Posted by *Iain McClatchie* on November 18, 2005, 7:29 pm

*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).*
Seems like the air-gapped foil-backed insulation's R-value is

dependent on the foil temp.

i^4 - (i-d)^4

= -4i^3d +6i^2d^2 -4id^3 +d^4

~= -4i^3d

So the air gap heat flow is just about linear with delta-T, but

goes up as the cube of ambient temp. 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.

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.

Posted by *nicksanspam* on November 18, 2005, 10:09 pm

*>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 i>>d.

*>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

Posted by *daestrom* on November 19, 2005, 6:15 pm

*>>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 i>>d.*

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

Posted by *daestrom* on November 19, 2005, 3:39 pm

*> 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).*

*> Seems like the air-gapped foil-backed insulation's R-value is*

*> dependent on the foil temp.*

*> i^4 - (i-d)^4*

*> = -4i^3d +6i^2d^2 -4id^3 +d^4*

*> ~= -4i^3d*

*> So the air gap heat flow is just about linear with delta-T, but*

*> goes up as the cube of ambient temp. 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.*

*> 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.*

Yes, you have an interesting point. The absolute temperature involved does

make a difference. But contriving a system where you can get the foil to be

coldest isn't easy. You can't just put it on the outside surface,

convection losses would over-shadow any radiant issues. If the air gap is

embedded in the wall, then it will be at some temperature between inside and

outside, depending on exactly where the air gap is in relation to other

materials. I suppose ideally we would put the air gap as close to the

lower temperature side of the construction as possible. That sounds like an

outer surface with a low emissivity foil, then protected from excessive

convection losses by some thin film to form the air gap. Not sure how

practical such a construction would be in a high wind situation though.

daestrom

Posted by *Iain McClatchie* on November 19, 2005, 10:33 pm

*Daestrom> I suppose ideally we would put the air gap as close to the*

*Daestrom> lower temperature side of the construction as possible. That*

*Daestrom> sounds like an outer surface with a low emissivity foil, then*

*Daestrom> protected from excessive convection losses by some thin*

*Daestrom> film to form the air gap. Not sure how practical such a*

*Daestrom> construction would be in a high wind situation though.*

I was thinking of foil-faced fiberglass in the wall, with the foil

facing the

plywood exterior sheath. Controlling the size of the air gap might be

tough, unless there was some sort of springy plastic spacer in there.

I suppose we could go all the way and use MLI like NASA does (many

many thin aluminized mylar sheets with vacuum gaps between, for

folks who haven't seen that TLA before). Seems MLI would work in

air reasonably well. I assume fiberglass is cheaper per R than MLI.

...but I wonder. To make code's R-19, we're going to have to use 6"

exterior walls with fiberglass. I presume that I could make R-19 with

4" MLI walls, no problem. (Or foam-filled, etc.) 4" construction is

cheaper than 6", and presumably the manufacturers have looked at

this issue.

>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).