A solar yard furnace
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nick pine
Jan 6, 2012, 4:52:23 AM1/6/12
to sunspace, james...@gmail.com, rdpi...@gmail.com
http://www.dbb-project.com/introduction/insulation.php says
Dynamic U-value Ud = VRhoaCa/(e^(VRhoaCaRs)-1) W/m^2-K, where
V is the air velocity in meters per second,
Rhoa is air density, 1.2 kg/m^3,
Ca is air's specific heat, 1000 J/(kg-K), and
Rs is the wall's static thermal resistance in m^2-K/W.
Using V = 1/3600 (1 meter per HOUR :-), and Rs = 5.7 m^2K/W (US R32),
Ud = 0.058 W/m^2, like a US R98 wall. V = 10 meters per hour makes
Ud = 1.7x10^-8 W/m^2K, ie a US R-value of 334 million :-)
This is exciting, no? We often try to make lots of air flow through
an air heater in order to keep temperatures and outdoor heat losses
low, but that uses lots of fan power and requires large air ducts
to move heat to a house. With "dynamic insulation" we might make
very hot air or water without much fan power or outdoor heat loss
and move it to a nearby house through smaller ducts.
My friend T just got her 100 year old house in New Jersey airsealed
and insulated with dense-packed cellulose in the formerly empty 2x4
wall cavities. It has no solar heat now, but it might have about
40'x8' of solar siding on the southeast and southwest walls.
With 100 F air inside for 6 hours per day, each square foot of R2
twinwall polycarbonate siding with 80% solar transmission would gain
0.8x980-6h(100-35)1ft^2/R2 = 589 Btu on an average 31.5 F December
day with a 35 F daytime temperature.
If the December gas bill says it used 65 therms at an average 40 F
outdoor temp and an 85% furnace efficiency and it used 600 kWh/mo
(68.2K Btu/day) of electricity indoors and 65x10^5/0.85 Btu
= 30d(24h(65-40)G-68.2K), the house conductance G = 539 Btu/h-F,
so it would need 24h(65-31.5)539-68.2K = 365K Btu of heat on
an average December day, which could come from 365K/589 = 620 ft^2
of solar siding on the southeast and southwest walls.
The house needs 1826K Btu for 5 cloudy days in a row, which could
come from 1826K/(140-80)/62 = 488 ft^3 of water cooling from 140
to 80 F in a 9'x18'x3' tall plywood box on the lawn with a $195
15'x24' folded EPDM liner from
http://www.pondliner.com/product/15_x_25_firestone_45_mil_epdm_pond_liner/Firestone_EPDM_Pond_Liners_15
At 140 F, with R30 insulation, it would lose 24h(140-31.5)486ft^2/R30
= 42K Btu on an average day. A 12'x20' twinwall roof with a 34 degree
slope would gain 12'x20'(0.8(980sin34+610cos34)-6h(140-35)/R2))
= 126K Btu/day. If the tank is 4' tall, the box would be 11' tall.
The box could provide about 126K-42K = 84K Btu of house heat
on an average December day at a rate of about 84K/6h = 14K Btu/h
in 14K/(140-70) = 200 cfm of 140 F air. A 20' duct (40' round trip)
with a 0.2 "H20 pressure drop with 0.2 = 0.1x40'/100'x200^2/D^5
would have a D = 6" diameter. It might be an underground foamboard
box with 2 6"x6" cavities.
(A more remote box might have hot and cool (80 F house return) tanks
and 2 car radiators, with one cooling 140 F air that exits the other
in order to provide cooler air near the twinwall and a near-infinite
mesh collector thermal resistance. This sounds complex for a heating
system, but it's child's play compared to what Google is up to, eg
video hangouts with tracking reindeer noses and antlers.)
With only 320 ft^2 of solar siding on the southwest wall, the box
would need to collect 42K+365K-320ft^2x589 = 219K Btu/day. It might
do that with a 16'x20' twinwall roof and a 17' box height.
Nick
Dynamic U-value Ud = VRhoaCa/(e^(VRhoaCaRs)-1) W/m^2-K, where
V is the air velocity in meters per second,
Rhoa is air density, 1.2 kg/m^3,
Ca is air's specific heat, 1000 J/(kg-K), and
Rs is the wall's static thermal resistance in m^2-K/W.
Using V = 1/3600 (1 meter per HOUR :-), and Rs = 5.7 m^2K/W (US R32),
Ud = 0.058 W/m^2, like a US R98 wall. V = 10 meters per hour makes
Ud = 1.7x10^-8 W/m^2K, ie a US R-value of 334 million :-)
This is exciting, no? We often try to make lots of air flow through
an air heater in order to keep temperatures and outdoor heat losses
low, but that uses lots of fan power and requires large air ducts
to move heat to a house. With "dynamic insulation" we might make
very hot air or water without much fan power or outdoor heat loss
and move it to a nearby house through smaller ducts.
My friend T just got her 100 year old house in New Jersey airsealed
and insulated with dense-packed cellulose in the formerly empty 2x4
wall cavities. It has no solar heat now, but it might have about
40'x8' of solar siding on the southeast and southwest walls.
With 100 F air inside for 6 hours per day, each square foot of R2
twinwall polycarbonate siding with 80% solar transmission would gain
0.8x980-6h(100-35)1ft^2/R2 = 589 Btu on an average 31.5 F December
day with a 35 F daytime temperature.
If the December gas bill says it used 65 therms at an average 40 F
outdoor temp and an 85% furnace efficiency and it used 600 kWh/mo
(68.2K Btu/day) of electricity indoors and 65x10^5/0.85 Btu
= 30d(24h(65-40)G-68.2K), the house conductance G = 539 Btu/h-F,
so it would need 24h(65-31.5)539-68.2K = 365K Btu of heat on
an average December day, which could come from 365K/589 = 620 ft^2
of solar siding on the southeast and southwest walls.
The house needs 1826K Btu for 5 cloudy days in a row, which could
come from 1826K/(140-80)/62 = 488 ft^3 of water cooling from 140
to 80 F in a 9'x18'x3' tall plywood box on the lawn with a $195
15'x24' folded EPDM liner from
http://www.pondliner.com/product/15_x_25_firestone_45_mil_epdm_pond_liner/Firestone_EPDM_Pond_Liners_15
At 140 F, with R30 insulation, it would lose 24h(140-31.5)486ft^2/R30
= 42K Btu on an average day. A 12'x20' twinwall roof with a 34 degree
slope would gain 12'x20'(0.8(980sin34+610cos34)-6h(140-35)/R2))
= 126K Btu/day. If the tank is 4' tall, the box would be 11' tall.
The box could provide about 126K-42K = 84K Btu of house heat
on an average December day at a rate of about 84K/6h = 14K Btu/h
in 14K/(140-70) = 200 cfm of 140 F air. A 20' duct (40' round trip)
with a 0.2 "H20 pressure drop with 0.2 = 0.1x40'/100'x200^2/D^5
would have a D = 6" diameter. It might be an underground foamboard
box with 2 6"x6" cavities.
(A more remote box might have hot and cool (80 F house return) tanks
and 2 car radiators, with one cooling 140 F air that exits the other
in order to provide cooler air near the twinwall and a near-infinite
mesh collector thermal resistance. This sounds complex for a heating
system, but it's child's play compared to what Google is up to, eg
video hangouts with tracking reindeer noses and antlers.)
With only 320 ft^2 of solar siding on the southwest wall, the box
would need to collect 42K+365K-320ft^2x589 = 219K Btu/day. It might
do that with a 16'x20' twinwall roof and a 17' box height.
Nick
nick pine
Jan 7, 2012, 7:20:43 AM1/7/12
to sunspace
> If the December gas bill says it used 65 therms at an average 40 F outdoor temp and an 85% furnace efficiency and it used 600 kWh/mo (68.2K Btu/day) of electricity indoors and 65x10^5/0.85 Btu
Oops. That shoulda been 65x10^5x0.85 = 5.25 million Btu...
> = 30d(24h(65-40)G-68.2K), the house conductance G = 539 Btu/h-F
> so it would need 24h(65-31.5)539-68.2K = 365K Btu of heat on an average December day, which could come from 365K/589 = 620 ft^2 of solar siding on the southeast and southwest walls.
> The house needs 1826K Btu for 5 cloudy days in a row, which could come from 1826K/(140-80)/62 = 488 ft^3 of water cooling from 140 to 80 F in a 9'x18'x3' tall plywood box on the lawn
>... A 12'x20' twinwall roof with a 34 degree slope would gain 12'x20'(0.8(980sin34+610cos34)-6h(140-35)/R2)) = 126K Btu/day. If the tank is 4' tall, the box would be 11' tall.
With a 60 degree slope, a 20' wide x 16' roof would gain
20'x16'(0.8(980sin60+610cos60)-6h(140-35)/R2)) = 195K Btu/day. If it
goes all the way to the ground, it would be 10.4' tall.
And it could provide 195K-42K = 153K Btu of house heat on an average
December day at a rate of 153K/6h = 25K Btu/h in 363 cfm of 140 F air.
A 20' duct (40' round trip) with a 0.2 "H20 pressure drop with 0.2 =
0.1x40'/100'x363^2/D^5 would have a D = 8" diameter.
Nick
nick pine
Jan 14, 2012, 4:40:03 PM1/14/12
to sunspace, li...@ecasavesenergy.org, jkpr...@verizon.net, jebar...@verizon.net, rdpi...@gmail.com
>... we might well solar heat an existing house without architectural changes or loss of floorspace by:
> 1. Measuring its thermal conductance by looking at fuel bills and indoor electrical use, or by heating it with electric space heaters with thermostats for a while and reading the house electric meter; and
>
> 2. Lowering that conductance with airsealing, more insulation, or seasonal foamboard window inserts; and
After blower door testing and airsealing the uninsulated 2x4 walls
with a dense pack cellulose insulation retrofit, Terri's 100-year-old
house (photo on page 8 of http://sbse.org/newsletter/issues/newsf08.pdf
) used 208 therms of natural gas at 95% efficiency (19.8 million Btu)
and 551 kWh of electricity totaling 21.6 million Btu in January of
2011, when the average outdoor temp was 32 F, so the house conductance
is now 21.6M/(65-32)/31d/24h = 881 Btu/h-F, so it would need
24h(65-36.6)881 = 600KBtu for an average 36.6 F December day near NYC.
Indoor electrical use provided about 61K Btu, leaving a need for 538K
Btu/day of solar heat in an average year, including gas water heating
energy.
> 3. Adding enough solar siding to heat it for 24 hours on an average December day, if possible; and
PVWATTS says the southeast and southwest walls receive 685 and 701 Btu/
ft^2 of sun on an average December day, with an average 40 F daytime
temp and a potential solar siding gain of 0.8x685-6h(70-40)1ft^2/R2 =
458 Btu/ft^2 on the southeast and 471 for the southwest. An 8'x40'
patch of SE siding would contribute 147K Btu/day, with 151K for SW.
> 4. Adding a solar yard furnace with enough glazing to heat the house on an average December day (if steps 2 and 3 won't accomplish that) and enough heat storage to keep the house warm and heat water for showers for 5 cloudy days in a row.
According to PVWATTS, an R2 twinwall yard furnace roof with 80% solar
transmission and a 60 degree tilt would receive 967 Btu/ft^2 on an
average December day. It could collect about 0.8x967-6h(70-40)1ft^2/R2
= 684 tu/ft^2 of heat.
With no solar siding, the yard furnace roof could be 538K/684 = 787
ft^2, eg 40'x20'. Adding 320 ft^2 of SE solar siding above the porch
roof would reduce this to (538K-147K)/684 = 572 ft^2, eg 32'x20'.
Adding 320 ft^2 of SW siding above the driveway would reduce the yard
furnace roof size to (538K-147K-151K)/684 = 351 ft^2, eg 24'x16'.
But Terri would like to enclose the whole south wall in a transparent
lean-to sunspace. It might have a $4500 40'x36' clear vinyl panel from
http://www.farmtek.com/farm/supplies/prod1;ft1_canopies_tents;pg105657_105659.html
that rolls down a quarter-cylindrical frame made with 6 $5 double
curved 1x3 beams on 8' centers with a 22' radius connecting the 2nd
floor eave with the ground 14' to the south of the porch. My first
sunspace had a $50 32'x24' piece of cloudy 4-year polyethylene
greenhouse film in lieu of vinyl.
Alternatively, Terri could have a sunspace like my current version,
with a low-slope 40'x12' corrugated polycarbonate roof extending 16'
to the south of the 2nd floor eave 8' above the flat porch roof, which
connects to a 40'x16' vertical wall made with 20 $100 4'x8' flat 10
mil HP92 polycarbonate double-glazed panels gently inflated with
welding argon via tire valves. The roof and wall would collect
0.8x40'x16'(510+685) = 612K Btu/day and lose 6h(70-40)640ft^2/R2 =
58K, for a net gain of 554K. Twinwall endwalls would collect more. It
would be easy to add more insulation to the outside of the enclosed
existing house walls.
This would enclose the existing porch and add 8'x40' of enclosed deck
roof floorspace and enclose 8'x40' on the ground, which might include
a picnic table with an awning and a December tomato and herb and red
bell pepper garden and a patio over a heat store, which might also
supply heat to a hot tub.
The house needs 5dx538K = 2.7 million Btu for 5 cloudy 36.6 F December
days with about 97% solar house heating in December. The sunspace
could have 2.7M/(140-80)/62 = 720 ft^3 of water cooling from 140 to 80
F in a well-insulated 8'x16'x6'-tall tank with a $300 folded 20'x32'
EPDM liner, dug 6' into the ground.
Nick
> 1. Measuring its thermal conductance by looking at fuel bills and indoor electrical use, or by heating it with electric space heaters with thermostats for a while and reading the house electric meter; and
>
> 2. Lowering that conductance with airsealing, more insulation, or seasonal foamboard window inserts; and
After blower door testing and airsealing the uninsulated 2x4 walls
with a dense pack cellulose insulation retrofit, Terri's 100-year-old
house (photo on page 8 of http://sbse.org/newsletter/issues/newsf08.pdf
) used 208 therms of natural gas at 95% efficiency (19.8 million Btu)
and 551 kWh of electricity totaling 21.6 million Btu in January of
2011, when the average outdoor temp was 32 F, so the house conductance
is now 21.6M/(65-32)/31d/24h = 881 Btu/h-F, so it would need
24h(65-36.6)881 = 600KBtu for an average 36.6 F December day near NYC.
Indoor electrical use provided about 61K Btu, leaving a need for 538K
Btu/day of solar heat in an average year, including gas water heating
energy.
> 3. Adding enough solar siding to heat it for 24 hours on an average December day, if possible; and
PVWATTS says the southeast and southwest walls receive 685 and 701 Btu/
ft^2 of sun on an average December day, with an average 40 F daytime
temp and a potential solar siding gain of 0.8x685-6h(70-40)1ft^2/R2 =
458 Btu/ft^2 on the southeast and 471 for the southwest. An 8'x40'
patch of SE siding would contribute 147K Btu/day, with 151K for SW.
> 4. Adding a solar yard furnace with enough glazing to heat the house on an average December day (if steps 2 and 3 won't accomplish that) and enough heat storage to keep the house warm and heat water for showers for 5 cloudy days in a row.
According to PVWATTS, an R2 twinwall yard furnace roof with 80% solar
transmission and a 60 degree tilt would receive 967 Btu/ft^2 on an
average December day. It could collect about 0.8x967-6h(70-40)1ft^2/R2
= 684 tu/ft^2 of heat.
With no solar siding, the yard furnace roof could be 538K/684 = 787
ft^2, eg 40'x20'. Adding 320 ft^2 of SE solar siding above the porch
roof would reduce this to (538K-147K)/684 = 572 ft^2, eg 32'x20'.
Adding 320 ft^2 of SW siding above the driveway would reduce the yard
furnace roof size to (538K-147K-151K)/684 = 351 ft^2, eg 24'x16'.
But Terri would like to enclose the whole south wall in a transparent
lean-to sunspace. It might have a $4500 40'x36' clear vinyl panel from
http://www.farmtek.com/farm/supplies/prod1;ft1_canopies_tents;pg105657_105659.html
that rolls down a quarter-cylindrical frame made with 6 $5 double
curved 1x3 beams on 8' centers with a 22' radius connecting the 2nd
floor eave with the ground 14' to the south of the porch. My first
sunspace had a $50 32'x24' piece of cloudy 4-year polyethylene
greenhouse film in lieu of vinyl.
Alternatively, Terri could have a sunspace like my current version,
with a low-slope 40'x12' corrugated polycarbonate roof extending 16'
to the south of the 2nd floor eave 8' above the flat porch roof, which
connects to a 40'x16' vertical wall made with 20 $100 4'x8' flat 10
mil HP92 polycarbonate double-glazed panels gently inflated with
welding argon via tire valves. The roof and wall would collect
0.8x40'x16'(510+685) = 612K Btu/day and lose 6h(70-40)640ft^2/R2 =
58K, for a net gain of 554K. Twinwall endwalls would collect more. It
would be easy to add more insulation to the outside of the enclosed
existing house walls.
This would enclose the existing porch and add 8'x40' of enclosed deck
roof floorspace and enclose 8'x40' on the ground, which might include
a picnic table with an awning and a December tomato and herb and red
bell pepper garden and a patio over a heat store, which might also
supply heat to a hot tub.
The house needs 5dx538K = 2.7 million Btu for 5 cloudy 36.6 F December
days with about 97% solar house heating in December. The sunspace
could have 2.7M/(140-80)/62 = 720 ft^3 of water cooling from 140 to 80
F in a well-insulated 8'x16'x6'-tall tank with a $300 folded 20'x32'
EPDM liner, dug 6' into the ground.
Nick
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