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Determine R value

  1. Feb 10, 2008 #1
    ... 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..
  2. jcsd
  3. Feb 10, 2008 #2


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    Staff: Mentor

    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?
  4. Feb 11, 2008 #3
    what Im going for ? Imagine you were given a task, to figure out a walls R value
    with no information on whats in it, and your not allowed to look.
    you have a wall with a proposed R value, you need to verify it ...

    How could you do it by measuring ?
  5. Feb 11, 2008 #4
    In practice, it is usually calculated from theoretical values and verified by thermography.
  6. Feb 11, 2008 #5
    problem is thermography doesnt 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 ...
  7. Feb 11, 2008 #6
    Right. It requires a reference.
  8. Feb 11, 2008 #7


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    Staff: Mentor

    Experimentally, you can take a heated room, shut off the heat, and watch the cooling profile. I'm actually doing that in a room right now...
  9. Feb 11, 2008 #8
    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 ?
  10. Feb 12, 2008 #9
    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'.
  11. Feb 12, 2008 #10


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    Staff: Mentor

    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.
  12. Feb 12, 2008 #11
    TVP, No this honestly isnt 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...
  13. Feb 12, 2008 #12
    Anomalies? What kind? Seems like the only such effects might be at the edges and a larger section minimizes that. What is this thing?
  14. Feb 12, 2008 #13
    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.....
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