Thermal Resistance due to Convection Heat Transfer

In summary, the conversation discusses the calculation of thermal resistance, thermal conductance, and percentage of total resistance for a typical wall construction in a single-family residence. It also includes a discussion on the effects of each wall-layer on heat flow and assumptions made in the calculations. The conversation also includes a second part that considers thermal resistances due to convection heat transfer on the inner and outer wall surfaces. The calculation procedures for both parts are provided.
  • #1
EmilyO89
4
0

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!
 
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  • #2
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
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
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
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:

Related to Thermal Resistance due to Convection Heat Transfer

1. What is thermal resistance due to convection heat transfer?

Thermal resistance due to convection heat transfer is a measure of how easily heat can be transferred through a material by convection. It is caused by the flow of a fluid (such as air or water) over a surface, which carries heat away from the surface and reduces the temperature difference between the surface and the surrounding fluid.

2. How is thermal resistance due to convection heat transfer calculated?

Thermal resistance due to convection heat transfer is calculated using the equation R = 1/hA, where R is the thermal resistance, h is the convective heat transfer coefficient, and A is the surface area. This equation is based on Newton's Law of Cooling, which states that the rate of heat transfer is directly proportional to the temperature difference between the surface and the surrounding fluid.

3. What factors influence thermal resistance due to convection heat transfer?

The main factors that influence thermal resistance due to convection heat transfer include the properties of the fluid (such as density and viscosity), the velocity of the fluid flow, the temperature difference between the surface and the fluid, and the surface roughness. Additionally, the geometry and orientation of the surface can also affect the convective heat transfer coefficient and thus impact the thermal resistance.

4. How does thermal resistance due to convection heat transfer impact heat transfer in different materials?

Thermal resistance due to convection heat transfer can have a significant impact on heat transfer in different materials. For example, materials with high thermal resistance due to convection may require more energy to maintain a certain temperature, while materials with low thermal resistance may lose heat more easily. This is why it is important to consider the thermal resistance of materials when designing efficient heating or cooling systems.

5. How can thermal resistance due to convection heat transfer be reduced?

There are several ways to reduce thermal resistance due to convection heat transfer. One way is to increase the surface area, as this will allow for more heat to be transferred through convection. Additionally, using materials with lower thermal resistance or increasing the fluid velocity can also help reduce thermal resistance. Improving the surface roughness or adding fins can also enhance heat transfer by promoting turbulence in the fluid flow.

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