Inflatable solar collectors?
6 views
Skip to first unread message
nick pine
Dec 23, 2011, 9:47:27 AM12/23/11
to sunspace, lbit...@yahoo.com
I see lots of inflatable snowmen and Santas and globes on holiday
lawns...
How about an inflatable solar collector, eg a 24' x 12' tall x 4'
thick double polyethylene film east-west wall containing some
evergreen shadecloth tree silhouettes attached to a black shadecloth
curtain?
The base could be a poly film air duct partly filled with water. We
could set it up on Thanksgiving with a few large brown turkeys and
roll it up after Valentine's day with a few large red hearts.
And inflate it with a fan that also moves air through a car radiator
that heats water in a loop through a heat store inside a nearby house.
Nick
lawns...
How about an inflatable solar collector, eg a 24' x 12' tall x 4'
thick double polyethylene film east-west wall containing some
evergreen shadecloth tree silhouettes attached to a black shadecloth
curtain?
The base could be a poly film air duct partly filled with water. We
could set it up on Thanksgiving with a few large brown turkeys and
roll it up after Valentine's day with a few large red hearts.
And inflate it with a fan that also moves air through a car radiator
that heats water in a loop through a heat store inside a nearby house.
Nick
nick pine
Dec 28, 2011, 5:39:50 AM12/28/11
to sunspace, rdpi...@gmail.com
Inflatables have more promise for air than water heating, IMO, altho
"solar siding" seems better for that purpose. Polyethylene film has
high IR transmittance, ie not much heat-trapping greenhouse effect,
compared to glass or polycarbonate, so a 140 F water heater would be
inefficient, but an air heater could double as an attached garage:
f ---
fsf f
fsf f \
S<-- fsf f 8'
fsf f \
/ fsf f
16'fsf 45 deg f
/ fsf.........................f ~14'
fsf . 9' f
fsf . f |
fsf . f 8'
fsf . f |
fsfoamboard . f
fsf 60 deg . f
gggggggggggggggggggggggggggggggggggggggggggggggground
| ~14' |
With 28' in the east-west direction, it needs 2x28'x16' = 896 ft^2
of film for the walls plus 265 for endwalls, totaling 1161 ft^2, eg
a $136 16'x75' 4-year roll from http://www.greenhousemegastore.com
In South Bend, the south wall would receive 580sin60+430cos60 = 717
Btu/ft^2 of sun on an average December day with a 32 F daytime temp,
ie 0.9x717x16'x28' = 289.1K Btu. With 70 F air inside for 6 hours,
it would lose 6h(70-32)16'x28'/R1 = 102.1K Btu to the outdoors, for
a net gain of (289.1K-102.1K)/6h = 31.2K Btu/h (140 F air inside
real R1 film for 6 hours would lose 6h(140-32)16'x28'/R1 = 290.1K
Btu/day, more than the solar gain.)
With a dark mesh screen ("s" above) and polyiso foamboard behind it,
the film might be the main south wall heat loss, with a net gain of
187K Btu/day. (A less expensive and less efficient version could
have film vs polyiso foamboard behind the mesh.)
With a few layers of gauzy mesh or felt to keep the surface near
the film cooler and increase its thermal resistance (a wall can't
lose much heat by convection to an airstream approaching the wall,
see equation 1 on page 4 of http://www.cibse.org/pdfs/8cimbabi.pdf),
we might collect and store higher temperature heat in water as well,
using a car radiator, despite the high IR loss of the polyethylene
film, ie keep a cloudy day heat store warm as well as solar heat
air for the house. The film's high IR loss can help prevent melting
the mesh or felt under stagnation conditions.
If 100 ft^3 of 140 F water in a 30" diameter x 25' polyethylene film
duct inside a 2'x2'x25' bench with 108 ft^2 of R20 surface (bags of
dry leaves?) loses 24h(140-29)108/20 = 14.4K Btu on an average day,
we can keep it warm with 14.4K/6h = 2400 Btu/h for 6 hours with 1000
cfm of air at T = 140+2400/1000 = 142 F with a 1000 Btu/h-F car
radiator. With C cfm of airflow and (T-140)C = 2400 Btu/h,
C = 2400/(T-140).
Heating C cfm from 70 to T (F) requires (T-70)C Btu/h. If the same
air flows into the mesh at 70 F and into the radiator at T (F) and
through the house and back into the mesh, and 31.2K Btu/h = (T-70)C
= (T-70)2400/(T-140), then T = 145.8 F and C = 411 cfm. (Or more,
since the radiator will be more efficient with less than 1000 cfm
of airflow.)
A 1000 Btu/h-F car radiator inside a house with a G Btu/h-F thermal
conductance can keep it 65 F on a 29 F day with a minimum water temp
Tm = 70+(65-29)G/1000. If 100 ft^3 of 140 F water cools to Tm over
5 cloudy days and 5x24h(65-29)G = (140-Tm)100x62.33, G = 96 Btu/h-F,
or more, with some help from indoor electrical use.
The garage might also supply 289.1K-102.1K-14.4K = 173K Btu of warm
air for the house on an average day, enough to warm a 200 Btu/h-F
house with no windows or electrical use, which would require another
100 ft^3 of water in the garage to stay warm for 5 cloudy days with
a solar heating fraction close to 100%.
Nick
"solar siding" seems better for that purpose. Polyethylene film has
high IR transmittance, ie not much heat-trapping greenhouse effect,
compared to glass or polycarbonate, so a 140 F water heater would be
inefficient, but an air heater could double as an attached garage:
f ---
fsf f
fsf f \
S<-- fsf f 8'
fsf f \
/ fsf f
16'fsf 45 deg f
/ fsf.........................f ~14'
fsf . 9' f
fsf . f |
fsf . f 8'
fsf . f |
fsfoamboard . f
fsf 60 deg . f
gggggggggggggggggggggggggggggggggggggggggggggggground
| ~14' |
With 28' in the east-west direction, it needs 2x28'x16' = 896 ft^2
of film for the walls plus 265 for endwalls, totaling 1161 ft^2, eg
a $136 16'x75' 4-year roll from http://www.greenhousemegastore.com
In South Bend, the south wall would receive 580sin60+430cos60 = 717
Btu/ft^2 of sun on an average December day with a 32 F daytime temp,
ie 0.9x717x16'x28' = 289.1K Btu. With 70 F air inside for 6 hours,
it would lose 6h(70-32)16'x28'/R1 = 102.1K Btu to the outdoors, for
a net gain of (289.1K-102.1K)/6h = 31.2K Btu/h (140 F air inside
real R1 film for 6 hours would lose 6h(140-32)16'x28'/R1 = 290.1K
Btu/day, more than the solar gain.)
With a dark mesh screen ("s" above) and polyiso foamboard behind it,
the film might be the main south wall heat loss, with a net gain of
187K Btu/day. (A less expensive and less efficient version could
have film vs polyiso foamboard behind the mesh.)
With a few layers of gauzy mesh or felt to keep the surface near
the film cooler and increase its thermal resistance (a wall can't
lose much heat by convection to an airstream approaching the wall,
see equation 1 on page 4 of http://www.cibse.org/pdfs/8cimbabi.pdf),
we might collect and store higher temperature heat in water as well,
using a car radiator, despite the high IR loss of the polyethylene
film, ie keep a cloudy day heat store warm as well as solar heat
air for the house. The film's high IR loss can help prevent melting
the mesh or felt under stagnation conditions.
If 100 ft^3 of 140 F water in a 30" diameter x 25' polyethylene film
duct inside a 2'x2'x25' bench with 108 ft^2 of R20 surface (bags of
dry leaves?) loses 24h(140-29)108/20 = 14.4K Btu on an average day,
we can keep it warm with 14.4K/6h = 2400 Btu/h for 6 hours with 1000
cfm of air at T = 140+2400/1000 = 142 F with a 1000 Btu/h-F car
radiator. With C cfm of airflow and (T-140)C = 2400 Btu/h,
C = 2400/(T-140).
Heating C cfm from 70 to T (F) requires (T-70)C Btu/h. If the same
air flows into the mesh at 70 F and into the radiator at T (F) and
through the house and back into the mesh, and 31.2K Btu/h = (T-70)C
= (T-70)2400/(T-140), then T = 145.8 F and C = 411 cfm. (Or more,
since the radiator will be more efficient with less than 1000 cfm
of airflow.)
A 1000 Btu/h-F car radiator inside a house with a G Btu/h-F thermal
conductance can keep it 65 F on a 29 F day with a minimum water temp
Tm = 70+(65-29)G/1000. If 100 ft^3 of 140 F water cools to Tm over
5 cloudy days and 5x24h(65-29)G = (140-Tm)100x62.33, G = 96 Btu/h-F,
or more, with some help from indoor electrical use.
The garage might also supply 289.1K-102.1K-14.4K = 173K Btu of warm
air for the house on an average day, enough to warm a 200 Btu/h-F
house with no windows or electrical use, which would require another
100 ft^3 of water in the garage to stay warm for 5 cloudy days with
a solar heating fraction close to 100%.
Nick
nick pine
Jan 1, 2012, 5:31:52 AM1/1/12
to sunspace
"jmygann" <jmygann@...> wrote:
>... hope you get a net gain.
Hope has nothing to do with it!
If a house has a measured conductance of 203 Btu/h-F including 32 cfm
of natural air leakage and 200 ft^2 of R2 windows, adding seasonal R10
foamboard inserts to 48 ft^2 of windows and reducing the air leaks to
10 cfm will lower the conductance to 161 Btu/h-F.
With an average 65 F indoor temp, it will need 24h(65-29)161 = 139K
Btu of heat on an average December day in South Bend, IN. With 300 kWh/
mo of indoor electricity, it will need 139K-34K = 105K Btu/day from
other sources, eg the sun.
If it ends up with 100 ft^2 of unshuttered unshaded south windows and
20 ft^2 on the east and west walls and 12 on the north wall and the
windows have 80% solar transmission, they will collect
0.8(100x580+20x260+20x280+12x170) = 57K Btu on an average December
day, so the house needs an additional 48K Btu/day of heat... 1 ft^2 of
R2 twinwall polycarbonate south solar siding with 80% transmission
with 100 F air inside would gain about 0.8x580-6h(100-32)1ft^2/R2 =
260 Btu/day, so it could have 48K/260 = 184 ft^2 of solar siding.
As an alternative, 1 ft^2 of twinwall over felt on an equilateral
solar shed near the house that provides warm air for the house might
gain 0.8x717-6h(70-32)1ft^2/R2 = 460 Btu/day, so it could have 104
ft^2 of shed twinwall, eg a shed with a 12'x12' south wall with a 60
degree tilt.
For near-100% solar heat in December, it needs to store 5daysx105K =
525K Btu in 525K/(140-75)/62 = 130 ft^3 of water cooling from 140 to
75 F. Keeping the water 140 F on an average day requires air with a
minimum 140 F temp, eg a sunspace containing 140 F air or solar siding
with 140 F air behind it or a solar shed with 140 F air inside.
Heating the water requires hot air. Hope is insufficient!
If the water lives in a 4'x12'x3' tall box in the shed with a 10'x18'
folded EPDM liner and 192 ft^2 of R32 surface and a 192/32 = 6 Btu/h-F
conductance and the box loses 24h(140-32)6 = 16K Btu/day and the house
can keep itself warm on an average December day and the shed does not
provide warm air for the house on an average December day and 1 ft^2
of twinwall gains 0.8x717-6h(140-32)1ft^2/R2 = 250 Btu/day, the shed
needs 16K/250 = 64 ft^2 of twinwall.
Nick
>... hope you get a net gain.
Hope has nothing to do with it!
If a house has a measured conductance of 203 Btu/h-F including 32 cfm
of natural air leakage and 200 ft^2 of R2 windows, adding seasonal R10
foamboard inserts to 48 ft^2 of windows and reducing the air leaks to
10 cfm will lower the conductance to 161 Btu/h-F.
With an average 65 F indoor temp, it will need 24h(65-29)161 = 139K
Btu of heat on an average December day in South Bend, IN. With 300 kWh/
mo of indoor electricity, it will need 139K-34K = 105K Btu/day from
other sources, eg the sun.
If it ends up with 100 ft^2 of unshuttered unshaded south windows and
20 ft^2 on the east and west walls and 12 on the north wall and the
windows have 80% solar transmission, they will collect
0.8(100x580+20x260+20x280+12x170) = 57K Btu on an average December
day, so the house needs an additional 48K Btu/day of heat... 1 ft^2 of
R2 twinwall polycarbonate south solar siding with 80% transmission
with 100 F air inside would gain about 0.8x580-6h(100-32)1ft^2/R2 =
260 Btu/day, so it could have 48K/260 = 184 ft^2 of solar siding.
As an alternative, 1 ft^2 of twinwall over felt on an equilateral
solar shed near the house that provides warm air for the house might
gain 0.8x717-6h(70-32)1ft^2/R2 = 460 Btu/day, so it could have 104
ft^2 of shed twinwall, eg a shed with a 12'x12' south wall with a 60
degree tilt.
For near-100% solar heat in December, it needs to store 5daysx105K =
525K Btu in 525K/(140-75)/62 = 130 ft^3 of water cooling from 140 to
75 F. Keeping the water 140 F on an average day requires air with a
minimum 140 F temp, eg a sunspace containing 140 F air or solar siding
with 140 F air behind it or a solar shed with 140 F air inside.
Heating the water requires hot air. Hope is insufficient!
If the water lives in a 4'x12'x3' tall box in the shed with a 10'x18'
folded EPDM liner and 192 ft^2 of R32 surface and a 192/32 = 6 Btu/h-F
conductance and the box loses 24h(140-32)6 = 16K Btu/day and the house
can keep itself warm on an average December day and the shed does not
provide warm air for the house on an average December day and 1 ft^2
of twinwall gains 0.8x717-6h(140-32)1ft^2/R2 = 250 Btu/day, the shed
needs 16K/250 = 64 ft^2 of twinwall.
Nick
nick pine
Jan 2, 2012, 9:08:45 AM1/2/12
to sunspace
> With a few layers of gauzy mesh or felt to keep the surface near
the film cooler and increase its thermal resistance (a wall can't
lose much heat by convection to an airstream approaching the wall,
see equation 1 on page 4 of http://www.cibse.org/pdfs/8cimbabi.pdf)
There's a typo in equation 1 above.
the film cooler and increase its thermal resistance (a wall can't
lose much heat by convection to an airstream approaching the wall,
see equation 1 on page 4 of http://www.cibse.org/pdfs/8cimbabi.pdf)
http://www.dbb-project.com/introduction/insulation.php says
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 :-)
Nick
Reply all
Reply to author
Forward
0 new messages