# polyiso vs styrofoam

wrote:

Well, the law of diminishing returns certainly applies to any insulation project. But you can certainly insulate with higher effective R values than 22. Don't see what you're getting at there. One foot thick fibreglass can give you about R-38, irrespective of the coatings or any 'facing' such as drywall or OSB over it.
Remember, a higher R-value *behind* the foil surface makes the foil surface temperature closer to the ambient temperature. And that reduces radiant losses as well.

As far as a foil barrier on the ambient side, it isn't really all that valuable in most circumstances. Yes, it would certainly reduce the heat gain from direct sunlight, and reduce radiant heat loss. But if you look at how much heat is lost to the environment due to simple convection, you will realize that even if radiant heat losses were cut to zero, it wouldn't reduce the total heat losses by a significant percentage in most cases. You won't get a lot of improvement for the 'buck'. And a lot of folks don't like the idea of living in an aluminum foil sided house.
Radiant losses can be a big issue if the temperature difference is large and/or you've already taken steps to reduce the other forms of heat loss (conduction/convection). Or if your goal is to reduce absorption from the sun.
daestrom
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What 1/2"? Why 0.5? Most materials are closer to 1.

= 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

Yes, with a + vs -, if i>>d.

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...

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

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
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wrote:

??
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).

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

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>

I'm sure you mean the 'He' boils at 4.2K ;-)
daestrom
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snipped-for-privacy@ece.villanova.edu wrote:

This is a bit offtrack, but about attic insulation...
From your calculations of a typical EeffHr (radiation coefficient) of .04555 this would give a maximum effective R value of 22 no matter how thick the insulation. Thicker would get you closer to it but there are diminishing returns.
Also, not knowing the Eeff of a paper barrier, it would certainly be higher than .05

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No. Add Hc to EeffHr and take the reciprocal to find the R above, for the foil only. THEN ADD THAT TO THE R-VALUE OF THE INSULATION ITSELF.
Nick
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Thanks Nick,
That was an old draft and I didn't mean to send it, you had answered this days ago...
Cheers, Jeff

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