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How Much Foam=A Thermal Break?

Stockwell | Posted in Energy Efficiency and Durability on

I am building in Western NC in Zone 4. In designing my “ideal wall”, I would like to include a thermal break. Since code does not require any amount of exterior foam for this Zone, what would constitute “enough”? I plan to use Zip-R sheathing, and would like to keep it as thin as possible despite all of Huber’s reassurances on it being an effective structural sheathing, even in the thicker versions. So, would the 1/2″ be enough of a thermal break? 1″? Is there a formula to figure this out or do I simply take the R5.5 of my 2×6 stud and add on the foam’s R value and say that is the R-value for the studded areas of the wall? Or is it more complicated since you are preventing the thermal bridge which must be worth some bonus points?

Thanks in advance.

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  1. Expert Member
    Dana Dorsett | | #1

    Thermal break is a relative term. There is no such thing as "...preventing a thermal bridge...", but a thermal break mitigates the thermal bridge, but it's a matter of degree.

    In studwalls it' usually referring to improving the performance of the thermally bridging framing. Most common framing species run about R1.2/inch, so a 3.5" deep 2x4 represents an R1.2/inch x 3.5" = R4.2 thermal bridge through the much higher R cavity insulation, with a disproportionately large heat loss for it's ~25% fraction of the wall surface (typical of 16" o.c. standard framing).

    But an inch of Type-II EPS sheet is also R4.2, which cuts the framing loss in half, even though it cuts the loss of the R11-R15 cavity by a much lesser amount. So it's something of a thermal break for the framing, not so much for the rest.

    With 2x6 framing you're looking at R1.2 x 5.5" =R6.6 for the framing fraction. To cut the framing fraction losses by roughly half would take an inch of polyisocyanurate, or 1.5" of EPS.

    But doubling the performance of the framing is an arbitrary criterion- greater amounts or lesser amounts could also be legitimately referred to as a thermal break. On the greater amount end of the scale, once the continuous insulation exceeds ~3/4 the R-value of the cavity insulation it becomes the primary insulation- the insulated studwall's net performance after thermal bridging is comparable to or less than the continuous layer. (eg: 2" of polyiso on a 2x4/R13 wall) In those situations it is much more than a mere thermal break, and isn't usually referred to as such.

    Whether R3 ZIP-R or R6 ZIP-R is used is really a matter of your performance goals. For a 2x6 /R20 wall the R3 ZIP-R reduces wall losses by about 16% compared to ZIP without the foam, R6 ZIP by about 29%.

  2. jwolfe1 | | #2

    I'm having trouble figuring out the equation to reduce heat loss when adding foam strips to rafters. If I have an 11.875 inch 2x12 cathedral roof rafter and add a strip (not a sheet) of 1 inch polyiso over the rafter face and then drywall to finish how much does that reduce the heat loss through that rafter? How much would be reduced with a 1.5 inch strip?

    Because the r value of the rafter (r14.25) is already much higher than a 2x4 (r4.2) or 2x6 (r6.6) and just an inch of polyiso is a much smaller proportion of of the rafter thickness (1 inch/11.875= 8.5% of the thickness vs 1 inch / 5.5 for a 2x6 = 18.2%) does it really even move the needle much to add a strip of 1 or 1.5 inch polyiso?

    The roof system will be 24 inch centers with no hips or valleys. There will be just a central ridge so the framing proportion is relatively low.


    1. Jon_R | | #3

      The software "THERM" does a good job. Simpler methods are probably good enough.

      1. jwolfe1 | | #4

        I'm unable to find THERM software and confirm that it is what I'm looking for. What I did see appears to have a significant learning curve. I'm really just looking for back of napkin estimates such as what was towards the end of post #1, not a full modeling breakdown.

        1. Jon_R | | #5

          Just add the R values. An inch of polyiso improves a R14.25 stud to about R20. 1.5" yields about R23.

          1. jwolfe1 | | #6

            So is the equation really as simple as having r15 insulation that is increased to r20 you end up with 20/15 -1 = 33.3% less heat loss in that area?

          2. Expert Member
            PETER Engle | | #7

            Yes, and you can simply compare the total area of rafter edges with the total area of cavities to calculate the total R value of the assembly.

          3. Expert Member
            NICK KEENAN | | #8

            Yes and no. If you are layering insulation, you can just add the R-Values. If you have a surface that has different insulation levels side-by-side, you have to take the weighted average of the inverse. Which is a mouthful but it's not that complicated. To make the math easier, people use U-values, which is the inverse of R-values: U=1/R. If you have a wall with different insulation levels, the U-value of the entire wall is equal to the the sum of the U-values of the sections, weighted by their share of the area.

            So U-Total=(U1*P1) + (U2*P2) .... (Un*Pn)

            Where U1= U value of section 1
            P1=percentage of the wall that is section 1

            So let's assume you have a 2x12 wall. The framing is 10% and has an R-value of 14.5. The bay is 90% and has an R-value of R-4 per inch or R-45 total.
            R14.5=U 0.07
            R45= U0.022
            So U-value of the assembly is:
            (10% * 0.07) + (90% * 0.022) = 0.027 which is equivalent to R37

            With an inch of polyiso, the studs are R20 or U 0.05. Total U-value of the assembly is:
            (10* 0.05) + (90% *.022) = 0.025 or R40.

            So that inch of foam increases the R-value of the assembly from R37 to R40, or about 7.5%.

            To figure out if that is worth it, you really have to have an energy model of the building.

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