Sample Question Temperature at interior of wall under Sheetrock.

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    Ryan Harvey

    Did the question provide any other information- R value of the sheetrock?, R value of the rest of the wall assembly? 

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    Daniel Burko

    It listed all of the individual r values of each piece of the assembly. So let’s say it’ all adds up to r21. I can figure out the btu loss but not what the temp is at the inside of the rock.

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    Scott Barber

    I came across a section of Ballast that covers this (chapter 16, pages 7-10 if you have Ballast 5.0). Essentially they use the formula below:

    delta Td = (Rc/Rt) x delta T

    where
    delta Td = temperature difference to reference point
    Rc = cumulative resistance (adding up all the R-values up to the reference point)
    Rt = total assembly resistance
    delta t = difference between inside and outside air

    So essentially you divide the R-value of the sheetrock by the R-value of the total assembly and multiply it by the total temperature difference. Then you can subtract that value from the outdoor air and get the temperature at that point.
    If it asks for the temperature at another point in the assembly you would have to combine the R-values of the materials up to that point and divide it by the total R-value of the assembly.

    Not sure if that all makes sense - I'd recommend reading that portion of Ballast or maybe doing a Google search to see if you can find a more comprehensive explanation. 

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    Daniel Burko

    Ok here is what I believe it is:

     

    Let’s say its 70 inside and 20 outside. Total R value of the assembly is R21 and total R value up to the Sheetrock is R19

    So, delta TD = (R19/R21) x 50

    = So the temp just under the rock would be 45 degrees?

    Almost doesn’t make sense that just under the Sheetrock with an R value of 2 (which is a lot) You would lose 15 degrees.  Since that is the warm side of the insulation.

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    Scott Barber (Edited )

    Maybe I didn't describe it well enough (I realized I described it as if the exterior was warmer). Looking at your formula, you would need to add 45.24 to 20, getting 65.24. 

    You could also calculate it this way, from the other side of the assembly:

    delta Td = (2/21) x 50 = 4.76
    So you would subtract 4.76 from 70 and get 65.24. 

    Here's a figure from Ballast (hope it's ok that I post it - I can take it down if needed) that may help clarify. Sorry, I couldn't figure out how to get it rotated...



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    Daniel Burko

    Agh ok.  So the way I did it, I just have to add the temp to the outdoor air temp.  Ok cool.  Thanks.

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    Sebastian Escobar

    This is exactly the explanation I needed. Thanks!

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    Jonathan Chertok

    hi guys,
    question. not totally sure if this is blindly obvious or more complicated but you don’t know til you ask...
    how do i know which side of the element the temperature is on. say i want to assign a value at the inside and outside of the airspace by labeling it.
    the number listed IN the airspace column is the value on the /outside/ (in reference to wall section) of the air space and the sheathing listing is the outside of the sheathing (inside face of air gap).
    is it by convention that the listed value in the column is the outside portion of the element in reference to the section (or just in reference to inside and outside i guess). seems like this could be dependent upon how you set up the hart or really easy to mix up if doing a hand calculation.
    thanks for any expertise (hard facts).

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