Walls are only about 60% insulated.

Many of us have a mental model of insulation as the nice fluffy stuff sandwiched between the inner and outer layers of our walls. The (thermally) ugly reality is that most walls contain lots of doors and windows, and that the wall area that is not doors and windows is full of wood and steel.

Here is a sketch (thanks to my newly acquired skills in Google Sketchup) of a typical section of wall for my Park City house.

Typical wall system in a U.S. stick-built house.

From the outside, the wall is:

• wood siding
• Tyvek housewrap
• OSB sheathing
• wall framing, with cavities filled with insulation (one cavity volume shown in pink)
• (sometimes a poly vapor barrier is inserted here, depending on the insulation strategy)
• drywall

Punched into the middle of this is a thermally challenged window assembly.

The pink stuff (whether pink or not, i.e., fiberglass, cellulose, spray foam) is a fantastic insulator. The problem is that, as you can see from the illustration, there isn’t very much wall cavity left to fill with insulation.

In my project, there is a shocking amount of wood and steel in my walls. In one section of wall, I have a 5 1/8″ x 28″ x 18′ glulam beam (a header over a bumped out section of the great room). My architects affectionately refer to this a “glu-wall.” I’ve resigned myself to filling my walls with wood and steel (to keep the house from falling down and to realize the architectural vision), but this reality smashes the ideal vision many of us have a well insulated wall.

The building code lives with the delusion. Code requires R-18 (or whatever…) but that is the measure of just the fluff that is put in the cavities of the wall. It ignores the overall thermal performance of the wall system factoring in the windows, doors, wood, and steel.

I’m working on a post that will do the full thermal analysis of alternative insulation strategies for this wall. But, here is a rough preview:

If the wall area is

25% window (~R3)
15% wood (~R7)
60% insulated cavity (~R19)

The net average R-value of the wall is just R13. Not so hot.

0.25 x 3 + 0.15 x 7 + 0.60 x 19 = 13

Whoa…but this is actually not how you add R values. (Thanks to commenter Ed for pointing this out.)

To add R values of areas arranged in parallel, calculate the “conduction coefficient” of the material first, U = 1/R. Then calculate the area-weighted average of the conduction, e.g., Unet = p1 x U1 + p2 x U2, where p1 and p2 are the fractions of the wall for each material. Then invert that value to get the net value of R for the system, Rnet = 1/Unet.

So, the U value of this wall is:

0.25 x 1/3 + 0.15 x 1/7 + 0.60 x 1/19 = 0.136

And therefore the net R value is just 1/0.136 = 7.3. Really not so hot.