Thermal Resistance due to Convection Heat Transfer

  • #1
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Homework Statement



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).

Homework 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

where,
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

The Attempt at a Solution



Part 1)

Siding

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

k=0.1 W/(m×℃)

Insulation

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

or…

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!
 

Answers and Replies

  • #2
stockzahn
Homework Helper
Gold Member
498
137
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?
 
  • #3
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0
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!!!
 
  • #4
stockzahn
Homework Helper
Gold Member
498
137
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.
 
  • #5
21,478
4,855
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.

Chet
 
Last edited:

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