Framing factors
Hatherly
| Posted in General Questions on
I am trying to calculate framing factors for walls. I read that until 1993 ASHRAE recommended using 15% for 16″oc walls and 12% for 24″oc. Then in 1993 they revised these to 25 and 22. Is this still what people are using? Also, how does this change for walls that use advanced framing techniques? Is there any resource that gives this information?
Thanks.
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ASHRAE has a 2003 publication regarding this. I have not had a chance to read it yet. The description says it is based on 180 case studies. Check their website store for "Characterization of Framing Factors for Wood-Framed Low-Rise Residential Buildings".
Also, a google search for that title brought up this document from the California Energy Commission:
energy.ca.gov/title24/2005standards/archive/documents/2001-11-14_workshop/2001-11-07_FRAMING_FACTORS.PDF
15% (@ 16" oc) is possible, but case studies have shown many houses closer to 25% due to over framing the walls. If this is for energy modeling/HERS index, then you'll have to make an assumption (high, low, or in the middle) and then verify it to your best judgement on site, like with the insulation grades. I generally use 20% and then adjust up or down depending on how the framing job looks; 15% for advanced framing.
I use 25% for 16" oc, 22% for 24" oc and 16% for OVE "advanced" framing.
The REM/Rate defaults appear to be 23% for 16" O.C. and 20% for 24" O.C., with nothing in the library for OVE.
Thanks so much everyone - this helps give me an idea where to go!
Here is a guide published by Oak Ridge Nation Lab since the ASHRAE link is a 404 now.
http://web.ornl.gov/sci/roofs+walls/research/detailed_papers/thermal_frame/
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I am curious about the translation of framing factors into real world transmissivity of energy flow.
I am using double frame walls with effectively zero thermal bridges. So, it feels to me like the framing factor for me is 0%.
I have real doubts about the Rem/ Rate calcs. validity in this modern world. They seem to be written in stone from a 1960's "cookbook of thermodynamics" approach. Somebody prove wrong here...
David,
Let me start by saying I don't know much about this. But whether your walls are separated by an insulated gap, or had continuous insulation on the exterior, you still have areas with insulation for the full depth of the wall, and areas where the insulation is interrupted by materials (wood) that has a much lower R value. Somehow that needs to be taken into account.
Framing factor isn't a scientific principle of heat flow so much as an input to get a numerical representative result (typically an assembly U-factor—I'm not at all familiar with the REM calcs). This can be calculated with double walls too. It'll depend on if studs are lined up, what size the gap is, size of studs...
Try: https://www.greenbuildingadvisor.com/article/the-fundamentals-of-series-and-parallel-heat-flow
There is a good article on this over on Energy Vanguard. It’s not as simple as you’d think if you want to model it properly, but there is a simple way to get close:
Let’s assume you are using insulation that is R5 per inch, and wood that is R1 per inch. Let’s assume you have a 10” thick double stud wall, with studs on 24” centers. You can model the wall as follows:
=== studs lined up gives R48.25 as shown below ===
With the studs lined up, look at the wall from above (“plan view”). Each area between studs is 10” of all insulation, so R5/inch x 10 inches gives R50 in the areas in the stud cavities. Since the studs line up, you have a 10” thick area of wall that is two studs at 3.5” each, so 7” of that 10” is wood at R1/inch with the remaining three inches filled with insulation at R5/inch. (R1/inch x 7 inches) + (R5/inch x 3 inches) = R22 in areas of the wall where the studs are.
Since the studs are 24” apart, think of each stud cavity as 24 inches wide with 1.5 inches of that made with stud. There are 16 “slices” of wall each 1.5” thick (24” divided by 1.5” gives 16”), and we use 1.5” for a slice because that’s the width of the stud and it makes the averaging easy. One of those 16 slices — the one where the stud is — is R22, the other 15 slices are full of insulation and are R50. We can now work out the average R value of the wall as ((1 slice x R22) + (15 slices x R50)) / 16 slices = R48.25 average R value over the entire stud cavity, so your wall here is R48.25.
=== studs offset gives R48.25 as shown below ===
With offset studs, each of the two of the “slices” described in the previous example have 3.5 inches of R1 and the remaining 6.5 inches full of insulation at R5/inch. That gives (R1/inch x 3.5 inches) + (R5/inch x 6.5 inches) = R36 in the area of the walls where the studs are.
Now we use the same “slice” method used in the previous example. Since we have offset studs, two of those 16 slices have studs in them. The average R value of the wall is ((2 slices x R36) + (14 slices x R50)) / 16 slices = R48.25
===
In reality, it’s a little more complex than that, and the Energy Vanguard article explains why. Offset studs are slightly more efficient (slightly higher R value) than “lined up” studs when modeled with the more complex method. I use the simple method I’ve described for most all comparisons though because it’s close enough to get you “in the ballpark” and allow you to make pretty good comparisons between different wall assemblies unless you want really high accuracy or have very oddball wall assemblies.
Bill
Bill, you may be including this in the "it's more complex than that" statement, but just to be clear: parallel flows should be calculated using U-values, not R-values.
It becomes a bigger deal when we use higher (and more accurate) framing fractions (as opposed to the 6.25% factor you used in your example). For example: if we use a framing factor of 20% instead, we'll get R-44.4 using your method, and R-39.85 using the U-value approach.
I won't profess to totally understand this, but it has to do with the mathematics and the fact that R-value is a reciprocal. Heat flow isn't simply proportional to area, since a highly conductive path is a sort of 'short circuit.' Averaging R-value doesn't properly account for this mathematically.
It also illustrates why in a super insulated assembly, thermal bridges can really undermine the overall assembly's performance by an even greater margin. For electrically minded folks, some similarities can be seen with a resistor in parallel circuit analysis. Adding a low resistance element to a circuit greatly increases total current, since the other high resistance paths don't restrict the flow apparent on that low resistance path. This is apparent in the equations that—like our heat flow equation—uses reciprocals of R (sort of like U-value).
1/Rtotal=1/R1+1/R2...1/Rn (notable difference being lack of an 'area' component)
I know you know this Bill, (and then some) I'm just expounding a bit.
When using the 6.25% in your example, and assessing the wall with studs aligned vs not aligned, using u-value shows a slight difference instead of an exact similarity: R-46.31 for lined up, R-47.68 for offset.
The 'more complex' statement re Energy Vanguard still exists with the U-value approach given things like air films and the 3-dimensional nature of heat flow (this being a 2-D analysis).
Yeah, I was including what you’re saying in the “it’s really more complicated” part. I used the simple method I describe because it’s easy for everyone to understand, and for basic wall assemblies it allows for reasonable approximations that give you a good idea of which is better and by how much. It’s a trade off of simple math that everyone can understand, but slightly reduced accuracy.
My simple method works best to compare assemblies that are very similar, things like “what if I use 2” of exterior rigid foam instead of 1”, or “what if I use R15 batts instead of R13 batts in my wall that also has 1” exterior rigid foam”. The error in the final numbers using the simple method get worse as the assemblies get more complex, which is one of the reason I used the double stud example the way I did since it shows the issue. You are usually within 10% or so, which is probably close enough for most purposes unless you’re really trying to squeeze out every last drop of performance.
Bill