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What is a reproducible test to measure the r-value of an exterior wall? Or even a specific spot on a specific wall? It needn't produce an absolutely certain result, but a good estimate with known level of confidence would be helpful.

This is more a science question than a question about a specific project.

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4 Answers 4

A thermographic camera would be an expensive way to measure heat loss. An inexpensive laser thermometer would give you readible difference between several points. Both methods require a temperature difference between sides of the wall. The most accurate measurents would require the greatest difference. It would be easier to see the difference between a 70 degree inside temp and a 25 degree outside temp.

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While I like the answer above I have to say that insofar as determining actual R-values you may be barking up the wrong tree, science-wise. R-values are assigned to building systems, but every wall/subfloor/attic has its own peculiarities that makes determining an R-value at any specific point a thorough crapshoot. For more on this subject I recommend checking out greenbuildingadvisor.com. –  Paul Jan 8 '13 at 22:01
    
As per the question"It needn't produce an absolutely certain result but an estimate" If the measurement difference between two points is significent the R value difference is significent –  mikes Jan 8 '13 at 22:30
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The temp difference alone will give you relative resistance of various components, not the actual R value. You also need to either know the heat flux of the section you are measuring or the actual conductance of a different section which will give you the base for the relative difference. –  bcworkz Jan 9 '13 at 3:02

You could cut out a section of the wall, place it in a box made of an extremely poor thermal conductor, add a known amount of heat to one side, and measure the resulting rate of change in temperature on the other side. This will allow you to calculate the wall's u-value, and the relationship u=1/r will give you the r-value.

You could also keep track of the temperatures at the internal and external wall surfaces as well as indoor air and outdoor air over time on a cold night while raising the interior temperature at a steady rate. You may need to ask the Physics site for help with the calculations and potential sources of error for this method.

Or you could estimate it by checking the type, thickness, and condition of the insulation and other materials and consulting a chart like this. As a general rule of thumb, you should subtract about 10-40% to account for thermal bridging at studs, imperfect installation, airflow, and other confounding factors. In a near-perfect wall, you might subtract 10% to account for studs, but if it is not properly air-sealed and there are insulation gaps and a couple damp spots, you might subtract 20-50% depending on the severity of the issues.

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Getting R-value is based on three things really.

  1. Materials
  2. Distance
  3. Sealing

Example - If you framed with 2x6s instead of 2x4s with just sheetrock - your wall would have more r-value because there would be a greater distance for air to travel to the other side.

Also materials are given an r-value based on their use in a best case scenario. Just because you have insulation that is r-13 doesn't mean you are getting r-13 returns after installing. Think about putting in one piece of insulation that is 12 inches wide to fix a 50 foot gap. Did it do anything measurable? Probably not.

Also r-value is a max measurement again. If the temperature outside is 70 degrees and it is 70 degree inside you could have 5 feet of materials with an r-value of 500 but in actuality it is providing an r-value of 0. r-value is measuring a change in temp from one side of a system to another. The more drastic the temperatures the greater need for more r-value and the greater r-value returns you will get.

So to answer your question without getting the CSI team together you will need a thermometer and a couple hours of steady weather (turn off your heating/air too). Optional piece of equipment would be a shoebox - to keep thermometer out of sun and drafts.

enter image description here

Notice that I didn't answer how much potential r-value a wall could have because that doesn't matter. You will have taped seems, outlets, studs, whatever. You want real world functionality. Take the temperature on outside wall, take the temperature inside, and take the temperature right by same wall in same location. This chart does work. Easiest to do on a window area but works everywhere.

To know your systems true r-value you will have to have extreme temps outside/inside. If your system really did perform at greater than r-15 which is about a 5 degree change over the course of a 100 degree temp conflict then you will have to have one of those -40 degree days with a toasty 80 degrees inside to get your true measurement. I have tested out very well insulated houses and they usually max out around r-15 to 20 range. Spray foam in some cases has out produced r-20.

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You need three temperature probes, a piece of material with known thickness and thermal conductivity (or known R-value) and a good measurement for the actual thickness of the wall. The inside and outside of the building need to be at different temperatures; the bigger the temperature difference, the cleaner the measurement.

Sandwich the wall with the known material and measure the temperatures as shown below.

enter image description here

The heat moving through the known material, per unit area, q'' = –k Δ*T*/Δ*x*, where k is the thermal conductivity of the known material (W/m/K), Δ*T* is T2–T1 and Δ*x* is the thickness of the known material.

At steady state, the amount of heat flowing through the mystery material has to be the same as through the known material, so you can find out k for the mystery material using k = –q'' Δ*x*/Δ*T*, where q'' is the heat flux you calculated before, Δ*x* is the thickness of the mystery material and Δ*T* is T3-T2.

R-value is Δ*x*/k, but you have to be careful about units. If k is in W/m/K, R will be in K·m^2/W. Multiply this by 5.678263 to get to h·ft^2·ºF/BTU, which are the usual units for R in the US.

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