How to find the R value of an exterior wall assembly

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... I need you folks !

Simply putting it, Id like to determine the R value of an exterior wall.

Im willing to treat the assembly as one and assign the R for the
complete assembly.

if anyone can come up with anything, Id love to hear it.

From what I have so far, comparing it to electricity, the problem may be figuring
the equivalent of I as E/I*R

thanks in advance..
 
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Heat transfer is just proportional to temperature difference: q=k(dT)

But I'm actually not sure what you're going for here. Are you trying to do it by experiment? Or just based on what you know of the materials the wall is built from?
 
what I am going for ? Imagine you were given a task, to figure out a walls R value
with no information on what's in it, and your not allowed to look.
Or
you have a wall with a proposed R value, you need to verify it ...

How could you do it by measuring ?
 
In practice, it is usually calculated from theoretical values and verified by thermography.
 
problem is thermography doesn't show R values, just differences in heat.

If heat were water and the the hull was a wall, how could you see how much water leaked in a given area while being on one side or the other ...
 
ghboom said:
problem is thermography doesn't show R values, just differences in heat.

If heat were water and the the hull was a wall, how could you see how much water leaked in a given area while being on one side or the other ...

Right. It requires a reference.
 
ok, Think of it like this...
Say you were given a brand new wall, A real honest wall in a new home. Instead of sheet rock, it had a new material called plasti-rock, then instead of
2x4 studs the wall has 5"x5" rails 24 OC made out of (pick any new material) and it was insulated with treated popcorn.
the exterior sheathing was made from tree bark, and the siding was rubber.

Your task is to find the R value of 5 sections, all measuring 1' square..

Could you (or anyone) figure out how to do it ?
 
ghboom said:
ok, Think of it like this...
Say you were given a brand new wall, A real honest wall in a new home. Instead of sheet rock, it had a new material called plasti-rock, then instead of
2x4 studs the wall has 5"x5" rails 24 OC made out of (pick any new material) and it was insulated with treated popcorn.
the exterior sheathing was made from tree bark, and the siding was rubber.

Your task is to find the R value of 5 sections, all measuring 1' square..

Could you (or anyone) figure out how to do it ?

Go back to Russ's answer previous.

However, the answer to the posed question, which I'm now certain is a homework question, is most likely no. Any testing which is compliant with ASTM C 236 or C 976 will require a specimen considerably larger than 1'x1'.
 
Actually, my previous post was not quite right - thermal inertia is difficult to estimate and has a large impact there. A better way would be to integrate over time and delta-T the heat provided by a heater to keep the room warm.
 
TVP, No this honestly isn't a Homework question.
The Reason for 1 ft sq, any larger can introduce
anomalys. A small section should be consistent.

Im Happy russ keyed back in, because the whole room is being
profiled, not a section measured.

R value per sq ft is a very meaningful value...
 
Anomalies? What kind? Seems like the only such effects might be at the edges and a larger section minimizes that. What is this thing?
 
Insulation values take a big hits for small gaps, areas of compression,
pipes in the bay etc...

If you have a 16" bay insulated well, close to perfect, then a sample from the
center is as good as that wall can be. That is the purpose of
my quest.

But to entertain the question further, If that 1 bay was perfect, and
the larger sample picked up pipes, 3 studs, compressions then the
perfect square foot took a large hit.

Determining the R value of a fake wall can show this more...
take a non standard 125' x 8' wall
(OA 1000 sq ft) built using 2x10 studs and single plates...
If the wall is built 16"oc

95 studs at 93" high R 7.5
32 studs for 1 top and 1 bottom plate at a full 8'
Cavity insulation, yet to be invented 9.5" R-100 fiber-stop1
1 large window 9' wide 5' high U= .35

whats the over all R value ?

Studs R7.5 U .13333 92 Sq ft UA=12.26
Plates R7.5 U .13333 2.6 Sq ft UA = .355555
Wall R100 U .01 860.4 sq ft UA = 8.604
window U .35 45 Sq ft UA = 15.75

1000 sq ft / 36.96 = Over all R27.7

The R 100 wall lost 72.3 on its over all R because of 16% performing poorly.

So accidently measuring areas that are already known bad performers cant
reliably say how well a product can perform...