midst of the post-arab oil embargo hysteria, TVA had
published tons of documents on solar heat. One I found particularly
useful was on how to fabricate a solar collector using (mostly)
ordinary building materials. The collector consisted of a frame of
1 or 2X4 pressure treated boards wrapped around a sheet of metal
corrugated roofing. The roofing is painted flat black (fancier
coatings were also listed in the paper but flat black paint is cheap
and works well. Behind the roofing are bats of fiberglass
insulation. Over the top is either glass or heavy plastic (I used
salvage glass). Along the top of the roofing is a length of copper
tubing drilled with holes so that water would spray into the grooves
of the roofing. AT the bottom is a gutter to collect the water.
The water is heated by being in direct contact with the black
collector.
I made a rough measure of the collection efficiency by simply
measuring the flow and temperature rise of water flowing through a
collector. Then using TVA's published insolation numbers for this
area, computed what I'd need on the worst case day I wanted the
system to work. based on that, I built sufficient collectors and
installed them in my large rural back yard facing south. These
would have fit on my roof but I'd have had to build a mount to angle
them due south.
For heat storage, I buried a 2000 gal septic tank in the back yard
and insulated it with several inches of styrofoam board that I
picked up salvage. No detailed computations done here - the 2k
gallon tank and the styrofoam were what I could find salvage. To
heat the house I modified a large condenser coil by converting it to
once-thru flow. copper tubing headers were soldered on each end
after all the elbows were removed. This was mounted in the same
duct work that served my homemade wood burning central furnace. A
solar cell, a mechanical time clock, some platinum RTD thermal
sensors (from the boneyard at TVA where I worked) and some discrete
logic/analog controls (no microprocessors yet.) drained the
collectors when the sun went down and/or when the outlet water
temperature dropped below the temperature of the water in the septic
tank. This prevented freezing and prevented diluting the hot water
with cooler water. I interconnected this control system with the
wood fired furnace control system so that if the water temperature
in the septic tank dropped below about 80 degrees, the furnace was
automatically lighted and took over heating. I later installed
water coils in the furnace to heat the water in addition to heating
air. The furnace fired only very seldom - a cord of wood lasted all
winter plus.
This was a case of "turn it on and see if it worked". It did. I
like to take that approach first and only if it doesn't work do the
formal engineering analysis. I ended up adding a few more
collectors as I could afford them. Total cost? I'm going to
estimate under $500 plus a lot of (fun) work plus the favor I owed a
friend for borrowing his back hoe. Figure 30 years of inflation and
with some clever scrounging, this system could still easily be built
for under $2k.
I believe that the system is still mostly in place but is not now
being used. I talked to the current owners of the house a couple of
years ago. They're still using the furnace but they've removed most
of the collectors to install a pool. It probably served this house
for 20+ years though.
This 'back of the match book' approach probably wouldn't work in
Minnesota but it worked fine here in the Sunny Mid South.
John
--
John De Armond
johngdDO...@bellsouth.net
http://personal.bellsouth.net/~johngd/
Cleveland, occupied TN
Victor Smith
Nick, I'm going to do this next week if I can verify that Rimol isn't
ripping me off on the $10.00 UPS charge for the poly.
Do you happen to know if that charge is for 4-day ground shipment?
TIA.
--Vic
Nick Pine
>The parabolic reflector bothers me. The problem is that if the parabola
>is accurate at all, there will be quite high temperatures if an
>efficient absorber is placed at the focus...
It's a linear parabola made with a dozen line segments softly focusing
sun onto a 4'x16' 160 F water cooled target in a 70 F room. I figured
the target would lose 8,640 Btu/h to the room air by all means. Perhaps
we should figure 0.1714x10^-8((460+160)^4-(460+70)^4)4'x16' = 7550 Btu/h
for radiation plus 0.27((160-70)/4')^0.25(160-70)4'x16' = 3387 Btu/h for
convection, which reduces the solar collection efficiency from 81% to
100(45.3K-10.9K)/45.3K = 76%.
Nick
Zentuck
>
> On or about 3/7/01 2:23 PM, Zentuck hit keys that wrote:
>
> > "Jeepers!" wrote:
> >
> >>
> >> What the hell did he say?
> >
> >
> > He basically explained how much energy is available to use. Then
> > explained the best ways to use it. Provided the numbers involved so
> > that the person wishing to do the project can do a cost benefit
> > analysis. By providing such a detailed post the builder can assess the
> > finished product and see if it is working at a reasonable efficiency.
> > Such a detailed post is a wonderful resource given freely by an engineer
> > that understands what is involved and how to quantify the project.
> >
> > For me it is refreshing to see detailed answers that can be evaluated
> > and checked. While I do find much value in the posts that have a less
> > technical nature they can not compare to this stuff.
> >
> > I thought it was clear and concise. But hey I like numbers that can be
> > crunched!
>
> That dictum is usually applied to hypotheses that run counter to accepted
> paradigms.
O My!
> --
> RANDOM THOUGHT FEED©
>
> I had a linguistics professor who said that it's man's ability to use
> language that makes him the dominant species on the planet. That may be, but
> I think there's one other thing that separates us from animals. We aren't
> afraid of vacuum cleaners.
Zentuck
snip
> --
> John De Armond
> johngdDO...@bellsouth.net
> http://personal.bellsouth.net/~johngd/
> Cleveland, occupied TN
What color was the insulation? This is a real question, I really want
to know if the insulation you used was yellow or pink?
BTW, Man o man I loved your post too!
--
Zen...@i-plus.net
Neon John
Zentuck wrote:
> What color was the insulation? This is a real question, I really want
> to know if the insulation you used was yellow or pink?
>
> BTW, Man o man I loved your post too!
Pink.
Well heck, they do make the stuff right up the road :-)
John
Zentuck
>
> Zentuck wrote:
>
> > What color was the insulation? This is a real question, I really want
> > to know if the insulation you used was yellow or pink?
> >
> > BTW, Man o man I loved your post too!
>
> Pink.
>
> Well heck, they do make the stuff right up the road :-)
>
> John
John,
Did you have trouble with outgassing of the insulation? I was told
years, perhaps decades, ago that the pink insulation would outgas and
fog the glass or plastic. Word was that the problem would exist as long
as the insulation was in place. Every commercial panel I installed or
had any dealings with had yellow insulation.
So did you notice a problem and if so how bad was it?
--
Zen...@i-plus.net
Sechler
news:3AA6AAB6...@bellsouth.net...
>
> This was a case of "turn it on and see if it worked". It did. I
> like to take that approach first and only if it doesn't work do the
> formal engineering analysis.
IMHO, a little preliminary engineering analysis with reasonable assumptions
helps one determine a starting point, an order of magnitude, from which a
prototype can be built, tested, and expanded upon. I'm sure the TVA's
documents listed "rule of thumbs" that were based on previous engineering
analysis and testing.
Joe
Neon John
I put the insulation BEHIND the corrugated metal. The metal is
RTVed to the wood frame to make it water-tight so the insulation is
not in the heated space. It simply insulated the back side of the
metal. I did not attempt to insulate the wooden sides.
I imagine the pink stuff would offgas. I used it to insulate the
walls of an annealing kiln for my glass shop (after stripping the
backing off, of course) and it smoked like HELL the first couple
hours of operation. When it quits smoking, the fiberglass is just
slightly on the grey side of white. The glass is stiff so it
obviously has some sort of coating, probably to make it less itchy.
When I need fiberglass insulation for a project now, I buy the stuff
in the pink plastic stocking. Inside the pink plastic is pure white
fiberglass wool without any apparent coating.
Neon John
Oh, absolutely. TVA spent $millions of tax dollars on this stuff.
Which meant I didn't have to re-invent the wheel.
Sechler
>
>
> Sechler wrote:
> >
> > "Neon John" <joh...@bellsouth.net> wrote in message
> > news:3AA6AAB6...@bellsouth.net...
> > >
> > > This was a case of "turn it on and see if it worked". It did. I
> > > like to take that approach first and only if it doesn't work do the
> > > formal engineering analysis.
> >
> > IMHO, a little preliminary engineering analysis with reasonable
assumptions
> > helps one determine a starting point, an order of magnitude, from which
a
> > prototype can be built, tested, and expanded upon. I'm sure the TVA's
> > documents listed "rule of thumbs" that were based on previous
engineering
> > analysis and testing.
> >
> > Joe
>
> Oh, absolutely. TVA spent $millions of tax dollars on this stuff.
> Which meant I didn't have to re-invent the wheel.
Yeah, but that kind of spending is true of any governmental agency. We're
talking a little well spent time.
Joe
Nick Pine
putting the heat storage tank in the basement. Less "roof" required,
10% more sun, the unpressurized target can be a lot simpler, the tank
can have less insulation, and the heat loss from the tank warms the
house, but... we need a pump.
NREL says a 1-axis EW concentrator can collect 729 Btu/ft^2 over 3 hours
in Phila on an average 30.4 F January day. With a 90% reflector and a
layer of R1 clear polycarbonate glazing with 90% solar transmission,
we might collect 0.9x0.9x729x16'x16' = 151K Btu/day with a 16'x16'x16'
tall open-sided box with a parabolic reflective north wall and roof.
The reflector would bounce dawn sun down onto a 4'x16' horizontal target
at the base of the north wall. The focus moves closer to the north wall
during the day until the reflector begins to shade the base of the wall
at a 45 degree sun elevation (at noon on April 4 and September 28, at
40 N latitude.) We would still have lots of hot water in summertime.
With a 4' strip near the north wall dedicated to solar collection and
12'x16' = 192 ft^2 of clear floorspace, the structure could also be
something else, eg a run-in shed for horses or cows or a place to park
a car temporarily when it's snowing.
The target might be a $10 16'x30" round UV polyethylene film greenhouse
air duct laid flat on the ground over black plastic film. It might heat
water for a 4'x12'x4' tall 1,024 gallon EPDM-lined tank in a basement
corner, with 9 inboard 4' studs on 2' centers and 1" of R7.2 foamboard
underneath inside 4' and 12' 2x4s attached flat on the floor. We might
put 3.5" fiberglass R11 insulation between the studs with a 1" layer of
foamboard over that and a 1" layer of foamboard between the tank and the
basement walls and another 2 layers over the tank. Its thermal conductance
would be 48ft^2/R17.2 = 2.79 for the floor plus 48ft^2/R14.4 = 3.33 for
the cover plus 64ft^2/R7.2 = 8.89 for the outside walls plus 64ft^2/R18.2
= 3.52 for the inside walls. G = 18.53 and RC = 8192Btu/F/18.53Btu/h-F
= 442 hours, or 18.4 days.
So far, we have a $10 duct (Griffin Greenhouse and Nursery Supplies,
978-851-4346) plus 64 ft^2 of tank OSB, $12 at $6/4x8 sheet. We've
also used a $72 20'x12' piece of EPDM for the tank liner, 64 ft^2 of
fiberglass worth $16, 272 ft^2 of foamboard at $85 and 52 ft of 2x4s
at $13. These materials subtotal $196.
The reflector needs 384 ft^2 of OSB and another 384 ft^2 of EPDM for its
roof ($187), and 9x24' = 216 ft of 2x4s on 2' centers with a dozen (11 :-)
kerfs in each bow. The lower and upper edges could use 32 ft of 2x4s. We
might support the open south wall with 3 16' posts with 2 2x4s attached at
right angles with deck screws, with some upper diagonal bracing. That's
344 ft of 2x4s worth $86. The bottom part of each post could be pressure-
treated wood bolted between the sidewalls of 2 vertical underground tires
filled with stone.
Gluing 384 ft^2 of Mylar ($0.09/ft^2 from http://www.snomo.com/mylar.html)
under the parabola adds $35. Grainger's 4PC89 120V 0.7A pump adds $135 and
$7/year at 10 cents/kWh, making the total materials cost $639.
The target has about 1.5x4'x16' = 96 Btu/h-F of thermal conductance to
outdoor air. We might collect 151K Btu/day of beam sun over 3 hours at
200 F and lose 3h(200-30)96 = 49K Btu, for an average daily hot water
production of 102K Btu at $639/(102K/3.41) = 2.1 cents per peak watt.
At T degrees F, the tank loses heat at a rate of (T-70)18.53Btu/h, and
hot water use removes heat at a constant 102K/24h = 4250 Btu/h, so dT/dt
= -((T-70)18.53+4250)/8192 = -0.002262T -0.3605 = -cT +d, so d/c = -159
and T = -159+(200+159)exp(-0.002262t) = 120 F when t = 111 hours, after
4.6 cloudy 30 F days in January.
Now how can we make it frugaler?
Nick
Chuck Simmons
Let's cover another base or two just to get the figures into the range
of what can be done. You give the transmission of the window material as
90%. However, transmisson for transparent materials is always given for
perpendicular incidence. At some angle of incidence, it actually goes to
zero. A cube shaped building is poor since reflection from the windows
could be 20% or more from each layer of multiple glazing. You have not
taken account of the spectral relationship of transmission. That can be
more or less important depending on the window material.
The reflector bothers me another way. Fresh aluminum on glass is about
95% reflective. It degrades to less than 90% in a year or two and is
often as low as 80% in five years time depending on various factors. 90%
is not a conservative estimate for sustained reflectivity of aluminum on
glass or mylar.
Your collector efficiency at the parabola focus needs attention. Flat
black paint, even the special paints used inside optical instruments,
reflect a bit. In optical instruments, surface roughening is used to
combat this but this increases scattering. There are ways to increase
absorbtion but at the expense of larger area which increases both
radiation and convection loss.
A couple of final things. Since there will be loss from unintercepted
solar radiation and heat loss into the air and surroundings, a heavy
north wall well insulated from the outside might be considered. A
masonry or rock wall comes to mind. This would capture some lost heat.
To avoid radiation loss at night, reflective shutters for use at night
would be a good idea as well.
dogsnus
snip
>For me it is refreshing to see detailed answers that can be evaluated
>and checked. While I do find much value in the posts that have a less
>technical nature they can not compare to this stuff.
>
>I thought it was clear and concise. But hey I like numbers that can be
>crunched!
IOW,
Not all rural folks are dumb.
BG!
Some of us are engineers, or retired engineers. Or better!
Just 'cause we like to retire, garden, farm,irrigate or grow
orchards, doesn't mean we're ignorant!
Terri
Nick Pine
>...You give the transmission of the window material as 90%. However,
>transmisson for transparent materials is always given for perpendicular
>incidence.
The cube walls would be 0.020" polycarbonate with an index of 1.6
and a max incidence angle of about 45 degrees in January. What's
the sun-power-weighted average transmittance?
>A cube shaped building is poor since reflection from the windows could
>be 20% or more from each layer of multiple glazing.
Say 10%, with this single layer at normal incidence.
>You have not taken account of the spectral relationship of transmission.
Polycarbonate's better than glass in that respect...
>The reflector bothers me another way. Fresh aluminum on glass is about
>95% reflective. It degrades to less than 90% in a year or two and is
>often as low as 80% in five years time depending on various factors. 90%
>is not a conservative estimate for sustained reflectivity of aluminum on
>glass or mylar.
I disagree. Nielsen sells this stuff to hydroponic growers and claims
the initial reflectance is more than 90% and it "lasts for years..."
Duane Johnson uses it for outdoor heliostats in Minnesota with lifetimes
in years. It would be indoors in this application, inside a building
that is likely to be quite dry. There are better materials for outdoor
use, eg 3M's old SA-85 film, which had a 0.25 mil layer of polyurethane
over the aluminum, but they sold that for more than $1/ft^2, vs $0.09.
>Your collector efficiency at the parabola focus needs attention. Flat
>black paint, even the special paints used inside optical instruments,
>reflect a bit.
The sun hits the polycarbonate target with air on one side and a few
inches of water on the other and a black surface beneath the water.
The water (what's its refractive index?) makes the water-polycarbonate
surface less reflective (1% vs 4%?) Polyethylene is cheaper and more
diffusive (what's its index?), and probably better with underwater PVs.
>Since there will be loss from unintercepted solar radiation and heat
>loss into the air and surroundings, a heavy north wall well insulated
>from the outside might be considered. A masonry or rock wall comes
>to mind. This would capture some lost heat.
Sounds pretty useless and expensive compared to making the target
as thin as possible, so it loses minimal heat at night.
Some people look for problems, and some people look for solutions.
Nick
Chuck Simmons
>
> Chuck Simmons <chr...@webaccess.net> wrote:
>
> >...You give the transmission of the window material as 90%. However,
> >transmisson for transparent materials is always given for perpendicular
> >incidence.
>
> The cube walls would be 0.020" polycarbonate with an index of 1.6
> and a max incidence angle of about 45 degrees in January. What's
> the sun-power-weighted average transmittance?
Essentially, I was asking you. However, with the angle of incidence at
45 degrees, crown glass (index of refraction near 1.5) will reflect at
least 10% at each of the air/glass and glass/air interface. A refractive
index of 1.6 makes this worse.
> >A cube shaped building is poor since reflection from the windows could
> >be 20% or more from each layer of multiple glazing.
>
> Say 10%, with this single layer at normal incidence.
You don't get normal incidence at any time during the day. That is the
point.
> >The reflector bothers me another way. Fresh aluminum on glass is about
> >95% reflective. It degrades to less than 90% in a year or two and is
> >often as low as 80% in five years time depending on various factors. 90%
> >is not a conservative estimate for sustained reflectivity of aluminum on
> >glass or mylar.
>
> I disagree. Nielsen sells this stuff to hydroponic growers and claims
> the initial reflectance is more than 90% and it "lasts for years..."
> Duane Johnson uses it for outdoor heliostats in Minnesota with lifetimes
> in years. It would be indoors in this application, inside a building
> that is likely to be quite dry. There are better materials for outdoor
> use, eg 3M's old SA-85 film, which had a 0.25 mil layer of polyurethane
> over the aluminum, but they sold that for more than $1/ft^2, vs $0.09.
Without seeing measurements on these materials, I have only experience
with deposited aluminum films on surfaces.
> >Your collector efficiency at the parabola focus needs attention. Flat
> >black paint, even the special paints used inside optical instruments,
> >reflect a bit.
>
> The sun hits the polycarbonate target with air on one side and a few
> inches of water on the other and a black surface beneath the water.
> The water (what's its refractive index?) makes the water-polycarbonate
> surface less reflective (1% vs 4%?) Polyethylene is cheaper and more
> diffusive (what's its index?), and probably better with underwater PVs.
The index of refraction of water is 1.33 near enouge. You have an
air/polycabonate interface and then a polycarbonate/water interface.
Taking your word for the 1.6 for polycarbonate, there will be
significant reflection at both interfaces. I would think that
eliminating both reflective interfaces would improve efficiency. I don't
think polyethelene is a good idea. Being more diffussive means it is a
scatterer making it probably more lossy yet.
> >Since there will be loss from unintercepted solar radiation and heat
> >loss into the air and surroundings, a heavy north wall well insulated
> >from the outside might be considered. A masonry or rock wall comes
> >to mind. This would capture some lost heat.
>
> Sounds pretty useless and expensive compared to making the target
> as thin as possible, so it loses minimal heat at night.
It is useless if the windows are not covered by a reflective material at
night.
> Some people look for problems, and some people look for solutions.
It is always best to investigate the problems before building something.
Better to solve problems first rather than be dissappointed.
Nick Pine
>> The cube walls would be 0.020" polycarbonate with an index of 1.6
>> and a max incidence angle of about 45 degrees in January. What's
>> the sun-power-weighted average transmittance?
>Essentially, I was asking you...
I invite you to answer that question.
>> >A cube shaped building is poor since reflection from the windows could
>> >be 20% or more from each layer of multiple glazing.
>>
>> Say 10%, with this single layer at normal incidence.
>
>You don't get normal incidence at any time during the day. That is the
>point.
Who cares? What's the sun-power-weighted average transmittance?
>> >...90% is not a conservative estimate for sustained reflectivity
>> >of aluminum on glass or mylar.
>> I disagree...
>Without seeing measurements on these materials, I have only experience
>with deposited aluminum films on surfaces.
You might get a sample and measure how the reflectance changes over
10 years inside a dry building.
>The index of refraction of water is 1.33 near enouge. You have an
>air/polycabonate interface and then a polycarbonate/water interface.
So the polycarb-water reflectance is 100((1.6-1.33)/(1.6+1.33)^2 = 0.85%.
>...I would think that eliminating both reflective interfaces would
>improve efficiency.
There are many things that could be done to improve efficiency, but
I'm more interested in cost-effectiveness, with oil at $1.50/gallon.
>I don't think polyethelene is a good idea. Being more diffussive means
>it is a scatterer making it probably more lossy yet.
The NRAES-33 Greenhouse Engineering book lists glass with an 88% solar
transmission, polycarbonate with 91-94% and polyethylene at 85%, but
polyethylene film is sooo cheap, $10 for a 4'x16' collector duct, and
scattering makes more uniform illumination for underwater PVs.
>> >Since there will be loss from unintercepted solar radiation and heat
>> >loss into the air and surroundings, a heavy north wall well insulated
>> >from the outside might be considered...
>> Sounds pretty useless and expensive compared to making the target
>> as thin as possible, so it loses minimal heat at night.
>It is useless if the windows are not covered by a reflective material at
>night.
But they wouldn't be. This structure would get cold inside at night,
and the heat in the target would be lost, while the tank stays warm.
Nick
Woodntya Likatano
Nick Pine wrote in message <9890jk$a...@acadia.ee.vill.edu>...
Chuck Simmons
>
> Chuck Simmons <chr...@webaccess.net> wrote:
>
> >> The cube walls would be 0.020" polycarbonate with an index of 1.6
> >> and a max incidence angle of about 45 degrees in January. What's
> >> the sun-power-weighted average transmittance?
>
> >Essentially, I was asking you...
>
> I invite you to answer that question.
>
> >> >A cube shaped building is poor since reflection from the windows could
> >> >be 20% or more from each layer of multiple glazing.
> >>
> >> Say 10%, with this single layer at normal incidence.
> >
> >You don't get normal incidence at any time during the day. That is the
> >point.
>
> Who cares? What's the sun-power-weighted average transmittance?
I don't know but with an air/polycarbonate boundary followed by a
polycarbonate/air boundary (the 20 mil separation makes it two distinct
boundaries), the maximum normal incidence aplitude transmission is about
80% at each making 64% for the sheet. This is about 80% power
transmission for the sheet.
> >The index of refraction of water is 1.33 near enouge. You have an
> >air/polycabonate interface and then a polycarbonate/water interface.
>
> So the polycarb-water reflectance is 100((1.6-1.33)/(1.6+1.33)^2 = 0.85%.
You forgot the air/polycarbonate interface which is 90% roughly using
the same formula (which is valid if the permeabilities of the two media
are the same - I don't know that they are). Anyway, that drops power
transmission to around 76.5%. Taking into account all of the reflection
losses including the walls of the structure with 20 mil polycarbonate, I
seem to get 61% of power transmitted to the black absorber. If it is 90%
efficient, I get 55% efficiency for the whole thing. Now if angle of
incidence on the walls and on your heat collector are taken into account
(remember, from your parabola, the angle of incidence will be nonzero
almost everywhere), it seems likely that 45% to 50% is kind of best
case.
My whole point was that terms that affect efficiency were left out.
Nick Pine
>> >> >A cube shaped building is poor since reflection from the windows could
>> >> >be 20% or more from each layer of multiple glazing.
>> >>
>> >> Say 10%, with this single layer at normal incidence.
>> >
>> >You don't get normal incidence at any time during the day. That is the
>> >point.
>>
>> Who cares? What's the sun-power-weighted average transmittance?
>I don't know but with an air/polycarbonate boundary followed by a
>polycarbonate/air boundary (the 20 mil separation makes it two distinct
>boundaries), the maximum normal incidence aplitude transmission is about
>80% at each making 64% for the sheet. This is about 80% power
>transmission for the sheet.
Sounds like those greenhouse engineering guys were confused when they
said a single layer of polycarbonate transmits 91-94% of the sun power in
the third (1994) revision of the NRAES-33 Greenhouse Engineering manual.
You can send your correction to the Northeast Regional Agricultural
Engineering Service, 152 Riley Robb Hall/Ithaca, NY 14853-5701, or notify
the authors directly at the University of Connecticut, University of
Delaware, University of the District of Colombia, University of Maine,
University of Maryland, University of Massachusetts, University of New
Hampshire, Rutgers, Cornell, Penn State, University of Vermont, and
West Virginia University. Perhaps they will invite you to give a lecture
about your discovery.
Nick
Chuck Simmons
>
> Chuck Simmons <chr...@webaccess.net> wrote:
>
> >> >> >A cube shaped building is poor since reflection from the windows could
> >> >> >be 20% or more from each layer of multiple glazing.
> >> >>
> >> >> Say 10%, with this single layer at normal incidence.
> >> >
> >> >You don't get normal incidence at any time during the day. That is the
> >> >point.
> >>
> >> Who cares? What's the sun-power-weighted average transmittance?
>
> >I don't know but with an air/polycarbonate boundary followed by a
> >polycarbonate/air boundary (the 20 mil separation makes it two distinct
> >boundaries), the maximum normal incidence aplitude transmission is about
> >80% at each making 64% for the sheet. This is about 80% power
> >transmission for the sheet.
>
> Sounds like those greenhouse engineering guys were confused when they
> said a single layer of polycarbonate transmits 91-94% of the sun power in
> the third (1994) revision of the NRAES-33 Greenhouse Engineering manual.
For further information see:
"Optics", Jenkins and White
"Optics", Hecht and Zajak
"Optics", Smith and Thomson
the above are elementary, more advanced are:
"Optical Physics", Lipson and Lipson
"On Light", Ditchburn
"Principles of Optics", Born and Wolf
These are all pretty standard books and are easy to find.
Nick Pine
>> >...with an air/polycarbonate boundary followed by a polycarbonate/air
>> >boundary... the maximum normal incidence aplitude transmission is about
>> >80% at each making 64% for the sheet. This is about 80% power
>> >transmission for the sheet...
>For further information see:
>
>"Optics", Jenkins and White...
Let's try again, Chuck. I don't have my Jenkins and White handy, but
Beckman and Duffie ("Solar Engineering of Thermal Processes," 1991, p 217)
say the power reflected from an air/n-index interface at normal incidence
is ((n-1)/(n+1))^2, and page 220 lists n = 1.60 for polycarbonate, which
makes the single-surface reflectance ((1.6-1)/(1.6+1))^2, 0.053, or 0.1065
for a sheet with air on each side, ie 0.893 power transmittance, right?
The NRAES guys measured 0.91 to 0.94. With a lower index polycarbonate?
And what's this nonsense about "aplitude"?
Nick
Chuck Simmons
The correct formula is (n1-n2)/(n1+n2) for amplitude (usually meaning
the elctric field amplitude but it equally applies to the magnetic
field). This formula is correct if the permeabilities are equal. To
optain power from amplitude, you square the amplitude. The formula must
be applied at each interface between materials of different refractive
index. With a polycarbonate sheet, there are two such interfaces. One on
each side.
As to a lower refractive index polycarbonate, that might be possible by
mixing in other materials. There are now ophthalmic plastics rivaling
ophthalmic crown glasses. Why not a lower index instead of pushing for
the highest possible? The low range for glass is below 1.5 and the high
range is above 1.8. Of course, glass is a very magical material in many
ways.
Amplitude is not nonsense. In the case of thin films, it is important
because power loses phase information. It is important other times. I
simply convert when I wish to know power.
I still object that you use numbers for normal incidence when the
geometries you present forbid normal incidence. In your 12' cube post,
there is hardly a time of day when any radiation on your "target" is at
normal incidence.