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Archived R-Value Dimensional Analysis

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data

    The ceiling of a single-family dwelling in a cold climate should have an R-value of 30. To give such insulation, how thick would a layer of (a) polyurethane foam and (b) silver have to be?
    per my book (W/mK)
    [tex] k_p = 0.024[/tex]
    [tex] k_s = 428[/tex]
    2. Relevant equations

    This is Thermal Resistance to Conduction (R-Value) and the equation is:
    [tex] R={\frac{L}{k}} [/tex]
    The 'R-Value' units per my book are
    [tex] {\frac {ft^2 F^{\circ} h}{Btu}}[/tex]


    3. The attempt at a solution

    I assumed that I would just solve for L and find the thickness of each material, although I ran into the problem with mixed units (F, Btu, Ft). The solutions manual simply lays down a dimensional analysis, although I'm having trouble following it and getting the same resulting unit.

    For Poly:

    [tex]
    ({\frac{0.024W}{mK}})
    ({\frac{30ft^2 F^{\circ} h}{Btu}})
    ({\frac{m}{3.281ft}})^2
    ({\frac{5C^{\circ}}{9F^{\circ}}})
    ({\frac{3600s}{h}})
    ({\frac{Btu}{1055J}})
    [/tex]

    EDIT
    Heh, as I was writing this, I figured it out. I wasn't thinking very carefully about which units cancel. So i'll finish anyways:

    After cancelling most of the obvious stuff, we end up with (taking out numbers for analysis):

    [tex]
    {\frac{Wm^2sC^{\circ}}{mKJ}}
    [/tex]
    Which is where I was stuck, so there are two things here:

    1. Rewrite C in K, and then they cancel
    2. Rewrite J as W=J/s and then cancel.

    We end up with m, which is the thickness (L) for the original question.

    Now, my question is: How was I to assume that the R-Value '30' was in US units instead of SI units as my book has no mention of this value in SI units? Wouldn't it just be 1 h·ft²·°F/Btu = 0.176110 K·m²/W? (per wikipedia)
     
  2. jcsd
  3. Sep 2, 2016 #2

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    Just the way things are.
     
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