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Homework Help: Thermal Resistance due to Convection Heat Transfer

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data

    1) A typical wall construction for a single-family residence might consist of 0.375 inches of outer sheathing (insulating board, hardwood siding, etc.), 3.5 inches of mineral fiber insulation and then a 0.375 inches inner surface (gypsum, etc.). Typical thermal conductivities for these materials are 0.1 W/m-C for the siding, 0.046 W/m-C for the insulation and 0.17 W/m-C for the inner surface. Determine the total thermal resistance (hr-ft2-F/Btu) and total wall conductance (Btu/hr-ft2-F). Also find the percent of total resistance for the three different materials. Provide step-by-step calculation procedures. Discuss/comment to include as a minimum things like assumptions, effects of each wall-layer on heat flow, etc (2.5 points).

    2) Part 1 neglected thermal resistances due to convection heat transfer on the inner and outer wall surfaces. Recalculate the total thermal resistance (along with percent) and the wall U-factor including convection heat transfer using an outer wall convection heat transfer coefficient of 6.0 Btu/hr-ft2-F (15 mph wind in the winter) and an inner wall value of 1.46 Btu/hr-ft2-F. Provide step-by-step calculation procedures. Discuss/comment to include as a minimum things like assumptions, comparisons, etc (2.5 points).

    2. Relevant equations

    Rth (thermal resistance ,R-value) =1/U-value=L/k (h∙ft2∙F/Btu)

    U-value (thermal conductance) = k/L (Btu/h∙ft2∙F)

    Fourier’s Law: qcond= kA (T1–T2)/L

    k=thermal conductivity(Btu∙in/h∙ft²∙F or W/m∙K)
    A=area through which conduction occurs
    T1= higher temperature
    T2= lower temperature
    L= thickness of material

    3. The attempt at a solution

    Part 1)


    0.375in thickness or 0.375in×2.54cm/1in×1m/100cm=0.009525 m

    k=0.1 W/(m×℃)


    3.5in thickness or 3.5in×2.54cm/1in×1m/100cm=0.0889 m

    k=0.46 W/(m×℃)

    Inner Surface

    0.375in thickness or 0.375in×2.54cm/1in×1m/100cm=0.009525 m

    k=0.17 W/(m×℃)

    Note: As the common SI unit for Thermal Resistance is (m^2×℃)/W it is necessary to convert the thickness value into meters.

    Thermal Resistance (R)

    〖R_th〗_siding=L/k=0.009525m/(0.1 W/m℃)=0.09525 (m^2×℃)/W

    〖R_th〗_insulation=L/k=0.0889m/(0.046 W/m℃)=1.9326 (m^2×℃)/W

    〖R_th〗_inner=L/k=0.009525m/(0.17 W/m℃)=0.056029 (m^2×℃)/W

    〖R_th〗_total=〖R_th〗_siding+ 〖R_th〗_insulation+ 〖R_th〗_inner=(0.09525 (m^2×℃)/W )+(1.9326 (m^2×℃)/W)+(0.056029 (m^2×℃)/W)=2.08 (m^2×℃)/W


    2.08 (m^2×℃)/W ×1W/(0.2931 Btu/h)×(10.76ft^2)/(1m^2 )×(9℉)/(5℃)=137 (hr×ft^2×℉)/Btu

    Percentage of R value per material:

    Siding: ((0.09525 (m^2×℃)/W)/(2.08 (m^2×℃)/W))×100=4.6%

    Insulation: ((1.9326 (m^2×℃)/W)/(2.08 (m^2×℃)/W))×100=92.9%

    Inner Surface: ((0.056029 (m^2×℃)/W)/(2.08 (m^2×℃)/W))×100=2.7%

    Note: Both the siding, and the inner wall surface have provide very little R-value. The insulation allows for thinner walls, as it would take an abundant amount of either siding or inner surface to achieve an acceptable level of thermal resistance.

    Wall Conductance (C)

    C=1/R=1/137 Btu/(hr×ft^2×℉)

    Note: Wall conductance is simply the inverse of the R-value, often expressed as U-value. It measures the rate of heat flow though a unit area of a material. This should not be confused with thermal conductivity (k), which is an inherent property of the material.

    Part 2) I have been staring at this for hours trying to understand what I'm not understanding! I know it must have something to do with the heat loss on the inside being equal to the heat gain on the outside, but I just cannot figure out how to get started on this problem without knowing the temperature difference.

    Any help, tips, or hints would be greatly appreciated!
  2. jcsd
  3. Sep 18, 2015 #2


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    Gold Member

    Regarding part 2): You don't need a temperature to calculate the resistance (or the u-value), you only would need it if you want to calculate the heat passing the wall. So you simply can calculate the total resistance/u-value with the given heat transfer coefficients and the thermal conductivities.

    Regarding part1): I didn't check it, because it seemed to be finished. Need something special with that?
  4. Sep 18, 2015 #3
    I think I understood the first part fine. I just included in in case anything was needed from it for part 2.

    I think I get the second part now...the use of different terms to refer to the same variable was throwing me off.

    I should just use 1/U=(1/h1)+(1/ktotal)+(1/h2). U will equal the U factor, and 1/U will equal the R factor (total thermal resistance)...correct?

    Thank you the help!!!
  5. Sep 18, 2015 #4


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    Gold Member

    You're welcome!

    Basically the formula is correct, but just to be sure about the (1/ktotal)-part. In your first post you defined k as the heat conductivity in W / (m⋅K). If ktotal has the same dimension, the equation isn't finished.
  6. Sep 18, 2015 #5
    I don't get the same answer as you for part 1. I get a conductance value of 0.08 ##\frac{BTU}{hr-ft^2-F}##.

    Your conversion factor from watts to BTU/hr looks flipped.

    Last edited: Sep 18, 2015
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