Taken from the SolarHeat yahoo group at:
http://tech.groups.yahoo.com/group/SolarHeat/message/27383
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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 = 1.7×10^-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 2×4
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
0.8×980-6h(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 65×10^5/0.85 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/Firest\
one_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 = 0.1×40′/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^2×589 = 219K Btu/day. It might
do that with a 16′x20′ twinwall roof and a 17′ box height.
Nick
