Sample Question Temperature at interior of wall under Sheetrock.
Hello All.
I had a sample question that I cannot figure out. In a wood frame wall section, the temperature outside is 20 degrees and the temperature inside is 70 degrees.
It asks for the temperature at the vapor barrier which is inside under the rock...
I am a little lost on this one.
I can draw it and post a pic if easier.

I came across a section of Ballast that covers this (chapter 16, pages 710 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 Rvalues up to the reference point)
Rt = total assembly resistance
delta t = difference between inside and outside airSo essentially you divide the Rvalue of the sheetrock by the Rvalue 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 Rvalues of the materials up to that point and divide it by the total Rvalue 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.

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.

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...

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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