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Calculating different in temperature using a material's conductivity

  1. May 24, 2012 #1
    this isn't really a problem, i'm trying to understand the table in this link:

    http://www.nrc-cnrc.gc.ca/eng/ibp/irc/cbd/building-digest-36.html [Broken]

    basically it shows a wall assembly, it's 70 degrees on the inside, and it gives a list of the materials and their thicknesses in the wall

    if we look at the gypsum plaster, we see it's conductivity (k) is 5 watts per meters kelvin, which i think means that for every square meter of this gypsum plaster, 5 watts of heat will pass through in an hour or something like that.

    anyway the next column shows it's conductance, which i think just expresses the conductivity of that material in a certain thickness, so you divide by the thickness. the next column, resistance, is the opposite of conductance. but then i don't understand how they jump from there and conclude that the temperature is going through drop 4 degrees through the gypsum plaster. i'm not sure how they get there.

    can anyone help explain this to me?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 24, 2012 #2
    basically i don't see how you can do this without knowing the area of the wall, which they don't give you in that article/table
  4. May 24, 2012 #3


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    Staff: Mentor

    That page is using British units, not metric. So the conductivity is given in Btu per square foot area per 1 inch thickness per hour per °F temperature differential.

    If the wall is uniform, then each square foot will pass heat at the same rate, and will have the same temperature differential. Once you calculate the temperature differential and heat flow for a representative square foot, multiply the heat flow by the total square footage of the wall to obtain the total heat flow.

    For materials sandwiched together the heat resistances add, so the net heat resistance of a representative square foot can be determined. Then, with the total temperature differential across the sandwich you can calculate the heat flow. This is analogous to taking a series string of resistors and placing a potential difference across it; you can determine the total current (heat) and individual potential drops (temperature changes) for each resistance (material heat resistance).
  5. May 24, 2012 #4
    I agree with Gneill. Thermal conductivity has been converted to RESISTANCE in the table because resistances add in series (Conductivity does not !!)

    great shame about the units !!!
  6. May 24, 2012 #5
    thank you guys so much for your help!

    to make sure i have this..

    1. so the first step is to make sure you have the resistance of each material in the wall assembly - you can get that from the conductance of a material, which is its conductivity / thickness in inches. you need the resistance because you can add them all up, which you can't do with conductivity (why, by the way?)

    2. then you simply take one material at a time, and divide it's individual resistance by the total overall resistance. so taking the concrete block (for example), i just take 2/11.32 = .1766%

    3. i take the overall thermal gradient (total heat loss from inside to outside of the assembly) and multiply it by this percentage, and round to the nearest degree. so using the concrete block again, i take 60 (total heat loss of the whole wall) * .1766% = 10.596 -> 11 degrees heat loss.

    is that accurate?

    my next question is that if i get a similar question with metric units, here's what would happen:

    1. conductivity is measured in watts per meters kelvin, but i still get that number from a table somewhere. it says that the conductivity for the cement mortar is 5 - but if that's british units, then i need to do some converting.. wikipedia says that 1 british unit is 1.730735 SI units, so cement mortar is now 8.65. right?

    2. conductance (C) is still C=k/n, except n would be in millimeters, right? so if were to use the cement mortar from the table, i would take 8.65 and divide by 6.35 (1/4" to mm). which gives me 1.362 conductance.

    3. resistance is still the reciprocal of that number, so i don't need to change too much around here.. 1/1.362 = .734 R

    4. i have to change farenheit to celcius (kelvin, i guess, but those are kinda the same, right?).. so 70 F is 21.11C and 10 F is -12.22C, so the new total temperature difference is 33.33C

    5. and finally, if i wanted to just get the individual potential drops, do the same math.. .734/total resistance, and then multiply that number by 33.33 to get the individual drop.

    is that right? do i have this thing down?
  7. Jun 2, 2012 #6
    new questions:

    1. if a 'layer' of a wall assembly contains two materials (ex, studs with insulation in between), do i average the two materials or add them up or what?

    2. i'm still not sure if i have the method right for doing this in imperial.. another example:

    calculating the thermal gradiant through 100 mm of brick with a conductivity of 1.21 w/mk

    to get the conductance (w/m2k), i would just take 1.21 / .1 = 12.1?

    is that correct? since it's in meters kelvin, 100 mm is .1 m and that's what i use?

    thanks in advance for any help on this.
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