Iain McClatchie wrote:

Daestrom Two surfaces spaced a modest 1/2 inch apart with
Daestrom emissivity/abortivity of about 0.5, with temperatures
Daestrom of 460 R and 461 R will have a net radiant flux of
Daestrom q' = emissivity*Stefan-Boltzmann * (461^4 - 460^4)
Daestrom = 0.34 BTU/hr-ft^2 (R-value of 3.0).

What 1/2"? Why 0.5? Most materials are closer to 1.

Seems like the air-gapped foil-backed insulation's R-value is
dependent on the foil temp.

i^4 -(i-d)^4
= i^4 -(i^2-2d+d^2)(i^2-2d+d^2)
= i^4 -(i^4-2i^2d+i^2d^2-2i^2d+4d^2-2d^3+i^2d^2-2d^3+d^4)
= 2i^2d-i^2d^2+2i^2d-4d^2+2d^3-i^2d^2+2d^3-d^4
= 4i^2d-2i^2d^2 -4d^2+4d^3 -d^4
~= -4i^3d

Yes, with a + vs -, if id.

So the air gap heat flow is just about linear with delta-T, but goes up
as the cube of ambient temp.

You've just reinvented the "linearized radiation conductance" :-)
G = 4x0.1714x10^-8Tm^3 Btu/h-F-ft^2, where Tm is the mean Rankine temp.
But air spaces also transfer heat by convection and conduction...

So while you got an R-value of 3.0 for the interior of a cold refrigerator,
the same foil against the interior of a house would see an R-value of 2.0.

That also depends on convection and conduction, which depend on the temp diff
and the direction of heatflow.

Seems like the R-value would be a lot higher if the foil faces
the cold side. If it's -30 F outside, R-value of that gap is 3.75.

Our local college keeps liquid helium for their electron microscope's
superconducting magnet in a Dewar vacuum flask surrounded by liquid
nitrogen, with insulation around that.

H2 boils at 4.2 K. N2 boils at 77.3 K. So 2 mirrors with e = 0.03 would lose
5.67x10^-8x0.03x40.75^3 = 0.000115 W/m^2C by radiation, ie 0.00002027
Btu/h-F-ft^2, ie US R49335, vs an R20 house wall.

Nick

wrote in message
...
Iain McClatchie wrote:

Daestrom Two surfaces spaced a modest 1/2 inch apart with
Daestrom emissivity/abortivity of about 0.5, with temperatures
Daestrom of 460 R and 461 R will have a net radiant flux of
Daestrom q' = emissivity*Stefan-Boltzmann * (461^4 - 460^4)
Daestrom = 0.34 BTU/hr-ft^2 (R-value of 3.0).

What 1/2"? Why 0.5? Most materials are closer to 1.


??

1/2" is just typical spacing between glazing on double-paned windows. It
doesn't factor into radiant transfer if the two surfaces are flat planes
whose area is spacing between them. Yes, the emissivity of many
materials are higher than 0.5. But the example (half of which has been
snipped here) compared the radiant heat loss with/without a low-e coating,
versus the conduction of having two films in direct contact. (remember Duane
had questioned why I said that emissivity is 'pretty much irrelavent' if
there is no air-gap).

Seems like the air-gapped foil-backed insulation's R-value is
dependent on the foil temp.

i^4 -(i-d)^4
= i^4 -(i^2-2d+d^2)(i^2-2d+d^2)
= i^4 -(i^4-2i^2d+i^2d^2-2i^2d+4d^2-2d^3+i^2d^2-2d^3+d^4)
= 2i^2d-i^2d^2+2i^2d-4d^2+2d^3-i^2d^2+2d^3-d^4
= 4i^2d-2i^2d^2 -4d^2+4d^3 -d^4
~= -4i^3d

Yes, with a + vs -, if id.


The absolute temperature of most building materials is the delta
temperature between said materials.

So the air gap heat flow is just about linear with delta-T, but goes up
as the cube of ambient temp.

You've just reinvented the "linearized radiation conductance" :-)
G = 4x0.1714x10^-8Tm^3 Btu/h-F-ft^2, where Tm is the mean Rankine temp.
But air spaces also transfer heat by convection and conduction...


The affects of convection and conduction were already discussed. Those
affects far outweigh the effects of radiant heat transfer in most building
applications. Only after reducing conduction with conventional insulation,
and controlling convection by properly designed air-gaps, does the use of
low-e coatings become a significant factor.

Interestingly, the spacing of glazing in double-paned windows is chosen to
minimize both convection currents and conduction. From a conduction
standpoint, a wider gap is better, but a wider gap allows stable currents to
set themselves up between the panes, bad from a convection standpoint.
Knowing the properties and pressure of the fill gas, one can choose a gap
that has the falling gas film on the cold pane interfere with the rising gas
film on the warm pane, and inhibit long-path (full height of pane)
circulation.

snip

Seems like the R-value would be a lot higher if the foil faces
the cold side. If it's -30 F outside, R-value of that gap is 3.75.

Our local college keeps liquid helium for their electron microscope's
superconducting magnet in a Dewar vacuum flask surrounded by liquid
nitrogen, with insulation around that.

H2 boils at 4.2 K. N2 boils at 77.3 K. So 2 mirrors with e = 0.03 would
lose
5.67x10^-8x0.03x40.75^3 = 0.000115 W/m^2C by radiation, ie 0.00002027
Btu/h-F-ft^2, ie US R49335, vs an R20 house wall.


I'm sure you mean the 'He' boils at 4.2K ;-)

daestrom