A pebble bed heat store
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nick pine
Feb 4, 2012, 6:44:38 AM2/4/12
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A 4'x8'x8'-tall plywood box with 8" blocks on the ground, then wire
mesh, then 4' of clean stone, then 8 3' vertical 55 gallon drums, then
4" of airspace at the top would have about 8x55x8.33 = 3665 Btu/F for
the drums.
If the stone has a 40% void ratio and weighs 150 lb/ft^3, it would add
0.16x0.6x150x4x4x8 = 3686 Btu/F.
Stone among the drums would add another 0.16x0.6x150x20.6ft^3 = 297
Btu/F, which does not seem worth adding, since that would increase the
vertical airflow resistance a lot. The drums might sit on 2x4s on
hollow block pillars so the bottoms don't block airflow into the stone
below.
The Dunkle and Ellul correlation (nice name :-) says dP = LG^2/
Rhoa(21+1750Mu/(GD)) Pascals, where L = 1.22m (4') is the length of
the bed in the flow direction. If Rhoa = 1.127 kg/m^3 and Mu = 1.9E-5
Pa-s at 40 C
(104 F) with equivalent pebble diameter D = 23.5 mm (0.92") and void
fraction 0.41, as in SETP example 3.16.1, 25 Pa (0.1" H20) = 1.22G^2/
(1.127x0.0235)(21+17501.9E-5/(0.0235G)) makes G = 0.131 = 1.127V,
which makes V = 0.116m/s, ie 22.8 feet per minute, ie 730 cfm for the
4'x8' bed face, if I did that right, which seems reasonable.
With 140 F (60 C) air, Rhoa = 1.059 kg/m^3 and Mu = 1.99E-5...
SETP equation 3.16.16 says the volumetric heat transfer coefficient Hv
= 650(G/D)^0.7 = 650(0.131/0.0235)^0.7 = 2164 W/m^3K, ie
2164x1.22x1.22x2.44 = 7859 W/K for the whole bed, ie 14893 Btu/h-F,
like 15 car radiators :-)
A flat 1"x300' 13-gallon plastic pipe spiral on top would with C =
13x8.33 = 108 Btu/F and G = 1.5x300xPi/12 = 118 Btu/h-F would have
time constant C/G = 0.92 hours. In a 140 F box, it would heat 13
gallons of 60 F water to 110 in -0.92ln((110-140)/(60-140)) = 0.9
hours. Two spirals in series with 26 vs 13 gallons would effectively
shorten this water heater recovery time.
Nick
mesh, then 4' of clean stone, then 8 3' vertical 55 gallon drums, then
4" of airspace at the top would have about 8x55x8.33 = 3665 Btu/F for
the drums.
If the stone has a 40% void ratio and weighs 150 lb/ft^3, it would add
0.16x0.6x150x4x4x8 = 3686 Btu/F.
Stone among the drums would add another 0.16x0.6x150x20.6ft^3 = 297
Btu/F, which does not seem worth adding, since that would increase the
vertical airflow resistance a lot. The drums might sit on 2x4s on
hollow block pillars so the bottoms don't block airflow into the stone
below.
The Dunkle and Ellul correlation (nice name :-) says dP = LG^2/
Rhoa(21+1750Mu/(GD)) Pascals, where L = 1.22m (4') is the length of
the bed in the flow direction. If Rhoa = 1.127 kg/m^3 and Mu = 1.9E-5
Pa-s at 40 C
(104 F) with equivalent pebble diameter D = 23.5 mm (0.92") and void
fraction 0.41, as in SETP example 3.16.1, 25 Pa (0.1" H20) = 1.22G^2/
(1.127x0.0235)(21+17501.9E-5/(0.0235G)) makes G = 0.131 = 1.127V,
which makes V = 0.116m/s, ie 22.8 feet per minute, ie 730 cfm for the
4'x8' bed face, if I did that right, which seems reasonable.
With 140 F (60 C) air, Rhoa = 1.059 kg/m^3 and Mu = 1.99E-5...
SETP equation 3.16.16 says the volumetric heat transfer coefficient Hv
= 650(G/D)^0.7 = 650(0.131/0.0235)^0.7 = 2164 W/m^3K, ie
2164x1.22x1.22x2.44 = 7859 W/K for the whole bed, ie 14893 Btu/h-F,
like 15 car radiators :-)
A flat 1"x300' 13-gallon plastic pipe spiral on top would with C =
13x8.33 = 108 Btu/F and G = 1.5x300xPi/12 = 118 Btu/h-F would have
time constant C/G = 0.92 hours. In a 140 F box, it would heat 13
gallons of 60 F water to 110 in -0.92ln((110-140)/(60-140)) = 0.9
hours. Two spirals in series with 26 vs 13 gallons would effectively
shorten this water heater recovery time.
Nick
John Canivan
Feb 4, 2012, 11:11:20 AM2/4/12
to suns...@googlegroups.com
Yes Nick this is a great way to store and exchange additional solar heat.
John
John
CJE
Feb 4, 2012, 1:19:42 PM2/4/12
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Weirdly enough I was just reading a carpentry textbook (1985 publish date) and saw that exact (or at least too close to tell the difference without close inspection) block and stone arrangement for storing heat. It was in the solar heating chapter of the book. I was actually looking for ideas in regard to building a complicated set of stairs when I stumbled across it.
stephen
----------
---- Original Message ----- From: "nick pine" <ni...@early.com>
> A 4'x8'x8'-tall plywood box with 8" blocks on the ground, then wire
nick pine
Feb 5, 2012, 2:58:29 AM2/5/12
to sunspace, zome...@zomeworks.com
CJE <cjec...@verizon.net> wrote:
> Weirdly enough I was just reading a carpentry textbook (1985 publish date) and saw that exact (or at least too close to tell the difference without close inspection) block and stone arrangement for storing heat.
Interesting. Can you post the book's author and title?
> Weirdly enough I was just reading a carpentry textbook (1985 publish date) and saw that exact (or at least too close to tell the difference without close inspection) block and stone arrangement for storing heat.
Rich Komp stores solar heat in his Maine hypocaust (hollow blocks
under a concrete slab.) Here's an older system...
http://factsanddetails.com/china.php?itemid=1089&catid=11&subcatid=71
says:
"Many Chinese houses are quite cold in the winters. It is not unusual
for people to wear thermal underwear and heavy coats inside their
houses throughout the winter. Many people in northern China sleep on
or around a kang, a traditional brick bed or concrete platform, built
over a stove, oven or fireplace which is heated with coal, wood or
animal dung and provides warmth in the winter. Kangs are usually
covered with cotton mattresses and colorfully embroidered quilts. They
serve as a couch, bed and work area."
http://en.wikipedia.org/wiki/Kang_bed-stove
http://www.google.com/imgres?imgurl=http://www.hedon.info/imgs/04bedstove550.gif\
&imgrefurl=http://www.hedon.info/
BP38_HouseholdEnergyInHighRegions&h=404&w=550&s\
z=10&tbnid=n86Cc37mtVbXDM:&tbnh=83&tbnw=113&prev=/search%3Fq%3Dkang
%2Bbed%2Bimag\
es%26tbm%3Disch%26tbo%3Du&zoom=1&q=kang+bed
+images&docid=MSoEPJ8IkOo79M&hl=en&sa\
=X&ei=bCouT-zGHIix0AHixf3VCg&sqi=2&ved=0CDUQ9QEwBg&dur=147
An 8' cubical plywood box full of rocks and drums could be a solar
guest kang in a sunspace. At 140 F on an average day, it would store 1
million Btu, the heat equivalent of about 10 gallons of oil or 10
therms of natural gas.
A 8'x8'x6'-tall plywood box with 4" blocks on the ground, then wire
mesh, then 28" of clean stone, then 16 3' vertical 55 gallon drums,
then 4" of airspace at the top would have about 16x55x8.33 = 7330 Btu/
F for the drums. If the stone has a 40% void ratio and weighs 150 lb/
ft^3, it would add 0.16x0.6x150x8x8x28/12 = 2150 Btu/F, enough to
store about (140-70)2150 = 151K Btu/day, ie 151K/6h = 25K Btu/h, eg
25K/(140-70) = 400 cfm of 140 F air from a 200 ft^2 deep mesh air
heater.
Dunkle and Ellul say dP = LG^2/Rhoair(21+1750Mu/(GD)) Pascals, where L
= 0.71m (28") is the length of the bed in the flow direction. If
Rhoair = 1.059 kg/m^3 and Mu = 1.99E-5 Pa-s at 60 C (140 F) with
equivalent pebble diameter D = 23.5 mm (0.92") and void fraction 0.41
and a 400cfm/64ft^2/196.7 = 0.0317 m/s air velocity and G =
0.0317x1.059 = 0.0336 kg/m^2-s mass velocity, dP = 0.71G^2/
(1.059x0.0235)(21+1750x1.99E-5/(0.0235G)) = 2.1 Pa, ie 0.008 "H20, if
I did that right.
SETP equation 3.16.16 says the volumetric heat transfer coefficient Hv
802x0.71x2.44^2 = 3391 W/K for the whole bed, ie 6426 Btu/h-F, like 6
car radiators :-)
Nick
nick pine
Feb 5, 2012, 5:42:42 AM2/5/12
to sunspace
"Brian" <bbkmski@...> wrote:
>I am located in Central Wisconsin.
The worst-case month for solar house heating in Eau Claire is
December, when 430 Btu/ft^2 of sun falls on the ground and 790 hits a
south wall on an average 16.8 F day with a 25.3 high. Brrr.
> I am planning around a 600sq.ft. solar water collector system.
On an average December day, a deep mesh solar air heater might gain
0.8sqrt(430^2+790^2) = 720 Btu/ft^2 and lose 6h(70-21)1ft^2/R2 = 147,
for a net for a net gain of 573.
> House is newer construction 1600 sq.ft., and heated shop area, of about 900 sq.ft.
With 2500 ft^2 of R40 ceiling and 2000 ft^2 of R20 walls and 100 ft^2
of R3 windows and 100 cfm of natural air leakage, it would have about
2500/40+2000/20+100/3+100 = 300 Btu/h-F of thermal conductance.
So it would need about 24h(65-16.8)300 = 342K Btu/day of heat. If 600
kWh/mo of indoor electrical use provides 68K Btu/day of that, it needs
274K Btu/day of solar heat, eg 274K/573 = 478 ft^2 of deep mesh air
heater, or less, with some solar siding on the house.
> The shed is approx. 150ft. away from the house.
Too far to move air back and forth...
> I would put a in ground tank in the shed, and another tank in the house.
With a bidirectional hose between them that's empty most of the time?
> Am I right on my tank sizes?
You could store 5x274K = 1.37 million Btu for 5 cloudy days in 1.37M/
(140-80)/62.33 = 366 ft^3 of water, eg 2740 gallons in 50 55 gallon
drums inside an 8'x32'x6'-tall box and collect 274K Btu/day in C =
274K/(140-80) = 4567 Btu/F of stone, ie 14.3 tons, about 7 cubic yards
in 2 14'x14'x6" layers above and below the drums...
http://wiki.answers.com/Q/What_does_one_cubic_yard_of_river_rock_weigh
The box could have 512 ft^2 of glazing, eg 8 $83 4'x16' sheets of
twinwall polycarbonate at a 60 degree angle, or maybe a double
inflated $30 16'x32' layer of 4-year 6 mil greenhouse polyethylene
film, if we can reduce the radiation loss from a deep mesh.
To collect hot water, sprinkle cool return water from the house over
the top of the upper layer of stones.
Nick
>I am located in Central Wisconsin.
The worst-case month for solar house heating in Eau Claire is
December, when 430 Btu/ft^2 of sun falls on the ground and 790 hits a
south wall on an average 16.8 F day with a 25.3 high. Brrr.
> I am planning around a 600sq.ft. solar water collector system.
On an average December day, a deep mesh solar air heater might gain
0.8sqrt(430^2+790^2) = 720 Btu/ft^2 and lose 6h(70-21)1ft^2/R2 = 147,
for a net for a net gain of 573.
> House is newer construction 1600 sq.ft., and heated shop area, of about 900 sq.ft.
With 2500 ft^2 of R40 ceiling and 2000 ft^2 of R20 walls and 100 ft^2
of R3 windows and 100 cfm of natural air leakage, it would have about
2500/40+2000/20+100/3+100 = 300 Btu/h-F of thermal conductance.
So it would need about 24h(65-16.8)300 = 342K Btu/day of heat. If 600
kWh/mo of indoor electrical use provides 68K Btu/day of that, it needs
274K Btu/day of solar heat, eg 274K/573 = 478 ft^2 of deep mesh air
heater, or less, with some solar siding on the house.
> The shed is approx. 150ft. away from the house.
Too far to move air back and forth...
> I would put a in ground tank in the shed, and another tank in the house.
With a bidirectional hose between them that's empty most of the time?
> Am I right on my tank sizes?
You could store 5x274K = 1.37 million Btu for 5 cloudy days in 1.37M/
(140-80)/62.33 = 366 ft^3 of water, eg 2740 gallons in 50 55 gallon
drums inside an 8'x32'x6'-tall box and collect 274K Btu/day in C =
274K/(140-80) = 4567 Btu/F of stone, ie 14.3 tons, about 7 cubic yards
in 2 14'x14'x6" layers above and below the drums...
http://wiki.answers.com/Q/What_does_one_cubic_yard_of_river_rock_weigh
The box could have 512 ft^2 of glazing, eg 8 $83 4'x16' sheets of
twinwall polycarbonate at a 60 degree angle, or maybe a double
inflated $30 16'x32' layer of 4-year 6 mil greenhouse polyethylene
film, if we can reduce the radiation loss from a deep mesh.
To collect hot water, sprinkle cool return water from the house over
the top of the upper layer of stones.
Nick
CJE
Feb 5, 2012, 9:32:47 AM2/5/12
to suns...@googlegroups.com
The book title is: Carpentry
Author: Leonard Koel
LOCN # 82-084716
ISBN # 0-8269-0728-8
American Technical Publishers - 1985
Solar Heat Storage - Page # 433
My daughter is down from Pace U. for the weekend, came into the library, saw me reading it, and asked:
"What are you reading?"
I held the book up so she could see the cover - it's a large book; like a textbook.
"Oh; car-pen-try - what's it about?"
"Building things out of wood."
"Oh. It looks like a textbook."
"Maybe it is - I bought it a long time ago - I forget why now."
"So you are just reading a textbook? For no reason?"
"It's very interesting. I am going to build a spiral staircase here. I thought it might have some ideas for doing that."
You must be the only person in the world reading a textbook just for fun.
We laughed about it for a while as she was coming in to do about the opposite - something with Facebook. <g>
stephen
-----------
Nathan Hurst
Feb 5, 2012, 12:46:20 PM2/5/12
to suns...@googlegroups.com
A textbook can be just as good as a murder thriller, scifi or
romance. And vastly more satisfying when you've read it.
romance. And vastly more satisfying when you've read it.
njh
nick pine
Feb 6, 2012, 10:15:55 AM2/6/12
to sunspace, jkpr...@verizon.net
A 6'x8'x8' tall 140 F box with 320 ft^2 of R32 surface would lose
24h(140-30)10 = 26K Btu/day in my 30 F sunspace. Behind 8'x16' of
sunspace double glazing, it would gain about 87K Btu on an average
January day.
When fully discharged, it would need at most 87K/(140-70) = 1243 Btu/F
of fast thermal mass, eg 1243/0.16 = 7768 pounds of stone, about 52
ft^3 in 2 6" layers above and below 24 55 gallon plastic water drums
stacked in 2 3x4 arrays.
I don't know how many drums are needed. This 1820 stone farmhouse
might catch on fire with active knob and tube wiring buried in
cellulose insulation running 10 1500W space heaters with thermostats.
The box could be up to 16' tall with more support and another 24 drums
on top. The house presently has about 400 cfm of natural air leakage.
With 2 cfm/ft^2 of glazing, the box would have 256 cfm of charging
airflow, which could come from a $70 Lasko 2155A 3-speed reversible
16" window fan at the bottom that can move up to 2470 cfm with 90
watts.
Dunkle and Ellul say dP = LG^2/Rhoair(21+1750Mu/(GD)) Pascals, where L
= 0.3048m (1') is the length of the bed in the flow direction. If
Rhoair = 1.059 kg/m^3 and V = 256cfm/4'/8' = 8 fpm (0.041 m/s) and G =
1.059V = 0.043 kg/m^2-s and Mu = 1.99E-5 Pa-s at 60 C (140 F) with
equivalent pebble diameter D = 23.5 mm (0.92"), then dP = 0.3048G^2/
(1.059x0.0235)(21+1750x1.99E-5/(0.0235G)) = 1.26 Pascals, ie 0.005
"H20 :-)
SETP equation 3.16.16 says the volumetric heat transfer coefficient Hv
= 650(G/D)^0.7 = 992 W/m^3K, ie 992x1x6x8x0.3048^3 = 1348 W/K for the
whole bed, ie 2555 Btu/h-F, like 2.6 car radiators. The air-stone temp
diff would be 15K/2555 = 6 F. The bottom might well have more stone.
On a cloudy day the fan could run harder, pushing house air into the
box bottom. At 0.1" H20, 25 Pa = 0.3048G^2/(1.059x0.0235)
(21+1750x1.99E-5/(0.0235G)) makes G = 0.278 = 1.059V, ie V = 0.263 m/
s, ie 45 fpm, ie 2144 cfm. At 1000 cfm, G = 0.112 and Hv = 1940 W/
m^3K, ie 2636 W/K for the whole bed, ie 5K Btu/F, like 5 car
radiators.
Nick
24h(140-30)10 = 26K Btu/day in my 30 F sunspace. Behind 8'x16' of
sunspace double glazing, it would gain about 87K Btu on an average
January day.
When fully discharged, it would need at most 87K/(140-70) = 1243 Btu/F
of fast thermal mass, eg 1243/0.16 = 7768 pounds of stone, about 52
ft^3 in 2 6" layers above and below 24 55 gallon plastic water drums
stacked in 2 3x4 arrays.
I don't know how many drums are needed. This 1820 stone farmhouse
might catch on fire with active knob and tube wiring buried in
cellulose insulation running 10 1500W space heaters with thermostats.
The box could be up to 16' tall with more support and another 24 drums
on top. The house presently has about 400 cfm of natural air leakage.
With 2 cfm/ft^2 of glazing, the box would have 256 cfm of charging
airflow, which could come from a $70 Lasko 2155A 3-speed reversible
16" window fan at the bottom that can move up to 2470 cfm with 90
watts.
Dunkle and Ellul say dP = LG^2/Rhoair(21+1750Mu/(GD)) Pascals, where L
Rhoair = 1.059 kg/m^3 and V = 256cfm/4'/8' = 8 fpm (0.041 m/s) and G =
1.059V = 0.043 kg/m^2-s and Mu = 1.99E-5 Pa-s at 60 C (140 F) with
equivalent pebble diameter D = 23.5 mm (0.92"), then dP = 0.3048G^2/
(1.059x0.0235)(21+1750x1.99E-5/(0.0235G)) = 1.26 Pascals, ie 0.005
"H20 :-)
SETP equation 3.16.16 says the volumetric heat transfer coefficient Hv
whole bed, ie 2555 Btu/h-F, like 2.6 car radiators. The air-stone temp
diff would be 15K/2555 = 6 F. The bottom might well have more stone.
On a cloudy day the fan could run harder, pushing house air into the
box bottom. At 0.1" H20, 25 Pa = 0.3048G^2/(1.059x0.0235)
(21+1750x1.99E-5/(0.0235G)) makes G = 0.278 = 1.059V, ie V = 0.263 m/
s, ie 45 fpm, ie 2144 cfm. At 1000 cfm, G = 0.112 and Hv = 1940 W/
m^3K, ie 2636 W/K for the whole bed, ie 5K Btu/F, like 5 car
radiators.
Nick
nick pine
Feb 21, 2012, 7:54:45 AM2/21/12
to sunspace
Here's the solar yard heat store I'm planning now:
12'
................... partial parts list
. . . . . . . . . .
. . . . 15 plastic 55 gallon drums
. . . . 2 10'x16' pieces of EPDM
. . . . 6 tons of 2" clean stone
. . . . 16 3'x4"-D PVC pipes
. . plan view . . 22.9' 2x12'x14' ft^2 wood pallets
. . . . 3x12'x14' ft^2 wire mesh
. . . . 2x12'x16' IR plastic film
. . . . 1x12'x16' 50% greenhouse shadecloth
. . . . 1x12'x16' weed barrier
. . . . . . . . . . 480 ft^2 R30 insulation
................... Grainger 10" fan
12'
................... .
. . / . .
. . 16'. east . <-- 2 12'x16' layers
. . / . elevation . of IR greenhouse film
. south . . . .
. elevation . 13.9' ......collar beam.... \
. . . . . . . . . . . . 16'
. . . . . . . . . . 8'. . . . \
. . . 6'. heat store . .
. . . . . 60.
.......................................................
|4.3'| 14' | 4.6'|
... 70 F air would inflate the space between the 2 layers of IR
greenhouse film and flow through 1/2" holes on a 1' grid in
the inner film into 3 N-S rows of 5 plastic 55 gallon drums in
a 12'x14'x6'H box which support 2 3'Wx4'Hx10'L water troughs
surrounded by 6 tons of rocks with about 500 ft^3 (4000 gallons)
of water, like this:
south south
^ ^
| 12' |
.................................
i insulation i
. . . . . . . . .
n n
. . D D D D D . .
s p p p p s
. . 4'x10'x3'H . . plan view
u u
. . EPDM tank . .
l p p p p l
. . D D D D D . . 14'
a p p p p a
. . 4'x10'x3'H . . east -->
t t
. . EPDM tank . .
i p p p p i p are 3'x4" pipes to
. . D D D D D . . connect the airspaces
o o above and below the drums.
. . . . . . . . .
n insulation n
.................................
south south
^ ^
| 6' |
.........................................
i insulation 8"
. . . . . . . . . . . --> down
n w | 4" w | 4" w | w f
. . D R U M . . 24"
s i a o i p o i p i o
. . 4'x10'x3'H . .
u r i f r a f r a r a 48"
. . EPDM tank . .
l e r s e l s e l e m
. . D R U M . . 24" 14'
a m g t m l t m l m b
. . 4'x10'x3'H . .
t e a o e e o e e e o 48"
. . EPDM tank . .
i s p n s t n s t s r
. . D R U M . . 24"
o h | e h | e h | h d
. . . . . . . . . . . --> down
n insulation 8"
.........................................
8" 4" 4" 5" 36" 4" 5" 2"
This would have about C = 3'x12'x14'x62.33 = 31.4K Btu/F of thermal
capacitance. Cooling it from 140 to 70 F would release (140-70)C
= 2.2 million Btu, equivalent to 2.2M/130K = 17 gallons of oil.
http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/bluebook/data/13739.SBF
says 620 Btu/ft^2 of sun falls on the ground and 1000 falls on south
walls on an average 30.4 F January day with a 37.9 high and a 30.4
average daytime temp near Philadelphia. On a clear day, 890 falls
on the ground and 1880 falls on south walls.
On an average day, a south wall with a 60 degree slope would receive
620cos60+1000sin60 = 1176 Btu/ft^2 of sun. This store could collect
0.8x1176x12'x16' = 180.6K Btu/day and lose 6h(70-34)12'x16'/R1 =
41.5K from the glazing, for a net gain of 139.1K Btu.
Storing 139K Btu over a 6 hour January solar collection day at a rate
of 139.1K/6h = 23.2K Btu/h with air cooling from 140 to 70 F requires
about 23.2K/(140-70) = 331 cfm of airflow, which could come from
a 26 watt Grainger 10" fan. If the entire stone bed warms from 70 to
140 F, it needs a 139.1K/(140-70) = 1987 Btu/F heat capacitance, eg
1987/0.16 = 12.4K lb of stone, ie 12.4K/3500 = 3.5 yd^3 or 100 ft^3
in 2 12x100ft^3/(12'x14')/2 = 4" layers above and below the drums.
A 12'x14'x6' tall 140 F box with 648 ft^2 of R30 surface and a 648/30
= 22 Btu/h-F thermal conductance would lose 24h(140-30)22 = 57K Btu
on an average 30 F January day near Phila, for a net gain of about
139.1K-57K = 82.1K, which should be plenty for hot water for showers
and trickle charging for cloudy days, since 5 cloudy days in a row
only happen 3% of the time, if cloudy days are like coin flips.
A yard furnace that heats a house on average days needs more glazing.
Dunkle and Ellul say dP = LG^2/Rhoair(21+1750Mu/(GD)) Pascals, where
L = 0.2032m (8/12') is the length of the bed in the flow direction.
If Rhoair = 1.059 kg/m^3 and V = 331cfm/12'/14' = 2 fpm (0.01 m/s)
and G = 1.059V = 0.011 kg/m^2-s and Mu = 1.99E-5 Pa-s at 60 C (140 F)
ie 0.000063 "H20 :-)
SETP equation 3.16.16 says the volumetric heat transfer coefficient
Hv = 650(G/D)^0.7 = 357 W/m^3K, ie 357x8/12x12x14x0.3048^3 = 1134 W/K
for the whole bed, ie 2148 Btu/h-F, like 2.1 car radiators. The air-
stone temp diff would be 15K/2148 = 6 F. The bottom layer could use
more stone.
A $70 Lasko 2155A 3-speed reversible 16" window fan can move up to
2470 cfm with 90 watts. On a cloudy day, it could push lots of house
air into the bottom. At 0.1" H20, 25 Pa = 0.2032G^2/(1.059x0.0235)
(21+1750x1.99E-5/(0.0235G)) makes G = 0.35 = 1.059V, ie V = 0.33 m/s,
ie 65 fpm, ie 65x12'x14' = 10.8K cfm. At 2000 cfm, 2000/12'/14' = 12
fpm, ie V = 0.06 m/s, so G = 0.064 and Hv = 1311 W/K, ie 4159 W/K
for the whole bed, ie 7.9K Btu/h-F, like 8 car radiators.
This heat store needs some dampers to avoid moving house air through
the poly film pillow on cloudy days.
Nick
12'
................... partial parts list
. . . . . . . . . .
. . . . 15 plastic 55 gallon drums
. . . . 2 10'x16' pieces of EPDM
. . . . 6 tons of 2" clean stone
. . . . 16 3'x4"-D PVC pipes
. . plan view . . 22.9' 2x12'x14' ft^2 wood pallets
. . . . 3x12'x14' ft^2 wire mesh
. . . . 2x12'x16' IR plastic film
. . . . 1x12'x16' 50% greenhouse shadecloth
. . . . 1x12'x16' weed barrier
. . . . . . . . . . 480 ft^2 R30 insulation
................... Grainger 10" fan
12'
................... .
. . / . .
. . 16'. east . <-- 2 12'x16' layers
. . / . elevation . of IR greenhouse film
. south . . . .
. elevation . 13.9' ......collar beam.... \
. . . . . . . . . . . . 16'
. . . . . . . . . . 8'. . . . \
. . . 6'. heat store . .
. . . . . 60.
.......................................................
|4.3'| 14' | 4.6'|
... 70 F air would inflate the space between the 2 layers of IR
greenhouse film and flow through 1/2" holes on a 1' grid in
the inner film into 3 N-S rows of 5 plastic 55 gallon drums in
a 12'x14'x6'H box which support 2 3'Wx4'Hx10'L water troughs
surrounded by 6 tons of rocks with about 500 ft^3 (4000 gallons)
of water, like this:
south south
^ ^
| 12' |
.................................
i insulation i
. . . . . . . . .
n n
. . D D D D D . .
s p p p p s
. . 4'x10'x3'H . . plan view
u u
. . EPDM tank . .
l p p p p l
. . D D D D D . . 14'
a p p p p a
. . 4'x10'x3'H . . east -->
t t
. . EPDM tank . .
i p p p p i p are 3'x4" pipes to
. . D D D D D . . connect the airspaces
o o above and below the drums.
. . . . . . . . .
n insulation n
.................................
south south
^ ^
| 6' |
.........................................
i insulation 8"
. . . . . . . . . . . --> down
n w | 4" w | 4" w | w f
. . D R U M . . 24"
s i a o i p o i p i o
. . 4'x10'x3'H . .
u r i f r a f r a r a 48"
. . EPDM tank . .
l e r s e l s e l e m
. . D R U M . . 24" 14'
a m g t m l t m l m b
. . 4'x10'x3'H . .
t e a o e e o e e e o 48"
. . EPDM tank . .
i s p n s t n s t s r
. . D R U M . . 24"
o h | e h | e h | h d
. . . . . . . . . . . --> down
n insulation 8"
.........................................
8" 4" 4" 5" 36" 4" 5" 2"
This would have about C = 3'x12'x14'x62.33 = 31.4K Btu/F of thermal
capacitance. Cooling it from 140 to 70 F would release (140-70)C
= 2.2 million Btu, equivalent to 2.2M/130K = 17 gallons of oil.
http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/bluebook/data/13739.SBF
says 620 Btu/ft^2 of sun falls on the ground and 1000 falls on south
walls on an average 30.4 F January day with a 37.9 high and a 30.4
average daytime temp near Philadelphia. On a clear day, 890 falls
on the ground and 1880 falls on south walls.
On an average day, a south wall with a 60 degree slope would receive
620cos60+1000sin60 = 1176 Btu/ft^2 of sun. This store could collect
0.8x1176x12'x16' = 180.6K Btu/day and lose 6h(70-34)12'x16'/R1 =
41.5K from the glazing, for a net gain of 139.1K Btu.
Storing 139K Btu over a 6 hour January solar collection day at a rate
of 139.1K/6h = 23.2K Btu/h with air cooling from 140 to 70 F requires
about 23.2K/(140-70) = 331 cfm of airflow, which could come from
a 26 watt Grainger 10" fan. If the entire stone bed warms from 70 to
140 F, it needs a 139.1K/(140-70) = 1987 Btu/F heat capacitance, eg
1987/0.16 = 12.4K lb of stone, ie 12.4K/3500 = 3.5 yd^3 or 100 ft^3
in 2 12x100ft^3/(12'x14')/2 = 4" layers above and below the drums.
A 12'x14'x6' tall 140 F box with 648 ft^2 of R30 surface and a 648/30
= 22 Btu/h-F thermal conductance would lose 24h(140-30)22 = 57K Btu
on an average 30 F January day near Phila, for a net gain of about
139.1K-57K = 82.1K, which should be plenty for hot water for showers
and trickle charging for cloudy days, since 5 cloudy days in a row
only happen 3% of the time, if cloudy days are like coin flips.
A yard furnace that heats a house on average days needs more glazing.
Dunkle and Ellul say dP = LG^2/Rhoair(21+1750Mu/(GD)) Pascals, where
If Rhoair = 1.059 kg/m^3 and V = 331cfm/12'/14' = 2 fpm (0.01 m/s)
and G = 1.059V = 0.011 kg/m^2-s and Mu = 1.99E-5 Pa-s at 60 C (140 F)
with equivalent pebble diameter D = 23.5 mm (0.92"), then dP =
0.2032G^2/(1.059x0.0235)(21+1750x1.99E-5/(0.0235G)) = 0.16 Pascals,
ie 0.000063 "H20 :-)
SETP equation 3.16.16 says the volumetric heat transfer coefficient
for the whole bed, ie 2148 Btu/h-F, like 2.1 car radiators. The air-
stone temp diff would be 15K/2148 = 6 F. The bottom layer could use
more stone.
A $70 Lasko 2155A 3-speed reversible 16" window fan can move up to
2470 cfm with 90 watts. On a cloudy day, it could push lots of house
air into the bottom. At 0.1" H20, 25 Pa = 0.2032G^2/(1.059x0.0235)
(21+1750x1.99E-5/(0.0235G)) makes G = 0.35 = 1.059V, ie V = 0.33 m/s,
ie 65 fpm, ie 65x12'x14' = 10.8K cfm. At 2000 cfm, 2000/12'/14' = 12
fpm, ie V = 0.06 m/s, so G = 0.064 and Hv = 1311 W/K, ie 4159 W/K
for the whole bed, ie 7.9K Btu/h-F, like 8 car radiators.
This heat store needs some dampers to avoid moving house air through
the poly film pillow on cloudy days.
Nick
nick pine
Feb 22, 2012, 5:23:54 AM2/22/12
to sunspace, jkpr...@verizon.net
> ... 70 F air would inflate the space between the 2 layers of IR greenhouse film and flow through 1/2" holes in the inner IR film...
The recommended inflation pressure for double polyethylene film
greenhouse covers is 0.25 "H20, ie 0.25/12x62.33 = 1.3 psf. What
should the 1/2" hole spacing be to inflate the films to 0.25" H20 if
331 cfm flows into a 12'x16' pillow at 331cfm/(12'x16') = 1.72 cfm/
ft^2?
One empirical chimney formula says cfm = 16.6Asqrt(HdT) for 2 A ft^2
vents with an H' vertical separation and a dT (F) temperature
difference between the top and bottom vents. With a 0.25" H20 HdT
bouyancy and H = 100' and a 70 F lower vent temp and 70 F air weighing
0.075 lb/ft^2, 1.3 psf = 100x0.075(530/530-530/(460+T)) makes T = 181
F and HdT = 100(181-70) = 11113 ft-F. If Q cfm = 16.6Asqrt(11113) and
A = 0.00136 in^2 for a 1/2" diameter hole, Q = 2.39 cfm, so each hole
should cover 2.39/1.72 = 1.39 ft^2, ie 200 in^2, ie a 14"x14" square.
So the inner glazing would have 12'x16'/1.39ft^2 = 138 holes. These
might be punched with a 1/2 steel pipe nipple with one end sharpened
with a grinder and a pipe cap screwed over the other end and a block
of wood under the poly film. It seems to me these holes would cool the
inner film sufficiently, since it would be absorbing longwave IR from
the mesh but it wouldn't absorb much incoming sun. A 3/8" hole would
allow better cooling, with Q = 1.34 cfm and 0.78 ft^2/hole and 246
holes on a 10.5" grid in the inner film, but that's more work.
Nick
The recommended inflation pressure for double polyethylene film
greenhouse covers is 0.25 "H20, ie 0.25/12x62.33 = 1.3 psf. What
should the 1/2" hole spacing be to inflate the films to 0.25" H20 if
331 cfm flows into a 12'x16' pillow at 331cfm/(12'x16') = 1.72 cfm/
ft^2?
One empirical chimney formula says cfm = 16.6Asqrt(HdT) for 2 A ft^2
vents with an H' vertical separation and a dT (F) temperature
difference between the top and bottom vents. With a 0.25" H20 HdT
bouyancy and H = 100' and a 70 F lower vent temp and 70 F air weighing
0.075 lb/ft^2, 1.3 psf = 100x0.075(530/530-530/(460+T)) makes T = 181
F and HdT = 100(181-70) = 11113 ft-F. If Q cfm = 16.6Asqrt(11113) and
A = 0.00136 in^2 for a 1/2" diameter hole, Q = 2.39 cfm, so each hole
should cover 2.39/1.72 = 1.39 ft^2, ie 200 in^2, ie a 14"x14" square.
So the inner glazing would have 12'x16'/1.39ft^2 = 138 holes. These
might be punched with a 1/2 steel pipe nipple with one end sharpened
with a grinder and a pipe cap screwed over the other end and a block
of wood under the poly film. It seems to me these holes would cool the
inner film sufficiently, since it would be absorbing longwave IR from
the mesh but it wouldn't absorb much incoming sun. A 3/8" hole would
allow better cooling, with Q = 1.34 cfm and 0.78 ft^2/hole and 246
holes on a 10.5" grid in the inner film, but that's more work.
Nick
nick pine
Feb 22, 2012, 6:29:38 AM2/22/12
to sunspace, jkpr...@verizon.net
Here's the solar yard heat store I'm planning now:
12'
................... partial parts list
. . . . . . . . . .
. . . . 15 plastic 55 gallon drums
. . . . 2 10'x16' pieces of EPDM
. . . . 6 tons of 1/2"-2" clean stone
12'
................... partial parts list
. . . . . . . . . .
. . . . 15 plastic 55 gallon drums
. . . . 2 10'x16' pieces of EPDM
. . . . 16 3'x4"-D perforated PVC pipes
. . plan view . . 22.9' 2x12'x14' ft^2 wood pallets
. . . . 3x12'x14' ft^2 wire mesh
. . . . 2x12'x16' IR plastic film
. . . . 1x12'x16' 50% greenhouse shadecloth
. . . . 1x12'x16' weed barrier
. . . . . . . . . . 480 ft^2 R30 insulation
................... Grainger 10" fan
12'
................... .
. . / . .
. . 16'. east . <-- 2 12'x16' layers
. . / . elevation . of IR greenhouse film
. south . . . .
. elevation . 13.9' ......collar beam.... \
. . . . . . . . . . . . 16'
. . . . . . . . . . 8'. . . . \
. . . 6'. heat store . .
. . . . . 60.
.......................................................
|4.3'| 14' | 4.6'|
. . . . 3x12'x14' ft^2 wire mesh
. . . . 2x12'x16' IR plastic film
. . . . 1x12'x16' 50% greenhouse shadecloth
. . . . 1x12'x16' weed barrier
. . . . . . . . . . 480 ft^2 R30 insulation
................... Grainger 10" fan
12'
................... .
. . / . .
. . 16'. east . <-- 2 12'x16' layers
. . / . elevation . of IR greenhouse film
. south . . . .
. elevation . 13.9' ......collar beam.... \
. . . . . . . . . . . . 16'
. . . . . . . . . . 8'. . . . \
. . . 6'. heat store . .
. . . . . 60.
.......................................................
|4.3'| 14' | 4.6'|
Storing 139K Btu over a 6 hour January solar collection day at a rate
of 139.1K/6h = 23.2K Btu/h with air cooling from 140 to 70 F requires
about 23.2K/(140-70) = 331 cfm of airflow, which could come from
a 26 watt Grainger 10" fan...
of 139.1K/6h = 23.2K Btu/h with air cooling from 140 to 70 F requires
about 23.2K/(140-70) = 331 cfm of airflow, which could come from
A $70 Lasko 2155A 3-speed reversible 16" window fan can move up to
2470 cfm with 90 watts. On a cloudy day, it could push lots of house
This heat store needs some dampers to avoid moving house air through
the poly film pillow on cloudy days.
. .
<-- to . airspace . <-- from
house . . collector
. . . . . . . . . . . . . . . . . . .
. .
. ^ | .
6'. | heat store v .
. .
. .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lasko .. house . Grainger .. collector
fan . . damper airspace . fan . . damper
--> . . . --> . .
..................................................................
| 14' |
To charge the heat store, the Grainger fan would move air out from
the pallet airspace under the stone at the bottom and back into
the airspace over the stone at the top, with the house damper closed.
This http://www.pexsupply.com/EWC-12X12ND-12-x-12-ND-Motorized-Damper
could be the $141 2-watt collector damper.
To discharge, close the collector damper and move house air with
the Lasko fan into the lower airspace and let hot air flow from
the upper airspace into the house via this $160 2-watt damper
http://www.pexsupply.com/EWC-18X18ND-18-x-18-ND-Motorized-Damper
Open both dampers and run both fans to do both at the same time.
Nick
nick pine
Feb 23, 2012, 11:18:57 AM2/23/12
to sunspace, jkpr...@verizon.net
With 2 layers of IR poly and a single layer of 50% shadecloth and a
black backwall...
20 S=1.714E-09'Stefan-Boltzmann constant
30 TA=34'ambient temp (F)
40 SUN=910/6'sun through outer poly film (Btu/ft^2-h)
50 TR=70'air heater inlet temp (F)
60 CFM=1'airflow/ft^2
70 C=10'relaxation damping cap
80 T1=50:T2=100:TS=100:TB=100'initial temps (F)
90 FOR I=1 TO 499'relaxation iterations
100 G1=(TR-T1)/(2/3+1/CFM)'outer poly air gain
105 GR1=(T2-T1)*G12R'rad gain
110 G1FLOW=(TA-T1)/1+G1+GR1'heatflow into outer poly
120 T1=T1+G1FLOW/C'new outer poly temp
130 T1R=INT(T1+.5)'round to print
140 T12=TR-G1/CFM'outer-inner poly air temp
150 T12R=INT(T12+.5)'round to print
160 G2=(T12-T2)/(2/3+1/CFM)'inner poly air gain
165 GR2=(T1-T2)*G12R+(TS-T2)*G2SR+(TB-T2)*G2BR'rad gain
170 G2FLOW=0*SUN+G2+GR2'heatflow into inner poly
180 T2=T2+G2FLOW/C'new inner poly temp
190 T2R=INT(T2+.5)'round to print
200 T2S=T12-G2/CFM'inner poly-shadecloth air temp
210 T2SR=INT(T2S+.5)'round to print
220 GS=(T2S-TS)/(2/3+1/CFM)'shadecloth air gain
225 GSR=(T2-TS)*G2SR+(TB-TS)*GSBR'rad gain
230 SCFLOW=.46*SUN+GS+GSR'heatflow into shadecloth
240 TS=TS+SCFLOW/C'new shadecloth temp
250 TSR=INT(TS+.5)'round to print
260 TSB=T12-GS/CFM'shadecloth-back wall air temp
270 TSBR=INT(TSB+.5)'round to print
280 GBW=(TSB-TB)/(2/3+1/CFM)'back wall air gain
285 GRBW=(T2-TB)*G2BR+(TS-TB)*GSBR'rad gain
290 BWFLOW=.54*SUN+GBW+GRBW'heatflow into back wall
300 TB=TB+BWFLOW/C'new back wall temp
310 TBR=INT(TB+.5)'round to print
320 TOUT=TSB-GBW/CFM'back wall outlet air temp
330 TOUTR=INT(TOUT+.5)'round to print
340 IF I>490 THEN PRINT 500+I"'";T1R;T12R;T2R;T2SR;TSR;TSBR;TBR;TOUTR
341 G12R=1*4*S*(460+(T1+T2)/2)^3'outer-inner poly radcon
344 G2SR=.5*4*S*(460+(T2+TS)/2)^3'inner poly-shadecloth radcon
345 G2BR=.5*4*S*(460+(T2+TB)/2)^3'inner poly-back wall radcon
346 GSBR=.5*4*S*(460+(TS+TB)/2)^3'shadecloth-back wall radcon
350 NEXT
T1 air T2 air Ts air Tb air output
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
... 167 Btu/h-ft^2 sun can heat 1 cfm of 70 F air to 150 F with
100(150-70)/167 = 48% efficiency. With 79 F R1 IR poly outer film
losing 79-34 = 45 Btu/h-ft^2 on a 34 F day, it's probably worth adding
another layer of 15 cent/ft^2 50% black greenhouse shadecloth to raise
the collection efficiency.
Nick
PS: PE Drew Gillett suggests folding the inner poly film in order to
drill or punch the multiple holes through several layers at once.
black backwall...
20 S=1.714E-09'Stefan-Boltzmann constant
30 TA=34'ambient temp (F)
40 SUN=910/6'sun through outer poly film (Btu/ft^2-h)
50 TR=70'air heater inlet temp (F)
60 CFM=1'airflow/ft^2
70 C=10'relaxation damping cap
80 T1=50:T2=100:TS=100:TB=100'initial temps (F)
90 FOR I=1 TO 499'relaxation iterations
100 G1=(TR-T1)/(2/3+1/CFM)'outer poly air gain
105 GR1=(T2-T1)*G12R'rad gain
110 G1FLOW=(TA-T1)/1+G1+GR1'heatflow into outer poly
120 T1=T1+G1FLOW/C'new outer poly temp
130 T1R=INT(T1+.5)'round to print
140 T12=TR-G1/CFM'outer-inner poly air temp
150 T12R=INT(T12+.5)'round to print
160 G2=(T12-T2)/(2/3+1/CFM)'inner poly air gain
165 GR2=(T1-T2)*G12R+(TS-T2)*G2SR+(TB-T2)*G2BR'rad gain
170 G2FLOW=0*SUN+G2+GR2'heatflow into inner poly
180 T2=T2+G2FLOW/C'new inner poly temp
190 T2R=INT(T2+.5)'round to print
200 T2S=T12-G2/CFM'inner poly-shadecloth air temp
210 T2SR=INT(T2S+.5)'round to print
220 GS=(T2S-TS)/(2/3+1/CFM)'shadecloth air gain
225 GSR=(T2-TS)*G2SR+(TB-TS)*GSBR'rad gain
230 SCFLOW=.46*SUN+GS+GSR'heatflow into shadecloth
240 TS=TS+SCFLOW/C'new shadecloth temp
250 TSR=INT(TS+.5)'round to print
260 TSB=T12-GS/CFM'shadecloth-back wall air temp
270 TSBR=INT(TSB+.5)'round to print
280 GBW=(TSB-TB)/(2/3+1/CFM)'back wall air gain
285 GRBW=(T2-TB)*G2BR+(TS-TB)*GSBR'rad gain
290 BWFLOW=.54*SUN+GBW+GRBW'heatflow into back wall
300 TB=TB+BWFLOW/C'new back wall temp
310 TBR=INT(TB+.5)'round to print
320 TOUT=TSB-GBW/CFM'back wall outlet air temp
330 TOUTR=INT(TOUT+.5)'round to print
340 IF I>490 THEN PRINT 500+I"'";T1R;T12R;T2R;T2SR;TSR;TSBR;TBR;TOUTR
341 G12R=1*4*S*(460+(T1+T2)/2)^3'outer-inner poly radcon
344 G2SR=.5*4*S*(460+(T2+TS)/2)^3'inner poly-shadecloth radcon
345 G2BR=.5*4*S*(460+(T2+TB)/2)^3'inner poly-back wall radcon
346 GSBR=.5*4*S*(460+(TS+TB)/2)^3'shadecloth-back wall radcon
350 NEXT
T1 air T2 air Ts air Tb air output
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
79 75 120 102 168 114 174 150
... 167 Btu/h-ft^2 sun can heat 1 cfm of 70 F air to 150 F with
100(150-70)/167 = 48% efficiency. With 79 F R1 IR poly outer film
losing 79-34 = 45 Btu/h-ft^2 on a 34 F day, it's probably worth adding
another layer of 15 cent/ft^2 50% black greenhouse shadecloth to raise
the collection efficiency.
Nick
PS: PE Drew Gillett suggests folding the inner poly film in order to
drill or punch the multiple holes through several layers at once.
nick pine
Feb 24, 2012, 5:19:18 AM2/24/12
to sunspace, jkpr...@verizon.net
With 2 layers of IR poly and a single layer of 50% shadecloth and a
black backwall, after fixing another bug...
20 S=1.714E-09'Stefan-Boltzmann constant
30 TA=34'ambient temp (F)
50 TR=70'air heater inlet temp (F)
60 CFM=1'airflow/ft^2
70 C=10'relaxation damping cap
80 T1=50:T2=100:TS=100:TB=100'initial temps (F)
90 FOR I=1 TO 499'relaxation iterations
100 G1=(TR-T1)/(2/3+1/CFM)'outer poly film air gain
60 CFM=1'airflow/ft^2
70 C=10'relaxation damping cap
80 T1=50:T2=100:TS=100:TB=100'initial temps (F)
90 FOR I=1 TO 499'relaxation iterations
105 GR1=(T2-T1)*G12R'rad gain
110 G1FLOW=(TA-T1)/1+G1+GR1'heatflow into outer poly
120 T1=T1+G1FLOW/C'new outer poly temp
130 T1R=INT(T1+.5)'round to print
140 T12=TR-G1/CFM'outer-inner poly air temp
150 T12R=INT(T12+.5)'round to print
160 G2=(T12-T2)/(2/3+1/CFM)'inner poly air gain
165 GR2=(T1-T2)*G12R+(TS-T2)*G2SR+(TB-T2)*G2BR'rad gain
170 G2FLOW=G2+GR2'heatflow into inner poly
110 G1FLOW=(TA-T1)/1+G1+GR1'heatflow into outer poly
120 T1=T1+G1FLOW/C'new outer poly temp
130 T1R=INT(T1+.5)'round to print
140 T12=TR-G1/CFM'outer-inner poly air temp
150 T12R=INT(T12+.5)'round to print
160 G2=(T12-T2)/(2/3+1/CFM)'inner poly air gain
165 GR2=(T1-T2)*G12R+(TS-T2)*G2SR+(TB-T2)*G2BR'rad gain
180 T2=T2+G2FLOW/C'new inner poly temp
190 T2R=INT(T2+.5)'round to print
200 T2S=T12-G2/CFM'inner poly-shadecloth air temp
210 T2SR=INT(T2S+.5)'round to print
220 GS=(T2S-TS)/(2/3+1/CFM)'shadecloth air gain
225 GSR=(T2-TS)*G2SR+(TB-TS)*GSBR'rad gain
230 GSFLOW=69+GS+GSR'heatflow into shadecloth
190 T2R=INT(T2+.5)'round to print
200 T2S=T12-G2/CFM'inner poly-shadecloth air temp
210 T2SR=INT(T2S+.5)'round to print
220 GS=(T2S-TS)/(2/3+1/CFM)'shadecloth air gain
225 GSR=(T2-TS)*G2SR+(TB-TS)*GSBR'rad gain
240 TS=TS+GSFLOW/C'new shadecloth temp
250 TSR=INT(TS+.5)'round to print
260 TSB=T2S-GS/CFM'shadecloth-back wall air temp
270 TSBR=INT(TSB+.5)'round to print
280 GB=(TSB-TB)/(2/3+1/CFM)'back wall air gain
285 GBR=(T2-TB)*G2BR+(TS-TB)*GSBR'rad gain
290 GBFLOW=69+GBW+GBR'heatflow into back wall
300 TB=TB+GBFLOW/C'new back wall temp
310 TBR=INT(TB+.5)'round to print
320 TOUT=TSB-GB/CFM'back wall outlet air temp
330 TOUTR=INT(TOUT+.5)'round to print
340 IF I>498 THEN PRINT "998'";T1R;T12R;T2R;T2SR;TSR;TSBR;TBR;TOUTR
341 G12R=1*4*S*(460+(T1+T2)/2)^3'outer-inner poly radcon
344 G2SR=.5*4*S*(460+(T2+TS)/2)^3'inner poly-shadecloth radcon
345 G2BR=.5*4*S*(460+(T2+TB)/2)^3'inner poly-back wall radcon
346 GSBR=.5*4*S*(460+(TS+TB)/2)^3'shadecloth-back wall radcon
350 NEXT
360 EFF=100*6*(TOUT-TR)/1000'solar collection efficiency (%)
344 G2SR=.5*4*S*(460+(T2+TS)/2)^3'inner poly-shadecloth radcon
345 G2BR=.5*4*S*(460+(T2+TB)/2)^3'inner poly-back wall radcon
346 GSBR=.5*4*S*(460+(TS+TB)/2)^3'shadecloth-back wall radcon
350 NEXT
365 LEFF=100*(1000-6*(T1-TA))/1000'glazing-loss-based efficiency
370 PRINT "999'";EFF;LEFF
T1 air T2 air Ts air Tb air
67.20852 68.66487
... 167 Btu/h-ft^2 sun can heat 1 cfm of 70 F air to 182 F with
100(173-70)/167 = 67% efficiency. With 86 F R1 IR poly outer film
losing 86-34 = 52 Btu/h-ft^2 on a 34 F day, it's probably worth adding
another layer of 15 cent/ft^2 50% black greenhouse shadecloth to raise
the collection efficiency.
Now the 67% air heating efficiency is close enough to the 69%
the collection efficiency.
efficiency based on glazing loss to say the difference is due to
rounding errors in these thousands of calculations, after which the
net flows into the layers (eg GBFLOW in line 290) are all on the order
of 10^-5 Btu/h-ft^2.
On to another layer of shadecloth...
Nick
nick pine
Feb 24, 2012, 6:44:09 AM2/24/12
to sunspace, jkpr...@verizon.net
With 2 IR poly layers and 2 50% black greenhouse shadecloth layers...
10 S=1.714E-09'Stefan-Boltzmann constant
20 TA=34'ambient temp (F)
30 SUN=910/6'sun through glazing (Btu/ft^2-h)
40 TR=70'air heater inlet temp (F)
50 CFM=1'airflow/ft^2
60 C=10'relaxation damping cap
70 T1=50:T2=100:TS=100:TB=100'initial temps (F)
80 FOR I=1 TO 499'relaxation iterations
90 G1=(TR-T1)/(2/3+1/CFM)'outer poly film air gain
100 GR1=(T2-T1)*G12R'rad gain
gain
180 G2FLOW=G2+GR2'heatflow into inner poly
190 T2=T2+G2FLOW/C'new inner poly temp
200 T2R=INT(T2+.5)'round to print
210 T2S1=T12-G2/CFM'inner poly-shade1 air temp
220 T2S1R=INT(T2S1+.5)'round to print
230 GS1=(T2S1-TS1)/(2/3+1/CFM)'shade1 air gain
240 GS1R=(T2-TS1)*GP21R+(TS2-TS1)*GS12R+(TB-TS1)*GS1BR'rad gain
250 GS1FLOW=69+GS1+GS1R'heatflow into shade1
260 TS1=TS1+GS1FLOW/C'new shade1 temp
270 TS1R=INT(TS1+.5)'round to print
280 TS12=T2S1-G2/CFM'shade1-shade2 air temp
290 TS12R=INT(TS12+.5)'round to print
300 GS2=(TS12-TS2)/(2/3+1/CFM)'shade2 air gain
310 GS2R=(T2-TS2)*GP22R+(TS1-TS2)*GS12R+(TB-TS2)*GS2BR'rad gain
320 GS2FLOW=34.5+GS2+GS2R'heatflow into shade2
330 TS2=TS2+GS2FLOW/C'new shade2 temp
340 TS2R=INT(TS2+.5)'round to print
350 TS2B=TS12-GS2/CFM'shade2-back wall air temp
360 TS2BR=INT(TS2B+.5)'round to print
370 GB=(TS2B-TB)/(2/3+1/CFM)'back wall air gain
380 GBR=(T2-TB)*GP2BR+(TS1-TB)*GS1BR+(TS2-TB)*GS2BR'rad gain
390 GBFLOW=34.5+GB+GBR'heatflow into back wall
400 TB=TB+GBFLOW/C'new back wall temp
410 TBR=INT(TB+.5)'round to print
420 TOUT=TS2B-GB/CFM'back wall outlet air temp
430 TOUTR=INT(TOUT+.5)'round to print
440 IF I>498 THEN PRINT
"998'";T1R;T12R;T2R;T2S1R;TS1R;TS12R;TS2R;TS2BR;TBR;TOUTR
450 GP12R=1*4*S*(460+(T1+T2)/2)^3'outer-inner poly radcon
460 GP21R=.5*4*S*(460+(T2+TS1)/2)^3'inner poly-shade1 radcon
470 GP22R=.25*4*S*(460+(T2+TS2)/2)^3'inner poly-shade2 radcon
480 GP2BR=.25*4*S*(460+(T2+TB)/2)^3'inner poly-back wall radcon
490 GS12R=.25*4*S*(460+(TS1+TS2)/2)^3'shade1-shade2 radcon
500 GS1BR=.25*4*S*(460+(TS1+TB)/2)^3'shade1-back wall radcon
510 GS2BR=.5*4*S*(460+(TS2+TB)/2)^3'shade2-back wall radcon
520 NEXT
530 EFF=100*6*(TOUT-TR)/1000'solar collection efficiency (%)
540 LEFF=100*(1000-6*(T1-TA))/1000'glazing loss-based efficiency
550 PRINT "999'";EFF;LEFF
T1 air T2 air Ts1 air Ts2 air Tb air outlet
48 57 179 130 211 203 221 214 223 219
89.49208 91.89999
... it looks like 167 Btu/h-ft^2 sun can heat 1 cfm of 70 F air to 219
F with about 90% efficiency :-)
The 89% air heating efficiency seems close enough to the 92% glazing
loss efficiency to say the difference is due to rounding errors in
thousands of relaxation calculations.
But it seems odd that T2 is so much warmer than T1...
This could be built more expensively and durably with an outer layer
of HP92W flat polycarbonate and an inner layer of perforated IR film
or polycarbonate.
Nick
10 S=1.714E-09'Stefan-Boltzmann constant
20 TA=34'ambient temp (F)
30 SUN=910/6'sun through glazing (Btu/ft^2-h)
40 TR=70'air heater inlet temp (F)
50 CFM=1'airflow/ft^2
60 C=10'relaxation damping cap
70 T1=50:T2=100:TS=100:TB=100'initial temps (F)
80 FOR I=1 TO 499'relaxation iterations
90 G1=(TR-T1)/(2/3+1/CFM)'outer poly film air gain
100 GR1=(T2-T1)*G12R'rad gain
110 G1FLOW=(TA-T1)/1+G1+GR1'heatflow into outer poly
120 T1=T1+G1FLOW/C'new outer poly temp
130 T1R=INT(T1+.5)'round to print
140 T12=TR-G1/CFM'outer-inner poly air temp
150 T12R=INT(T12+.5)'round to print
160 G2=(T12-T2)/(2/3+1/CFM)'inner poly air gain
170 GR2=(T1-T2)*G12R+(TS1-T2)*GP21R+(TS2-T2)*GP22R+(TB-T2)*GP2BR'rad
120 T1=T1+G1FLOW/C'new outer poly temp
130 T1R=INT(T1+.5)'round to print
140 T12=TR-G1/CFM'outer-inner poly air temp
150 T12R=INT(T12+.5)'round to print
160 G2=(T12-T2)/(2/3+1/CFM)'inner poly air gain
gain
180 G2FLOW=G2+GR2'heatflow into inner poly
190 T2=T2+G2FLOW/C'new inner poly temp
200 T2R=INT(T2+.5)'round to print
210 T2S1=T12-G2/CFM'inner poly-shade1 air temp
220 T2S1R=INT(T2S1+.5)'round to print
230 GS1=(T2S1-TS1)/(2/3+1/CFM)'shade1 air gain
240 GS1R=(T2-TS1)*GP21R+(TS2-TS1)*GS12R+(TB-TS1)*GS1BR'rad gain
250 GS1FLOW=69+GS1+GS1R'heatflow into shade1
260 TS1=TS1+GS1FLOW/C'new shade1 temp
270 TS1R=INT(TS1+.5)'round to print
280 TS12=T2S1-G2/CFM'shade1-shade2 air temp
290 TS12R=INT(TS12+.5)'round to print
300 GS2=(TS12-TS2)/(2/3+1/CFM)'shade2 air gain
310 GS2R=(T2-TS2)*GP22R+(TS1-TS2)*GS12R+(TB-TS2)*GS2BR'rad gain
320 GS2FLOW=34.5+GS2+GS2R'heatflow into shade2
330 TS2=TS2+GS2FLOW/C'new shade2 temp
340 TS2R=INT(TS2+.5)'round to print
350 TS2B=TS12-GS2/CFM'shade2-back wall air temp
360 TS2BR=INT(TS2B+.5)'round to print
370 GB=(TS2B-TB)/(2/3+1/CFM)'back wall air gain
380 GBR=(T2-TB)*GP2BR+(TS1-TB)*GS1BR+(TS2-TB)*GS2BR'rad gain
390 GBFLOW=34.5+GB+GBR'heatflow into back wall
400 TB=TB+GBFLOW/C'new back wall temp
410 TBR=INT(TB+.5)'round to print
420 TOUT=TS2B-GB/CFM'back wall outlet air temp
430 TOUTR=INT(TOUT+.5)'round to print
440 IF I>498 THEN PRINT
"998'";T1R;T12R;T2R;T2S1R;TS1R;TS12R;TS2R;TS2BR;TBR;TOUTR
450 GP12R=1*4*S*(460+(T1+T2)/2)^3'outer-inner poly radcon
460 GP21R=.5*4*S*(460+(T2+TS1)/2)^3'inner poly-shade1 radcon
470 GP22R=.25*4*S*(460+(T2+TS2)/2)^3'inner poly-shade2 radcon
480 GP2BR=.25*4*S*(460+(T2+TB)/2)^3'inner poly-back wall radcon
490 GS12R=.25*4*S*(460+(TS1+TS2)/2)^3'shade1-shade2 radcon
500 GS1BR=.25*4*S*(460+(TS1+TB)/2)^3'shade1-back wall radcon
510 GS2BR=.5*4*S*(460+(TS2+TB)/2)^3'shade2-back wall radcon
520 NEXT
530 EFF=100*6*(TOUT-TR)/1000'solar collection efficiency (%)
540 LEFF=100*(1000-6*(T1-TA))/1000'glazing loss-based efficiency
550 PRINT "999'";EFF;LEFF
T1 air T2 air Ts1 air Ts2 air Tb air outlet
48 57 179 130 211 203 221 214 223 219
89.49208 91.89999
... it looks like 167 Btu/h-ft^2 sun can heat 1 cfm of 70 F air to 219
F with about 90% efficiency :-)
The 89% air heating efficiency seems close enough to the 92% glazing
loss efficiency to say the difference is due to rounding errors in
thousands of relaxation calculations.
But it seems odd that T2 is so much warmer than T1...
This could be built more expensively and durably with an outer layer
of HP92W flat polycarbonate and an inner layer of perforated IR film
or polycarbonate.
Nick
nick pine
Mar 2, 2012, 6:19:05 PM3/2/12
to sunspace, jkpr...@verizon.net
>This would have about C = 3'x12'x14'x62.33 = 31.4K Btu/F of thermal capacitance. Cooling it from 140 to 70 F would release (140-70)C = 2.2 million Btu, equivalent to 2.2M/130K = 17 gallons of oil.
>On an average day, a south wall with a 60 degree slope would receive 620cos60+1000sin60 = 1176 Btu/ft^2 of sun. This store could collect... a net gain of 139.1K Btu/day.
>If the entire stone bed warms from 70 to 140 F, it needs a 139.1K/(140-70) = 1987 Btu/F capacitance, eg 1987/0.16 = 12.4K lb of stone...
If the heat store starts at 70 F and the stone is 140 F at the end of
the day and we wait a long time, the overall heat store temp will be
(140x1987+70x31.4K)/(1987+31.4K) = 74.16 F.
With series capacitance Cs = 1987x31.4K/(1987+34.1k) = 1869 Btu/F and
15x25ft^2 of drum surface and 168 ft^2 of EPDM tank surface and a G =
815 Btu/h-F conductance and time constant RC = Cs/G = 2.3 hours, the
overall temp after 24 hours would be 74.16+(70-74.16)e^(-24/2.3) =
74.1599 F. The stones would transfer almost all of their heat to the
water after 24 hours.
Nick
nick pine
Mar 3, 2012, 6:01:25 AM3/3/12
to sunspace, jkpr...@verizon.net
So 2 capacitors and a switch and a resistor walk into a bar...
/ R
----- ---www---
| T1 | T2
| |
--- C1 --- C2
--- ---
| |
| |
- -
If we want to cool 16 oz of hot coffee to make iced coffee without
diluting the hot coffee with a lot of ice, C1 might be 1 Btu/F and
T1 might be 212 F before we close the switch. C2 could be 4 Btu/F,
with 64 oz of T2 = 60 F cool tapwater before we close the switch,
ie put the coffee mug into the bowl of cool water.
The bottom ends of the capacitors are connected, so we really have
2 capacitors in series with capacitance Cs = 1x4/(1+4) = 0.8 Btu/F.
After a loooong time, the coffee and water temps would both be
(212x1+60x4)/(1+4) = 90.4 F.
A 3" diameter x 4" tall coffee mug with 0.31 ft^2 of side and bottom
surface surrounded by still water with a 30 Btu/h-F-ft^2 conductance
would have a thermal conductance G = 30x0.31 = 9.3 Btu/h-F with
time constant Cs/G = 0.8/9.3 = 0.086 hours, ie 5.2 minutes.
After 5.2 minutes, the coffee would be 90.4+(212-90.4)e^(-5.2/5.2)
= 135 F. After 10.4 minutes, it would be 90.4+(212-90.4)e^(-10.4/5.2)
= 107. After 15.6, 96.5 F. After 20.8, 92.6 F. After 26, 91.2.
We could speed up the cooling with by starting with ice in the water.
It takes 144 Btu to melt 1 pound of ice. Cooling 1 Btu/F of 212 F
coffee to 32 F requires (212-32)/144 = 1.25 pounds of ice. After 5.2
minutes, the coffee would be 32+(212-32)e^(-5.2/5.2) = 98.2 F. After
10.4, it would be 32+(212-32)e^(-10.4/5.2) = 56.4 F. After 26, 33.2.
This is 311-year-old physics, aka Newton's 1701 law of cooling:
http://en.wikipedia.org/wiki/Isaac_Newton
http://tuhsphysics.ttsd.k12.or.us/Research/ib01/IrlaTonl/index.html
Nick
/ R
----- ---www---
| T1 | T2
| |
--- C1 --- C2
--- ---
| |
| |
- -
If we want to cool 16 oz of hot coffee to make iced coffee without
diluting the hot coffee with a lot of ice, C1 might be 1 Btu/F and
T1 might be 212 F before we close the switch. C2 could be 4 Btu/F,
with 64 oz of T2 = 60 F cool tapwater before we close the switch,
ie put the coffee mug into the bowl of cool water.
The bottom ends of the capacitors are connected, so we really have
2 capacitors in series with capacitance Cs = 1x4/(1+4) = 0.8 Btu/F.
After a loooong time, the coffee and water temps would both be
(212x1+60x4)/(1+4) = 90.4 F.
A 3" diameter x 4" tall coffee mug with 0.31 ft^2 of side and bottom
surface surrounded by still water with a 30 Btu/h-F-ft^2 conductance
would have a thermal conductance G = 30x0.31 = 9.3 Btu/h-F with
time constant Cs/G = 0.8/9.3 = 0.086 hours, ie 5.2 minutes.
After 5.2 minutes, the coffee would be 90.4+(212-90.4)e^(-5.2/5.2)
= 135 F. After 10.4 minutes, it would be 90.4+(212-90.4)e^(-10.4/5.2)
= 107. After 15.6, 96.5 F. After 20.8, 92.6 F. After 26, 91.2.
We could speed up the cooling with by starting with ice in the water.
It takes 144 Btu to melt 1 pound of ice. Cooling 1 Btu/F of 212 F
coffee to 32 F requires (212-32)/144 = 1.25 pounds of ice. After 5.2
minutes, the coffee would be 32+(212-32)e^(-5.2/5.2) = 98.2 F. After
10.4, it would be 32+(212-32)e^(-10.4/5.2) = 56.4 F. After 26, 33.2.
This is 311-year-old physics, aka Newton's 1701 law of cooling:
http://en.wikipedia.org/wiki/Isaac_Newton
http://tuhsphysics.ttsd.k12.or.us/Research/ib01/IrlaTonl/index.html
Nick
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